Table of Contents Chapter R Review R.1 Real Numbers ............................................................................................................................ 1 R.2 Algebra Essentials ..................................................................................................................... 5 R.3 Geometry Essentials ................................................................................................................ 11 R.4 Polynomials ............................................................................................................................. 16 R.5 Factoring Polynomials ............................................................................................................ 23 R.6 Synthetic Division ................................................................................................................... 28 R.7 Rational Expressions ............................................................................................................... 30 R.8 nth Roots; Rational Exponents ................................................................................................ 40
Chapter 1 Equations and Inequalities 1.1 Linear Equations ...................................................................................................................... 50 1.2 Quadratic Equations................................................................................................................. 68 1.3 Complex Numbers; Quadratic Equations in the Complex Number System............................ 86 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations................................ 92 1.5 Solving Inequalities ............................................................................................................... 116 1.6 Equations and Inequalities Involving Absolute Value .......................................................... 127 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications ...... 137 Chapter Review ............................................................................................................................ 145 Chapter Test .................................................................................................................................. 153 Chapter Projects ............................................................................................................................ 155
Chapter 2 Graphs 2.1 The Distance and Midpoint Formulas ................................................................................... 156 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry ............................................. 169 2.3 Lines ...................................................................................................................................... 183 2.4 Circles .................................................................................................................................... 201 2.5 Variation ................................................................................................................................ 215 Chapter Review ............................................................................................................................ 221 Chapter Test .................................................................................................................................. 227 Cumulative Review ...................................................................................................................... 229 Chapter Project ............................................................................................................................. 231
Chapter 3 Functions and Their Graphs 3.1 Functions................................................................................................................................ 232 3.2 The Graph of a Function ........................................................................................................ 250 3.3 Properties of Functions .......................................................................................................... 259 3.4 Library of Functions; Piecewise-defined Functions .............................................................. 276 3.5 Graphing Techniques: Transformations ................................................................................ 288 3.6 Mathematical Models: Building Functions ........................................................................... 306 Chapter Review ............................................................................................................................ 314 Chapter Test .................................................................................................................................. 321 Cumulative Review ...................................................................................................................... 324 Chapter Projects ............................................................................................................................ 328
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Chapter 4 Linear and Quadratic Functions 4.1 Properties of Linear Functions and Linear Models ............................................................... 330 4.2 Building Linear Functions from Data .................................................................................... 341 4.3 Quadratic Functions and Their Properties ............................................................................. 347 4.4 Build Quadratic Models from Verbal Descriptions and from Data ...................................... 371 4.5 Inequalities Involving Quadratic Functions........................................................................... 378 Chapter Review ............................................................................................................................ 398 Chapter Test .................................................................................................................................. 406 Cumulative Review ...................................................................................................................... 408 Chapter Projects ............................................................................................................................ 411
Chapter 5 Polynomial and Rational Functions 5.1 Polynomial Functions ............................................................................................................ 414 5.2 Graphing Polynomial Functions; Models ............................................................................... 424 5.3 Properties of Rational Functions ........................................................................................... 440 5.4 The Graph of a Rational Function ......................................................................................... 450 5.5 Polynomial and Rational Inequalities .................................................................................... 506 5.6 The Real Zeros of a Polynomial Function ............................................................................. 527 5.7 Complex Zeros; Fundamental Theorem of Algebra .............................................................. 558 Chapter Review ............................................................................................................................ 567 Chapter Test .................................................................................................................................. 582 Cumulative Review ...................................................................................................................... 586 Chapter Projects ............................................................................................................................ 591
Chapter 6 Exponential and Logarithmic Functions 6.1 Composite Functions ............................................................................................................. 593 6.2 One-to-One Functions; Inverse Functions ............................................................................. 611 6.3 Exponential Functions ........................................................................................................... 634 6.4 Logarithmic Functions ........................................................................................................... 655 6.5 Properties of Logarithms ....................................................................................................... 677 6.6 Logarithmic and Exponential Equations................................................................................ 686 6.7 Financial Models ................................................................................................................... 707 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models ............................................................................................................... 715 6.9 Building Exponential, Logarithmic, and Logistic Models from Data ................................... 725 Chapter Review ............................................................................................................................ 729 Chapter Test .................................................................................................................................. 742 Cumulative Review ...................................................................................................................... 746 Chapter Projects ............................................................................................................................ 749
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Chapter 7 Trigonometric Functions 7.1 Angles, Arc Length, and Circular Motion ............................................................................. 752 7.2 Right Triangle Trigonometry ................................................................................................. 761 7.3 Computing the Values of Trigonometric Functions of Acute Angles ................................... 777 7.4 Trigonometric Functions of Any Angle ................................................................................ 790 7.5 Unit Circle Approach: Properties of the Trigonometric Functions ....................................... 806 7.6 Graphs of the Sine and Cosine Functions .............................................................................. 815 7.7 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions .................................... 837 7.8 Phase Shift; Sinusoidal Curve Fitting .................................................................................... 847 Chapter Review ............................................................................................................................ 860 Chapter Test .................................................................................................................................. 869 Cumulative Review ...................................................................................................................... 873 Chapter Projects ............................................................................................................................ 877
Chapter 8 Analytic Trigonometry 8.1 The Inverse Sine, Cosine, and Tangent Functions................................................................. 881 8.2 The Inverse Trigonometric Functions (Continued) ............................................................... 895 8.3 Trigonometric Equations ....................................................................................................... 907 8.4 Trigonometric Identities ........................................................................................................ 928 8.5 Sum and Difference Formulas ............................................................................................... 941 8.6 Double-angle and Half-angle Formulas................................................................................. 966 8.7 Product-to-Sum and Sum-to-Product Formulas .................................................................... 994 Chapter Review .......................................................................................................................... 1007 Chapter Test ................................................................................................................................ 1022 Cumulative Review .................................................................................................................... 1027 Chapter Projects .......................................................................................................................... 1033
Chapter 9 Applications of Trigonometric Functions 9.1 Applications Involving Right Triangles .............................................................................. 1037 9.2 The Law of Sines ................................................................................................................. 1045 9.3 The Law of Cosines ............................................................................................................. 1060 9.4 Area of a Triangle ................................................................................................................ 1072 9.5 Simple Harmonic Motion; Damped Motion; Combining Waves ........................................ 1082 Chapter Review .......................................................................................................................... 1092 Chapter Test ................................................................................................................................ 1098 Cumulative Review .................................................................................................................... 1101 Chapter Projects .......................................................................................................................... 1107
Chapter 10 Polar Coordinates; Vectors 10.1 Polar Coordinates............................................................................................................... 1111 10.2 Polar Equations and Graphs............................................................................................... 1120 10.3 The Complex Plane; De Moivre’s Theorem ...................................................................... 1149 10.4 Vectors ............................................................................................................................... 1163 10.5 The Dot Product ................................................................................................................. 1176 Chapter Review .......................................................................................................................... 1182 Chapter Test ................................................................................................................................ 1190 Cumulative Review .................................................................................................................... 1194 Chapter Projects .......................................................................................................................... 1196
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Chapter 11 Analytic Geometry 11.2 The Parabola ...................................................................................................................... 1199 11.3 The Ellipse ......................................................................................................................... 1215 11.4 The Hyperbola ................................................................................................................... 1232 11.5 Rotation of Axes; General Form of a Conic ...................................................................... 1252 11.6 Polar Equations of Conics ................................................................................................. 1265 11.7 Plane Curves and Parametric Equations ............................................................................ 1274 Chapter Review .......................................................................................................................... 1289 Chapter Test ................................................................................................................................ 1298 Cumulative Review .................................................................................................................... 1303 Chapter Projects .......................................................................................................................... 1305
Chapter 12 Systems of Equations and Inequalities 12.1 Systems of Linear Equations: Substitution and Elimination ............................................. 1309 12.2 Systems of Linear Equations: Matrices ............................................................................. 1332 12.3 Systems of Linear Equations: Determinants...................................................................... 1356 12.4 Matrix Algebra................................................................................................................... 1370 12.5 Partial Fraction Decomposition ......................................................................................... 1389 12.6 Systems of Nonlinear Equations ........................................................................................ 1407 12.7 Systems of Inequalities ...................................................................................................... 1435 12.8 Linear Programming .......................................................................................................... 1450 Chapter Review .......................................................................................................................... 1464 Chapter Test ................................................................................................................................ 1479 Cumulative Review .................................................................................................................... 1487 Chapter Projects .......................................................................................................................... 1491
Chapter 13 Sequences; Induction; the Binomial Theorem 13.1 Sequences .......................................................................................................................... 1493 13.2 Arithmetic Sequences ........................................................................................................ 1503 13.3 Geometric Sequences; Geometric Series ........................................................................... 1512 13.4 Mathematical Induction ..................................................................................................... 1524 13.5 The Binomial Theorem ...................................................................................................... 1533 Chapter Review .......................................................................................................................... 1540 Chapter Test ................................................................................................................................ 1544 Cumulative Review .................................................................................................................... 1547 Chapter Projects .......................................................................................................................... 1550
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Chapter 14 Counting and Probability 14.1 Counting ............................................................................................................................ 1553 14.2 Permutations and Combinations ........................................................................................ 1556 14.3 Probability.......................................................................................................................... 1561 Chapter Review .......................................................................................................................... 1568 Chapter Test ................................................................................................................................ 1570 Cumulative Review .................................................................................................................... 1571 Chapter Projects .......................................................................................................................... 1574
Appendix Graphing Utilities Section 1 The Viewing Rectangle ............................................................................................. 1577 Section 2 Using a Graphing Utility to Graph Equations ........................................................... 1578 Section 3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry ................. 1583 Section 5 Square Screens ........................................................................................................... 1585
vii Copyright © 2025 Pearson Education, Inc.
Chapter R Review Section R.1
16.
( A B) C 1, 3, 4,5, 9 2, 4, 6, 7,8 1,3, 4, 6
1. rational
4 1, 3, 4, 6
2. 4 5 6 3 4 30 3 31
1,3, 4, 6
3. Distributive
17. A 0, 2, 6, 7, 8
4. c 5. a
18. C 0, 2, 5, 7, 8, 9
6. b
19. A B 1, 3, 4, 5, 9 2, 4, 6, 7, 8 4 0, 1, 2, 3, 5, 6, 7, 8, 9
7. True 8. False; The Zero-Product Property states that if a product equals 0, then at least one of the factors must equal 0.
20. B C 2, 4, 6, 7, 8 1, 3, 4, 6 1, 2, 3, 4, 6, 7, 8 0, 5, 9
9. False; 6 is the Greatest Common Factor of 12 and 18. The Least Common Multiple is the smallest value that both numbers will divide evenly. The LCM for 12 and 18 is 36.
21. A B 0, 2, 6, 7, 8 0, 1, 3, 5, 9 0, 1, 2, 3, 5, 6, 7, 8, 9
22. B C 0, 1, 3, 5, 9 0, 2, 5, 7, 8, 9
10. True
0, 5, 9
11. A B 1, 3, 4,5, 9 2, 4, 6, 7,8
1, 2,3, 4, 5, 6, 7,8, 9
23. a.
12. A C 1, 3, 4,5, 9 1, 3, 4, 6 1, 3, 4, 5, 6, 9
13. A B 1, 3, 4,5, 9 2, 4, 6, 7,8 4
b.
6, 2,5
c.
d.
e.
14. A C 1, 3, 4,5, 9 1, 3, 4, 6 1, 3, 4 15.
( A B) C 1, 3, 4,5, 9 2, 4, 6, 7,8 1,3, 4, 6
24. a.
1, 2,3, 4,5, 6, 7,8,9 1,3, 4, 6 1, 3, 4, 6
2,5
1 6, , 1.333..., 2,5 2
1 6, , 1.333..., , 2,5 2
1
b.
0,1
c.
d.
5
5 , 2.060606... 2.06,1.25, 0,1 3
1 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
e. 25. a.
5 , 2.060606... 2.06,1.25, 0,1, 5 3
1
b.
0,1
c.
1 1 1 0,1, , , 2 3 4
d. None e.
1 1 1 0,1, , , 2 3 4
None
b.
1
c.
1.3, 1.2, 1.1, 1
d. e. 28. a.
c.
9.999
b.
9.998
36. a.
1.001
b.
1.000
37. a.
0.429
b.
0.428
38. a.
0.556
b.
0.555
39. a.
34.733
b.
34.733
40. a.
16.200
b.
16.200
45. 3 y 1 2 46. 2 x 4 6
None
47. x 2 6
None
48. 2 y 6
1 2, , 2 1, 2 2, , 2 1,
1 2
None
35. a.
44. 3 y 2 2
1.3, 1.2, 1.1, 1
b. None
0.053
43. x 2 3 4
b. None c.
b.
42. 5 2 10
d. None
27. a.
0.054
41. 3 2 5
26. a.
e.
34. a.
49.
x 6 2
50.
2 6 x
51. 9 4 2 5 2 7
52. 6 4 3 2 3 5
1 10.3 2
d.
2, 2
e.
53. 6 4 3 6 12 6 54. 8 4 2 8 8 0
1 2, 2, 10.3 2
55. 18 5 2 18 10 8
29. a.
18.953
b.
18.952
30. a.
25.861
b.
25.861
31. a.
28.653
b.
28.653
32. a.
99.052
b.
99.052
33. a.
0.063
b.
0.062
56. 100 10 2 100 20 80 57. 4
1 12 1 13 3 3 3
58. 2
1 4 1 3 2 2 2
2
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Section R.1: Real Numbers 59. 6 3 5 2 3 2 6 15 2 1 6 17 11 60. 2 8 3 4 2 3 2 8 3 6 3 2 8 18 3 2 10 3
70.
5 3 53 53 1 9 10 3 3 5 2 3 3 5 2 6
71.
6 10 2 3 5 2 2 3 5 2 4 25 27 5 5 3 9 5 5 3 9 45
72.
21 100 3 7 4 25 3 7 4 25 28 25 3 25 3 25 3
73.
3 2 15 8 23 4 5 20 20
74.
4 1 8 3 11 3 2 6 6
75.
7 4 49 32 81 8 7 56 56
76.
8 15 16 135 151 9 2 18 18
77.
5 1 10 3 13 18 12 36 36
78.
2 8 6 40 46 15 9 45 45
79.
5 8 25 64 39 13 24 15 120 120 40
80.
3 2 94 5 14 21 42 42
81.
3 2 98 1 20 15 60 60
82.
6 3 12 15 3 35 14 70 70
20 3 23
61. 4 9 5 6 7 3 4 14 42 3 56 42 3 14 3 17
62. 1 4 3 2 2 1 12 2 2 1 12 11
63. 10 6 2 2 8 3 2 10 6 4 5 2 10 2 5 2 10 7 2 10 14 4
64. 2 5 4 6 3 4 2 20 6 1 18 6 18 6 12
65.
1
1
5 3 2 2 2 1
66.
5 4 13 9 13 3
67.
4 8 12 6 53 2
68.
2 4 2 1 53 2
69.
3 10 3 2 5 3 2 5 2 5 21 5 3 7 5 3 7 7
5 18 5 27 5 9 3 5 9 3 15 83. 11 18 11 9 2 11 9 2 11 22 27 5 21 5 35 5 7 5 5 7 5 25 84. 2 21 2 7 3 2 7 3 2 6 35
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Chapter R: Review
85.
86.
1 4 17 4 17 21 1 3 7 21 21 21 21
98.
3x 1 x 5 3x 2 15 x x 5 3 x 2 14 x 5
2 4 1 2 22 2 2 2 2 2 3 5 6 3 5 3 2 3 5 3 2 3 15 2 5 2 10 2 10 2 12 3 5 15 15 15 15 15 43 4 3 4 53 5 3 5
99.
x 8 x 2 x 2 2 x 8 x 16 x 2 10 x 16
100.
x 4 x 2 x 2 2 x 4 x 8 x2 6 x 8
101. 3x( x 5k ) 3x 2 60 x
3 3 2 3 3 6 3 6 2 3 87. 2 4 8 1 4 8 4 8 4 2 8 12 3 12 3 15 8 8 8 8
3x 2 15 xk 3x 2 60 x 15 xk 60 x k 4
5 1 3 5 1 35 1 3 5 1 88. 3 6 2 1 6 2 3 2 2 3 2 2 5 1 5 1 4 2 2 2 2 2
102.
( x k )( x 3k ) x 2 4 x 12 x 2 3kx kx 3k 2 x 2 4 x 12 x 2 x(3k k ) 3k 2 x 2 4 x 12 x 2 x(3k k ) 3k 2 x 2 4 x 12
89. 6 x 4 6 x 24
x 2 x(2k ) 3k 2 x 2 4 x 12 2k 4
90. 4 2 x 1 8 x 4
k 2
91. x x 4 x 4 x 2
103. 2 x 3 x 2 x 3 x 2 3 x
92. 4 x x 3 4 x 2 12 x
5 x 5x
1 3 1 2 3x 2 3 93. 2 x 2 x 2 4 2 4 2 22 2 2 3x 2 3 x 1 2 2 2 2
104. 2 3 4 2 12 14 since multiplication comes before addition in the order of operations for real numbers.
2 3 4 5 4 20
1 2 1 3 2x 3 2 94. 3 x 3 x 3 3 6 3 6 3 3 2 3 2x 3 1 2x 3 3 2 2
95.
since operations inside parentheses come before multiplication in the order of operations for real numbers. 105. 2 3 4 2 12 24
x 2 x 4 x 2 4 x 2 x 8
2 3 2 4 6 8 48
2
x 6x 8
96.
106.
x 5 x 1 x 2 x 5 x 5 x2 6 x 5
97.
x 9 2 x 7 2 x 2 7 x 18 x 63
43 7 1 , but 25 7 4 3 4 5 3 2 20 6 26 13 2.6 2 5 10 10 10 5
107. Subtraction is not commutative; for example: 2 3 1 1 3 2 .
2 x 2 11x 63
4
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Section R.2: Algebra Essentials 108. Subtraction is not associative; for example: 5 2 1 2 4 5 2 1 . 109. Division is not commutative; for example: 2 3 . 3 2 110. Division is not associative; for example: 12 2 2 6 2 3 , but 12 2 2 12 1 12 .
111. The Symmetric Property implies that if 2 = x, then x = 2. 112. From the principle of substitution, if x 5 , then x x 5 5 x 2 25
Section R.2 1. variable 2. origin 3. strict 4. base; exponent (or power) 5. 1.2345678 103 6. d 7. a 8. b 9. True 10. False; the absolute value of a real number is nonnegative. 0 0 which is not a positive
x 2 x 25 5 2
x x 30
number.
113. There are no real numbers that are both rational and irrational, since an irrational number, by definition, is a number that cannot be expressed as the ratio of two integers; that is, not a rational number
Every real number is either a rational number or an irrational number, since the decimal form of a real number either involves an infinitely repeating pattern of digits or an infinite, nonrepeating string of digits. 114. The sum of an irrational number and a rational number must be irrational. Otherwise, the irrational number would then be the difference of two rational numbers, and therefore would have to be rational.
11. False; a number in scientific notation is expressed as the product of a number, x, 1 x 10 or 10 x 1 , and a power of 10. 12. True 13.
14.
116. Since 1 day = 24 hours, we compute 12997 541.5416 . 24 Now we only need to consider the decimal part of the answer in terms of a 24 hour day. That is,
0.5416 24 13 hours. So it must be 13 hours
later than 12 noon, which makes the time 1 a.m. CST.
15.
1 0 2
16. 5 6 17. 1 2 18. 3
5 2
19. 3.14
117. Answers will vary. 20.
3 4
1 3
115. Answers will vary.
2 1.41
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5 2
2 3 3 2
Chapter R: Review
21.
42. 3x y 3( 2) 3 6 3 3
1 0.5 2
43. 5 xy 2 5( 2)(3) 2 30 2 28
1 22. 0.33 3
44. 2 x xy 2( 2) ( 2)(3) 4 6 2
23.
2 0.67 3
45.
2( 2) 4 4 2x x y 2 3 5 5
24.
1 0.25 4
46.
x y 23 1 1 5 x y 2 3 5
47.
3x 2 y 3( 2) 2(3) 6 6 0 0 2 y 23 5 5
48.
2 x 3 2( 2) 3 4 3 7 3 3 3 y
29. x 1
49.
x y 3 ( 2) 1 1
30. x 2
50.
x y 3 ( 2) 5 5
31. Graph on the number line: x 2
51.
x y 3 2 3 2 5
52.
x y 3 2 3 2 1
53.
x 3 3 1 x 3 3
54.
y 2 2 1 y 2 2
55.
4 x 5 y 4(3) 5( 2)
25. x 0 26. z 0 27. x 2 28. y 5
32. Graph on the number line: x 4
33. Graph on the number line: x 1
34. Graph on the number line: x 7
12 10 22
35. d (C , D ) d (0,1) 1 0 1 1
22
36. d (C , A) d (0, 3) 3 0 3 3 37. d ( D, E ) d (1,3) 3 1 2 2
56.
3 x 2 y 3(3) 2( 2) 9 4 5 5
57.
4x 5 y
38. d (C , E ) d (0,3) 3 0 3 3
4(3) 5( 2) 12 10 12 10
39. d ( A, E ) d (3,3) 3 (3) 6 6
2 2
40. d ( D, B) d (1, 1) 1 1 2 2 41. x 2 y 2 2 3 2 6 4 6
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Section R.2: Algebra Essentials
be excluded from the domain because it causes division by 0.
58. 3 x 2 y 3 3 2 2 33 2 2 94
67.
13
59.
x2 1 x Part (c) must be excluded. The value x 0 must be excluded from the domain because it causes division by 0.
x2 1 60. x Part (c) must be excluded. The value x 0 must be excluded from the domain because it causes division by 0.
68.
x x x 9 ( x 3)( x 3) Part (a) , x 3 , must be excluded because it causes the denominator to be 0.
69.
61.
62.
63.
64.
65.
66.
2
x x 9 None of the given values are excluded. The domain is all real numbers. 2
x2 x 1 None of the given values are excluded. The domain is all real numbers. 2
x3 x3 x 2 1 ( x 1)( x 1) Parts (b) and (d) must be excluded. The values x 1, and x 1 must be excluded from the domain because they cause division by 0. x 2 5 x 10 x 2 5 x 10 3 x( x 1)( x 1) x x Parts (b), (c), and (d) must be excluded. The values x 0, x 1, and x 1 must be excluded from the domain because they cause division by 0.
9 x 2 x 1 9 x 2 x 1 x3 x x( x 2 1) Part (c) must be excluded. The value x 0 must
70.
4 x 5 x 5 must be exluded because it makes the denominator equal 0. Domain x x 5
6 x4 x 4 must be excluded sine it makes the denominator equal 0. Domain x x 4 x x4 x 4 must be excluded sine it makes the denominator equal 0. Domain x x 4 x2 x6 x 6 must be excluded sine it makes the denominator equal 0. Domain x x 6
5 5 5 71. C ( F 32) (32 32) (0) 0C 9 9 9 5 5 5 72. C ( F 32) (212 32) (180) 100C 9 9 9 5 5 5 73. C ( F 32) (77 32) (45) 25C 9 9 9 5 5 74. C ( F 32) ( 4 32) 9 9 5 (36) 9 20C
75. (9)2 (9)(9) 81 76. 42 (4) 2 16 77. 42
1 1 2 16 4
7 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
78. 42
1 1 16 42
94.
79. 36 34 36 4 32
4 x 2 ( y z ) 1 23 x 4 y
1 1 32 9
80. 42 43 42 3 41 4 81.
4 4 4 64
82.
2 2 2 8
83.
100 102 10
84.
2
36 6 6
85.
4 2 4 4
86.
3 3 3
87.
9 x 9 x 81x
88.
4 x 41x 41x
89.
x y x y x y xy
90.
x y x y x y xy
3 1
3 1
1 3
3
1 3
2 x 3 95. 1 3y
3
5 x 2 96. 2 6y
93.
3
97. 2 xy 1
8
2 1
2
2 1 2
2 2
3
1
3
3 3
4 2
3
7 2
(3) x y z
6 x2 2 5y
3
2 3
3
6 6
3 y 3 1 3 x 2 2
4
99. x 2 y 2 2 1 4 1 5 2
2
2
100. x 2 y 2 2 1 4 1 4 2
3 3
x 2 y 1 x 2 1 y1 2 x 3 y 1 3 2 xy x y ( 4) 2 y 5 ( x z )3
3
2x 2 2 4 y 1
98. 3x 1 y
2
1 2
1 3
3x3 32 x 6 9 x 6 2 2 2 2 y 4y 2y
5 y2 2 6x
3
4 2
2
2
2
216 x 125 y 5 y
2
4 2
3y 3 2x
63 x 2
x2 y5 y 91. 3 4 x 2 3 y 5 4 x 1 y1 x x y 92.
2
4 x 2 y 1 z 1 8x4 y 4 x 2 4 y 11 z 1 8 1 x 6 y 2 z 1 2 1 6 2 2x y z
101.
xy 2 2 1 2 2 4
102.
x y 2 2 1 12 1
103.
16 y 5 x3 z 3 27 x y 7 z 2
16 31 57 3 2 x y z 27 16 x 2 y 2 z1 27 16 x 2 z 27 y 2
2
2
2
x2 x 2 2
104.
x x 2
105.
x2 y 2
106.
x 2 y 2 x y 2 1 2 1 3
2
107. x y 21
8
Copyright © 2025 Pearson Education, Inc.
1 2
2 2 12
4 1 5
Section R.2: Algebra Essentials 108. y x 1 1
120. (8.11) 4 0.000
109. If x 2,
121. 454.2 4.542 102
2
2 x3 3 x 2 5 x 4 2 23 3 22 5 2 4 16 12 10 4 10
123. 0.013 1.3 102
If x 1, 3
2
122. 32.14 3.214 101
3
2
2 x 3 x 5 x 4 2 1 3 1 5 1 4
124. 0.00421 4.21 103 125. 32,155 3.2155 104
235 4 0
126. 21, 210 2.121 104
110. If x 1, 4 x3 3x 2 x 2 4 13 3 12 1 2 4 3 1 2
127. 0.000423 4.23 104 128. 0.0514 5.14 102
8
129. 6.15 104 61,500
If x 2, 4 x3 3x 2 x 2 4 23 3 22 2 2 32 12 2 2 44 4
(666) 4 666 4 111. 3 81 (222) 4 222 3
3 1 112. (0.1)3 (20)3 2 10 10 1 3 23 103 10 23 8
113. (8.2)6 304, 006.671 114. (3.7)5 693.440 115. (6.1) 3 0.004
130. 9.7 103 9700 131. 1.214 103 0.001214 132. 9.88 104 0.000988 133. 1.1 108 110, 000, 000 134. 4.112 102 411.2 135. 8.1 102 0.081 136. 6.453 101 0.6453 137. A lw 138. P 2 l w 139. C d
116. (2.2)5 0.019
140. A
1 bh 2
117. ( 2.8)6 481.890
141. A
3 2 x 4
118. (2.8)6 481.890 119. ( 8.11) 4 0.000
142. P 3x 4 143. V r 3 3
9 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
144. S 4 r 2
b.
209 volts is not acceptable.
145. V x3 153. a.
146. S 6 x 2 147. a.
b.
x 3 2.999 3 0.001 0.001 0.01 A radius of 2.999 centimeters is acceptable.
If x 1000, C 4000 2 x 4000 2(1000) 4000 2000 $6000 The cost of producing 1000 watches is $6000.
b.
x 3 2.89 3 0.11 0.11 0.01 A radius of 2.89 centimeters is not acceptable.
If x 2000, C 4000 2 x 4000 2(2000) 4000 4000 $8000 The cost of producing 2000 watches is $8000.
154. a.
x 98.6 97 98.6 1.6 1.6 1.5 97˚F is unhealthy.
b.
x 98.6 100 98.6 1.4
148. 210 80 120 25 60 32 5 $98 His balance at the end of the month was $98.
1.4 1.5 100˚F is not unhealthy.
149. We want the difference between x and 4 to be at least 6 units. Since we don’t care whether the value for x is larger or smaller than 4, we take the absolute value of the difference. We want the inequality to be non-strict since we are dealing with an ‘at least’ situation. Thus, we have x4 6
155. The distance from Earth to the Moon is about 4 108 400, 000, 000 meters. 156. The height of Mt. Everest is about 8848 8.848 103 meters. 157. The wavelength of visible light is about 5 107 0.0000005 meters.
150. We want the difference between x and 2 to be more than 5 units. Since we don’t care whether the value for x is larger or smaller than 2, we take the absolute value of the difference. We want the inequality to be strict since we are dealing with a ‘more than’ situation. Thus, we have x2 5 151. a.
x 220 209 220 11 11 8
158. The diameter of an atom is about 1 1010 0.0000000001 meters. 159. The diameter is about 0.0403 4.03 102 inches. 160. The tiniest motor is less than 0.00004 4 105 millimeters tall.
x 110 108 110 2 2 5
108 volts is acceptable. b.
x 110 104 110 6 6 5
104 volts is not acceptable. 152. a.
x 220 214 220 6 6 8
214 volts is acceptable. 10
Copyright © 2025 Pearson Education, Inc.
Section R.3: Geometry Essentials 161. 186, 000 60 60 24 365
1.86 10
5
6 10 2.4 10 3.65 10 1
2
1
2. A
2
586.5696 1010 5.865696 1012 There are about 5.9 1012 miles in one lightyear.
1 bh 2
3. C 2 r 4. similar 5. c
93, 000, 000 9.3 107 5 102 162. 186, 000 1.86 105 500 seconds 8 min. 20 sec. It takes about 8 minutes 20 seconds for a beam of light to reach Earth from the Sun.
163.
1 0.333333 ... 0.333 3 1 is larger by approximately 0.0003333 ... 3
164. 2 0.666666 ... 0.666 3 2 is larger by approximately 0.000666 ... 3 165.
13
19
34.06 10 3.406 10
6
7. True. 8. True. 62 82 36 64 100 102 9. False; the surface area of a sphere of radius r is given by V 4 r 2 . 10. True. The lengths of the corresponding sides are equal. 11. True. Two corresponding angles are equal. 12. False. The sides are not proportional.
5.24 10 6.5 10 5.24 6.5 10 10 6
6. b
13
13.
20
c 2 a 2 b2 52 122
1.62 104 1.62 104 166. 0.36 106 4.5 1010 4.5 1010 3.6 105
167. No. For any positive number a, the value
25 144 169 c 13 a is 2
14.
168. We are given that 1 x 2 10 . This implies that 1 x 10 . Since x 10 3.162 and x 3.142 , the number could be 3.15 or 3.16 (which are between 1 and 10 as required). The number could also be 3.14 since numbers such as 3.146 which lie between and 10 would equal 3.14 when truncated to two decimal places.
6 2 82 36 64 100 c 10
15.
1. right; hypotenuse
a 10, b 24, c 2 a 2 b2 102 242 100 576 676 c 26
169. Answers will vary. 170. Answers will vary. 5 < 8 is a true statement because 5 is further to the left than 8 on a real number line.
a 6, b 8, c 2 a 2 b2
smaller and therefore closer to 0.
Section R.3
a 5, b 12,
16.
a 4, b 3, c 2 a 2 b2 42 32 16 9 25 c 5
11 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
17.
a 7, b 24,
25. 62 32 42 36 9 16 36 25 false The given triangle is not a right triangle.
c 2 a 2 b2 7 2 242 49 576 625 c 25
18.
26. 7 2 52 42 49 25 16 49 41 false The given triangle is not a right triangle.
a 14, b 48, c2 a 2 b2 142 482 196 2304 2500 c 50
27. A l w 6 7 42 in 2 28. A l w 9 4 36 cm 2
19. 52 32 42 25 9 16 25 25 The given triangle is a right triangle. The hypotenuse is 5.
30. A
1 1 b h (4)(9) 18 cm 2 2 2
32. A r 2 (2) 2 4 ft 2 C 2 r 2 (2) 4 ft
21. 62 42 52 36 16 25 36 41 false The given triangle is not a right triangle. 2
1 1 b h (14)(4) 28 in 2 2 2
31. A r 2 (5) 2 25 m 2 C 2 r 2 (5) 10 m
20. 102 62 82 100 36 64 100 100 The given triangle is a right triangle. The hypotenuse is 10.
2
29. A
33. V l w h 6 8 5 240 ft 3 S 2lw 2lh 2wh 2 6 8 2 6 5 2 8 5 96 60 80
2
236 ft 2
22. 3 2 2 9 44 9 8 false The given triangle is not a right triangle.
34. V l w h 9 4 8 288 in 3 S 2lw 2lh 2wh 2 9 4 2 9 8 2 4 8
23. 252 7 2 242 625 49 576 625 625 The given triangle is a right triangle. The hypotenuse is 25.
72 144 64 280 in 2 4 3 4 500 r 53 cm3 3 3 3 S 4 r 2 4 52 100 cm 2
35. V
24. 262 102 242 676 100 576 676 676 The given triangle is a right triangle. The hypotenuse is 26.
4 3 4 r 33 36 ft 3 3 3 2 S 4 r 4 32 36 ft 2
36. V
12
Copyright © 2025 Pearson Education, Inc.
Section R.3: Geometry Essentials 43. Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get 8 x 4 2 8 2 x 4 4x In addition, corresponding angles must have the same angle measure. Therefore, we have A 90 , B 60 , and C 30 .
37. V r 2 h (9) 2 (8) 648 in 3 S 2 r 2 2 rh 2 9 2 9 8 2
162 144 306 in 2
38. V r 2 h (8) 2 (9) 576 in 3 S 2 r 2 2 rh 2 8 2 8 9 2
128 144 272 in 2
39. The diameter of the circle is 2, so its radius is 1. A r 2 (1) 2 square units 40. The diameter of the circle is 2, so its radius is 1. A 22 (1) 2 4 square units 41. The diameter of the circle is the length of the diagonal of the square. d 2 22 22 44 8 d 82 2 d 2 2 2 2 2 The area of the circle is: r
A r2
2 2 square units 2
42. The diameter of the circle is the length of the diagonal of the square. d 2 22 22 44 8 d 82 2 d 2 2 2 2 2 The area is: r
A
2 2 2 4 square units 2
2
44. Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get 6 x 12 16 6 16 x 12 8 x In addition, corresponding angles must have the same angle measure. Therefore, we have A 30 , B 75 , and C 75 . 45. Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get 30 x 20 45 30 45 x 20 135 x or x 67.5 2 In addition, corresponding angles must have the same angle measure. Therefore, we have A 60 , B 95 , and C 25 . 46. Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get 8 x 10 50 8 50 x 10 40 x In addition, corresponding angles must have the same angle measure. Therefore, we have A 50 , B 125 , and C 5 .
13 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review 54. Let x = the approximate distance from San Juan to Hamilton and y = the approximate distance from Hamilton to Fort Lauderdale. Using similar triangles, we get 1046 x 1046 y 58 53.5 58 57 1046 53.5 1046 57 x y 58 58 964.8 x 1028.0 y The approximate distance between San Juan and Hamilton is 965 miles and the approximate distance between Hamilton and Fort Lauderdale is 1028 miles.
47. The total distance traveled is 4 times the circumference of the wheel. Total Distance 4C 4( d ) 4 16
64 201.1 inches 16.8 feet
48. The distance traveled in one revolution is the circumference of the disk 4 . The number of revolutions = dist. traveled 20 5 1.6 revolutions circumference 4 49. Area of the border = area of EFGH – area of ABCD 102 62 100 36 64 ft 2 50. FG = 4 feet; BG = 4 feet and BC = 10 feet, so CG= 6 feet. The area of the triangle CGF is: 1 A (4)(6) 12 ft 2 2
55. Convert 20 feet to miles, and solve the Pythagorean Theorem to find the distance: 1 mile 20 feet 20 feet 0.003788 miles 5280 feet d 2 (3960 0.003788) 2 39602 30 sq. miles d 5.477 miles
51. Area of the window = area of the rectangle + area of the semicircle. 1 A (6)(4) 22 24 2 30.28 ft 2 2 Perimeter of the window = 2 heights + width + one-half the circumference. 1 P 2(6) 4 (4) 12 4 2 2 16 2 22.28 feet
d 20 ft
3960
52. Area of the deck = area of the pool and deck – area of the pool. A (13) 2 (10) 2 169 100
56. Convert 6 feet to miles, and solve the Pythagorean Theorem to find the distance: 1 mile 6 feet 6 feet 0.001136 miles 5280 feet d 2 (3960 0.001136) 2 39602 9 sq. miles d 3 miles
69 ft 2 216.77 ft 2
The amount of fence is the circumference of the circle with radius 13 feet. C 2(13) 26 ft 81.68 ft
d
53. We can form similar triangles using the Great Pyramid’s height/shadow and Thales’ height/shadow:
6 ft
3960
{
{
h
126 240
114
3960
2 3
This allows us to write h 2 240 3 2 240 160 h 3 The height of the Great Pyramid is 160 paces. 14
Copyright © 2025 Pearson Education, Inc.
3960
Section R.3: Geometry Essentials 57. Convert 100 feet to miles, and solve the Pythagorean Theorem to find the distance: 1 mile 100 feet 100 feet 0.018939 miles 5280 feet d 2 (3960 0.018939) 2 39602 150 sq. miles d 12.2 miles Convert 150 feet to miles, and solve the Pythagorean Theorem to find the distance: 1 mile 150 feet 150 feet 0.028409 miles 5280 feet d 2 (3960 0.028409) 2 39602 225 sq. miles d 15.0 miles
58. Given m 0, n 0 and m n ,
if a m 2 n 2 , b 2mn and c m 2 n 2 , then
a 2 b2 m2 n2
2mn 2
2
1 4
So A [(l w) 2 (l w) 2 ]
m 4 2m 2 n 2 n 4 4m 2 n 2 m 4 2m 2 n 2 n 4
2 2 2 and c m n
m 2m n n 2
4
2 2
4
a 2 b 2 c 2 a, b and c represent the sides of a right triangle. 59. V r 2 h (10)2 (4.5) 450 ft 3
So, 1ft 3 7.48052 gal so
450 ft 7.48052 gal/ft 10,575 gal 3
3
60. 10000(5.61458) 56145.8 ft 3 V r 2h 56145.8 (25) 2 h 56145.8 h 28.6 ft 625
61.
4 V r3 3 4 V2 (2r )3 3 4 8r 3 3 4 8 r 3 8V 3 If you double the radius the volume is 8 times the original volume. 63. Let l = length of the rectangle and w = width of the rectangle. Notice that (l w) 2 (l w) 2 [(l w) (l w)][(l w) (l w)] (2l )(2 w) 4lw 4 A
62.
A r2 A2 (2r ) 2
Since (l w) 2 0 , the largest area will occur when l – w = 0 or l = w; that is, when the rectangle is a square. But 1000 2l 2 w 2(l w) 500 l w 2l 250 l w The largest possible area is 2502 62500 sq ft. A circular pool with circumference = 1000 feet 500 yields the equation: 2 r 1000 r The area enclosed by the circular pool is: 2
5002 500 79577.47 ft 2 A r2 Thus, a circular pool will enclose the most area. 64. Consider the diagram showing the lighthouse at point L, relative to the center of Earth, using the radius of Earth as 3960 miles. Let P refer to the furthest point on the horizon from which the light is visible. Note also that 362 362 feet miles. 5280
4r 2 4 r 2 4 A If you double the radius, the area is four times the original area.
15 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
Let A refer to the airplane’s location. The distance from the plane to point P is d 2 . We want to show that d1 d 2 120 . Assume the altitude of the airplane is 10000 10,000 feet = miles. 5280
Apply the Pythagorean Theorem to CPL :
3960 2 d1 2 3960 5280 362
2
362 3960 2 d1 3960 5280 2
2
2
Apply the Pythagorean Theorem to CPA :
3960 362 3960 23.30 mi. 5280 Therefore, the light from the lighthouse can be seen at point P on the horizon, where point P is approximately 23.30 miles away from the lighthouse. Brochure information is slightly overstated. d1
2
3960 2 d 2 2 3960 5280 10000
2
2
d 2 2 3960 5280 3960 2 10000
2
10000 2 d 2 3960 3960 61Therefo 5280 122.49 miles. re, d1 d 2 23.30 122.49 145.79 120. The brochure information is slightly understated. Note that a plane at an altitude of 6233 feet could see the lighthouse from 120 miles away.
Verify the ship information: Let S refer to the ship’s location, and let x equal the height, in feet, of the ship. We need d1 d 2 40 . Since d1 23.30 miles we need d 2 40 23.30=16.70 miles. Apply the Pythagorean Theorem to CPS :
3960 2 16.7 2 3960 x 2
Section R.4
3960 2 16.7 2 3960 x
1. 4; 3
3960 2 16.7 2 3960 x
2. x 4 16
x 0.035 miles x 185.93 feet. The ship would have to be at least 186 feet tall to see the lighthouse from 40 miles away. Verify the airplane information:
3.
x
3
8
4. a 5. c 6. False; monomials cannot have negative degrees. 7. True 8. False; the dividend = (quotient)(divisor) + remainder 16
Copyright © 2025 Pearson Education, Inc.
Section R.4: Polynomials
9. 2x3 Monomial; Variable: x ; Coefficient: 2; Degree: 3 10. 4x 2 Monomial; Variable: x ; Coefficient: –4; Degree: 2 11.
8 8x 1 x
Not a monomial; when written in
24.
25. 2 y 3 2
Polynomial; Degree: 3
26. 10z 2 z
Polynomial; Degree: 2
k
the form ax , the variable has a negative exponent. 12. 2x 3
27.
Not a monomial; when written in the
form ax k , the variable has a negative exponent.
28.
13. 2xy 2 Monomial; Variables: x, y ; Coefficient: –2; Degree: 3 14. 5x 2 y 3 Monomial; Variables: x, y ; Coefficient: 5; Degree: 5 8x 15. 8 xy 1 y
29.
3x3 2 x 1 Not a polynomial; the x2 x 1 polynomial in the denominator has a degree greater than 0. ( x 2 6 x 8) (3 x 2 4 x 7) 4 x 2 2 x 15
Not a monomial; when written
2 x2 2 x 2 y 3 y3
30.
Not a monomial; when
( x3 3 x 2 2) ( x 2 4 x 4) x3 (3 x 2 x 2 ) ( 4 x) (2 4) x3 4 x 2 4 x 6
31. ( x3 2 x 2 5 x 10) (2 x 2 4 x 3) x3 2 x 2 5 x 10 2 x 2 4 x 3
n m
written in the form ax y , the exponent on the variable y is negative. 2
x2 5 Not a polynomial; the polynomial in x3 1 the denominator has a degree greater than 0.
( x 2 3x 2 ) (6 x 4 x) (8 7)
in the form ax n y m , the exponent on the variable y is negative. 16.
3 2 Not a polynomial; the variable in the x denominator results in an exponent that is not a nonnegative integer.
x3 ( 2 x 2 2 x 2 ) (5 x 4 x) (10 3) x3 4 x 2 9 x 7
2
17. x y Not a monomial; the expression contains more than one term. This expression is a binomial. 18. 3x 2 4 Not a monomial; the expression contains more than one term. This expression is a binomial. 19. 3x 2 5
Polynomial; Degree: 2
20. 1 4x
Polynomial; Degree: 1
21. 5
Polynomial; Degree: 0
22. –π
Polynomial; Degree: 0
x 2 3 x 4 x3 3x 2 x 5 x 3 ( x 2 3 x 2 ) (3x x) ( 4 5) x3 4 x 2 4 x 9
33.
6 x x x 5 x x 3x 5
3
4
3
6 x5 5 x 4 3 x 2 x
34.
10 x 8x 3x 2 x 6 5
2
3
2
10 x5 3x3 10 x 2 6
35.
5 Not a polynomial; the variable in the x denominator results in an exponent that is not a nonnegative integer.
23. 3x 2
32. ( x 2 3 x 4) ( x3 3x 2 x 5)
( x 2 6 x 4) 3(2 x 2 x 5) x 2 6 x 4 6 x 2 3x 15 7 x 2 3x 11
17 Copyright © 2025 Pearson Education, Inc.
2
Chapter R: Review
36.
37.
2( x 2 x 1) (5 x 2 x 2)
48.
(2 x 3)( x 2 x 1)
2 x2 2 x 2 5x2 x 2
2 x( x 2 x 1) 3( x 2 x 1)
7 x 2 3x
2 x3 2 x 2 2 x 3 x 2 3 x 3 2 x3 x 2 x 3
6( x3 x 2 3) 4(2 x 3 3 x 2 ) 6 x3 6 x 2 18 8 x3 12 x 2
49. ( x 2)( x 4) x 2 4 x 2 x 8
2 x3 18 x 2 18
38.
x2 6 x 8
8(4 x3 3x 2 1) 6(4 x3 8 x 2)
50. ( x 3)( x 5) x 2 5 x 3 x 15
32 x3 24 x 2 8 24 x3 48 x 12 3
x 2 8 x 15
2
8 x 24 x 48 x 4
39.
51. (2 x 7)( x 5) 2 x 2 7 x 10 x 35
x x 2 2 x 3x 5 x 1 2
2
2
2 x 2 17 x 35
x 2 x 2 2 x 2 3x 5 x2 1 2 x2 4 x 6
40.
52. (3x 1)(2 x 1) 6 x 2 3 x 2 x 1 6 x2 5x 1
x 1 4 x 5 x x 2 2
2
2
53. ( x 4)( x 2) x 2 2 x 4 x 8
x2 1 4 x2 5 x2 x 2 2 x 2 x 6
41.
7 y2 5 y 3 4 3 y2
x2 2x 8
54. ( x 4)( x 2) x 2 2 x 4 x 8 x2 2 x 8
7 y 2 35 y 21 12 4 y 2 11 y 2 35 y 9
42.
55. ( x 6)( x 3) x 2 6 x 3 x 18
8 1 y3 4 1 y y 2 y3 3
x 2 9 x 18
56. ( x 5)( x 1) x 2 x 5 x 5
2
8 8 y 4 4 y 4 y 4 y3
x2 6x 5
4 y 3 4 y 2 4 y 12
57. (2 x 3)( x 2) 2 x 2 4 x 3x 6
43. x 2 ( x 2 2 x 5) x 4 2 x3 5 x 2 2
3
5
3
44. 4 x ( x x 2) 4 x 4 x 8 x 2
3
5
45. 2 x (4 x 5) 8 x 10 x
2 x2 x 6
2
58. (2 x 4)(3x 1) 6 x 2 2 x 12 x 4
2
6 x 2 10 x 4
46. 5 x3 (3x 4) 15 x 4 20 x3 47.
59. ( 3 x 4)( x 2) 3x 2 4 x 6 x 8
3x 2 10 x 8
( x 1)( x 2 2 x 4) x( x 2 2 x 4) 1( x 2 2 x 4)
60. ( 3 x 1)( x 1) 3x 2 3 x x 1
x3 2 x 2 4 x x 2 2 x 4 3
3x 2 4 x 1
2
x 3x 2 x 4
61. ( x 5)(2 x 7) 2 x 2 10 x 7 x 35 2 x 2 17 x 35
18
Copyright © 2025 Pearson Education, Inc.
Section R.4: Polynomials
62. ( 2 x 3)(3 x ) 6 x 2 x 2 9 3x
83. ( x y ) 2 x 2 2 xy y 2
2 x 2 3x 9
84. ( x y ) 2 x 2 2 xy y 2
63. ( x 2 y )( x y ) x 2 xy 2 xy 2 y 2
85. ( x 2 y ) 2 x 2 2 x 2 y 2 y
x 2 xy 2 y 2 2
64. (2 x 3 y )( x y ) 2 x 2 xy 3 xy 3 y
x 2 4 xy 4 y 2
2
2 x 2 xy 3 y 2
86. (2 x 3 y ) 2 2 x 2 2 x 3 y 3 y 2
87. ( x 2)3 x3 3 x 2 2 3 x 22 23 x3 6 x 2 12 x 8
66. ( x 3 y )(2 x y ) 2 x 2 xy 6 xy 3 y 2
2 x 2 7 xy 3 y 2
2
88. ( x 1)3 x3 3 x 2 1 3 x 12 13 x3 3 x 2 3 x 1
67. ( x 7)( x 7) x 2 7 2 x 2 49 2
89. (2 x 1)3 (2 x)3 3(2 x) 2 (1) 3(2 x) 12 13
2
68. ( x 1)( x 1) x 1 x 1
8 x3 12 x 2 6 x 1
69. (2 x 3)(2 x 3) (2 x) 2 32 4 x 2 9
90. (3x 2)3 (3x)3 3(3 x) 2 (2) 3(3 x) 22 23 27 x3 54 x 2 36 x 8
70. (3x 2)(3x 2) (3x) 2 22 9 x 2 4 71. ( x 4) 2 x 2 2 x 4 42 x 2 8 x 16 2
2
2
2
72. ( x 5) x 2 x 5 5 x 10 x 25
4 x 2 11x 23 91. x 2 4 x3 3 x 2
11x 2
11x 22 x 23 x 1 23 x 46 45
75. (3x 4)(3x 4) (3x) 2 42 9 x 2 16 76. (5 x 3)(5 x 3) (5 x) 2 32 25 x 2 9 Check:
77. (2 x 3) 2 (2 x) 2 2(2 x)(3) 32
( x 2)(4 x 2 11x 23) ( 45)
4 x 2 12 x 9
78. (3x 4) (3 x) 2(3 x)(4) 4
x
2
74. ( x 5) 2 x 2 2 x 5 52 x 2 10 x 25
2
x 1
4 x3 8 x 2
73. ( x 4) 2 x 2 2 x 4 42 x 2 8 x 16
2
2
4 x 2 12 xy 9 y 2
65. ( 2 x 3 y )(3 x 2 y ) 6 x 2 4 xy 9 xy 6 y 2 6 x 2 13 xy 6 y 2
2
4 x3 11x 2 23 x 8 x 2 22 x 46 45
2
4 x3 3x 2 x 1
2
9 x 24 x 16
79. ( x y )( x y ) ( x) 2 y x 2 y 2 2
The quotient is 4 x 2 11x 23 ; the remainder is –45.
80. ( x 3 y )( x 3 y ) ( x) 2 3 y x 2 9 y 2 2
81. (3x y )(3x y ) (3x) 2 y 9 x 2 y 2 2
82. (3x 4 y )(3x 4 y ) (3x) 2 4 y 9 x 2 16 y 2 2
19 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
5 x 2 13 95. x 2 2 5 x 4 0 x3 3 x 2 x 1
3 x 2 7 x 15 92. x 2 3 x3 x 2 3
3x 6 x
x2
2
5x4
7 x2
10 x 2
x
13 x 2 x 1
7 x 2 14 x
13 x 2
x 27
15 x 2 15 x 30 32
Check:
x 2 5x 13 x 27 2
( x 2)(3 x 2 7 x 15) ( 32) 3
2
5 x 4 3x 2 x 1 The quotient is 5 x 2 13 ; the remainder is x 27 .
2
3x 7 x 15 x 6 x 14 x 30 32 3x3 x 2 x 2
The quotient is 3x 2 7 x 15 ; the remainder is –32.
5 x 2 11 96. x 2 5 x 4 0 x3 x 2 x 2 2
4x 3 93. x
2
5 x 4 10 x 2 13 x 2 26 x 27
Check:
2
5x4
4 x3 3x 2 x 1
10 x 2 11x 2 x 2
4 x3
11x 2
3x 2 x 1 3x 2
x 2 5x 11 x 20 2
Check: 2
3
2
5 x 4 10 x 2 11x 2 22 x 20
2
( x )(4 x 3) ( x 1) 4 x 3 x x 1
5x4 x2 x 2 The quotient is 5 x 2 11 ; the remainder is x 20 .
The quotient is 4 x 3 ; the remainder is x 1 .
3x 1 2
94. x 3 x3 x 2 x 2
2 x2 97. 2 x3 1 4 x5 0 x 4 0 x3 3 x 2 x 1
3
x2 x 2 x
22 x 20
Check:
x 1
3x
26
4 x5
2
2 x2 x2 x 1
x 2
Check:
2 x 1 2 x x x 1 3
Check:
( x 2 )(3 x 1) ( x 2) 3 x3 x 2 x 2 The quotient is 3 x 1 ; the remainder is x 2 .
2
2
4 x5 2 x 2 x 2 x 1 4 x5 3 x 2 x 1 The quotient is 2x 2 ; the remainder is x2 x 1 . 20
Copyright © 2025 Pearson Education, Inc.
Section R.4: Polynomials
x2 98. 3 x3 1 3 x5 0 x 4 0 x3 x 2 x 2 3x5 Check:
3x 1 x x 2 3x x x 2 2
3x x 1 x 23 x 19 169 x 179 2
2
3x 4 x3 x 2 2 x3 23 x 2 23 x
x2 x2
3
Check:
5
2
The quotient is x 2 ; the remainder is x 2 .
x 2 2 x 12
13 x 2 19 x 19 16 x 17 9 9 3x 4 x3 x 2 The quotient is x 2 2 x 1 ; the remainder is 3 9
16 17 x . 9 9 4 x 2 3x 3
99. 2 x 2 x 1 2 x 4 3 x3 0 x 2 x 1
101. x 1 4 x3 x 2 0 x 4
2 x 4 x3 x 2 4 x3 x 2 x
4 x3 4 x 2 3x 2
4 x3 2 x 2 2 x x 2 3x 1 1 1 x2 x 2 2 5 1 x 2 2 Check:
2x x 1 x 2x 1 5 x 1 2 2 2 4 3 2 3 2 1 2x 4x x x 2x x 2 2 x 2x 1 5 x 1 2 2 2 2
4
2
3
2 x 3x x 1 The quotient is x 2 2 x 12 ; the remainder is 5x 1 . 2 2
3x 2 3x 3x 4 3 x 3 7 Check:
( x 1)( 4 x 2 3 x 3) ( 7) 4 x3 3x 2 3 x 4 x 2 3 x 3 7 4 x3 x 2 4 The quotient is 4 x 2 3 x 3 ; the remainder is –7.
3 x3 3 x 2 3 x 5 102. x 1 3 x 4 0 x3 0 x 2 2 x 1
x2 2 x 1 3 9 2 4 3 100. 3 x x 1 3 x x 0 x 2 x 2
3 x 4 3 x3
3 x 4 x3 x 2 2 x3 x 2 x 2 x3 2 x 2 2 x 3 3 1 x2 5 x 2 3 3 2 1 1 x x1 3 9 9 16 x 17 9 9 21 Copyright © 2025 Pearson Education, Inc.
3 x3 3x3 3x 2
3x 2 2 x 3x 2 3x 5x 1 5 x 5 6
Chapter R: Review
Check:
Check:
( x 1)( 3 x3 3 x 2 3 x 5) ( 6)
( x 2 x 1)( x 2 x 1) ( 2 x 2)
3 x 4 3 x3 3 x 2 5 x 3x3 3x 2 3x 5 6
x 4 x3 x 2 x3 x 2 x x 2 x 1 2x 2 x4 x2 1 The quotient is x 2 x 1 ; the remainder is 2x 2 .
3 x 4 2 x 1 The quotient is 3 x3 3 x 2 3 x 5 ; the remainder is –6.
x 2 ax a 2
2
x x 1 2
4
3
2
4
3
2
105. x a x 3 0 x 2 0 x a3
103. x x 1 x 0 x x 0 x 1
x x x
x 3 ax 2 ax 2
x3 2 x 2
ax 2 a 2 x a 2 x a3
x3 x 2 x
a 2 x a3
x2 x 1
0
x2 x 1 2x 2
Check:
( x a)( x 2 ax a 2 ) 0
Check: 2
x3 ax 2 a 2 x ax 2 a 2 x a3
2
( x x 1)( x x 1) 2 x 2 4
3
2
3
2
x3 a3 The quotient is x 2 ax a 2 ; the remainder is 0.
2
x x x x x xx x 1 2x 2
x 4 ax3 a 2 x 2 a3 x a 4
x4 x2 1
106. x a x5 0 x 4 0 x3 0 x 2 0 x a 5
The quotient is x 2 x 1 ; the remainder is 2x 2 .
x5 ax 4 ax 4
2
x x 1 2
4
3
2
4
3
2
ax 4 a 2 x3
104. x x 1 x 0 x x 0 x 1 x x x
a 2 x3 a 2 x3 a 3 x 2
x3 2 x 2 3
a3 x 2
2
x x x
a3 x 2 a 4 x
x2 x 1
a 4 x a5
x2 x 1
a 4 x a5
2x 2
0
22
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Section R.5: Factoring Polynomials
of p2 x , the new polynomial will have degree
Check: 4
3
2 2
4
2 3
3
4
( x a)( x ax a x a x a ) 0 5
3 2
4
x ax a x a x a x ax a 2 x3 a 3 x 2 a 4 x a 5
x5 a 5 The quotient is x 4 ax 3 a 2 x 2 a3 x a 4 ; the remainder is 0.
107.
the degree of p1 x and p2 x .
4
(3 x 2k )(4 x 3k ) 12 x 2 kx 96 12 x 2 8kx 9kx 6k 2 12 x 2 kx 96
112. Answers will vary. 113. Answers will vary.
Section R.5 1. 3x x 2 x 2
2
kx 6k kx 96
2. prime
6k 2 96 k 2 16
3. c
k 4
4. b
108. The products ( x y )( x y ) and ( z w)( z w) will each result in a binomial that is the difference of squares. The product of those resulting binomials will have 4 terms. 109. When we multiply polynomials p1 x and p2 x , each term of p1 x will be multiplied
by each term of p2 x . So when the highestpowered term of p1 x multiplies by the highest powered term of p2 x , the exponents on the variables in those terms will add according to the basic rules of exponents. Therefore, the highest powered term of the product polynomial will have degree equal to the sum of the degrees of p1 x and p2 x . 110. When we add two polynomials p1 x and p2 x , where the degree of p1 x the degree
of p2 x , each term of p1 x will be added to each term of p2 x . Since only the terms with equal degrees will combine via addition, the degree of the sum polynomial will be the degree of the highest powered term overall, that is, the degree of the polynomial that had the higher degree. 111. When we add two polynomials p1 x and p2 x , where the degree of p1 x = the degree
5. d 6. c 7. True; x 2 4 is prime over the set of real numbers.
8. False; 3x3 2 x 2 6 x 4 3 x 2 x 2 2
9. 3x 6 3( x 2) 10. 7 x 14 7( x 2) 11. ax 2 a a ( x 2 1) 12. ax a a ( x 1) 13. x3 x 2 x x( x 2 x 1) 14. x3 x 2 x x( x 2 x 1) 15. 2 x 2 2 x 2 x( x 1) 16. 3x 2 3 x 3x( x 1) 17. 3x 2 y 6 xy 2 12 xy 3 xy ( x 2 y 4) 18. 60 x 2 y 48 xy 2 72 x3 y 12 xy (5 x 4 y 6 x 2 ) 19. x 2 1 x 2 12 ( x 1)( x 1)
23 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
41. 8 x3 27 (2 x)3 33
20. x 2 4 x 2 22 ( x 2)( x 2) 2
2
(2 x 3)(4 x 2 6 x 9)
2
21. 4 x 1 (2 x) 1 (2 x 1)(2 x 1)
42. 64 27 x 3 43 (3x)3
22. 9 x 2 1 (3x) 2 1 2 (3 x 1)(3 x 1)
(4 3x)(16 12 x 9 x 2 )
3 x 4 9 x 2 12 x 16
23. x 2 16 x 2 42 ( x 4)( x 4) 24. x 2 25 x 2 52 ( x 5)( x 5)
43. x 2 5 x 6 ( x 2)( x 3)
25. 25 x 2 4 (5 x 2)(5 x 2)
44. x 2 6 x 8 ( x 2)( x 4)
26. 36 x 2 9 9 4 x 2 1 9(2 x 1)(2 x 1)
45. x 2 7 x 6 ( x 6)( x 1)
27. x 2 2 x 1 ( x 1) 2
46. x 2 9 x 8 ( x 8)( x 1)
28. x 2 4 x 4 ( x 2) 2
47. x 2 7 x 10 ( x 2)( x 5)
29. x 2 4 x 4 ( x 2) 2
48. x 2 11x 10 ( x 10)( x 1)
30. x 2 2 x 1 ( x 1) 2
49. x 2 10 x 16 ( x 2)( x 8)
31. x 2 10 x 25 ( x 5) 2
50. x 2 17 x 16 ( x 16)( x 1)
32. x 2 10 x 25 ( x 5) 2
51. x 2 7 x 8 ( x 1)( x 8)
33. 4 x 2 4 x 1 (2 x 1) 2
52. x 2 2 x 8 ( x 2)( x 4)
34. 9 x 2 6 x 1 (3 x 1) 2
53. x 2 7 x 8 ( x 8)( x 1)
35. 16 x 2 8 x 1 (4 x 1) 2
54. x 2 2 x 8 ( x 4)( x 2)
36. 25 x 2 10 x 1 (5 x 1) 2
55. 2 x 2 4 x 3x 6 2 x( x 2) 3( x 2) ( x 2)(2 x 3)
37. x3 27 x3 33 ( x 3)( x 2 3 x 9) 3
3
3
56. 3x 2 3 x 2 x 2 3 x( x 1) 2( x 1) ( x 1)(3 x 2)
2
38. x 125 x 5 ( x 5)( x 5 x 25)
57. 5 x 2 15 x x 3 5 x( x 3) 1( x 3) ( x 3)(5 x 1)
39. x3 27 x3 33 ( x 3)( x 2 3 x 9) 40. 27 8 x3 33 (2 x)3 (3 2 x)(9 6 x 4 x 2 )
2 x 3 4 x 2 6 x 9
58. 3x 2 6 x x 2 3 x( x 2) 1( x 2) ( x 2)(3x 1)
24
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Section R.5: Factoring Polynomials
59. 6 x 2 21x 8 x 28 3 x(2 x 7) 4(2 x 7) (2 x 7)(3x 4) 60. 9 x 2 6 x 3 x 2 3x 3 x 2 1 3x 2 3 x 2 3x 1
76. Since b is -4 then we need half of -4 squared to be the last term in our trinomial. Thus 1 (4) 2; (2) 2 4 2 x 2 4 x 4 ( x 2) 2
77. Since b is 12 then we need half of 12 squared
61. 3 x 2 4 x 1 (3 x 1)( x 1)
to be the last term in our trinomial. Thus 1 ( 12 ) 14 ; ( 14 ) 2 161 2
62. 2 x 2 3 x 1 (2 x 1)( x 1)
x 2 12 x 161 ( x 14 )2
63. 2 z 2 9 z 7 (2 z 7)( z 1)
78. Since b is 13 then we need half of 13 squared to
64. 6 z 2 5 z 1 (3 z 1)(2 z 1)
be the last term in our trinomial. Thus 1 1 1 ( ) 16 ; ( 16 ) 2 36 2 3
65. 5 x 2 6 x 8 (5 x 4)( x 2)
1 x 2 13 x 36 ( x 16 ) 2
66. 3 x 2 10 x 8 (3 x 4)( x 2)
79. x 2 36 ( x 6)( x 6)
67. 5 x 2 6 x 8 (5 x 4)( x 2)
80. x 2 9 ( x 3)( x 3)
68. 3 x 2 10 x 8 (3 x 4)( x 2)
81. 2 8 x 2 2(1 4 x 2 ) 2 1 2 x 1 2 x
69. 5 x 2 22 x 8 (5 x 2)( x 4)
82. 3 27 x 2 3(1 9 x 2 ) 3 1 3x 1 3x
70. 3 x 2 14 x 8 (3 x 2)( x 4)
83. x 2 11x 10 ( x 1)( x 10)
71. 5 x 2 18 x 8 (5 x 2)( x 4)
84. x 2 5 x 4 ( x 4)( x 1)
72. 3 x 2 10 x 8 (3 x 2)( x 4)
85. x 2 10 x 21 x 7 x 3
73. Since b is 10 then we need half of 10 squared to be the last term in our trinomial. Thus 1 (10) 5; (5) 2 25 2
86. x 2 6 x 8 ( x 2)( x 4)
2
x 10 x 25 ( x 5)
74. Since b is 14 then we need half of 14 squared to be the last term in our trinomial. Thus 1 (14) 7; (7) 2 49 2 p 2 14 p 49 ( p 7) 2
75. Since b is -6 then we need half of -6 squared to be the last term in our trinomial. Thus 1 (6) 3; (3) 2 9 2
87. 4 x 2 8 x 32 4 x 2 2 x 8
2
88. 3x 2 12 x 15 3 x 2 4 x 5
89. x 2 4 x 16 is prime over the reals because there are no factors of 16 whose sum is 4. 90. x 2 12 x 36 ( x 6) 2 91. 15 2 x x 2 ( x 2 2 x 15) ( x 5)( x 3)
y 2 6 y 9 ( y 3) 2
25 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
92. 14 6 x x 2 ( x 2 6 x 14) is prime over the integers because there are no factors of –14 whose sum is –6.
106. x8 x5 x5 ( x3 1) x5 ( x 1)( x 2 x 1) 107. 16 x 2 24 x 9 4 x 3
93. 3x 2 12 x 36 3( x 2 4 x 12) 3( x 6)( x 2)
108. 9 x 2 24 x 16 3 x 4
110. 5 11x 16 x 2 (16 x 2 11x 5) (16 x 5)( x 1)
95. y 4 11y 3 30 y 2 y 2 ( y 2 11y 30)
y 2 ( y 5)( y 6)
111. 4 y 2 16 y 15 (2 y 5)(2 y 3)
96. 3 y 3 18 y 2 48 y 3 y ( y 2 6 y 16) 3 y ( y 2)( y 8)
112. 9 y 2 9 y 4 (3 y 4)(3 y 1)
97. 4 x 2 12 x 9 (2 x 3) 2
113. 1 8 x 2 9 x 4 (9 x 4 8 x 2 1) (9 x 2 1)( x 2 1)
98. 9 x 2 12 x 4 (3 x 2) 2
(3x 1)(3x 1)( x 2 1)
99. 6 x 2 8 x 2 2 3x 2 4 x 1
114. 4 14 x 2 8 x 4 2(4 x 4 7 x 2 2)
2 3x 1 x 1
2(4 x 2 1)( x 2 2)
2(2 x 1)(2 x 1)( x 2 2)
100. 8 x 2 6 x 2 2 4 x 2 3x 1
2 4 x 1 x 1
115. x( x 3) 6( x 3) ( x 3)( x 6)
9 ( x 9)( x 9) 2
2
2
116. 5(3 x 7) x(3 x 7) (3x 7)( x 5)
2
117. ( x 2) 2 5( x 2) ( x 2) ( x 2) 5
( x 3)( x 3)( x 2 9)
( x 2)( x 3)
1 ( x 1)( x 1)
102. x 4 1 x 2
2
2
2
109. 5 16 x 16 x 2 (16 x 2 16 x 5) (4 x 5)(4 x 1)
94. x3 8 x 2 20 x x ( x 2 8 x 20) x( x 10)( x 2)
101. x 4 81 x 2
2
2
2
118. ( x 1) 2 2( x 1) ( x 1) ( x 1) 2
( x 1)( x 1)( x 2 1)
( x 1)( x 3)
103. x 6 2 x3 1 ( x3 1) 2 ( x 1)( x x 1) 2
119.
2
( x 1) 2 ( x 2 x 1) 2
2 3 x 2 3 3 x 2 3 3x 2 9
3x 5 9 x 3x 7
104. x 6 2 x3 1 ( x3 1) 2 ( x 1)( x 2 x 1)
3x 2 3 27 3 3 x 2 33 3 x 5 9 x 2 12 x 4 9 x 6 9
2
2
( x 1) 2 ( x 2 x 1) 2
105. x 7 x5 x5 ( x 2 1) x5 ( x 1)( x 1) 26
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Section R.5: Factoring Polynomials
120.
5 x 13 1 3 5 x 1 13
127. 2 3 x 4 2 x 3 2 3 x 4 3 2
2 3x 4 3 x 4 2 x 3 3 2 3x 4 3 x 4 6 x 9
2 5 x 1 1 5 x 1 1 5 x 1 1
5 x 25 x 15 x 3
2 3x 4 9 x 13
5 x 25 x 2 10 x 1 5 x 1 1
128. 5 2 x 1 5 x 6 2 2 x 1 2 2
2
2 x 1 5 2 x 1 5 x 6 4
121. 3 x 2 10 x 25 4 x 5
2 x 110 x 5 20 x 24
3 x 5 4 x 5 2
2 x 1 30 x 19
x 5 3 x 5 4
129. 2 x 2 x 5 x 2 2 2 x 2 x 5 x
x 5 3 x 15 4
2x 2x 5 x
x 5 3 x 11
122.
2 x 3x 5
7 x 6 x 9 5 x 3 2
130. 3x 2 8 x 3 x3 8 x 2 3 8 x 3 8 x
7 x 3 5 x 3 2
x 2 24 x 9 8 x
x 3 7 x 3 5
x 2 32 x 9
x 3 7 x 21 5 x 3 7 x 16
131. 2 x 3 x 2 x 3 3 x 2 3
123. x3 2 x 2 x 2 x 2 ( x 2) 1 x 2
2
x 3 x 2 2 x 4 3x 9 2
( x 2)( x 1)( x 1)
x 3 x 2 5 x 5 2
124. x3 3 x 2 x 3 x 2 ( x 3) 1 x 3 ( x 3)( x 2 1) ( x 3)( x 1)( x 1)
5 x 3 x 2 x 1 2
132. 4 x 5 x 1 x 5 2 x 1 3
2
4
2 x 5 x 1 2 x 1 x 5 3
125. x 4 x3 x 1 x3 ( x 1) 1 x 1 ( x 1)( x3 1)
2 x 5 x 1 2 x 2 x 5
( x 1)( x 1)( x 2 x 1)
2 x 5 x 1 3 x 3
3 3
2 3 x 5 x 1 x 1 3
126. x x x 1 x ( x 1) 1 x 1 3
2
x 3 x 2 2 x 2 x 3 3
( x 2)( x 2 1)
4
2
3
6 x 5 x 1 x 1 3
( x 1)( x3 1) ( x 1)( x 1)( x 2 x 1) ( x 1) 2 ( x 2 x 1)
133.
4 x 32 x 2 4 x 3 4 4 x 3 4 x 3 8 x 4 x 3 4 x 3 8 x 4 x 312 x 3 3 4 x 3 4 x 1
27 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
Section R.6
134. 3x 2 3x 4 x3 2 3x 4 3 2
3x 2 3x 4 3x 4 2 x
1. quotient; divisor; remainder
3x 3x 4 3 x 4 2 x
2. 3 2 0 5 1
2
3x 2 3x 4 5 x 4
3. d
135. 2 3 x 5 3 2 x 1 3x 5 3 2 x 1 2 3
2
2
4. a
6 3 x 5 2 x 1 2 x 1 3 x 5 2
5. True
6 3 x 5 2 x 1 2 x 1 3 x 5 2
6. True
6 3 x 5 2 x 1 5 x 4 2
7. 2 1 7
136. 3 4 x 5 4 5 x 1 4 x 5 2 5 x 1 5 2
2
3
5
10
2 10 10
2 4 x 5 5 x 1 6 5 x 1 5 4 x 5 2
1 5 5 0
Quotient: x 2 5 x 5 Remainder: 0
2 4 x 5 5 x 1 30 x 6 20 x 25 2
2 4 x 5 5 x 1 50 x 31 2
137. x 4 x 2 x 4 (1 x 2 ) 138. 2 x 5 6 x 4 8 x 3 2 x 5 (1 3 x 4 x 2 )
8. 1 1
2 3 1 1 1 4
1
1 4 5
Quotient: x 2 x 4 Remainder: 5
139. x 2 ( x 1) x 1 ( x 1) x 2 [( x 1) x( x 1)] x 2 ( x 1 x 2 x)
2 1 9 33
9. 3 3
x 2 ( x 2 1)
140.
3 96
3 11 32
Quotient: 3x 11x 32 Remainder: 99
x( x 3) 1 4 x 2 ( x 3) 2 x( x 3) 2 [( x 3) 4 x] x( x 3) 2 (5 x 3)
10. 2 4
1
2
1
8 20 42
141. The possible factorizations are x 1 x 4 x2 5 x 4 or
4 10 21 43
Quotient: 4 x 2 10 x 21 Remainder: 43
x 2 x 2 x 2 4 x 4 , none of which equals x 2 4 .
11. 3 1
142. The possibile factorizations are
x 1 x 2 x 1 , neither of which equals 2
99 2
2
0 4 0 1 0 3 9 15 45 138
1 3
x2 x 1 .
5 15 4
46 138
Quotient: x 3x 5 x 2 15 x 46 Remainder: 138
143. Answers will vary. 144. Answers will vary.
28
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3
Section R.6: Synthetic Division
12. 2 1 0 1 0 2 2 4 10 20 1 2 5 10 22 3
4 5 2 8 Remainder = 8 ≠ 0. Therefore, x 2 is not a factor of 4 x3 3 x 2 8 x 4 .
2
Quotient: x 2 x 5 x 10 Remainder: 22 13. 1 4 4
0 3 0 4 4 1
1 0 5 1 2 2
4
2
2
4
3
1
1 5
20. 3 4
7 2
Quotient: 4 x 4 x x x 2 x 2 Remainder: 7
14. 1 1
1 1 6 6
6
1 1 6 6 6 16 4
3
2
Quotient: x x 6 x 6 x 6 Remainder: –16 15. 1.1 0.1
0
0.2
2
Quotient: 0.1x 0.11x 0.321 Remainder: –0.3531 0.2
0 0.21
24. 3 2
0
Quotient: x 2 x 4 x 2 8 x 16 Remainder: 0 18. 1 1
0 0
3
0 0
0 0 43 0 0 24 10 20 40 6 12 24
0 18 0 1 0 9 6 18 0 0 3 9
2 6 0 0 1 3 0 Remainder = 0. Therefore, x 2 is a factor of 2 x 6 18 x 4 x 2 9 .
1
1 1 1 1 1 1 1 1 1 1 4
3
4
3 10 20 3 6 12 0 Remainder = 0. Therefore, x 3 is a factor of 5 x 6 43 x3 24 .
17. 2 1 0 0 0 0 32 2 4 8 16 32 4
16 2
4 8 1 2 0 Remainder = 0. Therefore, x 2 is a factor of 4 x 4 15 x 2 4 .
0.441
1 2 4 8 16
0 15 0 4 8
23. 2 5
0.1 0.21 0.241 Quotient: 0.1x 0.21 Remainder: 0.241
0 7 21 0 0 21
2 0 0 7 0 Remainder = 0. Therefore, x 3 is a factor of 2 x 4 6 x3 7 x 21 .
0.3531
0.1 0.11 0.321 0.3531
16. 2.1 0.1
4 17 51 161 Remainder = 161 ≠ 0. Therefore, x 3 is not a factor of 4 x3 5 x 2 8 .
22. 2 4
0
0.11 0.121
5 0 8 12 51 153
21. 3 2 6 6
0 0 10
0 5
2 4 3 8 4 8 10 4
19.
0 2
Quotient: x x x x 1 Remainder: 0
29 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
25. 4 1
Yes, x y is a factor of
0 16 1 0 19 4 16 0 4 16
x 4 3 x3 y 3 x 2 y 2 xy 3 4 y 4 .
1 4 0 1 4 3 Remainder = 1 ≠ 0. Therefore, x 4 is not a factor of x5 16 x3 x 2 19 .
26. 4 1
32. Answers will vary.
Section R.7
0 16 0 1 0 16 4 16 0 0 4 16
1. lowest terms
1 4 0 0 1 4 0 Remainder = 0. Therefore, x 4 is a factor x 6 16 x 4 x 2 16 .
27.
2. Least Common Multiple 3. d 4. a
1 3 1 0 6 2 3 1 0 0 2 3
0 0 6
1 1 x3 5. True; 3 x 3x 1 1 x 5 5 x 5x 5( x 3) x 3 5x 3x x 5 3( x 5)
0
Remainder = 0; therefore x 13 is a factor of 3x 4 x3 6 x 2 .
28.
1 3 3
1 0 3 1 1 0
0 1
0 0
3 2
3
6. False; 2 x3 6 x 2 2 x 2 x 3 6 x 4 4 x3 2 x3 3x 2
Remainder = 2 0 ; therefore x 13 is not a 4
LCM 2 x3 x 3 3x 2
3
factor of 3x x 3 x 1 . 29. 2 1 2 2
3
5
x3 2 x 2 3x 5 17 x 2 4 x 11 x2 x2 a b c d 1 4 11 17 9
1
3h
2h 2h 2
1
8.
4 x 2 8 x 4 x( x 2) x 12 x 24 12( x 2) 3
9.
x 2 2 x x( x 2) x 3 x 6 3( x 2) 3
10.
15 x 2 24 x 3 x(5 x 8) 5 x 8 x 3x 2 3x 2
11.
24 x 2 24 x 2 4x 2 12 x 6 x 6 x(2 x 1) 2 x 1
12.
x 2 4 x 4 ( x 2) x 2 x 2 ( x 2)( x 2) x 2 x2 4
h3
h
3h
h 2 h3
3
h
h2
0
x3 3x 2 hx h 2 is the quotient and 0 is the remainder.
31. y 1
3( x 3) 3x 9 3 x 2 9 ( x 3)( x 3) x 3
8 22
1 4 11 17
30. h 1
7.
3y
3y2
y3
4 y4
y
2 y2
5 y3
4 y3
2y
5 y2
4 y3
0
30
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Section R.7: Rational Expressions
13.
y 7 y 7 y 2 49 2 3 y 18 y 21 3 y 2 6 y 7
22.
y 7 y 7 3 y 7 y 1
y7 3 y 1
14.
3 y 2 y 2 3 y 2 y 1 y 1 3 y 2 5 y 2 3 y 2 y 1 y 1
15.
x 2 4 x 12 ( x 6)( x 2) x 6 x 2 4 x 4 ( x 2)( x 2) x 2
16.
17.
23.
x ( x 1) x xx x ( 2)( 1) 2 2 x x x x x x2 2
24.
x 2 x 20 x 5 x 4 4 x 2 x x 5 x 4 1 x 4
25.
2 x 2 5 x 3 (2 x 1)( x 3) ( x 3) x 3 1 2x 1(2 x 1)
3( x 2) 3x 6 x x 2 19. 2 2 ( 2)( x x 2) 5x 5x x 4 3 5 x( x 2)
20.
21.
2
3 x 3 x 3x 2 x 6 x 10 2 2(3x 5) 4(3x 5) 2
3
4x x 64 2x x 2 16
2
2 x 2 x x 4 x 2 4 x 16
2 x x 4 x 4
2 x x 2 4 x 16
2 x x 1 2 x 1
6 x2 x 1 x 2 x 1
4 x 2 2 3 x 1 x 2
8 3x
x 2 4 x 12 x 2 4 x 32 x 2 2 x 48 x 2 10 x 16 x 6 x 2 x 8 x 4 x 8 x 6 x 8 x 2 x4 x 8
x2 x 6 x 2 25 x 2 4 x 5 x 2 2 x 15 x 2 x 3 x 5 x 5 x 5 x 1 x 5 x 3
x 2 x 3 x 5 x 5 x 1 x 3
6x 6x 2x 4 27. x 4 2 3x 9 x 4 3x 9 2x 4 2( x 2) 6x ( x 2)( x 2) 3( x 3) 4x ( x 2)( x 3) 2
x 4 x 4 x 16 4 x2 ( x 4)( x 4) 2x
3 2x 9 6 x 27 2 2 3 x 5x 4 x 18 5x 5 2 2x 9
26.
2 6 x 1 x 2 x 1
4 x 2 4 x 8 12 12 3x 12 6 x 3 x 62 x
2
x 5
18.
( x 1)( x 2 x 1) 12 x3 1 12 2(2 x 1) x 2 x 4 x 2 x( x 1)
x4
31 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
12 x 2 5 x 20 12 x x 16 28. 5 x 20 4 x 2 4 x2 2 x 16 12 x ( x 4)( x 4) 5( x 4) 4 x2 3( x 4) 5x
29.
8x x 1 8x x 1 10 x x 2 1 10 x x 1 8x x 1 x 1 x 1 10 x
31.
33.
x2 x2 12 x 4x 4 x x2 4 x 4 x2 4 x 4 12 x x2 12 x 4 x x 2 x 2
9 x3 3 x x 3 9 x3 x 3
3 x2
34.
2
9 x3
x 32
x 2 7 x 12 2 2 x 2 7 x 12 x 7 x 12 x x 12 x 2 x 12 x 2 7 x 12 x 2 x 12 x 2 x 12 ( x 3)( x 4) ( x 4)( x 3) ( x 3)( x 4) ( x 4)( x 3)
4 x 2 4 x 4 x x 16 4x 4 x 4x x 2 16 4 x x 4 x 4 4 x 4x 4 x x 4 4x
3 x 9 x3 3 x x 3 x 3
4 5 x 1
2
30.
32.
3 x 3 3 x 3 x 9x x2 9 3 x x2 9 9 x3
( x 3) 2 ( x 3) 2
x2 7 x 6 2 2 x2 x 6 x 7 x 6 x 5x 6 x2 5x 6 x2 x 6 x2 5x 6 2 x 5x 6 ( x 6)( x 1) ( x 2)( x 3) ( x 3)( x 2) ( x 6)( x 1) ( x 1)( x 2) ( x 2)( x 1)
5x2 7 x 6 2 5 x 2 7 x 6 2 x 2 13 x 20 35. 2 x2 3x 5 2 15 x 14 x 3 2 x 3 x 5 15 x 2 14 x 3 2 x 2 13 x 20 (5 x 3)( x 2) (2 x 5)( x 4) ( x 1)(2 x 5) (5 x 3)(3 x 1) ( x 2)( x 4) ( x 1)(3x 1)
x 4 2 4x
32
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Section R.7: Rational Expressions
9 x 2 3x 2 2 9 x 2 3 x 2 8 x 2 10 x 3 36. 12 x2 5 x 2 9 x 6 x 1 12 x 2 5 x 2 9 x 2 6 x 1 8 x 2 10 x 3 (3 x 2)(3x 1) (4 x 1)(2 x 3) (3x 2)(4 x 1) (3x 1)(3x 1) (4 x 1)(2 x 3) (4 x 1)(3x 1)
47.
7( x 1) 3( x 3) 7 3 x 3 x 1 ( x 3)( x 1) ( x 1)( x 3) 7 x 7 3x 9 ( x 1)( x 3) 4 x 16 ( x 1)( x 3) 4( x 4) ( x 1)( x 3)
2( x 5) 5( x 5) 2 5 x 5 x 5 ( x 5)( x 5) ( x 5)( x 5) 2 x 10 5 x 25 ( x 5)( x 5) 3 x 35 ( x 5)( x 5) 3x 35 ( x 5)( x 5)
37.
x 5 x5 2 2 2
38.
3 6 3 6 3 3 x x x x x
39.
4 x2 x 2 4 x 2 x 2 2x 3 2x 3 2x 3 2x 3
40.
3x2 9 3x2 9 3 x 3 2x 1 2x 1 2x 1 2x 1
41.
x 5 3x 2 x 5 3x 2 4 x 3 x4 x4 x4 x4
x2 x 2 x2 x 3 ( x 1)( x 1)
2 x 5 x 4 2 x 5 x 4 3x 1 3x 2 3x 2 3x 2 3x 2
3x2 2 x 3 ( x 1)( x 1)
42. 43.
44.
48.
2
3x 5 2 x 4 (3x 5) (2 x 4) 2x 1 2x 1 2x 1 3x 5 2 x 4 2x 1 x9 2x 1 5 x 4 x 1 (5 x 4) ( x 1) 3x 4 3x 4 3x 4 5x 4 x 1 3x 4 4x 5 3x 4
45.
4 x 4 x 4 x x2 2 x x2 x2 x2
46.
6 x 6 x x6 x 1 1 x x 1 x 1 x 1
49.
50.
51.
x( x 1) (2 x 3)( x 1) x 2x 3 x 1 x 1 ( x 1)( x 1) ( x 1)( x 1)
3 x( x 3) 2 x( x 4) 3x 2x x 4 x 3 ( x 4)( x 3) ( x 4)( x 3)
3x 2 9 x 2 x 2 8 x ( x 4)( x 3)
5x2 x ( x 4)( x 3)
x 5 x 1 ( x 4)( x 3)
x 3 x 4 ( x 3)( x 2) ( x 4)( x 2) x 2 x 2 ( x 2)( x 2) ( x 2)( x 2)
33 Copyright © 2025 Pearson Education, Inc.
x 2 5 x 6 ( x 2 6 x 8) ( x 2)( x 2)
x2 5x 6 x2 6 x 8 ( x 2)( x 2) (11x 2) 11x 2 or ( x 2)( x 2) ( x 2)( x 2)
Chapter R: Review
52.
2 x 3 2 x 1 (2 x 3)( x 1) (2 x 1)( x 1) x 1 x 1 ( x 1)( x 1) ( x 1)( x 1)
x 2 x 1 2 x 1
2 x 2 x 3 (2 x 2 x 1) ( x 1)( x 1) 2
2 x x x 2 x 1 3
53.
x x2 4
60. x 3 x 2 3 x x x 3
x 2 x x x x 2 x 1 x x 1 x 1 x 1 x x 1 Therefore, LCM x x 1 x 1 x x 1 .
61. x3 x x x 2 1 x x 1 x 1
3
2
2
x x x 1 x
4
3
Therefore, LCM x 2 x 2 . 3
x x x x 1
x3 x 2 1
63.
x 2 x 2 x 1 x 2
Therefore, LCM x 2 x 2 x 1 . x 2 x 12 x 3 x 4
x x 2 x 7 x 6 x 2 x 24 x x ( x 6)( x 1) ( x 6)( x 4) x( x 4) x( x 1) ( x 6)( x 1)( x 4) ( x 6)( x 4)( x 1) 2
x 2 8 x 16 x 4 x 4
Therefore, LCM x 3 x 4 . 2
64.
57. x3 x x x 2 1 x x 1 x 1 x x x x 1 2
Therefore, LCM x x 1 x 1 .
2
x2 4 x x2 x 5x ( x 6)( x 4)( x 1) ( x 6)( x 4)( x 1)
x x 1 x 3 x 2 5 x 24 x x 1 ( x 3) ( x 3)( x 8) x( x 8) x 1 ( x 3)( x 8) ( x 3)( x 8)
58. 3x 27 3 x 9 3 x 3 x 3 2
2
x 2 3
2
2
x3 2 x 2 x 2 x 2
4
55. x 2 4 x 2 x 2
56.
2
62. x 2 4 x 4 x 2
x3 x 2 1 4
2
2
2
x 1 x 1 x x 1 x 2 3 x x 1 x3 x 2 1 3
2
3
x x 2 x 2
x3 9 x x x 2 9 x x 3 x 3
Therefore, LCM x x 3 x 3 .
2 x2 2
54.
2
2 x2 4
2
Therefore, LCM x3 2 x 1 .
x 1 x2 x2 4 x 4 x x x2 4 2
2
x3
2
2x x 3 2x x 1 ( x 1)( x 1) 2 ( x 1)( x 1) 2 ( x 1)( x 1)
59. 4 x3 4 x 2 x x 4 x 2 4 x 1
x2 8x x 1 x2 7 x 1 ( x 3)( x 8) ( x 3)( x 8)
2 x x 15 2 x 5 x 3 2
Therefore, LCM 3 2 x 5 x 3 x 3 . 34
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Section R.7: Rational Expressions
65.
66.
67.
4x 2 x2 4 x2 x 6 4x 2 ( x 2)( x 2) ( x 3)( x 2) 4 x( x 3) 2( x 2) ( x 2)( x 2)( x 3) ( x 3)( x 2)( x 2)
2x 3 x4 x2 x 2 x2 2 x 8 2x 3 x4 ( x 2)( x 1) ( x 4)( x 2) ( x 4)( x 4) (2 x 3)( x 1) ( x 2)( x 1)( x 4) ( x 4)( x 2)( x 1)
4 x 2 12 x 2 x 4 ( x 2)( x 2)( x 3)
x 2 8 x 16 (2 x 2 5 x 3) ( x 2)( x 1)( x 4)
4 x 2 10 x 4 ( x 2)( x 2)( x 3)
x 2 3x 13 ( x 2)( x 1)( x 4)
2(2 x 2 5 x 2) ( x 2)( x 2)( x 3)
70.
3x x4 3x x4 x 1 x 2 2 x 1 ( x 1) ( x 1) 2
2x 3 x2 x 2 8 x 7 ( x 1) 2 2x 3 x2 ( x 1)( x 7) ( x 1) 2 (2 x 3)( x 1) ( x 2)( x 7) ( x 1)( x 7)( x 1) ( x 1) 2 ( x 7)
3 x( x 1) x4 ( x 1)( x 1) ( x 1) 2
3x 2 3x x 4 ( x 1) 2
2 x 2 x 3 ( x 2 5 x 14) ( x 1) 2 ( x 7)
3x 2 4 x 4 ( x 1) 2
x 2 6 x 11 ( x 1) 2 ( x 7)
3
2
x 1 x 1 x 1 x 1 3 x 1 2 x 1 x 12 x 12 2
68.
69.
2
2
2
5x 1
x 1 x 1 2
2
2
6
x 2 x 1 x 2 x 12 2 x 1 6 x 2 x 2 2 x 12 2
1 2 3 x x 2 x x3 x 2 1 2 3 x x x 1 x 2 x 1
3x 3 2 x 2
x 1 x 1
71.
x x 1 x 1 2 x x 1 3 x 1
x 2 x 1 x 1
x x 2 1 2 x 2 2 x 3x 3 x x 1 x 1 2
3
x x 2 x2 5x 3 x 2 x 1 x 1
x3 2 x 2 4 x 3 x 2 x 1 x 1
2 x 2 6 x 12
x 2 2 x 12 4 x 14
x 2 2 x 12 2 2 x 7 x 2 2 x 12 35 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
72.
x
x 1
2
x
x 1
2
2 x 1 x x3 x 2
4x2 1 4 x2 1 1 2 2 x x2 x2 x 76. 2 2 1 3 2 3 x 1 3 x 1 2 2 x x x2 x 4
2 x 1 x x 2 x 1
4 x2 1 x2 2 x2 3x 1 2 4x 1 2 3x 1
x3 2 x x 1 x 1 x 1 2
73.
x 2 x 1
2
x3 2 x x 2 2 x 1 x 2 1 x x 1 2
2
x 1 2x x 1 2x x 1 x x x x 77. x 1 3 x 1 x 1 3x 3 x 1 3 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 4x 2 x 2 2 x 1 x 1 x 1 x 1 2 x 2 x 1 2
x3 2 x3 4 x 2 2 x x 2 1 x 2 x 1
2
3x3 5 x 2 2 x 1 x 2 x 1
2
1( x h) 1 1 1 1 1 x h x h x h ( x h) x x( x h)
1 xxh h x( x h)
x 1 x 1 x x 1 x 1 x 1 x 1 78. x 1 2x x 1 x 1 2 x x x x x 1 x 1 x 1 x ( x 1) 2
h hx( x h) 1 x ( x h)
74.
1
1 1 1 h ( x h) 2 x 2
1( x h) 2 1 1 x2 h ( x h ) 2 x 2 x 2 ( x h) 2
x 4 x 3 79. x 2 x 1 x 1 ( x 4)( x 1) ( x 3)( x 2) ( x 2)( x 1) ( x 1)( x 2) x 1 x 2 5 x 4 ( x 2 5 x 6) ( x 2)( x 1) x 1 10 x 2 1 ( x 2)( x 1) x 1 2(5 x 1) ( x 2)( x 1) 2
1 x 2 ( x 2 2 xh h 2 ) h x 2 ( x h) 2
2 x h h2 hx 2 ( x h) 2 h( 2 x h)
hx 2 ( x h) 2 2x h 2 x ( x h) 2 2x h 2 x ( x h) 2 1 x 1 x 1 x x x x x 1 x x 1 75. 1 x 1 x 1 x x 1 x 1 1 x x x x 1
36
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Section R.7: Rational Expressions x2 x x 1 x 2 80. x3 x( x 1) ( x 2)( x 2) ( x 1)( x 2) ( x 2)( x 1) x3 x 2 4 x 4 ( x 2 x) ( x 2)( x 1) x3 5 x 4 1 ( x 2)( x 1) x 3 5 x 4 ( x 2)( x 1)( x 3)
5x 4 ( x 2)( x 1)( x 3)
x 2 x 1 2 x 1 x 81. 2x 3 x x 1 x ( x 2)( x 1) ( x 1)( x 2) ( x 2)( x 1) ( x 1)( x 2) (2 x 3)( x 1) x2 ( x 1)( x) x( x 1) x x2 x x2 ( x 2)( x 1) 2 2 x (2 x x 3) x( x 1) 2
2x 5 x x x 3 82. ( x 1) 2 x2 x 3 x3 x( x) (2 x 5)( x 3) x( x 3) x( x 3) x 2 ( x 3) ( x 3)( x 1) 2 ( x 3)( x 3) ( x 3)( x 3) 2 x 2 x 15 x 2 x( x 3) 3 2 x 3 x ( x3 x 2 5 x 3) ( x 3)( x 3) x 2 x 15 x( x 3) 2 4 x 5x 3 ( x 3)( x 3)
x 2 x 15 ( x 3)( x 3) x( x 3) 4 x 2 5 x 3
( x 2 x 15)( x 3) x(4 x 2 5 x 3)
83. 1
2
2x2 4 ( x 2)( x 1) 2 x x 3 x( x 1)
2( x 2 2) x( x 1) ( x 2)( x 1) ( x 2 x 3)
2 x( x 2 2) ( x 2)( x 2 x 3)
2 x ( x 2 2) ( x 2)( x 2 x 3)
84. 1
1 1 1 x
1 x 1 x x 1 x 1 x 1 x x 1 1 x 1 1
1 1 1 1 x
37 Copyright © 2025 Pearson Education, Inc.
1 1 1 x 1 x 1 1 x 1 x 1 x 1 x 1 1 x x x 1 x x 1 x 1
Chapter R: Review 3 x 1 2 2 2 x 1 3 x 1 3 x 1 x 1 85. 1 3 2 x 1 3 3 x 1 2 2 x 1 x 1 x 1 1
89.
2x x 1 x 1 x 1
x 2 x x 2 1 1 2
2
90.
2 3x 3 3 2 x 1 3 2 x 2
3 x 2 4 4 3 x2 x2 x2 86. 1 3 x 2 1 3 3 x 2 1 1 x2 x2 x2
91.
4 3 x 2 3 x 2
x 1 2
2
2
2
92.
2
2
2
2
2
x2 4
x 2 2 x 2 2
3x 1 2 x x 2 3 6 x 2 2 x 3x 2 3x 12 3x 12 3x2 2 x
2 x 5 3x 2 x3 2 6 x3 15 x 2 2 x3 2 x 5 2 2 x 5 2
4 x3 15 x 2
2 x 5 2 x 2 4 x 15 2 x 5 2
2 x 3 3 3x 5 2 6 x 9 6 x 10 3 x 5 2 3 x 5 2
x 1 3 3x 4 2 x 3x 3 6 x 8x x 1 x 1 2
93.
19 2
2
2
2
2
2
2
4 x 1 5 5 x 2 4 20 x 5 20 x 8 5 x 2 2 5 x 2 2
2
3x 12 x 3x 2 3x 12
4 3x 6 3 x 2
3x 5
2
2x x 4 x 4 x 4 x 4 x 4
3 x 2 3 x 2 x 1 x 1
2
x 1 x 1
4 3 x 2 x2 3 x 2 x2 4 3 x 2 x2 x2 3 x 2
88.
2
1
87.
x 1
x 2 x x 2 4 1
3x 1 2x 1 4 x 2 3
2
x2 1
2 3 x 1
2
2
2 3 x 1 x 1 3 2 x 1 x 1 2 3 x 1 x 1 x 1 3 2 x 1
2
3x 2 8 x 3
x 1 3 x 8 x 3 x 1 2
2
2
13
5 x 2 2
2
2
38
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3x 1 x 3
x 1 2
2
Section R.7: Rational Expressions
x 9 2 2 x 5 2 x 2 x 18 4 x 10 x x 9 x 9 2
94.
2
2
2
2
2
2
2 x 2 10 x 18
x 9 2 x 5 x 9 x 9 2
2
98.
A 7 12 x 1 x 2 x 3 x2 x 6 A( x 3) 7( x 2) 12 x 1 Ax 3 A 7 x 14 12 x 1 x( A 7) 3 A 14 12 x 1 so A 7 12
2
2
95.
2
A5 1 x 1 a 1, b 1, c 0 x x 1 1 x 1 1 1 1 x 1 x 1 1 x x x 1 x 2x 1 x 1 x 1 a 2, b 1, c 1
99. 1
1 1 1 (n 1) f R R 2 1 R R1 1 (n 1) 2 f R1 R2 R1 R2 (n 1) R2 R1 f f 1 R1 R2 (n 1) R2 R1
1
1
R1 R2 f (n 1) R2 R1
1
0.1(0.2) (1.5 1)(0.2 0.1) 0.02 0.02 2 meters 0.5(0.3) 0.15 15
1
1 1 x
1
f
96.
R R R1 R3 R1 R2 1 1 1 1 2 3 R R1 R2 R3 R1 R2 R3 R1 R2 R3 R R2 R3 R1 R3 R1 R2
2 x 1 x 1 3x 2 2x 1 2x 1 a 3, b 2, c 1
1
1 1
1
1 1
x 3 xk 3x 9k x x 3k 3 x 3k 97. ( x 3)( x 5) x 2 2 x 15 ( x 3)( x 5)
( x 5) x 12 x5 x 3k x 12
1 x
1 2x 1 1 x2 3 x 3 2 2x 1
3x 2 2 x 1 5 x 3 3x 2 3x 2 a 5, b 3, c 2
5 4 10 4 10 5 10 5 4 200 20 ohms 110 11
x 3 x 3k x 3k
1
1
2
1 x 1 1 x 1 2 x 2 1 x 1
If we continue this process, the values of a, b and c produce the following sequences: a :1, 2,3,5,8,13, 21,.... b :1,1, 2,3,5,8,13, 21,..... c : 0,1,1, 2,3,5,8,13, 21,..... In each case we have a Fibonacci Sequence, where the next value in the list is obtained from the sum of the previous 2 values in the list.
3k 12 k4
39 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review 100. Answers will vary.
21.
3
8 x 4 3 8 x3 x 2 x 3 x
22.
3
192 x5 3 64 3 x3 x 2 4 x 3 3 x 2
23.
4
243 4 81 3 3 4 3
24.
4
48 x5 4 16 x 4 3x 2 x 4 3x
25.
4
x12 y 8 4 x3
26.
5
x10 y 5 5 x 2
27.
4
x9 y 7 4 8 4 x y x2 y xy 3
28.
3
3 3xy 2 1 1 1 3 4 2 3 3 3 3 x 81x y 27 x 27 x
101. Answers will vary.
Section R.8 1. 9; 9 2. 4; 4 4 3. index
y x y 4
2 4
3 2
y x y 5
5
2
4. cube root 5. b 6. d 7. c 8. c 9. true 10. False; 4 3 3 3 4
11.
3
12.
4
13.
3
14.
3
3
29.
64 x 8 x
30.
9 x5 3 x 4 x 3x 2 x
31.
3
27 3 3
4
y 4
162 x9 y12 4 2 3 x x 2 4
3 4
3x 2 y3 4 2 x
16 4 24 2
32.
8 3 2 2 3
3
y y
40 x14 y10 3 5( 2)3 x 2 x 4
3
3 3
2 x 4 y 3 3 5 x 2 y
1 3 1 1 3
15.
8 42 2 2
16.
75 25 3 5 3
17.
700 100 7 10 7
18.
45 x3 9 5 x 2 x 3 x 5 x
19.
3
32 3 8 4 2 3 4
20.
3
54 3 27 2 3 3 2
33.
15 x 2 5 x 75 x 2 x 25 3 x 2 x 5 x 3 x
34.
5 x 20 x3 100 x 4 10 x 2
35.
5 9 5 9 3
2
2
3
2
5 3 92 5 3 81 5 3 3 3 15 3 3
36.
3 10 3 10 3
4
3
4
4
3 34 102 3 3 3 100 300 3 3
37.
3 6 2 2 6 12 6 4 3 12 3
40
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Section R.8: nth Roots; Rational Exponents
38.
5 8 3 3 15 24 30 6
49.
3
16 x 4 3 2 x 3 8 x3 2 x 3 2 x 2x 3 2x 3 2x
39. 3 2 4 2 3 4 2 7 2
2 x 1 3 2 x
40. 6 5 4 5 6 4 5 2 5
50.
4
32 x 4 2 x5 4 16 2 x 4 x 4 2 x 2 4 2x x 4 2x
41. 48 5 12 16 3 5 4 3
2 x 4 2 x or x 2 4 2 x
4 3 5 2 3 4 10 3
8 x3 3 50 x 4 x 2 2 x 3 25 2 x
51.
6 3
2 x 2 x 15 2 x 2 x 15 2 x
42. 2 12 3 27 2 4 3 3 9 3 4 3 9 3
52. 3x 9 y 4 25 y 9 x y 20 y
4 9 3
9 x 20 y
5 3
43.
3 3 3 1 3 3 3 3 3
53.
2
2 x 3 2 xy 3 x 3 2 xy 5 y 3 2 xy
2 3
2 x 3 x 5 y 3 2 xy
5 2 5 3 5 2 5 3 5 6 2
5 5 6
3
3
x 5 y 3 2 xy or x 5 y 3 2 xy
54. 8 xy 25 x 2 y 2 3 8 x3 y 3 8 xy 5 xy 2 xy
5 1 3
16 x 4 y 3x 3 2 xy 5 3 2 xy 4
3 8 x3 2 xy 3x 3 2 xy 5 3 y 3 2 xy
3 2 3 3
44.
3
8 5 2 xy 5 xy
3
45. 5 2 2 54 5 2 2 3 2 3
3
5 2 6 2 5 6 3 2
55.
1 1 2 2 2 2 2 2
56.
2 2 3 2 3 3 3 3 3
57.
3 3 5 15 5 5 5 5
58.
3 3 3 2 6 6 4 8 2 2 2 2 2 22
3 2
46. 9 3 24 3 81 9 2 3 3 3 3 3 18 3 3 3 3 3 18 3 3 3 15 3 3
47.
x 1 x 2 x 1 2
2
x 2 x 1
48.
x 5 x 2 x 5 5 2
2
2
x 2 5x 5
41 Copyright © 2025 Pearson Education, Inc.
Chapter R: Review
59.
3 5 2
3 5 2 25 2
3 5 2 23
67.
xh x xh x xh x xh x
x h 2 x x h x x h x
x h 2 x 2 xh x xhx
2 x h 2 x 2 xh h
or 5 3 6 23
2
7 2 74
2
7 2 or 3
14 2 2 3
2 5 2 5 23 5 23 5 23 5 23 5
3 1 3 1 2 3 3 2 3 3 2 3 3 2 3 3
5 2 1
xh xh xh xh xh xh xh xh
x h 2 x h x h x h x h x h
x h 2 x2 h2 x h xhxh
2 x 2 x 2 h2 2h
x x 2 h2 h 11 1 11 1 11 1 2 2 11 1 11 1 10 2 11 1 2 11 1
69.
62 3 3 3 3 95 3 12 9 3
63.
xh xh xh xh
68.
4 2 5 6 5 15 4 45 19 8 5 8 5 19 41 41
62.
xh x xh x
2 2 7 2 7 2 7 2 7 2
60.
61.
5 2 5 2 5 2 3
5 2 1 2 1 2 1
5 2 5 5 2 5 2 1
3 54
64.
70.
3 54 54 54
65.
66.
71.
5 5 3 4 53 4 3 2 2 32 34
5 11 1
18
3 5 43
6 6 5 43 5 43
6 15 6 15 6 15 15 15 6 15 6 15 9 90 15 3 10 15 9 3 10 5 3 10 5
2 2 3 3 2 3 3 3 9 39 33
3
5 43 5 43 5 43 25 43 3 3 5 43 3 5 43
3 5 12 3 5 12 11 5 16 3 5 12 11
42
Copyright © 2025 Pearson Education, Inc.
Section R.8: nth Roots; Rational Exponents
2 8
5 3 5 3 5 3 5 5 5 3 53 2 5 15 5 15
72.
80. 163/ 4 4 16 81. 1003/ 2
x c xc
x c x c xc x c xc x c x c
73.
x 7 1 x 8
75.
1 x c
x 2 x 2 x 2 x4 x4 x 2 x4 x 4 x 2
74.
82. 253/ 2
1 x 2
x 8 x 7 1
27 86. 8
1 x 7 1
1 4 x9
x 9
2 4
79.
2
4 2 8 3
89.
4
1
16
3
1 1 43 64
3
3
9 3 33 3 8 2 2 23 2
27 27 27 2 8 2 2 16 2 16 2 2
27 2 32 2
27 3 2 9 3 4 8 2 9 8
2 / 3
3/ 2
3
9 3 8 2 2
33
3
2
8 88. 27
1 1 23 8
3
1 163/ 2
x 25
4
3
1
3/2
x 25 4 x 9
78. 4
1
2/3
3/ 2
25 x
3
3
x 25 4
3/ 2
3
3
8 87. 9
100 10 1000
x 8
77. 82 / 3 3 8
3/ 2
x 8 x 7 1
4 x9 4 x9 4 x9 76. x 25 x 25 4 x 9 16 ( x 9) x 25 4 x 9
2
9 85. 8
3
25 5 125
84. 163/ 2
x 7 1 x 7 1 x 8 x 7 1 x 7 1
83. 43/ 2
3
2
3
27 27 8 2 2 16 2
27 2 27 2 32 16 2 2
27 8
2/3
2
27 3 2 9 3 4 8 2
1 1000 1/3 1000
1 90. 251/ 2 25
64 1/3 3 64 4 43 Copyright © 2025 Pearson Education, Inc.
3
1/3
1/2
3
1 1 1000 10
1 1 25 5
Chapter R: Review
2/3
64 91. 125
125 64
2/3
125 3 64
16 x y 99. xy
2 1/ 3 3/ 4
2
2 1/ 4
2
25 5 4 16 1 92. 813/4 81
3/ 4
1 4 81
96.
y
x y
4 8 3/ 4
1/ 3
4x y 100.
1 1/ 3 3/ 2
2
2 2/3
x2 / 3 y 2 / 3
xy 3/ 2
1/ 3
2/3
x 2 / 3 y1/ 3 x 2 / 3 y 4 / 3 x2 / 3 y2 / 3
x2 y
3/ 4
1/ 2
2 1/ 2
3/ 4
102.
1/ 4 13/ 2 1/ 4 1 3/ 4
y
x 1/ 4 y1/ 2
x 2 1 x
(1 x)1/ 2 x 2 2x (1 x)1/ 2 3x 2 (1 x)1/ 2
x1/ 4 y1/ 4 xy x3/ 2 y 3/ 4
x
y
x 2 1 x 1 x x 1/ 2 2 1 x 1/ 2 (1 x) (1 x)1/ 2
y x y 2 3/ 4
y
3/ 2 3/ 2
1/ 2
101.
x 2 / 3 y1 x 2 / 3 y
98.
3/ 2 1/ 2
8 x 3 y 1 8 3 x y
2 2/3
x2 / 3 y2 / 3
x1/ 4 y1/ 4 x 2
1/ 3 3/ 2
23 x 3/ 2 3/ 2 y1/ 2 3/ 2
x3 y 6
x y x y
1/ 2
3/ 2
x3/ 2 y 3/ 2 3
x 2 / 3 2 / 3 2 / 3 y1/ 3 4 / 3 2 / 3
xy 1/ 4 x 2 y 2
y
43/ 2 x 1
x
8 3/ 4
2 1/ 3
8 x5 / 4 y 3/ 4
4 x
xy 2
y
x
4 3/ 4
x y xy 97. 1/ 3
6 1/ 3
3/ 2 1/ 4
8 x5 / 4 y 3/ 4
94. x 2 / 3 x1/ 2 x 1/ 4 x 2 / 3 1/ 2 1/ 4 x11/12 x3
2 1/ 4
23 x3/ 2 1/ 4 y 1/ 4 1/ 2
93. x3/ 4 x1/ 3 x 1/ 2 x3 4 1/ 3 1/ 2 x 7 /12
3 6 1/ 3
3
1/ 3 3/ 4
x1/ 4 y1/ 2
3
1 1 3 27
x y
3/ 4
1/ 4
4
3
95.
y x y 16 x y 163/ 4 x 2
y1/ 2 x1/ 4
1 x 1 x x1/ 2 2 x1/ 2 x1/ 2 1/ 2 2x 2 x1/ 2 1 x 2 x 3x 1 1/ 2 2 x1/ 2 2x
44
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Section R.8: nth Roots; Rational Exponents
1/ 2
103. 2 x x 2 1
x2
2
2x x 1
1/ 2
1/ 2 1 2 x 1 2x 2 x3
106.
x 1 2 x x 1 x 1 x x 1 2 x x 1 x 2 x x 1 x x 1 x 1 2
1/ 2
2
2
1/ 2
3
2x 2x x
3
x 1 x 3x 2 x 1
1/ 2
2
3
1/ 2
2
1/ 2
2
2
1/ 2
104.
3
3x 2 x
x 1
x 11/ 3 x 13 x 12 / 3 , x 1 1/ 3
x 3 x 1
2/3
x 11/ 3 x 2/3 3 x 1 2 / 31/ 3 1 3 x 1 x 3 x 1 x 2/3 2/3 3 x 1 3 x 1
3 x 1
3 x 1
4x 3
105.
2/3
3x 3 x 2/3
1/ 2
1/ 2
x 1
24 3 8 x 1
2/3
1 1 x5 ,x 5 2 x5 5 4x 3
4x 3 x 5 2 x 5 5 4x 3 4x 3 5 4x 3 x 5 2 x 5 10 x 5 4 x 3 5 4 x 3 2 x 5 10 x 5 4 x 3 20 x 15 2 x 10 10 x 5 4 x 3 22 x 5 10 x 5 4 x 3
, x 2, x
2
8 3 8 x 1 3 x 2
1 8
2
2
2
3
24 3 x 2 3 8 x 1
2
8 8 x +1 x 2 24 3 x 2 8 x 1 2
2
64 x 8 x 2 24 3 x 2 8 x 1 2
2
65 x 6 24 3 x 2 8 x 1 2
2
1 x 1 x x 1 x 2 1 2 1 x x 107. 1 x 1 x 2 1 x 1 x x 2 1 x 1 x 2(1 x) x 1 2(1 x)1/ 2 1 x
4x 3 3 x 1
2
24 3 x 2 3 8 x 1 3
2
2
x2
8 3 8 x 1 3 8 x 1 3 x 2 3 x 2
3
1
2
3
2
3
2
8x 1
3 3 x 2
1/ 2
2
1/ 2 1/ 2
2
3
2 x 2(1 x)3/ 2
2 2x x 1 x 2 x2 1 108. x2 1 2 x2 x 1 x2 1 x2 1 2 x2 1 x2 x 1 x2 1 x 2 1 x2 1 x2 1 x2 x2 1 x2 1
1
x 1 2
45 Copyright © 2025 Pearson Education, Inc.
3/ 2
1
1 x 1 x 1 2
2
Chapter R: Review
109.
x 4 1/ 2 2 x x 4 1/ 2
x2
x 1 111.
x4
2
1/ 2 2x x 4 1/ 2 x 4 x4 1/ 2 2x 1/ 2 x 4 x 4 1/ 2 1/ 2 x 4 x 4 x4 x 4 2x x 4 1/ 2 x4 1 x 4 1/ 2 x 4 x 4
2
9 x 9 x 9 x
2 1/ 2
2 1/ 2
x
2
1
2
x 1
1/ 2
2
1/ 2
x2 x2 4
1/ 2
x2 4
2
1 9 x2
1/ 2
x2 4 x2
x 4 2
9
9 x
2 3/ 2
46
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1/ 2
1/ 2
2
1/ 2
2
1 1/ 2 9 x2 9 x2
2
x 4 x 4 x 4
9 x2 x2
1 x2
1/ 2 2 x2 x 4 1/ 2 x2 4 x2 4 x 2 4 1/ 2 x 2 4 1/ 2 x 2 1/ 2 2 x 4 x2 4
2 1/ 2
x2 x2 1 1 2 1/ 2 x x2 1
2
x2 9 x 2 1/ 2 2 1/ 2 9 x 2 9 x 9 x 2 1/ 2 9 x 2 1/ 2 x 2 2 1/ 2 9 x 9 x2
1/ 2
1/ 2
x 4 112.
, 3 x 3
1/ 2
2
2
2
1/ 2
9 x2
1/ 2
2
x
x 1 x 1 x x 1 1 x 1 x
x 4 3/ 2 x2 9 x2
2 1/ 2
, x 1 or x 1
x2 x2 1
4 x
9 x 110.
1/ 2
x2
x 4 3/ 2
x 2 x 2 1 1/ 2 x 2 1 1/ 2 1/ 2 2 x 1 2 x
x 4
x2 1
1/ 2
x2
1 x2 4
1 4 3/ 2 2 x 4 x 4 2
Section R.8: nth Roots; Rational Exponents
1 x2 2x x 2 x 113. ,x 0 2 1 x2
116.
1 x2 2 x 2 x x 2 x
1 x 1 x 2 x 2 x x
2 2
2
2 x
114.
2
1 x
2 2
1 x 4x 1 1 3x 2 2 2 2 x 1 x 2 x 1 x2
2 x 1 x2
23 x 1 x 1 x 1/ 3
2 2 / 3
3
, x 1, x 1
2 2/3
2 2/3
2 2/3
3
2 2/3
2 2/3
2 1/ 3 2 / 3
3
2 2/3
2
2 2/3
3
3
2 2 / 3 2 / 3
2
2 4/3
2 4/3
3 115. ( x 1)3/ 2 x ( x 1)1/ 2 2 3 ( x 1)1/ 2 x 1 x 2 5 ( x 1)1/ 2 x 1 2 1 ( x 1)1/ 2 5 x 2 2
1/ 2
2
2 x1/ 2 (3 x 4)( x 1)
118. 6 x1/ 2 2 x 3 x3/ 2 8 2 x1/ 2 3(2 x 3) 4 x
x 4 x 4 2x x 4 3 x 4 8 x x 4 3 x 12 8 x x 4 11x 12 4/3
2
2
1/ 3
2
1/ 3
2
2
1/ 3
2
120. 2 x 3x 4
1/ 3
2
4/3
2
2
x 2 4 3x 4
1/ 3
2 x 3x 4
3x 4 2 x
2 x 3x 4
5x 4
1/ 3 1/ 3
3
2 4/3
3
2 x 3x x 4
119. 3 x 2 4
1 x 2x 1 x 3 1 x 2 x 3 1 x 1 x 6 x 1 x 2x 1 3 1 x 1 x 6 x 1 x 2 x 6x 6x 2x 3 1 x 3 1 x 2x 3 2x 6x 4x 3 1 x 3 1 x 2 1/ 3
2 x1/ 2 10 x 9
2 x3 2 1/ 3 x x 2 1 2 / 3 3 1 x2
2 x1/ 2 3( x 2 x) 4 x 4 2
117. 6 x1/ 2 x 2 x 8 x3/ 2 8 x1/ 2
1
2
4 ( x 2 4) 4 / 3 x ( x 2 4)1/ 3 2 x 3 8 2 1/ 3 2 ( x 4) x 4 x 2 3 11 ( x 2 4)1/ 3 x 2 4 3 1/ 3 1 2 x 4 11x 2 12 3
121. 4 3 x 5
2 x 33/ 2 3 3x 5 4 / 3 2 x 31/ 2 1/ 3 1/ 2 3x 5 2 x 3 4 2 x 3 3 3 x 5 1/ 3
3x 5
2 x 31/ 2 8 x 12 9 x 15 1/ 3 1/ 2 3x 5 2 x 3 17 x 27 1/ 3
where x
47 Copyright © 2025 Pearson Education, Inc.
3 . 2
Chapter R: Review 122. 6 6 x 1
4 x 33/ 2 6 6 x 14 / 3 4 x 31/ 2 1/ 3 1/ 2 6 6 x 1 4 x 3 4 x 3 6 x 1 1/ 3
129.
2 3 4.89 3 5
130.
5 2 0.04 24
131.
3 35 2 2.15 3
132.
2 3 3 4 1.33 2
6 6 x 1
4 x 31/ 2 10 x 2 1/ 3 1/ 2 6 6 x 1 4 x 3 2 5 x 1 1/ 3 1/ 2 12 6 x 1 4 x 3 5 x 1 1/ 3
where x
3 . 4
3 1/ 2 x ,x 0 2 3 3 1/ 2 x1/ 2 2 x
123. 3x 1/ 2
3 2 3 x1/ 2 x1/ 2 6 3 x 3 x 2 1/ 2 2 x1/ 2 2x 2 x1/ 2
124. 8 x1/ 3 4 x 2 / 3 , x 0 4 8 x1/ 3 2 / 3 x
8 x1/ 3 x 2 / 3 4 8 x 4 4 2 x 1 2/3 x2 / 3 x x2 / 3
125.
2 1.41
126.
7 2.65
127.
3
V 40 12
b.
V 40 1
4 1.59
134. a.
128.
3
5 1.71
96 0.608 12 15, 660.4 gallons
133. a.
2
2
96 0.608 390.7 gallons 1
v 64 4 02 256 16 feet per second
b.
v 64 16 02 1024 32 feet per second
c.
v 64 2 42 144 12 feet per second
135. T 2
64 2 2 8.89 seconds 32
48
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Section R.8: nth Roots; Rational Exponents
16 1 2 2 32 2 2
136. T 2
2 4.44 seconds
137.
4
31
3
3 4 3 12
3 34
1
13
4 3
1
2
The quotient is x ( 3 4) x 4 3 The remainder is 1 138. 1 2 1
1
9
13
7
1 2
67 2 7
3 8
77 2
0
3
Yes, 1 2 is a factor of x 9 x 2 13 x 7 . 139. Answers may vary. One possibility follows: If a 5 , then
a2
5 2
25 5 a .
Since we use the principal square root, which is always non-negative, a if a 0 a2 a if a 0 which is the definition of a , so
a2 a .
49 Copyright © 2025 Pearson Education, Inc.
Chapter 1 Equations and Inequalities Section 1.1
14.
6 x 18 0 6 x 18 18 0 18 6 x 18 6 x 18 6 6 x 3 The solution set is {3}.
15.
2x 3 0 2x 3 3 0 3
1. Distributive 2. Zero-Product 3.
x x 4
4. False. Multiplying both sides of an equation by zero will not result in an equivalent equation. 5. identity
2x 3
6. linear; first-degree 7. False. The solution is
2x 3 2 2 3 x 2
8 . 3
3 The solution set is . 2
8. True 9. b 16.
10. d 11.
12.
13.
3x 4 0 3x 4 4 0 4
7 x 21 7 x 21 7 7 x3 The solution set is {3}.
3x 4 3x 4 3 3 4 x 3 4 The solution set is . 3
6 x 24 6 x 24 6 6 x 4 The solution set is {4}.
17.
3x 15 0 3x 15 15 0 15 3 x 15 3 x 15 3 3 x 5 The solution set is {5}.
1 7 x 4 20 1 7 4 x 4 4 20 28 7 x 20 5 7 The solution set is . 5
50 Copyright © 2025 Pearson Education, Inc.
Section 1.1: Linear Equations
18.
2 9 x 3 2 2 9 6 x 6 3 2 4 x 27
22.
5 y 6 6 18 y 6 5 y y 24 5 y y y 24 y 6 y 24
4 x 27 4 4 27 x 4
27 The solution set is . 4
19.
6 y 24 6 6 y 4 The solution set is {4}.
23.
3x 4 x
x 2x 3 x 2x 2x 3 2x
3x x 4
3 x 3
3x x x 4 x
3 x 3 3 3 x 1 The solution set is {1}.
2 x 4
20.
24.
2x x 1
2 x 9 9 5x 9
2x x x 1 x
2x 5x 9
x 1
2x 5x 5x 9 5x
x 1 1 1 x 1 The solution set is {1}.
3x 9
21.
2t 6 3 t 2t 6 6 3 t 6 2t 9 t 2t t 9 t t 3t 9 3t 9 3 3 t 3 The solution set is {3}.
3 2x 2 x 3 2x 3 2 x 3
2 x 9 5x
3x 9 3 3 x3 The solution set is {3}.
6 x 2x 9 6 x 6 2x 9 6
3x 4 4 x 4
2 x 4 2 2 x 2 The solution set is {2}.
5 y 6 18 y
25.
3 2n 4n 7 3 2n 3 4n 7 3 2n 4n 4 2n 4n 4n 4 4n 2n 4 2n 4 2 2 n 2 The solution set is {2}.
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Chapter 1: Equations and Inequalities 26.
6 2m 3m 1
30. 7 (2 x 1) 10 7 2 x 1 10 8 2 x 10 8 2 x 8 10 8 2 x 2 2 x 2 2 2 x 1 The solution set is {1}.
6 2m 6 3m 1 6 2m 3m 5 2m 3m 3m 5 3m 5m 5 5m 5 5 5 m 1 The solution set is {1}.
27.
28.
3(5 3 x) 8( x 1) 15 9 x 8 x 8 9 x 8 x 15 8 x 8 x 8 x 15 15 8 15 x 23 The solution set is {23}.
31.
3 1 1 x2 x 2 2 2 3 1 1 2 x 2 2 x 2 2 2 3x 4 1 x 3x 4 4 1 x 4 3 x 3 x
3(2 x) 2 x 1 6 3x 2 x 1 6 3x 6 2 x 1 6 3x 2 x 7 3 x 2 x 2 x 7 2 x 5 x 7 5 x 7 5 5 7 x 5 7 The solution set is . 5
3 x x 3 x x 4 x 3 4 x 3 4 4 3 x 4 3 The solution set is . 4
32.
29. 8 x (3 x 2) 3x 10 8 x 3x 2 3 x 10 5 x 2 3 x 10 5 x 2 2 3 x 10 2 5 x 3x 8 5 x 3x 3x 8 3x 2 x 8 2 x 8 2 2 x 4 The solution set is {4}.
1 2 x 2 x 3 3 2 1 3 x 3 2 x 3 3 x 6 2x x 2x 6 2x 2x 3x 6 3x 6 3 3 x2 The solution set is {2}.
52
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Section 1.1: Linear Equations
33.
34.
35.
36.
1 3 x 5 x 2 4 1 3 4 x 5 4 x 2 4 2 x 20 3 x 2 x 20 2 x 3x 2 x 20 x x 20 The solution set is {20}. 1 1 x 6 2 1 2 1 x 2 6 2 2 x 12 2 x 2 12 2 x 10 x 10 1 1 x 10 The solution set is {10}.
37.
38.
0.9t 1 t 0.9t t 1 t t 0.1t 1 1 0.1t 0.1 0.1 t 10 The solution set is {10}.
39.
x 1 x 2 2 3 7 x 1 x 2 21 21 2 7 3
7 x 1 3 x 2 42
7 x 7 3x 6 42 10 x 13 42 10 x 13 13 42 13 10 x 29 10 x 29 10 10 29 x 10 29 The solution set is . 10
2 1 1 p p 3 2 3 2 1 1 6 p 6 p 3 3 2 4p 3p 2 4 p 3p 3p 2 3p p2 The solution set is {2}. 1 1 4 p 2 3 3 1 1 4 6 p 6 2 3 3 3 2p 8 3 2p 3 83 2p 5 5 2p 2 2 5 p 2 5 The solution set is . 2
0.2m 0.9 0.5m 0.2m 0.5m 0.9 0.5m 0.5m 0.3m 0.9 0.3m 0.9 0.3 0.3 m 3 The solution set is {-3}.
40.
2x 1 16 3x 3 2x 1 3 16 3 3 x 3 2 x 1 48 9 x 2 x 49 9 x 2 x 49 2 x 9 x 2 x 49 7 x 49 7 x 7 7 x7 The solution set is {7}.
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Chapter 1: Equations and Inequalities
41.
42.
5 1 11 ( p 3) 2 (2 p 3) 8 4 16 10( p 3) 32 4(2 p 3) 11 10 p 30 32 8 p 12 11 10 p 2 8 p 1 10 p 8 p 2 8 p 8 p 1 2 p 2 2 1 2 2p 1 2 1 p 2 2 1 p 2 1 The solution set is . 2 1 3
44.
4 5 2y 5 2 y y 2y 8 10 y 5 8 10 y 8 5 8 10 y 3 10 y 3 10 10 3 y 10 3 Since y does not cause a denominator to 10 3 equal zero, the solution set is . 10
( w 1) 3 52 ( w 4) 152
5( w 1) 45 6( w 4) 2
45.
5w 5 45 6 w 24 2 5w 40 6 w 26 5w 6 w 40 6w 6 w 26 w 40 40 26 40 w 14 1 w 14 1 w 14 The solution set is 14 .
46. 43.
4 5 5 y 2y
2 4 3 y y 2 4 y y 3 y y 2 4 3y 6 3y 6 3y 3 3 2 y Since y = 2 does not cause a denominator to equal zero, the solution set is {2}.
1 2 3 2 x 4 1 2 3 4x 4x 2 x 4 2 x 8 3x 2 x 8 2 x 3x 2 x 8 x Since x = 8 does not cause any denominator to equal zero, the solution set is {8}. 3 1 1 x 3 6 3 1 1 6x 6x x 3 6 18 2 x x 18 2 x 2 x x 2 x 18 3 x 18 3 x 3 3 6x Since x 6 does not cause a denominator to equal zero, the solution set is {6}.
54
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Section 1.1: Linear Equations
47.
( x 7)( x 1) ( x 1) 2
48.
50.
x 2 x2 2x2 5x 2
x2 6x 7 x2 x2 2 x 1 x2 6x 7 2x 1 6x 7 7 2x 1 7 6x 2x 8 6x 2x 2x 8 2x 4x 8 4x 8 4 4 x2 The solution set is {2}.
x 2 x2 2 x2 2x2 5x 2 2 x2
2
x 5 x 2 x 5 x 5 x 2 5 x 6x 2 6x 2 1 x 6 6 3 1 The solution set is . 3
51.
p p 2 3 12 p 3 p 3 3 p 12 p 3
( x 2)( x 3) ( x 3) 2
p 3 p 3 3 p 12 p 3 p 3
x2 x 6 x2 6x 9
3 p 12 3 p 12 p4 3 3 The solution set is {4}.
x2 x 6 x2 x2 6 x 9 x2 x 6 6x 9 x 6 6 6x 9 6 x 6 x 15 x 6 x 6 x 15 6 x
49.
x(1 2 x) (2 x 1)( x 2)
x 6x 7 x 2x 1 2
52.
w(4 w2 ) 8 w3
7 x 15
4 w w3 8 w3
7 x 15 7 7 15 x 7 15 The solution set is . 7
4w w3 w3 8 w3 w3 4w 8 4w 8 4 4 w2 The solution set is {2}.
x(2 x 3) (2 x 1)( x 4) 2 x 2 3x 2 x 2 7 x 4 2 x 2 3x 2 x 2 2 x 2 7 x 4 2 x 2 3x 7 x 4 3 x 7 x 7 x 4 7 x 4x 4 4x 4 4 4 x 1 The solution set is {1}.
53.
2 x 3 x2 x2 x 2 3 x 2 x 2 x2 x2 x 3 x 2 2 x 3x 6 2 4x 6 2 4x 6 6 2 6 4x 8 4x 8 4 4 x2 Since x = 2 causes a denominator to equal zero, we must discard it. Therefore the original equation has no solution.
55 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
54.
2x 6 2 x3 x3 2x 6 2 x 3 x 3 x x 3 3 2 x 6 2 x 3 2 x 6 2 x 6 2 x 12 2 x 2 x 2 x 12 2 x 2 x 4 x 12 4 x 12 4 4 x 3 Since x = –3 causes a denominator to equal zero, we must discard it. Therefore the original equation has no solution.
55.
2x 4 3 2 x 4 x 4 x2 2x 4 3 x 2 x 2 x 2 x 2 x 2 2
2x 4 3 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 2x 4 3 x 2 2 x 4 3x 6 2 x 10 3 x 2 x 3 x 10 3 x 3 x 5 x 10 5 x 10 5 5 x2 Since x = 2 causes a denominator to equal zero, we must discard it. Therefore the original equation has no solution.
56
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Section 1.1: Linear Equations
x 4 3 x2 9 x 3 x2 9
56.
x
x 3 x 3
4 3 x 3 x 3 x 3
4 3 x x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 4 x 3 3 x 4 x 12 3 5 x 12 3 5 x 12 12 3 12 5 x 15 5 x 15 5 5 x3
Since x = 3 causes a denominator to equal zero, we must discard it. Therefore the original equation has no solution.
57.
x 3 x2 2 x 3 2 x 2 2 x 2 x2 2
59.
2x 3 x 2
7 2 3x 10 x 3 7 2 3 x 10 x 3 3 x 10 x 3 3 x 10 x3 7 x 3 2 3 x 10
7 x 21 6 x 20 7 x 21 6 x 6 x 20 6 x 21 x 20 21 x 21 25 20 21 x 41
2 x 3x 6 2 x 3x 3x 6 3x x 6 x 6 1 1 x 6 Since x = –6 does not cause any denominator to equal zero, the solution set is {6}.
58.
3x 2 x 1
Since x = 41 does not cause any denominator to equal zero, the solution is {41}. 60.
3x x 1 2 x 1 x 1 3x 2 x 2 3x 2 x 2 x 2 2 x x 2 Since x = –2 does not cause any denominator to equal zero, the solution set is {2}.
4 3 x4 x6 4 3 x 6 x 4 x 6 x 4 x4 x6 4 x 6 3 x 4 4 x 24 3x 12 4 x 24 4 x 3x 12 4 x 24 12 x 24 12 12 x 12 12 x Since x = –12 does not cause any denominator to equal zero, the solution set is {12}.
57 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
61.
6t 7 3t 8 4t 1 2t 4 6t 7 3t 8 4t 1 2t 4 4t 1 2t 4 t 4 1 2t 4
6t 7 2t 4 3t 8 4t 1
12t 24t 14t 28 12t 2 3t 32t 8 2
12t 2 10t 28 12t 2 29t 8 12t 2 10t 28 12t 2 12t 2 29t 8 12t 2 10t 28 29t 8 10t 28 29t 29t 8 29t 28 39t 8 28 39t 28 8 28 39t 20 39t 20 39 39 20 t 39 20 20 does not cause any denominator to equal zero, the solution set is . Since t 39 39
62.
8w 5 4 w 3 10w 7 5w 7 8w 5 4w 3 10 w 7 5w 7 10w 7 5w 7 w 10 7 5w 7
8w 5 5w 7 4w 310w 7
40w 56 w 25w 35 40 w2 28w 30w 21 2
40w2 81w 35 40 w2 58w 21 40 w2 81w 35 40 w2 40 w2 58w 21 40w2 81w 35 58w 21 81w 35 58w 58w 21 58w 139w 35 21 139w 35 35 21 35 139w 14 139w 14 139 139 14 w 139 14 14 does not cause any denominator to equal zero, the solution set is Since w . 139 139
58
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Section 1.1: Linear Equations
63.
4 3 7 x 2 x 5 x 5 x 2 3 7 4 x 5 x 2 x 5 x 2 x2 x 5 x 5 x 2 4 x 5 3 x 2 7
4 x 20 3 x 6 7 4 x 20 3 x 13 4 x 20 3x 3 x 13 3 x 7 x 20 13 7 x 20 20 13 20 7 x 7 7 x 7 7 7 x 1 Since x 1 does not cause any denominator to equal zero, the solution set is {1}.
64.
4 1 1 2 x 3 x 1 2 x 3 x 1 1 1 4 2 x 3 x 1 2 x 3 x 1 2x 3 x 1 2 x 3 x 1 4 x 1 1 2 x 3 1
4 x 4 2 x 3 1 2 x 7 1 2 x 7 7 1 7 2 x 6 2 x 6 2 2 x3 Since x 3 does not cause any denominator to equal zero, the solution set is {3}.
59 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
2 3 5 y3 y4 y6
65.
2 5 3 y 3 y 4 y 6 y 3 y 4 y 6 y3 y4 y6 2 y 4 y 6 3 y 3 y 6 5 y 3 y 4
2 y 2 6 y 4 y 24 3 y 2 6 y 3 y 18 5 y 2 4 y 3 y 12 2 y 2 2 y 24 3 y 2 9 y 18 5 y 2 y 12 2 y 2 4 y 48 3 y 2 27 y 54 5 y 2 5 y 60 5 y 2 31 y 6 5 y 2 5 y 60 5 y 2 31 y 6 5 y 2 5 y 2 5 y 60 5 y 2 31 y 6 5 y 60 31 y 6 5 y 5 y 60 5 y 36 y 6 60 36 y 6 6 60 6 36 y 66 36 y 66 36 36 11 y 6
Since y
66.
11 11 does not cause any denominator to equal zero, the solution set is . 6 6
5 4 3 5 z 11 2 z 3 5 z 4 5 3 5 z 11 2 z 3 5 z 5 z 11 2 z 3 5 z 5 z 11 2 z 3 5 z 5 2 z 3 5 z 4 5 z 11 5 z 3 5 z 11 2 z 3
5 10 z 2 z 15 3z 4 25 z 5 z 2 55 11z 3 10 z 2 15 z 22 z 33 2
5 2 z 2 13 z 15 4 5 z 2 36 z 55 3 10 z 2 37 z 33 10 z 2 65 z 75 20 z 2 144 z 220 30 z 2 111z 99 30 z 2 209 z 295 30 z 2 111z 99
30 z 2 209 z 295 30 z 2 30 z 2 111z 99 30 z 2 209 z 295 111z 99 209 z 295 209 z 111z 99 209 z 295 98 z 99 295 99 98 z 99 99 196 98 z 196 118 z 98 98 2z Since z 2 does not cause any denominator to equal zero, the solution set is {2}.
60
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Section 1.1: Linear Equations 10 x x4 x 2 9 x 2 3x x 2 3x 10 x x4 x 3 x 3 x x 3 x x 3
67.
10 x x4 x x 3 x 3 x x 3 x 3 x 3 x 3 x x 3 x x 3 x x x 4 x 3 10 x 3 x 2 x 2 3 x 4 x 12 10 x 30 x 2 x 2 7 x 12 10 x 30 x 2 x 2 7 x 12 10 x 30 7 x 12 10 x 30 7 x 12 12 10 x 30 12 7 x 10 x 42 7 x 10 x 10 x 10 x 42 3x 42 3x 42 3 3 x 14 Since x 6 does not cause any denominator to equal zero, the solution set is {14}.
x 1 x4 3 x 2 2 x x 2 x x 2 3x 2 x 1 x4 3 x x 2 x x 1 x 2 x 1
68.
x 1 x4 3 x x 2 x 1 x x 2 x 1 x x 2 x x 1 x 2 x 1 x 1 x 1 x 4 x 2 3x
x x x 1 x 2 x 4 x 8 3x x 2 x 1 x 6 x 8 3 x 2
2
2
2
x 2 2 x 1 x 2 6 x 8 3 x 2 x 1 6 x 8 3 x 4 x 7 3 x 4 x 7 4 x 3 x 4 x 7 x Since x 7 does not cause any denominator to equal zero, the solution set is {7}.
61 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
69.
21.3 19.23 65.871 21.3 21.3 21.3 19.23 3.2 x 65.871 65.871 65.871 21.3 3.2 x 19.23 65.871 21.3 1 1 3.2 x 19.23 65.871 3.2 3.2 3.2 x
21.3 1 x 19.23 5.91 65.871 3.2 The solution set is approximately {5.91}.
70.
19.1 0.195 83.72 19.1 19.1 19.1 0.195 6.2 x 83.72 83.72 83.72 19.1 6.2 x 0.195 83.72 19.1 1 1 6.2 x 0.195 6.2 83.72 6.2 6.2 x
19.1 1 x 0.195 0.07 83.72 6.2 The solution set is approximately {0.07}.
71.
18 x 2.4 2.11 18 18 18 x x 2.4 x 14.72 21.58 x 2.11 2.11 2.11 18 x 2.4 14.72 21.58 x 2.11 18 x 14.72 2.4 14.72 14.72 21.58 x 2.11 18 x 12.32 21.58 x 2.11 18 21.58 x 12.32 2.11 1 18 1 x 12.32 21.58 2.11 21.58 18 21.58 18 2.11 2.11 14.72 21.58 x
1 x 12.32 0.41 21.58 18 2.11 The solution set is approximately {0.41}.
62
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Section 1.1: Linear Equations
72.
21.2 14 x 20 2.6 2.32 21.2 14 14 14 18.63 x x x 20 x 2.6 2.32 2.32 2.32 21.2 14 18.63 x x 20 2.6 2.32 21.2 14 21.2 21.2 18.63x x 20 2.6 2.32 2.6 2.6 14 21.2 18.63x x 20 2.32 2.6 14 21.2 18.63 x 20 2.32 2.6 18.63x
21.2 1 1 14 18.63 x 20 14 14 2.6 18.63 2.32 18.63 2.32 2.32 21.2 1 x 20 0.94 2.6 18.63 14 2.32 The solution set is approximately {0.94}.
73.
ax b c, a 0 ax b b c b ax b c ax b c a a bc x a
74.
1 ax b, a 0 1 ax 1 b 1 ax b 1 ax b 1 a a b 1 1 b x a a
75.
76.
x x c, a 0, b 0, a b a b x x ab ab c a b bx ax abc (a b) x abc (a b) x abc ab ab abc x ab a b c, c 0 x x a b x xc x x a b cx a b cx c c ab x c
63 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 77. x 2a 16 ax 6a, if x 4 4 2a 16 a(4) 6a 4 2a 16 4a 6a 4 2a 16 2a 4a 12 4a 12 4 4 a3
81.
RF mv 2 RF mv 2 F F mv 2 R F
78. x 2b x 4 2bx, for x 2 2 2b 2 4 2b(2) 2 2b 2 4 4b 2 2b 2 4b 4 2b 4 b 2 b2 79.
82.
PV nRT PV nRT nR nR PV T nR
83.
1 1 1 R R1 R2 1 1 1 RR1 R2 RR1 R2 R R1 R2 R1 R2 RR2 RR1
a 1 r a S (1 r ) (1 r ) 1 r S Sr a S
S Sr S a S Sr a S Sr a S S S S a r S
R1 R2 R ( R2 R1 ) R1 R2 R ( R2 R1 ) R2 R1 R2 R1 R1 R2 R R2 R1
80.
mv 2 R mv 2 RF R R F
84.
v gt v0 v v0 gt
A P (1 r t ) A P P rt A P P rt A P P rt Pt Pt A P r Pt
v v0 gt g g v v0 v0 v t g g
85.
Amount in bonds Amount in CDs x
x 3000
Total 20, 000
x x 3000 20, 000 2 x 3000 20, 000 2 x 23, 000 x 11,500 $11,500 will be invested in bonds and $8500 will be invested in CD's.
64
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Section 1.1: Linear Equations
86.
Sean's Amount George's Amount x 3000
x
Total 10, 000
x x 3000 10, 000 2 x 3000 10, 000 2 x 13, 000 x 6500 Sean will receive $6500 and Jorge will receive $3500.
Dollars Hours Money per hour worked earned
87. Regular wage
x
40
40 x
Overtime wage
1.5 x
8
8(1.5 x)
40 x 8 1.5 x 910
89. Let x represent the score on the final exam. 80 83 71 61 95 x x 80 7 390 2 x 80 7 390 2 x 560 2 x 170 x 85 Camila needs a score of 85 on the final exam. 90. Let x represent the score on the final exam. Note: since the final exam counts for two-thirds of the overall grade, the average of the four test scores count for one-third of the overall grade. For a B, the average score must be 80. 1 86 80 84 90 2 x 80 3 4 3 1 340 2 x 80 3 4 3 85 2 x 80 3 3 85 2 3 x 3 80 3 3 85 2 x 240
40 x 12 x 910 52 x 910 910 x 17.50 52 Sandra’s regular hourly wage is $17.50. Dollars Hours Money per hour worked earned
88. Regular wage
x
40
40 x
Overtime wage
1.5 x
6
6(1.5 x)
Sunday wage
2x
4
4(2 x)
2 x 155 x 77.5 Ali needs a score of 78 to earn a B.
For an A, the average score must be 90. 1 86 80 84 90 2 x 90 3 4 3 1 340 2 x 90 3 4 3 85 2 x 90 3 3 85 2 3 x 3 90 3 3 85 2 x 270
40 x 6 1.5 x 4 2 x 1083 40 x 9 x 8 x 1083 57 x 1083 1083 19 x 57 Leigh’s regular hourly wage is $19.00.
2 x 185 x 92.5 Ali needs a score of 93 to earn an A.
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Chapter 1: Equations and Inequalities 7.50 x 4.50 5200 x 29,961
91. Let x represent the original price of the phone. Then 0.12x represents the reduction in the price of the phone. The new price of the phone is $572. original price reduction new price x 0.12 x 572 0.88 x 572 x 650 The original price of the phone was $650. The amount of the reduction (i.e., the savings) is 0.12($650) = $78.
7.50 x 23, 400 4.50 x 29,961 3.00 x 23, 400 29,961 3.00 x 6561 x 2187 There were 2187 adult patrons.
96. Let p represent the original price for the boots. Then, 0.30p represents the discounted amount. original price discount clearance price p 0.30 p 399 0.70 p 399 p 570 The boots originally cost $570.
92. Let x represent the original price of the car. Then 0.15x represents the reduction in the price of the car. The new price of the car is $8000. list price reduction new price x 0.15 x 18000 0.85 x 18000 x 21176.47 The list price of the car was $21,176.47. The amount of the reduction (i.e., the savings) is 0.15($21176.47) $3176.47 .
97. Let w represent the width of the rectangle. Then w 8 is the length. Perimeter is given by the formula P 2l 2 w. 2( w 8) 2 w 60 2w 16 2 w 60 4 w 16 60 4 w 44 w 11 Now, 11 + 8 = 19. The width of the rectangle is 11 feet and the length is 19 feet.
93. Let x represent the price the theater pays for the candy. Then 2.75x represents the markup on the candy. The selling price of the candy is $4.50. suppier price markup selling price x 2.75 x 4.50 3.75 x 4.50 x 1.20 The theater paid $1.20 for the candy.
98. Let w represent the width of the rectangle. Then 2w is the length. Perimeter is given by the formula P 2l 2 w. 2(2 w) 2w 42 4 w 2w 42 6w 42 w7 Now, 2(7) = 14. The width of the rectangle is 7 meters and the length is 14 meters.
94. Let x represent selling price for the new car. The dealer’s cost is 0.85($24, 000) $20, 400. The markup is $300. selling price = dealer’s cost + markup x 20, 400 300 $20, 700 At $300 over the dealer’s cost, the price of the care is $20,700. 95. Adults
Tickets sold x
Children 5200 x
Price per ticket 7.50
Money earned 7.50 x
4.50
4.50(5200 x)
99. We will let B be the calories from breakfast, L the calories from lunch and D the calories from dinner. So we have the following equations: B L 125 D 2 L 300 2025 B L D
Now we substitute the first two into the last one and solve for L. 66
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Section 1.1: Linear Equations 2025 ( L 125) L (2 L 300) 2025 4 L 175 2200 4 L L 550 Now we substitute L into the first two equations to get B and D. B 550 125 675 D 2(550) 300 800 So Herschel took in 675 calories from breakfast, 550 calories from lunch and 800 calories from dinner.
4 x 10 2 x 40 2 x 30 x 15 4 x 10 3 x 18 x8 2 x 40 3x 18 x 22 x 22 Since 22 is the largest of the numbers then the largest perimeter is: 4 22 10 2 22 40 3 22 18 266
100. We will let B be the calories from breakfast, L the calories from lunch, D the calories from dinner and S the calories from snacks. So we have the following equations: L 0.5 B
103.
D B 200
S B 120 E 700
1480 B L D E Now we substitute the first four into the last one and solve for B. 1480 B 0.5 B ( B 200) ( B 200) 700 1480 3.5B 620 2100 3.5 B B 600 Now we substitute B to get S. S B 120 600 120 480 So Tyshira took in 480 calories from snacks.
101.
Judy's Amount Tom's Amount Total x
2 x 3
18
2 x 18 3 5 x 18 3 3 x 18 5 x 10.80 Judy pays $10.80 and Tom pays $7.20. x
102. An isosceles triangle has three equal sides. Therefore: 4 x 10 2 x 40 3 x 18 . Solve each set separately:
3 11 1 1 4 x 3x 1 x 6 4 5 2 4 20 5 3 1 3 1 3 4 x x 1 x 4 10 5 80 2 5 Multiply both sides by the LCD 80 to clear fractions. 60 x 8 48 x 80 x 120 64 108 x 72 x 58 107 x 16 x
16 107
104. If a hexagon is inscribed in a circle then the sides of the hexagon are equal to the radius of the circle. Let the P = 6r be the perimeter of the hexagon. Let r be the radius of the circle. 6r r 10 5r 10 r2 Thus r = 2 inches is the radius of the circle where the perimeter of the hexagon is 10 inches more than the radius. 105. To move from step (6) to step (7), we divided both sides of the equation by the expression x 2 . From step (1), however, we know x = 2, so this means we divided both sides of the equation by zero. 106– 107. Answers will vary.
67 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
Section 1.2
14.
1. x 2 5 x 6 x 6 x 1
x 3 or x3 The solution set is {–3, 3}.
2. 2 x 2 x 3 2 x 3 x 1 3.
5 ,3 3
15.
z 3 or z2 The solution set is {–3, 2}.
2
1 5 5 25 25 5 ; ; x 2 5 x 5. 2 2 2 4 4 25 5 x 4 2
16.
2
7. False; a quadratic equation may have no real solutions.
17.
8. False; If x 2 p then x could also be negative. 9. b
1 or 2
x3
1 The solution set is , 3 2
2
x 9x 0 x x 9 0
18.
x 0 or x 9 0
3x 2 5 x 2 0 (3x 2)( x 1) 0 3x 2 0 or x 1 0
x 0 or x9 The solution set is {0, 9}.
x
2
x 4x 0 x( x 4) 0 x 0 or x 4 0
2 or 3
x 1
The solution set is 1,
x 0 or x 4 The solution set is {–4, 0}.
13.
2 x2 5x 3 0 (2 x 1)( x 3) 0 2x 1 0 or x 3 0
x
10. d
12.
v 2 7v 6 0 (v 6)(v 1) 0 v 6 0 or v 1 0 v 6 or v 1 The solution set is {–6, –1}
6. discriminant; negative
11.
z2 z 6 0 ( z 3)( z 2) 0 z 3 0 or z 2 0
4. True
x2 5x
x2 9 0 ( x 3)( x 3) 0 x 3 0 or x 3 0
19.
2 . 3
5 w2 180 0 5( w2 36) 0 5( w 6)( w 6) 0 w 6 0 or w 6 0 w 6 or w6 The solution set is {–6, 6}.
x 2 25 0 ( x 5)( x 5) 0 x 5 0 or x 5 0
x 5 or x5 The solution set is {–5, 5}.
68
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Section 1.2: Quadratic Equations
2 y 2 50 0
20.
25.
2( y 2 25) 0 2( y 5)( y 5) 0 y 5 0 or y 5 0 y 5 or y5 The solution set is {–5, 5}.
21.
6 p2 6 5 p 6 p2 5 p 6 0 (3 p 2)(2 p 3) 0 3p 2 0
or 2 p 3 0 2 3 p or p 3 2 2 3 The solution set is , . 3 2
x x 3 10 0 x 2 3x 10 0 ( x 2) x 5 0 x 2 0 or x 5 0 x 2 or x 5
26. 2(2u 2 4u ) 3 0 4u 2 8u 3 0
The solution set is 5, 2 .
(2u 1)(2u 3) 0 2u 1 0 or 2u 3 0
x( x 4) 12
22.
3 2 1 3 The solution set is , . 2 2 u
2
x 4 x 12 0 ( x 6)( x 2) 0 x 6 0 or x 2 0 x 6 or
x2
The solution set is 6, 2 . 23.
27.
4 x 2 9 12 x 4 x 2 12 x 9 0
The solution set is 24.
u
6 x 6 6 x 5 x x x 6x 5
6 x2 5x 6 0 (3x 2)(2 x 3) 0
3x 2 0
2x 3 0 2 3 x or x 3 2 Neither of these values causes a denominator to 2 3 equal zero, so the solution set is , . 3 2
3 . 2
25 x 2 16 40 x 25 x 2 40 x 16 0 (5 x 4) 2 0 5x 4 0 4 x 5
The solution set is
1 or 2
6x2 5x 6
2
(2 x 3) 0 2x 3 0 3 x 2
6( p 2 1) 5 p
28.
4 . 5
or
12 7 x 12 x x 7x x x
x 2 12 7 x x 2 7 x 12 0 ( x 3)( x 4) 0 x 3 0 or x 4 0 x 3 or x4
69 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
Neither of these values causes a denominator to equal zero, so the solution set is {3, 4}.
32. x 2 36 x 36
4 x 2 3 3 x 3 x x x 3
29.
x 6 The solution set is 6, 6 .
4 x 2 3 3 x x 3 x x 3 x x3 x x 3 4 x x 2 3 x 3 3
33.
x 1 4 x 1 2
4 x 2 8 x 3x 9 3
x 1 2 or x 1 2
4 x2 5x 6 0
x 3 or x 1 The solution set is 1, 3 .
4 x 3 x 2 0 4x 3 0
or x 2 0
3 or x2 4 Neither of these values causes a denominator to 3 equal zero, so the solution set is , 2 . 4
34.
x
x 2 1 x 2 1 or x 2 1 x 1 or x 3 The solution set is 3, 1 .
5 3 4 x4 x2 5 x 4 x 2 4 3 x 4 x 2 x2 x4 5 x 2 4 x 4 x 2 3 x 4
2
1 35. h 4 16 3 1 h 4 16 3 1 h 4 4 3 1 1 h 4 4 or h 4 4 3 3 1 1 h 0 or h 8 3 3 h 0 or h 24 The solution set is 24, 0 .
5 x 10 4 x 2 2 x 8 3x 12 5 x 10 4 x 2 8 x 32 3 x 12 0 4 x 2 6 x 10
0 2 2 x 2 3x 5
0 2 2 x 5 x 1 2x 5 0
or x 1 0
5 or x 1 2 Neither of these values causes a denominator to 5 equal zero, so the solution set is , 1 . 2 x
36.
31.
x 2 2 1 x2 1
30.
x 12 4
3 z 2 2 4 3z 2 4 3 z 2 2
x 2 25
3 z 2 2 or 3z 2 2 3 z 4 or 3z 0
x 25 x 5 The solution set is 5, 5 .
z
4 or 3
The solution set is
70
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z0
0,
4 . 3
Section 1.2: Quadratic Equations
2 1 x 0 3 3 2 1 2 x x 3 3 2 1 1 1 x2 x 3 9 3 9
x 2 4 x 21
37.
40. x 2
2
x 4 x 4 21 4
x 2 2 25 x 2 25 x 2 5
2
1 4 x 3 9
x 2 5 x 3 or x 7 The solution set is 7,3 .
38.
1 4 2 3 9 3 1 2 x 3 3 1 x or x 1 3 1 The solution set is 1, . 3 x
x 2 6 x 13 x 2 6 x 9 13 9
x 32 22
x 3 22 x 3 22 The solution set is
3 22, 3 22.
41.
1 3 39. x 2 x 0 2 16 1 3 x2 x 2 16 1 1 3 1 2 x x 2 16 16 16
1 0 2 1 1 x2 x 0 3 6 1 1 2 x x 3 6 1 1 1 1 x2 x 3 36 6 36 3x 2 x
2
2
1 7 x 6 36
1 1 x 4 4
1 1 1 4 4 2 1 1 x 4 2 3 1 x or x 4 4 1 3 The solution set is , . 4 4 x
x
1 7 6 36
x
1 7 6 6
1 7 6 1 7 1 7 The solution set is , . 6 6 x
42.
2 x 2 3x 1 0 3 1 x2 x 0 2 2 3 1 x2 x 2 2 3 9 1 9 x2 x 2 16 2 16
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Chapter 1: Equations and Inequalities
46. x 2 6 x 1 0 a 1, b 6, c 1
2
3 17 x 4 16 x
3 17 4 16
x
3 17 4 4 3 17 x 4
x
2 56 2 2 14 1 14 2 2
44. x 2 4 x 2 0 a 1, b 4, c 2
2
x
4 4 4(1)(2) 4 16 8 2(1) 2
4 8 4 2 2 2 2 2 2
48. 2 x 2 5 x 3 0 a 2, b 5, c 3
x
The solution set is 2 2, 2 2 .
5 25 24 5 1 5 1 4 4 4 5 1 5 1 x or x 4 4 4 6 x or x 4 4 3 x 1 or x 2 3 The solution set is , 1 . 2
( 4) ( 4) 2 4(1)(1) 4 16 4 2(1) 2
5 52 4(2)(3) 2(2)
45. x 2 4 x 1 0 a 1, b 4, c 1
4 20 4 2 5 2 5 2 2
( 5) ( 5) 2 4(2)(3) 2(2)
5 25 24 5 1 5 1 4 4 4 5 1 5 1 x or x 4 4 6 4 x or x 4 4 3 x or x 1 2 3 The solution set is 1, . 2
(2) (2) 2 4(1)(13) 2 4 52 2(1) 2
6 32 6 4 2 3 2 2 2 2
47. 2 x 2 5 x 3 0 a 2, b 5, c 3
The solution set is 1 14, 1 14 .
x
43. x 2 4 x 2 0 a 1, b 2, c 13
6 62 4(1)(1) 6 36 4 2(1) 2
The solution set is 3 2 2, 3 2 2 .
3 17 3 17 The solution set is , . 4 4
x
x
The solution set is 2 5, 2 5 .
72
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Section 1.2: Quadratic Equations
49. 4 y 2 y 2 0 a 4, b 1, c 2 y
4 x2 9 x 0 x(4 x 9) 0 x 0 or 4 x 9 0
(1) ( 1) 2 4(4)(2) 2(4)
1 1 32 1 31 8 8 No real solution.
x 0 or
0 4x2 5x 0 x(4 x 5) x 0 or 4 x 5 0
1 1 16 1 15 8 8 No real solution.
x 0 or
5 4
9x2 8x 5 0 a 9, b 8, c 5
5 . 4
55. 9t 2 6t 1 0 a 9, b 6, c 1
8 82 4(9)(5) 2(9)
t
8 64 180 8 244 18 18
8 2 61 4 61 18 9 4 61 4 61 The solution set is , . 9 9 2 x2 1 2 x
6 36 36 6 0 1 18 18 3 1 The solution set is . 3
56. 4u 2 6u 9 0 a 4, b 6, c 9 u
2x2 2 x 1 0 a 2, b 2, c 1
( 6) ( 6) 2 4(4)(9) 2(4)
6 36 144 6 108 8 8 No real solution.
2 22 4(2)(1) 2 4 8 x 2(2) 4 2 12 2 2 3 1 3 4 4 2 1 3 1 3 The solution set is , . 2 2
( 6) ( 6) 2 4(9)(1) 2(9)
x
The solution set is 0,
9 x2 8x 5
52.
9 . 4
54. 5 x 4 x 2
9 4
1 12 4(4)(1) t 2(4)
x
x
The solution set is 0,
50. 4t 2 t 1 0 a 4, b 1, c 1
51.
4x2 9 x
53.
57.
3 2 1 1 x x 0 4 4 2 1 3 2 1 4 x x 4 0 4 2 4 3x2 x 2 0 a 3, b 1, c 2
73 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
x
1
12 4 3 2 2 3
59.
1 1 24 1 25 1 5 6 6 6 1 5 1 5 x or x 6 6 4 6 x or x 6 6 2 x 1 or x 3 2 The solution set is ,1 . 3
5 x 2 3x 1 5 x 2 3x 1 0 a 5, b 3, c 1
3 9 20 3 29 10 10 3 29 3 29 The solution set is , . 10 10
60.
2 x 2 3x 9 0 a 2, b 3, c 9 x
32 4 5 1 2 5
2 2 x x3 0 3 2 3 x2 x 3 3 0 3
3
3
x
58.
5 2 1 x x 3 3 5 2 1 3 x x 3 3 3
3 2 1 x x 5 5 3 1 5 x2 x 5 5 5 3x2 5 x 1
32 4 2 9 2 2
3x2 5 x 1 0 a 3, b 5, c 1
3 9 72 3 81 3 9 4 4 4 39 39 x or x 4 4 6 12 x or x 4 4 3 x3 or x 2 3 The solution set is ,3 . 2
x
5
52 4 3 1 2 3
5 25 12 5 37 6 6 5 37 5 37 The solution set is , . 6 6
61.
2 x ( x 2) 3 2
2x 4x 3 0 a 2, b 4, c 3 x
4 42 4(2)(3) 4 16 24 2(2) 4
4 40 4 2 10 2 10 4 4 2 2 10 2 10 The solution set is , . 2 2
74
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Section 1.2: Quadratic Equations 3x( x 2) 1
62.
65.
2
3x 6 x 1 0 a 3, b 6, c 1 6 62 4(3)(1) 6 36 12 2(3) 6
x
3 x( x) ( x 2) 4 x 2 8 x 3x 2 x 2 4 x 2 8 x
6 48 6 4 3 3 2 3 6 6 3 3 2 3 3 2 3 The solution set is , . 3 3 4
63.
0 x2 9 x 2 a 1, b 9, c 2 x
1 1 0 x x2
9 81 8 9 73 2 2 Neither of these values causes a denominator to equal zero, so the solution set is 9 73 9 73 , . 2 2
4 x2 x 1 0 a 4, b 1, c 1 1 12 4 4 1 2 4
66.
1 1 16 1 17 8 8 Neither of these values causes a denominator to equal zero, so the solution set is 1 17 1 17 , . 8 8
64. 2
8 3 0 x x2
2 x 2 x 3 4 x 2 12 x 0 2 x 2 13 x 3 a 2, b 13, c 3 (13) (13) 2 4(2)(3) 2(2)
13 169 24 13 145 4 4 Neither of these values causes a denominator to equal zero, so the solution set is 13 145 13 145 , . 4 4
a 2, b 8, c 3
82 4 2 3 2 2
8 64 24 8 40 4 4 8 2 10 4 10 4 2 Neither of these values causes a denominator to equal zero, so the solution set is 4 10 4 10 , . 4 4
2 x( x) ( x 3) 4 x 2 12 x
2 x2 8x 3 0 8
2x 1 4 x3 x 2x 1 x 3 x x( x 3) 4 x( x 3)
x
8 3 x2 2 2 x2 0 x x
x
(9) (9) 2 4(1)(2) 2(1)
1 1 x2 4 2 x2 0 x x
x
3x 1 4 x2 x 1 3x x 2 x x( x 2) 4 x( x 2)
67. x 2 4.1x 2.2 0 a 1, b 4.1, c 2.2 x
4.1
4.12 4 1 2.2 2 1
4.1 16.81 8.8 4.1 8.01 2 2 x 3.47 or x 0.63 The solution set is 0.63, 3.47 .
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Chapter 1: Equations and Inequalities
72. x 2 x 2 0 a , b , c 2
68. x 2 3.9 x 1.8 0 a 1, b 3.9, c 1.8 x
3.9
3.9 2 4 11.8 2 1
x
3.9 15.21 7.2 3.9 8.01 2 2 x 0.53 or x 3.37 The solution set is 3.37, 0.53 .
69. x 2 3 x 3 0
73. 2 x 2 6 x 7 0 a 2, b 6, c 7
a 1, b 3, c 3 x
b 2 4ac (6) 2 4(2) 7 36 56 20
3 4 1 3 2
Since the b 2 4ac 0, the equation has no real solution.
2 1
3 3 12 3 15 2 2 x 1.07 or x 2.80 The solution set is 2.80, 1.07 .
74. x 2 4 x 7 0 a 1, b 4, c 7 b 2 4ac (4) 2 4(1) 7 16 28 12
Since the b 2 4ac 0, the equation has no real solution.
70. x 2 2 x 2 0 a 1, b 2, c 2 x
2
75. 9 x 2 30 x 25 0 a 9, b 30, c 25
2 4 1 2 2
b 2 4ac (30) 2 4(9) 25 900 900 0
2 1
Since b 2 4ac 0, the equation has one repeated real solution.
2 2 8 2 10 2 2 x 0.87 or x 2.29 The solution set is 2.29, 0.87 .
76. 25 x 2 20 x 4 0 a 25, b 20, c 4 b 2 4ac (20)2 4(25) 4 400 400 0
71. x 2 x 0 a , b 1, c x
1
2 4 2 2
2 8 2 x 0.44 or x 1.44 The solution set is 1.44, 0.44 .
3
Since b 2 4ac 0, the equation has one repeated real solution.
12 4 2
77. 3x 2 5 x 8 0 a 3, b 5, c 8
1 1 4 2 2 x 1.17 or x 0.85 The solution set is 0.85, 1.17 .
b 2 4ac (5) 2 4(3) 8 25 96 121
Since b 2 4ac 0, the equation has two unequal real solutions. 78. 2 x 2 3 x 7 0 a 2, b 3, c 7 b 2 4ac (3) 2 4(2) 7 9 56 65
Since b 2 4ac 0, the equation has two unequal real solutions. 76
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Section 1.2: Quadratic Equations
85. 2 z 6 z 2 0 6z2 z 2
79. x 2 5 0
0 3 z 2 2 z 1
x2 5 x 5
3z 2 0 or 2 z 1 0 2 1 z or z 3 2 1 2 The solution set is , . 2 3
The solution set is 5, 5 .
80. x 2 6 0 x2 6 x 6
86. 2 y 6 y 2
0 6 y2 y 2
The solution set is 6, 6 .
81.
0 3 y 2 2 y 1 3y 2 0
or 2 y 1 0 2 1 y or y 3 2 2 1 The solution set is , . 3 2
16 x 2 8 x 1 0
4 x 1 4 x 1 0
4x 1 0 1 x 4
The solution set is 82.
1 . 4
1 0 2 1 2 x2 2 x 2 0 2
3x 2 3x 2 0 3x 2 0 2 x 3
83.
2 x2 2 2 x 1 0 a 2, b 2 2, c 1
2 . 3
x
or 2 x 5 0 3 5 x or x 5 2 3 5 The solution set is , . 5 2
3x 4 2 x 5 0 3x 4 0 or 2 x 5 0 4 5 x or x 3 2 5 4 The solution set is , . 2 3
2(2)
5x 3 0
6 x 2 7 x 20 0
(2 2) (2 2) 2 4(2) 1
2 2 8 8 2 2 16 4 4 2 2 4 2 2 4 2 2 2 2 2 The solution set is , 2 2
10 x 2 19 x 15 0
5 x 3 2 x 5 0
84.
1 2
x2 2 x
9 x 2 12 x 4 0
The solution set is
x2 2 x
87.
1 2 x 2x 1 2
88.
1 2 x 2x 1 0 2 1 2 x 2 2 x 1 2 0 2 x2 2 2 x 2 0 a 1, b 2 2, c 2
77 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
x
(2 2) (2 2) 2 4(1) 2
90.
x2 x 1 0 a 1, b 1, c 1
2(1)
2 2 8 8 2 2 16 2 2 2 24 22 2 1
The solution set is
x
2(1)
x2 x 4 0 a 1, b 1, c 4 x
(1) (1) 2 4 1 1
1 1 4 1 5 2 2 1 5 1 5 The solution set is , . 2 2
2 2, 2 2.
x2 x 4
89.
x2 x 1
(1) (1) 2 4 1 4 2(1)
1 1 16 1 17 2 2 1 17 1 17 The solution set is , . 2 2
91.
x 2 7x 1 x 2 x 1 x2 x 2 x 2 7x 1 x 2 x 1 ( x 2)( x 1) 2 7x 1 x x 2 x 1 ( x 2)( x 1) ( x 2)( x 1) ( x 2)( x 1) x( x 1) 2( x 2) 7 x 1 x2 x 2 x 4 7 x 1 x 2 3x 4 7 x 1 x2 4 x 5 0 ( x 1)( x 5) 0 x 1 0 or x 5 0 x 1 or x5 The value x 1 causes a denominator to equal zero, so we disregard it. Thus, the solution set is {5}.
78
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Section 1.2: Quadratic Equations
92.
3x 1 4 7x x 2 x 1 x2 x 2 3x 1 4 7x x 2 x 1 ( x 2)( x 1) 1 4 7x 3x x 2 x 1 ( x 2)( x 1) ( x 2)( x 1) ( x 2)( x 1) 3x( x 1) ( x 2) 4 7 x 3x 2 3x x 2 4 7 x 3x 2 2 x 2 4 7 x 3x 2 5 x 2 0 (3 x 1)( x 2) 0 3x 1 0 or x 2 0 x
1 or 3
x 2
1 The value x 2 causes a denominator to equal zero, so we disregard it. Thus, the solution set is . 3
93. Since this is a right triangle then we can use the Pythagorean Theorem. So (2 x 3) 2 (2 x 5) 2 ( x 7) 2 4 x 2 12 x 9 4 x 2 20 x 25 x 2 14 x 49 12 x 9 x 2 6 x 74 0 x 2 18 x 65 0 ( x 5)( x 13)
x 5 0 or x 13 0 x 5 or
x 13
This means there are 2 possible that meet these requirements. Substituting x into the given sides gives: When x = 5: 5m, 12m, 13m When x = 13: 20m, 21m, 29m Thus there are 2 solutions. 94. Since this is a right triangle then we can use the Pythagorean Theorem. So (4 x 5) 2 (3 x 13)2 x 2 16 x 2 40 x 25 9 x 2 78 x 169 x 2 6 x 2 38 x 144 0 2(3x 2 19 x 72) 0 2(3 x 8)( x 9) 0
3x 8 0 or x 9 0 8 x or 3
x9
This means there are 2 possible solutions that meet these requirements. Substituting x into the given sides gives: When x = 9: 41m, 40m, 9m 8 When x = at least one side of the triangle 3 has a negative measurement which is impossible. Thus there is only 1 triangle possible 95. Let w represent the width of window. Then l w 2 represents the length of the window. Since the area is 143 square feet, we have: w( w 2) 143 w2 2w 143 0 ( w 13)( w 11) 0
w 13 or w 11 Discard the negative solution since width cannot be negative. The width of the rectangular window is 11 feet and the length is 13 feet. 96. Let w represent the width of window. Then l w 1 represents the length of the window. Since the area is 306 square centimeters, we have: w( w 1) 306 w2 w 306 0 ( w 18)( w 17) 0 w 18 or w 17 Discard the negative solution since width cannot
79 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 100. Let x = width of original sheet in feet. Length of sheet: 2x Length of box: 2 x 2 feet Width of box: x 2 feet Height of box: 1 foot V l wh
be negative. The width of the rectangular window is 17 centimeters and the length is 18 centimeters. 97. Let l represent the length of the rectangle. Let w represent the width of the rectangle. The perimeter is 26 meters and the area is 40 square meters. 2l 2 w 26 l w 13 so w 13 l l w 40 l (13 l ) 40
4 2 x 2 x 2 1 4 2x2 6 x 4 0 2x2 6 x 0 x 2 3x 0 x x 3 x 0 or x 3 Discard x 0 since that is not a feasible length for the original sheet. Therefore, the original sheet is 3 feet wide and 6 feet long.
13l l 2 40 l 2 13l 40 0 (l 8)(l 5) 0 l 8 or l 5 w5 w8 The dimensions are 5 meters by 8 meters.
101. a.
98. Let r represent the radius of the circle. Since the field is a square with area 1250 square feet, the length of a side of the square is 1250 25 2 feet. The length of the diagonal is 2r . Use the Pythagorean Theorem to solve for r :
(2r )2 25 2
25 2 2
When the ball strikes the ground, the distance from the ground will be 0. Therefore, we solve 96 80t 16t 2 0 16t 2 80t 96 0 t 2 5t 6 0
t 6 t 1 0 t 6 or t 1 Discard the negative solution since the time of flight must be positive. The ball will strike the ground after 6 seconds.
2
4r 2 1250 1250 4r 2 2500
b. When the ball passes the top of the building, it will be 96 feet from the ground. Therefore, we solve 96 80t 16t 2 96
r 2 625 r 25 The shortest radius setting for the sprinkler is 25 feet.
16t 2 80t 0 t 2 5t 0
99. Let x = length of side of original sheet in feet. Length of box: x 2 feet Width of box: x 2 feet Height of box: 1 foot V l wh
t t 5 0 t 0 or t 5 The ball is at the top of the building at time t 0 when it is thrown. It will pass the top of the building on the way down after 5 seconds.
4 x 2 x 2 1 4 x2 4 x 4 0 x2 4 x 0 x x 4 x 0 or x 4 Discard x 0 since that is not a feasible length for the original sheet. Therefore, the original sheet should measure 4 feet on each side.
80
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Section 1.2: Quadratic Equations 102. a.
To find when the object will be 15 meters above the ground, we solve 4.9t 2 20t 15 4.9t 2 20t 15 0 a 4.9, b 20, c 15 t
20 202 4 4.9 15 2 4.9
20 106 20 106 9.8 9.8 t 0.99 or t 3.09 The object will be 15 meters above the ground after about 0.99 seconds (on the way up) and about 3.09 seconds (on the way down).
b. The object will strike the ground when the distance from the ground is 0. Therefore, we solve 4.9t 2 20t 0 t 4.9t 20 0 t0
or
4.9t 20 0 4.9t 20
t 4.08 The object will strike the ground after about 4.08 seconds. 4.9t 2 20t 100
c.
4.9t 2 20t 100 0 a 4.9, b 20, c 100 20 20 4 4.9 100 2
t
2 4.9
20 1560 9.8 There is no real solution. The object never reaches a height of 100 meters.
103. Let x represent the number of centimeters the length and width should be reduced. 12 x = the new length, 7 x = the new width. The new volume is 90% of the old volume. (12 x)(7 x)(3) 0.9(12)(7)(3) 3x 2 57 x 252 226.8 3 x 2 57 x 25.2 0 2
x 19 x 8.4 0
(19) (19) 2 4(1)(8.4) 19 327.4 2(1) 2 x 0.45 or x 18.55 Since 18.55 exceeds the dimensions, it is discarded. The dimensions of the new chocolate bar are: 11.55 cm by 6.55 cm by 3 cm. x
104. Let x represent the number of centimeters the length and width should be reduced. 12 x = the new length, 7 x = the new width. The new volume is 80% of the old volume. (12 x)(7 x)(3) 0.8(12)(7)(3) 3x 2 57 x 252 201.6 3 x 2 57 x 50.4 0 x 2 19 x 16.8 0 x
2
( 19) ( 19) 4(1)(16.8) 2(1)
19 293.8 2
x 0.93 or x 18.07 Since 18.07 exceeds the dimensions, it is discarded. The dimensions of the new chocolate bar are: 11.07 cm by 6.07 cm by 3 cm.
105. Let x represent the width of the border measured in feet. The radius of the pool is 5 feet. Then x 5 represents the radius of the circle, including both the pool and the border. The total area of the pool and border is AT ( x 5) 2 .
The area of the pool is AP (5) 2 25 . The area of the border is AB AT AP ( x 5)2 25 . Since the concrete is 3 inches or 0.25 feet thick, the volume of the concrete in the border is
0.25 AB 0.25 ( x 5) 2 25
Solving the volume equation:
x 10 x 25 25 108
0.25 ( x 5) 2 25 27 2
x 2 10x 108 0 x
10 (10) 2 4()(108) 2()
31.42 1002 432 6.28 x 2.71 or x 12.71 Discard the negative solution. The width of the border is roughly 2.71 feet.
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Chapter 1: Equations and Inequalities 106. Let x represent the width of the border measured in feet. The radius of the pool is 5 feet. Then x 5 represents the radius of the circle, including both the pool and the border. The total area of the pool and border is AT ( x 5) 2 .
108. Let x = the width and 2x = the length of the patio. The height is 13 foot and the concrete
available is 8 27 216 cubic feet.. 1 V l w h x(2 x) 216 3
2 2 x 216 3 x 2 324 x 18 The dimensions of the patio are 18 feet by 36 feet.
The area of the pool is AP (5) 2 25 . The area of the border is AB AT AP ( x 5)2 25 . Since the concrete is 4 inches = 1 foot thick, the 3 volume of the concrete in the border is 1 1 A ( x 5) 2 25 3 B 3 Solving the volume equation: 1 ( x 5) 2 25 27 3
109. Let x = the length of a 12.9-inch iPad Pro in a 16:9 format. 9 Then x = the width of the iPad. The diagonal 16 of the 12.9-inch iPad is 9.7 inches, so by the Pythagorean theorem we have:
x 10 x 25 25 81 2
2
9 x 2 x 12.92 16 81 2 x2 x 166.41 256 81 2 256 x 2 x 256 166.41 256
x 2 10x 81 0 x
10 (10) 2 4()( 81) 2()
31.42 1002 324 6.28 x 2.13 or x 12.13 Discard the negative solution. The width of the border is approximately 2.13 feet.
256 x 2 81x 2 42600.96 337 x 2 42600.96 x 2 42600.96 337 x
107. Let x represent the width of the border measured in feet. The total area is AT (6 2 x )(10 2 x) . The area of the garden is AG 6 10 60 . The area of the border is AB AT AG (6 2 x)(10 2 x) 60 . Since the concrete is 3 inches or 0.25 feet thick, the volume of the concrete in the border is 0.25 AB 0.25 (6 2 x )(10 2 x) 60
42600.96 11.24 337
Since the length cannot be negative, the length of the iPad is 9 16
42600.96 inches and the width is 337
42600.96 6.32 inches. 337
iPad is
42600.96 9 16 337
Thus, the area of the
42600.96 71.11 square 337
inches. Let y = the length of a 14.4-inch 3:2 format 2 Microsoft Surface Pro. Then y = the width of 3 the Surface Pro. The diagonal of a 14.4-inch Surface Pro is 14.4 inches, so by the Pythagorean theorem we have:
Solving the volume equation: 0.25 (6 2 x)(10 2 x ) 60 27 60 32 x 4 x 2 60 108 4 x 2 32 x 108 0
2
2 y 2 y 14.42 3 4 y 2 y 2 207.36 9 4 9 y 2 y 2 9 207.36 9
2
x 8 x 27 0 2
8 8 4(1)( 27) 8 172 2(1) 2 x 2.56 or x 10.56 Discard the negative solution. The width of the border is approximately 2.56 feet. x
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Section 1.2: Quadratic Equations
9 y 2 4 y 2 1866.24
2
13 y 2 1866.24 y2
1866.24 13
1866.24 11.98 13 Since the length cannot be negative, the length of the Surface Pro is 11.98 inches and the width is y
2 1866.24 7.99 inches. Thus, the area of the 3 13 14.4-inch 3:2 format Surface Pro is 1866.24 2 3 13
10 y 2 y 82 16 100 2 y2 y 64 256 100 2 256 y 2 y 256 64 256 256 y 2 100 y 2 16384 356 y 2 16384 y2
16384 356
16384 6.78399 356 Since the length cannot be negative, the length of 16384 6.78399 inches and the the Fire is 356 y
1866.24 13
95.7 square inches. The Surface Pro format has the larger screen since its area is larger.
110. Let x = the length of a 8.3-inch iPad Mini in a 4:3 format. 3 Then x = the width of the iPad. The diagonal 4 of the 8.3-inch iPad is 8.3 inches, so by the Pythagorean theorem we have: 2
3 x 2 x 8.32 4 9 x 2 x 2 68.89 16 9 16 x 2 x 2 16 68.89 16
10 16384 4.240 inches. Thus, the area 16 356 of the Amazon Fire is 6.78399 4.240 28.8 square inches.
width is
The iPad Mini™ 4:3 format has the larger screen since its area is larger. 111. Let h be 1.1. Then 1.1 0.00025 x 2 0.04 x 0 0.00025 x 2 0.04 x 1.1 x
16 x 2 9 x 2 1102.24
2
0.04 (0.04) 4( 0.00025)( 1.1) 2( 0.00025)
35.3 ft or 124.7 ft 124.7 ft does not make sense in the context of the problem, so the answer is 35.3 ft.
25 x 2 1102.24 x 2 44.0896 x 44.0896 6.64 Since the length cannot be negative, the length of the iPad is 6.64 inches and the width is 3 6.32 4.98 inches. Thus, the area of the 4 iPad is (6.64)(4.98) 33.1 square inches. Let y = the length of a 8-inch 16:10 format 10 Amazon Fire HD 8™. Then y = the width of 16 the Fire. The diagonal of a 8-inch Fire is 8 inches, so by the Pythagorean theorem we have:
112. Since d is expressed in 1000’s we will set d = 12 and solve for x using the Quadratic Formula. d 0.33x 2 0.26 x 2.29 12 0.33x 2 0.26 x 2.29 0 0.33x 2 0.26 x 9.71 x
0.26 (0.26) 2 4(0.33)( 9.71) 2(0.33) 0.26 13.0772 0.66
x 5.085 or x 5.873 So the nearest year when the earnings were 12 occurred about 5 years after 2018 or 2023. The negative value -5.873 has no meaning.
83 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 113. We will set g = 2.97 and solve for h using the Quadratic Formula. g 0.0006 x 2 0.015 x 3.04
115. Let a be the age the individual is able to start saving money. Then we need to find where the models are equal. Solving these two equations together: 25a 2 2400a 30700 160a 7840
2.97 0.0006 x 2 0.015 x 3.04 0 0.0006 x 2 0.015 x 0.07
25a 2 2240a 38540 0
2
x
0.015 (0.015) 4( 0.0006)(0.07)
a
2( 0.0006)
0.015 0.000393 0.0012
a
x 29 or x 4.02
2
2240 (2240) 4(25)(38540) 2(25) 2240 1163600 50
2240 1078.7 50 2240 1078.7 2240 1078.7 or a a 50 50 or a 23.2 a 66.4 a
So the estimated numbers of hours worked by a student with a GPA of 2.97 is 29 hours. The value -4.02 has no meaning since it is negative. 114. Let x be the numbers of members in the fraternity and s be the share paid by each 1470 . If there are 7 member. Then s x members who cannot contribute then the share goes up by $5. So we have the following equation: 1470 or s 5 x 7 1470 s5 x7
Since x is the age to start saving, it makes sense that the answer is approximate at age 23. 1 n n 3 65 2 n n 3 130
116.
n 2 3n 130 0
n 13 n 10 0
Solving these two equations together:
n 13 or n 10 Since the number of sides cannot be negative, we discard the negative value. A polygon with 65 diagonals will have 13 sides.
s 5 x 7 1470 and s 1470 x 1470 x 5 x 7 1470
1 n n 3 80 2 n n 3 160
10290 5 x 35 1470 x 10290 5x 35 0 x 5 x 2 35 x 10290 0
1470
n 2 3n 160 0 a 1, b 3, c 160
n
5 x 2 35 x 10290 0
3
32 4 1 160 3 646 2 1 2
Neither solution is an integer, so there is no polygon that has 80 diagonals.
x 2 7 x 2058 0 ( x 42)( x 49) 0 x 42 or x 49
Since x is the number of members, it must be positive so the number of members is 49.
84
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Section 1.2: Quadratic Equations 117. The roots of a quadratic equation are x1
2
b 2 4ac 0 2
b b 4ac b b 4ac and x2 2a 2a
b b 2 4ac b b 2 4ac x1 x2 2a 2a b b 2 4ac b b 2 4ac 2a 2b 2a b a
k 2 4 1 4 0 k 2 16 0
k 4 k 4 0 k 4 or k 4 121. For ax 2 bx c 0 : b b 2 4ac 2a
x
For ax 2 bx c 0 :
118. The roots of a quadratic equation are b b 2 4ac b b 2 4ac x1 and x2 2a 2a 2 2 b b 4ac b b 4ac x1 x2 2a 2a
b 2
b 4ac b b 4ac 2
x
2a b b 2 4ac 2a
2
2
2a 2
2
4a 2
4ac 4a 2 c a
122. For ax 2 bx c 0 : x1
b b 2 4ac b b 2 4ac and x2 2a 2a
For cx 2 bx a 0 : x1*
119. In order to have one repeated solution, we need the discriminant to be 0. b 2 4ac 0 12 4 k k 0
1 4k 2 0 4k 2 1 1 k2 4 k k
b 2 4ac
b
1 2
b b 2 4 c a 2c
b b 4ac b b 2 4ac 2c b b 2 4ac
b 2 b 2 4ac
2
2c b b 4ac 2a
b b 2 4ac 1 x2 or k
b b 2 4ac 2c
2
1 4
1 2
120. In order to have one repeated solution, we need the discriminant to be 0.
85 Copyright © 2025 Pearson Education, Inc.
4ac
2c b b 4ac 2
Chapter 1: Equations and Inequalities
The first equation has the solution set 1
and b b 4 c a 2
x2*
2c
while the second equation has no solutions.
2
b b 4ac 2c
125. Answers will vary. Methods may include the quadratic formula, completing the square, graphing, etc. 126. Answers will vary. Knowing the discriminant allows us to know how many real solutions the equation will have. 127. Answers will vary. One possibility: Two distinct: x 2 3x 18 0 One repeated: x 2 14 x 49 0 No real: x 2 x 4 0
b b 2 4ac b b 2 4ac 2c b b 2 4ac
b 2 b 2 4ac
2
2c b b 4ac
4ac
2c b b 4ac 2
2a
b b 2 4ac 1 x1
128. Answers will vary.
123. If x = original width and y = original length, then 1 xy 1 or x . The ratio of side lengths is y x 1 . Folding along the longest side results y y2
Section 1.3
y 1 whose ratio is in sides of length x and y 2
1. Integers: 3, 0
y 2 2 y 1 2 y
Rationals: 3, 0,
2. True; the set of real numbers consists of all rational and irrational numbers.
Equating the ratios gives y2 1 2 2 y
3.
3 3 2 3 2 3 2 3 2 3
2 3 3 2 3
y4 2
4
y 2m
So 4 1 8 x 4 m. 2 2
124. a.
6 5
x 9 and x 3 are equivalent because 9 3.
c.
x 1 x 2 x 12 and x 2 x 1 are
2
2
43
3 2 3
x 2 9 and x 3 are not equivalent because they do not have the same solution set. In the first equation we can also have x 3 .
b.
3 2 3
4. real; imaginary; imaginary unit 5. False; the conjugate of 2 5i is 2 5i . 6. True; the set of real numbers is a subset of the set of complex numbers. 7. False; if 2 3i is a solution of a quadratic equation with real coefficients, then its conjugate, 2 3i , is also a solution.
not equivalent because they do not have the same solution set.
8. b 86
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Section 1.3: Complex Numbers; Quadratic Equations in the Complex Number System 9. a
25.
10. c 11. (2 3i ) (6 8i ) (2 6) (3 8)i 8 5i 12. (4 5i ) ( 8 2i ) (4 ( 8)) (5 2)i 4 7i 13. (3 2i ) (4 4i ) (3 4) (2 ( 4))i 7 6i
26.
14. (3 4i ) (3 4i ) (3 (3)) ( 4 ( 4))i 6 0i 6 15. (2 5i ) (8 6i ) (2 8) (5 6)i 6 11i 16. ( 8 4i ) (2 2i ) ( 8 2) (4 ( 2))i 10 6 i 17. 3(2 6i ) 6 18 i
27.
18. 4(2 8i ) 8 32 i 19. 3i (7 6i ) 21i 18i 2 21i 18(1) 18 21i
28.
20. 3i (3 4i ) 9i 12i 2 9i 12(1) 12 9i 21. (3 4i )(2 i ) 6 3i 8i 4i 2 6 5i 4(1) 10 5i 22. (5 3i )(2 i ) 10 5i 6i 3i 2 10 i 3(1) 13 i 23. ( 5 i )( 5 i ) 25 5i 5i i 2 25 (1) 26 24. ( 3 i )(3 i ) 9 3i 3i i 2 9 (1) 10
10 10 3 4i 30 40i 3 4i 3 4i 3 4i 9 12i 12i 16i 2 30 40i 30 40i 9 16(1) 25 30 40 i 25 25 6 8 i 5 5 13 13 5 12i 5 12i 5 12i 5 12i 65 156i 25 60i 60i 144i 2 65 156i 65 156i 25 144(1) 169 65 156 i 169 169 5 12 i 13 13 2 i 2 i i 2i i 2 i i i i 2 2i (1) 1 2i 1 2i 1 (1) 2 i 2 i i 2i i 2 2i 2i i 2i 2
29.
30.
2i (1) 1 2i 1 i 2(1) 2 2
6 i 6 i 1 i 6 6i i i 2 1 i 1 i 1 i 1 i i i2 6 7i (1) 5 7i 5 7 i 1 (1) 2 2 2 2 3i 2 3i 1 i 2 2i 3i 3i 2 1 i 1 i 1 i 1 i i i2 2 5i 3(1) 1 5i 1 5 i 1 (1) 2 2 2 2
1 3 1 1 3 3 2 i 2 i i 31. 2 2 4 2 2 4
87 Copyright © 2025 Pearson Education, Inc.
1 3 3 1 3 i (1) i 4 2 4 2 2
Chapter 1: Equations and Inequalities
2
46. 2i 4 (1 i 2 ) 2(1)(1 (1)) 2(0) 0
3 1 3 1 1 2 3 i 2 32. i i 4 2 2 2 2 4
i i i
i i i i i i
48. i 7 i 5 i 3 i i 2
2
0
7
1 1 1 1 38. i 23 23 22 1 22 2 11 i i i i (i ) i 1 1 1 i i i 2 i 11 (1) (1) i i i i i
5 (1) 5 1 5 6 3
2 2
i i i i
1 1 1 37. i 20 20 20 2 10 (i ) i i 1 1 1 10 1 (1)
39. i 6 5 i 2
3
(1)3 i (1) 2 i (1) i i
11
(1) 1
36. i14 i
3
49.
4 2i
50.
9 3i
51.
25 5i
52.
64 8i
53.
12 i 4 3 2 3i
54.
18 i 9 2 3 2i
55.
200 i 100 2 10 2i
56.
45 i 9 5 3 5i
57.
(3 4i )(4i 3) 12i 9 16i 2 12i
40. 4 i 3 4 i 2 i 4 (1) i 4 i 3
5
3
2
0
i (1) i i
2 7
2 2
1 1 1 1
34. (1 i ) 2 1 2i i 2 1 2i (1) 2i 11
2 3
(1) 4 (1)3 (1) 2 1
33. (1 i ) 2 1 2i i 2 1 2i (1) 2i
35. i 23 i 22 1 i 22 i i 2
4
47. i8 i 6 i 4 i 2 i 2
3 3 1 1 3 i (1) i 4 2 4 2 2
2
41. 6i 4i i (6 4i )
i 2 i (6 4(1)) 1 i (10) 10 i
9 16(1)
42. 4i 3 2i 2 1 4i 2 i 2i 2 1 4(1) i 2(1) 1 4i 2 1 3 4i
25 5i
58.
(4 3i )(3i 4) 12i 16 9i 2 12i
43. (1 i )3 (1 i )(1 i )(1 i ) (1 2i i 2 )(1 i ) (1 2i 1)(1 i ) 2i (1 i )
16 9(1) 25
2i 2i 2 2i 2(1)
5i
2 2i
44. (3i ) 4 1 81i 4 1 81(1) 1 82 45. i 7 (1 i 2 ) i 7 (1 (1)) i 7 (0) 0 88
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Section 1.3: Complex Numbers; Quadratic Equations in the Complex Number System
59. x 2 4 0
66. x 2 2 x 5 0 a 1, b 2, c 5
x 2 4
b 2 4ac ( 2) 2 4(1)(5) 4 20 16
x 4 x 2i The solution set is 2i, 2i .
x
The solution set is 1 2 i, 1 2i .
2
60. x 4 0 ( x 2)( x 2) 0 x 2 or x 2
67. 25 x 2 10 x 2 0 a 25, b 10, c 2
The solution set is 2, 2 .
b 2 4ac (10) 2 4(25)(2) 100 200 100 ( 10) 100 10 10i 1 1 i 50 50 5 5 1 1 1 1 i, i . The solution set is 5 5 5 5 x
61. x 2 16 0 x 4 x 4 0
x 4 or x 4 The solution set is 4, 4 .
b 2 4ac 62 4(10)(1) 36 40 4
x 25 5i
x
The solution set is 5i, 5i .
63. x 6 x 13 0 a 1, b 6, c 13, b 2 4ac ( 6) 2 4(1)(13) 36 52 16
2
x
64. x 4 x 8 0 a 1, b 4, c 8
(2) 16 2 4i 1 2 i 2(5) 10 5 5
The solution set is
2
b 4ac 4 4(1)(8) 16 32 16
1 2 1 2 i, i . 5 5 5 5
13x 2 1 6 x
70.
13x 2 6 x 1 0 a 13, b 6, c 1 b 2 4ac (6) 2 4(13)(1) 36 52 16
65. x 2 6 x 10 0 a 1, b 6, c 10 2
b 4ac ( 6) 4(1)(10) 36 40 4 x
3 1 3 1 i, i . 10 10 10 10
b 2 4ac 2 4(5)(1) 4 20 16
2
The solution set is 2 2i, 2 2i .
5x2 2 x 1 0 a 5, b 2, c 1
The solution set is 3 2i,3 2i .
4 16 4 4i 2 2i 2(1) 2
5x2 1 2 x
69.
( 6) 16 6 4i x 3 2i 2(1) 2
2
6 4 6 2i 3 1 i 2(10) 20 10 10
The solution set is
2
x
68. 10 x 2 6 x 1 0 a 10, b 6, c 1
62. x 2 25 0 x 2 25
2
( 2) 16 2 4i 1 2i 2(1) 2
( 6) 4 6 2i 3i 2(1) 2
x
( 6) 16 6 4i 3 2 i 2(13) 26 13 13
The solution set is
The solution set is 3 i, 3 i . 89 Copyright © 2025 Pearson Education, Inc.
3 2 3 2 i, i . 13 13 13 13
Chapter 1: Equations and Inequalities
71. x 2 x 1 0 a 1, b 1, c 1, 2
x 4 16 0
x 4 x 4 0 ( x 2)( x 2) x 4 0
2
b 4ac 1 4(1)(1) 1 4 3 x
2
1 3 1 3 i 1 3 i 2(1) 2 2 2
x 2 0 or x 2 0 or x 2 4 0 x 2 or x 2 or x 4 2i The solution set is 2, 2, 2i, 2i .
b 2 4ac (1) 2 4(1)(1) 1 4 3
x4 1 0
x 1 x 1 0 ( x 1)( x 1) x 1 0 2
1 3 1 3 i, i . The solution set is 2 2 2 2
73. x3 64 0
x4 1
76.
(1) 3 1 3 i 1 3 i 2(1) 2 2 2
2 2
x 1 0 or x 1 0 or x 2 1 0 x 1 or
( x 4) x 2 4 x 16 0
x 1 or x 2 1
x 1 or x 1 or x 1 i The solution set is 1, 1, i, i .
x4 0 x 4 or x 2 4 x 16 0 a 1, b 4, c 16
77.
4 48 4 4 3 i x 2 2 3i 2(1) 2
x 4 13x 2 36 0
x 9 x 4 0 2
b 2 4ac 42 4(1)(16) 16 64 48
2
x2 9 0
or x 2 4 0
x 2 9
or
x 2 4
x 9 or x 4 x 3i x 2i or The solution set is 3i, 3i, 2i, 2i .
The solution set is 4, 2 2 3i, 2 2 3i .
x 2 or x 2 4
x 2 or
72. x 2 x 1 0 a 1, b 1, c 1
74. x3 27 0
2 2
1 3 1 3 i, i . The solution set is 2 2 2 2
x
x 4 16
75.
( x 3) x 2 3 x 9 0
78.
x 3 0 x 3
x 4 3x 2 4 0
or x 3x 9 0 a 1, b 3, c 9
x 1 x 4 0 ( x 1)( x 1) x 4 0
b 2 4ac (3)2 4(1)(9) 9 36 27
x 1 0 or x 1 0 or x 2 4 0
2
2
x
2 2
(3) 27 3 3 3 i 3 3 3 i 2(1) 2 2 2
x 1 or
x 1 or x 2 4
x 1 or x 1 or x 4 2i The solution set is 1, 1, 2i, 2i .
3 3 3 3 3 3 i, i . The solution set is 3, 2 2 2 2
79. 3x 2 3 x 4 0 a 3, b 3, c 4
b 2 4ac ( 3) 2 4(3)(4) 9 48 39 The equation has two complex solutions that are conjugates of each other. 90
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Section 1.3: Complex Numbers; Quadratic Equations in the Complex Number System
80. 2 x 2 4 x 1 0 a 2, b 4, c 1
90. z w 3 4i (8 3i ) 3 4i 8 3i
b 2 4ac (4) 2 4(2)(1) 16 8 8 The equation has two unequal real number solutions.
81.
2 x 2 3x 4
5 7i 5 7i V 18 i 18 i 3 4i I 3 4i 3 4i 3 4i 54 72i 3i 4i 2 54 75i 4 9 16 9 12i 12i 16i 2 50 75i 2 3i 25 The impedance is 2 3i ohms.
91. Z
2 x2 3x 4 0 a 2, b 3, c 4 b 2 4ac 32 4(2)(4) 9 32 41 The equation has two unequal real solutions.
82.
x2 6 2x x2 2 x 6 0 a 1, b 2, c 6
92.
b 2 4ac (2) 2 4(1)(6) 4 24 20 The equation has two complex solutions that are conjugates of each other.
b 2 4ac ( 12) 2 4(9)(4) 144 144 0 The equation has a repeated real solution.
84. 4 x 2 12 x 9 0 a 4, b 12, c 9 b 2 4ac 122 4(4)(9) 144 144 0 The equation has a repeated real solution.
85. The other solution is 2 3i 2 3 i. 86. The other solution is 4 i 4 i. 87. z z 3 4i 3 4i 3 4i 3 4i 6
6 2i 8 6i 4i 3i
2
6 2i 6 2i 8 2i 3 11 2i
11 2i 11 2i 6 2i 6 2i 6 2i 6 2i 66 22i 12i 4i 2 66 10i 4 36 4 36 12i 12i 4i 2 70 10i 7 1 i 40 4 4 7 1 The total impedance is i ohms. 4 4
So, Z
83. 9 x 2 12 x 4 0 a 9, b 12, c 4
88. w w 8 3i 8 3i
1 1 1 1 1 (4 3i ) (2 i ) Z Z1 Z 2 2 i 4 3i (2 i )(4 3i )
93. z z (a b i ) (a b i ) a bi a bi 2a z z a b i (a b i) a b i (a b i) a bi a bi 2b i
8 3i (8 3i )
94. z a b i a b i a b i z
8 3i 8 3i 0 6i 6i
95. z w (a b i ) (c d i )
89. z z (3 4i )(3 4i )
(a c) (b d ) i
(3 4i )(3 4i )
(a c) (b d ) i ( a b i ) (c d i )
9 12i 12i 16i 2 9 16(1) 25
a bi c d i z w
91 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
This is only possible if v = 0 which then makes u = 0. Therefore, x 0 5 5 and y 0 5 5 , so x y 5 5 10
96. z w (a b i ) (c d i ) ac ad i bc i bd i 2
103. Answers will vary. A complex number is the sum or difference of two numbers (real and imaginary parts of the complex number) just as a binomial is the sum or difference of two monomial terms. We multiply two binomials by using the FOIL method, an approach we can also use to multiply two complex numbers.
z w a bi c d i (a b i )(c d i ) ac ad i bc i bd i 2 (ac bd ) (ad bc)i
a bi a bi 2
97.
104. Although the set of real numbers is a subset of the set of complex numbers, not all rules that work in the real number system can be used in the larger complex number system. The rule that allows us to write the product of two square roots as the square root of the product only works in the real number system. That is, a b ab only when a and b are real numbers. In the complex number system we must first convert the radicals to complex form. In this case this means we need to write 9 as
2
a 2abi b i a 2abi b i
a 2 2abi (bi ) 2 a 2 2abi (bi ) 2 2
2 2
Answers will vary.
100 – 102.
(ac bd ) (ad bc)i (ac bd ) (ad bc)i
2
2 2
a 2 2abi b 2 a 2 2abi b 2 a 2 b 2 ( a 2 b 2 ) 2(a 2 b 2 ) 0 a 2 b2 b a Any complex number of the form a ai or a ai will work.
1 9 9 1 3i . Then we can multiply to get
9 9 3i 3i 9i 2 9 1 9 .
98. Let u 3 2 in x3 2 0 so that x3 u 3 0 .
Then, ( x u )( x 2 ux u 2 ) 0 . From the first factor we find x u 3 2 . From the second factor, use the quadratic formula to get x
2
(u ) (u ) 4 1 u 2 1
Section 1.4
2
1. True
3 u 3u 2 u u 3 2 32 3 i i 2 2 2 2 2 The solution set is: 3 32 32 3 i 2, 2 2
2.
x x 3
3
3. 6 x3 2 x 2 2 x 2 3 x 1 4. False; you can also use the Quadratic Formula or completing the square.
99. ( x 5)( y 5) ( x y ) 2 ; let u x 5 (so x u 5 and v y 5 so y v 5 .
5. quadratic in form
Substituting gives uv (u v) 2 or
6. True
u 2 uv v 2 0 which is quadratic in u. Using the quadratic formula gives v v 2 4 1 v 2 v v 3 x . Since x 2 1 2 is a real number, u must also be a real number.
7. a 8. c
92
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Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
2t 1 1
9.
2t 1 1 2
x 2 2 x 1
x 2x 1
2
5
2t 1 1 2t 2 t 1 Check:
5
15.
5
2
x 2 2 x 1 x2 2 x 1 0
x 12 0
2(1) 1 1 1
x 1 0 x 1
The solution set is {1}. 3t 4 2
10.
3t 4 2 2
Check: 5 1 2 1 5 1 2 5 1 1 2
2
The solution set is {1}.
3t 4 4 3t 0
16.
t0 Check:
4
x 16 5 4
3(0) 4 4 2
x 2 16 5 4
2
4
x 2 16 25
The solution set is {0}.
11.
5
x2 9 x 3
3t 4 6 Since the principal square root is never negative, the equation has no real solution.
Check 3 : 4 3 16 4 9 16 4 25 5 2
Check 3 : 4 3 16 4 9 16 4 25 5 2
12.
5t 3 2 Since the principal square root is never negative, the equation has no real solution.
13.
3
x 2 8 x
1 2x 3 3
3
3 2
x 64 x 0 x x 64 0
1 2 x 27 2 x 26 x 13 Check:
x 0 or x 64 Check 0: 0 8 0 00
3 1 2( 13) 3 3 27 3 0
The solution set is {13}. 3
x 2 3 x
1 2x 1 3
3
x2 9 x
1 2x 1 2x 0 x0 Check:
x3 x
18.
1 2x 1 3
Check 64: 64 8 64 64 64
The solution set is 0, 64 .
1 2x 1 0 3
2
x 2 64 x
1 2x 3
14.
x8 x
17.
1 2x 3 0 3
The solution set is 3,3 .
x2 9 x 0 x x 9 0 x0
3 1 2(0) 1 3 1 1 0
The solution set is {0}.
93 Copyright © 2025 Pearson Education, Inc.
or x 9
2
Chapter 1: Equations and Inequalities
Check 9: 9 3 9 Check 0: 0 3 0 00 99 The solution set is 0,9 .
x2 2 x 1
2
15 2 x x
2
x2 4 x 4 0 ( x 2) 2 0 x 2
2
x 2 x 15 0 ( x 5)( x 3) 0 x 5 or x 3
Check: 2 2 ( 2) 1 2 2 The equation has no real solution.
15 2(5) 25 5 5
Check –5:
Check 3: 15 2(3) 9 3 3 Disregard x 5 as extraneous. The solution set is {3}.
20.
23.
12 x x 2
2
12 x x 2
x 2 2
2
2
Check:
12 ( 4) 16 4 4
Check 3: 12 3 9 3 3 Disregard x 4 as extraneous. The solution set is {3}.
21. x 2 x 1
x2 x 4
8 8 8 5 54 5 2 64 8 2 4 25 5 5 4 2 25 5 2 2 5 5 8 The solution set is . 5
x 2 x 12 0 ( x 4)( x 3) 0 x 4 or x 3
x2 2 x 1
x2 x 4 x 2
x2 x 4 x2 4 x 4 8 5 x 8 x 5
12 x x
Check –4:
2
x2 4x 4
15 2 x x 2
x 2 4( x 1)
15 2 x x
19.
x 2 x 1
22.
2
24.
2
x 4( x 1)
3 x x2 x 2
3 x x x 2 2
x2 4x 4 x2 4 x 4 0
2
2
3 x x2 x2 4 x 4 3x 1 1 x 3
( x 2) 2 0 x2 Check: 2 2 2 1 22 The solution set is {2}.
2
1 1 1 3 2 3 3 3 1 1 5 3 3 9 3 Since the principal square root is always a nonnegative number; x 13 does not check.
Check:
Therefore this equation has no real solution. 94
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Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
25. 3 3x 1 x
28.
1 x 3 x 2
3x 1 x 3
3x 1 ( x 3) 2
1 x x 5
1 x ( x 5) 2
2
3x 1 x 2 6 x 9
1 x x 2 10 x 25
0 x2 9 x 8 0 ( x 1)( x 8) x 1 or x 8 Check 1: 3 3(1) 1 3 4 5 1
0 x 2 11x 24 0 ( x 3)( x 8) x 3 or x 8 Check 3: 1 ( 3) 3 3 2 1 1
Check 8: 3 3(8) 1 3 25 8 8
Check 8:
Discard x 1 as extraneous. The solution set is {8}.
Discard x 8 as extraneous. The solution set is {-3}.
26. 2 12 2 x x
29.
12 2 x ( x 2) 2
3x 5 2 x 7
3x 5 2 x 7 2
2
12 2 x x 2 4 x 4 0 x 2x 8
2 x 16 4 x 7
( x 2)( x 4) 0
(2 x 16) 2 4 x 7
x 2 or x 4 Check 2: 2+ 12 2( 2) 2 16 6 2 Check 4: 2 12 2(4) 2 4 4 4 Discard x 2 as extraneous. The solution set is {4}. 3( x 10) 4 x 3( x 10) x 4
3( x 10) ( x 4)
2
4 x 2 64 x 256 16 x 112 4 x 2 80 x 144 0
x 2 20 x 36 0 ( x 2)( x 18) 0 x 2 or x 18 3(2) 5 2 7
Check 2:
1 9 1 3 2 2 Check 18:
0 x 2 5 x 14 0 ( x 7)( x 2) x 7 or x 2
Check 2:
4 x 2 64 x 256 16( x 7)
2
3x 30 x 2 8 x 16
Check 7:
2
3x 5 4 4 x 7 x 7
2
2
1 ( 8) 3 8 2 0 6
3x 5 x 7 2
12 2 x x 2
27.
2
3(18) 5 18 7
49 25 7 5 2 2 Discard x 2 as extraneous. The solution set is {18}.
3( 7 10) 4 9 4 1 7 3(2 10) 4 36 4 2 2
Discard x 7 as extraneous. The solution set is {2}.
95 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
30.
3x 7 x 2 1
2x 3 x 1 1
32.
2x 3 1 x 1
3x 7 1 x 2
3x 7 1 x 2 2
2 x 3 1 x 1 2
2
3x 7 1 2 x 2 x 2
2x 3 1 2 x 1 x 1
2x 4 2 x 2
x 1 2 x 1
x 2 x 2
( x 1) 2 2 x 1
( x 2) 2
x 2
2
x2 2x 3 0 ( x 1)( x 3) 0 x 1 or x 3
1 0 1 0 1 1
4 1 2 1 3 1
9 4 3 2 11
1 0 1 0 1 1 Discard x 1 as extraneous. The solution set is {2}.
The solution set is 1,3 . 32 x x
33.
3x 1 x 1 2 3x 1 2 x 1 2
2
2 x x 3 2
2
0 x 2 10 x 9 0 x 1 x 9 x 1 or x 9
x2 2x 1 4 x 4 x2 6 x 5 0 ( x 1)( x 5) 0 x 1 or x 5
Check 1:
3(1) 1 1 1
Check 9:
3 2 1 1
3 2 9 9
3 2 1
3 23 3
1 1
4 0 20 2 2 Check 5:
2
4 x x2 6x 9
2
4 x 2 8 x 4 16( x 1)
Check 1:
x 2
2 x x 3
2x 2 4 x 1
32 x
32 x x
3x 1 4 4 x 1 x 1
(2 x 2) 2 4 x 1
2(3) 3 3 1
Check 3:
3( 2) 7 2 2
3x 1 2 x 1
2(1) 3 1 1
Check –1:
3(1) 7 1 2
Check 2:
2
x2 2x 1 4 x 4
x 2 3x 2 0 ( x 1)( x 2) 0 x 1 or x 2 Check –1:
x 2 2 x 1 4( x 1)
x2 4 x 4 x 2
31.
2
3 3
11 Discard x 9 as extraneous. The solution set is {1}.
3(5) 1 5 1
16 4 4 2 2 2
The solution set is 1,5 .
96
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Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
5 x 2 1/ 3 2
37.
10 3 x x
34.
10 3 x x 2
5x 2 2 1/ 3 3
2
5x 2 8 5 x 10 x2
10 3 x x 3 x x 10
3 x x 10 2
Check: 5 2 2
1/ 3
2
0 x 2 29 x 100
2 x 1 1 1/ 3 3
or x 25
10 3 4 4
10 3 25 25
10 3 2 2
10 3 5 5
16 2
25 5
Check: 2 1 1
1/ 3
1/ 2 2
39.
x 9 2
1/ 2
2
2
x 2 16 x 16 4
4 9 25 5 Check 4: 4 9 25 5
x5 1/ 2
161/ 2 4
1/ 2 2
3x 5
1/ 2
The solution set is 4, 4 .
2 2
1/ 2
1/ 2
2
The solution set is {5}.
1/ 2
2
Check 4 :
Check: 3 5 1
3x 5
1
x 2 9 1/ 2 5 2 x 2 9 25
3 x 15
36.
1/ 3
5
3 x 1 16
1/ 2
1
The solution set is {1}.
55
3x 11/ 2 4
3x 1 4
3
2 x 1 1 2 x 2 x 1
Check 25:
42 Discard x 4 as extraneous. The solution set is {25}.
35.
2 x 11/ 3 1
38.
0 x 4 x 25
Check 4:
81/ 3 2
The solution set is {2}.
9 x x 2 20 x 100
x4
3
40.
2
x 16 2
1/ 2
9 2
x 2 16 1/ 2 9 2 x 2 16 81
3x 5 4 3x 9
x3
Check: 3 3 5
1/ 2
1/ 2
4
The solution set is {3}.
x 2 97
2
x 97
1/ 2
2 Check 97 : 97 16
Check
97 :
97
2
1/ 2
16
811/ 2 9
The solution set is 97, 97 .
97 Copyright © 2025 Pearson Education, Inc.
811/ 2 9
Chapter 1: Equations and Inequalities
41. x3/ 2 3x1/ 2 0
47.
x1/ 2 x 3 0
x 8 x 1 0
x1/ 2 0 or x 3 0 x 0 or x3 3/ 2 Check 0: 0 3 01/ 2 0 0 0 Check 3: 33/ 2 3 31/ 2 3 3 3 3 0 The solution set is 0,3 .
x3 8 0
3
48.
x0
49.
2
x 4 x 1 0
u 2 7u 12 0
u 3 u 4 0
x 2 4 0 or x 2 1 0 x 2 or x 1 The solution set is 2, 1,1, 2 .
u 3 0
x 5 or x 6 The solution set is 6, 5 .
x 5 x 5 0 2
50.
The solution set is 5, 5 .
2
u2 u 6 0
u 3 u 2 0
6 x 1 x 1 0 2
6 x2 1 0
or x 2 1 0
6 x 2 1 or x2 1 Not real or x 1 The solution set is 1,1 .
46.
or
u20
u 3
or
u 2
2x 5 3
or
2 x 5 2
x 1 or
x
2 x 3 x 4 0 2 x2 3 0
u 3 0
7 2
7 The solution set is , 1 . 2
2 x 4 5 x 2 12 0 2
2 x 5 2 2 x 5 6 0 Let u 2 x 5 so that u 2 2 x 5 .
6 x4 5x2 1 0
2
u 4
x 2 3 or x 2 4
x2 5 0
45.
or u 4 0
u 3 or
x 4 10 x 2 25 0
x 2 2 7 x 2 12 0 Let u x 2, so that u 2 x 2 .
2
x 5
x 3 1
x 2 or x 1 The solution set is 1, 2 .
x4 5x2 4 0
2
3
x3 8 or
Check 0: 03/ 4 9 01/ 4 0 0 0 Check 81: 813/4 9 811/ 4 27 27 0 The solution set is 0,81 .
44.
x 8 x 1 0
x3 8 0 or x3 1 0
x 81
2
x3 1
x 6 7 x3 8 0
3
0 or x1/ 2 9
x
or x3 1 0
x 2 or x 1 The solution set is 2,1 .
x1/ 4 x1/ 2 9 0 1/ 4
3
x3 8 or
x3/ 4 9 x1/ 4 0
42.
43.
x 6 7 x3 8 0
2
51.
or x 2 4 0
2 x 2 3 or x2 4 Not real or x 2 The solution set is 2, 2 .
4 x 9 2 10 4 x 9 25 0 2 Let u 4 x 9 so that u 2 4 x 9 .
98
Copyright © 2025 Pearson Education, Inc.
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations 3u 2 0
u 2 10u 25 0
2 or u 1 3 2 1 y or 1 y 1 3 5 y y2 or 3 5 The solution set is , 2 . 3
u 5 2 0
u
u 5 0 u 5 4x 9 5 4 x 14 x
7 2
The solution set is 52.
or u 1 0
7 . 2
55.
x 4x x 0
x 1 4 x 0
2 x 2 x 20 0 2
x 0 or 1 4 x 0
Let u 2 x so that u 2 x . 2
2
1 4 x 1 4 2 1 4
2
u u 20 0
x
u 5 u 4 0 u 5 0
or u 4 0
u 5 or or
Check: x 0 : 0 4(0) 0 0 00
x 2
The solution set is 2, 7 .
x 161 :
53. 2 s 1 5 s 1 3 2
Let u s 1 so that u 2 s 1 . 2
2
2u 5u 3
2u 1 u 3 0 or u 3 0
56. x 8 x 0 8 x x
1 or u u3 2 1 or s 1 3 s 1 2 3 or s s2 2 3 The solution set is , 2 . 2
8 x x 2
2
64 x x 2 0 x 2 64 x 0 x x 64
x 0 or x 64
54. 3 1 y 5 1 y 2 0 2
Let u 1 y so that u 2 1 y . 2
3u 2 u 1 0
1 1 0 16 16
2u 5u 3 0
3u 2 5u 2 0
161 4 161 161 0 1 4 161 14 0 16
00 1 The solution set is 0, . 16
2
2u 1 0
2
1 x 16
u4
2 x 5 or 2 x 4 x7
x
Check: x 0 : 0 8 0 0 00 x 64 : 64 8 64 0 64 64 0 The solution set is 0 .
99 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
60. z1/ 2 4t1/ 4 4 0 Let u z1/ 4 so that u 2 z1/ 2 . u 2 4u 4 0
57. x x 20 Let u x so that u 2 x. u 2 u 20
u 2 2 0
u 2 u 20 0
u 5 u 4 0
u2 0 u2
u 5 0 or u 4 0 u 5 or u4 x 5 or x 4 or not possible x 16
z1/ 4 2 z 16
Check: 161/ 2 4 16
1/ 4
48 4 0 00 The solution set is 16 .
Check: 16 16 20 16 4 20 The solution set is 16 . 58. x x 6 Let u x so that u 2 x. u2 u 6
61.
u2
u 3 u 2 0
2 or x 1 x 16 or x 1 Check: x 16 : 161/ 2 3 16
1/ 4
20
462 0 00
Check: 4 4 6 42 6 The solution set is 4 .
x 1: 11/ 2 3 1
1/ 4
20
1 3 2 0 00 The solution set is 1, 16 .
59. t1/ 2 2t1/ 4 1 0 Let u t1/ 4 so that u 2 t1/ 2 . u 2 2u 1 0
62. 4 x1/ 2 9 x1/ 4 4 0 Let u x1/ 4 so that u 2 x1/ 2 . 4u 2 9u 4 0
u 1 0 2
u 1 0 u 1
u
1/ 4
1 t 1 1/ 4
u 1 1/ 4
x
x 2 x4
Check: 11/ 2 2 1
or
1/ 4
u 3 0 or u 2 0 u 3 or u2
t
x1/ 2 3 x1/ 4 2 0 Let u x1/ 4 so that u 2 x1/ 2 . u 2 3u 2 0
u 2 u 1 0
u2 u 6 0
x 3 or not possible or
40
(9) (9) 2 4(4)(4) 9 17 2(4) 8
x1/ 4 1 0
9 17 8
9 17 x 8
1 2 1 0 00 The solution set is 1 .
100
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4
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
4
0 u 2 5u 6
9 17 Check x : 8 1/ 2
9 17 4 4 8
0 u 3 u 2 1/ 4
9 17 4 9 8
40
u 3
or u 2
2
or x 2 2
x 3
x 3 or x 2
2
9 17 9 17 4 9 4 0 8 8
Check:
x 3: 4 5 3
9 17 9 9 17 4 0 2
4
64
8
4
x 3: 4 5
2
4
2
15 6 3 4
4
4
10 6 2
1/ 4
x 2: 45
40
2 6 2 4
10 6 2 4
9 17 9 17 4 9 4 0 8 8
4
5x2 6 x
5x 6 x 4
2
4
4
5x2 6 x4 0 x4 5x2 6 Let u x 2 so that u 2 x 4 .
4 2 2 2
The solution set is
4 81 18 17 17 72 9 17 256 0 324 72 17 68 648 72 17 256 0 00 9 17 4 9 17 4 The solution set is , 8 . 8
4 2
2
2
63.
2
4
9 17 4 9 8
6 2
x 2: 4 5 2
9 17 Check x : 8
9 3 3 3
324 72 17 68 648 72 17 256 0 00
1/ 2
9 3
3 6 3
72 9 17 256 0 4 81 18 17 17 72 9 17 256 0
9 17 4 4 8
2
15 6 3
4
9 17 2 9 17 9 4 0 64 64 4 64 8 4 9 17
6 3
4
64.
2, 3 .
4 5x2 x
4 5x x 4
2
4
4
4 5x2 x4 0 x4 5x2 4 Let u x 2 so that u 2 x 4 . 0 u 2 5u 4 5 52 4(1)(4) 5 41 2 2 5 41 x2 2
u
x
5 41 2
101 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
Since 5 41 0, x
u 3 or
5 41 is not real. 2
5 41 is also 2 not real. Therefore, we have only one possible 5 41 : 2
Check x 4: 16 12 16 12 6
5 41 4 4 5 2
5 41 2
5 41 4 4 5 2
5 41 2
2
Check x 1:
12 3 1 12 3 1 6 1 3 1 3 6 4 4 6 66 The solution set is 4, 1 .
2
4 33 5
41
5 41 2
4 66 10
41
5 41 2
5 41 2
4
4 25 10
16 12 4 6 66
5 41
2
41 41 4
66. x 2 3x x 2 3 x 2 Let u x 2 3x so that u 2 x 2 3x. u2 u 2 u2 u 2 0 u 1 u 2 0
5 41 5 41 2
4
4
5 41 2
u 1 or
2
2
x 3x 1 or Not possible or
5 41 2
u2 2
x 3x 2 x 2 3x 4 x 2 3x 4 0 x 4 x 1 0 x 4 or x 1
5 41 The solution set is . 2 Check x 4:
65. x 2 3 x x 2 3 x 6 2
x 2 3x 4
4 2 3 4 4 2 3 4 6
2
8 5 5 41 4
x 3x 2 x2 3x 4 0 x 4 x 1 0 x 4 or x 1
5 41 : 2
Check x
2
x 3 x 3 or Not possible or
Since x is a fourth root, x
solution to check: x
u2
2
2
4 2 3 4 4 2 3 4 16 12 4
2
Let u x 3 x so that u x 3x.
42 2
u2 u 6
Check x 1:
u2 u 6 0 u 3 u 2 0
12 3 1 12 3 1 1 3 4 The solution set is 1, 4 .
102
Copyright © 2025 Pearson Education, Inc.
42 2
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
67.
1
x 1
2
1 2 x 1 2
Let u
1 1 so that u 2 . x 1 x 1 u2 u 2
2
u u 2 0
u 1 u 2 0 u 1
u2
or
3u 2 u 3 0
1 2 1 1 2
2 3 2 1 x 3 1 1 2 x 1 3 3 x 2 u
4 22 44 The solution set is 2, 1 . 2
68.
1
x 1
2
1 12 x 1 2
1 1 Let u so that u 2 . x 1 x 1 u 2 u 12 u 2 u 12 0
u 4 u 3 0 u 4 1 4 x 1 1 4 x 4 4x 3 3 x 4
or or or or or
69. 3x 2 7 x 1 6 0 Let u x 1 so that u 2 x 2 . 3u 2 7u 6 0
1 1 2 11
12 1
1 x 2: 2 2 2 1 2 1
1
2
Check:
1
1 1 x 4: 12 2 3 4 4 1 1 3 3 1 1 12 1 1 9 3 9 3 12 12 12 The solution set is 3 , 4 . 4 3
1 1 or 1 2 x 1 x 1 1 x 1 or 1 2x 2 or 2 x 1 x 2 1 x 2
1 x : 2
Check: 3 1 1 x : 12 2 3 4 3 1 1 4 4 1 1 12 1 1 4 16 16 4 12 12 12
u3 1 3 x 1 1 3x 3 4 3x 4 x 3
Check:
or
u3
or
x 1 3
or
x 3
or
x
1 1
1 3
2 1 3 x : 3 3 7 3 6 0 2 2 2 4 7 2 6 0 3 9 3 4 14 6 0 3 3 00
1 1 x : 3 3 3
2
1
1 7 6 0 3 3 9 7 3 6 0
27 21 6 0 00 3 1 The solution set is , . 2 3
103 Copyright © 2025 Pearson Education, Inc.
1
Chapter 1: Equations and Inequalities
70. 2 x 2 3 x 1 4 0 Let u x 1 so that u 2 x 2 . 2u 2 3u 4 0
Check x
2
3 41 4 3 41 1 x 4 u
or 1
2 64 3 8 3 41 4 3 41 0 128 72 24 41 4 9 6 41 41 0
3 41 4 3 41 1 x 4 u
or
2
1
1 3 41 3 41 1 or x 4 4 4 3 41 4 3 41 x or x 3 41 3 41 3 41 3 41
x 1
1
12 4 41 32 3 41 8
128 72 24 41 36 24 41 164 0 00 3 41 3 41 The solution set is , . 8 8
12 4 41 32 3 41 8
Check x
71. 2 x 2 / 3 5 x1/ 3 3 0 Let u x1/ 3 so that u 2 x 2 / 3 . 2u 2 5u 3 0
3 41 : 8 2
2u 1 u 3 0
1
3 41 3 41 2 3 4 0 8 8 64 8 3 2 4 0 2 3 41 3 41
1 2 1 1/ 3 x 2 u
x 12 1/ 3 3
0 128 72 24 41 4 9 6 41 41 0
2 64 3 8 3 41 4 3 41
1
3 41 3 41 2 3 4 0 8 8 64 8 3 2 4 0 2 3 41 3 41
(3) (3) 2 4(2)(4) 3 41 2(2) 4
u
3 41 : 8
2
x
128 72 24 41 36 24 41 164 0 00
1 8
or
u3
or
x1/ 3 3
or
x 3
or
x 27
3
1/ 3 3
3
1 1 Check x : 2
2/3
Check x 27: 2 27
5 27 3 0 2 9 5 3 3 0 18 15 3 0 33 0 00
8
2/3
1 8
The solution set is ,27 .
104
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1/ 3
1 5 3 0 8 8 1 1 2 5 3 0 4 2 1 5 3 0 2 2 33 0 00 1/ 3
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
u 2 3u 10
72. 3x 4 / 3 5 x 2 / 3 2 0 Let u x 2 / 3 so that u 2 x 4 / 3 .
u 2 3u 10 0
2
u 5 u 2 0
3u 5u 2 0
3u 1 u 2 0
u 5
1 u or u 2 3 1 x2 / 3 or x 2 / 3 2 3 3 3 3 3 1 x2 / 3 or x 2 / 3 2 3 1 x2 or x 2 8 27 1 x not real 27
2 5 5 3 3 5 3 10 Check v : 5 5 3 2 2 3 3
2/3
1 1 Check: 3 5 20 27 27 2/3 1/ 3 1 1 3 5 2 0 27 27 2 1 1 3 5 2 0 3 3 1 5 3 2 0 9 3 1 5 20 3 3 220 00 1 3 3 Note: 27 81 9
The solution set is
3 3 , . 9 9
25 15 9 3 10 1 1 9 3 25 15 10 10 10 2 3 4 4 Check v 4 : 10 4 2 4 2 16 12 10 4 2 4 6 10 10 10 5 The solution set is 4, . 3
2
y y 74. 6 y 1 16 y 1
2
3v v 73. 10 v2 v2
2
Let u
2
v v v 2 3 v 2 10
y y so that u 2 . y 1 y 1 u 2 6u 16
2
Let u
u2
v v or 5 2 v2 v2 v 5v 10 or v 2v 4 5 or v v 4 3
4/3
or
v v so that u 2 . v2 v2
u 2 6u 16 0
u 2 u 8 0 u 2 or u 8 y y 2 or 8 y 1 y 1 y 2 y 2 or y 8y 8 3y 2 or 7 y 8 2 8 y or y 3 7
105 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
2 2 2 2 3 Check y : 6 3 16 3 2 1 2 1 3 3 4 2 9 6 3 16 1 1 9 3
4 x3 3x 2 0 x 2 4 x 3 0 x 2 0 or 4 x 3 0 x0 4x 3 3 x 4
4 6 2 16 44
The solution set is 0,
2
8 8 8 7 Check y : 6 7 16 8 8 7 1 1 7 7 8 64 49 6 7 16 1 1 49 7 64 48 16 64 64 2 8 The solution set is , . 3 7
x5 4 x3 0
x3 x 2 x 2 0 x3 0 or x 2 0 or x 2 0 x0 x2 x 2 The solution set is 2, 0, 2 .
x3 9 x 0
x x 5 x 4 0
x x 3 x 3 0
x 0 or x 5 0 or x 4 0 x 5 x4 The solution set is 5, 0, 4 .
x 0 or x 3 0 x 3 0 x3 x 3 The solution set is 3, 0,3 .
80. 76.
x3 x 2 20 x 0 x x 2 x 20 0
x x 9 0
4
x3 x 2 4 0
79. 2
2
x3 6 x 2 7 x 0
x x 0
x x2 6x 7 0
x2 x2 1 0
x x 7 x 1 0
3 . 4
x5 4 x3
78.
75.
4 x3 3x 2
77.
x x 1 x 1 0 2
x 0 or x 7 0 or x 1 0 x 7 x 1 The solution set is 7, 0,1 .
2
x 0 or x 1 0 or x 1 0 x0 x 1 x 1 The solution set is 1, 0,1 .
81.
x3 x 2 x 1 0 x 2 x 1 1 x 1 0
x 1 x 2 1 0 x 1 x 1 x 1 0 x 1 0 or x 1 0 x 1 x 1 The solution set is 1,1 .
106
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Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
x3 4 x 2 x 4 0
87. 3( x 3) 3 9( x 3) 3 0
x 2 x 4 1 x 4 0
3( x 3) 3 [1 3( x 3)] 0
82.
1
1
x 4 x 2 1 0 x 4 x 1 x 1 0
3( x 3) 3 (1 3x 9) 0 1
3( x 3) 3 (3x 8) 0 1
1
( x 3) 0 x 3 0 x3 1
x3 3 x 2 16 x 48 0 x x 3 16 x 3 0
x 3 x 2 16 0 x 3 x 4 x 4 0
( x 2) 3 [4 ( x 2 4 x 4)] 0 ( x 2) 3 [4 x 2 4 x 4)] 0 ( x 2) 3 ( x 2 4 x) 0 ( x 2) 3 x( x 4) 0
x 2 x 3 1 x 3 0
x 3 x 2 1 0 x 3 x 1 x 1 0 x 3 0 or x 1 0 or x 1 0 x3 x 1 x 1 The solution set is 1,1,3 .
85. 2 x 4 3x 4 0 5
( x 2) 3 0 or x 0 or x 4 0 1 x 4 0 3 ( x 2) x 4 no solution The solution set is 4, 0 .
2 x 3x 0 x 3x x 2 x 3x 0 x 3x x 2 x 6 x 0 x 3x 2 x 5 x 0
89. x x 2 3 x
x 4 (2 3 x) 0
1/ 3
x 0 or 2 3x 0 x0 3x 2 2 x 3 2 The solution set is 0, . 3
2
1/ 3
2
4
1/ 3
2
4 x 1 x 0
2
1/ 3
2
x 3x
4/3
2
1/ 3
2
1
86.
8 ,3 . 3
( x 2) 3 [4 ( x 2) 2 ] 0
x3 3 x 2 x 3 0
1
4( x 2) 3 ( x 2) 1 0
88.
x 3 0 or x 4 0 or x 4 0 x3 x4 x 4 The solution set is 4,3, 4 .
1
3
The solution set is
2
84.
3x 8 0 3x 8 8 x 3
3( x 3) 3 0 or
x 4 0 or x 1 0 or x 1 0 x 4 x 1 x 1 The solution set is 4, 1,1 .
83.
4
2
0
or
2 x2 5x 0
x 2 3x 0
or
2 x2 5x 0
x x 3 0
or x 2 x 5 0
x 0 or x 3 or x 0 or x
x 1 (4 x 2 ) 0
or 4 x 2 0 x 1 0 1 x 2 4 0 x x2 4 no solution x 2 The solution set is 2, 2 .
5 The solution set is 0, , 3 . 2
107 Copyright © 2025 Pearson Education, Inc.
5 2
Chapter 1: Equations and Inequalities
90. 3x x 2 2 x
1/ 2
2 x2 2x
3/ 2
(4) (4)2 4(1)(2) 2 4 8 42 2 2 2 2 2
0
u
x 2 x 3x 2 x 2 x 0 x 2 x 3x 2 x 4 x 0 x 2 x 2 x x 0 x 2 x 0 or 2 x x 0 1/ 2
2
2
1/ 2
2
2
1/ 2
2
1/ 2
2
x 2x 0
or
2
x x 2 0
or x 2 x 1 0
2
2
u 2 2 1/ 2
x
2
3 0 02 2 0
3 0 0
2 0
3/ 2
2 02 2 0 1/ 2
3(2) (2) 2 2(2)
1/ 2
1 2
: 2 2 42 2 2 0 Check x 2 2
0
2 4 4
3/ 2
0
3/ 2
0
2 0
4 4 2 28 4 2 2 0 00 The solution set is
0
00
3 12
2 12 2
1 2
1/ 2
2
2 12 2
3 12 14 1
1 2
3/ 2
2 14 1
3/ 2
0
3 12 34
2 34
3/ 2
0
1/ 2
The solution set is 2, 0 .
, 2 2 0.34, 11.66 . 2
2
4 42 4(1)(2) 2(1) 4 8 4 2 2 2 2 2 2 u 2 2 u 2 2 or
u
0
1/ 2
2 2
92. x 2 / 3 4 x1/ 3 2 0 Let u x1/ 3 so that u 2 x 2 / 3 . u 2 4u 2 0
3(2) 0 2 0 0
Check x 12 :
2
2
3/ 2
1/ 2
2
4 4 2 28 4 2 2 0 00
0
3(2) 0
2
2
1/ 2
2 2
x
1/ 2 2
2
2 (2) 2 2(2)
3(2) 4 4
2 2
00
Check x 2 :
or
1/ 2
x 2 2 or x 2 2 x 2 2 or x 2 2 Check x 2 2 : 2 2 42 2 2 0
2x x 0
3/ 2
1/ 2
2 2
1/ 2 2
x 0 or x 2 or x 0 or x
Check x 0 :
u 2 2
or
x1/ 3 2 2
Not real
x1/ 3 2 2
or
or Check x 2 2 : x 2 2
3
x 2 2
3
3
1/ 2
91. x 4 x 2 0 Let u x1/ 2 so that u 2 x 2 . u 2 4u 2 0
2 2 3
2/3
1/ 3
3 4 2 2
20
2 2 4 2 2 2 0 2
4 4 2 28 4 2 2 0 00
108
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Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
3
4
Check x 2 2 :
2 2 3
2/3
1/ 3
3 4 2 2
20
2 2 4 2 2 2 0 2
4 4 2 28 4 2 2 0 00
The solution set is
2 2
, 2 2 39.80, 0.20 . 3
3
93. x 4 3 x 2 3 0 Let u x 2 so that u 2 x 4 . u 2 3u 3 0 u
3
3 4 1 3 3 15 2
2 1
2
3 15 2 3 15 2 x 2
3 15 2 3 15 2 or x 2
u
x
u
or
3 15 2
3 15 2 Not real
x
or
3 15 Check x : 2 4
2
2
3 15 3 15 3 3 0 2 2 3 3 3 15 3 2 3 15 15 3 0 4 2 18 2 45 3 45 3 0 4 2 9 45 3 45 3 0 2 2 9 45 3 45 3 0 2 33 0 00 Check x
3 15 : 2
2
3 15 3 15 3 3 0 2 2 3 3 3 15 3 2 3 15 15 3 0 4 2 18 2 45 3 45 3 0 4 2 9 45 3 45 3 0 2 2 9 45 3 45 3 0 2 33 0 00 The solution set is 3 15 3 15 , 1.03, 1.03 . 2 2
94. x 4 2 x 2 2 0 Let u x 2 so that u 2 x 4 . u 2 2u 2 0
u
2
2 4 1 2 2 10 2
2 1
2 10 2 2 10 x2 2
x
or
2 10 2
Check x
2
2 10 2 2 10 or x 2 2
u
3 15 3 15 3 3 0 2 2
2
3 15 3 15 3 3 0 2 2
u
or x
2 10 2 Not real
2 10 : 2 4
2
2 10 2 10 2 20 2 2 2
2 10 2 10 2 20 2 2 12 2 20 2 20 20 4 2 6 20 2 20 20 2 220 00
109 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
Check x
2
2 10 : 2 4
1 1 4 2 1 2 1 2 1 4 2 1 4 2 1 2 4
2
2 10 2 10 2 20 2 2 2
2 10 2 10 2 20 2 2 12 2 20 2 20 20 4 2 6 20 2 20 20 2 220 00 The solution set is 2 10 2 10 , 0.93, 0.93 . 2 2
1 1 4 2 2 2 2 2 1 1 4 2 2 2 The solution set is 1 1 4 2 1 1 4 2 , 1 1 2 2 1.85, 0.17 .
96. 1 r 2 1 r 2
Let u 1 r so that u 2 1 r . 2
2
u2 2 u
Let u 1 t so that u 1 t . 2
2
u2 u 2 0
u2 u u2 u 0
1 t
u
1 1 4 2 2 1 1 4 2 2
Check t 1
1 1 4 2 : 2
Check r 1
2
1 1 4 2 1 2 1 2 1 4 2 1 4 2 1 2 4
2
1 4 2 2
1 1 4 2 2 1 1 4 2 2
Check t 1
2
2
2 8 : 2
2 8 2 2 2 2 2 8 2 8 2 4 2
1 4 2 2
2 8 2 8 2 2
2 2 2 2 8 8 2 8 2 4 2
2 2 1 4 2 4 2 2 2 1 1 4 2 4 2 2
( ) ( ) 2 4( )(2) 2( )
2 8 2 2 8 1 r 2 2 8 r 1 2
(1) (1) 2 4( )( ) 1 1 4 2 2( ) 2
t 1
1 4 2 2
2 2 1 4 2 4 2 2 2 1 1 4 2 4 2
95. 1 t 1 t
u
1 4 2 2
2 8 4
2
2
1 1 4 2 : 2
110
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4 2 8 2
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
Check r 1
5a3 2a 2 45a 18 0
2
2 8 2 2 2 2 2 8 2 8 2 2 4
a 2 5a 2 9 5a 2 0
2 8 2
a 9 5a 2 0
2 8 2
2
a 3 a 3 5a 2 0 a 3 0 or a 3 0 or 5a 2 0 5a 2 a3 a 3 2 a 5 2 The solution set is 3, ,3 . 5
2 8 2 2 2 2 8 8 2 4 2 2 8 4 2
4 2 8 2
The solution set is 2 8 2 8 , 1 1 2 2 1.44, 0.44 . 97.
3x 2 7 x 20 0
3x 5 x 4 0 3x 5 0 3x 5
x
or x 4 0 x 4
5 3
5 The solution set is 4, . 3
98.
2 x 2 13x 21 0
2 x 7 x 3 0 2x 7 0 2x 7
x
or x 3 0 x3
7 2
7 The solution set is ,3 . 2
5a 3 45a 2a 2 18
99.
2 8 : 2
3 z 3 12 z 5 z 2 20
100.
3 z 3 5 z 2 12 z 20 0
z 2 3z 5 4 3z 5 0
z 4 3z 5 0 2
z 2 z 2 3z 5 0 z 2 0 or z 2 0 3z 5 0 3 z 5 z2 z 2 5 z 3 5 The solution set is 2, , 2 . 3
101. 4 w 3 w 3 4w 12 w 3 3w 15
w5 The solution set is 5 .
102. 6 k 3 2k 12 6k 18 2k 12 4k 6
k
3 2
3 The solution set is . 2
111 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
2
105.
2v v 8 103. 1 1 v v v Let u . Rewrite the equation: v 1
2x 5 x 1 2x 5 x 1
2 x 5 x 1 2
2
2x 5 x2 2 x 1
u 2 2u 8
x2 4 0
u 2 2u 8 0
x 2 x 2 0
u 2 u 4 0
x 2 or x 2 Check:
u 2 or u 4 Go back in terms of v and solve: v v 2 4 or v 1 v 1 v 2v 2 v 4v 4 v 2 5v 4 v 2 4 v 5 4 The solution set is 2, . 5
2 2 5 2 1
2 2 5 2 1
1 2 1
9 2 1
3 1 The solution set is 2 .
106.
11 T
3 x 1 2 x 6 3x 1 2 x 6
3x 1 2 x 6 2
2
y 6y 7 104. y 1 y 1 y . Rewrite the equation: Let u y 1
2
3 x 1 4 x 2 24 x 36 4 x 2 27 x 35 0
4 x 7 x 5 0
u 2 6u 7
4x 7 0
u 6u 7 0
4x 7
u 7 u 1 0
x
2
u 7 or u 1 Go back in terms of y and solve: y y 7 or 1 y 1 y 1 y y 1 y 7y 7 6 y 7 2y 1 7 1 y y 2 6 1 7 The solution set is , . 2 6
or x 5 0 x5
7 4
Check: 7 7 3 1 2 6 4 4 5 7 6 2 2 1 6 The solution set is 5 .
107.
3m 2 6m 1 3m 2 6m 1 0 a 3, b 6 , c 1
112
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3 5 1 2 5 6 4 10 6
6 6 T
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations
m
6 62 4 31
6 24 6
2 3
x 4 3x 2 4 0
6 2 6 3 6 6 3 3 6 3 6 The solution set is , . 3 3
x 1 x 4 0 2
4 4 3(1) 2 4 4 3
8 4 4 3 2 4 2
8 112 8 4 7 2 7 8 8 2 2 7 2 7 The solution set is , . 2 2
109.
k 2 k 12 0 k 4 k 3 0 k 4 x3 4 x3 x 3 4 x 12 3x 15
k 3 x3 3 or x3 or x 3 3 x 9 or 4x 6 6 3 x x5 or 4 2 Neither of these values causes a denominator to 3 equal zero, so the solution set is ,5 . 2
x4 5x2 6 0
x 2 x 3 0 2
x2 2
x2 3
x 2 Check: 4 5(
112.
2) 2 6 4 5(2) 6
4 2 2 4
4 5(
2
3) 6 4 5(3) 6
49 3 4 5(
3)2 6 4 5(3) 6
49 3 3
The solution set is
2, 3 .
k 2 3k 28 k 2 3k 28 0 k 4 k 7 0
44 2 4 5(
or
x 3
2) 2 6 4 5(2) 6
k 2 k 12
111.
5x2 6 x4
or x 2 3 0
4 1 1 1
The solution set is 1 .
5x 6 x
x2 2 0
4 4 3( 1) 2 4 4 3
41 1
2
2
x 2 4 no real solution
Check:
4
or x 2 4 0
x2 1 x 1
4 y2 8 y 3 0 a 4 , b 8 , c 3 8
2
x2 1 0
4 y2 8 y 3
y
4 3x2 x 4 3x2 x 4
108.
4
110.
k 4 or k 7 x3 x3 4 7 or x4 x4 x 3 4 x 16 or x 3 7 x 28 6 x 31 5 x 13 or 13 31 x x 2.6 or 5.17 5 6 Neither of these values causes a denominator to 13 31 equal zero, so the solution set is , . 5 6
113 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
s s 4. 4 1100
113. Solve the equation
16.5 2
l 32
l 16.5 2 32
s s 4 0 1100 4 s s 1100 1100 4 4 0 1100
2 16.5 l 2 32
s 275 s 4400 0
2
2
l 16.5 2 32
2
Let u s , so that u s.
u 2 275u 4400 0
2
16.5 l 32 220.7 2 The length was approximately 220.7 feet.
275 2752 4 1 4400 u 2 275 93, 225 2 u 15.1638 or u 290.1638 Since u s , it must be positive, so
3x 5 x 2 x 3
116.
3x 5 x 2 x 3 2
s u 2 15.1638 229.94
3x 5 2 3 x 5 x 2 x 2 x 3
The distance to the water's surface is approximately 229.94 feet.
4 x 3 2 3 x 5 x 2 x 3
2
2 3 x 2 x 10 3 x
2
LH 25 Let T 4 and H 10 , and solve for L.
114. T 4
L(10) 25 4 4 4L 44
4 4 4 4 L
2
4 3x 2 x 10 9 x 2 2
3 x 4 x 40 0
2
x
4
4 2 4 3 40
6 4 496 4 4 31 6 6 4 4 31 6 2 2 31 3
4
256 4 L 64 L The crushing load is 64 tons.
l 32 Let T 16.5 and solve for l.
115. T 2
Since x > 2, the negative solution is extraneous. 2 2 31 The solution set is . 3
114
Copyright © 2025 Pearson Education, Inc.
Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations a 1, b 3, c 9
117.
x
x 7 10 18 2
4 3
4 x 7 10 18 2 x 7 10 18 16
3
3
3
3 x 7 10 2
x 7 10 8
x 7 2 2
2
x7 4 x 11 The solution set is 11 7
12 x 10 3x 10 13x 10 0 14
9
4
1
x 10 0 x 0 1 1 To solve 12 x 13 x 2 3 0 , let u x 2 4
(4u 3)(3u 1) 0
123. Mya did not check her solutions and included the extraneous solution, x 1 . 2x 3 x 0
2x 3 x 2
1 3 1 u x 2 or x 2 4 3 9 1 x x 16 9 9 1 The solution set is 0, , 16 9 1
119.
2 1 3 1 3i 2 2 1 3i 3 3 3i The solution set is 3, 1, , . 2 2
2x 3 x
Then 12u 2 13u 3 0
12 4 11
122. Answers will vary.
x 10 12 x 13 x 2 3 0 4
1
121. Answers will vary. One example: x x 2 0.
9
2
x
120. Answers will vary. One example: x 1 1.
12 x 5 3 x 5 13 x 10
118.
32 4 1 9
2 3 27 3 3 3i 2 2 Also, a 1, b 1, c 1
4
4 3
3
z 6 28 z 3 27 0 ( z 3 27)( z 3 1) 0 ( z 3)( z 2 3 z 9)( z 1)( z 2 z 1) 0 z 3 0 or z 1 0 z 3 or z 1
2
2 x 3 x2
x2 2 x 3 0
x 3 x 1 0 x 3 or x 1
Check: 2 3 3 3 0
2 1 3 1 0
9 3 0
1 1 0
33 0
11 0
00 T The solution set is 3 .
20
115 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 19. a.
Section 1.5
35 33 53
1. x 2
68
b.
35 35 55
2. False.
2 0
3. closed interval
c.
35 3 3 3 5
4. multiplication properties (for inequalities)
9 15
5. True. This follows from the addition property for inequalities.
d.
35
2 3 2 5
6. True. This follows from the addition property for inequalities.
6 10
20. a.
7. True;. This follows from the multiplication property for inequalities.
2 1 2 3 1 3 54
8. False. Since both sides of the inequality are being divided by a negative number, the sense, or direction, of the inequality must be reversed. a b That is, . c c
b.
2 1 2 5 1 5
3 4
c.
9. True
2 1 3 2 3 1 63
10. False; either or both endpoints could be any real number.
d.
2 1
2 2 2 1
11. d
4 2
12. c 21. a.
13. Interval: 0, 2
4 3 4 3 3 3
Inequality: 0 x 2
70
14. Interval: 1, 2
b.
Inequality: 1 x 2
4 3 4 5 3 5
1 8
15. Interval: 2,
c.
Inequality: x 2
4 3 3 4 3 3
16. Interval: , 0
12 9
Inequality: x 0
d.
4 3
2 4 2 3
17. Interval: 0,3
8 6
Inequality: 0 x 3 18. Interval: 1,1
Inequality: 1 x 1 116
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Section 1.5: Solving Inequalities 22. a.
3 5 3 3 5 3
26. (–1, 5)
0 2
b.
27. [4, 6)
3 5 3 5 5 5
8 10
3 3 3 5 9 15 3 5
d.
28. (–2, 0)
3 5
c.
29.
3,
30.
, 5
2 3 2 5 6 10 2x 1 2
23. a.
2x 1 3 2 3 2x 4 5
31.
2x 1 2
b.
, 4
2x 1 5 2 5
2 x 4 3
32.
2x 1 2
c.
1,
3 2 x 1 3 2
6x 3 6 2 2 x 1 2 2
4 x 2 4
34. 1 x 2
1 2x 5
24. a.
33. 2 x 5
2x 1 2
d.
1 2x 3 5 3
35. 4 x 3
4 2x 8 1 2x 5
b.
1 2x 5 5 5
36. 0 x 1
4 2 x 0 1 2x 5
c.
3 1 2 x 3 5
37. x 4
3 6 x 15
1 2x 5
d.
2 1 2 x 2 5
38. x 2
2 4 x 10
25. [0, 4]
39. x 3
117 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 40. x 8
59.
x 1 5 x 11 5 1
x4 x x 4 or (, 4)
41. If x 5, then x 5 0. 42. If x 4, then x 4 0.
60.
43. If x 4, then x 4 0. 44. If x 6, then x 6 0. 45. If x 4, then 3 x 12.
x 6 1 x 6 6 1 6 x7 The solution set is x x 7 or (, 7) .
46. If x 3, then 2 x 6.
61. 3 5 x 7 5 x 10 x2 The solution set is x x 2 or [2, ) .
47. If x 6, then 2x 12. 48. If x 2, then 4 x 8. 49. If x 5, then 4 x 20. 50. If x 4, then 3 x 12.
62. 2 3x 5 3x 3 x 1 The solution set is x x 1 or [1, ) .
51. If 8 x 40, then x 5. 52. If 3 x 12, then x 4. 1 53. If x 3, then x 6. 2
63. 3x 7 2 3x 9 x3 The solution set is x x 3 or (3, ) .
1 54. If x 1, then x 4. 4
55. If 0 5 x, then 0
1 1 x 5
56. 0 4 x, then
1 1 0 4 x
57. 5 x 0, then
1 1 0 x 5
64. 2 x 5 1 2x 4 x 2
The solution set is x x 2 or ( 2, ) .
1 1 58. 0 x 10, then 0 10 x
118 Copyright © 2025 Pearson Education, Inc.
Section 1.5: Solving Inequalities 65. 3x 1 3 x 2x 4 x2 The solution set is x x 2 or [2, ) .
66. 2 x 2 3 x x5 The solution set is x x 5 or [5, ) .
70. 8 4(2 x) 2 x 8 8 4x 2x 4x 2x 6x 0 x0 The solution set is x x 0 or , 0 .
71.
67. 2( x 3) 8 2x 6 8 2 x 14 x 7
The solution set is x x 20 or (, 20) .
The solution set is x x 7 or ( 7, ) .
1 ( x 4) x 8 2 1 x2 x 8 2 1 x 10 2 x 20
68. 3(1 x) 12 3 3 x 12 3 x 15 x5 The solution set is x x 5 or ( , 5) .
72.
1 3x 4 ( x 2) 3 1 2 3x 4 x 3 3 9 x 12 x 2 8 x 14 7 x 4 7 7 The solution set is x x or ( , ) . 4 4
69. 4 3(1 x) 3 4 3 3x 3 3x 1 3
3x 2 2 x 3
73.
2 2 The solution set is x x or , . 3 3
x x 1 2 4 2x 4 x 3x 4 4 x 3
4 4 The solution set is x x or , . 3 3
119 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
74.
x x 2 3 6 2 x 12 x
79.
11 2 x 1
x 12 The solution set is x x 12 or 12, .
2x 1 0 4 12 2 x 1 0 3
11 1 The solution set is x x or 2 2 11 1 2 , 2 .
75. 0 3 x 7 5 7 3 x 12 7 x4 3
7 7 The solution set is x x 4 or , 4 . 3 3
3x 2 4 2 0 3x 2 8
The solution set is x 6 x 0 or 6, 0 .
78. 3 3 2 x 9 6 2x 6 3 x 3 The solution set is x 3 x 3 or 3, 3 .
1 81. 1 1 x 4 2 1 0 x3 2 0 x 6 or 6 x 0
2 3
2 3
2 2 The solution set is x x 3 or , 3 . 3 3
2 x2 3
2 2 The solution set is x x 2 or , 2 . 3 3
77. 5 4 3x 2 9 3x 2 2 3 x 3
2
2 3x 6
76. 4 2 x 2 10 2 2x 8 1 x 4 The solution set is x 1 x 4 or 1, 4 .
1
11 2
0
80.
11 1 x 2 2
82.
1 0 1 x 1 3 1 1 x 0 3 3 x 0 or 0 x 3 The solution set is x 0 x 3 or 0, 3 .
120
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Section 1.5: Solving Inequalities 83. ( x 2)( x 3) ( x 1)( x 1) 2
2
x x 6 x 1 x 6 1 x 5 x 5 The solution set is x x 5 or , 5 .
87.
2 4x 5 1 5 x 2 4 1 5 1 5 The solution set is x x or , . 2 4 2 4
84. ( x 1)( x 1) ( x 3)( x 4) x 2 1 x 2 x 12 1 x 12 x 11 x 11 The solution set is x x 11 or , 11 .
88.
1 x 1 2 3 2 3 2 3x 3 4 1 1 x 3 3 1 1 1 1 The solution set is x x or , . 3 3 3 3
2
4 x 2 3x 4 x 2 4 x 1
3x 4 x 1 x 1 x 1 The solution set is x x 1 or 1, .
86. x(9 x 5) (3 x 1)
89.
1 3
1 3
4 x 2 1 0 1 0 4x 2 4x 2 0
x
2
1 2
1 1 The solution set is x x or , . 2 2
9 x2 5x 9 x2 6 x 1 5 x 6 x 1
x 1 The solution set is x x 1 or , 1 .
5 4
1 2
1 3 x 1
85. x(4 x 3) (2 x 1)
1 x 1 3 2 3 4 6 4x 4 9
90.
2 x 11 0 1 0 2x 1 1 Since 0 , this means 2 x 1 0 . 2x 1 Therefore, 2x 1 0 1 x 2
121 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
1 1 The solution set is x x or , . 2 2
91.
1 4 x
1
1.7
0
Value of f
1 Positive
17 Negative
17 5
(, 143 )
( 143 , 14 )
( 14 , )
0
22 100
1
solution set is
Positive
3 or x 14 14
17 9
or x 53
or,
53 is not in the solution set because 72 is not in
the domain of f.
The solution set is 179 , 53 .
4 6 22 3 3 Negative Positive Negative
x x
x x
using interval notation, [ 179 , 53 ) . Note that
2 3 x 5 2 2 3 0 and x x 5 2 Since 0 , this means that x 0 . Therefore, x 2 3 x 5 2 3 5x 5x x 5 10 3 x 10 x 3 10 10 The solution set is x x or , . 3 3
93. 0
We want to know where f ( x) 0 , so the
or, using
interval notation, [ 143 , 14 ) . Note that 14 is not in the solution set because 14 is not in the domain of f. The solution set is 143 , 14 .
92.
2
We want to know where f ( x) 0 , so the
1 7 0 1 4x 1 7(1 4 x) 0 1 4x 6 28 x 0 1 4x The zeros and values where the expression is undefined are x 143 and x 14
solution set is
Number Chosen Conclusion
7
Interval Number Chosen Value of f Conclusion
( , 179 ) ( 179 , 53 ) ( 53 , )
Interval
2 3 x 5 3 2 3 0 (3x 5) 2 3(3x 5) 0 2(3x 5) 17 9 x 0 (3x 5) The zeros and values where the expression is undefined are x 179 and x 53 1
4 2 x 3 4 4 2 0 and x x 3 4 Since 0 , this means that x 0 . Therefore, x 4 2 x 3 4 2 3x 3x x 3 12 2 x 6 x
94. 0
122
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Section 1.5: Solving Inequalities
The solution set is x x 6 or 6, .
95. 0 2 x 4
1
1 2
1 1 0 2x 4 2 1 1 1 0 and 2x 4 2x 4 2 1 Since 0 , this means that 2 x 4 0 . 2x 4 Therefore, 1 1 2x 4 2 1 1 2( x 2) 2 1 1 2( x 2) 2( x 2) 2 2( x 2) 1 x2 3 x The solution set is x x 3 or 3, .
96. 0 3 x 6
1
1 3
1 1 3x 6 3 1 1 1 0 and 3x 6 3x 6 3 1 Since 0 , this means that 3x 6 0 . 3x 6 Therefore, 1 1 3x 6 3 1 1 3( x 2) 3 0
1 1 3( x 2) 3( x 2) 3 3( x 2) 1 x 2 1 x
The solution set is x x 1 or 1, .
97. If 1 x 1, then 1 4 x 4 1 4 3 x45 So, a 3 and b 5. 98. If 3 x 2, then 3 6 x 6 2 6 9 x 6 4 So, a 9 and b 4. 99. If 2 x 3, then 4(2) 4( x) 4(3) 12 4 x 8 So, a 12 and b 8. 100. If 4 x 0, then 1 1 1 4 2 x 2 0 2 1 2 x 0 2 So, a 2 and b 0. 101. If 0 x 4, then 2(0) 2( x) 2(4) 0 2x 8 0 3 2x 3 8 3 3 2 x 3 11 So, a 3 and b 11. 102. If 3 x 3, then 2(3) 2( x) 2(3) 6 2 x 6 6 1 2 x 1 6 1 7 1 2 x 5 5 1 2 x 7 So, a 5 and b 7. 103. If 3 x 0, then 3 4 x 4 0 4 1 x 4 4 1 1 1 x4 4 1 1 1 4 x4 1 So, a and b 1. 4
123 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 104. If 2 x 4, then 26 x6 46 4 x 6 2 1 1 1 4 x6 2 1 1 1 2 x6 4 1 1 So, a and b . 2 4
b. Let x = age at death. x 30 55.8 x 85.8 Therefore, the average life expectancy for a 30-year-old female in 2023 will be greater than or equal to 85.8 years. c.
112. V 20 T 80º T 120º
105. If 6 3 x 12, then 6 3 x 12 3 3 3 2 x4
V 120º 20 1600 V 2400 The volume ranges from 1600 to 2400 cubic centimeters, inclusive. 80º
22 x 2 42 4 x 2 16 So, a 4 and b 16.
113. Let P represent the selling price and C represent the commission. Calculating the commission: C 45, 000 0.25( P 900, 000) 45, 000 0.25 P 225, 000 0.25P 180, 000
106. If 0 2 x 6, then 0 2x 6 2 2 2 0 x3 02 x 2 32 0 x2 9 So, a 0 and b 9.
107.
108.
Calculate the commission range, given the price range: 900, 000 P 1,100, 000 0.25(900, 000) 0.25 P 0.25(1,100, 000) 225, 000 0.25 P 275, 000
3x 6 We need 3x 6 0 3x 6 x 2 To the domain is x x 2 or 2, .
225, 000 180, 000 0.25 P 180, 000 275, 000 180, 000
45, 000 C 95, 000
The agent's commission ranges from $45,000 to $95,000, inclusive.
8 2x We need 8 2 x 0 2 x 8 x 4 To the domain is x x 4 or 4, .
45, 000 95, 000 0.05 5% to 0.086 8.6%, 900, 000 1,100, 000
inclusive. As a percent of selling price, the commission ranges from 5% to 8.6%, inclusive. 114. Let C represent the commission. Calculate the commission range: 25 0.4(200) C 25 0.4(3000) 105 C 1225 The commissions are at least $105 and at most $1225.
109. 21 < young adult's age < 30 110. 40 ≤ middle-aged < 60 111. a.
By the given information, a female can expect to live 85.8 82.2 3.6 years longer.
Let x = age at death. x 30 52.2 x 82.2 Therefore, the average life expectancy for a 30-year-old male in 2023 will be greater than or equal to 82.2 years.
124
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Section 1.5: Solving Inequalities 115. Let W = weekly wages and T = tax withheld. Calculating the withholding tax range, given the range of weekly wages: 500 W 700 500 312.5 W 312.5 700 312.5 187.50 W 312.5 387.5 0.12(187.5) 0.12 W 312.5 0.12(387.5) 22.50 0.12 W 312.5 46.5 22.5 1100 0.12 W 312.5 1100 46.5 1100
1122.50 T 1146.50
The amount withheld varies from $1122.50 to $1146.50, inclusive. 116. Let x represent the length of time you should exercise on the last two days. 25 35 0 40 15 x 150 115 x 150 x 35 You will stay within the guidelines by exercising from 35 minutes total on the last two days. 117. Let x represent the amount paid for international minutes and y represent the number of international minutes. The range of the bills is $69.50 to $140.75. The rate plan is $60. Thus the range of costs of the international minutes is: 9.50 x 80.75 . The cost per min is $0.25. 9.50 80.75 y 0.25 0.25 38 y 323 The minutes varies from 38 to 323 minutes, inclusive. 118. Let C represent the amount paid for the fares. The range of the fare is $20.93 to $40.44. of miles. The number of miles is 23. 20.93 23 x 40.44 0.91 x 1.76 The cost per mile varies from $0.91 per mile to $1.76 per mile, inclusive. 119. You have already consumed 40 grams of fat. Let C represent the number of cookies. Then we have the following equation: 40 8C 64 8C 24 C3 You may eat up to 3 cookies and keep the total fat content of your meal not more than 64g.
120. You have already consumed 730 calories. Let x represent the number of apple sauce orders you can eat. Then we have the following equation: 730 50 x 830 50 x 100 x2 You may eat up to 2 orders of apple sauce and keep your calories below or equal to 830. 121. a.
Let T represent the score on the last test and G represent the course grade. Calculating the course grade and solving for the last test: 68 82 87 89 T G 5 326 T G 5 5G 326 T T 5G 326 Calculating the range of scores on the last test, given the grade range: 80 G 90 400 5G 450 74 5G 326 124 74 T 124 To get a grade of B, you need at least a 74 on the fifth test.
b. Let T represent the score on the last test and G represent the course grade. Calculating the course grade and solving for the last test: 68 82 87 89 2T G 6 326 2T G 6 163 T G 3 T 3G 163 Calculating the range of scores on the last test, given the grade range: 80 G 90 240 3G 270 77 3G 163 107 77 T 107 To get a grade of B, you need at least a 77 on the fifth test.
125 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 122. Let T represent the test scores of the people in the top 2.5%. T 1.96(12) 100 123.52 People in the top 2.5% will have test scores greater than 123.52. That is, T 123.52 or (123.52, ). 123. Since a b , a b and 2 2 a a a b and 2 2 2 2 ab and a 2 ab b. So, a 2
127. For 0 a b, h
a b 2 2 a b b b 2 2 2 2 ab b 2
a b 2b a b b a ab d b, . b 2 2 2 2 ab is equidistant from a and b. Therefore, 2
ab a 2
2
ab a
and
b2
and
b ab
ab
ab
ab . 2
h
2ab ab
2
Therefore, a ab b . 126. Show that
1ba h 2 ab
2
ab (geometric mean) 2 128. Show that h arithmetic mean 1 ( a b) 2 1 11 1 h 2 a b 2 1 1 ba h a b ab h ab 2 ab
125. If 0 a b, then b 2 ab 0
1
2ab a(a b) 2ab a ab ab 2 2ab a ab ab a 2 ab ab a (b a ) 0 ab Therefore, h a . 2ab b(a b) 2ab bh b ab ab 2 ab b 2ab b 2 ab ab ab b(b a) 0 ab Therefore, h b , and we have a h b .
ab b , so 2 a b 2a b a ab ab d a, and 2 a 2 2 2
and
1 1ba h h 2 ab
ha
124. From problem 123, a
ab a 2 0
1 11 1 h 2 a b
ab 1 ab a 2 ab b 2 2 2 1 a b 0, since a b. 2 ab . Therefore, ab 2
2
ab ab h 2 ab 1 2 ( a b)
129.
x5 3 3x 12 6 x 9 x 5 x 4 2x 3
3x 12 6 x 9 and
6x 9 x 5
3 3x
5 x 14
1 x
x
126
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14 4
Section 1.6: Equations and Inequalities Involving Absolute Value
14 . The solution 5 14 set, in interval notation, is 1, . 5
This is equivalent to 1 x
130. The largest value of 2 x 2 3 occurs at the largest value for x .
9.
3 x 15 3x 15 or 3 x 15 x 5 or x 5 The solution set is {–5, 5}.
10.
3 x 12 3x 12 or 3 x 12
2 5 x 9
x 4 or x 4 The solution set is {–4, 4}.
3 x 4 4 x 3
11.
3 x 4 or 4 x 3 The largest value for 2 x 2 3 is
2x 3 5 2 x 3 5 or 2 x 3 5 2 x 2 or
2(4) 2 3 32 3 29 . 131. Answers will vary
132. Answers will vary. One possibility: No solution: 4 x 6 2 x 5 2 x
12.
133. Since x 2 0 , we have x2 1 0 1 x2 1 1 Therefore, the expression x 2 1 can never be less than 5 .
3x 1 2 3x 1 2 or 3x 1 2
One solution: 3x 5 2 x 3 1 3 x 2 1
3x 3 or
3x 1
x 1 or
x
13.
1 4t 8 13 1 4t 5 1 4t 5 or 1 4t 5 4t 4 or
2 2
14.
1 2z 3
4. {x | 5 x 5}
z 1 or
z2
The solution set is 1, 2 .
5. True
8. a
1 2z 6 9 1 2 z 3 or 1 2 z 3 2 z 2 or 2 z 4
3. {5, 5}
7. d
t
Section 1.6
6. True
4t 6
3 2 3 The solution set is 1, . 2 t 1 or
2. True
1 3
1 The solution set is , 1 . 3
134. Answers will vary.
1.
2x 8
x 1 or x 4 The solution set is {–4, 1}.
15.
2x 8 2x 8 2x 8 or 2 x 8 x 4 or x4 The solution set is {–4, 4}.
127 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
16.
x 1
22.
x 1
x 1 x 1 1 or 1 2 3 2 3 3x 2 6 or 3 x 2 6 3x 8 or 3x 4 8 4 x x or 3 3 4 8 The solution set is , . 3 3
x 1 or x 1 The solution set is {–1, 1}.
17.
2 x 4 2x 4 x2 The solution set is {2}.
18. 3 x 9 3x 9 x3 The solution set is {3}.
19.
23.
8 x 3 7 21 x 8 21 21 x or x 8 8 21 21 The solution set is , . 8 8
24.
1 2 No solution, since absolute value always yields a non-negative number. u2
2 v 1
No solution, since absolute value always yields a non-negative number. 25. 5 4 x 4
4 x 1 4x 1 4 x 1 or 4 x 1 1 1 x or x 4 4
3 20. x 9 4 x 12
x 2 2 3 5
26. 5
1 x 3 2
1 x 2 2
x 2 x 2 2 or 2 3 5 3 5 5 x 6 30 or 5 x 6 30 5 x 24 or 5 x 36 24 36 x x or 5 5 36 24 The solution set is , . 5 5
1 1 The solution set is , . 4 4
x 12 or x 12 The solution set is {–12, 12}. 21.
x 1 1 2 3
1 x 2 2 1 1 x 2 or x 2 2 2 x 4 or x 4 The solution set is 4, 4 .
27.
x2 9 0 x2 9 0 x2 9 x 3 The solution set is 3, 3 .
128
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Section 1.6: Equations and Inequalities Involving Absolute Value
28.
x 2 16 0
33.
x 2 16 0 x 2 16 x 4 The solution set is 4, 4 .
29.
or
x2 2 x 3 0
or x 2 2 x 3 0
x 2 2 x 3
2 4 12 2 2 8 x 3 or x 1 or x no real sol. 2 The solution set is 1, 3 .
34.
x 2 x 12 x 2 x 12
or
x 2 x 12 0
or x 2 x 12 0
x 2 x 12
1 1 48 2 1 47 x 3 or x 4 or x no real sol. 2 The solution set is 4, 3 .
or
2x 1 1 3x 4 2x 1 2x 1 1 1 or 3x 4 3x 4 2 x 1 1 3x 4 or 2 x 1 1 3 x 4 2 x 1 3x 4
or
2 x 1 3 x 4
x 3
or
5 x 5
35.
x 2 3x x 2 2 x
x 2 3 x x 2 2 x or
x 2 3x x 2 2 x
3x 2 x
or
x 2 3x x 2 2 x
or x 2 x 1 1
5x 0
or
2 x2 x 0
or x 2 x 0
x0
or x (2 x 1) 0
x 1 x 2 0 or x x 1 0
x0
or x 0 or x
2
x x 1 1 x2 x 1 1 x2 x 2 0
x 1, x 2 or x 0, x 1
x 2 3x 2 2 x2 3x 2 2 2
x 3x 4
36. or x 2 3x 2 2
x 2 3 x 4 0 or x x 3 0
x 4 x 1 0
or
x 0, x 3
x 4, x 1
The solution set is 4, 3, 0,1 .
1 2
x2 2 x x2 6 x x 2 2 x x 2 6 x or
2
or x 3x 0
1 The solution set is , 0 . 2
The solution set is 2, 1, 0,1 . 32.
x 7
5 x 3 6 x 10
x 3 or x 1 Neither of these values cause the denominator to equal zero, so the solution set is 3, 1 .
x 3 x 4 0 or x
31.
or
x 3 x 1 0 or x
30.
5 x 3 6 x 10
11x 13 13 x7 x or 11 Neither of these values cause the denominator to 13 equal zero, so the solution set is , 7 . 11
x2 2 x 3 x2 2 x 3
5x 3 2 3x 5 5x 3 5x 3 2 or 2 3x 5 3x 5 5 x 3 2 3x 5 or 5 x 3 2 3 x 5
x2 2x x2 6x x2 2 x x2 6x
2 x 6 x
or
8 x 0 x0
or 2 x 2 4 x 0 or 2 x ( x 2) 0
x0 or x 0 or x 2 The solution set is 2, 0 .
129 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
37.
2x 8
43.
4 3t 2 4
4 x 4
2 3t 6
x 4 x 4 or 4,4
38.
3t 2 4
8 2x 8
2 t 2 3 2 2 t t 2 or , 2 3 3
3 x 15
15 3 x 15
39.
2 3
5 x 5 x 5 x 5 or 5,5
44.
2u 5 7 7 2u 5 7
12 2u 2
7 x 42
6 u 1
7 x 42 or 7 x 42 x 6 or x 6
u 6 u 1 or 6, 1
x x 6 or x 6 or , 6 6,
45.
2x 3 2 2 x 3 2 or 2 x 3 2
40.
2x 1
2x 6
2 x 6 or 2 x 6 x 3 or x 3
x x 3 or x 3 or , 3 3,
41.
x2 23 1 x 2 1 1 x 3
x 1 x 3 or 1,3 42.
x4 3 5
x4 2 2 x 4 2 6 x 2
x 6 x 2 or 6, 2
2x 5
46. 3 x 4 2 3x 4 2 or 3 x 4 2 3x 6 or 3 x 2 2 x 2 or x 3 2 2 x x 2 or x or , 2 , 3 3
x 2 1
or
1 5 or x x 2 2 1 5 1 5 x x or x or , , 2 2 2 2
130
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Section 1.6: Equations and Inequalities Involving Absolute Value
47.
1 4 x 7 2
51.
1 4x 5
4 x 5 1
5 1 4 x 5
4 x 4
6 4 x 4
This is impossible since absolute value always yields a non-negative number. The inequality has no solution.
4 6 x 4 4 3 x 1 2
1 x
or
x 1 x 32 or 1, 32
48.
3 2
52.
x 6
3 2
6 x 6 6 x 6 x | 6 x 6 or 6, 6
1 2 x 4 1 3 1 2 x 3 4 2 x 2 4 2 x 2 2 2 x 1 or 1 x 2 x 1 x 2 or 1,2
50.
x 4 2 x 4 2
1 2x 3
49.
4 x 5 1
53.
2 x 3 2x 3 2 x 3 or 2 x 3 3 3 x or x 2 2 3 3 3 3 x x or x or , , 2 2 2 2
5 2 x 7 5 2 x 7 or 5 2 x 7 2 x 12 or 2 x 2 x 6 or x 1 x x 1 or x 6 or , 1 6,
54.
x 2 1 x 2 1 or x 2 1 x 1 or x3 x 1 or x 3
x x 3 or x 1 or , 3 1,
2 3x 1
2 3 x 1 or 2 3 x 1 3 x 3 or 3 x 1 1 x 1 or x 3 1 1 x x or x 1 or , 1, 3 3
1 3
55. 3 2 x 5 21 2x 5 7 7 2 x 5 7 2 2 x 12 1 x 6
x 1 x 6 or 1, 6
131 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
56. 1 2 x 3 1 2x 3
62. 3 x 1
x 1
3 1 2x 3 4 2 x 2 2 x 1
x 1
x 1 x 2 or 1, 2
5 2
5 2
5 5 or x 1 2 2 7 3 x or x 2 2 7 3 7 3 x x or x or , , 2 2 2 2 x 1
57.
1 2
9 x 5
This is impossible since absolute value always yields a non-negative number. No solution.
58. 3x 0
This is impossible since absolute value always yields a non-negative number. No solution.
59. 5 x 1
64.
Absolute value yields a non-negative number, so this inequality is true for all real numbers, (, ).
7 x 4 9
This is impossible since absolute value always yields a non-negative number. No solution.
6 x 2
65.
This is impossible since absolute value always yields a non-negative number. No solution.
2x 3 1 1 3 2 2x 3 1 1 3 2 2x 3 1 6 1 6 6 1 2 3 6 2 2 x 3 3 6 6 4 x 6 3 6 6 4 x 3 6 9 4 x 3 9 3 x 4 4 9 3 9 3 x x or , 4 4 4 4 1
7 2x 0 3
Since the absolute value cannot be negative, the only possible solution would be: 7 2x 0 3 7 2x 0 2 x 7 7 x 2
61.
63. 8 4 x 13
Absolute value yields a non-negative number, so this inequality is true for all real numbers, (, ).
60.
66.
4 x 15 0 6 4 x 15 0 6
Since the absolute value cannot be negative, the only possible solution would be:
132
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Section 1.6: Equations and Inequalities Involving Absolute Value 4 x 15 0 6 4 x 15 0 4 x 15 15 x 4
67.
70.
3x 7 4 5 3x 7 3x 7 4 or 4 5 5 3x 7 20 or 3x 7 20 3x 13 or 3x 27 13 x or x 9 3 13 13 or x 9 or , 9, x x 3 3
68.
5 2x 8 9 5 2x 5 2x 8 or 8 9 9 5 2 x 72 or 5 2 x 72 2 x 77 or 2 x 67 77 67 x or x 2 2 67 77 67 77 or x or , , x x 2 2 2 2
2x 3 1 1 2 3 2x 3 1 2x 3 1 1 1 or 2 3 2 3 2x 3 1 2x 3 1 6 1 or 6 6 1 6 3 3 2 2 3(2 x 3) 2 6 or 3(2 x 3) 2 6 6 x 9 2 6 or 6x 9 2 6 6 x 7 6 or 6x 7 6 6x 1 or 6 x 13 1 13 x or x 6 6
1 13 1 13 x x or x or , , 6 6 6 6
71. A temperature x that differs from 98.6 F by at least 1.5F . x 98.6 1.5 x 98.6 1.5 or x 98.6 1.5 x 97.1 or x 100.1 The temperatures that are considered unhealthy are those that are less than 97.1˚F or greater than 100.1˚F, inclusive. 72. The length L must be within 0.0025 of 5.375 inches. L 5.375 0.0025 0.0025 L 5.375 0.0025 5.3725 L 5.3775 The lengths must be between 5.3725 and 5.3775 inches, inclusive.
73. The percentage must be within 3.9 percentage points of 44 percent. The inequality that represents this would be: x 44 3.9 1 69. 5 x 1 2
9 2 Absolute value yields a non-negative number, so this inequality is true for all real numbers, (, ). x 1
3.9 x 44 3.9 40.1 x 47.9 The actual percentage is likely to fall between 40.1% and 47.9%, inclusive.
74. The speed x varies from 707 mph by up to 55 mph. a.
x 707 55
133 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities b.
55 x 707 55 55 x 707 55 652 x 762 The speed of sound is between 652 and 762 miles per hour, depending on conditions.
75. x differs from 3 by less than x3
81.
2 x 4 2 6 x 2 12 2 x 4 15 2 x 3 7 a 15, b 7
1 . 2
82.
1 2
83.
x2 7 7 x 2 7 5 x 9 15 x 10 1 1 1 1 15 x 10 1 1 1 x 10 15 1 a 1, b 15
76. x differs from 4 by less than 1 x ( 4) 1 x 4 1 1 x 4 1 5 x 3
x 5 x 3 77. x differs from 3 by more than 2. x (3) 2 x3 2
84.
x 1 3
3 x 1 3 4 x 2 1 x 5 7 1 1 1 x5 7 1 1 1 7 x5 1 a , b 1 7
x 3 2 or x 3 2 x 5 or x 1 x | x 5 or x 1
78. x differs from 2 by more than 3. x2 3 x 2 3 or x 2 3 x 1 or x5 x | x 1 or x 5
85. Given that a 0, b 0, and that a b.
x 1 3
Note that b a
3 x 1 3 3 5 ( x 1) 5 3 5 2 x48 a 2, b 8
80.
x 3 1 1 x 3 1 2 x4 6 3x 12 7 3x 1 13 a 7, b 13
1 1 x 3 2 2 5 7 x 2 2 5 7 x x 2 2
79.
x4 2
Since ba
a b , show
b a b a .
a b means b a 0 , we have
b a b a 0 .
Therefore, b a 0 which means a b.
x2 5
86. Show that a a .
5 x 2 5 5 4 ( x 2) 4 5 4 9 x 2 1 a 9, b 1
We know 0 a . So if a < 0, then we have
134
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Section 1.6: Equations and Inequalities Involving Absolute Value
a 0 a which means a a . . Now, if
If x a , then x a 2 a 0 and
a 0, then a a . So a a .
x a 0 . Therefore, x a
which is a contradiction. So the solution set for
87. Prove a b a b .
x 2 a is
2
Note that a b a b a b . Case 1: If a b 0, then a b a b, so a b a b a b a b 2
a 2ab b 2
a b
2
x2 a 0
x a x a 0
2
If x a , then x a 0 and x a 2 a 0 .
by problem 86
Thus, a b a b 2
2
Therefore, x a
ab a b . a b a b a b a b
x a x a 0 , which is a contradiction.
a b a b
If x a , then x a 2 a 0 and
a 2 2ab b 2
x a 0 . Therefore, x a
by problem 86
So the solution set for x 2 a is
Thus, a b a b 2
2
2
a 2 a b b 2
91.
x2 1 1 x 1 1 x 1 The solution set is x 1 x 1 .
88. Prove a b a b . a a b b a b b by the Triangle
Inequality, so a a b b which means a b a b . Therefore, a b a b .
x a 0 .
x x< a or x a .
2
ab a b
92.
x2 4 4x 4 2 x 2 The solution set is x 2 x 2 .
89. Given that a > 0, x2 a x2 a 0
x a x a 0
93. x 2 9
If x a , then x a 0 and x a 2 a 0 . Therefore,
x a x a 0 , which is a contradiction.
If a x a , then 0 x a 2 a and 2 a x a 0 .
x a 0 .
If a x a , then 0 x a 2 a and 2 a x a 0 ..Therefore,
Case 2: If a b 0, then a b a b , so
a b
x a x a .
90. Given that a > 0, x2 a .
2
a 2 a b b
Therefore, x a
x a 0 ,
x a 0 .
x 9 or x 9 x 3 or x 3 The solution set is x x 3 or x 3 .
94. x 2 1 x 1 or x 1 x 1 or x 1 The solution set is x x 1 or x 1 .
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Chapter 1: Equations and Inequalities
95.
x 2 16
99. 3x 2 x 1 4
16 x 16 4 x 4 The solution set is x 4 x 4 .
96.
3x 2 x 1 4 or 3 x 2 x 1 4
3x 2 x 1 4 3x 4 2 x 1 2 x 1 3 x 4 or 2 x 1 3x 4 x 5 or 2 x 1 3x 4 x5 or 5x 3 3 x5 x or 5 or 3x 2 x 1 4 3x 4 2 x 1
x2 9 9x 9 3 x 3 The solution set is x 3 x 3 .
97. x 2 4 x 4 or x 4 x 2 or x 2 The solution set is x x 2 or x 2 .
2 x 1 3 x 4 or 2 x 1 3x 4 x 3 or 2 x 1 3 x 4 x 3 or 5 x 5 x 3 x 1 or 3 The values and 3 are extraneous. 5 The solution set is 1, 5 .
98. x 2 16 x 16 or x 16 x 4 or x 4 The solution set is x x 4 or x 4 .
100.
4 3 y 2 4 3y 2 3 y 6 or 3 y 2 2 y2 y 3 y is largest using x = 18 and y = 2, The value of x y 2 1 . so x 18 9
x 3x 2 2 x 3x 2 2 or x 3x 2 2
x 3x 2 2 3x 2 2 x 3x 2 2 x or 3x 2 2 x 4x 4 or 3 x 2 2 x or 2x 0 x 1 or x 1 x0 or x 3 x 2 2 3 x 2 2 x
102. Since x 0 for all real numbers, then x x 0 and 1 1 . This means 1 x 1 x
3x 2 2 x or 3 x 2 2 x 4x 0 or 3x 2 2 x x0 or 2x 4 x0 or x2 The value 2 is extraneous. The solution set is 0, 1 .
101.
x 3 1 x 2 1 x 1 1 0 2 1 x 1 1 1 2 1 x 1 1 1 1 x 2
1 1
2 x 5 x 13 or 2 x 5 ( x 13) 2 x 5 x 13 2 x 5 x 13 3 x 8 x 18 8 x 3
1 1 x 2 0 x 1
136
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Section 1.7: Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
Therefore 1 x 1 . The solution set in interval notation is 1,1 . 103 – 105.
Answers will vary.
15. Let W represent the work, F the force, and d the distance. Work equals force times distance: W Fd 16. Let K represent the kinetic energy, m the mass, and v the velocity. Kinetic energy is one-half the product of the mass and the square of the 1 velocity: K mv 2 2
Section 1.7 1. mathematical modeling
17. C total variable cost in dollars, x number of dishwashers manufactured: C 150 x
2. interest 3. uniform motion 4. False; the amount charged for the use of principal is the interest. 5. True; this is the uniform motion formula. 6. a
18. R total revenue in dollars, x number of dishwashers sold: R 250 x 19. Let x represent the amount of money invested in bonds. Then 5000 x represents the amount of money invested in CD's. Since the total interest is to be $500, we have: 0.15 x 0.07(5000 x) 500
100 0.15 x 0.07(5000 x) 500 100
7. b
15 x 7(5000 x ) 50, 000
8. c 9. Let A represent the area of the circle and r the radius. The area of a circle is the product of π times the square of the radius: A r 2 10. Let C represent the circumference of a circle and r the radius. The circumference of a circle is the product of π times twice the radius: C 2 r 11. Let A represent the area of the square and s the length of a side. The area of the square is the square of the length of a side: A s 2
15 x 35, 000 7 x 50, 000 8 x 35, 000 50, 000 8 x 15, 000 x 1875 $1875 should be invested in bonds at 15% and $3125 should be invested in CD's at 7%.
20. Let x represent the amount of money invested in bonds. Then 5000 x represents the amount of money invested in CD's. Since the total interest is to be $700, we have: 0.15 x 0.07(5000 x) 700
100 0.15 x 0.07(5000 x) 700 100
12. Let P represent the perimeter of a square and s the length of a side. The perimeter of a square is four times the length of a side: P 4s 13. Let F represent the force, m the mass, and a the acceleration. Force equals the product of the mass times the acceleration: F ma 14. Let P represent the pressure, F the force, and A the area. Pressure is the force per unit area: F P A
15 x 7(5000 x ) 70, 000 15 x 35, 000 7 x 70, 000 8 x 35, 000 70, 000 8 x 35, 000 x 4375 $4375 should be invested in bonds at 15% and $625 should be invested in CD's at 7%.
21. Let x represent the amount of money loaned at 4%. Then 12, 000 x represents the amount of money loaned at 5.5%. Since the total interest is to be $600, we have:
137 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 0.04 x 0.055(12, 000 x) 600
156 pounds of cashews must be added to the 60 pounds of almonds.
100 0.04 x 0.055(12, 000 x) 600 100 4 x 5.5(12, 000 x) 60, 000
26. Let x represent the number of caramels in the box. Then 30 x represents the number of cremes in the box. Revenue Cost Profit 12.50 0.25 x 0.45(30 x) 3.00
4 x 66, 000 5.5 x 60, 000 1.5 x 66, 000 60, 000 1.5 x 6, 000 x 4000
12.50 0.25 x 13.5 0.45 x 3.00
$4000 is loaned at 4%.
12.50 13.5 0.20 x 3.00
22. Let R represent the interest rate for the loan. Aditi will owe $16,350 at the end of two years. This would be $1350 in interest for the two years, or $675 per year. We have: R (15000) 675 675 R 15000 0.045 or 4.5% Her interest rate would be 4.5%.
12.50 13.50 0.20 x 3.00 1.00 0.20 x 3.00 0.20 x 4.00 x 20 The box should contain 20 caramels and 10 cremes.
27. Let r represent the speed of the current. Rate Time Distance
23. Let x represent the number of pounds of Earl Gray tea. Then 100 x represents the number of pounds of Orange Pekoe tea. 6 x 4(100 x) 5.50(100) 6 x 400 4 x 550 2 x 400 550 2 x 150 x 75 75 pounds of Earl Gray tea must be blended with 25 pounds of Orange Pekoe.
20 1 16 r 60 3 1 Downstream 16 r 15 60 4
Upstream
16 r 3 16 r 4
Since the distance is the same in each direction: 16 r 16 r 3 4 4(16 r ) 3(16 r ) 64 4r 48 3r 16 7 r 16 r 2.286 7 The speed of the current is approximately 2.286 miles per hour.
24. Let x represent the number of pounds of the first kind of coffee. Then 100 x represents the number of pounds of the second kind of coffee. 2.75 x 5(100 x) 4.10(100) 2.75 x 500 5 x 410 2.25 x 500 410
28. Let r represent the speed of the motorboat. Rate Time Distance Upstream r 3 5 5 r 3 Downstream r 3 2.5 2.5 r 3
2.25 x 90 x 40 40 pounds of the first kind of coffee must be blended with 60 pounds of the second kind of coffee.
The distance is the same in each direction: 5(r 3) 2.5(r 3) 5r 15 2.5r 7.5 2.5r 22.5 r 9 The speed of the motorboat is 9 miles per hour.
25. Let x represent the number of pounds of cashews. Then x 60 represents the number of pounds in the mixture. 9 x 4.50(60) 7.75( x 60) 9 x 270 7.75 x 465 1.25 x 195 x 156 138
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Section 1.7: Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications 29. Let r represent the speed of the current. Rate Time Distance 10 Upstream 15 r 10 15 r 10 Downstream 15 r 10 15 r Since the total time is 1.5 hours, we have: 10 10 1.5 15 r 15 r 10(15 r ) 10(15 r ) 1.5(15 r )(15 r ) 2
150 10r 150 10r 1.5(225 r ) 300 1.5(225 r 2 ) 200 225 r 2 r 2 25 0 (r 5)(r 5) 0 r 5 or r 5 Speed must be positive, so disregard r 5 . The speed of the current is 5 miles per hour.
30. Let r represent the rate of the slower car. Then r 10 represents the rate of the faster car. Rate Time Distance Slower car r 3.5 3.5r Faster car r 10 3 3 r 10 3.5 r 3(r 10) 3.5 r 3r 30 0.5 r 30 r 60 The slower car travels at a rate of 60 miles per hour. The faster car travels at a rate of 70 miles per hour. The distance is (70)(3) = 210 miles.
31. Let r represent Karen’s normal walking speed. Rate Time Distance 50 r 2.5 With walkway 50 r 2.5 50 Against walkway r 2.5 50 r 2.5 Since the total time is 48 seconds: 50 50 48 r 2.5 r 2.5 50(r 2.5) 50(r 2.5) 48(r 2.5)( r 2.5) 50r 125 50r 125 48(r 2 6.25) 100r 48r 2 300 0 48r 2 100r 300 0 12r 2 25r 75
r
(25) (25) 2 4(12)(75) 2(12)
25 4225 24 r 3.75 or r 1.67 Speed must be positive, so disregard r 1.67 . Karen’ normal walking speed is approximately 3.75 feet per second.
32. Let r represent the speed of the airport walkway. Rate Time Distance 280 Walking with 1.5 r 280 1.5 r 280 Standing still r 280 r Walking with the walkway takes 60 seconds less time than standing still on the walkway: 280 280 60 r 1.5 r 280r 280(1.5 r ) 60r ( r 1.5) 280r 420 280r 60r 2 90r 60r 2 90r 420 0 2r 2 3r 14 0 (2r 7)(r 2) 0 2r 7 0 or r 2 0 7 r or r2 2 7 . 2 The speed of the airport walkway is 2 meters per second.
Speed must be positive, so disregard r
33. Let w represent the width of a regulation doubles tennis court. Then 2w 6 represents the length. The area is 2808 square feet: w(2 w 6) 2808 2w2 6w 2808 2w2 6 w 2808 0 w2 3w 1404 0 ( w 39)( w 36) 0 w 39 0 or w 36 0 w 39 or w 36 The width must be positive, so disregard w 39 . The width of a regulation doubles tennis court is 36 feet and the length is 2(36) + 6 = 78 feet.
139 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 34. l length of the court and w width of the court
Time to do job Part of job done in one hour 1 Patrice 10 10 1 April t t 1 Together 6 6 1 1 1 10 t 6 3t 30 5 t 2t 30 t 15 April would take 15 hours to paint the rooms.
The width of the court is to be two feet less than 1 half of its length. Thus, w l 2 . The area of 2 the court is 880 square feet, so: 1 2 l 2 (l ) 880 1 2 l 2l 880 2 1 2 l 2l 880 0 2 l 2 4l 1760 0 (l 40)(l 44) 0 l 40 or l 44 Since the length cannot be negative then the length is 44 ft. The area is 880, so:
37. l length of the garden w width of the garden a.
44 w 880 880 44 w 20 w
The dimensions of the court are 44 ft by 20 feet. 35. Let t represent the time it takes to do the job together. Time to do job Part of job done in one minute 1 Trent 30 30 1 Lois 20 20 1 Together t t
The length of the garden is to be twice its width. Thus, l 2w . The dimensions of the fence are l 4 and w4 . The perimeter is 46 feet, so: 2(l 4) 2( w 4) 46 2(2w 4) 2( w 4) 46 4w 8 2 w 8 46 6w 16 46 6w 30 w5 The dimensions of the garden are 5 feet by 10 feet.
b. Area l w 5 10 50 square feet c.
1 1 1 30 20 t 2t 3t 60 5t 60 t 12 Working together, the job can be done in 12 minutes.
If the dimensions of the garden are the same, then the length and width of the fence are also the same (l 4) . The perimeter is 46 feet, so: 2(l 4) 2(l 4) 46 2l 8 2l 8 46 4l 16 46 4l 30 l 7.5 The dimensions of the garden are 7.5 feet by 7.5 feet.
36. Let t represent the time it takes April to do the job working alone.
d. Area l w 7.5(7.5) 56.25 square feet. 38. l length of the pond w width of the pond a.
The pond is to be a square. Thus, l w . The dimensions of the fenced area are w 6
140
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Section 1.7: Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
on each side. The perimeter is 100 feet, so: 4( w 6) 100 4 w 24 100 4 w 76 w 19 The dimensions of the pond are 19 feet by 19 feet. b. The length of the pond is to be three times the width. Thus, l 3w . The dimensions of the fenced area are w 6 and l 6 . The perimeter is 100 feet, so: 2( w 6) 2(l 6) 100 2( w 6) 2(3w 6) 100 2w 12 6 w 12 100 8w 24 100 8w 76 w 9.5 l 3(9.5) 28.5 The dimensions of the pond are 9.5 feet by 28.5 feet. c.
If the pond is circular, the diameter is d and the diameter of the circle with the pond and the deck is d 6 .
3
d
3
The perimeter is 100 feet, so: (d 6) 100 d 6 100 d 100 6 100 d 6 25.83 The diameter of the pond is 25.83 feet. d.
Time to run Time 100 yards
2
25.83 2 Area circle r = 524 ft . 2 The circular pond has the largest area. 2
39. Let t represent the time it takes for the defensive back to catch the tight end.
Distance
Tight End
12 sec
t
100 25 12 3
25 t 3
Def. Back
10 sec
t
100 10 10
10t
Since the defensive back has to run 5 yards farther, we have: 25 t 5 10t 3 25t 15 30t 15 5 t t 3 10t 30 The defensive back will catch the tight end at the 45 yard line (15 + 30 = 45). 40. Let x represent the number of highway miles traveled. Then 30, 000 x represents the number of city miles traveled. x 30, 000 x 900 40 25 x 30, 000 x 200 200 900 25 40 5 x 240, 000 8 x 180, 000 3 x 240, 000 180, 000 3 x 60, 000 x 20, 000 Therese is allowed to claim 20,000 miles as a business expense. 41. Let x represent the number of gallons of pure water. Then x 1 represents the number of gallons in the 60% solution. % gallons % gallons % gallons 0 x 1(1) 0.60( x 1) 1 0.6 x 0.6 0.4 0.6 x 4 2 x 6 3
Area square l w 19(19) 361 ft 2 .
Area rectangle l w 28.5(9.5) 270.75 ft 2 .
Rate
2 gallon of pure water should be added. 3
42. Let x represent the number of liters to be drained and replaced with pure antifreeze.
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Chapter 1: Equations and Inequalities
% liters % liters % liters 1 x 0.40(15 x) 0.60(15)
47. Let t represent the time it takes for Mike to catch up with Dan. Since the distances are the same, we have: 1 1 t (t 1) 6 9 3t 2t 2 t2 Mike will pass Dan after 2 minutes, which is a 1 distance of mile. 3
x 6 0.40 x 9 0.60 x 3 x5 5 liters should be drained and replaced with pure antifreeze.
43. Let x represent the number of ounces of water to be evaporated; the amount of salt remains the same. Therefore, we get 0.04(32) 0.06(32 x) 1.28 1.92 0.06 x 0.06 x 0.64 0.64 64 32 x 10 23 0.06 6 3 10 23 10.67 ounces of water need to be
48. Let t represent the time of flight with the wind. The distance is the same in each direction: 330 t 270(5 t ) 330 t 1350 270 t 600 t 1350 t 2.25 The distance the plane can fly and still return safely is 330(2.25) = 742.5 miles.
evaporated.
49. Let t represent the time the auxiliary pump needs to run. Since the two pumps are emptying one tanker, we have: 3 t 1 4 9 27 4t 36 4t 9 9 t 2.25 4 The auxiliary pump must run for 2.25 hours. It must be started at 9:45 a.m.
44. Let x represent the number of gallons of water to be evaporated; the amount of salt remains the same. 0.03(240) 0.05(240 x) 7.2 12 0.05 x 0.05 x 4.8 4.8 x 96 0.05 96 gallons of water need to be evaporated. 45. Let x represent the number of grams of pure gold. Then 60 x represents the number of grams of 12 karat gold to be used. 1 2 x (60 x) (60) 2 3 x 30 0.5 x 40 0.5 x 10 x 20 20 grams of pure gold should be mixed with 40 grams of 12 karat gold.
50. Let x represent the number of pounds of pure cement. Then x 20 represents the number of pounds in the 40% mixture. x 0.25(20) 0.40( x 20) x 5 0.4 x 8 0.6 x 3 30 5 6 5 pounds of pure cement should be added. x
46. Let x represent the number of atoms of oxygen. 2x represents the number of atoms of hydrogen. x 1 represents the number of atoms of carbon. x 2 x x 1 45 4 x 44 x 11 There are 11 atoms of oxygen and 22 atoms of hydrogen in the sugar molecule.
51. Let t represent the time for the tub to fill with the faucets on and the stopper removed. Since one tub is being filled, we have:
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Section 1.7: Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
t t 1 15 20 4t 3t 60 t 60 60 minutes is required to fill the tub.
52. Let t be the time the 5 horsepower pump needs to run to finish emptying the pool. Since the two pumps are emptying one pool, we have: t2 2 1 5 8 4(2 t ) 5 20 8 4t 5 20 4t 7 t 1.75 The 5 horsepower pump must run for an additional 1.75 hours or 1 hour and 45 minutes to empty the pool. 53. Let t represent the time spent running. Then 5 t represents the time spent biking. Rate Time Distance Run 6 t 6t Bike 25 5 t 25(5 t )
The total distance is 87 miles: 6t 25(5 t ) 87 6t 125 25t 87 19t 125 87 19t 38 t2 The time spent running is 2 hours, so the distance of the run is 6(2) 12 miles. The distance of the bicycle race is 25(5 2) 75 miles.
54. Let r represent the speed of the eastbound cyclist. Then r 5 represents the speed of the westbound cyclist. Rate Time Distance Eastbound r 6 6r Westbound r 5 6 6(r 5)
The total distance is 246 miles: 6r 6(r 5) 246 6r 6r 30 246 12r 30 246 12r 216 r 18
The speed of the eastbound cyclist is 18 miles per hour, and the speed of the westbound cyclist is 18 5 23 miles per hour. 100 meters/sec. In 9.81 seconds, 12 100 Burke will run (9.81) 81.75 meters. Bolt 12 would win by 100-81.75=18.25 meters.
55. Burke's rate is
56. A 2 r 2 2 r h . Since A 58.9 square inches and h 6.4 inches, 2 r 2 2 r (6.4) 58.9 2 r 2 12.8 r 58.9 0 2 r 2 12.8r 58.9 0 r
12.8 (12.8) 2 4(2)(58.9) . 2(2)
12.8 635.04 4 r 3.1 or r 9.5 The radius of the coffee can is 3.1 inches.
57. Let the individual times to complete the project be E for Elaine, B for Brian, and D for either daughter. Using the respective rates gives 1 1 1 1 1 1 1 1 2 1 , (or ), E B 2 E D D 2 E D 2 1 1 1 and . From the first two equations, B D 4 1 2 . Substituting into the third equation B D 2 1 1 3 1 gives D 12 hours. or D 4 D D 4 1 2 1 Then E 3 hours and E 12 2 1 1 1 B 6 hours. The combined rate B 12 4 of Elaine, Brian, and one of their daughters is 1 1 1 7 project per hour, so it will take 3 6 12 12 12 them hours to complete the project. 7
58. If x = liters of original solution, then there were originally 0.2x liters of salt and 0.8 liters of pure water. Over time, the solution loses 0.25(0.8 x ) 0.2 x liters of pure water. She adds 20 liters of salt so the total amount of salt is 143 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 0.2 x 20 liters. She also adds 10 liters of pure water, so the total amount of pure water is 0.8 x 0.2 x 10 0.6 x 10 liters. The resulting concentration is 33 1/3% which means 0.2 x 20 1 0.2 x 20 1 or or 0.2 x 20 0.6 x 10 3 0.8 x 30 3 0.6 x 60 0.8 x 30 x 150 . There were initially 150 liters of solution in the vat.
Computing the average speed: Distance 45t1 55 t2 Avg Speed Time t1 t2 11 45 t2 55 t2 55 t2 55 t2 9 11 11t2 9t2 t t 9 2 2 9 110t2 990 t2 20t2 20 t2 9 99 49.5 miles per hour 2 The average speed for the trip from Chicago to Miami is 49.5 miles per hour.
59. The speed of the train relative to the man is 30 – 4 = 26 miles per hour. The time is 5 5 1 5 sec min h h. 60 3600 720 d rt 1 26 720 26 miles 720 26 5280 190.67 feet 720 The freight train is about 190.67 feet long.
63. The time traveled with the tail wind was: 919 t 1.67091 hours . 550 Since they were 20 minutes 1 hour early, the 3 time in still air would have been: 1.67091 hrs 20 min 1.67091 0.33333 hrs
60. Answers will vary.
2.00424 hrs Thus, with no wind, the ground speed is 919 458.53 . Therefore, the tail wind is 2.00424 550 458.53 91.47 knots .
61. Let x be the original selling price of the shirt. Profit Revenue Cost 4 x 0.40 x 20 24 0.60 x x 40 The original price should be $40 to ensure a profit of $4 after the sale.
64. It is impossible to mix two solutions with a lower concentration and end up with a new solution with a higher concentration.
If the sale is 50% off, the profit is: 40 0.50(40) 20 40 20 20 0 At 50% off there will be no profit.
Algebraic Solution: Let x = the number of liters of 25% solution. % liters % liters % liters
62. Let t1 and t2 represent the times for the two segments of the trip. Since Atlanta is halfway between Chicago and Miami, the distances are equal. 45 t1 55 t2
0.25 x 0.48 20 0.58 20 x 0.25 x 9.6 10.6 0.58 x
0.33x 1
55 t 45 2 11 t1 t2 9 t1
x 3.03 liters (not possible)
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Chapter 1 Review Exercises
Chapter 1 Review
7.
x 8 3 6 x 24
1. 2
x 18 The solution set is {18}.
2. 2(5 3 x) 8 4 5 x 10 6 x 8 4 5 x
6x 2 4 5x x6 The solution set is {6}.
3.
1 1 3 x x 2 3 4 6 1 1 3 x 12 2 x 3 4 6 12 6x 2 9 2x 8 x 11 11 x 8 The solution set is 11 . 8
8.
x 6 x 1 5 5x 6x 6 6x Since x = 6 does not cause a denominator to equal zero, the solution set is {6}.
1 3x x 6 1 4 3 2 x x 1 3 6 1 12 4 3 2 12 3(1 3 x) 4( x 6) 6 3 9 x 4 x 24 6 13 x 27 27 x 13 The solution set is 27 . 13
4. (2 x 7) 2 20 2 x 7 20
9.
2 x 7 2 5
( x 1)(2 x 3) 3 2 x2 x 3 3
2 x 7 2 5 7 2 5 2 7 2 5 7 2 5 The solution set is , 2 2 x
2 x2 x 6 0 (2 x 3)( x 2) 0 x
3 or x 2 2
5. x(1 x) 6
3 . 2
10. 2 x 3 4 x 2
x x2 6
0 4 x2 2 x 3
0 x2 x 6 b 2 4ac 1 4 1 6 2
1 24 23 Therefore, there are no real solutions.
6.
The solution set is 2,
x(1 x) 6 x x2 6 x2 x 6 0 ( x 3)( x 2) 0 x 3 or x 2
x
(2) (2)2 4(4)(3) 2(4)
2 52 2 2 13 1 13 8 8 4 1 13 1 13 The solution set is , . 4 4
The solution set is 3, 2 .
145 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
x2 1 2
Check x 2:
2(2) 3 2 1 2 3
3 x 2 1 2 3
Check x 6:
2(6) 3 6 9 6 9 3
3
11.
3
The solution set is 2 .
x2 1 8 x2 9 x 3
Check x 3 : 3
(3) 2 1 2 3
Check x 3 : (3) 2 1 2
2 x 3 16
3
3
9 1 2
4
4
3
4
2 x 13
9 1 2
82 22 The solution set is 3,3 .
x
82 22
Check x
1 x3 3
4 2
1 x3 (3)2 2
13 2
13 : 2
13 4 4 2 3 13 3 16 2
The solution set is
1 x3 9 x3 8
17.
x 382
Check x 2 :
2x 3 2
2x 3 2
3
12.
4
16.
13 . 2
4 x2 x 6 x 1
4x x 6 x 1 2
2
1 2 3 3
2
4 x2 x 6 x 1
9 3 33 The solution set is 2 .
4 x2 5 x2
13. x( x 1) 2 0
5 4
x
x2 x 2 0 1 (1) 2 4(1)(2) 1 7 2(1) 2 No real solution. x
5 5 4 2
5 : 2
Check x 2
14.
4
5 5 5 4 6 + 1 2 2 2 0.34356 0.34356
2
x 5x 4 0
x 2 4 x 2 1 0 x 2 4 0 or x 2 1 0 x 2 or x 1 The solution set is 2, 1,1, 2 .
15.
Check x 2
5 : 2
5 5 5 6 4 1 2 2 2 The second solution is not possible because it makes the radicand negative. 5 The solution set is . 2
2x 3 x 3 2x 3 3 x 2 x 3 9 6 x x2 x 2 8 x 12 0 ( x 2)( x 6) 0 x 2 or x 6
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Chapter 1 Review Exercises
18.
2x 1 x 5 3
Check
2x 1 3 x 5
2x 1 3 x 5 2
1 1 : 2 2
6
Check 1: 1
2
1 7 2
6
8 64 56 8 0 3
7 1 8 1 7 8 0
The solution set is 1,
2x 1 9 6 x 5 x 5
3
1 . 2
x 5 6 x 5
x 5 2 6 x 5
2
x 2 m 2 2mx nx
21.
x 2 m 2 2mx n 2 x 2
x 2 10 x 25 36 x 5
2
x n x 2mx m 2 0
2 2
1 n x 2mx m 0
2
x 10 x 25 36 x 180
2
2
x 46 x 205 0
x 41 x 5 0
x
x 41 or x 5 Check x 41:
2 41 1 41 5 81 36 9 6 3
Check x 5 :
2
The solution set is 5, 41 . 19. 2 x1/ 2 3 0
x
2 x1/ 2 3
or
2 x1/ 2 2 32
x
4x 9 9 4
(2m) (2m) 2 4 1 n 2 m 2 2 1 n 2
2 m 4m 2 4 m 2 4 m 2 n 2 2 1 n 2
2 m 4m 2 n 2 2
m 1 n 1 n
3 3 2 3 3 3 0 2 9 The solution set is . 4
m 1 n 1 n
u 8 u 1 0 x
x
8 8
x
1 2
2
m 1 n m 1 n 1 n 1 n
m 1 n m 1 n 1 n 1 n
1/ 3
u 1
or
3
or x or
x
3 1/ 3
1 1
m m , , n 1, n 1. 1 n 1 n
or ax 2b 0 ax 2b 2b x a 9b 2b , a 0. The solution set is , 5a a
Let u x 3 so that u 2 x 6 . u 2 7u 8 0 or
2m 2mn
5ax 9b 0 5ax 9b 9b x 5a
20. x 6 7 x 3 8 0
u 8
22. 10a 2 x 2 2abx 36b 2 0 5a 2 x 2 abx 18b 2 0 5ax 9b ax 2b 0
1/ 2
9 2 4
3 1/ 3
2
The solution set is
9 Check x : 4
3
2
2 1 n 2 1 n 2 2m 1 n m 1 n 1 n2 2 1 n 2
2 5 1 5 5 9 0 3 0 3
x
2
1/ 3
x 1
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Chapter 1: Equations and Inequalities
25.
x 2 3x 7 x 2 3x 9 2 0
23.
2 3x 7
x 2 3x 7 x 2 3x 9 2
x 2 3x 7 x 2 3x 9 2 2
2 3x 7 3x 5
2
x 2 3x 7 x 2 3x 9 4 x 2 3x 9 4
x
2
6 x 6 4 x 3x 9
6 x 1 4 x2 3x 9 2
2 x3 3x 2 0 x 2 2 x 3 0
5 x 9 x 3 0
x2 0
or 2 x 3 0 3 x 0 or x 2 3 The solution set is 0, . 2
9 x or x 3 5
9 Check x : 5 2
x3
or
2 x3 3x 2
26.
5 x 2 6 x 27 0
2
9 3 9 7 9 3 9 9 2 5 5 5 5
27.
2 x3 5 x 2 8 x 20 0 x2 2 x 5 4 2 x 5 0
81 27 81 27 7 9 2 = 25 5 25 5
2 x 5 x2 4 0
81 135 175 81 135 225 2 25 25
2x 5 0
or x 2 4 0
2 x 5 or 5 x or 2
121 441 11 21 2 20 25 25 5 5
Check x 3 :
x2 4 x 2
5 The solution set is , 2, 2 . 2
32 3 3 7 32 3 3 9 2 997 999 2
28.
25 9 2 2 2 40 9 The solution set is . 5
24.
5 3
2
36 x 2 72 x 36 16 x 2 48 x 144 20 x 2 24 x 108 0
or 2 3 x 7 or 3x 9
5 The solution set is ,3 3
2
36 x 2 x 1 16 x 3x 9 2
2 3x 2 9
2x 3 7 2 x 3 7 or 2 x 3 7 2 x 4 or 2 x 10 x 2 or x 5 The solution set is {–5, 2}.
2x 3 x 2 5 2 2(2 x 3) 10(2) 5 x 4 x 6 20 5 x 14 x x 14 x x 14 or 14,
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Chapter 1 Review Exercises
29.
2x 3 7 4 36 2 x 3 28 33 2 x 31 33 31 x 2 2 31 33 x 2 2 31 33 31 33 x x or , 2 2 2 2 9
33. 2 2 3x 4 2 3x 2 2 2 3 x 2 4 3 x 0 4 x0 3 4 4 x 0 x or 0, 3 3
34. 1 2 3 x 4 2 3 x 5 3 3x 2 6 30. 12 24 3 3 x 72 21 3 x 69 7 x 23
2 3x 5 2 3x 5 or 2 3x 5 7 3x or 3 3x 7 x or 1 x 3 7 x 1 or x 3 7 7 x x 1 or x or , 1 , 3 3
x | 23 x 7 or 23, 7
31.
3x 4
1 2
1 1 3x 4 2 2 9 7 3x 2 2 3 7 x 2 6 3 7 3 7 x x or , 2 6 2 6
32.
35.
6 3 i 2 4i 6 2 3 4 i 4 7i
36. 4 3 i 3 5 2i 12 4i 15 6i 3 2i 37.
3 3 3i 9 3i 3 i 3 i 3 i 9 3i 3i i 2 9 3i 9 3 i 10 10 10
i 1 1 1
38. i 50 i 48 i 2 i 4
2x 5 9 2 x 5 9 or 2 x 5 9 2 x 4 or 2x 14 x 2 or x7
39.
x x 2 or x 7 or , 2 7,
12
2
12
2 3i 3 2 3i 2 2 3i 4 12i 9i 2 2 3i 5 12i 2 3i 10 15i 24i 36i 2 46 9i
149 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 0.08 x 0.05(70, 000 x) 5000
40. x 2 x 1 0 a 1, b 1, c 1,
100 0.08 x 0.05(70, 000 x) 5000 100
b 2 4ac 12 4 11 1 4 3 x
8 x 350, 000 5 x 500, 000 3x 350, 000 500, 000
1 3 1 3 i 1 3 i 2 1 2 2 2
3 x 150, 000 x 50, 000 $50,000 should be invested in bonds at 8% and $20,000 should be invested in CD's at 5%.
3 1 3 1 The solution set is i, i . 2 2 2 2
46. Using s v t , we have t 3 and v 1100 . Finding the distance s in feet: s 1100(3) 3300 The storm is 3300 feet away.
41. 2 x 2 x 2 0 a 2, b 1, c 2, b 2 4ac 12 4 2 2 1 16 17 x
1 17 1 17 2 2 4
47. 1600 I 3600 900 1600 2 3600 x x2 1 1 1600 900 3600 9 1 x2 16 4 3 1 x 4 2 The range of distances is from 0.5 meters to 0.75 meters, inclusive.
1 17 1 17 The solution set is , . 4 4
42.
x2 3 x x2 x 3 0 a 1, b 1, c 3, b 2 4ac 1 4 1 3 1 12 11 2
x
1 11 2 1
1 11 i 1 11 i 2 2 2
48. Let s represent the distance the plane can travel.
1 11 1 11 i, i . The solution set is 2 2 2 2
With wind Against wind Rate 250 30 280 250 30 220 (s / 2) ( s / 2) Time 280 220 s s Dist. 2 2 Since the total time is at most 5 hours, we have: s / 2 s / 2 5 280 220 s s 5 560 440 11s 14s 5(6160)
43. x(1 x ) 6 x2 x 6 0 a 1, b 1, c 6, b 2 4ac 12 4 1 6 1 24 23 x
1 23 2 1
1 23 i 1 23 i 2 2 2
23 1 23 1 The solution set is i, i . 2 2 2 2
25s 30,800 s 1232 The plane can travel at most 1232 miles or 616 miles one way and return 616 miles.
44. c 50, 000 95 x 45. Let x represent the amount of money invested in bonds. Then 70, 000 x represents the amount of money invested in CD's. Since the total interest is to be $5000, we have:
150
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Chapter 1 Review Exercises 49. Let t represent the time it takes the helicopter to reach the raft. Raft Helicopter Rate 5 90 t Time t Dist. 5t 90t
Since the total distance is 150 miles, we have: 5t 90t 150 95t 150 t 1.58 hours 1 hour and 35 minutes The helicopter will reach the raft in about 1 hour and 35 minutes. 50. Given that s 1280 32t 16t 2 , a.
The object hits the ground when s 0 . 0 1280 32t 16t 2 2
t 2t 80 0
t 10 t 8 0 t 10, t 8 The object hits the ground after 8 seconds. b. After 4 seconds, the object’s height is s 1280 32 4 16 4 896 feet.
52. Let t represent the time it takes the smaller pump to empty the tank. Small Pump Large Pump Time to do t t4 job alone Part of job 1 1 t t4 done in 1 hr Time on job 5 5 (hrs) Part of job 5 5 done by each t t 4 pump
Since the two pumps empty one tank, we have: 5 5 1 t t4 5(t 4) 5t t (t 4) 5t 20 5t t 2 4t t 2 14t 20 0 We can solve this equation for t by using the quadratic formula: t
(14) (14) 2 4(1)(20) 2(1)
14 116 14 2 29 2 2
2
51. Let t represent the time it takes Clarissa to complete the job by herself. Clarissa Shawna Time to do t t 5 job alone Part of job 1 1 t t 5 done in 1 day Time on job 6 6 (days) Part of job 6 6 done by each t t 5 person
Since the two people paint one house, we have: 6 6 1 t t 5 6(t 5) 6t t (t 5) 2
6t 30 6t t 5t
7 29 7 5.385 t 12.385 or t 1.615 (not feasible) It takes the small pump approximately 12.385 hours (12 hr 23 min) to empty the tank.
53. Let x represent the amount of water added. % salt Tot. amt. amt. of salt 10% 64 0.10 64 0% x 0.00 x 2% 64 x 0.02 64 x
0.10 64 0.00 x 0.02 64 x 6.4 1.28 0.02 x 5.12 0.02 x x 256 256 ounces of water must be added.
t 2 7t 30 0 (t 10)(t 3) 0 t 10 or t 3 It takes Clarissa 10 days to paint the house when working by herself.
151 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities 2 2 w 6 2 w 6 50
54. Consider the diagram 10
4w 12 2 w 12 50
w
6w 26 26 4 13 6 l 2w 8 23
w
w+2
By the Pythagorean Theorem we have w2 w 2 10 2
2
w2 w2 4 w 4 100
The painting is 8 23 inches by 4 13 inches.
2 w2 4w 96 0
The frame is 14 23 inches by 10 13 inches.
w2 2 w 48 0
57. Let x represent the amount Scott receives. Then 3 x represents the amount Alice receives and 4 1 x represents the amount Tricia receives. The 2 total amount is $900,000, so we have: 3 1 x x x 900, 000 4 2 3 1 4 x x x 4 900, 000 4 2 4 x 3x 2 x 3, 600, 000 9 x 3, 600, 000 x 400, 000 3 3 So, x 400, 000 300, 000 and 4 4 1 1 x 400, 000 200, 000 . 2 2 Scott receives $400,000, Alice receives $300,000, and Tricia receives $200,000.
w 8 w 6 0 w 8 or w 6 The width is 6 inches and the length is 6 + 2 = 8 inches.
55. Let x represent the amount of the 15% solution added. % acid tot. amt. amt. of acid 40% 60 0.40 60 15% x 0.15 x 25% 60 x 0.25 60 x
0.40 60 0.15 x 0.25 60 x 24 0.15 x 15 0.25 x 9 0.1x x 90 90 cubic centimeters of the 15% solution must be added, producing 150 cubic centimeters of the 25% solution.
56. a.
58. Let t represent the time it takes the older machine to complete the job by itself. Old copier New copier Time to do t t 1 job alone Part of job 1 1 t t 1 done in 1 hr Time on job 1.2 1.2 (hrs) Part of job 1.2 1.2 done by each t t 1 copier
Consider the following diagram: 4 s 6 50
4 s 24 50
s
4 s 26
s
s 6.5
The painting is 6.5 inches by 6.5 inches. s 6 12.5 , so the frame is 12.5 inches by 12.5 inches. b. Consider the following diagram:
Since the two copiers complete one job, we have:
w 2w
152
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Chapter 1 Test
Chapter 1 Test
1.2 1.2 1 t t 1 1.2(t 1) 1.2t t (t 1)
1.
1.2t 1.2 1.2t t 2 t t 2 3.4t 1.2 0 5t 2 17t 6 0 (5t 2)(t 3) 0 t 0.4 or t 3 It takes the old copier 3 hours to do the job by itself. (0.4 hour is impossible since together it takes 1.2 hours.)
59. Let rS represent Scott's rate and let rT represent Todd's rate. The time for Scott to run 95 meters is the same as for Todd to run 100 meters. 95 100 rS rT
x2 x 6 x2 x 6 0 ( x 3)( x 2) 0 x 3 0 or x 2 0 x 3 or x 2 The solution set is {2, 3}.
d S t rs t 0.95rT 0.95dT
3.
d S 0.95dT 0.95(105) 99.75 a.
d.
To end in a tie: 100 0.95(100 x) 100 95 0.95 x 5 0.95 x x 5.26 meters
e.
x 4 x 1 0 2
x2 4 0
b. Todd wins the race.
Todd wins by 0.25 meters.
x 4 3x2 4 0
2
The race does not end in a tie.
c.
x( x 1) 6
2.
rS 0.95rT
If Todd starts from 5 meters behind the start: dT 105
2x x 5 3 2 12 2x x 5 12 12 3 2 12 8x 6 x 5 2x 5 5 x 2 5 The solution set is . 2
x2 1 0
or
x 2 4 or x 2 1 Not real x 2 or The solution set is {2, 2}.
4.
2x 5 2 4 2x 5 2
2x 5 2 2
95 = 0.95(100) Therefore, the race ends in a tie.
60. We will use the formula for interest, I prt . Since she owed 27060 at the end of the loan she ad accumulated 3060 in interest and the principal is 24000. I prt 3060 (24000)r (3) 3060 r 3(24000) 0.0425 r The interest rate is 4.25 %.
2
2x 5 4 2x 9 9 x 2
Check:
9 2 5 2 4 2 95 2 4
42 4 22 4 44 9 The solution set is . 2
153 Copyright © 2025 Pearson Education, Inc.
Chapter 1: Equations and Inequalities
5.
2 x 3 7 10
9. 3x 4 8 8 3 x 4 8 12 3 x 4 4 4 x 3 4 4 x 4 x or 4, 3 3
2x 3 3 2 x 3 3 or 2 x 3 3 2 x 6 or 2x 0 x 3 or x0 The solutions set is {0, 3}. 3x3 2 x 2 12 x 8 0
6.
x 2 3x 2 4 3x 2 0
x 4 3x 2 0 2
10. 2 2 x 5 9
x 2 x 2 3x 2 0 x20
2x 5 7
or x 2 0 or 3x 2 0
x 2 or
x 2 or
x
2 x 5 7 or 2 x 5 7 2 x 2 or 2 x 12 x 1 or x6
2 3
2 The solution set is 2, , 2 . 3
x x 1 or x 6 or , 1 6, .
7. 3x 2 x 1 0 x
(1) (1) 2 4(3)(1) 2(3)
11.
1 11 (Not real) 6 This equation has no real solutions.
8.
2 2 3 i 6 2i 6 2i 3 i 3 i 3 i 9 3i 3i i 2 9 (1) 6 2i 3 i 3 1 i 10 5 5 5
12. 4 x 2 4 x 5 0
3x 4 6 2 3x 4 2 3 2 2 6 2 6 3 x 4 12 2 3 x 16 2 16 x 3 3 3
x
(4) (4) 2 4(4)(5) 2(4)
4 64 4 8i 1 i 8 8 2 1 1 i, i . This solution set is 2 2
2 16 2 16 x x or , 3 3 3 3
154
Copyright © 2025 Pearson Education, Inc.
Chapter 1 Projects 13. Let x represent the amount of the $8-per-pound coffee. Amt. of coffee Price Total $ (pounds) ($) 20 4 20 4 x 8 8 x 20 x 5 5 20 x 80 8 x 5 20 x 80 8 x 100 5 x 3 x 20 20 x 6 23 3 Add 6 23 pounds of $8/lb coffee to get 26 23
pounds of $5/lb coffee.
3. T = 0.15 board per second 1 0.15 0.2 p 2
0.2 p 2 0.15 1 0.03 p 0.3 1 0.03 p 0.7 p 23.3 parts per board Thus, only 23 parts per board will work.
For problems 4 – 6, C is requested, so solve for C first: n T Cnp L M
Cnp L M T n CnpT LT MT n CnpT n LT MT n LT MT C npT
Chapter 1 Projects Project I
4. T = 0.06, n = 3, p = 100, M = 1, L = 5 3 5 0.06 1 0.06 C 0.147 sec 3 100 0.06
Internet-based Project Project II n , n 3, L 5, M 1, C 0.2 Cnp L M 3 3 1 T 0.2 3 p 5 1 0.6 p 6 0.2 p 2
1. T
2. All of the times given in problem 1 were in seconds, so T = 0.1 board per second needs to used as the value for T in the equation found in problem 1. 1 0.1 0.2 p 2
5. T = 0.06, n = 3, p = 150, M = 1, L = 5 3 5 0.06 1 0.06 C 0.098 sec 3 150 0.06 6. T = 0.06, n = 3, p = 200, M = 1, L = 5 3 5 0.06 1 0.06 C 0.073 sec 3 200 0.06 7. As the number of parts per board increases, the tact time decreases, if all the other factors remain constant.
0.2 p 2 0.1 1 0.02 p 0.2 1 0.02 p 0.8 p 40 parts per board
155 Copyright © 2025 Pearson Education, Inc.
Chapter 2 Graphs (f) Quadrant IV
Section 2.1 1. 0 2.
5 3 8 8
3.
32 42 25 5
4. 112 602 121 3600 3721 612 Since the sum of the squares of two of the sides of the triangle equals the square of the third side, the triangle is a right triangle. 5.
1 bh 2
16. (a) Quadrant I (b) Quadrant III (c) Quadrant II (d) Quadrant I (e) y-axis (f) x-axis
6. true 7. x-coordinate or abscissa; y-coordinate or ordinate 8. quadrants 9. midpoint 10. False; the distance between two points is never negative. 11. False; points that lie in Quadrant IV will have a positive x-coordinate and a negative y-coordinate. The point 1, 4 lies in Quadrant II.
17. The points will be on a vertical line that is two units to the right of the y-axis.
x x y y2 12. True; M 1 2 , 1 2 2 13. b 14. a 15. (a) Quadrant II (b) x-axis (c) Quadrant III (d) Quadrant I (e) y-axis
156 Copyright © 2025 Pearson Education, Inc.
Section 2.1: The Distance and Midpoint Formulas
18. The points will be on a horizontal line that is three units above the x-axis.
28. d ( P1 , P2 )
6 ( 4) 2 2 (3) 2
102 52 100 25 125 5 5
29. d ( P1 , P2 )
2.3 (0.2) 2 1.1 (0.3) 2
2.52 0.82 6.25 0.64 6.89 2.62
30. d ( P1 , P2 )
(1.5) 2 (1.2) 2 2.25 1.44
19. d ( P1 , P2 ) (2 0) 2 (1 0) 2
3.69 1.92
22 12 4 1 5
31. d ( P1 , P2 ) (0 a) 2 (0 b) 2
20. d ( P1 , P2 ) (2 0) 2 (1 0) 2 (2) 2 12 4 1 5
(3) 2 12 9 1 10
22. d ( P1 , P2 )
2 (1) (2 1)
a 2 a 2 2a 2 a
23. d ( P1 , P2 ) (5 3) 2 4 4
d ( A, B )
2
24. d ( P1 , P2 )
2 1 4 0
3 4 9 16
25. d ( P1 , P2 )
4 (7) 2 (0 3)2
d ( B, C )
25 5
d ( A, C )
1 (2) 2 (0 5)2
12 (5) 2 1 25 26
112 ( 3) 2 121 9 130
4 2 2 2 (3) 2
22 52 4 25 29 27. d ( P1 , P2 ) (6 5) 2 1 (2)
1 12 (0 3)2
(2) 2 (3)2 4 9 13
2
2
1 (2) 2 (3 5)2
32 (2) 2 9 4 13
2
26. d ( P1 , P2 )
2
33. A (2,5), B (1,3), C (1, 0)
22 8 4 64 68 2 17
2
(a )2 (a )2
2
32 12 9 1 10
2
( a ) 2 ( b ) 2 a 2 b 2
32. d ( P1 , P2 ) (0 a ) 2 (0 a) 2
21. d ( P1 , P2 ) (2 1) 2 (2 1) 2
2
0.3 1.2 2 1.1 2.32
2
12 32 1 9 10
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Chapter 2: Graphs
Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:
d ( A, B)2 d ( B, C )2 d ( A, C )2
13 13 26 2
2
2
13 13 26 26 26
The area of a triangle is A
1 bh . In this 2
problem, A 1 d ( A, B) d ( B, C ) 2 1 13 13 1 13 2 2 13 2 square units
12 (2) 2 (3 5)2
2
2
200 200 400 400 400 1 The area of a triangle is A bh . In this 2 problem, 1 A d ( A, B ) d ( B, C ) 2 1 10 2 10 2 2 1 100 2 100 square units 2
d ( A, B)
6 ( 5) 2 (0 3)2
130 d ( B, C )
196 4 200 10 2
5 6 2 (5 0)2
(1) 2 52 1 25
10 12 2 (11 3)2
(2) (14) 2
26 d ( A, C )
2
4 196 200
5 ( 5) 2 (5 3)2
102 22 100 4
10 2 d ( A, C )
2
112 ( 3) 2 121 9
142 (2) 2
d ( B, C )
10 2 10 2 20
35. A ( 5,3), B (6, 0), C (5,5)
34. A (2, 5), B (12, 3), C (10, 11) d ( A, B )
d ( A, B)2 d ( B, C )2 d ( A, C )2
104
10 (2) (11 5) 2
2
2 26
122 (16) 2 144 256 400 20
Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:
Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: 158 Copyright © 2025 Pearson Education, Inc.
Section 2.1: The Distance and Midpoint Formulas
d ( A, C )2 d ( B, C )2 d ( A, B)2
d ( A, C )2 d ( B, C )2 d ( A, B)2
104 26 130
29 2 29 145
2
2
2
104 26 130 130 130 1 The area of a triangle is A bh . In this 2 problem, 1 A d ( A, C ) d ( B, C ) 2 1 104 26 2 1 2 26 26 2 1 2 26 2 26 square units
36. A (6, 3), B (3, 5), C (1, 5) d ( A, B)
3 (6) (5 3) 2
2
92 (8) 2 81 64 145 d ( B, C )
1 32 (5 (5))2
(4) 2 102 16 100 116 2 29 d ( A, C )
2
2
29 4 29 145 29 116 145 145 145 1 The area of a triangle is A bh . In this 2 problem, 1 A d ( A, C ) d ( B, C ) 2 1 29 2 29 2 1 2 29 2 29 square units
37. A (4, 3), B (0, 3), C (4, 2) d ( A, B) (0 4) 2 3 (3)
2
( 4)2 02 16 0 16 4 d ( B, C )
4 0 2 2 (3) 2
42 52 16 25 41
1 ( 6) (5 3) 2
2
2
52 22 25 4
d ( A, C ) (4 4) 2 2 (3)
2
02 52 0 25
29
25 5
Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:
Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:
159 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
d ( A, B)2 d ( A, C )2 d ( B, C )2 4 2 52
41
d ( A, B)2 d ( B, C )2 d ( A, C )2
2
42 22 2 5
1 bh . In this 2
problem, 1 A d ( A, B) d ( A, C ) 2 1 45 2 10 square units
d ( A, B) (4 4) 2 1 (3)
1 bh . In this problem, 2
1 d ( A, B) d ( B, C ) 2 1 42 2 4 square units
2
02 42 0 16 16 4
2 4 2 1 12
39. The coordinates of the midpoint are: x x y y ( x, y ) 1 2 , 1 2 2 2 35 4 4 , 2 2 8 0 , 2 2 (4, 0) 40. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2
(2) 2 02 4 0 4 2 d ( A, C ) (2 4) 2 1 (3)
The area of a triangle is A A
38. A (4, 3), B (4, 1), C (2, 1)
d ( B, C )
2
16 4 20 20 20
16 25 41 41 41
The area of a triangle is A
2
(2) 2 42 4 16 20
2 2 0 4 , 2 2 0 4 , 2 2 0, 2
41. The coordinates of the midpoint are: x x y y ( x, y ) 1 2 , 1 2 2 2 1 8 4 0 , 2 2 7 4 , 2 2 7 , 2 2
2 5
Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem:
160 Copyright © 2025 Pearson Education, Inc.
Section 2.1: The Distance and Midpoint Formulas
42. The coordinates of the midpoint are: x x y y ( x, y ) 1 2 , 1 2 2 2 2 4 3 2 , 2 2 6 1 , 2 2 1 3, 2 43. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2 7 9 5 1 , 2 2 16 4 , 2 2 (8, 2)
48. The new x coordinate would be 1 2 3 and the new y coordinate would be 6 4 10 . Thus the new point would be 3,10 49. a. If we use a right triangle to solve the problem, we know the hypotenuse is 13 units in length. One of the legs of the triangle will be 2+3=5. Thus the other leg will be: 52 b 2 132 25 b 2 169 b 2 144 b 12
44. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2 4 2 3 2 , 2 2 2 1 , 2 2 1 1, 2
Thus the coordinates will have an y value of 1 12 13 and 1 12 11 . So the points are 3,11 and 3, 13 . b. Consider points of the form 3, y that are a
distance of 13 units from the point 2, 1 . d
x2 x1 2 y2 y1 2
3 (2) 2 1 y 2
52 1 y 2
25 1 2 y y 2
45. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2 a0 b0 , 2 2 a b , 2 2
y 2 2 y 26
13 132
y 2 2 y 26
y 2 2 y 26
2
169 y 2 2 y 26 0 y 2 2 y 143
46. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2 a0 a0 , 2 2 a a , 2 2
47. The x coordinate would be 2 3 5 and the y coordinate would be 5 2 3 . Thus the new point would be 5,3 .
0 y 11 y 13 y 11 0
or y 13 0 y 11 y 13
Thus, the points 3,11 and 3, 13 are a distance of 13 units from the point 2, 1 .
161 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
50. a. If we use a right triangle to solve the problem, we know the hypotenuse is 17 units in length. One of the legs of the triangle will be 2+6=8. Thus the other leg will be:
d
x2 x1 2 y2 y1 2
4 x 2 3 0 2
82 b 2 17 2
16 8 x x 2 3
64 b 2 289
16 8 x x 2 9
b 2 225 b 15
2
x 2 8 x 25 6 x 2 8 x 25
Thus the coordinates will have an x value of 1 15 14 and 1 15 16 . So the points are 14, 6 and 16, 6 .
62
x 8x 25 2
2
36 x 2 8 x 25 0 x 2 8 x 11
b. Consider points of the form x, 6 that are
a distance of 17 units from the point 1, 2 . d
x2 x1 2 y2 y1 2
1 x 2 2 6
x 2 2 x 1 8
x
(8) (8) 2 4(1)(11) 2(1)
8 64 44 8 108 2 2 86 3 43 3 2 x 4 3 3 or x 4 3 3
2
2
x 2 2 x 1 64
Thus, the points 4 3 3, 0 and 4 3 3, 0 are
x 2 2 x 65
on the x-axis and a distance of 6 units from the point 4, 3 .
17 x 2 2 x 65 17 2
x 2 2 x 65
52. Points on the y-axis have an x-coordinate of 0. Thus, we consider points of the form 0, y that
2
are a distance of 6 units from the point 4, 3 .
289 x 2 2 x 65 0 x 2 2 x 224 0 x 14 x 16 x 14 0 or x 16 0 x 14 x 16 Thus, the points 14, 6 and 16, 6 are a
distance of 13 units from the point 1, 2 .
d
x2 x1 2 y2 y1 2
4 0 2 3 y 2
42 9 6 y y 2 16 9 6 y y 2
y 2 6 y 25
51. Points on the x-axis have a y-coordinate of 0. Thus, we consider points of the form x, 0 that are a
distance of 6 units from the point 4, 3 .
162 Copyright © 2025 Pearson Education, Inc.
Section 2.1: The Distance and Midpoint Formulas
6
y 2 6 y 25
62
y 6 y 25 2
x1 x2 2 3 x2 1 2 2 3 x2 x
2
36 y 2 6 y 25 0 y 2 6 y 11 y
Thus, P2 (1, 2) .
6 36 44 6 80 2 2 6 4 5 3 2 5 2 y 3 2 5 or y 3 2 5
x x y y2 56. M x, y 1 2 , 1 . 2 2 P2 x2 , y2 (7, 2) and ( x, y ) (5, 4) , so
x1 x2 2 x1 7 5 2 10 x1 7 x
Thus, the points 0, 3 2 5 and 0, 3 2 5
are on the y-axis and a distance of 6 units from the point 4, 3 . 53. a.
y1 y2 2 6 y2 4 2 8 6 y2 y
2 y2
1 x2
( 6) (6)2 4(1)( 11) 2(1)
and
To shift 3 units left and 4 units down, we subtract 3 from the x-coordinate and subtract 4 from the y-coordinate. 2 3,5 4 1,1
b. To shift left 2 units and up 8 units, we subtract 2 from the x-coordinate and add 8 to the y-coordinate. 2 2,5 8 0,13 54. Let the coordinates of point B be x, y . Using
the midpoint formula, we can write 1 x 8 y 2,3 2 , 2 . This leads to two equations we can solve. 1 x 8 y 2 3 2 2 1 x 4 8 y 6 x5 y 2 Point B has coordinates 5, 2 .
and
y1 y2 2 y1 (2) 4 2 8 y1 (2) y
6 y1
3 x1
Thus, P1 (3, 6) . 06 00 , 57. The midpoint of AB is: D 2 2 3, 0 04 04 The midpoint of AC is: E , 2 2 2, 2 64 04 , The midpoint of BC is: F 2 2 5, 2 d (C , D)
0 4 2 (3 4)2
( 4) 2 ( 1) 2 16 1 17 d ( B, E )
2 6 2 (2 0)2
( 4) 2 22 16 4 20 2 5 d ( A, F ) (2 0) 2 (5 0) 2
x x y y2 55. M x, y 1 2 , 1 . 2 2
22 52 4 25 29
P1 x1 , y1 (3, 6) and ( x, y ) (1, 4) , so
163 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs 58. Let P1 (0, 0), P2 (0, 4), P ( x, y ) d P1 , P2 (0 0) (4 0) 2
60. d ( P1 , P2 )
2
7 2 ( 2) 2
16 4
49 4
d P1 , P ( x 0) 2 ( y 0) 2
53
x y 4 2
2
2
d P2 , P ( x 0) ( y 4) 2
( 2)2 ( 7) 2
2
4 49
x 2 ( y 4) 2 4
53
x 2 ( y 4) 2 16 Therefore, y2 y 4
4 (1) 2 (5 4)2
d ( P1 , P3 )
52 ( 9) 2
2
25 81
y 2 y 2 8 y 16 8 y 16
106
y2 which gives x 2 22 16
Since d ( P1 , P2 ) d ( P2 , P3 ) d ( P1 , P3 ) , 2
2
2
the triangle is a right triangle. Since d P1 , P2 d P2 , P3 , the triangle is
x 2 12 x 2 3 Two triangles are possible. The third vertex is
2 3, 2 or 2 3, 2 .
isosceles. Therefore, the triangle is an isosceles right triangle. 61. d ( P1 , P2 )
59. d ( P1 , P2 ) ( 4 2) 2 (1 1) 2
0 ( 2) 2 7 (1) 2
22 82 4 64 68 2 17
( 6) 2 02
3 0 2 (2 7)2
d ( P2 , P3 )
36 6 d ( P2 , P3 )
4 6 2 (5 2)2
d ( P2 , P3 )
x y 16 2
6 (1) 2 (2 4)2
4 ( 4) (3 1) 2
0 ( 4) 2
32 ( 5) 2 9 25 2
34
2
3 (2) 2 2 (1) 2
d ( P1 , P3 )
16 4
52 32 25 9
( 6) 2 ( 4) 2
34 Since d ( P2 , P3 ) d ( P1 , P3 ) , the triangle is isosceles.
36 16
Since d ( P1 , P3 ) d ( P2 , P3 ) d ( P1 , P2 ) ,
52
the triangle is also a right triangle. Therefore, the triangle is an isosceles right triangle.
d ( P1 , P3 ) ( 4 2) 2 (3 1) 2
2
2 13
Since d ( P1 , P2 ) d ( P2 , P3 ) d ( P1 , P3 ) , 2
2
2
the triangle is a right triangle.
164 Copyright © 2025 Pearson Education, Inc.
2
2
Section 2.1: The Distance and Midpoint Formulas
65. a.
4 7 2 0 2 2
62. d ( P1 , P2 )
(11) 2 ( 2) 2
First: (90, 0), Second: (90, 90), Third: (0, 90) Y
121 4 125
(0,90)
(90,90)
5 5
4 ( 4) 2 (6 0)2
d ( P2 , P3 )
82 62 64 36
X
100 10
(0,0)
b. Using the distance formula:
4 7 2 6 2 2
d ( P1 , P3 )
(90,0)
d (310 90) 2 (15 90) 2
(3) 2 42 9 16
2202 (75)2 54025
25 5
5 2161 232.43 feet
Since d ( P1 , P3 ) d ( P2 , P3 ) d ( P1 , P2 ) , 2
2
2
c.
d (300 0) 2 (300 90)2
the triangle is a right triangle.
3002 2102 134100
63. Using the Pythagorean Theorem: 902 902 d 2 8100 8100 d
2
16200 d
2
Using the distance formula:
30 149 366.20 feet
66. a.
d 16200 90 2 127.28 feet
First: (60, 0), Second: (60, 60) Third: (0, 60) y
(0,60)
(60,60)
90
90 d
90
90
x (0,0)
64. Using the Pythagorean Theorem: 602 602 d 2
b. Using the distance formula: d (180 60) 2 (20 60) 2
3600 3600 d 2 7200 d 2
1202 ( 40) 2 16000
d 7200 60 2 84.85 feet 60
60
(60,0)
40 10 126.49 feet
c.
Using the distance formula: d (220 0) 2 (220 60)2
d
2202 1602 74000 60
60
20 185 272.03 feet
165 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
67. The Focus heading east moves a distance 60t after t hours. The Tesla heading south moves a distance 40t after t hours. Their distance apart after t hours is: d (60t ) 2 (45t ) 2 3600t 2 2025t 2 5625t 2 75t miles 60t
45t
68.
15 miles 5280 ft 1 hr 22 ft/sec 1 hr 1 mile 3600 sec 2
2018 2022 232.89 513.98 , 2 2 4040 746.87 , 2 2 2020, 373.44
71. For 2014 we have the ordered pair 2014,5645
and for 2022 we have the ordered pair 2022, 7951 . The midpoint is
4036 13596 , 2 2 2018, 6798
22t
69. a.
y y
year, $ 2014 2 2022 , 5645 2 7951
10000 484t 2 feet
100
x x
x, y 1 2 2 , 1 2 2
The estimate for 2020 is $373.44 billion. The estimate net sales of Amazon.com in 2020 is $12.62 billion off from the reported value of $386.44 billion.
d
d 1002 22t
70. Let P1 (2018, 232.89) and P2 (2022, 513.98) . The midpoint is:
d
Using the midpoint, we estimate the average credit card debt in 2018 to be $6,798. This is underestimate of the actual value.
The shortest side is between P1 (2.6, 1.5) and P2 (2.7, 1.7) . The estimate for the desired intersection point is: x1 x2 y1 y2 2.6 2.7 1.5 1.7 , 2 , 2 2 2 5.3 3.2 , 2 2 2.65, 1.6
72. Let P1 0, 0 , P2 a, 0 , and a 3a P3 , . Then 2 2
d P1 , P2
x2 x1 2 y2 y1 2
a 0 2 0 0 2
d P2 , P3
b. Using the distance formula: d (2.65 1.4) 2 (1.6 1.3) 2 (1.25) 2 (0.3) 2 1.5625 0.09 1.6525 1.285 units
166 Copyright © 2025 Pearson Education, Inc.
x2 x1 2 y2 y1 2
2 a 3a a 0 2 2
a2 a
a 2 3a 2 4 4
2
4a 2 a2 a 4
Section 2.1: The Distance and Midpoint Formulas
d P1 , P3
Since the lengths of the sides of the triangle formed by the midpoints are all equal, the triangle is equilateral. 73. Let P1 0, 0 , P2 0, s , P3 s, 0 , and
x2 x1 2 y2 y1 2
2 a 3a 0 0 2 2
2
P4 s, s be the vertices of the square.
a 2 3a 2 4a 2 a2 a 4 4 4 Since the lengths of the three sides are all equal, the triangle is an equilateral triangle. The midpoints of the saids are 0a 00 a , M P1P2 , 0 2 2 2 3 a 3a 3 a a 0 a M P2 P3 2, 2 4 , 4 2 2 3a a 0 0 2, 2 a, 3a M P1P3 2 2 4 4 Then,
d M P1 P2 , M P2 P3
2 3a a 3 a 0 4 2 4
a 3a 4 4 2
(0, )s
(,ss)
(,s 0) x
(0, 0)
The points P1 and P4 are endpoints of one diagonal and the points P2 and P3 are the endpoints of the other diagonal. 0s 0s s s M P1 P4 , , 2 2 2 2 0s s0 s s M P2 P3 , , 2 2 2 2 The midpoints of the diagonals are the same. Therefore, the diagonals of a square intersect at their midpoints.
2
74. Let P a, 2a . Then
a 5 2 2a 12 a 4 2 2a 4 2
2
a 5 2 2a 12 a 4 2 2a 4 2 5a 2 6a 26 5a 2 8a 32
a a 2 3a 2 16 16 2 d M P2 P3 , M P1 P3
y
6a 26 8a 32
2 3a 3a a 3 a 4 4 4 4
2a 6
2
a 3
Then P (3, 6) .
2
a 02 2
75. Arrange the parallelogram on the coordinate plane so that the vertices are P1 0, 0 , P2 (a, 0), P3 (a b, c) and P4 (b, c)
a a2 4 2
Then the lengths of the sides are:
a a 3a 0 d M P1 P2 , M P1 P3 4 2 4
2
a 3a 4 4 2
a a 2 3a 2 16 16 2
2
2
d ( P1 , P2 )
a 0 2 0 0 2
a2 a
d ( P2 , P3 )
(a b) a 2 c 0 2
b2 c2
167 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
d ( P3 , P4 )
b (a b)2 c c 2
78.
a2 a
and d ( P1 , P4 )
79.
b 0 2 c 0 2
b2 c2
80.
P1 and P3 are the endpoints of one diagonal, and P2 and P4 are the endpoints of the other diagonal. The lengths of the diagonals are d ( P1 , P3 )
( a b) 0 c 0 2
15 x 2 29 x 14 3 x 2 7 x 2 (3x 1)( x 2) 3x 2 11x 4 (3x 1)( x 4) x2 x4
81. 6( x 1)3 (2 x 7)7 6( x 1)3 (2 x 7)7
2
( x 1)3 (2 x 5)6 6(2 x 5) 5( x 1) ( x 1)3 (2 x 5)6 12 x 30 5 x 5
a 2 2ab b 2 c 2
( x 1)3 (2 x 5)6 7 x 35
and
7( x 1)3 (2 x 5)6 ( x 5)
d ( P2 , P4 ) (b a )2 c 0 . 2
82.
a 2 2ab b 2 c 2
Sum of the squares of the sides: a 2 ( b2 c 2 )2 a 2 ( b2 c 2 )2 2a 2 2b 2 2c 2
Sum of the squares of the diagonals:
a 2ab b c a 2ab b c 2
2
2
2
2
2
2
77. To find the domain, we know the denominator cannot be zero. 2x 5 0 2x 5 5 x 2
or x | x
83.
1 x 1 x3 x3 x( x 3) 1( x 3) 1( x 3)( x 3) x 2 3x x 3 x 2 9 3 x x 3 9 2 x 12
76. Answers will vary.
So the domain is all real numbers not equal to
3 x 2 7 x 20 0 (3x 5)( x 4) 0 (3x 5) 0 or ( x 4) 0 5 x or x 4 3 5 So the solution set is: ,4 3
2
2a 2 2b 2 2c 2
(4) 2 3(4)(7) 2 16 84 2 5(4) 2(7) 20 14 66 11 6 (5 x 2)(3x 7) 15 x 2 35 x 6 x 14
x6
5 2
5 . 2
168 Copyright © 2025 Pearson Education, Inc.
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
7 x 4 31
84.
8. True
7 x 4 31 or 7 x 4 31 7 x 27 27 x 7
9. False; the y-coordinate of a point at which the graph crosses or touches the x-axis is always 0. The x-coordinate of such a point is an x-intercept.
7 x 35 x5
The solution set is
10. False; a graph can be symmetric with respect to both coordinate axes (in such cases it will also be symmetric with respect to the origin). For example: x 2 y 2 1
27 ,5 7
85. 5( x 3) 2 x 6(2 x 3) 7 5 x 15 2 x 12 x 18 7 7 x 15 12 x 25 5 x 10 x2
11. d 12. c 13. y x 4 x 0 04 0
86. (7 3i )(1 2i) 7 14i 3i 6i 2 7 11i 6( 1) 7 11i 6 13 11i
4 (2) 4 2
1 14 1
00 1 0 4 16 2 The point (0, 0) is on the graph of the equation.
14. y x3 2 x 0 03 2 0
1 13 2 1
1 13 2 1
00 1 1 1 1 The points (0, 0) and (1, –1) are on the graph of the equation.
Section 2.2 1. 2 x 3 1 7 2 x 3 6 x 3 3 x 6 The solution set is 6 .
2. x 9 0 2
x2 9 x 9 3 The solution set is 3,3 .
3. intercepts 4. y 0
15. y 2 x 2 9 32 02 9
02 32 9
02 (3) 2 9
99 0 18 0 18 The point (0, 3) is on the graph of the equation.
16. y 3 x 1 23 1 1
13 0 1
03 1 1
82 00 11 The points (0, 1) and (–1, 0) are on the graph of the equation.
17. x 2 y 2 4 02 22 4
( 2) 2 22 4
44
84
(0, 2) and
2 2 4 2
2
2, 2 are on the graph of the
44
equation. 5. y-axis 6. 4 7.
3, 4
18. x 2 4 y 2 4 02 4 12 4 44
169 Copyright © 2025 Pearson Education, Inc.
22 4 02 4 44
22 4 12 4 2
54
Chapter 2: Graphs
The points (0, 1) and (2, 0) are on the graph of the equation. 19. y x 2 x-intercept: 0 x2 2 x
The intercepts are 4, 0 and 0,8 .
y-intercept: y 02 y2
The intercepts are 2, 0 and 0, 2 .
22. y 3x 9 x-intercept: y-intercept: 0 3x 9 y 30 9 3x 9 y 9 x3 The intercepts are 3, 0 and 0, 9 . 20. y x 6 x-intercept: 0 x6 6x
y-intercept: y 06 y 6
The intercepts are 6, 0 and 0, 6 .
23. y x 2 1 x-intercepts: 0 x2 1 x2 1
21. y 2 x 8 x-intercept: 0 2x 8 2 x 8 x 4
y-intercept: y 2 0 8
y-intercept: y 02 1 y 1
x 1 The intercepts are 1, 0 , 1, 0 , and 0, 1 .
y 8
170 Copyright © 2025 Pearson Education, Inc.
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
24. y x 2 9 x-intercepts: 0 x2 9 x2 9
The intercepts are 1, 0 , 1, 0 , and 0,1 . y-intercept: y 02 9 y 9
x 3 The intercepts are 3, 0 , 3, 0 , and 0, 9 .
y-intercept: 2 0 3 y 6
2x 6 x3
3y 6 y2
The intercepts are 3, 0 and 0, 2 .
25. y x 2 4 x-intercepts:
y-intercepts:
0 x 4
y 0 4
x2 4
y4
2
27. 2 x 3 y 6 x-intercepts: 2x 3 0 6
2
x 2 The intercepts are 2, 0 , 2, 0 , and 0, 4 .
26. y x 2 1 x-intercepts:
28. 5 x 2 y 10 x-intercepts: 5 x 2 0 10
y-intercept: 5 0 2 y 10
5 x 10 x2
2 y 10 y5
The intercepts are 2, 0 and 0,5 . y-intercept:
0 x2 1
y 0 1
x2 1
y 1
2
x 1
171 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
29. 9 x 2 4 y 36 x-intercepts:
32.
y-intercept:
9 x 2 4 0 36
9 0 4 y 36
9 x 2 36
4 y 36 y9
2
x2 4 x 2 The intercepts are 2, 0 , 2, 0 , and 0,9 .
33.
30. 4 x 2 y 4 x-intercepts:
34.
y-intercept:
4x2 0 4
4 0 y 4
4 x2 4
y4
2
x2 1 x 1 The intercepts are 1, 0 , 1, 0 , and 0, 4 .
35.
y 5
(c) = (5, 2)
(a) = (5, 2)
5
5
(b) = (5, 2) 5
31. 36.
172 Copyright © 2025 Pearson Education, Inc.
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
37.
43. a.
Intercepts: 2 , 0 , 0,1 , and 2 , 0
b. Symmetric with respect to the y-axis. 44. a.
Intercepts: 2, 0 , 0, 3 , and 2, 0
b. Symmetric with respect to the y-axis. 45. a.
Intercepts: 0, 0
b. Symmetric with respect to the x-axis.
38.
46. a.
Intercepts: 2, 0 , 0, 2 , 0, 2 , and 2, 0
b. Symmetric with respect to the x-axis, y-axis, and the origin. 47. a.
Intercepts: 2, 0 , 0, 0 , and 2, 0
b. Symmetric with respect to the origin. 48. a. 39.
Intercepts: 4, 0 , 0, 0 , and 4, 0
b. Symmetric with respect to the origin. 49. a.
x-intercepts: 2,1 , y-intercept 0
b. Not symmetric to x-axis, y-axis, or origin. 50. a.
x-intercepts: 1, 2 , y-intercept 0
b. Not symmetric to x-axis, y-axis, or origin. 51. a. Intercepts: none b. Symmetric with respect to the origin.
40.
52. a. Intercepts: none b. Symmetric with respect to the x-axis. 53.
41. a.
Intercepts: 1, 0 and 1, 0
b. Symmetric with respect to the x-axis, y-axis, and the origin. 42. a.
Intercepts: 0,1
b. Not symmetric to the x-axis, the y-axis, nor the origin
173 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
(0) 2 x 9 0 x 9 x9
54.
y2 0 9 y2 9 y 3
The intercepts are 9, 0 , 0, 3 and 0, 3 . Test x-axis symmetry: Let y y
y 2 x 9 y 2 x 9 same
Test y-axis symmetry: Let x x y 2 x 9 different
55.
Test origin symmetry: Let x x and y y .
y 2 x 9 y 2 x 9 different
Therefore, the graph will have x-axis symmetry. 59. y 3 x x-intercepts: y-intercepts: 3 0 x y 30 0 0x The only intercept is 0, 0 .
56.
Test x-axis symmetry: Let y y y 3 x different
57. y 2 x 16 x-intercepts: 02 x 16 16 x
y-intercepts: y 2 0 16 y 2 16 y 4
The intercepts are 16, 0 , 0, 4 and 0, 4 . Test x-axis symmetry: Let y y
y 2 x 16 y 2 x 16 same
Test y-axis symmetry: Let x x y 2 x 16 different Test origin symmetry: Let x x and y y .
y x 16 2
Test origin symmetry: Let x x and y y y 3 x 3 x y 3 x same
Therefore, the graph will have origin symmetry. 60. y 5 x x-intercepts: y-intercepts: 3 0 x y50 0 0x The only intercept is 0, 0 .
Test x-axis symmetry: Let y y y 5 x different
y x 16 different 2
Therefore, the graph will have x-axis symmetry. 58. y 2 x 9 x-intercepts:
Test y-axis symmetry: Let x x y 3 x 3 x different
Test y-axis symmetry: Let x x y 5 x 5 x different
y-intercepts:
174 Copyright © 2025 Pearson Education, Inc.
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
Test origin symmetry: Let x x and y y y 5 x 5 x
63. 25 x 2 4 y 2 100 x-intercepts:
y x same
25 0 4 y 2 100
25 x 2 100 x2 4 x 2
4 y 2 25 y2 5 y 5
Therefore, the is symmetric with respect to the origin. 61. x 2 y 9 0 x-intercepts: x2 9 0 x2 9
y-intercepts:
25 x 2 4 0 100 2
5
2
The intercepts are 2, 0 , 2, 0 , 0, 5 , and y-intercepts: 02 y 9 0 y9
x 3 The intercepts are 3, 0 , 3, 0 , and 0,9 .
Test x-axis symmetry: Let y y x y 9 0 different 2
Test y-axis symmetry: Let x x
x 2 y 9 0
0,5 . Test x-axis symmetry: Let y y 25 x 2 4 y 100 2
25 x 2 4 y 2 100 same
Test y-axis symmetry: Let x x 25 x 4 y 2 100 2
25 x 2 4 y 2 100 same
Test origin symmetry: Let x x and y y 25 x 4 y 100 2
x y 9 0 same 2
2
25 x 2 4 y 2 100 same
Test origin symmetry: Let x x and y y
x y 9 0 2
x 2 y 9 0 different
Therefore, the graph has y-axis symmetry. 62. x y 4 0 x-intercepts: y-intercept: 2 x 04 0 02 y 4 0 2 y 4 x 4 y 4 x 2 The intercepts are 2, 0 , 2, 0 , and 0, 4 . 2
Test x-axis symmetry: Let y y x2 y 4 0
Therefore, the graph has x-axis, y-axis, and origin symmetry. 64. 4 x 2 y 2 4 x-intercepts:
y-intercepts:
4 x 2 02 4
4 0 y2 4
4 x2 4
y2 4
x2 1
y 2
2
x 1 The intercepts are 1, 0 , 1, 0 , 0, 2 , and
0, 2 . Test x-axis symmetry: Let y y 4 x2 y 4 2
x 2 y 4 0 different
Test y-axis symmetry: Let x x
x 2 y 4 0
4 x 2 y 2 4 same
Test y-axis symmetry: Let x x 4 x y2 4 2
x 2 y 4 0 same
Test origin symmetry: Let x x and y y
x 2 y 4 0
4 x 2 y 2 4 same
Test origin symmetry: Let x x and y y 4x y 4
x y 4 0 different
2
2
Therefore, the graph has y-axis symmetry.
2
4 x 2 y 2 4 same
Therefore, the graph has x-axis, y-axis, and origin symmetry. 175 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
65. y x3 64 x-intercepts: 0 x3 64 x3 64
Test x-axis symmetry: Let y y y-intercepts: y 03 64 y 64
x4 The intercepts are 4, 0 and 0, 64 .
Test x-axis symmetry: Let y y y x3 64 different
Test y-axis symmetry: Let x x 3
y x3 64 different
Test origin symmetry: Let x x and y y y x 64 3
y x 64 different 3
Therefore, the graph has no symmetry.
y-intercepts: y 04 1 y 1
Test x-axis symmetry: Let y y y x 1 different 4
Test y-axis symmetry: Let x x y x 1 4
y x 4 1 same y x 1 4
y x 4 1 different
Therefore, the graph has y-axis symmetry.
0 x 4 x 2
2
y x 2 2 x 8 different
Test origin symmetry: Let x x and y y y x 2x 8 2
y x 2 2 x 8 different
68. y x 2 4 x-intercepts: 0 x2 4
y-intercepts: y 02 4 y4
x 2 4 no real solution The only intercept is 0, 4 . y x 2 4 different
Test y-axis symmetry: Let x x y x 4 2
y x 2 4 same
Test origin symmetry: Let x x and y y y x 4 2
y x 2 4 different
Therefore, the graph has y-axis symmetry. 69. y
Test origin symmetry: Let x x and y y
67. y x 2 2 x 8 x-intercepts: 0 x2 2x 8
y x 2x 8
Test x-axis symmetry: Let y y
x 1 x 1 The intercepts are 1, 0 , 1, 0 , and 0, 1 . 4
Test y-axis symmetry: Let x x
Therefore, the graph has no symmetry.
y x 64
66. y x 4 1 x-intercepts: 0 x4 1
y x 2 2 x 8 different
y-intercepts: y 02 2 0 8 y 8
4x
x 16 x-intercepts: y-intercepts: 4 0 0 4x y 2 0 0 2 16 0 16 x 16 4x 0 x0 The only intercept is 0, 0 . 2
Test x-axis symmetry: Let y y y
4x x 16 2
x 4 or x 2 The intercepts are 4, 0 , 2, 0 , and 0, 8 .
176 Copyright © 2025 Pearson Education, Inc.
different
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
Test y-axis symmetry: Let x x 4x y x 2 16 4x
y
x 16 2
different
Test origin symmetry: Let x x and y y y
4x
x 2 16
y y
x 2 16
x2 9 x-intercepts: y-intercepts: 3 x 03 0 y 2 0 0 2 0 9 9 x 9 x3 0 x0 The only intercept is 0, 0 .
Test x-axis symmetry: Let y y
4x x 2 16 4x
x3
71. y
y same y
Therefore, the graph has origin symmetry. x2 4 2x x-intercepts: x2 4 0 2x x2 4 0
x2 9 x3
different x2 9 Test y-axis symmetry: Let x x x
3
70. y
y
y-intercepts: 02 4 4 y 2 0 0 undefined
y
x 2 9 x3 x2 9
x 4 x 2 The intercepts are 2, 0 and 2, 0 .
y
Test x-axis symmetry: Let y y
y
x 4 different 2x 2
y
Test y-axis symmetry: Let x x
x 2 4 y 2x
different
Test origin symmetry: Let x x and y y
2
y
x3
x
3
x 2 9 x3 x2 9 x3 x2 9
same
Therefore, the graph has origin symmetry. 72. y
x4 1
x2 4 y different 2x
2 x5 x-intercepts:
Test origin symmetry: Let x x and y y
0
y-intercepts: 04 1 1 y 5 0 2 0
x4 1
2 x5 undefined x 4 1 no real solution There are no intercepts for the graph of this equation. Test x-axis symmetry: Let y y
x 2 4 y 2x x2 4 2 x x2 4 y same 2x
y
Therefore, the graph has origin symmetry.
y
x4 1 2 x5
177 Copyright © 2025 Pearson Education, Inc.
different
Chapter 2: Graphs
Test y-axis symmetry: Let x x
x 1
75. y x
4
y y
2x
5
x4 1 2 x5
different
Test origin symmetry: Let x x and y y y y y
x 4 1 5 2x x4 1 2 x5 x4 1 2 x5
76. y same
1 x
Therefore, the graph has origin symmetry. 73. y x3
77. If the point a, 4 is on the graph of y x 2 3x , then we have 4 a 2 3a
74. x y 2
0 a 2 3a 4 0 a 4 a 1 a 4 0 or a 1 0 a 4 a 1 Thus, a 4 or a 1 .
78. If the point a, 5 is on the graph of y x 2 6 x , then we have 5 a 2 6a 0 a 2 6a 5 0 a 5 a 1 a 5 0 or a 1 0 a 5 a 1 Thus, a 5 or a 1 .
178 Copyright © 2025 Pearson Education, Inc.
Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
79. For a graph with origin symmetry, if the point a, b is on the graph, then so is the point
y-intercepts:
0 y 0 0 y 2
a, b . Since the point 1, 2 is on the graph
2
y y 2 2
of an equation with origin symmetry, the point 1, 2 must also be on the graph.
2
2
y4 y2 y4 y2 0
80. For a graph with y-axis symmetry, if the point a, b is on the graph, then so is the point
y2 y2 1 0
a, b . Since 6 is an x-intercept in this case, the point 6, 0 is on the graph of the equation. Due to the y-axis symmetry, the point 6, 0 must
y 2 0 or
y2 1 0
y0
also be on the graph. Therefore, 6 is another xintercept. 81. For a graph with origin symmetry, if the point a, b is on the graph, then so is the point
2
2
y2 1 y 1
The intercepts are 0, 0 , 2, 0 , 0, 1 , and 0,1 . b. Test x-axis symmetry: Let y y
x y x x y
a, b . Since 4 is an x-intercept in this case, the point 4, 0 is on the graph of the equation. Due to the origin symmetry, the point 4, 0
2
2
2
2
x y x x y 2
2
2
2
2
2
same
Test y-axis symmetry: Let x x
must also be on the graph. Therefore, 4 is another x-intercept.
x y x x y 2
2
2
2
x y x x y
82. For a graph with x-axis symmetry, if the point a, b is on the graph, then so is the point
2
2
2
2
2
2
different
a, b . Since 2 is a y-intercept in this case, the point 0, 2 is on the graph of the equation. Due to the x-axis symmetry, the point 0, 2 must
Test origin symmetry: Let x x and y y
also be on the graph. Therefore, 2 is another yintercept.
different
83. a.
x y x x y 2
2
2
2
2
2
2
2
2
x x x 2
2
2
2
2
2
2
x 4 2 x3 0
2
2
2
2
2
Thus, the graph will have x-axis symmetry. 16 y 2 120 x 225 y-intercepts: 16 y 2 120 0 225
16 y 2 225 225 y2 16 no real solution
x 4 2 x3 x 2 x 2 x3 x 2 0 x3 0 or x0
2
2
x y x x y 84. a.
x-intercepts:
x 0 x x 0
x y x x y
x2 0 x2
179 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
x-intercepts: 2
16 0 120 x 225
Test y-axis symmetry: replace x by -x (( x) 2 y 2 ) 2 a 2 (( x) 2 y 2 )
0 120 x 225
( x 2 y 2 ) 2 a 2 ( x 2 y 2 ) equivalent
120 x 225 225 15 x 120 8
Test origin symmetry: replace x by -x and y by -y (( x) 2 ( y ) 2 ) 2 a 2 (( x) 2 ( y ) 2 )
15 The only intercept is , 0 . 8
( x 2 y 2 ) 2 a 2 ( x 2 y 2 ) equivalent
b. Test x-axis symmetry: Let y y 16 y 120 x 225 2
86. Let y = 0.
16 y 2 120 x 225 same
Test y-axis symmetry: Let x x 16 y 2 120 x 225 16 y 2 120 x 225 different
Test origin symmetry: Let x x and y y 16 y 120 x 225 2
16 y 2 120 x 225 different
Thus, the graph has x-axis symmetry.
( x 2 02 ax) 2 b 2 ( x 2 02 ) x 4 2ax3 a 2 x 2 b 2 x 2 0 x 2 ( x (a b) ( x (a b) 0 x 0 or x a b or x a b Let x = 0. (02 y 2 a 0) 2 b 2 (02 y 2 ) y 4 b2 y 2 0 y 2 ( y b)( y b) 0
x4 a2 ( x2 )
y 0, y b, y b So the intercepts are (0,0), (a-b,0), (a+b,0), (0,-b), (0, b). Test x-axis symmetry: replace y by -y
x4 a2 x2 0
x 2 ( y ) 2 ax b 2 x 2 ( y ) 2
85. Let y = 0. ( x 2 02 ) 2 a 2 ( x 2 02 )
2
x2 ( x2 a2 ) 0 x 0 2
The graph is symmetric by respect to the xaxis, the y-axis, and the origin.
or
( x 2 y 2 ax) 2 b 2 ( x 2 y 2 ) Equivalent
(x a ) 0 2
2
Test y-axis symmetry: replace x by -x
x 0 or x 2 a 2 x a, a Let x = 0. (02 y 2 ) 2 a 2 (02 y 2 )
2
( x) 2 y 2 a( x) b 2 ( x) 2 y 2 ( x 2 y 2 ax) 2 b 2 ( x 2 y 2 ) Not equivalent
Test origin symmetry: replace x by -x and y by -y
y 4 a 2 ( y 2 ) y4 a2 y2 0
2
( x) 2 ( y ) 2 a ( x) b 2 ( x) 2 ( y ) 2
y2 ( y2 a2 ) 0 y0
( x 2 y 2 ax) 2 b 2 ( x 2 y 2 ) No equivalent
(Note that the solutions to y 2 a 2 0 are not real)
The graph is symmetric with respect to the x-axis only.
So the intercepts are are (0,0), (a,0) and (-a,0). Test x-axis symmetry: Replace y by -y ( x 2 ( y )2 )2 a 2 ( x 2 ( y )2 ) ( x 2 y 2 ) 2 a 2 ( x 2 y 2 ) equivalent
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Section 2.2: Graphs of Equations in Two Variables; Intercepts; Symmetry
89. Answers will vary. One example:
87. a.
y
x
90. Answers will vary 91. Answers will vary 92. Answers will vary. Case 1: Graph has x-axis and y-axis symmetry, show origin symmetry. x, y on graph x, y on graph (from x-axis symmetry)
x, y on graph x, y on graph from y-axis symmetry Since the point x, y is also on the graph, the graph has origin symmetry. Case 2: Graph has x-axis and origin symmetry, show y-axis symmetry. x, y on graph x, y on graph
b.
Since
x 2 x for all x , the graphs of
y x 2 and y x are the same.
c.
For y
x , the domain of the variable 2
x is x 0 ; for y x , the domain of the variable x is all real numbers. Thus,
x x only for x 0. 2
d. For y x 2 , the range of the variable y is y 0 ; for y x , the range of the variable y is all real numbers. Also,
if x 0 . Otherwise,
x 2 x only
x2 x .
88. Answers will vary. A complete graph presents enough of the graph to the viewer so they can “see” the rest of the graph as an obvious continuation of what is shown.
from x-axis symmetry x, y on graph x, y on graph from origin symmetry Since the point x, y is also on the graph, the graph has y-axis symmetry. Case 3: Graph has y-axis and origin symmetry, show x-axis symmetry. x, y on graph x, y on graph
from y -axis symmetry x, y on graph x, y on graph from origin symmetry Since the point x, y is also on the graph, the graph has x-axis symmetry. 93. Answers may vary. The graph must contain the points 2,5 , 1,3 , and 0, 2 . For the
graph to be symmetric about the y-axis, the graph must also contain the points 2,5 and 1,3 (note that (0, 2) is on the y-axis).
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Chapter 2: Graphs
For the graph to also be symmetric with respect to the x-axis, the graph must also contain the points 2, 5 , 1, 3 , 0, 2 , 2, 5 , and
99.
1 3(4 x) 5 19 1 12 3 x 5 19 1 7 3 x 19 6 3 x 12 2 x 4 4 x 2 The solution set in interval notation is: 4, 2
100.
x2 8x 4 0
1, 3 . Recall that a graph with two of the symmetries (x-axis, y-axis, origin) will necessarily have the third. Therefore, if the original graph with y-axis symmetry also has xaxis symmetry, then it will also have origin symmetry. 94.
6 ( 2) 4 1 6 ( 2) 8 2
x 2 8 x 4 x 2 8 x 16 4 16
95. 3x 2 30 x 75
x 42 12
3( x 2 10 x 25) 3( x 5)( x 5) 3( x 5) 2
96.
x 4 12
2 x y 3x y (2 x ) ( y ) (3x ) ( y) 2 3 3
3
2
2 3
3 3
3 2
(8 x 6 )( y 9 )(9 x 6 )( y 2 )
x 4 12 2
42 3
The solution set is 4 2 3, 4 2 3 .
72 x y
12 11
97. 3 2 7
0
7
6
6 3 9
6
2 1 3 2
101. 3x 2 x 1 0 a 3, b 1, c 1 x
0
The quotient is 2 x x 2 3 x 2 and the remainder is 0.
1 12 4(3)(1) 1 1 12 1 13 2(3) 6 6
3
98.
2x 5 1 x 2x 5 x 1 2 x 5 x 1
2
2 x 5 x2 2 x 1 x2 4 x 2 Check each solution: 2(2) 5 1
4 5 1 9 1 3 1 2 2(2) 5 1 4 5 1
1 1 1 1 2 The solution set is {2}.
1 13 1 13 The solution set is , . 6 6 102. Let r be the radius of the pipe with the deposits and r + 2 be the original radius. Then the area of the cross section of the pipe with the deposits is A r 2 and the area of the cross section of the
original pipe is A (r 2) 2 . Since the area of the pipe with deposits is 30% less then r 2 .7 (r 2) 2 . Solving for r we have
r 2 .7 (r 2) 2 r 2 .7(r 2 4r 4) r 2 .7 r 2 2.8r 2.8
0 0.3r 2 2.8r 2.8 r
2.8 2.82 4(0.3)(2.8) 2(0.3)
2.8 11.2 10.2444 0.6
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Section 2.3: Lines
13. a.
The radius without the deposit is 10.2444 2 12.2444 and the diameter is 2(12.2444) 24.49 mm. 103.
196 ( 1)(196) 14i
Slope
1 0 1 20 2
b. If x increases by 2 units, y will increase by 1 unit. 14. a.
Slope
1 0 1 20 2
b. If x increases by 2 units, y will decrease by 1 unit.
Section 2.3
15. a.
1. undefined; 0
Slope
1 2 1 1 ( 2) 3
b. If x increases by 3 units, y will decrease by 1 unit.
2. 3; 2 x-intercept: 2 x 3(0) 6 2x 6 x3 y-intercept: 2(0) 3 y 6 3y 6 y2
16. a.
Slope
2 1 1 2 (1) 3
b. If x increases by 3 units, y will increase by 1 unit. 17.
3. True
Slope
y2 y1 0 3 3 2 x2 x1 4 2
4. False; the slope is 3 . 2 2 y 3x 5 3 5 y x 2 2 ?
5. True; 2 1 2 4 ?
2 24 4 4 True
6. m1 m2 ; y-intercepts; m1 m2 1
18. Slope
y2 y1 4 2 2 2 x2 x1 3 4 1
7. 2 8.
1 2
9. c 10. d 11. b 12. d 183 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
19. Slope
y2 y1 1 3 1 2 4 2 x2 x1 2 ( 2)
23. Slope
y2 y1 22 4 undefined. 0 x2 x1 1 (1)
20. Slope
y2 y1 3 1 2 x2 x1 2 ( 1) 3
24. Slope
y2 y1 2 0 2 undefined. x2 x1 2 2 0
21. Slope
y2 y1 1 (1) 0 0 2 (3) 5 x2 x1
y y 22 0 0 22. Slope 2 1 x2 x1 5 4 9
25. P 1, 2 ; m 3
26. P 2,1 ; m 4
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Section 2.3: Lines
27. P 2, 4 ; m
3 4
31. P 0,3 ; slope undefined
(note: the line is the y-axis) 28. P 1,3 ; m
2 5
29. P 1,3 ; m 0
32. P 2, 0 ; slope undefined
33. P 1, 2 ; m 3 ; y 2 3( x 1) 34. P 2,1 ; m 4 ; y 1 4( x 2) 3 3 35. P 2, 4 ; m ; y 4 ( x 2) 4 4
36. P 1,3 ; m
2 2 ; y 3 ( x 1) 5 5
37. P 1, 3 ; m 0 ; y 3 0 30. P 2, 4 ; m 0
38. P 2, 4 ; m 0 ; y 4 0 4 ; point: 1, 2 1 If x increases by 1 unit, then y increases by 4 units.
39. Slope 4
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Chapter 2: Graphs
Answers will vary. Three possible points are: x 1 1 2 and y 2 4 6
2, 6 x 2 1 3 and y 6 4 10
3,10 x 3 1 4 and y 10 4 14
4,14
1, 5 x 1 1 0 and y 5 2 7
2 ; point: 2,3 1 If x increases by 1 unit, then y increases by 2 units. Answers will vary. Three possible points are: x 2 1 1 and y 3 2 5
40. Slope 2
1,5 x 1 1 0 and y 5 2 7
0, 7 x 0 1 1 and y 7 2 9
0, 7 x 0 1 1 and y 7 2 9
1, 9 1 ; point: 4,1 1 If x increases by 1 unit, then y decreases by 1 unit. Answers will vary. Three possible points are: x 4 1 5 and y 1 1 0
44. Slope 1
5, 0
1,9
x 5 1 6 and y 0 1 1
3 3 ; point: 2, 4 2 2 If x increases by 2 units, then y decreases by 3 units. Answers will vary. Three possible points are: x 2 2 4 and y 4 3 7
41. Slope
4, 7 x 4 2 6 and y 7 3 10
6, 10 x 6 2 8 and y 10 3 13
8, 13 4 ; point: 3, 2 3 If x increases by 3 units, then y increases by 4 units. Answers will vary. Three possible points are: x 3 3 0 and y 2 4 6
42. Slope
0, 6 x 0 3 3 and y 6 4 10
3,10 x 3 3 6 and y 10 4 14
6,14
2 ; point: 2, 3 1 If x increases by 1 unit, then y decreases by 2 units. Answers will vary. Three possible points are: x 2 1 1 and y 3 2 5
43. Slope 2
6, 1 x 6 1 7 and y 1 1 2
7, 2 45. (0, 0) and (2, 1) are points on the line. 1 0 1 Slope 20 2 y -intercept is 0; using y mx b : 1 y x0 2 2y x 0 x 2y 1 x 2 y 0 or y x 2 46. (0, 0) and (–2, 1) are points on the line. 1 0 1 1 Slope 20 2 2 y -intercept is 0; using y mx b : 1 y x0 2 2 y x x 2y 0 1 x 2 y 0 or y x 2
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Section 2.3: Lines
47. (–1, 3) and (1, 1) are points on the line. 1 3 2 Slope 1 1 (1) 2 Using y y1 m( x x1 ) y 1 1( x 1) y 1 x 1 y x 2 x y 2 or y x 2 48. (–1, 1) and (2, 2) are points on the line. 2 1 1 Slope 2 (1) 3 Using y y1 m( x x1 ) 1 x (1) 3 1 y 1 ( x 1) 3 1 1 y 1 x 3 3 1 4 y x 3 3 y 1
1 4 x 3 y 4 or y x 3 3
49. y y1 m( x x1 ), m 2 y 3 2( x 3) y 3 2x 6 y 2x 3 2 x y 3 or y 2 x 3 50. y y1 m( x x1 ), m 1 y 2 1( x 1) y 2 x 1 y x 3 x y 3 or y x 3 1 51. y y1 m( x x1 ), m 2 1 y 2 ( x 1) 2 1 1 y2 x 2 2 1 5 y x 2 2 1 5 x 2 y 5 or y x 2 2
52. y y1 m( x x1 ), m 1 y 1 1( x (1)) y 1 x 1 y x2 x y 2 or y x 2 53. Slope = 3; containing (–2, 3) y y1 m( x x1 ) y 3 3( x ( 2)) y 3 3x 6 y 3x 9 3x y 9 or y 3 x 9
54. Slope = 2; containing the point (4, –3) y y1 m( x x1 ) y (3) 2( x 4) y 3 2x 8 y 2 x 11 2 x y 11 or y 2 x 11 1 55. Slope = ; containing the point (3, 1) 2 y y1 m( x x1 ) 1 ( x 3) 2 1 3 y 1 x 2 2 1 1 y x 2 2 y 1
x 2 y 1 or y
1 1 x 2 2
2 56. Slope = ; containing (1, –1) 3 y y1 m( x x1 ) 2 ( x 1) 3 2 2 y 1 x 3 3 2 1 y x 3 3
y (1)
2 x 3 y 1 or y
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2 1 x 3 3
Chapter 2: Graphs
57. Containing (1, 3) and (–1, 2) 2 3 1 1 m 1 1 2 2
62. x-intercept = 2; y-intercept = –1 Points are (2,0) and (0,–1) 1 0 1 1 m 02 2 2 y mx b 1 y x 1 2 1 x 2 y 2 or y x 1 2
y y1 m( x x1 ) 1 ( x 1) 2 1 1 y 3 x 2 2 1 5 y x 2 2 y 3
63. Slope undefined; containing the point (2, 4) This is a vertical line. x2 No slope-intercept form.
1 5 x 2 y 5 or y x 2 2
58. Containing the points (–3, 4) and (2, 5) 54 1 m 2 (3) 5 y y1 m( x x1 )
65. Horizontal lines have slope m 0 and take the form y b . Therefore, the horizontal line
1 y 5 ( x 2) 5 1 2 y 5 x 5 5 1 23 y x 5 5 x 5 y 23 or y
64. Slope undefined; containing the point (3, 8) This is a vertical line. x3 No slope-intercept form.
passing through the point 3, 2 is y 2 . 66. Vertical lines have an undefined slope and take the form x a . Therefore, the vertical line passing through the point 4, 5 is x 4 . 1 23 x 5 5
59. Slope = –3; y-intercept =3 y mx b y 3 x 3 3x y 3 or y 3x 3 60. Slope = –2; y-intercept = –2 y mx b y 2 x ( 2) 2 x y 2 or y 2 x 2 61. x-intercept = –4; y-intercept = 4 Points are (–4, 0) and (0, 4) 40 4 m 1 0 ( 4) 4 y mx b y 1x 4 y x4 x y 4 or y x 4
67. Parallel to y 2 x ; Slope = 2 Containing (–1, 2) y y1 m( x x1 ) y 2 2( x (1)) y 2 2x 2 y 2x 4 2 x y 4 or y 2 x 4
68. Parallel to y 3 x ; Slope = –3; Containing the point (–1, 2) y y1 m( x x1 ) y 2 3( x ( 1)) y 2 3 x 3 y 3x 1 3x y 1 or y 3 x 1
69. Parallel to x 2 y 5 ; 1 Slope ; Containing the point 0, 0 2 y y1 m( x x1 ) 1 1 ( x 0) y x 2 2 1 x 2 y 0 or y x 2 y0
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Section 2.3: Lines 70. Parallel to 2 x y 2 ; Slope = 2 Containing the point (0, 0) y y1 m( x x1 ) y 0 2( x 0) y 2x 2 x y 0 or y 2 x
76. Perpendicular to 2 x y 2 ; Containing the point (–3, 0) 1 Slope of perpendicular 2 y y1 m( x x1 ) 1 1 3 ( x (3)) y x 2 2 2 1 3 x 2 y 3 or y x 2 2 y0
71. Parallel to x 5 ; Containing (4,2) This is a vertical line. x 4 No slope-intercept form. 72. Parallel to y 5 ; Containing the point (4, 2) This is a horizontal line. Slope = 0 y2 1 x 4; Containing (1, –2) 2 Slope of perpendicular = –2 y y1 m( x x1 )
73. Perpendicular to y
y ( 2) 2( x 1) y 2 2x 2 y 2x 2 x y 0 or y 2 x
77. Perpendicular to x 8 ; Containing (3, 4) Slope of perpendicular = 0 (horizontal line) y4 78. Perpendicular to y 8 ; Containing the point (3, 4) Slope of perpendicular is undefined (vertical line). x 3 No slope-intercept form. 79. y 2 x 3 ; Slope = 2; y-intercept = 3
74. Perpendicular to y 2 x 3 ; Containing the point (1, –2) 1 Slope of perpendicular 2 y y1 m( x x1 ) 1 y ( 2) ( x 1) 2 1 1 1 3 y2 x y x 2 2 2 2 1 3 x 2 y 3 or y x 2 2
80. y 3 x 4 ; Slope = –3; y-intercept = 4
75. Perpendicular to x 2 y 5 ; Containing the point (0, 4) Slope of perpendicular = –2 y mx b y 2 x 4 2 x y 4 or y 2 x 4
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Chapter 2: Graphs
81.
82.
1 y x 1 ; y 2x 2 2 Slope = 2; y-intercept = –2
1 85. x 2 y 4 ; 2 y x 4 y x 2 2 1 Slope ; y-intercept = 2 2
1 1 x y 2; y x2 3 3 1 Slope ; y-intercept = 2 3
83. y
86. x 3 y 6 ; 3 y x 6 y Slope
1 x2 3
1 ; y-intercept = 2 3
1 1 x 2 ; Slope ; y-intercept = 2 2 2
87. 2 x 3 y 6 ; 3 y 2 x 6 y
Slope
84. y 2 x
2 ; y-intercept = –2 3
1 1 ; Slope = 2; y -intercept 2 2
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2 x2 3
Section 2.3: Lines
3 88. 3 x 2 y 6 ; 2 y 3 x 6 y x 3 2 3 Slope ; y-intercept = 3 2
92. y 1 ; Slope = 0; y-intercept = –1
93. y 5 ; Slope = 0; y-intercept = 5
89. x y 1 ; y x 1 Slope = –1; y-intercept = 1
94. x 2 ; Slope is undefined y-intercept - none
90. x y 2 ; y x 2 Slope = 1; y-intercept = –2
95. y x 0 ; y x Slope = 1; y-intercept = 0
91. x 4 ; Slope is undefined y-intercept - none
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Chapter 2: Graphs
y-intercept: 2 0 3 y 6
96. x y 0 ; y x Slope = –1; y-intercept = 0
3y 6 y2
The point 0, 2 is on the graph. y
b.
5
5
97. 2 y 3 x 0 ; 2 y 3x y
3 x 2
5
x
5
3 Slope ; y-intercept = 0 2
100. a.
x-intercept: 3x 2 0 6 3x 6 x2 The point 2, 0 is on the graph.
y-intercept: 3 0 2 y 6 2 y 6 y 3
The point 0, 3 is on the graph. 3 98. 3x 2 y 0 ; 2 y 3 x y x 2 3 Slope ; y-intercept = 0 2
y
b.
5
5
5
x
5
101. a.
99. a.
x-intercept: 4 x 5 0 40 4 x 40 x 10 The point 10, 0 is on the graph.
x-intercept: 2 x 3 0 6 2x 6 x3 The point 3, 0 is on the graph.
y-intercept: 4 0 5 y 40 5 y 40 y 8
The point 0,8 is on the graph.
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Section 2.3: Lines
b.
102. a.
b.
x-intercept: 6 x 4 0 24
104. a.
x-intercept: 5 x 3 0 18 5 x 18 18 x 5
6 x 24 x4 The point 4, 0 is on the graph.
18 The point , 0 is on the graph. 5
y-intercept: 6 0 4 y 24 4 y 24 y 6
y-intercept: 5 0 3 y 18 3 y 18 y6
The point 0, 6 is on the graph. b.
The point 0, 6 is on the graph. b.
103. a.
x-intercept: 7 x 2 0 21 7 x 21 x3 The point 3, 0 is on the graph.
y-intercept: 7 0 2 y 21 2 y 21 21 y 2
105. a.
1 1 x 0 1 2 3 1 x 1 2 x2 The point 2, 0 is on the graph.
x-intercept:
y-intercept:
21 The point 0, is on the graph. 2
1 1 0 y 1 2 3 1 y 1 3 y3
The point 0, 3 is on the graph.
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Chapter 2: Graphs
b.
106. a.
b.
2 0 4 3 x4 The point 4, 0 is on the graph.
x-intercept: x
108. a.
x-intercept: 0.3 x 0.4 0 1.2 0.3x 1.2 x 4 The point 4, 0 is on the graph.
y-intercept: 0.3 0 0.4 y 1.2
2 y4 3 2 y4 3 y 6
y-intercept: 0
0.4 y 1.2 y3
The point 0,3 is on the graph.
The point 0, 6 is on the graph.
b.
b.
107. a.
x-intercept: 0.2 x 0.5 0 1 0.2 x 1 x5 The point 5, 0 is on the graph.
y-intercept: 0.2 0 0.5 y 1 0.5 y 1 y 2
The point 0, 2 is on the graph.
109. The equation of the x-axis is y 0 . (The slope is 0 and the y-intercept is 0.) 110. The equation of the y-axis is x 0 . (The slope is undefined.) 111. The slopes are the same but the y-intercepts are different. Therefore, the two lines are parallel. 112. The slopes are opposite-reciprocals. That is, their product is 1 . Therefore, the lines are perpendicular. 113. The slopes are different and their product does not equal 1 . Therefore, the lines are neither parallel nor perpendicular. 114. The slopes are different and their product does not equal 1 (in fact, the signs are the same so the product is positive). Therefore, the lines are neither parallel nor perpendicular.
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Section 2.3: Lines 115. Intercepts: 0, 2 and 2, 0 . Thus, slope = 1. y x 2 or x y 2 116. Intercepts: 0,1 and 1, 0 . Thus, slope = –1. y x 1 or x y 1 1 117. Intercepts: 3, 0 and 0,1 . Thus, slope = . 3 1 y x 1 or x 3 y 3 3
118. Intercepts: 0, 1 and 2, 0 . Thus,
53 2 2 2 1 3 3 30 3 P2 1,3 , P3 1, 0 : m2 1 1 2
119. P1 2,5 , P2 1,3 : m1
Since m1 m2 1 , the line segments P1 P2 and P2 P3 are perpendicular. Thus, the points P1 , P2 , and P3 are vertices of a right triangle.
120. P1 1, 1 , P2 4,1 , P3 2, 2 , P4 5, 4 1 1
4 1 2 ; m24 3; 54 4 1 3 2 1 42 2 m34 ; m13 3 52 3 2 1 Each pair of opposite sides are parallel (same slope) and adjacent sides are not perpendicular. Therefore, the vertices are for a parallelogram.
121. P1 1, 0 , P2 2,3 , P3 1, 2 , P4 4,1 m12 m34
30 3 1 3 1 ; 1 ; m24 2 1 3 42 1 2 4 1
30 23 1 3 ; m23 ; 1 0 4 1 3 1 2 1 0 1 m34 3 ; m14 3 4 30 3 m12
d12
1 0 2 3 0 2 1 9 10
d 23
4 12 2 32 9 1 10
d34
3 4 2 1 2 2 1 9 10
d14
3 0 2 1 0 2 9 1 10
Opposite sides are parallel (same slope) and adjacent sides are perpendicular (product of slopes is 1 ). In addition, the length of all four sides is the same. Therefore, the vertices are for a square.
1 slope = . 2 1 y x 1 or x 2 y 2 2
m12
122. P1 0, 0 , P2 1,3 , P3 4, 2 , P4 3, 1
2 0 3 1 1 ; m13 1 1 3
Opposite sides are parallel (same slope) and adjacent sides are perpendicular (product of slopes is 1 ). Therefore, the vertices are for a rectangle.
123. Let x = number of miles driven, and let C = cost in dollars. Total cost = (cost per mile)(number of miles) + fixed cost C 0.60 x 39 When x = 110, C 0.60110 39 $105.00 .
When x = 230, C 0.60 230 39 $177.00 . 124. Let x = number of pairs of jeans manufactured, and let C = cost in dollars. Total cost = (cost per pair)(number of pairs) + fixed cost C 20 x 1200 When x = 400, C 20 400 1200 $9200 .
When x = 740, C 20 740 1200 $16, 000 . 125. Let x = number of miles driven annually, and let C = cost in dollars. Total cost = (approx cost per mile)(number of miles) + fixed cost C 0.28 x 6578 126. Let x = profit in dollars, and let S = salary in dollars. Weekly salary = (% share of profit)(profit) + weekly pay S 0.05 x 525
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Chapter 2: Graphs
127. a.
C 0.13 x 17 ; 0 x 1000
c.
slope
For 50 miles, C 0.13 50 17 $23.50
d.
For 120 miles, C 0.13 120 17 $32.60
e. 128. a.
129. (C , F ) (0, 32); (C , F ) (100, 212) 212 32 180 9 100 0 100 5 9 F 32 (C 0) 5 9 F 32 (C ) 5 5 C ( F 32) 9 If F 70 , then 5 5 C (70 32) (38) 9 9 C 21.1
b.
For every 1-mile in distance traveled, the cost will increase 13 cents.
130. a. b.
C 60 0.25 x
b.
131. a.
c.
d.
5 º C (º F 32) 9 5 K ( F 32) 273 9 5 160 273 K ºF 9 9 5 2297 K ºF 9 9
The y-intercept is (0, 30), so b = 30. Since the ramp drops 2 inches for every 25 inches 2 2 . Thus, of run, the slope is m 25 25 2 the equation is y x 30 . 25
b. Let y = 0.
For 20 minutes, C 60 0.25 20 $65.00
0
For 60 minutes,
2 x 30 25
2 x 30 25 25 2 25 x 30 2 25 2 x 375 The x-intercept is (375, 0). This means that the ramp meets the floor 375 inches (or 31.25 feet) from the base of the platform.
C 60 0.25 60 $75.00
e.
K º C 273
For every 1-minute increase in international calls the cost will increase by $0.25 (that is, 25 cents).
c.
No. From part (b), the run is 31.25 feet which exceeds the required maximum of 30 feet.
d. First, design requirements state that the maximum slope is a drop of 1 inch for each 1 12 inches of run. This means m . 12 196 Copyright © 2025 Pearson Education, Inc.
Section 2.3: Lines
Second, the run is restricted to be no more than 30 feet = 360 inches. For a rise of 30 inches, this means the minimum slope is 30 1 1 . That is, m . Thus, the 12 360 12 1 only possible slope is m . The 12 diagram indicates that the slope is negative. Therefore, the only slope that can be used to obtain the 30-inch rise and still meet design 1 requirements is m . In words, for 12 every 12 inches of run, the ramp must drop exactly 1 inch. 132. a.
Let x represent the percent of internet ad spending. Let y represent the percent of print ad spending. Then the points (0.19, 0.26) and (0.35, 0.16) are on the line. 16 26 10 0.625 . Using Thus, m 35 19 16 the point-slope formula we have y 26 0.625( x 19) y 26 0.625 x 11.875 y 0.625 x 37.875 0 0.625 x 37.875 37.875 0.625 x 60.6 x y-intercept: y 0.625(0) 37.875 37.875 The intercepts are (60.6, 0) and (0, 37.875).
b. x-intercept:
c.
60, 000 40, 000 200, 000 100, 000 20, 000 1 100, 000 5 1 A 40, 000 x 100, 000 5 1 A 40, 000 x 20, 000 5 1 A x 20, 000 5 slope
b. If x = 300,000, then 1 A 300, 000 20, 000 $80, 000 5 c.
Each additional box sold requires an additional $0.20 in advertising.
134. 2 x y C Graph the lines: 2x y 4 2x y 0 2x y 2 All the lines have the same slope, 2. The lines are parallel.
y-intercept: When Internet ads account for 0% of U.S. advertisement spending, print ads account for 37.875% of the spending. x-intercept: When Internet ads account for 60.6% of U.S. advertisement spending, print ads account for 0% of the spending.
d. Let x = 39.2. y 0.625(39.2) 37.875 13.4% 133. a.
( x2 , A2 ) (200, 000, 60, 000)
Let x = number of boxes to be sold, and A = money, in dollars, spent on advertising. We have the points ( x1 , A1 ) (100, 000, 40, 000);
135. Put each linear equation in slope/intercept form. x 2y 5 2 x 3 y 4 0 ax y 0 y ax 2 y x 5 3 y 2 x 4 1 5 2 4 y x y x 2 2 3 3
If the slope of y ax equals the slope of either of the other two lines, then no triangle is formed. 1 1 2 2 So, a a and a a . 3 3 2 2 Also if all three lines intersect at a single point,
197 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
then no triangle is formed. So, we find where 1 5 2 4 y x and y x intersect. 2 2 3 3 1 5 2 4 x x 2 2 3 3 7 7 x 6 6 x 1 1 5 (1) 2 2 2 The two lines intersect at (1, 2). If y ax also contains the point (1, 2), then 2 a 1 a 2 .
The three numbers are
1 2 , , and -2. 2 3
136. The slope of the line containing a, b and
b, a is ab 1 ba The slope of the line y x is 1. The two lines are perpendicular. The midpoint of (a, b) and (b, a) is ab ba M , . 2 2
Since the x and y coordinates of M are equal, M lies on the line y x . Note:
ab ba 2 2
137. The three midpoints are 0a 00 a ab 0c ab c , , , ,0, 2 2 2 2 2 2 2 0b 0c b c , and , . 2 2 2 2
ab c Line 1 from (0,0) to , 2 2 c 0 c 2 m ; ab 0 ab 2 c y0 ( x 0) ab c y x1 ab b c Line 2 from (a, 0) to , 2 2 c c 0 c 2 2 m b b 2a b 2a a 2 2 c y 0 ( x a) b 2a c y ( x a) b 2a a Line 3 from , 0 to (b, c) 2 c0 2c m a 2b a b 2 a 2c y0 x 2b a 2 a 2c y x 2b a 2 Find where line 1 and line 2 intersect: c c x ( x a) ab b 2a b 2a x xa ab b 2a a b x a ab 3a x a ab ab x ; 3 Substitute into line 1: c ab c y . ab 3 3
198 Copyright © 2025 Pearson Education, Inc.
Section 2.3: Lines
ab c So, line 1 and line 2 intersect at , . 3 3 ab c Show that line 3 contains the point , : 3 3
142. (d) The equation y 2 x 2 has slope 2 and yintercept (0, 2). The equation x 2 y 1 has 1 1 and y-intercept 0, . The lines 2 2 1 are perpendicular since 2 1 . One line 2 has a positive y-intercept and the other with a negative y-intercept.
slope
2c a b a 2c 2b a c So 2b a 3 2 2b a 6 3 ab c , . the three lines intersect at 3 3 y
143 – 145. Answers will vary.
138. Refer to Figure 47. Assume m1m2 1 . Then
146. No, the equation of a vertical line cannot be written in slope-intercept form because the slope is undefined.
d ( A, B)2 (1 1)2 (m1 m2 )2 (m1 m2 ) 2 m12 2m1m2 m2 2 m12 2(1) m2 2 m12 m2 2 2
147. No, a line does not need to have both an xintercept and a y-intercept. Vertical and horizontal lines have only one intercept (unless they are a coordinate axis). Every line must have at least one intercept.
d (O, B)2 (1 0)2 (m1 0)2 1 m12 , d (O, A)2 (1 0)2 (m2 0)2 1 m22
148. Two lines with equal slopes and equal y-intercepts are coinciding lines (i.e. the same).
Now,
So
d (O, B)2 d (O, A)2 1 m12 1 m22 m12 m2 2 2 d ( A, B )
2
By the converse of the Pythagorean Theorem, AOB is a right triangle with right angle at vertex O. Thus lines OA and OB are perpendicular. 139. (b), (c), (e) and (g) The line has positive slope and positive y-intercept. 140. (a), (c), and (g) The line has negative slope and positive y-intercept. 141. (c) The equation x y 2 has slope 1 and yintercept (0, 2). The equation x y 1 has slope 1 and y-intercept (0, 1). Thus, the lines are parallel with positive slopes. One line has a positive y-intercept and the other with a negative y-intercept.
149. Two lines that have the same x-intercept and yintercept (assuming the x-intercept is not 0) are the same line since a line is uniquely defined by two distinct points. 150. No. Two lines with the same slope and different xintercepts are distinct parallel lines and have no points in common. Assume Line 1 has equation y mx b1 and Line 2 has equation y mx b2 , b Line 1 has x-intercept 1 and y-intercept b1 . m b2 and y-intercept b2 . Line 2 has x-intercept m Assume also that Line 1 and Line 2 have unequal x-intercepts. If the lines have the same y-intercept, then b1 b2 .
b b b b b1 b2 1 2 1 2 m m m m b b But 1 2 Line 1 and Line 2 have the m m same x-intercept, which contradicts the original assumption that the lines have unequal x-intercepts. Therefore, Line 1 and Line 2 cannot have the same y-intercept.
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Chapter 2: Graphs
151. Yes. Two distinct lines with the same y-intercept, but different slopes, can have the same x-intercept if the x-intercept is x 0 . Assume Line 1 has equation y m1 x b and Line 2 has equation y m2 x b ,
Line 1 has x-intercept
b and y-intercept b . m1
155. h 2 a 2 b 2 82 152 16 225 289
h 289 17
156.
b Line 2 has x-intercept and y-intercept b . m2 Assume also that Line 1 and Line 2 have unequal slopes, that is m1 m2 . If the lines have the same x-intercept, then b b . m1 m2 b b m1 m2 m2 b m1b m2b m1b 0
Since we are assuming that m1 m2 , the only way that the two lines can have the same x-intercept is if b 0. 152. Answers will vary. y2 y1 4 2 6 3 x2 x1 1 3 4 2
It appears that the student incorrectly found the slope by switching the direction of one of the subtractions. 2
x 3 2 6
The solution set is: 3 2 6,3 2 6 . 2 x 5 7 10 3 2 x 5 3
b0 or m1 m2 0 m1 m2
x 2 y 3 154. 4 5 x y
x 3 2 6
2x 5 3
But m2 b m1b 0 b m1 m2 0
153. m
x 3 24
157.
x 3 2 25 49 x 3 2 24
x2 4 5 3 x y y 1 2 8 x y
2
2 2x 8 1 x 4 The solution set is: x |1 x 4 .
Interval notation: 1, 4
158. The radicand must be greater than or equal to 0 so: 2 8 x 0 3 2 x 8 3 3 x 8 2 x 12 The solution set is: x | x 12 .
Interval notation: ,12
2
2
x 2 y8 4 16 x y 1
200 Copyright © 2025 Pearson Education, Inc.
Section 2.4: Circles
2
7. d
2
9 1 3 3 159. 64 2 4 8
160.
161.
8. a
x2 ( x 4)( x 4) x 2 x 2 6 x 8 16 4 x ( x 2)( x 4) 4(4 x) ( x 4) 1 4(4 x ) 4
9. Center = (2, 1) Radius distance from (0,1) to (2,1)
x 1 x
10. Center = (1, 2) Radius distance from (1,0) to (1,2)
x 2 16
x 1 x
x 1 x x 1 x
x 1 2 x 1 x x x 1 x 2
2
2
2
x 1 2 x 1 x x x 1 2 x 1 x x x 1 x 2 x 1 2 x( x 1) x 5 x 6 0
162.
Equation: ( x 2) 2 ( y 1) 2 4
(1 1) 2 (2 0) 2 4 2
Equation: ( x 1) 2 ( y 2) 2 4 11. Center = midpoint of (1, 2) and (4, 2) 1 4 2 2 , 5, 2 2 2 2 Radius distance from 5 , 2 to (4,2) 2
2
5 9 3 4 (2 2)2 2 4 2
( x 3)( x 2) 0
x 3 0 or
(2 0) 2 (1 1) 2 4 2
x 2 0
2
5 9 Equation: x ( y 2) 2 2 4
x 3 or x 2 x 9 or x 4
12. Center = midpoint of (0, 1) and (2, 3) 0 2 1 3 , 1, 2 2 2 Radius distance from 1, 2 to (2,3)
The solution set is {9, 4}.
Section 2.4
2 12 (3 2)2
Equation: x 1 ( y 2) 2 2 2
1. add; 2.
2 1 10 25 2
13. ( x h) 2 ( y k ) 2 r 2
x 2 9 2
( x 0) 2 ( y 0) 2 22
x2 9 x 2 3 x 23 x 5 or x 1 The solution set is {1, 5}.
x2 y 2 4
General form: x 2 y 2 4 0
3. False. For example, x 2 y 2 2 x 2 y 8 0 is not a circle. It has no real solutions. 4. radius 5. True; r 2 9 r 3 6. False; the center of the circle
x 32 y 2 2 13 is 3, 2 . 201 Copyright © 2025 Pearson Education, Inc.
2
Chapter 2: Graphs
14. ( x h) 2 ( y k ) 2 r 2
17. ( x h) 2 ( y k ) 2 r 2
( x 0) 2 ( y 0) 2 32
( x 4) 2 ( y (3)) 2 52
x2 y 2 9
( x 4) 2 ( y 3) 2 25 General form: x 2 8 x 16 y 2 6 y 9 25
General form: x 2 y 2 9 0
x2 y 2 8x 6 y 0
15. ( x h) 2 ( y k ) 2 r 2 ( x 0) 2 ( y 2) 2 22
x 2 ( y 2) 2 4
General form: x y 4 y 4 4 2
2
x y 4y 0 2
2
18. ( x h) 2 ( y k ) 2 r 2 ( x 2) 2 ( y (3)) 2 42 ( x 2) 2 ( y 3) 2 16
General form: x 2 4 x 4 y 2 6 y 9 16 x2 y2 4 x 6 y 3 0
16. ( x h) 2 ( y k ) 2 r 2 ( x 1) 2 ( y 0)2 32 ( x 1) 2 y 2 9
General form: x 2 2 x 1 y 2 9 x2 y 2 2 x 8 0
19. ( x h) 2 ( y k ) 2 r 2 ( x 2 ) 2 ( y 1) 2 42 ( x 2) 2 ( y 1) 2 16
202 Copyright © 2025 Pearson Education, Inc.
Section 2.4: Circles
General form: x 2 4 x 4 y 2 2 y 1 16 x 2 y 2 4 x 2 y 11 0
22. ( x h) 2 ( y k ) 2 r 2
20. ( x h) 2 ( y k ) 2 r 2
( x 5 ) ( y (2)) 7 2
1 2
2
x 0 2 y 2
1 1 x y 2 4 2
General form: x 2 10 x 25 y 2 4 y 4 49
1 1 4 4 x2 y2 y 0
General form: x 2 y 2 y
x 2 y 2 10 x 4 y 20 0
21. ( x h) 2 ( y k ) 2 r 2 2
2
2
2
( x 5) 2 ( y 2) 2 49
1 1 2 x ( y 0) 2 2
1 2
2
2
1 1 2 x y 2 4 1 1 y2 4 4 x2 y 2 x 0
General form: x 2 x
203 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
23. ( x h) 2 ( y k ) 2 r 2 ( x 5) 2 ( y (1)) 2
b.
13
2
( x 5) 2 ( y 1) 2 13 General form: x 2 10 x 25 y 2 2 y 1 13
x 2 y 2 10 x 2 y 13 0
c.
x-intercepts: x 2 0 4 2
x2 4 x 4 2
y-intercepts: 0 y 2 4 2
y2 4 y 4 2
The intercepts are 2, 0 , 2, 0 , 0, 2 ,
24. ( x h) 2 ( y k ) 2 r 2
( x 3) 2 ( y 2) 2 2 5
and 0, 2 .
2
26. x 2 ( y 1) 2 1
( x 3) 2 ( y 2) 2 20
x 2 ( y 1) 2 12
General form: x 2 6 x 9 y 2 4 y 4 20 x2 y2 6 x 4 y 7 0
a.
Center:(0, 1); Radius 1
b.
c.
x-intercepts: x 2 (0 1) 2 1
25. x y 4 2
2
x y 2 2
a.
2
x2 1 1 2
x2 0
Center: (0, 0); Radius 2
x 0 0
y-intercepts: 0 ( y 1) 2 1 2
( y 1) 2 1 y 1 1 y 1 1 y 11
204 Copyright © 2025 Pearson Education, Inc.
Section 2.4: Circles y 2 or y 0
The intercepts are 0, 0 and 0, 2 .
c.
x-intercepts: x 1 0 1 2 2
x 12 12 2 x 12 1 2 x 12 1
27. 2 x 3 2 y 2 8 2
x 32 y 2 4 a.
2
x 1 1 x 1 1
Center: (3, 0); Radius 2
b.
x 1 1 x 0 or x 2
y-intercepts: 0 1 y 1 2 2
2
12 y 12 2 2 1 y 1 2 y 12 1 c.
y 1 1 y 1 1
x-intercepts: x 3 0 4 2
2
y 11 y 2 or y 0
x 32 4 x 3 4 x 3 2 x 3 2 x 5 or x 1
The intercepts are 2, 0 , 0, 0 , and 0, 2 . 29. x 2 y 2 2 x 4 y 4 0 x2 2 x y 2 4 y 4
y-intercepts: 0 3 y 2 4 2
( x 2 2 x 1) ( y 2 4 y 4) 4 1 4
32 y 2 4 9 y 4
( x 1) 2 ( y 2) 2 32
2
y 2 5 No real solution. The intercepts are 1, 0 and 5, 0 .
a.
Center: (1, 2); Radius = 3
b.
28. 3 x 1 3 y 1 6 2
2
x 12 y 12 2 a.
Center: (–1,1); Radius =
2
b. c.
x-intercepts: ( x 1) 2 (0 2) 2 32 ( x 1) 2 (2) 2 32
x 12 4 9 x 12 5 x 1 5 x 1 5
205 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
y-intercepts: (0 1) 2 ( y 2) 2 32
31.
(1) 2 ( y 2) 2 32 1 y 2 9 2
y 2 2 8
x2 y2 4 x 4 y 1 0 x2 4 x y 2 4 y 1 ( x 2 4 x 4) ( y 2 4 y 4) 1 4 4 ( x 2) 2 ( y 2) 2 32
Center: (–2, 2); Radius = 3 y b. a.
y2 8 y 2 2 2 y 22 2
0, 2 2 2 , and 0, 2 2 2 .
The intercepts are 1 5, 0 , 1 5, 0 ,
x 2 y 2 4 x 2 y 20 0
30.
x 4 x y 2 y 20 2
x
2
c.
( x 4 x 4) ( y 2 2 y 1) 20 4 1 2
x-intercepts: ( x 2) 2 (0 2) 2 32 ( x 2) 2 4 9
( x 2) 2 ( y 1) 2 52
( x 2) 2 5
a. Center: (–2,–1); Radius = 5 b.
x2 5 x 2 5
y-intercepts: (0 2) ( y 2) 2 32 2
4 ( y 2)2 9 ( y 2)2 5 y2 5
c.
2 5, 0 , 0, 2 5 , and 0, 2 5 .
( x 2) 2 1 25 x 2 24 x 2 2 6 x 2 2 6
32.
x2 y2 6 x 2 y 9 0 x 2 6 x y 2 2 y 9 2 ( x 6 x 9) ( y 2 2 y 1) 9 9 1 ( x 3) 2 ( y 1) 2 12
y-intercepts: (0 2) ( y 1) 2 52
a.
4 ( y 1) 25
b.
2
2
Center: (3, –1); Radius = 1
( y 1) 21 2
y 1 21
The intercepts are 2 5, 0 ,
x-intercepts: ( x 2) 2 (0 1) 2 52 ( x 2) 2 24
y 2 5
y 1 21
The intercepts are 2 2 6, 0 ,
2 2 6, 0 , 0, 1 21 , and 0, 1 21 . 206 Copyright © 2025 Pearson Education, Inc.
Section 2.4: Circles
c.
x-intercepts: ( x 3) 2 (0 1) 2 12
34.
( x 3) 1 1 2
x 32 0 x3 0 x3 y-intercepts: (0 3) 2 ( y 1) 2 12
1 0 2 1 x2 x y 2 y 2 1 1 1 1 1 2 2 x x y y 4 4 2 4 4 x2 y2 x y
2
2
1 1 2 x y 1 2 2
9 ( y 1) 1 2
y 12 8 a.
No real solution. The intercept only intercept is 3, 0 .
1 1 Center: , ; Radius = 1 2 2
b.
x y x 2 y 1 0 2
33.
2
x 2 x y 2 2 y 1 1 1 2 2 x x ( y 2 y 1) 1 1 4 4 2 2 1 1 2 x ( y 1) 2 2
a.
1 1 Center: , 1 ; Radius = 2 2
2
c.
b.
2
1 1 x-intercepts: x 0 12 2 2 2 1 1 x 1 2 4 2
2
c.
1 1 x-intercepts: x (0 1) 2 2 2 2 1 1 x 1 2 4
2
2
2
1 1 y-intercepts: 0 ( y 1) 2 2 2 1 1 2 y 1 4 4 y 12 0 y 1 0 y 1
The only intercept is 0, 1 .
2
1 1 y-intercepts: 0 y 12 2 2 2 1 1 y 1 4 2
1 3 x 2 4 No real solutions 2
1 3 x 2 4 1 3 x 2 2 1 3 x 2
2
2
1 3 y 2 4 1 3 y 2 2 1 3 y 2 1 3 1 3 , 0 , , 0 , The intercepts are 2 2 1 3 1 3 0, , and 0, . 2 2
207 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
2 x 2 2 y 2 12 x 8 y 24 0
35.
b.
x 2 y 2 6 x 4 y 12 x 2 6 x y 2 4 y 12 ( x 2 6 x 9) ( y 2 4 y 4) 12 9 4 ( x 3)2 ( y 2) 2 52
a.
Center: (3,–2); Radius = 5
b. c.
1 2 1 2 x 2 2
x-intercepts: ( x 2) 2 0 2
1 2 2 x2 2 x2
c.
x-intercepts: ( x 3) 2 (0 2) 2 52
x 2
x 3 4 25 x 32 21 2
1 2 1 2 4 y 2
y-intercepts: (0 2) 2 y 2
x 3 21 x 3 21
7 2 No real solutions. 2 , 0 and The intercepts are 2 2 2 , 0 . 2 2 y2
y-intercepts: (0 3) ( y 2) 2 52 2
9 y 2 25 2
y 2 2 16 y 2 4
y 2 4 y 2 or y 6
3 21, 0 ,
The intercepts are 3 21, 0 ,
0, 6 , and 0, 2 . 36. a.
2x2 2 y2 8x 7 0
Center: (–2, 0); Radius =
37.
2 x2 8x 2 y2 0 x2 4 x y2 0 x2 4 x 4 y 2 0 4
x 2 2 y 2 22 a. Center: 2, 0 ; Radius: r 2
2 x 2 8 x 2 y 2 7 7 x2 4x y2 2 7 2 2 ( x 4 x 4) y 4 2 1 2 2 ( x 2) y 2 2 2 ( x 2) 2 y 2 2
2 2
b.
2 2
208 Copyright © 2025 Pearson Education, Inc.
Section 2.4: Circles
c.
x-intercepts: x 2 0 22 2
2
39. Center at (0, 0); containing point (–2, 3).
2 0 2 3 0 2
( x 2)2 4
r
x 2 2 4
Equation: ( x 0)2 ( y 0) 2
x 2 2 x 2 2 x 0 or x 4
y-intercepts: 0 2 y 2 2
2
4 9 13
40. Center at (1, 0); containing point (–3, 2). r
4 y 4
3 12 2 0 2 16 4
2
Equation: ( x 1) 2 ( y 0) 2
y 0 y0 2
38. 3 x 2 3 y 2 12 y 0 x2 y 2 4 y 0
20
20 2 5 2
( x 1) 2 y 2 20
41. Endpoints of a diameter are (1, 4) and (–3, 2). The center is at the midpoint of that diameter: 1 (3) 4 2 , Center: 1,3 2 2
x2 y 2 4 y 4 0 4
Radius: r (1 (1)) 2 (4 3) 2 4 1 5
x2 y 2 4 2
a.
2
x 2 y 2 13
2
The intercepts are 4, 0 and 0, 0 .
13
Equation: ( x (1)) 2 ( y 3) 2
Center: 0, 2 ; Radius: r 2
b.
5
2
( x 1) 2 ( y 3) 2 5
42. Endpoints of a diameter are (4, 3) and (0, 1). The center is at the midpoint of that diameter: 4 0 3 1 , Center: 2, 2 2 2
Radius: r (4 2) 2 (3 2) 2 4 1 5 Equation: ( x 2) 2 ( y 2) 2
5
( x 2) 2 ( y 2) 2 5 c.
x-intercepts: x 2 0 2 4 2
43.
x2 4 4 x2 0
( x 2) 2 ( y (4)) 2 8
x0
y-intercepts: 02 y 2 4 2
y 2 2 4 y2 4 y 2 2 y 22 y 4 or y 0
The intercepts are 0, 0 and 0, 4 .
C 2 r 16 2 r r 8 2
( x 2) 2 ( y 4) 2 64
44.
A r2 49 r 2 r7 ( x (5)) 2 ( y 6) 2 7 ( x 5) 2 ( y 6) 2 49
209 Copyright © 2025 Pearson Education, Inc.
2
2
Chapter 2: Graphs
45. (c); Center: 1; Radius = 2
51. Consider the following diagram:
46. (d) ; Center: 3,3 ; Radius = 3 47. (b) ; Center: 1, 2 ; Radius = 2
(2,2)
48. (a) ; Center: 3,3 ; Radius = 3 49. The centers of the circles are: (4,-2) and (-1,5). 5 (2) 7 7 . Use the The slope is m 1 4 5 5 slope and one point to find the equation of the line. 7 y (2) ( x 4) 5 7 28 y2 x 5 5 5 y 10 7 x 28 7 x 5 y 18
Therefore, the path of the center of the circle has the equation y 2 . 52. Consider the following diagram:
(7,7)
50. Find the centers of the two circles: x2 y2 4x 6 y 4 0 ( x 2 4 x 4) ( y 2 6 y 9) 4 4 9 ( x 2) 2 ( y 3) 2 9
Therefore the path of the center of the circle has the equation x 7 .
Center: 2, 3 x2 y 2 6 x 4 y 9 0 ( x 2 6 x 9) ( y 2 4 y 4) 9 9 4 ( x 3) 2 ( y 2)2 4
Center: 3, 2 Find the slope of the line containing the centers: 1 2 (3) m 3 2 5 Find the equation of the line containing the centers: 1 y 3 ( x 2) 5 5 y 15 x 2 x 5 y 13 x 5 y 13 0
53. Let the upper-right corner of the square be the point x, y . The circle and the square are both
centered about the origin. Because of symmetry, we have that x y at the upper-right corner of the square. Therefore, we get x2 y 2 9 x2 x2 9 2x2 9 9 x2 2 9 3 2 2 2 The length of one side of the square is 2x . Thus, the area is x
2
2 3 2 A s 2 2 3 2 18 square units. 2
54. The area of the shaded region is the area of the circle, less the area of the square. Let the upperright corner of the square be the point x, y .
The circle and the square are both centered about 210 Copyright © 2025 Pearson Education, Inc.
Section 2.4: Circles
the origin. Because of symmetry, we have that x y at the upper-right corner of the square. Therefore, we get x 2 y 2 36 x 2 x 2 36
then the center of the building is 49.5 m above the ground, so the y-coordinate of the center is 49.5. The equation of the circle is given by x 2 ( y 49.5) 2 60.52 3660.25 58. Complete the square to find the equation of the circle representing the formula for the building. x 2 y 2 78 y 1521 1843 1521 3364
2 x 2 36 x 2 18 x3 2 The length of one side of the square is 2x . Thus,
the area of the square is 2 3 2
x 2 ( y 39) 2 582
72 square 2
units. From the equation of the circle, we have r 6 . The area of the circle is
r 2 6 36 square units. Therefore, the area of the shaded region is A 36 72 square units. 2
55. The diameter of the Ferris wheel was 250 feet, so the radius was 125 feet. The maximum height was 264 feet, so the center was at a height of 264 125 139 feet above the ground. Since the center of the wheel is on the y-axis, it is the point (0, 139). Thus, an equation for the wheel is:
x 0 2 y 139 2 1252 2 x 2 y 139 15, 625
Refer to figure. The y coordinate of the center is 39. The radius is 58. Thus the height of the building is 58 + 39 = 97 m. 59. Center at (2, 3); tangent to the x-axis. r 3 Equation: ( x 2) 2 ( y 3) 2 32 ( x 2) 2 ( y 3) 2 9
56. The diameter of the wheel is 520 feet, so the radius is 260 feet. The maximum height is 550 feet, so the center of the wheel is at a height of 550 260 290 feet above the ground. Since the center of the wheel is on the y-axis, it is the point (0, 290). Thus, an equation for the wheel is:
x 0 2 y 290 2 2602 2 x 2 y 290 67, 600
60. Center at (–3, 1); tangent to the y-axis. r 3 Equation: ( x 3) 2 ( y 1) 2 32 ( x 3) 2 ( y 1) 2 9
61. Center at (–1, 3); tangent to the line y = 2. This means that the circle contains the point (–1, 2), so the radius is r = 1. Equation: ( x 1) 2 ( y 3) 2 (1) 2 ( x 1) 2 ( y 3) 2 1
62. Center at (4, –2); tangent to the line x = 1. This means that the circle contains the point (1, –2), so the radius is r = 3. Equation: ( x 4) 2 ( y 2) 2 (3) 2
57.
( x 4) 2 ( y 2) 2 9
63. a.
Substitute y mx b into x 2 y 2 r 2 :
Refer to figure. Since the radius of the building is 60.5 m and the height of the building is 110 m, 211 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
x 2 (mx b) 2 r 2 x m x 2bmx b r 2
2 2
2
2
(1 m 2 ) x 2 2bmx b 2 r 2 0 This equation has one solution if and only if the discriminant is zero. (2bm) 2 4(1 m 2 )(b 2 r 2 ) 0 4b 2 m 2 4b 2 4r 2 4b 2 m 2 4m 2 r 2 0 4b 2 4r 2 4m 2 r 2 0 b2 r 2 m2 r 2 0 r 2 (1 m 2 ) b 2
b.
From part (a) we know (1 m 2 ) x 2 2bmx b 2 r 2 0 . Using the quadratic formula, since the discriminant is zero, we get: 2bm bm bmr 2 mr 2 x b 2(1 m 2 ) b 2 b2 2 r 2 mr y m b b
m r m r b r b b b b 2 2
2 2
2
2
mr 2 r 2 The point of tangency is , . b b
c.
The slope of the tangent line is m . The slope of the line joining the point of tangency and the center (0,0) is: r2 0 2 b r b 1 m mr 2 b mr 2 0 b The two lines are perpendicular.
2k 2 0h 2 k 2h The other tangent line is y 2 x 7 , and it has slope 2. 1 The slope from (h, k ) to (3, –1) is . 2 1 k 1 3h 2 2 2k 3 h 2k 1 h
h 1 2k Solve the two equations in h and k : 2 k 2(1 2k ) 2 k 2 4k 3k 0 k 0 h 1 2(0) 1 The center of the circle is (1, 0).
65. The slope of the line containing the center (0,0)
and 1, 2 2
is
2 2 0 2 2 .Then the slope of the tangent 1 0 1
2 . 4 2 2 So the equation of the tangent line is: 2 y2 2 x 1 4 2 2 y2 2 x 4 4 4y 8 2 2 x 2
line is
2 x 4y 9 2
64. Let (h, k ) be the center of the circle. x 2y 4 0 2y x 4 1 y x2 2 1 The slope of the tangent line is . The slope 2 from (h, k ) to (0, 2) is –2.
212 Copyright © 2025 Pearson Education, Inc.
Section 2.4: Circles
66. x 2 y 2 4 x 6 y 4 0
Then the slope of the tangent line is:
( x 4 x 4) ( y 6 y 9) 4 4 9 2
2
2 2 So, the equation of the tangent line is 2 ( x 3) y 2 2 3 4 2 3 2 y2 2 3 x 4 4 4 y 8 2 12 2 x 3 2
( x 2) ( y 3) 9 Center: (2, –3) The slope of the line containing the center and 2 2 3 (3) 2 2 2 2 3, 2 2 3 is 3 2 1 2
1
2
2 4
2 x 4 y 11 2 12
1 x x y y2 2 2 67. The center of the circle is 1 2 , 1 and the radius is 2 ( x1 x2 ) ( y1 y2 ) . Then the equation of 2 2 2
2
x x y y2 1 2 2 the circle is x 1 2 y 1 4 x1 x2 ( y1 y2 ) . Expanding, gives 2 2 x x( x1 x2 ) 2
x1 x2 2 4
y y ( y1 y2 ) 2
y1 y2 2 4
1
x12 2 x1 x2 x2 2 y12 2 y1 y2 y2 2
4
4 x 4 x1 x 4 x2 x x1 2 x1 x2 x2 4 y 4 y1 y 4 y2 y y1 2 y1 y2 y2 x1 2 x1 x2 x2 y1 2 y1 y2 y2 2
2
2
2
2
2
2
2
2
2
4 x 4 x1 x 4 x2 x 4 x1 x2 4 y 4 y1 y 4 y2 y 4 y1 y2 0 2
2
x x1 x x2 x x1 x2 y 2 y1 y y2 y y1 y2 0 2
x x x1 x2 x x1 y y y1 y2 y y1 0
x x1 x x2 y y1 y y2 0 2
2
d e d 2 e2 4 f 68. Complete the square to get x y . The slope of the line between the center 2 2 4 y0 2e x0 d2 d e m . So the slope of the tangent line is . tan , and the point of tangency x0 , y0 is m x0 d2 y0 2e 2 2 x d Therefore, the equation of the tangent line is y y0 0 2e ( x x0 ) which is equivalent to y0 2
( x x0 ) x0 d2 y y0 y0 2e 0
d d e e x x0 2 x0 y0 y y y0 2 y0 0 2 2 2 2 d e d e x0 x y0 y x y x0 2 y0 2 x0 y0 0 2 2 2 2 x0 x
Because x0 , y0 is on the circle, x0 2 y0 2 dx0 ey0 f 0 , and x0 2 y0 2
d e d e x0 y0 x0 y0 f Substituting this result gives 2 2 2 2
213 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
d e e d x0 y0 x0 y0 f 0 2 2 2 2 d e d e 2 2 x0 y0 x0 y0 x0 y0 f 0 2 2 2 2 x x0 y y0 x0 2 y0 2 d e f 0 2 2 x0 2 y0 2
69. (b), (c), (e) and (g) We need h, k 0 and 0, 0 on the graph. 70. (b), (e) and (g) We need h 0 , k 0 , and h r .
We need to check each possible solution: Check x 2 2( 2) 2 3( 2) 1 ( 2) 1 2(4) 6 1 1 no
71. Answers will vary.
Check x 1
72. The student has the correct radius, but the signs of the coordinates of the center are incorrect. The student needs to write the equation in the
standard form x h y k r 2 . 2
2
2(1) 2 3(1) 1 (1) 1 2 3 1 2 42
x 3 y 2 16 2
2
x 3 y 2 4 2
2
yes 2
Thus, h, k 3, 2 and r 4 . 73. A r 2 (13) 2 169 cm 2 C 2 r 2 (13) 26 cm
74. (3 x 2)( x 2 2 x 3) 3 x 3 6 x 2 9 x 2 x 2 4 x 6 3 x3 8 x 2 13 x 6
75.
2 x 2 3x 1 x 1 2 x 2 3x 1 x 1
2
2 x 2 3x 1 x 2 2 x 1 x2 x 2 0 ( x 2)( x 1) 0 x 2 or x 1
The solution is 1 76. Let t represent the time it takes to do the job together. of job done Time to do job Part in one minute 1 Aaron 22 22 1 Elizabeth 28 28 1 t Together t 1 1 1 22 28 t 14t 11t 308 25t 308 t 12.32 Working together, the job can be done in 12.32 minutes.
77. 9.57 105 0.0000957
214 Copyright © 2025 Pearson Education, Inc.
Section 2.5: Variation
78.
4 x 5x 7 x 2 2 x 2 3 8 x5 10 x 4 2 x3 11x 2 16 x 7 3
2
12 x3
8 x5
2(0.40) 1( X ) (2 X )0.75 0.8 X 1.5 0.75 X 0.25 X 0.7 X 2.8 You need to add 2.8 liters of pure antifreeze.
10 x 4 14 x3 11x 2 10 x 4
15 x 2 14 x3 4 x 2 16 x 21x
14 x3
4x 5x 7 2
4 x
6
2
5x 1 The quotient is 4 x3 5 x 2 7 x 2 ; the remainder is 5 x 1 . 4
3
4. c
3x 2 x 2 (4 x 5) 7(4 x 5) 3x 2 (4 x 5)( x 2 7) 4
405 x11 y 20 4 34 5 x8 x3 y 20
3x 2 y 5 4 5 x3
81. Since the area of the square is 64 square meters then each side must be 8 square meters. Thus the diagonal of the square is 642 642 128 8 2 square meters. Thus the diameter of the circle is also 8 2 square meters and the radius is 4 2 square meters. The circumference of the circle is:
C 2 r 2 4 2 8 2 m . The area of
the circle is:
A r2 4 2
32 m 2
2. False. If y varies directly with x, then y kx, where k is a constant.
2
3x 2 (4 x3 5 x 2 28 x 35)
80.
1. y kx
3. d
79. 12 x 15 x 84 x 105 x 5
Section 2.5
2
82. Let X be the amount of pure antifreeze to be added. 2L XL 2+XL 0.40 1.0 0.75 2(0.40) + 1(X) = (2 + X)0.75
5. y kx 2 10k 2 1 k 10 5 1 y x 5 6. v kt 16 2k 8k v 8t 7. V kx3 36 k (3)3 36 27 k 36 4 k 27 3 4 3 V x 3
8. A kx 2 4 k (2) 2 4 4k k A x2
215 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
9. y 4
k
13. M
x k
24
9
10
M
k d2 k
2 x
3
2
1 k 17 1 17 1 3 z x y2 17
k
26 k 52 122
26 k (169) 26 2 k 169 13 2 2 z x y2 13
15. T 2 2 2
4
18 k 8 3 2
3
T2
18 k 18 1 k
x d d 2
ka3 d2
k 23 42 k 8
16 k 4 2 k 8
12. T k 3 x d 2
T
9d 2
1 k 2 3
11. z k x 2 y 2
3
k 42
14. z k x3 y 2
52 k 10 25 k 250 250 F 2 d
x
9 16k 24 3 3 9 k 24 16 2
k 3 k 12 12 y x 4
10. F
kd 2
23
8a 3 d2
2 k 9 4
16. z 3 k x 2 y 2 x
3
2
2
8 k 97 8 97 8 2 z3 x y2 97 k
17. V
4 3 r 3
216 Copyright © 2025 Pearson Education, Inc.
Section 2.5: Variation
18. c 2 a 2 b 2
26.
19. p 2 l w 20. A
27. E kW
23.
k 32 Therefore, we have the linear equation v 32t. If t = 3 seconds, then v 32 3 96 ft/sec.
1 bh 2
21. F 6.67 10
22. T
2
11
3 k 20
mM 2 d
k
l
32
p kB
If W = 15, then E
0.00649 k Therefore we have the linear equation p 0.00649 B . If B 145000 , then p 0.00649 145000 $941.05 .
28.
p kB
16 k 1
k l k 256 48 k 12, 288 R
12, 288 . l
If R 576, then 12, 288 576 l 576l 12, 288 12, 288 64 l inches 576 3
0.00899 k Therefore we have the linear equation p 0.00899 B . If B 175000 , then p 0.00899 175000 $1573.25 . s kt 2
3 W. 20
3 15 2.25 . 20
Therefore, we have the equation R
8.99 k 1000
25.
3 20
Therefore, we have the linear equation E
6.49 k 1000
24.
v kt 64 k 2
29.
R kg 34.08 k 12
2
k 16 Therefore, we have equation s 16t 2 .
2.84 k Therefore, we have the linear equation R 2.84 g .
If t = 3 seconds, then s 16 3 144 feet.
If g 10.5 , then R 2.84 10.5 $29.82 .
2
If s = 64 feet, then 64 16t 2
30.
t2 4 t 2 Time must be positive, so we disregard t 2. It takes 2 seconds to fall 64 feet.
C kA 23.75 k 5 4.75 k Therefore, we have the linear equation C 4.75 A. If A 3.5 , then C 4.75 3.5 $16.63 .
217 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
31. D a.
k p
d2 If W 125, d 3960 then
D 156 , p 2.75 ;
D
429 143 bags of candy 3
and k 1,960, 200, 000
1,960, 200, 000 . d At the top of Mt. McKinley, we have d 3960 3.8 3963.8 , so 1,960, 200, 000 W 124.76 pounds. 3963.82 k
36. W
E 125 , L 5000 ; 125 5000k 125 1 k 5000 40
1 So, we have the equation E L . 40 1 b. 88 L 40 L 3520 likes 1 c. E (6723) 40 E $168.08
d2 k 55 39602 k 862, 488, 000
So, we have the equation W
d2
37. V r 2 h
39. I
k 600 150 k 90, 000
862, 488, 000
If d =3965, then 862, 488, 000 W 54.86 pounds. 39652
38. V
k P V 600, P 150 ;
33. V
3
r 2h
k
d2 If I 0.075, d 2 , then
So, we have the equation V If P 200 , then V
39602
So, we have the equation W
32. E kL a.
k
125
k 2.75 k 429 429 . So, D p 156
b.
k
35. W
90, 000 P
90, 000 450 cm3 . 200
0.075
k and k 0.3 . 22
So, we have the equation I If d 5, then I
k 34. i R k If i 30, R 8 , then 30 and k 240 . 8 240 . So, we have the equation i R 240 24 amperes . If R 10, then i 10
0.3 52
0.3 d2
.
0.012 foot-candles.
40. F kAv 2 11 k (20)(22) 2 11 9860k 11 1 k 9680 880
So, we have the equation F
218 Copyright © 2025 Pearson Education, Inc.
1 Av 2 . 880
.
Section 2.5: Variation
If A 47.125 and v 36.5 , then 1 F 47.125 36.5 2 71.34 pounds. 880 41.
If R = 1.44 and d = 3, then 1.24l 1.44 27(3) 2 1.24l 1.44 243 349.92 1.24l 349.92 l 282.2 feet 1.24
h ksd 3 36 k (75)(2)3 36 600k 0.06 k
So, we have the equation h 0.06sd 3 . If h 45 and s 125, then
44.
K kmv 2 1250 k (25)(10) 2 1250 2500k k 0.5
45 (0.06)(125)d 3 45 7.5d 3 6 d3
So, we have the equation K 0.5mv 2 . If m 25 and v 15, then
d 6 1.82 inches 3
K 0.5 25 15 2812.5 Joules 2
kT 42. V P k (300) 100 15 100 20k
45.
5k
So, we have the equation V
5T . P
0.6 pd . t If p 40, d 8, and t 0.50, then
So, we have the equation S
If V 80 and T 310, then 5(310) 80 P 80 P 1550 1550 P 19.375 atmospheres 80
S
46. 43.
R 1.24
kl d2 k 432
(4) 2 1.24 27 k
0.6(40)(8) 384 psi. 0.50
kwt 2 l k (4)(2)2 750 8 750 2k 375 k S
375wt 2 . l If l 10, w 6, and t 2, then
So, we have the equation S
1.24 k 27
So, we have the equation R
kpd t k (25)(5) 100 0.75 75 125k 0.6 k S
1.24l . 27 d 2
S
375(6)(2) 2 900 pounds. 10
47 – 50. Answers will vary.
219 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
51.
The solution set is 4 .
3x3 25 x 2 12 x 100 (3 x 25 x ) (12 x 100) 3
2
x 2 (3 x 25) 4(3 x 25)
57.
( x 2 4)(3 x 25) ( x 2)( x 2)(3 x 25)
52. 5 x2 5 x2 x 3 x 2 7 x 12 x 3 ( x 3)( x 4) x2 5( x 4) ( x 3)( x 4) ( x 3)( x 4) 5( x 4) ( x 2) ( x 3)( x 4) 5 x 20 x 2 ( x 3)( x 4) 6 x 18 ( x 3)( x 4) 6( x 3) 6 ( x 3)( x 4) ( x 4)
3 4 2
5 x 13 0 or 2 x 9 0 5 x 13 2 x 9 13 9 x x 5 2 9 13 The solution set is , . 2 5
4x 7 1
3
54. The term needed to rationalize the denominator is 7 2 .
56. 6 x 2( x 4) 24 6 x 2 x 8 24 4 x 16 x4
10 x 2 19 x 117 0 (5 x 13)(2 x 9) 0
3 4 x 7 3
8 2 5 125
(5)2 (2) 2 25 4 (3)(5) 2(2)(5 3) 15 (4)(2) 21 3 7
10 x 2 117 19 x
58.
59. 7 3 4 x 7 4
1 3 4 2
53. 25 25
55.
6 1 1 x 10 5 6 1 1 10 x 10 x x 10 5 60 x 2 x 3x 60 x 20 The solution set is 20
4 x 7 1 or 4 x 7 1 4x 6 4x 8 6 3 x x2 4 2 3 The solution set is , 2 . 2
60. 5( x 2) 9 x x 3(2 x 1) 7 5 x 10 9 x x 6 x 3 7 10 4 x 5 x 4 x 6 The solution is x | x 6 .
220 Copyright © 2025 Pearson Education, Inc.
Chapter 2 Review Exercises
4. y x 2 4
Chapter 2 Review Exercises
y
1. P1 0, 0 and P2 4, 2 a.
d P1 , P2
4 0 2 2 0 2
16 4 20 2 5
b. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2 04 02 4 2 , , 2,1 2 2 2 2
c.
slope
y 2 0 2 1 x 4 0 4 2
d. For each run of 2, there is a rise of 1. 2. P1 1, 1 and P2 2,3 a.
d P1 , P2
2 12 3 1
2
9 16 25 5
1 2 1 3 , 2 2 1 2 1 , ,1 2 2 2 slope
x
5. x-intercepts: 4, 0, 2 ; y-intercepts: 2, 0, 2 Intercepts: (4, 0), (0, 0), (2, 0), (0, 2), (0, 2) 6. 2 x 3 y 2 x-intercepts: y-intercepts: 2 2 x 3(0) 2(0) 3 y 2 2x 0 0 y2 y0 x0 The only intercept is (0, 0).
2 x 3( y ) 2 2 x 3 y 2 same Test y-axis symmetry: Let x x 2( x) 3 y 2
2( x) 3( y ) 2 2 x 3 y 2 different Therefore, the graph will have x-axis symmetry.
y 3 1 4 4 x 2 1 3 3
3. P1 4, 4 and P2 4,8
7. x 2 +4 y 2 =16
x-intercepts:
y-intercepts:
2
x +4 0 =16
0 2 +4 y 2 =16
0 144 144 12
x 2 16 x 4
4 y 2 16
d P1 , P2
4 4 2 8 4
b. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2 4 4 4 8 8 4 , , 4, 2 2 2 2 2
c.
2 x 3 y 2 different Test origin symmetry: Let x x and y y .
d. For each run of 3, there is a rise of 4.
a.
Test x-axis symmetry: Let y y
b. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2
c.
y 8 4 12 , undefined slope x 44 0
2
2
y2 4 y 2 The intercepts are (4, 0), (4, 0), (0, 2), and (0, 2).
Test x-axis symmetry: Let y y x 2 4 y =16 2
x 2 4 y 2 =16 same
d. An undefined slope means the points lie on a vertical line. There is no change in x. 221 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
Test y-axis symmetry: Let x x
Test y-axis symmetry: Let x x y ( x )3 ( x )
x 4 y =16 2
2
y x3 x
x 2 4 y 2 =16 same
different
Test origin symmetry: Let x x and y y .
Test origin symmetry: Let x x and y y .
x 2 4 y 2 =16
y ( x )3 ( x ) y x3 x y x3 x same
x 2 +4 y 2 =16
same
Therefore, the graph will have x-axis, y-axis, and origin symmetry.
10. x 2 x y 2 2 y 0
8. y x 4 +2 x 2 +1
x-intercepts: 0 x 4 +2 x 2 +1
Therefore, the graph will have origin symmetry.
x-intercepts: x 2 x (0) 2 2(0) 0 x2 x 0 x( x 1) 0 x 0, x 1
y-intercepts: y (0) 4 +2(0) 2 +1 1
0 x2 1 x2 1
x 1 0 x 2 1 no real solutions The only intercept is (0, 1). Test x-axis symmetry: Let y y 2
y-intercepts: (0) 2 0 y 2 2 y 0 y2 2 y 0 y ( y 2) 0 y 0, y 2 The intercepts are (1, 0), (0, 0), and (0, 2).
y x4 2 x2 1 y x 4 2 x 2 1 different
Test x-axis symmetry: Let y y
Test y-axis symmetry: Let x x
x 2 x ( y ) 2 2( y ) 0
y x 2x 1
x 2 x y 2 2 y 0 different Test y-axis symmetry: Let x x ( x) 2 ( x) y 2 2 y 0
4
2
y x4 2 x2 1
same
Test origin symmetry: Let x x and y y .
x 2 x y 2 2 y 0 different Test origin symmetry: Let x x and y y .
y x 2x 1 4
2
y x4 2 x2 1 y x4 2 x2 1
( x) 2 ( x) ( y ) 2 2( y ) 0 x 2 x y 2 2 y 0 different The graph has none of the indicated symmetries.
different
Therefore, the graph will have y-axis symmetry. 9. y x3 x
x-intercepts: 0 x3 x
11.
0 x x 1 2
2
x 2 2 y 32 16 12.
x 0, x 1, x 1
( x h) 2 ( y k ) 2 r 2
x 1 y 2 12 2
The intercepts are (1, 0), (0, 0), and (1, 0). Test x-axis symmetry: Let y y y x3 x y x 3 x different
x 2 y 32 42
y-intercepts: y (0)3 0 0
0 x x 1 x 1
( x h) 2 ( y k ) 2 r 2
2
x 12 y 2 2 1 13. x 2 y 1 4 2
x 2 y 1 22 Center: (0,1); Radius = 2 2
222 Copyright © 2025 Pearson Education, Inc.
Chapter 2 Review Exercises
y-intercepts: 0 1 y 2 32 2
2
1 y 2 9 2
y 2 2 8 y2 8 y 2 2 2 y 2 2 2
x-intercepts: x 2 0 1 4
The intercepts are 1 5, 0 , 1 5, 0 ,
2
0, 2 2 2 , and 0, 2 2 2 .
x 1 4 2
x 3 2
x 3
3x 2 3 y 2 6 x 12 y 0
15.
y-intercepts: 02 y 1 4
x2 y2 2 x 4 y 0
2
( y 1) 2 4 y 1 2 y 1 2 y 3 or y 1
3, 0 , 0, 1 ,
The intercepts are 3, 0 ,
x2 2x y2 4 y 0
x 2 x 1 y 4 y 4 1 4 x 1 y 2 5 2
2
2
2
Center: (1, –2) Radius =
2
5
and 0, 3 . x2 y 2 2 x 4 y 4 0
14.
x2 2 x y 2 4 y 4
x 2 x 1 y 4 y 4 4 1 4 2
2
x 12 y 2 2 32 Center: (1, –2) Radius = 3 x-intercepts: x 1 0 2 2
2
5
2
x 12 4 5 x 12 1 x 1 1 x 11 x 2 or x 0
y-intercepts: 0 1 y 2 2
x-intercepts: x 1 0 2 3 2
2
2
2
5
2
1 y 2 5 2
x 12 4 9 x 12 5
y 2 2 4 y 2 2 y 2 2 y 0 or y 4
x 1 5 x 1 5
The intercepts are 0, 0 , 2, 0 , and 0, 4 .
223 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs y y1 m x x1
16. Slope = –2; containing (3,–1) y y1 m x x1 y 1 2 x 6 y 2 x 5 or 2 x y 5
17. vertical; containing (–3,4) Vertical lines have equations of the form x = a, where a is the x-intercept. Now, a vertical line containing the point (–3, 4) must have an x-intercept of –3, so the equation of the line is x 3. The equation does not have a slopeintercept form. 18. y-intercept = –2; containing (5,–3) Points are (5,–3) and (0,–2) 1 1 2 (3) m 05 5 5 y mx b 1 y x 2 or x 5 y 10 5
21. Perpendicular to x y 2 x y 2 y x 2 The slope of this line is 1 , so the slope of a line perpendicular to it is 1. Slope = 1; containing (4,–3) y y1 m( x x1 ) y (3) 1( x 4) y3 x4 y x 7 or x y 7
19. Containing the points (3,–4) and (2, 1) 1 (4) 5 m 5 23 1 y y1 m x x1 y ( 4) 5 x 3 y 4 5 x 15 y 5 x 11 or 5 x y 11
20. Parallel to 2 x 3 y 4 2x 3y 4 3 y 2 x 4 3 y 2 x 4 3 3 2 4 y x 3 3 2 Slope ; containing (–5,3) 3
2 x (5) 3 2 y 3 x 5 3 2 10 y 3 x 3 3 2 19 y x or 2 x 3 y 19 3 3 y 3
y (1) 2 x 3
22. 4 x 5 y 20 5 y 4 x 20 4 y x4 5 4 slope = ; y-intercept = 4 5
x-intercept: Let y = 0. 4 x 5(0) 20 4 x 20 x 5
224 Copyright © 2025 Pearson Education, Inc.
Chapter 2 Review Exercises
23.
1 1 1 x y 2 3 6 1 1 1 y x 3 2 6 3 1 y x 2 2 3 1 slope = ; y -intercept 2 2 x-intercept: Let y = 0. 1 1 1 x (0) 2 3 6 1 1 x 2 6 1 x 3
25.
1 1 x y 2 2 3 x-intercept: 1 1 x (0) 2 2 3 1 x2 2 x4
The intercepts are 4, 0 and 0, 6 .
26. y x3
24. 2 x 3 y 12 x-intercept: 2 x 3(0) 12 2 x 12 x6
y-intercept: 1 1 (0) y 2 2 3 1 y2 3 y6
y-intercept: 2(0) 3 y 12 3 y 12 y 4
The intercepts are 6, 0 and 0, 4 .
27. y x
225 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
28. slope =
2 , containing the point (1,2) 3
mAB
40 4 2 4 ( 2) 2
mBC
54 1 8 4 12
mAC
50 5 1 8 2 10 2
1 1 , the sides AB 2 and AC are perpendicular and the triangle is a right triangle.
Since mAB mAC 2
29. Find the distance between each pair of points. d A, B (1 3) 2 (1 4) 2 4 9 13 d B ,C ( 2 1) 2 (3 1) 2 9 4 13
31. Endpoints of the diameter are (–3, 2) and (5,–6). The center is at the midpoint of the diameter: 3 5 2 6 Center: , 1, 2 2 2
Radius: r (1 (3)) 2 ( 2 2) 2
d A,C ( 2 3) 2 (3 4) 2 25 1 26
16 16
Since AB = BC, triangle ABC is isosceles.
32 4 2
30. Given the points A ( 2, 0), B ( 4, 4), and C (8, 5). a.
2
2
1 5 1 62 1 5 slope of AC 1 82 Therefore, the points lie on a line.
d A, B ( 4 ( 2)) 2 (4 0) 2
32. slope of AB
4 16 20 2 5 d B, C (8 ( 4)) 2 (5 4) 2
33.
144 1
p kB 854 k 130, 000
145
k
d A, C (8 ( 2)) 2 (5 0) 2 100 25
854 427 130, 000 65, 000
Therefore, we have the equation p
125 5 5 d A, B d A, C d B, C 2
20 125 145 2
2
x 12 y 2 2 32
Find the distance between each pair of points.
2
Equation: x 1 y 2 4 2
2
2
2
20 125 145
If B 165, 000 , then 427 p 165, 000 $1083.92 . 65, 000
145 145 The Pythagorean Theorem is satisfied, so this is a right triangle.
b. Find the slopes:
226 Copyright © 2025 Pearson Education, Inc.
427 B. 65, 000
Chapter 2 Test
4. y x 2 9 34. w
k d2
200
k 39602
k 200 39602 3,136,320, 000
Therefore, we have the equation 3,136,320, 000 . w d2 If d 3960 1 3961 miles, then 3,136,320, 000 w 199.9 pounds. 39612 35.
H ksd 135 k (7.5)(40) 135 300k k 0.45 So, we have the equation H 0.45sd . If s 12 and d 35, then H 0.45 12 35 189 BTU
5. y 2 x y
x
x2 9
1. d ( P1 , P2 )
5 (1) 2 1 32
62 4
y-intercept: (0) 2 y 9 y9
x 3 The intercepts are 3, 0 , 3, 0 , and 0, 9 .
Test x-axis symmetry: Let y y
2
x2 y 9
36 16
x 2 y 9 different
52 2 13
Test y-axis symmetry: Let x x
2. The coordinates of the midpoint are: x x y y2 ( x, y ) 1 2 , 1 2 2 1 5 3 (1) , 2 2 4 2 , 2 2 2, 1
3. a.
y2 x
6. x 2 y 9 x-intercepts: x2 0 9
Chapter 2 Test
x 2 y 9 x 2 y 9 same
Test origin symmetry: Let x x and y y
x 2 y 9 x 2 y 9 different
Therefore, the graph will have y-axis symmetry. 7. Slope = 2 ; containing (3, 4)
y y 2 1 3 4 m 2 1 3 x2 x1 5 (1) 6
b. If x increases by 3 units, y will decrease by 2 units.
y y1 m( x x1 ) y (4) 2( x 3) y 4 2 x 6 y 2 x 2
227 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
2 m . The line contains (1, 1) : 3 y y1 m( x x1 ) 2 y (1) ( x 1) 3 2 2 y 1 x 3 3 2 1 y x 3 3
8. ( x h) 2 ( y k ) 2 r 2
x 4 2 y (3) 2 52 x 4 2 y 32 25 General form: x 4 2 y 32 25 x 2 8 x 16 y 2 6 y 9 25 x y 8x 6 y 0 2
9.
2
x2 y2 4 x 2 y 4 0 x2 4 x y 2 2 y 4 ( x 2 4 x 4) ( y 2 2 y 1) 4 4 1 ( x 2) 2 ( y 1) 2 32
Center: (–2, 1); Radius = 3 y
3 . The line contains (0, 3) : 2 y y1 m( x x1 )
m
3 ( x 0) 2 3 y 3 x 2 3 y x3 2 y 3
11. Let R = the resistance, l = length, and r = radius. l Then R k 2 . Now, R = 10 ohms, when r l = 50 feet and r 6 103 inch, so 50 10 k 2 6 103
Perpendicular line Any line perpendicular to 2 x 3 y 6 has slope
6 10 7.2 10 k 10
x
3 2
6
50 Therefore, we have the equation l R 7.2 106 2 . r
10. 2 x 3 y 6 3 y 2 x 6 2 y x2 3
If l 100 feet and r 7 103 inch, then 100 14.69 ohms. R 7.2 106 2 7 103
Parallel line Any line parallel to 2 x 3 y 6 has slope
228 Copyright © 2025 Pearson Education, Inc.
Chapter 2 Cumulative Review
Chapter 2 Cumulative Review
2x 1 3 2
1. 3x 5 0 3x 5 5 x 3
2.
2
2x 1 9 2x 8 x4
5 The solution set is . 3
Check:
x x 12 0 x 4 x 3 0 x 4 or x 3
9 3? 3 3 True The solution set is 4 .
2
The solution set is 3, 4 . 3.
2x 1 3
6.
7.
x3 x 1 The solution set is 1,3 .
1 or x 3 2
x 2 4 x 22 2
x2 4x 4 x2 4 x 4 0
4. x 2 x 2 0 2
2
2 2 4 1 2 2 1
2 48 2 2 12 2 22 3 2 1 3
x2 4x 2
8.
1 The solution set is ,3 . 2
x
x 2 1 x 2 1 or x 2 1
2 x2 5x 3 0 2 x 1 x 3 0 x
2(4) 1 3?
x
4 42 4(1)(4) 4 16 16 2(1) 2
4 32 4 4 2 2 2 2 2 2
Check x 2 2 2 :
2 2 2 2 4 2 2 2 2? 4 8 2 8 8 8 2 2?
4 2 True
The solution set is 1 3, 1 3 . 5. x 2 2 x 5 0 x
2 22 4 1 5
Check x 2 2 2 :
2 2 2 2 4 2 2 2 2? 4 8 2 8 8 8 2 2?
2 1
2 4 20 2 2 16 2 No real solutions
4 2 True
The solution set is 2 2 2, 2 2 2 . 9. x 2 9 x 9 x 3i The solution set is 3i,3i .
229 Copyright © 2025 Pearson Education, Inc.
Chapter 2: Graphs
10. x 2 2 x 5 0 x
2
16. y x3 3x 1
2 2 4 1 5 2 4 20 2 1 2
2 16 2 4i 1 2i 2 2 The solution set is 1 2i, 1 2i .
a.
b.
11. 2 x 3 7 2 x 10 x5 x x 5 or ,5
c.
3,1 : 33 3 3 1 27 9 1 19 1 3,1 is not on the graph.
x 2 1 1 x 2 1 1 x 3 x 1 x 3 or 1,3
14.
2,3 : 2 3 3 2 1 8 6 1 3 2,3 is on the graph.
17. y x3
12. 1 x 4 5 5 x 1 x 5 x 1 or 5,1
13.
2, 1 : 2 3 3 2 1 8 6 1 1 2, 1 is on the graph.
18. The points (–1,4) and (2,–2) are on the line. 2 4 6 Slope 2 2 (1) 3 y y1 m( x x1 )
2 x 3
y 4 2 x 1
2 x 3 or 2 x 3 x 5 or x 1
x x 5 or x 1 or , 5 1,
y 4 2 x 1 y 2 x 2 4 y 2 x 2
19. Perpendicular to y 2 x 1 ; Contains 3,5 15. d P, Q
1 4 3 2 2
2
5 2 52
Slope of perpendicular = y y1 m( x x1 )
25 25 50 5 2 1 4 3 2 3 1 Midpoint , , 2 2 2 2
1 y 5 ( x 3) 2 1 3 y 5 x 2 2 1 13 y x 2 2
230 Copyright © 2025 Pearson Education, Inc.
1 2
Chapter 2 Project
20.
x2 y2 4x 8 y 5 0 x2 4 x y 2 8 y 5 ( x 2 4 x 4) ( y 2 8 y 16) 5 4 16 ( x 2) 2 ( y 4) 2 25 ( x 2) 2 ( y 4) 2 52 Center: (2,–4); Radius = 5
Chapter 2 Project Internet-based Project
231 Copyright © 2025 Pearson Education, Inc.
Chapter 3 Functions and Their Graphs 16. explicitly
Section 3.1 1.
17. a.
1,3
2. 3 2 5 2 2
1
2
3 4 5 2 12 10
1 2
43 or 21 12 or 21.5 2
4. 3 2 x 5 2 x 2 x 1 Solution set: x | x 1 or , 1
5.
b.
1 2
3. We must not allow the denominator to be 0. x 4 0 x 4 ; Domain: x x 4 .
Domain: {0,22,40,70,100} in C⁰ Range: {1.031, 1.121, 1.229, 1.305, 1.411} in kg/m3
c. {(0, 1.411), (22, 1.305), (40, 1.229), (70, 1.121), (100, 1.031)} 18. a.
Domain: {1.80, 1.78, 1.77} Range: {87.1, 86.9, 92.0, 84.1, 86.4}
b.
c. {(1.80, 87.1), (1.78, 86.9), (1.77, 83.0), (1.77, 84.1), (1.80, 86.4)}
52
6. radicals
19. Domain: {Elvis, Colleen, Kaleigh, Marissa} Range: {Jan. 8, Mar. 15, Sept. 17} Function
7. independent; dependent 8. a
20. Domain: {Bob, John, Chuck} Range: {Beth, Diane, Linda, Marcia} Not a function
9. c 10. False; g 0 11. False; every function is a relation, but not every relation is a function. For example, the relation x 2 y 2 1 is not a function. 12. verbally, numerically, graphically, algebraically 13. False; if the domain is not specified, we assume it is the largest set of real numbers for which the value of f is a real number. 14. False; if x is in the domain of a function f, we say that f is defined at x, or f(x) exists.
21. Domain: {20, 30, 40} Range: {200, 300, 350, 425} Not a function 22. Domain: {Less than 9th grade, 9th-12th grade, High School Graduate, Some College, College Graduate} Range: {$18,120, $23,251, $36,055, $45,810, $67,165} Function 23. Domain: {-3, 2, 4} Range: {6, 9, 10} Not a function
15. difference quotient 232 Copyright © 2025 Pearson Education, Inc.
Section 3.1: Functions 24. Domain: {–2, –1, 3, 4} Range: {3, 5, 7, 12} Function
34. Graph y x . The graph passes the VerticalLine Test. Thus, the equation represents a function.
25. Domain: {1, 2, 3, 4} Range: {3} Function 26. Domain: {0, 1, 2, 3} Range: {–2, 3, 7} Function
35. x 2 8 y 2 Solve for y : y 8 x 2
27. Domain: {-4, 0, 3} Range: {1, 3, 5, 6} Not a function
For x 0, y 2 2 . Thus, 0, 2 2 and
28. Domain: {-4, -3, -2, -1} Range: {0, 1, 2, 3, 4} Not a function
since a distinct x-value corresponds to two different y-values.
0, 2 2 are on the graph. This is not a function,
29. Domain: {–1, 0, 2, 4} Range: {-1, 3, 8} Function
36. y 1 2 x For x 0, y 1 . Thus, (0, 1) and (0, –1) are on the graph. This is not a function, since a distinct xvalue corresponds to two different y-values.
30. Domain: {–2, –1, 0, 1} Range: {3, 4, 16} Function
37. x y 2
31. Graph y 2 x 2 3 x 4 . The graph passes the Vertical-Line Test. Thus, the equation represents a function.
Solve for y : y x For x 1, y 1 . Thus, (1, 1) and (1, –1) are on the graph. This is not a function, since a distinct x-value corresponds to two different y-values. 38. x y 2 1
32. Graph y x3 . The graph passes the VerticalLine Test. Thus, the equation represents a function.
Solve for y : y 1 x For x 0, y 1 . Thus, (0, 1) and (0, –1) are on the graph. This is not a function, since a distinct xvalue corresponds to two different y-values. 39. Graph y 3 x . The graph passes the VerticalLine Test. Thus, the equation represents a function.
1 . The graph passes the Verticalx Line Test. Thus, the equation represents a function.
33. Graph y
233 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs 3x 1 . The graph passes the x2 Vertical-Line Test. Thus, the equation represents a function.
f.
40. Graph y
f x 1 3 x 1 2 x 1 4 2
3 x2 2 x 1 2x 2 4 3x2 6 x 3 2 x 2 4 3x2 8 x 1
41.
g.
f 2 x 3 2 x 2 2 x 4 12 x 2 4 x 4
h.
f x h 3 x h 2 x h 4
2
2
3 x 2 2 xh h 2 2 x 2h 4
Solve for y: y 2 x 3 or y (2 x 3)
3x 2 6 xh 3h 2 2 x 2h 4
For x 1, y 5 or y 5 . Thus, 1,5 and 44.
1, 5 are on the graph. This is not a function, since a distinct x-value corresponds to two different y-values. 42. x 2 4 y 2 1 Solve for y: x 2 4 y 2 1 2
2
4 y x 1 x2 1 y2 4
f x 2 x2 x 1
a.
f 0 2 0 0 1 1
b.
f 1 2 1 1 1 2
c.
f 1 2 1 1 1 4
d.
f x 2 x x 1 2x2 x 1
e.
f x 2 x 2 x 1 2 x 2 x 1
f.
f x 1 2 x 1 x 1 1
2
2
2
2
2 x2 2 x 1 x 1 1
x2 1 2 1 1 For x 2, y . Thus, 2, and 2 2 1 2, are on the graph. This is not a 2 function, since a distinct x-value corresponds to two different y-values.
2 x2 4x 2 x 2 x 2 3x 2
g.
f 2 x 2 2 x 2 x 1 8x2 2 x 1
h.
f x h 2( x h) 2 x h 1
2
2 x 2 2 xh h 2 x h 1 2 x 2 4 xh 2h 2 x h 1
f x 3x 2 2 x 4
a.
f 0 3 0 2 0 4 4
b.
f 1 3 1 2 1 4 3 2 4 1
2
45.
2
c.
f 1 3 1 2 1 4 3 2 4 3
d.
f x 3 x 2 x 4 3x 2 2 x 4
e.
f x 3x 2 2 x 4 3x 2 2 x 4
f x
a.
2
b.
2
2
y
43.
y 2x 3
c.
x 2
x 1
0 0 0 1 1 1 1 f 1 2 1 1 2 1 1 1 f 1 2 1 1 1 1 2
f 0
0
2
x
x
d.
f x
e.
x x f x 2 2 1 x x 1
234 Copyright © 2025 Pearson Education, Inc.
x 1 2
2
x 1
Section 3.1: Functions
f.
x 1
f x 1
x 1 1 x 1
x2 2 x 1 1 x 1
g. h.
46.
x2 2 x 2 2x 2x f 2x 2 2 2x 1 4x 1 f x h
f x
xh
x h 2 1
xh
48.
x 2 2 xh h 2 1
x2 1 x4 02 1 1 1 04 4 4
a.
f 0
b.
12 1 0 f 1 0 1 4 5
c.
f 1
12 1 1 4
f x x 4 x 4
e.
f x x 4 x 4
f.
f x 1 x 1 4
g.
f 2x 2x 4 2 x 4
h.
f x h x h 4
0 0 3
f x x2 x
a.
f 0 02 0 0 0
b.
f 1 12 1 2
c.
f 1
12 1 1 1 0 0
d.
f x
x 2 x
e.
f x
x x x x
f.
f x 1
x 12 x 1
d.
x2 1 f x x 4 x 4
e.
x2 1 x2 1 f x x 4 x4
f.
x 1 1 x 1 4
f x 1
4x2 1 2x 4
g.
f 2x
h.
x h 2 1 x 2 2 xh h 2 1 f x h xh4 x h 4
2x 4
g.
f 2x
2 x 2 2 x
h.
f x h
4 x2 2 x
x h 2 x h
x 2 2 xh h 2 x h
x2 2 x 1 1 x2 2 x x5 x5
2
x 2 3x 2
2
2 x 2 1
2
x2 x
x2 2 x 1 x 1
x 2 1
47.
d.
2
f x x 4
a.
f 0 0 4 0 4 4
b.
f 1 1 4 1 4 5
c.
f 1 1 4 1 4 5
49.
f x
2x 1 3x 5 2 0 1
0 1 1 05 5
a.
f 0
b.
f 1
c.
f 1
d.
f x
e.
2x 1 2x 1 f x 3x 5 3x 5
235 Copyright © 2025 Pearson Education, Inc.
3 0 5 2 1 1
3 1 5
2 1 3 3 35 2 2
2 1 1
3 1 5
2x 1
3 x 5
2 1 1 1 3 5 8 8
2x 1 2x 1 3 x 5 3x 5
Chapter 3: Functions and Their Graphs
f.
f x 1
g.
f 2x
h.
50.
2 2x 1
3 2x 5
2x 2 1 2x 3 3x 3 5 3x 2
2 x h 1
2 x 2h 1 3 x 3h 5
0 2
1 2
1
f 1 1
1 2
f x 1
2
1
1
1 2
2
1 3 4 4
56. h( x)
x 4 x 4 0
1 8 9 9
x 2 4 x 2 Domain: x x 2, x 2
1 1 0 1
57. F ( x)
x 2 2
x( x 2 1) 0
f.
f x 1 1
1
x 1 2
2x 2
f x h 1
2
2
1
1
x 0, x 2 1
Domain: x x 0
1
x 3
58. G ( x)
2
x 4x x 4x 0
1 4 x 1
x( x 2 4) 0
2
x 0, x 2 4 x 0, x 2
1
x h 2 2
Domain: x x 2, x 0, x 2
f ( x) 5 x 4
59. h( x ) 3 x 12 3x 12 0 3x 12 x4 Domain: x x 4
f ( x) x 2 2
Domain: x x is any real number 53.
f ( x)
x4 3
3
Domain: x x is any real number 52.
x2
x3 x x3 x 0
1
1
2x 2
2
e.
h.
x 2
x 16 x 16 0
1 1 f x 1 1 2 x 2 x 2 2
f 2x 1
x2 1 Domain: x x is any real number
x 2 16 x 4 Domain: x x 4, x 4 1
c.
x2
2
x 2 2
f 0 1
f ( x)
55. g ( x)
1
f 1 1
g.
54.
4x 1 6x 5
3 x h 5
b.
d.
51.
3 x 1 5
f x h
f x 1
a.
2 x 1 1
x 1
2 x2 8 Domain: x x is any real number
236
Copyright © 2025 Pearson Education, Inc.
Section 3.1: Functions
Also 3t 21 0
60. G ( x) 1 x 1 x 0 x 1 x 1 Domain: x x 1 61. p( x)
3t 21 0 3t 21 t7
Domain: t t 4, t 7
x 2x 3 1
z 3 z2 z 3 0
66. h( z )
2x 3 1 0 2x 3 1
z 3 Also z 2 0 z2 Domain: z z 3, z 2
2 x 3 1 or 2 x 3 1 2 x 4 2 x 2 x 2 x 1 Domain: x x 2, x 1
67. 62. p( x)
x 1 3x 1 4
3x 1 4 0
Domain: x x is any real number . 68. g (t ) t 2 3 t 2 7t
Domain: t t is any real number .
3x 1 4 3 x 1 4 or 3 x 1 4 3 x 3
3x 5 5 x 1 x 3 5 Domain: x x 1, x 3
63.
69. M (t ) 5
t 1 2
t 5t 14
t 2 5t 14 0 (t 2)(t 7) 0 t 2 0 or t 7 0 t 2 t 7 Domain: t t 2, x 7
x
f ( x)
f ( x) 3 5 x 4
x4
x4 0 x4 Domain: x x 4
70. N ( p ) 5
2 p 98 2
2 p 98 0 2( p 2 49) 0
x x2
64. q( x )
p 2
2( p 7)( p 7) 0 p 7 0 or p 7 0
x 2 0 x 2
p 7
Domain: p p 7, x 7
x 2
Domain: x x 2 65. P (t )
t4 3t 21
p7
71.
f ( x) 3x 4
a.
g ( x) 2 x 3
( f g )( x) 3 x 4 2 x 3 5 x 1
Domain: x x is any real number .
t4 0 t4
237 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
b.
( f g )( x) (3 x 4) (2 x 3) 3x 4 2 x 3 x7
Domain: x x is any real number . c.
e.
( f g )(3) 5(3) 1 15 1 14
f.
( f g )(4) 4 3 1
g.
( f g )(2) 6(2) 2 2 2 6(4) 2 2 24 2 2 20
( f g )( x) (3x 4)(2 x 3) 6 x 2 9 x 8 x 12
h.
6 x 2 x 12
Domain: x x is any real number . d.
73.
f 3x 4 ( x) 2x 3 g 2x 3 0 2x 3 x
b.
f.
( f g )(4) 4 7 11
g.
( f g )(2) 6(2) 2 2 12 24 2 12 10
h.
f 3(1) 4 3 4 7 7 (1) 2(1) 3 2 3 1 g
c.
Domain: x x is any real number . c.
( f g )( x) ( x 1)(2 x 2 ) 2 x3 2 x 2
Domain: x x is any real number . d.
g ( x) 3 x 2
( f g )( x) 2 x 1 3 x 2 5 x 1
( f g )( x) (2 x 1) (3 x 2) 2 x 1 3x 2 x 3 Domain: x x is any real number .
( f g )(3) 2(3) 2 3 1 2(9) 3 1 18 3 1 20
f.
( f g )(4) 2(4) 2 4 1 2(16) 4 1 32 4 1 29
g.
( f g )(2) 2(2)3 2(2) 2 2(8) 2(4) 16 8 8
( f g )( x) (2 x 1)(3 x 2) 6 x2 x 2 Domain: x x is any real number . f 2x 1 ( x) g x2 3 3x 2 0
h.
2 3 2 Domain: x x . 3 3x 2 x
74.
f x 1 ( x) 2 g 2x Domain: x x 0 .
e.
6 x 2 4 x 3x 2
d.
( f g )( x) ( x 1) (2 x 2 ) 2 x2 x 1
Domain: x x is any real number . b.
( f g )( x) x 1 2 x 2 2 x 2 x 1
x 1 2x2
( f g )(3) 5(3) 1 15 1 16
a.
g ( x) 2 x 2
Domain: x x is any real number .
3 2
e.
f ( x) 2 x 1
f ( x) x 1
a.
3 Domain: x x . 2
72.
f 2(1) 1 2 1 3 3 (1) g 3(1) 2 32 1
f 11 0 0 0 (1) 2 g 2(1) 2 2(1)
f ( x) 2 x 2 3
238
Copyright © 2025 Pearson Education, Inc.
g ( x) 4 x3 1
Section 3.1: Functions
a.
( f g )( x) 2 x 2 3 4 x3 1 3
b.
4x 2x 4 Domain: x x is any real number .
b.
2
c.
3
d.
4 x3 2 x 2 2
Domain: x x is any real number .
3x 5 x
f 2 x2 3 ( x) 3 4x 1 g 3 4x 1 0
e.
4 x3 1
f.
g.
( f g )(4) 4(4)3 2(4) 2 2
h. 76.
g.
a.
256 96 8 3 363
b.
75.
f 2(1) 3 2(1) 3 2 3 5 1 (1) g 4(1)3 1 4(1) 1 4 1 5
f ( x) x
a.
c.
Domain: x x 0 .
( f g )( x) x x
( f g )( x) x x x x
Domain: x x is any real number . d.
g ( x) 3 x 5
( f g )( x) x 3 x 5
( f g )( x) x x
Domain: x x is any real number .
2
h.
g ( x) x
Domain: x x is any real number .
( f g )(2) 8(2)5 12(2)3 2(2) 2 3 8(32) 12(8) 2(4) 3
f 1 1 1 1 (1) g 3(1) 5 3 5 2 2
f ( x) x
4(64) 2(16) 2 256 32 2 222
( f g )(2) 3(2) 2 5 2 6 2 5 2 2
( f g )(3) 4(3)3 2(3) 2 4 108 18 4 130
( f g )(4) 4 3(4) 5 2 12 5 5
4(27) 2(9) 4
f.
( f g )(3) 3 3(3) 5 3 95 3 4
3 1 1 2 x3 x 3 4 4 2 3 2 Domain: x x . 2
e.
f x ( x) 3x 5 g x 0 and 3 x 5 0 5 3 5 Domain: x x 0 and x . 3
( f g )( x) 2 x 2 3 4 x3 1
8 x5 12 x3 2 x 2 3 Domain: x x is any real number .
d.
( f g )( x) x (3x 5) 3x x 5 x
Domain: x x 0 .
( f g )( x) 2 x 3 4 x 1 2 x 2 3 4 x3 1
c.
( f g )( x) x (3 x 5) x 3 x 5
Domain: x x 0 .
2
x f ( x) x g Domain: x x 0 .
e.
( f g )(3) 3 3 3 3 6
f.
( f g )(4) 4 4 4 4 0
g.
( f g )(2) 2 2 2 2 4
239 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
h.
77.
1 1 f 1 (1) 1 1 g
f ( x) 1
a.
1 x
g ( x)
( f g )( x) 1
c.
( f g )( x) 1
1 x
x4 Domain: x 1 x 4 .
1 1 2 1 x x x d.
1 1 1 x x
Domain: x x 0 . c.
Domain: x 1 x 4 . e.
78.
( f g )(4) 1
g.
( f g )(2)
h.
f (1) 1 1 2 g
f ( x) x 1
a.
f.
( f g )(4) 4 1 4 4 3 0 3 0 3
g.
( f g )(2) (2)2 5(2) 4 4 10 4 2
2 5 3 3
f.
( f g )(3) 3 1 4 3 2 1 2 1
1 x 1 f x x x 1 x x 1 d. ( x) 1 1 x 1 g x x Domain: x x 0 . 1
( f g )(3) 1
f x 1 x 1 ( x) 4 x 4 x g x 1 0 and 4 x 0 x 1 and x 4 x4
11 1 1 ( f g )( x) 1 2 xx x x
Domain: x x 0 .
e.
h.
1 1 1 1 3 2 (2) 2 2 4 4
79.
f 1 1 0 0 0 (1) g 4 1 3
f ( x)
a.
2x 3 3x 2
g ( x)
3x 2 0
( f g )( x) x 1 4 x
3x 2 x 2 3 Domain: x x 2 . 3
x4
4x 3x 2
2x 3 4x 3x 2 3x 2 2x 3 4x 6x 3 3x 2 3x 2
( f g )( x)
g ( x) 4 x
x 1 0 and 4 x 0 x 1 and x 4
Domain: x 1 x 4 . b.
x 1 4 x
x2 5x 4 x 1 0 and 4 x 0 x 1 and x 4
Domain: x x 0 . b.
( f g )( x)
( f g )( x) x 1 4 x x 1 0 and 4 x 0 x 1 and x 4 x4
Domain: x 1 x 4 .
240
Copyright © 2025 Pearson Education, Inc.
Section 3.1: Functions
b.
2x 3 4x 3x 2 3x 2 2x 3 4x 2x 3 3x 2 3x 2
( f g )( x)
a.
x 1 0
2 3 2 Domain: x x . 3 3x 2 x
b.
c.
2 3 2 Domain: x x . 3
2 2 x 1 x x x0
( f g )( x) x 1 x 1 0
and
x 1 Domain: x x 1, and x 0 .
2x 3 f 3x 2 2 x 3 3x 2 2 x 3 ( x) 4 x 3x 2 4 x 4x g 3x 2 3x 2 0 and x 0
d.
f ( x) g x 1 0
x 1 x x 1 2 2 x and x 0
x 1 Domain: x x 1, and x 0 .
3x 2 2 3
2 Domain: x x and x 0 . 3
e.
( f g )(3) 3 1
2 2 2 8 4 2 3 3 3 3
f.
( f g )(4) 4 1
2 1 5 4 2
( f g )(3)
6(3) 3 18 3 21 3 3(3) 2 9 2 7
g.
( f g )(2)
f.
( f g )(4)
2(4) 3 8 3 5 1 3(4) 2 12 2 10 2
h.
f 1 11 2 (1) 2 2 g
g.
( f g )(2)
e.
h.
80.
and
x 1 Domain: x x 1, and x 0 .
2 2 x 3 4 x 8 x 12 x ( f g )( x) 3 x 2 3 x 2 (3x 2) 2 3x 2 0
x
2 x x0
( f g )( x) x 1 x 1 0
3x 2 x
d.
and
x 1 Domain: x x 1, and x 0 .
3x 2 0
c.
2 x x0
( f g )( x) x 1
8(2) 2 12(2)
3(2) 2 2 8(4) 24
6 2
2
32 24
4
2
f 2(1) 3 2 3 5 (1) 4(1) 4 4 g
f ( x) x 1
g ( x)
2 x
81.
56 7 16 2
f ( x) 3x 1
2 2 1 2 3 3 2 2
( f g )( x) 6
1 x 3 x 1 g ( x) 2 7 5 x g ( x) 2 7 g ( x) 5 x 2
6
241 Copyright © 2025 Pearson Education, Inc.
1 x 2
Chapter 3: Functions and Their Graphs
82.
f ( x)
1 x
f ( x h) f ( x ) h 3( x h) 2 2 (3 x 2 2) h 2 3 x 6 xh 3h 2 2 3 x 2 2 h 6 xh 3h 2 h 6 x 3h
f x 1 ( x) 2 g x x
1 x x 2 x g ( x) 1 1 x2 x g ( x) x x 1 x x 1 x2 x 1 x( x 1) x 1 x x 1 x 1 x 1
83.
87.
f ( x h) f ( x ) h ( x h) 2 ( x h) 4 ( x 2 x 4) h 2 2 x 2 xh h x h 4 x 2 x 4 h 2 xh h 2 h h 2x h 1
f ( x) 4 x 3
f ( x h) f ( x) 4( x h) 3 (4 x 3) h h 4 x 4h 3 4 x 3 h 4h 4 h
84.
f ( x) 3x 1 f ( x h) f ( x) 3( x h) 1 (3x 1) h h 3x 3h 1 3x 1 h 3h 3 h
85.
88.
f x 3x 2 2 x 6 f x h f x h 3 x h 2 2 x h 6 3 x 2 2 x 6 h
f ( x) x 2 4 f ( x h) f ( x ) h ( x h) 2 4 ( x 2 4) h 2 x 2 xh h 2 4 x 2 4 h 2 2 xh h h 2x h
86.
f ( x) x 2 x 4
3 x 2 2 xh h 2 2 x 2h 6 3x 2 2 x 6
h 3x 6 xh 3h 2h 3 x 2 6 xh 3h 2 2h h h 6 x 3h 2 2
f ( x) 3 x 2 2
242
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2
Section 3.1: Functions
89.
f ( x)
5 4x 3
91.
2
1 x3
1 1 f ( x h) f ( x ) x h 3 x 3 h h x 3 x 3 h x h 3 x 3 h x 3 x 3 h 1 x h 3 x 3 h 1 h x h x 3 3 h 1 x h 3 x 3
2
2 x 6 x 2hx 6h 2 x 6 x 2 xh
x h 3 x 3
h
1 6h x h 3 x 3 h
92. f ( x)
2x x3
2( x h) 2x f ( x h) f ( x ) x h 3 x 3 h h 2( x h)( x 3) 2 x x 3 h x h 3 x 3 h
5 5 f ( x h) f ( x) 4( x h) 3 4 x 3 h h 5(4 x 3) 5 4( x h) 3 4( x h) 3 4 x 3 h 20 x 15 20 x 15 20h 1 4( x h) 3 4 x 3 h 1 20h 4( ) 3 4 3 x h x h 20 4( x h) 3 4 x 3
90.
f ( x)
f ( x)
6 3 x 3 x h
5x x4
5( x h) 5x f ( x h) f ( x ) x h 4 x 4 h h 5( x h)( x 4) 5 x x 4 h x h 4 x 4 h
243 Copyright © 2025 Pearson Education, Inc.
5 x 20 x 5hx 20 h 5 x 20 x 5 xh 2
2
x h 4 x 4 h
1 20h x h 4 x 4 h
20 x h 4 x 4
Chapter 3: Functions and Their Graphs
93.
f x
x2
95.
f x h f x
94.
h
f ( x h) f ( x ) x h h h
2
2
h
x x 2 2 xh h 2 x x h 2
x h 2 x 2
2
h 1 2 h2 xh h x 2 x h 2
1 xh2 x2
1 h 2 x h h x 2 x h 2
f ( x) x 1 f x h f x
h x h 1 x 1 h x h 1 x 1 x h 1 x 1 h x h 1 x 1 x h 1 ( x 1) h h x h 1 x 1 h x h 1 x 1
x2 x h
x h 2 x 2
1 x2
x2 x h
h
h
2
xh2 x2 h 2 xh x2 xh2 x2 h xh2 x2 xh2 x2
1 x2 1
h
f x
96.
f x
2 x h x x h 2
2
2x h x2 x h
2
1 2
x 1
1 f ( x h) f ( x ) x h 1 h h 2
1 x 1
2
x2 1 x h 1
1
2
( x 1)( x h 1) 2
2
x h 1 x 1
2
h
x 1 x 2 2 xh h 2 1 ( x 1)( x h 1) 2
2
h 2 xh h 2
1 h ( x 2 1)( x h 2 1) h 2 x h 1 2 h ( x 1)( x h 2 1)
244
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2 x h ( x 1)( x h 1) 2
2
2x h ( x 1)( x h 1) 2
2
Section 3.1: Functions
97.
f ( x) 4 x 2
99.
f x h f x
0 x2 2 x 8 0 ( x 4)( x 2) x 4 0 or x 2 0 x4 or x 2
h 4 ( x h) 2 4 x 2
h 4 ( x h) 2 4 x 2
h
h
98.
4 ( x h) 2 4 x 2
100.
4 ( x h) 4 x 2
The solution set is: 2, 4
4 ( x h) 2 4 x 2
4 ( x h) 2 (4 x 2 )
2
4 ( x 2 2 xh h 2 ) (4 x 2 ) h
x h 1 x 1 2 xh h 2
h
x h 1 x 1 2 x h 4 ( x h) 2 4 x 2 4 ( x h) 2 4 x 2
101.
1
x2 xh2 x2 xh2
102.
x2 xh2
h x2 xh2 x2 xh2 x 2 ( x h 2) h( x 2) x h 2 ( x h 2) x 2 x2 xh2 h( x 2) x h 2 ( x h 2) x 2 h h( x 2) x h 2 ( x h 2) x 2 1 ( x 2) x h 2 ( x h 2) x 2
f ( x) 3x 2 Bx 4 and f (1) 12 : f (1) 3(1) 2 B (1) 4 12 3 B 4 B5
f ( x) 2 x3 Ax 2 4 x 5 and f (2) 5
f (2) 2(2)3 A(2) 2 4(2) 5 5 16 4 A 8 5 5 4 A 19 14 4 A 14 7 A 4 2
x2 f ( x h) f ( x ) h 1 1 xh2 x2 h
h x2 xh2
7 5 3 x 16 6 4 7 3 5 x 16 4 6 5 7 12 x 6 16 16 5 5 x 6 16 5 6 3 x 16 5 8
3 The solution set is: 8
(2 x h)
f x
11 x 2 2 x 3
103.
3x 8 and f (0) 2 2x A 3(0) 8 f (0) 2(0) A 8 2 A 2A 8 A 4 f ( x)
245 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
104.
2x B 1 and f (2) 3x 4 2 2(2) B f (2) 3(2) 4 1 4B 2 10 5 4B
b.
f ( x)
15 20 4.9 x 2 5 4.9 x 2 x 2 1.0204 x 1.01 seconds H x 10 :
B 1
10 20 4.9 x 2 10 4.9 x 2
105. Let x represent the length of the rectangle. x Then, represents the width of the rectangle 2 since the length is twice the width. The function x x2 1 2 for the area is: A( x ) x x 2 2 2
x 2 2.0408 x 1.43 seconds H x 5 : 5 20 4.9 x 2 15 4.9 x 2
106. Let x represent the length of one of the two equal sides. The function for the area is: 1 1 A( x ) x x x 2 2 2
x 2 3.0612 x 1.75 seconds
c.
107. Let x represent the number of hours worked. The function for the gross salary is: G ( x) 16 x
H 1 20 4.9 1
H x 0
0 20 4.9 x 2 20 4.9 x 2 x 2 4.0816
108. Let x represent the number of items sold. The function for the gross salary is: G ( x) 10 x 100 109. a.
H x 15 :
x 2.02 seconds
110. a.
H 1 20 13 1 20 13 7 meters 2
H 1.1 20 13 1.1 20 13 1.21 2
2
20 15.73 4.27 meters
20 4.9 15.1 meters H 1.1 20 4.9 1.1
2
H 1.2 20 13 1.2 20 13 1.44 2
20 4.9 1.21
20 18.72 1.28 meters
20 5.929 14.071 meters H 1.2 20 4.9 1.2
b.
2
H x 15 15 20 13 x 2
20 4.9 1.44
5 13 x 2
20 7.056 12.944 meters
x 2 0.3846 x 0.62 seconds H x 10 10 20 13 x 2 10 13 x 2 x 2 0.7692 x 0.88 seconds
246
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Section 3.1: Functions H x 5
2
c.
5 20 13x 2
2 8 5 8 5 2 2 A 4 1 3 3 3 3 9 3 3
15 13 x 2
2
x 1.1538 x 1.07 seconds
c.
L x L 113. R x x P x P
H x 0 0 20 13x 2 20 13x 2
114. T x V P x V x P x
x 2 1.5385
115. H x P I x P x I x
x 1.24 seconds
116. N x I T x I x T x
x 36, 000 111. C x 100 x 10 a.
117. a.
450 36, 000 C 450 100 10 450 100 45 80 $225
c.
600 36, 000 10 600 100 60 60 $220
d.
0.05 x 3 0.8 x 2 155 x 500
b.
118. a.
400 36, 000 10 400 100 40 90
P (20) 0.027(20) 2 6.530(20) 363.804
P (0) 0.027(0) 2 6.530(0) 363.804 363.804 In 2015 there are 363.804 million people.
119. a.
R (v) 2.2v; B (v) 0.05v 2 0.4 v 15 D (v ) R (v ) B (v )
8 2 1.26 ft 2 9
2.2v 0.05v 2 0.4 v 15 0.05v 2 2.6v 15
2
b.
P is the dependent variable; a is the independent variable
244.004 In 2015 there are 244.004 million people who are 20 years of age or older.
2
When 15 hundred smartphones are sold, the profit is $1836.25.
10.8 130.6 363.804
c.
1 4 8 4 2 2 1 1 A 4 1 3 3 9 3 3 3 3
2
$1836.25
$230
a.
3
P (15) 0.05(15) 0.8(15) 155(15) 500 168.75 180 2325 500
C 400 100
112. A x 4 x 1 x
1 3 3 1 1 A 4 1 2 2 2 4 2 2 2
1.2 x 2 220 x 0.05 x 3 2 x 2 65 x 500
b.
2
1.2 x 2 220 x 0.05 x 3 2 x 2 65 x 500
c.
C 600 100
P ( x) R( x) C ( x)
500 36, 000 10 500 100 50 72
C 500 100 $222
b.
8 5 1.99 ft 2 9
b.
2
D (60) 0.05(60) 2.6(60) 15
3 1.73 ft 2
180 156 15 321
247 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs c. 120. a.
The car will need 321 feet to stop once the impediment is observed.
c.
F a b 5 a b 2 5a 5b 2
h x 2x
Since 5a 5b 2 5a 2 5b 2 F a F b ,
h a b 2 a b 2a 2b
F x 5 x 2 does not have the property.
h a h b h x 2 x has the property.
b.
F x 5x 2
d.
g x x2
G x
1 x
G a b
g a b a b a 2 2ab b 2 2
Since a 2 2ab b 2 a 2 b 2 g a g b ,
G x
1 1 1 G a G b ab a b
1 does not have the property. x
g ( x) x 2 does not have the property.
121.
f ( x h) f ( x ) 3 x h 3 x h h 1
1
x h 3 x 3 h 1
1
2
1
1
2
2
1
1
2
x h 3 x 3 ( x h) 3 x 3 ( x h) 3 x 3
h
( x h) 3 x 3 ( x h) 3 x 3 h
2
1
1
2
( x h) 3 x 3 ( x h) 3 x 3
xhx h ( x h) 3 x 3 ( x h) 3 x 3 1 2
2
1
1
1
1
2
h h ( x h) 3 x 3 ( x h) 3 x 3 2
1
1
2
2
( x h) 3 x 3 ( x h) 3 x 3
122.
x4 2 f 3x 2 5x 4 x4 1. Solve 5x 4 x4 1 5x 4 x 4 5x 4
123. We need
x2 1 0 . Since x 2 1 0 for all 7 3x 1
real numbers x, we need 7 3 x 1 0 . 7 3x 1 0 3x 1 7 7 3 x 1 7
x2
2 x
Therefore, f 1 3(2) 2 10 2
8 3
8 8 The domain of f is x | 2 x , or 2, 3 3 in interval notation.
248
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Section 3.1: Functions 124. No. The domain of f is x x is any real number , but the domain of g is
x x 1 . 125.
3x x3 ( your age)
129. Let x represent the amount of the 7% fat hamburger added. % fat tot. amt. amt. of fat 20% 12 0.2012 7% x 0.07 x 15% 12 x 0.1512 x
0.2012 0.07 x 0.1512 x
126. Answers will vary.
2.4 0.07 x 1.8 0.15 x
127. ( x 12)2 y 2 16 x-intercept (y=0): ( x 12) 2 02 16
0.6 .08 x x 7.5 7.5 lbs. of the 7% fat hamburger must be added, producing 19.5 lbs. of the 15% fat hamburger.
( x 12) 2 16 ( x 12) 4
x 3 2 x 2 9 x 18 0
x 16, x 8 ( 16, 0), ( 8, 0) y-intercept (x=0): (0 12) 2 y 2 16
( x 3 2 x 2 ) (9 x 18) 0 x 2 ( x 2) 9( x 2) 0 ( x 2 9)( x 2) 0
(12) 2 y 2 16
( x 3)( x 3)( x 2) 0 ( x 3) 0 or ( x 3) 0 or ( x 2) 0
2
y 16 144 128 There are no real solutions so there are no yintercepts. Symmetry: ( x 12) 2 ( y ) 2 16 ( x 12) 2 y 2 16 This shows x-axis symmetry.
x 3, x 3, x 2 The solution set is: 3, 3, 2
131.
y 3( 1) 2 8 1 There is no solution so (-1,-5) is NOT a solution. y 3x2 8 x 48 16 32 So (4,32) is a solution. y 3x2 8 x y 3(9) 2 8 9 243 24 219 171 So (9,171) is NOT a solution.
a bx ac d a ac d bx a(1 c) d bx
128. y 3 x 2 8 x
y 3(4) 2 8 4
x 3 9 x 2 x 2 18
130.
x 12 4
a
132.
d bx 1 c
r kd 2 0.4 k (0.6) 2 10 k 9 Thus, 10 r (1.5) 2 9 2.5 kg m 2
249 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs 3. vertical 133. 3x 10 y 12 10 y 3x 12 3 6 y x 10 5
f 5 3
5.
f x ax 2 4 a 1 4 2 a 2 2
3 . The slope of a 10 10 perpendicular line would be . 3
6. False. The graph must pass the Vertical-Line Test in order to be the graph of a function.
(4 x 2 7) 3 (3 x 5) 8 x
7. False; e.g. y
The slope of the line is
134.
4.
(4 x 2 7) 2 12 x 2 21 (24 x 2 40 x) (4 x 2 7) 2 2
2
12 x 21 24 x 40 x 2
(4 x 7)
2
8. True
9. c 2
12 x 40 x 21
10. a
(4 x 2 7) 2
11. a.
12 x 2 40 x 21
b.
f (6) 0 since (6, 0) is on the graph. f (11) 1 since (11, 1) is on the graph.
Section 3.2
c.
f (3) is positive since f (3) 3.7.
d.
f (4) is negative since f (4) 1.
e.
f ( x) 0 when x 3, x 6, and x 10.
f.
f ( x) 0 when 3 x 6, and 10 x 11.
g.
The domain of f is x 6 x 11 or
2
1. x 4 y 16 x-intercepts:
6, 11 .
x 2 4 0 16 2
x 2 16
h.
x 4 4, 0 , 4, 0
The range of f is y 3 y 4 or
3, 4 .
y-intercepts:
0 4 y 2 16 2
4 y 2 16 y2 4 y 2 0, 2 , 0, 2
2. False;
f (0) 3 since (0,3) is on the graph. f ( 6) 3 since ( 6, 3) is on the graph.
(4 x 2 7) 2
135. Add the powers of x to obtain a degree of 7.
2
1 . x
x 2y 2 2 2 y 2 0 2y 0 y
i.
The x-intercepts are 3 , 6, and 10.
j.
The y-intercept is 3.
k.
The line y
l.
The line x 5 intersects the graph 1 time.
m.
f ( x) 3 when x 0 and x 4.
n.
f ( x) 2 when x 5 and x 8.
12. a.
1 intersects the graph 3 times. 2
f (0) 0 since (0, 0) is on the graph. f (6) 0 since ( 6, 0) is on the graph.
The point 2, 0 is on the graph. 250
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Section 3.2: The Graph of a Function
b.
f (2) 2 since (2, 2) is on the graph. f (2) 1 since (2, 1) is on the graph.
c.
Symmetry about y-axis.
16. Function
c.
f (3) is negative since f (3) 1.
d.
f (1) is positive since f (1) 1.0.
e.
f ( x) 0 when x 0, x 4, and x 6.
b. Intercepts: , 0 , , 0 , (0, 0)
f.
f ( x) 0 when 0 x 4.
c.
g.
The domain of f is x 4 x 6 or
4, 6 . h. The range of f is y 2 y 3 or 2, 3 . i.
The x-intercepts are 0, 4, and 6.
j.
The y-intercept is 0.
l.
The line x 1 intersects the graph 1 time.
m.
f ( x) 3 when x 5.
n.
f ( x) 2 when x 2.
Domain: x x 1 or x 1 ;
b. Intercepts: (1, 0), (1, 0)
17. Not a function since vertical lines will intersect the graph in more than one point. a.
Symmetry about the x-axis, y-axis and the origin
c.
Symmetry about the x-axis
18. Not a function since vertical lines will intersect the graph in more than one point.
Domain: x 2 x 2 ;
b. Intercepts: (2, 0)(2, 0)(0, 2)(0, 2) c.
Domain: x x is any real number ;
Symmetry about the x-axis, y-axis and the origin
19. Function
Domain: x 0 x 3 ; Range: y y <2
b. Intercepts: (1, 0) c.
Range: y y 0
None
20. Function a.
Domain: x 0 x 4 ; Range: y 0 y 3
b. Intercepts: (0,1)
None
b. Intercepts: (0, 0) c.
15. Function a.
Domain: x x 0 ; Range: y y is any real number
a.
14. Function
c.
Symmetry about the origin.
Range: y 2 y 2
Range: y y is any real number
a.
Range: y 1 y 1
a.
13. Not a function since vertical lines will intersect the graph in more than one point.
c.
Domain: x x ;
b. Intercepts: (0, 0)
k. The line y 1 intersects the graph 2 times.
a.
a.
Domain: x x ; Range: y 1 y 1
21. Function a.
b. Intercepts: , 0 , , 0 , (0,1) 2 2
None Domain: x x is any real number ; Range: y y 2
b. Intercepts: (–3, 0), (3, 0), (0,2)
251 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs c.
Symmetry about y-axis.
26.
22. Function a. Domain: x x 3 ;
a.
b.
b. Intercepts: (–3, 0), (2,0), (0,2) c. None
c.
Domain: x x is any real number ; Range: y y 3
Range: y y 5
x 3 x 5 0 x 0, x 5 3
f.
2
27.
f ( x)
f (1) 3 1 1 2 2 2
a.
f (2) 3 2 2 2 8 2
The point 2,8 is on the graph of f. c.
b.
Solve for x : 2 3 x 2 x 2 0 3x x
c.
0 x 3x 1 x 0, x 1 3 1 (0, –2) and , 2 are on the graph of f . 3
d. The domain of f is x x is any real number .
x-intercepts: f x =0 3 x 2 x 2 0
3x 2 x 1 0 x f.
x2 x6
3 2 5 14 36 3 The point 3,14 is not on the graph of f. f (3)
f (4)
42 6 3 46 2
The point 4, 3 is on the graph of f.
2
e.
y-intercept: f 0 3 0 5 0 0
None
The point 1, 2 is on the graph of f. b.
Domain: x x is any real number ;
f ( x) 3 x 2 x 2
a.
Solve for x : 2 3 x 2 5 x 3 x 2 5 x 2 0 3x 1 x 2 0 x 13 , x 2 (2, –2) and 1 , 2 on the graph of f . 3
x-intercepts: f x =0 3 x 2 5 x 0
e.
b. Intercepts: (–1, 0), (2,0), (0,4)
25.
2
d. The domain of f is x x is any real number .
None
24. Function
c.
f (2) 3 2 5 2 = 22
b. Intercepts: (1, 0), (3,0), (0,9)
a.
2
The point 2, 22 is on the graph of f.
23. Function
c.
f (1) 3 1 5 1 8 2
The point 1, 2 is not on the graph of f.
Range: y y 0
a.
f ( x) 3x 2 5 x
Solve for x : x2 2 x6 2 x 12 x 2 x 14 (14, 2) is a point on the graph of f .
d. The domain of f is x x 6 .
2 , x 1 3
e.
x-intercepts:
x2 0 x6 x 2 0 x 2 f x =0
y-intercept: f 0 =3 0 0 2 2 2
252
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Section 3.2: The Graph of a Function
f. y-intercept: f 0
28.
f ( x)
a.
b.
c.
c.
02 1 06 3
1
x 2 x4
2
12 2 3 1 4 5 3 The point 1, is on the graph of f. 5
(3 x 2 1)(4 x 2 1) 0
f (1)
3x 2 1 0 x
Solve for x : 1 x2 2 x 4 2x2 4 2 x4 0 2 x2 x 1 x 2 x 1 0 x 0 or x 2 1 1 1 0, and , are on the graph of f . 2 2 2
x-intercepts: x2 2 0 x2 2 0 x4 This is impossible, so there are no xintercepts. f x =0
f.
29.
y-intercept: 02 2 2 1 f 0 04 4 2
f ( x)
e.
a.
b.
12(1) 4 12 6 (1) 2 1 2 The point (–1,1) is on the graph of f. 12(3) 972 486 2 5 (3) 1 10 486 The point 3, is on the graph of f. 5 f (3)
12 x 4
x2 1 4 12 x 0 x 0
f.
3 3
30.
f ( x)
a.
0
y-intercept: f 0
f (1)
4
x-intercept: f x =0
c.
x2 1
3
d. The domain of f is x x is any real number .
b.
12 x 4
1
3 3 ,1 , ,1 are on the graph of f . 3 3
2
0 2 2 1 04 4 2 1 The point 0, is on the graph of f. 2 f (0)
12 x 4
x2 1 x 1 12 x 4 4 2 12 x x 1 0
2
d. The domain of f is x x 4 . e.
Solve for x :
12 0 2
4
0 1
0 0 0 1
2x x2
1 2 1 2 2 1 f 1 3 3 2 2 2 2 1 2 The point , is on the graph of f. 2 3 2(4) 8 4 42 2 The point 4, 4 is on the graph of f. f (4)
Solve for x : 2x 1 x 2 2x 2 x x2 (–2,1) is a point on the graph of f .
d. The domain of f is x x 2 . e.
x-intercept: 2x 0 2x 0 x2 x0
f x =0
f.
y-intercept: f 0
253 Copyright © 2025 Pearson Education, Inc.
0 0 02
Chapter 3: Functions and Their Graphs ( f g )(2) f (2) g (2) 2 1 3
5,13.2 , and 15,10 . The complete graph
b.
( f g )(4) f (4) g (4) 1 (3) 2
is given below.
c.
( f g )(6) f (6) g (6) 0 1 1
d.
( g f )(6) g (6) f (6) 1 0 1
e.
( f g )(2) f (2) g (2) 2(1) 2
f.
f f (4) 1 1 (4) g g (4) 3 3
31. a.
32. h x a.
136 x 2 v2
2.7 x 3.5
33. h x
We want h 15 10 .
136 15 v
2
2
2.7 15 3.5 10
30, 600 v2
h x
2.7 x 3.5 302 which simplifies to 34 2 h x x 2.7 x 3.5 225
c.
Using the velocity from part (b), 2 34 h 9 9 2.7 9 3.5 15.56 ft 225 The ball will be 15.56 feet above the floor when it has traveled 9 feet in front of the foul line.
x6
44 8
2
h 8
b.
h 12
c.
From part (a) we know the point 8,10.4 is
126 x 2
b.
v2
a.
34
v 2 900 v 30 ft/sec The ball needs to be thrown with an initial velocity of 30 feet per second.
44 x 2
8 6 282 2816 14 784 10.4 feet
44 12
2
12 6 282 6336 18 784 9.9 feet
on the graph and from part (b) we know the point 12,9.9 is on the graph. We could evaluate the function at several more values of x (e.g. x 0 , x 15 , and x 20 ) to obtain additional points. h 0
d. Select several values for x and use these to find the corresponding values for h. Use the results to form ordered pairs x, h . Plot the
44 0
h 15
points and connect with a smooth curve. 2 34 h 0 0 2.7 0 3.5 3.5 ft 225 2 34 h 5 5 2.7 5 3.5 13.2 ft 225 2 24 h 15 15 2.7 15 3.5 10 ft 225 Thus, some points on the graph are 0,3.5 ,
h 20
2
282
0 6 6
44 15
2
282 44 20 282
2
15 6 8.4 20 6 3.6
Some additional points are 0, 6 , 15,8.4 and 20,3.6 . The complete graph is given
254
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Section 3.2: The Graph of a Function
below.
Solve: 1 x 2 0
1 x 1 x 0 Case1: 1 x 0
1 x 0
and
x 1
x 1
and
(i.e. 1 x 1) 6
Case2: 1 x 0
d.
h 15
44 15
and
1 x 0
x 1 and
x 1
(which is impossible)
2
15 6 8.4 feet 282 No; when the ball is 15 feet in front of the foul line, it will be below the hoop. Therefore it cannot go through the hoop.
Therefore the domain of A is x 0 x 1 . b. Graphing A( x ) 4 x 1 x 2
In order for the ball to pass through the hoop, we need to have h 15 10 . 10 11
44 15 v
2
2
44 15
15 6
2
c.
v2 v 2 4 225
When x 0.7 feet, the cross-sectional area is maximized at approximately 1.9996 square feet. Therefore, the length of the base of the beam should be 1.4 feet in order to maximize the cross-sectional area.
v 2 900 v 30 ft/sec The ball must be shot with an initial velocity of 30 feet per second in order to go through the hoop.
34. A( x ) 4 x 1 x a.
35. h( x)
2
Domain of A( x ) 4 x 1 x 2 ; we know that x must be greater than or equal to zero, since x represents a length. We also need 1 x 2 0 , since this expression occurs under a square root. In fact, to avoid Area = 0, we require x 0 and 1 x 2 0 .
32 x 2 1302
a.
h(100)
b.
h(300)
c.
h(500)
x 32(100) 2
100 1302 320, 000 100 81.07 feet 16,900 32(300) 2
300 1302 2,880, 000 300 129.59 feet 16,900
255 Copyright © 2025 Pearson Education, Inc.
32(500) 2
500 1302 8, 000, 000 500 26.63 feet 16,900
Chapter 3: Functions and Their Graphs
The ball is about 26.63 feet high after it has traveled 500 feet. d.
Solving h( x)
32 x 2 1302
g.
The ball travels approximately 275 feet before it reaches its maximum height of approximately 131.8 feet.
x0
32 x 2
x0 1302 32 x x 1 0 1302 32 x x 0 or 1 0 1302 32 x 1 1302 2 130 32 x
h. The ball travels approximately 264 feet before it reaches its maximum height of approximately 132.03 feet.
1302 528.13 feet 32 Therefore, the golf ball travels 528.13 feet. x
e.
y1 150
32 x 2 1302
x 4000 36. W (h) m 4000 h
a.
2
h 14110 feet 2.67 miles ; 2
0 5
f.
4000 W (2.67) 120 119.84 4000 2.67 On Pike's Peak, Amy will weigh about 119.84 pounds.
600
Use INTERSECT on the graphs of 32 x 2 y1 x and y2 90 . 1302
b. Graphing: 120
150
0
0 119.5
600
c.
5 150
0
5
Create a TABLE:
The weight W will vary from 120 pounds to about 119.7 pounds.
600
5
The ball reaches a height of 90 feet twice. The first time is when the ball has traveled approximately 115.07 feet, and the second time is when the ball has traveled about 413.05 feet.
d. By refining the table, Amy will weigh 119.95 lbs at a height of about 0.83 miles
256
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Section 3.2: The Graph of a Function
(4382 feet).
c.
C 50 51, 000
It costs the company $51,000 to produce 50 computers in a day. d. The domain is q | 0 q 80 . This e.
37. C ( x) 100 a.
indicates that production capacity is limited to 80 computers in a day.
Yes, 4382 feet is reasonable. x 36000 10 x
C (480) 100
600 36000 C (600) 100 10 600 $220
x | x 0
c.
Graphing:
The graph is curved down and rises slowly at first. As production increases, the graph rises more quickly and changes to being curved up.
f.
The inflection point is where the graph changes from being curved down to being curved up.
480 36000 10 480
$223
b.
e.
39. a.
C 0 $50
It costs $50 if you use 0 gigabytes. b.
C 5 $50
It costs $50 if you use 5 gigabytes. c.
C 15 $150
It costs $90 if you use 15 gigabytes. d.
d. The domain is g | 0 g 30 . This
TblStart 0; Tbl 50
indicates that there are at most 30 gigabytes in a month. e. e.
The cost per passenger is minimized to about $220 when the ground speed is roughly 600 miles per hour.
The graph is flat at first and then rises in a straight line.
40. g (2) 5 f (2) 4
Since f (2) (2) 2 4(2) c 12 c we have
38.
a.
C 0 5000
This represents the fixed overhead costs. That is, the company will incur costs of $5000 per day even if no computers are manufactured. b.
12 c 4 5 3 12 c 9 3 12 c 27 c 15 f (3) 32 4 3 15 12
C 10 19, 000
It costs the company $19,000 to produce 10 computers in a day. 257 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs 48.
g (5) 52 n 25 n
41.
f ( g (5)) f (25 n) 25 n 2 4 so,
25 n 2 25 n 4 n 21
g (n) n 2 n (21) 2 (21) 420 .
42. Answers will vary. From a graph, the domain can be found by visually locating the x-values for which the graph is defined. The range can be found in a similar fashion by visually locating the y-values for which the function is defined.
If an equation is given, the domain can be found by locating any restricted values and removing them from the set of real numbers. The range can be found by using known properties of the graph of the equation, or estimated by means of a table of values.
49. a.
2 hours elapsed; Sobia was between 0 and 3 miles from home.
0.5 hours elapsed; Sobia was 3 miles from home. c. 0.3 hours elapsed; Sobia was between 0 and 3 miles from home. d. 0.2 hours elapsed; Sobia was at home. e. 0.9 hours elapsed; Sobia was between 0 and 2.8 miles from home. f. 0.3 hours elapsed; Sobia was 2.8 miles from home. g. 1.1 hours elapsed; Sobia was between 0 and 2.8 miles from home. h. The farthest distance Sobia is from home is 3 miles. i. Sobia returned home 2 times. b.
43. The graph of a function can have any number of x-intercepts. The graph of a function can have at most one y-intercept (otherwise the graph would fail the Vertical-Line Test). 44. Yes, the graph of a single point is the graph of a function since it would pass the Vertical-Line Test. The equation of such a function would be something like the following: f x 2 , where x 7.
45. (a) III; (b) IV; (c) I; (d) V; (e) II 46. (a) II; (b) V; (c) IV; (d) III; (e) I
50. a.
47.
Michael travels fastest between 7 and 7.4 minutes. That is, 7, 7.4 .
b. Michael's speed is zero between 4.2 and 6 minutes. That is, 4.2, 6 .
Between 0 and 2 minutes, Michael's speed increased from 0 to 30 miles/hour. d. Between 4.2 and 6 minutes, Michael was stopped (i.e, his speed was 0 miles/hour). e. Between 7 and 7.4 minutes, Michael was traveling at a steady rate of 50 miles/hour. f. Michael's speed is constant between 2 and 4 minutes, between 4.2 and 6 minutes, between 7 and 7.4 minutes, and between 7.6 and 8 minutes. That is, on the intervals (2, 4), (4.2, 6), (7, 7.4), and (7.6, 8). c.
258
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Section 3.3: Properties of Functions 51. Answers (graphs) will vary. Points of the form (5, y) and of the form (x, 0) cannot be on the graph of the function. 52. The only such function is f x 0 because it is
the only function for which f x f x . Any other such graph would fail the Vertical-Line Test.
60. The car traveling north travels a distance or 25t and the car traveling west travels a distance of 35t where t is the time of travel. Using the Pythagorean we have: 402 (35t ) 2 (25t ) 2 1600 1225t 2 625t 2 1600 1850t 2
53. Answers may vary. 54.
t 2 0.8649 t 0.93 hours Converting to minutes we have 0.93(60) 55.8 minutes
f ( x 2) ( x 2) 2 ( x 2) 3 ( x 2 4 x 4) x 2 3 x2 5x 9
55. d (1 3) 2 (0 ( 6)) 2 2
( 2) ( 6)
61. 3x 4 7 and 5 2 x 13 3x 3 2 x 8 x 1 x 4
2
4 36 40 2 10
The solution set is 4,1 .
2 x ( 6) 3 2 y4 x4 3 2 y x8 3
56. y 4
62.
(5 x 2 7 x 2) (8 x 10) 5 x 2 7 x 2 8 x 10 5 x 2 15 x 12
57. Since the function contains a cube root then the domain is:
63.
3,10
, 2
1 58. (12) 36 2
59.
Section 3.3
x 6 x 6 x6 x 6 1 x6 x 6 x 6 x 6
1. 2 x 5 2. slope
y 83 5 1 x 3 2 5
3. x-axis: y y
y 5x2 1 y 5x2 1 y 5 x 2 1 different
y-axis: x x y 5x 1 2
y 5 x 2 1 same
259 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
origin: x x and y y
19. Yes. The local maximum at x 2 is 10.
y 5x 1 2
y 5x2 1
20. No. There is a local minimum at x 5 ; the local minimum is 0.
y 5 x 2 1 different
21.
f has local maxima at x 2 and x 2 . The local maxima are 6 and 10, respectively.
22.
f has local minima at x 8, x 0 and x 5 . The local minima are –4, 0, and 0, respectively.
23.
f has absolute minimum of 4 at x = –8.
24.
f has absolute maximum of 10 at x = 2.
The equation has symmetry with respect to the y-axis only. y y1 m x x1
4.
y 2 5 x 3 y 2 5 x 3
5. y x 2 9 x-intercepts: 0 x2 9
25. a.
Intercepts: (–2, 0), (2, 0), and (0, 3).
b.
Domain: x 4 x 4 or 4, 4 ;
x 2 9 x 3
Range: y 0 y 3 or 0, 3 .
y-intercept:
Increasing: [–2, 0] and [2, 4]; Decreasing: [–4, –2] and [0, 2]. d. Since the graph is symmetric with respect to the y-axis, the function is even.
c.
y 0 9 9 2
The intercepts are 3, 0 , 3, 0 , and 0, 9 . 6. increasing
26. a.
Intercepts: (–1, 0), (1, 0), and (0, 2).
b. Domain: x 3 x 3 or 3, 3 ;
7. even; odd
Range: y 0 y 3 or 0, 3 .
8. True
Increasing: [–1, 0] and [1, 3]; Decreasing: [–3, –1] and [0, 1]. d. Since the graph is symmetric with respect to the y-axis, the function is even. c.
9. True 10. False; odd functions are symmetric with respect to the origin. Even functions are symmetric with respect to the y-axis.
27. a.
Intercepts: (0, 1).
11. c
b. Domain: x x is any real number ;
12. d
Range: y y 0 or 0, .
13. Yes
c.
14. No, it is increasing.
d. Since the graph is not symmetric with respect to the y-axis or the origin, the function is neither even nor odd.
15. No 16. Yes 17.
28. a.
Intercepts: (1, 0).
b. Domain: x x 0 or 0, ;
f is increasing on the intervals
Range: y y is any real number .
8, 2 , 0, 2 , 5, 7 . 18.
Increasing: (, ) ; Decreasing: never.
c.
f is decreasing on the intervals:
Increasing: [0, ) ; Decreasing: never.
10, 8 , 2, 0 , 2,5 . 260
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Section 3.3: Properties of Functions d. Since the graph is not symmetric with respect to the y-axis or the origin, the function is neither even nor odd.
d. Since the graph is not symmetric with respect to the y-axis or the origin, the function is neither even nor odd.
Intercepts: (, 0), (, 0), and (0, 0) .
33. a.
b. Domain: x x or , ;
b.
29. a.
c.
d. Since the graph is symmetric with respect to the origin, the function is odd. 30. a.
34. a. b.
35. a.
b.
31. a.
Range: y 1 y 2 or 1, 2 . c.
Increasing: 2, 3 ; Decreasing: 1, 1 ; Constant: 3, 1 and 1, 2
d. Since the graph is not symmetric with respect to the y-axis or the origin, the function is neither even nor odd. 32. a.
Intercepts: 2.3, 0 , 3, 0 , and 0, 1 .
b. Domain: x 3 x 3 or 3, 3 ;
Range: y 2 y 2 or 2, 2 . c.
Increasing: 3, 2 and 0, 2 ; Decreasing: 2, 3 ; Constant: 2, 0 .
2
.
f has a local minimum value of –1 at
36. a.
f has a local maximum value of 1 at x 0.
b.
f has a local minimum value of –1 both at x and x .
37.
f ( x) 4 x3 f ( x) 4( x)3 4 x3 f x
1 5 1 Intercepts: , 0 , , 0 , and 0, . 3 2 2
b. Domain: x 3 x 3 or 3, 3 ;
f has a local maximum value of 1 at
x . 2
Increasing: , 0 ; Decreasing: 0, .
d. Since the graph is symmetric with respect to the y-axis, the function is even.
f has a local minimum value of 0 at both
x
b. Domain: x x or , ;
c.
f has a local maximum value of 2 at x 0. x 1 and x 1.
Intercepts: , 0 , , 0 , and (0, 1) . 2 2
Range: y 1 y 1 or 1, 1 .
f has a local minimum value of 0 at both x 2 and x 2.
Range: y 1 y 1 or 1, 1 . Increasing: , ; 2 2 Decreasing: , and , . 2 2
f has a local maximum value of 3 at x 0.
Therefore, f is odd. 38.
f ( x) 2 x 4 x 2 f ( x) 2( x) 4 ( x) 2 2 x 4 x 2 f x
Therefore, f is even. 39. g ( x) 10 x 2 g ( x) 10 ( x) 2 10 x 2 g x
Therefore, g is even. 40. h( x) 3 x3 5 h( x) 3( x)3 5 3 x3 5
h is neither even nor odd.
41. F ( x) 3 4 x F ( x) 3 4 x 3 4 x F x
Therefore, F is odd.
261 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
51.
42. G ( x) x
f has an absolute minimum of 1 at x 1. f has an absolute maximum of 4 at x = 3.
G ( x) x G is neither even nor odd.
f has an local minimum value of 1 at x 1. f has an local maximum value of 4 at x = 3.
43.
f ( x) x x
52.
f ( x) x x x x
f has no absolute maximum.
f is neither even nor odd.
44.
f has an absolute minimum of 1 at x 0. f has no local minimum. f has no local maximum.
f ( x) 3 2 x 2 1
53.
f ( x) 3 2( x) 2 1 3 2 x 2 1 f x
f has an absolute minimum of 0 at x 0. f has no absolute maximum.
Therefore, f is even.
f has an local minimum value of 0 at x 0. f has an local minimum value of 2 at x 3.
1 2 x 8 1 1 g ( x) 2 g x 2 ( x) 8 x 8 Therefore, g is even.
45. g ( x)
f has an local maximum value of 3 at x 2.
54.
f has an absolute maximum of 4 at x 2. f has no absolute minimum. f has an local maximum value of 4 at x 2. f has an local minimum value of 2 at x 0.
x 46. h( x ) 2 x 1 x x 2 h x h( x ) 2 ( x) 1 x 1 Therefore, h is odd.
55.
f has no local maximum or minimum.
56.
f has no absolute maximum or minimum. f has no local maximum or minimum.
x3 3x2 9 ( x )3 x3 2 h x h( x ) 2 3( x) 9 3 x 9 Therefore, h is odd.
57.
47. h( x)
48. F ( x)
f has no absolute maximum or minimum.
f x x3 3 x 2 on the interval 2, 2
Use MAXIMUM and MINIMUM on the graph of y1 x3 3x 2 .
2x x
2( x ) 2 x F x x x Therefore, F is odd.
F ( x)
49.
f has an absolute maximum of 4 at x 1. f has an absolute minimum of 1 at x 5. f has an local maximum value of 3 at x 3.
local maximum: f ( 1) 4 local minimum: f (1) 0
f has an local minimum value of 2 at x 2.
50.
f has an absolute maximum of 4 at x 4.
f is increasing on: 2, 1 and 1, 2 ;
f has an absolute minimum of 0 at x 5.
f is decreasing on: 1,1
f has an local maximum value of 4 at x 4. f has an local minimum value of 1 at x 1.
262
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Section 3.3: Properties of Functions
58.
f x x3 3 x 2 5 on the interval 1,3
Use MAXIMUM and MINIMUM on the graph of y1 x3 3 x 2 5 .
local maximum: f (0) 5 local minimum: f (2) 1
f is increasing on: 1, 0 and 2,3 ;
local maximum: f (0) 0 local minimum: f ( 0.71) 0.25 ; f (0.71) 0.25
f is decreasing on: 0, 2 59.
f x x5 x3 on the interval 2, 2
f is increasing on: 0.71, 0 and 0.71, 2 ;
Use MAXIMUM and MINIMUM on the graph of y1 x5 x3 . 0.5
2
f is decreasing on: 2, 0.71 and 0, 0.71 61.
f x 0.2 x3 0.6 x 2 4 x 6 on the
interval 6, 4
2
Use MAXIMUM and MINIMUM on the graph of y1 0.2 x3 0.6 x 2 4 x 6 .
0.5 0.5
2
2
0.5
local maximum: f ( 0.77) 0.19 local minimum: f (0.77) 0.19
f is increasing on: 2, 0.77 and 0.77, 2 ; f is decreasing on: 0.77, 0.77 60.
f x x 4 x 2 on the interval 2, 2
Use MAXIMUM and MINIMUM on the graph of y1 x 4 x 2 .
local maximum: f (1.77) 1.91 local minimum: f ( 3.77) 18.89
f is increasing on: 3.77,1.77 ; f is decreasing on: 6, 3.77 and 1.77, 4
263 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
62.
f x 0.4 x3 0.6 x 2 3 x 2 on the
f ( 1.87) 0.95 , f (0.97) 2.65
f is increasing on: 1.87, 0 and 0.97, 2 ;
interval 4,5
f is decreasing on: 3, 1.87 and 0, 0.97
Use MAXIMUM and MINIMUM on the graph of y1 0.4 x3 0.6 x 2 3 x 2 .
64.
f x 0.4 x 4 0.5 x3 0.8 x 2 2 on the
interval 3, 2 Use MAXIMUM and MINIMUM on the graph of y1 0.4 x 4 0.5 x3 0.8 x 2 2 .
local maximum: f (2.16) 3.25 local minimum: f ( 1.16) 4.05
f is increasing on: 1.16, 2.16 ; f is decreasing on: 4, 1.16 and 2.16,5 63.
f x 0.25 x 4 0.3x3 0.9 x 2 3 on the
interval 3, 2 Use MAXIMUM and MINIMUM on the graph of y1 0.25 x 4 0.3x3 0.9 x 2 3 . local maxima: f ( 1.57) 0.52 , f (0.64) 1.87 local minimum: 0, 2 f (0) 2
f is increasing on: 3, 1.57 and 0, 0.64 ; f is decreasing on: 1.57, 0 and 0.64, 2 65.
f ( x) 2 x 2 4 a. Average rate of change of f from x 0 to x2 f 2 f 0 20
2 2 4 2 0 4 2
local maximum: f (0) 3 local minimum: 264
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4 4 2
2
2 8 4 2
Section 3.3: Properties of Functions b.
Average rate of change of f from x = 1 to x = 3:
c.
2 3 4 2 1 4 f 3 f 1 3 1 2 14 2 16 8 2 2 Average rate of change of f from x = 1 to x = 4: 2
2
2 4 4 2 1 4 f 4 f 1 4 1 3 28 2 30 10 3 3
66.
2
2
b. Average rate of change of g from x 1 to x 1: g 1 g 1 1 1
c.
3 1 33 4 3 7 13 4 1 7 2 22 4 18 9 2 2
f ( x) x3 1 a. Average rate of change of f from x = 0 to x = 2:
2 1 0 1 f 2 f 0 20 2 7 1 8 4 2 2 b. Average rate of change of f from x = 1 to x = 3:
c.
3
3
3 1 1 1 f 3 f 1 3 1 2 26 0 26 13 2 2 Average rate of change of f from x = –1 to x = 1: 3
3
1 1 1 1 f 1 f 1 2 1 1
3
3
0 2 2 1 2 2
67. g x x3 4 x 7 a.
Average rate of change of g from x 3 to x 2 : g 2 g 3
13 4 1 7 13 4 1 7 2 4 10 6 3 2 2 Average rate of change of g from x 1 to x 3: g 3 g 1
68. h x x 2 2 x 3 a.
Average rate of change of h from x 1 to x 1: h 1 h 1 1 1
12 2 1 3 12 2 1 3 2 2 6 4 2 2 2
b. Average rate of change of h from x 0 to x 2: h 2 h 0 20 2 2 2 2 3 0 2 2 0 3 2 3 3 0 0 2 2
2 3
2 3 4 2 7 33 4 3 7 1 7 8 15 15 1 1
265 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs c.
Average rate of change of h from x 2 to x 5: h 5 h 2
71. g x x 2 2 a.
52 5 2 2 5 3 2 2 2 2 3 3 18 3 15 5 3 3
69.
Therefore, the average rate of change of g from 2 to 1 is 1 . b. From (a), the slope of the secant line joining 2, g 2 and 1, g 1 is 1 .We use the
point-slope form to find the equation of the secant line: y y1 msec x x1
f x 5x 2
Average rate of change of f from 1 to 3: y f 3 f 1 13 3 10 5 x 3 1 3 1 2 Thus, the average rate of change of f from 1 to 3 is 5. b. From (a), the slope of the secant line joining 1, f 1 and 3, f 3 is 5. We use the
a.
y 2 1 x 2 y 2 x 2 y x
72. g x x 2 1 a.
point-slope form to find the equation of the secant line: y y1 msec x x1 y 3 5x 5
b. From (a), the slope of the secant line joining 1, g 1 and 2, g 2 is 1. We use the
y 5x 2
point-slope form to find the equation of the secant line: y y1 msec x x1
f x 4 x 1
a.
Average rate of change of g from 1 to 2: y g 2 g 1 52 3 1 x 2 1 2 1 3 Therefore, the average rate of change of g from 1 to 2 is 1.
y 3 5 x 1
70.
Average rate of change of g from 2 to 1: y g 1 g 2 1 2 3 1 x 1 2 1 2 3
Average rate of change of f from 2 to 5: y f 5 f 2 19 7 x 52 52 12 4 3 Therefore, the average rate of change of f from 2 to 5 is 4 .
y 2 1 x 1 y 2 x 1 y x3
73. h x x 2 2 x
Average rate of change of h from 2 to 4: y h 4 h 2 8 0 8 4 x 42 42 2 Therefore, the average rate of change of h from 2 to 4 is 4. b. From (a), the slope of the secant line joining 2, h 2 and 4, h 4 is 4. We use the
a.
b. From (a), the slope of the secant line joining 2, f 2 and 5, f 5 is 4 . We use the
point-slope form to find the equation of the secant line: y y1 msec x x1 y 7 4 x 2
point-slope form to find the equation of the secant line: y y1 msec x x1
y 7 4 x 8 y 4 x 1
y 0 4 x 2 y 4x 8
266
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Section 3.3: Properties of Functions f 2 (2)3 12(2) 8 24 16
74. h x 2 x 2 x a.
b.
Average rate of change from 0 to 3: y h 3 h 0 15 0 x 30 30 15 5 3 Therefore, the average rate of change of h from 0 to 3 is 5 . From (a), the slope of the secant line joining 0, h(0) and 3, h(3) is 5 . We use the point-slope form to find the equation of the secant line: y y1 msec x x1 y 0 5 x 0
So there is a local minimum value of 16 at x 2 . 77. F x x 4 8 x 2 9 a.
g x x3 27 x
Since F x F x , the function is even. b.
Since the function is even, its graph has y-axis symmetry. The second local maximum value is 25 and occurs at x 2 .
c.
Because the graph has y-axis symmetry, the area under the graph between x 0 and x 3 bounded below by the x-axis is the same as the area under the graph between x 3 and x 0 bounded below the x-axis. Thus, the area is 50.4 square units.
g x x 27 x
x3 27 x
g x
Since g x g x , the function is odd. b.
78. G x x 4 32 x 2 144 a.
Since g x is odd then it is symmetric
f x x 12 x
b.
Since the function is even, its graph has y-axis symmetry. The second local maximum is in quadrant II and is 400 and occurs at x 4 .
c.
Because the graph has y-axis symmetry, the area under the graph between x 0 and x 6 bounded below by the x-axis is the same as the area under the graph between x 6 and x 0 bounded below the x-axis. Thus, the area is 1612.8 square units.
3
3
x 12 x
x3 12 x
f x
Since f x f x , the function is odd. b.
2
Since G x G x , the function is even.
f x x3 12 x a.
4
G x
g 3 (3)3 27(3) 27 81 54
76.
G x x 32 x 144 x 4 32 x 2 144
about the origin so there exist a local maximum at x 3 . So there is a local maximum of 54 at x 3 .
2
F x
3
x3 27 x
4
x4 8x 9
y 5 x
75. a.
F x x 8x 9
Since f x is odd then it is symmetric about the origin so there exist a local maximum at x 3 .
267 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs 1
2500 x 2500 2 y1 0.3 x 21x 251 x
79. C x 0.3 x 2 21x 251 a.
b. Use MINIMUM. Rounding to the nearest whole number, the average cost is minimized when approximately 10 lawnmowers are produced per hour.
81. a.
c. 80. a.
The minimum average cost is approximately $239 per mower.
b.
C t .002t .039t .285t .766t .085 4
3
2
Graph the function on a graphing utility and use the Maximum option from the CALC menu. 1
c.
The concentration will be highest after about 2.16 hours. b. Enter the function in Y1 and 0.5 in Y2. Graph the two equations in the same window and use the Intersect option from the CALC menu.
avg. rate of change
P 2.5 P 0
2.5 0 0.18 0.09 2.5 0 0.09 2.5 0.036 gram per hour On average, the population is increasing at a rate of 0.036 gram per hour from 0 to 2.5 hours. avg. rate of change
P 6 P 4.5
6 4.5 0.50 0.35 6 4.5 0.15 1.5 0.1 gram per hour On average, the population is increasing at a rate of 0.1 gram per hour from 4.5 to 6 hours.
The average rate of change is increasing as time passes. This indicates that the population is increasing at an increasing rate.
82. a.
1
After taking the medication, the woman can feed her child within the first 0.71 hours (about 42 minutes) or after 4.47 hours (about 4hours 28 minutes) have elapsed.
268
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Section 3.3: Properties of Functions b.
83.
f ( x) x 2
a.
Average rate of change of f from x 0 to x 1: f 1 f 0 12 02 1 1 1 0 1 1
b. Average rate of change of f from x 0 to x 0.5 : f 0.5 f 0 0.5 0
c.
d.
avg. rate of change
P 2014 P 2012
2014 2012 9512 9273 2 239 2 $119.50 dollars/yr
avg. rate of change
c.
f 0.1 f 0 0.1 0
avg. rate of change
f.
The average rate of change is increasing.
g.
avg. rate of change
2016 2014 10409 9938 2 471 2 $235.50 dollars/yr
0.5
0.25 0.5 0.5
0.12 02 0.1
0.01 0.1 0.1
d. Average rate of change of f from x 0 to x 0.01 : f 0.01 f 0 0.01 0
e.
0.01 0.0001 0.01 0.01
f 0.001 f 0 0.001 0
f.
0.012 02
Average rate of change of f from x 0 to x 0.001 :
P 2018 P 2016
e.
0.52 02
Average rate of change of f from x 0 to x 0.1 :
P 2016 P 2014
2013 2011 9938 9512 2 426 2 $213.00 dollars/yr
0.0012 02
0.001 0.000001 0.001 0.001
Graphing the secant lines:
P 2021 P 2019
2021 2019 9177 10507 2 1330 2 $665.00 dollars/yr
The debt decreased. h. Quarantine from Covid19 most likely had a effect on spending habits. 269 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs d. Average rate of change of f from x 1 to x 1.01 : f 1.01 f 1 1.01 1
e.
1.001 1
g.
The secant lines are beginning to look more and more like the tangent line to the graph of f at the point where x 0 .
f ( x) x 2
a.
Average rate of change of f from x 1 to x 2: f 2 f 1 22 12 3 3 2 1 1 1
b. Average rate of change of f from x 1 to x 1.5 : f 1.5 f 1 1.5 1
c.
1.52 12 0.5
1.25 2.5 0.5
Average rate of change of f from x 1 to x 1.1 : f 1.1 f 1 1.1 1
1.12 12 0.1
0.0201 2.01 0.01
1.0012 12
0.001 0.002001 2.001 0.001
Graphing the secant lines:
h. The slopes of the secant lines are getting smaller and smaller. They seem to be approaching the number zero. 84.
0.01
Average rate of change of f from x 1 to x 1.001 : f 1.001 f 1
f.
1.012 12
0.21 2.1 0.1
270
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Section 3.3: Properties of Functions g.
The secant lines are beginning to look more and more like the tangent line to the graph of f at the point where x 1 .
c.
slope = 3 , we get the secant line: y 1 3 x 1
h. The slopes of the secant lines are getting smaller and smaller. They seem to be approaching the number 2. 85.
y 1 3 x 3 y 3 x 2
d. Graphing:
f ( x) 2 x 5
a.
f ( x h) f ( x ) h 2( x h) 5 2 x 5 2h 2 h h
msec
b. When x 1 : h 0.5 msec 2 h 0.1 msec 2 h 0.01 msec 2 as h 0, msec 2 c.
Using point 1, f 1 1, 1 and
The graph and the secant line coincide. 87.
Using the point 1, f 1 1, 7 and slope,
f ( x) x 2 2 x
a.
f ( x h) f ( x ) h 2 ( x h) 2( x h) ( x 2 2 x) h 2 2 x 2 xh h 2 x 2h x 2 2 x h 2 2 xh h 2h h 2x h 2
msec
m 2 , we get the secant line: y 7 2 x 1 y 7 2x 2 y 2x 5
d. Graphing:
The graph and the secant line coincide. 86.
c.
f ( x) 3 x 2
a.
f ( x h) f ( x ) msec h 3( x h) 2 (3x 2) 3h 3 h h
b. When x = 1, h 0.5 msec 3 h 0.1 msec 3 h 0.01 msec 3 as h 0, msec 3
b. When x = 1, h 0.5 msec 2 1 0.5 2 4.5 h 0.1 msec 2 1 0.1 2 4.1 h 0.01 msec 2 1 0.01 2 4.01 as h 0, msec 2 1 0 2 4
Using point 1, f 1 1,3 and slope = 4.01, we get the secant line: y 3 4.01 x 1 y 3 4.01x 4.01 y 4.01x 1.01
d. Graphing:
271 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
88.
b. When x = 1, h 0.5 msec 4 1 2 0.5 3 2
f ( x) 2 x 2 x
a.
f ( x h) f ( x ) h 2( x h) 2 ( x h) (2 x 2 x) h 2 2( x 2 xh h 2 ) x h 2 x 2 x h 2 2 2 x 4 xh 2h x h 2 x 2 x h 2 4 xh 2h h h 4 x 2h 1
msec
h 0.1 msec 4 1 2 0.1 3 1.2 h 0.01 msec 4 1 2 0.01 3 1.02 as h 0, msec 4 1 2 0 3 1
c.
Using point 1, f 1 1, 0 and slope = 1.02, we get the secant line: y 0 1.02 x 1 y 1.02 x 1.02
d. Graphing:
b. When x = 1, h 0.5 msec 4 1 2 0.5 1 6 h 0.1 msec 4 1 2 0.1 1 5.2 h 0.01 msec 4 1 2 0.01 1 5.02
as h 0, msec 4 1 2 0 1 5
c.
Using point 1, f 1 1,3 and
90.
slope = 5.02, we get the secant line: y 3 5.02 x 1 y 3 5.02 x 5.02 y 5.02 x 2.02
f ( x) x 2 3x 2 f ( x h) f ( x ) a. msec h 2
d. Graphing:
x h 3 x h 2 x 2 3x 2 2
2
2
( x 2 xh h ) 3 x 3h 2 x 3 x 2 h
x 2 xh h 3 x 3h 2 x 3 x 2 h
2
2
2
2 xh h 2 3h h 2 x h 3
89.
f ( x) 2 x 2 3 x 1 f ( x h) f ( x ) a. msec h 2
b. When x = 1, h 0.5 msec 2 1 0.5 3 0.5 h 0.1 msec 2 1 0.1 3 0.9 h 0.01 msec 2 1 0.01 3 0.99 as h 0, msec 2 1 0 3 1
2 x h 3 x h 1 2 x 2 3x 1 h 2
2
c.
2
2( x 2 xh h ) 3 x 3h 1 2 x 3 x 1 h 2
2
Using point 1, f 1 1, 0 and slope = 0.99, we get the secant line: y 0 0.99 x 1
2
2 x 4 xh 2h 3 x 3h 1 2 x 3 x 1 h
h
y 0.99 x 0.99
4 xh 2h 2 3h h 4 x 2h 3
272
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Section 3.3: Properties of Functions d. Graphing:
d. Graphing:
91.
f ( x)
a.
1 x
92. f ( x h) f ( x ) h 1 1 x x h xh x x h x h h x x h 1 h 1 x h x h x h x h
msec
f ( x)
a.
f ( x h) f ( x ) h 1 1 2 2 x h x h x 2 x h 2 x h 2 x 2 h x 2 x 2 2 xh h 2 1 2 2 h x h x
msec
1 x h x
b. When x = 1, 1 h 0.5 msec 1 0.5 1
1 10 0.909 1.1 11 1 h 0.01 msec 1 0.011
c.
1 100 0.990 1.01 101 1 1 msec 1 1 0 1 1
Using point 1, f 1 1,1 and 100 slope = , we get the secant line: 101 100 y 1 x 1 101 100 100 y 1 x 101 101 100 201 y x 101 101
2 xh h 2 1 x h 2 x 2 h 2 x h 2 x h 2 2 2 x 2 xh h 2 x 2 x h x
1 2 0.667 1.5 3 1 h 0.1 msec 1 0.11
as h 0,
1 x2
b. When x = 1, h 0.5 msec h 0.1 msec
2 1 0.5
1 0.5 1
2 2
2 1 0.1
1 0.1 1
h 0.01 msec
2 2
10 1.1111 9
210 1.7355 121
2 1 0.01
1 0.012 12
20,100 1.9704 10, 201 2 1 0 2 as h 0, msec 1 0 2 12
273 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
Using point 1, f 1 1,1 and
c.
95. Answers will vary. One possibility follows: y
slope = 1.9704 , we get the secant line: y 1 1.9704 x 1
2 (3, 0)
y 1 1.9704 x 1.9704
y 1.9704 x 2.9704
4
(1, 2) (0, 3)
d. Graphing: 5
(2, 6)
96. Answers will vary. See solution to Problem 89 for one possibility. 93.
f (2) 12 and f (1) 8, so
97. A function that is increasing on an interval can have at most one x-intercept on the interval. The graph of f could not "turn" and cross it again or it would start to decrease.
f (2) f (1) 12 4 2 (1) 3 f ( x) 4 3x 2 4 x 1 4
98. An increasing function is a function whose graph goes up as you read from left to right.
2
3x 4 x 5 0
y
4 42 4(3)(5) 4 76 x 2(3) 6
5
4 2 19 2 19 6 3
x
2 19 2.1 is outside the interval 3
x
2 19 .079 is in the interval. 3
3
3
5
2 19 . 3 x x 94. g ( x) 2 f 2 f . Since f is odd, 3 3 x x f f 3 3 x x g ( x) 2 f 2 f g ( x) 3 3 so g is odd. So g is odd. The only such number is
A decreasing function is a function whose graph goes down as you read from left to right. y 5
3
3
5
99. To be an even function we need f x f x
and to be an odd function we need f x f x . In order for a function be both 274
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Section 3.3: Properties of Functions
even and odd, we would need f x f x .
107.
This is only possible if f x 0 .
( x h) 2 ( y k ) 2 r 2 6 ( x 3) 2 ( y (2)) 2 2 3 ( x 3) 2 ( y 2) 2 2
100. The graph of y 5 is a horizontal line.
2
108.
x 2 2 3x 115x 3x 1 3 x 2 2 x 6 x x 2 3x 1 x 2 5 x 3x 1 6 x x 2 3x 1 5 x 10 x 3 x 1 6 x x 2 3x 1 8 x 10 x 1 3
2
The local maximum is y 5 and it occurs at each x-value in the interval. 101. Not necessarily. It just means f 5 f 2 .
The function could have both increasing and decreasing intervals. 102.
2
5
2
2
5
2
2
2
5
5
3
2
2
5
5
3
110.
3x 7 3 5 3x 7 8 3x 7 8 or 3 x 7 8 3 x 15 3x 1 1 x 5 x 3 1 The solution set is 5, . 3
540 36(15) 6 15
104. (4a b) 2 (4a b)(4a b)
16a 2 4ab 4ab b 2 16a 2 8ab b 2
105. C ( x) 0.80 x 40 111.
x6 7 x3 8 x6 7 x3 8 0 ( x3 8)( x3 1) 0 x3 8 0
or x3 1 0
x 3 8 x3 1 x 2 x 1 The solution set is 2,1 .
275 Copyright © 2025 Pearson Education, Inc.
3
2
3
x x y y2 109. 1 2 , 1 2 2 2 35 1 (4) 2 , 2
f ( x2 ) f ( x1 ) bb 0 x2 x1 x2 x1
k 106. S T k 33 10 k 330 330 S T 330 S 10 S 8.25 days
4
75 3 7 3 2 , 2 10 , 2
f (2) f (2) 0 0 0 4 2 2
103.
5
5
5
Chapter 3: Functions and Their Graphs
112.
3 y 2 D 3x 2 3 xy 2 3 x 2 y D 0
6. True
3 y 2 D 3 x 2 y D 3 x 2 3xy 2
7. False; the cube root function is odd and increasing on the interval , .
D 3 y 2 3x 2 y 3 x 2 3 xy 2 D
2 3 x 2 3 xy 2 xy 2 x 2 x y x y 2 x2 y y y x2 3 y 2 3x 2 y
8. False; the domain and range of the reciprocal function are both the set of real numbers except for 0. 9. b
Section 3.4 1. y
10. a
x
11. C 12. A 13. E 14. G 15. B 16. D 17. F
1 2. y x
3.
18. H 19.
f x x
20.
f x x2
y x3 8 y-intercept: Let x 0 , then y 0 8 8 . 3
x-intercept: Let y 0 , then 0 x3 8 x3 8 x2
The intercepts are 0, 8 and 2, 0 . 4.
, 0
5. piecewise-defined 276
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Section 3.4: Library of Functions; Piecewise-defined Functions
21.
f x x3
25.
f ( x) 3 x
22.
f x x
26.
f x 3
23.
1 f x x
27. a. b.
f (0) 4
c.
f (2) 3(3) 2 7
28. a.
24.
f x x
f ( 3) ( 3) 2 9
f 2 3 2 6
b.
f 1 0
c.
f 0 2 0 1 1
29. a.
2
f 2 2 2 4 0
b.
f 0 2 0 4 4
c.
f 1 2 1 4 6
d.
f 3 3 1 26
30. a.
f (1) ( 1)3 1
3
b.
f (0) 0 0
c.
f (1) 3(1) 2 5
3
277 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
d. 31.
f 3 3 3 2 11 if x 0 if x 0
2 x f ( x) 1
a.
33.
if x 1 2 x 3 f ( x) 3 x 2 if x 1 a. Domain: x x is any real number
b. x-intercept: none y-intercept: f 0 2 0 3 3
Domain: x x is any real number
The only intercept is 0,3 .
b. x-intercept: none y-intercept: f 0 1
c.
Graph:
The only intercept is 0,1 . c.
Graph:
d. Range: y y 1 ; 1, 34. d. Range: y y 0 ; , 0 0, 32.
a.
b.
if x 0 if x 0
3x f ( x) 4
if x 2 x 3 f ( x) if x 2 2 x 3 a. Domain: x x is any real number 2 x 3 0 2 x 3
Domain: x x is any real number
x
b. x-intercept: none y-intercept: f 0 4
x-intercepts: 3, y-intercept:
The only intercept is 0, 4 . c.
x3 0 x 3
3 2
3 2
f 0 2 0 3 3
3 The intercepts are 3, 0 , , 0 , and 2
Graph: y
0, 3 .
5
c.
Graph:
5
5
5
d. Range: y y 0 ; , 0 0, 278
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Section 3.4: Library of Functions; Piecewise-defined Functions d. Range: y y 1 ; , 1
35.
x 3 f ( x) 5 x 2
Graph:
if 2 x 1 if x 1 if x 1
a.
Domain: x x 2 ; 2,
b.
x3 0 x 3 (not in domain) x-intercept: 2
x 2 0 x 2 x2
y-intercept: f 0 0 3 3
c.
c.
d. Range: y y 5 ; , 5 37.
1 x f ( x) 2 x
if x 0 if x 0
The intercepts are 2, 0 and 0,3 .
a.
Domain: x x is any real number
Graph:
b.
1 x 0 x2 0 x 1 x0 x-intercepts: 1, 0
y-intercept:
f 0 02 0
The intercepts are 1, 0 and 0, 0 . c.
Graph:
d. Range: y y 4, y 5 ; , 4 5 36.
2 x 5 f ( x) 3 5 x
if 3 x 0 if x 0 if x 0
a.
Domain: x x 3 ; 3,
b.
2x 5 0 2 x 5 5 x 2
x-intercept:
5 x 0 x0 (not in domain of piece) 5 2
y-intercept: f 0 3
d. Range: y y is any real number 38.
1 if x 0 f ( x) x 3 x if x 0 a. Domain: x x is any real number
b.
5 The intercepts are , 0 and 0, 3 . 2
1 0 x (no solution) x-intercept: 0
3
x 0 x0
y-intercept: f 0 3 0 0 The only intercept is 0, 0 . 279 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs c.
Graph:
b.
2 x 0 x2
x 0 x0 (not in domain of piece)
no x-intercepts y-intercept: f 0 2 0 2 The intercept is 0, 2 . c.
y
d. Range: y y is any real number 39.
x f ( x) 3 x
a.
Graph: (3, 5)
5
if 2 x 0
(4, 2) (0, 2)
if x 0 5
Domain: x 2 x 0 and x 0 or
5
x | x 2, x 0 ; 2, 0 0, . 5
b. x-intercept: none There are no x-intercepts since there are no values for x such that f x 0 .
d. Range: y y 1 ; 1,
y-intercept: There is no y-intercept since x 0 is not in the domain. c.
41.
Graph:
2 x f ( x) x
a.
if 0 x 2 if 2 x 5 if 5
a.
Domain: x x 0 ; 0, .
b.
x2 0 x0 (not in domain of piece) x20 70 x 2 (not possible) (not in domain of piece) No intercepts.
d. Range: y y 0 ; 0, 40.
x2 f ( x) x 2 7
if 3 x 1 if x 1
Domain: x 3 x 1 and x 1 or
x | x 3, x 1 ; 3,1 1, .
280
Copyright © 2025 Pearson Education, Inc.
x
Section 3.4: Library of Functions; Piecewise-defined Functions c.
Graph:
43. Answers may vary. One possibility follows: if 1 x 0 x f ( x) 1 if 0 x 2 2 x 44. Answers may vary. One possibility follows: if 1 x 0 x f ( x) if 0 x 2 1
d. Range: y 0 y 7 ; 0, 7
42.
3x 5 if 3 x 0 if 0 x 2 f ( x) 5 2 x 1 if x 2
a.
Domain: x x 3 ; 3, .
b.
3x 5 0
50 x2 1 0 5 (not possible) x 2 1 x 3 (not possible)
x-intercept:
5 3
y-intercept: f 0 5 5 The intercepts are (0,5) and , 0 . 3
c.
Graph:
d. Range: 4,
45. Answers may vary. One possibility follows: if x 0 x f ( x) if 0 x 2 x 2 46. Answers may vary. One possibility follows: if 1 x 0 2 x 2 f ( x) if x 0 x 47. a.
f (1.7) int 2(1.7) int(3.4) 3
b.
f (2.8) int 2(2.8) int(5.6) 5
c.
f (3.6) int 2(3.6) int(7.2) 8
48. a.
1.2 f (1.2) int int(0.6) 0 2
b.
1.6 f (1.6) int int(0.8) 0 2
c.
1.8 f (1.8) int int( 0.9) 1 2
49. a.
b. The domain is 0, 6 . c.
Absolute max: f (2) 6 Absolute min: f (6) 2
281 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
For x 200 : C 100 1.5 x 50
50. a.
For x 200 : C 100 1.5 150 1.15( x 200) 325 1.15( x 200)
The cost function: if 0 x 50 2 x C 100 1.5( x 50) if 50 x 200 325 1.15( x 200) if x 200
b. The domain is 2, 2 .
d. Graph:
Absolute max: f ( 2) f (2) 3 Absolute min: none 0.13 x 17 if 0 x 3 51. C 0.4 x 23.5 if x 150 c.
a.
C 76 0.13 76 17 $26.88
b.
C 133 0.13 133 17 $34.29
c.
C 189 0.4 189 23.5 $52.10
3 3int 1 x 52. F x 6 4int 2 x 36 77
a.
0 x2 2 x4 4 x8 8 x 24
Taxes for $48,000: T 0.08(48000) $3840 b. Taxes for $132,000: T 0.08(75000) 0.16(57000) $15,120
54. a.
F 0.5 3 3int(1 0.5) 3
Parking for 0.5 hours costs $3. b.
c.
F 2 3 3int 1 2 6
Parking for 2 hours costs $6. c.
F 3.25 6 4int 2 3.25 14
Parking for 3.5 hours costs $14. d.
d. For 0 x 75, 000 : T 0.08 x
F 8 36
For 75, 000 x 200, 000 : T 0.08(75, 000) 0.16( x 75, 000) 6000 0.16( x 75, 000)
Parking for 8 hours costs $36. 53. a.
Taxes for $231,000: T 0.08(75000) 0.16(125000) 0.22(31000) $32,820
Charge for 2 dozen: C 2.00(24) $48.00
b.
Charge for 120 cupcakes: C 100 1.5( x 50) 100 1.5(120 50) $205
c.
For 0 x 50 : C 2x
For 200, 000 x 350, 000 : T 0.08(75, 000) 0.16(12,5000) 0.22( x 200, 000) 26000 0.22( x 200000) For x 350000 : T 59000 0.3( x 350000) 282
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Section 3.4: Library of Functions; Piecewise-defined Functions
The tax function: if 0.08 x 6000 0.16( x 75, 000) if T x 26, 000 0.22( x 200, 000) if 59, 000 0.3( x 350, 000) if
0 x 75, 000 75, 000 x 200, 000 200, 000 x 350, 000 x 350, 000
e.
55. For schedule X: 0.10 x 1100 0.12( x 11, 000) 5,175 0.22( x 44, 725) f ( x) 16, 290 0.24( x 95,375) 37,104 0.32( x 182,100) 52,832 0.35( x 231, 250) 174, 238 0.37( x 578,126) 56. For Schedule Y 1 : 0.10 x 2200 0.12( x 22, 000) 10, 294 0.22( x 89, 450) f ( x) 32,580 0.24( x 190, 750) 74, 208 0.32( x 364, 200) 105, 664 0.35( x 462,500) 186, 601 0.37( x 693, 750) 57. a.
if 0 x 11, 000 if 11,000 x 44, 725 if 44, 725 x 95,375 if 95,375 x 182,100 if 182,100 x 231, 250 if 231, 250 x 578,126 if x 578,126
if 0 x 22, 000 if 22, 000 x 89, 450 if 89, 450 x 190, 750 if 190, 750 x 364, 200 if 364, 200 x 462,500 if 462,500 x 693, 750 if x 693, 750
Let x represent the number of miles and C be the cost of transportation. if 0 x 100 0.50 x 0.50(100) 0.40( x 100) if 100 x 400 C ( x) if 400 x 800 0.50(100) 0.40(300) 0.25( x 400) 0.50(100) 0.40(300) 0.25(400) 0( x 800) if 800 x 960
283 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
0.50 x 10 0.40 x C ( x) 70 0.25 x 270
if 0 x 100 if 100 x 400 if 400 x 800 if 800 x 960
b. For hauls between 100 and 400 miles the cost is: C ( x) 10 0.40 x . c.
For hauls between 400 and 800 miles the cost is: C ( x) 70 0.25 x .
given by 9000 8250 5250 C s 3750 2250 1500
58. Let x = number of days car is used. The cost of renting is given by 185 if x 7 222 if 7 x 8 259 if 8 x 9 C x 296 if 9 x 10 333 if 10 x 11 370 if 11 x 14
if 720 s 739 if s 740
b. 725 is between 720 and 739 so the charge would be $2250. c.
59. a.
if s 659 if 660 s 679 if 680 s 699 if 700 s 719
670 is between 660 and 679 so the charge would be $8250.
60. Let x = the amount of the bill in dollars. The minimum payment due is given by x if 0 x 10 10 if 10 x 500 f x 30 if 500 x 1000 50 if 1000 x 1500 70 if x 1500
Let s = the credit score of an individual who wishes to borrow $300,000 with an 80% LTV ratio. The adverse market delivery charge is 284
Copyright © 2025 Pearson Education, Inc.
Section 3.4: Library of Functions; Piecewise-defined Functions 64. Use intervals 0,8 , 8,16 , 16,32 , 32,38
(exclude 0 and 38 since those would be the walls). Depth for the intervals 8,16 and 32,38 are constant (8 ft and 3 ft respectively). The other two are linear functions. On 0,8 the endpoint coordinates can be thought or as 0,3 and 8,8 . m W 10C
b.
W 33
c.
(10.45 10 15 15)(33 10) W 33 3C 22.04
d.
W 33 1.5958(33 10) 4C
e.
When 0 v 1.79 , the wind speed is so small that there is no effect on the temperature. When the wind speed exceeds 20, the wind chill depends only on the air temperature.
62. a. b.
(10.45 10 5 5)(33 10) 4C 22.04
W 10C
W 33
10.45 10 5 5 33 10 22.04
21C
c.
W 33
10.45 10 15 15 33 10 22.04
34C
d.
y
5 x3 8
On 16,32 the endpoint coordinates can be
61. a.
f.
83 5 80 8
W 33 1.5958 33 10 36C
63. Let x = the sales made and S x = the total
salary. For 0 x 250, 000 : S x $45, 000 0.04 x For 250, 000 x 500, 000 : S x 55, 000 0.06( x 250, 000)
thought of as 16,8 and 32,3 . 38 5 32 16 16 5 8 (16) b 16 13 b 5 y x 13 16 Therefore, 5 if 0 x 8 8 x3 if 8 x 16 8 d ( x) 5 x 13 if 16 x 32 16 3 if 32 x 38 m
65. The function f changes definition at 2 and the function g changes definition at 0. Combining these together, the sum function will change definitions at 0 and 2. On the interval , 0 : ( f g )( x ) f ( x ) g ( x ) (2 x 3) ( 4 x 1) . 2 x 4 On the interval 0, 2 :
f g ( x) f ( x) g ( x) (2 x 3) ( x 7) . 3x 4 On the interval 2, :
( f g )( x ) f ( x ) g ( x) ( x 2 5 x ) ( x 7)
For x 500, 000 : S x 70, 000 0.09( x 500, 000)
x2 6x 7 2 x 4 if x 0 So, f g ( x) 3x 4 if 0 x 2 x 2 6 x 7 if x 2
285 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
66. Each graph is that of y x 2 , but shifted vertically.
1 x is 4 the same as the graph of y x , but compressed
compressed vertically. The graph of y
vertically. The graph of y 5 x is the same as the graph of y x , but stretched vertically. 69. The graph of y x 2 is the reflection of the
graph of y x 2 about the x-axis.
2
If y x k , k 0 , the shift is up k units; if y x 2 k , k 0 , the shift is down k units. The
graph of y x 2 4 is the same as the graph of y x 2 , but shifted down 4 units. The graph of y x 2 5 is the graph of y x 2 , but shifted up 5 units.
The graph of y x is the reflection of the
67. Each graph is that of y x 2 , but shifted horizontally.
graph of y x about the x-axis.
Multiplying a function by –1 causes the graph to be a reflection about the x-axis of the original function's graph.
If y ( x k ) 2 , k 0 , the shift is to the right k units; if y ( x k ) 2 , k 0 , the shift is to the left k units. The graph of y ( x 4) 2 is the
70. The graph of y x is the reflection about the
same as the graph of y x 2 , but shifted to the
y-axis of the graph of y x .
left 4 units. The graph of y ( x 5) 2 is the
graph of y x 2 , but shifted to the right 5 units. y x
y x
68. Each graph is that of y x , but either
5
compressed or stretched vertically.
The same type of reflection occurs when graphing y 2 x 1 and y 2( x) 1 .
If y k x and k 1 , the graph is stretched vertically; if y k x and 0 k 1 , the graph is 286
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Section 3.4: Library of Functions; Piecewise-defined Functions
the points (1, 1) , (0, 0) , and (1, 1) . As n increases, the graph of the function increases at a greater rate for x 1 and is flatter around 0 for x 1.
The graph of y f ( x) is the reflection about the y-axis of the graph of y f ( x) . 71. The graph of y ( x 1)3 2 is a shifting of the
graph of y x3 one unit to the right and two units up. Yes, the result could be predicted.
74.
1 if x is rational f x 0 if x is irrational Yes, it is a function. Domain = x x is any real number or ,
Range = {0, 1} or y | y 0 or y 1 y-intercept: x 0 x is rational y 1 So the y-intercept is y 1 . 72. The graphs of y x n , n a positive even integer, are all U-shaped and open upward. All go through the points (1, 1) , (0, 0) , and (1, 1) . As n increases, the graph of the function is narrower for x 1 and flatter for x 1 .
x-intercept: y 0 x is irrational So the graph has infinitely many x-intercepts, namely, there is an x-intercept at each irrational value of x. f x 1 f x when x is rational;
f x 0 f x when x is irrational.
Thus, f is even. The graph of f consists of 2 infinite clusters of distinct points, extending horizontally in both directions. One cluster is located 1 unit above the x-axis, and the other is located along the x-axis. 75. Answers will vary. 73. The graphs of y x n , n a positive odd integer, all have the same general shape. All go through
76. ( x 3 y 5 )2 x 6 y 10
287 Copyright © 2025 Pearson Education, Inc.
x6 y10
Chapter 3: Functions and Their Graphs
77.
85. 3x3 y 2 x 2 y 2 18 x 12 y
x 2 y 2 6 y 16 x y 2 6 y 16 x 2 ( y 2 6 y 9) 16 9 x 2 ( y 3) 2 52
3x y 2 x y 18x 12 y
2
3
2 2
x 2 y 3x 2 y 6 3x 4 y
3x 2 y x 2 y 6
Center (h,k): (0, 3); Radius = 5 78. 4 x 5(2 x 1) 4 7( x 1) 4 x 10 x 5 4 7 x 7 6 x 5 7 x 3 x 8
Section 3.5
The solution set is: {8}
1. horizontal; right
79. Let x represent the amount of money invested in a mutual fund. Then 60, 000 x represents the amount of money invested in CD's. Since the total interest is to be $3700, we have: 0.08 x 0.03(60, 000 x) 3700
4. True; the graph of y f x is the reflection
100 0.08 x 0.03(60, 000 x) 3700100
about the x-axis of the graph of y f x .
2. y 3. False
8 x 3(60, 000 x) 370, 000 8 x 180, 000 3 x 370, 000
5. d
5 x 180, 000 370, 000 5 x 190, 000 x 38, 000 $38,000 should be invested in a mutual fund at 8% and $22,000 should be invested in CD's at 3%.
80. 2 1
3 0 6 2 2 4
1
1 6 2
6. b 7. B 8. E 9. H 10. D 11. I 12. A
2
Quotient: x x 2 Remainder: -2 81.
13. L 14. C
3 2i 2
15. F 16. J
82. 2 x 7 83.
5t 2
2
25t 2 7
2
17. G 25t 4 625t 7 4
25t 1 25t
3
18. K
19. y ( x 4)3
5t 2 1 25t 3
20. y ( x 4)3
84. The radicand cannot be negative so: x7 0 x 7 The domain is x | x 7 .
21. y x3 4 22. y x3 4 288
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Section 3.5: Graphing Techniques: Transformations 35. (c); To go from y f x to y 2 f x , we
23. y x x3 3
stretch vertically by a factor of 2. Multiply the y-coordinate of each point on the graph of y f x by 2. Thus, the point 1,3 would
24. y x3 25. y 5 x3
become 1, 6 . 3
1 3 1 26. y x x 4 64
27. y 2 x 8 x 3
28. y 29. (1)
36. (c); To go from y f x to y f 2 x , we
compress horizontally by a factor of 2. Divide the x-coordinate of each point on the graph of y f x by 2. Thus, the point 4, 2 would
3
become 2, 2 .
1 3 x 4
37.
y x 2
Using the graph of y x 2 , vertically shift downward 1 unit.
x 2 (3) y x 2 x 2 (2)
30. (1)
y
y x
(2)
y x3
(3)
y x3 2
31. (1)
f ( x) x 2 1
y3 x
(2)
y 3 x 4
(3)
y 3 x5 4
The domain is , and the range is 1, .
32. (1)
y x 2
(2)
y x 2
(3)
y ( x 3) 2 x 3 2
38.
f ( x) x 2 4
Using the graph of y x 2 , vertically shift upward 4 units.
33. (c); To go from y f x to y f x we
reflect about the x-axis. This means we change the sign of the y-coordinate for each point on the graph of y f ( x) . Thus, the point (3, 6) would become 3, 6 . 34. (d); To go from y f x to y f x , we
reflect each point on the graph of y f x
The domain is , and the range is 4, .
about the y-axis. This means we change the sign of the x-coordinate for each point on the graph of y f x . Thus, the point 3, 6 would become
3, 6 .
289 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
39. g ( x) 3x
Using the graph of y x , horizontally compress by a factor of 3.
The domain is 2, and the range is 0, .
The domain is 0, and the range is 0, . 40. g ( x) 3
42. h( x) x 1
Using the graph of y x , horizontally shift to the left 1 unit.
1 x 2
Using the graph of y 3 x , horizontally stretch by a factor of ½.
The domain is 1, and the range is 0, . The domain is , and the range is
43.
, .
f ( x) ( x 1)3 2
Using the graph of y x3 , horizontally shift to the right 1 unit y x 1 , then vertically 3
3 shift up 2 units y x 1 2 .
41. h( x)
x2
Using the graph of y x , horizontally shift to the left 2 units. The domain is , and the range is
, .
290
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Section 3.5: Graphing Techniques: Transformations
44.
f ( x) ( x 2)3 3
Using the graph of y x3 , horizontally shift to the left 2 units y x 2 , then vertically 3
3 shift down 3 units y x 2 3 .
The domain is 0, and the range is 0, . 47.
f ( x) 3 x
Using the graph of y 3 x , reflect the graph about the x-axis. The domain is , and the range is
, . 45. g ( x) 4 x
Using the graph of y x , vertically stretch by a factor of 4.
The domain is , and the range is
, . 48.
f ( x) x
Using the graph of y x , reflect the graph about the x-axis. The domain is 0, and the range is 0, . 1 x 2 Using the graph of y x , vertically compress
46. g ( x)
by a factor of
1 . 2
The domain is 0, and the range is , 0 .
291 Copyright © 2025 Pearson Education, Inc.
Chapter 3: Functions and Their Graphs
49.
f ( x) 2( x 1) 2 3
51. g ( x) 2 x 2 1 2
Using the graph of y x , horizontally shift to
Using the graph of y x , horizontally shift to
the left 1 unit y x 1 , vertically stretch
the right 2 units y x 2 , vertically stretch
by a factor of 2 y 2 x 1 , and then vertically shift downward 3 units y 2 x 12 3 .
shift upward 1 unit y 2 x 2 1 .
2
2
The domain is , and the range is 3, . 50.
by a factor of 2 y 2 x 2 , and vertically
The domain is 2, and the range is 1, . 52. g ( x) 3 x 1 3
Using the graph of y x , horizontally shift to the left 1 unit y x 1 , vertically stretch by a
f ( x) 3( x 2) 2 1
Using the graph of y x 2 , horizontally shift to 2
the right 2 units y x 2 , vertically
factor of 3 y 3 x 1 , and vertically shift downward 3 units y 3 x 1 3 .
stretch by a factor of 3 y 3 x 2 , and then vertically shift upward 1 unit y 3 x 2 2 1 . 2
The domain is , and the range is 3, . The domain is , and the range is 1, .
292 Copyright © 2025 Pearson Education, Inc.
Section 3.5: Graphing Techniques: Transformations
53. h( x ) x 2
Using the graph of y x , reflect the graph about the y-axis y x and vertically shift downward 2 units y x 2 .
The domain is , and the range is
, . 56.
Using the graph of y x , horizontally shift to
The domain is , 0 and the range is 2, . 54. h( x )
the right 1 unit y x 1 , reflect the graph
4 1 2 4 2 x x
Stretch the graph of y
f ( x) 4 x 1
about the x-axis y x 1 , and stretch vertically by a factor of 4 y 4 x 1 .
1 vertically by a factor x
1 4 of 4 y 4 and vertically shift upward 2 x x 4 units y 2 . x
The domain is 1, and the range is , 0 .
The domain is , 0 0, and the range is
, 2 2, . 55.
f ( x) ( x 1)3 1
Using the graph of y x3 , horizontally shift to the left 1 unit y x 1 , reflect the graph 3
about the x-axis y x 1 , and vertically 3
3 shift downward 1 unit y x 1 1 .
293
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Chapter 3: Functions and Their Graphs 57. g ( x) 2 1 x 2 1 x 2 x 1
Using the graph of y x , horizontally shift to the right 1 unit y x 1 , and vertically stretch by a factor or 2 y 2 x 1 .
The domain is , 0 0, and the range is
, 0 0, .
60.
The domain is , and the range is 0, .
f ( x) 3 x 1 3
Using the graph of f ( x) 3 x , horizontally shift to the right 1 unit y 3 x 1 , then vertically shift up 3 units y 3 x 1 3 .
58. g ( x) 4 2 x 4 ( x 2)
Using the graph of y x , reflect the graph about the y-axis y x , horizontally shift to the right 2 units y x 2 , and vertically stretch by a factor of 4 y 4 x 2 .
The domain is , and the range is
, . 61. a.
F ( x) f ( x) 3 Shift up 3 units.
The domain is , 2 and the range is 0, . 59. h( x)
1 2x
Using the graph of y by a factor of
1 , vertically compress x
1 . 2
294 Copyright © 2025 Pearson Education, Inc.
Section 3.5: Graphing Techniques: Transformations
b.
G ( x) f ( x 2) Shift left 2 units.
f.
g ( x) f ( x) Reflect about the y-axis.
c.
P ( x) f ( x) Reflect about the x-axis.
g.
h( x) f (2 x)
d.
e.
Compress horizontally by a factor of
H ( x) f ( x 1) 2 Shift left 1 unit and shift down 2 units.
Q( x)
62. a.
F ( x) f ( x) 3 Shift up 3 units.
1 f ( x) 2
Compress vertically by a factor of
1 . 2
b.
G ( x) f ( x 2) Shift left 2 units.
295
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1 . 2
Chapter 3: Functions and Their Graphs
c.
d.
e.
P ( x) f ( x) Reflect about the x-axis.
Q( x)
1 f ( x) 2
g ( x) f ( x) Reflect about the y-axis.
h( x) f (2 x)
Compress horizontally by a factor of
H ( x) f ( x 1) 2 Shift left 1 unit and shift down 2 units.
Compress vertically by a factor of
f.
g.
1 . 2
63. a.
F ( x) f ( x) 3 Shift up 3 units.
b.
G ( x) f ( x 2) Shift left 2 units.
c.
P ( x) f ( x) Reflect about the x-axis.
296 Copyright © 2025 Pearson Education, Inc.
1 . 2
Section 3.5: Graphing Techniques: Transformations
d.
H ( x) f ( x 1) 2 Shift left 1 unit and shift down 2 units.
e.
Q( x)
1 f ( x) 2
64. a.
F ( x) f ( x) 3 Shift up 3 units.
b.
G ( x) f ( x 2) Shift left 2 units.
c.
P ( x) f ( x) Reflect about the x-axis.
d.
H ( x) f ( x 1) 2 Shift left 1 unit and shift down 2 units.
1 Compress vertically by a factor of . 2
f.
g ( x) f ( x) Reflect about the y-axis.
g.
h( x) f (2 x)
Compress horizontally by a factor of
1 . 2
297
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Chapter 3: Functions and Their Graphs
e.
Q( x)
1 unit.
1 f ( x) 2
Compress vertically by a factor of
1 . 2
66.
f ( x) x 2 6 x
f ( x) ( x 2 6 x 9) 9
f.
g.
g ( x) f ( x) Reflect about the y-axis.
f ( x) ( x 3) 2 9
Using f ( x) x 2 , shift right 3 units and shift down 9 units.
h( x) f (2 x)
Compress horizontally by a factor of 12 .
67.
f ( x) x 2 8 x 1
f ( x) x 2 8 x 16 1 16 f ( x) x 4 15 2
Using f ( x) x 2 , shift right 4 units and shift down 15 units.
65.
f ( x) x 2 2 x f ( x) ( x 2 2 x 1) 1 f ( x) ( x 1) 2 1
Using f ( x) x 2 , shift left 1 unit and shift down 68.
f ( x) x 2 4 x 2
f ( x) x 2 4 x 4 2 4 f ( x) x 2 2 2
298 Copyright © 2025 Pearson Education, Inc.
Section 3.5: Graphing Techniques: Transformations
Using f ( x) x 2 , shift left 2 units and shift down 2 units.
71.
f x 3x 2 12 x 17
3 x 4 x 4 17 12 3 x 2 4 x 17 2
3 x 2 5 2
Using f x x 2 , shift left 2 units, stretch vertically by a factor of 3, reflect about the xaxis, and shift down 5 units.
69.
f x 2 x 2 12 x 19
2 x 6 x 9 19 18 2 x 2 6 x 19 2
2 x 3 1 2
Using f x x 2 , shift right 3 units, vertically stretch by a factor of 2, and then shift up 1 unit. 72.
f x 2 x 2 12 x 13
2 x 6 x 9 13 18 2 x 2 6 x 13 2
2 x 3 5 2
Using f x x 2 , shift left 3 units, stretch vertically by a factor of 2, reflect about the xaxis, and shift up 5 units. 70.
f x 3x 2 6 x 1
3 x 2 x 1 1 3 3 x2 2 x 1 2
3 x 1 2 2
Using f x x 2 , shift left 1 unit, vertically stretch by a factor of 3, and shift down 2 units. 73. a.
The graph of y f x 2 is the same as the graph of y f x , but shifted 2 units to the left. Therefore, the x-intercepts are 7 and 1.
b. The graph of y f x 2 is the same as
the graph of y f x , but shifted 2 units to
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Chapter 3: Functions and Their Graphs
c.
the right. Therefore, the x-intercepts are 3 and 5.
x-axis. Therefore, we can say that the graph of y f x must be decreasing on the
The graph of y 4 f x is the same as the
interval 1,5 .
graph of y f x , but stretched vertically
d. The graph of y f x is the same as the
by a factor of 4. Therefore, the x-intercepts are still 5 and 3 since the y-coordinate of each is 0.
graph of y f x , but reflected about the y-axis. Therefore, we can say that the graph of y f x must be decreasing on the
d. The graph of y f x is the same as the
interval 5,1 .
graph of y f x , but reflected about the y-axis. Therefore, the x-intercepts are 5 and 3 . 74. a.
76. a.
the graph of y f x , but shifted 2 units to the left. Therefore, the graph of f x 2 is
The graph of y f x 4 is the same as the graph of y f x , but shifted 4 units to
decreasing on the interval 4,5 .
the left. Therefore, the x-intercepts are 12 and 3 .
b. The graph of y f x 5 is the same as
the graph of y f x , but shifted 5 units to
b. The graph of y f x 3 is the same as
the right. Therefore, the graph of f x 5
the graph of y f x , but shifted 3 units to the right. Therefore, the x-intercepts are 5 and 4. c.
is decreasing on the interval 3,12 . c.
The graph of y 2 f x is the same as the by a factor of 2. Therefore, the x-intercepts are still 8 and 1 since the y-coordinate of each is 0.
d. The graph of y f x is the same as the
x-axis. Therefore, we can say that the graph of y f x must be increasing on the interval 2, 7 . d. The graph of y f x is the same as the
graph of y f x , but reflected about the
graph of y f x , but reflected about the
y-axis. Therefore, we can say that the graph of y f x must be increasing on the
y-axis. Therefore, the x-intercepts are 8 and 1 .
interval 7, 2 .
The graph of y f x 2 is the same as the graph of y f x , but shifted 2 units to
The graph of y f x is the same as the graph of y f x , but reflected about the
graph of y f x , but stretched vertically
75. a.
The graph of y f x 2 is the same as
77. a.
y f ( x)
the left. Therefore, the graph of f x 2 is increasing on the interval 3, 3 . b. The graph of y f x 5 is the same as
the graph of y f x , but shifted 5 units to the right. Therefore, the graph of f x 5 is increasing on the interval 4,10 . c.
The graph of y f x is the same as the graph of y f x , but reflected about the 300 Copyright © 2025 Pearson Education, Inc.
Section 3.5: Graphing Techniques: Transformations
b.
y f x
b. The graph of y 2 f x 2 1 is the
graph of y f x but shifted right 2 units, stretched vertically by a factor of 2, reflected about the x-axis, and shifted up 1 unit. Thus, the point 1,3 becomes the point 3, 5 . c.
The graph of y f 2 x 3 is the graph of
y f x but shifted left 3 units and
78. a.
horizontally compressed by a factor of 2. Thus, the point 1,3 becomes the point
To graph y f ( x) , the part of the graph
1,3 .
for f that lies in quadrants III or IV is reflected about the x-axis.
80. a.
The graph of y g x 1 3 is the graph of y g x but shifted left 1 unit and down 3 units. Thus, the point 3,5 becomes the point 4, 2 .
b. The graph of y 3 g x 4 3 is the graph
of y g x but shifted right 4 units, stretched vertically by a factor of 3, reflected about the x-axis, and shifted up 3 units. Thus, the point 3,5 becomes the point 1, 12 .
b. To graph y f x , the part of the graph
for f that lies in quadrants II or III is replaced by the reflection of the part in quadrants I and IV reflected about the yaxis.
c.
The graph of y g 3x 9 is the graph of
y f x but shifted left 9 units and horizontally compressed by a factor of 3. Thus, the point 3,5 becomes the point
4,5 . 81. a. f ( x) int( x) Reflect the graph of y int( x) about the yaxis.
79. a.
The graph of y f x 3 5 is the graph of y f x but shifted left 3 units and down 5 units. Thus, the point 1,3 becomes the point 2, 2 . b. g ( x) int( x) 301
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Chapter 3: Functions and Their Graphs
Reflect the graph of y int( x) about the xaxis.
83. a. f ( x) x 3 3
Using the graph of y x , horizontally shift to the right 3 units y x 3 and vertically shift downward 3 units y x 3 3 .
82. a. f ( x) int( x 1) Shift the graph of y int( x) right 1 unit.
1 bh 2 1 (6)(3) 9 2 The area is 9 square units.
b. A
84. a. f ( x) 2 x 4 4 b. g ( x) int(1 x) int( ( x 1))
Using the graph of y x , horizontally shift to
Using the graph of y int( x) , reflect the
the right 4 units y x 4 , vertically stretch
graph about the y-axis y int( x) ,
by a factor of 2 and flip on the x-axis y 2 x 4 , and vertically shift upward 4 units y 2 x 4 4 .
horizontally shift to the right 1 unit y int( ( x 1)) .
1 bh 2 1 (4)(4) 8 2 The area is 8 square units.
b. A
302 Copyright © 2025 Pearson Education, Inc.
Section 3.5: Graphing Techniques: Transformations f 6 0.12 6 1.29 6 9.96 2
85. a.
From the graph, the thermostat is set at 72F during the daytime hours. The thermostat appears to be set at 65F overnight.
b.
To graph y T t 2 , the graph of T t is
13.38 The estimated hours of daylight in July is 13.38. f 11 0.12 11 1.29 11 9.96 2
shifted down 2 units. This change will lower the temperature in the house by 2 degrees.
9.63 The estimated hours of daylight in December is 13.38. b. x represents the number of months after July. c. January corresponds to: F (6) f 6 6 f (0) f (0) 9.96
d. Orlando is in the Northern Hemisphere and Alice Springs is in the Southern Hemisphere; therefore, they have “opposite” seasons in each month. For example, it is winder in Orlando in January while it is summer in Alice Spring. c.
To graph y T t 1 , the graph of T t
9 87. F C 32 5
should be shifted left one unit. This change will cause the program to switch between the daytime temperature and overnight temperature one hour sooner. The home will begin warming up at 5 a.m. instead of 6 a.m. and will begin cooling down at 8 p.m. instead of 9 p.m.
9 F ( K 273) 32 5 Shift the graph 273 units to the right.
86. a.
f 0 0.12 0 1.29 0 9.96 9.96 2
The estimated hours of daylight in January is 9.96.
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Chapter 3: Functions and Their Graphs
88. a.
T 2
90. y x 2 c
l g
If c 0, y x 2 . If c 3, y x 2 3; shift up 3 units. If c 2, y x 2 2; shift down 2 units.
b.
T1 2
l 1 l2 ; T2 2 ; g g
T3 2
l 3 g
91.
f ( x 5) is a shift right 5 units; increasing on
2,8 and 16, 24 ; decreasing of 8,16 . f (2 x 5) compresses horizontally by a factor of
c.
d.
As the length of the pendulum increases, the period increases. T1 2
2l 3l 4l ; T2 2 ; T3 2 g g g
½; increasing on 1, 4 and 8,12 ; decreasing on
4,8 . f (2 x 5) reflects about the x-axis; increasing on 4,8 ; decreasing on 1, 4 and 8,12 . 3 f (2 x 5) stretches vertically by a factor or 3 but does not affect increasing/decreasing. Therefore 3 f (2 x 5) is increasing on 4,8 .
e.
If the length of the pendulum is multiplied by k , the period is multiplied by k .
89. y ( x c) 2 If c 0, y x 2 . If c 3, y ( x 3) 2 ; shift right 3 units. If c 2, y ( x 2) 2 ; shift left 2 units.
92. Write the general normal density as x 2 1 1 exp . Starting f ( x) 2 2 with the standard normal density, x2 1 exp , stretch/compress f ( x) 2 2 horizontally by a factor of to get x 2 1 exp f ( x) 2 2 1 multiply all the x-coordinates by , then shift the graph horizontally units (left if 0
and right if 0 ) to get
304 Copyright © 2025 Pearson Education, Inc.
Section 3.5: Graphing Techniques: Transformations
f ( x)
x 2 1 . Then exp 2 2
stretch/compress vertically by a factor of
the second region would also be units.
1
96. The range of f ( x) x 2 is 0, . The graph of g ( x) f ( x) k is the graph of f shifted up k
to
units if k > 0 and shifted down k units if k < 0,
get
so the range of g is k , .
x 2 1 1 exp f ( x) 2 2 1 multiply all the y -coordinates by . 1. Stretch/compress horizontally by a factor or (stretch if 1 ) 2. Shift horizontally units (left if 0 and
97. The domain of g ( x)
3.Stretch/compress vertically by a factor of
x is 0, . The graph
of g ( x k ) is the graph of g shifted k units to the right, so the domaine of g is k , . 98. 3x 5 y 30 5 y 3x 30 3 y x6 5 3 The slope is and the y-intercept is -6. 5
right if 0 ).
(compress if 1 )
16 square 3
1
13.1 13.1 8.4214 . The 7 2 total distance is 26.2 mile. Thus the average 26.2 speed is 3.11 mph . 8.4214
99. The total time run is
93. The graph of y 4 f ( x) is a vertical stretch of the graph of f by a factor of 4, while the graph of y f (4 x) is a horizontal compression of the
graph of f by a factor of 14 .
100. W kT 7 k4 7 k 4 7 7 W T (9) 15.75 gal 4 4
94. The graph of y f ( x) 2 will shift the graph of y f ( x) down by 2 units. The graph of y f ( x 2) will shift the graph of y f ( x) to the right by 2 units. 95. The graph of y x is the graph of y x but reflected about the y-axis. Therefore, our region is simply rotated about the y-axis and does not change shape. Instead of the region being bounded on the right by x 4 , it is bounded on the left by x 4 . Thus, the area of
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Chapter 3: Functions and Their Graphs
z 3 216 z 6 3
101. y 2 x 4 x-intercepts: (0) 2 x 4 0 x4 x 4
106.
y-intercepts: y2 0 4 y2 4 y 2
The intercepts are 4, 0 , 0, 2 and 0, 2 . Test x-axis symmetry: Let y y
( z 6)( z 2 6 z 36)
Section 3.6 1. a.
y x 4
graph of y x 2 8 , we have:
y 2 x 4 same
d ( x) x 2 ( x 2 8) 2 x 4 15 x 2 64
Test y-axis symmetry: Let x x y 2 x 4 different Test origin symmetry: Let x x and y y .
y x 4 2
b.
d (0) 04 15(0) 2 64 64 8
c.
d (1) (1) 4 15(1) 2 64 1 15 64 50 5 2 7.07
y 2 x 4 different
Therefore, the graph will have x-axis symmetry.
So the domain is: x | x 7, x 2 103. 16t 2 96t 200 88 16t 2 96t 112 0 16(t 2 6t 7) 0 16(t 7)(t 1) 0 t 7, t 1 Since t represents time the only answer that is reasonable is 7 seconds.
105.
e.
2. a.
d is smallest when x 2.74 or when x 2.74 .
The distance d from P to (0, –1) is d x 2 ( y 1) 2 . Since P is a point on
16 x5 y 6 z 3 8 2 x3 x 2 y 6 z 2 xy 2 3 2 x 2 z
f ( x h) f ( x ) h 3( x h) 2 2( x h) 1 (3x 2 2 x 1) h 3( x 2 2 xh h 2 ) 2 x 2h 1 3 x 2 2 x 1 h 3x 2 6 xh 3h 2 2 x 2h 1 3x 2 2 x 1 h 6 xh h 2 2h h(6 x h 2) 6x h 2 h h
d.
102. The denominator must not be zero. x 2 5 x 14 0 ( x 7)( x 2) 0 x 7, x 2
3
The distance d from P to the origin is d x 2 y 2 . Since P is a point on the
2
104.
3
the graph of y x 2 8 , we have: d ( x) x 2 ( x 2 8 1) 2
x2 x2 7
x 13x 49 2
4
2
b.
d (0) 04 13(0) 2 49 49 7
c.
d (1) (1)4 13(1) 2 49 37 6.08
306 Copyright © 2025 Pearson Education, Inc.
Section 3.6: Mathematical Models: Building Functions 10
d.
–4
e.
b.
8
4
0
–5
d is smallest when x 2.55 or when x 2.55 .
c.
5
0
d is smallest when x 1 or x 1 .
d. 3. a.
The distance d from P to the point (1, 0) is d ( x 1) 2 y 2 . Since P is a point on
d (1)
the graph of y x , we have: d ( x) ( x 1) 2
x x x 1 2
c.
d is smallest when x 12 .
d.
d ( x)
4. a.
2
0
1
2
5. By definition, a triangle has area 1 A b h, b base, h height. From the figure, 2 we know that b x and h y. Expressing the area of the triangle as a function of x , we have: 1 1 1 A( x ) xy x x3 x 4 . 2 2 2
2
0
12 1
2
where x 0 . b.
12 1 2; d (1) 1
6. By definition, a triangle has area 1 A b h, b=base, h height. Because one 2 vertex of the triangle is at the origin and the other is on the x-axis, we know that b x and h y. Expressing the area of the triangle as a function of x , we have: 1 1 9 1 A( x ) xy x 9 x 2 x x3 . 2 2 2 2
12 1 3 1 2 2 2
7. a.
The distance d from P to the origin is
A( x ) xy x 16 x 2
d x 2 y 2 . Since P is a point on the
b. Domain: x 0 x 4
1 graph of y , we have: x
c.
The area is largest when x 2.31 . 30
2
1 1 d ( x) x 2 x 2 2 x x 0
0
307
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4
Chapter 3: Functions and Their Graphs
9. a.
In Quadrant I, x 2 y 2 4 y 4 x 2 A( x) (2 x)(2 y ) 4 x 4 x 2
d. The largest area is
2
A(2.31) 2.31 16 2.31
b.
p ( x) 2(2 x) 2(2 y ) 4 x 4 4 x 2
c.
Graphing the area equation: 10
24.63 square
units. 8. a.
A( x) 2 xy 2 x 4 x 2
0
b.
p( x) 2(2 x) 2( y ) 4 x 2 4 x 2
c.
Graphing the area equation:
2
0
4
The area is largest when x 1.41 . d. Graphing the perimeter equation: 0
0
12
2
0
2
0
The area is largest when x 1.41 . d. Graphing the perimeter equation: 10
The perimeter is largest when x 1.41 . 10. a. 0
0
2
b. 11. a.
The perimeter is largest when x 1.79 . e.
The largest area is A(1.41) 2 1.41 4 1.412 4 square
A(r ) (2r )(2r ) 4r 2
p (r ) 4(2r ) 8r C circumference, A total area, r radius, x side of square C 2r 10 4 x r 52 x Total Area area square + area circle x 2 r 2
A( x ) x 2 52 x
units. The largest perimeter is p (1.79) 4 1.79 2 4 1.792 8.94
units. 308 Copyright © 2025 Pearson Education, Inc.
2
x2
25 20 x 4 x 2
Section 3.6: Mathematical Models: Building Functions b. Since the lengths must be positive, we have: 10 4 x 0 and x 0 4 x 10 and x 0 x 2.5 and x 0 Domain: x 0 x 2.5 c.
c.
The area is smallest when x 2.08 meters. 8
0
The total area is smallest when x 1.40 meters.
3.33
0
8
0
2.5
0
13. a.
Since the wire of length x is bent into a circle, the circumference is x . Therefore, C ( x) x .
b. Since C x 2 r , r
x . 2
2
12. a.
x2 x . A( x) r 2 4 2
C circumference, A total area, r radius, x side of equilateral triangle
C 2r 10 3x r
14. a.
10 3x 2
The height of the equilateral triangle is Total Area area triangle area circle
3 x. 2
b. Since P x 4s, s
3 2 10 3 x x 4 2
1 x , we have 4
2
1 1 A( x ) s 2 x x 2 . 16 4
1 3 x x r2 2 2
A( x)
Since the wire of length x is bent into a square, the perimeter is x . Therefore, p( x) x .
15. a.
2
A area, r radius; diameter 2r A(r ) (2r )(r ) 2r 2
3 2 100 60 x 9 x 2 x 4 4
b.
p perimeter p(r ) 2(2r ) 2r 6r
b. Since the lengths must be positive, we have: 10 3x 0 and x 0 3x 10 and x 0 10 x and x 0 3 10 Domain: x 0 x 3
16. C circumference, r radius; x length of a side of the triangle
Since ABC is equilateral, EM 309
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3x . 2
Chapter 3: Functions and Their Graphs
d 2 3 40t
3x 3x OE r 2 2
Therefore, OM
2 x 3x r In OAM , r 2 2 2
2
d1 2 30t
d
2
x 3 x 2 3 rx r 2 4 4 3 rx x 2
r2
x 3 Therefore, the circumference of the circle is x 2 3 C ( x ) 2 r 2 x 3 3
b. The distance is smallest at t 0.07 hours.
r
20. r radius of cylinder, h height of cylinder, V volume of cylinder
17. Area of the equilateral triangle 1 3 3 2 A x x x 2 2 4
2
x2 . 3 Area inside the circle, but outside the triangle: 3 2 A( x) r 2 x 4 x2 3 2 3 2 x x 3 4 3 4
h2 h2 h r 2 R2 r 2 R2 r 2 R2 4 4 2 2 V r h
From problem 16, we have r 2
h2 h2 V (h) R 2 h h R 2 4 4
21.
r radius of cylinder, h height of cylinder, V volume of cylinder H H h R r Hr R H h
By similar triangles:
18. d 2 d12 d 2 2 d 2 30t 40t 2
Hr RH Rh
2
d t 900 t 1600 t 2500 t 50 t 2
2
2
d2 =40t d1=30t
d
Rh RH Hr RH Hr H R r R R H R r H R r r2 V (r ) r 2 h r 2 R R h
22. a. 19. a.
d
2
d12 d 2 2
d 2 2 30t 3 40t 2
d t
2
2 30t 2 3 40t 2
The total cost of installing the cable along the road is 500 x . If cable is installed x miles along the road, there are 5 x miles between the road to the house and where the cable ends along the road.
4 120t 900t 2 9 240t 1600t 2 2500t 2 360t 13
310 Copyright © 2025 Pearson Education, Inc.
Section 3.6: Mathematical Models: Building Functions House
23. a.
time on land is given by
d
2
Town x
P
d (5 x) 2 22
T ( x)
C ( x) 500 x 700 x 2 10 x 29 Domain: x 0 x 5
x2 4 12 x d1 12 x 5 3 5 3
b. Domain: x 0 x 12
C (1) 500 1 700 1 10 1 29 2
500 700 20 $3630.50
c.
T (4)
C (3) 500 3 700 32 10 3 29
1500 700 8 $3479.90
d.
T (8)
e.
12–x
d1 x 2 22 x 2 4 The total time for the trip is:
25 10 x x 2 4 x 2 10 x 29 The total cost of installing the cable is:
d.
12 x . 5
d1
2
x
5x
c.
d1 . The 3
Island Box
b.
The time on the boat is given by
3000
24. a.
Using MINIMUM, the graph indicates that x 2.96 miles results in the least cost.
12 4 42 4 5 3 8 20 3.09 hours 5 3
12 8 82 4 5 3 4 68 3.55 hours 5 3
Let A amount of material , x length of the base , h height , and V volume . 10 V x 2 h 10 h 2 x Total Area A Area base 4 Area side x 2 4 xh 10 x2 4 x 2 x 40 x2 x 40 2 A x x x 40 1 40 41 ft 2 1
b.
A 1 12
c.
A 2 22
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40 4 20 24 ft 2 2
Chapter 3: Functions and Their Graphs
d.
y1 x 2
26. Consider the diagrams shown below.
40 x
100
10
0
0
The amount of material is least when x 2.71 ft. e.
The largest area is A 2.71 2.712
25.
40 22.1 ft 2 2.71
a.
length = 24 2x ; width = 24 2x ; height = x V ( x) x(24 2 x)(24 2 x) x(24 2 x) 2
b.
V (3) 3(24 2(3)) 2 3(18) 2 3(324) 972 in 3 .
c.
V (10) 10(24 2(10))2 10(4) 2 10(16) 160 in 3 .
d.
y1 x(24 2 x )2
0
2
1 1 1 3 V h r 2h h h h . 3 3 4 48
27. a.
1100
0
There is a pair of similar triangles in the diagram. This allows us to write r 4 r 1 1 r h h 16 h 4 4 Substituting into the volume formula for the conical portion of water gives
12
Use MAXIMUM.
The total cost is the sum of the shipment cost, storage cost, and product cost. Since each shipment will contain x units, there are 600/x shipments per year, each costing $15. 600 9000 So the shipment cost is 15 = . x x The storage cost for the year is given as 1.60 x. The product costs is 600(4.85) 2910. So, the total cost is C ( x)
The volume is largest when x 4 inches. e.
The largest volume is V (4) 4(24 2(4)) 2 1024 in 3 312 Copyright © 2025 Pearson Education, Inc.
9000 1.60 x 2910. x
Section 3.6: Mathematical Models: Building Functions b.
31.
10 14 4 x 10 x 4(14) 10 x 56 x 5.6
32. x u 1 u 1 u 1 y u 11 u 33.
The retailer should order 75 drives per order for a minimum yearly cost of $3150. 28.
x5 2
1
x 3
3x 3
2 x 3 5 2 2x 3 3
2 x 3 3 or 2 x 3 3 2 x 0 or 2x 6 x 0 or
3x 2
3x 3 3x 3 x 5 3x 2
3x 3 4x 5 2
34. 3x 2 4
x3
3x 2 4
No solution since a square root cannot be negative.
29. In order for the 16-foot long Ford Fusion to pass the 50-foot truck, the Ford Fusion must travel the length of the truck and the length of itself in the time frame of 5 seconds. Thus the Fusion must travel an additional 66 feet in 5 seconds.
35. Since the graph is symmetric is symmetric about the origin then (3, -2) is symmetric to (-3, 2). 36.
Convert this to miles-per-hour. 5 5 1 5 sec min hr hr. 60 3600 720 66 66 ft mi 5280
v
2.6t d2
vd 2 2.6t
E P
E P
2
vd 2 E 2.6t P
66
distance 5280 1 9 mph time 720
v2 d 4 E 2 P 6.76t 2 4 Pv d E 6.76t 2 6.76t 2 E P 2 4 v d
Since the truck is traveling 55 mph, the Fusion must travel 55 + 9 = 64 mph. 30. m
2
3x 3
The solution set is 0,3 .
speed=
x5
y2 y1 6 ( 2) 8 4 1 3 x2 x1 2
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Chapter 3: Functions and Their Graphs
3x 2 7 x 4 x 2
37. 2
3x 11x 2 0 b 2 4ac (11) 2 4(3)(2) 121 24 97
d.
3x 3 x f ( x) 2 x 1 x2 1
e.
f ( x 2)
3( x 2) ( x 2) 2 1 3 x 2 3x 6 2 x 4x 4 1 x 4x 3
f.
Chapter 3 Review Exercises 1. a.
Domain {8, 16, 20, 24}
7.
Range {$6.30, $12.32, $13.99} b. {(8,$6.30), (16,$13.99), (20,$12.32), (24,$13.99)} c.
d.
f (2 x)
2
3(2 x ) 6x 2 2 (2 x) 1 4 x 1
f ( x) x 2 4
a.
f (2) 22 4 4 4 0 0
b.
f (2)
c.
f ( x) ( x) 2 4 x 2 4
d.
f ( x) x 2 4
e.
f ( x 2) ( x 2) 2 4
2 2 4
44 0 0
x2 4 x 4 4 x2 4 x
f.
f (2 x) (2 x) 2 4 4 x 2 4
4 x2 1 2 x2 1
2. This relation represents a function. Domain = {–1, 2, 4}; Range = {0, 3}.
8.
3. Domain {2,4}; Range {-1,1,2} Not a function
x2 4 x2 22 4 4 4 0 0 4 4 22
a.
f (2)
b.
f (2)
2 2 4 4 4 0 0 4 4 2 2
c.
f ( x)
( x) 2 4 x 2 4 x2 ( x) 2
a.
3(2) 6 6 f (2) 2 2 4 1 3 (2) 1
d.
x2 4 4 x2 x2 4 f ( x) 2 x2 x2 x
b.
3(2) 6 6 f (2) 2 (2) 2 1 4 1 3
e.
f ( x 2)
c.
3( x) 3 x f ( x) 2 2 ( x) 1 x 1
4. not a function; domain [-1, 3]; range [-2, 2] 5. function; domain: all real numbers; range 3, 6.
f ( x)
f ( x)
3x x 1 2
314
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( x 2) 2 4 x 2 4 x 4 4 ( x 2) 2 ( x 2) 2 x2 4 x x x 4 ( x 2) 2 ( x 2) 2
Chapter 3 Review Exercises
f.
f (2 x)
9.
(2 x) 2 4 4 x 2 4 (2 x) 2 4 x2
14.
x 1
4 x2 1
2
4 x2
x2
x 8 Domain: x x 8
x x 9 The denominator cannot be zero: x2 9 0 f ( x)
2
15.
g ( x) 3 x 1
2 x 3x 1 2 x 3 Domain: x x is any real number
x 3 or 3
Domain: x x 3, x 3
( f g )( x) f x g ( x) 2 x 3 x 1
f ( x) 2 x The radicand must be non-negative: 2 x 0
2 x 3x 1 4 x 1 Domain: x x is any real number
x2 Domain: x x 2 or , 2
( f g )( x) f ( x) g x 2 x 3 x 1
x 11. g ( x) x The denominator cannot be zero: x0
6 x 2 3x 2 x 3x 2 5 x 2
Domain: x x is any real number
Domain: x x 0
f x 2 x f g ( x) g x 3x 1 3x 1 0
x 12. f ( x) 2 x 2x 3 The denominator cannot be zero: x2 2 x 3 0
1 3 1 Domain: x x 3 3 x 1 x
x 3 x 1 0 x 3 or 1 Domain: x x 3, x 1
13.
f ( x) 2 x
( f g )( x) f x g ( x)
( x 3)( x 3) 0
10.
x x8 The radicand must be non-negative and not zero: x8 0 f ( x)
16.
f ( x) 3x 2 x 1
g ( x) 3x
( f g )( x) f x g ( x)
x 1 x2 4 The denominator cannot be zero: x2 4 0 f ( x)
3x2 x 1 3x 3x2 4 x 1 Domain: x x is any real number
x 2 x 2 0
( f g )( x) f x g ( x)
x 2 or 2 Also, the radicand must be non-negative: x 1 0
3x 2 x 1 3x 3x 2 2 x 1 Domain: x x is any real number
x 1 Domain: 1, 2 2,
315
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Chapter 3: Functions and Their Graphs ( f g )( x) f ( x) g x
18.
f ( x) 2 x 2 x 1 f x h f x h
3x 2 x 1 3x 9 x3 3x 2 3 x Domain: x x is any real number
2
f x 3x 2 x 1 f g ( x) g x 3x 3x 0 x 0
x 1 1 g ( x) x 1 x f g x f x g x ( )( ) ( ) f ( x)
x 1 1 x x 1 1 x 1 x 1 x x x 1 2
h
2
2 x 2 xh h
2
x h 1 2x x 1
19. a.
Range:
2
x x x 1 x 2x 1 x x 1 x x 1
y 3 y 3 ; 3, 3
b. Intercept: 0, 0 c.
f 2 1
( f g )( x) f x g ( x)
d.
f x 3 when x = –4
e.
f ( x) 0 when 0 x 3
x 1 1 x x 1 1 x 1 x 1 x x x 1 2
2
Domain: x 4 x 3 ; 4, 3
Domain: x x 0, x 1
h 2 x 2 4 xh 2h 2 x h 1 2 x 2 x 1 h 4 xh 2h 2 h h 4 x 2h 1 h h 4 x 2h 1
Domain: x x 0
17.
2 x h x h 1 2 x 2 x 1
x | 0 x 3
2
x x x 1 x 1 x x 1 x x 1
f.
To graph y f x 3 , shift the graph of f horizontally 3 units to the right.
Domain: x x 0, x 1 x 1 x 1 1 ( f g )( x) f ( x) g x x x 1 x x 1 Domain: x x 0, x 1 x 1 f x x 1 x 1 x x( x 1) f g ( x) g x 1 x 1 1 x 1 x Domain: x x 0, x 1
316
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Chapter 3 Review Exercises
g.
4 x2 1 x4 4 ( x) 2 4 x 2 g ( x) g ( x) 1 ( x) 4 1 x 4 g is even.
1 To graph y f x , stretch the graph of 2 f horizontally by a factor of 2.
22. g ( x)
23. G ( x) 1 x x3 G ( x ) 1 ( x ) ( x )3 1 x x 3 G ( x) or G ( x ) G is neither even nor odd.
h.
To graph y f x , reflect the graph of f
x 1 x2 x x f ( x) f ( x) 1 ( x) 2 1 x 2 f is odd.
24.
f ( x)
25.
f x 2 x 3 5 x 1 on the interval 3,3
vertically about the y-axis.
Use MAXIMUM and MINIMUM on the graph of y1 2 x3 5 x 1 . 20
20. a.
3
Domain: , 4 Range: ,3
20
Decreasing: 2, 2
e.
The graph has no symmetry.
f.
The function is neither.
g.
x-intercepts: 3, 0 , 0, 0 , 3, 0 ;
Use MAXIMUM and MINIMUM on the graph of y1 2 x 4 5 x3 2 x 1 . 20
2
f ( x) x3 4 x
20
3 2 10 20
3
f ( x) ( x) 4( x) x 4 x
20
f x 2 x 4 5 x3 2 x 1 on the interval 2,3
26.
y-intercept: (0,0) 3
3
f is decreasing on: 0.91, 0.91 .
Local minimum is 1 at x 2 ; Local maximum is 1 at x 2
d. No absolute minimum; Absolute maximum is 3 at x 4
21.
3 3
local maximum value: 4.04 when x 0.91 local minimum value: 2.04 when x 0.91 f is increasing on: 3, 0.91 and 0.91,3 ;
b. Increasing: , 2 and 2, 4 ; c.
20
3 10
x3 4 x f ( x) 2
f is odd.
3 10
local maximum: 1.53 when x 0.41 317
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Chapter 3: Functions and Their Graphs
local minimal values: 0.54 when x 0.34 , 3.56 when x 1.80 f is increasing on: 0.34, 0.41 and 1.80, 3 ;
33.
f ( x) x
34.
f ( x) x
f is decreasing on: 2, 0.34 and 0.41, 1.80 . 27.
f ( x) 8 x 2 x
a.
b.
c.
28.
f (2) f (1) 8(2) 2 2 [8(1) 2 1] 2 1 1 32 2 (7) 23 f (1) f (0) 8(1) 2 1 [8(0) 2 0] 1 0 1 8 1 0 7 f (4) f (2) 8(4) 2 4 [8(2) 2 2] 42 2 128 4 (30) 94 47 2 2
f ( x) 2 5 x
f (3) f (2) 2 5 3 2 5 2 3 2 32 2 15 2 10 1 13 8 5
29.
f ( x) 3x 4 x 2
35. F ( x ) x 4 . Using the graph of y x ,
2 2 f (3) f (2) 3 3 4 3 3 2 4 2 3 2 3 2 9 36 6 16 1 27 10 17
vertically shift the graph downward 4 units.
30. Refer to question 29 for the slope. y 10 17( x 2) y 10 17 x 34 y 17 x 24
Intercepts: (–4,0), (4,0), (0,–4) Domain: x x is any real number
31. The graph does not pass the Vertical-Line Test and is therefore not a function.
Range: y y 4 or 4,
32. The graph passes the Vertical-Line Test and is therefore a function.
36. g ( x) 2 x . Reflect the graph of y x
about the x-axis and vertically stretch the graph by a factor of 2.
318
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Chapter 3 Review Exercises
39. h( x ) ( x 1) 2 2 . Using the graph of y x 2 , horizontally shift the graph to the right 1 unit and vertically shift the graph up 2 units.
Intercepts: (0, 0) Domain: x x is any real number Range: y y 0 or , 0
Intercepts: (0, 3) Domain: x x is any real number Range: y y 2 or 2,
37. h( x) x 1 . Using the graph of y x , horizontally shift the graph to the right 1 unit.
40. g ( x) 2( x 2)3 8
Using the graph of y x3 , horizontally shift the graph to the left 2 units, vertically stretch the graph by a factor of 2, reflect about the x-axis, and vertically shift the graph down 8 units.
y
x
2 3 4, 0 Intercept: (1, 0) Domain: x x 1 or 1,
5
Range: y y 0 or 0, 38.
f ( x) 1 x ( x 1) . Reflect the graph of
y x about the y-axis and horizontally shift the graph to the right 1 unit.
Intercepts: (0,–24), 2 3 4, 0 3.6, 0 Domain: x x is any real number Range: y y is any real number 41.
3x f ( x) x 1
a.
if 2 x 1 if x 1
Domain: x x 2 or 2,
b. Intercept: 0, 0
Intercepts: (1, 0), (0, 1) Domain: x x 1 or , 1 Range: y y 0 or 0,
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Chapter 3: Functions and Their Graphs c.
Graph:
44. a.
x 2 h 10 h
10 x2
A( x ) 2 x 2 4 x h 10 2 x2 4 x 2 x 40 2 x2 x
d. Range: y | y 6 or 6,
42.
x f ( x) 1 3x
a.
A(1) 2 12
c.
A(2) 2 22
40 8 20 28 ft 2 2
d. Graphing:
if 4 x 0
50
if x 0 if x 0
Domain: x x 4 or 4, 0
b. Intercept: (0, 1) c.
40 2 40 42 ft 2 1
b.
5
0
Graph:
The area is smallest when x 2.15 feet. 45. a.
Consider the following diagram: P(x,y)
y
y 10 x
d. Range: y y 4, y 0 43.
2
x
Ax 5 and f (1) 4 6x 2 A(1) 5 4 6(1) 2 A5 4 4 A 5 16 f ( x)
The area of the rectangle is A xy . Thus, the area function for the rectangle is: A( x ) x(10 x 2 )
A 11
b.
The maximum area is roughly: A(1.83) (1.83)3 10(1.83) 12.17 square units
The maximum value occurs at the vertex:
320
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Chapter 3 Chapter Test
can never equal 0. This means that x 2 . Domain: x | x 2
Chapter 3 Test 1. a.
2,5 , 4, 6 , 6, 7 , 8,8
g 1
This relation is a function because there are no ordered pairs that have the same first element and different second elements. Domain: 2, 4, 6,8
x4 x 2 5 x 36 The function tells us to divide x 4 by
4. h x
Range: 5, 6, 7,8 b.
1,3 , 4, 2 , 3,5 , 1, 7
x 2 5 x 36 . Since division by 0 is not defined, we need to exclude any values which make the denominator 0. x 2 5 x 36 0
This relation is not a function because there are two ordered pairs that have the same first element but different second elements.
x 9 x 4 0
Domain: 3,1, 4
x 9 or x 4 Domain: x | x 9, x 4
Range: 2,3,5, 7 c.
This relation is not a function because the graph fails the Vertical-Line Test.
(note: there is a common factor of x 4 but we must determine the domain prior to simplifying)
Domain: 1,
h 1
Range: x x is any real number d. This relation is a function because it passes the Vertical-Line Test. Domain: x x is any real number
5. a.
Range: y | y 2 or 2, 2.
f x 4 5x
The function tells us to take the square root of 4 5x . Only nonnegative numbers have real square roots so we need 4 5 x 0 . 4 5x 0 4 5x 4 0 4 5 x 4 5 x 4 5 5 4 x 5 4 4 Domain: x x or , 5 5
1 4 5 1 1 5 1 36 40 8 2
To find the domain, note that all the points on the graph will have an x-coordinate between 5 and 5, inclusive. To find the range, note that all the points on the graph will have a y-coordinate between 3 and 3, inclusive. Domain: x | 5 x 5 or 5, 5 Range: y | 3 y 3 or 3, 3
b. The intercepts are 0, 2 , 2, 0 , and 2, 0 .
x-intercepts: 2, 2 y-intercept: 2 c.
f 1 is the value of the function when x 1 . According to the graph, f 1 3 .
d. Since 5, 3 and 3, 3 are the only
points on the graph for which y f x 3 , we have f x 3 when x 5 and x 3 .
f 1 4 5 1 4 5 9 3
3. g x
1 2 1 1 1 2 1
e.
x2 x2
To solve f x 0 , we want to find xvalues such that the graph is below the xaxis. The graph is below the x-axis for values in the domain that are less than 2 and greater than 2. Therefore, the solution set is x | 5 x 2 or 2 x 5 . In
The function tells us to divide x 2 by x 2 . Division by 0 is undefined, so the denominator 321
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Chapter 3: Functions and Their Graphs
keep the part for which x 1 .
interval notation we would write the solution set as 5, 2 2,5 . 6.
f x x 4 2 x3 4 x 2 2
We set Xmin = 5 and Xmax = 5. The standard Ymin and Ymax will not be good enough to see the whole picture so some adjustment must be made.
b. To find the intercepts, notice that the only piece that hits either axis is y x 4 . y x4 y x4 y 04 0 x4 y 4 4x
The intercepts are 0, 4 and 4, 0 . c.
To find g 5 we first note that x 5 so we must use the first “piece” because 5 1 . g 5 2 5 1 10 1 9
d. To find g 2 we first note that x 2 so we
We see that the graph has a local maximum value of 0.86 (rounded to two places) when x 0.85 and another local maximum value of 15.55 when x 2.35 . There is a local minimum value of 2 when x 0 . Thus, we have Local maxima: f 0.85 0.86
must use the second “piece” because 2 1 . g 2 2 4 2 8. a. The average rate of change from 3 to 4 is
given by f 4 f 3 43
f 2.35 15.55
Local minima: f 0 2
3 4 3 4 4 3 3 3 3 4 2
The function is increasing on the intervals 5, 0.85 and 0, 2.35 and decreasing on the
43 40 22 18 18 43 1
intervals 0.85, 0 and 2.35,5 . 7. a.
2
x 1 2 x 1 f x x 1 x4 To graph the function, we graph each “piece”. First we graph the line y 2 x 1 but only keep the part for which x 1 . Then we plot the line y x 4 but only
b.
y 40 18( x 4) y 40 18 x 72 y 18 x 32
9. a.
( f g )( x) 2 x 2 1 3 x 2 2
2 x 1 3x 2 2 x 2 3x 3
b.
( f g )( x) 2 x 2 1 3 x 2 3
6 x 4 x 2 3x 2
322
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Chapter 3 Chapter Test
c.
y
f x h f x
2 x 2 xh h 1 2 x 1 2 x h 1 2 x2 1 2
2
2
2
x
2 x 2 4 xh 2h 2 1 2 x 2 1
4 xh 2h 2
10. a.
y 2 x 1
3
3
The basic function is y x so we start with the graph of this function. y
The last step is to shift this graph up 3 units to obtain the graph of y 2 x 1 3 . 3
y x3
y
x
x
Next we shift this graph 1 unit to the left to
y 2 x 1 3 3
obtain the graph of y x 1 . 3
y x 1
3
y
b. The basic function is y x so we start
with the graph of this function.
y
y x
x
x
Next we reflect this graph about the x-axis to obtain the graph of y x 1 . 3
Next we shift this graph 4 units to the left to obtain the graph of y x 4 .
y
y
y x4
x
y x 1
3
Next we stretch this graph vertically by a factor of 2 to obtain the graph of
x
Next we shift this graph up 2 units to obtain
y 2 x 1 . 3
323
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Chapter 3: Functions and Their Graphs b. If the rink is 90 feet wide, then we have x 90 .
the graph of y x 4 2 . y
y x4 2
V 90
2
The volume of ice is roughly 3460.29 ft 3 .
902 10 90 90 3460.29 3 3 24
x
13.
f x ( x ) 2 7 x 2 7 same The function is even.
11. a. f ( x h) f ( x) ( x h) 2 3( x h) ( x 2 3 x) h h 2 2 x 2 xh h 3 x 3h x 2 3x h 2 xh h 2 3h h h(2 x h 3) 2x h 3 h
12. a.
Let x = width of the rink in feet. Then the length of the rectangular portion is given by 2 x 20 . The radius of the semicircular x portions is half the width, or r . 2 To find the volume, we first find the area of the surface and multiply by the thickness of the ice. The two semicircles can be combined to form a complete circle, so the area is given by A l w r2 x 2 x 20 x 2 2
2 x 20 x
1.
3 x 8 10 3x 8 8 10 8 3 x 18 3x 18 3 3 x6 The solution set is 6 .
2.
3x 2 x 0 x 3 x 1 0 x 0 or 3x 1 0 3x 1 1 3 1 The solution set is 0, . 3 x
2
3.
x2
x2 8x 9 0
x 9 x 1 0
4 We have expressed our measures in feet so we need to convert the thickness to feet as well. 1 ft 2 1 2 in ft ft 12 in 12 6 Now we multiply this by the area to obtain the volume. That is, 1 x2 V x 2 x 2 20 x 6 4 V x
Chapter 3 Cumulative Review
x 9 0 or x 1 0 x9 x 1 The solution set is 1,9 .
x 2 10 x x 2 3 3 24
324
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Chapter 3 Cumulative Review
4.
6 x2 5x 1 0
4 Interval notation: , 3
3x 1 2 x 1 0 3x 1 0 or 2 x 1 0 3x 1 2x 1 1 2 1 1 The solution set is , . 3 2 x
5.
1 3
x
8.
3 2 x 5 3 2 2x 8 1 x 4 Solution set: x |1 x 4
2x 3 4 2 x 3 4 or 2 x 3 4 2 x 7 2x 1
6.
7 2
Interval notation: 1, 4
1 2 7 1 The solution set is , . 2 2 x
x
9.
2x 3 2 2
4x 1 7 4 x 1 7 or 4 x 1 7 4 x 8 4x 6 3 x 2 x 2 3 Solution set: x | x 2 or x 2 3 Interval notation: , 2 , 2
2x 3 2 2
2x 3 4 2x 1 x
2x 5 3
1 2
Check: ? 1 2 3 2 2 ?
1 3 2
10. a.
?
4 2 22 T 1 The solution set is . 2
x2 x1 y2 y1 2
2
3 2 5 3
3 2 5 3
2
2
2
52 2 25 4 2
7. 2 3x 6 3x 4 x
d
29 4 3
4 Solution set: x | x 3
325
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2
Chapter 3: Functions and Their Graphs
b.
c.
x x y y2 M 1 2 , 1 2 2 2 3 3 5 , 2 2 1 , 4 2 m
12. x y 2
x, y 2 2 x 2 4 4, 2 2 1 x 1 1 1, 1 x 02 0 0 0, 0 x 12 1 1 1,1 2 x2 4 2 4, 2 y
y2 y1 5 3 2 2 x2 x1 5 5 3 2
x y2
11. 3 x 2 y 12 x-intercept: 3x 2 0 12 3 x 12 x4 The point 4, 0 is on the graph.
y-intercept: 3 0 2 y 12
13. x 2 y 3 16 2
2 y 12 y 6
This is the equation of a circle with radius r 16 4 and center at 0,3 . Starting at the
The point 0, 6 is on the graph.
center we can obtain some points on the graph by moving 4 units up, down, left, and right. The corresponding points are 0, 7 , 0, 1 ,
4,3 , and 4,3 , respectively.
326
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Chapter 3 Cumulative Review
The graph of the equation has y-axis symmetry.
14. y x x 0 1 4
x, y y 0 0 0, 0 y 1 1 1,1 y 4 2 4, 2
16. First we find the slope: 84 4 1 m 8 2 6 2
y x
Next we use the slope and the given point 6,8 in the point-slope form of the equation of a line: y y1 m x x1 1 x 6 2 1 y 8 x 3 2 1 y x5 2 y 8
17. 15. 3 x 2 4 y 12 x-intercepts: 3x 2 4 0 12
f x x 2 3 2
Starting with the graph of y x 2 , shift the graph 2 units to the left y x 2 and down 3 2
2 units y x 2 3 .
3 x 2 12 x2 4 x 2 y-intercept: 3 0 4 y 12 2
4 y 12 y 3
The intercepts are 2, 0 , 2, 0 , and 0, 3 . Check x-axis symmetry: 3x 2 4 y 12 3x 2 4 y 12 different
Check y-axis symmetry: 3 x 4 y 12 2
3 x 2 4 y 12 same Check origin symmetry: 3 x 4 y 12 2
3x 2 4 y 12 different
327
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Chapter 3: Functions and Their Graphs
18.
f x
Project II
1 x
1 x y x 1 1 y 1 1 1 y 1 1 1 1 y 2 2
1. Silver: C x 20 0.16 x 200 0.16 x 12
x, y
20 C ( x) 0.16 x 12
1, 1
0 x 200 x 200
Gold: C x 50 0.08 x 1000 0.08 x 30
1,1
50.00 0 x 1000 C ( x) 0.08 x 30 x 1000
1 2, 2
Platinum: C x 100 0.04 x 3000 0.04 x 20 C ( x) 100.00 0 x 3000 0.04 x 20 x 3000
Cost (dollars)
C(x) 300
2 x if x 2 19. f x if x 2 x Graph the line y 2 x for x 2 . Two points
Silver Gold
200 Platinum
100 0
1000
2000
3000
4000 x
K-Bytes
on the graph are 0, 2 and 2, 0 .
3. Let y = #K-bytes of service over the plan minimum.
Graph the line y x for x 2 . There is a hole in the graph at x 2 .
Silver: 20 0.16 y 50 0.16 y 30 y 187.5 Silver is the best up to 187.5 200 387.5 K-bytes of service. Gold: 50 0.08 y 100 0.08 y 50 y 625 Gold is the best from 387.5 K-bytes to 625 1000 1625 K-bytes of service. Platinum: Platinum will be the best if more than 1625 K-bytes is needed. 4. Answers will vary.
Chapter 3 Projects Project I – Internet-based Project – Answers will vary
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Chapter 3 Projects Project III
6. C(4.5) = 100(4.5) + 140 4 (5 4.5) 2
1.
$738.62 The cost for the Steven’s cable would be $738.62.
Possible route 1
Driveway 2 miles
7. 5000(738.62) = $3,693,100 State legislated 5000(695.96) = $3,479,800 cheapest cost It will cost the company $213,300 more.
Cable box
5 miles
Possible route 2
Highway
House
2.
$140/mile L
4 (5 x )2
Project IV
2 miles
1. A r 2
Cable box
2. r 2.2t
5 miles $10 0/mile
C ( x) 100 x 140 L
3. r 2.2 2 4.4 ft
C ( x) 100 x 140 4 (5 x)
3.
2
r 2.2 2.5 5.5 ft
x C x
4. A (4.4) 2 60.82 ft 2
0 100 0 140 4 25 $753.92
A (5.5)2 95.03 ft 2
1 100 1 140 4 16 $726.10
5. A (2.2t ) 2 4.84 t 2
2 100 2 140 4 9 $704.78
6. A 4.84 (2) 2 60.82 ft 2
3 100 3 140 4 4 $695.98
A 4.84 (2.5) 2 95.03 ft 2
4 100 4 140 4 1 $713.05 5 100 5 140 4 0 $780.00
The choice where the cable goes 3 miles down the road then cutting up to the house seems to yield the lowest cost. 4. Since all of the costs are less than $800, there would be a profit made with any of the plans.
A(2.5) A(2) 95.03 60.82 68.42 ft/hr 2.5 2 0.5
8.
A(3.5) A(3) 186.27 136.85 98.84 ft/hr 3.5 3 0.5
9. The average rate of change is increasing. 10. 150 yds = 450 ft r 2.2t 450 t 204.5 hours 2.2
C(x ) dollars
7.
11. 6 miles = 31680 ft Therefore, we need a radius of 15,840 ft. 15,840 t 7200 hours 2.2
x miles
Using the MINIMUM function on a graphing calculator, the minimum occurs at x 2.96 . C(x) dollars
x miles
The minimum cost occurs when the cable runs for 2.96 mile along the road. 329
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Chapter 4 Linear and Quadratic Functions 11. a
Section 4.1 1. From the equation y 3x 1 , we see that the yintercept is 1 . Thus, the point 0, 1 is on the graph. We can obtain a second point by choosing a value for x and finding the corresponding value for y. Let x 1 , then y 3 1 1 1 . Thus, the point 1,1 is also on the graph. Plotting the two
12. d 13.
f x 2x 3
a.
Slope = 2; y-intercept = 3
b. Plot the point (0, 3). Use the slope to find an additional point by moving 1 unit to the right and 2 units up.
points and connecting with a line yields the graph below.
y y 3 5 2 2 2. m 2 1 x2 x1 1 2 3 3
3.
c.
average rate of change = 2
d.
increasing
14. g x 5 x 4
f (2) 4(2) 3 5 f (4) 4(4) 3 13 y f (4) f (2) 13 (5) 8 4 x 42 42 2
a.
Slope = 5; y-intercept = 4
b. Plot the point (0, 4) . Use the slope to find an additional point by moving 1 unit to the right and 5 units up.
4. 60 x 900 15 x 2850 75 x 900 2850 75 x 3750 x 50 The solution set is {50}. 5.
f 2 7.5 2 15 0
6. True
c.
average rate of change = 5
d.
increasing
7. slope; y-intercept 8. positive 9. True 10. False. The y-intercept is 8. The average rate of change is 2 (the slope). 330 Copyright © 2025 Pearson Education, Inc.
Section 4.1: Properties of Linear Functions and Linear Models 15. h x 3x 4 a. Slope = 3 ; y-intercept = 4 b. Plot the point (0, 4). Use the slope to find an additional point by moving 1 unit to the right and 3 units down.
c. d.
c.
average rate of change =
d.
increasing
1 4
2 18. h x x 4 3 2 a. Slope = ; y-intercept = 4 3 b. Plot the point (0, 4). Use the slope to find an additional point by moving 3 units to the right and 2 units down.
average rate of change = 3 decreasing
16. p x x 6 a. Slope = 1 ; y-intercept = 6 b. Plot the point (0, 6). Use the slope to find an additional point by moving 1 unit to the right and 1 unit down.
c.
average rate of change =
d.
decreasing
2 3
19. F x 4 a. Slope = 0; y-intercept = 4 b. Plot the point (0, 4) and draw a horizontal line through it.
c. d. 17.
average rate of change = 1 decreasing
1 x 3 4 1 a. Slope = ; y-intercept = 3 4 b. Plot the point (0, 3) . Use the slope to find an additional point by moving 4 units to the right and 1 unit up. f x
c. d.
average rate of change = 0 constant
331 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions 20. G x 2 a. b.
Slope = 0; y-intercept = 2 Plot the point (0, 2) and draw a horizontal line through it.
23.
x
y f x
2
8
1
3
0
0
Avg. rate of change = 3 8 1 2 0 3 0 1
5 5 1
3 3 1
y x
1 1 2 0 This is not a linear function, since the average rate of change is not constant. c. d. 21.
average rate of change = 0 constant
x
y f x
2
4
24.
Avg. rate of change =
y x
1
1
1 4 3 3 1 2 1
0
2
2 1 3 3 0 1 1
1
5
5 2 1 0
3 3 1
22.
8
x
y f x
2
1 4
1
1 2
0
1
Avg. rate of change =
y x
12 14 14 1 1 2 1 4 1 12 12 1 0 1 1 2
1 2 2 4 This is not a linear function since the average rate of change is not constant.
y f x
2
4
1
0
0
4
40 4 4 0 1 1
1
8
84 4 4 1 0 1
Avg. rate of change = 0 4
1 2
y x
4 4 1
12 8 4 4 2 1 1 This is a linear function with slope = 4, since the average rate of change is constant at 4.
8 5
3 3 2 1 1 This is a linear function with slope = –3, since the average rate of change is constant at –3.
2
x
25.
2
12
x
y f x
2
26
1
4
0
2
Avg. rate of change = 4 26 1 2
2 4 0 1
y x
22 22 1
6 6 1
1 –2 2 –10 This is not a linear function, since the average rate of change is not constant.
332 Copyright © 2025 Pearson Education, Inc.
Section 4.1: Properties of Linear Functions and Linear Models
26.
x
y f x
2
4
1
3.5
0
3
1
2.5
Avg. rate of change = 3.5 4 1 2
3 3.5 0 1
2.5 3 1 0 2 2.5
y x
29.
a.
27.
2
g x 2 x 5
f x 0 4x 1 0
0.5 0.5 1
0.5 0.5 1
0.5 0.5 1
x
b.
y f x
2
8
1
8
88 0 0 1 2 1
0
8
88 0 0 0 1 1
1
8
88 0 0 1 0 1
Avg. rate of change =
f x 0 x
1 4
1 1 The solution set is x x or , . 4 4
x
1 4
4x 1 0
0.5 0.5 2 1 1 This is a linear function, since the average rate of change is constant at 0.5
2
f x 4 x 1;
c.
f x g x 4 x 1 2 x 5 6x 6
y x
x 1
d.
f x g x 4 x 1 2 x 5 6x 6 x 1 The solution set is x x 1 or , 1 .
e.
88 0 0 2 1 1 This is a linear function with slope = 0, since the average rate of change is constant at 0.
28.
2
8
x
y f x
2
0
1
1
1 0 1 1 1 2 1
4
4 1 3 3 0 1 1
0
Avg. rate of change =
y x
1 9 2 16 This is not a linear function, since the average rate of change is not constant.
30.
f x 3 x 5;
a.
g x 2 x 15
f x 0 3x 5 0 x
b.
5 3
f x 0 3x 5 0 x
5 3
5 5 The solution set is x x or , . 3 3
333 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
c.
f x g x
32. a.
3x 5 2 x 15 5 x 10
b. The point (15, 60) is on the graph of y g ( x) , so the solution to g ( x) 60 is x 15 .
x2
d.
The point (5, 20) is on the graph of y g ( x) , so the solution to g ( x) 20 is x 5 .
f x g x
c.
3x 5 2 x 15 5 x 10 x2 The solution set is x x 2 or 2, .
e.
The point (15, 0) is on the graph of y g ( x) , so the solution to g ( x) 0 is x 15 .
d. The y-coordinates of the graph of y g ( x) are above 20 when the x-coordinates are smaller than 5. Thus, the solution to g ( x) 20 is
x x 5 or (, 5) . e.
The y-coordinates of the graph of y f ( x) are below 60 when the x-coordinates are larger than 15 . Thus, the solution to g ( x) 60 is
x x 15 or [15, ) . f. 31. a.
The point (40, 50) is on the graph of y f ( x) , so the solution to f ( x) 50 is x 40 .
b. The point (88, 80) is on the graph of y f ( x) , so the solution to f ( x) 80 is x 88 . c.
The point (40, 0) is on the graph of y f ( x) , so the solution to f ( x) 0 is x 40 .
d. The y-coordinates of the graph of y f ( x) are above 50 when the x-coordinates are larger than 40. Thus, the solution to f ( x) 50 is
x x 40 or (40, ) . e.
f.
The y-coordinates of the graph of y f ( x) are between 0 and 80 when the x-coordinates are between 40 and 88. Thus, the solution to 0 f ( x) 80 is x 40 x 88 or (40, 88) .
x 15 x 15 or (15, 15) . 33. a.
f x g x when their graphs intersect.
Thus, x 4 . b.
f x g x when the graph of f is above
the graph of g. Thus, the solution is x x 4 or (, 4) . 34. a.
f x g x when their graphs intersect.
Thus, x 2 . b.
The y-coordinates of the graph of y f ( x) are below 80 when the x-coordinates are smaller than 88. Thus, the solution to f ( x) 80 is x x 88 or (, 88] .
The y-coordinates of the graph of y f ( x) are between 0 and 60 when the xcoordinates are between 15 and 15. Thus, the solution to 0 f ( x) 60 is
f x g x when the graph of f is below
or intersects the graph of g. Thus, the solution is x x 2 or , 2 . 35. a.
f x g x when their graphs intersect.
Thus, x 6 . b.
g x f x h x when the graph of f is
above or intersects the graph of g and below the graph of h. Thus, the solution is x 6 x 5 or 6, 5 .
334 Copyright © 2025 Pearson Education, Inc.
Section 4.1: Properties of Linear Functions and Linear Models
36. a.
f x g x when their graphs intersect.
Thus, x 7 . b.
g x f x h x when the graph of f is
above the graph of g and below or intersects the graph of h. Thus, the solution is x 4 x 7 or 4, 7 .
b.
40. a. b.
37. C x 2.5 x 85 a.
C 40 2.5 40 85 $185 .
b.
Solve C x 2.5 x 85 245 2.5 x 85 245 2.5 x 160 160 64 miles x 2.5
c.
39. a.
Solve C x 2.5 x 85 150 2.5 x 85 150 2.5 x 65 65 326 miles x 2.5
a.
b.
c.
c.
16.64 64 minutes 0.26
Solve C x 0.26 x 5 50 0.26 x 5 50 0.26 x 45 45 x 173 minutes 0.26
d. The number of minutes cannot be negative, so x 0 . If there are 30 days in the month, then the number of minutes can be at most 30 24 60 43, 200 . Thus, the implied domain for C is {x | 0 x 43, 200} or [0, 43200] .
Solve S p D p . 600 50 p 1200 25 p 75 p 1800 1800 24 p 75 S 24 600 50 24 600
Solve D p S p . 1800 75 p 1800 p 75 24 p The demand will exceed supply when the price is less than $24 (but still greater than $0). That is, $0 p $24 .
b. Solve C x 0.26 x 5 21.64
x
152.4 2.47 x 54.10 2.47 x 98.3 x 39.8 cm
1200 25 p 600 50 p
C 50 0.26 50 5 $18 .
0.26 x 16.64
H (46.8) 2.47(46.8) 54.1 169.7 cm
Thus, the equilibrium price is $24, and the equilibrium quantity is 600 T-shirts.
38. C x 0.26 x 5
0.26 x 5 21.64
175.3 2.89 x 78.1 2.89 x 97.2 x 33.6 cm
41. S p 600 50 p; D p 1200 25 p
d. The number of miles driven cannot be negative, so the implied domain for C is {x | x 0} or [0, ) .
a.
H (37.1) 2.89(37.1) 78.1 185.3 cm
The price will eventually be increased.
42. S p 2000 3000 p; D p 10000 1000 p a.
Solve S p D p . 2000 3000 p 10000 1000 p 4000 p 12000 12000 3 4000 S 3 2000 3000 3 7000 p
Thus, the equilibrium price is $3, and the equilibrium quantity is 7000 hot dogs.
335 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions b. Solve D p S p .
e.
0.12 x 11000 1100 3155
10000 1000 p 2000 3000 p 12000 4000 p
c. 43. a.
11000, 44725 . b.
c.
d.
0.12 x 1320 1100 3155 0.12 x 220 3155 0.12 x 3375 x 28,125 A single filer with an adjusted gross income of $28,125 will have a tax bill of $3155.00.
12000 p 4000 3 p The demand will be less than the supply when the price is greater than $3. The price will eventually be decreased.
We are told that the tax function T is for adjusted gross incomes x between $11,000 and $44,725 (inclusive). Thus, the domain is x 11, 000 x 44, 725 or
44. a.
b.
The independent variable is payroll, p. The payroll tax only applies if the payroll exceeds $230 million. Thus, the domain of T is p | p 230 or (230, ) . T 288 0.20 288 230 11.6
The balance tax for the New York Mets was $11.6 million.
T 20, 000 0.12 20, 000 11000 1100 2180 If a single filer’s adjusted gross income is $20,000, then his or her tax bill will be $2180.00. The independent variable is adjusted gross income, x. The dependent variable is the tax bill, T.
We must solve T x 3155 .
c.
Evaluate T at p 230 and 288 million. T 230 0.20 230 230 0 million T 288 0.20 288 230 11.6 million
Thus, the points 230 million, 0 million and
288 million, 11.6 million are on the graph.
Evaluate T at x 11000 and 44725 . T 11, 000 0.12 11, 000 11, 000 1100 1100 T 44, 725 0.12 44, 725 11, 000 1100 5147 Thus, the points 11, 000,1100 and
44, 725,5147 are on the graph. d.
We must solve T p 6.7 . 0.20 p 230 6.7 0.20 p 46 6.7 0.20 p 52.7 p 263.5 If the luxury tax is $6.7 million, then the payroll of the team is $263.5 million.
336 Copyright © 2025 Pearson Education, Inc.
Section 4.1: Properties of Linear Functions and Linear Models 45. R x 8 x; C x 4.5 x 17,500 a.
c.
The graph of V ( x) 1000 x 3000
d.
V (2) 1000(2) 3000 1000 The computer’s book value after 2 years will be $1000.
e.
Solve V ( x) 2000 1000 x 3000 2000 1000 x 1000 x 1 The computer will have a book value of $2000 after 1 year.
Solve R x C x . 8 x 4.5 x 17,500 3.5 x 17,500 x 5000 The break-even point occurs when the company sells 5000 units.
b. Solve R x C x 8 x 4.5 x 17,500 3.5 x 17,500 x 5000 The company makes a profit if it sells more than 5000 units.
46. R ( x) 12 x; C ( x) 10 x 15, 000 a. Solve R ( x) C ( x) 12 x 10 x 15, 000 2 x 15, 000 x 7500 The break-even point occurs when the company sells 7500 units.
48. a.
the age in years of the machine and y = the value in dollars of the machine. So we have the points 0,120000 and 10, 0 . The
b. Solve R ( x) C ( x) 12 x 10 x 15, 000 2 x 15, 000 x 7500 The company makes a profit if it sells more than 7500 units. 47. a.
Consider the data points ( x, y ) , where x = the age in years of the computer and y = the value in dollars of the computer. So we have the points (0,3000) and (3, 0) . The slope formula yields: y 0 3000 3000 m 1000 x 30 3 The y-intercept is (0,3000) , so b 3000 . Therefore, the linear function is V ( x ) mx b 1000 x 3000 .
Consider the data points x, y , where x =
slope formula yields: y 0 120000 120000 m 12000 x 10 0 10 The y-intercept is 0, 120000 , so b 120, 000 . Therefore, the linear function is V x mx b 12, 000 x 120, 000 .
b. The age of the machine cannot be negative, and the book value of the machine will be $0 after 10 years. Thus, the implied domain for V is {x | 0 x 10} or [0, 10]. c.
The graph of V x 12, 000 x 120, 000
b. The age of the computer cannot be negative, and the book value of the computer will be $0 after 3 years. Thus, the implied domain for V is {x | 0 x 3} or [0, 3].
337 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
d.
V 4 12000 4 120000 72000
c.
The graph of C x 90 x 1805
d.
The cost of manufacturing 14 bicycles is given by C 14 90 14 1805 $3065 .
e.
Solve C x 90 x 1805 3780 90 x 1805 3780 90 x 1975 x 21.94 So approximately 21 bicycles could be manufactured for $3780.
The machine’s value after 4 years is given by $72,000. e.
49. a.
Solve V x 72000 . 12000 x 120000 72000 12000 x 48000 x4 The machine will be worth $72,000 after 4 years. Let x = the number of bicycles manufactured. We can use the cost function C x mx b , with m = 90 and b = 1800. Therefore C x 90 x 1800
b.
The graph of C x 90 x 1800
51. a.
c.
The cost of manufacturing 14 bicycles is given by C 14 90 14 1800 $3060 .
b.
d (2.4) 5.5(2.4) 13.2 cm
c.
19.8 5.5w w 3.6 kg
52. a.
d. Solve C x 90 x 1800 3780 90 x 1800 3780 90 x 1980 x 22 So 22 bicycles could be manufactured for $3780. 50. a.
b.
First we find the equation of the function given two ordered pairs: (1.5,9), (2.5,5) 59 4 2.5 1.5 y 9 4( x 1.5)
m
y 9 4 x 6 y 4 x 15 d ( w) 4 w 15
b.
The new daily fixed cost is 100 1800 $1805 20
d ( w) 5.5w
c.
d (0) 4(0) 15 15 cm
0 4 w 15 4 w 15
Let x = the number of bicycles manufactured. We can use the cost function C x mx b , with m = 90 and b = 1805.
w 3.75 lb
Therefore C x 90 x 1805
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Section 4.1: Properties of Linear Functions and Linear Models h h1 m( s s1 ) h 0 0.6( s 20) h 0.6s 12 Using function notation, we have h( s ) 0.6s 12 .
9 53. Solving F C 32 for C gives 5 5 C F 32 , and solving R F 459.67 for 9 F gives F R 459.67 . Then
d. The number of sodas cannot be negative, so s 0 . Likewise, the number of hot dogs cannot be negative, so, h( s ) 0 . 0.6 s 12 0 0.6 s 12 s 20 Thus, the implied domain for h(s) is {s | 0 s 20} or [0, 20] .
5 F 32 9 5 R 459.67 32 9 5 R 491.67 9 5 R 273.15 9
C
e.
So C ( R)
5 R 273.15 . 9
54. a.
f.
b.
c.
h Avg. rate of change = s
s
h
20
0
15
3
30 3 0.6 15 20 5
10
6
63 3 0.6 10 15 5
5
9
96 3 0.6 5 10 5
Since each input (soda) corresponds to a single output (hot dogs), we know that number of hot dogs purchased is a function of number of sodas purchased. Also, because the average rate of change is constant at 0.6 hot dogs per soda, the function is linear. From part (b), we know m 0.6 . Using ( s1 , h1 ) (20, 0) , we get the equation:
g.
If the number of hot dogs purchased increases by $1, then the number of sodas purchased decreases by 0.6. s-intercept: If 0 hot dogs are purchased, then 20 sodas can be purchased. h-intercept: If 0 sodas are purchased, then 12 hot dogs may be purchased.
55. The graph shown has a positive slope and a positive y-intercept. Therefore, the function from (d) and (e) might have the graph shown. 56. The graph shown has a negative slope and a positive y-intercept. Therefore, the function from (b) and (e) might have the graph shown. 57. A linear function f x mx b will be odd
provided f x f x . That is, provided m x b mx b .
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mx b mx b b b 2b 0 b0
Chapter 4: Linear and Quadratic Functions
So a linear function f x mx b will be odd provided b 0 .
61.
A linear function f x mx b will be even provided f x f x . That is, provided m x b mx b . mx b mx b mxb mx 0 2mx m0 So, yes, a linear function f x mx b cab be
f (3) f (1) 3 1 12 ( 2) 2 14 2 7
62.
even provided m 0 . 58. Answers may vary. x 2 4 x y 2 10 y 7 0
59.
( x 2 4 x 4) ( y 2 10 y 25) 7 4 25 ( x 2) 2 ( y 5) 2 62
Center: (2, 5) ; Radius = 6
63.
f ( x) x 2 10 x 7 ( x 2 10 x 25) 7 25 2
( x 5) 18
64. g ( x) 3x 2 15 x 13 3( x 2 5 x) 13 25 75 3 x 2 5 x 13 4 4 2
5 23 3x 2 4
60.
2x B x3 2(5) B f (5) 8 53 10 B 8 2 16 10 B B6
f ( x)
65. 4(0) 2 9 y 72 y 8 y-intercept: (0,8) 4 x 2 9(0) 72 4 x 2 72 x 2 18 x 18 3 2
The x-intercepts are: 3 2, 0 , 3 2, 0
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Section 4.2: Building Linear Models from Data 66. Since the radicand must not be negative then: x20 x 2 Also, x 4 0 x 4 The domain is: x | x 2, x 4 67.
No, the relation is not a function because an input, 1, corresponds to two different outputs, 5 and 12. 2. Let x1 , y1 1, 4 and x2 , y2 3, 8 . m
f (2) f (1) 2 (1) 5 11 3 6 3 2 y 5 2( x 2) y 5 2 x 4 y 2 x 9 m
y2 y1 8 4 4 2 x2 x1 3 1 2
y y1 m x x1 y 4 2 x 1 y 4 2x 2 y 2x 2
3. scatter plot 4. True 5. Linear relation, m 0 6. Nonlinear relation
68.
7. Linear relation, m 0 8. Linear relation, m 0 9. Nonlinear relation 10. Nonlinear relation 11. a.
Local maximum: f (1) 4 Local minimum: f (4.33) 14.52 Increasing: 2,1 , 4.33,8 Decreasing: 1, 4.33
b.
y 6 2( x 4)
Section 4.2 1.
Answers will vary. We select (4, 6) and (8, 14). The slope of the line containing these points is: 14 6 8 m 2 84 4 The equation of the line is: y y1 m( x x1 ) y 6 2x 8 y 2x 2
y
c.
x
341 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions d.
Using the LINear REGression program, the line of best fit is: y 2.0357 x 2.3571
e.
r 0.996
13. a.
f. b.
12. a.
b.
Answers will vary. We select (5, 2) and (11, 9). The slope of the line containing 92 7 these points is: m 11 5 6 The equation of the line is: y y1 m( x x1 ) 7 ( x 5) 6 7 35 y2 x 6 6 7 23 y x 6 6 y2
Answers will vary. We select (–2, –4) and (2, 5). The slope of the line containing these points is: 5 ( 4) 9 m . 2 ( 2) 4 The equation of the line is: y y1 m( x x1 ) 9 y ( 4) ( x ( 2)) 4 9 9 9 1 y4 x y x 4 2 4 2
c.
d
Using the LINear REGression program, the line of best fit is: y 2.2 x 1.2
e.
r 0.976
f.
c.
d.
Using the LINear REGression program, the line of best fit is: y 1.1286 x 3.8619
e.
r 0.991
14. a.
f. b.
Answers will vary. We select (–2, 7) and (2, 0). The slope of the line containing 07 7 7 . these points is: m 2 (2) 4 4 The equation of the line is:
342 Copyright © 2025 Pearson Education, Inc.
Section 4.2: Building Linear Models from Data y y1 m( x x1 ) 7 y 7 ( x ( 2)) 4 7 7 y7 x 4 2 7 7 y x 4 2
c.
c.
d.
Using the LINear REGression program, the line of best fit is: y 3.8613 x 180.2920
e.
r 0.957
f. d.
Using the LINear REGression program, the line of best fit is: y 1.8 x 3.6
e.
r 0.988
f.
15. a.
b.
16. a.
b.
Answers will vary. We select (–20,100) and (–10,140). The slope of the line containing these points is: 140 100 40 m 4 10 20 10
The equation of the line is: y y1 m( x x1 )
Selection of points will vary. We select (–30, 10) and (–14, 18). The slope of the line containing these points is: 18 10 8 1 m 16 2 14 30 The equation of the line is: y y1 m( x x1 ) 1 x (30) 2 1 y 10 x 15 2 1 y x 25 2 y 10
c.
y 100 4 x (20) y 100 4 x 80 y 4 x 180
d.
Using the LINear REGression program, the line of best fit is: y 0.4421x 23.4559
343 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions e.
r 0.944
18. a.
f.
b. c.
17. a.
Linear with positive slope. Answers will vary. We will use the points (200, 2.5) and (500, 5.8) . 5.8 2.5 3.3 0.011 500 200 300 N N1 m w w1
m
N 2.5 0.011 w 200 N 2.5 0.011w 2.2 N 0.011w 0.3
b. Linear with positive slope. c. Answers will vary. We will use the points (39.52, 210) and (66.45, 280) .
d.
280 210 70 2.5993316 66.45 39.52 26.93 y 210 2.5993316( x 39.52) y 210 2.5993316 x 102.7255848 y 2.599 x 107.288
m
d. e.
f.
e.
f.
x 62.3 : y 2.599(62.3) 107.274 269 We predict that a candy bar weighing 62.3 grams will contain 269 calories. If the weight of a candy bar is increased by one gram, then the number of calories will increase by 2.599.
19. a.
N (450) 0.011(450) 0.3 5.25 We predict that the length of a tornado that is 450 yards wide will be 6.25 miles. For each 1-yard increase in the width of a tornado, the length of the tornado increases by 0.011 mile, on average.
The independent variable is the number of hours spent playing video games and cumulative grade-point average is the dependent variable because we are using number of hours playing video games to predict (or explain) cumulative grade-point average.
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Section 4.2: Building Linear Models from Data 85 1.1857 p 1231.8279 1146.8279 1.1857 p 967 p A hurricane with a wind speed of 85 knots would have a pressure of approximately 967 millibars.
b.
21. c.
Using the LINear REGression program, the line of best fit is: G (h) 0.0942h 3.2763
d.
If the number of hours playing video games in a week increases by 1 hour, the cumulative grade-point average decreases 0.09, on average. G (8) 0.0942(8) 3.2763 2.52 We predict a grade-point average of approximately 2.52 for a student who plays 8 hours of video games each week. 2.40 0.0942(h) 3.2763 2.40 3.2763 0.0942h 0.8763 0.0942h 9.3 h
e.
f.
The data do not follow a linear pattern so it would not make sense to find the line of best fit. 22. a.
A student who has a grade-point average of 2.40 will have played approximately 9.3 hours of video games. 20. a.
The relation appears to be linear. b.
b.
Using the LINear REGression program, the line of best fit is:
w( p) 1.1857 p 1231.8279
c.
d. e.
For each 10-millibar increase in the atmospheric pressure, the wind speed of the tropical system decreases by 1.1857 knots, on average. w(990) 1.1857(990) 1231.8279 58 knots To find the pressure, we solve the following equation:
Using the LINear REGression program, the line of best fit is: y 4.94 x 50.60 .
c. For every 1 hour increase in study time the text score increases by 4.94%. d.
S (9) 4.94(9) 50.6 95.06 The test score about 95.06%.
e.
85 4.94 x 50.60 The student should study 34.4 4.94 x x7 about 7 hours.
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Chapter 4: Linear and Quadratic Functions
23. Using the ordered pairs (1, 5) and (3, 8) , the line of best fit is y 1.5 x 3.5 .
x2 4x 3
31.
x2 4x 3 0 a 1, b 4, c 3 x
4 16 12 2 4 28 2 2 2 7
The correlation coefficient is r 1 . This is reasonable because two points determine a line.
24. A correlation coefficient of 0 implies that the data do not have a linear relationship. 25. The y-intercept would be the calories of a candy bar with weight 0 which would not be meaningful in this problem. 26. G (0) 0.0942(0) 3.2763 3.2763 . The approximate grade-point average of a student who plays 0 hours of video games per week would be 3.28. 27. m
3 5 8 2 3 ( 1) 4 y y1 m x x1
y 5 2 x 1 y 5 2 x 2 y 2 x 3 or 2x y 3
x 2 25 x 5 So the domain is: x | x 5, 5
29.
32. 5(2 x 7) 6 x 10 3( x 9) 10 x 35 6 x 10 3x 27 4 x 35 3x 17 7 x 52 52 x 7 52 The solution set is , 7 f ( x)
( x) 2 5 2( x) 2
x2 f ( x) 5 2 x2 The function is even.
34. 3(0) 8 y 6 8 y 6
f ( x) 5 x 8 and g ( x) x 2 3x 4
( g f )( x) ( x 2 3 x 4) (5 x 8) x 2 3x 4 5 x 8 x 2 8 x 12
30. Since y is shifted to the left 3 units we would use y ( x 3) 2 . Since y is also shifted down 4
units, we would use y ( x 3) 2 4 .
The solution set is 2 7, 2 7 .
33.
28. The domain would be all real numbers except those that make the denominator zero. x 2 25 0
(4) (4) 2 4(1)(3) 2(1)
6 3 8 4 3 The y-intercept is 0, . 4 3x 8(0) 6 3x 6 x2 The x-intercept is 2, 0 . y
346 Copyright © 2025 Pearson Education, Inc.
Section 4.3: Quadratic Functions and Their Properties
35.
6 x 1 6 x 1 x 35 6 x 1 36 ( x 1) 35 x x 35 6 x 1 x 35 6 x 1
36.
x 35
x 35 6 x 1
4x
1
2
1 6 x 1
4x
3(2 x)
3(2 x) 1 1 1 (2 x) 2 (2 x) 2 (2 x ) 2 4 x 3(2 x) 4 x 6 3x 7x 6 1 1 1 (2 x) 2 (2 x ) 2 (2 x) 2 2
7 1 7 49 6. 3 x 2 x 3 2 3 36 7 49 3 x2 x 3 36 7 3 x 6
2
7. parabola 8. axis (or axis of symmetry) 9.
b 2a
10. True; a 2 0 . 11. True;
Section 4.3 1. y x 2 9 To find the y-intercept, let x 0 : y 02 9 9 . To find the x-intercept(s), let y 0 :
13. a 14. d 15. C
x 9
16. E
x 9 3 The intercepts are (0, 9), (3, 0), and (3, 0) .
2.
12. True
x2 9 0 2
17. F 18. A
2 x2 7 x 4 0
19. G
2 x 1 x 4 0 2 x 1 0 or x 4 0 2 x 1 or
x 4
1 or 2
x 4
x
b 4 2 2a 2 1
20. B 21. H 22. D
1 The solution set is 4, . 2
23. a.
Vertex: 3, 2
Axis of symmetry: x 3 b. concave up
2
25 1 3. (5) 4 2
4. right; 4 5. The discriminant is (5) 2 4(2)(8) 89 so there are two real solutions.
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Chapter 4: Linear and Quadratic Functions c.
27. a.
Vertex: 6,3
Axis of symmetry: x 6 b. concave up c.
24. a.
Vertex: 4, 1
Axis of symmetry: x 4 b. concave down c.
28. a.
Vertex: 1, 3
Axis of symmetry: x 1 b. concave up c.
25. a.
Vertex: 3,5
Axis of symmetry: x 3 b. concave down c.
29. a.
1 7 Vertex: , 2 6
Axis of symmetry: x
1 2
b. concave down c.
26. a.
Vertex: 1, 4
Axis of symmetry: x 1 b. concave up c. 30. a.
Vertex: 5, 0
Axis of symmetry: x 5 b. concave down
348 Copyright © 2025 Pearson Education, Inc.
Section 4.3: Quadratic Functions and Their Properties c.
33.
f ( x) ( x 2) 2 2
Using the graph of y x 2 , shift left 2 units, then shift down 2 units.
31.
1 2 x 4 Using the graph of y x 2 , compress vertically f ( x)
1 by a factor of . 4
32.
34.
f ( x) ( x 3) 2 10
Using the graph of y x 2 , shift right 3 units, then shift down 10 units.
f ( x) 2 x 2 4
Using the graph of y x 2 , stretch vertically by a factor of 2, then shift up 4 units.
35.
f ( x) x 2 4 x 2 ( x 2 4 x 4) 2 4 ( x 2) 2 2
Using the graph of y x 2 , shift left 2 units, then shift down 2 units.
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Chapter 4: Linear and Quadratic Functions
36.
f ( x) x 2 6 x 1 ( x 2 6 x 9) 1 9 ( x 3) 2 10
Using the graph of y x 2 , shift right 3 units, then shift down 10 units.
39.
f ( x) x 2 2 x
x2 2x
( x 2 2 x 1) 1 ( x 1) 2 1
37.
Using the graph of y x 2 , shift left 1 unit, reflect across the x-axis, then shift up 1 unit.
2
f ( x) 2 x 4 x 1
2 x2 2x 1 2( x 2 2 x 1) 1 2 2( x 1) 2 1
Using the graph of y x 2 , shift right 1 unit, stretch vertically by a factor of 2, then shift down 1 unit.
40.
f ( x) 2 x 2 6 x 2
2 x 2 3x 2 9 9 2 x 2 3 x 2 4 2 2
3 13 2 x 2 2
38.
f ( x) 3 x 2 6 x
2
3 x 2x
3 units, 2 reflect about the x-axis, stretch vertically by a 13 units. factor of 2, then shift up 2
Using the graph of y x 2 , shift right
2
3( x 2 x 1) 3 3( x 1) 2 3
Using the graph of y x 2 , shift left 1 unit, stretch vertically by a factor of 3, then shift down 3 units.
350 Copyright © 2025 Pearson Education, Inc.
Section 4.3: Quadratic Functions and Their Properties
41.
1 2 x x 1 2 1 x2 2 x 1 2 1 2 1 x 2x 1 1 2 2 1 3 2 x 1 2 2 Using the graph of y x 2 , shift left 1 unit, f ( x)
1 , then shift 2
3 units. 2
For f ( x) x 2 2 x , a 1 , b 2 , c 0. Since a 1 0 , the graph is concave up. The x-coordinate of the vertex is x
compress vertically by a factor of down
43. a.
b (2) 2 1 . 2a 2(1) 2
The y-coordinate of the vertex is b f f (1) (1) 2 2(1) 1 2 1. 2a Thus, the vertex is (1, 1) . The axis of symmetry is the line x 1 . b. The discriminant is b 2 4ac (2) 2 4(1)(0) 4 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: x2 2 x 0 x( x 2) 0 x 0 or x 2 The x-intercepts are –2 and 0 . The y-intercept is f (0) 0 . c.
42.
2 2 4 x x 1 3 3 2 2 x 2x 1 3 2 2 x2 2 x 1 1 3 3 2 5 2 x 1 3 3 Using the graph of y x 2 , shift left 1 unit, f ( x)
compress vertically by a factor of down
5 unit. 3
2 , then shift 3
d.
The domain is (, ) . The range is [1, ) .
e.
Decreasing on , 1 . Increasing on 1, .
f.
f ( x) 0 on , 2 0, f ( x) 0 on 2, 0
351 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
44. a.
b.
For f ( x) x 2 4 x , a 1 , b 4 , c 0 . Since a 1 0 , the graph is concave up. The x-coordinate of the vertex is b (4) 4 x 2. 2a 2(1) 2 The y-coordinate of the vertex is b f f (2) (2) 2 4(2) 4 8 4. 2a Thus, the vertex is (2, 4) . The axis of symmetry is the line x 2 . The discriminant is: b 2 4ac (4) 2 4(1)(0) 16 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: x2 4 x 0 x( x 4) 0 x 0 or x 4. The x-intercepts are 0 and 4. The y-intercept is f (0) 0 .
b.
The y-coordinate of the vertex is b f f (3) (3) 2 6(3) 2a 9 18 9. Thus, the vertex is (3, 9) . The axis of symmetry is the line x 3 . The discriminant is: b 2 4ac (6) 2 4(1)(0) 36 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: x2 6 x 0 x( x 6) 0 x 0 or x 6. The x-intercepts are 6 and 0 . The y-intercepts are f (0) 0 .
c.
c.
Increasing on , 3 .
f ( x) 0 on 6, 0 f ( x) 0 on , 6 0,
Decreasing on , 2 . 46. a.
f ( x) 0 on , 0 4, f ( x) 0 on 0, 4
45. a.
e.
f.
Increasing on 2, . f.
The domain is (, ) . The range is (, 9] . Decreasing on 3, .
d.. The domain is (, ) . The range is [4, ) . e.
d.
For f ( x) x 2 6 x , a 1 , b 6 , c 0 . Since a 1 0, the graph is concave down. The x-coordinate of the vertex is b (6) 6 x 3. 2a 2(1) 2
For f ( x) x 2 4 x, a 1, b 4 , c 0 . Since a 1 0 , the graph is concave down. The x-coordinate of the vertex is b 4 4 x 2. 2a 2(1) 2 The y-coordinate of the vertex is b f f (2) 2a
352 Copyright © 2025 Pearson Education, Inc.
(2) 2 4(2) 4.
Section 4.3: Quadratic Functions and Their Properties
The x-intercepts are found by solving: x2 2 x 8 0 ( x 4)( x 2) 0 x 4 or x 2. The x-intercepts are 4 and 2 . The y-intercept is f (0) 8 .
Thus, the vertex is (2, 4) . The axis of symmetry is the line x 2 . b.
The discriminant is: b 2 4ac 42 4(1)(0) 16 0, so the graph has two x-intercepts. The x-intercepts are found by solving: x2 4 x 0 x( x 4) 0 x 0 or x 4. The x-intercepts are 0 and 4. The y-intercept is f (0) 0 .
c.
c.
d. The domain is (, ) . The range is [9, ) . e.
Decreasing on , 1 . Increasing on 1, .
d.
The domain is (, ) . The range is (, 4] .
e.
Increasing on , 2 .
f.
f ( x) 0 on 4, 2
Decreasing on 2, . f.
48. a.
f ( x) 0 on 0, 4 f ( x) 0 on , 0 4,
47. a.
f ( x) 0 on , 4 2,
For f ( x) x 2 2 x 8 , a 1 , b 2 , c 8 . Since a 1 0 , the graph is concave up. The x-coordinate of the vertex is b 2 2 x 1 . 2a 2(1) 2 The y-coordinate of the vertex is b f f (1) (1) 2 2(1) 8 2a 1 2 8 9. Thus, the vertex is (1, 9) . The axis of symmetry is the line x 1 .
b. The discriminant is: b 2 4ac 22 4(1)(8) 4 32 36 0 , so the graph has two x-intercepts.
For f ( x) x 2 2 x 3, a 1, b 2, c 3. Since a 1 0 , the graph is concave up. The x-coordinate of the vertex is b (2) 2 x 1. 2a 2(1) 2 The y-coordinate of the vertex is b f f (1) 12 2(1) 3 4. 2a Thus, the vertex is (1, 4) . The axis of symmetry is the line x 1 .
b.
The discriminant is: b 2 4ac ( 2) 2 4(1)( 3) 4 12 16 0 ,
so the graph has two x-intercepts. The x-intercepts are found by solving: x2 2 x 3 0 ( x 1)( x 3) 0 x 1 or x 3.
353 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
The x-intercepts are 1 and 3 . The y-intercept is f (0) 3 .
c.
c.
d.
The domain is (, ) . The range is [4, ) .
e.
Decreasing on , 1 . Increasing on 1, .
f.
f ( x) 0 on , 1 3, f ( x) 0 on 1,3
49. a.
2
For f ( x) x 2 x 1 , a 1 , b 2 , c 1 . Since a 1 0 , the graph is concave up. The x-coordinate of the vertex is b 2 2 x 1 . 2a 2(1) 2 The y-coordinate of the vertex is b f f (1) 2a 2
(1) 2(1) 1 1 2 1 0. Thus, the vertex is (1, 0) . The axis of symmetry is the line x 1 .
b. The discriminant is: b 2 4ac 22 4(1)(1) 4 4 0 , so the graph has one x-intercept. The x-intercept is found by solving: x2 2 x 1 0 ( x 1) 2 0 x 1. The x-intercept is 1 . The y-intercept is f (0) 1 .
d.
The domain is (, ) . The range is [0, ) .
e.
Decreasing on , 1 . Increasing on 1, .
f.
f ( x) 0 on , 1 1, f ( x) 0 has no solution.
50. a.
For f ( x) x 2 6 x 9 , a 1 , b 6 , c 9 . Since a 1 0 , the graph is concave up. The x-coordinate of the vertex is b 6 6 x 3 . 2a 2(1) 2 The y-coordinate of the vertex is b f f (3) 2a
(3) 2 6(3) 9 9 18 9 0. Thus, the vertex is (3, 0) . The axis of symmetry is the line x 3 . b. The discriminant is: b 2 4ac 62 4(1)(9) 36 36 0 , so the graph has one x-intercept. The x-intercept is found by solving: x2 6 x 9 0 ( x 3) 2 0 x 3. The x-intercept is 3 . The y-intercept is f (0) 9 .
354 Copyright © 2025 Pearson Education, Inc.
Section 4.3: Quadratic Functions and Their Properties
c.
d.
The domain is (, ) . The range is [0, ) .
e.
Decreasing on , 3 . Increasing on 3, .
f.
The domain is (, ) . The range is
e.
1 1 Decreasing on , . Increasing on , .
f.
f ( x) 0 on ,
52. a.
For f ( x) 2 x 2 x 2 , a 2 , b 1 , c 2 . Since a 2 0 , the graph is concave up. The x-coordinate of the vertex is b (1) 1 x . 2a 2(2) 4 The y-coordinate of the vertex is
b.
The axis of symmetry is the line x 1 . 4 The discriminant is: b 2 4ac (1) 2 4(2)(2) 1 16 15 , so the graph has no x-intercepts. The y-intercept is f (0) 2 .
4
4
For f ( x) 4 x 2 2 x 1 , a 4 , b 2 , c 1 . Since a 4 0 , the graph is concave up. The x-coordinate of the vertex is b (2) 2 1 x . 2a 2(4) 8 4 The y-coordinate of the vertex is 2
b 1 1 1 f f 4 2 1 a 2 4 4 4 1 1 3 1 . 4 2 4 Thus, the vertex is 1 , 3 . 4 4 The axis of symmetry is the line x 1 . 4
2
b 1 1 1 f f 2 2 2a 4 4 4 1 1 15 2 . 8 4 8 1 15 Thus, the vertex is , . 4 8
15 , . 8
f ( x) 0 has no solution.
f ( x) 0 on , 3 3, f ( x) 0 has no solution.
51. a.
d.
b.
The discriminant is: b 2 4ac (2) 2 4(4)(1) 4 16 12 , so the graph has no x-intercepts. The y-intercept is f (0) 1 .
c.
c.
355 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
d.
The domain is (, ) .
e.
The range is 3 , . 4 e.
f.
Decreasing on , 1 . 4 Increasing on 1 , . 4
f.
f ( x) 0 on ,
For f ( x) 2 x 2 2 x 3 , a 2 , b 2 , c 3 . Since a 2 0 , the graph is concave down. The x-coordinate of the vertex is b (2) 2 1 x . 2a 2(2) 4 2 The y-coordinate of the vertex is
54. a. For f ( x) 3 x 2 3 x 2 , a 3 , b 3 , c 2 . Since a 3 0 , the graph is concave down. The x-coordinate of the vertex is b 3 3 1 x . 2a 2(3) 6 2 The y-coordinate of the vertex is 2
b 1 1 1 f f 3 3 2 2a 2 2 2 3 3 5 2 . 4 2 4 1 5 Thus, the vertex is , . 2 4 1 The axis of symmetry is the line x . 2 The discriminant is: b 2 4ac 32 4(3)(2) 9 24 15 , so the graph has no x-intercepts. The y-intercept is f (0) 2 .
2
b 1 1 1 f f 2 2 3 a 2 2 2 2 1 5 1 3 . 2 2 1 5 Thus, the vertex is , . 2 2 1 The axis of symmetry is the line x . 2
b.
f ( x) 0 has no solution. f ( x) 0 on ,
f ( x) 0 has no solution.
53. a.
1 Increasing on , . 2 1 Decreasing on , . 2
b.
The discriminant is: b 2 4ac 22 4(2)(3) 4 24 20 , so the graph has no x-intercepts. The y-intercept is f (0) 3 .
c.
c. d.
The domain is (, ) . 5 The range is , . 4
d.
The domain is (, ) . 5 The range is , . 2
e.
1 Increasing on , . 2 1 Decreasing on , . 2
356 Copyright © 2025 Pearson Education, Inc.
Section 4.3: Quadratic Functions and Their Properties
f.
f ( x) 0 has no solution.
f.
f ( x) 0 on ,
55. a.
For f ( x) 3 x 2 6 x 2 , a 3 , b 6 , c 2 . Since a 3 0 , the graph is concave up. The x-coordinate of the vertex is b 6 6 x 1 . 2a 2(3) 6 The y-coordinate of the vertex is b f f (1) 3(1) 2 6(1) 2 2a 3 6 2 1. Thus, the vertex is (1, 1) . The axis of symmetry is the line x 1 .
3 3 f ( x) 0 on , 1 , 1 3 3 3 3 f ( x) 0 on 1 , 1 3 3
56. a.
For f ( x) 2 x 2 5 x 3 , a 2 , b 5 , c 3 . Since a 2 0 , the graph is concave up. The x-coordinate of the vertex is b 5 5 x . 2a 2(2) 4 The y-coordinate of the vertex is b 5 f f 2a 4 2
b.
5 5 2 5 3 4 4 25 25 3 8 4 1 . 8 5 1 Thus, the vertex is , . 4 8
The discriminant is: b 2 4ac 62 4(3)(2) 36 24 12 , so the graph has two x-intercepts. The x-intercepts are found by solving: 3x 2 6 x 2 0 x
b b 2 4ac 2a
6 12 6 2 3 3 3 6 6 3 3 3 and 1 . The x-intercepts are 1 3 3 The y-intercept is f (0) 2 .
The axis of symmetry is the line x
b.
c.
The discriminant is: b 2 4ac 52 4(2)(3) 25 24 1 , so the graph has two x-intercepts. The x-intercepts are found by solving: 2 x2 5x 3 0 (2 x 3)( x 1) 0 3 x or x 1. 2 3 The x-intercepts are and 1 . 2 The y-intercept is f (0) 3 .
c.
d.
The domain is (, ) . The range is 1, .
e.
Decreasing on , 1 . Increasing on 1, .
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5 . 4
Chapter 4: Linear and Quadratic Functions
d.
The domain is (, ) . 1 The range is , . 8
e.
f.
57. a.
5 Decreasing on , . 4 5 Increasing on , . 4 3 f ( x) 0 on , 1, 2 3 f ( x) 0 on , 1 2
For f ( x) 4 x 2 6 x 2 , a 4 , b 6 , c 2 . Since a 4 0 , the graph is concave down. The x-coordinate of the vertex is 3 b (6) 6 x . 2a 2(4) 8 4 The y-coordinate of the vertex is
d. The domain is (, ) . 17 The range is , . 4
e.
f.
2
b.
b 3 3 3 f f 4 6 2 a 2 4 4 4 9 9 17 2 . 4 2 4 3 17 Thus, the vertex is , . 4 4 3 The axis of symmetry is the line x . 4 The discriminant is: b 2 4ac (6) 2 4(4)(2) 36 32 68 , so the graph has two x-intercepts. The x-intercepts are found by solving: 4 x 2 6 x 2 0
x
58. a.
For f ( x) 3 x 2 8 x 2, a 3, b 8, c 2 . Since a 3 0 , the graph is concave up. The x-coordinate of the vertex is b (8) 8 4 x . 2a 2(3) 6 3 The y-coordinate of the vertex is 2
b 4 4 4 f f 3 8 2 2 a 3 3 3 16 32 10 2 . 3 3 3 4 10 Thus, the vertex is , . 3 3
c.
3 17 3 17 f ( x) 0 on , 4 4 3 17 3 17 f ( x) 0 on , 4 , 4
b b 2 4ac (6) 68 2a 2(4)
6 68 6 2 17 3 17 8 8 4 3 17 3 17 The x-intercepts are and . 4 4 The y-intercept is f (0) 2 .
3 Decreasing on , . 4 3 Increasing on , . 4
The axis of symmetry is the line x b.
4 . 3
The discriminant is: b 2 4ac (8) 2 4(3)(2) 64 24 40 , so the graph has two x-intercepts. The x-intercepts are found by solving: 3x 2 8 x 2 0
358 Copyright © 2025 Pearson Education, Inc.
Section 4.3: Quadratic Functions and Their Properties
x
b b 2 4ac (8) 40 2a 2(3)
8 40 8 2 10 4 10 6 6 3 4 10 4 10 The x-intercepts are and . 3 3 The y-intercept is f (0) 2 .
1 a 0 1 2 2
1 a 1 2 2
1 a 2 1 a The quadratic function is f x x 1 2 x 2 2 x 1 . 2
60. Consider the form y a x h k . From the 2
graph we know that the vertex is 2,1 so we have h 2 and k 1 . The graph also passes through the point x, y 0,5 . Substituting
c.
these values for x, y, h, and k, we can solve for a: 5 a 0 2 1 2
5 a 2 1 2
5 4a 1 4 4a 1 a The quadratic function is f x x 2 1 x2 4 x 5 . 2
d. The domain is (, ) .
61. Consider the form y a x h k . From the
10 The range is , . 3
graph we know that the vertex is 3,5 so we
e.
f.
2
have h 3 and k 5 . The graph also passes through the point x, y 0, 4 . Substituting
4 Decreasing on , . 3 4 Increasing on , . 3
these values for x, y, h, and k, we can solve for a: 4 a 0 (3) 5 2
4 a 3 5 2
4 10 4 10 , f ( x) 0 on , 3 3 4 10 4 10 , f ( x) 0 on 3 3
59. Consider the form y a x h k . From the 2
graph we know that the vertex is 1, 2 so we
4 9a 5 9 9a 1 a The quadratic function is
f x x 3 5 x 2 6 x 4 . 2
62. Consider the form y a x h k . From the 2
graph we know that the vertex is 2,3 so we
have h 1 and k 2 . The graph also passes through the point x, y 0, 1 . Substituting
have h 2 and k 3 . The graph also passes through the point x, y 0, 1 . Substituting
these values for x, y, h, and k, we can solve for a:
these values for x, y, h, and k, we can solve for a:
359 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions 1 a 0 2 3 2
1 a 2 3 2
1 4a 3 4 4a 1 a The quadratic function is
f x x 2 3 x2 4 x 1 . 2
63. Consider the form y a x h k . From the 2
graph we know that the vertex is 1, 3 so we have h 1 and k 3 . The graph also passes through the point x, y 3,5 . Substituting these values for x, y, h, and k, we can solve for a: 5 a 3 1 (3) 2
66. For f ( x) 2 x 2 12 x, a 2, b 12, c 0, . Since a 2 0, the graph opens down, so the vertex is a maximum point. The maximum b 12 12 3. occurs at x 2a 2(2) 4 The maximum value is f (3) 2(3) 2 12(3) 18 36 18 . 67. For f ( x) 2 x 2 12 x 3, a 2, b 12, c 3. Since a 2 0, the graph opens up, so the vertex is a minimum point. The minimum occurs at b 12 12 x 3. The minimum value is 2a 2(2) 4 f (3) 2(3) 2 12(3) 3 18 36 3 21 .
68. For f ( x) 4 x 2 8 x 3, a 4, b 8, c 3. Since a 4 0, the graph opens up, so the vertex is a minimum point. The minimum occurs at b (8) 8 x 1. The minimum value is 2a 2(4) 8
5 a 2 3 2
5 4a 3 8 4a 2a The quadratic function is
f (1) 4(1) 2 8(1) 3 4 8 3 1 .
f x 2 x 1 3 2 x 2 4 x 1 . 2
64. Consider the form y a x h k . From the 2
graph we know that the vertex is 2, 6 so we have h 2 and k 6 . The graph also passes through the point x, y 4, 2 . Substituting these values for x, y, h, and k, we can solve for a: 2 a 4 (2) 6 2
2 a 2 6 2
2 4a 6 8 4a 2 a The quadratic function is f x 2 x 2 6 2 x 2 8 x 2 .
69. For f ( x) x 2 6 x 1 , a 1, b 6 , c 4 . Since a 1 0, the graph opens down, so the vertex is a maximum point. The maximum occurs b 6 6 3 . The maximum value at x 2a 2(1) 2
is f (3) (3) 2 6(3) 1 9 18 1 8 . 70. For f ( x) 2 x 2 8 x 3 , a 2, b 8, c 3. Since a 2 0, the graph opens down, so the vertex is a maximum point. The maximum 8 8 b occurs at x 2 . The 2a 2( 2) 4 maximum value is f (2) 2(2) 2 8(2) 3 8 16 3 11 .
2
65. For f ( x) 3 x 2 24 x, a 3, b 24, c 0 . Since a 3 0, the graph opens up, so the vertex is a minimum point. The minimum b 24 24 occurs at x 4. 2a 2(3) 6 The minimum value is f (4) 3(4) 2 24(4) 48 96 48 .
71. For f ( x) 5 x 2 20 x 3 , a 5, b 20, c 3. Since a 5 0, the graph opens down, so the vertex is a maximum point. The maximum occurs b 12 12 2 . The maximum value at x 2a 2(3) 6
is f (2) 5(2)2 20(2) 3 20 40 3 23 . 72. For f ( x) 4 x 2 4 x , a 4, b 4, c 0. Since a 4 0, the graph opens up, so the vertex
360 Copyright © 2025 Pearson Education, Inc.
Section 4.3: Quadratic Functions and Their Properties
is a minimum point. The minimum occurs at b ( 4) 4 1 x . The minimum value is 2a 2(4) 8 2
75. a and d.
2
1 1 1 f 4 4 1 2 1 . 2 2 2
73. Use the form f ( x) a ( x h) 2 k . The vertex is (0, 2) , so h = 0 and k = 2. f ( x) a( x 0) 2 2 ax 2 2 . Since the graph passes through (1, 8) , f (1) 8 . f ( x) ax 2 2 8 a (1) 2 2 8 a2 6a f x 6 x2 2 .
b.
a 6, b 0, c 2
f ( x) g ( x) 2x 1 x2 4 0 x2 2x 3 0 ( x 1)( x 3)
74. Use the form f ( x) a ( x h) 2 k . The vertex is (1, 4) , so h 1 and k 4 .
x 1 0 or x 3 0 x 1 x3
f ( x) a ( x 1) 2 4 . Since the graph passes through (1, 8) , f (1) 8 .
The solution set is {1, 3}.
8 a (1 1) 2 4 8 a (2) 2 4 8 4a 4 12 4a 3 a f ( x) 3( x 1) 2 4
c.
f (1) 2(1) 1 2 1 3 g (1) (1) 2 4 1 4 3 f (3) 2(3) 1 6 1 5 g (3) (3) 2 4 9 4 5 Thus, the graphs of f and g intersect at the points (1, 3) and (3, 5) .
76. a and d.
3( x 2 2 x 1) 4 3 x 2 6 x 3 4 3 x 2 6 x 1 a 3, b 6, c 1
361 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
b.
f ( x) g ( x) 2 x 1 x 2 9 0 x2 2 x 8 0 ( x 4)( x 2)
78. a and d.
x 4 0 or x 2 0 x 4 x2
The solution set is {4, 2}. c.
f (4) 2(4) 1 8 1 7 g (4) (4) 2 9 16 9 7 f (2) 2(2) 1 4 1 5 g (2) (2) 2 9 4 9 5 Thus, the graphs of f and g intersect at the points 4, 7 and 2, 5 .
b.
77. a and d.
f x g x x2 9 2 x 1 0 x2 2x 8 0 x 4 x 2 x 4 0 or x 2 0 x 4 x2 The solution set is {4, 2}.
c.
f 4 4 9 16 9 7 2
g 4 2 4 1 8 1 7
b.
f x g x
f 2 2 9 4 9 5 2
g 2 2 2 1 4 1 5
x 2 4 2 x 1 0 x 2x 3 2
Thus, the graphs of f and g intersect at the points 4, 7 and 2, 5 .
0 x 1 x 3 x 1 0 or x 3 0 x 1 x3 The solution set is {1, 3}.
c.
79. a and d.
f 1 1 4 1 4 3 2
g 1 2 1 1 2 1 3 f 3 3 4 9 4 5 2
g 3 2 3 1 6 1 5
Thus, the graphs of f and g intersect at the points 1, 3 and 3, 5 .
362 Copyright © 2025 Pearson Education, Inc.
Section 4.3: Quadratic Functions and Their Properties f x g x
b.
81. a.
x 2 5 x x 2 3x 4
1( x (3))( x 1)
0 2 x2 2 x 4
( x 3)( x 1) x 2 2 x 3 For a 2 : f ( x) 2( x (3))( x 1)
0 x2 x 2 0 x 1 x 2 x 1 0 or x 2 0 x 1 x2 The solution set is {1, 2}.
c.
For a 1: f ( x) a ( x r1 )( x r2 )
2( x 3)( x 1) 2( x 2 2 x 3) 2 x 2 4 x 6 For a 2 : f ( x) 2( x (3))( x 1)
f 1 1 5 1 1 5 6 2
2( x 3)( x 1)
g 1 1 3 1 4 1 3 4 6 2
2( x 2 2 x 3) 2 x 2 4 x 6 For a 5 : f ( x) 5( x (3))( x 1)
f 2 2 5 2 4 10 6 2
g 2 22 3 2 4 4 6 4 6
Thus, the graphs of f and g intersect at the points 1, 6 and 2, 6 . 80. a and d.
5( x 3)( x 1) 5( x 2 2 x 3) 5 x 2 10 x 15
b. The x-intercepts are not affected by the value of a. The y-intercept is multiplied by the value of a . c.
The axis of symmetry is unaffected by the value of a . For this problem, the axis of symmetry is x 1 for all values of a.
d. The x-coordinate of the vertex is not affected by the value of a. The y-coordinate of the vertex is multiplied by the value of a . e. f x g x
b.
82. a.
x 7 x 6 x2 x 6 2
0 2 x2 6 x 0 2 x x 3 2 x 0 or x 3 0 x0 x3 The solution set is {0, 3}.
c.
f 0 0 7 0 6 6 2
g 0 02 0 6 6 f 3 3 7 3 6 9 21 6 6
The x-coordinate of the vertex is the mean of the x-intercepts. For a 1: f ( x) 1( x (5))( x 3) ( x 5)( x 3) x 2 2 x 15 For a 2 : f ( x) 2( x (5))( x 3) 2( x 5)( x 3) 2( x 2 2 x 15) 2 x 2 4 x 30 For a 2 : f ( x) 2( x (5))( x 3) 2( x 5)( x 3)
2
g 3 32 3 6 9 3 6 6
Thus, the graphs of f and g intersect at the points 0, 6 and 3, 6 .
2( x 2 2 x 15) 2 x 2 4 x 30 For a 5 : f ( x) 5( x (5))( x 3) 5( x 5)( x 3) 5( x 2 2 x 15) 5 x 2 10 x 75
363 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions b. The x-intercepts are not affected by the value of a. The y-intercept is multiplied by the value of a . c.
d.
The axis of symmetry is unaffected by the value of a . For this problem, the axis of symmetry is x 1 for all values of a.
d. The x-coordinate of the vertex is not affected by the value of a. The y-coordinate of the vertex is multiplied by the value of a . e.
83. a.
The x-coordinate of the vertex is the mean of the x-intercepts. x
4 b 2 2a 2 1
y f 2 2 4 2 21 25 2
84. a.
The vertex is 2, 25 .
2
The vertex is 1, 9 .
x 4 x 21 0 2
x2 2x 8 0
x3 0
or
x 7
x 4 x 2 0
x3
x40
The x-intercepts of f are (7, 0) and (3, 0).
x2
The x-intercepts of f are (4, 0) and (2, 0).
x 4 x 21 21 2
f x 8
c.
x2 4x 0
x 2 x 8 8 2
x x 4 0 x 0 or
x2 0
or
x 4
f x 21
c.
f x 0
b.
x 7 x 3 0 x7 0
2 b 1 2a 2 1
y f 1 1 2 1 8 9
f x 0
b.
x
x2 2x 0
x40
x x 2 0
x 4
x 0 or
The solutions f x 21 are 4 and 0.
x20 x 2
Thus, the points 4, 21 and 0, 21 are
The solutions f x 8 are 2 and 0. Thus,
on the graph of f.
the points 2, 8 and 0, 8 are on the graph of f. d.
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Section 4.3: Quadratic Functions and Their Properties
85. h( x) a.
32 x 2 8 2 x 200 x x 200 625 (50) 2
f.
8 , b 1, c 200. 625 The maximum height occurs when b 1 625 x 39.1 feet 2a 2 8 / 625 16 a
The maximum height is 2
625 8 625 625 h 16 200 16 625 16 7025 219.5 feet. 32
c.
8 2 x x 200 0 625
x
1 12 4 8 / 625 (200) 2 8 / 625
b.
1 11.24 x 0.0256
32 x 2 2 2 x x x 2 625 (100) 2 a , b 1, c 0. 625 The maximum height occurs when b 1 625 x 156.25 feet 2a 2 2 / 625 4 h( x )
The maximum height is 2
x 91.90 or x 170 Since the distance cannot be negative, the projectile strikes the water approximately 170 feet from the base of the cliff. d.
e.
86. a.
Solving when h( x ) 0 :
12 4 8 / 625 100
1 6.12 0.0256 2 8 / 625 x 57.57 or x 135.70 Since the distance cannot be negative, the projectile is 100 feet above the water when it is approximately 135.7 feet from the base of the cliff. x
from base of the cliff. b.
8 2 x x 200 100 625 8 2 x x 100 0 625
Using the MAXIMUM function
Using the ZERO function
625 2 625 625 h 4 4 625 4 625 78.125 feet 8
c.
Solving when h( x) 0 : 2 2 x x0 625 2 x x 1 0 625 2 x 0 or x 1 0 625
2 x 625 625 x 0 or x 312.5 2 Since the distance cannot be zero, the projectile lands 312.5 feet from where it was fired. d. x 0 or
e.
1
Using the MAXIMUM function
365 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
The maximum revenue is 1 2 R 2900 2900 2900 2900 2 4205000 8410000 $4, 205, 000
Using the ZERO function
f.
Solving when h x 50 :
2 2 x x 50 625
2 2 x x 50 0 625
x
1
12 4 2 / 625 50
89. a.
b. The minimum marginal cost is 2 b f f 70 70 140 70 7400 2a 4900 9800 7400 $2500 90. a.
2 2 / 625
1 0.36 1 0.6 0.0064 0.0064 x 62.5 or x 250 The projectile is 50 feet above the ground 62.5 feet and 250 feet from where it was fired.
1 2 1 p 2900 p , a , b 2900, c 0. 2 2 1 Since a 0, the graph is a parabola that is 2 concave down, so the vertex is a maximum point. The maximum occurs at b 2900 2900 p 2900 . Thus, the 2a 2 1/ 2 1
88. R ( p )
C ( x) 5 x 2 200 x 4000 , a 5, b 200, c 4000. Since a 5 0, the graph is concave up, so the vertex is a minimum point. The minimum marginal cost b 200 200 20 , occurs at x 2a 2(5) 10 20,000 smartphones manufactured.
87. R ( p ) 4 p 2 4000 p , a 4, b 4000, c 0. Since a 4 0 the graph is a parabola that is concave down, so the vertex is a maximum point. b 4000 500 . The maximum occurs at p 2a 2( 4) Thus, the unit price should be $500 for maximum revenue. The maximum revenue is R (500) 4(500) 2 4000(500) 1000000 2000000 $1, 000, 000
C ( x) x 2 140 x 7400 , a 1, b 140, c 7400. Since a 1 0, the graph is concave up, so the vertex is a minimum point. The minimum marginal cost b (140) 140 70 , occurs at x 2a 2(1) 2 70,000 digital music players produced.
b. The minimum marginal cost is 2 b f f 20 5 20 200 20 4000 2a 2000 4000 4000 $2000 91. a.
R ( x) 75 x 0.2 x 2 a 0.2, b 75, c 0 The maximum revenue occurs when b 75 75 x 187.5 2a 2 0.2 0.4
The maximum revenue occurs when x 187 or x 188 watches. The maximum revenue is: R (187) 75 187 0.2 187 $7031.20 2
R (188) 75 188 0.2 188 $7031.20
unit price should be $2900 for maximum revenue. 366 Copyright © 2025 Pearson Education, Inc.
2
Section 4.3: Quadratic Functions and Their Properties
b.
P( x) R x C x
b.
75 x 0.2 x 2 32 x 1750
0 200 1.1v 0.06v 2 x 1.1 1.1 4 0.06 200 2
0.2 x 2 43 x 1750
c.
2 0.06
P ( x ) 0.2 x 2 43 x 1750 a 0.2, b 43, c 1750
x
b 43 43 107.5 2a 2 0.2 0.4
The maximum profit occurs when x 107 or x 108 watches. The maximum profit is: 2
$561.20 P (108) 0.2 108 43 108 1750 2
94. a.
0 0.000092 x 2 0.020741x 0 x(0.000092 x 0.020741) x 0 or x 225 A width of 0 has no meaning so the width is 225 feet.
R ( x) 9.5 x 0.04 x 2
b.
118.75 119 boxes of candy The maximum revenue is: 2
P( x) R x C x 0.04 x 2 8.25 x 250
c.
P( x) 0.04 x 2 8.25 x 250 a 0.04, b 8.25, c 250 The maximum profit occurs when b 8.25 8.25 x 2a 2 0.04 0.08
96. Vertex: (3, 5); Center: 3,1 d (3 3) 2 (1 5) 2 (6) 2 (4) 2 36 16
P(103) 0.04 103 8.25 103 250 2
93. a.
d (v) 1.1v 0.06v
52 2 13
97.
d. Answers will vary.
b .020741 112.723 2a 2 0.000092
95. We are given: V ( x) kx(a x) kx 2 akx . The reaction rate is a maximum when: b ak ak a x . 2a 2(k ) 2k 2
103.125 103 boxes of candy The maximum profit is: $175.39
x
y (112.723) 0.000092(112.723) 2 0.020741(112.723) 213.37 1.17 feet or 14 inches
R(119) 9.5 119 0.04 119 $564.06 9.5 x 0.04 x 2 1.25 x 250
0.12
The 1.1v term might represent the reaction time.
a 0.04, b 9.5, c 0 The maximum revenue occurs when b 9.5 9.5 x 2a 2 0.04 0.08
b.
0.12 1.1 7.015
c.
$561.20
d. Answers will vary.
1.1 49.21
v 49 or v 68 Disregard the negative value since we are talking about speed. So the maximum speed you can be traveling would be approximately 49 mph.
P (107) 0.2 107 43 107 1750
92. a.
200 1.1v 0.06v 2
x 2 10 x y 2 8 y 32 x 2 10 x 25 y 2 8 y 16 32 25 16
2
x 5 y 4 9 2
d (45) 1.1(45) 0.06(45) 49.5 121.5 171 ft. 2
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2
center: (5, 4)
Chapter 4: Linear and Quadratic Functions
d ( x)2 2 x 2 8 x 16 .
3( x 2 2 x) 1 3( x 2 2 x 1) 1 3
Since a = 2 > 0, it has a minimum. The xcoordinate of the minimum point of d ( x) will 2
3( x 1) 4 vertex: (1, 4) 2
also provides the x-coordinate of the minimum b (8) 8 2 . So, 2 is point of d ( x) : x 2a 2(2) 4 the x-coordinate of the point on the line y that is closest to the point (4, 1). The y-coordinate is y = 2 + 1 = 3. Thus, the point is (2, 3) is the point on the line y = x + 1 that is closest to (4, 1).
d (1 (5)) 2 (4 (4)) 2 (6) 2 (8) 2 36 64 100 10 2
98. If x is even, then ax and bx are even and ax 2 bx is even, which means that ax 2 bx c is odd. If x is odd, then ax 2 and bx are odd and ax 2 bx is even, which means that ax 2 bx c is odd. In either case f ( x) is odd. 99. Let (x, y) be a point on the line y = x. Then the distance from (x, y) to the point (3, 1) is d
x 3 y 1 . Since y = x, 2
d ( x)
2
x 32 x 12
x2 6x 9 x2 2 x 1 2 x 2 8 x 10
d ( x)2 2 x 2 8 x 10 . Because a = 2 > 0, it has a minimum. The xcoordinate of the minimum point of d ( x) also 2
provides the x-coordinate of the minimum point of b (8) 8 2 . So, 2 is the xd ( x) : x 2a 2(2) 4 coordinate of the point on the line y = x that is closest to the point (3, 1). Since y = x, the ycoordinate is also 2. So the point (2, 2) is the point on the line y = x that is closest to (3, 1). 100. Let (x, y) be a point on the line y = x + 1. Then the distance from (x, y) to the point (4, 1) is d
x 4 y 1 . Since y = x + 1, 2
d ( x)
2
x 4 2 ( x 1) 12
x 2 8 x 16 x 2 2 x 2 8 x 16
101. The derivative f x 3x 2 14 x 5 is a
quadratic function, so its graph is a parabola. Note that a = 3, b = 14 , and c = 5 . Because a 3 0 , the parabola is concave up. The x14 7 coordinate of the vertex is h . The 23 3 y-coordinate is 2
64 7 7 7 k f 3 14 5 . So, the 3 3 3 3 7 64 vertex is , ,which lies below the x-axis. 3 3 So the graph of f is below the x-axis when x is between the x-intercepts, and the graph of f is above the x-axis when x is outside the xintercepts. The find the x-intercepts, solve f ( x) 0 : 3x 2 14 x 5 0 (3 x 1)( x 5) 0 1 x or x 5. 3 1 The x-intercepts are and 5. So f ( x) 0 for 3 1 on the interval ,5 , and f ( x) 0 on the 3 1 interval , , 5, . Then 3 f ( x) x3 7 x 2 5 x 35 is increasing on
1 1 , 3 , 5, and decreasing on 3 ,5 . 2 102. The second derivative f x 3x 14 x 5 is a
quadratic function, so its graph is a parabola. Note that a = 36, b = -48, and c = 0. Because a 36 0 , the parabola is concave up. The x48 2 coordinate of the vertex is h . The 2 36 3 y-coordinate is
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Section 4.3: Quadratic Functions and Their Properties
2
2 2 2 k f 36 48 16 . So, the 3 3 3 2 vertex is , 16 ,which lies below the x-axis. 3 So the graph of f is below the x-axis when x is between the x-intercepts, and the graph of f is above the x-axis when x is outside the xintercepts. The find the x-intercepts, solve f ( x) 0 : 36 x 2 48 x 0 12 x(3 x 4) 0 4 x 0 or x . 3 4 The x-intercepts are 0 and . So f ( x) 0 for 3 4 on the interval 0, , and f ( x) 0 on the 3 4 interval , 0 , , . Then 3 f ( x) 3 x 4 8 x3 6 x 1 is concave up on
4 4 , 0 , 3 , and concave down on 0, 3 .
103. Answers will vary. 104. y x 2 2 x 3 ; y x 2 2 x 1 ; y x 2 2 x
Each member of this family will be a parabola with the following characteristics: (i) is concave up since a > 0; b 2 (ii) vertex occurs at x 1 ; 2a 2(1) (iii) There is at least one x-intercept since b 2 4ac 0 . 105. y x 2 4 x 1 ; y x 2 1 ; y x 2 4 x 1
Each member of this family will be a parabola with the following characteristics: (i) is concave up since a > 0 (ii) y-intercept occurs at (0, 1). 106. The graph of the quadratic function f x ax 2 bx c will not have any
x-intercepts whenever b 2 4ac 0 . 107. By completing the square on the quadratic function f x ax 2 bx c we obtain the 2
b b2 equation y a x c . We can then 2a 4a draw the graph by applying transformations to the graph of the basic parabola y x 2 , which is concave up. When a 0 , the basic parabola will either be stretched or compressed vertically. When a 0 , the basic parabola will either be stretched or compressed vertically as well as reflected across the x-axis. Therefore, when a 0 , the graph of f ( x) ax 2 bx c will open up, and when a 0 , the graph of f x ax 2 bx c will open down. 108. No. We know that the graph of a quadratic function f x ax 2 bx c is a parabola with
vertex 2ba , f 2ba
. If a > 0, then the vertex
is a minimum point, so the range is f b , . If a < 0, then the vertex is a 2a
maximum point, so the range is , f 2ba . Therefore, it is impossible for the range to be , .
369 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions 109. Two quadratic functions can intersect 0, 1, or 2 times.
A parallel line would have the same slope, 5 m . 7 5 ( y 3) ( x 14) 7 5 y 3 x 10 7 5 y x7 7
110. x 2 4 y 2 16 To check for symmetry with respect to the xaxis, replace y with –y and see if the equations are equivalent. x 2 4( y ) 2 16 x 2 4 y 2 16 So the graph is symmetric with respect to the xaxis. To check for symmetry with respect to the yaxis, replace x with –x and see if the equations are equivalent. ( x) 2 4 y 2 16 2
Range: 3, 4, 5, , 6, 7 The relation is a function.
2
x 4 y 16 So the graph is symmetric with respect to the yaxis. To check for symmetry with respect to the origin, replace x with –x and y with –y and see if the equations are equivalent. ( x)2 4( y ) 2 16 So the graph is x 2 4 y 2 16 symmetric with respect to the origin. 111. 27 x 5 x 3 6 x 24 x4 So the solution set is: , 4 or x | x 4 . 2
115. Domain: 1, 2,3, 4,5
116.
f (7) 3(7) 2 25(7) 28 147 189 28 0
3 23 117. g x 12 x 12 8 2 32 x 88 x
118.
4 x2 (3 x 5)
2
2
x 10 x y 4 y 20 ( x 2 10 x 25) ( y 2 4 y 4) 20 25 4 ( x 5) 2 ( y 2) 2 32
Center: (5, 2) ; Radius = 3
8 x (3 x 5) 3 3
4 x2 2
(3 x 5) 3
8 x(3x 5)
2
(3 x 5) 3
2
(3 x 5) 3
4 x 2 24 x 2 40 x 2
(3x 5) 3
28 x 2 40 x 2
(3 x 5) 3
x 5 x c 5c x 5x c 5c 2
119.
4 x 2 8 x(3 x 5)
2
112. x y 10 x 4 y 20 0
1
2
2
xc
2
xc x c 5 x 5c xc ( x c)( x c) 5( x c) xc ( x c ) ( x c ) 5 xc xc5 2
113. To reflect a graph about the y-axis, we change f ( x ) to f ( x) so to reflect y x about the y-
axis we change it to y x . 114. 5 x 7 y 35 7 y 5 x 35 5 y x5 7
370 Copyright © 2025 Pearson Education, Inc.
2
2
Section 4.4: Build Quadratic Models from Verbal Descriptions and from Data
Section 4.4 1. A r 2 2. Use LIN REGression to get y 1.7826 x 4.0652 3. a.
R ( p) p 6 p 600 6 p 2 600 p
b. The quantity sold price cannot be negative, so p 0 . Similarly, the price should be positive, so p 0 . 6 p 600 0 6 p 600 p 100 Thus, the implied domain for R is { p | 0 p 100} or 0, 100 . c.
6 x 2 600 x 12600 0
b 600 600 $50 p 2a 2 6 12
x 2 100 x 2100 0 ( x 30)( x 70) 0
d. The maximum revenue is
x 30, x 70
R (50) 6(50) 2 600(50)
The company should charge between $30 and $70.
15000 30000 $15, 000
e.
4. a.
x 6(50) 600 300
b. The quantity sold price cannot be negative, so p 0 . Similarly, the price should be positive, so p 0 . 3 p 360 0 3 p 360 p 120 Thus, the implied domain for R is { p | 0 p 120} or 0, 120 .
f.
c. g.
R ( p ) p 3 p 360 3 p 2 360 p
Graph R 6 x 2 600 x and R 12600 . Find where the graphs intersect by solving 12600 6 x 2 600 x .
p
b 360 360 $60 2a 2 3 6
d. The maximum revenue is R (60) 3(60) 2 360(60) 10800 21600 $10,800
e.
x 3(60) 360 180
371 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions f.
R (10) 5(10) 2 100(10) 500 1000 $500
e.
x 5(10) 100 50
g.
f.
graph
g.
Graph R 5 x 2 100 x and R 480 . Find where the graphs intersect by solving 480 5 x 2 100 x .
Graph R 3 p 2 360 p and R 9600 . Find where the graphs intersect by solving 9600 3 p 2 360 p .
3x 2 360 x 9600 0 x 2 30 x 3200 0 ( x 40)( x 80) 0 x 40, x 80
5. a.
The company should charge between $40 and $80.
5 x 2 100 x 480 0
R ( p ) p 5 p 100 5 p 2 100 p
( x 8)( x 12) 0
b. The quantity sold price cannot be negative, so p 0 . Similarly, the price should be positive, so p 0 . 5 p 100 0 5 p 100 p 20 Thus, the implied domain for R is { p | 0 p 20} or 0, 20 . c.
p
b 100 100 $10 2a 2 5 10
x 2 20 x 96 0 x 8, x 12
The company should charge between $8 and $12. 6.
a. R ( p ) p 20 p 500 20 p 2 500 p
b. The quantity sold price cannot be negative, so p 0 . Similarly, the price should be positive, so p 0 .
d. The maximum revenue is 372 Copyright © 2025 Pearson Education, Inc.
Section 4.4: Build Quadratic Models from Verbal Descriptions and from Data 20 p 500 0 20 p 500 p 25 Thus, the implied domain for R is { p | 0 p 25} or 0, 25 .
c.
p
b 500 500 $12.50 2a 2 20 40
d. The maximum revenue is R (12.5) 20(12.5) 2 500(12.5) 3125 6250 $3125
e.
x 20(12.5) 500
20 x 2 500 x 3000 0 x 2 25 x 150 0 ( x 10)( x 15) 0 x 10, x 15
The company should charge between $10 and $15. 7. a. Let w width and l length of the rectangular area. Solving P 2 w 2l 400 for l : 400 2w l 200 w . 2 Then A( w) (200 w) w 200 w w2 w2 200 w
250
b 200 200 100 yards 2a 2(1) 2
b.
w
c.
A(100) 1002 200(100)
f.
10000 20000 10, 000 yd 2
8. a.
Let x = width and y = width of the rectangle. Solving P 2 x 2 y 3000 for y: 3000 2 x 1500 x. 2 Then A( x) (1500 x) x y
g.
Graph R 20 x 2 500 x and R 3000 . Find where the graphs intersect by solving 3000 20 x 2 500 x .
1500 x x 2 x 2 1500 x.
b 1500 1500 750 feet 2a 2( 1) 2
b.
x
c.
A(750) 7502 1500(750) 562500 1125000 562,500 ft 2
9. Let x = width and y = length of the rectangle. Solving P 2 x y 4000 for y: y 4000 2 x . Then A( x) (4000 2 x) x 4000 x 2 x 2 2 x 2 4000 x b 4000 4000 x 1000 meters 2a 2( 2) 4 maximizes area.
373 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
A(1000) 2(1000) 2 4000(1000) . 2000000 4000000 2, 000, 000 The largest area that can be enclosed is 2,000,000 square meters.
10. Let x = width and y = length of the rectangle. 2 x y 2000 y 2000 2 x Then A( x) (2000 2 x) x
2000 x 2 x 2 x 2000 x b 2000 2000 x 500 meters 2a 2( 2) 4 maximizes area. A(500) 2(500) 2 2000(500) 500, 000 1, 000, 000 500, 000 The largest area that can be enclosed is 500,000 square meters. 2
y 0.001875(100) 2 18.75 meters . y
–200
(0,0)
(0,25)
(–60,0)
(0,0)
10
20
40
(60,0)
2
11. Locate the origin at the point where the cable touches the road. Then the equation of the parabola is of the form: y ax 2 , where a 0. Since the point (200, 75) is on the parabola, we can find the constant a : 75 Since 75 a (200) 2 , then a 0.001875 . 2002 When x 100 , we have:
(–200,75)
25 . The equation of the parabola is: 602 25 h( x ) 2 x 2 25 . 60 a
(200,75)
100
x 200
12. Locate the origin at the point directly under the highest point of the arch. Then the equation of the parabola is of the form: y ax 2 k , where a > 0. Since the maximum height is 25 feet, when x 0, y k 25 . Since the point (60, 0) is on the parabola, we can find the constant a : Since 0 a(60) 2 25 then
At x 10 : 25 25 (10) 2 25 25 24.3 ft. 36 602 At x 20 : 25 25 h(20) 2 (20) 2 25 25 22.2 ft. 9 60 At x 40 : 25 100 h(40) 2 (40) 2 25 25 13.9 ft. 9 60 h(10)
13. a.
Let x = the depth of the gutter and y the width of the gutter. Then A xy is the crosssectional area of the gutter. Since the aluminum sheets for the gutter are 12 inches wide, we have 2 x y 12 . Solving for y : y 12 2 x . The area is to be maximized, so: A xy x(12 2 x) 2 x 2 12 x . This equation is a parabola opening down; thus, it has a maximum b 12 12 when x 3. 2a 2( 2) 4 Thus, a depth of 3 inches produces a maximum cross-sectional area.
b. Graph A 2 x 2 12 x and A 16 . Find where the graphs intersect by solving 16 2 x 2 12 x .
374 Copyright © 2025 Pearson Education, Inc.
Section 4.4: Build Quadratic Models from Verbal Descriptions and from Data
2 x 2 12 x 16 0 x2 6 x 8 0 ( x 4)( x 2) 0 x 4, x 2
The graph of A 2 x 2 12 x is above the graph of A 16 where the depth is between 2 and 4 inches. 14. Let x width of the window and y height of the rectangular part of the window. The x perimeter of the window is: x 2 y 20. 2 40 2 x x Solving for y : y . 4 The area of the window is: 2
40 2 x x 1 x A( x) x 2 2 4 x 2 x 2 x 2 10 x 2 4 8 1 x 2 10 x. 2 8 This equation is a parabola opening down; thus, it has a maximum when 10 b 10 x 5.6 feet 2a 1 2 1 2 8 4 40 2(5.60) (5.60) y 2.8 feet 4 The width of the window is about 5.6 feet and the height of the rectangular part is approximately 2.8 feet. The radius of the semicircle is roughly 2.8 feet, so the total height is about 5.6 feet.
15. Let x the width of the rectangle or the diameter of the semicircle and let y the length of the x rectangle. The perimeter of each semicircle is . 2 The perimeter of the track is given x x by: y y 1500 . 2 2 Solving for x : x 2 y 1500 x 1500 2 y 1500 2 y x
This equation is a parabola opening down; thus, it has a maximum when 1500 b 1500 375. y 4 2a 2 2 1500 2(375) 750 Thus, x 238.73 The dimensions for the rectangle with maximum 750 area are 238.73 meters by 375 meters. 16. Let x = width of the window and y = height of the rectangular part of the window. The perimeter of the window is: 3x 2 y 16 16 3 x y 2 The area of the window is 3 2 16 3 x A( x ) x x 2 4 3 2 3 2 x x 2 4 3 3 2 x 8x 2 4 8x
This equation is a parabola opening down; thus, it has a maximum when 8 b x 2a 3 3 2 2 4
8
16 3.75 ft. 6 3
3 2 The window is approximately 3.75 feet wide. 16 48 16 3 16 6 3 8 24 6 3 y 2 2 6 3 The height of the equilateral triangle is 3 16 8 3 feet, so the total height is 2 6 3 6 3 3
8
24 8 3 5.62 feet. 6 3 6 3
The area of the rectangle is: 2 2 1500 1500 2 y A xy y y y. 375 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions d. The maximum height will be: h(139.4) 2 0.0037(139.4) 1.0318(139.4) 5.6667
17. a.
77.6 feet
e.
From the graph, the data appear to follow a quadratic relation with a 0 . b. Using the QUADratic REGression program
19. a.
y 0.0082 x 2 0.746 x 13.469
c.
b 0.746 45.5 2a 2(0.0082) The speed at which the gas mileage is greatest is about 45.5 miles per hour. x
d.
From the graph, the data appear to be linearly relation. b. Using the LINear REGression program y 24.520 x 311.104 c.
y (25) 24.520(25) 311.104 301.89 The numbers of subscribers in 2025 will be about 301.89 million.
18. a.
20. a.
From the graph, the data appear to follow a quadratic relation with a 0 . b. Using the QUADratic REGression program
From the graph, the data appear to be linearly related with m 0 . b. Using the LINear REGression program
h( x ) 0.0037 x 2 1.0318 x 5.6667
c.
b 1.0318 139.4 2a 2(0.0037) The ball will travel about 139.4 feet before it reaches its maximum height. x
C ( x) 0.233 x 2.037
c.
C (80) 0.233(80) 2.037 16.6 When the temperature is 80F , there will be about 16.6 chirps per second.
376 Copyright © 2025 Pearson Education, Inc.
Section 4.4: Build Quadratic Models from Verbal Descriptions and from Data
So
21. a.
2 h y0 4 y1 y2 16 4 8 16 3 3 2 128 64 3 3
area
From the graph, the data appear to follow a quadratic relation with a 0 .
25. Note that h = 4. Then y0 (4) 2 3(4) 5 9, y1 02 3 0 5 5, and y2 42 3 4 5 33 So 4 h area y0 4 y1 y2 9 4 5 33 3 3 4 248 62 3 3
b. Using the QUADratic REGression program y 0.0671x 2 2.696a 500.971
c.
y (74.5) 0.0671(74.5)2 2.696(74.5) 500.971 672 The reaction time is about 672 milliseconds.
22. Substitute each point into the equation: y0 a (h) 2 b(h) c ah 2 bh c;
26. Note that h = 1. Then y0 (1) 2 (1) 4 2, y1 02 0 4 4, and y2 12 1 4 4 So 1 h area y0 4 y1 y2 2 4 4 4 3 3 1 22 22 3 3
y1 a (0) 2 b(0) c c; y2 a (h) 2 b(h) c ah 2 bh c Then: y0 4 y1 y2 ah 2 bh c 4c ah 2 bh c 2ah 2 6c.
So, Area
h h 2ah 2 6c y0 4 y1 y2 . 3 3
23. Note that h = 1. Then y0 5(1) 2 8 3,
27. Answers will vary. One possibility follows: If the price is $140, no one will buy the calculators, thus making the revenue $0. 28. d ( x2 x1 ) 2 ( y2 y1 ) 2
y1 5 02 8 8,
((1) 4) 2 (5 (7)) 2
and y2 5 12 8 3 So 1 h area y0 4 y1 y2 3 4 8 3 3 3 1 38 38 3 3
(5)2 (12) 2 25 144 169 13
29.
( x h) 2 ( y k ) 2 r 2 ( x ( 6)) 2 ( y 0) 2 ( 7) 2 ( x 6) 2 y 2 7
24. Note that h = 2. Then y0 2(2) 2 8 16, y1 2 02 8 8, and y2 2 22 8 16
377 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
30. x
(8) 82 4(5)( 3) 8 64 60 2(5) 10
36.
4 31 4 31 , So the zeros are: 5 5
31. 5(0) 7 y 140 7 y 140 y 20 5 x 7(0) 140 5 x 140 x 28
The x-intercept is 28, 0 and the y-intercept is
0, 20
37. 4( x 1)5 ( x 7)3 5( x 1) 4 ( x 7) 4
32. 2 3x 7 9 21
( x 1) 4 ( x 7)3 4( x 1) 5( x 7)
2 3x 7 30
( x 1) 4 ( x 7)3 4 x 4 5 x 35
3x 7 15 3 x 7 15 or 3x 7 15 3 x 8 3x 22 8 22 x x 3 3 8 22 The solution set is , 3 3 7 19 15 3 12 21
1 4
7
3 x 1
3 3 f ( x h) f ( x ) x h 1 x 1 h h x 1 x 1 h x h 1 x 1 h x 1 x 1 h 1 x h 1 1 h 1 h x h 1 x 1 h 1 x h 1 x 1
8 124 8 2 31 4 31 10 10 5
33. 3 1
f ( x)
6 2
The quotient is x 4 x 7 8 and the remainder is 6. 34. The denominator cannot be 0. x3 16 x 0 x( x 2 16) 0 x( x 4)( x 4) 0 x 0, 4, 4 The solution set is x | x 4, 0, 4 .
( x 1) 4 ( x 7)3 9 x 31
Section 4.5 1. 3x 2 7 3 x 9 x 3 The solution set is x | x 3 or 3, . 2.
2, 7 represents the numbers between 2 and 7, including 7 but not including 2 . Using inequality notation, this is written as 2 x 7 .
3. a.
f ( x) 0 when the graph of f is above the x-
axis. Thus, x x 2 or x 2 or, using interval notation, , 2 2, .
35. y 2 9 ( x 3) 2 4
378 Copyright © 2025 Pearson Education, Inc.
Section 4.5: Inequalities Involving Quadratic Functions
b.
intersects the x-axis. Thus, x 2 x 2 or, using interval notation, 2, 2 .
4. a.
g ( x) 0 when the graph of g is below the
x-axis. Thus, x x 1 or x 4 or, using
interval notation, , 1 4, . b.
f ( x) 0 when the graph of f is below or
g ( x) 0 when the graph of f is above or
intersects the x-axis. Thus, x 1 x 4
The graph is below the x-axis for 2 x 5 . Since the inequality is strict, the solution set is x 2 x 5 or, using interval notation,
2, 5 .
or, using interval notation, 1, 4 . 5. a.
g ( x) f x when the graph of g is above
or intersects the graph of f. Thus x 2 x 1 or, using interval notation,
2, 1 . b.
f ( x) g x when the graph of f is above
the graph of g. Thus, x x 2 or x 1 or, using interval notation, , 2 1, . 6. a.
f ( x) g x when the graph of f is below
the graph of g. Thus, x x 3 or x 1 or, using interval notation, , 3 1, . b.
f ( x) g x when the graph of f is above
or intersects the graph of g. Thus, x 3 x 1 or, using interval notation,
3, 1 . 7. x 2 3 x 10 0 We graph the function f ( x) x 2 3 x 10 . The intercepts are y-intercept: f (0) 10 2
x-intercepts: x 3x 10 0 ( x 5)( x 2) 0 x 5, x 2 b (3) 3 The vertex is at x . Since 2a 2(1) 2 49 3 3 49 f , the vertex is , . 4 4 2 2
8. x 2 3x 10 0 We graph the function f ( x) x 2 3 x 10 . The intercepts are y-intercept: f (0) 10
x-intercepts: x 2 3x 10 0 ( x 5)( x 2) 0 x 5, x 2 b (3) 3 . Since The vertex is at x 2a 2(1) 2 49 3 3 49 f , the vertex is , . 4 4 2 2
The graph is above the x-axis when x 5 or x 2 . Since the inequality is strict, the solution set is x x 5 or x 2 or, using interval notation, , 5 2, . 9. x 2 4 x 0 We graph the function f ( x) x 2 4 x . The intercepts are y-intercept: f (0) 0
x-intercepts: x 2 4 x 0 x( x 4) 0 x 0, x 4
379 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
The vertex is at x
b (4) 4 2 . Since 2a 2(1) 2
The vertex is at x
b (0) 0 . Since 2a 2(1)
f (0) 9 , the vertex is 0, 9 .
The graph is above the x-axis when x 0 or x 4 . Since the inequality is strict, the solution set is x x 0 or x 4 or, using interval
The graph is below the x-axis when 3 x 3 . Since the inequality is strict, the solution set is x 3 x 3 or, using interval notation,
f 2 4 , the vertex is 2, 4 .
3, 3 .
notation, , 0 4, . 10. x 2 8 x 0 We graph the function f ( x) x 2 8 x . The intercepts are y-intercept: f (0) 0
12. x 2 1 0 We graph the function f ( x) x 2 1 . The intercepts are y-intercept: f (0) 1 x2 1 0
x-intercepts:
x-intercepts: x 2 8 x 0 x( x 8) 0 x 0, x 8 b (8) 8 The vertex is at x 4 . 2a 2(1) 2
( x 1)( x 1) 0 x 1, x 1 b (0) 0 . Since The vertex is at x 2a 2(1)
Since f 4 16 , the vertex is 4, 16 .
f (0) 1 , the vertex is 0, 1 .
The graph is above the x-axis when x 8 or x 0 . Since the inequality is strict, the solution set is x x 8 or x 0 or, using interval
The graph is below the x-axis when 1 x 1 . Since the inequality is strict, the solution set is x 1 x 1 or, using interval notation,
1, 1 .
notation, , 8 0, . 11. x 2 9 0 We graph the function f ( x) x 2 9 . The intercepts are y-intercept: f (0) 9
x-intercepts:
x2 9 0 ( x 3)( x 3) 0 x 3, x 3
x 2 x 12
13.
x x 12 0 We graph the function f ( x) x 2 x 12 . y-intercept: f (0) 12 2
x-intercepts:
x 2 x 12 0 ( x 4)( x 3) 0 x 4, x 3
The vertex is at x 380 Copyright © 2025 Pearson Education, Inc.
b (1) 1 . Since 2a 2(1) 2
Section 4.5: Inequalities Involving Quadratic Functions
49 1 1 49 f , the vertex is , . 2 4 4 2
The vertex is at x
49 5 5 49 f , the vertex is , . 8 8 4 4
The graph is above the x-axis when x 4 or x 3 . Since the inequality is strict, the solution set is x x 4 or x 3 or, using interval
1 x 3. 2 Since the inequality is strict, the solution set is 1 x x 3 or, using interval notation, 2
The graph is below the x-axis when
notation, , 4 3, . 14.
x 2 7 x 12 x 2 7 x 12 0 We graph the function f ( x) x 2 7 x 12 . y-intercept: f (0) 12
x-intercepts: x 2 7 x 12 0 ( x 4)( x 3) 0 x 4, x 3 The vertex is at x
1 , 3 . 2
16.
b (7) 7 . Since 2a 2(1) 2
1 7 1 1 f , the vertex is , . 4 2 2 4
b (5) 5 . Since 2a 2(2) 4
6 x2 6 5x 6 x2 5x 6 0 We graph the function f ( x) 6 x 2 5 x 6 . The intercepts are y-intercept: f (0) 6
x-intercepts:
6x2 5x 6 0 (3x 2)(2 x 3) 0
2 3 x ,x 3 2 b (5) 5 The vertex is at x . Since 2a 2(6) 12
The graph is below the x-axis when 4 x 3 . Since the inequality is strict, the solution set is x | 4 x 3 or, using interval notation,
169 5 5 169 f , the vertex is , . 24 24 12 12
4, 3 . 15.
2 x2 5x 3 2 x2 5x 3 0 We graph the function f ( x) 2 x 2 5 x 3 . The intercepts are y-intercept: f (0) 3
x-intercepts:
2x2 5x 3 0 (2 x 1)( x 3) 0
2 3 x . 3 2 Since the inequality is strict, the solution set is
The graph is below the x-axis when
1 x , x3 2
381 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
2 3 x x or, using interval notation, 3 2 2 3 , . 3 2
2
17. x x 1 0 We graph the function f ( x) x 2 x 1 . The intercepts are y-intercept: f (0) 1
x-intercepts: x
(1) (1) 2 4(1)(1) 2(1)
The graph is always above the x-axis. Thus, the solution is all real numbers or using interval notation, , . 19.
1 3 (not real) 2 Therefore, f has no x-intercepts.
4 x2 9 6 x 4 x2 6 x 9 0 We graph the function f ( x) 4 x 2 6 x 9 . y-intercept: f (0) 9 x-intercepts: x
The vertex is at x
b (1) 1 . Since 2a 2(1) 2
(6) ( 6) 2 4(4)(9) 2(4)
6 108 (not real) 8 Therefore, f has no x-intercepts. b (6) 6 3 . Since The vertex is at x 2a 2(4) 8 4
1 3 1 3 f , the vertex is , . 2 4 2 4
3 27 3 27 , the vertex is , f . 4 4 4 4
The graph is never below the x-axis. Thus, there is no real solution.
18. x 2 x 4 0 We graph the function f ( x) x 2 2 x 4 . y-intercept: f (0) 4 2
x-intercepts: x
(2) (2) 2 4(1)(4) 2(1)
2 12 (not real) 2 Therefore, f has no x-intercepts. b (2) 1 . Since The vertex is at x 2a 2(1)
The graph is never below the x-axis. Thus, there is no real solution. 20.
25 x 2 16 40 x 25 x 2 40 x 16 0 We graph the function f ( x) 25 x 2 40 x 16 .
y-intercept: f (0) 16 x-intercepts: 25 x 2 40 x 16 0
f 1 3 , the vertex is 1,3 .
(5 x 4) 2 0 5x 4 0 x
382 Copyright © 2025 Pearson Education, Inc.
4 5
Section 4.5: Inequalities Involving Quadratic Functions
The vertex is at x
b (40) 40 4 . 2a 2(25) 50 5
22. 2 2 x 2 3 x 9 4 x 2 6 x 9 4 x2 6 x 9 0 We graph the function f ( x) 4 x 2 6 x 9 . y-intercept: f (0) 9
4 4 Since f 0 , the vertex is , 0 . 5 5
x-intercepts: x
6 108 (not real) 8 Therefore, f has no x-intercepts. b (6) 6 3 . Since The vertex is at x 2a 2(4) 8 4
The graph is never below the x-axis. Thus, there is no real solution. 21.
6 x 2 1 5 x
3 27 3 27 f , the vertex is , . 4 4 4 4
6 x2 6 5x 6 x2 5x 6 0 We graph the function f ( x) 6 x 2 5 x 6 . y-intercept: f (0) 6
x-intercepts:
6x2 5x 6 0 (3x 2)(2 x 3) 0
2 3 x ,x 3 2 b (5) 5 . Since The vertex is at x 2a 2(6) 12 169 5 5 169 f , the vertex is , . 12 24 24 12
The graph is always above the x-axis. Thus, the solution set is all real numbers or, using interval notation, , . 23.
f ( x) x 2 1; g ( x) 3x 3 f ( x) 0
a.
x 1 0 ( x 1)( x 1) 0 x 1; x 1 2
Solution set: 1, 1 . b.
The graph is above the x-axis when x
2 or 3
3 . Since the inequality is strict, solution set 2 2 3 is x x or x or, using interval 3 2 x
2 3 notation, , , . 3 2
(6) ( 6) 2 4(4)(9) 2(4)
g ( x) 0 3x 3 0 3x 3 x 1 Solution set: 1 . f ( x) g ( x)
c.
x 1 3x 3 x 3x 4 0 ( x 4)( x 1) 0 x 4; x 1 2
2
Solution set: 1, 4 .
383 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
d.
f ( x) 0
We graph the function f ( x) x 1 . y-intercept: f (0) 1 2
x-intercepts:
x2 1 0 ( x 1)( x 1) 0 x 1, x 1
The vertex is at x
The graph of p is above the x-axis when x 1 or x 4 . Since the inequality is strict, the solution set is x x 1 or x 4 or, using interval
b (0) 0 . Since 2a 2(1)
f (0) 1 , the vertex is (0, 1).
notation, , 1 4, . g.
The graph is above the x-axis when x 1 or x 1 . Since the inequality is strict, the solution set is x x 1 or x 1 or, using
x-intercepts: x 2 2 0 x2 2
interval notation, (, 1) (1, ) . e.
f ( x) 1 x2 1 1 x2 2 0 We graph the function p( x) x 2 2 . The intercepts of p are y-intercept: p(0) 2
x 2 b (0) The vertex is at x 0 . Since 2a 2(1)
g ( x) 0 3x 3 0 3x 3 x 1 The solution set is x x 1 or, using
p(0) 2 , the vertex is (0, 2).
interval notation, , 1 . f.
f ( x) g ( x) x 2 1 3x 3 x 2 3x 4 0 We graph the function p( x) x 2 3x 4 . The intercepts of p are y-intercept: p(0) 4
x-intercepts:
The graph of p is above the x-axis when x 2 or x 2 . Since the inequality is not strict, the solution set is
x x 2 or x 2 or, using interval
x 2 3x 4 0 ( x 4)( x 1) 0 x 4, x 1
The vertex is at x
b (3) 3 . Since 2a 2(1) 2
25 3 3 25 , the vertex is , . p 4 4 2 2
notation, , 2 2, . 24.
f ( x) x 2 3;
a.
g ( x ) 3x 3
f ( x) 0 x 3 0 x2 3 2
x 3
Solution set: 3, 3 .
384 Copyright © 2025 Pearson Education, Inc.
Section 4.5: Inequalities Involving Quadratic Functions
b.
g x 0 3x 3 0 3 x 3 x 1 Solution set: {1}.
x-intercepts: x 2 3 x 0 x( x 3) 0 x 0; x 3 b (3) 3 3 . The vertex is at x 2a 2(1) 2 2
c.
f ( x) g ( x) x 3 3 x 3 0 x 2 3x 0 x( x 3) x 0; x 3 Solution set: 0, 3 .
3 9 3 9 Since p , the vertex is , . 2 4 2 4
2
d.
f ( x) 0
The graph of p is above the x-axis when 0 x 3 . Since the inequality is strict, the solution set is x 0 x 3 or, using
We graph the function f ( x) x 2 3 . y-intercept: f (0) 3 x-intercepts: x 2 3 0 x2 3 x 3 b (0) 0 . Since The vertex is at x 2a 2(1) f (0) 3 , the vertex is (0, 3).
The graph is above the x-axis when 3 x 3 . Since the inequality is strict,
x 3 x 3 or,
g.
f ( x) 1 x2 3 1 x2 2 0 We graph the function p( x) x 2 2 . The intercepts of p are y-intercept: p(0) 2
x-intercepts: x 2 2 0 x2 2
the solution set is
interval notation, (0, 3) .
x 2 b (0) 0 . Since The vertex is at x 2a 2(1) p(0) 2 , the vertex is (0, 2).
using interval notation, 3, 3 . e.
g ( x) 0 3x 3 0 3 x 3 x 1 The solution set is x x 1 or, using
interval notation, 1, . f.
f ( x) g ( x) x 2 3 3x 3 x 2 3x 0 We graph the function p( x) x 2 3 x . The intercepts of p are y-intercept: p(0) 0
The graph of p is above the x-axis when 2 x 2 . Since the inequality is not
strict, the solution set is
x 2 x 2
or, using interval notation, 2, 2 .
385 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
25.
f ( x) x 2 1;
a.
g ( x) 4 x 1
e.
f x 0 x2 1 0 1 x2 0 1 x 1 x 0
1 The solution set is x x or, using 4 1 interval notation, , . 4
x 1; x 1
Solution set: 1, 1 . b.
g x 0 4x 1 0 4 x 1 1 x 4
f.
f x g x
x-intercepts: x 2 4 x 0 x( x 4) 0 x 0; x –4
x2 1 4x 1 0 x2 4x 0 x x 4 x 0; x 4
The vertex is at x
Solution set: 4, 0 . d.
b (4) 4 2 . 2a 2(1) 2
Since p(2) 4 , the vertex is (2, 4).
f x 0
We graph the function f ( x) x 2 1 . y-intercept: f (0) 1
x 1 0 x2 1 0 ( x 1)( x 1) 0 x 1; x 1 b (0) 0 . Since The vertex is at x 2a 2(1)
x-intercepts:
f x g x x2 1 4x 1 x2 4 x 0 We graph the function p( x) x 2 4 x . The intercepts of p are y-intercept: p(0) 0
1 Solution set: . 4
c.
g x 0 4x 1 0 4 x 1 1 x 4
2
f (0) 1 , the vertex is (0, 1).
The graph is above the x-axis when 1 x 1 . Since the inequality is strict, the solution set is x 1 x 1 or, using
The graph of p is above the x-axis when 4 x 0 . Since the inequality is strict, the solution set is x 4 x 0 or, using
interval notation, 4, 0 . g.
f ( x) 1 x2 1 1 x2 0 We graph the function p( x) x 2 . The
vertex is at x
b (0) 0 . Since 2a 2(1)
p(0) 0 , the vertex is (0, 0). Since a 1 0 , the parabola opens downward.
interval notation, 1, 1 .
386 Copyright © 2025 Pearson Education, Inc.
Section 4.5: Inequalities Involving Quadratic Functions
The graph is above the x-axis when 2 x 2 . Since the inequality is strict, the solution set is x 2 x 2 or, using
The graph of p is never above the x-axis, but it does touch the x-axis at x = 0. Since the inequality is not strict, the solution set is {0}. 26.
f ( x) x 2 4; g ( x) x 2 a. f ( x) 0 2 x 4 0 x2 4 0 ( x 2)( x 2) 0 x 2; x 2
interval notation, 2, 2 . e.
interval notation, 2, . f.
Solution set: 2, 2 . b.
g ( x) 0 x 2 0 2 x Solution set: 2 .
c.
f ( x) g ( x) x2 4 x 2 0 x2 x 6 0 x 3 x 2 x 3; x 2 Solution set: 2, 3 .
d.
x2 x 6 0 x2 x 6 0 ( x 2)( x 3) 0 x 2; x 3
The vertex is at x
b (1) 1 1 . 2a 2(1) 2 2
1 25 1 25 , the vertex is , . Since p 2 4 2 4
The graph of p is above the x-axis when 2 x 3 . Since the inequality is strict, the solution set is x 2 x 3 or, using
x 4 0 x2 4 0 ( x 2)( x 2) 0 x 2; x 2 2
The vertex is at x
f ( x) g ( x) x2 4 x 2 x2 x 6 0 We graph the function p( x) x 2 x 6 . The intercepts of p are y-intercept: p(0) 6
x-intercepts:
f ( x) 0 x2 4 0 We graph the function f ( x) x 2 4 . y-intercept: f (0) 4
x-intercepts:
g ( x) 0 x 2 0 x 2 x 2 The solution set is x x 2 or, using
b (0) 0 . Since 2a 2(1)
f (0) 4 , the vertex is (0, 4).
interval notation, (2, 3) . g.
f ( x) 1 x2 4 1 x2 3 0 We graph the function p( x) x 2 3 . The intercepts of p are y-intercept: p(0) 3
x-intercepts: x 2 3 0 x2 3 x 3
387 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
The vertex is at x
b (0) 0 . Since 2a 2(1)
f (0) 4 , the vertex is (0, 4).
The graph of p is above the x-axis when 3 x 3 . Since the inequality is not
The graph is above the x-axis when x 2 or x 2 . Since the inequality is strict, the solution set is x x 2 or x 2 or,
x 3 x 3
or, using interval notation, 3, 3 . f x x 4; 2
using interval notation, , 2 2, . e.
g x x 4
g ( x) 0 x 4 0 We graph the function g ( x) x 2 4 . 2
2
f x 0
a.
y-intercept: g (0) 4
x2 4 0 x 2 x 2 0 x 2; x 2
x-intercepts:
Solution set: 2, 2 . g x 0
b.
x 4 0 x2 4 0 x 2 x 2 0
f ( x) g ( x) x2 4 x2 4 2 x2 8 0 2 x 2 x 2 0
The graph is below the x-axis when x 2 or x 2 . Since the inequality is not strict, the solution set is x x 2 or x 2 or,
x 2; x 2
Solution set: 2, 2 .
using interval notation, , 2 2, .
f ( x) 0 x 4 0 We graph the function f ( x) x 2 4 . 2
y-intercept: f (0) 4 x-intercepts:
b (0) 0 . Since 2a 2(1)
g (0) 4 , the vertex is (0, 4).
x 2; x 2 Solution set: 2, 2 .
d.
x2 4 0 x2 4 0 ( x 2)( x 2) 0 x 2; x 2
The vertex is at x
2
c.
b (0) 0 . Since 2a 2(1)
p(0) 3 , the vertex is (0, 3).
strict, the solution set is
27.
The vertex is at x
x2 4 0 ( x 2)( x 2) 0 x 2; x 2
f.
f ( x) g ( x) x 4 x2 4 2 x2 8 0 We graph the function p( x) 2 x 2 8 . 2
y-intercept: p(0) 8 x-intercepts:
388 Copyright © 2025 Pearson Education, Inc.
2 x2 8 0 2( x 2)( x 2) 0 x 2; x 2
Section 4.5: Inequalities Involving Quadratic Functions
The vertex is at x
b (0) 0 . Since 2a 2(2)
b.
p(0) 8 , the vertex is (0, 8).
g ( x) 0 x2 1 0 x2 1 0 x 1 x 1 0 x 1; x 1
Solution set: 1, 1 .
c.
f ( x) g ( x) x 2 x 1 x2 1 2 x2 2 x 0 2 x x 1 0 2
The graph is above the x-axis when x 2 or x 2 . Since the inequality is strict, the solution set is x x 2 or x 2 or,
x 0, x 1 Solution set: 0, 1 .
using interval notation, (, 2) (2, ) . g.
f ( x) 1 2 x 4 1 x2 5 0 We graph the function p( x) x 2 5 .
d.
y-intercept: f (0) 1 x-intercepts: x 2 2 x 1 0
y-intercept: p(0) 5
x 1 0 2
x-intercepts: x 2 5 0 x2 5
x 1 0 x 1 b (2) 2 1. The vertex is at x 2a 2(1) 2
x 5 b (0) 0 . Since The vertex is at x 2a 2(1) p(0) 5 , the vertex is (0, 5).
Since f (1) 0 , the vertex is (1, 0).
The graph of p is above the x-axis when x 5 or x 5 . Since the inequality is not strict, the solution set is
The graph is above the x-axis when x 1 or x 1 . Since the inequality is strict, the solution set is x x 1 or x 1 or, using
x x 5 or x 5 or, using interval
notation, , 5 5, . 28.
f x x 2 2 x 1; g x x 2 1
a.
f ( x) 0 x 2x 1 0 We graph the function f ( x) x 2 2 x 1 . 2
f ( x) 0 x2 2x 1 0 ( x 1) 2 0 x 1 0 x 1 Solution set: 1 .
interval notation, , 1 1, . e.
g ( x) 0 x2 1 0 We graph the function g ( x) x 2 1 .
y-intercept: g (0) 1 x-intercepts:
389 Copyright © 2025 Pearson Education, Inc.
x2 1 0 x2 1 0 ( x 1)( x 1) 0 x 1; x 1
Chapter 4: Linear and Quadratic Functions
The vertex is at x
f.
b (0) 0 . Since 2a 2(1)
The vertex is at x
g (0) 1 , the vertex is (0, 1).
Since p (1) 1 , the vertex is (1, 1).
The graph is below the x-axis when x 1 or x 1 . Since the inequality is not strict, the solution set is x x 1 or x 1 or,
The graph of p is above the x-axis when x 0 or x 2 . Since the inequality is not strict, the solution set is x x 0 or x 2
using interval notation, , 1 1, .
or, using interval notation, , 0 2, .
f ( x) g ( x) x 2 x 1 x2 1 2 x2 2 x 0 We graph the function p( x) 2 x 2 2 x . 2
29.
f x x 2 x 2;
a.
x-intercepts: 2 x 2 2 x 0 2 x( x 1) 0 x 0; x 1 The vertex is at x
b.
x 2; x 1
Solution set: 2, 1 . c.
interval notation, , 0 1, . f ( x) 1 x2 2x 1 1 x2 2 x 0 We graph the function p( x) x 2 2 x .
f ( x) g ( x) x x 2 x2 x 2 2 x 0 x0 Solution set: 0 . 2
The graph is above the x-axis when x 0 or x 1 . Since the inequality is strict, the solution set is x x 0 or x 1 or, using
g ( x) 0 x x2 0 x 2 x 1 0 2
1 1 1 1 Since p , the vertex is , . 2 2 2 2
x-intercepts: x 2 2 x 0 x( x 2) 0 x 0; x 2
f ( x) 0 x x2 0 x 2 x 1 0 x 2, x 1
Solution set: 1, 2 .
b (2) 2 1 . 2a 2(2) 4 2
y-intercept: p(0) 0
g x x2 x 2
2
y-intercept: p(0) 0
g.
b (2) 2 1 . 2a 2(1) 2
d.
f ( x) 0 x x2 0 We graph the function f ( x) x 2 x 2 . 2
y-intercept: f (0) 2 x2 x 2 0 ( x 2)( x 1) 0 x 2; x 1 b (1) 1 . Since The vertex is at x 2a 2(1) 2
x-intercepts:
9 1 9 1 f , the vertex is , . 2 4 2 4
390 Copyright © 2025 Pearson Education, Inc.
Section 4.5: Inequalities Involving Quadratic Functions
x-intercepts: x 2 x 3 0 x
1
1 4 1 3 2 1 2
1 1 12 1 13 2 2 x 1.30 or x 2.30 b (1) 1 . Since The vertex is at x 2a 2(1) 2
The graph is above the x-axis when x 1 or x 2 . Since the inequality is strict, the solution set is x x 1 or x 2 or, using
13 1 13 1 p , the vertex is , . 4 2 4 2
interval notation, , 1 2, . e.
g ( x) 0 x2 x 2 0 We graph the function g ( x) x 2 x 2 .
y-intercept: g (0) 2 x2 x 2 0 ( x 2)( x 1) 0 x 2; x 1 b (1) 1 . Since The vertex is at x 2a 2(1) 2
x-intercepts:
The graph of p is above the x-axis when 1 13 1 13 or x . Since the x 2 2 inequality is not strict, the solution set is 1 13 1 13 or x x x or, using 2 2 interval notation, 1 13 1 13 , . , 2 2
7 1 7 1 f , the vertex is , . 4 2 4 2
The graph is below the x-axis when 2 x 1 . Since the inequality is not strict, the solution set is x 2 x 1 or, using
30.
f ( x) x 2 x 1; g ( x) x 2 x 6 a. f ( x) 0 2 x x 1 0 x2 x 1 0
interval notation, 2, 1 . f.
interval notation, , 0 . g.
x
f ( x) g ( x) 2 x x 2 x2 x 2 2 x 0 x0 The solution set is x x 0 or, using
y-intercept: p(0) 3
1 4 1 1 2 1 2
1 1 4 1 5 2 2 1 5 1 5 Solution set: , . 2 2 g ( x) 0 2 x x 6 0 x2 x 6 0 x 3 x 2 0
b.
f ( x) 1 x x2 1 x2 x 3 0 We graph the function p( x) x 2 x 3 . 2
1
x 3; x 2 Solution set: 2, 3 .
391 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
c.
f ( x) g ( x) x x 1 x2 x 6 2 x 5 0 2 x 5 5 x 2 5 Solution set: . 2
x2 x 6 0 x2 x 6 0 x 3 x 2 0 x 3; x 2 b (1) 1 1 . The vertex is at x 2a 2(1) 2 2
x-intercepts:
2
d.
1 25 1 25 , the vertex is , . Since f 2 4 2 4
f ( x) 0 x2 x 1 0 We graph the function f ( x) x 2 x 1 . y-intercept: f (0) 1
x-intercepts: x 2 x 2 0 x2 x 2 0
The graph is below the x-axis when x 2 or x 3 . Since the inequality is not strict, the solution set is x x 2 or x 3 or,
(1) (1) 4(1)(1) 2(1) 2
x
1 1 4 1 5 2 2 x 1.62 or x 0.62 b (1) 1 1 . The vertex is at x 2a 2(1) 2 2
using interval notation, , 2 3, . f.
1 5 1 5 Since f , the vertex is , . 2 4 2 4
f ( x) g ( x) x x 1 x2 x 6 2 x 5 5 x 2 The solution set is x x 52 or, using 2
interval notation, , 52 . g.
f ( x) 1 x x 1 1 x2 x 0 We graph the function p( x) x 2 x . y-intercept: p(0) 0 2
The graph is above the x-axis when 1 5 1 5 x . Since the inequality 2 2
is strict, the solution set is
x-intercepts:
1 5 1 5 x x or, using interval 2 2
1 5 1 5 , . 2 2
notation, e.
g ( x) 0 2 x x 6 0 We graph the function g ( x) x 2 x 6 . y-intercept: g (0) 6
x2 x 0 x x 1 0
x 0; x 1 The vertex is at x
b (1) 1 1 . 2a 2(1) 2 2
1 1 1 1 Since p , the vertex is , . 2 4 2 4
392 Copyright © 2025 Pearson Education, Inc.
Section 4.5: Inequalities Involving Quadratic Functions
The vertex of p is at x
The graph of p is above the x-axis when 1 x 0 . Since the inequality is not strict, the solution set is x 1 x 0 or, using interval notation, 1, 0 . 31. The domain of the expression f ( x) x 2 16
includes all values for which x 2 16 0 . We graph the function p( x) x 2 16 . The intercepts of p are y-intercept: p(0) 6 x 2 16 0
x-intercepts:
( x 4)( x 4) 0 x 4, x 4
The vertex of p is at x
b (0) 0 . Since 2a 2(1)
1 1 1 1 Since p , the vertex is , . 6 12 6 12
The graph of p is above the x-axis when 1 0 x . Since the inequality is not strict, the 3 1 solution set of x 3 x 2 0 is x 0 x . 3
1 Thus, the domain of f is also x 0 x or, 3 1 using interval notation, 0, . 3 33. a.
p(0) 16 , the vertex is 0, 16 .
b.
The graph of p is above the x-axis when x 4 or x 4 . Since the inequality is not strict, the solution set of x 2 16 0 is x | x 4 or x 4 . Thus, the domain of f is also x | x 4 or x 4 or, using interval notation, , 4 4, . 32. The domain of the expression f x x 3 x 2
includes all values for which x 3 x 2 0 . We graph the function p( x) x 3x 2 . The intercepts of p are y-intercept: p(0) 6 x-intercepts:
b (1) 1 1 . 2a 2(3) 6 6
The ball strikes the ground when s (t ) 80t 16t 2 0 . 80t 16t 2 0 16t 5 t 0 t 0, t 5 The ball strikes the ground after 5 seconds. Find the values of t for which 80t 16t 2 96 16t 2 80t 96 0 We graph the function f (t ) 16t 2 80t 96 . The intercepts are
y-intercept: f (0) 96 t-intercepts: 16t 2 80t 96 0 16(t 2 5t 6) 0 16(t 2)(t 3) 0 t 2, t 3 The vertex is at t
b (80) 2.5 . 2a 2(16)
Since f 2.5 4 , the vertex is 2.5, 4 .
x 3x 2 0 x(1 3 x) 0
1 x 0, x . 3 393 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
35. a.
4 p p 1000 0 p 0, p 1000 Thus, the revenue equals zero when the price is $0 or $1000.
The graph of f is above the t-axis when 2 t 3 . Since the inequality is strict, the solution set is t | 2 t 3 or, using interval notation, 2, 3 . The ball is more than 96 feet above the ground for times between 2 and 3 seconds. 34. a.
R ( p) 4 p 2 4000 p 0
The ball strikes the ground when s (t ) 96t 16t 2 0 . 96t 16t 2 0 16t 6 t 0 t 0, t 6 The ball strikes the ground after 6 seconds.
b. Find the values of t for which 96t 16t 2 128 2 16t 96t 128 0 We graph f (t ) 16t 2 96t 128 . The intercepts are y-intercept: f (0) 128
b. Find the values of p for which 4 p 2 4000 p 800, 000 4 p 2 4000 p 800, 000 0
We graph f ( p ) 4 p 2 4000 p 800, 000 . The intercepts are y-intercept: f (0) 800, 000 p-intercepts: 4 p 2 4000 p 800000 0 p 2 1000 p 200000 0 p
1000
1000 4 1 200000 2 1 2
1000 200000 2 1000 200 5 2 500 100 5 p 276.39; p 723.61 .
The vertex is at p
b (4000) 500 . 2a 2(4)
t-intercepts: 16t 2 96t 128 0 16(t 2 6t 8) 0 16(t 4)(t 2) 0 t 4, t 2 b (96) 3 . Since The vertex is at t 2a 2(16)
Since f 500 200, 000 , the vertex is
f 3 16 , the vertex is 3, 16 .
The graph of f is above the t-axis when 2 t 4 . Since the inequality is strict, the solution set is t 2 t 4 or, using interval
500, 200000 .
The graph of f is above the p-axis when 276.39 p 723.61 . Since the inequality is strict, the solution set is p 276.39 p 723.61 or, using interval
notation, 276.39, 723.61 . The revenue is more than $800,000 for prices between $276.39 and $723.61.
notation, 2, 4 . The ball is more than 128 feet above the ground for times between 2 and 4 seconds. 394 Copyright © 2025 Pearson Education, Inc.
Section 4.5: Inequalities Involving Quadratic Functions
36. a.
R( p)
1 2 p 1900 p 0 2
1 p p 3800 0 2 p 0, p 3800 Thus, the revenue equals zero when the price is $0 or $3800.
b.
2000c 24.3845 24.3845c 200 24.3845c 2 2000c 224.3845 0 We graph f (c ) 24.3845c 2 2000c 224.3845 . The intercepts are y-intercept: f (0) 224.3845 c-intercepts: 24.3845c 2 2000c 224.3845 0 c
2000
2000 4 24.3845 224.3845 2 24.3845 2
2000 3,978,113.985 48.769 c 0.112 or c 81.907
1 2 p 1900 p 1200000 0 2 p 2 3800 p 2400000 0 p 800 p 3000 0 p 800; p 3000
The vertex is at p
2
Find the values of p for which 1 p 2 1900 p 1200000 2 1 2 p 1900 p 1200000 0 2 1 We graph f ( p) p 2 1900 p 1200000 . 2 The intercepts are y-intercept: f (0) 1, 200, 000 p-intercepts:
2
9.81 2000 c(2000) 1 c 2 200 2 897 2000c 24.3845 1 c 2 200
The vertex is at b (2000) c 41.010 . Since 2a 2(24.3845) f 41.010 40, 785.273 , the vertex is
b (1900) 1900 . 2a 2(1/ 2)
41.010, 40785.273 .
Since f 1900 605, 000 , the vertex is
1900, 605000 .
The graph of f is above the c-axis when 0.112 c 81.907 . Since the inequality is strict, the solution set is c 0.112 c 81.907 or, using interval
The graph of f is above the p-axis when 800 p 3000 . Since the inequality is strict, the solution set is p 800 p 3000 or, using interval
notation, 800, 3000 . The revenue is more than $1,200,000 for prices between $800 and $3000.
g x 37. y cx 1 c 2 2 v
a.
2
notation, 0.112, 81.907 . b.
Since the round is to be on the ground y 0 . Note, 75 km = 75,000 m. So, x 75, 000, v 897, and g 9.81 .
2
9.81 75, 000 c(75, 000) 1 c 2 0 2 897 75, 000c 34, 290.724 1 c 2 0
2
Since the round must clear a hill 200 meters high, this mean y 200 . Now x 2000, v 897, and g 9.81 .
75, 000c 34, 290.724 34, 290.724c 0 34, 290.724c 2 75, 000c 34, 290.724 0 We graph f (c) 34,290.724c 2 75,000c 34,290.724 .
395 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
The intercepts are y-intercept: f (0) 34, 290.724 c-intercepts: 34, 290.724c 2 75, 000c 34, 290.724 0 (75, 000)
c=
75, 000 4 34, 290.724 34, 290.724 2 34, 290.724 2
75,000 921,584,990.2 68,581.448 c 0.651 or c 1.536
It is possible to hit the target 75 kilometers away so long as c 0.651 or c 1.536 . 38. Note that v = 25 mph =
110 ft/sec. We solve 3
1 2 w 2 kx v for x when k = 9450, g = 32.2, 2 2g
and v =
110 . 3
1 4000 110 (9450) x 2 2 2(32.2) 3
40. ( x 2) 2 0
We graph the function f ( x) ( x 2) 2 . y-intercept: f (0) 4 x-intercepts: ( x 2) 2 0 x2 0 x2 The vertex is the vertex is 2, 0 .
The graph is above the x-axis when x 2 or x 2 . Since the inequality is strict, the solution set is x x 2 or x 2 . Therefore, the given
inequality has exactly one real number that is not a solution, namely x 2 .
2
4725 x 2 83,505.86611 x 2 17.67319918 x 4.204 since x > 0. To the nearest tenth, the spring must be able to compress at least 4.3 feet.
39. ( x 4) 0 2
41. Solving x 2 x 1 0 We graph the function f ( x) x 2 x 1 . y-intercept: f (0) 1
x-intercepts: b 2 4ac 12 4 11 3 , so f has no x-intercepts. The vertex is at x
We graph the function f ( x) ( x 4) 2 . y-intercept: f (0) 16 x-intercepts: ( x 4) 2 0 x4 0 x4 The vertex is the vertex is 4, 0 .
b (1) 1 . Since 2a 2(1) 2
1 3 1 3 f , the vertex is , . 2 4 2 4
The graph is always above the x-axis. Thus, the solution is the set of all real numbers or, using interval notation, (, ) .
The graph is never below the x-axis. Since the inequality is not strict, the only solution comes from the x-intercept. Therefore, the given inequality has exactly one real solution, namely x 4.
42. Solving x 2 x 1 0 We graph the function f ( x) x 2 x 1 . y-intercept: f (0) 1
x-intercepts: b 2 4ac (1) 2 4(1)(1) 3 , so f has no x-intercepts.
396 Copyright © 2025 Pearson Education, Inc.
Section 4.5: Inequalities Involving Quadratic Functions
The vertex is at x
b.
b (1) 1 . Since 2a 2(1) 2
1 3 1 3 f , the vertex is , . 2 4 2 4
The graph is never below the x-axis. Thus, the inequality has no solution. That is, the solution set is { } or . 43. The x-intercepts are included when the original inequality is not strict (when it contains an equal sign with the inequality). 44. Since the radical cannot be negative we determine what makes the radicand a nonnegative number. 10 2 x 0 2 x 10 x5 So the domain is: x | x 5 . 45.
( x) ( x) 2 9 x 2 f ( x) x 9 Since f x f ( x) then the function is odd. f x
46. a.
2 x6 3 2 6 x 3 x9 2 y (0) 6 3 6 The intercepts are: 9, 0 , 0, 6 0
47.
d kt 203 3.5k k 58 so d 58t
48. a 1, b 6, c 8 x
6 62 4(1)(8) 2(1)
6 36 32 2 6 68 6 2 17 2 2 3 17
The solution set is 3 17, 3 17
4(0) 2 25 02 1 25 25 1
49. y
4 x 2 25 x2 1 0 4 x 2 25 25 0 x2 4 25 2 x 4 5 x 2 0
5 5 The x-intercepts are , 0 , , 0 and the y 2 2 intercept is 0, 25 .
397 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
right and 2 units up.
50. ( g f )( x) (3x 4) ( x 2 2 x 7) 3x 4 x 2 2 x 7 x2 x 3
51. ( f g )( x) ( x 2 2 x 7)(3x 4) 3 x3 4 x 2 6 x 2 8 x 21x 28 3 x3 2 x 2 29 x 28
52. f ( x h) f ( x) 3( x h) 2 5( x h) (3x 2 5 x) h h 2 2 2 3x 6 xh 3h 5 x 5h 3 x 5 x h 2 6 xh 3h 5h h(6 x 3h 5) h h 6 x 3h 5
53. 4
4
5
5 x (2 x 7) 8 x (2 x 7) (2 x 7)
4
3
(2 x 7)
8
4 ; y-intercept = 6 5
Slope =
b.
average rate of change =
c.
Plot the point (0, 6) . Use the slope to find an additional point by moving 5 units to the right and 4 units up.
d.
increasing
8
4 5
x (2 x 7) 10 x 35 8 x 4
4 x6 5
a.
3
(2 x 7)
increasing
2. h( x)
x (2 x 7) 5(2 x 7) 8 x
3
x (2 x 7) 5(2 x 7) 8 x 4
8
d.
3
(2 x 7)
8
x 2 x 35 4
(2 x 7)
5
Chapter 4 Review Exercises 1.
f x 2x 5
3. G x 4
a.
Slope = 2; y-intercept = 5
b.
average rate of change = 2
c.
Plot the point (0, 5) . Use the slope to find an additional point by moving 1 unit to the
a.
Slope = 0; y-intercept = 4
b.
average rate of change = 0
398 Copyright © 2025 Pearson Education, Inc.
Chapter 4 Review Exercises
7.
constant
d. 4.
x
y f x
–1
–2
0
3
1
8
2
13
3
18
Avg. rate of change = 3 2 0 1
x
y f x
–1
–3
0
4
5 5 1
83 5 5 1 0 1 13 8 5 5 2 1 1 18 13 5 5 3 2 1
1
7
2 3
6 1
Avg. rate of change = 4 3 0 1
f ( x) ( x 4) 2
Using the graph of y x 2 , shift the graph 4 units right, then reflect about the x-axis.
y x
This is a linear function with slope = 5, since the average rate of change is constant at 5. 5.
shift down 4 units.
Plot the point (0, 4) and draw a horizontal line through it.
c.
y x
8.
f ( x) 3( x 2) 2 1
Using the graph of y x 2 , stretch vertically by a factor of 3, then shift 2 units left, then reflect about the x-axis, then shift 1 unit up.
7 7 1
74 3 3 1 0 1
This is not a linear function, since the average rate of change is not constant. 6.
9. a.
f ( x) ( x 1) 2 4
Using the graph of y x 2 , shift left 1 unit, then
f ( x) ( x 2) 2 2 x2 4 x 4 2 x2 4 x 6 a 1, b 4, c 6. Since a 1 0, the graph is concave up. The x-coordinate of b 4 4 the vertex is x 2. 2a 2(1) 2
399 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
The y-coordinate of the vertex is b f f (2) (2) 2 4 2 6 2 . 2a Thus, the vertex is (2, 2). The axis of symmetry is the line x 2 . The discriminant is: b 2 4ac (4) 2 4 1 (6) 8 0 , so the
1 2 x 16 0 4 x 2 64 0 x 2 64 x 8 or x 8 The x-intercepts are –8 and 8. The y-intercept is f (0) 16 .
graph has no x-intercepts. The y-intercept is f (0) 6 .
b. c.
The domain is (, ) . The range is [2, ) .
The domain is (, ) . The range is [16, ) .
c.
Decreasing on , 0 . Increasing on 0, .
Decreasing on , 2 . Increasing on 2, .
10. a.
b.
f ( x)
11. a.
1 2 x 16 4
1 1 , b 0, c 16. Since a 0, the 4 4 graph is concave up. The x-coordinate of b 0 0 0. the vertex is x 1 2a 1 2 2 4 The y-coordinate of the vertex is 1 b f f (0) (0) 2 16 16 . 4 2a Thus, the vertex is (0, –16). The axis of symmetry is the line x 0 . The discriminant is: 1 b 2 4ac (0) 2 4 (16) 16 0 , so 4 the graph has two x-intercepts. The x-intercepts are found by solving: a
f ( x) 4 x 2 4 x a 4, b 4, c 0. Since a 4 0, the graph is concave down. The x-coordinate of the vertex is b 4 4 1 x . 2a 2( 4) 8 2 The y-coordinate of the vertex is 2
b 1 1 1 f f 4 4 2a 2 2 2 1 2 1 1 Thus, the vertex is , 1 . 2 1 The axis of symmetry is the line x . 2 The discriminant is: b 2 4ac 42 4( 4)(0) 16 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: 4x2 4 x 0 4 x( x 1) 0 x 0 or x 1 The x-intercepts are 0 and 1. The y-intercept is f (0) 4(0) 2 4(0) 0 .
400 Copyright © 2025 Pearson Education, Inc.
Chapter 4 Review Exercises
b.
The domain is (, ) .
b.
The range is , 1 . c.
12. a.
1 The range is , . 2
1 Increasing on , 2 1 Decreasing on , . 2 f ( x)
The domain is (, ) .
c.
9 2 x 3x 1 2
9 9 a , b 3, c 1. Since a 0, the 2 2 graph is concave up. The x-coordinate of the vertex is 3 3 b 1 x . 2a 9 3 9 2 2 The y-coordinate of the vertex is 2
b 1 9 1 1 f f 3 1 2a 3 2 3 3 1 1 1 1 2 2 1 1 Thus, the vertex is , . 3 2 1 The axis of symmetry is the line x . 3 The discriminant is: 9 b 2 4ac 32 4 (1) 9 18 9 0 , 2 so the graph has no x-intercepts. The y9 2 intercept is f 0 0 3 0 1 1 . 2
13. a.
1 Decreasing on , . 3 1 Increasing on , . 3 f ( x) 3x 2 4 x 1 a 3, b 4, c 1. Since a 3 0, the graph is concave up. The x-coordinate of the b 4 4 2 vertex is x . 2a 2(3) 6 3 The y-coordinate of the vertex is 2
b 2 2 2 f f 3 4 1 2a 3 3 3 4 8 7 1 3 3 3 2 7 Thus, the vertex is , . 3 3 2 The axis of symmetry is the line x . 3 The discriminant is: b 2 4ac (4) 2 4(3)(1) 28 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: 3x 2 4 x 1 0 . x
b b 2 4ac 4 28 2a 2(3) 4 2 7 2 7 6 3
The x-intercepts are
401 Copyright © 2025 Pearson Education, Inc.
2 7 1.55 and 3
Chapter 4: Linear and Quadratic Functions
2 7 0.22 . 3
16.
a 3, b 12, c 4. Since a 3 0, the graph is concave down, so the vertex is a maximum point. The maximum occurs at 12 12 b x 2. 2a 2(3) 6 The maximum value is 2 b f f 2 3 2 12 2 4 2 a 12 24 4 16
2
The y-intercept is f (0) 3(0) 4(0) 1 1 .
b.
17. x 2 6 x 16 0 We graph the function f ( x) x 2 6 x 16 . The intercepts are y-intercept: f (0) 16
The domain is (, ) .
x-intercepts: x 2 6 x 16 0 ( x 8)( x 2) 0 x 8, x 2 b (6) 3 . Since The vertex is at x 2a 2(1)
7 The range is , . 3
c.
14.
15.
f ( x) 3x 2 12 x 4
2 Decreasing on , 3 2 Increasing on , . 3
f (3) 25 , the vertex is 3, 25 .
f ( x) 3 x 2 6 x 4
a 3, b 6, c 4. Since a 3 0, the graph is concave up, so the vertex is a minimum point. The minimum occurs at b 6 6 x 1. 2a 2(3) 6 The minimum value is 2 b f f 1 3 1 6 1 4 2a 36 4 1
The graph is below the x-axis when 8 x 2 . Since the inequality is strict, the solution set is x | 8 x 2 or, using interval notation,
f ( x) x 2 8 x 4
a 1, b 8, c 4. Since a 1 0, the graph is concave down, so the vertex is a maximum point. The maximum occurs at b 8 8 x 4. 2a 2(1) 2 The maximum value is 2 b f f 4 4 8 4 4 2a 16 32 4 12
8, 2 . 18.
3x 2 14 x 5 3x 2 14 x 5 0 We graph the function f ( x) 3 x 2 14 x 5 . The intercepts are y-intercept: f (0) 5
x-intercepts: 3x 2 14 x 5 0 (3x 1)( x 5) 0 1 x , x5 3 b (14) 14 7 . The vertex is at x 2a 2(3) 6 3
402 Copyright © 2025 Pearson Education, Inc.
Chapter 4 Review Exercises
64 7 64 7 Since f , the vertex is , . 3 3 3 3
21. a. b.
c.
0.01x 25, 000 100, 000 0.01x 75, 000 x 7,500, 000 Orlando’s sales would have to be $7,500,000 in order to earn $100,000.
The graph is above the x-axis when x
19. Use the form f ( x) a ( x h) 2 k . The vertex is (2, 4) , so h 2 and k 4 .
S (1, 000, 000) 0.01(1, 000, 000) 25, 000 10, 000 25, 000 35, 000 In 2005, Orlando’s salary was $35,000.
1 or 3 x 5 . Since the inequality is not strict, the 1 solution set is x x or x 5 or, using 3 1 interval notation, , 5, . 3
S ( x) 0.01x 25, 000
d.
22. a.
0.01x 25, 000 150, 000 0.01x 125, 000 x 12,500, 000 Orlando’s sales would have to be more than $12,500,000 in order for his salary to exceed $150,000.
If x 1500 10 p, then p
1500 x . 10
R ( p ) px p (1500 10 p ) 10 p 2 1500 p
f ( x) a ( x 2) 2 4 .
Since the graph passes through (0, 16) , f (0) 16 . 16 a (0 2) 2 4
b. Domain: p 0 p 150 c.
16 a ( 2) 2 4 12 4a 3 a f ( x) 3( x 2) 2 4
p
b 1500 1500 $75 2a 2 10 20
d. The maximum revenue is R(75) 10(75) 2 1500(75) 56250 112500 $56, 250
3( x 2 4 x 4) 4 3x 2 12 x 12 4 3x 2 12 x 16
20. Use the form f ( x) a ( x h) 2 k . The vertex is (1, 2) , so h 1 and k 2 .
e.
x 1500 10(75) 1500 750 750
f.
Graph R 10 p 2 1500 p and R 56000 .
f ( x) a ( x 1) 2 2 .
Since the graph passes through (1, 6) , f (1) 6 . 6 a(1 1) 2 2 6 a(2) 2 2 6 4a 2 4 4a 1 a f ( x) 1( x 1) 2 2 ( x 2 2 x 1) 2 x2 2 x 3
Find where the graphs intersect by solving 56000 10 p 2 1500 p . 10 p 2 1500 p 56000 0 p 2 150 p 5600 0 ( p 70)( p 80) 0 p 70, p 80 The company should charge between $70 and $80.
403 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions 23. Since there are 200 feet of border, we know that 2 x 2 y 200 . The area is to be maximized, so A x y . Solving the perimeter formula for y : 2 x 2 y 200 2 y 200 2 x y 100 x The area function is: A( x ) x(100 x) x 2 100 x The maximum value occurs at the vertex: b (100) 100 50 x 2 2a 2(1) The pond should be 50 feet by 50 feet for maximum area.
25. Consider the diagram
d
x Let d diameter of the semicircles width of the rectangle Let x length of the rectangle 100 outside dimension length 100 2 x 2 circumference of a semicircle 100 2 x circumference of a circle 100 2 x d
24. Consider the diagram
100 d 2 x 100 d x 2 1 50 d x 2
x
y Total amount of fence = 3x 2 y 10, 000 y
10, 000 3 x 3 5000 x 2 2
3 Total area enclosed = x y x 5000 x 2 3 3 A x 5000 x x 2 x 2 5000 x is a 2 2 3 quadratic function with a 0 . 2 So the vertex corresponds to the maximum value for this function. The vertex occurs when b 5000 5000 x . 2a 2 3 / 2 3
The maximum area is: 2
3 5000 5000 5000 A 5000 2 3 3 3 3 25, 000, 000 25, 000, 000 2 9 3 12,500, 000 25, 000, 000 3 3 12,500, 000 3 4,166, 666.67 square meters
We need an expression for the area of a rectangle in terms of a single variable. Arectangle x d 1 50 d d 2 1 2 50d d 2 1 This is a quadratic function with a 0 . 2 Therefore, the x-coordinate of the vertex represents the value for d that maximizes the area of the rectangle and the y-coordinate of the vertex is the maximum area of the rectangle. The vertex occurs at 50 50 50 b d 2a 1 2 2 This gives us 1 1 50 x 50 d 50 50 25 25 2 2 Therefore, the side of the rectangle with the 50 feet and the other side semicircle should be should be 25 feet. The maximum area is 50 1250 25 397.89 ft 2 .
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Chapter 4 Review Exercises
26. C ( x) 4.9 x 2 617.4 x 19, 600 ; a 4.9, b 617.4, c 19, 600. Since a 4.9 0, the graph is concave up, so the vertex is a minimum point. a. The minimum marginal cost occurs at 617.40 617.40 b x 63 . 2a 2(4.9) 9.8 Thus, 63 golf clubs should be manufactured in order to minimize the marginal cost. b. The minimum marginal cost is C 63 4.9 63 617.40 63 19600 2
0 a(10) 2 10 1 0.10 10 10 The equation of the parabola is: 1 y x 2 10 10 At x 8 : 1 y (8) 2 10 6.4 10 3.6 feet 10 a
10
2
29. a.
$151.90
27. The area function is: A( x ) x(10 x) x 2 10 x The maximum value occurs at the vertex: b 10 10 x 5 2a 2(1) 2 The maximum area is: A(5) (5)2 10(5) 25 50 25 square units
x
b. Yes, the two variables appear to have a linear relationship. c.
Using the LINear REGression program, the line of best fit is: y 1.3902 x 1.1140
d.
y 1.39017 26.5 1.11395 37.95 mm
xx
x
28. Locate the origin at the point directly under the highest point of the arch. Then the equation is in the form: y ax 2 k , where a 0 . Since the maximum height is 10 feet, when x 0, y k 10 . Since the point (10, 0) is on the parabola, we can find the constant:
405 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions 30. a.
c.
The slope is negative, so the graph is decreasing.
d.
Plot the point (0, 3) . Use the slope to find an additional point by moving 1 unit to the right and 4 units down.
The data appear to be quadratic with a < 0. b. The maximum revenue occurs at b 411.88 A 2a 2(7.76) 411.88 $26.5 thousand 15.52 c.
The maximum revenue is b R R 26.53866 2a
2.
f ( x) 3x 2 2 x 8
y-intercept: f (0) 8 x-intercepts:
3x 2 2 x 8 0 (3x 4)( x 2) 0
7.76 26.5 411.88 26.5 942.72 2
$6408 thousand
d. Using the QUADratic REGression program, the quadratic function of best fit is: y 7.76 x 2 411.88 x 942.72 .
x
4 ; x2 3
The intercepts are (0, 8), , 0 , and (2, 0) . 4 3
3. G ( x) 2 x 2 4 x 1
y-intercept: G (0) 1 x-intercepts: 2 x 2 4 x 1 0 a 2, b 4, c 1
e.
x
2 b b 2 4ac 4 4 4 2 1 2a 2 2
4 24 4 2 6 2 6 4 4 2 2 6 The intercepts are (0, 1) , , 0 , and 2
2 6 , 0 . 2
Chapter 4 Test 1.
f x 4 x 3
a. Slope = 4 ; y-intercept = 3. b. average rate of change = 4 406 Copyright © 2025 Pearson Education, Inc.
Chapter 4 Test f ( x) g ( x)
4. a.
b. The x-coordinate of the vertex is b 12 12 x 2. 2a 2 3 6
x 2 3x 5 x 3 x2 2x 3 0 ( x 1)( x 3) 0 x 1 0 or x 3 0 x 1 or x3
The y-coordinate of the vertex is 2 b f f 2 3 2 12 2 4 2a 12 24 4 8 Thus, the vertex is 2, 8 .
The solution set is 1, 3 . b.
c.
The axis of symmetry is the line x 2 .
d. The discriminant is: b 2 4ac 12 4 3 4 96 0 , so 2
the graph has two x-intercepts. The xintercepts are found by solving: 3x 2 12 x 4 0 . x
b b 2 4ac (12) 96 2a 2(3) 12 4 6 6 2 6 6 3
The x-intercepts are c.
x | 1 < x 3 ; 1,3
62 6 0.37 and 3
62 6 3.63 . The y-intercept is 3
f (0) 3(0) 2 12(0) 4 4 .
e. 5.
f ( x) x 3 2 2
Using the graph of y x 2 , shift right 3 units, then shift down 2 units. y
6. a.
7.
x
f ( x) 3 x 2 12 x 4 a 3, b 12, c 4. Since a 3 0, the graph is concave up.
f ( x) 2 x 2 12 x 3 a 2, b 12, c 3. Since a 2 0, the graph is concave down, so the vertex is a maximum point. The maximum occurs at b 12 12 x 3. 2a 2(2) 4
407 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
c.
The maximum value is f 3 2 3 12 3 3 18 36 3 21 . 2
8. x 2 10 x 24 0 We graph the function f ( x) x 2 10 x 24 . The intercepts are y-intercept: f (0) 24
x-intercepts: x 2 10 x 24 0 ( x 4)( x 6) 0 x 4, x 6 b (10) 10 5. The vertex is at x 2a 2(1) 2
p
b 10000 10000 $500 2a 2 10 20
d. The maximum revenue is R(500) 10(500) 2 10000(500) 2500000 15000000 $2,500, 000 e.
x 10000 10(500) 10000 5000 5000
f.
Graph R 10 p 2 10000 p and R 1600000 . Find where the graphs intersect by solving 56000 10 p 2 1500 p .
Since f (5) 1 , the vertex is (5, 1).
10 p 2 10000 p 1600000 0 p 2 1000 p 160000 0 ( p 200)( p 800) 0 p 200, p 800 The company should charge between $200 and $800.
The graph is above the x-axis when x 4 or x 6 . Since the inequality is not strict, the solution set is x x 4 or x 6 or, using
interval notation, , 4 6, . 9. a. b.
c.
Chapter 4 Cumulative Review
C (m) 0.15m 129.50 C (860) 0.15(860) 129.50 129 129.50 258.50 If 860 miles are driven, the rental cost is $258.50. C (m) 213.80 0.15m 129.50 213.80 0.15m 84.30 m 562 The rental cost is $213.80 if 562 miles were driven.
10. a.
If x 10000 10 p, then
1. P 1,3 ; Q 4, 2
Distance between P and Q: d P, Q
2
2
(5) 2 (5) 2 25 25 50 5 2 Midpoint between P and Q: 1 4 3 2 3 1 , , 1.5, 0.5 2 2 2 2
2. y x3 3x 1 a.
10000 x . p 10
4 1 2 3
2, 1 : 1 2 3 2 1 3
1 8 6 1
R ( p ) px p (10000 10 p) 10 p 2 10000 p
1 1 Yes, 2, 1 is on the graph.
b. Domain: p 0 p 1000
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Chapter 4 Cumulative Review
b.
2,3 : 3 2 3 2 1 3
Slope of perpendicular =
3 8 6 1
y y1 m( x x1 )
33 Yes, 2,3 is on the graph.
c.
1 2
1 x 3 2 1 3 y 5 x 2 2 1 13 y x 2 2 y 5
3,1 : 1 3 3 3 1 3
1 27 9 1 1 35 No, 3,1 is not on the graph.
3. 5 x 3 0 5 x 3 3 x 5
3 3 The solution set is x x or , . 5 5 6. x 2 y 2 4 x 8 y 5 0 x2 4x y2 8 y 5
4. (–1,4) and (2,–2) are points on the line. 2 4 6 Slope 2 2 1 3
( x 2 4 x 4) ( y 2 8 y 16) 5 4 16 ( x 2) 2 ( y 4) 2 25 ( x 2) 2 ( y 4) 2 52 Center: (2, –4) Radius = 5
y y1 m x x1
y 4 2 x 1 y 4 2 x 1 y 4 2 x 2 y 2 x 2
7. Yes, this is a function since each x-value is paired with exactly one y-value. 8. 5. Perpendicular to y 2 x 1 ; Containing (3, 5)
f ( x) x 2 4 x 1 a.
f (2) 2 2 4 2 1 4 8 1 3
b.
f ( x) f 2 x 2 4 x 1 3 x2 4x 2
c.
f ( x) x 4 x 1 x 2 4 x 1
d.
f ( x) x 2 4 x 1 x 2 4 x 1
2
409 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions
e.
f ( x 2) x 2 4 x 2 1 2
12.
f ( x)
x2 4 x 4 4 x 8 1
f ( x)
f ( x h) f x
Therefore, f is neither even nor odd.
h
x h 4 x h 1 x 2 4 x 1 2
x x2 f x or f x 2 x 1 2 x 1 2
x2 3
f.
x2 2x 1
13.
f x x 3 5 x 1 on the interval 4, 4
Use MAXIMUM and MINIMUM on the graph of y1 x3 5 x 1 .
h x 2 2 xh h 2 4 x 4h 1 x 2 4 x 1 h 2 xh h 2 4h h h 2x h 4 2x h 4 h
3z 1 6z 7 The denominator cannot be zero: 6z 7 0
9. h( z )
6z 7 7 6 7 Domain: z z 6 z
Local maximum is 5.30 and occurs at x 1.29 ; Local minimum is –3.30 and occurs at x 1.29 ; f is increasing on 4, 1.29 or 1.29, 4 ;
f is decreasing on 1.29,1.29 .
10. Yes, the graph represents a function since it passes the Vertical Line Test. 11.
f ( x)
a.
b. c.
14.
x x4
1 1 1 1 , so 1, is not on 1 4 5 4 4 the graph of f. f (1)
2 2 1, so 2, 1 is a 2 4 2 point on the graph of f. f (2)
Solve for x: x 2 x4 2x 8 x x 8 So, (8, 2) is a point on the graph of f.
f ( x) 3x 5; g ( x) 2 x 1 a. f ( x) g ( x) 3x 5 2 x 1 3x 5 2 x 1 x 4
b.
f x g x 3x 5 2 x 1 3x 5 2 x 1 x 4 The solution set is x x 4 or 4, .
15. a.
Domain: x | 4 x 4 or 4, 4 Range: y | 1 y 3 or 1, 3
b.
Intercepts: 1, 0 , 0, 1 , 1, 0 x-intercepts: 1, 1 y-intercept: 1
410 Copyright © 2025 Pearson Education, Inc.
Chapter 4 Projects c.
point is multiplied by 2.
The graph is symmetric with respect to the y-axis.
d. When x 2 , the function takes on a value of 1. Therefore, f 2 1 . e.
The function takes on the value 3 at x 4 and x 4 .
f.
f x 0 means that the graph lies below
the x-axis. This happens for x values between 1 and 1. Thus, the solution set is x | 1 x 1 or 1, 1 . g.
Since the graph is symmetric about the yaxis, the function is even.
j.
The graph of y f x 2 is the graph of y f x but shifted up 2 units.
k. The function is increasing on the interval 0, 4 .
Chapter 4 Projects Project I – Internet-based Project
Answers will vary. h.
The graph of y f x is the graph of
Project II
y f x but reflected about the y-axis.
a.
1000 m/sec
kg
b. The data would be best fit by a quadratic function.
i.
The graph of y 2 f x is the graph of y f x but stretched vertically by a
factor of 2. That is, the coordinate of each
y 0.085 x 2 14.46 x 1069.52
411 Copyright © 2025 Pearson Education, Inc.
Chapter 4: Linear and Quadratic Functions 1000
These results seem reasonable since the function fits the data well.
m/sec
kg
c.
s0 = 0m
Type
Weight kg
Velocity m/sec
MG 17
10.2
905
2 v0 t s0 2 s (t ) 4.9t 2 639.93t Best. (It goes the highest)
MG 131
19.7
710
s (t ) 4.9t 2 502.05t
MG 151
41.5
850
s (t ) 4.9t 2 601.04t
MG 151/20
42.3
695
s (t ) 4.9t 2 491.44t
MG/FF
35.7
575
s (t ) 4.9t 2 406.59t
MK 103
145
860
s (t ) 4.9t 2 608.11t
MK 108
58
520
s (t ) 4.9t 2 367.70t
WGr 21
111
315
s (t ) 4.9t 2 222.74t
Type
Weight kg
Velocity m/sec
MG 17
10.2
905
2 v0 t s0 2 s (t ) 4.9t 2 639.93t 200 Best. (It goes the highest)
MG 131
19.7
710
s (t ) 4.9t 2 502.05t 200
MG 151
41.5
850
s (t ) 4.9t 2 601.04t 200
MG 151/20
42.3
695
s (t ) 4.9t 2 491.44t 200
MG/FF
35.7
575
s (t ) 4.9t 2 406.59t 200
MK 103
145
860
s (t ) 4.9t 2 608.11t 200
MK 108
58
520
s (t ) 4.9t 2 367.70t 200
WGr 21
111
315
s (t ) 4.9t 2 222.74t 200
Type
Weight kg
Velocity m/sec
MG 17
10.2
905
2 v0 t s0 2 s (t ) 4.9t 2 639.93t 30 Best. (It goes the highest)
MG 131
19.7
710
s (t ) 4.9t 2 502.05t 30
MG 151
41.5
850
s (t ) 4.9t 2 601.04t 30
MG 151/20
42.3
695
s (t ) 4.9t 2 491.44t 30
MG/FF
35.7
575
s (t ) 4.9t 2 406.59t 30
Equation in the form: s (t ) 4.9t 2
s0 = 200m
Equation in the form: s (t ) 4.9t 2
s0 = 30m
Equation in the form: s (t ) 4.9t 2
412 Copyright © 2025 Pearson Education, Inc.
Chapter 4 Projects
MK 103
145
860
s (t ) 4.9t 2 608.11t 30
MK 108
58
520
s (t ) 4.9t 2 367.70t 30
WGr 21
111
315
s (t ) 4.9t 2 222.74t 30
Notice that the gun is what makes the difference, not how high it is mounted necessarily. The only way to change the true maximum height that the projectile can go is to change the angle at which it fires. Project III a.
e. x
1 2
3
4
5
f.
y y2 y1 3 (1) 2 1 x x2 x1
1
3
5 15 33 59
2
3
4
14 6 2 10 18 26
x
2 1 0 1
y
23
2 y x2
y y2 y1 5 (3) 2 1 x x2 x1
9
2
3
4
3 5 15 33 59 8 8
8
8
8
The second differences are all the same.
y are the same. x
g. The paragraph should mention at least two observations: 1. The first differences for a linear function are all the same. 2. The second differences for a quadratic function are the same.
Median Income ($)
0
9
As x increases,
y y2 y1 1 1 2 1 x x2 x1
c.
1
23
y increases. This makes sense x because the parabola is increasing (going up) steeply as x increases.
y y2 y1 1 3 2 1 x x2 x1
All of the values of
2
y y x
y 2 x 5 3 1 1 3 5
b.
x
Age Class Midpoint
I 30633 9548 d. 2108.50 x 10 I 37088 30633 645.50 x 10 I 41072 37088 398.40 x 10 I 34414 41072 665.80 x 10 I 19167 34414 1524.70 x 10 I values are not all equal. The data are These x not linearly related.
Project IV a. – i.
Answers will vary , depending on where the CBL is located above the bouncing ball.
j.
The ratio of the heights between bounces will be the same.
413 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
Chapter 5 Polynomial and Rational Functions Section 5.1 1.
15.
degree 3. Leading term: x3 ; Constant term: none
2, 0 , 2, 0 , and 0,9 x-intercepts: let y 0 and solve for x
16.
9 x 2 4 0 36
2 3x 2 2 3 2 x is a polynomial 5 5 5 3 function of degree 2. Leading term: x 2 ; 5 2 Constant term: 5
17. g ( x)
x2 4 x 2
y-intercepts: let x 0 and solve for y 9 0 4 y 36 2
4 y 36 y9
2. Yes; it has the form an x n an 1 x n 1 ... a1 x a0 where each ai is a real number and n is a positive integer.; degree 3 3. down; 4
1 x is a polynomial function of 2 1 degree 1. Leading term: x ; Constant term: 3 2
18. h( x) 3
19.
4. True: f ( x) 0 indicates that y 0 which indicates that the point is an x-intercept. 5. b; c
20.
6. smooth; continuous
1,1 , 0, 0 , and 1,1
9. a. r is a real zero of a polynomial function f . b. r is an x-intercept of the graph of f .
10. turning points 11. y 3x 4 12. ; 13. b 14. d
1 1 x 1 is not a polynomial x function because it contains a negative exponent. f ( x) 1
f ( x) x( x 1) x 2 x is a polynomial
function of degree 2. Leading term: x 2 ; Constant term: 0
7. b
c. x r is a factor of f .
f ( x) 5 x 2 4 x 4 is a polynomial function of
degree 4. Leading term: 4 x 4 ; Constant term: 0
9 x 2 36
8.
f ( x) 4 x x3 is a polynomial function of
21. g ( x) x 2/3 x1 3 2 is not a polynomial function because it contains a fractional exponent. 22. h( x ) x
x 1 x x
1/ 2
is not a
polynomial function because it contains fractional exponents. 1 is a polynomial function 2 of degree 4. Leading term: 5 x 4 ; Constant term: 1 2
23. F ( x ) 5 x 4 x3
x2 5
x 1 5 x 3 is not a polynomial x3 function because it contains a negative exponent.
24. F ( x )
414 Copyright © 2025 Pearson Education, Inc.
Section 5.1: Polynomial Functions
25. G ( x) 2( x 1) 2 ( x 2 1) 2( x 2 2 x 1)( x 2 1) 2( x 4 x 2 2 x3 2 x x 2 1) 2( x 4 2 x3 2 x 2 2 x 1) 2 x 4 4 x3 4 x 2 4 x 2 is a polynomial function of degree 4. Leading term: 2 x 4 ; Constant term: 2
26
G ( x) 3x 2 ( x 2)3 3 x 2 ( x3 6 x 2 12 x 8) 3x5 18 x 4 36 x3 24 x 2 is a polynomial function of degree 5. Leading term: 3 x5 ; Constant term: 0
27.
f ( x) ( x 1)
30.
4
Using the graph of y x 4 , shift the graph vertically up 2 units.
Using the graph of y x 4 , shift the graph horizontally, 1 unit to the left.
28.
31.
f ( x) ( x 2)5
Using the graph of y x5 , shift the graph horizontally to the right 2 units.
29.
f ( x) x 4 2
1 4 x 2 Using the graph of y x 4 , compress the graph f ( x)
1 vertically by a factor of . 2
f ( x) x5 3
Using the graph of y x5 , shift the graph vertically, 3 units down. 415 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
32.
f ( x) 3x5
35.
Using the graph of y x5 , stretch the graph vertically by a factor of 3.
33.
34.
Using the graph of y x5 , shift the graph horizontally, 1 unit to the right, and shift vertically 2 units up.
f ( x) x5
Using the graph of y x5 , reflect the graph about the x-axis.
f ( x) ( x 1)5 2
36.
f ( x) ( x 2) 4 3
Using the graph of y x 4 , shift the graph horizontally left 2 units, and shift vertically down 3 units.
f ( x) x 4
Using the graph of y x 4 , reflect the graph about the x-axis.
37.
f ( x) 2( x 1)4 1
Using the graph of y x 4 , shift the graph horizontally, 1 unit to the left, stretch vertically by a factor of 2, and shift vertically 1 unit up.
416 Copyright © 2025 Pearson Education, Inc.
Section 5.1: Polynomial Functions
38.
1 ( x 1)5 2 2 Using the graph of y x5 , shift the graph horizontally 1 unit to the right, compress 1 vertically by a factor of , and shift vertically 2 down 2 units. f ( x)
41.
f ( x) a x (1) ( x 1)( x 3)
For a 1 :
f ( x) ( x 1)( x 1)( x 3) x 2 1 ( x 3) x3 3x 2 x 3
42.
f ( x) a x ( 2) ( x 2)( x 3)
For a 1 :
f ( x) ( x 2)( x 2)( x 3) x 2 4 ( x 3) 3
2
x 3x 4 x 12
43.
f ( x) a x (5) ( x 0)( x 6)
For a 1 : f ( x) ( x 5)( x)( x 6)
x( x 5)( x 6) x 2 5 x ( x 6) 3
2
2
x 6 x 5 x 30 x
39.
x3 x 2 30 x
f ( x) 4 ( x 2)5 ( x 2)5 4
Using the graph of y x5 , shift the graph horizontally, 2 units to the right, reflect about the x-axis, and shift vertically 4 units up.
44.
f ( x) a x ( 4) ( x 0)( x 2)
For a 1 : f ( x) ( x 4)( x)( x 2)
x( x 4)( x 2) x 2 4 x ( x 2) 3
2
2
x 2 x 4 x 8x x3 2 x 2 8 x
45.
f ( x) a x ( 5) x (2) ( x 3)( x 5)
For a 1 : f ( x) ( x 5)( x 2)( x 3)( x 5)
x 2 7 x 10 x 2 8 x 15 4
40.
4
2
3
2
2
x 8 x 15 x 7 x 56 x 105 x 10 x 80 x 150
4
f ( x) 3 ( x 2) ( x 2) 3
x 4 x3 31x 2 25 x 150
4
Using the graph of y x , shift the graph horizontally, 2 units to the left, reflect about the x-axis, and shift vertically 3 units up.
3
46.
f ( x) a x ( 3) x (1) ( x 2)( x 5)
For a 1 : f ( x) ( x 3)( x 1)( x 2)( x 5)
x 2 4 x 3 x 2 7 x 10 4
3
2
3
x 7 x 10 x 4 x 28 x 2 40 x 3x 2 21x 30 x 4 3 x3 15 x 2 19 x 30
417 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
47.
f ( x) a x (1) x 3
For a 1 : f ( x) ( x 1)( x 3) 2
52.
2
x 1 x 2 6 x 9
5 5 5 5 15 a 5 1 2 6 2 2 2 2
735 a 16 16 a 49
15
x3 6 x 2 9 x x 2 6 x 9 x3 5 x 2 3 x 9
48.
16 ( x 5)( x 1)( x 2)( x 6) 49 16 ( x 4 2 x3 31x 2 32 x 60) 49 16 4 32 3 496 2 512 960 x x x x 49 49 49 49 49
f ( x) a x ( 2) x 4 2
f ( x)
For a 1 : f ( x) ( x 2) 2 ( x 4)
x2 4 x 4 x 4 3
x 4 x 4 x 2 16 x 4 x 16
2
53.
x3 12 x 16
49.
36 12a a3
36 a(4)(1)(3)
f ( x) 3( x 3)( x 1)( x 4)
36 12a a3
54.
f ( x) 3( x 2)( x 3)( x 5)
f ( x) a ( x 4)( x 1)( x 2) 16 a (0 4)(0 1)(0 2) 16 8a
3x 18 x 3 x 90
50.
f ( x) a ( x 3)( x 1)( x 4) 36 a (0 3)(0 1)(0 4)
f ( x) a( x 2)( x 3)( x 5) 36 a(2 2)(2 3)(2 5)
3
f ( x) a ( x 5)( x 1)( x 2)( x 6)
2
a 2
f ( x) ax( x 2)( x 2)
f ( x) 2( x 4)( x 1)( x 2)
16 a 4 (4 2)(4 2) 16 48a 1 3 1 f ( x) x( x 2)( x 2) 3 a
55.
45 a 2 1 2 1 2
2
45 9a a5
1 4 x3 x 3 3
51.
f ( x) a ( x 1) 2 ( x 1) 2
f ( x) 5( x 1) 2 ( x 1) 2 5 x 4 10 x 2 5
f ( x) ax( x 2)( x 1)( x 3) 1 1 1 1 63 a 2 1 3 2 2 2 2 63 63 a 16 a 16
56.
f ( x) ax( x 1) 2 ( x 3) 2 48 a (1) 1 1 1 3 2
2
48 16a a 3 f ( x) 3 x( x 1) 2 ( x 3) 2
f ( x) 16 x( x 2)( x 1)( x 3) 16 x 4 32 x 3 80 x 2 96 x
3x 5 12 x 4 6 x3 36 x 2 27 x
418 Copyright © 2025 Pearson Education, Inc.
Section 5.1: Polynomial Functions
57.
f ( x) a ( x 5) 2 ( x 2)( x 4)
62. a.
128 a (3 5) 2 (3 2)(3 4)
is: 3, with multiplicity one. x 2 4 0 has no real solution.
128 64a a 2
b. The graph crosses the x-axis at 3 c.
f ( x) 2( x 5) 2 ( x 2)( x 4)
58.
values of x .
64 a 2 (2 4)(2 2) 3
64 64a a 1
2
63. a.
f ( x) x3 ( x 4)( x 2)
The real zeros of f ( x) 3( x 7)( x 3) 2 are: 7, with multiplicity one; and –3, with multiplicity two.
b. The graph crosses the x-axis at 7 (odd multiplicity) and touches it at –3 (even multiplicity). c.
1 (even 2 multiplicity) and crosses the x-axis at -4 (odd multiplicity).
c.
n 1 3 1 2
values of x . 2
64. a.
The real zeros of f ( x) 4( x 4)( x 3)3 are: –4, with multiplicity one; and –3, with multiplicity three.
b. The graph crosses the x-axis at –4 and at –3 (odd multiplicities).
1 (even 3 multiplicity), and crosses the x-axis at 1 (odd multiplicity).
n 1 4 1 3
c.
values of x .
2
2
The real zeros of f ( x) 7 x 4 ( x 5) is: 5, with multiplicity three. x 2 4 0 has no real solution.
b. The graph crosses the x-axis at 5 (odd multiplicity). c.
n 1 7 1 6
d. The function resembles y 4 x 7 for large
values of x .
3
1 3 The real zeros of f ( x) x x 1 3 1 are: , with multiplicity two; and 1, with 3 multiplicity 3.
b. The graph touches the x-axis at
d. The function resembles y 4 x 4 for large
61. a.
n 1 5 1 4
d. The function resembles y 2 x5 for large
values of x .
c.
1 3 The real zero of f ( x) 2 x x 4 2 1 are: , with multiplicity two; -4 with 2 multiplicity 3.
b. The graph touches the x-axis at
d. The function resembles y 3x3 for large
60. a.
n 1 7 1 6
d. The function resembles y 2 x 7 for large
f ( x) ax3 ( x 4)( x 2)
59. a.
The real zeros of f ( x) 2 x 3 ( x 2 4)3
n 1 5 1 4
d. The function resembles y x5 for large
values of x . 65. a.
The real zeros of f ( x) ( x 5)3 ( x 4) 2 are: 5, with multiplicity three; and –4, with multiplicity two.
b. The graph crosses the x-axis at 5 (odd multiplicity) and touches it at –4 (even multiplicity). c.
n 1 5 1 4
d. The function resembles y x5 for large
values of x . 419 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
66. a.
The real zeros of f ( x) x 3
( x 2) 2
4
are: 3 , with multiplicity two; and 2, with multiplicity four. b. The graph touches the x-axis at 3 and at 2 (even multiplicities). c.
n 1 6 1 5 6
d. The function resembles y x for large
2 1 2 x 2 9 x 2 7 has no real 2 zeros. 2 x 2 9 0 and x 2 7 0 have no real solutions.
f ( x)
b. The graph neither touches nor crosses the xaxis. c.
n 1 6 1 5
d. The function resembles y x 6 for large
values of x . 68. a.
3
f ( x) 2 x 2 3 has no real zeros. 2
x 3 0 has no real solutions.
b. The graph neither touches nor crosses the xaxis. c.
n 1 6 1 5
d. The function resembles y 2 x 6 for large
values of x . 69. a.
The real zeros of f ( x) 2 x 2 ( x 2 2) are: 2 and 2 with multiplicity one; and 0, with multiplicity two.
b. The graph touches the x-axis at 0 (even multiplicity) and crosses the x-axis at 2 and 2 (odd multiplicities). c.
n 1 4 1 3
d. The function resembles y 2 x 4 for large
values of x . 70. a.
The real zeros of f ( x) 4 x( x 2 3) are: 3,
n 1 3 1 2
d. The function resembles y 4 x3 for large
values of x . 71. The graph could be the graph of a polynomial function.; zeros: 1, 1, 2 ; min degree = 3 72. The graph could be the graph of a polynomial.; zeros: 1, 2 ; min degree = 4 73. The graph cannot be the graph of a polynomial.; not continuous at x 1
values of x . 67. a.
c.
3 and 0, with multiplicity one.
b. The graph crosses the x-axis at 3 , and 0 (odd multiplicities).
3
74. The graph cannot be the graph of a polynomial.; not smooth at x 0 75. The graph crosses the x-axis at x 0 , x 1 , and x 2 . Thus, each of these zeros has an odd multiplicity. Using one for each of these multiplicities, a possible function is f ( x) ax( x 1)( x 2) . Since the y-intercept is 0, we know f (0) 0 . Thus, a can be any positive constant. Using a 1 , the function is f ( x) x( x 1)( x 2) . 76. The graph crosses the x-axis at x 0 and x 2 and touches at x 1 . Thus, 0 and 2 each have odd multiplicities while 1 has an even multiplicity. Using one for each odd multiplicity and two for the even multiplicity, a possible function is f ( x) ax( x 1) 2 ( x 2) . Since the y-intercept is 0, we know f (0) 0 . Thus, a can be any positive constant. Using a 1 , the function is f ( x) x( x 1) 2 ( x 2) . 77. The graph crosses the x-axis at x 1 and x 2 and touches it at x 1 . Thus, 1 and 2 each have odd multiplicities while 1 has an even multiplicity. Using one for each odd multiplicity and two for the even multiplicity, a possible funtion is f ( x) ax( x 1) 2 ( x 2) . Since the y-
intercept is 1, we know f (0) 1 . Thus, a(0 1)(0 1)(0 2) 1 a(1)(1)(2) 1 2a 1 1 a 2 1 The function is f ( x) ( x 1)( x 1) 2 ( x 2) . 2
420 Copyright © 2025 Pearson Education, Inc.
Section 5.1: Polynomial Functions
The graph crosses the x-axis at x 1 , x 1 , and x 2 . Thus, each of these zeros has an odd multiplicity. Using one for each of these multiplicities, a possible funtion is f ( x) a( x 1)( x 1)( x 2) . Since the yintercept is 1 , we know f (0) 1 . Thus, a(0 1)(0 1)(0 2) 1 a (1)(1)(2) 1 2a 1 1 a 2 1 The function is f ( x) ( x 1)( x 1)( x 2) . 2 79. The graph crosses the x-axis at x 4 and x 3 and touches it at x 1 . Thus, 4 and 3 each have odd multiplicities while 1 has an even multiplicity. Using one for each odd multiplicity and two for the even multiplicity, a possible funtion is f ( x) a ( x 4)( x 1) 2 ( x 3) . We know f (1) 8 . Thus,
f (2) 50 . Thus,
78.
a(1 4)(1 1) 2 (1 3) 8 a(5)(2) 2 ( 2) 8 40a 8 a 0.2 The function is f ( x) 0.2( x 4)( x 1) 2 ( x 3) .
80. The graph crosses the x-axis at x 4 and x 1 and touches it at x 1 . Thus, 4 and 1 each have odd multiplicities while 1 has an even multiplicity. Using one for each odd multiplicity and two for the even multiplicity, a possible funtion is f ( x) a ( x 4)( x 1)( x 1) 2 . We
a(2)(2 3) 2 (2 3) 2 50 a(2)(5) 2 (1) 2 50 50a 50 a 1 The function is f ( x) x( x 3) 2 ( x 3) 2 .
82. The graph touches the x-axis at x 0 and crosses it at x 3 and x 1 and x 2 . Thus, 3 and 1 and 2 each have odd multiplicities while 0 has an even multiplicity. Using one for each odd multiplicity and two for the even multiplicity, a possible funtion is f ( x) ax 2 ( x 3)( x 1)( x 2) . We know f ( 2) 16 . Thus, a(2) 2 (2 3)(2 1)(2 2) 16 a (2) 2 (1)(1)(4) 16 16a 16 a 1
The function is f ( x) x 2 ( x 3)( x 1)( x 2) . 83. a. 0 ( x 3) 2 ( x 2) x 3, x 2 b. The graph is shifted to the left 3 units so the xintercepts would be x 3 3 6 and x 2 3 1 84. a. 0 ( x 2)( x 4)3 x 2, x 4 b. The graph is shifted to the right 2 units so the xintercepts would be x 2 2 0 and x 4 2 6 85.
know f (3) 8 . Thus,
f ( x) 3( x 1)( x 1) ( x 3)( x 1)
2
3( x 1)( x 1)( x 3) 2 ( x 1) 2
a(3 4)(3 1)(3 1) 2 8 a( 1)(2)(4) 2 8 32a 8 a 0.25 The function is f ( x) 0.25( x 4)( x 1)( x 1) 2 .
3( x 1)( x 1)3 ( x 3) 2 The zeros are 1 with multiplicity 1, -3 with multiplicity 2 and -1 with multiplicity 3.
86. The power function is found by multiplying the leading terms from all the factors. 3
81. The graph crosses the x-axis at x 0 and touches it at x 3 and x 3 . Thus, 3 and 3 each have even multiplicities while 0 has an odd multiplicity. Using one for each odd multiplicity and two for the even multiplicity, a possible funtion is f ( x) ax( x 3) 2 ( x 3) 2 . We know
1 g ( x) 4 x 2 (4 5 x) 2 (2 x 3) x 1 2 So multiply 1 1 4 x 2 (5 x) 2 (2 x)( x)3 (4 x 2 )(25 x 2 )(2 x)( x3 ) 2 8 8 25 x
421 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
The power function is y 25 x8 . 87. The graph of a polynomial function will always have a y-intercept since the domain of every polynomial function is the set of real numbers. Therefore f 0 will always produce a y-
coordinate on the graph. A polynomial function might have no x-intercepts. For example, f ( x) x 2 1 has no x-intercepts since the equation x 2 1 0 has no real solutions. The degree will be even because the ends of the graph go in the same direction. b. The leading coefficient is positive because both ends go up. c. The function appears to be symmetric about the y-axis. Therefore, it is an even function. d. The graph touches the x-axis at x 0 . Therefore, x n must be a factor, where n is even and n 2 . e. There are six zeros with odd multiplicity and one with even multiplicity. Therefore, the minimum degree is 6(1) 1(2) 8 .
88. a.
f.
Answers will vary.
89. f ( x) x3 bx 2 cx d a.
True since every polynomial function has exactly one y-intercept, in this case (0, d ) .
b. True, a third degree polynomial will have at most 3 x-intercepts since the equation x3 bx 2 cx d 0 will have at most 3 real solutions. c. True, a third degree polynomial will have at least one x-intercept since the equation x3 bx 2 cx d 0 will have at least one real solution. d. True, since f has degree 3 and the leading coefficient 1. e. False, since f ( x) x b x c x d 3
2
x3 bx 2 cx d
90.
1 is smooth but not continuous; x g ( x) x is continuous but not smooth. f ( x)
91. Answers will vary , f ( x) ( x 2)( x 1) 2 and g ( x) ( x 2)3 ( x 1) 2 are two such polynomials.
92. Answers will vary, one such polynomial is f ( x) x 2 ( x 1)(4 x)( x 2) 2 93. Answers will vary. 94. Answers will vary. One possibility: 1 3 f ( x) 5( x 1)3 ( x 2) x x 2 5 95. We need to put the equation in standard form. 5x 2 y 6 2 y 5 x 6 5 y x3 2
Since we are looking for a perpendicular line, the 2 new slope must be m . 5 2 y y1 ( x x1 ) 5 2 y 3 ( x 2) 5 2 4 y3 x 5 5 2 11 y x 5 5 96. The denominator cannot be zero so the domain is: x | x 5 97. x
(8) (8) 2 4(4)( 3) 2(4)
8 64 48 8 112 8 8
8 4 7 2 7 8 2
f ( x). (unless b d 0)
f.
True only if d 0 , otherwise the statement is false.
b b 2 4ac 2a
422 Copyright © 2025 Pearson Education, Inc.
Section 5.1: Polynomial Functions
So the zeros are: 98.
2 7 2 7 , 2 2
102.
5x 3 7
1 1 ( x h) 2 ( x 2 2) 3 3 h 1 1 ( x h) 2 ( x 2) 3 3 h 1 1 1 x h2 x2 3 3 3 h 1 h 1 3 3 h
f ( x h) f ( x ) h
5 x 3 7 or 5 x 3 7 5 x 4
5 x 10
4 x 5
x2
4 So the solution set is , 2 5 99. Since the function is linear and the slope, -3, is negative the function is decreasing. 100. y ( x 2) 2 5
103.
x (2) 3 2 x2 6 x 8
101. Graph:
y4 5 2 y 4 10 y 14
The endpoint is (8, 14)
4x 7 We see the graph is increasing on: , 0 1,
2
3
104. x 1 4 x 7 x 2 0 x 5
4 x3
4x 2
7x 4x 5 7 x 2
+7 4x 2
The quotient is 4 x 7 and the remainder is 4x 2 .
423 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
Section 5.2 1.
6.
2, 0 and 0,10
f ( x) x( x 2) 2
Step 1: Degree is 3. The function resembles y x3 for large values of x .
x-intercepts: let y 0 and solve for x 0 5 x 10 10 5 x x 2
Step 2: y-intercept: f (0) 0(0 2) 2 0
x-intercepts: solve f ( x) 0 0 x( x 2) 2
y-intercepts: let x 0 and solve for y y 5(0) 10 y 10
x 0, 2
Step 3: Real zeros: 2 with multiplicity two, 0 with multiplicity one. The graph touches the x-axis at x 2 and crosses the x-axis at x 0 . Step 4: 3 1 2
2. 7 x3 3. Local maximum value 6.48 at x 0.67; Local minimum value 3 at x 2 .
Step 5:
f ( 3) 3; f ( 1) 1; f (1) 9 :
2
4. 0.337 x 2.311x 0.216 5.
f ( x) x 2 ( x 3)
Step 1: Degree is 3. The function resembles y x3 for large values of x . Step 2: y-intercept: f (0) 02 (0 3) 0 x-intercepts: solve f ( x) 0 0 x 2 ( x 3) x 0, x 3
Step 3: Real zeros: 0 with multiplicity two, 3 with multiplicity one. The graph touches the x-axis at x 0 and crosses the x-axis at x 3 . Step 4: 3 1 2 Step 5:
f ( 1) 4; f (2) 4; f (4) 16
7.
f ( x) ( x 4) 2 (1 x)
Step 1: Degree is 3. The function resembles y x3 for large values of x . Step 2: y-intercept: f (0) (0 4) 2 (1 0) 16 x-intercepts: solve f ( x) 0 0 ( x 4) 2 (1 x) x 4, 1
Step 3: Real zeros: 4 with multiplicity two, 1 with multiplicity one. The graph touches the x-axis at x 4 and crosses the x-axis at x 1 Step 4: 3 1 2 Step 5:
f ( 5) 6; f ( 2) 12; f (2) 36
424 Copyright © 2025 Pearson Education, Inc.
Section 5.2: Graphing Polynomial Functions; Models
Step 5:
8.
f ( 3) 250; f ( 1) 54; f (3) 10
f ( x) ( x 1)( x 3) 2
Step 1: Degree is 3. The function resembles y x3 for large values of x .
10.
Step 2: y-intercept: f (0) (0 1)(0 3)2 9
x-intercepts: solve f ( x) 0 0 ( x 1)( x 3) 2
1 Step 2: y-intercept: f (0) (0 4)(0 1)3 2 2 x-intercepts: solve f ( x) 0
x 1, 3
Step 3: Real zeros: 3 with multiplicity two, 1 with multiplicity one. The graph touches the x-axis at x 3 and crosses it at x 1 .
1 0 ( x 4)( x 1)3 2 x 4, 1
Step 3: Real zeros: 4 with multiplicity one, 1 with multiplicity three. The graph crosses the x-axis at x 4 and x 1 .
Step 4: 3 1 2 Step 5:
1 f ( x) ( x 4)( x 1)3 2 Step 1: Degree is 4. The function resembles 1 y x 4 for large value of x . 2
f ( 4) 5; f ( 1) 8; f (2) 25
Step 4: 4 1 3 Step 5:
9.
f ( 5) 108; f ( 3) 32; f (3) 28
f ( x) 2( x 2)( x 2)3 Step 1: Degree is 4. The function resembles y 2 x 4 for large values of x .
Step 2: y-intercept: f (0) 2(0 2)(0 2)3 32 x-intercepts: solve f ( x) 0 0 2( x 2)( x 2)3 x 2, 2
Step 3: Real zeros: 2 with multiplicity one, 2 with multiplicity three. The graph crosses the x-axis at x 2 and x 2 . Step 4: 4 1 3
11.
f ( x) ( x 1) x 2 ( x 4)
Step 1: Degree is 3. The function resembles y x3 for large values of x . Step 2: y-intercept: f (0) (0 1) 0 2 (0 4)
425 Copyright © 2025 Pearson Education, Inc.
8
Chapter 5: Polynomial and Rational Functions
x-intercepts: solve f ( x) 0 0 ( x 1) x 2 ( x 4)
13.
x 1, 2, 4
Step 1: Degree is 3. The function resembles y x3 for large values of x .
Step 3: Real zeros: 4 with multiplicity one, 1 with multiplicity one, 2 with multiplicity one. The graph crosses the x-axis at x 4, 1, 2 .
Step 2: y-intercept: f (0) (0) 1 0 (2 0) 0 x-intercepts: solve f ( x) 0
Step 4: 3 1 2 Step 5:
f ( x) x 1 x (2 x)
0 x 1 x (2 x)
f ( 5) 28; f ( 2) 8; f (1) 10;
x 0,1, 2
f (3) 28
Step 3: Real zeros: 0 with multiplicity one, 1 with multiplicity one, 2 with multiplicity one. The graph crosses the x-axis at x 0,1, 2 . Step 4: 3 1 2 Step 5: Graphing by hand;
12.
f ( x) x 1 ( x 4)( x 3)
Step 1: Degree is 3. The function resembles y x3 for large values of x Step 2: y-intercept: f (0) 0 1 (0 4)(0 3) 12 x-intercepts: solve f ( x) 0 0 x 1 ( x 4)( x 3) x 1, 4, 3
Step 3: Real zeros: 4 with multiplicity one, 1 with multiplicity one, 3 with multiplicity one. The graph crosses the x-axis at x 4, 1, 3 .
f ( 5) 48; f ( 2) 30; f (2) 6; f (4) 24
f ( x) (3 x) 2 x ( x 1)
Step 1: Degree is 3. The function resembles y x3 for large values of x . Step 2: y-intercept: f (0) (3 0) 2 0 (0 1) 6 x-intercepts: solve f ( x) 0 0 (3 x) 2 x ( x 1)
x 3, 2, 1
Step 4: 3 1 2 Step 5:
14.
Step 3: Real zeros: 3 with multiplicity one, 2 with multiplicity one, 1 with multiplicity one. The graph crosses the x-axis at x 3, 2, 1 . Step 4: 3 1 2 Step 5: Graphing by hand;
426 Copyright © 2025 Pearson Education, Inc.
Section 5.2: Graphing Polynomial Functions; Models
Step 5:
15.
17.
f ( x) ( x 1) 2 ( x 2) 2
f ( 3) 49; f (2) 64; f (5) 49
f ( x) 2( x 1)2 ( x 2 16)
Step 1: Degree is 4. The graph of the function resembles y 2 x 4 for large values of
Step 1: Degree is 4. The graph of the function resembles y x 4 for large values of x .
x .
Step 2: y-intercept: f (0) (0 1) 2 (0 2) 2 4 x-intercepts: solve f ( x) 0
Step 2: y-intercept: f (0) 2(0 1) 2 (02 16) 32 x-intercept: solve f ( x) 0
( x 1) 2 ( x 2) 2 0 x 1 or x 2
2( x 1) 2 ( x 2 16) 0
Step 3: Real zeros: 1 with multiplicity two, 2 with multiplicity two. The graph touches the x-axis at x 1 and x 2 .
x 1, 4, 4 Step 3: Real zeros: 1 with multiplicity two, -4 with multiplicity one and 4 with multiplicity one. The graph touches the xaxis at x 1 and crosses the x-axis at x 4 and x 4 .
Step 4: 4 1 3 Step 5: f ( 2) 16; f (1) 4; f (3) 16
Step 4: 4 1 3 Step 5: Graphing by hand:
16.
f ( x) ( x 4) 2 ( x 2) 2
Step 1: Degree is 4. The graph of the function resembles y x 4 for large values of x . Step 2: y-intercept: f (0) (0 4) 2 (0 2) 2 64 x-intercept: solve f ( x) 0 ( x 4) 2 ( x 2) 2 0
x 4 or x 2 Step 3: Real zeros: 2 with multiplicity two, 4 with multiplicity two. The graph touches the x-axis at x 2 and x 4 .
18.
f ( x) x 1 ( x 3) 3
Step 1: Degree is 4. The graph of the function resembles y x 4 for large values of x . Step 2: y-intercept: f (0) 0 1 0 3 3
Step 4: 4 1 3 427 Copyright © 2025 Pearson Education, Inc.
3
x-intercept: solve f ( x) 0 ( x 1)3 ( x 3) 0 x 1 or x 3
Chapter 5: Polynomial and Rational Functions
Step 3: Real zeros: 1 with multiplicity three, 3 with multiplicity one. The graph crosses the x-axis at x = –1 and x = 3.
Step 2: y-intercept: f (0) (0 2) 2 (0 2)(0 4) 32 x-intercept: solve f ( x) 0
Step 4: 4 1 3
( x 2) 2 ( x 2)( x 4) 0
x 2 or x 2 or x 4 Step 3: Real zeros: 2 with multiplicity two, 2 with multiplicity one, 4 with multiplicity one. The graph touches the x-axis at x 2 , and crosses it at x 2 and x 4 . Step 4: 4 1 3 Step 5: f ( 5) 147; f ( 3) 25; f ( 1) 27;
Step 5: Graphing by hand;
f (3) 35
19.
f ( x) 5 x x 2 4 ( x 3)
Step 1: Degree is 4. The graph of the function resembles y 5 x 4 for large values of x . Step 2: y-intercept:
f (0) 5(0) 02 4 0 3 0
x-intercept: solve f ( x) 0 2
5 x( x 4)( x 3) 0
x 0 or x 2, 2 or x 3
Step 3: Real zeros: 3 with multiplicity one, 2 with multiplicity one, 0 with multiplicity one and 2 with multiplicity one. The graph crosses the x-axis at x = –3 , x = - 2, x = 0 and x = 2. Step 4: 4 1 3 Step 5: f ( 4) 240; f ( 2.5) 14.1; f ( 1) 30; f (1) 60; f (3) 450
21.
f ( x) x 2 ( x 2)( x 2 3)
Step 1: Degree is 5. The graph of the function resembles y x5 for large values of x . Step 2: y-intercept: f (0) 02 (0 2)(02 3) 0 x-intercept: solve f ( x) 0 x 2 ( x 2)( x 2 3) 0 x 0 or x 2 Note: x 2 3 0 has no real solution.
Step 3: Real zeros: 0 with multiplicity two, 2 with multiplicity one. The graph touches the x-axis at x 0 and crosses it x 2 . Step 4: 5 1 4 Step 5: f ( 1) 12; f (1) 4; f (3) 108
20.
f ( x) ( x 2) 2 ( x 2)( x 4)
Step 1: Degree is 4. The graph of the function resembles y x 4 for large values of x . 428 Copyright © 2025 Pearson Education, Inc.
Section 5.2: Graphing Polynomial Functions; Models
22.
Step 4:
f ( x) x 2 ( x 2 1)( x 4)
Step 1: Degree is 5. The graph of the function resembles y x5 for large values of x . Step 2: y-intercept: f (0) 02 (02 1)(0 4) 0 x-intercept: Solve f ( x) 0 2
Step 5: 2 turning points; local maximum: (–0.80, 0.57); local minimum: (0.66, –0.99)
2
x ( x 1)( x 4) 0 x 0 or x 4 Note: x 2 1 0 has no real solution.
Step 6: Graphing by hand
Step 3: Real zeros: 0 with multiplicity two, 4 with multiplicity one. The graph touches the x-axis at x 0 and crosses it at x 4 . Step 4: 5 1 4 Step 5: f ( 5) 650; f ( 3) 90; f ( 2) 40; f (1) 10
Step 7: Domain: , ; Range: , Step 8: Increasing on , 0.80 and 0.66, ; decreasing on 0.80, 0.66 24.
23.
f ( x) x3 0.2 x 2 1.5876 x 0.31752
Step 1: Degree = 3; The graph of the function resembles y x3 for large values of x .
f ( x) x3 0.8 x 2 4.6656 x 3.73248
Step 1: Degree = 3; The graph of the function resembles y x3 for large values of x . Step 2: Graphing utility
Step 2: Graphing utility
Step 3: x-intercepts: –2.16, 0.8, 2.16; y-intercept: 3.73248 Step 4: Step 3: x-intercepts: –1.26, –0.20, 1.26; y-intercept: –0.31752 Step 5: 2 turning points; local maximum: (–1.01, 6.60); local minimum: (1.54, –1.70)
429 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
Step 7: Domain: , ; Range: ,
Step 6: Graphing by hand
Step 8: Increasing on , 2.21 and 0.50, ; decreasing on 2.21, 0.50 . 26.
f ( x) x3 2.91x 2 7.668 x 3.8151
Step 1: Degree = 3; The graph of the function resembles y x3 for large values of x . Step 7: Domain: , ; Range: ,
Step 2: Graphing utility
Step 8: Increasing on , 1.01 and 1.54, ; decreasing on 1.01,1.54 25.
f ( x) x3 2.56 x 2 3.31x 0.89
Step 1: Degree = 3; The graph of the function resembles y x3 for large values of x . Step 2: Graphing utility
Step 3: x-intercepts: –0.9, 4.71; y-intercept: –3.8151 Step 4:
Step 3: x-intercepts: –3.56, 0.50; y-intercept: 0.89 Step 4:
Step 5: 2 turning points; local maximum: (–0.9, 0); local minimum: (2.84, –26.16) Step 6: Graphing by hand
Step 5: 2 turning points; local maximum: (–2.21, 9.91); local minimum: (0.50, 0) Step 6: Graphing by hand
Step 7: Domain: , ; Range: , Step 8: Increasing on , 0.9 and 2.84, ; decreasing on 0.9, 2.84 .
430 Copyright © 2025 Pearson Education, Inc.
Section 5.2: Graphing Polynomial Functions; Models
27.
Step 2: Graphing utility
f ( x) x 4 2.5 x 2 0.5625
Step 1: Degree = 4; The graph of the function resembles y x 4 for large values of x . Step 2: Graphing utility
Step 3: x-intercepts: –3.90, –1.82, 1.82, 3.90; y-intercept: 50.2619 Step 4: Step 3: x-intercepts: –1.5, –0.5, 0.5,1.5; y-intercept: 0.5625 Step 4: Step 5: 3 turning points: local maximum: (0, 50.26); local minima: (–3.04, –35.30), (3.04, –35.30) Step 5: 3 turning points: local maximum: (0, 0.5625); local minima: (–1.12, –1), (1.12, –1)
Step 6: Graphing by hand
Step 6: Graphing by hand
Step 7: Domain: , ; Range: 35.30, Step 8: Increasing on 3.04, 0 and 3.04, ; Step 7: Domain: , ; Range: 1, Step 8: Increasing on 1.12, 0 and 1.12, ; decreasing on , 1.12 and 0,1.12 28.
f ( x) x 4 18.5 x 2 50.2619
Step 1: Degree = 4; The graph of the function resembles y x 4 for large values of x .
431 Copyright © 2025 Pearson Education, Inc.
decreasing on , 3.04 and 0, 3.04
Chapter 5: Polynomial and Rational Functions
29.
f x 2 x 4 x3 5 x 4
30.
Step 1: Degree = 4; The graph of the function resembles y 2 x 4 for large values of
f x 1.2 x 4 0.5 x 2 3x 2
Step 1: Degree = 4; The graph of the function resembles y 1.2 x 4 for large values of x .
x .
Step 2: Graphing utility
Step 2: Graphing utility:
Step 3: x-intercepts: –1.07, 1.62; y-intercept: –4 Step 4:
Step 5: 1 turning point; local minimum: (–0.42, –4.64) Step 6: Graphing by hand
Step 7: Domain: , ; Range: 4.64, Step 8: Increasing on 0.42, ; decreasing on , 0.42
Step 3: x-intercepts: –1.47, 0.91; y-intercept: 2 Step 4:
Step 5: 1 turning point: local maximum: (–0.81, 3.21) Step 6: Graphing by hand
Step 7: Domain: , ; Range: , 3.21 Step 8: Increasing on , 0.81 ; decreasing on 0.81,
432 Copyright © 2025 Pearson Education, Inc.
Section 5.2: Graphing Polynomial Functions; Models
31.
f ( x) 4 x x3 x( x 2 4) x( x 2)( x 2)
Step 1: Degree is 3. The function resembles y x 3 for large values of x . Step 2: y-intercept: f (0) 4(0) 03 0 x-intercepts: Solve f ( x) 0 0 x( x 2)( x 2) x 0, 2, 2 Step 3: Real zeros: 0 with multiplicity one, 2 with multiplicity one, 2 with multiplicity one. The graph crosses the x-axis at x 0, x 2, and x 2. Step 4: 3 1 2 Step 5:
f (3) 15; f (1) 3; f (1) 3; f (3) 15
33.
f ( x) x3 x 2 12 x x( x 2 x 12) x( x 4)( x 3)
Step 1: Degree is 3. The function resembles y x3 for large values of x . Step 2: y-intercept: f (0) 03 02 12(0) 0 x-intercepts: Solve f ( x) 0 0 x( x 4)( x 3) x 0, 4, 3
32.
f ( x) x x3 x( x 2 1) x( x 1)( x 1)
Step 1: Degree is 3. The function resembles y x3 for large values of x .
Step 3: Real zeros: 0 with multiplicity one, 4 with multiplicity one, 3 with multiplicity one. The graph crosses the x-axis at x 0, x 4, and x 3. Step 4: 3 1 2 Step 5:
Step 2: y-intercept: f (0) 0 03 0 x-intercepts: Solve f ( x) 0 0 x( x 1)( x 1) x 0, 1, 1 Step 3: Real zeros: 0 with multiplicity one, 1 with multiplicity one, 1 with multiplicity one. The graph crosses the x-axis at x 0, x 1, and x 1 . Step 4: 3 1 2 Step 5:
f (2) 6; f ( 12 ) 83 ; f ( 12 ) 83 ; f (2) 6
433 Copyright © 2025 Pearson Education, Inc.
f (5) 40; f (2) 20; f (2) 12; f (4) 32
Chapter 5: Polynomial and Rational Functions
34.
Step 4: 3 1 2 Step 5: f (7) 630; f (4) 192; f (1) 30 f (1) 42; f (3) 270
f ( x) x3 2 x 2 8 x x( x 2 2 x 8) x( x 4)( x 2)
Step 1: Degree is 3. The function resembles y x3 for large values of x . Step 2: y-intercept: f (0) 03 2(0) 2 8(0) 0 x-intercepts: Solve f ( x) 0 0 x( x 4)( x 2) x 0, 4, 2 Step 3: Real zeros: 0 with multiplicity one, 4 with multiplicity one, 2 with multiplicity one. The graph crosses the x-axis at x 0, x 4, and x 2.
f ( x) 4 x3 10 x 2 4 x 10 2(2 x3 5 x 2 2 x 5 2 x 2 (2 x 5) 1(2 x 5)
Step 4: 3 1 2 Step 5:
36.
f (5) 35; f (2) 16; f (1) 5; f (3) 21
2(2 x 5)( x 2 1) 2(2 x 5)( x 1)( x 1)
Step 1: Degree is 3. The function resembles y 4 x3 for large values of x . Step 2: y-intercept: f (0) 4(0)3 10(0) 2 4(0) 10 10
35.
f ( x) 2 x 4 12 x3 8 x 2 48 x 2 x( x3 6 x 2 4 x 24) 2 x x 2 ( x 6) 4( x 6) 2 x( x 6)( x 2 4) 2 x( x 6)( x 2)( x 2)
Step 1: Degree is 4. The function resembles y 2 x 4 for large values of x .
x-intercepts: solve f ( x) 0 0 2(2 x 5)( x 1)( x 1) 5 x , 1, 1 2 5 Step 3: Real zeros: with multiplicity one, 2 1 with multiplicity one, 1 with multiplicity one. The graph crosses the 5 x-axis at x , x 1, and x 1. 2 Step 4: 3 1 2 Step 5: f (3) 16; f (2) 6; f ( 12 ) 9; f ( 23 ) 20
Step 2: y-intercept: f (0) 2(0) 4 12(0)3 8(0)2 48(0) 0 x-intercepts: Solve f ( x) 0 0 2 x( x 6)( x 2)( x 2) x 0, 6, 2, 2 Step 3: Real zeros: 0 with multiplicity one, 6 with multiplicity one, 2 with multiplicity one, 2 with multiplicity one. The graph crosses the x-axis at x 0, x 6, x 2, and x 2. 434 Copyright © 2025 Pearson Education, Inc.
Section 5.2: Graphing Polynomial Functions; Models
37.
Step 3: Real zeros: 0 with multiplicity two, 2 with multiplicity one, 2 with multiplicity one, 5 with multiplicity one. The graph touches the x-axis at x 0 and crosses it at x 2, x 2, and x 5 .
f ( x) x5 x 4 x3 x 2 x 2 ( x3 x 2 x 1) x 2 x 2 ( x 1) 1( x 1) x 2 ( x 1)( x 2 1) x 2 ( x 1)( x 1)( x 1)
Step 4: 5 1 4 Step 5: Graphing by hand:
x ( x 1) ( x 1) 2
2
Step 1: Degree is 5. The graph of the function resembles y x5 for large values of x . Step 2: y-intercept: f (0) (0)5 (0) 4 (0)3 (0) 2 0 x-intercept: Solve f ( x) 0 x 2 ( x 1) 2 ( x 1) 0 x 0, 1, 1
Step 3: Real zeros: 0 with multiplicity two, 1 with multiplicity two, 1 with multiplicity one. The graph touches the x-axis at x 0 and x 1 , and crosses it at x 1 . Step 4: 5 1 4 Step 5: f (1.5) 1.40626; f (0.54) 0.10; f (0.74) 0.43; f (1.2) 1.39392
38.
f ( x) x5 5 x 4 4 x3 20 x 2 x 2 ( x3 5 x 2 4 x 20) x 2 x 2 ( x 5) 4( x 5)
39.
f ( x) 15 x 5 80 x 4 80 x3 5 x3 (3x 2 16 x 16) 5 x3 (3x 4)( x 4) Step 1: The graph of the function resembles y 15 x5 for large values of x .
Step 2: y-intercept: f (0) 15(0)5 80(0) 4 80(0)3 0 x-intercept: solve f ( x) 0 5 x3 (3 x 4)( x 4) 0 4 x 0, , 4 3 Step 3: Real zeros: 0 with multiplicity three, 4 with multiplicity one, -4 with 3 multiplicity one. The graph crosses the 4 x-axis at x 0, x , and x 4 . 3 Step 4: 5 1 4 Step 5: f (4.2) 637.157; f (3) 675;
x 2 ( x 5)( x 2 4) x 2 ( x 5)( x 2)( x 2)
Step 1: Degree is 5. The graph of the function resembles y x 5 for large values of x . Step 2: y-intercept: f (0) (0)5 5(0) 4 4(0)3 20(0) 2 0 x-intercept: solve f ( x) 0 x 2 ( x 2)( x 2)( x 5) 0 x 0, 2, 2, 5
435 Copyright © 2025 Pearson Education, Inc.
f (1) 15; f (1) 175
Chapter 5: Polynomial and Rational Functions
40.
Step 5:
f ( x) 3x 6 6 x 5 12 x 4 24 x3 3
f (5.5) 9.975; f (3) 22.4; f (2) 8.4; f (4.5) 0.475; f (6) 4.4
3x ( x 2 x 4 x 8) 3
2
3x 3 ( x 2)( x 2) 2
Step 1: The graph of the function resembles y 3x 6 for large values of x . Step 2: y-intercept: f (0) 3(0) 6 6(0)5 12(0)4 24(0)3 0 x-intercept: solve f ( x) 0 3x3 ( x 2)( x 2) 2 0 x 0, 2, 2 Step 3: Real zeros: 0 with multiplicity three, 2 with multiplicity one, -2 with multiplicity two. The graph crosses the x-axis at x 0, and x 2 and touches at x 2 . Step 4: 6 1 5 Step 5: f (3) 405; f (1) 9; f (1) 27; f (2.5) 474.609
41.
1 3 4 2 x x 5 x 20 5 5 1 ( x 5)( x 4)( x 5) 5 Step 1: The graph of the function resembles 1 y x5 for large values of x . 5 Step 2: y-intercept: 1 4 f (0) (0)3 (0) 2 5(0) 20 20 5 5 x-intercept: solve f ( x) 0 1 ( x 5)( x 4)( x 5) 0 5 x 5, 4,5 f ( x)
42.
3 3 15 2 x x 6 x 15 2 4 3 (2 x 5)( x 2)( x 2) 2 Step 1: The graph of the function resembles 3 y x5 for large values of x . 2 Step 2: y-intercept: 3 15 f (0) (0)3 (0) 2 6(0) 15 15 2 4 x-intercept: solve f ( x) 0 f ( x)
1 (2 x 5)( x 2)( x 2) 0 5 5 x , 2, 2 2 5 Step 3: Real zeros: with multiplicity one, 2 2 with multiplicity one, -2 with multiplicity one. The graph crosses the 5 x-axis at x , x 4 and x 5 . 2 Step 4: 3 1 2 Step 5: f (2.5) 16.875; f (1) 15.75; f (1) 6.75; f (2.25) 0.398; f (3) 3.75
Step 3: Real zeros: -5 with multiplicity one, 4 with multiplicity one, 5 with multiplicity one. The graph crosses the x-axis at x 5, x 4 and x 5 . Step 4: 3 1 2 436 Copyright © 2025 Pearson Education, Inc.
Section 5.2: Graphing Polynomial Functions; Models 43. a.
Graphing, we see that the graph may be a cubic relation.
44. a,c.
c. b. P (t ) 0.0158t 3 0.4122t 2 2.9305t 8.7022 P (13) 0.0158(15)3 0.4122(15) 2 2.9305(15) 8.7022 13.2%
b. The cubic function of best fit is H ( x) 0.2096 x3 3.4167 x 2 16.5404 x 1.6667
45. a.
c.
Graphing, we see that the graph may be a cubic relation.
For the decade 1961-1970, we have x 5 . H (5) 0.2096(5)3 3.4167(5)2 16.5404(5) 1.6667 25 The model predicts that about 25 major hurricanes struck the Atlantic Basin between 1961 and 1970.
b.
T T (12) T (9) 43.1 41.0 2.1 0.7 x 12 9 12 9 3 The average rate of change in temperature from 9 a.m. to noon was 0.7F per hour.
c.
d.
T T (21) T (15) 45 48.9 3.9 0.65 x 21 15 21 15 6
The average rate of change in temperature from 9 a.m. to noon was 1F per hour. d.
The cubic function of best fit is T ( x) 0.0079 x3 0.2930 x 2 2.6481x 47.6857
e.
For the decade 2011 to 2020 we have x 10 . H (10) 0.2096(11)3 3.4167(11)2 16.5404(11) 1.6667 49 The model predicts that approximately 49 major hurricanes will strike the Atlantic Basin between 2021 and 2030. The prediction does not seem to be reasonable. It appears to be too high.
At 5 p.m. we have x 17 . T 17 0.0079 17 0.2930 17 3
2
2.648117 47.6857 48.32 The predicted temperature at 5 p.m. is 48.32F .
437 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions e.
c. f.
The y-intercept is approximately 47.7F . The model predicts that the midnight temperature was 47.7F .
46. a. T (r ) 500(1 r )(1 r ) 500(1 r ) 500 Deposit 3 Account value of deposit 1
Account value of deposit 2
500(1 2r r 2 ) 500(1 r ) 500 500 1000r 500r 2 500 500r 500 500r 2 1500r 1500
b.
F (0.05) 500(.05)3 2000(.05) 2 3000(.05) 2000 2155.06 The value of the account at the beginning of the fourth year will be $2155.06.
47. a.
d. The values of the polynomial function get closer to the values of f. The approximations near 0 are better than those near 1 or 1. 48. a.
P3 (0.6) (0.6) 2 (1 2(1 0.6)) 0.648
b.
P5 (0.6) (0.6)3 (1 3(1 0.6) 6(1 0.6) 2 ) 0.68256
c.
b.
438 Copyright © 2025 Pearson Education, Inc.
Section 5.2: Graphing Polynomial Functions; Models
the y-axis gives f ( x 4) 0 on (, b 4)
d. Using MAXIMUM on the graph we find that the maximum is: E (0.7236) 0.10733 . The maximum edge is 0.10733 when the probability of winning a set is 0.7236.
f. The graph of f is decreasing on , c .
e.
E (0.5) 0 is the edge if both players have the same chance of winning a set.
50. 2 3x 1 4 10
f.
E (1) 0 is the edge if the better player has a 100% chance of winning a set.
3x 1 3
Since b and c are positive, the zeros are 0, b, and -c. The x-intercepts are 0, b, and -c. The y-intercept is 0. The leading term will be y ax x 2 x 2 ax5 . The value of a is positive, so the leading coefficient will be negative. Since the degree is odd, this means as x the graph will increase without bound ( as x f ( x) ) and as x the graph will decrease without bound ( f ( x) as x ). The multiplicity of c and 0 is 2 (even), so the graph will touch the x-axis at -c and 0. The multiplicity of b is 1 (odd), so the graph will cross the x-axis at b. There are at most 4 turning points, but the actual turning points cannot be determined exactly.
49.
2 3x 1 6 3 x 1 3 or 3x 1 3
51. y 52.
1 x 3
f ( x) 2 x 2 7 x 3 x
b 7 7 2a 4 4 2
25 7 7 7 f 2 7 3 8 4 4 4 7 25 The vertex is , 4 8
f k c 1.47 104 k 150 4.9 104 k 0.000005 S
2.45 105 S 0.000005 140 3 8.75 10
b. There is a local maximum on the intervals (c, 0) and (0, b)
e. f ( x 4) is a transformation of f that is shifted right 4 units and then reflected about the y-axis. Starting with (b, ) , going right 4 units gives (b 4, ) and then reflecting about
x
2 4 , , 3 3
53.
d. f ( x) 0 on the interval (b, ) [where the graph is below the x-axis]
3x 4
2 x 3
4 3 2 4 The solution set is x | x or x or 3 3
a.
c. Both -c and 0 yield a local minimum value of 0.
3 x 2
3
54.
17 1 1 1 f 2 7 1 4 2 2 2
55. x 4 0 x4 The domain is 4, . 56.
f (1) f (2) 2 (7) 9 3 1 (2) 3 3
439 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
3 units.
x 2 4 x y 2 2 y 11
57.
x 2 4 x 4 y 2 2 y 1 11 4 1
5
( x 2) 2 ( x 1) 2 16 The center is (2,1) and the radius is 4.
58. g ( x)
x 3 x 3 x 3 ( x ) ( x ) ( x ) x x 3 x
y
3
3
5
3
x g ( x) x3 x
5
x
5
The function g is even.
5. False
59. The amount of interest is $2500.
6. horizontal asymptote
I Pr t 2500 5000(0.08)t t 6.25 years
7. vertical asymptote 8. proper 9. True 10. False, a graph cannot intersect a vertical asymptote.
Section 5.3
11. y 0
1. True
12. True
2. Quotient: 3x 3 ; Remainder: 2 x 2 3 x 3 3x 3 x3 x 2 1 3x 4 0 x3 x 2 0 x 0
13. d
(3x 4 3x3
4x , the denominator, q( x) x 7 , x7 has a zero at 7. Thus, the domain of R( x) is all
15. In R ( x)
3 x)
3x 3 x 2 3 x (3x3 3x 2 3) 2 x 2 3x 3
3. y
14. a
real numbers except 7. x | x 7 5x2 , the denominator, q( x) 3 x , 3 x has a zero at –3. Thus, the domain of R ( x) is all
1 x
16. In R ( x)
real numbers except –3. x | x 3 17. In H ( x)
4 x2 , the denominator, ( x 2)( x 4)
q( x) ( x 2)( x 4) , has zeros at 2 and –4. Thus, the domain of H ( x) is all real numbers
except –4 and 2. x | x 4, x 2 4. Using the graph of y x 2 , stretch vertically by a factor of 2, then shift left 1 unit, then shift down
18. In G ( x)
6 , the denominator, ( x 3)(4 x)
q( x) ( x 3)(4 x) , has zeros at –3 and 4.
440 Copyright © 2025 Pearson Education, Inc.
Section 5.3: Properties of Rational Functions
Thus, the domain of G ( x) is all real numbers except –3 and 4. x | x 3, x 4 3 x( x 1) , the denominator, 2 x 2 5 x 12 q( x) 2 x 2 5 x 12 (2 x 3)( x 4) , has zeros at
19. In F ( x)
3 and 4 . Thus, the domain of F ( x) is all real 2 3 3 numbers except and 4 . x | x , x 4 2 2
x(1 x) , the denominator, 3x 2 5 x 2 q( x) 3 x 2 5 x 2 (3x 1)( x 2) , has zeros at
20. In Q( x)
1 and – 2 . Thus, the domain of Q( x) is all real 3 1 1 numbers except –2 and . x | x 2, x 3 3
25. In R ( x)
3( x 2 x 6) , the denominator, 5( x 2 4)
q( x) 5( x 2 4) 5( x 2)( x 2) , has zeros at 2 and –2. Thus, the domain of R( x) is all real numbers except –3 and 3. x | x 2, x 2 26. In F ( x)
2( x 2 4) , the denominator, 3( x 2 4 x 4)
q( x) 3( x 2 4 x 4) 3( x 2) 2 , has a zero at –2. Thus, the domain of F ( x) is all real
numbers except –2. x | x 2 27. a.
Domain: x x 2 ; Range: y y 1
b. Intercept: (0, 0) c.
Horizontal Asymptote: y 1
d. Vertical Asymptote: x 2 e.
Oblique Asymptote: none
x 21. In R ( x) 3 , the denominator, x 64 q( x) x 3 4 ( x 4)( x 2 4 x 16) , has a zero
28. a.
at 2 ( x 2 4 x 4 has no real zeros). Thus, the domain of R( x) is all real numbers except 4.
b. Intercept: (0, 2) c.
x | x 4
d. Vertical Asymptote: x 1
x , the denominator, 4 x 1 q( x) x 4 1 ( x 1)( x 1)( x 2 1) , has zeros at
22. In R ( x)
–1 and 1 ( x 2 1 has no real zeros). Thus, the domain of R( x) is all real numbers except –1
e. 29. a.
Domain: x x 1 ; Range: y y 0 Horizontal Asymptote: y 0
Oblique Asymptote: none Domain: x x 0 ; Range: all real numbers
b.
Intercepts: (–1, 0) and (1, 0)
and 1. x | x 1, x 1
c.
Horizontal Asymptote: none
d.
Vertical Asymptote: x 0
3x x , the denominator, x2 9 q( x) x 2 9 , has no real zeros. Thus, the domain of H ( x) is all real numbers.
e.
Oblique Asymptote: y 2 x
23. In H ( x)
2
x3 , the denominator, x4 1 q( x) x 4 1 , has no real zeros. Thus, the domain of G ( x) is all real numbers.
24. In G ( x)
30. a.
Domain: x x 0 ; Range: y y 2 or y 2
b. Intercepts: none c.
Horizontal Asymptote: none
d. Vertical Asymptote: x 0 e.
Oblique Asymptote: y x
441 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
31. a.
Domain: x x 2, x 2 ; Range: y y 0 or y 1
c.
b.
Intercept: (0, 0)
c.
Horizontal Asymptote: y 1
d.
Vertical Asymptotes: x 2, x 2
e.
Oblique Asymptote: none
32. a.
Range: y | y 3 Vertical asymptote: x 0 Horizontal asymptote: y 3
35. a. R x
1
x 1
2
; Using the function, y
1 , x2
shift the graph horizontally 1 unit to the right.
Domain: x x 1, x 1 ; Range: all real numbers
b. Intercept: (0, 0) c.
Horizontal Asymptote: y 0
d. Vertical Asymptotes: x 1, x 1 e.
Oblique Asymptote: none
1 1 ; Using the function, y , x x shift the graph vertically 2 units up.
33. a. F x 2
b. Domain: x | x 1
Range: y | y 0 c.
Vertical asymptote: x 1 Horizontal asymptote: y 0
1 3 1 3 ; Using the function y , x x x stretch the graph vertically by a factor of 3.
36. a. Q( x)
b. Domain: x | x 0
Range: y | y 2 c.
Vertical asymptote: x 0 Horizontal asymptote: y 2
1 1 ; Using the function y 2 , 2 x x shift the graph vertically 3 units up.
34. a. Q( x) 3
b. Domain: x | x 0
Range: y | y 0 c.
Vertical asymptote: x 0 Horizontal asymptote: y 0
37. a. H ( x)
2 1 2 ; Using the function x 1 x 1
1 , shift the graph horizontally 1 unit to the x left, reflect about the x-axis, and stretch y
b. Domain: x | x 0 442 Copyright © 2025 Pearson Education, Inc.
Section 5.3: Properties of Rational Functions
vertically by a factor of 2.
units to the left, and reflect about the x-axis.
b. Domain: x | x 1
b. Domain: x | x 2
Range: y | y 0 c.
Range: y | y 0
Vertical asymptote: x 1 Horizontal asymptote: y 0
c.
Vertical asymptote: x 2 Horizontal asymptote: y 0
1 1 1 ; Using the function y , x x 1 shift the graph horizontally 1 unit to the right, and shift vertically 1 unit up.
40. a. R( x) 38. a. G ( x)
1 2 ; Using the 2 2 2 ( x 2) ( x 2)
1 , shift the graph horizontally 2 x2 units to the left, and stretch vertically by a factor of 2. function y
b. Domain: x | x 1
Range: y | y 1 c. b. Domain: x | x 2
Range: y | y 0 c.
Vertical asymptote: x 2 Horizontal asymptote: y 0
39. a. R ( x)
1 1 ; Using the 2 x2 4 x 4 x 2
function y
1 , shift the graph horizontally 2 x2
Vertical asymptote: x 1 Horizontal asymptote: y 1
41. a. G ( x) 1
2 2 1 ; 2 ( x 3) ( x 3) 2
1 2 1 ( x 3) 2
1 , shift the graph right 3 x2 units, stretch vertically by a factor of 2, and shift
Using the function y
443 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
vertically 1 unit up.
vertically 1 unit up.
b. Domain: x | x 3
b. Domain: x | x 0
Range: y | y 1 c.
Vertical asymptote: x 3 Horizontal asymptote: y 1
1 1 2 ; Using the x 1 x 1 1 function y , shift the graph left 1 unit, reflect x about the x-axis, and shift vertically up 2 units.
42. a. F ( x) 2
Range: y | y 1 c.
Vertical asymptote: x 0 Horizontal asymptote: y 1
x4 4 1 1 4 1 ; Using the x x x 1 function y , reflect about the x-axis, stretch x vertically by a factor of 4, and shift vertically 1 unit up.
44. a. R ( x)
b. Domain: x | x 1
Range: y | y 2 c.
Vertical asymptote: x 1 Horizontal asymptote: y 2
x2 4 4 1 1 2 4 2 1 ; Using 2 x x x 1 the function y 2 , reflect about the x-axis, x stretch vertically by a factor of 4, and shift
43. a. R ( x)
b. Domain: x | x 0
Range: y | y 1 c.
Vertical asymptote: x 0 Horizontal asymptote: y 1
3x ; The degree of the numerator, x4 p( x) 3x, is n 1 . The degree of the denominator, q( x) x 4, is m 1 . Since
45. R( x)
3 3 is a horizontal 1 asymptote. The denominator is zero at x 4 , so x 4 is a vertical asymptote.
n m , the line y
444 Copyright © 2025 Pearson Education, Inc.
Section 5.3: Properties of Rational Functions 3x 5 ; The degree of the numerator, x6 p( x) 3 x 5, is n 1 . The degree of the denominator, q( x) x 6, is m 1 . Since
46. R ( x)
3 3 is a horizontal 1 asymptote. The denominator is zero at x 6 , so x 6 is a vertical asymptote.
n m , the line y
47. H ( x)
x3 8 x2 5x 6
x 2 x2 2 x 4 x 2 x 3
x2 2x 4 , where x 2,3 x 3 The degree of the numerator in lowest terms is n 2 . The degree of the denominator in lowest terms is m 1 . Since n m 1 , there is an oblique asymptote. Dividing: x5 x 3 x2 2 x 4
x 2 3x 5x 4 19 19 H ( x) x 5 , x 2,3 x 3 Thus, the oblique asymptote is y x 5 . The denominator in lowest terms is zero at x 3 so x 3 is a vertical asymptote. x3 1 x 2 5 x 14
x3 ; The degree of the numerator, x 1 p( x) x3 , is n 3 . The degree of the
49. T ( x)
4
denominator, q( x) x 4 1 is m 4 . Since n m , the line y 0 is a horizontal asymptote. The denominator is zero at x 1 and x 1 , so x 1 and x 1 are vertical asymptotes. 4x2 ; The degree of the numerator, x3 1 p( x) 4 x 2 , is n 2 . The degree of the
50. P ( x)
denominator, q( x) x3 1 is m 3 . Since n m , the line y 0 is a horizontal asymptote. The denominator is zero at x 1 , so x 1 is a vertical asymptote. 2 x 2 5 x 12 (2 x 3)( x 4) 3 x 2 11x 4 (3 x 1)( x 4) 2x 3 1 , where x , 4 3x 1 3 The degree of the numerator in lowest terms is n 1 . The degree of the denominator in lowest 2 terms is m 1 . Since n m , the line y is a 3 horizontal asymptote. The denominator in 1 1 lowest terms is zero at x , so x is a 3 3 vertical asymptote.
51. Q( x)
5 x 15
48. G ( x)
39 x 71 , x 2, 7 x 2 5 x 14 Thus, the oblique asymptote is y x 5 . The denominator is zero at x 2 and x 7 , so x 2 and x 7 are vertical asymptotes. G ( x) x 5
x 1 3
x 2 x 7
The degree of the numerator, p( x) x3 1, is n 3 . The degree of the denominator, q( x) x 2 5 x 14, is m 2 . Since n m 1 , there is an oblique asymptote. Dividing: x5 2 x 5 x 14 x3 1
x3 5 x 2 14 x
5 x 14 x 1 5 x 2 25 x 70 2
39 x 71
x2 6 x 5 ( x 5)( x 1) 2 2 x 7 x 5 (2 x 5)( x 1) x5 5 , where x , 1 2x 5 2 The degree of the numerator in lowest terms is n 1 . The degree of the denominator in lowest 1 terms is m 1 . Since n m , the line y is a 2 horizontal asymptote. The denominator in 5 5 lowest terms is zero at x , so x is a 2 2 vertical asymptote.
52. F ( x)
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Chapter 5: Polynomial and Rational Functions
6 x 2 19 x 7 (3 x 1)(2 x 7) 3x 1 3x 1 1 2 x 7, where x 3 The degree of the numerator in lowest terms is n 1 . Since R in lowest terms is linear there is 1 no oblique asymptote. Since is not in the 3 1 domain of R, but is not a real zero of the 3 denominator of R in lowest terms, there is no vertical asymptote to the graph of R.
53. R ( x)
8 x 2 26 x 7 (4 x 1)(2 x 7) 4x 1 4x 1 1 2 x 7, where x 4 The degree of the numerator in lowest terms is n 1 . The degree of the denominator in lowest terms is m 0 . Since R in lowest terms is linear there is no oblique asymptote. The denominator of R(x) in lowest terms is 1, so there is no vertical asymptote.
54. R ( x)
55. G ( x)
x 4 1 ( x 2 1)( x 2 1) x2 x x( x 1)
denominator in lowest terms is zero at x 0 , so x 0 is a vertical asymptote. 57. g h
x 16 ( x 4)( x 4) 56. F ( x ) 2 x( x 2) x 2x
2
( x 4)( x 2)( x 2) x( x 2) 2
( x 2 4)( x 2) , where x 0, 2 x The degree of the numerator in lowest terms is n 3 . The degree of the denominator in lowest terms is m 1 . Since n m 2 , there is no horizontal asymptote or oblique asymptote. The
2
3.99 1014
b.
g 443
6.374 10 0 6
2
9.8208 m/s 2
3.99 1014
6.374 10 443 6
2
9.8195 m/s 2
c.
g 8848
3.99 1014
6.374 10 8848 6
2
9.7936 m/s 2
d.
g h
3.99 1014
6.374 10 h 6
2
3.99 1014 0 as h h2 Thus, the h-axis is the horizontal asymptote.
g h
3.99 1014
6.374 10 h 6
2
0 , to solve this
equation would require that 3.99 1014 0 , which is impossible. Therefore, there is no height above sea level at which g 0. In other words, there is no point in the entire universe that is unaffected by the Earth’s gravity!
2
6
g 0
( x 2 1)( x 1)( x 1) x( x 1)
4
6.374 10 h
a.
e.
( x 2 1)( x 1) , where x 0, 1 x The degree of the numerator in lowest terms is n 3 . The degree of the denominator in lowest terms is m 1 . Since n m 2 , there is no horizontal asymptote or oblique asymptote. The denominator in lowest terms is zero at x 0 , so x 0 is a vertical asymptote.
3.99 1014
58. P t a.
50 1 0.5t 2 0.01t
P 0
50 1 0 20
50 25 insects 2
b. 5 years = 60 months; 50 1 0.5 60 1550 P 60 2 0.01 60 2.6 596 insects
c.
P t
50 1 0.5t
50 0.5t
2500 2 0.01t 0.01t as t Thus, y 2500 is the horizontal asymptote. The area can sustain a maximum population of 2500 insects.
446 Copyright © 2025 Pearson Education, Inc.
Section 5.3: Properties of Rational Functions
59. a.
Rtot
10 R2 10 R2
p (4) 4.097560976 p '(4) p(4.097560976) x2 4.097560976 p '(4.097560976) 4.094906 x1 4
p(4.094906) 4.094904 p '(4.094906) Since x1 and x2 are the same to 4 decimal places, the zero is approximately x 4.0949 . x3 4.094906
b. Horizontal asymptote: y Rtot 10 As the value of R2 increases without bound, the total resistance approaches 10 ohms, the resistance of R1 . c.
Rtot 17
61. a. Dividing: 2 x 1 2x 3 2 x 2
R1 R2 R1 R2
5
R1 2 R1 R( x) 2
R1 2 R1
Solving graphically, let Y1 17 and
x 2 x .
Y2 2 x x
5 1 5 2 x 1 x 1
b.
We would need R1 103.5 ohms. c. 60. a.
p (3) (3)3 7(3) 40 34 p(5) (5)3 7(5) 40 50
b.
c.
p is continuous and p(3) 0 p(5) , so there must be at least one real zero in the interval 3,5 . 2
p '( x) 3 x 7 . Start with x0 4 .
The only real zero of the denominator is 1, so the line x 1 is a vertical asymptote. As 2x 3 2x x , 2 so the line y 2 x 1 x is a horizontal asymptote.
62. a. Dividing: 3 2 x 7 6 x 16 6 x 21 -5 R ( x) 3
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5 2x 7
Chapter 5: Polynomial and Rational Functions b.
denominator. However, if the numerator has a higher degree, there is no horizontal asymptote. 66. Answers will vary. If x 4 is a vertical asymptote, then x 4 is a zero of the denominator. If x 4 is a zero of a polynomial, then ( x 4) is a factor of the polynomial. Therefore, if x 4 is a vertical asymptote of a rational function, then ( x 4) must be a factor of the denominator.
c. The only real zero of the denominator is
7 , so 2
7 is a vertical asymptote. As 2 6 x 16 6 x x , 3 so the line 2x 7 2x y 3 is a horizontal asymptote.
the line x
67. The equation of a vertical line through the point (5, 3) is x 5 . 68.
63. Answers will vary. We want a rational function n x where n and d such that r x 2 x 1 d x
are polynomial functions and the degree of n x is less than the degree of d x . We could let n x 1 and d x x 1 . Then our function is
69. 2 x3 xy 2 4 Test x-axis symmetry: Let y y 2 x3 x y 4 2
1 r x 2x 1 . Getting a common x 1 denominator yields 2 x 1 x 1 1 r x x 1 x 1 2 x2 x 2 x 1 1 x 1 2 2 x 3x 2 x 1
2 x3 xy 2 4 same
Test y-axis symmetry: Let x x
2 x ( x) y 2 4 3
2 x3 xy 2 4 not the same Test origin symmetry: Let x x and y y
2 x ( x) y 4 3
2
Therefore, the graph will have x-axis symmetry.
x 1
65. A rational function cannot have both a horizontal and oblique asymptote. To have an oblique asymptote, the degree of the numerator must be exactly one larger than the degree of the
2
2 x3 xy 2 4 not the same
Therefore, one possibility is r x 2 x 3x 2 . 64. Answers will vary. With rational functions, the only way to get a non-zero horizontal asymptote is if the degree of the numerator equals the degree of the denominator. In such cases, the horizontal asymptote is the ratio of the leading coefficients.
2 x (3x 7) 1 2 5 4 6 14 x x 1 2 5 5 4 6 14 x x 2 1 5 4 5 19 1 x 20 5 1 20 4 x 5 19 19
70.
f ( x) 3x 2; g ( x) x 2 2 x 4 3 x 2 x 2 2 x 4 0 x2 x 6 0 ( x 2)( x 3) x 2 or x 3 f (2) 3(2) 2 4
f ( 3) 3( 3) 2 11 So the intersection points are: (2, 4), ( 3,11)
448 Copyright © 2025 Pearson Education, Inc.
Section 5.3: Properties of Rational Functions 71. x 0 06 6 y 3 02 2 y0 x6 0 x2 0 x6 x6
5 x 2 13 x 6 0 (5 x 2)( x 3) 0 2 and x 3 5 2 f ( x) 0 at ,3 5
x
The x-intercept is (6, 0). The y-intercept is (0, 3)
74.
f ( x)
3
x 2
( x) 6
3 x x2 6
3
x f ( x) x 6 The function is odd.
72. The local maximum value is f (3) 19 .
75.
3 2
x 9
2
2 3 2 x 3 x 3 x 3 x 3
3
2 x 3
x 3 x 3 x 3 x 3 3 2 x 3 x 3 x 3 73. 5 x 2 13 x 6 0 (5 x 2)( x 3) 0 2 x and x 3 5
3 2x 6
x 3 x 3
9 2x x2 9
76. 3 (2 x 4) 5 x 13 3 2 x 4 5 x 13 2 x 1 5 x 13 7 x 14 x2 The solution set is x | x 2 or , 2
449 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
Section 5.4 2. false: sometimes the rational function may only have a hole at the undefined point.
1. y-intercept: f (0)
0 1 1 1 02 4 4 4 2
3. c 4. True
x-intercepts: Set the numerator equal to 0 and solve for x.
5. a.
x 1 0 ( x 1)( x 1) 0 x 1 or x 1
x | x 2
2
b.
0 x( x 2) 2 x 0 or x 2
But x = 2 makes the function undefined so the only x-intercept is 0.
1 The intercepts are 0, , (1, 0) , and (1, 0) . 4 6. a
In problems 7–44, we will use the terminology: R ( x)
p ( x) , where the degree of p( x) n and the degree of q ( x)
q( x) m .
7. R ( x)
Step 1:
x 1 x( x 4)
p ( x) x 1; q( x) x( x 4) x 2 4 x; n 1; m 2
Domain: x x 4, x 0 Since 0 is not in the domain, there is no y-intercept.
Step 2 & 3:The function is in lowest terms. The x-intercept is the zero of p ( x) : x 1 with odd multiplicity. Plot the point 1, 0 . The graph will cross the x-axis at this point. Step 4:
R( x)
x 1 is in lowest terms. x( x 4)
The vertical asymptotes are the zeros of q( x) : x 4 and x 0 . Plot these lines using dashes. The multiplicity of 0 and -4 are odd so the graph will approach plus or minus infinity on either side of the asymptotes. Step 5:
Since n m , the line y 0 is the horizontal asymptote. Solve R x 0 to find intersection points: x 1 0 x( x 4) x 1 0 x 1 R ( x) intersects y 0 at (–1, 0). Plot the point 1, 0 and the line y 0 using dashes.
450
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Section 5.4: The Graph of a Rational Function
Step 6:
Steps 7: Graphing
8. R ( x)
Step 1:
x ( x 1)( x 2)
p ( x) x; q ( x) ( x 1)( x 2) x 2 x 2; n 1; m 2
Domain: x x 2, x 1 The y-intercept; R (0) 0
Step 2 & 3:The function is in lowest terms. The x-intercept is the zero of p ( x) . 0 with odd multiplicity. Plot the point 0, 0 . The graph will cross the x-axis at this point. Step 4:
R( x)
x is in lowest terms. ( x 1)( x 2)
The vertical asymptotes are the zeros of q( x) : x 2 and x 1 . Graph these asymptotes using dashed lines. The multiplicity of -2 and 1 are both odd so the graph will approach plus or minus infinity on either side of the asymptotes. Step 5:
Since n m , the line y 0 is the horizontal asymptote. Solve to find intersection points: x 0 ( x 1)( x 2) x0 R ( x) intersects y 0 at (0, 0). Plot the point 0, 0 and the line y 0 using dashes.
451
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Chapter 5: Polynomial and Rational Functions
Step 6:
Steps 7: Graphing:
9. R ( x)
Step 1:
3 x 3 3( x 1) 2 x 4 2( x 2)
p ( x) 3 x 3; q ( x) 2 x 4; n 1; m 1
Domain: x x 2 The y-intercept is R (0)
Step 2 & 3: R( x)
3 0 3
2 0 4
3 3 . Plot the point 0, . 4 4
3 x 1 is in lowest terms. The x-intercept is the zero of p ( x) , x 1 with odd multiplicity. 2 x 2
Plot the point 1, 0 . The graph will cross the x-axis at this point. Step 4:
Step 5:
R( x)
3x 3 3 x 1 is in lowest terms. 2x 4 2 x 2
The vertical asymptote is the zero of q( x) : x 2 . Graph this asymptote using a dashed line. The multiplicity of -2 is odd so the graph will approach plus or minus infinity on either side of the asymptote. 3 Since n m , the line y is the horizontal asymptote. 2 Solve to find intersection points:
452
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Section 5.4: The Graph of a Rational Function 3x 3 3 2x 4 2 2 3 x 3 3 2 x 4 6x 6 6x 4 02 R ( x) does not intersect y
3 3 . Plot the line y with dashes. 2 2
Step 6:
Steps 7: Graphing:
10. R ( x)
Step 1:
2x 4 2 x 2 x 1 x 1
p( x) 2 x 4; q ( x) x 1; n 1; m 1
Domain: x x 1 The y-intercept is R (0)
2(0) 4 4 4 . Plot the point 0, 4 . 0 1 1
Step 2 & 3:R is in lowest terms. The x-intercept is the zero of p( x) : x 2 with odd multiplicity. Plot the point 2, 0 . The graph will cross the x-axis at this point. Step 4:
2x 4 2 x 2 is in lowest terms. x 1 x 1 The vertical asymptote is the zero of q( x) : x 1 . Graph this asymptote using a dashed line. The multiplicity of 1 is odd so the graph will approach plus or minus infinity on either side of the asymptote. R( x)
453
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Chapter 5: Polynomial and Rational Functions
Since n m , the line y 2 is the horizontal asymptote. Solve to find intersection points: 2x 4 2 x 1 2 x 4 2 x 1
Step 5:
2x 4 2x 1 0 5 R ( x) does not intersect y 2 . Plot the line y 2 with dashes.
Step 6:
Steps 7: Graphing:
11. R ( x)
Step 1:
3 2
x 4
3
x 2 x 2
p ( x) 3; q ( x) x 2 4; n 0; m 2
Domain: x x 2, x 2 The y-intercept is R (0)
3 2
0 4
3 3 3 . Plot the point 0, . 4 4 4
Step 2 & 3:R is in lowest terms. The x-intercepts are the zeros of p( x) . Since p x is a constant, there are no xintercepts. Step 4:
3
is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 2 and x 2 . x 4 Graph each of these asymptotes using dashed lines. The multiplicity of -2 and 2 is odd so the graph will approach plus or minus infinity on either side of the asymptotes. R( x)
2
454
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Section 5.4: The Graph of a Rational Function
Since n m , the line y 0 is the horizontal asymptote. Solve to find intersection points:
Step 5:
3
0
2
x 4
3 0 x2 4
30 R ( x) does not intersect y 0 . Plot the line y 0 with dashes.
Step 6:
Steps 7: Graphing:
12. R ( x)
Step 1:
6 2
x x6
6 ( x 3)( x 2)
p ( x) 6; q ( x) x 2 x 6; n 0; m 2
Domain: x x 2, x 3 The y-intercept is R (0)
6 2
0 06
6 1 . Plot the point 0, 1 . 6
Step 2 & 3:R is in lowest terms. The x-intercepts are the zeros of p( x) . Since p x is a constant, there are no xintercepts. Step 4:
6 is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 2 and x 3 x2 x 6 Graph each of these asymptotes using dashed lines. The multiplicity of -2 and 3 is odd so the graph will approach plus or minus infinity on either side of the asymptotes. R( x)
455
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Chapter 5: Polynomial and Rational Functions
Step 5:
Since n m , the line y 0 is the horizontal asymptote. Solve to find intersection points: 6 2
x x6
0
6 0 x2 x 6
60 R ( x) does not intersect y 0 .
Step 6:
Steps 7: Graphing:
13. P ( x)
Step 1:
x4 x2 1 2
x 1
p ( x) x 4 x 2 1; q ( x) x 2 1; n 4; m 2
Domain: x x 1, x 1 The y-intercept is P (0)
04 02 1 02 1
1 1 . Plot the point 0, 1 . 1
x4 x2 1
is in lowest terms. The x-intercept is the zero of p ( x) . Since p x is never 0, x2 1 there are no x-intercepts.
Step 2 & 3: P ( x)
Step 4:
x4 x2 1
is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 1 and x 1 . x2 1 Graph each of these asymptotes using dashed lines. The multiplicity of -1 and 1 is odd so the graph will approach plus or minus infinity on either side of the asymptotes. P( x)
456
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Section 5.4: The Graph of a Rational Function
Since n m 1 , there is no horizontal or oblique asymptote.
Step 5: Step 6:
Steps 7: Graphing:
14. Q( x)
Step 1:
x4 1 2
x 4
( x 2 1)( x 1)( x 1) ( x 2)( x 2)
p ( x ) x 4 1; q( x) x 2 4; n 4; m 2
Domain: x x 2, x 2 The y-intercept is Q(0)
04 1 2
0 4
1 1 1 . Plot the point 0, . 4 4 4
( x 2 1)( x 1)( x 1) is in lowest terms. The x-intercepts are the zeros of p( x) : –1 and 1 both ( x 2)( x 2) with odd multiplicity. Plot the points 1, 0 and 1, 0 . The graph crosses the x-axis at both points.
Step 2 & 3: Q( x)
Step 4:
Step 5:
Q( x)
x4 1 2
( x 2 1)( x 1)( x 1) is in lowest terms. ( x 2)( x 2)
x 4 The vertical asymptotes are the zeros of q( x) : x 2 and x 2 . Graph each of these asymptotes using dashed lines. The multiplicity of -2 and 2 is odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Since n m 1 , there is no horizontal asymptote and no oblique asymptote.
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Chapter 5: Polynomial and Rational Functions
Step 6:
Steps 7: Graphing:
15. H ( x)
Step 1:
x3 1 x2 9
x 1 x 2 x 1 x 3 x 3
p( x) x3 1; q( x) x 2 9; n 3; m 2
Domain: x x 3, x 3 03 1
1 1 1 . Plot the point 0, . 02 9 9 9 9 Step 2 & 3: H ( x) is in lowest terms. The x-intercept is the zero of p ( x) : 1 with odd multiplicity.
The y-intercept is H (0)
Plot the point 1, 0 . The graph will cross the x-axis at this point. Step 4:
H ( x) is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 3 and x 3 . Graph each of these asymptotes using dashed lines. The multiplicity of -3 and 3 is odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: x 9x 1 2 x 9 x3 0 x 2 0 x 1 H ( x) x 2 x 9 x3 9x 9x 1 The oblique asymptote is y x . Graph the asymptote with a dashed line. Solve to find intersection points:
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Section 5.4: The Graph of a Rational Function
x3 1
x x2 9 x3 1 x3 9 x 1 9 x x
1 9
1 1 The oblique asymptote intersects H ( x) at , . 9 9
Step 6:
Steps 7: Graphing:
16. G ( x)
Step 1:
x3 1 2
x 2x
( x 1)( x 2 x 1) x( x 2)
p ( x) x3 1; q ( x) x 2 2 x; n 3; m 2
Domain: x x 2, x 0 There is no y-intercept since G (0)
Step 2 & 3: G ( x)
03 1 2
0 2(0)
1 . 0
x3 1
is in lowest terms. The x-intercept is the zero of p( x) : –1 with odd multiplicity. x2 2x Plot the point 1, 0 . The graph will cross the x-axis at this point.
Step 4:
x3 1
is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 2 and x 0 . x2 2x Graph each of these asymptotes using dashed lines. The multiplicity of -2 and 0 is odd so the graph will approach plus or minus infinity on either side of the asymptotes. G ( x)
459
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Chapter 5: Polynomial and Rational Functions
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: x2 4x 1 2 3 x 2x x 0x2 0 x 1 G ( x) x 2 2 x 2x x3 2 x 2 2x2
1
2
2x 4x 4x 1 The oblique asymptote is y x 2 . Graph this asymptote with a dashed line. Solve to find intersection points: x3 1 x2 x2 2 x x3 1 x3 4 x 1 4x 1 x 4 1 9 The oblique asymptote intersects G ( x) at , . 4 4
Step 6:
Steps 7:
Graphing:
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Section 5.4: The Graph of a Rational Function
17. R ( x)
Step 1:
x2 x2 x 6
x2 ( x 3)( x 2)
p( x) x 2 ; q ( x) x 2 x 6; n 2; m 2
Domain: x x 3, x 2 The y-intercept is R (0)
Step 2 & 3: R ( x)
02 2
0 06
0 0 . Plot the point 0, 0 . 6
x2
is in lowest terms. The x-intercept is the zero of p ( x) : 0 with even multiplicity. x2 x 6 Plot the point 0, 0 . The graph will touch the x-axis at this point. R( x)
x2
Step 4:
is in lowest terms. The vertical asymptotes are the zeros of q( x) : x2 x 6 x 3 and x 2 . Graph each of these asymptotes using dashed lines. The multiplicity of -3 and 2 is odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote with a dashed line. Solve to find intersection points: x2 1 x2 x 6 x2 x2 x 6 0 x6 x6 R ( x) intersects y 1 at (6, 1).
Step 6:
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Chapter 5: Polynomial and Rational Functions
Steps 7: Graphing:
18. R ( x)
Step 1:
x 2 x 12 2
x 4
( x 4)( x 3) p ( x) x 2 x 12; q( x) x 2 4; n 2; m 2 ( x 2)( x 2)
Domain: x x 2, x 2 The y-intercept is R (0)
02 0 12 02 4
12 3 . Plot the point 0, 3 . 4
( x 4)( x 3) is in lowest terms. The x-intercepts are the zeros of p ( x) : –4 and 3 each with ( x 2)( x 2) odd multiplicity. Plot the points 4, 0 and 3, 0 . The graph will cross the x-axis at these point.
Step 2 & 3: R ( x)
R( x)
x 2 x 12
Step 4:
is in lowest terms. x2 4 The vertical asymptotes are the zeros of q( x) : x 2 and x 2 . Graph each of these asymptotes using a dashed line. The multiplicity of -2 and 2 is odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x 2 x 12 1 x2 4 x 2 x 12 x 2 4 x 8 R ( x) intersects y 1 at (8, 1).
462
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Section 5.4: The Graph of a Rational Function
Step 6:
Steps 7: Graphing:
19. G ( x)
Step 1:
x x2 4
x ( x 2)( x 2)
p( x) x; q ( x) x 2 4; n 1; m 2
Domain: x x 2, x 2 The y-intercept is G (0)
0 2
0 4
0 0 . Plot the point 0, 0 . 4
x is in lowest terms. The x-intercept is the zero of p( x) : 0 with odd multiplicity. x 4 Plot the point 0, 0 . The graph will cross the x-axis at this point.
Step 2 & 3: G ( x)
Step 4:
Step 5:
2
x is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 2 and x 2 . x 4 Graph each of these asymptotes using a dashed line. The multiplicity of -2 and 2 is odd so the graph will approach plus or minus infinity on either side of the asymptote. G ( x)
2
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x 0 x2 4 x0 G ( x) intersects y 0 at (0, 0).
463
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Chapter 5: Polynomial and Rational Functions
Step 6:
Steps 7: Graphing:
20. G ( x)
Step 1:
3x x2 1
3x ( x 1)( x 1)
p( x) 3 x; q( x) x 2 1; n 1; m 2
Domain: x x 1, x 1 The y-intercept is G (0)
Step 2 & 3: G ( x)
3(0) 2
0 1
0 0 . Plot the point 0, 0 . 1
3x
is in lowest terms. The x-intercept is the zero of p ( x) : 0 with odd multiplicity. x 1 Plot the point 0, 0 . The graph will cross the x-axis at this point.
Step 4:
Step 5:
2
3x is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 1 and x 1 x 1 Graph each of these asymptotes using a dashed line. The multiplicity of -1 and 1 is odd so the graph will approach plus or minus infinity on either side of the asymptotes. G ( x)
2
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: 3x 0 x2 1 3x 0 x0 G ( x) intersects y 0 at (0, 0).
464
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Section 5.4: The Graph of a Rational Function
Step 6:
Steps 7: Graphing:
21. R ( x)
Step 1:
3 ( x 1)( x 2 4)
3 ( x 1)( x 2)( x 2)
p( x) 3; q( x) ( x 1)( x 2 4); n 0; m 3
Domain: x x 2, x 1, x 2 The y-intercept is R (0)
Step 2 & 3: R ( x)
R( x)
3 ( x 1)( x 2 4)
3 2
(0 1)(0 4)
3 3 . Plot the point 0, . 4 4
is in lowest terms. There is no x-intercept.
3
Step 4:
is in lowest terms. ( x 1)( x 2 4) The vertical asymptotes are the zeros of q( x) : x 2, x 1, and x 2 . Graph each of these asymptotes using a dashed line. The multiplicity of -2, 1 and 2 is odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Step 5:
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote with a dashed line. Solve to find intersection points: 3 0 ( x 1)( x 2 4) 30 R ( x) does not intersect y 0 .
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Chapter 5: Polynomial and Rational Functions
Step 6:
Steps 7: Graphing:
22. R ( x)
Step 1:
4 2
( x 1)( x 9)
4 ( x 1)( x 3)( x 3)
p ( x) 4; q ( x) ( x 1)( x 2 9);
n 0; m 3
Domain: x x 3, x 1, x 3 The y-intercept is R (0)
Step 2 & 3: R ( x)
R( x)
4 ( x 1)( x 2 9)
4 2
(0 1)(0 9)
4 4 4 . Plot the point 0, . 9 9 9
is in lowest terms. There is no x-intercept.
4
Step 4:
is in lowest terms. ( x 1)( x 2 9) The vertical asymptotes are the zeros of q( x) : x 3, x 1, and x 3 Graph each of these asymptotes using a dashed line. The multiplicity of -3, -1 and 3 is odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Step 5:
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote with a dashed line. Solve to find intersection points: 4 0 ( x 1)( x 2 9) 4 0 R ( x) does not intersect y 0 .
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Section 5.4: The Graph of a Rational Function
Step 6:
Steps 7: Graphing:
23. H ( x)
Step 1:
x2 1 4
x 16
( x 1)( x 1) 2
( x 4)( x 2)( x 2)
p ( x) x 2 1; q ( x) x 4 16; n 2; m 4
Domain: x x 2, x 2 The y-intercept is H (0)
Step 2 & 3: H ( x)
02 1 4
0 16
1 1 1 . Plot the point 0, . 16 16 16
2
x 1
is in lowest terms. The x-intercepts are the zeros of p( x) : –1 and 1 each with odd x 4 16 multiplicity. Plot 1, 0 and 1, 0 . The graph will cross the x-axis at these points. H ( x)
x2 1
Step 4:
is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 2 and x 2 x 4 16 Graph each of these asymptotes using a dashed line. The multiplicity of -2 and 2 is odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Step 5:
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x2 1 0 x 4 16 x2 1 0 x 1 H ( x) intersects y 0 at (–1, 0) and (1, 0). 467
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Chapter 5: Polynomial and Rational Functions
Step 6:
Steps 7: Graphing:
24. H ( x)
Step 1:
x2 4 x4 1
x2 4
p ( x) x 2 4; q( x) x 4 1;
( x 2 1)( x 1)( x 1)
n 2; m 4
Domain: x x 1, x 1 The y-intercept is H (0)
Step 2 & 3: H ( x)
H ( x)
x2 4 x4 1
02 4 4
0 1
4 4 . Plot the point 0, 4 . 1
is in lowest terms. There are no x-intercepts.
x2 4
Step 4:
is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 1 and x 1 x4 1 Graph each of these asymptotes using a dashed line. The multiplicity of -1 and 1 is odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Step 5:
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x2 4 0 x4 1 x2 4 0 no real solution H ( x) does not intersect y 0 .
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Section 5.4: The Graph of a Rational Function
Step 6:
Steps 7: Graphing:
25. F ( x)
x 2 3x 4 ( x 1)( x 4) x2 x2
Step 1:
Domain: x x 2 The y-intercept is F (0)
p( x) x 2 3 x 4; q( x) x 2; n 2; m 1
02 3(0) 4 4 2 . Plot the point 0, 2 . 02 2
x 2 3x 4 is in lowest terms. The x-intercepts are the zeros of p ( x) : –1 and 4 each with odd x2 multiplicity. Plot 1, 0 and 4, 0 . The graph will cross the x-axis at these points.
Step 2 & 3: F ( x)
Step 4:
x 2 3x 4 is in lowest terms. The vertical asymptote is the zero of q( x) : x 2 x2 Graph this asymptote using a dashed line. The multiplicity of -2 is odd so the graph will approach plus or minus infinity on either side of the asymptote. F ( x)
469
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Chapter 5: Polynomial and Rational Functions
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: x5 6 2 x 2 x 3x 4 F ( x) x 5 x2 x2 2 x 5x 4 5 x 10 6 The oblique asymptote is y x 5 . Graph this asymptote using a dashed line. Solve to find intersection points: x 2 3x 4 x 5 x2 x 2 3x 4 x 2 3 x 10 4 10 The oblique asymptote does not intersect F ( x) .
Step 6:
Steps 7: Graphing:
470
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Section 5.4: The Graph of a Rational Function
26. F ( x)
Step 1:
x 2 3 x 2 ( x 2)( x 1) x 1 x 1
p ( x) x 2 3 x 2; q ( x) x 1; n 2; m 1
Domain: x x 1 The y-intercept is F (0)
02 3(0) 2 2 2 . Plot the point 0, 2 . 0 1 1
x2 3x 2 is in lowest terms. The x-intercepts are the zeros of p( x) : –2 and –1 each with x 1 odd multiplicity. Plot 2, 0 and 1, 0 . The graph will cross the x-axis at these points.
Step 2 & 3: F ( x)
Step 4:
Step 5:
x2 3x 2 is in lowest terms. The vertical asymptote is the zero of q( x) : x 1 x 1 Graph this asymptote using a dashed line. The multiplicity of 1 is odd so the graph will approach plus or minus infinity on either side of the asymptote. F ( x)
Since n m 1 , there is an oblique asymptote. Dividing: x4 6 x 1 x 2 3x 2 F ( x) x 4 1 x x2 x 4x 2 4x 4 6 The oblique asymptote is y x 4 . Graph this asymptote using a dashed line. Solve to find intersection points: x2 3x 2 x4 x 1 x2 3x 2 x 2 3x 4 2 4 The oblique asymptote does not intersect F ( x ) .
Step 6:
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Chapter 5: Polynomial and Rational Functions
Steps 7: Graphing:
27. R ( x)
Step 1:
x 2 x 12 ( x 4)( x 3) x4 x4
p( x) x 2 x 12; q( x) x 4; n 2; m 1
Domain: x x 4 The y-intercept is R (0)
02 0 12 12 3 . Plot the point 0,3 . 04 4
x 2 x 12 is in lowest terms. The x-intercepts are the zeros of p( x) : –4 and 3 each with odd x4 multiplicity . Plot 4, 0 and 3, 0 . The graph will cross the x-axis at these points.
Step 2 & 3: R ( x)
Step 4:
Step 5:
x 2 x 12 is in lowest terms. The vertical asymptote is the zero of q( x) : x 4 x4 Graph this asymptote using a dashed line. The multiplicity of 4 is odd so the graph will approach plus or minus infinity on either side of the asymptote. R( x)
Since n m 1 , there is an oblique asymptote. Dividing: x5 x 4 x 2 x 12
R( x) x 5
x2 4 x 5 x 12
8 x4
5 x 20 8
The oblique asymptote is y x 5 . Graph this asymptote using a dashed line. Solve to find intersection points: x 2 x 12 x5 x4 x 2 x 12 x 2 x 20 12 20 The oblique asymptote does not intersect R ( x) .
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Section 5.4: The Graph of a Rational Function
Step 6:
Steps 7: Graphing:
28. R ( x)
x 2 x 12 ( x 4)( x 3) x5 x5
Step 1:
Domain: x x 5 The y-intercept is R (0)
p( x) x 2 x 12; q ( x) x 5; n 2; m 1
02 0 12 12 12 . Plot the point 0, . 05 5 5
x 2 x 12 is in lowest terms. The x-intercepts are the zeros of p( x) : –3 and 4 each with odd x5 multiplicity. Plot 3, 0 and 4, 0 . The graph will cross the x-axis at these points.
Step 2 & 3: R ( x )
Step 4:
x 2 x 12 is in lowest terms. The vertical asymptote is the zero of q( x) : x 5 x5 Graph this asymptote using a dashed line. The multiplicity of -5 is odd so the graph will approach plus or minus infinity on either side of the asymptote. R( x)
473
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Chapter 5: Polynomial and Rational Functions
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: x6 x 5 x x 12 2
R( x) x 6
x2 5x
18 x5
6 x 12 6 x 30 18
The oblique asymptote is y x 6 . Graph this asymptote using a dashed line. Solve to find intersection points: x 2 x 12 x6 x5 x 2 x 12 x 2 x 30 12 30 The oblique asymptote does not intersect R ( x) . Step 6:
Steps 7: Graphing:
474
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Section 5.4: The Graph of a Rational Function
29. F ( x )
x 2 x 12 ( x 4)( x 3) x2 x2
Step 1:
Domain: x x 2 The y-intercept is F (0)
p( x) x 2 x 12; q( x) x 2; n 2; m 1
02 0 12 12 6 . Plot the point 0, 6 . 02 2
x 2 x 12 is in lowest terms. The x-intercepts are the zeros of p( x) : –4 and 3 each with odd x2 multiplicity. Plot 4, 0 and 3, 0 . The graph will cross the x-axis at these points.
Step 2 & 3: F ( x )
Step 4:
Step 5:
x 2 x 12 is in lowest terms. The vertical asymptote is the zero of q( x) : x 2 x2 Graph this asymptote using a dashed line. The multiplicity of -2 is odd so the graph will approach plus or minus infinity on either side of the asymptote. F ( x)
Since n m 1 , there is an oblique asymptote. Dividing: x 1 x 2 x 2 x 12
F ( x) x 1
x2 2 x
10 x2
x 12 x 2 10
The oblique asymptote is y x 1 . Graph this asymptote using a dashed line. Solve to find intersection points: x 2 x 12 x 1 x2 x 2 x 12 x 2 x 2 12 2 The oblique asymptote does not intersect F ( x ) . Step 6:
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Chapter 5: Polynomial and Rational Functions
Steps 7: Graphing:
30. G ( x)
Step 1:
x 2 x 12 ( x 3)( x 4) x 1 x 1
p ( x ) x 2 x 12; q( x) x 1; n 2; m 1
Domain: x x 1 The y-intercept is F (0)
02 0 12 12 12 . Plot the point 0, 12 . 0 1 1
x 2 x 12 is in lowest terms. The x-intercepts are the zeros of p( x) : –3 and 4 each with odd x 1 multiplicity. Plot 3, 0 and 4, 0 . The graph will cross the x-axis at these points.
Step 2 & 3: G ( x)
Step 4:
Step 5:
x 2 x 12 is in lowest terms. The vertical asymptote is the zero of q( x) : x 1 x 1 Graph this asymptote using a dashed line. The multiplicity of -1 is odd so the graph will approach plus or minus infinity on either side of the asymptote. G ( x)
Since n m 1 , there is an oblique asymptote. Dividing: x2 x 1 x x 12 2
x2 x
G ( x) x 2
10 x 1
2 x 12 2x 2 10 The oblique asymptote is y =x 2. Graph this asymptote using a dashed line. Solve to find intersection points: x 2 x 12 x2 x 1 x 2 x 12 x 2 x 2 12 2 The oblique asymptote does not intersect G ( x) .
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Section 5.4: The Graph of a Rational Function
Step 6:
Steps 7: Graphing:
31. R ( x)
Step 1:
x( x 1) 2 ( x 3)3
p ( x) x( x 1) 2 ; q( x) ( x 3)3 ; n 3; m 3
Domain: x x 3 The y-intercept is R (0)
Step 2 & 3: R ( x)
0(0 1) 2 (0 3)
3
0 0 . Plot the point 0, 0 . 27
x( x 1) 2
is in lowest terms. The x-intercepts are the zeros of p( x) : 0 with odd multiplicity ( x 3)3 and 1 with even multiplicity. Plot 0, 0 . The graph will cross the x-axis at this point.
Plot 1, 0 . The graph will touch the x-axis at this point. x( x 1) 2
Step 4:
is in lowest terms. The vertical asymptote is the zero of q( x) : x 3 ( x 3)3 Graph this asymptote with a dashed line. The multiplicity of -2 is odd so the graph will approach plus or minus infinity on either side of the asymptote.
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote with a dashed line. Solve to find intersection points:
R( x)
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Chapter 5: Polynomial and Rational Functions
x( x 1)2 ( x 3)3
1
x3 2 x 2 x x3 9 x 2 27 x 27 0 11x 2 26 x 27 b 2 4ac 262 4 11 27 512 no real solution R ( x) does not intersect y 1 .
Step 6:
Steps 7: Graphing:
32. R ( x)
Step 1:
( x 1)( x 2)( x 3) x ( x 4) 2
p( x) ( x 1)( x 2)( x 3); q ( x) x( x 4) 2 ;
n 3; m 3
Domain: x x 0, x 4 There is no y-intercept since R (0)
Step 2 & 3: R ( x)
(0 1)(0 2)(0 3) 0(0 4)
2
6 . 0
( x 1)( x 2)( x 3)
is in lowest terms. The x-intercepts are the zeros of p( x) : –2, 1, and 3 x( x 4) 2 each with odd multiplicity. Plot 2, 0 , 1, 0 and 3, 0 . The graph will cross the x-axis at these points.
478
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Section 5.4: The Graph of a Rational Function
R( x)
( x 1)( x 2)( x 3)
Step 4:
is in lowest terms. x( x 4) 2 The vertical asymptotes are the zeros of q( x) : x 0 and x 4 Graph each of these asymptotes with a dashed line. The multiplicity of 0 is odd so the graph will approach plus or minus infinity on either side of the asymptote. The multiplicity of 4 is even so the graph will approach the same infinity on either side of the asymptote.
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote with a dashed line. Solve to find intersection points: ( x 1)( x 2)( x 3) 1 x( x 4) 2 ( x 2 x 2)( x 3) x( x 2 8 x 16) x3 2 x 2 5 x 6 x3 8 x 2 16 x 6 x 2 21x 6 0 2 x2 7 x 2 0 x
7 49 4(2)(2) 7 33 2(2) 4
7 33 7 33 R ( x) intersects y 1 at , 1 and , 1 . 4 4
Step 6:
Steps 7: Graphing:
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Chapter 5: Polynomial and Rational Functions
33. R ( x)
Step 1:
x 2 x 12 2
x x6
( x 4)( x 3) x 4 ( x 3)( x 2) x 2
p ( x) x 2 x 12; q( x) x 2 x 6;
n 2; m 2
Domain: x x 2, x 3 The y-intercept is R (0)
02 0 12 2
0 06
12 2 . Plot the point 0, 2 . 6
x4 , x 3 . Note: R x is still undefined at both 3 and 2 . x2 The x-intercept is the zero of y x 4 : –4 with odd multiplicity.
Step 2 & 3:In lowest terms, R ( x )
Plot 4, 0 . The graph will cross the x-axis at this point. x4 , x 3 . The vertical asymptote is the zero of f x x 2 : x 2 ; x2 Graph this asymptote using a dashed line. The multiplicity of -2 is odd so the graph will approach plus or minus infinity on either side of the asymptote. Note: x 3 is not a vertical asymptote because the 7 reduced form must be used to find the asymptotes. The graph has a hole at 3, . 5
Step 4:
In lowest terms, R ( x )
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x 2 x 12 1 x2 x 6 x 2 x 12 x 2 x 6 2x 6 x3 R ( x) does not intersect y 1 because R ( x) is not defined at x 3 .
Step 6:
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Section 5.4: The Graph of a Rational Function
Steps 7: Graphing:
34. R ( x)
Step 1:
x 2 3 x 10 2
x 8 x 15
( x 5)( x 2) x 2 ( x 5)( x 3) x 3
p( x) x 2 3x 10;
q( x) x 2 8 x 15;
n 2; m 2
Domain: x x 5, x 3 The y-intercept is R (0)
Step 2 & 3:In lowest terms, R ( x )
02 3(0) 10 2
0 8(0) 15
2 2 10 . Plot the point 0, . 3 15 3
x2 , x 5 . The x-intercept is the zero of y x 2 : 2 with odd x3
multiplicity. Note: –5 is not a zero because reduced form must be used to find the zeros. Plot the point 2,0 . The graph will cross the x-axis at this point. x2 , x 5 . The vertical asymptote is the zero of f x x 3 : x 3 ; x3 Graph this asymptote using a dashed line. The multiplicity of -3 is odd so the graph will approach plus or minus infinity on either side of the asymptote. Note: x 5 is not a vertical asymptote because reduced form must be used to find the asymptotes. The graph has a hole at 5, 3.5 .
Step 4:
In lowest terms, R ( x )
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x 2 3 x 10 1 x 2 8 x 15 x 2 3 x 10 x 2 8 x 15 5 x 25 x 5 R ( x) does not intersect y 1 because R( x) is not defined at x 5 .
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Chapter 5: Polynomial and Rational Functions
Step 6:
Steps 7: Graphing:
35. R ( x)
Step 1:
6 x2 7 x 3 2
2x 7x 6
(3 x 1)(2 x 3) 3x 1 (2 x 3)( x 2) x2
p ( x) 6 x 2 7 x 3;
q( x) 2 x 2 7 x 6; n 2; m 2
3 Domain: x x , x 2 2
The y-intercept is R (0)
Step 2 & 3:In lowest terms, R ( x )
6(0) 2 7(0) 3 2
2(0) 7(0) 6
1 3 1 . Plot the point 0, . 2 6 2
3x 1 3 1 , x . The x-intercept is the zero of y 3 x 1 : with odd 3 2 x2
multiplicity. 3 Note: x is not a zero because reduced form must be used to find the zeros. 2 1 Plot the point , 0 . The graph will cross the x-axis at this point. 3 Step 4:
3x 1 3 , x . The vertical asymptote is the zero of f x x 2 : x 2 ; 2 x2 Graph this asymptote using a dashed line. The multiplicity of 2 is odd so the graph will approach plus or minus infinity on either side of the asymptote. 3 Note: x is not a vertical asymptote because reduced form must be used to find the asymptotes. 2 3 The graph has a hole at , 11 . 2
In lowest terms, R ( x )
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Section 5.4: The Graph of a Rational Function
Step 5:
Since n m , the line y 3 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: 6 x2 7 x 3 3 2 x2 7 x 6 6 x 2 7 x 3 6 x 2 21x 18 14 x 21 3 x 2 3 R ( x) does not intersect y 3 because R ( x) is not defined at x . 2
Step 6:
Steps 7: Graphing:
36. R ( x )
Step 1:
8 x 2 26 x 15 2
2 x x 15
(4 x 3)(2 x 5) 4 x 3 (2 x 5)( x 3) x3
p ( x) 8 x 2 26 x 15; q ( x) 2 x 2 x 15; n 2; m 2
5 Domain: x x , x 3 2 8 0 26 0 15 2
The y-intercept is R (0)
Step 2 & 3:In lowest terms, R( x)
2 0 0 15 2
15 1 . Plot the point 0, 1 . 15
4x 3 5 3 , x . The x-intercept is the zero of y 4 x 3 : with odd x 3 2 4
multiplicity. 483
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Chapter 5: Polynomial and Rational Functions
5 is not a zero because reduced form must be used to find the zeros. 2 3 Plot the point , 0 . The graph will cross the x-axis at this point. 4
Note:
4x 3 5 , x . The vertical asymptote is the zero of f x x 3 : x 3 ; 2 x 3 Graph this asymptote using a dashed line. The multiplicity of 3 is odd so the graph will approach plus or minus infinity on either side of the asymptote. 5 Note: x is not a vertical asymptote because reduced form must be used to find the asymptotes. 2 5 14 The graph has a hole at , . 2 11
Step 4:
In lowest terms, R ( x)
Step 5:
Since n m , the line y 4 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: 8 x 2 26 x 15 4 2 x 2 x 15 8 x 2 26 x 15 8 x 2 4 x 60 30 x 75 5 x 2 5 R ( x) does not intersect y 4 because R ( x) is not defined at x . 2
Step 6:
Steps 7: Graphing:
484
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Section 5.4: The Graph of a Rational Function
37. R ( x)
Step 1:
x 2 5 x 6 ( x 2)( x 3) x2 x3 x3
p( x) x 2 5 x 6; q( x) x 3;
n 2; m 1
Domain: x x 3 The y-intercept is R (0)
02 5(0) 6 6 2 . Plot the point 0, 2 . 03 3
Step 2 & 3:In lowest terms, R ( x ) x 2, x 3 . The x-intercept is the zero of y x 2 : –2 with odd multiplicity. Note: –3 is not a zero because reduced form must be used to find the zeros. Plot the point 0, 2 . The graph will cross the x-axis at this point. Step 4:In lowest terms, R ( x ) x 2, x 3 . There are no vertical asymptotes. Note: x 3 is not a vertical asymptote because reduced form must be used to find the asymptotes. The graph has a hole at 3, 1 . Step 5:
Since n m 1 there is a possibility of an oblique asymptote. However, R in lowest terms, y x 2 is a linear function and therefore the graph has no oblique asymptotes.
Step 6:
Steps 7: Graphing:
485
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Chapter 5: Polynomial and Rational Functions
38. R ( x )
Step 1:
x 2 x 30 ( x 6)( x 5) x5 x6 x6
p( x) x 2 x 30; q( x) x 6; n 2; m 1
Domain: x x 6 The y-intercept is R (0)
02 (0) 30 30 5 . Plot the point 0, 5 . 06 6
Step 2 & 3:In lowest terms, R ( x ) x 5, x 6 . The x-intercept is the zero of y x 5 : 5 with odd multiplicity. Note: –6 is not a zero because reduced form must be used to find the zeros. Plot the point 5, 0 . The graph will cross the x-axis at this point. Step 4:
In lowest terms, R ( x ) x 5, x 6 . There are no vertical asymptotes. Note: x 6 is not a vertical asymptote because reduced form must be used to find the asymptotes. The graph has a hole at 6, 11 .
Step 5:
Since n m 1 there is a possibility of an oblique asymptote. However, R in lowest terms, y x 5 is a linear function and therefore the graph has no oblique asymptotes.
Step 6:
Steps 7: Graphing:
486
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Section 5.4: The Graph of a Rational Function
39. H ( x)
Step 1:
3x 6 4 x2
3x 6 ( x 2 4)
3( x 2) ( x 2)( x 2)
Domain: x x 2, x 2 The y-intercept is H (0)
Step 2 & 3: H ( x)
Step 4:
Step 5:
p ( x ) 3x 6; q( x) 4 x 2 ; n 1; m 2
H ( x)
3x 6 4 x
2
3x 6 2
3(0) 6 40
2
3 3 6 . Plot the point 0, . 4 2 2
3 is in lowest terms. The x-intercept is the zero of p( x) : none x2
3 is in lowest terms so R ( x) x 2 . The vertical asymptotes are the zeros of x2
4 x R ( x) : x 2 . Graph the asymptote using a dashed line. The multiplicity of -2 is odd so the graph will approach plus or minus infinity on either side of the asymptote.
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: 3x 6 0 4 x2 3x 6 0 x2 The function is not defined at x 2 so there is no interection.
Step 6:
Steps 7: Graphing:
487
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Chapter 5: Polynomial and Rational Functions
40. H ( x)
Step 1:
2 2x x2 1
2( x 1) ( x 1)( x 1)
p ( x) 2 2 x; q ( x ) x 2 1; n 1; m 2
Domain: x x 1, x 1 The y-intercept is H (0)
2 2(0) 2
0 1
2 2 . Plot the point 0, 2 . 1
2( x 1) is in lowest terms. The possible x-intercept is the zero of p ( x) : -1 but H ( 1) is ( x 1)( x 1) not defined.
Step 2 & 3: H ( x)
Step 4:
2 is in lowest terms so R ( x ) x 1 . The vertical asymptotes are the zeros of R( x) : x 1 x 1 . There is a hole at 1, 1
H ( x)
Graph the asymptote using a dashed line. The multiplicity of -1 is odd so the graph will approach plus or minus infinity on either side of the asymptote. Step 5:
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: 2 2x 0 x2 1 2 2x 0 x 1 H ( x) is not defined at x 1 so there is no intersection.
Step 6:
Steps 7: Graphing:
488
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Section 5.4: The Graph of a Rational Function
41. F ( x )
Step 1:
x2 5x 4 2
x 2x 1
( x 1)( x 4) ( x 1)
2
x4 x 1
p ( x) x 2 5 x 4;
q( x) x 2 2 x 1; n 2; m 2
Domain: x x 1 The y-intercept is R (0)
02 5(0) 4 2
0 2(0) 1
4 4 . Plot the point 0, 4 . 1
x4 , x 1 . The x-intercept is the zero of y x 4 : 4 with odd multiplicity. x 1 Note: –5 is not a zero because reduced form must be used to find the zeros. Plot the point 4,0 . The graph crosses the x-axis at this point.
Step 2 & 3:In lowest terms, F ( x )
x4 , x 1 . The vertical asymptote is the zero of f x x 1 : x 1 ; Graph x 1 this asymptote using a dashed line. The multiplicity of 1 is odd so the graph will approach plus or minus infinity on either side of the asymptote.
Step 4:
In lowest terms, F ( x )
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x2 5x 4 1 x2 2 x 1 x2 5x 4 x2 2 x 1 3x 3 x 1 F ( x ) does not intersect y 1 because R ( x) is not defined at x 1 .
Step 6:
Steps 7: Graphing:
489
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Chapter 5: Polynomial and Rational Functions
42. F ( x )
Step 1:
x 2 2 x 15 2
x 6x 9
( x 5)( x 3) x 5 ( x 3)( x 3) x 3
p ( x) x 2 2 x 15;
q( x) x 2 6 x 9;
n 2; m 2
Domain: x x 3 The y-intercept is F (0)
Step 2 & 3:In lowest terms, F ( x)
02 2(0) 15 2
0 6(0) 9
5 5 15 . Plot the point 0, . 3 9 3
x5 , x 3 . The x-intercept is the zero of y x 5 : 5 with odd x3
multiplicity. Note: –3 is not a zero because reduced form must be used to find the zeros. Plot the point 5,0 . The graph crosses the x-axis at this point. x5 , x 3 . The vertical asymptote is the zero of f x x 3 : x 3 ; x3 Graph this asymptote using a dashed line. The multiplicity of -3 is odd so the graph will approach plus or minus infinity on either side of the asymptote.
Step 4:
In lowest terms, F ( x)
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x 2 2 x 15 1 x2 6 x 9 x 2 2 x 15 x 2 6 x 9 8 x 24 x 3 R ( x) does not intersect y 1 because F ( x ) is not defined at x 3 .
Step 6:
Steps 7: Graphing:
490
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Section 5.4: The Graph of a Rational Function
43. G ( x)
x ( x 2) 2
p( x) x; q ( x) ( x 2) 2 ; n 1; m 2
Step 1:
Domain: x x 2 .
Step 2:
G ( x)
Step 3:
The y-intercept is G (0)
x ( x 2) 2
is in lowest terms. (0) (0 2)
2
0 0 . Plot the point 0, 0 . 4
The x-intercept is the zero of p( x) : 0 with odd multiplicity. Plot the point 0, 0 . The graph crosses the x-axis at this point. Step 4:
Step 5:
x is in lowest terms. The vertical asymptote is the zero of q( x) : x 2 . Graph this ( x 2) 2 asymptote. The multiplicity of -2 is even so the graph will approach the same infinity on both sides of the asymptote. G ( x)
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x 0 ( x 2) 2 x0 G ( x) intersects y 0 at (0, 0) .
Step 6:
Steps 7: Graphing:
491
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Chapter 5: Polynomial and Rational Functions
44. G ( x)
2 x ( x 1) 2
p ( x ) 2 x; q ( x) ( x 1) 2 ; n 1; m 2
Step 1:
Domain: x x 1 .
Step 2:
G ( x)
Step 3:
The y-intercept is G (0)
2 x ( x 1) 2
is in lowest terms. 2 (0) (0 1)
2
2 2 . Plot the point 0, 2 . 1
The x-intercept is the zero of p( x) : 2 with odd multiplicity. Plot the point 2, 0 . The graph crosses the x-axis at this point. Step 4:
Step 5:
2 x is in lowest terms. The vertical asymptote is the zero of q( x) : x 1 . Graph this ( x 1) 2 asymptote. The multiplicity of 1 is even so the graph will approach the same infinity on both sides of the asymptote. G ( x)
Since n m , the line y 0 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: 2 x 0 ( x 1) 2 2 x 0 x2 G ( x) intersects y 0 at (2, 0) .
Step 6:
Steps 7: Graphing:
492
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Section 5.4: The Graph of a Rational Function
45.
f ( x) x
1 x2 1 x x
Step 1:
Domain: x x 0
p( x) x 2 1; q ( x) x; n 2; m 1
There is no y-intercept because 0 is not in the domain. Step 2 & 3: f ( x)
Step 4:
Step 5:
x2 1 is in lowest terms. There are no x-intercepts since x 2 1 0 has no real solutions. x
x2 1 is in lowest terms. The vertical asymptote is the zero of q( x) : x 0 Graph this x asymptote using a dashed line. The multiplicity of 0 is odd so the graph will approach plus or minus infinity on either side of the asymptote. f ( x)
Since n m 1 , there is an oblique asymptote. x 1 f ( x) x Dividing: x x2 1 x 2 The oblique asymptote is y =x. x 1 Graph this asymptote using a dashed line. Solve to find intersection points: x2 1 x x x2 1 x2 1 0 The oblique asymptote does not intersect f ( x ) .
Step 6:
493
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Chapter 5: Polynomial and Rational Functions
Steps 7: Graphing :
46.
f ( x) 2 x
Step 1:
9 2x2 9 x x
p( x) 2 x 2 9; q ( x) x; n 2; m 1
Domain: x x 0 There is no y-intercept because 0 is not in the domain.
Step 2 & 3: f ( x)
Step 4:
Step 5:
2 x2 9 is in lowest terms. There are no x-intercepts since 2 x 2 9 0 has no real solutions. x
2 x2 9 is in lowest terms. The vertical asymptote is the zero of q( x) : x 0 x Graph this asymptote using a dashed line. The multiplicity of 0 is odd so the graph will approach plus or minus infinity on either side of the asymptote. f ( x)
Since n m 1 , there is an oblique asymptote. Dividing: 2x x 2x2 9 2x2
f ( x) 2 x
9 x
9 The oblique asymptote is y 2 x . Graph this asymptote using a dashed line. Solve to find intersection points: 2 x2 9 2x x 2 x2 9 2 x2 90 The oblique asymptote does not intersect f ( x ) .
494
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Section 5.4: The Graph of a Rational Function
Step 6:
Steps 7: Graphing:
47.
p( x) x 1; q( x) x; n 3; m 1 .
1 x3 1 x 1 x x 1 f ( x) x x x x 2
Step 1:
2
3
Domain: x x 0 There is no y-intercept because 0 is not in the domain. x3 1 is in lowest terms. The x-intercept is the zero of p ( x) : –1 with odd multiplicity. x Plot the point 1, 0 . The graph crosses the x-axis at this point.
Step 2 & 3: f ( x)
Step 4:
Step 5:
x3 1 is in lowest terms. The vertical asymptote is the zero of q( x) : x 0 x Graph this asymptote using a dashed line. The multiplicity of 0 is odd so the graph will approach plus or minus infinity on either side of the asymptote. f ( x)
Since n m 1 , there is no horizontal or oblique asymptote.
Step 6:
495
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Chapter 5: Polynomial and Rational Functions
Steps 7: Graphing:
48.
f ( x) 2 x 2
Step 1:
2 x 2 x 2x 4 16 2 x3 16 2 x 8 x x x x 3
2
p( x) 2 x 16; q( x) x; n 3; m 1 3
Domain: x x 0 There is no y-intercept because 0 is not in the domain. 2 x3 16 is in lowest terms. The x-intercept is the zero of p( x) : 2 with odd multiplicity. x Plot the point. 2, 0 . The graph crosses the x-axis at this point.
Step 2 & 3: f ( x)
Step 4:
Step 5:
2 x3 16 is in lowest terms. The vertical asymptote is the zero of q( x) : x 0 x Graph this asymptote using a dashed line. The multiplicity of 0 is odd so the graph will approach plus or minus infinity on either side of the asymptote. f ( x)
Since n m 1 , there is no horizontal or oblique asymptote.
Step 6:
Interval
, 2
2, 0
0,
Number Chosen
3
1
1
Value of f f 3 16 f 1 14 f 1 18 Location of Graph Above x-axis Below x-axis Above x-axis Point on Graph
3,16
1, 14
496
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1,18
Section 5.4: The Graph of a Rational Function
Steps 7: Graphing:
49.
f x x
Step 1:
1
x4 1
x
3
3
x
p x x 4 1; q x x3 ; n 4; m 3
Domain: x x 0 There is no y-intercept because 0 is not in the domain.
Step 2 & 3: f x
x4 1
f x
x4 1
x3
is in lowest terms. There are no x-intercepts since x 4 1 0 has no real solutions.
Step 4:
is in lowest terms. The vertical asymptote is the zero of q( x) : x 0 x3 Graph this asymptote using a dashed line. The multiplicity of 0 is odd so the graph will approach plus or minus infinity on either side of the asymptote.
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: x x3 x 4 1 x4
f ( x) x
1 x3
1 The oblique asymptote is y x. Graph this asymptote using a dashed line. Solve to find intersection points: x4 1 x x3 x4 1 x4 1 0 The oblique asymptote does not intersect f ( x ) .
497
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Chapter 5: Polynomial and Rational Functions
Step 6:
Steps 7: Graphing:
50.
f ( x) 2 x
Step 1:
9
2 x4 9
x
x3
3
p ( x) 2 x 4 9; q ( x) x3 ; n 4; m 3
Domain: x x 0 There is no y-intercept because 0 is not in the domain.
Step 2 & 3: f ( x)
2 x4 9 x3
is in lowest terms. There are no x-intercepts since 2 x 4 9 0 has no real solutions.
2 x4 9
Step 4:
is in lowest terms. The vertical asymptote is the zero of q( x) : x 0 x3 Graph this asymptote using a dashed line. The multiplicity of 0 is odd so the graph will approach plus or minus infinity on either side of the asymptote.
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: 2x 9 3 f ( x) 2 x 3 x 2 x4 9 x 2 x4 9 The oblique asymptote is y 2 x . Graph this asymptote using a dashed line. Solve to find intersection points:
f ( x)
498
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Section 5.4: The Graph of a Rational Function
2 x4 9
2x x3 2 x4 9 2 x4
90 The oblique asymptote does not intersect f ( x ) .
Step 6:
Steps 7: Graphing:
51. One possibility: R ( x)
2
x 4
52. One possibility: R ( x)
53. One possibility: R ( x )
55.
x2
x 2
x 1
x 1 x 3 x 2 a ( x 1) 2 ( x 2) 2
(Using the point 0,1 leads to a 4 / 3 .) Thus, R( x)
x 1 x 3 x 2 43 ( x 1) 2 ( x 2) 2
54. One possibility: R ( x )
.
The likelihood of your ball not being chosen increases very quickly and approaches 1 as the number of attendees, x, increases.
3( x 2)( x 1) 2 ( x 3)( x 4) 2
499
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Chapter 5: Polynomial and Rational Functions
56. Begin with the graph of f ( x)
b. Graphing:
1 . Changing to x
1 flips the graph about the x-axis. x 1 Changing to f ( x) shifts the graph to x6 the right by 6 units. f ( x)
c. 59. a.
57. a.
The degree of the numerator is 1 and the degree of the denominator is 2. Thus, the horizontal asymptote is C 0 . The concentration of the drug decreases to 0 as time increases.
b. Graphing:
Using MAXIMUM, the concentration is highest after t 5 minutes. The cost of the project is the sum of the cost for the parallel side, the two other sides, and the posts. A xy 1000 xy 1000 y x If the length of a perpendicular side is x feet, 1000 the length of the parallel side is y x feet. Thus, 1000 C x 2 8 x 5 4 25 x 5000 16 x 100 x
b. The domain is x 0 . Note that x is a length so it cannot be negative. In addition, if x 0 , there is no rectangle (that is, the area is 0 square feet). c.
58. a.
c.
Using MAXIMUM, the concentration is highest after t 0.71 hours.
The degree of the numerator is 1 and the degree of the denominator is 2. Thus, the horizontal asymptote is C 0 . The concentration of the drug decreases to 0 as time increases.
C x 16 x
5000 100 x
d. Using MINIMUM, the dimensions of cheapest cost are about 17.7 feet by 56.6 feet (longer side parallel to river).
500
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Section 5.4: The Graph of a Rational Function
2 53 feet and 3 3 1000 3000 y feet . 53 / 3 53
Note: x 17
60. a.
b.
Thus, y
10, 000
, so x2 10, 000 2 S ( x) 4 x 2x x2 40, 000 2 x2 x 3 2 x 40, 000 x
772.4 45 f ' vs 600 772.4 v s 727.4 600 772.4 v s
b. Graphing:
727.4 620 600 772.4 v s 436, 440 620 772.4 vs
620 772.4 vs 436, 440 436, 440 620 436, 440 vs 772.4 68.5 620 If f ' 620 Hz , the speed of the ambulance is roughly 68.5 miles per hour.
c.
772.4 vs
c.
y
d. The surface area is a minimum when x 21.54 inches. 10, 000 y 21.54 inches 21.544 2
436, 440 772.4 x
The dimensions of the box are: 21.54 in. by 21.54 in. by 21.54 in. e.
62. a.
Answers will vary. One possibility is to save costs or reduce weight by minimizing the material needed to construct the box. The surface area is the sum of the areas of the five sides. S xy xy xy xy x 2 4 xy x 2 The volume is x x y x 2 y 5000.
436, 440 and Y2 620 , then find 772.4 x the intersection point.
d. Let Y1
5000 , so x2 5000 S ( x) 4 x 2 x 2 x 20, 000 x2 x 3 x 20, 000 x
Thus, y
The graph agrees with our direct calculation. 61. a.
Using MINIMUM, the minimum surface area (amount of cardboard) is about 2784.95 square inches.
The surface area is the sum of the areas of the six sides. S xy xy xy xy x 2 x 2 4 xy 2 x 2 The volume is x x y x 2 y 10, 000 . 501
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Chapter 5: Polynomial and Rational Functions b. Graphing:
64. a.
100 r 2 h h
100 r2
A(r ) 2 r 2 2 rh
c.
100 2 r 2 2 r 2 r 200 2 r 2 r
Using MINIMUM, the minimum surface area (amount of cardboard) is about 1392.48 square inches.
A(3) 2 32
c.
A(4) 2 42
Answers will vary. One possibility is to save costs or reduce weight by minimizing the material needed to construct the box.
d.
A(5) 2 52
500 r 2 h
e.
Graphing:
d. The surface area is a minimum when x 21.54 . 5000 y 10.78 21.54 2
The dimensions of the box are: 21.54 in. by 21.54 in. by 10.78 in. e.
63. a.
200 3 200 18 123.22 square feet 3
b.
h
200 4 32 50 150.53 square feet
200 5 50 40 197.08 square feet
500 r2
C (r ) 6(2 r 2 ) 4(2 rh) 500 12 r 2 8 r 2 r 4000 12 r 2 r
Using MINIMUM, the area is smallest when r 2.52 feet.
b. Graphing: 65. a.
P ( x) P (0.64)
Using MINIMUM, the cost is least for r 3.76 cm.
502
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x 4 (8 x3 28 x 2 34 x 15) 2 x2 2 x 1 (.64) 4 (8(.64)3 28(.64)2 34(.64) 15) 2(.64) 2 2(.64) 1 (.64) 4 (2.611648) 0.8126 0.5392
Section 5.4: The Graph of a Rational Function b.
67. a.
P (0.62)
7 3, , we can remove the discontinuity by 5 defining
(.62) 4 (8(.62)3 28(.62)2 34(.62) 15) 2(.62) 2 2(.62) 1
(.62) 4 (2.776576) 0.7759 0.5288 A player serving, with probability 0.62 of winning a point on a serve, has probability 0.7759 of winning the game.
x 2 x 12 if x 3 2 x x6 7 if x 3 5
c.
Graph P ( x)
Since R( x) is undefined and has a hole at
x 4 (8 x3 28 x 2 34 x 15)
b. Since R( x) is undefined and has a hole at
2 x2 2 x 1 and y 0.9 and find the intersection.
3 , 11 , we can remove the discontinuity be 2 defining 6 x2 7 x 3 3 if x 2 2 2x 7 x 6 3 11 if x 2
68. a. Since R ( x) is undefined and has a hole at
5,3.5 , we can remove the discontinuity by
The intersection is x 0.7
defining
d.
x 2 3 x 10 if x 5 2 x 8 x 15 3.5 if x 5
b. Since R( x) is undefined and hole at 5 14 , , we can remove the discontinuity be 2 11 defining
The P values seems to be approaching 1. 66. a.
32(t 2) t 5 32(0 2) 64 N (0) =12.8 words per min 05 5
8 x 2 26 x 15 5 if x 2 2 2 x x 15 5 14 if x 11 2
N (t )
32(7 2) 288 =24 words per min 75 12
b.
N (7)
c.
The function has the same degree in the numerator and denominator so there is a 32 horizontal asymptote at y ) 32 . The 1 number of words per minute seems to be approaching 32 as the number days increases.
69. y
x2 1 x 1
503
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Chapter 5: Polynomial and Rational Functions
y
x3 1 x 1
70. y
x2 x 1
y
x4 1 x 1
y
x4 x 1
y
x5 1 x 1
y
x6 x 1
y
x8 x 1
x = 1 is not a vertical asymptote because of the following behavior: When x 1 : y
x 2 1 x 1 x 1 x 1 x 1 x 1
x3 1 x 1 x x 1 y x2 x 1 x 1 x 1 y
2
All four graphs have a vertical asymptote at x2 x 1. y has an oblique asymptote at x 1 y x 1.
x2 1 x2 1 x4 1 x 1 x 1
x 1 x 1 x 1 2
x 1 x x x 1 3
y
71. Answers will vary. One example is
2
R x
x5 1 ( x 4 x3 x 2 x 1)( x 1) x 1 x 1 4 3 2 x x x x 1
2 x 3 x 2
2
x 13
.
72. Answers will vary. One example is
In general, the graph of xn 1 y , n 1, an integer, x 1 will have a “hole” with coordinates 1, n .
R x
3 x 2 x 1
2
x 5 x 6 2
73. Answers will vary. 504
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.
Section 5.4: The Graph of a Rational Function 83. 2 x 6 y 7 6 y 2 x 7 2 7 y x 6 6 1 7 y x 3 6
74. Answers will vary. 75. Answers will vary. 76. (4 x3 7 x 1) (5 x 2 9 x 3)
4 x3 7 x 1 5 x 2 9 x 3 4 x3 5 x 2 2 x 2 77.
3x x2 3x 1 x 5 3x ( x 5) ( x 2)(3 x 1)
The slopes are opposite reciprocals so the lines are perpendicular. 84. x x 7 5
3x 2 15 x 3 x 2 5 x 2
x7 5 x
15 x 5 x 2 20 x 2 x
x 7 x5
1 10
x 7 x 2 10 x 25 0 x 2 11x 18 0 ( x 2)( x 9)
78. The maximum value occurs at b 6 9 x 2 2a 2( 3 ) 2
x 2,9 Now check solution in the original problem. 2 27 23 5
2
2 9 9 9 f 6 5 2 2 3 2
9 97 94 5 The solution set is 9 .
2 81 54 17 5 3 4 2 2
85.
b 12 2 79. The vertex occurs at x 2a 2(3)
x2 4 x2 0 2 4 x
f 2 3 2 12 2 7 2
3 4 24 7 5
x2 4 x2 0 4 x2 2 4 x
The vertex is 2, 5 80. y x 4
x2 4 x2 0
81. g (3) 5(3) 9 15 9 6 82.
x2 4 x2 0 2 2 4 x
2 x2 4 x2 4
f ( x 2) ( x 2)2 3( x 2) 2 x 2 4 x 4 3x 6 2
x 2
x x4 2
505
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Chapter 5: Polynomial and Rational Functions
set is x x 0 or 1 x 2 or, using
Section 5.5
interval notation (, 0] [1, 2] .
1. 3 4 x 5 4 x 2
6. The x-intercepts of the graph of f are 1 , 1, and 2. a. The graph of f is below the x-axis (so f is negative) for 1 x 1 or x 2 . Therefore, the solution set is x 1 x 1 or x 2 or,
1 x 2 1 The solution set is x x or, using interval 2 1 notation, , . 2
using interval notation, (1, 1) (2, ) . b. The graph of f is above the x-axis (so f is positive) for x 1 or 1 x 2 . Since the inequality is not strict, we include 0, 1, and 2 in the solution set. Therefore, the solution set is x x 1 or 1 x 2 or, using interval notation (, 1] [1, 2] .
x 2 5 x 24
2.
7. The x-intercept of the graph of f is 0. a. The graph of f is below the x-axis (so f is negative) for 1 x 0 or x 1 . Therefore, the solution set is x 1 x 0 or x 1 or,
x 2 5 x 24 0 ( x 3)( x 8) 0 f ( x) x 2 5 x 24 ( x 3)( x 8) x 3, x 8 are the zeros of f.
using interval notation, (1, 0) (1, ) .
Interval (, 3) (3,8) (8, ) Number 4 0 9 Chosen 24 Value of f 12 12 Conclusion Positive Negative Positive
b. The graph of f is above the x-axis (so f is positive) for x 1 or 0 x 1 . Since the inequality is not strict, we include 0 in the solution set. Therefore, the solution set is x x 1 or 0 x 1 or, using interval
notation (, 1) [0, 1) .
The solution set is x | 3 x 8 or, using interval notation, 3, 8 .
8. The x-intercepts of the graph of f are 1 and 3. a. The graph of f is above the x-axis (so f is positive) for x 1 or 1 x 1 or x 3 . Therefore, the solution set is x x 1 or 1 x 1 or x 3 or, using
3. c
interval notation, (, 1) (1, 1) (3, ) .
4. False. The value 3 is not in the domain of f, so it must be restricted from the solution. The solution set would be x | x 0 or x 3 .
b. The graph of f is below the x-axis (so f is negative) for 1 x 2 or 2 x 3 . Since the inequality is not strict, we include 1 and 3 in the solution set. Therefore, the solution set is x 1 x 2 or 2 x 3 or, using
5. The x-intercepts of the graph of f are 0, 1, and 2. a. The graph of f is above the x-axis (so f is positive) for 0 x 1 or x 2 . Therefore, the solution set is x 0 x 1 or x 2 or,
interval notation [1, 2) (2, 3] . 9. We graphed f ( x) x 2 ( x 3) in Problem 81 of Section 5.1. The graph is reproduced below.
using interval notation, (0, 1) (2, ) . b. The graph of f is below the x-axis (so f is negative) for x 0 or 1 x 2 . Since the inequality is not strict, we include 0, 1, and 2 in the solution set. Therefore, the solution 506
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Section 5.5: Polynomial and Rational Inequalities
positive) for x 1 . Since the inequality is not strict, we include 1 in the solution set. Therefore, the solution set is x x 1 or, using interval notation ,1 .
12. We graphed f ( x) ( x 1)( x 3) 2 in Problem 84 of Section 5.1. The graph is reproduced below.
From the graph, we see that f is below the x-axis (so f is negative) for x 0 or 0 x 3 . Thus, the solution set is x x 0 or 0 x 3 or, using interval notation (, 0) (0, 3) .
From the graph, we that f is above the x-axis (so f is positive) for x 1 . Therefore, the solution set is x x 1 or, using interval notation (1, ) .
10. We graphed f ( x) x( x 2) 2 in Problem 82 of Section 5.1. The graph is reproduced below.
13. We graphed f ( x) 2( x 2)( x 2)3 in Problem 85 of Section 5.1. The graph is reproduced below.
From the graph, we see that f is below the x-axis (so f is negative) for x 2 or 2 x 0 . Since the inequality is not strict, we include 2 and 0 in the solution set. Therefore, the solution set is x x 0 or, using interval notation
From the graph, we see that f is below the x-axis (so f is negative) for x 2 or x 2 . Since the inequality is not strict, we include 2 and 2 in the solution set. Therefore, the solution set is x x 2 or x 2 or, using interval notation
(, 0] .
11. We graphed f ( x) ( x 4) 2 (1 x) in Problem 83 of Section 5.1. The graph is reproduced below.
(, 2] [2, ) .
From the graph, we that f is above the x-axis (so f is 507
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Chapter 5: Polynomial and Rational Functions
1 14. We graphed f ( x) ( x 4)( x 1)3 in Problem 2 86 of Section 5.1. The graph is reproduced below.
R is negative) for x 2 or 0 x 1 . Thus, the solution set is x x 2 or 0 x 1 or, using interval notation (, 2) (0, 1) . 3x 3 in Problem 9 of 2x 4 Section 5.3. The graph is reproduced below.
17. We graphed R ( x)
From the graph, we see that f is below the x-axis (so f is negative) for x 4 or x 1 . Therefore, the solution set is x x 4 or x 1 or, using interval notation (, 4) (1, ) . x 1 in Problem 7 of x( x 4) Section 5.3. The graph is reproduced below.
15. We graphed R ( x)
From the graph, we that R is below the x-axis (so R is negative) for 2 x 1 . Since the inequality is not strict, we include 1 in the solution set. Therefore, the solution set is x 2 x 1 or, using interval notation (2, 1] . 2x 4 in Problem 10 of x 1 Section 5.3. The graph is reproduced below.
18. We graphed R ( x)
From the graph, we that R is above the x-axis (so R is positive) for 4 x 1 or x 0 . Therefore, the solution set is x 4 x 1 or x 0 or, using interval
notation (4, 1) (0, ) . x in Problem 8 ( x 1)( x 2) of Section 5.3. The graph is reproduced below.
16. We graphed R ( x)
From the graph, we that R is above the x-axis (so R is positive) for x 2 or x 1 . Since the inequality is not strict, we include 2 in the solution set. Therefore, the solution set is x x 2 or x 1 or, using interval notation (, 2] (1, ) .
19. ( x 4) 2 ( x 6) 0 f ( x) ( x 4) 2 ( x 6) x 4, x 6 are the zeros of f .
From the graph, we that R is below the x-axis (so 508 Copyright © 2025 Pearson Education, Inc.
Section 5.5: Polynomial and Rational Inequalities
Interval Number
(, 6)
(6, 4)
(6, )
7
0
7
121
96
117
Chosen Value of f Conclusion
The solution set is x x 8 or, using interval notation, , 8 . 23.
2 x3 8 x 2 0
Negative Positive Positive
The solution set is x x 6 or, using interval
2x2 x 4 0
notation, , 6 .
f x 2 x3 8 x 2
x 0, x 4 are the zeros of f.
20. ( x 5)( x 2) 2 0
Interval Number Chosen Value of f Conclusion
f ( x) ( x 5)( x 2) 2 x 5, x 2 are the zeros of f .
(, 2)
(2, 5)
(5, )
3
0
6
Value of f
8
20
64
Conclusion
Negative
Interval Number Chosen
notation, 5, . 3
21. x 4 x 0 x 2 ( x 4) 0
(, 0)
(0, 4)
(4, )
1
1
5
5
3
25
Chosen Value of f
Negative Negative Positive
The solution set is x x 4 or, using interval notation, 4, . x3 8 x 2 0
24.
6 10 50 Negative Positive Positive
3x3 15 x 2 0
Interval Number Chosen Value of f Conclusion
(, 5)
(5, 0)
(0, )
6
1
1
12 18 108 Negative Positive Positive
The solution set is x | x 5 or, using interval notation, , 5 . 25. ( x 2)( x 4)( x 6) 0 f ( x) ( x 2)( x 4)( x 6) x 2, x 4, x 6 are the zeros of f .
f ( x) x3 8 x 2 x 2 x 8
Number
x – 8, x 0 are the zeros of f .
Chosen
Value of f
1
3 x3 15 x 2
Interval
Conclusion
1
using interval notation, 4, 0 0, .
x 2 ( x 8) 0
Chosen
5
x 0, x 5 are the zeros of f.
x 0, x 4 are the zeros of f .
Interval Number
(0, )
f x 3 x3 15 x 2
f ( x) x3 4 x 2 x 2 x 4
Conclusion
(4, 0)
3x2 x 5 0
2
Interval Number
(, 4)
The solution set is x | 4 x 0 or x 0 or,
Negative Positive
The solution set is x x 5 or, using interval
22.
2 x3 8 x 2
( , 2)
( 2, 4)
(4, 6)
(6, )
3
3
5
7
Value of f
63
15
7
48
Conclusion
Negative
Positive
Negative
Positive
(, 8)
(8, 0)
(0, )
9
1
1
The solution set is x x 2 or 4 x 6 or,
81
7
9
using interval notation, , 2 4, 6 .
Negative Positive Positive
509 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions 26. ( x 1)( x 2)( x 3) 0 f ( x) ( x 1)( x 2)( x 3) x 1, x 2, x 3 are the zeros of f .
x4 x2 0 x 2 ( x 2 1) 0
( , 3)
( 3, 2)
( 2, 1)
( 1, )
4
2.5
1.5
0
Value of f
6
0.375
0.375
6
Conclusion
Negative
Positive
Negative
Positive
Interval Number Chosen
x 2 ( x 1)( x 1) 0 f ( x) x 2 ( x 1)( x 1) x 1 , x 0, x 1 are the zeros of f ( , 1)
( 1, 0)
(0, 1)
(1, )
2
0.5
0.5
2
Value of f
12
0.1875
0.1875
12
Conclusion
Positive
Negative
Negative
Positive
Interval Number
The solution set is x x 3 or 2 x 1 or, using interval
Chosen
notation, , 3 2, 1 . 27.
x4 x2
29.
x 3 4 x 2 12 0
The solution set is x x 1 or x 1 or,
x( x 2)( x 6) 0
using interval notation, , 1 1, .
3
2
f ( x) x 2 x 3 x
30. x 4 9 x 2
x 2, x 0, x 6 are the zeros of f . ( , 2)
( 2, 0)
(0, 6)
(6, )
3
1
1
7
Value of f
27
7
15
63
Conclusion
Negative
Positive
Negative
Positive
Interval Number Chosen
x4 9 x2 0 x 2 ( x 2 9) 0 x 2 ( x 3)( x 3) 0 f ( x) x 2 ( x 3)( x 3) x 0, x 3, x 3 are the zeros of f
The solution set is x 2 x 0 or x 6 or,
Interval
using interval notation, 2, 0 6, .
Number
( , 3)
( 3, 0)
(0, 3)
(3, )
4
1
1
4
Value of f
112
8
8
112
Conclusion
Positive
Negative
Negative
Positive
Chosen
28. x3 2 x 2 3x 0 2
x( x 2 x 3) 0 x( x 3)( x 1) 0 f ( x) x( x 3)( x 1) x 0, x 3, x 1 are the zeros of f Interval Number Chosen
The solution set is x – 3 x 0 or 0 x 3 or, using interval notation, 3, 0 0,3 .
( , 3)
( 3, 0)
(0, 1)
(1, )
4
1
0.5
2
Value of f
20
4
0.875
10
Conclusion
Negative
Positive
Negative
Positive
The solution set is x 3 x 0 or x 1 or, using interval notation, 3, 0 1, .
x4 1
31.
x4 1 0 ( x 2 1)( x 2 1) 0 ( x 1)( x 1)( x 2 1) 0 f ( x) ( x 1)( x 1)( x 2 1) x 1, x 1 are the zeros of f ; x 2 1 has no real solution Interval
( , 1)
( 1,1)
(1, )
Number Chosen
2
0
2
Value of f
15
1
15
Conclusion
Positive
Negative
Positive
510 Copyright © 2025 Pearson Education, Inc.
Section 5.5: Polynomial and Rational Inequalities
The solution set is x x 1 or x 1 or, using interval notation, , 1 1, .
35.
x3 1
32.
x3 1 0 ( x 1)( x 2 x 1) 0
Interval Number Chosen
f ( x) ( x 1)( x 2 x 1)
x 1 is a zero of f; x 2 x 1 has no real solution. Interval
( , 1)
(1, )
Number Chosen
0
2
Value of f
1
7
Conclusion
Negative
Positive
Conclusion
notation, 1, .
(1,1)
(1, )
2
0
2
1
1
3
3
Positive
Negative Positive
The solution set is x x 1 or x 1 or, using interval notation, , 1 1, . 36.
33. 3( x 2 2) 2( x 1) 2 x 2 3x 2 6 2( x 2 2 x 1) x 2 3x 2 6 2 x 2 4 x 2 x 2 3x 2 6 3x 2 4 x 2 6 4 x 2 4x 6 2 4x 8 x2 The solution set is x x 2 or, using interval
x 3 0 x 1 x 3 f ( x) x 1 The zeros and values where f is undefined are x 1 and x 3 . Interval (, 1) (1, 3) (3, ) Number 2 0 4 Chosen Value of f 5 0.2 3 Conclusion Positive Negative Positive
The solution set is x x 1 or x 3 or,
notation, (, 2) .
using interval notation, , 1 3, .
( x 3)( x 2) x 2 3x 5 x 2 2 x 3x 6 x 2 3x 5 x 2 x 6 x 2 3x 5 x 6 3x 5 4 x 6 5 4 x 11 11 x 4 11 The solution set is x x or, using 4
11 interval notation, , . 4
(, 1)
Value of f
The solution set is x x 1 or, using interval
34.
x 1 0 x 1 x 1 f ( x) x 1 The zeros and values where f is undefined are x 1 and x 1 .
37.
( x 2)( x 2) 0 x ( x 2)( x 2) f ( x) x The zeros and values where f is undefined are x 2, x 0 and x 2 . ( , 2)
( 2, 0)
(0, 2)
(2, )
3
1
1
3
Value of f
1.67
3
3
1.67
Conclusion
Negative
Positive
Negative
Positive
Interval Number Chosen
The solution set is x x 2 or 0 x 2 or, using interval notation, , 2 0, 2 . 511 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
38.
( x 3)( x 2) 0 x 1 ( x 3)( x 2) f ( x) x 1 The zeros and values where f is undefined are x 2, x 1 and x 3 . ( , 2)
( 2, 1)
(1, 3)
(3, )
3
0
2
4
Value of f
1.5
6
4
2
Conclusion
Negative
Positive
Negative
Positive
Interval Number Chosen
Interval
x2 4
41.
0
x 4 The zeros and values where f is undefined are ( , 2)
( 2, 2)
(2, 3)
(3, )
3
0
2.5
4
Value of f
0.03125
0.8
6.25
12.8
Conclusion
Positive
Positive
Negative
Positive
x4 1 x2 x4 1 0 x2 x 4 ( x 2) 0 x2 6 0 x2 6 f ( x) x2 The value where f is undefined is x 2 . Interval
(, 2)
(2, )
Number Chosen
0
3
3
6
Negative Positive
notation, , 2 . 42.
Value of f
7.2
2.25
.11
.083
Conclusion
Positive
Negative
Positive
Positive
using interval notation, , 2 2, . ( x 5) 2
0 x2 4 ( x 5) 2 0 ( x 2)( x 2)
f ( x)
3
The solution set is x x 2 or, using interval
The solution set is x x 2 or x 2 or,
40.
0
Conclusion
x 2, x 2 and x 3 .
Chosen
3
Value of f
2
Number
6
using interval notation, , 2 2, .
( x 3) 2
Interval
(2, )
The solution set is x x 2 or x 2 or,
( x 3) 2 0 ( x 2)( x 2) f ( x)
( 2, 2)
Number
using interval notation, , 2 1,3 . ( x 3) 2
( 5, 2)
Chosen
The solution set is x x 2 or 1 x 3 or,
39.
( , 5)
( x 5) 2
x2 4 The zeros and values where f is undefined are x 5, x 2 and x 2 .
x2 1 x4 x2 1 0 x4 x 2 ( x 4) 0 x4 6 0 x4 6 f ( x) x4 The value where f is undefined is x 4 . Interval
(, 4)
(4, )
Number Chosen
0
5
Value of f
1.5
6
Conclusion
Negative Positive
The solution set is x x 4 or, using interval notation, 4, .
512 Copyright © 2025 Pearson Education, Inc.
Section 5.5: Polynomial and Rational Inequalities
43.
3x 5 2 x2 3x 5 2 0 x2 3x 5 2( x 2) 0 x2 x9 0 x2 x9 f ( x) x2 The zeros and values where f is undefined are
45.
x 2 and x 9 . Interval Number
(, 2)
(2, 9)
(9, )
3
0
10
Chosen Value of f
12
Conclusion
Positive
1 12 Negative Positive 4.5
The solution set is x 2 x 9 or, using
We want to know where f ( x) 0 , so the
interval notation, 2,9 .
solution set is x x 3 or x 7 or, using
x4 1 44. 2x 4 x4 1 0 2x 4 x 4 2x 4 0 2x 4 x8 0 2 x 2
f ( x)
interval notation, (, 3) [7, ) . Note that 3 is not in the solution set because 3 is not in the domain of f. 46.
x 8 2 x 2
The zeros and values where f is undefined are x 8 and x 2 . Interval Number Chosen Value of f Conclusion
x 1 2 x 3 x 1 2 0 x3 x 1 2( x 3) 0 x3 x 1 2 x 6) 0 x 3 x 7 0 x 3 x 7 f ( x) x 3 The zeros and values where f is undefined are x 3 and x 7 . Interval (, 3) (3, 7) (7, ) Number 1 5 8 Chosen 1 Value of f 3 1 5 Conclusion Negative Positive Negative
(, 8)
(8, 2)
(2, )
9
3
0
2.5
2
1 14 Positive
Negative Positive
x 1 2 x2 x 1 20 x2 x 1 2( x 2) 0 x2 x 1 2x 4 0 x2 3x 3 0 x2 3( x 1) 0 x2 3( x 1) f ( x) x2 The zeros and values where f is undefined are x 2 and x 1 .
The solution set is x 8 x 2 or, using interval notation, 8, 2 .
513 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
The zeros and values where f is undefined are x 7, x 1, and x 3 .
Interval (, 2) (2, 1) (1, ) 3 Number 3 1 Chosen 2 Value of f 6 3 2 Conclusion Positive Negative Positive
Interval Number
We want to know where f ( x) 0 , so the
Chosen
solution set is x x 2 or x 1 or, using
Value of f
interval notation, (, 2) [1, ) . Note that 2 is not in the solution set because 2 is not in the domain of f. 47.
49.
0
4
14
22
2
2
3
5
Positive
Negative
Positive
77
Negative
x 2 (3 x)( x 4) ( x 5)( x 1)
The zeros and values where f is undefined are x 5, x 4, x 3, x 0 and x 1 . Interval
Number Chosen
, 5
6
Value of f
Negative
Negative
(5, )
5, 4
4.5
0
2.5
4
6
4, 3
3.5
1
1
(3, 0)
1
0.75
6
36
Negative
Positive
(0, 1)
0.5
(1, )
2
5
10
18
3
Negative
Positive
using interval notation, , 2 3,5 . 5 3 x 3 x 1 5 3 0 x 3 x 1 5 x 5 3x 9 0 ( x 3)( x 1) 2 x 7 ( x 3)( x 1)
0
2 x 7 ( x 3)( x 1)
Conclusion
216 7 243 44 49 108
(3, 5)
The solution set is x x 2 or 3 x 5 or,
f ( x)
2
(2, 3)
Chosen
48.
8
( , 2)
Number
Conclusion
(3, )
x 2 (3 x)( x 4) 0 ( x 5)( x 1)
f ( x)
The zeros and values where f is undefined are x 2, x 3, and x 5 .
Value of f
( 1, 3)
The solution set is x 7 x 1 or x 3 or,
x5 ( x 2)(3x 9)
Interval
( 7, 1)
using interval notation, 7, 1 3, .
1 2 x 2 3x 9 1 2 0 x 2 3x 9 3x 9 2( x 2) 0 ( x 2)(3 x 9) x5 0 ( x 2)(3x 9) f ( x)
Conclusion
( , 7)
63 44 120 7
Positive
Positive
Negative Positive
The solution set is x x 5 or 4 x 3 or x 0 or x 1 or, using interval notation, , 5 4, 3 0 1, . 50.
x( x 2 1)( x 2) 0 ( x 1)( x 1) f ( x)
x( x 2 1)( x 2) ( x 1)( x 1)
The zeros and values where f is undefined are x 1, x 0, x 1 and x 2 .
514 Copyright © 2025 Pearson Education, Inc.
Section 5.5: Polynomial and Rational Inequalities
( , 1)
Interval Number
( 1, 0)
(0,1)
(2, )
(1, 2)
2
0.5
0.5
1.5
3
Value of f
40 3
25 12
1.25
1.95
3.75
Conclusion
Positive
Negative
Positive
Negative
Positive
Chosen
, 1
2,
interval notation, , 1 0,1 2, .
, 1 2 1 ,1 2
1, 3 3,
Number
Value of f
Conclusion
1
32
Positive
0
27
Negative
Chosen
0
16
Negative
1
1/ 2
Positive
3
1/ 4
Negative
6x 5
53.
6 x
6 0 x 6x2 5x 6 0 x (2 x 3)(3 x 2) 0 x (2 x 3)(3 x 2) f ( x) x The zeros and values where f is undefined are
2
5/7
Positive
4
1 / 7
Negative
Interval Number Chosen Value of f Conclusion
2
2 x 3 3 x 2
3
6x 5
using interval notation, 1 ,1 (3, ) . 2
2 2 , , 0 3 3
1
0.5
3 0, 2
3 , 2
1
2
4 4 5 5 Negative Positive Negative Positive
We want to know where f ( x) 0 , so the 0
2 3 solution set is x x or 0 x or, 3 2 2 3 using interval notation, , 0, . 3 2
2 x 3 3 x 2 0 x 1 x 2 x 1 f x
Positive
2 3 x , x 0 and x . 3 2
x3 1
512 / 7
The solution set is x 1 x 2 or x 2 or,
The solution set is x 1 x 1 or x 3 or,
52.
2
using interval notation, 1, 2 (2, ) . 3
(3 x) (2 x 1) ( x 1)( x 2 x 1) The zeros and values where f is undefined are 1 x 3, x , and x 1 . 2 Interval
Conclusion
3
f ( x)
Value of f
Chosen
2 1, 3 2 , 2 3
The solution set is x x 1 or 0 x 1 or x 2 or, using
(3 x)3 (2 x 1) 51. 0 x3 1 (3 x)3 (2 x 1) 0 ( x 1)( x 2 x 1)
Number
Interval
2 x 3 3 x 2 x 1 x 2 x 1
x
54.
The zeros and values where f is undefined are 2 x 2, x , and x 1 . 3
12 7 x
12 7 0 x x 2 7 x 12 0 x ( x 3)( x 4) 0 x
x
515 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
solution set is x x 0 or 3 x 4 or, using
( x 3)( x 4) ; The zeros and values x where f is undefined are x 0, x 3 and x 4 .
f ( x)
Interval
( , 0)
(0, 3)
(3, 4)
(4, )
1
1
3.5
5
20
6
Number Chosen Value of f
1
interval notation, , 0 3,4 .
0.4
14 Conclusion
Negative
Positive
Negative
Positive
We want to know where f ( x) 0 , so the
55. a.
R( x)
Step 1:
x2 5x 6 2
x 4x 4
( x 6)( x 1) ( x 2)( x 2)
p( x) x 2 5 x 6;
q( x) x 2 4; n 2; m 2
Domain: x x 2 The y-intercept is R (0)
Step 2 & 3:In lowest terms, R ( x )
(0) 2 5(0) 6 2
(0) 4(0) 4
6 3 3 . Plot the point 0, . 4 2 2
( x 6)( x 1) , x 2 . The x-intercepts are the zeros of y x 6 and ( x 2)( x 2)
y x 1 : 6,1 ;
Step 4:
In lowest terms, R ( x )
( x 6)( x 1) , x 2 . The vertical asymptote is the zero of f x x 2 : ( x 2)( x 2)
x 2; Graph this asymptote using a dashed line.
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: x2 5x 6 1 x2 4 x 4 x2 5x 6 x2 4x 4 5 x 6 4 x 4 9 x 10 10 x 9
516 Copyright © 2025 Pearson Education, Inc.
Section 5.5: Polynomial and Rational Inequalities
Steps 6 & 7: Graphing:
( x 6)( x 1) 0 ( x 2)( x 2)
b.
The zeros and values where f is undefined are x 6, x 1, and x 2 . Interval
Number Chosen
, 6
6
6, 1
4.5
1, 2
3.5
(2, )
2
Value of f
Conclusion
216 7 243 44 49 108 120 7
Positive Negative Positive Positive
The solution set is x x 6 or 1 x 2 or x 2 or, using interval notation,
, 6 1, 2 2, 56. a.
R( x)
Step 1:
2 x2 9 x 9 2
x 4
(2 x 3)( x 3) ( x 2)( x 2)
p( x) 2 x 2 9 x 9;
q ( x) x 2 4; n 2; m 2
Domain: x x 2, 2 The y-intercept is R (0)
Step 2 & 3:In lowest terms, R ( x)
2(0) 2 9(0) 9 2
(0) 4
9 9 9 . Plot the point 0, . 4 4 4
(2 x 3)( x 3) , x 2, 2 . The x-intercepts are the zeros of y 2 x 3 and ( x 2)( x 2)
3 y x 3 : , 3 ; 2
Step 4:
In lowest terms, R ( x)
(2 x 3)( x 3) , x 2, 2 . The vertical asymptotes are the zeros of ( x 2)( x 2)
f x x 2 and f x x 2 : x 2 and x 2 ;
Graph these asymptotes using dashed lines.
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Chapter 5: Polynomial and Rational Functions
Step 5:
Since n m , the line y 1 is the horizontal asymptote. Graph this asymptote using a dashed line. Solve to find intersection points: 2 x2 9 x 9 1 x2 4 x2 9 x 9 x2 4 9 x 13 13 x 9
Steps 6 & 7: Graphing:
(2 x 3)( x 3) 0 ( x 2)( x 2)
b.
3 The zeros and values where f is undefined are x , x 3, x 2 and x 2 . 2 Interval
, 3 3, 2
Number Chosen
Value of f
Conclusion
4
5
12
Positive
2.5
49
Negative
2, 3 2
1.75
2
3
Positive
3 , 2 2
0
94
Negative
(2, )
3
54
Positive
5
3 The solution set is x x 3 or 2 x or x 2 or, using interval notation, 2 3 , 3 2, 2, 2
57. a.
R( x)
( x 4)( x 2 2 x 3) ( x 4)( x 3)( x 1) ( x 4)( x 1) ( x 3)( x 2) ( x 3)( x 2) ( x 2)
p( x) x3 2 x 2 11x 12;
q ( x) x 2 x 6; n 3; m 2
518 Copyright © 2025 Pearson Education, Inc.
Section 5.5: Polynomial and Rational Inequalities
Step 1:
Domain: x x 2,3 The y-intercept is R (0)
Step 2 & 3:In lowest terms, R ( x)
(0)3 2(0) 2 11(0) 12 2
(0) (0) 6
12 2 . Plot the point 0, 2 . 6
( x 4)( x 1) , x 2 . The x-intercepts are the zeros of y x 4 and ( x 2)
y x 1 : 4, 1 ; Note: x 3 is not a zero because reduced form must be used to find the zeros.
Step 4:
In lowest terms, R ( x)
( x 4)( x 1) , x 2 . The vertical asymptote is the zero of f x x 2 : ( x 2)
x 2 ; Graph this asymptote using a dashed line.
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: x3 x 2 x 6 x3 2 x 2 11x 12 3
x x
2
3x
6x 2
5x 12
2
3x 18
3x
G ( x) x 3
2 x 6 x2 x 6
2x 6 The oblique asymptote is y x 3 . Graph this asymptote with a dashed line.
Steps 6 & 7: Graphing:
b.
( x 3)( x 4)( x 1) 0 ( x 3)( x 2)
The zeros and values where f is undefined are x 4, x 2, and x 1 .
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Chapter 5: Polynomial and Rational Functions
Interval
Number Chosen
Value of f
Conclusion
, 4
5
Negative
4, 2
4 3
3
2
Positive
2, 1
1.5
( 1, )
0
5 2 2
Negative Positive
The solution set is x 4 x 2 or 1 x 3 or x 3 or, using interval notation,
4, 2 1,3 3, 58. a.
R( x)
x3 6 x 2 9 x 4 2
x x 20
( x 1)( x 1)( x 4) ( x 1)( x 1) ( x 5)( x 4) ( x 5)
p( x) x3 6 x 2 9 x 4;
q ( x) x 2 x 20; n 3; m 2
Step 1:
Domain: x x 5, 4 The y-intercept is R (0)
(0)3 6(0) 2 9(0) 4 2
(0) (0) 20
4 1 1 . Plot the point 0, . 20 5 5
( x 1)( x 1) , x 5 . The x-intercept is the zero of y x 1 : 1 ; ( x 5) Note: x 4 is not a zero because reduced form must be used to find the zeros.
Step 2 & 3:In lowest terms, R ( x)
Step 4:
In lowest terms, R ( x)
( x 1)( x 1) , x 5 . The vertical asymptote is the zero of f x x 5 : ( x 5)
x 5 ; Graph this asymptote using a dashed line.
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: x7 x 2 x 20 x3 6 x 2 9 x 4 x3 x 2 20 x
G ( x) x 3
2 x 6
x2 x 6
7 x 2 29 x 4 7 x 2 7 x 140 36 x 144 The oblique asymptote is y x 7 . Graph this asymptote with a dashed line.
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Section 5.5: Polynomial and Rational Inequalities
Steps 6 & 7: Graphing:
( x 1)( x 1)( x 4) 0 ( x 5)( x 4)
b.
The zeros and values where f is undefined are x 5, x 1, and x 4 . Interval
Number Chosen
, 5
Value of f
Conclusion
49
Negative
6
5,1
0
1, 4
3
(4, )
5
1 5 1 2 8 5
Positive Positive Positive
The solution set is x 5 x 4 or x 4 or, using interval notation, 5, 4 4, 59. Let x be the positive number. Then x3 4 x 2
60. Let x be the positive number. Then x3 x
x3 4 x 2 0
x3 x 0
x 2 ( x 4) 0
x x 1 x 1 0
f ( x) x ( x 4)
f x x x 1 x 1
x 0 and x 4 are the zeros of f .
x 1, x 0, and x 1 are the zeros of f.
2
Interval Number Chosen Value of f Conclusion
(, 0)
(0, 4)
(4, )
1
1
5
Number
5
3
25
Chosen
Negative Negative Positive
Since x must be positive, all real numbers greater than 4 satisfy the condition. The solution set is x x 4 or, using interval notation,
4, .
( , 1)
( 1, 0)
(0,1)
(1, )
2
1 / 2
1/ 2
2
Value of f
6
0.375
0.375
6
Conclusion
Negative
Positive
Negative
Positive
Interval
Since x must be positive, all real numbers between (but not including) 0 and 1 satisfy the condition. The solution set is x 0 x 1 or, using interval notation, 0,1 .
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Chapter 5: Polynomial and Rational Functions
Interval
61. The domain of f ( x) x 4 16 consists of all real numbers x for which x 4 16 0
(2, )
5
0
3
Chosen
x 4 x 2 x 2 0 p( x) x 4 x 2 x 2 2
Value of R
7
Conclusion
Positive
1 1 7 2 Negative Positive
The domain of f will be where R ( x) 0 . Thus,
2
the domain of f is x x 4 or x 2 or,
x – 2 and x 2 are the zeros of p . (, 2)
(2, 2)
(2, )
3
0
3
Value of p
65
16
65
Conclusion
Positive
Chosen
(4, 2)
Number
( x 2 4)( x 2 4) 0
Interval Number
(, 4)
using interval notation, , 4 2, . 64. The domain of f ( x)
x 1 includes all x4
values for which x 1 0 x4 x 1 R( x) x4 The zeros and values where the expression is undefined are x – 4 and x 1 .
Negative Positive
The domain of f will be where p( x) 0 . Thus, the domain of f is x x 2 or x 2 or, using interval notation, , 2 2, . 62. The domain of f ( x) x3 3 x 2 consists of all real numbers x for which x3 3x 2 0
Interval Number Chosen
(, 4)
(4,1)
(1, )
5
0
2
x 2 ( x 3) 0
Value of R
6
p( x) x 2 ( x 3)
1 4
1 6
Conclusion Positive Negative Positive
x 0 and x 3 are the zeros of p .
The domain of f will be where R ( x) 0 . Thus,
Interval Number
(, 0)
(0,3)
(3, )
1
1
4
4
2
16
Chosen Value of p Conclusion
the domain of f is x x 4 or x 1 or, using interval notation, , 4 1, . f x g x
65.
x 4 1 2 x 2 2
Negative Negative Positive
The domain of f will be where p( x) 0 . Thus, the domain of f is x x 0 or x 3 or, using interval notation, 0 3, . 63. The domain of f ( x)
values for which
x 3 x 1 0 x 3 x 1 x 1 0 h x x 3 x 1 x 1 2
2
2
x2 includes all x4
x2 0. x4
x2 x4 The zeros and values where R is undefined are x – 4 and x 2 . R( x)
x4 2 x2 3 0
2
x 1 and x 1 are the zeros of h. Interval (, 1) (1,1) (1, ) Number 2 0 2 Chosen Value of h 21 21 3 Conclusion Positive Negative Positive f x g x if 1 x 1 . That is, on the
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Section 5.5: Polynomial and Rational Inequalities
interval 1,1 .
f x g x
67.
x 4 4 3x2 x4 3x 2 4 0
x 4 x 1 0 x 2 x 2 x 1 0 h x x 2 x 2 x 1 2
2
2
2
x 2 and x 2 are the zeros of h. Interval (, 2) (2, 2) (2, ) Number 0 3 3 Chosen Value of h 50 50 4 Conclusion Positive Negative Positive
f x g x
66.
x4 1 x 1
x4 x 0
x x 1 x x 1 0 h x x x 1 x x 1 x x3 1 0
f x g x if 2 x 2 . That is, on the
2
interval 2, 2 .
2
x 0 and x 1 are the zeros of h. Interval (, 0) (0,1) (1, ) Number 1/ 2 2 1 Chosen Value of h 2 14 7 /16 Conclusion Positive Negative Positive f x g x if 0 x 1 . That is, on the
interval 0,1 . f x g x
68.
x4 2 x2 x4 x2 2 0
x 2 x 1 0 x 2 x 1 x 1 0 h x x 2 x 1 x 1 2
2
2
2
x 1 and x 1 are the zeros of h. Interval (, 1) (1,1) (1, ) Number 2 0 2 Chosen 2 Value of h 18 18 Conclusion Positive Negative Positive f x g x if 1 x 1 . That is, on the
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Chapter 5: Polynomial and Rational Functions
interval 1,1 .
69. R ( x)
x 4 16
2
x 9
( x 4)( x 4) ( x 3)( x 3)
x2 9 x2 9 x4
16
4
x 9x
2
9 x 2 16 9 x 2 81 65 The x-intercepts are where x 4 0 or x 4 0 . The vertical asymptotes are x = 3, x = -3. So we need to test a number in each interval. Interval (, 4) 4, 3 3,3 3, 4 (4, ) 5 3.5 test value 0 3.5 5 x4 x4 x 3 x3 Sign of expression
So R(x) >0 on the intervals: , 4 3,3 4, 70. R ( x)
x3 8 x 2 25
( x 2)( x 2 2 x 4) ( x 5)( x 5)
The x-intercept are where x 2 0 . The vertical asymptotes are x = 5, x = -5. So we need to test a number in each interval. Interval Number Chosen Conclusion
(, 5)
(5, 2)
(2, 5)
(5, )
6
0
3
6
Negative Positive Negative Positive
So R(x)>0 on the intervals: 5, 2 (5, )
524 Copyright © 2025 Pearson Education, Inc.
Section 5.5: Polynomial and Rational Inequalities
71. We need to solve C ( x) 100 . 80 x 5000 100 x 80 x 5000 100 x 0 x x 5000 20 x 0 x 20(250 x) 0 x 20 250 x f x x The zeros and values where the expression is undefined are x 0 and x 250 . Interval (, 0) (0, 250) (250, ) Number 1 1 260 Chosen Value of f 5020 4980 10 /13 Conclusion Negative Positive Negative
The number of bicycles produced cannot be negative, so the solution is x x 250 or, using interval notation, 250, . The company must produce at least 250 bicycles each day to keep average costs to no more than $100. 72. We need to solve C x 100 . 80 x 6000 100 x 80 x 6000 100 x 0 x x 6000 20 x 0 x 20 300 x 0 x 20 300 x f x x The zeros and values where the expression is undefined are x 0 and x 300 . Interval (, 0) (0,300) (300, ) Number 1 1 310 Chosen Value of f 6020 5980 20 / 31 Conclusion Negative Positive Negative
The number of bicycles produced cannot be negative, so the solution is x x 300 or, using interval notation, 300, . The company
must produce at least 300 bicycles each day to keep average costs to no more than $100. K 16
73. a.
2 150 S 42 S2 300S 12, 600 S2 300 S 12, 600 S2
16 16
16 0
300 S 12, 600 16S 2
0 S2 Solve 16 S 2 300S 12, 600 0 and S 2 0 . The zeros and values where the left-hand side is undefined are S 0 , S 39 , S 20 . Since the stretch cannot be negative, we only consider cases where S 0. Interval Test Value
(0, 39) 1
(39, ) 40
Left side
12884
0.625
Conclusion Positive Negative
The cord will stretch less than 39 feet. b. The safe height is determined by the minimum clearance (3 feet), the free length of the cord (42 feet), and the stretch in the cord (39 feet). Therefore, the platform must be at least 3 42 39 84 feet above the ground for a 150-pound jumper. 74. Let r = the distance between Earth and the object in kilometers. Then 384, 400 r = the distance between the object and the moon. We want mmoon mobj mearth mobj G G 2 r2 384, 400 r mmoon
384, 400 r mmoon
2
mearth
mearth r2
0
384, 400 r r 2 r 2 mmoon 384, 400 r mearth 0 2 r 2 384, 400 r 2
2
The zeros and values where the left-hand side is undefined are r 0 , r 432,353 , r 346, 022 , and r 384, 400 . Since the
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Chapter 5: Polynomial and Rational Functions
distance from Earth to the object will be greater than 0 but less than the distance to the moon, we can exclude some of these values. Interval Test Value
(0, 346022) (346022,384400) 100, 000 350, 000
Left side
6 1014
1.3 1013
Conclusion
Negative
Positive
2k 9 2 3k so k
So y 82.
The gravitational force on the object due to the moon will be greater than the force due to the Earth when the object is more than 346,022 kilometers from Earth.
9x2 1 1 9 1 x (the negative solution will not work) 3 1 The domain is , 3 x2
x 3 x 3 83. f 4 3 4 4 x 3 3 x
84.
78. Answers will vary, for example, x 2 0 has no real solution and x 2 0 has exactly one real solution. 85.
4 3 4 So the solution is: , 3 6 x 8 x
80.
2
1
LC 1 2 LC 2 LC 1 1 C L 2
x y 2x 9 x y 2
2
2
2
2
Substitute -x for x:
( x) y 2( x) 9 ( x) y x y 2 x 9 x y not the same 2
2
6 x 4 y 4 3 x3 y 5 18 x 2 y 6 2 4
f g 3x 1 3x 1 9x2 1 0
77. Answers will vary. One example: x 5 0 x3
79. 9 2 x 4 x 1
2 x 3
9 x2 1
75. x 4 1 5 has no solution because the quantity x 4 1 is never negative. ( x 4 1 1 ) 76. No, the student is not correct. For example, x 5 is in the solution set, but does not satisfy the original inequality. 5 4 1 1 0 5 3 8 8 When multiplying both sides of an inequality by a negative, we must switch the direction of the inequality. Since we do not know the sign of x 3 , we cannot multiply both sides of the inequality by this quantity.
2 3
2
2
2
2
2
2
2
2
not symmetric to y-axis (or origin) Substitute -y for y:
2
3 x y (2 x xy 6 y ) 3 x 2 y 4 ( x 2 y )(2 x 3 y )
x ( y ) 2 x 9 x ( y) x y 2 x 9 x y the same 2
81. Since we have y varying directly with x then y k x . When x = 9, then y = 2 so we have
2
2
2
2
2
symmetric to x-axis
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2
2
2
2
Section 5.6: The Real Zeros of a Polynomial Function 86. The turning points are 0, 4 and 1.33, 2,81
3. Using synthetic division: 3 3 5 0 7 4 9 12 36 3
129
4 12 43 3
125 2
Quotient: 3x 4 x 12 x 43 Remainder: 125 4. x 2 x 3 0 x
1 12 4 1 3 2 1
1 1 12 1 13 2 2 1 13 1 13 , The solution set is . 2 2
5. a
5 x 2 3 2 x 2 11x 1
87.
6.
3x 2 11x 4 0 (3x 1)( x 4) 0
7. b
1 x ,x 4 3 1 The solution set is , 4 3
8. False; every polynomial function of degree 3 with real coefficients has at most three real zeros. 9. 0. 10. True
3 2x 2 88. 4 x 1 8 x 2 4 x 5
11.
6x 5 3 6x 2 13 2
12.
3 13 and the remainder is . 2 2
Section 5.6 f 1 2 1 1 2 1 3 2
f ( x) 4 x 3 5 x 2 8; c 3
f ( 3) 4( 3)3 5( 3) 2 8 108 45 8 161 0 Thus, –3 is not a zero of f and x 3 is not a factor of f .
13.
1.
f ( x) 4 x3 3 x 2 8 x 4; c 2
f (2) 4(2)3 3(2) 2 8(2) 4 32 12 16 4 8 0 Thus, 2 is not a zero of f and x 2 is not a factor of f .
8x2 2 x
The quotient is 2 x
f c
f ( x) 5 x 4 20 x3 x 4; c 2
f (2) 5(2) 4 20(2)3 (2) 4 80 160 2 4 82 Thus, 2 is not a zero of f and x 2 is not a factor of f .
2. 6 x 2 x 2 3x 2 2 x 1
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Chapter 5: Polynomial and Rational Functions
14.
f x 4 x 4 15 x 2 4 ; c 2
1 1 is not a zero of f and x is not a 3 3 factor of f.
Thus,
f 2 4 2 15 2 4 64 60 4 0 4
2
Thus, 2 is a zero of f and x 2 is a factor of f . 15.
6
21.
3
f ( x) 2 x 129 x 64; c 4 6
The maximum number of zeros is the degree of the polynomial, which is 7. Examining f x 4 x 7 x3 x 2 2 , there are
3
f (4) 2( 4) 129( 4) 64 8192 8256 64 0 Thus, –4 is a zero of f and x 4 is a factor of f .
16.
three variations in sign; thus, there are three positive real zeros or there is one positive real zero. Examining
f x 2 x 6 18 x 4 x 2 9 ; c 3
f x 4 x x x 2 , 7
f 3 2 3 18 3 3 9 6
4
2
1458 1458 9 9 0 Thus, –3 is a zero of f and x 3 is a factor of f .
17.
6
4
22.
f ( x) 4 x 64 x x 15; c 4 2
16,384 16,384 16 15 1 0 Thus, –4 is not a zero of f and x 4 is not a factor of f .
18.
f x 5x 2 x 6 x 5 , 4
2
4096 4096 16 16 0 Thus, –4 is a zero of f and x 4 is a factor of f .
19.
f ( x) 2 x 4 x3 2 x 1; c 4
23.
1 2
two variations in sign; thus, there are two positive real zeros or no positive real zeros. Examining
3
f ( x) 3x 4 x3 3x 1; c 4
3
1 3
1 1 1 1 f 3 3 1 3 3 3 3 1 1 11 2 0 27 27
f x 8x6 7 x2 x 5
The maximum number of zeros is the degree of the polynomial, which is 6. Examining f x 8 x 6 7 x 2 x 5 , there are
1 1 1 1 f 2 2 1 2 2 2 2 1 1 11 0 8 8 1 1 Thus, is a zero of f and x is a factor of f . 2 2
20.
2
5x4 2 x2 6 x 5 there is one variation in sign; thus, there is one negative real zero.
f 4 4 16 4 4 16 4
f x 5x4 2 x2 6 x 5
one variation in sign; thus, there is one positive real zero. Examining
f x x 6 16 x 4 x 2 16 ; c 4 6
2
The maximum number of zeros is the degree of the polynomial, which, is 4. Examining f x 5 x 4 2 x 2 6 x 5 , there is
f (4) 4 4 64 4 4 15 4
3
4 x7 x3 x 2 2 there are two variations in sign; thus, there are two negative real zeros or no negative real zeros.
2
6
f x 4 x 7 x3 x 2 2
f x 8x 7 x x 5 , 6
2
8x6 5x 2 x 5 there are two variations in sign; thus, there are two negative real zeros or no negative real zeros.
24.
f x 3 x5 4 x 4 2
The maximum number of zeros is the degree of the polynomial, which is 5. Examining f x 3 x5 4 x 4 2 , there is one variation in sign; thus, there is one positive real zero. Examining
528 Copyright © 2025 Pearson Education, Inc.
Section 5.6: The Real Zeros of a Polynomial Function f x 3 x 4 x 2 ,
f x x 5 x 2 x 4 5 x3 2 ,
3x5 4 x 4 2 there is no variation in sign; thus, there are no negative real zeros.
there is one variation in sign; thus, there is one negative real zero.
5
25.
4
4
29.
f x 2 x3 5 x 2 x 7
no variations in sign; thus, there are no positive real zeros. Examining
two variations in sign; thus, there are two positive real zeros or no positive real zeros. Examining
f x x x x x 1 , 5
f x 2 x 5 x x 7 , 2
30.
f x x3 x 2 x 1
f x x x x 1 , 2
f x x x x x x 1 5
2
x x x 1 there are two variations in sign; thus, there are two negative real zeros or no negative real zeros.
27.
f x x4 x2 1
The maximum number of zeros is the degree of the polynomial, which is 4. Examining f x x 4 x 2 1 , there are two
31.
f x x 4 5 x3 2
2
f x x6 1
in sign; thus, there is one positive real zero. Examining f x x 1 x 6 1 , there is 6
Examining f x x x 1
28.
3
The maximum number of zeros is the degree of the polynomial, which is 6. Examining f x x6 1 , there is one variation
2
x 4 x 2 1 , there are two variations in sign; thus, there are two negative real zeros or no negative real zeros.
4
x5 x 4 x3 x 2 x 1 there is no variation in sign; thus, there are no negative real zeros.
variations in sign; thus, there are two positive real zeros or no positive real zeros. 4
f x x5 x 4 x3 x 2 x 1
there are five variations in sign; thus, there are five positive real zeros or three positive real zeros or there is one positive real zero. Examining
variation in sign; thus, there is one positive real zero. Examining 3
2
The maximum number of zeros is the degree of the polynomial, which is 5. Examining f x x5 x 4 x3 x 2 x 1 ,
The maximum number of zeros is the degree of the polynomial, which is 3. Examining f x x3 x 2 x 1 , there is one
3
4
x5 x 4 x 2 x 1 there are three variations in sign; thus, there are three negative real zeros or there is one negative real zero.
2 x3 5 x 2 x 7 there is one variation in sign; thus, there is one negative real zero.
26.
f x x5 x 4 x 2 x 1
The maximum number of zeros is the degree of the polynomial, which is 5. Examining f x x5 x 4 x 2 x 1 , there are
The maximum number of zeros is the degree of the polynomial, which is 3. Examining f x 2 x3 5 x 2 x 7 , there are
3
3
one variation in sign; thus, there is one negative real zero. 32.
f x x6 1
The maximum number of zeros is the degree of the polynomial, which is 6. Examining f x x6 1 , there is no variation
The maximum number of zeros is the degree of the polynomial, which is 4. Examining f x x 4 5 x3 2 , there is one
in sign; thus, there are no positive real zeros.
variation in sign; thus, there is one positive real zero. Examining
no variation in sign; thus, there are no negative real zeros.
Examining f x x 1 x6 1 , there is 6
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Chapter 5: Polynomial and Rational Functions
33.
f x 3x 4 3x3 x 2 x 1
40.
f ( x) 4 x3 x 2 x 6 p must be a factor of 6: p 1, 2, 3, 6 q must be a factor of –4: q 1, 2, 4 The possible rational zeros are: p 1 1 3 3 1, 2, , , 3, , , 6 q 2 4 2 4
41.
f ( x) 2 x5 x3 2 x 2 12 p must be a factor of 12: p 1, 2, 3, 4, 6, 12 q must be a factor of 2: q 1, 2 The possible rational zeros are: p 1 3 1, 2, 4, , 3, , 6, 12 q 2 2
42.
f ( x) 3 x5 x 2 2 x 18 p must be a factor of 18: p 1, 2, 3, 6, 9, 18 q must be a factor of 3: q 1, 3 The possible rational zeros are: p 1 2 1, , 2, , 3, 6, 9 18 q 3 3
43.
f ( x) 6 x 4 2 x3 x 2 20 p must be a factor of 20: p 1, 2, 4, 5, 10, 20 q must be a factor of 6: q 1, 2, 3, 6 The possible rational zeros are: p 1 1 2 1 4 5 1, 2, , , , , 4, , 5, , q 2 3 3 6 3 2 5 5 10 20 , , 10, , 20, 3 6 3 3
44.
f ( x) 6 x3 x 2 x 10 p must be a factor of 10: p 1, 2, 5, 10 q must be a factor of –6: q 1, 2, 3, 6 The possible rational zeros are: p 1 1 1 2 5 1, , , , 2, , 5, , q 2 3 6 3 2 5 5 10 , , 10, 3 6 3
45.
f x x3 2 x 2 5 x 6
p must be a factor of 1: p 1 q must be a factor of 3: q 1, 3
The possible rational zeros are: 34.
p 1 1, q 3
f x x5 x 4 2 x 2 3
p must be a factor of 3: p 1, 3 q must be a factor of 1: q 1 The possible rational zeros are: 35.
p 1, 3 q
f x x5 2 x 2 8 x 5
p must be a factor of –5: p 1, 5 q must be a factor of 1: q 1 The possible rational zeros are: 36.
p 1, 5 q
f x 2 x5 x 4 x 2 1
p must be a factor of 1: p 1 q must be a factor of 2: q 1, 2 The possible rational zeros are: 37.
p 1 1, q 2
f x 9 x3 x 2 x 3
p must be a factor of 3: p 1, 3 q must be a factor of –9: q 1, 3, 9 The possible rational zeros are: p 1 1 1, 3, , 9 3 q 38.
f x 6 x4 x2 2
p must be a factor of 2: p 1, 2 q must be a factor of 6: q 1, 2, 3, 6 The possible rational zeros are: p 1 1 2 1 1, 2, , , , q 2 3 3 6 39.
f ( x) 6 x 4 x 2 9 p must be a factor of 9: p 1, 3, 9 q must be a factor of 6: q 1, 2, 3, 6 The possible rational zeros are: p 1 1 1 3 9 1, , , , 3, , 9, q 2 3 6 2 2
Step 1:
f (x) has at most 3 real zeros.
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Section 5.6: The Real Zeros of a Polynomial Function
Step 2: By Descartes’ Rule of Signs, there is one positive real zero.
x 2 3x 4 . Thus,
f x x 2 x 5x 6 3
3
2
x 2x 5x 6 thus, there are two negative real zeros or no negative real zeros.
Step 3: Possible rational zeros: p 1, 2, 3, 6; q 1; p 1, 2, 3, 6 q
2 5 6 3 3 6 1 1 2 0
x 3 x 1 x 2
The real zeros are –3, –1, and 2, each of multiplicity 1. 46.
f x x3 8 x 2 11x 20
Step 1:
f (x) has at most 3 real zeros.
Step 2: By Descartes’ Rule of Signs, there is one positive real zero. f x x 8 x 11 x 20 , 3
f x 2 x3 x 2 2 x 1
f (x) has at most 3 real zeros.
Step 2: By Descartes’ Rule of Signs, there are three positive real zeros or there is one positive real zero. f ( x) 2( x)3 ( x) 2 2( x ) 1 2 x3 x 2 2 x 1 thus, there are no negative real zeros.
Since the remainder is 0, x (3) x 3 is a factor. The other factor is the quotient: x2 x 2 . Thus, f x x 3 x 2 x 2
x 5 x 4 x 1
Step 1:
31
The real zeros are –5, –4, and 1, each of multiplicity 1. 47.
Step 4: Using synthetic division: We try x 3 :
f x x 5 x 2 3x 4
2
2
x3 8 x 2 11x 20 thus, there are two negative real zeros or no negative real zeros.
Step 3: Possible rational zeros: p 1, 2, 4, 5, 10, 20 q 1
Step 3: Possible rational zeros: p 1 q 1, 2 1 p 1, q 2 Step 4: Using synthetic division: We try x 1 : 1 2 1 2 1 2 1 3 2 1 3 2 x 1 is not a factor 1 We try x : 2 1 2 1 2 1 2 1 0 1 2
0 2 0
1 is a factor and the quotient is 2 x 2 2 . 2 Thus, 1 f x 2 x3 x 2 2 x 1 x 2 x 2 2 2 . 1 2 x x2 1 2 x
p 1, 2, 4, 5, 10, 20 q
Step 4: Using synthetic division: We try x 5 :
51
8 11 20 5 15 20 1 3 4 0 Since the remainder is 0, x (5) x 5 is a
Since x 2 1 0 has no real solutions, the only 1 real zero is x , of multiplicity 1. 2
factor. The other factor is the quotient: 531 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
48.
f ( x) 2( x)3 4( x) 2 10 x 20 ,
f ( x) 2 x3 x 2 2 x 1
Step 1:
f (x) has at most 3 real zeros.
2 x3 4 x 2 10 x 20 thus, there is one negative real zeros.
Step 2: By Descartes’ Rule of Signs, there are no positive real zeros.
Step 3: Possible rational zeros: p 1, 2, 5, 10; q 1; p 1, 2, 5, 10 q
f x 2x x 2x 1 3
2
2 x3 x 2 2 x 1 thus, there are three negative real zeros or there is one negative real zero.
Step 4: Using synthetic division: We try x 2 : 2 1 2 5 10 2 0 10
Step 3: Possible rational zeros: p 1; q 1, 2; 1 p 1, q 2
1 0 5 0 Since the remainder is 0, x 2 is a factor. The other factor is the quotient: x 2 5 . We can find the remaining real zeros by solving x2 5 0
Step 4: Using synthetic division: We try x 1 : 1 2 1 2 1 2 1 3
x2 5
2 1 3 2
x 5
x 1 is not a factor
2
0
2
real zeros are 2, multiplicity 1. 50.
Since x 2 1 0 has no real solutions, the only 1 real zero is x , of multiplicity 1. 2 f x 2 x 3 4 x 2 10 x 20
2 x3 2 x 2 5 x 10
Step 1:
3 x3 2 x 2 5 x 10
1 x is a factor and the quotient is 2 x 2 2 2 1 f ( x) 2 x3 x 2 2 x 1 x 2 x 2 2 2 1 2 x x2 1 2
49.
5 , and 5 , each of
f x 3 x3 6 x 2 15 x 30
0
Thus, f ( x) 2 x 2 x 5 x 5 . The
1 We try x : 2 1 2 1 2 1 2 1 0 1
f (x) has at most 3 real zeros.
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros.
Step 1:
f (x) has at most 3 real zeros.
Step 2: By Descartes’ Rule of Signs, there is one positive real zero. f ( x) 3( x)3 6( x) 2 15 x 30 , 3 x3 6 x 2 15 x 30 thus, there are two negative real zeros or no negative real zeros.
Step 3: Possible rational zeros: p 1, 2, 5, 10; q 1; p 1, 2, 5, 10 q Step 4: Using synthetic division: We try x 2 : 2 1 2 5 10 2 0 10 1
0
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5
0
Section 5.6: The Real Zeros of a Polynomial Function
Since the remainder is 0, x 2 is a factor. The other factor is the quotient: x 2 5 . We can find the remaining real zeros by solving x2 5 0 x2 5
The real zeros are
52.
Thus, f ( x) 3 x 2 x 5 x 5 . The real zeros are 2 , multiplicity 1. 51.
4
3
f ( x) 2 x 4 x3 5 x 2 2 x 2
5 , and 5 , each of
f x 2 x x 5x 2 x 2 4
f (x) has at most 4 real zeros.
f x 2 x x 7 x 3 x 3 2
2 x 4 x3 7 x 2 3x 3 thus, there are two negative real zeros or no negative real zeros.
Step 4: Using synthetic division: We try x 1 : 1 2 1 5 2 2 2 1 4 2
Step 3: Possible rational zeros: p 1, 3; q 1, 2; p 1 3 , 1, , 3 q 2 2
2 1 4 2 0 x 1 is a factor and the quotient is 2 x3 x 2 4 x 2 . Factoring by grouping gives 2 x3 x 2 4 x 2 x 2 2 x 1 2 2 x 1
Step 4: Using synthetic division: We try x 1 : 1 2 1 7 3 3 2 1 6 3
2 x 1 x 2 2
2 x3 x 2 6 x 3 . Factoring by grouping gives 2 x3 x 2 6 x 3 x 2 2 x 1 3 2 x 1 2 x 1 x 2 3
x 2 1 2 x x 1 x 2 x 2 2
f ( x) 2 x 1 x 1 x 2
Set each of these factors equal to 0 and solve: x2 3 0 2x 1 0 2x 1 x2 3 1 x 3 x 2 Thus,
1 2 x x 1 x 3 x 3 2
f ( x) 2 x 1 x 1 x 3 x 3
Set each of these factors equal to 0 and solve: x2 2 0 2x 1 0 2 x 1 x2 2 1 x 2 x 2 Thus,
2 1 6 3 0 x 1 is a factor and the quotient is
2
Step 3: Possible rational zeros: p 1, 2; q 1, 2; p 1 , 1, 2, 3 q 2
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros. 3
3
2 x 4 x3 5 x 2 2 x 2 thus, there are two negative real zeros or no negative real zeros.
2
4
f (x) has at most 4 real zeros.
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros.
f ( x) 2 x x 7 x 3 x 3
Step 1:
3 , and 3 , each
of multiplicity 1.
Step 1:
x 5
1 , 1 , 2
1 The real zeros are , 1, 2 of multiplicity 1
53.
2 , and 2 , each
f ( x) x 4 x3 3 x 2 x 2
Step 1:
f (x) has at most 4 real zeros.
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Chapter 5: Polynomial and Rational Functions
Step 4: Using synthetic division: We try x 2 : 21 1 6 4 8 2 6 0 8 1 3 0 4 0
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros. f x x x 3 x x 2 thus, 4
3
2
x 4 x3 3x 2 x 2 there are two negative real zeros or no negative real zeros.
x 2 is a factor and the quotient is x3 3x 2 4 .
Step 3: Possible rational zeros: p 1, 2; q 1; p 1, 2 q
We try x 1 on x3 3x 2 4 11 3 0 4 1 4 4 1 4 4 0
Step 4: Using synthetic division: We try x 2 :
x 1 is a factor and the quotient is x 2 4 x 4 . Thus,
21
f x x 2 x 1 x 2 4 x 4
1 3 1 2 2 2 2 2 1 1 1 1 0
x 2 x 1 x 2
x3 x 2 x 1 .
55.
2
We try x 1 on x x x 1 11 1 1 1 1 2 1 1 2 1 0
2
The real zeros are –2, –1, each of multiplicity 1, and 1, of multiplicity 2. 3
2
54. f ( x ) x x 6 x 4 x 8 Step 1: f (x) has at most 4 real zeros.
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros. f x x x 6x 4x 8 4
3
f (x) has at most 4 real zeros.
Step 2: By Descartes’ Rule of Signs, there are no positive real zeros.
f ( x) x 2 x 1 x 2 2 x 1
4
f ( x) 4 x 4 5 x3 9 x 2 10 x 2
Step 1:
x 1 is a factor and the quotient is x 2 2 x 1 . Thus, x 2 x 1 x 1
2
The real zeros are –2, –1, each of multiplicity 1, and 2, of multiplicity 2.
x 2 is a factor and the quotient is
3
2
x 4 x3 6 x 2 4 x 8 thus, there are two negative real zeros or no negative real zeros.
Step 3: Possible rational zeros: p 1, 2, 4, 8; q 1; p 1, 2, 4, 8 q
f x 4 x 5 x 9 x 10 x 2 4
3
2
4 x 4 5 x3 9 x 2 10 x 2 thus, there are four negative real zeros or two negative real zeros or no negative real zeros.
Step 3: Possible rational zeros: p 1, 2; q 1, 2, 4; p 1 1 , , 1, 2 q 4 2 Step 4: Using synthetic division: We try x 1 : 1 4 5 9 10 2 4 1 8 2 4 1 8 2 0 x 1 is a factor and the quotient is 4 x3 x 2 8 x 2 . Factoring by grouping gives 4 x3 x 2 8 x 2 x 2 4 x 1 2 4 x 1
4 x 1 x 2 2
Set each of these factors equal to 0 and solve:
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Section 5.6: The Real Zeros of a Polynomial Function
f ( x) 3 x 1 x 1 x 2 2
4x 1 0
x2 2 0
4 x 1
x 2 2
1 x 4
x 2 no real sol.
Thus,
f ( x) 4 x 1 x 1 x 2 2
1 4 x x 1 x 2 2 4 1 The real zeros are and 1 , each of 4 multiplicity 1.
56.
1 3 x x 1 x 2 2 3 1 The real zeros are and 1 , each of 3 multiplicity 1.
57. x 4 x3 2 x 2 4 x 8 0 The solutions of the equation are the zeros of f x x 4 x3 2 x 2 4 x 8 .
f (x) has at most 4 real zeros.
Step 1:
f ( x) 3x 4 4 x3 7 x 2 8 x 2 Step 1: f (x) has at most 4 real zeros.
Step 2: By Descartes’ Rule of Signs, there are three positive real zeros or there is one positive real zero.
Step 2: By Descartes’ Rule of Signs, there are no positive real zeros.
f x x x 2x 4x 8
f x 3x 4 x 7 x 8x 2 4
3
2
3x 4 4 x3 7 x 2 8 x 2 thus, there are four negative real zeros or two negative real zeros or no negative real zeros.
4
3
2
x 4 x3 2 x 2 4 x 8 thus, there is one negative real zero.
Step 3: Possible rational zeros: p 1, 2, 4, 8; q 1; p 1, 2, 4, 8 q
Step 3: Possible rational zeros: p 1, 2; q 1, 3; p 1 2 , , 1, 2 q 3 3
Step 4: Using synthetic division: We try x 1 :
Step 4: Using synthetic division: We try x 1 : 1 3 4 7 8 2 3 1 6 2
x 1 is a factor and the quotient is x3 2 x 2 4 x 8 .
11 1 2 4 8 1 2 4 8 1 2 4 8 0
3 1 6 2 0 x 1 is a factor and the quotient is 3x3 x 2 6 x 2 . Factoring by grouping gives 3x3 x 2 6 x 2 x 2 3x 1 2 3x 1
3 x 1 x 2 2
Set each of these factors equal to 0 and solve: 3x 1 0 x2 2 0 3x 1 x 2 2 1 x 2 x 3 no real sol. Thus,
We try x 2 on x3 2 x 2 4 x 8
21 2 2 1 0
4 8 0 8 4 0
x 2 is a factor and the quotient is x 2 4 .
Thus, f x x 1 x 2 x 2 4 . 2
Since x 4 0 has no real solutions, the solution set is 1, 2 .
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Chapter 5: Polynomial and Rational Functions
58. 2 x3 3 x 2 2 x 3 0 Solve by factoring: x 2 (2 x 3) (2 x 3) 0
x
3 2 2 Since x 1 0 has no real solutions, the 3 solution set is . 2 x
59. 3x3 4 x 2 7 x 2 0 The solutions of the equation are the zeros of f x 3 x3 4 x 2 7 x 2 .
f (x) has at most 3 real zeros.
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros. f x 3x 4 x 7 x 2 3
2 8 2 2 2 2 1 2 2 2 The solution set is 1 2, 1 2, . 3
(2 x 3) x 2 1 0
Step 1:
2 4 4(1)(1) 2(1)
2
60. 2 x3 3x 2 3x 5 0 The solutions of the equation are the zeros of f x 2 x3 3 x 2 3 x 5 .
Step 1:
f (x) has at most 3 real zeros.
Step 2: By Descartes’ Rule of Signs, there is one positive real zero. f x 2 x 3 x 3 x 5 3
2
3 x3 4 x 2 7 x 2 thus, there is one negative real zero.
2 x3 3 x 2 3x 5 thus, there are two negative real zeros or no negative real zeros.
Step 3: Possible rational zeros: p 1, 2; q 1, 3 p 1 2 1, 2, , q 3 3
Step 3: Possible rational zeros: p 1, 5; q 1, 2 p 1 5 1, 5, , q 2 2
Step 4: Using synthetic division: 2 We try x : 3 2 3 4 7 2 3 2 4 2
Step 4: Using synthetic division: 5 We try x : 2 5 2 3 3 5 2 5 5 5
3 6 3
2
0
2
2
0
5 is a factor. The other factor is the 2 quotient: 2 x 2 2 x 2 . Thus, 5 f ( x) x 2 x 2 2 x 2 2 5 2 x x2 x 1 2 x
x
2 is a factor. The other factor is the quotient 3
3x 2 6 x 3 . Thus, 2 f ( x) x 3x 2 6 x 3 3 2 3 x x2 2 x 1 3 Using the quadratic formula to solve x2 2 x 1 0 :
Since x 2 x 1 0 has no real solutions, the 5 solution set is . 2
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Section 5.6: The Real Zeros of a Polynomial Function
thus, there are two negative real zeros or no negative real zeros.
61. 3x3 x 2 15 x 5 0 Solving by factoring: x 2 (3 x 1) 5(3 x 1) 0
Step 3: Possible rational zeros: p 1, 2, 3, 6; q 1; p 1, 2, 3, 6 q
(3 x 1) x 5 x 5 0 (3x 1) x 2 5 0
The solution set is 5, 3
5,
1 . 3
2
62. 2 x 11x 10 x 8 0 The solutions of the equation are the zeros of f x 2 x3 11x 2 10 x 8 .
Step 1:
f (x) has at most 3 real zeros.
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros. f x 2 x 11 x 10 x 8 3
2
2 x3 11x 2 10 x 8 thus, there is one negative real zero.
Step 3: Possible rational zeros: p 1, 2, 4, 8; q 1, 2 p 1 1, 2, 4, 8, q 2
f (x) has at most 4 real zeros.
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros. f x x 4 x 2 x x 6 x 4 4 x3 2 x 2 x 6
x 2 is a factor and the quotient is x 2 x 1 .
Thus, f ( x) x 3 x 2 x 2 x 1 .
64. x 4 2 x3 10 x 2 18 x 9 0 The solutions of the equation are the zeros of f x x 4 2 x3 10 x 2 18 x 9
Step 1:
f (x) has at most 4 real zeros.
Step 2: By Descartes’ Rule of Signs, there are four positive real zeros or two positive real zeros or no positive real zeros. 3
2
x 4 2 x3 10 x 2 18 x 9 Thus, there are no negative real zeros.
63. x 4 4 x3 2 x 2 x 6 0 The solutions of the equation are the zeros of f x x 4 4 x3 2 x 2 x 6 .
3
We try x 2 on x3 x 2 x 2 21 1 1 2 2 2 2 1 1 1 0
4
1 The solution set is , 2, 4 . 2
4
x 3 is a factor and the quotient is x3 x 2 x 2 .
f x x 2 x 10 x 18 x 9
x 4 2 x 1 x 2
Step 1:
4 2 1 6 3 3 3 6 1 1 1 2 0
Since x x 1 0 has no real solutions , the solution set is 3, 2 .
4 2 11 10 8 8 12 8 2 3 2 0 x 4 is a factor. The other factor is the quotient: 2 x 2 3 x 2 . Thus,
31
2
Step 4: Using synthetic division: We try x 4 :
f ( x) x 4 2 x 2 3x 2
Step 4: Using synthetic division: We try x 3 :
2
Step 3: Possible rational zeros: p 1, 3, 9; q 1 p 1, 3, 9 q Step 4: Using synthetic division: We try x 1 : 11 2 10 18 9 1 1 9 9 1 1 9 9 0 x 1 is a factor and the quotient is x3 x 2 9 x 9 .
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Chapter 5: Polynomial and Rational Functions
We try x 1 on x3 x 2 9 x 9 11 1 9 9 1 0 9 1 0 9 0
3 2 x 3 x 2 0 2 x3 3x 2 6 x 4 0 2 The solutions of the equation are the zeros of f ( x) 2 x3 3x 2 6 x 4 .
66. x3
x 1 is a factor and the quotient is x 2 9 . Thus, f ( x) x 1
2
x 9 . 2
2
Since x 9 0 has no real solutions, the solution set is 1 . 2 2 8 x x 1 0 3 x3 2 x 2 8 x 3 0 3 3 The solutions of the equation are the zeros of f ( x) 3x3 2 x 2 8 x 3 .
65. x3
f (x) has at most 3 real zeros.
Step 1:
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros. f ( x) 3( x)3 2( x) 2 8( x) 3 , 3x3 2 x 2 8 x 3 thus, there is one negative real zero.
Step 3: To find the possible rational zeros: p 1, 3; q 1, 3 p 1 1, 3, q 3 Step 4: Using synthetic division: 1 We try x : 3 1 3 2 8 3 3 1 1 3 3 9
3 x
0
1 is a factor. The other factor is the 3 2
quotient: 3x 3x 9 . Thus, 1 f ( x) x 3 x 2 3 x 9 3 1 x 3 x 2 x 3 3
3 x 1 x 2 x 3 2
Step 1:
f (x) has at most 3 real zeros.
Step 2: By Descartes’ Rule of Signs, there is one positive real zero. f ( x) 2( x)3 3( x) 2 6( x) 4 2 x3 3 x 2 6 x 4 thus, there are two negative real zeros or no negative real zeros.
Step 3: To find the possible rational zeros: p 1, 2, 4; q 1, 2 p 1 1, , 2, 4 q 2 Step 4: Using synthetic division: We try x 1 : 1 2 3 6 4 2 5 11 2 5 11 7 x 1 is not a factor
We try x 12 1 2
2
3 6 4 1 2 4
2
4 8
0
x 12 is a factor Thus,
1 f ( x) x 2 x 2 4 x 8 2 1 2 x x2 2 x 4 2
Since x 2 2 x 4 0 has no real solutions, the 1 solution set is . 2 67. 2 x 4 19 x3 57 x 2 64 x 20 0 The solutions of the equation are the zeros of f ( x) 2 x 4 19 x3 57 x 2 64 x 20 .
Step 1:
Since x x 3 0 has no real solutions, the 1 solution set is . 3
f (x) has at most 4 real zeros.
Step 2: By Descartes’ Rule of Signs, there are four positive real zeros or two positive real zeros or no positive real zeros.
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Section 5.6: The Real Zeros of a Polynomial Function
f ( x) 2 x 19 x 57 x 64 x 20 4
3
2
2 x 4 19 x3 57 x 2 64 x 20 Thus, there are no negative real zeros.
68. 2 x 4 x3 24 x 2 20 x 16 0 The solutions of the equation are the zeros of f ( x) 2 x 4 x3 24 x 2 20 x 16 .
f (x) has at most 4 real zeros.
Step 3: To find the possible rational zeros: p 1, 2, 4, 5, 10, 20; q 1, 2; p 1 5 1, , 2, 4, 5, , 10, 20 q 2 2
Step 1:
Step 4: Using synthetic division: We try x 1 : 1 2 19 57 64 20 2 17 40 24
2 x 4 x3 24 x 2 20 x 16 thus, there are two negative real zeros or no negative real zeros.
1 We try x : 2 1 2 19 57 64 20 2 1 9 24 20 48 40
0
3
2
2 5 14 8 0 x 2 is a factor, and the other factor is the quotient 2 x3 5 x 2 14 x 8 .
3
1 7 10 0 x 2 is a factor, and the other factor is the quotient x 2 7 x 10 . Thus,
x 9 x 24 x 20 x 2 x 7 x 10
x 2 x 2 x 5 1 2 f ( x) 2 x x 2 x 5 2 1 The solution set is , 2,5 . 2
2
Now try x 4 as a factor of 2 x 5 x 14 x 8 . 4 2 5 14 8 8 12 8
2
Thus, f ( x) x 2 2 x3 5 x 2 14 x 8 .
Now try x – 2 as a factor of x3 9 x 2 24 x 20 . 2 1 9 24 20 2 14 20
2
4
Step 4: Using synthetic division: We try x 2 : 2 2 1 24 20 16 4 10 28 16
1 x is a factor and the quotient is 2 2 x3 18 x 2 48 x 40 . Thus, 1 f ( x) x 2 x3 18 x 2 48 x 40 2 1 2 x x3 9 x 2 24 x 20 2
3
f ( x) 2 x x 24 x 20 x 16
Step 3: To find the possible rational zeros: p 1, 2, 4, 8, 16; q 1, 2; p 1 1, , 2, 4, 8, 16 q 2
2 17 40 24 4 x 1 is not a factor
2 18
Step 2: By Descartes’ Rule of Signs, there are two positive real zeros or no positive real zeros.
2 3 2 0 x 4 is a factor, and the other factor is the quotient 2 x 2 3 x 2 . Thus,
2 x3 5 x 2 14 x 8 x 4 2 x 2 3 x 2
x 4 2 x 1 x 2
f ( x) x 2 x 4 2 x 1 x 2 x 2 x 4 2 x 1 2
1 The solution set is 4, , 2 . 2
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Chapter 5: Polynomial and Rational Functions
69.
f x x 4 3x 2 4 r
1 2 1 2
coeff of q(x)
remainder
1 2 2 6 2 1 2 0 1 2 2 6 2 1 2 0
1 1 1 1
For r = 2, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 2. For r = -2, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -2. The upper bound is 2 and the lower bound is -2.
70.
f x x 4 5 x 2 36 r
coeff of q(x)
remainder
1 2
1 1
1 2
4 1
4 2
40 40
3 1 2 3
1 1 1 1
3 1 2 3
4 4 1 4
12 0 4 40 2 40 12 0
For r = 3, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 3. For r = -3, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -3. The upper bound is 3 and the lower bound is -3.
71.
f x x 4 x3 x 1 r
1 1
coeff of q(x)
1 1
2 0
2 0
remainder
1 1
0 0
For r = 1, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 1. For r = -1, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -1. The upper bound is 1 and the lower bound is -1.
72.
f x x 4 x3 x 1 r
1 1
coeff of q(x)
1 1
0 2
0 2
remainder
1 1
0 0
For r = 1, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 1. For r = -1, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -1. The upper bound is 1 and the lower bound is -1.
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Section 5.6: The Real Zeros of a Polynomial Function
73.
f x 3 x 4 3 x3 x 2 12 x 12 r
coeff of q(x)
remainder
1
3
6
5
7
5
2 1
3 3
9 0
17 1
22 11
56 23
2
3
3
5
22
56
For r = 2, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 2. For r = -2, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -2. The upper bound is 2 and the lower bound is -2.
74. f x 3 x 4 3 x3 5 x 2 27 x 36 r
coeff of q(x)
1 2 1 2 3
3 3 3 3 3
0 3 6 9 12
5 1 1 13 31
remainder
22 29 26 1 66
14 22 62 38 162
For r = 2, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 2. For r = -3, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -3. The upper bound is 2 and the lower bound is -3.
75.
f x 4 x5 x 4 2 x3 2 x 2 x 1 r
coeff of q(x)
1 1
4 4
3 5
5 7
remainder
3 9
4 10
3 11
For r = 1, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 1. For r = -1, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -1. The upper bound is 1 and the lower bound is -1.
76.
1 1 1 1 1 f x 4 x 5 x 4 x3 x 2 2 x 2 4 x5 x 4 x3 x 2 x 4 4 4 2 2 r
1 1
coeff of q(x)
4 4
5 3
6 4
remainder
7 3
5 1
3 3
For r = 1, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 1. For r = -1, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -1. The upper bound is 1 and the lower bound is -1.
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Chapter 5: Polynomial and Rational Functions
77.
f x x 4 3 x3 4 x 2 2 x 9 x 4 3 x3 4 x 2 2 x 9 r
coeff of q(x)
remainder
1 2 3
1 1 1
2 1 0
2 2 4
4 6 14
5 3 33
1 2
1 1
4 5
8 14
6 26
3 43
For r = 3, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 3. For r = -2, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -2. The upper bound is 3 and the lower bound is -2.
78.
f x 4 x5 5 x3 9 x 2 3 x 12 4 x5 5 x3 9 x 2 3 x 12 r
1 2 1 2
coeff of q(x)
4 4 4 4
4 8 4 8
1 11 1 11
remainder
10 13 8 31
13 23 5 59
1 58 7 106
For r = 2, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 2. For r = -2, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -2. The upper bound is 2 and the lower bound is -2.
79.
f x 8 x 4 2 x 2 5 x 1;
0,1
82.
f (0) 1 0 and f (1) 10 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
80.
f ( x) x 4 8 x3 x 2 2;
1, 0
f ( x) 2 x3 6 x 2 8 x 2;
5, 4
f (5) 58 0 and f ( 4) 2 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
3, 2
f (3) 42 0 and f ( 2) 5 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
83.
f (1) 6 0 and f (0) 2 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
81.
f ( x) 3x3 10 x 9;
f ( x) x5 x 4 7 x3 7 x 2 18 x 18; 1.4, 1.5 f (1.4) 0.1754 0 and f (1.5) 1.4063 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
84.
f ( x) x5 3 x 4 2 x3 6 x 2 x 2;
1.7, 1.8
f (1.7) 0.35627 0 and f (1.8) –1.021 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
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Section 5.6: The Real Zeros of a Polynomial Function
85. 8 x 4 2 x 2 5 x 1 0;
f 0.67 0.6535; f 0.66 0.5458
0 r 1
Consider the function f x 8 x 2 x 5 x 1
f 0.66 0.5458; f 0.65 0.4410
Subdivide the interval [0,1] into 10 equal subintervals: [0,0.1]; [0.1,0.2]; [0.2,0.3]; [0.3,0.4]; [0.4,0.5]; [0.5,0.6]; [0.6,0.7]; [0.7,0.8]; [0.8,0.9]; [0.9,1] f 0 1; f 0.1 0.5192
f 0.65 0.4410; f 0.64 0.3390
f 0.1 0.5192; f 0.2 0.0672
f 0.61 0.0495; f 0.60 0.0416
4
2
f 0.2 0.0672; f 0.3 0.3848
So f has a real zero on the interval [0.2,0.3]. Subdivide the interval [0.2,0.3] into 10 equal subintervals: [0.2,0.21]; [0.21,0.22]; [0.22,0.23]; [0.23,0.24]; [0.24,0.25]; [0.25,0.26];[0.26,0.27]; [0.27,0.28]; [0.28,0.29]; [0.29,0.3] f 0.2 0.0672; f 0.21 0.02264 f 0.21 0.02264; f 0.22 0.0219
So f has a real zero on the interval [0.21,0.22], therefore r 0.21 , correct to two decimal places.
f 0.64 0.3390; f 0.63 0.2397 f 0.63 0.2397; f 0.62 0.1433
f 0.62 0.1433; f 0.61 0.0495
So f has a real zero on the interval [–0.61, –0.6], therefore r 0.60 , correct to two decimal places.
87. 2 x3 6 x 2 8 x 2 0;
5 r 4
Consider the function f x 2 x3 6 x 2 8 x 2 Subdivide the interval [–5, –4] into 10 equal subintervals: [–5, –4.9]; [–4.9, –4.8]; [–4.8, –4.7]; [–4.7, –4.6]; [–4.6, –4.5]; [–4.5, –4.4]; [–4.4, –4.3]; [–4.3, –4.2]; [–4.2, –4.1]; [–4.1, –4] f 5 58; f 4.9 50.038 f 4.9 50.038; f 4.8 42.544 f 4.8 42.544; f 4.7 35.506
86. x 4 8 x3 x 2 2 0;
1 r 0
Consider the function f x x 4 8 x3 x 2 2 Subdivide the interval [–1, 0] into 10 equal subintervals: [–1, –0.9]; [–0.9, –0.8]; [–0.8, –0.7]; [–0.7, –0.6]; [–0.6, –0.5]; [–0.5, –0.4]; [–0.4, –0.3]; [–0.3, –0.2]; [–0.2, –0.1]; [–0.1,0] f 1 6; f 0.9 3.9859 f 0.9 3.9859; f 0.8 2.3264 f 0.8 2.3264; f 0.7 0.9939
f 0.7 0.9939; f 0.6 0.0416
So f has a real zero on the interval [–0.7, –0.6]. Subdivide the interval [–0.7, –0.6] into 10 equal subintervals: [–0.7, –0.69]; [–0.69, –0.68]; [–0.68, –0.67]; [–0.67, –0.66]; [–0.66, –0.65]; [–0.65, –0.64]; [–0.64, –0.63]; [–0.63, –0.62]; [–0.62, –0.61]; [–0.61, –0.6] f 0.7 0.9939; f 0.69 0.8775 f 0.69 0.8775; f 0.68 0.7640 f 0.68 0.7640; f 0.67 0.6535
f 4.7 35.506; f 4.6 28.912
f 4.6 28.912; f 4.5 22.75 f 4.5 22.75; f 4.4 17.008 f 4.4 17.008; f 4.3 11.674
f 4.3 11.674; f 4.2 6.736 f 4.2 6.736; f 4.1 2.182 f 4.1 2.182; f 4 2
So f has a real zero on the interval [–4.1, –4]. Subdivide the interval [–4.1, –4] into 10 equal subintervals: [–4.1, –4.09]; [–4.09, –4.08]; [–4.08, –4.07]; [–4.07, –4.06]; [–4.06, –4.05]; [–4.05, –4.04]; [–4.04, –4.03]; [–4.03, –4.02]; [–4.02, –4.01]; [–4.01, –4] f 4.1 2.182; f 4.09 1.7473 f 4.09 1.7473; f 4.08 1.3162
f 4.08 1.3162; f 4.07 0.8889 f 4.07 0.8889; f 4.06 0.4652 f 4.06 0.4652; f 4.05 0.0453
f 4.05 0.4653; f 4.04 0.3711
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Chapter 5: Polynomial and Rational Functions
Subdivide the interval [1.1,1.2] into 10 equal subintervals: [1.1,1.11]; [1.11,1.12]; [1.12,1.13]; [1.13,1.14]; [1.14,1.15]; [1.15,1.16];[1.16,1.17]; [1.17,1.18]; [1.18,1.19]; [1.19,1.2] f 1.1 0.359; f 1.11 0.2903
So f has a real zero on the interval [–4.05, –4.04], therefore r 4.04 , correct to two decimal places.
88. 3x3 10 x 9 0;
3 r 2
Consider the function f x 3 x3 10 x 9
f 1.11 0.2903; f 1.12 0.2207
Subdivide the interval [–3, –2] into 10 equal subintervals: [–3, –2.9]; [–2.9, –2.8]; [–2.8, –2.7]; [–2.7, –2.6]; [–2.6, –2.5]; [–2.5, –2.4]; [–2.4, –2.3]; [–2.3, –2.2]; [–2.2, –2.1]; [–2.1, –2] f 3 42; f 2.9 35.167
f 1.12 0.2207; f 1.13 0.1502 f 1.13 0.1502; f 1.14 0.0789
f 1.14 0.0789; f 1.15 0.0066 f 1.15 0.0066; f 1.16 0.0665
f 2.9 35.167; f 2.8 28.856
So f has a real zero on the interval [1.15,1.16], therefore r 1.15 , correct to two decimal places.
f 2.8 28.856; f 2.7 23.049 f 2.7 23.049; f 2.6 17.728 f 2.6 17.728; f 2.5 12.875
f 2.5 12.875; f 2.4 8.472 f 2.4 8.472; f 2.3 4.501 f 2.3 4.501; f 2.2 0.944
f 2.2 0.944; f 2.1 2.217
So f has a real zero on the interval [–2.2, –2.1]. Subdivide the interval [–2.2, –2.1] into 10 equal subintervals: [–2.2, –2.19]; [–2.19, –2.18]; [–2.18, –2.17]; [–2.17, –2.16]; [–2.16, –2.15]; [–2.15, –2.14]; [–2.14, –2.13]; [–2.13, –2.12]; [–2.12, –2.11]; [–2.11, –2.1] f 2.2 0.944; f 2.19 0.6104 f 2.19 0.6104; f 2.18 0.2807 f 2.18 0.2807; f 2.17 0.0451
So f has a real zero on the interval [–2.18, –2.17], therefore r 2.17 , correct to two decimal places.
89.
f x x3 x 2 x 4 f 1 1; f 2 10
So f has a real zero on the interval [1,2]. Subdivide the interval [1,2] into 10 equal subintervals: [1,1.1]; [1.1,1.2]; [1.2,1.3]; [1.3,1.4]; [1.4,1.5]; [1.5,1.6]; [1.6,1.7]; [1.7,1.8]; [1.8,1.9]; [1.9,2] f 1 1; f 1.1 0.359
90.
f x 2 x4 x2 1 f 0 1; f 1 2
So f has a real zero on the interval [0,1]. Subdivide the interval [0,1] into 10 equal subintervals: [0,0.1]; [0.1,0.2]; [0.2,0.3]; [0.3,0.4]; [0.4,0.5]; [0.5,0.6]; [0.6,0.7]; [0.7,0.8]; [0.8,0.9]; [0.9,1] f 0 1; f 0.1 0.9898 f 0.1 0.9898; f 0.2 0.9568 f 0.2 0.9568; f 0.3 0.8938 f 0.3 0.8938; f 0.4 0.7888
f 0.4 0.7888; f 0.5 0.625 f 0.5 0.625; f 0.6 0.3808 f 0.6 0.3808; f 0.7 0.0298
f 0.7 0.0298; f 0.8 0.4592
So f has a real zero on the interval [0.7,0.8]. Subdivide the interval [0.7,0.8] into 10 equal subintervals: [0.7,0.71]; [0.71,0.72]; [0.72,0.73]; [0.73,0.74]; [0.74,0.75]; [0.75,0.76];[0.76,0.77]; [0.77,0.78]; [0.78,0.79]; [0.79,0.8] f 0.7 0.298; f 0.71 0.0123 So f has a real zero on the interval [0.7,0.71], therefore r 0.70 , correct to two decimal places.
f 1.1 0.359; f 1.2 0.368
So f has a real zero on the interval [1.1,1.2].
544 Copyright © 2025 Pearson Education, Inc.
Section 5.6: The Real Zeros of a Polynomial Function
91.
therefore r 2.53 , correct to two decimal places.
f x 2 x 4 3 x3 4 x 2 8 f 2 16; f 3 37
92.
So f has a real zero on the interval [2,3]. Subdivide the interval [2,3] into 10 equal subintervals: [2,2.1]; [2.1,2.2]; [2.2,2.3]; [2.3,2.4]; [2.4,2.5]; [2.5,2.6]; [2.6,2.7]; [2.7,2.8]; [2.8,2.9]; [2.9,3] f 2 16; f 2.1 14.5268
f 2 4; f 3 43
So f has a real zero on the interval [2,3]. Subdivide the interval [2,3] into 10 equal subintervals: [2,2.1]; [2.1,2.2]; [2.2,2.3]; [2.3,2.4]; [2.4,2.5]; [2.5,2.6]; [2.6,2.7]; [2.7,2.8]; [2.8,2.9]; [2.9,3] f 2 4; f 2.1 1.037
f 2.1 14.5268; f 2.2 12.4528
f 2.2 12.4528; f 2.3 9.6928 f 2.3 9.6928; f 2.4 6.1568
f 2.1 1.037; f 2.2 2.264
f 2.4 6.1568; f 2.5 1.75
So f has a real zero on the interval [2.1,2.2].
f 2.5 1.75; f 2.6 3.6272
Subdivide the interval [2.1,2.2] into 10 equal subintervals: [2.1,2.11]; [2.11,2.12]; [2.12,2.13]; [2.13,2.14]; [2.14,2.15]; [2.15,2.16];[2.16,2.17]; [2.17,2.18]; [2.18,2.19]; [2.19,2.2] f 2.1 1.037; f 2.11 0.7224
So f has a real zero on the interval [2.5,2.6]. Subdivide the interval [2.5,2.6] into 10 equal subintervals: [2.5,2.51]; [2.51,2.52]; [2.52,2.53]; [2.53,2.54]; [2.54,2.55]; [2.55,2.56];[2.56,2.57]; [2.57,2.58]; [2.58,2.59]; [2.59,2.6] f 2.5 1.75; f 2.51 1.2576
f 2.11 0.7224; f 2.12 0.4044 f 2.12 0.4044; f 2.13 0.0830
f 2.51 1.2576; f 2.52 0.7555
f 2.13 0.0830; f 2.14 0.2418
f 2.52 0.7555; f 2.53 0.2434
So f has a real zero on the interval [2.13,2.14], therefore r 2.13 , correct to two decimal places.
f 2.53 0.2434; f 2.54 0.2787
So f has a real zero on the interval [2.53,2.54],
93.
f x 3x3 2 x 2 20
f ( x) x3 2 x 2 5 x 6 ( x 3)( x 1)( x 2) x-intercepts: –3, –1, 2; Near 3 : f x x 3 3 1 3 2 10 x 3
Near 1 : f x 1 3 x 1 1 2 6 x 1 Near 2: f x 2 3 2 1 x 2 15 x 2 Plot the point 3, 0 and show a line with positive slope there. Plot the point 1, 0 and show a line with negative slope there. Plot the point 2, 0 and show a line with positive slope there. y-intercept: f 0 03 2 0 5 0 6 6 ; 2
The graph of f crosses the x-axis at x = –3, –1 and 2 since each zero has multiplicity 1. Interval
, 3
3, 1
1, 2
2,
Number Chosen Value of f Location of Graph Point on Graph
–4 –18
–2 4
0 –6
3 24
Below x-axis 4, 18
Above x-axis 2, 4
Below x-axis 0, 6
Above x-axis 3, 24
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Chapter 5: Polynomial and Rational Functions
94.
f ( x) x3 8 x 2 11x 20 ( x 5)( x 4)( x 1) x-intercepts: –5, –4, 1; Near 5 : f x x 5 5 4 5 1 6 x 5
Near 4 : f x 4 5 x 4 4 1 5 x 4 Near 1: f x 1 5 1 4 x 1 30 x 1 Plot the point 5, 0 and show a line with positive slope there. Plot the point 4, 0 and show a line with negative slope there. Plot the point 1, 0 and show a line with positive slope there. y-intercept: f 0 03 8 0 11 0 20 20 2
The graph of f crosses the x-axis at x = –5, –4 and 1 since each zero has multiplicity 1. Interval
, 5
5, 4
4,1
1,
Number Chosen Value of f Location of Graph Point on Graph
–6 –14
–4.5 1.375
0 –20
2 42
Below x-axis 6, 14
Above x-axis 4.5,1.375
Below x-axis 0, 20
Above x-axis 2, 42
546 Copyright © 2025 Pearson Education, Inc.
Section 5.6: The Real Zeros of a Polynomial Function
95.
1 f x 2 x3 x 2 2 x 1 x 2 x 2 2 2
x-intercept:
1 ; 2
Near
2 5 1 1 1 1 : f x x 2 2 x 2 2 2 2 2
1 Plot the point , 0 and show a line with positive slope there. 2
y-intercept: f 0 2 0 02 2 0 1 1 3
The graph of f crosses the x-axis at x = Interval Number Chosen Value of f Location of Graph Point on Graph
96.
1 since the zero has multiplicity 1. 2
1 , 2 0 –1
1 , 2 1 2
Below x-axis 0, 1
Above x-axis 1, 2
1 f ( x) 2 x3 x 2 2 x 1 x 2 x 2 2 2 1 x-intercept: 2 2 5 1 1 1 1 Near : f x x 2 2 x 2 2 2 2 2 1 Plot the point , 0 and show a line with positive slope there. 2
y-intercept: f 0 2 0 02 2 0 1 1 3
The graph of f crosses the x-axis at x =
1 since the zero has multiplicity 1. 2
547 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
Interval Number Chosen Value of f Location of Graph Point on Graph
97.
1 , 2 –1 –2
1 , 2 0 1
Below x-axis 1, 2
Above x-axis 0,1
f ( x) x 4 x 2 2 x 1 x 1 x 2 2
x-intercepts: –1, 1
Near 1 : f x x 1 1 1 1 2 6 x 1
2
Near 1: f x 1 1 x 1 12 2 6 x 1 Plot the point 1, 0 and show a line with negative slope there. Plot the point 1, 0 and show a line with positive slope there. y-intercept: f 0 04 02 2 2 The graph of f crosses the x-axis at x = –1 and 1 since each zero has multiplicity 1.
Interval
, 1
1,1
1,
Number Chosen Value of f Location of Graph Point on Graph
–2 18
0 –2
2 18
Above x-axis 2,18
Below x-axis 0, 2
Above x-axis 2,18
548 Copyright © 2025 Pearson Education, Inc.
Section 5.6: The Real Zeros of a Polynomial Function
98.
f x x 4 3 x 2 4 x 2 x 2 x 2 1
x-intercepts: –2, 2
Near 2 : f x x 2 2 2 2 1 20 x 2
2
Near 2: f x 2 2 x 2 22 1 20 x 2 Plot the point 2, 0 and show a line with negative slope there. Plot the point 2, 0 and show a line with positive slope there. y-intercept: f 0 04 3 0 4 4 2
The graph of f crosses the x-axis at x = –2 and 2 since each zero has multiplicity 1.
99.
Interval
, 2
2, 2
2,
Number Chosen Value of f Location of Graph Point on Graph
–3 50
0 –4
3 50
Above x-axis 3,50
Below x-axis 0, 4
Above x-axis 3,50
f x 4 x 4 7 x 2 2 2 x 1 2 x 1 x 2 2
1 1 x-intercepts: , 2 2 2 1 1 9 1 Near : f x 2 x 1 2 1 2 2 x 1 2 2 2 2
Near
1 2 9 1 1 : f x 2 1 2 x 1 2 2 x 1 2 2 2 2
1 Plot the point , 0 and show a line with negative slope there. 2 1 Plot the point , 0 and show a line with positive slope there. 2
y-intercept: f 0 4 0 7 0 2 2 4
2
The graph of f crosses the x-axis at x =
1 1 and since each zero has multiplicity 1. 2 2
549 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
Interval Number Chosen Value of f Location of Graph Point on Graph
100.
1 , 2 –1 9
1 1 , 2 2 0 –2
1 , 2 1 9
Above x-axis 1, 9
Below x-axis 0, 2
Above x-axis 1,9
f x 4 x 4 15 x 2 4 2 x 1 2 x 1 x 2 4
1 1 x-intercepts: , 2 2 2 1 1 17 1 Near : f x 2 x 1 2 1 4 2 x 1 2 2 2 2
Near
1 2 17 1 1 : f x 2 1 2 x 1 4 2 x 1 2 2 2 2
1 Plot the point , 0 and show a line with negative slope there. 2 1 Plot the point , 0 and show a line with positive slope there. 2
y-intercept: f 0 4 0 15 0 4 4 4
2
The graph of f crosses the x-axis at x =
Interval Number Chosen Value of f Location of Graph Point on Graph
1 1 and since each zero has multiplicity 1. 2 2
1 , 2 –1 15
1 1 , 2 2 0 –4
1 , 2 1 15
Above x-axis 1,15
Below x-axis 0, 4
Above x-axis 1,15
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Section 5.6: The Real Zeros of a Polynomial Function
101.
f ( x) x 4 x3 3x 2 x 2 ( x 2)( x 1)( x 1) 2 x-intercepts: –2, –1, 1
Near 2 : f x x 2 2 1 2 1 9 x 2 2
Near 1 : f x 1 2 x 1 1 1 4 x 1 2
Near 1: f x 1 2 1 1 x 1 6 x 1 2
2
Plot the point 2, 0 and show a line with negative slope there. Plot the point 1, 0 and show a line with positive slope there. Plot the point 1, 0 and show a parabola opening up there. y-intercept: f 0 04 03 3 0 0 2 2 2
The graph of f crosses the x-axis at x = –2 and –1 since each zero has multiplicity 1. The graph of f touches the x-axis at x = 1 since the zero has multiplicity 2.
102.
Interval
, 2
2, 1
1,1
1,
Number Chosen Value of f Location of Graph Point on Graph
–3 32
–1.5 –1.5625
0 2
2 12
Above x-axis 3, 32
1.5, 1.5625
Below x-axis
Above x-axis 0, 2
Above x-axis 2,12
f ( x) x 4 x3 6 x 2 4 x 8 ( x 2)( x 1)( x 2) 2 x-intercepts: –2, –1, 2
Near 2 : f x x 2 2 1 2 2 16 x 2 2
Near 1 : f x 1 2 x 1 1 2 9 x 1 2
Near 2: f x 2 2 2 1 x 2 12 x 2 2
2
551 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
Plot the point 2, 0 and show a line with negative slope there. Plot the point 1, 0 and show a line with positive slope there. Plot the point 2, 0 and show a parabola opening up there. y-intercept: f 0 04 03 6 0 4 0 8 8 2
The graph of f crosses the x-axis at x = –2 and –1 since each zero has multiplicity 1. The graph of f touches the x-axis at x = 2 since the zero has multiplicity 2.
103.
Interval
, 2
2, 1
1, 2
2,
Number Chosen Value of f Location of Graph Point on Graph
–3 50
–1.5 –3.0625
0 8
3 20
Above x-axis 3, 50
1.5, 3.0625
Below x-axis
Above x-axis 0,8
Above x-axis 3, 20
f ( x) 4 x5 8 x 4 x 2 ( x 2)
x-intercepts: Near
2 x 1 2 x 1 2 x 1 2
2 2 , ,2 2 2
2 2 2 : f x 2 2 1 2 2 2
2 2 Near : f x 2 2 2
Near 2: f x x 2
2 2 2 x 1 2 1 2 2
2 2 2 2 2 x 1 2 1 2 1 2 2
2 4 2 x 1
2 4 2 x 1
2 2 1 2 2 1 2 2 1 63 x 2 2
2 , 0 and show a line with positive slope there. Plot the point 2 2 Plot the point , 0 and show a line with negative slope there. 2 Plot the point 2, 0 and show a line with positive slope there.
y-intercept: f 0 4 0 8 0 0 2 2 5
4
The graph of f crosses the x-axis at x
2 2 and x 2 since each zero has multiplicity 1. , x 2 2
552 Copyright © 2025 Pearson Education, Inc.
Section 5.6: The Real Zeros of a Polynomial Function
2 , 2 –1 –9
2 2 , 2 2 0 2
2 , 2 2 1 –3
Below x-axis 1, 9
Above x-axis 0, 2
Below x-axis 1, 3
Interval Number Chosen Value of f Location of Graph Point on Graph
104.
f ( x) 4 x5 12 x 4 x 3 ( x 3)
x-intercepts: 3,
3 323 Above x-axis 3,323
2 x 1 2 x 1 2 x 1 2
2 2 , 2 2
Near 3 : f x x 3 Near
2,
2 3 1 2 3 1 2 3 1 323 x 3 2
2 2 2 : f x 3 2 1 2 2 2
2 2 Near : f x 3 2 2
2 2 2 x 1 2 1 2 2
2 2 2 2 2 x 1 2 1 2 1 2 2
2 6 2 x 1
2 6 2 x 1
Plot the point 3, 0 and show a line with positive slope there.
2 Plot the point , 0 and show a line with negative slope there. 2 2 Plot the point , 0 and show a line with positive slope there. 2
y-intercept: f 0 4 0 12 0 0 3 3 5
4
The graph of f crosses the x-axis at x
2 2 and x 3 since each zero has multiplicity 1. , x 2 2
553 Copyright © 2025 Pearson Education, Inc.
Chapter 5: Polynomial and Rational Functions
105.
–4 –1023
2 3, 2 –2 63
2 2 , 2 2 0 –3
2 , 2 1 12
Below x-axis 4, 1023
Above x-axis 2, 63
Below x-axis 0, 3
Above x-axis 1,12
Interval
, 3
Number Chosen Value of f Location of Graph Point on Graph
f ( x) 3 x 3 16 x 2 3 x 10 will factor into
f ( x) ( x 1)(3x 2)( x 5) . Solving ( x 1)(3x 2)( x 5) 0 we get 2 , x 5 . f ( x 3) would shift the 3 graph left by three units and thus would shift the zeros left by three units. So the zeros would be x 1 3 4 2 7 x 3 3 3 x 5 3 8 x 1, x
106.
107. From the Remainder and Factor Theorems, x 2 is a factor of f if f 2 0 .
2 3 k 2 2 k 2 2 0 8 4k 2k 2 0 2k 10 0 2k 10 k 5
108. From the Remainder and Factor Theorems, x 2 is a factor of f if f 2 0 .
2 4 k 2 3 k 2 2 1 0
f ( x) 4 x3 11x 2 26 x 24 will factor into
16 8k 4k 1 0
f ( x) ( x 2)(4 x 3)( x 4) . Solving ( x 2)(4 x 3)( x 4) 0 we get 3 , x 4 . f ( x 2) would shift the 4 graph right by two units and thus would shift the zeros right by two units. So the zeros would be x 2 2 0 3 11 x 2 4 4 x 42 6
12k 17 0 12k 17 k
x 2, x
17 12
109. From the Remainder Theorem, we know that the remainder is f 1 2 1
20
8 1 1 2 2 8 1 2 7 10
The remainder is –7.
110. From the Remainder Theorem, we know that the remainder is 554 Copyright © 2025 Pearson Education, Inc.
Section 5.6: The Real Zeros of a Polynomial Function f 1 3 1 1 1 2 1 1 17
9
5
The remainder is 1.
111. We want to prove that x c is a factor of x n c n , for any positive integer n. By the Factor Theorem, x c will be a factor of f x provided f c 0 . Here, f x x n c n , so that f c c n c n 0 . Therefore, x c is a factor of x n c n .
112. We want to prove that x c is a factor of x n c n , if n 1 is an odd integer. By the Factor Theorem, x c will be a factor of f x provided f c 0 . Here, f x x c , so n
n
that f c c c n c n c n 0 if n 1 n
is an odd integer. Therefore, x c is a factor of x n c n if n 1 is an odd integer.
113. x3 8 x 2 16 x 3 0 has solution x 3 , so x 3 is a factor of f x x3 8 x 2 16 x 3 . Using synthetic division 3 1 8 16 3 3 15 3 1 5 1 0 Thus,
f x x 3 8 x 2 16 x 3 x 3 x 2 5 x 1 .
Solving x 2 5 x 1 0 5 25 4 5 21 2 2 The sum of these two roots is 5 21 5 21 10 5. 2 2 2 x
x3 x 2 294 x3 x 2 294 0 The solutions to this equation are the same as the real zeros of f x x3 x 2 294 .
By Descartes’ Rule of Signs, we know that there is one positive real zero. p 1, 2, 3, 6, 7, 14, 21, 42, 49, 98, 147, 294 q 1 The possible rational zeros are the same as the values for p. p 1, 2, 3, 6, 7, 14, 21, 42, 49, 98, q 147, 294 Using synthetic division: 71 1 0 294 7 42 294 1 6 42 0 7 is a zero, so the length of the edge of the original cube was 7 inches.
x 18x 72 ( x 4) 2 x 2
3
x3 14 x 2 288 2 x3
f x x3 5 x 2 5 x 2 x 2 x 2 3x 1 .
Solving x 2 3x 1 0 ,
115. Let x be the length of a side of the original cube. After removing the 1-inch slice, one dimension will be x 1 . The volume of the new solid will be: ( x 1) x x . Solve the volume equation: ( x 1) x x 294
116. Let x be the length of a side of the original 3 cube. The volume is x . The dimensions are changed to x 6, x 12, and x 4 . The volume of the new solid will be (x 6)(x 12)(x 4) . Solve the volume equation: ( x 6)( x 12)( x 4) 2 x3
114. x3 5 x 2 5 x 2 0 has solution x 2 , so x 2 is a factor of f x x3 5 x 2 5 x 2 . Using synthetic division 2 1 5 5 2 2 6 2 1 3 1 0 Thus,
3 9 4 3 13 . 2 2 The sum of these two roots is 3 13 3 13 6 3 . 2 2 2 x
x3 14 x 2 288 0 The solutions to this equation are the same as the real zeros of f x x3 14 x 2 288 .
By Descartes’ Rule of Signs, we know that there are two positive real zeros or no positive real zeros.
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Chapter 5: Polynomial and Rational Functions p 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 72, 96, 144, 288 q 1 The possible rational zeros are the same as the values for p: p 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, q 32, 36, 48, 72, 96, 144, 288 Using synthetic division: 61 14 0 288 6 48 288 1 8 48 0 Therefore, 6 is a zero; the other factor is 2 x 8x 48 x 12 x 4 . The other zeros are 12 and –4. The length of the edge of the original cube was 6 cm or 12 cm.
118.
f x 8x4 2 x2 5x 1
f x x n an 1 x n 1 an 2 x n 2 ... a1 x a0 ;
117.
where an 1 , an 2 ,...a1 , a0 are integers. If r is a real zero of f , then r is either rational or irrational. We know that the rational roots of f must be of the form
p where p is a divisor of q
a0 and q is a divisor of 1. This means that p p . q Therefore, r is an integer or r is irrational. q 1 . So if r is rational, then r
0 r 1
We begin with the interval [0,1]. f 0 1; f 1 10 Let mi = the midpoint of the interval being considered. So m1 0.5 n mn f mn
New interval
1
0.5
f 0.5 1.5 0
[0,0.5]
2
0.25
f 0.25 0.15625 0
[0,0.25]
3
0.125
f 0.125 0.4043 0
[0.125,0.25]
4
0.1875
f 0.1875 0.1229 0
[0.1875,0.25]
5
0.21875
f 0.21875 0.0164 0
[0.1875,0.21875]
6
0.203125
f 0.203125 0.0533 0
[0.203125,0.21875]
7
0.2109375
f 0.2109375 0.0185 0
[0.2109375,0.21875]
8
0.21484375
f 0.21484375 0.00105 0
[0.21484375,0.21875]
9
0.216796875
f 0.216796875 0.007655271 0
[0.21484375,0.2167968]
10
0.2158203125
f 0.216796875 0.007655271 0
[0.21484375,0.21582031]
11
0.21533203125
f 0.216796875 0.007655271 0
[0.21484375,0.21533203]
Since the endpoints of the new interval at Step 11 agree to three decimal places, r = 0.215, correct to three decimal places.
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Section 5.6: The Real Zeros of a Polynomial Function 119. b p be a rational zero of the polynomial f ( x ) = an x n + an -1 x n -1 + an - 2 x n - 2 + ... + a1 x + a0 where q an , an -1 , an - 2 ,...a1 , a0 are integers . Suppose also that p and q have no common factors other than 1 and – 1. Then
120. Let
n
æ pö æ pö æ pö f ç ÷ = an ç ÷ + an -1 ç ÷ èqø èqø èqø Þ
1 qn
(a p + a p q + a n
n
n -1
n -1
n -1
æ pö + an - 2 ç ÷ èqø
n-2 p
n-2
æ pö + ... + a1 ç ÷ + a0 = 0 èqø
)
n-2 2
q + ... + a1 pq n -1 + a0 q n = 0
Þ an p n + an -1 p n -1q + an - 2 p n - 2 q 2 + ... + a1 pq n -1 + a0 q n = 0 Þ an p n + an -1 p n -1q + an - 2 p n - 2 q 2 + ... + a1 pq n -1 = -a0 q n
Because p is a factor of the left side of this equation, p must also be a factor of a0 q n . Since p is not a factor of q, p must be a factor of a0 . Similarly, q must be a factor of an 121. Let y1 = x3 , y2 = 1 - x 2 , and
124.
f ( x ) = y1 - y2 = ( x3 ) - (1 - x 2 ) = x3 + x 2 - 1.
Note that f ( x) = 0 where y1 = y2 , that is, where y1 and y2 intersect. Since f is a polynomial function and f (0) = -1 and f (1) = 1 are of opposite sign, the Intermediate Value Theorem guarantees there is at least one real number c in the interval (0, 1) where f (c) = 0 . That is there is at least one real number c in the interval where the graphs of y1 = x3 and y2 = 1 - x 2 intersect. 122.
1 is not in the list of possible rational 3 zeros, it is not a zero of f .
123.
3 is not in the list of possible rational 5 zeros, it is not a zero of f (x).
Since
125.
f ( x ) = 2 x3 + 3x 2 - 6 x + 7 By the Rational Zero Theorem, the only possible p 1 7 rational zeros are: = ±1, ±7, ± , ± . q 2 2
Since
f ( x ) = 2 x 6 - 5 x 4 + x3 - x + 1 By the Rational Zero Theorem, the only possible p 1 rational zeros are: = ±1, ± . q 2
f ( x ) = x 7 + 6 x5 - x 4 + x + 2 By the Rational Zero Theorem, the only possible p rational zeros are: = ±1, ±2 . q 2 is not in the list of possible rational 3 zeros, it is not a zero of f .
Since
126.
f ( x) = -3 x 2 + 30 x - 4
( ) = -3 ( x - 10 x + 25 ) - 4 + 75 = -3 ( x - 5 ) + 71 = -3 x 2 - 10 x - 4 2
f ( x ) = 4 x3 - 5 x 2 - 3x + 1 By the Rational Zero Theorem, the only possible
rational zeros are:
p 1 1 = ±1, ± , ± . q 2 4
1 is not in the list of possible rational 3 zeros, it is not a zero of f .
Since
2
127.
[ 3,8)
128. 2 x - 5 y = 3 -5 y = -2 x + 3 2 3 y = x5 5
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2
Chapter 5: Polynomial and Rational Functions
1 129. a , b 2, c 9 3
4.
25 4(1) 29
1 (2) (2)2 4 (9) 3 x 1 2 3
5. one 6. 3 4i
7. True
2 4 12 2 8 2 2 3 3
8. False; would also need 3 5i 9. Since complex zeros appear in conjugate pairs, 4 i , the conjugate of 4 i , is the remaining zero of f .
2 2i 2 3 3i 2 2 3
130.
3, 2 and 5,
131.
, 3 and 2,5
5 2i 5 2i 25 10i 10i 4i 2
10. Since complex zeros appear in conjugate pairs, 3 i , the conjugate of 3 i , is the remaining zero of f . 11. Since complex zeros appear in conjugate pairs, i , the conjugate of i , and 3 i , the conjugate of 3 i , are the remaining zeros of f .
132. -5 and -1
12. Since complex zeros appear in conjugate pairs, 2 i , the conjugate of 2 i , is the remaining zero of f .
133. (5, 0), (1, 0), (0,3) 134. (3, 2), (2, 6), and (5,1)
13. Since complex zeros appear in conjugate pairs, i , the conjugate of i , and 5i , the conjugate of 5i , are the remaining zeros of f .
135. Absolute minimum: f (3) 2 ; no absolute maximum
14. Since complex zeros appear in conjugate pairs, i , the conjugate of i , is the remaining zero. 15. Since complex zeros appear in conjugate pairs, i , the conjugate of i , is the remaining zero.
Section 5.7 1.
16. Since complex zeros appear in conjugate pairs, 2 i , the conjugate of 2 i , and i , the conjugate of i , are the remaining zeros of f .
3 2i 3 5i 3 3 2i 5i 3i
3 2i 3 5i 9 15i 6i 10i 2 9 21i 10 1
17. Since complex zeros appear in conjugate pairs, 4 9i , the conjugate of 4 9i , and 7 2i , the conjugate of 7 2i , are the remaining zeros.
1 21i
2. The zeros of f x are the solutions to the
18. Since complex zeros appear in conjugate pairs, i , the conjugate of i , 3 2i , the conjugate of 3 2i , and 2 i , the conjugate of 2 i , are the remaining zeros of f .
2
equation x 2 x 2 0 . x2 2 x 2 0 a 1, b 2, c 2 x
2 22 4 1 2 2 1
2 4 2 2i 1 i 2 2
For 19–24, we will use an 1 as the lead coefficient of the polynomial. Also note that
x a b i x a b i x a b i x a b i 2 2 x a b i
The solution set is {1 i, 1 i} .
3. 3 4i 558
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Section 5.7: Complex Zeros; Fundamental Theorem of Algebra
19. Since 3 2i is a zero, its conjugate 3 2i is also a zero of f .
20. Since 1 2i and i are zeros, their conjugates 1 2i and i are also zeros of f .
f ( x) ( x 4)( x 4) x (3 2i ) x (3 2i )
f ( x) ( x i )( x (i )) x (1 2i ) x (1 2i )
x 8 x 16 x 6 x 9 4i x 8 x 16 x 6 x 13
( x i )( x i ) ( x 1) 2i ( x 1) 2i
x 8 x 16 ( x 3) 2i ( x 3) 2i 2 2
2
2
2
x 2 x 1 4i x 1 x 2 x 5 x2 i 2
2
2
2
2
2
x 4 2 x3 5 x 2 1x 2 2 x 5
x 4 6 x 3 13 x 2 8 x3 48 x 2
104 x 16 x 2 96 x 208
x 4 2 x3 6 x 2 2 x 5
x 4 14 x 3 77 x 2 200 x 208
21. Since i is a zero, its conjugate i is also a zero, and since 1 i is a zero, its conjugate 1 i is also a zero of f . f ( x) ( x 2)( x i )( x i ) x (1 i ) x (1 i )
( x 2) x 1 x 2 x 1 i x 2 x x 2 x 2 x 2
( x 2) x 2 i 2 ( x 1) i ( x 1) i 2
3
2
2
2
2
x5 2 x 4 2 x3 2 x 4 4 x3 4 x 2 x3 2 x 2 2 x 2 x 2 4 x 4 x5 4 x 4 7 x3 8 x 2 6 x 4 22. Since i is a zero, its conjugate i is also a zero; since 4 i is a zero, its conjugate 4 i is also a zero; and since 2 i is a zero, its conjugate 2 i is also a zero of f . f ( x) ( x i )( x i ) x (4 i ) x (4 i ) x (2 i ) x (2 i )
( x 4) i ( x 4) i ( x 2) i ( x 2) i x 1 x 8 x 16 i x 4 x 4 i x 1 x 8 x 17 x 4 x 5 x 8 x 17 x x 8 x 17 x 4 x 5 x 8 x 18 x 8 x 17 x 4 x 5 2
x i
2
2
2
2
2
2
2
2
2
4
3
2
4
3
2
2
2
2
x 6 4 x 5 5 x 4 8 x 5 32 x 4 40 x3 18 x 4 72 x3 90 x 2 8 x3 32 x 2 40 x 17 x 2 68 x 85
x 6 12 x 5 55 x 4 120 x3 139 x 2 108 x 85
23. Since i is a zero, its conjugate i is also a zero. f ( x) ( x 3)( x 3)( x i ) x i
x 6 x 9 x 1 x2 6 x 9 x2 i 2 2
2
x 4 x 2 6 x3 6 x 9 x 2 9 x 4 6 x3 10 x 2 6 x 9
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Chapter 5: Polynomial and Rational Functions
24. Since 1 i is a zero, its conjugate 1 i is also a zero of f .
4 x2 7 x 2 x 2 16 4 x 4 7 x3 62 x 2 112 x 32
f ( x) ( x 1) x (1 i ) x (1 i ) 3
4x4
x 3 x 3x 1 x 2 x 1 i x 3 x 3x 1 x 2 x 2
x 3 x 3x 1 ( x 1) i ( x 1) i 3
2
3
2
2
3
2
2
64 x 2 7 x3 2 x 2 112 x
2
7 x3
3 x3 6 x 2 6 x x 2 2 x 2 4
3
1 and 2 . 4 1 The zeros of f are 4i, 4i, 2, . 4
Thus, ( x 3i )( x 3i ) x 2 9 is a factor of f . Using division to find the other factor:
28. Since 3i is a zero, its conjugate 3i is also a zero of h . x 3 i and x 3 i are factors of h .
x5 x 2 9 x3 5 x 2 9 x 45
Thus, ( x 3i )( x 3i ) x 2 9 is a factor of h . Using division to find the other factor:
9x 2
45
5x2
45
5x
3x 2 5 x 2 x 2 9 3x 4 5 x3 25 x 2 45 x 18 3x 4
x 5 is a factor, so the remaining zero is 4. The zeros of f are 5, 3i, 3i .
27 x 2 5 x3 2 x 2 45 x 5 x3
26. Since 5i is a zero, its conjugate 5i is also a zero of g . x 5 i and x 5 i are factors of g . 2
Thus, ( x 5 i )( x 5 i ) x 25 is a factor of g . Using division to find the other factor:
45 x 2 x2
18
2
18
2x
3x 2 5 x 2 (3 x 1)( x 2)
x3 x 2 25 x3 3x 2 25 x 75 x3
1 and 2 . 3 1 The zeros of h are 3i, 3i, 2, . 3 The remaining zeros are
25 x 3x 2
75
2
75
3x
32
The remaining zeros are
25. Since 3i is a zero, its conjugate 3i is also a zero of f . x 3i and x 3i are factors of f .
x
32
2
4 x 2 7 x 2 (4 x 1)( x 2)
2
x 5 x 11x 13x 8 x 2
3
2 x2 2x
x5 2 x 4 2 x3 3 x 4 6 x3 6 x 2 5
112 x
29. Since 2 5i is a zero, its conjugate 2 5i is also a zero of h . x (2 5i ) and x (2 5i ) are factors of h . Thus,
x 3 is a factor, so the remaining zero is –3. The zeros of g are –3, 5 i, 5 i .
27. Since 4i is a zero, its conjugate 4i is also a zero of f . x 4i and x 4i are factors of f .
( x (2 5i ))( x (2 5i )) (( x 2) 5i )(( x 2) 5i )
x 2 4 x 4 25i 2
2
Thus, ( x 4i )( x 4i ) x 16 is a factor of f . Using division to find the other factor:
x 2 4 x 29
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Section 5.7: Complex Zeros; Fundamental Theorem of Algebra
is a factor of h . Using division to find the other factor:
3
2
3 x 2 x 21x 14 2
5
x 4
4
4 x 2 3x 24
3x
5
12 x
x 2 4 x 29 4 x 4 7 x3 23 x 2 15 x 522 4
3
4 x 4 x 29 x
2
3
2 x 21x 6 x 4
2
2x
3x3 6 x 2 15 x 3
3
3 x 2 x 9 x 6 x 84 x 56
3
4
8x
2
2
3
2
21x 14 x 84 x
2
3 x 12 x 87 x
21x3
2
18 x 72 x 522 2
18 x 72 x 522
12 x
14 x 2
56
2
56
14 x
2
x 3 x 18 ( x 3)( x 6) The remaining zeros are –3 and 6. The zeros of h are 2 5i, 2 5i, 3, 6 .
30. Since 1 3i is a zero, its conjugate 1 3i is also a zero of f . x (1 3 i ) and x (1 3 i ) are factors of f . Thus, ( x (1 3i ))( x (1 3i )) (( x 1) 3i )(( x 1) 3i ) is a factor of f . (( x 1) 3i )(( x 1) 3i ) x 2 2 x 1 9i 2 x 2 2 x 10 Using division to find the other factor: x2 5x 6 x 2 2 x 10 x 4 7 x3 14 x 2 38 x 60
3x3 2 x 2 21x 14 x 2 (3x 2) 7(3 x 2) (3 x 2)( x 2 7)
x 7
(3 x 2) x 7
2 The remaining zeros are , 7, and 7 . 3 2 The zeros of h are 2i, 2i, 7, 7, . 3
32. Since 3 i is a zero, its conjugate 3i is also a zero of g . x 3 i and x 3 i are factors of g . Thus, ( x 3 i )( x 3 i ) x 2 9 is a factor of g . Using division to find the other factor: 2 x3 3x 2 23 x 12
x 4 2 x3 10 x 2 5 x3 4 x 2 38 x
x 2 9 2 x 5 3 x 4 5 x3 15 x 2 207 x 108
2 x5
5 x3 10 x 2 50 x
18 x3
6 x 2 12 x 60
3x 4 23x3 15 x 2
6 x 2 12 x 60
3 x 4
27 x 2 23x3 12 x 2 207 x
x 2 5 x 6 ( x 1)( x 6) The remaining zeros are –1 and 6. The zeros of f are 1 3i, 1 3i, 1, 6 .
31. Since 2i is a zero, its conjugate 2i is also a zero of h . x 2i and x 2i are factors of h . Thus, ( x 2i )( x 2i ) x 2 4 is a factor of h . Using division to find the other factor:
23x3
207 x 12 x 2
108
2
108
12 x
0
Using the Rational Root theorem, we see that 3 is a potential rational zero. 3 2 3
23
12
6
27
12
9
4
0
2
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Chapter 5: Polynomial and Rational Functions x 3 is a factor. The remaining factor is
2 1 8
2 x 2 9 x 4 (2 x 1)( x 4) .
2 12
1 The zeros of g are 3i, 3i, 3, , 4 . 2
33.
1 3 2
x
1 3 1 3 i and i 2 2 2 2
36.
f ( x) x3 13 x 2 57 x 85
f ( x ) has 3 complex zeros.
Step 1:
Step 2: By Descartes Rule of Signs, there are no positive real zeros.
2
The solutions of x 1 0 are x i . The zeros are: 1, 1, i, i .
f ( x) ( x)3 13( x) 2 57( x) 85 , thus,
f x x 1 x 1 x i x i
x 3 13x 2 57 x 85 there are three negative real zeros or there is one negative real zero.
f ( x) x3 8 x 2 25 x 26
Step 1:
Step 3:
Possible rational zeros:
p 1, 5, 17, 85; q 1;
f ( x ) has 3 complex zeros.
p 1, 5, 17, 85 q
Step 2: By Descartes Rule of Signs, there are three positive real zeros or there is one positive real zero.
Step 4:
f ( x) ( x)3 8( x) 2 25( x) 26 , thus,
We try x 5 :
3
2
x 8 x 25 x 26 there are no negative real zeros. Step 3:
5 1
Using synthetic division:
13
57
85
5 40 85
Possible rational zeros:
1
p 1, 2, 13, 26; q 1;
8
17
0
x 5 is a factor. The other factor is the
p 1, 2, 13, 26 q
Step 4:
.
f x x 2 x 3 2i x 3 2i
( x 1)( x 1) x 2 1
35.
2(1)
6 16 6 4i 3 2i 2 2 The zeros are 2, 3 2i, 3 2i .
f ( x) x 4 1 x 2 1 x 2 1
( 6) ( 6) 2 4(1)(13)
1 3 1 3 The zeros are: 1, i, i. 2 2 2 2 1 3 1 3 f x x 1 x i x i 2 2 2 2
34.
0
The solutions of x 2 6 x 13 0 are:
2
2 1
13
26
x 2 is a factor. The other factor is the quotient: x 2 6 x 13 .
of x 2 x 1 0 are: 1 1 4 1 1
6
1
f ( x) x3 1 ( x 1) x 2 x 1 The solutions
x
25 26
quotient: x 2 8 x 17 . The solutions of x 2 8 x 17 0 are:
Using synthetic division:
We try x 2 :
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Section 5.7: Complex Zeros; Fundamental Theorem of Algebra
x
8 82 4 117
2 1
x3 x 2 25 x 25 x 2 ( x 1) 25( x 1)
8 4
( x 1) x 2 25
2
8 2i 4i 2 The zeros are 5, 4 i, 4 i .
( x 1)( x 5i )( x 5i )
The zeros are 3, 1, 5 i, 5 i . f x x 3 x 1 x 5i x 5i
f x x 5 x 4 i x 4 i
37.
40.
f ( x) x 4 5 x 2 4 x 2 4 x 2 1
( x 2i )( x 2i )( x i )( x i )
f ( x) x 4 13 x 2 36 x 2 4 x 2 9
f ( x) x 4 3 x3 19 x 2 27 x 252
( x 2i )( x 2i )( x 3i )( x 3i )
Step 2: By Descartes Rule of Signs, there are three positive real zeros or there is one positive real zero. 4
f x x 3i x 2i x 2i x 3i f ( x) x 4 2 x3 22 x 2 50 x 75
f ( x ) has 4 complex zeros.
Step 1:
Step 2: By Descartes Rule of Signs, there is 1 positive real zero. f ( x) ( x) 4 2( x)3 22( x)2 50( x) 75 4
3
2
x 2 x 22 x 50 x 75
Thus, there are three negative real zeros or there is one negative real zero. Step 3:
Possible rational zeros:
p 1, 3, 5, 15, 25, 75; q 1; p 1, 3, 5, 15, 25, 75 q
Step 4:
Using synthetic division:
We try x 3 : 3 1
3
2
f ( x) ( x ) 3( x) 19( x ) 27( x) 252 4
3
2
x 3 x 19 x 27 x 252
The zeros are: 3 i, 2 i, 2 i, 3 i .
39.
f ( x ) has 4 complex zeros.
Step 1:
The zeros are: 2i, i, i, 2i .
38.
Thus, there is 1 negative real zero. Step 3:
Possible rational zeros:
p 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252; q 1;
The possible rational zeros are the same as the values of p . Step 4:
Using synthetic division:
We try x 7 : 7 1
3 19
27
252
7
28
63
252
1 4
9
36
0
x 7 is a factor. The other factor is the quotient: x3 4 x 2 9 x 36 x 2 ( x 4) 9( x 4)
2 22
50 75
3
3
75
75
1 1
25
25
0
2
( x 4) x 9
( x 4)( x 3 i )( x 3 i )
The zeros are 7, 4, 3 i, 3 i .
x 3 is a factor. The other factor is the quotient: x3 x 2 25 x 25 .
f x x 7 x 4 x 3i x 3i
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.
Chapter 5: Polynomial and Rational Functions
41.
The solutions of x 2 4 x 13 0 are:
f ( x) 3x 4 x3 9 x 2 159 x 52 f ( x ) has 4 complex zeros.
x
Step 2: By Descartes Rule of Signs, there are three positive real zeros or there is one positive real zero.
Step 1:
3
2
Thus, there is 1 negative real zero. Possible rational zeros:
42.
p 1, 2, 4, 13, 26, 52;
4 36 4 6i 2 3i 2 2
f ( x) 2 x 4 x3 35 x 2 113 x 65
f ( x ) has 4 complex zeros.
Step 1:
q 1, 3;
Step 2: By Descartes Rule of Signs, there are two positive real zeros or no positive real zeros.
p 1, 2, 4, 13, 26, 52, q 1 2 4 13 26 52 , , , , , 3 3 3 3 3 3
Step 4:
.
The zeros are 4,
3 x x 9 x 159 x 52
Step 3:
2(1)
1 , 2 3 i, 2 3 i . 3 1 f x 3 x 4 x x 2 3i x 2 3i 3
f ( x ) 3( x ) 4 ( x )3 9( x) 2 159( x ) 52 4
( 4) ( 4) 2 4(1)(13)
f ( x) 2( x) 4 ( x )3 35( x) 2 113( x) 65 2 x 4 x3 35 x 2 113 x 65
Using synthetic division:
Thus, there are two negative real zeros or no negative real zeros.
We try x 4 : 1 9
159 52
Step 3:
12
52
172
52
p 1, 5, 13, 65; q 1, 2;
3 13
43
13
0
p 1 5 13 65 1, 5, 13, 65, , , , q 2 2 2 2
4 3
x 4 is a factor and the quotient is 3x3 13 x 2 43x 13 .
We try x
Step 4:
12
39
Using synthetic division:
We try x 5 :
1 on 3x3 13 x 2 43x 13 : 3
5 2
1 3 13 43 13 3 1 4 13 3
Possible rational zeros:
1 35 113 10
55
100 65
2 11
20
13
2 x 3 11x 2 20 x 13
1 is a factor and the quotient is 3 3x 2 12 x 39 . x
0
x 5 is a factor and the quotient is
0
3x 2 12 x 39 3 x 2 4 x 13
65
We try x
1 on 2 x 3 11x 2 20 x 13 : 2
1 2 11 2 1
20 13
2 12
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6
13
26
0
Section 5.7: Complex Zeros; Fundamental Theorem of Algebra
g ( x) a( x 2 1)( x 2 4)
1 is a factor and the quotient is 2 2 x 2 12 x 26 . x
2 x 2 12 x 26 2 x 2 6 x 13
4 a(02 1)(02 4) 4 a(1)(4)
a 1 g ( x) ( x 2 1)( x 2 4)
The solutions of x 2 6 x 13 0 are: x
6 62 4(1)(13) 2(1) 6 16 6 4i 3 2 i 2 2
1 , 3 2 i, 3 2 i . 2 1 f x 2 x 5 x x 3 2i x 3 2i 2
f ( x) 2 x3 14 x 2 bx 3 3
2
0 2(2) 14(2) b(2) 3 0 16 26 2b 3 43 2 so
( f g )(1) f (1) g (1) 15 (10) 25
45. a.
f ( x) x 4 1 x4 2 x2 1 2 x2
x 1 2 x x 1 2 x x 2 x 1 x 2 x 1 2 2 4(1)(1) b. x 2
x2 1 2 x2 2
2
2
2
2
b
2(1)
43 x3 2 g ( x) x3 cx 2 8 x 30 0 (3 i )3 (3 i ) 2 c 8(3 i ) 30 0 (18 26i ) (8 6i )c (24 8i ) 30 0 18 26i 8 6ic 24 8i 30 0 18 24 30 8c 26i 6ic 8i c 3 3
2 2 2 2 2 2 2 x , 2 2 2 2 and
f ( x) 2 x3 14 x 2
x
2
2 4(1)(1) 2
2(1)
2 2 2 2 2 2 2 x , 2 2 2 2 The zeros are 2 2 2 2 2 2 2 2 i, i, i, i 2 2 2 2 2 2 2 2
2
So g ( x) x 3 x 8 x 30 43 13 f (1) 2 14 3 2 2 g (1) 1 3 8 30 20 13 ( f g )(1) 20 130 2
44.
f (1) 15 g (1) 10 so
.
The zeros are 5,
43.
( x 4 5 x 2 4)
f ( x) ( x (3 i ))( x (3 i ))( x 2)( x 2) ( x 3 i )( x 3 i )( x 2)( x 2) x 4 6 x3 6 x 2 24 x 40 g ( x) a( x i )( x i )( x 2i )( x 2i )
46. If the coefficients are real numbers and 2 i is a zero, then 2 i would also be a zero. This would then require a polynomial of degree 4. 47. Three zeros are given. If the coefficients are real numbers, then the complex zeros would also have their conjugates as zeros. This would mean
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Chapter 5: Polynomial and Rational Functions
that there are 5 zeros, which would require a polynomial of degree 5.
54. A r 2 (3) 2
48. If the coefficients are real numbers, then complex zeros must appear in conjugate pairs. We have a conjugate pair and one real zero. Thus, there is only one remaining zero, and it must be real because a complex zero would require a pair of complex conjugates.
9 ft 2 28.274 ft 2 C 2 r 2 (3) 6 ft 18.850 ft x 3x 2 3x 2 x 3x 2 x 1 x 1 x 1 x We cannot use any values of x that makes any denominator zero so the domain is: x | x 1, x 0
49. One of the remaining zeros must be 4 i , the conjugate of 4 i . The third zero is a real number. Thus, the fourth zero must also be a real number in order to have a degree 4 polynomial.
55. ( f / g )( x)
50. a. f ( x) (3 i )2 2(3 i )i 10 (8 6i ) 6i 2 10 0
56.
x x5
y 3 5 y3
x 5 y 3 2 y x 5 3
2
2
b. f ( x) (3 i ) 2(3 i )i 10 (8 6i ) 6i 2 10 12i 4 0
57. The domain of the radical must be nonnegative and the domain of the denominator cannot be zero. So the domain is: x | x 0 or 0,
c. The coefficients are not real numbers. 51.
58. 3(0) y 2 12 y 2 12 y 12 2 3
The y-intercepts are : 0, 2 3 , 0, 2 3 . 3x 02 12 3 x 12
52.
x4 The x-intercept is: 4, 0
3 x 5 3 x 25 x 22
59.
53. (2 x 5)(3 x 2 x 4) (2 x)3 x 2 (2 x) x (2 x)4
x 3 x 3 x 3 x 3 x 9 x7 x 3 x 7 x 3
5(3x ) 5( x) 5( 4) 2
6 x3 2 x 2 8 x 15 x 2 5 x 20 6 x3 13 x 2 13 x 20
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x9
x 7 x 3
Chapter 5 Review Exercises 60.
6.
f ( x) ( x 1) 4
Using the graph of y x 4 , shift right 1 unit, then reflect about the x-axis.
f ( x h) f ( x) (( x h)3 8) ( x3 8) h (( x h)3 8) ( x3 8) h 3 2 x 3 x h 3xh 2 h3 8 x3 8 h 2 2 3 3 x h 3xh h h 2 h(3 x 3xh h 2 ) 3x 2 3 xh h 2 h
7.
f ( x) ( x 1) 4 2
Using the graph of y x 4 , shift right 1 unit, then shift up 2 units.
Chapter 5 Review Exercises 1.
2.
3.
f ( x) 4 x5 3 x 2 5 x 2 is a polynomial of degree 5. 3 x5 is a rational function. It is not a 2x 1 polynomial because there are variables in the denominator. f ( x)
f ( x) 3 x 2 5 x1/ 2 1 is not a polynomial
because the variable x is raised to the
1 power, 2
8.
f ( x) x( x 2)( x 4)
Step 1:
Degree is 3. The function resembles y x3 for large values of x .
Step 2:
y-intercept: f 0 0 0 2 0 4 0
which is not a nonnegative integer. 4.
f ( x) 3 is a polynomial of degree 0.
5.
f ( x) ( x 2)3
x-intercepts: solve f ( x) 0 : x( x 2)( x 4) 0 x 0 or x 2 or x 4
Using the graph of y x3 , shift left 2 units. Step 3:
The graph crosses the x-axis at x = 4 , x = 2 and x = 0 since each zero has multiplicity 1.
Step 4:
The polynomial is of degree 3 so the graph has at most 3 1 2 turning points.
Step 5:
f ( 5) 15; f ( 3) 3; f ( 1) 3; f (1) 15
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Chapter 5 Polynomial and Rational Functions
Step 4: Step 5:
9.
f ( x) ( x 2) 2 ( x 4)
Step 1:
Degree is 3. The function resembles y x3 for large values of x .
Step 2:
y-intercept: f 0 0 2 0 4 16 2
x-intercepts: solve f ( x) 0 : Step 3:
Step 4:
Step 5:
( x 2) 2 ( x 4) 0 x 2 or x 4 The graph crosses the x-axis at x = –4 since this zero has multiplicity 1. The graph touches the x-axis at x = 2 since this zero has multiplicity 2.
11.
f ( x) x 1 x 3 x 1 2
Step 1:
Degree is 4. The function resembles y x 4 for large values of x .
Step 2:
y-intercept: f 0 0 1 0 3 0 1 2
3 x-intercepts: solve f ( x) 0 :
The polynomial is of degree 3 so the graph has at most 3 1 2 turning points. f ( 5) 49; f ( 2) 32; f (3) 7
x 12 x 3 x 1 0 Step 3:
Step 4:
Step 5:
10.
the x-axis at x 0 since it has multiplicity 1. The polynomial is of degree 3 so the graph has at most 3 1 2 turning points. f ( 1) 6; f (1) 2; f (3) 18
f ( x) 2 x3 4 x 2 2 x 2 x 2
Step 1:
Degree is 3. The function resembles y 2 x3 for large values of x .
Step 2:
x-intercepts: 0, 2; y-intercept: 0
Step 3:
The graph crosses x-axis at x 2 since it has multiplicity 2 and touches 568 Copyright © 2025 Pearson Education, Inc.
x 1 or x 3 or x 1 The graph crosses the x-axis at x 3 and x 1 since each zero has multiplicity 1.The graph touches the x-axis at x = 1 since this zero has multiplicity 2. The polynomial is of degree 4 so the graph has at most 4 1 3 turning points. f ( 4) 75; f ( 2) 9; f (2) 15
Chapter 5 Review Exercises x2 is in lowest terms. ( x 3)( x 3) x 9 The denominator has zeros at –3 and 3. Thus, the domain is x x 3, x 3 . The degree of
12. R ( x)
x2 2
the numerator, p( x) x 2, is n 1 . The degree of the denominator q( x) x 2 9, is m 2 . Since n m , the line y 0 is a horizontal asymptote. Since the denominator is zero at –3 and 3, x = –3 and x = 3 are vertical asymptotes. x2 4 13. R ( x) is in lowest terms. The x2 denominator has a zero at 2. Thus, the domain is x x 2 . The degree of the numerator, p( x) x 2 4, is n 2 . The degree of the
denominator, q( x) x 2, is m 1 . Since n m 1 , there is an oblique asymptote.
x2 x 2 x2
4
2
x 2x 2x 4 2x 4 8
R( x) x 2
8 x2
Thus, the oblique asymptote is y x 2 . Since the denominator is zero at 2, x = 2 is a vertical asymptote. 14. R ( x )
x2 3x 2
x 2
2
x 2 x 1 x 1 is in x2 x 2 2
lowest terms. The denominator has a zero at –2. Thus, the domain is x x 2 . The degree of
Dividing:
the numerator, p ( x) x 2 3 x 2, is n 2 . The degree of the denominator,
(cont on next column)
q( x) x 2 x 2 4 x 4, is m 2 . Since 2
1 n m , the line y 1 is a horizontal 1 x 1 x2 is zero at 2 , x 2 is a vertical asymptote.
asymptote. Since the denominator of y
15. R ( x)
Step 1:
2x 6 x
p ( x ) 2 x 6; q ( x) x; n 1; m 1
Domain: x x 0 There is no y-intercept because 0 is not in the domain. 2 x 6 2 x 3 is in lowest terms. x x
Step 2:
R( x)
Step 3:
The x-intercept is the zero of p( x) : 3 Near 3: R x
Step 4:
2 x 3 . Plot the point 3, 0 and show a line with positive slope there. 3
2 x 6 2 x 3 is in lowest terms. The vertical asymptote is the zero of q( x) : x 0 . x x Graph this asymptote using a dashed line. The multiplicity of 0 is odd so the graph will approach plus or minus infinity on either side of the asymptote. R( x)
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Chapter 5 Polynomial and Rational Functions
Step 5:
Since n m , the line y
2 2 is the horizontal asymptote. Solve to find intersection points: 1
2x 6 2 x 2x 6 2x 6 0 R ( x) does not intersect y 2 . Plot the line y 2 with dashes.
Step 6:
Steps 7:
16. H ( x)
Graphing:
x2 x( x 2)
p( x) x 2; q( x) x( x 2) x 2 2 x; n 1; m 2
Step 1:
Domain: x x 0, x 2 .
Step 2:
H ( x)
Step 3:
There is no y-intercept because 0 is not in the domain.
x2 is in lowest terms. x( x 2)
The x-intercept is the zero of p( x) : –2 Near 2 : H x Step 4:
1 x 2 . Plot the point 2, 0 and show a line with positive slope there. 8
x2 is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 0 and x 2 . x( x 2) Graph these asymptotes using dashed lines. The multiplicity of 0 and 2 are odd so the graph will approach plus or minus infinity on either side of the asymptotes. H ( x)
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Chapter 5 Review Exercises
Since n m , the line y 0 is the horizontal asymptote. Solve to find intersection points: x2 0 x( x 2) x20 x 2 H ( x) intersects y 0 at (–2, 0). Plot the line y 0 using dashes.
Step 5:
Step 6:
Steps 7: Graphing:
17. R ( x)
x2 x 6 2
x x6
( x 3)( x 2) ( x 3)( x 2)
p( x) x 2 x 6; q ( x) x 2 x 6;
Step 1:
Domain: x x 2, x 3 .
Step 2:
R( x)
Step 3:
The y-intercept is R (0)
x2 x 6 x2 x 6
is in lowest terms. 02 0 6 2
0 06
6 1 . Plot the point 0,1 . 6
The x-intercepts are the zeros of p ( x) : –3 and 2. Near 3 : R x Near 2: R x
5 x 3 . Plot the point 3, 0 and show a line with negative slope there. 6
5 x 2 . Plot the point 2, 0 and show a line with negative slope there. 4
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Chapter 5 Polynomial and Rational Functions
Step 4:
Step 5:
R( x)
x2 x 6
is in lowest terms. The vertical asymptotes are the zeros of q( x) : x2 x 6 x 2 and x 3 . Graph these asymptotes with dashed lines. The multiplicity of -2 and 3 are odd so the graph will approach plus or minus infinity on either side of the asymptotes. 1 Since n m , the line y 1 is the horizontal asymptote. Solve to find intersection points: 1 x2 x 6 1 x2 x 6 x2 x 6 x2 x 6 2x 0 x0 R ( x) intersects y 1 at (0, 1). Plot the line y 1 using dashes.
Step 6:
Steps 7: Graphing:
18. F ( x)
x3 x2 4
x3 x 2 x 2
p( x) x3 ; q( x) x 2 4; n 3; m 2
Step 1:
Domain: x x 2, x 2 .
Step 2:
F ( x)
Step 3:
The y-intercept is F (0)
x3 x2 4
is in lowest terms. 03 2
0 4
0 0 . Plot the point 0, 0 . 4
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Chapter 5 Review Exercises
The x-intercept is the zero of p( x) : 0. 1 Near 0: F x x3 . Plot the point 0, 0 and indicate a cubic function there (left tail up and right 4 tail down). x3
Step 4:
F ( x)
is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 2 and x 2 . x2 4 Graph these asymptotes using dashed lines. The multiplicity of -2 and 2 are odd so the graph will approach plus or minus infinity on either side of the asymptotes.
Step 5:
Since n m 1 , there is an oblique asymptote. Dividing: x x3 4x 2 x 2 x 4 x3 2 x 4 x 4 4x x3 4x The oblique asymptote is y x . Solve to find intersection points: x3 x2 4
x
x3 x3 4 x 4x 0 x0 F ( x ) intersects y x at (0, 0). Plot the line y x using dashed lines.
Step 6:
Steps 7: Graphing:
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Chapter 5 Polynomial and Rational Functions
19. R ( x)
2 x4 ( x 1) 2
p( x) 2 x 4 ; q( x) ( x 1) 2 ; n 4; m 2
Step 1:
Domain: x x 1 .
Step 2:
R ( x)
Step 3:
The y-intercept is R (0)
2x4 ( x 1) 2
is in lowest terms. 2(0) 4 (0 1)
2
0 0 . Plot the point 0, 0 . 1
The x-intercept is the zero of p( x) : 0. Near 0: R x 2 x 4 . Plot the point 0, 0 and show the graph of a quartic opening up there. R ( x)
2x4
Step 4:
is in lowest terms. The vertical asymptote is the zero of q( x) : x 1 . ( x 1) 2 Graph this asymptote using a dashed line. The multiplicity of 1 is even so the graph will approach plus or minus infinity on the same side of the asymptote.
Step 5:
Since n m 1 , there is no horizontal asymptote and no oblique asymptote.
Step 6:
Steps 7: Graphing:
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Chapter 5 Review Exercises
20. G ( x)
x2 4 2
x x2
( x 2)( x 2) x 2 ( x 2)( x 1) x 1
Step 1:
Domain: x x 1, x 2 .
Step 2:
In lowest terms, G ( x)
Step 3:
The y-intercept is G (0)
p ( x) x 2 4; q( x) x 2 x 2;
x2 , x2. x 1 02 4 2
0 02
4 2 . Plot the point 0, 2 . 2
The x-intercept is the zero of y x 2 : –2; Note: 2 is not a zero because reduced form must be used to find the zeros. Near 2 : G x x 2 . Plot the point 2, 0 and show a line with negative slope there. Step 4:
Step 5:
x2 , x 2 . The vertical asymptote is the zero of f ( x) x 1 : x 1 ; x 1 Graph this asymptote using a dashed line. Note: x 2 is not a vertical asymptote because reduced form 4 must be used to find the asymptotes. The graph has a hole at 2, . The multiplicity of -1 is odd so 3 the graph will approach plus or minus infinity on either side of the asymptote. In lowest terms, G ( x)
1 Since n m , the line y 1 is the horizontal asymptote. Solve to find intersection points: 1 2 x 4 1 2 x x2 x2 4 x2 x 2 x2 G ( x) does not intersect y 1 because G ( x) is not defined at x 2 . Plot the line y 1 using dashes.
Step 6:
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Chapter 5 Polynomial and Rational Functions
Steps 7: Graphing:
x3 x 2 4 x 4
21. 3
( , 4)
( 4, 1)
( 1, 1)
(1, )
5
2
0
2
Value of f
24
6
4
18
Conclusion
Negative
Positive
Negative
Positive
Interval
2
x x 4x 4 0
Number
x x 1 4 x 1 0
Chosen
2
x 4 x 1 0 2
x 2 x 2 x 1 0 f ( x) x 2 x 2 x 1
The solution set is x 4 x 1 or x 1 , or,
x 2, x 1, and x 2 are the zeros of f . ( , 2)
( 2, 1)
( 1, 2)
(2, )
3
3 / 2
0
3
Value of f
10
0.875
4
20
Conclusion
Negative
Positive
Negative
Positive
Interval Number Chosen
The solution set is x | x 2 or 1 x 2 , or, using interval notation, , 2 1, 2 .
x3 4 x 2 x 4
22.
using interval notation, 4, 1 1, .
x3 4 x 2 x 4 0 x 2 x 4 1 x 4 0
x 1 x 4 0 2
23.
Chosen
x 4, x 1, and x 1 are the zeros of f .
2x 6 2 1 x 2x 6 2 0 1 x 2 x 6 2(1 x) 0 1 x 4x 8 0 1 x 4 x 2 f ( x) 1 x The zeros and values where the expression is undefined are x 1, and x 2 . Interval Number
x 1 x 1 x 4 0 f ( x) x 1 x 1 x 4
Value of f Conclusion
, 1
1, 2
2,
0
1.5
3
8
4
2
Negative Positive Negative
The solution set is x x 1 or x 2 , or,
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Chapter 5 Review Exercises
using interval notation, ,1 2, .
24.
interval notation, , 4 2, 4 6, .
( x 2)( x 1) 0 x3 ( x 2)( x 1) f ( x) x3 The zeros and values where the expression is undefined are x 1, x 2, and x 3 . Interval
, 1
1, 2
2, 3
(3, )
0
1.5
2.5
4
2
1
Number Chosen Value of f
3
6
Conclusion
Negative
Positive
3 2
Negative
26.
25.
x 2 8 x 12 x 2 16 f ( x)
f ( x) 8 x3 3 x 2 x 4
10 So R 10 and g is not a factor of f .
27.
f ( x) x 4 2 x3 15 x 2
Since g ( x) x 2 then c 2 . From the Remainder Theorem, the remainder R when f ( x ) is divided by g ( x ) is f (c ) :
Positive
f (2) (2) 4 2(2)3 15(2) 2 16 2(8) 30 2 0 So R 0 and g is a factor of f .
28. 4 12
x 2 8 x 12
x 2 16 ( x 2)( x 6) 0 ( x 4)( x 4) The zeros and values where the expression is undefined are x 4, x 2, x 4, and x 6 . Interval
Number Chosen
, 4
5
4, 2
0
2, 4
3
4, 6
5
6,
7
Value of f
77 9 3 4 3 7 1 3 5 33
8 3 1 4
6
0
f (1) 8(1)3 3(1) 2 1 4
using interval notation, 1, 2 3, .
Since g ( x) x 1 then c 1 . From the Remainder Theorem, the remainder R when f ( x ) is divided by g ( x ) is f (c ) :
The solution set is x 1 x 2 or x 3 , or,
12
0
8
48
192
736 2944 11,776 47,104
48
184
736 2944 11,776 47,105
0
0
0
1
f (4) 47,105
29.
f x 12 x8 x7 8 x 4 2 x3 x 3
The maximum number of zeros is the degree of the polynomial, which is 8. Examining f x 12 x8 x 7 8 x 4 2 x3 x 3 ,
Conclusion
there are four variations in signs; thus, there are four, two or no positive real zeros. Examining
Positive
f x 12 x x 8( x) 4 2( x)3 x 3 8
Negative
7
12 x8 x 7 8 x 4 2 x3 x 3
, there are two variations in sign; thus, there are two negative real zeros or no negative real zeros.
Positive Negative Positive
The solution set is x x 4 or 2 x 4 or x 6 , or, using
30.
f x 6 x5 x 4 5 x3 x 1
The maximum number of zeros is the degree of the polynomial, which is 5. Examining f x 6 x5 x 4 5 x3 x 1 , there is one variation in sign; thus, there is one positive real zero.
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Chapter 5 Polynomial and Rational Functions
Examining
The zeros are –2, of multiplicity 1 and
f x 6 x x 5 x ( x) 1 , 5
4
3
6 x5 x 4 5 x3 x 1 there are two variations in sign; thus, there are two negative real zeros or no negative real zeros. 31. a0 3 , a8 12 p 1, 3 q 1, 2, 3, 4, 6, 12 p 1 3 1 1 3 1 1 1, 3, , , , , , , q 2 2 3 4 4 6 12
32.
f ( x) x3 3x 2 6 x 8
10
Using synthetic division: We try x 2 : 2 1 4 9 20 20 4
10 20
Thus, f ( x) ( x 2)( x 2) x 5 ( x 2) x 5 2
2
Thus, f ( x) ( x 2) x 2 5 x 4 . ( x 2)( x 1)( x 4) The zeros are –2, 1, and 4, each of multiplicity 1. f ( x) 4 x3 4 x 2 7 x 2
Possible rational zeros: p 1, 2; q 1, 2, 4; p 1 1 1, 2, , 2 4 q
2
Since x 2 5 0 has no real solutions, the only zero is 2, of multiplicity 2. 35. 2 x 4 2 x3 11x 2 x 6 0 The solutions of the equation are the zeros of f ( x) 2 x 4 2 x3 11x 2 x 6 . Possible rational zeros: p 1, 2, 3, 6; q 1, 2; p 1 3 1, 2, 3, 6, , q 2 2
Using synthetic division: We try x 3 : 3 2 2 11 1 6 6
Using synthetic division: We try x 2 : 2 4 4 7 2 8 8 2
12 3
6
2 4 1 2 0 x 3 is a factor and the quotient is 2x3 4 x 2 x 2 2 x 2 x 2 1 x 2
4 4 1 0 x 2 is a factor. The other factor is the quotient: 4 x2 4 x 1 .
Possible rational zeros: p 1, 2, 4, 5, 10, 20; q 1; p 1, 2, 4, 5, 10, 20 q
x 2 x2 5
1 5 4 0 x 2 is a factor. The other factor is the quotient: x 2 5 x 4 .
33.
f ( x) x 4 4 x3 9 x 2 20 x 20
x3 2 x 2 5 x 10 x 2 x 2 5 x 2
8
34.
1 2 5 10 0 x 2 is a factor and the quotient is
Using synthetic division: We try x 2 : 2 1 3 6 8 2
multiplicity 2.
2
Possible rational zeros: p 1, 2, 4, 8; q 1; p 1, 2, 4, 8 q
1 , of 2
Thus, f ( x) x 2 4 x 2 4 x 1 . x 2 (2 x 1)(2 x 1)
Thus, f ( x) ( x 3)( x 2) 2 x 1 . x 2 2 x2 1 2
Since 2 x 2 1 0 has no real solutions, the solution set is 3, 2 .
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Chapter 5 Review Exercises
For r = 5, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 5. For r = -3, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -3. The upper bound is 5 and the lower bound is -3.
36. 2 x 4 7 x3 x 2 7 x 3 0 The solutions of the equation are the zeros of f ( x) 2 x 4 7 x3 x 2 7 x 3 . Possible rational zeros: p 1, 3; q 1, 2; p 1 3 1, 3, , q 2 2 39.
Using synthetic division: We try x 3 : 3 2 7 1 7 3 6 3
6
2 x 1 x 1
Thus,
2
40.
x 3 2 x 1 x 1 x 1
1 The solution set is 3, 1, , 1 . 2
37.
41.
f x x3 x 2 4 x 2 r
coeff of q(x)
1
1
0
2 3 1
1 1 1
1 2 2 2 2 8 2 2 0
2
1
3
4 2
2
f x 2 x3 7 x 2 10 x 35 r
coeff of q(x)
1
2
5
15
20
5 1 2
2 2 2
3 5 5 10 11 12
60 45 11
3
2
13
remainder
52
29
f x x3 x 2
So by the Intermediate Value Theorem, f has a zero on the interval [1,2]. Subdivide the interval [1,2] into 10 equal subintervals: [1,1.1]; [1.1,1.2]; [1.2,1.3]; [1.3,1.4]; [1.4,1.5]; [1.5,1.6]; [1.6,1.7]; [1.7,1.8]; [1.8,1.9]; [1.9,2] f 1 2; f 1.1 1.769
For r = 3, the last row of synthetic division contains only numbers that are positive or 0, so we know there are no zeros greater than 3. For r = -2, the last row of synthetic division results in alternating positive (or 0) and negative (or 0) values, so we know that there are no zeros less than -2. The upper bound is 3 and the lower bound is -2. 38.
0, 1
f 1 2; f 2 4
remainder
2
f ( x) 8 x 4 4 x3 2 x 1;
f (0) 1 0 and f (1) 1 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
f ( x) ( x 3) 2 x 1 x 1 2
0, 1
f (0) 1 0 and f (1) 1 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
3
2 1 2 1 0 x 3 is a factor and the quotient is 2x3 x 2 2 x 1 x 2 2 x 1 1 2 x 1 .
f ( x) 3x3 x 1;
f 1.1 1.769; f 1.2 1.472
f 1.2 1.472; f 1.3 1.103 f 1.3 1.103; f 1.4 0.656 f 1.4 0.656; f 1.5 0.125
f 1.5 0.125; f 1.6 0.496
So f has a real zero on the interval [1.5,1.6]. Subdivide the interval [1.5,1.6] into 10 equal subintervals: [1.5,1.51]; [1.51,1.52]; [1.52,1.53]; [1.53,1.54]; [1.54,1.55]; [1.55,1.56];[1.56,1.57]; [1.57,1.58]; [1.58,1.59]; [1.59,1.6] f 1.5 0.125; f 1.51 0.0670 f 1.51 0.0670; f 1.52 0.0082
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Chapter 5 Polynomial and Rational Functions f 1.52 0.0082; f 1.53 0.0516
So f has a real zero on the interval [1.52,1.53], therefore the zero is 1.52, correct to two decimal places.
44. Since complex zeros appear in conjugate pairs, i , the conjugate of i , and 1 i , the conjugate of 1 i , are the remaining zeros of f .
f x x i x i x 1 i x 1 i x 4 2 x3 3x 2 2 x 2
42.
45.
f x 8 x 4 4 x3 2 x 1 f 0 1; f 1 1 ,
f ( x) x3 3 x 2 6 x 8 .
Possible rational zeros: p 1, 2, 4, 8; q 1; p 1, 2, 4, 8 q
So by the Intermediate Value Theorem, f has a zero on the interval [0,1]. Subdivide the interval [0,1] into 10 equal subintervals: [0,0.1]; [0.1,0.2]; [0.2,0.3]; [0.3,0.4]; [0.4,0.5]; [0.5,0.6]; [0.6,0.7]; [0.7,0.8]; [0.8,0.9]; [0.9,1] f 0 1; f 0.1 1.2032
Using synthetic division: We try x 1 : 1 1 3 6 8
f 0.1 1.2032; f 0.2 1.4192
1 2 8
f 0.2 1.4192; f 0.3 1.6432
1
2 8
0
x 1 is a factor and the quotient is x 2 2 x 8 Thus,
f 0.3 1.6432; f 0.4 1.8512 f 0.4 1.8512; f 0.5 2
f ( x) ( x 1) x 2 2 x 8 ( x 1)( x 4)( x 2) .
f 0.5 2; f 0.6 2.0272
The complex zeros are 1, 4, and –2, each of multiplicity 1.
f 0.6 2.0272; f 0.7 1.8512 f 0.7 1.8512; f 0.8 1.3712 f 0.8 1.3712; f 0.9 0.4672
f 0.9 0.4672; f 1 1
So f has a real zero on the interval [0.9,1]. Subdivide the interval [0.9,1] into 10 equal subintervals: [0.9,0.91]; [0.91,0.92]; [0.92,0.93]; [0.93,0.94]; [0.94,0.95]; [0.95,0.96];[0.96,0.97]; [0.97,0.98]; [0.98,0.99]; [0.99,1] f 0.9 0.4672; f 0.91 0.3483 f 0.91 0.3483; f 0.92 0.2236 f 0.92 0.2236; f 0.93 0.0930
f 0.93 0.0930; f 0.94 0.0437
So f has a real zero on the interval [0.93,0.94], therefore the zero is 0.93, correct to two decimal places. 43. Since complex zeros appear in conjugate pairs, 4 i , the conjugate of 4 i , is the remaining zero of f . f x x 6 x 4 i x 4 i
46.
f ( x) 4 x3 4 x 2 7 x 2 .
Possible rational zeros: p 1, 2; q 1, 2, 4; 1 1 p 1, , , 2 2 4 q Using synthetic division: We try x 2 : 2 4 4 7 2 8 8 2 4
4
1
0
x 2 is a factor and the quotient is 4x 2 4 x 1 . Thus,
f ( x) ( x 2) 4 x 2 4 x 1
.
x 2 2 x 1 2 x 1
x 2 2 x 1 4 x 2 x 12 2
2
The complex zeros are –2, of multiplicity 1, and 1 , of multiplicity 2. 2
x3 14 x 2 65 x 102
580 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Review Exercises
47.
f ( x) x 2 x 3
f ( x) x 4 4 x 3 9 x 2 20 x 20 . Possible rational zeros: p 1, 2, 4, 5, 10, 20; q 1; p 1, 2, 4, 5, 10, 20 q
The complex zeros are 2, –3,
2 2 i , and i, 2 2
each of multiplicity 1. 49.
10 20
f ( x) x3 2.37 x 2 4.68 x 6.93
Step 1: Degree = 3; The graph of the function resembles y x3 for large values of x .
1 2 5 10 0 x 2 is a factor and the quotient is x3 2 x 2 5 x 10 .
2 x 2 x 3 x 22 i x 22 i
Using synthetic division: We try x 2 : 2 1 4 9 20 20 2 4
2 x i 2 x i
Step 2: Graphing utility
Thus, f ( x) ( x 2) x3 2 x 2 5 x 10 . 3
2
We can factor x 2 x 5 x 10 by grouping. x3 2 x 2 5 x 10 x 2 x 2 5 x 2
x 2 x 5i x 5i f ( x) ( x 2) x 5i x 5i x 2 x2 5
2
The complex zeros are 2, of multiplicity 2, and 5i and 5i , each of multiplicity 1. 48.
Step 3: x-intercepts: -1.93, 1.14, 3.16; y-intercept: 6.93 Step 4:
f ( x) 2 x 4 2 x3 11x 2 x 6 .
Possible rational zeros: p 1, 2, 3, 6; q 1, 2; p 1 3 1, , 2, 3, 6 2 2 q
Step 5: 2 turning points; local maximum: (–0.69, 8.70); local minimum: (2.27, –4.21)
Using synthetic division: We try x 2 : 2 2 2 11 1 6 4
12
2
Step 6: Graphing by hand
6
2 6 1 3 0 x 2 is a factor and the quotient is 2 x3 6 x 2 x 3 .
Thus, f ( x) ( x 2) 2 x3 6 x 2 x 3 . 3
2
We can factor 2 x 6 x x 3 by grouping. 2 x 3 6 x 2 x 3 2 x 2 x 3 x 3
x 3 2 x i 2 x i x 3 2 x 2 1
Step 7: Domain: , ; Range: , Step 8: Increasing on , 0.69 and 2.27, ; decreasing on 0.69, 2.27
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Chapter 5 Polynomial and Rational Functions
50. a.
250 r 2 h
h
250 r
2
then shift down 2 units.
;
250 A(r ) 2 r 2 2rh 2r 2 2r 2 r 500 2r 2 r
b.
c.
500 3 500 18 223.22 square cm 3 500 A(5) 2 52 5 50 100 257.08 square cm
A(3) 2 32
The maximum number of real zeros is the degree, n 3 . b. g x 2 x3 5 x 2 28 x 15
2. a.
We list all integers p that are factors of a0 15 and all the integers q that are factors of a3 2 . p : 1, 3, 5, 15 q : 1, 2
d. Use MINIMUM on the graph of 500 y1 2x 2 x
p
Now we form all possible ratios q :
The area is smallest when the radius is approximately 3.41 cm. 51. a.
c.
p 1 3 5 15 : , 1, , , 3, 5, , 15 q 2 2 2 2 If g has a rational zero, it must be one of the 16 possibilities listed.
c.
We can find the rational zeros by using the fact that if ris a zero of g, then g r 0 . That is, we evaluate the function for different values from our list of rational zeros. If we get g r 0 , we have a zero.
b.
P (t ) 2.206t 3 21.346t 2 67.698t 250.486 3
2
P (10) 2.206(8) 21.346(8) 67.698(8) 250.486 555.4 The predicted median sale price for January 2023 is approximately $555,100.
Then we use long division to reduce the polynomial and start again on the reduced polynomial. We will start with the positive integers: g 1 2 1 5 1 28 1 15 , 3
2
2 5 28 15 36 g 3 2 3 5 3 28 3 15 3
2
54 45 84 15 0 So, we know that 3 is a zero. This means that x 3 must be a factor of g. Using
Chapter 5 Test 1.
long division we get 4
f ( x) ( x 3) 2
Using the graph of y x 4 , shift right 3 units, 582 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Test
2 x3 6 x 2
The power function that the graph of g resembles for large values of x is given by
f.
2 x 2 11x 5 x 3 2 x3 5 x 2 28 x 15
the term with the highest power of x. In this case, the power function is y 2 x3 . So, the graph of g will resemble the graph of y 2 x3 for large values of x .
11x 2 28 x
11x 2 33 x
g.
5 x 15 5 x 15 0
Thus, we can now write
g x x 3 2 x 2 11x 5
We could first evaluate the function at several values for x to help determine the scale. Putting all this information together, we obtain the following graph:
The quadratic factor can be factored so we get: g x x 3 2 x 1 x 5 To find the remaining zeros of g, we set the last two factors equal to 0 and solve. 2x 1 0 x5 0 2 x 1 x 5 1 x 2 1 Therefore, the zeros are 5 , , and 3. 2 Notice how these rational zeros were all in the list of potential rational zeros. d. The x-intercepts of a graph are the same as the zeros of the function. In the previous 1 part, we found the zeros to be 5 , , and 2 1 3. Therefore, the x-intercepts are 5 , , 2 and 3.
To find the y-intercept, we simply find g 0 . g 0 2 0 5 0 28 0 15 15 3
2
So, the y-intercept is 15 . e.
Whether the graph crosses or touches at an x-intercept is determined by the multiplicity. Each factor of the polynomial occurs once, so the multiplicity of each zero is 1. For odd multiplicity, the graph will cross the x-axis at the zero. Thus, the graph crosses the xaxis at each of the three x-intercepts.
3.
x 3 4 x 2 25 x 100 0 x 2 x 4 25 x 4 0
x 4 x 2 25 0
x 4 0 or x 2 25 0 x4
x 2 25 x 25
x 5i The solution set is 4, 5i, 5i .
4.
3 x3 2 x 1 8 x 2 4 3x3 8 x 2 2 x 3 0 If we let the left side of the equation be f x ,
then we are simply finding the zeros of f. We list all integers p that are factors of a0 3 and all the integers q that are factors of a3 3 . p : 1, 3 ; q : 1, 3 p
Now we form all possible ratios q : p 1 : , 1, 3 q 3
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Chapter 5 Polynomial and Rational Functions
the denominator is 1. Therefore, the graph has the horizontal asymptote y 12 2 .
It appears that there is a zero near x 1 . f 1 3 1 8 1 2 1 3 0 3
2
Therefore, x=1 is a zero and x 1 is a factor of f x . We can reduce the polynomial expression
by using synthetic division. 1 3 8
2
3
3 5 3 3 5 3
0
Thus, f x x 1 3 x 2 5 x 3 . We can find the remaining zeros by using the quadratic formula. 3x 2 5 x 3 0 a 3, b 5, c 3 x
5
Asymptotes: Since the function is in lowest terms, the graph has one vertical asymptote, x 1 . The degree of the numerator is one more than the degree of the denominator so the graph will have an oblique asymptote. To find it, we need to use long division (note: we could also use synthetic division in this case because the dividend is linear). x 1 x 1 x2 2 x 3
x2 x
5 2 4 3 3 2 3
x3
5 25 36 5 61 6 6 5 61 5 61 Thus, the solution set is 1, , . 6 6
5. We start by factoring the numerator and denominator. 2 x 2 14 x 24 2 x 3 x 4 g x 2 x 10 x 4 x 6 x 40
The domain of f is x | x 10, x 4 . In lowest terms, g x
x2 2 x 3 x 1 Start by factoring the numerator. x 3 x 1 r x x 1 The domain of the function is x | x 1 .
6. r x
2 x 3
with x 4 . x 10 The graph has one vertical asymptote, x 10 , since x 10 is the only factor of the denominator of g in lowest terms. The graph is still undefined at x 4 , but there is a hole in the graph there instead of an asymptote.
Since the degree of the numerator is the same as the degree of the denominator, the graph has a horizontal asymptote equal to the quotient of the leading coefficients. The leading coefficient in the numerator is 2 and the leading coefficient in
x 1 4 The oblique asymptote is y x 1 .
7. From problem 6 we know that the domain is x | x 1 and that the graph has one vertical
asymptote, x 1 , and one oblique asymptote, y x 1. x-intercepts: To find the x-intercepts, we need to set the numerator equal to 0 and solve the resulting equation. x 3 x 1 0 x 3 0 or x 1 0 x 3 x 1 The x-intercepts are 3 and 1. The points 3, 0 and 1, 0 are on the graph.
y-intercept: 02 2 0 3 r 0 3 0 1 The y-intercept is 3 . The point 0, 3 is on the graph.
584 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Test
Test for symmetry:
x 2 x 3 x 2x 3 x 1 x 1 Since r x r x , the graph is not symmetric 2
r x
2
with respect to the y-axis. Since r x r x , the graph is not symmetric with respect to the origin. Behavior near the asymptotes: To determine if the graph crosses the oblique asymptote, we solve the equation r x x 1
The zeros of the numerator and denominator, 3 , 1 , and 1, divide the x-axis into four subintervals. , 3 , 3, 1 , 1,1 , 1, We can check a point in each subinterval to determine if the graph is above or below the xaxis. Interval , 3 3, 1 1,1 1, Number 0 3 5 2 Value of r 3 3 3 3 Location below above below above Point 5, 3 2,3 0, 3 3,3
get f x x 2 x x 3 i x 3 i
4
2
3
2
y x 1
3
2
x 6 x 10 x 2 x 12 x 20 x x 4 4 x3 2 x 2 20 x
9. Since the domain excludes 4 and 9, the denominator must contain the factors x 4
and x 9 . However, because there is only one vertical asymptote, x 4 , the numerator must also contain the factor x 9 . The horizontal asymptote, y 2 , indicates that the degree of the numerator must be the same as the degree of the denominator and that the ratio of the leading coefficients needs to be 2. We can accomplish this by including another factor in the numerator, x a , where a 4 , along with a factor of 2. Therefore, we have r x
2 x 9 x a
x 4 x 9
.
If we let a 1 , we get 2 x 9 x 1 2 x 2 20 x 18 . r x x 4 x 9 x 2 13x 36
5
(3, 0) (1, 0) (0, 3)
x 1
where a is any real number. If we let a 1 , we
2
3 1 false The result is a contradiction so the graph does not cross the oblique asymptote.
5
f x a x 2 x 0 x 3 i x 3 i
x2 2 x x 3 i x 3 i
x2 2 x 3 x2 2 x 1
5
factor of the polynomial. This allows us to write the following function:
x 2 x x 6 x 10
x2 2 x 3 x 1, x 1 x 1
y
8. Since the polynomial has real coefficients, we can apply the Conjugate Pairs Theorem to find the remaining zero. If 3 i is a zero, then its conjugate, 3 i , must also be a zero. Thus, the four zeros are 2 , 0, 3 i , and 3 i . The Factor Theorem says that if f c 0 , then x c is a
5
x
10. Since we have a polynomial function and polynomials are continuous, we simply need to show that f a and f b have opposite signs
(where a and b are the endpoints of the interval). f 0 2 0 3 0 8 8 2
f 4 2 4 3 4 8 36 2
Since f 0 8 0 and f 4 36 0 , the Intermediate Value Theorem guarantees that there is at least one real zero between 0 and 4. 585 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Polynomial and Rational Functions
11.
x2 2 x 3 We note that the domain of the variable consists of all real numbers except 3. Rearrange the terms so that the right side is 0. x2 2 0 x 3 x2 For f x 2 , we find the zeros of f and x3 the values of x at which f is undefined. To do this, we need to write f as a single rational expression. x2 2 f x x 3 x2 x 3 2 x 3 x 3 x 2 2x 6 x3 x 8 x3 The zero of f is x 8 and f is undefined at x 3 . We use these two values to divide the real number line into three subintervals.
Interval Num. ch osen Value of f C onclus ion
,3
Interval Number
2.5
1
1
Value of f
8
0.625
2
12
Conclusion
Negative
Positive
Negative
Positive
The solution set is , 3 2,0
Chapter 5 Cumulative Review 1. P 1, 3 , Q 4, 2 d P ,Q
4 12 2 32 5 2 12
25 1
26
2.
x2 x x2 x 0 x( x 1) 0
8,
f ( x) x 2 x
0 4 9 8 4 1 3 6 negative pos itive negative
x 0, x 1 are the zeros of f .
Interval
(, 0)
(0, 1)
(1, )
1
0.5
2
2
0.25
2
Number
Since we want to know where f x is negative,
Chosen
we conclude that values of x for which x 3 or x 8 are solutions. The inequality is strict so the solution set is x | x 3 or x 8 . In
Conclusion Positive Negative Positive
Value of f
The solution set is x x 0 or x 1 or
interval notation we write ,3 or 8, .
, 0 or 1, in interval notation.
x3 7 x 2 2 x 2 6 x
12.
4
Chosen
3,8
(, 3) (3, 2) (2, 0) (0, )
x3 5 x 2 6 x 0 x( x 3)( x 2) 0
f x x ( x 3)( x 2)
x 0, x 3, x 2 are the zeros of f .
3.
x 2 3x 4 x 2 3x 4 0
x 4 x 1 0 f x x 2 3x 4 x 1, x 4 are the zeros of f .
586 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Cumulative Review
(, 1)
(1, 4)
(4, )
2
0
5
Value of f
6
4
6
Conclusion
Positive
Interval Number Chosen
Negative Positive
The solution set is x 1 x 4 or 1, 4 in interval notation. 6. y x3
4. Slope –3, Containing the point (–1, 4)
Using the point-slope formula yields: y y1 m x x1 y 4 3 x 1 y 4 3 x 3 y 3 x 1
7. This relation is not a function because the ordered pairs (3, 6) and (3, 8) have the same first element, but different second elements.
Thus, f x 3 x 1 .
8.
x3 6 x 2 8 x 0
x x2 6 x 8 0 x x 4 x 2 0 x 0 or x 4 or x 2
The solution set is 0, 2, 4 . 5. Parallel to y 2 x 1 ; Slope 2, Containing the point (3, 5) Using the point-slope formula yields: y y1 m x x1 y 5 2 x 3 y 5 2x 6 y 2x 1
9. 3x 2 5 x 1 3 2x 3 x 2 3 x 2 3 3 The solution set is x x or , in 2 2 interval notation.
587 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Polynomial and Rational Functions
10.
Origin: Replace x by x and y by y :
x2 4x y2 2 y 4 0
y x 9 x 3
( x 2 4 x 4) ( y 2 2 y 1) 4 4 1
y x3 9 x
( x 2) 2 ( y 1) 2 9
which is equivalent to y x3 9 x . Therefore, the graph is symmetric with respect to origin.
( x 2) 2 ( y 1) 2 32 Center: (–2, 1) Radius 3
12. 3x 2 y 7 2 y 3 x 7 3 7 y x 2 2 3 . Every line that is 2 perpendicular to the given line will have slope 2 . Using the point 1, 5 and the point-slope 3 formula yields: y y1 m x x1
The given line has slope
2 x 1 3 2 2 y 5 x 3 3 2 17 y x 3 3 y 5
13. Not a function, since the graph fails the VerticalLine Test, for example, when x 0 . 11. y x 3 9 x
14.
3
x-intercepts: 0 x 9 x
0 x x2 9
0 x x 3 x 3 x 0, 3, and 3 0, 0 , 3, 0 , 3, 0
y-intercepts: y 03 9 0 0 0, 0
f ( x) x 2 5 x 2
a.
f (3) 32 5(3) 2 9 15 2 22
b.
f ( x) x 5 x 2 x 2 5 x 2
c.
f ( x) x 2 5 x 2 x 2 5 x 2
d.
f (3x) 3x 5 3 x 2 9 x 2 15 x 2
Test for symmetry: 3
x-axis: Replace y by y : y x 9 x , which is not equivalent to y x 3 9 x .
e.
2
2
f x h f x h
y-axis: Replace x by x : y x 9 x 3
3
x 9x
which is not equivalent to y x 3 9 x .
x h 2 5 x h 2 x 2 5 x 2 h
x 2 xh h 5 x 5h 2 x 2 5 x 2 h 2 2 xh h 5h h 2x h 5
2
588 Copyright © 2025 Pearson Education, Inc.
2
Chapter 5 Cumulative Review
15.
f ( x)
a. b.
x5 x 1
Domain
17.
a 2, b 4, c 1. Since a 2 0, the graph is concave up.
x x 1 .
25 7 = =7 6; 2 1 1 2,6 is not on the graph of f . f (2)
The point 2, 7 is on the graph. c.
35 8 = =4; 3 1 2 3,4 is on the graph of f . f (3)
The y-coordinate of the vertex is 2 b f f 1 2 1 4 1 1 1 . 2a
The discriminant is: b 2 4ac 4 4 2 1 8 0 , so the graph 2
has two x-intercepts. The x-intercepts are found by solving: 2 x2 4 x 1 0
x 5 9x 9 14 8 x 14 7 8 4 7 Therefore, ,9 is on the graph of f . 4 x
x
4 8 2 2
42 2 2 2 4 2
f ( x ) is a rational function since it is in the
The x-intercepts are
p( x) form . q( x)
16.
The x-coordinate of the vertex is b 4 x 1. 2a 2 2
Thus, the vertex is (1, –1). The axis of symmetry is the line x 1 .
d. Solve for x x5 9 x 1 x 5 9 x 1
e.
f ( x) 2 x 2 4 x 1
2 2 2 2 and . 2 2
The y-intercept is f (0) 1 .
f x 3 x 7
The graph is a line with slope –3 and y-intercept (0, 7).
589 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Polynomial and Rational Functions
18.
f x x2 3x 1 average rate of change of f from 1 to 2 :
f 2 f 1 2 1
11 5 6 msec 1
f 2 11 so the point 2,11 is on the graph.
Using this point and the slope m 6 , we can obtain the equation of the secant line: y y1 m x x1 y 11 6 x 2
d. Range: y y 5 or ,5
y 11 6 x 12 y 6x 1
19. a.
22.
x-intercepts: 5, 0 ; 1, 0 ; 5, 0 ;
2
Using the graph of y x 2 , shift left 1 unit, vertically stretch by a factor of 3, reflect about the x-axis, then shift up 5 units.
y-intercept: (0,–3) b. The graph is not symmetric with respect to the origin, x-axis or y-axis. c.
f ( x) 3 x 1 5
The function is neither even nor odd.
d. f is increasing on , 3 and 2, ; f is
decreasing on 3, 2 ; e.
f has a local maximum value at x 3, and
the local maximum is f 3 5 . f.
f has a local minimum value at x 2, and
the local minimum is f 2 6 . 20.
f ( x)
23.
f ( x) x 2 5 x 1
a.
5x
x2 9 5 x 5 x 2 f x , therefore f ( x) 2 x 9 x 9 f is an odd function.21. 2 x 1 f ( x) 3 x 4
a.
( f g )( x) x 2 5 x 1 4 x 7 x2 9 x 6 The domain is: x x is a real number .
b.
if 3 x 2
f x x2 5x 1 f ( x) 4 x 7 g x g
7 The domain is: x x . 4
if x 2
Domain: x x 3 or 3,
g ( x ) 4 x 7
24. a.
R( x) x p 1 x x 150 10 1 2 x 150 x 10
1 b. x-intercept: , 0 2 y-intercept: (0,1)
c.
590 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Projects
b.
1 (100) 2 150(100) 10 1000 15, 000
R (100)
b. x 2 bx c 0 ( x r1 )( x r2 ) 0 x 2 r1 x r2 x r1r2 0
$14, 000
c.
x 2 (r1 r2 ) x r1r2 0
1 2 x 150 x is a quadratic 10 1 function with a 0 , the vertex will 10 be a maximum point. The vertex occurs b 150 when x 750 . 2a 2 1 / 10
b (r1 r2 )
Since R ( x )
c r1r2
c.
The maximum revenue is given by
f ( x) ( x 2)( x 2 3 x 4) f ( x) ( x 2)( x 4)( x 1) zeros: 2 . 4. 1 sum 2 4 1 1 , product (2)(4)(1) 8 sum of double products 2(4) (2)(1) 4(1) 8 2 4 10
1 R (750) (750) 2 150(750) . 10 56, 250 112, 500
The coefficient of x 2 is the negative sum. The coefficient of x is the sum of the double products. The constant term is the negative product.
Thus, the revenue is maximized when x 750 units sold.
$56, 250
d.
f ( x) x3 x 2 10 x 8
d.
1 (750) 150 75 150 $75 is 10 the selling price that maximizes the revenue. p
f ( x) x3 bx 2 cx d f ( x) ( x r1 )( x r2 )( x r3 ) f ( x) ( x 2 (r1 r2 ) x r1r2 )( x r3 ) f ( x) x3 (r1 r2 r3 ) x 2 (r1r2 r1r3 r2 r3 ) x r1r2 r3 b (r1 r2 r3 )
c r1r2 r1r3 r2 r3
Chapter 5 Projects
d r1r2 r3
Project I – Internet-based Project
e.
Answers will vary
f ( x) x 4 bx3 cx 2 dx e f ( x) ( x r1 )( x r2 )( x r3 )( x r4 ) f ( x) ( x3 (r1 r2 r3 ) x 2 (r1r2 r1r3 r2 r3 ) x r1r2 r3 )( x r3 ) f ( x) x 4 (r1 r2 r3 r4 ) x3
Project II
r1r2 r1r3 r2 r3 r1r4 r2 r4 r3 r4 x 2
a. x 2 8 x 9 0 ( x 9)( x 1) 0
(r1r2 r4 r1r3 r4 r2 r3 r4 r1r2 r3 ) x r1r2 r3 r4
sum 9 1 8 , product 9 1 9
b (r1 r2 r3 r4 )
x 9 or x 1
c r1r2 r1r3 r2 r3 r1r4 r2 r4 r3 r4 d (r1r2 r4 r1r3 r4 r2 r3 r4 r1r2 r3 ) e r1r2 r3 r4
f. The coefficients are sums, products, or sums of products of the zeros. If f ( x) x n an 1 x n 1 an 2 x n 2 ... a1 x a0 , 591 Copyright © 2025 Pearson Education, Inc.
Chapter 5 Polynomial and Rational Functions
then: an 1 will be the negative of the sum of the zeros. an 2 will be the sum of the double products. a1 will be the negative (if n is even) or positive (if n is odd) of the sum of (n-1) products. a0 will be the negative (if n is odd) or positive (if n is even) product of the zeros. These will always hold. These would be useful if you needed to multiply a number of binomials in x c form together and you did not want to have to do the multiplication out. These formulas would help same time.
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Chapter 6 Exponential and Logarithmic Functions Section 6.1 1.
10. a.
f 3 4 3 5 3 2
b.
f g 2 f g 2 f 3 11
c.
g f 2 g f 2 g 1 0
d.
g f 3 g f 3 g 1 0
e.
g g 1 g g 1 g 0 1
f.
f f 3 f f 3 f 1 7
11. a.
g f 1 g f (1) g 1 4
4 9 15 36 15 21
2.
f 3x 4 2 3x
2
4 2 9 x2 4 18 x 2
3.
f ( x)
x2 1 2
x 25 x 2 25 0 x 5 x 5 0 x 5, x 5
b.
g f 0 g f (0) g 0 5
c.
f g 1 f g (1) f 3 1
d.
f g 4 f g (4) f 2 2
12. a.
g f 1 g f (1) g 1 3
b.
g f 5 g f (5) g 1 4
c.
f g 0 f g (0) f 5 1
d.
f g 2 f g (2) f 2 2
Domain: x x 5, x 5 4. composite function; f g ( x) 5. False: ( f g )( x) f ( g (4)) f ( 4 9) f ( 13) 2
( 13) 13
6. c 13.
7. a
f ( x) 2 x
a.
8. False. The domain of f g ( x) is a subset of
f g 1 f g 1 f 0 5
( f g )(4) f ( g (4))
f (49) 2(49) 98
f g 1 f g 1 f 0 1
b.
f g 1 f g 1 f 0 1
c.
g f 1 g f 1 g 3 8
d.
g f 0 g f 0 g 1 0
e.
g g 2 g g 2 g 3 8
f.
f f 1 f f 1 f 3 7
f 3(4) 2 1
the domain of g ( x). 9. a.
g ( x) 3x 2 1
b.
( g f )(2) g ( f (2)) g (2 2) g (4) 3(4) 2 1 48 1 49
593 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
c.
d.
( f f )(1) f ( f (1)) f (2(1)) f (2) 2(2) 4 ( g g )(0) g ( g (0))
b.
( g f )(2) g ( f (2)) g (8(2)2 3) g (29) 1 3 (29) 2 2 841 3 2 835 2
g 3(0) 2 1 g (1) 3(1) 2 1 4
14.
f ( x) 3x 2
a.
c.
( f f )(1) f ( f (1)) f (8(1) 2 3) f (5)
g ( x) 2 x 2 1
( f g )(4) f ( g (4))
8(5) 2 3 197
f 2(4) 2 1
d.
f (31) 3(31) 2 95
b.
( g g )(0) g ( g (0))
1 g 3 (0) 2 2 g (3) 1 3 (3) 2 2 9 3 2 3 2
( g f )(2) g ( f (2)) g (3(2) 2) g (8) 2(8) 2 1 128 1 127
c.
( f f )(1) f ( f (1)) f 3(1) 2
16.
f (5) 3(5) 2 17
d.
( g g )(0) g ( g (0))
f ( x) 2 x 2 g ( x) 1 3 x 2 a. ( f g )(4) f ( g (4))
f 1 3(4) 2 f ( 47) 2( 47) 2 4418
g 2(0) 2 1
b.
g (1)
( g f )(2) g ( f (2)) g (2(2) 2 ) g (8)
2(1) 2 1 1
1 3(8) 2 1 192 191
1 15. f ( x) 8 x 2 3 g ( x) 3 x 2 2 a. ( f g )(4) f ( g (4)) 1 f 3 (4) 2 2 f (5)
c.
( f f )(1) f ( f (1))
f 2(1) 2 f (2)
2
2(2) 2 8
8(5) 3 197
594
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Section 6.1: Composite Functions
d.
( g g )(0) g ( g (0))
g 1 3(0)
2
c.
( f f )(1) f ( f (1))
1 1 f 2 f
g (1) 1 3(1) 2 1 3 2
17.
f ( x) x
a.
d.
g ( x) 5 x
( f g )(4) f ( g (4)) f 5(4) f (20) 20
19.
2 5
b.
( g f )(2) g ( f (2)) g
2
5 2
c.
( f f )(1) f ( f (1)) f
1
f (1) 1 1
d.
( g g )(0) g ( g (0)) g 3(0) g (0) 3(0) 0
f ( x) x
a.
b.
1 2
( g f )(2) g ( f (2)) g 2
( g g )(0) g ( g (0)) g 5(0)
g (2) 1 2 2 9 1 13
f ( x) x 1 g ( x) 3 x a. ( f g )(4) f ( g (4)) f 3(4) f (12)
c.
( f f )(1) f ( f (1)) f1 f (1) 1
12 1
1
13
b.
g ( x)
x 9 ( f g )(4) f ( g (4)) 1 f 2 4 9 1 f 25 1 25 1 25
g (0) 5(0) 0
18.
2 1
d.
( g f )(2) g ( f (2))
2 1 g 3 g
( g g )(0) g ( g (0)) 1 g 2 0 9 g 19
3 3
1
1 2 9 9
1 81 730 730 81
595
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Chapter 6: Exponential and Logarithmic Functions
20.
f ( x) x 2
a.
g ( x)
3
21.
x2 2
( f g )(4) f ( g (4))
a.
f 2 4 2 3 f 18 1 f 6 1 2 6 11 6 11 6
3 g ( x) 3 x x 1 ( f g )(4) f ( g (4))
f ( x)
3
b.
f 3
b.
3 4 1
31 1
( g f )(2) g ( f (2))
c.
( f f )(1) f ( f (1))
g 22
3 f 11 3 f 2 3 3 1 2 3 5 2 6 5
( f f )(1) f ( f (1))
f 1 2 f (1) 1 2 1
d.
3
( g f )(2) g ( f (2)) 3 g 2 1 3 g 3 g 1
g (0) 3 2 0 2 3 2
c.
4
d.
( g g )(0) g ( g (0))
( g g )(0) g ( g (0))
g 30
3 g 2 0 2 3 g 2 3 2 3 2 2 3 17 4 12 17
g (0) 30 0
596
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Section 6.1: Composite Functions
22.
a.
23.
2 x 1 ( f g )(4) f ( g (4))
f ( x) x3/ 2
g ( x)
The domain of g is x x is any real number . a.
( f g )( x) f ( g ( x)) f (4 x) 2(4 x) 3 8x 3 Domain: x x is any real number .
b.
( g f )( x) g ( f ( x)) g (2 x 3) 4(2 x 3) 8 x 12 Domain: x x is any real number .
c.
( f f )( x) f ( f ( x)) f (2 x 3) 2(2 x 3) 3 4x 6 3 4x 9
3/ 2
2 5
3
8 125 2 2 5 5 5 5 2 10 25
b.
( g f )(2) g ( f (2))
g 23/ 2 g
3
c.
Domain: x x is any real number .
2
g 2 2
d.
2 2 2 1
or
4 2 2 7
( f f )(1) f ( f (1))
f 13/ 2
24.
f 1
( g g )( x) g ( g ( x)) g (4 x) 4(4 x) 16 x Domain: x x is any real number .
f ( x) x
g ( x) 2 x 4
The domain of f is x x is any real number .
13/ 2 1
d.
g ( x) 4 x
The domain of f is x x is any real number .
2 f 4 1 2 f 5 2 5
f ( x) 2 x 3
The domain of g is x x is any real number . a.
( g g )(0) g ( g (0)) 2 g 0 1 g (2) 2 2 1 2 3
( f g )( x) f ( g ( x)) f (2 x 4) (2 x 4) 2x 4
Domain: x x is any real number . b.
( g f )( x) g ( f ( x)) g ( x) 2( x) 4 2 x 4 Domain: x x is any real number .
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Chapter 6: Exponential and Logarithmic Functions
c.
d.
25.
( f f )( x) f ( f ( x)) f ( x) ( x ) x Domain: x x is any real number .
a.
x 4 1 x2 5 Domain: x x is any real number .
b.
x2 2 x 1 4 x2 2 x 5 Domain: x x is any real number .
g ( x) x 2
c.
( f g )( x) f ( g ( x))
d.
3x 2 1 Domain: x x is any real number .
x2 4
4
( g f )( x) g ( f ( x)) g (3 x 1)
2
4
x 8 x 2 16 4 x 4 8 x 2 20 Domain: x x is any real number .
2
9x 6x 1 Domain: x x is any real number .
27.
( f f )( x) f ( f ( x)) f (3 x 1) 3(3x 1) 1 9x 3 1 9x 4
f ( x) x 2
g ( x) x 2 4
The domain of f is x x is any real number . The domain of g is x x is any real number . a.
( f g )( x) f ( g ( x))
Domain: x x is any real number .
f x2 4
( g g )( x) g ( g ( x))
x2 4
x 2 2
b.
x4 Domain: x x is any real number . f ( x) x 1
2
x 4 8 x 2 16 Domain: x x is any real number .
g x2
26.
( g g )( x) g ( g ( x)) g x2 4
(3 x 1) 2
d.
( f f )( x) f ( f ( x)) f ( x 1) ( x 1) 1 x2
Domain: x x is any real number .
f x2
c.
( g f )( x) g ( f ( x)) g ( x 1) ( x 1) 2 4
The domain of g is x x is any real number .
b.
2
The domain of f is x x is any real number . a.
f x2 4
( g g )( x) g ( g ( x)) g (2 x 4) 2(2 x 4) 4 4x 8 4 4 x 12 Domain: x x is any real number .
f ( x) 3 x 1
( f g )( x) f ( g ( x))
( g f )( x) g ( f ( x))
x 4 g x2
2 2
g ( x) x 2 4
x4 4 Domain: x x is any real number .
The domain of f is x x is any real number . The domain of g is x x is any real number . 598
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Section 6.1: Composite Functions
c.
( f f )( x) f ( f ( x))
d.
x f x
( g g )( x) g ( g ( x))
2 2 x 3 3 2 4 x 12 x 9 3 g 2x2 3
2
2 2
x4 Domain: x x is any real number .
d.
( g g )( x) g ( g ( x))
4
8 x 4 24 x 2 18 3
x2 4
4 2
29.
x 4 8 x 2 16 4
f ( x) x 2 1
f ( x)
a.
g ( x) 2 x 2 3
2x2 3
1
f 2x2 3
2
4 x 4 12 x 2 9 1 4 x 4 12 x 2 10 Domain: x x is any real number .
b.
( g f )( x) g ( f ( x))
b.
2 x 1 3 2 x 2 x 1 3 2
g x 1 2
2
4
2
2 x4 4 x2 2 3 2 x4 4 x2 5 Domain: x x is any real number .
c.
( f f )( x) f ( f ( x))
( f g )( x) f ( g ( x))
( g f )( x) g ( f ( x)) 3 g x 1 2 3 x 1 2( x 1) 3 Domain x x 1
f x2 1
2 x
2 f x 3 2 1 x 3 2 x x 3x 2 x Domain x x 0, x 2 .
The domain of g is x x is any real number . ( f g )( x) f ( g ( x))
g ( x)
x x 0 .
The domain of f is x x is any real number . a.
3 x 1
The domain of f is x x 1 . The domain of g is
x 4 8 x 2 20 Domain: x x is any real number .
28.
2
8 x 4 24 x 2 21 Domain: x x is any real number .
g x2 4
2
2
2
x2 1 1 x4 2x2 1 1 x4 2x2 2 Domain: x x is any real number .
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Chapter 6: Exponential and Logarithmic Functions
c.
( f f )( x) f ( f ( x))
c.
d.
30.
2 2 2x x ( g g )( x) g ( g ( x)) g 2 x 2 x Domain x x 0 .
f ( x)
1 x3
g ( x)
d.
2 x
The domain of f is x x 3 . The domain of g is x x 0 . a.
( f g )( x) f ( g ( x))
31.
2 f x 1 1 2 3x 2 3 x x x x or 2 3x 3x 2 2 Domain x x 0, x . 3
b.
( f f )( x) f ( f ( x)) 1 f x3 1 1 1 1 3x 9 3 x3 x3 x3 3x 10 10 Domain x x , x 3 . 3
3 f x 1 3 3 3 3 ( x 1) 1 x 1 x 1 3( x 1) 4 x Domain x x 1, x 4 .
( g g )( x) g ( g ( x)) 2 g x 2 2x 2 2 x x Domain x x 0 .
f ( x)
x x 1
g ( x)
4 x
The domain of f is x x 1 . The domain of g is x x 0 . a.
( f g )( x) f ( g ( x)) 4 f x 4 4 4 x x 4 x 4 x 4 1 x x 4 4 x Domain x x 4, x 0 .
( g f )( x) g ( f ( x)) 1 g x3 2( x 3) 2 1 1 x3 2( x 3)
b.
Domain x x 3 .
( g f )( x) g ( f ( x)) x g x 1 4 x x 1 4( x 1) x Domain x x 0, x 1 .
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Section 6.1: Composite Functions
c.
( f f )( x) f ( f ( x))
c.
x f x 1 x x x x 1 x 1 x 1 x x ( x 1) 1 1 x 1 x 1 x 1 x Domain x x 1 .
d.
32.
x f x3 x x x3 x3 4x 9 x 3 x3 x3 x 4x 9 9 Domain x x 3, x . 4
( g g )( x) g ( g ( x)) 4 g x 4x 4 4 4 x x Domain x x 0 .
f ( x)
x x3
g ( x)
d.
33.
2 2 2x x ( g g )( x) g ( g ( x)) g 2 x 2 x Domain x x 0 .
f ( x) x
is x x is any real number . a.
g is x x 0 .
( f g )( x) f ( g ( x)) f 2 x 5 2 x 5 5 Domain x x . 2
( f g )( x) f ( g ( x)) 2 f x 2 2 x x 2 2 3x 3 x x 2 2 3x
b.
( g f )( x) g ( f ( x)) g
Domain c.
x 2 x 5
x x 0 .
( f f )( x) f ( f ( x)) f
x
x
x1/ 2
2 Domain x x , x 0 . 3 b.
g ( x) 2 x 5
The domain of f is x x 0 . The domain of g
2 x
The domain of f is x x 3 . The domain of a.
( f f )( x) f ( f ( x))
1/ 2
x1/ 4 4x
( g f )( x) g ( f ( x)) x g x3 2 x x3 2( x 3) x Domain x x 3, x 0 .
Domain x x 0 . d.
( g g )( x) g ( g ( x)) g 2 x 5 2(2 x 5) 5 4 x 10 5 4 x 15 Domain x x is any real number .
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Chapter 6: Exponential and Logarithmic Functions
34.
f ( x) x 2
b.
g ( x) 1 2 x
The domain of f is x x 2 . The domain of g
g x2 7
is x x is any real number .
x2 7 7
a.
( f g )( x) f ( g ( x)) f 1 2 x
x2 x
Domain x x is any real number .
1 2x 2 2 x 1
c.
1 Domain x x . 2
b.
x 2 d.
f
x 2
x2 2
Now,
36.
a.
x7
( f g )( x) f ( g ( x))
x 2 x 2 4 f
2
x24 x2 Domain x x 2 .
The domain of g is x x 7 . ( f g )( x) f ( g ( x))
g ( x) x 2
The domain of g is x x 2 .
The domain of f is x x is any real number .
x7 7 0
f ( x) x 2 4
g ( x) x 7
x7 7
The domain of f is x x is any real number .
( g g )( x) g ( g ( x)) g 1 2 x
f
x7
x7 7 x 7 49 x 56 Domain x x 56 .
1 2(1 2 x) 1 2 4x 4x 1 Domain x x is any real number .
a.
( g g )( x) g ( g ( x)) g
x2 2 0
f ( x) x 2 7
7
x 4 14 x 2 56 Domain x x is any real number .
x2 2 x2 4 x6 Domain x x 6 .
d.
x2 7
2
x 4 14 x 2 49 7
( f f )( x) f ( f ( x))
Now,
x x 2 .
Domain
f x2 7
1 2 x 2
c.
( f f )( x) f ( f ( x))
( g f )( x) g ( f ( x)) g
35.
( g f )( x) g ( f ( x))
b.
( g f )( x) g ( f ( x))
g x2 4
x2 4 2
x7 7 2
x2 2
x77 x Domain x x 7 .
Domain x x is any real number .
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Section 6.1: Composite Functions
c.
( f f )( x) f ( f ( x))
2
f x 4
x2 4
b.
4 2
x 4 8 x 2 16 4 x 4 8 x 2 20 Domain x x is any real number .
d.
x 5 2( x 1) x 5 2 x 2 x 5 3( x 1) x 5 3 x 3 3x 3 3x 3 or 2 x 8 2x 8
( g g )( x) g ( g ( x)) g
Now,
x 2
x2 2
Now, 2 x 8 0, so x 4. Also, from the domain of f, we know x 1 . Domain of g f : x x 4, x 1 .
x2 2 0
x2 2 x2 4 x6 Domain x x 6 .
37.
x5 ( g f )( x) g ( f ( x)) g x 1 x5 x 5 2 ( x 1) 2 x 1 x 1 x 5 x 5 3 ( x 1) 3 x 1 x 1
f ( x)
x5 x 1
g ( x)
c.
x5 ( f f )( x) f ( f ( x)) f x 1 x 5 x5 5 ( x 1) 5 x 1 x 1 x 5 x5 1 1 ( x 1) x 1 x 1 x 5 5( x 1) x 5 5 x 5 x 5 1( x 1) x 5 x 1 4 x 10 2(2 x 5) 2x 5 x2 2x 4 2( x 2)
x2 x3
The domain of f is x x 1 . The domain of g is
x x 3 . a.
x2 ( f g )( x) f ( g ( x)) f x3 x 2 5 ( x 3) x2 5 x3 x3 x2 x 2 1 ( x 3) 1 x3 x3 x 2 5( x 3) x 2 5 x 15 x2 x3 x 2 1( x 3) 4 x 17 4 x 17 or 2x 1 2x 1 1 Now, 2 x 1 0, so x . Also, from the 2 domain of g, we know x 3 . 1 Domain of f g : x x , x 3 . 2
Now, x 2 0, so x 2. Also, from the domain of f, we know x 1 . Domain of f f : x x 1, x 2 . d.
x2 ( g g )( x) g ( g ( x)) g x 3 x2 x2 2 ( x 3) 2 x 3 x3 x2 x2 3 3 ( x 3) x3 x3 x 2 2( x 3) x 2 2 x 6 x 2 3( x 3) x 2 3 x 9 3x 4 3x 4 or 2 x 11 2 x 11 11 Now, 2 x 11 0, so x . Also, from the 2 domain of g, we know x 3 . 11 Domain of g g : x x , x 3 . 2
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Chapter 6: Exponential and Logarithmic Functions
38.
f ( x)
2x 1 x2
g ( x)
Now, x 8 0, so x 8. Also, from the domain of f, we know x 2 . Domain of f g : x x 2, x 8 .
x4 2x 5
The domain of f is x x 2 . The domain of g 5 is x x . 2
a.
c.
x4 ( f g )( x) f ( g ( x)) f 2x 5 x4 2 1 2x 5 x4 2 2x 5 x4 1 (2 x 5) 2 2x 5 x4 2 (2 x 5) 2x 5 2( x 4) 1(2 x 5) x 4 2(2 x 5) 2x 8 2x 5 x 4 4 x 10 13 13 or 3x 14 3 x 14
Domain of f f : x x 2 .
14 . Also, from the 3 5 domain of g, we know x . 2 5 14 Domain of f g : x x , x . 2 3
Now, 3x 14 0, so x
b.
2x 1 ( f f )( x) f ( f ( x)) f x2 2x 1 2 1 x2 2x 1 2 x2 2x 1 2 x 2 1 ( x 2) 2x 1 2 ( x 2) x2 2(2 x 1) 1( x 2) 2 x 1 2( x 2) 4 x 2 x 2 3x x 2x 1 2x 4 3 From the domain of f, we know x 2 .
d.
x4 ( g g )( x) g ( g ( x)) g 2x 5 x4 4 2 x5 x4 2 5 2x 5 x4 4 (2 x 5) 2x 5 x4 5 (2 x 5) 2 2x 5 x 4 4(2 x 5) 2( x 4) 5(2 x 5) x 4 8 x 20 2 x 8 10 x 25 9 x 16 9 x 16 or 8 x 33 8 x 33
2x 1 ( g f )( x) g ( f ( x)) g x2 2x 1 4 x2 2x 1 2 5 x2 2x 1 4 ( x 2) x2 2x 1 2 x 2 5 ( x 2) 2 x 1 4( x 2) 2(2 x 1) 5( x 2) 2x 1 4x 8 4 x 2 5 x 10 6x 9 6x 9 or x 8 x 8
33 . Also, from the 8 5 domain of g, we know x . 2 5 33 Domain of f g : x x , x . 2 8
Now, 8 x 33 0, so x
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Section 6.1: Composite Functions 45. ( f g )( x) f ( g ( x)) 1 f ( x b) a 1 a ( x b) b a xbb x ( g f )( x) g ( f ( x)) g ax b 1 (ax b) b a 1 (ax) a x
1 1 39. ( f g )( x) f ( g ( x)) f x 2 x x 2 2 1 ( g f )( x) g ( f ( x)) g (2 x) (2 x) x 2 1 1 40. ( f g )( x) f ( g ( x)) f x 4 x x 4 4 1 ( g f )( x) g ( f ( x)) g (4 x) (4 x) x 4
x x x ( g f )( x) g ( f ( x)) g x x x
41. ( f g )( x) f ( g ( x)) f
3
3
3
3
3
3
42. ( f g )( x) f ( g ( x)) f x 5 x 5 5 x ( g f )( x) g ( f ( x)) g x 5 x 5 5 x
x 1 1 46. ( f g )( x) f ( g ( x)) f 1 x 1 x 1 x x 1 1 ( g f )( x) g ( f ( x)) g 1 x 1 x 1 x
43. ( f g )( x) f ( g ( x)) 1 f ( x 6) 9 1 9 ( x 6) 6 9 x66 x ( g f )( x) g ( f ( x)) g 9x 6
47. H ( x) (2 x 3) 4 Answers may vary. One possibility is f ( x) x 4 , g ( x) 2 x 3
48. H ( x) 1 x 2
1 (9 x 6) 6 9 1 (9 x) 9 x
3
Answers may vary. One possibility is f ( x) x3 , g ( x) 1 x 2 49. H ( x) x 2 1 Answers may vary. One possibility is f ( x) x , g ( x) x 2 1
44. ( f g )( x) f ( g ( x)) 1 f (4 x) 3 1 4 3 (4 x) 3 44 x x ( g f )( x) g ( f ( x)) g 4 3x 1 4 (4 3 x) 3 1 (3x ) 3 x
50. H ( x) 1 x 2 Answers may vary. One possibility is f ( x) x , g ( x) 1 x 2 51. H ( x) 2 x 1
Answers may vary. One possibility is f ( x) x , g ( x) 2 x 1
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Chapter 6: Exponential and Logarithmic Functions
When x 0 , ( f g )(0) 68 .
52. H ( x) 2 x 2 3
Solving: 3(2 0 a) 2 7 68
Answer may vary. One possibility is f ( x) x , g ( x) 2 x 2 3 53.
3
2
f ( x) 2 x 3x 4 x 1
3a 2 7 68 3a 2 75 0 3(a 5)(a 5) 0 a 5 or a 5
g ( x) 2
( f g )( x) f ( g ( x)) f (2) 2(2)3 3(2) 2 4(2) 1 16 12 8 1 11
( g f )( x) g ( f ( x)) g 2 x3 3 x 2 4 x 1 2
54.
55.
x 1 , x 1 x 1 ( f f )( x) f ( f ( x)) x 1 f x 1 x 1 1 x 1 x 1 1 x 1 x 1 x 1 x 1 x 1 ( x 1) x 1 2x x 1 2 x 1 2x x 1 x 1 2 x, x 1
57. a.
( f g )( x) f ( g ( x)) f (cx d ) a (cx d ) b acx ad b
b.
( g f )( x) g ( f ( x)) g (ax b) c(ax b) d acx bc d
c.
Since the domain of f is the set of all real numbers and the domain of g is the set of all real numbers, the domains of both f g and g f are all real numbers.
d.
( f g )( x) ( g f )( x) acx ad b acx bc d ad b bc d Thus, f g g f when ad b bc d .
f ( x)
f ( x) 2 x 2 5
58. a.
( f g )( x) f ( g ( x)) f (mx) a (mx) b c(mx ) d amx b cmx d
b.
( g f )( x) g ( f ( x)) ax b g cx d ax b m cx d m(ax b) cx d
c.
To find the domain of f g , we first recognize that the domain of g is the set of all real numbers. This means that the only restrictions are those that cause zero in the denominator of the final result in part (a).
g ( x) 3 x a
( f g )( x) f ( g ( x)) f (3 x a ) 2(3 x a ) 2 5
When x 0 , ( f g )(0) 23 . Solving: 2(3 0 a ) 2 5 23 2a 2 5 23 2a 2 18 0 2(a 3)(a 3) 0 a 3 or a 3
56.
f ( x) 3x 2 7
g ( x) 2 x a
( f g )( x) f ( g ( x)) f (2 x a ) 3(2 x a ) 2 7
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Section 6.1: Composite Functions cmx d 0 cmx d d x cm
4 3 2 r r (t ) t 3 , t 0 3 3 2 V (r (t )) V t 3 3
60. V (r )
d Thus the domain of f g is x x . cm
3
4 2 3 t 3 3 4 8 t9 3 27 32 9 t 81 32 Thus, V (t ) t 9 . 81
To find the domain of g f , we first recognize that the domain of f is d x x and the domain of g is the set c of all real numbers. Thus, the domain of d g f is also x x . c d.
61.
( f g )( x) ( g f )( x) amx b m(ax b) cmx d cx d amx b amx bm cmx d cx d (amx bm)(cmx d ) (amx b)(cx d ) Now, this equation will only be true if m 1 . Thus, f g g f when m 1.
59. S (r ) 4r 2
r (t )
N (t ) 100t 5t 2 , 0 t 10 C ( N ) 15, 000 8000 N
C ( N (t )) C 100t 5t 2
15, 000 8000 100t 5t 2
15, 000 800, 000t 40, 000t 2
Thus, C (t ) 15, 000 800, 000t 40, 000t 2 . 62. A(r ) r 2
r (t ) 200 t
2 3 t , t0 3
A r (t ) A 200 t 200 t
40, 000t 2
Thus, A(t ) 40, 000t .
2 S (r (t )) S t 3 3
63.
2
2 4 t 3 3 4 4 t 6 9 16 6 t 9 16 Thus, S (t ) t 6 . 9
1 p x 100, 0 x 400 4 1 x 100 p 4 x 4(100 p ) x 600 25 4(100 p) 600 25 2 100 p 600, 0 p 100 25 2 100 p 600, 0 p 100 . Thus, C ( p) 25 C
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Chapter 6: Exponential and Logarithmic Functions K C F .
1 p x 200, 0 x 1000 5
64.
5 K C F F 32 273 9 5 F 32 273 9 5 160 F 273 9 9 5 2297 5 F 2297 F or 9 9 9
1 x 200 p 5 x 5(200 p ) x 400 10 5(200 p ) 400 10 1000 5 p 400, 0 p 200 10 1000 5 p 400, 0 p 200 . Thus, C ( p) 10 C
2
65. V r h
b. 69. a.
h 2r 2
V (r ) r (2r ) 2r
3
5 80 2297 9
299.7 kelvins
f p p 200
b.
g p 0.80 p
c.
f g p f g p 0.80 p 200 0.80 p 200 This represents the final price when the rebate is issued on the sale price.
1 66. V r 2 h h 2r 3 1 2 V (r ) r 2 (2r ) r 3 3 3
67.
K C 80
g f p g f p 0.80 p 200
f x = the number of Euros bought for x dollars;
a.
f x 0.9101x
0.80 p 160 This represents the final price when the sale price is calculated after the rebate is given.
b.
g x 128.4337 x
Appling the 20% first is a better deal since a larger portion will be removed up front.
c.
g f x g f x g 0.9101x 128.4337 0.9101x
g x = the number of yen bought for x Euros
70. G (h) 20h ; N (G (h)) 0.8G (h) ( N G )(h) (0.8)(20h) 16h
The net salary for working h hours is given by ( N G )(h) 16h
116.8875104 x
d.
g f 1000 116.8875104 1000
71.
116,887.5104 yen
68. a.
f g ( x) ( x 2 4)3 ( x 2 4)2 16( x 2 4) 16 x 6 11x 4 24 x x 2 ( x 4 11x 2 24)
5 Given C F F 32 and 9 K C C 273 , we need to find
x 2 ( x 2 3)( x 2 8) x 2 0 or ( x 2 3) 0 or ( x 2 8) 0 x0
x 3
x 8 2 2
The zeros are 0, 3, 3, 2 2, 2 2 .
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Section 6.1: Composite Functions
72.
( g f )( x) g f x
f g ( x) 2( x 5)3 3( x 5)2 8( x 5) 12 3
g f x since f is odd
2
2 x 27 x 112 x 147
g f x
( x 7)( x 3)(2 x 7) x 7 0 or x 3 0 or 2 x 7 0 x 7
( g f )( x) So, g f is even.
7 x 3
x 3
( f g )( x) f g x
7 The zeros are 7, , 3 . 2
73.
since g is even
f g x since g is even
f x ax b; g ( x ) bx a
f g x since f is odd
f (1) 8; f ( g (20)) g ( f (20)) 14
( f g )( x)
We will solve as a system of equations. The first equation is f 1 a (1) b a b 8 . The
So, f g is even. 76. ( f g )( x) f ( g ( x)) f (ax b)
second is:
f (20b a) g (20a b) 14
(ax b) 2 5(ax b) c
a(20b a ) b b(20a b) a 14 2
a 2 x 2 2abx b 2 5ax 5b c
2
20ab a b 20ab b a 14
a 2 x 2 2ab 5a x b 2 5b c
a 2 b b 2 a 14
We know ( f g )( x) 4 x 2 22 x 31 , so
Now we substitute from the first equation, a 8b.
a 2 4, 2ab 5a 22, b 2 5b c 31 . Now,
(8 b) 2 b b 2 (8 b) 14 2
a 2 4 a 2 or a 2. If a = 2, then 2(2)b 5(2) 22 4b 10 22 4b 12 b3
2
64 16b b b b 8 b 14 14b 70 b5
Substituting back into the first equation to solve for a gives a 3 . So the product is:
(3) 2 5(3) c 31 9 15 c 31 c7 If a 2 : 2(2)b 5(2) 22 4b 10 22 4b 32 b 8 (8) 2 5(8) c 31 64 40 c 31 c7 So a 2, b 3, c 7 or a 2, b 8, c 7
ab (3)(5) 15
74. Since f and g are odd functions, f x f x
and g x g x Then ( f g )( x) f g x f g x f g x
since g is odd since f is odd
( f g )( x) So, f g is odd.
75. Since f is an odd function, we know that f x f x . Since g is an even function, g x g x . Also,
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Chapter 6: Exponential and Logarithmic Functions ( f g )( x) 3 x 8 x 5 2 x 13
77. h(2) 2 7 9 3 1 f (h(2)) g (3) 3 1 f ( g (h(2)) f 3 1 6 7 2 7 5 3 ( f g h)(2) 5
78. From
Domain: all real numbers ( f g )( x) (3 x 8)( x 5) 3 x 2 7 x 40
Domain: all real numbers f 3x 8 g ( x) x 5
x 3, x 3 .
From ( g h)( x)
Domain: x | x 5
x3 x3 2
,
80. 2 x 5 x 2 0 (2 x 1)( x 2) 0
x3 2 0
2 x 1 0 or 1 x 2 1 x 4
x3 2 x3 4 x 1
From ( f g h)( x)
x3 x3 2
x3 4
1 x3 4 x32
,
x2 x4
1 The real zeros are , 4 4
4 0
x 2 6 x 5 ( x 5)( x 1) x3 x3 where the domain is x | x 3
x 3 2 0
81. R ( x)
5 x3 8 25( x 3) 64 25 x 75 64 25 x 11 11 x 25 So the domain of ( f g h)( x) is
The degree of the numerator in lowest terms is n 2 . The degree of the denominator in lowest terms is m 1 . Since n m , There is no horizontal asymptote. The denominator in lowest terms is zero at x 3 , so x 3 is a vertical asymptote. x9 x 3 x 6x 5
11 x | x 3, x , x 1 or 25 11 11 3, 25 25 ,1 1, .
79.
x 2 0
2
x 2 3x
9x 5 9 x 27
f x 3 x 8; g ( x) x 5
32 32 R ( x) x 9 , x3 x3 Thus, the oblique asymptote is y x 9 .
( f g )( x) 3 x 8 x 5 4x 3
Domain: all real numbers
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Section 6.2: One-to-One Functions; Inverse Functions
82.
b 2 3 2 2a 3 1 f (3) (3) 2 2(3) 5 8 3 The vertex is (3,8) . The axis of symmetry is x 3 and since a is negative the graph is concave down.
86.
x
52 (15) 2 250 25 10 5 10
87.
83. x 2 6 x 7 0 We graph the function f ( x) x 2 6 x 7 . The intercepts are y-intercept: f (0) 7
x-intercepts:
(2 (3)) 2 (7 8) 2
x2 6x 7 0 ( x 7)( x 1) 0 x 7, x 1
88.
3(c 1) 3( x 1) 3 3 x c x 1)(c 1) ( 1)( 1) ( x 1 c 1 xc xc 3(c 1) 3( x 1) 3c 3 x ( x 1)(c 1) ( x 1)(c 1) xc xc 1 3(c x) 3 x c ( x 1)(c 1) ( x 1)(c 1)
x3 (9 x 2 ) x(9 x 2 )
1
1
1 2 2 x (9 x 2 ) 2 0
2 ( x 2 2 x (9 x 2 )) 0
x(9 x 2 )
1
x 2 (9 x 2 )
2 (3 x 2 18 x ) 0 1
x 2 0 or (9 x 2 ) x0
2 ( x 18) 0
1
2 0 or ( 3 x 2 18) 0
x 3
3 x 2 18 x 6
But substituting x 3 into the original equation gives an undefined situation so the solution set is
0, 6, 6
The graph is below the x-axis for 1 x 7 . Since the inequality is inclusive, the solution set is x 1 x 7 or, using interval notation,
1, 7 . 84.
Section 6.2
a 2 b2 c2 (1) 2 b 2 22
1.
2
1 b 4 b2 3
2. The function f ( x) x 2 is increasing on the
b 3
85.
x2 x2 f 4 2 x22 x . 4 4
interval 0, . It is decreasing on the interval
, 0 .
x 2 3 x 7 2 x 3 x2 5x 4 0
3. The function is not defined when x 2 3 x 18 0 . Solve: x 2 3x 18 0 ( x 6)( x 3) 0 x 6 or x 3 The domain is {x | x 6, x 3} .
( x 4)( x 1) 0 x 4, x 1 g (4) 2(4) 3 11 g (1) 2(1) 3 5
The points of intersection are: (4,11), (1,5) . 611
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Chapter 6: Exponential and Logarithmic Functions
4.
1 1 x 1 x 1 x x x2 x 2 1 1 x 1 1 x 2 2 x2 x x x2
19. The function is one-to-one because there are no two distinct inputs that correspond to the same output. 20. The function is one-to-one because there are no two distinct inputs that correspond to the same output.
2 1 x x x 1 x 2
x2 1 x x (1 x)(1 x) x ; x 0,1, 1 (1 x)
5.
21. The function f is one-to-one because every horizontal line intersects the graph at exactly one point. 22. The function f is one-to-one because every horizontal line intersects the graph through at most one point.
f x1 f x2
23. The function f is not one-to-one because there are horizontal lines (for example, y 1 ) that intersect the graph at more than one point.
6. one-to-one 7. 3 8. y x 9.
24. The function f is not one-to-one because there are horizontal lines (for example, y 1 ) that intersect the graph at more than one point.
4,
25. The function f is one-to-one because every horizontal line intersects the graph through at most one point.
10. True 11. a
26. The function f is not one-to-one because the horizontal line y 2 intersects the graph at more than one point.
12. d 13. The function is one-to-one because there are no two distinct inputs that correspond to the same output.
27. To find the inverse, interchange the elements in the domain with the elements in the range:
14. The function is one-to-one because there are no two distinct inputs that correspond to the same output. 15. The function is not one-to-one because there are two different inputs, 20 Hours and 25 Hours, that correspond to the same output, $380. 16. The function is not one-to-one because there are two different inputs, John and Chuck, that correspond to the same output, Phoebe.
Domain: {49.7, 43.8, 4.2, 61.9, 12.8} Range: {Atlanta, Boston, Las Vegas, Miami, Los Angeles}
17. The function is not one-to-one because there are two distinct inputs, 2 and 3 , that correspond to the same output. 18. The function is one-to-one because there are no two distinct inputs that correspond to the same output.
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Section 6.2: One-to-One Functions; Inverse Functions 28. To find the inverse, interchange the elements in the domain with the elements in the range:
33. Interchange the entries in each ordered pair: {(1, 2), (2, 3), (0, 10), (9,1), (4, 2)} Domain: {1, 2, 0, 9, 4} Range: {2, 3, 10, 1, 2} 34. Interchange the entries in each ordered pair: {(8, 2), (1, 1), (0, 0), (1,1), (8, 2)} Domain: {8, 1, 0, 1, 8} Range: {2, 1, 0, 1, 2}
Domain: {$937, $679, $786, $761, $700} (in millions)
35. Graphing the inverse:
Range: {Star Wars: The Force Awakens, Avengers: Endgame, Avatar, Black Panther, Avengers: Infinity War} 29. To find the inverse, interchange the elements in the domain with the elements in the range: Monthly Cost of Life Insurance $213 $328 $532
Age 30 40 50
36. Graphing the inverse:
Domain: {$213, $328, $532} Range: {30, 40, 50} 30. To find the inverse, interchange the elements in the domain with the elements in the range: Unemployment Rate 3.3% 5.3% 3.6% 4.8%
State Virginia Nevada Tennessee Texas 37. Graphing the inverse:
Domain: {3.8%, 5.3%, 3.6%, 4.8%} Range: {Virginia, Nevada, Tennessee, Texas} 31. Interchange the entries in each ordered pair: {(5, 3), (9, 2), (2, 1), (11, 0), (5,1)} Domain: {5, 9, 2, 11, 5} Range: {3, 2, 1, 0, 1} 32. Interchange the entries in each ordered pair: {(2, 2), (6, 1), (8, 0), (3,1), (9, 2)} Domain: {2, 6, 8, 3, 9} Range: {2, 1, 0, 1, 2}
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Chapter 6: Exponential and Logarithmic Functions
Thus, f and g are inverses of each other.
38. Graphing the inverse: 42.
f ( x) 3 2 x;
1 g ( x) ( x 3) 2
1 f g ( x) f ( x 3) 2 1 3 2 ( x 3) 2 3 ( x 3) x g f ( x) g (3 2 x)
39. Graphing the inverse:
1 (3 2 x) 3 2 1 ( 2 x) 2 x
Thus, f and g are inverses of each other.
43. 40. Graphing the inverse:
41.
f ( x) 3x 4;
y=x
f ( x) 4 x 8;
g ( x)
x 2 4
x f g ( x) f 2 4 x 4 2 8 4 x 88 x g f ( x) g (4 x 8) 4x 8 2 4 x22
1 g ( x) ( x 4) 3
x
Thus, f and g are inverses of each other.
1 f g ( x) f ( x 4) 3 1 3 ( x 4) 4 3 ( x 4) 4 x
44.
f ( x) 2 x 6;
g ( x)
1 x 3 2
1 f g ( x) f x 3 2 1 2 x 3 6 x 6 6 2 x
g f ( x) g (3 x 4) 1 (3 x 4) 4 3 1 (3 x) x 3
614
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Section 6.2: One-to-One Functions; Inverse Functions g f ( x) g (2 x 6)
g f ( x) g x x
1 (2 x 6) 3 x 3 3 2 x
Thus, f and g are inverses of each other. 49. f ( x)
Thus, f and g are inverses of each other. 45.
f ( x) x3 8;
4x 3 f g ( x) f , x 2 2 x 4x 3 2 3 2 x 4x 3 4 2 x 4x 3 2 2 x 3 (2 x) 4x 3 4 (2 x ) 2 x 2(4 x 3) 3(2 x) 4 x 3 4(2 x) 8 x 6 6 3x 4x 3 8 4x 5x 5 x
g ( x) 3 x 8
x 8 x 8 8
f g ( x) f
3
3
3
x 88 x g f ( x) g ( x3 8) 3 ( x3 8) 8 3 x3 x
Thus, f and g are inverses of each other. 46.
f ( x) ( x 2) 2 , x 2;
f g ( x) f
g ( x) x 2
x 2
x 2 2 x
2x 3 g f ( x) g , x 4 x4 2x 3 4 3 x4 2x 3 2 x4 2x 3 4 x 4 3 ( x 4) 2x 3 2 ( x 4) x4 4(2 x 3) 3( x 4) 2( x 4) (2 x 3) 8 x 12 3 x 12 2x 8 2x 3 5x 5 x Thus, f and g are inverses of each other.
2
2
x
g f ( x) g ( x 2) 2
2
( x 2) 2 x22 x Thus, f and g are inverses of each other.
47.
1 f ( x) ; x
g ( x)
1 x
x 1 1 f g ( x) f 1 x 1 1 x x x 1 1 g f ( x) g 1 x 1 x 1 x Thus, f and g are inverses of each other.
48.
f ( x) x;
2x 3 4x 3 ; g ( x) x4 2 x
g ( x) x
f g ( x) f x x
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Chapter 6: Exponential and Logarithmic Functions
50.
f ( x)
x 5 ; 2x 3
g ( x)
3x 5 1 2x
1 3x 5 f g ( x) f , x 2 1 2x 3x 5 5 1 2x 3x 5 2 3 1 2x 3x 5 5 (1 2 x) x 1 2 3x 5 2 1 2 x 3 (1 2 x) 3 x 5 5(1 2 x) 2(3 x 5) 3(1 2 x ) 3 x 5 5 10 x 6 x 10 3 6 x 13 x 13 x
b. domain of f range of f 1 all real numbers c/ range of f domain of f 1 all real numbers c.
3 x5 g f ( x) g , x 2 2x 3 x5 3 5 2x 3 x5 1 2 2x 3 x5 3 2 x 3 5 (2 x 3) x 5 1 2 2 x 3 (2 x 3) 3( x 5) 5(2 x 3) 1(2 x 3) 2( x 5) 3x 15 10 x 15 2 x 3 2 x 10 13 x 13 x Thus, f and g are inverses of each other.
51. a.
1 1 Verifying: f f 1 ( x) f x 3 x x 3 3 1 f 1 f ( x) f 1 (3x) (3x) x 3
f ( x) 4 x y 4x x 4 y Inverse x y 4 1 1 f ( x) x 4 Verifying: 1 1 f f 1 ( x) f x 4 x x 4 4 1 f 1 f ( x) f 1 ( 4 x) ( 4 x) x 4 b. domain of f range of f 1 all real numbers
52. a.
range of f domain of f 1 all real numbers
c.
f ( x) 3 x y 3x x 3 y Inverse x y 3 1 f 1 ( x) x 3
616
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Section 6.2: One-to-One Functions; Inverse Functions f ( x) 4 x 2 y 4x 2 x 4y 2 4y x 2 x2 y 4 x 1 y 4 2 x 1 1 f ( x) 4 2
53. a.
b. domain of f range of f 1 all real numbers
Inverse
range of f domain of f 1 all real numbers
c.
Verifying:
x 1 x 1 f f 1 ( x) f 4 2 4 2 4 2 x22 x 4x 2 1 f 1 f ( x) f 1 4 x 2 4 2 1 1 x x 2 2 b. domain of f range of f 1 all real numbers
f ( x) x3 1
55. a.
y x3 1 x y3 1
Inverse
3
y x 1
range of f domain of f 1 all real numbers
y 3 x 1
c.
f
1
( x) 3 x 1
Verifying: f f 1 ( x) f
x 1 x 1 1 3
x 1 1 x
3
3
f 1 f ( x) f 1 x3 1 3 x3 1 1 3 x3 x
b. domain of f range of f 1 all real numbers range of f domain of f 1 all real numbers
f ( x) 1 3x y 1 3x x 1 3y 3y 1 x 1 x y 3 1 x 1 f ( x) 3
54. a.
c.
Inverse
Verifying: 1 x 1 x f f 1 ( x ) f 1 3 3 3 1 (1 x) x
f 1 f ( x) f 1 (1 3x)
1 (1 3 x) 3 x x 3 3
617
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Chapter 6: Exponential and Logarithmic Functions
56. a.
f ( x) x3 1
b. domain of f range of f 1 x | x 0
y x3 1
range of f domain of f 1 x | x 4
x y3 1
Inverse
c.
3
y x 1 y 3 x 1 f 1 ( x ) 3 x 1
Verifying:
f f 1 ( x ) f
x 1 x 1 1 3
3
3
x 1 1 x
f
1
f ( x) f 1 x3 1 3 x3 1 1 3 x3 x
b. domain of f range of f 1 all real numbers range of f domain of f
1
f ( x ) x 2 9, x 0
58. a.
all real numbers
y x 2 9, x 0
c.
x y 2 9, y 0
Inverse
2
y x 9, x 9 y x 9, x 9 f
1
( x) x 9, x 9
x 9 x 9 9
Verifying: f f 1 ( x) f
2
x 99 x
f 1 f ( x) f 1 x 2 9
f ( x ) x 2 4, x 0
57. a.
2
y x 4, x 0 2
x y 4, y 0
Inverse
y x 4, x 4 y x 4, x 4 f
x 4 x 4 4
Verifying: f f 1 ( x) f
1 b. domain of f range of f x | x 0 range of f domain of f 1 x | x 9
2
x44 x
f 1 f ( x) f 1 x 2 4
2
x, x 0
( x) x 4, x 4
x 9 9
x2 x
2
1
x 4 4 2
x2 x x, x 0
618
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Section 6.2: One-to-One Functions; Inverse Functions c.
a. f ( x)
60.
3 x 3 x y xy 3 3 y x 3 f 1 ( x) x Verifying:
3 x
y
f ( x)
59. a.
4 x 4 x y xy 4 4 y x 4 f 1 ( x ) x Verifying:
4 x
3 3 x f f 1 ( x) f 3 x 3 x 3 x 3 3 x f 1 f ( x) f 1 3 x 3 x 3 x b. domain of f range of f 1 x | x 0
y
Inverse
Inverse
range of f domain of f 1 x | x 0
c.
4 4 x f f 1 ( x) f 4 x x 4 4 x 4 4 x f 1 f ( x) f 1 4 x x 4 4 x b. domain of f range of f 1 x | x 0
range of f domain of f 1 x | x 0 61.
c.
a. f ( x)
1 x2
1 x2 1 x Inverse y2 xy 2 x 1 xy 2 x 1 2x 1 y x x 1 2 f 1 ( x) x y
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Chapter 6: Exponential and Logarithmic Functions
Verifying:
Verifying: 4 4 2x f f 1 ( x) f x 4 2x 2 x 4 x 4x 4x x 4 2x 2x 4 4 2x x 2 x 4 4 2 4 x2 f 1 f ( x) f 1 4 x2 x2 4 4 2 x 2 ( x 2) 4( x 2) 2(4) 4 4 ( x 2) x2 4x 8 8 4x x 4 4 b. domain of f range of f 1 x | x 2
1 2x 1 f f 1 ( x) f x 2x 1 2 x 1 x x 2x 1 2x 2x 1 2 x x x x 1 1 2 1 1 x2 f 1 f ( x) f 1 1 x2 x2 1 2 x 2 1 ( x 2) 1 ( x 2) x2 2 ( x 2) x x 1 1 domain of f range of f 1 x | x 2 b. range of f domain of f 1 x | x 0
range of f domain of f 1 x | x 0
c.
c.
62.
2 3 x 2 y 3 x 2 x Inverse 3 y x(3 y ) 2 3 x xy 2 xy 2 3 x 2 3x y x 2 3x 1 f ( x) x
63. a.
4 x2 4 y x2 4 x Inverse y2 x( y 2) 4 xy 2 x 4 xy 4 2 x 4 2x y x 4 2x 4 1 or f 1 x 2 f ( x) x x
a.
f ( x)
f ( x)
620
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Section 6.2: One-to-One Functions; Inverse Functions
Verifying:
Verifying:
2 2 3x f f 1 ( x) f x 3 2 3x x 2 x 2x 2 3x 3x 2 3x 3 x x 2x x 2
f
1
4 4 f 1 f ( x) f 1 2 4 2 x 2 x 2 x 2 4 2 2 x 4 22 x x
2 2 3 3 x f ( x) f 2 3 x 2 2 3 3 x (3 x) 2 (3 x) 3 x 2(3 x ) 3(2) 6 2 x 6 2 2 2x x 2 1
2 3 x
b. domain of f range of f 1 x | x 2
range of f domain of f 1 x | x 0 3x x2 3x y x2 3y x y2 x y 2 3 y
65. a. f ( x)
b. domain of f range of f 1 x | x 3 range of f domain of f 1 x | x 0
64.
4 4 f f 1 ( x) f 2 4 x 22 x x 4 4 4 x 4 4 4 22 x x
Inverse
xy 2 x 3 y xy 3 y 2 x y x 3 2 x
4 a. f ( x ) 2 x 4 y 2 x 4 x Inverse 2 y x(2 y ) 4 2 x xy 4 xy 2 x 4
2 x x 3 2 x 1 f x x 3 y
Verifying: 2 x f f 1 ( x) f x 3 2 x 3 2 x ( x 3) 3 x 3 x 3 2 x 2 x 2 2 ( x 3) x 3 x3 6 x 6 x x 2 x 2 x 6 6
2x 4 y x 4 y 2 x 4 f 1 ( x) 2 x
621
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Chapter 6: Exponential and Logarithmic Functions
3x f 1 f ( x) f 1 x2 3x 2 3 x ( x 2) 2 x 2 x 2 3x 3x 3 3 ( x 2) x2 x2 6 x 6 x x 3 x 3 x 6 6
f
1
b. domain of f range of f 1 x | x 2 range of f domain of f 1 x | x 3 2x x 1 2x y x 1 2y x y 1 x y 1 2 y f ( x)
66. a.
b. domain of f range of f 1 x | x 1
range of f domain of f 1 x | x 2 2x 3x 1 2x y 3x 1 2y Inverse x 3 y 1 3xy x 2 y 3xy 2 y x y (3x 2) x x y 3x 2 x 1 f ( x) 3x 2
Inverse
f ( x)
67. a.
xy x 2 y xy 2 y x y x 2 x x x2 x 1 f x x2 y
Verifying: x 2 x x2 f f 1 ( x) f x x2 1 x2 x 2 x 2 ( x 2) x 1 ( x 2) x2 2 x x ( x 2) 2 x 2 x
2x x 1 f ( x) f x 1 2x 2 x 1 2x ( x 1) x 1 2x 2 ( x 1) x 1 2 x 2 x 2 x 2 2 x 2 x 1 2 x
Verifying: x 2 x 3x 2 f f 1 ( x) f 3x 2 3 x 1 3x 2 x 2 3 x 2 (3x 2) x 3 3 x 2 1 (3 x 2) 2x 2x x 3x (3 x 2) 2
622
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Section 6.2: One-to-One Functions; Inverse Functions
f
1
2x 2x 1 3 x f ( x) f 3x 1 3 2 x 2 3x 1 2x (3 x 1) 3x 1 2x 3 3x 1 2 (3 x 1) 2x 3(2 x) 2(3x 1) 2x 2x x 6x 6x 2 2
3x 1 f 1 f ( x) f 1 x 1 1 3x 1 3x 1 3 3 x x 1 x x 3x 1 3x 3x 1 3 x x x x 1
b. domain of f range of f 1 x | x 0 range of f domain of f 1 x | x 3
1 b. domain of f range of f x | x 3 2 range of f domain of f 1 x | x 3 1
3x 4 2x 3 3x 4 y 2x 3 3y 4 x 2y 3 x(2 y 3) 3 y 4 2 xy 3 x 3 y 4 2 xy 3 y 3x 4 y (2 x 3) 3x 4 3x 4 y 2x 3 x4 3 f 1 ( x) 2x 3 Verifying:
3x 1 x 3x 1 y x 3y 1 x Inverse y xy (3 y 1) xy 3 y 1 xy 3 y 1 y ( x 3) 1 1 y x3 1 1 f ( x) x3 f ( x)
68. a.
Inverse
3x 4 3 4 3x 4 2x 3 f f 1 ( x) f 2 x 3 2 3x 4 3 2x 3 3x 4 3 2 x 3 4 (2 x 3) 3x 4 2 2 x 3 3 (2 x 3) 3(3 x 4) 4(2 x 3) 2(3 x 4) 3(2 x 3) 9 x 12 8 x 12 17 x x 6x 8 6x 9 17
Verifying: 1 f f 1 ( x) f x3 1 3 3 1 1 x 3 x3 1 1 x 3 x 3 3 x3 1 x 3 1 3 ( x 3) x
f ( x)
69. a.
623
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Chapter 6: Exponential and Logarithmic Functions
3x 4 3 4 x 3 4 2x 3 1 1 f f ( x) f 2 x 3 2 3x 4 3 2x 3 3x 4 3 2 x 3 4 (2 x 3) 3x 4 2 2 x 3 3 (2 x 3) 3(3 x 4) 4(2 x 3) 2(3 x 4) 3(2 x 3) 9 x 12 8 x 12 17 x x 6x 8 6x 9 17
2x 3 f 1 f ( x) f 1 x4 2x 3 4 3 x4 2x 3 2 x4 4 2 x 3 3 x 4 2 x 4 2 x 3
3 b. domain of f range of f 1 x | x 2 3 range of f domain of f 1 x | x 2
b. domain of f range of f 1 x | x 4
8 x 12 3 x 12 2x 8 2x 3 11x 11 x
range of f domain of f 1 x | x 2 2x 3 x2 2x 3 y x2 2y 3 x Inverse y2 xy 2 x 2 y 3 xy 2 y 2 x 3 y ( x 2) 2 x 3 2x 3 y x2 2x 3 f 1 ( x) x2
2x 3 x4 2x 3 y x4 2y 3 x Inverse y4 x( y 4) 2 y 3 xy 4 x 2 y 3 xy 2 y 4 x 3 y ( x 2) (4 x 3) (4 x 3) 4 x 3 y 2 x x2 4 3 x f 1 ( x) 2 x
a.
70.
f ( x)
Verifying: 2x 3 2 3 2x 3 x2 f f 1 ( x) f 2x 3 x2 2 x2 2x 3 2 x 2 3 ( x 2) 2x 3 2 ( x 2) 2 x 2(2 x 3) 3( x 2) 2 x 3 2( x 2) 4 x 6 3x 6 x x 2 x 3 2 x 4 1
Verifying: 4x 3 2 3 4 3 x 2 x 1 f f ( x) f 4x 3 2 x 4 2 x 2(4 x 3) 3(2 x) 4 x 3 4(2 x) 8 x 6 6 3x 4x 3 8 4x 11x 11 x
f ( x)
71. a.
624
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Section 6.2: One-to-One Functions; Inverse Functions 3 x 4 f 1 f ( x) f 1 x2 3x 4 2 4 x2 3 x 4 3 x2 2(3 x 4) 4( x 2) 3 x 4 3( x 2) 6x 8 4x 8 3 x 4 3 x 6 10 x 10 x
2x 3 2 3 x2 2x 3 2 x2 2x 3 2 x 2 3 ( x 2) 2x 3 2 ( x 2) x2 2(2 x 3) 3( x 2) 2 x 3 2( x 2) 4 x 6 3 x 6 x x 1 2x 3 2x 4
2x 3 f 1 f ( x) f 1 x2
b. domain of f range of f 1 x | x 2
b. domain of f range of f 1 x | x 2
range of f domain of f 1 x | x 2 3 x 4 x2 3 x 4 y x2 3 y 4 x y2 x( y 2) 3 y 4 xy 2 x 3 y 4 xy 3 y 2 x 4 y ( x 3) 2 x 4 2x 4 y x3 2x 4 1 f ( x) x3
range of f domain of f 1 x | x 3
f ( x)
72. a.
73.
f ( x) y
Inverse x
x2 4
2 x2 x2 4 2 x2 y2 4
, x0
, x0
2 y2
, y0
1 2 1 2 2 x 2 xy y 4, 2 1 2 y 2 x 1 4, x 2 1 2 y 1 2 x 4, x 2 4 1 y2 x , 1 2x 2 4 1 y , x 1 2x 2 2 1 y , x 2 1 2x 2 1 f 1 x , x 2 1 2x 2 xy 2 y 2 4,
Verifying: 2x 4 f f 1 ( x) f x3 2x 4 3 4 x3 2x 4 2 x3 3(2 x 4) 4( x 3) 2 x 4 2( x 3) 6 x 12 4 x 12 2x 4 2x 6 10 x 10 x
a.
625
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x
Inverse
Chapter 6: Exponential and Logarithmic Functions
Verifying:
Verifying: 2
2
2 4 2 1 2x 1 f f ( x) f 2 1 2x 2 2 1 2x 4 4 4 (1 2 x) 4 1 2x x 1 2 4 4 2 2 (1 2 x) 1 2x 1 2x 4 4(1 2 x) 4 4 8 x 8 x x 2(4) 8 8
3 3 3 x 1 3 1 f f ( x) f 2 3x 1 3 3 3x 1 3 3 3 (3 x 1) 3 3 1 x 3x 1 3 3 3 3 (3x 1) 3x 1 3x 1 3 3(3x 1) 3(3) 3 9x 3 9 9x 9 x
x2 4 f 1 f ( x) f 1 2 2x
2 x2 4 1 2 2 2x 2
2
4 x2 4 11 2 1 2 x x 2 2 |x| 2 2 2 4 2 |x| x
x2 3 3 f 1 f ( x) f 1 2 2 x 3 3x 1 3 2 3x
x x, x 0
b. domain of f range of f 1 x | x 0 range of f domain of f
74. a.
f ( x) y x
x2 3 3x 2 x2 3 3x 2 y2 3 3 y2
,
1
x0
,
y0
x 3 1 x2
3 3 1 2 1 x
x2 3 3 x 2 3 3 2 x x x, x 0
b. domain of f range of f 1 x | x 0 1 range of f domain of f 1 x | x 3 Inverse
75. a.
2
f ( x) x 3 4 2
y x3 4
1 3 xy y 3, x 3 1 2 2 3xy y 3, x 3 1 y 2 3 x 1 3, x 3 3 1 2 y , x 3x 1 3 3 1 y , x 3x 1 3 3 1 f 1 x , x 3x 1 3 2
2
1 x | x 2
x0
,
3
2
2
x y3 4
Inverse
2 3
y x4 3
y ( x 4) 2 , x 4 f 1 ( x) ( x 4) 2 , x 4 3
Verifying: f f 1 ( x) f ( x 4) 2
3
3
( x 4) 2
4 2 3
x44 x
626
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Section 6.2: One-to-One Functions; Inverse Functions
f 1 f ( x) f 1 x 3 4
2
f 1 f ( x) f 1 3 x5 2
x 4 4 2 3
3 2
5
x,since x 0
x
3 2
2 3
range of f domain of f
76. a.
x | x 4
78. a.
3 2
f ( x ) 5 x3 13
y x 5
y 5 x3 13
3 2
x y 5
Inverse
x 5 y 3 13
3 2
y x5
3
y 3 x5 13
f 1 ( x) ( x 5) 3 , x 5 2
Inverse
5
y x 13
2
y ( x 5) 3 , x 5
range of f domain of f 1 all real numbers
3
Verifying: f f
( x) f ( x 5)
( x 5)
2 3
2 3
f 1 ( x) 3 x5 13
5 3 2
f 1 f ( x) f 1 x 2 5 3
3
5
5
3
5
x 5 5 3 2
3
5 x5 13 13 5 x5 x
2 3
f 1 f ( x) f 1 5 x3 13
x,since x 0
x
3 2
x 13 x 13 13
Verifying: f f 1 ( x) f
x 5 5 x,since x 5
2 3
b. domain of f range of f 1 all real numbers
f ( x) x 2 5
1
x5 2
5 x5 2 2 5 x5 x
b. domain of f range of f 1 x | x 0 1
3
2 3
3
b. domain of f range of f 1 x | x 0
x 13 2 5
5
3
3 x 3 13 13 3 x3 x
range of f domain of f 1 x | x 5
b. domain of f range of f 1 all real numbers 77. a.
3
5
range of f domain of f 1 all real numbers
f ( x) x 2 y 3 x5 2 x 5
5
y 2
3
f ( x ) 19 ( x 1) 2 2
79. a.
Inverse
y 19 ( x 1) 2 2
3
y x 2
x 19 ( y 1) 2 2
y 5 x3 2
x 2 19 ( y 1) 2
f 1 ( x) 5 x3 2
Verifying: f f
1
Inverse
9( x 2) 1 y
x 2 x 2 2 5
( x) f
5
3
3
y 3 x 2 1, x 2 f
5
3
3 x3 2 2 3 x3 x
627
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1
( x ) 3 x 2 1, x 2
Chapter 6: Exponential and Logarithmic Functions
f 1 f ( x) f 1 2 x 3 5
Verifying: f f 1 ( x) f 3 x 2 1
2
19 (3 x 2 1) 1 2 2
19 (9( x 2) 2
2
2 x3 3 2 ( x 3) 3 x
( x 2) 2 x f
f ( x)
f 1 19 ( x 1) 2 2
3
( x 1) 2 2 1
3
1 ( x 1) 2 1 3 13 9
1 9
b. domain of f range of f 1 x | x 3
2
range of f domain of f 1 x | x 5
( x 1) 1 81. a.
( x 1) 1 x,since x 1
b. domain of f range of f 1 x | x 1 range of f domain of f 1 x | x 2
c.
y 2 x3 5 x 2 y 3 5
Inverse
d. Because the ordered pair (1, 2) is on the graph, f 1 (2) 1 .
2
x5 y3 2
82. a. 2
x5 y 3, x 5 2 f
2
x5 ( x) 3, x 5 2
1 Because the ordered pair 2, is on the 2 1 graph, f (2) . 2
b. Because the ordered pair (1, 0) is on the graph, f (1) 0 .
Verifying:
c.
x 5 2 f f 1 ( x) f 3 2
Because the ordered pair (0,1) is on the graph, f 1 (1) 0 .
x5 2 y3
1
Because the ordered pair (1, 0) is on the graph, f (1) 0 .
b. Because the ordered pair (1, 2) is on the graph, f (1) 2 .
f ( x) 2 x 3 5
80. a.
2
2 x 3 5 5 3 2
19 (3 x 2 2
1
Because the ordered pair (1, 0) is on the graph, f 1 (0) 1 .
d. Because the ordered pair (0, 1) is on the
x 5 2 2 3 3 5 2
graph, f 1 (1) 0 . 83. Since f 7 13 , we have f 1 13 7 ; the input
2
x5 2 5 2 x5 2 5 2 ( x 5) 5 x,since x 5
of the function is the output of the inverse when the output of the function is the input of the inverse. 84. Since g 5 3 , we have g 1 3 5 ; the input
of the function is the output of the inverse when the output of the function is the input of the inverse. 85. Since the domain of a function is the range of the inverse, and the range of the function is the domain of the inverse, we get the following for f 1 :
Domain: 2, 628
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Range: 5,
Section 6.2: One-to-One Functions; Inverse Functions 93. If a, b is on the graph of f, then b, a is on the
86. Since the domain of a function is the range of the inverse, and the range of the function is the domain of the inverse, we get the following for f 1 :
Domain: 5,
graph of f 1 . Since the graph of f 1 lies in quadrant I, both coordinates of (a, b) are positive, which means that both coordinates of (b, a) are
Range: 0,
positive. Thus, the graph of f 1 must lie in quadrant I.
87. Since the domain of a function is the range of the inverse, and the range of the function is the domain of the inverse, we get the following for g 1 :
Domain: 0,
94. If a, b is on the graph of f, then b, a is on the
Range: , 0
graph of f 1 . Since the graph of f lies in quadrant II, a must be negative and b must be positive. Thus, (b, a) must be a point in quadrant IV, which
88. Since the domain of a function is the range of the inverse, and the range of the function is the domain of the inverse, we get the following for g 1 :
Domain: 0,8
means the graph of f 1 lies in quadrant IV.
Range: 0,15
95. Answers may vary. One possibility follows: f ( x) x , x 0 is one-to-one.
89. Since f x is increasing on the interval 0,5 , it is
one-to-one on the interval and has an inverse, f 1 x . In addition, we can say that f 1 x is
f ( x ) x, x 0 y x, x 0
Thus,
increasing on the interval f 0 , f 5 .
f 1 ( x) x, x 0
96. Answers may vary. One possibility follows: f ( x) x 4 , x 0 is one-to-one.
90. Since f x is decreasing on the interval 0,5 , it is
one-to-one on the interval and has an inverse, f 1 x . In addition, we can say that f 1 x is
f ( x) x 4 , x 0
Thus,
y x4 , x 0
decreasing on the interval f (5), f (0) . 91.
92.
x y4
f ( x ) mx b, m 0 y mx b x my b Inverse x b my 1 y x b m 1 1 f ( x) x b , m 0 m
y x, x 0 f
97. a.
f ( x) r 2 x 2 , 0 x r y r 2 x2 x r2 y2
Inverse
4
b.
Inverse
( x) 4 x , x 0
d 6.97r 90.39 d 90.39 6.97r d 90.39 r 6.97 Therefore, we would write d 90.39 r d 6.97 r d r
6.97r 90.39 90.39
6.97 6.97r 90.39 90.39 6.97 r 6.97 6.97 r
x2 r 2 y 2 y 2 r 2 x2 y r 2 x2
1
d 90.39 d r d 6.97 90.39 6.97 d 90.39 90.39 d
f 1 ( x) r 2 x 2 , 0 x r
629
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Chapter 6: Exponential and Logarithmic Functions
c.
98. a.
300 90.39 56.01 6.97 If the distance required to stop was 300 feet, the speed of the car was roughly 56 miles per hour. r 300
c.
H C 2.15C 10.53 H 2.15C 10.53 H 10.53 2.15C H 10.53 C 2.15 H 10.53 C H 2.15
b.
d.
99. a.
2.15C 10.53 10.53
b.
26 10.53 16.99 2.15 The head circumference of a child who is 26 inches tall is about 17 inches. C 26
9 F C 32 5 9 F 32 C 5
5 9 C F C C 32 32 9 5 5 9 C C 9 5 95 F C F F 32 32 59 F 32 32 F
6 feet = 72 inches W 72 50 2.3 72 60
c.
50 2.3 12 50 27.6 77.6
The ideal weight of a 6-foot man is 77.6 kilograms. b.
80 88 168 73.04 2.3 2.3 The height of a man who is at his ideal weight of 80 kg is roughly 73 inches. h 80
5 F 32 C 9 Therefore, we would write 5 C F F 32 9
2.15 2.15C 10.53 10.53 2.15 2.15C C 2.15
c.
50 2.3 h 60 88
2.3 50 2.3h 138 88 2.3h h 2.3 2.3 W 88 60 W h W 50 2.3 2.3 50 W 88 138 W
100. a.
H 10.53 H C H 2.15 10.53 2.15 H 10.53 10.53 H C H C
h W h
101. a.
W 50 2.3 h 60 W 50 2.3h 138 W 88 2.3h W 88 h 2.3 Therefore, we would write W 88 h W 2.3
b.
C 70
5 5 70 32 38 21.1C 9 9
From the restriction given in the problem statement, the domain is g | 41, 775 g 89, 075 or 41775,89075 . T 41775 4807.5 0.22 41775 41775 4807.50 T 89075 4807.5 0.22 89075 41775
15, 213.50
Since T is linear and increasing, we have that the range is T | 4807.5 T 15, 213.5 or
4807.50,15213.50 .
630
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Section 6.2: One-to-One Functions; Inverse Functions
c.
T 4807.5 0.22 g 41, 775 T 4807.5 0.22 g 41, 775
H 4.9t 2 100
T 4807.5 g 41, 775 0.22 T 4807.5 41, 775 g 0.22 Therefore, we would write T 4807.5 g T 41, 775 0.22 Domain: T | 4807.5 T 15, 213.5
4.9t 2 100 H 100 H t2 4.9 100 H t 4.9 100 H . 4.9 (Note: we only need the principal square root since we know t 0 )
Therefore, we would write t H
Range: g | 41, 775 g 89, 075 102. a.
b.
H 100 4.9t 2
b.
100 H H t H 100 4.9 4.9 100 H 100 4.9 4.9 100 100 H H
From the restriction given in the problem statement, the domain is g | 19, 400 g 78,950 or 19400, 78950 . T 19, 400 1940 0.12 19, 400 19, 400 1940 T 78,950 1940 0.12 78,950 19, 400 9086 Since T is linear and increasing, we have that the range is T |1940 T 9086 or
t H t
1940, 9086 . c.
T 1940 0.12 g 19, 400
c.
T 1940 0.12 g 19, 400 T 1940 g 19, 400 0.12 T 1940 19, 400 g 0.12 T 1940 19, 400 . We would write g T 0.12 Domain: T | 1940 T 9086
104. a.
4.9 2
4.9t t2 t 4.9
(since t 0)
100 80 2.02 4.9 It will take the rock about 2.02 seconds to fall 80 meters. l T (l ) 2 32.2 t 80
T 2
l 32.2
T l 2 32.2
Range: g |19, 400 g 78,950 103. a.
100 100 4.9t 2
2
2
l T 32.2 2
The graph of H is symmetric about the y-axis. Since t represents the number of seconds after the rock begins to fall, we know that t 0 . The graph is strictly decreasing over its domain, so it is one-to-one.
T l 32.2 2
2
2
T l (T ) 32.2 , T 0 2 2
b.
3 l 3 32.2 7.34 2 A pendulum whose period is 3 seconds will be about 7.34 feet long.
631
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Chapter 6: Exponential and Logarithmic Functions
105.
f ( x) 3x 4, x 0 y 3 x 4, x 0. Interchange x and y: x 3 y 4, y 0 Note : If y 0, then x 4 so x4 y, x 4 x 4 3 y, x 4 3 x4 f 1 ( x) ,x 4 3 Putting the pieces together we have x3 2 , x 3 f 1 ( x ) x4,x 4 3
ax b ax b y cx d cx d Interchange x and y, and then solve for y: ay b x cy d x(cy d ) ay b cxy dx ay b cxy ay b dx y (cx a) b dx b dx y cx a dx b So f 1 ( x) cx a Now, f f 1 f ( x)
c. The domain of f 1 is ,3 4, , the
range of f. The range is f 1 is , , the domain of f.
ax b dx b cx d cx a ax b cx a dx b cx d
108. Yes. In order for a one-to-one function and its inverse to be equal, its graph must be symmetric about the line y x . One such example is the
acx 2 a 2 x bcx ab cdx 2 d 2 x bcx bd
function f x
acx 2 a 2 bc x ab cdx 2 d 2 bc x bd
So ac cd ; a 2 bc d 2 bc; ab bd , which means a d .
1 . x
109. Answers will vary. 110. Answers will vary. One example is 1 , if x 0 f x x x, if x 0
106. h( x) ( f g )( x) f ( g ( x)) . y f ( g ( x)) Interchange x and y, and then solve for y: x f ( g ( y )) f 1 ( x ) g ( y ) g 1 ( f 1 ( x)) y
So, h 1 ( x) g 1 ( f 1 ( x)) ( g 1 f 1 )( x) 107. a. The domain of f is , . From y 2 x 3 ,
x < 0, then y < 3, From y 3x 4 , x 0, then y 4. The range of f is
,3 4, . b. Consider the piece f ( x) 2 x 3, x 0 y 2 x 3, x 0 Interchange x and y: x 2 y 3, y 0 Note : If y 0 then x 3 so x3 y, x 3 x 3 2 y, x 3 2 x3 f 1 ( x) ,x 3 2 Now consider the piece
This function is one-to-one since the graph passes the Horizontal Line Test. However, the function is neither increasing nor decreasing on its domain. 111. No, not every odd function is one-to-one. For example, f ( x) x3 x is an odd function, but it is not one-to-one. 112. C 1 (800, 000) represents the number of cars manufactured for $800,000. 632
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Section 6.2: One-to-One Functions; Inverse Functions 113. If a horizontal line passes through two points on a graph of a function, then the y value associated with that horizontal line will be assigned to two different x values which violates the definition of one-to-one.
6 x 2 11x 2 ( x 2)(6 x 1) 2 x2 x 6 ( x 2)(2 x 3) 6x 1 3 , where x , 2 2x 3 2
118. R ( x )
3 Domain: x | x , 2 2 The degree of the numerator in lowest terms is n 1 . The degree of the denominator in lowest 6 terms is m 1 . Since n m , the line y 3 is 2 a horizontal asymptote. The denominator in lowest 3 3 terms is zero at x , so x is a vertical 2 2 asymptote.
114. Answers may vary. 115.
f ( x h) f ( x ) 3( x h) 2 7( x h) (3x 2 7 x) 3( x 2 2 xh h 2 ) 7 x 7 h 3x 2 7 x 3 x 2 6 xh 3h 2 7 x 7 h 3x 2 7 x 6 xh 3h 2 7h
116. x
b b 2 4ac 2a
119. ( x (3)) 2 ( y 5) 2 7 2
(5) (5) 2 4(3)(1) 2(3)
( x 3) 2 ( y 5) 2 49
5 25 12 5 13 6 6 So the zeros and the x-intercepts are: 5 13 5 13 , 6 6 5 5 The x value of the vertex is . The y 2(3) 6 value of the vertex is
120. 3x 6 y 5 6 y 3x 5 1 5 y x 2 6 The slope of a perpendicular line would be 2 . y 1 2( x 4) y 1 2 x 8 y 2 x 7
2
13 5 5 5 f 3 5 1 . The 12 6 6 6 5 13 vertex is , . Since the first term is 6 12 positive the graph is concave up and thus the vertex represents a minimum. 117. We start with the graph of y x . The graph will
121.
f ( x)
3( x) 5( x)3 7( x) 3x 5 x 3 7 x 3 x
(5 x3 7 x) 3x 3 f ( x) 5x 7 x The function is even.
be shifted horizontally to the left by 2 units. The graph will be reflected on the x-axis. Then the graph will be shifted vertically by 3 units upward. 122.
2 x 2 yD xD y 2 yD xD y 2 x D(2 y x) y 2 x y 2x D 2y x
633
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Chapter 6: Exponential and Logarithmic Functions 123.
f (4) f (0) 3(4) 5 3(0) 5 40 4 (12 5) (0 5) 4 7 (5) 4 12 4 3
4. f (4) f (2) (3(4) 2 2(4) 1) (3(2) 2 2(2) 1) 42 2 (39) (7) 32 16 2 2
124.
f ( x) 2 x 3 f x h f x
5. True
h
7. x 2 4 x 3 0
2 x 2h 3 2 x 3 2 x 2h 3 2 x 3 h 2 x 2h 3 2 x 3 2 x 2h 3 (2 x 3)
6. line; 3 ; 10
2 x 2h 3 2 x 3 h
h h
2 x 2h 3 2 x 3 2h 2 x 2h 3 2 x 3
x
b 4 2 2a 2(1)
f (2) (2) 2 4(2) 3 1
The vertex is 2, 1 . Graphing the function shows that the graph is increasing on 2, and
decreasing on , 2
2 2 x 2h 3 2 x 3
Section 6.3
2 4 ; 3 31 19
1. 43 64 ; 82 / 3 3 8 2.
2
2
2
1 1 ;1; 2 ; ;1;3 3 2 1 1 b. 2;1; ; 3;1; 3 2 c. y-axis d. decreasing; increasing e. Horizontal line y = 1
8. a.
2
x 2 3x 4 0 ( x 4)( x 1) 0 x 4 0 or x 1 0 The solution set is 4,1 .
9. a. b. c. d. e. f.
3. False. To obtain the graph of y ( x 2)3 , we 3
would shift the graph of y x to the right 2 units.
horizontally; right f ( x) 2 x 3 ; horizontally; left vertically; up f ( x) 2 x 3 ; y 3 ; vertically; down stretched; 1; 2; 4 decreasing
10. exponential function; growth factor; initial value 11. a
634
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Section 6.3: Exponential Functions 12. True
5 26. 158 6
13. True 14.
1 1, , 0,1 , 1, a a
8.63
32.758
27. e1.2 3.320 28. e 1.3 0.273
15. 4
29. 125e0.025(7) 149.952
16. False; for example, the point 1,3 is on the first
graph and 1, 13 is on the other.
30. 83.6e 0.157(9.5) 18.813
17. b
31. 23.14 8.815
b.
23.141 8.821
c.
23.1415 8.824
d.
2 8.825
20. a.
22.7 6.498
b.
22.71 6.543
c.
22.718 6.580
d.
2e 6.581
21. a.
–1
3
0
6
63 3 0 1
12 2 6
1
12
12 6 6 1 0
18 3 12 2
3.14
c.
3.1412.718 22.440
d.
e 22.459
22. a.
32.
2.73.1 21.738
b.
2.713.14 22.884
c.
2.7183.141 23.119
d.
e 23.141
23. (1 0.04)6 1.265 0.09 24. 1 12 1 25. 8.4 3
6 2 3
Not an exponential function since the ratio of consecutive terms is not constant.
22.217
b.
f x
2 18 3 30 Not a linear function since the average rate of change is not constant.
3.12.7 21.217 2.71
f x 1
y f x
18. c 19. a.
y x
x
x
y g x
–1
2
0
5
1
8
2
11
3
14
y x
52 3 0 1
g x 1 g x 5 2 8 5
85 3 1 0 11 8 3 2 1 14 11 3 3 2
Not an exponential function since the ratio of consecutive terms is not constant.
24
1.196
The average rate of change is a constant, 3. Therefore, this is a linear function. In a linear function the average rate of change is the slope m. So, m 3 . When x 0 we have y 5 so the y-
2.9
0.347
635
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Chapter 6: Exponential and Logarithmic Functions
function is C 1 . Therefore, the exponential function that models the data is
intercept is b 5 . The linear function that models this data is g x mx b 3x 5 . 33.
x
y H x
–1
5 4
32 .
F x Ca x 1 32
H x 1
y x
5 4 5 / 4
x
y f x
–1
54
0
18
1
6
Not a linear function since the average rate of change is not constant.
2
2
The ratio of consecutive outputs is a constant, 4. This is an exponential function with growth factor a 4 . The initial value of the exponential function is C 1 . Therefore, the exponential function that
3
2 3
1 14
0
5
3 0 1 4
1
20
4 1 3 1 0
2
80
3
320
20 4 5 80 4 20 320 4 80
35.
–1
y F x
F x
1 23
0
1
0 1
1
3 2
3 1 2
2
9 4
3
27 8
1 1 0 2
1 3
3 / 2 1
18 1 54 3 3 32
0 1
3 2
63 3 1 0
6 1 18 3 2 1 6 3 2/3 1 2 3
x
1 models the data is f x Ca x 18 . 3
1 3 2 / 3 2
2 3
f x
The ratio of consecutive outputs is a constant, 1/3. This is an exponential function with growth factor 1 a . The initial value of the exponential function 3 is C 18 . Therefore, the exponential function that
F x 1
y x
f x 1
y x
Not a linear function since the average rate of change is not constant.
x
x
x
H x
models the data is H x Ca x 5 4 5 4 x .
34.
x
3 2
36.
9 / 4 3 3 / 2 2 27 / 8 3 9 / 4 2
x
y g x
–1
6
0
1
1
0
y x
g x 1 g x 1 6
1 6 5 0 1
0 0 1
0 1 1 1 0
2 3 3 10 Not a linear function since the average rate of change is not constant.
Not a linear function since the average rate of change is not constant. The ratio of consecutive outputs is a constant, 32 .
Not an exponential function since the ratio of consecutive terms is not constant.
This is an exponential function with growth factor a 32 . The initial value of the exponential 636
Copyright © 2025 Pearson Education, Inc.
Section 6.3: Exponential Functions
x
y H x
–1
2
37.
0
4
H x 1
y x
42 2 0 1
39. B
H x
40. F
4 2 2
41. D
6 3 4 2
43. A
42. H
44. C
64 2 1 0 86 2 2 8 2 1 10 8 2 3 10 3 2 Not an exponential function since the ratio of consecutive terms is not constant.
1
6
45. E 46. G 47.
Using the graph of y 2 x , shift the graph up 1 unit. Domain: All real numbers Range: { y | y 1} or (1, ) Horizontal Asymptote: y 1 y-interecpt: 2
The average rate of change is a constant, 2. Therefore, this is a linear function. In a linear function the average rate of change is the slope m. So, m 2 . When x 0 we have y 4 so the yintercept is b 4 . The linear function that models this data is H x mx b 2 x 4 . 38.
x
y F x
–1
1 2
0
1 4
1
1 8
2
1 16
3
1 32
f ( x) 2 x 1
F x 1
y x
F x
1/ 4 1 1/ 2 2 1 12 4
0 1 1 14 8
1 1 0 8
1 4
1/ 8 1 1/ 4 2
48.
f x 3x 2
Using the graph of y 3x , shift the graph down 2 units. Domain: All real numbers Range: y | y 2 or 2,
1/16 1 1/ 8 2 1/ 32 1 1/16 2
Horizontal Asymptote: y 2
Not a linear function since the average rate of change is not constant. The ratio of consecutive outputs is a constant, 12 . This is an exponential function with growth factor a 12 . The initial value of the exponential function is C 14 . Therefore, the exponential function that models the data is F x Ca x 14 12 . x
637
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Chapter 6: Exponential and Logarithmic Functions y-intercept: 1
49.
51.
f x 3x 1
1 f x 3 2
x
x
1 Using the graph of y , vertically stretch the 2 graph by a factor of 3. That is, for each point on the graph, multiply the y-coordinate by 3. Domain: All real numbers Range: y | y 0 or 0,
Using the graph of y 3x , shift the graph right 1 unit. Domain: All real numbers Range: y | y 0 or 0, Horizontal Asymptote: y 0
y-intercept: 13
Horizontal Asymptote: y 0
y-intercept: 3
50.
f ( x) 2 x 2
Using the graph of y 2 x , shift the graph left 2 units. Domain: All real numbers Range: { y | y 0} or (0, ) Horizontal Asymptote: y 0 y-intercept: 4
52.
1 f x 4 3
x
x
1 Using the graph of y , vertically stretch the 3 graph by a factor of 4. That is, for each point on the graph, multiply the y-coordinate by 4. Domain: All real numbers Range: y | y 0 or 0,
Horizontal Asymptote: y 0
638
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Section 6.3: Exponential Functions
y-intercept: 4
55.
f ( x) 2 4 x 1
Using the graph of y 4 x , shift the graph to the right one unit and up 2 units. Domain: All real numbers Range: { y | y 2} or (2, ) Horizontal Asymptote: y 2
y-intercept: 94
53.
f ( x) 3 x 2
Using the graph of y 3x , reflect the graph about the y-axis, and shift down 2 units. Domain: All real numbers Range: { y | y 2} or ( 2, ) Horizontal Asymptote: y 2 y-intercept: 1
56.
f ( x) 1 2 x 3
Using the graph of y 2 x , shift the graph to the left 3 units, reflect about the x-axis, and shift up 1 unit. Domain: All real numbers Range: { y | y 1} or (, 1) Horizontal Asymptote: y 1 y-intercept: 7
54.
f ( x ) 3 x 1
Using the graph of y 3x , reflect the graph about the x-axis, and shift up 1 unit. Domain: All real numbers Range: { y | y 1} or ( , 1) Horizontal Asymptote: y 1 y-intercept: 0 57.
f ( x ) 2 3x / 2
Using the graph of y 3x , stretch the graph horizontally by a factor of 2, and shift up 2 units. Domain: All real numbers Range: { y | y 2} or (2, ) Horizontal Asymptote: y 2 y-intercept: 3
639
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Chapter 6: Exponential and Logarithmic Functions
58.
60.
f ( x) 1 2 x / 3
Using the graph of y e x , reflect the graph about the x-axis. Domain: All real numbers Range: { y | y 0} or (, 0) Horizontal Asymptote: y 0 y-intercept: 1
Using the graph of y 2 x , stretch the graph horizontally by a factor of 3, reflect about the yaxis, reflect about the x-axis, and shift up 1 unit. Domain: All real numbers Range: { y | y 1} or (, 1) Horizontal Asymptote: y 1 y-intercept: 0
61. 59.
f ( x ) e x
f ( x) e x
f ( x) e x 2
Using the graph of y e x , shift the graph 2 units to the left. Domain: All real numbers Range: { y | y 0} or (0, ) Horizontal Asymptote: y 0 y-intercept: 7.39
Using the graph of y e x , reflect the graph about the y-axis. Domain: All real numbers Range: { y | y 0} or (0, ) Horizontal Asymptote: y 0 y-intercept: 1
640
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Section 6.3: Exponential Functions
62.
f ( x) e x 1
Using the graph of y e x , shift the graph down 1 unit. Domain: All real numbers Range: { y | y 1} or (1, ) Horizontal Asymptote: y 1 y-intercept: 0
65.
63.
f ( x) 2 e x / 2
Using the graph of y e x , reflect the graph about the y-axis, stretch horizontally by a factor of 2, reflect about the x-axis, and shift up 2 units. Domain: All real numbers Range: { y | y 2} or (, 2) Horizontal Asymptote: y 2 y-intercept: 1
f ( x) 5 e x
Using the graph of y e x , reflect the graph about the y-axis, reflect about the x-axis, and shift up 5 units. Domain: All real numbers Range: { y | y 5} or (, 5) Horizontal Asymptote: y 5 y-intercept: 4
66.
f ( x) 7 3e 2 x
Using the graph of y e x , reflect the graph about 1 , 2 stretch vertically by a factor of 3, reflect about the x-axis, and shift up 7 units. Domain: All real numbers Range: { y | y 7} or (, 7) Horizontal Asymptote: y 7
the y-axis, shrink horizontally by a factor of
64.
f ( x) 9 3e x
Using the graph of y e x , reflect the graph about the y-axis, stretch vertically by a factor of 3, reflect about the x-axis, and shift up 9 units. Domain: All real numbers Range: { y | y 9} or (, 9) Horizontal Asymptote: y 9 y-intercept: 6
641
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Chapter 6: Exponential and Logarithmic Functions
67. 6 x 65 We have a single term with the same base on both sides of the equation. Therefore, we can set the exponents equal to each other: x 5 . The solution set is 5 .
73.
32 x 5 9 32 x 5 32 2x 5 2 2x 7 7 x 2
68. 5 x 56 We have a single term with the same base on both sides of the equation. Therefore, we can set the exponents equal to each other: x 6 . The solution set is 6 .
7 The solution set is . 2 1 5 5 x 3 51 x 3 1 x 4 The solution set is 4 .
74. 5 x 3
69. 2 x 16 2 x 24 x 4 x 4 The solution set is 4 .
3
3x 9 x
75.
70. 3 x 81
3
3x 32
3 x 34 x 4 x 4
3
3x 32 x x3 2 x
The solution set is 4 .
x3 2 x 0
x x2 2 0 x 0 or x 2 2 0
x
1 1 71. 25 5
x2 2 x 2
x
1 1 2 5 5 x
x
1 1 5 5 x2 The solution set is 2 .
76.
2
4x 2x 2 2 x
2 2
x
2
22 x 2 x x
1 1 72. 4 64
2 x2 x 2 x2 x 0 x 2 x 1 0
x
1 1 3 4 4 x
The solution set is 2, 0, 2 . 2
x 0 or 2 x 1 0 2x 1 1 x 2 1 The solution set is 0, . 2
3
1 1 4 4 x3 The solution set is 3 .
642
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Section 6.3: Exponential Functions
8 x 11 162 x
77.
2
3 x 11
24
81.
2 2 2
2x
2 x
23 x 33 28 x 3x 33 8 x 33 11x 3 x The solution set is 3 .
3
2 x 15
2 2
22 x x 28 x2 2 x 8 x2 2 x 8 0 x 4 x 2 0 x 4 0 or x 2 0 x 4 x2 The solution set is 4, 2 .
3 x
3
2 x 30
3 33 x 2 x 30 3x 30 5 x 6x The solution set is 6 .
82.
2
2 2
34 x 3 x 31 3x 2 4 x 1
2x
3x2 4 x 1 0 3x 1 x 1 0
3x 7 36 x x2 7 6 x
3x 1 0 or x 1 0 3 x 1 x 1 1 x 3 1 The solution set is 1, . 3
2
x 6x 7 0
x 7 x 1 0 x 7 0 or x 1 0 x7 x 1 The solution set is 1, 7 .
83.
2
5 x 8 1252 x 2
5 x 8 53
1
34 x 33 x 31
2
80.
2 3 x
2 2x
2
3x 7 33
2
92 x 27 x 31
3 3 3
3x 7 27 2 x
79.
4 2
x2
22 x 2 x 28
9 x 15 27 x
78.
2
4 x 2 x 162
2x
2
5 x 8 56 x x2 8 6 x x2 6 x 8 0 x 4 x 2 0
e 2 x e5 x 12 2 x 5 x 12 3x 12 x 4 The solution set is 4 .
84. e3 x e 2 x 3x 2 x 4x 2 1 x 2
x 4 0 or x 2 0 x4 x2 The solution set is 2, 4 .
1 The solution set is . 2
643
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Chapter 6: Exponential and Logarithmic Functions
3
2 1 e x e3 x 2 e
85.
90. If 5 x 3 , then 5 x
e
e
x 3x 2 2
x 3x 2 0 x 2 x 1 0
5 3 5
86.
x
2
x 2
2
91. If 9 x 25 , then 32
x 2 0 or x 1 0 x2 x 1 The solution set is 1, 2 .
3x 5
92. If 23 x
e e e x2
12
2
e 4 x e x e12
x 1 , then 2 3 103 1000 3 2x 103
2 x 10
2
e 4 x x e12 x 2 4 x 12
93. We need a function of the form f x k a p x ,
x 2 4 x 12 0 x 6 x 2 0
with a 0, a 1 . The graph contains the points 1 1, , 0,1 , 1,3 , and 2,9 . In other words, 3 1 f 1 , f 0 1 , f 1 =3 , and f 2 9 . 3
x 6 0 or x 2 0 x 6 x2 The solution set is 6, 2 . x
x 2
42 x
Therefore, f 0 k a p 0
7
87. If 4 7 , then 4
2
1 k a0 1 k 1 1 k
1
72 1 42 x 49
and f 1 a p 1
3 ap Let’s use a 3, p 1. Then f x 3x . Now we
3
88. If 2 x 3 , then 2 x
2
22 x
2
need to verify that this function yields the other 1 known points on the graph. f 1 31 ; 3
1
32 x 1 22 9 1 x 4 9
x
89. If 3
1
33 1 53 x 27
3x2
2
4 x
3
53 x
2
e x e3 x e 2 x2
3
f 2 32 9
So we have the function f x 3x .
2 x 2
2 , then 3
32 x
94. We need a function of the form f x k a p x ,
2
with a 0, a 1 . The graph contains the points 1 1, , 0,1 , and 1,5 . In other words, 5
1
22 1 32 x 4
644
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Section 6.3: Exponential Functions
f 1
Therefore, f 0 k a p 0
1 , f 0 1 , and f 1 5 . Therefore, 5
f 0 k a
1 k a 0 1 k 1 1 k
p 0
1 k a0 1 k 1 1 k
and f 1 a p 1
e a p
and f 1 a p 1
e ap Let’s use a e, p 1 . Then f x e x . Now
p
5a Let’s use a 5, p 1. Then f x 5x . Now we
we need to verify that this function yields the other known points on the graph. 1 f 1 e1 e
need to verify that this function yields the other known point on the graph. 1 f 1 51 5
f 2 e 2
So we have the function f x 5 . x
So we have the function f x e x .
95. We need a function of the form f x k a p x ,
97. We need a function of the form f x k a p x b ,
with a 0, a 1 . The graph contains the points 1 1, , 0, 1 , 1, 6 , and 2, 36 . In other 6 1 words, f 1 , f 0 1 , f 1 = 6 , and 6 f 2 36 .
with a 0, a 1 and b is the vertical shift of 2 units upward. The graph contains the points 0, 3 , and 1,5 . In other words, f 0 1 and f 1 3 . We can assume the graph has the same
shape as the graph of f x k a p x . The reference (unshifted) graph would contain the points 0,1 , and 1,3 .
Therefore, f 0 k a and f 1 a . p 1
p 0
6 a p 1 k a 0 1 k 1 6 ap 1 k Let’s use a 6, p 1. Then f x 6 x .
Therefore, f 0 k a and f 1 a p 0
3 ap 1 k a0 1 k 1 1 k Let’s use a 3, p 1 . Then f x 3x . To shift
Now we need to verify that this function yields the other known points on the graph. 1 f 1 61 ; f 2 62 36 6
the graph up by 2 units we would have f x 3x 2 . Now we need to verify that this
So we have the function f x 6 x .
function yields the other known points on the graph. f 0 30 2 3
96. We need a function of the form f x k a p x ,
f 1 31 2 5
with a 0, a 1 . The graph contains the points 1 2 1, , 0, 1 , 1, e , and 2, e . In other e 1 words, f 1 , f 0 1 , f 1 = e , and e
p 1
So we have the function f x 3x 2 . 98. We need a function of the form f x k a p x b ,
with a 0, a 1 and b is the vertical shift of 3 units downward. The graph contains the points 0, 2 , and 2,1 . In other words, f 0 2
f 2 e 2 .
and f 2 1 . We can assume the graph has the 645
Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
same shape as the graph of f x k a p x . The
b.
reference (unshifted) graph would contain the points 0,1 , and 2, 4 . p 0 p 2 Therefore, f 0 k a and f 2 a
1 a2 p 4 1 ap 2
1 k a0 1 k 1 1 k
1 1x Let’s use a , p 1 . Then f x . To shift 2 2 the graph down by 3 units we would have 1x f x 3 . Now we need to verify that this 2 function yields the other known points on the graph. 10 f 0 3 2 2 1 2 f 2 3 43 1 2
101. a.
b.
g x 66 4 x 64 4 x 43 x3 The point 3, 66 is on the graph of g.
f 4 24 16
102. a.
The point 4,16 is on the graph of f.
100. a.
1 9 2 4 4
4 x 2 66
x
b.
g 1 41 2
9 The point 1, is on the graph of g. 4
1 So we have the function f x 3 . 2
99. a.
1 9 1 3x 9 1 x 3 2 3 3x 32 x 2 1 The point 2, is on the graph of f. 9 f x
1 16 1 x 2 16 1 2x 4 2 x 2 24 x 4 1 The point 4, is on the graph of f. 16 f x
b.
1 14 3 5 5 14 The point 1, is on the graph of g. 15 g 1 51 3
g x 122 5 x 3 122 5 x 125 5 x 53 x3 The point 3,122 is on the graph of g. 6
103. a.
f 4 3 81 4
The point 4,81 is on the graph of f.
6 1 H 6 4 2 4 60 2 The point 6, 60 is on the graph of H.
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Section 6.3: Exponential Functions
b.
H x 12
x
c.
x
1 4 12 2
1 3 0 3 x
1 3 3
x
1 16 2
2
x
3 3 1 x
4
2 x 4 x 4 The point 4,12 is on the graph of H.
3 x 31 x 1 x 1 The zero of F is x 1 .
x
c.
1
1 4 0 2
105.
x
e x f x x e
if x 0 if x 0
1 4 2
2 2 1 x
2
2 x 22 x 2 x 2 The zero of H is x 2 . 5
104. a.
b.
5 1 F 5 3 3 3 240 3 The point 5, 240 is on the graph of F.
Domain: , Range: y | y 1 or 1, Intercept: 0,1
F x 24 x
1 3 24 3 x
1 27 3 3 x 33 x 3 x 3 The point 3, 24 is on the graph of F.
647
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Chapter 6: Exponential and Logarithmic Functions
106.
e x f x x e
109. p(n) 100(0.97) n
if x 0 if x 0
a.
p (10) 100(0.97)10 74% of light
b.
p(25) 100(0.97) 25 47% of light
c.
Each pane allows only 97% of light to pass through.
110. p(h) 760e 0.145 h a.
p(2) 760e0.145(2)
760e0.290 568.68 mm of Hg
Domain: , Range: y | 0 y 1 or 0,1
b.
p(10) 760e0.145(10)
Intercept: 0,1 107.
x e f x x e
760e1.45 178.27 mm of Hg
if x 0
111. p ( x) 25, 495(0.90) x
if x 0
a.
p(3) 25, 495(0.90)3 $18,586
b. p(9) 25, 495(0.90)9 $9877 c. As each year passes, the sedan is worth 90% of its value the previous year. 112. A(n) A0 e 0.35 n a.
A(3) 100e 0.35(3)
100e 1.05 34.99 square millimeters
Domain: , Range: y | 1 y 0 or 1, 0
b.
Intercept: 0, 1 e f x x e
x
108.
A(10) 100e0.35(10)
100e3.5 3.02 square millimeters
if x 0 if x 0
113. Since the number of bacteria doubles every minute, half of the container is full one minute before it is full. Thus, it takes 59 minutes to fill the container. 114. P (t ) 30(1.149)t a. 30 b. P (5) 30(1.149)5 60 c.
P (10) 30(1.149)10 120
d. P (15) 30(1.149)15 241 e. The population appears to be doubling every 5 years.
Domain: , Range: y | y 1 or , 1 Intercept: 0, 1 648
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Section 6.3: Exponential Functions 115. D h 5e 0.4 h
b.
D 1 5e 0.4(1) 5e 0.4 3.35
F 30 1 e 0.15(30) 1 e4.5 0.989
The probability that a car will arrive within 30 minutes of 5:00 p.m. is 0.989.
After 1 hours, 3.35 milligrams will be present.
As t , F t 1 e0.15 t 1 0 1
D 6 5e0.4(6) 5e 2.4 0.45 milligrams
c.
After 6 hours, 0.45 milligrams will be present.
d. Graphing the function:
116. N P 1 e0.15 d
1000 1 e 362
N 3 1000 1 e0.15(3) 0.45
After 3 days, 362 students will have heard the rumor.
e.
117. F t 1 e0.1t a.
F 6 0.60 , so 6 minutes are needed for the
probability to reach 60%.
F 10 1 e0.1(10) 1 e 1 0.632
The probability that a car will arrive within 10 minutes of 12:00 p.m. is 0.632. b.
F 40 1 e0.1(40) 1 e4 0.982
The probability that a car will arrive within 40 minutes of 12:00 p.m. is 0.982. c.
As t , F t 1 e0.1t 1 0 1
119. P ( x)
d. Graphing the function:
e.
a.
b.
F 7 0.50 , so about 7 minutes are needed
2015 e20 0.0516 or 5.16% 15! The probability that 15 cars will arrive between 5:00 p.m. and 6:00 p.m. is 5.16%. P (15)
2020 e20 0.0888 or 8.88% 20! The probability that 20 cars will arrive between 5:00 p.m. and 6:00 p.m. is 8.88%. P (20)
for the probability to reach 50%. 120. P ( x) a.
118. F t 1 e0.15 t a.
b.
F 15 1 e0.15(15) 1 e2.25 0.895
The probability that a car will arrive within 15 minutes of 5:00 p.m. is 0.895.
20 x e20 x!
4 x e4 x!
45 e4 0.1563 or 15.63% 5! The probability that 5 people will arrive within the next minute is 15.63%. P 5
48 e4 0.0298 or 2.98% 8! The probability that 8 people will arrive within the next minute is 2.98%. P 8
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Chapter 6: Exponential and Logarithmic Functions
4221
4221
121. R 10 T 459.4 D 459.4
4221
4221
2 70.95%
a.
R 10 50 459.4 41 459.4
b.
4221 4221 2 R 10 68 459.4 59 459.4 72.62%
c.
b.
500 1 e
0.183
500 1 e
L 60 500 1 e 0.0061(60) 0.366
c.
See the graph at the end of the solution.
d.
I2
0.5 120 1 e 10 24 1 e0.25 5 5.31 amperes after 0.5 second 5
I2
153 The student will learn about 153 words after 60 minutes.
123. I
a.
0.3 120 1 e 10 24 1 e 0.15 5 3.34 amperes after 0.3 second 5
84 The student will learn about 84 words after 30 minutes.
b.
Therefore, as,
10 t 5
t , I1
L 30 500 1 e0.0061(30)
10 t As t , e 5 0 .
120 1 e 12 1 0 12 , 10 which means the maximum current is 12 amperes.
4221 4221 2 R 10 T 459.4 T 459.4 102 100%
122. L t 500 1 e0.0061 t a.
10 1 120 I1 1 e 5 12 1 e2 10.38 10 amperes after 1 second
2
5 1 120 I2 1 e 10 24 1 e0.5 5 9.44 amperes after 1 second
R t E 1 e L R 10 0.3 120 1 e 5 12 1 e0.6 5.41 I1 10 amperes after 0.3 second
e.
5 t As t , e 10 0 .
t , I1
Therefore, as,
10 t 120 1 e 5 24 1 0
24 ,
5 which means the maximum current is 24 amperes.
0.5 120 1 e 5 12 1 e1 7.59 I1 10 amperes after 0.5 second 10
f.
See the graph that follows.
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Section 6.3: Exponential Functions
t
124. I
E RC e R
a.
0
120 2000 1 120 0 I1 e e 0.06 2000 2000 amperes initially. 1000
120 2000 1 120 1/ 2 e 0.0364 e 2000 2000 amperes after 1000 microseconds I1
t
3000
120 2000 1 120 1.5 e 0.0134 I1 e 2000 2000 amperes after 3000 microseconds
125. Since the growth rate is 3 then a 3 . So we have
b. The maximum current occurs at t 0 . Therefore, the maximum current is 0.06 amperes. c.
f ( x) C 3x
So
f (6) C 36 12 C 36 12 C 36
Graphing the function:
12
37 36 12 3 36
f (7)
So f (7) 36 1 1 1 1 ... 2! 3! 4! n! 1 1 1 n 4; 2 2.7083 2! 3! 4! 1 1 1 1 1 n 6; 2 2.7181 2! 3! 4! 5! 6! 1 1 1 1 1 1 1 n 8; 2 2! 3! 4! 5! 6! 7! 8! 2.7182788 1 1 1 1 1 1 1 1 1 n 10; 2 2! 3! 4! 5! 6! 7! 8! 9! 10! 2.7182818
126. 2
t
0
120 1000 2 120 0 e e 0.12 d. I 2 1000 1000 amperes initially. 1000
120 1000 2 120 1/ 2 e 0.0728 e 1000 1000 amperes after 1000 microseconds I2
e 2.7182818
3000
127. 2 1 3
120 1000 2 120 1.5 I2 e e 0.0268 1000 1000 amperes after 3000 microseconds e.
The maximum current occurs at t 0 . Therefore, the maximum current is 0.12 amperes.
f.
Graphing the functions:
2 1 2.5 e 11 2 1 2.8 e 1 1 22
651
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Chapter 6: Exponential and Logarithmic Functions 2 1
2.7 e
130.
1 1
f ( x) a x f ( x) a x
2 2 33
131.
2 1
1 f ( x)
f ( x) a x f ( x) a x a x
1 1 2 2
1 x x e e 2 f ( x) sinh( x) 1 e x e x 2 1 e x e x 2 sinh x f ( x) Therefore, f ( x) sinh x is an odd function.
132. sinh x
3 3
a.
44
2.717770035 e
1 1 2 2 3 3 4 4 55
b. Let Y1
2 1
a
x
f ( x)
2.721649485 e
2 1
1
2.718348855 e
1 x x e e . 2
1 1 2 2
3 3 4 4
5 5 66 e 2.718281828
128.
a.
f ( x) a x f ( x h) f ( x ) a x h a x h h a x ah a x h x a ah 1 h h 1 a a x h
129.
b. Let Y1
f ( x) a x
1 x e e x 2 f ( x) cosh( x) 1 e x e x 2 1 x e e x 2 cosh x f ( x) Thus, f ( x) cosh x is an even function.
133. cosh x
1 x e e x . 2
f ( A B ) a A B a A a B f ( A) f ( B )
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Section 6.3: Exponential Functions
(cosh x) 2 (sinh x) 2
c.
2
e x e x e x e x 2 2
2u 2 3u 20 0 (2u 5)(u 4) 0
2
5 or u 4 2 1x 1x 5 2 3 or 2 3 4 2 u
e2 x 2 e2 x e2 x 2 e2 x 4 4 e2 x 2 e2 x e2 x 2 e2 x 4 4 4 1
134.
not possible
1x
2x 1
The solution set is 6
f ( x) 2 f (1) 2
2 1 22 1 4 1 5
137 - 138. Answers will vary. 139. Given the function f x a x , with a 1 ,
23 1 28 1 256 1 257
If x 0 , the graph becomes steeper as a increases. If x 0 , the graph becomes less steep as a increases.
f (3) 2
f (4) 2
24 1 216 1 65,536 1 65,537
25
f (5) 2 1 232 1 4, 294,967, 296 1 4, 294,967, 297 641 6, 700, 417
x 3 5 x 2 4 x 20 0
( x 2)( x 2)( x 5) 0 We graph the function f ( x) x 3 5 x 2 4 x 20 . The intercepts are y-intercept: f (0) 20
x
32 1 4 3x 9 0 x
x
32 12 3x 27 0 Let u 3x. u 2 12u 27 0 (u 3)(u 9) 0 u 3 or u 9
x 3 5 x 2 4 x 20 0 ( x 2)( x 2)( x 5) 0 x 2, x 2, x 5 The graph is below the x-axis when x 5 or 2 x 2 . So the solution set is , 5 2, 2 .
x-intercepts:
Then, 3x 3 or 3x 9 x 1 or x 2 The solution set is 1, 2 2 x 1
23
141.
1
3 2 3 20 0
2x
x3 5 x 2 4 x 20
140.
3(32 1 4 3x 9) 3 0
136.
2 3 22 1 x2 3 x6
1
22 1 24 1 16 1 17 f (2) 2
135.
1x
or 2 3 4
1
2 2 3 3 2 3 20 0 1x
Let u 2 3 .
x 1 1 x2 x 1 1 0 x2 x 1 x 2 0 x2 3 0 x2 3 f ( x) x2 The value where f is undefined is x 2 .
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Chapter 6: Exponential and Logarithmic Functions
Interval
( , 2)
(2, )
Number Chosen
0
5
1.5
1
Value of f Conclusion
x 3 or x 1
Negative Positive
The solution set is x x 2 or, 2, . 142. Use the form f ( x) a ( x h) 2 k . The vertex is (3, 5) , so h 3 and k 5 . f ( x) a ( x 3) 2 5 .
Since the graph passes through (2, 3) , f (2) 3 .
Domain: (, ) . Range: [4, ) .
c.
Decreasing on , 1 ; increasing on
1, .
3 a(2 3) 2 5 3 a( 1) 2 5
144. 13x (5 x 6) 2 x (8 x 27) 8 x 6 6 x 27 14 x 21 21 3 x 14 2
3 a5 2 a f ( x) 2( x 3) 2 5 2( x 2 6 x 9) 5 2 x 2 12 x 18 5
3 The solution is 2
2 x 2 12 x 13
143. a.
b.
145. ( x 0)2 ( y 0) 2 12
f x x2 2 x 3
x2 y2 1
a 1, b 2, c 3. Since a 1 0, the graph is concave up. The x-coordinate of the vertex is b 2 1 . x 2a 2(1) The y-coordinate of the vertex is b f f (1) (1) 2 2 1 3 4 . 2a Thus, the vertex is (–1, –4). The axis of symmetry is the line x 1 . The discriminant is:
146.
x 16 x 48 0
( x 12)( x 4) 0 x 12 0
or
x 12
x 4
x 144
x 16
The solution set is 144,16 147. I prt 12000(0.035)(2.5) $1050
b 2 4ac 2 4 1 3 16 0 , so the 2
graph has two x-intercepts. The x-intercepts are found by solving: x2 2 x 3 0
148.
x 3 x 1 0
x 4 5 x 2 6 2 x 2 12 x 4 7 x 2 18 0
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x 40
Section 6.4: Logarithmic Functions The x-intercepts are (3, 0) and (3, 0).
The intercepts are y-intercept: f (0) 6
x2 x 6 0
x-intercepts:
x 2 x 3 0 x 2, x 3 b (1) 1 The vertex is at x . Since 2a 2(1) 2
25 1 1 25 f , the vertex is , . 4 4 2 2
The graph is above the x-axis when x 2 or x 3 . Since the inequality is strict, the solution set is x x 2 or x 3 or, using interval notation, , 2 3, .
We see that the graph is less than or equal to zero at {x | 3 x 3} or 3,3 149.
f ( x) 2 x 2 7 x
2.
f ( x h) f ( x ) h 2( x h) 2 7( x h) (2 x 2 7 x) h 2 2 2 x 4 xh 2h 7 x 7 h 2 x 2 7 x h 2 4 xh 2h 7h h 4 x 2h 7
x 1 0 x4 x 1 f x x4 f is zero or undefined when x 1 or x 4 . Interval
(, 4)
(4,1)
(1, )
Test Value
5
0
2
Value of f
6
Conclusion
positive
1 1 4 6 negative positive
The solution set is x x 4 or x 1 or, using interval notation, , 4 1, .
Section 6.4 1. a.
b.
3x 7 8 2 x 5 x 15 x3 The solution set is x x 3 . x2 x 6 0 We graph the function f ( x) x 2 x 6 .
3. a. b. c. d.
decreasing; increasing -1; 0; 1 -1; 0; 1 1; 0; -1
4. a. b. c. d. e. f.
horizontally; right horizontally; left x5 vertically; up y 2 increasing
655 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
5, ; 5,
27. ln 4 x is equivalent to e x 4 .
x x 0 or (0, )
28. ln x 4 is equivalent to e4 x .
g. 5.
29. log 2 1 0 since 20 1 .
1 6. , 1 , 1, 0 , a,1 a
30. log8 8 1 since 81 8 .
7. 1 31. log 7 49 2 since 7 2 49 .
y
8. False. If y log a x , then x a .
1 1 32. log3 2 since 32 . 9 9
9. True 10. a
1 33. log1/5 125 3 since 5
11. c 12. b 13. 9 32 is equivalent to 2 log 3 9 .
1 34. log1/ 3 9 2 since 3
14. 16 42 is equivalent to 2 log 4 16 .
35. log 10
53 125 .
2
32 9 .
1 since 101/ 2 10 . 2
15. a 2 1.6 is equivalent to 2 log a 1.6 . 36. log 3 100
16. a3 2.1 is equivalent to 3 log a 2.1 .
3
2 since 102/3 1001/3 3 100 . 3
17. 2 x 7.2 is equivalent to x log 2 7.2 .
37. log 2 4 4 since
2 4 .
18. 3x 4.6 is equivalent to x log3 4.6 .
38. log 3 9 4 since
3 9 .
19. e x 8 is equivalent to x ln 8 . 39. ln e
20. e2.2 M is equivalent to 2.2 ln M .
4
4
1 since e1/ 2 e . 2
40. ln e3 3 since e3 e3 .
21. log 2 8 3 is equivalent to 23 8 .
41.
1 1 22. log3 2 is equivalent to 32 . 9 9
f ( x) ln( x 3) requires x 3 0 . x3 0 x3
The domain of f is x x 3 or 3, .
23. log a 3 6 is equivalent to a 6 3 .
42. g ( x) ln( x 1) requires x 1 0 . x 1 0 x 1
24. logb 4 2 is equivalent to b 2 4 . 25. log3 2 x is equivalent to 3x 2 .
The domain of g is x x 1 or 1, .
x
26. log 2 6 x is equivalent to 2 6 .
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Section 6.4: Logarithmic Functions
The domain of g is x x 5 or 5, .
43. F ( x ) log 2 x 2 requires x 2 0 . 2
x 0 for all x 0 .
The domain of F is x x 0 . 44. H ( x) log 5 x3 requires x3 0 . x3 0 for all x 0 .
The domain of H is x x 0 or 0, . 45.
x 1 x 1 0. 49. g ( x) log5 requires x x x 1 p x is zero or undefined when x x 1 or x 0 .
x x f x 3 2 log 4 5 requires 5 0 . 2 2 x 5 0 2 x 5 2 x 10 The domain of f is x x 10 or 10, .
46. g x 8 5ln 2 x 3 requires 2 x 3 0 . 2x 3 0 2 x 3 3 x 2 3 3 The domain of g is x x or , . 2 2
(, 1)
(1, )
Test Value
2
0
Value of p
1
1
Conclusion negative positive
The domain of f is x x 1 or 1, . 1 1 0. 48. g ( x) ln requires x 5 x 5 1 p x is undefined when x 5 . x5 Interval Test Value
(,5) 4
(5, ) 6
Value of p
1
1
(, 1)
(1, 0)
(0, )
Test Value
2
1 2
1
1
2
Value of p Conclusion
1 2 positive
negative positive
The domain of g is x x 1 or x 0 ;
, 1 0, . x x 50. h( x) log 3 0. requires x 1 x 1 x p x is zero or undefined when x 1 x 0 or x 1 .
1 1 47. f ( x) ln 0. requires x 1 x 1 1 p x is undefined when x 1 . x 1 Interval
Interval
Interval
(, 0)
(0,1)
(1, )
Test Value
1
1 2
2
1 1 2 2 Conclusion positive negative positive Value of p
The domain of h is x x 0 or x 1 ;
, 0 1, . 51.
f ( x) ln x requires ln x 0 and x 0
ln x 0 x e0 x 1 The domain of h is x x 1 or 1, .
52. g ( x)
1 requires ln x 0 and x 0 ln x ln x 0
x e0 x 1
Conclusion negative positive
657 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
The domain of h is x x 0 and x 1 ;
62. If the graph of f ( x) log a x contains the point 1 1 1 , 4 , then f log a 4 . Thus, 2 2 2 1 log a 4 2 1 a 4 2 1 1 a4 2 a4 2
0,1 1, . 5 53. ln 0.511 3
54.
ln 5 0.536 3 10 3 30.099 0.04
ln
55.
a 21/ 4 1.189
63.
2 3 4.055 56. 0.1 ln
57.
ln 4 ln 2 2.303 log 4 log 2
58.
log15 log 20 0.434 ln15 ln 20
59.
2 ln 5 log 50 53.991 log 4 ln 2
60.
3log 80 ln 5 1.110 log 5 ln 20
64.
61. If the graph of f ( x) log a x contains the point (2, 2) , then f (2) log a 2 2 . Thus, log a 2 2 a2 2 a 2 Since the base a must be positive by definition, we have that a 2 .
65.
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Section 6.4: Logarithmic Functions 66.
e.
The domain of the inverse found in part (d) is all real numbers. Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 4, .
f.
Shift the graph of y e x down 4 units.
67. B 68. F 69. D 70. H 71. A 72. C
76.
f ( x) ln( x 3)
Domain: (3, )
73. E
a.
74. G
b. Using the graph of y ln x , shift the graph 3 units to the right.
75.
f ( x) ln( x 4)
a.
Domain: ( 4, )
b. Using the graph of y ln x , shift the graph 4 units to the left.
c. d. c.
Range: (, ) Vertical Asymptote: x 4 f ( x) ln( x 4) y ln( x 4) x ln( y 4)
d.
y 4 ex x
y e 4 f
1
( x) e x 4
Range: (, ) Vertical Asymptote: x 3 f ( x) ln( x 3) y ln( x 3) x ln( y 3) y 3 e
Inverse
x
y ex 3 f 1 ( x) e x 3
Inverse
e.
The domain of the inverse found in part (d) is all real numbers. Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 3, .
659 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
f.
77.
Using the graph of y e x , shift the graph 3 units up.
f.
f ( x) 2 ln x
a.
78.
Domain: (0, )
d.
c. d.
Inverse
y e x2
Range: (, ) Vertical Asymptote: x 0 f ( x) ln( x) y ln( x) x ln( y ) x ln( y )
Inverse
y e x
f 1 ( x) e x 2
e.
Domain: ( , 0)
b. Using the graph of y ln x , reflect the graph about the y-axis, and reflect about the x-axis.
Range: (, ) Vertical Asymptote: x 0 f ( x) 2 ln x y 2 ln x x 2 ln y x 2 ln y
f ( x) ln( x)
a.
b. Using the graph of y ln x, shift up 2 units.
c.
Using the graph of y e x , shift the graph 2 units to the right.
y e x f 1 ( x) e x
The domain of the inverse found in part (d) is all real numbers.
e.
Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 0, .
The domain of the inverse found in part (d) is all real numbers. Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is , 0 .
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Section 6.4: Logarithmic Functions
f.
e.
Using the graph of y e x , reflect the graph about the y-axis, and reflect about the x-axis.
Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 0, . f.
79.
The domain of the inverse found in part (d) is all real numbers.
Using the graph of y e x , reflect the graph about the y-axis, and reflect about the x-axis.
f ( x) ln(2 x) 3
a.
Domain: (0, )
b. Using the graph of y ln x , compress the
graph horizontally by a factor of shift down 3 units.
1 , and 2
80.
f ( x) 2 ln( x 1)
a.
Domain: (1, )
b. Using the graph of y ln x , shift the graph to the left 1 unit, reflect about the x-axis and stretch vertically by a factor of 2.
c. d.
Range: (, ) Vertical Asymptote: x 0 f ( x) ln(2 x) 3 y ln(2 x) 3 x ln(2 y ) 3 x 3 ln(2 y )
c. Inverse
d.
x 3
2y e 1 y e x 3 2 1 x 3 1 f ( x) e 2
Range: (, ) Vertical Asymptote: x 1 f ( x) 2 ln( x 1) y 2 ln( x 1) x 2 ln( y 1) x ln( y 1) 2 y 1 e x / 2
Inverse
y e x / 2 1 f 1 ( x) e x / 2 1
e.
The domain of the inverse found in part (d) is all real numbers.
661 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 1, . f.
81.
Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 4, .
Using the graph of y e x , reflect the graph about the y-axis, stretch horizontally by a factor of 2, and shift down 1 unit.
f.
f ( x) log x 4 2
a.
82.
Domain: (4, )
f ( x)
a.
b. Using the graph of y log x , shift the graph 4 units to the right and 2 units up.
Using the graph of y 10 x , shift the graph 2 units to the right and 4 units up.
1 log x 5 2
Domain: (0, )
b. Using the graph of y log x , compress the
graph vertically by a factor of 5 units down.
c.
Range: (, ) Vertical Asymptote: x 4
d.
f ( x) log x 4 2
c.
y log x 4 2
x log y 4 2
Range: (, ) Vertical Asymptote: x 0
Inverse
x 2 log y 4 y 4 10 x 2 y 10 x 2 4 f 1 ( x) 10 x 2 4
e.
The domain of the inverse found in part (d) is all real numbers.
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1 , and shift it 2
Section 6.4: Logarithmic Functions
d.
1 log x 5 2 1 y log x 5 2 1 x log y 5 2 1 x 5 log y 2 2( x 5) log y f ( x)
Inverse
y 102( x 5) f
e.
1
( x) 10
c.
2( x 5)
The domain of the inverse found in part (d) is all real numbers.
d.
Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 0, . f.
Using the graph of y 10 x , shift the graph 5 units to the left, and compress horizontally 1 by a factor of . 2 e.
Range: (, ) Vertical Asymptote: x 0 1 log 2 x 2 1 y log 2 x 2 1 x log 2 y 2 2 x log 2 y
f ( x)
Inverse
2 y 102 x 1 y 102 x 2 1 f 1 ( x ) 102 x 2 The domain of the inverse found in part (d) is all real numbers.
Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 0, . f. 83.
1 , and 2 1 compress vertically by a factor of . 2
graph horizontally by a factor of
1 log 2 x 2 Domain: (0, )
f ( x)
a.
Using the graph of y 10 x , compress the
b. Using the graph of y log x , compress the 1 , and 2 1 compress vertically by a factor of . 2
graph horizontally by a factor of
663 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
84.
f ( x) log(2 x)
a.
85.
Domain: (, 0)
a.
b. Using the graph of y log x , reflect the graph across the y-axis and compress horizontally by a factor of 12 .
c. d.
c.
Range: (, ) Vertical Asymptote: x 2
d.
f ( x) 3 log 3 x 2 y 3 log 3 x 2
x 3 log 3 y 2
Inverse
y 2 3x 3 y 3x 3 2 f 1 ( x) 3x 3 2
The domain of the inverse found in part (d) is all real numbers.
e.
1
Since the domain of f is the range of f , we can use the result from part (a) to say that the range of f 1 is , 0 . f.
Inverse
x 3 log3 y 2
2 y 10 x 1 f 1 ( x) 10 x 2
e.
Domain: (2, )
b. Using the graph of y log3 x , shift 2 units to the left, and shift up 3 units.
Range: (, ) Vertical Asymptote: x 0 f ( x) log(2 x) y log(2 x) x log(2 y )
f ( x) 3 log 3 x 2
The domain of the inverse found in part (d) is all real numbers. Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 2, .
Using the graph of y 10 x , reflect the graph across the x-axis and compress vertically by a factor of 12 .
f.
Using the graph of y 3x , shift 3 units to the right, and shift down 2 units.
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Section 6.4: Logarithmic Functions
86.
f ( x) 2 log 3 x 1
a.
87.
Domain: (1, )
f ( x) e x 2 3
a.
b. Using the graph of y log3 x , shift 1 unit to the left, reflect the graph about the x-axis, and shift 2 units up.
b. Using the graph of y e x , shift the graph two units to the left, and shift 3 units down.
c. c. d.
Range: (, ) Vertical Asymptote: x 1
Domain: (, )
Range: (3, ) Horizontal Asymptote: y 3 f ( x) e x 2 3
d.
f ( x) 2 log3 x 1
y ex2 3
y 2 log3 x 1
x 2 log3 y 1
x e y2 3
Inverse
x 2 log 3 y 1
x3 e y 2 ln x 3
y 1 32 x
1
2 x log3 y 1
y ln x 3 2
f
y 32 x 1
e.
f 1 ( x) 32 x 1
e.
The domain of the inverse found in part (d) is all real numbers. Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is 1, .
f.
Inverse
y2
Using the graph of y 3x , reflect the graph about the y-axis, shift 2 units to the right, and shift down 1 unit.
( x) ln x 3 2
For the domain of f 1 we need x3 0 x 3 So the domain of the inverse found in part (d) is 3, . Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is , .
665 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
f.
Using the graph of y ln x , shift 3 units to the left, and shift down 2 units.
e.
For the domain of f 1 we need x2 0 3 x2 0 x2 The domain of the inverse found in part (d) is 2, .
Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is , . f. 88.
f ( x) 3e x 2
a.
Using the graph of y ln x , shift 3 units to the left, and shift down 2 units.
Domain: (, )
b. Using the graph of y e x , stretch the graph vertically by a factor of 3, and shift 2 units up.
89.
f ( x) 2 x / 3 4
a. c.
d.
Range: (2, ) Horizontal Asymptote: y 2
Domain: (, )
b. Using the graph of y 2 x , stretch the graph horizontally by a factor of 3, and shift 4 units up.
f ( x) 3e x 2
y 3e x 2 x 3e y 2 x 2 3e x2 ey 3
Inverse
y
x2 y ln 3 x2 f 1 ( x) ln 3
c.
Range: (4, ) Horizontal Asymptote: y 4
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Section 6.4: Logarithmic Functions
d.
c.
f ( x) 2 x / 3 4 y2
x/3
4
x2
y/3
4
Inverse
Range: (, 0) Horizontal Asymptote: y 0 f ( x) 3x 1
d.
x 4 2y/3 y log 2 x 4 3 y 3log 2 x 4
y 3x 1 x 3 y 1
x 3 y 1 log3 x
f 1 ( x) 3log 2 x 4
e.
y log3 x 1
1
For the domain of f we need x4 0 x4 The domain of the inverse found in part (d) is 4, .
f
e.
Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is , . f.
90.
Using the graph of y log 2 x , shift 4 units to the right, and stretch vertically by a factor of 3.
f ( x) 3x 1
a.
Inverse
y 1
1
( x) log3 x 1
For the domain of f 1 we need x 0 x0 The domain of the inverse found in part (d) is , 0 . Since the domain of f is the range of f 1 , we can use the result from part (a) to say that the range of f 1 is , .
f.
Using the graph of y log3 x , reflect the graph across the y-axis, and shift down 1 unit.
91. log3 x 2
Domain: (, ) x
b. Using the graph of y 3 , shift the graph to the left 1 unit, and reflect about the x-axis.
x 32 x9 The solution set is 9 .
92. log5 x 3 x 53 x 125 The solution set is 125 .
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Chapter 6: Exponential and Logarithmic Functions 93. log 2 (3 x 4) 5
99. log 4 64 x 5
4 x 64
3x 4 2 3 x 4 32 3 x 28 28 x 3
4 x 43 x3 The solution set is 3 .
100. log5 625 x
28 The solution set is . 3
5 x 625 5 x 54 x4 The solution set is 4 .
94. log3 (3x 2) 2 3 x 2 32 3x 2 9 3 x 11 11 x 3
101. log3 243 2 x 1 32 x 1 243 32 x 1 35 2x 1 5 2x 4 x2 The solution set is 2 .
11 The solution set is . 3
95. log x 16 2 x 2 16 x 4 ( x 4, base is positive)
102. log 6 36 5 x 3
The solution set is 4 .
65 x 3 36
1 96. log x 3 8 1 x3 8 1 x 2
65 x 3 6 2 5x 3 2 5 x 1 1 x 5 1 The solution set is . 5
1 The solution set is . 2
103. e3 x 10 3 x ln10 ln10 x 3
97. ln e x 5 e x e5 x5 The solution set is 5 .
ln10 The solution set is . 3
98. ln e 2 x 8 e 2 x e8 2x 8 x 4
The solution set is 4 .
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Section 6.4: Logarithmic Functions
104. e 2 x
109. log 2 8 x 6
1 3
8 x 26
1 2 x ln 3
2 2 3 x
23 x 26 3x 6 x 2 The solution set is 2 .
2 x ln 31 2 x ln 3 2 x ln 3 ln 3 x 2
110. log3 3x 1
ln 3 The solution set is . 2
105.
3x 31 x 1 The solution set is 1 .
e2 x 5 8 2 x 5 ln 8 2 x 5 ln 8 5 ln 8 x 2
111.
5 ln 8 . 2
1 ln13 . 2
112. 8 102 x 7 3 3 102 x 7 8
107. log 7 x 2 4 2 x2 4 72
2 x 7 log
x 2 4 49 x 45 3 5
The solution set is 3 5, 3 5 .
1 3 x 7 log 2 8
x 2 x 4 52 x 2 x 4 25 x 2 x 21 0 12 4 1 21 2 1
3 8
1
3
The solution set is 7 log . 2 8
108. log5 x 2 x 4 2
1
3 8
2 x 7 log
x 2 45
x
7 5
The solution set is 5ln .
The solution set is
7 5
7 5 0.2 x 5 ln 5 7 x 5ln 5
e 2 x 1 13 2 x 1 ln13 2 x 1 ln13 1 ln13 1 ln13 x 2 2
5e0.2 x 7 7 e0.2 x 5 0.2 x ln
The solution set is 106.
6
1 85 2
1 85 1 85 , . 2 2
The solution set is
669 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
113. 2 102 x 5 5 102 x 2
log3 2 x 1 2 0 log3 2 x 1 2
5 2 x log 2
2 x 1 32 2x 1 9
5 2
x 2 log
2x 8 x4 The zero of G is x 4 .
5 x 2 log 2
5 2
The solution set is 2 log .
116. a.
5 4
x 1 ln
b.
5 4
G x log3 2 x 1 2
log 2 x 1 3 1 log 2 x 1 2
x 1 22 x 1 4 x3 The point 3, 1 is on the graph of F.
G 40 log3 2 40 1 2
F x 0
d.
log 2 x 1 3 0
log3 81 2 42 2 The point 40, 2 is on the graph of G.
c.
F x 1
c.
We require that 2 x 1 be positive. 2x 1 0 2 x 1 1 x 2 1 1 Domain: x | x or , 2 2 b.
F 7 log 2 7 1 3
log 2 8 3 33 0 The point 7, 0 is on the graph of F.
5 4
The solution set is 1 ln . 115. a.
F x log 2 x 1 3
We require that x 1 be positive. x 1 0 x 1 Domain: x | x 1 or 1,
114. 4e x 1 5 5 e x 1 4 x 1 ln
G x 0
d.
log 2 x 1 3 x 1 23 x 1 8 x7 The zero of G is x 7 .
G x 3
log3 2 x 1 2 3 log3 2 x 1 5
2 x 1 35 2 x 1 243 2 x 242 x 121 The point 121,3 is on the graph of G.
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Section 6.4: Logarithmic Functions
117.
ln x if x 0 f x if x 0 ln x
120.
Domain: x | x 0 ; 0,
Domain: x | x 0
Range: y | y 0 ; , 0
Range: ,
Intercept: 1, 0
Intercepts: 1, 0 , 1, 0 118.
ln x if 0 x 1 f x if x 1 ln x
if x 1 ln x f x 1 x 0 x ln if
121. pH log10 H a.
pH log10 0.1 (1) 1
b.
pH log10 0.01 (2) 2
c.
pH log10 0.001 (3) 3
d.
As the H decreases, the pH increases.
e.
3.5 log10 H 3.5 log10 H H 10 3.5 3.16 104 0.000316
Domain: x | x 0 ; , 0 Range: y | y 0 ; 0, Intercept: 1, 0 119.
ln x if 0 x 1 f x if x 1 ln x
f.
7.4 log10 H 7.4 log10 H H 10 7.4 3.981 108 0.00000003981
122. H p1 log p1 p2 log p2 pn log pn p1 log p1 p2 log p2 pn log pn
a.
H 0.593log 0.593 0.126 log 0.126
0.007 log 0.007 0.059 log 0.059
Domain: x | x 0 ; 0,
0.002 log 0.002 0.189 log 0.189
Range: y | y 0 ; 0, Intercept: 1, 0
0.023log 0.023 0.5154
671 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
b.
H max log 7 0.8451
c.
E
124. A A0 e 0.35 n a.
H 0.5154 0.6098 H max 0.8451
0.5 e 0.35n ln 0.5 0.35n
d. (answers may vary)
t
Race
Proportion
White Black or African American
0.720
American Indian/ Native Alaskan
0.009
Asian
0.050
Native Hawaiian/Pacific
0.002
b.
0.130
Some other race
0.060
Two or More Races
0.029
10 100e 0.35 n 0.1 e 0.35n ln 0.1 0.35n t
ln 0.1
6.58 0.35 About 6.58 days, or 6 days and 14 hours.
125. F (t ) 1 e 0.1t a.
0.5 1 e 0.1t 0.5 e 0.1t
H 0.720 log 0.720 0.130 log 0.130
0.5 e 0.1t ln 0.5 0.1t
0.009 log 0.009 0.050 log 0.050 0.002 log 0.002 0.060 log 0.060
0.029 log 0.029 0.4247 The United States appears to be growing more diverse.
t
ln 0.5
6.93 0.1 Approximately 6.93 minutes.
b.
0.8 1 e 0.1t 0.2 e 0.1t
123. p 760e 0.145 h
b.
ln 0.5
1.98 0.35 Approximately 2 days.
Islander
a.
50 100e 0.35 n
0.2 e 0.1t ln 0.2 0.1t
320 760e 0.145 h 320 e 0.145h 760 320 ln 0.145h 760 320 ln 760 5.97 h 0.145 Approximately 5.97 kilometers.
t
ln 0.2
16.09 0.1 Approximately 16.09 minutes.
c.
It is impossible for the probability to reach 100% because e 0.1t will never equal zero; thus, F (t ) 1 e 0.1t will never equal 1.
667 760e 0.145 h 667 e 0.145h 760 667 ln 0.145h 760 667 ln 760 0.90 h 0.145 Approximately 0.90 kilometers.
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Section 6.4: Logarithmic Functions
126. F (t ) 1 e 0.15 t a.
0.50 1 e
129. I
0.15 t
0.5 e
R / L t
0.5 e0.15t ln 0.5 0.15t t
ln 0.5
4.62 0.15 Approximately 4.62 minutes, or 4 minutes and 37 seconds.
b.
0.80 1 e 0.15 t 0.2 e0.15t
ln 0.2
10.73 0.15 Approximately 10.73 minutes, or 10 minutes and 44 seconds. D 5e 0.4 h 2 5e0.4 h 0.4 e0.4 h ln 0.4 0.4h h
ln 0.4
2.29
0.4 Approximately 2.29 hours, or 2 hours and 17 minutes.
128.
N P 1 e0.15 d
450 1000 1 e 0.45 1 e
0.15 d
Substituting E 12 , R 10 , L 5 , and I 1.0 , we obtain: 12 10 / 5 t 1.0 1 e 10 10 1 e 2t 12 1 e 2 t 6 2t ln 1/ 6 t
ln 1/ 6
0.8959 2 It takes approximately 0.8959 second to obtain a current of 1.0 ampere.
Graphing:
0.15 d
0.55 e0.15 d 0.55 e0.15 d ln 0.55 0.15 d d
ln 7 /12
0.2695 2 It takes approximately 0.2695 second to obtain a current of 0.5 ampere.
0.2 e ln 0.2 0.15t
127.
Substituting E 12 , R 10 , L 5 , and I 0.5 , we obtain: 12 10 / 5 t 0.5 1 e 10 5 1 e2t 12 7 e 2 t 12 2t ln 7 /12 t
0.15t
t
E R/L t 1 e R
ln 0.55
3.99 0.15 Approximately 4 days.
673 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
130. L(t ) A 1 e k t
20 200 1 e k (5)
a.
0.1 1 e
134. Intensity of car: x 70 10 log 12 10 x 7 log 12 10 x 107 12 10 x 105
5 k
5 k
e 0.9 5 k ln 0.9 ln 0.9 k 0.0211 5 ln 0.9 (10) 5 L(10) 200 1 e
b.
200 1 e2ln 0.9
Intensity of truck is 10 105 104 . 104 L 104 10 log 12 10
38 words L(15) 200 1 e
ln 0.9 (15) 5
c.
200 1 e3ln 0.9
10 log 108
10 8 80 decibels 125,892 135. M (125,892) log 8.1 103
54 words ln 0.9 t 5 180 200 1 e
d.
0.9 1 e e
ln 0.9 t 5
50,119 136. M (50,119) log 7.7 103
ln 0.9 t 5
0.1
137. R ekx a. 1.4 e k (0.03)
ln 0.9 t ln 0.1 5
ln 0.1 t ln 0.9 109.27 minutes
1.4 e0.03 k ln 1.4 0.03 k
5
107 131. L 107 10 log 12 10
k
10 log 105
b.
10 5 50 decibels
c.
10 log 1011
0.03
11.216
R e11.216(0.17) e1.90672 6.73 100 e11.216 x
x
10 11 110 decibels
0.41 percent
e.
Answers will vary.
10 log 109
10 9 90 decibels
11.216
5 e11.216 x ln 5 11.216 x ln 5 x 0.14 percent 11.216 At a percent concentration of 0.14 or higher, the driver should be charged with a DUI.
ln 100
d.
103 133. L 103 10 log 12 10
ln 1.4
100 e11.216 x ln 100 11.216 x
101 132. L 101 10 log 12 10
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Section 6.4: Logarithmic Functions 138. L( x) x ln x
141.
log 3 92 x 3 x 2 1 2
a.
b.
c.
L(0.1) (0.1) ln(0.1) 0.230 If you reject 10% of the first individuals you date, the probability of finding the ideal mate is 0.230. L(0.6) (0.6) ln(0.6) 0.306 If you reject 60% of the first individuals you date, the probability of finding the ideal mate is 0.306. Since you can only reject between 0 and 100 percent of the individuals (not 100 since you would not find a mate), then the implied domain is (0, 1].
d.
3x 1 92 x 3 2
3x 1 32
2 x 3
x 2 1 2(2 x 3) x2 1 4x 6 x2 4 x 5 0 ( x 5)( x 1) 0 x 5 or x 1 The solution set is 1,5 .
142. No. Explanations will vary. 143. If the base of a logarithmic function equals 1, we would have the following: f x log 1 x
f 1 x 1x 1 for every real number x. In other words, f 1 would be a constant function and, therefore, f 1 would not be oneto-one.
e.
L is maximized at x = 0.368 and the highest probability is 0.368.
139. log 6 (log 2 x) 1 61 log 2 x 26 x x 64 The solution set is 64 .
140. log 2 (log 4 (log3 x)) 0 20 log 4 (log 3 x) 1 log 4 (log 3 x) 41 log 3 x 34 x x 81
The solution set is 81 .
675 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
144. New Old e R t
Age Depriciation rate 5 38, 000 21, 200e R 5 38, 000 e5 R 21, 200
Age Depriciation rate R 1 1 38, 000 36, 600e 38, 000 eR 36, 600
38, 000 ln 5R 21, 200 38, 000 ln 21, 200 R 0.1167 11.7% 5
38, 000 R ln 0.03754 3.8% 36, 600 2
R 2 38, 000 32, 400e 38, 000 e2 R 32, 400 38, 000 ln 2R 32, 400
Answers will vary. 145. g ( x) 4 x 4 37 x 2 9 (4 x 2 1)( x 2 9)
38, 000 ln 32, 400 R 0.07971 8% 2 3
(2 x 1)(2 x 1)( x 3)( x 3) 0 (2 x 1)(2 x 1)( x 3)( x 3) Setting each factor to 0, we get: 1 1 x , x , x 3, x 3 2 2 So the zeros of the function and the x-intercepts 1 1 are: , ,3, 3 2 2 1 1 f (1) f 2 91 9 2 146. 1 1 1 1 2 2 93 6 12 1 1 2 2
R 3 38, 000 28, 750e 38, 000 e3 R 28, 750 38, 000 ln 3R 28, 750
38, 000 ln 28, 750 R 0.0930 9.3% 3 4
R 4 38, 000 25, 400e 38, 000 e4 R 25, 400 38, 000 ln 4R 25, 400
147.
f (1) 4(1)3 2(1) 2 7 427
38, 000 ln 24,500 R 0.1007 10.1% 4
5 f (2) 4(2)3 2(2) 2 7 4(8) 2(4) 7 32 8 7 17 Since f(b) and f(a) are of opposite signs there is at least one real zero of f between a and b.
148. The remaining root must be the conjugate of (3 i ) which is (3 i ) . A polynomial could be (a = 1): f ( x) ( x 1)( x 2)( x (3 i ))( x (3 i )) x 4 7 x 3 14 x 2 2 x 20
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Section 6.5: Properties of Logarithms 149. a 2, b 7, c 1 x
3. r
(7) (7) 2 4(2)(1) 2(2)
4. log a M ; log a N 5. log a M ; log a N
7 49 8 4 7 57 4
6. r log a M
7 57 7 57 The solution set is , 4 4
150.
5 4 1 5 80 8 8 5 y 1 ( x 0) 8 5 y x 1 8 8 y 5 x 8 5x 8 y 8
x3 8. False: ln( x 3) ln(2 x) ln 2x
9. False: log 2 3 x 4 log 2 3 4 log 2 x
m
10. False 11. b 12. b 13. log 7 7 29 29
151. 2 x 17 45 or 2 x 17 45 2 x 62 2 x 28 x 31 x 14 The solution set is 31,14
14. log 2 213 13 15. ln e 4 4 16. ln e 2 2
152. H (30.9) 2.90(30.9) 61.53
153.
7. 7
89.61 61.53 151.1 cm
17. 9log9 13 13
f ( x) f 2
18. eln 8 8
x2
x3 8 x2 ( x 2)( x 2 2 x 4) x2 x2 2x 4
19. log8 2 log8 4 log8 4 2 log8 8 1 20. log 6 9 log 6 4 log 6 9 4 log 6 36
154. ( x 5) 4 7( x 3)6 ( x 3)7 4( x 5)3
log 6 62
( x 5) ( x 3) ( x 5) 7 ( x 3) 4 3
6
( x 5) ( x 3) 7 x 35 4 x 12 3
2
6
( x 5)3 ( x 3)6 11x 23
21. log 5 35 log 5 7 log 5
35 log 5 5 1 7
22. log8 16 log8 2 log8
16 log8 8 1 2
Section 6.5 1. 0 2. M 677 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
23. log 2 6 log 6 8 log 6 8log2 6
30. ln
log 6 2
3 log 2 6
log 6 23log2 6 3
2 ln 2 ln 3 a b 3
31. ln1.5 ln
3 ln 3 ln 2 b a 2
32. ln 0.5 ln
1 ln1 ln 2 0 a a 2
log 6 2log2 6 log 6 63 3
24. log 3 8 log8 9 log8 9log3 8
33. ln 8 ln 23 3 ln 2 3a
log8 32
34. ln 27 ln 33 3 ln 3 3b
log3 8
log8 32 log3 8
35. ln 5 6 ln 61/ 5 1 ln 6 5 1 ln 2 3 5 1 ln 2 ln 3 5 1 a b 5
log3 82
log8 3
log8 82 2
25. 4log4 6 log4 5 4
log 4
6 5
6 5
26. 5log5 6 log5 7 5log5 (67) 5log5 42 42 27. e
log 2 16
1/ 4
e
36. ln 4
Let a log e2 16, then e 2 16. a
e 2 a 16 e2 a 42
e
2 a 1/ 2
42
1/ 2
ea 4 a ln 4
Thus, e 28. e
log 2 16 e
37. log 6 36 x log 6 36 log 6 x 2 log 6 x
eln 4 4 .
log 2 9 e
38. log 3
Let a log e2 9, then e 9. 2 a
e2a 9 1/ 2
40. log 7 x 5 5log 7 x
32
1/ 2
41. ln ex ln e ln x 1 ln x
ea 3 a ln 3
Thus, e
log 2 9 e
x x log 3 2 log 3 x log 3 32 log 3 x 2 9 3
39. log 5 y 6 6 log 5 y
e 2 a 32
e2 a
2 2 ln 3 3 1 2 ln 4 3 1 ln 2 ln 3 4 1 a b 4
eln 3 3 .
42. ln
29. ln 6 ln(2 3) ln 2 ln 3 a b
e ln e ln x 1 ln x x
x 43. ln x ln x ln e x ln x x e
678
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Section 6.5: Properties of Logarithms 44. ln xe x ln x ln e x ln x x
1/ 3
53.
45. log a u 2 v3 log a u 2 log a v3 2 log a u 3log a v a 46. log 2 2 log 2 a log 2 b 2 log 2 a 2 log 2 b b
47. ln x 2 1 x ln x 2 ln 1 x ln x 2 ln(1 x)1/ 2 1 2 ln x ln(1 x) 2
48. ln x 1 x 2 ln x ln 1 x 2 ln x ln 1 x 2
1/ 2
1 ln x ln 1 x 2 2 x3 3 49. log 2 log 2 x log 2 ( x 3) x 3 3log 2 x log 2 ( x 3) 3 x2 1 50. log 5 2 x 1 log 5 x 2 1
1/ 3
x( x 2) log x( x 2) log( x 3) 2 51. log 2 x ( 3) log x log( x 2) 2 log( x 3) x x 1 52. log log x3 x 1 log( x 2) 2 2 ( x 2) log x3 log( x 1)1/ 2 2 log( x 2) 1 3log x log( x 1) 2 log( x 2) 2
1 ( x 2)( x 1) ln 3 ( x 4) 2 1 ln( x 2)( x 1) ln( x 4) 2 3 1 ln( x 2) ln( x 1) 2 ln( x 4) 3 1 1 2 ln( x 2) ln( x 1) ln( x 4) 3 3 3 2/3
x 4 2 54. ln 2 x 1 2 2 x 4 ln 2 3 x 1
2 2 ln x 4 ln x 2 1 3 2 2 ln x 4 ln x 1 x 1 3 2 2 ln x 4 ln x 1 ln x 1 3 4 2 2 ln x 4 ln x 1 ln x 1 3 3 3
55. ln
log 5 ( x 2 1)
1 log 5 x 2 1 log 5 x 2 1 3 1 log 5 x 2 1 log 5 x 1 x 1 3 1 log 5 x 2 1 log 5 x 1 log 5 x 1 3
3
x2 x 2 ln 2 ( x 4)
5 x 1 3x ( x 4)3
ln 5 x 1 3 x ln( x 4)3 ln 5 ln x ln 1 3 x 3ln x 4 ln 5 ln x ln 1 3x
1/ 2
3ln x 4
1 ln 5 ln x ln 1 3x 3ln x 4 2 5x2 3 1 x 56. ln 2 4( x 1)
ln 5 x 2 3 1 x ln 4( x 1) 2 ln 5 ln x 2 ln 1 x
1/ 3
ln 4 ln x 1 2
1 ln 5 2 ln x ln 1 x ln 4 2 ln x 1 3
57. 3log 5 u 4 log 5 v log 5 u 3 log 5 v 4
679 Copyright © 2025 Pearson Education, Inc.
log 5 u 3 v 4
Chapter 6: Exponential and Logarithmic Functions x x 1 2 63. ln ln ln x 1 x x 1 x x 1 2 ln ln x 1 x 1 x
58. 2 log 3 u log 3 v log 3 u 2 log 3 v u log 3 v 2
x 59. log 3 x log 3 x3 log 3 3 x x1/ 2 log 3 3 x
x 1 ln x 2 1 x 1 x 1 ln x 1 x 2 1 x 1 ln x x x ( 1)( 1)( 1)
log 3 x 5 / 2 1 log 3 5 / 2 x
1 ln 2 ( x 1) ln( x 1) 2 2 ln( x 1)
1 1 1 1 60. log 2 log 2 2 log 2 2 x x x x 1 log 2 3 x
x2 2 x 3 x2 7 x 6 log 64. log 2 x 4 x2
61. log 4 x 2 1 5log 4 x 1 log 4 x 2 1 log 4 x 1
x2 2x 3 x2 4 log 2 x 7x 6 x 2
5
x2 1 log 4 5 x 1 x 1 x 1 log 4 5 x 1 x 1 log 4 4 x 1
( x 3)( x 1) x2 log ( x 2)( x 2) ( x 6)( x 1) ( x 3)( x 1) log ( x 2)( x 6)( x 1) 4 65. 8log 2 3x 2 log 2 log 2 4 x
62. log x 2 3x 2 2 log 2 x 1 log x 2 3 x 2 log 2 x 1
2
log 2
x 2 3x 2 log x 12 x 2 x 1 log x 12 2 x log x 1
3x 2 log 4 log x log 4 8
2
2
2
log 2 (3 x 2) log 2 4 log 2 x log 2 4 4
log 2 (3 x 2) 4 log 2 x log 2 x(3 x 2) 4
66. 21log 3 3 x log 3 9 x 2 log 3 9 log 3 x1/ 3 log 3 9 log 3 x 2 log 3 9 21
log 3 x 7 log 3 x 2 log 3 x 7 x 2 log 3 x 9
680
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Section 6.5: Properties of Logarithms
1 67. 2 log a 5 x 3 log a (2 x 3) 2
76. log 5 8
log a 5 x 3 log a (2 x 3)1/ 2
log 8 log 5
2.584
2
68.
ln e 0.874 ln
log a 25 x 6 log a 2 x 3
77. log e
25 x 6 log a 2x 3
78. log 2
1 1 log x 3 1 log x 2 1 3 2
79. y log 4 x
log x3 1
1/ 3
log x 2 1
1/ 2
3
ln x log x or y ln 4 log 4
log x 1 x 1 3
ln 2 0.303 ln
2
69. 2 log 2 x 1 log 2 x 3 log 2 x 1 log 2 x 1 log 2 x 3 log 2 x 1 2
x 1 log 2 x 1 x 3 x 12 log 2 x 3 x 1
2
log 2
80. y log 5 x
ln x log x or y ln 5 log 5
70. 3log 5 3x 1 2 log 5 2 x 1 log 5 x log 5 3 x 1 log 5 2 x 1 log 5 x 3
2
3x 1 log 5 log 5 x 2 2 x 1 3x 13 log 5 2 x 2 x 1 3
71. log 3 21
81. y log 2 ( x 2)
log 21 2.771 log 3
82. y log 4 ( x 3)
log 71 log 71 3.880 log 1/ 3 log 3
74. log1/ 2 15
log15 log15 3.907 log 1/ 2 log 2
log 2
ln( x 3) log( x 3) or y ln 4 log 4
73. log1/ 3 71
log 7
log18 72. log 5 18 1.796 log 5
75. log 2 7
ln( x 2) log( x 2) or y ln 2 log 2
5.615
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Chapter 6: Exponential and Logarithmic Functions
83. y log x 1 ( x 1)
ln( x 1) log( x 1) or y ln( x 1) log( x 1)
d.
f h x f h x log 2 4 x or log 2 4 log 2 x 2 log 2 x
Domain: x | x 0 or 0,
e.
f h 8 log 2 4 8 log 2 32 5 or 2 log 2 8 2 3 5
84. y log x 2 ( x 2)
ln( x 2) log( x 2) or y ln( x 2) log( x 2)
87. ln y ln x ln C ln y ln xC
y Cx
88. ln y ln x C y xC
85.
89. ln y ln x ln( x 1) ln C
f x ln x ; g x e x ; h x x 2
a.
b.
ln y ln x x 1 C
f g x f g x ln e x x Domain: x | x is any real number or ,
y Cx x 1
90. ln y 2 ln x ln x 1 ln C x 2C ln y ln x 1 Cx 2 y x 1
g f x g f x eln x x Domain: x | x 0 or 0, (Note: the restriction on the domain is due to the domain of ln x )
c. d.
e. 86.
f g 5 5
91. ln y 3x ln C
[from part (a)]
f h x f h x ln x 2 Domain: x | x 0 or , 0 0,
ln y ln e3 x ln C ln y ln Ce3 x y Ce3 x
f h e ln e 2 ln e 2 1 2 2
92. ln y 2 x ln C
f x log 2 x ; g x 2 ; h x 4 x x
a.
b.
ln y ln e 2 x ln C ln y ln Ce 2 x
f g x f g x log 2 2 x x Domain: x | x is any real number or ,
y Ce 2 x
g f x g f x 2log2 x x Domain: x | x 0 or 0, (Note: the restriction on the domain is due to the domain of log 2 x )
c.
f g 3 3
[from part (a)] 682
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Section 6.5: Properties of Logarithms 93. ln y 3 4 x ln C ln y 3 ln e
4 x
97. log 2 3 log 3 4 log 4 5 log 5 6 log 6 7 log 7 8
ln C
ln y 3 ln Ce4 x
log 3 log 4 log 5 log 6 log 7 log 8 log 2 log 3 log 4 log 5 log 6 log 7
log 8 log 23 log 2 log 2 3log 2 log 2 3
y 3 Ce4 x y Ce4 x 3
94. ln y 4 5 x ln C ln y 4 ln e5 x ln C ln y 4 ln Ce5 x
log 4 log 6 log 8 log 2 log 4 log 6 log 8 log 2
98. log 2 4 log 4 6 log 6 8
y 4 Ce5 x y Ce5 x 4
log 23 log 2 3log 2 log 2 3
1 1 95. 3ln y ln 2 x 1 ln x 4 ln C 2 3 ln y 3 ln 2 x 1
ln x 4
1/ 2
1/ 3
ln C
C 2 x 11/ 2 ln y 3 ln 1/ 3 x 4
99. log 2 3 log 3 4 log n n 1 log n 1 2
C 2 x 1
1/ 2
y3
x 4
1/ 3 1/ 3
C 2 x 11/ 2 y 1/ 3 x 4 3
y
log 2 log 2 1
C 2 x 1
1/ 6
x 4
1/ 9
100. log 2 2 log 2 4 . . . log 2 2n
1 1 96. 2 ln y ln x ln x 2 1 ln C 2 3 ln y 2 ln x1/ 2 ln x 2 1
1/ 3
C x 2 1 ln y 2 ln x1/ 2
1/ 3
C x 2 1
1/ 3
y 2
log 3 log 4 log n 1 log 2 log 2 log 3 log n log n 1
x1/ 2 1/ 2
C x 2 11/ 3 y x1/ 2
log 2 2 log 2 22 log 2 2n 1 2 3 n n!
ln C
log a x x 2 1 x x 2 1 log a x 2 x 2 1 log a x 2 x 2 1 log a 1 0
C x 2 1
1/ 6
y
101. log a x x 2 1 log a x x 2 1
x1/ 4
683 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
x x 1 log x x 1 log x x 1 x x 1
102. log a
107.
a
1 1 f log a x x log a 1 log a x
a
log a x x 1
log a x
log a x x 1
f ( x)
log a 1 0
108.
103. 2 x ln 1 e
2 x
2 x
2 x
2x
109. If A log a M and B log a N , then a A M
2 x
and a B N . aA M log a log a B N a
f ( x h) f ( x) log a ( x h) log a x h h x h log a x h 1 h log a 1 h x
log a a A B A B log a M log a N 1 110. log a log a N 1 N 1 log a N
1
h h log a 1 , h 0 x
105.
log a N ,
f ( x) log a x means that x a f ( x ) . Now, raising both sides to the 1 power, we f ( x) f ( x) 1 1 . obtain x 1 a f ( x ) a 1 a f ( x)
1 x 1 means that log1/ a x 1 f ( x) . a Thus, log1/ a x 1 f ( x)
111. log a b
log b b log b a
1 log b a
112. log a m
log a m log a a
a 1
log a m 1
log a a 2
log m log a m 1 a 1 2 log a a 2
log1/ a x f ( x) f ( x) log1/ a x
106.
f x log a x
f x log a x log a x f ( x)
ln 1 e ln e e 1 ln e ln e 1 2 x ln e 1 2x
2x
104.
f ( x) log a x
2 log a m log a m 2
f ( AB) log a ( AB) log a A log a B f ( A) f ( B)
113. log an b m
log a b m m log a b log a a n n log a a
m log a b m log a b n n
684
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Section 6.5: Properties of Logarithms
The zeros are: 1.78,1.29,3.49 .
114. log 2 3 log 3 4 log 4 5 log n (n 1) 10 ln(n 1) ln 3 ln 4 ln 5 10 ln 2 ln 3 ln 4 ln n ln(n 1) 10 ln 2 log 2 (n 1) 10
120. The discriminant is b 2 4ac . So b 2 4ac ( 28) 2 4(4)(49) 784 784 0 Since the discriminant is zero then there is a repeated real solution (double root).
n 1 210 n 1023
115. Y1 log x 2
Y2 2 log x
121.
y log x 2
y 2 log x
The domain of Y1 log a x is x x 0 . The 2
domain of Y2 2 log a x is x x 0 . These two domains are different because the logarithm property log a x n n log a x holds only when log a x exists. 116. Answers may vary. One possibility follows: Let a 2 , x 8 , and r 3 . Then
f ( x) 5 x5 44 x 4 116 x3 95 x 2 4 x 4 . Possible rational zeros: p 1, 2, 4; q 1, 5;
p 1 2 4 1, 2, 4, , , q 5 5 5 Using synthetic division: We try 2 : 2 5 44 116 95 4 4 10 68 96 2 4 5 34 48 1 2 0 2 is a zero and we try it again: 2 5
5 24
log 2 8 3log 2 8 and, in general, r log a x r log a x . 3
117. Answers may vary. One possibility follows: Let x 4 and y 4 . Then log 2 ( x y ) log 2 (4 4) log 2 8 3 . But log 2 x log 2 y log 2 4 log 2 4 2 2 4 . Thus, log 2 (4 4) log 2 4 log 2 4 and, in general, log 2 ( x y ) log 2 x log 2 y . 118. No. log 3 (5) does not exist. The argument of a logarithm must be nonnegative. 119. Using a graphing utility set Y 1 x3 3 x 2 4 x 8 and Y 2 0 . Graph both and use the graphing utility to find the intersection points. You can also use the SOLVE function on the graphing utility to find the zeros.
0
0
2
1
0
So 2 is a repeated root. We now try 1 5 5
3
r log a x 3log 2 8 3 3 9 . Thus,
48 1 2
10 48
log a x log 2 8 33 27 . But r
34
24
1 5
0 1
1 5 1 5 25 5 0 We can use the quadratic formula on 5 x 2 25 x 5 0 . x2 5x 1 0
x
b b 2 4ac 2a
5 52 4(1)(1) 2(1)
5 25 4 2 5 21 2
The zeros are 2 , of multiplicity 2, 5 21 , each of multiplicity 1. 2
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1 , 5
Chapter 6: Exponential and Logarithmic Functions
122.
f ( x) 2 x ( x 2) . Use the graph of
127.
f ( x) x . The graph would be reflected about the y-axis and shifted horizontally by 2 units to the right.
128.
f (3) f (1) 33 (1)3 3 (1) 3 (1) 27 1 4 28 7 4 f ( x)
Domain: , 2
Range: 0,
3 x 5 x 2 3x 4 3
x
f ( x)
Section 6.6 1.
4 x 1 32
x 1 8
x 2 7 x 30 0 ( x 3)( x 10) 0 x 3 0 or x 10 0 x 3 or x 10 The solution set is {3, 10} .
8 x 1 8 9 x 1 7 The solution set is x | 9 x 1 7 or 9, 7 .
2. Let u x 3 . Then ( x 3) 2 4( x 3) 3 0
b 4 4 2a 2( 12 )
u 2 4u 3 0 (u 1)(u 3) 0
1 f (4) (4) 2 4(4) 5 13 2
u 1 0 or u 3 0 u 1 or u 3 Back substituting u x 3 , we obtain x 3 1 or x 3 3 x 2 or x0 The solution set is {2, 0} .
The vertex is (4,13) and the graph is concave down since a is negative. 126.
5 x 2 3x4
The function is odd.
124. 4 x 1 9 23
x
3 ( x)
123. The radicand must be non-negative: 3 5x 0 5 x 3 3 x 5 3 The solution set is , 5
125.
5( x) 2 3( x) 4
x 2 10 x y 2 4 y 35 ( x 2 10 x 25) ( y 2 4 y 4) 35 25 4 ( x 5) 2 ( y 2) 2 64
3. x3 x 2 5 Using INTERSECT to solve: y1 x3 ; y2 x 2 5
The center is 5, 2 and the radius is 8.
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Section 6.6: Logarithmic and Exponential Equations
8. log3 (3x 1) 2
Thus, x 1.43 , so the solution set is {1.43} . 4. x3 2 x 2 0 Using ZERO to solve: y1 x3 2 x 2
Thus, x 1.77 , so the solution set is {1.77} . 5. log 4 x 2 x 42 x 16 The solution set is 16 .
3 x 1 32 3x 1 9 3 x 10 10 x 3 10 3
The solution set is . 9. log 4 ( x 12) log 4 7 x 12 7 x 5 The solution set is 5 . 10. log5 (2 x 11) log 5 3 2 x 11 3 2 x 8 x 4 The solution set is 4 . 11. log 4 x 3 x 43 x 64 or x 64 The solution set is 64, 64 .
6. log ( x 6) 1 x 6 101 x 6 10 x4 The solution set is 4 .
7. log 2 (5 x) 4 5 x 24 5 x 16 16 x 5
12. log 2 x 7 4 x 7 24 x 7 16 or x 7 16 x 23 x 9 The solution set is 9, 23 .
13. log5 2 x 1 log5 13 x 7 13
16 5
The solution set is .
2 x 1 13 or 2x 1 13 x7 x 6 The solution set is 6, 7 .
14. log9 3x 4 log 9 5 x 12 3x 4 5 x 12 or 3x 4 5 x 12 2 x 16 8 x 8 x8 x 1 The solution set is 1,8 .
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Chapter 6: Exponential and Logarithmic Functions
15.
19. 2 log 6 ( x 5) log 6 9 2 2 log 6 ( x 5) log 6 9 2
1 log 7 x 3log 7 2 2 log 7 x1/ 2 log 7 23
log 6 ( x 5) 2 log 6 9 2
x1/ 2 8 x 64 The solution set is 64 .
log 6 9( x 5) 2 2 2
9( x 5) 62 36 ( x 5) 2 9 2 ( x 5) 4 x5 2 x 3 The solution set is 3 .
16. 2 log 4 x log 4 9 log 4 x 2 log 4 9 x 2 9 1 9 x2 1 x2 9
20. 2 log3 ( x 20) log3 25 2 log3 ( x 20) 2 log3 25 2
1 x 3 1 Since log 4 is undefined, the solution set is 3 1 . 3
( x 20) 2 2 25 ( x 20) 2 32 25 ( x 20) 2 225 x 20 15 x 20 15 x 5 or x 35 Since log3 35 20 log3 15 is undefined, log3
17. 3log 2 x log 2 27 log 2 x3 log 2 27 1 x3 27 1 1 x3 27 1 x 3
the solution set is 5 . 21. log x log( x 15) 2 log x( x 15) 2
1 The solution set is . 3
x( x 15) 102 x 2 15 x 100 0 ( x 20)( x 5) 0 x 20 or x 5
18. 2 log 5 x 3log5 4 log5 x 2 log5 43
Since log 20 is undefined, the solution set is
x 2 64 x 8 Since log 5 8 is undefined, the solution set is
5 . 22. log x log( x 21) 2 log x( x 21) 2
8 .
x( x 21) 102 x 2 21x 100 0 ( x 4)( x 25) 0 x 4 or x 25 Since log 4 is undefined, the solution set is
25 . 688
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Section 6.6: Logarithmic and Exponential Equations
23.
Since log 6 6 4 log 6 2 is undefined,
log(7 x 6) 1 log( x 1) log(7 x 6) log( x 1) 1 7x 6 log 1 x 1 7x 6 101 x 1 7 x 6 10( x 1) 7 x 6 10 x 10 3 x 16 16 16 x 3 3
the solution set is 1 . 27.
log8 ( x 6) 1 log8 ( x 4) log8 ( x 6) log8 ( x 4) 1 log8 ( x 6)( x 4) 1
( x 6)( x 4) 81 x 2 4 x 6 x 24 8 x 2 10 x 16 0 ( x 8)( x 2) 0 x 8 or x 2 Since log8 8 6 log8 2 is undefined, the
16 3
The solution set is .
solution set is 2 .
24. log(2 x) log( x 3) 1 2x log 1 x 3 2x 101 x 3 2 x 10( x 3) 2 x 10 x 30 8 x 30 30 15 x 4 8
28.
log 5 ( x 3) 1 log5 ( x 1) log5 ( x 3) log5 ( x 1) 1 log 5 ( x 3)( x 1) 1
( x 3)( x 1) 51 x 2 x 3x 3 5 x2 2 x 8 0 ( x 4)( x 2) 0 x 4 or x 2 Since log 5 4 3 log 5 1 is undefined, the
15 4
The solution set is .
solution set is {2}.
25. log 2 ( x 7) log 2 ( x 8) 1 log 2 ( x 7)( x 8) 1
29. ln x ln( x 2) 4 ln x( x 2) 4
( x 7)( x 8) 21
x( x 2) e 4
x 2 8 x 7 x 56 2
x 2 2 x e4 0
x 2 15 x 54 0 ( x 9)( x 6) 0 x 9 or x 6 Since log 2 9 7 log 2 2 is undefined,
the solution set is 6 . 26. log 6 ( x 4) log 6 ( x 3) 1 log 6 ( x 4)( x 3) 1
( x 4)( x 3) 61 x 2 3 x 4 x 12 6 x2 7 x 6 0 ( x 6)( x 1) 0 x 6 or x 1
x
2 22 4(1)( e 4 ) 2(1)
2 4 4e 4 2
2 2 1 e4 2
1 1 e 4 x 1 1 e 4 or x 1 1 e4 8.456 6.456 Since ln 8.456 is undefined, the solution set
is 1 1 e4 6.456 .
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Chapter 6: Exponential and Logarithmic Functions 32. log 2 x 1 log 2 x 7 3
30. ln( x 1) ln x 2
log 2 x 1 x 7 3
x 1 ln 2 x x 1 2 e x x 1 e2 x
x 1 x 7 23
x2 7 x x 7 8 x2 8x 1 0
e2 x x 1
x
x e2 1 1 x
1
8 82 4(1)(1) 2(1)
8 68 2 8 2 17 2 4 17
0.157
e2 1 1 The solution set is 2 0.157 . e 1
31. log9 x 8 log9 x 7 2
x 4 17 or x 4 17 8.123 0.123 Since log 2 8.123 1 log 2 7.123 is
log9 x 8 x 7 2
x 8 x 7 92
undefined, the solution set is
x 2 8 x 7 x 56 81
4 17 0.123.
x 2 15 x 25 0
15 (15)2 4(1)(25) 2 15 325 2 15 5 13 2
x
Since log3 16.514 8 log 3 8.514 is undefined, the solution set is 15 5 13 0.854 . 2
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Section 6.6: Logarithmic and Exponential Equations 36. log a x log a ( x 2) log a ( x 4)
33. log1/ 3 ( x 2 x) log1/ 3 ( x 2 x) 1
log a x( x 2) log a ( x 4) x( x 2) x 4
x x log1/ 3 2 1 x x 2
x2 x
1 x2 x 3
x2 2x x 4
1
x2 x
3 x2 x x2 x 3 x2 x
2
2
x x 3x 3x 2
2 x 4 x 0 2 x x 2 0
x 2 3x 4 0 ( x 4)( x 1) 0 x 4 or x 1 Since log a (1) is undefined, the solution set is
4 .
37. 2 log5 ( x 3) log 5 8 log 5 2 log5 ( x 3) 2 log 5 8 log 5 2
2 x 0 or x 2 0 x 0 or x2 Since each of the original logarithms are not defined for x 0 , but are defined for x 2 , the solution set is 2 .
log 5
( x 3)2 log 5 2 8 ( x 3)2 2 8 ( x 3)2 16
x 2 6 x 9 16
2
34. log 4 ( x 9) log 4 ( x 3) 3 x 9 log 4 3 x3 ( x 3)( x 3) 43 x3 x 3 64 x 67 Since each of the original logarithms is defined for x 67 , the solution set is 67 . 2
35. log a ( x 1) log a ( x 6) log a ( x 2) log a ( x 3) x 1 x2 log a log a x6 x3 x 1 x 2 x6 x3 x 1 x 3 x 2 x 6 x 2 2 x 3 x 2 4 x 12 2 x 3 4 x 12 9 2x 9 x 2 Since each of the original logarithms is defined for 9 9 x , the solution set is . 2 2
x2 6x 7 0 ( x 7)( x 1) 0 x 7 or x 1 Since log5 ( 4) is undefined, the solution set is
7 .
38. log3 x 2 log 3 5 log 3 ( x 1) 2 log 3 10 log3 x log3 52 log 3 ( x 1) log3 102 x ( x 1) log 3 25 100 x x 1 25 100 4x x 1 3x 1 1 x 3 1 The solution set is . 3 log3
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Chapter 6: Exponential and Logarithmic Functions 39. 2 log 6 ( x 2) 3log 6 2 log 6 4 2
1 42. log( x 1) log 2 3
3
log 6 ( x 2) log 6 2 log 6 4
1
log 6 ( x 2) 2 log 6 (8)(4)
log( x 1) log 2 3
( x 2) 2 32
1
x 1 23
2
x 4 x 4 32
1 x 2 3 1 3 2 1 2.260
x 2 4 x 28 0 x
4 42 4(1)( 28) 2(1)
The solution set is
4 128 2 48 2 2 2 4 2
43.
3
(log3 x) 2 3(log3 x) 10 (log 3 x) 2 3(log 3 x) 10 0 (log 3 x 5)(log3 x 2) 0 log3 x 5 or log3 x 2
x 2 4 2 or x 2 4 2
The solution set is 0.123 .
x 243
40. 3(log 7 x log 7 2) 2 log 7 4 3log 7 x 3log 7 2 2 log 7 4
1 The solution set is , 243 . 9
log 7 x3 log 7 23 log 7 42 x3 log 7 42 8 x3 16 8 x3 128
44.
ln x 3 ln x 2 0 ( ln x 2)( ln x 1) 0 ln x 2 ln x =4
x 4 3 2 5.040
or
ln x 1 ln x 1
x e4 54.600 x e 2.718
The solution set is 4 3 2 .
The solution set is e, e4 .
41. 2 log13 ( x 2) log13 (4 x 7)
45. 2 x 5 8
2
log13 ( x 2) log13 (4 x 7)
2 x 5 23 x 5 3 x 8 The solution set is 8 .
( x 2) 2 (4 x 7) x2 4 x 4 4 x 7 x2 3 0 x2 3
46. 5 x 25
x 3 1.732
x 32 1 x 9
x 35
0.123
8.123
log 7
2 1 .
5 x 5 2 x 2 x 2 The solution set is 2 .
The solution set is 3, 3 .
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Section 6.6: Logarithmic and Exponential Equations
47. 2 x 10 log10 x log 2 10 3.322 log 2 The solution set is 1 log 2 10 3.322 . log 2
40.2 x
ln14 2.402 ln 3 The solution set is ln14 log3 14 2.402 . ln 3 x log3 14
53.
ln 1.2 ln 8
0.088
50. 2 x 1.5 x log 2 1.5
54. log1.5 0.585 log 2
The solution set is ln1.5 log 2 1.5 0.585 . ln 2
2
2 x 1 51 2 x
ln 2 x 1 ln 51 2 x
( x 1) ln 2 (1 2 x) ln 5 x ln 2 ln 2 ln 5 2 x ln 5 x ln 2 2 x ln 5 ln 5 ln 2 x(ln 2 2 ln 5) ln 5 ln 2 x ln 2 ln 25 ln 52
51. 5 23 x 8 3x
(1 2 x) ln 3 x ln 4 ln 3 2 x ln 3 x ln 4 ln 3 2 x ln 3 x ln 4 ln 3 x(2 ln 3 ln 4) ln 3 x 0.307 2 ln 3 ln 4 ln 3 The solution set is 0.307 . 2 ln 3 ln 4
The solution set is ln 1.2 log8 1.2 0.088 . ln 8
x log 2 1.5
31 2 x 4 x
ln 31 2 x ln 4 x
x
x log8 1.2
2 3
2 0.2 x log 4 3 log 4 (2 / 3) ln 2 / 3 x 1.462 0.2 0.2 ln 4 The solution set log (2 / 3) ln 2 / 3 is 4 1.462 . 0.2 0.2 ln 4
48. 3x 14
49. 8 1.2 x log8 1.2
52. 0.3 40.2 x 0.2
x ln 50 ln 52
8 5
8 3 x log 2 5 1 8 ln 8 / 5 x log 2 0.226 3 3ln 2 5 The solution set is 1 8 ln 8 / 5 log 2 0.226 . 3 5 3ln 2
ln 52
0.234 ln 50 ln 5 The solution set is 2 0.234 . ln 50
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x
Chapter 6: Exponential and Logarithmic Functions x ln 0.3 2 x ln 1.7 ln 1.7 ln 0.3
x
3 1 x 7 5
55.
x
x ln 0.3 2 ln 1.7 ln 1.7 ln 0.3
3 ln ln 71 x 5 x ln 3 / 5 (1 x) ln 7
x
x ln 3 / 5 x ln 7 ln 7
x ln 3 / 5 ln 7 ln 7
ln 1 x ln e x (1 x) ln x ln x ln x ln x x ln ln x(1 ln ) ln x 0.534 1 ln ln The solution set is 0.534 . 1 ln
ln 7 1.356 ln 3 / 5 ln 7
1 x
4 3
5x 1 x
4 ln ln 5 x 3 (1 x) ln 4 / 3 x ln 5
60.
ln 4 / 3 x ln 4 / 3 x ln 5
ln 4 / 3 x ln 5 ln 4 / 3
ln 43 x 0.152 ln 20 3 ln 4 The solution set is 3 0.152 . 20 ln 3 x
1.2 (0.5)
57.
22 x 2 x 12 0
61.
2 2 12 0 2 3 2 4 0 x 2
x
x
ln1.2 x ln(0.5) x x ln 1.2 x ln 0.5
x ln 1.2 x ln 0.5 0
58.
e x 3 x ln e x 3 ln x x 3 x ln 3 x ln x 3 x(ln 1) 3 x 20.728 ln 1 3 The solution set is 20.728 . ln 1
ln 4 / 3 x ln 5 x ln 4 / 3 ln 4 / 3 x ln 20 3
ln 0.51 0.297 ln(2.89 / 3)
1 x e x
59.
ln 7 The solution set is 1.356 . ln 3 / 5 ln 7
56.
ln 0.3 2 ln 1.7
ln 0.51 The solution set is 0.297 . ln(2.89 / 3)
x ln 3 / 5 ln 7 x ln 7
x
ln 1.7 ln 0.3
x
x
2x 3 0
or
2x 3
or
2x 4 0 2x 4
x ln 1.2 ln 0.5 0
ln 2 x ln 3
x0 The solution set is 0 .
x ln 2 ln 3 ln 3 x 1.585 ln 2 ln 3 The solution set is 1.585 . ln 2
0.31 x 1.7 2 x 1
ln 0.31 x ln 1.7 2 x 1
(1 x) ln 0.3 (2 x 1) ln 1.7
ln 0.3 x ln 0.3 2 x ln 1.7 ln 1.7
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No solution
Section 6.6: Logarithmic and Exponential Equations
(we ignore the first solution since 4 x is never negative)
32 x 3x 2 0
62.
3 3 2 0 3 13 2 0 x 2
x
x
The solution set is log 4 2 7
x
3x 1 0 or 3x 2 0 3x 1 or x0
63.
3
3 3 3 1 0 3 3 3 1 0
3x 2 No solution
3
4 0
x
x 2
x
u 2 3u 1 0 a 1, b 3, c 1
3 3 3 4 0 3 13 4 0 x
x
2 x
Let u 3x .
x 1
x 2
9 x 3x 1 1 0
66.
The solution set is 0 . 2x
x
u
3x 1 0
or
3x 4 0
3x 1 x0
or
3x 4 No solution
3
32 4 11 3 5 2 1 2
2 2 2 12 0 2 2 2 6 0 x
x
Therefore, we get 3 5 3x 2 3 5 x log3 2 The solution set is 3 5 3 5 , log 3 log3 2 2
2x 2 0
or
2x 6 0
0.876, 0.876 .
or
x
The solution set is 0 . 22 x 2 x 2 12 0
64.
x 2
x
2
2 2 x 1
x
2 6 No solution
67.
65. 16 x 4 x 1 3 0
4 4 4 3 0 4 4 4 3 0 2 x
x
x 2
x
25 x 8 5 x 16
5 8 5 16 5 8 5 16
The solution set is 1 .
2 x
x
x 2
x
Let u 5 x . u 2 8u 16 u 2 8u 16 0
Let u 4 x . u 2 4u 3 0 a 1, b 4, c 3 u
0.315 .
u 4 2 0
4 42 4 1 3 2 1
u4 Therefore, we get 5x 4 x log 5 4
4 28 2
The solution set is log5 4 0.861 .
4 2 7 2 7 2 Therefore, we get
4 x 2 7 or 4 x 2 7
x log 4 2 7
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Chapter 6: Exponential and Logarithmic Functions
71. 4 x 10 4 x 3 Multiply both sides of the equation by 4 x .
36 x 6 6 x 9
68.
6 6 6 9 0 6 6 6 9 0 6 3 0 2 x
x
x 2
x
4 10 4 4 3 4 4 10 3 4 4 3 4 10 0 4 5 4 2 0 x 2
x
6 3 x log 6 3
The solution set is log 6 3 0.613 .
x
x 2
x
x 2
4 80 not real 6 The equation has no real solution.
11 7 5 0 2 7 11 7 5 0 x
x
x 2
x
x
x
x x
or 3x 2 0
3x 7 x log3 7
70. 2 49 11 7 5 0
x 2
x
3x 7 0
x
x
4 x 2
x 2
2 3
x
or 4 x 2 0
3 14 3 3 5 3 3 14 5 3 3 5 3 14 0 3 7 3 2 0
4 42 4 3 8
2 72
x
x
72. 3x 14 3 x 5 Multiply both sides of the equation by 3x .
Let u 2 . 3u 2 4u 8 0 a 3, b 4, c 8
x
x
x
The solution set is log 4 5 1.161 .
x
u
x
4x 5 x log 4 5
42 8 0 32 4 2 8 0 x
x 2
4x 5 0
69. 3 4 x 4 2 x 8 0 3 22
x
x 2
2
x
x
3x 2
The solution set is log3 7 1.771 . 73.
x
Let u 7 . 2u 2 11u 5 0 2u 1 u 5 0
e x e x 1 2 e x e x 2 e x ( e x e x ) 2e x e 2 x 1 2e x
2u 1 0 or u 5 0 2u 1 u 5 1 u 2 Therefore, we get 1 7x or 7 x 5 2 Since 7 x 0 for all x, the equation has no real solution.
( e x ) 2 2e x 1 0 (e x 1) 2 0 ex 1 0 ex 1 x0 The solution set is {0} .
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Section 6.6: Logarithmic and Exponential Equations
e x e x 3 2 e x e x 6
74.
76.
e x ( e x e x ) 6e x
e
x
( e ) 6e 1 0 6 (6) 4(1)(1) 2(1)
2x
ex
6 32 6 4 2 3 2 2 2 2
x ln 3 2 2
4 20 2 4 2 5 2 5 2 x ln 2 5 or x ln 2 5
or x ln 3 2 2
ln 3 2 2 , ln 3 2 2 1.763,1.763 .
(e x ) 2 4e x 1 0 ( 4) ( 4) 2 4(1)(1)
4
2(1)
4 20 4 2 5 2 5 2 2 x ln 2 5 or x ln 2 5
x ln(0.236) or x 1.444 Since ln(0.236) is undefined, the solution set is
ln 2 5 1.444.
1.444.
77. log5 ( x 1) log 4 ( x 2) 1 Using INTERSECT to solve: y1 ln( x 1) / ln(5) ln( x 2) / ln(4) y2 1
e 2 x 1 4e x
x ln 4.236 or x 1.444
is ln 2 5
e x (e x e x ) 4e x
Since ln 4.236 is undefined, the solution set
e x e x 2 2 e x e x 4
ex
4 42 4(1)(1) 2(1)
x 1.763 or x 1.763 The solution set is
75.
1 4e x
( e x ) 2 4e x 1 0
2
ex
e x e x e x 4e x
e 2 x 1 6e x x 2
e x e x 2 2 e x e x 4
0
5
–4
Thus, x 2.79 , so the solution set is {2.79} . 78. log 2 ( x 1) log 6 ( x 2) 2 Using INTERSECT to solve: y1 ln( x 1) / ln(2) ln( x 2) / ln(6) y2 2 4
0
15
–4
Thus, x 12.15 , so the solution set is {12.15} .
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Chapter 6: Exponential and Logarithmic Functions 83. ln x x Using INTERSECT to solve: y1 ln x; y2 x
79. e x x x
Using INTERSECT to solve: y1 e ; y2 x 2
2 3
–3
–2
4
–2
–2
Thus, x 0.57 , so the solution set is {0.57} .
Thus, x 0.57 , so the solution set is {0.57} .
2x
80. e x 2 Using INTERSECT to solve: y1 e 2 x ; y2 x 2 3
84. ln(2 x) x 2 Using INTERSECT to solve: y1 ln(2 x); y2 x 2
3
–3
2
–3
4
0
2
–1
–1
4
–4
Thus, x 1.98 or x 0.45 , so the solution set is {1.98, 0.45} .
Thus, x 1.16 , so the solution set is {1.16} . 85. ln x x3 1 Using INTERSECT to solve: y1 ln x; y2 x3 1
81. e x x 2 Using INTERSECT to solve: y1 e x ; y2 x 2
2
3
2
–2 –3
4
–2
4
3
–2
–1
Thus, x 0.70 , so the solution set is {0.70} . 82. e x x3 Using INTERSECT to solve: y1 e x ; y2 x3 12
–2
Thus, x 0.39 or x 1 , so the solution set is {0.39, 1} . 86. ln x x 2 Using INTERSECT to solve: y1 ln x; y2 x 2
120
2 –2
–3
2
3 –4
-3
-4
6
–6
Thus, x 1.86 or x 4.54 , so the solution set is {1.86, 4.54} .
Thus, x 0.65 , so the solution set is {0.65} .
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Section 6.6: Logarithmic and Exponential Equations
87. e x ln x 4 Using INTERSECT to solve: y1 e x ln x; y2 4
log 2 x 3 3 x 3 23 x38 x5 The solution set is {5}. The point 5, 3 is
5
–2
f x 3
91. a.
on the graph of f.
4
g x 4
b.
–2
log 2 3x 1 4
Thus, x 1.32 , so the solution set is {1.32} .
3 x 1 24 3 x 1 16 3 x 15 x5 The solution set is {5}. The point 5, 4 is
x
88. e ln x 4 Using INTERSECT to solve: y1 e x ln x; y2 4 6
6
on the graph of g. –1
2
–1
–1
2
log 2 x 3 log 2 3x 1
–1
x 3 3x 1 2 2x 1 x The solution set is {1}, so the graphs intersect when x 1 . That is, at the point 1, 2 .
Thus, x 0.05 or x 1.48 , so the solution set is {0.05, 1.48} . 89. e x ln x Using INTERSECT to solve: y1 e x ; y2 ln x 2
–2
d. 4
–2
f x g x
c.
f g x 7 log 2 x 3 log 2 3 x 1 7 log 2 x 3 3x 1 7 x 3 3x 1 27 3 x 2 10 x 3 128
Thus, x 1.31 , so the solution set is {1.31} . 90. e x ln x Using INTERSECT to solve: y1 e x ; y2 ln x
3x 25 0 or x 5 0 3 x 25 x5 25 x 3
2
–1
3x 2 10 x 125 0 3x 25 x 5 0
The solution set is 5 .
2
–2
Thus, x 0.57 , so the solution set is {0.57}
699 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
e.
f g x 2 log 2 x 3 log 2 3x 1 2 log 2
d.
x3 2 3x 1 x3 22 3x 1 x 3 4 3x 1
x 2 4 x 5 27 x 2 4 x 32 0 x 8 x 4 0
x 3 12 x 4 1 11x 1 x 11 1 The solution set is . 11
92. a.
f g x 3 log3 x 5 log3 x 1 3 log 3 x 5 x 1 3 x 5 x 1 33
x 8 0 or x 4 0 x 8 x4
The solution set is 4 . e.
f x 2
f g x 2 log3 x 5 log3 x 1 2
log3 x 5 2
log3
x 5 32 x5 9 x4 The solution set is {4}. The point 4, 2 is
x 5 9x 9 14 8 x 7 x 4 7 The solution set is . 4
on the graph of f. b.
g x 3
log3 x 1 3 x 1 33 x 1 27 x 28 The solution set is {28}. The point 28,3
93. a. y
is on the graph of g. c.
x5 2 x 1 x5 32 x 1 x 5 9 x 1
f x 3 x 1 g x 2 x 2
8
f x g x
(0.710, 6.541)
log3 x 5 log 3 x 1 x 5 x 1 5 1 False This is a contradiction, so the equation has no solution. The graphs do not intersect.
4
2
700
Copyright © 2025 Pearson Education, Inc.
2
x
Section 6.6: Logarithmic and Exponential Equations
b.
f x g x 3
x 1
2
c.
x | x 2.513 or 2.513, .
ln 2
ln 3
x 1
Based on the graph, f x g x for x 2.513 . The solution set is
x2 x2
x 1 ln 3 x 2 ln 2 x ln 3 ln 3 x ln 2 2 ln 2 x ln 3 x ln 2 2 ln 2 ln 3 x ln 3 ln 2 2 ln 2 ln 3
95. a., b.
2 ln 2 ln 3 0.710 ln 3 ln 2 2 ln 2 ln 3 f 6.541 ln 3 ln 2 The intersection point is roughly 0.710, 6.541 . x
c.
Based on the graph, f x g x for x 0.710 . The solution set is x | x 0.710 or 0.710, .
f x g x
c.
3x 10 x log3 10
94. a.
The intersection point is log 3 10,10 . 96. a., b.
b.
f x g x 5 x 1 2 x 1
ln 5 x 1 ln 2 x 1
x 1 ln 5 x 1 ln 2
c.
x ln 5 ln 5 x ln 2 ln 2 x ln 5 x ln 2 ln 5 ln 2 x ln 5 ln 2 ln 5 ln 2
f x g x 2 x 12 x log 2 12
ln 5 ln 2 x 2.513 ln 5 ln 2 ln 5 ln 2 f 11.416 ln 5 ln 2 The intersection point is roughly 2.513,11.416 .
The intersection point is log 2 12,12 .
701 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions 97. a., b.
99. a. y
f x 2x 4
Using the graph of y 2 x , shift the graph down 4 units.
f x 2 x 1
4 1 ,2 2 2
2
g x 2 x 2 2
2
x
f x g x
c.
b.
2 x 1 2 x 2 x 1 x 2 2x 1 1 x 2 1 f 21/ 2 1 23/ 2 2 2 2 1 The intersection point is , 2 2 . 2
f x 0 2x 4 0 2x 4 2 x 22 x2 The zero of f is x 2 .
c.
Based on the graph, f x 0 when x 2 . The solution set is x | x 2 or , 2 .
98. a., b.
100. a.
g x 3x 9
Using the graph of y 3x , shift the graph down nine units.
c.
f x g x
b.
3 x 1 3x 2 x 1 x 2 2 x 3 3 x 2
g x 0 3x 9 0 3x 9 3x 32 x2 The zero of g is x 2 .
1 3 3 f 33/ 2 1 31/ 2 3 3 2 3 3 The intersection point is , . 2 3
c.
Based on the graph, g x 0 when x 2 . The solution set is x | x 2 or 2, .
702
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Section 6.6: Logarithmic and Exponential Equations
101. a.
329 1.007
t 2019
According to the model, the population of the
415
world will reach 9 billion people at the beginning of the year 2033.
415 1.007 329 t 2019 415 ln 1.007 ln 329 415 (t 2019) ln 1.007 ln 329 ln 415 / 329 t 2019 ln 1.007 t 2019
t
ln 415 / 329
b.
329 1.007
t 2019
12.5 7.71 t 2019 12.5 ln 1.011 ln 7.71 (t 2019) ln 1.011 ln 12.5 / 7.71
t 2019
2019
t
470 329 t 2019 470 ln 1.007 ln 329 470 (t 2019) ln 1.007 ln 329 ln 470 / 329 t 2019 ln 1.007
7.711.011
103. a.
2019
12, 000 t log 0.82 log 19, 200 log 12, 000 / 19, 200 t log 0.82 2.4 According to the model, the car will be worth
$12,000 after about 2.4 years. 19, 200 0.82 9, 000 t
9, 000 19, 200 9, 000 t log 0.82 log 19, 200
9, 000 t log 0.82 log 19, 200 log 9, 000 / 19, 200 t log 0.82
ln 9 / 7.71 ln 1.011
ln 1.011
t
0.82 t
9 1.011 7.71 t 2019 9 ln 1.011 ln 7.71 (t 2019) ln 1.011 ln 9 / 7.71
ln 9 / 7.71
19, 200 0.82 12, 000 12, 000 19, 200 12, 000 t log 0.82 log 19, 200
b.
t 2019
t
ln 1.011
0.82 t
ln 470 / 329
9
t 2019
ln 12.5 / 7.71
2063
2019 ln 1.007 2070 According to the model, the population of the U.S. will reach 470 million people in the beginning of the year 2070.
102. a.
ln 1.011
world will reach 12.5 billion people at the beginning of the year 2063.
1.007 t 2019
t 2019
ln 12.5 / 7.71
According to the model, the population of the
470
t
12.5
1.011t 2019
ln 1.007 2052 According to the model, the population of the U.S. will reach 415 million people around the beginning of the year 2052.
b.
t 2019
7.711.011
2019 2033
3.8 According to the model, the car will be worth
$9,000 after about 3.8 years.
703 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
c.
19, 200 0.82 3, 000 t
c.
7,500 19, 705 7,500 t log 0.848 log 19, 705
3, 000 t log 0.82 log 19, 200 log 3, 000 / 19, 200 t log 0.82
7,500 t log 0.848 log 19, 705 log 7,500 / 19, 705 t 5.9 log 0.848
0.848t
9.4
According to the model, the car will be worth
According to the model, the car will be worth
$7,500 after about 5.9 years.
$3,000 after about 9.4 years. 105. a.
19, 705 0.848 14, 000 t
14, 000 19, 705 14, 000 t log 0.848 log 18, 705
0.848t
14, 000 t log 0.848 log 18, 705 log 14, 000 / 18, 705 t 2.1 log 0.848
y (50) 0.09 ln(50) 1.889 2.24
b.
y (50) 0.048ln(50) 2.337 2.15
c.
2.5 0.09 ln x 1.889 0.611 0.09 ln x 0.611 ln x 0.09 0.611
x e 0.09 $887.93
d.
According to the model, the car will be worth
$14,000 after about 2.1 years. b.
t
3, 000 19, 200 3, 000 t log 0.82 log 19, 200
0.82 t
104. a.
19, 705 0.848 7,500
19, 705 0.848 10, 000 t
2.5 0.048ln x 2.337 0.163 0.048ln x 0.611 ln x 0.048 0.611
10, 000 19, 705 10, 000 t log 0.848 log 19,507
0.848t
x e 0.048 $0.03
e.
10, 000 t log 0.848 log 19, 705 log 10, 000 / 19, 705 t 4.1 log 0.848
0.09 ln x 1.889 0.048ln x 2.337 0.138ln x 0.448 0.448 ln x 0.138 0.448 x e 0.138 $25.70
According to the model, the car will be worth
The price where the ratings are equal is $25.70.
$10,000 after about 4.1 years.
y 0.09 ln(25.70) 1.889 2.18
The rating of the wine at this price is 2.18. 106. The domain of the variable is x > 0.
704
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Section 6.6: Logarithmic and Exponential Equations log 2 ( x 1) log 4 x 1
for both x
log 2 x 1 log 2 ( x 1) log 2 4 log 2 x 1 log 2 ( x 1) 2 2 log 2 ( x 1) log 2 x 2
1 , 4 . 4
109. The domain of the variable is x > 0. ln x 2 (ln x) 2
log 2 ( x 1) 2 log 2 x 2
2 ln x (ln x ) 2 0 ln x(2 ln x) 0 ln x 0 or 2 ln x 0
( x 1) 2 log 2 2 x ( x 1) 2 22 x x2 2 x 1 4 x ( x 1) 2 0 x 1 0 x 1 Since each of the original logarithms is defined for x 1 , the solution set is 1 . 2 x
2
2 2x 2 x 2 2 2x 3
e0 x
ln x 2
1 x
e2 x
The solution set is 1, e 2 .
x2 2x 1 0
107.
1 and x 4 , the solution set is 4
110. The domain of the variable is x > 0. log x 2 log 3 log x log 3 log x (log 3) 2 x 10
log 3
2
10 3log 3 Since we squared both sides of the equation, we must check. Since log 3 log 3
1/ 3
1 (2 x) 2 23 2x
log 3
log 3
log 3 log 3 log 3 log
1 (2 x) x 2 3 2 x 3x 2
3 2 log 3 2
, the solution set is 3log 3 . 111. The domain of the variable is x 0 . Note that x log 4 x 4log 4 ( x Then,
3x 2 x 2 0 (3x 2)( x 1) 0 2 x or x 1 3 2 The solution set is 1, . 3
log 4 x
)
2
4(log4 x )(log4 x ) 4(log4 x ) .
2
4(log 4 x ) x log 4 x 128 0 2
2
4(log4 x ) 4(log 4 x ) 128 0 2
2 4(log 4 x ) 128 0 2
2 4(log 4 x ) 128
108. The domain of the variable is x > 0. log x log 2 x 2 4 log 2 x log 2 x 4
2
4(log 4 x ) 64 2
4(log 4 x ) 43
log 2 x 2 4
(log 4 x) 2 3
log 2 x 2 or log 2 x 2 x 22 or x 22 1 x or x 4 4 Since each of the original logarithms is defined
log 4 x 3
x 4 3
The solution set is 4 3 , 4 3
705 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
the function would be the most restrictive of these so the domain of f ( x ) is x | x 1 .
112. Solution A: change to exponential expression; square root method; meaning of ; solve.
Solution B: log a M r r log a M ; divide by 2; change to exponential expression; solve.
117.
x 5 x 7
The power rule log a M r r log a M only applies when M 0 . In this equation, M x 1 . Now, x 2 causes M 2 1 3 . Thus, if we use the power rule, we lose the valid solution x 2 . 113.
3
x x7 5
( x 5) 2 x 7 x 2 10 x 25 x 7 x 2 11x 18 0 ( x 9)( x 2) 0 x 9, x 2 Check: 9 4 5 correct 2 2 0 false The solution set is 9
2
f ( x) 4 x 3x 25 x 6 p 1, 2, 3, 6
q 1, 2, 4 The possible rational zeros are: p 1 1 3 3 1, 2, , , 3, , , 6 q 2 4 2 4 Using synthetic division: We try 2 : 2 4 3 25 6 8 22 6 4 11 3 0
118.
y
2.16
k x2 k
52 k 2.16 25 54 y
54 6 9 x 5 x2 x2 x( x 2) 5( x 2) ( x 2)( x 2)
119. ( f g )( x)
f ( x) ( x 2)(4 x 2 11x 3) 0
( x 2)(4 x 1)( x 3) 0 1 So the solution set is: 3, 2, . 4 114. Since the x elements are not repeated and the y elements are not repeated the ordered pairs represent a function that is one-to-one. x5 x5 3 x x3 115. ( f g )( x) x5 x 5 2( x 3) 2 x3 x3 x3 x5 x5 3 x x3 x 5 2 x 6 x 11 x3 x3 x5 x 5 x 3 x 3 x 11 x 11 The domain would be any x that works in g(x) or ( f g )( x) so the domain is: x | x 3, x 11
116.
domain of
x 2 2 x 5 x 10 ( x 2)( x 2)
x 2 7 x 10 ( x 2)( x 2)
120. Center: (2, 3) vertex: (6,9) d (6 2) 2 (9 (3)) 2 (4) 2 (12) 2 16 144 160 4 10
121.
f ( x) x 3 x 1
The domain of
f (b) f (a ) log 2 16 log 2 4 ba 16 4 42 2 1 16 4 12 6
x 3 is x | x 3 and the
x 1 is x | x 1 . The domain of
706
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Section 6.7: Financial Models 9. P $900, r 0.03, n 2, t 2.5
x6 x x6 x 6 x6 x
122.
r A P 1 n
x 6 x x 6 x 6 x 6 x x6 x
6 6
x6 x
6 x6 x
r A P 1 n
x6 x
0.03 900 1 2
(2)(2.5)
$969.56
10. P $300, r 0.12, n 12, t 1.5
1
nt
nt
0.12 300 1 12
(12)(1.5)
$358.84
11. P $1200, r 0.05, n 365, t 3
r A P 1 n
nt
0.05 1200 1 365
(365)(3)
$1394.13
12. P $700, r 0.06, n 365, t 2 r A P 1 n
Section 6.7
nt
0.06 700 1 365
1. P $500, r 0.06, t 6 months 0.5 year I Prt (500)(0.06)(0.5) $15.00
13. P $1000, r 0.11, t 2
2. P $5000, t 9 months 0.75 year, I $500 500 5000r (0.75) 500 r (5000)(0.75) 2 2 40 1 % 13 % 100% 15 15 3 3 1 The per annum interest rate was 13 % . 3
14. P $400, r 0.07, t 3
3. principal
5. effective rate of interest
0.04 100 1 4
(4)(2)
$108.29
r A P 1 n
0.06 50 1 12
r P A 1 n
nt
0.06 100 1 12
( 12)(2)
$88.72
16. A $75, r 0.08, n 4, t 3 nt
0.08 75 1 4
( 4)(3)
$59.14
nt
0.015 1500 1 365
( 365)(2.5)
$1444.79
18. A $800, r 0.07, n 12, t 3.5
8. P $50, r 0.06, n 12, t 3 nt
15. A $100, r 0.06, n 12, t 2
r P A 1 n
7. P $100, r 0.04, n 4, t 2 r A P 1 n
A Per t 400e(0.07)(3) $493.47
17. A $1700, r 0.015, n 365, t 2.5
6. a
nt
$789.24
A Pe r t 1000e(0.11)(2) $1246.08
r P A 1 n
4. I; Prt; simple interest
(365)(2)
(12)(3)
$59.83
r P A 1 n
nt
0.07 800 1 12
707 Copyright © 2025 Pearson Education, Inc.
( 12)(3.5)
$626.61
Chapter 6: Exponential and Logarithmic Functions I Prt 0.04081 P r 1 .04081 r The effective interest rate is 4.081%.
19. A $750, r 0.025, n 4, t 2 r P A 1 n
nt
0.025 750 1 4
Thus, ( 4)(2)
$713.53
20. A $300, r 0.03, n 365, t 4 r P A 1 n
26. Suppose P dollars are invested for 1 year at 6%.
nt
0.03 300 1 365
Compounded continuously yields: A Pe(0.06)(1) 1.06184 P The interest earned is I 1.06184 P P 0.06184 P I Prt Thus, 0.06184 P P r 1 0.06184 r The effective interest rate is 6.184%.
( 365)(4)
$266.08
21. A $120, r 0.05, t 3.25 P Ae r t 120e( 0.05)(3.25) $102.00
22. A $800, r 0.08, t 2.5
27. 6% compounded quarterly:
P Ae r t 800e( 0.08)(2.5) $654.98
(4)(1)
0.06 A 10, 000 1 4
23. Suppose P dollars are invested for 1 year at 5%.
$10, 613.64
6 14 % compounded annually:
Compounded quarterly yields:
A 10, 000 1 0.0625 $10, 625 1
4 1
0.05 A P 1 1.05095 P . 4 The interest earned is I 1.05095 P P 0.05095 P I Prt Thus, 0.05095 P P r 1 0.05095 r The effective interest rate is 5.095%.
6 14 % compounded annually is the better deal.
28. 9% compounded quarterly: 0.09 A 10, 000 1 4
(4)(1)
$10,930.83
9 14 % compounded annually:
A 10, 000 1 0.0925 $10,925 1
24. Suppose P dollars are invested for 1 year at 6%.
9% compounded quarterly is the better deal.
Compounded monthly yields:
29. 9% compounded monthly:
12 1
0.06 A P 1 1.06168 P . 12 The interest earned is I 1.06168 P P 0.06168 P I Prt Thus, 0.06168 P P r 1 0.06168 r The effective interest rate is 6.168%.
(12)(1)
0.09 A 10, 000 1 12 8.8% compounded daily:
$10,938.07
365
0.088 A 10, 000 1 $10,919.77 365 9% compounded monthly is the better deal.
30. 8% compounded semiannually:
25. Suppose P dollars are invested for 1 year at 4%.
(2)(1)
0.08 A 10, 000 1 $10,816 2 7.9% compounded daily:
Compounded continuously yields: A Pe(0.04)(1) 1.04081P The interest earned is I 1.04081P P 0.04081P
365
0.079 A 10, 000 1 $10,821.95 365 7.9% compounded daily is the better deal.
708
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Section 6.7: Financial Models
31. 2 P P 1 1r
3(1)
b.
2 e0.08t ln 2 0.08t ln 2 8.66 t 0.08 It will take about 8.66 years to double.
2 P P 1 r
3
2 (1 r )3 3
2 1 r
r 3 2 1 0.25992 The required rate is 25.992%.
32. 2 P P 1 1r
2 P P 1 r
12t
36. a.
6(1)
12t
ln 3 ln 1.005
12t
ln 3 12t ln 1.005
2 1 r
r 6 2 1 0.12246 The required rate is 12.246%.
33. 3P P 1 1r
3P P 1 r 3 (1 r ) 5
t
b.
5
3 1 r
37. Since the effective interest rate is 7%, we have: I Prt I P 0.07 1 I 0.07 P Thus, the amount in the account is A P 0.07 P 1.07 P
10(1)
3P P 1 r
10
3 (1 r )10 10
3 1 r
Let x be the required interest rate. Then,
r 10 3 1 0.11612 The required rate is 11.612%.
r 1.07 P P 1 4
12t
35. a.
3P Pe0.06t 3 e0.06t ln 3 0.06t ln 3 18.31 t 0.06 It will take about 18.31 years to triple.
5
3P P 1 1r
ln 3 18.36 12 ln 1.005
It will take about 18.36 years to triple.
5(1)
r 5 3 1 0.24573 The required rate is 24.573%.
34.
0.06 3P P 1 12 3 1.005
6
2 (1 r )6 6
2 P Pe0.08t
0.08 2 P P 1 12
4
12t
0.08 2 1 12
12t
4 1
0.08 ln 2 ln 1 12 0.08 ln 2 12t ln 1 12 ln 2 t 8.69 0.08 12 ln 1 12 It will take about 8.69 years to double.
r 1.07 1 4 r 4 1.07 1 4 r 4 1.07 1 4 r 4 4 1.07 1 0.06823
Thus, an interest rate of 6.823% compounded quarterly has an effective interest rate of 7%. 38. Since the effective interest rate is 6%, we have: I Prt I P 0.06 1 I 0.06 P
709 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
Thus, the amount in the account is A P 0.06 P 1.06 P
41. 25, 000 10, 000e0.06 t 2.5 e0.06 t ln 2.5 0.06 t ln 2.5 t 15.27 0.06 It will take about 15.27 years (or 15 years, 3 months).
Let x be the required interest rate. Then, 1.06 P Pe( r )(1) 1.06 er r ln(1.06) 0.05827 Thus, an interest rate of 5.827% compounded continuously has an effective interest rate of 6%.
39.
0.04 150 100 1 12
42. 80, 000 25, 000e0.07 t
12 t
3.2 e0.07 t ln 3.2 0.07 t ln 3.2 t 16.62 0.07 It will take about 16.62 years (or 16 years, 7 months).
1.5 1.003333 ln1.5 12t ln 1.003333 ln1.5 t 10.15 12 ln 1.003333 12 t
Compounded monthly, it will take about 10.15 years (or 121.85 months).
43. A 90, 000(1 0.03)5 $104,335 The house will cost $104,335 in five years.
150 100e0.04 t
44. A 200 1 0.0125 $215.48 6
1.5 e0.04 t ln1.5 0.04 t ln1.5 t 10.14 0.04 Compounded continuously, it will take about 10.14 years (or 121.64 months).
40.
0.025 175 100 1 12
Her bill will be $215.48 after 6 months. 45. P 15, 000e( 0.05)(3) $12,910.62 Jerome should ask for $12,910.62. ( 12)(0.5)
0.03 $2955.39 46. P 3, 000 1 12 John should save $2955.39.
12 t
1.75 1.002083
12 t
47. A 15(1 0.15)5 15(1.15)5 $30.17 per share for a total of about $3017.
ln1.75 12t ln 1.002083 ln1.75 t 22.41 12 ln 1.002083
48. 850, 000 650, 000 1 r 85 3 1 r 65 85 3 1 r 65 r 3 1.3077 1 0.0935 The annual return is approximately 9.35%. 3
Compounded monthly, it will take about 22.41 years (or 268.94 months). 175 100e0.025 t 1.75 e0.025 t ln1.75 0.025 t ln1.75 t 22.38 0.025 Compounded continuously, it will take about 22.38 years (or 268.62 months).
49. 5.6% compounded continuously: A 1000e(0.056)(1) $1057.60 Jim will not have enough money to buy the computer.
710
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Section 6.7: Financial Models
5.9% compounded monthly:
b. 5.5% compounded monthly:
12
0.055 A 100, 000 1 12
0.059 A 1000 1 $1060.62 12 The second bank offers the better deal.
50. 6.8% compounded continuously for 3 months: Amount on April 1: A 1000e(0.068)(0.25) $1017.15
5.25% compounded monthly for 1 month: Amount on May 1 0.0525 A 1017.15 1 12
(12)(1/12)
$1021.60
51. Will: 9% compounded semiannually: 0.09 A 2000 1 2
(2)(20)
$11, 632.73
(12)(5)
$131,570
I $131,570 $100, 000 $31, 570
5.25% compounded continuously: A 100, 000e(0.0525)(5) $130, 018 I $130, 018 $100, 000 $30, 018 Thus, simple interest at 6% is the best option since it results in the least interest. c.
55. Graph the following two functions and find the intersection. 0.05 A 1000 1 12
(12) t
; $1000 invested at 5%
A 2000 1 0.04 ; $2000 invested at 4% t
Henry: 8.5% compounded continuously: A 2000e(0.085)(20) $10,947.89 Will has more money after 20 years. 52. Value of $1000 compounded continuously at 10% for 3 years: A 1000e(0.10)(3) $1349.86
April will have more money if she takes the $1000 now and invests it. 53. a.
Let x = the year, then the average annual cost C of a 4-year private college is by the function C ( x) 38, 070(1.021) x 2021 . C (2041) 38070(1.021) 2041 2021 38070(1.033) 20 57, 690 In 2041, the average annual cost at a 4-year private college will be about $57,690.
b.
A Pe
rt
57690 Pe0.02(18) 57690 P 0.02(18) $40, 249 e An investment of $40,249 in 2023 would pay for the cost of college at a 4-year private college in 2041.
54. P 100, 000; t 5 a. Simple interest at 6% per annum: A 100, 000 100, 000(0.06)(5) $130, 000 I $130, 000 $100, 000 $30, 000
The two account balances will be approximately equal after 64.9 years. 56. From 2021 to 2030 would be 9 years, so t = 9. The federal debt (in millions) would be:
F 30000 1 0.055 30000 1.055 . For t = t
t
9: F 30000 1.055 48572.8282 million. 9
The U.S. population (in millions) would be: P 332 1 0.007 332 1.007 . For t = 9: t
t
P 332 1.007 353.5113 . 9
The per capita debt in 2030 will be 485728282 $137, 401 . 353.5113 57. P 1000, r 0.03, n 2 A
1000 $942.60 (1 0.03)2
58. P 1000, r 0.08, n 3 A
1000 $793.83 (1 0.08)3
711 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions 59. P 1000, A 950, n 2
64. A $80, 000, r 0.06, n 1, t 17
950 10002
0.06 P 80, 000 1 1
(1 r )
1.05263 (1 r ) 2 1.05263 1 r
0.045 P 10, 000 1 1
25, 000 8 1 r 15,334.65
2
1.07527 1 r
8
r 1 1.07527 r 0.0370 or r 2.037 Disregard r 1.9644 . The inflation rate was 3.7%.
25, 000 1 r 15,334.65
25, 000 1 15,334.65 r 0.063 The annual rate of return is about 6.3%. r8
61. Let A=1 and P=2. 2 1 n (1 0.02)
2 (1.02) n
t
b.
t
c.
r mP P 1 n
ln 2 35.003 n ln(1.02)
The purchasing power will be half in 35.003 years. 62. Let A=1 and P=2. 2 1 n (1 0.04)
2 (1.04) n ln 2 ln(1.04) n n ln(1.04) ln 2 ln 2 17.7 n ln(1.04)
ln 2 0.06 1 ln 1 1 ln 2 11.90 years ln 1.06
67. a.
ln 2 ln(1.02) n n ln(1.02) ln 2
ln 3 0.05 4 ln 1 4 ln 3 22.11 years 4 ln 1.0125
nt
nt
r m 1 n r ln m n t ln 1 n ln m t r n ln 1 n
The purchasing power will be half in 17.7 years.
b.
$6439.28
25, 000 15,334.65 1 r 8
930 10002 (1 r )
A $10, 000, r 0.05, n 12, t 20 0.05 P 10, 000 1 12
( 1)(10)
66. A $25, 000, P 15,334.65, n 1, t 8
60. P 1000, A 930, n 2
63. a.
$29, 709.15
65. A $10, 000, r 0.045, n 1, t 10
r 1 1.05263 r 0.0259783 or r 2.026 Disregard r 2.026 . The inflation rate was 2.6%.
1.07527 (1 r )
17
( 12)(20)
$3686.45
A $10, 000, r 0.05, t 20 P 10, 000e( 0.05)(20) $3678.79
712
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Section 6.7: Financial Models
t
b.
35
c.
The CPI will reach 300 about 17 years after 2013, or in the year 2030.
ln 4500 ln1000 26.16 years 0.0575
68. a.
ln 30, 000 ln 2000 r ln 30, 000 ln 2000 r 35 0.0774 r 7.74%
70. CPI 0 234.2, r 2.8%, n 5
A Per t A er t P A ln r t P ln A ln P r t ln A ln P t r
3.1 2 CPI 0 CPI 0 1 100
5
2.8 CPI 234.2 1 268.9 100 In 5 years, the CPI index will be about 268.9.
71. r 3.1%
2 1.031
ln 2 22.7 ln1.031 It will take about 22.7 years for the CPI index to double.
72. CPI 0 100, CPI 456.5, r 5.57
r 251.11 232.96 1 100
1
b.
5.57 456.5 100 1 100 456.5 100 1.0557
5
4.565 1.0557
5
251.11 r 5 100 232.96 251.11 r 5 1 100 232.96 251.11 r 100 5 1 1.51% 232.96
CPI 0 229.39, CPI 300, r 1.23 1.51 300 232.96 1 100 300 1.51 1 232.96 100
n
n log1.031 2
69. a. CPI 0 232.96, CPI 251.11, n 2018 2013 5
251.11 r 1 232.96 100
n
73-75. 76.
n
Answers will vary.
f ( x) 6 x3 3 x 2 2 x 11; c 1 f (1) 6(1)3 3(1) 2 2(1) 11 6 3 2 11 0 Thus, 1 is a zero of f and x 1 is a factor of f .
n
77. n
n
n log1.0558 4.565 ln 4.565 28.0 years ln1.0558 The yeas that was used as the base period for the CPI was about 28 years before 1995, or the year 1967.
n
300 1.51 ln ln 1 232.96 100 300 1.51 ln n ln 1 232.96 100 300 ln 232.96 n 17 years 1.51 ln 1 100
n
x x2 y x y2 x( y 2) y xy 2 x y xy y 2 x f ( x)
713 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions y ( x 1) 2 x 2x y x 1 2x f 1 ( x) x 1 78. x5 x 4 15 x3 21x 2 16 x 20 0 Step 1: f (x) has at most 5 real zeros.
80. 2 x 4 6 x3 50 x 2 150 x 2 x( x3 6 x 2 25 x 75) 2 x[ x 2 ( x 3) 25( x 3)] 2 x( x 3)( x 2 25) 2 x( x 3)( x 5)( x 5)
81.
f ( g ( x)) 5(3x 1) 2 4(3 x 1) 8
Step 2: Possible rational zeros: p 1, 2, 4, 5, 10, 20; q 1; p 1, 2, 4, 5, 10, 20 q
5(9 x 2 6 x 1) 12 x 4 8 45 x 2 30 x 5 12 x 4 8 45 x 2 18 x 7
Step 3: Using synthetic division: We try x 2 : 2 1 1 15 21 16 20 2 6 18 6 20
82. Domain ,
Vertex:
1 3 9 3 10 0 x 2 is a factor and the quotient is x 4 3 x3 9 x 2 3 x 10 . We try x 2 again on x 4 3 x3 9 x 2 3 x 10 2 1 3 9 3 10
2 10 2
0 1
83.
10
2 x2 5x 4 x7 The degree of the numerator, p( x) 2 x 2 5 x 4, is n 2 . The degree of the f ( x)
denominator, q( x) x 7, is m 1 . Since n m 1 , there is an oblique asymptote. Dividing: 2x 9 x 7 2 x2 5x 4
2 x 2 14 x
0
x 5 is a factor and the quotient is x 1 . So the real zeros of f ( x) are 2,5.
9x 4 9 x 63
2
59 59 , x7 G ( x) 2 x 9 x7 Thus, the oblique asymptote is y 2 x 9 . The denominator is zero at x 7 , so x 7 is a vertical asymptotes. There are no horizontal asymptotes.
Thus, f ( x) x 2 x 5 x 2 1 . 2
8 2 2(2)
f (2) 9 The graph is concave down so this is a maximum so the range is ,9
1 5 1 5 0 x 2 is again a factor and the quotient is x3 5 x 2 x 5 . We try x 5 again on x3 5 x 2 x 5 5 1 5 1 5 5 0 5 1
x
79. log 2 ( x 3) 2 log 2 ( x 3) log 2 ( x 3) log 2 ( x 3) 2 ( x 3) ( x 3) 2 x 3 x2 6 x 9 x2 7 x 6 0 ( x 6)( x 1) 0
x 6 or x 1 But x = 1 will not work since we cannot take the log of a negative number so the solution set is 6 . 714
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Section 6.8: Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models d. Find t when N 1700 : 1700 1000e0.01t
84. The points on the line are: (3, 6) and (5, 2) .
1.7 e0.01t ln1.7 0.01t ln1.7 53.1 hours t 0.01
2 (6) 8 4 m 53 2 y 2 4( x 5) y 2 4 x 20
e.
y 4 x 18
85.
Find t when N 2000 : 2000 1000e0.01t 2 e0.01t ln 2 0.01t ln 2 t 69.3 hours 0.01
f ( x h) f ( x) 3( x h) 5 (3x 5) h h 3x 3h 5 3 x 5 3h 3 h h
3. A(t ) A0 e0.0244 t 500e0.0244 t a. decay rate: k = 0.0244 2.44%
Section 6.8 1. P (t ) 500e0.02 t a.
P (0) 500e
0.02 0
b.
A(10) 500e
c.
Find t when A 400 :
P (10) 500e
0.02 10
0.8 e 0.0244 t ln 0.8 0.0244 t ln 0.8 t 9.1 years 0.0244
611 insects
d. Find t when P 800 :
d. Find t when A 250 :
800 500e0.02 t
250 500e 0.0244 t
1.6 e0.02 t ln1.6 0.02 t ln1.6 23.5 days t 0.02
e.
0.5 e 0.0244 t ln 0.5 0.0244 t ln 0.5 t 28.4 years 0.0244
Find t when P 1000 : 1000 500e0.02 t
4. A(t ) A0 e0.087 t 100e0.087 t
2 e0.02 t ln 2 0.02 t ln 2 34.7 days t 0.02
N (0) 1000e
0.01 0
1000 bacteria
b. growth rate: k = 0.01 = 1 % c.
N (4) 1000e
a.
decay rate: k 0.087 8.7%
b.
A(9) 100e
c.
Find t when A 70 :
0.087 9
45.7 grams
70 100e 0.087 t
2. N (t ) 1000e0.01t a.
391.7 grams
400 500e 0.0244 t
500 insects
b. growth rate: k = 0.02 = 2 % c.
0.0244 10
0.01 4
1041 bacteria
0.7 e 0.087 t ln 0.7 0.087 t ln 0.7 t 4.1 days 0.087
715 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions d. Find t when A 50 :
If N 0 10, 000 and t 2 , then
50 100e 0.087 t
ln 2 2
P (2) 10, 000e 1.5 25,198 The population 2 years from now will be 25,198.
0.087 t
0.5 e ln 0.5 0.087 t ln 0.5 t 7.97 days 0.087
5. a.
8. a.
b. If N (t ) 800, 000, N 0 900, 000, and t 2022 2020 2 , then 800, 000 900, 000e k (2) 8 e2k 9 8 ln 2k 9 ln 8 / 9 k 2 If t 2024 2020 4 , then
N (t ) N 0 e k t
b. If N (t ) 1800, N 0 1000, and t 1 , then 1800 1000e k (1) 1.8 e k k ln1.8
If t 3 , then N (3) 1000e mosquitoes. c.
ln1.8 3
5832
Find t when N (t ) 10, 000 : 10, 000 1000e
ln1.8 t
ln 8 / 9 4 2 P (4) 900, 000e 711,111
10 e ln10 ln1.8 t ln1.8 t
t
6. a.
The population in 2024 will be 711,111.
ln10 3.9 days ln1.8
kt
9. Use A A0 e and solve for k : 0.5 A0 A0 ek (1690)
N (t ) N 0 e k t
0.5 e1690 k ln 0.5 1690k ln 0.5 k 1690 A When 0 10 and t 50 :
b. If N (t ) 800, N 0 500, and t 1 , then 800 500e k (1) 1.6 e k k ln1.6
If t 5 , then N (5) 500e bacteria c.
ln1.6 5
5243
ln 0.5 50
A 10e 1690
Find t when N (t ) 20, 000 : 20, 000 500e
0.5 A0 A0 e
40 e ln 40 ln1.6 t
7. a.
9.797 grams
10. Use A A0 ek t and solve for k :
ln1.6 t
ln1.6 t
t
N (t ) N 0 e k t , k 0
0.5 e
k 1.3109
1.3109 k
ln 0.5 1.3 109 k ln 0.5 k 1.3 109
ln 40 7.85 hours ln1.6
N (t ) N 0 e k t
When A0 10 and t 100 :
b. Note that 18 months = 1.5 years, so t = 1.5. 2 N 0 N 0 e k (1.5)
ln 0.5 100 9
A 10e 1.310
1.5 k
2e ln 2 1.5k ln 2 k 1.5
9.999999467 grams
When A0 10 and t 1000 : ln 0.5 1000 9 A 10e 1.310 9.999994668 grams
716
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Section 6.8: Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
11. Use A A0 e kt and solve for k : half-life = 5730 years
13. a.
Using u T (u0 T )e k t with t 5 , T 70 , u0 450 , and u 300 :
0.5 A0 A0 ek (5730)
300 70 (450 70)e k (5)
0.5 e5730 k ln 0.5 5730k ln 0.5 k 5730
230 380e5k 230 e5 k 380 23 ln 5k 38 1 23 k ln 0.1004 5 38
Solve for t when A 0.3 A0 : ln 0.5 t
0.3 A0 A0 e 5730 ln 0.5 t
T 70, u0 450, u 135 :
0.3 e 5730 ln 0.5 ln 0.3 t 5730 5730 t ln 0.3 9953 ln 0.5 The tree died approximately 9953 years ago.
135 70 (450 70)e ln 23/ 38
65 380e 65 e 380
12. Use A A0 ekt and solve for k : half-life = 5730 years 0.5 A0 A0 ek (5730)
5
5
t
t
t
65 ln 23 / 38 ln t 5 380 5 65 t ln 18 minutes ln 23 / 38 380
0.5 A0 A0 ek (5730) 0.5 e5730 k ln 0.5 5730k ln 0.5 k 5730
The temperature of the pan will be 135˚F at about 5:18 p.m. b.
T 70, u0 450, u 160 : ln 23/ 38 t 5 160 70 (450 70)e
Solve for t when A 0.7 A0 : 0.7 A0 A0 e
ln 23/ 38
5
ln 23/ 38
ln 23/ 38 t 5 90 380e
ln 0.5 t 5730
ln 23/ 38 t 5
ln 0.5 t
90 e 380
0.7 e 5730 ln 0.5 ln 0.7 t 5730 5730 t ln 0.7 2949 ln 0.5 The fossil is about 2949 years old.
90 ln 23 / 38 ln t 5 380 5 90 t ln 14.3 minutes ln 23 / 38 380
c.
The pan will be 160˚F after about 14.3 minutes. As time passes, the temperature of the pan approaches 70˚F.
717 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
14. a.
Using u T (u0 T )e kt with t 2 , T 38, u0 72 , and u 60 : 60 38 (72 38)ek (2)
15. Using u T (u0 T )e kt with t 3 , T 35 , u0 8 , and u 15 :
22 34e 2 k 22 e2k 34 22 ln 2k 34 ln 22 / 34 k 2
15 35 (8 35)e k (3) 20 27e3k 20 e3k 27 20 ln 3k 27 ln 20 / 27 k 3
T 38, u0 72, t 7
ln 22 / 34 7 2
u 38 (72 38)e
ln 22 / 34 7 2 u 38 34e 45.41º F
After 7 minutes the thermometer will read about 45.41˚F. b. Find t when u 39º F ln 22 / 34 t 2 39 38 (72 38)e ln 22 / 34 t 2 1 34e ln 22 / 34 t 2 e
1 34 1 ln 22 / 34 ln t 2 34 2 1 ln 16.2 t ln 22 / 34 34
The thermometer will read 39 degrees after about 16.2 minutes. c.
T 38, u0 72, u 45 : ln 22 / 34
45 38 (72 38)e ln 22 / 34
7 34e 7 e 34
2
ln 22 / 34 2
d. As time passes, the temperature gets closer to 38˚F.
2
t
At t 5 : ln 20 / 27 5 3 u 35 (8 35)e 18.63C
After 5 minutes, the thermometer will read approximately 18.63C . At t 10 : ln 20 / 27 10 3 u 35 (8 35)e 25.1C
After 10 minutes, the thermometer will read approximately 25.1C 16. Using u T (u0 T )e kt with t 10 , T 70 , u0 28 , and u 35 :
35 70 (28 70)e k (10) 35 42e10 k 35 e10 k 42 35 ln 10k 42 ln 35 / 42 k 10
At t 30 :
t
ln 35 / 42 30 10 u 70 (28 70)e 45.69F
t
7 ln 22 / 34 ln t 2 34 2 7 t ln 7.26 minutes ln 22 / 34 34
After 30 minutes, the temperature of the stein will be approximately 45.69 F .
The thermometer will read 45 F after about 7.26 minutes. 718 Copyright © 2025 Pearson Education, Inc.
Section 6.8: Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models When A 0 0.25 and t 30 :
Find the value of t so that the u = 45˚F: ln 35 / 42 t 10 45 70 (28 70)e
ln 0.6 30 A 0.25e 17 0.10
ln 35/ 42 t 10 25 42e
After 30 minutes, approximately 0.10 M of dinitrogen pentoxide will remain.
ln 35 / 42
Find t when A 0.01 :
t 25 10 e 42 25 ln 35 / 42 ln t 10 42
t
ln 0.6 t
0.01 0.25 e 17 ln 0.6 t 0.04 e 17
10 25 ln 28.46 ln 35 / 42 42
ln 0.6 ln 0.04 t 17 17 ln 0.04 107 t ln 0.6 It will take approximately 107 minutes until 0.01 M of dinitrogen pentoxide remains.
The temperature of the stein will be 45 F after about 28.46 minutes. 17. Use A A0 e kt and solve for k: 2.2 2.5ek (24)
19. Use A A0 e kt and solve for k:
0.88 e24 k ln 0.88 24k ln 0.88 k 24
0.36 0.40e k (30) 0.9 e30 k ln 0.9 30k ln 0.9 k 30
When A 0 2.5 and t 72 : ln 0.88 72 A 2.5e 24 1.70
Note that 2 hours = 120 minutes. When A 0 0.40 and t 120 :
After 3 days (72 hours), the amount of free chlorine will be 1.70 parts per million.
ln 0.9 120 A 0.40e 30 0.26
Find t when A 1 :
After 2 hours, approximately 0.26 M of sucrose will remain.
ln 0.88 t 1 2.5 e 24
Find t when A 0.10 :
ln 0.88 t 0.4 e 24
ln 0.9 t
0.10 0.40e 30
ln 0.88 ln 0.4 t 24 24 ln 0.4 172 t ln 0.88 Ben will have to shock his pool again after 172 hours (or 7.17 days) when the level of free chlorine reaches 1.0 parts per million.
ln 0.9 t 0.25 e 30
ln 0.9 ln 0.25 t 30 30 t ln 0.25 395 ln 0.9 It will take approximately 395 minutes (or 6.58 hours) until 0.10 M of sucrose remains.
18. Use A A0 e kt and solve for k: 0.15 0.25e k (17) 0.6 e17 k ln 0.6 17k ln 0.6 k 17
719 Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
20. Use A A0 ekt and solve for k : 15 25e
k (10)
298 1 3.439e 0.3785 x
b.
Y1
c.
We need to find t such that P = 280 298 295 1 3.439e 0.3785t 298 1 3.439e0.3785 t 295 298 1 3.439e0.3785 t 295
0.6 e10 k ln 0.6 10k ln 0.6 k 10
When A0 25 and t 24 : ln 0.6 24 A 25e 10 7.34
There will be about 7.34 kilograms of salt left after 1 day. Find t when A 0.5 A0 : ln 0.6 t 0.5 25 e 10 ln 0.6 t
0.02 e 10 ln 0.6 ln 0.02 t 10 10 t ln 0.02 76.6 ln 0.6 It will take about 76.6 hours (about 3.19 days) until ½ kilogram of salt is left.
298 1 295 298 3.439e0.3785 t 1 295 3.439e 0.3785t 0.01017 3.439e0.3785 t
e 0.3785t 0.01017 3.439
0.3785 t ln 0.01017 3.439
21. Use A A0 ekt and solve for k : 0.5 A0 A0 e
k (8)
t
8k
0.5 e ln 0.5 8k ln 0.5 k 8
0.3785
15.4
Thus, t 15 . Now, 2010 15 2025 . The number of U.S. smartphone users should reach 295 million in early 2025.
Find t when A 0.1A0 :
23. a.
ln 0.5 t 0.1A0 A0 e 8 ln 0.5 t 8
0.1 e
ln 0.5 ln 0.1 t 8 8 t ln 0.1 26.6 ln 0.5 The farmers need to wait about 26.6 days before using the hay.
22. a.
ln 0.01017 3.439
50.9249 3.2 1 14.9863e 1.0404 (0) In 2010, about 3.2% of households owned a tablet computer. P (0)
50.9249 1 14.9863e1.0404 t
b.
Y1
c.
t 2018 2010 8 50.9249 P (11) 50.7 1 14.9863e 1.0404 (8)
Growth rate = 0.3785.
720 Copyright © 2025 Pearson Education, Inc.
Section 6.8: Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
in the United States would be approaching 0.
In 2018, about 50.7% of households owned a tablet computer. d. We need to find t such that P = 47 50.9249 Y1 ; Y2 47 1 14.9863e 1.0404 x
25. a.
Y1
b.
Thus, t 5 . Now, 2010 5 2015 . The percentage of households that owned a tabled computer reached 47% during 2015. 24. a.
b.
c.
14, 656, 248 W (0) 13,839, 705 1 0.059e0.057 (0) In 1910, there were about 13,839,705 farm workers. Y1
113.3198 1 0.115e0.0912 x
113.3198 78 1 0.115e0.0912 (15) In a room of 15 people, the probability that no two people share the same birthday is about 78% or 0.78. We need to find n such that P = 10. 113.3198 Y1 ; Y2 10 1 0.115e0.0912 x P (15)
14, 656.248 1 0.059e0.057 x
Thus, t 49.3 . The probability falls below 10% when 50 people are in the room. d. As n , 1 0.115e0.0912 n . Thus, P (n) 0 . This means that as the number of people in the room increases, the more likely it will be that two will share the same birthday.
t 2010 1910 100 14, 656, 248 W (100) 786,567 1 0.059e0.057(100) In 2010, there were about 787.56 workers. d. We need to find t such that W = 10,000,000. 14, 656, 248 Y1 ; Y2 10, 000, 000 1 0.059e0.057 x
c.
26. a.
As t , e 0.162 t 0. Thus, P (t ) 500 . The carrying capacity is 500 bald eagles.
b. Growth rate = 0.162 = 16.2%. c.
500 9.68 1 82.33e 0.162 (3) After 3 years, the population is almost 10 bald eagles. P (3)
Thus, t 36.2 . Now, 1910 36.2 1946.2 . There were 10,000 workers in 1946. e.
As t , 1 0.059e0.057 t . Thus, W (t ) 0 . No, it is not reasonable to use this model to predict the number of workers in 2060 because the number of workers left
721
Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions d. We need to find t such that P = 300: 500 300 1 82.33e 0.162t 300 1 82.33e0.162 t 500 5 3 5 0.162 t 82.33e 1 3 2 e 0.162 t 3 82.33 2 0.162 t ln 3 82.33 t 29.7
1 82.33e0.162t
d.
e.
431
y
176 1 7.91e 0.017(100) The number of invasive species present in 2000 was approximately 176. We need to find t such that P = 175. 431 Y1 ; Y2 175 1 7.91e 0.071 x
Thus, t 29.7 . The bald eagle population will be 300 in approximately 29.7 years. e.
We need to find t such that 1 P 500 250 : 2
Thus, t 99 . Now, 1900 99 1999 . There were 175 invasive species in 1999.
500 1 82.33e 0.162t 250 1 82.33e 0.162 t 500 250
1 82.33e 0.162t 2 82.33e 0.162 t 2 1 1 82.33 1 0.162 t ln 82.33 t 27.2 e 0.162 t
28. a.
P(0)
86.1
27.6 1 2.12e0.361(0) In 2008, 27.6% of people in the U. S. had a social media profile.
b. The growth rate of the percentage of people in the U.S. who have a social media profile is 36.1% . c.
Y1
d.
y
86.1 1 2.12e 0.361t
Thus, t 27.2 . The bald eagle population will reach one-half of its carrying capacity after about 27.2 years. 27. a.
P (0)
431
48 1 7.91e 0.017(0) In 1900 the number of invasive species present in the Great Lakes was approximately 48.
b. The growth rate of invasive species is 1.7% . c.
Y1
431 1 7.91e 0.017 x
e.
86.1
79.6 1 2.12e 0.361(9) The percentage of people in the U.S. had a social media profile in 2017 was approximately 79.6. We need to find t such that P = 69.3. 86.1 Y1 ; Y2 69.3 1 2.12e0.361t
722 Copyright © 2025 Pearson Education, Inc.
Section 6.8: Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models c. t
80, 000 25, 000(2) 8 t
80, 000 (2) 8 25, 000 t
ln 3.2 ln(2) 8
Thus, t 6 . Now, 2008 6 2014 . 29. a.
t ln 3.2 ln(2) 8
n = 20; P0 50
ln 3.2
P (t ) 50(3)t / 20
b.
ln 2 47 20
t8
t 47 , then P (47) 50(3) 661 The population 47 days from now will be 661. t
25, 000 eln 2
25, 000e0.087t P t 25, 000e0.087t
t 20
ln 14 ln 3
31. m 48
y 50 3
3 y 1 ( x 4) 2 3 y 1 x 6 2 3 y f ( x) x 7 2
t /20
50 eln 3
t / 20
50eln 3(t / 20) 50e0.055t
32. The slope of f ( x) is 5 . The slope of g ( x) is
P t 50e0.055t
30. a.
1 . 5
They are not equal so they are not parallel functions. They are not opposite reciprocals so they are not perpendicular.
n = 8; P0 25, 000 P (t ) 25, 000(2)t /8
b.
5 1 6 3 84 4 2
( y y1 ) m( x x1 )
The population will reach 700 in 48 days. d.
t /8
25, 000eln 2(t /8)
t 20
t ln 14 ln(3) 20
t 20
13.42
t /8
t
700 (3) 20 50
ln 3
ln 2
y 25, 000 2
d.
700 50(3) 20
ln 14
ln 3.2
The population will reach 80,000 in 13.42 years.
c.
ln 14 ln(3)
t 8
3
t 3 , then P (3) 25, 000(2) 8 32, 421 The population 3 years from now will be 32,421.
723
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Chapter 6: Exponential and Logarithmic Functions
The y-intercept is 1. The x-intercepts are 5 17 5 17 , 4 4
2 12 x2 y x y 33. ln ln z z 1 ln x 2 y 2 ln z
37.
1 ln x 2 ln y 2 ln z
x2 2 x 1 x2 2 x2 2 x
1 2 ln x ln y ln z 2
2 x 1 2 x2 2 x 2 x2 1
34. The denominator cannot be zero. x 2 2 x 8 ( x 4)( x 2)
x2
x 4, 2
1 2
x
The domain is x | x 4, x 2
1 2 2 2
2 2 The solution set is , . 2 2
35. 3x 1 2 x 3 x 3 x 4 (3 x 1)( x 4) (2 x 3)( x 3) ( x 3)( x 4) ( x 3)( x 4) (3 x 1)( x 4) (2 x 3)( x 3) ( x 3)( x 4)
( g f )( x)
36.
x 1 x 2 x x 1 ( x 1)( x 1) x 2 2 x( x 1)
38. Using Linear Regression on the data gives: y 1.0714 x 3.9048 . The coefficient is r 0.985 . 39.
(3 x 2 11x 4) (2 x 2 9 x 9) ( x 3)( x 4) 3 x 11x 4 2 x 9 x 9 2
2
( x 3)( x 4)
f (0) 2(0) 2 5(0) 1 1 f ( x) 0 2 x 2 5 x 1 x
(5) (5) 2 4(2)(1) 2(2)
5 25 8 5 17 4 4
x 2 x 13 2
( x 3)( x 4)
Local minima: f (1.37) 5.85 and f (1) 1 Local maximum: f (0.37) 0.65 Increasing: [1.37, 0.37] [1,3] Decreasing: [3, 1.37] [0.37,1]
724 Copyright © 2025 Pearson Education, Inc.
Section 6.9: Building Exponential, Logarithmic, and Logistic Models from Data 40.
f. 10 x
5 x(2 x 3)
13
2
3(2 x 3) 3
11.1287e
2 2 3(2 x 3) 3 3(2 x 3) 3 10 x 15 x(2 x 3) 2
3(2 x 3) 3 10 x 30 x 45 2
3(2 x 3) 3 40 x 45 2
3(2 x 3) 3
0.2464 t
1000000 1000000 0.2464 t e 11.1287 1000000 0.2464t ln 11.1287 1000000 ln 11.1287 46 days t 0.2464 46 days after February 20 would be April 6.
15 x(2 x 3)
10 x
We need to find t when N 1000000 :
2. a.
Section 6.9 1. a. b. Using EXPonential REGression on the data
yields: y 0.1948 1.7354 c.
y 0.1948 1.7354
x
0.1948 e
b. Using EXPonential REGression on the data
yields: y 11.1287 1.2794 c.
y 11.1287 1.2794
11.1287e
ln 1.2794
ln 1.7354
x
0.1948e
ln 1.7354 x
x
A t 0.1948e
x
11.1287 e
x
x
0.5512 t
d.
Y1 0.1948e
0.5512 t
e.
A 9 0.1948e
ln 1.2794 x
N t 11.1287e0.2464t
d.
Y1 11.1287e0.2464t 0.5512 (12)
145.27 billion
dollars f.
e.
k 0.5512 55.12% is the exponential growth rate. It represents the rate at which Tesla’s revenue is increasing.
N 7 11.1287e0.2464(45) 727615 bacteria
725
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Chapter 6: Exponential and Logarithmic Functions
Therefore, the predicted percent of U.S. citizens on social networking sites in 2022 is 87.6%.
3. a.
e.
90 9.3274 28.9155ln x 80.6726 28.9155ln x 80.6726 ln x 28.9155 80.6726
x e 28.9155 14.8 For 2008 + 14.8 = 2023, the function predicts that the 90% of U.S. citizens will be on social networking sites.
b. Using lnREGression on the data yields: y 371.7593 42.2140 ln x c.
Y1 371.7593 42.2140 ln x
5. a.
d. Note that 2008 is represented by t 28 . y 371.7593 42.2140 ln(67) 194 billion pounds. e.
It is over by 4 billion pounds.
4. a.
b. Using LOGISTIC REGression on the data 656.6308 yields: y 1 7.4868e 0.0171x c.
b. Let x 1 correspond to 2008, x 2 correspond to 2009, x 3 correspond to 2010, etc. Using LNREGression on the data yields: y 9.3274 28.9155ln x c.
Y1 9.3274 28.9155ln x
y 9.3274 28.9155ln(15) 87.6%
Y1
656.6308 1 7.4868e 0.0171x
d. As x , 7.4868e 0.0171x 0 , which means 1 7.4868e 0.0171x 1 , so 656.6308 y 656.6308 1 7.4868e0.0171x Therefore, the carrying capacity of the United States is approximately 656.6 million people. e.
d.
Let x 0 correspond to 1900, x 10 correspond to 1910, x 20 correspond to 1920, etc.
The year 2022 corresponds to x = 122, so 656.6308 y 1 7.4868e 0.0171(122) 340 million people
726 Copyright © 2025 Pearson Education, Inc.
Section 6.9: Building Exponential, Logarithmic, and Logistic Models from Data
f.
Find x when y 350 656.6308 350 1 7.4868e 0.0171x
656.6308 350 1 7.4868e
e. 0.0171x
656.6308 1 7.4868e 0.0171x 350 656.6308 1 7.4868e 0.0171x 350 0.8761 7.4868e 0.0171x
f.
We need to find x when y 9 : 13.0401
9 1 1.09943e 0.0256 x 13.0401 9 1 1.09943e 0.0256 x
0.8761 e 0.0171x 7.4868 0.8761 ln 0.0171x 7.4868 0.8761 ln 7.4868 x 0.0171 x 125.46 Therefore, the United States population will be 350,000,000 in the year 2026.
6. a.
Therefore, the carrying capacity of the world is approximately 13.0401 billion people. The year 2025 corresponds to x = 24 so 13.0401 8.18 . y 1 1.09943e 0.0256(24) In 2025, the population of the world was approximately 8.18 billion people.
13.0401 9 9.89487e
0.0256 x
13.0401 9 9.89487e 0.0256 x 4.0401 9.89487e 0.0256 x 4.0401 e 0.0256 x 9.89487 4.0401 ln 0.0256 x 9.89487 4.0401 ln 9.89487 35.0 x 0.0256 Therefore, the world population will be 9 billion in approximately the year 2036.
Let x = 0 correspond to 2001, x = 1 correspond to 2002, etc.
7. a.
b. Using LOGISTIC REGression on the data 13.0401 yields: y 1 1.09943e 0.0256 x c.
b. Based on the shape of the graph, a cubic model might best describe the data.
13.0401 Y1 : 1 1.09943e0.0256 x
c.
Using CubicReg, the cubic model is y 0.0607 x3 0.5530 x 2 4.1380 x 13.1566
d.
d. As x , 1.09943e 0.0256 x 0 , which means 1 1.09943e 0.0256 x 1 , so 13.0401 y 13.0401 1 1.09943e 0.0256 x 727
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Chapter 6: Exponential and Logarithmic Functions
e.
d.
y (11) 0.0607(16)3 0.5530(16) 2 4.1380(16) 13.1566 186.4 The model predicts an online advertising revenue of $186.4 billion in 2021.
8. a. e.
y 115.5779 0.9012
30
5.1 The model predicts the expected percentage of a 30-foot putt to be made is 5.1%.
b. Based on the “upside down U-shape” of the graph, a quadratic model with a 0 would best describe the data. c.
Using QUADratic REGression, the quadratic model is y 0.0311x 2 3.4444 x 118.2493 .
10. The graph crosses the x-axis at x 3 and x 2 and touches it at x 1 . Thus, 3 and 2 each have odd multiplicities while 1 has an even multiplicity. Using one for each odd multiplicity and two for the even multiplicity, a possible funtion is f ( x) a ( x 3)( x 1) 2 ( x 2) . The yintercept appears to be negative so we can use (0,-2) as a typical y-intercept. So, a(0 3)(0 1) 2 (0 2) 2
d.
a (3)(1)( 2) 2 6a 2 1 3 The a posible function is 1 f ( x) ( x 3)( x 1) 2 ( x 2) . 3 a
e.
y 0.0311 35 3.4444 35 118.2493 2
201 The model predicts a total cholesterol of 201 for a 35-year-old male.
11.
a2 b2 c2 (1) 2 b 2 (2) 2 1 b2 4
9. a.
b2 3 b 3
12. The equation represents a circle with center (3, 0) and radius 5. b. Based on the graph, an exponential model would best describe the data. c.
Using EXPonential REGression, the model is y 115.5779 0.9012 . x
115.5449e 0.1040 x
728 Copyright © 2025 Pearson Education, Inc.
Chapter 6 Review Exercises
x x y y2 7 1 5 (9) 13. 1 2 , 1 , 2 2 2 2 6 4 , 2 2 3, 2
14.
18.
f ( x) x 4
15. x
(4) (4) 2 4(3)(5) 2(3)
19.
4 16 60 4 76 6 6 4 2 19 2 19 6 3
2 19 2 19 The solution set is , . 3 3
16.
2 1 2 1 f f 3 3 8 3 8 3 2 1 1 3 3 3 42 2 1 1 3 3 6 f (2) (2)4 2(2)3 5(2) 1 16 16 10 1 21 f (1) (1) 4 2(1)3 5(1) 1 1 2 5 1 3 Since f (2) is negative and f (1) is positive and f ( x) is continuous on (2, 1) , then there exists a number c in (2, 1) such that f (c) 0 .
x 1
0 x 25 x 1 0 ( x 5)( x 5) 2
f ( x)
From the graph we see that the zero is approximately 1.32 .
x 1
x 2 25 The zeros and values where f is undefined are x 1, x 5 and x 5 . ( , 5)
( 5, 1)
( 1, 5)
(5, )
6
3
0
6
Value of f
0.45
0.125
0.04
0.64
Conclusion
Negative
Positive
Negative
Positive
Interval Number Chosen
Chapter 6 Review Exercises 1.
f ( x) 3x 5 g ( x) 1 2 x 2 a. ( f g )(2) f ( g (2))
f 1 2(2) 2
The solution set is x 5 x 1, x 5 or,
f (7) 3(7) 5 26
using interval notation, 5, 1 5, . 17. b.
f (3) 3(3)5 7(3) 4 27(3)3 67(3) 2 36 729 567 729 603 36 0 f (3) 0 . Therefore, x 3 is a factor of 3x5 7 x 4 27 x3 67 x 2 36 .
( g f )( 2) g ( f ( 2)) g 3( 2) 5 g (11) 1 2(11) 2 241
729
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Chapter 6: Exponential and Logarithmic Functions
c.
d.
( f f )(4) f ( f (4)) f 3(4) 5 f (7) 3(7) 5 16
b.
3e2 2 3 2 2 e
( g g )(1) g ( g (1))
c.
( f f )(4) f ( f (4))
f e4
g (1) 1 2(1) 2 1
ee
d.
f ( x) x 2 g ( x) 2 x 2 1 a. ( f g )(2) f ( g (2))
g e2
g 1 2(1) 2
2.
( g f )( 2) g ( f ( 2))
( g g )(1) g ( g (1)) g 3(1) 2 g (5) 3(5) 2 17
f 2(2) 2 1 f (9) 92
4.
11
b.
2 2
g (0) 2(0) 2 1 1
c.
( f f )(4) f ( f (4))
4 2 f 6 f
d.
62
( g g )(1) g ( g (1))
g 2(1) 2 1 g (3) 2(3) 2 1 19
3.
f ( x) e x
a.
g ( x) 3x 2 ( f g )(2) f ( g (2)) f 3(2) 2
f (4)
f ( x) 2 x
g ( x) 3 x 1
The domain of f is x x is any real number .
( g f )( 2) g ( f ( 2)) g
4
The domain of g is x x is any real number . ( f g )( x) f ( g ( x)) f (3 x 1) 2 (3x 1) 2 3x 1 1 3x Domain: x x is any real number . ( g f )( x) g ( f ( x)) g (2 x) 3(2 x) 1 6 3x 1 7 3x Domain: x x is any real number . ( f f )( x) f ( f ( x)) f (2 x) 2 (2 x) 22 x x Domain: x x is any real number .
e4
730 Copyright © 2025 Pearson Education, Inc.
Chapter 6 Review Exercises
f g x f g x
( g g )( x) g ( g ( x)) g (3x 1) 3(3x 1) 1 9x 3 1 9x 4 Domain: x x is any real number .
5.
1 1 1 x f x 1 1 x 1 1 x 1 x x 1 x 1 1 x x
g ( x) 1 x x 2
f ( x) 3x
The domain of f is x x 0 .
Domain x x 0, x 1 .
The domain of g is x x is any real number . ( f g )( x) f ( g ( x))
g f x g f x 1 x 1 x 1 g x 1 x 1 x 1 x 1
f 1 x x2
3 1 x x2
Domain x x 1, x 1
3 3x 3x2
f f x f f x
Domain: x x is any real number .
x 1
( g f )( x) g ( f ( x)) g
3x
1 3x
3x
x 1
2
x 1 1 ( x 1) x 1 x 1 1 ( x 1) x 1
1 3x 3x
Domain: x x 0 .
( f f )( x) f ( f ( x)) f
3x 3 3x
Domain: x x 0 .
x 1 x 1 x 1 x 1
2x 2
x
Domain x x 1 .
( g g )( x) g ( g ( x))
g 1 x x2
g g x g g x g 1
1 1 x x2 1 x x2
x
2
Domain x x 0 .
1 1 x x 2 1 2 x 3x 2 2 x3 x 4 2
3
1 x 1 x
4
3 3x 4 x 2 x x Domain: x x is any real number .
6.
1
x 1 x 1 f x 1 x 1 1
7. a.
x 1 1 g ( x) x 1 x The domain of f is x x 1 .
The function is one-to-one because there are no two distinct inputs that correspond to the same output.
b. The inverse is 2,1 , 5,3 , 8,5 , 10, 6 .
f ( x)
The domain of g is x x 0 .
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Chapter 6: Exponential and Logarithmic Functions 8. The function f is one-to-one because every horizontal line intersects the graph with at most one point.
Domain of f = Range of f 1 = All real numbers except
2 . 5
Range of f = Domain of f 1 = All real numbers except 12.
9.
1 f ( g ( x)) 5 x 2 10 x 10 10 x 5 1 g ( f ( x)) (5 x 10) 2 x 2 2 x 5
4 4 1 x 4 4(1 x) 1 x 10. f ( g ( x)) 4 1 x 4 1 x 4 4 4x 4x x 4 4 x 4 4x g ( f ( x)) x 4 x x ( x 4) 1 x 4x 4x x xx4 4
1 x 1 1 y x 1 1 x Inverse y 1 x( y 1) 1 xy x 1 xy x 1 x 1 y x x 1 1 f ( x) x Domain of f = Range of f 1 = All real numbers except 1 Range of f = Domain of f 1 = All real numbers except 0 f ( x)
f ( x) x 2
13.
y x2 x
Thus f and g are inverses of each other. 11.
2x 3 5x 2 2x 3 y 5x 2 2y 3 x 5y 2 x(5 y 2) 2 y 3 5 xy 2 x 2 y 3 5 xy 2 y 2 x 3 y (5 x 2) 2 x 3 2x 3 y 5x 2 2x 3 1 f ( x) 5x 2
2 . 5
y2
2
x y2
f ( x)
f
1
Inverse x0
2
y x 2
x0
2
x0
( x) x 2
Domain of f = Range of f 1 x | x 2 or 2, Range of f = Domain of f 1 x | x 0 or 0,
Inverse
14.
f ( x ) x1/ 3 1 y x1/ 3 1 x y1/ 3 1 1/ 3
y
Inverse
x 1
y ( x 1)3 f 1 ( x ) ( x 1)3
Domain of f = Range of f 1 = All real numbers or ,
732 Copyright © 2025 Pearson Education, Inc.
Chapter 6 Review Exercises
Range of f = Domain of f 1
uv 2 2 23. log3 log 3 uv log 3 w w log 3 u log 3 v 2 log 3 w log 3 u 2 log 3 v log 3 w
= All real numbers or , 15. a. b.
f 4 34 81
g 9 log3 9 log 3 32 2
24. log 2 a 2 b
1 9
c.
f 2 32
d.
1 1 g log3 log3 33 3 27 27
2
1/ 2
1/ 2
1 2 log x log x3 1 2
f x log(3x 2) requires: 3x 2 0 2 x 3 2 2 Domain: x x or , 3 3
2
26.
2 ln(2 x 3) 2 ln( x 1) 2 ln( x 2)
p( x) x 2 3 x 2 0 x 2 x 1 0
(2, )
log 4 x 6 x1/ 4
3 Test Value 0 3 2 1 Value of p 2 2 4 Conclusion positive negative positive
Thus, the domain of H x log 2 x 2 3 x 2
3 1/ 2 1 27. 3log 4 x 2 log 4 x log 4 x 2 log 4 x1/ 2 2 log 4 x 6 log 4 x1/ 4
x 2 and x 1 are the zeros of p . (1, 2)
2x 3 ln 2 x 3x 2 2x 3 2 ln 2 x 3x 2 2 ln(2 x 3) ln ( x 1)( x 2)
2 ln(2 x 3) ln( x 1) ln( x 2)
19. H x log 2 x 2 3 x 2 requires
(,1)
2
25. log x 2 x3 1 log x 2 log x3 1
17. log5 u 13 is equivalent to 513 u
Interval
2
2
1 4 2 log 2 a log 2 b 2 8log 2 a 2 log 2 b
4
2
16. 52 z is equivalent to 2 log5 z
18.
4 log a b 4 log a log b
25 / 4
log 4 x 25 log 4 x 4
is x x 1 or x 2 or ,1 2, . 1 20. log 2 log 2 23 3log 2 2 3 8
21. ln e 2 2 22. 2log 20.4 0.4
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Chapter 6: Exponential and Logarithmic Functions x 1 x 2 28. ln ln ln x 1 x x 1 x 1 x 2 ln ln x 1 x x 1 x 1 ln x2 1 x 1 x 1 1 ln x 1 ( x 1)( x 1) 1 ln ( x 1) 2
b. Using the graph of y 2 x , shift the graph horizontally 3 units to the right.
ln( x 1)2 2 ln( x 1)
c.
d. 29.
1 1 1 ln x 2 1 4 ln ln( x 4) ln x 2 2 2 1/ 2 1 ln ln x( x 4) 2 1/ 2 x2 1 ln 1 x( x 4) 1/ 2 16 16 x 2 1 ln x( x 4)
ln x 2 1
30. log 4 19
f ( x ) 2 x 3 y 2 x 3 x 2 y 3 Inverse y 3 log 2 x y 3 log 2 x
4
1/ 2
Range: (0, ) Horizontal Asymptote: y 0
f 1 ( x) 3 log 2 x
e.
Range of f = Domain f 1 : (0, ) Domain of f = Range of f 1 : (, )
f.
Using the graph of y log 2 x , shift the graph vertically 3 units up.
ln19 2.124 ln 4
31. Y1 log3 x
ln x ln 3
32.
f ( x) 2 x 3
a.
Domain: (, )
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Chapter 6 Review Exercises
33.
f ( x) 1 3 x
a.
34.
a.
Domain: (, )
y
x
c.
c.
Range: (1, ) Horizontal Asymptote: y 1
d.
Inverse
y
x 1 3 y log3 x 1 f
e.
( x) log 3 x 1
1
x 1 0 x 1
e.
Range of f = Domain f 1 : (0, ) Domain of f = Range of f 1 : (, )
Range of f = Domain f 1 : (1, ) f.
Domain of f = Range of f 1 : (, ) f.
Using the graph of y log3 x , shift the graph horizontally to the right 1 unit, and reflect vertically about the x-axis. y x
Using the graph of y ln x , stretch horizontally by a factor of 3, and shift vertically up 2 units.
x
Inverse
y 2 ln 3x f ( x ) 2 ln 3x
y log 3 x 1
1
f ( x ) 3e x 2 x 3e y 2 x e y 2 3 y 2 ln 3x
y 1 3 x x 1 3 y
Range: (0, ) Horizontal Asymptote: y 0 y 3e x 2
f ( x) 1 3 x
d.
Domain: (, )
b. Using the graph of y e x , shift the graph two units horizontally to the right, and stretch vertically by a factor of 3.
b. Using the graph of y 3x , reflect the graph about the y-axis, and shift vertically 1 unit up. y
f ( x) 3e x 2
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Chapter 6: Exponential and Logarithmic Functions
35.
f ( x)
a.
down 3 units.
1 ln x 3 2
Domain: (3, )
b. Using the graph of y ln x , shift the graph to the left 3 units and compress vertically by a factor of 1 . 2
86 3 x 4
36.
2
3 63 x
c.
d.
218 9 x 22 18 9 x 2 9 x 16 16 x 9
Range: (, ) Vertical Asymptote: x 3 1 ln x 3 2 1 y ln x 3 2 1 x ln y 3 2 2 x ln y 3
The solution set is 16 .
f ( x)
9
2
3x x 3
37.
2
3x x 31/ 2 1 x2 x 2 2 x2 2 x 1 0
Inverse
y 3 e2 x y e2 x 3 f 1 ( x) e 2 x 3
e.
22
x
2 22 4(2)(1) 2(2)
2 12 2 2 3 1 3 4 4 2
Using the graph of y e x , compress horizontally by a factor of 12 , and shift
2
Domain of f = Range of f 1 : (3, ) f.
The solution is 1 3 , 1 3 {1.366, 0.366} .
Range of f = Domain f 1 : (, ) 38.
log x 64 3 x 3 64
x
3 1/ 3
641/ 3
x 3
1 64
1 4
The solution set is 1 . 4
736 Copyright © 2025 Pearson Education, Inc.
2
Chapter 6 Review Exercises
39.
5 x 3x 2
ln 5 x ln 3x 2
42.
2
3 2 x2 5x 0 2 x2 5x 3 0 (2 x 1)( x 3) 1 or x 3 x 2 1 The solution set is 3, . 2
2
2 2x
x 2 12
4x
x 2 12
5
5
5x
23 22 x 5 x
252 x 5 x 12
5 5
x2
2
23 22
x ln 5 ( x 2) ln 3 x ln 5 x ln 3 2 ln 3 x ln 5 x ln 3 2 ln 3 x(ln 5 ln 3) 2 ln 3 2 ln 3 x 4.301 ln 5 ln 3 2 ln 3 The solution set is 4.301 . ln 5 ln 3
40.
2
8 4 x 25 x
43.
2 x 5 10 x
ln 2 x 5 ln10 x ln 2 x ln 5 ln10 x x ln 2 ln 5 x ln10 ln 5 x ln10 x ln 2 ln 5 x(ln10 ln 2) ln 5 x ln10 ln 2 ln 5 ln 5 1 x 10 ln 5 ln 2 The solution set is 1 .
4 x x 2 12 x 2 4 x 12 0 ( x 6)( x 2) 0 x 6 or x 2
The solution set is 2, 6 . 41. log3 x 2 2 x 2 32 x2 9 x 2 92 x 2 81 x 83
44. log 6 ( x 3) log 6 ( x 4) 1 log 6 ( x 3)( x 4) 1
( x 3)( x 4) 61
Check: log3 83 2 log3 81 log3 9 2 The solution set is 83 .
x 2 7 x 12 6 x2 7 x 6 0 ( x 6)( x 1) 0 x 6 or x 1
Since log 6 (6 3) log 6 (3) is undefined, the solution set is 1 .
45. e1 x 5 1 x ln 5 x 1 ln 5 x 1 ln 5 0.609 The solution set is 1 ln 5 0.609 .
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Chapter 6: Exponential and Logarithmic Functions
9 x 4 3x 3 0
46.
x
x 2
x
log 2 x 2 1 0
log 2 x 2 1
x 2 21 1 x2 2 5 x 2 Based on the graph drawn in part (a), 5 f x 0 when x . The solution set is 2 5 5 x | x or , . 2 2
x
Let u 3 . u 2 4u 3 0 a 1, b 4, c 3 u
(4) (4) 2 4(1)(3) 2(1) 4 28 4 2 7 2 7 2 2
or 3x 2 7
3x 2 7
The solution set is log 2 7 0.398 3x can't be negative
x log3 2 7
3
47. a.
f x 0
d.
3 4 3 3 0 3 4 3 3 0 2 x
e.
f x log 2 x 2 1 y log 2 x 2 1
.
x log 2 y 2 1
f x log 2 x 2 1
y 2 2 x 1
Using the graph of y log 2 x , shift the graph right 2 units and up 1 unit. y x
y 2 x 1 2 f 1 x 2 x 1 2
y x
f x log 2 x 2 1
x
f 1 ( x) 2 x 1 2 f x log 2 x 2 1
y x
b.
Inverse
x 1 log 2 y 2
yx
f 6 log 2 6 2 1
log 2 4 1 2 1 3
The point 6,3 is on the graph of f. c.
48. P 25e0.1d a. P 25e0.1(4) 25e0.4 37.3 watts
f x 4
log 2 x 2 1 4
b.
log 2 x 2 3
x 2 23 x2 8 x 10 The solution set is {10}. The point 10, 4
50 25e0.1d 2 e0.1d ln 2 0.1d ln 2 6.9 decibels d 0.1
is on the graph of f.
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Chapter 6 Review Exercises 49. L 9 5.1log d a. L 9 5.1log 3.5 11.77 b.
53.
0.5 A0 A0 ek (5730)
14 9 5.1log d 5 5.1log d 5 log d 0.9804 5.1 d 100.9804 9.56 inches
50. a.
n
b.
n
0.5 e5730 k ln 0.5 5730k ln 0.5 k 5730 ln 0.5 t
0.05 A0 A0 e 5730
log10, 000 log 90, 000 9.85 years log(1 0.20)
ln 0.5 t
0.05 e 5730 ln 0.5 ln 0.05 t 5730 ln 0.05 24, 765 t ln 0.5 5730
log 0.5i log i
log(1 0.15) 0.5i log i log 0.5 4.27 years log 0.85 log 0.85
0.04 51. In 18 years, A 10, 000 1 2 10, 000(1.02)36 $20,398.87
The man died approximately 24,765 years ago.
(2) (18)
54. Using u T (u0 T )e kt , with t 5 , T 70 , u0 450 , and u 400 . 400 70 (450 70)ek (5) 330 380e5 k 330 e5 k 380 330 ln 5k 380 ln 330 / 380 k 5
The effective interest rate is computed as follows: 0.04 When t = 1, A 10, 000 1 2 10, 000(1.02) 2 $10, 404
(2) (1)
10, 404 10, 000 404 0.0404 , so 10, 000 10, 000 the effective interest rate is 4.04%.
Find time for temperature of 150˚F:
Note,
ln 330 / 380 t 5 150 70 (450 70)e
In order for the bond to double in value, we have the equation: A 2 P .
ln 330 / 380 t 5 80 380e
2t
0.04 10, 000 1 20, 000 2
ln 330 / 380
t 80 5 e 380 80 ln 330 / 380 ln t 5 380 80 ln 380 55.22 t ln 330 / 380 5 The temperature of the skillet will be 150˚F after approximately 55.22 minutes (or 55 minutes, 13 seconds).
1.02 2t 2 2t ln1.02 ln 2 ln 2 17.5 years t 2 ln1.02 nt
A A0 ekt
r 0.04 52. P A 1 85, 000 1 2 n $41, 668.97
2(18)
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Chapter 6: Exponential and Logarithmic Functions 55. P0 7, 714,576,923 , k 0.011 , and t 2024 2019 5 P P0 ekt 7, 714,576,923e0.011 (5) 8,150, 763,844 people
d. Find t such that P (t ) 0.75 . 0.8 0.75 1 1.67e 0.16 t 0.8 0.75 1 1.67e 0.16 t
0.8 1 1.67e0.16 t 0.75 0.8 1 1.67e 0.16 t 0.75 0.8 1 0.75 e0.16 t 1.67 0.8 0.75 1 ln 0.16t 1.67 0.8 0.75 1 ln t 1.67 20.13 0.16 Note that 2006 20.13 2026.13 , so 75% of new cars will have GPS in 2026.
A A0 e kt
56.
0.5 A0 A0 ek (5.27) 0.5 e5.27 k ln 0.5 5.27k ln 0.5 k 5.27 ln 0.5 (20)
In 20 years: A 100e 5.27 In 40 years:
57. a.
7.204 grams
ln 0.5 (40) A 100e 5.27 0.519 grams
0.8
0.8 0.3 1 1.67 1 1.67e In 2006, about 30% of cars had a GPS. P (0)
0.16 (0)
b. The maximum proportion is the carrying capacity, c = 0.8 = 80%. c.
Y1
58. a.
0.8 1 1.67e0.16 x
1
b. Using EXPonential REGression on the data 0
0
yields: y 0.0903 1.3384
30
c.
y 0.0903 1.3384
x
x
0.0903 e
ln 1.3384
x
0.0903e
ln 1.3384 x
N t 0.0903e0.2915t
d.
Y1 0.0903e0.2915 x
e.
N 7 0.0903e
740 Copyright © 2025 Pearson Education, Inc.
0.2915 7
0.69 bacteria
Chapter 6 Review Exercises f.
We need to find t when N 0.75 : 0.0903e
c.
0.2915 t
0.75 0.75 0.2915 t e 0.0903 0.75 0.2915t ln 0.0903 0.75 ln 0.0903 t 7.26 hours 0.2915
In reality, all 50 people living in the town might catch the cold.
b. Using LnREGression on the data yields: y 18.903 7.096 ln x where y = wind chill and x = wind speed. c.
e.
Y1 18.903 7.096 ln x
Find t when C 10 . 46.9292 10 1 21.2733e 0.7306t 46.9292 10 1 21.2733e 0.7306t
46.9292 1 21.2733e 0.7306t 10 46.9292 1 21.2733e 0.7306t 10 3.69292 21.2733e 0.7306t 3.69292 e 0.7306t 21.2733 3.69292 ln 0.7306t 21.2733 3.69292 ln 21.2733 t 0.7306 t 2.4
d. If x = 23, then y 18.903 7.096 ln 23 3o F .
60. a.
d. As t , 21.2733e0.7306t 0 , which means 1 21.2733e0.7306t 1 , so 46.9292 C 46.9292 1 21.2733e 0.7306t Therefore, according to the function, a maximum of about 47 people can catch the cold.
46.93 1 21.273e 0.7306t
59. a.
Y1
The data appear to have a logistic relation. b. Using LOGISTIC REGression on the data yields: 46.93 C 1 21.273e 0.7306t
Therefore, after approximately 2.4 days (during the 10th hour on the 3rd day), 10 people had caught the cold.
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Chapter 6: Exponential and Logarithmic Functions
f.
Find t when C 46 . 46.9292 46 1 21.2733e 0.7306t 46.9292 46 1 21.2733e 0.7306t
46.9292 1 21.2733e 0.7306t 46 46.9292 1 21.2733e 0.7306t 46 0.0202 21.2733e 0.7306t 0.0202 e 0.7306t 21.2733 0.0202 e 0.7306t 21.2733 0.0202 ln 0.7306t 21.2733 0.0202 ln 21.2733 t 0.7306 t 9.5 Therefore, after approximately 9.5 days (during the 12th hour on the 10th day), 46 people had caught the cold.
b.
g f 2 g f 2 2 2 g 2 2 g 0
2(0) 5 5
c.
f g 2 f g 2 f 2(2) 5 f 1
2. a.
1 2 3 3 1 2 1
Graph y 4 x 2 3 : y
x
The function is not one-to-one because it fails the horizontal line test. A horizontal line (for example, y 4 ) intersects the graph twice. b. Graph y x 3 5 : y
Chapter 6 Test 1.
x2 g ( x) 2 x 5 x2 The domain of f is x x 2 .
x
f ( x)
The domain of g is all real numbers. a.
f g x f g x f 2 x 5
The function is one-to-one because it passes the horizontal line test. Every horizontal line intersects the graph at most once.
(2 x 5) 2 (2 x 5) 2 2x 7 2x 3 3 Domain x x . 2
742 Copyright © 2025 Pearson Education, Inc.
Chapter 6 Chapter Test
3.
9. log10000 x
2 3x 5 2 y 3x 5 2 x 3y 5 x(3 y 5) 2 3 xy 5 x 2 3xy 5 x 2 5x 2 y 3x 5x 2 1 f ( x) 3x f ( x)
10 x 10000 104 x4
log 2 8log 2 5 log 2 x log 2 5log 2 8 log 2 x 3log 2 5 log 2 x log 2 53 log 2 x x 53 125
11. ln e7 7 ln e 7
Domain of f = x | x 53 , range of f =
12.
y | y 0 ; domain of f 1 = x | x 0 , range of f
1
=
y | y 53
8log 2 5 x
10.
Inverse
f ( x) 4 x 1 2
a.
Domain: (, )
b. Using the graph of y 4 x , shift the graph 1 unit to the left, and shift 2 units down.
4. If the point (3, 5) is on the graph of f, then the
point (5, 3) must be on the graph of f 1 . 5. 3x 243 3x 35 x5 The solution set is 5
6. logb 16 2
c.
b 2 16 b 16 4 Since the base of a logarithm must be positive, the only viable solution is b 4 . The solution set is 4
d.
Range: (2, ) Horizontal Asymptote: y 2 f ( x ) 4 x 1 2 y 4 x 1 2 x 4 y 1 2
Inverse
y 1
x2 4 y 1 log 4 ( x 2) y log 4 ( x 2) 1
7. log5 x 4 x 54 x 625 The solution set is 625
f 1 ( x) log 4 ( x 2) 1
e. 8. log 6
1 x 36 1 1 6x 62 36 62 x 2
Range of f = Domain f 1 : (2, ) Domain of f = Range of f 1 : (, )
f.
Using the graph of y log 4 x , shift the graph 2 units to the left, and shift down 1 unit.
743
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Chapter 6: Exponential and Logarithmic Functions
right 1 unit, and shift up 2 units.
13.
14. 5 x 2 125
f ( x) 1 log 5 x 2
a.
Domain: (2, )
b. Using the graph of y log5 x , shift the graph to the right 2 units, reflect vertically about the y-axis, and shift up 1 unit.
5 x 2 53 x23 x 1 The solution set is {1} .
15. log( x 9) 2 x 9 102 x 9 100 x 91 The solution set is {91} .
16. 8 2e x 4 2e x 4
c.
Range: (, ) Vertical Asymptote: x 2
d.
f ( x ) 1 log5 x 2
e x 2 x ln 2 x ln 2 0.693 The solution set is { ln 2} {0.693} .
x 1 log5 y 2
x x3 0
x
y 2 51 x
e.
( x) 5
(1) (1) 2 4(1)(3) 1 13 2(1) 2
1 13 1 13 , The solution set is 2 2
y 51 x 2 f
x 3 x6 2
1 x log5 y 2
1 x
2
Inverse
x 1 log 5 y 2
1
17. log x 2 3 log x 6
y 1 log5 x 2
2
Range of f = Domain f 1 : (, )
1.303, 2.303 .
Domain of f = Range of f 1 : (2, ) f.
Using the graph of y 5 x , reflect the graph horizontally about the y-axis, shift to the
744 Copyright © 2025 Pearson Education, Inc.
Chapter 6 Chapter Test
7 x 3 e x
18.
ln 0.68 t 2 50e 30
ln 7 x 3 ln e x ( x 3) ln 7 x x ln 7 3ln 7 x x ln 7 x 3ln 7 x(ln 7 1) 3ln 7 3ln 7 3ln 7 x 6.172 ln 7 1 1 ln 7 3ln 7 The solution set is 6.172 . 1 ln 7
ln 0.68 t
0.04 e 30 ln 0.68 ln 0.04 t 30 ln 0.04 250.39 t ln 0.68 30 There will be 2 mg of the substance remaining after about 250.39 days.
19. log 2 x 4 log 2 x 4 3
22. a.
log 2 x 4 x 4 3
log 2 x 2 16 3
Note that 8 months =
P 1000 , r 0.05 , n 12 , and t
x 2 16 23
0.05 So, A 1000 1 12
2
x 16 8 x 2 24
(12) (2 / 3)
b.
8
Because x 2 6 results in a negative arguments for the original logarithms, the only viable solution is x 2 6 . That is, the solution
3 year. Thus, 4 3 A 1000 , r 0.05 , n 4 , and t . So, 4
Note that 9 months =
set is 2 6 4.899 . 4 x3 20. log 2 2 x 3x 18 22 x3 log 2 ( x 3)( x 6)
0.05 1000 A0 1 4
(4) (3/ 4)
1000 A0 1.0125 1000 A0 $963.42 1.0125 3 3
log 2 22 x3 log 2 ( x 6)( x 3) log 2 2 log 2 x3 log 2 ( x 6) log 2 ( x 3) 2
c.
2 3log 2 x log 2 ( x 6) log 2 ( x 3)
21.
2 . 3
0.05 1000 1 12 $1033.82
x 24 2 6
2 year. Thus, 3
r 0.06 and n 1 . So,
0.06 2 A0 A0 1 1 2 A0 A0 (1.06)t
A A0 ekt 34 50ek (30)
(1)t
2 (1.06)t
30 k
0.68 e ln 0.68 30k ln 0.68 k 30
ln 2 11.9 ln1.06 It will take about 11.9 years to double your money under these conditions. t log1.06 2
ln 0.68 t
Thus, the decay model is A 50e 30 . We need to find t when A 2 :
745
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Chapter 6: Exponential and Logarithmic Functions
23. a.
I 80 10 log 12 10 I 8 log 12 10 8 log I log1012 8 log I (12) 8 log I 12 4 log I
c.
8
2
2 x 4 xh 2h 3 x 3h 1
3. x 2 y 2 1 2
a.
b. Let n represent the number of people who must shout. Then the intensity will be n 104 . If D 125 , then n 104 125 10 log 12 10 12.5 log n 108
2
2
2 x 2 2 xh h 2 3 x 3h 1
I 104 0.0001 If one person shouts, the intensity is 104 watts per square meter. Thus, if two people shout at the same time, the intensity will be 2 104 watts per square meter. Thus, the loudness will be 2 104 10 log 2 108 83 D 10 log 12 10 decibels
125 10 log n 108
f x h 2 x h 3 x h 1
2
1 1 1 1 1 1 1 1 ; , is 4 4 2 2 2 2 2 not on the graph. 2
b.
2 1 3 1 3 1 3 1 ; , is on 4 4 2 2 2 2 the graph.
4. 3 x 2 4 x 5 3 x 6 4 x 20 26 x The solution set is 26 . 5. 2 x 4 y 16
x-intercept: 2 x 4 0 16 2 x 16 x 8
y-intercept: 2 0 4 y 16 4 y 16 y 4
12.5
n 10 10
n 104.5 31, 623 About 31,623 people would have to shout at the same time in order for the resulting sound level to meet the pain threshold.
6. a.
Chapter 6 Cumulative Review 1. The graph represents a function since it passes the Vertical Line Test.
The function is not a one-to-one function since the graph fails the Horizontal Line Test. 2.
f ( x) x 2 2 x 3 ; a 1, b 2, c 3.
Since a 1 0, the graph is concave down. The x-coordinate of the vertex is b 2 2 x 1. 2a 2 1 2
f x 2 x 2 3x 1
a.
f 3 2 3 3 3 1 18 9 1 10
b.
f x 2 x 3 x 1 2 x 2 3x 1
2
2
746 Copyright © 2025 Pearson Education, Inc.
Chapter 6 Cumulative Review f 4 8
The y-coordinate of the vertex is b f f 1 2a 12 2 1 3 1 2 3 2 Thus, the vertex is 1, 2 .
a 4 b 4 24 8 2
16a 4b 24 8 16a 4b 32 4a b 8 Replacing b with 8a in this equation yields 4a 8a 8 4a 8 a2 So b 8a 8 2 16 .
The axis of symmetry is the line x 1 .
The discriminant is: b 2 4ac 22 4 1 3 4 12 8 0 .
Therefore, we have the function f x 2 x 2 16 x 24 .
The graph has no x-intercepts. The y-intercept is f (0) 02 2(0) 3 3 . 8.
f ( x) 3( x 1)3 2
Using the graph of y x3 , shift the graph 1 unit to the left, stretch vertically by a factor of 3, and shift 2 units down.
b. The graph of f ( x) x 2 2 x 3 indicates
that f ( x) 0 for all values of x. Thus, the solution to f ( x) 0 is (, ) .
9.
7. Given that the graph of f ( x) ax 2 bx c has
f ( x) x 2 2
2 x3
2 ( f g )( x) f x 3
vertex 4, 8 and passes through the point b 4 , f 4 8 , 2a and f 0 24 . Notice that
2
0, 24 , we can conclude
2 2 x 3 4 2 x 3 2
f 0 24 a 0 b 0 c 24 2
The domain of f is x x is any real number .
c 24 Therefore, f x ax 2 bx c ax 2 bx 24 .
Furthermore,
g ( x)
The domain of g is x x 3 . So, the domain of f g x is x x 3 .
b 4 , so that b 8a , and 2a
( f g )(5)
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Copyright © 2025 Pearson Education, Inc.
4
5 3
2
2
4 2
2
2
4 23 4
Chapter 6: Exponential and Logarithmic Functions
10.
Range of g: 2,
f ( x) 3x 4 15 x3 12 x 2 60 x 3
Horizontal Asymptote for g: y 2
2
3x( x 5 x 4 x 20) 3x x ( x 5) 4( x 5) 2
g ( x ) 3x 2
b.
y 3x 2
3x( x 5)( x 2 4) 3x( x 5)( x 2)( x 2)
x 3y 2
x2 3 y log3 ( x 2)
a. Degree is 4. The function resembles y 3x 4
for large values of x .
g 1 x log3 x 2
b. y-intercept: f (0) 3(0) 4 15(0)3 12(0) 2 60(0) 0 x-intercepts: Solve f ( x) 0 0 3 x( x 5)( x 2)( x 2) x 0,5, 2, 2 c. Real zeros: 0 with multiplicity one, 5 with multiplicity one, 2 with multiplicity one, 2 with multiplicity one. The graph crosses the x-axis at x 0, x 5, x 2, and x 2. d.
4 1 3
e.
Graph
Inverse
y
Domain of g 1 : (2, ) Range of g 1 : (, ) Vertical Asymptote for g 1 : x 2 c. 12.
See part a. 4 x 3 82 x
2
2 x 3
23
2x
2 2 x 6 26 x 2x 6 6x 6 4 x 6 3 x 4 2 3 The solution set is . 2
13. log3 ( x 1) log3 (2 x 3) log 9 9 log3 ( x 1)(2 x 3) 1
( x 1)(2 x 3) 31
11. a.
2 x2 x 3 3
g ( x ) 3x 2
2 x2 x 6 0 2 x 3 ( x 2) 0
Using the graph of y 3x , shift up 2 units.
3 or x 2 2 3 1 Since log3 1 log 3 is 2 2 undefined the solution set is 2 . x
Domain of g: ,
748 Copyright © 2025 Pearson Education, Inc.
Chapter 6 Projects
14. a.
b.
log3 x 2 0
100 70 (200 70)e30 k
x 2 30 x 2 1 x 1 The solution set is 1 .
30 130e30 k 30 e30 k 130 30 30k ln 130 1 30 k ln 0.04888 30 130
log3 x 2 0 x 2 30 x 2 1 x 1 The solution set is x x 1 or 1, .
c.
u1 (t ) 70 130e 0.04888t
Container 2: u0 = 200ºF, T = 60ºF, u(25)=110ºF, t = 25 mins. 100 60 (200 60)e25k
log3 x 2 3 x 2 33 x 2 27 x 25 The solution set is 25 .
15. a.
50 140e25k 50 e25k 140 50 25k ln 140 50 ln 140 0.04118 k 25 u2 (t ) 60 140e0.04118t
Container 3: u0 = 200ºF, T = 65ºF, u(20)=120ºF, t = 20 mins.
b. Logarithmic: y 49.293 10.563ln x c.
Answers will vary 100 65 (200 65)e 20 k 55 135e20 k 55 e20 k 135 55 20k ln 135 55 ln 135 0.04490 k 20 u3 (t ) 65 135e 0.04490t
Chapter 6 Projects Project I – Internet-based Project - Answers will vary Project II a. Newton’s Law of Cooling: u (t ) T (u0 T )e kt , k < 0 Container 1: u0 = 200ºF, T = 70ºF, u(30)=100ºF, t = 30 mins.
b. We need time for each of the problems, so solve for t first then substitute the specific values for each container:
749
Copyright © 2025 Pearson Education, Inc.
Chapter 6: Exponential and Logarithmic Functions
u T (u0 T )e kt u T (u0 T )e kt
u T e kt u0 T
f. Since all three containers are within seconds of each other in cooling and staying warm, the cost would have an effect. The cheaper one would be the best recommendation.
u T ln u0 T u T kt ln t k u0 T
Container 1: 130 70 ln 200 70 t 15.82 minutes 0.04888 Container 2: 130 60 ln 200 60 t 16.83 minutes 0.04118 Container 3: 130 65 ln 200 65 t 16.28 minutes 0.04490
Project III Solder Joint X=ln(εp) Fatigue Strain, εp Cycles, Nf 0.01 4.605 10, 000 0.035 3.352 1000
9.210 6.908
2.303 0.916 0.405
4.605 2.303 0
0.1 0.4 1.5
c. Container 1: 110 70 ln 130 70 t 8.295 0.04888 It will remain between 110º and 130º for about 8.3 minutes.
1.
Container 2: 110 60 ln 130 60 t 8.171 0.04118 It will remain between 110º and 130º for about 8.17 minutes
2.
Container 3: 110 65 ln 130 65 t 8.190 0.04490 It will remain between 110º and 130º for about 8.19 minutes. d. All three graphs basically lie on top of each other.
100 10 1
The shape becomes exponential.
3.
The shape became linear.
e. Container 1 would be the best. It cools off the quickest but it stays in a warm beverage range the longest. 750 Copyright © 2025 Pearson Education, Inc.
Y=ln(Nf)
Chapter 6 Projects
4.
Y 1.84 X 0.63
5. Y 1.84 X 0.63 ln( Nf ) 1.84 ln( p ) 0.63
ln( Nf ) ln ( p ) 1.84 ln(e0.63 )
ln( Nf ) ln ( p)
1.84
(e ) 0.63
Nf ( p) 1.84 (e0.63 ) Nf e
0.63
( p )
1.84
6. Nf e0.63 (0.02) 1.84 Nf 2510.21 cycles Nf e0.63 ( p ) 1.84 3000 e0.63 ( p) 1.84 3000 ( p ) 1.84 e0.63 3000 p 0.63 e p 0.018
1 1.84
7. Nf e0.63 ( p ) 1.84 Nf 1.88( p ) 1.84 Nf ( p ) 1.84 1.88
p 0.53Nf
1 1.84
p 0.53Nf
.543
p 1.41 Nf
p 1.41 Nf
.543
p 1.41 3000 p 0.018
.543
.543
751
Copyright © 2025 Pearson Education, Inc.
Chapter 7 Trigonometric Functions 16.
Section 7.1 1. C 2 r ; A r 2
17.
2. d r t 3. standard position 4. central angle
18.
5. d 6. r ;
1 2 r 2
19.
7. b 8.
s ; t t
20.
9. True 10. False; r 21.
11.
12.
22. 13.
14.
15.
_
23. 30 30
radian radian 180 6
24. 120 120
2 radian radians 180 3
25. 495 495
11 radian radians 180 4
26. 330 330
11 radian radians 180 6
752 Copyright © 2025 Pearson Education, Inc.
Section 7.1: Angles, Arc Length, and Circular Motion radian radian 180 3
44.
radian radian 180 6
45.
17 17 180 degrees 204 15 15
29. 540 540
radian 3 radians 180
46.
3 3 180 degrees 135 4 4
30. 270 270
3 radian radians 180 2
47. 17 17
27. 60 60 28. 30 30
31. 240 240
4 radian radians 180 3
32. 225 225 33. 90 90
5 radian radians 180 4
radian radians 180 2
34. 180 180
radian radians 180
35.
180 degrees 60 3 3
36.
5 5 180 degrees 150 6 6
37. 38.
39.
2 2 180 degrees 120 3 3
40. 4 4
41.
42.
180 degrees 720
3 3 180 degrees 27 20 20 5 5 180 degrees 75 12 12
43.
180 degrees 90 2 2
17 radian radian 0.30 radian 180 180
radian 180 73 radians 180 1.27 radians
48. 73 73
49. 40 40
radian 180
2 radian 9 0.70 radian
radian 180 17 radian 60 0.89 radian
50. 51 51
13 13 180 degrees 390 6 6
9 9 180 degrees 810 2 2
180 degrees 180
51. 125 125
radian 180
25 radians 36 2.18 radians
52. 350 350
radian 180
35 radians 18 6.11 radians
53. 3.14 radians 3.14
54. 0.75 radian 0.75
753 Copyright © 2025 Pearson Education, Inc.
180 degrees 179.91º
180 degrees 42.97º
Chapter 7: Trigonometric Functions
55. 7 radians 7
56. 3 radians 3
180 degrees 401.07º
180 degrees 171.89º
57. 9.28 radians 9.28
58.
2 radians 2
180 degrees 531.70º
180 degrees 81.03º
1 1 1 º 59. 40º10 ' 25" 40 10 25 60 60 60 (40 0.1667 0.00694)º 40.17º 1 1 1 º 60. 61º 42 ' 21" 61 42 21 60 60 60 (61 0.7000 0.00583)º 61.71º 1 1 1º 61. 50º14 '20" 50 14 20 60 60 60 (50 0.2333 0.00556)º 50.24º 1 1 1 º 62. 73º 40 ' 40" 73 40 40 60 60 60 (73 0.6667 0.0111)º 73.68º 1 1 1 º 63. 9º 9 '9" 9 9 9 60 60 60 (9 0.15 0.0025)º 9.15º 1 1 1 º 64. 98º 22 ' 45" 98 22 45 60 60 60 (98 0.3667 0.0125)º 98.38º
65. 40.32º 40º 0.32º 40º 0.32(60 ') 40º 19.2 ' 40º 19 ' 0.2 ' 40º 19 ' 0.2(60") 40º 19 ' 12" 40º19 '12" 66. 61.24º 61º 0.24º 61º 0.24(60 ') 61º 14.4 ' 61º 14 ' 0.4 ' 61º 14 ' 0.4(60") 61º 14 ' 24" 61º14 ' 24" 67. 18.255º 18º 0.255º 18º 0.255(60 ') 18º 15.3' 18º 15' 0.3' 18º 15' 0.3(60") 18º 15' 18" 18º15'18" 68. 29.411º 29º 0.411º 29º 0.411(60 ') 29º 24.66 ' 29º 24 ' 0.66 ' 29º 0.66(60") 29º 24 ' 39.6" 29º 24 ' 40" 69. 19.99º 19º 0.99º 19º 0.99(60 ') 19º 59.4 ' 19º 59 ' 0.4 ' 19º 59 ' 0.4(60") 19º 59 ' 24" 19º 59 ' 24"
754 Copyright © 2025 Pearson Education, Inc.
Section 7.1: Angles, Arc Length, and Circular Motion 70. 44.01º 44º 0.01º 44º 0.01(60 ') 44º 0.6 ' 44º 0 ' 0.6 ' 44º 0 ' 0.6(60") 44º 0 ' 36" 44º 0 '36" 71. r 10 meters; s r 10
1 radian 2 1 1 100 21 A r 2 10 =25 m 2 2 2 4 2
79. r 10 meters;
80. r 6 feet; 2 radians A
1 radian; 2
1 5 meters 2
2 radian; s 8 feet; 3 s r s 8 r 12 feet 2 / 3
73.
r 12 2 3 3.464 feet 1 radian; A 6 cm 2 4 1 A r 2 2 1 21 6 r 2 4 1 2 6 r 8 48 r 2
82.
1 74. radian; s 6 cm; 4 s r
s
6 24 cm 1 / 4
75. r 10 miles; s 9 miles; s r s 9 0.9 radian r 10
r 48 4 3 6.928 cm
76. r 6 meters; s 8 meters; s r s 8 4 1.333 radians r 6 3 77. r 2 inches; 30º 30 s r 2
radian; 180 6
1.047 inches 6 3
78. r 3 meters; 120º 120 s r 3
1 radian; A 2 ft 2 3 1 A r 2 2 1 2 1 2 r 2 3 1 2 2 r 6 12 r 2
81.
72. r 6 feet; 2 radian; s r 6 2 12 feet
r
1 2 1 2 r 6 2 =36 ft 2 2 2
83. r 5 miles; A 3 mi 2 1 A r 2 2 1 2 3 5 2 25 3 2 6 0.24 radian 25
2 radians 180 3
2 2 6.283 meters 3 755 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 91. r 6 inches In 15 minutes, 15 1 rev 360º 90º radians 60 4 2 s r 6 3 9.42 inches 2
84. r 6 meters; A 8 m 2 1 A r 2 2 1 2 8 6 2 8 18 8 4 0.444 radian 18 9 85. r 2 inches; 30º 30 A
radian 180 6
1 2 1 2 r 2 1.047 in 2 2 2 6 3
86. r 3 meters; 120º 120 A
2 radians 180 3
1 2 1 2 2 2 r 3 =3 9.425 m 2 2 3
87. r 2 feet;
radians 3 2 s r 2 2.094 feet 3 3 1 1 2 2 A r 2 2 = 2.094 ft 2 2 2 3 3
88. r 4 meters;
6
radian
2 2.094 meters 6 3 1 1 4 2 A r 2 4 4.189 m 2 2 2 6 3
s r 4
89. r 12 yards; 70º 70
7 radians 180 18
7 14.661 yards 18 1 1 2 7 2 A r 2 12 28 87.965 yd 2 2 18
s r 12
90. r 9 cm; 50º 50
5 radian 180 18
5 7.854 cm 18 1 1 2 5 45 A r 2 9 35.343 cm 2 = 2 2 4 18
In 25 minutes, 25 5 5 rev 360º 150º radians 60 12 6 5 s r 6 5 15.71 inches 6 92. r 40 inches; 20º s r 40
radian 9
40 13.96 inches 9 9
radian 180 4 1 1 2 A r 2 4 2 6.28 m 2 2 2 4
93. r 4 m; 45º 45
radians 180 3 1 2 1 2 3 A r 3 4.71 cm 2 2 2 3 2
94. r 3 cm; 60º 60
3 radians 180 4 1 1 675 2 3 A r 2 30 1060.29 ft 2 2 2 2 4
95. r 30 feet; 135º 135
96. r 15 yards; A 100 yd 2 1 A r 2 2 1 2 100 15 2 100 112.5 100 8 0.89 radian 112.5 9
or
8 180 160 50.93 9
s r 9
756 Copyright © 2025 Pearson Education, Inc.
Section 7.1: Angles, Arc Length, and Circular Motion 1 2 1 2 r1 r2 2 120 2 2 3 1 2 1 2 (34) 2 (9) 2 3 2 3 2
102. r 6.5 m; 22 rev/min 44 rad/min v r (6.5) 44 m/min 286 898.5 m/min m 1km 60 min v 286 53.9 km/hr min 1000m 1hr
1 2 1 2 (1156) (81) 3 2 3 2
103. r
97. A
222 m; 14 rev/min 28 rad/min 2 v r (111) 28 m/min 3108 m/min
(1156) (81) 3 3
m 1km 60 min min 1000m 1hr 585.8 km/hr
v 3108
1156 81 3 3
1075 1125.74 in 2 3 1 2 1 25 r1 r2 2 125 2 2 36 1 25 1 2 25 (30) 2 (6) 36 2 36 2
98. A
1 25 1 25 (900) (36) 36 2 36 2
25 25 (450) (18) 36 36 11250 450 36 36
10800 300 942.48 in 2 36
99. r 5 cm; t 20 seconds;
1 / 3
1 radian 3
1 1 1 radian/sec 20 3 20 60 t 1 s r 5 1 / 3 5 1 cm/sec v 20 3 20 12 t t
100. r 30 feet 1 rev 2 0.09 radian/sec 70 sec 70 sec 35 rad 6 ft 2.69 feet/sec v r 30 feet 35 sec 7 sec 101. r 25 feet; 13 rev/min 26 rad/min v r 25 26 ft./min 650 2042.0 ft/min ft. 1mi 60 min v 650 23.2 mi/hr min 5280ft 1hr
104. r 4 m; 8000 rev/min 16000 rad/min v r (4) 16000 m/min 64000 cm/min cm 1m 1km 60 min v 64000 min 100cm 1000m 1hr 120.6 km/hr 105. d 26 inches; r 13 inches; v 35 mi/hr 35 mi 5280 ft 12 in. 1 hr v hr mi ft 60 min 36,960 in./min v 36,960 in./min r 13 in. 2843.08 radians/min 2843.08 rad 1 rev min 2 rad 452.5 rev/min 106. r 15 inches; 3 rev/sec 6 rad/sec v r 15 6 in./sec 90 282.7 in/sec in. 1ft 1mi 3600sec v 90 16.1 mi/hr sec 12in. 5280ft 1hr 107. r 860 feet; 4
6 4.1; 60
4.1 0.07156 180 0.07156(860) 61.54 feet
108. r 920 feet; 1
42 1.7; 60
1.7 0.02967 180 0.02967(920) 27.30 feet
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Chapter 7: Trigonometric Functions 109. r 3429.5 miles
1 rev/day 2 radians/day v r 3429.5
radians/hr 12
898 miles/hr 12
110. r 3033.5 miles
1 rev/day 2 radians/day v r 3033.5
radians/hr 12
794 miles/hr 12
111. r 2.39 105 miles 1 rev/27.3 days 2 radians/27.3 days radians/hr 12 27.3 v r 2.39 105 2292 miles/hr 327.6 112. r 9.29 107 miles 1 rev/365 days 2 radians/365 days radians/hr 12 365 v r 9.29 107
66, 633 miles/hr 4380
113. r1 2 inches; r2 8 inches; 1 3 rev/min 6 radians/min Find 2 : v1 v2 r11 r22
pulleys is the same, we have: v1 v2 r11 r22 r11 r22 r21 r21 r1 2 r2 1
115. r 4 feet; 10 rev/min 20 radians/min v r 4 20 ft 80 min 80 ft 1 mi 60 min min 5280 ft hr 2.86 mi/hr 116. d 26 inches; r 13 inches; 480 rev/min 960 radians/min v r 13 960 in 12480 min 12480 in 1 ft 1 mi 60 min min 12 in 5280 ft hr 37.13 mi/hr v r 80 mi/hr 12 in 5280 ft 1 hr 1 rev 13 in 1 ft 1 mi 60 min 2 rad 1034.26 rev/min 117. d 8.5 feet;
2(6) 82
r 4.25 feet;
v 9.55 mi/hr
v 9.55 mi/hr 4.25 ft r 9.55 mi 1 5280 ft 1 hr 1 rev hr 4.25 ft mi 60 min 2 31.47 rev/min
12 2 8 1.5 radians/min 1.5 rev/min 2 3 rev/min 4
114. r1 rotates at 1 rev/min , so v1 r11 . r2 rotates at 2 rev/min , so v2 r22 . Since the linear speed of the belt connecting the
118. Let t represent the time for the earth to rotate 90 miles. t 24 90 2(3559) 90(24) 0.0966 hours 5.8 minutes t 2(3559)
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Section 7.1: Angles, Arc Length, and Circular Motion 119. The earth makes one full rotation in 24 hours. The distance traveled in 24 hours is the circumference of the earth. At the equator the circumference is 2(3960) miles. Therefore, the linear velocity a person must travel to keep up with the sun is: s 2(3960) v 1037 miles/hr t 24 120. Find s, when r 3960 miles and 1'. 1 degree radians 0.00029 radian 1' 60 min 180 degrees s r 3960(0.00029) 1.15 miles Thus, 1 nautical mile is approximately 1.15 statute miles. 121. We know that the distance between Alexandria and Syene to be s 500 miles. Since the measure of the Sun’s rays in Alexandria is 7.2 , the central angle formed at the center of Earth between Alexandria and Syene must also be 7.2 . Converting to radians, we have 7.2 7.2 s r
25
25
radian . Therefore,
25
500
12,500
C 2 r 2
3979 miles
12,500
25, 000 miles. The radius of Earth is approximately 3979 miles, and the circumference is approximately 25,000 miles.
122. a.
123. Large area: A r 2 (9) 2 81 243 3 We need ¾ of this area. 81 . 4 4
A r2
Small area: (3) 2 9 9 1 We need ¼ of the small area. 9 4 4
The total area is:
243 9 252 63 4 4 4
square feet.
500 r r
180
1 2 1 R r 2 R 2 r 2 2 2 2 96 2002 1902 2 180 4 3900 3267.3 15 The area of the warning track is about 3267.3 square feet. A
The length of the outfield fence is the arc length subtended by a central angle 96 with r 200 feet. s r 200 96
335.10 feet 180 The outfield fence is approximately 335.1 feet long. b. The area of the warning track is the difference between the areas of two sectors with central angle 96 . One sector with r 200 feet and the other with r 190 feet.
2 is one124. BAE is a right angle so arc BE fourth of the circumference C of the circle so C 8 . First we find the radius of the circle: C 2 r 8 2 r r4
The area of the circle is A r 2 (4) 2 16 . The area of the sector of the circle is 4 . The area of the rectangle is lw (4)(4 7) 44 . So the area of the rectangle that is outside the circle is 44 4 square units. 125. Since 50 feet = 600 inches, moving 600 inches equates to an are length of 600 inches for a circle with radius 15 inches (the radius of the wheel). 600 15( w ) so w 40 radians . The wheel and rear cog have the same angle of rotation so sc rc w 1.8(40) 72 inches for the chain on the cog wheel. The chain needs to have the same arc length on the pedal drive wheel, so
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Chapter 7: Trigonometric Functions 72 (5.2) p or p 13.84615 radians.
5 x 2 2 5 14 x
134.
Dividing by 2 (for 1 revolution) gives 13.84615 / (2 ) 2.2 revolutions.
5 x 2 14 x 3 0 (5 x 1)( x 3) 0 x
126. Answers will vary. 127. If the radius of a circle is r and the length of the arc subtended by the central angle is also r, then the measure of the angle is 1 radian. Also, 180 1 radian degrees .
1
135. Shift to the left 3 units would give y x 3 .
Reflecting about the x-axis would give y x 3 . Shifting down 4 units would result in y x 3 4 .
radians 128. Note that 1 1 0.017 radian 180 180 and 1 radian 57.296 . radians Therefore, an angle whose measure is 1 radian is larger than an angle whose measure is 1 degree.
129. Linear speed measures the distance traveled per unit time, and angular speed measures the change in a central angle per unit time. In other words, linear speed describes distance traveled by a point located on the edge of a circle, and angular speed describes the turning rate of the circle itself. 130. This is a true statement. That is, since an angle measured in degrees can be converted to radian measure by using the formula 180 degrees radians , the arc length formula
can be rewritten as follows: s r
133.
1 So the solution set is 3, 5
1 revolution 360
131 – 132. Answers will vary. f ( x) 3x 7
0 3x 7 3 x 7 x
7 3
1 or x 3 5
180
r .
3 x 2 12 3( x 2)( x 2) 3( x 2) x 5 x 14 ( x 2)( x 7) ( x 7) The vertical asymptote is: x 7 As f(x) go to then the graph behaves like 3 x 2 12 3x 2 3 so the horizontal x 2 5 x 14 x 2 asymptote is y 3 . 137. 2 x y 5 y 2x 5
136.
2
1 A perpendicular line would have slope . So 2 the line containing the point (-1, 4) is: 1 y 4 ( x 1) 2 1 1 y4 x 2 2 1 7 y x 2 2 The other point would be: 1 7 c (2) 2 2 7 5 1 2 2
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Section 7.2: Right Triangle Trigonometry b. -3, 1, 4, 7 c. The values are the same. d. , 3 , 1, 4 , 7,
138. 2 x 3 5 8 2 x3 3 3 2 9 x3 4 21 x 4
x3
x 4 9 x3 3x 2 89 x 84 0
e.
( x 3)( x 1)( x 4)( x 7) 0 f ( x) x 4 9 x3 3 x 2 89 x 84 x 3, x 1, x 4, x 7 are the zeros of f .
21 The solution set is 4
( , 3)
( 3,1)
(1, 4)
(4, 7)
(7, )
4
0
2
6
8
Value of f
440
84
50
90
308
Conclusion
Pos
Neg
Pos
Neg
Pos
Interval Number
139. The denominator cannot be zero. x2 9 0
Chosen
x 9 x 3 x 9 0 2
2
The solution set is x x 3 or 1 x 4 or x 7 or, using
x 9 x 3 So the domain is. {x | x 3, x 3} . 2
140.
interval notation, , 3 1, 4 7, .
f ( x h) f ( x ) h 2( x h)3 5 ( x3 5) h
Section 7.2
2 x3 3 x 2 h 3 xh 2 h3 5 2 x3 5
h 2 x 6 x h 6 xh 2h3 5 2 x3 5 h 2 2 3 6 x h 6 xh 2h h 6 x 2 6 xh 2h 2 3
2
The intervals are the same.
f.
f ( x) 2 x3 5
1. c 2 a 2 b 2 62 102 36 100 136 c 136 2 34
2
3 141. (3x 2) (3x 2)(3 x 2)(3 x 2) (9 x 2 12 x 4)(3 x 2)
27 x3 54 x 2 36 x 8
142. a. (3,395.65); (1, 39.55); (4, 69.8);(7, 104.35)
2.
f (5) 3(5) 7 15 7 8
3.
32 16 2 4 2 45 9 5 3 5
4.
10 x 8 3 30 8 x 30 15 x 8 4
5. complementary 6. cosine 7. 62o 8. c 9. True 761 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 10. a
16. opposite = 3; adjacent = 3; hypotenuse = ? (hypotenuse) 2 32 32 18
11. True
hypotenuse 18 3 2
3 12. False; tan cot cot 5 2 5 10
13. opposite = 5; adjacent = 12; hypotenuse = ? (hypotenuse) 2 52 122 169 hypotenuse 169 13 opp 5 hyp 13 sin csc hyp 13 opp 5 adj 12 hyp 13 cos sec hyp 13 adj 12 opp 5 adj 12 tan cot adj 12 opp 5
14. opposite = 3; adjacent = 4, hypotenuse = ? (hypotenuse) 2 32 42 25 hypotenuse 25 5 opp 3 sin hyp 5 adj 4 cos hyp 5 opp 3 tan adj 4
hyp 5 csc opp 3 hyp 5 sec adj 4 adj 4 cot opp 3
15. opposite = 2; adjacent = 3; hypotenuse = ? (hypotenuse) 2 22 32 13
hypotenuse 13 sin
adj 3 3 13 3 13 hyp 13 13 13 13 opp 2 tan adj 3
csc
opp 3 3 2 2 hyp 3 2 3 2 2 2
adj 3 3 2 2 hyp 3 2 3 2 2 2 opp 3 tan 1 adj 3 cos
csc
hyp 3 2 2 opp 3
hyp 3 2 2 adj 3 adj 3 cot 1 opp 3
sec
17. adjacent = 2; hypotenuse = 4; opposite = ? (opposite) 2 22 42 (opposite) 2 16 4 12 opposite 12 2 3
opp 2 3 3 hyp 4 2 adj 2 1 cos hyp 4 2
sin
tan
opp 2 3 3 adj 2
hyp 4 4 3 2 3 opp 2 3 2 3 3 3 hyp 4 2 sec adj 2 csc
opp 2 2 13 2 13 hyp 13 13 13 13
cos
sin
cot
adj 2 2 3 3 opp 2 3 2 3 3 3
hyp 13 opp 2
hyp 13 adj 3 adj 3 cot opp 2
sec
762 Copyright © 2025 Pearson Education, Inc.
Section 7.2: Right Triangle Trigonometry 18. opposite = 3; hypotenuse = 4; adjacent = ? 32 (adjacent) 2 42 (adjacent) 2 16 9 7
adj 7 hyp 4
opp 3 3 7 3 7 adj 7 7 7 7 hyp 4 csc opp 3 tan
sec
hyp 4 4 7 4 7 adj 7 7 7 7
adj 7 cot opp 3
19. opposite =
2 ; adjacent = 1; hypotenuse = ?
(hypotenuse) 2
2
2
(hypotenuse) 2 22
3 ; hypotenuse = ?
3 7 2
hypotenuse 7
adjacent 7 opp 3 sin hyp 4 cos
20. opposite = 2; adjacent =
12 3
sin
opp 2 2 7 2 7 hyp 7 7 7 7
cos
adj 3 3 7 21 hyp 7 7 7 7
tan
opp 2 2 3 2 3 adj 3 3 3 3
csc
hyp 7 opp 2
sec
hyp 7 7 3 21 adj 3 3 3 3
cot
adj 3 opp 2
21. opposite = 1; hypotenuse = 12 (adjacent) 2
5
5 ; adjacent = ?
2
(adjacent) 2 5 1 4
hypotenuse 3
adjacent 4 2
opp 2 2 3 6 sin hyp 3 3 3 3
sin
cos
adj 1 1 3 3 hyp 3 3 3 3
tan
opp 2 2 adj 1
csc
hyp 3 3 2 6 opp 2 2 2 2
sec
hyp 3 3 adj 1
cot
adj 1 1 2 2 opp 2 2 2 2
opp 1 1 5 5 hyp 5 5 5 5
adj 2 2 5 2 5 hyp 5 5 5 5 opp 1 tan adj 2 cos
csc
hyp 5 5 opp 1
hyp 5 adj 2 adj 2 2 cot opp 1
sec
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Chapter 7: Trigonometric Functions
22. adjacent = 2; hypotenuse = (opposite) 2 2
2
5
5 ; opposite = ?
2
opposite 1 1 opp 1 1 5 5 hyp 5 5 5 5
hyp 5 5 opp 1
cot
hyp 5 adj 2 adj 2 2 cot opp 1
sec
cos
cos
5 3
2 sin 2 3 2 2 5 2 5 3 tan cos 5 5 3 5 5 5 5 3
3 2
1 1 1 2 2 sin 1 2
1 1 2 2 cos 3 3 3 2 1 1 3 3 cot tan 3 3 3 3
sec
1 1 1 3 3 tan 3 3 3 3
2 25. sin ; 3
1 sin 1 2 1 1 3 3 tan 2 cos 3 3 2 3 3 3 3 2
csc
1 2
1 1 2 2 3 2 3 sin 3 3 3 3 3 2 1 1 1 2 2 sec cos 1 2
adj 2 2 5 2 5 hyp 5 5 5 5 opp 1 tan adj 2
1 23. sin ; 2
cos
csc
cos
csc
3 ; 2
3 sin 3 2 tan 2 3 1 cos 2 1 2
(opposite) 5 4 1 2
sin
24. sin
3 3 3 3
2 3 3
3 3 3 3
csc
1 1 3 sin 2 2 3
1 1 3 5 3 5 cos 5 5 5 5 3 1 1 5 5 5 5 5 cot tan 2 5 2 5 5 10 2 5
sec
1 26. sin ; 3
cos
2 2 3
1 sin 1 3 1 2 2 tan 3 cos 2 2 3 2 2 2 2 2 4 3
csc
1 1 1 3 3 sin 1 3
sec
1 1 3 3 2 3 2 cos 2 2 2 2 2 2 2 4 3
cot
1 1 4 4 2 4 2 2 2 tan 2 2 2 2 2 4
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Section 7.2: Right Triangle Trigonometry
27. sin
2 corresponds to the right triangle: 2
c
b 2
c
a 2
b 2
a
sin
opp 2 hyp 2
tan
opp adj
csc
hyp 2 2 2 2 opp 2 2 2
sec
hyp 2 2 2 2 adj 2 2 2
cot
adj opp
Using the Pythagorean Theorem: a2
2 2 2
2
a2 4 2 2 a 2 So the triangle is: c
b 2
a 2
tan
opp adj
sec
hyp 2 2 2 2 adj 2 2 2
csc
hyp 2 2 2 2 opp 2 2 2
cot
adj opp
2 2
1 sin 2 1 3 1 sin 2 1 9 8 sin 2 9 sin
1
b
tan
sin 2 3 2 2 2 3 1 2 2 cos 3 1 3
csc
1 1 3 3 2 3 2 sin 2 3 2 2 2 2 2 2 4
sec
1 1 1 3 3 cos 13
cot
1 1 1 2 2 tan 2 2 2 2 2 4
a 2
Using the Pythagorean Theorem: 2
2
b2 4 2 2 b 2 So the triangle is:
8 2 2 9 3
(Note: sin must be positive since is acute.)
c
2 2
1
2
2
1
2 28. cos corresponds to the right triangle: 2
b2
2
1 3 Using the Pythagorean Identities: sin 2 cos 2 1
adj 2 hyp 2 2
1
2
29. cos
cos
2
2
3 4 Using the Pythagorean Identities: sin 2 cos 2 1
30. sin
2
3 2 cos 1 4
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Chapter 7: Trigonometric Functions
3 cos 2 1 16 13 cos 2 16 13 13 cos 16 4
1 corresponds to the right triangle: 2
32. cot
c
b
(Note: cos must be positive since is acute.) tan
sin cos
csc
1 1 4 4 3 4 3 3 sin 3 3 3 3 4
sec
1 1 4 4 13 4 13 13 cos 13 13 13 13 4
cot
cos sin
31. tan
b
3 4 13 4
13 4 3 4
3 13
13 3
3 13
13 3
c
Using the Pythagorean Theorem: c 2 12 22 5 c 5 So, the triangle is:
3 3
39 13
c 5
c 5 So the triangle is: c 5
b
39 3
a
sin
opp 2 2 5 2 5 hyp 5 5 5 5
adj 1 1 5 5 hyp 5 5 5 5 opp 2 tan 2 adj 1
cos
csc
hyp 5 opp 2
sec
hyp 5 5 adj 1
tan 2 1 32
tan 2 32 1 8
sin
opp 1 1 5 5 hyp 5 5 5 5
cos
adj 2 2 5 2 5 hyp 5 5 5 5
csc
hyp 5 5 opp 1
hyp 5 adj 2 adj 2 2 cot opp 1
Using the Pythagorean Theorem: c 2 12 22 5
33. sec 3 Using the Pythagorean Identities: tan 2 1 sec2
a
sec
13
1 corresponds to the right triangle: 2
a
b
13
a
tan 8 2 2 (Note: tan must be positive since is acute.) 1 1 cos sec 3 sin tan , so cos 1 2 2 sin tan cos 2 2 3 3 csc
1 1 3 3 2 3 2 2 2 sin 4 2 2 2 2 2 3
cot
1 1 1 2 2 tan 2 2 2 2 2 4
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Section 7.2: Right Triangle Trigonometry 34. csc 5 Using the Pythagorean Identities: cot 2 1 csc 2
2
25 16 5 tan 2 1 1 9 9 3 16 4 9 3 (Note: tan must be positive since is acute.) 1 1 3 cos 5 sec 3 5 tan
cot 2 1 52 cot 2 52 1 24 cot 24 2 6 (Note: cot must be positive since is acute.) 1 1 sin csc 5 cos cot , so sin
tan
sin , so cos
sin tan cos
1 2 6 cos cot sin 2 6 5 5
csc
1 1 5 4 sin 5 4
tan
1 1 1 6 6 cot 2 6 2 6 6 12
cot
1 1 3 4 tan 3 4
sec
1 1 5 5 6 5 6 cos 2 5 6 2 6 2 6 6 12
35. tan 2 Using the Pythagorean Identities: sec 2 tan 2 1 sec 2
2 1 3 2
sec 3 (Note: sec must be positive since is acute.)
37. csc 2 corresponds to the right triangle: c
b
a
Using the Pythagorean Theorem: a 2 12 22 a2 1 4 a2 4 1 3
1 1 1 3 3 cos sec 3 3 3 3 sin tan , so cos
a 3 So the triangle is:
3 6 sin tan cos 2 3 3
b
1 1 3 3 6 3 6 6 csc 6 sin 6 2 6 6 6 3
1 1 1 2 2 cot tan 2 2 2 2 5 3 Using the Pythagorean Identities: tan 2 1 sec2
36. sec
5 tan 2 1 3
2
4 3 4 3 5 5
c a 3
sin
opp 1 hyp 2
cos
adj 3 hyp 2
tan
opp 1 1 3 3 adj 3 3 3 3
sec
hyp 2 2 3 2 3 adj 3 3 3 3
cot
adj 3 3 opp 1
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Chapter 7: Trigonometric Functions 38. cot 2 corresponds to the right triangle:
b
sin 70º tan 70º tan 70º 0 , using cos 70º sin the identity tan cos
43. tan 70º
c a
Using the Pythagorean Theorem: c 2 12 22 1 4 5 c 5 So the triangle is:
b
45. sin 38º cos 52º sin 38º sin(90º 52º ) sin 38º sin 38º 0 using the identity cos sin 90
c 5 a
sin
opp 1 1 5 5 hyp 5 5 5 5
adj 2 2 5 2 5 hyp 5 5 5 5 opp 1 tan adj 2
cos
csc
hyp 5 5 opp 1
sec
hyp 5 adj 2
cos 25º cot 25º cot 25º 0 , using the sin 25º cos identity cot sin
44. cot 25º
46. tan12º cot 78º tan12º tan(90º 78º ) tan12º tan12º 0 using the identity cot tan 90 47.
48.
39. sin 2 26º cos 2 26º 1 , using the identity
cos13º sin(90º 13º ) sin 77º 1 sin 77º sin 77º sin 77º using the identity cos sin 90 cos 40º sin(90º 40º ) sin 50º 1 sin 50º sin 50º sin 50º using the identity cos sin 90
49. 1 cos 2 15º cos 2 75º 1 cos 2 15º sin 2 (90º 75º )
sin 2 cos 2 1
1 cos 2 15º sin 2 (15º )
40. sec 2 28º tan 2 28º 1 , using the identity
1 cos 2 15º sin 2 (15º )
tan 2 1 sec 2
1 41. sin 80º csc80º sin 80º 1 , using the sin 80º 1 identity csc sin 42. tan10º cot10º tan10º
identity cot
1 tan
1 1 0
using the identities cos sin 90 and sin 2 cos 2 1 .
50. 1 tan 2 5º csc2 85º sec 2 5º csc 2 85º sec 2 5º sec 2 (90º 85º )
1 1 , using the tan10º
sec2 5º sec2 5º 0
using the identities 1 tan 2 sec 2 and csc sec 90
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Section 7.2: Right Triangle Trigonometry
51. tan 20º
cos 70º sin(90º 70º ) tan 20º cos 20º cos 20º sin 20º tan 20º cos 20º tan 20º tan 20º
0 using the identities cos sin 90 and sin tan . cos
52. cot 40º
sin 50º cos(90º 50º ) cot 40º sin 40º sin 40º cos 40º cot 40º sin 40º cot 40º cot 40º
0 using the identities sin cos 90 and
cot
cos . sin
55. cos 35º sin 55º cos 55º sin 35º cos 35º cos(90º 55º ) sin(90º 55º ) sin 35º cos 35º cos 35º sin 35º sin 35º cos 2 35º sin 2 35º 1 using the identities sin cos 90 , cos sin 90 , and sin 2 cos 2 1 .
56. sec 35º csc 55º tan 35º cot 55º sec 35º sec(90º 55º ) tan 35º tan(90º 55º ) sec 35º sec 35º tan 35º tan 35º sec 2 35º tan 2 35º (1 tan 2 35º ) tan 2 35º 1 using the identities csc sec 90 , cot tan 90 , and 1 tan 2 sec 2
57. Given: sin 30o sin10º sec80º cos10º cos10º sin10º sec80º
53. tan10º sec80º cos10º
1 2
1 2
a.
cos 60o sin 90o 60o sin 30o
sin10º csc 90º 80º
b.
sin10º csc10º 1 sin10º sin10º 1
3 1 cos 30 1 sin 30 1 4 2
c.
csc
d.
sec
using the identities tan
sin , cos
sec csc 90 , and csc
1 . sin
cos 25º csc65º sin 25º sin 25º cos 25º csc65º
54. cot 25º csc65º sin 25º
cos 25º sec 90º 65º cos 25º sec 25º 1 cos 25º cos 25º 1
using the identities cot
cos , sin
csc sec 90 , and sec
1 . cos
2
2
o
2
o
1 1 csc 30o 2 o 1 6 sin 30 2 csc csc csc 30o 2 3 6 2 3 3 2
58. Given: sin 60o
3 2
a.
cos 30o sin 90o 30o sin 60o
b.
3 1 cos 2 60o 1 sin 2 60o 1 2 4
c.
sec
d.
csc
2
1 1 1 2 3 2 3 6 cos cos30o 3 3 3 3 6 2 2 3 sec sec sec30o 3 3 2 3 6
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Chapter 7: Trigonometric Functions 59. Given: tan 7 a. sec2 1 tan 2 1 7 2 1 49 50
1 1 tan 7
b.
cot
c.
cot tan 7 2
d.
63. Given: sin 38o 0.62 a. cos 38o ? sin 2 38o cos 2 38o 1 cos 2 38o 1 sin 2 38o cos 38o 1 sin 2 38o 1 (0.62) 2
csc2 1 cot 2 1
0.78
1 1 1 50 1 2 1 49 49 tan 2 7
60. Given: sec 3 1 1 a. cos sec 3 b.
tan sec 1 3 1 9 1 8
c.
csc 90º sec 3
d.
sin 2 1 cos 2
2
2
2
1 1 1 8 1 1 2 1 2 9 9 sec 3
61. Given: csc 4 1 1 a. sin csc 4 cot csc 1 4 1 16 1 15
c.
sec(90º ) csc 4
d.
1 1 16 1 sec 1 tan 1 2 15 15 cot
2
2
tan 38o
sin 38o 0.62 0.79 cos 38o 0.78
c.
cot 38o
cos 38o 0.79 1.27 sin 38o 0.62
d.
sec 38o
1 1 1.28 o 0.78 cos 38
e.
csc 38o
1 1 1.61 o 0.62 sin 38
f.
sin 52o cos 90o 52o cos 38o 0.78
g. h.
b.
2
b.
2
cos 52 sin 90 52 sin 38 0.62 tan 52 cot 90 52 cot 38 1.27 o
o
o
o
o
o
o
o
64. Given: cos 21o 0.93 a. sin 21o ? sin 2 21o cos 2 21o 1
sin 2 21o 1 cos 2 21o
2
sin 21o 1 cos 2 21o 1 (0.93) 2
62. Given: cot 2 1 1 a. tan cot 2
0.37
b.
tan 21o
sin 21o 0.37 0.40 cos 21o 0.93
b.
csc 2 cot 2 1 22 1 4 1 5
c.
tan cot 2 2
c.
cot 21o
cos 21o 0.93 2.51 sin 21o 0.37
d.
sec2 1 tan 2
d.
sec 21o
1 1 1.08 o 0.93 cos 21
e.
csc 21o
1 1 2.70 o 0.37 sin 21
f.
sin 69o cos 90o 69o cos 21o 0.93
1
1 1 1 5 1 2 1 2 4 4 cot 2
g. h.
cos 69 sin 90 69 sin 21 0.37 tan 69 cot 90 69 cot 21 2.51
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o
o
o
o
o
o
o
o
Section 7.2: Right Triangle Trigonometry 65. Given: sin 0.3 sin cos sin sin 0.3 0.3 0.6 2 66. Given: tan 4 tan tan tan cot 2 1 1 17 tan 4 tan 4 4
72.
4 x 2 4 ( 2 tan )2 4 4 tan 2 4(1 tan 2 ) 4sec 2 2sec
73.
x 2 36 ( 6sec ) 2 36 36sec 2 36 36(sec 2 1)
67. The equation sin cos 2 30o will be true
when 90o 2 30o
60o 2 3 60
36 tan 2 6 tan 2
74.
o
9 100 x 2 81 100 sec 81 10
20o
100
68. The equation tan cot 45o will be true
when 90 45 o
o
81sec 2 81 81(sec 2 1)
45 o
81tan 2 9 tan
2 45o
22.5o 69.
75. First we set up a ratio to find the relationship 17 , we know between the sides. Since tan 10 the ratio of the sides. The width of the droplet is 4 mm so we setup the following ratio and solve for x:
25 x 2 25 (5sin ) 2 25 25sin 2 25(1 sin 2 )
17 4 10 x x 2.353
25cos 2 5cos
70.
4 16 9 x 16 9 sin 3
2
2
We now solve for side r (the hypotenuse). 42 2.3532 r 2
16 16 9 sin 2 6
r 2 21.5366 r 4.64 mm
16 16sin 2 16(1 sin 2 ) 16 cos 2 4 cos
71.
81 sec 2 81 100
49 x 2 49 (7 tan ) 2 49 49 tan 2 49(1 tan 2 ) 49sec2 7 sec
771 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
76. cos
5 1.5 7 L1
b.
1.5 7 1.5 5 5 7 2.1 cm
L1
1.52 SD 2 2.12
SD 2 2.12 1.52 SD 1.47 cm
77. a.
Z 2 X 2 R2 Z
X 2 R 2 4002 6002
520, 000 200 13 721.1 ohms The impedance is about 721.1 ohms.
b.
sin
X 400 2 13 13 Z 200 13
cos
R 600 3 13 Z 200 13 13
tan
X 400 2 R 600 3
csc
13 Z 200 13 400 2 X
79. a.
T
1500 500 5 5 10 minutes 300 100
b.
T
500 1500 5 15 20 minutes 100 100
c.
13 Z 200 13 600 3 R R 600 3 cot X 400 2
sec
78. a.
X R 5 X 12 588 5 588 X or X 245 ohms 12 The inductive reactance is 245 ohms. tan
Z X 2 R 2 2452 5882 637 The impedance is 637 ohms.
X 245 5 Z 637 13 R 588 12 cos Z 637 13 Z 637 13 csc X 245 5 Z 637 13 sec R 588 12 R 588 12 cot X 245 5
sin
d.
500 500 , so x . tan x 500 , so sin distance in sand 500 . distance in sand sin 1500 x distance in sand T 300 100 500 500 1500 tan sin 300 100 5 5 5 3 tan sin 1 1 5 1 3 tan sin tan
500 1 , so we can consider the 1500 3 triangle: tan
10
772 Copyright © 2025 Pearson Education, Inc.
3
Section 7.2: Right Triangle Trigonometry 80. a.
5 5 3 tan sin 5 5 5 1 1 3 3 10
T 5
that hall that is 4 feet wide. Then, cos so x
5 5 5 10 15.8 minutes
e.
f.
Consider the length of the line segment in two sections, x, the portion across the hall that is 3 feet wide and y, the portion across
1000 feet along the paved path leaves an additional 500 feet in the direction of the path, so the angle of the path across the sand is 45˚. 5 5 T 5 3 tan 45º sin 45º 5 5 5 3 1 2 2 5 10 5 3 2 10.4 minutes 5 5 with the 3 tan x sin x calculator in DEGREE mode.
Let Y1 5
3 4 4 and sin , so y . cos sin y
Thus, L( ) x y
3 4 . cos sin
b. Answers will vary. 81. cos 0.6 h
30 h
30 50 yds 0.6
302 d 2 502 d 2 502 302 d 40 yds
The receiver would be tackled 40 yards from the 50-yard line or at the 10-yard line. x . Consider a right triangle 1 with acute angle .
82. Rewrite as tan
x
0
90
Use the MINIMUM feature:
1
The hypotenuse is given by c 1 x 2 . sin
0
90
The time is least when the angle is approximately 70.5˚. The value of x for this 500 angle is x 177 feet . The tan 70.53º least time is approximately 9.7 minutes. g.
Answers will vary.
3 , x
opposite x x 1 x2 hypotenuse 1 x2 1 x2
adjacent 1 1 x2 hypotenuse 1 x2 1 x2 adjacent 1 cot opposite x cos
sec
hypotenuse 1 x2 1 x2 adjacent 1
csc
hypotenuse 1 x2 x opposite
773 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
83. a.
Since OA OC 1 , OAC is isosceles. Thus, OAC OCA. . Now OAC OCA AOC 180º OAC OCA (180º ) 180º OAC OCA
Area of OBD
c.
Area OBD Area OBC Area OBC 1 1 sin sin 2 2 2cos sin sin sin sin sin cos 1 1 sin cos
2 OAC OAC
b.
sin cos
c.
tan
2
CD OC OD OC
CD
2
CD
1 OD
1
OD
CD CD CD sin AD AO OD 1 OD 1 cos
84. Let h be the height of the triangle and b be the base of the triangle. h sin , so h a sin a 1 b cos 2 , so b 2a cos a 1 1 A bh (2a cos )(a sin ) a 2 sin cos 2 2
1 1 tan 2 1 tan 2 sin 2 cos
b.
1 OC AC 2 AC 1 OC 2 1 1 1 cos sin 2 1 sin cos 2
87. a.
Area OAC
b.
Area OCB
h x tan x h h (1 x) (1 x) tan n 1 x x tan 1 x tan n
1 OC BC 2 OC BC 1 2 OB OB OB 2 1 2 OB cos sin 2 1 2 OB sin cos 2
85. h x
x tan tan n x tan n
c.
1 BD OA 2 1 BD 1 2 BD 1 OB OB 2
Area OAB
x tan x tan n tan n x tan tan n tan n x
tan n
tan tan n
86. a.
Area of OBC
1 1 1 sin sin 2 2
d.
1 OB sin( ) 2
OC OA OC OB cos OB OC OC cos 1 OB
774 Copyright © 2025 Pearson Education, Inc.
Section 7.2: Right Triangle Trigonometry e.
Area OAB Area OAC Area OCB 1 OB sin( ) 2 1 1 2 sin cos OB sin cos 2 2 cos sin( ) cos
sin cos cos tan cos
89. sin
cos cos cos tan cos
cos 2 sin cos cos 2 cos cos sin( ) sin cos sin cos cos cos sin( ) sin cos cos sin sin cos
88. Let x be the distance from O to the first circle. a and From the diagram, we have sin xa b sin . x 2a b a b Therefore, x a x 2a b xb ab xa 2a 2 ab xb xa 2a 2 x(b a ) 2a 2 2a 2 ba a Then we write, sin xa a a 2 2a 2 2a ab a 2 a ba ba a a (b a ) a(b a) 2 a ab a 2 ab a(b a ) ba ba ba
x
ba cos 1 sin 1 ba
2
b 2 2ab a 2 1 2 b 2ab a 2
sin cos
sin sin 2 cos 2 1 sin 2 tan 2 1 sin 2
sin 2 1 cos 2
sin 2 1 1 sin 2 sin 2 2 1 sin 2 sin 2 1 1 sin 1 sin 2 sin 2
sin 2 sin 4 sin 2 1 sin 2 sin 4 3sin 2 1 0 Using the quadratic formula: 3 5 sin 2 2 3 5 2
sin
But
3 5 3 5 . 1 . So, sin 2 2 cos 2 A cos 2 B cos 2 C
90. a.
(cos sin C ) 2 (sin sin C ) 2 cos 2 C cos 2 sin 2 C sin 2 sin 2 C cos 2 C sin 2 C (cos 2 sin 2 ) cos 2 C sin 2 C (1) cos 2 C sin 2 C cos 2 C 1 sin 2 A sin 2 B sin 2 C
b.
1 cos 2 A 1 cos 2 B 1 cos 2 C 3 (cos 2 A cos 2 B cos 2 C ) 3 1 2
b 2 2ab a 2 b 2 2ab a 2 b 2 2ab a 2 ab 4ab 2 ab ab ab ( a b) 2 2
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Chapter 7: Trigonometric Functions
91. cos 2 10 cos 2 20 cos 2 30 cos 2 40 cos 2 50 cos 2 60 cos 2 70 cos 2 80 sin 80 sin 2 70 sin 2 60 sin 2 50 cos 2 50 2
cos 2 60 cos 2 70 cos 2 80 sin 2 80 cos 2 80 sin 2 70 cos 2 70 sin 2 60 c os 2 60 sin 2 50 cos 2 50 1111 4
c
a
95. The argument for the ln function must be a positive number : 5x 2 0 5 x 2 x
2 5
2 2 So the domain is: x | x or , 5 5
92. Consider the right triangle:
b
94. Answers will vary.
If is an acute angle in this triangle, then b a 0, b 0 and c 0 . So sin 0 . c Also, since a 2 b 2 c 2 , we know that: 0 b2 c 2 0bc b Thus, 0 1 c Therefore, 0 sin 1 .
96. Since 4 3i is a zero, its conjugate 4 3i is also a zero of h . x (4 3i ) and x (4 3i ) are factors of h . Thus, ( x (4 3i ))( x (4 3i )) (( x 4) 3i )(( x 4) 3i )
x 2 8 x 16 9i 2 x 2 8 x 25
is a factor of h . Using division to find the other factor: x 2 3 x 10 x 2 8 x 25 x 4 5 x3 9 x 2 155 x 250 x 4 8 x3 25 x 2 3 x3 34 x 2 155 x
93. Consider the right triangle:
3 x3 24 x 2 75 x c
b a
10 x 2 80 x 250 10 x 2 80 x 250
If is an acute angle in this triangle, then: a a 0, b 0 and c 0 . So cos 0 . c 2 2 2 Also, since a b c , we know that: 0 a2 c2 0ac a Thus, 0 1 . c So we now know that 0 cos 1 which 1 1 implies that: cos 1 sec 1
x 2 3 x 10 ( x 2)( x 5) The remaining zeros are 2 and 5 . The zeros of h are 4 3i, 4 3i, 2, 5 .
97. Using the Remainder Theorem: P( 2) 8( 2) 4 2( 2)3 ( 2) 8 8(16) 2( 8) 2 8 128 16 2 8 134 98. The outer circle would have area: A r2 (15) 2 225 ft 2 The garden has area:
776 Copyright © 2025 Pearson Education, Inc.
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles A r2
2
(12) 2 144 ft 2 So the sidewalk has area: 225 144 81 ft 2 .
2
x2 1 x4 4 2 2 4 x 4x x4 4 2 1 g ( x) 4 x2
(7) (7) 2 4(3)(9) 99. x 2(3)
100.
x4 1 1 16 2 x 4 x2 1 x2 1 2 2 4 x 4 x
1 g ( x)
7 157 6
1 since 2x 1 1 2x2 1 f ( g ( x)) 2 2 1 2 2 x2 1 x 1 f ( x)
x2 3 x 3
101.
Section 7.3 1. elevation 2. depression
x x6 0 2
3. True
( x 3)( x 2) 0 x 3, x 2 The graphs intersect at -3 and 2.
4. False 5. b 6. c 2 2 2 cos 45 2 tan 45 1
7. sin 45
g ( x) f ( x) at 3, 2 .
102. int( x 3) 2 int( x ) 5
8. sin 30
3 2 3 tan 30 3
The interval is 5, 4 .
cos 30
103. The x-value is moved to the right by 3 units and the y-value is shrunk by a value of ½. So the new point is (6, 2) . 104.
csc 30 2
x2 1 x2 1 2 2 4 x 4 x
sec 30
x4 1 1 1 x4 1 1 16 4 4 x 4 16 2 x 4
cot 30 3
g ( x) 2
1 2
9.
csc 45 2 sec 45 2 cot 45 1 3 2 1 cos 60 2
sin 60
tan 60 3 csc 60
2 3 3
f 60o sin 60o
777 Copyright © 2025 Pearson Education, Inc.
2 3 3
sec 60 2 cot 60 3 2
3 3
2
Chapter 7: Trigonometric Functions
10. g 60o cos 60o
11.
1 2
24. tan
60o 1 o o f f 30 sin 30 2 2
2
14. g 60o cos 60o
3 3 4 2 2
12 14 2
4
11 2 2
26. 4 tan 2
2
2
cot
2 3 12 4 11 25. csc 5 5 5 5 3 9 3 3 3
2
4 2
60o 3 o o 12. g g 30 cos 30 2 2
13. f 60o sin 60o
3
4
3 4 3 7 2
27. 2
o
2
o
28. sec 2 60o tan 2 45o 2 1 4 1 3 2
15. 2 f 60o 2sin 60o 2
3 3 2
2
17.
1 1 2
3
sin 60 2 3 1 3
f 60o
2
o
2
2
2 2
4
18.
1
cos 60 2 1 1 1 2
o
2
2
2 2
4
2 2 2 2 2 2 2 2
19. 4 cos 45º 2sin 45º 4
o
2
o
2
3 30. 1 tan 30 csc 45 1 3 3 1 2 9 2 3 2
g 60o
2
3 1 2 29. 1 cos 30 cos 60 1 2 2 3 1 1 4 4 0 2
16. 2 g 60o 2 cos 60o 2
2
3 3 3 3 3 sin 60 cos 30 2 2 4 4 2 2
o
2
o
2
2
31. Set the calculator to degree mode: sin 28º 0.47 .
2
20. 2sin 45º 4 cos 30º 2
2 4 3 22 3 2 2
1 21. 6 tan 45º 8cos 60º 6 1 8 6 4 2 2
22. sin 30º tan 60º
23. sec
6
2 csc
4
32. Set the calculator to degree mode: cos14º 0.97 .
1 3 3 2 2 2 3 2 2 3
778 Copyright © 2025 Pearson Education, Inc.
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles 33. Set the calculator to degree mode: tan 36º 0.73 .
34. Set the calculator to degree mode: 1 cot 70º 0.36 . tan 70º
40. Set the calculator to radian mode: 1 5.67. cot 18 tan 18
35. Set the calculator to degree mode: 1 sec 53º 1.66 . cos 53º
41. Set the calculator to radian mode: 1 sec 1.02 18 cos 18
36. Set the calculator to degree mode: 1 csc55º 1.22 . sin 55º
37. Set the calculator to radian mode: sin
39. Set the calculator to radian mode: 8 tan 9.51 . 15
42. Set the calculator to radian mode: 5 1 csc 1.07. 13 sin 5 13
0.43 . 7 43. Set the calculator to radian mode: sin 4 0.76 .
38. Set the calculator to radian mode: cos
0.92 . 8 44. Set the calculator to radian mode: tan 1 1.56 .
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Chapter 7: Trigonometric Functions 45. Set the calculator to degree mode: sin 4º 0.07 .
53. ( f h) f h 6 6 3 sin 2 sin 3 2 6
46. Set the calculator to degree mode: tan1º 0.02 .
54. ( g p )(60) g p (60) 3 60 cos cos 30 2 2
55. ( p g )(45) p g (45) cos 45 2 1 1 2 2 cos 45 2 2 2 4
47. Set the calculator to radian mode: tan 0.9 1.26 .
48. Set the calculator to radian mode: tan 0.1 0.10 .
56. (h f ) h f 6 6 1 2 sin 2 1 2 6
57. a. 49. ( f g )(30) f (30) g (30) sin 30 cos 30
1 3 1 3 2 2 2
2 , is on the graph of f 1 . b. The point 2 4
50. ( f g )(60) f (60) g (60) sin 60 cos 60
c.
3 1 3 1 2 2 2
2 The point , 2 is on the graph of 4 y f x 3 . 4
2 2 4 2 1 2 2 4 4 2
52. ( f g ) f g 3 3 3 sin cos 3 3 3 1 3 2 2 4
f 3 f 3 4 4 2 sin 3 2 1 3
51. ( f g ) f g 4 4 4 sin cos 4 4
2 f sin 2 4 4 2 The point , is on the graph of f. 4 2
58. a.
3 g cos 6 6 2 3 The point , is on the graph of g. 6 2
3 , is on the graph of g 1 . b. The point 2 6
780 Copyright © 2025 Pearson Education, Inc.
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles
c.
Case 2: 25º , b 5
2 g 2 g (0) 6 6 2 cos(0)
c
2 1
25º
2 Thus, the point , 2 is on the graph of 6 y 2g x . 6
cos 25º c
5 c
5 5 5.52 in. cos 25º 0.9063
b. There are two possible cases because the given side could be adjacent or opposite the given angle.
59. c 8, 35º
a
a
62. a.
35º
Case 1:
, a3 8
c
b
a 8 a 8sin 35º
b 8 b 8cos 35º
8(0.5736) 4.59 in.
8(0.8192) 6.55 in.
sin 35º
cos 35º
_ 8
b
3 sin 8 c 3 3 c 7.84 m. 0.3827 sin 8
60. c 10, 40º
a
Case 2:
, b3 8
c
a
40º
b
a 10 a 10sin 40º
b 10 b 10 cos 40º
10(0.6428)
10(0.7660)
6.43 cm.
7.66 cm.
cos 40º
sin 40º
61. a.
_ 8
Case 1: 25º , a 5 c
b. There are two possible cases because the given side could be adjacent or opposite the given angle.
25º
b
sin 25º c
3 cos 8 c 3 3 c 3.25 m. 0.9239 cos 8
5 c
63. Use the formula R
5 5 11.83 in. sin 25º 0.4226
2v0 2 sin cos with g
g 32.2ft/sec 2 ; 45º ; v0 100 ft/sec :
2 100 sin 45º cos 45º 2
R
781 Copyright © 2025 Pearson Education, Inc.
32.2
310.56 feet
Chapter 7: Trigonometric Functions
Use the formula H
v0 2 sin 2 with 2g
g 32.2ft/sec 2 ; 45º ; v0 100 ft/sec : H
1002 sin 2 45º 77.64 feet 2(32.2)
64. Use the formula R
2v0 2 sin cos with g
g 9.8 m/sec 2 ; 30º ; v0 150 m/sec : 2 150 sin 30º cos 30º 2
R
g
Use the formula H
H
v0 sin with 2g 2
2
2
150 sin 30º 22,500(0.5) 286.99 m 2(9.8) 19.6
2v sin cos 65. Use the formula R 0 with g g 9.8 m/sec ; 25º ; v0 500 m/sec : 2
2 500 sin 25º cos 25º 2
g
19,541.95 m
v 2 sin 2 Use the formula H 0 with 2g g 9.8 m/sec 2 ; 25º ; v0 500 m/sec : H
5002 sin 2 25º 2278.14 m 2(9.8)
66. Use the formula R
a.
t
b.
t
c.
t
2 10 32sin 30º cos 30º 2 10 32sin 45º cos 45º 2 10 32sin 60º cos 60º
2v0 2 sin cos with g
g 32.2ft/sec 2 ; 50º ; v0 200 ft/sec :
2 200 sin 50º cos 50º
x cos
g
1.20 seconds
When 30º :
g 32.2ft/sec 2 ; 50º ; v0 200 ft/sec : 2002 sin 2 50º 364.49 ft 2(32.2)
x cos 30º 16 0.5 2 cos 2 30º 1 4.897 in.
When 45º : x cos 45º 16 0.5 2 cos 2 45º 1 4.707 in.
69. tan 35º
AC
70. tan 40º
AC
100 AC 100 tan 35º 100(0.7002) 70.02 feet
100 AC 100 tan 40º 100(0.8391) 83.91 feet
71. Let x = the height of the Eiffel Tower. x tan 78.232º 205 x 205 tan 78.232º 205(4.8001) 984.02 feet 72. Let x = the distance to the shore. 100 ft
1223.36 ft
v 2 sin 2 Use the formula H 0 with 2g
H
1.12 seconds
16 0.5(2 cos 2 1) .
2
R
1.20 seconds
68. Use the formula 2
2
R
g 32 ft/sec 2 and a 10 feet :
1988.32 m
g 9.8 m/sec 2 ; 30º ; v0 150 m/sec : 2
2a with g sin cos
67. Use the formula t
30º
x
tan 30º x
100 x
100 100 300 100 3 173.21 feet tan 30º 3 3 3
782 Copyright © 2025 Pearson Education, Inc.
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles 73. Let x = the distance to the base of the plateau.
tan 54º 50 m
x
60º x
50 50 50 3 28.87 meters tan 60º 3 3
74. Let x = the distance up the building feet 70º
x sin 70º 22 x 22sin 70º 22(0.9397) 20.67 feet
75. We construct the figure below:
32º
h h 100 tan 61º tan 54º
h
tan(61) tan(54)
h 100 tan 61
h
100 tan 61
580.61 tan 61 1 tan 54 Thus, the height of the balloon is approximately 580.61 feet.
77. Let h represent the height of Lincoln's face.
23º
ft y
x
tan 32º x
h x 100 h ( x 100) tan 61º
tan(61) h 1 100 tan 61 tan(54)
x
32º
h tan 54º
tan 61º
50 tan 60º x x
h x
500 x
500 tan 32º
h 23º
b
tan 23º y
500 tan 23º
Distance = x y 500 500 tan 32º tan 23º
32º
feet
b 800 b 800 tan 32º 499.90 tan 32º
bh 800 b h 800 tan 35º 560.17 tan 35º
1978.09 feet
76. Let h = the height of the balloon. 54º
500 y
35º
Thus, the height of Lincoln’s face is: h (b h) b 560.17 499.90 60.27 feet 78. Let h represent the height of tower above the Sky Pod.
61º
h h
h
b
61º 54º
ft
x
20.1º
x
783 Copyright © 2025 Pearson Education, Inc.
24.4º
feet
Chapter 7: Trigonometric Functions
b 4000 b 4000 tan 20.1º 1463.79
sin 50º
tan 20.1º
z
1 1.3054 mi sin 50º
bh 4000 b h 4000 tan 24.4º 1814.48
tan 40º
Thus, the height of tower above the Sky Pod is: h (b h) b 1814.48 1463.79 350.69 feet
a
tan 24.4º
79. Let x = the length of the guy wire. 10 ft
x 45º
x
1 b
1 0.8391 mi tan 50º
a yb 3 y 3 a b 3 1.1918 0.8391 0.9691 mi The length of the highway is about: 1.5557 0.9691 1.3054 3.83 miles .
190 ft
sin 45º
1 a
1 1.1918 mi tan 40º
tan 50º b
1 z
190 x
190 190 380 190 2 268.70 ft sin 45º 2 2 2
82. Let x = the distance from George at which the camera must be set in order to see his head and feet. x
80. Let h = the height of the monument.
20º
ft
h
tan 20º
35.1º
ft
h tan 35.1º 789 h 789 tan 35.1º 789(0.7028) 554.52 ft
81. Let x, y, and z = the three segments of the highway around the bay (see figure). y
x
4 x
4 10.99 feet tan 20º
If the camera is set at a distance of 10 feet from George, his feet will not be seen by the lens. The camera would need to be moved back about 1 additional foot (11 feet total). 83. a.
We label the diagram as follows: 8 mi
z
mi
mi
x 140º 40º
50º 130º
a
mi
b x
The length of the highway x y z sin 40º x
1 x
1 1.5557 mi sin 40º
s
1 mi
s 8 2x
1 mi x
1 1 and sin , so x s 1 1 x and s . Also, note that tan sin distance distance rate time , so time . rate
Note that tan
784 Copyright © 2025 Pearson Education, Inc.
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles
Then, distance on sand time on sand rate on sand 1 2 2s 2 sin 3 3 3sin distance on road and time on road rate on road x 8 2x 1 8 4 1 tan 1 1 1 4 4 tan So, total time time on sand time on road
d.
e.
2 1 1 3sin 4 tan 2 1 1 3sin 4 tan
T
b.
2 1 3sin 30º 4 tan 30º 2 1 1 1 1 3 4 2 3
T (30º ) 1
4 3 1.9 hr 3 4 Sally is on the paved road for 1 1 0.57 hr . 4 tan 30º
f.
1
c.
2 1 3sin 60o 4 tan 60o 2 1 1 3 4 3 3 2 4 1 1 1.63 hr 3 3 4 3 Sally is on the paved road for 1 1 0.86 hr . 4 tan 60o T (60º ) 1
2 1 . 3sin 90º 4 tan 90º But tan 90º is undefined, so we can’t use the function formula for this path. However, the distance would be 2 miles in the sand and 8 miles on the road. The total 2 5 time would be: 1 1.67 hours. The 3 3 path would be to leave the first house walking 1 mile in the sand straight to the road. Then turn and walk 8 miles on the road. Finally, turn and walk 1 mile in the sand to the second house. T (90º ) 1
1 1 1 , so x 4 . Thus, tan 1/ 4 4 the Pythagorean Theorem yields: s 2 x 2 12 tan
s x 2 1 42 1 17 Total time time on sand time on road
2 1 3sin 45º 4 tan 45º 2 1 1 1 1 4 3 2
2 s 8 2 x 2 17 8 2 4 3 8 3 8 2 17 8 8 2 17 0 3 8 3 2 17 2.75 hrs 3 The path would be to leave the first house and walk in the sand directly to the bridge. Then cross the bridge (approximately 0 miles on the road), and then walk in the sand directly to the second house.
T (45º ) 1
T
2 2 1 1.69 hr 3 4 Sally is on the paved road for 1 1 0.75 hr . 4 tan 45o 1
g.
Let Y1 1
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2 1 3sin x 4 tan x
Chapter 7: Trigonometric Functions
Then r h cot R R cos cos cos sin R R cos sin 1 Thus, V r 2 h 3
0
90 Use the MINIMUM feature:
2
1 R R cos R R cos 3 sin cos
0
90 The time is least when 67.98º . The least time is approximately 1.62 hour. Sally’s time on the paved road is 1 1 1 1 0.90 hour. 4 tan 4 tan 67.98o
84. a.
R R cos 3sin 2 cos
3
b. When 30º : V 30º
2 2 cos 30º
3
3sin 2 30º cos 30º
251.4 cm3
When 45º : V 45º
We label the diagram as follows:
2 2 cos 45º
3
3sin 2 45º cos 45º
117.9 cm3
When 60º : V 60º
c.
Let Y1
2 2 cos 60º
3
3sin 2 60º cos 60º
2 2 cos x
75.4 cm3
3
3(sin x) 2 cos x
.
h h , so r h cot . r tan Consider the smaller triangle in the figure. R From this, sin 90o . Since hR
Note that tan
sin 90 cos , we have that: o
R hR R hR cos R R R cos h R cos cos cos
0
90
Using a slant angle of approximately 70.5o will yield the minimum volume 67.0 cm3 . 85. Refer to the diagram below
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Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles
If we let x = length shown in the figure, we see 3 that, tan and x 3 tan 52 x 3 x 2.34 ft. tan 52 Now we add 1 foot. 1 2.34 3.34 ft The player should hit the top cushion at a point that is 3.34 feet from upper left corner. 86. a.
x x 15 tan . And by the 15 Pythagorean Theorem: tan
L2 2 y 2 102 L2 100 y 2 Using this information we have: y 2 (30 x) 2 900 60 x x 2
L2 100 (900 60 x x 2 )
The distance between the buildings is the length of the side adjacent to the angle of elevation in a right triangle.
1000 60 x x 2 Since x 15 tan then: L2 1000 60(15 tan ) (15 tan ) 2 ) 1000 900 tan 225 tan 2 25(40 36 tan 9 tan 2 )
1776
34
5 40 36 tan 9 tan 2
x
opposite and we know the adjacent angle measure, we can use the tangent to find the distance. Let x = the distance between the buildings. This gives us 1776 tan 34 x 1776 x tan 34 x 2633 The office building is about 2633 feet from the base of the tower.
Since tan
b. Let y = the difference in height between 1 WTC and the office building. Together with the result from part (a), we get the following diagram
Since L L1 L2 then L L1 L2
15 5 40 36 tan 9 tan 2 cos
15 b. L(45) cos(45) 5 40 36 tan(45) 9 tan 2 (45)
21.213 5 40 36 9 21.213 18.028 39.2 ft. x 15 tan 15 tan(45) 15 ft.
c.
y
20 2633
opposite tan adjacent y tan 20 2633 y 958 The 1 WTC is about 958 feet taller than the office building. Therefore, the office building is 1776 958 818 feet tall.
87. a.
The minimum occurs when 50.2 . d. Using the minimum function we find that the minimum length for the wire is 39.05 feet and occurs when 50.2 . Thus x 15 tan(50.2) 18 so it is anchored 18 feet from the 15-ft pole.
Let L L1 L2 and y 30 x . We have L1
15 . Also we have cos 787 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
cos 1
91. cos1o cos 2o ... cos 45o csc 46o ... csc89o
cos1o csc89o cos 2o csc88o ...
cos 1
0.5
0.1224
0.2448
0.4
0.0789
0.1973
0.2
0.0199
0.0997
0.1
0.0050
0.0050
0.01
0.00005
0.0050
0.001
0.0000
0.0005
0.0001
0.0000
0.00005
0.00001
0.0000
0.000005
88.
cos 1
0.5
0.9589
0.4
0.3894
0.9735
0.2
0.1987
0.9933
0.1
0.0998
0.9983
0.01
0.0100
1.0000
0.001
0.0010
1.0000
0.0001
0.0001
1.0000
1 1 1
o
o
2 2 2 2
o
cot1 cot 89 cot1 cot 90 1 cot1 tan1 o
o
o
o
o
o
o
1
cot 2 cot 88 cot 2 cot 90 2 cot 2 tan 2 o
o
o
o
o
o
o
1 and so on. This result holds for each pair in our product. Since we know that cot 45o 1 , our product can be rewritten as: 1 1 1 ... 1 1 . Therefore, cot1o cot 2o cot 3o ... cot 89o 1 .
93. tan1o tan 2o tan 3o ... tan 89o tan1o tan 89o tan 2o tan 88o ...
o
1 1 1
sin1 csc1 sin 2 csc 2 sin 44 csc 44 sin 45 o
o
Now each set of parentheses contains a pair of complementary angles. For example, using cofunction properties, we have:
o
sin1 csc1
cot 44 cot 46o cot 45o
2 2 2 2
sin 44 sec 46o sin 45o o
o
cot1o cot 89o cot 2o cot 88o ...
sin1o sec89o sin 2o sec88o ...
o
o
92. We can rearrange the order of the terms in this product as follows: cot1o cot 2o cot 3o ... cot 89o
90. sin1o sin 2o ... sin 45o sec 46o ... sec89o
o
approaches 1 as approaches 0.
o
cos 44 sec 44o cos 45o
0.00001 0.00001 1.0000
sin
o
cos1 sec1 cos 2 sec 2 ...
0.4794
sin
sin
cos 44 csc 46o cos 45o
approaches 0 as approaches 0.
89.
tan 44o tan 46o tan 45o
tan 44 cot 44 tan 45
tan1 cot1 tan 2 cot 2 o
o
o
o
1 1 1 1 1
94 – 96. Answers will vary.
788 Copyright © 2025 Pearson Education, Inc.
o
o
Section 7.3: Computing the Values of Trigonometric Functions of Acute Angles
97. x 2 5 ( x 3)( x 3) 4 x 5 x 9 4 2
2
5 5 This is true for all real numbers so the solution set is all real numbers.
98. We need to use completing the square to put the function in the form f ( x ) a ( x h) 2 k f ( x) 2 x 2 12 x 13 144 144 2 x 2 6 x 13 2 2 4( 2) 4( 2) 2 2( x 2 6 x 9) 13 18
So the vertex is (3,5) and the axis of symmetry is x 3. ex4 6 ln e x 4 ln 6
2
( x 4) ln 6 x ln 6 4 So the solution set is: ln 6 4 . f ( x) x3 9 x 2 3x 27 ( x 9)( x 2 3)
The factor ( x 2 3) has no real roots so the solution set is: 9 . x2 x5 2 x 2 x 5 2 x 2 x 5 4 x 5 4 3 4 x 5
b 28 7 2a 12 3 7 when: The y-value is 3
104. The x-value is
7 7 7 6 28 c 3 3 3
( x 4) ln e ln 6
2 4 x 5 1
42 x 5 x 21 The solution set is 21
2( x 3) 2 5
101.
103. log 4 ( x 5) 2 16 x 5
2( x 2 6 x 9) 5
100.
89 The solution set is 16
102. (x + 6) moves the graph 6 units to the left so the zeros are 8 and 3 .
2( x 2 6 x) 13
99.
3 x5 4 9 x5 16 89 x 16
7 49 196 6 c 3 9 3 7 294 196 c 3 9 3 105 c 35 3
105. 3(0) 5 y 15 5 y 15 y3 3 x 5(0) 15 3 x 15 x 5
The intercepts are 0,3 , 5, 0 .
789 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
106.
3 2 x 5x 1 2 f x h f x
f x
h 2 3 3 2 2 x h 5 x h 1 2 x 5 x 1 h 3 2 3 x 2 xh h 2 5 x 5h 1 x 2 5 x 1 2 2 h 3 2 3 3 3 x 3 xh h 2 5h x 2 3 xh h 2 5h 2 2 2 2 h h 3 3x h 5 2
b 4 r 5 b 4 4 tan = a 3 3 r 5 5 sec = a 3 3 sin
12.
a 3 3 r 5 5 a 3 3 cot b 4 4 r 5 csc b 4
cos
5, 12 : a 5, b 12 r a 2 b 2 52 122 25 144 169 13
Section 7.4 1. tangent, cotangent 2. coterminal b 12 12 = r 13 3 b 12 12 tan = a 5 5 r 13 sec a 5
3. x-axis
sin
4. False: sin182º cos 92º 5. True 6. True 7. 360º ; 2 8. b 9.
2
and
3 2
10. d 11.
3, 4 : a 3, b 4 r a 2 b2
3 42 9 16 2
25 5
790 Copyright © 2025 Pearson Education, Inc.
a 5 r 13 a 5 5 cot b 12 12 r 13 13 csc b 12 12
cos
Section 7.4: Trigonometric Functions of Any Angle
13.
2, 3 : a 2, b 3
15.
r a 2 b 2 22 3 4 9 13 2
sin
r a 2 b2
b 3 13 3 13 = r 13 13 13
sin
a 2 13 2 13 cos = r 13 13 13 b 3 a 2 3 2 tan = cot a 2 b 3 2 3 r r 13 13 13 sec csc a b 3 2 3
14.
1, 2 : a 1, b 2 r a b 2
sin
2
2
b 2 5 2 5 = r 5 5 5
2
2 2
=
2
2 2
a 3 2 2 = r 3 2 2 2 b 3 a 3 tan =1 cot 1 a 3 b 3 r 3 2 r 3 2 sec = 2 csc = 2 a b 3 3
2, 2 : a 2, b 2 r a 2 b 2 22 2 8 2 2 2
sin
a 1 5 5 cos = r 5 5 5 b 2 a 1 1 tan =2 cot a 1 b 2 2 r r 5 5 5 sec = 5 csc = a 1 b 2 2
b 3 r 3 2
3 3 18 3 2
cos
16.
1 2 1 4 5 2
3, 3 : a 3, b 3
b 2 r 2 2
2 2
=
2 2
a 2 2 2 = r 2 2 2 2 b 2 a 2 tan = 1 cot 1 a 2 b 2 r 2 2 r 2 2 sec = 2 csc = 2 a b 2 2 cos
791 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
3 1 3 1 , : a , b 2 2 2 2
17.
2
3 1 2 3 1 r a 2 b 2 1 1 4 4 2 2
1 b 2 1 sin = r 1 2
csc
r 1 2 b 1 2
3 a 3 cos 2 r 1 2 1 b 1 3 3 tan 2 = = a 3 3 3 3 2 r 1 2 3 2 3 sec = = a 3 3 3 3 2 3 a 2 cot = 3 1 b 2
2 2 2 2 , , b 19. : a 2 2 2 2 2
2
2 1 3 1 3 r a b 1 1 4 4 2 2 2
2
2 2 r 2 2
1 3 1 3 18. , : a , b 2 2 2 2 2
3 3 b 2 sin 1 2 r 1 1 a 2 cos = 1 2 r 3 b 2 tan = 3 a 1 2 1 2 3 2 3 r csc = = 3 b 3 3 3 2 1 r 2 sec a 1 2 1 3 a 2 1 3 cot = = 3 b 3 3 3 2
2 2 1 1 4 4
2 2 2 2 b 2 a 2 sin cos = 1 2 1 2 r r 2 2 b 2 a 2 tan = 1 cot = 1 a b 2 2 2 2 1 2 2 r sec = 2 a 2 2 2 2 1 2 2 r csc = 2 b 2 2 2 2
792 Copyright © 2025 Pearson Education, Inc.
Section 7.4: Trigonometric Functions of Any Angle 25. csc 315º csc(360º (315)º ) csc 45º 2
2 2 2 2 , , b 20. : a 2 2 2 2 2
2
2 2 r 2 2
26. sec 540º sec(360º 180º ) sec180º 1
2 2 1 1 4 4
27. cot 405º cot(405º 360º ) cot 45º 1 28. sec 420º sec(360º 60º ) sec 60º 2
2 2 2 1 2 2 2 a 2 cos = 1 2 r 2 b 2 tan =1 a 2 2 r 1 2 csc b 2 2 2 r 1 2 sec a 2 2 2 2 a 2 cot =1 b 2 2 b sin r
23 23 2 2 29. cos cos 6 6 23 24 cos 6 6 cos 6 3 2
30. sin
2 2 2 2
9 8 sin 4 4 4 sin 2 4 sin 4 2 2
= 2
31. tan 19 tan(19 20) tan 0 = 2
32. csc
21. sin 405º sin(360º 45º ) sin 45º
2 2
22. cos 420º cos(360º 60º ) cos 60º
1 2
23. tan 390º tan(360º 30º ) tan 30º
3 3
24. sin 390º sin(360º 30º ) sin 30º
1 2
9 8 csc 2 2 2 csc 4 2 csc 2 2 2 csc 2 1
33. Since sin 0 for points in quadrants I and II, and cos 0 for points in quadrants II and III, the angle lies in quadrant II. 34. Since sin 0 for points in quadrants III and IV, and cos 0 for points in quadrants I and IV, the angle lies in quadrant IV.
793 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 35. Since sin 0 for points in quadrants III and IV, and tan 0 for points in quadrants II and IV, the angle lies in quadrant IV. 36. Since cos 0 for points in quadrants I and IV, and tan 0 for points in quadrants I and III, the angle lies in quadrant I. 37. Since cos 0 for points in quadrants I and IV, and cot 0 for points in quadrants II and IV, the angle lies in quadrant IV. 38. Since sin 0 for points in quadrants III and IV, and cot 0 for points in quadrants I and III, the angle lies in quadrant III. 39. Since sec 0 for points in quadrants II and III, and tan 0 for points in quadrants I and III, the angle lies in quadrant III. 40. Since csc 0 for points in quadrants I and II, and cot 0 for points in quadrants II and IV, the angle lies in quadrant II.
5 is in quadrant II, so the reference angle 6 5 . is 6 6
48.
8 is in quadrant II. Note that 3 8 2 2 , so the reference angle is 3 3 2 . 3 3
49.
7 is in quadrant IV, so the reference angle 4 7 . is 2 4 4
50.
51. 165o is in quadrant III. Note that 165o 360o 195o , so the reference angle is 195o 180o 15o .
41. 30o is in quadrant IV, so the reference angle is 30o .
52. 240o is in quadrant II. Note that 240o 360o 120o , so the reference angle is 180o 120o 60o .
42. 60o is in quadrant IV, so the reference angle is 60o .
53.
43. 120o is in quadrant II, so the reference angle is 180o 120o 60o . 44. 210o is in quadrant III, so the reference angle is 210o 180o 30o . 45. 320o is in quadrant IV, so the reference angle is 360o 320o 40o . 46. 330o is in quadrant IV, so the reference angle is 360o 330o 30o .
5 47. is in quadrant II, so the reference angle 8 5 3 . is 8 8
5 is in quadrant III. Note that 7 5 9 2 , so the reference angle is 7 7 9 2 . 7 7
7 is in quadrant II. Note that 6 7 5 2 , so the reference angle is 6 6 5 . 6 6
54.
55. 440o is in quadrant I. Note that 440o 360o 80o , so the reference angle is 80o . 56. 490o is in quadrant II. Note that 490o 360o 130o , so the reference angle is 180o 130o 50o .
794 Copyright © 2025 Pearson Education, Inc.
Section 7.4: Trigonometric Functions of Any Angle 15 is in quadrant IV. Note that 4 15 7 2 , so the reference angle is 4 4 7 2 . 4 4
57.
19 is in quadrant III. Note that 6 19 7 2 , so the reference angle is 6 6 7 . 6 6
58.
1 59. sin150o sin 30o , since 150o has 2 reference angle 30o in quadrant II. 3 , since 210o has 2 reference angle 30o in quadrant III.
60. cos 210o cos 30o
1 , since 510o has 2 reference angel 30o in quadrant II.
61. sin 510o sin 30o
1 , since 600o has 2 reference angel 60o in quadrant III.
62. cos 600o cos 60o
2 , since 45o has 2 reference angel 45o in quadrant IV.
63. cos 45o sin 45o
3 64. sin 240 sin 60 , since 240o 2 has reference angel 60o in quadrant II.
o
o
65. sec 240o sec 60o 2 , since 240o has reference angle 60o in quadrant III. 2 3 66. csc 300 csc 60 , since 300o 3 o
o
68. tan 225o tan 45o 1 , since 225o has reference angle 45o in quadrant III. 69. sin
3 2 3 sin , since has 4 4 2 4
reference angle 70. cos
in quadrant II.
3
in quadrant II.
13 2 13 cos , since has 4 4 2 4
reference angle 72. tan
4
2 1 2 cos , since has 3 3 2 3
reference angle
71. cos
4
in quadrant III.
8 8 tan 3 , since has 3 3 3
reference angle
3
in quadrant II.
2 3 2 73. sin , since sin 3 2 3 3
has reference angle
3
in quadrant III.
74. cot cot 3 , since has 6 6 6 reference angle
75. tan
6
in quadrant IV.
14 14 tan 3 , since has 3 3 3
reference angle
76. sec
3
in quadrant II.
11 11 sec 2 , since has 4 4 4
reference angle
4
in quadrant II.
has reference angle 60o in quadrant IV.
77. sin 8 sin(0 8) sin 0 0
67. cot 330o cot 30o 3 , since 330o has reference angle 30o in quadrant IV.
78. cos 2 cos(0 2) cos 0 1 79. tan 7 tan( 6) tan 0
795 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 80. cot(5) cot( 4) cot( ) , which is undefined 81. sec(3) sec( 4) sec( ) 1 5 3 4 1 82. csc csc 2 2 12 , in quadrant II 13 Since is in quadrant II, sin 0 and csc 0 , while cos 0 , sec 0 , tan 0 , and cot 0 . If is the reference angle for , then 12 . Now draw the appropriate triangle sin 13 and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
83. sin
4 4 5 tan sec 5 3 3 5 3 csc cot 4 4 Finally, assign the appropriate signs to the values of the other trigonometric functions of . 4 4 5 sin tan sec 5 3 3 5 3 csc cot 4 4 sin
4 85. cos , in quadrant III 5 Since is in quadrant III, cos 0, sec 0, sin 0 and csc 0, while tan 0 and cot 0.
If is the reference angle for , then cos 5 12 13 tan sec 13 5 5 13 5 csc cot 12 12 Finally, assign the appropriate signs to the values of the other trigonometric functions of . 5 12 13 cos tan sec 13 5 5 13 5 csc cot 12 12 cos
3 84. cos , in quadrant IV 5 Since is in quadrant IV, cos 0 and sec 0, while sin 0, csc 0, tan 0, and cot 0.
If is the reference angle for , then cos Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
3 . 5
4 . 5
Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
3 5 5 csc 3
sin
3 4 4 cot 3 tan
sec
5 4
Finally, assign the appropriate signs to the values of the other trigonometric functions of .
796 Copyright © 2025 Pearson Education, Inc.
Section 7.4: Trigonometric Functions of Any Angle
3 5 5 csc 3
sin
3 4 4 cot 3 tan
sec
the other trigonometric functions of .
5 4
5 , in quadrant III 13 Since is in quadrant III, cos 0, sec 0, sin 0 and csc 0, while tan 0 and cot 0.
86. sin
If is the reference angle for , then sin
12 13 5 tan csc 13 5 12 13 12 sec cot 12 5 Finally, assign the appropriate signs to the values of the other trigonometric functions of . 12 13 5 tan cos csc 13 5 12 13 12 sec cot 12 5 cos
5 . 13
Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
4 , 270º 360º (quadrant IV) 5 Since is in quadrant IV, sin 0, csc 0, tan 0, cot 0, cos 0, and sec 0.
88. cos 12 13 13 csc 5 cos
5 12 12 cot 5 tan
sec
13 12
If is the reference angle for , then cos
Finally, assign the appropriate signs to the values of the other trigonometric functions of . 12 13 13 sec 12 cos
5 12 12 cot 5 tan
csc
13 5
5 , 90º 180º , so in quadrant II 13 Since is in quadrant II, cos 0, sec 0, tan 0, and cot 0, while sin 0 and csc 0. If is the reference angle for , then 5 sin . Now draw the appropriate triangle and 13 use the Pythagorean Theorem to find the values of
4 . 5
Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
87. sin
3 3 5 tan sec 5 4 4 5 4 csc cot 3 3 Finally, assign the appropriate signs to the values of the other trigonometric functions of . 3 3 5 tan sec sin 5 4 4 5 4 csc cot 3 3 sin
797 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
other trigonometric functions of .
1 89. cos , 180o 270o (quadrant III) 3 Since is in quadrant III, cos 0, sec 0, sin 0, csc 0, tan 0, and cot 0.
If is the reference angle for , then cos
1 . 3
Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of . cos
5 3
tan
2 5
5 5
2 5 5
sec
3
cot
5 2
5
5 5
3 5 5
3 2 Finally, assign the appropriate signs to the values of the other trigonometric functions of . csc
sin
2 2 3
csc
3 2 2
2
3 2 4 2
2 2 1 2 2 tan cot 2 2 1 4 2 2 2 3 sec 3 1 Finally, assign the appropriate signs to the values of the other trigonometric functions of . sin
2 2 3
tan 2 2 sec 3
csc cot
3 2 4
2 4
5 3
2 5 5 3 csc 2 tan
sec cot
3 5 5
5 2
2 91. sin , tan 0 (quadrant II) 3 Since is in quadrant II, cos 0, sec 0, tan 0 and cot 0, while sin 0 and csc 0.
If is the reference angle for , then sin Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
2 90. sin , 180o 270o (quadrant III) 3 Since is in quadrant III, cos 0, sec 0, sin 0, csc 0, tan 0, and cot 0. If is the reference angle for , then sin
cos
2 . 3
Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the
798 Copyright © 2025 Pearson Education, Inc.
2 . 3
Section 7.4: Trigonometric Functions of Any Angle
cos
5 3
tan
2
5
2 5 5
sec
3
cot
5 2
5
5 5
3 5 5
5 5 3 csc 2 Finally, assign the appropriate signs to the values of the other trigonometric functions of . cos
5 3
sec
3 5 5
tan
2 5 5
cot
5 2
csc
3 2
sin
15 4
csc
tan 15
cot
sec 4
4 15 15
15 15
93. sec 2, sin 0 (quadrant IV) Since is in quadrant IV, sin 0, csc 0, tan 0 and cot 0, while cos 0 and sec 0. If is the reference angle for , then sec 2 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
1 92. cos , tan 0 (quadrant III) 4 Since is in quadrant III, cos 0, sec 0, sin 0 and csc 0, while tan 0 and cot 0.
If is the reference angle for , then cos
1 . 4
Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of . 3 3 1 3 2 3 1 3 3 csc cot 3 3 3 3 3 3 Finally, assign the appropriate signs to the values of the other trigonometric functions of . sin
3 2 2
cos
1 2
3 1 cos 2 2 2 3 csc 3
sin
sin
15 4
csc
4
tan
15 15 1
cot
1
15 15
15 15 15 15
4 15 15
15 15
4 4 1 Finally, assign the appropriate signs to the values of the other trigonometric functions of . sec
tan
tan 3 cot
3 3
94. csc 3, cot 0, (quadrant II) Since is in quadrant II, cos 0, sec 0, tan 0 and cot 0, while sin 0 and csc 0. If is the reference angle for , then csc 3 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the
799 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
other trigonometric functions of .
Finally, assign the appropriate signs to the values of the other trigonometric functions of . 3 4 4 sin cos cot 5 5 3 5 5 sec csc 3 4 4 , cos 0 (quadrant III) 3 Since is in quadrant III, cos 0, sec 0, sin 0, csc 0, tan 0, and cot 0.
96. cot
1 3
cos
2 2 3
cot
2 2 2 2 1
If is the reference angle for , then cot
3 2 4 2 2 2 Finally, assign the appropriate signs to the values of the other trigonometric functions of .
Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
sin tan
1 2 2
3
sec
sin
2 2
2
2 4
1 3
cos
2 4 3 2 sec 4
4 . 3
2 2 3
cot 2 2
tan
3 95. tan , sin 0 (quadrant III) 4 Since is in quadrant III, cos 0, sec 0, sin 0, csc 0, tan 0, and cot 0.
If is the reference angle for , then tan Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
3 4 3 tan cos 5 5 4 5 5 sec csc 3 4 Finally, assign the appropriate signs to the values of the other trigonometric functions of . 3 4 3 tan sin cos 5 5 4 5 5 sec csc 3 4 sin
3 . 4
1 97. tan , sin 0 (quadrant II) 3 Since is in quadrant II, cos 0, sec 0, tan 0 and cot 0, while sin 0 and csc 0.
If is the reference angle for , then tan 3 5 5 csc 3
sin
4 5 5 sec 4 cos
cot
4 3
Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the
800 Copyright © 2025 Pearson Education, Inc.
1 . 3
Section 7.4: Trigonometric Functions of Any Angle
other trigonometric functions of .
sin
1
cos
3
sin
10 10
10
10 10 10
3 2 2 3 csc 3
cos
sin
10 10
csc
10 10 1
3 10 10
sec
10 3
10 10 3 cot 3 1 Finally, assign the appropriate signs to the values of the other trigonometric functions of .
1 2
tan 3 cot
3 3
99. csc 2, tan 0 in quadrant III Since is in quadrant III, cos 0, sec 0, sin 0 and csc 0, while tan 0 and cot 0. If is the reference angle for , then csc 2 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
csc 10
3 10 10 cot 3
cos
sec
10 3
98. sec 2, tan 0 (quadrant III) Since is in quadrant III, cos 0, sec 0, sin 0 and csc 0, while tan 0 and cot 0. If is the reference angle for , then sec 2 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of .
sin
tan
1 2 1
3
cos
3 3
3 3
sec
3 2 2
3
3 3
2 3 3
3 3 1 Finally, assign the appropriate signs to the values of the other trigonometric functions of . cot
1 2 3 tan 3
sin
3 2 2 3 sec 3 cos
cot 3
3 3 1 3 2 3 1 3 3 csc cot 3 3 3 3 3 3 Finally, assign the appropriate signs to the values of the other trigonometric functions of . sin
3 2 2
cos
1 2
tan
100. cot 2, sec 0 (quadrant IV) Since is in quadrant IV, cos 0 and sec 0, while sin 0, csc 0, tan 0 and cot 0. If is the reference angle for , then cot 2 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the
801 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
other trigonometric functions of .
b.
F (120) csc120 csc 60
2 3 3
2 3 The point 120, is on the graph of F. 3
c.
sin
1 5
2
5 5
5 5
csc
5 5 1
2 5 5 cos sec 5 2 5 5 1 tan 2 Finally, assign the appropriate signs to the values of the other trigonometric functions of . sin
5 5
2 5 5 1 tan 2
cos
5
csc 5 sec
b.
The point 315, 2 is on the graph of G. c.
7 has reference angle in 6 6 quadrant III. 7 3 7 cos a. g cos 6 6 2 6 7 3 , The point is on the graph of g. 2 6
103. Note:
b.
c.
G (315) sec 315 sec 45 2
h(315) tan 315 tan 45 1
The point 315, 1 is on the graph of h. 102. Note: 120 has reference angle 60 in quadrant II. 1 a. g (120) cos120 cos 60 2 1 The point 120, is on the graph of g. 2
3 3
3 The point 120, is on the graph of H. 3
5 2
101. Note: 315 has reference angle 45 in quadrant IV. 2 a. f (315) sin 315 sin 45 2 2 The point 315, is on the graph of f. 2
H (120) cot120 cot 60
7 7 csc 2 F csc 6 6 6 7 The point , 2 is on the graph of F. 6
Note: 315 has reference angle 45 in quadrant I. H (315) cot(315) cot 45 1 The point 315, 1 is on the graph of H.
7 has reference angle in 4 4 quadrant IV. 7 2 7 sin a. f sin 4 4 4 2 7 2 , The point is on the graph of f. 4 2
104. Note:
b.
c.
7 7 sec 2 G sec 4 4 4 7 The point , 2 is on the graph of G. 4
Note: 225 has reference angle 45 in quadrant II. F (225) csc(225) csc 45 2 The point 225, 2 is on the graph of F.
802 Copyright © 2025 Pearson Education, Inc.
Section 7.4: Trigonometric Functions of Any Angle 105. Since f sin 0.2 is positive, must lie
112. tan 40º tan140º tan 40º tan 180º 40º
either in quadrant I or II. Therefore, must lie either in quadrant III or IV. Thus, f sin 0.2 106. Since g cos 0.4 is positive, must lie
either in quadrant I or IV. Therefore, must lie either in quadrant II or III. Thus, g cos 0.4 . 107. Since F tan 3 is positive, must lie
either in quadrant I or III. Therefore, must also lie either in quadrant I or III. Thus, F tan 3 . 108. Since G cot 2 is negative, must lie
either in quadrant II or IV. Therefore, must also lie either in quadrant II or IV. Thus, G cot 2 . 1 1 1 5 , then csc sin 1 5 5 Since csc 0 , must lie in quadrant I or II. This means that csc must lie in quadrant
109. Given sin
III or IV with the same reference angle as . Since cosecant is negative in quadrants III and IV, we have csc 5 . 1 1 3 2 , then sec cos 2 2 3 3 Since sec 0 , must lie in quadrant I or IV. This means that csc must lie in quadrant
110. Given cos
II or III with the same reference angle as . Since secant is negative in quadrants II and III, 3 we have sec . 2 111. sin 40º sin130º sin 220º sin 310º sin 40º sin 40º 90º sin 40º 180º
3 3 0
113. sin1º sin 2º sin 3º ... sin 357º sin 358º sin 359º sin1º sin 2º sin 3º sin(360º 3º ) sin(360º 2º ) sin(360º 1º ) sin1º sin 2º sin 3º sin(3º ) sin( 2º ) sin(1º ) sin1º sin 2º sin 3º sin 3º sin 2º sin1º sin 180º 0
114. cos1º cos 2º cos 3º cos 357º cos 358º cos 359º cos1º cos 2º cos 3º ... cos(360º 3º ) cos(360º 2º ) cos(360º 1º ) cos1º cos 2º cos 3º ... cos(3º ) cos( 2º ) cos(1º ) cos1º cos 2º cos 3º ... cos 3º cos 2º cos1º 2 cos1º 2 cos 2º 2 cos 3º ... 2 cos178º 2 cos179º cos180º 2 cos1º 2 cos 2º 2 cos 3º ... 2 cos(180º 2º ) 2 cos(180º 1º ) cos 180º 2 cos1º 2 cos 2º 2 cos 3º ... 2 cos 2º 2 cos1º cos180º cos180º 1
115. a.
R
322 2 sin 2 60º cos 2 60º 1 32
32 2 0.866 ( 0.5) 1 16.6 ft
b. Let Y1
322 2 sin 2 x cos 2 x 1 32
sin 40º 270º sin 40 sin 40 sin 40 sin 40 0
45
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90
Chapter 7: Trigonometric Functions c.
Using the MAXIMUM feature, we find:
45
2 2 3 128 2 sin cos ln tan L 4 3 32 3 3 611.98 ft
90
R is largest when 67.5º .
116. L r 2 2 cos 920 2 2 cos
b. Let Y1
2 x 1282 2 sin x cos x ln tan 32 4
Y2 550
30
920(0.10467) 96.30 ft
117. a.
180 100 80
b.
180 135 45
c.
Answers may vary.
118. a. 2 128 6 2 L sin cos ln tan 4 6 32 6 6 2
2 1 3 3 512 ln tan 2 4 4
The angles required for the object to have a path length of 550 feet are:
0.687 (38.8º ) or 1.364 (78.2º ) c.
Using the MAXIMUM feature, we find:
1 3 512 ln tan 466.93 ft 2 4 6
L is largest when 0.986 56.5º . 2 128 4 sin cos 2 ln tan L 4 4 32 4 4 587.67 ft 2
Let Y1
1282 sin x cos x 32.2
(Refer to Section
7.3). Using the MAXIMUM feature, we find:
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Section 7.4: Trigonometric Functions of Any Angle
125.
2x 16 x 3 16 4 2 x 4 x 3
42 4 x 3 2 x 4x 5 2x 22( x 5) 2 x
R is largest when 0.785 45º . This is
2 x 10 x
x 10 The solutions et is 10
larger than the angle needed to get the maximum range. 119.
tan 3 sec 2
2
126.
tan 9 6sec sec 2
2
D(32) 3500 (32) 2 3500 1024 2476 The numbers of units sold is 2,476,000.
6sec 9 sec tan 2
2
6sec 9 1 10 5 1 6 3 cos 3 cos 5 4 7 So, sin and sin cos 5 5 sec
127.
2sin 2 3cos 2 3sin cos 1 sin 2 sin 2 1 3cot 2 3cot csc2
120.
1 2 cot 2 3cot 2 cot 2 3cot 1 0
128.
(2 cot 1)(cot 1) 0
b a b
2
12n 25n 2 144n 2
12n (5n) (12n) 2
12n 169n 2
122 – 124. Answers will vary.
2
12n 12 13n 13
9 1 f (12) 2 2 10 5 f (12) 2 f ( g ( x)) x 2 8 x 19 ( x 4) 2 3
Given that f ( x) x 2 3 then g ( x) x 4 .
121. The point (a, b) (5n, 12n) is in quadrant IV 2
3 f (12) f (0) , 8 12 0 3 f (12) 12 8 12 3 12 f (12) 12 8
x 2 8 x 16 3
1 cot or cot 1 2
sin
D( x) 3500 x 2
x f ( x) x 1 . Any number than makes the 129. x2 g x2 1 denominators zero would not be in the domain so the domain is: x | x 1,1, 2 .
130.
f ( x) 3( x) ( x) 2 5 3 x x 2 5 f ( x) The function is odd.
805 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
3 1 3 1 11. P , ; a , b 2 2 2 2
131. (6 x3 44 x 2 70 x)( x 2 5) 0 2
2
2 x(3x 22 x 35)( x 5) 0 2 x(3 x 7)( x 5)( x 2 5) 0 7 x 0, x , x 5, x 5 3 7 The solution set is 0, , 5, 5, 5 . 3
132. The domain is , . 133. The power function that resembles f(x) at large 2 values of x is y 32 x 7 6 x 7 . 3 134. ( y 2) 2 3 y2 3 y 2 3
The solution set is 2 3, 2 3 .
1 3 cos t 2 2 1 1 3 3 1 2 tan t 2 3 3 2 3 3 3 2 1 2 csc t 1 2 1 1 2 1 2 2 3 2 3 sec t 1 3 3 3 3 3 2 3 3 2 cot t 2 3 1 2 1 2 sin t
3 1 3 1 , ; a , b 12. P 2 2 2 2 sin t
Section 7.5 1. (0, 0); 1; x 2 y 2 1 2.
x x 4
3. even 4. 2 , 5. b; a 6.
b a ; r r
7. 0.2, 0.2
1 2
cos t
3 2
1 3 1 2 1 3 tan t 2 3 3 2 3 3 3 2 1 2 csc t 1 2 1 1 2
2 3 2 3 2 1 3 3 3 3 3 2 3 3 2 2 cot t 3 1 2 1 2
sec t
1
8. True 9. b 10. a
806 Copyright © 2025 Pearson Education, Inc.
Section 7.5: Unit Circle Approach; Properties of the Trigonometric Functions
2 2 2 2 13. P , , b ; a 2 2 2 2
5 2 5 2 , ; a , b 15. P 3 3 3 3
2 2 cos t 2 2 2 2 2 2 tan t cot t 1 1 2 2 2 2 1 2 2 2 csc t 1 2 2 2 2 2 2 1 2 2 2 sec t 1 2 2 2 2 2 2
sin t
sin t
2 3
cos t
5 3
2 2 3 2 5 2 5 tan t 3 = 5 5 3 5 5 5 3 1 3 3 csc t 1 2 2 2 3 sec t
3 3 5 3 5 1 5 5 5 5 5 3
1
5 5 3 5 cot t 3 2 3 2 2 3
2 2 2 2 14. P , , b ; a 2 2 2 2
2 2 cos t 2 2 2 2 tan t 2 1 cot t 2 1 2 2 2 2 1 2 2 2 csc t 1 2 2 2 2 2 2 1 2 2 2 sec t 1 2 2 2 2 2 2
sin t
5 2 5 5 2 5 , , b 16. P ; a 5 5 5 5
2 5 5 cos t 5 5 2 5 2 5 5 tan t 5 = 2 5 5 5 5 1 5 5 5 csc t 1 2 2 5 2 5 5 5 1 5 5 sec t 1 5 5 5 5 5 5 5 5 1 5 cot t 2 2 5 5 2 5 5 sin t
17. For the point (3, 4) , x 3, y 4, r x 2 y 2 9 16 25 5 4 5 5 csc 4
sin
807 Copyright © 2025 Pearson Education, Inc.
3 5 5 sec 3
cos
4 3 3 cot 4
tan
Chapter 7: Trigonometric Functions 18. For the point (5, –12), x 5, y 12, r x y 25 144 169 13 2
2
12 13 13 csc 12
5 13 13 sec 5
sin
12 5 5 cot 12
cos
tan
19. For the point (–2, 3), x 2, y 3,
sin
2
3
13
13 13 2
cos
13
13 13
3 13 13
csc
2 13 13
3 tan 2
2 2
24. cos 420º cos(360º 60º ) cos 60º
1 2
25. tan 405º tan(180º 180º 45º ) tan 45º 1 26. sin 390º sin(360º 30º ) sin 30º
r x y 4 9 13 2
23. sin 405º sin(360º 45º ) sin 45º
13 3
13 2 2 cot 3
sec
27. csc 390º csc(360º 30º ) csc 30º 2 28. sec 540º sec(360º 180º ) sec180º 1 29. cot 390º cot(180º 180º 30º ) cot 30º 3 30. sec 420º sec(360º 60º ) sec 60º 2
20. For the point (2, –4), x 2, y 4, 31. cos
r x 2 y 2 4 16 20 2 5 sin cos
4
5
2 5 5 2
5
2 5 5 4 tan 2 2
2 5 5
csc
2 5 5 4 2
5 5
sec
5
5 5 5 2 1 cot 4 2
r x2 y 2 1 1 2 sin cos
1
2
2
2
1
2
2 2 1 tan 1 1
2 2
csc
2 2 1
2 2
sec
2 2 1
1 cot 1 1
22. For the point (–3, 1), x 3, y 1,
r x 2 y 2 9 1 10 sin
1 10 10 10 10 10
3 10 3 10 10 10 10 1 1 tan 3 3
cos
csc
10 10 1
sec
10 10 3 3
cot
3 3 1
37 cos 6 6 6 cos 3 2 6 cos 6
5
21. For the point (–1, –1), x 1, y 1,
1 2
32. sin
3 2
9 2 sin 2 sin 4 4 2 4
33. tan 15 tan(0 15) tan 0 0 34. csc
9 csc 4 2 2 csc 2 2 2 csc 2 1
808 Copyright © 2025 Pearson Education, Inc.
Section 7.5: Unit Circle Approach; Properties of the Trigonometric Functions
35. csc
21 5 csc 4 4 4
47. tan tan 1 4 4
5 csc 2 2 4 5 csc 4 2
36. cot
48. sin() sin 0
2 49. cos cos 4 2 4 3 50. sin sin 3 2 3
17 cot 4 4 4 cot 2 2 4 cot 4 1
51. tan() tan 0 3 3 52. sin sin (1) 1 2 2 53. csc csc 2 4 4
19 tan 6 tan 3 37. tan 3 3 3
54. sec() sec 1
25 sec 4 38. sec 6 6
2 3 55. sec sec 6 3 6
sec 2 2 6 sec 6
2 3 56. csc csc 3 3 3
2 3 3
57. sin cos 5 sin cos 4
39. sin( 60º ) sin 60º
40. cos(30º ) cos 30º
0 cos 1
3 2
7 5 5 58. tan cot tan cot 3 2 6 6 2 5 tan cot 6 2 3 0 3
3 2
41. tan(60º ) tan 60º 3 42. sin(135º ) sin135º
2 2
43. sec( 45º ) sec 45º 2
3 3
59. sec csc sec csc 3 2 3 2 2 1 1
44. csc(30º ) csc 30º 2 45. sin(90º ) sin 90º 1 46. cos( 270º ) cos 270º 0 809
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Chapter 7: Trigonometric Functions
60. tan 6 cos
61.
9 tan(0 6) cos 2 4 4 tan 0 cos 4 2 0 2 2 2
9 9 sin tan 4 4 9 9 sin tan 4 4 sin 2 tan 2 4 4 sin tan 4 4 2 2 2 1, or 2 2
69. The range of the sine function is the set of all real numbers between –1 and 1, inclusive. That is, the interval 1,1 .
74. The range of the cosecant function is the set of all real number greater than or equal to 1 and all real numbers less than or equal to –1. That is, the interval , 1 1, . 75. The sine function is odd because sin( ) sin . Its graph is symmetric with respect to the origin.
2 2 2
64. The domain of the cosine function is the set of all real numbers. That is, , .
66.
f ( ) csc is not defined for numbers that are multiples of .
73. The range of the secant function is the set of all real number greater than or equal to 1 and all real numbers less than or equal to –1. That is, the interval , 1 1, .
63. The domain of the sine function is the set of all real numbers. That is, , .
f ( ) tan is not defined for numbers that are
odd multiples of
68.
. 2
72. The range of the cotangent function is the set of all real numbers. That is, , .
65.
odd multiples of
71. The range of the tangent function is the set of all real numbers. That is, , .
3 sin 4 2
2 (1) 2 2 1, or 2
f ( ) sec is not defined for numbers that are
70. The range of the cosine function is the set of all real numbers between –1 and 1, inclusive. That is, the interval 1,1 .
17 3 62. cos sin 4 2 17 3 cos sin 4 2 3 cos 2 2 sin 2 4
cos
67.
. 2
f ( ) cot is not defined for numbers that are multiples of .
76. The cosine function is even because cos( ) cos . Its graph is symmetric with respect to the y-axis. 77. The tangent function is odd because tan( ) tan . Its graph is symmetric with respect to the origin. 78. The cotangent function is odd because cot( ) cot . Its graph is symmetric with respect to the origin. 79. The secant function is even because sec( ) sec . Its graph is symmetric with respect to the y-axis.
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Section 7.5: Unit Circle Approach; Properties of the Trigonometric Functions 80. The cosecant function is odd because csc( ) csc . Its graph is symmetric with respect to the origin.
88. a.
f (a ) f (a ) (3) 3
b.
f ( a ) f ( a ) f ( a 4 ) f (a) f (a) f (a) 3 (3) (3) 9
81. If sin 0.3 , then sin sin 2 sin 4 0.3 0.3 0.3 0.9
89. a.
82. If cos 0.2 , then cos cos 2 cos 4
b.
0.2 0.2 0.2 0.6
83. If tan 3 , then tan tan tan 2
90. a.
333 9
b.
84. If cot 2 , then cot cot cot 2 2 2 2
91. a.
6
85. a. b.
86. a. b.
87. a. b.
f (a) f (a)
1 3
f (a) f (a 2) f (a 4) f (a ) f (a ) f (a) 4 ( 4) ( 4) 12 f (a ) f (a ) 2 f (a) f (a 2) f (a 4) f (a ) f (a ) f (a) 222 6
When t 1 , the coordinate on the unit circle is approximately (0.5, 0.8) . Thus, sin1 0.8
f (a) f (a 2) f (a 4) f (a ) f (a ) f (a ) 1 1 1 3 3 3 1 f (a) f (a)
f (a) f (a) 4
cos1 0.5
tan1
0.8 1.6 0.5
1 1.3 0.8 1 sec1 2.0 0.5 0.5 cot1 0.6 0.8 csc1
Using a calculator on RADIAN mode: sin1 0.8 csc1 1.2 cos1 0.5 sec1 1.9 tan1 1.6 cot1 0.6
1 4
f (a) f (a 2) f (a 2) f (a ) f (a ) f (a) 1 1 1 4 4 4 3 4
b. When t 5.1 , the coordinate on the unit circle is approximately (0.4, 0.9) . Thus,
f (a ) f (a ) 2
sin 5.1 0.9
f ( a ) f ( a ) f ( a 2 ) f (a) f (a) f (a) 222 6
cos 5.1 0.4 tan 5.1
0.9 2.3 0.4
1 1.1 0.9 1 sec 5.1 2.5 0.4 0.4 cot 5.1 0.4 0.9 csc 5.1
Using a calculator on RADIAN mode: sin 5.1 0.9 csc 5.1 1.1 811
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Chapter 7: Trigonometric Functions cos 5.1 0.4 tan 5.1 2.4
92. a.
sec 5.1 2.6 cot 5.1 0.4
When t 2 , the coordinate on the unit circle is approximately (0.4, 0.9) . Thus, sin 2 0.9
94. Suppose there is a number p, 0 p 2 , for
which cos( p) cos for all . If
1 1.1 0.9 1 sec 2 2.5 0.4 0.4 cot 2 0.4 0.9 csc 2
cos 2 0.4 tan 2
3 But p . Thus, sin 1 sin 1, 2 2 or 1 1 . This is impossible. Therefore, the smallest positive number p for which sin( p ) sin for all is p 2 .
0.9 2.3 0.4
Using a calculator on RADIAN mode: sin 2 0.9 csc 2 1.1 cos 2 0.4 sec 2 2.4 tan 2 2.2 cot 2 0.5
then cos p cos 0 ; so that p . 2 2 If =0 , then cos 0 p cos 0 . But p .
Thus, cos 1 cos 0 1, or 1 1. This is impossible. Therefore, the smallest positive number p for which cos( p) cos for all is p 2 . 95.
96. b. When t 4 , the coordinate on the unit circle is approximately (0.6, 0.8) . Thus, sin 4 0.8 cos 4 0.7 tan 4
0.8 1.1 0.7
1 1.3 0.8 1 sec 4 1.4 0.7 0.7 cot 4 0.9 0.8 csc 4
Set the calculator on RADIAN mode: csc 4 1.3 sin 4 0.8 cos 4 0.7 sec 4 1.5 tan 4 1.2 cot 4 0.9
then sin 0 p sin p sin 0 0 ; so that p . If
then sin p sin . 2 2 2
1 : since cos has period cos 2 , so does f ( ) sec . f ( ) sec
1 : since sin has period sin 2 , so does f ( ) csc . f ( ) csc
97. If P (a, b) is the point on the unit circle corresponding to , then Q (a, b) is the point on the unit circle corresponding to . b b tan . If there Thus, tan( ) a a exists a number p, 0 p , for which tan( p ) tan for all , then if 0 , tan( p ) tan(0) 0. But this means that p is a multiple of . Since no multiple of exists in the interval (0, ) , this is impossible. Therefore, the fundamental period of f ( ) tan is . 98.
93. Suppose there is a number p, 0 p 2, for which sin( p ) sin for all . If 0 ,
, 2
1 : Since tan has period tan , so does f ( ) cot . f ( ) cot =
99. Let P ( x, y ) be the point on the unit circle that corresponds to an angle t. Consider the equation y tan t a . Then y ax . Now x 2 y 2 1 , x
812 Copyright © 2025 Pearson Education, Inc.
Section 7.5: Unit Circle Approach; Properties of the Trigonometric Functions
so x 2 a 2 x 2 1 . Thus, x y
1 1 a
2
(20) P (20) 40 cos 110 6 40 cos 10.4720 110
and
a
; that is, for any real number a, 1 a2 there is a point P ( x, y ) on the unit circle for which tan t a . In other words, tan t , and the range of the tangent function is the set of all real numbers.
90 (30) P (30) 40 cos 110 6 40 cos 15.7080 110 70
100. Let P ( x, y ) be the point on the unit circle that corresponds to an angle t. Consider the equation x cot t a . Then x ay . Now x 2 y 2 1 , y
so a 2 y 2 y 2 1 . Thus, y x
1 1 a2
1 104. cos 2 sin ;cos 2 sin 2 1 9 Substitute x cos ; y sin and solve these simultaneous equations for y. 1 x2 y ; x2 y 2 1 9 x2 1 y 2 1 (1 y 2 ) y 9 10 y2 y 0 9 9 y 2 9 y 10 0
and
a
; that is, for any real number a, 1 a2 there is a point P ( x, y ) on the unit circle for which cot t a . In other words, cot t , and the range of the cotangent function is the set of all real numbers. sin 0 sin tan . cos 0 cos Since L is parallel to M, the slope of L tan .
101. The slope of M is
a 9, b 9, c 10 y
2 (2.5 1.25) 102. V (2.5) 250sin 2650 5 250sin 1.5708 2650
(9) (9) 2 4(9)( 10) 18
9 81 360 9 441 9 21 18 18 18 Since the point is in Quadrant II then 3 21 12 2 y and 18 18 3 2 4 5 2 x2 1 1 3 9 9
2900 mL 2 (10 1.25) V (10) 250sin 2650 5 250sin 10.9956 2650 2400 mL
x
2 (17 1.25) V (17) 250sin 2650 5 250sin 19.7920 2650
5 5 9 3
41 ; cos 2 sin 2 1 49 Substitute x cos ; y sin and solve these simultaneous equations for y.
105. cos sin 2
2852.25 mL (10) 103. P (10) 40 cos 110 6 40 cos 5.2360 110 130
813
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Chapter 7: Trigonometric Functions 108 – 111. Answers will vary.
41 2 ; x y2 1 49 y 2 1 x2
x y2
112. a 3, b 10, c 5
41 x (1 x ) 49 8 x2 x 0 49
x
2
10 100 60 6 10 40 10 2 10 5 10 6 6 3
a 1, b 1, c x
8 49
( 1) ( 1) 2 4(1)( 498 )
The zeros are:
2
1 1 32 49 2
1
81 49
2
1 97 8 1 or 2 7 7
Since the point is in quadrant III then x
1 7
1 . Using the point slope formula and 4 the point ( 3, 7) :
which is
2
( y y1 ) m( x x1 )
48 4 3 49 7
106. Since sin(4 ) cos(2 ) and 0 4
know 4 2
. So
2
1 ( x 3) 4 1 3 y7 x 4 4 1 31 y x 4 4 y 7
we
. 12 2 sin(8 ) cos(4 ) 2 sin cot 2 3 3 2
3 3 2 2 3 5 3 5 3 12 2 6 6 cos 7 8sin
107.
r P A 1 n
114.
1 sin 2 49 112 sin 64 sin 2
65sin 2 112sin 48 0 (13sin 12)(5sin 4) 0
1800 r 1 1500 4 12
12 5 (cos ) or 13 13 4 3 sin (cos ) extraneous 5 5
43
12
r 6 1 5 4 12
sin
nt
r 1800 1500 1 4
cos 2 49 112 sin 64 sin 2
sin cos
5 10 5 10 , 3 3
113. The slope of the line perpendicular to y 4 x 5 is the opposite reciprocal of -4
1 48 1 and y 2 1 1 7 49 49
y
( 10) ( 10) 2 4(3)(5) 2(3)
6 r 1 5 4
6 4 12 1 r 5
12 5 7 13 13 13
814 Copyright © 2025 Pearson Education, Inc.
r .0612 or 6.12%
Section 7.6: Graphs of the Sine and Cosine Functions
115.
We see from the graphs that the first derivative is positive and the second derivative is negative on the interval [0, 1].
5x 2 x3 The vertical asymptote is where the denominator is undefined: x 3 f ( x)
120.
5 1
The horizontal asymptote is y 5 .
4 x 2 5 x 10 7 x3 4 x 2 7 x3 4 x 2 x 8
3
116. ( x 7) 5 8 5
( x 7) 8 3 x7 2
1
1 121. 4 log 3 x log 3 y log 3 x 4 log 3 y 2 2 1 log 3 x 4 y 2 log 3 x 4 y
5
x 32 7 39
The solution set is {39} . 117.
4 x 2 5 x 10 (7 x3 4 x 2)
f (7) f (3) 142 (22) 73 4 120 30 4
Section 7.6 1. y 3 x 2
118. The functions intersect at:
Using the graph of y x 2 , vertically stretch the graph by a factor of 3.
x 2 21 4 x x 2 4 x 21 0 ( x 7)( x 3) 0 x 7, x 3
2. y 2 x
The graph of x 2 is below the graph of 21 4 x on the interval (7,3) .
Using the graph of y x , compress horizontally by a factor of
119. Graph the first and second derivative.
3. a. b. c. d. e.
1 0 -1 0 1,1
815
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1 . 2
Chapter 7: Trigonometric Functions
origin; odd 1 2 ; ; 1 2 2
f. g.
g.
The graph of y cos x crosses the y-axis at the point (0, 1), so the y-intercept is 1. b. The graph of y cos x is decreasing for 0 x. c. The smallest value of y cos x is 1 .
14. a.
4. a. -1 b. 0; 1 c. 1,1 d. y-axis; even 3 2 e. ; ;0 2 2 5. 1;
d. e.
2
3 11 11 when x , , , 2 6 6 6 6 g. The x-intercepts of cos x are 2k 1 , k an integer x | x 2 15. y 5sin x cos x
f.
2 6 3
8. True 9. False; The period is
2
This is in the form y A sin( x) where A 5 2
and 1 . Thus, the amplitude is A 5 5 and the period is T
10. True 11. d
2
2 2 . 1
16. y 3cos x This is in the form y A cos( x) where A 3
12. d 13. a.
3 , 2 2 cos x 1 when x 2, 0, 2;
cos x 0 when x
cos x 1 when x , .
6. 3; 7. 3;
The x-intercepts of sin x are x | x k , k an integer
and 1 . Thus, the amplitude is A 3 3
The graph of y sin x crosses the y-axis at the point (0, 0), so the y-intercept is 0.
and the period is T
b. The graph of y sin x is increasing for
2
2 2 . 1
17. y 3cos(4 x) This is in the form y A cos( x) where A 3 and 4 . Thus, the amplitude is
x . 2 2
c.
The largest value of y sin x is 1.
A 3 3 and the period is
d.
sin x 0 when x 0, , 2 .
T
e.
f.
3 , ; 2 2 3 sin x 1 when x , . 2 2
2 . 4 2
1 18. y sin x 2 This is in the form y A sin( x) where A 1
sin x 1 when x
sin x
2
1 . Thus, the amplitude is A 1 1 2 2 2 and the period is T 1 4 .
and
1 5 7 11 when x , , , 2 6 6 6 6
816
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2
Section 7.6: Graphs of the Sine and Cosine Functions 19. y 6sin( x) This is in the form y A sin( x) where A 6
9 3 9 3 24. y cos x cos x 5 2 5 2
and . Thus, the amplitude is A 6 6 and the period is T
2
This is in the form y A cos( x) where A
2 2.
3 . Thus, the amplitude is 2 9 9 A and the period is 5 5 2 2 4 T . 3 3
and
20. y 3cos(3x) This is in the form y A cos( x) where A 3
and 3 . Thus, the amplitude is A 3 3 and the period is T
2
2 . 3
2
25. F
1 7 21. y cos x 7 2 This is in the form y A cos( x) where
26. E 27. A
1 7 and . Thus, the amplitude is 7 2 1 1 A and the period is 7 7 2 2 4 . T 7 7 2 A
28. I 29. H 30. B 31. C
4 2 22. y sin x 3 3
32. G
This is in the form y A sin( x) where A
4 3
33. J 34. D
2 4 4 . Thus, the amplitude is A 3 3 3 2 2 2 3 . and the period is T
and
35. Comparing y 4 cos x to y A cos x , we
find A 4 and 1 . Therefore, the amplitude 2 is 4 4 and the period is 2 . Because 1 the amplitude is 4, the graph of y 4 cos x will lie between 4 and 4 on the y-axis. Because the period is 2 , one cycle will begin at x 0 and end at x 2 . We divide the interval 0, 2
3
10 2 10 2 x sin x sin 9 5 9 5 This is in the form y A sin( x) where
23. y
10 2 and . Thus, the amplitude is 9 5 10 10 and the period is A 9 9 2 2 T 5. A
9 5
into four subintervals, each of length
2 by 4 2
finding the following values: 3 , and 2 0, , , 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 4 cos x , we multiply the y-coordinates of the five key points for y cos x by A 4 . The five key points are
2 5
817
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Chapter 7: Trigonometric Functions 3 , 0 , , 4 , , 0 , 2 , 4 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
0, 4 ,
direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, , and the range is 3,3 . 37. Comparing y 4sin x to y A sin x , we
From the graph we can determine that the domain is all real numbers, , and the
find A 4 and 1 . Therefore, the amplitude 2 is 4 4 and the period is 2 . Because 1 the amplitude is 4, the graph of y 4sin x will lie between 4 and 4 on the y-axis. Because the period is 2 , one cycle will begin at x 0 and end at x 2 . We divide the interval 0, 2
range is 4, 4 . 36. Comparing y 3sin x to y A sin x , we
find A 3 and 1 . Therefore, the amplitude 2 is 3 3 and the period is 2 . Because 1 the amplitude is 3, the graph of y 3sin x will lie between 3 and 3 on the y-axis. Because the period is 2 , one cycle will begin at x 0 and end at x 2 . We divide the interval 0, 2 into four subintervals, each of length
2 by 4 2 3 finding the following values: 0, , , , 2 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 4sin x , we multiply the y-coordinates of
into four subintervals, each of length
2 by 4 2
finding the following values: 3 , and 2 0, , , 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 3sin x , we multiply the y-coordinates of the five key points for y sin x by A 3 . The five key
the five key points for y sin x by A 4 . The five key points are 3 0, 0 , , 4 , , 0 , , 4 , 2 , 0 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
3 points are 0, 0 , ,3 , , 0 , , 3 , 2 2 2 , 0
We plot these five points and fill in the graph of the curve. We then extend the graph in either
818
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Section 7.6: Graphs of the Sine and Cosine Functions
between 1 and 1 on the y-axis. Because the
From the graph we can determine that the domain is all real numbers, , and the
period is
range is 4, 4 .
2
, one cycle will begin at x 0 and
. We divide the interval 0, 2 2 /2 into four subintervals, each of length 4 8 by finding the following values: 3 , and 0, , , 8 4 8 2 These values of x determine the x-coordinates of the five key points on the graph. The five key points are 3 0,1 , , 0 , , 1 , , 0 , ,1 8 4 8 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
end at x
38. Comparing y 3cos x to y A cos x , we
find A 3 and 1 . Therefore, the amplitude 2 is 3 3 and the period is 2 . Because 1 the amplitude is 3, the graph of y 3cos x will lie between 3 and 3 on the y-axis. Because the period is 2 , one cycle will begin at x 0 and end at x 2 . We divide the interval 0, 2 into four subintervals, each of length
2 by 4 2
finding the following values: 3 , and 2 0, , , 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 3cos x , we multiply the y-coordinates of the five key points for y cos x by A 3 . The five key points are 3 0, 3 , , 0 , ,3 , , 0 , 2 , 3 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. y 5 ( , 3) ( , 3) 3 ( ––– , 0) 2 x 2 2 , 0) ( –– 2 (0, 3) (2 , 3) 5
From the graph we can determine that the domain is all real numbers, , and the range is 1,1 . 40. Comparing y sin 3 x to y A sin x , we
find A 1 and 3 . Therefore, the amplitude 2 is 1 1 and the period is . Because the 3 amplitude is 1, the graph of y sin 3 x will lie
From the graph we can determine that the domain is all real numbers, , and the
between 1 and 1 on the y-axis. Because the 2 , one cycle will begin at x 0 and period is 3 2 2 . We divide the interval 0, end at x 3 3 2 / 3 into four subintervals, each of length 4 6 by finding the following values:
range is 3,3 . 39. Comparing y cos 4 x to y A cos x , we
find A 1 and 4 . Therefore, the amplitude 2 is 1 1 and the period is . Because the 4 2 amplitude is 1, the graph of y cos 4 x will lie 819
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Chapter 7: Trigonometric Functions
2 , , , and 6 3 2 3 These values of x determine the x-coordinates of the five key points on the graph. The five key points are 2 0, 0 , ,1 , , 0 , , 1 , , 0 6 3 2 3 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. 0,
3 , 1 , , 0 , ,1 , , 0 4 2 4 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. y 2 3 , 1) ( ––– , 1) ( , 0) ––– ( 4 4
0, 0 ,
2
2 x
(0, 0)
, 0) (–– 2
, 1) 2 (–– 4
From the graph we can determine that the domain is all real numbers, , and the range is 1,1 . 42. Since cosine is an even function, we can plot the equivalent form y cos 2 x .
Comparing y cos 2 x to y A cos x , we
From the graph we can determine that the domain is all real numbers, , and the
find A 1 and 2 . Therefore, the amplitude 2 is 1 1 and the period is . Because the 2 amplitude is 1, the graph of y cos 2 x will lie
range is 1,1 . 41. Since sine is an odd function, we can plot the equivalent form y sin 2 x .
between 1 and 1 on the y-axis. Because the period is , one cycle will begin at x 0 and end at x . We divide the interval 0, into
Comparing y sin 2 x to y A sin x , we find A 1 and 2 . Therefore, the 2 amplitude is 1 1 and the period is . 2 Because the amplitude is 1, the graph of y sin 2 x will lie between 1 and 1 on the
four subintervals, each of length
4
4
by finding
the following values: 3 , and 0, , , 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y cos 2 x , we multiply the y-coordinates of
y-axis. Because the period is , one cycle will begin at x 0 and end at x . We divide the interval 0, into four subintervals, each of length
by finding the following values:
the five key points for y cos x by A 1 .The five key points are 3 0,1 , , 0 , , 1 , , 0 , ,1 4 2 4 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
3 0, , , , and 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y sin 2 x , we multiply the y-coordinates of
the five key points for y sin x by A 1 .The five key points are 820
Copyright © 2025 Pearson Education, Inc.
Section 7.6: Graphs of the Sine and Cosine Functions
direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, , and the
From the graph we can determine that the domain is all real numbers, , and the
range is 1,1 .
range is 2, 2 .
1 43. Comparing y 2sin x to y A sin x , 2 1 we find A 2 and . Therefore, the 2 2 amplitude is 2 2 and the period is 4 . 1/ 2 Because the amplitude is 2, the graph of 1 y 2sin x will lie between 2 and 2 on the 2 y-axis. Because the period is 4 , one cycle will begin at x 0 and end at x 4 . We divide the interval 0, 4 into four subintervals, each
of length
1 44. Comparing y 2 cos x to y A cos x , 4 1 we find A 2 and . Therefore, the 4 2 amplitude is 2 2 and the period is 8 . 1/ 4 Because the amplitude is 2, the graph of 1 y 2 cos x will lie between 2 and 2 on 4 the y-axis. Because the period is 8 , one cycle will begin at x 0 and end at x 8 . We divide the interval 0,8 into four subintervals,
4 by finding the following 4
each of length
8 2 by finding the following 4
values: 0, , 2 , 3 , and 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 y 2sin x , we multiply the y-coordinates of 2 the five key points for y sin x by A 2 . The five key points are 0, 0 , , 2 , 2 , 0 , 3 , 2 , 4 , 0
values: 0, 2 , 4 , 6 , and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 y 2 cos x , we multiply the y-coordinates 4 of the five key points for y cos x by A 2 .The five key points are 0, 2 , 2 , 0 , 4 , 2 , 6 , 0 , 8 , 2
We plot these five points and fill in the graph of the curve. We then extend the graph in either
We plot these five points and fill in the graph of the curve. We then extend the graph in either
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Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
direction to obtain the graph shown below.
direction to obtain the graph shown below. y ( 2 , 0) (0, 2) (8 , 2) 2 ( 8 , 2) (2 , 0) (6 , 0) x 8 4 4 8 ( 4 , 2)
2
(4 , 2)
From the graph we can determine that the domain is all real numbers, , and the
From the graph we can determine that the domain is all real numbers, , and the
range is 2, 2 .
1 1 range is , . 2 2
1 45. Comparing y cos 2 x to y A cos x , 2 1 we find A and 2 . Therefore, the 2 1 1 2 amplitude is and the period is . 2 2 2 1 Because the amplitude is , the graph of 2 1 1 1 y cos 2 x will lie between and on 2 2 2 the y-axis. Because the period is , one cycle will begin at x 0 and end at x . We divide the interval 0, into four subintervals, each of
length
4
2 16 . Because the amplitude is 4, the 1/ 8 1 graph of y 4sin x will lie between 4 8 and 4 on the y-axis. Because the period is 16 , one cycle will begin at x 0 and end at x 16 . We divide the interval 0,16 into
by finding the following values:
3 , , , and 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 y cos 2 x , we multiply the y-coordinates 2 of the five key points for y cos x by 0,
1 46. Comparing y 4sin x to y A sin x , 8 1 we find A 4 and . Therefore, the 8 amplitude is 4 4 and the period is
four subintervals, each of length
16 4 by 4
finding the following values: 0, 4 , 8 , 12 , and 16 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 y 4sin x , we multiply the y-coordinates 8 of the five key points for y sin x by A 4 . The five key points are 0, 0 , 4 , 4 , 8 , 0 , 12 , 4 , 16 , 0
1 A .The five key points are 2 1 1 3 1 0, , , 0 , , , , 0 , , 2 4 2 2 4 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either
We plot these five points and fill in the graph of the curve. We then extend the graph in either
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Copyright © 2025 Pearson Education, Inc.
Section 7.6: Graphs of the Sine and Cosine Functions
direction to obtain the graph shown below.
direction to obtain the graph shown below. y 5
(12 , 4) (0, 0)
16
x (16 , 0)
8
(8 , 0) 5
(4 , 4)
From the graph we can determine that the domain is all real numbers, , and the range is 4, 4 .
From the graph we can determine that the domain is all real numbers, , and the range is 1,5 .
47. We begin by considering y 2sin x . Comparing y 2sin x to y A sin x , we find A 2
48. We begin by considering y 3cos x . Comparing
and 1 . Therefore, the amplitude is 2 2
y 3cos x to y A cos x , we find A 3
2 2 . Because the 1 amplitude is 2, the graph of y 2sin x will lie between 2 and 2 on the y-axis. Because the period is 2 , one cycle will begin at x 0 and end at x 2 . We divide the interval 0, 2
and the period is
into four subintervals, each of length
and 1 . Therefore, the amplitude is 3 3 2 2 . Because the 1 amplitude is 3, the graph of y 3cos x will lie between 3 and 3 on the y-axis. Because the period is 2 , one cycle will begin at x 0 and end at x 2 . We divide the interval 0, 2
and the period is
2 by 4 2
finding the following values: 3 , and 2 0, , , 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 2sin x 3 , we multiply the y-coordinates of the five key points for y sin x by A 2 and then add 3 units. Thus, the graph of y 2sin x 3 will lie between 1 and 5 on the yaxis. The five key points are 3 0, 3 , ,5 , ,3 , ,1 , 2 ,3 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either
into four subintervals, each of length
2 by 4 2
finding the following values: 3 , and 2 0, , , 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 3cos x 2 , we multiply the y-coordinates of the five key points for y cos x by A 3 and then add 2 units. Thus, the graph of y 3cos x 2 will lie between 1 and 5 on the y-axis. The five key points are 3 0, 5 , , 2 , , 1 , , 2 , 2 , 5 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either
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Chapter 7: Trigonometric Functions
direction to obtain the graph shown below. y
direction to obtain the graph shown below.
3 (0, 2) 2
3 ( –– , 3) 2
(2, 2) 2 x 3 (–– , 3) 2 1 (–– , 3) 2
(1, 8)
9
From the graph we can determine that the domain is all real numbers, , and the
From the graph we can determine that the domain is all real numbers, , and the
range is 8, 2 .
range is 1,5 .
50. We begin by considering y 4sin x . 2 Comparing y 4sin x to y A sin x , 2
49. We begin by considering y 5cos x .
Comparing y 5cos x to y A cos x , we find A 5 and . Therefore, the amplitude 2 2 . Because the is 5 5 and the period is
we find A 4 and
amplitude is 5, the graph of y 5cos x will
2
. Therefore, the
2 4. /2 Because the amplitude is 4, the graph of y 4sin x will lie between 4 and 4 on 2 the y-axis. Because the period is 4 , one cycle will begin at x 0 and end at x 4 . We divide the interval 0, 4 into four subintervals, each of amplitude is 4 4 and the period is
lie between 5 and 5 on the y-axis. Because the period is 2 , one cycle will begin at x 0 and end at x 2 . We divide the interval 0, 2 into four subintervals, each of length
2 1 by 4 2
finding the following values: 1 3 0, , 1, , and 2 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 5cos x 3 , we multiply the y-coordinates
4 1 by finding the following values: 4 0, 1, 2, 3, and 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 4sin x 2 , we multiply the y2 coordinates of the five key points for y sin x by A 4 and then subtract 2 units. Thus, the graph of y 4sin x 2 will lie between 6 2 and 2 on the y-axis. The five key points are 0, 2 , 1, 2 , 2, 2 , 3, 6 , 4, 2
length
of the five key points for y cos x by A 5 and then subtract 3 units. Thus, the graph of y 5cos x 3 will lie between 8 and 2 on the y-axis. The five key points are 1 3 0, 2 , , 3 , 1, 8 , , 3 , 2, 2 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either
We plot these five points and fill in the graph of the curve. We then extend the graph in either 824
Copyright © 2025 Pearson Education, Inc.
Section 7.6: Graphs of the Sine and Cosine Functions
direction to obtain the graph shown below.
0, 4 , , 2 , 3, 4 , ,10 , 6, 4
3 9 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, , and the range is 6, 2 .
From the graph we can determine that the domain is all real numbers, , and the
51. We begin by considering y 6sin x . 3 Comparing y 6sin x to y A sin x , 3
we find A 6 and
3
range is 2,10 . 52. We begin by considering y 3cos x . 4 Comparing y 3cos x to y A cos x , 4
. Therefore, the
2 6. /3 Because the amplitude is 6, the graph of y 6sin x will lie between 6 and 6 on the 3 y-axis. Because the period is 6, one cycle will begin at x 0 and end at x 6 . We divide the interval 0, 6 into four subintervals, each of
amplitude is 6 6 and the period is
we find A 3 and
4
. Therefore, the
2 8. /4 Because the amplitude is 3, the graph of y 3cos x will lie between 3 and 3 on 4 the y-axis. Because the period is 8, one cycle will begin at x 0 and end at x 8 . We divide the interval 0,8 into four subintervals, each of amplitude is 3 3 and the period is
6 3 by finding the following values: 4 2 3 9 0, , 3, , and 6 2 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 6sin x 4 , we multiply the y3 coordinates of the five key points for y sin x by A 6 and then add 4 units. Thus, the graph of y 6sin x 4 will lie between 2 and 3 10 on the y-axis. The five key points are
length
8 2 by finding the following values: 4 0, 2, 4, 6, and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 3cos x 2 , we multiply the y4 coordinates of the five key points for y cos x by A 3 and then add 2 units. Thus, the graph
length
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Chapter 7: Trigonometric Functions
on the y-axis. The five key points are 3 0,5 , , 2 , ,5 , ,8 , ,5 4 2 4 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
of y 3cos x 2 will lie between 1 and 4 5 on the y-axis. The five key points are 0, 1 , 2, 2 , 4,5 , 6, 2 , 8, 1
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
y
0,5
3 4 ,8 ,5 2
,5 ,2 4
From the graph we can determine that the domain is all real numbers, , and the
From the graph we can determine that the domain is all real numbers, , and the
range is 2,8 .
range is 1,5 .
54. y 2 4 cos 3x 4 cos 3x 2
53. y 5 3sin 2 x 3sin 2 x 5
We begin by considering y 4 cos 3 x .
We begin by considering y 3sin 2 x .
Comparing y 4 cos 3 x to y A cos x ,
Comparing y 3sin 2 x to y A sin x ,
we find A 4 and 3 . Therefore, the 2 amplitude is 4 4 and the period is . 3 Because the amplitude is 4, the graph of y 4 cos 3 x will lie between 4 and 4 on
we find A 3 and 2 . Therefore, the 2 amplitude is 3 3 and the period is . 2 Because the amplitude is 3, the graph of y 3sin 2 x will lie between 3 and 3 on the
2 , one cycle 3 2 will begin at x 0 and end at x . We 3 2 divide the interval 0, into four 3 2 / 3 by subintervals, each of length 4 6 finding the following values: 2 0, , , , and 3 6 3 2 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 4 cos 3x 2 , we multiply the ythe y-axis. Because the period is
y-axis. Because the period is , one cycle will begin at x 0 and end at x . We divide the interval 0, into four subintervals, each of length
4
by finding the following values:
3 , , , and 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for y 3sin 2 x 5 , we multiply the y0,
coordinates of the five key points for y sin x by A 3 and then add 5 units. Thus, the graph of y 3sin 2 x 5 will lie between 2 and 8 826
Copyright © 2025 Pearson Education, Inc.
Section 7.6: Graphs of the Sine and Cosine Functions
coordinates of the five key points for y cos x by A 4 and then adding 2 units. Thus, the graph of y 4 cos 3x 2 will lie between 2
3 3 9 , , , and 3 4 2 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 5 2 y sin x , we multiply the y3 3
0,
and 6 on the y-axis. The five key points are 2 0, 2 , , 2 , , 6 , , 2 , , 2 6 3 2 3 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
coordinates of the five key points for y sin x 5 by A .The five key points are 3 3 5 3 9 5 0, 0 , , , , 0 , , , 3, 0 4 3 2 4 3 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, , and the range is 2, 6 . 55. Since sine is an odd function, we can plot the 5 2 equivalent form y sin x . 3 3
From the graph we can determine that the domain is all real numbers, , and the
5 2 x to Comparing y sin 3 3 5 2 y A sin x , we find A and . 3 3 5 5 Therefore, the amplitude is and the 3 3 2 3 . Because the amplitude is period is 2 / 3 5 5 2 , the graph of y sin x will lie 3 3 3
5 5 range is , . 3 3
56. Since cosine is an even function, we consider the 9 3 x . Comparing equivalent form y cos 5 2
9 3 y cos x to y A cos x , we find 5 2 9 3 and . Therefore, the amplitude is 5 2 2 4 9 9 and the period is . Because 5 5 3 / 2 3 9 the amplitude is , the graph of 5 9 9 9 3 y cos x will lie between and 5 5 5 2 A
5 5 and on the y-axis. Because the 3 3 period is 3 , one cycle will begin at x 0 and end at x 3 . We divide the interval 0,3 into
between
four subintervals, each of length
3 by finding 4
the following values:
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Chapter 7: Trigonometric Functions
3 57. We begin by considering y cos x . 2 4
4 , one 3 4 cycle will begin at x 0 and end at x . We 3 4 divide the interval 0, into four subintervals, 3 4/3 1 by finding the following each of length 4 3 values: 1 2 4 0, , , 1 , and 3 3 3 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 9 3 y cos x , we multiply the y-coordinates 5 2
on the y-axis. Because the period is
of the five key points for y cos x by A
3 Comparing y cos x to 2 4 3 y A cos x , we find A and . 4 2 3 3 Therefore, the amplitude is and the 2 2 2 3 8 . Because the amplitude is , period is 2 /4 3 the graph of y cos x will lie between 2 4 3 3 and on the y-axis. Because the period is 2 2 8, one cycle will begin at x 0 and end at x 8 . We divide the interval 0,8 into four
9 . 5
8 2 by finding the 4 following values: 0, 2, 4, 6, and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 3 1 y cos x , we multiply the y2 4 2 coordinates of the five key points for y cos x
subintervals, each of length
9 3 x will lie Thus, the graph of y cos 5 2
between
9 9 and on the y-axis. The five key 5 5
points are 9 1 2 9 4 9 0, , , 0 , , , 1, 0 , , 5 3 3 5 3 5 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
3 1 and then add unit. Thus, the 2 2 3 1 graph of y cos x will lie between 2 4 2 1 and 2 on the y-axis. The five key points are 1 1 0, 1 , 2, , 4, 2 , 6, , 8, 1 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
by A
y
( 4, 2)
2
From the graph we can determine that the domain is all real numbers, , and the
8 4
9 9 range is , . 5 5
( 8, 1)
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Copyright © 2025 Pearson Education, Inc.
2
1 (2, –– ) 2 (4, 2) 1 (6, –– ) 2 x 4 8
(0, 1) (8, 1)
Section 7.6: Graphs of the Sine and Cosine Functions
From the graph we can determine that the domain is all real numbers, , and the
y 2.5
range is 1, 2 .
3 (0, –– ) 2 3 (8, –– ) 2
1 58. We begin by considering y sin x . 2 8 1 Comparing y sin x to y A sin x , 2 8
(4, 1)
3 (16, –– ) 2
x 16 12 8 4 0.5
1 and . Therefore, the 8 2 1 1 amplitude is and the period is 2 2 2 1 16 . Because the amplitude is , the /8 2 1 1 graph of y sin x will lie between 2 8 2
4
8 12 16
From the graph we can determine that the domain is all real numbers, , and the
we find A
range is 1, 2 . 59.
A 3; T ;
2 2 2 T
y 3sin(2 x)
1 on the y-axis. Because the period is 16, 2 one cycle will begin at x 0 and end at x 16 . We divide the interval 0,16 into four
60.
and
subintervals, each of length
(12, 2)
( 4, 2)
A 2; T 4;
2 2 1 T 4 2
1 y 2sin x 2
16 4 by finding 4
61.
the following values: 0, 4, 8, 12, and 16 These values of x determine the x-coordinates of the five key points on the graph. To obtain the ycoordinates of the five key points for 1 3 y sin x , we multiply the y2 8 2
A 3; T 2; y 3sin(x)
62.
2 2 T 2
A 4; T 1; y 4sin(2 x)
2 2 2 T 1
63. The graph is a cosine graph with amplitude 5 and period 8. 2 Find : 8
coordinates of the five key points for y sin x 1 3 and then add units. Thus, the 2 2 1 3 graph of y sin x will lie between 2 8 2 1 and 2 on the y-axis. The five key points are 3 3 3 0, , 4,1 , 8, , 12, 2 , 16, 2 2 2 We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
by A
8 2
2 8 4
The equation is: y 5cos x . 4
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Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 64. The graph is a sine graph with amplitude 4 and period 8π. 2 Find : 8
Find :
2 1 8 4
69. The graph is a reflected sine graph with 4 . amplitude 1 and period 3 4 2 Find : 3 4 6 6 3 4 2 3 The equation is: y sin x . 2
65. The graph is a reflected cosine graph with amplitude 3 and period 4π. 2 Find : 4
4 2 2 1 4 2 1 The equation is: y 3cos x . 2
70. The graph is a reflected cosine graph with amplitude π and period 2π. 2 Find : 2
66. The graph is a reflected sine graph with amplitude 2 and period 4. 2 Find : 4
2 2 2 1 2 The equation is: y cos x .
4 2 2 4 2 The equation is: y 2sin x . 2
71. The graph is a reflected cosine graph, shifted up 3 1 unit, with amplitude 1 and period . 2 3 2 Find : 2 3 4 4 3 4 The equation is: y cos x 1 . 3
3 and 4
period 1. Find : 1
5 The equation is: y cos x . 2
1 The equation is: y 4sin x . 4
67. The graph is a sine graph with amplitude
2
2 2 2 2
8 2
2
2
2
3 The equation is: y sin 2 x . 4
68. The graph is a reflected cosine graph with 5 amplitude and period 2. 2
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Section 7.6: Graphs of the Sine and Cosine Functions 72. The graph is a reflected sine graph, shifted down 1 4 . 1 unit, with amplitude and period 2 3 4 2 Find : 3 4 6 6 3 4 2 1 3 The equation is: y sin x 1 . 2 2
Find :
2
2 2 2 The equation is: y 4sin 2 x .
77.
73. The graph is a sine graph with amplitude 3 and period 4. 2 Find : 4
f / 2 f 0
/20
sin / 2 sin 0
/2
1 0 2 /2
The average rate of change is
4 2 2 4 2
78.
The equation is: y 3sin x . 2
f / 2 f 0
/20
2
cos / 2 cos 0
/2 0 1 2 /2
The average rate of change is
74. The graph is a reflected cosine graph with amplitude 2 and period 2. 2 Find : 2
79.
f / 2 f 0
/20
2 2 2 2 The equation is: y 2 cos( x) .
75. The graph is a reflected cosine graph with 2 . amplitude 4 and period 3 2 2 Find : 3 2 6 6 3 2 The equation is: y 4 cos 3 x .
.
2
.
cos 2 cos(2 0) 2 /2 cos( ) cos(0) 1 1 /2 /2 2 4 2
f / 2 f (0) /20
The average rate of change is
76. The graph is a sine graph with amplitude 4 and period π.
2
1 1 sin sin 0 2 2 2 /2 sin / 4 sin 0 /2 2 2 2 2 2 /2 2
The average rate of change is
80.
.
831
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4
.
Chapter 7: Trigonometric Functions
81.
f g x sin 4 x
g f x cos 2 x
g f x 4 sin x 4sin x
82.
84.
f g x 3 sin x 3sin x
g f x sin 3x
f g x cos
1 x 2
85. 1 2
g f x cos x
83.
1 cos x 2
f g x 2 cos x 2 cos x
832
Copyright © 2025 Pearson Education, Inc.
Section 7.6: Graphs of the Sine and Cosine Functions 90. I (t ) 120sin(30 t ), t 0 2 2 1 Period: T second 30 15 Amplitude: A 120 120 amperes
86.
87. y sin x , 2 x 2
V (t )
2
91. a.
P (t )
R
V0 sin 2ft R 2 2 V sin 2ft 0 R V0 2 sin 2 2ft R
88. y cos x , 2 x 2
2
b. The graph is the reflected cosine graph translated up a distance equivalent to the 1 , so 4 f . amplitude. The period is 2f
The amplitude is
The equation is: V2 V2 P t 0 cos 4ft 0 2R 2R 2 V 0 1 cos 4ft 2R
89. I t 220sin(60 t ), t 0 2 1 second 60 30 Amplitude: A 220 220 amperes Period: T
2
1 V0 2 V0 2 . 2 R 2R
c.
92. a.
Comparing the formulas: 1 sin 2 2ft 1 cos 4ft 2 Since the tunnel is in the shape of one-half a sine cycle, the width of the tunnel at its base is one-half the period. Thus, 2 T 2(28) 56 or . 28 The tunnel has a maximum height of 15 feet so we have A 15 . Using the form y A sin( x) , the equation for the sine
833
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Chapter 7: Trigonometric Functions
is 23.65 + 51.75 = 75.4. The highest average monthly temperature is 75.4 .
curve that fits the opening is x y 15sin . 28
b. The magnitude is 23.65 and the vertical shift is 51.75 so the lowest the function reaches is 23.65 + 51.75 = 28.1. The lowest average monthly temperature is 28.1 .
b. Since the shoulders are 7 feet wide and the road is 14 feet wide, the edges of the road correspond to x 7 and x 21 . 7 15 2 10.6 15sin 15sin 2 28 4
c.
21 3 15 2 10.6 15sin 15sin 28 2 4 The tunnel is approximately 10.6 feet high at the edge of the road. 93. a.
period
The magnitude is 20 and the vertical shift is 100 so the highest the function will go is 100 + 20 = 120. This is the individual’s systolic pressure.
96. a.
2 6 The period is . Therefore, the 7 3 7
c.
rate in beats per minute would be 1 beat 60 sec 70 beats per min . 6 sec 1 min 7
The time between the shortest and longest dayes would be half the period. Thus the time between the days would be: 2
365
365 period 365 182.5 days. 2 2
The magnitude is 2.91 and the vertical shift is 2.97 so the highest the function will reach is 2.91+2.97 = 5.88. The height of the water at high tide is 5.88 ft.
97. a.
The amplitude is 100 and the vertical shift is 105 so the highest the wheel would go is 100 + 105 = 205 ft.
b. The lowest the seat would so is 100 + 105 = 5 ft. c.
The time between high and low tide would be half of the period. Thus the time between the tides would be: 2 2(149) 12.42 24 24 149 period 12.42 6.21 hrs. 2 2
period
95. a.
The magnitude is 1.615 and the vertical shift is 12.135 so the highest the function reaches is 1.615 + 12.135 = 13.75. The longest day would have 13.75 hours of daylight.
period
b. The magnitude is 2.91 and the vertical shift is 2.97 so the lowest reaches will be 2.91 + 2.97 = 0.06. The height of the water at low tide is 0.06 ft. c.
12
b. The magnitude is 1.615 and the vertical shift is 12.135 so the highest the function reaches is 1.615 + 12.135 = 10.52. The shortest day would have 10.52 hours of daylight.
heart beats once every 76 seconds. The heart
94. a.
2
6 period 12 6 mo. 2 2
b. The magnitude is 20 and the vertical shift is 100 so the lowest the function will go is 20 + 100 = 80. This is the individual’s diastolic pressure. c.
The time between the high and low temperatures would be half the period. Thus the time between the temperatures would be:
The period of the function is 2 period 100 sec . In 5 minutes (300 50 seconds) the wheel would make 3 revolutions.
d. The radius of the wheel is 100 so d s r 100(2 ) 200 d 200 ft 1 mile 60sec 60 min t 100sec 5280 ft 1min 1 hour 4.3 mph
v
The magnitude is 23.65 and the vertical shift is 51.75 so the highest the function reaches
834
Copyright © 2025 Pearson Education, Inc.
Section 7.6: Graphs of the Sine and Cosine Functions 98. a.
The cosine function has a max and min or 1 and 1 respectively so the average of the cos function is 0. So the average of d (t ) 50 cos( t 39) is 0. Adding 60 to the function will increase the average value by 60. So, d (t ) 50 cos( t 39) 60 will have an average value of 60 miles.
b. The period is P c.
2
d.
78 minutes
39
1 rev 2 rad rad/min 78 min 1 rad 39 rad 60 min r 50 mi 241.66 mph 39 min 1 hr
Physical potential peaks at 15 days after the 20th birthday, with minimums at the 3rd and 26th days. Emotional potential is 50% at the 17th day, with a maximum at the 10th day and a minimum at the 24th day. Intellectual potential starts fairly high, drops to a minimum at the 13th day, and rises to a maximum at the 29th day.
100. The graph of y A sin( Bx C ) D oscillates
between A D and A D . So to lie below d.
1 hr d rt (241.66 mph)(78 min) 60 min 314.125 miles
the x-axis completely, A D 0 or D A . 101. The y-intercept: A cos( BC ) A x-intercepts: 0 A cos B( x C ) A
1.8 mi 314.125 miles x gal 1 gal 1.8 x 314.125
0 cos B ( x C ) 1 1 cos B ( x C )
x 174.53 gal
99. a.
1 cos B ( x C )
2 ; 23 2 Emotional potential: ; 28 14 2 Intellectual potential: 33
Physical potential:
So, B ( x C ) (2k 1) , k an integer. (2k 1) x C B (2k 1) x C B The intercepts are (0, A cos( BC ) A) and
b.
(2k 1) C , 0 , k an integer. B
102 – 106. Answers will vary.
2 #1: P t 50sin t 50 23 # 2 : P t 50sin t 50 14 2 #3 : P t 50sin t 50 33
c.
No.
835
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Chapter 7: Trigonometric Functions
107.
f ( x h) f ( x ) h ( x h) 2 5( x h) 1 ( x 2 5 x 1) h 2 2 ( x 2 xh h ) (5 x 5h) 1 x 2 5 x 1 h 2 2 x 2 xh h 5 x 5h 1 x 2 5 x 1 h 2 xh h 2 5h h(2 x h 5) 2x h 5 h h
0.04 111. 2 P P 1 4
0.04 2 1 4
4t
4t
0.08 ln 2 ln 1 12 0.04 ln 2 4t ln 1 4 ln 2 t 17.4 0.04 4 ln 1 4 It will take about 17.4 years to double.
108. We need to use completing the square to put the function in the form f ( x ) a ( x h) 2 k
112. e3 x 7 3x ln 7 ln 7 x 0.649 3
f ( x) 3 x 2 12 x 7 3( x 2 4 x) 7 144 144 3 x 2 4 x 7 3 4( 3) 4( 3) 2
4t
2
113. Dividing:
3( x 2 4 x 4) 7 12
2x 7 2 x 2 4 x 3 4 x3 +6 x 2 3 x 1
3( x 2 4 x 4) 5 3( x 2) 2 5 So the vertex is (2,5) .
4 x3 8 x 2 6 x
109. The y-intercept is: y 3 0 2 1
14 x 9 x 1 14 x 2 28 x 21
y 6 1 5
19 x 20 19 x 20
2
0,5
G ( x) 2 x 7
2 x2 4 x 3 Thus, the oblique asymptote is y 2 x 7 .
The x-interecpts are: 0 3 x 2 1
19 19 . Since the 2(6) 12 graph is concave up the graph is decreasing on 19 12 , .
1 x2 3 1 1 x 2 or x 2 3 3 5 7 or x x 3 3 5 7 , 0 , , 0 3 3
114. The vertex occurs at
1 3 x2
115.
, 2 ,
110. 3x 2(5 x 16) 3 x 4(8 x) 3 x 10 x 32 3 x 32 4 x 8 x 64 x 8
The solution set is: 8 .
836
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4 3
Section 7.7: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions 116. log( x 3) log( x 3) log( x 4) x3 log log( x 4) x3 x3 x4 x3 x 3 ( x 4)( x 3) x 3 ( x 4)( x 3) x 3 x 2 x 12
0 x2 2 x 9 a 1, b 2, c 9
11. sec x 1 when x 2, 0, 2; sec x 1 when x , 3 , ; 2 2 3 csc x 1 when x , 2 2
12. csc x 1 when x
13. y sec x has vertical asymptotes when
(2) (2) 2 4(1)(9) x 2(1)
10. y csc x has no y-intercept.
2 40 2 2 10 1 10 2 2
But 1 10 would result is taking a logarithm of a negative expression and thus cannot be used so the solution set is 1 10 .
x
3 3 . , , , 2 2 2 2
14. y csc x has vertical asymptotes when x 2, , 0, , 2 . 15. y tan x has vertical asymptotes when x
3 3 . , , , 2 2 2 2
16. y cot x has vertical asymptotes when x 2, , 0, , 2 . 17. y 3 tan x ; The graph of y tan x is stretched vertically by a factor of 3.
Section 7.7 1. x 4 2. True 3. a.
,
b. odd; origin c. d. odd; 2 e. 1, 1, 0 4. y-axis; x = odd multiples of
k The domain is x x , k is an odd integer . 2 The range is the set of all real number or (, ) .
2
18. y 2 tan x ; The graph of y tan x is stretched vertically by a factor of 2 and reflected about the
5. b 6. True 7. The y-intercept of y tan x is 0. 8. y cot x has no y-intercept. 9. The y-intercept of y sec x is 1.
837 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
x-axis.
21. y tan x ; The graph of y tan x is 2 2 horizontally compressed by a factor of .
k The domain is x x , k is an odd integer . 2 The range is the set of all real number or (, ) .
19. y 4 cot x ; The graph of y cot x is stretched vertically by a factor of 4.
The domain is x x does not equal an odd integer . The range is the set of all real number or (, ) . 1 22. y tan x ; The graph of y tan x is 2 horizontally stretched by a factor of 2.
The domain is x x k , k is an integer . The range is the set of all real number or (, ) .
20. y 3cot x ; The graph of y cot x is stretched vertically by a factor of 3 and reflected about the x-axis.
The domain is x x k , k is an odd integer . The range is the set of all real number or (, ) . 1 23. y cot x ; The graph of y cot x is 4 horizontally stretched by a factor of 4.
The domain is x x k , k is an integer . The range is the set of all real number or (, ) . The domain is x x 4k , k is an integer . The range is the set of all real number or (, ) . 838 Copyright © 2025 Pearson Education, Inc.
Section 7.7: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
24. y cot x ; The graph of y cot x is 4 4 horizontally stretched by a factor of .
The domain is x x k , k is an integer . The 1 1 range is y y or y . 2 2
27. y 3csc x ; The graph of y csc x is vertically stretched by a factor of 3 and reflected about the x-axis.
The domain is x x 4k , k is an integer . The range is the set of all real number or (, ) . 25. y 2sec x ; The graph of y sec x is stretched vertically by a factor of 2.
The domain is x x k , k is an integer . The range is y y 3 or y 3 . 28. y 4sec x ; The graph of y sec x is vertically stretched by a factor of 4 and reflected about the x-axis.
k The domain is x x , k is an odd integer . 2
The range is y y 2 or y 2 .
1 csc x ; The graph of y csc x is vertically 2 1 compressed by a factor of . 2
26. y
k The domain is x x , k is an odd integer . 2
The range is y y 4 or y 4 .
1 29. y 4sec x ; The graph of y sec x is 2 horizontally stretched by a factor of 2 and
839 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
vertically stretched by a factor of 4.
The domain is x x does not equal an integer . The domain is x x k , k is an odd integer . The range is y y 4 or y 4 . 1 csc 2 x ; The graph of y csc x is 2 1 horizontally compressed by a factor of and 2 1 vertically compressed by a factor of . 2
30. y
The range is y y 2 or y 2 . 32. y 3sec x ; The graph of y sec x is 2 2 horizontally compressed by a factor of ,
vertically stretched by a factor of 3, and reflected about the x-axis.
The domain is x x does not equal an odd integer . k The domain is x x , k is an integer . The 2 1 1 range is y y or y . 2 2
31. y 2 csc x ; The graph of y csc x is
horizontally compressed by a factor of
1
The range is y y 3 or y 3 . 1 33. y tan x 1 ; The graph of y tan x is 4 horizontally stretched by a factor of 4 and shifted up 1 unit.
,
vertically stretched by a factor of 2, and reflected about the x-axis.
840 Copyright © 2025 Pearson Education, Inc.
Section 7.7: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
2 35. y sec x 2 ; The graph of y sec x is 3 3 and horizontally compressed by a factor of 2 shifted up 2 units.
The domain is x x 2k , k is an odd integer . The range is the set of all real number or (, ) . 34. y 2 cot x 1 ; The graph of y cot x is vertically stretched by a factor of 2 and shifted down 1 unit.
3 The domain is x x k , k is an odd integer . 4
The range is y y 1 or y 3 .
3 36. y csc x ; The graph of y csc x is 2 2 horizontally compressed by a factor of . 3
The domain is x x k , k is an integer . The range is the set of all real number or (, ) .
2 The domain is x x k , k is an integer . The 3
range is y y 1 or y 1 .
841 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
1 tan 2 x 2 ; The graph of y tan x is 2 horizontally compressed by a factor of 2, 1 vertically compressed by a factor of , and 2 shifted down 2 units.
37. y
39. y 2 csc x 1 ; The graph of y csc x is 2 2 horizontally compressed by a factor of ,
vertically stretched by a factor of 2, and shifted down 1 unit.
The domain is x x k , k is an odd integer . 4 The range is the set of all real number or (, ) .
1 38. y 3cot x 2 ; The graph of y cot x is 2 horizontally stretched by a factor of 2, vertically stretched by a factor of 3, and shifted down 2 units.
The domain is x x 2 k , k is an integer . The range is the set of all real number or (, ) .
The domain is x x k , k is an integer . The range is y y 3 or y 1 . 1 40. y 3sec x 1 ; The graph of y sec x is 4 horizontally stretched by a factor of 4, vertically stretched by a factor of 3, and shifted up 1 unit.
The domain is x x 2 k , k is an odd integer . The range is y y 2 or y 4 .
842 Copyright © 2025 Pearson Education, Inc.
Section 7.7: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
3 f f 0 0 tan / 6 tan 0 6 41. 3 /6 /6 0 6 3 6 2 3 3 2 3 . The average rate of change is
45.
f g x tan 4 x
42.
2 3 f f 0 1 sec / 6 sec 0 6 3 /6 /6 0 6 2 3 3 6 2 3 2 3 3
The average rate of change is
2 3 2 3
.
f f 0 tan 2 / 6 tan 2 0 6 43. /6 0 6 3 0 6 3 /6 6 3 . The average rate of change is
g f x 4 tan x 4 tan x
46.
f g x 2sec
f f 0 sec 2 / 6 sec 2 0 6 44. /6 0 6 2 1 6 /6 6 The average rate of change is .
843 Copyright © 2025 Pearson Education, Inc.
1 x 2
Chapter 7: Trigonometric Functions
g f x 2 csc
1 x 2
1 2
g f x 2sec x sec x
47.
f g x 2 cot x 2 cot x 49.
g f x cot 2 x
50.
48.
1 2
f g x 2 csc x csc x
51. a.
Consider the length of the line segment in two sections, x, the portion across the hall that is 3 feet wide and y, the portion across that hall that is 4 feet wide. Then, 3 4 and sin cos x y 3 4 x y cos sin
844 Copyright © 2025 Pearson Education, Inc.
Section 7.7: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
in the rotation of the beacon when the beam of light being cast on the wall changes from one side of the beacon to the other.
Thus, L x y b. Let Y1
c.
3 4 3sec 4 csc . cos sin c.
3 4 . cos x sin x
2
d.
Use MINIMUM to find the least value:
2
e.
L is least when 0.83 .
d.
L
3 4 9.86 feet . cos 0.83 sin 0.83
Note that rounding up will result in a ladder that won’t fit around the corner. Answers will vary. 52. a.
b.
d t 10 tan( t )
d t 10 tan( t ) is undefined at t
t
d t 10 tan( t )
0 0.1 0.2 0.3 0.4
0 3.2492 7.2654 13.764 30.777
d (0.1) d (0) 3.2492 0 32.492 0.1 0 0.1 0 d (0.2) d (0.1) 7.2654 3.2492 40.162 0.2 0.1 0.2 0.1 d (0.3) d (0.2) 13.764 7.2654 64.986 0.3 0.2 0.3 0.2 d (0.4) d (0.3) 30.777 13.764 170.13 0.4 0.3 0.4 0.3 The first differences represent the average rate of change of the beam of light against the wall, measured in feet per second. For example, between t 0 seconds and t 0.1 seconds, the average rate of change of the beam of light against the wall is 32.492 feet per second.
53.
1 and 2
3 , or in general at 2 k k is an odd integer . At these t 2 instances, the length of the beam of light approaches infinity. It is at these instances t
845 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
Time to do job Part of job done in one hour 1 Hazel t t 1 Gwyneth t2 t 2 1 Together 2.4 2.4 1 1 1 t t 2 2.4 24(t 2) 24t 10 t (t 2) 24t 48 24t 10t 2 20t 0 10t 2 28t 48
Yes, the two functions are equivalent.
0 5t 2 14t 24 0 (t 4)(5t 6) t 4 0 or 5t 6 0 6 t 4 or t 5
54. We need tan x 0 so the angle x needs to be in Quadrant I or Quadrant III. The domain is 2k 1 , k an integer . The range x | k x 2 is the set of all real numbers. The graph of y tan x has vertical asymptotes when cos x 0 , so f ( x) log(tan x) will have these
asymptotes at every integer multiple of is, at x
k , where k an integer . 2
2
6 Time must be positive, so disregard t . 5 Hazel takes 4 hours to complete the job alone. Gwyneth 4 + 2 = 6 hours to complete the job alone.
. That
55. We need sin 0 or x k , so the domain is x | x k , k an integer . Since 0 sin x 1 ,
we have ln sin x 0 so the range is
y | y 0 or , 0 . The function
58.
2
9 x 1 3x 5 2
32( x 1) 3x 5 2( x 1) x 2 5 2 x 2 x2 5 x2 2 x 3 0
f ( x) ln sin x will have vertical asymptotes at
( x 3)( x 1) 0
where sin 0 . That is, at x k , where k an integer .
So x 3 or x 1 . The solution set is: 3, 1 .
56. We use the difference in cubes formula: 125 p 3 8q 6 (5 p)3 (2q 2 )3
59. The slope is
(5 p 2q 2 )(25 p 2 10 pq 2 4q 4 )
57. Let t represent the time it takes Hazel to complete the paint job alone. Then t 2 represents the time it takes Gwyneth to complete the paint job alone.
846 Copyright © 2025 Pearson Education, Inc.
1 and the y-intercept is (0, 3) . 4
Section 7.8: Phase Shift; Sinusoidal Curve Fitting 60. The log function must be positive so we need x4 0. x x4 f ( x) x The zeros and values where f is undefined are x 0 and x 4 . Interval Number
(, 0)
(0, 4)
(4, )
2
1
5
3
3
0.2
Chosen Value of f
Conclusion Positive Negative Positive
65. The argument of 4 5 x 2 3 cannot be negative solve 5 x 2 0 to find the domain. 5x 2 0 5x 2 x
2 5
2 The domain is , . 5
Section 7.8 1. phase shift
The domain is x x 0 or x 4 or, using interval notation, , 0 4, . 61.
62.
3 1 4 3 2 g (4) 3(4) 7 5 f (3)
2
2
x 3x c 3c f ( x ) f (c ) xc xc
x c 3x 3c 2
2
xc ( x c)( x c) 3( x c) xc ( x c)[ x c 3] xc3 xc
63. The y-intercept occurs when x = 0: y
2(0) 2 0 6 6 2 03 3
2. False 3. y 4sin(2 x ) Amplitude:
A 4 4
Period:
T
Phase Shift: 2
2 x 2 x 6 (2 x 3)( x 2) 0 2 x 3 0 or x 2 0 3 or x 2 2
3 The intercepts are: 0, 2 , ,0 , 2, 0 . 2
64.
2 2
Interval defining one cycle: 3 , T 2 , 2 Subinterval width: T 4 4 Key points: 3 5 3 , 0 , , 4 , , 0 , , 4 , , 0 2 4 4 2
The x-intercept occurs when y = 0:
x
2
x 2 2 x 26 ( x 2 2 x 1) 26 1 ( x 1) 2 25
847 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 4. y 3sin(3 x ) Amplitude:
A 3 3
Period:
T
2
Phase Shift: 3
2 3
Interval defining one cycle: , T 3 , Subinterval width: T 2 / 3 4 4 6
6. y 3cos 2 x Amplitude:
Key points: 2 5 , 0 , , 3 , , 0 , 0 , ,3 , 3 2 3 6
A 3 3
2 2 Phase Shift: 2 2 Interval defining one cycle: , T 2 , 2 Subinterval width: T 4 4 Key points: ,3 , , 0 , 0, 3 , , 0 , ,3 2 4 4 2 Period:
T
2
5. y 2 cos 3 x 2 Amplitude: A 2 2 2 3 2 Phase Shift: 3 6 Interval defining one cycle: , T 6 , 2 Subinterval width: T 2 / 3 4 4 6 Key points: , 2 , 0, 0 , , 2 , , 0 , , 2 6 6 3 2 Period:
T
2
7. y 3sin 2 x 2 Amplitude: A 3 3 Period:
T
2
2 2
Phase Shift: 2 2 4 Interval defining one cycle: 3 , T 4 , 4
848 Copyright © 2025 Pearson Education, Inc.
Section 7.8: Phase Shift; Sinusoidal Curve Fitting
Subinterval width: T 4 4 Key points: 3 , 0 , 0, 3 , , 0 , ,3 , , 0 4 4 2 4
9. y 4 cos(4 x ) 2 Amplitude:
A 4 4
Period:
T
2
Phase Shift: 4
2 4 2
Interval defining one cycle: , T 4 , 4 2 4 , 4 Subinterval width:
T 4 8
Key points: , 2 , , 2 , 0, 6 , , 2 , 4 8 8 ,2 4
8. y 2 cos 2 x 2 Amplitude: A 2 2 Period:
T
2
2 2
2 Phase Shift: 2 4 Interval defining one cycle: 5 , T 4 , 4 Subinterval width: T 4 4 Key points: 3 5 , 2 , , 0 , , 2 , , 0 , , 2 4 2 4 4
10. y 4sin(4 x ) 3 Amplitude:
A 4 4
Period:
T
Phase Shift:
4
2
2 4 2
Interval defining one cycle: , T 4 , 4 2 4 , 4 Subinterval width: T 4 8 Key points: ,3 , ,8 , 0,3 , , 2 , ,3 4 8 8 4
849 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
Interval defining one cycle: 2 2 , T ,1 T 1 Subinterval width: 4 4 Key points: 2 1 2 1 2 3 2 ,6 , , 4 , , 2 , , 4 , 4 2 4 2 1 , 6 11. y 4sin(x 2) 5 Amplitude:
A 4 4
2 2 2 2 Phase Shift: Interval defining one cycle: 2 2 , T , 2 Subinterval width: T 2 1 4 4 2 Key points: 2 1 2 2 , 5 , , 1 , 1 , 5 , 2 Period:
T
2
2 3 2 , 9 , 2 , 5 2
13. y 3cos(x 2) 5 Amplitude:
A 3 3
Period:
T
Phase Shift:
2
2
2 2
Interval defining one cycle: 2 2 , T , 2 Subinterval width: T 2 1 4 4 2 Key points: 2 1 2 2 3 2 ,8 , ,5 , 1 , 2 , ,5 , 2 2 12. y 2 cos(2x 4) 4 Amplitude:
2 2 ,8
A 2 2
2 1 2 2 4 Phase Shift: 2 Period:
T
2
850 Copyright © 2025 Pearson Education, Inc.
Section 7.8: Phase Shift; Sinusoidal Curve Fitting
15. y 3sin 2 x 3sin 2 x 2 2 3sin 2 x 2 Amplitude:
A 3 3
Period:
T
2
2 2
2 Phase Shift: 2 4 14. y 2sin(2x 4) 1 Amplitude:
A 2 2
2 1 2 4 2 Phase Shift: 2 Interval defining one cycle: 2 2 , T ,1 Subinterval width: T 1 4 4 Key points: 2 1 2 1 2 , 1 , ,1 , , 1 , 4 2 Period:
T
2
Interval defining one cycle: 5 , T 4 , 4 Subinterval width: T 4 4 Key points: 3 5 , 0 , ,3 , , 0 , , 3 , , 0 4 2 4 4
2 3 2 , 3 , 1 , 1 4 16. y 3cos 2 x 3cos 2 x 2 2 3cos 2 x 2 Amplitude:
A 3 3
Period:
T
2
2 2
2 Phase Shift: 2 4 Interval defining one cycle: 5 , T 4 , 4 Subinterval width: 851 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions T 4 4 Key points: 3 5 , 3 , , 0 , ,3 , , 0 , , 3 4 2 4 4
2) Horizontally compress by a factor of y cot 2 x
3) Vertically compress by a factor of
1 2
1 2
1 y 2 cot 2 x
17. y 2 tan 4 x
Begin with the graph of y tan x and apply the following transformations: 1) Shift right units y tan x 2) Horizontally compress by a factor of y tan 4 x 3) Vertically stretch by a factor of 2 y 2 tan 4 x
1 4
19. y 3csc 2 x 4 Begin with the graph of y csc x and apply the following transformations: 1) Shift right
units y csc x 4 4
2) Horizontally compress by a factor of y csc 2 x 4 3) Vertically stretch by a factor of 3 y 3csc 2 x 4
1 cot 2 x 2 Begin with the graph of y cot x and apply the following transformations:
18. y
1) Shift right units y cot x
852 Copyright © 2025 Pearson Education, Inc.
1 2
Section 7.8: Phase Shift; Sinusoidal Curve Fitting
1 20. y sec 3x 2 Begin with the graph of y sec x and apply the following transformations: 1) Shift right units y sec x 1 2) Horizontally compress by a factor of 3 y sec 3 x 1 3) Vertically compress by a factor of 2 1 y 2 sec 3 x
, 1 , ,1 , 0,3 , ,1 , , 1 4 8 8 4
22. y 2sin 4 x 2 2 2sin 4 x 2 2 2sin 4 x 2 2 Amplitude:
2 4 2 2 Phase Shift: 4 2 Interval defining one cycle: , T 2 , 0 Subinterval width: T 4 8 Key points: 3 , 2 , , 0 , , 2 , , 4 , 2 8 4 8 Period:
21. y 2 cos 4 x 1 3cos 4 x 1 3cos 4 x 1 Amplitude:
A 2 2
Period:
T
2
2 4 2
4 Interval defining one cycle: , T 4 , 4 2 4 , 4 Subinterval width: T Phase Shift:
A 3 3
0, 2
4 8 Key points:
853 Copyright © 2025 Pearson Education, Inc.
T
2
Chapter 7: Trigonometric Functions
23.
1 2 2 2 1 2 2 2 T 1 Assuming A is positive, we have that y A sin( x ) 2sin(2 x 1) A 2; T ;
1 2sin 2 x 2
24.
A 3; T
27. I t 120sin 30 t , t 0 3 2 2 1 Period: T second 30 15 Amplitude: A 120 120 amperes
1 Phase Shift: second 3 30 90
; 2 2
2 2 2 4 T 4 2 8 Assuming A is positive, we have that y A sin( x ) 3sin(4 x 8)
3sin 4 x 2
25.
1 3 2 2 2 1 2 T 3 3 3 A 3; T 3;
3
28. I t 220sin 60 t , t 0 6 2 2 1 Period: T second 60 30 Amplitude: A 220 220 amperes
1 Phase Shift: second 6 60 360
1 2 2 3 3 9 Assuming A is positive, we have that 2 2 y A sin( x ) 3sin x 3 9 2 1 3sin x 3 3
26.
2 2 2 2 2 T 2 4 A 2; T ;
29. a.
Assuming A is positive, we have that y A sin( x ) 2sin(2 x 4) 2sin 2 x 2
854 Copyright © 2025 Pearson Education, Inc.
Section 7.8: Phase Shift; Sinusoidal Curve Fitting
b.
33 16 17 8.5 2 2 33+16 49 Vertical Shift: 24.5 2 2 2 2 5 5 Phase shift (use y 16, x 6): Amplitude: A
2 16 8.5sin 6 24.5 5 12 8.5 8.5sin 5
30. a.
b.
12 1 sin 5 12 2 5 29 10 11 2 Thus, y 8.5sin x 24.5 or 10 5
79.8 36.0 43.8 21.9 2 2 79.8+36.0 115.8 57.9 Vertical Shift: 2 2 2 12 6 Phase shift (use y 36.0, x 1): Amplitude: A
36.0 21.9sin 1 57.9 6 21.9 21.9sin 6 1 sin 6
2 11 y 8.5sin x 24.5 . 5 4
c.
2 6 2 3
2 Thus, y 21.9sin x 57.9 or 6 3 y 21.9sin x 4 57.9 . 6
c.
d.
y 9.52sin 1.229 x 2.976 24.230
e. d.
y 21.68sin 0.516 x 2.124 57.81
e.
855 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 31. a.
b.
32. a.
75.4 28.1 47.3 23.65 2 2 75.4+28.1 103.5 51.75 Vertical Shift: 2 2 2 12 6 Phase shift (use y 28.1, x 1): Amplitude: A
b.
77.0 32.9 44.1 22.05 2 2 77.0+32.9 109.9 54.95 Vertical Shift: 2 2 2 12 6 Phase shift (use y 32.9, x 1): Amplitude: A
28.1 23.65sin 1 51.75 6 23.65 23.65sin 6
22.05 22.05sin 6
1 sin 6
1 sin 6
2 6 2 3
e.
2 6 2 3
2 Thus, y 23.65sin x 51.75 or 6 3
2 Thus, y 22.05sin x 54.95 or 6 3
y 23.65sin x 4 51.75 . 6
y 22.05sin x 4 54.95 . 6
c.
d.
32.9 22.05sin 1 54.95 6
c.
y 24.25sin(0.493 x 1.927) 51.61
d.
y 21.73sin(0.518 x 2.139) 54.82
e.
856 Copyright © 2025 Pearson Education, Inc.
Section 7.8: Phase Shift; Sinusoidal Curve Fitting 33. a.
b.
1.17 + 12.4167 = 13.6167 hours which is at 1:35 p.m.
24 3.72 4.235 10.27 6.035sin 149 24 3.72 6.035 6.035sin 149
5.73 0.15
5.88 2.94 2 2 5.73 ( 0.15) 5.58 Vertical Shift: 2.79 2 2 2 24 12.4167 6.20835 149 Phase shift (use y 5.73, x 1.17): Ampl: A
89.28 1 sin 149 89.28 2 149 0.312
24 1.17 2.79 5.73 2.94sin 149 24 1.17 2.94 2.94sin 149 28.08 1 sin 149 28.08 2 149 0.979
c.
24 Thus, y 2.94sin x 0.979 2.79 149 24 or y 2.94sin x 1.934 2.79 . 149
c.
34. a. b.
24 Thus, y 6.035sin x 0.312 4.235 149 24 or y 6.035sin x 0.616 4.235 . 149
35. a.
24 y 2.94sin 17 1.934 2.79 149 2.33 feet
24 y 6.035sin 15 0.616 4.235 149 9.30 feet 13.75 10.52 1.615 2 13.75 10.52 Vertical Shift: 12.135 2 2 365 Phase shift (use y 13.75, x 172): Amplitude: A
2 13.75 1.615sin 172 12.135 365
3.72 + 12.4167 = 16.1367 hours which is at 4:08 p.m.
2 1.615 1.615sin 172 365
10.27 (1.80) 12.07 6.035 2 2 10.27 (1.80) 8.47 Vertical Shift: 4.235 2 2 2 24 12.4167 6.20835 149 Phase shift (use y 10.27, x 3.72):
344 1 sin 365
Ampl: A
344 2 365 1.3900 2 Thus, y 1.615sin x 1.39 12.135 or 365 2 y 1.615sin x 80.75 12.135 . 365
b.
2 y 1.615sin 91 80.75 12.135 365 12.42 hours
857 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions c.
c.
d. The actual hours of sunlight on April 1, 2022 were 12.43 hours. This is very close to the predicted amount of 12.42 hours.
d. The actual hours of sunlight on April 1, 2022 were 12.75 hours. This is very close to the predicted amount of 12.71 hours.
36. a.
15.27 9.07 3.1 2 15.27 9.07 Vertical Shift: 12.17 2 2 365 Phase shift (use y 15.27, x 172): Amplitude: A
37. a.
2 15.27 3.1sin 172 12.17 365
2 19.37 6.96sin 172 12.41 365
2 3.1 3.1sin 172 365
2 6.96 6.96sin 172 365
344 1 sin 365
344 1 sin 365
344 2 365 1.39
344 2 365 1.39
2 Thus, y 3.1sin x 1.39 12.17 or 365
2 Thus, y 6.96sin x 1.39 12.41 or 365
2 y 3.1sin x 80.75 12.17 . 365
b.
2 y 3.1sin 91 1.39 12.17 365 12.71 hours
19.37 5.45 6.96 2 19.37 5.45 Vertical Shift: 12.41 2 2 365 Phase shift (use y 19.37, x 172): Amplitude: A
2 y 6.96sin x 80.75 12.41 . 365
b.
2 y 6.96sin 91 1.39 12.41 365 13.63 hours
858 Copyright © 2025 Pearson Education, Inc.
Section 7.8: Phase Shift; Sinusoidal Curve Fitting d. The actual hours of sunlight on April 1, 2022 were 12.38 hours. This is very close to the predicted amount of 12.35 hours.
c.
d. The actual hours of sunlight on April 1, 2022 was 13.37 hours. This is close to the predicted amount of 13.63 hours. 38. a.
13.42 10.83 1.295 2 13.42 10.83 Vertical Shift: 12.125 2 2 365 Phase shift (use y 13.42, x 172): Amplitude: A
2 13.42 1.295sin 172 12.125 365 2 172 1.295 1.295sin 365 344 1 sin 365 344 2 365 1.39 2 Thus, y 1.295sin x 1.39 12.125 . 365
b.
2 y 1.295sin 91 1.39 12.125 365 12.35 hours
c.
39. The coaster car travels from a high of 106 ft to a low of 4 ft, so the amplitude is A = 51 and the vertical shift is B = 55. The car moves from a high point to a low point (1/2 period) in 1.8 seconds, so the period T = 3.6 seconds. 2 2 5 T 3.6 9 Assuming the car starts at the top of a hill, there is a phase shift to the left of 0.9 seconds. 5 y 51sin t 0.9 55 9 5 t 51sin 55 2 9 40 – 41. Answers will vary. 42.
4x 9 2 4y 9 x 2 2x 4 y 9
f ( x)
2x 9 4 y 2x 9 y 4 2x 9 f 1 ( x) 4
43. 0.25(0.4 x 0.8) 3.7 1.4 x 0.1x 0.2 3.7 1.4 x 1.5 x 3.5 35 7 x 15 3 7 The solution set is 3 44. (8 x 15 y ) 2 (8 x 15 y )(8 x 15 y ) 64 x 2 120 xy 120 xy 225 y 2 64 x 2 240 xy 225 y 2
859 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
51. log 2 (8 x 2 y 5 ) log 2 8 log 2 x 2 log 2 y 5
45. d ( x2 x1 ) 2 ( y2 y1 ) 2
3 2 log 2 x 5log 2 y
(10 4) 2 (3 ( 1)) 2 (6) 2 (4) 2 36 16 52 4(13) 2 13
46. 3x 4 5 x 7 or 3x 4 5 x 7 2 x 11 8x 3 11 3 x x 2 8
1. 135 135
2. 15 15
3 11 The solution set is , . 8 2
3.
47. u x 4 x u4 1
1
sin t
5. tan
x opp a hypo
adj 2 opp 2 hypo 2
radian radian 180 12
5 5 180 degrees 450 2 2
u 2 4u 2
48.
3 radian radians 180 4
2 2 180 degrees 120 3 3
4.
y (u 4) u (u 4)u 2 3
Chapter 7 Review Exercises
1 1 sin 1 4 6 2 2
6. 3sin 45º 4 tan
adj 2 x 2 a 2 adj a 2 x 2 adj cos t hypo
a x a 2
7. 6 cos
2
3 2 2 tan 6 2 3 4 3 2
3 2 2 3
49. The amount of fencing can be represented by 2l 2 w 54 . The length can be represented by l 2w 3 . Thus, 2l 2 w 54 l 2w 3 2(2w 3) 2 w 54 4 w 6 2 w 54 6w 6 54 6 w 48 w 8 ft l 2(8) 3 19 ft ( x 5)( x 5) ( x 5) . Since the (x + 3) ( x 5)( x 3) ( x 3) factor does not cancel, there is a vertical asymptote at x 3 .
50. R ( x)
2 3 3 2 4 3 3 4 6 2 3 2 3
5 5 8. sec cot sec cot 2 1 3 3 4 3 4
9. tan sin 0 0 0 10. cos 540º tan( 405º ) 1 ( 1) 1 1 0 11. sin 2 20º
1 sin 2 20º cos 2 20º 1 sec 20º 2
12. sec 50º cos 50º
13.
1 cos 50º 1 cos 50º
sin 50º sin 50º 1 cos 40 sin 50
860 Copyright © 2025 Pearson Education, Inc.
Chapter 7 Review Exercises
14.
sin 40º cos 50
4 and 0 , so lies in quadrant I. 5 2 Using the Pythagorean Identities: cos 2 1 sin 2
sin 40 1 sin 40
20. sin
15. sin 400º sec 50º sin 400º sec 50º 1 cos 50º sin 40º sin 40º 1 cos 50º sin 40º sin 40º 360º
16. sin
17 5 cos sin cos 6 6 6 6 3 1 3 2 2 4
17. 3sin110 csc110 3sin110 3
1 sin110
sin110 3(1) 3 sin110
18. 2 cos 2 2 4 4 3 cot 2 3 sin 2 2 sin 2 2 sin 2 2 3 3 3 2
1 2 4 4 cos 2 3 2 2 3 sin 2 3 2 1 15 4 4 4 5 3 3 4 4 cos 70 sin 20 1 csc 2 (135) 4 2 sin 20 sin ( 135) sin 20 1 1 4 4 2 1 2 2 42 2 2
9 3 25 5 Note that cos must be positive since lies in 3 quadrant I. Thus, cos . 5 sin 54 4 5 4 tan cos 53 5 3 3 cos
csc
1 1 5 5 1 sin 54 4 4
sec
1 1 5 5 3 1 cos 5 3 3
cot
1 1 3 3 1 tan 43 4 4
12 and sin 0 , so lies in quadrant III. 5 Using the Pythagorean Identities: sec 2 tan 2 1
21. tan
2
19. 4
2
16 9 4 cos 2 1 1 25 25 5
2
144 169 12 sec 2 1 1 25 25 5 169 13 25 5 Note that sec must be negative since lies in 13 quadrant III. Thus, sec . 5 1 1 5 cos sec 135 13 sec
tan
sin , so cos
12 5 12 5 13 13 1 1 13 12 csc sin 13 12
sin tan cos
cot
1 1 5 tan 125 12
861 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
5 and tan 0 , so lies in quadrant II. 4 Using the Pythagorean Identities: tan 2 sec 2 1
22. sec
2
2
25 9 5 tan 2 1 1 16 16 4 tan
25 144 5 cos 2 1 1 169 169 13
9 3 16 4
sin , so cos
3 4 3 sin tan cos . 4 5 5 1 1 5 csc sin 35 3 1 1 4 3 cot tan 4 3 3 and lies in quadrant III. 5 Using the Pythagorean Identities: sin 2 1 cos 2
23. cos
2
9 16 3 sin 2 1 1 25 25 5 16 4 25 5 Note that sin must be negative because lies 4 in quadrant III. Thus, sin . 5 sin 45 4 5 4 tan cos 35 5 3 3 sin
csc
1 1 5 4 sin 5 4
sec
1 1 5 cos 53 3
cot
1 1 3 4 tan 3 4
144 12 169 13 Note that cos must be positive because lies 12 in quadrant IV. Thus, cos . 13 sin 135 5 13 5 12 tan cos 13 12 12 13 cos
3 Note that tan 0 , so tan . 4 1 1 4 cos sec 54 5
tan
5 3 and 2 (quadrant IV) 13 2 Using the Pythagorean Identities: cos 2 1 sin 2
24. sin
csc
1 1 13 5 sin 13 5
sec
1 1 13 cos 12 12 13
cot
1 1 12 5 tan 12 5
1 and 180º 270º (quadrant III) 3 Using the Pythagorean Identities: sec 2 tan 2 1
25. tan
2
1 10 1 sec 2 1 1 3 9 9 10 10 9 3 Note that sec must be negative since lies in sec
quadrant III. Thus, sec cos
tan
1 sec
1
10 3
3 10
10 . 3
10 10
3 10 10
sin , so cos
1 3 10 10 sin tan cos 3 10 10 1 1 10 10 csc sin 10 10 10 1 1 3 cot tan 13
862 Copyright © 2025 Pearson Education, Inc.
Chapter 7 Review Exercises 3 (quadrant III) 2 Using the Pythagorean Identities: cot 2 csc 2 1
26. csc 4 and
factor of
1 . 4
cot 2 4 1 16 1 15 2
cot 15 Note that cot must be positive since lies in quadrant III. Thus, cot 15 . 1 1 1 sin csc 4 4 cos , so cot sin 15 1 cos cot sin 15 4 4
tan
1 1 15 15 cot 15 15 15
sec
1 1 4 15 4 15 cos 15 15 15 15 4
Domain: , Range: 2, 2 29. y 3cos(2 x) The graph of y cos x is stretched vertically by a factor of 3, reflected across the x-axis, and 1 compressed horizontally by a factor of . 2
(quadrant II) 2 Using the Pythagorean Identities: csc 2 1 cot 2
27. cot 2 and
csc 2 1 2 1 4 5 2
csc 5 Note that csc must be positive because lies in quadrant II. Thus, csc 5 .
1 1 5 5 csc 5 5 5 cos , so cot sin 5 2 5 . cos cot sin 2 5 5 1 1 1 tan cot 2 2
sin
sec
Domain: , Range: 3,3
1 1 5 5 cos 2 5 5 2 2 5
28. y 2sin(4 x) The graph of y sin x is stretched vertically by a factor of 2 and compressed horizontally by a 863 Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 30. y tan( x ) The graph of y tan x is shifted units to the left.
32. y cot x 4
The graph of y cot x is shifted
4
units to the
left.
k Domain: x | x , k is an odd integer 2 Range: ,
31. y 2 tan(3x ) The graph of y tan x is stretched vertically by a factor of 2, reflected across the x-axis, and 1 compressed horizontally by a factor of . 3
Domain: x | x k , k is an integer 4 Range: , 33. y 4sec 2 x The graph of y sec x is stretched vertically by a factor of 4 and compressed horizontally by a 1 factor of . 2 y 4 3 2 1
(⫺, 4)
3
⫺––– 4 , ⫺4 ⫺ –– 2
( Domain: x | x k , k is an integer 6 3 Range: ,
)
⫺ –– 4
⫺2 ⫺3 ⫺4 ⫺5
(, 4)
(0, 4)
–– 4
3 ––– 4
x
(––2 , ⫺4)
k Domain: x | x , k is an odd integer 4 Range: y | y 4 or y 4 34. y csc x 4
The graph of y csc x is shifted
864 Copyright © 2025 Pearson Education, Inc.
4
units to the
Chapter 7 Review Exercises
stretched vertically by a factor of 5.
left.
Domain: x | x k , k is an integer 4 Range: y | y 1 or y 1
3 Domain: x | x k 3 , k is an integer 4 Range: ,
35. y 4sin 2 x 4 2
37. y sin(2 x)
The graph of y sin x is shifted left 4 units,
Amplitude = 1 1 ; Period =
1 compressed horizontally by a factor of , 2 stretched vertically by a factor of 4, and shifted down 2 units.
2 2
38. y 2 cos(3 x)
Amplitude = 2 2 ; Period = 39. y 4sin(3 x) Amplitude:
A 4 4
Period:
T
2
2 3
0 Phase Shift: 0 3
Domain: , Range: 6, 2
x 36. y 5cot 3 4 The graph of y cot x is shifted right 4 units, stretched horizontally by a factor of 3, and
865 Copyright © 2025 Pearson Education, Inc.
2 2 3 3
Chapter 7: Trigonometric Functions
1 40. y cos x 2 2 Amplitude: A 1 1 Period:
T
2
2 4 1 2
Phase Shift: 2 1
2 42. y cos x 6 3 2 2 3 3 2 2 T Period: 2 6 Phase Shift: Amplitude:
A
2
1 3 41. y sin x 2 2
1 1 Amplitude: A 2 2 2 2 4 Period: T 3 3 2 2 Phase Shift: 3 3 2
43. The graph is a cosine graph with amplitude 5 and period 8π. 2 Find : 8
8 2 2 1 8 4 1 The equation is: y 5cos x . 4
44. The graph is a reflected sine graph with amplitude 7 and period 8. 2 Find : 8
8 2 2 8 4 The equation is: y 7 sin x . 4
45. opposite = 3; hypotenuse = 7; adjacent = ? 32 (adjacent) 2 7 2 (adjacent)2 49 9 40 adjacent 40 2 10 sin
opp 3 hyp 7
866 Copyright © 2025 Pearson Education, Inc.
cos
adj 2 10 hyp 7
Chapter 7 Review Exercises
tan
opp 3 3 10 3 10 adj 2 10 2 10 10 20
sec
hyp 7 7 10 7 10 adj 2 10 2 10 10 20
csc
hyp 7 opp 3
cot
50. The point P (2, 5) is on a circle of radius r (2) 2 52 4 25 29 with the center at the origin. So, we have x 2 , y 5 , and r 29 . Thus, sin t
adj 2 10 opp 3
46. Set the calculator to radian mode: sin
8
cos t
0.38 .
y 5 5 29 ; r 29 29
x 2 29 y 5 2 ; tan t . x 2 r 29 29
51. The domain of y sec x is
x x odd multiple of . 2 The range of y sec x is y y 1 or y 1 . The period is 2 .
Set the calculator to degree mode: 1 sec10o 1.02 . cos10o
52. a. b.
63.18o 0.18o (0.18)(60 ') 10.8' 0.8' (0.8)(60") 48"
53. r 2 feet, 30º or
6
1.05 feet 6 3 1 1 2 A r 2 2 1.05 square feet 2 2 6 3
s r 2
48. The angle is in Quadrant I so the reference angle is 750 2(360) 30 . 1 2 2 49. P , 3 3
54. In 30 minutes: r 8 inches, 180º or s r 8 8 25.13 inches
1 3 2 3 2 2 2 ; csc t 4 3 2 2 2 2 2 3
In 20 minutes: r 8 inches, 120º or s r 8
1 1 cos t ; sec t 3 3 1 3 2 2 2 2 3 2 2 ; tan t 3 3 1 1 3
cot t
20 35 32.34o 60 3600
Thus, 63.18o 63o10 '48"
47. Terminal side of in Quadrant III implies sin 0 csc 0 cos 0 sec 0 tan 0 cot 0
sin t
32o 20 '35" 32
1 2 2 4 2 2 2
867
Copyright © 2025 Pearson Education, Inc.
2 16 16.76 inches 3 3
2 3
Chapter 7: Trigonometric Functions
55. v 180 mi/hr ;
1 mile 2 1 r 0.25 mile 4
d
v 180 mi/hr r 0.25 mi 720 rad/hr 720 rad 1 rev hr 2 rad 360 rev hr 114.59 rev/hr
59. Let x = the distance the boat is from shore (see figure). Note that 1 mile = 5280 feet. ft ft
56. Since there are two lights on opposite sides and the light is seen every 5 seconds, the beacon makes 1 revolution every 10 seconds: 1 rev 2 radians radians/second 10 sec 1 rev 5 57. Let x = the length of the lake, and let y = the distance from the edge of the lake to the point on the ground beneath the balloon (see figure).
x
65º
y
500 x 500 x tan 65º
tan 65º
500 x y 500 x y tan 25º
tan 25º
x
x
5º
1454 x 5280 1454 x 5280 tan(5) 1454 x 5280 tan(5) 16, 619.30 5280 11,339.30 Thus, the boat is approximately 11,339.30 feet, 11,339.30 2.15 miles, from shore. or 5280 tan(5)
60. I (t ) 220sin 30t , t 0 6 2 1 a. Period second 30 15 b. The amplitude is 220 amperes. c. The phase shift is: 1 1 second 6 30 6 30 180
ft 25º
58. Let x = the distance across the river. x tan(25) 50 x 50 tan(25) 23.32 Thus, the distance across the river is 23.32 feet.
500 y tan 25º 500 500 tan 25º tan 65º
1072.25 233.15 839.10 Thus, the length of the lake is approximately 839.10 feet.
868 Copyright © 2025 Pearson Education, Inc.
Chapter 7 Test d.
c.
y 19.81sin 0.543 x 2.296 75.66
d.
61. a. e.
95 55 40 20 2 2 95+55 150 75 Vertical Shift: 2 2 2 12 6 Phase shift (use y 55, x 1):
b.
Amplitude: A
Chapter 7 Test
55 20sin 1 75 6
1. 260 260 1 degree 260
20 20sin 6
1 sin 6
radian
180
260 13 radian radian 180 9
2. 400 400 1 degree
2 6 2 3
400
180
radian
400 20 radian radian 180 9
2 Thus, y 20sin x 75 , or 6 3
3. 13 13 1 degree 13
y 20sin x 4 75 . 6
4.
8
radian
5.
8
180
radian
13 radian 180
1 radian
180 degrees 22.5 8
9 9 1 radian radian 2 2 9 180 degrees 810 2
869
Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions
6.
3 3 1 radian radian 4 4 3 180 degrees 135 4
7. sin
6
8. cos
16. Set the calculator to radian mode: 28 1 cot 2.747 28 9 tan 9
1 2
5 3 5 3 cos cos 2 cos 4 4 4 4 3 3 cos cos 0 4 4
9. cos 120 cos 120
1 2
10. tan 330 tan 150 180 tan 150 11. sin
2
tan
3 3
19 3 sin tan 4 4 2 4 sin
3 tan 1 1 2 2 4 2
3 2 12. 2sin 2 60 3cos 45 2 3 2 2 3 3 2 3 3 2 3 1 2 2 2 2 2 2 4
13. Set the calculator to degree mode: sin17 0.292
17. To remember the sign of each trig function, we primarily need to remember that sin is positive in quadrants I and II, while cos is positive in quadrants I and IV. The sign of the other four trig functions can be determined directly from sine and cosine by knowing sin 1 1 tan , sec , csc , and cos cos sin cos cot . sin
in QI in QII in QIII in QIV
sin cos tan sec csc cot + + +
18. Because f ( x) sin x is an odd function and
since f (a) sin a
3 , then 5
3 f (a ) sin(a ) sin a . 5
5 and in quadrant II. 7 Using the Pythagorean Identities: 2 25 24 5 cos 2 1 sin 2 1 1 7 49 49
19. sin 14. Set the calculator to radian mode: cos
2 0.309 5
24 2 6 49 7 Note that cos must be negative because lies cos
15. Set the calculator to degree mode: 1 sec 229 1.524 cos 229
in quadrant II. Thus, cos
2 6 . 7
tan
5 sin 5 7 6 5 6 27 6 cos 7 7 2 6 6 12
csc
1 1 7 5 sin 7 5
870 Copyright © 2025 Pearson Education, Inc.
Chapter 7 Test
sec
1 1 7 6 7 6 2 6 cos 7 12 2 6 6
csc
1 1 13 sin 12 12 13
cot
1 1 12 6 2 6 5 6 tan 12 5 5 6 6
cot
1 1 5 12 tan 5 12
22. The point 2, 7 lies in quadrant I with x 2
2 3 and 2 (in quadrant IV). 2 3 Using the Pythagorean Identities: 2 4 5 2 sin 2 1 cos 2 1 1 9 9 3
20. cos
and y 7 . Since x 2 y 2 r 2 , we have r 22 7 2 53 . So, y 7 7 53 7 53 . sin 53 r 53 53 53
5 5 9 3 Note that sin must be negative because lies sin
23. The point 5,11 lies in quadrant II with x 5 and y 11 . Since x 2 y 2 r 2 , we
5 . 3
have r
tan
sin 35 5 3 5 2 cos 3 2 2 3
cos
csc
1 1 3 5 3 5 5 sin 3 5 5 5
in quadrant IV. Thus, sin
and y 3 . Since x 2 y 2 r 2 , we have r 62 3 45 3 5 . So, 2
1 1 2 5 2 5 5 tan 2 5 5 5
tan
y A sin x , we see that
A2,
169 13 25 5 Note that sec must be negative since lies in 13 quadrant II. Thus, sec . 5 1 1 5 cos sec 135 13
1 , and . The graph is a sine 3 6
curve with amplitude A 2 , period
sec
T
2
2 6 , and phase shift 1/ 3
x 6 . The graph of y 2sin 1/ 3 2 3 6
will lie between 2 and 2 on the y-axis. One
sin , so cos
sin tan cos
y 3 1 x 6 2
x 25. Comparing y 2sin to 3 6
12 and (in quadrant II) 2 5 Using the Pythagorean Identities: 2 144 169 12 1 sec 2 tan 2 1 1 25 25 5
21. tan
tan
x 5 5 146 5 146 . r 146 146 146 146
24. The point 6, 3 lies in quadrant IV with x 6
1 1 3 2 sec cos 3 2 cot
5 2 112 146 . So,
period will begin at x x
12 5 12 5 13 13
2
and end at 2 13
6 . We divide the 2 2 13
interval , into four subintervals, each of 2 2
871
Copyright © 2025 Pearson Education, Inc.
Chapter 7: Trigonometric Functions 6 3 . 4 2 7 7 13 2 , 2 , 2 , 2 , 2 ,5 , 5 , 2 The five key points on the graph are 7 13 , 0 , 5 , 2 , ,0 , 0 , 2 , 2 , 2 2 2 We plot these five points and fill in the graph of the sine function. The graph can then be extended in both directions.
length
27. For a sinusoidal graph of the form y A sin x , the amplitude is given by A , the period is given by
2
, and the phase
. Therefore, we have A 3 , 3 3 , and 3 . The equation
shift is given by
4
4
3 . for the graph is y 3sin 3 x 4
y (⫺4, 2)
11 , (⫺ –––– 2 0(
3 2 1
28. The area of the walk is the difference between the area of the larger sector and the area of the smaller shaded sector.
(2, 2) x
( ––2 , 0( (5, ⫺2) 7 , ( ––– 2 0(
5 , (⫺ ––– 2 0(
(⫺, ⫺2)
26. y tan x 2 4 Begin with the graph of y tan x , and shift it 4
units to the left to obtain the graph of
y tan x . Next, reflect this graph about 4 the y-axis to obtain the graph of y tan x . 4 Finally, shift the graph up 2 units to obtain the graph of y tan x 2 . 4 y tan x 2 4
y
x
k
3 ft
The area of the walk is given by 1 1 A R 2 r 2 , 2 2
r 3 r
R2 r 2 2 where R is the radius of the larger sector and r is the radius of the smaller sector. The larger radius is 3 feet longer than the smaller radius because the walk is to be 3 feet wide. Therefore, R r 3 , and
A
2
2
2
2
r 6r 9 r 2
2
6r 9 2 The shaded sector has an arc length of 25 feet 5 and a central angle of 50 radians . The 18 s 25 90 radius of this sector is r 5 feet .
2
l
50
4
2
W a
18
Thus, the area of the walk is given by 5 90 5 540 9 A 18 6 9 2 36 5 2 2 75 ft 78.93 ft 4
872 Copyright © 2025 Pearson Education, Inc.
Chapter 7 Cumulative Review 29. To throw the hammer 83.19 meters, we need v2 s 0 g v0 2 83.19 m 9.8 m/s 2 2 v0 815.262 m 2 / s 2 v0 28.553 m/s Linear speed and angular speed are related according to the formula v r . The radius is r 190 cm 1.9 m . Thus, we have 28.553 r 28.553 1.9 15.028 radians per second radians 60 sec 1 revolution 15.028 sec 1 min 2 radians 143.5 revolutions per minute (rpm) To throw the hammer 83.19 meters, Adrian must have been swinging it at a rate of 143.5 rpm upon release. 30. Let x = the distance to the base of the statue.
h 50 tan 32º h tan 40º h 50 0.6249 h 0.8391 h 0.7447h 31.245
0.2553h 31.245 h 122.39 feet The building is roughly 122.4 feet tall.
Chapter 7 Cumulative Review
2 x 1 x 1 0 x
2. Slope 3 , containing (–2,5) Using y y1 m( x x1 )
20º
x
x
305 x
y 5 3 x (2) y 5 3( x 2) y 5 3 x 6 y 3 x 1
305 305 837.98 feet tan 20º 0.3640
The ship is about 838 feet from the base of the Statue of Liberty.
3. radius = 4, center (0,–2) Using x h y k r 2 2
40º
tan 40º x
x 2 y 2 16 2
4. 2 x 3 y 12 This equation yields a line. 2 x 3 y 12 3 y 2 x 12 2 y x4 3 2 The slope is m and the y-intercept is 4 . 3 Let y 0 : 2 x 3(0) 12 2 x 12 x6
x
h x
h tan 40º
tan 32º
2
2
h 32º
2
x 0 y 2 42
31. Let h = the height of the building and let x = the distance from the building to the first sighting.
feet
1 or x 1 2
1 The solution set is 1, . 2
feet
tan 20º
2 x2 x 1 0
1.
h x 50
h ( x 50) tan 32º
873
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Chapter 7: Trigonometric Functions
The x-intercept is 6.
5. x 2 y 2 2 x 4 y 4 0 x 2x 1 y 4 y 4 4 1 4 2
2
7. a.
y x2
b.
y x3
c.
y ex
d.
y ln x
x 1 y 2 9 2 2 x 1 y 2 32 2
2
This equation yields a circle with radius 3 and center (1,–2).
6. y ( x 3) 2 2
Using the graph of y x 2 , horizontally shift to the right 3 units, and vertically shift up 2 units.
874 Copyright © 2025 Pearson Education, Inc.
Chapter 7 Cumulative Review
e.
10. y 3sin(2 x)
y sin x
A 3 3
Amplitude:
2 2 0 Phase Shift: 0 2
f.
8.
Period:
T
y tan x
11. tan
f ( x) 3x 2 y 3x 2 x 3 y 2 Inverse x 2 3y x2 y 3 x2 1 f 1 ( x) x 2 3 3
6
csc
3 1 3 2 6 2
3 3 2 63 3 2 3
12. We need a function of the form y A b x , with b 0, b 1 . The graph contains the points
0, 2 and 1, 6 . Therefore, 2 A b0 .
9. Since sin cos 1 , then 2
4
3cos
2
2 A 1
sin14 cos14 3 1 3 2 o 2
o 2
A2
And y 2b x 6 2b1 b3
So we have the function y 2 3x . 13. The graph is a cosine graph with amplitude 3 and period 12. 2 Find : 12
12 2
2 12 6
The equation is: y 3cos x . 6
875
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Chapter 7: Trigonometric Functions
14. a.
Given points (2, 3) and (1, 6) , we compute the slope as follows: y y 6 3 9 3 slope 2 1 x2 x1 1 2 3
So 3 8a 6 a 9 9a a 1 Thus, b 2a 2 1 2
Using y y1 m( x x1 ) :
Therefore, we have the function
and c 3 8a 3 8 1 5
y 3 3 x 2
f x x 2 2 x 5 x 1 6 . 2
y 3 3 x 2 y 3 x 6 3 y 3 x 3
y-intercept: f 0 02 2 0 5 5 x-intercepts: 0 x 2 2 x 5
The linear function is f x 3 x 3 .
x
Slope: m 3 ; y-intercept: f 0 3 0 3 3
2 2 4 1 5 2 1
2 4 20 2 24 2 2 6 2 2 2 1 6 1.45 or 3.45
x-intercept: 0 3 x 3 3x 3 x 1 Intercepts: 1, 0 , 0, 3
Intercepts: 0, 5 , 1 6, 0 , 1 6, 0
b. Given that the graph of f x ax bx c 2
has vertex (1, –6) and passes through the b 1, point (–2,3), we can conclude 2a f 2 3 , and f 1 6 . b 1 2a b 2a Also note that f 2 3
Notice that
a 2 b 2 c 3 4a 2b c 3 f 1 6 2
c.
If f x ae x contains the points (–2,3) and (1,–6), we would have the equations f 2 ae 2 3 and f 1 ae1 6 . Note that ae 2 3 3 a 2 3e 2 e 6 1 But ae 6 a e 6 Since 3e 2 , there is no exponential e function of the form f x ae x that contains the points (–2,3) and (1,–6).
a 1 b 1 c 6 a b c 6 Replacing b with 2a in these equations yields: 4a 2 2a c 3 c 3 8a and a 2a c 6 c 6 a 2
2
15. a. A polynomial function of degree 3 whose x-intercepts are –2, 3, and 5 will have the form f ( x) a( x 2)( x 3)( x 5) , since the x-intercepts correspond to the zeros of the function. Given a y-intercept of 5, we
876 Copyright © 2025 Pearson Education, Inc.
Chapter 7 Projects f (0) 5 a(0 2)(0 3)(0 5) 5 02
have
f (0) 5 a(0 2)(0 3)(0 5) 5
30a 10 a
5 1 30a 5 a 30 6 Therefore, we have the function 1 f ( x) ( x 2)( x 3)( x 5) . 6
1 3
Therefore, we have the function 1 ( x 2)( x 3)( x 5) f ( x) 3 x2 ( x 2)( x 2 8 x 15) 3( x 2) ( x 3 6 x 2 x 30) 3( x 2) 3 x 6 x 2 x 30 x 3 6 x 2 x 30 6 3x 3 x 6
b. A rational function whose x-intercepts are 2 , 3, and 5 and that has the line x 2 as a vertical asymptote will have the form a ( x 2)( x 3)( x 5) f ( x) , since the xx2 intercepts correspond to the zeros of the numerator, and the vertical asymptote corresponds to the zero of the denominator. Given a y-intercept of 5, we have
Chapter 7 Projects Project I – Internet-based Project Project II 1. November 15: High tide: 11:18 am and 11:15 p.m. November 19: low tide: 7:17 am and 8:38 p.m. 2. The low tide was below sea level. It is measured against calm water at sea level.
877
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Chapter 7: Trigonometric Functions
3.
Nov
14 0-24 15 24-48 16 48-72 17 72-96 18 96-120 19 120-144 20 144-168
Low Tide
Low Tide
High Tide
High Tide
Time
Ht (ft) t
Time
Ht (ft) t
Time
Ht (ft)
t
Time
6:26a
2.0
6.43
4:38p
1.4 16.63
9:29a
2.2
9.48
11:14p 2.8
23.23
6:22a
1.6
30.37
5:34p
1.8 41.57
11:18a 2.4
35.3
11:15p 2.6
47.25
6:28a
1.2
54.47
6:25p
2.0 66.42
12:37p 2.6
60.62
11:16p 2.6
71.27
6:40a
0.8
78.67
7:12p
2.4 91.2
1:38p
2.8
85.63
11:16p 2.6
95.27
6:56a
0.4
102.93
7:57p
2.6 115.95
2:27p
3.0
110.45
11:14p 2.8
119.23
7:17a
0.0
127.28
8:38p
2.6 140.63
3:10p
3.2
135.17
11:05p 2.8
143.08
7:43a -0.2
151.72
3:52p
3.4
159.87
Ht (ft) t
similar.
4. The data seems to take on a sinusoidal shape (oscillates). The period is approximately 12 hours. The amplitude varies each day: Nov 14: 0.1, 0.7 Nov 15: 0.4, 0.4 Nov 16: 0.7, 0.3 Nov 17: 1.0, 0.1 Nov 18: 1.3, 0.1 Nov 19: 1.6, 0.1 Nov 20: 1.8 5. Average of the amplitudes: 0.66 Period : 12 Average of vertical shifts: 2.15 (approximately) There is no phase shift. However, keeping in mind the vertical shift, the amplitude y A sin Bx D A 0.66
12 B
2 B
7. Find the high and low tides on November 21 which are the min and max that lie between t 168 and t 192 . Looking at the graph of the equation for part (5) and using MAX/MIN for values between t 168 and t 192 :
Low tides of 1.49 feet when t = 178.2 and t = 190.3.
D 2.15
0.52 6 Thus, y 0.66sin 0.52 x 2.15
High tides of 2.81 feet occur when t = 172.2 and t = 184.3.
(Answers may vary) 6. y 0.848sin 0.52 x 1.25 2.23
The two functions are not the same, but they are
878 Copyright © 2025 Pearson Education, Inc.
Chapter 7 Projects
Looking at the graph for the equation in part (6) and using MAX/MIN for values between t = 168 and t = 192:
5. t
1 5 9 45 , t , t ,…, t 4 f0 4 f0 4 f0 4 f0
6. M = 0 1 0 P = 0 π 0
A low tide of 1.38 feet occurs when t = 175.7 and t 187.8 .
7. S0 (t ) 1sin(2 f 0 t 0) , S1 (t ) 1sin(2 f 0 t ) 8. [0, 4 T0 ] S0 [4 T0 , 8 T0 ] S1 [ 8 T0 , 12 T0 ] S0
A high tide of 3.08 feet occurs when t = 169.8 and t 181.9 .
8. The low and high tides vary because of the moon phase. The moon has a gravitational pull on the water on Earth.
Project IV 1. Lanai:
s
0
s (t ) 0 1
396
1 2 f0
3 4 f0
1 f0
0
1
0
2. s r 65 s 0.0164 r 3960
4. Let f 0 1 =1. Let 0 x 12 , with x 0.5 . Label the graph as 0 x 12T0 , and each tick
mark is at x
i
t
3.
1 4 f0
3,370 ft
2 1 2 f 0 f0
3.
1 . 2 f0
12 12T0 __ f0
3960 cos(0.164) 3960 h 3960 0.9999(3960 h) h 0.396 miles 0.396 5280 2090 feet
879
Copyright © 2025 Pearson Education, Inc.
}
1. s (t ) 1sin 2 f 0 t 2. T0
mi
Lanai
39 60
Peak of Lanaihale
m 65
Project III
Oahu
s=
39 60 m i
Oahu
0m
i
Lanai
Chapter 7: Trigonometric Functions 4. Maui: Oahu
s Maui
}
i
39 60 m i
Peak of Haleakala
0m 11 s
39 60 m i
Oahu
10,023 ft
396
i 0m
Maui
s 40 0.0101 r 3960 3960 cos(0.0101) 3960 h 3960 0.9999(3960 h) h 0.346 miles h 0.346 x5280 2090 feet
5. Kamakou, Haleakala, and Lanaihale are all visible from Oahu.
s 110 0.0278 r 3960 3960 cos(0.278) 3960 h 3960 0.9996(3960 h) h 1.584 miles h 1.584 5280 8364 feet
Project V
Answers will vary.
Hawaii: Oahu
s mi
}
39 60
Hawaii
i 0m 19
Peak of Mauna Kea
s
39 60 m i
Oahu
13,796 ft
396
0
mi
Hawaii
s 190 0.0480 r 3960 3960 cos(0.480) 3960 h 3960 0.9988(3960 h) h 4.752 miles h 4.752 5280 25, 091 feet
Molokai:
}
Peak of Kamakou
mi
Molokai
i
s 39 60 m
Oahu
0 s
39 60 m i
Oahu
4,961 ft
396
i 0m
Molokai
880 Copyright © 2025 Pearson Education, Inc.
Section 8 Analytic Trigonometry Section 8.1 1. Domain: x x is any real number ; Range: y 1 y 1 2.
3,
3. True
whose sine equals 1 . sin 1, 2 2 2 sin 1 1 2 14. cos 1 1
3 1 4. 1; ; ; 1 2 2
We are finding the angle , 0 , whose cosine equals 1 . cos 1, 0
5. x sin y
6. 0 x
cos
7. True
1
1
15. tan 1 0
8. True
We are finding the angle ,
9. True
, whose 2 2
tangent equals 0.
10. d
tan 0,
11. sin 1 0 We are finding the angle ,
, whose 2 2
sine equals 0. sin 0,
0
0
2 2
1
tan 0 0
16. tan 1 1
2 2
We are finding the angle , tangent equals 1 .
sin 1 0 0
12. cos 1 1 We are finding the angle , 0 , whose cosine equals 1. cos 1, 0 0 cos 1 1 0 13. sin 1 1
We are finding the angle ,
tan 1, 4 tan 1 (1) 4
, 2 2
881 Copyright © 2025 Pearson Education, Inc.
2 2
, whose 2 2
Section 8: Analytic Trigonometry
17. sin 1
3 20. sin 1 2
2 2
We are finding the angle , 2 . 2 2 sin , 2 4 2 sin 1 2 4
, whose 2 2
We are finding the angle ,
sine equals
18. tan 1
sine equals
2 2
We are finding the angle , tangent equals 3 , 3 6
tan
tan 1
3 , 2 3 3 sin 1 3 2
2 2
3 21. cos 1 2 We are finding the angle , 0 , whose
, whose 2 2
3 . 2 3 cos , 2 5 6 3 5 cos 1 2 6
3 . 3
3 . 2
sin
3 3
, whose 2 2
cosine equals 2 2
3 3 6
19. tan 1 3
0
2 22. sin 1 2
We are finding the angle , , whose 2 2 tangent equals 3 . tan 3, 2 2 3 1 tan 3 3
We are finding the angle , sine equals
2 . 2
2 , 2 4 2 sin 1 4 2 sin
882 Copyright © 2025 Pearson Education, Inc.
, whose 2 2
2 2
Section 8.1: The Inverse Sine, Cosine, and Tangent Functions
2 23. cos 1 2 We are finding the angle , 0 , whose 2 . 2 2 cos , 2 4 2 cos 1 4 2
1 . 2 1 sin , 2 6 1 sin 1 2 6
sine equals
cosine equals
0
27. sin 1 0.1 0.10 28. cos 1 0.6 0.93 29. tan 1 5 1.37
1 24. cos 1 2 We are finding the angle , 0 , whose 1 cosine equals . 2 1 cos , 0 2 2 3 1 2 cos 1 2 3
30. tan 1 0.2 0.20
7 0.51 8
32. sin 1
1 0.13 8
34. tan 1 ( 3) 1.25 35. sin 1 ( 0.12) 0.12
We are finding the angle ,
, whose 2 2
36. cos 1 ( 0.44) 2.03
3 . 3
3 , 3 6 3 tan 1 3 6 tan
31. cos 1
33. tan 1 ( 0.4) 0.38
3 25. tan 1 3
tangent equals
2 2
2 2
2 1.08 3
38. sin 1
3 0.35 5
4 39. cos 1 cos follows the form of the equation 5 4 f 1 f x cos 1 cos x x . Since is 5
in the interval 0, , we can apply the equation
1 26. sin 1 2
We are finding the angle ,
37. cos 1
4 4 . directly and get cos 1 cos 5 5
, whose 2 2
883 Copyright © 2025 Pearson Education, Inc.
Section 8: Analytic Trigonometry
40. sin 1 sin follows the form of the 10
so sine is negative. The reference angle of
equation f 1 f x sin 1 sin x x . Since
and we want to be in quadrant IV so sine 8 will still be negative. Thus, we have 9 sin . Since is in the interval sin 8 8 8
is in the interval , , we can apply 10 2 2 the equation directly and get sin 1 sin . 10 10
2 , 2 , we can apply the equation above and 9 1 get sin 1 sin sin sin . 8 8 8
3 41. tan 1 tan follows the form of the 8
11 44. sin 1 sin follows the form of the 4
equation f 1 f x tan 1 tan x x . Since 3 is in the interval , , we can apply 8 2 2 the equation directly and get 3 3 . tan 1 tan 8 8
11 is not 4
11 11 sin . The angle is in quadrant II 4 4 11 so sine is positive. The reference angle of is 4 3 and we need to be in quadrant I so sine 4 will still be positive. Thus, we have 3 sin . Since sin is in the interval 4 4 4
sin
3 is in the interval , , we can apply 7 2 2 the equation directly and get 3 3 . sin 1 sin 7 7 9 43. sin 1 sin follows the form of the 8
in the interval , . We need to find an 2 2 angle in the interval , for which 2 2
cannot use the formula directly since
equation f 1 f x sin 1 sin x x . Since
equation f 1 f x sin 1 sin x x , but we
3 42. sin 1 sin follows the form of the 7
9 is 8
2 , 2 , we can apply the equation above and
equation f 1 f x sin 1 sin x x , but we
11 sin 1 sin . get sin 1 sin 4 4 4
9 cannot use the formula directly since is not 8 in the interval , . We need to find an 2 2
5 45. cos 1 cos follows the form of the 3
angle in the interval , for which 2 2 9 9 sin . The angle is in quadrant III sin 8 8
equation f 1 f x cos 1 cos x x , but we cannot use the formula directly since
5 is 3
not in the interval 0, . We need to find an 884
Copyright © 2025 Pearson Education, Inc.
Section 8.1: The Inverse Sine, Cosine, and Tangent Functions
4 4 tan is in quadrant tan . The angle 5 5 II so tangent is negative. The reference angle of 4 is and we want to be in quadrant IV 5 5 so tangent will still be negative. Thus, we have 4 tan tan . Since is in the 5 5 5
angle in the interval 0, for which 5 5 is in cos cos . The angle 3 3 5 quadrant I so the reference angle of is . 3 3 5 Thus, we have cos cos . Since is 3 3 3 in the interval 0, , we can apply the equation
interval , , we can apply the equation 2 2 above and get 4 1 tan 1 tan tan tan . 5 5 5
above and get 5 1 cos 1 cos cos cos . 3 3 3 7 46. cos 1 cos follows the form of the 6
10 follows the form of the 48. tan 1 tan 9
equation f 1 f x cos 1 cos x x , but
above and get 5 5 7 cos 1 cos cos 1 cos . 6 6 6
is in the interval , , we can apply 9 2 2 the equation above and get 10 tan 1 tan . tan 1 tan 9 9 9
4 47. tan 1 tan follows the form of the 5
equation f 1 f x tan 1 tan x x , but we cannot use the formula directly since
10 9
10 10 is in tan . The angle tan 9 9 quadrant II so tangent is negative. The reference 10 angle of is and we want to be in 9 9 quadrant IV so tangent will still be negative. 10 Thus, we have tan tan . Since 9 9
in the interval 0, , we can apply the equation
is not in the interval , . We need to find 2 2 an angle in the interval , for which 2 2
7 7 cos cos . The angle is in quadrant 6 6 III so the we need an angle in the desired interval 7 . whose cosine is equal to the cosine of 6 5 7 5 Thus, we have cos cos . Since is 6 6 6
we cannot use the formula directly since
angle in the interval 0, for which
equation f 1 f x tan 1 tan x x , but
7 is 6 not in the interval 0, . We need to find an
we cannot use the formula directly since
4 is 5
2 49. tan 1 tan follows the form of the 3
not in the interval , . We need to find an 2 2 angle in the interval , for which 2 2
equation f 1 f x tan 1 tan x x . but we cannot use the formula directly since 885
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2 is not 3
Section 8: Analytic Trigonometry
in the interval , . We need to find an angle 2 2 in the interval , for which 2 2
we cannot use the formula directly since
4 not in the interval 0, . We need to find an cos cos . The angle is in 4 4
quadrant IV so the reference angle of
3
equation f
3
.
cannot use the formula directly since
4 is 3 not in the interval 0, . We need to find an
we cannot use the formula directly since
3 is not 4
in the interval , . We need to find an 2 2 angle in the interval , for which 2 2
angle in the interval 0, for which 4 4 is in quadrant cos cos . The angle 3 3
3 sin sin . The reference angle of 4
4 is . We want 3 3 the angle to be in quadrant II and the cosine to be 2 4 negative. Thus, we have cos cos . 3 3
3 is and we want to be in quadrant IV 4 4 so sine will still be negative. Thus, we have 3 sin sin . Since is in the 4 4 4
III so the reference angle of
2 is in the interval 0, , we can apply 3 the equation above and get 2 2 4 cos 1 cos cos 1 cos . 3 3 3
interval , , we can apply the equation 2 2 above and get 3 sin 1 sin sin 1 sin . 4 4 4
Since
51. cos 1 cos follows the form of the 4
4
equation f 1 f x sin 1 sin x x , but we
f x cos cos x x , but
3 52. sin 1 sin follows the form of the 4
1
is
above and get cos 1 cos cos 1 cos . 4 4 4
4 50. cos 1 cos follows the form of the 3 1
4
in the interval 0, , we can apply the equation
above and get tan 1 tan 2 tan 1 tan .
Thus, we have cos cos . Since is 4 4 4
interval , , we can apply the equation 2 2 3
is
angle in the interval 0, for which
2 2 tan is in tan . The angle 3 3 quadrant III so tangent is positive. The reference 2 angle of is and we want to be in 3 3 quadrant I so tangent will still be positive. Thus, 2 we have tan is in the tan . Since 3 3 3
equation f 1 f x cos 1 cos x x , but
886
Copyright © 2025 Pearson Education, Inc.
Section 8.1: The Inverse Sine, Cosine, and Tangent Functions
apply the equation directly and get 2 2 cos cos 1 . 3 3
53. tan 1 tan follows the form of the 2
equation f 1 f x tan 1 tan x x . We need to find an angle in the interval , 2 2 for which tan tan . In this case, 2
and get tan tan 1 4 4 .
58. tan tan 1 2 follows the form of the equation
f f 1 x tan tan 1 x x . Since 2 is a
3 54. tan tan follows the form of the 2 1
real number, we can apply the equation directly
also be undefined.
f f 1 x tan tan 1 x x . Since 4 is a
tan is undefined so tan 1 tan would 2 2
57. tan tan 1 4 follows the form of the equation
real number, we can apply the equation directly
and get tan tan 1 2 2 .
equation f 1 f x tan 1 tan x x . We
59. Since there is no angle such that cos 1.2 , the quantity cos 1 1.2 is not defined. Thus,
need to find an angle in the interval , 2 2 3 for which tan tan . The reference 2
cos cos 1 1.2 is not defined.
3 is . Thus, we have 2 2 3 tan tan . In this case, tan is 2 2 2
60. Since there is no angle such that sin 2 , the quantity sin 1 2 is not defined. Thus,
undefined so tan 1 tan would also be 2 undefined.
61. tan tan 1 follows the form of the equation
angle of
sin sin 1 2 is not defined.
f f 1 x tan tan 1 x x . Since is a
real number, we can apply the equation directly
1 55. sin sin 1 follows the form of the equation 4 1 is in f f 1 x sin sin 1 x x . Since 4 the interval 1,1 , we can apply the equation
62. Since there is no angle such that sin 1.5 , the quantity sin 1 1.5 is not defined. Thus,
63.
x 5sin y 2 5sin y x 2 x2 sin y 5 x2 f 1 x y sin 1 5
equation f f 1 x cos cos 1 x x . Since
f x 5sin x 2 y 5sin x 2
2 56. cos cos 1 follows the form of the 3
sin sin 1 1.5 is not defined.
1 1 directly and get sin sin 1 . 4 4
and get tan tan 1 .
2 is in the interval 1,1 , we can 3
887
Copyright © 2025 Pearson Education, Inc.
Section 8: Analytic Trigonometry
The domain of f x equals the range of f 1 ( x ) and is
2
x
x 2 cos 3 y
or , in 2 2 2
cos 3 y
x 3 y cos 1 2 1 x y cos 1 f 1 x 3 2
interval notation. To find the domain of f 1 x we note that the argument of the inverse sine x2 and that it must lie in the function is 5 interval 1,1 . That is,
The domain of f x equals the range of
x2 1 5 5 x 2 5 3 x 7 The domain of f 1 x is x | 3 x 7 , or
f 1 ( x ) and is 0 x
1
y 2 tan x 3 x 2 tan y 3 2 tan y x 3 x3 tan y 2 x3 y tan 1 f 1 x 2 The domain of f x equals the range of f 1 ( x)
2
x
2
, or 0, in interval 3
2, 2 in interval notation. Recall that the domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is 2, 2 . 66.
f x 3sin 2 x
y 3sin 2 x x 3sin 2 y
or , in interval 2 2
sin 2 y
notation. To find the domain of f 1 x we note that the argument of the inverse tangent function can be any real number. Thus, the domain of f 1 x is all real numbers, or , in
x 3
2 y sin 1
x 3
1 x y sin 1 f 1 x 2 3 The domain of f x equals the range of
interval notation. Recall that the domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is , . 65.
3
that the argument of the inverse cosine function x and that it must lie in the interval 1,1 . is 2 That is, x 1 1 2 2 x 2 2 x 2 The domain of f 1 x is x | 2 x 2 , or
f x 2 tan x 3
and is
notation. To find the domain of f 1 x we note
3, 7 in interval notation. Recall that the domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is also 3, 7 .
64.
x 2
f 1 ( x) and is
f x 2 cos 3 x
4
x
, or , in 4 4 4
interval notation. To find the domain of f 1 x
y 2 cos 3 x
we note that the argument of the inverse sine x and that it must lie in the interval function is 3 888
Copyright © 2025 Pearson Education, Inc.
Section 8.1: The Inverse Sine, Cosine, and Tangent Functions
The domain of f x equals the range of
1,1 . That is, x 1 1 3 3 x 3 The domain of f 1 x is x | 3 x 3 , or
f 1 ( x) and is 2 x 2 , or 2, 2 in
interval notation. To find the domain of f 1 x we note that the argument of the inverse cosine function is x 1 and that it must lie in the interval 1,1 . That is, 1 x 1 1 0 x2 1 The domain of f x is x | 0 x 2 , or
3,3 in interval notation. Recall that the domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is 3,3 .
67.
0, 2 in interval notation. Recall that the domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is 0, 2 .
f x tan x 1 3 y tan x 1 3 x tan y 1 3 tan y 1 x 3
69.
f x 3sin 2 x 1 y 3sin 2 x 1
y 1 tan 1 x 3
x 3sin 2 y 1
y 1 tan 1 x 3 1 tan 1 x 3 f 1 x
sin 2 y 1
(note here we used the fact that y tan 1 x is an odd function). The domain of f x equals the range of f 1 ( x) and is 1
2
x
2
x 3 x 2 y sin 1 1 3
2 y 1 sin 1
1 , or
1 x 1 y sin 1 f 1 x 2 3 2
1 , 1 in interval notation. To find the 2 2
The domain of f x equals the range of
domain of f 1 x we note that the argument of
1 1 f 1 ( x) and is x , or 2 4 2 4 1 1 2 4 , 2 4 in interval notation. To find
the inverse tangent function can be any real number. Thus, the domain of f 1 x is all real numbers, or , in interval notation. Recall
the domain of f 1 x we note that the argument
that the domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is , . 68.
x 3
of the inverse sine function is
x and that it must 3
lie in the interval 1,1 . That is,
f x cos x 2 1
x 1 3 3 x 3 The domain of f 1 x is x | 3 x 3 , or 1
y cos x 2 1 x cos y 2 1 cos y 2 x 1
3,3 in interval notation. Recall that the domain of a function equals the range of its inverse and the range of a function equals the
y 2 cos 1 x 1 y cos 1 x 1 2 f 1 ( x)
889
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Section 8: Analytic Trigonometry
domain of its inverse. Thus, the range of f is 3,3 . 70.
72. 2 cos 1 x cos 1 x
f x 2 cos 3 x 2
2
x cos
0 2 The solution set is {0} .
y 2 cos 3x 2
x 2 cos 3 y 2 x 2
73. 3cos 1 2 x 2
x 3 y 2 cos 1 2 x 3 y cos 1 2 2
cos 1 2 x
cos 3 y 2
2 x cos
2 3
1 2 1 x 4
2x
1 x 2 y cos 1 f 1 x 3 2 3
The domain of f x equals the range of
1 The solution set is . 4
2 2 f ( x ) and is x , or 3 3 3 2 2 3 , 3 3 in interval notation. To find the 1
74. 6sin 1 3x sin 1 3x
domain of f 1 x we note that the argument of
6 3 x sin 6 1 3x 2 1 x 6 1 The solution set is . 6
x and that it must 2 lie in the interval 1,1 . That is,
the inverse cosine function is
x 1 2 2 x 2 The domain of f 1 x is x | 2 x 2 , or 1
2, 2 in interval notation. Recall that the domain of a function equals the range of its inverse and the range of a function equals the domain of its inverse. Thus, the range of f is 2, 2 .
75. 3 tan 1 x tan 1 x
3
x tan
71. 4sin x 1
sin 1 x
2 3
3
3
The solution set is
3 .
4
76. 4 tan 1 x
2 4 2 2 The solution set is . 2 x sin
tan 1 x
4
x tan 1 4 The solution set is {1} .
890
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Section 8.1: The Inverse Sine, Cosine, and Tangent Functions 81. Note that 2118 21.3 .
77. 4 cos 1 x 2 2 cos 1 x 2 cos 1 x 2 0
a.
2 cos 1 x 2 cos 1 x x cos 1 The solution set is {1} .
b.
78. 5sin 1 x 2 2sin 1 x 3 3sin 1 x sin 1 x
c.
3
3 x sin 2 3 3 The solution set is . 2
b.
c.
a.
cos 1 tan 23.5 180 tan 29.75 180 D 24 1 13.92 hours or 13 hours, 55 minutes
b.
cos 1 tan 0 180 tan 29.75 180 D 24 1 12 hours
c.
cos 1 tan 22.8 180 tan 29.75 180 D 24 1 13.85 hours or 13 hours, 51 minutes
83. a.
80. Note that 4045 40.75 . a.
b.
c.
cos 1 tan 0 180 tan 21.3 180 D 24 1 12 hours cos 1 tan 22.8 180 tan 21.3 180 D 24 1 13.26 hours or 13 hours, 15 minutes
82. Note that 6110 61.167 .
79. Note that 2945 29.75 . a.
cos 1 tan 23.5 180 tan 21.3 180 D 24 1 13.30 hours or 13 hours, 18 minutes
b.
cos 1 tan 23.5 180 tan 40.75 180 D 24 1 14.93 hours or 14 hours, 56 minutes
c.
cos 1 tan 0 180 tan 40.75 180 D 24 1 12 hours
cos 1 tan 23.5 180 tan 61.167 180 D 24 1 18.96 hours or 18 hours, 57 minutes cos 1 tan 0 180 tan 61.167 180 D 24 1 12 hours cos 1 tan 22.8 180 tan 61.167 180 D 24 1 18.64 hours or 18 hours, 38 minutes
cos 1 tan 23.5 180 tan 0 180 D 24 1 12 hours cos 1 tan 0 180 tan 0 180 D 24 1 12 hours cos 1 tan 22.8 180 tan 0 180 D 24 1 12 hours
d. There are approximately 12 hours of daylight every day at the equator.
cos 1 tan 22.8 180 tan 40.75 180 D 24 1 14.83 hours or 14 hours, 50 minutes
84. Note that 6630 66.5 . a.
cos 1 tan 23.5 180 tan 66.5 180 D 24 1 24 hours
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Section 8: Analytic Trigonometry
b.
cos 1 tan 0 180 tan 66.5 180 D 24 1 12 hours
34 6 86. x tan 1 tan 1 . x x 34 6 a. 10 tan 1 tan 1 42.6 10 10 If you sit 10 feet from the screen, then the viewing angle is about 42.6 . 34 6 15 tan 1 tan 1 44.4 15 15 If you sit 15 feet from the screen, then the viewing angle is about 44.4 . 34 6 20 tan 1 tan 1 42.8 20 20 If you sit 20 feet from the screen, then the viewing angle is about 42.8 .
cos tan 22.8 180 tan 66.5 180 D 24 1 22.02 hours or 22 hours, 1 minute 1
c.
Therefore, a person atop Cadillac Mountain will see the first rays of sunlight about 3.35 minutes sooner than a person standing below at sea level.
d. The amount of daylight at this location on the winter solstice is 24 24 0 hours. That is, on the winter solstice, there is no daylight. In general, for a location at 6630 ' north latitude, it ranges from around-the-clock daylight to no daylight at all. 85. Let point C represent the point on the Earth’s axis at the same latitude as Cadillac Mountain, and arrange the figure so that segment CQ lies along the x-axis (see figure).
b. Let r = the row that result in the largest viewing angle. Looking ahead to part (c), we see that the maximum viewing angle occurs when the distance from the screen is about 14.3 feet. Thus, 5 3(r 1) 14.3 5 3r 3 14.3 3r 12.3 r 4.1 Sitting in the 4th row should provide the largest viewing angle.
y
P
D (x,y )
s
C
2710 mi
x 2710 Q (2 71 0 ,0 )
c.
At the latitude of Cadillac Mountain, the effective radius of the earth is 2710 miles. If point D(x, y) represents the peak of Cadillac Mountain, then the length of segment PD is 1 mile 1530 ft 0.29 mile . Therefore, the 5280 feet point D( x, y ) (2710, y ) lies on a circle with radius r 2710.29 miles. We now have x 2710 cos r 2710.29 2710 cos 1 0.01463 radians 2710.29 Finally, s r 2710(0.01463) 39.64 miles ,
Set the graphing calculator in degree mode 34 6 and let Y1 tan 1 tan 1 : x x 90
0 Use MAXIMUM: 90
2 (2710) 39.64 , so 24 t 24(39.64) t 0.05587 hours 3.35 minutes 2 (2710)
and
0
892
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Section 8.1: The Inverse Sine, Cosine, and Tangent Functions 90. Here we have 1 2118' , 1 15750 ' , 2 3747 ' , and 2 14458' . Converting minutes to degrees gives 1 21.3 ,
The maximum viewing angle will occur when x 14.3 feet. a 0 ; b 3 ; The area is: tan 1 b tan 1 a tan 1 3 tan 1 0
87. a.
b.
a
3
3
144 . Substituting these values, and 1 157 56 , 2 37 47 , and 60
0
r 3960 , into our equation gives d 5518 miles. The distance from Honolulu to Melbourne is about 5518 miles. (remember that S and W angles are negative)
square units
3 ; b 1 ; The area is: 3
91. Let 1 sin 1 x and 2 cos 1
3 tan b tan a tan 1 tan 3 1
1
1
1
cos 1 1 x 2 and tan 2
4 6 5 square units 12
1 x2
b.
3
3
x2
7 16 7 7 16 4
0
x
square units
7 7 The solution set is , 4 4
92. Let 1 cos 1 tan 1
square units 3 89. Here we have 1 4150 ' , 1 8737 ' , 2 2118' , and 2 15750 ' . Converting minutes to degrees gives
157 . Substituting these values, and
1 41 56 , 1 87 37 , 2 21.3 , and 60 2
3 4
16 x 2 7
1 1 a ; b ; The area is: 2 2 1 1 sin 1 b sin 1 a sin 1 sin 1 2 2 6 6
3 . So, 4
9 16 16 16 x 2 9
3 ; The area is: 2 3 1 sin 1 b sin 1 a sin 1 sin 0 2
4 . Then 5
1 x2
a 0; b
88. a.
29 30
2
5 6
r 3960 , into our equation gives d 4250 miles. The distance from Chicago to Honolulu is about 4250 miles. (remember that S and W angles are negative)
893
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x and 2 tan 1 u . Then r
r 2 x2 u and sin 2 . So, x u2 1
Section 8: Analytic Trigonometry
r 2 x2 u x u2 1
96.
r x u 2 2 x u 1 2
2
2
r u r x u x x u
2
2
2
2
2
2
2
2
2
2
2
2
2
12
( x 2 3)
32
1
x(2 x 1) 2
(2 x 1)
12
( x 2 3)
32
( x 2 3) x(2 x 1)
(2 x 1)
12
( x 2 3)
32
( x 2 x 3)
e4 x 3 2
2
r 2 x2 u 2 2 x2 r 2
2
ln e 4 x ln 3
4 x ln 3
r 2 x2 u2 2 x2 r 2 u
( x 2 3)
97. e 4 x 7 10
r x 2x u r u 2
12
2
r x u 1 x u 2
(2 x 1)
x
r 2 x2 2 x2 r 2
ln 3 4
ln 3 The solution set is 4
93. 3x 2 5 9
98. The circumference of a circle is given by C 2 r . Thus,
3x 2 4 2 3x 2 3 2 So the solution is: , 2 3
C 2 (10) 20 inches 336 rev 1 min 336(20 in) 1 ft 1 mi 60 min 1 min 12 in 5280 ft 1 hr 19.912 20 mph
94. The function f is one-to-one because every horizontal line intersects the graph at exactly one point.
3 3 1 99. sin cos 3 3 2 2 4 24 , in quadrant I 25 Solve for sin : sin 2 cos 2 1
100. cos
sin 2 1 cos 2
95.
sin 1 cos 2 Since is in quadrant I, sin 0 .
f ( x) 1 2 x y 1 2x x 1 2
2
49 7 24 sin 1 cos 2 1 25 625 25
y
x 1 2y log 2 ( x 1) log 2 2
7 sin 25 7 25 7 tan cos 24 25 24 24 25
y
log 2 ( x 1) y log 2 2 log 2 ( x 1) y f 1 ( x) log 2 ( x 1)
894
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Section 8.2: The Inverse Trigonometric Functions (Continued)
cotangent equals
1 1 25 csc sin 7 7 25 1 1 25 sec cos 24 24 25 1 1 24 cot 7 tan 7 24
cot 3,
4
6
12
1
cot 1
3 3 12 4 3
11. csc 1 (1) , 2 2 0 , whose cosecant equals 1 . csc 1, , 0 2 2 2 1 csc (1) 2
We are finding the angle ,
Section 8.2 1. Domain: x x odd integer multiples of , 2
Range: y y 1 or y 1
12. csc 1 2 , 2 2 0 , whose cosecant equals 2 . csc 2, , 0 2 2 4 1 csc 2 4
We are finding the angle ,
2. True
1 5
6 3 6
10. cot 1 1 We are finding the angle , 0 , whose cotangent equals 1. cot 1, 0 4 cot 1 1 4
12
3.
0
101. Quadrant II f f tan tan 4 6 4 6 102.
3.
5 5
4. x sec y , 1 , 0 , 5. cosine 6. False 7. True
13. sec 1
8. True
2 3 3
We are finding the angle , 0 ,
9. cot 1 3 We are finding the angle , 0 , whose
895
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, 2
Section 8: Analytic Trigonometry
whose secant equals
2 3 . 3
2 3 , 3 6
sec
sec 1
14. sec
1
0 ,
2 3 , 3 3 2 3 csc 1 3 3 csc
2
2 3 3 6
17. sec 1 2
We are finding the angle , 0 , whose secant equals 2 . 0 ,
sec 2,
0 ,
3 4 3 sec 1 2 4
2
, 2
2
2 3 2 1 sec 2 3
18. cot 1 1
We are finding the angle , 0 , whose cotangent equals -1. cot 1, 0 3 4 3 1 cot 1 4
3 15. cot 1 3 We are finding the angle , 0 , whose 3 . 3
3 , 3 2 3 3 2 cot 1 3 3 cot
whose secant equals 2 .
, 2
cotangent equals
, 0 2 2
We are finding the angle , 0 ,
2
sec 2,
0
19. csc 1 2
, 2 2 0 , whose cosecant equals 2 . csc 2, , 0 2 2 4 1 csc 2 4
We are finding the angle ,
2 3 16. csc 1 3
, 2 2 2 3 0 , whose cosecant equals . 3
We are finding the angle ,
20. sec 1 1
We are finding the angle , 0 ,
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, 2
Section 8.2: The Inverse Trigonometric Functions (Continued)
whose secant equals 1 . sec 1,
0 sec
1
1 24. sec 1 (3) cos 1 3 We seek the angle , 0 , whose cosine
0 , 2
1 1 equals . Now cos , lies in 3 3 quadrant II. The calculator yields 1 cos 1 1.91 , which is an angle in 3 quadrant II, so sec1 3 1.91 .
1 0
1 4 We seek the angle , 0 , whose cosine
21. sec 1 4 cos 1
1 1 . Now cos , so lies in quadrant 4 4 1 I. The calculator yields cos 1 1.32 , which is 4 1 an angle in quadrant I, so sec 4 1.32 .
equals
1 25. csc 1 3 sin 1 3
We seek the angle , 22. csc 1 5 sin 1
2
2
, whose sine
1 1 equals . Now sin , so lies in 3 3 quadrant IV. The calculator yields 1 sin 1 0.34 , which is an angle in 3 quadrant IV, so csc1 3 0.34 .
1 5
We seek the angle ,
, whose sine 2 2 1 1 equals . Now sin , so lies in 5 5 1 quadrant I. The calculator yields sin 1 0.20 , 5 which is an angle in quadrant I, so csc1 5 0.20 .
1 26. cot 1 tan 1 ( 2) 2 We seek the angle , 0 , whose tangent equals 2 . Now tan 2 , so lies in quadrant II. The calculator yields tan 1 2 1.11 , which is an angle in
1 2 We seek the angle , 0 , whose tangent
23. cot 1 2 tan 1
quadrant IV. Since lies in quadrant II, 1.11 2.03 . Therefore, 1 cot 1 2.03 . 2
1 1 . Now tan , so lies in 2 2 1 quadrant I. The calculator yields an 1 0.46 , 2 which is an angle in quadrant I, so cot 1 2 0.46 .
equals
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Section 8: Analytic Trigonometry
4 3 30. sec 1 cos 1 3 4
1 27. cot 1 5 tan 1 5 We seek the angle , 0 , whose tangent
1
We are finding the angle , 0 ,
, 2 3 3 whose cosine equals . Now cos , so 4 4 lies in quadrant II. The calculator yields 3 cos 1 2.42 , which is an angle in 4 4 quadrant II, so sec 1 2.42 . 3
1
. Now tan , so lies in 5 5 quadrant II. The calculator yields 1 tan 1 0.42 , which is an angle in 5 quadrant IV. Since is in quadrant II, 0.42 2.72 . Therefore, equals
cot 1 5 2.72 .
3 2 31. cot 1 tan 1 2 3 We are finding the angle , 0 , whose
1 28. cot 8.1 tan 8.1 We seek the angle , 0 , whose tangent 1
1
2 2 . Now tan , so 3 3 lies in quadrant II. The calculator yields 2 tan 1 0.59 , which is an angle in 3 quadrant IV. Since is in quadrant II, 3 0.59 2.55 . Thus, cot 1 2.55 . 2
tangent equals
1 1 . Now tan , so lies in 8.1 8.1 quadrant II. The calculator yields 1 tan 1 0.12 , which is an angle in 8.1 quadrant IV. Since is in quadrant II,
equals
0.12 3.02 . Thus, cot 1 8.1 3.02 .
3 2 29. csc 1 sin 1 2 3
We seek the angle ,
2
1 32. cot 1 10 tan 1 10 We are finding the angle , 0 , whose
2
, 0,
1 . Now tan , so 10 10 lies in quadrant II. The calculator yields 1 tan 1 0.306 , which is an angle in 10 quadrant IV. Since is in quadrant II,
tangent equals
2 2 whose sine equals . Now sin , so 3 3 lies in quadrant IV. The calculator yields 2 sin 1 0.73 , which is an angle in 3 3 quadrant IV, so csc1 0.73 . 2
1
0.306 2.84 . So, cot 1 10 2.84 .
898
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Section 8.2: The Inverse Trigonometric Functions (Continued)
2 33. cos sin 1 2
1 equals . 2
Find the angle , equals
1 sin , 2 6
, whose sine 2 2
2 . 2
1 37. sec cos 1 2 Find the angle , 0 , whose cosine
equals
1 34. sin cos 1 2 Find the angle , 0 , whose cosine
1 , 2 3
0
1 38. cot sin 1 2
Find the angle ,
1 sin , 2 2 2 6 1 cot sin 1 cot 3 2 6
3 . 2
3 , 2
0
5 6
39. csc tan 1 1
Find the angle ,
3 5 3 tan cos 1 tan 2 6 3
, whose tangent 2 2
equals 1. tan 1,
1 36. tan sin 1 2
Find the angle ,
, whose sine 2 2
1 equals . 2
3 35. tan cos 1 2 Find the angle , 0 , whose cosine
cos
0
1 sec cos 1 sec 2 2 3
1 3 sin cos 1 sin 2 3 2
equals
1 . 2 cos
1 . 2 1 cos , 2 3
2 2
3 1 tan sin 1 tan 3 2 6
2 sin , 2 2 2 4 1 2 2 cos sin cos 4 2 2
equals
, whose sine 2 2
4
csc tan 1 1 csc
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4
2 2
2
Section 8: Analytic Trigonometry
40. sec tan 1 3
Find the angle , equals
1 equals . 2
, whose tangent 2 2
1 sin , 2 6
3. tan 3,
2 2
2 3
3 44. csc cos 1 2 Find the angle , 0 , whose cosine
41. sin tan 1 (1)
Find the angle , equals 1 .
, whose tangent 2 2
tan 1,
cos
2 2
3 2
0
5 6
3 5 2 csc cos 1 csc 2 6
2 sin tan 1 (1) sin 4 2 3 42. cos sin 1 2
5 2 45. cos 1 sin cos 1 4 2 Find the angle , 0 , whose cosine
, whose sine 2 2
2 . 2
equals
3 . equals 2
cos
3 sin , 2 2 2 3 1 3 1 cos sin cos 2 3 2
2 , 2
0
3 4 5 3 cos 1 sin 4 4
1 2 46. tan 1 cot tan 1 3 3
1 43. sec sin 1 2
Find the angle ,
3 . 2
equals
4
Find the angle ,
2 2
1 2 3 sec sin 1 sec 3 2 6
3
sec tan 1 3 sec
Find the angle ,
, whose sine 2 2
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, whose tangent 2 2
Section 8.2: The Inverse Trigonometric Functions (Continued)
equals
1 3
Since is in quadrant I, x 2 2 .
.
tan
1 3
,
1 y 1 2 2 tan sin 1 tan 3 4 x 2 2 2
2 2
1 50. tan cos 1 3
6 2 tan 1 cot 3 6
1 1 . Since cos and 0 , 3 3 is in quadrant I, and we let x 1 and r 3 . Solve for y: 1 y2 9 Let cos 1
3 7 47. sin 1 cos sin 1 6 2 Find the angle , , whose sine 2 2 3 . equals 2 3 sin , 2 2 2 3 7 sin 1 cos 3 6
y2 8 y 8 2 2
Since is in quadrant I, y 2 2 . 1 y 2 2 tan cos 1 tan 2 2 3 1 x 1 51. sec tan 1 2
Let tan 1
1 1 . Since tan and 2 2
, is in quadrant I, and we let 2 2 x 2 and y 1 . Solve for r: 22 1 r 2
48. cos 1 tan cos 1 1 4 Find the angle , 0 , whose cosine equals 1 . cos 1, 0 4 cos 1 tan 4
r2 5 r 5
is in quadrant I. 1 r 5 sec tan 1 sec x 2 2
1 49. tan sin 1 3
2 52. cos sin 1 3
1 1 Let sin 1 . Since sin and 3 3 , is in quadrant I, and we let 2 2 y 1 and r 3 . Solve for x: x2 1 9
Let sin 1
2 2 . Since sin and 3 3
, is in quadrant I, and we let 2 2 y 2 and r 3 . Solve for x: x2 2 9
x2 8
x2 7
x 8 2 2
x 7
901
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Section 8: Analytic Trigonometry
Since is in quadrant I, x 7 . 2 x 7 cos sin 1 cos 3 r 3
Since is in quadrant IV, r 10 . y sin tan 1 (3) sin r 10 3 10 3 10 10 10
2 53. cot sin 1 3
3 56. cot cos 1 3 3 3 Let cos 1 . Since cos 3 and 3 0 , is in quadrant II, and we let x 3 and r 3 . Solve for y: 3 y2 9
2 2 Let sin 1 and . Since sin 3 3 , is in quadrant IV, and we let 2 2 y 2 and r 3 . Solve for x: x2 2 9 x2 7
y2 6
x 7
y 6
Since is in quadrant IV, x 7 . x 2 7 2 14 cot sin 1 cot y 2 2 2 3
Since is in quadrant II, y 6 . 3 x cot cos 1 cot y 3
54. csc tan 1 ( 2) Let tan 1 ( 2) . Since tan 2 and
, is in quadrant IV, and we let 2 2 x 1 and y 2 . Solve for r: 1 4 r2
6
1 2
2 2
2 2
2 5 57. sec sin 1 5
Let sin 1
r 5 2
2 5 2 5 . Since sin and 5 5
, is in quadrant I, and we let 2 2 y 2 5 and r 5 . Solve for x: x 2 20 25
r 5
Since is in quadrant IV, r 5 . csc tan 1 ( 2) csc
3
r 5 5 y 2 2
55. sin tan 1 (3) Let tan 1 (3) . Since tan 3 and
x2 5 x 5
, is in quadrant IV, and we let 2 2 x 1 and y 3 . Solve for r: 1 9 r2
Since is in quadrant I, x 5 . 2 5 r 5 sec sin 1 5 sec 5 x 5 1 58. csc tan 1 2
r 2 10 r 10
Let tan 1
902
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1 1 . Since tan and 2 2
Section 8.2: The Inverse Trigonometric Functions (Continued)
, is in quadrant I, and we let 2 2 x 2 and y 1 . Solve for r: 22 1 r 2
64. Let cos 1 u so that cos u , 0 , 1 u 1 . Then, sin tan cos 1 u tan cos
r 5 2
r 5 is in quadrant I. 1 r 5 5 csc tan 1 csc 2 y 1
2
1
2
u . Then,
2
,
62. Let cos 1 u so that cos u , 0 , 1 u 1 . Then,
sin cos 1 u sin sin 2
63. Let sin 1 u so that sin u , 1 u 1 . Then,
sin
sin cos
cos 2 u
sec 2
sec 2 1 sec 2
u2 1 u
67. Let csc1 u so that csc u ,
1 cos 2 1 u 2
1
66. Let cot 1 u so that cot u , 0 , u . Then, 1 sin cot 1 u sin sin 2 csc2 1 1 2 1 cot 1 u2
1 1 sec sec 2 1 1 1 tan 2 1 u2
cos tan 1 u cos
tan sin 1 u tan
1 u2 u
sin sec 1 u sin sin 2 1 cos 2
61. Let tan 1 u so that tan u ,
, u 1 . Then,
7 1 2 60. cos 1 sin cos 1 6 2 3
sin 2 1 cos 2 cos cos
65. Let sec 1 u so that sec u , 0 and
3 2 59. sin 1 cos sin 1 4 2 4
2
2
,
u 1 . Then,
2
2
cos csc1 u cos cos
,
sin 1 sin 2
sin cot sin sin
cot cot 2 csc 2 1 csc csc csc
u2 1 u
68. Let sec1 u so that sec u , 0 and
1 u2
2
, u 1 . Then,
cos sec 1 u cos
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1 1 sec u
Section 8: Analytic Trigonometry
69. Let cot 1 u so that cot u , 0 , u . Then, 1 1 tan cot 1 u tan cot u
73. g 1 f cos 1 sin 4 4
2 3 cos 1 2 4
70. Let sec1 u so that sec u , 0 and
2
74.
, u 1 . Then,
tan sec 1 u tan tan 2 sec2 1 u 2 1 ( if u 0; if u 0)
3 3 75. h f 1 tan sin 1 5 5 3 3 Let sin 1 . Since sin and 5 5
12 12 71. g f 1 cos sin 1 13 13 12 12 and Let sin 1 . Since sin 13 13
x 2 (3) 2 52
x 2 122 132
x 2 9 25
x 2 144 169
x 2 16 x 16 4 Since is in quadrant IV, x 4 . 3 3 h f 1 tan sin 1 5 5
x 2 25 x 25 5 Since is in quadrant I, x 5 . 12 x 5 12 g f 1 cos sin 1 cos 13 13 r 13
72.
, is in quadrant IV, and we let 2 2 y 3 and r 5 . Solve for x:
, is in quadrant I, and we let 2 2 y 12 and r 13 . Solve for x:
5 5 f 1 g sin 1 cos 6 6 3 sin 1 2 3
tan
5 5 f g 1 sin cos 1 13 13 5 5 Let cos 1 . Since cos and 13 13 0 , is in quadrant I, and we let x 5 and r 13 . Solve for y:
3 y 3 4 4 x
4 4 76. h g 1 tan cos 1 5 5 4 4 Let cos 1 . Since cos and 5 5 0 , is in quadrant II, and we let x 4 and r 5 . Solve for y: (4) 2 y 2 52
52 y 2 132 25 y 2 169
16 y 2 25
y 2 144
y2 9
y 144 12 Since is in quadrant I, y 12 .
y 9 3 Since is in quadrant II, y 3 .
5 y 12 5 f g 1 sin cos 1 sin 13 r 13 13
4 4 h g 1 tan cos 1 5 5 3 y 3 tan x 4 4
904
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Section 8.2: The Inverse Trigonometric Functions (Continued) x 1 and r 4 . Solve for y: (1) 2 y 2 42
12 12 77. g h 1 cos tan 1 5 5 12 12 and Let tan 1 . Since tan 5 5
1 y 2 16 y 2 15
y 15
, is in quadrant I, and we let 2 2 x 5 and y 12 . Solve for r:
Since is in quadrant II, y 15 . 1 1 h g 1 tan cos 1 4 4
r 2 52 122 r 2 25 144 169
tan
r 169 13 Now, r must be positive, so r 13 . 12 12 x 5 g h 1 cos tan 1 cos 5 r 13 5
78.
2 2 82. h f 1 tan sin 1 5 5 2 2 Let sin 1 . Since sin and 5 5
5 5 f h 1 sin tan 1 12 12 5 5 and Let tan 1 . Since tan 12 12
, is in quadrant IV, and we let 2 2 y 2 and r 5 . Solve for x:
, is in quadrant I, and we let 2 2 x 12 and y 5 . Solve for r:
y 15 15 x 1
x 2 (2)2 52 x 2 4 25
r 2 122 52
x 2 21
r 2 144 25 169
x 21
Since is in quadrant IV, x 21 . 2 2 h f 1 tan sin 1 5 5
r 169 13 Now, r must be positive, so r 13 . 5 5 y 5 f h 1 sin tan 1 sin 12 12 13 r
tan
79. g 1 f cos 1 sin 3 3
83. a.
3 5 cos 1 2 6 80. g 1 f cos 1 sin 6 6 1 2 cos 1 2 3
b.
y 2 21 2 21 x 21
Since the diameter of the base is 45 feet, we 45 have r 22.5 feet. Thus, 2 22.5 cot 1 31.89 . 14
cot 1
r h
r r h cot h Here we have 31.89 and h 17 feet.
cot
1 1 81. h g 1 tan cos 1 4 4 1 1 Let cos 1 . Since cos and 4 4 0 , is in quadrant II, and we let
Thus, r 17 cot 31.89 27.32 feet and the diameter is 2 27.32 54.64 feet.
905
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Section 8: Analytic Trigonometry
c.
From part (b), we get h
86.
r . cot
22 61 feet. 2 r 61 h 37.96 feet. cot 22.5 /14 Thus, the height is 37.96 feet.
The radius is
84. a.
b.
c.
cot sec
cot sec
Since the diameter of the base is 6.68 feet, , 3.34 cot 1 50.14 4
Let sec 1
2 is in quadrant I. So
r cot 1 h r r cot h h cot Here we have 50.14 and r 4 feet. 4 Thus, h 4.79 feet. The cot 50.14
cot
6 39
5 3 6
5 3 5 3 . Then sec . 6 6
Because 0
, and because sec 0 ,
6 39 2 39 39 13
87. Let 1 cos 1 x . Then cos 1 x . Because 0 , 1 is in quadrant I and
bunker will be 4.79 feet high.
sin 1 1 x 2 . Let 2 tan 1 1 x 2 . Then
4.22 54.88 6
tan 2 1 x 2 0 , so 2 is in quadrant I. So
TG cot 1
sec 2 1 tan 2 2 2 x 2
cot
88 – 89. Answers will vary. 90.
( x 2 4)( x 2 25) 0
2x
x 2 4 0 or x 2 25 0 x 2 or x 5i So the complex zeros are: 2, 2, 5i,5i
2 y gt 2
2x 2 y gt 2 The artillery shell begins at the origin and lands at the coordinates 6175, 2450 . Thus,
91.
cot
1
f ( x ) ( x )3 ( x ) 2 ( x ) x3 x 2 x f ( x) So the function is not even. f ( x ) ( x )3 ( x ) 2 ( x )
2 2450 32.2 2.27 2
cot 1
f ( x) 4 x 4 21x 2 100 4 x 4 21x 2 100 0
cot 1
2 6175
( x3 x 2 x) f ( x) So the function is not odd.
2.437858 22.3
The artilleryman used an angle of elevation of 22.3 . b.
1
From part (a) we have USGA 50.14 . For steep bunkers, a larger angle of repose is required. Therefore, the Tour Grade 50/50 sand is better suited since it has a larger angle of repose. 85. a.
3 3 1 sin tan cot sec 3 6 3 2
1
sec
v0 t x
x sec 6175 sec 22.30325 t 2.27 2940.23 ft/sec
v0
906
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Section 8.3: Trigonometric Equations 7 92. 315 radians 180 4
93. 75 s r
97.
5 12
5 6 12
94. a.
5 7.85 in. 2
x
2 B 3 B 6 B6 and C 1 6 C6 y 4sin(6 x 6)
98.
5 (10) 2(2) 2
1 x2 1 c2 1 x2 1 c2 xc 1 x2 1 c2
2
31 5 5 y 2 10 3 2 2 2 5 31 The vertex is , 2 2 b. Since the leading coefficient is negative, the parabola is concave down. c. Since the graph has a maximum then the 5 graph is increasing on , and 2 5 decreasing on , 2 2 95. log 5 ( x 16) 2
2
1 x 1 c
x c
2
2
x c
2
1 x 1 c x c 1 x 1 c 2
x c
x c
2
2
2
2
2
( x c )( x c )
1 x 1 c x c 1 x 1 c 2
2
( x c )
2
2
( x c)
1 x 1 c 1 x 1 c 2
2
2
2
99. 4 4 3 f 1 f sin sin 3 2 3 2 2 4 4 5 3 2 3 2 6
52 x 2 16 25 x 2 16 25 x 2 16
x2 9 x 3 The solution set is 3,3 .
3 3 6 5
f x 3 96. Any number that would make the x7 g x4 argument of the square root negative or any denominator zero would not be in the domain. So the domain is: {x | x 3, x 4, x 7} or
3 36 5
Section 8.3
3, 4 4, 7 7, . 1. sec 2
2.
12
tan 2
2 1 , 2 2
907
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tan 2 1 tan 2 15 15 15 1
Section 8: Analytic Trigonometry 9. True
4 x2 x 5 0
3.
4 x 5 x 1 0
10. False, 2 is outside the range of the sin function.
4 x 5 0 or x 1 0 x
5 or 4
11. d
x 1
12. a
5 The solution set is 1, . 4
13. 2sin 3 2 2sin 1 1 sin 2 7 11 2k or 2k , k is any integer 6 6 7 11 On 0 2 , the solution set is , . 6 6
4. x 2 x 1 0 x
1
1 4 1 1 2 1 2
1 1 4 2 1 5 2
1 2 1 1 cos 2 1 cos 2 5 2k or 2k , k is any integer 3 3 5 On 0 2 , the solution set is , . 3 3
14. 1 cos
1 5 1 5 , . 2 2
The solution set is 5.
(2 x 1) 2 3(2 x 1) 4 0
(2 x 1) 1(2 x 1) 4 0 2 x(2 x 5) 0 2 x 0 or 2 x 5 0 5 x 0 or x 2 5 The solution set is 0, . 2
15. 2sin 1 0 2sin 1 1 sin 2 7 11 2k or 2k , k is any integer 6 6 7 11 On 0 2 , the solution set is , . 6 6
6. 5 x3 2 x x 2 Let y1 5 x3 2 and y2 x x 2 . Use INTERSECT to find the solution(s):
16. cos 1 0 cos 1 2k , k is any integer
In this case, the graphs only intersect in one location, so the equation has only one solution. Rounding as directed, the solution set is 0.76 .
On the interval 0 2 , the solution set is .
7. False because of the circular nature of the functions. 8. True 908
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Section 8.3: Trigonometric Equations 17. tan 1 0 tan 1 3 k , k is any integer 4
22. 4sin 3 3 3 4sin 2 3 2 3 3 4 2 4 5 2k or 2k , k is any integer 3 3 4 5 On 0 2 , the solution set is , . 3 3 sin
3 7 On 0 2 , the solution set is , . 4 4
18.
3 cot 1 0 3 cot 1 cot
1 3
3 3
23. 4 cos 2 1 1 cos 2 4
2 k , k is any integer 3
cos
2 5 On 0 2 , the solution set is , . 3 3
2 k or k , k is any integer 3 3 On the interval 0 2 , the solution set is 2 4 5 , , . , 3 3 3 3
19. 4sec 6 2 4sec 8 sec 2 2 4 2k or 2k , k is any integer 3 3 2 4 On 0 2 , the solution set is , . 3 3
24. tan 2
1 3 3 3 5 k or k , k is any integer 6 6 On the interval 0 2 , the solution set is 5 7 11 , , , . 6 6 6 6
On 0 2 , the solution set is . 2
25. 2sin 2 1 0 2sin 2 1 1 sin 2 2
21. 3 2 cos 2 1 3 2 cos 3 1 2
1 3
tan
20. 5csc 3 2 5csc 5 csc 1 2k , k is any integer 2
cos
1 2
2 2
1 2 2 2 3 k or k , k is any integer 4 4 On the interval 0 2 , the solution set is 3 5 7 , , , . 4 4 4 4 sin
3 5 2k or 2k , k is any integer 4 4 3 5 On 0 2 , the solution set is , . 4 4
909
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Section 8: Analytic Trigonometry
On the interval 0 2 , the solution set is 3 7 11 15 , , , . 8 8 8 8
26. 4 cos 2 3 0 4 cos 2 3 3 cos 2 4
3 2 2 3 2 3 4 2k or 2k 2 3 2 3 4 4k 8 4k or , 9 3 9 3 k is any integer On the interval 0 2 , the solution set is 4 8 16 , , . 9 9 9
31. sec
3 cos 2 5 k or k , k is any integer 6 6 On the interval 0 2 , the solution set is 5 7 11 , , , . 6 6 6 6
27. sin 3 1 3 2k 2 2k , k is any integer 2 3 On the interval 0 2 , the solution set is 7 11 , , . 2 6 6 3
2 3 3 2 5 k , k is any integer 3 6 5 3k , k is any integer 4 2
32. cot
5 On 0 2 , the solution set is . 4
28. tan 3 2
33. cos 2 1 2 2 2k 2 3 2k 2 2 3 k , k is any integer 4 3 7 On 0 2 , the solution set is , . 4 4
k , k is any integer 3 2 2 k , k is any integer 3 2 On 0 2 , the solution set is . 3 2
29. cos 2
1 2
2 4 2k or 2 2k 3 3 2 k or k , k is any integer 3 3 On the interval 0 2 , the solution set is 2 4 5 , , . , 3 3 3 3 2
34. sin 3 1 18 3 2k 18 2 4 3 2k 9 4 2k , k is any integer 27 3 On the interval 0 2 , the solution set is 4 22 40 , , . 27 27 27
30. tan 2 1 3 k , k is any integer 4 3 k , k is any integer 8 2
2
910
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Section 8.3: Trigonometric Equations 35. tan 1 2 3 k 2 3 4 k 2 12 2k , k is any integer 6 11 On 0 2 , the solution set is . 6
40. cos
5 7 2k or 2k , k is any 6 6 integer. Six solutions are 5 7 17 19 29 31 , , , . , , 6 6 6 6 6 6
41. cos 0 3 2k or = 2k , k is any 2 2 integer 3 5 7 9 11 . , , Six solutions are , , , 2 2 2 2 2 2
1 36. cos 3 4 2 5 2k or 2k 3 4 3 3 4 3 7 23 2k or 2k 3 12 3 12 7 23 6k or 6k , 4 4 k is any integer. 7 On 0 2 , the solution set is . 4
37. sin
2 2 3 2k , k is any 2k or 4 4 integer 3 9 11 17 19 . , , Six solutions are , , , 4 4 4 4 4 4
42. sin
43.
1 2
3 cot 0 cot 3
5 2k , k is any 2k or 6 6 integer. Six solutions are 5 13 17 25 29 , , , . , , 6 6 6 6 6 6
k , k is any integer 6 Six solutions are 7 13 19 25 31 , , . , , , 6 6 6 6 6 6
38. tan 1 k , k is any integer 4
Six solutions are
39. tan
3 2
44. 2 3 csc 0 2 csc 3
5 9 13 17 21 . , , , , , 4 4 4 4 4 4
2 2k , k is any 2k or 3 3 integer 2 7 8 13 14 , , , , . Six solutions are , 3 3 3 3 3 3
3 3
5 k , k is any integer 6 Six solutions are 5 11 17 23 29 35 , , , , . , 6 6 6 6 6 6
45. cos 2 2
1 2
2 4 2k or 2 2k , k is any integer 3 3
911
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Section 8: Analytic Trigonometry 51. tan 5
2 k , k is any integer k or 3 3 2 4 5 7 8 , , , , . Six solutions are , 3 3 3 3 3 3
tan 1 5 1.37 1.37 or 1.37 4.51 .
The solution set is 1.37, 4.51 .
46. sin 2 1
52. cot 2 1 tan 2
3 2 2k , k is any integer 2 3 k , k is any integer 4 Six solutions are 3 7 11 15 19 23 , , . , , , 4 4 4 4 4 4
1 0.46 or 0.46 3.61 . The solution set is 0.46, 3.61 .
tan 1 0.46 2
53. cos 0.9
3 47. sin 2 2 4 5 2k or 2k , k is any integer 2 3 2 3 8 10 4k or 4k , k is any 3 3 integer. Six solutions are 8 10 20 22 32 34 , . , , , , 3 3 3 3 3 3
cos 1 0.9 2.69 2.69 or 2 2.69 3.59 . The solution set is 2.69, 3.59 . 54. sin 0.2
sin 1 0.2 0.20
0.20 2 or 0.20 . 6.08 3.34 The solution set is 3.34, 6.08 .
1 2 3 k , k is any integer 2 4 3 2k , k is any integer 2 Six solutions are 3 7 11 15 19 23 , , . , , , 2 2 2 2 2 2
48. tan
55. sec 4 1 cos 4 1 1.82 or 2 1.82 4.46 . The solution set is 1.82, 4.46 .
cos 1 1.82 4
56. csc 3 1 sin 3
49. sin 0.4
sin 1 0.4 0.41 0.41 or 0.41 2.73 . The solution set is 0.41, 2.73 .
1 0.34 2 or 0.34 . 5.94 3.48 The solution set is 3.48, 5.94 .
sin 1 0.34 3
50. cos 0.6
cos1 0.6 0.93 0.93 or 2 0.93 5.36 .
The solution set is 0.93, 5.36 .
912
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Section 8.3: Trigonometric Equations 57. 5 tan 9 0 5 tan 9 9 tan 5
61.
2 cos 2 cos 0 cos (2 cos 1) 0 cos 0 3 , 2 2
2 cos 1 0 2 cos 1 1 cos 2 2 4 , 3 3 2 4 3 The solution set is , , , . 3 2 2 3
9 1.064 or 1.064 2 2.08 5.22 The solution set is 2.08, 5.22 .
tan 1 1.064 5
58. 4 cot 5 5 cot 4 4 tan 5
62.
or
sin 2 1 0 (sin 1)(sin 1) 0 sin 1 0 or sin 1 3 2
sin 1 0 sin 1 2 3 The solution set is , . 2 2
4 0.675 or 0.675 2 . 2.47 5.61 The solution set is 2.47, 5.61 .
tan 1 0.675 5
63.
59. 3sin 2 0 3sin 2 2 sin 3
2sin 2 sin 1 0 (2sin 1)(sin 1) 0 2sin 1 0 or sin 1 0 2sin 1 sin 1 1 sin 2 2 7 11 , 6 6 7 11 The solution set is , , . 6 2 6
2 0.73 or 0.73 2.41 . The solution set is 0.73, 2.41 .
sin 1 0.73 3
60. 4 cos 3 0 4 cos 3 3 cos 4
64.
2 cos 2 cos 1 0 (cos 1)(2 cos 1) 0 cos 1 0 or cos 1
2 cos 1 0 2 cos 1 1 cos 2 5 , 3 3 5 The solution set is , , . 3 3
3 2.42 or 2 2.42 3.86 . The solution set is 2.42, 3.86 .
cos 1 2.42 4
913
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Section 8: Analytic Trigonometry 65. (tan 1)(sec 1) 0 tan 1 0 or sec 1 0 tan 1 sec 1 5 0 , 4 4 5 The solution set is 0, , . 4 4
sin 2 6 cos 1 1 cos 2 6 cos 6 cos 2 6 cos 5 0
cos 5 cos 1 0 cos 5 0 or cos 5 (not possible)
1 66. (cot 1) csc 0 2 csc
2 cos 2 3cos 1 0
2 cos 1 cos 1 0
1 cos cos 1 cos
2 cos 1 0 1 cos 2
1 2 cos 2 1 cos 2 cos 2 cos 0
cos 2 cos 1 0
or
2 cos 1 0
3
cos
, 2 2
1 2
cos sin sin 1 cos tan 1 5 , 4 4
1 sin sin sin 0 2
1 2sin 2 sin 0
5 The solution set is , . 4 4
2sin 2 sin 1 0
2sin 1 sin 1 0 sin
or 1 2
, 3 3
cos sin
cos 2 sin 2 sin 0
2sin 1 0
cos 1 0 cos 1 0
71. cos sin
2 4 , 3 3 2 4 3 The solution set is , , , . 3 2 2 3
2
5
or
5 The solution set is 0, , . 3 3
68.
2 2 cos 2 3 3cos
2
cos 0
2sin 2 3 1 cos
2 1 cos 2 3 1 cos
sin 2 cos 2 1 cos 2
2sin 2 3 1 cos
70.
3 7 The solution set is , . 4 4
67.
cos 1 0 cos 1
The solution set is .
1 0 2 1 csc 2 (not possible)
cot 1 0 or cot 1 3 7 , 4 4
sin 2 6 cos 1
69.
sin 1 0 sin 1
7 11 , 6 6 7 11 The solution set is , , . 6 2 6
2
914
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Section 8.3: Trigonometric Equations
72.
73.
74.
cos sin 0
1 sin 2 cos 2
75.
cos sin 0
1 sin 2(1 sin 2 )
cos sin 0 sin cos sin 1 cos tan 1 3 7 , 4 4 3 7 The solution set is , . 4 4
1 sin 2 2sin 2 2sin 2 sin 1 0 (2sin 1)(sin 1) 0 2sin 1 0 or sin 1 0 1 sin 1 sin 2 3 5 2 , 6 6 5 3 The solution set is , , . 2 6 6
tan 2sin sin 2sin cos sin 2sin cos 0 2sin cos sin 0 sin (2 cos 1) 2 cos 1 0 or sin 0 1 0, cos 2 5 , 3 3 5 The solution set is 0, , , . 3 3
sin 2 2 cos 2
76.
1 cos 2 2 cos 2 cos 2 2 cos 1 0
cos 1 0 2
cos 1 0 cos 1 The solution set is .
77.
2sin 2 5sin 3 0
2sin 3 sin 1 0 2sin 3 0 or sin 1 0 3 sin (not possible) 2 2 The solution set is . 2 2 78. 2 cos 7 cos 4 0
tan cot 1 tan tan tan 2 1 tan 1 3 5 7 , , , 4 4 4 4 3 5 7 , , The solution set is , . 4 4 4 4
2 cos 1 cos 4 0 2 cos 1 0
or cos 4 0 1 cos 4 sin 2 (not possible) 2 4 , 3 3 2 4 The solution set is , . 3 3
915
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Section 8: Analytic Trigonometry
3(1 cos ) sin 2
79.
83.
tan 2 1 tan 0 This equation is quadratic in tan . The discriminant is b 2` 4ac 1 4 3 0 . The equation has no real solutions.
3 3cos 1 cos 2 cos 2 3cos 2 0
cos 1 cos 2 0 cos 1 0 or cos 2 0 cos 1 cos 2 0 (not possible)
84.
The solution set is 0 . 4(1 sin ) cos 2
80.
4 4sin 1 sin 2 sin 2 4sin 3 0
sin 1 sin 3 0 sin 1 0 sin 1 3 2
or
sin 3 0 sin 3 (not possible)
3 The solution set is . 2
sec tan cot 1 sin cos cos cos sin 1 sin 2 cos 2 cos sin cos 1 1 cos sin cos sin cos 1 cos sin 1 2 Since sec and tan do not exist, the 2 2 equation has no real solutions.
85. x 5cos x 0 Find the zeros (x-intercepts) of Y1 x 5cos x :
3 sec 2 3 sec2 1 sec 2 2sec2 2 3sec tan 2
81.
sec 2 tan 0
2sec 3sec 2 0 (2sec 1)(sec 2) 0 2sec 1 0 or sec 2 0 1 sec 2 sec 2 5 , (not possible) 3 3 5 The solution set is , . 3 3
2
x 1.31, 1.98, 3.84
86. x 4sin x 0 Find the zeros (x-intercepts) of Y1 x 4sin x :
csc2 cot 1
82.
1 cot cot 1 2
cot cot 0 cot (cot 1) 0 2
cot 0 or cot 1 3 5 , , 2 2 4 4 5 3 The solution set is , , , . 2 4 2 4
916
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Section 8.3: Trigonometric Equations
90. sin x cos x x Find the intersection of Y1 sin x cos x and Y2 x :
x 1.26
x 2.47, 0, 2.47
91. x 2 2 cos x 0
Find the zeros (x-intercepts) of Y1 x 2 2 cos x :
87. 22 x 17 sin x 3 Find the intersection of Y1 22 x 17 sin x and Y2 3 :
x 1.02, 1.02
92. x 2 3sin x 0
x 0.52
Find the zeros (x-intercepts) of Y1 x 2 3sin x :
88. 19 x 8cos x 2 Find the intersection of Y1 19 x 8cos x and Y2 2 :
x 1.72, 0
93. x 2 2sin 2 x 3 x
x 0.30
Find the intersection of Y1 x 2 2sin 2 x and Y2 3 x :
89. sin x cos x x Find the intersection of Y1 sin x cos x and Y2 x :
x 0, 2.15
x 1.26
917
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Section 8: Analytic Trigonometry
On the interval 0, 2 , the zeros of f are
94. x 2 x 3cos(2 x)
2 4 5 , , , . 3 3 3 3
Find the intersection of Y1 x and Y2 x 3cos(2 x) : 2
f x 0
98.
2 cos 3 x 1 0
2 cos 3 x 1
cos 3 x
2 4 2k or 3x 2k 3 3 2 2k 4 2k x or x , 9 3 9 3 k is any integer On the interval 0, , the zeros of f are
x 0.62, 0.81
3x
95. 6sin x e x 2, x 0
Find the intersection of Y1 6sin x e x and Y2 2 :
1 2
2 4 8 , , . 9 9 9
99. a.
f x 0 3sin x 0
sin x 0 x 0 2k or x 2k , k is any integer On the interval 2 , 4 , the zeros of f are
x 0.76, 1.35
96. 4 cos(3 x) e x 1, x 0
2, , 0, , 2, 3, 4 .
Find the intersection of Y1 4 cos(3 x) e x and Y2 1 :
b.
f x 3sin x
x 0.31
97.
f x 0 4sin 2 x 3 0
c.
4sin 2 x 3 sin 2 x
3 4
3 3 4 2 2 x k or x k , k is any integer 3 3 sin x
3 2 3 3sin x 2 1 sin x 2 f x
x
6
918
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2k or x
5 2k , k is any integer 6
Section 8.3: Trigonometric Equations
On the interval 2 , 4 , the solution set is
7 5 x interval 2 , 4 is x 6 6 5 7 17 19 or x or x . 6 6 6 6
11 7 5 13 17 , , , , , . 6 6 6 6 6 6
d. From the graph in part (b) and the results of 3 part (c), the solutions of f x on the 2 11 7 interval 2 , 4 is x x 6 6 or
101.
a.
5 13 17 x or x . 6 6 6 6
b.
2 cos x 0 3 2k or x 2k , k is any 2 2 integer On the interval 2 , 4 , the zeros of f are
b.
f x 4 4 tan x 4 tan x 1 Graphing y1 tan x and y2 1 on the
cos x 0 x
f x 4 4 tan x 4 tan x 1 x x k , k is any integer 4
f x 0
100. a.
f x 4 tan x
interval , , we see that y1 y2 for 2 2 x or , . 2 4 2 4
3 3 5 7 . , , , , , 2 2 2 2 2 2
f x 2 cos x
2
2
102.
f x cot x
a. c.
cot x 3
f x 3
5 k , k is any integer x x 6
2 cos x 3 cos x
f x 3
3 2
b.
f x 3 cot x 3
5 7 2k or x 2k , k is any 6 6 integer On the interval 2 , 4 , the solution set is x
1 and y2 3 on the tan x interval 0, , we see that y1 y2 for
7 5 5 7 17 19 , , , , , . 6 6 6 6 6 6
0 x
Graphing y1
d. From the graph in part (b) and the results of part (c), the solutions of f x 3 on the
919
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5 5 or 0, . 6 6
Section 8: Analytic Trigonometry
f x g x
b.
x 2 cos 3 4 2 x 2 cos 1 2 x 1 cos 2 2 x x 5 2k or 2k 2 3 2 3 2 10 x 4k or x 4k , 3 3 k is any integer 2 10 On 0, 4 , the solution set is , . 3 3
103. a, d.
f x 3sin 2 x 2 ; g x
7 2
c.
From the graph in part (a) and the results of part (b), the solution of f x g x on
0, 4 is x
f x g x
b.
7 2 3 3sin 2 x 2 1 sin 2 x 2
3sin 2 x 2
2x
6
f x 4 cos x ; g x 2 cos x 3
5 2k 6 5 x k , 12
2k or 2 x
k or 12 k is any integer x
105. a, d.
2 10 2 10 x or , . 3 3 3 3
5 On 0, , the solution set is , . 12 12
c.
From the graph in part (a) and the results of part (b), the solution of f x g x on
b.
4 cos x 2 cos x 3
5 5 0, is x x or , . 12 12 12 12
104. a, d.
f x g x 6 cos x 3 cos x
x f x 2 cos 3 ; g x 4 2
3 1 6 2
2 4 2k or x 2k , 3 3 k is any integer 2 4 On 0, 2 , the solution set is , . 3 3 x
920
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Section 8.3: Trigonometric Equations c.
For k 0 , t 0 sec. 3 For k 1 , t 0.43 sec. 7 6 For k 2 , t 0.86 sec. 7 The blood pressure will be 100 mmHg after 0 seconds, 0.43 seconds, and 0.86 seconds.
From the graph in part (a) and the results of part (b), the solution of f x g x on
0, 2 is x
106. a, d.
2 4 2 4 x , or . 3 3 3 3
f x 2sin x ; g x 2sin x 2
b. Solve P t 120 on the interval 0,1 . 7 100 20sin t 120 3 7 20sin t 20 3
b.
7 sin t 1 3
f x g x
7 t 2 k , k is any integer 3 2 3 2k 12 , k is any integer t 7 We need 3 2k 12 0 1 7 0 2k 12 73
2sin x 2sin x 2 4sin x 2 sin x
2 1 4 2
5 2k or x 2k , 6 6 k is any integer 5 On 0, 2 , the solution set is , . 6 6 x
c.
12 2k 11 6 11 14 k 12
From the graph in part (a) and the results of part (b), the solution of f x g x on
0, 2 is x
3 0.21 sec 14 The blood pressure will be 120mmHg after 0.21 sec .
For k 0 , t
5 5 x or , . 6 6 6 6
7 107. P t 100 20sin t 3 a. Solve P t 100 on the interval 0,1 .
c.
Solve P t 105 on the interval 0,1 . 7 100 20sin t 105 3 7 20sin t 5 3
7 100 20sin t 100 3 7 20sin t 0 3
7 3 sin t 3 4 7 3 t sin 1 3 4
7 sin t 0 3 7 t k , k is any integer 3
t
3
t k , k is any integer
3 3 sin 1 7 4
On the interval 0,1 , we get t 0.03
7
seconds, t 0.39 seconds, and t 0.89 seconds. Using this information, along with
We need 0 73 k 1 , or 0 k 7 . 3 921
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Section 8: Analytic Trigonometry
the results from part (a), the blood pressure will be between 100 mmHg and 105 mmHg for values of t (in seconds) in the interval 0, 0.03 0.39, 0.43 0.86, 0.89 .
t
108. h t 125sin 0.157t 125 2 a. Solve h t 125sin 0.157t 125 125 2 on the interval 0, 40 .
2
c.
k , k is any integer
0.157t k
t
k
2
For k 1, t
sin 0.157t 0 2 Graphing y1 sin 0.157 x and y2 0 2 on the interval 0, 40 , we see that y1 y2 for
2 30 seconds . 0.157
10 x 30 .
2 50 seconds . 0.157 So during the first 40 seconds, an individual on the Ferris wheel is exactly 125 feet above the ground when t 10 seconds and again when t 30 seconds .
For k 2, t
125sin 0.157t 0 2
2
, k is any integer
125sin 0.157t 125 125 2
, k is any integer
2 10 seconds . For k 0, t 0.157
0.157
Solve h t 125sin 0.157t 125 125 2 on the interval 0, 40 .
2 , k is any integer 0.157
0
2k
20 seconds . 0.157 2 60 seconds . For k 1, t 0.157 4 For k 2, t 100 seconds . 0.157 So during the first 80 seconds, an individual on the Ferris wheel is exactly 250 feet above the ground when t 20 seconds and again when t 60 seconds .
sin 0.157t 0 2 0.157t
For k 0, t
125sin 0.157t 125 125 2 125sin 0.157t 0 2
2k , k is any integer 2 2 0.157t 2k , k is any integer
0.157t
b. Solve h t 125sin 0.157t 125 250 2 on the interval 0,80 .
So during the first 40 seconds, an individual on the Ferris wheel is more than 125 feet above the ground for times between about 10 and 30 seconds. That is, on the interval 10 x 30 , or 10, 30 .
125sin 0.157t 125 250 2 125sin 0.157t 125 2 sin 0.157t 1 2 922
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Section 8.3: Trigonometric Equations
the interval 0, 20 , we see that y1 y2 for
109. d x 70sin 0.65 x 150 a.
d 0 70sin 0.65 0 150
0 x 6.06 , 8.44 x 15.72 , and 18.11 x 20 .
70sin 0 150
150 miles
b. Solve d x 70sin 0.65 x 150 100 on
the interval 0, 20 . 70sin 0.65 x 150 100
So during the first 20 minutes in the holding pattern, the plane is more than 100 miles from the airport before 6.06 minutes, between 8.44 and 15.72 minutes, and after 18.11 minutes.
70sin 0.65 x 50 sin 0.65 x
5 7
5 0.65 x sin 1 2 k 7 5 sin 1 2 k 7 x 0.65 3.94 2 k 5.94 2 k x or x , 0.65 0.65 k is any integer 3.94 0 5.94 0 For k 0 , x or x 0.65 0.65 6.06 min 8.44 min
d. No, the plane is never within 70 miles of the airport while in the holding pattern. The minimum value of sin 0.65x is 1 . Thus,
the least distance that the plane is from the airport is 70 1 150 80 miles. 110. R 672sin 2 a.
interval 0, . 2 672sin 2 450
3.94 2 5.94 2 or x 0.65 0.65 15.72 min 18.11 min
For k 1 , x
450 225 672 336 225 2 sin 1 2k 336
sin 2
For k 2 , 3.94 4 5.94 4 or x x 0.65 0.65 25.39 min 27.78 min
225 sin 1 2k 336 2 0.7337 2k 2.408 2k or , 2 2 k is any integer
So during the first 20 minutes in the holding pattern, the plane is exactly 100 miles from the airport when x 6.06 minutes , x 8.44 minutes , x 15.72 minutes , and x 18.11 minutes . c.
Solve R 672sin 2 450 on the
the interval 0, 20 .
0.7337 0 2.408 0 or 2 2 0.36685 1.204
70sin 0.65 x 150 100
21.02
For k 0 ,
Solve d x 70sin 0.65 x 150 100 on
70sin 0.65 x 50 sin 0.65 x
For k 1 ,
5 7
Graphing y1 sin 0.65 x and y2
5 on 7
0.7337 2 2
68.98
or
2.408 2 2
3.508
4.3456
200.99
248.98
So the golfer should hit the ball at an angle of either 21.02 or 68.98 . 923
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Section 8: Analytic Trigonometry
b. Solve R 672sin 2 540 on the interval 0, . 2 672sin 2 540 540 135 672 168 135 2 sin 1 2k 168
sin 2
So, the golf ball will travel at least 480 feet if the angle is between about 22.79 and 67.21 . d. No; since the maximum value of the sine function is 1, the farthest the golfer can hit the ball is 672 1 672 feet.
135 sin 1 2k 168 2 0.9333 2k 2.2083 2k or , 2 2 k is any integer
111. Find the first two positive intersection points of Y1 x and Y2 tan x .
26.74
63.26
0.9330 2 2
3.608
or
2.2083 2
4.246
112. a.
Solve R 672sin 2 480 on the
Let L be the length of the ladder with x and y being the lengths of the two parts in each hallway. L x y cos
interval 0, . 2 672sin 2 480
x
480 672 5 sin 2 7 sin 2
L( )
Graphing y1 sin 2 x and y2
3 x
4 y
sin
3 cos
y
4 sin
3 4 3sec 4 csc cos sin
3sec tan 4 csc cot 0 3sec tan 4 csc cot
5 on the 7
sec tan 4 csc cot 3 4 tan 3 3
interval 0, and using INTERSECT, we 2 see that y1 y2 when 0.3978 x 1.1730 radians, or 22.79 x 67.21 .
4 1.10064 3 47.74º
tan 3
The first two positive solutions are x 2.03 and x 4.91 .
2
206.72 243.28 So the golfer should hit the ball at an angle of either 26.74 or 63.26 .
c.
0.9330 0 2.2083 0 or 2 2 0.46665 1.10415
For k 0 ,
For k 1 ,
2
2
b.
L 47.74º
3 4 cos 47.74º sin 47.74º
9.87 feet
924
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Section 8.3: Trigonometric Equations
c.
3 4 and use the cos x sin x MINIMUM feature:
Graph Y1
114. a.
An angle of 47.74 minimizes the length at L 9.87 feet .
2 42.4º or 137.6º
21.2º
d. For this problem, only one minimum length exists. This minimum length is 9.87 feet, and it occurs when 47.74 . No matter if we find the minimum algebraically (using calculus) or graphically, the minimum will be the same. 113. a.
107
(40) 2 sin(2 ) 9.8 110 9.8 sin(2 ) 0.67375 402 2 sin 1 0.67375 110
or 68.8º
b. The maximum distance will occur when the angle of elevation is 45 : (40) 2 sin 2(45) R 45 163.3 9.8 The maximum distance is approximately 163.3 meter
(34.8) 2 sin 2
c.
9.8 107(9.8) 0.8659 sin 2 (34.8) 2
Let Y1
(40) 2 sin(2 x) : 9.8
2 sin 1 0.8659 2 60º or 120º
30º or 60º
b. Notice that the answers to part (a) add up to 90 . The maximum distance will occur when the angle of elevation is 90 2 45 : (34.8)2 sin 2 45 123.6 R 45 9.8 The maximum distance is 123.6 meters. c.
Let Y1
d.
(34.8) 2 sin(2 x) 9.8
115.
sin 40 1.33 sin 2 1.33sin 2 sin 40
sin 40 0.4833 1.33 2 sin 1 0.4833 28.90
sin 2
d.
925
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Section 8: Analytic Trigonometry
116.
121. Here we have n1 1.33 and n2 1.52 . n1 sin B n2 cos B
sin 50 1.66 sin 2 1.66sin 2 sin 50
sin B n2 cos B n1
sin 50 0.4615 1.66 2 sin 1 0.4615 27.48
sin 2
tan B
117. Calculate the index of refraction for each: v1 sin 1 1 2 v2 sin 2 sin10º 1.2477 10º 8º sin 8º sin 20º 1.2798 20º 15º 30 ' 15.5º sin15.5º sin 30º 1.3066 30º 22º 30 ' 22.5º sin 22.5º sin 40º 1.3259 40º 29º 0 ' 29º sin 29º sin 50º 1.3356 50º 35º 0 ' 35º sin 35º sin 60º 1.3335 60º 40º 30 ' 40.5º sin 40.5º sin 70º 70º 45º 30 ' 45.5º 1.3175 sin 45.5º sin 80º 80º 50º 0 ' 50º 1.2856 sin 50º
n2 n1
B tan 1
n2 1.52 tan 1 48.8 n1 1.33
122. If is the original angle of incidence and is sin n2 . The sin angle of incidence of the emerging beam is also 1 , and the index of refraction is . Thus, is n2 the angle of refraction of the emerging beam. The two beams are parallel since the original angle of incidence and the angle of refraction of the emerging beam are equal.
the angle of refraction, then
Yes, these data values agree with Snell’s Law. The results vary from about 1.25 to 1.34. 118.
v1 2.998 108 1.56 v2 1.92 108 The index of refraction for this liquid is about 1.56.
123.
3sin 3 cos 0 cos cos 3 tan 3 0
119. Calculate the index of refraction: sin 1 sin 40º 1.47 1 40º , 2 26º ; sin 2 sin 26º
tan
120. The index of refraction of crown glass is 1.52. sin 30º 1.52 sin 2
3 3
5 k , where k is any integer 6
5 k , where k is any integer | 6
1.52sin 2 sin 30
124. Substitute x 2 3 to get
sin 30 0.3289 sin 2 1.52 2 sin 1 0.3352 19.20
8 4 3 2 tan 2 cot 3 tan 3 cot 0 (1)
Substitute x 2 3 to get 8 4 3 2 tan 2 cot 3 tan 3 cot 0 (2)
The angle of refraction is about 19.20 .
Subtract equation (1) from equation (2) to get 8 3 2 3 tan 2 3 cot 0 .
926
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Section 8.3: Trigonometric Equations 130. y 2sin 2 x
So, tan cot 4 sin cos 4 cos sin sin 2 cos 2 4sin cos 1 4sin cos 1 sin cos 4
Amplitude:
A 2 2
Period:
T
Phase Shift:
2 2
2
2 2
125. Answers will vary. 126. Since the range of y sin x is 1 y 1 , then y 5sin x x cannot be equal to 3 when x 4 or x since you are multiplying the result by 5 and adding x. 127. 6 x y x log 6 y 128. x
( 9) ( 9) 2 4(2)(8) 2(2)
131. First find the inverse of the function. 1 y e x 1 3 2 1 x 1 y 3 e 2 2( y 3) e x 1
9 81 64 4 9 17 4
ln 2( y 3) ln e x 1
9 17 9 17 , So the zeros are . 4 4
ln 2 ln( y 3) x 1 ln 2 ln( y 3) 1 x y ln 2 ln( x 3) 1 Since the argument of the ln function must be positive then the domain of the inverse function is x | x 3 or 3, .
10 3 10 , cos 10 10 10 sin 10 10 10 1 tan cos 3 10 10 3 10 3 10
129. sin
csc
132. s r 15(36) 180 3 9.425 cm
1 1 10 10 1 10 sin 10 10 10 10
sec
1 1 10 10 10 cos 3 10 3 10 10 3 10
cot
1 3 tan
133. ax 3 y 10 3 y ax 10 10 a y x 3 3 a 2 3 a6
927
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Section 8: Analytic Trigonometry
134.
10. b
3( x) 5 ( x)2 3x 5 x2 3x f ( x) 5 x2 The function is odd. f ( x)
5 y 8 ( x 1) 2 5 5 y 8 x 2 2 5 21 y x 2 2
12. cot sec
cos 1 1 sin cos sin
14.
1 f (1) f cos 1 1 cos 1 1 2 2 1 4 1 1 2
sin 1 1 cos sin cos
13.
1 8 f (4) f (1) 2 135. m 4 1 3 15 15 5 2 3 6 2
136.
11. tan csc
0 1 2
3
15.
3 2 1 3 2
Section 8.4 16.
1. True
cos 1 sin cos 1 sin 1 sin 1 sin 1 sin 2 cos 1 sin cos 2 1 sin cos sin 1 cos sin 1 cos 1 cos 1 cos 1 cos 2 sin 1 cos sin 2 1 cos sin sin cos cos sin cos sin sin 2 sin cos cos cos sin sin cos 2 sin sin cos cos 2 cos sin sin cos 2 2 sin cos sin cos cos sin sin cos 1 sin cos 1 1 1 cos v 1 cos v 1 cos v 1 cos v 1 cos v 1 cos v 2 1 cos 2 v 2 sin 2 v
2. True
3. identity; conditional 4. 1 5. 0 6. True 7. False, you need to work with one side only. 8. True 9. c 928
Copyright © 2025 Pearson Education, Inc.
Section 8.4: Trigonometric Identities
17.
18.
sin cos sin cos 1
sin cos 25. cos (tan cot ) cos cos sin sin 2 cos 2 cos cos sin
sin cos 2 sin 2sin cos cos 2 1 sin cos sin 2 cos 2 2sin cos 1 sin cos 1 2sin cos 1 sin cos 2sin cos sin cos 2
1 cos cos sin 1 sin csc cos sin 26. sin (cot tan ) sin sin cos cos 2 sin 2 sin sin cos
tan 1 tan 1 sec2 tan tan 2 tan 1 sec2 tan 2 tan 1 2 tan sec2 tan 2 sec 2 tan sec2 tan 2 tan tan 2 2
19.
1 sin sin cos 1 cos sec 1 cos 2 u tan u 1 cos 2 u
27. tan u cot u cos 2 u tan u
3sin 2 4sin 1 3sin 1 sin 1 sin 2 2sin 1 sin 1 sin 1
sin 2 u 1 cos 2 u sin u 1 cos 2 u
28. sin u csc u cos 2 u sin u
3sin 1 sin 1
20.
cos 1 cos 1 cos 2 1 cos cos 1 cos 2 cos
sin 2 u
29. (sec 1)(sec 1) sec2 1 tan 2
cos 1 cos
30. (csc 1)(csc 1) csc2 1 cot 2
21. csc cos
1 cos cos cot sin sin
22. sec sin
1 sin sin tan cos cos
31. (sec tan )(sec tan ) sec2 tan 2 1 32. (csc cot )(csc cot ) csc2 cot 2 1 33. cos 2 (1 tan 2 ) cos 2 sec2 1 cos 2 cos 2 1
23. 1 tan 2 ( ) 1 ( tan ) 2 1 tan 2 sec2 24. 1 cot 2 ( ) 1 ( cot ) 2 1 cot 2 csc2
929
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Section 8: Analytic Trigonometry
34. (1 cos 2 )(1 cot 2 ) sin 2 csc2 1 sin 2 2 sin 1 35. (sin cos ) (sin cos ) 2
1 cos u sin u sin u 1 cos u 1 cos u sin u 1 cos u
40. csc u cot u
2
sin 2 2sin cos cos 2
sin 2 u sin u (1 cos u ) sin u 1 cos u
sin 2sin cos cos 2
2
2sin 2 cos 2 2
2(sin 2 cos 2 ) 2 1 2
41. 3sin 2 4 cos 2 3sin 2 3cos 2 cos 2 3(sin 2 cos 2 ) cos 2 3 1 cos 2
36. tan 2 cos 2 cot 2 sin 2
3 cos 2
sin 2 cos 2 cos 2 sin 2 2 cos sin 2 sin 2 cos 2 1
42. 9sec 2 5 tan 2 4sec2 5sec2 5 tan 2 4sec2 5(sec2 tan 2 ) 4sec2 5 1 5 4sec2
37. sec4 sec2 sec 2 (sec2 1) (tan 2 1) tan 2
43. 1
tan 4 tan 2
38. csc4 csc2 csc2 (csc2 1) (cot 2 1) cot 2 cot 4 cot 2 1 sin u cos u cos u 1 sin u 1 sin u cos u 1 sin u
39. sec u tan u
1 cos 2 u sin u (1 cos u )
44. 1
1 sin 2 u cos u (1 sin u )
cos 2 u cos u (1 sin u ) cos u 1 sin u
cos 2 1 sin 2 1 1 sin 1 sin (1 sin )(1 sin ) 1 1 sin 1 1 sin 1 1 sin sin sin 2 1 cos 2 1 1 cos 1 cos (1 cos )(1 cos ) 1 1 cos 1 (1 cos ) 1 1 cos cos
1 1 1 tan v cot v 45. 1 tan v 1 1 cot v 1 1 cot v cot v 1 1 cot v cot v cot v 1 cot v 1
930
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Section 8.4: Trigonometric Identities
1 1 csc v 1 sin v 46. 1 csc v 1 1 sin v 1 1 sin v sin v 1 1 sin v sin v 1 sin v 1 sin v
1 1 cos 1 sec 50. 1 cos 1 1 sec 1 sec sec 1 sec sec 1 sec 1 sec
1 sec sin cos sin 47. 1 csc cos cos sin sin sin cos cos tan tan 2 tan
51.
48.
1 sin v cos v (1 sin v) 2 cos 2 v cos v 1 sin v cos v(1 sin v)
1 2sin v sin 2 v cos 2 v cos v(1 sin v)
1 2sin v 1 cos v(1 sin v) 2 2sin v cos v(1 sin v) 2(1 sin v) cos v(1 sin v) 2 cos v 2sec v
csc 1 csc 1 csc 1 cot cot csc 1 csc 2 1 cot (csc 1) cot 2 cot (csc 1) cot csc 1
52.
cos v 1 sin v cos 2 v (1 sin v) 2 1 sin v cos v cos v(1 sin v)
1 1 1 sin csc 49. 1 sin 1 1 csc csc 1 csc csc 1 csc csc 1 csc csc csc 1 csc 1 csc 1
cos 2 v 1 2sin v sin 2 v cos v(1 sin v)
2 2sin v cos v(1 sin v) 2(1 sin v) cos v(1 sin v) 2 cos v 2sec v
1 sin sin sin 53. 1 sin cos sin cos sin 1 cos 1 sin 1 1 cot
931
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Section 8: Analytic Trigonometry
54. 1
sin 2 1 cos 2 1 1 cos 1 cos 1 cos 1 cos 1 1 cos 1 1 cos
57.
cos
55.
(sec tan ) 2 sec2 2sec tan tan 2 1 1 sin sin 2 2 cos cos cos 2 cos 2 1 2sin sin 2 cos 2 (1 sin )(1 sin ) 1 sin 2 (1 sin )(1 sin ) (1 sin )(1 sin ) 1 sin 1 sin
58.
56. (csc cot ) 2 csc 2 2 csc cot cot 2 1 1 cos cos 2 2 2 sin sin sin 2 sin 1 2 cos cos 2 sin 2 (1 cos )(1 cos ) 1 cos 2 (1 cos )(1 cos ) (1 cos )(1 cos ) 1 cos 1 cos
cos sin 1 tan 1 cot cos sin sin cos 1 1 cos sin cos sin cos sin sin cos cos sin 2 cos sin 2 cos sin sin cos cos 2 sin 2 cos sin (cos sin )(cos sin ) cos sin sin cos cot tan 1 tan 1 cot cos sin sin cos sin cos 1 1 cos sin cos sin sin cos cos sin sin cos cos sin 2 cos sin 2 sin (cos sin ) cos (sin cos )
cos 2 cos sin 2 sin sin cos (sin cos )
sin 3 cos3 sin cos (sin cos )
(sin cos )(sin 2 sin cos cos 2 ) sin cos (sin cos )
sin 2 sin cos cos 2 sin cos 2 sin cos cos 2 sin sin cos sin cos sin cos sin cos 1 cos sin 1 tan cot
932
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Section 8.4: Trigonometric Identities
59. tan
cos sin cos 1 sin cos 1 sin sin (1 sin ) cos 2 cos (1 sin )
62.
sin sin 2 cos 2 cos (1 sin ) sin 1 cos (1 sin ) 1 cos sec
sin 2 cos 2 2sin 1 sin 2 2sin cos cos 2 1 sin 2 (1 sin 2 ) 2sin 1 2sin cos 1 1 2sin 2 2sin 2sin cos 2sin (sin 1) 2sin cos sin 1 cos sin cos tan cot cos sin 63. tan cot sin cos cos sin sin 2 cos 2 sin cos sin 2 cos 2 cos sin sin 2 cos 2 1 2 sin cos 2
tan sec 1 tan sec 1 tan (sec 1) tan (sec 1) tan (sec 1) tan (sec 1)
sin 2 cos 2 sin cos sin cos 1 (sin cos ) 2 1
1 (sin cos ) sin cos cos 2 60. 2 2 cos sin (cos 2 sin 2 ) 1 cos 2 sin cos 2 sin 1 cos 2 tan 1 tan 2
61.
sin cos 1 sin cos 1 (sin cos ) 1 (sin cos ) 1 (sin cos ) 1 (sin cos ) 1
tan 2 2 tan (sec 1) sec2 2sec 1 tan 2 (sec2 2sec 1)
1 cos 2 sec cos cos cos 64. sec cos 1 cos 2 cos cos
sec 2 1 2 tan (sec 1) sec2 2sec 1 sec2 1 sec 2 2sec 1 2 2sec 2sec 2 tan (sec 1) 2sec 2 2sec (sec 1) 2 tan (sec 1) 2sec 2 2(sec 1)(sec tan ) 2(sec 1) tan sec
1 cos 2 cos 2 1 cos cos 1 cos 2 1 cos 2 sin 2 1 cos 2
933
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Section 8: Analytic Trigonometry
1 sec 68. cos 1 sec 1 1 cos 1 cos cos 1 cos 1 1 cos 1 cos 1 cos 1 cos 1 cos 2 1 cos sin 2
sin u cos u tan u cot u u sin u 1 cos 65. 1 sin u cos u tan u cot u cos u sin u sin 2 u cos 2 u u sin u 1 cos sin 2 u cos 2 u cos u sin u sin 2 u cos 2 u 1 1 sin 2 u cos 2 u 1 sin 2 u (1 cos 2 u ) sin 2 u sin 2 u 2sin 2 u
69.
sin u cos u tan u cot u 2 cos 2 u cos u sin u 2 cos 2 u 66. sin u cos u tan u cot u cos u sin u sin 2 u cos 2 u u sin u 2 cos 2 u cos sin 2 u cos 2 u cos u sin u sin 2 u cos 2 u 2 cos 2 u 1 sin 2 u cos 2 u 1
70.
1 sin sec tan cos cos 67. cot cos cos cos sin 1 sin cos cos cos sin sin 1 sin sin cos cos (1 sin ) sin 1 cos cos tan sec
1 tan 2 1 tan 2 1 tan 2 1 2 1 tan 1 tan 2 1 tan 2 1 tan 2 1 tan 2 1 tan 2 2 2 2 1 tan sec 2 1 2 2 sec 2 cos 2 1 cot 2 1 cot 2 2 cos 2 2 cos 2 2 2 1 cot csc 1 cot 2 2 cos 2 csc 2 csc 2 cos 2 2 sin 2 sin 2 cos 2 1 sin 2 2 sin cos 2 2 cos 2 sin 2 cos 2 1
71.
sec csc sec csc sec csc sec csc sec csc 1 1 csc sec sin cos
934
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Section 8.4: Trigonometric Identities
72.
sin 2 tan cos 2 cot sin sin 2 cos cos 2 cos sin 2 sin cos sin cos 2 cos sin cos sin 2 sin cos sin sin cos cos 2 sin cos sin (sin cos 1) sin cos cos (cos sin 1)
76.
1 2sin sin 2 (1 2sin sin 2 ) 1 sin 2 4sin cos 2 sin 1 4 cos cos 4 tan sec
77.
sin 2 cos 2 tan 2
1 cos cos 1 cos 2 cos sin 2 cos sin sin cos sin tan
73. sec cos
78.
sec sec 1 sin 1 sin 1 sin 1 sin sec (1 sin ) 1 sin 2 sec (1 sin ) cos 2 1 1 sin cos cos 2 1 sin cos3 1 sin (1 sin )(1 sin ) 1 sin (1 sin )(1 sin ) (1 sin ) 2 1 sin 2 (1 sin ) 2 cos 2
sin cos cos sin sin 2 cos 2 sin cos 1 sin cos 1 1 cos sin sec csc
74. tan cot
75.
1 sin 1 sin 1 sin 1 sin (1 sin ) 2 (1 sin ) 2 (1 sin )(1 sin )
1 sin cos
2
sin 1 cos cos (sec tan ) 2
1 1 1 sin 1 sin 1 sin 1 sin (1 sin )(1 sin ) 2 1 sin 2 2 cos 2 2sec 2
935
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2
Section 8: Analytic Trigonometry
79.
(sec v tan v) 2 1 csc v(sec v tan v)
82.
sec2 v 2sec v tan v tan 2 v 1 csc v(sec v tan v)
sec 2 v 2sec v tan v sec2 v csc v(sec v tan v)
2sec 2 v 2sec v tan v csc v(sec v tan v) 2sec v(sec v tan v) csc v (sec v tan v) 2sec v csc v 1 2 cos v 1 sin v 1 sin v 2 cos v 1 sin v 2 cos v 2 tan v
80.
81.
83.
84.
sec2 v tan 2 v tan v 1 tan v sec v sec v sin v 1 cos v 1 cos v cos v sin v cos v 1 cos v cos v sin v sin cos sin cos cos sin sin cos sin cos cos cos sin sin sin cos 11 cos sin 2 2 sin cos cos sin 1 cos sin sec csc
85.
sin cos cos sin sin cos sin cos cos sin sin sin cos cos cos sin 1 1 sin cos cos 2 sin 2 cos sin 1 cos sin sec csc sin 3 cos3 sin cos (sin cos )(sin 2 sin cos cos 2 ) sin cos sin 2 cos 2 sin cos 1 sin cos sin 3 cos3 1 2 cos 2 (sin cos )(sin 2 sin cos cos 2 ) 1 cos 2 cos 2 (sin cos )(sin 2 cos 2 sin cos ) sin 2 cos 2 (sin cos )(1 sin cos ) (sin cos )(sin cos ) 1 1 sin cos cos 1 sin cos cos 1 sin cos sin 1 cos sec sin tan 1
cos 2 sin 2 cos 2 sin 2 1 tan 2 sin 2 1 cos 2 2 cos sin 2 cos 2 sin 2 cos 2 cos 2 sin 2 cos 2
936
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cos 2 cos 2 sin 2
Section 8.4: Trigonometric Identities
86.
cos sin sin 3 cos sin sin 3 sin sin sin sin cot 1 sin 2
90.
cot cos 2
1 2cos cos 2 2sin (1 cos ) sin 2 1 2cos cos 2 sin 2 1 2cos cos 2 2sin (1 cos ) 1 cos 2 1 2cos cos 2 (1 cos 2 )
2
87.
2 cos 2 (sin 2 cos 2 ) (2 cos 2 1)2 cos 4 sin 4 (cos 2 sin 2 )(cos 2 sin 2 )
(cos 2 sin 2 ) 2 (cos 2 sin 2 )(cos 2 sin 2 )
2 2 cos 2sin (1 cos ) 2 cos 2 cos 2 2(1 cos ) 2sin (1 cos ) 2 cos (1 cos ) 2(1 cos )(1 sin ) 2 cos (1 cos ) 1 sin cos 1 sin cos cos sec tan
cos 2 sin 2 cos 2 sin 2 cos 2 sin 2
1 sin 2 sin 2 1 2sin 2
88.
89.
1 cos sin 1 cos sin (1 cos ) sin (1 cos ) sin (1 cos ) sin (1 cos ) sin
1 2 cos 2 1 cos 2 cos 2 sin cos sin cos 2 sin cos 2 sin cos sin 2 cos 2 sin cos sin cos sin cos cos sin tan cot
91. (a sin b cos ) 2 (a cos b sin ) 2 a 2 sin 2 2ab sin cos b 2 cos 2 a 2 cos 2 2ab sin cos b 2 sin 2 a 2 (sin 2 cos 2 ) b 2 (sin 2 cos 2 ) a 2 b2
1 sin cos 1 sin cos (1 sin ) cos (1 sin ) cos (1 sin ) cos (1 sin ) cos
92. (2a sin cos ) 2 a 2 (cos 2 sin 2 ) 2 4a 2 sin 2 cos 2
a 2 cos 4 2cos 2 sin 2 sin 4
1 2sin sin 2 2cos (1 sin ) cos 2 1 2sin sin 2 cos 2 1 2sin sin 2 2cos (1 sin ) (1 sin 2 ) 1 2sin sin 2 (1 sin 2 ) 2 2sin 2cos (1 sin ) 2sin 2sin 2 2(1 sin ) 2cos (1 sin ) 2sin (1 sin ) 2(1 sin )(1 cos ) 2sin (1 sin ) 1 cos sin
a cos 2cos sin sin a cos sin
a 4sin cos cos 2cos sin sin 2
2
2
4
2
2
a 2 1
2
a2
937
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2
4
2
2
2
2
2
4
2
4
Section 8: Analytic Trigonometry
93.
tan tan tan tan 1 1 cot cot tan tan tan tan tan tan tan tan
100. ln sec tan ln sec tan ln sec tan sec tan ln sec2 tan 2 ln tan 2 1 tan 2 ln 1
tan tan (tan tan ) tan tan
0
tan tan
101.
sin x
94. (tan tan )(1 cot cot ) (cot cot )(1 tan tan ) tan tan tan cot cot tan cot cot cot cot cot tan tan cot tan tan tan tan cot cot cot cot tan tan 0
sin x cos x
sin 2 x cos x 1 cos 2 x cos x 1 cos 2 x cos x cos x sec x cos x
g x
95. (sin cos ) 2 (cos sin )(cos sin ) sin 2 2sin cos cos 2 cos 2 sin 2
102.
2sin cos 2 cos 2
f x cos x cot x cos x
2 cos (sin cos )
cos x sin x
cos 2 x sin x 1 sin 2 x sin x 1 sin 2 x sin x sin x csc x sin x
96. (sin cos ) 2 (cos sin )(cos sin ) sin 2 2sin cos cos 2 cos 2 sin 2 2sin cos 2cos 2 2cos (sin cos )
97. ln sec ln
f x sin x tan x
1 1 ln cos ln cos cos
g x
sin 98. ln tan ln ln sin ln cos cos
99. ln 1 cos ln 1 cos ln 1 cos 1 cos ln 1 cos 2 ln sin 2 2 ln sin
938
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Section 8.4: Trigonometric Identities
103.
f
1 sin cos cos 1 sin 1 sin 1 sin cos 1 sin
cos cos 1 sin cos
1 sin 2 cos 2 cos 1 sin
cos 1 sin
0 cos 1 sin
1200 1 1 cos 2
f tan sec
1200 1 sin 2 cos
4 A2 cos 2 2 A cos
2
109. Let sin 1 ( x) . Then x sin . So, x sin sin( ) because the sine function is
odd. This means sin 1 x , and sin 1 x . So, sin 1 ( x) sin 1 x .
cos 1 sin g
1 1 110. Let tan 1 . Then tan . So x x 1 x cot . This means cot 1 x . So, tan 1 cot 1 x tan 1 x
16 16 tan 2 16 1 tan 2 4 sec 2 4sec 2
csc 1 sec tan
4 A2 1 sin 2
3
4 A2 1 sin 1 sin
cos 2 cos 1 sin
Since sec 0 for
cos
3
csc sec csc 1 sec tan 4 A2 csc sec 1 tan 4 A2 1 1 csc sec
sin 1 cos cos 1 sin cos 1 sin 1 sin cos 1 sin 1 sin 2 cos 1 sin
106.
1 2 cos 2 cos cos 2
108. I t 4 A2
105.
1200
g
1 2 cos 2 cos cos 2 cos 2
11 cos 1 sin
1 2 1 cos cos 2
1200
1 sin 2 cos 2
0
104.
107. 1200sec 2sec2 1 1200
2
.
111. Answers will vary.
9sec 2 9 9 sec 2 1
112. sin 2 cos 2 1 tan 2 1 sec2 1 cot 2 csc 2
3 tan 2 3 tan 3 Since tan 0 for . 2
939
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Chapter 8: Analytic Trigonometry 1 2 1 r (8) 2 (54) 2 2 180 48 2 m 30.159 m 2 5
113 – 114. Answers will vary.
120. A
115. Since a is negative then the graph opens up so the function has a maximum value. To find the maximum value we can find the vertex. x
121. tup tdown 6 hr
b 120 20 2a 2( 3)
dup
2
f (20) 3(20) 120(20) 50 1250
tup
8(r 1) 8(r 1) 6(r 2 1)
x 1 f ( x) 116. ; g ( x) 3 x 4 x2 (3x 4) 1 ( f g )( x) (3 x 4) 2 3x 3 3x 6 3( x 1) 3( x 2) x 1 x2
8r 8 8r 8 6r 2 6 6r 2 16r 6 0 3r 2 8r 3 0 (3r 1)(r 3) 0 r 3
So, if no current, d rt 3(6) 18 mi 122. If the angle is in quadrant III then x 8 and y 5. Solving for r we get:
117. For the point ( 12,5) , x 12 , y 5 ,
118.
f / 2 f 0
/20
64 25 r 2
13 csc 5 13 sec 12 12 cot 5
Since the secant function is negative in Quadrant 89 r . III, the answer is sec 8 x 123. x 2 y 2 12 x 4 y 31 0 x 2 12 x y 2 4 y 31
/2
0 1 2 /2
8 (4) 5 3 2
r 89
cos / 2 cos 0
The average rate of change is
119.
(8) 2 (5) 2 r 2
x 2 y 2 144 25 169 13
5 sin 13 12 cos 13 5 tan 12
d down 6 tdown
8 8 6 r 1 r 1
The vertex is (20,1250) so the maximum value of the function is 1250.
r
x 2 12 x 36 y 2 4 y 16 31 36 4
x 6 y 2 9 2
2
.
2
122 82 144 64 208 16(13) 4 13
940 Copyright © 2025 Pearson Education, Inc.
2
Section 8.5: Sum and Difference Formulas
124.
f g ( x)
12. d
x3 4 . x6
13. cos165º cos 120º 45º cos120º cos 45º sin120º sin 45º
The domain of f is x | x 4 . The domain of g is x | x x 6 . x3 4 0 x6 x3 4 x6 x 3 4 x 6
sin 60º cos 45º cos 60º sin 45º
x9 The domain of f g is x | 6 x 9
Section 8.5 2
2
32 42 9 16 25 5
2.
3 5
3. a. b.
96 3 3 93
1 1 2 2
12 6 3 6
1
5. congruent 2 2 1 ; ; 2 2 3 3
7. a.
b.
2 3
16. tan195º tan(135º 60º ) tan135º tan 60º 1 tan135º tan 60º 1 3 1 (1) 3
1 3 1 3 1 3 1 3
1 2 3 3 1 3
4 2 3 2
8. False 9. False 10. True
15. tan15º tan(45º 30º ) tan 45º tan 30º 1 tan 45º tan 30º 3 1 3 3 3 3 1 1 3 3 3 3 3 3 3 3 3
2 1 2 2 2 4
4. y 4, r 5, x 3 (Quadrant 2) 3 x cos r 5
6.
3 2 1 2 2 2 2 2 1 6 2 4
3 x 27
5 2 1 3
14. sin105º sin 60º 45º
x 3 4 x 24
1.
1 2 3 2 2 2 2 2 1 2 6 4
2 3
11. a 941 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
17. sin
5 3 2 sin 12 12 12 sin cos cos sin 4 6 4 6 2 3 2 1 2 2 2 2 1 6 2 4
18. sin
21. sin
3 2 sin 12 12 12 sin cos cos sin 4 6 4 6 2 3 2 1 2 2 2 2 1 6 2 4
17 15 2 sin 12 12 12 5 5 sin cos cos sin 4 6 4 6 2 3 2 1 2 2 2 2 1 6 2 4
22. tan
19 15 4 tan 12 12 12 5 tan tan 4 3 5 1 tan tan 4 3
7 4 3 19. cos cos 12 12 12 cos cos sin sin 3 4 3 4 1 2 3 2 2 2 2 2 1 2 6 4
20. tan
1 3 1 3 1 3 1 3
1 2 3 3 1 3
42 3 2
12
42 3 2
1
12
12
1 cos cos sin sin 4 3 4 3 1 2 1 2 3 2 2 2 2 1 2 6 4 4 2 6 2 6 2 6
1 3 1 3 1 3 1 3
1
23. sec 12 cos cos 3 4
1 1 3
1 2 3 3 1 3
1 1 3
1 3
1 3
2 3
7 3 4 tan 12 12 12 tan tan 4 3 1 tan tan 4 3
4 2 4 6 26 4 2 4 6 4 6 2
2 3
942 Copyright © 2025 Pearson Education, Inc.
Section 8.5: Sum and Difference Formulas
5
5
1
24. cot cot 12 tan 5 12
29.
12 1 3 2 tan 12 12 1 tan tan 4 6 1 tan tan 4 6 1 tan 4 tan 6 tan tan 4 6 1 1 1 3 3 1 3 1 3
tan 20º tan 25º tan 20º 25º 1 tan 20º tan 25º tan 45º 1
30.
tan 40º tan10º tan 40º 10º 1 tan 40º tan10º tan 30º
31. sin
3 3
7 7 7 cos cos sin sin 12 12 12 12 12 12 6 sin 12 sin 2 1
3 1 3 1 3 1 3 1
32. cos
3 3 3 1 3 1 42 3 2
5 7 5 7 5 7 cos sin sin cos 12 12 12 12 12 12 12 cos 12 cos 1
2 3
25. sin 20º cos10º cos 20º sin10º sin(20º 10º ) sin 30º 1 2
33. cos
5 5 5 cos sin sin cos 12 12 12 12 12 12 4 cos 12 cos 3 cos 3 1 2
26. sin 20º cos80º cos 20º sin 80º sin(20º 80º ) sin( 60º ) sin 60º
3 2
27. cos 70º cos 20º sin 70º sin 20º cos(70º 20º ) cos 90º 0 28. cos 40º cos10º sin 40º sin10º cos(40º 10º ) cos 30º
34. sin
5 5 5 cos cos sin sin 18 18 18 18 18 18 6 sin 18 sin 3
3 2
943 Copyright © 2025 Pearson Education, Inc.
3 2
Chapter 8: Analytic Trigonometry 3 35. sin , 0 5 2
sin( ) sin cos cos sin
c.
3 2 5 4 5 5 5 5 5
2 5 cos , 0 5 2 y
6 54 5 25
y
10 5 25
2 5, y
2 5 5
y
(x, 3) 5
2 5
x
3
5
x
x
x 2 32 52 , x 0
tan( )
d.
x 2 25 9 16, x 0
3 1 4 2 3 1 1 4 2 5 4 5 8 2
x4
4 cos , 5
tan
3 4
2 5 y 5 , y 0 2
2
2
y 2 25 20 5, y 0 y 5 sin
a.
5 5 1 , tan 5 2 2 5
5 , 0 5 2 4 sin , 0 5 2
36. cos
sin( ) sin cos cos sin 3 2 5 4 5 5 5 5 5
y
6 54 5 25
b.
y
5, y 5
2 5 25
8 5 3 5 25
11 5 25
x
x
5
(x, )
x
5
5 y 5 , y 0
4 2 5 3 5 5 5 5 5
y
cos( ) cos cos sin sin
tan tan 1 tan tan
2
2
2
y 2 25 5 20, y 0 y 20 2 5 sin
2 5 , 5
tan
2 5 5
2
x 2 ( 4) 2 52 , x 0 x 2 25 16 9, x 0 x3 3 cos , 5
944 Copyright © 2025 Pearson Education, Inc.
tan
4 4 3 3
Section 8.5: Sum and Difference Formulas
a.
sin( ) sin cos cos sin
2 5 3 5 4 5 5 5 5
b.
6 54 5 25
2 5 25
sin
4 , 5
3 3 5 5
y 3 sin
sin( ) sin cos cos sin
a.
3 , 2
tan
3 3 1
sin( ) sin cos cos sin 4 1 3 3 5 2 5 2
b.
43 3 10
cos( ) cos cos sin sin 3 1 4 3 5 2 5 2
2 5 5
4 2 3 4 1 2 3 10 3 5 3 2
cos
y 2 4 1 3, y 0
2 5 3 5 4 5 5 5 5
d.
1
12 y 2 22 , y 0
11 5 25
tan tan tan( ) 1 tan tan
x
r 5
3 5 8 5 25
6 54 5 25 10 5 25
y
r 2 (3) 2 42 25
5 3 2 5 4 5 5 5 5
c.
(1, y) 2
r
4
cos( ) cos cos sin sin
y
y
c.
3 4 3 10
sin( ) sin cos cos sin 4 1 3 3 5 2 5 2
4 37. tan , 3 2 1 cos , 0 2 2
945 Copyright © 2025 Pearson Education, Inc.
43 3 10
x
Chapter 8: Analytic Trigonometry tan tan 1 tan tan 4 3 3 4 1 3 3
cos( ) cos cos sin sin
b.
tan( )
d.
3 5 1 12 13 2 13 2
sin( ) sin cos cos sin
c.
43 3 3 3 4 3 3 43 3 3 4 3 3 4 3 3 4 3
3 12 1 5 13 2 13 2 tan( )
d.
48 25 3 39 25 3 48 39
5 3 , 12 2 1 3 sin , 2 2
r
540 507 3 180 1296 75 720 507 3 1221 240 169 3 407
x
y x
(x, )
5 3 , 13 2 tan 3, 2
39. sin
r 2 (12) 2 (5) 2 169 r 13 5 5 12 12 , cos 13 13 13 13 x 2 (1)2 22 , x 0
sin
3 tan 3 3
y
r
x
1
x
x 2 52 132 , x 0 x 2 169 25 144, x 0
sin( ) sin cos cos sin
x 12
3 12 1 5 13 2 13 2
3
x
x 3 3 cos , 2
5
x 4 1 3, x 0
1, 3
y
(x, )
2
a.
tan tan 1 tan tan
y x
5 3 12 26
5 3 54 3 12 3 12 5 3 36 5 3 1 12 3 36 15 12 3 36 5 3 36 5 3 36 5 3
38. tan
12 3 5 26
cos
5 3 12 12 5 3 26 26
946 Copyright © 2025 Pearson Education, Inc.
12 12 , 13 13
tan
5 12
Section 8.5: Sum and Difference Formulas 1 , 0 2 2 1 sin , 0 3 2
2
r 2 (1) 2 3 4
40. cos
r2 sin
a.
3 1 1 , cos 2 2 2
y
sin( ) sin cos cos sin
b.
5 12 3 5 12 3 or 26 26
c.
y
x
x
3
3 2
3 3 , tan 2 1
3
x 2 12 32 , x 0
5 1 12 3 13 2 13 2
d.
sin
x 2 9 1 8. x 0
sin( ) sin cos cos sin
3
(1, y)
y 2 4 1 3, y 0
12 1 5 3 13 2 13 2 12 5 3 26
(x, 1)
12 y 2 22 , y 0
cos( ) cos cos sin sin
x
y
2
5 1 12 3 13 2 13 2
y
5 12 3 26
x 82 2 cos
a.
tan tan 1 tan tan 5 3 12 5 1 3 12
2 2 , 3
2 4
3 2 2 1 1 2 3 2 3
1 2 2
sin( ) sin cos cos sin
tan( )
tan
5 12 3 12 12 5 3 12 5 12 3 12 5 3 12 5 3 12 5 3
b.
1 2 6 6
cos( ) cos cos sin sin 3 1 1 2 2 2 3 2 3
c.
32 2 6
sin( ) sin cos cos sin
240 169 3 69
3 2 2 1 1 2 3 2 3
947 Copyright © 2025 Pearson Education, Inc.
1 2 6 6
Chapter 8: Analytic Trigonometry
d.
tan( )
tan tan 1 tan tan
d.
2 3 4
42
1 3
4 3 2 4 4 6 4 4 3 2 4 6 4 6 4 6
tan tan 4 tan 4 1 tan tan 4 1 1 2 2 1 1 1 2 2 1 2 2 2 2 2 2 1 2 2 2 2 1 2 2 1 2 2 1 2 2 1
16 3 4 2 4 18 12 16 6
8 4 2 1 8 1 4 2 7
18 3 16 2 10 9 3 8 2 5
42. cos
1 41. sin , in quadrant II 3
a.
1 cos 1 sin 1 3
1 1 4
1 9
1
8 9
2 2 3
15 16
15 4
sin sin cos cos sin 6 6 6 1 3 2 2 1 3 2 3 2
c.
sin 1 cos 2
2
1
b.
a.
2
1 , in quadrant IV 4
b.
3 2 2 2 2 3 6 6
1 16
sin sin cos cos sin 6 6 6 15 3 1 1 4 2 4 2
cos cos cos sin sin 3 3 3 2 2 1 1 3 3 2 3 2
c.
2 2 3 6
2
1 3 5 8
cos cos cos sin sin 3 3 3 1 1 15 3 4 2 4 2
948 Copyright © 2025 Pearson Education, Inc.
1 3 5 8
Section 8.5: Sum and Difference Formulas
d.
tan tan 4 tan 4 1 tan tan 4 15 1 1 15 1
44. From the solution to Problem 43, we have 2 2 1 3 , sin , and sin , cos 3 2 2 1 cos . Thus, 3 g cos cos cos sin sin
1 15 1 15 1 15 1 15
1 2 15 15 1 15 16 2 15 14 8 15 7
3 1 1 2 2 3 2 3 2
43. lies in quadrant I . Since x 2 y 2 4 , r 4 2 . Now, ( x, 1) is on the circle, so x 2 12 4
45. From the solution to Problem 43, we have 2 2 1 3 sin , cos , sin , and 3 2 2 1 cos . Thus, 3 g cos cos cos sin sin
x 2 4 12
3 1 1 2 2 3 2 3 2
x 4 12 3 y 1 x 3 and cos . 2 r 2 r lies in quadrant IV . Since x 2 y 2 1 ,
Thus, sin
1 r 1 1 . Now, , y is on the circle, so 3 2
1 2 y 1 3 1 y2 1 3
2
sin cos cos sin
2
y 23 2 2 2 and r 1 3 x 1 1 cos 3 . Thus, r 1 3 f sin
1 1 3 2 2 3 2 3 2
Thus, sin
sin cos cos sin 1 1 3 2 2 3 2 3 2
3 2 2 32 2 6 6 6
46. From the solution to Problem 43, we have 2 2 1 3 sin , cos , sin , and 2 3 2 1 cos . Thus, 3 f sin
8 2 2 1 y 1 9 3 3
3 2 2 32 2 6 6 6
1 2 6 1 2 6 6 6 6
47. From the solution to Problem 43, we have 2 2 1 3 sin , cos , sin , and 3 2 2 1 cos . Thus, 3
1 2 6 1 2 6 6 6 6
949 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry 49. sin sin cos cos sin 2 2 2 1 cos 0 sin cos
1 sin 1 3 and tan 2 cos 3 3 3 2 2 2 sin 3 2 2 . Finally, tan 1 cos 3 tan tan h tan 1 tan tan
50. cos cos cos sin sin 2 2 2 0 cos 1 sin sin
51. sin sin cos cos sin
3 2 2 3 3 1 2 2 3 3 2 2 3 3 2 6 3 1 3 3 6 2 3 2 6 3 2 6 3 2 6
0 cos 1 sin sin
52. cos cos cos sin sin 1 cos 0 sin cos
53. sin sin cos cos sin 0 cos 1 sin
3 3 6 2 18 2 24 3
sin
9 6 6 6 6 24
54. cos cos cos sin sin
27 3 24 2 8 2 9 3 15 5
1 cos 0 sin cos
48. From the solution to Problem 47, we have 3 and tan 2 2 . Thus, tan 3 tan tan h tan 1 tan tan
tan tan 1 tan tan 0 tan 1 0 tan tan 1 tan
55. tan
3 2 2 3 3 1 2 2 3 3 2 2 3 3 2 6 3 1 3 3 6 2 3 2 6 3 2 6 3 2 6
tan 2 tan 1 tan 2 tan 0 tan 1 0 tan tan 1 tan
56. tan 2
3 3 6 2 18 2 24 3
3 3 3 57. sin sin cos cos sin 2 2 2 1 cos 0 sin cos
9 6 6 6 6 24 27 3 24 2 8 2 9 3 15 5
950
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Section 8.5: Sum and Difference Formulas 3 3 3 58. cos cos cos sin sin 2 2 2 0 cos (1) sin sin
65.
sin( ) sin cos cos sin sin( ) sin cos cos sin sin cos cos sin cos cos sin cos cos sin cos cos sin cos cos sin cos cos cos cos sin cos cos sin cos cos cos cos tan tan tan tan
59. sin( ) sin( ) sin cos cos sin sin cos cos sin 2sin cos 60. cos( ) cos( ) cos cos sin sin cos cos sin sin 2 cos cos 61.
62.
63.
64.
66.
cos( ) cos cos sin sin cos( ) cos cos sin sin
sin( ) sin cos cos sin sin cos sin cos sin cos cos sin sin cos sin cos 1 cot tan
cos cos sin sin cos cos cos cos sin sin cos cos cos cos sin sin cos cos cos cos cos cos sin sin cos cos cos cos 1 tan tan 1 tan tan
sin( ) sin cos cos sin cos cos cos cos sin cos cos sin cos cos cos cos tan tan cos( ) cos cos sin sin cos cos cos cos cos cos sin sin cos cos cos cos 1 tan tan
cos( ) sin( ) cos cos sin sin sin cos cos sin cos cos sin sin sin sin sin cos cos sin sin sin cos cos sin sin sin sin sin sin sin cos cos sin sin sin sin sin cot cot 1 cot cot
67. cot( )
cos( ) cos cos sin sin sin cos sin cos cos cos sin sin sin cos sin cos cot tan
951 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry 71. sin( )sin( ) sin cos cos sin sin cos cos sin
cos( ) sin( ) cos cos sin sin sin cos cos sin cos cos sin sin sin sin sin cos cos sin sin sin cos cos sin sin sin sin sin sin sin cos cos sin sin sin sin sin cot cot 1 cot cot
68. cot( )
69. sec( )
sin 2 cos 2 cos 2 sin 2 sin 2 (1 sin 2 ) (1 sin 2 )sin 2 sin 2 sin 2 sin 2 sin 2 sin 2 sin 2 sin 2 sin 2
72. cos( )cos( ) cos cos sin sin cos cos sin sin cos 2 cos 2 sin 2 sin 2 cos 2 (1 sin 2 ) (1 cos 2 )sin 2 cos 2 cos 2 sin 2 sin 2 cos 2 sin 2 cos 2 sin 2
73. sin( k ) sin cos k cos sin k
1
(sin )(1) k (cos )(0)
cos( )
(1) k sin , k any integer
1 cos cos sin sin 1 sin sin cos cos sin sin sin sin 1 1 sin sin cos cos sin sin sin sin sin sin csc csc cot cot 1
70. sec( )
74. cos( k ) cos cos k sin sin k (cos )(1) k (sin )(0) (1) k cos , k any integer 1 75. sin sin 1 cos 1 0 sin 2 6 2 2 sin 3
1
3 2
3 76. sin sin 1 cos 1 1 sin 0 2 3 sin 3 3 2
cos( )
1 cos cos sin sin 1 cos cos cos cos sin sin cos cos 1 1 cos cos cos cos sin sin cos cos cos cos sec sec 1 tan tan
3 4 77. sin sin 1 cos 1 5 5 3 4 Let sin 1 and cos 1 . is in 5 5
952
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Section 8.5: Sum and Difference Formulas
quadrant I; is in quadrant II. Then sin
3 , 5
4 0 , and cos , . 5 2 2 cos 1 sin 2 2
9 16 4 3 1 1 5 25 25 5 sin 1 cos 2
2
16 4 1 1 25 5
9 3 25 5
3 4 sin sin 1 cos 1 sin 5 5 sin cos cos sin 3 4 4 3 5 5 5 5 12 12 25 25 24 25
4 3 , 0 , and tan , 5 4 2 0 . 2
cos 1 sin 2 9 3 25 5
sec 1 tan 2
sin cos cos sin 4 4 3 3 5 5 5 5 16 9 25 25 25 25 1 4 5 79. cos tan 1 cos 1 3 13 4 5 Let tan 1 and cos 1 . is in 3 13
2
, and cos
5 , 0 . 13 2
sec 1 tan 2 2
16 4 1 1 9 3 cos
25 5 9 3
3 5
sin 1 cos 2 2
9 16 4 3 1 1 25 25 5 5 sin 1 cos 2 2
25 144 12 5 1 1 169 169 13 13
2
9 25 5 3 1 1 4 16 16 4
cos
9 3 25 5
3 4 sin sin 1 tan 1 5 4 sin
0
sin
2
2
16 4 1 1 25 5
quadrant I; is in quadrant I. Then tan
3 4 78. sin sin 1 tan 1 5 4 3 4 Let sin 1 and tan 1 . is in 4 5 quadrant IV; is in quadrant I. Then
16 4 1 1 5 25
sin 1 cos 2
4 5
953 Copyright © 2025 Pearson Education, Inc.
4 , 3
Chapter 8: Analytic Trigonometry
5 3 81. cos sin 1 tan 1 13 4 3 1 5 and tan 1 . is in Let sin 13 4
4 5 cos tan 1 cos 1 3 13 cos cos cos sin sin
quadrant I; is in quadrant I. Then sin
3 5 4 12 5 13 5 13 15 48 33 65 65 65
0
3 , and tan , 0 . 4 2 2
cos 1 sin 2
5 3 80. cos tan 1 sin 1 12 5 5 3 Let tan 1 and sin 1 . is in 12 5 quadrant I; is in quadrant IV. Then
2
25 144 12 5 1 1 13 169 169 13 sec 1 tan 2 2
9 25 5 3 1 1 16 16 4 4
3 5 tan , 0 , and sin , 5 12 2 0. 2
cos
sec 1 tan 2
4 5
sin 1 cos 2
2
25 169 13 5 1 1 12 144 144 12
cos
5 , 13
2
16 4 1 1 25 5
12 13
9 3 25 5
5 3 cos sin 1 tan 1 13 4 cos
sin 1 cos 2 2
cos cos sin sin 12 4 5 3 13 5 13 5 48 15 65 65 63 65
144 25 5 12 1 1 13 169 169 13 cos 1 sin 2 2
9 16 4 3 1 1 25 25 5 5 5 3 cos tan 1 sin 1 12 5 cos
4 12 82. cos tan 1 cos 1 3 13 4 12 Let tan 1 and cos 1 . is in 3 13
cos cos sin sin 12 4 5 3 48 15 33 13 5 13 5 65 65 65
quadrant I; is in quadrant I. Then tan 0
12 , and cos , 0 . 13 2 2
954
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4 , 3
Section 8.5: Sum and Difference Formulas
sec 1 tan 2 2
16 4 1 1 9 3 cos
25 5 9 3
3 tan sin 1 tan 5 6 1 3 tan sin 5 6 1 3 1 tan sin tan 5 6 3 3 4 3 3 3 1 4 3 9 3 12 12 3 3 12 9 3 12 3 3 12 3 3 12 3 3
3 5
sin 1 cos 2 2
9 16 4 3 1 1 25 25 5 5 sin 1 cos 2 2
144 25 5 12 1 1 169 169 13 13
108 75 3 36 144 27 144 75 3 117 48 25 3 39
4 12 cos tan 1 cos 1 3 13 cos cos cos sin sin 3 12 4 5 5 13 5 13 36 20 65 65 16 65 3 83. tan sin 1 5 6 3 Let sin 1 . is in quadrant I. Then 5 3 sin , 0 . 5 2 cos 1 sin 2 2
9 16 4 3 1 1 25 25 5 5 3 sin 5 3 5 3 = tan cos 4 5 4 4 5
3 84. tan cos 1 4 5 3 Let cos 1 . is in quadrant I. Then 5 3 cos , 0 . 5 2 sin 1 cos 2 2
9 16 4 3 1 1 5 25 25 5 4 sin 5 4 5 4 = tan cos 3 5 3 3 5
3 tan tan cos 1 3 4 5 tan cos 1 5 1 3 4 1 tan tan cos 4 5 4 1 1 1 3 1 3 3 4 7 3 7 7 1 1 3 3
955 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
4 85. tan sin 1 cos 1 1 5 1 4 and cos 11 ; is in Let sin 5 4 quadrant I. Then sin , 0 , and 5 2
4 sin cos 1 sin 1 1 5 1 4 tan cos sin 1 1 5 tan cos 1 4 sin 1 1 5 sin cos
cos 1 , 0 . So, cos 1 1 0 .
cos 1 sin 2 2
16 4 1 1 25 5
3 4 (0) (1) 5 5 4 3 (0) (1) 5 5 4 4 5 3 3 5
9 3 25 5
4 sin 5 4 5 4 = tan cos 3 5 3 3 5 1 4 tan sin cos 1 1 5 4 tan sin 1 tan cos 1 1 5 4 1 tan sin 1 tan cos 1 1 5 4 4 0 4 3 3 4 1 0 1 3 3
87. cos cos 1 u sin 1 v
sin 1 ,
2
2
sin 1 cos 2 1 u 2 cos 1 sin 2 1 v 2
. So, sin 1
1
2
2
cos cos 1 u sin 1 v cos( ) cos cos sin sin u 1 v2 v 1 u 2
88. sin sin 1 u cos 1 v
Let sin u and cos 1 v . Then 1
.
, and 2 2 cos v, 0 . 1 u 1 , 1 v 1
sin u ,
sin 1 cos 2 16 4 1 1 5 25
Let cos 1 u and sin 1 v . Then cos u, 0 , and sin v, 2 2 1 u 1 , 1 v 1
4 86. tan cos 1 sin 1 1 5 1 4 and sin -1 1 ; is in Let cos 5 4 quadrant I. Then cos , 0 , and 2 5
sin cos cos sin cos cos sin sin
9 3 25 5
cos 1 sin 2 1 u 2
sin 1 cos 2 1 v 2
3 sin 5 3 5 3 = , but tan is tan cos 4 5 4 4 2 5 undefined. Therefore, we cannot use the sum formula for tangent. Rewriting using sine and cosine, we obtain:
sin sin 1 u cos 1 v sin( ) sin cos cos sin uv 1 u 2 1 v 2
956
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Section 8.5: Sum and Difference Formulas
89. sin tan 1 u sin 1 v
sin 1 cos 2
Let tan 1 u and sin 1 v . Then
1
, and 2 2 sin v, . 2 2 u , 1 v 1 tan u,
sec tan 2 1 u 2 1
1
cos
u2 1
u2 11 u2 1 u2 u 1 u 2
u2 1
sec tan 2 1 v 2 1
cos 1 sin 2 1 v 2
1
cos
v 1 2
sin 1 cos 2
sin 1 cos 2
1 2 u 1
1
1
u2 11 u2 1 2
u 1 u
u
2
u2 1
sin tan 1 u sin 1 v
u 1
Let tan u and tan 1 v . Then tan u, , and 2 2 tan v, . 2 2 u , v sec tan 2 1 u 2 1
cos
1
u 1 2
v2 1
cos cos sin sin
2
1
v2 v 1 v 2
cos( )
u 1 v v
v2 1 1 v2 1
cos tan 1 u tan 1 v
2
90. cos tan 1 u tan 1 v
1 v 1 2
sin( ) sin cos cos sin u 1 1 v2 v 2 u 1 u2 1
1 u2 1
1
1
u 1 v 1 1 uv 2
2
u u 1 2
v v 1 2
u 2 1 v2 1
91. tan sin 1 u cos 1 v
Let sin u and cos 1 v . Then 1
, and 2 2 cos v, 0 . 1 u 1 , 1 v 1
sin u ,
cos 1 sin 2 1 u 2
tan
sin u cos 1 u2
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Chapter 8: Analytic Trigonometry
sec tan 1 u cos 1 v
sin 1 cos 2 1 v 2
tan
sec( )
sin 1 v2 cos v
tan sin 1 u cos 1 v tan( )
1 u2
1
u 1 u2
1 v2 v
uv 1 u 2 1 v 2 v 1 u2 v 1 u 2 u 1 v2 v 1 u
92. sec tan 1 u cos 1 v
uv 1 u
u2 1
2
1 v
2
v 1 u u 1 v
2
2
2
, and 2 2 cos v, 0 . u , 1 v 1 tan u,
sec tan 2 1 u 2 1
1 u2 1
sin 1 cos 2 1
u2 1 v u 1 v2
93. sin 3 cos 1 Divide each side by 2: 1 3 1 sin cos 2 2 2 Rewrite in the difference of two angles form 1 3 using cos , sin , and : 3 2 2 1 sin cos cos sin 2 1 sin( ) 2 5 or 6 6 5 3 6 3 6 7 2 6
Let tan 1 u and cos 1 v . Then
cos
cos( )
1 cos cos sin sin 1 u 1 v 1 v2 2 2 u 1 u 1 1 v u 1 v2 u2 1 u2 1 1 v u 1 v2
1 v2 v
1
tan tan 1 tan tan u
1 u 1 2
u2 11 u2 1
7 The solution set is , . 2 6
2
u u2 1 u u2 1
sin 1 cos 2 1 v 2
958
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Section 8.5: Sum and Difference Formulas
94.
3 sin cos 1 Divide each side by 2: 3 1 1 sin cos 2 2 2 Rewrite in the sum of two angles form using 1 3 , sin , and : cos 6 2 2 1 sin cos cos sin 2 1 sin( ) 2 5 or 6 6 5 or 6 6 6 6 2 0 or 3 2 The solution set is 0, . 3
95. sin cos 2 Divide each side by 2 : 1 1 sin cos 1 2 2 Rewrite in the sum of two angles form using 1 1 , sin , and : cos 4 2 2 sin cos cos sin 1 sin( ) 1 2 4 2 4 The solution set is . 4
96. sin cos 2
Divide each side by 2 : 1 1 sin cos 1 2 2 Rewrite in the sum of two angles form using 1 1 , sin , and : cos 4 2 2 sin cos sin cos 1 sin( ) 1 3 2 3 4 2 7 4
The solution set is 74 . 97.
tan 3 sec sin 1 3 cos cos sin 3 cos 1 sin 3 cos 1 Divide each side by 2: 1 3 1 sin cos 2 2 2 Rewrite in the difference of two angles form 1 3 using cos , sin , and : 3 2 2 1 sin cos cos sin 2 1 sin( ) 2 5 or 6 6 5 3 6 3 6 11 2 6 6 But since is not in the domain of the tangent 2 11 function then the solution set is . 6
959 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
98.
cot csc 3 cos 1 3 sin sin cos 1 3 sin
101.
3 sin cos 1 Divide each side by 2: 3 1 1 sin cos 2 2 2 Rewrite in the sum of two angles form using 1 3 , sin , and : cos 6 2 2 1 sin cos cos sin 2 1 sin( ) 2 7 11 or 6 6 7 11 or 6 6 6 6 5 or 3 But since is not in the domain of the cotangent function then the solution set is 5 . 3
99. sin sin 1 v cos 1 v
102.
f ( x h) f ( x ) h sin( x h) sin x h sin x cos h cos x sin h sin x h cos x sin h sin x sin x cos h h cos x sin h sin x 1 cos h h sin h 1 cos h cos x sin x h h f ( x h) f ( x ) h cos( x h) cos x h cos x cos h sin x sin h cos x h sin x sin h cos x cos h cos x h sin x sin h cos x 1 cos h h sin h 1 cos h sin x cos x h h
cos sin v sin cos v
sin sin 1 v cos cos 1 v
1
1
v v 1 v2 1 v2 v2 1 v2 1
100. cos sin 1 v cos 1 v
sin sin v sin cos v
cos sin 1 v cos cos 1 v
1
1
1 v2 v v 1 v2 0
960
Copyright © 2025 Pearson Education, Inc.
Section 8.5: Sum and Difference Formulas
103. a.
1
1
1
1
1
1
1 tan tan 1 tan 2 tan tan 3
tan tan 1 tan 2 tan 3 tan tan 1 tan 2 tan 3
tan tan 1 1 tan 1 2 tan tan 1 3 1
1
1
3 3 1 2 3 3 1 tan tan 1 tan tan 2 3 3 0 1 1 2 1 0 2 3 1 1 9 10 tan tan 1 tan tan 2 3 1 3 1 1 3 1 1 1 2 1 tan tan 1 tan tan 2 tan tan 1 1 tan tan 1 2 1
1
1
1
1
1
b. From the definition of the inverse tangent function we know 0 tan 1 1
, 0 tan 1 2
, and 2 2 3 3 0 tan 1 3 . Thus, 0 tan 1 1 tan 1 2 tan 1 3 . On the interval 0, , tan 0 if and only if 2 2 2
. Therefore, from part (a), tan 1 1 tan 1 2 tan 1 3 .
sin t sin t cos cos t sin
104. cos sin 2 t sin sin t cos t sin t cos sin t sin cos t sin t sin t
105. a.
3 2 1200 tan 12 12 12 3 2 tan tan tan tan 4 6 12 12 1200 1200 3 2 tan 1 tan 1 tan tan 12 12 4 6 3 1 3 1200 1 1 33
A 12(10) 2 tan
3 3
1200 33 3 1200 3
1200
3 3 3 3
3 3 3 3 3 3 3 3
12 6 3 1200 2 3 6 2400 1200 3 1200
961 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
b.
675 3 2 A 3(15) 2 cot 675cot 12 12 12 tan 3 2 12 12
675 675 675 675 3 3 3 3 675 3 1 3 3 3 3 3 3 tan tan 4 6 1 1 33 33 3 3 3 1 tan tan 4 6 3 3 3 3 12 6 3 675 675 675 2 3 6 3 3 3 3
1350 675 3 cm3
106. a.
A 3 52 cot a 2 75cot 12 12 75
3 2 75 3 2
150 75 3 cm 2
b. We will use one of the small triangles to compute radius (see figure).
tan 15
5
2
r 5 r tan 15 2 5 r 2 3 2
2r 2 3 5 r
c.
5 42 3
10 5 3 A r2 2
10 5 3 cm 2 2
10 5 3 10 5 3 2 2 100 50 3 50 3 75 4
175 100 3 cm 2 4
962
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Section 8.5: Sum and Difference Formulas
d.
175 100 3 4 600 300 3 175 100 3 cm 2 4
150 75 3
107. Note that 2 1 .
Then tan tan 2 1
tan 2 tan 1 m m1 2 1 tan 2 tan 1 1 m2 m1
108. Let tan 1 e v . Then tan e v , so 1 cot v ev . Because 0 , we know 2 e
that e v 0 , which means cot 1 ev cot 1 cot tan 1 e v . 109. Let sin 1 v and cos 1 v . Then sin v cos , and since sin cos , cos cos . If 2 2 v 0 , then 0 , so that and 2 2 both lie in the interval 0, . If v 0 , then 2 0 , so that and both lie in 2 2 the interval , . Either way, 2 cos cos implies , or 2 2 . Thus, sin 1 v cos 1 v . 2 2
110. Let tan 1 v and cot 1 v . Then tan v cot , and since tan cot , cot cot . If 2 2 v 0 , then 0 , so that and 2 2
both lie in the interval 0, . If v 0 , then 2 0 , so that and both lie in 2 2 the interval , . Either way, 2 cot cot implies , or 2 2 . Thus, tan 1 v cot 1 v . Note 2 2 1 that v 0 since cot 0 is undefined.
1 111. Let tan 1 and tan 1 v . Because v v 0, , 0 . Then tan
1 1 cot , and since v tan
tan cot , cot cot . 2 2 Because v 0 , 0 . So and 2 2 both lie in the interval 0, . Then 2 cot cot implies or 2 2 . Thus, 2 1 tan 1 tan 1 v, if v 0 . v 2
963 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry 112. sin( )sin( )sin( ) sin cos cos sin sin cos cos sin sin cos cos sin cos cos cos sin sin cos sin sin cos sin sin cos sin sin sin cos cos cos cos cos cos sin 3 sin sin sin sin sin s in sin sin sin
sin 3 sin cot cot sin cot cot sin cot cot sin 3 sin sin sin cot cot cot cot cot cot
cos cos cos cos cos cos sin 3 sin sin sin sin sin sin sin sin sin sin( ) sin( ) sin( ) sin 3 sin sin sin sin sin sin sin sin sin sin(180º ) sin(180º ) sin(180º ) sin 3 sin sin sin sin sin sin sin sin sin sin sin sin sin 3 sin sin sin sin sin sin sin sin sin sin 3
113. 2cot 2
115. If formula (7) is used, we obtain tan tan 2 tan . However, this is 2 1 tan tan 2 impossible because tan is undefined. Using 2 formulas (3a) and (3b), we obtain sin 2 . tan 2 cos 2 cos sin cot
1
tan
2 tan tan 1 tan tan 2 1 tan tan tan tan
2 1 x 1 x 1 x 1 x 1
2 1 x2 1
x 1 x 1 2x2 2 2 x
116. x 2 5 x 1 2 x 2 11x 4
114. The first step in the derivation,
tan tan 2 , is impossible tan 2 1 tan tan 2
because tan
3x 2 16 x 5 0 (3x 1)( x 5) 0 3x 1 0 or x 5 0 1 x x 5 3
is undefined. 2
964
Copyright © 2025 Pearson Education, Inc.
Section 8.5: Sum and Difference Formulas
For x
1 3
120.
1 2 x x2 4 1 x2 4 x 2 4 1 2 x 4x 4 2 1 4 1 2 x 2 3 4
f ( x)
2
1 1 y 5 1 3 3 1 5 5 1 9 3 9 For x 5
y 5 5 5 1 2
25 25 1 1 The intersection points are: 1 5 , , 5,1 3 9
117.
121.
The solution set is 6 .
radians
122.
log 7 x3 log 7 y 2 log 7 z 5
2
x3 y 2 log 7 5 z
123.
sec 5 Note that sec must be positive since lies in quadrant IV. Thus, sec 5 . 1 1 5 5 sec 5 5 5 sin tan , so cos 5 2 5 . sin tan cos 2 5 5 cos
log 7 x 3 y 2 log 7 z 5
sec2 2 1 4 1 5
1
3log 7 x 2 log 7 y 5log 7 z log 7 x3 log 7 y 2 log 7 z 5
119. tan 2 and 270 360 (quadrant IV) Using the Pythagorean Identities: sec2 tan 2 1
1 sin
8 x 4 42 x 9 3( x 4) 2(2 x 9) 3 x 12 4 x 18 x 6 x6
4 1 2 1 A r (6) 2 2 2 4 36 9 14.14 m 2 8 2
csc
23( x 4) 22(2 x 9)
17 180 510 6
118. 45
5
2 5 2 5 5 1 1 1 cot tan 2 2
2 x y 3x y 2 x y 3 x y 2
3
4
4
2
2
5
4
3
4
2
5
2
2
144 x18 y14
124.
3x 2 2 x 3 1 3x 2 1 2 x 3
3x 2 1 2 x 3
2
3x 2 1 2 2 x 3 2 x 3 x 2 2x 3
5 2
x 2 4(2 x 3) x 2 8 x 12 x 2 8 x 12 0 ( x 6)( x 2) 0 x 6, x 2
The solution set is 2, 6
965 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
125.
6x
6x
3
( x 3)
1
8( x 3) 4 4
( x 3)
1
4
8( x 3) ( x 3)
1
c.
4
sin
2
6 x 8( x 3)
1
( x 3) 4 6 x 8 x 24
d.
1
( x 3) 4 14 x 24 , x 3 1 ( x 3) 4
cos
2
2
2
4 5
1 5 1 1 10 10 2 10 10 10 10
1 cos 2 1 2
4 5
9 5 9 3 10 3 10 2 10 10 10 10
2 tan 1 tan 2 3 3 3 2 24 4 2 2 2 9 7 7 3 1 1 16 16 4
tan 2
f.
The angle is in QI so
1. sin 2 , 2 cos 2 , 2sin 2
1
e.
Section 8.6
2.
1 cos 2
1 45 1 cos tan 1 cos 1 54 2
3. sin 4. True
5. False, only the first one is equivalent.
1 5 9 5
1 1 9 3
3 10. cos , 0 . Thus, 0 , which 2 4 5 2
6. False, you cannot add the arguments or tan. 7. b
means
lies in quadrant I. 2 x 3, r 5
8. c
3 9. sin , 0 . Thus, 0 , which 2 4 5 2 means
32 y 2 52 , y 0 y 2 25 9 16, y 0
y4
lies in quadrant I. 2 y 3, r 5
So, sin
x 2 32 52 , x 0 x 2 25 9 16, x 0 x4
4 So, cos . 5
4 . 5
a.
4 3 24 sin(2 ) 2sin cos 2 5 5 25
b.
cos(2 ) cos 2 sin 2 2
a.
3 4 24 sin(2 ) 2sin cos 2 5 5 25
b.
cos(2 ) cos 2 sin 2 2
2
9 16 7 3 4 5 5 25 25 25
2
16 9 7 4 3 25 25 25 5 5
966
Copyright © 2025 Pearson Education, Inc.
Section 8.6: Double-angle and Half-angle Formulas
c.
sin
2
1 cos 2
1
d.
cos
2
2
2
2 5 1 1 5 5 2 5 5 5 5
1 2
3 5
8 5 4 2 52 5 2 5 5 5 5
1 cos 2
3 1 5 2
d.
cos
2
8 5 4 2 52 5 2 5 5 5 5
1 cos 2
3 1 5 2
2 tan tan 2 1 tan 2 4 8 8 2 24 3 3 3 2 16 7 7 4 1 1 9 9 3
The angle is in QI so
f.
sin
1 cos 2
e.
3 5
c.
2 1 1 5 5 5 2 5 5 5 5 2 tan 1 tan 2 4 8 8 2 24 3 3 3 2 16 7 7 4 1 1 9 9 3
e.
tan 2
f.
The angle is in QII so
1 53 1 cos tan 1 cos 1 53 2
2 5 8 5
1 1 4 2
4 3 3 11. tan , . Thus, , 2 2 4 3 2 which means
2 x 3, y 4
1 53 1 cos tan 1 cos 1 53 2
lies in quadrant II.
r 2 (3) 2 ( 4) 2 9 16 25
12. tan
r 5
4 3 sin , cos 5 5 a.
b.
cos(2 ) cos sin 2
2
2
2 x 2, y 1
4 3 24 2 5 5 25 2
lie in quadrant II.
r 2 ( 2) 2 (1) 2 4 1 5 r 5 5 2 2 5 , cos 5 5 5 5 sin(2 ) 2sin cos
sin
9 16 7 3 4 25 25 25 5 5
4 2
1 3 3 , . Thus, , 2 2 2 2 4
which means
sin(2 ) 2sin cos
8 5 2 5
a.
967 Copyright © 2025 Pearson Education, Inc.
1
5 2 5 4 2 5 5 5
Chapter 8: Analytic Trigonometry
b.
cos(2 ) cos 2 sin 2 2
2 5 5 5 5 20 5 15 3 25 25 25 5
c.
which means
2
2
sin
5 2 5 10
a.
y 3
3 3
sin(2 ) 2sin cos 3 6 2 3 3
b.
52 5 10
c.
52 5
d.
52 5
3 6 3 2 3 6 6
6 1 3 1 cos cos 2 2 2
5 2 5 5 2 5 5 2 5 5 2 5
2
6 1 3 1 cos sin 2 2 2
1 25 1 cos tan 1 cos 2 1 25
5 2 5
cos(2 ) cos 2 sin 2 2
2 tan 1 tan 2 1 2 1 1 4 2 2 1 3 3 1 1 1 4 4 2
2 18 6 2 2 2 9 9 3
6 3 3 3 6 3 3 1 9 9 9 3
e. tan 2
5 2 5
2
y2 9 6 3
2 5 1 1 cos 5 cos 2 2 2
The angle is in QII so
lies in quadrant I.
6 y 3
52 5 5 2
f.
2 x 6, r 3
5 2 5 5 2
6 , . Thus, , 3 2 4 2 2
13. cos
2 5 1 1 cos 5 sin 2 2 2
d.
2
25 20 5 20 45 20 5 25 20 5
94 5
968
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3 6 3 2 3 6 6
Section 8.6: Double-angle and Half-angle Formulas 2 tan 1 tan 2 2 2 2 2 2 2 2 2 1 1 2 1 1 2 2 2
tan 2
e.
1 36 1 cos tan 1 cos 2 1 36
3 6 3 3 6 3
3 6 3 6 9 6 6 6 96 3 6 3 6
15 6 6 5 2 6 3
which means
2 y 3, r 3
x 3
d.
3 6
3 2
2
3 6 6
2 tan 1 tan 2 2 2 2 2 2 2 2 2 1 1 2 1 1 2 2 2
e.
tan 2
f.
The angle is in QII so 1 1 cos tan 1 cos 2 1
6 cos 3 sin(2 ) 2sin cos 3 6 2 3 3 2 18 6 2 2 2 9 9 3 cos(2 ) cos 2 sin 2 6 3 3 3 6 3 3 1 9 9 9 3
6 3
1
3 6 3 2
lies in quadrant II.
2
3 6 6
1 cos cos 2 2
x 6
b.
3 6 3 2
3 6
x2 9 3 6
a.
2
3 3 3 , , 2 . Thus, 3 2 4 2
14. sin
2
1 cos sin 2 2
The angle is in QI so
f.
c.
6 3
1
2
969 Copyright © 2025 Pearson Education, Inc.
6 3
6 3
3 6 3 3 6 3
3 6 3 6 3 6 3 6
96 6 6 15 6 6 96 3
52 6
3 6 3 6
Chapter 8: Analytic Trigonometry
15. sec 3, sin 0 , so 0
16. csc 5, cos 0 , so
, which means lies in quadrant I. 2 4 2 1 cos , x 1 , r 3 . 3 0
. Thus, 2
3 , which means lies in quadrant II. 2 2 2 4 1 5 sin , r 5, y 1 5 5
12 y 2 32
x 2 (1) 2
y 9 1 8 2
2 2 3
cos
a.
2 2 1 4 2 sin(2 ) 2sin cos 2 3 3 9
b.
cos(2 ) cos 2 sin 2
a.
2 5
2
d.
cos
2
e.
b.
1 3
2
2
2 5 5 5 5 20 5 15 3 25 25 25 5
2 3 1 1 3 3 2 3 3 3 3
1 cos 2 1 2
tan 2
1 3
c. 4 3 2
2 3
2 3
6 3 3
2 tan 1 tan 2
4 2 4 2 1 8 7 1 2 2 2
d.
52 5 10
52 5 5 2
1 13 1 cos tan 1 cos 1 13 2
52 5 5 2
2 5 1 5 1 cos cos 2 2 2
The angle is in QI so
2 3 4 3
2
2 5 1 5 1 cos sin 2 2 2
3
2 2 2
f.
2 5 5
cos(2 ) cos 2 sin 2
1 cos 2 1
sin(2 ) 2sin cos
2
sin
2
5 2 5 4 2 5 5 5
2 1 8 7 1 2 2 9 9 9 3 3
c.
5
x2 5 1 4 x 2
y 82 2
sin
3 . Thus, 2
1 2 2 2
970
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52 5 10
Section 8.6: Double-angle and Half-angle Formulas
e.
2 tan 1 tan 2 1 2 1 1 4 2 2 1 3 3 1 1 1 4 4 2
tan 2
The angle is in QII so
f.
1 1 cos tan 1 cos 2 1 2
d.
2 5 1 5 1 cos cos 2 2 2
2 5
5
5 2 5
5 2 5
17. cot 2, sec 0 , so
52 5 52 5
tan 2
f.
The angle is in QI so
. Thus, 2
r 2 ( 2) 2 12 4 1 5 5 2 2 5 , cos sin 5 5 5 5 sin(2 ) 2sin cos 5 2 5 20 4 2 5 25 5 5
b.
cos(2 ) cos sin 2
2
2 5 5 5 5 20 5 15 3 25 25 25 5
c.
52 5 5 2 52 5 10
5 2 5
52 5 52 5
5 2 5 5 2 5 5 2 5 5 2 5
25 40 5 20 25 20
45 40 5 5
94 5
2
2 5 1 5 1 cos sin 2 2 2
5 2 5
1
2
1 25 1 cos tan 1 cos 2 1 25
r 5
a.
52 5 10
2 tan 1 tan 2 1 2 1 1 4 2 2 1 3 3 1 1 1 4 4 2
e.
94 5
lies in quadrant I. , which means 4 2 2 2 x 2, y 1
52 5 5 2
18. sec 2, csc 0 , so
3 2 . Thus, 2
3 , which means lies in quadrant II. 2 4 2 1 cos , x 1, r 2 2 2 2 1 y 22 y2 4 1 3 y 3 sin
971 Copyright © 2025 Pearson Education, Inc.
3 2
Chapter 8: Analytic Trigonometry
a.
sin(2 ) 2sin cos
a.
sin 2 2sin cos
3 1 3 2 2 2 2
b.
3 10 10 2 10 10 6 3 10 5
cos(2 ) cos 2 sin 2 2
2 3 1 3 1 1 4 4 2 2 2
c.
cos
2
1 2
tan 2
f.
2
1 2
1 2 1 1 2 4 2
1 cos 2
c.
1 3 2 2 3 3 2 2 4 2
1 cos sin 2 2
2 3 3 1 3 1 3 2 3
2
The angle is in QII so
d.
1 12 1 cos tan 1 cos 1 12 2
19. tan 3, sin 0 , so
10
1 10 10 2 5
1 cos cos 2 2
3 2 . Thus, 2
e.
tan 2
r 10
10 10 20
1 3 3 3
r 2 12 (3) 2 1 9 10 3
1
10 10 2
10 10 10 2
3 , which means lies in quadrant II. 4 2 2 x 1, y 3
sin
10 10 2
10 10 10 2
2 tan 1 tan 2
1
1
1 2 3 2
2
10 3 10 10 10 10 90 80 4 100 100 100 5
1 cos sin 2 2
e.
cos(2 ) cos 2 sin 2
d.
b.
3 10 1 10 , cos 10 10 10
972
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2 tan 1 tan 2 2 3 1 3
2
10 10 20
1 10 10 2 5
6 3 6 1 9 8 4
Section 8.6: Double-angle and Half-angle Formulas
The angle is in QII so
f.
1 1 cos tan 1 cos 2 1 10 1
10 1
1 10
c.
1 10
3 10 1 1 cos 10 sin 2 2 2 10 3 10 10 2
10 10 10 10
10 10 10 10 10 10 10 10
100 20 10 10 110 20 10 100 10 90
11 2 10 3
d.
20. cot 3, cos 0 , so
r 10
cos
3
a.
10
10
10 , 10
3 10 10
b.
cos(2 ) cos 2 sin 2 2
3 10 10 10 10 90 10 80 4 100 100 100 5
2
1 10 3 10 2 5
10 3 10 20
1 10 3 10 2 5
2 tan 1 tan 2 1 2 2 3 3 3 2 8 4 1 1 9 3
e.
tan 2
f.
The angle is in QII so
sin 2 2sin cos 10 3 10 6 3 2 10 10 10 5
10 3 10 10 2
3 . Thus, 2
r 2 (3) 2 (1) 2 9 1 10 1
10 3 10 20
3 10 1 1 cos 10 cos 2 2 2
3 which means is in quadrant II. 2 2 4 2 x 3, y 1
sin
3 1 10 1 cos tan 1 cos 2 1 3
10 3 10 3
10
10 3 10 10 3 10
10 3 10 10 3 10 10 3 10 10 3 10
100 60 10 90 190 60 10 100 90 10
19 6 10
973 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry 45 21. sin 22.5 sin 2
9 9 1 cos 4 9 4 24. tan tan 9 2 8 1 cos 4
1 cos 45 2 1 2
2 2
2 2 4
2 2 2 2 2 1 2 1
2 2 2
45 22. cos 22.5 cos 2
2 2 2 2 2 2 2 2
1 cos 45 2 1 2
2 2
2 2 2
2 2 4
2 2 2
2
1 2 330 25. cos165 cos 2
2 2 2 2 2 1 2 1
2 2 2 2 2 2 2 2
2 2
2 2
2 1
7 7 1 cos 4 7 4 tan 23. tan 7 8 2 1 cos 4
1 cos 330 2 1
3 2 2 3 2 3 2 4 2
1 cos 390 390 26. sin195 sin 2 2
2
2
2 2 2
2
2 1
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3 2
2
2 3 4
2 3 2
1 2
974
1
Section 8.6: Double-angle and Half-angle Formulas
27. sec
15 1 1 15 15 8 cos 4 8 cos 2 1 15 1 cos 4 2 1 2 1 2 2 1 2 2 4 2 2 2
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 2 2
2 2
28. csc
1 7 1 7 8 sin 7 8 sin 4 2 1 7 1 cos 4 2 1 2 1 2 2 1 2 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 2
2 2 2
2 2
2 2
2 2 2 2
2
2 2
2
2 2
29. sin sin 4 2 8 1 cos 4 2
1
2 2 2 2 2 2 2 4 2
3 3 30. cos cos 4 2 8 3 1 cos 4 2 2 1 2 2
975 Copyright © 2025 Pearson Education, Inc.
2 2 4
2 2 2
Chapter 8: Analytic Trigonometry
31. lies in quadrant II. Since x 2 y 2 5 , r 5 . Now, the point (a, 2) is on the circle, so
34. Note: Since lies in quadrant II,
a 2 22 5
quadrant I. Therefore, sin
a 52 2
2
f 2 sin 2 2sin cos
2 5 5 20 4 2 25 5 5 5
32. From the solution to Problem 31, we have 2 5 5 sin and cos . 5 5
Thus, g 2 cos 2 cos 2 sin 2
5 . 5
5 5 5 2 5 5 10
10 5 5
10
2
5 2 5 5 5 5 20 15 3 25 25 25 5
33. Note: Since lies in quadrant II,
quadrant I. Therefore, cos
2
2
35. lies in quadrant II. Since x 2 y 2 5 , r 5 . Now, the point (a, 2) is on the circle, so a 2 22 5 a 2 5 22 a 5 2 2 1 1 (a is negative because lies in quadrant II.) b 2 2 . Thus, tan a 1 h 2 tan 2
must lie in
is positive. From the
solution to Problem 31, we have cos
5 . 5
1 cos Thus, g cos 2 2 2
5 1 5 2
is positive. From the
5 1 5 2
a 1 5 cos . Thus, r 5 5
must lie in
1 cos Thus, f sin 2 2 2
b 2 2 5 and Thus, sin 5 r 5
2
2
solution to Problem 31, we have cos
a 5 2 2 1 1 (a is negative because lies in quadrant II.)
2
2 tan 1 tan 2 2 2 1 2
5 5 5 2 5 5 10
10 5 5
10
976
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2
4 4 4 1 4 3 3
Section 8.6: Double-angle and Half-angle Formulas 36. From the solution to Problem 31, we have 2 5 5 sin and cos . Thus, 5 5 5 1 1 cos 5 h tan 2 sin 2 5 2 5 5 5 5 2 5 5 5 5 2 5
5 5 2 5
5 5
5 5 5 10
5 1 1 5 2 2
38. From the solution to Problem 37, we have 1 15 and cos . Thus, sin 4 4 f 2 sin 2 2sin cos 15 1 15 2 8 4 4
39. Note: Since lies in quadrant III,
quadrant II. Therefore, sin
1 b2 1 4
2
2 1 15 4 4 1 15 14 7 16 16 16 8
is positive. From 1 . 4
1 cos 2
1 1 4 2
5 4 5 5 2 10 10 2 8 8 2 16 4
40. Note: Since lies in quadrant III,
quadrant II. Therefore, cos 2
must lie in
Thus, f sin 2 2
2
15 15 1 b 1 16 4 4 (b is negative because lies in quadrant III.) 1 a 1 Thus, cos 4 and r 1 4 15 b 4 15 . Thus, sin r 1 4 2 g 2 cos 2 cos sin 2
2
2
the solution to Problem 37, we have cos
37. lies in quadrant III. Since x 2 y 2 1 , 1 r 1 1 . Now, the point , b is on the 4 circle, so 2 1 2 b 1 4
2
2
must lie in
is negative. From
the solution to Problem 37, we have cos
1 . 4
Thus, g cos 2 2
1 cos 2
1 1 4 2 3 3 3 2 6 6 4 2 8 8 2 16 4
977 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
41. From the solution to Problem 37, we have 1 15 and cos . Thus, sin 4 4 1 cos h tan 2 sin 2
43. sin 4 sin 2
5
15
2
44. sin 4 sin 2 2
15
2sin 2 cos 2
15
5 15 15
2(2sin cos ) 1 2sin 2
4sin cos 1 2sin 2
15 3
cos 4sin 8sin 3
1 cos(2 ) 1 cos(2 ) 2 2 1 1 cos 2 (2 ) 4 1 1 cos(4 ) 1 4 2 1 1 1 cos(4 ) 4 2 2 1 1 cos(4 ) 8 8
45. sin 2 cos 2
2
2
15 15 1 b 1 16 4 4 (b is negative because lies in quadrant III.) 15 b 4 15 . Thus, tan 1 a 4 h 2 tan 2
1 r 1 1 . Now, the point , b is on the 4 circle, so 2 1 2 b 1 4
cos 4sin 1 2sin 2
42. lies in quadrant III. Since x 2 y 2 1 ,
1 b2 1 4
2
1 cos 2 2 1 1 2 cos 2 cos 2 2 4 1 1 1 cos 2 cos 2 2 4 2 4 1 1 1 1 cos 4 cos 2 4 2 4 2 1 1 1 1 cos 2 cos 4 4 2 8 8 3 1 1 cos 2 cos 4 8 2 8
1 1 4 15 4 5 4 15 4 5 15
2 tan 1 tan 2
15 2 15 2 15 15 14 1 15 7 1 15 2
2
978
Copyright © 2025 Pearson Education, Inc.
Section 8.6: Double-angle and Half-angle Formulas
46. sin 4 cos 4 sin 4 cos 4
49. We use the result of problem 44 to help solve this problem: sin 5 sin(4 )
2
2
1 1 cos(4 ) 8 8 1 2 1 cos(4 ) 64 1 1 2 cos(4 ) cos 2 (4 ) 64 1 cos 8 1 1 2 cos(4 ) 64 2 1 1 2 4 cos(4 ) 1 cos 8 64 2 1 3 4 cos(4 ) cos 8 128 3 1 1 cos 8 cos(4 ) 128 32 128
sin 4 cos cos 4 sin
2
sin 1 8sin 2 cos 2
5
sin 8sin 3 1 sin 2
5sin 12sin 8sin 8sin 8sin 3
5
3
5
16sin 5 20sin 3 5sin
50. We use the results from problems 44 and 46 to help solve this problem: cos(5 ) cos(4 ) cos 4 cos sin 4 sin
8cos 4 8cos 2 1 cos
2 cos cos 2 cos 2 cos 3
cos 4sin 8sin 3 sin 8cos 8cos cos 5
3
4 cos sin 2 8cos sin 4
48. cos 4 cos 2 2
8cos5 8cos3 cos 4 cos (1 cos 2 ) 8cos (1 cos 2 ) 2
2 cos 2 (2 ) 1
2 4 cos 4 cos 1 1 2
2 2 cos 2 1 1 4
2
8cos 4 8cos 2 2 1 8cos 8cos 1 4
4sin 12sin 3 8sin 5
2 cos cos 2 1 cos cos 4 cos3 3cos
2 cos cos 2sin cos
4sin 12sin 8sin
2
3
3
3
2
2
sin 1 2 2sin cos
2 cos 2 1 cos 2sin cos sin 3
3
1 sin 4sin 8sin
cos 2 cos sin 2 sin 3
cos 4sin 8sin 1 2sin 2 2 sin 2
47. cos(3 ) cos(2 )
cos 4sin 8sin 3 cos cos 2(2 ) sin
2
8cos5 8cos3 cos 4 cos 4 cos3 8cos (1 2 cos 2 cos 4 ) 8cos5 4 cos3 3cos 8cos 16 cos3 8cos5 16 cos5 20 cos3 5cos
51. cos 4 sin 4 cos 2 sin 2 1 cos 2 cos 2
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cos sin 2
2
Chapter 8: Analytic Trigonometry cos sin cot tan sin cos 52. cot tan cos sin sin cos cos 2 sin 2 cos sin cos 2 sin 2 sin cos cos 2 sin 2 sin cos sin cos cos 2 sin 2 2 2 cos sin 1 cos 2
53. cot(2 )
54. cot(2 )
1 1 2 tan tan(2 ) 1 tan 2 1 tan 2 2 tan 1 1 2 cot 2 cot cot 2 1 2 cot 2 cot cot 2 1 cot 2 cot 2 2 cot 1 2 cot
1 1 cos(2 ) 2 cos 2 1 1 2 1 sec 2 1 2 sec2 sec2 sec2 2 sec2
56. csc 2
1 1 sin 2 2sin cos 1 1 1 2 cos sin 1 sec csc 2
57. cos 2 (2u ) sin 2 (2u ) cos 2(2u ) cos(4u ) 58. (4sin u cos u )(1 2sin 2 u ) 2(2sin u cos u )(1 2sin 2 u ) 2sin 2u cos 2u sin 2 2u sin 4u
59.
1 1 2 tan tan(2 ) 1 tan 2 1 tan 2 2 tan 1 1 tan 2 2 tan tan
55. sec(2 )
1 cot tan 2
cos(2 ) cos 2 sin 2 1 sin(2 ) 1 2sin cos (cos sin )(cos sin ) cos 2 sin 2 2sin cos (cos sin )(cos sin ) (cos sin )(cos sin ) cos sin cos sin cos sin sin cos sin sin cos sin sin sin cos sin sin sin cot 1 cot 1
980
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Section 8.6: Double-angle and Half-angle Formulas
1 4sin 2 cos 2 4 1 2 2sin cos 4 2 1 sin 2 4 1 2 sin 2 4
60. sin 2 cos 2
64. tan
v 1 cos v 1 cos v csc v cot v 2 sin v sin v sin v
1 cos 1 cos 65. 1 cos 1 tan 2 1 2 1 cos 1 cos (1 cos ) 1 cos 1 cos 1 cos 1 cos 2 cos 1 cos 2 1 cos 2 cos 1 cos 1 cos 2 cos 1 tan 2
1 1 2 61. sec2 1 cos 1 cos 2 cos 2 2 2 1 1 2 62. csc2 2 sin 2 1 cos 1 cos 2 2 1 1 v 63. cot 2 2 tan 2 v 1 cos v 1 cos v 2 1 cos v 1 cos v 1 1 sec v 1 1 sec v sec v 1 sec v sec v 1 sec v sec v 1 sec v sec v sec v 1 sec v 1 sec v 1
66.
2
1
sin 3 cos3 sin cos
sin cos sin 2 sin cos cos 2
sin cos sin sin cos cos 2 1 sin 2 cos 2 2sin cos 2 1 1 sin 2 2 2
67.
sin(3 ) cos(3 ) sin 3 cos cos 3 sin sin cos sin cos sin(3 ) sin cos sin 2 sin cos 2sin cos sin cos 2
981 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry cos sin cos sin cos sin cos sin cos sin cos sin cos sin cos sin 2
68.
2
cos 2 2 cos sin sin 2 cos 2 2 cos sin sin 2
cos sin cos 2 2 cos sin sin 2 cos 2 2 cos sin sin 2 cos 2 sin 2 4 cos sin cos 2
2
2
2(2sin cos ) cos 2 2sin 2 cos 2
2 tan 2
69. tan 3 tan(2 ) 2 tan 2 tan tan tan 3 tan 2 tan 2 tan 3 tan tan 3 1 tan 2 3 tan tan 3 1 tan 2 1 tan 2 2 2 2 1 tan 2 tan 1 2 tan tan 1 tan 2 tan 1 tan 1 3 tan 1 3 tan 2 2 1 tan 1 tan 2
70. tan tan( 120º ) tan( 240º ) tan tan120º tan tan 240º tan 1 tan tan120º 1 tan tan 240º tan 3 tan 3 tan 1 tan 3 1 tan 3
tan
tan 3 1 3 tan
tan 3 1 3 tan
tan 1 3 tan tan 3 1 3 tan tan 3 1 3 tan 2
1 3 tan tan 3 tan 3 tan 3 tan 2 3 3 tan tan 3 tan 2 3 3 tan 1 3 tan 2 3 tan 3 9 tan 1 3 tan 2
2
3 3 tan tan 3
1 3 tan 3 tan 3 (from Problem 69) 2
982
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Section 8.6: Double-angle and Half-angle Formulas
71.
1 ln 1 cos 2 ln 2 2 1 1 cos 2 ln 2 2
2 cos 2 1 cos 2 cos 2 cos 1 0 (2 cos 1)(cos 1) 0
1 cos 2 1/ 2 ln 2
ln sin 2
2 cos 1 0
1 ln 1 cos 2 ln 2 2 1 1 cos 2 ln 2 2
76.
sin(2 ) cos 2sin cos cos 2sin cos cos 0 (cos )(2sin 1) 0 cos 0 or 2sin 1 cos 0 1 sin 2 3 , 5 2 2 , 6 6 5 3 The solution set is , , , . 2 6 2 6
77.
sin(2 ) sin(4 ) 0 sin(2 ) 2sin(2 ) cos(2 ) 0 sin(2 ) 1 2 cos(2 ) 0
1 cos 2 1/ 2 ln 2
ln cos 2
1/ 2
ln cos
73.
cos 2 6sin 2 4 1 2sin 2 6sin 2 4 4sin 2 3 3 sin 2 4 3 2 2 4 5 , , , 3 3 3 3 2 4 5 , , The solution set is , . 3 3 3 3 sin
74.
cos 1 0 cos 1 0
1 2 2 4 , 3 3 2 4 , The solution set is 0, . 3 3
ln sin
72.
or
cos
1/ 2
cos(2 ) cos
75.
cos 2 2 2sin 2 1 2sin 2 2 2sin 2 1 2 (not possible) The equation has no real solution.
sin(2 ) 0
1 2 cos(2 ) 0 1 cos(2 ) 2 2 0 2k or 2 2k or k k 2 2 4 2 2k or 2 2k 3 3 2 k k 3 3 On the interval 0 2 , the solution set is 4 3 5 2 , , , , 0, , , . 3 2 3 3 2 3
983 Copyright © 2025 Pearson Education, Inc.
or
Chapter 8: Analytic Trigonometry 78.
sin(2 ) 2sin 0 cos(2 ) sin 2 2sin cos 2 0 cos 2 2sin cos 2sin (2 cos 2 1) 0
2 cos 1 2 cos (2 ) 1 0 2
2
2 cos 2 1 2 cos(2 ) cos(2 ) 1 0
2 cos 2 2 2 cos 2 ( ) 1 2 cos 2 ( ) 1 2 0
2 cos 1 2 4 cos 4 cos 1 1 0 2
4
2sin 2 cos cos 1 0
2
2sin cos 2 cos 2 1 0
2 cos 1 8cos 8cos 2 1 0 2
4
2
2
8cos 6 cos 0 4
2
2sin (2 cos 1)(cos 1) 0 2 cos 1 0 or 2sin 0 1 sin 0 cos 2 0, 5 , 3 3
4 cos 4 3cos 2 0
cos 2 4 cos 2 3 0 cos ( ) 0 or 4 cos 3 0 3 cos 0 or cos 2 4 3 cos 2 3 5 7 11 or , , , , 2 2 6 6 6 6 2
tan(2 ) 2sin 0
81.
cos(2 ) cos(4 ) 0
2
cos 1 0 cos 1 5 The solution set is 0, , , . 3 3
On the interval 0 2 , the solution set is 5 7 3 11 , , , , , . 6 2 6 6 2 6
tan(2 ) 2 cos 0
82.
sin(2 ) 2 cos 0 cos(2 )
3 sin cos(2 )
79.
or
sin 2 2 cos cos 2
3 sin 1 2sin 2
cos 2
2sin sin 2 0 This equation is quadratic in sin . The discriminant is b 2` 4ac 1 16 15 0 . The equation has no real solutions. 2
0
2sin cos 2 cos (1 2sin 2 ) 0
2 cos 2sin sin 1 0 2 cos sin 1 2sin 2 0 2
80.
cos(2 ) 5cos 3 0
2 cos (2sin 1)(sin 1) 0
2 cos 2 1 5cos 3 0
2 cos 0 or 2sin 1 0 or cos 0 1 sin 2 3 , 7 11 2 2 , 6 6 sin 1 0 sin 1 2 7 3 11 The solution set is , , , . 2 6 2 6
2 cos 2 5cos 2 0 (2 cos 1)(cos 2) 0 2 cos 1 or cos 2 1 (not possible) cos 2 2 4 , 3 3 2 4 The solution set is , . 3 3
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Section 8.6: Double-angle and Half-angle Formulas
1 3 83. sin 2sin 1 sin 2 sin 2 3 2 6 3 2 3 84. sin 2sin 1 sin 2 sin 2 3 2 3 3 3 85. cos 2sin 1 1 2sin 2 sin 1 5 5 3 1 2 5 18 1 25 7 25
2
4 4 86. cos 2 cos 1 2 cos 2 cos 1 1 5 5 2
4 2 1 5 32 1 25 7 25 3 87. tan 2 cos 1 5 3 Let cos 1 . lies in quadrant II. 5 3 . Then cos , 5 2 5 sec 3 tan sec2 1 2
25 16 4 5 1 1 9 9 3 3
2 tan 3 tan 2 cos 1 tan 2 5 1 tan 2 4 2 3 2 4 1 3 8 3 9 16 9 1 9 24 9 16 24 7 24 7 3 2 tan tan 1 4 1 3 88. tan 2 tan 3 4 1 tan 2 tan 1 4 3 2 4 2 3 1 4 3 16 2 9 16 1 16 24 16 9 24 7 4 89. sin 2 cos 1 5 4 . is in quadrant I. Let cos 1 5 4 Then cos , 0 . 5 2
985 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
sin 1 cos 2 2
16 4 1 1 25 5
cos 1 sin 2 2
9 16 4 3 1 1 25 25 5 5
9 3 25 5
3 1 1 cos 2 sin 1 cos 2 5 2 2
4 sin 2 cos 1 sin 2 5 3 4 24 2sin cos 2 5 5 25
1 cos 2
4 90. cos 2 tan 1 3 4 Let tan 1 . is in quadrant IV. 3 4 Then tan , 0 . 3 2
3 93. sec 2 tan 1 4 3 Let tan 1 . is in quadrant I. 4 3 Then tan , 0 . 4 2
sec tan 2 1 2
16 4 1 1 9 3
cos
sec tan 2 1
25 5 9 3
2
9 25 5 3 1 1 4 16 16 4
3 5
cos
4 cos 2 tan 1 cos 2 2 cos 2 1 3 2 3 2 1 5 18 1 25 7 25
3 1 91. sin 2 cos 1 5 2
4 9 55 9 2 2 10
1
4 5
3 1 sec 2 tan 1 sec 2 4 cos 2 1 2 cos 2 1 1 2 4 2 1 5 1 32 1 25 1 7 25 25 7
3 1 cos cos 1 1 3 5 5 2 2 2 5 2 1 5
3 94. csc 2sin 1 5 3 Let sin 1 . is in quadrant IV. 5 3 Then sin , 0 . 5 2
3 1 92. cos sin 1 5 2 3 Let sin 1 . is in quadrant I. Then 5 3 sin , 0 . 5 2 2
986
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Section 8.6: Double-angle and Half-angle Formulas
3 cos 1 sin 2 1 5 1
9 25
16 25
97.
2
x
4 5
3
, ,
5 . 3
f x 0 cos 2 x cos x 0 2 cos 2 x 1 cos x 0 2 cos 2 x cos x 1 0 2 cos x 1 cos x 1 0
x
5
or cos x 1 0 cos x 1 x
, 3 3
The zeros on 0 x 2 are
3
, ,
5 . 3
, . 2 2
f x 0 2sin x sin 2 x 0 2sin 2 x 2sin x cos x 0 2sin x(sin x cos x) 0 or sin x cos x 0 sin x cos x 5 x , 4 4 5 The zeros on 0 x 2 are 0, , , . 4 4
99.
f x 0 sin 2 x cos x 0 2sin x cos x cos x 0 cos x (2sin x 1) 0 cos x 0 or 2sin x 1 0 sin x 1 sin x 3 2 x , 7 11 2 2 x , 6 6 7 3 11 , , . The zeros on 0 x 2 are , 2 6 2 6
100.
f x 0 cos 2 x 5cos x 2 0 2 2 cos x 1 5cos x 2 0
, 3 3
3
2sin x 0 sin x x 0,
5
The zeros on 0 x 2 are 0,
2 cos x 1 0 1 cos x 2
, 2 2
2
or 2 cos x 1 0 1 cos x 2 x
96.
98.
f x 0 sin 2 x sin x 0 2sin x cos x sin x 0 sin x 2 cos x 1 0
sin x 0 x 0,
3
The zeros on 0 x 2 are
1 3 csc 2sin 1 csc 2 sin 2 5 1 2sin cos 1 3 4 2 5 5 1 24 25 25 24 95.
f x 0 cos 2 x sin 2 x 0 cos 2 x sin 2 x sin 2 x 0 cos 2 x 0 cos x 0
2 cos 2 x 5cos x 3 0 (2 cos x 1)(cos x 3) 0 2 cos x 1 0 or cos x 3 0 1 cos x 3 cos x no sol 2 2 4 , x 3 3 2 4 The zeros on 0 x 2 are , . 3 3
987 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
101. a.
2 cos 1 0
1 cos 4 A 8 12 tan 1152 8 sin 4 2 2 2 1 2 1152 2 1152 2 2 2 2
b.
A(60º ) 16sin 60º cos 60º 1
2 2 2 2 2 1152 1152 2 2 2
16
b.
2 1 =1152 2 1152 in
c.
103. a.
D
1W 2
csc cot W 2 D csc cot csc cot
2 2 2 1 2 162 2 162 2 2 2 2
1 cos 1 cos sin sin sin
tan
2
Therefore, W 2 D tan
2 2 2 2 2 162 162 2 2 2
2
.
b. Here we have D 15 and W 6.5 .
2 2 2 2 2 2 162 162 2 2 2
102. a.
The maximum area is approximately 20.78 in.2 when the angle is 60˚.
1 cos 4 162 sin 4
2 1 =162 2 162 cm
Graph Y1 16sin x cos x 1 and use the MAXIMUM feature:
2
1 2 A 2 9 cot 162 8 sin 4 1 cos 4
162
31 1 2 2
12 3 in 2 20.78 in 2
2 2 2 2 2 2 1152 1152 2 2 2 1152
or cos 1 0
1 cos 1 cos 2 180º 60º , 300º On the interval 0º 90º , the solution is 60˚.
2
6.5 2 15 tan
2 13 tan 2 60 13 tan 1 2 60 1 13 24.45 2 tan 60
2
cos(2 ) cos 0 , 0º 90º 2 cos 2 1 cos 0 2 cos 2 cos 1 0 (2 cos 1)(cos 1) 0
988
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Section 8.6: Double-angle and Half-angle Formulas
104.
I x sin cos I y sin cos I xy cos 2 sin 2
1 sin 2 I xy cos 2 2
I x I y sin cos I xy cos 2 sin 2
Ix I y
105. a.
Ix I y 2
322 2 sin(2 67.5º ) cos(2 67.5º ) 1 32
32 2 sin 135º cos 135º 1 2 2 32 2 1 2 2
sin 2 I xy cos 2
2 1 32 2 2 feet 18.75 feet 32 2
v02 2 cos (sin cos ) 16 v2 2 0 (cos sin cos 2 ) 16 v2 2 1 0 (2 cos sin 2 cos 2 ) 16 2 v2 2 1 cos 2 0 sin 2 2 32 2
R ( )
v02 2 sin 2 1 cos 2 32 v2 2 0 sin 2 cos 2 1 32
b.
c. R
sin(2 ) cos(2 ) 0
Divide each side by 2 : 1 1 sin(2 ) cos(2 ) 0 2 2 Rewrite in the sum of two angles form using 1 1 cos and sin and : 4 2 2 sin(2 ) cos cos(2 ) sin 0 sin(2 ) 0 2 0 k 2 0 k 4 2 k 4 k 8 2 3 67.5º 8
322 2 sin(2 x) cos(2 x) 1 and 32 use the MAXIMUM feature:
d. Graph Y1
The angle that maximizes the distance is 67.5˚, and the maximum distance is 18.75 feet. 1 1 106. y sin(2 x) sin(4 x) 2 4 1 1 sin(2 x) sin(2 2 x) 2 4 1 1 sin(2 x) 2sin(2 x) cos(2 x) 2 4 1 1 sin(2 x) sin(2 x) cos(2 x) 2 2 1 1 sin(2 x) sin(2 x) 2 cos 2 ( x) 1 2 2
1 1 sin(2 x) sin(2 x) cos 2 ( x) sin(2 x) 2 2 2 sin(2 x) cos ( x)
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Chapter 8: Analytic Trigonometry
107. Let b represent the base of the triangle. h b/2 cos sin 2 s 2 s h s cos
b 2 s sin
2
cos 2 sin 2 cos 2 sin 2 cos 2 sin 2 cos 2 2 cos sin 2 cos 2 1 tan 2 4 1 tan 2 4 4 4 tan 2 4 4 tan 2
110. cos 2 cos 2 sin 2
2
1 A bh 2 1 2s sin s cos 2 2 2 s 2 sin
2
cos
2
1 s 2 sin 2
108. sin
y x y; cos x 1 1
a.
A 2 xy 2 cos sin 2sin cos
b.
2sin cos sin(2 )
c.
The largest value of the sine function is 1. Solve: sin 2 1 2
d.
111.
4
45
2 2 y sin 4 2 4 2 The dimensions of the largest rectangle are 2 . 2 by 2 x cos
112.
2sin cos 2 cos 1 sin 2 cos 1 cos 2 2 tan sec 2 2 tan 4 1 tan 2 4 4(2 tan ) 4 (2 tan ) 2 4x 4 x2
109. sin 2 2sin cos
4 2 tan
2
4 x2 4 x2
2
2
1 1 sin 2 x C cos 2 x 2 4 1 1 C cos 2 x sin 2 x 4 2 1 cos 2 x 2sin 2 x 4 1 1 2sin 2 x 2sin 2 x 4 1 (1) 4 1 4
4 2 tan
1 1 cos 2 x C cos 2 x 2 4 1 1 C cos 2 x cos 2 x 4 2 1 1 2 2 cos x 1 cos 2 x 4 2 1 1 1 cos 2 x cos 2 x 2 4 2 1 4
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Section 8.6: Double-angle and Half-angle Formulas 113. If z tan , then 2 2 tan 2z 2 1 z2 2 1 tan 2 2 tan 2 2 sec 2 2 tan cos 2 2 2
2sin 2 cos 2 2 cos 2 2sin cos 2 2 sin 2 2 sin 114. If z tan , then 2 1 tan 2 1 z2 2 1 z2 2 1 tan 2 1 cos 1 1 cos 1 cos 1 1 cos 1 cos (1 cos ) 1 cos 1 cos 1 cos 1 cos 1 cos (1 cos ) 1 cos 1 cos 2 cos 2 cos
115.
f ( x) sin 2 x
1 cos 2 x
2 Starting with the graph of y cos x , compress horizontally by a factor of 2, reflect across the xaxis, shift 1 unit up, and shrink vertically by a factor of 2.
116. g ( x) cos 2 x
1 cos 2 x
2 Starting with the graph of y cos x , compress horizontally by a factor of 2, reflect across the xaxis, shift 1 unit up, and shrink vertically by a factor of 2.
1 cos 12 12 sin 117. sin 24 2 2
6 2
1 1 4
82
2
1 1 2 8
6 2
6 2 8 2 6 2 16
4
6 2 2 4 6 2
2 4
4
4
1 cos 12 12 cos cos 24 2 2 1 1 4
991 Copyright © 2025 Pearson Education, Inc.
82
6 2 2
1 1 2 8
6 2
6 2 8 2 6 2 16
2 4 6 2 4
4
2 4 6 2 4
Chapter 8: Analytic Trigonometry
1 cos 8 8 cos cos 16 2 2
1 cos 4 4 118. cos cos 8 2 2
1 2
2 2
2 2 4
2 2 2 1 cos 8 8 sin sin 16 2 2
1
2 2 2 2
1
2 2 2 2
2 2 2 2
2 2 2 4
2 2 2 2
119. sin 3 sin 3 ( 120º ) sin 3 ( 240º ) sin 3 sin cos 120º cos sin 120º sin cos 240º cos sin 240º 3
3
3
3
1 1 3 3 sin sin cos sin cos 2 2 2 2 1 sin 3 sin 3 3 3 sin 2 cos 9sin cos 2 3 3 cos3 8 1 sin 3 3 3 sin 2 cos 9sin cos 2 3 3 cos3 8 3
1 3 3 9 3 3 sin 3 sin 3 sin 2 cos sin cos 2 cos3 8 8 8 8 1 3 3 9 3 3 sin 3 sin 2 cos sin cos 2 cos3 8 8 8 8 3 9 3 3 3 3 2 2 3 sin sin cos sin 3sin 1 sin sin 3sin 3sin 3 4 4 4 4 3 3 (from Example 2) 4sin 3 3sin sin 3 4 4
992
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2 2 2 4
Section 8.6: Double-angle and Half-angle Formulas 120. tan tan 3 3
3 tan
3
tan 3
1 3 tan
a tan
3 tan
3a tan
3
tan
2
3
(from problem 69)
3
tan 3 tan 2 3 3 1 3 tan 2
3
3
3
3a 1 tan 2 tan 2 tan
121.
3
a tan 1 3 tan 2 3 3 3 3 3 tan 2 a 1 3 tan 2 3 3 tan 3
3 tan 2 2
2
1 ( x x1 ) 2 1 y ( 3) ( x 2) 2 1 y 3 x 1 2 1 y x4 2 b 6 124. Vertex: x 3 2a 2( 1) y y1
3
3
3
a 3a tan 2
f (3) (3) 2 6(3) 7 16 ; (3,16)
x-intercepts: 0 x 2 6 x 7 0 x2 6 x 7 0 ( x 7)( x 1) x 7 or x 1 y-intercepts: y (0) 2 6(0) 7 y7
3
a3 a 3
a 3 3a 1
a 3 3a 1
cos(2 x) (2m 1) sin x m 1 0 (1 2sin 2 x) (2m 1) sin x m 1 0
3 1 2 4 125. sin cos 3 3 2 2
2sin 2 x (2m 1) sin x m 0 2sin 2 x (2m 1) sin x m 0 which is in quadratic form. For this equation to have exactly one real solution,
(2m 1)2 4(2)( m) 0
126. Amplitude: 2; Period:
4m 2 4m 1 8m 0 4m 2 4m 1 0 (2m 1) 2 0
So m
3 1 2 2
1 . 2
122. Answers will vary. 123. Since the line is perpendicular the slope would 1 be m . 2
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2
2
4
3 1 2
Chapter 8: Analytic Trigonometry
127.
f ( x) a x (5) ( x (2))( x 2)
For a 1 :
130. Vertex 1: 4 2 x 2 f (2) 5 (2, 5) Vertex 2: 6 3 x 2 f (3) 7 (3, 7)
f ( x) ( x 5)( x 2)( x 2) x 2 7 x 10 ( x 2)
x3 5 x 2 4 x 20
128.
3 y 2y 5 x(2 y 5) 3 y 2 xy 5 x 3 y 2 xy y 5 x 3 y (2 x 1) 5 x 3 5x 3 y f 1 ( x) 2x 1 x
129.
2
x7
3
d (3 2) 2 (7 (5)) 2
(5)2 (12) 2 25 144 169 13
131.
x2
x7
ln 2 ln 3x 2 x 7 ln 2 x 2 ln 3
132. 6 x 5 xD 5 y 4 yD 3 4 D 0 5 xD 4 yD 4 D 6 x 5 y 3 D 5 x 4 y 4 6 x 5 y 3
x ln 2 7 ln 2 x ln 3 2 ln 3 x ln 2 x ln 3 2 ln 3 7 ln 2 x(ln 2 ln 3) 2 ln 3 7 ln 2 x
2 ln 3 7 ln 2 ln 2 ln 3
f (b) f (a) log 2 16 log 2 4 ba 16 4 42 2 1 16 4 12 6
ln 9 ln128 ln 2 ln 3
6 x 5 y 3 5 x 4 y 4 6x 5 y 3 5x 4 y 4
D
6.548
2 ln 3 7 ln 2 The solution set is 6.548 . ln 2 ln 3
Section 8.7 1. sin(195) cos(75) sin(150 45) cos(30 45) sin(150 45) cos(30 45) sin150 cos 45 cos150 sin 45 cos 30 cos 45 sin 30 sin 45 1 2 3 2 3 2 1 2 2 2 2 2 2 2 2 2 2 12 4 36 12 6 6 2 16 16 16 16 4 4 4 4
2 3 2 6 2 3 3 1 3 3 2 3 4 3 1 1 3 1 16 16 16 16 8 8 8 8 8 8 4 2 2 2
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Section 8.7: Product-to-Sum and Sum-to-Product Formulas 2. cos(285) cos(195) cos(240 45) cos(240 45) cos(240 45) cos(240 45) cos 240 cos 45 sin 240 sin 45 cos 240 cos 45 sin 240 sin 45 cos 240 cos 45 sin 240 sin 45 2
2
2
2
2
2
2
2 3 2 1 2 3 2 1 2 2 2 2 2 4 4 4 4
1 1 3 4 8 8
3. sin(285) sin(75) sin(240 45) sin(30 45) sin(240 45) sin(30 45) sin 240 cos 45 cos 240 sin 45 sin 30 cos 45 cos 30 sin 45 3 2 1 2 1 2 3 2 2 2 2 2 2 2 2 2 12 36 4 12 6 2 2 6 16 16 16 16 4 4 4 4
2 3 6 2 2 3 3 3 1 3 2 3 4 3 1 1 3 1 16 16 16 16 8 8 8 8 8 8 4 2 2 2
4. sin(75) sin(15) sin(45 30) sin(45 30) sin(45) cos(30) cos(45) sin(30) sin(45) cos(30) cos(45) sin(30) 2sin(45) cos(30) 2 3 6 2 2 2 2
5. cos(255) cos(195) cos(225 30) cos(225 30) cos(225) cos(30) sin(225) sin(30) cos(225) cos(30) sin(225) sin(30) 2sin(225) sin(30) 2 1 2 2 2 2 2
6. sin(255) sin(15) sin(135 120) sin(135 120) sin(135) cos(120) cos(135) sin(120) sin(135) cos(120) cos(135) sin(120) sin(135) cos(120) cos(135) sin(120) sin(135) cos(120) cos(135) sin(120) 2 cos(135) sin(120) 2 3 6 2 2 2 2
995
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Chapter 8: Analytic Trigonometry
1 cos(4 2 ) cos(4 2 ) 2 1 cos 2 cos 6 2
1 cos(3 4 ) cos(3 4 ) 2 1 cos( ) cos 7 2 1 cos cos 7 2
7. sin(4 ) sin(2 )
14. cos(3 ) cos(4 )
1 cos(4 2 ) cos(4 2 ) 2 1 cos(2 ) cos 6 2
8. cos(4 ) cos(2 )
15. sin
3 1 3 3 cos sin sin 2 2 2 2 2 2 2
1 sin(4 2 ) sin(4 2 ) 2 1 sin 6 sin 2 2
9. sin(4 ) cos(2 )
1 sin 2 sin 2
5 1 5 5 sin sin 16. sin cos 2 2 2 2 2 2 2 1 sin 3 sin( 2 ) 2 1 sin 3 sin 2 2
1 cos(3 5 ) cos(3 5 ) 2 1 cos( 2 ) cos 8 2 1 cos 2 cos 8 2
10. sin(3 ) sin(5 )
4 2 4 2 17. sin(4 ) sin(2 ) 2sin cos 2 2 2sin cos 3
1 cos(3 5 ) cos(3 5 ) 2 1 cos( 2 ) cos 8 2 1 cos 2 cos 8 2
11. cos(3 ) cos(5 )
4 2 4 2 18. sin(4 ) sin(2 ) 2sin cos 2 2 2sin 3 cos 2 4 2 4 19. cos(2 ) cos(4 ) 2 cos cos 2 2 2 cos 3 cos( )
1 12. sin(4 ) cos(6 ) sin(4 6 ) sin(4 6 ) 2 1 sin 10 sin( 2 ) 2 1 sin 10 sin 2 2
2 cos 3 cos 5 3 5 3 20. cos(5 ) cos(3 ) 2sin sin 2 2 2sin 4 sin
1 cos( 2 ) cos( 2 ) 2 1 cos( ) cos 3 2 1 cos cos 3 2
13. sin sin(2 )
3 3 21. sin sin(3 ) 2sin cos 2 2 2sin 2 cos( ) 2sin 2 cos
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Section 8.7: Product-to-Sum and Sum-to-Product Formulas 3 3 22. cos cos(3 ) 2 cos cos 2 2 2 cos 2 cos( )
4 2 4 2 2sin cos 2 2 cos(4 ) cos(2 ) 2sin(3 ) cos 2 cos(3 ) cos sin(3 ) cos(3 ) tan(3 )
sin(4 ) sin(2 ) 27. cos(4 ) cos(2 )
2 cos 2 cos 3 3 3 2sin 2 2 sin 2 2 23. cos cos 2 2 2 2 2sin sin 2 2sin sin 2
2sin sin
3 3 2sin sin 2 2 3 3 2sin cos 2 2 2sin(2 ) sin( ) 2sin cos(2 ) ( sin ) sin(2 ) sin cos(2 ) tan(2 )
cos cos(3 ) 28. sin(3 ) sin
2
3 3 2 2 2 2 3 24. sin sin 2sin cos 2 2 2 2 2sin cos 2
3 3 2sin sin 2 2 3 3 2sin cos 2 2 2sin(2 ) sin( ) 2sin(2 ) cos( ) ( sin ) cos tan
cos cos(3 ) 29. sin sin(3 )
2sin cos 2 3 3 2sin cos 2 2 2sin(2 ) 2sin(2 ) cos( ) 2sin(2 ) cos( )
sin sin(3 ) 25. 2sin(2 )
cos
5 5 2sin sin 2 2 5 5 2sin cos 2 2 2sin(3 ) sin( 2 ) 2sin(3 ) cos( 2 ) ( sin 2 ) cos 2
cos cos(5 ) 30. sin sin(5 )
3 3 2 cos cos cos cos(3 ) 2 2 26. 2 cos(2 ) 2 cos(2 ) 2 cos(2 ) cos( ) 2 cos(2 ) cos( ) cos
tan 2
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Chapter 8: Analytic Trigonometry 31. sin sin sin(3 )
35.
3 3 sin 2sin cos 2 2 sin 2sin(2 ) cos( )
4 8 4 8 2sin cos 2 2 4 8 4 8 2sin cos 2 2 2sin(6 ) cos( 2 ) 2sin( 2 ) cos(6 )
cos 2sin(2 ) sin 1 cos 2 cos cos(3 ) 2 cos cos cos(3 )
sin(6 ) cos(2 ) sin(2 ) cos(6 ) tan(6 ) cot(2 )
32. sin sin 3 sin(5 ) 3 5 3 5 sin 2sin cos 2 2
sin 2sin(4 ) cos( ) cos 2sin(4 ) sin
36.
1 cos 2 cos 3 cos(5 ) 2 cos cos 3 cos(5 )
33.
tan(6 ) tan(2 )
cos(4 ) cos(8 ) cos(4 ) cos(8 ) 4 8 4 8 2sin sin 2 2 4 8 4 8 2 cos cos 2 2 2sin(6 )sin(2 ) 2 cos(6 ) cos(2 ) sin(6 ) sin( 2 ) cos(6 ) cos(2 ) tan(6 ) tan(2 )
sin(4 ) sin(8 ) cos(4 ) cos(8 ) 4 8 4 8 2sin cos 2 2 4 8 4 8 2 cos cos 2 2 2sin(6 ) cos( 2 ) 2 cos(6 ) cos( 2 )
tan(2 ) tan(6 ) 2sin cos 2 2 2sin cos 2 2 sin cos 2 2 cos sin 2 2 tan cot 2 2
sin(6 ) cos(6 ) tan(6 )
34.
sin(4 ) sin(8 ) sin(4 ) sin(8 )
sin sin 37. sin sin
sin(4 ) sin(8 ) cos(4 ) cos(8 ) 4 8 4 8 2sin cos 2 2 4 8 4 8 2sin sin 2 2 2sin( 2 ) cos(6 ) 2sin(6 ) sin( 2 ) cos(6 ) sin(6 ) cot(6 )
998
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Section 8.7: Product-to-Sum and Sum-to-Product Formulas
2 cos cos 2 2 2sin sin 2 2 cos cos 2 2 sin sin 2 2 cot cot 2 2
2sin cos 2 2 2sin sin 2 2 cos 2 sin 2 cot 2
cos cos 38. cos cos
sin sin 40. cos cos
2sin cos 2 2 2 cos cos 2 2 sin 2 cos 2 tan 2
sin sin 39. cos cos
41. 1 cos(2 ) cos(4 ) cos(6 ) cos 0 cos(6 ) cos(2 ) cos(4 ) 0 6 0 6 2 4 2 4 2 cos cos 2 cos cos 2 2 2 2 2 cos(3 ) cos(3 ) 2 cos(3 ) cos( ) 2 cos 2 (3 ) 2 cos(3 ) cos 2 cos(3 ) cos(3 ) cos 3 3 2 cos(3 ) 2 cos cos 2 2 2 cos(3 ) 2 cos(2 ) cos 4 cos cos(2 ) cos(3 )
42. 1 cos(2 ) cos(4 ) cos(6 ) cos 0 cos(6 ) cos(4 ) cos(2 ) 0 6 0 6 2 4 2 4 2sin sin 2sin sin 2 2 2 2 2sin(3 ) sin(3 ) 2sin(3 ) sin( ) 2sin 2 (3 ) 2sin(3 ) sin 2sin(3 ) sin(3 ) sin 3 3 2sin(3 ) 2sin cos 2 2 2sin(3 ) 2sin cos(2 ) 4sin cos(2 ) sin(3 )
999
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Chapter 8: Analytic Trigonometry
43. sin 4 cos 2 sin 2
cos 2
2
2
1 cos(2 ) 1 cos(2 ) 2 2 1 2 1 cos(2 ) 1 cos(2 ) 8 1 1 cos(2 ) 1 cos2 (2 ) 8 1 1 cos(4 ) 1 cos(2 ) 1 8 2 1 1 cos(2 ) 2 1 cos(4 ) 16 1 1 cos(2 ) 1 cos(4 ) 16 1 1 cos(2 ) cos(4 ) cos(4 ) cos(2 ) 16 1 1 1 cos(2 ) cos(4 ) cos(2 ) cos(6 ) 16 2 1 2 2 cos(2 ) 2 cos(4 ) cos(2 ) cos(6 ) 32 1 2 cos(2 ) 2 cos(4 ) cos(6 ) 32 1 1 1 1 cos(2 ) cos(4 ) cos(6 ) 16 32 16 32
44. sin 2 cos 4 sin 2 cos 2
2
2
1 cos(2 ) 1 cos(2 ) 2 2 1 2 1 cos(2 ) 1 cos(2 ) 8
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Section 8.7: Product-to-Sum and Sum-to-Product Formulas
1 1 cos 2 (2 ) 1 cos(2 ) 8 1 1 cos(4 ) 1 1 cos(2 ) 8 2 1 2 1 cos(4 ) 1 cos(2 ) 16 1 1 cos(4 ) 1 cos(2 ) 16 1 1 cos(2 ) cos(4 ) cos(4 ) cos(2 ) 16 1 1 1 cos(2 ) cos(4 ) cos(2 ) cos(6 ) 16 2 1 2 2 cos(2 ) 2 cos(4 ) cos(2 ) cos(6 ) 32 1 2 cos(2 ) 2 cos(4 ) cos(6 ) 32 1 1 1 1 cos(2 ) cos(4 ) cos(6 ) 16 32 16 32
45. sin 6 sin 2
3
3
1 cos(2 ) 2 1 3 1 cos(2 ) 8 1 1 2 cos cos 2 (2 ) 1 cos(2 ) 8 1 1 cos(4 ) 1 2 cos 2 1 cos(2 ) 8 2 1 2 4 cos(2 ) 1 cos(4 ) 1 cos(2 ) 16 1 3 4 cos(2 ) cos(4 ) 1 cos(2 ) 16 1 3 3cos(2 ) 4 cos(2 ) 4 cos 2 (2 ) cos(4 ) cos(4 ) cos(2 ) 16 1 1 cos(4 ) 1 cos(4 ) cos(2 ) cos(6 ) 3 7 cos(2 ) 4 16 2 2 1 6 14 cos(2 ) 4 4 cos(4 ) 2 cos(4 ) cos(2 ) cos(6 ) 32 1 10 15cos(2 ) 6 cos(4 ) cos(6 ) 32 5 15 3 1 cos(2 ) cos(4 ) cos(6 ) 16 32 16 32
1001
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Chapter 8: Analytic Trigonometry
46. cos6 cos 2
3
3
1 cos(2 ) 2 1 3 1 cos(2 ) 8 1 1 2 cos cos 2 (2 ) 1 cos(2 ) 8 1 1 cos(4 ) 1 2 cos 2 1 cos(2 ) 8 2 1 2 4 cos(2 ) 1 cos(4 ) 1 cos(2 ) 16 1 3 4 cos(2 ) cos(4 ) 1 cos(2 ) 16 1 3 3cos(2 ) 4 cos(2 ) 4 cos 2 (2 ) cos(4 ) cos(4 ) cos(2 ) 16 1 1 cos(4 ) 1 cos(4 ) cos(2 ) cos(6 ) 3 7 cos(2 ) 4 16 2 2 1 6 14 cos(2 ) 4 4 cos(4 ) 2 cos(4 ) cos(2 ) cos(6 ) 32 1 10 15cos(2 ) 6 cos(4 ) cos(6 ) 32 5 15 3 1 cos(2 ) cos(4 ) cos(6 ) 16 32 16 32
47.
sin(2 ) sin(4 ) 0 sin(2 ) 2sin(2 ) cos(2 ) 0 sin(2 ) 1 2 cos(2 ) 0 sin(2 ) 0
cos(2 ) cos(4 ) 0
48.
2 4 2 4 2 cos cos 0 2 2 2 cos(3 ) cos( ) 0
1 2 cos(2 ) 0 1 cos(2 ) 2 2 0 2k or 2 2k or k k 2 2 4 2 2k or 2 2k 3 3 2 k k 3 3 On the interval 0 2 , the solution set is 4 3 5 2 , , , , 0, , , . 3 2 3 3 2 3 or
2 cos 3 cos 0 cos(3 ) 0 or cos 0 3 2k or 3 2k or 2 2 2k 2k 6 3 2 3 3 2k or 2k 2 2 On the interval 0 2 , the solution set is 5 7 3 11 , , , , , . 6 2 6 6 2 6 3
1002
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Section 8.7: Product-to-Sum and Sum-to-Product Formulas cos(4 ) cos(6 ) 0
49.
4 6
2sin
4 6
sin 0 2 2 2sin(5 ) sin( ) 0
2sin
4 6 0 cos 2 2 2sin( ) cos(5 ) 0
2sin 5 sin 0
2sin cos(5 ) 0
sin(5 ) 0 or sin 0
cos(5 ) 0 or sin 0
5 0 2k or
5 2k
0 2k or 2k
or
2k 2k 5 5 5 0 2k or 2k On the interval 0 2 , the solution set is 6 7 8 9 2 3 4 0, , , , , , , , , . 5 5
5
5
5
5
5
or
3 5 2k or 5 2k 2 2 2k 3 2k 10 5 10 5 On the interval 0 2 , the solution set is 11 13 3 17 19 3 7 9 , , , , 0, , , , , , , .
51. a.
sin(4 ) sin(6 ) 0
50.
4 6
5
10 10 2 10 10
10
10
2
10
10
y sin 2 (852)t sin 2 (1209)t 2 (852)t 2 (1209)t 2 (852)t 2 (1209)t 2sin cos 2 2 2sin(2061 t ) cos(357 t )
2sin(2061 t ) cos(357 t )
b. Because sin 1 and cos 1 for all , it follows that sin(2061 t ) 1 and cos(357 t ) 1 for all
values of t. Thus, y 2sin(2061 t ) cos(357 t ) 2 1 1 2 . That is, the maximum value of y is 2. c.
Let Y1 2sin(2061 x) cos(357 x) .
52. a.
y sin 2 (941)t sin 2 (1477)t 2 (941)t 2 (1477)t 2 (941)t 2 (1477)t 2sin cos 2 2 2sin(2418 t ) cos(536 t )
2sin(2418 t ) cos(536 t )
b.
Because sin 1 and cos 1 for all , it follows that sin(2418 t ) 1 and cos(2418 t ) 1 for all values of t. Thus, y 2sin(2418 t ) cos(536 t ) 2 1 1 2 . That is, the maximum value of y is 2.
1003
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Chapter 8: Analytic Trigonometry
c.
Let Y1 2sin(2418 x) cos(536 x) .
2 2 53. I u I x cos I y sin 2 I xy sin cos
cos 2 1 1 cos 2 Ix Iy I xy 2sin cos 2 2 I cos 2 I x I y I y cos 2 x I xy sin 2 2 2 2 2 Ix I y Ix I y cos 2 I xy sin 2 2 2 I v I x sin 2 I y cos 2 2 I xy sin cos 1 cos 2 cos 2 1 Ix Iy I xy 2sin cos 2 2 I x I x cos 2 I y cos 2 I y I xy sin 2 2 2 2 2 Ix I y Ix I y cos 2 I xy sin 2 2 2
54. a.
Since and v0 are fixed, we need to maximize sin cos .
1 sin sin 2 1 sin 2 sin 2 This quantity will be maximized when sin 2 1 . So, sin cos
1 2v02 1 sin v 2 1 sin v02 1 sin v02 0 2 Rmax g 1 sin 1 sin g 1 sin g cos 2 g 1 sin 2
b.
Rmax
50
2
9.8 1 sin 35
598.24
The maximum range is about 598 meters.
1004
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Section 8.7: Product-to-Sum and Sum-to-Product Formulas 55. Add the sum formulas for sin( ) and sin( ) and solve for sin cos : sin( ) sin cos cos sin sin( ) sin cos cos sin sin( ) sin( ) 2sin cos sin cos
1 sin( ) sin( ) 2
59. sin 2 sin 2 sin 2
56. 2sin cos 2 2 1 2 sin sin 2 2 2 2 2
2 2 2 2 2sin cos sin 2 2 2 2sin( ) cos( ) 2sin cos
2 2 sin sin 2 2 sin sin
2sin( ) cos( ) 2sin cos
sin sin
2sin 2 cos cos 2 2 2 2 4sin cos cos 2 2
2sin cos( ) 2sin cos 2sin cos( ) cos
Thus, sin sin 2sin cos . 2 2 57. 2 cos cos 2 2 1 2 cos cos 2 2 2 2 2 2 2 cos cos 2 2 cos cos
4sin cos cos 2 2 4sin sin sin 4sin sin sin
Thus, cos cos 2 cos cos . 2 2 58. 2sin sin 2 2 1 2 cos cos 2 2 2 2 2 2 2 cos cos 2 2 cos cos Thus, cos cos 2sin sin . 2 2
1005
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Chapter 8: Analytic Trigonometry sin sin sin cos cos cos sin cos cos sin cos cos sin cos cos cos cos cos cos (sin cos cos sin ) sin cos cos cos cos cos cos sin( ) sin cos cos cos sin( ) sin cos cos cos cos cos cos cos cos cos sin sin cos cos sin (cos cos cos ) cos cos cos cos cos cos
60. tan tan tan
sin cos ( ) cos cos cos cos cos
sin cos( ) cos cos
sin cos cos sin sin cos cos cos cos cos
cos cos cos
sin (sin sin ) tan tan tan cos cos cos
Note that ,sin( ) sin cos ( ) cos( ) .
61.
x 1
f ( x ) 3sin x 5 y 3sin x 5 x 3sin y 5 x 5 3sin y x5 sin y 3 x 5 sin 1 y 3 x 5 f 1 ( x ) sin 1 3
x5
27 9 33( x 1) 32( x 5) 3( x 1) 2( x 5) 3 x 3 2 x 10 x 13 The solution set is 13 .
62. Amplitude: 5 2 Period: 4 2 Phase Shift: 4 4
The domain of sin 1 (u ) is 1,1 so x 5 1 1 3 8 x 2 Range of f = Domain of f 1 8, 2
63. cos csc 1 5 7
7 , let r 7 and y 5 . 5 2 2 Solve for x: x 2 25 49
Since csc ,
Range of f 1 = , 2 2
x 2 24
3 65. tan
x 2 6
6
Since is in quadrant I, x 2 6 . 7 x 2 6 . Thus, cos csc 1 cos
5
r
66.
7
64. We find the inverse function by switching the x and y variables and solving for y.
3
1 f ( x) x 2 2 x 2 3 1 2 x 6x 2 3 1 2 x 6x 9 2 3 3 2 1 x 3 5 3
1006
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Chapter 8 Review Exercises 67. The circumference of the large flywheel is C 2 r 2 (3) 6 in . The circumference of the small flywheel is C 2 (1.25) 2.5 in . The velocity of the large flywheel is 2000(6 in) v 12000 in/min . The velocity of 1 min
68.
Chapter 8 Review Exercises 1. Domain: x | 1 x 1 Range: y | y 2 2
the small flywheel is 12000 in 1 rev v 4800 rpm 1 min 2.5 in
2. Domain: x | 1 x 1
1 A bh 2 2 A bh
3. Domain: x | x
Range: y | 0 y
Range: y | y 2 2
2A h b
4. Domain: x | x 1
69. Find the points of intersection.
Range: y | 0 y , y 2
2 x 2 3x 4 x 3 2x2 7 x 3 0
5. Domain: x | x 1
(2 x 1)( x 3) 0 1 x ,3 2
Range: y | y , y 0 2 2 6. Domain: x | x
Range: y | 0 y 7. sin 1 1 1 f ( x) g ( x) on the interval ,3 2
70.
Find the angle , equals 1. sin 1,
2 f ( x) x 9 3 2 2 ( x h) 9 x 9 f ( x h) f ( x ) 3 3 h h 2 2 2 x h9 x9 3 3 3 h 2 h 2 3 3 h
2
, whose sine 2 2
2 2
Thus, sin 1 1
. 2
8. cos 1 0 Find the angle , 0 , whose cosine equals 0. cos 0, 0 2 Thus, cos 1 0 . 2
1007 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
9. tan 1 1
Find the angle , equals 1. tan 1,
4
tan 3,
, whose tangent 2 2
equals 2 . sec 2, 4
1 10. sin 2 , whose sine 2 2
. 4
14. cot 1 1
Find the angle , 0 , whose cotangent equals 1 . cot 1, 0 3 4 3 . Thus, cot 1 1 4
3 11. cos 1 2 Find the angle , 0 , whose cosine
3 15. sin 1 sin follows the form of the 8
equation f 1 f x sin 1 sin x x . Since
3 . 2 3 cos , 0 2 5 6 3 5 Thus, cos 1 . 2 6
equals
3 is in the interval , , we can apply 8 2 2 the equation directly and get 3 3 . sin 1 sin 8 8
3 16. cos 1 cos follows the form of the equation 4 3 f 1 f x cos 1 cos x x . Since is 4 in the interval 0, , we can apply the equation
Find the angle ,
0
Thus, sec 1 2
1 equals . 2 1 sin , 2 2 2 6 1 Thus, sin 1 . 6 2
. 3
13. sec 1 2 Find the angle , 0 , whose secant
1
12. tan 1 3
3
Thus, tan 1 3
. 4
Find the angle ,
2 2
2 2
Thus, tan 1 1
, whose tangent 2 2
equals 3 .
3 3 . directly and get cos 1 cos 4 4
1008
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Chapter 8 Review Exercises
2 17. tan 1 tan follows the form of the 3
equation f 1 f x tan 1 tan x x but we cannot use the formula directly since
2 is not 3
8 19. sin 1 sin follows the form of the 9
equation f 1 f x sin 1 sin x x , but we cannot use the formula directly since
8 is not 9
in the interval , . We need to find an 2 2 angle in the interval , for which 2 2
in the interval , . We need to find an 2 2 angle in the interval , for which 2 2
2 2 is in quadrant tan tan . The angle 3 3 II so tangent is negative. The reference angle of 2 is and we want to be in quadrant IV 3 3 so tangent will still be negative. Thus, we have 2 tan tan . Since is in the 3 3 3
8 8 is in sin sin . The angle 9 9 quadrant III so sine is negative. The reference 8 is and we want to be in angle of 9 9 quadrant IV so sine will still be negative. Thus, 8 we have sin sin . Since is 9 9 9
interval , , we can apply the equation 2 2 above and get 2 1 tan 1 tan tan tan . 3 3 3
in the interval , , we can apply the 2 2 equation above and get 8 sin 1 sin sin 1 sin . 9 9 9
15 18. cos 1 cos follows the form of the 7
equation f
1
f x cos cos x x , but 1
15 is we cannot use the formula directly since 7
not in the interval 0, . We need to find an angle in the interval 0, for which 15 15 is in cos cos . The angle 7 7 15 is . quadrant I so the reference angle of 7 7 15 is Thus, we have cos cos . Since 7 7 7
in the interval 0, , we can apply the equation above and get 15 1 cos 1 cos cos cos . 7 7 7
20. sin sin 1 0.9 follows the form of the equation
f f 1 x sin sin 1 x x . Since 0.9 is in
the interval 1,1 , we can apply the equation
directly and get sin sin 1 0.9 0.9 .
21. cos cos 1 0.6 follows the form of the equation
f f 1 x cos cos 1 x x . Since 0.6 is
in the interval 1,1 , we can apply the equation
directly and get cos cos 1 0.6 0.6 .
22. tan tan 1 5 follows the form of the equation
f f 1 x tan tan 1 x x . Since 5 is a
real number, we can apply the equation directly
and get tan tan 1 5 5 .
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Chapter 8: Analytic Trigonometry 23. Since there is no angle such that cos 1.6 , the quantity cos 1 1.6 is not defined. Thus,
28. sin cot 1 4 3
3 4
cos cos 1 1.6 is not defined.
Since cot , 0 , is in quadrant I. Let x 3 and y 4 . Solve for r: 9 16 r 2
2 1 1 1 24. sin cos sin 3 2 6
r 2 25 r 5 3 y 4 Thus, sin cot 1 sin . 4 5 r
3 1 1 25. cos tan cos 1 4
Find the angle ,
4
29. tan sin 1 5
3 26. tan sin 1 2
4 Since sin , , let y 4 and 5 2 2 2 r 5 . Solve for x: x 16 25
, whose sine 2 2
x2 9
3 . equals 2 3 sin , 2 2 2 3 3 So, sin 1 . 2 3 3 Thus, tan sin 1 tan 3 . 3 2
x 3
Since is in quadrant IV, x 3 . 4 y 4 4 Thus, tan sin 1 tan
30.
5
x
3
3
f x 2sin 3 x y 2sin 3x x 2sin 3 y x sin 3 y 2 x 3 y sin 1 2
3 27. sec tan 1 3
1 x y sin 1 f 1 x 3 2
Find the angle , , whose tangent is 2 2 3 3
The domain of f x equals the range of f 1 x and is
3 , 3 2 2 6 3 . So, tan 1 3 6 1 3 2 3 . Thus, sec tan sec 3 3 6 tan
6
x
, or , in 6 6 6
interval notation. To find the domain of f 1 x we note that the argument of the inverse sine x and that it must lie in the interval function is 2 1,1 . That is, x 1 2 2 x 2 The domain of f 1 x is x | 2 x 2 , or 1
2, 2 in interval notation. Recall that the 1010
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Chapter 8 Review Exercises
domain of a function is the range of its inverse and the domain of the inverse is the range of the function. Therefore, the range of f x is
33. Let csc 1 u so that csc u ,
2
2
and 0 , u 1 . Then,
2, 2 . 31. f x cos x 3
tan csc1 u tan tan 2
y cos x 3
sec 2 1
x cos y 3 x 3 cos y
3 x cos y y cos 1 3 x f 1 x
The domain of f x equals the range of
1 csc 1 2
u u u2 1 1 sin 2 tan 1 sin 2 cos 2
34. tan cot sin 2 tan
f 1 x and is 0 x , or 0, in interval
notation. To find the domain of f 1 x we note that the argument of the inverse cosine function is 3 x and that it must lie in the interval 1,1 . That is, 1 3 x 1
35. sin 2 (1 cot 2 ) sin 2 csc 2 1 sin 2 2 1 sin 36. 5cos 2 3sin 2 2 cos 2 3cos 2 3sin 2 2 cos 2 3 cos 2 sin 2
4 x 2
2 cos 2 3 1 3 2 cos 2
4 x2 2 x4 The domain of f 1 x is x | 2 x 4 , or
37.
2, 4 in interval notation. Recall that the domain of a function is the range of its inverse and the domain of the inverse is the range of the function. Therefore, the range of f x is
1 cos sin (1 cos ) 2 sin 2 sin 1 cos sin (1 cos ) 1 2cos cos 2 sin 2 sin (1 cos ) 1 2cos 1 sin (1 cos ) 2 2cos sin (1 cos ) 2(1 cos ) sin (1 cos ) 2 2csc sin
2, 4 .
32. Let sin 1 u so that sin u , 1 u 1 . Then,
2
2
,
cos sin 1 u cos cos 2 1 sin 2 1 u 2
1 cos cos cos 38. 1 cos sin cos sin cos 1 1 sin 1 tan 1 cos
1011 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
1 csc sin 39. sin 1 csc 1 1 sin sin 1 sin 1 1 1 sin 1 sin 1 sin 1 sin 1 sin 2 1 sin cos 2
43.
44.
1 sin sin 1 sin 2 sin cos 2 sin cos cos sin cos cot
40. csc sin
41.
1 sin cos (1 sin ) sec cos (1 sin )
cos 1 sin 2 1 sin cos cos 2
cos( ) cos cos sin sin cos sin cos sin cos cos sin sin cos sin cos sin cos sin sin cos cot tan cos( ) cos cos sin sin cos cos cos cos cos cos sin sin cos cos cos cos 1 tan tan
45. (1 cos ) tan
2
(1 cos )
46. 2 cot cot 2 2
sin sin 1 cos
cos cos 2 sin sin 2
2 cos cos 2 sin 2 sin 2sin cos
cos 2 sin 2 sin 2 2 cos sin 2 sin 2 sin 2 2 cot 1
1 sin 1 sin
47. 1 8sin 2 cos 2 1 2 2sin cos 1 2sin 2 2 cos 2 2 cos 4
1 sin cos3 1 sin
cos sin sin cos cos 2 sin 2 sin cos 1 sin 2 sin 2 sin cos 1 2sin 2 sin cos
42. cot tan
48.
sin 3 cos sin cos 3 sin 2
2
sin 3 sin 2 sin 2
sin 2 1
1012
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Chapter 8 Review Exercises
2 4 2 4 2sin cos sin 2 sin 4 2 2 49. cos 2 cos 4 2cos 2 4 cos 2 4 2 2 2sin 3 cos 2cos 3 cos sin 3 cos 3 tan 3
50.
cos(2 ) cos(4 ) tan tan(3 ) cos(2 ) cos(4 ) 2sin(3 ) sin( ) tan tan(3 ) 2 cos(3 ) cos( ) 2sin(3 )sin tan tan(3 ) 2 cos(3 ) cos tan(3 ) tan tan tan(3 ) 0
51. sin165º sin 120º 45º sin120º cos 45º cos120º sin 45º 3 2 1 2 2 2 2 2 6 2 4 4 1 6 2 4
52. tan105º tan 60º 45º tan 60º tan 45º 1 tan 60º tan 45º 3 1 1 3 1 3 1 1 3 1 3 1 3
53. cos
5 3 2 cos 12 12 12 cos cos sin sin 4 6 4 6 2 3 2 1 2 2 2 2 6 2 4 4 1 6 2 4
2 3 54. sin sin 12 12 12 sin cos cos sin 6 4 6 4 1 2 3 2 2 2 2 2 2 6 4 4 1 2 6 4
55. cos80º cos 20º sin 80º sin 20º cos 80º 20º cos 60º 1 2 56. sin 70º cos 40º cos 70º sin 40º sin 70º 40º sin 30º 1 2
1 2 3 3 1 3 42 3 2 2 3
1013 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
2 1 cos 1 4 4 2 57. tan tan 8 2 2 1 cos 1 4 2
sin( ) sin cos cos sin 4 12 3 5 5 13 5 13 48 15 63 65 65
d.
tan( )
e.
4 3 24 sin(2 ) 2sin cos 2 5 5 25
f.
cos(2 ) cos 2 sin 2
2 2 2 2
tan tan 1 tan tan 4 5 3 12 4 5 1 3 12 11 11 9 33 12 14 12 14 56 9
2 2 2 2 2 2 2 2
2 2
2
4 2 2 2 2 2 2 2 2 2 2 1
2 5 5 1 1 cos 2 5 4 sin 4 58. sin 8 2 2 2
2
2
12 5 144 25 119 13 13 169 169 169
2 2 4 2 2 2
4 5 59. sin , 0 ; sin , 5 2 13 2 3 4 12 5 cos , tan , cos , tan , 5 3 13 12 0 , 2 4 4 2 2 a. sin( ) sin cos cos sin 4 12 3 5 5 13 5 13 48 15 33 65 65 b.
c.
g.
sin
h.
cos
1 cos 2 2 12 1 13 2 25 25 5 5 26 13 2 26 26 26
2
1 cos 2 3 8 1 5 5 2 2
4 2 2 5 5 5 5
3 3 12 3 60. sin , ; cos , 2 5 2 13 2 4 3 5 5 cos , tan , sin , tan , 5 4 13 12 3 3 , 2 2 4 4 2
cos( ) cos cos sin sin 3 12 4 5 5 13 5 13 36 20 56 65 65
1014
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Chapter 8 Review Exercises
a.
sin( ) sin cos cos sin 3 12 4 5 5 13 5 13 36 20 65 16 65
b.
cos( ) cos cos sin sin 4 12 3 5 5 13 5 13 48 15 65 63 65
c.
sin( ) sin cos cos sin 3 12 4 5 5 13 5 13 36 20 65 56 65
d.
tan( )
tan tan 1 tan tan
h.
f.
sin(2 ) 2sin cos 3 4 24 2 5 5 25
cos( ) cos cos sin sin 4 5 3 12 5 13 5 13 20 36 65 65 16 65
c.
sin( ) sin cos cos sin 3 5 4 12 5 13 5 13 15 48 65 65 33 65
d.
tan( )
2
12 5 13 13 144 25 119 169 169 169
g.
sin
2
1 cos 2 12 1 1 13 13 2 2
1 cos 2 2 4 1 5 2 1 1 1 10 5 2 10 10 10
b.
cos(2 ) cos 2 sin 2 2
3 3 12 61. tan , ; tan , 0 4 2 5 2 3 4 12 5 sin , cos , sin , cos , 5 5 13 13 3 , 0 2 2 4 2 4 a. sin( ) sin cos cos sin 3 5 4 12 5 13 5 13 15 48 65 65 63 65
3 5 1 1 16 16 4 12 3 3 5 21 3 21 63 1 4 12 16
e.
cos
1 1 26 26 26 26
1015 Copyright © 2025 Pearson Education, Inc.
tan tan 1 tan tan 3 12 4 5 3 12 1 4 5 63 63 5 63 20 4 20 4 16 5
Chapter 8: Analytic Trigonometry
e.
3 4 24 sin(2 ) 2sin cos 2 5 5 25
f.
cos(2 ) cos 2 sin 2 2
d.
tan( )
2
25 144 119 5 12 169 13 13 169 169
g.
h.
sin
cos
9 3 8 2 23 8 2 9 3 23
1 cos 2 2 4 1 5 2 1 1 1 10 5 2 10 10 10
e.
3 1 3 sin(2 ) 2sin cos 2 2 2 2
f.
cos(2 ) cos 2 sin 2 2
2 1 8 7 1 2 2 3 9 9 9 3
3 0; sec 3, 2 2 2 3 1 sin , cos , tan 3, 2 2 2 2 1 sin , cos , tan 2 2, 3 3 3 0, 4 2 4 2 a. sin( ) sin cos cos sin 3 1 1 2 2 2 3 2 3 32 2 6
g.
sin
h.
cos
2
62. sec 2,
b.
cos( ) cos cos sin sin 1 1 3 2 2 2 3 2 3 1 2 6 6
c.
sin( ) sin cos cos sin 3 1 1 2 2 2 3 2 3 32 2 6
1 3 2 2 3 2 2
3 2 2 1 2 6 1 2 6 1 2 6
1 cos 2 2 5 8 1 13 13 4 2 2 13 2 2 13 13 13
tan tan 1 tan tan
1 cos 2 1 2 1 3 3 1 1 3 2 2 3 3 3
1 cos 2 2 1 3 1 2 2 2 2
3 3 4 2
2 3 2 3 63. sin , ; cos , 3 2 3 2 5 2 5 5 cos , tan , sin , 3 5 3 5 3 3 tan , , 2 2 2 4 2 2 4 a. sin( ) sin cos cos sin 5 5 2 2 3 3 3 3 4 5 9 9 1
1016
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Chapter 8 Review Exercises
b.
c.
d.
cos( ) cos cos sin sin 5 2 2 5 3 3 3 3 2 5 2 5 9 9 0
3 5 3 2
sin( ) sin cos cos sin 5 5 2 2 3 3 3 3 4 5 9 9 1 9 tan( )
e.
sin(2 ) 2sin cos 5 4 5 2 2 9 3 3
f.
cos(2 ) cos 2 sin 2 2
2 5 4 5 1 2 9 9 9 3 3
2
1 cos 2 2 1 3 2
3 5 6
6 3 5
6 6 3 5 6
quadrant I; is in quadrant I. Then sin
4 5 5 5 10 11 9 5 10 ; Undefined 0
sin
3 1 64. cos sin 1 cos 1 5 2 3 1 Let sin 1 and cos 1 . is in 2 5
tan tan 1 tan tan
2 5 5 2 5 2 5 5 1 5 2
g.
h.
5 1 3 1 cos cos 2 2 2
5 3 5 30 2 6 6
0
3 , 5
1 , and cos , 0 . 2 2 2
cos 1 sin 2 2
9 16 4 3 1 1 25 25 5 5 sin 1 cos 2 2
1 3 3 1 1 1 4 4 2 2 3 1 cos sin 1 cos 1 cos 5 2 cos cos sin sin 4 1 3 3 5 2 5 2 4 3 3 43 3 10 10 10 5 4 65. sin cos 1 cos 1 13 5 4 1 5 and cos 1 . is in Let cos 13 5
quadrant I; is in quadrant I. Then cos 0
4 , and cos , 0 . 2 5 2
1017 Copyright © 2025 Pearson Education, Inc.
5 , 13
Chapter 8: Analytic Trigonometry
1 3 tan sin 1 tan 1 tan 2 4 tan tan 1 tan tan
sin 1 cos 2 2
25 144 12 5 1 1 13 169 169 13 sin 1 cos 2
3 3 3 4 3 3 1 3 4
2
16 9 3 4 1 1 5 25 25 5 4 1 5 sin cos cos 1 sin 13 5 sin cos cos sin 12 4 5 3 13 5 13 5 48 15 33 65 65 65
4 3 9 12 3 3 1 12 9 4 3 12 3 3 12 3 3 12 3 3
3 1 66. tan sin 1 tan 1 2 4 1 3 Let sin 1 and tan 1 . is in 4 2 quadrant IV; is in quadrant I. Then,
tan
1 3
48 25 3 39 48 25 3 39
4 67. cos tan 1 (1) cos 1 5
cos 1 sin 2 1 1 1 1 2 4
144 75 3 117
1 3 sin , 0 , and tan , 2 4 2 0 . 2 2
4
Let tan 1 (1) and cos 1 . is in 5
3 3 4 2
quadrant IV; is in quadrant II. Then tan 1,
3 3
2
4 0 , and cos , 5 2
.
sec 1 tan 2 1 (1) 2 2 cos
1 2
2 2
sin 1 cos 2 2
2 1 1 2 1 1 2 2 2 2 sin 1 cos 2 2
16 9 3 4 1 1 25 25 5 5
1018
Copyright © 2025 Pearson Education, Inc.
Chapter 8 Review Exercises
4 cos tan 1 (1) cos 1 cos 5 cos cos sin sin 2 4 2 3 2 5 2 5 4 2 3 2 10 10 2 10
3 68. sin 2 cos 1 5 3 Let cos 1 . is in quadrant II. Then 5 3 cos , . 5 2 sin 1 cos 2 2
9 16 4 3 1 1 5 25 25 5
3 sin 2cos 1 sin 2 5 2sin cos 24 4 3 2 25 5 5
4 69. cos 2 tan 1 3 4 Let tan 1 . is in quadrant I. Then 3 4 tan , 0 . 3 2 sec tan 2 1 2
16 4 1 1 3 9 3 cos 5 4 cos 2 tan 1 cos 2 3 2 cos 2 1 2
25 5 9 3
70. cos
5 2k , k is any integer 3 3 5 On 0 2 , the solution set is , . 3 3
2k or
71. tan 3 0 tan 3 2 k , k is any integer 3 On the interval 0 2 , the solution set is 2 5 , . 3 3
72. sin(2 ) 1 0 sin(2 ) 1 3 2k 2 3 k , k is any integer 4 On the interval 0 2 , the solution set is 3 7 , 4 4 2
73. tan 2 0 2 0 k
k , where k is any integer 2 On the interval 0 2 , the solution set is 3 0, , , . 2 2
74. sec 2 4 sec 2 1 cos 2
7 3 9 2 1 2 1 25 5 25
1 2
+k
or
2 +k , 3
3 where k is any integer On the interval 0 2 , the solution set is 2 4 5 , , . , 3 3 3 3
1019 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry 75. 0.2sin 0.05 Find the intersection of Y1 0.2sin and Y2 0.05 :
4sin 2 1 4 cos
79.
4 1 cos 2 1 4 cos 4 4 cos 1 4 cos 2
4 cos 4 cos 3 0 2
2 cos 1 2 cos 3 0
0, 1 2 2 4 , 3 3 On 0 2 , the solution set is 4 2 , , . 0, 3 3
3 , 2 2
5 On 0 2 , the solution set is 0, , . 6 6
2sin 2 3sin 1 0 (2sin 1)(sin 1) 0
6
sin( )
sin 1 0 sin 1
5 ,
3
, 4 4 3 3
,
81. sin cos 1 Divide each side by 2 : 1 1 1 sin cos 2 2 2 Rewrite in the difference of two angles form 1 1 , sin , and : where cos 4 2 2 1 sin cos cos sin 2
6
or
2 2
4 2 4
0
1 sin 2
sin
On 0 2 , the solution set is , ,
or cos 1
2sin 1 0
or 2sin 2 0
(2sin 1)(cos 1) 0
78.
cos 0
cos (2sin 1) 1(2sin 1) 0
6
sin 2 2 cos
cos 2sin 2 0
2sin cos cos 2sin 1 0
,
3
2sin cos 2 cos 0
sin(2 ) cos 2sin 1 0
5
,
2sin cos 2 cos
cos
3
80.
or sin 0
1 2
5
5 . 3 3
sin (1 2 cos ) 0
sin
3 2 (not possible) cos
On 0 2 , the solution set is ,
sin 2sin cos 0
77.
1 2
sin sin(2 ) 0
1 2 cos 0
or 2 cos 3 0
cos
On the interval 0 2 , x 0.25 or x 2.89 The solution set is 0.25, 2.89 . 76.
2 cos 1 0
2
6 5 . 6 2 6
On 0 2 , the solution set is , ,
1020
Copyright © 2025 Pearson Education, Inc.
2 2
. 2
Chapter 8 Review Exercises 3 or 4 4 3 or 4 4 4 4 or 2 On 0 2 , the solution set is , .
2
quadrant II. The calculator yields 1 tan 1 0.24 , which is an angle in 4
quadrant IV. Since lies in quadrant II, 0.24 2.90 . Therefore, cot 1 4 2.90 .
82. sin 1 0.7 0.78 87. 2 x 5cos x Find the intersection of Y1 2 x and Y2 5cos x :
83. tan 1 2 1.11
x 1.11 The solution set is 1.11 .
84. cos 1 0.2 1.77
1 85. sec1 3 cos 1 3 We seek the angle , 0 , whose cosine
1 1 equals . Now cos , so lies in 3 3 1 quadrant I. The calculator yields cos 1 1.23 , 3 which is an angle in quadrant I, so sec1 3 1.23 .
88. 2sin x 3cos x 4 x Find the intersection of Y1 2sin x 3cos x and Y2 4 x :
x 0.87 . The solution set is 0.87 .
89. sin x ln x Find the intersection of Y1 sin x and Y2 ln x :
1 86. cot 1 4 tan 1 4 We seek the angle , 0 , whose tangent 1 1 equals . Now tan , so lies in 4 4
x 2.22 The solution set is 2.22 .
1021 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry 90. 3sin 1 x sin 1 x
Verifying equality: 1 6 2 6 2 4 4 2 3 2 4
3 x sin 3
3 2
3 The solution set is . 2
91.
2 cos 1 x
x cos
4
2 cos 1 x 0
2
3 1
2 3 1 4
2 cos 1 x 4 cos 1 x
cos 1 x
2
0 2 The solution set is {0}.
92. Using a half-angle formula: 30 sin15 sin 2 1 cos 30 2
2 3 2 3 1 16
2 42 3
16
22 2 3
16
2 3 4
2 3 2
93. Given the value of cos , the most efficient Double-angle Formula to use is cos 2 2 cos 2 1 .
3 1 2 2 3 2 3 2 4 2 Note: since 15º lies in quadrant I, we have sin15 0 .
Using a difference formula: sin15 sin(45 30) sin(45) cos(30) cos(45) sin(30) 2 3 2 1 2 2 2 2 6 2 6 2 1 4 4 4 4
2
Chapter 8 Test 2 1. Let sec1 . We seek the angle , such 3
6 2
that 0 and
2
, whose secant equals
2 . The only value in the restricted range with 3 2 2 is . Thus, sec1 a secant of . 6 3 3 6
1022
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Chapter 8 Test
2 2. Let sin 1 . We seek the angle , such 2 2 , whose sine equals . The 2 2 2 only value in the restricted range with a sine of 2 2 is . Thus, sin 1 . 4 4 2 2
that
3. Let tan 1 3 . We seek the angle , such
that
, whose tangent equals 3 . The
2 2 only value in the restricted range with a tangent of
3 is
3
. Thus, tan
1
3 3 .
4. Let cos 1 0 . We seek the angle , such that 0 , whose cosine equals 0 . The only value
in the restricted range with a cosine of 0 is Thus, cos 1 0
2
2
.
.
, whose cotangent equals 1 . The only 2 2 value in the restricted range with a cotangent of 1
is
. Thus, cot 1 1
4
4
, whose cosecant equals 2 . The only 2 2 value in the restricted range with a cosecant of 2
is
6
. Thus, csc 1 2
6
.
11 7. sin 1 sin follows the form of the equation 5
7 8. tan tan 1 follows the form 3
f f 1 x tan tan 1 x x . Since the
domain of the inverse tangent is all real numbers, we can directly apply this equation to get 7 7 tan tan 1 . 3 3
9. cot csc1 10
r 10 , , let y 2 2
r 10 and y 1 . Solve for x: x 2 12
10
2
x 2 1 10 x2 9
.
6. Let csc 1 2 . We seek the angle , such that
11 above and get sin 1 sin . 5 5
Since csc 1
5. Let cot 1 1 . We seek the angle , such that
11 is in quadrant I. The reference angle of 5 11 11 is and sin sin . Since is in 5 5 5 5 5 the interval , , we can apply the equation 2 2
f 1 f x sin 1 sin x x , but because
x3 is in quadrant I.
Thus, cot csc 1 10 cot
x 3 3. y 1
3 10. Let cos 1 . 4
3 sec cos 1 sec 4 1 cos
11 is not in the interval , , we cannot 5 2 2 directly use the equation. We need to find an angle in the interval 11 2 , 2 for which sin 5 sin . The angle 1023 Copyright © 2025 Pearson Education, Inc.
1 1 3 cos cos 4 1 3 4 4 3
Chapter 8: Analytic Trigonometry 11. sin 1 0.382 0.39 radian
sin cos cos sin 2 cos 2 cos cos sin 2 cos 2 cos 1 cos sec
16. sin tan cos sin
1 12. sec1 1.4 cos 1 0.78 radian 1.4
17. tan cot
sin cos cos sin
sin 2 cos 2 sin cos sin cos sin 2 cos 2 sin cos 1 sin cos 2 2sin cos 2 sin 2
13. tan 1 3 1.25 radians
1 14. cot 1 5 tan 1 0.20 radian 5
2 csc 2
18. 15.
sin cos cos sin sin sin cos cos sin cos cos sin sin cos cos sin cos cos cos cos sin cos cos sin sin cos cos sin cos cos sin cos cos sin
csc cot sec tan csc cot csc cot sec tan csc cot csc2 cot 2 sec tan csc cot
1 sec tan csc cot
sec tan sec tan csc cot sec tan
1
1
cos cos
sec tan
sec tan csc cot 2
sin tan tan
2
sec tan csc cot
1024
Copyright © 2025 Pearson Education, Inc.
cos cos sin cos cos sin
Chapter 8 Test
19.
22. tan 75 tan 45 30
sin 3 sin 2 sin cos 2 cos sin 2
sin cos 2 sin 2 cos 2sin cos sin cos sin 2sin cos 2 3sin cos 2 sin 3 3sin 1 sin 2 sin 3 2
3
3sin 3sin 3 sin 3 3sin 4sin 3
sin cos tan cot cos sin 20. tan cot sin cos cos sin sin 2 cos 2 cos sin sin 2 cos 2 sin cos sin 2 cos 2 sin 2 cos 2 cos 2 1 2 cos 2 1
3 1 23. sin cos 1 2 5 1 3 Let cos . Since 0 (from the 5 2
range of cos 1 x ),
1 2 cos 2
21. cos15 cos 45 30 cos 45 cos 30 sin 45 sin 30 2 3 2 1 2 2 2 2 2 3 1 4 6 2 1 or 6 2 4 4
tan 45 tan 30 1 tan 45 tan 30 3 1 3 3 1 1 3 3 3 3 3 3 3 3 3 3 3 3 3 96 3 3 32 3 12 6 3 6 2 3
1 cos 1 sin 2 2 3 1 cos cos 1 1 53 5 2 2 1 5 5 5 6 24. tan 2sin 1 11 6 6 and lies in Let sin 1 . Then sin 11 11 y 6 quadrant I. Since sin , let y 6 and r 11 r 11 , and solve for x: x 2 62 112 x 2 85 x 85
1025 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
tan
26. Let 75 , 15 .
y 6 6 85 x 85 85
Since sin cos
6 85 2 2 tan 85 tan 2 2 1 tan 2 6 85 1 85
1 sin sin , 2
1 sin 90 sin 60 2 1 3 1 2 3 1 2 3 2 2 4 4
sin 5 cos15
12 85 12 85 85 85 36 85 49 1 85 12 85 49
27.
sin 75 sin15 75 15 75 15 2sin cos 2 2 2 3 6 2sin 45 cos 30 2 2 2 2
2 3 25. cos sin 1 tan 1 3 2 2 3 Let sin 1 and tan 1 . Then 3 2 2 3 sin and tan , and both and 3 2 y 2 lie in quadrant I. Since sin 1 , let r1 3
28.
cos 65 cos 20 sin 65 sin 20 cos 65 20 cos 45
4sin 2 3 3 sin 2 4 3 sin 2 On the interval 0, 2 , the sine function takes
Thus, cos
x1 5 . 3 r1
on a value of
Since tan
y2 3 , let x2 2 and y2 3 . x2 2
sine takes on a value of
y2 3 . r2 13 Therefore, cos cos cos sin sin
2 5 6 3 13
2 13
30. 3cos tan 2 3sin tan sin 0 3sin cos 1 0 sin 3 cos
Thus, sin
3 2 when or . The 3 3 2
3 4 when and 3 2 5 2 4 5 . The solution set is . , , , 3 3 3 3 3
Solve for x2 : 22 32 r2 2 4 9 r2 2 r2 2 13 r2 13
5 2 2 3 3 13 3 13
2 2
29. 4sin 2 3 0
y1 2 and r1 3 . Solve for x1 : x12 22 32 x12 4 9 x12 5 x1 5
sin 0 or
1 3 0 cos
1 3 On the interval 0, 2 , the sine function takes cos
5 3 39
on a value of 0 when 0 or . The cosine 1026
Copyright © 2025 Pearson Education, Inc.
Chapter 8 Cumulative Review
1 in the second and 3 1 third quadrants when cos 1 and 3 1 cos 1 . That is 1.911 and 4.373 . 3 The solution set is 0,1.911, , 4.373 .
function takes on a value of
31.
cos 2 2sin cos sin 2 0
cos sin 2sin cos 0 2
2
cos 2 sin 2 0 sin 2 cos 2 tan 2 1 The tangent function takes on the value 1 3 k . Thus, we need when its argument is 4 3 k 2 4 3 k 8 2
3 4k 8 On the interval 0, 2 , the solution set is
3 7 11 15 , , , 8 8 8
Chapter 8 Cumulative Review 1. 3x 2 x 1 0 b b 2 4ac 2a
x
.
sin cos1 cos sin1 cos cos cos tan cos1 sin1 1
1 13 1 13 , The solution set is . 6 6
tan cos1 1 sin1 1 sin1 cos1
1 sin1 Therefore, tan 1 0.285 or cos1 1 sin1 tan 1 3.427 cos1 The solution set is 0.285,3.427 .
4sin 2 7 sin 2 0 Let u sin . Then,
2 3
1 1 12 6 1 13 6
sin cos1 cos sin1 cos
33. 4sin 2 7 sin 2
1 12 4 3 1
32. sin 1 cos
tan
4u 1 0 or u 2 0 4u 1 u 2 1 u 4 Substituting back in terms of , we have 1 or sin 2 sin 4 The second equation has no solution since 1 sin 1 for all values of . Therefore, we only need to find values of 1 between 0 and 2 such that sin . These will 4 occur in the first and second quadrants. Thus, 1 1 sin 1 0.253 and sin 1 2.889 . 4 4 The solution set is 0.253, 2.889 .
2. Line containing points (2,5) and (4, 1) : m
y2 y1 1 5 6 1 x2 x1 4 2 6
Using y y1 m( x x1 ) with point (4, 1) , y (1) 1 x 4 y 1 1 x 4 y 1 x 4 y x 3 or x y 3
4u 2 7u 2 0
4u 1 u 2 0 1027 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
Distance between points (2,5) and (4, 1) : d
x2 x1 y2 y1
4 2 1 52
2
the right 3 units and vertically up 2 units.
2
2
62 6 72 36 2 6 2 2
Midpoint of segment with endpoints (2,5) and (4, 1) : x1 x2 y1 y2 2 4 5 1 2 , 2 2 , 2 1, 2
5. y 3e x 2
Using the graph of y e x , stretch vertically by a factor of 3, and shift down 2 units.
3. 3x y 2 9
x-intercept: 3x 02 9 ; 3, 0 3x 9 x3 y-intercepts: 3 0 y 2 9 ; 0, 3 , 0,3 y2 9 y 3
Tests for symmetry: x-axis: Replace y with y : 3x y 9 2
3x y 2 9 Since we obtain the original equation, the graph is symmetric with respect to the x-axis.
6. y cos x 1 2 Using the graph of y cos x , horizontally shift
y-axis: Replace x with x : 3 x y 2 9 3x y 2 9 Since we do not obtain the original equation, the graph is not symmetric with respect to the y-axis.
to the right unit.
Origin: Replace x with x and y with y : 3 x y 9 2
3x y 2 9 Since we do not obtain the original equation, the graph is not symmetric with respect to the origin.
4. y x 3 2
Using the graph of y x , shift horizontally to
1028
Copyright © 2025 Pearson Education, Inc.
2
units, and vertically shift down 1
Chapter 8 Cumulative Review
7. a.
y x3 y
x
Inverse function: y 3 x y
c.
y sin x ,
y
x
x
2
2 , 1
2
2 , 1
x
Inverse function: y sin 1 x y
b.
1, 2
y ex y e
1 1, e
x
Inverse function: y ln x y e
x
1 e , 1
1029 Copyright © 2025 Pearson Education, Inc.
1, 2
x
Chapter 8: Analytic Trigonometry d.
y cos x , 0 x
1 2 2 sin(2 ) 2sin cos 2 3 3
c.
y
2 , 0
2
2 2 1 2 8 1 7 3 3 9 9 9
Inverse function: y cos 1 x
e.
y
cos(2 ) cos 2 sin 2
d. x
0, 2
4 2 9
3 3 , we have that . 2 2 2 4 1 1 Thus, lies in quadrant II and sin 0 . 2 2
Since
2 2 1 3 1 cos 1 sin 2 2 2
x
f.
Since
3 2 2 3 2 2 3 2 6
1 1 lies in quadrant II, cos 0 . 2 2
2 2 1 3 1 cos 1 cos 2 2 2 32 2 3 2 2 3 2 6
1 3 8. sin , , so lies in quadrant III. 3 2 a. In quadrant III, cos 0 1 cos 1 sin 2 1 3 1
b.
9. cos tan 1 2
2
Let tan 1 2 . Then tan
1 8 9 9
y 2 , x 1
. Let x 1 and y 2 . 2 2 Solve for r: r 2 x 2 y 2
2 2 3
r 2 12 22
1 sin 3 tan cos 2 2 3 1 3 1 2 3 2 2 2 2 4
r2 5 r 5 is in quadrant I.
cos tan 1 2 cos
1030
Copyright © 2025 Pearson Education, Inc.
x 1 1 5 5 r 5 5 5 5
Chapter 8 Cumulative Review 1 1 3 10. sin , ; cos , 3 2 3 2 a. Since , we know that lies in 2 quadrant II and cos 0 . cos 1 sin 2 2
1 8 1 1 1 9 9 3
1 2 3 The smaller of the two numbers is 3. Thus, every zero of f must lie between 3 and 3.
2
Use synthetic division with –1:
cos(2 ) cos sin 2
2
2
2 2 1 8 1 7 3 3 9 9 9 2
cos( ) cos cos sin sin 2 2 1 1 2 2 3 3 3 3 2 2 2 2 4 2 9 9 9 3 3 Since , we have that . 2 2 2 4
2
lies in quadrant II and sin
2
1 1 1 cos 3 sin 2 2 2
4 3 2
1 Max 0.5 , 1 , 1 , 2 , 0.5
1 8 2 2 1 9 9 3
Thus,
Possible rational zeros: p 1 1, p 1; q 1, 2; q 2 Using the Bounds on Zeros Theorem: f ( x) 2 x5 0.5 x 4 2 x3 x 2 x 0.5
Max 1, 5 5
3 , we know that lies in 2 quadrant III and sin 0 .
1 1 3
e.
f ( x ) has at most 5 real zeros.
Max 1, 0.5 1 1 2 0.5
sin 1 cos
d.
a.
a4 0.5, a3 2, a2 1, a1 1, a0 0.5
2
c.
f ( x) 2 x5 x 4 4 x3 2 x 2 2 x 1
2 2 3
b.
11.
4 2 2 6 6 6 6 3 6
0.
1 2 1 4 2
2 1
2
1 3
1
2 3 1 3 1
0
3
Since the remainder is 0, x 1 x 1 is a factor. The other factor is the quotient: 2 x 4 3 x3 x 2 3x 1 . Use synthetic division with 1 on the quotient: 1 2 3 1 3
1
2 1 2
1
2 1 2 1
0
Since the remainder is 0, x 1 is a factor. The other factor is the quotient: 2 x3 x 2 2 x 1 . Factoring: 2 x3 x 2 2 x 1 x 2 2 x 1 1 2 x 1
2 x 1 x 2 1
2 x 1 x 1 x 1
Therefore, f x 2 x 1 x 1 x 1 2
2
2 2 1 2 x x 1 x 1 2
1031 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
The real zeros are 1 and 1 (both with 1 multiplicity 2) and (multiplicity 1). 2 1 b. x-intercepts: 1, , 1 2 y-intercept: 1 1 The intercepts are (0, 1) , (1, 0) , , 0 , 2 and (1, 0) c.
g.
f resembles the graph of y 2 x5 for large
f is increasing on , 1 , 0.29, 0.69 ,
and 1, . f is decreasing on 1, 0.29
x .
and 0.69,1 .
d. Let Y1 2 x 5 x 4 4 x 3 2 x 2 2 x 1 12.
f ( x) 2 x 2 3 x 1 ; g ( x) x 2 3x 2
f ( x) 0
a.
2 x 3x 1 0 2
(2 x 1)( x 1) 0 1 x or x 1 2
e.
f.
The solution set is 1,
Four turning points exist. Use the MAXIMUM and MINIMUM features to locate local maxima at 1, 0 , 0.69, 0.10
1 . 2
f ( x) g ( x)
b.
and local minima at 1, 0 , 0.29, 1.33 .
2 x 3x 1 x 2 3x 2
To graph by hand, we determine some additional information about the intervals between the xintercepts:
x2 1 0 ( x 1)( x 1) 0 x 1 or x 1 The solution set is 1, 1 .
Interval Test number Value of f Location Point
, 1 1, 0.5 0.5,1
1,
2
2
45
0 1
Below Below x-axis x-axis 2, 45 0, 1
0.7 0.1
2
f ( x) 0
c.
2 x 3x 1 0 2
27
Above Above x-axis x-axis 0.7, 0.1 2, 27
f is above the x-axis for 0.5,1 and
(2 x 1)( x 1) 0
f ( x) 2 x 1 x 1
The zeros of f are x
1, , and below the x-axis for , 1 and 1,0.5 .
1032
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1 and x 1 2
Chapter 8 Projects
Interval
, 1 1, 2 2 , 1
1
Test number
2
0.75
0
Value of f
3
0.125
1
Conclusion
Positive Negative Positive
e. If x 1 , the resulting equation is y 0.00421sin(68.3 2.68t ) . To graph, let Y1 0.00421sin(68.3 2.68 x) .
1 The solution set is , 1 , . 2
f ( x) g ( x)
d.
f. Note: (kx t ) (kx t ) 2kx 2t and (kx t ) (kx t ) .
2 x 3x 1 x2 3x 2 2
x2 1 0 ( x 1)( x 1) 0
y1 y2 ym sin(kx t ) ym sin( kx t ) ym [sin(kx t ) sin(kx t )]
p x x 1 x 1
The zeros of p are x 1 and x 1 . Interval Test number
, 1
1,1
1,
2
0
2
Value of p
3
1
3
Conclusion
Positive Negative Positive
The solution set is , 1 1, .
g. ym 0.0045 , 2.5 , 0.09 , f 2.3 Let x 1 : 2 0.09 f 2.3 k 2 200 4.6 14.45 k 69.8 9
0.0045sin(69.8 14.45t )
y2 ym sin(kx t )
Project II
0.0045sin(69.8 1 14.45t 2.5)
a. Amplitude = 0.00421 m
0.0045sin(72.3 14.45t )
b. 2.68 radians/sec
d.
2kx 2 wt cos 2 ym sin 2 2
0.0045sin(69.8 1 14.45t )
Project I – Internet-based Project
f
2kx 2 wt ym 2sin cos 2 2
y1 ym sin(kx t )
Chapter 8 Projects
c.
2
2.68 0.4265 vibrations/sec 2
2 2 0.09199 m k 68.3
2kx 2t y1 y2 2 ym sin cos 2 2 2 69.8 1 2 14.45t 2.5 2.5 2 0.0045sin cos 2 2 142.1 28.9t 0.009sin cos(1.25) 2 0.009sin 71.05 14.45t cos(1.25)
h. Let Y1 0.0045sin(69.8 14.45 x) , Y2 0.0045sin(72.3 14.45 x) , and
Y3 0.009sin 71.05 14.45 x cos(1.25) .
1033 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
Project III
y1 y1 y2
y
a.
h
y2
x
h
i. ym 0.0045 , 0.4 , 0.09 , f 2.3 Let x 1 : 2 0.09 f 2.3 k 2 200 4.6 14.45 k 69.8 9 y1 0.0045sin(69.8 14.45t ) y2 ym sin(kx t )
b. Let Y1 1
0.0045sin(70.2 14.45t )
c. Let Y1 1
2kx 2t y1 y2 2 ym sin cos 2 2 2 69.8 1 2 14.45t 0.4 0.4 2 0.0045sin cos 2 2 140 28.9t 0.009sin cos(0.2) 2 0.009sin 70 14.45t cos(0.2)
sin(17 x) 4 sin x sin(3x ) ... 3 17 1
d. Let Y1 1
Let Y1 0.0045sin(69.8 14.45 x) , Y2 0.0045sin(70.2 14.45 x) , and
Y3 0.009sin 70 14.45 x cos(0.2) .
y1 y2 y1
4 sin x
1
sin(3x ) sin(37 x) ... 3 37
y2
e. The best one is the one with the most terms.
0.0045sin(69.8 1 14.45t 0.4)
4 sin x sin(3x ) sin(5 x ) sin(7 x) 3 5 7 1
j. The phase shift causes the amplitude of y1 y2 to increase from 0.009 cos(1.25) 0.003 to 0.009 cos(0.2) 0.009 .
1034
Copyright © 2025 Pearson Education, Inc.
Chapter 8 Projects
The shape looks like a sinusoidal graph.
Project IV a.
f ( x) sin x (see table column 2)
x 0
f ( x) 0 1 2 2 2 3 2
6
4
3
1
2 2 3 3 4 5 6
3 2 2 2 1 2 0 1 2 2 2 3 2
7
g ( x) 0.954
h( x ) k ( x) 0.311 0.749
m( x ) 6.085
0.791
0.703
2.437
4.011
0.607
1.341 1.387
3.052
0.256
0.978
0.588
1.243
0.256 0.670 0.063
0.413
0.607 0.703
0.153
8.507
0.791 0.623
2.380
6.822
0.954
0.594
2.695
0
0.954
0.311 0.817 1.536
0.791 6 5 0.607 4 4 0.256 3 3 0.256 1 2 5 3 0.607 3 2 7 2 0.791 4 2 11 1 0.954 6 2 2 1
0.117 0.013 5.248
b. g ( x)
1.341
1.387
0.978
0.588 1.243
0.670
0.063
0.703
0.306
Rounding a, b, c, and d to the nearest tenth, we have that y sin( x 1.8) . Barring error due to rounding and approximation, this looks like y cos x g xi 1 g xi (see table column 4) xi 1 xi
d. h( x)
The shape is sinusoidal. It looks like an upsidedown sine wave.
3.052
0.705
0.623
Rounding a, b, c, and d to the nearest tenth, we have that y 0.5sin(6.4 x ) . e. k ( x)
h xi 1 h xi (see table column 5) xi 1 xi
f xi 1 f xi (see table column 3) xi 1 xi
This curve is losing its sinusoidal features, although it still looks like one. It takes on the features of an upside-down cosine curve
c.
. Rounding a, b, c, and d to the nearest tenth, we have that y 0.8sin(1.1x) 0.3 . Note: The rounding error is getting greater and greater.
1035 Copyright © 2025 Pearson Education, Inc.
Chapter 8: Analytic Trigonometry
f. m( x)
k xi 1 k xi (see table column 6) xi 1 xi
The sinusoidal features are gone.
Rounding a, b, c, and d to the nearest tenth, we have that y 2.1sin(5.1x 1.5) 0.6 . g. It would seem that the curves would be less “involved,” but the rounding error has become incredibly great that the points are nowhere near accurate at this point in calculating the differences.
1036
Copyright © 2025 Pearson Education, Inc.
Chapter 9 Applications of Trigonometric Functions Section 9.1
b c 5 sin 20º c sin B
1. a 652 632 4225 3969 256 16 2. False: sin 52 cos 38
c
3. s r (5)(2.7) 13.5 ft.
5 5 14.62 sin 20º 0.3420
A 90º B 90º 20º 70º
1 4. tan , 0 90 2 1 tan 1 26.6o 2
12. b 4, B 10º b a 4 tan 10º a tan B
1 5. sin , 0 90 2 1 sin 1 30 2 tan 1, 0 90 1 tan 1 45 2
a
4 4 22.69 tan 10º 0.1763
b c 4 sin 10º c sin B
y 3 r 5 y 3 tan x 4
c
6. sin
4 4 23.04 sin 10º 0.1736
A 90º B 90º 10º 80º
13. a 6, B 40º
7. b
b a b tan 40º 6 b 6 tan 40º 6 (0.8391) 5.03 tan B
8. direction; bearing 9. True 10. False
a c 6 cos 40º c cos B
11. b 5, B 20º b a 5 tan 20º a tan B
a
c
5 5 13.74 tan 20º 0.3640
6 6 7.83 0.7660 cos 40º
A 90º B 90º 40º 50º
1037 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions 17. a 5, A 25º
14. a 7, B 50º tan B
b a b tan 50º 7 b 7 tan 50º 7 (1.1918) 8.34
b a b cot 25º 5 b 5cot 25º 5 2.1445 10.72
a c 7 cos 50º c
c a c csc 25º 5 c 5csc 25º 5 2.3662 11.83
cot A
cos B
c
csc A
7 7 10.89 cos 50º 0.6428
B 90 A B 90º A 90 25 65
A 90 B 90 50 40
18. a 6, A 40º
15. b 7, A 14º a b a tan 14º 7 a 7 tan 14º 7 (0.2493) 1.75
a b b cot 40º 6 b 6 cot 40º 6 1.1918 7.15
b c 7 cos 14º c
c a c csc 45º 6 c 6 csc 40º 6 1.5557 9.33
cot A
tan A
csc A
cos A
c
7 7 7.21 cos 14º 0.9703
B 90 A 90 40 50
B 90º A 90º 14º 76º
19. c 9, B 20º b c b sin 20º 9 b 9sin 20º 9 (0.3420) 3.08
16. b 6, A 20º
sin B
a b a tan 20º 6 a 6 tan 20º 6 (0.3640) 2.18 tan A
a c a cos 20º 9 a 9 cos 20º 9 (0.9397) 8.46 cos B
b c 6 cos 20º c cos A
c
6 6 6.39 cos 20º 0.9397
A 90 A 90 20 70
20. c 10, A 40º
B 90º A 90º 20º 70º
a c a sin 40º 10 a 10sin 40º 10 (0.6428) 6.43 sin A
1038 Copyright © 2025 Pearson Education, Inc.
Section 9.1: Applications Involving Right Triangles
b c b cos 40º 10 b 10 cos 40º 10 (0.7660) 7.66
B 90º A 90º 48.2º 41.8º
cos A
25. c 5, a 2 sin A
2 A sin 1 23.6 5 B 90º A 90º 23.6º 66.4º The two angles measure about 23.6 and 66.4º .
B 90 A 90 40 50
21. a 5, b 3 c 2 a 2 b 2 52 32 25 9 34 c 34 5.83
2 5
26. c 3, a 1 B 90º A 90º 19.5º 70.5º The two angles measure about 19.5 and 70.5º .
a 5 b 3 5 A tan 1 59.0 3
tan A
27. c 8, 35º
B 90 A 90 59.0 31.0
22. a 2, b 8
a
35º
b
c 2 a 2 b 2 22 82 4 64 68
a 2 1 b 8 4 1 A tan 1 14.0 4
tan A
B 90 A 90 14.0 76.0
a 8 a 8sin 35º
b 8 b 8cos 35º
8(0.5736)
8(0.8192)
4.59 in.
6.55 in.
sin 35º
c 68 8.25
cos 35º
28. c 10, 40º
23. a 3, c 11
c 2 a 2 b2
a
40º
b 2 c 2 a 2 112 32 121 9 112
b
b 112 10.58
a 10 a 10sin 40º
b 10 b 10 cos 40º
10(0.6428)
10(0.7660)
6.43 cm.
7.66 cm.
sin 40º
a 3 c 11 3 A sin 1 15.8º 11
sin A
cos 40º
B 90 A 90 15.8 74.2
29.
24. b 4, c 6 c 2 a 2 b2 a 2 c 2 b 2 62 42 36 16 20 a 20 4.47
300 6 50 A tan 1 6 80.5º The angle of elevation of the sun is about 80.5º . tan A
b 4 2 c 6 3 2 A tan 1 48.2º 3
cos A
1039 Copyright © 2025 Pearson Education, Inc.
300 feet A 50 feet
Chapter 9: Applications of Trigonometric Functions 30. opposite side = 10 feet, adjacent side = 35 feet 10 tan 35 10 tan 1 15.9º 35 31. a.
b.
x 1.44 ft.
34. a.
Let x represent the distance the truck traveled in the 1 second time interval. 30 tan 15º x 30 30 x 111.96 feet tan 15º 0.2679
tan 20º
x
b.
35. a.
30 x
4.22 5.9 1012 24.898 1012 2.4898 1013 Proxima Centauri is about 2.4898 1013 miles from Earth.
30 30 82.42 feet tan 20º 0.3640
b. Construct a right triangle using the sun, Earth, and Proxima Centauri as shown. The hypotenuse is the distance between Earth and Proxima Centauri. b Sun Proxima Centauri
A ticket is issued for traveling at a speed of 60 mi/hr or more. 60 mi 5280 ft 1hr 88 ft/s. hr mi 3600 sec 30 If tan , the trooper should issue a 88 30 ticket. Now, tan 1 18.8 , so a ticket 88 is issued if 18.8º .
a
Parallax
c Earth
a 9.3 107 c 2.4898 1013 sin 0.000003735 sin
32. If the camera is to be directed to a spot 6 feet above the floor 12 feet from the wall, then the “side opposite” the angle of depression is 3 feet. (see figure) 12 3 1 tan A A 12 4 3 1 A tan 1 14.0 4 The angle of depression should be about 14.0 . 33. a.
x 5 5sin 5.7 x sin 5.7
x 0.50 ft.
The truck is traveling at 82.42 ft/s, or 82.42 ft 1 mile 3600 sec 56.2 mi/h . sec 5280 ft hr c.
tan c k tan c 0.1
c tan 1 0.1 5.7
The truck is traveling at 111.96 ft/s, or 111.96 ft 1 mile 3600 sec 76.3 mi/h . sec 5280 ft hr b.
x 5 5sin16.7 x sin16.7
sin 1 0.000003735 0.000214 The parallax of Proxima Centauri is 0.000214 . 36. a.
11.14 5.9 1012 65.726 1012 6.5726 1013 61 Cygni is about 6.5726 1013 miles from Earth.
tan c s tan c 0.3
b. Construct a right triangle using the sun, Earth, and 61 Cygni as shown. The hypotenuse is the distance between Earth and 61 Cygni.
c tan 1 0.3 c 16.7
1040
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Section 9.1: Applications Involving Right Triangles
b
Sun
a
61 Cygni
39. Let y = the height of the embankment.
51
Parallax
c
.8
y
o
36 2’ 362 ' 36.033
Earth
a 9.3 107 sin c 6.5726 1013 sin 0.000001415
y 51.8 y 51.8sin 36.033 30.5 meters
sin 36.033
sin 1 0.000001415 0.000081 The parallax of 61 Cygni is 0.000081 .
The embankment is about 30.5 meters high. 40. a.
37. Begin by finding angle BAC : (see figure) 40º
i
B m
D
Let h = the height of the building, a = the height of the antenna, and x = the distance between the surveyor and the base of the building.
m
a
i
E
2760
C
95 25
1 tan 2 0.5 63.4º DAC 40º 63.4º 103.4º
o
34
EAC 103.4º 90º 13.4º Now, 90º 13.4º 76.6º The control tower should use a bearing of S76.6˚E.
80º
x 2595 x 2595 cos 34 2151 The surveyor is located approximately 2151 feet from the building.
b.
C
c.
B
CMA 80
30 2 15 AMB tan 1 2 63.4º 80º 63.4º 16.6º The bearing of the ship from port is S16.6ºE . tan AMB
x
cos 34
38. Find AMB and subtract from 80˚ to obtain (see figure). M
h
h 2595 h 2595 sin 34 1451 The building is about 1451 feet high. sin 34
Let = the angle of inclination from the surveyor to the top of the antenna. 2151 cos 2760 2151 38.8 cos 1 2760
1041 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions
d.
10 6 4 15 15 1 4 A tan 14.9º 15 The angle of elevation from the player’s eyes to the center of the rim is about 14.9º .
h a 1451 a 2760 2760 2760 sin 1451 a a 2760 sin 1451 sin
44. tan A
2151 1451 a 2760sin cos 1 2760 a 278
45. A line segment drawn from the vertex of the angle through the centers of the circles will bisect (see figure). Thus, the angle of the two
The antenna is about 278 feet tall. 41. Let x = the distance between the buildings.
right triangles formed is
1451 10.3
x
o
2
.
x 1451 x 1451 7984 x tan 10.3 The two buildings are about 7984 feet apart. tan
42. tan
Let x = the length of the segment from the vertex of the angle to the smaller circle. Then the hypotenuse of the smaller right triangle is x 2 , and the hypotenuse of the larger right triangle is x 2 2 4 x 8 . Since the two right triangles are similar, we have that: x 2 x8 2 4 4( x 2) 2( x 8) 4 x 8 2 x 16 2x 8 x4 Thus, the hypotenuse of the smaller triangle is 2 1 4 2 6 . Now, sin , so we have 2 6 3
2070 630
2070 630 2070 tan 1 630 2070 1 67 tan 1 cot 630 55 33.69 Let x = the distance between the Arch and the boat on the Missouri side. x tan 630 x 630 tan 630 tan 33.69
tan 1
that:
x 420 Therefore, the Mississippi River is approximately 2070 420 1650 feet wide at the St. Louis riverfront.
46. a.
43. The height of the beam above the wall is 46 20 26 feet. 26 tan 2.6 10 tan 1 2.6 69.0º The pitch of the roof is about 69.0º .
1 sin 1 3 1 2 sin 1 38.9 3 2
3960 cos 2 3960 h
b.
d 3960
c.
3960 d cos 7920 3960 h
1042
Copyright © 2025 Pearson Education, Inc.
Section 9.1: Applications Involving Right Triangles
d.
e.
3960 2500 cos 7920 3960 h 3960 0.9506 3960 h 0.9506(3960 h) 3960 3764 0.9506h 3960 0.9506h 196 h 206 miles 3960 3960 d cos 7920 3960 300 4260 d 3960 cos 1 7920 4260 3960 d 7970 cos 1 4260 2990 miles
47. Extend the tangent line until it meets a line extended through the centers of the pulleys at P. Let x d ( P, B ), y the distance from P to the center of the smaller circle, and x d ( A, B ) . . Using similar triangles gives
The arc length, s1 , where the belt touches the top half of the smaller pulley is 2.5(1.4033) , 3.51 inches. The length of the belt is about: 2(11.30 + 3.51 + 23.66) = 76.94 inches. 48. Let x = the hypotenuse of the larger right triangle and z = length of its third side. Then 24 x is the hypotenuse of the smaller triangle and let y be its third side. The two triangles are similar so
6.5 x 20 which yields 24 x . 2.5 24 x 3 cos
3 8
cos 1
6.5 3 cos 1 1.1864 rad x 8
1.1864 1.9552 rad z 6.5 tan 16.07 in; y 2.5 tan 6.18 in;
24 y 6.5 which yields y 15 . y 2.5
Use the Pythagorean Theorem twice to find x and z: x 2 2.52 152 x 14.79 ( z 14.79) 2 6.52 24 15
2
z 23.66 2.5 0.1667 cos 15 2.5 cos 1 1.4033 radians 15 1.4033 1.7383 radians
The arc length, s2 , where the belt touches the top half of the larger pulley is 6.5(1.7383) , 11.30 inches.
Distance between points of tangency = z y 22.25 The arc length, s2 , where the belt touches the top half of the larger pulley is: 6.5 1.9552 12.71 in. The arc length, s2 , where the belt touches the bottom half of the smaller pulley is 2.5 1.9552 4.89 in. The distance between the points of tangency is z y 16.07 6.18 22.25 inches. The length of the belt is about: 2s2 2( z y ) 2s1 2(12.71 + 4.89 + 22.25) = 79.69 in. 49. Let = the central angle formed by the top of the lighthouse, the center of the Earth and the point P on the Earth’s surface where the line of sight from the top of the lighthouse is tangent to 362 miles. the Earth. Note also that 362 feet 5280
1043 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions
Let = the central angle formed by 40 nautical o 40 2 miles then. 60 3 3960 cos( ) 3960 x 3960 cos (2 / 3) 0.33715 3960 x 3960 x cos (2 / 3) 0.33715 3960 3960 x
cos 1
3960 o 0.33715 3960 362 / 5280
x
Verify the airplane information: Let = the central angle formed by the plane, the center of the Earth and the point P. 3960 cos 1 1.77169 3960 10, 000 / 5280 Note that d d tan 1 and tan 2 3690 3690 d1 3960 tan d 2 3960 tan
3960 cos (2 / 3) 0.33715 3960 3960 cos (2 / 3) 0.33715
0.06549 miles ft 0.06549 mi 5280 mi 346 feet Therefore, a ship that is 346 feet above sea level can see the lighthouse from a distance of 40 nautical miles.
50. 3 1
So, d1 d 2 3960 tan 3960 tan
1
2
21
19
3
3
15 18
3
5
6
0
1
Since the remainder is 0 then ( x 3) is a factor
3960 tan(0.33715) 3960 tan(1.77169) 146 miles To express this distance in nautical miles, we express the total angle in minutes. That is,
of x 4 2 x3 21x 2 19 x 3 . 51. sin 12 sin 4 6
0.33715o 1.77169o 60 126.5
sin 4 cos 6 cos 4 sin 6
nautical miles. Therefore, a plane flying at an altitude of 10,000 feet can see the lighthouse 120 miles away.
2 3 2 1 2 2 2 2 6 2 4 4 1 6 2 4
Verify the ship information:
1044
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Section 9.2: The Law of Sines
52.
53.
Section 9.2
f (5) f (4) 52 0.236 54 1 f (4.5) f (4) 4.5 2 0.243 4.5 4 0.5 f (4.1) f (4) 4.1 2 0.248 4.1 4 0.1
1. sin A cos B cos A sin B 2. sin A A
2sin 2 sin 1 0 2sin 1 0 or sin 1 0
7 11 , , 2 6 6
.
8 x 42
8. False
42 21 5.25 8 4
(e 4 x 2e 2 x 1) 1 (e 4 x 2e 2 x 1)
57. Using the Remainder Theorem we find P (2) 2(2) 4 3(2)3 (2) 7 65 The remainder is 65.
58.
5 , 6 6
9. False: You must have at least one angle opposite one side.
(e 2 x 1) 2 (2e x ) 2 (e 4 x 2e2 x 1) (4e2 x ) (e 2 x 1) 2 (e 4 x 2e 2 x 1)
.
sin A sin B sin C a b c
7. d
55. Since the polynomial has 4 roots the remaining zero would be the conjugate of 3 5 which is 3 5 . 56.
5 6
5. a
x 14 3 8 8 x 14(3) x
or
4. sin 1 (0.76) 49.5
6. 54.
6
3. sin(40) 0.64 sin(80) 0.98
1 or sin 1 2 7 11 , , 2 6 6
sin
The solution set is
(2sin 1)(sin 1) 0
So the solution set is
1 2
.
( x h) 2 ( y k ) 2 r 2 ( x (4)) 2 ( y 0) 2 ( 5)2
10. ambiguous case 11. c 5, B 45º , C 95º A 180º B 180º 45º 95º 40º sin A sin C a c sin 40º sin 95º a 5 5sin 40º a 3.23 sin 95º sin B sin C b c sin 45º sin 95º b 5 5sin 45º b 3.55 sin 95º
12. c 4, A 45º , B 40º C 180º A B 180º 45º 40º 95º
( x 4) 2 y 2 5
59. The domain is all real numbers or , .
1045 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions 15. b 7, A 40º , B 45º C 180º A B 180º 40º 45º 95º
sin A sin C a c sin 45º sin 95º 4 a 4sin 45º a 2.84 sin 95º
sin A sin B a b sin 40º sin 45º a 7 7 sin 40º a 6.36 sin 45º
sin B sin C b c sin 40º sin 95º 4 b 4sin 40º b 2.58 sin 95º
sin C sin B c b sin 95º sin 45º 7 c 7 sin 95º c 9.86 sin 45º
13. b 3, A 50º , C 85º B 180º A C 180º 50º 85º 45º sin A sin B a b sin 50º sin 45º 3 a 3sin 50º a 3.25 sin 45º
16. c 5, A 10º , B 5º C 180º A B 180º 10º 5º 165º sin A sin C a c sin10º sin165º 5 a 5sin10º a 3.35 sin165º
sin C sin B c b sin 85º sin 45º 3 c 3sin 85º c 4.23 sin 45º
sin B sin C b c sin 5º sin165º 5 b 5sin 5º b 1.68 sin165º
17. b 2, B 40º , C 100º A 180º B C 180º 40º 100º 40º
14. b 10, B 30º , C 125º A 180º B C 180º 30º 125º 25º
sin A sin B a b sin 40º sin 40º 2 a 2sin 40º a 2 sin 40º
sin A sin B a b sin 25º sin 30º 10 a 10sin 25º a 8.45 sin 30º
sin C sin B c b sin100º sin 40º 2 c 2sin100º c 3.06 sin 40º
sin C sin B c b sin125º sin 30º 10 c 10sin125º c 16.38 sin 30º
1046
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Section 9.2: The Law of Sines 18. b 6, A 100º , B 30º C 180º A B 180º 100º 30º 50º
21. B 64º , C 47º , b 6 A 180º B C 180º 64º 47º 69º
sin A sin B a b sin100º sin 30º 6 a 6sin100º a 11.82 sin 30º
sin A sin B a b sin 69º sin 64º a 6 6sin 69º a 6.23 sin 64º
sin C sin B c b sin 50º sin 30º 6 c 6sin 50º c 9.19 sin 30º
sin C sin B c b sin 47º sin 64º c 6 6sin 47º c 4.88 sin 64º
19. A 55º , B 25º , a 4 C 180º A B 180º 55º 25º 100º
22. A 70º , B 60º , c 4 C 180º A B 180º 70º 60º 50º
sin A sin B a b sin 55º sin 25º 4 b 4sin 25º b 2.06 sin 55º
sin A sin C a c sin 70º sin 50º 4 a 4sin 70º a 4.91 sin 50º
sin C sin A c a sin100º sin 55º 4 c 4sin100º c 4.81 sin 55º
sin B sin C b c sin 60º sin 50º 4 b 4sin 60º b 4.52 sin 50º
20. A 50º , C 20º , a 3 B 180º A C 180º 50º 20º 110º
23. A 110º , C 30º , c 3 B 180º A C 180º 110º 30º 40º
sin A sin B a b sin 50º sin110º 3 b 3sin110º b 3.68 sin 50º
sin A sin C a c sin110º sin 30º a 3 3sin110º a 5.64 sin 30º
sin C sin A c a sin 20º sin 50º 3 c 3sin 20º c 1.34 sin 50º
sin C sin B c b sin 30º sin 40º 3 b 3sin 40º b 3.86 sin 30º
1047 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions 24. B 10º , C 100º , b 2 A 180º B C 180º 10º 100º 70º
27. a 3, b 2, A 50º sin B sin A b a sin B sin 50º 2 3 2sin 50º sin B 0.5107 3
sin A sin B a b sin 70º sin10º 2 a 2sin 70º a 10.82 sin10º
B sin 1 0.5107
sin C sin B c b sin100º sin10º 2 c 2sin100º c 11.34 sin10º
B 30.7º or B 149.3º The second value is discarded because A B 180º . C 180º A B 180º 50º 30.7º 99.3º sin C sin A c a sin 99.3º sin 50º 3 c 3sin 99.3º c 3.86 sin 50º One triangle: B 30.7º , C 99.3º , c 3.86
25. A 40º , B 40º , c 2 C 180º A B 180º 40º 40º 100º sin A sin C a c sin 40º sin100º 2 a 2sin 40º a 1.31 sin100º
28. b 4, c 3, B 40º sin B sin C b c sin 40º sin C 4 3 3sin 40º 0.4821 sin C 4
sin B sin C b c sin 40º sin100º 2 b 2sin 40º b 1.31 sin100º
C sin 1 0.4821 C 28.8º or C 151.2º The second value is discarded because B C 180º . A 180º B C 180º 40º 28.8º 111.2º
26. B 20º , C 70º , a 1 A 180º B C 180º 20º 70º 90º sin A sin B a b sin 90º sin 20º 1 b 1sin 20º b 0.34 sin 90º
sin B sin A b a sin 40º sin111.2º 4 a 4sin111.2º a 5.80 sin 40º One triangle: A 111.2º , C 28.8º , a 5.80
sin C sin A c a sin 70º sin 90º c 1 1sin 70º c 0.94 sin 90º
1048
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Section 9.2: The Law of Sines 29. b 9, c 4, B 115º
31. a 7, b 14, A 30º sin B sin A b a sin B sin 30º 14 7 14sin 30º sin B 1 7
sin B sin C b c sin115º sin C 9 4 4sin115º sin C 0.4028 9 C sin 1 0.4028
B sin 1 1
C 23.8º or C =156.2º The second value is discarded because B C 180º . A 180º B C 180º 115º 23.8º 41.2º sin B sin A b a sin115º sin 41.2º 9 a 9sin 41.2º a 6.55 sin115º One triangle: A 41.2º , C 23.8º , a 6.55
B 90º There is one solution. C 180º A B 180º 30º 90º 60º sin C sin A c a sin 60º sin 30º c 7 7 sin 60º c 12.12 sin 30º
One triangle: B 90º , C 60º , c 12.12 32. b 2, c 3, B 40º
30. a 2, c 1, A 120º
sin B sin C b c sin 40º sin C 2 3 3sin 40º 0.9642 sin C 2 C sin 1 0.9642
sin C sin A c a sin C sin120º 1 2 1sin120º 0.4330 sin C 2
C sin 1 0.4330 C 25.7º or C =154.3º The second value is discarded because A C 180º . B 180º A C 180º 120º 25.7º 34.3º
C1 74.6º or C2 105.4º For both values, B C 180º . Therefore, there are two triangles. A1 180º B C1 180º 40º 74.6º 65.4º sin B sin A1 b a1
sin B sin A b a sin 34.3º sin120º b 2 2sin 34.3º b 1.30 sin120º
sin 40º sin 65.4º 2 a1
One triangle: B 34.3º , C 25.7º , b 1.30
A2 180º B C2 180º 40º 105.4º 34.6º
a1
2sin 65.4º 2.83 sin 40º
sin B sin A2 b a2 sin 40º sin 34.6º 2 a2 a2
1049 Copyright © 2025 Pearson Education, Inc.
2sin 34.6º 1.77 sin 40º
Chapter 9: Applications of Trigonometric Functions
Two triangles: A1 65.4º , C1 74.6º , a1 2.83 or A2 34.6º , C2 105.4º , a2 1.77
8sin125º 2.1844 3 There is no angle A for which sin A 1 . Thus, there is no triangle with the given measurements. sin A
33. b 4, c 6, B 20º
36. b 4, c 5, B 95º
sin B sin C b c sin 20º sin C 4 6 6sin 20º sin C 0.5130 4 C sin 1 0.5130
sin C sin B c b sin C sin 95º 5 4 5sin 95º 1.2452 sin C 4 There is no angle C for which sin C 1 . Thus, there is no triangle with the given measurements.
C1 30.9º or C2 149.1º For both values, B C 180º . Therefore, there are two triangles. A1 180º B C1 180º 20º 30.9º 129.1º
37. a 7, c 3, C 12º sin A sin C a c sin A sin12º 7 3 7 sin12º sin A 0.4851 3 A sin 1 0.4851
sin B sin A1 b a1 sin 20º sin129.1º 4 a1 a1
4sin129.1º 9.07 sin 20º
A1 29.0º or A2 151.0º For both values, A C 180º . Therefore, there are two triangles.
A2 180º B C2 180º 20º 149.1º 10.9º sin B sin A2 b a2
B1 180º A1 C 180º 29º 12º 139º
sin 20º sin10.9º 4 a2 a2
sin B1 sin C b1 c
4sin10.9º 2.20 sin 20º
sin139º sin12º 3 b1
Two triangles: A1 129.1º , C1 30.9º , a1 9.07 or A2 10.9º , C2 149.1º , a2 2.20
b1
3sin139º 9.47 sin12º
B2 180º A2 180º 151º 12º 17º
34. a 3, b 7, A 70º
sin B2 sin C b2 c
sin B sin 70º 7 3 7 sin 70º sin B 2.1926 3 There is no angle B for which sin B 1 . Thus, there is no triangle with the given measurements.
sin17º sin12º b2 3
b2
3sin17º 4.22 sin12º
Two triangles: A1 29.0º , B1 139.0º , b1 9.47 or A2 151.0º , B2 17.0º , b2 4.22
35. a 8, c 3, C 125º sin C sin A c a sin125º sin A 3 8
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Section 9.2: The Law of Sines 38. b 4, c 5, B 40º
41. Let h = the height of the plane and x = the distance from Q to the plane (see figure).
sin B sin C b c sin 40º sin C 4 5 5sin 40º 0.8035 sin C 4 C sin 1 0.8035
C1 53.5º or C2 126.5º For both values, B C 180º . Therefore, there are two triangles. A1 180º B C1 180º 40º 53.5º 86.5º sin B sin A1 b a1
h h x 793.07 h 793.07 sin 25º 335 .16 feet The plane is about 335.16 feet high.
4sin 86.5º 6.21 sin 40º
A2 180º B C2 180º 40º 126.5º 13.5º sin B sin A2 b a2
42. Let h = the height of the bridge, x = the distance from C to point A (see figure).
A
sin 40º sin13.5º a2 4
a2
sin 50º sin105º x 1000 1000sin 50º x 793.07 feet sin105º sin 25º
sin 40º sin 86.5º 4 a1
a1
PRQ 180º 50º 25º 105º
B
ft 69.2º
65.5º
h
x
4sin13.5º 1.45 sin 40º
Two triangles: A1 86.5º , C1 53.5º , a1 6.21 or A2 13.5º , C2 126.5º , a2 1.45 39. QPR 180º 25º 155º PQR 180º 155º 15º 10º Let c represent the distance from P to Q. sin15º sin10º 1000 c 1000sin15º c 1490.48 feet sin10º 40. From Problem 39, we have that the distance from P to Q is 1490.48 feet. Let h represent the distance from Q to D. h sin 25º 1490.48 h 1490.48sin 25º 629.90 feet
C ACB 180º 69.2º 65.5º 45.3º
sin 65.5º sin 45.3º 880 x 880sin 65.5º x 1126.57 feet sin 45.3º h h x 1126.57 h 1126.57 sin 69.2º 1053.15 feet The bridge is about 1053.15 feet high. sin 69.2º
43. Let A = the angle opposite the road. Then angle A 180º 50º 43º 87º . Let x equal the side opposite the 50º angle and y equal the side opposite the 43º angle. Using the Law of Sines, we can solve for x and y.
Equation 1:
1051 Copyright © 2025 Pearson Education, Inc.
sin 87º sin 50º 200 x 200sin 50º x 153.42 ft. sin 87º
Chapter 9: Applications of Trigonometric Functions
Set the two equations equal to each other and solve for h.
sin 87º sin 43º 200 y 200sin 43º y 136.59 ft. sin 87º
Equation 2:
h sin 60º h sin 70º 40 sin 30º sin 20º sin 60º sin 70º h 40 sin 30º sin 20º 40 h sin 60º sin 70º sin 30º sin 20º 39.39 feet
44. Let x = the distance between the runners. First we must calculate h, the distance between runner A and the helicopter. Then we will calculate the two remaining angles in the left triangle (see figure), angle B, the supplementary angle to runner B and C, the top angle. Then we will use the Law of Siner to solve for x.
h 38º
C B
The height of the tree is about 39.39 feet. 46. Let x = the length of the new ramp (see figure).
1700
45º
x
x
1700 Solve for h: sin 38º h 1700 h 2761.26 sin 38º
12º 168º
sin 7º sin135º x 2761.26 2761.26sin 7º x sin135º 475.90 feet
47. Note that KOS 57.7 29.6 28.1 (See figure) 79.4o K
The distance between the runners is about 475.90 feet.
S
29.6o
iles 1m . 1 46
45. Let h = the height of the tree, and let x = the distance from the first position to the center of the tree (see figure).
ft
Equation 2:
o
O From the diagram we find that KSO 180 79.4 57.7 42.9 and OKS 180 28.1 42.9 109.0 . We can use the Law of Sines to find the distance between Oklahoma City and Kansas City, as well as the distance between Kansas City and St. Louis. sin 42.9 sin109.0 OK 461.1 461.1sin 42.9 OK 332.0 sin109.0
x
Using the Law of Sines twice yields two equations relating x and h. Equation 1:
57.7
28.1o
10º 60º h 30º
h
Using the Law of Sines: sin162º sin12º 10 x 10sin162º x 14.86 feet sin12º The new ramp is about 14.86 feet long.
Solve for B,C: B 180º 45º 135º C 180º 38º 135º 7º
20º
ft 18º
sin 30º sin 60º h x h sin 60º x sin 30º sin 20º sin 70º h x 40 h sin 70º x 40 sin 20º
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Section 9.2: The Law of Sines sin 28.1 sin109.0 KS 461.1 461.1sin 28.1 KS 229.7 sin109.0
sin 55º sin 65º 150 a 150sin 55º a 135.58 miles sin 65º sin 60º sin 65º 150 b 150sin 60º b 143.33 miles sin 65º Station Able is about 143.33 miles from the ship, and Station Baker is about 135.58 miles from the ship.
Therefore, the total distance using the connecting flight is 332.0 229.7 561.7 miles. Using the connecting flight, Catalina would receive 561.7 461.1 100.6 more frequent flyer miles. 48. The time of the actual trip was: 50 300 350 t 1.4 hour 250 250
b.
Q mi
P
50. Consider the figure below.
10º
mi
R
C
RQ 300, PR 50, P 10º Solve the triangle: sin10º sin Q 300 50 50sin10º 0.0289414 sin Q 300 Q sin 1 0.0289414 1.65845º
R 180º 10º 1.65845º 168.34155º sin10º sin168.34155º PQ 300 300sin168.34155º PQ 349.115 sin10º 349.115 1.396459 hour 250 The trip should have taken 1.396459 hour but, because of the incorrect course, took 1.4 hour. Thus, the trip took 0.003541 hour, or about 12.7 seconds, longer. t
49. a.
Find C ; then use the Law of Sines (see figure):
B 60º
a C
mi 55º
a 135.6 0.68 hours 200 r min 0.68 hr 60 41 minutes hr
t
b
A C 180º 60º 55º 65º
b a
c B A We are given that A 49.8974 , B 180 49.9312 130.0688 , and c 300 km . Using angles A and B, we can find C 180 130.0688 49.8974 0.0338 Using the Law of Sines, we can determine b and a. sin B sin C b c c sin B b sin C 300 sin130.0688 sin 0.0338 389,173.319 sin A sin C a c c sin A a sin C 300 sin 49.8974 sin 0.0338 388,980.139
1053 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions
At the time of the measurements, the moon was about 389,000 km from Earth.
A 140º
51. Let h = the perpendicular distance from R to PQ, (see figure). R ft
D
0.125 mi
C
mi
E
0.125 mi
B 135º h 60º
sin 40º sin 95º BC 2 2sin 40º BC 1.290 mi sin 95º sin 45º sin 95º 2 AC 2sin 45º AC 1.420 mi sin 95º BE 1.290 0.125 1.165 mi
Q
P ft sin R sin 60º 123 184.5 123sin 60º 0.5774 sin R 184.5 R sin 1 0.5774 35.3º RPQ 180º 60º 35.3º 84.7º h 184.5 h 184.5sin 84.7º 183.72 feet
AD 1.420 0.125 1.295 mi For the isosceles triangle, 180º 95º CDE CED 42.5º 2 sin 95º sin 42.5º DE 0.125 0.125sin 95º DE 0.184 miles sin 42.5º The approximate length of the highway is AD DE BE 1.295 0.184 1.165 2.64 mi.
sin 84.7º
52. Let AOP sin sin15º 9 3 9sin15º 0.7765 sin 3 sin 1 0.7765 50.94º or 180º 50.94º 129.06º
54. Let PR = the distance from lighthouse P to the ship, QR = the distance from lighthouse Q to the ship, and d = the distance from the ship to the shore. From the diagram, QPR 75º and PQR 55º , where point R is the ship.
A 114.06º or A 35.94º sin114.06º sin15º a 3 3sin114.06º a 10.58 inches sin15º sin 35.94º sin15º or a 3 3sin 35.94º a 6.80 inches sin15º The approximate distance from the piston to the center of the crankshaft is either 6.80 inches or 10.58 inches.
P 15º
mi
d 35º
R
Q a.
53. A 180º 140º 40º ; B 180º 135º 45º ; C 180º 40º 45º 95º
Use the Law of Sines: sin 50º sin 55º 3 PR 3sin 55º PR 3.21 miles sin 50º The ship is about 3.21 miles from lighthouse P.
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Section 9.2: The Law of Sines b. Use the Law of Sines: sin 50º sin 75º QR 3 3sin 75º QR 3.78 miles sin 50º The ship is about 3.78 miles from lighthouse Q.
Use the Law of Sines: sin 90º sin 75º 3.2 d 3.2sin 75º d 3.10 miles sin 90º The ship is about 3.10 miles from the shore.
c.
Using the Law of Sines: sin(65 40) sin 25 88 L 88sin 25 L 38.5 inches sin105 The awning is about 38.5 inches long. 56. The tower forms an angle of 95˚ with the ground. Let x be the distance from the ranger to the tower. B 45º
ft
A
95º
55. Determine other angles in the figure:
x
40º 5º
C
ABC 180º 95º 40º 45º
50º 40º 65º
sin 40º sin 45º 100 x 100sin 45º x 110.01 feet sin 40º The ranger is about 110.01 feet from the tower.
88 in 65º 65º
57. Let h = height of the pyramid, and let x = distance from the edge of the pyramid to the point beneath the tip of the pyramid (see figure).
h 40.3º
46.27º
ft
ft
x
Using the Law of Sines twice yields two equations relating x and y: sin 46.27º sin(90º 46.27º ) Equation 1: h x 100 ( x 100) sin 46.27º h sin 43.73º x sin 46.27º 100sin 46.27º h sin 43.73º h sin 43.73º 100sin 46.27º sin 46.27º sin 40.3º sin(90º 40.3º ) h x 200 x 200 sin 40.3º h sin 49.7º x
Equation 2:
x sin 40.3º 200sin 40.3º h sin 49.7º h sin 49.7º 200sin 40.3º x sin 40.3º Set the two equations equal to each other and solve for h.
1055 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions h sin 43.73º 100sin 46.27º h sin 49.7º 200sin 40.3º sin 46.27º sin 40.3º h sin 43.73º sin 40.3º 100sin 46.27º sin 40.3º h sin 49.7º sin 46.27º 200sin 40.3º sin 46.27º h sin 43.73º sin 40.3º h sin 49.7º sin 46.27º 100sin 46.27º sin 40.3º 200sin 40.3º sin 46.27º
100sin 46.27º sin 40.3º 200sin 40.3º sin 46.27º sin 43.73º sin 40.3º sin 49.7º sin 46.27º 449.36 feet The current height of the pyramid is about 449.36 feet. 59. Using the Law of Sines: 58. Let h = the height of the aircraft, and let x = the h
distance from the first sensor to a point on the ground beneath the airplane (see figure).
Mercury B
x 5º 70º 15º
ft
20º
h
Earth
Equation 2:
C
Sun
149,600,000
x
sin15º sin B 57,910, 000 149, 600, 000 149, 600, 000 sin15º sin B 57,910, 000 14,960 sin15º 5791 14,960 sin15º o B sin 1 41.96 5791 or
Using the Law of Sines twice yields two equations relating x and h. Equation 1:
15º
57,910,000
sin 20º sin 70º h x h sin 70º x sin 20º
sin15º sin 75º h x 700 h sin 75º x 700 sin15º
B 138.04o
Set the equations equal to each other and solve for h.
C 180o 41.96o 15º 123.04o or
h sin 70º h sin 75º 700 sin 20º sin15º sin 70º sin 75º h 700 sin 20º sin15º 700 h sin 70º sin 75º sin 20º sin15º 710.97 feet
C 180o 138.04o 15º 26.96o
sin15º sin C 57,910, 000 x 57,910, 000 sin C x sin15 57,910, 000 sin123.04o sin15º 187,564,951.5 km or 57,910, 000 sin 26.96o x sin15º 101, 439,834.5 km So the possible distances between Earth and Mercury are approximately 101,440,000 km and 187,600,000 km.
The height of the aircraft is about 710.97 feet.
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Section 9.2: The Law of Sines 60. Using the Law of Sines:
62.
V enus B x
Earth
10 º
108,200,000 C
Sun
149,600,000
sin10º sin B 108, 200, 000 149, 600, 000 149, 600, 000 sin10º sin B 108, 200, 000 1496 sin10º 1082 1496 sin10º o B sin 1 13.892 1082
or
B 166.108o
C 180 13.892 10 156.108 or C 180 166.108 10 13.892 sin10º sin C x 108, 200, 000 108, 200, 000 sin C x sin10º 108, 200, 000 sin156.108 x sin10º 252,363, 760.4 km or 108, 200, 000 sin 3.892 x sin10º 42, 293, 457.3 km So the approximate possible distances between Earth and Venus are 42,300,000 km and 252,400,000 km.
61. Since there are 36 equally spaced cars, each car 360 10 . The angle between is separated by 36 the radius and a line segment connecting 170 consecutive cars is 85 (see figure). If 2 we let r = the radius of the wheel, we get sin10 sin 85 r 22 22sin 85 126 r sin10 The length of the diameter of the wheel is approximately d 2r 2 126 252 feet.
63.
a b a b sin A sin B c c c sin C sin C sin A sin B sin C A B A B 2sin cos 2 2 C C 2sin cos 2 2 C A B sin cos 2 2 2 C sin C cos 2 C A B cos cos 2 2 C C sin cos 2 2 A B 1 cos cos A B 2 2 C 1 sin sin C 2 2
a b a b sin A sin B sin A sin B c c c sin C sin C sin C A B A B 2sin cos 2 2 C sin 2 2 A B cos A B 2sin 2 2 C C 2sin cos 2 2 A B C sin cos 2 2 2 C C sin cos 2 2 A B C sin sin 2 2 C C sin cos 2 2 A B 1 sin sin 2 A B 2 1 C cos cos C 2 2
1057 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions
64. a
b sin A b sin 180º ( B ) sin B sin B b sin( B C ) sin B b sin B cos C cos B sin C sin B b sin C b cos C cos B sin B b cos C c cos B
67 – 69. Answers will vary. 70. Let x be the distance from the second surveyor to the bottom of the mountain. Then x + 900 is the distance from the first surveyor to the bottom of the mountain. Then we have the following equations: tan 35
h h and tan 47 x 900 x
Solve for x in the second equation:
a b a b 65. c ab ab c 1 sin ( A B ) 2 1 cos C 2 1 cos ( A B ) 2 1 sin C 2 1 1 sin ( A B ) sin C 2 2 1 1 cos C cos ( A B) 2 2 1 1 tan ( A B) tan C 2 2 1 1 tan ( A B) tan ( A B ) 2 2 A B 1 tan ( A B) tan 2 2 2 1 A B tan ( A B) cot 2 2 1 tan ( A B ) 2 1 tan ( A B ) 2
h and substitute into the first equation tan 47 and solve for h: x
h
tan 35
h tan 47 h tan 35 900 h tan 47 900
630.187 0.653h h 0.347h 630.187 h 1816
So the height of the mountain is approximately 1816 2 1818 meters . 71. 0 3 x3 4 x 2 27 x 36 Possible rational zeros: p 1, 2, 3, 6; q 1, 3; 1 2 p 1, 2, 3, 6, , q 3 3
Using synthetic division: We try x 3 : 3 3
4 9
27 36 15 36
3
5
12
0
Since the remainder is 0, x (3) x 3 is a factor. The other factor is the quotient: 3x 2 5 x 12 .
66. Since PQR and PP ' R are inscribed angles intersecting the same arc, they are congruent. Therefore, b sin B sin PQR sin PP ' R 2r sin B 1 sin A sin C b a c 2r (from the Law of Sines).
Thus, 0 x 3 3 x 2 5 x 12
x 3 3 x 4 x 3 4 The zeros are 3, ,3 . 3
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Section 9.2: The Law of Sines
72. d
x2 x1 2 y2 y1 2
2 (1) 2 1 (7) 2
32 6 2
y
3 and 2 contains the point (-2, -5). The equation of the line is y y1 m( x x1 )
78. The slope of the perpendicular line is m
9 36 45 3 5 6.71
3 ( x (2)) 2 3 y 5 ( x 2) 2 3 y5 x3 2 3 y x2 2
7 73. tan cos 1 8 7 Since cos , 0 , let x 7 and 8 r 8 . Solve for y: 49 y 2 64 y 2 15 y 15
Since is in quadrant II, y 15 . 15 15 7 y . Thus, tan cos 1 8 x 7 7
y (5)
79. h( x) 5( x)3 4( x) 1 5 x 3 4 x 1 (5 x3 4 x 1) h( x) nor h( x)
1 74. y 4sin x 2
Therefore f ( x) is neither even nor odd.
The graph of y sin x is stretched vertically by a factor of 4 and stretched horizontally by a factor of 2 .
80.
1 ( x 6) 4 x 0 3 1 x 2 4x 0 3 13 x2 0 3 13 x2 3 6 x 13
75. log a 100 0.2 x 76.
3 1 . 9 3
The solution set is x | x
f (b) f (a) (e 2(3) 3ln 3) (e 2(1) 3ln1) 3 1 ba 6 e 3ln 3 e2 199.668 2
77. The horizontal asymptote will be constructed using the coefficients of the highest power of the variable in the numerator and denominator,
1059 Copyright © 2025 Pearson Education, Inc.
6 6 or , . 13 13
Chapter 9: Applications of Trigonometric Functions
Section 9.3
c 2 a 2 b 2 2ab cos C
x2 x1 y2 y1 2
1. d
a 2 b2 c2 2ab 2.052 32 42 2.7975 cos C 2(2.05)(3) 12.3 cos C
2
2 2 45 or 4
2. cos
2.7975 C cos 1 103.1º 12.3
The solution set is 45 or 4
B 180º A C 180º 30º 103.1º 46.9º
3. Cosines
11. a 2, b 3, C 95º
4. a
c 2 a 2 b 2 2ab cos C
5. b
c 2 22 32 2 2 3cos 95º 13 12 cos 95º
6. False: Use the Law of Cosines
c 13 12 cos 95º 3.75
7. False
a 2 b 2 c 2 2bc cos A
8. True
b2 c 2 a 2 2bc 32 3.752 22 19.0625 cos A 2(3)(3.75) 22.5
cos A
9. a 2, c 4, B 45º b 2 a 2 c 2 2ac cos B b 2 22 42 2 2 4 cos 45º
19.0625 A cos 1 32.1º 22.5
2 20 16 2 20 8 2
B 180º A C 180º 32.1º 95º 52.9º
12. a 2, c 5, B 20º
b 20 8 2 2.95
b 2 a 2 c 2 2ac cos B
a 2 b 2 c 2 2bc cos A
b 2 22 52 2 2 5cos 20º 29 20 cos 20º
2bc cos A b 2 c 2 a 2
b 29 20 cos 20º 3.19
b2 c 2 a 2 2bc 2.952 42 22 20.7025 cos A 2(2.95)(4) 23.6 cos A
a 2 b 2 c 2 2bc cos A cos A
b2 c 2 a 2 2bc
29 20 cos 20º 5 2 0.97681 cos A
20.7025 A cos 1 28.7º 23.6
2
2
2
2( 29 20 cos 20º )(5)
C 180º A B 180º 28.7º 45º 106.3º
A cos 1 0.97681 12.4º
10. b 3, c 4, A 30º
C 180º A B 180º 12.4º 20º 147.6º
a 2 b 2 c 2 2bc cos A
13. a 6, b 5, c 8
a 2 32 42 2 3 4 cos 30º
a 2 b 2 c 2 2bc cos A
3 25 24 2
cos A
25 12 3
b 2 c 2 a 2 52 82 62 53 2bc 2(5)(8) 80
53 A cos 1 48.5º 80
a 25 12 3 2.05
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Section 9.3: The Law of Cosines
b 2 a 2 c 2 2ac cos B
b 2 a 2 c 2 2ac cos B
a 2 c2 b2 2ac 2 6 82 52 75 cos B 2(6)(8) 96 cos B
cos B
23 B cos 1 44.0º 32
75 B cos 1 38.6º 96
C 180º A B 180º 68.0º 44.0º 68.0º
C 180º A B 180º 48.5º 38.6º 92.9º
c 2 32 42 2 3 4 cos 40º 25 24 cos 40º
a b c 2bc cos A 2
2
b 2 c 2 a 2 52 4 2 82 23 cos A 2bc 2(5)(4) 80 23 A cos 1 125.1º 80 a 2 c 2 b 2 82 42 52 55 2ac 2(8)(4) 64
55 B cos 1 30.8º 64
a b c 2bc cos A cos A
2
2
2
cos B
2
2
2
61 B cos 32.1º 72 1
a 2 b 2 c 2 2bc cos A a 20 16 cos 75º 3.98 b 2 a 2 c 2 2ac cos B
16. a 4, b 3, c 4
cos B
a 2 b 2 c 2 2bc cos A b 2 c 2 a 2 32 42 42 9 2bc 2(3)(4) 24
9 A cos 1 68.0º 24
19. b 2, c 4, A 75º a 2 22 42 2 2 4 cos 75º 20 16 cos 75º
C 180º A B 180º 127.2º 32.1º 20.7º
cos A
b 2 c 2 a 2 1.032 12 22 1.9391 2bc 2(1.03)(1) 2.06
C 180º A B 180º 160.3º 10º 9.7º
a c b 9 4 6 61 2ac 2(9)(4) 72 2
cos A
1.9391 A cos 1 160.3º 2.06
b 2 a 2 c 2 2ac cos B 2
b 2 22 12 2 2 1cos10º 5 4 cos10º
2
29 A cos 1 127.2º 48 2
B 180º A C 180º 48.6º 40º 91.4º
a 2 b 2 c 2 2bc cos A
b c a 6 4 9 29 2bc 2(6)(4) 48 2
13.6049 A cos 1 48.6º 20.56
b 5 4 cos10º 1.03
2
2
b 2 c 2 a 2 42 2.57 2 32 13.6049 2bc 2(4)(2.57) 20.56
b 2 a 2 c 2 2ac cos B
15. a 9, b 6, c 4 2
a 2 b 2 c 2 2bc cos A
18. a 2, c 1, B 10º
C 180º A B 180º 125.1º 30.8º 24.1º
2
c 25 24 cos 40º 2.57
cos A
b 2 a 2 c 2 2ac cos B cos B
17. a 3, b 4, C 40º c 2 a 2 b 2 2ab cos C
14. a 8, b 5, c 4 2
a 2 c 2 b 2 42 42 32 23 2ac 2(4)(4) 32
a 2 c 2 b 2 3.982 42 22 26.84 2ac 2(3.98)(4) 31.84
26.84 B cos 1 32.5º 31.84 C 180º A B 180º 75º 32.5º 72.5º
1061 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions 20. a 6, b 4, C 60º
a 2 b 2 c 2 2bc cos A
c 2 a 2 b 2 2ab cos C
cos A
c 2 62 42 2 6 4 cos 60º 28
5.2441 A cos 1 55.1º 9.16
c 28 5.29 a 2 b 2 c 2 2bc cos A
B 180º A C 180º 55.1º 70º 54.9º
b 2 c 2 a 2 42 5.292 62 7.9841 cos A 2bc 2(4)(5.29) 42.32
24. a 3, c 2, B 90º
7.9841 A cos 1 79.1º 42.32
b 2 a 2 c 2 2ac cos B b 2 32 22 2 3 2 cos 90º 13
B 180º A C 180º 79.1º 60º 40.9º
b 13 3.61
21. a 5, c 3, B 105º
a 2 b 2 c 2 2bc cos A
b 2 a 2 c 2 2ac cos B
cos A
b 2 52 32 2 5 3cos105º 34 30 cos105º
C 180º A B 10º 56.3º 90º 33.7º
c 2 a 2 b 2 2ab cos C a 2 b 2 c 2 52 6.462 32 57.7316 2ab 2(5)(6.46) 64.6
25. a 20, b 29, c 21 a 2 b 2 c 2 2bc cos A
57.7316 C cos 26.7º 64.6 1
cos A
A 180º B C 180º 105º 26.7º 48.3º
a 2 b 2 c 2 2bc cos A
b 2 a 2 c 2 2ac cos B
a 4 1 2 4 1cos120º 21 2
2
cos B
a 21 4.58
a 2 c 2 b 2 202 212 292 0 2ac 2(20)(21)
B cos 1 0 90º
c 2 a 2 b 2 2ab cos C cos C
b 2 c 2 a 2 292 212 202 882 2bc 2(29)(21) 1218
882 A cos 1 43.6º 1218
22. b 4, c 1, A 120º 2
b 2 c 2 a 2 ( 13) 2 22 32 0.55470 2bc 2( 13)(2)
A cos 1 0.55470 56.3º
b 13 12 cos105º 6.46
cos C
b 2 c 2 a 2 22 2.292 22 5.2441 2bc 2(2)(2.29) 9.16
C 180º A B 180º 43.6º 90º 46.4º
a 2 b 2 c 2 4.582 42 12 35.9764 2ab 2(4.58)(4) 36.64
26. a 4, b 5, c 3
35.9764 C cos 10.9º 36.64 1
a 2 b 2 c 2 2bc cos A cos A
B 180º A C 180º 120º 10.9º 49.1º
b 2 c 2 a 2 52 32 42 0.6 2bc 2(5)(3)
A cos 1 0.6 53.1º
23. a 2, b 2, C 70º c 2 a 2 b 2 2ab cos C
b 2 a 2 c 2 2ac cos B
c 2 22 22 2 2 2 cos 70º 8 8cos 70º
cos B
c 8 8cos 70º 2.29
a 2 c 2 b 2 42 32 52 0 2ac 2(4)(3)
B cos 1 0 90º C 180º A B 180º 53.1º 90º 36.9º
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Section 9.3: The Law of Cosines 27. a 2, b 2, c 2
30. a 4, b 3, c 6
a b c 2bc cos A 2
cos A
2
a 2 b 2 c 2 2bc cos A
2
b 2 c 2 a 2 22 22 22 0.5 2bc 2(2)(2)
A cos 1 0.5 60º a 2 c 2 b 2 22 22 22 0.5 2ac 2(2)(2)
B cos 1 0.5 60º C 180º A B 180º 60º 60º 60º
28. a 3, b 3, c 2 b 2 c 2 a 2 32 22 32 1 2bc 2(3)(2) 3
1 A cos 1 70.53º 3
cos B
a c b 3 2 3 1 2ac 2(3)(2) 3 2
2
2
2
2
1 B cos 1 70.53º 3
29. a 6, b 11, c 12 112 122 a 2 112 122 62 229 2bc 2(11)(12) 264
229 A cos 1 29.84º 264
cos B
6 12 11 6 12 11 59 2ac 144 2 6 12 2
a 2 b 2 c 2 2bc cos A
2
47 b 2 c 2 a 2 132 32 152 2bc 2(13)(3) 78
47 A cos 1 127.1º 78 b 2 a 2 c 2 2ac cos B a 2 c 2 b 2 152 32 132 65 2ac 2(15)(3) 90
65 B cos 1 43.8º 90
32. a 9, b 7, c 10 a 2 b 2 c 2 2bc cos A cos A
b 2 a 2 c 2 2ac cos B 2
31. a 15, b 13, c 3
C 180o A B 180o 127.1o 43.8o 9.1o
a 2 b 2 c 2 2bc cos A
2
a 2 c 2 b 2 42 62 32 43 2ac 48 2 4 6
43 B cos 1 26.4º 48
cos B
C 180º A B 180º 70.53º 70.53º 38.9º
cos A
cos B
cos A
b 2 a 2 c 2 2ac cos B 2
b 2 a 2 c 2 2ac cos B
C 180o A B 180o 36.3o 26.4o 117.3o
a 2 b 2 c 2 2bc cos A cos A
b 2 c 2 a 2 32 62 42 29 2bc 2(3)(6) 36
29 A cos 1 36.3º 36
b 2 a 2 c 2 2ac cos B cos B
cos A
2
2
59 o B cos 1 65.81 144
68 A cos 1 60.9º 140 b 2 a 2 c 2 2ac cos B cos B
C 180o A B 180o 29.8o 65.8o 84.4o
b 2 c 2 a 2 7 2 102 92 68 2bc 2(7)(10) 140
a 2 c 2 b 2 92 102 7 2 132 2ac 2(9)(10) 180
132 B cos 1 42.8º 180 C 180o 60.94o 42.83o 76.2o
1063 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions 33. B 20 , C 75 , b 5 sin B sin C b c b sin C 5sin 75 c 14.12 sin B sin 20
c 2 a 2 b 2 2ab cos C a2 b2 c2 2ab 2 6 82 92 19 2 6 8 96
cos C
A 180 20 75 85
C cos 1
sin A sin B a b b sin A 5sin 85 a 14.56 sin B sin 20
19 78.6 96
36. a 14 , b 7 , A 85 sin 85 sin B 14 7 sin 85 sin B 0.49810 2 B sin 1 0.49810 29.9 or 150.1 The second value is discarded since A B 180 . Therefore, B 29.9 . C 180 29.9 85 65.1 sin 85 sin 65.1 14 c 14 sin 65.1 c 12.75 sin 85
34. A 50 , B 55 , c 9 C 180 50 55 75 sin C sin B c b c sin B 9sin 55 b 7.63 sin C sin 75 sin C sin A c a c sin A 9sin 50 a 7.14 sin C sin 75
35. a 6 , b 8 , c 9 a 2 b 2 c 2 2bc cos A
37. B 35 , C 65 , a 15 A 180 35 65 80
b2 c2 a 2 2bc 2 8 92 62 109 2 8 9 144
sin A sin C a c a sin C 15sin 65 c 13.80 sin A sin 35
109 40.8 144
sin A sin B a b a sin B 15sin 35 b 8.74 sin A sin 80
cos A
A cos 1
b 2 a 2 c 2 2ac cos B a2 c2 b2 2ac 2 6 9 2 82 53 2 6 9 108
cos B
B cos 1
53 60.6 108
1064
Copyright © 2025 Pearson Education, Inc.
Section 9.3: The Law of Cosines 38. a 4 , c 5 , B 55 2
2
10sin10 0.578827 3 B sin 1 0.578827 35.4 or 144.6
sin B
2
b a c 2ac cos B b 2 42 52 2 4 5 cos 55 b 2 41 40 cos 55 b 41 40 cos 55 4.25 sin 55
sin A 4 41 40 cos 55 4sin 55 sin A 0.77109 41 40 cos 55
A sin 1 0.77109 50.5 or 129.5 We discard the second value since A B 180 . Therefore, A 50.5 . C 180 55 50.5 74.5
sin C sin B c b c sin B 8sin 52 6.30 b sin C sin 90
12.29
c2 32 102 2 3 10 cos 25.4 7.40
C 180 65 72 43 sin 43 sin 72 7 c 7 sin 43 5.02 c sin 72
sin C sin A c a c sin A 8sin 38 4.93 a sin C sin 90
40. A 73º , C 17º , a 20 B 180º A C 180º 73º 17º 90º
sin C sin A c a sin17º sin 73º c 20 20sin17º c 6.11 sin 73º
c1 32 102 2 3 10 cos134.6
42. A 65 , B 72 , b 7 sin 72 sin 65 7 a 7 sin 65 6.67 a sin 72
39. A 38 , B 52 , c 8 C 180 38 52 90
sin A sin B a b sin 73º sin 90º 20 b 20sin 90º b 20.91 sin 73º
Since both values yield A B 180 , there are two triangles. B1 35.4 and B2 144.6 C1 180 10 35.4 134.6 C2 180 10 144.6 25.4 Using the Law of Cosines we get
43. b 5 , c 12 , A 60 a 2 b 2 c 2 2bc cos A a 2 52 122 2 5 12 cos 60 109 a 109 10.44 sin 60
sin B 5 109 5sin 60 sin B 0.414751 109
B sin 1 0.414751 24.5 or 155.5
We discard the second value because it would give A B 180 . Therefore, B 24.5 . C 180 60 24.5 95.5
41. a 3 , b 10 , A 10 sin10 sin B 3 10
1065 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions 44. a 10 , b 10 , c 15 a 2 b 2 c 2 2bc cos A
sin A sin130º 100 227.56 100sin130º sin A 227.56 100sin130º A sin 1 19.7º 227.56 Since the angle of the triangle is 19.7˚, the pilot should fly at a bearing of N 19.7 E .
b2 c2 a 2 2bc 2 10 152 102 225 3 2 10 15 300 4
cos A
A cos 1
3 41.4 4
47. After 10 hours the ship will have traveled 150 nautical miles along its altered course. Use the Law of Cosines to find the distance from Barbados on the new course. a 600, b 150, C 20º
b 2 a 2 c 2 2ac cos B a 2 c2 b2 2ac 2 10 152 102 225 3 2 10 15 300 4
cos B
B cos
1 3
4
c 2 a 2 b 2 2ab cos C 6002 1502 2 600 150 cos 20º 382,500 180, 000 cos 20º
41.4
c 382,500 180, 000 cos 20º
c 2 a 2 b 2 2ab cos C
461.9 nautical miles
a 2 b2 c2 2ab 2 10 102 152 25 1 2 10 10 200 8
cos C
a.
1 C cos 1 97.2 8
124,148.39 A cos 1 153.6º 138,570 The captain needs to turn the ship through an angle of 180 153.6 26.4 .
45. Find the third side of the triangle using the Law of Cosines: a 150, b 35, C 110º c 2 a 2 b 2 2ab cos C 1502 352 2 150 35cos110º 23, 725 10,500 cos110º
b.
c 23, 725 10,500 cos110º 165 The ball is approximately 165 yards from the center of the green.
46. a.
Use the Law of Cosines to find the angle opposite the side of 600: b2 c 2 a 2 cos A 2bc 2 150 461.92 6002 124,148.39 cos A 2(150)(461.9) 138,570
48. a.
The angle inside the triangle at Sarasota is 180 50 130 . Use the Law of Cosines to find the third side: a 150, b 100, C 130º
461.9 nautical miles 30.8 hours are 15 knots required for the second leg of the trip. (The total time for the trip will be about 40.8 hours.) t
After 15 minutes, the plane would have flown 220(0.25) = 55 miles. Find the third side of the triangle: a 55, b 330, 10º c 2 a 2 b 2 2ab cos C
c 2 a 2 b 2 2ab cos C
552 3302 2 55 330 cos10º 111,925 36,300 cos10º
1502 1002 2 150 100 cos130º 32,500 30, 000 cos130º
c 111,925 36,300 cos10º 276
c 32,500 30, 000 cos130º 227.56 mi
Find the measure of the angle opposite the 330-mile side:
b. Use the Law of Sines to find the angle inside the triangle at Ft. Myers: 1066
Copyright © 2025 Pearson Education, Inc.
Section 9.3: The Law of Cosines
a 2 c2 b2 2ac 2 55 2762 3302 29, 699 2(55)(276) 30,360
The pitcher needs to turn through an angle of about 92.8˚ to face first base.
cos B
50. a.
Find x in the figure: 3rd
29, 699 B cos 168.0º 30,360 The pilot should turn through an angle of 180 168.0 12.0 .
2nd
1
2nd
60 .5
ft
y
Home
45
90 ft
1st
60 ft
x 2 462 602 2(46)(60) cos 45º 2 5716 5520 5716 2760 2 2 x 5716 2760 2 42.58 feet It is about 42.58 feet from the pitching rubber to first base.
b. Use the Pythagorean Theorem to find y in the figure: 602 602 (46 y ) 2
Find x in the figure: 3rd
60 ft x
45
Home
b. If the total trip is to be done in 90 minutes, and 15 minutes were used already, then there are 75 minutes or 1.25 hours to complete the trip. The plane must travel 276 miles in 1.25 hours: 276 r 220.8 miles/hour 1.25 The pilot must maintain a speed of 220.8 mi/hr to complete the trip in 90 minutes. 49. a.
46
ft
y
7200 (46 y ) 2
90 ft x
46 y 7200 84.85 y 38.85 feet It is about 38.85 feet from the pitching rubber to second base.
1st
x 2 60.52 902 2(60.5)(90) cos 45º 2 11, 760.25 10,980 2
c.
11, 760.25 5445 2 x 11, 760.25 5445 2 63.7 feet It is about 63.7 feet from the pitching rubber to first base. b. Use the Pythagorean Theorem to find y in the figure: 902 902 (60.5 y ) 2 16, 200 (60.5 y )
Find B in the figure by using the Law of Cosines: 462 42.582 602 329.0564 cos B 2(46)(42.58) 3917.36 329.0564 B cos 1 85.2º 3917.36 The pitcher needs to turn through an angle of 85.2˚ to face first base.
51. a.
Find x by using the Law of Cosines:
2
60.5 y 16, 200 127.3
c.
Find B in the figure by using the Law of Cosines: 60.52 63.7 2 902 382.06 cos B 2(60.5)(63.7) 7707.7
500 ft x y
250 ft
y 66.8 feet It is about 66.8 feet from the pitching rubber to second base.
100 ft
382.06 B cos 1 92.8º 7707.7
1067 Copyright © 2025 Pearson Education, Inc.
80
10
Chapter 9: Applications of Trigonometric Functions
x 2 5002 1002 2(500)(100) cos80º
x 2 30.12 51.42 2(30.1)(51.4) cos89.2º
260, 000 100, 000 cos80º
3547.97 3094.28cos89.2º x 3547.97 3094.28cos89.2º 59.2 mm
x 260, 000 100, 000 cos80º 492.58 ft
b. The length of 59.2 mm is very close to the average male length of 59.4 mm.
The guy wire needs to be about 492.58 feet long.
54. a.
b. Use the Pythagorean Theorem to find the value of y:
Find x by using the Law of Cosines:
y 2 1002 2502 72,500 y 269.26 feet The guy wire needs to be about 269.26 feet long. 52. Find x by using the Law of Cosines:
x y
x 2 48.82 62.22 2(48.8)(62.2) cos89º
500 ft
6250.28 6070.72 cos89º x 6250.28 6070.72 cos89º 78.4 mm
95 85 5
b. The length of 78.4 mm is less than 80 so this indicates a typical female.
x 2 5002 1002 2(500)(100) cos85º 260, 000 100, 000 cos85º
55. a. Begin by finding angle NP by using the Law of Cosines:
x 260, 000 100, 000 cos85º 501.28 feet The guy wire needs to be about 501.28 feet long.
Find y by using the Law of Cosines: y 2 5002 1002 2(500)100 cos 95º 260, 000 100, 000 cos 95º y 260, 000 100, 000 cos 95º 518.38 feet The guy wire needs to be about 518.38 feet long.
53. a.
Find x by using the Law of Cosines:
242 202 82 2(20)(8) cos NP 242 202 82 2(20)(8) cos NP 242 202 82 cos NP 2(20)(8) 0.35 cos NP NP 110.49
Now we can find the distance from the ball to the goalie:
1068
Copyright © 2025 Pearson Education, Inc.
Section 9.3: The Law of Cosines
G 2 202 42 2(20)(4) cos110.49
56. Consider the figure:
G 2 416 160 cos110.49 G 2 470.01 G 21.73 yd
Now use the Law of Cosines to solve for the two angles, and . 42 202 21.732 2(20)(21.73) cos 42 202 21.732 2(20)(21.73) cos 42 202 21.732 cos 2(20)(21.73) 9.9
First find the angle at the near post by using the Law of Cosines.
4 24 21.73 2(24)(21.73) cos 2
2
2
4 24 21.73 2(24)(21.73) cos 2
2
2
4 24 21.73 cos 2(24)(21.73) 8.3 2
2
2
The total of the angles is 9.9 8.3 18.2 . Thus equal angles 18.2 would be 9.1 . The angle of the 2 goalie needs to make with the goal on the near post side is 180 9.1 110.49 60.41 . Using the Law of Sines gives: sin 60.41 sin 9.1 20 x x sin 60.41 20sin 9.1 x
302 242 82 2(24)(8) cos NP 302 242 82 cos NP 2(24)(8) NP 132.6
Now find the angle at the ball using the Law of Cosines.
b. The distance from the ball to the goalie is 21.73 yards as calculated in part a. c.
302 242 82 2(24)(8) cos NP
82 242 302 2(24)(30) cos B 82 242 302 2(24)(30) cos B 82 242 302 cos B 2(24)(30) B 11.32
Half of the angle (angle ) is 5.66 . Thus angle 180 90 5.66 84.34 and angle 132.6 84.34 48.26 . We can now solve for x. x 24 x 24sin 5.66 2.37 yd
sin 5.66
20sin 9.1 3.64 yd sin 60.41
Consider the smaller triangle:
The goalie needs to move 4 – 3.64 = 0.36 yds.
Use the Law of Cosines to solve for d:
1069 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions
d 2 2.37 2 42 2(2.37)(4) cos 48.26
c 2 a 2 b 2 2ab cos C 102 102 2 10 10 cos 45
d 2 8.994 d 3.0 yds
2 100 100 200 2
57. Find x by using the Law of Cosines:
200 100 2 100 2 2 x
c 100 2 2 10 2 2 7.65
400 ft 90 ft
45
The footings should be approximately 7.65 feet apart.
x 2 4002 902 2(400)(90) cos 45º
60. Use the Law of Cosines:
168,100 36, 000 2
A
x 168,100 36, 000 2 342.33 feet It is approximately 342.33 feet from dead center to third base.
L
r
O
58. Find x by using the Law of Cosines:
B
x
L2 x 2 r 2 2 x r cos x 2 2 x r cos r 2 L2 0
x
Using the quadratic formula:
280 ft 60 ft
45
x
x 2 2802 602 2(280)(60) cos 45º
x
82, 000 16,800 2 x 82, 000 16,800 2 241.33 feet It is approximately 241.33 feet from dead center to third base.
x x
59. Use the Law of Cosines:
2r cos 4r 2 cos 2 4 r 2 L2
2
2r cos 4 r cos 2 r 2 L2 2
2 2r cos 2 r cos 2 r 2 L2 2 2
x r cos r 2 cos 2 L2 r 2
45o
61. d 2 r 2 r 2 2 r r cos
10
10
2r cos (2r cos )2 4(1)(r 2 L2 ) 2(1)
2r 2 2r 2 cos 2r 2 (1 cos ) 1 cos 4r 2 2
c
2r
1 cos 2
2r sin 2 d s r so 2r sin r or 2sin . 2 2
Since sin 2sin cos and cos 1 , Then 2 2 2 1070
Copyright © 2025 Pearson Education, Inc.
Section 9.3: The Law of Cosines
sin 2sin
0 .
62. cos
2
. Therefore, sin for
C 1 cos C 2 2
1
a2 b2 c2 2ab 2
2ab a 2 b 2 c 2 4ab
( a b) 2 c 2 4ab
63. sin
64.
65 – 69. Answers will vary. 70. Domain: x x 3
y-intercept: R (0)
a b c a b c 2 s (2s c c) 4ab
4 s( s c) 4ab
1 2 vertical asymptotes: x 1 The multiplicity of 1 is odd so the graph will approach plus or minus infinity on either side of the asymptote. Since the degree of the numerator and denominator are equal then the horizontal asymptote is: y 2 .
s( s c) ab
C 1 cos C 2 2
1
a 2 b2 c 2 2ab 2
2ab a 2 b 2 c 2 4ab
(a 2 2ab b 2 c 2 ) 4ab
2(0) 1 1 1 3 03 3
x-interecpt: x
4ab
cos A cos B cos C a b c b2 c 2 a 2 a 2 c 2 b2 a 2 b2 c2 2bca 2acb 2abc 2 2 2 2 2 2 2 b c a a c b a b2 c2 2abc 2 2 2 a b c 2abc
( a b) 2 c 2
4ab
(a b c)(a b c ) 4ab
(a b c)(b c a ) 4ab
(2s 2b)(2s 2a ) 4ab
4( s b)( s a) 4ab
( s a )( s b) ab
4 x 3x 1
71.
ln 4 x ln 3x 1 x ln 4 ( x 1) ln 3 x ln 4 x ln 3 ln 3 x ln 4 x ln 3 ln 3 x(ln 4 ln 3) ln 3 ln 3 x 3.819 ln 4 ln 3
The solution is
1071 Copyright © 2025 Pearson Education, Inc.
ln 3 3.819 . ln 4 ln 3
Chapter 9: Applications of Trigonometric Functions
5 72. cos ; x 5, r 7 7 2 6 tan ; x 5, y 2 6 5
1
x
x
4 3 ln 3 x 2 4 3
76.
sin
y 2 6 x 5 5 6 ;cot 7 12 r y 2 6
csc
7 7 6 7 r r ;sec 12 5 y 2 6 x
x
1 2
x
1
x
43 x
2
2
1
Find :
2
2
ln 3 x
x
1
2
2
1 4 3x ln 3 x 2 1
x 2
x
2
1 4 3x ln 3 x 2 3
x 2
73. The graph is a reflected sine graph with
amplitude 3 and period
2
77.
.
4x 3 x 1 4 x 3 x 1 or 4 x 3 x 1 3x 4 5x 2 4 2 x x 3 5 The solution set is 2 4 , 5 3 ,
2 2
2
4 4 The equation is: y 3sin 4 x .
8 78. 96 180 15
74. Swap the variables x and y and solve for y: A 5y 2 x(5 y 2) A 5 xy 2 x A 5 xy A 2 x A 2x f 1 ( x ) y 5x x
79. The x value of the vertex would be b 2 , a 0 . This means the x value 2a 2a would be negative. The graphs is concave down and the y-intercept is 5. The only option is for the vertex to be in Quadrant II. Since the vertex is in Quadrant II and the graph is concave down, there are 2 x-intercepts.
75. F (b) F (a ) F (2) F (1) 23 13 3(2) C 3(1) C 3 3 8 1 6C 3C 3 3 8 1 6C 3C 3 3 8 1 6C 3C 3 3 2 3
Section 9.4 1 1. K bh 2
2. False; cos 2 3. K 4.
2
1 cos 2
1 ab sin C 2
s ( s a)( s b)( s c) ;
1072
Copyright © 2025 Pearson Education, Inc.
1 (a b c) 2
Section 9.4: Area of a Triangle
5. K
16. a 4, b 3, c 4
1 (3)(4) sin 90 6 2
1 1 11 a b c 2 4 3 4 2 2 K s ( s a)( s b)( s c)
s
6. True 7. c
11 3 5 3 2 2 2 2
8. c 9. a 2, c 4, B 45º K
17. a 3, b 4, C 50º
1 1 ac sin B (2)(4) sin 45º 2.83 2 2
1 1 K bc sin (3)(4) sin 30º 3 2 2
1 1 ab sin C (3)(4) sin 50º 4.60 2 2
K
1 1 ac sin B (2)(1)sin10º 0.17 2 2
19. b 1, c 8, A 75º
11. a 5, b 7, C 94º 1 1 ab sin C (5)(7) sin 94º 17.46 2 2
12. a 2, c 5, B 20º
1 1 K bc sin A (1)(8) sin 75º 3.86 2 2
20. a 6, b 4, C 60º
1 1 K ac sin B (2)(5) sin 20º 1.71 2 2
K
1 1 ab sin C (6)(4) sin 60º 10.39 2 2
21. a 3, c 2, B 115º
13. a 7, b 4, c 9 1 1 a b c 2 7 4 9 10 2 K s ( s a)( s b)( s c)
s
K
18. a 2, c 1, B 10º
10. b 3, c 4, A 30º
K
495 5.56 16
10 3 6 1 180 13.42
14. a 8, b 5, c 4 1 1 17 a b c 8 5 4 2 2 2 K s ( s a)( s b)( s c)
s
1071 17 1 7 9 8.18 16 2 2 2 2
15. a 9, b 6, c 4
K
1 1 ac sin B (3)(2) sin115º 2.72 2 2
22. b 4, c 1, A 120º 1 1 K bc sin A (4)(1) sin120º 1.73 2 2
23. a 12, b 35, c 37 1 1 a b c 2 12 35 37 42 2 K s ( s a)( s b)( s c)
s
42 30 7 5
44100 210
24. a 4, b 5, c 3
1 1 19 s a b c 9 6 4 2 2 2 K s ( s a )( s b)( s c) 1463 19 1 7 11 9.56 16 2 2 2 2
1 1 a b c 4 5 3 6 2 2 K s ( s a)( s b)( s c)
s
6 2 1 3 36 6
1073 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions 25. a 4, b 4, c 4
1 1 b sin C K bc sin A b sin A 2 2 sin B
1 1 s a b c 4 4 4 6 2 2 K s ( s a)( s b)( s c)
6 2 2 2
b 2 sin A sin C 2sin B 1 1 c sin A K ac sin B c sin B 2 2 sin C c 2 sin A sin B 2sin C
48 6.93
26. a 3, b 3, c 2 1 1 a b c 2 3 3 2 4 2 K s ( s a)( s b)( s c)
s
31. A 40º , B 20º , a 2 C 180º A B 180º 40º 20º 120º
4 11 2 8 2.83
K
27. a 11, b 14, c 20
32. A 50º , C 20º , a 3 B 180º A C 180º 50º 20º 110º
1 1 45 a b c 2 11 14 20 2 2 K s ( s a)( s b)( s c)
s
K
45 23 17 5 5498.4375 74.15 2 2 2 2
a 2 sin B sin C 32 sin110º sin 20º 1.89 2sin A 2sin 50º
33. B 70º , C 10º , b 5 A 180º B C 180º 70º 10º 100º
28. a 4, b 3, c 6
K
1 1 13 a b c 4 3 6 2 2 2 K s ( s a)( s b)( s c)
s
13 5 7 1 2 2 2 2
a 2 sin B sin C 22 sin 20º sin120º 0.92 2sin A 2sin 40º
b 2 sin A sin C 52 sin100º sin10º 2.27 2sin B 2sin 70º
34. A 70º , B 60º , c 4 C 180º A B 180º 70º 60º 50º
455 5.33 16
K
c 2 sin A sin B 42 sin 70º sin 60º 8.50 2sin C 2sin 50º
35. A 110º , C 30º , c 3 B 180º A C 180º 110º 30º 40º
sin A sin B . a b a sin B Solving for b, so we have that b . Thus, sin A 1 1 a sin B K ab sin C a sin C 2 2 sin A a 2 sin B sin C 2sin A
29. From the Law of Sines we know
K
c 2 sin A sin B 32 sin110º sin 40º 5.44 2sin C 2sin 30º
36. B 10º , C 100º , b 2 A 180º B C 180º 10º 100º 70º K
sin A sin B sin C , we have that a b c b sin C c sin A and a . Thus, c sin B sin C
b 2 sin A sin C 22 sin 70º sin100º 10.66 2sin B 2sin10º
30. From
37. Area of a sector
1 2 r where is in radians. 2
7 180 18 1 7 112 2 ASector 82 ft 2 18 9
70
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Copyright © 2025 Pearson Education, Inc.
Section 9.4: Area of a Triangle
1 8 8sin 70º 32sin 70º ft 2 2 112 ASegment 32sin 70º 9.03 ft 2 9 ATriangle
38. Area of a sector
42. Begin by adding a diagonal to the diagram.
1 2 r where is in radians. 2
2 180 9 1 2 25 2 ASector 52 in 2 9 9 1 25 ATriangle 5 5sin 40º sin 40º in 2 2 2 25 25 ASegment sin 40º 0.69 in 2 9 2
40
39. Find the area of the lot using Heron's Formula: a 100, b 50, c 75 1 1 225 a b c 2 100 50 75 2 2 K s ( s a)( s b)( s c)
s
225 25 125 75 2 2 2 2 52, 734,375 16 1815.46 Cost $31815.46 $5446.38
40. Diameter of canvas is 24 feet; radius of canvas is 12 feet; angle is 260˚. 1 Area of a sector r 2 where is in radians. 2 13 260 180 9 1 936 2 13 ASector 12 104 326.73 ft 2 2 9 9 41. Find the area of the lot using Heron's Formula: a 8.05, b 4.55, c 8.75 s
1 1 a b c 8.05 4.55 8.75 10.675 2 2
d 24897.29 157.8 Using Heron’s formula for the lower triangle, 138 38 157.8 166.9 2 Alower 166.9(28.9)(128.9)(9.1) s
5657811.7 2378.6 sq. feet Total Area 5997.1 2378.6 8376 sq. feet
43. Divide home plate into a rectangle and a triangle. 12”
12” 17”
8.5”
8.5”
17”
ARectangle lw (17)(8.5) 144.5 in 2 Using Heron’s formula we get ATriangle s ( s a )( s b)( s c) 1 1 a b c 12 12 17 20.5 2 2 Thus, ATriangle (20.5)(20.5 12)(20.5 12)(20.5 17) s
(20.5)(8.5)(8.5)(3.5)
K s ( s a)( s b)( s c)
1 (140)(86)(sin 85) 5997.1 sq. feet 2 By the law of cosines, d 2 (140) 2 (86) 2 2(140)(86)(cos85) 19600 7396 2098.71 24897.29 Aupper
10.675 2.625 6.125 1.925
5183.9375 72.0 sq. in.
330.3754 18.18 m 2
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Chapter 9: Applications of Trigonometric Functions
So, ATotal ARectangle ATriangle
Find the area of the three triangles:
144.5 72.0
1 35 80 47.072 81.036 2 s1 a1 81.036 35 46.036
s1
2
216.5 in The area of home plate is about 216.5 in 2 .
s1 b1 81.036 80 1.036
44. Find the area of the shaded region by subtracting the area of the triangle from the area of the semicircle. Area of the semicircle 1 1 25 ASemicircle r 2 (5) 2 in 2 2 2 2 The triangle is a right triangle. Find the other leg: 82 b 2 102
s1 c1 81.036 47.072 33.964 K1 81.036(46.036)(1.036)(33.964) 362.307 ft 2 1 40 45.961 47.072 66.5165 2 s2 a2 66.5165 40 26.5165 s2
s2 b2 66.5165 45.961 20.5555
b 2 100 64 36
s2 c2 66.5165 47.072 19.4445
b 36 6 1 ATriangle 8 6 24 in 2 2 AShaded region 12.5 24 15.27 in 2
K 2 66.5165(26.5165)(20.5555)(19.4445) 839.6247 ft 2 1 (45 20 45.961) 55.4805 2 s3 a3 55.4805 45 10.4805
s3
45. The area is the sum of the area of a triangle and a sector. 1 1 ATriangle r r sin( ) r 2 sin( ) 2 2 1 2 ASector r 2
s3 b3 55.4805 20 35.4805 s3 c3 55.4805 45.961 9.5195 K 3 55.4805(10.4805)(35.4805)(9.5195) 443.1626 ft 2 The approximate area of the lake is 362.307 839.6247 443.1626 1645.1 ft 2
1 2 1 r sin r 2 2 2 1 2 r sin( ) 2 1 r 2 sin cos cos sin 2 1 2 r 0 cos (1) sin 2 1 2 r sin 2
K
47. Use Heron’s formula: a 87 , b 190 , c 173 1 1 s a b c 87 190 173 225 2 2 K s s a s b s c
225 225 87 225 190 225 173 225 138 35 52
46. Use the Law of Cosines to find the lengths of the diagonals of the polygon. x 2 352 802 2 35 80 cos15º 7625 5600 cos15º
56,511, 000 7517.4 The building covers approximately 7517.4 square feet of ground area.
x 7625 5600 cos15º 47.072 feet The interior angle of the third triangle is: 180 100 80 . y 2 452 202 2 45 20 cos80º
48. Use Heron’s formula: a 1028 , b 1046 , c 965 1 1 s a b c 1028 1046 965 1519.5 2 2
2425 1800 cos80º y 2425 1800 cos80º 45.961 feet
1076
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Section 9.4: Area of a Triangle K s s a s b s c
The area of the Bermuda Triangle is approximately 442,816 square miles.
1519.5 491.5 473.5 554.5 442,816
49. Letting d 0 gives K ( s a )( s b)( s c)( s 0) abc(0) cos 2 s ( s a)( s b)( s c ) where s
1 1 (a b c 0) (a b c) . 2 2
sin 75 sin 30 , where d is the distance from a vertex to the center. So d a a sin 75 a sin(90 15) a cos15 a d . The area of the triangle is sin 30 sin(2 15) 2sin15 cos15 2sin15
50. a. Using the Law of Sines
2
Ki
1 1 a a 2 cos15 a 2 d d sin 30 2sin15 cos15 cot15, i 1, 2, . Since there are 12 2 2 2sin15 4sin15 4
a2 cot15 3a 2 cot15. The radius of the 4 a 1 inscribed circle is perpendicular to each side at its midpoint. So, tan15 2 , or a 2r tan15. Using K bh r 2 1 1 2 gives the area of each triangle as K i ar 2r tan(15)r r tan15. Since there are 12 congruent triangles, the 2 2
congruent triangles, the area of the dodecagon is K 12 Ki 12
total area of the dodecagon is K 12 K i 12r 2 tan b.
51. a.
b.
K
12
.
n a2 cot or K n r 2 tan 4 n n 1 OC AC 2 AC 1 OC 2 1 1 1 cos sin 2 1 sin cos 2
Area OAC
1 OC BC 2 OC BC 2 1 OB 2 OB OB
c.
Area OCB
2 1 OB cos sin 2 2 1 OB sin cos 2
1 BD OA 2 1 BD 1 2 BD 1 OB 2 OB
Area OAB
d.
e.
1 OB sin( ) 2
OC OA OC OB cos OB cos 1 OC OC OB Area OAB Area OAC Area OCB 2 1 1 1 OB sin( ) sin cos OB sin cos 2 2 2
1077 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions
cos cos 2 sin( ) sin cos sin cos cos cos 2
A3
cos cos sin cos sin cos cos cos sin( ) sin cos cos sin sin( )
52. a.
Area of OBC
1 sin 1 1 sin 2 2
b.
Area of OBD
1 tan sin 1 tan 2 2 2 cos
c.
Area OBD Area OBC Area OBC 1 1 sin sin 2 2 2 cos sin sin cos sin sin sin sin sin cos 1 1 sin cos
1 (10)(90) sin 40.49º 292.19 ft 2 2
The angle for the sector A2 is 90º 40.49º 49.51º . A2
2 1 90 49.51 3499.66 ft 2 2 180
Since the cow can go in either direction around the barn, both A2 and A3 must be doubled. Thus, the total grazing area is: 23,561.94 2(3499.66) 2(292.19) 31,145 ft 2 54. We begin by dividing the grazing area into five regions: three sectors and two triangles (see figure). Region A1 is a sector representing three-fourths of a circle with radius 100 feet: 2 3 Thus, A1 100 7500 23,561.9 ft 2 . 4 To find the areas of regions A2 , A3 , A4 , and A5 , we first position the rectangular barn on a rectangular coordinate system so that the lower right corner is at the origin. The coordinates of the corners of the barn must then be O(0, 0) , P (20, 0) , Q(20,10) , and R (0,10) .
53. The grazing area must be considered in sections. Region A1 represents three-fourth of a circle with radius 100 feet. Thus, 2 3 A1 100 7500 23,561.94 ft 2 4
T
Angles are needed to find regions A2 and A3 : (see the figure)
y
A2
C
R D B
10 ft
90 ft
10 ft Q P O 20 ft A1
A
In ABC , CBA 45º , AB 10, AC 90 . Find BCA : sin CBA sin BCA 90 10 sin 45º sin BCA 90 10 10sin 45º 0.0786 sin BCA 90 10sin 45º BCA sin 1 4.51º 90
S
90 ft 80 ft
A3 A4
A5
U
x
Now, region A2 is a sector of a circle with center Q(20,10) and radius 90 feet. The equation of the circle then is ( x 20) 2 ( y 10) 2 902 . Likewise, region A5 is a sector of a circle with center O(0, 0) and radius 80 feet. The equation
BAC 180º 45º 4.51º 130.49º
of this circle then is x 2 y 2 802 . We use a graphing calculator to find the intersection point
DAC 130.49º 90º 40.49º
1078
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Section 9.4: Area of a Triangle
S of the two sectors. Let Y1 802 x 2 and
Thus, the total grazing area is: 23,561.9 4220.5 454 288.5 2448.6
Y2 902 ( x 20)2 10 .
30,973 ft 2
55. a.
Area: 1 1 s a b c 9 10 17 18 2 2
The approximate coordinates are S (57.7,55.4) . Now, consider QRS (i.e. region A3 ). The “base” of this triangle is 20 feet and the “height” is approximately 55.4 10 45.4 feet (the ycoordinate of the intersection point minus the side of the barn). Thus, the area of region A3 is 1 A3 20 45.4 454 ft 2 . Likewise, consider 2 ORS (i.e. region A4 ). The “base” of this triangle is 10 feet and the “height” is about 57.7 feet (the x-coordinate of the intersection point). 1 Thus, A4 10 57.7 288.5 ft 2 . 2
K s s a s b s c 18 18 9 18 10 18 17 18 9 8 1 1296 36 Since the perimeter and area are numerically equal, the given triangle is a perfect triangle.
b. Perimeter: P a b c 6 25 29 60
Area: 1 1 s a b c 6 25 29 30 2 2
To find the area of sectors A2 and A5 , we must determine their angles: TQS and SOU ,
K s s a s b s c
respectively. Now, we know A3 454 ft 2 . Also, we know A3 Thus,
30 30 6 30 25 30 29
1 90 20sin SQR . 2
30 24 5 1
1 90 20sin SQR 454 2 sin SQR 0.5044
SQR 0.5287 rad. Since TQR is a right angle, we have TQS
2
.5287 1.0421 rad. So,
1 2 1 r 902 1.0421 4220.5 ft 2 . 2 2 1 Similarly, 80 10sin SOR 288.5 2 sin SOR 0.72125 A2
Perimeter: P a b c 9 10 17 36
3600 60 Since the perimeter and area are numerically equal, the given triangle is a perfect triangle. 2K 2K 1 h a , so h1 . Similarly, h 2 a b 2 1 2K and h 3 . Thus, c 1 1 1 a b c h1 h 2 h 3 2 K 2 K 2 K
56. K
SOR 0.8056 rad.
So, SOU A5
2
.8056 0.7652 rad. and
1 802 0.7652 2448.6 ft 2 2
1079 Copyright © 2025 Pearson Education, Inc.
a b c 2s 2K 2K s K
Chapter 9: Applications of Trigonometric Functions
1 1 ah and K ab sin C , which 2 2 means h b sin C . From the Law of Sines, we sin A sin B a sin B know , so b . Therefore, a b sin A a sin B sin C a sin B . h sin C sin A sin A
57. We know K
C cos C 2 60. cot 2 sin C 2
( s b)( s c) ( s a )( s c) bc ac ( s a)( s b) r ab c ( s b)( s c) ( s a )( s c) ab r bc ac ( s a)( s b) c
58. C is on the unit circle so the distance from the origin to C is 1. Therefore: 1 ac sin 2 1 1 1 sin 105 2 1 sin 45 60 2 1 sin 45 cos 60 cos 45 sin 60 2 1 2 1 2 3 1 2 6 2 2 2 2 2 2 4 4
A
2 6 or 8
2 1 3
A B sin 2 2 r ( s a )( s b) ab
c sin
c ab( s a )( s b)( s c) 2 r abc 2 ( s a)( s b)
c ( s c) 2 c s c r r c c2 sc r
61. cot
8
59. Using the formula from Problem 57 with OPQ A B sin 2 2 . Now, gives r sin POQ c sin
A B POQ 180 2 2 and C 1 180 180 C 90 2 2 C C sin 90 cos 2 2 A B c sin sin 2 2 . So, r C cos 2
A B C s a s b s c cot cot 2 2 2 r r r s a s b s c r 3s (a b c) r 3s 2s r s r
62. K Area POQ Area POR Area QOR 1 1 1 rc rb ra 2 2 2 1 r a b c 2 rs
Now, K s ( s a)( s b)( s c) , so rs s ( s a)( s b)( s c) s ( s a)( s b)( s c) s ( s a )( s b)( s c) s
r
63 – 65. Answers will vary.
1080
Copyright © 2025 Pearson Education, Inc.
Section 9.4: Area of a Triangle
66. For f ( x) 3 x 2 12 x 5 , a 3, b 12, c 5. Since a 3 0, the graph opens down, so the vertex is a maximum point. The b 12 12 2. maximum occurs at x 2a 2(3) 6 The maximum value is f (2) 3(2) 2 12(2) 5 12 24 5 17 . 67.
x 1 0 ( x 3)( x 3) The zeros and values where the expression is undefined are x 1, x 3, and x 3 . Interval
, 3
( 3, 1)
( 1, 3)
(3, )
Number Chosen
4
2
0
4
Value of f
3 7
1 5
1 9
5 16
Conclusion
Negative
Positive
Negative
Positive
69. csc sin
1 sin sin 1 sin 2 sin
cos 2 sin cos cos sin cos cot
70. x 2 25 0 x 2 25 so x 5 or x 5 The domain is , 5 5, .
71. Using the Pythagorean Theorem, we have w2 l 2 122
The solution set is x x 3 or 1 x 3 ,
144 w2 l 2
or, using interval notation, , 3 1,3 .
l 144 w
. The perimeter is
2
P 2w 2l 2 w 2 144 w2 7 2 7 2 , ;x ,y 68. P 3 3 3 3 2
2 7 r 3 3
72.
2
2
7 2 1 9 9
2 7 y x sin t ; cos t 3 3 r r csc t
3 3 2 r ; 2 y 2
sec t
3 3 7 r 7 x 7
2 2 14 y tan t 3 7 x 7 7 3 x cot t y
7 3 7 14 2 2 2 3
f ( x) 2 x3 5 x 2 13 x 6 p must be a factor of 6: p 1, 2, 3, 6 q must be a factor of 2: q 1, 2 The possible rational zeros are: p 1 3 , , 1, 2, 3, 6 q 2 2
73. (5 x 7) 5 0.05 0.05 (5 x 7) 5 0.05 0.05 5 x 12 0.05 11.95 5 x 12.05 2.39 x 2.41 The solution set is 2.39, 2.41
x( x 7) 18
74. 2
x 7 x 18 0 ( x 9)( x 2) 0 x 9, x 2 The solution set is 2,9 .
75.
f (1) 3(1) 4 7(1) 2 2 2 . The point on the line is (1, -2). The slope of the tangent line is
1081 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions 15. d 5sin(3t ) a. Simple harmonic
f (1) 12(1)3 14(1) 2 . The equation of the tangent line is: y y1 m( x x1 ) y (2) 2( x 1) y 2 2 x 2 y 2 x
b. 5 meters
2
3 oscillation/second 2
π seconds 1 oscillation/second d. c.
2 0.105 rad/sec 60
17. d 8cos(2t ) a. Simple harmonic
12
d.
b. 4 meters
2 1. 5 5 ; 4 2
3.
2 seconds 3
16. d 4sin(2t ) a. Simple harmonic
Section 9.5
2.
c.
b. 8 meters
c. 1 second d. 1 oscillation/second
6
x y 7 sin 6
18. d 5cos t 2 a. Simple harmonic
4. simple harmonic; amplitude 5. simple harmonic; damped
b. 5 meters
6. True
c.
4 seconds
7. d 5cos t
d.
1 oscillation/second 4
2 8. d 10 cos t 3
1 19. d 9sin t 4 a. Simple harmonic
2 9. d 7 cos t 5
b. 9 meters
10. d 4 cos 4 t
c.
8 seconds
11. d 5sin t
d.
1 oscillation/second 8
20. d 2 cos(2t ) a. Simple harmonic
2 12. d 10sin t 3
b. 2 meters
2 13. d 7 sin t 5
14. d 4sin 4 t
c.
π seconds
d.
1 oscillation/second
1082
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Section 9.5: Simple Harmonic Motion; Damped Motion; Combining Waves 25. d t e t / 2 cos t
21. d 3 7 cos(3t ) a. Simple harmonic b. 7 meters c.
2 second 3
d.
3 oscillations/second 2
22. d 4 3sin(t ) a. Simple harmonic
26. d t e t / 4 cos t
b. 3 meters c.
2 seconds
d.
1 oscillation/second 2
23. d t e t / cos 2t
27.
f x x cos x
28.
f x x cos 2 x
24. d t e t / 2 cos 2t
1083 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions
29.
f x x sin x
33.
f x sin x sin 2 x
f x cos 2 x cos x
30.
f x x cos x
34.
31.
f x sin x cos x
35. a.
f x sin 2 x sin x 1 cos(2 x x) cos(2 x x) 2 1 cos( x) cos(3 x) 2
b.
32.
f x sin 2 x cos x
36. a.
F x sin 3x sin x 1 cos(3x x) cos(3x x) 2 1 cos(2 x) cos(4 x) 2
1084 Copyright © 2025 Pearson Education, Inc.
Section 9.5: Simple Harmonic Motion; Damped Motion; Combining Waves b.
39. a.
H x 2sin 3 x cos x 1 2 sin(3x x ) sin(3 x x) 2 sin(4 x) sin(2 x)
b.
37. a.
G x cos 4 x cos 2 x 1 cos(4 x 2 x) cos(4 x 2 x) 2 1 cos(2 x) cos(6 x) 2
b.
40. a.
g x 2sin x cos 3 x 1 2 sin( x 3 x) sin( x 3x) 2 sin(4 x) sin(2 x) sin(4 x) sin(2 x)
38. a.
h x cos 2 x cos x 1 cos(2 x x) cos(2 x x) 2 1 cos( x) cos(3 x) 2
b.
41. a.
2 2 (0.7) 2 d 10e0.7t / 2(25) cos t 5 4(25) 2 42 0.49 d 10e0.7t / 50 cos t 25 2500
b.
1085 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions
42. a.
46. a.
2 (0.6) 2 d 18e 0.6 t / 2(30) cos t 2 4(30) 2 2 0.36 d 18e 0.6 t / 60 cos t 4 3600
2 2 (0.7) 2 d 5e 0.7 t / 2(10) cos t 3 4(10) 2 2 4 0.49 d 5e 0.7 t / 20 cos t 9 400
b.
b.
b.
43. a.
b.
2 2 (0.75) 2 cos d 15e t 6 4(20) 2 2 0.5625 d 15e0.75 t / 40 cos t 9 1600 0.75 t / 2(20)
47. a.
Damped motion with a bob of mass 20 kg and a damping factor of 0.7 kg/sec.
b. 20 meters leftward
c.
44. a.
2 2 (0.65) 2 d 16e0.65 t / 2(15) cos t 5 4(15) 2 42 0.4225 d 16e0.65 t / 30 cos t 25 900
d. The displacement of the bob at the start of the second oscillation is about 18.33 meters.
b.
e.
The displacement of the bob approaches zero, since e0.7 t / 40 0 as t .
48. a.
Damped motion with a bob of mass 20 kg and a damping factor of 0.8 kg/sec.
45. a.
2 2 (0.8) 2 d 5e0.8 t / 2(10) cos t 3 4(10) 2 2 4 0.64 d 5e0.8 t / 20 cos t 9 400
b. 20 meters leftward
c.
d. The displacement of the bob at the start of the second oscillation is about 18.10 meters. 1086 Copyright © 2025 Pearson Education, Inc.
Section 9.5: Simple Harmonic Motion; Damped Motion; Combining Waves 51. a.
Damped motion with a bob of mass 15 kg and a damping factor of 0.9 kg/sec.
b. 15 meters leftward e.
49. a.
Damped motion with a bob of mass 40 kg and a damping factor of 0.6 kg/sec.
d. The displacement of the bob at the start of the second oscillation is about 12.53 meters.
b. 30 meters leftward c.
c.
The displacement of the bob approaches zero, since e 0.8 t / 40 0 as t .
d. The displacement of the bob at the start of the second oscillation is about 28.47 meters.
e.
The displacement of the bob approaches zero, since e0.9 t / 30 0 as t .
52. a.
Damped motion with a bob of mass 25 kg and a damping factor of 0.8 kg/sec.
b. 10 meters leftward e.
50. a.
Damped motion with a bob of mass 35 kg and a damping factor of 0.5 kg/sec.
d. The displacement of the bob at the start of the second oscillation is 9.53 meters.
b. 30 meters leftward c.
c.
The displacement of the bob approaches zero, since e 0.6 t / 80 0 as t .
e.
d. The displacement of the bob at the start of the second oscillation is about 29.16 meters.
The displacement of the bob approaches zero, since e0.8 t / 50 0 as t .
53. The maximum displacement is the amplitude so we have a 0.80 . The frequency is given by
e.
520 . Therefore, 1040 and the 2 motion of the diaphragm is described by the equation d 0.80 cos 1040 t . f
The displacement of the bob approaches zero, since e 0.5 t / 70 0 as t .
1087 Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions t 0, 2 . The graph of V touches the graph
54. If we consider a horizontal line through the center of the wheel as the equilibrium line, then 390 the amplitude is a 195 . The wheel 18 2 min to complete 1 rev so 1 min = 1 revolution so the period is 18 2 1 and . We want the rider to 9 1 18 be at the lowest position at time t 0 . Since a cos t peaks at t 0 if a 0 and is at its
of y e t / 3 when t 1, 3 . c.
0 t 3 . To do so, we consider the graphs
of y 0.4, y =e t / 3 cos t , and y 0.4 . On the interval 0 t 3 , we can use the INTERSECT feature on a calculator to determine that y e t / 3 cos t intersects y 0.4 when t 0.35, t 1.75 , and
lowest if a 0 , we select a cosine model and need a 195 . Using the model d a cos t b , we have
t 2.19, y e t / 3 cos t intersects y 0.4 when t 0.67 and t 1.29 and the graph shows that 0.4 e t / 3 cos t 0.4 when t 3 .
d 195cos wt b . When t 0 , the rider
should be 10 feet above the ground. That is, 10 195cos 0 b
Therefore, the voltage V is between –0.4 and 0.4 on the intervals 0.35 t 0.67 , 1.29 t 1.75 , and 2.19 t 3 .
10 195 b 205 b Therefore, the equation that describes the rider’s t motion is h(t ) 205 195cos . 9
440 . Therefore, 880 and the 2
329.63 . Therefore, 659.26 and 2 the movement of the tuning fork is described by the equation d 0.025sin 659.26 t .
58. a.
1 1 Let Y1 sin 2 sin 4 x . 2 4 1
t
V
56. The maximum displacement is the amplitude so we have a 0.025 . The frequency is given by
57. a.
movement of the tuning fork is described by the equation d 0.01sin 880 t .
f
55. The maximum displacement is the amplitude so we have a 0.01 . The frequency is given by
f
We need to solve the inequality 0.4 e t / 3 cos t 0.4 on the interval
0
1
b. On the interval 0 t 3 , the graph of V touches the graph of y e t / 3 when 1088 Copyright © 2025 Pearson Education, Inc.
Section 9.5: Simple Harmonic Motion; Damped Motion; Combining Waves
b. Let Y1
1 1 1 sin 2 sin 4 x sin 8 x . 2 4 8
c.
1 1 Let Y1 sin 2 sin 4 x 2 4 1 1 sin 8 x sin 16 x 8 16
61. The sound emitted by touching * is y sin 2 941 t sin 2 1209 t .
Let Y1 sin 2 941 x sin 2 1209 x .
d. f x
59.
1 1 1 sin 2 x sin 4 x sin 8 x 2 4 8 1 1 sin 16 x sin 32 x 16 32
Y1 2.35 sin x
sin(3 x ) 3
sin(5 x ) 5
sin(3 x ) 7
62 – 63. CBL Experiments 64. a.
sin(9 x )
y3 y1 y2 3cos( 1t ) 3cos( 2 t ) 3 cos( 1t ) cos( 2 t )
9
2 2 t 3cos 1 t 3 2 cos 1 2 2 2 2 6 cos 1 t 3cos 1 t 2 2 y3 0 if t
60.
Y1 1.6 cos x
1 9
cos(3 x)
1 25
cos(5 x)
1 49
1 2
or t
1 2
. Since
, y first equals 0 at 1 2 1 2 3 t seconds. 1 2
cos(7 x)
b.
1 t
2 2 2 2 , 2 ; 20 10 T1 19 T2
2 19 10
1089
Copyright © 2025 Pearson Education, Inc.
1 190 4.87 sec 20 19 39 190
Chapter 9: Applications of Trigonometric Functions
No, the waves are not in tune since there is an interference pattern present.
c.
1 67. y sin x x
65. Let Y1
1 y 2 sin x x
sin x . x
1 y 3 sin x x
As x approaches 0,
sin x approaches 1. x
66. y x sin x
Possible observation: As x gets larger, the graph 1 of y n sin x gets closer to y 0 . x 68. Answers will vary.
y x 2 sin x
69. We fill swap the x and y and solve for y. x3 y x4 y 3 x y4 x( y 4) y 3
xy 4 x y 3 xy y 4 x 3 y ( x 1) 4 x 3 4x 3 y x 1 4x 3 1 f ( x) x 1
y x3 sin x
70. log 7 x 3log 7 y log 7 ( x y )
Possible observations: The graph lies between the bounding curves y x, y x 2 , y x3 , respectively, touching them at odd multiples of . The x-intercepts of each graph are the 2 multiples of .
log 7 x log 7 y 3 log 7 ( x y ) log 7 ( xy 3 ) log 7 ( x y ) xy 3 log 7 x y
1090
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Section 9.5: Simple Harmonic Motion; Damped Motion; Combining Waves 71. log( x 1) log( x 2) 1
73. g ( f ( x)) g ( 3 5 x )
log ( x 1)( x 2) 1
10 ( x 1)( x 2) x 2 x 12 0 ( x 4)( x 3) 0 So x 4 or x 3 But x 3 will not work since we cannot take the log of a negative number. The solution set is: 4 4 72. cos , 0 . Thus, 0 , which 5 2 2 4
the Pythorean Theorem, x 5, r 7, y 2 6
lies in quadrant I. 2 x 4, r 5
csc
4 2 y 2 52 , y 0 y 2 25 16 9, y 0 y3
So, sin
cos
2
sin
2
3 3 and tan . 5 4
1 2
4 5
3 y 1 ( x 3) 2 3 9 y 1 x 2 2 3 11 y x 2 2
9 5 9 3 3 10 3 10 2 10 10 10 10
1 x 2 ln x 2 x x 76. 0 2 x2
1 cos 2 1 2
4 5
1 5 1 1 10 10 2 10 10 10 10
The numerator must be zero. 1 x 2 ln x 2 x 0 x x ln x 2 x 0 x(1 2 ln x) 0 1 2 ln x 0 2 ln x 1 1 ln x 2
1 45 1 cos tan 1 cos 1 54 2
1 5 9 5
r 7 7 6 y 12 2 6
75. The slope of the perpendicular line would be 3 m and a point on the line is (3, 1). 2 The normal line is: ( y y1 ) m( x x1 )
1 cos 2
2
3 5 x 7 10 5 x The radicand cannot be negative so 3 5x 0 5 x 3 3 x 5 3 The domain is , . 5 5 74. cos , tan 0 (quadrant IV) 7 Since is in quadrant IV and using
10 x 2 x 2
means
3 5x 7
1 1 9 3
1
e 2 x
1091
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Chapter 9: Applications of Trigonometric Functions
Note that the x=0 answer will not work in the
b 2 c 5 2 B sin 1 23.6º 5 A 90º B 90º 23.6º 66.4º
sin B
original equation so the solution set is e
1
2
77. The function is moved to the left 3 units and is stretched by a factor. This does not change the range so the range is [5,8] . 78. x 2 (5 x 3)( x 2) 0
3. A 50º , B 30º , a 1 C 180º A B 180º 50º 30º 100º
f ( x) x 2 (5 x 3)( x 2)
3 x 0, x , x 2 are the zeros of f . 5 ( , 2)
( 2, 0)
(0, )
3 5
( , )
4
1
0.5
1
Value of f
736
8
0.3125
6
Conclusion
Positive
Negative
Negative
Positive
Interval Number Chosen
sin A sin B a b sin 50º sin 30º 1 b 1sin 30º b 0.65 sin 50º
3 5
sin C sin A c a sin100º sin 50º 1 c 1sin100º c 1.29 sin 50º
3 The solution set is x 2 x or, using 5 3 interval notation, 2, . 5
4. A 100º , c 2, a 5 sin C sin A c a sin C sin100º 2 5 2sin100º 0.3939 sin C 5 2sin100º C sin 1 5 C 23.2º or C 156.8º The value 156.8º is discarded because A C 180º . Thus, C 23.2º . B 180º A C 180º 100º 23.2º 56.8º
Chapter 9 Review Exercises 1. c 10, B 20º b c b sin 20º 10 b 10sin 20º 3.42 sin B
a c a cos 20º 10 a 10 cos 20º 9.40
sin B sin A b a sin 56.8º sin100º 5 b 5sin 56.8º b 4.25 sin100º
cos B
A 90º B 90º 20º 70º
2. b 2, c 5 c2 a 2 b2 a 2 c 2 b 2 52 22 25 4 21 a 21 4.58
1092
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Chapter 9 Review Exercises 5. a 2.4, b 7.4, c 7
sin C sin B c b sin 63.8º sin 80º c 5 5sin 63.8º 4.55 c sin 80º
a b c 2bc cos A 2
2
cos A
2
98 b 2 c 2 a 2 7.42 7 2 2.42 2bc 2(7.4)(7) 103.6
98 A cos 1 18.9º 103.6
8. a 2, b 3, c 1
b 2 a 2 c 2 2ac cos B
a 2 b 2 c 2 2bc cos A
a 2 c 2 b 2 2.42 7 2 7.42 0 cos B 2ac 2(2.4)(7) 33.6
cos A
0 B cos 1 90.0º 33.6 C 180º A B 180º 18.9º 90.0º 71.1º
b 2 c 2 a 2 32 12 22 1 2bc 2(3)(1)
A cos 1 1 0º No triangle exists with an angle of 0˚.
9. a 10, b 7, c 8
6. a 3, c 1, B 100º
a 2 b 2 c 2 2bc cos A
b 2 a 2 c 2 2ac cos B
cos A
32 12 2 3 1cos100º 10 6 cos100º
b 2 c 2 a 2 7 2 82 102 13 2bc 2(7)(8) 112
b 10 6 cos100º 3.32
13 A cos 1 83.33º 112
a 2 b 2 c 2 2bc cos A
b 2 a 2 c 2 2ac cos B
cos A
b2 c 2 a 2 2bc
cos B
( 10 6 cos100º ) 2 12 32 2( 10 6 cos100º )(1)
a 2 c 2 b 2 102 82 7 2 115 2ac 2(10)(8) 160
115 B cos 1 44.05º 160 C 180º 83.33º 44.05º 52.6º
0.45771
A cos 1 0.45771 62.8º C 180º A B 180º 62.8º 100º 17.2º
10. a 1, b 3, C 40º c 2 a 2 b 2 2ab cos C
7. a 3, b 5, B 80º
12 32 2 1 3cos 40º
sin A sin B a b sin A sin 80º 3 5 3sin 80º sin A 0.5909 5 3sin 80º A sin 1 5 A 36.2º or A 143.8º The value 143.8º is discarded because A B 180º . Thus, A 36.2º .
10 6 cos 40º c 10 6 cos 40º 2.32 a 2 b 2 c 2 2bc cos A cos A
b 2 c 2 a 2 32 2.322 12 13.3824 2bc 2(3)(2.32) 13.92
13.3824 A cos 1 16.1º 13.92 B 180º A C 180º 16.1º 40º 123.9º
C 180º A B 180º 36.2º 80º 63.8º
1093
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Chapter 9: Applications of Trigonometric Functions 11. a 5, b 3, A 80º
13. a 3, A 10º , b 4 sin B sin A b a sin B sin10º 4 3 4sin10º sin B 0.2315 3 4sin10º B sin 1 3 B1 13.4º or B2 166.6º For both values, A B 180º . Therefore, there are two triangles.
sin B sin A b a sin B sin 80º 3 5 3sin 80º sin B 0.5909 5 3sin 80º B sin 1 5 B 36.2 or B 143.8 The value 143.8 is discarded because A B 180º . Thus, B 36.2 .
C1 180º A B1 180º 10º 13.4º 156.6º
C 180º A B 180º 80º 36.2º 63.8º
sin A sin C1 a c1
sin C sin A c a sin 63.8º sin 80º c 5 5sin 63.8º c 4.55 sin 80º
12. a 1, b
sin10º sin156.6º 3 c1 c1
C2 180º A B2 180º 10º 166.6º 3.4º
1 4 , c 2 3
sin A sin C2 a c2
a 2 b 2 c 2 2bc cos A 2
sin10º sin 3.4º 3 c2
2
1 4 12 2 3 b c a 1.03 cos A 2bc 1.33 1 4 2 2 3 1.03 A cos 1 39.57º 1.33 2
2
3sin156.6º 6.86 sin10º
2
c2
Two triangles: B1 13.4º , C1 156.6º , c1 6.86 or B2 166.6º , C2 3.4º , c2 1.02
b 2 a 2 c 2 2ac cos B 2
3sin 3.4º 1.02 sin10º
14. a 4, A 20º , B 100º C 180º A B 180º 20º 100º 60º
2
4 1 1 3 a2 c2 b2 2 91 cos B 2ac 96 4 2(1) 3 91 B cos 1 18.57º 96 2
sin A sin B a b sin 20º sin100º b 4 4sin100º 11.52 b sin 20º sin C sin A c a sin 60º sin 20º c 4 4sin 60º 10.13 c sin 20º
C 180º A B 180º 39.57º 18.57º 121.9º
1094
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Chapter 9 Review Exercises 20. a 4, b 3, c 5
15. c 5, b 4 , A 70º
1 1 (a b c ) (4 3 5) 6 2 2 K s ( s a)( s b)( s c )
a b c 2bc cos A 2
2
2
s
a 2 42 52 2 4 5cos 70º 41 40 cos 70º a 41 40 cos 70º 5.23
a 2 b2 c 2 2ab 5.232 42 52 18.3529 2(5.23)(4) 41.48
21. a 4, b 2, c 5
cos C
1 1 (a b c) (4 2 5) 5.5 2 2 K s ( s a)( s b)( s c) s
18.3529 C cos 1 64.0º 41.48 B 180º A C 180º 70º 64.0º 46.0º
22. A 50º , B 30º , a 1 C 180º A B 180º 50º 30º 100º
a 2 b 2 c 2 2bc cos A
K
b 2 c 2 a 2 632 162 652 0 2bc 2(63)(16) 2016
ft
b 2 a 2 c 2 2ac cos B
a 2 c 2 b 2 652 162 632 512 2ac 2(65)(16) 2080
ft
900 4100 900 sin 1 12.7º 4100 The trail is inclined about 12.7º from the lake to the hotel. sin
512 B cos 1 75.7º 2080 C 180º A B 180º 75.7º 90.0º 14.3º
17. a 3, c 1, C 110º sin C sin A c a sin110º sin A 1 3 3sin110º sin A 2.8191 1 No angle A exists for which sin A 1 . Thus, there is no triangle with the given measurements.
24. Let h = the height of the helicopter, x = the distance from observer A to the helicopter, and AHB (see figure).
H x 25º
h 40º
A B ft 180º 40º 25º 115º
18. a 2, b 3, C 40º K
a 2 sin B sin C 12 sin 30º sin100º 0.32 2sin A 2sin 50º
23. Let = the inclination (grade) of the trail. The “rise” of the trail is 4100 5000 900 feet (see figure).
0 A cos 1 90º 2016
cos B
5.5 1.5 3.5 0.5 14.4375 3.80
16. a 65, b 63, c 16
cos A
6 2 31 36 6
c 2 a 2 b 2 2ab cos C
1 1 ab sin C (2)(3) sin 40º 1.93 2 2
sin 40º sin115º x 100 100sin 40º x 70.92 feet sin115º
19. b 4, c 10, A 70º 1 1 K bc sin A (4)(10) sin 70º 18.79 2 2
1095
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Chapter 9: Applications of Trigonometric Functions a 72, b 200, C 15º
h h x 70.92 h 70.92sin 25º 29.97 feet The helicopter is about 29.97 feet high. sin 25º
c 2 a 2 b 2 2ab cos C c 2 722 2002 2 72 200 cos15º 45,184 28,800 cos15º
25. 180º 120º 60º ; 180º 115º 65º ; 180º 60º 65º 55º
A 120º
D
mi
E
c 131.8 miles The sailboat is about 131.8 miles from the island.
b. Find the measure of the angle opposite the 200 side: a 2 c2 b2 cos B 2ac 2 72 131.82 2002 17, 444.76 cos B 2(72)(131.8) 18,979.2
0.25 mi C 0.25 mi
B 115º
17, 444.76 B cos 1 156.802º 18,979.2 The sailboat should turn through an angle of 180 156.8 23.2 to correct its course.
sin 60º sin 55º BC 3 3sin 60º BC 3.17 mi sin 55º
c.
sin 65º sin 55º AC 3 3sin 65º AC 3.32 mi sin 55º BE 3.17 0.25 2.92 mi
27. Find the lengths of the two unknown sides of the middle triangle: x 2 1002 1252 2 100 125 cos 50º
AD 3.32 0.25 3.07 mi
For the isosceles triangle, CDE CED
180º 55º 62.5º 2
25, 625 25, 000 cos 50º x 25, 625 25, 000 cos 50º 97.75 feet
sin 55º sin 62.5º DE 0.25 0.25sin 55º 0.23 miles DE sin 62.5º
y 2 702 502 2 70 50 cos100º 7400 7000 cos100º y 7400 7000 cos100º 92.82 feet
The length of the highway is 2.92 3.07 0.23 6.22 miles. 26. a.
Find the areas of the three triangles: 1 K1 (100)(125) sin 50º 4787.78 ft 2 2 1 K 2 (50)(70) sin100º 1723.41 ft 2 2 1 s (50 97.75 92.82) 120.285 2 K 3 120.285 70.285 22.535 27.465
After 4 hours, the sailboat would have sailed 18(4) 72 miles. Find the third side of the triangle to determine the distance from the island:
ST
mi
15º
B
mi
The original trip would have taken: 200 t 11.11 hours . The actual trip takes: 18 131.8 t 4 11.32 hours . The trip takes 18 about 0.21 hour, or about 12.6 minutes longer.
BWI c
2287.47 ft 2
The approximate area of the lake is 4787.78 + 1723.41 + 2287.47 = 8798.67 ft 2 .
1096
Copyright © 2025 Pearson Education, Inc.
Chapter 9 Review Exercises 28. Construct a diagonal. Find the area of the first triangle and the length of the diagonal:
40 4 10 AMB tan 1 4 76.0º
tan AMB
ft A 100º ft
d
80º 76.0º 4.0º The bearing is S4.0E .
B
40º
31. d 3cos t 2
ft
1 K1 50 100 sin 40º 1606.969 ft 2 2 2 d 502 1002 2 50 100 cos 40º
32. d 6sin(2 t ) a. Simple harmonic
12,500 10, 000 cos 40º
b. 6 feet
d 12,500 10, 000 cos 40º 69.566914 feet
Use the Law of Sines to find B : sin B sin100º 20 69.566914 20sin100º sin B 69.566914 20sin100º B sin 1 16.447º 69.566914
c.
seconds
d.
1 oscillation/second
33. d 2 cos( t ) a. Simple harmonic b. 2 feet
A 180º 100º 16.447º 63.553º
Find the area of the second triangle: 1 K 2 20 69.566914 sin 63.553º 622.865 ft 2 2
c.
2 seconds
d.
1 oscillation/second 2
34. a.
The cost of the parcel is approximately 100(1606.969 622.865) $222,983.40 .
42 0.5625 d 15e0.75 t / 80 cos t 25 6400
29. To find the area of the segment, we subtract the area of the triangle from the area of the sector. 1 1 ASector r 2 62 50 15.708 in 2 2 2 180 1 1 A Triangle ab sin 6 6sin 50º 13.789 in 2 2 2 ASegment 15.708 13.789 1.92 in 2
b.
25
–15
Damped motion with a bob of mass 20 kg and a damping factor of 0.6 kg/s. b. 15 meters leftward 15 c.
35. a.
80º
15 0
30. Find angle AMB and subtract from 80˚ to obtain . M
2 2 (0.75) 2 d 15e0.75 t / 2(40) cos t 5 4(40) 2
25
0
–15 B
d. The displacement of the bob at the start of the second oscillation is about 13.92 meters leftward.
1097
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Chapter 9: Applications of Trigonometric Functions
Note that 5 miles = 26,400 feet. 600 1 26400 44 1 A tan 1 1.3 44
tan A
It approaches zero, since e 0.6 t / 40 0 as t .
e.
The angle of depression from the balloon to the airport is about 1.3 .
36. y 2sin x cos 2 x , 0 x 2
3. Use the law of cosines to find a: a 2 b 2 c 2 2bc cos A (17) 2 (19) 2 2(17)(19) cos 52 289 361 646(0.616) 252.064 a 252.064 15.88
Use the law of sines to find B . a b sin A sin B 15.88 17 sin 52 sin B 17 sin B (sin 52) 0.8436 15.88
Since b is not the longest side of the triangle, we know that B 90 . Therefore, B sin 1 (0.8436) 57.5
Chapter 9 Test 1. Let = the angle formed by the ground and the ladder. ft
C 180 A B 180 52 57.5 70.5
4. Use the Law of Sines to find b : ft
a b sin A sin B 12 b sin 41 sin 22 12 sin 22 b 6.85 sin 41
A
Then sin A
10.5 12
10.5 A sin 1 61.0 12 The angle formed by the ladder and ground is about 61.0 .
C 180 A B 180 41 22 117 Use the Law of Sines to find c: a c sin A sin C 12 c sin 41 sin117 12 sin117 c 16.30 sin 41
2. Let A = the angle of depression from the balloon to the airport. mi A MSFC
ft AP
1098
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Chapter 9 Test 5. Use the law of cosines to find A . a 2 b 2 c 2 2bc cos A
sin B sin A b a sin B sin 40º 7 3 7 sin 40º sin B 1.4998 3 There is no angle B for which sin B 1 . Therefore, there is no triangle with the given measurements.
82 (5) 2 (10) 2 2(5)(10) cos A 64 25 100 100 cos A 100 cos 61 cos A 0.61 A cos 1 (0.61) 52.4
Use the law of sines to find B . a b sin A sin B 8 5 sin 52.41 sin B 5 sin B sin 52.4 8 sin B 0.495 Since b is not the longest side of the triangle, we have that B 90 . Therefore B sin 1 (0.495) 29.7
8. a 8, b 4, C 70º c 2 a 2 b 2 2ab cos C c 2 82 42 2 8 4 cos 70º 80 64 cos 70º c 80 64 cos 70º 7.62 a 2 b 2 c 2 2bc cos A cos A
b 2 c 2 a 2 42 7.622 82 10.0644 2bc 2(4)(7.62) 60.96
10.0644 A cos 1 80.5º 60.96
C 180 A B 180 52.4 29.7
B 180º A C 180º 80.5º 70º 29.5º
97.9
9. a 8, b 4, C 70º
6. A 55º , C 20º , a 4 B 180º A C 180º 55º 20º 105º Use the law of sines to find b.
1 ab sin C 2 1 (8)(4) sin 70 15.04 square units 2
K
sin A sin B a b sin 55º sin105º 4 b 4sin105º b 4.72 sin 55º
10. a 8, b 5, c 10 1 1 a b c 8 5 10 11.5 2 2 K s ( s a)( s b)( s c)
s
Use the law of sines to find c. sin C sin A c a sin 20º sin 55º 4 c 4sin 20º c 1.67 sin 55º
11.5(11.5 8)(11.5 5)(11.5 10) 11.5(3.5)(6.5)(1.5) 392.4375 19.81 square units
11. We can find the area of the shaded region by subtracting the area of the triangle from the area of the semicircle. Since triangle ABC is a right triangle, we can use the Pythagorean theorem to find the length of the third side.
7. a 3, b 7, A 40º Use the law of sines to find B
1099
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Chapter 9: Applications of Trigonometric Functions sin A sin 40 OB AB sin 70 sin 40 AB 5 5sin 40 AB 3.420 sin 70 Now, AB is the diameter of the semicircle, so the 3.420 radius is 1.710 . 2 1 1 2 ASemicircle r 2 1.710 4.593 sq. units 2 2
a 2 b2 c 2 2
2
a 6 8
2
a 2 64 36 28 a 28 2 7 The area of the triangle is 1 1 A bh 2 7 6 6 7 square cm . 2 2 The area of the semicircle is 2 1 1 A r 2 4 8 square cm . 2 2 Therefore, the area of the shaded region is 8 6 7 9.26 square centimeters .
1 ab sin(O) 2 1 (5)(5)(sin 40) 8.035 sq. units 2
ATriangle
12. Begin by adding a diagonal to the diagram. 5
72
11 d
7
ATotal ASemicircle ATriangle 4.593 8.035 12.63 sq. units
8
15. Using Heron’s formula: 5 x 6 x 7 x 18 x s 9x 2 2 K 9 x(9 x 5 x)(9 x 6 x)(9 x 7 x)
1 (5)(11)(sin 72) 26.15 sq. units 2 By the law of cosines, d 2 (5) 2 (11) 2 2(5)(11)(cos 72) 25 121 110(0.309) 112.008 Aupper
9 x 4 x 3x 2 x 216 x 4
d 112.008 10.58 Using Heron’s formula for the lower triangle,
6 6 x2
Thus, (6 6) x 2 54 6
7 8 10.58 12.79 2 Alower 12.79(5.79)(4.79)(2.21) s
x2 9 x3 The sides are 15, 18, and 21.
783.9293 28.00 sq. units Total Area 26.15 28.00 54.15 sq. units
16. Since we ignore all resistive forces, this is simple harmonic motion. Since the rest position t 0
13. Use the law of cosines to find c: c 2 a 2 b 2 2ab cos C
is the vertical position d 0 , the equation will
(4.2) 2 (3.5) 2 2(4.2)(3.5) cos 32
have the form d a sin(t ) .
17.64 12.25 29.4(0.848) 4.9588
42
c 4.9588 2.23 Madison will have to swim about 2.23 miles.
5 feet a
14. Since OAB is isosceles, we know that 180 40 A B 70 2 Then, 1100
Copyright © 2025 Pearson Education, Inc.
Chapter 9 Cumulative Review
Now, the period is 6 seconds, so 2 6
3.
f x x 2 3x 4
f will be defined provided
2
g x x 2 3x 4 0 .
6
x 2 3x 4 0 x 4 x 1 0
radians/sec 3 From the diagram we see a sin 42 5 a 5 sin 42
x 4, x 1 are the zeros.
t Thus, d 5 sin 42 sin or 3 t d 3.346 sin . 3
Interval
Test Number
g ( x) Pos./Neg.
x 1
2
6
Positive
1 x 4
0
4
Negative
4 x
5
6
Positive
The domain of f x x 2 3x 4 is
x x 1 or x 4 . 4. y 3sin x
Chapter 9 Cumulative Review
Amplitude:
A 3 3
Period:
T
Phase Shift:
0 0
2
3x 1 4 x
1. 2
3x 4 x 1 0 3x 1 x 1 0 x
2
2
1 or x 1 3
1 The solution set is ,1 . 3
2. Center (5, 1) ; Radius 3
x h 2 y k 2 r 2 x (5) 2 y 12 32 x 52 y 12 9
5. y 2 cos 2 x 2 cos 2 x 2 Amplitude: A 2 2 2 2
Period:
T
Phase Shift:
2
1101
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Chapter 9: Applications of Trigonometric Functions
6. tan 2,
3 2 , so lies in quadrant IV. 2
f.
5
a.
b.
c.
3 2 , so sin 0 . 2 2 5 2 5 sin 5 5 5
1 cos 1 cos 2 2
7. a.
sin(2 ) 2sin cos
cos(2 ) cos sin 2
2
5 2 5 5 5 5 20 25 25 15 3 25 5
e.
y ex , 0 x 4
2 5 5 2 5 5 20 4 25 5
d.
1 2
5 5
5 5 5 2
3 2 , so cos 0 2 1 5 5 cos 5 5 5
2
3 2 2 3 1 4 2 1 1 Since lies in Quadrant II, cos 0 . 2 2
b.
2
5 5 10
y sin x , 0 x 4
c.
y e x sin x , 0 x 4
5 1 1 cos 1 5 sin 2 2 2
3 2 2 3 1 4 2 1 1 Since lies in Quadrant II, sin 0 . 2 2
5 5 5 2
d.
y 2 x sin x , 0 x 4
1102
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5 5 10
Chapter 9 Cumulative Review 8. a.
yx
e.
y ex
b.
y x2
f.
y ln x
c.
y x
g.
y sin x
h.
y cos x
d.
y x3
1103
Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions i.
y tan x
10. 3x5 10 x 4 21x3 42 x 2 36 x 8 0 Let f x 3 x5 10 x 4 21x3 42 x 2 36 x 8 0 f ( x ) has at most 5 real zeros.
Possible rational zeros: p 1, 2, 4, 8; q 1, 2, 3; p 1 1 2 4 8 1, , , 2, , 4, , 8, 2 3 3 3 3 q
Using the Bounds on Zeros Theorem: 10 8 f ( x) 3 x5 x 4 7 x3 14 x 2 12 x 3 3 10 8 a4 , a3 7, a2 14, a1 12, a0 3 3
9. a 20, c 15, C 40o sin C sin A c a sin 40o sin A 15 20 20sin 40o sin A 15 20sin 40o A sin 1 15
8 Max 1, 12 3 Max 1, 39 39
o
o
o
o
sin C sin B1 c b1
sin 40o sin 81.01o 15 b1 15sin 81.01o sin 40
o
o
o
o
o
o
Using synthetic division with 1: 1 3 10 21 42 36 8 3 7 14 28 8
sin 40o sin18.99o 15 b2 b2
From the graph it appears that there are x1 intercepts at ,1, and 2. 3
sin C sin B2 c b2
15sin18.99o
23.05
B2 180 A2 C 180 40 121.01 18.99
3
The smaller of the two numbers is 15. Thus, every zero of f lies between –15 and 15. Graphing using the bounds: (Second graph has a better window.)
B1 180 A1 C 180 40 58.99 81.01
b1
10
8 10 1 Max , 12 , 14 , 7 , 3 3 1 14 15
A1 58.99o or A2 121.01o For both values, A C 180º . Therefore, there are two triangles. o
14 7
3
7.59
7 14 28
8
0
Two triangles: A1 59.0o , B1 81.0o , b1 23.05
Since the remainder is 0, x 1 is a factor. The other factor is the quotient: 3x 4 7 x3 14 x 2 28 x 8 .
A2 121.0o , B2 19.0o , b2 7.59 .
Using synthetic division with 2 on the quotient:
sin 40o
or
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Chapter 9 Cumulative Review
2 3 7 14 28 8 6 2 24 8 3 1
4
12
R ( x)
0
x5 0 or x 3 0 x 5 x3
Since n m , the line y 2 is the horizontal asymptote. 26 R ( x) intersects y 2 at , 2 , since: 11 2 2x 7x 4 2 x 2 2 x 15 2 x 2 7 x 4 2 x 2 2 x 15
0 12 0 1 is a factor. The 3
other factor is the quotient:
3x 2 12 3 x 2 4 3 x 2i x 2i .
Factoring,
2
1 . The imaginary 3
Graphing utility:
zeros are 2i and 2i. Therefore, over the complex numbers, the equation 3x5 10 x 4 21x3 42 x 2 36 x 8 0 has solution 1 set 2i, 2i, , 1, 2 . 3 11. R ( x)
2 x2 7 x 4
2
2 x 7 x 4 2 x 4 x 30 11x 26 26 x 11
1 f ( x) 3( x 1) x 2 x x 2i x 2i 3
The real zeros are 1,2, and
which
x 2 2 x 15 0 x 5 x 3 0
1 3 1 12 4 3 1 0 4
2 x2 7 x 4
The vertical asymptotes are the zeros of q( x) :
1 Using synthetic division with on the quotient: 3
Since the remainder is 0, x
( x) 2 2( x) 15 x 2 2 x 15 is neither R( x) nor R ( x) , so there is no symmetry.
Since the remainder is 0, x 2 is a factor. The other factor is the quotient: 3x3 x 2 12 x 4 .
3
2( x) 2 7( x) 4
Graph by hand:
(2 x 1)( x 4) ( x 3)( x 5)
Interval
Location Point
p( x) 2 x 2 7 x 4; q ( x) x 2 2 x 15; n 2; m 2
Test Value number of f
, 5
6
12.22
Above x-axis
6,12.22
Domain: x x 5, x 3
5, 0.5
1
0.3125
Below x-axis
1, 0.3125
0.5,3
0
0.27
Above x-axis
0, 0.27
3, 4
3.5
0.94
Below x-axis
3.5, 0.94
4,
5
0.55
Above x-axis
5, 0.55
2
x 2 x 15
R is in lowest terms. The x-intercepts are the zeros of p ( x) : 2 x2 7 x 4 0 2 x 1 x 4 0 2x 1 0 or x 4 0 1 x x4 2 The y-intercept is 2 02 7 0 4 4 4 R (0) 2 . 0 2 0 15 15 15
1105
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Chapter 9: Applications of Trigonometric Functions
c.
f x g x 4 x 4 x 2 5 x 24 0 x 2 x 28 x
1 12 4(1)(29) 2(1)
1 117 1 3 13 2 2 1 3 13 1 3 13 , . 2 2
The solution set is 3x 12
12.
d.
ln 3x ln 12
4x 5 0 4 x 5 5 x 4
x ln 3 ln 12 x
ln 12 ln 3
2.26
5 5 The solution set is x x or , . 4 4
The solution set is {2.26}. 13. log3 x 8 log3 x 2
2
x 5 x 24 0 x 8 x 3 0
x 8 x 32 x2 8x 9
x 8, x 3 are the zeros.
x2 8x 9 0 x 9 x 1 0 x 9 or x 1 x 9 is extraneous because it makes the original logarithms undefined. The solution set is 1 .
Interval Test number 9 , 8
8,3 3,
g ( x) Pos./Neg. 12 Positive
0
24
Negative
4
12
Positive
The solution set is x 8 x 3 or 8,3 .
f ( x) 4 x 5 ; g ( x) x 2 5 x 24
a.
g x 0
e.
log 3 x 8 x 2
14.
f x 0
f.
y f ( x) 4 x 5 The graph of f is a line with slope 4 and yintercept 5.
g.
y g ( x) x 2 5 x 24 The graph of g is a parabola with y-intercept 24 and x-intercepts 8 and 3. The xcoordinate of the vertex is
f x 0 4x 5 0 4 x 5 5 x 4 5 The solution set is . 4
b.
f x 13 4 x 5 13 4x 8 x2 The solution set is 2 .
1106
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Chapter 9 Projects
x
C
b 5 5 2.5 . 2a 2 1 2
The y-coordinate of the vertex is b y f f (2.5) 2a (2.5) 2 5(2.5) 24 30.25
a
Q
O
sin a
The vertex is 2.5, 30.25 . 2.
CQ , OC 2 CQ 2 OQ 2 OQ
For OPQ , we have that ( PQ ) 2 (OP ) 2 (OQ ) 2 2(OQ )(OP ) cos c
For CPQ , we have that ( PQ) 2 (CQ) 2 (CP ) 2 2(CQ )(CP ) cos C
3.
0 (OP ) 2 (OQ) 2 2(OQ)(OP ) cos c (CQ) 2 (CP) 2 2(CQ )(CP ) cos C 2(OQ)(OP ) cos c (OQ) 2 (CQ) 2 (OP) 2 (CQ) 2 2(CQ)(CP) cos C
4.
Chapter 9 Projects
(OC ) 2 (OP ) 2 (CP ) 2
Thus, 2(OQ)(OP) cos c OQ 2 CQ 2 OP 2
Project I C
1.
CQ 2 2(CQ)(CP) cos C 2(OQ)(OP) cos c OC 2 OC 2 2(CQ)(CP ) cos C
a
O
c
From part a: (OC ) 2 (OQ) 2 (CQ) 2
a b
B
Q
b c
A
P
5.
Triangles OCP and OCQ are plane right triangles.
cos c
2OC 2 2(CQ)(CP ) cos C 2(OQ)(OP ) 2(OQ)(OP )
cos c
(CQ)(CP ) cos C OC 2 (OQ)(OP ) (OQ)(OP )
C
O
b
sin b
(CQ)(CP) cos C OC 2 (OQ)(OP ) (OQ)(OP ) OC OC CQ CP cos C OQ OP OQ OP cos a cos b sin a sin b cos C
cos c
P
CP , OC 2 CP 2 OP 2 OP
1107
Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions Project II 1.
4.
Lewiston and Clarkston
Putting this on a circle of radius 1 in order to apply the Law of Cosines from A:
45 N 42.5
45
x 282 sin 48.5 sin 89.5 282sin 48.5 x 211.2 sin 89.5
0
111.5W
y 282 sin 42 sin 89.5 282sin 42 y 188.8 sin 89.5
113.5W O 113.5 111.3 2.2
cos c cos 45 cos 42.5 sin 45 sin 42.5 cos 2.2 cos c 0.99870 c 2.93
Using a plane triangle, they traveled 211.2 188.7 399.9 miles.
s rc 3960 2.93 202.5 180
The mileage by using spherical triangles and that by using a plane triangle are relatively close. The total mileage is basically the same in each case. This is because compared to the surface of the Earth, these three towns are very close to each other and the surface can be approximated very closely by a plane.
It is 202.5 miles from Great Falls to Lemhi. N
43.5
45
117.0W
46.5 N 45 N
c 43.5
Lemhi
45
Project III
0
1.
113.5W O 117.0 113.5 3.5
cos c cos 45 cos 43.5 sin 45 sin 43.5 cos 3.5 cos c 0.99875 c 2.87
f1 sin( t ) 1 f3 sin(3 t ) 3 1 f5 sin(5 t ) 5 1 f 7 sin(7 t ) 7 1 f9 sin(9 t ) 9
s rc 3960 2.87 198.4 180 It is about 198.4 miles from Lemhi to Lewiston and Clarkston. 3.
x
180 48.5 42 89.5
47.5 N
c
Lewiston/Clarkton
48.5
Great Falls
Great Falls
42.5
Lemhi
2.
42
y
N 45
282 miles
They traveled 202.5 198.4 miles just to go from Great Falls to Lewiston and Clarkston.
1108
Copyright © 2025 Pearson Education, Inc.
Chapter 9 Projects
2.
f1 sin( t )
11.67 184.5 11.67 sin 1 4 184.5
d. sin
1 f1 f3 sin( t ) sin(3 t ) 3 1 1 f1 f3 f5 sin( t ) sin(3 t ) sin(5 t ) 3 5 f1 f3 f5 f 7 1 1 1 sin( t ) sin(3 t ) sin(5 t ) sin(7 t ) 3 5 7 f1 f3 f5 f 7 f9 1 1 sin( t ) sin(3 t ) sin(5 t ) 3 5 1 1 sin(7 t ) sin(9 t ) 7 9
e.
f. The angles are relatively small and for part (d), where 4 was acquired, it was arrived at by rounding. g. Answers will vary.
3. If one graphs each of these functions, one observes that with each iteration, the function becomes more square. 4.
h sin 86 184.5 h 184.5sin 86 184.1 ft
Project V a. Answers will vary.
1 f1 f13 f3 sin( t ) cos(2 t ) sin(3 t ) 2 By adding in the cosine term, the curve does not become as flat. The waves at the “tops” and the “bottoms” become deeper.
Mountain
b.
50 yd
Project IV Rock
a.
6
184 .5
ft
70 yd 84
40
13 ft
Palm Tree
b. Let h = the height of the tower with a lean of 6 .
184.5
ft
6
h 84
sin sin 40 70 50 70sin 40 sin 0.8999 50 70sin 40 sin 1 64 50
h sin 84 184.5 h 184.5sin 84 183.5 feet
13 ft
c. 180 64 116
2 ft 3
1 84.5
ft
c. 13 ft 16 in 11 ft 8 in 11
11 _23 ft
1109
Copyright © 2025 Pearson Education, Inc.
Chapter 9: Applications of Trigonometric Functions Mountain 50 yd
d.
Rock
116 x
70 yd 40
d L sin(180 ) sin( ) L sin(180 ) d sin( ) L sin(180) cos( ) cos180 sin( ) d sin( ) d
Palm Tree
The third angle (i.e., at the rock) is 24 . x 70 sin 24 sin116 70sin 24 x 32 sin116 The treasure is about 32 yards away from the palm tree.
d
L (0) cos( ) (1) sin( ) sin( )
L sin( ) sin( ) y l y l sin
sin
y
L sin sin sin( )
Project VI highest point
a.
home plate
b. d
fence
ce stan l di a m 22.5 in i l =m
10
410’
147.5 32.5
x
d 410 sin147.5 sin 22.5 410sin147.5 d sin 22.5 d 575.7 feet
c.
y = height
nc e ista d l im a min l d= x L
l 410 sin10 sin 22.5 410sin10 l sin 22.5 l 186.0 feet
y = height
d L sin(180 ) sin( ) L sin(180 ) d sin( ) l L sin sin( ) L sin l sin( )
1110
Copyright © 2025 Pearson Education, Inc.
Chapter 10 Polar Coordinates; Vectors Section 10.1
15. C 16. C
1.
17. B 18. D 19. A The point lies in quadrant IV.
x2 x1 2 y2 y1 2
2.
20. D 21.
2
6 3. 9 2
4. 22.
5.
23.
b a
6.
24.
4
7. pole, polar axis
25.
8. r cos ; r sin 9. b 10. d 11. True
26.
12. False 13. A 14. B 1111 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors 34.
27.
28.
35.
29.
30.
a.
r 0, 2 0
4 5, 3
b.
r 0, 0 2
5 5, 3
c.
r 0, 2 4
8 5, 3
a.
r 0, 2 0
5 4, 4
b.
r 0, 0 2
7 4, 4
c.
r 0, 2 4
11 4, 4
a.
r 0, 2 0
2, 2
b.
r 0, 0 2
2,
c.
r 0, 2 4
2, 2
36.
31.
32.
37.
33.
1112 Copyright © 2025 Pearson Education, Inc.
Section 10.1: Polar Coordinates 38.
42.
a.
r 0, 2 0
3,
b.
r 0, 0 2
3, 0
c.
r 0, 2 4
3, 3
39.
r 0, 2 0
3 1, 2
b.
r 0, 0 2
3 1, 2
c.
r 0, 2 4
5 1, 2
a.
r 0, 2 0
2,
b.
r 0, 0 2
2, 0
c.
r 0, 2 4
2, 3
r 0, 2 0
5 2, 3
b.
r 0, 0 2
4 2, 3
c.
r 0, 2 4
7 2, 3
3 0 0 2 y r sin 3sin 3 1 3 2
43. x r cos 3cos
Rectangular coordinates of the point 3, are 2 0, 3 .
40.
a.
a.
3 40 0 2 3 y r sin 4sin 4 (1) 4 2
44. x r cos 4 cos
3 Rectangular coordinates of the point 4, are 2 0, 4 .
41.
45. x r cos 2 cos 0 2 1 2 y r sin – 2sin 0 – 2 0 0 Rectangular coordinates of the point – 2, 0 are
2, 0 . a.
r 0, 2 0
5 3, 4
b.
r 0, 0 2
7 3, 4
c.
(r 0, 2 4
11 3, 4
46. x r cos 3cos 3(1) 3 y r sin 3sin 3 0 0 Rectangular coordinates of the point 3, are
3, 0 .
1113 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
47. x r cos 6 cos y r sin 6sin
2 52. x r cos 6 cos 6 3 2 4 2
5 3 6 3 3 6 2
5 1 6 3 6 2
2 y r sin 6sin 6 3 2 4 2
5 Rectangular coordinates of the point 6, 6
Rectangular coordinates of the point 6, 4
are 3 3, 3 .
5 1 5 5 3 2 2 5 3 5 3 5 y r sin 5sin 3 2 2 5 Rectangular coordinates of the point 5, 3
53. x r cos 2 cos( ) 2 1 2
48. x r cos 5cos
y r sin – 2sin( ) – 2 0 0
Rectangular coordinates of the point – 2, are 2, 0 . 54. x r cos 3cos 3 0 0 2 y r sin – 3sin 3(1) 3 2 Rectangular coordinates of the point – 3, 2 are 0, 3 .
5 5 3 are , . 2 2
49. x r cos 2 cos
3 2 2 2 4 2
3 2 2 2 4 2 3 Rectangular coordinates of the point 2, 4 y r sin 2sin
are
11 7.5( 0.3420) 2.57 18 11 y r sin 7.5sin 7.5(0.9397) 7.05 18 11 Rectangular coordinates of the point 7.5, 18 are about 2.57, 7.05 .
55. x r cos 7.5cos
2, 2 .
50. x r cos 2 cos
2 1 2 1 3 2
2 3 2 3 3 2 2 Rectangular coordinates of the point 2, 3 y r sin 2sin
are 3 2, 3 2 .
91 3.1( 0.9994) 3.10 90 91 y r sin 3.1sin 3.1( 0.0349) 0.11 90 91 Rectangular coordinates of the point 3.1, 90 are about 3.10, 0.11 .
56. x r cos 3.1cos
are 1, 3 . 3 5 3 51. x r cos 5cos 5 6 2 2 1 5 y r sin 5sin 5 6 2 2
57. x r cos 6.3cos 3.8 6.3( 0.7910) 4.98 y r sin 6.3sin 3.8 6.3( 0.6119) 3.85
Rectangular coordinates of the point 5, 6 5 3 5 , . are 2 2
Rectangular coordinates of the point 6.3, 3.8 are about 4.98, 3.85 .
1114 Copyright © 2025 Pearson Education, Inc.
Section 10.1: Polar Coordinates 58. x r cos 8.1cos 5.2 8.1(0.4685) 3.79
64. The point (3, 3) lies in quadrant II.
y r sin 8.1sin 5.2 8.1( 0.8835) 7.16
r x 2 y 2 (3) 2 32 3 2
Rectangular coordinates of the point 8.1, 5.2
y 3 Polar coordinates of the point 3, 3 are
are about 3.79, 7.16 . 59. r x 2 y 2 32 02 9 3
3 3 2, . 4
y 0 tan 1 tan 1 tan 1 0 0 x 3 Polar coordinates of the point (3, 0) are (3, 0) .
r x 2 y 2 52 5 3
2
10, . 3 3 1 , lies in quadrant III. 66. The point 2 2
61. r x 2 y 2 (1) 2 02 1 1 y 0 The point lies on the negative x-axis, so . Polar coordinates of the point (1, 0) are 1, .
tan 1 tan 1 tan 1 0 0 x 1
2
62. r x 2 y 2 02 ( 2) 2 4 2 2
tan 1 tan 1 x 0 2 is undefined, . 0 2
. 2 Polar coordinates of the point (0, 2) are 2, . 2
The point lies on the negative y-axis, so
63. The point (1, 1) lies in quadrant IV. 2
2
2
1 1
2 3 1 r x 2 y 2 1 1 2 2 1 y 2 tan 1 1 11 tan 1 tan 1 6 x 3 3 2 7 The point lies in quadrant III, so 6 6 3 1 Polar coordinates of the point , are 2 2 7 1, . 6
67. The point (1.3, 2.1) lies in quadrant IV. r x 2 y 2 1.32 ( 2.1) 2 6.1 2.47
2
r x y 1 (1) 2
y 2.1 The polar coordinates of the point 1.3, 2.1 are
tan 1 tan 1 1.02 1.3 x
tan tan tan (1) 4 x 1 Polar coordinates of the point (1, 1) are 1 y
Polar coordinates of the point 5,5 3 are
Polar coordinates of the point (0, 2) are 2, . 2
Since
2
tan 1
2 is undefined, 0 2
y
100 10
5 3 1 3 tan 5 3
tan 1 tan 1 x 0 Since
65. The point 5,5 3 lies in quadrant I.
60. r x 2 y 2 02 22 4 2 y
tan 1 tan 1 tan 1 (1) x 3 4
1
2.47, 1.02 .
2, . 4
1115
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors 68. The point (0.8, 2.1) lies in quadrant III.
75.
r x 2 y 2 ( 0.8)2 ( 2.1) 2 5.05 2.25
r 2 2sin cos 1
1 2.1
1 y
2 xy 1 2(r cos )(r sin ) 1
tan tan 1.21 x 0.8
r 2 sin 2 1
Since the point lies in quadrant III, 1.21 1.93 . The polar coordinates of the point 0.8, 2.1 are
4x2 y 1
76.
4(r cos ) 2 r sin 1 4r 2 cos 2 r sin 1 1 r 3 cos 2 sin 4
2.25, 1.93 . 69. The point (8.3, 4.2) lies in quadrant I. r x 2 y 2 8.32 4.22 86.53 9.30 y 4.2 The polar coordinates of the point 8.3, 4.2 are
tan 1 tan 1 0.47 x 8.3
9.30, 0.47 .
77.
x4 r cos 4
78.
y 3 r sin 3 r cos
79.
70. The point (2.3, 0.2) lies in quadrant II. 2
2
2
r r cos 2
2
x y2 x
2
r x y ( 2.3) 0.2 5.33 2.31
x2 x y 2 0 1 1 x2 x y 2 4 4 2 1 1 2 x y 2 4
0.2 y Since the point lies in quadrant II, 0.09 3.05 . The polar coordinates of the point 2.3, 0.2
tan 1 tan 1 0.09 2.3 x
are 2.31, 3.05 . 71.
r r sin r 2
2 x2 2 y 2 3
2
x y 2 y x2 y 2
2 x2 y2 3 2r 2 3 3 6 2 2
r 2r sin r cos 2
2
x y2 2 y x x2 y 2 2 y x 0
72. x 2 y 2 x r 2 r cos r cos
r sin cos
82.
r r sin r cos 2
2
x y2 y x
2
x 4y
x2 x y 2 y 0 1 1 1 1 x2 x y 2 y 4 4 4 4 2 2 1 1 1 x y 2 2 2
r cos 2 4r sin r 2 cos 2 4r sin 0
74.
r 2sin cos
81.
3 r or r 2 2
73.
r sin 1
80.
y2 2x
r sin 2 2r cos r 2 sin 2 2r cos 0
1116
Copyright © 2025 Pearson Education, Inc.
Section 10.1: Polar Coordinates r2
83.
3 3 cos r (3 cos ) 3 3r r cos 3 r
86.
2
r 4 x2 y 2 4 r4
84.
3 x2 y2 x 3
2
r 16
3 x2 y 2 x 3
2
x y 2 16
2
9 x 9 y x2 6 x 9
4 1 cos r (1 cos ) 4 r r cos 4 r
85.
2
8x2 6 x 9 y2 9 0 64 x 2 48 x 72 y 2 72 0 3 64 x 2 x 72 y 2 72 4 9 2 3 9 64 x x 72 y 2 72 64 4 64 64
x2 y 2 x 4 x2 y 2 x 4 x 2 y 2 x 2 8 x 16
2
y2 8 x 2
87. a.
9 x2 y2 x2 6 x 9
3 64 x 72 y 2 81 8
For this application, west is a negative direction and north is positive. Therefore, the rectangular coordinate is (10, 36) .
b. The distance r from the origin to (10, 36) is r x 2 y 2 ( 10) 2 (36) 2 1396 2 349 37.36 . y Since the point (10, 36) lies in quadrant II, we use 180 tan 1 . Thus, x 36 1 18 180 tan 1 180 tan 105.5 . 10 5
18 The polar coordinate of the point is 2 349, 180 tan 1 37.36, 105.5 . 5 c.
For this application, west is a negative direction and south is also negative. Therefore, the rectangular coordinate is (3, 35) .
d. The distance r from the origin to (3, 35) is r x 2 y 2 (3) 2 (35) 2 1234 35.13 . y Since the point (3, 35) lies in quadrant III, we use 180 tan 1 . Thus, x 35 1 35 180 tan 1 180 tan 265.1 3 3 35 The polar coordinate of the point is 1234, 180 tan 1 35.13, 265.1 . 3
88. Rewrite the polar coordinates in rectangular form: Since P1 r1 , 1 and P2 r2 , 2 , we have that x1 , y1 r1 cos 1 , r1 sin 1 and
1117
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
x1 , y1 r2 cos 2 , r2 sin 2 . d
x2 x1 2 y2 y1 2 r2 cos 2 r1 cos 1 2 r2 sin 2 r1 sin 1 2
r22 cos 2 2 2r1r2 cos 2 cos 1 r12 cos 2 1 r22 sin 2 2 2r1r2 sin 2 sin 1 r12 sin 2 1
r12 cos 2 1 sin 2 1 r22 cos 2 2 sin 2 2 2r1r2 cos 2 cos 1 sin 2 sin 1 r12 r22 2r1r2 cos 2 1
89. a.
At 10:15 a.m., 80, 25 . At 10:25 a.m.,
90.
110, 5 . b. At 10:15 a.m.:
so 72.50,33.81 .
x 80 cos 25 72.50 y 80sin 25 33.81
At 10:25 a.m., x 110 cos(5) 109.58 y 110sin(5) 9.59 So, 109.58, 9.59 .
See the figure. 1 180 24 156 , so radar station A is located at (150,156) on the second system. Using the Law of Cosines,
c.
rate=
distance time
342.5 mph
109.58 72.52 9.59 33.812 1 h 6
r2 1002 1502 2 100 150 cos 56 125.4 . Use the Law of Sines to find the measure of Angle B in triangle ABC: sin B sin 56 100sin 56 sin B 100 125.4 125.4 1 100sin 56 B sin 41.4 125.4
Then 2 1 B 156 41.4 114.6 , so
radar station C is located at 125.4,114.6 on the second system. 91. x r cos and y r sin 92 – 93. Answers will vary. 94. log 4 ( x 3) log 4 ( x 1) 2 x 3 log 4 2 x 1 42
1118
Copyright © 2025 Pearson Education, Inc.
x3 x 1
Section 10.1: Polar Coordinates
value becomes y 16 5 11 . Thus the final point is (0, 11) .
16( x 1) x 3 16 x 16 x 3 15 x 19 19 x 15
98. z w (2 5i )(4 i )
The solution set is:
8 2i 20i 5i 2 8 18i 5
19 . 15
13 18i
99. 4sin cos 1 2(2sin cos ) 1
95. f x 2 x3 6 x 2 7 x 8
Examining f x 2 x3 6 x 2 7 x 8 , there is
2sin 2 1
two variations in sign; thus, there are 2 or 0 positive real zeros. Examining
sin 2
5 2 n 6 5 n or 2 n 12 12 5 13 17 . The solution set is , , , 12 12 12 12 2
f x 2 x 6 x 7 x 8 , 3
2
2 x3 6 x 2 7 x 8 there is one variation in sign; thus, there is one negative real zero.
1 2
6
2 n or 2
x x y y2 96. 1 2 , 1 2 2
100. A 65º , B 37º , c 10 C 180º A B 180º 65º 37º 78º
1 3 7 2 2 , 2 2
sin A sin C a c sin 65º sin 78º a 10 10sin 65º a 9.27 sin 78º
5 2 , 9 5 , 9 2 2 4 2
sin B sin C b c sin 37º sin 78º b 10 10sin 37º b 6.15 sin 78º
97. We move the graph horizontally left 3 units so the x value becomes x 3 3 0 . We stretch the graph by a factor of 2 and flip on the x axis to the y value becomes y 2(8) 16 . Then we shift the graph vertically 5 units so the y
101. sin( ) sin cos cos sin sin
7 3 sin 12 4 6
3 3 cos cos sin 4 6 4 6 2 3 2 1 2 2 2 2 sin
1119
Copyright © 2025 Pearson Education, Inc.
6 2 6 2 4 4 4
Chapter 10: Polar Coordinates; Vector
102.
5 x 2 3e3 x e3 x 10 x
5x
2 2
15. r 4 The equation is of the form r a, a 0 . It is a circle, center at the pole and radius 4. Transform to rectangular form: r4
5 xe3 x ( x 3 2) 25 x 4
5 xe3 x (3 x 2) 25 x 4 e3 x (3x 2) 5 x3
r 2 16 x 2 y 2 16
103. sin 5 x sin x(sin 2 x) 2 sin x(1 cos 2 x) 2 sin x(1 2 cos 2 x cos 4 x) sin x 2sin x cos 2 x cos 4 x sin x
Section 10.2 1.
4, 6
16. r 2 The equation is of the form r a, a 0 . It is a circle, center at the pole and radius 2. Transform to rectangular form: r2
2. cos A cos B sin A sin B 3.
x (2) 2 y 52 32 x 2 2 y 5 2 9
r2 4 x2 y2 4
4. odd, since sin( x) sin x . 5.
2 2
6.
1 2
7. polar equation 8. False. They are sufficient but not necessary. 9. 10. 11. 2n ; n 12. True 13. c 14. b
1120
Copyright © 2025 Pearson Education, Inc.
Section 10.2: Polar Equations and Graphs
3 The equation is of the form . It is a line, passing through the pole making an angle of 3 with the polar axis. Transform to rectangular form: 3 tan tan 3 y 3 x y 3x
17.
19. r sin 4 The equation is of the form r sin b . It is a horizontal line, 4 units above the pole. Transform to rectangular form: r sin 4 y4
4 The equation is of the form . It is a line, passing through the pole making an angle of 3 or with the polar axis. Transform to 4 4 rectangular form: 4 tan tan 4 y 1 x y x
18.
20. r cos 4 The equation is of the form r cos a . It is a vertical line, 4 units to the right of the pole. Transform to rectangular form: r cos 4 x4
1121
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vector r 2 cos
21. r cos 2 The equation is of the form r cos a . It is a vertical line, 2 units to the left of the pole. Transform to rectangular form: r cos 2 x 2
r 2r cos 2
2
x y2 2x x2 2 x y 2 0 ( x 1) 2 y 2 1 center (1, 0) ; radius 1
22. r sin 2 The equation is of the form r sin b . It is a horizontal line, 2 units below the pole. Transform to rectangular form: r sin 2 y 2
24. r 2sin The equation is of the form r 2a sin , a 0 . It is a circle, passing through the pole, and center on the line . Transform to rectangular form: 2 r 2sin r 2 2r sin x2 y 2 2 y x2 y 2 2 y 0 x 2 ( y 1) 2 1 center (0, 1) ; radius 1
23. r 2 cos The equation is of the form r 2a cos , a 0 . It is a circle, passing through the pole, and center on the polar axis. Transform to rectangular form:
1122
Copyright © 2025 Pearson Education, Inc.
Section 10.2: Polar Equations and Graphs 27. r sec 4 The equation is a circle, passing through the pole, center on the polar axis and radius 2. Transform to rectangular form: r sec 4 1 r 4 cos r 4 cos
25. r 4sin The equation is of the form r 2a sin , a 0 . It is a circle, passing through the pole, and center on the line . Transform to rectangular form: 2 r 4sin r 2 4r sin x2 y 2 4 y 2
r 2 4r cos
2
x y 4y 0 2
x2 y2 4 x
2
x ( y 2) 4 center (0, 2) ; radius 2
x2 4 x y 2 0 ( x 2) 2 y 2 4 center (2, 0) ; radius 2
26. r 4 cos The equation is of the form r 2a cos , a 0 . It is a circle, passing through the pole, and center on the polar axis. Transform to rectangular form: r 4 cos
28. r csc 8 The equation is a circle, passing through the pole, center on the line and radius 4. 2 Transform to rectangular form: r csc 8 1 r 8 sin r 8sin
r 2 4r cos x2 y 2 4 x x2 4 x y 2 0 ( x 2) 2 y 2 4 center (2, 0) ; radius 2
r 2 8r sin x2 y2 8 y x2 y 2 8 y 0 x 2 ( y 4) 2 16 center (0, 4) ; radius 4
1123
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vector
r 2 4r cos x2 y 2 4 x x2 4 x y 2 0 ( x 2) 2 y 2 4 center (2, 0) ; radius 2
29. r csc 2 The equation is a circle, passing through the pole, center on the line and radius 1. 2 Transform to rectangular form: r csc 2 1 r 2 sin r 2sin
31. E 32. A
r 2 2r sin
33. F
x2 y 2 2 y 2
2
x y 2y 0
34. B
x 2 ( y 1) 2 1 center (0, 1) ; radius 1
35. H 36. G 37. D 38. C 39. r 2 2 cos The graph will be a cardioid. Check for symmetry:
Polar axis: Replace by . The result is r 2 2 cos( ) 2 2 cos . The graph is symmetric with respect to the polar axis.
30. r sec 4 The equation is a circle, passing through the pole, center on the polar axis and radius 2. Transform to rectangular form: r sec 4 1 r 4 cos r 4 cos 1124
Copyright © 2025 Pearson Education, Inc.
Section 10.2: Polar Equations and Graphs r 1 sin( )
: Replace by . 2 r 2 2 cos( ) 2 2 cos( ) cos sin( ) sin
The line
1 sin cos cos sin
2 2( cos 0) 2 2 cos The test fails.
1 (0 sin ) 1 sin The graph is symmetric with respect to the line . 2
The pole: Replace r by r . r 2 2 cos . The test fails.
The pole: Replace r by r . r 1 sin . The test fails.
Due to symmetry with respect to the polar axis, assign values to from 0 to .
Due to symmetry with respect to the line
r 2 2 cos
0
4
6 3 2 2 3 5 6
assign values to from
2 3 3.7
3
2
3
6
2 3 0.3
0
0
2
6 0
1
r 1 sin
3
3 0.1 2 1 2 1 3 2 3 1.9 1 2
1
2
40. r 1 sin The graph will be a cardioid. Check for symmetry:
Polar axis: Replace by . The result is r 1 sin( ) 1 sin . The test fails. The line
: Replace by . 2
1125
Copyright © 2025 Pearson Education, Inc.
2
to . 2 2
, 2
Chapter 10: Polar Coordinates; Vector 41. r 3 3sin The graph will be a cardioid. Check for symmetry: Polar axis: Replace by . The result is r 3 3sin( ) 3 3sin . The test fails. : Replace by . 2 r 3 3sin( )
The line
: Replace by . 2 r 2 2 cos( )
3 3(0 sin ) 3 3sin The graph is symmetric with respect to the line . 2 The pole: Replace r by r . r 3 3sin . The test fails. Due to symmetry with respect to the line , 2 assign values to from to . 2 2 2 3 6 0
2 2( cos 0) 2 2 cos The test fails.
The pole: Replace r by r . r 2 2 cos . The test fails. Due to symmetry with respect to the polar axis, assign values to from 0 to .
3
3 3 5.6 2 9 2 3 3 2
3
r 2 2 cos
0
0
6 3 2 2 3 5 6
6
6 3 2
2 2 cos cos sin sin
r 3 3sin
Polar axis: Replace by . The result is r 2 2 cos( ) 2 2 cos . The graph is symmetric with respect to the polar axis. The line
3 3 sin cos cos sin
42. r 2 2 cos The graph will be a cardioid. Check for symmetry:
3 3 0.4 2
2 3 0.3 1 2 3 2 3 3.7
0
1126
Copyright © 2025 Pearson Education, Inc.
4
Section 10.2: Polar Equations and Graphs 44. r 2 cos The graph will be a limaçon without an inner loop. Check for symmetry:
43. r 2 sin The graph will be a limaçon without an inner loop. Check for symmetry: Polar axis: Replace by . The result is r 2 sin( ) 2 sin . The test fails.
Polar axis: Replace by . The result is r 2 cos( ) 2 cos . The graph is symmetric with respect to the polar axis.
: Replace by . 2 r 2 sin( )
The line
: Replace by . 2 r 2 cos( )
The line
2 sin cos cos sin
2 cos cos sin sin
2 (0 sin ) 2 sin The graph is symmetric with respect to the line . 2
2 ( cos 0) 2 cos The test fails.
The pole: Replace r by r . r 2 sin . The test fails.
The pole: Replace r by r . r 2 cos . The test fails.
Due to symmetry with respect to the line assign values to from
2 3 6 0
Due to symmetry with respect to the polar axis, assign values to from 0 to .
to . 2 2
r 2 sin
2
3 1.1 2 3 2 2 5 2
2
r 2 cos
0
1
6 3 2 2 3
1
6 3 2
, 2
5 6
3 2.9 2
2
3 1.1 2 3 2 2 5 2
2
3
1127
Copyright © 2025 Pearson Education, Inc.
3 2.9 2 3
Chapter 10: Polar Coordinates; Vector 45. r 4 2 cos The graph will be a limaçon without an inner loop. Check for symmetry:
46. r 4 2sin The graph will be a limaçon without an inner loop. Check for symmetry:
Polar axis: Replace by . The result is r 4 2 cos( ) 4 2 cos . The graph is symmetric with respect to the polar axis.
Polar axis: Replace by . The result is r 4 2sin( ) 4 2sin . The test fails. : Replace by . 2 r 4 2sin( )
The line
The line : Replace by . 2 r 4 2 cos( )
4 2 sin cos cos sin
4 2 cos cos sin sin
4 2(0 sin ) 4 2sin The graph is symmetric with respect to the line . 2
4 2( cos 0) 4 2 cos The test fails.
The pole: Replace r by r . r 4 2 cos . The test fails.
The pole: Replace r by r . r 4 2sin . The test fails.
Due to symmetry with respect to the polar axis, assign values to from 0 to .
0 6 3 2 2 3 5 6
Due to symmetry with respect to the line
r 4 2 cos 2
assign values to from
4 3 2.3
2 3 6 0
r 4 2sin
3 4 5 4 3 5.7
6 3 2
6
2 4 3 2.3 3 4 5 4 3 5.7
1128
Copyright © 2025 Pearson Education, Inc.
6
to . 2 2
, 2
Section 10.2: Polar Equations and Graphs 47. r 1 2sin The graph will be a limaçon with an inner loop. Check for symmetry:
48. r 1 2sin The graph will be a limaçon with an inner loop. Check for symmetry:
Polar axis: Replace by . The result is r 1 2sin( ) 1 2sin . The test fails.
Polar axis: Replace by . The result is r 1 2sin( ) 1 2sin . The test fails.
: Replace by . 2 r 1 2sin( )
: Replace by . 2 r 1 2sin( )
The line
The line
1 2 sin cos cos sin
1 2 sin cos cos sin
1 2(0 sin ) 1 2sin The graph is symmetric with respect to the line . 2
1 2(0 sin ) 1 2sin The graph is symmetric with respect to the line . 2
The pole: Replace r by r . r 1 2sin . The test fails.
The pole: Replace r by r . r 1 2sin . The test fails.
Due to symmetry with respect to the line assign values to from
2 3 6 0
6 3 2
, 2
Due to symmetry with respect to the line
to . 2 2
assign values to from
r 1 2sin
1
2 3 6 0
1 3 0.7 0 1
6 3 2
2 1 3 2.7 3
r 1 2sin 3 1 3 2.7 2 1 0 1 3 0.7
1129
Copyright © 2025 Pearson Education, Inc.
1
to . 2 2
, 2
Chapter 10: Polar Coordinates; Vector 49. r 2 3cos The graph will be a limaçon with an inner loop. Check for symmetry:
50. r 2 4 cos The graph will be a limaçon with an inner loop. Check for symmetry:
Polar axis: Replace by . The result is r 2 3cos( ) 2 3cos . The graph is symmetric with respect to the polar axis.
Polar axis: Replace by . The result is r 2 4 cos( ) 2 4 cos . The graph is symmetric with respect to the polar axis.
: Replace by . 2 r 2 3cos( )
The line
: Replace by . 2 r 2 4 cos( )
The line
2 3 cos cos sin sin
2 4 cos cos sin sin
2 3( cos 0) 2 3cos The test fails.
2 4( cos 0) 2 4 cos The test fails.
The pole: Replace r by r . r 2 3cos . The test fails.
The pole: Replace r by r . r 2 4 cos . The test fails.
Due to symmetry with respect to the polar axis, assign values to from 0 to .
r 2 3cos
0
1
6 3 2 2 3
3 3 2 0.6 2 1 2
5 6
Due to symmetry with respect to the polar axis, assign values to from 0 to . r 2 4 cos
0
6
6 3 2 2 3 5 6
2 7 2 2
3 3 4.6 2 5
2 2 3 5.5 4 2 0 2 2 3 1.5
1130
Copyright © 2025 Pearson Education, Inc.
2
Section 10.2: Polar Equations and Graphs 51. r 3cos(2 ) The graph will be a rose with four petals. Check for symmetry:
52. r 2sin(3 ) The graph will be a rose with three petals. Check for symmetry:
Polar axis: Replace by . r 3cos(2( )) 3cos( 2 ) 3cos(2 ) . The graph is symmetric with respect to the polar axis.
Polar axis: Replace by .
: Replace by . 2 r 3cos 2( )
The line
r 2sin 3( ) 2sin(3 ) 2sin 3 . The
test fails. : Replace by . 2 r 2sin 3( )
The line
2sin(3 3 )
3cos(2 2 )
2 sin 3 cos 3 cos 3 sin 3
3 cos 2 cos 2 sin 2 sin 2
2 0 sin 3
3(cos 2 0) 3cos 2 The graph is symmetric with respect to the line . 2
2sin 3
The graph is symmetric with respect to the line . 2 The pole: Replace r by r . r 2sin 3 .
The pole: Since the graph is symmetric with respect to both the polar axis and the line
, 2
The test fails. Due to symmetry with respect to the line
it is also symmetric with respect to the pole. Due to symmetry, assign values to from 0 to . 2 r 3cos 2 3 3 6 2 0 4 3 3 2 3 2
assign values to from
2 3 4 6
r 2sin 3
2 0 2 1.4
0
1131
Copyright © 2025 Pearson Education, Inc.
2 0
, 2
to . 2 2 r 2sin 3 2 6 2 1.4 4 0 3 2 2
Chapter 10: Polar Coordinates; Vector 54. r 3cos(4 ) The graph will be a rose with eight petals. Check for symmetry:
53. r 4sin(5 ) The graph will be a rose with five petals. Check for symmetry: Polar axis: Replace by .
Polar axis: Replace by . r 3cos(4( )) 3cos( 4 ) 3cos(4 ) . The graph is symmetric with respect to the polar axis.
r 4sin 5( ) 4sin(5 ) 4sin 5 .
The test fails. The line : Replace by . 2 r 4sin 5( ) 4sin(5 5 )
: Replace by . 2 r 3cos 4( )
The line
4 sin 5 cos 5 cos 5 sin 5
3cos(4 4 )
4 0 sin 5
3 cos 4 cos 4 sin 4 sin 4
4sin 5 The graph is symmetric with respect to the line . 2 The pole: Replace r by r . r 4sin 5 .
3(cos 4 0) 3cos 4 The graph is symmetric with respect to the line . 2
The test fails.
The pole: Since the graph is symmetric with
Due to symmetry with respect to the line , 2 assign values to from to . 2 2 r 4 sin 5 r 4sin 5 2 3 4 6
0
4 2 3 3.5 2 2 2.8 2
6 4 3 2
respect to both the polar axis and the line it is also symmetric with respect to the pole. Due to symmetry, assign values to from 0 to . 2 r 3cos 4 0 2 3 6 2 3 4 3 3 2 3 2
2 2 2 2.8 2 3 3.5 4
0
1132
Copyright © 2025 Pearson Education, Inc.
, 2
Section 10.2: Polar Equations and Graphs
55. r 2 9 cos(2 ) The graph will be a lemniscate. Check for symmetry:
56. r 2 sin(2 ) The graph will be a lemniscate. Check for symmetry: Polar axis: Replace by .
Polar axis: Replace by .
r 2 sin(2( )) sin( 2 ) sin(2 ) . The test fails. The line : Replace by . 2 2 r sin 2( )
r 2 9 cos(2( )) 9 cos( 2 ) 9 cos(2 ) . The graph is symmetric with respect to the polar axis. : Replace by . 2 r 2 9 cos 2( )
The line
sin(2 2 ) sin 2 cos 2 cos 2 sin 2
9 cos(2 2 ) 9 cos 2 cos 2 sin 2 sin 2
0 sin 2
9(cos 2 0) 9 cos 2 The graph is symmetric with respect to the line . 2 The pole: Since the graph is symmetric with respect to both the polar axis and the line , 2 it is also symmetric with respect to the pole. Due to symmetry, assign values to from 0 to . 2
r 9 cos 2
0
3
6 4 3 2
sin 2 The test fails. The pole: Replace r by r . (r ) 2 sin 2 r 2 sin 2
The graph is symmetric with respect to the pole. Due to symmetry, assign values to from 0 to .
r sin 2
0
0
6 3 2 2 3 5 6
3 2 2.1 2 0 undefined undefined
1133
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3 2
3 2 0
undefined undefined 0
Chapter 10: Polar Coordinates; Vector 57. r 2 The graph will be a spiral. Check for symmetry:
58. r 3 The graph will be a spiral. Check for symmetry:
Polar axis: Replace by . r 2 . The test fails.
Polar axis: Replace by . r 3 . The test fails.
: Replace by . 2 r 2 . The test fails.
: Replace by . r 3 . 2 The test fails.
The line
The line
The pole: Replace r by r . r 2 . The test fails.
The pole: Replace r by r . r 3 . The test fails.
r 2
r 3
0.1
0.03
0.3
2 4 0
4 2 3 2 2
2 4 0
0.6 1
4 2
1.7 3.0 8.8
3 2 2
26.2 77.9
0.2 0.4 1 2.4 5.6 31.5 117.2 995
1134
Copyright © 2025 Pearson Education, Inc.
Section 10.2: Polar Equations and Graphs 59. r 1 cos The graph will be a cardioid. Check for symmetry:
60. r 3 cos The graph will be a limaçon without an inner loop. Check for symmetry:
Polar axis: Replace by . The result is r 1 cos( ) 1 cos . The graph is symmetric with respect to the polar axis.
Polar axis: Replace by . The result is r 3 cos( ) 3 cos . The graph is symmetric with respect to the polar axis. : Replace by . 2 r 3 cos( )
The line
The line : Replace by . 2 r 1 cos( ) 1 (cos cos sin sin )
3 cos cos sin sin
3 ( cos 0) 3 cos The test fails.
1 ( cos 0) 1 cos The test fails.
The pole: Replace r by r . r 3 cos . The test fails.
The pole: Replace r by r . r 1 cos . The test fails.
Due to symmetry, assign values to from 0 to .
Due to symmetry, assign values to from 0 to .
0
r 1 cos 0
0
r 3 cos 4
3 3 3.9 6 2 7 3 2 3 2 2 5 3 2 5 3 3 2.1 6 2 2
3 1 0.1 6 2 1 3 2 1 2 2 3 3 2 5 3 1 1.9 6 2 2
1135
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vector 61. r 1 3cos The graph will be a limaçon with an inner loop. Check for symmetry:
62. r 4 cos(3 ) The graph will be a rose with three petals. Check for symmetry:
Polar axis: Replace by . The result is r 1 3cos( ) 1 3cos . The graph is symmetric with respect to the polar axis.
Polar axis: Replace by . r 4 cos(3( )) 4 cos( 3 ) 4 cos(3 ) . The graph is symmetric with respect to the polar axis.
: Replace by . 2 r 1 3cos( )
: Replace by . 2 r 4 cos 3
The line
The line
1 3 cos cos sin sin
4 cos 3 3
1 3( cos 0) 1 3cos The test fails.
4 cos 3 cos 3 sin 3 sin 3 4 cos 3 0 4 cos 3
The pole: Replace r by r . r 1 3cos . The test fails.
The test fails. The pole: Replace r by r . r 4 cos 3 .
Due to symmetry, assign values to from 0 to .
0
The test fails.
r 1 3cos 2
Due to symmetry, assign values to from 0 to .
3 3 1 1.6 6 2 1 3 2 1 2 2 5 3 2 5 3 3 1 3.6 6 2 4
r 4 cos 3
0
4
6 3 2 2 3 5 6
1136
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0 4 0 4 0 4
Section 10.2: Polar Equations and Graphs 63. The graph is a cardioid whose equation is of the form r a b cos . The graph contains the point (6, 0) , so we have 6 a b cos 0 6 a b(1) 6 ab
The graph contains the point 3,
3 a b cos
5 a b sin 5 a b 1
, so we have 2
2
3 a b(0) 3a Substituting a 3 into the first equation yields: 6 ab 6 3b 3b Therefore, the graph has equation r 3 3cos .
66. The graph is a limaçon with inner loop whose equation is of the form r a b sin , where 0 a b . The graph contains the point 1, 0 ,
so we have 1 a b sin 0 1 a b 0 1 a
64. The graph is a cardioid whose equation is of the form r a b cos . The graph contains the point (6, ) , so we have
The graph contains the point 5, , so we have 2
6 a b cos 6 a b(1) 6 a b
5 a b sin 5 a b 1
3 a b cos
2
5 ab Substituting a 4 into the second equation yields: 5 ab 5 4b 1 b Therefore, the graph has equation r 4 sin .
The graph contains the point 3,
2
5 ab Substituting a 1 into the second equation yields: 5 ab 5 1 b 4b Therefore, the graph has equation r 1 4sin .
, so we have 2
2
3 a b(0) 3a Substituting a 3 into the first equation yields: 6 a b 6 3b b 3 Therefore, the graph has equation r 3 3cos .
67. r 8cos The equation is of the form r 2a cos , a 0 . It is a circle, passing through the pole, and center on the polar axis. r 2sec 2 r cos r cos 2 The equation is of the form r cos a . It is a vertical line, 2 units to the right of the pole.
65. The graph is a limaçon without inner loop whose equation is of the form r a b sin , where 0 b a . The graph contains the point 4, 0 ,
so we have 4 a b sin 0 4 a b 0 4a The graph contains the point 5, , so we have 2
1137
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vector 68. r 8sin The equation is of the form r 2a sin , a 0 . It is a circle, passing through the pole, and center on the line . 2 r 4 csc 4 r sin r sin 4 The equation is of the form r sin b . It is a horizontal line, 4 units above the pole.
Use substitution to find the point(s) of intersection: 8cos 2sec 2 8cos cos 1 2 cos 4 1 cos 2 2 4 5 , , , for 0 2 3 3 3 3 1 If , r 8cos 8 4 . 3 3 2 2 2 1 , r 8cos 8 4 . If 3 3 2 4 4 1 , r 8cos 8 4 . If 3 3 2 5 5 1 , r 8cos 8 4 . If 3 3 2 The points of intersection are 4, and 3 5 4, . 3
Use substitution to find the point(s) of intersection: 8sin 4 csc 4 8sin sin 1 sin 2 2 2 sin 2 3 5 7 , , , for 0 2 4 4 4 4 2 If , r 8sin 8 4 2 . 2 4 4 2 3 3 8 , r 8sin If 4 2. 2 4 4 5 2 5 8 , r 8sin If 4 2 . 2 4 4 7 2 7 8 , r 8sin If 4 2 . 2 4 4 The points of intersection are 4 2, and 4 3 4 2, . 4 1138
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Section 10.2: Polar Equations and Graphs 69. r sin The equation is of the form r 2a sin , a 0 . It is a circle, passing through the pole, and center on the line . 2 r 1 cos The graph will be a limaçon without an inner loop. Check for symmetry: Polar axis: Replace by . The result is r 1 cos( ) 1 cos . The graph is symmetric with respect to the polar axis. The line : Replace by . 2 r 1 cos( )
Use substitution to find the point(s) of intersection: sin 1 cos sin cos 1
sin cos 2 12 sin 2 2sin cos cos 2 1 1 2sin cos 1 2sin cos 0 sin cos 0 sin 0 or cos 0 3 0, or , for 0 2 2 2 If 0 , r sin 0 0 (doesn’t check). If , r sin 0 .
1 cos cos sin sin
If
0
, r sin 1 . 2 2 3 3 , r sin 1 (doesn’t check). If 2 2 The points of intersection are 0, and 1, . 2
1 ( cos 0) 1 cos The test fails. The pole: Replace r by r . r 1 cos . The test fails. Due to symmetry, assign values to from 0 to .
70. r 3 The equation is of the form r a, a 0 . It is a circle, center at the pole and radius 3.
r 1 cos 2
r 2 2 cos The graph will be a limaçon without an inner loop. Check for symmetry: Polar axis: Replace by . The result is r 2 2 cos( ) 2 2 cos . The graph is symmetric with respect to the polar axis. The line : Replace by . 2 r 2 2 cos( )
3 1 1.9 6 2 3 3 2 1 2 2 1 3 2 5 3 1 0.1 6 2 0
2 2 cos cos sin sin
2 2( cos 0) 2 2 cos The test fails. The pole: Replace r by r . r 2 2 cos . The test fails. Due to symmetry, assign values to from 0 to .
1139
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vector
0 6 3 2 2 3 5 6
71. r 1 sin The graph will be a cardioid. Check for symmetry: Polar axis: Replace by . The result is r 1 sin( ) 1 sin . The test fails.
r 2 2 cos 4 2 3 3.7 3
: Replace by . 2 r 1 sin( ) 1 sin( ) cos cos( ) sin
The line
2 1
1 (0 sin ) 1 sin The graph is symmetric with respect to the line . 2 The pole: Replace r by r . r 1 sin . The test fails. Due to symmetry with respect to the line , assign values to from to . 2 2 2
2 3 0.3 0
Use substitution to find the point(s) of intersection: 3 2 2 cos 1 2 cos 1 cos 2 5 for 0 2 , 3 3 1 If , r 2 2 cos 2 2 3 . 3 3 2 5 5 1 , r 2 2 cos 2 2 3 . If 3 3 2 The points of intersection are 3, and 3 5 3, . 3
r 1 sin
0
2
3
1
6
0 6
3
3 0.1 2 1 2
1 3 2 3 1 1.9 2
2
2
r 1 cos The graph will be a limaçon without an inner loop. Check for symmetry: Polar axis: Replace by . The result is r 1 cos( ) 1 cos . The graph is symmetric with respect to the polar axis. The line : Replace by . 2 r 1 cos( ) 1 cos cos sin sin
1 ( cos 0) 1 cos The test fails.
1140
Copyright © 2025 Pearson Education, Inc.
Section 10.2: Polar Equations and Graphs 72. r 1 cos The graph will be a limaçon without an inner loop. Check for symmetry: Polar axis: Replace by . The result is r 1 cos( ) 1 cos . The graph is symmetric with respect to the polar axis. The line : Replace by . 2 r 1 cos( )
The pole: Replace r by r . r 1 cos . The test fails. Due to symmetry, assign values to from 0 to .
0
r 1 cos 2
3 1 1.9 6 2 3 3 2 1 2 2 1 3 2 5 3 0.1 1 6 2
1 cos cos sin sin
1 ( cos 0) 1 cos The test fails. The pole: Replace r by r . r 1 cos . The test fails. Due to symmetry, assign values to from 0 to .
0
0
r 1 cos 2
3 1 1.9 6 2 3 3 2 1 2 2 1 3 2 5 3 0.1 1 6 2 0
Use substitution to find the point(s) of intersection: 1 sin 1 cos sin cos sin 1 cos tan 1 5 for 0 2 , 4 4
r 3cos The equation is of the form r 2a cos , a 0 . It is a circle, passing through the pole, and center on the polar axis.
2 , r 1 sin 1 1.7 . 4 2 4 5 5 2 , r 1 sin If 1 0.3 . 4 4 2 2 , and The points of intersection are 1 2 4 2 5 , 1 . 2 4
If
Use substitution to find the point(s) of intersection: 1141
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vector 1 cos 3cos 1 2 cos 1 cos 2 5 , for 0 2 3 3 1 3 If , r 1 cos 1 . 3 3 2 2 5 5 1 3 , r 1 cos 1 . If 3 3 2 2
Due to symmetry, assign values to from 0 to .
0 6
3 3 5 The points of intersection are , and , . 2 3
73. r
2 3
2 Check for symmetry: 1 cos
Polar axis: Replace by . The result is r
2 2 . 1 cos 1 cos
2 1 cos undefined
r
2 1 3 2
3 2 2 3 5 6
1 3 2
1
14.9
4 2 4 3 2
The graph is symmetric with respect to the polar axis. The line : Replace by . 2 2 r 1 cos 2 1 cos cos sin sin 2 1 ( cos 0) 2 1 cos The test fails. 2 . The pole: Replace r by r . r 1 cos The test fails.
1142
Copyright © 2025 Pearson Education, Inc.
1.1
Section 10.2: Polar Equations and Graphs
74. r
2 1 2 cos
75. r
Check for symmetry:
Polar axis: Replace by . The result is 2 2 . The graph is 1 2 cos( ) 1 2 cos symmetric with respect to the polar axis. The line : Replace by . 2 2 r 1 2 cos 2 1 2 cos cos sin sin
2 2 1 2 cos 0 1 2 cos
r
The test fails.
The pole: Replace r by r . r
2 . 1 2 cos
The pole: Replace r by r . r
The test fails. Due to symmetry, assign values to from 0 to .
0 6 3 2 2 3 5 6
1 1 . The graph is 3 2 cos 3 2 cos
symmetric with respect to the polar axis. The line : Replace by . 2 1 r 3 2 cos 1 3 2 cos cos sin sin 1 1 3 2 cos 0 3 2 cos
The test fails.
Check for symmetry:
Polar axis: Replace by . The result is
r
1 3 2 cos
The test fails. Due to symmetry, assign values to from 0 to .
2 1 2 cos 2 2 2.7 1 3
r
0 6 3 2 2 3 5 6
undefined 2 1 2 1 3 2 3
0.7
1 3 2 cos 1 1 0.8 3 3 1 2 1 3 1 4 1 0.2 3 3 1 5
r
1143
Copyright © 2025 Pearson Education, Inc.
1 . 3 2 cos
Chapter 10: Polar Coordinates; Vector
76. r
1 1 cos
Check for symmetry:
Polar axis: Replace by . The result is r
1 1 . The graph is 1 cos 1 cos
symmetric with respect to the polar axis. The line r
: Replace by . 2
1 1 cos
77. r , 0 Check for symmetry:
1
1 cos cos sin sin 1 1 cos 0
Polar axis: Replace by . r . The test fails.
1 1 cos The test fails.
: Replace by . r . 2 The test fails.
The line
The pole: Replace r by r . r
1 . 1 cos
The pole: Replace r by r . r . The test fails.
The test fails. Due to symmetry, assign values to from 0 to .
0 6
1 1 cos undefined
r
1 1 3 2
7.5
2 3 2 2 2 2 3 3 5 1 0.5 6 1 3 2
r
0
0
6 3 2
0.5 6 1.0 3 1.6 2 3.1
3 2 2
3 4.7 2 2 6.3
1 2
1144
Copyright © 2025 Pearson Education, Inc.
Section 10.2: Polar Equations and Graphs
78. r
3
1 2, 0 sin Check for symmetry:
79. r csc 2
Check for symmetry:
Polar axis: Replace by . r
3 . The
Polar axis: Replace by . r csc( ) 2 csc 2 . The test fails.
test fails. The line r
: Replace by . 2
: Replace by . 2 r csc 2
The line
3 . The test fails.
The pole: Replace r by r . r
3
. The test
1 2 sin cos cos sin 1 2 0 cos 1 sin 1 2 sin csc 2 The graph is symmetric with respect to the line . 2
fails. 3
r
0
undefined
6 3 2
18 5.7 9 2.9 6 1.9 3 1.0 2 0.6 3 0.5 2
3 2 2
1 2 sin
The pole: Replace r by r . r csc 2 . The test fails. Due to symmetry, assign values to from 0 to
0 6 4 3 2
r csc 2 undefined 0 2 2 0.6 2 3 2 0.8 3
1145
Copyright © 2025 Pearson Education, Inc.
1
. 2
Chapter 10: Polar Coordinates; Vector 80. r sin tan Check for symmetry:
2 2 Check for symmetry:
81. r tan ,
Polar axis: Replace by . r sin( ) tan( ) ( sin )( tan ) sin tan The graph is symmetric with respect to the polar axis.
Polar axis: Replace by . r tan( ) tan . The test fails. The line
: Replace by . 2 r sin tan
: Replace by . 2
The line
r tan( )
tan tan sin cos cos sin 1 tan tan tan sin 1 sin tan The test fails.
r sin tan
0
0
6 3 2 2 3
1 3 0.3 2 3 3 2
5 6
1 tan tan
tan tan 1
The test fails. The pole: Replace r by r . r tan . The test fails.
The pole: Replace r by r . r sin tan . The test fails. Due to symmetry, assign values to from 0 to .
tan tan
3 4 6 0
r tan
3 1.7 1
6 4 3
undefined 3 2 1 3 0.3 2 3 0
1146
Copyright © 2025 Pearson Education, Inc.
3 0.6 3 0 3 0.6 3 1 3 1.7
Section 10.2: Polar Equations and Graphs
82. r cos
83. Convert the equation to rectangular form: r sin a ya The graph of r sin a is a horizontal line a units above the pole if a 0 , and |a| units below the pole if a 0 .
2 Check for symmetry:
Polar axis: Replace by .
r cos cos . The graph is symmetric 2 2 with respect to the polar axis.
84. Convert the equation to rectangular form: r cos a xa The graph of r cos a is a vertical line a units to the right of the pole if a 0 and |a| units to the left of the pole if a 0 .
The line : Replace by . 2 r cos cos 2 2 2 cos sin
2
cos
2
sin
2
sin
2
85. Convert the equation to rectangular form: r 2a sin , a 0
2 The test fails.
r 2 2a r sin
The pole: Replace r by r . r cos
2
x 2 y 2 2ay
. The
x 2 y 2 2ay 0
test fails.
x 2 ( y a)2 a 2 Circle: radius a, center at rectangular coordinates (0, a ).
Due to symmetry, assign values to from 0 to .
r cos
0
1
6
86. Convert the equation to rectangular form: r 2a sin , a 0
0.97
r 2 2a r sin
3
3 0.87 2
2 2 3 5 6
2 0.71 2 1 2
2
x 2 y 2 2ay x 2 y 2 2ay 0 x 2 ( y a)2 a 2 Circle: radius a, center at rectangular coordinates (0, a).
0.26
87. Convert the equation to rectangular form: r 2a cos , a 0
0
r 2 2a r cos x 2 y 2 2ax x 2 2ax y 2 0 ( x a)2 y 2 a 2 Circle: radius a, center at rectangular coordinates (a, 0).
1147
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Chapter 10: Polar Coordinates; Vector 88. Convert the equation to rectangular form: r 2a cos , a 0
1 ab sin C . Let 2 A (r2 , 2 ), B (r1 , 1 ), and C (0, 0). Then
92. The area of triangle ABC is K
r 2 2a r cos
a r1 , b r2 , and the measure of angle
x 2 y 2 2ax x 2 2ax y 2 0
C 2 1 . So K
( x a)2 y 2 a 2 Circle: radius a, center at rectangular coordinates ( a, 0) .
93. Answers will vary. 94. Answers will vary.
Reading the graph we obtain 5 knots. Reading the graph we obtain 6 knots. Reading the graph we obtain 10 knots. Reading the graph we obtain approximately 80 to 150 . Reading the graph we obtain approximately 9 knots that occurs at approximately 90 to 100 .
89. a. b. c. d. e.
95.
90. First we will calculate the distance of the speaker from the microphone. We use the Pythagorean Theorem. 2
2
r (6) (10)
1 r1r2 sin( 2 1 ) . 2
5 1 x3 5 5 1( x 3) x 8 1 0 0 0 x3 x3 x3 x 8 f ( x) x3 The zeros and values where the expression is undefined are x 8 and x 3 . Interval Number
, 3
3, 8
8,
0
5
10
Chosen
2
8 3 2 3 2 7 Negative Positive Negative
r 2 36 100
Value of f
r 2 136 11.99 meters
Conclusion
Now we need the angle that the speaker makes with the microphone. y 6 tan x 10 6 tan 1 0.5404 rad 10 We now calculate the distance of the microphone from the boundary of the optimal pickup region using the angle calculated above.
The solution set is x 3 x 8 , or, using
interval notation, 3,8 . 96.
97. y 2sin(5 x) Ampl 2 2
Period
r 8 8sin 0.5405 12.1 meters This means that the speaker will not be outside the optimal pickup range.
r 2 cos 2 sin 2 r 2 r 2 r 2 cos 2 r 2 sin 2
r r cos r sin x y x y 2 2
2
2 2
2
2
2 5
x3 is in lowest x x 12 ( x 3)( x 4) terms. The denominator has zeros at –3 and 4. The degree of the numerator is n = 1 and the degree of the denominator is m 2 . Since n m , the line y 0 is a horizontal asymptote. Since the denominator is zero at 4, x 4 is a vertical asymptotes. Since the factor ( x 3) cancels, x 3 is not an asymptote.
98. R ( x)
r 2 cos(2 )
91.
7 180 420 3
2
x3
2
2
99. P (1) 3(1)5 2(1)3 7(1) 5 3 2 7 5 3 By the Remainder Theorem, the remainder is 3. 1148
Copyright © 2025 Pearson Education, Inc.
Section 10.3: The Complex Plane; De Moivre’s Theorem
100.
s
Section 10.3
1 (6 11 13) 15 2
1. 4 3i
K 15(15 6)(15 11)(15 13) 15(9)(4)(2)
2. a.
sin A cos B cos A sin B
b.
cos A cos B sin A sin B
1080 6 30 32.86 sq units
101.
32 x 3 91 x 2 x 3
3
3
2 x 3 2(1 x)
4. e7 , e12
2x 3 2 2x
5. real; imaginary
4x 5
x
3 1 , 2 2
3.
2(1 x )
5 4
6. magnitude, modulus, argument
The solution set is
7. r1r2 e 1
i 2
5 . 4
8. false; z n r n e
i n
6 x 7 x 20 2
102.
9. three
6 x 7 x 20 0 2
10. True
(2 x 5)(3 x 4) 0 5 4 x ,x 2 3
11. c
12. a
5 4 The solution set is , 2 3
13. r x 2 y 2 12 12 2
103. Since f (2) (2)3 4(2) 2 5 3 , the point on the line is (-2, 3). The slope is f (2) 3(2) 2 8(2) 4 . The equation of the line is:
tan
y 1 x
4 The polar form of z 1 i is z r cos i sin 2 cos i sin 4 4
y y1 m x x1 y 3 4( x 2) y 3 4 x 8
i
2e 4
y 4 x 5
104. cos3 x cos x cos 2 x cos x(1 sin 2 x) cos x sin 2 x cos x
1149
Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
14. r x 2 y 2 (1)2 12 2 y 1 x 3 4 The polar form of z 1 i is tan
3
2e 4
15. r x 2 y 2
2
y 3 3 x 1 5 3 The polar form of z 1 3i is 5 5 i sin z r cos i sin 2 cos 3 3 i
5
2e 3
3 1 4 2 2
2
y 1 3 tan x 3 3
42
tan
3 3 i sin z r cos i sin 2 cos 4 4 i
16. r x 2 y 2 12 3
17. r x 2 y 2 02 (3) 2 9 3 y 3 x 0 3 2 The polar form of z 3 i is tan
11 6
3 3 z r cos i sin 3 cos i sin 2 2
The polar form of z 3 i is 11 11 z r cos i sin 2 cos i sin 6 6 i
i
3
3e 2
11
2e 6
18. r x 2 y 2 ( 2) 2 02 4 2 tan
y 0 0 x 2
1150 Copyright © 2025 Pearson Education, Inc.
Section 10.3: The Complex Plane; De Moivre’s Theorem The polar form of z 2 is z r cos i sin 2 cos i sin 2ei
21. r x 2 y 2 32 ( 4) 2 25 5 y 4 3 x 5.356 The polar form of z 3 4i is z r cos i sin 5 cos 5.356 i sin 5.356 tan
19. r x 2 y 2 42 ( 4) 2 32 4 2 y 4 1 4 x 7 4 The polar form of z 4 4i is tan
5ei 5.356
7 7 i sin z r cos i sin 4 2 cos 4 4 i
7
4 2e 4
22. r x 2 y 2 22
3 7 2
y 3 x 2 0.714 The polar form of z 2 3 i is tan
20. r x 2 y 2
9 3 9 324 18 2
2
7 cos 0.714 i sin 0.714
y 9 3 3 x 9 3
tan
7ei 0.714
z r cos i sin
6 The polar form of z 9 3 9 i is
z r cos i sin 18 cos i sin 6 6 . i
18e 6
1151 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
23. r x 2 y 2 ( 2) 2 32 13 tan
y 3 3 x 2 2
1 2 2 3 i sin 25. 2 cos 2 2 2 i 3 3 1 3 i
2.159 The polar form of z 2 3i is
7 7 3 1 26. 3 cos i sin 3 i 6 6 2 2
z r cos i sin 13 cos 2.159 i sin 2.159
13ei 2.159
3 3 3 i 2 2
7 2 i 7 7 2 27. 4e 4 4 cos i i sin 4 4 4 2 2
2 2 2 2 i 5 i 5 5 3 1 28. 2e 6 2 cos i sin 2 i 6 6 2 2
3i
24. r x y 2
tan
2
5 1 6 2
2
y 1 5 x 5 5
5.863 The polar form of z 5 i is z r cos i sin 6 cos 5.863 i sin 5.863 6ei 5.863
3 3 29. 3 cos i sin 3 0 1i 3 i 2 2 30. 4 cos i sin 4 0 1i 4 i 2 2
31. 7ei 7 cos i sin 7 1 0i 7 i 32. 3e 2 3 cos i sin 3 0 1i 2 2 3i
5 5 33. 0.2 cos i sin 0.2 0.1736 0.9848 i 9 9 0.035 0.197 i
1152 Copyright © 2025 Pearson Education, Inc.
Section 10.3: The Complex Plane; De Moivre’s Theorem 10 10 34. 0.4 cos i sin 0.4 0.9397 0.3420 i 9 9 0.376 0.137 i i 35. 2e 18 2 cos i sin 2 0.985 0.174i 18 18 1.970 0.347 i
36. 3e
i
10
3 cos i sin 3 0.951 0.309i 10 10 2.853 0.927i
2 2 37. z w 2 cos i sin 4 cos i sin 9 9 9 9 2e
i
2 9
4e
i
8e
9
2 i 9 2
8e
i
3
8 cos i sin 3 3 i
2
e
e
2 5 i 9 3
e
11 11 cos i sin 9 9 i
2 3
z e 5 e w i e 9 cos
9
2 5 i 9 3
i sin
7
i
e 9
13
i
12e
i
10
4 10 9
i
14
14 14 12 cos i sin 9 9 i
4
4
10
2
z 2e 9 2 i 10 e 9 9 w 6 i 6e 9
11 i 9
i
i
i
41. z w 2e 8 2e 10 2 2e 8 10 i
9
4e 40 9 9 4 cos i sin 40 40
i z 2e 8 2 i 8 10 e e 40 i w 2 2e 10 cos i sin 40 40 3
9
i
3 9
42. z w 4e 8 2e 16 4 2e 8 16
3
i
13 3 i 2 18
20 i 2 9
i
12e 9
9
39. z w 3e 18 4e 2 3 4e
4
i
40. z w 2e 9 6e 9 2 6e 9
i
i
7
3 i 3 i 2 3 i 11 e 9 e 9 e 9 4 4 4 3 11 11 cos i sin 4 9 9
i
13 3
1 i 4 e 3 3 1 4 4 cos i sin 3 3 3
2 2 5 5 38. z w cos i sin cos 9 i sin 9 3 3 e
13 3
2
2
5 i 9
13
1 i 1 i 2 e 3 e 3 3 3
z 2e 9 2 i 9 9 1 i 9 e e 4 2 w i 4e 9 1 cos i sin 2 9 9
2 i 3
i
z 3e 18 3 i 18 2 3 i 18 2 e e 3 i w 4 4 4e 2
i
20
12e 9
12e
2 i 9
15
8e 16
15 15 8 cos i sin 16 16
2 2 12 cos i sin 9 9
1153 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
i
3
3 9
44. z 1 i
3
z 4e 8 4 i 4 i 9 e 8 16 e 16 w 2 2 2e 16 2e
3 i 2 16
2e
i
r 12 (1) 2 2
1 1 1 7 4 7 i 7 7 4 i sin 2 z 2 e 4 4
tan
29 16
29 29 2 cos i sin 16 16
43. z 2 2i
w 1 3 i
r 22 22 8 2 2 2 tan 1 2
w 3 i
3 1 4 2 2
i
z w 2 2e 4 2e
11 i 6
11
i
2 2 2e 4 i
25
4 2e 12
6
2
i
i
4 2e 12
4 2 cos i sin 12 12 i
11
z 2 2e 4 2 2 i 4 6 e 11 w 2 i 2e 6 19 i 12 19 i 2 12
25
4 2e 12
7
i
i
5
z w 2e 4 2e 3
1 3 3 3 11 6 11 i 11 11 w 2 cos i sin 2e 6 6 6
2e
3 3 1 5 3 5 i 5 5 i sin 2e 3 w 2 cos 3 3
2
tan
2e
2
tan
4
i z 2 2 cos i sin 2 2e 4 4 4
r
42
r 12 3
i
7 5 3
2 2e 4 i
41
2 2e 12
i
2
i
i
7
7
5
z 2e 4 2 i 4 3 e 5 i w 2 2e 3 2 i 12 e 2 2 cos i sin 2 12 12 3
i 29 2 2 i sin 45. 4 cos 4e 9 9 3
4 e 3
2 i 9
i
i
5
17
2 2e 12 17 17 2 2 cos i sin 12 12
4 e 3
2 i 3 9
2
64e 3
2e 12 5 5 2 cos i sin 12 12
41
2 2e 12
2 2 64 cos i sin 3 3 1 3 64 i 2 2 32 32 3 i
1154 Copyright © 2025 Pearson Education, Inc.
3
Section 10.3: The Complex Plane; De Moivre’s Theorem
3
i 4 4 4 i sin 3e 9 46. 3 cos 9 9 4 4 i 3 i 33 e 9 27e 3
3
5
1 i 2 2 2 1 50. cos i sin e 5 5 5 2 2 5
2 1 1 1 i (5 ) 1 e 5 e i 2 e i 0 2 32 32 32 1 cos 2 i sin 2 32 1 1 0 i 32 1 32
4 4 27 cos i sin 3 3 1 3 27 i 2 2
27 27 3 i 2 2
4
5
i 47. 2 cos i sin 2e 10 10 10
3 i 3 3 51. 5e 16 5 cos i sin 16 16 4 3 3 5 cos 4 i sin 4 16 16
5
3 3 25 cos i sin 4 4
32 cos i sin 2 2
2 2 i 25 2 2
32 0 1 i 32 i
4 i 5 5 5 48. 2 cos i sin 2 e 16 16 16
5
i
5
4e 16 4e 4
2 2 i 4 2 2 i
5
6
3 e
i
27e 3
27 cos i sin 3 3 1 3 i 27 2 2
27 27 3 i 2 2
6
6 5 i 5 5 52. 3e 18 3 cos i sin 18 18 6 5 5 3 cos 6 i sin 6 18 18 5 5 27 cos i sin 3 3
5
i 49. 3 cos i sin 3e 18 18 18
4
1 3 i 27 2 2
2 2 2 2 i 4e 4
i (6 ) 18
3 i 25 2 25 2 i 25e 4 2 2
5 5 4 cos i sin 4 4
6
4
i 5 i 25 e 10 32e 2
i 4
5
6
i 27 27 3 i 27e 3 2 2
53. 1 i r 12 (1) 2 2 1 1 1 7 4 7 7 1 i 2 cos i sin 4 4 tan
1155 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
5
7 7 (1 i )5 2 cos i sin 4 4 5 7 7 2 cos 5 i sin 5 4 4 35 35 4 2 cos i sin 4 4 3 3 4 2 cos i sin 4 4 2 2 i 4 2 2 2
i
3
3 i r
3 (1) 4 2 2
1
tan
6
27 cos 34.006 i sin 34.006 27 0.8519 0.5237i
23 14.142i 27ei 2.590
56. 1 5i
r 12 5
5 5 1 5.133 1 5 i 6 cos 5.133 i sin 5.133
1296 0.9753 0.2208i
1264 286.217i 1296ei 3.364 6
11 11 2 cos i sin 6 6 11 11 26 cos 6 i sin 6 6 6
57. 1 i r 12 12 2 1 tan 1 1
4
i 1 i 2 cos i sin 2e 4 4 4
64 1 0 i 64 64ei 2 i
2 (1) 3 2
tan
1 2
2
2
1296 cos 41.064 i sin 41.064
64 cos i sin
r
6
8
3 3
64 cos11 i sin11
55.
6
8
3 11 6 3 i 2 cos 330º i sin 330º
3 i
6
1 5 i 6 cos 8 5.133 i sin 8 5.133
2
6
tan
4 4 i 4 2e 4
54.
2 i 3 cos 5.668 i sin 5.668 3 cos 6 5.668 i sin 6 5.668
2 2
5.668
2 i 3 cos 5.668º i sin 5.668º
1156 Copyright © 2025 Pearson Education, Inc.
Section 10.3: The Complex Plane; De Moivre’s Theorem
i
The three complex cube roots of 1 i 2e 4 are: zk 3
r 42 4 3
2 i k 3
tan
2 i 0 i 3 6 2e 12
6 2 cos i sin 12 12
5
2 3 i 1 i 3 6 2e 4
i
1 5 5 1 i 2 k i k 4 8e 12 2
2 17 i 2 i 2e 12 3 6 2e 12
zk 4 8e 4 3
17 17 2 cos i sin 12 12
z0
6
5 1 11 i 1 i 4 8e 12
3 1 4 2
4
1
3 tan 3 3 11 6
z2 4
11
i 11 11 6 3 i 2 cos i sin 2e 6 6
The four complex fourth roots of are: 1 11 11 2 k k i i 4 2e 24 2
z0
11 11 k 0 i i 4 2e 24 2 4 2e 24
11 11 4 2 cos i sin 24 24 11 23 k 1 i i 2 4 2e 24
z1 4 2e 24
23 23 4 2 cos i sin 24 24 z2
4
11 35 k 2 i i 2e 24 2 4 2e 24
35 35 4 2 cos i sin 24 24 z3
4
5 1 5 i 0 i 8e 12 2 4 8e 12
z1 4 8e 12 2
2
zk 4 2e 4 6
4
5 5 4 8 cos i sin 12 12
3 i r
5
The four complex fourth roots of 4 4 3 i 8e 3 are:
3 3 6 2 cos i sin 4 4
2
2
i 5 5 3 4 4 3 i 8 cos i sin 8e 3 3
z1 6 2e 12
58.
64 8
4 3 3 4 5 3
z0 6 2e 12
z2
1 i 2 k 2 e 3 4
6 2e 12
6
59. 4 4 3 i
11 11 8 cos i sin 12 12 4
5 1 17 i 2 i 8e 12 2 4 8e 12
17 17 8 cos i sin 12 12 5 1 23 i 3 i 4 8e 12
z3 4 8e 12 2 i
11
3 i 2e 6
4
23 23 8 cos i sin 12 12
60. 8 8i r ( 8) 2 ( 8) 2 8 2
8 1 8 5 4
tan
5
i 5 5 4 8 8i 8 2 cos i sin 8 2e 4 4
i
5
The three complex cube roots of 8 8i 8 2e 4 are:
11 47 k 2 i i 2e 24 2 4 2e 24
47 47 4 2 cos i sin 24 24
1157 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors 62. 8
1 5 5 2 i 2 k i k 2 6 2e 12 3
zk 3 8 2 e 3 4
z0
r ( 8) 2 02 8
5 2 5 i 0 i 2 6 2e 12 3 2 6 2e 12
0 0 8 180º
tan
5 5 2 6 2 cos i sin 12 12
8 8 cos180º i sin180º 8ei
5 2 13 i 1 i 3 2 6 2e 12
z1 2 6 2e 12
The three complex cube roots of 8 8ei are:
13 13 2 6 2 cos i sin 12 12 z2 2
6
1 i 2 k
zk 3 8e 3
5 2 7 i 2 i 2e 12 3 2 6 2e 4
2 i 0 i 3 2e 3
z0 2e 3
7 7 2 6 2 cos i sin 4 4
2 cos i sin 3 3
61. 16 i
z1
r 02 16 256 16 2
16 tan 0 270º 16 i 16 cos 270º i sin 270º
3 3 i 0 i 2 2e 8
z0 2e 8
3 3 2 cos i sin 8 8 3 7 i 1 i 2 2e 8
z1 2e 8
7 7 2 cos i sin 8 8 z2
2 cos i sin 2 5 i 2 i 3 2e 3
5 5 2 cos i sin 3 3
3 16e 2 i
i
1 3 3 i 2 k i k 2e 8 2
2 i 1 2e 3 3 2ei
z2 2e 3
3
The four complex fourth roots of 16 i 16e 2 are: zk 4 16e 4 2
2 i k 3
2e 3
63. i r 02 12 1 1 1 tan 0
2 i
i 1(cos 90º i sin 90º ) e 2 The five complex fifth roots of i
i 1 cos 90º i sin 90º e 2 are:
3 11 i 2 i 2e 8 2 2e 8
11 11 2 cos i sin 8 8 3 15 i 3 i 2 2e 8
z3 2e 8
15 15 2 cos i sin 8 8
1158 Copyright © 2025 Pearson Education, Inc.
Section 10.3: The Complex Plane; De Moivre’s Theorem
1 3 2 i 2
1 2 i 2 k i k 1e 10 5
zk 5 1e 5 2
2 i 0 i 5 e 10
z0 1e 10 cos z1
10
i sin
2
i sin
1 3 2 i 3
10
z4
2
2 9 i 2 i 1e 10 5 e 10
cos
i
2 i 4
i
13
17
The four complex fourth roots of unity are: i
z1
z1
3 3 i sin 10 10
i 0 1 i e 2 e 2 cos 90º i sin 90º
0 1i i
3 i i e 2
cos
3 i 1e 2
i
z2
i 0 2 e 2 ei cos180º i sin180º
1 0 i 1
are:
3 i 0 3 i e 2 e 2
cos 270º i sin 270º 0 1i i The complex fourth roots of unity are: 1, i , 1, i . z3
1 3 2 7 i 1 i 1e 5 10 5 e 10
cos
i 0 k 2
1e
i 0 0 2 ei 0 1
1 3 tan 0 2
z0
0 2 k
z0 e
r 02 (1) 2 1 1
3 2 3 i 0 i 1e 10 5 e 10
1
zk 4 1e 4
64. i
zk
19 19 i sin 10 10
r 12 02 1 1 0 tan 0 1 0º 1 0 i 1 cos 0º i sin 0º ei 0
z4 1 e 10 5 e 10 17 17 cos i sin 10 10
1 3 3 2 i 2 k i k e 10 5 5 1e 5 2
3
65. 1 1 0 i
z3 1 e 10 5 e 10 13 13 cos i sin 10 10
The five complex fifth roots of
i
1 3 2 19 i 4 i 1 e 5 10 5 e 10
cos
9 9 i sin 10 10
2 i 3
11
z3 1e 5 10 5 e 2 3 3 cos i sin 2 2
2 i 1 i 1e 10 5 e 2
cos z2
i
z2 1 e 5 10 5 e 10 11 11 cos i sin 10 10
7 7 i sin 10 10
1159 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors 66. 1 1 0 i r 12 02 1 1 0 tan 0 1 0 1 0 i ei 0 The six complex sixth roots of unity are: i
1
zk 6 1e 6
0 2 k
i k
1e 3
i 0
z0 e 3 ei 0 cos 0º i sin 0º 1 0i 1 z1
i 1 i e 3 e 3 cos 60º i sin 60º
1 3 i 2 2 i 2
i
2
z2 e 3 e 3 cos120º i sin120º
i 3 e 3 ei cos180º i sin180º
1 0 i 1 z4
If w 0 , there are n distinct nth roots of w, given by the formula: 2k 2k zk n r cos i sin , n n n n where k 0, 1, 2, ... , n 1 zk n r for all k
1 3 i 2 2 z3
67. Let w r cos i sin be a complex number.
4 i 4 i e3 e 3
68. Since zk n r for all k, each of the complex nth roots lies on a circle with center at the origin and radius n w n r , where w is the original
cos 240º i sin 240º
complex number.
1 3 i 2 2 i 5
i
5
z5 e 3 e 3 cos 300º i sin 300º 1 3 i 2 2 The complex sixth roots of unity are: 1 3 1 3 1 3 1 3 1, i, i, 1, i, i . 2 2 2 2 2 2 2 2
69. Examining the formula for the distinct complex nth roots of the complex number w r cos i sin , 2k 2k zk n r cos i sin , where k 0, 1, 2, ... , n 1 , we see that the zk are spaced apart by an n n n n 2 angle of . n
1160 Copyright © 2025 Pearson Education, Inc.
Section 10.3: The Complex Plane; De Moivre’s Theorem 70. Let z1 r1 cos 1 i sin 1 and z2 r2 cos 2 i sin 2 . Then r cos 1 i sin 1 z1 r1ei1 i 1 z2 r2 e 2 r2 cos 2 i sin 2
r1 cos 1 i sin 1
cos 2 i sin 2 r2 cos 2 i sin 2 cos 2 i sin 2
r cos 1 cos 2 i cos 1 sin 2 i sin 1 cos 2 sin 1 sin 2 1 r2 cos 2 2 sin 2 2
r cos 1 cos 2 sin 1 sin 2 i sin 1 cos 2 cos 1 sin 2 1 1 r2 r r i 1 cos 1 2 i sin 1 2 1 e 1 2 r2 r2
71. By the periodicity of the sine and cosine functions, we know cos cos( 2k ) and sin sin( 2k ) ,
where k is any integer. Then, rei r (cos i sin ) r cos 2k i sin 2k re
i 2 k
, k any
integer. 72. Let r 1 and . Then rei 1ei cos i sin 1 i 0 1 . So, ei 1 1 1 0 . 73. Assume the theorem is true for true for n 1 . For n = 0: z 0 r 0 ei (0 ) r 0 cos(0 ) i sin(0 ) 1 1 cos 0 i sin 0 1 1 1 0 1 1 True
For negative integers:
r e r cos(n ) i sin(n ) with n 1
z n z n
1
n i ( n ) 1
1 r cos(n ) i sin(n ) n
1
n
cos(n ) i sin(n ) cos(n ) i sin(n ) n cos( n ) i sin( n ) r cos(n ) i sin(n ) r (cos 2 (n ) sin 2 (n ) 1
n
r n cos( n ) i sin(n ) r n ei ( n )
74. a.
z a0 a1 0.1 0.4i 0.05 0.48i 0.5 0.8i 0.11 1.6i 0.9 0.7i 0.58 0.56i 1.1 0.1i 0.1 0.12i 0 1.3i 1.69 1.3i 1 1i
b.
1 3i
a2 0.13 0.35i 2.05 1.15i 0.87 1.35i 1.10 0.76i 1.17 3.09i 7 7 i
a3 a4 0.01 0.31i 0.01 0.395i 3.37 3.92i 3.52 25.6i 1.95 1.67i 0.13 7.21i 0.11 0.068i 1.09 0.085i 8.21 5.92i 32.46 98.47i 1 97i
9407 193i
a5 0.06 0.4i 641.7 180.8i 52.88 2.56i 0.085 0.85i 8643.6 6393.7i
a6 0.06 0.35i 379073 232071i 2788.5 269.6i 1.10 0.086i 33833744 110529134.4i
88454401 3631103i
7.8 1015 6.4 1014 i
z1 and z4 are in the Mandlebrot set. a6 for the complex numbers not in the set have very large components.
1161 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
z 0.1 0.4i 0.5 0.8i 0.9 0.7i 1.1 0.1i 0 1.3i 1 1i
c.
z a6 0.4 0.4 0.9 444470 1.1 2802 1.1 1.1 1.3 115591573 1.4 7.8 1015
The numbers which are not in the Mandlebrot set satisfy this condition. The numbers which are in the Mandlebrot set satisfy the condition an 2 .
e x cos y 7 and e x sin y 0 By the Zero-Product Property, e x 0 or sin y 0
e x sin 2k 1 6 . If k is even, then 2 sin 2k 1 1 , and e x 1 6 x ln 6 . If 2 k is odd, then sin 2k 1 1 , and 2 x x e 1 6 e 6 . So,
e x 0 or sin y 0 y k , k an integer.
x yi ln 6 i 2k 1
If y k , then we have e x cos(k ) 7 . If k is even, then cos(k ) 1 , and we have
equivalently,
e x yi 7
75.
x
yi
e e 7 x
e (cos y i sin y ) 7 e x cos y ie x sin y 7
x yi ln 6 i 4k 1
x
e 1 7 x ln 7 . If k is odd, then
77. A
cos( k ) 1 , and we have e x 1 7 e x 7 . , So
2
2
equivalently, x yi ln 7 i 2k , k an integer.
79.
3
24 x 2 y 5 3 8 3 x 2 y 3 y 2 2 y 3 3x 2 y 2
e x e yi 6i
80. Since a = 5 is positive, then the graph opens up and the function has a maximum value. The x 12 12 6 b value is x 2a 2(5) 10 5
e x (cos y i sin y ) 6i e x cos y ie x sin y 6i e x cos y 0 and e x sin y 6
2
6 6 6 f 5 12 4 5 5 5
By the Zero-Product Property, e x 0 or cos y 0 e 0 or cos y 0 y 2k 1 x
2
, k an integer .
1 1 ab sin C (8)(11) sin(113) 40.50 2 2
e x yi 6i
If y 2k 1
, k an even integer , or
4 78. 240 180 3
x yi ln 7 ik , k an even integer, or
76.
, k an integer. , then
2
, k an integer.
36 72 5 4 25 5 16 5
1162 Copyright © 2025 Pearson Education, Inc.
Section 10.4: Vectors 81. a 6, b 8, c 12
86.
16sec 2 x 16 16(sec2 x 1)
a 2 b 2 c 2 2bc cos A cos A
16 tan 2 x 4 tan x
8 12 6 172 b c a 2bc 2(8)(12) 192 2
2
2
2
2
2
172 A cos 1 26.4º 192 b 2 a 2 c 2 2ac cos B cos B
a 2 c 2 b 2 62 122 82 116 2ac 144 2 6 12
116 B cos 1 36.3º 144
Section 10.4 1. vector 2. 0
C 180o A B 180o 36.3o 26.4o 117.3o
3. unit 4. position
82. 3log a x 2 log a y 5log a z log a x 3 log a y 2 log a z 5 log a
x3 y 2 z5
5. horizontal, vertical 6. resultant 7. True
83. log 5 x 4 2 52 x 4
8. False
25 x 4
9. a
252 x 4
10. b
625 x 4 x 621
11. v w
The soluition set is 621 . 84. ( f g )( x) f ( g ( x)) 3(5 x3 ) 2 4(5 x 3 ) 3(25 x 6 ) 4(5 x3 ) 75 x 6 20 x3 3 85. The line would have slope . 2 2 f (6) (6) 5 1 so the line contains the 3 point 6, 1 .
12. u v
13. 3v
3 y (1) ( x 6) 2 3 y 1 x 9 2 3 y x 8 2
1163 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors 14. 2w
19. True 20. False
K G F
21. False
C F E D
22. True 23. False
DE HG
24. False
C H G F
25. True 15. v w
26. True 27. P (0, 0), Q (3, 4) v (3 0)i (4 0) j 3i 4 j 28. P (0, 0), Q (3, 5) v (3 0)i (5 0) j 3i 5 j
16. u v
29. P (3, 2), Q (5, 6) v (5 3)i (6 2) j 2i 4 j 30. P (3, 2), Q (6, 5) v 6 (3) i (5 2) j 9i 3 j
17. 3v u 2w
31. P ( 2, 1), Q (6, 2) v 6 ( 2) i 2 (1) j 8i j 32. P (1, 4), Q (6, 2) v 6 ( 1) i (2 4) j 7i 2 j 33. P (1, 0), Q (0, 1) v (0 1)i (1 0) j i j 34. P (1, 1), Q (2, 2) v (2 1)i (2 1) j i j 35. For v 3i 4 j , v 32 ( 4) 2 25 5 .
18. 2u 3v w
36. For v 5i 12 j , v (5) 2 122 169 13 .
37. For v i j , v 12 (1) 2 2 . 38. For v i j , v (1) 2 ( 1) 2 2 . 1164 Copyright © 2025 Pearson Education, Inc.
Section 10.4: Vectors
39. For v 2i 3 j , v ( 2) 2 32 13 .
51. u
40. For v 6i 2 j , v 62 22 40 2 10 . 41.
x 2 y 2 cos 2 sin 2 1 1
42.
x 2 y 2 12 cot 2 1 cot 2
52. u
csc 2 csc
43. 2 v 3w 2 3i 5 j 3 2i 3 j 6i 10 j 6i 9 j j
44. 3v 2w 3 3i 5 j 2 2i 3j 9i 15 j 4i 6 j 13i 21j
45.
vw
53. u
3i 5 j 2i 3j
5i 8 j 52 ( 8) 2 89
46.
vw
54. u
3i 5 j 2i 3j
v 3i 4 j 3i 4 j 2 v 3i 4 j 3 ( 4) 2 3i 4 j 25 3i 4 j 5 3 4 i j 5 5
v 5i 12 j 5i 12 j 5i 12 j v (5) 2 122 5i 12 j 169 5i 12 j 13 5 12 i j 13 13
ij v ij ij 2 2 v ij 2 1 (1) 1 1 i j 2 2 2 2 i j 2 2 v 2i j v 2i j
2i j 2
2 (1)
v w 3i 5 j 2i 3j
32 ( 5) 2 ( 2) 2 32 34 13
48.
55.
1 1 v 3v (4) 3(4) 14 2 2
56.
3 3 3 v (2) 4 4 2
v w 3i 5 j 2i 3 j 2
2
2
2
3 ( 5) ( 2) 3 34 13 5i
49. u
v 5i v 5i
50. u
v 3 j 3 j 3j j v 3j 3 09
25 0
i 2 j 12 ( 2) 2 5
47.
2
5i i 5
1165 Copyright © 2025 Pearson Education, Inc.
2i j 5 2 5
i
1 5
j
2 5 5 i j 5 5
Chapter 10: Polar Coordinates; Vectors
57. Let v ai bj . We want v 4 and a 2b . v a 2 b2
x
2b 2 b 2 5b 2
4 84 2 4 2 21 2 2 21
5b 2 4 5b 2 16 16 b2 5 b
16 4 4 5 5 5 5
60. P (3, 1), Q ( x, 4), v x (3) i (4 1) j ( x 3)i 3j
8 5 4 5 8 5 4 5 i j or v i j 5 5 5 5
v ( x 3) 2 32
x2 6 x 9 9
58. Let v ai bj . We want v 3 and a b .
x 2 6 x 18 Solve for x:
v a 2 b 2 b 2 b 2 2b 2
x 2 6 x 18 5
2b 2 3
x 2 6 x 18 25
2b 2 9 9 b2 2 b
x2 6 x 7 0 ( x 7)( x 1) 0 x 7 or x 1 The solution set is {7, 1}.
9 3 3 2 2 2 2
3 2 2 3 2 3 2 3 2 3 2 v i j or v i j 2 2 2 2
ab
59. v 2i j , w xi 3j,
61.
5 cos 60º i sin 60º j 1 3 j 5 i 2 2 5 5 3 j i 2 2
v w 2i j xi 3j
(2 x)i 2 j
2
x 4x 4 4 2
x 4x 8
Solve for x: x2 4 x 8 5 x 2 4 x 8 25
v 5, 60º v v cos i sin j
vw 5
(2 x) 2 22
The solution set is 2 21, 2 21 .
4 5 8 5 a 2b 2 5 5 v
4 16 4(1)(17) 2(1)
62.
v 8, 45º v v cos i sin j 8 cos 45º i sin 45º j 2 2 8 i j 2 2 4 2i 4 2 j
x 2 4 x 17 0
1166 Copyright © 2025 Pearson Education, Inc.
Section 10.4: Vectors
63.
v 14, 120º
69. v 3 3i 3j 3 3i j
v v cos i sin j
14 cos 120º i sin 120º j 1 3 14 i j 2 2 7i 7 3j
64.
tan 1
The angle is in quadrant III, thus, 225 . 71. v 4i 2 j 2 2i j
v cos i sin j
1 1 tan 1 26.6 2 2 The angle is in quadrant IV, thus, 333.4 . tan
v 25, 330º
72. v 6i 4 j 2 3i 2 j
v v cos i sin j 25 cos 330º i sin 330º j
v cos i sin j
2 2 tan 1 33.7 3 3 The angle is in quadrant IV, thus, 326.3 .
3 1 25 i j 2 2 25 3 25 i j 2 2
tan
73. v i 5 j v cos i sin j
v 15, 315º
tan 1 5 78.7 The angle is in quadrant III, thus, 258.7 . tan 5
v v cos i sin j
15 cos 315º i sin 315º j 2 2 15 2 15 2 15 i j i j 2 2 2 2
67. v 3i 3j 3 i j
v cos i sin j
74. v i 3 j v cos i sin j
tan 1 3 71.6 The angle is in quadrant II, thus, 108.4 . tan 3
75. F 40 cos 30º i sin 30º j
tan 1 tan 1 1 45
The angle is in quadrant I, thus, 45 . 68. v i 3 j v cos i sin j
tan 1
1
tan 1 tan 1 1 45
1 3 3 i j 2 2 3 3 3 i j 2 2
tan 3
1 30 3 3 The angle is in quadrant II, thus, 150 . tan
v cos i sin j
v 3, 240º
3 cos 240º i sin 240º j
66.
v cos i sin j
70. v 5i 5 j 5 i j
v v cos i sin j
65.
3 60
3 1 40 i j 2 2 20 3 i 20 j
76. F 100 cos 20º i sin 20º j
The angle is in quadrant I, thus, 60 .
1167 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
77. F1 40 cos(30º )i sin(30º ) j
tan
3 1 40 i j 20 3 i 20 j 2 2
83.5
F2 60 cos( 45º )i sin( 45º ) j
F F1 F2 20 3 i 20 j 30 2 i 30 2 j
20 3 30 2 i 20 30 2 j
80. Let v a = the velocity of the plane in still air, v w = the velocity of the wind, and v g = the
velocity of the plane relative to the ground. a.
78. F1 30 cos 45º i sin 45º j 2 2 i j 15 2 i 15 2 j 30 2 2 1 3 j 35 i 35 3j 70 i 2 2
b.
15 2 35 i 15 2 35 3 j
79. Let v a = the velocity of the plane in still air, v w = the velocity of the wind, and v g = the
velocity of the plane relative to the ground. v a 550 j v w 100(cos 45i sin 45 j)
550 j 50 2i 50 2 j
50 2i 550 50 2 j
c.
The speed of the plane relative to the ground is: vg
50 2 550 50 2 2
c.
The speed of the plane relative to the ground is: vg
50 2 500 50 2 2
2
189289.3219 435.1 To find the direction, find the angle between v g and the x-axis and consider a convenient
The plane is traveling with a ground speed of 435.1 mph in an approximate direction of 80.6 degrees west of south ( S 80.6 W ).
v g va vw
50 2 500 i 50 2 j
vector such as due south. 50 2 tan 50 2 500 9.4
2 2 100 i j 2 2 50 2i 50 2 j
b.
v g va vw 500i 50 2i 50 2 j
F F1 F2 15 2 i 15 2 j (35) i 35 3j
a.
v a 500i v w 100(cos 315i sin 315 j) 2 2 100 i j 2 2 50 2i 50 2 j
F2 70 cos 120º i sin 120º j
50 2
The plane is traveling with a ground speed of 624.7 mph in an approximate direction of 6.5 degrees east of north ( N 6.5 E ).
2 2 60 i j 30 2 i 30 2 j 2 2
550 50 2
2
81. Let v a = the velocity of the plane in still air, v w = the velocity of the wind, and v g = the velocity
of the plane relative to the ground. v g va v w
390281.7459 624.7 To find the direction, find the angle between v g and the x-axis.
1168 Copyright © 2025 Pearson Education, Inc.
Section 10.4: Vectors va 500 cos 45º i sin 45º j
v g va v w 300 i 300 3 j 20 2 i 20 2 j
2 2 i j 500 2 2 250 2 i 250 2 j
vg
1 3 j 60 i 2 2 30i 30 3 j
The speed of the plane relative to the ground is:
30 250 2 250 2 30 3 2
2
269,129.1 518.8 km/hr
To find the direction, find the angle between v g and a convenient vector such as due east, i . j component tan i component
250 2 30 3 250 2 30
1.2533 51.5º The plane is traveling with a ground speed of about 518.8 km/hr in a direction of 38.6º east of north N38.6E .
82. Let v a = the velocity of the plane in still air, v w = the velocity of the wind, and v g = the velocity
of the plane relative to the ground. v g va v w va 600 cos 60º i sin 60º j 1 3 j 300 i 300 3 j 600 i 2 2 v w 40 cos 45º i sin 45º j 2 2 i j 40 2 2 20 2 i 20 2 j
2
2
and a convenient vector such as due east, i . j component tan i component
250 2 30 i 250 2 30 3 j
300 20 2 300 3 20 2
To find the direction, find the angle between v g
250 2 i 250 2 j 30i 30 3 j
407,964 638.7 km/hr
v g va v w
vg
The speed of the plane relative to the ground is:
v w 60 cos 120º i sin 120º j
300 20 2 i 300 3 20 2 j
300 3 20 2 300 20 2
1.6689 59.1º The plane is traveling with a ground speed of about 638.7 km/hr in a direction of about 30.9 degrees east of south ( S30.9E ).
83. Let F1 be the force of gravity and F2 be the force required to hold the weight on the ramp. F Then sin10 2 F1 sin10
700 F1
700 sin10 4031
F1
So the combined weight of the boat and its trailer is 4031 lbs. 84. Let F1 be the force of gravity and F2 be the force required to hold the weight on the ramp. F Then sin15 2 F1 sin15
1200 F1
1200 sin15 4636
F1
So the weight of the car is 4636 lbs.
1169 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
85. Let the positive x-axis point downstream, so that the velocity of the current is vc 3i . Let v w = the velocity of the boat in the water, and v g =
the velocity of the boat relative to the land. Then v g v w v c and v g k since the boat is going directly across the river. The speed of the boat is v w 20 ; we need to find the direction. Let v w a i b j , so
v w a 2 b 2 20
ij v w 40 i j i j 40 40 i j 20 2 i j 2 11 v g va v w 250 cos i 250sin j 20 2 i 20 2 j ai Examining the j components:
250sin 20 2 0
a 2 b 2 400
250sin 20 2
Since v g v w v c ,
20 2 0.11314 250 6.5 The heading of the plane should be about N83.5˚E, that is, about 6.5 north of east. sin
k j a i b j 3i (a 3) i b j a3 0 a 3 k b
Examining the i components: 250 cos 6.5º i 20 2 i a i 276.7 a The speed of the plane relative to the ground is about 276.7 miles per hour.
a 2 b 2 400 9 b 2 400
b 2 391 k b 391 19.8
v w 3i 391j and v g 391j
87. a.
Find the angle between v w and j : cos
vw j vw j 3 0 391(1)
391 0.9887 20
20 02 12 8.6º The heading of the boat needs to be about 8.6º upstream. The velocity of the boat directly across the river is about 19.8 kilometers per hour. The time to cross the river is: 0.5 t 0.025 hours or 19.8 0.5 t 60 1.52 minutes . 19.8
v a v a cos 73.74º i sin 73.74º j
v a 120 cos i sin j v a 120i To find we use the law of sines: sin 73.74 sin 120 20 9.21 90 9.21 73.74 7.05 So the plane should head N 7.05 E .
86. Let v a = the velocity of the plane in still air, v w = the velocity of the wind, and v g = the velocity of the plane relative to the ground. v g va v w
b.
v a 250 cos i sin j
1170 Copyright © 2025 Pearson Education, Inc.
Section 10.4: Vectors
73.74 9.21 82.95 v a 120 cos 83.20º 20 120sin 82.95º 124.05 25mi Travel time 0.2 hr = 12 min 124.05mph 2
88.
2
89. Let F1 be the tension on the left cable and F2 be the tension on the right cable. Let F3 represent the force of the weight of the box. F1 F1 cos 155º i sin 155º j F1 0.9063i 0.4226 j
F2 F2 cos 40º i sin 40º j F2 0.7660i 0.6428 j F3 1000 j For equilibrium, the sum of the force vectors must be zero. F1 F2 F3 0.9063 F1 i 0.4226 F1 j 0.7660 F2 i 0.6428 F2 j 1000 j
0.9063 F1 0.7660 F2 i
0.4226 F1 0.6428 F2 1000 j
a. Find : tan
0 Set the i and j components equal to zero and solve: 0.9063 F1 0.7660 F2 0 0.4226 F1 0.6428 F2 1000 0
1 2
Solve the first equation for F2 and substitute the result into the second equation to solve the system: 0.9063 F2 F1 1.1832 F1 0.7660 0.4226 F1 0.6428 1.1832 F1 1000 0
1 26.57 2 So 90 63.43 and 180 63.43 116.57 Using Law of Sines, Find angle A: sin116.57 sin A 10 2 2sin116.57 A sin 1 10 10.30 The boat must head 10.30 26.37 36.87 left of perpendicular to the shore. b.
tan 1
v a 10 cos 126.57º 2 10sin 126.57º 8.954 km/h 2
The boat must travel
5 km.
5 km 0.25 hr (15 min) Time: 8.945 km / h
1.1832 F1 1000 F2 1.1832 845.2 1000
F1 845.2
The tension in the left cable is about 845.2 pounds and the tension in the right cable is about 1000 pounds.
2
90. Let F1 be the tension on the left cable and F2 be the tension on the right cable. Let F3 represent the force of the weight of the box. F1 F1 cos 145º i sin 145º j F1 0.8192i 0.5736 j
F2 F2 cos 50º i sin 50º j F2 0.6428i 0.7660 j F3 800 j
1171 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
For equilibrium, the sum of the force vectors must be zero. F1 F2 F3 0.8192 F1 i 0.5736 F1 j 0.6428 F2 i 0.7660 F2 j 800 j
0.8192 F1 0.6428 F2 i
0.5736 F1 0.7660 F2 800 j
0 Set the i and j components equal to zero and solve: 0.8192 F1 0.6428 F2 0 0.5736 F1 0.7660 F2 800 0
Solve the first equation for F2
and substitute the
result into the second equation to solve the system: 0.8192 F2 F1 1.2744 F1 0.6428 0.5736 F1 0.7660 1.2744 F1 800 0 1.5498 F1 800 F1 516.2 F2 1.2744 516.2 657.8
The tension in the left cable is about 516.2 pounds and the tension in the right cable is about 657.8 pounds. 91. Let F1 be the tension on the left end of the rope and F2 be the tension on the right end of the rope. Let F3 represent the force of the weight of the tightrope walker. F1 F1 cos 175.8º i sin 175.8º j F1 0.99731i 0.07324 j
F2 F2 cos 3.7º i sin 3.7º j F2 0.99792i 0.06453 j F3 150 j For equilibrium, the sum of the force vectors must be zero. F1 F2 F3 0.99731 F1 i 0.07324 F1 j 0.99792 F2 i 0.06453 F2 j 150 j
0.99731 F1 0.99792 F2 i
0.07324 F1 0.06453 F2 150 j
0 Set the i and j components equal to zero and solve:
0.99731 F1 0.99792 F2 0 0.07324 F1 0.06453 F2 150 0 Solve the first equation for F2 and substitute the
result into the second equation to solve the system: 0.99731 F2 F1 0.99939 F1 0.99792 0.07324 F1 0.06453 0.99939 F1 150 0 0.13773 F1 150 F1 1089.1 F2 0.99939(1089.1) 1088.4
The tension in the left end of the rope is about 1089.1 pounds and the tension in the right end of the rope is about 1088.4 pounds. 92. Let F1 be the tension on the left end of the rope and F2 be the tension on the right end of the rope. Let F3 represent the force of the weight of the tightrope walker. F1 F1 cos 176.2º i sin 176.2º j F1 0.99780i 0.06627 j
F2 F2 cos 2.6º i sin 2.6º j F2 0.99897i 0.04536 j F3 135 j For equilibrium, the sum of the force vectors must be zero. F1 F2 F3 0.99780 F1 i 0.06627 F1 j 0.99897 F2 i 0.04536 F2 j 135 j
0.99780 F1 0.99897 F2 i
0.06627 F1 0.04536 F2 135 j
0
Set the i and j components equal to zero and solve: 0.99780 F1 0.99897 F2 0 0.06627 F1 0.04536 F2 135 0 Solve the first equation for F2 and substitute the result into the second equation to solve the system: 0.99780 F2 F1 0.99883 F1 0.99897 0.06627 F1 0.04536 0.99883 F1 135 0
1172 Copyright © 2025 Pearson Education, Inc.
Section 10.4: Vectors T F2 W2 sin 35 0 FN2 W2 cos 35 T W2 sin 35 F2 Tensions are equal so: F1 W2 sin 35 F2 FN1 W2 sin 35 FN2 W1 W2 sin 35 W2 cos 35
0.11158 F1 135
F1 1209.9 F2 0.99883(1209.9) 1208.4
The tension in the left end of the rope is about 1209.9 pounds and the tension in the right end of the rope is about 1208.4 pounds.
W2 sin 35 cos 35
W1
93.
W1
100 sin 35 0.6 cos 35 0.6
13.68 lb
96.
First find FN : FN (20 lbs )(cos 20) 18.7939 lbs Now find the force causing the bos to slide down the incline. Fd (20 lbs )(sin 20) 6.8404 lbs The frictional force f is f FN . Therefore f Fd right at the moment the box begins to slide. So FN Fd F 6.8404 lbs 0.36 d FN 18.7939 lbs
R F1 F2 800(cos10i sin10 j) 710(cos 35i sin 35 j) R
2
(800 cos 10 710 cos 35 ) (800 sin 10 710 sin 35 )
1474.3 N
Direction is 800sin10 710sin 35 tan 800 cos10 710 cos 35 0.3988
94.
tan 1 (0.3988) 21.7 97. F1 3000i Tg (3 lbs )(sin )
To keep box from sliding: T Tg 2 3sin 2 sin 3 2 sin 1 41.8 3
95. Left box: FN1 W1 T = F1 Right box:
F2 2000 cos 45º i sin 45º j 2 2 i j 2000 2 2 1000 2 i 1000 2 j
F F1 F2 3000i 1000 2 i 1000 2 j
3000 1000 2 i 1000 2 j
F
3000 1000 2 1000 2 2
2
4635.2 The monster truck must pull with a force of
1173 Copyright © 2025 Pearson Education, Inc.
2
Chapter 10: Polar Coordinates; Vectors
approximately 4635.2 pounds in order to remain unmoved. 98. a.
a' a v 3, 0 3, 2 0, 2 , b' b v 1, 2 3, 2 2, 4 ,
F1 7000i
c' c v 3, 1 3, 2 6, 1 , and
F2 5500 cos 40º i sin 40º j
d' d v 1,3 3, 2 4, 1 .
5500 0.766044i 0.642788 j
The vertices of the new parallelogram A ' B ' C ' D ' are (0, 2), (2, 4), (6, 1), and (4, 1).
4213.24 i 3535.33j
F F1 F2 7000i 4213.24 i 3535.33 j 11, 213.24 i 3535.33j F (11, 213.24) 2 (3535.33) 2 11, 757.4 The farmer will not be successful in removing the stump. The two tractors will have a combined pull of only about 11,757.4 pounds, which is less than the 6 tons needed.
b.
F1 7000i F2 5500 cos 25º i sin 25º j
5500 0.906308i 0.422618 j 4984.69 i 2324.40 j
F F1 F2 7000i 4984.69 i 2324.40 j 11,984.69 i 2324.40 j F (11,984.69) 2 (2324.40) 2 12, 208.0 The farmer will be successful in removing the stump. The two tractors will have a combined pull of about 12,208 pounds, which is more than the 6 tons needed.
99. a.
Let u 3, 1 . Then u' u v 3, 1 4,5 1, 4 .
The new coordinate will be (1, 4). b.
100. a.
b.
1 1 3 v 3, 2 ,1 . Then 2 2 2 1 3 9 a' a v 3, 0 ,1 ,1 , 2 2 2 1 3 5 b' b v 1, 2 ,1 , 1 , 2 2 2 1 3 3 c' c v 3, 1 ,1 , 2 , and 2 2 2 1 3 1 d' d v 1,3 ,1 , 4 . 2 2 2 The vertices of the new parallelogram 9 5 3 A ' B ' C ' D ' are , 1 , , 1 , , 2 , 2 2 2 1 and , 4 . 2
101. The given forces are: F1 3i; F2 i 4 j; F3 4i 2 j; F4 4 j A vector v a i b j needs to be added for equilibrium. Find vector v a i b j : F1 F2 F3 F4 v 0 3i (i 4 j) (4i 2 j) (4 j) (ai bj) 0 0i 2 j (ai bj) 0 ai (2 b) j 0 a 0; 2 b 0 b2 Therefore, v 2 j .
Let a 3, 0 , b 1, 2 , c 3,1 , and d 1,3 . Then
1174 Copyright © 2025 Pearson Education, Inc.
Section 10.4: Vectors
1 . 2 1 cos , 0 2 3 1 So, cos 1 2 3
equals
102. Let x = Bill’s force. We have FBill xi, FTyrone 240 cos 30i 240 sin 30 j 207.846i 120 j,
1 Thus, tan cos 1 tan 3 . 3 2 3 cos(6 x 3 ) 2 3 2 Amplitude = ; Period = 2 6 3 3 Phase Shift = 6 2
110. y
FChuck 110 cos 25 i 25 sin 30 j 99.6939i 46.4880 j Then F FBill FAdam FChuck ( x 307.5400)i 73.5120 j . F
2
( x 307.5400) 73.5120 2
( x 307.5400) 73.5120
2
2
500
500
2
x 187.0 lb
103. Let = direction angle of the boulder due east. We have FBill 200i, FAdam 240 cos 30i 240 sin 30 j 207.846i 120 j, FChuck 110 cos 25 i 25 sin 30 j 99.6939i 46.4880 j
111.
x2 x1 2 y2 y1 2 7 5 2 1 8 2
F FBill FAdam FChuck (507.5400)i 73.5120 j . So
144 81
73.5120 8.2 north of east. 507.5400
225 15
tan 1
112.
104 – 106. Answers will vary. 107.
12 2 9 2
3
x 2 y 2 20 x 4 y 55 0 ( x 2 20 x 100) ( y 2 4 y 4) 55 100 4 ( x 10) 2 ( y 2) 2 49
x2 3 x 2 33 x 2 27 x 29
113.
0 x3 2 x 2 9 x 18 0 ( x3 2 x 2 ) (9 x 18)
The solution set is 29 . 3
2
f (0) 03 2(0) 2 9(0) 18 18
0 x 2 ( x 2) 9( x 2) 2
108. 3x 12 x 36 x 3x( x 4 x 12) 3x( x 6)( x 2) 1 109. tan cos 1 2 Find the angle , 0 , whose cosine
0 ( x 2 9)( x 2) 0 ( x 3)( x 3)( x 2) x 3, x 3, x 2
The x-intercepts are -3, 3, -2. The y-intercept is 18.
1175 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
7. d 8. b
4( x 5) 2 9 53
114. 2
4( x 10 x 25) 9 53
9. v i j, w i j a. v w 1(1) (1)(1) 1 1 0
4 x 2 40 x 100 44 0 4 x 2 40 x 56 0 a 4, b 40, c 56 x
cos
c.
The vectors are orthogonal.
(40) (40) 2 4(4)(56) 2(4)
40 704 40 8 11 8 8 5 11
10. v i j, w i j a. v w 1(1) 1(1) 1 1 0
The solution set is 5 11,5 11 115.
f ( x) f (3) x 4 34 x 4 81 x3 x 3 x 3 ( x 2 9)( x 2 9) x3 ( x 3)( x 3)( x 2 9) x3 ( x 3)( x 2 9) x3 3x 2 9 x 27
116. ( f g )( x) 25 5sin
cos
c.
The vectors are orthogonal.
11. v 2i j, w i 2 j a. v w 2(1) 1(2) 2 2 0 b.
c.
4. parallel
0
0 2
2
12 (2) 2
2 1
0 0 5
The vectors are orthogonal.
12. v 2i 2 j, w i 2 j a. v w 2(1) 2(2) 2 4 6 vw v w
6
b.
cos
c.
The vectors are neither parallel nor orthogonal.
1. c 2 a 2 b 2 2ab cos C
3. orthogonal
vw v w
5 5 90º
Section 10.5
2. dot product
cos
25 25sin 2 25cos 2 5cos
vw 0 0 v w 12 12 (1) 2 12 90º
b.
2
25(1 sin 2 )
vw 0 0 2 v w 1 (1) 2 12 12 90º
b.
2
2
2 2 12 22 6 3 3 10 10 2 2 5 10 18.4º
13. v 3 i j, w i j a.
v w 3 (1) (1)(1) 3 1
5. True 6. False
1176 Copyright © 2025 Pearson Education, Inc.
Section 10.5: The Dot Product
b.
cos
vw v w 3 1
3 1
3 (1) 1 1 2
2
2
2
3 1
b.
cos
vw v w
c.
1 3
4 2 105º
c.
1 3 2 2
3 1 (1) 2
c.
2
2
The vectors are neither parallel nor orthogonal.
vw 50 2 2 v w 3 4 (6) 2 (8) 2 50 50 1 50 25 100 180º
cos
1 Note that v w and 180º , so the 2 vectors are parallel.
cos
vw v w
vw v w
0 2
1 0 90º
c.
2
2
0 (3)
2
0 0 0 1 3 3
The vectors are orthogonal.
19. v i a j, w 2i 3 j Two vectors are orthogonal if the dot product is zero. Solve for a: vw 0 1(2) (a )(3) 0 2 3a 0 3a 2 2 a 3 20. v i j, w i b j Two vectors are orthogonal if the dot product is zero. Solve for b: vw 0 1(1) 1(b) 0 1 b 0 b 1 21. v 2i 3 j, w i j
75
32 ( 4) 2 92 ( 12) 2 75 75 1 25 225 75 0º
c.
cos
2 6 4
16. v 3i 4 j, w 9i 12 j a. v w 3(9) ( 4)(12) 27 48 75 b.
02 12
The vectors are orthogonal.
b.
15. v 3i 4 j, w 6i 8 j a. v w 3(6) 4(8) 18 32 50 b.
4 0
2
18. v i , w 3 j a. v w 1(0) 0(3) 0 0 0
1 3 12
0 2
0 0 0 4 1 4 90º
14. v i 3 j, w i j v w 1(1) 3(1) 1 3
vw v w
The vectors are neither parallel nor orthogonal.
a.
cos
b.
6 2 4 4 2 2 2 o 75
c.
17. v 4i , w j a. v w 4(0) 0(1) 0 0 0
1 Note that v w and 0º , so the 3 vectors are parallel.
v1
vw
w
2
w
2(1) (3)(1)
12 (1) 2
2
i j
5 5 5 i j i j 2 2 2
1 1 5 5 v 2 v v 1 2i 3 j i j i j 2 2 2 2
1177 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
27. Let 4ai 3aj be the parallel vector.
22. v 3i 2 j, w 2i j v1
vw
w
2
w
3(2) 2(1)
2 1 2
2
2
15 (4a ) 2 (3a) 2
2i j
225 16a 2 9a 2
4 8 4 2i j i j 5 5 5 8 4 v 2 v v1 3i 2 j i j 5 5 7 14 i j 5 5
225 25a 2 a2 9 a 3 We can use the positive answer so the vector with magnitude 15 is 4(3)i 3(3) j 12i 9 j .
28. Let 12ai 9aj be the parallel vector.
23. v i j, w i 2 j v1
vw
w
2
w
5 (12a ) 2 (9a) 2
1(1) (1)(2)
12 2 2
2
i 2 j
1 1 2 i 2 j i j 5 5 5 1 2 6 3 v 2 v v1 i j i j i j 5 5 5 5
24. v 2i j, w i 2 j v1
vw
w
2
w
2(1) (1)( 2)
1 ( 2) 2
2
2
i 2 j
4 4 8 i 2 j i j 5 5 5 4 8 6 3 v 2 v v 1 2i j i j i j 5 5 5 5
25. v 3i j, w 2i j v1
vw
w
2
w
3( 2) 1(1)
( 2) (1) 2
2
2
2i j
7 14 7 2i j i j 5 5 5 14 7 1 2 v 2 v v1 3i j i j i j 5 5 5 5
26. v i 3 j, w 4i j v1
vw
w
2
w
1(4) (3)(1)
4 (1) 2
2
2
4i j
7 28 7 4i j i j 17 17 17 7 28 v 2 v v1 i 3 j i j 17 17 11 44 i j 17 17
25 144a 2 81a 2 25 225a 2 25 a2 225 1 a 3 1 If a then 3 1 1 1 12 i 9 j 4i 3 j . If a then 3 3 3 1 1 12 i 9 j 4i 3 j 3 3
29. F 3 cos 60º i sin 60º j 1 3 3 3 3 j i j 3 i 2 2 2 2 3 3 3 W F AB i j 6i 2 2 3 3 3 (6) 0 9 ft-lb 2 2 30. F 20 cos 30º i sin 30º j 3 1 i j 10 3i 10 j 20 2 2 W F AB 10 3i 10 j 100i
10 3(100) 10 0 1000 3 1732 ft-lb
31. a.
I (0.02) 2 (0.01) 2
0.0005 0.022 The intensity of the sun’s rays is about 0.022 watts per square centimeter.
1178 Copyright © 2025 Pearson Education, Inc.
Section 10.5: The Dot Product
A (300) 2 (400) 2 250, 000 500
The area of the solar panel is 500 square centimeter. b.
Fd 10º Fp F
W I A (0.02i 0.01j) (300i 400 j)
(0.02)(300) (0.01)(400) 6 (4) 10 10
This means 10 watts of energy is collected. c.
To collect the maximum number of watts, I and A should be parallel with the solar panels facing the sun.
32. a.
R (0.75) 2 (1.75) 2 3.625 1.90
About 1.90 inches of rain fell. A (0.3) 2 (1) 2 1.09 1.04
The area of the opening of the gauge is about 1.04 square inches. b.
V R A (0.75i 1.75 j) (0.3i j)
(0.75)(0.3) (1.75)(1) 0.225 (1.75) 1.525 1.525
This means the gauge collected 1.525 cubic inches of rain. c.
To collect the maximum volume of rain, R and A should be parallel and oriented in opposite directions.
33. Split the force into the components going down the hill and perpendicular to the hill. Fd 8º
Fp
F
Fd F sin 8º 5300sin 8º 737.6 Fp F cos8º 5300 cos8º 5248.4
Fd F sin10º 4500sin10º 781.4 Fp F cos10º 4500 cos10º 4431.6
The force required to keep the Silverado from rolling down the hill is about 781.4 pounds. The force perpendicular to the hill is approximately 4431.6 pounds. 35. We must determine the component force going down the ramp. Fd F sin 20º 250sin 20º 85.5 Timmy must exert about 85.5 pounds of force to hold the piano in position. 36. We must determine the angle if the force of the boulder is F 5000 pounds and the component force going down the hill is Fd 1000 pounds. Fd F sin 1000 5000sin 1000 sin 0.2 5000 sin 1 (0.2) 11.5 The angle of inclination of the hill is about 11.5. W 2, AB 4i 37. W F AB F cos i sin j 2 cos i sin j 4i 2 4 cos 1 cos 2 60º
The force required to keep the Explorer from rolling down the hill is about 737.6 pounds. The force perpendicular to the hill is approximately 5248.4 pounds. 34. Split the force into the components going down the hill and perpendicular to the hill.
1179 Copyright © 2025 Pearson Education, Inc.
Chapter 10: Polar Coordinates; Vectors
38. Let u a1i b1 j , v a2 i b2 j , and w a3i b3 j . u v w
a1i b1 j a2 i b2 j a3i b3 j a1i b1 j a2 i a3i b2 j b3 j
a1i b1 j (a2 a3 )i (b2 b3 ) j a1 (a2 a3 ) b1 (b2 b3 ) a1a2 a1a3 b1b2 b1b3 a1a2 b1b2 a1a3 b1b3
43.
w v v w w v v w 2 w vv w v vw 2 w v vw v ww 2 2 w vv v ww 2 2 2 2 w v v w 0
a1i b1 j a2 i b2 j a1i b1 j a3i b3 j
Therefore, the vectors are orthogonal. 44. ( v w ) w v w w w
uv uw
v w w vw
2
39. Since 0 0 i 0 j and v =a i +b j , we have that 0 v 0 a 0b 0 .
vw
40. Let v x i y j . Since v is a unit vector, we
0 Therefore, the vectors are orthogonal.
have that v x 2 y 2 1 , or x 2 y 2 1 . If is the angle between v and i , then xi yj i vi x . Now, cos 1 1 v i
cos 60
2
y 2 1 cos 2
0.5
y 2 sin 2 y sin Thus, v cos i sin j .
2
i
1 0 2
2
i 2
a(1) b(0) 12
2
2
6.5(16 y 2 ) 16 40 y 25 y 2
v1 the projection of v onto i
a i bi i
26 16 y 2
0.5 26 16 y 4 5 y
42. Let v a i b j . The projection of v onto i is
i 2
u v 0.5 u v 4 5y
0.5 26 16 y 2 4 5 y
41. If v a1i b1 j cos i sin j and w a2 i b2 j cos i sin j , then cos( ) v w a1a2 b1b2 cos cos sin sin
v1 i
2
u v 4 5 y, u 26, v 16 y 2
cos y 1
w
45. u i 5 j and v 4i yj
x2 y 2 1 2
w
2
i ai
18.5 y 2 40 y 88 0 Using a solving utility we find two solutions: y 3.515 or y 1.353 . The y 3.515 will not work so the solution is y 1.353
46. u xi 2 j and v 7i 3 j u v 7 x 6, u
v 2 v v1 ai bj ai bj
Since v i ai bj i a 1 b 0 a and v j ai bj j a 0 b 1 b , v v1 v 2 v i i v j j.
1180 Copyright © 2025 Pearson Education, Inc.
x 2 4, v 58
Section 10.5: The Dot Product
cos 30 0.8660
u v 0.8660 u v 7x 6
51. Answers will vary. 52.
Average rate of change of f from x 3 to x2
x 2 4 58
0.8660 58 x 2 4 7 x 6
0.8660 58 x 4 7 x 6 2
2
g 2 g 3 2 3
2
23 5 2 2 27 33 5 3 2 27 5 15 45 60 12 5 5
43.5( x 2 4) 49 x 2 84 x 36 5.5 x 2 84 x 138 0 Using a solving utility we find two solutions: x 16.769 or x 1.496 . The x 1.496 will not work so the solution is y 16.769
f x x3 5 x 2 27
53. 5cos 60 2 tan
4
47. Since we want orthogonal vectors then their dot product is zero.
2 x 2 24 0 2 x 2 24
If u a1i b1 j and v a2 i b2 j , then since
55. If the rectangle is 19 inches long and we cut out x from each end then the resulting length would be 19 – 2x. If the rectangle is 13 inches wide and we cut out x from each end then the resulting width would be 13 – 2x. Turning up the ends would make the height of the box x. Thus the volume of the box would be V x(19 2 x)(13 2 x) 4 x3 64 x 2 247 x
u v ,
56.
x 2 12 x 12 2 3 48. If F is orthogonal to AB , then F AB 0 . So, W F AB 0 .
a12 b12 u
and
2
v
2
2
2
2
2
0
b. The legs of the angle can be made to correspond to vectors u v and u v . 2
2
u v u v u v u v (u u u v v u v v) (u u u v v u v v) 2(u v) 2(u v) 4(u v)
x4
x ln 7 ln 7 ln 3 ( x 4) ln 2 x ln 7 ln 7 ln 3 x ln 2 4 ln 2 x ln 7 x ln 2 ln 3 4 ln 2 ln 7 x(ln 7 ln 2) ln 3 4 ln 2 ln 7 ln 3 4 ln 2 ln 7 x 4.634 ln 7 ln 2 The solution set is {4.634}.
a b
u v u v
( x 1) ln 7 ln 3 ln 2
u v u v a12 b12
7 x 1 3 2 x 4 ln 7 x 1 ln 3 2 x 4
a2 2 b2 2
(a1 a2 )(a1 a2 ) (b1 b2 )(b1 b2 )
50.
5 9 2 2 2
54. (1 sin 2 )(1 tan 2 ) (cos 2 )(sec 2 ) 1 cos 2 cos 2 1
uv 0 2 x x 3 8 0
49. a.
1 5 2(1) 2
57.
f ( x) 3 x 4 9
1181 Copyright © 2025 Pearson Education, Inc.
58.
f ( x)
2 x2 5 2
x 2 x 15
4 2. 2, 3
2 x2 5 ( x 3)( x 5)
Since the function is undefined at x = -3 and 5, then the vertical asymptotes are x = -3 and x = 5. The horizontal asymptote is obtained by using the coefficients of the highest powered terms in the numerator and denominator. So the horizontal asymptote is y 2. 59. Using the sum formula we have: cos80 cos 70 sin 80 sin 70 cos(80 70) 3 cos(150) 2
60.
x
f (9)
12 12 b 9 2a 2 4 2 3 3
4 4 1; y 2sin 3 3 3 4 are 1, 3 . Rectangular coordinates of 2, 3 x 2 cos
3. 3, P 2
2 2 (9) 12(9) 10 44 3
The vertex is 9, 44 and since a is positive the graph is concave up. 1
61. ( f g )( x)
3
3 tan 2 9 2 1
3
9(tan 2 1 2 1
1
9 tan 9 2
3
2
1
9sec 2
3
2
x 3cos 0; y 3sin 3 2 2 Rectangular coordinates of 3, are (0, 3) . 2
4. The point (3, 3) lies in quadrant II. r x 2 y 2 (3) 2 32 3 2
27 sec3
3 y Polar coordinates of the point (3, 3) are
tan 1 tan 1 tan 1 (1) x 4 3 3 3 2, 4 or 3 2, 4 .
Chapter 10 Review Exercises 1. 3, 6
5. The point (0, –2) lies on the negative y-axis. r x 2 y 2 02 ( 2) 2 2 2 2 is undefined, so ; . 0 2 2 2 Polar coordinates of the point (0, –2) are y
tan 1 tan 1 x 0
x 3cos
3 3 3 ; y 3sin 6 2 6 2
3 3 3 , . Rectangular coordinates of 3, are 6 2 2
2, 2 or 2, 2 .
1182 Copyright © 2025 Pearson Education, Inc.
Chapter 10 Review Exercises 6. The point 3, 4 lies in quadrant I.
9. a.
r x 2 y 2 32 42 5 4 y tan 1 tan 1 0.93 3 x
4
tan tan 4 y 1 x yx x y 0
4 tan 1 4.07 3 Polar coordinates of the point (3, 4) are (5, 0.93) or (5, 4.07) .
7. a.
b. The graph is a line through the point 0, 0
with slope 1.
r 2sin r 2 2r sin x2 y 2 2 y 2 x y2 2 y 0 x2 y 2 2 y 1 1 x 2 y 1 12 2
b. The graph is a circle with center (0,1) and radius 1.
10. a.
r 2 4r sin 8r cos 5 x2 y2 4 y 8x 5 x 2 8 x 16 y 2 4 y 4 5 16 4
x 4 y 2 25 2
2
b. The graph is a circle with center 4, 2 and
radius 5 . 8. a.
r 5 r 2 25 x 2 y 2 52
b. The graph is a circle with center 0, 0 and
radius 5.
11. r 4 cos The graph will be a circle with radius 2 and center (2, 0) . Check for symmetry:
Polar axis: Replace by . The result is r 4 cos( ) 4 cos . 1183
Copyright © 2025 Pearson Education, Inc.
The graph is symmetric with respect to the polar axis. : Replace by . 2 r 4 cos( )
The line
4(cos cos sin sin )
3 3(sin cos cos sin ) 3 3(0 sin ) 3 3sin The graph is symmetric with respect to the line . 2
4( cos 0) 4 cos The test fails.
The pole: Replace r by r . r 4 cos . The test fails. Due to symmetry with respect to the polar axis, assign values to from 0 to .
: Replace by . 2 r 3 3sin( )
The line
r 4 cos 4
0 2 3 3.5 6 2 3 0 2 2 2 3 5 2 3 3.5 6 4
The pole: Replace r by r . r 3 3sin . The test fails.
Due to symmetry with respect to the line assign values to from
2 3 6 0
r 3 3sin 6 3
6 3 2
3 3 5.6 2 9 2 3 3 2
3
12. r 3 3sin The graph will be a cardioid. Check for symmetry:
Polar axis: Replace by . The result is r 3 3sin( ) 3 3sin . The test fails.
1184 Copyright © 2025 Pearson Education, Inc.
3 3 0.4 2 0
to . 2 2
, 2
Chapter 10 Review Exercises 13. r 4 cos The graph will be a limaçon without inner loop. Check for symmetry:
14. r x 2 y 2 (1) 2 (1) 2 2 y 1 tan 1 x 1 5 4 The polar form of z 1 i is 5 5 . i sin z r cos i sin 2 cos 4 4
Polar axis: Replace by . The result is r 4 cos( ) 4 cos . The graph is symmetric with respect to the polar axis. : Replace by . 2 r 4 cos( )
The line
15. r x 2 y 2 42 (3) 2 25 5 y 3 tan x 4 5.64 The polar form of z 4 3 i is
4 (cos cos sin sin ) 4 ( cos 0) 4 cos The test fails.
z r cos i sin 5 cos 5.64 i sin 5.64 .
The pole: Replace r by r . r 4 cos . The test fails.
5 i 5 5 3 1 16. 2e 6 2 cos i sin 2 i 6 6 2 2
Due to symmetry with respect to the polar axis, assign values to from 0 to .
r 4 cos
0
3
3 i
3 4 3.1 6 2 7 3 2 4 2 2 9 3 2 5 3 4 4.9 6 2 5
1 2 2 3 17. 3 cos i sin 3 i 3 3 2 2 3 3 3 i 2 2
1185
Copyright © 2025 Pearson Education, Inc.
18. 35 i 35 35 i sin 0.1 0.9848 0.1736 i 0.1e 18 0.1 cos 18 18 0.10 0.02i
71 71 i sin 21. z w 5 cos i sin cos 18 18 36 36
i
i
i
71
71
i
73
5 cos i sin 36 36 5 cos i sin i 18 18 5e 18 z 72 72 72 w i sin cos e 36 36 36 72 i
23 i
i
5e 18 36 5e 12 5e 12
5 cos i sin 12 12 3
4 4 5 5 i cos i sin 19. z w cos 9 9 18 18 i
4
i
i
4 5 18
1 1e 9 e 9 e 9 cos
i
13
e 18
13 13 i sin 18 18
4 4 4 i z cos 9 i sin 9 e 9 5 5 5 w i sin cos e 18 18 18 e
4 5 i 9 18
cos
6
e
i sin
20. z w 3 cos
i
i
6
i
i
9 5
3 2e 5 e 9 6e 5
6ei 2 6ei 0
6 cos0 i sin 0 6 9 9 9 i 3 cos i sin 5 5 3e 5 z w 2 cos i sin 2e 5 5 5 9
3 i 3 i 8 e 5 5 e 5 2 2 3 8 8 cos i sin 2 5 5
3
5 i 5 5 23. 2 cos i sin 2e 8 8 8 5 4 i 4 i 2 e 8 4e 2 4 cos i sin 2 2 4(0 1 i ) 4i
6
9 9 i sin 2 cos i sin 5 5 5 5
9
i 22. 3 cos i sin 3e 9 9 9 i 3 i 33 e 9 27e 3 27 cos i sin 9 9 1 3 27 i 2 2 27 27 3 i 2 2 4
i
5e 18 e 36 5e 18 36 5e 36 5e 36
24. 1 3 i
r 12 3
2 2
3 3 1 5 3
tan
1186 Copyright © 2025 Pearson Education, Inc.
4
Chapter 10 Review Exercises 5 5 1 3 i 2 cos i sin 3 3
1 3 i 2e 6
i
5 3
2 2 z0 3 cos 0 i sin 0 3 3
6
3 cos 0º i sin 0º 3
i 6 5 26 e 3 64ei10 64ei 2 64ei 0
2 2 z1 3 cos 1 i sin 1 3 3
64 cos 0 i sin 0
2 2 3 3 3 i 3 cos i sin 3 3 2 2
64 0 i 64
i
2
3e 3
25. 3 4i
2 2 z2 3 cos 2 i sin 2 3 3
r 32 42 5 4 3 0.9273
tan
4 4 3 3 3 i 3 cos i sin 3 3 2 2
3 4i 5 cos 0.9273 i sin 0.9273 5ei 0.9273
5 e
(3 4i ) 4 54 ei 0.9273 4 i 3.7092
5 e 4
4 i 40.9273
i
4
3e 3
27.
625 cos 3.7092 i sin 3.7092 625 0.8432 0.5376i 527 336i
28.
26. 27 0 i r 27 2 02 27 0 0 tan 27 0º 27 0i 27 cos 0º i sin 0º
The three complex cube roots of 27 0i 27 cos 0º i sin 0º are: 0º 2 k 0º 2 k zk 3 27 cos i sin 3 3 3 3
29. P (1, 2), Q (3, 6) v (3 1)i 6 ( 2) j 2i 4 j
2 k 2 k 3 cos i sin 3 3
v 22 ( 4)2 20 2 5
30. P (0, 2), Q (1, 1) v (1 0)i 1 ( 2) j i 3 j v (1) 2 32 10
31. v w 2i j 4i 3 j 2i 2 j 1187
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32. 4 v 3w 4 2i j 3 4i 3j
37. v i 3 j v cos i sin j
8i 4 j 12i 9 j 20i 13 j
33.
v 2i j ( 2) 2 12 5
34.
v w 2i j 4i 3 j ( 2) 2 12 42 (3)2 5 5 7.24
2i j 5
cos
vw v w
11 ( 2) 12 42 (3) 2 2
11 11 5 5 5 5 169.7º
2 5 5 i j 5 5
36. Let v x i y j,
v 3 , with the direction o
angle of v equal to 60 . v 3
39. v i 3j, w i j v w 1(1) (3)(1) 1 3 4 cos
x2 y 2 3
x2 y 2 9
vw v w
4 1 (3)2 (1) 2 12 2
o
The angle between v and i equals 60 . Thus, xi yj i x x vi cos 60o . 3 1 3 1 3 v i We also conclude that v lies in Quadrant I. x cos 60o 3 1 x 2 3 3 x 2 x2 y 2 9
2
v x i y j
2 5
2 5 5
40. v 2i 3 j, w 4i 6 j v w 2 (4) 3 (6) 8 18 26
cos
vw v w 26 2 3 2
2
4 6 2
2
26 26 26 1 26 13 52 676
cos 1 1 180o Thus, the vectors are parallel.
27 3 3 4 2
Since v lies in Quadrant I, y
10 2 153.4º
9 36 9 27 3 y2 9 9 4 4 4 2
4
2
3 2 2 y 9
y
38. v 2i j, w 4i 3 j v w 2(4) 1(3) 8 3 11
2i j 2i j v 2i j v ( 2) 2 12
35. u
3 3 tan 1 3 60 1 The angle is in quadrant II, thus, 120 . tan
3 3 . So, 2
3 3 3 i j. 2 2
1188 Copyright © 2025 Pearson Education, Inc.
Chapter 10 Review Exercises
44. v 2i 3 j, w 3i j The projection of v onto w is given by: 2i 3j 3i j vw v1 w 3i j 2 2 w 32 12
41. v 2i 2 j, w 3i 2 j v w 2 (3) 2 (2) 6 4 10
cos
vw v w
2 22 3 22
10 10 0.9806 8 13 104
2
2
v w 3 (4) 2 (6) 12 12 0
Thus, the vectors are orthogonal.
vw 2
w
swimmer is v w 2 , and the heading is directly across the river, so v w 2 j . Then
v g v w v c 2 j 5i 5i 2 j
vw
v g 22 52 29 5.39 mi/hr
2
2i j 4i 3j
4 3 2
2
25
Since the river is 1 mile wide, it takes the swimmer about 0.5 hour to cross the river. The swimmer will end up (0.5)(5) 2.5 miles downstream.
4i 3j
2
2 4 1 3
9 27 9 3i j i j 10 10 10
= the velocity of the swimmer relative to the land. Then v g v w v c . The speed of the
and v 2 v v1 w
3i j
45. Let the positive x-axis point downstream, so that the velocity of the current is v c 5i . Let v w = the velocity of the swimmer in the water and v g
43. v 2i j, w 4i 3 j The decomposition of v into 2 vectors v1 and v 2 so that v1 is parallel to w and v 2 is w
10
9 27 2i 3 j i j 10 10 27 9 2i 3 j i j 10 10 7 21 i j 10 10
42. v 3i 2 j, w 4i 6 j
perpendicular to w is given by: v1
2 3 31
v 2 v v1
10 o 11.31 104 Thus, the vectors are neither parallel nor orthogonal.
cos 1
v1
10
4i 3j
46. Let F1 = the tension on the left cable, F2 = the tension on the right cable, and F3 = the force of the weight of the box. F1 F1 cos 140º i sin 140º j
1 4 3 4i 3j i j 5 5 5
v 2 v v1 4 3 2i j i j 5 5 4 3 2i j i j 5 5 6 8 i j 5 5
F1 0.7660i 0.6428 j
F2 F2 cos 30º i sin 30º j F2 0.8660i 0.5000 j F3 2000 j
For equilibrium, the sum of the force vectors must be zero.
1189 Copyright © 2025 Pearson Education, Inc.
F1 F2 F3
down the street is about 697.2 pounds. The force perpendicular to the street is approximately 7969.6 pounds.
0.766044 F1 i 0.642788 F1 j 0.866025 F2 i 0.5 F2 j 2000 j
0.766044 F1 0.866025 F2 i
0.642788 F1 0.5 F2 2000 j
0
Set the i and j components equal to zero and solve: 0.766044 F1 0.866025 F2 0 0.642788 F1 0.5 F2 2000 0
Chapter 10 Test 1 – 3.
Solve the first equation for F2 and substitute the result into the second equation to solve the system: 0.766044 F2 F1 0.884552 F1 0.866025
0.642788 F1 0.5 0.884552 F1 2000 0
1.085064 F1 2000 F1 1843.21 lb
4. x 2 and y 2 3
F2 0.884552(1843.21) 1630.41 lb
r x 2 y 2 (2) 2 (2 3) 2 16 4 and
The tension in the left cable is about 1843.21 pounds and the tension in the right cable is about 1630.41 pounds.
y 2 3 3 2 x Since x, y is in quadrant I, we have 3 .
tan
47. F 5 cos 60º i sin 60º j
The point (2, 2 3) in rectangular coordinates is
1 3 5 5 3 5 i j i j 2 2 2 2 AB 20i 5 5 3 W F AB i j 20i 2 2 5 3 5 (20) (0) 50 ft-lb 2 2
48. Split the force into the components going down the hill and perpendicular to the hill.
r7
5.
x y 7 2
2
x y 7 2
2
2
2
x 2 y 2 49 Thus the equation is a circle with center (0, 0) and radius = 7.
Fd 5º
the point 4, 3 in polar coordinates.
Fp
F
Fd F sin 5º 8000sin 5º 697.2 Fp F cos 5º 8000 cos 5º 7969.6
The force required to keep the van from rolling
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Chapter 10 Test y 3 4
y
2
4
x
3 4
7 4
3 2
7.
7 4
3 2
The line 2 : Replace with : r 2 (cos cos sin sin ) 5
4
r 2 ( cos ) 5 r 2 cos 5 Since the resulting equation is not the same as the original, the graph may or may not be
0
5 4
r 2 cos 5 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the polar axis.
x
5 4
x
r 2 cos( ) 5
4
8. r 2 cos 5 Polar axis: Replace with : r 2 cos( ) 5
y
3 4
0
6. tan 3 sin 3 cos r sin 3 r cos y 3 or y 3x x Thus the equation is a straight line with m = 3 and b = 0. 2
2
0
5 4
symmetrical with respect to the line
7 4
2
.
The pole: Replacing r with r : (r ) 2 cos 5
3 2
r 2 cos 5 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the pole. Note: Since we have now established symmetry about the pole and the polar axis, it can be shown that the graph must also be symmetric about the line 2 .
r sin 2 8sin r r 2 sin 2 8r sin r 2 y 2 8 y x2 y 2 8 y x 2 or 4 2 y x 2
The graph is a parabola with vertex (0, 0) and focus (0, 2) in rectangular coordinates.
9. r 5sin cos 2 The polar axis: Replace with : r 5sin( ) cos 2 ( ) r 5( sin ) cos 2 r 5sin cos 2 Since the resulting equation is the not same as
1191 Copyright © 2025 Pearson Education, Inc.
the original, the graph may or may not be symmetrical with respect to the polar axis. The line : Replace with : 2 r 5sin( ) cos 2 ( )
22 22 22 5 i 180 i sin )] 3e 12. w [3(cos 180 180 22 i 5
22 i 5
i
11
35 e 180 35 e 180 243e 18 11 11 243 cos i sin 18 18
r 5(sin cos cos sin )( cos ) 2 r 5(0 cos (1) sin )(cos 2 ) r 5sin cos 2 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the line . 2 The pole: Replacing r with r : r 5sin cos 2
5
5
13. Let w 8 8 3 i ; then w (8) 2 (8 3) 2 64 192 256 16
so we can write w in polar form as 2 i 1 3 2 2 w 16 i 16(cos i sin ) 16e 3 3 3 2 2
r 5sin cos 2 Since the resulting equation is not the same as the original, the graph may or may not be symmetrical with respect to the pole.
Using De Moivre’s Theorem, the three distinct 3rd roots of w are 1 2 2 2 i i 2 k k 3 2 3 2e 9
zk 3 16e 3 3
10.
where k 0, 1, and 2 . So we have
17 11 11 17 z w 2 cos i sin i sin 3 cos 36 36 90 90
2 2 2 0 i i 3 2 3 2e 9
17 11 i i 2e 36 3e 90
z0 2 3 2e 9
17 11 107 i i 2 3 e 36 90 6e 180
z1 2 3 2e 9
107 107 6 cos i sin 180 180
z2 2 3 2e 9
i
11
2 2 8 i 1 i 3 2 3 2e 9 2 2 14 i 2 i 3 2 3 2e 9
11 17
w 3e 90 3 i 11. e 90 36 17 z 2 i 2e 36 7
33
3 i 20 3 i 20 e e 2 2 3 [cos( 3320 ) i sin( 3320 )] 2
Since 720 has the same terminal side as 3320 , we can write
w 3 (cos 3320 i sin 3320 ) z 2
14. v 8 2 3 2 , 2 2 7 2 5 2, 5 2 15.
v
16. u
5 2 5 2 100 10 2
2
v 1 5 2, 5 2 v 10
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2 2 , 2 2
Chapter 10 Test
17. From Exercise 16, we can write 2 2 v v u 10 , 2 2
23. We first calculate the angle .
10 cos 315, sin 315
16
16 F1
F2
10(cos 315 i sin 315 j) Thus, the angle between v and i is 315 . 1200 lbs
18. From Exercise 17, we can write v 10 cos 315 i sin 315 j 10
F3
2 2 i j 5 2 i 5 2 j 2 2
19. v1 2 v 2 v 3 4, 6 2 3, 6 8, 4
tan 1 2 63.435 .
4 2(3) (8), 6 2(6) 4
Thus the three force vectors in the problem can be written as F1 F1 (cos116.565 i sin116.565 j)
4 6 8, 6 12 4 6, 10 6i 10 j
F1 (0.44721i 0.89443j)
20 – 21. If ij is the angle between v i and v j , then cos ij cos 12
vi v j vi v j
F2 F2 (cos 63.435 i sin 63.435 j) F2 (0.44721i 0.89443 j)
. Thus,
4(3) 6(6) 52 45 4(8) 6(4)
Using the right triangle in the sketch, we 16 conclude tan 2 so that 8
48
F3 1200 j Since the system is in equilibrium, we need F1 F2 F3 0 i 0 j .
8
6 65 65 8 1 cos 13 52 80 8 65 65 4(10) 6(15) 130 cos 14 1 52 325 130 (3)(8) (6)(4) 0 cos 23 0 60 45 80 (3)10 (6)15 120 8 cos 24 45 325 15 65 65 20 1 (8)10 (4)15 cos 34 20 65 65 80 325 The vectors v i and v j are parallel if cos ij 1
This means F1 (0.44721) F2 (0.44721) 0 0 and F1 (0.89443) F2 (0.89443) 1200 0 . The first equation gives F1 F2 ; if we call
this common value F and substitute into the second equation we get 2 F (0.89443) 1200 1200 670.82 2(0.89443) Thus, the cable must be able to endure a tension of approximately 670.82 lbs.
F
and are orthogonal if cos ij 0 . Thus, vectors v1 and v 4 are parallel and v 2 and v 3 are orthogonal.
22. The angle between vectors v1 and v 2 is v v 48 cos 1 1 2 cos 1 v v 52 45 . 2 1 cos 1 (0.992278) 172.87
1193 Copyright © 2025 Pearson Education, Inc.
Chapter 10 Cumulative Review 1.
2
ln e x 9 ln 1 x2 9 0 x 3 x 3 0 x 3 or x 3 The solution set is 3,3 .
2. The line containing point (0, 0), making an angle of 30o with the positive x-axis has polar
3
x2 y3 2 x4
which is not equivalent to x 2 y 3 2 x 4 . y-axis: Replace x by x :
x 2 y 3 2 x 4 x 2 y3 2 x 4
which is equivalent to x 2 y 3 2 x 4
6
3 tan tan 6 3
Origin: Replace x by x and y by y :
x 2 y 3 2 x 4
3 y So, x 3 y
5. x 2 y 3 2 x 4 Test for symmetry: x-axis: Replace y by y : x2 y 2 x4
. 6 Using rectangular coordinates: y tan 1 x y tan x
f x ln 1 2 x f will be defined provided 1 2 x 0 . 1 2x 0 1 2x 1 x 2 1 1 The domain of is x x or , . 2 2
2
e x 9 1
equation
4.
x2 y3 2 x4
3 x 3
3. The circle with center point (h, k) = (0,1) and radius r = 3 has equation:
x h 2 y k 2 r 2 x 0 2 y 12 32 2 x 2 y 1 9
which is not equivalent to x 2 y 3 2 x 4 . Therefore, the graph is symmetric with respect to the y-axis. 6. y ln x ln x, ln x, ln x, ln x,
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when ln x 0 when ln x 0 when x 1 when 0<x 1
Chapter 10 Cumulative Review
10. Graphing x 3 and y 4 using rectangular coordinates: x 3 yields a vertical line passing through the point 3, 0 .
7. y sin x sin x, sin x, sin x, sin x,
when sin x 0 when sin x 0 when 0 x when x 2
y 4 yields a horizontal line passing through
the point 0, 4 .
11. Graphing r 2 and
using polar coordinates: 3 r 2 yields a circle, centered at 0, 0 , with radius
when x 0 sin x, 8. y sin x sin x, when x 0
= 2.
3
yields a line passing through the point 0, 0 ,
forming an angle of
3
with the positive x-axis.
1 9. sin 1 2
We are finding the angle , 1 sine equals . 2 1 sin , 2 6 1 sin 1 6 2
, whose 2 2
12. y 4 cos( x) is of the form y a cos( x) Amplitude: a 4 4
2 2
Period:
2
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2
2
Chapter 10 Projects
2 2 Magnitude 1 0.25 i 2 2
Project I
2
2. The aircraft will fall (lose altitude).
If h is high enough such that cos(2 h) 0 and sin(2 h) 1 , then
3. The aircraft will speed up. 4. The aircraft will slow down. 5. W 700 lb
Magnitude 1 0.25 0 i
2.205 kg 1543.5 kg 1 lb
6. F 1543.5 kg 9.80 m/s 2 15,126.3 kg m/s 15,126.3 N
1 0.25 1.03 2
If h is high enough such that cos(2 h) 0 and sin(2 h) 1 , then
Magnitude 1 0.25 0 i 1.03
2
If h is high enough such that cos(2 h) 1 and sin(2 h) 0 , then
7. The lift force of the Wright brother’s plane must have exceeded 15,126.3 newtons in order for it to have gotten off the ground.
Magnitude 1 0.25 1 0.75 (min).
Because sine and cosine oscillate between 1 and 1, the magnitude will oscillate between a maximum and a minimum. 1.25 SWR 1.67 .75
2.205 kg 8. W 2400 lb 5292 kg 1 lb
F 5292 kg 9.80 m/s 2
51,861.6 kg m/s 2 51,861.6 N The lift force of the Cessna 172P must exceed 51,861.6 newtons in order to get off the ground.
9. W 560, 000 lb
2.205 kg 1, 234,800 kg 1 lb
2
2 2 1 1.19 8 8
1. The aircraft will rise (gain altitude).
F 1, 234,800 kg 9.80 m/s 2
2
12,101, 040 kg m/s 12,101, 040 N The lift force of the Boeing 787 must exceed 5,510,295 newtons in order to get off the ground.
Project II
2. The distance between two consecutive minima will be the period of cosine (the value of cosine 2 1. is 1 once per cycle): Period 2 3. The distance between two consecutive maxima will be the period of cosine (the value of cosine 2 is 1 once per cycle): Period 1. 2 4. The distance between a minimum and a maximum is half a period: 0.5(1) 0.5 . 5. The pager must be sensitive enough to receive a minimum signal strength of 0.75. 6.
1. If h 0 , then cos(2 h) 1 and sin(2 h) 0 ,
Imaginary axis
then Magnitude 1 0.25(1) 1.25 (max). If h is high enough such that cos(2 h) and sin(2 h)
2 2
2 , then 2
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Real axis
Chapter 10 Projects
Since the magnitude of a complex number is the distance from (0, 0) to the point, z, the minimum is 0.75 and the maximum is 1.25
Project III
a. A 100e0.05 , 0 30
g. Doubling the interest rate causes the time to double to be less than half. 0.5 19.7 9.85, not 7.5
Project IV
A.
b. Approximately 19.7 years.
a. ei cos i sin 1 i (0) 1
Therefore, ei 1 0 .
b.
c. Approximately 26.1 years.
eix e ix cos x i sin x cos( x) i sin( x) 2i 2i cos x i sin x cos x i sin x 2i sin x 2i 2i sin x eix e ix cos x i sin x cos( x) i sin( x) 2 2 cos x i sin x cos x i sin x 2 cos x 2 2 cos x
c. sin(1 i )
d. A 100e0.10 , 0 30
e 1 ei (1) e e i (1) 2i e 1 (cos1 i sin1) e cos( 1) i sin( 1)
2i 0.1988 i (0.3096) 1.4687 2.2874i 2i 1.2699 2.597i i 1.2985 0.6350i 2i i
e. Approximately 7.5 years
d. a bi r1 (cos x i sin x) r1eix ,
ei (1i ) ei (1 i ) e 1i e1i 2i 2i
r1 a 2 b 2
c di r2 (cos y i sin y ) r2 eiy ,
f. Approximately 13.5 years.
r2 c 2 d 2 (a bi )(c di ) r1eix r2 eiy r1r2 eix eiy r1r2 eix iy r1r2 ei ( x y )
r1r2 cos( x y ) i sin( x y )
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three points of intersection, however, the “missing” point is complex so it will not show up on the graph.
a bi r1eix r1 ix iy r1 i ( x y ) e e c di r2 eiy r2 r2 r 1 cos( x y ) i sin( x y ) r2
B. a. 1.
x u iv
i (u iv)3 8 0 i (u 3 3u 2 (iv) 3u (iv) 2 (iv)3 ) 8 0 i (u 3 3iu 2 v 3uv 2 iv3 ) 8 0
2.
v3 3u 2 v 8 0
8 v3 u 3v
3
b. ix 8 0
2
3
x3 8i 0
2
u 3uv 0 2
x3 8i 0
2
u (u 3v ) 0 2
x3 8i 8i 8(0 i ) 8(cos 90 i sin 90)
2
u 0 or u 3v 0
If u 2 3v 2 0 , then u 2 3v 2 0
90 360 90 360 zk 3 8 cos k i sin k 3 3 3 3 2 cos 30 120k i sin 30 120k
u 2 3v 2 8 v3 3v 3 9v 8 v3
3v 2
z0 2 cos 30 i sin 30
8v3 8
3 1 2 i 3 i 2 2 z1 2 cos 30 120 i sin 30 120
v3 1 v 1
If u 0 and v 2 , then x 2i . If v 1 , then u 3 . Thus, x 3 i or x 3 i
3.
(v 3 3u 2 v 8) (u 3 3uv 2 )i 0
8 v3 and u 3v 3v
Graph u
3 1 2 i 3 i 2 2 z1 2 cos 30 240 i sin 30 240 2 0 1i 2i
These solutions do match those in part (a). If u 0 , then 8 v3 3v 3 8v 0 v 2 0
u 0 or u 2 3v 2 0
4.
The points of intersection are (1, 1.73) and (1, 1.73) , which matches up to the second set of intersection points. There should be
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Chapter 11 Analytic Geometry Section 11.1
12. (3, 2)
Not applicable
13. d
Section 11.2
14. False; y 2
x2 x1 2 y2 y1 2
1.
15. B; the graph has a vertex h, k 0, 0 and opens up. Therefore, the equation of the graph has the form x 2 4ay . The graph passes
2
4 2. 4 2
3.
through the point 2,1 so we have
2 2 4a 1
x 4 2 9
4 4a
x 4 3 x43 or x 4 3
x 1 or
1 a Thus, the equation of the graph is x 2 4 y .
x 7
The solution set is {7, 1} .
16. G; the graph has vertex h, k 1,1 and opens to the left. Therefore, the equation of the graph
4. (2, 5)
has the form y 1 4a x 1 . 2
5. 3, up
17. E; the graph has vertex h, k 1,1 and opens
6. (3, 5) ; x 3
to the right. Therefore, the equation of the graph has the form y 1 4a x 1 . 2
7. parabola; axis of symmetry 8. latus rectum
18. D; the graph has vertex h, k 0, 0 and opens
9. a. 6 b. right c. 3; 3 d. (1, 2)
down. Therefore, the equation of the graph has the form x 2 4ay . The graph passes through the point 2, 1 so we have
2 2 4a 1
e. ( y 2) 2 12( x 1) f. (3, 2) g. 10 h. left i. ( y 2) 2 20( x 3)
4 4a a 1 Thus, the equation of the graph is x 2 4 y .
19. H; the graph has vertex h, k 1, 1 and
10. a. (0, 0);(0, 2); 2 b. down c. (0, 0); (0, 2); 2 d. up e. 2; (1,3); ( x 1) 2 8( y 3); y 1
opens down. Therefore, the equation of the graph has the form x 1 4a y 1 . 2
20. A; the graph has vertex h, k 0, 0 and opens
to the right. Therefore, the equation of the graph has the form y 2 4ax . The graph passes
11. c 1199
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Chapter 11: Analytic Geometry
through the point 1, 2 so we have
24. The focus is (0, 2) and the vertex is (0, 0). Both lie on the vertical line x 0 . a = 2 and since (0, 2) is above (0, 0), the parabola opens up. The equation of the parabola is: x 2 4ay
2 2 4a 1 4 4a 1 a
x2 4 2 y
Thus, the equation of the graph is y 2 4 x .
x2 8 y
21. C; the graph has vertex h, k 0, 0 and opens
Letting y 2, we find x 2 16 or x 4 . The points (–4, 2) and (4, 2) define the latus rectum.
to the left. Therefore, the equation of the graph has the form y 2 4ax . The graph passes through the point 1, 2 so we have
2 2 4a 1 4 4a 1 a Thus, the equation of the graph is y 2 4 x .
22. F; the graph has vertex h, k 1, 1 and
opens up; further, a 1 . Therefore, the equation of the graph has the form x 1 4 y 1 . 2
23. The focus is (4, 0) and the vertex is (0, 0). Both lie on the horizontal line y 0 . a = 4 and since (4, 0) is to the right of (0, 0), the parabola opens to the right. The equation of the parabola is: y 2 4ax
25. The focus is (0, –3) and the vertex is (0, 0). Both lie on the vertical line x 0 . a = 3 and since (0, –3) is below (0, 0), the parabola opens down. The equation of the parabola is: x 2 4ay
y2 4 4 x
x2 4 3 y
2
y 16 x
x 2 12 y
Letting x 4, we find y 2 64 or y 8 . The points (4, 8) and (4, –8) define the latus rectum.
Letting y 3, we find x 2 36 or x 6 . The points 6, 3 and 6, 3 define the latus rectum.
1200
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Section 11.2: The Parabola 26. The focus is (–4, 0) and the vertex is (0, 0). Both lie on the horizontal line y 0 . a = 4 and since (–4, 0) is to the left of (0, 0), the parabola opens to the left. The equation of the parabola is: y 2 4ax
x 2 4ay x 2 4 1 y x2 4 y
Letting y –1, we find x 2 4 or x 2 . The points (–2, –1) and (2, –1) define the latus rectum.
2
y 44 x y 2 16 x
Letting x 4, we find y 2 64 or y 8 . The points (–4, 8) and (–4, –8) define the latus rectum.
1 and the vertex is (0, 0). 2 1 1 1 The focus is 0, . a and since 0, is 2 2 2 above (0, 0), the parabola opens up. The equation of the parabola is: x 2 4ay 1 x2 4 y 2 x2 2 y 1 Letting y , we find x 2 1 or x 1 . 2 1 1 The points 1, and 1, define the latus 2 2 rectum.
29. The directrix is y
27. The focus is (–2, 0) and the directrix is x 2 . The vertex is (0, 0). a = 2 and since (–2, 0) is to the left of (0, 0), the parabola opens to the left. The equation of the parabola is: y 2 4ax y2 4 2 x y 2 8x
Letting x – 2, we find y 2 16 or y 4 . The points (–2, 4) and (–2, –4) define the latus rectum.
28. The focus is (0, –1) and the directrix is y 1 . The vertex is (0, 0). a = 1 and since (0, –1) is below (0, 0), the parabola opens down. The equation of the parabola is: 1201
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Chapter 11: Analytic Geometry
1 and the vertex is (0, 0). 2 1 1 1 The focus is , 0 . a and since , 0 is 2 2 2 to the right of (0, 0), the parabola opens to the right. The equation of the parabola is: y 2 4ax
2 1 3 , 3 define the latus rectum.
30. The directrix is x
1 y2 4 x 2 2 y 2x 1 Letting x , we find y 2 1 or y 1 . The 2 1 1 points , 1 and , 1 define the latus 2 2 rectum.
32. Vertex: (0, 0). Since the axis of symmetry is horizontal, the parabola opens left or right. Since (2, 3) is to the right of (0, 0), the parabola opens to the right. The equation has the form y 2 4ax . Substitute the coordinates of (2, 3) into the equation to find a : 32 4a 2 9 8a 9 a 8 9 The equation of the parabola is: y 2 x . The 2 9 9 focus is , 0 . Letting x , we find 8 8 81 9 9 9 y2 or y . The points , and 16 4 8 4 9 9 , define the latus rectum. 8 4
31. Vertex: (0, 0). Since the axis of symmetry is vertical, the parabola opens up or down. Since (2, 3) is above (0, 0), the parabola opens up. The equation has the form x 2 4ay . Substitute the coordinates of (2, 3) into the equation to find a : 2 2 4a 3 4 12a 1 a 3 4 The equation of the parabola is: x 2 y . The 3 1 1 focus is 0, . Letting y , we find 3 3 x2
4 2 2 1 or x . The points , and 9 3 3 3
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Section 11.2: The Parabola 33. The vertex is (2, –3) and the focus is (2, –5). Both lie on the vertical line x 2 . a 5 3 2 and since (2, –5) is below
rectum.
(2, –3), the parabola opens down. The equation of the parabola is:
x h 4a y k 2 x 2 4 2 y 3 2 x 2 8 y 3 2
Letting y 5 , we find
x 2 16
35. The vertex is (–1, –2) and the focus is (0, –2). Both lie on the horizontal line y 2 .
2
x 2 4 x 2 or x 6 The points (–2, –5) and (6, –5) define the latus rectum.
a 1 0 1 and since (0, –2) is to the right of
(–1, –2), the parabola opens to the right. The equation of the parabola is:
y k 2 4a x h
y 2 4 1 x 1 2
y 2 2 4 x 1 Letting x 0 , we find
y 2 2 4 y 2 2 y 4 or y 0 The points (0, –4) and (0, 0) define the latus rectum.
34. The vertex is (4, –2) and the focus is (6, –2). Both lie on the horizontal line y 2 . a 4 6 2 and since (6, –2) is to the right of
(4, –2), the parabola opens to the right. The equation of the parabola is:
y k 2 4a x h
y 2 4 2 x 4 2
y 2 2 8 x 4 Letting x 6 , we find
y 2 2 16 y 2 4 y 6 or y 2 The points (6, –6) and (6, 2) define the latus
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Chapter 11: Analytic Geometry 38. The directrix is x 4 and the focus is (2, 4). This is a horizontal case, so the vertex is (–1, 4). a = 3 and since (2, 4) is to the right of x 4 , the parabola opens to the right. The equation of the parabola is: ( y k ) 2 4a ( x h)
36. The vertex is (3, 0) and the focus is (3, –2). Both lie on the horizontal line x 3 . a 2 0 2
and since (3, –2) is below of (3, 0), the parabola opens down. The equation of the parabola is:
x h 2 4a y k x 32 4 2 y 0 x 32 8 y
( y 4) 2 4 3 ( x (1)) ( y 4) 2 12( x 1)
Letting x 2 , we find ( y 4) 2 36 or y 4 6 . So, y 2 or y 10 . The points (2, –2) and (2, 10) define the latus rectum.
Letting y 2 , we find
x 32 16 x 3 4 x 1 or x 7 The points (–1, –2) and (7, –2) define the latus rectum.
39. The directrix is x 1 and the focus is (–3, –2). This is a horizontal case, so the vertex is (–1, –2). a = 2 and since (–3, –2) is to the left of x 1 , the parabola opens to the left. The equation of the parabola is: ( y k ) 2 4a ( x h)
37. The directrix is y 2 and the focus is (–3, 4). This is a vertical case, so the vertex is (–3, 3). a = 1 and since (–3, 4) is above y 2 , the parabola opens up. The equation of the parabola is: ( x h) 2 4a ( y k )
( y ( 2)) 2 4 2 ( x (1))
( x (3)) 2 4 1 ( y 3)
( y 2) 2 8( x 1)
( x 3) 2 4( y 3)
Letting x 3 , we find ( y 2) 2 16 or y +2 4 . So, y 2 or y 6 . The points (–3, 2) and (–3, –6) define the latus rectum.
Letting y 4 , we find ( x 3) 2 4 or x 3 2 . So, x 1 or x 5 . The points (–1, 4) and (–5, 4) define the latus rectum.
1204
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Section 11.2: The Parabola 40. The directrix is y 2 and the focus is (–4, 4). This is a vertical case, so the vertex is (–4, 1). a = 3 and since (–4, 4) is above y 2 , the parabola opens up. The equation of the parabola is: ( x h) 2 4a ( y k )
42. The equation y 2 8 x is in the form y 2 4ax where 4a 8 or a 2 . Thus, we have: Vertex: (0, 0) Focus: (2, 0) Directrix: x 2
( x (4)) 2 4 3 ( y 1) ( x 4) 2 12( y 1)
Letting y 4 , we find ( x 4) 2 36 or x 4 6 . So, x 10 or x 2 . The points (–10, 4) and (2, 4) define the latus rectum.
43. The equation y 2 16 x is in the form y 2 4ax where 4a 16 or a 4 . Thus, we have: Vertex: (0, 0) Focus: (–4, 0) Directrix: x 4
41. The equation x 2 4 y is in the form x 2 4ay where 4a 4 or a 1 . Thus, we have: Vertex: (0, 0) Focus: (0, 1) Directrix: y 1
44. The equation x 2 4 y is in the form x 2 4ay where 4a 4 or a 1 . Thus, we have: Vertex: (0, 0) Focus: (0, –1) Directrix: y 1
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Chapter 11: Analytic Geometry
45. The equation ( y 2)2 8( x 1) is in the form ( y k ) 2 4a ( x h) where 4a 8 or a 2 ,
h 1, and k 2 . Thus, we have: Vertex: (–1, 2); Focus: (1, 2); Directrix: x 3
48. The equation ( y 1) 2 4( x 2) is in the form ( y k ) 2 4a ( x h) where – 4a 4 or a 1 , h 2, and k 1 . Thus, we have: Vertex: (2, –1); Focus: 1, 1
Directrix: x 3 2
46. The equation ( x 4) 16( y 2) is in the form ( x h) 2 4a ( y k ) where 4a 16 or a 4 ,
h 4, and k 2 . Thus, we have: Vertex: (–4, –2); Focus: (–4, 2) Directrix: y 6
49. The equation ( y 3) 2 8( x 2) is in the form ( y k ) 2 4a ( x h) where 4a 8 or a 2 , h 2, and k 3 . Thus, we have: Vertex: (2, –3); Focus: (4, –3) Directrix: x 0
47. a.
The equation ( x 3) 2 ( y 1) is in the
form ( x h) 2 4a ( y k ) where – 4a 1 or a
1 , h 3, and k 1 . Thus, 4
we have: Vertex: (3, –1); Focus: 3, 5 ; 4
Directrix: y 3 4
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Section 11.2: The Parabola
50. The equation ( x 2) 2 4( y 3) is in the form ( x h) 2 4a ( y k ) where 4a 4 or a 1 ,
h 2, and k 3 . Thus, we have: Vertex: (2, 3); Focus: (2, 4); Directrix: y 2
53. Complete the square to put in standard form: x2 8x 4 y 8 x 2 8 x 16 4 y 8 16 ( x 4) 2 4( y 2)
The equation is in the form ( x h) 2 4a ( y k ) where 4a 4 or a 1 , h 4, and k 2 . Thus, we have: Vertex: (–4, –2); Focus: (–4, –1) Directrix: y 3
51. Complete the square to put in standard form: y2 4 y 4x 4 0 y 2 4 y 4 4 x
y 2 2 4 x The equation is in the form ( y k ) 2 4a( x h) where 4a 4 or a 1 , h 0, and k 2 . Thus, we have: Vertex: (0, 2); Focus: (–1, 2); Directrix: x 1
54. Complete the square to put in standard form: y 2 2 y 8x 1 y 2 2 y 1 8x 1 1
y 12 8 x
52. Complete the square to put in standard form: x2 6 x 4 y 1 0
The equation is in the form ( y k ) 2 4a ( x h) where 4a 8 or a 2 , h 0, and k 1 . Thus, we have: Vertex: (0, 1); Focus: (2, 1) Directrix: x 2
x2 6 x 9 4 y 1 9
x 32 4 y 2 The equation is in the form ( x h) 2 4a ( y k ) where 4a 4 or a 1 , h 3, and k 2 . Thus, we have: Vertex: (–3, –2); Focus: (–3, –1) Directrix: y 3
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Chapter 11: Analytic Geometry
57. Complete the square to put in standard form: x2 4 x y 4
55. Complete the square to put in standard form: y2 2 y x 0
x2 4 x 4 y 4 4
y2 2 y 1 x 1
( x 2) 2 y 8
y 12 x 1
The equation is in the form ( x h) 2 4a ( y k )
The equation is in the form ( y k ) 2 4a ( x h) where 4a 1 or a
where 4a 1 or a
1 , h 1, and k 1 . 4
we have:
Thus, we have:
31 Vertex: (2, –8); Focus: 2, 4 33 Directrix: y 4
3 Vertex: (–1, –1); Focus: , –1 4 5 Directrix: x 4
56. Complete the square to put in standard form: x2 4 x 2 y
58. Complete the square to put in standard form: y 2 12 y x 1
x2 4 x 4 2 y 4
y 2 12 y 36 x 1 36
x 2 2 y 2 2
( y 6) 2 ( x 37) The equation is in the form ( y k ) 2 4a ( x h) where
The equation is in the form ( x h) 2 4a ( y k ) where 4a 2 or a
1 , h 2, and k 8 . Thus, 4
1 , h 2, and k 2 . 2
4a 1 or a
Thus, we have:
1 , h 37, and k 6 . Thus, 4
we have:
3 Vertex: (2, –2); Focus: 2, 2 5 Directrix: y 2
Vertex: (37, –6); Focus:
147 , – 6 ; 4
Directrix: x 1208
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149 4
Section 11.2: The Parabola
65. ( y 0) 2 c( x ( 2)) y 2 c( x 2) 12 c(0 2) 1 2c c y2
1 2
1 ( x 2) 2
66. ( y 0) 2 c( x 1) y 2 c x 1 12 c 0 1
59. ( y 1) 2 c( x 0)
1 c
( y 1) 2 cx
c 1 2 y x 1
(2 1) 2 c(1) 1 c ( y 1) 2 x
67. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 cy . The point (300, 80) is a point on the parabola. Solve for c and find the equation: 3002 c(80) c 1125
2
60. ( x 1) c( y 2) (2 1) 2 c(1 2) 1 c c 1 ( x 1) 2 ( y 2)
x 2 1125 y
61. ( y 1) 2 c( x 2) (0 1) 2 c(1 2)
y
1 c c 1
(300,80) (150,h)
( y 1) 2 ( x 2)
x
(0,0)
80 h
62. ( x 0) 2 c( y (1)) x 2 c( y 1)
Since the height of the cable 150 feet from the center is to be found, the point (150, h) is a point on the parabola. Solve for h: 1502 1125h 22,500 1125h 20 h The height of the cable 150 feet from the center is 20 feet.
22 c(0 1) 4 c x 2 4( y 1)
63. ( x 0) 2 c( y 1) x 2 c y 1 22 c 2 1 4c x 4 y 1 2
68. Set up the problem so that the vertex of the parabola is at (0, 10) and it opens up. Then the equation of the parabola has the form: x 2 c( y 10) . The point (200, 100) is a point on the parabola. Solve for c and find the equation:
64. ( x 1) 2 c( y (1))
x 12 c y 1 (0 1) 2 c(1 1) 1 2c c ( x 1) 2
1 2
1 ( y 1) 2
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Chapter 11: Analytic Geometry 2002 c 100 10
is about 25 – 0.69 = 24.31 feet. To find the height of the bridge 30 feet from the center the point (30, y) is a point on the parabola. Solve for y: 302 144 y 900 144 y 6.25 y The height of the bridge 30 feet from the center is 25 – 6.25 = 18.75 feet. To find the height of the bridge, 50 feet from the center, the point (50, y) is a point on the parabola. Solve for y: 502 144 y 2500 144 y y 17.36 The height of the bridge 50 feet from the center is about 25 – 17.36 = 7.64 feet.
40, 000 90c 444.44 c x 2 444.44 y 10 (–200,100)
(200,100)
(0,10)
h 50
Since the height of the cable 50 feet from the center is to be found, the point (50, h) is a point on the parabola. Solve for h: 502 444.44 h 10
70. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens down. Then the equation of the parabola has the form: x 2 cy . The points (50, –h) and (40, –h+10) are points on the parabola. Substitute and solve for c and h: 502 c( h) 402 c( h 10) ch 2500 1600 ch 10c
2500 444.44h 4444.4 6944.4 444.44h 15.625 h The height of the cable 50 feet from the center is about 15.625 feet.
69. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens down. Then the equation of the parabola has the form: x 2 cy . The point (60, –25) is a point on the parabola. Solve for c and find the equation: 602 c( 25) c 144
y
x h
(40,–h+10) 10
x 2 144 y
(50,–h)
1600 2500 10c
y
1600 2500 0c 900 10c x
90 c 90h 2500 h 27.78 The height of the bridge at the center is about 27.78 feet.
25 (60,–25)
To find the height of the bridge 10 feet from the center the point (10, y) is a point on the parabola. Solve for y: 102 144 y 100 144 y 0.69 y The height of the bridge 10 feet from the center
71. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 4ay . Since the parabola is 10 feet across and 4 feet deep, the points (5, 4) and (–5, 4) are on the parabola. Substitute and solve for a :
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Section 11.2: The Parabola
(2.5, y) on the parabola. Solve for y: x 2 8 y
25 16 a is the distance from the vertex to the focus. Thus, the receiver (located at the focus) is 25 1.5625 feet, or 18.75 inches from the base 16 of the dish, along the axis of the parabola. 52 4a (4) 25 16a a
2.52 8 y 6.25 8 y y 0.78125 feet The depth of the searchlight should be 0.78125 feet.
72. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 4ay . Since the parabola is 6 feet across and 2 feet deep, the points (3, 2) and (–3, 2) are on the parabola. Substitute and solve for a : 9 32 4a(2) 9 8a a 8 a is the distance from the vertex to the focus. Thus, the receiver (located at the focus) is 9 1.125 feet, or 13.5 inches from the base of 8 the dish, along the axis of the parabola.
76. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 4ay . a is the distance from the vertex to the focus (where the source is located), so a = 2. Since the depth is 4 feet, there is a point (x, 4) on the parabola. Solve for x: x 2 8 y x 2 8 4 x 2 32 x 4 2 The width of the opening of the searchlight should be 8 2 11.31 feet. 77. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 4ay . Since the parabola is 20 inches across and 6 inches deep, the points (10, 6) and (–10, 6) are on the parabola. Substitute and solve for a : 102 4a(6) 100 24a a 4.17 feet The heat will be concentrated about 4.17 inches from the base, along the axis of symmetry.
73. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 4ay . Since the parabola is 4 inches across and 1 inch deep, the points (2, 1) and (–2, 1) are on the parabola. Substitute and solve for a : 22 4a (1) 4 4a a 1 a is the distance from the vertex to the focus. Thus, the bulb (located at the focus) should be 1 inch from the vertex.
78. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 4ay . Since the parabola is 4 inches across and 3 inches deep, the points (2, 3) and (–2, 3) are on the parabola. Substitute and solve for a : 22 4a (3) 4 12a 1 a inch 3 The collected light will be concentrated 1/3 inch from the base of the mirror along the axis of symmetry.
74. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 4ay . Since the focus is 1 inch from the vertex and the depth is 2 inches, a 1 and the points ( x, 2) and ( x, 2) are on the parabola. Substitute and solve for x : x 2 4(1)(2) x 2 8 x 2 2
The diameter of the headlight is 4 2 5.66 inches. 75. Set up the problem so that the vertex of the parabola is at (0, 0) and it opens up. Then the equation of the parabola has the form: x 2 4ay . a is the distance from the vertex to the focus (where the source is located), so a = 2. Since the opening is 5 feet across, there is a point 1211
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Chapter 11: Analytic Geometry
Imagine placing the Arch along the x-axis with the peak along the y-axis. Since the Arch is 630 feet high and is 630 feet wide at its base, we would have the points 315, 0 , 0, 630 , and 315, 0 . The
79. a.
81. Cy 2 Dx 0
Cy Dx D y2 x C This is the equation of a parabola with vertex at (0, 0) and whose axis of symmetry is the x-axis. D , 0 . The directrix is The focus is 4C D . The parabola opens to the right if x 4C D D 0 and to the left if 0 . C C
equation of the parabola would have the form y ax 2 c . Using the point 0, 630 we have
630 a 0 c 2
630 c The model then becomes y ax 2 630 .
Next, using the point 315, 0 we get 0 a 315 630
82. Ax 2 Dx Ey F 0
2
630 315 a
a.
2
2 a 2 315 315
315
2
2 D D2 1 x 2 A A Ey F 4 A 2 D F D2 E x y 2A A E 4 AE
x 2 630 , we get
2
D D 2 4 AF E x 2 A A y 4 AE This is the equation of a parabola with D D 2 4 AF vertex at , and whose 4 AE 2A
x Width (ft) Height (ft), model 567 283.5 119.7 478 239 267.3 308 154 479.4
axis of symmetry is parallel to the y-axis.
No; the heights computed by using the model do not fit the actual heights.
c. 2
80. Ax Ey 0
If E 0 , then: D D2 D2 A x 2 x 2 Ey F A 4A 4A
Thus, the equation of the parabola with the same given dimensions is 2 2 y x 630 . 315 630
A0
Ax 2 Dx Ey F
630
b. Using y
C 0, D 0
2
b. If E 0 , then D D 2 4 AF 2A D 2 If D 4 AF 0 , then x is a single 2A vertical line. Ax 2 Dx F 0 x
A 0, E 0
E y A This is the equation of a parabola with vertex at (0, 0) and axis of symmetry being the y-axis. E The focus is 0, . The directrix is 4 A E E . The parabola opens up if 0 and y 4A A E down if 0 . A Ax 2 Ey x 2
c.
If E 0 , then Ax 2 Dx F 0 x
D D 2 4 AF 2A
If D 2 4 AF 0 , then x x
D
2
D 4 AF and 2A 2
D D 4 AF are two vertical lines. 2A
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Section 11.2: The Parabola d. If E 0 , then
84. y 2 2 y 8 x 1 0 2
D D 4 AF 2A If D 2 4 AF 0 , there is no real solution. The graph contains no points. Ax 2 Dx F 0 x
2
83. Cy Dx Ey F 0 a.
y 2 2 y 1 8x
( y 1) 2 8 x
The equation is in the form
y k 2 4 x( x h) where
C0
4a 8 so a 2, h 0, k 1.
If D 0 , then:
The vertex is (0,1) and the focus is (2,1). Letting x = 2 gives
Cy 2 Ey Dx F
E E2 E2 C y2 y Dx F C 4C 4C 2
y 12 8 2 where y 1 4
2
E E2 1 y Dx F C 2C 4C 2
y 5 or y 3
The latus rectum endpoints are (2,5) and (2,-3). The distance between the vertex and one of the endpoints is
E F E D y 2C C x D 4CD 2
2 E E 2 4CF D y 2C C x 4CD This is the equation of a parabola with E 2 4CF E vertex at , , and whose 4 2 CD C
d
x2 x1 2 y2 y1 2
2 0 2 5 12
85. x 9 y 2 36
x-intercepts:
axis of symmetry is parallel to the x-axis.
y-intercepts:
x 9(0) 2 36
b. If D 0 , then
x 36
E E 2 4CF Cy Ey F 0 y 2C E is a single If E 2 4CF 0 , then y 2C horizontal line. 2
c.
36 9 x 2
The intercepts are: (0, 2), (0, 2), ( 36, 0) f ( y ) 9 y 2 36 f ( y ) 9( y ) 2 36
E E 2 4CF 2C
9 y 2 36 f ( y )
2
If E 4CF 0 , then
So the function is symmetric with respect to the x-axis.
E E 2 4CF y and 2C E E 2 4CF are two horizontal 2C lines.
86.
y
4 x 1 8 x 1 22( x 1) 23( x 1) 2( x 1) 3( x 1)
d. If D 0 , then Cy 2 Ey F 0 y
0 9 x 2 36 4 x2 x 2
If D 0 , then Cy 2 Ey F 0 y
20 2 5
2 x 2 3x 3
E E 2 4CF 2C
x 5 x5
If E 2 4CF 0 , then there is no real solution. The graph contains no points.
The solution set is 5 .
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Chapter 11: Analytic Geometry
5 87. tan , is in quadrant II 8 Solve for sec : sec2 1 tan 2
y 2 10 2 10 3 tan cos 1 tan 3 7 x 3
sec 1 tan 2 Since is in quadrant II, sec 0 .
89. d ( x2 x1 ) 2 ( y2 y1 ) 2 2
1 2 (3) 5 2 3
sec 1 tan 2 2
25 89 89 5 1 1 8 64 64 8 cos
2
11 11 3 2
2
121 121 9 4
1 1 8 8 89 sec 89 89 89 8
sin 1 cos 2
2
90.
8 89 64 1 1 89 89
2
1573 11 13 36 6 ( x h) 2 ( y k ) 2 r 2
( x (12)) 2 ( y 7) 2
25 5 5 89 89 89 89
6
2
( x 12) 2 ( y 7) 2 6
csc
1 1 89 89 sin 5 89 5 89 5 89
cot
1 1 8 tan 5 5 8
91.
f (b) f (a) ln(5 3) ln(1 3) ba 5 1 ln 8 ln 4 5 1 8 ln ln 2 4 4 4
1 cos , let 92. Using the identity cos 2 2
3 88. tan cos 1 7 3 3 Let cos 1 . Since cos and 7 7
34. Then
17. So 2 1 cos 34 cos17 . 2
, is in quadrant II, and we let 2 x 3 and r 7 . Solve for y: ( 3) 2 y 2 49
y 2 40 y 40 2 10
Since is in quadrant II, y 2 2 .
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Section 11.3: The Ellipse
93.
3 a 2 c2 b2 5; 3; 4 x2 y 2 h. 1 ; the denominator of the 16 25
e. f. g.
x2 5x 2 4 x2 5x 6 x2 5x 6
or x 2 5 x 6
x2 5x 6 0
x2 5x 6 0
( x 6)( x 1) 0
( x 2)( x 3) 0
x 2 -term is b 2 and the denominator of the y 2 -term is a 2
x 6, x 1 x 2, x 3 The solution set is 1, 2,3, 6 .
10. a. 5; 3; parallel to x-axis b. 5; 4; parallel to y-axis ( x h) 2 ( y k ) 2 1 c. a2 b2
Section 11.3 1. d
4 2 2 2 5 2
11. (0, 5); (0,5)
22 32 13
12. 5; 3; x
2
9 3 2. 4 2
3. x-intercepts:
13.
14. a
02 16 4 x 2
15. C; the major axis is along the x-axis and the vertices are at 4, 0 and 4, 0 .
4 x 2 16
x2 4 x 2 2, 0 , 2, 0
y-intercepts: y 2 16 4 0
16. D; the major axis is along the y-axis and the vertices are at 0, 4 and 0, 4 .
2
17. B; the major axis is along the y-axis and the vertices are at 0, 2 and 0, 2 .
y 2 16 y 4 0, 4 , 0, 4
18. A; the major axis is along the x-axis and the vertices are at 2, 0 and 2, 0 .
The intercepts are 2, 0 , 2, 0 , 0, 4 , and
0, 4 . 4.
2,5 ; change x to x : 2 2 2
19.
5. left 1; down 4 6.
2, 3 6, 3
x 2 2 y 3 12 2
x 2 2 y 32 1
x2 y 2 1 25 4 The center of the ellipse is at the origin. a 5, b 2 . The vertices are (5, 0) and (–5, 0). Find the value of c: c 2 a 2 b 2 25 4 21 c 21
7. ellipse 8. b 9. a.
2,3
b. 10 c. 5 d. 4 1215
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Chapter 11: Analytic Geometry
The foci are
21, 0 and 21, 0 .
y2 1 16 The center of the ellipse is at the origin. a 4, b 1 . The vertices are (0, 4) and (0, –4). Find the value of c: c 2 a 2 b 2 16 1 15
22. x 2
c 15
The foci are 0, 15 and 0, 15
20.
x2 y 2 1 9 4 The center of the ellipse is at the origin. a 3, b 2 . The vertices are (3, 0) and (–3, 0). Find the value of c: c2 a 2 b2 9 4 5 c 5
The foci are
21.
5, 0 and 5, 0 .
23. 4 x 2 y 2 16 Divide by 16 to put in standard form: 4 x 2 y 2 16 16 16 16 x2 y 2 1 4 16 The center of the ellipse is at the origin. a 4, b 2 . The vertices are (0, 4) and (0, –4). Find the value of c: c 2 a 2 b 2 16 4 12
x2 y 2 1 9 25 The center of the ellipse is at the origin. a 5, b 3 . The vertices are (0, 5) and (0, –5). Find the value of c: c 2 a 2 b 2 25 9 16 c4 The foci are (0, 4) and (0, –4).
c 12 2 3
The foci are 0, 2 3 and 0, 2 3 .
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Section 11.3: The Ellipse
24. x 2 9 y 2 18 Divide by 18 to put in standard form: x 2 9 y 2 18 18 18 18 x2 y2 1 18 2 The center of the ellipse is at the origin.
a 3 2, b 2 . The vertices are 3 2, 0
26. 4 y 2 9 x 2 36 Divide by 36 to put in standard form: 4 y 2 9 x 2 36 36 36 36 x2 y2 1 4 9 The center of the ellipse is at the origin. a 3, b 2 . The vertices are (0, 3) and (0, –3). Find the value of c: c2 a2 b2 9 4 5 c 5
and 3 2, 0 . Find the value of c: 2
2
2
c a b 18 2 16 c4 The foci are (4, 0) and (–4, 0).
27. 25. 4 y 2 x 2 8 Divide by 8 to put in standard form: 4 y2 x2 8 8 8 8 x2 y2 1 8 2 The center of the ellipse is at the origin. a 8 2 2, b 2 .
x2 y2 1 16 16
This is a circle whose center is at (0, 0) and radius = 4. The focus of the ellipse is 0, 0 and the vertices are 4, 0 , 4, 0 , 0, 4 , 0, 4 .
the value of c: c2 a 2 b2 8 2 6 The foci are
x 2 y 2 16
The vertices are 2 2, 0 and 2 2, 0 . Find
c 6
The foci are 0, 5 and 0, 5 .
6, 0 and 6, 0 .
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Chapter 11: Analytic Geometry 31. Center: (0, 0); Focus: (0, –4); Vertex: (0, 5); Major axis is the y-axis; a 5; c 4 . Find b:
x2 y 2 1 4 4 This is a circle whose center is at (0, 0) and radius = 2. The focus of the ellipse is 0, 0 and
28. x 2 y 2 4
b 2 a 2 c 2 25 16 9 b3
the vertices are 2, 0 , 2, 0 , 0, 2 , 0, 2 .
Write the equation:
29. Center: (0, 0); Focus: (3, 0); Vertex: (5, 0); Major axis is the x-axis; a 5; c 3 . Find b: 2
2
x2 y 2 1 9 25
32. Center: (0, 0); Focus: (0, 1); Vertex: (0, –2); Major axis is the y-axis; a 2; c 1 . Find b:
2
b a c 25 9 16 b4 x2 y 2 Write the equation: 1 25 16
b2 a 2 c 2 4 1 3 b 3
Write the equation:
x2 y 2 1 3 4
33. Foci: (±2, 0); Length of major axis is 6. Center: (0, 0); Major axis is the x-axis; a 3; c 2 . Find b:
30. Center: (0, 0); Focus: (–1, 0); Vertex: (3, 0); Major axis is the x-axis; a 3; c 1 . Find b: b2 a 2 c 2 9 1 8
b2 a 2 c2 9 4 5 b 5
b2 2 x2 y 2 Write the equation: 1 9 8
Write the equation:
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x2 y 2 1 9 5
Section 11.3: The Ellipse 34. Foci: (0, ±2); length of the major axis is 8. Center: (0, 0); Major axis is the y-axis; a 4; c 2 . Find b :
37. Foci: (0, ±3); x-intercepts are ±2. Center: (0, 0); Major axis is the y-axis; c 3; b 2 . Find a :
b 2 a 2 c 2 16 4 12 b 2 3
Write the equation:
a 2 b 2 c 2 4 9 13 a 13
x2 y 2 1 12 16
Write the equation:
35. Focus: 4, 0 ; Vertices: 5, 0 and 5, 0 ;
38. Vertices: (±4, 0); y-intercepts are ±1. Center: (0, 0); Major axis is the x-axis; a 4; b 1 . Find c: c 2 a 2 b 2 16 1 15
Center: 0, 0 ; Major axis is the x-axis. a 5 ; c 4 . Find b: b 2 a 2 c 2 25 16 9 b 3
Write the equation:
a 15
x2 y 2 1 25 9
Write the equation:
36. Focus: (0, –4); Vertices: (0, ±8). Center: (0, 0); Major axis is the y-axis; a 8; c 4 . Find b:
x2 y2 1 16
39. Center: (0, 0); Vertex: (0, 4); b 1 ; Major axis is the y-axis; a 4; b 1 .
Write the equation: x 2
b 2 a 2 c 2 64 16 48 b 4 3
Write the equation:
x2 y 2 1 4 13
x2 y 2 1 48 64
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y2 1 16
Chapter 11: Analytic Geometry 40. Vertices: (±5, 0); c 2 ; Major axis is the xaxis; a 5; Find b :
45. The equation
b 2 a 2 c 2 25 4 21 x2 y 2 1 25 21
2
( y k )2
1 (major axis parallel b a2 to the y-axis) where a 3, b 2, h 3, and k 1 . Solving for c:
form
b 21
Write the equation:
( x h) 2
( x 3) 2 ( y 1) 2 1 is in the 4 9
c2 a 2 b2 9 4 5 c 5 Thus, we have: Center: (3, –1)
3, 1 5 , 3, 1 5
Foci:
Vertices: (3, 2), (3, –4)
41. Center: 1,1
Major axis: parallel to x-axis Length of major axis: 4 2a a 2 Length of minor axis: 2 2b b 1 ( x 1) 2 ( y 1) 2 1 4
46. The equation
42. Center: 1, 1
( y 1) 2 1 4
43. Center: 1, 0
Major axis: parallel to y-axis Length of major axis: 4 2a a 2 Length of minor axis: 2 2b b 1 ( x 1) 2
2
( y k )2
1 a b2 (major axis parallel to the x-axis) where a 3, b 2, h 4, and k 2 . Solving for c: c2 a 2 b2 9 4 5 c 5 Thus, we have: Center: (–4, –2); Vertices: (–7, –2), (–1, –2)
form
Major axis: parallel to y-axis Length of major axis: 4 2a a 2 Length of minor axis: 2 2b b 1 ( x 1) 2
( x h) 2
( x 4) 2 ( y 2) 2 1 is in the 9 4
Foci: 4 5, 2 , 4 5, 2
y2 1 4
44. Center: 0,1
Major axis: parallel to x-axis Length of major axis: 4 2a a 2 Length of minor axis: 2 2b b 1 x2 ( y 1) 2 1 4
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Section 11.3: The Ellipse 47. Divide by 16 to put the equation in standard form: ( x 5) 2 4( y 4) 2 16 ( x 5) 2 4( y 4) 2 16 16 16 16 ( x 5) 2 ( y 4) 2 1 16 4 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the a2 b2 x-axis) where a 4, b 2, h 5, and k 4 . Solving for c: c 2 a 2 b 2 16 4 12 c 12 2 3 Thus, we have: Center: (–5, 4)
Foci: 5 2 3, 4 , 5 2 3, 4
49. Complete the square to put the equation in standard form: x2 4x 4 y2 8 y 4 0 ( x 2 4 x 4) 4( y 2 2 y 1) 4 4 4 ( x 2) 2 4( y 1) 2 4 ( x 2) 2 4( y 1) 2 4 4 4 4 2 ( x 2) ( y 1) 2 1 4 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the a2 b2 x-axis) where a 2, b 1, h 2, and k 1 .
Vertices: (–9, 4), (–1, 4)
Solving for c: c 2 a 2 b 2 4 1 3 c 3 Thus, we have: Center: (–2, 1) Foci:
Vertices: (–4, 1), (0, 1)
48. Divide by 18 to put the equation in standard form: 9( x 3) 2 ( y 2) 2 18 9( x 3) 2 ( y 2) 2 18 18 18 18 ( x 3) 2 ( y 2) 2 1 2 18 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the b2 a2 y-axis) where a 3 2, b 2, h 3, and k 2 . Solving for c: c 2 a 2 b 2 18 2 16 c 4 Thus, we have: Center: (3, –2) Foci: (3, 2), (3, –6)
Vertices:
2 3,1 , 2 3,1
50. Complete the square to put the equation in standard form: x 2 3 y 2 12 y 9 0 x 2 3( y 2 4 y 4) 9 12 x 2 3( y 2)2 3 x 2 3( y 2)2 3 3 3 3 2 x ( y 2)2 1 3
3, 2 3 2 , 3, 2 3 2
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Chapter 11: Analytic Geometry
The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the a2 b2 x-axis) where a 3, b 1, h 0, and k 2 . Solving for c: c2 a 2 b2 3 1 2 c 2 Thus, we have: Center: (0, 2)
2, 2 , 2, 2 Vertices: 3, 2 , 3, 2
Foci:
52. Complete the square to put the equation in standard form: 4 x2 3 y2 8x 6 y 5 4( x 2 2 x) 3( y 2 2 y ) 5 4( x 2 2 x 1) 3( y 2 2 y 1) 5 4 3 4( x 1) 2 3( y 1) 2 12 4( x 1) 2 3( y 1) 2 12 12 12 12 2 2 ( x 1) ( y 1) 1 3 4 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the b2 a2 y-axis) where a 2, b 3, h 1, and k 1 .
51. Complete the square to put the equation in standard form: 2 x2 3 y 2 8x 6 y 5 0 2( x 2 4 x) 3( y 2 2 y ) 5
Solving for c: c 2 a 2 b 2 4 3 1 c 1 Thus, we have: Center: (–1, 1) Foci: (–1, 0), (–1, 2) Vertices: (–1, –1), (–1, 3)
2( x 2 4 x 4) 3( y 2 2 y 1) 5 8 3 2( x 2) 2 3( y 1) 2 6 2( x 2) 2 3( y 1) 2 6 6 6 6 2 2 ( x 2) ( y 1) 1 3 2 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the a2 b2 x-axis) where a 3, b 2, h 2, and k 1 .
Solving for c: c 2 a 2 b 2 3 2 1 c 1 Thus, we have: Center: (2, –1) Foci: (1, –1), (3, –1) Vertices:
53. Complete the square to put the equation in standard form: 9 x 2 4 y 2 18 x 16 y 11 0
2 3, 1 , 2 3, 1
9( x 2 2 x) 4( y 2 4 y ) 11 9( x 2 2 x 1) 4( y 2 4 y 4) 11 9 16 9( x 1) 2 4( y 2) 2 36
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Section 11.3: The Ellipse
9( x 1) 2 4( y 2) 2 36 36 36 36 ( x 1) 2 ( y 2) 2 1 4 9 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the b2 a2 y-axis) where a 3, b 2, h 1, and k 2 . Solving for c: c2 a 2 b2 9 4 5 c 5 Thus, we have: Center: 1, 2
55. Complete the square to put the equation in standard form: 4x2 y2 4 y 0 4 x2 y 2 4 y 4 4
1, 2 5 , 1, 2 5
Foci:
Vertices:
4 x 2 ( y 2) 2 4
1,1 , 1, 5
4 x 2 ( y 2) 2 4 4 4 4 2 y ( 2) x2 1 4 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the b2 a2 y-axis) where a 2, b 1, h 0, and k 2 . Solving for c: c2 a 2 b2 4 1 3 c 3 Thus, we have: Center: (0,–2)
54. Complete the square to put the equation in standard form: x 2 9 y 2 6 x 18 y 9 0
Foci:
( x 2 6 x) 9( y 2 2 y ) 9 2
0, 2 3 , 0, 2 3
Vertices: (0, 0), (0, –4)
2
( x 6 x 9) 9( y 2 y 1) 9 9 9 ( x 3) 2 9( y 1) 2 9 ( x 3) 2 9( y 1) 2 9 9 9 9 ( x 3) 2 ( y 1) 2 1 9 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the a2 b2 x-axis) where a 3, b 1, h 3, and k 1 .
Solving for c: c 2 a 2 b 2 9 1 8 c 2 2 Thus, we have: Center: (–3, 1)
Foci: 3 2 2, 1 , 3 2 2, 1 Vertices: (0, 1), (–6, 1)
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Chapter 11: Analytic Geometry 56. Complete the square to put the equation in standard form: 9 x 2 y 2 18 x 0
58. Center: (–3, 1); Vertex: (–3, 3); Focus: (–3, 0); Major axis parallel to the y-axis; a 2; c 1 . Find b: b2 a 2 c 2 4 1 3 b 3
9( x 2 2 x 1) y 2 9 9( x 1) 2 y 2 9
Write the equation:
9( x 1) 2 y 2 9 9 9 9 y2 ( x 1) 2 1 9 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the b2 a2 y-axis) where a 3, b 1, h 1, and k 0 . Solving for c: c2 a 2 b2 9 1 8 c 2 2 Thus, we have: Center: (1, 0)
Foci: 1, 2 2 , 1, 2 2
59. Vertices: (4, 3), (4, 9); Focus: (4, 8); Center: (4, 6); Major axis parallel to the y-axis; a 3; c 2 . Find b:
b2 a 2 c2 9 4 5 b 5
Vertices: (1, 3), (1, –3)
Write the equation:
57. Center: (2, –2); Vertex: (7, –2); Focus: (4, –2); Major axis parallel to the x-axis; a 5; c 2 . Find b: b 2 a 2 c 2 25 4 21 b 21
Write the equation:
( x 3) 2 ( y 1) 2 1 3 4
( x 4) 2 ( y 6) 2 1 5 9
60. Foci: (1, 2), (–3, 2); Vertex: (–4, 2); Center: (–1, 2); Major axis parallel to the x-axis; a 3; c 2 . Find b: b2 a 2 c2 9 4 5 b 5
( x 2) 2 ( y 2) 2 1 25 21
Write the equation:
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( x 1) 2 ( y 2) 2 1 9 5
Section 11.3: The Ellipse 61. Foci: (5, 1), (–1, 1); Length of the major axis = 8; Center: (2, 1); Major axis parallel to the x-axis; a 4; c 3 . Find b: b 2 a 2 c 2 16 9 7 b 7
Write the equation:
( x 2) 2 ( y 1) 2 1 16 7
64. Center: (1, 2); Focus: (1, 4); contains the point (2, 2); Major axis parallel to the y-axis; c 2 . The equation has the form: ( x 1) 2 ( y 2) 2 1 b2 a2 Since the point (2, 2) is on the curve: 1 0 2 1 2 b a 1 1 b2 1 b 1 b2 Find a : a2 b2 c2 1 4 5 a 5
62. Vertices: (2, 5), (2, –1); c 2 ; Center: (2, 2); Major axis parallel to the y-axis; a 3; c 2 . Find b: b2 a 2 c2 9 4 5 b 5
Write the equation:
( x 2) 2 ( y 2) 2 1 5 9
Write the equation: ( x 1) 2
63. Center: (1, 2); Focus: (4, 2); contains the point (1, 3); Major axis parallel to the x-axis; c 3 . The equation has the form: ( x 1) 2 ( y 2) 2 1 a2 b2 Since the point (1, 3) is on the curve: 0 1 2 1 2 a b 1 1 b2 1 b 1 b2 Find a : a 2 b 2 c 2 1 9 10 a 10
Write the equation:
( y 2) 2 1 5
65. Center: (1, 2); Vertex: (4, 2); contains the point (1, 5); Major axis parallel to the x-axis; a 3 . The equation has the form: ( x 1) 2 ( y 2) 2 1 a2 b2 Since the point (1, 5) is on the curve: 0 32 1 9 b2 9 1 b2 9 b 3 2 b Solve for c: c 2 a 2 b 2 9 9 0 . Thus, c 0 .
( x 1) 2 ( y 2) 2 1 10
Write the equation: 1225
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( x 1) 2 ( y 2) 2 1 9 9
Chapter 11: Analytic Geometry 68. Rewrite the equation: y 9 9 x2 y 2 9 9 x2 , 2
2
2
2
9 x y 9, x y 1, 1 9
y0 y0 y0
66. Center: (1, 2); Vertex: (1, 4); contains the point (1 3,3) ; Major axis parallel to the y-axis; a2.
The equation has the form:
( x 1) 2 2
( y 2) 2
b a2 Since the point (1 3,3) is on the curve: 3 1 1 4 b2 1 1 b2 4 b 2 2 4 b
1
69. Rewrite the equation: y 64 16 x 2 y 2 64 16 x 2 ,
( x 1) 2 ( y 2) 2 Write the equation: 1 4 4 Solve for c: c 2 a 2 b 2 4 4 0 . Thus, c 0 .
2
2
2
2
16 x y 64, x y 1, 4 64
y0 y0 y0
67. Rewrite the equation: y 16 4 x 2 y 2 16 4 x 2 ,
70. Rewrite the equation:
y0
4 x 2 y 2 16,
y0
x2 y2 1, 4 16
y0
y 4 4x2 y 2 4 4 x2 , 2
2
2
2
4 x y 4, x y 1, 1 4
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y0 y0 y0
Section 11.3: The Ellipse 71. The center of the ellipse is (0, 0). The length of the major axis is 20, so a 10 . The length of half the minor axis is 6, so b 6 . The ellipse is situated with its major axis on the x-axis. The x2 y2 1. equation is: 100 36
74. Assume that the half ellipse formed by the gallery is centered at (0, 0). Since the distance between the foci is 100 feet and Jim is 6 feet from the nearest wall, the length of the gallery is 112 feet. 2a 112 or a 56 . The distance from the center to the foci is 50 feet, so c 50 . Find the height of the gallery which is b : b 2 a 2 c 2 3136 2500 636
72. The center of the ellipse is (0, 0). The length of the major axis is 30, so a 15 . The length of half the minor axis is 10, so b 10 . The ellipse is situated with its major axis on the x-axis. The x2 y2 1. equation is: 225 100 The roadway is 12 feet above the axis of the ellipse. At the center ( x 0 ), the roadway is 2 feet above the arch. At a point 5 feet either side of the center, evaluate the equation at x 5 :
b 636 25.2 The ceiling will be 25.2 feet high in the center.
75. Place the semi-elliptical arch so that the x-axis coincides with the water and the y-axis passes through the center of the arch. Since the bridge has a span of 120 feet, the length of the major axis is 120, or 2a 120 or a 60 . The maximum height of the bridge is 25 feet, so x2 y2 1. b 25 . The equation is: 3600 625 The height 10 feet from the center: y2 102 1 3600 625 y2 100 1 625 3600 3500 y 2 625 3600 y 24.65 feet The height 30 feet from the center: y2 302 1 3600 625 y2 900 1 625 3600 2700 y 2 625 3600 y 21.65 feet The height 50 feet from the center: y2 502 1 3600 625 y2 2500 1 625 3600 1100 2 y 625 3600 y 13.82 feet
52 y2 1 225 100 y2 25 200 1 100 225 225 200 9.43 225 The vertical distance from the roadway to the arch is 12 9.43 2.57 feet. At a point 10 feet either side of the center, evaluate the equation at x 10 : y 10
102 y 2 1 225 100 y2 100 125 1 100 225 225 125 7.45 225 The vertical distance from the roadway to the arch is 12 7.45 4.55 feet. At a point 15 feet either side of the center, the roadway is 12 feet above the arch. y 10
73. Assume that the half ellipse formed by the gallery is centered at (0, 0). Since the hall is 100 feet long, 2a 100 or a 50 . The distance from the center to the foci is 25 feet, so c 25 . Find the height of the gallery which is b : b 2 a 2 c 2 2500 625 1875 b 1875 43.3 The ceiling will be 43.3 feet high in the center.
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Chapter 11: Analytic Geometry 76. Place the semi-elliptical arch so that the x-axis coincides with the water and the y-axis passes through the center of the arch. Since the bridge has a span of 100 feet, the length of the major axis is 100, or 2a 100 or a 50 . Let h be the maximum height of the bridge. The equation is: x2 y2 2 1. 2500 h The height of the arch 40 feet from the center is 10 feet. So (40, 10) is a point on the ellipse. Substitute and solve for h :
282
132 1 400 a 784 169 231 1 2 400 400 a 231a 2 313600 2
a 2 1357.576 a 36.845 The span of the bridge is 73.69 feet.
79. Because of the pitch of the roof, the major axis will run parallel to the direction of the pitch and the minor axis will run perpendicular to the direction of the pitch. The length of the major axis can be determined from the pitch by using the Pythagorean Theorem. The length of the minor axis is 8 inches (the diameter of the pipe).
402 102 1 2500 h 2 102 2
1
1600 9 2500 25
h 9h 2 2500
50 16.67 3 The height of the arch at its center is 16.67 feet. h
77. If the x-axis is placed along the 100 foot length and the y-axis is placed along the 50 foot length, x2 y2 1. the equation for the ellipse is: 502 252 Find y when x = 40: 402 y 2 1 502 252 1600 y2 1 625 2500 9 2 y 625 25 y 15 feet To get the width of the ellipse at x 40 , we need to double the y value. Thus, the width 10 feet from a vertex is 30 feet.
2(5) = 10
2(4) = 8
The length of the major axis is
8 10 164 2 41 inches. 2
2
80. The length of the football gives the length of the major axis so we have 2a 11.125 or a 5.5625 . At its center, the prolate spheroid is a circle of radius b. This means 2 b 28.25 28.25 b 2 2
78. Place the semi-elliptical arch so that the x-axis coincides with the major axis and the y-axis passes through the center of the arch. Since the height of the arch at the center is 20 feet, b 20 . The length of the major axis is to be found, so it is necessary to solve for a . The equation is:
4 4 28.25 ab 2 5.5625 471 . 3 3 2 The football contains approximately 471 cubic inches of air.
Thus,
81. Since the mean distance is 93 million miles, a 93 million. The length of the major axis is 186 million. The perihelion is 186 million – 94.5 million = 91.5 million miles. The distance from the center of the ellipse to the sun (focus) is 93 million – 91.5 million = 1.5 million miles.
x2
y2 1. a 2 400 The height of the arch 28 feet from the center is to be 13 feet, so the point (28, 13) is on the ellipse. Substitute and solve for a :
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Section 11.3: The Ellipse
Therefore, c 1.5 million. Find b: b2 a 2 c2
93 10
b2 a 2 c 2
483.8 106
1.5 10
6 2
6 2
8.64675 10 8646.75 10
x2
b 92.99 106 The equation of the orbit is: x2 y2 1 2 2 93 106 92.99 106
We can simplify the equation by letting our units for x and y be millions of miles. The equation then becomes: x2 y2 1 8649 8646.75
483.2 106
5448.5 106
2
1
897.5 10 2
6 2
2.8880646 1019
13.5 10
6 2
b 5374.07 106 The equation of the orbit of Pluto is: x2 y2 1 2 2 5448.5 106 5374.07 106
b 141.36 106 The equation of the orbit is: x2 y2 1 2 2 142 106 141.36 106
y2
84. The mean distance is 4551 million + 897.5 million = 5448.5 million miles. The aphelion is 5448.5 million + 897.5 million = 6346 million miles. Since a 5448.5 106 and c 897.5 106 , we can find b: b2 a 2 c 2
1.998175 1016
x2 y2 1 234, 062.44 233, 524.2
82. Since the mean distance is 142 million miles, a 142 million. The length of the major axis is 284 million. The aphelion is 284 million – 128.5 million = 155.5 million miles. The distance from the center of the ellipse to the sun (focus) is 142 million – 128.5 million = 13.5 million miles. Therefore, c 13.5 million. Find b: b2 a 2 c2 142 10
483.8 10
6 2
We can simplify the equation by letting our units for x and y be millions of miles. The equation then becomes:
6 2
6 2
b 483.2 106 The equation of the orbit of Jupiter is:
12
2
2.335242 1017
15
23.2 10
We can simplify the equation by letting our units for x and y be millions of miles. The equation then becomes: x2 y2 1 29, 686,152.25 28,880, 646
We can simplify the equation by letting our units for x and y be millions of miles. The equation then becomes:
c 0.75 . Perihelion: a – c = 5 a c 0.75a a 0.75a 5 0.25a 5 a 20 ac 5 20 c 5 c 15 So the aphelion is a + c = 35 million mi.
x2 y2 1 20,164 19, 981.75
85. e
83. The mean distance is 507 million – 23.2 million = 483.8 million miles. The perihelion is 483.8 million – 23.2 million = 460.6 million miles.
Since a 483.8 106 and c 23.2 106 , we can find b:
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Chapter 11: Analytic Geometry
86.
d ( M ,V ) d ( M , B)
x2 y 2 1 49 4
( 2 5 ( 5)) (0 y )
Area of rectangle: (2x)(2y) = 4xy
( 2 5 ( 5)) (0 y ) (0 ( 5)) (2 y )
y 2 1
2
2
2
2
2
(0 ( 5)) (2 y ) 2
x2 49
20 20 5 25 y 2 25 4 4 y y 2
x2 49
5 54 y
A 8x 1
2
16 20 5 4 y
90. The center of the ellipse is the midpoint of the vertices. So the center is (4,0). So h = 4, k = 0 and a = 2. One focus is at 4 c 4 3 , so c 3 . Then b 2 a 2 c 2 4 3 1 b 1.
On the calculator, set Y1 equal to this equation and find the max. The max y is 28 so the max are is 28 m2. 87. Let a 324 18 and b 100 10. . Then c2 a 2 b2 324 100 224
The equation of the ellipse is
x 4 2 4
y 2 1.
The points of intersection satisfy
x 4 2
c 224 4 14 The vertices on the major axis are (18, 0) and (18, 0) and the vertices on the minor axis are (10, 0) and (10, 0) . The foci are
4
y2 x 2 y2 2
x 4 4 x 2 2
2
x 2 8 x 16 4 x 2 16 x 16 3x 2 8 x 0
(4 14, 0) and (4 14, 0) . We are looking for the distance between the foci which is: c 2 a 2 b2
x(3 x 8) 0 x 0 or x 83
324 100
Substitute each x into either equation to solve for y.
224 2c 2 4 14 8 14 29.93 cm
x 0; (0 2) 2 y 2 1 y 2 3 (no real solution)
88. Given that the length of the major axis is 20 then the coordinates of the vertices are (10, 0) and (10, 0) and a 10 . The length of the minor axis is 9 so the vertices are 9 9 9 0, and 0, and b . So the 2 2 2
equation of the ellipse is
2
2
8 8 x ; 2 y2 1 3 3 5 y2 9 5 y 3
x2 y2 1. 100 20.25
The points of intersection are 8 8 5 5 , and , . 3 3 3 3
89. The ellipse x 2 5 y 2 20 can be written as x2 y 2 1 , so the vertices are at 20 4 ( 2 5, 0) and (2 5, 0) . The endpoints of the minor axis are at (0, 2) and (0, 2) . V (2 5, 0) and B (0, 2) and M ( 5, y )
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Section 11.3: The Ellipse 93. Answers will vary.
Ax 2 Cy 2 F 0
91. a.
Ax 2 Cy 2 F 2
94.
2
Ax Cy 1 F F x2 y2 1 ( F / A) ( F / C ) where A 0, C 0, F 0 , and F / A and F / C are positive. F F If A C , then . So, this is the A C equation of an ellipse with center at (0, 0).
0 ( x 5)2 12 12 ( x 5) 2 12 x 5 x 5 12 5 2 3
The zeros are x 5 2 3 and 5 2 3 . The x-intercepts are 5 2 3 and 5 2 3 . 95.
b. If A C , the equation becomes: F Ax 2 Ay 2 F x 2 y 2 A This is the equation of a circle with center at (0, 0) F and radius of . A
2 2 is a horizontal 1 asymptote. The denominator is zero at x 5 , so x 5 is a vertical asymptote. The domain is x | x 5 .
Ax 2 Cy 2 Dx Ey F
A x2
D E x C y2 y F A C 2
96. F 80 cos 50º i sin 50º j
80 0.6428i 0.7660 j 51.423i 61.284 j W F AB 51.423i 61.284 j 12i
2
D E D2 E 2 F A x C y 2A 2C 4 A 4C
51.426(12) 61.284 0 617.1 ft-lb
where A C 0 . D2 E 2 F . Let U 4 A 4C
a.
97. a 14, A 52º b cot 52º 14 b 14 cot 52º 14 0.7813 10.94
If U is of the same sign as A (and C ) , then 2
2
x D y E 2 A 2C 1 U U A C
c 14 c 14 csc 52º 14 1.2690 17.77
csc 52º
This is the equation of an ellipse with center at D E , . 2 A 2C
B 90 A 90 52 38
b. If U 0 , the graph is the single point
98. 2 3 tan 5 x 7 9
D E , . 2 A 2C
c.
2x 3 ; The degree of the numerator, x5 p( x) 2 x 3, is n 1 . The degree of the denominator, q( x) x 5, is m 1 . Since f ( x)
n m , the line y
Ax 2 Cy 2 Dx Ey F 0, A 0, C 0
92.
f ( x) ( x 5)2 12
2 3 tan 5 x 2
If U is of the opposite sign as A (and C ) , this graph contains no points since the left side always has the opposite sign of the right side.
tan 5 x 5x x
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1 3
6
30
5
k
Chapter 11: Analytic Geometry
On the interval 0 7 13 , , . 30 30 30
2
Section 11.4 , the solution set is
52 5 2
3x 2 14 x 8 (3x 2)( x 4) (3x 2) ( x 4)( x 3) ( x 3) x 2 x 12 Now we can evaluate. (3(4) 2) 10 10 R (4) ((4) 3) 7 7
3. x-intercepts: 02 9 4 x 2 4 x 2 9 9 x 2 (no real solution) 4
2 x 1 ln 8 2 x ln 8 1
y-intercepts: y 2 9 4 0
ln 8 1 0.5397 2
2
y2 9
0, 3 , 0,3 The intercepts are 0, 3 and 0,3 . y 3
f ( x h) f ( x) 2( x h) 2 7( x h) (2 x 2 7 x) 101. h h 2 2 2 2 x 4 xh 2h 7 x 7h 2 x 7 x h 2 4 xh 2h 7h h h(4 x 2h 7) h 4 x 2h 7
4. True; the graph of y 2 9 x 2 is a hyperbola with its center at the origin. 5. right 5; down 4 x2 9 ; p x x 2 9, q x x 2 4 x2 4 The vertical asymptotes are the zeros of q .
6. y
As h approaches 0 the expressions becomes 4 x 2(0) 7 4 x 7
q x 0 x2 4 0
x 102. log 3 1 4 2 x 34 1 2 x 81 1 2 x 82 2 x 164
x2 4 x 2, x 2 The lines x 2, and x 2 are the vertical asymptotes. The degree of the numerator, p( x) x 2 9 , is n 2 . The degree of the
denominator, q ( x) x 2 4 , is m 2 . 1 Since n m , the line y 1 is a horizontal 1 asymptote. Since this is a rational function and there is a horizontal asymptote, there are no oblique asymptotes.
103. ( x 3) 2 20 x 3 20 x 3 20
7. hyperbola
3 2 5
25 25 50 5 2
25 5 2. 2 4
e 2 x 1 8
x
2
2
99. R( x)
100.
2 32 1 4
1. d
The solution set is 3 2 5, 3 2 5
8. transverse axis
9. b 1232
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Section 11.4: The Hyperbola
10.
2, 4 ; 2, 2
11.
2, 6 ; 2, 4
12. a.
20. D; the hyperbola opens up and down, and has vertices at 0, 1 . Thus, the graph has an
equation of the form y 2
1, 1
21. Center: (0, 0); Focus: (3, 0); Vertex: (1, 0); Transverse axis is the x-axis; a 1; c 3 . Find the value of b: b2 c 2 a 2 9 1 8
4 As a changes, the distance between the vertices changes. The value of a represents the distance from the center to the vertices. d. 4 x2 y 2 e. 1 9 7 f. 5 y 2 x2 g. 1 9 16
b. c.
13. a. b. c. d.
x2 1. b2
b 82 2
Write the equation: x 2
y2 1. 8
x2 c 1 ( y k ) 2 ( x h) 2 1 a2 b2 ( x h) 2 ( y k ) 2 1 a2 b2
14. c
22. Center: (0, 0); Focus: (0, 5); Vertex: (0, 3); Transverse axis is the y-axis; a 3; c 5 . Find the value of b: b 2 c 2 a 2 25 9 16 b4 y 2 x2 1. Write the equation: 9 16
15. 2; 3; x 4 4 16. y x ; y x 9 9
17. B; the hyperbola opens to the left and right, and has vertices at 1, 0 . Thus, the graph has an
equation of the form x 2
y2 1. b2
18. C; the hyperbola opens up and down, and has vertices at 0, 2 . Thus, the graph has an
equation of the form
y 2 x2 1. 4 b2
23. Center: (0, 0); Focus: (0, –6); Vertex: (0, 4) Transverse axis is the y-axis; a 4; c 6 . Find the value of b: b 2 c 2 a 2 36 16 20
19. A; the hyperbola opens to the left and right, and has vertices at 2, 0 . Thus, the graph has an
equation of the form
x2 y 2 1. 4 b2
b 20 2 5
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Chapter 11: Analytic Geometry
Write the equation:
26. Focus: (0, 6); Vertices: (0, –2), (0, 2) Center: (0, 0); Transverse axis is the y-axis; a 2; c 6 . Find the value of b: b 2 c 2 a 2 36 4 32 b 4 2
y 2 x2 1. 16 20
Write the equation:
y 2 x2 1. 4 32
24. Center: (0, 0); Focus: (–3, 0); Vertex: (2, 0) Transverse axis is the x-axis; a 2; c 3 . Find the value of b: b2 c 2 a 2 9 4 5 b 5
Write the equation:
27. Vertices: (0, –6), (0, 6); asymptote: y 2 x ; Center: (0, 0); Transverse axis is the y-axis; a 6 . Find the value of b using the slope of the a 6 asymptote: 2 2b 6 b 3 b b Find the value of c: c 2 a 2 b 2 36 9 45
x2 y 2 1. 4 5
c3 5
Write the equation:
25. Foci: (–5, 0), (5, 0); Vertex: (3, 0) Center: (0, 0); Transverse axis is the x-axis; a 3; c 5 . Find the value of b: b 2 c 2 a 2 25 9 16 b 4
Write the equation:
x2 y 2 1. 9 16
1234
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y 2 x2 1. 36 9
Section 11.4: The Hyperbola 28. Vertices: (–4, 0), (4, 0); asymptote: y 2 x ; Center: (0, 0); Transverse axis is the x-axis; a 4 . Find the value of b using the slope of the b b asymptote: 2 b 8 a 4 Find the value of c: c 2 a 2 b 2 16 64 80
30. Foci: (0, –2), (0, 2); asymptote: y x ; Center: (0, 0); Transverse axis is the y-axis; c 2 . Using the slope of the asymptote: a 1 b a b a b Find the value of b: b2 c2 a2
Write the equation:
c 2
a2 b2 c2
c4 5 2
2
y x 1. 16 64
b 2 b 2 4 2b 2 4 b2 2 b 2 a 2
(a b)
Write the equation:
29. Foci: (–4, 0), (4, 0); asymptote: y x ; Center: (0, 0); Transverse axis is the x-axis; c 4 . Using the slope of the asymptote: b 1 b a b a . a Find the value of b: b2 c2 a2 a2 b2 c2 c 4
31.
b 2 b 2 16 2b 2 16 b 2 8
Write the equation:
x2 y 2 1 25 9 The center of the hyperbola is at (0, 0). a 5, b 3 . The vertices are 5, 0 and
5, 0 . Find the value of c:
b 82 2 a 82 2
y 2 x2 1. 2 2
c 2 a 2 b 2 25 9 34 c 34
( a b)
The foci are
x2 y 2 1. 8 8
34, 0 and 34, 0 .
The transverse axis is the x-axis. The asymptotes 3 3 are y x; y x . 5 5
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Chapter 11: Analytic Geometry
32.
y 2 x2 1 16 4 The center of the hyperbola is at (0, 0). a 4, b 2 . The vertices are (0, 4) and (0, –4). Find the value of c: c 2 a 2 b 2 16 4 20 c 20 2 5
34. 4 y 2 x 2 16 Divide both sides by 16 to put in standard form: 4 y 2 x 2 16 y 2 x2 1 16 16 16 4 16 The center of the hyperbola is at (0, 0). a 2, b 4 . The vertices are 0, 2 and
0, 2 . Find the value of c:
The foci are 0, 2 5 and 0, 2 5 .
c 2 a 2 b 2 4 16 20 c 20 2 5
The transverse axis is the y-axis. The asymptotes are y 2 x; y 2 x .
The foci are 0, 2 5 and 0, 2 5 .
The transverse axis is the y-axis. The asymptotes 1 1 are y x and y x . 2 2
33. 4 x 2 y 2 16 Divide both sides by 16 to put in standard form: 4 x 2 y 2 16 x2 y 2 1 16 16 16 4 16 The center of the hyperbola is at (0, 0). a 2, b 4 . The vertices are (2, 0) and (–2, 0). Find the value of c: c 2 a 2 b 2 4 16 20 c 20 2 5
35. y 2 9 x 2 9 Divide both sides by 9 to put in standard form: y 2 9 x2 9 y2 x2 1 9 9 9 9 The center of the hyperbola is at (0, 0). a 3, b 1 . The vertices are (0, 3) and (0, –3). Find the value of c: c 2 a 2 b 2 9 1 10
The foci are 2 5, 0 and 2 5, 0 .
c 10
The transverse axis is the x-axis. The asymptotes are y 2 x; y 2 x .
The foci are 0, 10 and 0, 10 .
The transverse axis is the y-axis. The asymptotes are y 3x; y 3x .
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Section 11.4: The Hyperbola
36. x 2 y 2 4 Divide both sides by 4 to put in standard form: x2 y 2 4 x2 y 2 1. 4 4 4 4 4 The center of the hyperbola is at (0, 0). a 2, b 2 . The vertices are 2, 0 and
38. 2 x 2 y 2 4 Divide both sides by 4 to put in standard form: x2 y 2 1. 2 4 The center of the hyperbola is at (0, 0). a 2, b 2 .
2, 0 . Find the value of c: 2
2
The vertices are
2
c a b 44 8 c 8 2 2
Find the value of c: c 2 a 2 b2 2 4 6
The foci are 2 2, 0 and 2 2, 0 .
The foci are
The transverse axis is the x-axis. The asymptotes are y x; y x .
Find the value of c: c 2 a 2 b2 1 1 2 c 2
The foci are
c 2 a 2 b 2 25 25 50
6, 0 and 6, 0 .
39. The center of the hyperbola is at (0, 0). a 1, b 1 . The vertices are 1, 0 and 1, 0 .
0, 5 . Find the value of c:
c 6
The transverse axis is the x-axis. The asymptotes are y 2 x; y 2 x .
37. y 2 x 2 25 Divide both sides by 25 to put in standard form: y 2 x2 1. 25 25 The center of the hyperbola is at (0, 0). a 5, b 5 . The vertices are 0, 5 and
c 50 5 2
2, 0 and 2, 0 .
2, 0 and 2, 0 .
The transverse axis is the x-axis. The asymptotes are y x; y x .
The foci are 0,5 2 and 0, 5 2 .
The equation is: x 2 y 2 1 .
The transverse axis is the y-axis. The asymptotes are y x; y x .
40. The center of the hyperbola is at (0, 0). a 1, b 1 . The vertices are 0, 1 and 0, 1 .
Find the value of c: c 2 a 2 b2 1 1 2 c 2
The foci are 0, 2 and 0, 2 . The transverse axis is the y-axis. The asymptotes are y x; y x . The equation is: y 2 x 2 1 .
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Chapter 11: Analytic Geometry 41. The center of the hyperbola is at (0, 0). a 6, b 3 .
44. Center: (–3, 1); Focus: (–3, 6); Vertex: (–3, 4); Transverse axis is parallel to the y-axis; a 3; c 5 . Find the value of b:
The vertices are 0, 6 and 0, 6 . Find the
b 2 c 2 a 2 25 9 16 b 4 ( y 1) 2 ( x 3) 2 1. Write the equation: 9 16
value of c: c 2 a 2 b 2 36 9 45 c 45 3 5
The foci are 0, 3 5 and 0,3 5 . The transverse axis is the y-axis. The asymptotes are y 2 x; y 2 x . The equation is:
y 2 x2 1. 36 9
42. The center of the hyperbola is at (0, 0). a 2, b 4 .
The vertices are 2, 0 and 2, 0 . Find the value of c: c 2 a 2 b 2 4 16 20 c 20 2 5
45. Center: (–3, –4); Focus: (–3, –8); Vertex: (–3, –2); Transverse axis is parallel to the y-axis; a 2; c 4 . Find the value of b: b 2 c 2 a 2 16 4 12
The foci are 2 5, 0 and 2 5, 0 . The transverse axis is the x-axis. The asymptotes are y 2 x; y 2 x . The equation is:
b 12 2 3
x2 y 2 1. 4 16
Write the equation:
43. Center: (4, –1); Focus: (7, –1); Vertex: (6, –1); Transverse axis is parallel to the x-axis; a 2; c 3 . Find the value of b: b2 c 2 a 2 9 4 5 b 5 ( x 4) 2 ( y 1) 2 Write the equation: 1. 4 5
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( y 4) 2 ( x 3) 2 1. 4 12
Section 11.4: The Hyperbola 46. Center: (1, 4); Focus: (–2, 4); Vertex: (0, 4); Transverse axis is parallel to the x-axis; a 1; c 3 . Find the value of b:
48. Focus: (–4, 0); Vertices: (–4, 4), (–4, 2); Center: (–4, 3); Transverse axis is parallel to the y-axis; a 1; c 3 . ind the value of b: b2 c 2 a 2 9 1 8
b2 c 2 a 2 9 1 8 b 82 2
Write the equation: ( x 1) 2
b 82 2
( y 4) 2 1. 8
Write the equation: ( y 3) 2
47. Foci: (3, 7), (7, 7); Vertex: (6, 7); Center: (5, 7); Transverse axis is parallel to the x-axis; a 1; c 2 . Find the value of b: b2 c 2 a 2 4 1 3
49. Vertices: (–1, –1), (3, –1); Center: (1, –1); Transverse axis is parallel to the x-axis; a 2 . 3 Asymptote: y 1 x 1 2 Using the slope of the asymptote, find the value of b: b b 3 b3 a 2 2 Find the value of c: c 2 a 2 b 2 4 9 13
b 3
Write the equation: ( x 5) 2
( x 4) 2 1. 8
( y 7) 2 1. 3
c 13
Write the equation:
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( x 1) 2 ( y 1) 2 1. 4 9
Chapter 11: Analytic Geometry 50. Vertices: (1, –3), (1, 1); Center: (1, –1); Transverse axis is parallel to the y-axis; a 2 . 3 Asymptote: y 1 x 1 2 Using the slope of the asymptote, find the value of b: a 2 3 4 3b 4 b b b 2 3 Find the value of c: 16 52 52 2 13 c 2 a 2 b2 4 c 9 9 9 3 ( y 1) 2 9( x 1) 2 Write the equation: 1. 4 16
51.
52.
( y 3) 2 ( x 2) 2 1 4 9 The center of the hyperbola is at (2, –3). a 2, b 3 . The vertices are (2, –1) and (2, –5). Find the value of c: c 2 a 2 b 2 4 9 13 c 13
Foci: 2, 3 13 and 2, 3 13
Transverse axis: x 2 , parallel to the y-axis 2 Asymptotes: y 3 ( x 2); 3 2 y 3 ( x 2) 3
53. ( y 2) 2 4( x 2) 2 4 Divide both sides by 4 to put in standard form: ( y 2) 2 ( x 2) 2 1 . 4 The center of the hyperbola is at (–2, 2). a 2, b 1 . The vertices are (–2, 4) and (–2, 0). Find the value of c: c2 a2 b2 4 1 5 c 5
( x 2) 2 ( y 3) 2 1 4 9 The center of the hyperbola is at (2, –3). a 2, b 3 . The vertices are (0, –3) and (4, –3). Find the value of c: c 2 a 2 b 2 4 9 13 c 13
Foci: 2 13, 3 and 2 13, 3 . Transverse axis: y 3 , parallel to x-axis.
Foci:
2, 2 5 and 2, 2 5 .
Transverse axis: x 2 , parallel to the y-axis. Asymptotes: y 2 2( x 2); y 2 2( x 2) .
3 ( x 2); 2 3 y 3 ( x 2) 2
Asymptotes: y 3
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Section 11.4: The Hyperbola
54. ( x 4) 2 9( y 3) 2 9 Divide both sides by 9 to put in standard form: ( x 4) 2 ( y 3) 2 1 . 9 The center of the hyperbola is (–4, 3). a 3, b 1 . The vertices are (–7, 3) and (–1, 3). Find the value of c: c 2 a 2 b 2 9 1 10 c 10
Foci: 4 10, 3 and 4 10, 3
56. ( y 3) 2 ( x 2) 2 4 Divide both sides by 4 to put in standard form: ( y 3) 2 ( x 2) 2 1 . The center of the 4 4 hyperbola is at (–2, 3). a 2, b 2 . The vertices are (–2, 5) and (–2, 1). Find the value of c: c 2 a 2 b2 4 4 8 c 8 2 2
1 1 Asymptotes: y 3 ( x 4), y 3 ( x 4) 3 3
55. ( x 1)2 ( y 2) 2 4 Divide both sides by 4 to put in standard form: ( x 1) 2 ( y 2) 2 1. 4 4 The center of the hyperbola is (–1, –2). a 2, b 2 . The vertices are (–3, –2) and (1, –2). Find the value of c: c 2 a 2 b2 4 4 8 c 8 2 2
Transverse axis: x 2 , parallel to the y-axis. Asymptotes: y 3 x 2; y 3 ( x 2)
Transverse axis: y 3 , parallel to the x-axis.
Foci: 1 2 2, 2 and 1 2 2, 2
Foci: 2, 3 2 2 and 2, 3 2 2
57. Complete the squares to put in standard form: x2 y 2 2 x 2 y 1 0 ( x 2 2 x 1) ( y 2 2 y 1) 1 1 1 ( x 1) 2 ( y 1) 2 1 The center of the hyperbola is (1, –1). a 1, b 1 . The vertices are (0, –1) and (2, –1). Find the value of c: c 2 a 2 b2 1 1 2 c 2
Foci: 1 2, 1 and 1 2, 1 .
Transverse axis: y 2 , parallel to the x-axis. Asymptotes: y 2 x 1; y 2 ( x 1)
Transverse axis: y 1 , parallel to x-axis. Asymptotes: y 1 x 1; y 1 ( x 1) .
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Chapter 11: Analytic Geometry 58. Complete the squares to put in standard form: y2 x2 4 y 4 x 1 0
60. Complete the squares to put in standard form: 2 x2 y2 4x 4 y 4 0
( y 2 4 y 4) ( x 2 4 x 4) 1 4 4
2( x 2 2 x 1) ( y 2 4 y 4) 4 2 4
( y 2) 2 ( x 2) 2 1 The center of the hyperbola is (2, 2). a 1, b 1 . The vertices are 2,1 and 2,3 .
2( x 1) 2 ( y 2) 2 2 ( y 2) 2 1 2 The center of the hyperbola is (–1, 2). a 1, b 2 . The vertices are (–2, 2) and (0, 2). Find the value of c: c 2 a 2 b2 1 2 3 c 3 ( x 1)2
Find the value of c: c2 a2 b2 1 1 2 c 2
Foci: 2, 2 2 and 2, 2 2 .
Transverse axis: x 2 , parallel to the y-axis. Asymptotes: y 2 x 2; y 2 ( x 2) .
1 3, 2 and 1 3, 2 .
Foci:
Transverse axis: y 2 , parallel to the x-axis. Asymptotes: y 2 2( x 1); y 2 2( x 1) .
59. Complete the squares to put in standard form: y2 4x2 4 y 8x 4 0 ( y 2 4 y 4) 4( x 2 2 x 1) 4 4 4 ( y 2) 2 4( x 1) 2 4 ( y 2) 2 ( x 1) 2 1 4 The center of the hyperbola is (–1, 2). a 2, b 1 . The vertices are (–1, 4) and (–1, 0). Find the value of c: c2 a2 b2 4 1 5 c 5
61. Complete the squares to put in standard form: 4 x 2 y 2 24 x 4 y 16 0 4( x 2 6 x 9) ( y 2 4 y 4) 16 36 4 4( x 3) 2 ( y 2) 2 16 ( x 3) 2 ( y 2) 2 1 4 16 The center of the hyperbola is (3, –2). a 2, b 4 . The vertices are (1, –2) and (5, –2). Find the value of c: c 2 a 2 b 2 4 16 20
Foci: 1, 2 5 and 1, 2 5 .
Transverse axis: x 1 , parallel to the y-axis. Asymptotes: y 2 2( x 1); y 2 2( x 1) .
c 20 2 5
Foci: 3 2 5, 2 and 3 2 5, 2 . Transverse axis: y 2 , parallel to x-axis. Asymptotes: y 2 2( x 3); y 2 2( x 3) 1242
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Section 11.4: The Hyperbola 63. Complete the squares to put in standard form: y 2 4 x 2 16 x 2 y 19 0 ( y 2 2 y 1) 4( x 2 4 x 4) 19 1 16 ( y 1) 2 4( x 2) 2 4 ( y 1) 2 ( x 2) 2 1 4 The center of the hyperbola is (–2, 1). a 2, b 1 . The vertices are (–2, 3) and (–2, –1). Find the value of c: c2 a 2 b2 4 1 5
62. Complete the squares to put in standard form: 2 y 2 x2 2 x 8 y 3 0
c 5
2( y 2 4 y 4) ( x 2 2 x 1) 3 8 1
Foci:
2( y 2) 2 ( x 1) 2 4 ( y 2) 2 ( x 1) 2 1 2 4 The center of the hyperbola is (1,–2). a 2, b 2 .
Vertices: 1, 2 2 and 1, 2 2
2, 1 5 and 2, 1 5 .
Transverse axis: x 2 , parallel to the y-axis. Asymptotes: y 1 2( x 2); y 1 2( x 2) .
Find the value of c: c2 a2 b2 2 4 6 c 6
Foci: 1, 2 6 and 1, 2 6 . Transverse axis: x 1 , parallel to the y-axis. 2 ( x 1); 2 2 y2 ( x 1) 2
Asymptotes: y 2
64. Complete the squares to put in standard form: x2 3 y 2 8x 6 y 4 0 ( x 2 8 x 16) 3( y 2 2 y 1) 4 16 3 ( x 4) 2 3( y 1) 2 9 ( x 4) 2 ( y 1) 2 1 9 3 The center of the hyperbola is (–4, –1). a 3, b 3 . The vertices are (–7, –1) and (–1, –1). Find the value of c: c 2 a 2 b 2 9 3 12 c 12 2 3
Foci: 4 2 3, 1 and 4 2 3, 1 . Transverse axis: y 1 , parallel to x-axis. Asymptotes: 3 3 y 1 ( x 4); y 1 ( x 4) 3 3 1243
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Chapter 11: Analytic Geometry 67. Rewrite the equation: y 25 x 2 y 2 25 x 2 , 2
2
2
2
x y 25, y x 1, 25 25
y0 y0 y0
65. Rewrite the equation: y 16 4 x 2 y 2 16 4 x 2 , y 2 4 x 2 16, 2
2
y x 1, 16 4
y0 y0 y0
68. Rewrite the equation: y 1 x 2 y 2 1 x 2 , 2
2
x y 1,
y0 y0
66. Rewrite the equation: y 9 9 x2 y 2 9 9x2 , y 2 9 x 2 9, 2
2
y x 1, 9 1
y0 y0 y0
69.
( x 3) 2 y 2 1 4 25 The graph will be a hyperbola. The center of the hyperbola is at (3, 0). a 2, b 5 . The vertices are (5, 0) and (1, 0). Find the value of c: c 2 a 2 b 2 4 25 29 c 29
Foci: 3 29, 0 and 3 29, 0
Transverse axis is the x-axis. 5 5 Asymptotes: y ( x 3); y ( x 3) 2 2
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Section 11.4: The Hyperbola
72. y 2 12( x 1) The graph will be a parabola. The equation is in the form ( y k ) 2 4a( x h) where – 4a 12 or a 3 , h 1, and k 0 . Thus, we have: Vertex: (1, 0) ; Focus: (4, 0) ; Directrix: x 2
70.
( y 2) 2 ( x 2) 2 1 16 4 The graph will be a hyperbola. The center of the hyperbola is at (2, –2). a 4, b 2 . The vertices are (2, 2) and (2, –6). Find the value of c: c 2 a 2 b 2 16 4 20 c 20 2 5
Foci: 2, 2 2 5 and 2, 2 2 5
73. The graph will be an ellipse. Complete the square to put the equation in standard form: 25 x 2 9 y 2 250 x 400 0
Transverse axis: x 2 , parallel to the y-axis Asymptotes: y 2 2( x 2); y 2 2( x 2)
(25 x 2 250 x) 9 y 2 400 25( x 2 10 x) 9 y 2 400 25( x 2 10 x 25) 9 y 2 400 625 25( x 5) 2 9 y 2 225 25( x 5) 2 9 y 2 225 225 225 225 ( x 5) 2 y 2 1 9 25 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the b2 a2 y-axis) where a 5, b 3, h 5, and k 0 .
71. x 2 16( y 3) The graph will be a parabola. The equation is in the form ( x h) 2 4a ( y k ) where 4a 16 or a 4 , h 0, and k 3 . Thus, we have: Vertex: (0, 3); Focus: (0, 7) ; Directrix: y 1
Solving for c: c 2 a 2 b 2 25 9 16 c 4 Thus, we have: Center: (5, 0) Foci: (5, 4), (5, –4) Vertices: (5, 5), (5, –5)
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Chapter 11: Analytic Geometry 74. The graph will be an ellipse. Complete the square to put the equation in standard form: x 2 36 y 2 2 x 288 y 541 0 ( x 2 2 x) (36 y 2 288 y ) 5 41 ( x 2 2 x) 36( y 2 8 y ) 5 41 ( x 2 2 x 1) 36( y 2 8 y 16) 541 1 576 ( x 1) 2 36( y 4) 2 36 ( x 1) 2 36( y 4) 2 36 36 36 36 ( x 1) 2 ( y 4) 2 1 36 The equation is in the form ( x h) 2 ( y k ) 2 1 (major axis parallel to the a2 b2 x-axis) where a 6, b 1, h 1, and k 4 . Solving for c: c 2 a 2 b 2 36 1 35 c 35 Thus, we have: Center: (1, 4)
Foci:
76. The graph will be a hyperbola. Complete the squares to put in standard form: 9 x 2 y 2 18 x 8 y 88 0 9( x 2 2 x 1) ( y 2 8 y 16) 88 9 16 9( x 1) 2 ( y 4) 2 81 ( x 1) 2 ( y 4) 2 1 9 81 The center of the hyperbola is (1, 4) . a 9, b 81 . The vertices are (2, 4) and (4, 4) . Find the value of c: c 2 a 2 b 2 9 81 90
1 35, 4 , 1 35, 4
Vertices: (7, 4) , (5, 4)
c 90 3 10
Foci: 1 3 10, 4 and 1 3 10, 4 . Transverse axis: x 2 , parallel to the y-axis. Asymptotes: y 4 3( x 1); y 4 3( x 1) .
75. The graph will be a parabola. Complete the square to put the equation in standard form: x 2 6 x 8 y 31 0 x 2 6 x 8 y 31 x 2 6 x 9 8 y 31 9 ( x 3) 2 8 y 40 ( x 3) 2 8( y 5) The equation is in the form ( x h) 2 4a ( y k ) where 4a 8 or a 2 , h 3, and k 5 . Thus, we have: Vertex: (3, 5) ; Focus: (3, 3) ; Directrix: y 7
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Section 11.4: The Hyperbola 77. First note that all points where a burst could take place, such that the time difference would be the same as that for the first burst, would form a hyperbola with A and B as the foci. Start with a diagram: (x, y)
B
78. First note that all points where the strike could take place, such that the time difference would be the same as that for the first strike, would form a hyperbola with A and B as the foci. Start with a diagram: (x, y)
B
A 2 miles
A 1 mile
Assume a coordinate system with the x-axis containing BA and the origin at the midpoint of BA . The ordered pair x, y represents the location of
Assume a coordinate system with the x-axis containing BA and the origin at the midpoint of BA . The ordered pair x, y represents the location
the fireworks. We know that sound travels at 1100 feet per second, so the person at point A is 1100 feet closer to the fireworks display than the person at point B. Since the difference of the distance from x, y to A and from x, y to B is the
of the lightning strike. We know that sound travels at 1100 feet per second, so the person at point A is 2200 feet closer to the lightning strike than the person at point B. Since the difference of the distance from x, y to A and from x, y
constant 1100, the point x, y lies on a hyperbola
to B is the constant 2200, the point x, y lies on
whose foci are at A and B. The hyperbola has the equation x2 y2 1 a 2 b2 where 2a 1100 , so a 550 . Because the distance between the two people is 2 miles (10,560 feet) and each person is at a focus of the hyperbola, we have 2c 10,560 c 5280 b 2 c 2 a 2 52802 5502 27,575,900 The equation of the hyperbola that describes the location of the fireworks display is y2 x2 1 5502 27,575,900 Since the fireworks display is due north of the individual at A, we let x 5280 and solve the equation for y. y2 52802 1 5502 27,575,900
a hyperbola whose foci are at A and B. The hyperbola has the equation x2 y2 1 a2 b2 where 2a 2200 , so a 1100 . Because the distance between the two people is 1 mile (5,280 feet) and each person is at a focus of the hyperbola, we have 2c 5, 280 c 2, 640
b 2 c 2 a 2 26402 11002 5, 759, 600 The equation of the hyperbola that describes the location of the lightning strike is y2 x2 1 11002 5, 759, 600 Since the lightning strike is due north of the individual at A, we let x 2640 and solve for y. y2 26402 1 2 5, 759, 600 1100 y2 4.76 5, 759, 600
y2 91.16 27,575,900
y 2 27, 415, 696 y 5236 The lightning strikes 5236 feet (approximately 0.99 miles) due north of the person at point A.
2
y 2,513,819, 044 y 50,138 Therefore, the fireworks display was 50,138 feet (approximately 9.5 miles) due north of the person at point A.
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Chapter 11: Analytic Geometry 79. To determine the height, we first need to obtain the equation of the hyperbola used to generate the hyperboloid. Placing the center of the hyperbola at the origin, the equation of the x2 y2 hyperbola will have the form 2 2 1 . a b The center diameter is 200 feet so we have 200 a 100 . We also know that the base 2 diameter is 400 feet. Since the center of the hyperbola is at the origin, the points 200, 360
80. First note that all points where an explosion could take place, such that the time difference would be the same as that for the first detonation, would form a hyperbola with A and B as the foci. Start with a diagram:
(1200, 0) A
200 2 360 2 1 b2 100 2
3
b2 3602
(a, 0)
(1200, 0) B
Since A and B are the foci, we have 2c 2400 c 1200 Since D1 is on the transverse axis and is on the hyperbola, then it must be a vertex of the hyperbola. Since it is 300 feet from B, we have a 900 . Finally, b 2 c 2 a 2 12002 9002 630, 000 Thus, the equation of the hyperbola is y2 x2 1 810, 000 630, 000
hyperbola (recall the center is 360 feet above ground). Therefore,
3602
D1 (0, 0)
(1200, y)
2400 ft
and 200, 360 must be on the graph of our
4
D2
1
b2 b 2 43, 200
The point 1200, y needs to lie on the graph of the hyperbola. Thus, we have
b 43, 200 120 3 The equation of the hyperbola is x2 y2 1 10, 000 43, 200
1200 2 810, 000
At the top of the tower we have x
300 150 . 2
y2 1 630, 000
7 y2 630, 000 9
y 2 490, 000 y 700 The second explosion should be set off 700 feet due north of point B.
y2 1502 1 10, 000 43, 200 y2 1.25 43, 200
81. a.
y 2 54000 y 232.4 The height of the tower is approximately 232.4 360 592.4 feet.
Since the particles are deflected at a 45 angle, the asymptotes will be y x .
b. Since the vertex is 10 cm from the center of the hyperbola, we know that a 10 . The b slope of the asymptotes is given by . a Therefore, we have b b 1 1 b 10 10 a Using the origin as the center of the hyperbola, the equation of the particle path would be x2 y2 1 , x 0 100 100 1248
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Section 11.4: The Hyperbola 82. Assume the origin lies at the center of the hyperbola. From the equation we know that the hyperbola has a transverse axis that is parallel to the y-axis. The foci of the hyperbola are located at 0, c
84. Let x = 32 and solve for y. y2 x2 1 484 100 (32) 2 y 2 1 484 100 1024 y 2 1 484 100 256 y 2 1 121 100 y2 256 1 100 121 13500 y2 121 13500 y 121
c 2 a 2 b 2 9 16 25 or c 5 Therefore, the foci of the hyperbola are at 0, 5 and 0,5 .
If we assume the parabola opens up, the common focus is at 0,5 . The equation of our parabola will be x 2 4a y k . The focal length of the parabola is given as a 6 . We also know that the distance focus of the parabola is located at 0, k a 0,5 . Thus, ka 5 k 6 5 k 1 and the equation of our parabola becomes x 2 4 6 y 1
30 15 11 Therefore the width is 30 15 30 15 60 15 21.13 mi 11 11 11 y
x 2 24 y 1 or y
x2 y2 1. a2 b2 If the eccentricity is close to 1, then c a and b 0 . When b is close to 0, the hyperbola is very narrow, because the slopes of the asymptotes are close to 0. If the eccentricity is very large, then c is much larger than a and b is very large. The result is a hyperbola that is very wide, because the slopes of the asymptotes are very large.
1 2 x 1 . 24
85. Assume
83. Since the distance between the vertices is 18, the vertices are (0,9) and (0, 9) giving a 9 . The foci are 4 inches from the vertices so the foci are (0,13) and (0, 13) giving c 13 . Solving for b 2 we have b2 c 2 a 2 . 132 92 88
y2
So the equation of the hyperbola is: y 2 x2 1 81 88
2
x2
1 , the opposite is true. When the a b2 eccentricity is close to 1, the hyperbola is very wide because the slopes of the asymptotes are close to 0. When the eccentricity is very large, the hyperbola is very narrow because the slopes of the asymptotes are very large.
For
86. If a b , then c 2 a 2 a 2 2a 2 . Thus, c2 c 2 or 2 . The eccentricity of an 2 a a equilateral hyperbola is 2 .
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Chapter 11: Analytic Geometry
87.
89. Put the equation in standard hyperbola form: Ax 2 Cy 2 F 0 A C 0, F 0
x2 y 2 1 (a 2, b 1) 4 This is a hyperbola with horizontal transverse axis, centered at (0, 0) and has asymptotes: 1 y x 2 x2 y2 1 (a 1, b 2) 4 This is a hyperbola with vertical transverse axis, 1 centered at (0, 0) and has asymptotes: y x . 2 Since the two hyperbolas have the same asymptotes, they are conjugates.
Ax 2 Cy 2 F Ax 2 Cy 2 1 F F y2 x2 1 F F A C Since F / A and F / C have opposite signs, this is a hyperbola with center at (0, 0). The transverse axis is the x-axis if F / A 0 and the y-axis if F / A 0 .
90. Ax 2 Cy 2 Dx Ey F 0 , A C 0 .
A x D C y E D E F 2A 2C 4 A 4C A x2 D x C y2 E y F A C 2
2
2
2
2 2 Let U D E F . 4 A 4C a. If U 0 , then 2
88.
2
2
y2
a 2 x 2 b2 1 2 2 b x
2
x D y E 2 A 2C 1 U U A C U U and having opposite signs. This with C A is the equation of a hyperbola whose center E D , is . 2 A 2C
y x 2 1 2 a b Solve for y: y2 x2 1 a2 b2 x2 y 2 a 2 1 2 b
b. If U 0 , then
A x D 2A
ax b 2 y 1 b x2
C y 2EC 0 y 2EC CA x 2DA A y E x D 2C 2A C 2
2
2
b2 gets x2 close to 0, so the expression under the radical gets closer to 1. Thus, the graph of the hyperbola a a gets closer to the lines y x and y x . b b These lines are the asymptotes of the hyperbola.
As x or as x , the term
2
which is the graph of two intersecting lines, E D containing the point , , with 2 A 2C slopes
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A . C
Section 11.4: The Hyperbola 94. r 6sin
1 91. y sin 3x 5 2 1 1 Amplitude: A 2 2 2 2 Period: T 3 Phase Shift: 3 3 Vertical Shift: B 5
r 2 6r sin x2 y 2 6 y x2 y2 6 y 0 x2 y 2 6 y 9 9 x 2 ( y 3) 2 9 The graph will be a circle with radius 3 and center 0,3 .
c 2 a 2 b 2 2ab cos C
Check for symmetry: Polar axis: Replace by . The result is r 6sin( ) 6sin . The test fails.
c 2 7 2 102 2 7 10 cos100º 149 140 cos100º
The line
92. a 7, b 10, C 100º
: Replace by . 2 r 6sin( )
c 149 140 cos100º 13.16 a 2 b 2 c 2 2bc cos A 102 13.162 7 2 102 13.162 7 2 224.1856 cos A 2bc 2(10)(13.16) 263.2 224.1856 31.6º A cos 1 263.2 B 180º A C 180º 31.6º 100º 48.4º
93.
12, 3 x r cos 12 cos 3
6(sin cos cos sin ) 6(0 sin ) 6sin The graph is symmetric with respect to the line . 2 The pole: Replace r by r . r 6sin . The test fails. Due to symmetry with respect to the line , assign values to from to . 2 2 2
1 12 6 2
2 3 6 0
r 6sin
y r sin 12sin 3 3 12 6 3 2
6
6 3 5.2 2 3 0
6
The rectangular coordinates are 6, 6 3 .
3 2
3 6 3 5.2 2
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6
Chapter 11: Analytic Geometry
(2 x 3) 2 x 2 5 x(2 x) 1
97.
4 x 2 12 x 9 x 2 10 x 5 x 2 1 5 x 2 12 x 9 10 x 5 x 2 1 12 x 9 10 x 1 2 x 8 x 4
The solution set is 4 .
95.
x x y y2 3 (2) 8 5 98. 1 2 , 1 , 2 2 2 2 1 3 , 2 2
y 3e x 1 4 x 3e y 1 4 x 4 3e y 1
99. Since sin
x4 e y 1 3 x4 ln y 1 3 x4 1 y ln 1 f ( x) 3
opposite x , then the adjacent hypotenuse 4
side is 16 x 2 . cos
adjacent 16 x 2 hypotenuse 4
96.
Section 11.5 1. sin A cos B cos A sin B 2. 2sin cos
The area of the left side of the semicircle is 1 A r2 4 1 2 9 3 4 4 The area of the triangle under the line in Quadrant II is 1 A bh 2 1 9 33 2 2 The area between the curves is 9 9 2.57 units 2 4 2
3.
1 cos 2
4.
1 cos 2
5. cot 2
AC B
6. d 7. B 2 4 AC 0 8. c 9. True 10. False; cot 2 1252
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AC B
Section 11.5: Rotation of Axes; General Form of a Conic
11. x 2 4 x y 3 0 A 1 and C 0; AC (1)(0) 0 . Since AC 0 , the equation defines a parabola.
21. x 2 4 xy y 2 3 0 A 1, B 4, and C 1; A C 1 1 0 cot 2 0 B 4 4 2 2 4 x x cos y sin 4 4 2 2 2 x y x y 2 2 2 2 2 y x sin y cos x y 4 4 2 2 2 x y 2
12. 2 y 2 3 y 3x 0 A 0 and C 2; AC (0)(2) 0 . Since AC 0 , the equation defines a parabola. 13. 6 x 2 3 y 2 12 x 6 y 0 A 6 and C 3; AC (6)(3) 18 . Since AC 0 and A C , the equation defines an ellipse.
14. 2 x 2 y 2 8 x 4 y 2 0 A 2 and C 1; AC (2)(1) 2 . Since AC 0 and A C , the equation defines an ellipse.
22. x 2 4 xy y 2 3 0 A 1, B 4, and C 1; A C 1 1 0 cot 2 0 B 4 4 2 2 4 2 2 x x cos y sin x y 4 4 2 2 2 x y 2 2 2 y x sin y cos x y 4 4 2 2 2 x y 2
15. 3 x 2 2 y 2 6 x 4 0 A 3 and C 2; AC (3)( 2) 6 . Since AC 0 , the equation defines a hyperbola. 16. 4 x 2 3 y 2 8 x 6 y 1 0 A 4 and C 3; AC (4)( 3) 12 . Since AC 0 , the equation defines a hyperbola.
17. 2 y 2 x 2 y x 0 A 1 and C 2; AC (1)(2) 2 . Since AC 0 , the equation defines a hyperbola. 18. y 2 8 x 2 2 x y 0 A 8 and C 1; AC ( 8)(1) 8 . Since AC 0 , the equation defines a hyperbola.
23. 5 x 2 6 xy 5 y 2 8 0 A 5, B 6, and C 5; AC 55 0 cot 2 0 B 6 6 2 2 4 2 2 x x cos y sin x y 4 4 2 2 2 x y 2 2 2 y x sin y cos x y 4 4 2 2 2 x y 2
19. x 2 y 2 8 x 4 y 0 A 1 and C 1; AC (1)(1) 1 . Since AC 0 and A C , the equation defines a circle.
20. 2 x 2 2 y 2 8 x 8 y 0 A 2 and C 2; AC (2)(2) 4 . Since AC 0 and A C , the equation defines a circle.
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Chapter 11: Analytic Geometry
24. 3x 2 10 xy 3 y 2 32 0
26. 11x 2 10 3 xy y 2 4 0
A 3, B 10, and C 3; AC 33 0 cot 2 0 B 10 10 2 2 4 2 2 x x cos y sin x y 4 4 2 2 2 x y 2 2 2 y x sin y cos x y 4 4 2 2 2 x y 2
A 11, B 10 3, and C 1; A C 11 1 10 3 B 3 10 3 10 3 2 3 6 3 1 x x cos y sin x y 6 6 2 2 1 3 x y 2 3 1 y x sin y cos x y 6 6 2 2 1 x 3 y 2
cot 2
25. 13 x 2 6 3 xy 7 y 2 16 0
2
6 3 A C 13 7 B 3 6 3 6 3
cot 2
2 3 3
A C 4 1 3 3 ; cos 2 B 4 4 5
3 1 5 sin 2
1 3 x x cos y sin x y 3 3 2 2 1 x 3 y 2 3 1 y x sin y cos x y 3 3 2 2 1 3x y 2
27. 4 x 2 4 xy y 2 8 5 x 16 5 y 0 A 4, B 4, and C 1;
A 13, B 6 3, and C 7; cot 2
4 2 2 5 ; 5 5 5
3 1 5 1 1 5 cos 2 5 5 5
x x cos y sin
5 x 2 y 5
y x sin y cos
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5 2 5 x y 5 5
5 2 x y 5
2 5 5 x y 5 5
Section 11.5: Rotation of Axes; General Form of a Conic
28. x 2 4 xy 4 y 2 5 5 y 5 0 A 1, B 4, and C 4; cot 2
29. 25 x 2 36 xy 40 y 2 12 13 x 8 13 y 0 A 25, B 36, and C 40;
A C 1 4 3 3 ; cos 2 B 4 4 5
3 1 5 sin 2
cot 2
4 2 2 5 ; 5 5 5
sin
3 1 5 1 1 5 cos 2 5 5 5
cos
5 2 5 x y 5 5
x x cos y sin
y x sin y cos
2 5 5 x y 5 5
5 13 9 3 3 13 2 13 13 13
1
3 13 2 13 x y 13 13
13 3 x 2 y 13
y x sin y cos
5 2 x y 5
5 13 4 2 2 13 ; 2 13 13 13
1
x x cos y sin
5 x 2 y 5
A C 25 40 5 5 ; cos 2 B 36 12 13
2 13 3 13 x y 13 13
13 2 x 3 y 13
30. 34 x 2 24 xy 41 y 2 25 0 A 34, B 24, and C 41; cot 2
A C 34 41 7 7 ; cos 2 B 24 24 25 7 25 2
7 1 9 3 16 4 25 ; cos sin 25 5 2 25 5 4 3 1 x x cos y sin x y 4 x 3 y 5 5 5 3 4 1 y x sin y cos x y 3 x 4 y 5 5 5 1
31. x 2 4 xy y 2 3 0 ; 45º 2
y
2
2 2 2 2 x y 4 x y x y x y 3 0 2 2 2 2 1 2 1 2 2 2 2 x 2 xy y 2 x y x 2 xy y 2 3 0 2 2 1 2 1 1 1 x xy y 2 2 x2 2 y 2 x2 xy y 2 3 2 2 2 2 2 3 x y 2 3
x 2 y 2 1 1 3 Hyperbola; center at the origin, transverse axis is the x -axis,
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x'
3
y'
(1, 0) –3
(–1, 0)
–3
x
Chapter 11: Analytic Geometry
vertices (±1, 0). 32. x 2 4 xy y 2 3 0 ; 45º 2
y 3
2
2 2 2 2 x y 4 x y x y x y 3 0 2 2 2 2 1 2 1 x 2 xy y 2 2 x2 y 2 x2 2 xy y 2 3 0 2 2 1 2 1 1 1 x xy y 2 2 x2 2 y 2 x2 xy y 2 3 2 2 2 2 2 x 3 y 2 3
y'
x'
(0, 1)
x
–3
(0, –1)
3
–3
y 2 x 2 1 1 3 Hyperbola; center at the origin, transverse axis is the y -axis, vertices (0, ±1).
33. 5 x 2 6 xy 5 y 2 8 0 ; 45º 2
y 3
2
2 2 2 2 5 x y 6 x y x y 5 x y 8 0 2 2 2 2 5 2 5 x 2 xy y 2 3 x2 y 2 x2 2 xy y 2 8 0 2 2 5 2 5 2 5 5 x 5 xy y 3 x2 3 y 2 x2 5 xy y 2 8 2 2 2 2 2 8 x 2 y 2 8
y'
x'
(0, 2)
x –3
3 (0, –2)
x 2 y 2 1 1 4 Ellipse; center at the origin, major axis is the y -axis, vertices (0, ±2).
–3
34. 3x 2 10 xy 3 y 2 32 0 ; 45º 2
y
2
2 2 2 2 3 x y 10 x y x y 3 x y 32 0 2 2 2 2 3 2 3 x 2 xy y 2 5 x2 y 2 x2 2 xy y 2 32 0 2 2 3 2 3 2 3 3 2 x 3 xy y 5 x 5 y 2 x2 3xy y 2 32 2 2 2 2 2 2 x 8 y 2 32
y 2 x 2 1 4 16 Hyperbola; center at the origin, transverse axis is the y -axis, vertices (0, ±2).
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x'
5
y' (0, 2)
x
–5
(0, –2)
–5
5
Section 11.5: Rotation of Axes; General Form of a Conic
35. 13x 2 6 3 xy 7 y 2 16 0 ; 60º
2
1 1 1 13 x 3 y 6 3 x 3 y 2 2 2 13 4
2
3x y 7
1
y
2
3 x y 16 0
(2, 0) y'
x 2 3xy 3 y 2 3x 2 xy 3 y 4 3x 2 3xy y 16 2
13 4
x 2
3 3
2
13 3
x y
2
39 4
y 2
2
9 2
2
x 2 3 3 x y
9 2
y 2
7
21 4
2
x 2
2
7 3 2
7
x y
4
x –3
y 2 16
3 (–2, 0)
4 x 16 y 16 2
2
x 2
Ellipse; center at the origin, major axis is the x -axis, vertices (±2, 0).
4
y 2 1
x'
3
–3
1
36. 11x 2 10 3 xy y 2 4 0 ; 30º 1 11 2
2
2
3x y 10 3 12 3x y 12 x 3 y 12 x 3 y 4 0
11 5 3 1 3x2 2 3xy y 2 3 x2 2 xy 3 y 2 x2 2 3 xy 3 y 2 4 4 2 4 33 2 11 3 11 2 15 2 15 2 1 2 3 3 x xy y x 5 3xy y x xy y 2 4 4 2 4 2 2 4 2 4 2 16 x 4 y 2 4
y 1
y'
x'
1, 0 2
x –1
1 – 1, 0
2
–1
4 x 2 y 2 1 x '2 y '2 1 1 1 4 Hyperbola; center at the origin, transverse axis is the x -axis, vertices 12 , 0 .
37. 4 x 2 4 xy y 2 8 5 x 16 5 y 0 ; 63.4º 2
y 4
2
5 5 5 5 4 x 2 y 4 x 2 y 2 x y 2 x y 5 5 5 5 5 5 8 5 x 2 y 16 5 2 x y 0 5 5 4 2 4 1 x 4 xy 4 y 2 2 x2 3xy 2 y 2 4 x2 4 xy y 2 5 5 5 8 x 16 y 32 x 16 y 0 4 2 16 16 2 8 2 12 8 4 4 1 x xy y x xy y 2 x2 xy y 2 40 x 0 5 5 5 5 5 5 5 5 5 5 y 2 40 x 0
y 2 8 x Parabola; vertex at the origin, axis of symmetry is the x ' axis, focus at (2, 0).
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y'
x'
(2, 0)
–4
4
–4
x
Chapter 11: Analytic Geometry
38. x 2 4 xy 4 y 2 5 5 y 5 0 ; 63.4º 2
2
5 5 5 5 x 2 y 4 x 2 y 2 x y 4 2 x y 5 5 5 5 5 5 5 2 x y 5 0 5 1 2 4 4 x 4 xy 4 y 2 2 x2 3 xy 2 y 2 4 x2 4 xy y 2 5 5 5 10 x 5 y 5 0 1 2 4 4 8 12 8 16 16 4 x xy y 2 x2 xy y 2 x2 xy y 2 5 5 5 5 5 5 5 5 5 10 x 5 y 5 0
5 x2 10 x 5 y 5 0 x 2 2 x 1 y 1 1 ( x 1)2 y 1 Parabola; vertex at (–1, 0), axis of symmetry parallel to the y -axis; focus at x ', y ' 1, . 4
39. 25 x 2 36 xy 40 y 2 12 13 x 8 13 y 0 ; 33.7º 2
13 13 13 13 25 3x 2 y 36 3x 2 y 2 x 3 y 40 2 x 3 y 13 13 13 13 13 13 12 13 3x 2 y 8 13 2 x 3 y 0 13 13 25 36 40 9 x2 12 xy 4 y 2 6 x2 5 xy 6 y 2 4 x2 12 xy 9 y 2 13 13 13 36 x 24 y 16 x 24 y 0
225 2 300 100 2 216 2 180 216 2 x xy y x xy y 13 13 13 13 13 13 160 2 480 360 2 x xy y 52 x 0 13 13 13
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2
Section 11.5: Rotation of Axes; General Form of a Conic
13x 2 52 y 2 52 x 0 x 2 4 x 4 y 2 0
x 2 2 4 y 2 4 ( x 2) 2 y 2 1 4 1 Ellipse; center at (2, 0), major axis is the x -axis, vertices (4, 0) and (0, 0). y
5
y'
x' (2, 1) (4, 0) x –3 –1
(2, –1)
40. 34 x 2 24 xy 41y 2 25 0 ; 36.9º 2
2
1 1 1 1 34 4 x 3 y 24 4 x 3 y 3x 4 y 41 3x 4 y 25 0 5 5 5 5 34 24 41 16 x 2 24 x y 9 y 2 12 x2 7 xy 12 y 2 9 x2 24 xy 16 y 2 25 25 25 25 544 2 816 306 2 288 2 168 288 2 369 2 984 656 2 x x y y x xy y x xy y 25 25 25 25 25 25 25 25 25 25 25 x 2 50 y 2 25
x 2 2 y 2 1
Ellipse; center at the origin, major axis is the x -axis, vertices (±1, 0). y 2 y' x' (1, 0) –2
x 2
(–1, 0) –2
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x '2 y ' 2 1 1 1 2
Chapter 11: Analytic Geometry
41. 16 x 2 24 xy 9 y 2 130 x 90 y 0 A 16, B 24, and C 9; cot 2
A C 16 9 7 7 cos 2 B 24 24 25
7 25 2
7 1 9 3 25 16 4 36.9o sin ; cos 25 5 2 25 5 4 3 1 x x cos y sin x y 4 x 3 y 5 5 5 3 4 1 y x sin y cos x y 3 x 4 y 5 5 5 1
2
2
1 1 1 1 16 4 x 3 y 24 4 x 3 y 3 x 4 y 9 3 x 4 y 5 5 5 5 1 1 130 4 x 3 y 90 3x 4 y 0 5 5 16 24 2 2 2 2 16 x 24 xy 9 y 12 x 7 xy 12 y 25 25 9 9 x2 24 xy 16 y 2 104 x 78 y 54 x 72 y 0 25 256 2 384 144 2 288 2 168 288 2 x xy y x xy y 25 25 25 25 25 25 81 2 216 144 2 x xy y 50 x 150 y 0 25 25 25 25 x 2 50 x 150 y 0
x 2 2 x 6 y ( x 1) 2 6 y 1 1 ( x 1) 2 6 y 6 4 1 Parabola; vertex 1, , focus 1, ; axis of symmetry parallel to the y ' axis. 3 6
y'
y 5
1, 1 6
x' x
–5
1, – 4 3
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Section 11.5: Rotation of Axes; General Form of a Conic
42. 16 x 2 24 xy 9 y 2 60 x 80 y 0 A 16, B 24, and C 9; cot 2
7 A C 16 9 7 cos 2 24 24 25 B
7 25 2
7 1 9 3 25 16 4 36.9 ; cos sin 25 5 2 25 5 4 3 1 x x cos y sin x y 4 x 3 y 5 5 5 3 4 1 y x sin y cos x y 3 x 4 y 5 5 5 1
2
2
1 1 1 1 16 4 x 3 y 24 4 x 3 y 3x 4 y 9 3x 4 y 5 5 5 5 1 1 60 4 x 3 y 80 3x 4 y 0 5 5 16 24 16 x 2 24 xy 9 y 2 12 x2 7 xy 12 y 2 25 25 9 9 x 2 24 xy 16 y 2 48 x 36 y 48 x 64 y 0 25 256 2 384 144 2 288 2 168 288 2 x xy y x xy y 25 25 25 25 25 25 81 2 216 144 2 x xy y 100 y 0 25 25 25 25 x2 100 y 0
x 2 4 y Parabola; vertex (0, 0), axis of symmetry is the y ' axis, focus (0, –1).
y'
y 2
x'
x –2
2 (0, –1) –2
43. A 1, B 3, C 2
B 2 4 AC 32 4(1)( 2) 17 0 ; hyperbola
44. A 2, B 3, C 4
B 2 4 AC (3) 2 4(2)(4) 23 0 ; ellipse
45. A 1, B 7, C 3
B 2 4 AC (7) 2 4(1)(3) 37 0 ; hyperbola
46. A 2, B 3, C 2
B 2 4 AC (3) 2 4(2)(2) 7 0 ; ellipse
47. A 9, B 12, C 4
B 2 4 AC 122 4(9)(4) 0 ; parabola
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Chapter 11: Analytic Geometry
48. A 10, B 12, C 4
B 2 4 AC 122 4(10)(4) 16 0 ; ellipse
49. A 10, B 12, C 4
B 2 4 AC (12) 2 4(10)(4) 16 0 ; ellipse
50. A 4, B 12, C 9
B 2 4 AC 122 4(4)(9) 0 ; parabola
51. A 3, B 2, C 1
B 2 4 AC ( 2) 2 4(3)(1) 8 0 ; ellipse
52. A 3, B 2, C 1
B 2 4 AC 22 4(3)(1) 8 0 ; ellipse
21 21 x 21 y 2 171 y 324 0 2 A C 4 21 17 17 A 4, B 4 21, and C 21; cot 2 cos 2 B 4 21 4 21 4 21
53. 4 x 2 4 21xy
17 2 cot 1 47.58 4 21 23.6
54. 20 x 2 10 xy
19 48 x 89 y 2 89 y 0 2 5
A 20, B 10, and C 89; cot 2
A C 20 89 69 69 cos 2 B 10 10 10
69 2 cot 1 8.25 10 4.1
55. A A cos 2 B sin cos C sin 2 B B (cos 2 sin 2 ) 2(C A)(sin cos ) C A sin 2 B sin cos C cos 2 D D cos E sin E D sin E cos F F
56. A C A cos 2 B sin cos C sin 2 A sin 2 B sin cos C cos 2
A cos 2 sin 2 C sin 2 cos 2 A(1) C (1) A C
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Section 11.5: Rotation of Axes; General Form of a Conic
57. B '2 [ B(cos 2 sin 2 ) 2(C A) sin cos ]2 [ B cos 2 ( A C ) sin 2 ]2 B 2 cos 2 2 2 B( A C ) sin 2 cos 2 ( A C ) 2 sin 2 2 4 A ' C ' 4[ A cos 2 B sin cos C sin 2 ][ A sin 2 B sin cos C cos 2 ]
1 cos 2 1 cos 2 1 cos 2 sin 2 B 4 A 1 cos 2 B sin 2 C A C 2 2 2 2 2 2 [ A(1 cos 2 ) B(sin 2 ) C (1 cos 2 )][ A(1 cos 2 ) B (sin 2 ) C (1 cos 2 )] [( A C ) B sin 2 ( A C ) cos 2 ][( A C ) ( B sin 2 ( A C ) cos 2 )]
( A C ) 2 [ B sin 2 ( A C ) cos 2 ]2 ( A C ) 2 [ B 2 sin 2 2 2 B( A C ) sin 2 cos 2 ( A C ) 2 cos 2 2 ] B '2 4 A ' C ' B 2 cos 2 2 2 B ( A C ) sin 2 cos 2 ( A C ) 2 sin 2 2 ( A C ) 2 B 2 sin 2 2 2 B( A C ) sin 2 cos 2 ( A C ) 2 cos 2 2 B 2 (cos 2 2 sin 2 2 ) ( A C ) 2 (cos 2 2 sin 2 2 ) ( A C ) 2 B 2 ( A C ) 2 ( A C ) 2 B 2 ( A2 2 AC C 2 ) ( A2 2 AC C 2 ) B 2 4 AC
58. Since B 2 4 AC B 2 4 AC for any rotation (Problem 55), choose so that B 0 . Then B 2 4 AC 4 AC .
a.
If B 2 4 AC 4 AC 0 then AC 0 . Using the theorem for identifying conics without completing the square, the equation is a parabola.
b. If B 2 4 AC 4 AC 0 then AC 0 . Thus, the equation is an ellipse (or circle). c.
If B 2 4 AC 4 AC 0 then AC 0 . Thus, the equation is a hyperbola.
59. d 2 ( y2 y1 ) 2 ( x2 x1 ) 2 x2 sin y2 cos x1 sin y1 cos x2 cos y2 sin x1 cos y1 sin 2
x2 x1 sin y2 y1 cos x2 x1 cos y2 y1 sin 2
2
x2 x1 sin 2 2 x2 x1 y2 y1 sin cos y2 y1 cos 2 2
2
x2 x1 cos 2 2 x2 x1 y2 y1 sin cos y2 y1 sin 2 2
2
x2 x1 sin 2 cos 2 y2 y1 cos 2 sin 2 2
x2 x1 y2 y1 2
2
2
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2
Chapter 11: Analytic Geometry z r cos i sin 29 cos 291.8º i sin 291.8º
60. x1/ 2 y1/ 2 a1/ 2 1/ 2
y
1/ 2
a
1/ 2
x
y a1/ 2 x1/ 2
1/ 2 1/ 2
y a 2a
x
(4 x 1) 3(2 x 3) 2 (2 x 3) 8(4 x 1) 8 x 2
2
67. x
2a1/ 2 x1/ 2 (a x) y 2
4ax (a x) 2 y (a x) y
61 – 62. Answers will vary. 63. a 7, b 9, c 11 a b c 2bc cos A cos A
9 11 7 153 b c a 2bc 2(9)(11) 198 2
2
2
2
68.
2(4 x 2 1)7 (2 x 3) 2 (12 x 2 3 64 x 2 96 x)
2(2 x 3)2 3 52 x 2 96 x
(4 x 2 1)16 (4 x 2 1)16 (4 x 2 1)9 2(2 x 3) 2 52 x 2 3 96 x (4 x 2 1)9
f ( x) 4e x 1 5
Using the graph of y 4e x 1 , shift the graph down 5 units. Horizontal Asymptote: y 5 69. log 5 x log 5 ( x 4) 1
b 2 a 2 c 2 2ac cos B a 2 c 2 b 2 7 2 112 92 89 2ac 2(7)(11) 154
log 5 ( x( x 4)) 1 5 x ( x 4)
89 B cos 54.7º 154 1
5 x2 4 x x2 4x 5 0
C 180o A B 180o 39.4o 54.7 o 85.9o
( x 5)( x 1) 0 x 5 or x 1 We cannot use the negative answer since the domain of the log function must be positive. So the solutions set is 5 .
1 64. A ab sin C 2 1 (14)(11) sin 30 2 38.5 65. xy 1
(r cos )(r sin ) 1 r 2 cos sin 1
66. r
2
2
153 39.4º A cos 1 198
cos B
7
2(4 x 2 1)7 (2 x 3) 2 3(4 x 2 1) 32 x(2 x 3)
2
2
2
(4 x 2 1)8
B 2 4 AC ( 2) 2 4(1)(1) 4 4 0 The graph of the equation is part of a parabola.
2
3
(4 x 2 1)8 6 (4 x 2 1)8 (2 x 3) 2 64 x(2 x 3)3 (4 x 2 1)7
0 x 2 2 xy y 2 2ax 2ay a 2
2
2
2
4ax a 2 2ax x 2 2ay 2 xy y 2
2
8
x 2 y 2 22 ( 5) 2 29 y 5 x 2 291.8º The polar form of z 2 5i is tan
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Section 11.6: Polar Equations of Conics 70. The minimum occurs when x 1 0 2 x 25 2 x x 25 2
r 2 x 2 y 2 , we obtain x 2 y 2 6 x . Move all the variable terms to the left side of the equation and complete the square in x. x2 6 x y2 0
1 2
x 6x 9 y 9 2
2 x x 2 25
x 32 y 2 9
4 x x 25 2
2
3. conic; focus; directrix
3 x 2 25
4. parabola; hyperbola; ellipse
25 5 3 3 From substitution we see that the positive answer only is acceptable. The minimum value is x
5. b 6. True
2
5 1 5 25 2 8 0 3 3
7. e 1; p 1 ; parabola; directrix is perpendicular to the polar axis and 1 unit to the right of the pole.
5 25 3 25 4 0 2 3
8. e 1; p 3 ; parabola; directrix is parallel to the polar axis and 3 units below the pole.
100 5 4 0 3 2 3
9.
10 5 4 8.33 3 2 3
71. Find the inverse of g(x). x y7 2 x2
2
y7
( x 2) 2 y 7 ( x 2) 2 7 y g 1 ( x) g 1 (3) (3 2)2 7 12 7 8
4 4 2 3sin 3 2 1 sin 2 2 3 1 sin 2 3 4 ep 2, e ; p 2 3 Hyperbola; directrix is parallel to the polar axis 4 and units below the pole. 3 r
2 ; ep 2, e 2; p 1 1 2 cos Hyperbola; directrix is perpendicular to the polar axis and 1 unit to the right of the pole.
72. As x goes to the term 4e x 1 tends to be 0. Thus 4e x 1 5 tends to 5 . So the horizontal asymptote would be y 5 .
10. r
Section 11.6 1. r cos ; r sin 2. r 6 cos Begin by multiplying both sides of the equation by r to get r 2 6r cos . Since r cos x and 1265
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Chapter 11: Analytic Geometry
11.
12.
3 1 sin ep 3, e 1, p 3 Parabola; directrix is parallel to the polar axis 3 3 3 units below the pole; vertex is , . 2 2
3 3 4 2 cos 1 4 1 cos 2 3 4 ; 1 1 cos 2 3 1 3 ep , e ; p 4 2 2 Ellipse; directrix is perpendicular to the polar 3 axis and units to the left of the pole. 2 r
14. r
6 6 1 8 2sin 8 1 sin 4 3 4 1 1 sin 4 3 ep 4
r
8 4 3sin 8 2 r 3 1 3 sin 4 1 sin 4 4 3 8 ep 2, e , p 4 3
15. r
1 e ; p3 4 Ellipse; directrix is parallel to the polar axis and 3 units above the pole.
Ellipse; directrix is parallel to the polar axis
8 3
units above the pole; vertices are 8 3 , and 8, . 7 2 2
1 1 cos ep 1, e 1, p 1 Parabola; directrix is perpendicular to the polar axis 1 unit to the right of the pole; vertex is 1 , 0 . 2
13. r
1 8 32 so the center is at Also: a 8 2 7 7 32 3 24 3 24 24 8 7 , 2 7 , 2 , and c 7 0 7 so that the second focus is at 24 24 3 48 3 7 7 , 2 7 , 2 Directrix
8 7, 2
y
(2, )
(2, 0)
3 8, 2
1266 Copyright © 2025 Pearson Education, Inc.
Polar x Axis
Section 11.6: Polar Equations of Conics
10 5 4 cos 10 2 r 4 1 4 cos 5 1 cos 5 5 4 5 ep 2, e , p 5 2 Ellipse; directrix is perpendicular to the polar 5 axis units to the right of the pole; vertices are 2 10 , 0 and 10, . 9 1 10 50 so the center is at Also: a 10 2 9 9 50 40 40 40 10 9 , 9 , , and c 9 0 9 so that the second focus is at 40 40 80 9 9 , 9 , .
16. r
2, 2
y
12 4 8sin 12 3 r 4 1 2sin 1 2sin
18. r
3 2 Hyperbola; directrix is parallel to the polar axis 3 units above the pole; vertices are 2 3 1, and 3, . 2 2 1 Also: a 3 1 1 so the center is at 2 3 1 1, 2 2, 2 [or 2, 2 ], and c 2 0 2 so that the second focus is at 3 2 2, 2 4, 2 [or 4, 2 ]. ep 3, e 2, p
Directrix
Polar x Axis
(10, )
3 2, 2
10 ,0 9
9 3 6 cos 9 3 r 3 1 2 cos 1 2 cos
17. r
3 2 Hyperbola; directrix is perpendicular to the polar 3 axis units to the left of the pole; vertices are 2 3, 0 and 1, . ep 3, e 2, p
1 3 1 1 so the center is at 2 1 1, 2, [or 2, 0 ], and c 2 0 2
Also: a
so that the second focus is at 2 2, 4, [or 4, 0 ].
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Chapter 11: Analytic Geometry
8 2 sin 8 4 r 1 1 1 sin 2 1 sin 2 2 1 ep 4, e , p 8 2 Ellipse; directrix is parallel to the polar axis 8 units below the pole; vertices are 8 3 8, and , . 2 3 2 1 8 16 so the center is at Also: a 8 2 3 3 8 8 16 8 8 3 , 2 3 , 2 , and c 3 0 3 so that 8 8 16 the second focus is at , , . 3 3 2 3 2
8 8 16 16 3 3 , 0 3 , 0 [or 3 , ].
19. r
6 3 2sin 6 2 r 2 2 3 1 sin 1 sin 3 3
21. r 3 2sin 6 r
2 , p3 3 Ellipse; directrix is parallel to the polar axis 3 units below the pole; vertices are 6 3 6, and , . 2 5 2 1 6 18 so the center is at Also: a 6 2 5 5 12 12 18 12 6 5 , 2 5 , 2 , and c 5 0 5 so that the second focus is at 12 12 24 5 5 ,2 5 ,2. ep 2, e
8 2 4 cos 8 4 r 2 1 2 cos 1 2 cos
20. r
ep 4, e 2, p 2 Hyperbola; directrix is perpendicular to the polar axis 2 units to the right of the pole; vertices are 4 , 0 and 4, . 3 1 4 4 Also: a 4 so the center is at 2 3 3 4 4 8 8 3 3 , 0 3 , 0 [or 3 , ], and 8 8 c 0 so that the second focus is at 3 3
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Section 11.6: Polar Equations of Conics
6 4, 0 2, 0 , and c 2 0 2 so that the second focus is at 2 2, 0 4, 0 .
2 2 cos 2 1 r 1 1 1 cos 2 1 cos 2 2
22. r 2 cos 2 r
1 , p2 2 Ellipse; directrix is perpendicular to the polar axis 2 units to the left of the pole; vertices are 2 2, 0 and , . 3 1 2 4 Also: a 2 so the center is at 2 3 3 4 2 2 2 2 3 , 0 3 , 0 , and c 3 0 3 so that 2 2 4 the second focus is at , 0 , 0 . 3 3 3 ep 1, e
1 3 3 3csc sin sin 24. r 1 1 sin csc 1 1 sin sin 3 sin 3 sin 1 sin 1 sin ep 3, e 1, p 3 Parabola; directrix is parallel to the polar axis 3 3 3 units below the pole; vertex is , . 2 2
1 6 6 6sec cos cos 23. r 2 cos 2sec 1 1 2 1 cos cos 6 6 cos cos 2 cos 2 cos 6 3 r 1 1 2 1 cos 1 cos 2 2
1 1 cos r r cos 1
25. r
1 ep 3, e , p 6 2 Ellipse; directrix is perpendicular to the polar axis 6 units to the left of the pole; vertices are 6, 0 and 2, .
Also: a
r 1 r cos r 2 (1 r cos ) 2 x 2 y 2 (1 x) 2
1 6 2 4 so the center is at 2
x2 y2 1 2 x x2 y2 2x 1 0
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Chapter 11: Analytic Geometry
3 1 sin r r sin 3
30. r
26. r
r 3 r sin r (3 r sin ) 2 2
12 4 8sin 4r 8r sin 12 4r 12 8r sin r 3 2r sin r 2 (3 2r sin ) 2
x 2 y 2 (3 y ) 2
x 2 y 2 (3 2 y ) 2
x2 y 2 9 6 y y2
x 2 y 2 9 12 y 4 y 2
x2 6 y 9 0
x 2 3 y 2 12 y 9 0
8 27. r 4 3sin 4r 3r sin 8 4r 8 3r sin
31. r
16r 2 (8 3r sin ) 2
8 2 sin 2r r sin 8 2r 8 r sin 4r 2 (8 r sin ) 2
16( x 2 y 2 ) (8 3 y ) 2
4( x 2 y 2 ) (8 y )2
16 x 2 16 y 2 64 48 y 9 y 2
4 x 2 4 y 2 64 16 y y 2
16 x 2 7 y 2 48 y 64 0
4 x 2 3 y 2 16 y 64 0
10 28. r 5 4 cos 5r 4r cos 10 5r 10 4r cos
32. r
8 2 4 cos 2 r 4 r cos 8 2 r 8 4 r cos
25r 2 (10 4r cos ) 2 25( x 2 y 2 ) (10 4 x)2
r 4 2r cos
25 x 2 25 y 2 100 80 x 16 x 2
r (4 2 r cos ) 2 2
x 2 y 2 (4 2 x) 2
9 x 2 25 y 2 80 x 100 0
x 2 y 2 16 16 x 4 x 2
29. r
9 3 6 cos 3 r 6 r cos 9 3 r 9 6 r cos r 3 2r cos
3x 2 y 2 16 x 16 0
33. r (3 2sin ) 6 3 r 2 r sin 6 3 r 6 2 r sin
r 2 (3 2 r cos ) 2 2
2
x y (3 2 x)
9 r 2 (6 2 r sin ) 2
2
9( x 2 y 2 ) (6 2 y ) 2
x 2 y 2 9 12 x 4 x 2
9 x 2 9 y 2 36 24 y 4 y 2
3x 2 y 2 12 x 9 0
9 x 2 5 y 2 24 y 36 0
1270
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Section 11.6: Polar Equations of Conics 34. r (2 cos ) 2 2r r cos 2 2r 2 r cos
ep 1 e cos 4 e ; p3 5 12 12 5 r 4 1 cos 5 4 cos 5
39. r
4r 2 (2 r cos ) 2 4( x 2 y 2 ) (2 x)2 4 x2 4 y 2 4 4 x x2 3x 2 4 y 2 4 x 4 0
ep 1 e sin 2 e ; p3 3 2 6 r 2 3 2sin 1 sin 3
6sec 2sec 1 6 r 2 cos 2r r cos 6 2r 6 r cos
40. r
r
35.
4r 2 (6 r cos ) 2
ep 1 e sin e 6; p 2 12 r 1 6sin
4( x 2 y 2 ) (6 x) 2
41. r
4 x 2 4 y 2 36 12 x x 2 3x 2 4 y 2 12 x 36 0
36. r
3csc csc 1
ep 1 e cos e 5; p 5 25 r 1 5cos
42. r
3 r 1 sin r r sin 3 r 3 r sin r (3 r sin ) 2 2
43. Consider the following diagram:
x 2 y 2 (3 y ) 2
P = ( r, )
x2 y 2 9 6 y y2 2
x 6y 9 0
ep 1 e sin e 1; p 1
r
d( D, P)
r
37. r
Pole O (Focus F)
1 1 sin
Q p
d ( F , P ) e d ( D, P ) d ( D, P ) p r cos r e( p r cos ) r ep er cos r e r cos ep r (1 e cos ) ep ep r 1 e cos
ep 1 e sin e 1; p 2 2 r 1 sin
38. r
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Directrix D
Polar Axis
Chapter 11: Analytic Geometry 44. Consider the following diagram:
46. r
Directrix D
1 0.206 cos At aphelion, the greatest distance from the sun, cos 1 .
d( D, P) P = ( r, )
p
r
r
Polar Axis
Q
Pole O (Focus F)
r
4 107
4 107
1.155 1 0.967 cos At aphelion, the greatest distance from the sun, cos 1 . 1.155 1.155 r 1 0.967(1) 0.794 35 AU At perihelion, the shortest distance from the sun, cos 1 . 1.155 1.155 r 1 0.967(1) 1.967 0.587 AU
47. r
d (D , P) Q
6 107
6 107
P = ( r, )
(3.442)107 (3.442)107 1 0.206(1) 1.206
2.854 107 miles
45. Consider the following diagram:
Pole O (Focus F)
(3.442)107 (3.442)107 1 0.206(1) 0.794
4.335 107 miles At perihelion, the shortest distance from the sun, cos 1 .
d ( F , P ) e d ( D, P ) d ( D, P ) p r sin r e( p r sin ) r ep er sin r e r sin ep r (1 e sin ) ep ep r 1 e sin
r
3.442 107
Polar Axis
p Directrix D
d ( F , P ) e d ( D, P ) d ( D, P ) p r sin r e( p r sin ) r ep er sin r e r sin ep r (1 e sin ) ep ep r 1 e sin
1272
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Section 11.6: Polar Equations of Conics
M (1 e) m(1 e) M Me m me e e M m M m e( M m ) e M m M m M 1 M m M ( M m M m) p M m M m M m M (2m) 2mM M m M m
0.8 0.8 gives r 0.4 so 2 1 sin 2 2 the vertex (peak) is 0.4 inches above the focus. Since the focus is at the pole (0,0), the vertex is 0, a 0, 0.4 . The parabola is concave down
48. Letting
so x h 4a y k . So, 2
x 2 1.6 y 0.4 . The water hits the ground
when x 4 , so 42 1.6 y 0.4 10 y 0.4 y 9.6 The water hits the ground 9.6 inches below the focus. This is, the base of the tank is 9.6 inches below the focus. The puncture is at the vertex, so the puncture is 9.6 + 0.4 = 10 inches above the base of the tank.
49. Letting
2
gives r
51.
GM e 1 1 GM e 1 ; Using cos 2 r r0 r0 v0 r0 v0 2 r
ep and the fact that e 1 , if the 1 e cos
graph is a parabola, r
250 125 so 1 sin 2
p 1 1 cos 1 cos . or 1 cos r p p p
Comparing to the giving trajectory equation, it must be that 1 1
the vertex is 125 meters above the focus. Since the focus is at the pole (0,0), the vertex is 0, a 0, 125 . The parabola is concave up
p
r0
GM e and r0 v0 2
1 GM e . Equating the two expressions gives p r0 v0 2
so we have x h 4a y k . So, 2
1 GM e GM e 1 2GM e . So and 2 r0 r0 v0 2 r r0 v0 r0 v0 2 0
x 2 500 y 125 . The right end of the board is
located at x 2.5 , so
v0 2
2.5 500 y 125 2
0.0125 y 125 y 124.9875 The vertex is at (0, -125) and the right end of the board is at (0, -124.9875) so the displacement at the center is 0.0125 meters.
2GM e . Using the positive solution, the r0
escape velocity is ve
2GM e . r0
52. a 7, b 8, c 10 1 1 s a b c 7 8 10 12.5 2 2 K s ( s a)( s b)( s c)
ep M (1 e) so p ; 1 e e ep m(1 e) so p m 1 e e
50. M
12.5 5.5 4.5 2.5
27.81
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773.4375
Chapter 11: Analytic Geometry
1 53. y 4 cos x 5
60.
Ampl: 4 4; Period:
2 1
f ( x) a ( x 3) 2 8 10
5 a ((0) 3) 2 8 1 3 9a a 3 1 2 f ( x) ( x 3) 8 3
5
54. 2 cos 2 x cos x 1 0 (2 cos x 1)(cos x 1) 0 (2 cos x 1) 0 or (cos x 1) 0 1 cos x or cos x 1 2 x
f ( x ) a ( x h) 2 k
61. The figure is a trapezoid so we can find the length of each end and use the area formula for a trapezoid. The left side has a length or 3 (the yint) and the right side has a length of 1 f (8) (8) 3 7. The area of the trapezoid is 2 1 1 (b1 b2 )h (3 7)8 5(8) 40 sq units 2 2
5
, , 3 3
5 The solution set is , , . 3 3
55. For v 10i 24 j , v (10) 2 ( 24) 2 676 26 .
56.
Section 11.7
s r 7 14 r 12 12 24 14 ft r 7
1.
3 3;
2 4 2
2. plane curve; parameter 3. b
57. First find k. 1 e k (15) 2 1 ln 15k 2 1 ln k 2 0.04621 15
4. a 5. False; for example: x 2 cos t , y 3sin t define the same curve as in problem 3. 6. True 7. x(t ) 3t 2, y (t ) t 1, 0 t 4 x 3( y 1) 2 x 3y 3 2 x 3 y 1 x 3y 1 0
0.4 e 0.04621t ln 0.4 0.04621t ln 0.4 k 19.83 yrs 0.04621
58. Constant: none Decreasing: 1, 0
Increasing: 2, 1 and 0, 59.
2 5 k 6 5 k 12 k
12 5
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Section 11.7: Plane Curves and Parametric Equations 8. x(t ) t 3, y (t ) 2t 4, 0 t 2 y 2( x 3) 4 y 2x 6 4 y 2 x 10 2 x y 10 0
11. x(t ) t 2 4, y (t ) t 2 4, t y ( x 4) 4 y x 8 For t 0 the movement is to the left. For 0 t the movement is to the right.
9. x(t ) t 2, y (t ) t , t 0
12. x(t ) t 4, y (t ) t 4, t 0 y x 4 4 x 8
y x2
13. x(t ) 3t 2 , y (t ) t 1 , t x 3( y 1) 2
10. x(t ) 2t , y (t ) 4t , t 0 x2 y 4 2 x 2 , x 0 2
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Chapter 11: Analytic Geometry
17. x(t ) t , y (t ) t 3/2 , t 0
14. x(t ) 2t 4, y (t ) 4t 2 , t
2
2 x4 y 4 x 4 2
y x2 y x3
15. x(t ) 2et , y (t ) 1 et , t 0 x y 1 2 2y 2 x
18. x(t ) t 3/ 2 1, y (t ) t , t 0
x y2
3/ 2
1
x y3 1
19. x(t ) 2 cos t , y (t ) 3sin t , 0 t 2
16. x(t ) et , y (t ) e t , t 0
y x 1
3/ 2
x cos t 2
1 x
2
2
y sin t 3
x y 2 2 cos t sin t 1 2 3 x2 y 2 1 4 9
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Section 11.7: Plane Curves and Parametric Equations 20. x(t ) 2 cos t , y (t ) 3sin t , 0 t x cos t 2 2
22. x(t ) 2 cos t , y (t ) sin t , 0 t
y sin t 3
x cos t 2
2
x y 2 2 cos t sin t 1 2 3 x2 y 2 1 4 9
2
2
y0
x2 y2 1 4
y sin t 3
x y 1 4 9
4
sec2 t 1 tan 2 t
2
2
x 0, y 0
23. x(t ) sec t , y (t ) tan t , 0 t
x y 2 2 cos t sin t 1 2 3 2
y sin t
2 x 2 2 y cos t sin t 1 2
21. x(t ) 2 cos t , y (t ) 3sin t , t 0 x cos t 2
2
x2 1 y2 x2 y 2 1
1 x 2, 0 y 1
y0
24. x(t ) csc t , y (t ) cot t ,
t 4 2
csc2 t 1 cot 2 t x2 1 y2 x2 y 2 1
1 x 2, 0 y 1
y
2,1
1
(1, 0)
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1
x
Chapter 11: Analytic Geometry 36. x(t ) t 1, y (t ) t 2; 0 t 4
25. x(t ) sin 2 t , y (t ) cos 2 t , 0 t 2 2
2
sin t cos t 1 x y 1
37. x(t ) 3cos t , y (t ) 2sin t ; 0 t 2 38. x(t ) cos t 1 , y (t ) 4sin t 1 ; 0 t 2 2 2
39. Since the motion begins at (2, 0), we want x = 2 and y = 0 when t = 0. For the motion to be clockwise, we must have x positive and y negative initially. x 2 cos t , y 3sin t
26. x(t ) t 2 , y (t ) ln t , t 0 y ln x 1 y ln x 2
2
2
x 2 cos t , y 3sin t , 0 t 2
40. Since the motion begins at (0, 3), we want x = 0 and y = 3 when t = 0. For the motion to be counter-clockwise, we need x negative and y positive initially. x 2sin t , y 3cos t 2
27. x(t ) t , y (t ) 4t 1; x(t )
28. x(t ) t , y (t ) 8t 3; x(t )
x 2sin 2t , y 3cos 2t , 0 t 1
t 1 , y (t ) t 4
41. Since the motion begins at (0, 3), we want x = 0 and y = 3 when t = 0. For the motion to be clockwise, we need x positive and y positive initially. x 2sin t , y 3cos t
3t , y (t ) t 8
29. x(t ) t , y (t ) t 2 1
2
x(t ) t 3 , y (t ) t 6 1
x(t ) t 3 , y (t ) 2 t 6 1
42. Since the motion begins at (2, 0), we want x = 2 and y = 0 when t = 0. For the motion to be counter-clockwise, we need x positive and y positive initially. x 2 cos t , y 3sin t
x(t ) 3 t , y t
32. x(t ) t , y (t ) t 4 1
1 2
x 2sin 2t , y 3cos 2t , 0 t 1
30. x(t ) t , y (t ) 2 t 2 1
31. x(t ) t , y t 3
1 2
x(t ) t 3 , y (t ) t12 1 3
2
2 3 2 2 x 2 cos t , y 3sin t , 0 t 3 3 3
2
33. x(t ) t , y (t ) t 3 , t 0; x(t ) t , y (t ) t , t 0 2
34. x(t ) t , y (t ) t 2 , t 0; x(t ) t , y (t ) t , t 0
3
35. x(t ) t 2, y (t ) t ; 0 t 5 1278
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Section 11.7: Plane Curves and Parametric Equations
43. C1
44. C1
C2
C2
C3 C3
C4 C4
45. x(t ) t sin t , y (t ) t cos t 7
–7
11 –5
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Chapter 11: Analytic Geometry 46. x(t ) sin t cos t , y (t ) sin t cos t 2
c.
3
–3
–2
The maximum height occurs at the vertex of the quadratic function. b 50 t 1.5625 seconds 2a 2(16) Evaluate the function to find the maximum height: 16(1.5625) 2 50(1.5625) 6 45.0625 The maximum height is 45.0625 feet.
d. We use x 3 so that the line is not on top of the y-axis. 50
47. x(t ) 4sin t 2sin(2t ) y (t ) 4 cos t 2 cos(2t ) 3.5 –7.5
7.5
0 –6.5
50. a.
48. x(t ) 4sin t 2sin(2t ) y (t ) 4 cos t 2 cos(2t ) 6
–9
5
0
Use equations (1): x(t ) 40 cos 90º t 0 1 32 t 2 40sin 90º t 5 2 16t 2 40t 5
y (t ) 9
b. The ball is in the air until y 0 . Solve: 16t 2 40t 5 0
–6
49. a.
t
Use equations (1): x(t ) 50 cos 90º t 0
40 1920 0.12 or 2.62 32 The ball is in the air for about 2.62 seconds. (The negative solution is extraneous.)
1 32 t 2 50sin 90º t 6 2 16t 2 50t 6
y (t )
c.
b. The ball is in the air until y 0 . Solve: 16t 2 50t 6 0 t
40 402 4(16)(5) 2(16)
50 502 4(16)(6) 2(16)
50 2884 32 0.12 or 3.24 The ball is in the air for about 3.24 seconds. (The negative solution is extraneous.)
The maximum height occurs at the vertex of the quadratic function. b 40 t 1.25 seconds 2a 2(16) Evaluate the function to find the maximum height: 16(1.25) 2 40(1.25) 5 30 The maximum height is 30 feet.
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Section 11.7: Plane Curves and Parametric Equations d. We use x 3 so that the line is not on top of the y-axis. 40
0
52. Let y1 1 be the bus’s path and y2 5 be Jodi’s path. a. Bus: Using the hint, 1 x1 (t ) (3)t 2 1.5 t 2 2 y1 (t ) 1
5
0
Jodi: x2 (t ) 5(t 2) y2 (t ) 3
51. Let y1 1 be the train’s path and y2 5 be Bill’s path. a.
b. Jodi will catch the bus if x1 x2 .
Train: Using the hint, 1 x1 (t ) (2)t 2 t 2 2 y1 (t ) 1
1.5 t 2 5(t 2) 1.5 t 2 5t 10 1.5 t 2 5 t 10 0 Since b 2 4ac (5) 2 4(1.5)(10) , 25 60 35 0 the equation has no real solution. Thus, Jodi will not catch the bus.
Bill: x2 (t ) 5(t 5) y2 (t ) 5
b. Bill will catch the train if x1 x2 .
c.
t 2 5(t 5) t 2 5t 25 t 2 5 t 25 0 Since b 2 4ac (5) 2 4(1)(25) , 25 100 75 0 the equation has no real solution. Thus, Bill will not catch the train.
c.
y
53. a.
10 t7
5
y (t )
t 10 t 5
Use equations (1): x(t ) 145cos 20º t
50
Bill t 10
100
1 32 t 2 145sin 20º t 5 2
b. The ball is in the air until y 0 . Solve: 16 t 2 145sin 20º t 5 0
Train x
t
145sin 20º
145sin 20º 2 4(16)(5) 2( 16)
0.10 or 3.20 The ball is in the air for about 3.20 seconds. (The negative solution is extraneous.)
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Chapter 11: Analytic Geometry c.
Find the horizontal displacement: x 145cos 20º 3.20 436 feet
e.
x 125cos 40º t y 16 t 2 125sin 40º t 3 120
d. The maximum height occurs at the vertex of the quadratic function. b 145sin 20º t 1.55 seconds 2a 2(16) Evaluate the function to find the maximum height:
500
0 60
16 1.55 145sin 20º (1.55) 5 43.43 2
55. a.
The maximum height is about 43.43 feet.
Use equations (1): x(t ) 40 cos 45º t 20 2t 1 9.8 t 2 40sin 45º t 300 2 4.9t 2 20 2t 300
y (t )
e. 250
b. The ball is in the air until y 0 . Solve: 0
4.9t 2 20 2t 300 0
440 –50
54. a.
t
Use equations (1): x(t ) 125cos 40º t
b. The ball is in the air until y 0 . Solve:
c.
16 t 2 125sin 40º t 3 0 t
2
2( 4.9)
Find the horizontal displacement:
x 20 2 11.23 317.6 meters
125 sin 40º 2 4(16)(3)
2(16) 0.037 or 5.059
d. The maximum height occurs at the vertex of the quadratic function. 20 2 b t 2.89 seconds 2a 2( 4.9) Evaluate the function to find the maximum height:
The ball is in the air for about 5.059 sec. (The negative solution is extraneous.) c.
20 2 4( 4.9)(300)
5.45 or 11.23 The ball is in the air for about 11.23 seconds. (The negative solution is extraneous.)
y (t ) 16 t 2 125sin 40º t 3
125 sin 40º
20 2
Find the horizontal displacement: x 125cos 40º 5.059 484.41 feet
4.9 2.89 20 2 2.89 300 2
340.8 meters
d. The maximum height occurs at the vertex of the quadratic function. b 125sin 40º t 2.51 seconds 2a 2(16) Evaluate the function to find the maximum height:
e. 350
16 2.51 125 sin 40º 2.51 3 103.87 2
The maximum height is about 103.87 feet.
0
0
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400
Section 11.7: Plane Curves and Parametric Equations 56. a.
Use equations (1): x(t ) 40 cos 45º t 20 2t
57. a.
1 1 y (t ) 9.8 t 2 40sin 45º t 300 2 6 4.9 2 t 20 2 t 300 6
b. The ball is in the air until y 0 . Solve:
4.9 2 t 20 2t 300 0 6
2
d B
b. Let d represent the distance between the cars. Use the Pythagorean Theorem to find
Find the horizontal displacement:
(0,0)
P
4.9 20 2 20 2 4 (300) 6 t 4.9 2 6 8.51 or 43.15 The ball is in the air for about 43.15 seconds. (The negative solution is extraneous.)
c.
At t 0 , the Camry is 5 miles from the intersection (at (0, 0)) traveling east (along the x-axis) at 40 mph. Thus, x 40t 5 , y 0 , describes the position of the Camry as a function of time. The Impala, at t 0 , is 4 miles from the intersection traveling north (along the y-axis) at 30 mph. Thus, x 0 , y 30t 4 , describes the position of the Impala as a function of time.
the distance: d (40t 5) 2 (30t 4) 2 .
x 20 2 43.15 1220.5 meters
c.
d. The maximum height occurs at the vertex of the quadratic function. b 20 2 t 17.32 seconds 2a 4.9 2 6 Evaluate the function to find the maximum height: 4.9 17.32 2 20 2 17.32 300 6 544.9 meters
Note this is a function graph not a parametric graph.
d. The minimum distance between the cars is 0.2 miles and occurs at 0.128 hours (7.68 min). e.
e. 700
y
0
1300
5
–100
Impala
t 0.25
t 0.125
t 0.25 Camry
t0 t 0.125 t0
1283
Copyright © 2025 Pearson Education, Inc.
5
x
Chapter 11: Analytic Geometry 58. a.
At t 0 , the Boeing 747 is 550 miles from the intersection (at (0, 0)) traveling west (along the x-axis) at 600 mph. Thus, x 600t 550 , y 0 , describes the position of the Boeing 747 as a function of time. The Cessna, at t 0 , is 100 miles from the intersection traveling south (along the y-axis) at 120 mph. Thus, x 0 , y 120t 100 describes the position of the Cessna as a function of time.
c.
Note this is a function graph not a parametric graph.
b.
C
d. The minimum distance between the planes is 9.8 miles and occurs at 0.913 hours (54.8 min). e.
d
y
B
t 0 200
Let d represent the distance between the planes. Use the Pythagorean Theorem to find the distance: 2
t 56
t 56 t0
500
Boeing 747
2
d ( 600t 550) (120t 100) . Cessna
59. a.
1 We start with the parametric equations x(t ) v0 cos t and y (t ) gt 2 v0 sin t h . 2 2 We are given that 45 , g 32 ft/sec , and h 3 feet. This gives us 2 1 2 v0 t 3 v0 t ; y (t ) 32 t 2 v0 sin 45 t 3 16t 2 2 2 2 where v0 is the velocity at which the ball leaves the bat. x(t ) v0 cos 45 t
b.
Letting v0 90 miles/hr 132 ft/sec , we have 2 132 t 3 16t 2 66 2 t 3 2 66 2 33 2 2.9168 sec The height is maximized when t 2 16 16 y (t ) 16t 2
y 16 2.9168 66 2 2.9168 3 139.1 2
The maximum height of the ball is about 139.1 feet. c.
From part (b), the maximum height is reached after approximately 2.9168 seconds. 2 x(t ) v0 cos t 132 2.9168 272.25 2 The ball will reach its maximum height when it is approximately 272.25 feet from home plate. 1284
Copyright © 2025 Pearson Education, Inc.
x
(0,0)
Section 11.7: Plane Curves and Parametric Equations d. The ball will reach the left field fence when x 310 feet. x v0 cos t 66 2 t 310 66 2 t t
310
3.3213 sec 66 2 Thus, it will take about 3.3213 seconds to reach the left field fence.
y 16 3.3213 66 2 3.3213 3 136.5 2
Since the left field fence is 37 feet high, the ball will clear the Green Monster by 136.5 37 99.5 feet. 60. a.
x(t ) v0 cos t , t
y (t ) v0 sin t 16t 2
x v0 cos
x x y v0 sin 16 v0 cos v0 cos 16 y (tan ) x 2 x2 v0 cos 2
2
y is a quadratic function of x; its graph is a parabola with a
16 , b tan , and c 0 . v0 cos 2 2
y0
b.
v0 sin t 16t 2 0 t v0 sin 16t 0 t 0 or v0 sin 16t 0 v sin t 0 16
c.
v sin v0 sin(2 ) x(t ) v0 cos t v0 cos 0 feet 32 16
d.
x y
2
v0 cos t v0 sin t 16t 2 16t 2 v0 cos t v0 sin t 0 t 16t v0 cos v0 sin 0 t 0 or 16t v0 cos v0 sin 0 v0 sin v0 cos v0 sin cos 16 16 v0 At t sin cos : 16 t
2 v v x v0 cos 0 sin cos 0 cos sin cos 16 16
y
v0 2 cos 2 sin cos 16
(recall we want x y )
1285
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Chapter 11: Analytic Geometry
2
v 2 v2 x y 2 0 cos sin cos 0 2 cos sin cos 16 16 Note: since must be greater than 45 ( sin cos 0 ), thus absolute value is not needed. _________________________________________________________________________________________________ 2
2
61. x x2 x1 t x1 ,
62. a.
y y2 y1 t y1 , t
x(t ) cos3 t , y (t ) sin 3 t , 0 t 2
1
x x1 t x2 x1
–1.5
x x1 y y2 y1 y1 x2 x1 y y y y1 2 1 x x1 x2 x1
1.5
–1
This is the two-point form for the equation of a line. Its orientation is from x1, y1 to x2 , y2 .
b.
y
cos 2 t sin 2 t x1/ 3
2
1/ 3 2
x2 / 3 y2 / 3 1
63. The line connecting R,0 and x, y and slope m
y0 y so y m( x R) . The line intersects the x ( R) x R
circle so substitute this expression for y in the equation for the circle x 2 y 2 R 2 : x 2 m( x R ) R 2 2
x 2 m 2 ( x R )2 R 2 x 2 m 2 ( x 2 2 xR R 2 ) R 2 x 2 m 2 x 2 2m2 xR m 2 R 2 R 2 (1 m 2 ) x 2 2m 2 xR R 2 ( m2 1) 0
Using the quadratic formula gives x
2m 2 R (2m 2 R )2 4(1 m 2 ) R 2 (m 2 1) 2
2(1 m )
2m 2 R 4m 4 R 2 4 R 2 (m 4 1) 2(1 m 2 )
2m 2 R 4 R 2 2m 2 R 2 R m 2 R R 2(1 m 2 ) 2(1 m 2 ) (1 m 2 )
So x
m 2 R R R (m 2 1) m 2 R R R (1 m 2 ) The first value is for the fixed point, so the second R or x 2 2 (1 m ) (1 m ) (1 m 2 ) (1 m 2 )
value is for the other point. x ( m)
R (1 m 2 ) Using this for x in y m( x R) gives (1 m 2 )
R Rm 2 R Rm 2 R Rm 2 2 R 2mR R m y m m 2 2 1 1 1 m2 1 m2 m m 2mR , m . y ( m) 1 m2
Thus x( m)
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R (1 m 2 ) , (1 m 2 )
Section 11.7: Plane Curves and Parametric Equations
64. The slope of the line containing (1,1) and (x,y) is m
y 1 , so m x 1 y 1 or y mx m 1 . Substitute this x 1
expression for y in y x 2 . mx m 1 x 2 x 2 mx (m 1) 0 . Using the quadratic formula gives x
( m) ( m)2 4 1 m 1) m m 2 4m 4 m ( m 2)2 m (m 2) 2(1) 2 2 2
So x
m (m 2) m (m 2) 2m 2 1 or x m 1 2 2 2
The first value is from the fixed point, so the second value is from the other point. Thus, x(m) m 1 . Substituting this expression for x in y x 2 gives y (m 1)2 . Thus x(m) m 1 , y (m 1)2 , m .
65 – 66. Answers will vary.
69.
67. 3x 4 y 8 4 y 3 x 8 3 y x2 4
cos 1
3960 3960 248
0.345032 radians s1 r 3960(0.345032) 1366.3 miles Total distance s1 s2 x 68. y 2 cos 2 x sin , 2
1366.3 1366.3 2733 miles 0 x 2
70. d 2 cos(4 t ) a. Simple harmonic b. 2 meters c.
seconds 2
d.
2 oscillation/second
1287 Copyright © 2025 Pearson Education, Inc.
Chapter 11: Analytic Geometry 74. log5 (7 x) log 5 (3 x 5) log 5 (24 x) log5 (7 x )(3 x 5) log 5 (24 x)
71. Dividing: 11 2x 2 2x 1 4 x2 9 x 7
(7 x)(3x 5) 24 x 16 x 3x 2 35 24 x
4x 2x 2
3 x 2 8 x 35 0
11x 7
(3 x 7)( x 5) 0
11 11x 2
7 x , x 5 3
25 2 11 25 P( x) 2 x 2 2 2 x 1
We cannot use the negative value since we cannot take the log of a negative value. So the 7 solution set is 3 .
Thus, the oblique asymptote is y 2 x
72.
f ( x)
11 . 2
75.
1 x3
1 1 f ( x h) f ( x ) x h 3 x 3 h h x 3 x 3 h x h 3 x 3 h x 3 x 3 h 1 x h 3 x 3 h 1 h x h 3 x 3 h 1 x h 3 x 3
As h 0
1
1
x h 3 x 3 x 0 3 x 3
1 3 1 3 (2) 1 (0) 1 f (b) f (a) 4 4 20 ba 2 1 0 1 2 3 1 1 2 3 2 c 1 4 4 c2 3
c
1
x 32
1 2 3 2 2 2 2 2 1 4
2 3
2 3 3
The only answer that is in the interval 0, 2 is 2 3 c 3 .
73. cos 285º cos 240º 45º cos 240º cos 45º sin 240º sin 45º
4 3
2 6 6 4 2 1288 Copyright © 2025 Pearson Education, Inc.
Chapter 11 Review Exercises
Chapter 11 Review Exercises
a 2, b 8 2 2 . Find the value of c: c 2 a 2 b 2 2 8 10
1. y 2 16 x This is a parabola. a4 Vertex: (0, 0) Focus: (–4, 0) Directrix: x 4 2.
c 10 Center: (0, 0)
2, 0 , 2, 0 Foci: 10, 0 , 10, 0 Vertices:
x2 y2 1 25 This is a hyperbola. a 5, b 1 . Find the value of c: c 2 a 2 b 2 25 1 26
Asymptotes: y 2 x; y 2 x 6. x 2 4 x 2 y This is a parabola. Write in standard form: x2 4 x 4 2 y 4
c 26 Center: (0, 0) Vertices: (5, 0), (–5, 0)
Foci:
26, 0 , 26, 0
Asymptotes: y 3.
( x 2) 2 2( y 2) 1 a 2 Vertex: (2, –2) 3 Focus: 2, 2 5 Directrix: y 2
1 1 x; y x 5 5
y 2 x2 1 25 16 This is an ellipse. a 5, b 4 . Find the value of c: c 2 a 2 b 2 25 16 9 c3 Center: (0, 0) Vertices: (0, 5), (0, –5) Foci: (0, 3), (0, –3)
7. y 2 4 y 4 x 2 8 x 4 This is a hyperbola. Write in standard form: ( y 2 4 y 4) 4( x 2 2 x 1) 4 4 4 ( y 2) 2 4( x 1) 2 4 ( y 2) 2 ( x 1) 2 1 4 1 a 2, b 1 . Find the value of c:
4. x 2 4 y 4 This is a parabola. Write in standard form: x2 4 y 4 x 2 4( y 1) a 1 Vertex: (0, 1) Focus: (0, 0)
c2 a 2 b2 4 1 5 c 5 Center: (1, 2) Vertices: (1, 0), (1, 4)
Foci: 1, 2 5 , 1, 2 5 Directrix:
y2
Asymptotes: y 2 2( x 1); y 2 2( x 1)
5. 4 x 2 y 2 8 This is a hyperbola. Write in standard form: x2 y 2 1 2 8 1289 Copyright © 2025 Pearson Education, Inc.
Chapter 11: Analytic Geometry
8. 4 x 2 9 y 2 16 x 18 y 11 This is an ellipse. Write in standard form: 4 x 2 9 y 2 16 x 18 y 11 4( x 2 4 x 4) 9( y 2 2 y 1) 11 16 9 2
2
2
2
4( x 2) 9( y 1) 36 ( x 2) ( y 1) 1 9 4 a 3, b 2 . Find the value of c:
11. Parabola: The focus is (–2, 0) and the directrix is x 2 . The vertex is (0, 0). a = 2 and since (–2, 0) is to the left of (0, 0), the parabola opens to the left. The equation of the parabola is: y 2 4ax y2 4 2 x y 2 8x
c 2 a 2 b2 9 4 5 c 5 Center: (2, 1); Vertices: (–1, 1), (5, 1)
Foci: 2 5, 1 , 2 5, 1
9. 4 x 2 16 x 16 y 32 0 This is a parabola. Write in standard form: 4( x 2 4 x 4) 16 y 32 16
12. Hyperbola: Center: (0, 0); Focus: (0, 4); Vertex: (0, –2); Transverse axis is the y-axis; a 2; c 4 . Find b: b 2 c 2 a 2 16 4 12
4( x 2) 2 16( y 1) ( x 2) 2 4( y 1)
a 1 Vertex: (2, –1); Focus: 2, 2 ;
b 12 2 3
Directrix: y 0
Write the equation:
10. 9 x 2 4 y 2 18 x 8 y 23 This is an ellipse. Write in standard form: 9( x 2 2 x 1) 4( y 2 2 y 1) 23 9 4 9( x 1) 2 4( y 1) 2 36 ( x 1)2 ( y 1) 2 1 4 9 a 3, b 2 . Find the value of c: c2 a 2 b2 9 4 5
c 5 Center: (1, –1) Vertices: (1, –4), (1, 2)
Foci:
1, 1 5 , 1, 1 5
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y 2 x2 1 4 12
Chapter 11 Review Exercises 13. Ellipse: Foci: (–3, 0), (3, 0); Vertex: (4, 0); Center: (0, 0); Major axis is the x-axis; a 4; c 3 . Find b:
Write the equation:
( x 2) 2 ( y 3) 2 1 1 3
b 2 a 2 c 2 16 9 7 b 7
Write the equation:
x2 y 2 1 16 7
16. Ellipse: Foci: (–4, 2), (–4, 8); Vertex: (–4, 10); Center: (–4, 5); Major axis is parallel to the yaxis; a 5; c 3 . Find b: b 2 a 2 c 2 25 9 16
14. Parabola: The focus is (2, –4) and the vertex is (2, –3). Both lie on the vertical line x 2 . a = 1 and since (2, –4) is below (2, –3), the parabola opens down. The equation of the parabola is: ( x h) 2 4a ( y k )
b4 2
Write the equation:
( x 4) ( y 5) 2 1 16 25
( x 2) 2 4 1 ( y (3)) ( x 2) 2 4( y 3)
17. Hyperbola: Center: (–1, 2); a 3; c 4 ; Transverse axis parallel to the x-axis; Find b: b 2 c 2 a 2 16 9 7 b 7
Write the equation:
15. Hyperbola: Center: (–2, –3); Focus: (–4, –3); Vertex: (–3, –3); Transverse axis is parallel to the x-axis; a 1; c 2 . Find b: b2 c 2 a 2 4 1 3 b 3
1291 Copyright © 2025 Pearson Education, Inc.
( x 1) 2 ( y 2) 2 1 9 7
Chapter 11: Analytic Geometry 18. Hyperbola: Vertices: (0, 1), (6, 1); Asymptote: 3 y 2 x 9 0 ; Center: (3, 1); Transverse axis is parallel to the x-axis; a 3 ; The slope of the 2 Find b: asymptote is ; 3 b b 2 3b 6 b 2 a 3 3
Write the equation:
20. x 2 2 y 2 4 x 8 y 2 0 A 1 and C 2; AC (1)(2) 2 . Since AC 0 and A C , the equation defines an ellipse. 21. 9 x 2 12 xy 4 y 2 8 x 12 y 0 A 9, B 12, C 4
( x 3) 2 ( y 1) 2 1 9 4
B 2 4 AC (12) 2 4(9)(4) 0 Parabola 22. 4 x 2 10 xy 4 y 2 9 0 A 4, B 10, C 4 B 2 4 AC 102 4(4)(4) 36 0 Hyperbola
23. x 2 2 xy 3 y 2 2 x 4 y 1 0 A 1, B 2, C 3 B 2 4 AC ( 2) 2 4(1)(3) 8 0 Ellipse
19. y 2 4 x 3 y 8 0 A 0 and C 1; AC (0)(1) 0 . Since AC 0 , the equation defines a parabola.
_________________________________________________________________________________________________ 24. 2 x 2 5 xy 2 y 2
9 0 2
AC 22 0 2 B 5 2 4 2 2 2 x x 'cos y 'sin x ' y' x ' y ' 2 2 2 2 2 2 y x 'sin y 'cos x ' y' x ' y ' 2 2 2 A 2, B 5, and C 2; cot 2
2
2
2 2 2 2 9 2 x ' y ' 5 x ' y ' x ' y ' 2 x ' y ' 0 2 2 2 2 2 5 9 x '2 2 x ' y ' y '2 x '2 ' y '2 x '2 2 x ' y ' y '2 0 2 2 2 2 x y 9 2 1 2 9 ' ' x ' y ' 9 x '2 y '2 9 1 2 2 2 1 9
Hyperbola; center at (0, 0), transverse axis is the x'-axis, vertices at x ', y ' 1, 0 ; foci at x ', y ' 10, 0 ;
1292 Copyright © 2025 Pearson Education, Inc.
Chapter 11 Review Exercises
asymptotes: y ' 3x ' .
25. 6 x 2 4 xy 9 y 2 20 0 A 6, B 4, and C 9; cot 2
AC 69 3 3 cos 2 B 4 4 5
3 3 1 5 1 5 4 2 5 1 5 63.4º sin ; cos 2 5 5 2 5 5 5 2 5 5 x ' y' x ' 2 y ' 5 5 5 2 5 5 5 y x 'sin y 'cos x ' y' 2 x ' y ' 5 5 5
x x 'cos y 'sin
2
2
5 5 5 5 6 x ' 2 y ' 4 x ' 2 y ' 2 x ' y ' 9 2 x ' y ' 20 0 5 5 5 5 6 2 4 9 2 2 2 2 2 x ' 4 x ' y ' 4 y ' 2 x ' 3 x ' y ' 2 y ' 4 x ' 4 x ' y ' y ' 20 0 5 5 5 6 2 24 24 2 8 2 12 8 2 36 2 36 9 x' x ' y ' y ' x ' x ' y ' y ' x ' x ' y ' y '2 20 5 5 5 5 5 5 5 5 5 2 2 x ' y ' 10 x '2 5 y '2 20 1 2 4
Ellipse; center at the origin, major axis is the y'-axis, vertices at x ', y ' 0, 2 ; foci at x ', y ' 0, 2 .
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Chapter 11: Analytic Geometry
26. 4 x 2 12 xy 9 y 2 12 x 8 y 0 A 4, B 12, and C 9; cot 2
5 AC 49 5 cos 2 12 12 13 B
5 5 1 1 4 2 13 9 3 13 13 13 sin ; cos 33.7º 2 13 13 2 13 13 x x 'cos y 'sin
3 13 2 13 13 x ' y' 3x ' 2 y ' 13 13 13
y x 'sin y 'cos
2 13 3 13 13 x ' y' 2 x ' 3 y ' 13 13 13
2
13 13 13 13 4 3x ' 2 y ' 12 3x ' 2 y ' 2 x ' 3 y ' 9 2 x ' 3 y ' 13 13 13 13
2
13 13 12 3x ' 2 y ' 8 2 x ' 3 y ' 0 13 13 4 12 9 9 x '2 12 x ' y ' 4 y '2 6 x '2 5 x ' y ' 6 y '2 4 x '2 12 x ' y ' 9 y '2 13 13 13
36 13 24 13 16 13 24 13 x ' y ' x ' y' 0 13 13 13 13 36 2 48 16 2 72 2 60 72 2 36 2 108 81 2 x' x ' y ' y' x ' x ' y ' y' x' x ' y ' y ' 4 13x ' 0 13 13 13 13 13 13 13 13 13
13 y '2 4 13 x ' 0
y '2
4 13 x' 13
Parabola; vertex at the origin, focus at x ', y ' 13 , 0 . 13
4 1 cos ep 4, e 1, p 4 Parabola; directrix is perpendicular to the polar axis 4 units to the left of the pole; vertex is 2, .
27. r
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Chapter 11 Review Exercises
6 3 2 sin 1 1 sin 2 1 ep 3, e , p 6 2 Ellipse; directrix is parallel to the polar axis 6 units below the pole; vertices are 3 6, and 2, . Center at 2, 2 ; other 2 2
28. r
focus at 4, . 2
4 1 cos r r cos 4
30. r
r 4 r cos r (4 r cos ) 2 2
x 2 y 2 (4 x) 2 x 2 y 2 16 8 x x 2 y 2 8 x 16 0
31. r
8 4 8cos 4r 8r cos 8 4r 8 8r cos r 2 2r cos r 2 (2 2r cos ) 2 x 2 y 2 (2 2 x )2 x2 y 2 4 8x 4 x2
3x 2 y 2 8 x 4 0
8 2 4 8cos 1 2 cos ep 2, e 2, p 1 Hyperbola; directrix is perpendicular to the polar axis 1 unit to the right of the pole; vertices are 2 4 , 0 and 2, . Center at 3 , 0 ; other 3 8 focus at , . 3
29. r
32. x(t ) 4t 2, y (t ) 1 t , t
x 4(1 y ) 2 x 4 4y 2 x 4y 2
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Chapter 11: Analytic Geometry 33. x(t ) 3sin t , y (t ) 4 cos t 2, 0 t 2 x y2 sin t , cos t 3 4 sin 2 t cos 2 t 1 2
2
x y2 1 3 4
36.
x y cos t ; sin t 4 3 2 2 4 4 2 4 2 x y cos t ; sin t 4 2 3 2 x(t ) 4 cos t ; y (t ) 3sin t , 0 t 4 2 2
x 2 ( y 2) 2 1 9 16
37. Write the equation in standard form: x2 y 2 4 x 2 9 y 2 36 1 9 4 The center of the ellipse is (0, 0). The major axis is the x-axis. a 3; b 2; c 2 a 2 b 2 9 4 5 c 5. For the ellipse: Vertices: (–3, 0), (3, 0);
Foci: 34. x(t ) sec 2 t , y (t ) tan 2 t , 0 t
4
5, 0 , 5, 0
For the hyperbola: Foci: (–3, 0), (3, 0); Vertices:
5, 0 , 5, 0 ;
Center: (0, 0) a 5; c 3; b2 c 2 a 2 9 5 4 b 2
The equation of the hyperbola is:
38. Let ( x, y ) be any point in the collection of points. The distance from
tan 2 t 1 sec 2 t y 1 x
35. Answers will vary. One example: y 2 x 4 x(t ) t , y (t ) 2t 4 x(t )
4t , y (t ) t 2
x2 y 2 1 5 4
( x, y ) to (3, 0) ( x 3) 2 y 2 . The distance from 16 16 ( x, y ) to the line x is x . 3 3 Relating the distances, we have: 3 16 ( x 3) 2 y 2 x 4 3 ( x 3) 2 y 2 x2 6 x 9 y 2
1296 Copyright © 2025 Pearson Education, Inc.
9 16 x 16 3
2
9 2 32 256 x x 16 3 9
Chapter 11 Review Exercises
16 x 2 96 x 144 16 y 2 9 x 2 96 x 256 7 x 2 16 y 2 112 7 x 2 16 y 2 1 112 112 x2 y2 1 16 7 The set of points is an ellipse.
39. Locate the parabola so that the vertex is at (0, 0) and opens up. It then has the equation: x 2 4ay . Since the light source is located at the focus and is 1 foot from the base, a 1 . Thus, x 2 4 y . The width of the opening is 2, so the point (1, y) is located on the parabola. Solve for y: 12 4 y 1 4 y y 0.25 feet The mirror is 0.25 feet, or 3 inches, deep. 40. Place the semi-elliptical arch so that the x-axis coincides with the water and the y-axis passes through the center of the arch. Since the bridge has a span of 60 feet, the length of the major axis is 60, or 2a 60 or a 30 . The maximum height of the bridge is 20 feet, so b 20 . The x2 y2 1. 900 400 The height 5 feet from the center: 52 y2 1 900 400 25 y2 1 400 900 875 y 2 400 y 19.72 feet 900 The height 10 feet from the center: 102 y2 1 900 400 y2 100 1 400 900 800 y 2 400 y 18.86 feet 900
equation is:
The height 20 feet from the center: 202 y 2 1 900 400 y2 400 1 400 900 500 y 2 400 y 14.91 feet 900 41. First note that all points where an explosion could take place, such that the time difference would be the same as that for the first detonation, would form a hyperbola with A and B as the foci. Start with a diagram: N
D2
(1000, 0)
D1
A
(0, 0) ( a, 0)
(1000, y)
(1000, 0) B
2000 feet
Since A and B are the foci, we have 2c 2000 c 1000 Since D1 is on the transverse axis and is on the hyperbola, then it must be a vertex of the hyperbola. Since it is 200 feet from B, we have a 800 . Finally, b 2 c 2 a 2 10002 8002 360, 000 Thus, the equation of the hyperbola is y2 x2 1 640, 000 360, 000 The point 1000, y needs to lie on the graph of the hyperbola. Thus, we have
1000 2 640, 000
y2 1 360, 000
y2 9 360, 000 16 y 2 202,500
y 450 The second explosion should be set off 450 feet due north of point B.
1297 Copyright © 2025 Pearson Education, Inc.
Chapter 11: Analytic Geometry 42. a.
Train: x1 3 t 2 ; 2 Let y1 1 for plotting convenience. Mary: x2 6(t 2);
Evaluate the function to find the maximum height: 16(1.43) 2 80 sin 35º (1.43) 6 38.9 ft d. Find the horizontal displacement: x 80 cos 35º (2.99) 196 feet
Let y2 3 for plotting convenience.
b.
Mary will catch the train if x1 x2 . 3 t 2 6(t 2) 2 3 t 2 6t 12 2 t2 4t 8 0 Since b 2 4ac ( 4) 2 4(1)(8) 16 32 16 0 the equation has no real solution. Thus, Mary will not catch the train.
c.
50
50
44. Answers will vary.
Chapter 11 Test t 3
t2
t4
Mary
t3
t2
10
1.
t4
20
Train
b. The ball is in the air until y 0 . Solve: 16 t 80sin 35º t 6 0 2
t
y2 1 4 9 Rewriting the equation as
2
Use equations (1) in Section 11.7. x 80 cos 35º t
80 sin 35º
x 12
x 1 y 0 2 1 , we see that this is the
x
1 y (32) t 2 80sin 35º t 6 2
80 sin 35º 2 4(16)(6) 2(16)
45.886 2489.54 32 0.1253 or 2.9932 The ball is in the air for about 2.9932 seconds. (The negative solution is extraneous.)
c.
e.
y 5
43. a.
(or roughly 65.3 yards)
The maximum height occurs at the vertex of the quadratic function. b 80sin 35º t 1.43 seconds 2a 2(16)
22 32 equation of a hyperbola in the form
x h 2 y k 2
1 . Therefore, we have a2 b2 h 1 , k 0 , a 2 , and b 3 . Since a 2 4
and b 2 9 , we get c 2 a 2 b 2 4 9 13 , or c 13 . The center is at 1, 0 and the transverse axis is the x-axis. The vertices are at h a, k 1 2, 0 , or 3, 0 and 1, 0 .
The foci are at h c, k 1 13, 0 , or
1 13, 0 and 1 13, 0 . The asymptotes are y 0 y
3 3 x 1 , or y x 1 and 2 2
3 x 1 . 2
1298 Copyright © 2025 Pearson Education, Inc.
Chapter 11 Test
focus is a 1.5 . Because the focus lies above the vertex, we know the parabola opens upward. As a result, the form of the equation is
2. 8 y x 1 4 2
Rewriting gives ( x 1) 2 8 y 4
x h 2 4a y k where h, k 1,3 and a 1.5 . Therefore,
1 ( x 1) 2 8 y 2 1 x 1 4(2) y 2 This is the equation of a parabola in the form 2
x h 2 4a y k . Therefore, the axis of symmetry is parallel to the y-axis and we have 1 h, k 1, and a 2 . The vertex is at 2 1 h, k 1, , the axis of symmetry is x 1 , 2 1 3 the focus is at h, k a 1, 2 1, , 2 2 and the directrix is given by the line y k a ,
the equation is
x 12 4 1.5 y 3 x 12 6 y 3 The points h 2a, k , that is 4, 4.5 and 2, 4.5 , define the lattice rectum; the line y 1.5 is the directrix.
5 or y . 2
3. 2 x 2 3 y 2 4 x 6 y 13 Rewrite the equation by completing the square in x and y. 2 x 2 3 y 2 4 x 6 y 13
2 x 2 4 x 3 y 2 6 y 13
vertex 0, 4 is a 4 . Then,
2
b 2 a 2 c 2 42 32 16 9 7 The form of the equation is
2 x 1 3 y 1 18 2
2
x h 2 y k 2
x 1 y 1 1 2
2
9 6 This is the equation of an ellipse with center at 1,1 and major axis parallel to the x-axis.
Since a 2 9 and b 2 6 , we have a 3 , b 6 , and c 2 a 2 b 2 9 6 3 , or
Since the center, focus, and vertex all lie on the line x 0 , the major axis is the y-axis. The distance from the center 0, 0 to a focus 0,3 is c 3 . The distance from the center 0, 0 to a
2 x 2 x 1 3 y 2 y 1 13 2 3 2 x 2 2 x 3 y 2 2 y 13
2
5. The center is h, k 0, 0 so h 0 and k 0 .
c 3 . The foci are h c, k 1 3,1 or
1 3,1 and 1 3,1 . The vertices are at h a, k 1 3,1 , or 4,1 and 2,1 . 4. The vertex 1,3 and the focus 1, 4.5 both
lie on the vertical line x 1 (the axis of symmetry). The distance a from the vertex to the
1 b2 a2 where h 0 , k 0 , a 4 , and b 7 . Thus, we get x2 y 2 1 7 16 To graph the equation, we use the center h, k 0, 0 to locate the vertices. The major
axis is the y-axis, so the vertices are a 4 units above and below the center. Therefore, the vertices are V1 0, 4 and V2 0, 4 . Since c 3 and the major axis is the y-axis, the foci are 3 units above and below the center. Therefore, the foci are F1 0,3 and
1299 Copyright © 2025 Pearson Education, Inc.
Chapter 11: Analytic Geometry
F2 0, 3 . Finally, we use the value b 7
and F2 2, 2 2 3 . The asymptotes are given
to find the two points left and right of the center:
by the lines a y k x h . Therefore, the asymptotes b are 2 y2 x 2 2 2
7, 0 and 7, 0 .
y
2 x 2 2 2
6. The center h, k 2, 2 and vertex 2, 4 both
lie on the line x 2 , the transverse axis is parallel to the y-axis. The distance from the center 2, 2 to the vertex 2, 4 is a 2 , so the other vertex must be 2, 0 . The form of the equation is
y k 2 x h 2
7. 2 x 2 5 xy 3 y 2 3 x 7 0 is in the form
1 a2 b2 where h 2 , k 2 , and a 2 . This gives us
Ax 2 Bxy Cy 2 Dx Ey F 0 where A 2 , B 5 , and C 3 .
y 2 2 x 2 2
B 2 4 AC 52 4 2 3
1 4 b2 Since the graph contains the point
25 24
x, y 2 10,5 , we can use this point to
1 Since B 2 4 AC 0 , the equation defines a hyperbola.
determine the value for b.
5 2 2 4
2 10 2 1 2
8. 3 x 2 xy 2 y 2 3 y 1 0 is in the form
b2 9 10 1 4 b2 5 10 4 b2
Ax 2 Bxy Cy 2 Dx Ey F 0 where A 3 , B 1 , and C 2 . B 2 4 AC 1 4 3 2 2
1 24
b2 8
23 Since B 4 AC 0 , the equation defines an ellipse. 2
b2 2 Therefore, the equation becomes
y 2 2 x 2 2 4
8
1
Since c 2 a 2 b 2 4 8 12 , the distance from the center to either focus is c 2 3 . Therefore, the foci are c 2 3 units above and
below the center. The foci are F1 2, 2 2 3
1300 Copyright © 2025 Pearson Education, Inc.
Chapter 11 Test
9. x 2 6 xy 9 y 2 2 x 3 y 2 0 is in the form 2
2
Ax Bxy Cy Dx Ey F 0 where A 1 , B 6 , and C 9 .
B 2 4 AC 6 4 1 9 2
36 36
2
0 Since B 2 4 AC 0 , the equation defines a parabola.
10. 41x 2 24 xy 34 y 2 25 0 Substituting x x 'cos y 'sin and y x 'sin y 'cos transforms the equation into one that represents a rotation through an angle . To eliminate the x ' y ' term in the new
equation, we need cot 2
AC . That is, we B
need to solve 41 34 cot 2 24 7 cot 2 24 Since cot 2 0 , we may choose
4 4 4 3 3 3 41 x ' y ' 24 x ' y ' x ' y ' 5 5 5 5 5 5 2
3 4 34 x ' y ' 25 5 5 Multiply both sides by 25 and expand to obtain
41 9 x '2 24 x ' y ' 16 y '2 24 12 x '2 7 x ' y ' 12 y '2
2
34 16 x ' 24 x ' y ' 9 y '
2
625
625 x '2 1250 y '2 625 x '2 2 y ' 2 1 x 2 y 2 1 1 1 2 1 2 2 2 This is the equation of an ellipse with center at 0, 0 in the x ' y ' plane . The vertices are at Thus: a 1 1 and b
90 2 180 , or 45 90 .
1, 0 and 1, 0 in the x ' y ' plane .
( 7, 24)
1 1 2 c 2 2 2 2 The foci are located at , 0 in the 2 x ' y ' plane . In summary: c2 a 2 b2 1
7 2 24 2 25
24
3 4 4 3 x ' y ' and y x ' y ' 5 5 5 5 Substituting these values into the original equation and simplifying, we obtain 41x 2 24 xy 34 y 2 25 0 x
2 7
xy plane
xy plane
center
(0, 0)
(0, 0)
We have cot 2 7 so it follows that 24 7 cos 2 . 25
vertices
(1, 0)
7 1 1 cos 2 25 16 4 sin 2 2 25 5
minor axis
2 0, 2
cos
1 cos 2 2
7 1 25 2
9 3 25 5
intercepts
foci
3 With these values, the rotation formulas are
cos 1 53.13 5
1301 Copyright © 2025 Pearson Education, Inc.
2 , 0 2
3 4 3 4 , , , 5 5 5 5 2 2 3 2 , , 5 10 2 2 3 2 , 5 10 3 2 2 2 , , 5 10 3 2 2 2 10 , 5
Chapter 11: Analytic Geometry
The graph is given below.
11. r
ep 3 1 e cos 1 2 cos
3 . Since e 1 , this is 2 the equation of a hyperbola. 3 r 1 2 cos r 1 2 cos 3
Therefore, e 2 and p
r 2r cos 3 Since x r cos and x 2 y 2 r 2 , we get r 2r cos 3
r 2r cos 3 r 2r cos 3 2
x y 2 x 3 2
2
2
To find the rectangular equation for the curve, we need to eliminate the variable t from the equations. We can start by solving equation (1) for t. x 3t 2 3t x 2 x2 t 3 Substituting this result for t into equation (2) gives x2 y 1 , 2 x 25 3 13. We can draw the parabola used to form the reflector on a rectangular coordinate system so that the vertex of the parabola is at the origin and its focus is on the positive y-axis. y
2
2
( 2, 1.5)
x 2 y 2 4 x 2 12 x 9
ft (2, 1.5)
3x 2 12 x y 2 9
3 x 4 x 4 y 9 12
2
3 x 2 4 x y 2 9 2
3 x 2 y2 3 2
x 2
1
12.
y2 1 3
x
0 x 3 0 2 2 1
x 3 1 2 1
4
x 3 4 2 10
9
x 3 9 2 25
x
The form of the equation of the parabola is x 2 4ay and its focus is at 0, a . Since the point 2,1.5 is on the graph, we have 22 4a 1.5
x(t ) 3t 2 (1) y (t ) 1 t (2) t
2
2
2
2
0
x, y y 1 0 1 2,1 y 1 1 0 1, 0 y 1 4 1 10, 1 y 1 9 2 25, 2 y
4 6a a2 3 The microphone should be located 23 feet (or 8
inches) from the base of the reflector, along its axis of symmetry.
1302 Copyright © 2025 Pearson Education, Inc.
Chapter 11 Cumulative Review y
9 x3 12 x 2 11x 2 .
ft
2
( 2, 1.5)
2
Using synthetic division on the quotient and x = 2: 2 9 12 11 2 18 12 2
x
0
2
2
F 0, 23
Thus, f ( x) ( x 5) 9 x3 12 x 2 11x 2 .
(2, 1.5)
9 6 1 0 Since the remainder is 0, 2 is a zero for f. So x 2 is a factor; thus,
.
f ( x) ( x 5) x 2 9 x 2 6 x 1
( x 5) x 2 3 x 1 3 x 1
Chapter 11 Cumulative Review
1.
1 is also a zero for f (with 3 multiplicity 2). Solution set: 5, 1 , 2 . 3
Therefore, x
f x h f x h
3 x +h +5 x h 2 3 x +5 x 2 2
2
3 x 2 xh h
2
2
6 x x2
3.
0 x2 x 6
h
x2 x 6 0
5 x 5h 2 3x 2 5 x 2
x 3 x 2 0
h 2 2 3 x 6 xh 3h 5 x 5h 2 3 x 2 5 x 2 h 6 xh 3h 2 5h 6 x 3h 5 h
2. 9 x 4 +33 x3 71x 2 57 x 10=0 There are at most 4 real zeros. Possible rational zeros: p 1, 2, 5, 10; q 1, 3, 9; p 1 1 2 2 1, , , 2, , , 5, 3 9 3 9 q 5 5 10 10 , , 10, , 3 9 3 9 Graphing y1 9 x 4 +33x3 71x 2 57 x 10 indicates that there appear to be zeros at x = –5 and at x = 2. Using synthetic division with x = –5: 5 9 33 71 57 10
f ( x) x 2 x 6 x 3, x 2 are the zeros of f .
Interval Test Value
, 3
3, 2
2,
4
0
3
Value of f
6
6
6
Conclusion
Positive
Negative Positive
The solution set is x 3 x 2 , or 3, 2 . 4.
f x 3x 2
a.
Domain: , ; Range: 2, .
b.
f x 3x 2 y 3x 2 x 3y 2 x2 3
Inverse
y
log3 x 2 y f 1 x log3 x 2
55
10
Domain of f 1 = range of f = 2, .
9 12 11 2
0
Range of f 1 = domain of f = , .
45
60
Since the remainder is 0, –5 is a zero for f. So x ( 5) x 5 is a factor. The other factor is the quotient: 1303 Copyright © 2025 Pearson Education, Inc.
Chapter 11: Analytic Geometry
5.
f x log 4 x 2
a.
x h 2 4a y k
f x log 4 x 2 2
x 12 4ay 0 12 4a 2
x 2 42 x 2 16
1 8a
x 18 The solution set is 18 .
b.
1 8 1 2 x 1 2 y or y 2 x 12 a
f x log 4 x 2 2 x 2 42
and x 2 0
x 2 16
and x 2
x 18
and x 2
e.
This graph is a hyperbola with center 0, 0 and vertices 0, 1 , containing the point
3, 2 . y k 2 x h 2
2 x 18
2,18
a2
6. a.
This graph is a line containing points 0, 2 and 1, 0 . y 0 2 2 2 x 1 0 1 using y y1 m x x1
2 2 32 1
y 0 2 x 1 y 2 x 2 or 2 x y 2 0 b. This graph is a circle with center point (2, 0) and radius 2.
x h 2 y k 2 r 2 x 2 2 y 0 2 22 x 2 2 y 2 4
0, 2 . a2
f.
b2
x 0 2 y 0 2 32
22
x2 y2 1 9 4
d. This graph is a parabola with vertex 1, 0
and y-intercept 0, 2 .
1 b2 4 9 1 1 b2 9 4 2 1 b 9 3 2 b 3b 2 9
This is the graph of an exponential function with y-intercept 0,1 , containing the point
1, 4 .
1 1
b2 3 The equation of the hyperbola is: y 2 x2 1 1 3
This graph is an ellipse with center point (0, 0); vertices 3, 0 and y-intercepts
x h 2 y k 2
1
y 2 x2 1 1 b2
slope
c.
b2
y A bx
y-intercept 0,1 1 A b0 A 1 A 1 point 1, 4 4 b1 b Therefore, y 4 x .
1304 Copyright © 2025 Pearson Education, Inc.
Chapter 11 Projects 7. sin 2 0.5 2k 6 5 or 2 2k 6 where k is any integer. 2
k 12 , 5 k 12
6
.
9. Using rectangular coordinates, the circle with center point (0, 4) and radius 4 has the equation:
x h 2 y k 2 r 2 x 0 2 y 4 2 42 2
11. cot 2 1, where 0o 90o 2
8. The line containing point (0,0), making an angle of 30o with the positive x-axis has polar
equation:
The domain is 3 k , where k is any integer . x x 4
k k , where k is any 4 8 2
integer. On the interval 0o 90o , the solution is 22.5o . 8 x tan t 5
12. x(t ) 5 tan t sec2 t 1 tan 2 t
2
2
y (t ) 5sec 2 t 5 1 tan 2 t x x 5 1 5 5 5
2
x y 8 y 16 16 x2 y 2 8 y 0 Converting to polar coordinates: r 2 8r sin 0
x2 5. 5
The rectangular equation is y
r 2 8r sin r 8sin
Chapter 11 Projects Project I – Internet-based Project Project II 1. Figure: 37.1 10 6
Sun
(0, 0)
3 10. f x sin x cos x f will be defined provided sin x cos x 0 . sin x cos x 0
4458.0 10 6
4495.1 10 6
c 37.2 106 b2 a 2 c 2
sin x cos x
b 2 4495.1 106
sin x 1 cos x tan x 1 3 k , k is any integer x 4
b 4494.9 106 x2
37.110 2
6 2
y2 1 (4495.1 x 106 )2 (4494.9 x 106 ) 2
1305 Copyright © 2025 Pearson Education, Inc.
Chapter 11: Analytic Geometry
The focal length is 0.03125 m. 1 1 z x2 y2 8 8
2. Figure: 1467.7 10 6 4445.8 10 6
2.
Sun
(0, 0)
5913.5 10 6
Target
x
77381.2 106 4445.8 106 11827 106
a 0.5 11827 106 5913.5 106 c 1467.7 10
6
2
b 5913.5 10
1467.7 10
6 2
6 2
b 5728.5 106 x2 (5913.5 x 106 ) 2
y2 (5728.5 x 106 ) 2
1
4. Shift Pluto's distance Neptune's distance 1467.7 106 37.1 106 1430.6 106 ( x 1430.6 x 106 ) 2 (5913.5 x 106 ) 2
y2 (5728.5 x 106 ) 2
1
5. Yes. One must adjust the scale accordingly to see it. 6.
6
0
9.551
11.78
0.5
1.950
0
9.948
5.65
0.125
0.979
T3
0
9.928
5.90
0.125
1.021
T4
0
9.708
11.89
0.5
2.000
T5
9.510
9.722
11.99
11.31
0
T6
9.865
9.917
5.85
12.165
0
T7
9.875
9.925
5.78
12.189
0
T8
9.350
9.551
11.78
10.928
0
1 2 1 2 x y , y = Rsinθ, x = Rcosθ 8 8
Project IV
Figure 1
(0, 0)
x1 , 70
6
x2 , 70 280 ft
280 ft ( 525, 350)
(525, 350) 1050 ft
1.
Project III 1. As an example, T1 will be used. (Note that any of the targets will yield the same result.) z 4ax 2 4ay 2
2
x 4ay (525) 2 4a(350) 272625 1400a a 196.875 x 2 787.5 y
0.5 16a a
y
T1
6
7. No, The timing is different. They do not both pass through those points at the same time.
0.5 4a(0) 2 4a(2) 2
Z
4. T1 through T4 do not need to be adjusted. T5 must move 11.510 m toward the y-axis and the z coordinate must move down 10.81 m. T6 must move 10.865 m toward the y-axis and the z coordinate must move down 12.04 m. T7 must move 8.875 toward the y-axis and z must move down 12.064m. T8 must move 7.35 m toward the y-axis and z must move down 10.425 m.
4431.6 10 , 752.6 10 , 4431.6 10 , 752.6 10 6
T2
z
3. The two graphs are being graphed with the same center. Actually, the sun should remain in the same place for each graph. This means that the graph of Pluto needs to be adjusted.
R
1 32
1306 Copyright © 2025 Pearson Education, Inc.
Chapter 11 Projects 2.
Let y 70 . (The arch needs to be 280 ft high. Remember the vertex is at (0, 0), so we must measure down to the arch from the x-axis at the point where the arch’s height is 280 ft.) x 2 787.5(70) x 2 55125 x 234.8 The channel will be 469.6 ft wide.
3.
Figure 2 (0, 350) 280 ft
280 ft (0, 0)
( 525, 0)
x2
(525, 0)
1050 ft
x2 a2
y2 b2
1
1. 4t 2 sec 2 t , 2
1 t tan t ,
0t 0t
4
4 For the x-values, t = 1.99, which is not in the domain [0, π/4]. Therefore there is no t-value that allows the two x values to be equal.
2. On the graphing utility, solve these in parametric form, using a t-step of π/32. It appears that the two graphs intersect at about (1.14, 0.214). However, for the first pair, t = 0.785 at that point. That t-value gives the point (2, 1) on the second pair. There is no intersection point.
4. x1 4t 2
x2 (280) 2 1 275625 122500 (280) 2 x 2 1 275625 122500
y1 1 t D=R; R = R 1 y1 t
x 4(1 y ) 2 x 4y 2 x2 sec 2 t
x 2 99225 x 315 The channel will be 630 feet wide.
5.
Project V
3. Since there were no solutions found for each method, the “solutions” matched.
x2 y2 1 275625 122500
4.
channel doesn’t shrink in width in a flood as fast as a parabola.
If the river rises 10 feet, then we need to look for how wide the channel is when the height is 290 ft. For the parabolic shape: x 2 787.5(60) x 2 47250 x 217.4 There is still a 435 ft wide channel for the ship. For the semi-ellipse: x2 (290) 2 1 275625 122500 (290) 2 x 2 1 275625 122500 x 2 86400 x 293.9 The ship has a 588-ft channel. A semiellipse would be more practical since the
1 tan 2 t sec 2 t 1 y x
y2 tan 2 t
D=[1,2], R=[0,1]
x y 1 x 4 y 1 x y 1 5y 0 y0 x 1 The t-values that go with those x, y values are not the same for both pairs. Thus, again, there is no solution.
5. x : t
3/ 2
ln t
3
y : t 2t 4 Graphing each of these and finding the intersection: There is no intersection for the x-values, so there is no intersection for the system. Graphing the two parametric pairs: The parametric equations show an intersection point. However, the t-value that gives that point for
1307 Copyright © 2025 Pearson Education, Inc.
Chapter 11: Analytic Geometry
each parametric pair is not the same. Thus there is no solution for the system. Putting each parametric pair into rectangular coordinates: x1 ln t t e x D = R, R (0, ) y1 t 3 e3 x x2 t 2 t x 3 3
2
D [0, ), R [4, )
y2 2t 4 2 x 3 4 Then solving that system: y e3 x 2 y 2 x 3 4 This system has an intersection point at (0.56, 5.39). However, ln t = 0.56, gives t = 1.75 and 2
t 0.56
2/3
7. Efficiency depends upon the equations. Graphing the parametric pairs allows one to see immediately whether the t-values will be the same for each pair at any point of intersection. Sometimes, solving for t, as was done in the first method is easy and can be quicker. It leads straight to the t-values, so that allow the method to be efficient. If the two graphs intersect, one must be careful to check that the t-values are the same for each curve at that point of intersection.
0.68 . Since the t-values are not
the same, the point of intersection is false for the system. 6. x: 3 sin t = 2 cos t tan t = 2/3 t = 0.588 y: 4 cos t +2 = 4 sin t by graphing, the solution is t = 1.15 or t = 3.57. Neither of these are the same as for the x-values, thus the system has no solution. Graphing parametrically: If the graphs are done simultaneously on the graphing utility, the two graphs do not intersect at the same t-value. Tracing the graphs shows the same thing. This backs up the conclusion reached the first way. x1 3sin t y1 4 cos t 2 x y2 cos t 3 4 2 2 sin t cos t 1 sin t
x 2 ( y 2) 2 1 9 16 D=[-3, 3], R=[-2, 6] x2 2 cos t
y2 4sin t
x y sin t 2 4 cos 2 t sin 2 t 1 cos t
x2 y 2 1 4 16 D=[-2, 2], R=[-4,4] Solving the system graphically: x 1.3 , y 3.05 . However, the t-values associated with these values are not the same. Thus there is no solution. (Similarly with the symmetric pair in the third quadrant.)
1308 Copyright © 2025 Pearson Education, Inc.
Chapter 12 Systems of Equations and Inequalities Section 12.1
2(2) (1) 4 1 5 5(2) 2(1) 10 2 8 Each equation is satisfied, so x 2, y 1 , or (2, 1) , is a solution of the system of equations.
1. 3x 4 8 x 4x 4 x 1 The solution set is 1 . 2. a.
3 x 2 y 2 10. x 7 y 30 Substituting the values of the variables: 3( 2) 2(4) 6 8 2 ( 2) 7(4) 2 28 30 Each equation is satisfied, so x 2, y 4 , or ( 2, 4) , is a solution of the system of equations.
3x 4 y 12
x-intercept: 3x 4 0 12 3x 12 x4 y-intercept: 3 0 4 y 12 4 y 12 y3
b.
3x 4 y 4 11. 1 1 2 x 3 y 2 Substituting the values of the variables: 1 3(2) 4 2 6 2 4 1 (2) 3 1 1 3 1 2 2 2 2
Each equation is satisfied, so x 2, y 1 , or 2 1 , is a solution of the system of equations. 2, 2
3x 4 y 12 4 y 3x 12
3 y x3 4 3 A parallel line would have slope . 4
3. false; It is inconsistent. 4. consistent; independent 5. (3, 2) 6. consistent; dependent 7. b
2x 1 y 0 2 12. 3 x 4 y 19 2 Substituting the values of the variables, we obtain: 1 1 2 2 2 2 1 1 0 3 1 4 2 3 8 19 2 2 2 Each equation is satisfied, so x 1 , y 2 , or 2 1 , 2 , is a solution of the system of equations. 2
8. a
2 x y 5 9. 5 x 2 y 8 Substituting the values of the variables:
1309 Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities 3 2 3 2 2 2 6 6 4 4 2 3 2 2 2 6 2 10 5 2 2 2 3 2 10 4 6 8 Each equation is satisfied, so x 2 , y 2 ,
x y 3 13. 1 2 x y 3 Substituting the values of the variables, we obtain: 4 1 3 1 2 (4) 1 2 1 3 Each equation is satisfied, so x 4, y 1 , or (4, 1) , is a solution of the system of equations.
z 2 , or (2, 2, 2) is a solution of the system of equations. 5z 6 4x 5 y z 17 18. x 6 y 5 z 24
x y 3 14. 3x y 1 Substituting the values of the variables: 2 5 2 5 3 3 2 5 6 5 1 Each equation is satisfied, so x 2, y 5 , or ( 2, 5) , is a solution of the system of equations.
Substituting the values of the variables: 4 4 5 2 16 10 6 5 3 2 15 2 17 4 6 3 5 2 4 18 10 24 Each equation is satisfied, so x 4 , y 3 , z 2 , or (4, 3, 2) , is a solution of the system of equations.
3x 3 y 2 z 4 15. x y z 0 2 y 3z 8 Substituting the values of the variables: 3(1) 3(1) 2(2) 3 3 4 4 1 (1) 2 1 1 2 0 2(1) 3(2) 2 6 8 Each equation is satisfied, so x 1, y 1, z 2 ,
x y 8 19. x y 4 Solve the first equation for y, substitute into the second equation and solve: y 8 x x y 4
x (8 x) 4 x 8 x 4
or (1, 1, 2) , is a solution of the system of equations.
2 x 12 x6 Since x 6, y 8 6 2 . The solution of the system is x 6, y 2 or using ordered pairs (6, 2) .
z 7 4x 16. 8 x 5 y z 0 x y 5 z 6 Substituting the values of the variables: 4 2 1 8 1 7 8 2 5 3 1 16 15 1 0 2 3 5 1 2 3 5 6 Each equation is satisfied, so x 2 , y 3 , z 1 , or (2, 3, 1) , is a solution of the system of equations.
3x 3 y 2 z 4 17. x 3 y z 10 5 x 2 y 3 z 8
Substituting the values of the variables: 1310
Copyright © 2025 Pearson Education, Inc.
Section 12.1: Systems of Linear Equations: Substitution and Elimination
x 2 y 7 20. x y 3 Solve the first equation for x, substitute into the second equation and solve: x 7 2 y x y 3 (7 2 y ) y 3 7 y 3 4 y Since y 4, x 7 2(4) 1 . The solution of the system is x 1, y 4 or using an ordered pair, (1, 4) .
x 3y 5 22. 2 x 3 y 8 Add the equations: x 3 y 5 2 x 3 y 8 3 x 1 Substitute and solve for y: 1 3 y 5 3x
3y 6 y2 The solution of the system is x 1, y 2 or using ordered pairs (1, 2) .
5 x y 21 21. 2 x 3 y 12 Multiply each side of the first equation by 3 and add the equations to eliminate y: 15 x 3 y 63 2 x 3 y 12 51 x3 Substitute and solve for y: 5(3) y 21 15 y 21 y 6 y 6 The solution of the system is x 3, y 6 or 17 x
using ordered an pair 3, 6 .
1311
Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities
24 3x 23. x 2 y 0 Solve the first equation for x and substitute into the second equation: x8 x 2 y 0 8 2y 0 2y 8 y 4 The solution of the system is x 8, y 4 or
using ordered pairs (8, 4) 3x 6 y 2 25. 5 x 4 y 1 Multiply each side of the first equation by 2 and each side of the second equation by 3, then add to eliminate y: 6 x 12 y 4 15 x 12 y 3 7 1 x 3 Substitute and solve for y: 3 1/ 3 6 y 2 21x
4 x 5 y 3 24. 2y 8 Solve the second equation for y and substitute into the first equation: 4 x 5 y 3 y4
1 6y 2 6y 1 y
1 1 The solution of the system is x , y or 3 6 1 1 using ordered pairs , . 3 6
4 x 5(4) 3 4 x 20 3 4 x 23 x
23 4
The solution of the system is x
1 6
23 , y 4 or 4
23 using ordered pairs , 4 . 4
1312
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Section 12.1: Systems of Linear Equations: Substitution and Elimination
2 2 x 4 y 26. 3 3x 5 y 10 Multiply each side of the first equation by 5 and each side of the second equation by 4, then add to eliminate y: 10 10 x 20 y 3 12 x 20 y 40
110 3 5 x 3 Substitute and solve for y: 3 5 / 3 5 y 10 22 x
x y 5 28. 3x 3 y 2 Solve the first equation for x, substitute into the second equation and solve: x y 5 3x 3 y 2
5 5 y 10 5 y 5 y 1
5 The solution of the system is x , y 1 or 3 5 using ordered pairs , 1 . 3
3( y 5) 3 y 2 3 y 15 3 y 2 0 17 This equation is false, so the system is inconsistent.
2x y 1 27. 4 x 2 y 3 Solve the first equation for y, substitute into the second equation and solve: y 1 2x 4 x 2 y 3 4 x 2(1 2 x) 3 4x 2 4x 3 0 1 This equation is false, so the system is inconsistent.
1313
Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities
2 x y 0 29. 4 x 2 y 12 Solve the first equation for y, substitute into the second equation and solve: y 2x 4 x 2 y 12
The solution of the system is x 1, y 4 using ordered pairs 1, . 3
4 x 2(2 x) 12 4 x 4 x 12 8 x 12 3 x 2 3 3 Since x , y 2 3 2 2
The solution of the system is x
4 or 3
x 2y 4 31. 2 x 4 y 8 Solve the first equation for x, substitute into the second equation and solve: x 4 2 y 2 x 4 y 8
3 , y 3 or 2
3 using ordered pairs ,3 . 2
2(4 2 y ) 4 y 8 8 4y 4y 8 00 These equations are dependent. The solution of the system is either x 4 2 y , where y is any real 4 x , where x is any real number. 2 Using ordered pairs, we write the solution as ( x, y) x 4 2 y, y is any real number or as
number or y
4 x , x is any real number . ( x, y ) y 2
3 x 3 y 1 30. 8 4 x y 3 Solve the second equation for y, substitute into the first equation and solve: 3x 3 y 1 8 y 3 4 x 8 3x 3 4 x 1 3 3x 8 12 x 1 9 x 9 x 1 8 8 4 Since x 1, y 4(1) 4 . 3 3 3
3x y 7 32. 9 x 3 y 21 Solve the first equation for y, substitute into the second equation and solve: y 3x 7 9 x 3 y 21 9 x 3(3x 7) 21 9 x 9 x 21 21 00 These equations are dependent. The solution of the system is either y 3 x 7 , where x is any real
y7 , where y is any real number. 3 Using ordered pairs, we write the solution as
number is x
1314
Copyright © 2025 Pearson Education, Inc.
Section 12.1: Systems of Linear Equations: Substitution and Elimination
( x, y) y 3x 7, x is any real number or as
2 x 3 y 6 1 x y 2 1 2 y 3y 6 2 2y 1 3y 6 5y 5 y 1
y7 , y is any real number . ( x, y ) x 3 2 x 3 y 1 33. 10 x y 11 Multiply each side of the first equation by –5, and add the equations to eliminate x: 10 x 15 y 5 10 x y 11
Since y 1, x 1
1 3 . The solution of the 2 2
3 system is x , y 1 or using ordered pairs 2 3 , 1 . 2
16 y 16 y 1 Substitute and solve for x: 2 x 3(1) 1
2 x 3 1 2x 2 x 1 The solution of the system is x 1, y 1 or using ordered pairs (1, 1).
1 x y 2 36. 2 x 2 y 8 Solve the second equation for x, substitute into the first equation and solve: 1 x y 2 2 x 2 y 8 1 (2 y 8) y 2 2 y 4 y 2 2y 6 y 3 Since y 3, x 2(3) 8 6 8 2 . The solution of the system is x 2, y 3 or using ordered pairs (2, 3) .
3x 2 y 0 34. 5 x 10 y 4 Multiply each side of the first equation by 5, and add the equations to eliminate y: 15 x 10 y 0 5 x 10 y 4 4 1 x 5 Substitute and solve for y: 5 1/ 5 10 y 4 20 x
1 10 y 4 10 y 3 3 y 10
1 1 2 x 3 y 3 37. 1 x 2 y 1 4 3 Multiply each side of the first equation by –6 and each side of the second equation by 12, then add to eliminate x: 3 x 2 y 18 3x 8 y 12
1 3 or The solution of the system is x , y 5 10 1 3 using ordered pairs , . 5 10 2 x 3 y 6 35. 1 x y 2 Solve the second equation for x, substitute into the first equation and solve:
10 y 30 y 3
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Chapter 12: Systems of Equations and Inequalities 3 2 / 3 6 y 7
Substitute and solve for x: 1 1 x (3) 3 2 3 1 x 1 3 2 1 x2 2 x4
2 6y 7 6 y 5 y
5 6
The solution of the system is x
The solution of the system is x 4, y 3 or using ordered pairs (4, 3).
2 5 , y or 3 6
2 5 using ordered pairs , . 3 6
3 1 3 x 2 y 5 38. 3 x 1 y 11 4 3
2 x y 1 40. 1 3 x 2 y 2 Multiply each side of the second equation by 2, and add the equations to eliminate y: 2 x y 1 2 x y 3
Multiply each side of the first equation by –54 and each side of the second equation by 24, then add to eliminate x: 18 x 81 y 270 18 x 8 y 264
4x
89 y 534 y 6
2 1 x 2
1 Substitute and solve for y: 2 y 1 2 1 y 1 y 2 y2
Substitute and solve for x: 3 1 x (6) 11 4 3 3 x 2 11 4 3 x9 4 x 12
1 The solution of the system is x , y 2 or 2 1 using ordered pairs , 2 . 2
The solution of the system is x 12, y 6 or using ordered pairs (12, 6).
1 1 x y 8 41. 3 5 0 x y
3x 6 y 7 39. 5 x 2 y 5 Multiply each side of the second equation by -3 then add the equations to eliminate y: 3 x 6 y 7 15 x 6 y 15
Rewrite letting u
1 1 , v : x y
u v8 3u 5v 0 Solve the first equation for u, substitute into the second equation and solve: u 8 v 3u 5v 0
12 x
8 2 x 3 Substitute and solve for x:
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Section 12.1: Systems of Linear Equations: Substitution and Elimination 3(8 v) 5v 0
x y 9 43. 2 x z 13 3 y 2 z 7
24 3v 5v 0 8v 24 v3
Multiply each side of the first equation by –2 and add to the second equation to eliminate x: 2 x 2 y 18
1 1 Since v 3, u 8 3 5 . Thus, x , u 5 1 1 y . The solution of the system is v 3 1 1 1 1 x , y or using ordered pairs , . 5 3 5 3
2y z 5 Multiply each side of the result by 2 and add to the original third equation to eliminate z: 3y 2z 7 4 y 2 z 10
4 3 x y 0 42. 6 3 2 x 2 y
Rewrite letting u
y 3 y3 Substituting and solving for the other variables: 3(3) 2 z 7 2 x (1) 13 2 x 12 2 z 2 z 1 x6 The solution is x 6, y 3, z 1 or using ordered triples (6,3, 1) .
1 1 , v : x y
4u 3v 0 3 6u 2 v 2 Multiply each side of the second equation by 2, and add the equations to eliminate v: 4u 3v 0 12u 3v 4 16u 4 4 1 u 16 4 Substitute and solve for v: 1 4 3v 0 4 1 3v 0
2x y 4 44. 2 y 4 z 0 3 x 2 z 11
Multiply each side of the first equation by 2 and add to the second equation to eliminate y: 4x 2 y 8 2 y 4z 0 4x
4z 8
1 and add to 2 the original third equation to eliminate z: 2x 2z 4 3x 2 z 11 5x 15 x 3 Substituting and solving for the other variables: 2(3) y 4 3(3) 2 z 11 6 y 4 9 2 z 11 y2 2z 2 z 1 The solution is x 3, y 2, z 1 or using ordered triples (3, 2, 1) .
Multiply each side of the result by
3v 1 v
z 13
2x
1 3
1 1 4, y 3 . The solution of the u v system is x 4, y 3 or using ordered pairs (4, 3). Thus, x
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Chapter 12: Systems of Equations and Inequalities 66 x 77 z 77 66 x 90 z 42
x 2 y 3z 7 45. 2 x y z 4 3x 2 y 2 z 10 Multiply each side of the first equation by –2 and add to the second equation to eliminate x; and multiply each side of the first equation by 3 and add to the third equation to eliminate x: 2 x 4 y 6 z 14 2x y z
13z 35 35 13 Substituting and solving for the other variables: 35 6 x 7 7 13 245 6x 7 13 336 6x 13 56 x 13 z
4
5 y 5 z 10 3x 6 y 9 z 21 3x 2 y 2 z 10 4 y 7 z 11
Multiply each side of the first result by
4 and 5
56 35 2 y 3 0 13 13 112 105 y 0 13 13
add to the second result to eliminate y: 4 y 4 z 8 4 y 7 z 11 3z 3 z 1 Substituting and solving for the other variables:
y
56 7 35 , y , z or 13 13 13 56 7 35 using ordered triples , , . 13 13 13
The solution is x
x 2(1) 3(1) 7 x23 7 x2 The solution is x 2, y 1, z 1 or using ordered triples (2, 1, 1) . y 1 2 y 1
x y z 1 47. 2 x 3 y z 2 3x 2 y 0
2 x y 3z 0 46. 2 x 2 y z 7 3x 4 y 3z 7 Multiply each side of the first equation by –2 and add to the second equation to eliminate y; and multiply each side of the first equation by 4 and add to the third equation to eliminate y: 4 x 2 y 6 z 0
Add the first and second equations to eliminate z: x y z 1 2x 3y z 2 3x 2 y 3 Multiply each side of the result by –1 and add to the original third equation to eliminate y: 3x 2 y 3
2 x 2 y z 7 6x
3x 2 y 0
7z 7
0 3 This equation is false, so the system is inconsistent.
8 x 4 y 12 z 0 3x 4 y 3z 7 11x
7 13
15 z 7
Multiply each side of the first result by 11 and multiply each side of the second result by 6 to eliminate x: 1318
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Section 12.1: Systems of Linear Equations: Substitution and Elimination x (4 z 3) z 1
2x 3y z 0 48. x 2 y z 5 3x 4 y z 1
x 4z 3 z 1 x 5z 3 1 x 5z 2 The solution is {( x, y , z ) x 5 z 2, y 4 z 3 ,
Add the first and second equations to eliminate z; then add the second and third equations to eliminate z: 2x 3y z 0 x 2 y z 5 x y
z is any real number}. 2x 3y z 0 50. 3x 2 y 2 z 2 x 5 y 3z 2
5
x 2 y z 5
Multiply the first equation by 2 and add to the second equation to eliminate z; multiply the first equation by 3 and add to the third equation to eliminate z: 4x 6 y 2z 0
3x 4 y z 1 2x 2 y
6
Multiply each side of the first result by –2 and add to the second result to eliminate y: 2 x 2 y 10
3x 2 y 2 z 2 7x 4y
2x 2 y 6 0 2 This equation is false, so the system is inconsistent.
6 x 9 y 3z 0 x 5 y 3z 2 7x 4y
x y z 1 49. x 2 y 3 z 4 3x 2 y 7 z 0
2
2
Multiply each side of the first result by –1 and add to the second result to eliminate y: 7 x 4 y 2 7x 4y 2
Add the first and second equations to eliminate x; multiply the first equation by –3 and add to the third equation to eliminate x: x y z 1
0 0 The system is dependent. If y is any real
x 2 y 3z 4
4 2 y . 7 7 Solving for z in terms of x in the first equation: z 2x 3y
number, then x
y 4z 3 3x 3 y 3 z 3
4y 2 2 3y 7 8 y 4 21 y 7 13 y 4 7 4 2 The solution is ( x, y, z ) x y , 7 7 13 4 z y , y is any real number . 7 7
3x 2 y 7 z 0 y 4 z 3
Multiply each side of the first result by –1 and add to the second result to eliminate y: y 4z 3 y 4 z 3 0 0 The system is dependent. If z is any real number, then y 4 z 3 . Solving for x in terms of z in the first equation:
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Chapter 12: Systems of Equations and Inequalities
the third equation to eliminate z: x y z 6
2 x 2 y 3z 6 51. 4 x 3 y 2 z 0 2 x 3 y 7 z 1 Multiply the first equation by –2 and add to the second equation to eliminate x; add the first and third equations to eliminate x: 4 x 4 y 6 z 12 4x 3 y 2z
3x 2 y z 5 4x y
6 x 4 y 2 z 10 x 3 y 2 z 14 7x y
0
2 x 2 y 3z 6 2x 3y 7z 1
x 1 Substituting and solving for the other variables: 4(1) y 1 3(1) 2(3) z 5 y 3 3 6 z 5 y3 z 2 The solution is x 1, y 3, z 2 or using ordered triplets (1, 3, 2) .
y4z 7 0 19 This result is false, so the system is inconsistent. 3x 2 y 2 z 6 52. 7 x 3 y 2 z 1 2 x 3 y 4 z 0
x y z 4 54. 2 x 3 y 4 z 15 5 x y 2 z 12
Multiply the first equation by –1 and add to the second equation to eliminate z; multiply the first equation by –2 and add to the third equation to eliminate z: 3x 2 y 2 z 6 7 x 3 y 2z 1
Multiply the first equation by –3 and add to the second equation to eliminate y; add the first and third equations to eliminate y: 3x 3 y 3z 12 2 x 3 y 4 z 15 x
z 3 z x 3 x y z 4 5 x y 2 z 12
7
6 x 4 y 4 z 12 4 x y
4
3x 3
y 4z 7 Multiply each side of the first result by –1 and add to the second result to eliminate y: y 4 z 12
2x 3y 4z
Multiply each side of the first result by –1 and add to the second result to eliminate y: 4 x y 1 7x y 4
y 4 z 12
4x y
1
0
12
6x
Add the first result to the second result to eliminate y: 4x y 7 4 x y 12
z 8
Substitute and solve: 6 x ( x 3) 8 6x x 3 8 5x 5 x 1 z x 3 1 3 2
0 19 This result is false, so the system is inconsistent.
y 12 5 x 2 z 12 5(1) 2( 2) 3 The solution is x 1, y 3, z 2 or using ordered triplets (1, 3, 2) .
x y z 6 53. 3 x 2 y z 5 x 3 y 2 z 14
Add the first and second equations to eliminate z; multiply the second equation by 2 and add to 1320
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Section 12.1: Systems of Linear Equations: Substitution and Elimination 2 x 8 y 6 z 16 x y 6z 1
x 2 y z 3 55. 2 x 4 y z 7 2 x 2 y 3z 4 Add the first and second equations to eliminate z; multiply the second equation by 3 and add to the third equation to eliminate z: x 2y z 3 2x 4 y z 7 3x 2 y 10
3x 9 y
Multiply each side of the second result by 1/ 3 and add to the first result to eliminate y: 4x 3y 4 x 3y 5 9 x3 Substituting and solving for the other variables: 3 3 y 5 3y 8 8 y 3 3x
6 x 12 y 3z 21 2 x 2 y 3z 4 4 x 10 y 17
Multiply each side of the first result by –5 and add to the second result to eliminate y: 15 x 10 y 50 4 x 10 y 17
8 3 6z 1 3 2 6z 3 1 z 9
11x
33 x 3 Substituting and solving for the other variables: 3(3) 2 y 10 9 2 y 10 2 y 1 1 y 2
8 3
The solution is x 3, y , z
1 or using 9
8 1
ordered triplets 3, , . 3 9
1 3 2 z 3 2 3 1 z 3 z 1
57. Let l be the length of the rectangle and w be the width of the rectangle. Then: l 2w and 2l 2w 90
Solve by substitution: 2(2 w) 2w 90 4 w 2w 90 6w 90 w 15 feet l 2(15) 30 feet The floor is 15 feet by 30 feet.
z 1 1 The solution is x 3, y , z 1 or using 2 1 ordered triplets 3, , 1 . 2 x 4 y 3z 8 56. 3x y 3 z 12 x y 6z 1
58. Let l be the length of the rectangle and w be the width of the rectangle. Then: l w 50 and 2l 2 w 3000
Add the first and second equations to eliminate z; multiply the first equation by 2 and add to the third equation to eliminate z: x 4 y 3z 8
Solve by substitution: 2( w 50) 2w 3000 2 w 100 2w 3000 4w 2900 w 725 meters l 725 50 775 meters
3x y 3z 12 4x 3 y
15
4
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Chapter 12: Systems of Equations and Inequalities
The dimensions of the field are 775 meters by 725 meters.
x y 14 0.30 x 0.65 y 0.40(14) Solve the first equation for y: y 14 x Solve by substitution: 0.30 x 0.65(14 x) 5.6
59. Let x = the number of Starlink launches and y = the number of other launches. Then: 1 x y 61 and x y 10 2 Solve by substitution: 1 y 31 y 61 2 3 y 51 2 y 34 In 2022 there were 27 Starlink launches and 34 other launches.
0.3x 9.1 0.65 x 5.6 0.35 x 3.5 x 10 y 14 10 4 The chemist needs 10 liters of the 30% solution and 4 liters of the 65% solution.
63. Let s = the price of a smartphone and t = the price of a tablet. Then: s t 1665 340 s 250t 486000 Solve the first equation for t: t 1665 s Solve by substitution: 340 s 250(1665 s ) 486000 340 s 416250 250 s 486000 90 s 69750 s 775 t 1665 775 890 The price of the smartphone is $775.00 and the price of the tablet is $890.00.
60. Let x = the number of adult tickets sold and y = the number of senior tickets sold. Then: x y 325 15 x 13 y 4445 Solve the first equation for y: y 325 x Solve by substitution: 15 x 13(325 x) 4445 15 x 4225 13 x 4445 2 x 220 x 110 y 325 110 215 There were 110 adult tickets sold and 215 senior citizen tickets sold.
64. Let x = the amount invested in B bonds. Let y = the amount invested in a CD. a. Then x y 5000 represents the total investment. 0.15 x 0.07 y 500 represents the earnings on the investment.
61. Let x = the number of pounds of cashews. Let y = is the number of pounds in the mixture. The value of the cashews is 5x . The value of the peanuts is 1.50(30) = 45. The value of the mixture is 3y . Then x 30 y represents the amount of mixture. 5 x 45 3 y represents the value of the mixture.
Solve by substitution: 0.15(5000 y ) 0.07 y 500 750 0.15 y 0.07 y 500 0.08 y 250 y 3125 x 5000 3125 1875 Thus, $1875 should be invested in B Bonds and $3125 in a CD.
Solve by substitution: 5 x 45 3( x 30) 2 x 45 x 22.5 So, 22.5 pounds of cashews should be used in the mixture.
b. Then x y 5000 represents the total investment. 0.15 x 0.07 y 650 represents the earnings on the investment.
62. Let x = the number of liters of 30% solution and y = the number liters of 65% solution. Then:
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Section 12.1: Systems of Linear Equations: Substitution and Elimination
Solve by substitution: 0.15(5000 y ) 0.07 y 650 750 0.15 y 0.07 y 650 0.08 y 100 y 1250 x 5000 1250 3750 Thus, $3750 should be invested in B Bonds and $1250 in a CD.
67. Let x = the number of $25-design. Let y = the number of $45-design. Then x y = the total number of sets of dishes. 25 x 45 y = the cost of the dishes.
Setting up the equations and solving by substitution: x y 200 25 x 45 y 7400 Solve the first equation for y, the solve by substitution: y 200 x 25 x 45(200 x ) 7400 25 x 9000 45 x 7400 20 x 1600 x 80 y 200 80 120 Thus, 80 sets of the $25 dishes and 120 sets of the $45 dishes should be ordered.
65. Let x = the plane’s average airspeed and y = the average wind speed. Rate
Time Distance
With Wind
x y
3
600
Against
x y
4
600
( x y )(3) 600 ( x y )(4) 600
Multiply each side of the first equation by
1 , 3
68. Let x = the cost of a hot dog. Let y = the cost of a soft drink. Setting up the equations and solving by substitution: 10 x 5 y 35.00 7 x 4 y 25.25
1 multiply each side of the second equation by , 4 and add the result to eliminate y x y 200 x y 150 2 x 350 x 175
10 x 5 y 35.00 2x y 7
175 y 200 y 25 The average airspeed of the plane is 175 mph, and the average wind speed is 25 mph.
y 7 2x 7 x 4(7 2 x ) 25.25 7 x 28 8 x 25.25 x 2.75
66. Let x = the average wind speed and y = the distance. Rate
Time Distance
With Wind 150 x
2
y
150 x
3
y
Against
x 2.75 y 7 2(2.75) 1.50 A single hot dog costs $2.75 and a single soft drink costs $1.50.
69. Let x = the cost per package of bacon. Let y = the cost of a carton of eggs. Set up a system of equations for the problem: 3x 2 y 13.45 2 x 3 y 11.45 Multiply each side of the first equation by 3 and each side of the second equation by –2 and solve by elimination:
(150 x)(2) y (150 x)(3) y Solve by substitution: (150 x)(2) (150 x)(3) 300 2 x 450 3x 5 x 150 x 30 Thus, the average wind speed is 30 mph.
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Chapter 12: Systems of Equations and Inequalities 9 x 6 y 40.35
Multiplying each equation by 10 yields 2 x 4 y 400 6 x 4 y 600
4 x 6 y 22.90
5x
17.45
x 3.49
Subtracting the bottom equation from the top equation yields
Substitute and solve for y: 3(3.49) 2 y 13.45 10.47 2 y 13.45 2 y 2.98 y 1.49 A package of bacon costs $3.49 and a carton of eggs cost $1.49. The refund for 2 packages of bacon and 2 cartons of eggs will be 2($3.49) + 2($1.49) = $9.96.
2 x 4 y 6 x 4 y 400 600 2 x 6 x 200 4 x 200 x 50 2 50 4 y 400 100 4 y 400 4 y 300 300 75 4 So 50 mg of compound 1 should be mixed with 75 mg of compound 2. y
70. Let x = Pamela’s average speed in still water. Let y = the speed of the current. Rate
Time Distance
Downstream
x y
3
15
Upstream
x y
5
15
72. Let x = the # of units of powder 1. Let y = the # of units of powder 2. Setting up the equations and solving by substitution:
Set up a system of equations for the problem: 3( x y ) 15 5( x y ) 15 Multiply each side of the first equation by
0.2 x 0.4 y 12 vitamin B12 0.3 x 0.2 y 12 vitamin E
1 , 3
Multiplying each equation by 10 yields
1 multiply each side of the second equation by , 5 and add the result to eliminate y: x y 5 x y 3
2 x 4 y 120 6 x 4 y 240
Subtracting the bottom equation from the top equation yields 2 x 4 y 6 x 4 y 120 240
2x 8
4 x 120
x4 4 y 5
2 30 4 y 120
y 1
x 30
60 4 y 120
Pamela's average speed is 4 miles per hour and the speed of the current is 1 mile per hour.
4 y 60 60 15 4 So 30 units of powder 1 should be mixed with 15 units of powder 2. y
71. Let x = the # of mg of compound 1. Let y = the # of mg of compound 2. Setting up the equations and solving by substitution: 0.2 x 0.4 y 40 vitamin C 0.3 x 0.2 y 30 vitamin D
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Section 12.1: Systems of Linear Equations: Substitution and Elimination
The system of equations is: a bc 2 a bc –4 4a 2b c 4
73. y ax 2 bx c At (–1, 4) the equation becomes: 4 a (–1) 2 b(1) c 4 a bc
Multiply the first equation by –1 and add to the second equation; multiply the first equation by – 1 and add to the third equation to eliminate c:
At (2, 3) the equation becomes: 3 a(2) 2 b(2) c 3 4a 2b c
a b c 2 ab c –4
At (0, 1) the equation becomes: 1 a(0) 2 b(0) c 1 c
4a 2b c 4 3a 3b
2 b 1
2b
The system of equations is: a bc 4 4a 2b c 3 c1
a b c 2
Substitute and solve: a (1) 2 a3
6
ab 2 c a b 4 3 (1) 4 6
Substitute c 1 into the first and second equations and simplify: 4a 2b 1 3 a b 1 4 4a 2b 2 a b 3 a b3
The solution is a 3, b 1, c 6 . The equation is y 3 x 2 x 6 0.06Y 5000r 240 75. 0.06Y 6000r 900 Multiply the first equation by 1 , the add the result to the second equation to eliminate Y. 0.06Y 5000r 240
Solve the first result for a, substitute into the second result and solve: 4(b 3) 2b 2 4b 12 2b 2 6b 10 5 b 3 5 4 a 3 3 3 4 5 The solution is a , b , c 1 . The 3 3 4 5 equation is y x 2 x 1 . 3 3
0.06Y 6000r 900 11000r 660 r 0.06 Substitute this result into the first equation to find Y. 0.06Y 5000(0.06) 240 0.06Y 300 240 0.06Y 540 Y 9000 The equilibrium level of income and interest rates is $9000 million and 6%.
74. y ax 2 bx c At (–1, –2) the equation becomes: 2 a (1) 2 b(1) c a b c 2
0.05Y 1000r 10 76. 0.05Y 800r 100 Multiply the first equation by 1 , the add the result to the second equation to eliminate Y. 0.05Y 1000r 10
At (1, –4) the equation becomes: 4 a(1) 2 b(1) c a b c 4
0.05Y 800r 100 1800r 90
At (2, 4) the equation becomes: 4 a (2) 2 b(2) c 4a 2b c 4
r 0.05 Substitute this result into the first equation to find Y.
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Chapter 12: Systems of Equations and Inequalities 0.05Y 1000(0.05) 10 0.05Y 50 10 0.05Y 60
8I1 4 6 I 2
8 4 I1 10 I 2
8I1 6 I 2 4
4 I1 10 I 2 8
Y 1200 The equilibrium level of income and interest rates is $1200 million and 5%.
Multiply both sides of the first result by –2 and add to the second result to eliminate I1 : 8 I1 20 I 2 16
I 2 I1 I 3 77. 5 3I1 5I 2 0 10 5I 7 I 0 2 3
8 I1 6 I 2
4
26 I 2 12 12 6 26 13 Substituting and solving for the other variables: 6 4 I1 10 8 13 60 4 I1 8 13 44 4 I1 13 11 I1 13 11 6 17 I 3 I1 I 2 13 13 13 11 6 17 The solution is I1 , I 2 , I 3 . 13 13 13
Substitute the expression for I 2 into the second and third equations and simplify: 5 3I1 5( I1 I 3 ) 0
I2
8 I1 5I 3 5 10 5( I1 I 3 ) 7 I 3 0 5I1 12 I 3 10
Multiply both sides of the first result by 5 and multiply both sides of the second result by –8 to eliminate I1 : 40 I1 25I 3 25 40 I1 96 I 3 80 71I 3 55 I3
equation and simplify: 8 4( I1 I 2 ) 6 I 2
55 71
Substituting and solving for the other variables: 55 8I1 5 5 71 275 8I1 5 71 80 8 I1 71 10 I1 71
79. Let x = the number of orchestra seats. Let y = the number of main seats. Let z = the number of balcony seats. Since the total number of seats is 500, x y z 500 . Since the total revenue is $64,250 if all seats are sold, 150 x 135 y 110 z 64, 250 . If only half of the orchestra seats are sold, the revenue is $56,750. 1 So, 150 x 135 y 110 z 56, 750 . 2 Thus, we have the following system: x y z 500 150 x 135 y 110 z 64, 250 75 x 135 y 110 z 56, 750
10 55 65 I2 71 71 71 10 65 55 . The solution is I1 , I 2 , I 3 71 71 71 I 3 I1 I 2 78. 8 4 I 3 6 I 2 8I 4 6 I 2 1 Substitute the expression for I 3 into the second
Multiply each side of the first equation by –110 and add to the second equation to eliminate z; multiply each side of the third equation by –1 and add to the second equation to eliminate z:
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Section 12.1: Systems of Linear Equations: Substitution and Elimination 110 x 110 y 110 z 55, 000
y 2x
3x z 405
150 x 135 y 110 z 64, 250
2(110)
3(110) z 405
40x 25 y
220
330 z 405
9250
z 75 There were 110 adults, 220 children, and 75 senior citizens that bought tickets.
150 x 135 y 110 z 64, 250
75 x 135 y 110 z 56, 750 75 x
7500
81. Let x = the number of servings of chicken. Let y = the number of servings of corn. Let z = the number of servings of 2% milk.
x 100
Substituting and solving for the other variables: 40(100) 25 y 9250 100 210 z 500 4000 25 y 9250 310 z 500 25 y 5250 z 190 y 210
Protein equation: 30 x 3 y 9 z 66 Carbohydrate equation: 35 x 16 y 13 z 94.5 Calcium equation: 200 x 10 y 300 z 910 Multiply each side of the first equation by –16 and multiply each side of the second equation by 3 and add them to eliminate y; multiply each side of the second equation by –5 and multiply each side of the third equation by 8 and add to eliminate y: 480 x 48 y 144 z 1056 105 x 48 y 39 z 283.5
There are 100 orchestra seats, 210 main seats, and 190 balcony seats. 80. Let x = the number of adult tickets. Let y = the number of child tickets. Let z = the number of senior citizen tickets. Since the total number of tickets is 405, x y z 405 . Since the total revenue is $3315, 11x 6 y 9 z 3315 . Twice as many children's tickets as adult tickets are sold. So, y 2 x . Thus, we have the following system: y z 405 x 11x 6.50 y 9 z 3315 y 2x
375 x
175 x 80 y
65 z 472.5
1600 x 80 y 2400 z 7280 1425 x
2335 z 6807.5
Multiply each side of the first result by 19 and multiply each side of the second result by 5 to eliminate x: 7125 x 1995 z 14, 677.5 7125 x 11, 675 z 34, 037.5
Substitute for y in the first two equations and simplify: x (2 x) z 405 3 x z 405
9680 z 19,360 z2
Substituting and solving for the other variables: 375 x 105(2) 772.5 375 x 210 772.5 375 x 562.5 x 1.5
11x 6.50(2 x) 9 z 3315 24 x 9 z 3315 Multiply the first result by –9 and add to the second result to eliminate z: 27 x 9 z 3645 24 x 9 z 3315 3 x
105 z 772.5
30(1.5) 3 y 9(2) 66 45 3 y 18 66
330
3y 3
x 110
y 1
The dietitian should serve 1.5 servings of chicken, 1 serving of corn, and 2 servings of 2% milk. 1327
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Chapter 12: Systems of Equations and Inequalities 4 x 3 y 3 z 13.05
82. Let x = the amount in Treasury bills. Let y = the amount in Treasury bonds. Let z = the amount in corporate bonds.
5 x 3 y 4 z 15.80 z 2.75
x
Since the total investment is $2000, x y z 2000
x 2.75 z
Substitute and solve for y in terms of z: 5 2.75 z 3 y 4 z 15.80
Since the total income is to be $139, 0.05 x 0.07 y 0.10 z 139
13.75 3 y z 15.80
The investment in Treasury bills is to be $300 more than the investment in corporate bonds. So, x 300 z
3 y z 2.05 1 41 z 3 60 Solutions of the system are: x 2.75 z , 1 41 y z . 3 60 Since we are given that 0.60 z 0.90 , we choose values of z that give two-decimal-place values of x and y with 1.75 x 2.25 and 0.75 y 1.00 . The possible values of x, y, and z are shown in the table. y
Substitute for x in the first two equations and simplify: (300 z ) y z 2000 y 2 z 1700 5(300 z ) 7 y 10 z 13900 7 y 15 z 12400
Multiply each side of the first result by –7 and add to the second result to eliminate y: 7 y 14 z 11,900 7 y 15 z 12, 400
x
y
z
z 500 x 300 z 300 500 800 y 2 z 1700
2.13
0.89
0.62
2.10
0.90
0.65
y 2(500) 1700
2.07
0.91
0.68
y 1000 1700 y 700 Asako should invest $800 in Treasury bills, $700 in Treasury bonds, and $500 in corporate bonds.
2.04
0.92
0.71
2.01
0.93
0.74
1.98
0.94
0.77
1.95
0.95
0.80
1.92
0.96
0.83
1.89
0.97
0.86
1.86
0.98
0.89
83. Let x = the price of 1 hamburger. Let y = the price of 1 order of fries. Let z = the price of 1 drink.
We can construct the system 8 x 6 y 6 z 26.10 10 x 6 y 8 z 31.60
84. Let x = the price of 1 hamburger. Let y = the price of 1 order of fries. Let z = the price of 1 drink We can construct the system 8 x 6 y 6 z 26.10 10 x 6 y 8 z 31.60 3 x 2 y 4 z 10.95
A system involving only 2 equations that contain 3 or more unknowns cannot be solved uniquely. 1 Multiply the first equation by and the 2 1 second equation by , then add to eliminate y: 2
Subtract the second equation from the first equation to eliminate y: 8 x 6 y 6 z 26.10 10 x 6 y 8 z 31.60 2 x 2 z 5.5 1328
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Section 12.1: Systems of Linear Equations: Substitution and Elimination
Multiply the third equation by –3 and add it to the second equation to eliminate y: 10 x 6 y 8 z 31.60 9 x 6 y 12 z 32.85 x 4 z 1.25 Multiply the second result by 2 and add it to the first result to eliminate x: 2x 2 z 5.5 2x 8 z 2.5
In 15 hours they complete 1 entire job, so 1 1 15 1 . y z
10 z 8 z 0.8 Substitute for z to find the other variables: x 4(0.8) 1.25 x 3.2 1.25 x 1.95 3(1.95) 2 y 4(0.8) 10.95 5.85 2 y 3.2 1.095 2 y 1.9 y 0.95
in 1 hour
1 1 1 y z 15 Beth Solana Edie
Beth Dan Edie
in 1 hour
z
1 x
1 y
1 z
In 10 hours they complete 1 entire job, so 1 1 1 10 1 x y z
in 1 hour
z
1 y
1 z
1 x
1 y
1 z
1 30 x 30
Substitute x = 30 into the third equation: 12 12 4 1 30 y z . 12 4 3 y z 5 Now consider the system consisting of the last result and the second original equation. Multiply the second original equation by –12 and add it to the last result to eliminate y:
Dan Edie y
Part of job done
1 x
1 1 1 1 x y z 10 Hours to do job Part of job done
z
We have the system 1 1 1 1 x y z 10 1 1 1 y z 15 12 12 4 1 y z x Subtract the second equation from the first equation: 1 1 1 1 x y z 10 1 1 1 y z 15
We can use the following tables to organize our work: y
y
12 12 4 1 x y z
85. Let x = Beth’s time working alone. Let y =Solana’s time working alone. Let z = Edie’s time working alone.
x
x
With all 3 working for 4 hours and Beth and Solana working for an additional 8 hours, they complete 1 entire job, so 1 1 1 1 1 4 8 1 x y z x y
Therefore, one hamburger costs $1.95, one order of fries costs $0.95, and one drink costs $0.80.
Hours to do job Part of job done
Hours to do job
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Chapter 12: Systems of Equations and Inequalities
Multiply the first equation by -c and add to the second equation.
12 12 12 y z 15 12 4 3 y z 5
2 2 acx bcy c z ac bc c 2 2 2 a x b y c z ac ab bc
8 3 z 15 z 40
a(a c) x b(b c) y ab c 2 (2)
Multiply (2) by b and (3) by –(a-c) then add.
Plugging z = 40 to find y: 12 4 3 y z 5 12 4 3 y 40 5 12 1 y 2 y 24 Working alone, it would take Beth 30 hours, Solana 24 hours, and Edie 40 hours to complete the job.
ab(a c) x b 2 (b c) y ab 2 bc 2 2 2 2 ab(a c) x bc(a c) abc a c b c a c b(b 2 bc ac c 2 ) y a (b 2 bc ac c 2 ) y
a b
Substitute y
a bc ac b abx ac bc ac abx bc
ax by a b 86. 2 2 abx b y b ab
abx bc c a
Multiply the first equation by b and add.
x
abx b 2 y ab b 2 2 2 abx b y b ab
Substitute x
2abx 2b 2 b x a
Substitute x
a into equation (3): b
c a , y into equation (1): a b
c a a b cz a b c a b c a cz a b c cz b
b into equation (1): a
z
b by a b a b by a b by a a y b a
b c
c a b The solution set is , , a b c
88. – 90.
Answers will vary.
b a So the solution is , . a b ax by cz a b c (1) 2 2 2 87. a x b y c z ac ab bc (2) (3) abx byc bc ac
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Section 12.1: Systems of Linear Equations: Substitution and Elimination 91.
10 have sin sin . Since is in the 9 9 9
interval , , we can apply the equation 2 2 above and get 10 sin 1 sin . sin 1 sin 9 9 9
3 1 4 2 2
94. r x 2 y 2 92. a.
y 1 3 x 3 3
tan
3
5 6 The polar form of z 3 i is
3
3
3
z r cos i sin 2 cos 56 i sin 56 2e 6
3
3
3
2
4 2 x 3 2 x3 5 2 x3 5 3x 2 2 x 3 3
4
2 2 x 3 x 5 4 x 20 6 x 9 x 2 2 x 3 x 510 x 9 x 20
2 2 x 3 x3 5 4 x3 5 3 x 2 2 x 3 3
i 5
2
.
2
95. A B 6,12,18, 24,30
b. 1 2
3 x 5
12
12 3 x 5
3 x 3
12
12
12 x 3
32
96. Since the length of the major axis is 20 then the vertices are (10,0) and (-10,0) and a = 10. The length of the minor axis is 12 so the minor vertices are (0,6) and (0,-6) and b = 6. So the x2 y2 1. formula for the ellipse is 100 36
3 x 5 2 1
x 3 3 x 3 3x 5 3
2
x 3 3x 9 3x 5 12 3 x 5 x 3 14 7 3 x 5 x 3 12 3 x 5
2
12
3
1
3
1
2
3
2
2
2
i
10 93. sin sin follows the form of the 9
i
5
i
7
31
2
62e 4
1
7
97. z w 6e 4 2e 6
2
i
12e 12
equation f 1 f x sin 1 sin x x , but
5 6
i
31
12e 12 i
7
12e 12
i 7 7 7 12 cos i sin 12e 12 12 12
10 we cannot use the formula directly since 9 is not in the interval , . We need to find 2 2
i
7
7 5
7 5
i z 6e 4 6 i 4 6 e 3e 4 6 5 i w 2 2e 6
an angle in the interval , for which 2 2 10 10 sin . The angle sin is in 9 9 quadrant II so sine is positive. The reference 10 angle of is and we want to be in 9 9 quadrant I so sine will still be positive. Thus, we
i
11
3e 12
i 11 11 11 3 cos i sin 3e 12 12 12
98. A $5000, r 0.04, n 12, t 1.5 r P A 1 n
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nt
0.04 5000 1 12
( 12)(1.5)
$4709.29
Chapter 12: Systems of Equations and Inequalities
2x 3y 6 0 9. 4 x 6 y 2 0 Write the system in standard form and then write the augmented matrix for the system of equations: 2x 3 y 6 3 6 2 4 x 6 y 2 4 6 2
1 1 f (b) f (a) cos ( 12 ) cos ( 12 ) 1 ( 12 ) ba 2
99.
3
2 3 1 3
100. Find the length of the sides of the triangle.
S1:
(3 0) 2 (9 5) 2 9 16 5
S2:
(12 0) 2 (0 5) 2 144 25 13
9x y 0 10. 3x y 4 0 Write the system in standard form and then write the augmented matrix for the system of equations: 9 x y 0 9 1 0 3 x y 4 3 1 4
S3: (12 3) 2 (0 9) 2 81 81 9 2 Now use the area formula: 1 S = (5 13 9 2) 15.364 2 K=
11. Writing the augmented matrix for the system of equations: 0.01x 0.03 y 0.06 0.01 0.03 0.06 0.10 0.20 0.13 x 0.10 y 0.20 0.13
15.364(15.364 5)(15.364 13)(15.364 9 2) 31.5 square units
12. Writing the augmented matrix for the system of equations: 3 3 3 3 4 4 3 x 2 y 4 3 2 4 1 2 1 1 x 1 y 2 4 3 3 3 3 4
Section 12.2 1. matrix 2. augmented
13. Writing the augmented matrix for the system of equations: x y z 10 1 1 1 10 3 3 0 5 3 3 5 x y x y 2z 2 1 1 2 2
3. third; fifth 4. b 5. True
14. Writing the augmented matrix for the system of equations: 5 x y z 0 5 1 1 0 5 1 1 0 5 x y 2x 3z 2 2 0 3 2
6. c 7. Writing the augmented matrix for the system of equations: x 5y 5 5 5 1 3 6 4 x 3 y 6 4
15. Writing the augmented matrix for the system of equations: 1 1 2 x yz 2 1 3 2 0 2 3 x 2 y 2 5 x 3 y z 1 5 3 1 1
8. Writing the augmented matrix for the system of equations: 3x 4 y 7 3 4 7 4 x 2 y 5 4 2 5
1332
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Section 12.2: Systems of Linear Equations: Matrices
1 3 4 3 x 3 y 4 z 3 21. 3 5 6 6 3 x 5 y 6 z 6 5 3 4 6 5 x 3 y 4 z 6
2 x 3 y 4 z 0 16. x 5 z 2 0 x 2 y 3z 2 Write the system in standard form and then write the augmented matrix for the system of equations: 0 2 x 3 y 4 z 0 2 3 4 5 z 2 1 0 5 2 x x 2 y 3z 2 1 2 3 2
R2 3r1 r2 1 3 4 3 5 6 5 3 4 1 3(1) 3 5
17. Writing the augmented matrix for the system of equations: 1 1 1 10 x y z 10 2 x y 2 z 1 2 1 2 1 3 4 0 5 3 x 4 y 5 4 x 5 y z 0 4 5 1 0
3 6 6 3 4 3 3(3) 5 3(4) 6 3(3) 6 3 4 6
1 3 4 3 0 4 6 3 5 3 4 6
R3 5r1 r3 1 3 4 3 0 4 6 3 5 3 4 6 3 4 3 1 0 6 3 4 5(1) 5 5(3) 3 5(4) 4 5(3) 6 1 3 4 3 0 4 6 3 0 12 24 21
18. Writing the augmented matrix for the system of equations: 1 1 2 1 5 x y 2z w 5 x 3 y 4 z 2 w 2 1 3 4 2 2 3 x y 5 z w 1 3 1 5 1 1 1 3 2 x 3 y 2 19. 2 5 5 2 x 5 y 5
1 3 3 5 x 3 y 3 z 5 22. 4 5 3 5 4 x 5 y 3 z 5 3 2 4 6 3x 2 y 4 z 6
R2 2r1 r2 1 3 2 1 3 2 2 5 5 2(1) 2 2( 3) 5 2( 2) 5 1 3 2 0 1 9
R2 4r1 r2 1 3 3 5 4 5 3 5 3 2 4 6 3 5 3 1 4(1) 4 4( 3) 5 4(3) 3 4(5) 5 2 4 6 3
1 3 3 x 3 y 3 20. 2 5 4 2 x 5 y 4 R2 2r1 r2
1 3 3 5 0 17 9 25 3 2 4 6
1 1 3 3 3 3 2 5 4 2(1) 2 2(3) 5 2(3) 4 1 3 3 0 1 2
1333
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Chapter 12: Systems of Equations and Inequalities R3 3r1 r3
1 3 4 6 x 3 y 4 z 6 24. 6 5 6 6 6 x 5 y 6 z 6 1 1 4 6 x y 4 z 6
1 3 3 5 4 5 3 5 3 2 4 6 3 3 5 1 0 9 17 25 3(1) 3 3(3) 2 3(3) 4 3(5) 6
R2 6r1 r2 1 3 4 6 6 5 6 6 1 1 4 6 1 3 4 6 6(1) 6 6(3) 5 6(4) 6 6( 6) 6 1 4 6 1 1 3 4 6 0 13 30 30 1 1 4 6
1 3 3 5 0 17 9 25 0 11 13 9
1 3 2 6 x 3 y 2 z 6 23. 2 5 3 4 2 x 5 y 3z 4 3 6 4 6 3x 6 y 4 z 6
R3 r1 r3
R2 2r1 r2 1 3 2 6 2 5 3 4 3 6 4 6 1 2 3 6 2(1) 2 2(3) 5 2(2) 3 2(6) 4 4 6 6 3 1 3 2 6 0 1 1 8 3 6 4 6
R3 3r1 r3
3 4 6 1 3 4 6 1 6 5 6 6 6 6 5 6 1 1 4 6 1 1 3 1 4 4 6 6 1 3 4 6 6 5 6 6 0 2 0 0
5 3 1 2 5 x 3 y z 2 25. 2 5 6 2 2 x 5 y 6 z 2 4 1 4 6 4 x y 4 z 6
R1 2r2 r1
1 3 2 6 0 1 1 8 3 6 4 6 2 3 6 1 1 8 0 1 3(1) 3 3(3) 6 3(2) 4 3(6) 6 1 3 2 6 0 1 1 8 0 15 10 12
5 3 1 2 2 5 6 2 4 1 4 6 2(2) 5 2( 5) 3 2(6) 1 2(2) 2 2 6 5 2 1 4 6 4 1 7 11 2 2 5 6 2 4 6 4 1
R3 2r2 r3
1 7 11 2 2 5 6 2 4 6 4 1 1 7 2 11 2 6 5 2 2(2) (4) 2( 5) 1 2(6) 4 2(2) 6 1 7 11 2 2 5 6 2 2 0 9 16
1334
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Section 12.2: Systems of Linear Equations: Matrices
3 1 2 4 x 3 y z 2 5 2 6 3x 5 y 2 z 6 3 6 4 6 3x 6 y 4 z 6
x 2 z 1 31. y 4 z 2 00 Consistent; x 1 2 z y 2 4 z z is any real number
4
26. 3
R1 r2 r1
4 3 1 2 3 5 2 6 3 6 4 6 (3) 4 ( 5) 3 (2) 1 (6) 2 2 6 3 5 3 6 4 6 1 2 3 4 3 5 2 6 3 6 4 6
or {( x, y, z ) | x 1 2 z , y 2 4 z , z is any real number} x 4z 4 32. y 3 z 2 00
Consistent; x 4 4z y 2 3z z is any real number
R3 r2 r3
1 2 3 4 3 5 2 6 3 6 4 6 2 3 4 1 2 6 3 5 3 (3) 5 (6) 2 4 6 6 1 2 3 4 3 5 2 6 0 11 6 12
or {( x, y, z ) | x 4 4 z , y 2 3 z , z is any real number} x1 1 33. x2 x4 2 x 2x 3 4 3 Consistent; x1 1 x2 2 x4 x3 3 2 x4 x4 is any real number or {( x1 , x2 , x3 , x4 ) | x1 1, x2 2 x4 , x3 3 2 x4 , x4 is any real number}
x 5 27. y 1 Consistent; x 5, y 1, or using ordered pairs (5, 1) .
x 4 28. y 0 Consistent; x 4, y 0, or using ordered pairs ( 4, 0) .
x1 1 34. x2 2 x4 2 x 3x 0 4 3
x 1 29. y 2 0 3
Consistent; x1 1 x2 2 2 x4 x3 3x4 x4 is any real number or {( x1 , x2 , x3 , x4 ) | x1 1, x2 2 2 x4 , x3 3 x4 , x4 is any real number}
Inconsistent x 0 30. y 0 0 2
Inconsistent
1335
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Chapter 12: Systems of Equations and Inequalities
x1 4 x4 2 35. x2 x3 3x4 3 00 Consistent; x1 2 4 x4 x2 3 x3 3 x4 x , x are any real numbers 3 4
x1 1 x2 2 x3 3 x4 0 or (1, 2,3, 0)
x y 8 39. x y 4 Write the augmented matrix: 1 1 8 1 8 1 R2 r1 r2 1 1 4 0 2 4 8 1 1 R2 12 r2 0 1 2 0 6 1 R1 r2 r1 0 1 2 The solution is x 6, y 2 or using ordered pairs (6, 2).
or {( x1 , x2 , x3 , x4 ) | x1 2 4 x4 , x2 3 x3 3x4 , x3 and x4 are any real numbers} x1 1 36. x2 2 x 2x 3 4 3 Consistent; x1 1 x2 2 x3 3 2 x4 x4 is any real number
x 2 y 5 40. x y 3 Write the augmented matrix: 1 2 5 1 2 5 R2 r1 r2 1 1 3 0 1 2 1 2 5 R2 r2 0 1 2 0 1 1 R1 2r2 r1 0 1 2 The solution is x 1, y 2 or using ordered pairs (1, 2).
or {( x1 , x2 , x3 , x4 ) | x1 1, x2 2, x3 3 2 x4 , x4 is any real number} x1 x4 2 x 2 x4 2 37. 2 x3 x4 0 00 Consistent; x1 2 x4 x2 2 2 x4 x3 x4 x4 is any real number
3 x 6 y 4 41. 5 x 4 y 5 Write the augmented matrix: 3 6 4 1 2 43 5 4 5 5 5 4
or {( x1 , x2 , x3 , x4 ) | x1 2 x4 , x2 2 2 x4 , x3 x4 , x4 is any real number} x1 1 x2 2 38. x3 3 x4 0
Consistent;
1 43 2 35 3 0 14
R2 5r1 r2
1 43 2 5 0 6 1
R2 141 r2
0 1 0 1
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R1 13 r1
1 3 3 4
R1 2r2 r1
Section 12.2: Systems of Linear Equations: Matrices
1 5 The solution is x , y or using ordered pairs 3 6 1 5 , . 3 6 3x 3 y 3 42. 8 4 x 2 y 3 Write the augmented matrix: 3 3 3 1 1 1 R1 1 r1 8 3 8 4 2 3 4 2 3
2 x 3 y 6 45. 1 x y 2 Write the augmented matrix: 2 3 6 1 3 3 2 R1 12 r1 1 1 1 1 2 1 1 2 3 3 2 R2 r1 r2 1 0 5 5 2 2 1 3 3 2 R2 52 r2 0 1 1
1 1 1 R2 4r1 r2 0 2 43 1 1 1 R2 12 r2 2 0 1 3 0 1 0 1
1 3 2 3
1 2 The solution is x , y or using ordered 3 3 1 2 pairs , . 3 3
46. 2 x y 2 1
x 2 y 8
x 2y 4 43. 2 x 4 y 8 Write the augmented matrix: 1 2 4 1 2 4 R2 2r1 r2 2 4 8 0 0 0 This is a dependent system. x 2y 4 x 4 2y The solution is x 4 2 y, y is any real number or {( x, y ) | x 4 2 y, y is any real number}
-1 7 1 3 3 0 0 0 This is a dependent system.
3 R1 32 r2 r1 1 0 2 0 1 1 3 3 The solution is x , y 1 or , 1 . 2 2
R1 r2 r1
3x y 7 44. 9 x 3 y 21 Write the augmented matrix: 1 7 3 1 7 1 3 3 9 3 21 9 3 21
3x y 7 3x 7 y The solution is y 3x 7, x is any real number or {( x, y ) | y 3x 7, x is any real number}
Write the augmented matrix: 1 2 4 1 2 1 2 8 8 1 2 1 2
R1 2r1
2 4 R r r 1 1 2 2 0 4 12 1 2 4 R2 14 r2 0 1 3 0 2 1 R1 2r2 r1 0 1 3 The solution is x 2, y 3 or (2, 3) .
3x 5 y 3 47. 15 x 5 y 21 Write the augmented matrix:
R1 13 r1 R2 9 r1 r2
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Chapter 12: Systems of Equations and Inequalities
3 5 3 1 5 1 3 R1 13 r1 15 5 21 15 5 21 5 3 1 1 R2 15r1 r2 0 30 6 53 1 1 R2 30 r2 1 1 0 1 5 0 4 3 R1 53 r2 r1 1 0 1 1 5
1 0 3 8 2 3 0 1 2 2 4 0 0 0 1 0 3 8 2 0 1 32 2 1 0 0 0
4 2 x y 50. 2 y 4 z 0 3x 2 z 11 Write the augmented matrix: 2 0 4 1 0 4 0 2 3 0 2 11
2 x y 1 48. 1 3 x 2 y 2
1 0 1 0
12 R1 12 r1 3 2 12 12 R 1r r 2 1 2 1 2 0 12 R1 12 r2 r1 1 2
R3 14 r3
1 0 0 8 R1 32 r3 r1 0 1 0 2 R2 3 r3 r2 0 0 1 0 2 The solution is x 8, y 2, z 0 or (8, 2, 0).
4 1 4 1 The solution is x , y or , . 3 5 3 5
2 1 1 1 1 2 3 1 1 12 1 2 2
R1 r2 r1 R3 2r2 r3
1 1 The solution is x , y 2 or , 2 . 2 2
1 1 0 2 2 0 2 0 4 0 2 11 3
R1 12 r1
1 0 0 1 0 0 1 0 0
0 2 0 R3 3r1 r3 4 2 32 2 5 1 0 2 2 0 R2 12 r2 1 2 32 2 5 0 1 2 1 R1 2 r2 r1 0 1 2 R3 3 r2 r3 2 0 5 5 1 0 1 2 0 R3 15 r3 0 1 2 0 0 1 1 1 0 0 3 R1 r3 r1 0 1 0 2 0 0 1 1 R2 2r3 r2 The solution is x 3, y 2, z 1 or (3, 2, 1) .
6 x y 3 z 16 49. 2 x 2y z 4 Write the augmented matrix: 1 1 0 6 2 0 3 16 0 2 1 4 1 1 0 6 0 2 3 4 R2 2r1 r2 0 2 1 4 1 1 0 6 0 1 32 2 R2 12 r2 1 4 0 2
1 2
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Section 12.2: Systems of Linear Equations: Matrices 51. Write the augmented matrix: 1 4 2 9 1 1 4 3 3 3 7 2 1 4 2 9 0 13 5 31 0 5 1 11 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0
2 9 5 31 1 13 13 1 11 5 1 5
4
1 3 0 2 2 2 1 7 3 4 3 7
R2 3r1 r2 R3 2r1 r3 1 R2 13 r2 R3 1 r3 5
2
R1 12 r1
1 1 32 0 2 0 3 2 7 3 0 11 7 2 2
R2 2r1 r2 R3 3r1 r3
1 0 0 1 0 0
9 5 31 13 1 13 R3 r2 r3 12 12 0 65 65 2 9 4 5 31 65 R3 12 r3 1 13 13 0 1 1 6 7 0 13 13 5 31 13 13 1 R1 4r2 r1 0 1 1 0 0 1 5 R3 13 r3 r2 0 1 2 R1 6 r3 r1 13 0 1 1 4
1 1 23 0 2 2 2 1 7 3 4 3 7
1 2
32
11 2
3 2
1 23
0
76
1
23
0 13 6
0 73 7 7 6 73 35 6
7 1 0 76 6 0 1 23 73 35 1 13 0 0
R2 13 r2 R1 12 r2 r1 R3 11 r2 r3 2
R3 136 r3
56 1 0 0 13 R1 76 r3 r1 7 0 1 0 13 R2 2 r3 r2 35 3 0 0 1 13 56 7 35 or The solution is x , y , z 13 13 13 56 7 35 , , . 13 13 13
The solution is x 1, y 2, z 1 or (1, 2, 1) .
2 x y 3z 0 52. 2 x 2 y z 7 3x 4 y 3z 7
Write the augmented matrix:
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Chapter 12: Systems of Equations and Inequalities
2x 2 y 2z 2 53. 2 x 3 y z 2 3x 2 y 0 Write the augmented matrix: 2 2 2 2 3 1 2 2 3 0 0 2 1 1 1 1 2 3 1 2 R1 12 r1 3 2 0 0 1 1 1 1 R2 2r1 r2 0 5 3 0 R3 3r1 r3 0 5 3 3 1 1 1 1 0 5 3 0 R3 r2 r3 0 0 0 3
1 1 1 1 2 3 4 1 3 2 7 0 1 1 1 1 1 2 3 4 R1 r1 3 2 7 0 1 1 1 1 R r r 0 1 4 3 2 1 2 R 3r1 r3 0 1 4 3 3 1 0 5 2 R r r 0 1 4 3 1 2 1 R r2 r3 0 0 0 0 3 The matrix in the last step represents the system x 5z 2 x 5z 2 y 4 z 3 or, equivalently, y 4 z 3 0 0 00 The solution is x 5 z 2 , y 4 z 3 , z is any
There is no solution. The system is inconsistent.
real number or {( x, y, z ) | x 5 z 2, y 4 z 3, z is any real number}.
2x 3 y z 0 54. x 2 y z 5 3x 4 y z 1
2x 3y z 0 56. 3x 2 y 2 z 2 x 5 y 3z 2 Write the augmented matrix: 2 3 1 0 3 2 2 2 1 5 3 2
Write the augmented matrix: 2 3 1 0 2 1 5 1 3 4 1 1 1 2 1 5 Interchange 2 3 1 0 r and r 1 2 3 4 1 1 1 2 1 5 R2 2r1 r2 0 1 1 10 R 3r r 1 3 3 0 16 2 2 1 2 1 5 0 1 1 10 R3 2r2 r3 0 0 0 4
1 5 3 2 3 2 2 2 2 3 1 0 1 5 3 2 0 13 7 4 0 13 7 4
Interchange r1 and r3 R2 3r1 r2 R3 2r1 r3
1 5 3 2 R3 r2 r3 7 4 0 1 13 13 1 R2 13 r2 0 0 0 0 1 0 4 6 13 13 7 4 0 1 13 R1 5 r2 r1 13 0 0 0 0 The matrix in the last step represents the system
There is no solution. The system is inconsistent. x y z 1 55. x 2 y 3 z 4 3x 2 y 7 z 0 Write the augmented matrix:
1340
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Section 12.2: Systems of Linear Equations: Matrices
3 2 2 6 7 3 2 1 2 3 4 0 1 2 2 2 3 3 7 3 2 1 2 3 4 0
4 6 x 13 z 13 y 7 z 4 13 13 0 0 or, equivalently, 4 6 x 13 z 13 y 7 z 4 13 13 0 0
1 2 3 5 0 3 0 5 3
4 6 7 4 z , y z , 13 13 13 13 4 6 z is any real number or ( x, y, z ) x z , 13 13
The solution is x
y
Write the augmented matrix: 2 2 3 6 4 3 2 0 2 3 7 1
3 1 1 3 2 0 1 4 12 7 0 1 4
R2 4r1 r2 R3 2r1 r3
R2 7 r1 r2 R3 2r1 r3
x y z 6 59. 3 x 2 y z 5 x 3 y 2 z 14 Write the augmented matrix: 1 1 1 6 1 5 3 2 3 2 14 1
2 x 2 y 3z 6 57. 4 x 3 y 2 z 0 2 x 3 y 7 z 1
R1 12 r1
2 15 4
2 1 2 2 3 3 5 83 15 0 R3 r2 r3 3 0 0 19 0 There is no solution. The system is inconsistent.
7 4 z , z is any real number . 13 13
1 1 3 3 2 4 3 2 0 2 3 7 1
2 3 83 8 3
R1 13 r1
1 1 1 6 0 5 4 23 0 2 1 8
1 0 5 9 2 R1 r2 r1 0 1 4 12 R3 r2 r3 0 19 0 0 There is no solution. The system is inconsistent.
1 1 1 4 0 1 5 0 2 1
6 23 5 8
1 0 1 5 4 0 1 5 3 0 0 5
7 5 23 5 65
7 1 0 1 5 5 23 4 0 1 5 5 0 0 1 2
R2 3r1 r2 R3 r1 r3
R2 15 r2 R1 r2 r1 R3 2r2 r3
R3 53 r3
1 0 0 1 R1 15 r3 r1 0 1 0 3 R2 4 r3 r2 0 0 1 2 5 The solution is x 1, y 3, z 2 , or (1, 3, 2) .
3x 2 y 2 z 6 58. 7 x 3 y 2 z 1 2 x 3 y 4 z 0 Write the augmented matrix:
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Chapter 12: Systems of Equations and Inequalities
x y z 4 60. 2 x 3 y 4 z 15 5 x y 2 z 12
Write the augmented matrix: 1 1 1 4 4 15 2 3 5 1 2 12
1 0 1 13 4 4 3 1 0 1 8 8 11 0 0 11 4 4 1 0 1 13 4 4 3 1 0 1 8 8 1 1 0 0
1 1 1 4 0 1 2 7 0 6 7 32
1 0 0 3 1 0 1 0 2 0 0 1 1
1 1 1 4 7 0 1 2 0 6 7 32 1 0 1 3 0 1 2 7 0 0 5 10 1 0 1 3 0 1 2 7 0 0 1 2
R2 2r1 r2 R3 5r1 r3
R1 2r2 r1 R3 6r2 r3
R3 114 r3 R1 14 r3 r1 R2 3 r3 r2 8
1 2
1 2
The solution is x 3, y , z 1 or 3, , 1 .
R2 r2
x 4 y 3z 8
62. 3x y 3z 12
R1 r2 r1 R3 6r2 r3
x y 6z 1
Write the augmented matrix: 1 4 3 8 3 1 3 12 1 1 6 1
R3 15 r3
1 4 3 8 R2 3r1 r2 0 13 12 36 R3 r1 r3 0 3 9 9 1 4 3 8 36 1 13 0 R2 13 r2 1 12 13 9 9 0 3 9 40 13 13 1 0 R1 4r2 r1 36 13 0 1 12 13 R3 3r2 r3 0 0 81 9 13 13
1 0 0 1 R1 r3 r1 3 0 1 0 R2 2r3 r2 0 0 1 2 The solution is x 1, y 3, z 2 or (1, 3, 2) .
x 2 y z 3 61. 2 x 4 y z 7 2 x 2 y 3z 4
Write the augmented matrix: 1 2 1 3 1 7 2 4 2 2 3 4 1 2 1 3 0 8 3 1 0 6 5 2
R2 2r1 r2 R3 2r1 r3
1 2 1 3 3 1 0 1 8 8 0 6 5 2
R2 18 r2
1 0 0 1 0 0
40 13 36 13 R3 13 r 1 81 3 1 0 1 9 0 0 3 R1 9 r3 r1 13 1 0 83 R3 12 r3 r2 13 1 0 1 9 8 1 8 1 The solution is x 3, y , z or 3, , . 3 9 3 9
0
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9 13 12 13
Section 12.2: Systems of Linear Equations: Matrices
2 3x y z 3 63. 2 x y z 1 8 4x 2 y 3 Write the augmented matrix: 3 1 1 2 3 2 1 1 1 8 4 2 0 3
1 1 0 1 2 1 1 1 1 2 1 8 3
1 1 1 2 3 3 9 1 1 2 1 8 0 3 4 2
R1 13 r1
1 1 3 5 0 3 2 0 3
R2 2r1 r2 R3 4r1 r3
13 5 3 4 3
2 9 5 9 16 9
2 1 1 1 3 3 9 1 0 1 1 3 16 4 0 23 3 9 1 1 0 0 3 1 0 1 1 3 2 0 0 2 1 1 0 0 3 1 0 1 1 3 0 0 1 1
1 0 0 1 3 2 0 1 0 3 0 0 1 1
R2 53 r2 R1 13 r2 r1 R3 2 r2 r3 3
1 1 0 1 0 3 1 1 0 1 1 5 3
R2 2r1 r2 R3 r1 r3
1 1 0 1 0 1 1 5 3 0 3 1 1
Interchange r2 and r3
1 0 0 1 0 0
1 0 1 1 1 5 3 0 4 4 1 0 1 1 1 5 3 0 1 1
1 0 0 1 0 0
1 0 1 1 0 2 3 0 1 1 0 0 1 3 1 0 2 3 0 1 1
R3 3r2 r3
R3 14 r3 R2 r2 r3
R1 r1 r2
1 2 1 2 The solution is x , y , z 1 or , , 1 . 3 3 3 3
R3 12 r3
x y z w 4 2x y z 0 65. 3x 2 y z w 6 x 2 y 2 z 2 w 1 Write the augmented matrix: 1 1 1 1 4 1 0 0 2 1 3 2 1 1 6 1 2 2 2 1
R2 r3 r2
1 2 1 2 The solution is x , y , z 1 or , , 1 . 3 3 3 3 x y 1 64. 2 x y z 1 8 x 2y z 3 Write the augmented matrix:
1 1 1 1 4 0 3 1 2 8 0 1 2 4 6 0 3 3 1 5
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R2 2r1 r2 R3 3r1 r3 R r r 4 1 4
Chapter 12: Systems of Equations and Inequalities
1 1 1 1 4 0 1 2 4 6 0 3 1 2 8 0 3 3 1 5
Interchange r2 and r3
1 0 0 0
1 3
4 2 1 4 1 1 3 2 3 0 4 7
R2 r1 r2 R3 2r1 r3 R4 2 r1 r4
1 1 1 1 4 0 6 1 2 4 0 3 1 2 8 0 3 3 1 5
R2 r2
1 0 0 0
1 4 1 1 3 2 3 2 1 4 3 0 4 7
Interchange r2 and r3
1 0 0 0
0 1 3 2 6 1 2 4 0 5 10 10 0 3 13 13
R1 r2 r1 R3 3 r2 r3 R 3r r 4 2 4
1 0 0 0
6 4 1 1 3 2 0 5 10 10 0 3 13 13
R1 r2 r1 R3 3 r2 r3 R 3 r r 4 2 4
1 0 0 0
0 1 3 2 6 1 2 4 0 1 2 2 0 3 13 13
R3 15 r3
1 0 0 0
6 1 1 2 0 5 10 10 0 0 35 35
R4 3 r3 5r4
1 0 0 0
0 0 1 0 1 0 0 2 0 1 2 2 0 0 7 7 0 0 1 0 1 0 0 2 0 1 2 2 0 0 1 1
1 0 0 0
1
0
0
1
1
1
2
2
4 3
1 0 2 4 6 R4 15 r3 0 1 1 3 2 R4 1 r4 0 0 1 2 2 35 0 0 0 1 1 Write the matrix as the corresponding system: x 2 z 4w 6 y z 3w 2 z 2w 2 w 1 Substitute and solve: z 2(1) 2
R1 r3 r1 R2 2 r3 r2 R 3 r r 3 4 4
R4 17 r4
1 0 0 0 1 0 1 0 0 2 R1 r4 r1 0 0 1 0 0 R3 2 r4 r3 0 0 0 1 1 The solution is x 1, y 2, z 0, w 1 or (1, 2, 0, 1).
z2 2 z0 y 0 3(1) 2 y 3 2 y 1
x y z w 4 x 2 y z 0 66. 2x 3 y z w 6 2 x y 2 z 2w 1 Write the augmented matrix: 1 1 1 1 4 1 0 0 1 2 2 3 1 1 6 2 1 2 2 1
x 2(0) 4(1) 6 x04 6 x2 The solution is x 2 , y 1 , z 0 , w 1 or (2, 1, 0, 1).
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Section 12.2: Systems of Linear Equations: Matrices
x y z 5 69. x 3 2 y 2z 0 Write the augmented matrix: 1 1 1 5 3 2 2 0
x 2y z 1 67. 2 x y 2 z 2 3x y 3z 3 Write the augmented matrix: 1 2 1 1 2 1 2 2 3 1 3 3 1 2 1 1 R2 2r1 r2 0 5 0 0 R3 3r1 r3 0 5 0 0 1 2 1 1 0 5 0 0 R3 r2 r3 0 0 0 0
1 1 1 5 R2 3r1 r2 0 5 5 15 1 1 1 5 R2 15 r2 0 1 1 3 1 0 0 2 R1 r2 r1 0 1 1 3
The matrix in the last step represents the system x2 x 2 or, equivalently, y z 3 y z 3 Thus, the solution is x 2 , y z 3 , z is any real number or {( x, y, z ) | x 2, y z 3, z is any real number}.
The matrix in the last step represents the system x 2y z 1 5y 0 00
2x y z 4 70. x y 3z 1 Write the augmented matrix: 2 1 1 4 1 1 3 1
Substitute and solve: 5 y 0 x 2(0) z 1 y0 z 1 x The solution is y 0, z 1 x, x is any real number or {( x, y, z ) | y 0, z 1 x, x is any real number}.
1 1 3 1 2 1 1 4 1 1 3 1 0 3 5 6 1 1 3 1 5 3 2 0 1
x 2y z 3 68. 2 x y 2 z 6 x 3 y 3z 4 Write the augmented matrix: 1 2 1 3 2 1 2 6 1 3 3 4 1 2 1 3 R2 2r1 r2 0 5 4 0 R r r 1 3 3 0 5 4 1 1 2 1 3 0 5 4 0 R3 r2 r3 0 0 0 1 There is no solution. The system is inconsistent.
interchange r and r 1 2
R2 2r1 r2
R2 13 r2
1 0 4 1 3 R1 r2 r1 5 0 1 3 2 The matrix in the last step represents the system 4 4 x 3 z 1 x 1 3 z or, equivalently, y 5 z 2 y 2 5 z 3 3 4 5 Thus, the solution is: x 1 z , y 2 z , z 3 3
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Chapter 12: Systems of Equations and Inequalities
x 8 z 7 y 7 z 7 z 19 18 Substitute and solve: 19 y 7 7 18 7 y 18
4 is any real number or ( x, y, z ) x 1 z , 3 5 y 2 z , z is any real number . 3 2 x 3 y z 3 x yz 0 71. x y z 0 x y 3 z 5 Write the augmented matrix: 2 3 1 3 1 1 1 0 1 1 1 0 1 1 3 5 1 1 1 0 2 3 1 3 interchange 1 1 1 0 r1 and r2 1 1 3 5
19 x 8 7 18 13 x 9 13 7 19 Thus, the solution is x , y , z or 9 18 18 13 7 19 , , . 9 18 18 x 3y z 1 2 x y 4 z 0 72. x 3 y 2z 1 x 2y 5 Write the augmented matrix: 1 3 1 1 2 1 4 0 1 3 2 1 1 2 0 5 1 3 1 1 R2 2r1 r2 0 5 6 2 R r1 r2 0 0 1 0 3 R4 r1 r2 0 1 1 4
1 1 1 0 0 5 1 3 R2 2r1 r2 R3 r1 r3 0 0 0 0 0 2 4 5 R4 r1 r4 1 1 1 0 0 5 1 3 interchange 0 2 4 5 r3 and r4 0 0 0 0 1 1 1 0 0 1 7 7 R2 2r3 r2 0 2 4 5 0 0 0 0 1 0 8 7 0 1 7 7 R1 r2 r1 0 0 18 19 R3 2r2 r3 0 0 0 0
1 3 1 1 0 1 1 4 0 0 1 0 0 5 6 2 1 0 2 13 0 1 1 4 0 0 1 0 0 0 1 22
0 8 7 1 7 7 R3 = 181 r3 0 1 19 18 0 0 0 The matrix in the last step represents the system 1 0 0 0
1 0 0 0
0 1 0 0
0 13 0 4 1 0 0 22
interchange r2 and r4
R1 3r2 r1 R4 5r2 r4
R1 2r3 r1 R2 r3 r2 R r r 4 3 4
There is no solution. The system is inconsistent.
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Section 12.2: Systems of Linear Equations: Matrices
4 x y 5 74. 2 x y z w 5 zw4 Write the augmented matrix: 4 1 0 0 5 2 1 1 1 5 0 0 1 1 4
4x y z w 4 73. x y 2 z 3w 3 Write the augmented matrix: 4 1 1 1 4 1 1 2 3 3 1 1 2 3 3 4 1 1 1 4
interchange r1 and r2
1 1 2 3 3 R2 4r1 r2 0 5 7 13 8 The matrix in the last step represents the system x y 2 z 3w 3 5 y 7 z 13w 8 The second equation yields 5 y 7 z 13w 8 5 y 7 z 13w 8 7 13 8 y z w 5 5 5 The first equation yields x y 2 z 3w 3 x 3 y 2 z 3w Substituting for y: 13 8 7 x 3 z w 2 z 3w 5 5 5 3 2 7 x z w 5 5 5 3 2 7 Thus, the solution is x z w , 5 5 5 7 13 8 y z w , z and w are any real numbers or 5 5 5 3 2 7 7 13 8 ( x, y, z, w) x z w , y z w , 5 5 5 5 5 5
1 1 0 0 5 4 4 2 1 1 1 5 0 0 1 1 4 1 1 0 0 5 4 4 0 12 1 1 152 0 0 1 1 4 1 0 0 1 0 0 1 0 0
R1 14 r1 R2 2r1 r2
0 54 1 2 2 15 R2 2r2 0 1 1 4 0 12 12 5 1 2 2 15 R1 14 r2 r1 0 1 1 4 0 0 1 3 R1 12 r3 r1 1 0 4 7 R2 2r3 r2 0 1 1 4 The matrix in the last step represents the system x w 3 x 3 w or, equivalently, y w 4 7 y 7 4 w z w 4 z 4 w The solution is x 3 w , y 7 4 w , z 4 w , 14
0
w is any real number or ( x, y, z , w) | x 3 w, y 7 4 w, z 4 w, w is any real number}
z and w are any real numbers .
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Chapter 12: Systems of Equations and Inequalities 75. Each of the points must satisfy the equation y ax 2 bx c . (1, 2) : 2 abc (2, 7) : 7 4a 2b c (2, 3) : 3 4a 2b c Set up a matrix and solve: 1 1 1 2 4 2 1 7 4 2 1 3 1 1 1 2 R2 4r1 r2 0 6 3 15 R3 4r1 r3 0 2 3 11 1 1 1 2 5 1 R2 16 r2 0 1 2 2 0 2 3 11 1 1 0 12 2 R1 r2 r1 5 1 0 1 2 2 R 2r r 2 3 3 0 0 2 6
1 0 0 1 0 0
0 1
1 2 1 2
0 1 0 0 1 0 0 1
1 0 0 1 0 0
0 13 1 0
4 3
1
1 3 4 3
2
1 R2 6 r2 R3 r2 r3
R1 r2 r1 1 R3 5 r3
1 0 0 1 R1 13 r3 r1 0 1 0 4 R1 4 r3 r2 0 0 1 3 2 The solution is a 1, b – 4, c 2 ; so the
equation is y x 2 4 x 2 .
77. Each of the points must satisfy the equation f ( x) ax3 bx 2 cx d . f (3) 112 : 27 a 9b 3c d 112 f (1) 2 : a b c d 2 f (1) 4 : abcd 4 8a 4b 2c d 13 f (2) 13 : Set up a matrix and solve: 27 9 3 1 112 2 1 1 1 1 1 1 1 1 4 13 8 4 2 1 1 1 1 1 4 2 Interchange 1 1 1 1 27 9 3 1 112 r3 and r1 13 8 4 2 1 1 1 1 1 4 R2 r1 r2 0 0 2 2 2 R 27 r r 1 3 0 36 24 28 4 3 R4 8 r1 r4 0 4 6 7 19
12 5 R3 12 r3 2 3 2 R1 12 r3 r1 1 R2 1 r3 r2 2 3
The solution is a 2, b 1, c 3 ; so the equation is y 2 x 2 x 3 . 76. Each of the points must satisfy the equation y ax 2 bx c . (1, 1) : 1 a b c (3, 1) : 1 9a 3b c (– 2,14) : 14 4a 2b c Set up a matrix and solve: 1 1 1 1 3 1 1 9 4 2 1 14 1 1 1 1 0 6 8 8 0 6 3 18
1 1 43 43 0 5 10 1 1
R2 9r1 r2 R 4r r 1 3 3
1 1 1 1 4 0 0 1 1 1 0 36 24 28 4 0 4 6 7 19
R2 12 r2
1 0 0 0
R1 r2 r1 R3 36 r2 r3 R 4r r 2 4 4
3 1 1 1 0 24 8 40 0 6 3 15 0
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1 0
0
Section 12.2: Systems of Linear Equations: Matrices
1 0 0 0
3 1 1 1 5 5 0 1 3 3 0 6 3 15
0
1 0 0 0
0 0 1 0
1 0 0 0
0
1 0
1 3
R3 241 r3
1 0 0 0
14 3
1 R1 r3 r1 5 1 R 6 r3 r4 0 1 3 3 4 0 0 5 25 1 14 1 0 0 3 3 0 1 0 1 1 R4 15 r4 5 1 0 0 1 3 3 0 0 0 5 1 1 0 0 0 1 3 R1 3 r4 r1 4 0 0 0 1 R r r 2 4 2 0 0 1 0 0 1 R3 3 r4 r3 5 0 0 0 1 The solution is a 3, b 4, c 0, d 5 ; so the 1
1 0 0 0
1 0 0 0 1 0 0 0
equation is f ( x) 3 x 3 4 x 2 5 .
1 1 1 1 4 2 9 3
5 1 3 1 1 10 1 15
R2 r1 r2 R 8 r r 1 3 3 R 27 r r 1 4 4
1 5 1 1 1 0 0 1 1 4 0 12 6 9 30 0 18 24 26 120
R2 12 r2
0 1 0 0 0 0 1 0 0 1 0 0
4 1 4 R1 r3 r1 R 24 r3 r4 12 3 4 20 120 1 4 2 1 4 1 r R4 20 4 12 3 1 6 1 2
1 1 0 0 2 0 1 0 0 0 0 1 6 0 0 0
R1 1 r4 r1 2 R2 r4 r2 R3 1 r4 r3 2
79. Let x = the number of servings of salmon steak. Let y = the number of servings of baked eggs. Let z = the number of servings of acorn squash. Protein equation: 30 x 15 y 3 z 78 Carbohydrate equation: 20 x 2 y 25 z 59 Vitamin A equation: 2 x 20 y 32 z 75 Set up a matrix and solve: 30 15 3 78 20 2 25 59 2 20 32 75
Interchange r3 and r1
1 5 1 1 1 0 0 8 2 2 0 12 6 9 30 0 18 24 26 120
1 0
R3 16 r3
equation is f ( x) x3 2 x 2 6 .
4 2 1 10 3 1 1 1 5 1 1 1 9 3 1 15
1 1 8 27
0 0
R1 r2 r1 R3 12 r2 r3 R 18 r r 4 2 4
The solution is a 1, b 2, c 0, d 6 ; so the
78. Each of the points must satisfy the equation f ( x) ax3 bx 2 cx d . f ( 2) 10 : 8a 4b 2c d 10 f (1) 3 : abcd 3 f (1) 5 : abcd 5 27a 9b 3c d 15 f (3) 15 : Set up a matrix and solve: 8 1 1 27
1 0 1 0 1 1 4 0 6 3 18 0 24 8 48 0 0 1 1 0 1 1 4 1 0 3 1 2 0 24 8 48 0
2 20 32 75 20 2 25 59 30 15 3 78
Interchange r3 and r1
1 10 16 37.5 59 R1 12 r1 20 2 25 30 15 3 78 1 10 16 37.5 R2 20r1 r2 0 198 295 691 0 285 477 1047 R3 30r1 r3
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Chapter 12: Systems of Equations and Inequalities
1 10 16 37.5 0 198 295 691 3457 3457 0 66 66 0 1 10 16 37.5 0 198 295 691 0 0 1 1 Substitute z 1 and solve: 198 y 295(1) 691 198 y 396 y2
The dietitian should provide 1 serving of pork chops, 2 servings of corn on the cob, and 2 servings of 2% milk.
R3 9566 r2 r3
81. Let x = the amount loaned from Bank One. Let y = the amount loaned from Bank Two. Let z = the amount loaned from Bank Three. Total load equation: x y z 10, 000 Annual interest equation: 0.02 x 0.03 y 0.04 z 330 Condition on interest equation: x 0.5 z x 0.5 z 0 Set up a matrix and solve:
66 r3 R3 3457
x 10(2) 16(1) 37.5 x 36 37.5 x 1.5
1 1 1 10,000 330 0.02 0.03 0.04 1 0 0.5 0
The dietitian should serve 1.5 servings of salmon steak, 2 servings of baked eggs, and 1 serving of acorn squash. 80. Let x = the number of servings of pork chops. Let y = the number of servings of corn on the cob. Let z = the number of servings of 2% milk. Protein equation: 23x 3 y 9 z 47 Carbohydrate equation: 16 y 13 z 58 Calcium equation: 10 x 10 y 300 z 630 Set up a matrix and solve: 23 3 9 47 0 16 13 58 10 10 300 630 1 1 30 63 0 16 13 58 23 3 9 47 1 30 63 1 13 29 0 1 16 8 0 20 681 1402 467 475 1 0 16 8 13 29 0 1 8 16 0 0 2659 1402 4 1 0 0 1 0 0
0 1
467 16 13 16
0
1
1 1 0 2 0 1 2 0 0
475 8 29 8
Interchange 1 r3 and r1 10 R3 23r1 r3 R2 1 r2 16
1 1 1 10,000 0 0.01 0.02 130 0 1 1.5 10,000
R2 0.02 r1 r2 R3 r1 r3
1 1 1 10,000 13,000 0 1 2 0 1 1.5 10,000
R2 100 r2
1 0 1 3000 0 1 2 13,000 0 0 0.5 3000
R1 r2 r1 R3 r2 r3
1 0 1 3000 0 1 2 13,000 0 0 1 6000
R3 2 r3
1 0 0 3000 0 1 0 1000 0 0 1 6000
R1 r3 r1 R2 2r3 r2
Carletta should borrow $3000 from Bank One, $1000 from Bank Two, and $6000 from Bank Three.
R1 r2 r1 R3 20r2 r3
82. Let x = the fixed delivery charge; let y = the cost of each tree, and let z = the hourly labor charge. 1st subdivision: x 250 y 166 z 7520 2nd subdivision: x 200 y 124 z 5945 3rd subdivision: x 300 y 200 z 8985 Set up a matrix and solve:
4 r R3 2659 3
2 R1 467 r3 r1 16 R 13 r r 2 3 2 16
1 250 166 7520 1 200 124 5945 1 300 200 8985
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Section 12.2: Systems of Linear Equations: Matrices
1 0 4 48 0 1 2 15 0 0 1 10
1 250 166 7520 R2 r1 r2 0 50 42 1575 0 50 34 1465 R3 r3 r1 1 250 166 7520 1 R2 50 r2 0 1 0.84 31.5 R 1 r 0 1 0.68 29.3 3 50 3 1 0 44 355 0 1 0.84 31.5 0 0 0.18 2.2 1 0 44 355 0 1 0.84 31.5 0 0 1 13.75 1 0 0 250 0 1 0 19.95 0 0 1 13.75
1 0 0 8 R1 4r3 r1 0 1 0 5 R2 2r3 r2 0 0 1 10 The company should produce 8 Deltas, 5 Betas, and 10 Sigmas.
R1 r1 250 r2 R3 r2 r3
84. Let x = the number of cases of orange juice produced; let y = the number of cases of grapefruit juice produced; and let z = the number of cases of tomato juice produced. Sterilizing equation: 9 x 10 y 12 z 398 Filling equation: 6 x 4 y 4 z 164 Labeling equation: x 2 y z 58 Set up a matrix and solve: 9 10 12 398 6 4 4 164 1 2 1 58 1 2 1 58 Interchange 6 4 4 164 9 10 12 398 r1 and r3 1 2 58 1 R2 6 r1 r2 0 8 2 184 0 8 R3 9r1 r3 3 124 1 2 1 58 23 R2 18 r2 0 1 14 0 8 3 124
R3 0.161 r3 R1 r1 44r3 R2 r2 0.84 r3
The delivery charge is $250 per job, the cost for each tree is $19.95, and the hourly labor charge is $13.75. 83. Let x = the number of Deltas produced. Let y = the number of Betas produced. Let z = the number of Sigmas produced. Painting equation: 10 x 16 y 8 z 240 Drying equation: 3 x 5 y 2 z 69 Polishing equation: 2 x 3 y z 41 Set up a matrix and solve: 10 16 8 240 3 5 2 69 2 3 1 41 1 3 2 1 0 0
1 2 33 5 2 69 3 1 41 1 2 33 2 4 30 1 3 25
R1 3r2 r1
1 0 1 12 2 0 1 14 23 0 0 5 60 1 0 1 12 2 0 1 14 23 0 0 1 12 1 0 0 6 0 1 0 20 0 0 1 12
R2 3r1 r2 R3 2r1 r3
1 1 2 33 0 1 2 15 0 1 3 25
R2 12 r2
1 0 4 48 0 1 2 15 0 0 1 10
R1 r1 r2 R3 r3 r2
R3 r3
R1 2r2 r1 R3 8 r2 r3
R3 15 r3 R1 12 r3 r1 1 R2 4 r3 r2
The company should prepare 6 cases of orange juice, 20 cases of grapefruit juice, and 12 cases of tomato juice. 1351
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Chapter 12: Systems of Equations and Inequalities 85. Rewrite the system to set up the matrix and solve: I I I I I I I I 0 200 I 40 80 I 0 200 I 80 I 40 360 20 I 80 I 40 0 80 I 20 I 400 80 70 I 200 I 0 200 I 70 I 80 1
2
1
4
3
1
2
2
3
1
2
4
3
1
1 0 0 0
4
2
2
3
1
4
0 1 1 1 1 0 0 40 200 80 0 80 20 0 400 0 0 70 80 200 1 0 1 1 1 0 280 200 200 40 0 80 20 0 400 80 0 200 200 270 1 0 1 1 1 5 5 1 0 7 7 1 7 0 80 20 0 400 80 0 200 200 270 2 27 17 1 0 7 5 1 57 7 7 0 1 0 0 540 7 400 7 2720 7 400 890 360 7 7 0 0 7 27 57
1 0 0 0
7 1 20 0 1 27 400 890 0 7 7
1 0 0 0
0 0 27 5 27 1 0 20 0 1 27 2290 0 0 27
0
2
7
5
2
17 1 7 136 27 360 7
35
27
101
27
136 9160
27 27
R2 200r1 r2 R4 200 r1 r4
1 r2 R2 = 280
R1 r1 r2 R r r 80 2 3 3 R r 200r 4 4 2
2
0 0 27 5 27 1 0 20 0 1 27 0 0 1
35
27
101
27
27 R4 2290 r4 4 1 0 0 0 1 R r 2 r 1 1 27 4 0 1 0 0 3 R2 r2 5 27 r4 0 0 1 0 8 R3 r3 20 27 r4 0 0 0 1 4 The solution is I1 1 , I 2 3 , I 3 8 , I 4 4 . 136
27
86. Rewrite the system to set up the matrix and solve: I1 I 3 I 2 I1 I 2 I 3 0 6 I1 3I 3 24 24 6 I1 3I 3 0 12 24 6 I 6 I 0 6 I 6 I 36 1 2 1 2 0 1 1 1 6 0 3 24 6 6 0 36 0 1 1 1 0 6 9 24 0 12 6 36 0 1 1 1 3 0 1 4 2 0 12 6 36
R3 5407 r3
1 0 12 4 0 1 32 4 0 0 12 12
R1 2 7 r3 r1 5 R2 7 r3 r2 R 400 r r 4 7 3 4
1 0 12 4 0 1 32 4 0 0 1 1 1 0 0 3.5 0 1 0 2.5 0 0 1 1
R2 6r1 r2 R3 6 r1 r3
R2 16 r2
R1 r2 r1 R3 12 r2 r3
R3 121 r3 R1 12 r3 r1 R2 3 r3 r2 2
The solution is I1 3.5, I 2 2.5, I 3 1 .
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Section 12.2: Systems of Linear Equations: Matrices 87. a. Let x = the number of bananas needed; let y = the number of oranges needed; and let z = the number of papayas needed. Fruit equation: x y z 7 Orange/banana equation: y 2 x Cost equation: x 1.5 y 2.5 z 10.5 Set up a matrix and solve: 1 7 1 1 0 0 1 2 1 1.5 2.5 10.5 1 1 1 7 0 R3 10r3 2 1 0 10 15 25 105 1 1 1 7 R2 2 r1 r2 0 3 2 14 0 5 15 35 R3 10r1 r3 1 1 1 7 2 14 R2 13 r2 0 1 3 3 0 5 15 35 1 0 1 73 3 14 R1 r2 r1 0 1 23 3 R3 5r2 r3 35 0 0 35 3 3 7 1 0 1 3 3 14 3 0 1 23 3 R3 35 r3 0 0 1 1
1 1 1 12 0 0 1 2 0.5 1.5 2.5 18 1 1 1 12 0 2 1 0 5 15 25 180
R3 10r3
1 1 1 12 0 3 2 24 0 10 20 120
R2 2 r1 r2 R3 5r1 r3
1 1 1 12 2 0 1 3 8 0 10 20 120
R2 13 r2
1 0 0 1 0 0
1 3 2 3 40 3
4 8 40
1 0 1 4 3 0 1 23 8 0 0 1 3
R1 r2 r1 R3 10r2 r3
R3 403 r3
1 0 0 3 1 R1 3 r3 r1 0 1 0 6 2 0 0 1 3 R2 3 r3 r2 You should use 3 bananas, 6 oranges and 3 papaya.
88. a. Let a = the first coefficient. Let b = the second coefficient. Let c = the third coefficient. Height equation: h(t ) at 2 bt c Using the given heights at given times we have:
1 0 0 2 1 R1 3 r3 r1 0 1 0 4 2 0 0 1 1 R2 3 r3 r2 You should use 2 bananas, 4 oranges and 1 papaya.
2
1 1 a b c 24 2 2
b. Fruit equation: x y z 12 Orange/banana equation: y 2 x Cost equation: 0.5 x 1.5 y 2.5 z 18 Set up a matrix and solve:
a 1 b 1 c 37 2
a 3 b 3 c 9 2
or 1 1 a b c 24 2 2 a b c 37 9a 3b c 9
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Chapter 12: Systems of Equations and Inequalities
Set up a matrix and solve: 1 1 4 2 1 24 1 1 1 37 9 3 1 9 1 2 4 96 1 1 1 37 R1 4r1 9 3 1 9 1 96 2 4 0 1 3 59 0 15 35 855
89. Let x = the amount of supplement 1. Let y = the amount of supplement 2. Let z = the amount of supplement 3. 0.20 x 0.40 y 0.30 z 40 Vitamin C 0.30 x 0.20 y 0.50 z 30 Vitamin D Multiplying each equation by 10 yields 2 x 4 y 3z 400 3x 2 y 5 z 300
1 0 0 1 0 0
96 59 R2 r2 1 15 35 855 0 2 22 R1 2r2 r1 1 3 59 R 15r2 r3 0 10 30 3
1 0 0 1 0 0
0 2 22 1 3 59 0 3 1 0 0 16 1 0 50 0 1 3
2
Set up a matrix and solve:
R2 r1 r2 R3 9r1 r3
2 4 3 400 3 2 5 300 1 2 32 200 3 2 5 300
4 3
1 2 0 4
3 2 1 2
R1 12 r1
200 300
R2 r2 3r1
1 2 3 200 2 1 0 1 8 75
R2 14 r2
R3 101 r3
1 0 7 50 4 1 0 1 75 8
R1 r1 2r2
R1 2r3 r1 R2 3r3 r2
The matrix in the last step represents the system x 74 z 50 1 y 8 z 75
The matrix in the last step represents the coefficients so the solution is h(t ) 16t 2 50t 3 . b.
h(0) 16(0) 2 50(0) 3 3 ft
c.
Use the quadratic formula to solve 0 16t 2 50t 3 t
50 (50) 2 4(16)(3) 2(16)
50 2692 32
7 Therefore the solution is x 50 z , 4 1 y 75 z , z is any real number. 8
Possible combinations:
50 2692 50 2692 or t t 32 32 1.8849 101.8845 t -0.059 t 3.184 32 32 The negative number for time does not exist so the egg hits the ground is 3.184 seconds.
Supplement 1
Supplement 2
Supplement 3
50mg
75mg
0mg
36mg
76mg
8mg
22mg
77mg
16mg
8mg
78mg
24mg
1354
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Section 12.2: Systems of Linear Equations: Matrices
Possible combinations:
90. Let x = the amount of powder 1. Let y = the amount of powder 2. Let z = the amount of powder 3. 0.20 x 0.40 y 0.30 z 12 Vitamin B12 0.30 x 0.20 y 0.40 z 12 Vitamin E Multiplying each equation by 10 yields 2 x 4 y 3 z 120 3x 2 y 4 z 120
Set up a matrix and solve: 2 4 3 120 3 2 4 120 2 4 3 120 0 4 0.5 60
Powder 1
Powder 2
Powder 3
30 units
15 units
0 units
20 units
14 units
8 units
10 units
13 units
16 units
0 units
12 units
24 units
91 – 93. Answers will vary. 94.
x 2 3x 6 2 x x2 5x 6 0 We inspect the graph of the function f ( x) x 2 5 x 6 . y-intercept: f (0) 6
R2 r2 32 r1
2 0 2.5 60 R1 r1 r2 0 4 0.5 60 The matrix in the last step represents the system 2 x 2.5 z 60 4 y 0.5 z 60 Thus, the solution is x 30 1.25 z , y 15 0.125 z , z is any real number.
x2 5x 6 0 ( x 6)( x 1) 0 x 6, x 1 The graph is below the x-axis when 1 x 6 . Since the inequality is strict, the solution set is x | 1 x 6 or, using interval notation,
x-intercepts:
1, 6 . 95.
R( x)
2
2x x 1 x2 2x 1
(2 x 1)( x 1) ( x 1)( x 1)
Domain: x x 1 . R ( x)
p ( x) 2 x 2 x 1; q ( x) x 2 2 x 1;
2 x2 x 1
is in lowest terms. x2 2 x 1 2 02 0 1 1 The y-intercept is f (0) 2 1 . Plot the point 0, 1 . 0 20 1 1 1 The x-intercepts are the zeros of p ( x) : 1 and . 2 2 x2 x 1 R( x) 2 is in lowest terms. The vertical asymptotes are the zeros of q( x) : x 2x 1 x 1 . Graph this asymptote with dashed lines. 2 Since n m , the line y 2 is the horizontal asymptote. Solve to find intersection points: 1 Plot the line y 2 using dashes. Graph:
1355
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Chapter 12: Systems of Equations and Inequalities
f (b) f (a) sin 1 (1) sin 1 (1) ba 1 (1)
102.
( ) 2 2 2 2
103.
f x
1 x2
f ( x h) f ( x ) h
1 2 x
1
x h
2
h x x h
96. Since any real number can be evaluated in the equation then the domain is x | x is any real number or , .
2
x2 x h
2
2
h
97. cos 1 ( 0.75) 2.42 radians
x x 2 2 xh h 2
3
x2 x h
2
3
18 x 4 y 5 2 x 8 x3 98. 3 9 4 27 y12 27 x y 3 y
h 1 2 xh h2 h x 2 x h 2
99. Focus: (4, 0),(4,10); Vertices: (4,1), (4,9); Center: (4, 5); Transverse axis is parallel to the y-axis; a 4; c 5 . Find the value of b: b 2 c 2 a 2 25 16 9
1 h 2x h h x 2 x h 2
b 9 3
4
i 5 5 5 100. 6 cos i sin 6 e 12 12 12
36e
36e
i
2x h x x h 2
2
5 12
Section 12.3 1. ad bc 2.
1 3 i 36 2 2
3. False; If ad=bc, the the det = 0.
nt
x x h
4
5 5 36 cos i sin 3 3
18 18 3 i 36e
2x h 2
( y 5) 2 ( x 4) 2 1. 16 9
Write the equation:
5 i 4 12
2
5 i 3
5 3 3 4
4. False; The solution cannot be determined.
r .036 A P 1 2700 1 101. n 12 $3007.44
(12)(3)
5. False; See Thm (14) 6. a 7.
6 4 6(3) (1)(4) 18 4 22 1 3
1356 Copyright © 2025 Pearson Education, Inc.
2
Section 12.3: Systems of Linear Equations: Determinants
8.
9.
10.
8 3 4
2
3 1 4
2
8(2) 4(3) 16 12 28
14.
3 4(1) (3)(0) 9 1(1) 8(0) 4 1(3) 8(4)
3(2) 4(1) 6 4 2
3(4) 9(1) 4(35) 12 9 140 119
4 2 4(3) (5)(2) 12 10 2 5 3 x y 8 15. x y 4
3
11.
4 2 5 5 5 3 1 1 1 4 1 2 1 1 2 2 2 1 1 2 1 2 2
D 1 1 1 1 2 1 1
3 1 ( 2) 2(5) 4 1( 2) 1(5)
2 1(2) 1( 1)
Dx
3( 8) 4( 7) 2(3) 24 28 6
1 3 2 5 6 5 6 1 3 ( 2) 6 1 5 1 1 8 3 8 2 2 3 8 2 3 11(3) 2( 5) 3 6(3) 8(5)
1 8 4 12
4 1
x 2 y 5 16. x y 3
2 6(2) 8(1) 1(13) 3(58) 2(4) 13 174 8
D 1 2 1 2 3 1 1
169
13.
8
8 Dy 1 48 4 1 4 Find the solutions by Cramer's Rule: Dy 4 D 12 x x y 6 2 D 2 D 2 The solution is (6, 2).
10
12.
3 9 4 4 0 1 0 1 4 1 4 0 3 (9) 4 3 8 8 3 1 1 8 3 1
Dx
4 1 2 0 6 0 6 1 (1) 2 6 1 0 4 1 3 4 1 4 1 3 1 3 4
5
2 5 6 11 3 1
5 Dy 1 35 2 1 3 Find the solutions by Cramer's Rule: Dy 2 2 D 11 11 x x y D D 3 3 3 3
4 1(4) 0(3) 1 6(4) 1(0) 2 6( 3) 1(1) 4( 4) 1(24) 2(17)
11 2 , . 3 3
16 24 34 26
The solution is
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Chapter 12: Systems of Equations and Inequalities
5 x y 13 17. 2 x 3 y 12
4 x 5 y 3 20. 2y 4
D
5 1 15 2 17 2 3
5 D 4 8 0 8 0 2
Dx
13 1 39 12 51 12 3
Dx
Dy
5 13
3 Dy 4 16 0 16 0 4 Find the solutions by Cramer's Rule: Dy 16 D 26 13 x x y 2 D 8 D 8 4
60 26 34
2 12 Find the solutions by Cramer's Rule: Dy 34 D 51 x x 3 y 2 D 17 D 17 The solution is (3, 2). x 3y 5 18. 2 x 3 y 8 D 1
3
Dx
5
3
8 3
15 (24) 9
5 Dy 1 8 10 18 2 8 Find the solutions by Cramer's Rule: Dy 18 D 9 1 2 x x y 9 D 9 D The solution is (1, 2) . 24 3 x 19. x 2 y 0 D
3 0 1 2
5
6 (20) 26
4 2
13 The solution is , 2 . 4 4 x 6 y 42 21. 7 x 4 y 1
3 6 9
2 3
3
D 4 7
6
Dx 42 1
4
16 (42) 58
6 4
168 6 174
Dy 4 42 4 (294) 290 7 1 Find the solutions by Cramer's Rule: Dy 290 D 174 3 5 x x y D D 58 58 The solution is (3,5) . 2 x 4 y 16 22. 3x 5 y 9
60 6
0 Dx 24 48 0 48 0 2 3 24 Dy 0 24 24 1 0 Find the solutions by Cramer's Rule: Dy 24 D 48 8 4 x x y 6 6 D D The solution is (8, 4) .
D
2 4 10 12 22 3 5
Dx
16 4 80 36 44 9 5
2 16 18 48 66 3 9 Find the solutions by Cramer's Rule: Dy 66 D 44 2 3 x x y D 22 D 22 The solution is (2, 3). Dy
1358 Copyright © 2025 Pearson Education, Inc.
Section 12.3: Systems of Linear Equations: Determinants
2 x 3 y 1 27. 10 x 10 y 5
3x 2 y 4 23. 6 x 4 y 0
3 2 12 (12) 0 6 4 Since D 0 , Cramer's Rule does not apply. D
D
3 Dx 1 10 (15) 5 5 10
x 2 y 5 24. 4x 8 y 6
2 1 10 (10) 20 10 5 Find the solutions by Cramer's Rule: Dy 20 2 D 5 1 x x y D 50 10 D 50 5 1 2 The solution is , . 10 5 Dy
D 1 2 8 8 0 4 8 Since D 0 , Cramer's Rule does not apply. 2x 4 y 2 25. 3 x 2 y 3 4
D 2 3 Dx
2 3
3x 2 y 0 28. 5 x 10 y 4
4 12 16
2 4 2
4 12 8
2 6 6 12 Dy 2 3 3 Find the solutions by Cramer's Rule: Dy 12 3 D 8 1 x x y D 16 2 D 16 4 1 3 The solution is , . 2 4
D Dx Dy
4 2 3 3 8 3
3 2 30 (10) 40 5 10
Dx
0 2 0 (8) 8 4 10
3 0 12 0 12 5 4 Find the solutions by Cramer's Rule: Dy 12 3 D 8 1 x x y D 40 5 D 40 10 1 3 The solution is , . 5 10 2 x 3 y 6 29. 1 x y 2
6 12 6 68 2
D 2 3 2 3 5 1 1
8 12 4
Dx
2
3 3
D
Dy
3x 3 y 3 26. 8 4 x 2 y 3
3 3
2 3 20 (30) 50 10 10
4 83 Find the solutions by Cramer's Rule: Dy 4 2 D 2 1 x x y D 6 3 D 6 3
Dy
6
3
1 2
1
2 6
6
3 15 2 2
1 6 5 1 1 2 Find the solutions by Cramer's Rule:
1 2 The solution is , . 3 3
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Chapter 12: Systems of Equations and Inequalities
15 Dx 2 3 x 2 D 5 3 2
Dy
5 y 1 D 5
The solution is , 1 .
2 x y 1 32. 1 3 x 2 y 2 D
2 1 1
1 x y 2 30. 2 x 2 y 8
Dx
1
Dy
1 1 1 2 D 2 1 2 1 4 8 4 Dx 2 8 2 1
Dy 2 2 4 (2) 6 1 8 Find the solutions by Cramer's Rule: Dy D 6 4 x x 2 y 3 D 2 D 2
The solution is (2, 3) .
1 2
1 1 3 2
1 2
2 1 1
3 2
11 2 1 3 1 2 2 3 1 4
Find the solutions by Cramer's Rule: Dy 4 D 1 x x y 2 D 2 D 2 1
The solution is , 2 . 2 x y z 6 33. 3 x 2 y z 5 x 3 y 2 z 14
1 1 1 D 3 2 1 3 2 1
3x 5 y 3 31. 15 x 5 y 21 D
3 5 15 (75) 90 15 5
Dx
3 5 15 (105) 120 21 5
3 3 63 45 18 15 21 Find the solutions by Cramer's Rule: D y 18 1 D 120 4 x x y 90 3 D D 90 5 Dy
4 1
The solution is , . 3 5
1 1 3 1 (1) 3 2 1 2 3 2 3 1 2 1 1(4 3) 1( 6 1) 1(9 2) 1 7 11 3 6 1 1 Dx 5 2 1 3 2 14 1 1 5 1 (1) 5 2 6 2 3 2 3 14 2 14 6(4 3) 1(10 14) 1(15 28) 6 4 13 3 1 6 Dy 3 5
1
1 1 14 2
1 6 3 1 (1) 3 5 14 2 1 2 1 14 1(10 14) 6( 6 1) 1(42 5) 1
5
4 42 47 9
1360 Copyright © 2025 Pearson Education, Inc.
Section 12.3: Systems of Linear Equations: Determinants
1 1 4 Dz 2 3 15 5 1 12
1 1 6 Dz 3 2 5 1 1
3 14
2 5
1
3 5
6
1 3 15 (1) 2 15 ( 4) 2 3 5 12 5 1 12 1 1(36 15) 1(24 75) 4(2 15) 21 99 68 10 Find the solutions by Cramer's Rule: Dy 15 D 5 1 3 x x y 5 D 5 D D 10 2 z z D 5 The solution is (1, 3, 2) .
3 2
3 14 3 1 14 1 1( 28 15) 1(42 5) 6(9 2) 13 47 66 6 Find the solutions by Cramer's Rule: Dy 9 D 3 x x 1 y 3 D 3 D 3 D 6 z z 2 D 3 The solution is (1, 3, 2) .
x 3 y z 2 35. 2 x 6 y z 5 3x 3 y 2 z 5
x y z 4 34. 2 x 3 y 4 z 15 5 x y 2 z 12
1 3 1 1 D 2 6 3 3 2
1 1 1 D 2 3 4 5 1 2
1
6
1 3 2 1 (1) 2 6 3 2 3 3 3 2 1(12 3) 2( 4 3) 1(6 18) 9 3 12 24
4 (1) 2 4 1 2 3 1 3 5 2 5 1 2 1 1(6 4) 1( 4 20) 1(2 15) 2 24 17 5
3 1 2 Dx 5 6 1 5 3 2
1 4 1 Dx 15 3 4 12 1 2
1 3 5 1 (1) 5 6 2 6 3 2 5 2 5 3 2(12 3) 3(10 5) 1(15 30) 18 15 15 48
4 (1) 15 4 1 15 3 4 3 1 2 12 2 12 1 4(6 4) 1(30 48) 1(15 36) 8 18 21 5 1 4 Dy 2 15 4 5 12 2 1
Dy
1 2 1 2 5 1 3 5 2
2 1 (1) 2 5 3 2 3 5 5 2 1(10 5) 2(4 3) 1(10 15) 1
15
4 ( 4) 2 4 1 2 15 1 5 2 5 12 12 2 1(30 48) 4( 4 20) 1(24 75)
5
525
18 96 99
8
15
1361
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1 (2)
Chapter 12: Systems of Equations and Inequalities
1 3 2 Dz 2 6 5 3 3 5
1 4 8 Dz 3 1 12 1
1 6 5 3 2 5 (2) 2 6 3 5 3 3 3 5 1(30 15) 3(10 15) 2(6 18) 15 15 24 24 Find the solutions by Cramer's Rule: Dy D 8 1 48 x x y 2 D D 24 3 24 D 24 z z 1 D 24 1 The solution is 2, , 1 . 3
x 4 y 3z 8 36. 3x y 3 z 12 x y 6 z 1
1
1
1 12
1 4
3 12
( 8)
3 1
1 1 1 1 1 1(1 12) 4(3 12) 8(3 1)
1
13 36 32 9 Find the solutions by Cramer's Rule: x
Dx 243 3 81 D
z
9 1 Dz D 81 9
y
Dy D
216 8 81 3
8 1
The solution is 3, , . 3 9
x 2 y 3z 1 37. 3x y 2 z 0 2x 4 y 6z 2
1 4 3 D 3 1 3 1 1 6 1 1 3 4 3 3 (3) 3 1 1 6 1 6 1 1 1( 6 3) 4(18 3) 3(3 1) 9 60 12 81
3 1 2 D 3 1 2 6 2 4 1 2 ( 2) 3 2 3 3 1 6 6 4 2 2 4 1(6 8) 2(18 4) 3(12 2) 1
2 44 42
8 4 3 Dx 12 1 3 1 1 6
0 Since D 0 , Cramer's Rule does not apply.
8 1 3 4 12 3 (3) 12 1 1 6 1 6 1 1 8( 6 3) 4(72 3) 3(12 1) 72 276 39 243
x y 2z 5 4 38. 3 x 2 y 2 x 2 y 4 z 10 D
1 8 3 Dy 3 12 3 1 1 6
1 1 3 2 2
1 12 3 ( 8) 3 3 (3) 3 12 1 6 1 6 1 1 1(72 3) 8(18 3) 3(3 12) 69 120 27 216
2 0
2 4
0 3 0 3 2 1 2 (1) 2 2 4 2 4 2 2 1( 8 0) 1(12 0) 2(6 4) 8 12 20 0 Since D 0 , Cramer's Rule does not apply.
1362 Copyright © 2025 Pearson Education, Inc.
Section 12.3: Systems of Linear Equations: Determinants
x 2y z 0 39. 2 x 4 y z 0 2 x 2 y 3z 0 D
1 0 3 Dy 3 0 3 0 1 0
1 2 1 2 4 1 2 2 3
6
1 4 0 Dz 3 1 0 0 1 1 0
1 2 2 1 (1) 2 4 1 4 2 3 2 2 3 2 1(12 2) 2( 6 2) 1(4 8) 10 8 4 22
[By Theorem (12)]
Find the solutions by Cramer's Rule: Dy D 0 0 0 0 x x y D 81 D 81 D 0 0 z z D 81 The solution is (0, 0, 0).
0 2 1 Dx 0 4 1 0 0 2 3
[By Theorem (12)]
1 0 1 1 0 Dy 2 0 2 0 3
[By Theorem (12)]
x 2 y 3z 0 41. 3x y 2 z 0 2x 4 y 6z 0
2 0 Dz 2 4 0 0 2 2 0
[By Theorem (12)]
3 1 2 D 3 1 2 6 2 4
1
1 2 ( 2) 3 2 3 3 1 4 6 6 2 2 4 1(6 8) 2(18 4) 3(12 2) 1
Find the solutions by Cramer's Rule: Dy D 0 0 x x 0 y 0 D 22 D 22 D 0 z z 0 D 22 The solution is (0, 0, 0).
2 44 42 0 Since D 0 , Cramer's Rule does not apply. x y 2z 0 42. 3 x 2 y 0 2x 2 y 4z 0
x 4 y 3z 0 40. 3x y 3z 0 x y 6z 0 1 4 3 D 3 1 3 1 1 6
D
1 1 3 2 2
2 0
2 4
0 3 0 3 2 1 2 (1) 2 2 4 2 4 2 2 1( 8 0) 1(12 0) 2(6 4)
3 3 3 3 1 1 1 4 (3) 1 6 1 6 1 1 1( 6 3) 4(18 3) 3(3 1)
8 12 20
9 60 12
0 Since D 0 , Cramer's Rule does not apply.
81 0 4 3 Dx 0 1 3 0 0 1 6
[By Theorem (12)]
[By Theorem (12)]
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Chapter 12: Systems of Equations and Inequalities
43.
x
y
u
v w 4
1 2
z
x y z 47. Let u v w 4 1 2 3
3
By Theorem (11), the value of a determinant changes sign if any two rows are interchanged. 1 2 3 Thus, u v w 4 . x y z
44.
y
u
v w 4 3
2v
y
z
2w 2
2 x 3
y 6 z 9 v
[Theorem (14)]
w
x3
y 6 z 9
1
2
3
u
v
w
[Theorem (11)]
x3
y 6 z 9
u
v
w
1
2
3
2(1)(1) x
y
z
2(1)(1) u
v
w
1
2
3
x [Theorem (14)]
x y z 3( 1) u v w [Theorem (11)] 1 2 3 3(4) 12
[Theorem (11)]
[Theorem (15)] ( R1 3r3 r1 )
zx
x
y
u 1
v wu u v w 2 2 1 2 3
x
1 2 3 1 2 3 xu y v z w x y z u v w u v w
u
x
y
z
[Theorem (15)] (C3 c1 c3 )
z 3
[Theorem (15)] R2 r2 r3
z
1
2
3
2x
2y
2z
u 1 v 2 w 3 [Theorem (11)]
v w
y z v w 4
y
49. Let u v w 4 1 2 3
2
x y z ( 1)(1) u v w [Theorem (11)] 1 2 3
1 2
z
4
46. Let u v w 4 1 2 3
x y ( 1) 1 2
y
48. Let u v w 4 1 2 3
z
x u
3
2(1)(1)(4) 8
x y z x y z 3 6 9 3 1 2 3 u v w u v w
y
3
1
2(1)
45. Let u v w 4 . 1 2 3
x
2u
z
By Theorem (14), if any row of a determinant is multiplied by a nonzero number k, the value of the determinant is also changed by a factor of k. x y z x y z Thus, u v w 2 u v w 2(4) 8 . 2 4 6 1 2 3 x
2
y 6 z 9
u
x
1 2
1 x3
1
2
3
x
y
z
[Theorem (14)]
u 1 v 2 w 3 2(1)
x
y
z
1
2
3
u 1 v 2 w 3
3
1364 Copyright © 2025 Pearson Education, Inc.
[Theorem (11)]
Section 12.3: Systems of Linear Equations: Determinants
x
y
53. Solve for x:
z
2(1)(1) u 1 v 2 w 3 [Theorem (11)] 1
2
x
y
z
2(1)(1) u
v
w
1
2
3
x
1 1 4 3 2 2 1 2 5
3 [Theorem (15)] ( R2 r3 r2 )
x
2(1)(1)(4) 8 x
y
3 2 3 1 4 2 1 4 2 2 5 1 5 1 2 x 15 4 20 2 8 3 2 11x 22 11 2 11x 13 13 x 11
z
50. Let u v w 4 1 2 3 x3
y6
z 9
54. Solve for x:
3u 1 3v 2 3w 3 1
2 x
y
z
3u 1 3v 2 3w 3 1 x
2 y
3
z
2
x
y
3 u
[Theorem (15)] (R1 =3r3 r1 )
3
[Theorem (15)]
3u 3v 3w 1
3 2 4 5 0 1 x 0 1 2
3
( R2 r3 r2 )
3
5 5 1 1 x 2 4 0 0 2 0 1 1 2 3 2 x 5 2 2 4 1 0
x
6 x 15 4 4 0 6x 7 0 6x 7
z
v w
1 2
[Theorem (14)]
3
x
3(4) 12
7 6
55. Solve for x: 3 x 2 0 7 1 x 6 1 2
51. Solve for x: x
x 5 3 4 3x 4 x 5
0 2 1 0 3 1 x 7 x x 6 2 6 1 1 2 x 2 x 2 2 3 1 6 x 7
x 5 x 5
2 x 2 4 3 18 x 7
52. Solve for x:
2 x 2 18 x 0
x
1 2 3 x
2x x 9 0 x 0 or x 9
2
x 3 2 x2 1 0 ( x 1)( x 1) 0 x 1 0 or x 1 0 x 1
or
x 1
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Chapter 12: Systems of Equations and Inequalities 56. Solve for x: x
1 2 1 x 3 4 x 0 1 2 x
x 3
3 x 1 1 2 1 4 x 0 2 0 1 1 2 x 2 x 3 1 2 2 1 4 x 2 x 2 3 x 2 2 4 x 2 x2 x 0 x 2 x 1 0
x
y 1 y1 1 0 y2 1
y1 1 x 1 x y 1 1 1 y2 1 x2 1 x2
y1 0 y2
x( y1 y2 ) y x1 x2 ( x1 y2 x2 y1 ) 0
x( y1 y2 ) y x2 x1 x2 y1 x1 y2
y x2 x1 x2 y1 x1 y2 x( y2 y1 ) y x2 x1 y1 ( x2 x1 ) x2 y1 x1 y2 x( y2 y1 ) y1 ( x2 x1 )
x2 x1 ( y y1 )
x( y2 y1 ) x2 y1 x1 y2 y1 x2 y1 x1
x2 x1 ( y y1 ) ( y2 y1 ) x ( y2 y1 ) x1 x2 x1 ( y y1 ) ( y2 y1 )( x x1 ) ( y2 y1 ) ( x x1 ) x2 x1 This is the 2-point form of the equation for a line. ( y y1 )
x x2 x3
y 1 y2 1 0 y3 1
If the point ( x1 , y1 ) is on the line containing ( x2 , y2 ) and ( x3 , y3 ) [the points are collinear], x1 then x2 x3
y1 1 y2 1 0 . y3 1
x1 Conversely, if x2 x3
1 x 0 or x 2
57. Expanding the determinant: x x1 x2
58. Any point ( x, y ) on the line containing ( x2 , y2 ) and ( x3 , y3 ) satisfies:
y1 1 y2 1 0 , then ( x1 , y1 ) is y3 1
on the line containing ( x2 , y2 ) and ( x3 , y3 ) , and the points are collinear. 59. If the vertices of a triangle are (2, 3), (5, 2), and (6, 5), then: 2 5 6 1 D 3 2 5 2 1 1 1 3 5 3 2 1 2 5 2 5 6 1 1 1 1 1 1 2 1 2(2 5) 5(3 5) 6(3 2) 2 1 2(3) 5(2) 6(1) 2 1 6 10 6 2 5 The area of the triangle is 5 5 square units. 6 8 6 1 60. D1 8 4 2 10 2 1 1 1 10 10 1 6 6 1 D2 6 8 2 25 2 1 1 1 25 25
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Section 12.3: Systems of Linear Equations: Determinants
1 1 6 1 D3 3 6 2 15.5 2 1 1 1
Now set this expression equal to 0. Then complete the square to obtain the standard form.
15.5 15.5
20 x 2 120 x 20 y 2 80 y 240
Total area =10 25 15.5 50.5 square units
x 2 6 x y 2 4 y 12
61. A
1 2
6 8 1 6
1
6
1 3
1
3
6
2
240 120 x 80 y 20 x 2 20 y 2 0
6 2
8
4
8 4
6 8
x 2 6 x 9 y 2 4 y 4 12 9 4
x 3 2 y 2 2 25
1 368 3 6 2 18 2416 64 24 2 1 28 9 16 40 40 50.5 square units 2
62.
0
0
8 1
2
8
0 1
10
4
4 1
4
10 6 1
2
8
1
2
8
0
8 10
4
1 1 10
4
4
4
10 1
4
64. Expanding the determinant:
10 6
x
7
3
6
5
3
2
y x y
74 18 40
7
3
6
1
1
1
5
3
2 x 5
3
2
y 7
3
1
2
6 x y
2
( y z )( x y )( x z )
by s ad bc 0 , and the system is . cx dy t s The solution of the system is y , b
t dy t d b tb sd . Using Cramer’s c c bc 0 b bc , Rule, we get D c d s b Dx sd tb , t d 0 s Dy 0 sc sc , so c t D ds tb td sd x x and D bc bc Dy sc s y , which is the solution. Note D bc b that these solutions agree if d 0.
1 1
7 3 6
5 3 2
74 18 40
240 x 120 y 80 x 2 y 2 20 2
240 120 x 80 y 20 x 20 y
z
2
65. If a 0, then b 0 and c 0 since
1
2
y
( y z ) x( x y ) z ( x y )
74 18 40 1
2
s
2
1
y 1 y2 1 y2 x 1 z 1 z2 1 z2
x
2
74 18 40
z 1
( y z ) x 2 xy xz yz
1 576 96 cubic units 6
63.
z
x 2 ( y z ) x( y z )( y z ) yz ( y z )
8(24) 384 576
2
y 1
2
x ( y z ) x( y z ) 1( y z z 2 y )
8 (12) 48 84) 1 32 352
2
y
2
1 2(24 40) 8(60 16)
2
x 1
2
x2
10 1 10 4 4 1 8 2 8 1 4 1 4 10 10 1 10 4 4 4 1 2 8 10 6 4 6 8 2(4 10) 8(10 4) 1(100 16)
1
x2
2
1367
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Chapter 12: Systems of Equations and Inequalities
If b 0, then a 0 and d 0 since s ax . ad bc 0 , and the system is cx dy t The solution of the system is x
s , a
t cx at cs y . Using Cramer’s Rule, we d ad a 0 s 0 get D ad , Dx sd , and c d t d a s D sd s Dy at cs , so x x c t D ad a Dy at cs , which is the solution. and y D ad Note that these solutions agree if c 0. If c 0, then a 0 and d 0 since ax by s . ad bc 0 , and the system is dy t The solution of the system is y
t , d
s by sd tb . Using Cramer’s Rule, we a ad a b s b ad , Dx sd tb , get D t d 0 d a s D sd tb at , so x x and and Dy 0 t D ad Dy at t y , which is the solution. Note D ad d that these solutions agree if b 0. x
Dy
at cs cs at , which is the D bc bc solution. Note that these solutions agree if a 0. y
66.
a13 a23 a33
a12 a22 a32
a11 a21 a31
a13
a22 a32
a21 a a12 23 a31 a33
a11 ( a23a32 a22a33 ) a13a22a31 a13a21a32 a12a23a31 a12a21a33 a11a23a32 a11a22 a33 a11a22a33 a11a23a32 a12a21a33 a12a23a31 a13a21a32 a13a22a31 a11 (a22a33 a23a32 ) a12 ( a21a33 a23a31 ) a13 (a21a32 a22a31 ) a11
a22 a32
s b 0 tb tb , and t 0 a s c
t
at cs , so x
a23 a a a a a12 21 23 a13 21 22 a33 a31 a33 a31 a32
a a23 a a a a a11 22 a12 21 23 a13 21 22 a a a a a 32 33 31 33 31 a32 a11 a12 a13 a21 a22 a23 a31 a32 a33
t The solution of the system is x , c s ax cs at y . Using Cramer’s Rule, we b bc a b 0 bc bc , get D c 0
Dy
a22 a32
a13 ( a22a31 a21a32 ) a12 (a23a31 a21a33 )
If d 0, then b 0 and c 0 since ax by s ad bc 0 , and the system is . t cx
Dx
a21 a a11 23 a31 a33
Dx tb t and D bc c
1368 Copyright © 2025 Pearson Education, Inc.
Section 12.3: Systems of Linear Equations: Determinants
67.
a11 a12 ka21 ka22 a31 a32
a13 ka23 a33
(a11 ka21 )
(a13 ka23 )
ka21 (a12 a33 a32 a13 ) ka22 (a11a33 a31a13 )
(a13 ka23 )(a21a32 a22a31 )
a13 a23 a33
a11 ( a22a33 a23a32 ) ka21 (a22a33 a23a32 ) a12 (a21a33 a23a31 ) ka22 ( a21a33 a23a31 ) a13 (a21a32 a22a31 ) ka23 ( a21a32 a22a31 ) a11 ( a22a33 a23a32 ) ka21a22a33
68. Set up a 3 by 3 determinant in which the first column and third column are the same and evaluate:
ka21a23a32 a12 ( a21a33 a23a31 ) ka22a21a33 ka22a23a31 a13 (a21a32 a22a31 ) ka23a21a32
a11 a12 a21 a22 a31 a32
a11 a21 a31
a a11 22 a32
a21 a a a a a12 21 21 a11 21 22 a31 a31 a31 a31 a32
ka23a22a31 a11 ( a22a33 a23a32 ) a12 (a21a33 a23a31 ) a13 (a21a32 a22a31 ) a11
a11 (a22 a31 a32 a21 ) a12 (a21a31 a31a21 )
a22 a32
a23 a a a a a12 21 23 a13 21 22 a33 a31 a33 a31 a32
a11 a12 a21 a22 a31 a32
a11 (a21a32 a31a22 ) a11a22 a31 a11a32 a21 a12 (0) a11a21a32 a11 a31a22 0 a11 ka21 a12 ka22 a21 a22 a31 a32
a21 a22 a31 a32
(a12 ka22 )( a21a33 a23a31 )
k [ a21 (a12 a33 a32 a13 ) a22 (a11a33 a31a13 ) a11 a12 a23 (a11a32 a31a12 )] k a21 a22 a31 a32
a23 a a ( a12 ka22 ) 21 23 a33 a31 a33
(a11 ka21 )(a22a33 a23a32 )
ka23 (a11a32 a31a12 )
69.
a22 a32
a13 ka23 a23 a33
a13 a23 a33
70. v ( x2 x1 )i ( y2 y1 ) j (5 ( 4))i ( 1 3) j 9i 4 j v 92 ( 4)2 81 16 97
71.
f ( x) 2 x3 5 x 2 x 10 p must be a factor of 10: p 1, 2, 5, 10 q must be a factor of 2: q 1, 2 The possible rational zeros are: p 1 5 , , 1, 2, 5, 10 2 2 q
1369
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Chapter 12: Systems of Equations and Inequalities
72.
f ( x) ( x 1) 2 4 . Using the graph of y x 2 , horizontally shift the graph to the left 1 unit and vertically shift the graph down 4 units.
d (5 3) 2 (2 2) 2 (8) 2 (4)2 64 16 80 4 5
77.
2 x 5 (2 x 5)(2 x 5)(2 x 5) 3
(4 x 2 20 x 25)(2 x 5) 8 x3 60 x 2 150 x 125 x 7 10 x 7 10 x x 7 10
78.
73.
tan 42 cot 48 cot(90 42) cot 48 cot(48) cot 48 0
74. r x y ( 5) 5 5 2 2
2
2
2
y 5 1 x 5 3 4 The polar form of z 5 5i is tan
z r cos i sin 3 3 5 2 cos i sin 4 4 i
3
5 2e 4
y 3 log 5 ( x 1)
75.
x 7 10 x 7 10 x 7 100 x x 93
x
x 7 10
2 so the perpendicular slope 5 5 would be m . The y value of the point on 2 2 the line is y (10) 7 3 so a point on the 5 line is (10,3) . Using the point-slope formula we have:
79. The slope is
5 ( x 10) 2 5 y 3 x 25 2 5 y x 22 2
y 3
y 3 log 5 ( x 1) 5 y 3 x 1 5 y 3 1 x y f 1 5 x 3 1
76. x
(12) 3 2(2)
y 2(3) 12(3) 20 2 2
The vertex of f(x) is (3, 2) x
(30) 5 2(3)
y 3(5) 2 30(5) 77 2
The vertex of g(x) is (5, 2)
x 7 10
Section 12.4 1. square 2. true 3. false 4. inverse 5. true 6. A1 B 7. a 1370 Copyright © 2025 Pearson Education, Inc.
Section 12.4: Matrix Algebra 8. d
4 1 0 3 5 6 2 1 2 6 2 3 0(4) 3(6) (5)( 2) 0(1) 3(2) ( 5)(3) 1(1) 2(2) 6(3) 1(4) 2(6) 6( 2)
15. AC
0 0 3 5 4 1 9. A B 1 2 6 2 3 2 5 0 0 4 3 1 1 ( 2) 2 3 6 ( 2)
28 9 4 23
4 4 5 1 5 4 0 0 3 5 4 1 10. A B 1 2 6 2 3 2 5 0 0 4 3 1 1 ( 2) 2 3 6 ( 2)
4 1
0
4
1
16. BC 6 2 2 3 2 2 3 4(1) 1(2) 0(3) 4(4) 1(6) 0( 2) 3(2) ( 2)(3) 2(4) 3(6) ( 2)( 2) 2(1) 22 6 14 2
4 2 5 3 1 8 0 3 5 11. 4 A 4 1 2 6 4 0 4 3 4(5) 4 6 4 1 4 2
4 1 0 3 5 17. CA 6 2 1 2 6 2 3 4(3) 1(2) 4(5) 1(6) 4(0) 1(1) 6(0) 2(1) 6(3) 2(2) 6( 5) 2(6) 2(0) 3(1) 2(3) 3(2) 2(5) 3(6)
0 12 20 24 4 8
1 14 14 2 22 18 3 0 28
0 4 1 12. 3B 3 2 3 2 3 0 3 4 3 1 3 2 3 3 3 2
4 1 0 4 1 18. CB 6 2 2 3 2 2 3 4(1) 1(3) 4(0) 1( 2) 4(4) 1( 2) 6(1) 2(3) 6(0) 2( 2) 6(4) 2( 2) 2(4) 3( 2) 2(1) 3(3) 2(0) 3( 2)
12 3 0 6 9 6 0 0 3 5 4 1 13. 3 A 2 B 3 2 2 3 2 1 2 6 0 0 9 15 8 2 4 6 4 3 6 18
14 7 2 20 12 4 14 7 6
8 7 15 7 0 22
19. Since the number of columns in A does not match the number of rows in B then AB is not defined.
0 0 3 5 4 1 4 14. 2 A 4 B 2 1 2 6 2 3 2 0 0 6 10 16 4 2 4 12 8 12 8
20. Since the number of columns in B does not match the number of rows in A then BA is not defined.
16 10 10 4 6 16
1371
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Chapter 12: Systems of Equations and Inequalities
4 21. C ( A B) 6 2 4 6 2
1 0 3 5 4 1 0 2 1 2 6 2 3 2 3 1 4 4 5 2 1 5 4 3
15 21 16 22 34 22 11 7 22
4 1 4 1 0 0 3 5 4 1 6 2 6 2 1 2 6 2 3 2 2 3 2 3 1 14 14 14 7 2 2 22 18 20 12 4 3 0 28 14 7 6 13 7 12 18 10 14 17 7 34
4 1 0 0 3 5 4 1 2 3 2 6 2 1 2 6 2 3 4 1 4 4 5 6 2 1 5 4 2 3
22. ( A B)C
50 3 18 21
4 1 0 3 5 1 0 23. AC 3I 2 6 2 3 0 1 1 2 6 2 3 28 9 3 0 4 23 0 3 25 9 4 20
26. AC BC 4 1 4 1 0 0 3 5 4 1 6 2 2 3 2 6 2 1 2 6 2 3 2 3 28 9 22 6 4 23 14 2 50 3 18 21
27. a11 2(2) ( 2)(3) 2 a12 2(1) ( 2)(1) 4 a13 2(4) ( 2)(3) 2 a14 2(6) ( 2)(2) 8 a21 1(2) 0(3) 2 a22 1(1) 0(1) 1
4 1 1 0 0 0 3 5 0 1 0 24. CA 5I 3 6 2 5 1 2 6 2 3 0 0 1 1 14 14 5 0 0 2 22 18 0 5 0 3 0 28 0 0 5 6 14 14 2 27 18 3 0 33
25. CA CB
a23 1(4) 0(3) 4 a24 1(6) 0(2) 6
2 2 2 1 4 6 2 4 2 8 1 0 3 1 3 2 2 1 4 6 28. a11 4( 6) 1(2) 22 a12 4(6) 1(5) 29 a13 4(1) 1(4) 8 a14 4(0) 1(1) 1 a21 2( 6) 1(2) 10 a22 2(6) 1(5) 17 a23 2(1) 1(4) 6 a24 2(0) 1(1) 1 4 1 6 6 1 0 22 29 8 1 2 1 2 5 4 1 10 17 6 1
1372 Copyright © 2025 Pearson Education, Inc.
Section 12.4: Matrix Algebra
1
2
3
1
2 1 35. A 1 1 Augment the matrix with the identity and use row operations to find the inverse: 2 1 1 0 1 1 0 1
2
29. 1 0 0 1 4 2 4 1(2) 2(0) 3(4) 1(1) 2(1) 3(2) 0(1) ( 1)( 1) 4(2) 0(2) (1)(0) 4(4) 5 14 9 16
1 2 1 0 1 0 1 0
1 1 2 8 1 30. 3 2 3 6 0 0 5 1(2) (1)(3) 1(8) ( 1)(6) 1( 1) (1)(0) (3)(2) 2(3) (3)(8) 2(6) (3)(1) 2(0) 0(2) 5(3) 0(8) 5(6) 0(1) 5(0)
1 Interchange 1 1 0 r1 and r2 1 0 1 R2 2r1 r2 1 1 2 1 0 1 R2 r2 1 1 2 1 0
1 1 R1 r2 r1 1 1 2 1 1 Thus, A1 . 1 2
2 1 1 0 12 3 15 30 0
31. Since the number of columns in the first matrix does not match the number of rows in the second then the operation is undefined.
0
3 1 36. A 2 1 Augment the matrix with the identity and use row operations to find the inverse: 3 1 1 0 2 1 0 1
32. Since the number of columns in the first matrix does not match the number of rows in the second then the operation is undefined.
1 0 1 1 R1 r2 r1 2 1 0 1 1 0 1 1 R2 2r1 r2 0 1 2 3 1 1 Thus, A1 . 2 3
1 0 1 1 3 33. 2 4 1 6 2 3 6 1 8 1 1(1) 0(6) 1(8) 1(3) 0(2) 1(1) 2(1) 4(6) 1(8) 2(3) 4(2) 1( 1) 3(1) 6(6) 1(8) 3(3) 6(2) 1(1) 9 2 34 13 47 20
6 5 37. A 2 2 Augment the matrix with the identity and use row operations to find the inverse:
4 2 3 2 6 1 2 1 1 34. 0 1 0 1 0 2 4(2) ( 2)(1) 3(0) 4(6) ( 2)( 1) 3(2) 0(2) 1(1) 2(0) 0(6) 1(1) 2(2) 1(2) 0(1) 1(0) 1(6) 0(1) 1(2) 6 32 1 3 2 4
1373
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Chapter 12: Systems of Equations and Inequalities
6 5 1 0 2 2 0 1 2 2 0 1 6 5 1 0
Interchange r1 and r2
1 2 2 0 0 1 1 3
R2 3r1 r2
1 1 0 12 0 1 1 3
R1 12 r1 R2 r2
1 0 1 52 R1 r2 r1 3 0 1 1 1 52 Thus, A1 . 3 1 1 4 38. A 6 2 Augment the matrix with the identity and use row operations to find the inverse: 1 1 0 4 6 2 0 1 1 14 14 0 0 1 6 2 1 14 14 0 3 1 1 0 2 2
R1 14 r1 R2 6r1 r2
1 14 14 0 R2 2r2 1 3 2 0 1 0 1 12 1 R1 4 r2 r 0 1 3 2 1 12 Thus, A1 . 3 2
2 1 39. A where a 0. a a Augment the matrix with the identity and use row operations to find the inverse: 2 1 1 0 a a 0 1 1 1 1 2 2 0 a a 0 1
R1 12 r1
1 1 1 0 2 2 R2 a r1 r2 1 1 0 2 a 2 a 1 1 1 12 0 2 R2 a2 r2 2 0 1 1 a 1 0 1 1a R1 12 r2 r1 2 0 1 1 a
1 1a Thus, A1 . 2 1 a b 3 40. A where b 0. b 2 Augment the matrix with the identity and use row operations to find the inverse: b 3 1 0 b 2 0 1 b 3 1 0 R2 r1 r2 0 1 1 1 1 1 b3 0 b R1 b1 r1 0 1 1 1 1 b3 b1 0 R2 r2 0 1 1 1 3 1 0 b2 b R1 b3 r2 r1 1 1 0 1
3 2 b . Thus, A1 b 1 1
1374 Copyright © 2025 Pearson Education, Inc.
Section 12.4: Matrix Algebra
1 1 1 41. A 0 2 1 2 3 0 Augment the matrix with the identity and use row operations to find the inverse: 1 1 1 1 0 0 0 2 1 0 1 0 2 3 0 0 0 1
1 0 2 1 0 0 5 1 1 0 1 2 0 2 2 0 0 1 1 1 1 2 2 2 1 0 2 1 0 0 1 1 0 1 52 0 2 2 0 0 1 1 1 2
1 12 0 12 2 52
0 0 1
Thus, A
R1 r2 r1 R3 5 r2 r3
1 1 1 0 0 1 0 1 4 3 1 0 0 2 1 3 0 1 1 1 1 0 0 1 1 4 3 1 0 0 0 2 1 3 0 1
1 3 3 2 2 1 . 4 5 2
R2 3 r1 r2 R3 3 r1 r3
R2 r2
1 0 3 2 1 0 0 1 4 R3 17 r3 3 1 0 3 0 0 72 17 1 7 3 1 1 0 0 75 7 7 R1 3r3 r1 9 1 0 1 0 74 7 7 R 4r3 r2 3 2 1 2 0 0 1 7 7 7
1 0 2 42. A 1 2 3 1 1 0 Augment the matrix with the identity and use row operations to find the inverse: 1 0 2 1 0 0 1 2 3 0 1 0 1 1 0 0 0 1 2 1 0 0 1 0 5 1 1 0 0 2 0 1 2 1 0 1 2 1 0 0 1 0 5 1 1 0 0 1 2 2 2 0 1 2 1 0 1
R1 2r3 r1 5 R2 2 r3 r2
1 1 1 43. A 3 2 1 3 1 2 Augment the matrix with the identity and use row operations to find the inverse: 1 1 1 1 0 0 3 2 1 0 1 0 3 1 2 0 0 1
1 1 0 1 12 0 2 1 1 0 2 0 R3 2 r3 0 1 2 0 0 1 4 5 2 3 3 1 1 0 0 R1 12 r3 r1 0 1 0 2 2 1 R2 1 r3 r2 2 0 0 1 4 5 2
1
R3 2 r3
1 0 0 3 2 4 0 1 0 3 2 5 0 0 1 1 1 2 3 2 4 1 Thus, A 3 2 5 . 1 2 1
1 1 1 1 0 0 0 2 1 0 1 0 R3 2 r1 r3 0 5 2 2 0 1 1 1 0 0 1 1 1 1 2 0 12 0 R2 12 r2 0 0 5 2 2 0 1 1 1 0 2 0 1 12 0 0 1 2
R3 r2 r3
75 Thus, A1 97 3 7
R2 r1 r2 R3 r1 r3
R2 12 r2 1375
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1 7 1 7 72
3 7 74 . 1 7
Chapter 12: Systems of Equations and Inequalities
3 3 1 44. A 1 2 1 2 1 1 Augment the matrix with the identity and use row operations to find the inverse: 3 3 1 1 0 0 1 2 1 0 1 0 2 1 1 0 0 1 1 2 3 3 2 1 1 2 0 3 0 5
1 0 1 0 Interchange 1 1 0 0 r and r2 1 0 0 1 1 1 0 1 0 R2 3 r1 r2 2 1 3 0 R 2 r1 r3 1 0 2 1 3
0 1 0 1 2 1 2 1 0 1 3 3 1 0 R2 13 r2 0 2 1 0 5 1 2 1 0 1 0 1 3 3 R 2r2 r1 2 1 1 0 1 1 0 3 3 R3 5 r2 r3 7 5 0 0 3 1 3 3 2 1 0 1 0 1 3 3 2 1 0 1 1 0 R3 73 r3 3 3 5 9 3 0 0 1 7 7 7
3 1 0 0 7 1 0 1 0 7 0 0 1 5 7
74
3 7 Thus, A1 17 75
74
1 7 9 7
1 7 9 7
1 7 72 3 7
1 7 72 . 3 7
R1 1 r3 r1 3 R 2r r 2 3 3 2
2 x y 1 45. x y 3 Rewrite the system of equations in matrix form: 2 1 x 1 , X , B A 1 1 y 3
Find the inverse of A and solve X A1 B : 1 1 From Problem 35, A1 , so 1 2 1 1 1 4 X A1 B . 1 2 3 7 The solution is x 4, y 7 or (4, 7) . 3x y 8 46. 2 x y 4 Rewrite the system of equations in matrix form: 3 1 x 8 , X , B A 2 1 y 4
Find the inverse of A and solve X A1 B : 1 1 From Problem 36, A1 , so 2 3 1 1 8 12 X A1 B . 2 3 4 28 The solution is x 12, y 28 or (12, 28) . 2 x y 0 47. x y 5 Rewrite the system of equations in matrix form: 2 1 x 0 , X , B A 1 1 y 5
Find the inverse of A and solve X A1 B : 1 1 From Problem 35, A1 , so 1 2 1 1 0 5 X A1 B . 1 2 5 10 The solution is x 5, y 10 or (5, 10) .
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Section 12.4: Matrix Algebra
3x y 4 48. 2 x y 5 Rewrite the system of equations in matrix form: 3 1 x 4 , X , B A 2 1 y 5
6 x 5 y 13 51. 2 x 2 y 5 Rewrite the system of equations in matrix form: 6 5 x 13 , X , B A 2 2 y 5
Find the inverse of A and solve X A1 B : 1 1 From Problem 36, A1 , so 2 3
Find the inverse of A and solve X A1 B : 1 52 From Problem 37, A1 , so 3 1 1 52 13 12 X A1 B . 3 5 2 1
1 1 4 9 X A1 B . 2 3 5 23 The solution is x 9, y 23 or (9, 23) .
1 1 The solution is x , y 2 or , 2 . 2 2
6x 5 y 7 49. 2 x 2 y 2 Rewrite the system of equations in matrix form: 6 5 x 7 , X , B A 2 2 y 2
4 x y 5 52. 6 x 2 y 9 Rewrite the system of equations in matrix form: 1 4 x 5 A , X y , B 9 6 2
Find the inverse of A and solve X A1 B : 1 52 From Problem 37, A1 , so 3 1 1 52 7 2 X A1 B . 3 2 1 1
Find the inverse of A and solve X A1 B : 1 12 From Problem 38, A1 , so 3 2 1 12 5 12 X A1 B . 3 2 9 3
The solution is x 2, y 1 or (2, 1) .
1 1 The solution is x , y 3 or , 3 . 2 2
4 x y 0 50. 6 x 2 y 14 Rewrite the system of equations in matrix form: 1 4 x 0 , X , B A 6 2 y 14
2 x y 3 53. a0 ax ay a Rewrite the system of equations in matrix form: 2 1 x 3 , X , B A a a y a
Find the inverse of A and solve X A1 B : 1 12 From Problem 38, A1 , so 3 2 1 12 0 7 X A1 B . 3 2 14 28
Find the inverse of A and solve X A1 B : 1 1a From Problem 39, A1 , so 2 a 1
The solution is x 7, y 28 or (7, 28) .
1 1a 3 2 X A1 B . 2 a a 1 1 The solution is x 2, y 1 or (2, 1) .
1377
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Chapter 12: Systems of Equations and Inequalities
bx 3 y 2b 3 54. b0 bx 2 y 2b 2 Rewrite the system of equations in matrix form: b 3 x 2b 3 , X , B A b 2 y 2b 2
Find the inverse of A and solve X A1 B : 3 2 b From Problem 40, A1 b , so 1 1 3 2 2b 3 2 b X A1 B b . 1 1 2b 2 1 The solution is x 2, y 1 or (2, 1).
7 2 x y 55. a a0 ax ay 5 Rewrite the system of equations in matrix form: 7 2 1 x A , X , B a a a y 5
Find the inverse of A and solve X A1 B : 1 1a From Problem 39, A1 , so 2 a 1 1 1a 7 a2 X A1 B a 3 . 2 1 a 5 a
The solution is x
2 3 2 3 , y or , . a a a a
bx 3 y 14 56. b0 bx 2 y 10 Rewrite the system of equations in matrix form: b 3 x 14 A , X , B b 2 y 10
x yz 4 57. 2y z 1 2 x 3 y 4 Rewrite the system of equations in matrix form: 1 1 1 x 4 A 0 2 1 , X y , B 1 2 3 0 z 4
Find the inverse of A and solve X A1 B : 1 3 3 1 From Problem 41, A 2 2 1 , so 4 5 2 1 4 5 3 3 X A1 B 2 2 1 1 2 . 4 5 2 4 3 The solution is x 5, y 2, z 3 or (5, 2,3) . 2z 6 x 58. x 2 y 3z 5 x y 6 Rewrite the system of equations in matrix form: 1 0 2 x 6 A 1 2 3 , X y , B 5 1 1 0 z 6
Find the inverse of A and solve X A1 B : 3 2 4 From Problem 42, A1 3 2 5 , so 1 1 2 3 2 4 6 4 X A1 B 3 2 5 5 2 . 1 1 2 6 1 The solution is x 4, y 2, z 1 or (4, 2, 1) .
Find the inverse of A and solve X A1 B : 3 2 b From Problem 40, A1 b , so 1 1 3 2 2 14 b X A1 B b b . 1 1 10 4 2 2 The solution is x , y 4 or , 4 . b b
1378 Copyright © 2025 Pearson Education, Inc.
Section 12.4: Matrix Algebra
x yz 2 2y z 2 59. 2 x 3 y 1 2 Rewrite the system of equations in matrix form: 2 1 1 1 x A 0 2 1 , X y , B 2 1 2 3 0 z 2
x y z 9 61. 3 x 2 y z 8 3x y 2 z 1
Rewrite the system of equations in matrix form: 1 1 1 x 9 A 3 2 1 , X y , B 8 3 1 2 z 1 Find the inverse of A and solve X A1 B : 3 1 75 7 7 1 From Problem 43, A1 97 74 , so 7 3 2 1 7 7 7 3 34 1 75 7 7 9 7 9 85 1 1 4 7 8 7 . X A B 7 7 3 12 2 1 1 7 7 7 7
Find the inverse of A and solve X A1 B : 1 3 3 1 From Problem 41, A 2 2 1 , so 4 5 2 1 2 12 3 3 X A1 B 2 2 1 2 12 4 5 2 12 1 1 1 The solution is x , y , z 1 or 2 2 1 1 , , 1 . 2 2
The solution is x
34 85 12 or ,y ,z 7 7 7
34 85 12 , , . 7 7 7 3x 3 y z 8 62. x 2 y z 5 2x y z 4 Rewrite the system of equations in matrix form: 3 3 1 x 8 A 1 2 1 , X y , B 5 2 1 1 z 4
2z 2 x 3 60. x 2 y 3 z 2 2 x y Rewrite the system of equations in matrix form: 1 0 2 x 2 A 1 2 3 , X y , B 32 1 1 0 z 2
Find the inverse of A and solve X A1 B : 1 73 74 7 1 2 , so From Problem 44, A1 17 7 7 5 9 3 7 7 7
1
Find the inverse of A and solve X A B : 3 2 4 1 From Problem 42, A 3 2 5 , so 1 1 2
73 X A1 B 17 5 7
3 2 4 2 1 1 X A B 3 2 5 32 1 . 1 1 2 2 12 1 1 The solution is x 1, y 1, z or 1, 1, . 2 2
74 1 7 9 7
1 87 7 8 72 5 75 . 17 3 4 7 7
8 5 17 or The solution is x , y , z 7 7 7 8 5 17 , , . 7 7 7
1379
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Chapter 12: Systems of Equations and Inequalities x y z 2 7 63. 3 x 2 y z 3 10 3x y 2 z 3 Rewrite the system of equations in matrix form: 2 1 1 1 x A 3 2 1 , X y , B 73 10 3 1 2 z 3
Find the inverse of A and solve X A1 B : 3 1 75 7 7 1 1 From Problem 43, A 97 74 , so 7 3 2 1 7 7 7 5 3 1 1 7 2 3 7 7 9 1 1 4 X A B 7 7 73 1 . 7 3 2 2 1 10 3 3 7 7 7 1 2 2 1 The solution is x , y 1, z or , 1, . 3 3 3 3 3x 3 y z 1 64. x 2 y z 0 2x y z 4
4 2 65. A 2 1 Augment the matrix with the identity and use row operations to find the inverse: 4 2 1 0 2 1 0 1 1 0 4 2 1 R2 2 r1 r2 1 0 0 1 2 1 1 1 2 0 1 4 R1 4 r1 1 0 0 1 2 There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse. 3 12 66. A 6 1 Augment the matrix with the identity and use row operations to find the inverse: 3 12 1 0 6 1 0 1 3 12 1 0 R2 2r1 r2 0 0 2 1 1 16 13 0 R1 13 r1 0 2 1 0 There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse.
Rewrite the system of equations in matrix form: 3 3 1 x 1 A 1 2 1 , X y , B 0 2 1 1 z 4 Find the inverse of A and solve X A1 B : 1 73 74 7 1 2 , so From Problem 44, A1 17 7 7 5 9 3 7 7 7 1 73 74 1 7 1 1 1 1 2 7 0 1 . X A B 7 7 5 9 3 4 1 7 7 7 The solution is x 1, y 1, z 1 or (1, 1, 1) .
15 3 67. A 10 2 Augment the matrix with the identity and use row operations to find the inverse: 15 3 1 0 10 2 0 1 1 0 15 3 2 0 0 3 1
R2 23 r1 r2
1 15 151 0 R1 151 r1 2 0 0 3 1 There is no way to obtain the identity matrix on the left; thus, there is no inverse.
1380 Copyright © 2025 Pearson Education, Inc.
Section 12.4: Matrix Algebra
3 0 68. A 4 0 Augment the matrix with the identity and use row operations to find the inverse: 3 0 1 0 4 0 0 1 3 0 0 0
1 0 1
4 3
5
Augment the matrix with the identity and use row operations to find the inverse: 1 3 1 0 0 1 2 4 1 0 1 0 7 1 0 0 1 5 1 3 1 0 0 1 R2 2 r1 r2 7 2 1 0 0 6 R 5 r1 r3 0 12 14 5 0 1 3 0 0 1 1 3 1 7 1 0 1 6 3 16 0 R2 16 r2 0 12 14 5 0 1
R2 34 r1 r2
1 0 13 0 R1 13 r1 4 1 0 0 3 There is no way to obtain the identity matrix on the left; thus, there is no inverse.
1 1 3 69. A 1 4 7 1 2 5 Augment the matrix with the identity and use row operations to find the inverse: 1 1 1 0 0 3 1 4 7 0 1 0 2 5 0 0 1 1 2 5 0 0 1 1 1 4 7 0 1 0 3 1 1 1 0 0 2 5 0 0 1 1 0 6 12 0 1 1 0 7 14 1 0 3 1 2 5 0 0 1 0 1 2 0 16 16 0 7 14 1 0 3
1 3 1 7 1
1
70. A 2 4
1 0 11 6 0 1 76 0 0 0
2 3 1 3
0 0 2 1
1 6 1 6
1
R1 r2 r1 R3 12 r2 r3
There is no way to obtain the identity matrix on the left; thus, there is no inverse. 61 12 25 71. A 18 12 7 3 4 1
Interchange r1 and r3 R2 r1 r2 R3 3 r1 r3
R2 16 r2 Thus, A
1 2 1 0 1 0 3 3 R1 2r2 r1 1 1 0 1 2 0 6 6 R3 7 r2 r3 7 11 0 0 0 1 6 6 There is no way to obtain the identity matrix on the left; thus, there is no inverse.
1
0.05 0.01 0.01 0.01 0.01 0.02 0.02 0.01 0.03
4 18 3 72. A 6 20 14 10 25 15
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Chapter 12: Systems of Equations and Inequalities
0.26 0.29 0.20 1.63 1.20 2.53 1.80 1.84
Thus, A1 1.21
21 18 44 2 10 15 73. A 21 12 12 4 8 16
25
61
3
4
76. A 18 12
12 15 7 ; B 3 1 12
Enter the matrices into a graphing utility and use A1B to solve the system. The result is shown below:
6 5 4 9
Thus, the solution to the system is x 4.56 , y 6.06 , z 22.55 or (4.56, 6.06, 22.55) . 0.01 0.02 0.04 0.01 0.02 0.05 0.03 0.03 Thus, A1 . 0.02 0.01 0.04 0.00 0.06 0.07 0.06 0.02
25 61 12 21 77. A 18 12 7 ; B 7 3 2 1 4
22 3 5 16 21 17 4 8 74. A 2 8 27 20 15 3 10 5
Thus, the solution to the system is x 1.19 , y 2.46 , z 8.27 or (1.19, 2.46, 8.27) . 25
61
3
4
78. A 18 12 0.04 0.01 0.02 0.02 Thus, A1 0.04 0.02 0.05 0.02
0.00 0.03 0.01 0.01 . 0.04 0.06 0.00 0.09
12 25 7 ; B 10 1 4
Thus, the solution to the system is x 2.05 , y 3.88 , z 13.36 or (2.05, 3.88, 13.36) .
25 61 12 10 75. A 18 12 7 ; B 9 3 12 1 4
Enter the matrices into a graphing utility and use A1B to solve the system. The result is shown below:
Thus, the solution to the system is x 4.57 , y 6.44 , z 24.07 or (4.57, 6.44, 24.07) .
79.
2 x 3 y 11 5 x 7 y 24 Multiply each side of the first equation by 5, and each side of the second equation by 2 . Then add the equations to eliminate x: 10 x 15 y 55 10 x 14 y 48 y7 Substitute and solve for x:
1382 Copyright © 2025 Pearson Education, Inc.
Section 12.4: Matrix Algebra 2 x 3 7 11
1 0 0 4 R1 2r2 r1 0 1 0 2 5 0 0 1 2
2 x 21 11 2 x 10
x 5 The solution of the system is x 5, y 7 or
The solution is x 4, y 2, z
using ordered pairs 5, 7 .
5 5 or 4,2, . 2 2
2 x 3 y z 2 3z 6 82. 4 x 6 y 2z 2
2 x 8 y 8 80. x 7 y 13 Multiply each side of the second equation by 2 . Then add the equations to eliminate x: 2x 8 y 8 2 x 14 y 26
Write the augmented matrix: 2 4 0 1 4 0
6 y 18
y 3 Substitute and solve for x: x 7 3 13 x 21 13
3 1 2 0 3 6 6 2 2 3 1 2 1 R r / 2 2 1 1 0 3 6 6 2 2
3 12 1 1 2 0 6 5 10 R2 4r1 r2 0 6 2 2
x8 The solution of the system is x 8, y 3 or
using ordered pairs 8, 3 .
1 0 0 1 0 0 1 0 0 1 0 0
x 2 y 4z 2 81. 3x 5 y 2 z 17 4x 3y 22
Write the augmented matrix: 1 2 4 2 3 5 17 2 4 3 0 22
1 2 4 2 0 1 10 23 R2 3r1 r2 5 16 30 R3 4r1 r3 0 1 2 4 2 0 1 10 23 R2 r2 0 5 16 30 1 2 4 2 0 1 10 23 0 0 34 85 R 5r r 3 2 3
3 2
1 6 3 2
1 0 3 2
1 0 3 2
1 0
1 56 53 R2 r2 / 6 2 2 12 1 56 53 3 12 R3 6r2 r3 12 1 56 53 1 4 R3 r3 / 3 0 1 R1 12 r3 r1 5 0 3 R2 56 r3 r2 1 4 12
32 R 3 r r 1 1 0 0 1 2 2 5 0 3 1 0 0 0 1 4
1 2 4 2 0 1 10 23 5 1 2 R3 r3 / 34 0 0 1 2 0 8 R1 4r3 r1 0 1 0 2 R2 10r3 r2 5 0 0 1 2
3 2
5 3
The solution is x , y , z 4 or 3 5 , , 4 . 2 3
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Chapter 12: Systems of Equations and Inequalities
5x y 4 z 2
2x 3y z 4
85. 3x 2 y z 3
83. x 5 y 4 z 3
7 x 13 y 4 z 17
5y z 6
Write the augmented matrix:
Write the augmented matrix:
5 1 4 2 1 5 4 3 7 13 4 17
2 3 1 4 3 2 1 3 0 5 1 6 3 1 2 2 2 R1 r1 / 2 1 3 2 1 3 0 5 1 6
1 5 4 3 R1 r2 5 1 4 2 R2 r1 7 13 4 17 1 5 4 3 0 24 16 17 R2 5r1 r2 0 48 32 38 R 7 r r 3 1 3
3 1 2 2 2 1 0 52 12 3 R2 3r1 r2 0 5 1 6 1 3 2 2 2 1 6 1 2 1 5 5 R2 5 r2 0 6 0 5 1
1 5 4 3 2 17 0 1 3 24 R2 r2 / 24 0 48 32 38 1 5 4 3 2 17 0 1 3 24 0 4 R3 48r2 r3 0 0
The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is , or . 84. 3x 2 y z 2
2 x y 6 z 7 2 x 2 y 14 z 17
1 0 0 1 0 0
2 65 1 0 0 0 R3 5r2 r3 1 1 3 0 5 5 R1 r2 r1 2 6 1 1 5 5 0 0 0 32
1 2 15
Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is 1 1 1 6 x z , y z , and z is any real 5 5 5 5 number. That is, 1 1 1 6 x, y, z | x z , y z , 5 5 5 5
Write the augmented matrix: 3 2 1 2 6 7 2 1 2 2 14 17 1 1 7 9 R1 r1 r2 6 7 2 1 2 2 14 17
z is any real number
1 1 7 9 0 1 20 25 R2 2r1 r2 0 0 0 1 R 2r r 3 1 3
86. 4 x 3 y 2 z 6
The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is , or .
3 x y z 2 x 9y z 6
Write the augmented matrix: 4 3 2 6 3 1 1 2 1 9 1 6
1384 Copyright © 2025 Pearson Education, Inc.
Section 12.4: Matrix Algebra
$62.00 in interest, and Stephanie’s loans accrued $50.30 in interest.
1 9 1 6 R1 r3 3 1 1 2 4 3 2 6 R r 3 1
c.
1 9 6 1 0 26 4 20 R2 3r1 r2 0 39 6 30 R 4r r 3 1 3 1 9 0 1 0 39 1 0 0 1 0 0
9 1 0
1 2 13
6
10 13 R2 r2 /
6 30
1 2 13
26
7062.00 6350.30 Jamal’s loan balance after one month was $7062.00, and Stephanie’s loan balance was $6350.30.
6
10 13
0 R3 39r2 r3 5 12 0 13 13 R1 9r2 r1 10 2 13 13 1 0 0 0 0
89. a.
Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is 5 12 2 10 x z , y z , and z is any real 13 13 13 13 number. That is, 5 12 2 10 x, y, z | x z , y z , 13 13 13 13
b.
88. a. b.
The rows of the 2 by 3 matrix represent stainless steel and aluminum. The columns represent 10-gallon, 5-gallon, and 1-gallon. 500 350 400 The 2 by 3 matrix is: . 700 500 850 500 700 The 3 by 2 matrix is: 350 500 . 400 850
b. The 3 by 1 matrix representing the amount of 15 material is: 8 . 3
z is any real number
87. a.
4000 3000 1 0.011 A(C B ) 2500 3800 1 0.006 4000 3000 1.011 2500 3800 1.006 4000(1.011) 3000(1.006) 2500(1.011) 3800(1.006)
6 9 148.00 A ; B 407.30 3 12
c.
6 9 148.00 AB 3 12 407.30 6(148.00) 9(407.30) 4553.70 3(148.00) 12(407.30) 5331.60 Nikki’s total tuition is $4553.70, and Joe’s total tuition is $5331.60.
The days usage of materials is: 15 500 350 400 11,500 700 500 850 8 17, 050 3 Thus, 11,500 pounds of stainless steel and 17,050 pounds of aluminum were used that day.
d. The 1 by 2 matrix representing cost is: 0.10 0.05 .
4000 3000 0.011 A ; B 0.006 2500 3800
e.
4000 3000 0.011 AB 2500 3800 0.006 4000(0.011) 3000(0.006) 62.00 2500(0.011) 3800(0.006) 50.30 After one month, Jamal’s loans accrued
The total cost of the day’s production was: 11,500 0.10 0.05 2002.50 . 17, 050 The total cost of the day’s production was $2002.50.
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Chapter 12: Systems of Equations and Inequalities 90. a.
The rows of the 2 by 3 matrix represent the location. The columns represent the type of car sold. The 2 by 3 matrix for January is: 400 250 50 450 200 140 . The 2 by 3 matrix for
1 0 0 1 0 0
350 100 30 February is: . 350 300 100
750 350 80 800 500 240
c.
b.
M EK
1
d. Multiplying to find the profit at each location: 100 750 350 80 143,500 800 500 240 150 203, 000 . 200
91. a.
R2 r2
R2 r2 r3
1 0 0 1 0 1 0 1 0 1 1 1 R1 r1 r2 0 0 1 0 1 1 1 0 1 1 Thus, K 1 1 1 . 0 1 1
b. Adding the matrices: 400 250 50 350 100 30 450 200 140 350 300 100
100 The 3 by 1 matrix representing profit: 150 . 200
1 0 0 1 0 1 1 1 2 0 0 1 0 1 1 1 0 0 1 0 1 0 1 1 1 0 1 0 1 1
47 34 33 1 0 1 44 36 27 1 1 1 47 41 20 0 1 1 13 1 20 8 9 19 6 21 14
The city location has a two-month profit of $143,500. The suburban location has a twomonth profit of $203,000.
because a11 47(1) 34(1) 33(0) 13
2 1 1 K 1 1 0 1 1 1 Augment the matrix with the identity and use row operations to find the inverse: 2 1 1 1 0 0 1 1 0 0 1 0 1 1 1 0 0 1
a13 47(1) 34(1) 33(1) 20
1 2 1 1 0 0
1 0 0 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 0 1 1 1 2 0 0 1 0 1 1
Interchange r and r 1 2 R2 2 r1 r2 R r r 1 3 3
a12 47(0) 34(1) 33(1) 1 a21 44(1) 36(1) 27(0) 8 a22 44(0) 36(1) 27(1) 9 a23 44(1) 36(1) 27(1) 19 a31 47(1) 41(1) 20(0) 6 a32 47(0) 41(1) 20(1) 21 a33 47(1) 41(1) 20(1) 14
c.
13 M ; 1 A; 20 T ; 8 H ; 9 I ; 19 S ; 6 F ; 21 U ; 14 N The message: Math is fun.
é0 ê ê1 ê 92. a. A = ê 0 ê ê0 ê ê0 ë
1 0 0 0 1
1 0 0 0 0
0 1 1 0 0
0ù ú 1ú ú 0ú ú 1úú 0úû
b. The diagonal entries are all zero indicating that no page has a link to itself.
1386 Copyright © 2025 Pearson Education, Inc.
Section 12.4: Matrix Algebra
é 1 0 0 2 1ù ê ú ê 0 2 1 0 1ú ê ú c. A2 = ê 0 0 0 0 1ú ; each i, j entry ê ú ê 0 1 0 0 0ú ê ú ê 1 0 0 1 1ú ë û indicates the number of ways to get from page i to page j in exactly two clicks. é0 ê ê1 ê 93. A = ê1 ê ê0 ê ê0 ë
1 0 0 0 1
1 0 0 1 0
a 2 b2 2ab
0 2 1 1 0
0 2 2 0 0
a 2 a b2 0 2ab b 0
é2 ê ê5 ê 2 3 A + A + A = ê4 ê ê2 ê ê1 ë
4 3 2 2 3
a
1 into equation (1); 2 1 1 1 1 b2 0 b2 b 4 2 4 2
5 2 2 3 2
2 5 4 2 1
3 1 1 1 2
4 0 0 2 2
0 4 3 1 0
2ù ú 2ú ú 1ú ú 2úú 1úû
Substitute b 0 into equation (1); a 2 a 0 a (a 1) a 0 or a 1
So, 1 1 1 1 a , b ; a , b ; a 0, b 0; a 1, b 0 2 2 2 2
3ù ú 4ú ú 2ú ú 3úú 2úû
96. Answers will vary. 97. Since the product is found by multiplying the components from the columns of the first matrix by the components in the rows of the second matrix and then adding those products, then the number of columns in the first must equal the number of rows in the second. 98. For real numbers this you multiply both sides by the multiplicative inverse of c, 1c , c not equal 0.
b. The largest number in row 1 (page 1) is 5 which corresponds to page 3.
Since c times 1c results in 1, the multiplicative identity, then a = b. For matrices, this would hold true as long as C has an inverse. Then you
AB BA b d a b a b a c c d c d a c b d
So we have a b a c ab bd c d a c c d b d
1 or b 0 2
Substitute a
Yes, all pages can reach every other page within 3 clicks.
94.
(1) (2)
Solve equation (2); 2ab b 0 (2a 1)b 0
é0 1ù ú ê ú ê4 1 ú 3 ê 1ú A = ê 3 ú ê ê1 0úú ê ê0 1úû ë
2 0 0 1 1
2ab b 0 0 a 2 a b 2 0 0
So,
a. é2 ê ê0 ê 2 A = ê0 ê ê1 ê ê1 ë
2ab a b 0 0 a 2 b 2 b a 0 0
a 2 a b2 2ab b
0ù ú 1ú ú 0ú ú 1úú 0úû
0 1 1 0 0
A2 A 0
95.
would multiply both sides by C
AC BC
ACC 1 BCC 1 AI BI A B
b c 0 (1) a d 0 (2) (3) a d 0 (3) (4) b c 0 (4) (1) (2)
99. If the inverse of A exists:
AX 0
So, a d and b c.
A1 AX A1 0 IX 0 X 0 1387
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1
to get:
Chapter 12: Systems of Equations and Inequalities
If the inverse of A does not exist, then A is singular and you would not be able to multiply A by its inverse to give the identity inverse and X would have no solution. 100.
105.
x 1 x 3 4 x 3 x 1 4 x 3 x 3 x 3 x 3 x 3 x 3
f ( x) ax x 3 x ( 2) 2
3
For a 1 : f ( x) x3 ( x 3) 2 ( x 2)
x3 x 2 6 x 9 x 2 6
5
4
x 4 x 3 x 18 x
x 1 x 3 4 x 3 x 3 x 3
x 2 4 x 3 4 x 12 x 3 x 3
3
101. v w ( 2)(2) ( 1)(1) 4 1 5 vw 5 5 cos 1 v w 5 5 5
x2 8x 9 2
x 9
x 9 x 1 x 3 x 3
106. The radical cannot be negative and the denominator cannot be zero so: 10 2 x 0 x3 0 2 x 10 and x 3 x5 So the domain is: x | x 5, x 3 .
cos 1 ( 1) 180 102.
5x x x2 x2 5 x( x 2) x( x 2) 5 x 2 10 x x 2 2 x
107.
4 x 2 12 x 0 4 x( x 3) 0
3x 4 12 x3 108 x 2 432 x 3x( x3 4 x 2 36 x 144)
4 x 0 or x 3 0
3x x 4 ( x 2 36)
x 0 or x3 The solution set is 0,3 .
3x x 4 ( x 6)( x 6)
103. Let csc 1 u so that csc u ,
2
2
u 1 . Then,
cos csc1 u cos cos
sin cot sin sin
cot cot 2 csc2 1 csc csc csc
u2 1 u
1 i 3 104. 8e 3 8 cos i sin 8 i 3 3 2 2
,
108. Graph the functions. The area enclosed by the 1 1 semi-circle is: r 2 (2) 2 2 . The area 2 2 1 1 enclosed by the triangle is: bh (4)(2) 4 . 2 2 The total area is 2 4 10.28 square units . 109.
f g ( x)
44 3 i
25( 52 sec x) 2 4 2 sec x 5
4 25( 25 sec 2 x ) 4 2 sec x 5
4(sec2 x 1) x
4sec2 x 4 2 sec x 5
4 tan 2 x 2 tan x 2sin x 5cos x 2 2 sec x sec x cos x 2 5 5
5sin x
1388 Copyright © 2025 Pearson Education, Inc.
Section 12.5: Partial Fraction Decomposition x 1
Section 12.5 2
x 2 x 15 x3 + x 2 12 x 9
1. True
x3 2 x 2 15 x x 2 3x 9
2. True
x 2 2 x 15 5x 6 The proper rational expression is:
3x 12
3( x 4) 3 3. 2 x 16 x 4 x 4 x 4
x3 x 2 12 x 9
4. True 5. The rational expression
2
x 2 x 15
x
is proper, since x2 1 the degree of the numerator is less than the degree of the denominator.
10. The rational expression
5x 2
x 5 x2 4
10 x 2 7 x 10 x 2 25 x 18 x 3 18 x 45 42 The proper rational expression is:
is improper, so
x2 4 x2 5
6 x3 5 x 2 7 x 3 42 3x 2 5 x 9 2x 5 2x 5
x2 4 9 The proper rational expression is: x2 5 9 1 2 2 x 4 x 4 3x 2 2 x2 1
11. The rational expression
is improper, so
12. The rational expression
13. The rational expression
3x 2 3 1 The proper rational expression is: 3x 2 2 1 3 2 2 x 1 x 1
x 2 2 x 15
x3 12 x 2 9 x
is 9x2 x4 proper, since the degree of the numerator is less than the degree of the denominator.
x 2 1 3x 2 2
x3 x 2 12 x 9
5x2 7 x 6
is proper, x x3 since the degree of the numerator is less than the degree of the denominator.
perform the division: 3
9. The rational expression
6 x3 5 x 2 7 x 3 is 2x 5
6 x3 15 x 2
perform the division: 1
8. The rational expression
x 2 x 15
6 x3 5 x 2 7 x 3
2x 5
2
7. The rational expression
2
improper, so perform the division: 3x 2 5 x 9
is proper, since x3 1 the degree of the numerator is less than the degree of the denominator.
6. The rational expression
5x 6
x 1
5 x3 2 x 1 x2 4
so perform the division: 5x x 2 4 5 x3
2x 1
3
20 x 22 x 1 The proper rational expression is: 5x
is
5 x3 2 x 1
improper, so perform the division:
2
x 4
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5x
22 x 1 x2 4
is improper,
Chapter 12: Systems of Equations and Inequalities
3x 4 x 2 2
14. The rational expression
3
x 8
is improper,
so perform the division: 3x x3 8 3x 4 x 2 3x
4
4 4 4 x( x 1) x x 1
2 24 x
x 2 24 x 2 The proper rational expression is: 3x 4 x 2 2 3
x 8
3x
18. Find the partial fraction decomposition 3x A B ( x 2)( x 1) x 2 x 1
x 2 24 x 2 x3 8
Multiplying both sides by ( x 2)( x 1) , we obtain: 3x A( x 1) B( x 2)
15. The rational expression x( x 1) x2 x is improper, so 2 ( x 4)( x 3) x x 12 perform the division: 1
Let x 1 , then 3(1) A(0) B(3) 3B 3 B 1 Let x 2 , then 3(– 2) A(3) B (0) 3 A 6 A2 3x 2 1 ( x 2)( x 1) x 2 x 1
x 2 x 12 x 2 x 0 x 2 x 12 2 x 12 The proper rational expression is: x( x 1) 2 x 12 2( x 6) 1 2 1 ( x 4)( x 3) ( x 4)( x 3) x x 12 2
2 x ( x 4)
16. The rational expression
Let x 1 , then 4 A(0) B B4 Let x 0 , then 4 A(1) B (0) A 4
2
x 1
3
2x 8x 2
x 1
improper, so perform the division:
19. Find the partial fraction decomposition: 1 2
is
x ( x 1)
1 A( x 2 1) ( Bx C ) x
x 2 1 2 x3 8 x
Let x 0 , then 1 A(02 1) ( B (0) C )(0)
3
2x 2x 6x
A 1
Let x 1 , then
The proper rational expression is: 2
x 1
A Bx C x x2 1
1 A Bx C 2 x( x 2 1) 2 x( x 1) 2 x x 1 x( x 1)
2x
2 x( x 2 4)
2x
6x 2
x 1
1 A(12 1) ( B (1) C )(1) 1 2A B C 1 2(1) B C B C 1
Let x 1 , then
17. Find the partial fraction decomposition: 4 A B x ( x 1) x x 1
4 B A x( x 1) x ( x 1) x x 1
x( x 1)
4 A( x 1) Bx
1 A((1) 2 1) ( B(1) C )(1) 1 A(1 1) ( B C )(1) 1 2A B C 1 2(1) B C B C 1
Solve the system of equations: B C 1 B C 1 2 2B B 1
1390 Copyright © 2025 Pearson Education, Inc.
Section 12.5: Partial Fraction Decomposition 1 C 1 C0
1 2
x( x 1)
Let x 1 , then 1 A(1 2) B (1 1) 1 A A 1 Let x 2 , then 2 A(2 2) B(2 1) 2B x 1 2 ( x 1)( x 2) x 1 x 2
1 x x x2 1
20. Find the partial fraction decomposition: 1 ( x 1)( x 2 4)
A Bx C x 1 x2 4
22. Find the partial fraction decomposition: 3x A B ( x 2)( x 4) x 2 x 4
Multiplying both sides by ( x 1)( x 2 4) , we 2
obtain: 1 A( x 4) ( Bx C )( x 1) 1 A(5) ( B (1) C )(0)
Let x 1 , then
Multiplying both sides by ( x 2)( x 4) , we obtain: 3x A( x 4) B( x 2)
5A 1 1 5 Let x 1 , then 1 A(12 4) ( B (1) C )(1 1)
A
Let x 2 , then 3( 2) A( 2 4) B ( 2 2) 6 6A A 1 Let x 4 , then 3(4) A(4 4) B (4 2) 12 6 B B2 3x 1 2 ( x 2)( x 4) x 2 x 4
1 5 A ( B C )(2) 1 5 1/ 5 2 B 2C 1 1 2 B 2C 0 2 B 2C 0 BC Let x 0 , then 1 A(02 4) ( B (0) C )(0 1)
23. Find the partial fraction decomposition: x2 A B C 2 2 x 1 ( x 1) ( x 1) x 1 ( x 1)
1 4A C 1 4 1/ 5 C 4 C 5 1 C 5 1
Multiplying both sides by ( x 1) 2 ( x 1) , we obtain: x 2 A( x 1)( x 1) B ( x 1) C ( x 1) 2
Since B C 0 , we have that B
B 1 2
( x 1)( x 4)
1 5
x 1
Let x 1 , then 12 A(1 1)(1 1) B (1 1) C (1 1) 2
1 0 5
1 A(0)(2) B (2) C (0) 2
1 5
1 2B
1 x 1 52 5
B
x 4
1 2
Let x 1 , then (1) 2 A(1 1)(1 1) B (1 1) C (1 1) 2
21. Find the partial fraction decomposition: x A B ( x 1)( x 2) x 1 x 2
1 A(2)(0) B (0) C (2) 2 1 4C
Multiplying both sides by ( x 1)( x 2) , we obtain: x A( x 2) B( x 1)
C
1 4
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Chapter 12: Systems of Equations and Inequalities
Let x 0 , then 02 A(0 1)(0 1) B (0 1) C (0 1) 2 0 A B C A BC 1 1 3 A 2 4 4 x2 2
( x 1) ( x 1)
3 4
1 2
x 1 ( x 1)
2
1 4
x 1
24. Find the partial fraction decomposition: x 1 x 2 ( x 2)
A B C x x2 x 2
Let x 2 , then
1 A 22 2(2) 4 ( B (2) C )(2 2) 1 12 A 1 A 12 Let x 0 , then
1 A 02 2(0) 4 ( B (0) C )(0 2) 1 4 A 2C 1 4 1/12 2C 2C 23
Multiplying both sides by x 2 ( x 2) , we obtain:
C 1 3
x 1 Ax( x 2) B ( x 2) Cx 2
Let x 1 , then
Let x 0 , then 0 1 A(0)(0 2) B (0 2) C (0) 2 1 2B 1 B 2
1 A 12 2(1) 4 ( B (1) C )(1 2)
Let x 2 , then 2 1 A(2)(2 2) B (2 2) C (2) 2 3 4C 3 C 4 Let x 1 , then 2 A B C A B C 2 3 1 3 A 2 2 4 4 3 3 1 x 1 4 22 4 x x2 x 2 ( x 2) x
25. Find the partial fraction decomposition: 1 x3 8
1
( x 2)( x 2 2 x 4) 1 A Bx C ( x 2)( x 2 2 x 4) x 2 x 2 2 x 4
Multiplying both sides by ( x 2)( x 2 2 x 4) , we obtain: 1 A( x 2 2 x 4) ( Bx C )( x 2)
1 7A B C 1 7 1/12 B B
1 3
1 12
1 x3 8
1 12
x2 1 12
x2
1 x1 12 3
x2 2x 4 1 x4 12
x2 2x 4
26. Find the partial fraction decomposition: 2x 4 2x 4 A Bx C 2 3 2 x 1 ( x 1)( x x 1) x 1 x x 1
Multiplying both sides by ( x 1)( x 2 x 1) , we obtain: 2 x 4 A( x 2 x 1) ( Bx C )( x 1) Let x 1 , then
2(1) 4 A 12 1 1 B (1) C (1 1) 6 3A A2 Let x 0 , then
2(0) 4 A 02 0 1 ( B (0) C )(0 1) 4 A C 4 2C C 2
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Section 12.5: Partial Fraction Decomposition
Let x 1 , then
Let x 2 , then
2
2(1) 4 A (1) (1) 1 ( B (1) C )(1 1)
C (2 1) 2 (2 1) D(2 1) 2 4 9 A 9 B 3C D 9 A 3C 4 9 B D
2 A 2 B 2C 2 2 2 B 2( 2) 2B 4 B 2 2x 4 x3 1
1 1 3 9 A 3C 4 9 4 4 2 1 3A C 2 Solve the system of equations: AC 1 2 1 3A C 2 4A 1 A 1 4 3 C 1 4 2 C 1 4 1 1 1 14 x2 4 4 4 ( x 1) 2 ( x 1) 2 x 1 ( x 1) 2 x 1 ( x 1) 2
2x 2 2 x 1 x2 x 1
27. Find the partial fraction decomposition: x2 2
( x 1) ( x 1)2
A B C D x 1 ( x 1)2 x 1 ( x 1) 2
Multiplying both sides by ( x 1)2 ( x 1)2 , we obtain: x 2 A( x 1)( x 1)2 B ( x 1)2 C ( x 1) 2 ( x 1) D ( x 1) 2
Let x 1 , then 12 A(1 1)(1 1) 2 B(1 1) 2 C (1 1) 2 (1 1) D(1 1) 2 1 4B 1 B 4 Let x 1 , then
28. Find the partial fraction decomposition: x 1 x 2 ( x 2) 2
(1) 2 A(1 1)(1 1) 2 B (1 1) 2
A B C D x x 2 x 2 ( x 2) 2
Multiplying both sides by x 2 ( x 2)2 , we obtain:
C (1 1) 2 (1 1) D(1 1) 2
x 1 Ax( x 2) 2 B ( x 2)2 Cx 2 ( x 2) Dx 2
1 4D 1 D 4 Let x 0 , then 2
22 A(2 1)(2 1) 2 B (2 1) 2
Let x 0 , then 0+1 A(0)(0 2) 2 B(0 2) 2 2
0 A(0 1)(0 1) B (0 1) 2
C (0) 2 (0 2) D(0)2
2
C (0 1) (0 1) D(0 1) 0 A B C D AC B D 1 1 1 AC 4 4 2
1 4B 2
1 4 Let x 2 , then B
2 1 A(2)(2 2) 2 B (2 2) 2 C (2) 2 (2 2) D(2)2 3 4D 3 D 4
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Chapter 12: Systems of Equations and Inequalities
Let x 1 , then 1 1 A(1)(1 2) 2 B (1 2) 2 C (1) 2 (1 2) D(1) 2 2 A B C D AC 2 B D 1 3 AC 2 1 4 4 Let x 3 , then 3 1 A(3)(3 2) 2 B (3 2)2 C (3)2 (3 2) D (3)2 4 3 A B 9C 9 D 3 A 9C 4 B 9 D 1 3 3 A 9C 4 9 3 4 4 A 3C 1 Solve the system of equations: AC 1 A 3C 1 A C 1
Let x 0 , then 0 3 A(0 1) 2 B (0 2)(0 1) C (0 2) 3 A 2 B 2C 3 5 2 B 2( 4) 2 B 10 B5 x 3 5 5 4 2 x 2 x 1 ( x 1) 2 ( x 2)( x 1) 30. Find the partial fraction decomposition: x2 x A B C 2 x 2 x 1 ( x 1)2 ( x 2)( x 1)
Multiplying both sides by ( x 2)( x 1) 2 , we obtain: x 2 x A( x 1) 2 B( x 2)( x 1) C ( x 2) Let x 2 , then ( 2) 2 ( 2) A( 2 1) 2 B ( 2 2)( 2 1) C ( 2 2)
A C 1
2 9A
A 3C 1
A
C 1 3C 1 4C 2 1 C 2 1 1 A C 1 1 2 2 3 1 1 12 x 1 2 4 4 x 2 ( x 2) 2 x x 2 x 2 ( x 2) 2
29. Find the partial fraction decomposition: x 3
2 9
Let x 1 , then 12 1 A(1 1) 2 B (1 2)(1 1) C (1 2) 2 3C 2 C 3 Let x 0 , then 02 0 A(0 1) 2 B (0 2)(0 1) C (0 2) 0 A 2 B 2C 2 B A 2C
A B C x 2 x 1 ( x 1)2
obtain: x 3 A( x 1) 2 B ( x 2)( x 1) C ( x 2)
2 2 14 2 9 3 9 7 B 9
Let x 2 , then
x2 x
( x 2)( x 1)
2
2B
Multiplying both sides by ( x 2)( x 1) 2 , we
2 3 A( 2 1) 2 B( 2 2)( 2 1) C ( 2 2) 5 A A 5
Let x 1 , then 1 3 A(1 1) 2 B(1 2)(1 1) C (1 2) 4 C C 4
( x 2)( x 1)
2
2 9
x2
7 9
2 3
x 1 ( x 1)2
31. Find the partial fraction decomposition: x4 2
2
x ( x 4)
A B Cx D x x2 x2 4
Multiplying both sides by x 2 ( x 2 4) , we obtain: x 4 Ax( x 2 4) B( x 2 4) (Cx D) x 2
1394 Copyright © 2025 Pearson Education, Inc.
Section 12.5: Partial Fraction Decomposition
Let x 0 , then 0 4 A(0)(0 4) B (0 4) C (0) D (0) 4 4B B 1 2
2
Multiply both sides by ( x 1) 2 ( x 2 2) :
2
10 x 2 2 x A( x 1)( x 2 2) B( x 2 2) (Cx D)( x 1) 2
Let x 1 , then 10(1) 2 2(1) A(1 1)(12 2) B(12 2)
Let x 1 , then 1 4 A(1)(12 4) B (12 4) (C (1) D)(1) 2 5 5 A 5B C D 5 5A 5 C D 5A C D 0
C (1) D (1 1) 2 12 3B B4 Let x 0 , then 10(0) 2 2(0) A(0 1)(02 2) B(02 2)
Let x 1 , then 1 4 A(1)((1) 2 4) B ((1) 2 4) (C (1) D)(1)2 3 5 A 5 B C D 3 5 A 5 C D 5 A C D 2
C (0) D (0 1) 2 0 2 A 2 B D 0 2 A 8 D 2A D 8 D 2A 8 Let x 1 , then
Let x 2 , then 2 4 A(2)(22 4) B(22 4) (C (2) D)(2)2 6 16 A 8B 8C 4 D 6 16 A 8 8C 4 D 16 A 8C 4 D 2 Solve the system of equations: 5A C D 0 5 A C D 2
10(1) 2 2(1) A(1 1)((1) 2 2) B ((1) 2 2) (C (1) D)(1 1) 2 8 6 A 3B 4C 4 D 8 6 A 12 4C 4 D 6 A 4C 4 D 4 Let x 2 , then
2D 2 D 1 5A C 1 0 C 1 5A
10(2) 2 2(2) A(2 1)(22 2) B(22 2) (C (2) D )(2 1)2 44 6 A 6 B 2C D 44 6 A 24 2C D 6 A 2C D 20 Solve the system of equations (Substitute for D): D 2A 8 6 A 4C 4 D 4 6 A 4C 4(2 A 8) 4 2 A 4C 28 A 2C 14
16 A 8(1 5 A) 4(1) 2 16 A 8 40 A 4 2 24 A 6 1 A 4 5 1 1 C 1 5 1 4 4 4 1 1 x4 1 x 1 4 2 42 2 2 x ( x 4) x x x 4 1 1 1 x 4 4 2 42 x x x 4
6 A 2C D 20 6 A 2C 2 A 8 20 8 A 2C 28
32. Find the partial fraction decomposition: 10 x 2 2 x A B Cx D 2 2 2 2 ( x 1) ( x 2) x 1 ( x 1) x 2
Add the equations and solve:
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Chapter 12: Systems of Equations and Inequalities A 2C 14
x2 2 x 3
8 A 2C 28 9A 42 14 A 3 2C A 14 C
( x 1)( x 2 2 x 4)
14 3
( x 1) 2 ( x 2 2)
4 x 1 ( x 1)
2 3
x 1
1 x1 3 3
x2 2 x 4 1 ( x 1) 3 2
x 2x 4
Multiplying both sides by x( x 2 3 x 3) ( x 1)( x 2 2 x 4) , we obtain:
4 14 D 2A 8 2 8 3 3 10 x 2 2 x
x 1
34. Find the partial fraction decomposition: x 2 11x 18 A Bx C 2 2 x x( x 3x 3) x 3x 3
14 28 14 3 3
14 3
2 3
x 2 11x 18 A( x 2 3 x 3) ( Bx C ) x
143 x 43 2 2 x 2
Let x 0 , then
02 11(0) 18 A 02 3(0) 3 B(0) C (0) 18 3 A A 6 Let x 1 , then
33. Find the partial fraction decomposition: x2 2x 3 A Bx C 2 2 1 x ( x 1)( x 2 x 4) x 2x 4
Multiplying both sides by ( x 1)( x 2 2 x 4) , we obtain: x 2 2 x 3 A( x 2 2 x 4) ( Bx C )( x 1)
12 11(1) 18 A 12 3(1) 3 B (1) C (1) 28 7 A B C 28 7(6) B C B C 14 Let x 1 , then
(1)2 11(1) 18 A (1) 2 3(1) 3
B (1) C (1)
Let x 1 , then (1) 2 2(1) 3 A((1) 2 2(1) 4) ( B (1) C )(1 1) 2 3A 2 A 3 Let x 0 , then
6 A B C 6 6 B C B C 0 Add the last two equations and solve: B C 14
02 2(0) 3 A(02 2(0) 4) ( B(0) C )(0 1)
2B
3 4A C 3 4 2 / 3 C C
1 3
Let x 1 , then 2
1 2(1) 3 A(1 2(1) 4) ( B(1) C )(1 1) 6 7 A 2 B 2C 6 7 2 / 3 2 B 2 1/ 3 2 B 6 143 32 32 B 13
B C 0 14 B7 B C 14 7 C 14 C7 x 2 11x 18
2
2
x( x 3x 3)
6 7x 7 2 x x 3x 3
35. Find the partial fraction decomposition: x A B (3x 2)(2 x 1) 3x 2 2 x 1
Multiplying both sides by (3 x 2)(2 x 1) , we obtain: x A(2 x 1) B(3 x 2) 1396 Copyright © 2025 Pearson Education, Inc.
Section 12.5: Partial Fraction Decomposition
Multiplying both sides by ( x 3)( x 1) , we obtain: x A( x 1) B( x 3)
Let x 1 , then 2 1 A 2 1/ 2 1 B 3 1/ 2 2 2 1 7 B 2 2 1 B 7 2 Let x , then 3 2 A 2 2 / 3 1 B 3 2 / 3 2 3 2 7 A 3 3 2 A 7
Let x 1 , then 1 A(1 1) B(1 3) 1 4B 1 B 4 Let x 3 , then 3 A(3 1) B(3 3) 3 4 A 3 A 4 x 2
x 2x 3
3 4
x3
1 4
x 1
38. Find the partial fraction decomposition: x2 x 8
x2 x 8 ( x 1)( x 2 5 x 6) ( x 1)( x 2)( x 3) A B C x 1 x 2 x 3
2 1 x 7 7 (3x 2)(2 x 1) 3x 2 2 x 1
Multiplying both sides by ( x 1)( x 2)( x 3) , we obtain:
36. Find the partial fraction decomposition: 1 A B (2 x 3)(4 x 1) 2 x 3 4 x 1
x 2 x 8 A( x 2)( x 3) B ( x 1)( x 3)
Multiplying both sides by (2 x 3)(4 x 1) , we obtain: 1 A(4 x 1) B(2 x 3)
C ( x 1)( x 2)
Let x 1 , then (1) 2 ( 1) 8 A(1 2)(1 3)
3 Let x , then 2 3 3 1 A 4 1 B 2 3 2 2 1 7 A
A
B (1 1)(1 3) C ( 1 1)(1 2) 6 2 A A 3
Let x 2 , then
1 7
( 2) 2 ( 2) 8 A( 2 2)( 2 3) B ( 2 1)( 2 3) C ( 2 1)( 2 2) 2 B B2
1 , then 4 1 1 1 A 4 1 B 2 3 4 4 7 1 B 2 2 B 7 2 17 1 7 (2 x 3)(4 x 1) 2 x 3 4 x 1
Let x
Let x 3 , then
(3) 2 ( 3) 8 A(3 2)( 3 3) B(3 1)(3 3) C (3 1)(3 2) 4 2C C2 x2 x 8 2
( x 1)( x 5 x 6)
37. Find the partial fraction decomposition: x x A B x 2 2 x 3 ( x 3)( x 1) x 3 x 1 1397
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3 2 2 x 1 x 2 x 3
Chapter 12: Systems of Equations and Inequalities 39. Find the partial fraction decomposition: x 2 2 x 3 Ax B Cx D 2 ( x 2 4) 2 x 4 ( x 2 4) 2 2
2
Multiplying both sides by ( x 4) , we obtain:
obtain: 7 x 3 A( x 3)( x 1) Bx( x 1) Cx( x 3) Let x 0 , then 7(0) 3 A(0 3)(0 1) B(0)(0 1) C (0)(0 3) 3 3 A
x 2 2 x 3 ( Ax B)( x 2 4) Cx D x 2 2 x 3 Ax3 Bx 2 4 Ax 4 B Cx D x 2 2 x 3 Ax3 Bx 2 (4 A C ) x 4 B D A0;
A 1
Let x 3 , then 7(3) 3 A(3 3)(3 1) B (3)(3 1) C (3)(3 3)
B 1;
24 12 B
4A C 2 4(0) C 2 C2 x2 2 x 3 ( x 2 4) 2
4B D 3 4(1) D 3 D 1 1 x2 4
2x 1 ( x 2 4) 2
B2
Let x 1 , then 7(1) 3 A(1 3)(1 1) B(1)(1 1) C (1)(1 3) 4 4C C 1 7x 3 3
2
x 2 x 3x 40. Find the partial fraction decomposition: x3 1 Ax B Cx D 2 2 2 2 ( x 16) x 16 ( x 16) 2
Multiplying both sides by ( x 2 16) 2 , we obtain: x3 1 ( Ax B)( x 2 16) Cx D
1 2 1 x x 3 x 1
42. Find the partial fraction decomposition: x3 1 x3 1 x5 x 4 x 4 ( x 1) A B C D E 2 3 4 x x x 1 x x
x3 1 Ax3 Bx 2 16 Ax 16 B Cx D
Multiplying both sides by x 4 ( x 1) , we obtain:
x3 1 Ax3 Bx 2 (16 A C ) x 16 B D
x3 1 Ax3 ( x 1) Bx 2 ( x 1) Cx( x 1) D( x 1) Ex 4
A 1; B 0 ;
Let x 0 , then
16 A C 0 16(1) C 0 C 16
04 1 A 03 (0 1) B 02 (0 1) C 0(0 1) D(0 1) E 04 1 D D 1
16 B D 1 16(0) D 1 D 1 x3 1 ( x 2 16)
2
Let x 1 , then x x 2 16
16 x 1
14 1 A 13 (1 1) B 12 (1 1) C 1(1 1)
( x 2 16) 2
D (1 1) E 14
41. Find the partial fraction decomposition: 7x 3 7x 3 3 2 x 2 x 3 x x( x 3)( x 1) A B C x x 3 x 1 Multiplying both sides by x( x 3)( x 1) , we
2E
Let x 1 , then (1) 4 1 A(1)3 (1 1) B( 1) 2 (1 1) C (1)( 1 1) D (1 1) E (1) 4 0 2 A 2 B 2C 2 D E
0 2 A 2 B 2C 2(1) 2 2 A 2 B 2C 4 A B C 2
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Section 12.5: Partial Fraction Decomposition
Let x 2 , then 4
x3 4 x 2 5 x 2 ( x 2)( x 2 2 x 1)
3
2
2 1 A 2 (2 1) B 2 (2 1) C 2(2 1)
( x 2)( x 1) 2
4
D (2 1) E 2 9 8 A 4 B 2C D 16 E
Find the partial fraction decomposition: x2
9 8 A 4 B 2C (1) 16(2)
x3 4 x 2 5 x 2
8 A 4 B 2C 22 4 A 2 B C 11
Let x 2 , then C (2)(2 1) D (2 1) E (2) 4
7 24 A 12 B 6C 3D 16 E 7 24 A 12 B 6C 3(1) 81(2) 42 24 A 12 B 6C 14 8 A 4 B 2C
Solve the system of equations by using a matrix equation: A B C 2 2B C 11 A 4 8 A 4 B 2C 14
1 1 1 A 2 4 2 1 B 11 8 4 2 C 14
x2 3
x 4 x2 5x 2 1
2 2 11 1 14 1
x3 1
5
4
x x
3
1 2
( x 3)( x 1) 2
Find the partial fraction decomposition: x2 1 x3 x 2 5 x 3
5 2 1
0
2
43. Perform synthetic division to find a factor: 2 4
4 3 1 x 2 x 1 ( x 1) 2
x x 5 x 3 ( x 1)( x 2 2 x 3)
x ( x 1) 2 2 1 1 1 2 3 4 x x x 1 x x
2 1 4
5 3 2 3
11 1 1
So, A 2 , B 1 , and C 1 . Thus, 4
( x 2)( x 1) 2 A B C x 2 x 1 ( x 1) 2
44. Perform synthetic division to find a factor:
1 2 3 x3 1
x2
Multiplying both sides by ( x 2)( x 1) 2 , we obtain: x 2 A( x 1) 2 B ( x 2)( x 1) C ( x 2) Let x 2 , then 22 A(2 1) 2 B (2 2)(2 1) C (2 2) 4 A Let x 1 , then 12 A(1 1) 2 B (1 2)(1 1) C (1 2) 1 C C 1 Let x 0 , then 02 A(0 1) 2 B (0 2)(0 1) C (0 2) 0 A 2 B 2C 0 4 2 B 2(1) 2 B 6 B 3
(22)4 1 A( 2)3 (2 1) B (2)2 (2 1)
A 1 1 1 B 4 2 1 C 8 4 2
x2 1
( x 3)( x 1) 2 A B C x 3 x 1 ( x 1) 2
2
Multiplying both sides by ( x 3)( x 1) 2 , we
0
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Chapter 12: Systems of Equations and Inequalities
obtain: x 2 1 A( x 1) 2 B( x 3)( x 1) C ( x 3) Let x 3 , then (3) 2 1 A(3 1) 2 B (3 3)(3 1) C (3 3) 10 16 A 5 A 8 Let x 1 , then 12 1 A(1 1) 2 B(1 3)(1 1) C (1 3) 2 4C 1 C 2
2
x x 5x 3
32 B D 0 32(0) D 0 D0
256 A 16C E 0 256(0) 16(1) E 0 E 16
256 B 16 D F 0 256(0) 16(0) F 0 F 0 2
( x 16)
5 1 1 3B 3 8 2 9 3B 8 3 B 8 x2 1
32 A C 1 32(0) C 1 C 1
x3
Let x 0 , then 02 1 A(0 1) 2 B (0 3)(0 1) C (0 3) 1 A 3B 3C
3
A 0; B 0 ;
3
x 2
( x 16)
2
16 x 2
( x 16)3
46. Find the partial fraction decomposition: x2 Ax B Cx D Ex F 2 2 2 2 3 2 x 4 ( x 4) ( x 4) ( x 4)3
Multiplying both sides by ( x 2 4)3 , we obtain: x 2 ( Ax B )( x 2 4) 2 (Cx D)( x 2 4) Ex F x 2 ( Ax B )( x 4 8 x 2 16) Cx3 Dx 2 4Cx 4 D Ex F x 2 Ax5 Bx 4 8 Ax3 8 Bx 2 16 Ax 16 B
5 8
x3
3 8
Cx3 Dx 2 4Cx 4 D Ex F
1 2
x 2 Ax5 Bx 4 (8 A C ) x3 (8 B D) x 2 (16 A 4C E ) x (16 B 4 D F )
x 1 ( x 1) 2
45. Find the partial fraction decomposition: x3 Ax B Cx D Ex F 2 2 2 2 3 2 x 16 ( x 16) ( x 16) ( x 16)3
A 0; B 0 ;
8A C 0 8(0) C 0 C0
Multiplying both sides by ( x 2 16)3 , we obtain:
8B D 1
16 A 4C E 0
x3 ( Ax B )( x 2 16) 2 (Cx D)( x 2 16) Ex F
8(0) D 1
16(0) 4(0) E 0
D 1
E0
3
16 B 4 D F 0 16(0) 4(1) F 0
4
2
3
2
x ( Ax B )( x 32 x 256) Cx Dx 16Cx 16 D Ex F x3 Ax5 Bx 4 32 Ax3 32 Bx 2 256 Ax 256 B Cx3 Dx 2 16Cx 16 D Ex F
F 4 x 2
2
( x 4)
3
x3 Ax5 Bx 4 (32 A C ) x3 (32 B D) x 2 (256 A 16C E ) x (256 B 16 D F )
1400 Copyright © 2025 Pearson Education, Inc.
1 2
( x 4)
2
4 2
( x 4)3
Section 12.5: Partial Fraction Decomposition 47. Find the partial fraction decomposition: 4 4 A B 2 x 2 5 x 3 ( x 3)(2 x 1) x 3 2 x 1
49. Find the partial fraction decomposition: 2x 3 2x 3 x 4 9 x 2 x 2 ( x 3)( x 3) A B C D 2 x x x 3 x 3
Multiplying both sides by ( x 3)(2 x 1) , we obtain: 4 A(2 x 1) B( x 3)
Multiplying both sides by x 2 ( x 3)( x 3) , we obtain: 2 x 3 Ax( x 3)( x 3) B ( x 3)( x 3)
1 , then 2 1 1 4 A 2 1 B 3 2 2 7 4 B 2 8 B 7 Let x 3 , then 4 A(2(3) 1) B (3 3)
Let x
Cx 2 ( x 3) Dx 2 ( x 3)
Let x 0 , then 2 0 3 A 0(0 3)(0 3) B (0 3)(0 3) C 02 (0 3) D 02 (0 3) 3 9 B 1 3 Let x 3 , then 2 3 3 A 3(3 3)(3 3) B (3 3)(3 3)
4 7A A 4 2 x2 5x 3
4 7
x3
B
4 7
87
C 32 (3 3) D 32 (3 3)
2x 1
9 54C
48. Find the partial fraction decomposition: 4x 4x A B 2 2 x 3x 2 ( x 2)(2 x 1) x 2 2 x 1
C
Let x 3 , then 2(3) 3 A(3)(3 3)(3 3) B (3 3)(3 3)
Multiplying both sides by ( x 2)(2 x 1) , we obtain: 4 x A(2 x 1) B ( x 2)
C (3) 2 (3 3)
1 Let x , then 2 1 1 1 4 A 2 1 B 2 2 2 2
D(3)2 (3 3) 3 54 D D
5 B 3 4 B 5 Let x 2 , then 4( 2) A(2( 2) 1) B ( 2 2) 2
C 12 (1 3) D 12 (1 3) 5 8 A 8B 4C 2 D 5 8 A 8 1/ 3 4 1/ 6 2 1/18 8 2 1 5 8A 3 3 9 16 8A 9 2 A 9 1 1 2 1 2x 3 9 23 6 18 4 2 x x3 x3 x 9x x
8 5
4x 2 x 2 3x 2
8 5
x2
1 18
Let x 1 , then 2 1 3 A 1(1 3)(1 3) B (1 3)(1 3)
8 5 A A
1 6
4 5
2x 1
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Chapter 12: Systems of Equations and Inequalities 50. Find the partial fraction decomposition: x2 9 x2 9 x 4 2 x 2 8 ( x 2 2)( x 2)( x 2) A B Cx D x 2 x 2 x2 2
51.
x3 x 2 3 x 2 3x 4 Dividing:
x3 3x 2 4 x
Multiplying both sides by (x 2 2)( x 2)( x 2) , we obtain: x 2 9 A( x 2 2)( x 2) B ( x 2)( x 2 2) (Cx D)( x 2)( x 2)
2x 4x 3 2 x 2 6 x 8
13 24 A 13 24 Let x 2 , then A
B ( 2 2)(( 2)2 2) (C ( 2) D)( 2 2)( 2 2) 13 24 B
1 5 B 1 B 5 Let x 4 , then 10(4) 11 A(( 4) 1) B ( 4 4) 51 5 A 51 A 5 51 15 10 x 11 5 x 2 3x 4 x 4 x 1
13 24 Let x 0 , then B
02 9 A(02 2)(0 2) B (0 2)(02 2) (C (0) D )(0 2)(0 2) 9 4 A 4B 4D
x2
x2
10 x 11
Multiplying both sides by ( x 4)( x 1) , we obtain: 10 x 11 A( x 1) B( x 4) Let x 1 , then 10 1 11 A 1 1 B 1 4
( 2) 2 9 A(( 2) 2 2)( 2 2)
13 24
x2
, x 4,1 x 3x 4 x 2 3x 4 Find the partial fraction decomposition: 10 x 11 10 x 11 A B 2 1 x x x x ( 4)( 1) 4 x 3x 4 2
(C (2) D )(2 2)(2 2)
x4 2 x2 8
2
x x 3
13 24
10 x 11 3
22 9 A(22 2)(2 2) B (2 2)(22 2)
x2 9
2
Let x 2 , then
13 13 9 4 4 4D 24 24 14 4D 3 7 D 6 Let x 1 , then 12 9 A(12 2)(1 2) B (1 2)(12 2) (C (1) D)(1 2)(1 2) 10 9 A 3B 3C 3D 39 13 7 10 3C 8 8 2 3C 0 C0
x2 x3 x 2 0 x 3
x 2 3x 4
Thus,
52.
x3 x 2 3 x 2 3x 4
x2
51 5
x4
x3 3x 2 1 x2 5x 6 Dividing:
x 8 x3 3x 2 0 x 1
x2 5x 6
76 x2 2
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x3 5 x 2 6 x
8x2 6 x 1 8 x 2 40 x 48
34x 49
15 x 1
Section 12.5: Partial Fraction Decomposition
x3 3x 2 1
x 8
34 x 49
Since x 2 4 is irreducible then we cannot go any further.
, x 2, 3
x2 5x 6 x2 5x 6 Find the partial fraction decomposition: 34 x 49 34 x 49 A B 2 x 5 x 6 ( x 2)( x 3) x 2 x 3
55.
Multiplying both sides by ( x 2)( x 3) , we obtain: 34 x 49 A( x 3) B ( x 2) Let x 3 , then 34 3 49 A 3 3 B 3 2
53.
2
x 5x 6
x 8
x 4 4 x3 4 x 2
19 53 x3 x2
Multiplying both sides by ( x 2) 2 , we obtain: 11x 32 A( x 2) B 11x 32 Ax 2 A B 11x 32 Ax (2 A B) Since the coefficient of x is A then A 11 . Let A 11 , then 32 2 A B 32 2(11) B 10 B 11x 32 11 10 x 2 4 x 4 x 2 ( x 2) 2
x x2 1
x
x x2 1
Since x 2 1 is irreducible then we cannot go any further. 54.
Thus, x 4 5 x3 x 4
x3 x
x2 4x 4
x2 4 Dividing: x x 0 x 4 x 0 x2 x 2
3
x3 0 x 2 4 x
3x x3 x 2
x 4
x
11x 32 4 3 x 5x x 4 11x 32 x2 4 x 7 2 , 2 x 4x 4 x 4x 4 x 2 Find the partial fraction decomposition: 11x 32 11x 32 A B 2 x 4 x 4 ( x 2)( x 2) x 2 ( x 2) 2
x x 0 x 1 x 0 x2 0 x
x3
7x 2 17 x 4 7 x 2 28 x 28
3
2
x2 1 Dividing:
x3 0 x 2 x
4x 9x x 4 x3 16 x 2 16 x 3
x3
2
x2 4 x 7 x 4 0 x3 5 x 2 x 4
x 4x 4
19 A 19 34 x 49 53 2 x x 2 3 x 5x 6 x3 3 x 2 1
x2 4 x 4 Dividing: 2
53 B 53 B Let x 2 , then 34 2 49 A 2 3 B 2 2
Thus,
x 4 5 x3 x 4
3 x x2 4
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x2 4 x 7
11 10 x 2 ( x 2) 2
Chapter 12: Systems of Equations and Inequalities
56.
x 4 x3 x 2
57.
x2 2 x 1 Dividing: x 2x 1
x 4 2 x3 x 2
x5 0 x 4 2 x3 0 x 2 x
3x3 x 2 x 3 x3 6 x 2 3 x
4
3
x x x2
6x 3 6x 3
Multiplying both sides by ( x 1) 2 , we obtain: 6 x 3 A( x 1) B 6 x 3 Ax A B 6 x 3 Ax ( A B) Since the coefficient of x is A then A 6 . Let A 6 , then 3 A B 3 1(6) B 3 B 6x 3 6 3 2 x 2 x 1 x 1 ( x 1) 2
x2 2 x 1
3
2
2x x
2
x 2 3x 5 2 , x 1 x2 2 x 1 x 2x 1 Find the partial fraction decomposition: 6x 3 6x 3 A B 2 x 2 x 1 ( x 1)( x 1) x 1 ( x 1) 2
Thus, x 4 x3 x 2
x 2x x x 2 x 4 0 x3 2 x 2 0 x 1 4
5x 2 4 x 2 5 x 2 10 x 5
x4 2 x2 1 Dividing:
x 1 x 4 0 x3 2 x 2 0 x 1 x5 x 4 0 x3 x 2 0 x 2
x 2 3x 5 x 4 3 x3 0 x 2 x 2
2
x5 x 4 x 2 2
6 3 x 2 3x 5 x 1 ( x 1) 2
3
x5 x 4 x 2 2
x 1
x 1
2 x3 x 2 x 1
, x4 2 x2 1 x4 2 x2 1 x 1, 1 Find the partial fraction decomposition: 2 x3 x 2 x 1 A B C D 2 2 2 x 1 x 1 ( x 1) ( x 1) ( x 1) ( x 1) 2 Multiplying both sides by ( x 1) 2 ( x 1) 2 , we obtain: 2 x3 x 2 x 1 A( x 1)( x 1) 2 B( x 1) 2 C ( x 1) 2 ( x 1) D ( x 1) 2 Let x 1 , then 2(1)3 (1) 2 (1) 1 A(1 1)(1 1) 2 B (1 1) 2 C (1 1) 2 (1 1) D(1 1) 2 1 4B 1 B 4
Let x 1 , then 2(1)3 (1) 2 (1) 1 A(1 1)(1 1) 2 B (1 1) 2 C (1 1) 2 (1 1) D(1 1) 2 3 4D 3 D 4
Let x 0 , then 2(0)3 (0) 2 (0) 1 A(0 1)(0 1) 2 1 3 (0 1) 2 C (0 1) 2 (0 1) (0 1) 2 4 4 1 3 1 A C 4 4 0 AC
1404 Copyright © 2025 Pearson Education, Inc.
Section 12.5: Partial Fraction Decomposition
Let x 2 , then 2(2)3 (2) 2 (2) 1 A(2 1)(2 1) 2 1 3 (2 1)2 C (2 1) 2 (2 1) (2 1) 2 4 4 1 27 19 3 A 9C 4 4 12 3 A 9C
9 3 A C 1 3B D So 9 3(7) C C 12 and 1 3(1) D D 2
0 AC 12 3 A 9C AC 12 3 A 9 A 12 12 A 1 A and 1 C
( x 2 3)( x 2 3)
7 x3 x 2 9 x 1
Thus,
e
1 3 1 1 4 4 x 1 ( x 1) 2 x 1 ( x 1) 2 x4 2 x2 1 x5 x 4 x 2 2 2
x 2x 1
x 1
1 x 1
4
x 6x 9
2x
x
e 2 A
u2
So
u 1
x x 4 0 x 3 6 x 2 0 x 9 x5 0x 4 x3 x 2 0 x 1
x5 0 x 4 6 x3 0 x 2 9 x
7 x x 9x 1 3
x
2
7 x 3 x 2 9 x 1
, x 6x 9 x4 6 x2 9 x 1, 1 Find the partial fraction decomposition: 7 x3 x 2 9 x 1 Ax B Cx D 2 Multiply ( x 2 3)( x 2 3) x 3 ( x 2 3) 2 2
x
12 x 2 ( x 2 3) 2
7 x 1 2
x 3
12 x 2 ( x 2 3) 2
3u 2
u u2
3u (u 2)(u 1)
3u A(u 1) B (u 2)
3u 2
u u 2
u e x to get
x4 6x2 9 Dividing:
4
B
x5 x3 x 2 1
x5 x3 x 2 1
2
If u 1, then B 1;if u 2 then A 2 .
1 3 1 4 2 4 x 1 ( x 1) 2 ( x 1)
58.
x2 3
59. Let u e x . Then
2 x3 x 2 x 1
4
7 x 1
x5 x3 x 2 1
3e x
Thus,
2 1 . Back substitute u 2 u 1
e
2x
3e x ex 2
2 ex 2
1 ex 1
.
60. Let u 3 x . Then 2 2 2 A B C x 3 x u 3 u u (u 1)(u 1) u u 1 u 1 2 A(u 1)(u 1) Bu (u 1) Cu (u 1) If u = 0, then A = -2; if u = 1, the C = 1; if u = -1 then B = 1. 2 2 1 1 So we have 3 . Back u u u u 1 u 1
substitute u 3 x to get 2 2 1 1 3 3 3 . 3 x x x x 1 x 1
ing both sides by ( x 2 3) 2 , we obtain: 7 x3 x 2 9 x 1 ( Ax B )( x 2 3) (Cx D) Ax3 Bx 2 3 Ax 3B Cx D Ax3 Bx 2 x(3 A C ) x (3B D) Then A 7 and B 1 .
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Chapter 12: Systems of Equations and Inequalities
r 61. A P 1 n
nt
0.18 8400 4200 1 365 2 1.000493
x 4 8 y 3 y 3 log8 ( x 4)
365t
y log8 ( x 4) 3
365t
ln 2 ln 1.000493
f
365t
ln 2 365ln 1.000493
3.85 years
62.
1
( x) log8 ( x 4) 3
67. Focus: (0, 13); Vertices: (0, 5), (0, 13); Center: (0, 0); Transverse axis is parallel to the y-axis; a 5; c 13 . Find the value of b: b 2 c 2 a 2 169 25 144 b 12 12 y 2 x2 Write the equation: 1. 25 144
ln 2 365t ln 1.000493 t
x 8 y 3 4
66.
f ( x) x 4; g ( x) x 2 3 x f ( 3) 3 4 1 g (1) 12 3(1) 2
68.
1 cos 308 cos 52 1 cos 308 cos(360 52) 1 cos 308 1 cos 308
63. sec 52 cos 308
5 64. 1, 4
5 4 x 1 x 2 5 3 9 x 2 5 6 x 5 6 6 The solutions set is x | x or , . 5 5
69. 2 x 4 xD 4 y 2 yD D 2 x 4 y D 4 xD 2 yD 2 x 4 y D(1 4 x 2 y ) 2x 4 y D 1 4x 2 y 70. From the calculator we see that the intersection occurs at the point (3, 4) . The slope of the
x 1 cos
5 2 5 2 ; y 1 sin 4 2 4 2
Rectangular coordinates of 1,
5 are 4
2 2 , . 2 2
65.
f ( x)
3( x) 2
( x) 10 function is odd.
3x 2
x 10
f ( x ) . The
perpendicular line would be 1 ( x 3) 2 1 3 y4 x 2 2 1 11 y x 2 2 y4
1406 Copyright © 2025 Pearson Education, Inc.
1 . 2
Section 12.6: Systems of Nonlinear Equations
Section 12.6
y 2 x2 1
3.
x2 y 2 1
1. y 3 x 2 The graph is a line. x-intercept: 0 3x 2 3x 2 2 x 3
x2 y 2 1 12 12 The graph is a hyperbola with center (0, 0), transverse axis along the x-axis, and vertices at (1, 0) and (1, 0) . The asymptotes are y x and y x . y
y-intercept: y 3 0 2 2
y
2 3 ,0
2
(0, 2)
x
x
4.
2
x2 4 y 2 4 x2 4 y 2 4 4 4 2 x y2 1 4 x2 y 2 1 22 12 The graph is an ellipse with center (0, 0) , major
2. y x 2 4 The graph is a parabola. x-intercepts: 0 x2 4 x2 4 x 2, x 2
axis along the x-axis, vertices at (2, 0) and (2, 0) . The graph also has y-intercepts at (0, 1) and (0,1) .
2
y-intercept: y 0 4 4 The vertex has x-coordinate: 0 b x 0. 2a 2 1
y
The y-coordinate of the vertex is
y 0 2 4 4 .
1407 Copyright © 2025 Pearson Education, Inc.
x
Chapter 12: Systems of Equations and Inequalities
y 36 x 2 7. y 8 x
2 y x 1 5. y x 1
(0, 1) and (1, 2) are the intersection points.
(2.59, 5.41) and (5.41, 2.59) are the intersection points.
Solve by substitution: x2 1 x 1
Solve by substitution: 36 x 2 8 x
2
x x0
36 x 2 64 16 x x 2
x( x 1) 0
2 x 2 16 x 28 0
x 0 or x 1
x 2 8 x 14 0
y 1 y2 Solutions: (0, 1) and (1, 2)
8 64 56 2 8 2 2 2 4 2
x
2 y x 1 6. y 4 x 1
If x 4 2, y 8 4 2 4 2 Solutions: 4 2, 4 2 and 4 2, 4 2 If x 4 2, y 8 4 2 4 2
y 4 x 2 8. y 2 x 4
(0, 1) and (4, 17) are the intersection points. Solve by substitution: x2 1 4 x 1 x2 4 x 0 x( x 4) 0 x 0 or x 4 y 1 y 17 Solutions: (0, 1) and (4, 17)
(–2, 0) and (–1.2, 1.6) are the intersection points. 1408 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations
Solve by substitution:
y x 10. y 6 x
2
4 x 2x 4 4 x 2 4 x 2 16 x 16 5 x 2 16 x 12 0 ( x 2)(5 x 6) 0 x 2 or x y0
or y
6 5
8 5
6 8 Solutions: 2, 0 and , 5 5 y x 9. y 2 x
(4, 2) is the intersection point. Solve by substitution: x 6 x x 36 12 x x 2 x 2 13x 36 0 ( x 4)( x 9) 0 x4
or x 9
y 2 or y 3 Eliminate (9, –3); we must have y 0 . Solution: (4, 2) x 2 y 11. 2 x y 2 y
(1, 1) is the intersection point. Solve by substitution: x 2 x x 4 4 x x2 x2 5x 4 0 ( x 4)( x 1) 0 x4
or x 1
y 2 or y =1 Eliminate (4, –2); we must have y 0 . Solution: (1, 1)
(0, 0) and (8, 4) are the intersection points. Solve by substitution: 2 y y2 2 y y2 4 y 0 y ( y 4) 0 y 0 or y =4 x 0 or x =8 Solutions: (0, 0) and (8, 4)
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Chapter 12: Systems of Equations and Inequalities
y x 1 12. 2 y x 6 x 9
x2 y2 8 14. 2 2 x y 4 y 0
(2, 1) and (5, 4) are the intersection points.
(–2, –2) and (2, –2) are the intersection points.
Solve by substitution: x2 6 x 9 x 1
Substitute 8 for x 2 y 2 in the second equation. 8 4y 0 4y 8 y 2
x 2 7 x 10 0 ( x 2)( x 5) 0 x 2 or x =5
x 8 ( 2) 2 2
y 1 or y =4 Solutions: (2, 1) and (5, 4)
Solution: (–2, –2) and (2, –2) y 3x 5 15. 2 2 x y 5
x2 y 2 4 13. 2 2 x 2 x y 0
(1, –2) and (2, 1) are the intersection points.
(–2, 0) is the intersection point. Substitute 4 for x 2 y 2 in the second equation. 2x 4 0 2x 4 x 2 y 4 ( 2) 2 0
Solution: (–2, 0)
Solve by substitution: x 2 (3 x 5) 2 5 x 2 9 x 2 30 x 25 5 10 x 2 30 x 20 0 x 2 3x 2 0 ( x 1)( x 2) 0 x 1 or x 2 y 2 y 1 Solutions: (1, –2) and (2, 1)
1410 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations
x 2 y 2 16 18. 2 x 2 y 8
2 2 x y 10 16. y x2
(–3.46, 2), (0, –4), and (3.46, 2) are the intersection points. Substitute 2 y 8 for x 2 in the first equation.
(1, 3) and (–3, –1) are the intersection points.
2 y 8 y 2 16
Solve by substitution: x 2 ( x 2) 2 10
y2 2 y 8 0 ( y 4)( y 2) 0 y 4 or y 2
x 2 x 2 4 x 4 10 2x2 4 x 6 0
x 2 0 or x 2 12
2( x 3)( x 1) 0 x 3 or x 1 y 1 y3 Solutions: (–3, –1) and (1, 3)
x0
x 2 3
Solutions: (0, – 4), 2 3, 2 , 2 3, 2
xy 4 19. 2 2 x y 8
2 2 x y 4 17. 2 y x 4
(–2, –2) and (2, 2) are the intersection points. Solve by substitution: 2
(–1, 1.73), (–1, –1.73), (0, 2), and (0, –2) are the intersection points.
4 x2 8 x 16 x2 2 8 x x 4 16 8 x 2 x 4 8 x 2 16 0
Substitute x 4 for y 2 in the first equation:
x2 x 4 4 x2 x 0 x ( x 1) 0 x0
or x 1
y2 4
y2 3
y 2
y 3
( x 2 4) 2 0
Solutions: (0, – 2), (0, 2), 1, 3 , 1, 3
x2 4 0 x2 4 x 2 or x 2 y 2 or y 2 Solutions: (–2, –2) and (2, 2)
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Chapter 12: Systems of Equations and Inequalities
xy 1 22. y 2x 1
2 x y 20. xy 1
(1, 1) is the intersection point.
(–1, –1) and (0.5, 2) are the intersection points.
Solve by substitution: 1 x2 x 3 x 1 x 1
Solve by substitution: x(2 x 1) 1 2 x2 x 1 0 ( x 1)(2 x 1) 0
y 1
1 Solutions: (–1, –1) and , 2 2
x y 4 21. y x2 9 2
1 2 y2
x 1 or x
y (1) 2 1 Solution: (1, 1) 2
y x 2 4 23. y 6 x 13
No solution; Inconsistent. Solve by substitution: x 2 ( x 2 9) 2 4
(3, 5) is the intersection point.
x 2 x 4 18 x 2 81 4
Solve by substitution: x 2 4 6 x 13
x 4 17 x 2 77 0 17 289 4(77) x 2 17 19 2 There are no real solutions to this expression. Inconsistent. 2
x2 6 x 9 0 ( x 3) 2 0 x3 0 x3 y (3) 2 4 5 Solution: (3,5)
1412 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations
24. x 2 y 2 10 xy 3
If x 2 2 :
y
If x 2 2 :
y
If x 1:
y
4 2 2 4
2
2 2
2
4 4 1 4 y 4 1
If x 1:
Solutions:
2 2, 2 , 2 2, 2 , (1, 4), (1, 4)
26. Solve the second equation for y, substitute into the first equation and solve: x 2 y 2 21 x y 7 y 7 x
(1, 3), (3, 1), (–3, –1), and (–1, –3) are the intersection points. Solve by substitution: 2 x 2 3x 10 x 2 92 10 x 4 x 9 10 x 2
4
x 2 7 x 21 2
x 2 49 14 x x 2 21 14 x 70 x5 y 75 2
2
x 10 x 9 0 ( x 2 9)( x 2 1) 0 ( x 3)( x 3)( x 1)( x 1) 0 x 3 or x –3 or x 1 or x 1 y 1 y 1 y 3 y –3 Solutions: (3, 1), (–3, –1), (1, 3), (–1, –3)
Solution: (5, 2) 27. Substitute the first equation into the second equation and solve: y 3x 2 2 2 3x y 4
25. Solve the second equation for y, substitute into the first equation and solve: 2 x 2 y 2 18 4 xy 4 y x
3x 2 3x 2 4 2
3x 2 9 x 2 12 x 4 4 12 x 2 12 x 0 12 x x 1 0
2
4 2 x 18 x 16 2 x 2 2 18 x 4 2 x 16 18 x 2 2
12 x 0 or x 1 0 x 0 or x 1 If x 0 : y 3(0) 2 2 If x 1:
Solutions: (0, 2), 1, 1
2 x 4 18 x 2 16 0
28. Solve the second equation for x and substitute into the first equation and solve: x 2 4 y 2 16 2y x 2 x 2y 2
x4 9 x2 8 0
x 8 x 1 0 2
y 3 1 2 1
2
x2 8
or
x2 1
x 8= 2 2
or
x 1
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Chapter 12: Systems of Equations and Inequalities
(2 y 2) 2 4 y 2 16 4 y 2 8 y 4 4 y 2 16 8 y 12 3 y 2 x 2 32 2 5
Solutions:
5, 32
29. Solve the first equation for y, substitute into the second equation and solve: x y 1 0 y x 1 2 2 x y 6 y x 5 x 2 ( x 1)2 6( x 1) x 5 x 2 x 2 2 x 1 6 x 6 x 5 2 x2 5x 0 x(2 x 5) 0 x 0 or x 5 2 y (0) 1 1 If x 0 : 5 5 7 y 1 If x : 2 2 2 5 7 Solutions: (0, 1), 2 , 2
30. Solve the second equation for y, substitute into the first equation and solve: 2 x 2 xy y 2 8 4 xy 4 y x 2
4 4 2 x2 x 8 x x 16 2x2 4 2 8 x 2 x 4 16 12 x 2 x4 6 x2 8 0
x 4 x 2 0 ( x 2)( x 2) x 2 x 2 0 2
2
x 2 or x 2 or x 2 or x 2 y2 y 2 Solutions:
y2 2
(2, 2), 2, 2 2 ,
y 2 2
2, 2 2 , (2, 2)
31. Solve the second equation for y, substitute into the first equation and solve:
9 x 2 8 xy 4 y 2 70 3 3 x 2 y 10 y x 5 2 2
3 3 9 x 2 8 x x 5 4 x 5 70 2 2 9 x 2 12 x 2 20 x 9 x 2 40 x 100 70 30 x 2 60 x 100 70 3x 2 6 x 3 0 (3x 1)( x 3) 0
1 or x 3 3 1 31 9 y 5 If x : 3 23 2 3 1 If x 3 : y (3) 5 2 2 1 9 1 Solutions: , , 3, 3 2 2 x
2 y 2 3xy 6 y 2 x 4 0 32. 2x 3y 4 0 Solve the second equation for x, substitute into the first equation and solve: 2x 3y 4 0 2x 3y 4 3y 4 x 2 3y 4 3y 4 2 2 y 3 y 6y 2 4 2 2 9 2 y2 y2 6 y 6 y 3y 4 4 2 5 y 2 15 y 0 2 5 y 2 30 y 0 5 y ( y 6) 0 y 0 or y 6 If y 0 :
x
If y 6 :
x
3 0 4 2 36 4
2 Solutions: (–2, 0), (7, 6)
2
7
33. Multiply each side of the second equation by 4 and add the equations to eliminate y:
1414 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations
If x 1 : 2
2 2 x2 4 y 2 7 x 4 y 7 2 4 2 12 x 2 4 y 2 124 3x y 31
13x 2
2
9 3 1 3 5 y 2 12 y 2 y 4 2 2 If x 1 : 2
117 2
x 9 x 3 2 2 2 If x 3 : 3(3) y 31 y 4 y 2
2
9 3 1 3 5 y 2 12 y 2 y 2 4 2 Solutions: 1 3 1 3 1 3 1 3 , , , , , , , 2 2 2 2 2 2 2 2
If x 3 : 3(3) 2 y 2 31 y 2 4 y 2 Solutions: (3, 2), (3, –2), (–3, 2), (–3, –2) 34. 3x 2 2 y 2 5 0 2 2 2 x y 2 0
36. x 2 3 y 2 1 0 2 2 2 x 7 y 5 0
2 2 3x 2 y 5 2 2 2 x y 2 Multiply each side of the second equation by –2 and add the equations to eliminate y: 3x 2 2 y 2 5
x 2 3 y 2 1 2 2 2 x 7 y 5
Multiply each side of the first equation by –2 and add the equations to eliminate x: 2 x 2 6 y 2 2
4 x 2 2 y 2 4 x 2 1
2 x 2 7 y 2 5
x2 1 x 1
y 2 3
If x 1:
y2 3
2(1) 2 y 2 2 y 2 4 y 2 If x 1:
y 3 If y 3 :
2(1)2 y 2 2 y 2 4 y 2 Solutions: (1, 2), (1, –2), (–1, 2), (–1, –2)
x2 3
2
2
If y 3 :
35. 7 x 2 3 y 2 5 0 2 2 3x 5 y 12
x2 3 3
1 x 8 x 2 2 2
2
Solutions:
7 x 2 3 y 2 5 2 2 3x 5 y 12 Multiply each side of the first equation by 5 and each side of the second equation by 3 and add the equations to eliminate y: 35 x 2 15 y 2 25
2 2, 3 , 2 2, 3 , 2 2, 3 , 2 2, 3
37. Multiply each side of the second equation by 2 and add the equations to eliminate xy: 2 x 2 2 xy 10 x 2 xy 10 2 2 6 x 2 2 xy 4 3x xy 2
9 x 2 15 y 2 36 44 x 2 11 x2
3 1 x 8 x 2 2
7 x2
1 4
14 2
x 2
1 x 2
x 2
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Chapter 12: Systems of Equations and Inequalities
If x 2 :
4x2 2 y 2 4
2 2 y 2
x 2 2 y 2 8
3
2
2 y 4 y
4 2
y2 2
If x 2 :
2 y 2
2 y 4 y
3 2
2
4
y 2 2
2
2, 2 2 , 2, 2 2
Solutions:
38. 5 xy 13 y 2 36 0 xy 7 y 2 6 5 xy 13 y 2 36 2 xy 7 y 6 Multiply each side of the second equation by –5 and add the equations to eliminate xy: 5 xy 13 y 2 36 5 xy 35 y 2 30 y2 3 y 3
If y 3 :
3 7 3 6 3 x 15 x
15 3
x 5 3
If y 3 :
x 3 7 3
6 2
3 x 15 x
Solutions:
2 x2 3 y 2 6 5 y 2 2 2 y2 5 No real solution. The system is inconsistent.
8 x 2 2 y 2 48
2
40. y 2 x 2 4 0 2 2 2 x 3 y 6 x 2 y 2 4 2 2 2 x 3 y 6 Multiply each side of the first equation by 2 and add the equations to eliminate x: 2 x 2 2 y 2 8
41. x 2 2 y 2 16 2 2 4 x y 24 Multiply each side of the second equation by 2 and add the equations to eliminate y: x 2 2 y 2 16
22 y 2 66
x
5 x 2 4 4 x2 5 No real solution. The system is inconsistent.
15 3
x5 3
5 3, 3 , 5 3, 3
39. 2 x 2 y 2 2 2 2 x 2 y 8 0 2 x 2 y 2 2 2 2 x 2 y 8 Multiply each side of the first equation by 2 and add the equations to eliminate y:
9 x 2 64 64 x2 9 8 x 3 8 If x : 3 2 80 8 2 2 2 y 16 2 y 3 9 y2
40 2 10 y 9 3
8 If x : 3 2 8 80 2 2 2 y 16 2 y 9 3 y2
40 2 10 y 9 3
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Section 12.6: Systems of Nonlinear Equations
Solutions: 8 2 10 8 2 10 8 2 10 , , , , , , 3 3 3 3 3 3
2 5 43. 2 2 3 0 x y 3 1 7 x 2 y 2
8 2 10 , 3 3
2 5 x 2 y 2 3 3 1 7 x 2 y 2 Multiply each side of the second equation by 2 and add the equations to eliminate y: 5 2 2 3 2 x y 6 2 2 14 2 x y
42. 4 x 2 3 y 2 4 2 2 2 x 6 y 3
Multiply each side of the first equation by 2 and add the equations to eliminate y: 8x2 6 y 2 8 2 x 2 6 y 2 3 10 x 2 5 1 x2 2 x If x
11
11 x2 x2 1 x 1 If x 1: 3 1 1 1 2 7 2 4 y2 2 4 y y (1) 1 y 2 If x 1: 3 1 1 1 2 7 2 4 y2 2 4 y y (1) 1 y 2 1 1 1 1 Solutions: 1, , 1, , 1, , 1, 2 2 2 2
2 2
2 : 2 2
2 2 2 4 3 y 4 3 y 2 2 2 6 y 3 3 2 If x : 2
y2
2
2 2 2 4 3 y 4 3 y 2 2 2 6 y 3 3 Solutions: 2 6 2 6 2 6 2 , 3 , 2 , 3 , 2 , 3 , y2
3 2 44. 2 2 1 0 y x 6 7 20 x 2 y 2
2 6 , 3 2
3 2 x 2 y 2 1 6 7 2 x 2 y 2
Multiply each side of the first equation by –3 and add the equations to eliminate x:
1417
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Chapter 12: Systems of Equations and Inequalities
6 2
y 6
x2
9 2
y 7
y2 2 y2
46. Add the equations to eliminate y: 1 1 1 x4 y 4 1 1 4 x4 y4
3 2 1
2 x4
2
y 2
x4
y 2 If y 2 :
2 x
2
3
2
2
1
2 x
2
1 2
2
3
2 : 5
1 1 2 2 2 x2 x 2
2 5
x4 If x 4
x 2 4 x 2 If y 2 :
5
1 2 4 5
2
y4
2
x 4 x 2 Solutions:
2, 2 , 2, 2 , 2, 2 , 2, 2
6 1 45. 4 4 6 y x 2 2 19 x 4 y 4
If x 4
Multiply each side of the first equation by –2 and add the equations to eliminate x: 2 12 12 x4 y4 2 2 19 x4 y4 14 4 7 y y 4 2 There are no real solutions. The system is inconsistent.
4
1 y
4
4
1 y
4
3 2
2 2 y 4 3 3
2 : 5
y4
2 5
1 2 4 5
4
1 y4
4
1 y4
3 2
2 2 y 4 3 3
Solutions: 2 2 2 2 2 2 4 , 4 , 4 , 4 , 4 , 4 , 5 3 5 3 5 3 2 2 4 , 4 3 5 47. x 2 3 xy 2 y 2 0 x 2 xy 6
Subtract the second equation from the first to eliminate the x 2 term. 4 xy 2 y 2 6 2 xy y 2 3 Since y 0 , we can solve for x in this equation to get y2 3 x , y0 2y Now substitute for x in the second equation and solve for y.
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Section 12.6: Systems of Nonlinear Equations
x 2 xy 6
49. y 2 y x 2 x 2 0 x2 y 1 0 y
2
y2 3 y2 3 y6 2y 2y
Multiply each side of the second equation by –y and add the equations to eliminate y: y 2 y x2 x 2 0
y4 6 y2 9 y2 3 6 2 4 y2 y 4 6 y 2 9 2 y 4 6 y 2 24 y 2 4
y2 y
2
3 y 12 y 9 0 4
x2 2 x 0
2
y 4y 3 0
x x 2 0
y 3 y 1 0 2
2
x 0 or x 2 If x 0 :
Thus, y 3 or y 1 . If y 1: x 2 1 2 If y 1: x 2(1) 2 If y 3 :
x 3
If y 3 :
x 3
y 2 y 02 0 2 0 y 2 y 2 0 ( y 2)( y 1) 0 y 2 or y 1 If x 2 :
Solutions: (2, 1), (–2, 1),
y 2 y 22 2 2 0 y 2 y 0 y ( y 1) 0 y 0 or y 1 Note: y 0 because of division by zero. Solutions: (0, –2), (0, 1), (2, –1)
3, 3 , 3, 3
2 2 x xy 2 y 0 48. xy x 6 0 Factor the first equation, solve for x, substitute into the second equation and solve: x 2 xy 2 y 2 0 ( x 2 y )( x y ) 0 x 2 y or x y Substitute x 2 y and solve: xy x 6 0 (2 y ) y 2 y 6
50. x3 2 x 2 y 2 3 y 4 0 y2 y x2 0 x2 Multiply each side of the second equation by x 2 and add the equations to eliminate x: x3 2 x 2 y 2 3 y 4 0 x3 2 x 2 y 2 y
2( y 2 y 3) 0
1 12 4(1)(3) (No real solution) 2(1)
If y 1: x3 2 x 2 12 3 1 4 0 x3 2 x 2 0 x 2 ( x 2) 0 x 0 or x 2 Note: x 0 because of division by zero. Solution: (2, 1)
Substitute x y and solve: xy x 6 0 y y ( y ) 6 y2 y 6 0
51. Rewrite each equation in exponential form: log x y 3 y x3 5 log x (4 y ) 5 4 y x
( y 3)( y 2) 0 y 3 or y =2 If y 3 :
x3
If y 2 :
x 2
0
4y 4 0 4y 4 y 1
2 y2 2 y 6 0
y
x20
Substitute the first equation into the second and solve:
Solutions: (3, –3), (– 2, 2) 1419
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Chapter 12: Systems of Equations and Inequalities 54. Rewrite each equation in exponential form:
4 x3 x5
ln x 5ln y
x5 4 x3 0
log 2 x 3 2 log 2 y
x3 ( x 2 4) 0
If x 2 : y 23 8 Solution: (2, 8)
52. Rewrite each equation in exponential form: 3 log x (2 y ) 3 2 y x 2 log x (4 y ) 2 4 y x Multiply the first equation by 2 then substitute the first equation into the second and solve: 2 x3 x 2 2
2x x 0 x (2 x 1) 0 1 1 x or x 0 2 2 The base of a logarithm must be positive, thus x 0. x 2 0 or x
2
1 1 1 4y y 4 16 2 1 1 Solution: , 2 16 1 : 2
x y 5 So we have the system 2 x 8 y Therefore we have 8 y 2 y5
8 y 2 y5 0
y 0 or 8 y 3 0 y 2
Since ln y is undefined when y 0 , the only solution is y 2 . If y 2 :
x y 5 x 25 32
x2 x y 2 3 y 2 0 55. y2 y x 1 0 x x 1 2 y 3 2 1 2 2 2 2 2 x 12 y 12 12
53. Rewrite each equation in exponential form: 4
ln x 4 ln y x e4ln y eln y y 4 log3 x 2 2 log 3 y x 32 2log3 y 32 32log3 y 32 3log3 y 9 y 2 2
x y 4 So we have the system 2 x 9 y Therefore we have : 9 y 2 y 4 9 y 2 y 4 0 y 2 (9 y 2 ) 0 y 2 (3 y )(3 y ) 0 y 0 or y 3 or y 3
Since ln y is undefined when y 0 , the only solution is y 3 . If y 3 :
y 2 8 y3 0
Solution: 32, 2
2
If x
x 23 2log 2 y 23 22log 2 y 23 2log 2 y 8 y 2 2
x3 0 or x 2 4 x 0 or x 2 The base of a logarithm must be positive, thus x 0 and x 2 .
3
5
x e5ln y eln y y 5
x y 4 x 34 81
Solution: 81, 3
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Section 12.6: Systems of Nonlinear Equations 3.1
56. y 2 y x 2 x 2 0 x2 y 1 0 y
–4.7
x 1 2 y 1 2 5 2 2 2 2 9 1 x y 2 4
4.7
–3.1 Solution: x 1.65, y 0.89 or (–1.65, –0.89)
60. Graph: y1 2 x3 ; y2 2 x3 ;
y3 4 / x 2 Use INTERSECT to solve: 3.1
–4.7
57. Graph: y1 x (2 / 3); y2 e ( x) Use INTERSECT to solve: 3.1
4.7
–3.1 Solution: x 1.37, y 2.14 or (–1.37, 2.14)
61. Graph: y1 4 12 x 4 ; y2 4 12 x 4 ; –4.7
y3 2 / x ; y4 2 / x Use INTERSECT to solve:
4.7
-3.1 Solution: x 0.48, y 0.62 or (0.48, 0.62)
58. Graph: y1 x (3 / 2); y2 e ( x) Use INTERSECT to solve: 3.1
–4.7
4.7
–3.1 Solution: x 0.65, y 0.52 or (0.65, 0.52)
Solutions: x 0.58, y 1.86; x 1.81, y 1.05; x 1.81, y 1.05; ; x 0.58, y 1.86 or (0.58, 1.86), (1.81, 1.05), (1.81, –1.05), (0.58, –1.86)
59. Graph: y1 3 2 x 2 ; y2 4 / x3 Use INTERSECT to solve:
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Chapter 12: Systems of Equations and Inequalities
62. Graph: y1 4 6 x 4 ; y2 4 6 x 4 ; y3 1/ x Use INTERSECT to solve:
Solution: x 1.90, y 0.64; x 0.14, y 2.00 or (1.90, 0.64), (0.14, –2.00) 65. Solve the first equation for x, substitute into the second equation and solve: x 2y 0 x 2y 2 2 ( x 1) ( y 1) 5 ( 2 y 1) 2 ( y 1) 2 5
Solutions: x 0.64, y 1.55; x 1.55, y 0.64; x 0.64, y 1.55; ; x 1.55, y 0.64 or (0.64, 1.55), (1.55, 0.64), (–0.64, –1.55), (–1.55, –0.64)
4 y2 4 y 1 y2 2 y 1 5 5 y2 2 y 3 0 (5 y 3)( y 1) 0 3 y =0.6 or y 1 5 6 x = 1.2 or x 2 5 6 3 The points of intersection are , , (2, –1) . 5 5
63. Graph: y1 2 / x; y2 ln x Use INTERSECT to solve: 3.1
–4.7
4.7
–3.1 Solution: x 2.35, y 0.85 or (2.35, 0.85) 64. Graph: y1 4 x 2 ; y2 4 x 2 ;
y3 ln x Use INTERSECT to solve:
66. Solve the first equation for x, substitute into the second equation and solve: x 2y 6 0 x 2y 6 2 2 ( x 1) ( y 1) 5 ( 2 y 6 1)2 ( y 1) 2 5 4 y 2 20 y 25 y 2 2 y 1 5 5 y 2 22 y 21 0 (5 y 7)( y 3) 0 7 or y 3 5 16 x or x 0 5 The points of intersection are y
1422 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations
16 7 , , (0, 3) . 5 5
( x 2) 2 x 6 4
x2 4 x 4 x 6 4 x2 5x 6 0 ( x 2)( x 3) 0 x 2 or x 3 If x 2 : ( y 1)2 2 6 y 1 2 y 1 or y 3 2
If x 3 : ( y 1) 3 6 y 1 3 y 1 3 The points of intersection are:
3, 1 3 , 3, 1 3 , ( 2, 1), ( 2, 3) .
67. Complete the square on the second equation. y2 4 y 4 x 1 4 ( y 2)2 x 3 Substitute this result into the first equation. ( x 1) 2 x 3 4 x2 2 x 1 x 3 4 x2 x 0 x( x 1) 0 x 0 or x 1 If x 0 :
( y 2) 2 0 3
69. Solve the first equation for x, substitute into the second equation and solve: 4 y 3 x x2 6x y2 1 0
y 2 3 y 2 3 If x 1: ( y 2) 2 1 3 y 2 2 y 2 2 The points of intersection are:
0, 2 3 , 0, 2 3 , 1, – 4 , 1, 0 .
4 x3 4 x 3 y 4 x 3 y y
2
4 4 2 3 6 3 y 1 0 y y 16 24 24 9 18 y 2 1 0 2 y y y
68. Complete the square on the second equation, substitute into the first equation and solve: ( x 2) 2 ( y 1) 2 4 2 y 2 y x 5 0 y2 2 y 1 x 5 1
16 y2
( y 1) 2 x 6 1423
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y2 8 0
Chapter 12: Systems of Equations and Inequalities
The points of intersection are: (0, 2), (–4, –2).
16 y 4 8 y 2 0 y 4 8 y 2 16 0 ( y 2 4) 2 0 y2 4 0 y2 4 y 2 4 If y 2 : x 3 5 2 4 If y 2 : x 3 1 2 The points of intersection are: (1, –2), (5, 2).
71. Let x and y be the two numbers. The system of equations is: x y2 x y 2 2 2 x y 10 Solve the first equation for x, substitute into the second equation and solve:
y 2 2 y 2 10 y 2 4 y 4 y 2 10
70. Substitute the first equation into the second equation and solve: 4 y x2 x2 4 x y 2 4 0 2
4 x2 4x 4 0 x2 4 x2 4 x 4 x2
2
x 4 x 4 x 4 x 4 16
( x 2) 2 x 2 4 x 4 16
2
2
4
3
3
72. Let x and y be the two numbers. The system of equations is: x y 7 x 7 y 2 2 x y 21 Solve the first equation for x, substitute into the second equation and solve:
14 y 28
x 2 x 2 8 x 16 0 2
If y 3 : x 3 2 1 If y 1: x 1 2 3 The two numbers are 1 and 3 or –1 and –3.
49 14 y y 2 y 2 21
2
x 8 x 16 x 0
y 3 y 1 0 y 3 or y 1
7 y 2 y 2 21
2
x 8 x 16 x 16 16 4
y2 2 y 3 0
y 2 x 72 5 The two numbers are 2 and 5.
2
x ( x 4) 0 x 0 or x 4 y2
y 2
1424 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations
73. Let x and y be the two numbers. The system of equations is: 4 xy 4 x y x2 y 2 8 Solve the first equation for x, substitute into the second equation and solve:
75. Let x and y be the two numbers. The system of equations is: x y xy 1 1 x y 5 Solve the first equation for x, substitute into the second equation and solve: x xy y y x 1 y y x 1 y
2
4 2 y 8 y 16 y2 8 y2
1
16 y 4 8 y 2
1 y 1 5 y y 2 y 5 y 2 y 5y
y 4 8 y 2 16 0
y 4 0 2
2
y2 4 y 2
6y 2
4 4 2; If y 2 : x 2 2 2 The two numbers are 2 and 2 or –2 and –2. If y 2 : x
y
2
1
100 y 4 21 y 2
y 4 y 25 0 or
1
1
y y 3 y 1
y 4 21y 2 100 0
y2 4
1
1 x 3 3 1 2 1 3 3 2
76. Let x and y be the two numbers. The system of equations is: x y xy 1 1 x y 3 Solve the first equation for x, substitute into the second equation and solve: xy x y y x y 1 y x y 1
10 2 y 21 y 100 y 2 21 2 y
2
1 3
The two numbers are 1 and 1 . 2 3
74. Let x and y be the two numbers. The system of equations is: 10 xy 10 x y x 2 y 2 21 Solve the first equation for x, substitute into the second equation and solve:
2
1
y y 5 1 y
y 1 1 3 y y y2 3 y y 2 3y
y 2 25 (no real solution)
y 2 x 10 5 2 10 If y 2 : x 5 2 The two numbers are 2 and 5 or –2 and –5. If y 2 :
2 y 2 1 1 1 1 2 The two numbers are 1 and 1 . 2 y 1 x
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Chapter 12: Systems of Equations and Inequalities
a 2 77. b 3 a b 10 a 10 b Solve the second equation for a , substitute into the first equation and solve: 10 b 2 b 3 3(10 b) 2b 30 3b 2b 30 5b b6a4 a b 10; b a 2
The ratio of a b to b a is 10 5 . 2 a 4 78. b 3 a b 14 a 14 b Solve the second equation for a , substitute into the first equation and solve: 14 b 4 b 3 3 14 b 4b
80. Let 2x = the side of the first square. Let 3x = the side of the second square.
2 x 2 3x 2 52 4 x 2 9 x 2 52 13 x 2 52 x2 4 x 2 Note that we must have x 0 . The sides of the first square are (2)(2) = 4 feet and the sides of the second square are (3)(2) = 6 feet.
81. Let x = the radius of the first circle. Let y = the radius of the second circle. 2 x 2 y 12 2 2 x y 20 Solve the first equation for y, substitute into the second equation and solve: 2 x 2 y 12 x y 6 y 6 x
42 3b 4b
x 2 y 2 20
42 7b
x 2 y 2 20
b6 a8 a b 2; a b 14
x 2 (6 x) 2 20
The ratio of a b to a b is 2 1 . 14 7 79. Let x = the width of the rectangle. Let y = the length of the rectangle. 2 x 2 y 16 xy 15 Solve the first equation for y, substitute into the second equation and solve. 2 x 2 y 16 2 y 16 2 x y 8 x x 8 x 15
x 2 36 12 x x 2 20 2 x 2 12 x 16 0 x 2 6 x 8 0 ( x 4)( x 2) 0 x 4 or x 2 y2 y4 The radii of the circles are 2 centimeters and 4 centimeters.
82. Let x = the length of each of the two equal sides in the isosceles triangle. Let y = the length of the base. The perimeter of the triangle: x x y 18 Since the altitude to the base y is 3, the Pythagorean theorem produces another equation. 2
8 x x 2 15 x 2 8 x 15 0
x 5 x 3 0 x 5 or x 3 The dimensions of the rectangle are 3 inches by 5 inches.
y2 y 2 2 9 x2 3 x 2 4 Solve the system of equations: 2 x y 18 y 18 2 x 2 y 2 9 x 4
1426 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations 84. Let v1 , v2 , v3 the speeds of runners 1, 2, 3. Let t1 , t2 , t3 the times of runners 1, 2, 3. Then by the conditions of the problem, we have the following system: 5280 v1 t1 5270 v2 t1 5260 v3 t1 5280 v2 t2 Distance between the second runner and the third runner after t2 seconds is:
Solve the first equation for y, substitute into the second equation and solve.
18 2 x 2
9 x2 4 324 72 x 4 x 2 9 x2 4 81 18 x x 2 9 x 2 18 x 90 x 5 y 18 2 5 8
The base of the triangle is 8 centimeters.
v t 5280 v3 t2 5280 v3 t1 2 2 v2 t1 5280 5280 5260 5270 10.02 The second place runner beats the third place runner by about 10.02 feet.
83. The tortoise takes 9 + 3 = 12 minutes or 0.2 hour longer to complete the race than the hare. Let r = the rate of the hare. Let t = the time for the hare to complete the race. Then t + 0.2 = the time for the tortoise and r 0.5 = the rate for the tortoise. Since the length of the race is 21 meters, the distance equations are: 21 r t 21 r t r 0.5 t 0.2 21
85. Let x = the width of the cardboard. Let y = the length of the cardboard. The width of the box will be x 4 , the length of the box will be y 4 , and the height is 2. The volume is V ( x 4)( y 4)(2) . Solve the system of equations: 216 y xy 216 x 2( x 4)( y 4) 224 Solve the first equation for y, substitute into the second equation and solve. 216 2 x 8 4 224 x 1728 32 224 432 8 x x 432 x 8 x 2 1728 32 x 224 x
Solve the first equation for r, substitute into the second equation and solve: 21 0.5 t 0.2 21 t 4.2 21 0.5t 0.1 21 t 4.2 10t 21 0.5t 0.1 10t 21 t 210t 42 5t 2 t 210t 5t 2 t 42 0
5t 14 t 3 0 5t 14 0 5t 14
or t 3 0 t 3
8 x 2 240 x 1728 0
14 2.8 5 t 3 makes no sense, since time cannot be negative. Solve for r: 21 r 7.5 2.8 The average speed of the hare is 7.5 meters per hour, and the average speed for the tortoise is 7 meters per hour.
x 2 30 x 216 0
t
x 12 x 18 0 x 12 0 or x 18 0 x 12 x 18 The cardboard should be 12 centimeters by 18 centimeters.
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Chapter 12: Systems of Equations and Inequalities 86. Let x = the width of the cardboard. Let y = the length of the cardboard. The area of the cardboard is: xy 216
The volume of the tube is: V r 2 h 224 x where h y and 2r x or r . 2 Solve the system of equations: 216 xy 216 y x 2 2 x y 224 x y 224 2 4 Solve the first equation for y, substitute into the second equation and solve. x 2 216 x 224 4 216 x 896 896 x 13.03 216
216 216 216 y 16.57 896 896 x 2
216
The cardboard should be about 13.03 centimeters by 16.57 centimeters. 87. Find equations relating area and perimeter: 2 2 x y 4500 3x 3 y ( x y ) 300 Solve the second equation for y, substitute into the first equation and solve: 4 x 2 y 300 2 y 300 4 x y 150 2 x x 2 (150 2 x) 2 4500 x 2 22,500 600 x 4 x 2 4500 5 x 2 600 x 18, 000 0
x 2 120 x 3600 0 ( x 60) 2 0
x 60 0 x 60 y 150 2(60) 30 The sides of the squares are 30 feet and 60 feet.
88. Let x = the length of a side of the square. Let r = the radius of the circle. The area of the square is x 2 and the area of the circle is r 2 . The perimeter of the square is 4 x and the circumference of the circle is 2 r . Find equations relating area and perimeter: x 2 r 2 100 4 x 2 r 60 Solve the second equation for x, substitute into the first equation and solve: 4 x 2 r 60 4 x 60 2 r 1 x 15 r 2 2
1 2 15 r r 100 2 1 225 15 r 2 r 2 r 2 100 4 1 2 2 r 15 r 125 0 4 1 b 2 4ac (15) 2 4 2 (125) 4 1 2252 500 2 4 1002 500 0 Since the discriminant is less than zero, it is impossible to cut the wire into two pieces whose total area equals 100 square feet.
89. Solve the equation: m 2 4(2m 4) 0 m 2 8m 16 0
m 4 2 0 m4 Use the point-slope equation with slope 4 and the point (2, 4) to obtain the equation of the tangent line: y 4 4( x 2) y 4 4 x 8 y 4 x 4
90. Solve the system: 2 2 x y 10 y mx b Solve the system by substitution:
1428 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations b 3 m 3 2 1 The equation of the tangent line is y 2 x 1 .
x 2 mx b 10 2
x 2 m 2 x 2 2bmx b 2 10 0
1 m x 2bmx b 10 0 2
2
92. Solve the system: x 2 y 5 y mx b Solve the system by substitution: x 2 mx b 5 x 2 mx b 5 0 Note that the tangent line passes through (–2, 1). Find the relation between m and b: 1 m 2 b b 2m 1
2
Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 m(1) b b 3 m There is one solution to the quadratic if the discriminant is zero.
2bm 2 4 m2 1 b 2 10 0
4b 2 m 2 4b 2 m 2 40m 2 4b 2 40 0
Substitute into the quadratic to eliminate b: x 2 mx 2m 1 5 0
40m 2 4b 2 40 0 Substitute for b and solve:
x 2 mx 2m 4 0
40m 2 4 3 m 40 0 2
Find when the discriminant equals 0:
40m 2 4m 2 24m 36 40 0
m 2 4 1 2m 4 0
36m 2 24m 4 0
m 2 8m 16 0
2
9m 6m 1 0
m 4 2 0
3m 1 0 2
m4 0
3m 1
m4 b 2m 1 2 4 1 9
1 m 3
The equation of the tangent line is y 4 x 9 .
1 10 b 3 m 3 3 3 The equation of the tangent line is 1 10 y x . 3 3
93. Solve the system: 2 x 2 3 y 2 14 y mx b Solve the system by substitution: 2 x 2 3 mx b 14 2
91. Solve the system: 2 y x 2 y mx b Solve the system by substitution: x 2 2 mx b x 2 mx 2 b 0 Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 m(1) b b 3 m Substitute into the quadratic to eliminate b: x 2 mx 2 (3 m) 0 x 2 mx (m 1) 0 Find when the discriminant equals 0:
2 x 2 3m 2 x 2 6mbx 3b 2 14
3m 2 x 6mbx 3b 14 0 2
2
2
Note that the tangent line passes through (1, 2). Find the relation between m and b: 2 m(1) b b 2 m Substitute into the quadratic to eliminate b: (3m 2 2) x 2 6m(2 m) x 3(2 m) 2 14 0 (3m 2 2) x 2 (12m 6m 2 ) x (3m 2 12m 2) 0 Find when the discriminant equals 0:
m 2 4 1 m 1 0 m 2 4m 4 0
m 2 2 0 m2 0 m2
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Chapter 12: Systems of Equations and Inequalities
(12m 6m 2 ) 2 4(3m 2 2)(3m 2 12m 2) 0 144m 2 96m 16 0 9m 2 6 m 1 0 (3m 1) 2 0 3m 1 0 1 m 3 1 7 b 2m 2 3 3
1 7 The equation of the tangent line is y x . 3 3 94. Solve the system: 3 x 2 y 2 7 y mx b Solve the system by substitution: 3x mx b 7 2
2
95. Solve the system: 2 2 x y 3 y mx b Solve the system by substitution: x 2 mx b 3 2
x 2 m 2 x 2 2mbx b 2 3
1 m x 2mbx b 3 0 2
2
2
Note that the tangent line passes through (2, 1). Find the relation between m and b: 1 m(2) b b 1 2m Substitute into the quadratic to eliminate b: (1 m 2 ) x 2 2m(1 2m) x (1 2m) 2 3 0 (1 m 2 ) x 2 ( 2m 4m 2 ) x 1 4m 4m 2 3 0 (1 m 2 ) x 2 ( 2m 4m 2 ) x ( 4m 2 4m 4) 0 Find when the discriminant equals 0:
2m 4m 4 1 m 4m 4m 4 0 2 2
2
4m 2 16m3 16m 4 16m 4 16m3 16m 16 0
3 x 2 m 2 x 2 2mbx b 2 7
4m 2 16m 16 0
m 3 x 2mbx b 7 0 2
2
2
Note that the tangent line passes through (–1, 2). Find the relation between m and b: 2 m(1) b b m 2 There is one solution to the quadratic if the discriminant equals 0.
2bm 2 4 m2 3 b2 7 0
4b 2 m 2 4b 2 m 2 28m 2 12b 2 84 0 28m 2 12b 2 84 0 7 m 2 3b 2 21 0 Substitute for b and solve: 7m 2 3 m 2 21 0
m 2 4m 4 0
m 2 2 0 m2 The equation of the tangent line is y 2 x 3 .
96. Solve the system: 2 2 2 y x 14 y mx b Solve the system by substitution: 2 mx b x 2 14 2
2m 2 x 2 4mbx 2b 2 x 2 14
2
2m 1 x 4mbx 2b 14 0 2
7 m 2 3m 2 12m 12 21 0
2
2
Note that the tangent line passes through (2, 3). Find the relation between m and b: 3 m(2) b b 3 2m There is one solution to the quadratic if the discriminant equals 0.
4m 2 12m 9 0
2m 3 2 0 2m 3 3 m 2 b m2
2
4bm 2 4 2m2 1 2b 2 14 0
3 7 2 2 2
16b 2 m 2 16b 2 m 2 112m 2 8b 2 56 0
3 7 The equation of the tangent line is y x . 2 2
112m 2 8b 2 56 0 14m 2 b 2 7 0 Substitute for b and solve:
1430 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations 98.
14m 2 3 2m 7 0 2
14m 2 4m 2 12m 9 7 0
x 2 y 2 a 2 b2 2 2 2 2 2 2 2 2 2 2 b x a y a b (1) a b a b x y a b bx ay a b (2) a b ab
2
18m 12m 2 0 2 3m 1 0 2
3m 1 1 m 3 1 7 b 3 2m 3 2 3 3
Square equation (2) and subtract equation (1). b 2 x 2 2abxy a 2 y 2 a 2 2ab b 2 b2 x2
The equation of the tangent line is y
So, xy 1 y Substitute
b r1 r2 a rr c 1 2 a Substitute and solve:
b2
2ab
1 x
1 into equation (2). x
1 ab x bx 2 a (a b) x bx a
bx 2 (a b) x a 0
b r1 r2 a b c r2 r2 a a b c r2 2 r2 0 a a 2 ar2 br2 c 0
Using the quadratic formula we have: x
b b 2 4ac r2 2a b r1 r2 a b b 2 4ac b a 2a
(a b) (a b) 2 4ba 2b
a b a 2 2ab b 2 4ba 2b 2 a b a 2 2ab b 2 a b (a b) 2b 2b a b a b
2b If a > b, then a b a b and x
b b 2 4ac 2b 2a 2a
b b 2 4ac 2a The solutions are:
a b ( a b) a b a b abab or 2b 2b 2b 2a a 2b = or 1 2b b 2b
If a < b, then a b a b and
r1
2abxy
1 7 x . 3 3
97. Solve for r1 and r2 :
a2 y2 a2
a b (b a) 2b a 1 or b x
b b 2 4ac b b 2 4ac and r2 2a 2a
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Chapter 12: Systems of Equations and Inequalities
a 1 If x 1 , then y 1 . If x , then b 1 1 b a b y . So we have, (1,1), , . . a a b a b Because we squared equation (2) at the start, we need to check. 12 12 1 1 Check (1,1) : 2 2 2 2 a b a b 2 2 b a a 2 b2 2 2 2 2 2 2 a b a b a b 1 1 b a ab a b ab ab ab So (1,1) checks.
2 2 a b Check , : b 2 a 2 b 2 a2 b a b b a a a 2
a b
1 b
2
b 2
1
a
2
a2
b2
a 2 b2 a 2b2
b a
1 1 ab a b b a ab a b So , checks. b a 2l 2 w P 99. lw A Solve the first equation for l , substitute into the second equation and solve. 2l P 2 w l
P w 2
w
P2 4 A 4
2 P 2
If w
P P 2 16 A then 4
P P P 2 16 A P P 2 16 A 2 4 4 Assuming l w , the solution is: l
P P 2 16 A P P 2 16 A and l 4 4 Note: To show that the solutions are real, we can verify P 2 16 A 0 as follows: P 2 16 A (2l 2w) 2 16lw w
4l 2 8lw 4 w2 16lw 4l 2 8lw 4 w2 4(l w) 2 0
100. Solve the system for l and b : P b 2l b P 2l 2 b2 l2 h 4 Solve the first equation for b , substitute into the second equation and solve. 4h 2 b 2 4l 2 4h 2 P 2 4 Pl 4l 2 4l 2 4h 2 P 2 4 Pl l b P
P 16 A
2
P P P 2 16 A P P 2 16 A 2 4 4
2
P 2
P2 16 A 4 4
2
4h 2 P 2 4P
4h 2 P 2 P 2 4h 2 2P 2P
101. Consider the circle with equation
x h 2 y k 2 r 2 and the third degree
2
2
l
P P 2 16 A then 4
4h 2 P 2l 4l 2
P w w A 2 P w w2 A 2 P w2 w A 0 2 P 2
If w
P P 2 16 A 4
polynomial with equation y ax3 bx 2 cx d . Substituting the second equation into the first equation yields
x h 2 ax3 bx 2 cx d k r 2 . 2
In order to find the roots for this equation we can expand the terms on the left hand side of the 1432 Copyright © 2025 Pearson Education, Inc.
Section 12.6: Systems of Nonlinear Equations
2
3
2
polynomial, and ax bx cx d k
In order to get a volume equal to 9 cubic feet,
a.
equation. Notice that x h yields a 2nd degree
we solve 10 2 x x 9. 2
yields a 2
10 2 x 2 x 9
th
6 degree polynomial. Therefore, we need to find the roots of a 6th degree equation, and the Fundamental Theorem of Algebra states that there will be at most six real roots. Thus, the circle and the 3rd degree polynomial will intersect at most six times. Now consider the circle with equation
100 40 x 4 x x 9 2
100 x 40 x 2 4 x3 9 So we need to solve the equation 4 x3 40 x 2 100 x 9 0 .
x h 2 y k 2 r 2 and the polynomial of
Graphing y1 4 x3 40 x 2 100 x 9 on a calculator yields the graph
degree n with equation y a0 a1 x a2 x 2 a3 x3 ... an x n . Substituting the first equation into the first equation yields
80
x h 2 a0 a1 x a2 x 2 a3 x3 ... an x n k r 2 2
2
In order to find the roots for this equation we can expand the terms on the left hand side of the equation.
40
80
Notice that x h yields a 2 degree polynomial, 2
nd
and a0 a1 x a2 x 2 a3 x3 ... an x n k
2
2
yields a polynomial of degree 2n. Therefore, we need to find the roots of an equation of degree 2n, and the Fundamental Theorem of Algebra states that there will be at most 2n real roots. Thus, the circle and the nth degree polynomial will intersect at most 2n times.
40 80
2
102. Since the area of the square piece of sheet metal is 100 square feet, the sheet’s dimensions are 10 feet by 10 feet. Let x the length of the cut.
40
x
10 – 2x 10 – 2x
The graph indicates that there are three real zeros on the interval [0, 6]. Using the ZERO feature of a graphing calculator, we find that the three roots shown occur at x 0.093 , x 4.274 and x 5.632 . But we’ve already noted that we must have 0<x <5 , so the only practical values for the sides of the square base are x 0.093 feet and x 4.274 feet.
x x
10
10
b. Answers will vary.
The dimensions of the box are: length 10 2 x; width 10 2 x; height x . Note that each of these expressions must be positive. So we must have x 0 and 10 2 x 0 x 5, that is, 0 x 5 . So the volume of the box is given by V length width height
103. 7 x 2 6 x 8 0
10 2 x 10 2 x x
x
2 b b 2 4ac 6 6 4(7)( 8) 2a 2(7)
6 260 6 2 65 3 65 14 14 7
3 65 3 65 , The solution set is 7 7
10 2 x x 2
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Chapter 12: Systems of Equations and Inequalities
104.
y y1 m( x x1 )
y-intercept: f (0) 21 x-intercepts: x 2 4 x 21 0 ( x 3)( x 7) 0 x 3, x 7
2 y ( 7) ( x 10) 5 2 2 y7 x4 y x3 5 5
The vertex is at x
24 and cos 0 , so lies in quadrant 7 III. Using the Pythagorean Identities: csc 2 1 cot 2
105. cot
f 2 25 , the vertex is 2, 25 .
The graph is below the x-axis when 3 x 7 . Since the inequality is strict, the solution set is x 3 x 7 or, using interval notation,
2
625 24 csc 2 1 7 49
3, 7 .
625 25 49 7 Note that csc must be negative because lies in 25 quadrant III. Thus, csc . 7 1 1 7 sin csc 257 25 csc
109. Reflecting about the x-axis would be f ( x) 25 x 2 . Shifting 4 units to the right
would give f ( x) 25 ( x 4) 2 . 110.
f ( x 3) 2( x 3) 2 8( x 3) 7 2( x 2 6 x 9) 8 x 24 7
cos , so cot sin cos cot sin
2 x 2 12 x 18 8 x 31 2 x 2 20 x 49
24 7 24 7 25 25
111.
1 1 7 24 tan cot 24 7 1 1 25 sec cos 24 24 25
106. The hotel is 1200 feet above the lake. The hypotenuse of the triangle is 4420. So opp 1200 sin hypo 4420
f ( x)
3x x 8
3( x h) 3x f ( x h) f ( x ) x h 8 x 8 h h 3( x h)( x 8) 3 x x 8 h x h 8 x 8 h
1200 15.8 4420
107.
3 x 24 x 3hx 24 h 3 x 24 x 3 xh 2
2
x h 8 x 8
sin 1
The inclination of the trail is 15.8 .
1 24h x h 8 x 8 h 24 x h 8 x 8
( x h) 2 ( y k ) 2 r 2 ( x (3)) 2 ( y 4) 2 102 ( x 3) 2 ( y 4) 2 100
108.
b (4) 2 . Since 2a 2(1)
x 2 4 x 21 x 2 4 x 21 0 We graph the function f ( x) x 2 4 x 21 . The intercepts are
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h
Section 12.7: Systems of Inequalities
112.
f ( x)
(2 x 5)9 3 3x 9(2 x 5)8 2 (2 x 5)9
2
3. x 2 y 2 9 The graph is a circle. Center: (0, 0) ; Radius: 3
3(2 x 5)8 (2 x 5) 18 x
(2 x 5)18 3(5 16 x) 15 48 x 48 x 15 (2 x 5)10 (2 x 5)10 (2 x 5)10
4. y x 2 4 The graph is a parabola. x-intercepts: 0 x 2 4
Section 12.7 1. 3x 4 8 x 4x 4 x 1 x x 1 or ,1
x 2 4, no x intercepts y-intercept: y 02 4 4 The vertex has x-coordinate: b 0 x 0. 2a 2 1
2. 3x 2 y 6 The graph is a line. x-intercept: 3x 2 0 6
The y-coordinate of the vertex is y 02 4 4 .
3x 6 x2 y-intercept: 3 0 2 y 6
2 y 6 y 3
5. True 6.
x2 4 5 x2 9 0 ( x 3)( x 3) 0
The x-intercepts are 3 and -3. The graph is below the x-axis when 3 x 3 . Since the inequality includes the equal sign, the solution set is x 3 x 3 or, using interval notation, 3, 3 . 7. dashes; solid 8. half-planes 1435 Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities 9. False, see example 7b. 10. b 11. x 0 Graph the line x 0 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (2, 0). Since 2 ≥ 0 is true, shade the side of the line containing (2, 0).
12. y 0 Graph the line y 0 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2 ≥ 0 is true, shade the side of the line containing (0, 2).
13. x 4 Graph the line x 4 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (5, 0). Since 5 ≥ 0 is true, shade the side of the line containing (5, 0).
14. y 2 Graph the line y 2 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (5, 0). Since 0 ≤ 2 is true, shade the side of the line containing (5, 0).
15. 2 x y 6 Graph the line 2 x y 6 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6 is false, shade the opposite side of the line from (0, 0).
16. 3x 2 y 6 Graph the line 3x 2 y 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6 is true, shade the side of the line containing (0, 0).
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Section 12.7: Systems of Inequalities
17. x 2 y 2 1
20. y x 2 2
Graph the circle x 2 y 2 1 . Use a dashed line since the inequality uses >. Choose a test point not on the circle, such as (0, 0). Since 02 02 1 is false, shade the opposite side of the circle from (0, 0).
18. x 2 y 2 9 2
2
Graph the circle x y 9 . Use a solid line since the inequality uses . Choose a test point not on the circle, such as (0, 0). Since 02 02 9 is true, shade the same side of the circle as (0, 0).
2
19. y x 1
Graph the parabola y x 2 1 . Use a solid line since the inequality uses . Choose a test point not on the parabola, such as (0, 0). Since 0 02 1 is false, shade the opposite side of the parabola from (0, 0).
Graph the parabola y x 2 2 . Use a dashed line since the inequality uses >. Choose a test point not on the parabola, such as (0, 0). Since 0 02 2 is false, shade the opposite side of the parabola from (0, 0).
21. xy 4 Graph the hyperbola xy 4 . Use a solid line since the inequality uses . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 4 is false, shade the opposite side of the hyperbola from (0, 0).
22. xy 1 Graph the hyperbola xy 1 . Use a solid line since the inequality uses . Choose a test point not on the hyperbola, such as (0, 0). Since 0 0 1 is true, shade the same side of the hyperbola as (0, 0).
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Chapter 12: Systems of Equations and Inequalities
x y 2 23. 2 x y 4 Graph the line x y 2 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 2 is true, shade the side of the line containing (0, 0). Graph the line 2 x y 4 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
3x y 6 24. x 2y 2 Graph the line 3 x y 6 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6 is false, shade the opposite side of the line from (0, 0). Graph the line x 2 y 2 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2 is true, shade the side of the line containing (0, 0). The overlapping region is the solution.
2x y 4 25. 3 x 2 y 6 Graph the line 2 x y 4 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4 is true, shade the side of the line containing (0, 0). Graph
the line 3x 2 y 6 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6 is true, shade the side of the line containing (0, 0). The overlapping region is the solution.
4x 5 y 0 26. 2 x y 2 Graph the line 4 x 5 y 0 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (2, 0). Since 4(2) 5(0) 0 is false, shade the opposite side of the line from (2, 0). Graph the line 2 x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
2 x 3 y 0 27. 3x 2 y 6 Graph the line 2 x 3 y 0 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 3). Since 2(0) 3(3) 0 is true, shade the side of the line containing (0, 3). Graph the line 3x 2 y 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 2(0) 6 is true, shade the side of the line containing (0, 0).
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Section 12.7: Systems of Inequalities
The overlapping region is the solution.
4 x y 2 28. x 2y 2 Graph the line 4 x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 4(0) 0 2 is false, shade the opposite side of the line from (0, 0). Graph the line x 2 y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
The overlapping region is the solution.
x 4 y 8 30. x 4 y 4 Graph the line x 4 y 8 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 8 is true, shade the side of the line containing (0, 0). Graph the line x 4 y 4 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
x 2y 6 29. 2 x 4 y 0 Graph the line x 2 y 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6 is true, shade the side of the line containing (0, 0). Graph the line 2 x 4 y 0 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2(0) 4(2) 0 is false, shade the opposite side of the line from (0, 2).
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Chapter 12: Systems of Equations and Inequalities
2 x y 2 31. 2 x y 2 Graph the line 2 x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2 is true, shade the side of the line containing (0, 0). Graph the line 2 x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
2 x 3 y 6 33. 2 x 3 y 0 Graph the line 2 x 3 y 6 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6 is false, shade the opposite side of the line from (0, 0). Graph the line 2 x 3 y 0 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 2). Since 2(0) 3(2) 0 is false, shade the opposite side of the line from (0, 2). Since the regions do not overlap, the solution is an empty set.
x 4 y 4 32. x 4 y 0 Graph the line x 4 y 4 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 4(0) 4 is true, shade the side of the line containing (0, 0). Graph the line x 4 y 0 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (1, 0). Since 1 4(0) 0 is true, shade the side of the line containing (1, 0). The overlapping region is the solution.
2 x y 0 34. 2 x y 2 Graph the line 2 x y 0 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (1, 0). Since 2(1) 0 0 is true, shade the side of the line containing (1, 0). Graph the line 2 x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
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Section 12.7: Systems of Inequalities
2 2 x y 9 35. x y 3 Graph the circle x 2 y 2 9 . Use a solid line since the inequality uses . Choose a test point not on the circle, such as (0, 0). Since 02 02 9 is true, shade the same side of the circle as (0, 0). Graph the line x y 3 . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 0 3 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
x y 9 36. x y 3 Graph the circle x 2 y 2 9 . Use a solid line since the inequality uses ≥. Choose a test point not on the circle, such as (0, 0). Since 02 02 9 is false, shade the opposite side of the circle as (0, 0). Graph the line x y 3 . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 3 is true, shade the same side of the line as (0, 0). The overlapping region is the solution. 2
2
2 y x 4 37. y x 2 Graph the parabola y x 2 4 . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 0 02 4 is true, shade the same side of the parabola as (0, 0). Graph the line y x 2 . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 0 0 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.
y 2 x 38. y x Graph the parabola y 2 x . Use a solid line since the inequality uses . Choose a test point not on the parabola, such as (1, 2). Since 22 1 is false, shade the opposite side of the parabola from (1, 2). Graph the line y x . Use a solid line since the inequality uses . Choose a test point not on the line, such as (1, 2). Since 2 1 is true, shade the same side of the line as (1, 2). The overlapping region is the solution.
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Chapter 12: Systems of Equations and Inequalities
39. x 2 y 2 16 2 y x 4
Graph the circle x 2 y 2 16 . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0, 0) . Since 02 02 16 is true, shade the side of the circle containing (0, 0) . Graph the parabola y x 2 4 . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0, 0) . Since 0 02 4 is true, shade the side of the parabola that contains (0, 0) . The overlapping region is the solution.
40. x 2 y 2 25 2 y x 5
Graph the circle x 2 y 2 25 . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0, 0) . Since 2
2
0 0 25 is true, shade the side of the circle containing (0, 0) . Graph the parabola y x 2 5 . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0, 0) . Since 0 02 5 is false, shade the side of the parabola opposite that which contains the point (0, 0) . The overlapping region is the solution.
xy 4 41. 2 y x 1 Graph the hyperbola xy 4 . Use a solid line since the inequality uses . Choose a test point not on the parabola, such as (0, 0). Since 0 0 4 is false, shade the opposite side of the hyperbola from (0, 0). Graph the parabola y x 2 1 . Use a solid line since the inequality uses . Choose a test point not on the parabola, such as (0, 0). Since 0 02 1 is false, shade the opposite side of the parabola from (0, 0).The overlapping region is the solution.
y x 2 1 42. 2 y x 1 Graph the parabola y x 2 1 . Use a solid line since the inequality uses . Choose a test point not on the parabola, such as (0, 0). Since
0 02 1 is true, shade the same side of the
parabola as (0, 0). Graph the parabola y x 2 1 . Use a solid line since the inequality uses . Choose a test point not on the parabola, such as (0, 0). Since 0 02 1 is true, shade the same side of the parabola as (0, 0). The overlapping region is the solution.
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Section 12.7: Systems of Inequalities
x0 0 y 43. 2 x y 6 x 2 y 6 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line 2 x y 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 6 is true, shade the side of the line containing (0, 0). Graph the line x 2 y 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 6 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:
The x-axis and y-axis intersect at (0, 0). The intersection of x 2 y 6 and the y-axis is (0, 3). The intersection of 2 x y 6 and the x-axis is (3, 0). To find the intersection of x 2 y 6 and 2 x y 6 , solve the system:
x0 0 y 44. x y 4 2 x 3 y 6 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x y 4 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 4 is false, shade the opposite side of the line from (0, 0). Graph the line 2 x 3 y 6 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded. Find the vertices: The intersection of x y 4 and the y-axis is (0, 4). The intersection of x y 4 and the xaxis is (4, 0). The two corner points are (0, 4), and (4, 0).
x 2y 6 2 x y 6 Solve the first equation for x: x 6 2 y . Substitute and solve: 2(6 2 y ) y 6 12 4 y y 6
=
12 3 y 6 3 y 6 y2 x 6 2(2) 2 The point of intersection is (2, 2). The four corner points are (0, 0), (0, 3), (3, 0), and (2, 2).
x0 y0 45. x y 2 2 x y 4 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). Graph the line 2 x y 4 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 4 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.
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Chapter 12: Systems of Equations and Inequalities
Find the vertices: The intersection of x y 2 and the x-axis is (2, 0). The intersection of 2 x y 4 and the yaxis is (0, 4). The two corner points are (2, 0), and (0, 4).
x0 y0 46. x y6 3 2 x y 2 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line 3 x y 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 6 is true, shade the side of the line containing (0, 0). Graph the line 2 x y 2 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of x 0 and y 0 is (0, 0). The intersection of 2 x y 2 and the x-axis is (1, 0). The intersection of 2 x y 2 and the yaxis is (0, 2). The three corner points are (0, 0), (1, 0), and (0, 2).
x0 y 0 47. x y 2 2 x 3 y 12 3x y 12
Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2 is false, shade the opposite side of the line from (0, 0). Graph the line 2 x 3 y 12 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 12 is true, shade the side of the line containing (0, 0). Graph the line 3x y 12 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) 0 12 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of x y 2 and the y-axis is (0, 2). The intersection of x y 2 and the xaxis is (2, 0). The intersection of 2 x 3 y 12 and the y-axis is (0, 4). The intersection of 3x y 12 and the x-axis is (4, 0). To find the intersection of 2 x 3 y 12 and 3 x y 12 , solve the system: 2 x 3 y 12 3 x y 12 Solve the second equation for y: y 12 3 x . Substitute and solve: 2 x 3(12 3x) 12 2 x 36 9 x 12 7 x 24 24 x 7 72 12 24 y 12 3 12 7 7 7 24 12 The point of intersection is , . 7 7 The five corner points are (0, 2), (0, 4), (2, 0), 24 12 (4, 0), and , . 7 7
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Section 12.7: Systems of Inequalities
x0 y0 48. x y 1 x y 7 2 x y 10 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x y 1 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 1 is false, shade the opposite side of the line from (0, 0). Graph the line x y 7 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 7 is true, shade the side of the line containing (0, 0). Graph the line 2 x y 10 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
x 0 y 0 49. x y 2 x y 8 2 x y 10 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2 is false, shade the opposite side of the line from (0, 0). Graph the line x y 8 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8 is true, shade the side of the line containing (0, 0). Graph the line 2 x y 10 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of x y 1 and the y-axis is (0, 1). The intersection of 2 x y 10 and the yaxis is (0, 10). To find the intersection of 2 x y 10 and x y 7 , solve the system:
Find the vertices: The intersection of x y 2 and the y-axis is (0, 2). The intersection of x y 2 and the x-axis is (2, 0). The intersection of x y 8 and the yaxis is (0, 8). The intersection of 2 x y 10 and the x-axis is (5, 0). To find the intersection of x y 8 and 2 x y 10 , solve the system:
2 x y 10 x y 7 Solve the second equation for y: y 7 x . Substitute and solve: 2 x 7 x 10 x3 y 73 4 The point of intersection is (3, 4). The five corner points are (0, 1), (1, 0), (0, 7), (5, 0) and (3, 4).
x y 8 2 x y 10 Solve the first equation for y: y 8 x . Substitute and solve: 2 x 8 x 10 x2 y 82 6 The point of intersection is (2, 6). The five corner points are (0, 2), (0, 8), (2, 0), (5, 0), and (2, 6).
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Chapter 12: Systems of Equations and Inequalities
x0 y0 50. x y 2 x y 8 x 2 y 1 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2 is false, shade the opposite side of the line from (0, 0). Graph the line x y 8 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8 is true, shade the side of the line containing (0, 0). Graph the line x 2 y 1 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of x y 2 and the y-axis is (0, 2). The intersection of x y 2 and the xaxis is (2, 0). The intersection of x y 8 and the y-axis is (0, 8). The intersection of x y 8 and the x-axis is (8, 0). The four corner points are (0, 2), (0, 8), (2, 0), and (8, 0).
x 0 y 0 51. x 2 y 1 x 2 y 10 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x 2 y 1 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1 is false, shade the opposite side of the line from (0, 0). Graph the line x 2 y 10 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 10 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.
Find the vertices: The intersection of x 2 y 1 and the y-axis is (0, 0.5). The intersection of x 2 y 1 and the x-axis is (1, 0). The intersection of x 2 y 10 and the y-axis is (0, 5). The intersection of x 2 y 10 and the x-axis is (10, 0). The four corner points are (0, 0.5), (0, 5), (1, 0), and (10, 0).
x 0 y 0 x 2 y 1 52. x 2 y 10 x y 2 x y 8 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x 2 y 1 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 2(0) 1 is false, shade the opposite side of the line from (0, 0). Graph the line x 2 y 10 . Use a solid line since the inequality uses ≤. Choose a
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Section 12.7: Systems of Inequalities
test point not on the line, such as (0, 0). Since 0 2(0) 10 is true, shade the side of the line containing (0, 0). Graph the line x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2 is false, shade the opposite side of the line from (0, 0). Graph the line x y 8 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 8 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of x y 2 and the y-axis is (0, 2). The intersection of x y 2 and the xaxis is (2, 0). The intersection of x 2 y 10 and the y-axis is (0, 5). The intersection of x y 8 and the x-axis is (8, 0). To find the intersection of x y 8 and x 2 y 10 , solve the system: x y 8 x 2 y 10 Solve the first equation for x: x 8 y . Substitute and solve: (8 y ) 2 y 10 y2 x 82 6 The point of intersection is (6, 2). The five corner points are (0, 2), (0, 5), (2, 0), (8, 0), and (6, 2).
53. The system of linear inequalities is: x4 x y 6 x 0 y 0
54. The system of linear inequalities is: y5 x y 2 x6 x 0 y 0 55. The system of linear inequalities is: x 20 y 15 x y 50 x y 0 x 0 56. The system of linear inequalities is: y 6 x 5 3 4 x y 12 2x y 8 x 0 57. a.
Let x = the amount invested in Treasury bills, and let y = the amount invested in corporate bonds. The constraints are: x y 5000 because the total investment cannot exceed $5000. x 3500 because the amount invested in Treasury bills must be at least $3500. y 1000 because the amount invested in corporate bonds must not exceed $1000. y 0 because a non-negative amount must be invested. The system is x y 5000 x 3500 y 1000 y0
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Chapter 12: Systems of Equations and Inequalities b. Graph the system.
58. a.
59. a.
The corner points are (3500, 0), (3500, 1000), (4000, 1000), (5000, 0) and (3500,1500). Let x = the # of standard model trucks, and let y = the # of deluxe model trucks. The constraints are: x 0, y 0 because a non-negative number of trucks must be manufactured. 2 x 3 y 80 because the total painting hours worked cannot exceed 80. 3 x 4 y 120 because the total detailing hours worked cannot exceed 120. The system is x 0 y 0 2 x 3 y 80 3x 4 y 120
Let x = the # of packages of the economy blend, and let y = the # of packages of the superior blend. The constraints are: x 0, y 0 because a non-negative # of packages must be produced. 4 x 8 y 75 16 because the total amount of “A grade” coffee cannot exceed 75 pounds. (Note: 75 pounds = (75)(16) ounces.) 12 x 8 y 120 16 because the total amount of “B grade” coffee cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) Simplifying the inequalities, we obtain: 4 x 8 y 75 16
12 x 8 y 120 16
x 2 y 75 4
3x 2 y 120 4
x 2 y 300
3x 2 y 480
The system is: x 0 y 0 x 2 y 300 3x 2 y 480 b. Graph the system.
b. Graph the system.
The corner points are (0, 0), (0, 150), (90, 105), (160, 0). 60. a. 80 The corner points are 0, 0 , 0. , 40, 0 . 3
Let x = the # of lower-priced packages, and let y = the # of quality packages. The constraints are: x 0, y 0 because a non-negative # of packages must be produced. 8 x 6 y 120 16 because the total amount of peanuts cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) 4 x 6 y 90 16 because the total amount of cashews cannot exceed 90 pounds. (Note: 90 pounds = (90)(16) ounces.) Simplifying the inequalities, we obtain:
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Section 12.7: Systems of Inequalities 8 x 6 y 120 16 4 x 3 y 120 8 4 x 3 y 960
4 x 6 y 90 16 2 x 3 y 90 8 2 x 3 y 720
62.
x y 4 y x 4 y x 4 and y x 4
Graph y x 4 and shade above; graph
The system is 4 x 3 y 960 2 x 3 y 720 x 0; y 0
y x 4 and shade below.
y x 2 3 y x 2 3 and y x 2 3
Graph y x3 3 and shade above; graph
b. Graph the system.
y x 2 3 and shade below. The intersection is
the solution to the system as shown in the figure.
The corner points are (0, 0), (0, 240), (120, 160), (240, 0). 61. a.
Let x = the # of microwaves, and let y = the # of printers. The constraints are: x 0, y 0 because a non-negative # of items must be shipped. 30 x 20 y 1600 because a total cargo weight cannot exceed 1600 pounds. 2 x 3 y 150 because the total cargo volume cannot exceed 150 cubic feet. Note that the inequality 30 x 20 y 1600 can be simplified: 3x 2 y 160 . The system is: 3x 2 y 160 2 x 3 y 150 x 0; y 0
b. Graph the system.
63. 2( x 1) 2 8 0 2( x 1) 2 8 ( x 1) 2 4 ( x 1) 4 ( x 1) 2i x 1 2i
The solution set is 1 2i, 1 2i . 64.
3r sin 3r 2 r sin 3( x 2 y 2 ) y 2 3x 3 y 2 y 0 1 x2 y 2 y 0 3 1 1 1 2 2 x y y 3 36 36 2 2 1 1 x2 y 6 6 1 The graph is a circle with center 0, and
radius The corner points are (0, 0), (0, 50), (36, 26), 160 ,0 . 3
1449 Copyright © 2025 Pearson Education, Inc.
1 . 6
6
Chapter 12: Systems of Equations and Inequalities
71. ( f g )( x)
2
x 2 5
x20 x2 and x 2 5 0 x 2 25 x 23
So the domain is x | x 2, x 23 . 65.
f ( x) 6 x3 5 x 6;
72.
1, 2
f ( 1) 5 0 and f (2) 28 0 The value of the function is positive at one endpoint and negative at the other. Since the function is continuous, the Intermediate Value Theorem guarantees at least one zero in the given interval.
66.
2 cos 2 cos 1 0 (2 cos 1)(cos 1) 0 2 cos 1 0
or cos 1 0
67. x 2 4 x 3 x 18 5 5 x 15 1 x 3 3 x 1
Section 12.8 1. objective function 2. True 3. z x y
Or x | 3 x 1
69.
1 3 ,5 . The graph is above the x-axis and thus 1 f ( x ) is increasing over the intervals , 3 and 5, .
0
2 4 On 0 2 , the solution set is 0, , . 3 3
nt
.0325 7500 1 365
1 and 5 . From the graph of 3 f ( x) we see that the graph is below the x-axis and thus f ( x) is decreasing over the interval
intercepts are
cos 1
1 cos 2 2 4 , 3 3
r 68. A P 1 n
f ( x) 3 x 2 14 x 5 (3 x 1)( x 5) The x-
365(5)
$8823.30
Vertex Value of z x y z 03 3 (0, 3) z 06 6 (0, 6) z 5 6 11 (5, 6) z 52 7 (5, 2) z 40 4 (4, 0) The maximum value is 11 at (5, 6), and the minimum value is 3 at (0, 3).
h ks 3 150 k (12)3 k 0.0868 h 0.0868(6)3 18.75 horsepower
70. log 5 x y 5 y x 1450 Copyright © 2025 Pearson Education, Inc.
Section 12.8: Linear Programming 4. z 2 x 3 y
8. z 7 x 5 y
Vertex Value of z 2 x 3 y z 2(0) 3(3) 9 (0, 3) (0, 6) z 2(0) 3(6) 18 (5, 6) z 2(5) 3(6) 28 (5, 2) z 2(5) 3(2) 16 z 2(4) 3(0) 8 (4, 0) The maximum value is 28 at (5, 6), and the minimum value is 8 at (4, 0).
5. z x 10 y Vertex Value of z x 10 y z 0 10(3) 30 (0, 3) z 0 10(6) 60 (0, 6) z 5 10(6) 65 (5, 6) z 5 10(2) 25 (5, 2) z 4 10(0) 4 (4, 0) The maximum value is 65 at (5, 6), and the minimum value is 4 at (4, 0).
Vertex Value of z 7 x 5 y (0, 3) z 7(0) 5(3) 15 (0, 6) z 7(0) 5(6) 30 (5, 6) z 7(5) 5(6) 65 (5, 2) z 7(5) 5(2) 45 (4, 0) z 7(4) 5(0) 28 The maximum value is 65 at (5, 6), and the minimum value is 15 at (0, 3).
9. Maximize z 2 x y subject to x 0, y 0, x y 6, x y 1 . Graph the constraints. y
(0,6)
x+y=6
(0,1)
6. z 10 x y
(6,0)
x
(1,0)
Vertex Value of z 10 x y z 10(0) 3 3 (0, 3) z 10(0) 6 6 (0, 6) z 10(5) 6 56 (5, 6) z 10(5) 2 52 (5, 2) z 10(4) 0 40 (4, 0) The maximum value is 56 at (5, 6), and the minimum value is 3 at (0, 3).
7. z 5 x 7 y Vertex Value of z 5 x 7 y (0, 3) z 5(0) 7(3) 21 (0, 6) z 5(0) 7(6) 42 (5, 6) z 5(5) 7(6) 67 (5, 2) z 5(5) 7(2) 39 (4, 0) z 5(4) 7(0) 20 The maximum value is 67 at (5, 6), and the minimum value is 20 at (4, 0).
x+y=1
The corner points are (0, 1), (1, 0), (0, 6), (6, 0). Evaluate the objective function: Vertex Value of z 2 x y z 2(0) 1 1 (0, 1) z 2(0) 6 6 (0, 6) z 2(1) 0 2 (1, 0) z 2(6) 0 12 (6, 0) The maximum value is 12 at (6, 0). 10. Maximize z x 3 y subject to x 0, y 0 , x y 3 x 5, y 7 . Graph the constraints. y (0,7) y = 7
(5,7)
x=5 (0,3)
(3,0) x+y=3
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(5,0)
x
Chapter 12: Systems of Equations and Inequalities
The corner points are (0, 3), (3, 0), (0, 7), (5, 0), (5, 7). Evaluate the objective function: Vertex Value of z x 3 y z 0 3(3) 9 (0, 3) z 3 3(0) 3 (3, 0) z 0 3(7) 21 (0, 7) z 5 3(0) 5 (5, 0) z 5 3(7) 26 (5, 7) The maximum value is 26 at (5, 7). 11. Minimize z 2 x 5 y subject to x 0, y 0, x y 2, x 5, y 3 . Graph the constraints. y
x=5
(0,3)
(5,3)
y=3
(5,0)
x+y=8
(0,2) (8,0) x
(3,0) 2x + 3y = 6
The corner points are (0, 2), (3, 0), (0, 8), (8, 0). Evaluate the objective function: Vertex Value of z 3 x 4 y z 3(0) 4(2) 8 (0, 2) z 3(3) 4(0) 9 (3, 0) (0, 8) z 3(0) 4(8) 32 (8, 0) z 3(8) 4(0) 24 The minimum value is 8 at (0, 2). 13. Maximize z 3 x 5 y subject to x 0, y 0, x y 2, 2 x 3 y 12, 3x 2 y 12 . Graph the constraints.
(0,2) (2,0)
y (0,8)
x
x+y=2
y
The corner points are (0, 2), (2, 0), (0, 3), (5, 0), (5, 3). Evaluate the objective function: Vertex Value of z 2 x 5 y (0, 2) z 2(0) 5(2) 10 (0, 3) z 2(0) 5(3) 15 z 2(2) 5(0) 4 (2, 0) (5, 0) z 2(5) 5(0) 10 (5, 3) z 2(5) 5(3) 25 The minimum value is 4 at (2, 0). 12. Minimize z 3 x 4 y subject to x 0 , y 0, 2 x 3 y 6, x y 8 . Graph the constraints.
3x + 2y = 12 (0,4) (2.4,2.4) (0,2) (2,0)
2x + 3y = 12 (4,0) x
x+y=2
To find the intersection of 2 x 3 y 12 and 3x 2 y 12 , solve the system: 2 x 3 y 12 3x 2 y 12 3 Solve the second equation for y: y 6 x 2 Substitute and solve:
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Section 12.8: Linear Programming
3 2 x 3 6 x 12 2 9 2 x 18 x 12 2 5 x 6 2 12 x 5 3 12 18 12 y 6 6 2 5 5 5
Vertex Value of z 5 x 3 y z 5(0) 3(2) 6 (0, 2) (0, 8) z 5(0) 3(8) 24 (2, 0) z 5(2) 3(0) 10 (5, 0) z 5(5) 3(0) 25 (2, 6) z 5(2) 3(6) 28 The maximum value is 28 at (2, 6).
The point of intersection is 2.4, 2.4 . The corner points are (0, 2), (2, 0), (0, 4), (4, 0), (2.4, 2.4). Evaluate the objective function: Vertex Value of z 3x 5 y z 3(0) 5(2) 10 (0, 2) z 3(0) 5(4) 20 (0, 4) z 3(2) 5(0) 6 (2, 0) z 3(4) 5(0) 12 (4, 0) (2.4, 2.4) z 3(2.4) 5(2.4) 19.2 The maximum value is 20 at (0, 4). 14. Maximize z 5 x 3 y subject to x 0, y 0, x y 2, x y 8, 2 x y 10 . Graph the constraints. y 2x + y = 10 (0,8) (2,6)
x+y=8 (0,2) (2,0)
(5,0)
x
15. Minimize z 5 x 4 y subject to x 0, y 0, x y 2, 2 x 3 y 12, 3 x y 12 . Graph the constraints. y
3x + y = 12
(0,4) 24
7 , 12 7 (0,2)
2x + 3y = 12 x
(4,0)
(2,0) x+y=2
To find the intersection of 2 x 3 y 12 and 3x y 12 , solve the system: 2 x 3 y 12 3 x y 12 Solve the second equation for y: y 12 3 x Substitute and solve: 2 x 3(12 3 x) 12 2 x 36 9 x 12 7 x 24 x 24 7
y 12 3 24 12 72 12 7 7 7
x+y=2
To find the intersection of x y 8 and 2x y 10 , solve the system: x y 8 2 x y 10 Solve the first equation for y: y 8 x . Substitute and solve: 2 x 8 x 10 x2 y 82 6 The point of intersection is (2, 6). The corner points are (0, 2), (2, 0), (0, 8), (5, 0), (2, 6). Evaluate the objective function:
The point of intersection is 24 , 12 . 7
7
The corner points are (0, 2), (2, 0), (0, 4), (4, 0), 24 12 , . Evaluate the objective function: 7 7
Vertex
Value of z 5 x 4 y
(0, 2)
z 5(0) 4(2) 8
(0, 4)
z 5(0) 4(4) 16
(2, 0)
z 5(2) 4(0) 10
(4, 0)
z 5(4) 4(0) 20
247 , 127 z 5 247 4 127 24 The minimum value is 8 at (0, 2).
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Chapter 12: Systems of Equations and Inequalities 16. Minimize z 2 x 3 y subject to x 0, y 0, x y 3, x y 9, x 3 y 6 . Graph the constraints. y
17. Maximize z 5 x 2 y subject to x 0, y 0, x y 10, 2 x y 10, x 2 y 10 . Graph the constraints. y
(0,9)
(0,10) x + y = 10
x+y=9
(0,3)
103 , 103
, 3 2
3 2
(6,0)
x+y=3
(10,0) x
x (9,0)
x + 3y = 6
2x + y = 10
x + 2y = 10
To find the intersection of x y 3 and x 3 y 6 , solve the system:
To find the intersection of 2 x y 10 and x 2 y 10 , solve the system:
x y 3 x 3y 6 Solve the first equation for y: y 3 x . Substitute and solve: x 3(3 x ) 6 x 9 3x 6
2 x y 10 x 2 y 10 Solve the first equation for y: y 10 2 x . Substitute and solve: x 2(10 2 x) 10 x 20 4 x 10 3 x 10 10 x 3 20 10 10 y 10 2 10 3 3 3 The point of intersection is (10/3 10/3). The corner points are (0, 10), (10, 0), (10/3, 10/3). Evaluate the objective function:
2 x 3 x y 3
3 2
3 3 2 2
3 3 The point of intersection is , . 2 2 The corner points are (0, 3), (6, 0), (0, 9), (9, 0), 3 3 , . Evaluate the objective function: 2 2
Vertex
Value of z 2 x 3 y
(0, 3) (0, 9)
z 2(0) 3(3) 9 z 2(0) 3(9) 27
(6, 0)
z 2(6) 3(0) 12
(9, 0)
z 2(9) 3(0) 18
Vertex
Value of z 5 x 2 y
(0, 10) (10, 0)
z 5(0) 2(10) 20 z 5(10) 2(0) 50
10 10 10 10 70 23 13 , z 5 2 3 3 3 3 3
The maximum value is 50 at (10, 0).
3 3 3 3 15 , z 2 4 2 2 2 2 2 15 3 3 The minimum value is at , . 2 2 2
18. Maximize z 2 x 4 y subject to x 0, y 0, 2 x y 4, x y 9 . Graph the constraints.
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Section 12.8: Linear Programming y 32 8 24 The point of intersection is (8, 24). The corner points are (0, 0), (0, 32), (20, 0), (8, 24). Evaluate the objective function: Vertex Value of P 70 x 50 y
y (0,9)
x+y=9 (0,4)
(2,0)
(9,0) x 2x + y = 4
The corner points are (0, 9), (9, 0), (0, 4), (2, 0). Evaluate the objective function: Vertex Value of z 2 x 4 y (0, 9) z 2(0) 4(9) 36 (9, 0) z 2(9) 4(0) 18 (0, 4) z 2(0) 4(4) 16 z 2(2) 4(0) 4 (2, 0) The maximum value is 36 at (0, 9). 19. Let x = the number of downhill skis produced, and let y = the number of cross-country skis produced. The total profit is: P 70 x 50 y . Profit is to be maximized, so this is the objective function. The constraints are: x 0, y 0 A positive number of skis must be produced. 2 x y 40 Manufacturing time available. x y 32 Finishing time available. Graph the constraints.
(0, 0)
P 70(0) 50(0) 0
(0, 32)
P 70(0) 50(32) 1600
(20, 0) P 70(20) 50(0) 1400 (8, 24) P 70(8) 50(24) 1760 The maximum profit is $1760, when 8 downhill skis and 24 cross-country skis are produced.
With the increase of the manufacturing time to 48 hours, we do the following: The constraints are: x 0, y 0
A positive number of skis must be produced. 2 x y 48 Manufacturing time available. x y 32 Finishing time available. Graph the constraints. y 2x + y = 48 (0,32)
(16,16) x + y = 32 (24,0)
x
(0,0)
To find the intersection of x y 32 and 2 x y 48 , solve the system:
y 2x + y = 40
x y 32 2 x y 48 Solve the first equation for y: y 32 x . Substitute and solve: 2 x (32 x) 48
(0,32) (8,24)
x + y = 32 (20,0)
x
(0,0)
To find the intersection of x y 32 and 2 x y 40 , solve the system: x y 32 2 x y 40 Solve the first equation for y: y 32 x . Substitute and solve: 2 x (32 x) 40
x8
x 16 y 32 16 16 The point of intersection is (16, 16). The corner points are (0, 0), (0, 32), (24, 0), (16, 16). Evaluate the objective function: Vertex Value of P 70 x 50 y (0, 0)
P 70(0) 50(0) 0
(0, 32)
P 70(0) 50(32) 1600
(24, 0) P 70(24) 50(0) 1680 (16, 16) P 70(16) 50(16) 1920
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Chapter 12: Systems of Equations and Inequalities
The maximum profit is $1920, when 16 downhill skis and 16 cross-country skis are produced. 20. Let x = the number of acres of soybeans planted , and let y = the number of acres of wheat planted. The total profit is: P 180 x 100 y . Profit is to be maximized, so this is the objective function. The constraints are: x 0, y 0 A non-negative number of acres must be planted. x y 70 Acres available to plant. 60 x 30 y 1800 Money available for preparation. 3 x 4 y 120 Workdays available. Graph the constraints.
The maximum profit is $5520, when 24 acres of soybeans and 12 acres of wheat are planted. With the increase of the preparation costs to $2400, we do the following: The constraints are: x 0, y 0 A non-negative number of acres must be planted. x y 70 Acres available to plant. 60 x 30 y 2400 Money available for preparation. 3 x 4 y 120 Workdays available. Graph the constraints. y x + y = 70
y x + y = 70
(0,30) 3x + 4y = 120
60x + 30y = 1800
60x + 30y = 2400
(0,30)
(0,0)
(24,12) 3x + 4y = 120 (0,0)
(30,0)
x
To find the intersection of 60 x 30 y 1800 and 3 x 4 y 120 , solve the system: 60 x 30 y 1800 3x 4 y 120 Solve the first equation for y: 60 x 30 y 1800 2 x y 60 y 60 2 x Substitute and solve: 3x 4(60 2 x) 120 3x 240 8 x 120 5 x 120 x 24 y 60 2(24) 12 The point of intersection is (24, 12). The corner points are (0, 0), (0, 30), (30, 0), (24, 12). Evaluate the objective function: Vertex Value of P 180 x 100 y (0, 0) P 180(0) 100(0) 0 (0, 30) P 180(0) 100(30) 3000 (30, 0) P 180(30) 100(0) 5400 (24, 12) P 180(24) 100(12) 5520
(40,0)
x
The corner points are (0, 0), (0, 30), (40, 0). Evaluate the objective function: Vertex
Value of P 180 x 100 y
(0, 0)
P 180(0) 100(0) 0
(0, 30)
P 180(0) 100(30) 3000
(40, 0) P 180(40) 100(0) 7200 The maximum profit is $7200, when 40 acres of soybeans and 0 acres of wheat are planted.
21. Let x = the number of rectangular tables rented, and let y = the number of round tables rented. The cost for the tables is: C 28 x 52 y . Cost is to be minimized, so this is the objective function. The constraints are: x 0, y 0 A non-negative number of tables must be used. x y 35 Maximum number of tables. 6 x 10 y 250 Number of guests. x 15 Rectangular tables available. Graph the constraints.
1456 Copyright © 2025 Pearson Education, Inc.
Section 12.8: Linear Programming y
x
xy xy
x
The corner points are (0, 25), (0, 35), (15, 20), (15, 16). Evaluate the objective function: Vertex Value of C 28 x 52 y (0, 25) C 28(0) 52(25) 1300 (0, 35) C 28(0) 52(35) 1820 (15, 20) C 28(15) 52(20) 1460 (15, 16) C 28(15) 52(16) 1252 Kathleen should rent 15 rectangular tables and 16 round tables in order to minimize the cost. The minimum cost is $1252.00. 22. Let x = the number of buses rented, and let y = the number of vans rented. The cost for the vehicles is: C 975 x 350 y . Cost is to be minimized, so this is the objective function. The constraints are: x 0, y 0 A non-negative number of buses and vans must be used. 40 x 8 y 320 Number of regular seats. x 3 y 36 Number of special seats. Graph the constraints. y xy
Substitute and solve: 40(3 y 36) 8 y 320 120 y 1440 8 y 320 112 y 1120 y 10 x 3(10) 36 30 36 6 The point of intersection is (6, 10). The corner points are (0, 40), (6, 10), and (36, 0). Evaluate the objective function: Vertex
(0, 40) C 975(0) 350(40) 14, 000 (6, 10)
C 975(6) 350(10) 9350
(36, 0) C 975(36) 350(0) 35,100 The college should rent 6 buses and 10 vans for a minimum cost of $9350.00.
23. Let x = the amount invested in junk bonds, and let y = the amount invested in Treasury bills. The total income is: I 0.09 x 0.07 y . Income is to be maximized, so this is the objective function. The constraints are: x 0, y 0 A non-negative amount must be invested. x y 20, 000 Total investment cannot exceed $20,000. x 12, 000 Amount invested in junk bonds must not exceed $12,000. y 8000 Amount invested in Treasury bills must be at least $8000. a. y x Amount invested in Treasury bills must be equal to or greater than the amount invested in junk bonds. Graph the constraints. (0,20000) x + y = 20000
Value of C 975 x 350 y
y=x
(10000,10000)
x xy
(0,8000)
To find the intersection of 40 x 8 y 320 and x 3 y 36 , solve the system: 40 x 8 y 320 x 3 y 36 Solve the second equation for x: x 3 y 36 x 3 y 36
(8000,8000)
y = 8000
x = 12000
The corner points are (0, 20,000), (0, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function:
1457 Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities
Value of I 0.09 x 0.07 y I 0.09(0) 0.07(20000) 1400 (0, 8000) I 0.09(0) 0.07(8000) 560 (8000, 8000) I 0.09(8000) 0.07(8000) 1280 (10000, 10000) I 0.09(10000) 0.07(10000) 1600
x 10 Time used on machine 1. y 10 Time used on machine 2. 60 x 40 y 240 8-inch pliers to be produced. 70 x 20 y 140 6-inch pliers to be produced. Graph the constraints.
The maximum income is $1600, when $10,000 is invested in junk bonds and $10,000 is invested in Treasury bills.
12 , 214
Vertex (0, 20000)
b.
(0,10) (10,10) (0,7)
yx
Amount invested in Treasury bills must not exceed the amount invested in junk bonds. Graph the constraints.
y=x (10000,10000) y = 8000 (12000,8000)
x = 12000
The corner points are (12,000, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function: Vertex (12000, 8000)
70x + 20y = 140
60x + 40y = 240
To find the intersection of 60 x 40 y 240 and 70x 20 y 140 , solve the system:
x + y = 20000
(8000,8000)
(10,0)
(4,0)
Value of I 0.09 x 0.07 y I 0.09(12000) 0.07(8000)
1640 I 0.09(8000) 0.07(8000) 1280 (10000, 10000) I 0.09(10000) 0.07(10000) 1600 (8000, 8000)
The maximum income is $1640, when $12,000 is invested in junk bonds and $8000 is invested in Treasury bills. 24. Let x = the number of hours that machine 1 is operated, and let y = the number of hours that machine 2 is operated. The total cost is: C 50 x 30 y . Cost is to be minimized, so this is the objective function. The constraints are: x 0, y 0 A positive number of hours must be used.
60 x 40 y 240 70 x 20 y 140 Divide the first equation by 2 and add the result to the second equation: 30 x 20 y 120 70 x 20 y 140
20
40 x
20 1 40 2 Substitute and solve: 1 60 40 y 240 2 30 40 y 240
x
40 y 210 210 21 1 5 40 4 4 1 1 The point of intersection is , 5 . 2 4
y
The corner points are (0, 7), (0, 10), (4, 0), 1 1 (10, 0), (10, 10), , 5 . Evaluate the 2 4 objective function:
1458 Copyright © 2025 Pearson Education, Inc.
Section 12.8: Linear Programming
Vertex (0, 7) (0, 10) (4, 0) (10, 0) (10, 10)
Value of C 50 x 30 y C 50(0) 30(7) 210 C 50(0) 30(10) 300 C 50(4) 30(0) 200 C 50(10) 30(0) 500 C 50(10) 30(10) 800
1 1 1 1 , 5 P 50 30 5 182.50 2 4 2 4
The minimum cost is $182.50, when machine 1 1 1 is used for hour and machine 2 is used for 5 4 2 hours. 25. Let x = the number of pounds of ground beef, and let y = the number of pounds of ground pork. The total cost is: C 2.25 x 1.35 y . Cost is to be minimized, so this is the objective function. The constraints are: x 0, y 0 A positive number of pounds must be used. x 200 Only 200 pounds of ground beef are available. y 50 At least 50 pounds of ground pork must be used. 0.75 x 0.60 y 0.70( x y ) (1) Leanness condition x y 180 (2) (Note that the equation (1) will simplify to 1 y x and equation (2) will simplify to 2 y x 180 ) Graph the constraints.
Vertex Value of C 2.25 x 1.35 y (120, 60) C 2.25(120) 1.35(60) 351.00 (200, 50) C 2.25(200) 1.35(50) 517.50 (200, 100) C 2.25(200) 1.35(100) 585.00 C 2.25(130) 1.35(50) 360.00
(130, 50)
The minimum cost is $351.00, when 120 pounds of ground beef and 60 pounds of ground pork are used. 26. Let x = the number of gallons of regular, and let y = the number of gallons of premium. The total profit is: P 0.75 x 0.90 y . Profit is to be maximized, so this is the objective function. The constraints are: x 0, y 0 A positive number of gallons must be used. 1 y x At least one gallon of premium 4 for every 4 gallons of regular. 5 x 6 y 3000 Daily shipping weight limit. 24 x 20 y 16(725) Available flavoring. 12 x 20 y 16(425) Available milk-fat (Note: the last two inequalities simplify to 6 x 5 y 2900 and 3x 5 y 1700 .) Graph the constraints. xy
y
xy xy
y
y
(200,100) (120.60)
1 x 4
x
(100,50) (130.50)
(200,50)
The corner points are (0, 0), (400, 100), (0, 340). Evaluate the objective function:
x
The corner points are (120, 60), (200, 50), (200, 100), (130. 50). Evaluate the objective function:
Vertex Value of P 0.75 x 0.90 y (0, 0) P 0.75(0) 0.90(0) 0 (400, 100) P 0.75(400) 0.90(100) 390 (0, 340) P 0.75(0) 0.90(340) 306
Mom and Pop should produce 400 gallons of regular and 100 gallons of premium ice cream. The maximum profit is $390.00.
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Chapter 12: Systems of Equations and Inequalities 27. Let x = the number of racing skates manufactured, and let y = the number of figure skates manufactured. The total profit is: P 10 x 12 y . Profit is to be maximized, so this is the objective function. The constraints are: x 0, y 0
A positive number of skates must be manufactured. 6 x 4 y 120 Only 120 hours are available for fabrication. x 2 y 40 Only 40 hours are available for finishing. Graph the constraints. y
(0,20)
28. Let x = the amount placed in the AAA bond. Let y = the amount placed in a CD. The total return is: R 0.08 x 0.04 y . Return is to be maximized, so this is the objective function. The constraints are: x 0, y 0 A positive amount must be invested in each. x y 5000 Total investment cannot exceed $5000. x 2000 Investment in the AAA bond cannot exceed $2000. y 1500 Investment in the CD must be at least $1500. y x Investment in the CD must exceed or equal the investment in the bond. Solve each simultaneous equation and find the intersections.
The corner points are (0, 5000), (0, 1500), (1500, 1500), (2000, 2000), (2000, 3000). Evaluate the objective function:
(10,15)
(20,0)
Vertex (0, 5000)
x
(0,0)
To find the intersection of 6 x 4 y 120 and x +2y 40 , solve the system: 6 x 4 y 120 x 2 y 40 Solve the second equation for x: x 40 2 y Substitute and solve: 6(40 2 y ) 4 y 120
240 12 y 4 y 120 8 y 120 y 15 x 40 2(15) 10 The point of intersection is (10, 15). The corner points are (0, 0), (0, 20), (20, 0), (10, 15). Evaluate the objective function: Vertex Value of P 10 x 12 y P 10(0) 12(0) 0 (0, 0) (0, 20)
P 10(0) 12(20) 240
(20, 0)
P 10(20) 12(0) 200
(10, 15) P 10(10) 12(15) 280 The maximum profit is $280, when 10 racing skates and 15 figure skates are produced.
(0, 1500)
Value of R 0.08 x 0.04 y R 0.08(0) 0.04(5000) 200 R 0.08(0) 0.04(1500)
60 R 0.08(1500) 0.04(1500) 180 (2000, 2000) R 0.08(2000) 0.04(2000) (1500, 1500)
(2000, 3000)
240 R 0.08(2000) 0.04(3000) 280
The maximum return is $280, when $2000 is invested in a AAA bond and $3000 is invested in a CD. 29. Let x = the number of metal fasteners, and let y = the number of plastic fasteners. The total cost is: C 9 x 4 y . Cost is to be minimized, so this is the objective function. The constraints are: x 2, y 2 At least 2 of each fastener must be made. x y 6 At least 6 fasteners are needed. 4 x 2 y 24 Only 24 hours are available. Graph the constraints.
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Section 12.8: Linear Programming
Vertex (0, 47.5) (0, 60)
y (2,8)
Value of C 1.40 x 1.12 y C 1.40(0) 1.12(47.5) 53.20 C 1.40(0) 1.12(60) 67.20
C 1.40(60) 1.12(0) 84.00 (60, 0) (58.75, 0) C 1.40(58.75) 1.12(0) 82.25 C 1.40(15) 1.12(25) 49.00 (15, 25)
(2,4) (5,2) (4,2) x
The corner points are (2, 4), (2, 8), (4, 2), (5, 2). Evaluate the objective function: Vertex Value of C 9 x 4 y (2, 4) C 9(2) 4(4) 34 (2, 8) C 9(2) 4(8) 50 (4, 2) C 9(4) 4(2) 44 (5, 2) C 9(5) 4(2) 53 The minimum cost is $34, when 2 metal fasteners and 4 plastic fasteners are ordered. 30. Let x = the amount of “Gourmet Dog,” and let y = the amount of “Chow Hound.” The total cost is: C 1.40 x 1.12 y . Cost is to be minimized, so this is the objective function. The constraints are: x 0, y 0 A non-negative number of cans must be purchased. 20 x 35 y 1175 At least 1175 units of vitamins per month. 75 x 50 y 2375 At least 2375 calories per month. x y 60 Storage space for 60 cans. Graph the constraints.
The minimum cost is $49, when 15 cans of "Gourmet Dog" and 25 cans of “Chow Hound” are purchased. 31. Let x = the number of first class seats, and let y = the number of coach seats. Using the hint, the revenue from x first class seats and y coach seats is Fx Cy, where F C 0. Thus, R Fx Cy is the objective function to be maximized. The constraints are: 8 x 16 Restriction on first class seats. 80 y 120 Restriction on coach seats. a.
x 1 Ratio of seats. y 12 The constraints are: 8 x 16 80 y 120 12x y Graph the constraints.
y (0,60)
(0,47.5)
x + y = 60
The corner points are (8, 96), (8, 120), and (10, 120). Evaluate the objective function:
20x+35y=1175 (15,25)
75x+50y=2375
(60,0) x (58.75,0)
The corner points are (0, 47.5), (0, 60), (60, 0), (58.75, 0), (15, 25). Evaluate the objective function:
Vertex Value of R Fx Cy (8, 96) R 8F 96C (8, 120) R 8 F 120C (10, 120) R 10 F 120C
Since C 0, 120C 96C , so 8F 120C 8 F 96C. Since F 0, 10 F 8F , so 10 F 120C 8 F 120C. Thus, the maximum revenue occurs when the
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Chapter 12: Systems of Equations and Inequalities
aircraft is configured with 10 first class seats and 120 coach seats. b.
x 1 y 8 The constraints are: 8 x 16 80 y 120 8x y Graph the constraints.
Corner Points (0, 0) (0, 9) (1, 13) (3,14) (7,12) (8,11) (9,7) (8,2) (4,0)
The corner points are (8, 80), (8, 120), (15, 120), and (10, 80). Evaluate the objective function: Vertex Value of R Fx Cy (8, 80) R 8 F 80C (8, 120) R 8F 120C (15, 120) R 15 F 120C (10, 80) R 10 F 80C
Since F 0 and C 0, 120C 96C , the maximum value of R occurs at (15, 120). The maximum revenue occurs when the aircraft is configured with 15 first class seats and 120 coach seats. c.
Answers will vary.
z 10 x 4 y z 10 0 4 0 0 z 10 0 4 9 36 z 10 1 4 13 62 z 10 3 4 14 86 z 10 7 4 12 118 z 10 8 4 11 124 z 10 9 4 7 118 z 10 8 4 2 88 z 10 4 4 0 40
The maximum is 124 occurs when x = 8 and y = 11. 33. Answers will vary. 2m 2/5 m1/5 1
34.
2m 2/5 m1/5 1 0 (2m1/5 1)(m1/5 1) 0 (2m1/5 1) 0 or (m1/5 1) 0
32. The figure shows the graph of the constraints with the corner points labeled. The table shows the value of the objective function at each corner point.
2m1/5 1
or m1/5 1
1 2
or m1/5 1
m1/5
1 m 2 m
1 32
5
or m 15 or m 1
1 The solution set is ,1 . 32
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Section 12.8: Linear Programming 35. y tan x 2
The graph of y tan x is shifted
2 right and reflected across the x-axis.
r 39. P A 1 n
nt
0.04 15000 1 365
units to the
365(3)
$13,303.89
40. Focus: (0, 3) ; Vertices (0, 5) Center: (0, 0); Major axis is the y-axis; a 5; c 3 . Find b: b 2 a 2 c 2 25 9 16 b 4
Write the equation:
41.
Domain: x | x k , k is an integer Range: , 36.
y7 5y 2 x7 f 1 y 5x 2 x
ln 0.5 63t r 0.011 Find t when A 75 and A0 200 : 75 200e 0.011t
3
1
ln 0.375 0.011t
1
15 x 2 ( x 7) 2 (3 x 14)
ln 0.375 89.1 years 0.011
37. The slope would be the same so the slope is m 3.
43. 2 x 6 0 2x 6 x3 The function is concave down on the interval ,3
y 1 3( x ( 2))
2x 6 0
y 1 3( x 2)
2x 6
y 1 3x 6 y 3x 7 x y 16
1
42. 30 x 2 ( x 7) 2 15 x3 ( x 7) 2 15 x 2 ( x 7) 2 (2( x 7) x)
0.375 e 0.011t
38.
5 xy y 2 x 7 x(5 y 2) 7 y
1 2e63t
y y1 m( x x1 )
2x 7 5x 1 y (5 x 1) 2 x 7 y
5 xy 2 x 7 y
A(t ) A0 e rt
t
x2 y 2 1 16 25
x3 The function is concave up on the interval 3, .
y 3x 4 x (3 x 4) 16 4 x 20 x5 y 15 4 11 The two numbers are 5 and 11.
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Chapter 12: Systems of Equations and Inequalities
Chapter 12 Review Exercises 2x y 5 1. 5 x 2 y 8 Solve the first equation for y: y 2 x 5 . Substitute and solve: 5 x 2(2 x 5) 8 5 x 4 x 10 8 9 x 18 x2 y 2(2) 5 4 5 1 The solution is x 2, y 1 or (2, 1) .
3 x 4 y 4 2. 1 x 3 y 2 1 2 Substitute into the first equation and solve: 1 3 3 y 4 y 4 2 3 9y 4y 4 2 5 5y 2 1 y 2 1 1 x 3 2 2 2 1 1 The solution is x 2, y or 2, . 2 2
Solve the second equation for x: x 3 y
x 2y 4 0 3. 3 x 2 y 4 0 Solve the first equation for x: x 2 y 4 Substitute into the second equation and solve: 3(2 y 4) 2 y 4 0 6 y 12 2 y 4 0 8 y 8 y 1 x 2(1) 4 2 The solution is x 2, y 1 or (2, 1) .
y 2x 5 4. x 3y 4 Substitute the first equation into the second equation and solve: x 3(2 x 5) 4 x 6 x 15 4 5 x 11 11 x 5 3 11 y 2 5 5 5 11 3 11 3 The solution is x , y or , . 5 5 5 5 x 3y 4 0 5. 1 3 4 2 x 2 y 3 0 Multiply each side of the first equation by 3 and each side of the second equation by 6 and add: 3 x 9 y 12 0 3x 9 y 8 0 40 There is no solution to the system. The system is inconsistent.
2 x 3 y 13 0 6. 0 3x 2 y Multiply each side of the first equation by 2 and each side of the second equation by 3, and add to eliminate y: 4 x 6 y 26 0 0 9 x 6 y 13 x 26 0 13 x 26 x2 Substitute and solve for y: 3(2) 2 y 0 2y 6 y3 The solution is x 2, y 3 or (2, 3). 2 x 5 y 10 7. 4 x 10 y 20 Multiply each side of the first equation by –2 and add to eliminate x:
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Chapter 12 Review Exercises
4 x 10 y 20 4 x 10 y 20 0 0 The system is dependent. 2 x 5 y 10 5 y 2 x 10 2 y x2 5 2 The solution is y x 2 , x is any real number 5 2 or ( x, y ) y x 2, x is any real number . 5 x 2y z 6 8. 2 x y 3z 13 3x 2 y 3 z 16
Multiply each side of the first equation by –2 and add to the second equation to eliminate x; 2 x 4 y 2 z 12 2 x y 3 z 13 5 y 5 z 25 yz 5 Multiply each side of the first equation by –3 and add to the third equation to eliminate x: 3x 6 y 3 z 18 3x 2 y 3z 16 8 y 6 z 34 Multiply each side of the first result by 8 and add to the second result to eliminate y: 8 y 8 z 40 8 y 6 z 34 2z 6 z 3 Substituting and solving for the other variables: y (3) 5 x 2(2) (3) 6 x43 6 y2 x 1 The solution is x 1, y 2, z 3 or (1, 2, 3) . 2 x 4 y z 15 9. x 2 y 4 z 27 5 x 6 y 2 z 3
equation by 2, and then add to eliminate x: 2 x 4 y z 15 2 x 4 y 8 z 54 8 y 9 z 69 Multiply the second equation by 5 and add to the third equation to eliminate x: 5 x 10 y 20 z 135 5 x 6 y 2 z 3 16 y 18 z 138 Multiply both sides of the first result by 2 and add to the second result to eliminate y: 16 y 18 z 138 16 y 18 z 138 00 The system is dependent. 16 y 18 z 138 18 z 138 16 y 9 69 y z 8 8 Substituting into the second equation and solving for x: 69 9 x 2 z 4 z 27 8 8 9 69 x z 4 z 27 4 4 7 39 x z 4 4 9 69 7 39 , z is The solution is x z , y z 8 8 4 4 7 39 any real number or ( x, y, z ) x z , 4 4 y
9 69 z , z is any real number . 8 8
x 4 y 3 z 15 10. 3x y 5 z 5 7 x 5 y 9 z 10
Multiply the first equation by 3 and then add the second equation to eliminate x: 3x 12 y 9 z 45 3x y 5 z 5 11y 4 z 40 Multiply the first equation by 7 and add to the
Multiply the first equation by 1 and the second 1465 Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities
third equation to eliminate x: 7 x 28 y 21z 105
3 4 0 4 3 5 1 1 1 2 5 2 4(3) 3(1) 0(5) 4(4) 3(5) 0(2) 1(3) 1(1) 2(5) 1( 4) 1(5) 2(2)
16. BC
7 x 5 y 9 z 10 33 y 12 z 115 115 3 Multiply the first result by 1 and adding it to the second result: 11y 4 z 40 115 11y 4 z 3 11 y 4 z
5 3 The system has no solution. The system is inconsistent. 0
3x 2 y 8 11. x 4 y 1 x 2 y 5z 2 12. 5 x 3z 8 2 x y 0 1 0 3 4 13. A C 2 4 1 5 1 2 5 2 1 3 0 ( 4) 4 4 2 1 4 5 3 9 1 5 2 2 4 4 1 0 6 1 6 0 6 0 14. 6 A 6 2 4 6 2 6 4 12 24 1 2 6(1) 6 2 6 12
9 31 6 3
4 6 17. A 1 3 Augment the matrix with the identity and use row operations to find the inverse: 4 6 1 0 1 3 0 1 1 3 0 1 Interchange 4 6 1 0 r1 and r2 3 0 1 1 R2 4r1 r2 0 6 1 4 1 0 1 0
1
3 0 1 16
2 3
1 0 2 1 16
1 2 3
1 Thus, A1 12 6
R2 16 r2 R1 3r2 r1
1 . 2 3
1 3 3 18. A 1 2 1 1 1 2 Augment the matrix with the identity and use row operations to find the inverse:
1 0
0 4 3 1 2 1 2 1(3) 0(1) 1(0) 0( 2) 1(4) 0(1) 2(4) 4(1) 2( 3) 4(1) 2(0) 4( 2) 1(4) 2(1) 1( 3) 2(1) 1(0) 2(2)
15. AB 2 4 1
0 4 3 12 2 8 2 5 4
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Chapter 12 Review Exercises
1 3 3 1 0 0 1 2 1 0 1 0 1 1 2 0 0 1 3 3 1 0 0 1 0 1 2 1 1 0 0 4 1 1 0 1 3 3 1 0 0 1 0 1 2 1 1 0 0 4 1 1 0 1 3 0 1 0 3 2 0 1 2 1 1 0 0 0 7 3 4 1 1 0 3 2 3 0 1 1 0 0 1 2 3 1 74 71 7 0 0
R2 r1 r2 R3 r1 r3
R2 r2 R1 3 r2 r1 R3 4 r2 r3
R3 17 r3
1 0 0 5 7 1 0 1 0 7 3 0 0 1 7
9 7 1 7 74
3 7 R1 3 r3 r1 72 R2 2 r3 r2 1 7
75 Thus, A1 17 3 7
9 7 1 7 74
3 7 72 . 1 7
3x 2 y 1 20. 10 x 10 y 5 Write the augmented matrix: 3 2 1 10 10 5
3 2 1 1 16 2
R2 3r1 r2
16 2 1 3 2 1
Interchange r1 and r2
16 2 1 0 50 5 1 16 2 1 0 1 10
R2 3r1 r2
0 1 0 1
2 5
1 10
The solution is x
R2 501 r2 R1 16r2 r1 2 1 2 1 ,y or , . 5 10 5 10
5 x 6 y 3 z 6 21. 4 x 7 y 2 z 3 3 x y 7 z 1
4 8 19. A 1 2 Augment the matrix with the identity and use row operations to find the inverse: 4 8 1 0 1 2 0 1 1 2 0 1 Interchange 4 8 1 0 r1 and r2 1 2 0 1 R2 4r1 r2 0 0 1 4 1 2 0 1 R1 r1 0 1 4 0 There is no inverse because there is no way to obtain the identity on the left side. The matrix is singular.
Write the augmented matrix: 5 6 3 6 4 7 2 3 3 1 7 1 1 1 1 9 4 7 2 3 R1 r2 r1 3 1 7 1 1 9 1 1 R2 4r1 r2 0 11 2 39 0 2 4 26 R3 3r1 r3 1 1 1 9 1 R2 11 r2 39 2 0 1 11 11 R3 1 r3 2 2 13 0 1 9 60 1 0 11 11 R1 r2 r1 39 2 0 1 11 11 R3 r2 r3 104 24 0 0 11 11
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Chapter 12: Systems of Equations and Inequalities
9 1 0 11 2 0 1 11 0 0 1
1 0 0 0 1 0 0 0 1
60 11 39 11 13 3
9 13 3 13 3
R3 11 r 24 3
9 R1 11 r3 r1 R2 2 r3 r2 11
The solution is x 9, y
13 13 ,z or 3 3
13 13 9, , . 3 3 2 x y z 5 22. 4 x y 3z 1 8x y z 5
1 0 0 1 0 0
0 2 1 4 1 53 3 0 1 34 0 0 12 1 0 23 0 1 34
R2 13 r2 R3 3r2 r3
R3 81 r3 R1 2r3 r1 4 R2 3 r3 r2
1 2 3 The solution is x , y , z or 2 3 4 1 2 3 , , . 2 3 4
Write the augmented matrix: 2 1 1 5 4 1 3 1 8 1 1 5 2 5 1 1 0 3 5 9 0 3 5 15
R2 2r1 r2 R3 4r1 r3
5 1 1 12 2 2 5 1 3 3 0 0 3 5 15
R1 12 r1 R2 1 r2 3
1 0 13 1 R1 12 r2 r1 5 3 0 1 3 R3 3 r2 r3 0 6 0 0 There is no solution; the system is inconsistent. 2z 1 x 3 23. 2 x 3 y 4x 3 y 4z 3
Write the augmented matrix: 1 1 0 2 2 3 0 3 4 3 4 3 1 1 0 2 0 3 4 5 0 3 4 1
1 0 2 1 5 4 0 1 3 3 0 3 4 1 1 0 2 1 5 4 0 1 3 3 0 0 8 6
R2 2r1 r2 R3 4r1 r3
x y z 0 24. x y 5 z 6 2 x 2 y z 1
Write the augmented matrix: 1 1 1 0 1 1 5 6 2 2 1 1 1 0 1 1 0 0 6 6 0 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 1 0 1 0 0 1 1 0 0 0 0
R2 r1 r2 R3 2r1 r3
R2 16 r2 R1 r2 r1 R3 r2 r3
The system is dependent. x y 1 z 1 The solution is x y 1 , z 1 , y is any real number or ( x, y, z ) x y 1, z 1, y is any real number .
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Chapter 12 Review Exercises
x y z t 1 2 x y z 2t 3 25. x 2 y 2 z 3t 0 3x 4 y z 5t 3 Write the augmented matrix: 1 1 1 1 1 2 1 1 2 3 1 2 2 3 0 1 5 3 3 4 1 1 1 1 1 R2 2r1 r2 0 3 1 4 1 R r1 r3 0 1 1 2 1 3 R 3 r1 r4 8 6 4 0 1 4 1 1 1 1 1 0 1 1 2 1 0 3 1 4 1 8 6 0 1 4 1 1 1 1 1 0 1 1 2 1 0 3 1 4 1 0 1 4 8 6 1 2 1 0 0 0 1 1 2 1 0 0 2 2 2 0 0 5 10 5 1 0 0 1 2 0 1 1 2 1 0 0 1 1 1 0 0 1 2 1 1 0 0 1 2 0 1 0 1 0 0 0 1 1 1 0 0 0 1 2
26.
27.
3 4 1 3
3(3) 4(1) 9 4 5
1 4 0 1 6 1 2 2 6 1 2 6 1 4 0 1 3 4 3 4 1 4 1 3 1(6 6) 4(3 24) 0(1 8) 1(0) 4( 27) 0(9) 0 108 0 108
28.
2 1 3 0 1 5 1 5 0 5 0 1 (3) 1 2 6 0 2 0 2 6 2 6 0 2(0 6) 1(0 2) 3(30 0) 2(6) 1(2) 3(30)
Interchange r2 and r3
R2 r2 R1 r2 r1 R3 3 r2 r3 R r r 4 2 4 R3 12 r3 R4 1 r4 5
R2 = r3 r2 R4 = r3 r4
1 0 0 0 4 R1 r4 r1 0 1 0 0 2 R2 r4 r2 0 0 1 0 3 R r r 4 3 3 0 0 0 1 2 The solution is x 4, y 2, z 3, t 2 or (4, 2, 3, 2) .
12 2 90 100 x 2y 4 29. 3x 2 y 4 Set up and evaluate the determinants to use Cramer’s Rule: D
1 2 1(2) 3(2) 2 6 8 3 2
Dx
4 2 4(2) 4(2) 8 8 16 4 2
Dy =
1 4 1(4) 3(4) 4 12 8 3 4
The solution is x
D Dx 16 8 1 2, y y D D 8 8
or (2, 1) . 2 x 3 y 13 0 30. 3x 2 y 0 Write the system is standard form: 2 x 3 y 13 3 x 2 y 0 Set up and evaluate the determinants to use Cramer’s Rule: 2 3 D 4 9 13 3 2 Dx
13 3 26 0 26 0 2
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Chapter 12: Systems of Equations and Inequalities
Dy 2 13 0 39 39 3 0
The solution is x y
Dy D
y
Dx 26 2, 13 D
39 3 or (2, 3). 13
Dy D
D 30 20 3 or 2, z z 10 10 D
(1, 2, 3) .
32. Let
x
y
8.
a b
2x y 2 8 16 by Theorem (14). 2a b The value of the determinant is multiplied by k when the elements of a column are multiplied by k.
Then
x 2y z 6 31. 2 x y 3z 13 3 x 2 y 3 z 16
Set up and evaluate the determinants to use Cramer’s Rule: 1 2 1 D 2 1 3 3 2 3 1
33. Let
1 3 6 2(3 6) ( 1)(4 3) 3 6 1 10 6 2 1 Dx 13 1 3 16 2 3
y
8.
y
x
8 by Theorem (11). The b a value of the determinant changes sign when any 2 columns are interchanged.
Then
2 1 1 3 1 3 2 (1) 3 2 2 3 2 3
x
a b
34. Find the partial fraction decomposition: 6 B A x( x 4) x( x 4) x ( x 4) x x 4 6 A( x 4) Bx
1 3 13 3 13 1 6 2 (1) 2 3 16 3 16 2
6 A(4 4) B (4)
Let x =4, then
4B 6
18 18 10 10 1 6 1 Dy 2 13 3 3 16 3 1
3 2 6 A(0 4) B (0)
B
6 3 6 2(39 48) (1)(26 16)
Let x = 0, then
4 A 6 A
2 3 2 13 13 3 6 (1) 3 3 3 16 16 3
1 39 48 6(6 9) (1)( 32 39) 9 18 7 20 1 2 6 Dz 2 1 13 3 2 16
3 2
3 3 6 2 2 x( x 4) x x4
35. Find the partial fraction decomposition: x4 A B C 2 2 x 1 x ( x 1) x x
2 13 2 1 1 13 1 2 6 3 16 3 2 2 16 116 26 2(32 39) 6(4 3) 10 14 6 30 D 10 The solution is x x 1 , 10 D
Multiply both sides by x 2 ( x 1) x 4 Ax( x 1) B ( x 1) Cx 2 Let x 1 , then
1 4 A(1)(1 1) B(1 1) C (1) 2 3 C C 3 Let x 0 , then
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Chapter 12 Review Exercises
0 4 A(0)(0 1) B(0 1) C (0) 2
1 5 1 B 10
2B
4 B B4 Let x 2 , then 2 4 A(2)(2 1) B (2 1) C (2) 2 2 2 A B 4C 2 A 2 4 4(3) 2A 6 A3 x4 3 4 3 2 2 x 1 x ( x 1) x x
36. Find the partial fraction decomposition: x A Bx C ( x 2 9)( x 1) x 1 x 2 9
Multiply both sides by ( x 1)( x 2 9) . x A( x 2 9) ( Bx C )( x 1) Let x 1 , then
1 A 1 9 B 1 C 1 1 2
1 A(10) ( B C )(0) 1 10 A 1 A 10 Let x 0 , then
1 1 9 x 10 10 2 10 ( x 2 9)( x 1) x 1 x 9
37. Find the partial fraction decomposition: x3 Ax B Cx D 2 2 2 x 4 ( x 2 4) 2 ( x 4)
Multiply both sides by ( x 2 4) 2 . x3 ( Ax B )( x 2 4) Cx D x3 Ax3 Bx 2 4 Ax 4 B Cx D x3 Ax3 Bx 2 (4 A C ) x 4 B D A 1; B 0 4A C 0 4(1) C 0 C 4 4B D 0 4(0) D 0 D0 x3 ( x 2 4) 2
0 A 02 9 B 0 C 0 1 0 9A C 1 0 9 C 10 9 C 10
Let x 1 , then 1 A 12 9 B 1 C 1 1 1 A(10) ( B C )(2) 1 10 A 2 B 2C 1 9 1 10 2 B 2 10 10 9 1 1 2 B 5
x
x x2 4
4x ( x 2 4) 2
38. Find the partial fraction decomposition: x2 x2 ( x 2 1)( x 2 1) ( x 2 1)( x 1)( x 1) A B Cx D x 1 x 1 x2 1
Multiply both sides by ( x 1)( x 1)( x 2 1) . x 2 A( x 1)( x 2 1) B ( x 1)( x 2 1) (Cx D)( x 1)( x 1) Let x 1 , then 12 A(1 1)(12 1) B (1 1)(12 1) (C (1) D)(1 1)(1 1) 1 4A 1 4 Let x 1 , then A
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Chapter 12: Systems of Equations and Inequalities
(1)2 A(1 1)((1) 2 1) B (1 1)((1)2 1) (C (1) D)(1 1)(1 1) 1 4B 1 4 Let x 0 , then 02 A(0 1)(02 1) B (0 1)(02 1)
40. Multiply each side of the second equation by 2 and add the equations to eliminate xy: 2 xy y 2 10 2 xy y 2 10 2 2 2 xy 6 y 2 4 xy 3 y 2 7 y 2 14
B
(C (0) D)(0 1)(0 1)
y2 2 y 2 If y 2 :
2 2 10 2 2 x 8 x 2 2 2
0 A B D
2x
1 1 D 4 4 1 D 2 Let x 2 , then 22 A(2 1)(22 1) B (2 1)(22 1) (C (2) D)(2 1)(2 1) 4 15 A 5B 6C 3D
If y 2 :
0
1 1 1 4 15 5 6C 3 4 4 2 15 5 3 6C 4 4 4 2 6C 0 C0 1 1 1 x2 4 4 22 2 2 x x 1 1 x 1 x 1 x 1
39. Solve the first equation for y, substitute into the second equation and solve: 2 x y 3 0 y 2 x 3 2 2 x y 5 x 2 ( 2 x 3) 2 5 x 2 4 x 2 12 x 9 5 5 x 2 12 x 4 0 (5 x 2)( x 2) 0 2 or x 2 5 11 y y 1 5 2 11 Solutions: , , (2, 1) . 5 5 x
2x 2 2
10 2 2 x 8 2
x 2 2
Solutions:
2 2, 2 , 2 2, 2
41. Substitute into the second equation into the first equation and solve: x 2 y 2 6 y x2 3 y 3y y2 6 y y2 3 y 0 y ( y 3) 0 y 0 or y 3 x 2 3(0) x 2 0 x 0
If y 0 :
If y 3 : x 2 3(3) x 2 9 x 3 Solutions: (0, 0), (–3, 3), (3, 3).
42. Factor the second equation, solve for x, substitute into the first equation and solve: 3x 2 4 xy 5 y 2 8 2 2 x 3 xy 2 y 0 x 2 3 xy 2 y 2 0 ( x 2 y )( x y ) 0 x 2 y or x y Substitute x 2 y and solve: 3 x 2 4 xy 5 y 2 8 3( 2 y ) 2 4( 2 y ) y 5 y 2 8 12 y 2 8 y 2 5 y 2 8 9 y2 8 y2
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8 2 2 y 9 3
Chapter 12 Review Exercises y
Substitute x y and solve:
3x 2 4 xy 5 y 2 8
x y 12
3( y ) 2 4( y ) y 5 y 2 8 3y2 4 y2 5 y2 8
x
4 y2 8
y2 2 y 2
2 2 If y : 3
2 2 4 2 x 2 3 3
2 2 : 3
2 2 4 2 x 2 3 3
If y
If y 2 :
45. y x 2
Graph the parabola y x 2 . Use a solid curve since the inequality uses . Choose a test point not on the parabola, such as (0, 1). Since 0 12 is false, shade the opposite side of the parabola from (0, 1).
x 2
y
If y 2 : x 2 Solutions: 4 2 2 2 4 2 2 2 , , , , 2, 2 , 3 3 3 3
y x2
2, 2
x 2 3x y 2 y 2 43. x 2 x y 1 0 y Multiply each side of the second equation by –y and add the equations to eliminate y: x 2 3x y 2 y 2 x2 x y2 y 0 2 x 2 x 1 If x 1: 12 3(1) y 2 y 2
y2 y 0 y ( y 1) 0 y 0 or y 1 Note that y 0 because that would cause division by zero in the original system. Solution: (1, –1)
44. 3 x 4 y 12 Graph the line 3 x 4 y 12 . Use a solid line since the inequality uses . Choose a test point not on the line, such as (0, 0). Since 3 0 4 0 12 is true, shade the side of the line
containing (0, 0).
x
2 x y 2 46. x y 2 Graph the line 2 x y 2 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 2 is true, shade the side of the line containing (0, 0). Graph the line x y 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 0 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. y 5
x+y=2 x –5
5
–2x + y = 2 –5
The graph is unbounded. Find the vertices: To find the intersection of x y 2 and 2 x y 2 , solve the system: x y 2 2 x y 2
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Chapter 12: Systems of Equations and Inequalities
Solve the first equation for x: x 2 y . Substitute and solve: 2(2 y ) y 2 4 2y y 2 3y 6 y2 x 22 0 The point of intersection is (0, 2). The corner point is (0, 2). x0 y0 47. x y 4 2 x 3 y 6 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x y 4 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 4 is true, shade the side of the line containing (0, 0). Graph the line 2 x 3 y 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 3(0) 6 is true, shade the side of the line containing (0, 0). y 8
x+ y=4 (0, 2) (0, 0) (3, 0) –2 2x + 3y = 6
x 8
The overlapping region is the solution. The graph is bounded. Find the vertices: The x-axis and yaxis intersect at (0, 0). The intersection of 2 x 3 y 6 and the y-axis is (0, 2). The intersection of 2 x 3 y 6 and the x-axis is (3, 0). The three corner points are (0, 0), (0, 2), and (3, 0). x0 0 y 48. 2 x y 8 x 2 y 2 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line 2 x y 8 . Use a solid
line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≤ 8 is true, shade the side of the line containing (0, 0). Graph the line x 2 y 2 . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) ≥ 2 is false, shade the opposite side of the line from (0, 0). y 9
(0, 8)
2x + y = 8 (0, 1) x + 2y = 2
(4, 0)
–1–1
x 9
(2, 0)
The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of x 2 y 2 and the y-axis is (0, 1). The intersection of x 2 y 2 and the x-axis is (2, 0). The intersection of 2 x y 8 and the y-axis is (0, 8). The intersection of 2 x y 8 and the xaxis is (4, 0). The four corner points are (0, 1), (0, 8), (2, 0), and (4, 0). 49. Graph the system of inequalities: 2 2 x y 16 x y 2
Graph the circle x 2 y 2 16 .Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since 02 02 16 is true, shade the side of the circle containing (0, 0). Graph the line x y 2 . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 0 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. y 5
x2+ y2 = 16 x
–5
5
–5
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x+y=2
Chapter 12 Review Exercises 50. Graph the system of inequalities: 2 y x xy 4
Graph the parabola y x 2 . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (1, 2). Since 2 12 is false, shade the opposite side of the parabola from (1, 2). Graph the hyperbola xy 4 . Use a solid line since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (1, 2). Since 1 2 4 is true, shade the same side of the hyperbola as (1, 2). The overlapping region is the solution. y 5
y = x2
–5 5 x xy = 4 –5
51. Maximize z 3x 4 y subject to x 0 , y 0 , 3x 2 y 6 , x y 8 . Graph the constraints. y (0,8)
(0,3)
(2,0)
52. Minimize z 3 x 5 y subject to x 0 , y 0 , x y 1 , 3x 2 y 12 , x 3 y 12 . Graph the constraints. y
(0,4)
127 , 247
(0,1)
(4,0)
x
(1,0)
To find the intersection of 3x 2 y 12 and x 3 y 12 , solve the system: 3x 2 y 12 x 3 y 12 Solve the second equation for x: x 12 3 y Substitute and solve: 3(12 3 y ) 2 y 12 36 9 y 2 y 12 7 y 24 24 y 7 72 12 24 x 12 3 12 7 7 7 12 24 The point of intersection is , . 7 7 The corner points are (0, 1), (1, 0), (0, 4), (4, 0), 12 24 , . 7 7
(8,0) x
The corner points are (0, 3), (2, 0), (0, 8), (8, 0). Evaluate the objective function: Vertex Value of z 3 x 4 y (0, 3) z 3(0) 4(3) 12 (0, 8) z 3(0) 4(8) 32 z 3(2) 4(0) 6 (2, 0) (8, 0) z 3(8) 4(0) 24 The maximum value is 32 at (0, 8).
Evaluate the objective function: Vertex Value of z 3x 5 y z 3(0) 5(1) 5 (0, 1) z 3(0) 5(4) 20 (0, 4) z 3(1) 5(0) 3 (1, 0) z 3(4) 5(0) 12 (4, 0) 12 24 12 24 156 , z 3 5 7 7 7 7 7 The minimum value is 3 at (1, 0).
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Chapter 12: Systems of Equations and Inequalities
2 x 5 y 5 53. 4 x 10 y A Multiply each side of the first equation by –2 and eliminate x: 4 x 10 y 10 4 x 10 y A 0 A 10 If there are to be infinitely many solutions, the result of elimination should be 0 = 0. Therefore, A 10 0 or A 10 .
2 x 5 y 5 54. 4 x 10 y A Multiply each side of the first equation by –2 and eliminate x: 4 x 10 y 10 4 x 10 y A 0 A 10 If the system is to be inconsistent, the result of elimination should be 0 = any number except 0. Therefore, A 10 0 or A 10 .
55. y ax 2 bx c At (0, 1) the equation becomes: 1 a (0) 2 b(0) c c 1
At (1, 0) the equation becomes: 0 a(1) 2 b(1) c 0 abc abc 0 At (–2, 1) the equation becomes: 1 a( 2) 2 b( 2) c 1 4a 2b c 4a 2b c 1 The system of equations is: a bc 0 4a 2b c 1 c1 Substitute c 1 into the first and second equations and simplify: 4a 2b 1 1 a b 1 0 a b 1 4a 2b 0 a b 1 Solve the first equation for a, substitute into the second equation and solve:
4(b 1) 2b 0 4b 4 2b 0 6b 4 b a
2 3
2 1 1 3 3
1 2 The quadratic function is y x 2 x 1 . 3 3
56. Let x = the number of pounds of coffee that costs $6.00 per pound, and let y = the number of pounds of coffee that costs $9.00 per pound. Then x y 100 represents the total amount of coffee in the blend. The value of the blend will be represented by the equation: 6 x 9 y 6.90(100) . Solve the system of equations: x y 100 6 x 9 y 690 Solve the first equation for y: y 100 x . Solve by substitution: 6 x 9(100 x) 690 6 x 900 9 x 690 3x 210 x 70 y 100 70 30 The blend is made up of 70 pounds of the $6.00per-pound coffee and 30 pounds of the $9.00per-pound coffee. 57. Let x = the number of small boxes, let y = the number of medium boxes, and let z = the number of large boxes. Oatmeal raisin equation: x 2 y 2 z 15 Chocolate chip equation: x y 2 z 10 Shortbread equation: y 3 z 11 x 2 y 2 z 15 x y 2 z 10 y 3 z 11
Multiply each side of the second equation by –1 and add to the first equation to eliminate x: x 2 y 2 z 15 x y 2 z 10 y 3 z 11
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y 5
Chapter 12 Review Exercises
Substituting and solving for the other variables: 5 3z 11 x 5 2(2) 10 x 9 10 3z 6 x 1 z2 Thus, 1 small box, 5 medium boxes, and 2 large boxes of cookies should be purchased. 58. a.
Let x = the number of lower-priced packages, and let y = the number of quality packages. 8 x 6 y 120(16) Peanut inequality: 4 x 3 y 960 Cashew inequality: 4 x 6 y 72(16) 2 x 3 y 576 The system of inequalities is:
59. Let x = the speed of the boat in still water, and let y = the speed of the river current. The distance from Chiritza to the Flotel Orellana is 100 kilometers. Rate Time Distance trip downstream x y 5 / 2 100 trip downstream x y 3 100
The system of equations is: 5 ( x y ) 100 2 3( x y ) 100 Multiply both sides of the first equation by 6, multiply both sides of the second equation by 5, and add the results. 15 x 15 y 600 15 x 15 y 500
x0 y0 4 x 3 y 960 2 x 3 y 576
30 x 1100 1100 110 x 30 3 110 3 3 y 100 3 110 3 y 100 10 3 y 10 y 3 The speed of the boat is 110 / 3 36.67 km/hr ; the speed of the current is 10 / 3 3.33 km/hr .
b. Graphing:
To find the intersection of 2 x 3 y 576 and 4 x 3 y 960 , solve the system: 4 x 3 y 960 2 x 3 y 576 Subtract the second equation from the first: 4 x 3 y 960 2 x 3 y 576 2 x 384 x 192 Substitute and solve: 2(192) 3 y 576 3 y 192 y 64 The corner points are (0, 0), (0, 192), (240, 0), and (192, 64).
60. Let x = the number of hours for Bruce to do the job alone, let y = the number of hours for Bryce to do the job alone, and let z = the number of hours for Marty to do the job alone. Then 1/x represents the fraction of the job that Bruce does in one hour. 1/y represents the fraction of the job that Bryce does in one hour. 1/z represents the fraction of the job that Marty does in one hour. The equation representing Bruce and Bryce working together is: 1 1 1 3 0.75 x y 4 / 3 4
The equation representing Bryce and Marty working together is: 1 1 1 5 0.625 y z 8 / 5 6 The equation representing Bruce and Marty
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Chapter 12: Systems of Equations and Inequalities
working together is: 1 1 1 3 0.375 x z 8 / 3 8
y
Solve the system of equations: x 1 y 1 0.75 1 1 y z 0.625 1 1 x z 0.375 Let u x 1 , v y 1 , w z 1
(20,40)
(60,40)
(20,30) (60,15) (35,15) x
u v 0.75 v w 0.625 u w 0.375
Solve the first equation for u: u 0.75 v . Solve the second equation for w: w 0.625 v . Substitute into the third equation and solve: (0.75 v) (0.625 v) 0.375 2v 1 v 0.5 u 0.75 0.5 0.25 w 0.625 0.5 0.125 Solve for x, y, and z : x 4, y 2, z 8 (reciprocals) Bruce can do the job in 4 hours, Bryce in 2 hours, and Marty in 8 hours. 61. Let x = the number of gasoline engines produced each week, and let y = the number of diesel engines produced each week. The total cost is: C 450 x 550 y . Cost is to be minimized; thus, this is the objective function. The constraints are: 20 x 60 number of gasoline engines needed and capacity each week. 15 y 40 number of diesel engines needed and capacity each week. x y 50 number of engines produced to prevent layoffs. Graph the constraints.
The corner points are (20, 30), (20, 40), (35, 15), (60, 15), (60, 40) Evaluate the objective function: Vertex Value of C 450 x 550 y (20, 30) C 450(20) 550(30) 25,500 (20, 40) C 450(35) 550(40) 31, 000 (35, 15) C 450(35) 550(15) 24, 000 (60, 15) C 450(60) 550(15) 35, 250 60, 40 C 450 60 550 40 49, 000 The minimum cost is $24,000, when 35 gasoline engines and 15 diesel engines are produced. The excess capacity is 15 gasoline engines, since only 20 gasoline engines had to be delivered. 62. Answers will vary.
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Chapter 12 Test
Chapter 12 Test 1. 2 x y 7 4x 3y 9 Substitution: We solve the first equation for y, obtaining y 2x 7 Next we substitute this result for y in the second equation and solve for x. 4x 3 y 9 4x 3 2x 7 9 4 x 6 x 21 9 10 x 30 30 x 3 10 We can now obtain the value for y by letting x 3 in our substitution for y. y 2x 7
y 2 3 7 6 7 1 The solution of the system is x 3 , y 1 or (3, 1) . Elimination: Multiply each side of the first equation by 2 so that the coefficients of x in the two equations are negatives of each other. The result is the equivalent system 4 x 2 y 14 4x 3y 9 We can replace the second equation of this system by the sum of the two equations. The result is the equivalent system 4 x 2 y 14 5 y 5 Now we solve the second equation for y. 5 y 5 5 y 1 5 We back-substitute this value for y into the original first equation and solve for x. 2 x y 7 2 x 1 7 2 x 6 6 3 2 The solution of the system is x 3 , y 1 or (3, 1) . x
1 2. x 2 y 1 3 5 x 30 y 18 We choose to use the method of elimination and multiply the first equation by 15 to obtain the equivalent system 5 x 30 y 15 5 x 30 y 18
We replace the second equation by the sum of the two equations to obtain the equivalent system 5 x 30 y 15 03 The second equation is a contradiction and has no solution. This means that the system itself has no solution and is therefore inconsistent. x y 2 z 5 (1) 3. 3 x 4 y z 2 (2) 5 x 2 y 3z 8 (3)
We use the method of elimination and begin by eliminating the variable y from equation (2). Multiply each side of equation (1) by 4 and add the result to equation (2). This result becomes our new equation (2). x y 2z 5 4 x 4 y 8 z 20 3x 4 y z 2
3 x 4 y z 2 7x
7 z 18 (2)
We now eliminate the variable y from equation (3) by multiplying each side of equation (1) by 2 and adding the result to equation (3). The result becomes our new equation (3). x y 2z 5 2x 2 y 4 z 10 5 x 2 y 3z 8
5x 2 y 3 z 8 7x
7 z 18 (3)
Our (equivalent) system now looks like x y 2 z 5 (1) 7 z 18 (2) 7 x 7 x 7 z 18 (3) Treat equations (2) and (3) as a system of two equations containing two variables, and eliminate the x variable by multiplying each side of equation (2) by 1 and adding the result to equation (3). The result becomes our new equation (3).
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Chapter 12: Systems of Equations and Inequalities 7 x 7 z 18
7 x 7 z 18
7 x 7 z 18
7 x 7 z 18
0 0 (3) We now have the equivalent system x y 2 z 5 (1) 7 z 18 (2) 7 x 0 0 (3)
This is equivalent to a system of two equations with three variables. Since one of the equations contains three variables and one contains only two variables, the system will be dependent. There are infinitely many solutions. We solve equation (2) for x and determine that 18 x z . Substitute this expression into 7 equation (1) to obtain y in terms of z. x y 2z 5 18 z y 2z 5 7 18 z y 2z 5 7 17 y z 7 17 7 18 17 The solution is x z , y z , 7 7 18 z is any real number or ( x, y, z ) x z , 7 17 y z , z is any real number . 7 y z
3x 2 y 8 z 3 (1) 4. x 23 y z 1 (2) 6 x 3 y 15 z 8 (3) We start by clearing the fraction in equation (2) by multiplying both sides of the equation by 3. 3x 2 y 8 z 3 (1) 3x 2 y 3z 3 (2) 6 x 3 y 15 z 8 (3)
We use the method of elimination and begin by eliminating the variable x from equation (2). The coefficients on x in equations (1) and (2) are negatives of each other so we simply add the two equations together. This result becomes our new
equation (2). 3x 2 y 8 z 3 3 x 2 y 3z 3 5 z 0 (2)
We now eliminate the variable x from equation (3) by multiplying each side of equation (1) by 2 and adding the result to equation (3). The result becomes our new equation (3). 3x 2 y 8 z 3 6 x 4 y 16 z 6 6 x 3 y 15 z 8 6 x 3 y 15 z 8 7 y 31z 14 (3)
Our (equivalent) system now looks like 3x 2 y 8 z 3 (1) 5 z 0 (2) 7 y 31z 14 (3) We solve equation (2) for z by dividing both sides of the equation by 5 . 5 z 0 z0 Back-substitute z 0 into equation (3) and solve for y. 7 y 31z 14 7 y 31(0) 14 7 y 14 y 2 Finally, back-substitute y 2 and z 0 into equation (1) and solve for x. 3x 2 y 8 z 3 3x 2(2) 8(0) 3 3x 4 3 3x 1 1 x 3 The solution of the original system is 1 1 x , y 2 , z 0 or , 2, 0 . 3 3 5. 4 x 5 y z 0 2 x y 6 19 x 5 y 5 z 10
We first check the equations to make sure that all variable terms are on the left side of the equation and the constants are on the right side. If a variable is missing, we put it in with a coefficient of 0. Our system can be rewritten as
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Chapter 12 Test 3 3 matrix. 4 6 1 2 5 CB 1 3 0 3 1 1 8
4x 5 y z 0 2 x y 0 z 25 x 5 y 5 z 10
The augmented matrix is 0 4 5 1 2 1 0 25 1 5 5 10
4( 2) 6 3 4 5 6 1 4 1 6 0 1 1 ( 3)0 1( 2) ( 3)3 1 5 ( 3)1 ( 1)1 8 0 ( 1)( 2) 8 3 ( 1)5 8 1
6. The matrix has three rows and represents a system with three equations. The three columns to the left of the vertical bar indicate that the system has three variables. We can let x, y, and z denote these variables. The column to the right of the vertical bar represents the constants on the right side of the equations. The system is 3 x 2 y 4 z 6 3x 2 y 4 z 6 x 8z 2 or 1x 0 y 8 z 2 2 x 1y 3 z 11 2 x y 3 z 11 1 1 4 6 7. 2 A C 2 0 4 1 3 3 2 1 8 2 2 4 6 6 4 0 8 1 3 1 11 6 4 1 8 5 12 1 8. A 3C 0 3 1 0 3
1 4 6 4 3 1 3 2 1 8 1 12 18 11 19 5 4 3 9 3 2 3 24 6 22
9. Here we are taking the product of a 3 2 matrix and a 2 3 matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (2 in both cases), the operation can be performed and will result in a
4 10 26 1 11 2 1 26 3
10. Here we are taking the product of a 2 3 matrix and a 3 2 matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (3 in both cases), the operation can be performed and will result in a 2 2 matrix. 1 1 1 2 5 BA 0 4 0 3 1 3 2 1 1 2 0 5 3 1 1 2 4 5 2 0 1 3 0 1 3 0 1 3 4 1 2 16 17 3 10
11. We first form the matrix 3 2 1 0 A | I2 5 4 0 1
Next we use row operations to transform A | I 2 into reduced row echelon form. 1 23 13 0 3 2 1 0 5 4 0 1 5 4 0 1
1 23 13 0 5 0 23 3 1 1 23 13 0 5 3 0 1 2 2 1 0 2 1 5 3 0 1 2 2 2 Therefore, A1 5 2
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R 13 r 1
1
R2 5r1 r2
R 32 r 2
2
R 23 r r 1
1 3 . 2
2
1
Chapter 12: Systems of Equations and Inequalities 12. We first form the matrix 1 1 1 1 0 0 B | I 3 2 5 1 0 1 0 2 3 0 0 0 1 Next we use row operations to transform B | I 3
6 3 12 2 1 2 2 1 2 6 3 12
R1 r2 R2 r1
1 12 1 6 3 12
R1 12 r1
1 12 1 0 6 18 1 12 1 3 0 1
into reduced row echelon form. 1 1 1 1 0 2 5 1 0 1 2 3 0 0 0 1 1 1 1 0 7 3 2 0 5 2 2 1 0 0 1 0 0
1 0 0 1 0 0
1
1
1
73
0 0 1
0 0 R2 2r1 r2 1 0 R 2r1 r3 0 1 3 1 0 0 72 71 0 R2 71 r2 2 0 1
5
2
0
4 7 73 1 7
5 7 72 74
1 7 1 7 75
4 7 73
5 7 72
1 7 1 7
1 0
0 1 0 0 1 0
0 0 1
0 0 1 4 5 7 0 3 3 4 0 2 2 3 1 4 5 7
R1 r2 r1 R3 5r2 r3
R3 7r3 R1 74 r3 r1 R 3r r 2 7 3 2
3 3 4 1 Thus, B 2 2 3 4 5 7
13. 6 x 3 y 12 2 x y 2 We start by writing the augmented matrix for the system. 6 3 12 2 1 2 Next we use row operations to transform the augmented matrix into row echelon form.
1 0 12 0 1 3
R2 6r1 r2
R2 16 r2
R2 12 r2 r1
The solution of the system is x 12 , y 3 or
12 , 3
1 14. x y 7 4 8 x 2 y 56 We start by writing the augmented matrix for the system. 1 14 7 8 2 56
Next we use row operations to transform the augmented matrix into row echelon form. 1 14 7 8 2 56 R2 8R1 r2 1 14 7 0 0 0 The augmented matrix is now in row echelon form. Because the bottom row consists entirely of 0’s, the system actually consists of one equation in two variables. The system is dependent and therefore has an infinite number of solutions. Any ordered pair satisfying the 1 equation x y 7 , or y 4 x 28 , is a 4 solution to the system.
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Chapter 12 Test
15. x 2 y 4 z 3 2 x 7 y 15 z 12 4 x 7 y 13z 10
16. 2 x 2 y 3 z 5 x y 2z 8 3x 5 y 8 z 2
We start by writing the augmented matrix for the system. 1 2 4 3 2 7 15 12 4 7 13 10
We start by writing the augmented matrix for the system. 2 2 3 5 1 1 2 8 3 5 8 2
Next we use row operations to transform the augmented matrix into row echelon form. 1 2 4 3 2 7 15 12 4 7 13 10
Next we use row operations to transform the augmented matrix into row echelon form. 2 2 3 5 1 1 2 8 3 5 8 2
1 0 0 1 0 0
3 3 7 6 1 3 2 2 4 3 1 3 2 3 7 6 2
4
1 1 2 8 2 2 3 5 3 5 8 2 8 1 1 2 0 4 7 11 0 8 14 26
R2 2r1 r2 R3 4r1 r3 R2 r3 R3 r2
1 2 4 3 0 1 3 2 R3 3r2 r3 0 0 2 0 1 2 4 3 0 1 3 2 R3 12 r3 0 0 1 0 The matrix is now in row echelon form. The last row represents the equation z 0 . Using z 0 we back-substitute into the equation y 3z 2 (from the second row) and obtain y 3 z 2 y 3 0 2
1 1 2 8 0 1 7 11 4 4 0 8 14 26
x 2 2 4 0 3 x 1 The solution is x 1 , y 2 , z 0 or (1, 2, 0) .
R2 2r1 r2 R3 3r1 r3
R2 14 r2
1 1 2 8 0 1 7 11 R3 8r2 r3 4 4 4 0 0 0 The last row represents the equation 0 4 which is a contradiction. Therefore, the system has no solution and is be inconsistent.
17.
y 2 Using y 2 and z 0 , we back-substitute into the equation x 2 y 4 z 3 (from the first row) and obtain x 2 y 4 z 3
R1 r2 R2 r1
18.
2 5
2 7 5 3 14 15 29
3
7
2
4
1
4
0
1
2
4
4
0
2
6
2 4
(4)
1
0
1 4
6
1
4
1 2
2 4(4) 2(0) 4 1(4) (1)(0) 6 1(2) (1)4 2(16) 4(4) 6(6) 32 16 36 12
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Chapter 12: Systems of Equations and Inequalities
19. 4 x 3 y 23 3 x 5 y 19 The determinant D of the coefficients of the variables is 4 3 D 4 5 3 3 20 9 29 3 5 Since D 0 , Cramer’s Rule can be applied. 23 3 Dx 23 5 319 58 19 5 Dy
4 23 3
4 19 23 3 145
19
Dx 58 2 D 29 Dy 145 y 5 D 29 The solution of the system is x 2 , y 5 or (2, 5) . x
20. 4 x 3 y 2 z 15 2 x y 3 z 15 5 x 5 y 2 z 18
The determinant D of the coefficients of the variables is 4 3 2 D 2 1 3 5 5 2 4
1
3
5
2
3
2 3 5
2
2
2
1
5
5
5
2
1
3
5
2
3
15
15 3 18
2
2
9
2
15 3 18
2
15
2 3 5
2
2
2 15 5
18
4 30 54 15 4 15 2 36 75 4 24 15 11 2 39 9 4
3
15
Dz 2
1
15
5
5
18
1
15
5
18
3
4
2 15 5
18
15
2
1
5
5
4 18 75 3 36 75 15 10 5 4 57 3 39 15 5 36 Dy D 9 9 1 , x x 1, y D 9 D 9 D 36 4 z z 9 D The solution of the system is x 1 , y 1 ,
x 2 3x 4 0 ( x 1)( x 4) 0 x 1 or x 4 Back substitute these values into the second equation to determine y: x 1 : y 2 9(1) 9
15
1
y 3
18
5
x 4 : y 2 9(4) 36
15 2 15 3 30 54 2 75 18 15 13 3 24 2 57
4
18
3x 2 9 x 12 0
9 Since D 0 , Cramer’s Rule can be applied. 15 3 2 18
5
Substitute 9x for y 2 into the first equation and solve for x: 3x 2 9 x 12
52 33 10
3
2
2 2 3x y 12 21. y2 9x
4 13 3 11 2 5
1
15
z 4 or (1, 1, 4) .
4 2 15 3 4 15 2 10 5
Dx 15
4
Dy 2 15 3
y 36 (not real) The solutions of the system are (1, 3) and (1, 3) .
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Chapter 12 Test
2 y 2 3x 2 5 22. y x 1 y x 1 Substitute x 1 for y into the first equation and solve for x:
24.
x 3
2
The denominator contains the repeated linear factor x 3 . Thus, the partial fraction decomposition takes on the form 3x 7 A B 2 2 x 3 x 3 x 3
2 x 1 3x 2 5 2
3x 7
2 x 2 2 x 1 3x 2 5 2 x 2 4 x 2 3x 2 5
Clear the fractions by multiplying both sides by
x 3 . The result is the identity 3x 7 A x 3 B 2
x 4x 3 0 2
x2 4x 3 0 ( x 1)( x 3) 0 x 1 or x 3 Back substitute these values into the second equation to determine y: x 1 : y 11 2 x 3 : y 3 1 4 The solutions of the system are (1, 2) and (3, 4) .
or 3x 7 Ax 3 A B We equate coefficients of like powers of x to obtain the system 3 A 7 3 A B Therefore, we have A 3 . Substituting this result into the second equation gives 7 3A B
x 2 y 2 100 23. 4 x 3 y 0
7 3 3 B 2
2 B Thus, the partial fraction decomposition is 2 3x 7 3 . 2 2 x 3 x x 3 3
2
Graph the circle x y 100 . Use a solid curve since the inequality uses . Choose a test point not on the circle, such as (0, 0). Since 02 02 100 is true, shade the same side of the circle as (0, 0); that is, inside the circle. Graph the line 4 x 3 y 0 . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 1). Since 4(0) 3(1) 0 is false, shade the opposite side of the line from (0, 1). The overlapping region is the solution.
25.
4x2 3
x x2 3
2
The denominator contains the linear factor x and the repeated irreducible quadratic factor x 2 3 . The partial fraction decomposition takes on the form 4 x2 3 A Bx C Dx E 2 2 2 2 x x 3 x x 3 x2 3
We clear the fractions by multiplying both sides
by x x 2 3
to obtain the identity
2
4 x2 3 A x2 3
x x 3 Bx C x Dx E 2
2
Collecting like terms yields 4 x 2 3 A B x 4 Cx 3 6 A 3B D x 2 3C E x 9 A
Equating coefficients, we obtain the system
1485 Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities 2x 3y 2
A B 0 C0 6 A 3 B D 4 3C E 0 9 A 3
2 2 x 3 3 Test the point 0, 0 . y
2x 3 y 2
1 From the last equation we get A . 3 Substituting this value into the first equation 1 gives B . From the second equation, we 3 know C 0 . Substituting this value into the fourth equation yields E 0 . 1 1 Substituting A and B into the third 3 3 equation gives us 6 13 3 13 D 4
2 0 30 2 ? 0 2 false The point 0, 0 is not a solution. Thus, the
graph of the inequality 2 x 3 y 2 includes the 2 2 x . 3 3 Because the inequality is non-strict, the line is also part of the graph of the solution. The overlapping shaded region (that is, the shaded region in the graph below) is the solution to the system of linear inequalities. half-plane below the line y
2 1 D 4 D5 Therefore, the partial fraction decomposition is 4x2 3 x x 3 2
2
1 3
x
1 x 3
5x
x 3 x 3 2 2
2
26. x 0 y 0 x 2 y 8 2 x 3 y 2
The inequalities x 0 and y 0 require that the graph be in quadrant I. x 2y 8 1 y x4 2 Test the point 0, 0 . x 2y 8
0 2 0 8 ? 0 8 false The point 0, 0 is not a solution. Thus, the
graph of the inequality x 2 y 8 includes the 1 half-plane above the line y x 4 . Because 2 the inequality is non-strict, the line is also part of the graph of the solution.
The graph is unbounded. The corner points are 4, 2 and 8, 0 . 27. The objective function is z 5 x 8 y . We seek the largest value of z that can occur if x and y are solutions of the system of linear inequalities x 0 2 x y 8 x 3 y 3 2x y 8 x 3 y 3 y 2 x 8 3 y x 3 1 y x 1 3 The graph of this system (the feasible points) is shown as the shaded region in the figure below. The corner points of the feasible region are 0,1 , 3, 2 , and 0,8 .
1486 Copyright © 2025 Pearson Education, Inc.
Chapter 12 Cumulative Review y 8
Substituting this result into y z 2.5 (from the second row) gives y z 2.5 y 6 2.5 y 8.5 Substituting z 6 into x 3 z 42.5 (from the first row) gives x 3z 42.5
x 3 y 3
x 3 6 42.5
x
4
x 24.5 Thus, flare jeans cost $24.50, camisoles cost $8.50, and t-shirts cost $6.00.
2x y 8
Corner point, x, y Value of obj. function, z
0,1 3, 2 0,8
z 5 0 8 1 8
z 5 3 8 2 31
z 5 0 8 8 64
From the table, we can see that the maximum value of z is 64, and it occurs at the point 0,8 . 28. Let j = unit price for flare jeans, c = unit price for camisoles, and t = unit price for t-shirts. The given information yields a system of equations with each of the three women yielding an equation. 2 j 2c 4t 90 (Megan) 3t 42.5 (Paige) j j 3c 2t 62 (Kara)
We can solve this system by using matrices. 2 2 4 90 1 1 2 45 1 0 3 42.5 1 0 3 42.5 R1 12 r1 1 3 2 62 1 3 2 62
1 1 2 45 0 1 1 2.5 0 2 0 17
R2 r1 r2 R3 r1 r3
1 1 2 45 0 1 1 2.5 0 2 0 17
R2 r2
1 0 3 42.5 0 1 1 2.5 0 0 2 12
R1 r2 r1 R3 2r2 r3
1 0 3 42.5 0 1 1 2.5 R3 12 r3 6 0 0 1 The last row represents the equation z 6 .
Chapter 12 Cumulative Review 1. 2 x 2 x 0 x 2 x 1 0 x 0 or 2 x 1 0 2x 1 1 2 1 The solution set is 0, . 2 x
2.
3x 1 4
3x 1 4 2
2
3 x 1 16 3x 15 x5 Check: 35 1 4 15 1 4 16 4 44 The solution set is 5 .
3. 2 x3 3x 2 8 x 3 0 The graph of Y1 2 x3 3x 2 8 x 3 appears to have an x-intercept at x 3 .
Using synthetic division:
1487 Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities
32
3 6
2
3
8 9
Thus, g is an odd function and its graph is symmetric with respect to the origin.
3 3
1
0
8.
x 2 y 2 2 x 4 y 11 0
Therefore, 2 x 3x 8 x 3 0 3
2
x 2 2 x y 2 4 y 11
x 3 2 x 3x 1 0 x 3 2 x 1 x 1 0 2
( x 2 2 x 1) ( y 2 4 y 4) 11 1 4 ( x 1) 2 ( y 2) 2 16 Center: (1,–2); Radius: 4
1 or x 1 2 1 The solution set is 1, ,3 . 2 x x1 4. 3 9
x 3 or x
3x 32
x 1
3x 32 x 2 x 2x 2 x 2 The solution set is 2 .
9.
f ( x ) 3x 2 1
Using the graph of y 3x , shift the graph horizontally 2 units to the right, then shift the graph vertically upward 1 unit.
5. log 3 x 1 log 3 2 x 1 2 log 3 x 1 2 x 1 2
x 1 2 x 1 32 2 x2 x 1 9 2 x 2 x 10 0
2 x 5 x 2 0 5 or x 2 2 Since x 2 makes the original logarithms 5 undefined, the solution set is . 2 x
6.
3x e
x ln 3 1 1 0.910 ln 3 1 0.910 . The solution set is ln 3 x
2 x3 x4 1
g ( x)
10.
f ( x)
5 x2
5 x2 5 x Inverse y2 x( y 2) 5 xy 2 x 5 xy 5 2 x 5 2x 5 y 2 x x 5 Thus, f 1 ( x) 2 x y
ln 3x ln e
7. g ( x)
Domain: (, ) Range: (1, ) Horizontal Asymptote: y 1
2x
3
x 1 4
2 x3 g x x4 1
Domain of f = {x | x 2} Range of f = { y | y 0}
1488 Copyright © 2025 Pearson Education, Inc.
Chapter 12 Cumulative Review
Domain of f 1 = {x | x 0} Range of f 11. a.
1
y 3x 6 The graph is a line. x-intercept: 0 3x 6
3x 6 x 2
b.
= { y | y 2} .
y
e.
y x
y-intercept: y 3 0 6 6
x2 y 2 4 The graph is a circle with center (0, 0) and radius 2.
f. y e x
g.
c.
1 x
d.
y ln x
y x3
1489 Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities
h.
2 x2 5 y 2 1 The graph is an ellipse. x2 y2 1 1 2
j.
x2 2 x 4 y 1 0 x2 2 x 1 4 y 4 y ( x 1) 2 1 y ( x 1) 2 4
1 5
2
2
x y 1 2 5 2 5
12.
i.
x 3y 1 The graph is a hyperbola x2 y 2 1 1 1 2
f ( x) x3 3x 5
a.
2
3 2
Let Y1 x3 3x 5 .
2
x y 1 1 3 3
The zero of f is approximately 2.28 .
b.
f has a local maximum of 7 at x 1 and a local minimum of 3 at x 1 . c.
f is increasing on the intervals , 1
and 1, .
1490 Copyright © 2025 Pearson Education, Inc.
Chapter 12 Projects Project II
Chapter 12 Projects
a. 2 2 2 2 16 codewords.
Project I – Internet-based Project 1. 80% = 0.80 40% = 0.40 20% = 0.20
18% = 0.18 50% = 0.50 60% = 0.60
2% = 0.02 10% = 0.10 20% = 0.20
0.80 0.18 0.02 2. 0.40 0.50 0.10 0.20 0.60 0.20
3. 0.80 0.18 0.02 1.00 0.40 0.50 0.10 1.00 0.20 0.60 0.20 1.00 The sum of each row is 1 (or 100%). These represent the three possibilities of educational achievement for a parent of a child, unless someone does not attend school at all. Since these are rounded percents, chances are the other possibilities are negligible. 2
0.8 0.18 0.02 2 4. P 0.4 0.5 0.1 0.2 0.6 0.2 0.716 0.246 0.038 0.54 0.382 0.078 0.44 0.456 0.104
Grandchild of a college graduate is a college graduate: entry (1, 1): 0.716. The probability is 71.6% 5. Grandchild of a high school graduate finishes college: entry (2,1): 0.54. The probability is 54%. 6. grandchildren → k = 2. v (2) v (0) P 2 0.716 0.246 0.038 [0.317 0.565 0.118] 0.54 0.382 0.078 0.44 0.456 0.104 [0.583992 0.34762 0.068388]
College: 58.4% High School: 34.8% Elementary: 6.8% 7. The matrix totally stops changing at 0.64885496 0.29770992 0.05343511 30 P 0.64885496 0.29770992 0.05343511 0.64885496 0.29770992 0.05343511
b. v uG u will be the matrix representing all of the 4-digit information bit sequences. 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 u 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 (Remember, this is mod two. That means that you only write down the remainder when dividing by 2. ) v uG 0 0 0 0 0 0 0 0 v 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 1 0 1 1 0 1 1 1 0 0 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 0 1 0 1 0 1 0 1 1 0 0 0 1 1 1 1 1 1 1
1491 Copyright © 2025 Pearson Education, Inc.
Chapter 12: Systems of Equations and Inequalities c. Answers will vary, but if we choose the 6th row and the 10th row: 0101101 1001001 1102102 → 1100100 (13th row) d. v uG VH uGH 0 0 0 0 GH 0 0 0 0
0 0 0 0
1 0 1 e. rH [0 1 0 1 0 0 0] 1 1 0 0
1 1 0 1 0 1 0
1 1 1 0 0 0 1
[1 0 1]
error code: 0010 000 r : 0101 000 0111 000 This is in the codeword list. Project III
1 1 1 1 1 1 1 a. AT 3 4 5 6 7 8 9 b. B ( AT A) 1 AT Y 2.357 B 2.0357
c. y 2.0357 x 2.357 d. y 2.0357 x 2.357
Project IV
Answers will vary.
1492 Copyright © 2025 Pearson Education, Inc.
Chapter 13 Sequences; Induction; the Binomial Theorem Section 13.1 1.
f 2
1 1 2 2 1 , a2 , 1 2 3 22 4 2 3 3 4 4 2 a3 , a4 , 3 2 5 42 6 3 5 5 a5 52 7
17. a1
3 1 2 2 1 1 ; f 3 3 3 2 2
2. True 3. sequence
2 1 1 3 2 2 1 5 , b2 , 2 1 2 22 4 2 3 1 7 2 4 1 9 b3 , b4 , 23 6 24 8 2 5 1 11 b5 25 10
18. b1
4. True 5. True 6. b 7. summation
19. c1 (1)11 (12 ) 1, c2 (1) 2 1 (22 ) 4,
8. b
c3 (1)31 (32 ) 9, c4 (1) 41 (42 ) 16,
9. 10! 10 9 8 7 6 5 4 3 2 1 3, 628,800
c5 (1)51 (52 ) 25
10. 9! 9 8 7 6 5 4 3 2 1 362,880
1 20. d1 (1)11 1, 2 1 1 2 2 d 2 (1) 2 1 , 3 2 2 1
9! 9 8 7 6! 9 8 7 504 11. 6! 6!
12.
13.
14.
12! 12 11 10! 12 11 132 10! 10!
3 3 d3 (1)31 , 2 3 1 5 4 4 d 4 (1) 4 1 , 7 2 4 1
4!11! 4 3 2 1 11 10 9 8 7! 7! 7! 4 3 2 1 11 10 9 8 190, 080
5 5 d5 (1)51 2 5 1 9
5! 8! 5 4 3! 8! 3! 3! 5 4 8 7 6 5 4 3 2 1 806, 400
21. s1 s3
15. s1 1, s2 2, s3 3, s4 4, s5 5
s5
16. s1 12 1 2, s2 22 1 5, s3 32 1 10,
31
3 32 9 , s2 2 , 5 2 3 2 3 7 1
33 23 3 35 5
2 3
s4 42 1 17, s5 52 1 26
1493 Copyright © 2025 Pearson Education, Inc.
27 24 81 , s4 4 , 11 2 3 19
243 35
Chapter 13: Sequences; Induction; the Binomial Theorem
1
2
16 4 4 4 22. s1 , s2 , 9 3 3 3 3
4
64 256 4 4 s3 , s4 , 27 81 3 3 5
4 1024 s5 243 3
23. t1
(1)1 1 1 , (1 1)(1 2) 2 3 6
t2
(1) 2 1 1 , (2 1)(2 2) 3 4 12
30. Each term is equal to a fraction with the numerator equal to a power of 2 and the denominator equal to a power of 3. Both powers are equal to the term number. Since the powers are the same, we can use rules for exponents to write each term as a power of 2 . 3 2 an 3
(1)3 1 1 t3 , (3 1)(3 2) 4 5 20 t4
(1) 4 1 1 , (4 1)(4 2) 5 6 30
t5
(1)5 1 1 (5 1)(5 2) 6 7 42
24. a1 a4
an 1
31 3 32 9 33 27 3, a2 , a3 9, 1 1 2 2 3 3 34 81 35 243 , a5 4 4 5 5
26. c1 c4
1
1
2
1
21 42 2
2
n 1
32. The terms appear to alternate between whole numbers and fractions. If we write the whole numbers as fractions (e.g. 1 1 , 3 3 , etc.), we 1 1 see that each term consists of a 1 and the term number. When n is odd, the numerator is n and the denominator is 1. When n is even, the numerator is 1 and the denominator is n. This alternating behavior occurs if we have a power that alternates sign. The alternating sign is
obtained by using 1 2
1 2 3 9 , c2 2 1, c3 3 , 2 8 2 2
4
n
31. The terms form an alternating sequence. Ignoring the sign, each term always contains a 1. The sign alternates by raising 1 to a power. Since the first term is positive, we use n 1 as the power.
1 2 3 , b2 2 , b3 3 , e e e e 4 5 b4 4 , b5 5 e e
25. b1
29. Each term is a fraction with the numerator equal to 1 and the denominator equal to a power of 2. The power is equal to one less than the term number. 1 an n 1 2
16 52 25 1, c5 5 16 32 2
1 an n
n1
. Thus, we get
n1
33. The terms (ignoring the sign) are equal to the term number. The alternating sign is obtained by
27. Each term is a fraction with the numerator equal to the term number and the denominator equal to one more than the term number. n an n 1 28. Each term is a fraction with the numerator equal to 1 and the denominator equal to the product of the term number and one more than the term number. 1 an n n 1
using 1 an 1
n1
n 1
.
n
34. Here again we have alternating signs which will
be taken care of by using 1
n 1
. The rest of the
term is twice the term number. an 1
n 1
2n
35. a1 2, a2 3 2 5, a3 3 5 8, a4 3 8 11, a5 3 11 14
1494 Copyright © 2025 Pearson Education, Inc.
Section 13.1: Sequences 36. a1 3, a2 4 3 1, a3 4 1 3, a4 4 3 1, a5 4 1 3
48. a1 2, a2
37. a1 2, a2 2 ( 2) 0, a3 3 0 3, a4 4 3 7, a5 5 7 12
2 2 2 , a 5 2
38. a1 1, a2 2 1 1, a3 3 1 2, a4 4 2 2, a5 5 2 3
a4
39. a1 4, a2 3 4 12, a3 3 12 36, a4 3 36 108, a5 3 108 324
49.
2 2 2 2 2
n
(k 2) 3 4 5 6 7 n 2
k 1
40. a1 2, a2 2, a3 ( 2) 2,
50.
a4 2, a5 ( 2) 2
n
(2k 1) 3 5 7 9 2n 1
k 1
3 3 1 41. a1 3, a2 , a3 2 , 2 3 2 1 1 1 1 a4 2 , a5 8 4 8 5 40
51.
k2 1 9 25 49 n2 2 8 18 32 2 2 2 2 2 k 1 2
52.
k 1 4 9 16 25 36 n 1
n
n
2
2
k 1
42. a1 2, a2 2 3( 2) 4,
53.
a3 3 3( 4) 9, a4 4 3(9) 23, a5 5 3( 23) 64
54.
43. a1 1, a2 2, a3 2 1 2, a4 2 2 4, a5 4 2 8
55.
44. a1 1, a2 1, a3 1 3 1 2, a4 1 4 2 9, a5 2 5 9 47
56.
n
1
1 1 3 9
k 1
k 0 3
1 1 n 27 3
k
n
3 9 3 3 1 2 4 2 k 0 2
n 1
1
1 1 3 9
k 1
k 0 3
n
1 1 n 27 3
n 1
(2k 1) 1 3 5 7 2(n 1) 1
k 0
45. a1 A, a2 A d , a3 ( A d ) d A 2d ,
1 3 5 7 (2n 1)
a4 ( A 2d ) d A 3d , a5 ( A 3d ) d A 4d
57.
n
(1)k ln k ln 2 ln 3 ln 4 (1)n ln n
k 2
46. a1 A, a2 rA, a3 r (rA) r 2 A,
2 2 , 2
2 , a3 2
a4 r r 2 A r 3 A, a5 r r 3 A r 4 A
58.
n
(1)k 1 2k
k 3
(1) 4 23 (1)5 24 (1)6 25 (1) n 1 2n
47. a1 2, a2 2 2 , a3 2 2 2 ,
23 24 25 26 (1) n 1 2n
a4 2 2 2 2 ,
8 16 32 64 ... (1) n 1 2n
a5 2 2 2 2 2
20
59. 1 2 3 20 k k 1
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Chapter 13: Sequences; Induction; the Binomial Theorem
8
60. 13 23 33 83 k 3
74.
k 1
61.
26
26
26
26
26
k 1
k 1
k 1
k 1
k 1
3k 7 3k 7 3 k 7 26 26 1 3 7 26 2 1053 182 871
13 k 1 2 3 13 2 3 4 13 1 k 1 k 1 12
62. 1 3 5 7 2(12) 1 (2k 1)
75.
k 1
k 2 4 k 2 4 16
16
16
k 1
k 1
k 1
1 1 1 1 6 1 63. 1 (1)6 6 (1) k k 3 9 27 3 k 0 3
64.
2 3
65. 3
4 9
8 27
11 111 2
11 k 1 2 (1) 3 3 k 1
( 1)
6 1496 64 1560
k
76.
14
14
k 0
k 1
14
k 1
77.
60
n
or (a k 1 d ) k 1
n
68. a a r a r 2 a r n 1 a r k 1 k 1
72.
73.
78.
40
5 5 5 40 5 200 5 5
40
40
40 40 1
k 1
2
24
24
24 24 1
k 1
k 1
2
20
20
20
20
20
k 1
k 1
k 1
k 1
k 1
20 41 820
( k ) k
7
k 8
k 1 k 1 40 40 1 7 7 1 3 2 2 3 820 28 2376
20
20
4
k 5
k 1
k 1
k 8
k3 k3 k3 2
20 20 1 4 4 1 2 2
300
(5k 3) (5k ) 3 5 k 3 20 20 1 5 3 20 2 1050 60 1110
40
40
3k 3 k 3 k k
50
k
k 1 k 1 60 60 1 9 9 1 2 2 2 2 1830 45 3570
8 8 8 50(8) 400 8 8
79.
9
k 10
40 times
50 times
60
60
2k 2 2k 2 k k
k 10
k 0
71.
14 14 1 2 14 1
6 4 1015 64 955
67. a ( a d ) ( a 2d ) ( a nd ) (a kd )
k 1
k 1
4
n
70.
14
k2 4
n 3k 32 33 3n n k 1 k 2 3
k 1
4 16
k 2 4 02 4 k 2 4
n k 1 2 3 n 66. 2 3 n k e e e e k 1 e
69.
16 16 1 2 16 1
2
2102 102 44, 000
80.
24
24
3
k 4
k 1
k 1
k3 k3 k3 2
24 24 1 3 3 1 2 2 3002 62 89,964
1496 Copyright © 2025 Pearson Education, Inc.
2
4 14
Section 13.1: Sequences 81. B1 1.01B0 100 1.01(3000) 100 $2930 Jane’s balance is $2930 after making the first payment.
84. p1 0.9 p0 15 0.9(250) 15 240 p2 0.9 p1 15 0.9(240) 15 231 There are 231 tons of pollutants after two years.
82. p1 1.03 p0 20 1.03(2000) 20 2080 p2 1.03 p1 20 1.03(2080) 20 2162.4 There are approximately 2162 trout in the pond after 2 months.
85. a1 1, a2 1, a3 2, a4 3, a5 5, a6 8, a7 13, a8 21, an an 1 an 2 a8 a7 a6 13 8 21 After 7 months there are 21 mature pairs of rabbits.
83. B1 1.005B0 534.47 1.005(18,500) 534.47 $18, 058.03 Phil’s balance is $18,058.03 after making the first payment.
1 5 1 5 1 5 1 5 2 5 1 86. a. u 1
1
1
21 5
2 5
u2
2 5
1 5 1 5 1 2 5 5 1 2 5 5 4 5 1 2
2
22 5
4 5
4 5
b. Set A 1 5, B 1 5 . un 2
An 2 B n 2
2n 2 5
A 2 An B 2 B n 2n 2 5
1 5 A 1 5 B 2
2
n
n
2n 2 5
3 5 A 3 5 B n
n
2n 1 5
1 5 A 1 5 B 2 A 2B n
n
n
n
2n 1 5
An 1 B n 1
2n 1 5 un 1 un
c.
An B n 2n 5
Since u1 1, u2 1, un 2 un 1 un , un is the Fibonacci sequence.
1497
Copyright © 2025 Pearson Education, Inc.
Chapter 13: Sequences; Induction; the Binomial Theorem 87. 1, 1, 2, 3, 5, 8, 13 This is the Fibonacci sequence. 88. a. b.
u1 1, u2 1, u3 2, u4 3, u5 5, u6 8, u7 13, u8 21, u9 34, u10 55 , u11 89 u3 2 u u u2 1 u4 3 5 8 1, 2, 1.5, 5 1.67, 6 1.6, u1 1 u2 1 u3 2 u4 3 u5 5 u7 13 u8 21 u9 34 1.625, 1.615, 1.619, u6 8 u7 13 u8 21 u10 55 u 89 1.618, 11 1.618 u9 34 u10 55
c. d.
1.618 (the exact value is
1 5 ) 2
u3 2 u1 1 u2 1 u4 3 1, 0.5, 0.667, 0.6, u2 1 u3 2 u4 3 u5 5 u5 5 u6 8 u 13 0.625, 0.615, 7 0.619, u6 8 u7 13 u8 21 u8 21 u9 34 u 55 0.618, 0.618, 10 0.618 u9 34 u10 55 u11 89
e.
0.618 (the exact value is
89. a.
f 1.3 e1.3 1.3 k!
4
k
k 0
b.
7
f 1.3 e1.3 1.3 k! k 0
c.
k
2 ) 1 5
1.30 0!
1.31 1!
...
1.34 4!
3.630170833
1.30 1.31 1.37 ... 3.669060828 0! 1! 7!
f 1.3 e1.3 3.669296668
d. It will take n 12 to approximate f 1.3 e1.3 correct to 8 decimal places.
1498 Copyright © 2025 Pearson Education, Inc.
Section 13.1: Sequences
3
2.4 2.4 2.4 2.4 2.4 0.824 k
0
90. a.
f 2.4 e2.4
b.
f 2.4 e2.4
c.
f 2.4 e2.4 0.0907179533
k 0 6
k 0
0!
k!
2.4 k!
1
k
2.4 0!
1!
0
2.4
1
1!
2
3
2!
...
3!
2.4 6!
6
0.1602688
d. It will take n 17 to approximate f 2.4 e2.4 correct to 8 decimal places.
91. a.
a1 0.4 , a2 0.4 0.3 22 2 0.4 0.3 0.7 , a3 0.4 0.3 23 2 0.4 0.3 2 1.0 , a4 0.4 0.3 24 2 0.4 0.3 4 1.6 , a5 0.4 0.3 25 2 0.4 0.3 8 2.8 ,
a6 0.4 0.3 26 2 0.4 0.3 16 5.2 , a7 0.4 0.3 27 2 0.4 0.3 32 10.0 , a8 0.4 0.3 28 2 0.4 0.3 64 19.6 The first eight terms of the sequence are 0.4, 0.7, 1.0, 1.6, 2.8, 5.2, 10.0, and 19.6. b. Except for term 5, which has no match, Bode’s formula provides excellent approximations for the mean distances of the planets from the sun. c.
The mean distance of Ceres from the Sun is approximated by a5 2.8 , and that of Uranus is a8 19.6 .
d.
a9 0.4 0.3 29 2 0.4 0.3 128 38.8
a10 0.4 0.3 210 2 0.4 0.3 256 77.2 e.
Pluto’s distance is approximated by a9 , but no term approximates Neptune’s mean distance from the sun.
f.
a11 0.4 0.3 211 2 0.4 0.3 512 154
According to Bode’s Law, the mean orbital distance of Eris will be 154 AU from the sun.
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Chapter 13: Sequences; Induction; the Binomial Theorem
1 4 1 4 4 1 4 4 1 4 4 92. a1 4 , a2 (4) , a3 , a4 , a5 , 5 5 5 5 25 5 25 125 5 125 625 1 4 4 a5 0.00128 5 625 3125 93. a.
Let I 0 represent the intensity. Then the nth term would be an 0.95n I 0 . 0.02 0.95n
b.
log 0.02 log 0.95n log 0.02 n log 0.95 n
94. To show that S 1
Let
log 0.02 77 log 0.95
1 2 3 ... n 1 n 2
n n 1
2 3 ... n 1 n, we can reverse the order to get
S n n 1 n 2 +...+
2
+ 1,
now add these two lines to get
2S 1 n 2 n 1 3 n 2 ...... n 1 2 n 1
So we have 2S 1 n 1 n 1 n .... n 1 n 1 n n 1 2S n n 1 S
95.
n n 1 2
5 We begin with an initial guess of a0 2 . 1 5 2 2.25 2 2 1 5 a2 2.25 2.236111111 2 2.25 1 5 a3 2.236111111 2 2.236111111 2.236067978 1 5 a4 2.236067978 2 2.236067978 2.236067977 1 5 a5 2.236067977 2 2.236067977 2.236067977 a1
For both a5 and the calculator approximation, we obtain
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5 2.236067977 .
Section 13.1: Sequences
96.
97.
8 We begin with an initial guess of a0 3 . a1
1 8 1 8 a0 3 2.833333333 2 3 a0 2
a2
1 8 a1 2 a1
1 21 4 4.625 2 4 1 21 a2 4.625 4.58277027 2 4.625 1 21 a3 4.58277027 2 4.58277027 4.582575699 1 21 a4 4.582575699 2 4.582575699 4.582575695 1 21 a5 4.582575695 2 4.582575695 4.582575695 a1
1 8 2.833333333 2 2.2.833333333 2.828431373
a3
1 8 a2 2 a2
1 8 2.828431373 2 2.828431373 2.828427125
a4
21 We begin with an initial guess of a0 4 .
1 8 a3 2 a3
8 1 2.828427125 2.828427125 2 2.828427125
a5
For both a5 and the calculator approximation,
1 8 a4 2 a4
we obtain
1 8 2.828427125 2 2.828427125 2.828427125
98.
89 We begin with an initial guess of a0 9 .
1 89 5 9.444444444 2 5 1 89 a2 9.444444444 2 9.444444444 9.433986928 1 89 a3 9.433986928 2 9.433986928 9.433981132 1 89 a4 9.433981132 2 9.433981132 9.433981132 1 89 a5 9.433981132 2 9.433981132 9.433981132 a1
For both a5 and the calculator approximation, we obtain
21 4.582575695 .
8 2.828427125 .
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Chapter 13: Sequences; Induction; the Binomial Theorem
For both a5 and the calculator approximation, 89 9.433981132 .
we obtain
a1n
r a2 n
a2
a2 n
= n
r n a2 n a2 n
rn
r r r r
99. u1 1 and un 1 un (n 1) : So u1 1
n factors
u2 u1 (1 1) 1 2 3
a a a a a 1 2 3 n 1 a2 a3 a4 an 1 an 1
u3 u2 (2 1) 3 3 6 u4 u3 (3 1) 6 4 10
103 - 104. Answers will vary.
u5 u4 (4 1) 10 5 15 u6 u5 (5 1) 15 6 21
105.
u7 u6 (6 1) 21 7 28
100. Note that: un 1 un (n 1) 1 2 3
; (n 2) (n 1) n (n 1) Reverse the order and add: So adding these together we have un 1 1 2 3 (n 1) n
r A P 1 n
nt
0.03 2500 1 12
12(2)
$2654.39
106. r x 2 y 2 (1) 2 (1) 2 2 y 1 (n 1) tan 1 x 1 un 1 (n 1) n (n 1) 3 2 1 225º 2(un 1 ) (n 2) (n 2) (n 2) (n 2) (n 2) (n 2) The polar form of z 1 i is 2(un 1 ) (n 1)(n 2) z r cos i sin 2 cos 225º i sin 225º . (n 1)(n 2) un 1 107. v w (2)(1) ( 1)(2) 2 2 ( 2) (n 1)(n 2) (n)(n 1) 0 and un 101. un 1 2 2 so 108. The vertex is (–3, 4) and the focus is (1, 4). Both (n 1)(n 2) (n)(n 1) lie on the horizontal line y 4 . a 3 1 4 un 1 un + 2 2 and since (1, 4) is to the right of (–3, 4), the (n 1)(n 2) (n)(n 1) parabola opens to the right. The equation of the parabola is: 2 n 2 3n 2 n 2 n 2 2n 2 4n 2 2 2 2(n 2n 1) n 2 2n 1 (n 1) 2 2 102. Let r be the common ratio so a a1 a2 a = n 1 r . Then 1 r and a2 a3 a a2
y k 2 4a x h y 4 2 4 4 x (3) y 4 2 16 x 3
109. Since the degree of the denominator is higher than the degree of the numerator the horizontal asymptote is y 0 .
a1 r a2 .
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Section 13.2: Arithmetic Sequences 110. A B C 180 B A4 C 3B 11 Setting up a system of equations gives: A B C 180 2 A B 4 3B C 11 The coefficient matrix would be: 1 1 1 180 2 1 0 4 . Solving this system gives 0 3 1 11 1 0 0 23 0 1 0 42 0 0 1 115 So the three angles are A 23, B 42, and C 115 .
x 2 4 x 12 0 ( x 2)( x 6) 0 x 2, 6 So the critical numbers are 2, 6 .
Section 13.2 1. arithmetic 2. False; the sum of the first and last terms equals twice the sum of all the terms divided by the number of terms. 3. 12 a1 (5 1)5 a1 8 So a6 8 (6 1)5 17
10 r r 10, x 1, y 9 3 111. sec so 1 x tan 3
4. True
r 10, x 3, y 1 1 10 r so sec 1 x 3 tan 3 1 8 3 f (b) f (a) 3 3 ba 10 2 10 10 3 3
5. d 6. c 7. d sn sn 1 (n 4) (n 1 4) (n 4) (n 3) n 4n3 1 The difference between consecutive terms is constant, therefore the sequence is arithmetic. s1 1 4 5, s2 2 4 6, s3 3 4 7,
8 8 10 2 10 20 5 2 10
112. f (a 1) 5(a 1) 2 2(a 1) 9 16
8. d sn sn 1
5(a 2 2a 1) 2a 2 9 16
(n 5) (n 1 5) n 5 n 6
5a 2 10a 5 2a 7 16
n5n6 1 The difference between consecutive terms is constant, therefore the sequence is arithmetic. s1 1 5 4, s2 2 5 3, s3 3 5 2,
5a 2 8a 4 0 (5a 2)(a 2) 0 a
s4 4 4 8
2 or a 2 5
s4 4 5 1
9. d an an 1
113. The function is undefined when the denominator is equal to zero: x2 0 x2 However the function is not defined at x = 2 so it is not a critical number. The function is equal to zero when the numerator is equal to zero:
2n 5 (2(n 1) 5) 2n 5 2n 2 5
2n 5 2n 7 2 The difference between consecutive terms is
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Chapter 13: Sequences; Induction; the Binomial Theorem
constant, therefore the sequence is arithmetic. a1 2 1 5 3, a2 2 2 5 1, a3 2 3 5 1, a4 2 4 5 3
10. d bn bn 1
3n 1 (3(n 1) 1) 3n 1 3n 3 1
3n 1 3n 2 3 The difference between consecutive terms is constant, therefore the sequence is arithmetic. b1 3 1 1 4, b2 3 2 1 7, b3 3 3 1 10, b4 3 4 1 13
11. d cn cn 1
14. d tn tn 1 2 1 2 1 n (n 1) 3 4 3 4 1 2 1 2 1 n n 4 3 4 3 4 2 1 2 1 1 1 n n 3 4 3 4 4 4 The difference between consecutive terms is constant, therefore the sequence is arithmetic. 2 1 11 2 1 7 t1 1 , t2 2 , 3 4 12 3 4 6 2 1 17 2 1 5 t3 3 , t4 4 3 4 12 3 4 3
15. d sn sn 1
6 2n (6 2(n 1))
ln 3n ln 3n 1
6 2n 6 2n 2 6 2n 6 2n 2 2 The difference between consecutive terms is constant, therefore the sequence is arithmetic. c1 6 2 1 4, c2 6 2 2 2, c3 6 2 3 0, c4 6 2 4 2
n ln 3 n 1 ln 3 ln 3 (n n 1) ln 3 n n 1 ln 3 The difference between consecutive terms is constant, therefore the sequence is arithmetic.
s ln 3 3ln 3 , s ln 3 4 ln 3 s1 ln 31 ln 3 , s2 ln 32 2 ln 3 ,
12. d an an 1
4 2n (4 2(n 1))
3
4 2n 4 2n 2 4 2n 4 2n 2 2 The difference between consecutive terms is constant, therefore the sequence is arithmetic. a1 4 2 1 2, a2 4 2 2 0, a3 4 2 3 2, a4 4 2 4 4
13. d tn tn 1 1 1 1 1 n (n 1) 2 3 2 3 1 1 1 1 1 n n 3 2 3 2 3 1 1 1 1 1 1 n n 2 3 2 3 3 3 The difference between consecutive terms is constant, therefore the sequence is arithmetic. 1 1 1 1 1 1 t1 1 , t2 2 , 2 3 6 2 3 6 1 1 1 1 1 5 t3 3 , t4 4 2 3 2 2 3 6
3
4
4
16. d sn sn 1 eln n eln( n 1) n n 1 1
The difference between consecutive terms is constant, therefore the sequence is arithmetic. s1 eln1 1, s2 eln 2 2, s3 eln 3 3, s4 eln 4 4
17. an a1 (n 1)d 2 (n 1)3 2 3n 3 3n 1 a51 3 51 1 152
18. an a1 (n 1)d 2 (n 1)4 2 4n 4 4n 6 a51 4 51 6 198
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Section 13.2: Arithmetic Sequences 19. an a1 (n 1)d
27. a1 3, d 3 3 6, an a1 (n 1)d a90 3 (90 1)(6) 3 89(6)
8 (n 1)(7) 8 7n 7 15 7n a51 15 7 51 342
3 534 531
28. a1 5, d 0 5 5, an a1 (n 1)d a80 5 (80 1)(5) 5 79(5)
20. an a1 (n 1)d
5 395 390
6 (n 1)( 2) 6 2n 2 8 2n a51 8 2 51 94
5 1 2 , an a1 (n 1)d 2 2 1 83 a80 2 (80 1) 2 2
29. a1 2, d
21. an a1 (n 1)d 0 (n 1)
30. a1 2 5, d 4 5 2 5 2 5,
1 2
an a1 (n 1)d a70 2 5 (70 1)2 5
1 1 n 2 2 1 n 1 2 1 a51 51 1 25 2
2 5 69 2 5
2 5 138 5 140 5
31. a8 a1 7 d 8 a20 a1 19d 44 Solve the system of equations by subtracting the first equation from the second: 12d 36 d 3 a1 8 7(3) 8 21 13 an an 1 3 Recursive formula: a1 13
22. an a1 (n 1)d 1 1 (n 1) 3 1 1 1 n 3 3 4 1 n 3 3 4 1 4 51 47 a51 51 3 3 3 3 3
nth term: an a1 n 1 d
13 n 1 3 13 3n 3 3n 16
23. an a1 (n 1)d
32. a4 a1 3d 3 a20 a1 19d 35 Solve the system of equations by subtracting the first equation from the second: 16d 32 d 2 a1 3 3(2) 3 6 3 an an 1 2 Recursive formula: a1 3
2 (n 1) 2 2 2n 2 2n a51 51 2
24. an a1 (n 1)d 0 (n 1) n 1
nth term: an a1 n 1 d
a51 51 50
3 n 1 2
25. a1 2, d 2, an a1 (n 1)d a100 2 (100 1)2 2 99(2) 2 198 200
3 2n 2
26. a1 1, d 2, an a1 (n 1)d a80 1 (80 1)2 1 79(2) 1 158 157
33. a9 a1 8d 5 a15 a1 14d 31 Solve the system of equations by subtracting the first equation from the second:
2n 5
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Chapter 13: Sequences; Induction; the Binomial Theorem 6d 36 d 6 a1 5 8(6) 5 48 53 Recursive formula: a1 53
an an 1 6
nth term: an a1 n 1 d
53 n 1 6 53 6n 6
37. a14 a1 13d 1 a18 a1 17 d 9 Solve the system of equations by subtracting the first equation from the second: 4d 8 d 2 a1 1 13( 2) 1 26 25 an an 1 2 Recursive formula: a1 25
nth term: an a1 n 1 d
6n 59
25 n 1 2
34. a8 a1 7 d 4 a18 a1 17 d 96 Solve the system of equations by subtracting the first equation from the second: 10d 100 d 10 a1 4 7(10) 4 70 74 an an 1 10 Recursive formula: a1 74
nth term: an a1 n 1 d
74 n 1 10 74 10n 10
25 2n 2 27 2n
38. a12 a1 11d 4 a18 a1 17 d 28 Solve the system of equations by subtracting the first equation from the second: 6d 24 d 4 a1 4 11(4) 4 44 40 an an 1 4 Recursive formula: a1 40
nth term: an a1 n 1 d
84 10n
40 n 1 4
35. a15 a1 14d 0 a40 a1 39d 50 Solve the system of equations by subtracting the first equation from the second: 25d 50 d 2 a1 14( 2) 28 an an 1 2 Recursive formula: a1 28
nth term: an a1 n 1 d
40 4n 4 4n 44
39. Sn
n n n a1 an 1 2n 1 2n n2 2 2 2
40. Sn
n n a1 an 2 2n n n2 n n 1 2 2
41. Sn
n n n a1 an 7 2 5n 9 5n 2 2 2
28 n 1 2 28 2n 2 30 2n
36. a5 a1 4d 2 a13 a1 12d 30 Solve the system of equations by subtracting the first equation from the second: 8d 32 d 4 a1 2 4(4) 18 an an 1 4 Recursive formula: a1 18
n n a1 an 1 4n 5 2 2 n 4n 6 2n 2 3n 2 n 2n 3
42. Sn
nth term: an a1 n 1 d
18 n 1 4 18 4n 4 4n 22
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Section 13.2: Arithmetic Sequences 48. a1 7 , d 1 7 6 , an a1 n 1 d
43. a1 2, d 4 2 2, an a1 (n 1)d 70 2 (n 1)2 70 2 2n 2 70 2n n 35 n 35 Sn a1 an 2 70 2 2 35 72 35 36 2 1260
299 7 n 1 6 306 6 n 1 51 n 1 52 n n 52 Sn a1 an 7 299 2 2 26 292 7592
49. a1 4 , d 4.5 4 0.5 , an a1 n 1 d
44. a1 1, d 3 1 2, an a1 (n 1)d 59 1 (n 1)2 59 1 2n 2 60 2n n 30 n 30 Sn a1 an 1 59 15 60 900 2 2
100 4 n 1 0.5 96 0.5 n 1 192 n 1 193 n n 193 Sn a1 an 4 100 2 2 193 104 10, 036 2
45. a1 9, d 5 (9) 4, an a1 (n 1)d 39 9 n 1 4
1 1 50. a1 8 , d 8 8 , an a1 n 1 d 4 4 1 50 8 n 1 4
39 9 4n 4 39 4n 13 n 13 n 13 13 Sn a1 an 9 39 30 195 2 2 2
1 n 1 4 168 n 1 42
46. a1 2, d 5 2 3, an a1 (n 1)d 41 2 n 1 3
169 n n 169 Sn a1 an 8 50 2 2 169 58 4901 2
41 2 3n 3 42 3n n 14 n 14 Sn a1 an 2 41 7 43 301 2 2
51. a1 4 1 9 5 , a80 4 80 9 311
47. a1 93 , d 89 93 4 , an a1 n 1 d
S80
287 93 n 1 4
80 5 311 40 306 12, 240 2
52. a1 3 2 1 1 , a90 3 2 90 177
380 4 n 1 380 4n 4 384 4n
S90
90 1 177 45 176 7920 2
96 n n 96 Sn a1 an 93 (287) 2 2 48 194 9312
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Chapter 13: Sequences; Induction; the Binomial Theorem
1 11 1 1 2 , a100 6 2 100 44 2 100 11 S100 44 2 2
n 2(11) (n 1)(3) 2 n 1092 22 3n 3 2 2194 n 19 3n
53. a1 6
1092
77 50 1925 2
2194 19n 3n 2
1 1 5 1 1 163 54. a1 1 , a80 80 3 2 6 3 2 6 80 5 163 S80 40 28 1120 2 6 6
3n 2 19n 2184 0 (3n 91)(n 24) 0 So n 24 .
60. d 4, a1 78, and S 702 n 2(78) (n 1)(4) 2 n 702 156 4n 4 2 1404 n 160 4n 702
55. a1 14 , d 16 14 2 , an a1 n 1 d
a120 14 120 1 2 14 119 2 252 S120
120 14 252 60 266 15,960 2
56. a1 2 , d 1 2 3 , an a1 n 1 d
a46 2 46 1 3 2 45 3 133 S46
46 2 133 23 131 3013 2
57. Find the common difference of the terms and solve the system of equations: (2 x 1) ( x 3) d x 2 d (5 x 2) (2 x 1) d 3 x 1 d 3x 1 x 2 2 x 3 3 x 2 58. Find the common difference of the terms and solve the system of equations: (3 x 2) (2 x) d x 2 d (5 x 3) (3x 2) d 2 x 1 d 2x 1 x 2 x 1 59. d 3, a1 11, and S 1092
1404 160n 4n 2 4n 2 160n 1404 0 n 2 40n 351 0 (n 13)(n 27) 0 So n 13 or n 27 .
61. The total number of seats is: S 25 26 27 25 29 1
This is the sum of an arithmetic sequence with d 1, a1 25, and n 30 . Find the sum of the sequence: 30 S30 2(25) (30 1)(1) 2 15(50 29) 15(79) 1185 There are 1185 seats in the theater. 62. a1 35 , d 37 35 2 , an a1 n 1 d
a27 35 27 1 2 35 26 2 87 27 27 35 87 2 122 1647 2 The amphitheater has 1647 seats. S27
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Section 13.2: Arithmetic Sequences 63. The total number of seats is:
S 15 17 19 15 39 2
n
This is the sum of an arithmetic sequence with d 2, a1 15, and n 40 . Find the sum of the sequence: 40 S40 2(15) (40 1)(2) 2 20(30 78) 20(108) 2160 The corner section has 2160 seats.
49 4001 49 63.25 2 2 n 7.13 or n 56.13 It takes about 8 years to have an aggregate salary of at least $280,000. The aggregate salary after 8 years will be $319,200.
66. Find n in an arithmetic sequence with a1 10, d 4, Sn 2040 . n Sn 2a1 (n 1)d 2 n 2040 2(10) (n 1)4 2 4080 n 20 4n 4
64. The number of bricks required decreases by 2 on each successive step. This is an arithmetic sequence with a1 100, d 2, and n 30 . a.
49 492 4(1)( 400) 2(1)
The number of bricks for the top step is: a30 a1 (n 1)d 100 (30 1)( 2) 100 29( 2) 100 58
4080 n(4n 16)
42 42 bricks are required for the top step. b. The total number of bricks required is the sum of the sequence: 30 S 100 42 15(142) 2130 2 2130 bricks are required to build the staircase.
4080 4n 2 16n 1020 n 2 4n n 2 4n 1020 0 (n 34)(n 30) 0 n 34 or n 30 There are 30 rows in the corner section of the stadium.
67. The lighter colored tiles have 20 tiles in the bottom row and 1 tile in the top row. The number decreases by 1 as we move up the triangle. This is an arithmetic sequence with a1 20, d 1, and n 20 . Find the sum:
65. The yearly salaries form an arithmetic sequence with a1 35, 000, d 1400, Sn 280, 000 . Find the number of years for the aggregate salary to equal $280,000. n Sn 2a1 (n 1)d 2 n 280, 000 2(35, 000) (n 1)1400 2 280, 000 n 35, 000 700n 700
20 2(20) (20 1)(1) 2 10(40 19) 10(21) 210 There are 210 lighter tiles. S
The darker colored tiles have 19 tiles in the bottom row and 1 tile in the top row. The number decreases by 1 as we move up the triangle. This is an arithmetic sequence with a1 19, d 1, and n 19 . Find the sum:
280, 000 n(700n 34,300) 280, 000 700 n 2 34,300 n 400 n 2 49 n n 2 49 n 400 0
19 2(19) (19 1)(1) 2 19 19 (38 18) (20) 190 2 2 are 190 darker tiles. S
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Chapter 13: Sequences; Induction; the Binomial Theorem 68. a. We are given that a1 57, d 95 . The formula for the sequence would be an a1 (n 1)d an 57 (n 1)95
57 95n 95 95n 38
The predictive formula would be an 95n 38 . b.
a10 95(10) 38 912 min.
912 minutes after 12:57 a.m. would be 3:12 p.m. c.
We need the number of hours to be less than 24. This is 1440 minutes. We must add multiples of 1h 35m (1.583h) until we get to the end of the day. Using 0.95 for the time of 12:57 a.m. and solving for n gives: 24 15.16 . It would erupt 15 times. So 1.583 0.95 (15 1)(1.583) 0.95 22.162 23.112
23.112 would be approximately 11:07 p.m. 69. The air cools at the rate of 5.5 F per 1000 feet. Since n represents thousands of feet, we have d 5.5 . The ground temperature is 67F so we have T1 67 5.5 61.5 . Therefore,
Tn 61.5 n 1 5.5 5.5n 67 or 67 5.5n
After the parcel of air has risen 5000 feet, we have T5 61.5 5 1 5.5 39.5 . The parcel of air will be 39.5F after it has risen 5000 feet. 70. If we treat the length of each rung as the term of an arithmetic sequence, we have a1 49 , d 2.5 , and an 24 .
To find the total material required for the rungs, we need the sum of their lengths. Since there are 11 rungs, we have 11 11 S11 49 24 73 401.5 2 2 It would require 401.5 feet of material to construct the rungs for the ladder. 71. Let an 2 (n 1) 7 7 n 5 and bn 5 (n 1) 6 6n 1.
Solving 7 n 5 6n 1 gives n 4 , so the fourth term in each sequence is the same. Subsequent pairs occur every 6 terms for an , the sequence with d 7 , and every 7 terms for bn , the sequence with d 6 . Thus, the number of pairs in limited by bn and the restriction of 100 terms. After the fourth term, 96 there are int 13 terms that appear in both 7 sequences of 14 terms in common. n (a1 an ) and 2 a2 n a1 (2n 1)d an nd , so
72. Sn
2n 2n (a1 a2 n ) (a1 an nd ) 2 2 n 2 (a1 an ) n 2 d 2Sn n 2 d 2 S2 n 2Sn n 2 d n2 d 2 C (a constant). Sn Sn Sn Rearranging gives n2 d C 2 Sn . n C 2 2a1 (n 1)d 2 From which 2nd C 2 nd 2a1 d . S2 n
an a1 n 1 d
Because this equation is an identity in n, 2 (C 2) C 4 and
24 49 n 1 2.5
2a1 d 0 a1
25 2.5 n 1 10 n 1 11 n Therefore, the ladder contains 11 rungs.
d . 2
So an a1 (n 1)d
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d 2n 1 (n 1)d d . 2 2
Section 13.2: Arithmetic Sequences é2 0 1 0ù ê ú ê 3 -1 0 1ú ë û é 1 0 1 0ù 2 ú R =1r ê ê3 -1 0 1ú ( 1 2 1 ) ë û 1 é1 0 0ùú 2 êê 3 ú ( R2 = - 3r1 + r2 ) ëê 0 -1 - 2 1ûú é1 0 1 0ùú 2 êê 3 ú ( R2 = -1r2 ) êë 0 1 2 -1úû é1 0ùú Thus, A-1 = êê 23 ú. ëê 2 -1ûú
73. Answers will vary. Both increase (or decrease) at a constant rate, but the domain of an arithmetic sequence is the set of natural numbers while the domain of a linear function is the set of all real numbers. 74. Answers will vary. 0.153 75. re 1 12 16.42%
12
1 0.1642
76. v ( x2 x1 )i ( y2 y1 ) j (3 ( 1))i ( 4 2) j 4i 6 j
79. Find the partial fraction decomposition: 3x 3x A Bx C 2 3 2 x 1 ( x 1)( x x 1) x 1 x x 1
77. 25 x 2 4 y 2 100 x2 y 2 1 4 25 The center of the ellipse is at the origin. a 2, b 5 . The vertices are (0, 5) and (0, –5). Find the value of c: c 2 b 2 a 2 25 4 21 c 21
Multiplying both sides by ( x 1)( x 2 x 1) , we obtain: 3x A( x 2 x 1) ( Bx C )( x 1)
Let x 1 , then 3(1) A 12 1 1 B(1) C (1 1) 3 3A A 1
The foci are 0, 21 and 0, 21 .
Let x 0 , then
3(0) A 02 0 1 ( B (0) C )(0 1) 0 1C C 1
Let x 1 , then
3(1) A (1) 2 (1) 1 ( B (1) C )(1 1) 3 A 2 B 2C 3 1 2 B 2(1) 2 2 B B 1
2 0 78. A 3 1 Augment the matrix with the identity and use row operations to find the inverse:
x 1 1 2 x 1 x 1 x x 1 1 x 1 x 1 x2 x 1 3x
3
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Chapter 13: Sequences; Induction; the Binomial Theorem
u 2 2u 1 0
2 2 2 5 1 cos 5 4 5 sin( 4 ) 80. sin 8 2 2 2
u 2 2u 1 Complete the square u 2 2u 1 1 1
2 2 2
u 12 2
2 2 2 5 1 cos 5 4 5 cos( 4 ) cos 8 2 2 2
u 1 2 u 1 2 x2 1 2
2 2 2
x 1 2 The radicand cannot be negative so the zeros are:
0, 1 2 , 1 2 . 2
sin 2
5 5 2 2 2 2 cos 2 8 8 2 2 2 2 2 2 4 4 1 2 1 2 2 2 4 2 4 2
2
83. 12 x 2 6 y 2 24 x 24 y 0 A 12, C 6 AC 72 Since AC<0 the equation represents a hyperbola.
x 2 6 x 9 x 2 2 x 15 7 6 x 9 2 x 8 8 x 17
81. (x – 1) moves the function right by 1 unit so the domain of 2 g ( x 1) would be 3,11 . 82.
x 1 2 x x 1 4 x 0 2 x x 1 x 1 2 x 0 2 x x 1 2 x 2 x 0 4
2
4
4
17 8
The solution set is
17 8
2
4
2
2 x x 4 2 x 2 1 0 2 x x 4 2 x 2 1 0 2 x 0 or x 4 2 x 2 1 0
Let u x 2 . Then
x
3
2
( x 3) 2 ( x 3)( x 5) 7
84.
Section 13.3 0.04 1. A 1000 1 2
22
0.05 2. P 10, 000 1 12
3. geometric 4.
a 1 r
5. b 6. True
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$1082.43 121
$9513.28
Section 13.3: Geometric Sequences; Geometric Series 7. False; the common ratio can be positive or negative (or 0, but this results in a sequence of only 0s).
2n 11 4 2n 13. r n 1 2n ( n 1) 2 2n 1 2 4 The ratio of consecutive terms is constant, therefore the sequence is geometric. 211 20 1 c1 2 22 , 4 4 2
8. True 4n 1
4n 1 n 4 4n The ratio of consecutive terms is constant, therefore the sequence is geometric. s1 41 4, s2 42 16,
9. r
s3 43 64, s4 44 256 (5) n 1
(5) n 1 n 5 (5) n The ratio of consecutive terms is constant, therefore the sequence is geometric. s1 (5)1 5, s2 (5) 2 25,
10. r
n 1
1 3 n 1 n 1 2 1 11. r n 2 2 1 3 2 The ratio of consecutive terms is constant, therefore the sequence is geometric. 2
3
4
3 3 1 1 a3 3 , a4 3 2 8 2 16
231 22 2 1, 4 2
c4
241 23 2 2 4 2
15. r
n 1
33 27 34 81 3, d 4 9 9 9 9 9
n 1 7 4 n 7 4
n 1 n 4 71/4
7 4
The ratio of consecutive terms is constant, therefore the sequence is geometric. e1 71/4 , e2 71/2 , e3 73/4 2, e4 7
5 n 1 n 5 2 5 12. r n 2 2 5 2 The ratio of consecutive terms is constant, therefore the sequence is geometric.
32( n 1)
32 n 2 2 n 32 9 32 n The ratio of consecutive terms is constant, therefore the sequence is geometric. f1 321 9, f 2 322 34 81,
16. r
2
25 5 5 5 b1 , b2 , 4 2 2 2 3
c3
d3
3 3 1 1 a1 3 , a2 3 , 2 4 2 2
1
221 21 1 2 21 , 4 2 2
3n 1 9 3n 1 14. r n 3n 1 n 3 n 3 3 9 The ratio of consecutive terms is constant, therefore the sequence is geometric. 31 1 32 9 d1 , d 2 1, 9 3 9 9
s3 (5)3 125, s4 (5) 4 625
1
c2
4
625 5 125 5 b3 , b4 8 16 2 2
f3 323 36 729, f 4 324 38 6561
1513
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Chapter 13: Sequences; Induction; the Binomial Theorem
3n 11 n 1 n n 2 3 2 17. r 3n 1 3n 1 2n 1 n 2
1 24. a5 1 3
3 2 The ratio of consecutive terms is constant, therefore the sequence is geometric. 311 30 1 32 1 31 3 , t2 2 2 , t1 1 2 2 4 2 2 2 t3
3
2
3
4 1
2
3
3
9 3 3 27 , t4 4 4 8 16 2 2 2 3
2n 1 3n 11 3n 1 2n 1 18. r 2n 3n 2n 3n 1
8 24 16 16 u3 31 2 , u4 41 3 9 27 3 3 3 3 8
3 a 3 3
5 1
25. a5 3
1 26. a5 0
1 an 0
20. a5 2 451 2 44 2 256 512
4
1 0 0
n 1
0
1 a7 1 2
7 1
6
1 1 64 2
28. a1 1, r 3, n 8 a8 1 381 37 2187
29. a1 1, r 1, n 9 15 1
a15 1 1
1
8 1
21. a5 5(1)51 5(1) 4 5 1 5
1 2 1(512) 512 9
0.4 0.1 0.00000004 7
a7 0.1 107 1 0.110 100, 000 6
22. a5 6( 2)51 6( 2) 4 6 16 96
1 an 0 7
1
32. a1 0.1, r 10, n 7
n 1
n 1
5 1
14
31. a1 0.4, r 0.1, n 8
a8 0.4 0.1
n 1
an 6 ( 2)
4
n
10 1
an 2 3
an 5 (1)
3 3 9 9 3 3
n 1
5 1
n 1
3
n
a10 1 2
n 1
1 23. a5 0 7
1 3
30. a1 1, r 2, n 10
19. a5 2 351 2 34 2 81 162
an 2 4
n 1
4
1 27. a1 1, r , n 7 2
2 3n 1 n 2n 1 n 31 2 3 The ratio of consecutive terms is constant, therefore the sequence is geometric. 21 2 2 22 4 u1 11 0 2, u2 2 1 , 1 3 3 3 3 23
1 1 1 81 3
1 an 1 3
3n ( n 1) 2n ( n 1) 3 21
31
5 1
4
1 0 0 7
n 1
0
33. a1 6 , r
18 3 , an a1 r n 1 6
an 6 3n 1
34. a1 5 , r an 5 2n 1
1514 Copyright © 2025 Pearson Education, Inc.
10 2 , an a1 r n 1 5
Section 13.3: Geometric Sequences; Geometric Series
1 1 , an a1 r n 1 3 3
35. a1 3 , r
1 an 3 3
36. a1 4 , r 1 an 4 4
n 1
1 3
40.
n2
1
r 3 81 1 3
1 , an a1 r n 1 4
n 1
1 4
r3
243 a1 3
5
1 Therefore, an 3 3
Therefore, an (3)n 1
41.
38. an a1 r n 1
1 3
n2
.
1 , r2 4 1 r n 1 1 2n 1 n Sn a1 1 2 1 4 1 2 4 r 1 n 2 1 4 a1
42.
n 1
.
a4 a1 r 4 1 r 3 r2 a2 a1 r 2 1 r
3 1 , r 3 9 3 1 r n 1 1 3n 1 1 3n Sn a1 1 r 3 1 3 3 2 1 1 1 3n 3n 1 6 6 a1
1575 225 7 r 225 15
r2
43. a1
2 2 , r 3 3
2 n 1 1 r n 2 3 Sn a1 1 r 3 1 2 3
an a1 r n 1 7 a1 152 1 7 15a1 7 15
Therefore, an
n 1
2 1
1 Therefore, an 21 3
a1
31
1 1 a1 3 9 3 a1
6 1
1 a1
39.
1 1 27 3
1 1 a1 3 3
243 243a1
1 7 a1 3 1 7 a1 3 21 a1
1 1 3 81 27
an a1 r n 1
n2
37. an a1 r n 1 243 a1 3
a6 a1 r 6 1 r 5 r3 a3 a1 r 31 r 2
2 n 1 2 n 2 3 2 1 1 3 3 3
7 15n 1 7 15n 2 . 15
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Chapter 13: Sequences; Induction; the Binomial Theorem
44.
a1 4, r 3
51. Using the sum of the sequence feature:
1 rn 1 3n 1 3n Sn a1 4 4 1 r 1 3 2
2 1 3n 2 3n 1
45. a1 1, r 2 1 rn 1 2n n Sn a1 1 1 2 1 r 1 2
46. a1 2, r
52. Using the sum of the sequence feature:
3 5
3 n 3 n 1 1 1 rn 5 5 2 Sn a1 2 2 3 1 r 1 5 5
1 3 Since r 1, the series converges.
53. a1 1, r
S
3 n 5 1 5
47. Using the sum of the sequence feature:
2 3 Since r 1, the series converges.
54. a1 2, r
S
48. Using the sum of the sequence feature:
a1 1 1 3 1 r 1 2 2 1 3 3
a1 2 2 6 2 1 1 r 1 3 3
1 2 Since r 1, the series converges.
55. a1 8, r
S
49. Using the sum of the sequence feature:
a1 8 8 16 1 r 1 1 1 2 2
1 3 Since r 1, the series converges.
56. a1 6, r
S
50. Using the sum of the sequence feature:
a1 6 6 9 1 r 1 2 1 3 3
1516 Copyright © 2025 Pearson Education, Inc.
Section 13.3: Geometric Sequences; Geometric Series
1 4 Since r 1, the series converges.
2 3 Since r 1, the series converges.
57. a1 2, r
S
65. a1 6, r
a1 2 2 8 1 r 1 5 5 1 4 4
58. a1 1, r
S
3 4
1 2 Since r 1, the series converges.
66. a1 4, r
Since r 1, the series converges. S
a1 1 1 4 1 r 3 7 7 1 4 4
S
3 2 Since r 1 , the series diverges.
59. a1 8 , r
60. a1 9 , r
a1 6 6 18 1 r 2 5 5 1 3 3
67.
4 3
Since r 1 , the series diverges.
a1 4 4 8 1 r 1 3 3 1 2 2 k 1
k
2 2 2 2 3 3 3 3 3 2 3 k 1 k 1 k 1 2 a1 2 , r 3 Since r 1, the series converges.
S
a1 2 2 6 2 1 1 r 1 3
1 61. a1 5, r 4 Since r 1, the series converges. S
3
k 1
k
3 3 33 3 68. 2 2 4 4 4 k 1 k 1 k 1 2 4 3 3 a1 , r 2 4 Since r 1, the series converges.
a1 5 5 20 1 r 1 3 3 1 4 4
1 62. a1 8, r 3 Since r 1, the series converges. S
k 1
S
69.
a1 8 8 12 1 r 1 2 1 3 3
k 1
3 3 a1 3 2 2 4 6 1 r 1 3 1 2 4 4
n 2 d (n 1 2) (n 2) n 3 n 2 1 The difference between consecutive terms is constant. Therefore the sequence is arithmetic.
1 , r 3 2 Since r 1 , the series diverges.
50
50
50
k 1
k 1
k 1
S50 (k 2) k 2
63. a1
50(50 1) 2(50) 1275 100 1375 2
3 2 Since r 1 , the series diverges.
64. a1 3 , r
70.
2n 5 d 2(n 1) 5 (2n 5) 2n 2 5 2n 5 2 The difference between consecutive terms is constant. Therefore the sequence is arithmetic.
1517
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Chapter 13: Sequences; Induction; the Binomial Theorem
50
50
50
k 1
k 1
k 1
S50 (2k 5) 2 k 5 50(50 1) 2 5(50) 2550 250 2300 2
71.
4n Examine the terms of the sequence: 4, 2
16, 36, 64, 100, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric. 72.
5n 1 Examine the terms of the sequence: 2
6, 21, 46, 81, 126, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric. 2 73. 3 n 3 2 2 d 3 (n 1) 3 n 3 3 2 2 2 2 3 n 3 n 3 3 3 3 The difference between consecutive terms is constant. Therefore the sequence is arithmetic. 50 2 50 2 50 S50 3 k 3 k 3 k 1 3 k 1 k 1 2 50(50 1) 3(50) 150 850 700 3 2 3 74. 8 n 4 3 3 d 8 (n 1) 8 n 4 4 3 3 3 3 8 n 8 n 4 4 4 4 The difference between consecutive terms is constant. Therefore the sequence is arithmetic. 50 3 50 3 50 S50 8 k 8 k 4 k 1 4 k 1 k 1 3 50(50 1) 8(50) 4 2 400 956.25 556.25
75. 1, 3, 6, 10, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric. 76. 2, 4, 6, 8, ... The common difference is 2. The difference between consecutive terms is constant. Therefore the sequence is arithmetic. 50 50 50(50 1) S50 2k 2 k 2 2550 2 k 1 k 1 2 n 77. 3 n 1
2 n 1 n 2 3 2 r n 3 3 2 3 The ratio of consecutive terms is constant. Therefore the sequence is geometric. 2 1 k 50 2 3 2 S50 3 1 2 k 1 3 3
50
1.999999997
5 n 78. 4 n 1
5 n 1 n 5 4 5 r n 4 4 5 4 The ratio of consecutive terms is constant. Therefore the sequence is geometric. 5 1 50 5 4 5 S50 5 4 4 k 1 1 4
50
k
350,319.6161
79. –1, 2, –4, 8, ... 2 4 8 r 2 1 2 4 The ratio of consecutive terms is constant. Therefore the sequence is geometric. 50 1 (2)50 S50 1 (2) k 1 1 1 (2) k 1 3.752999689 1014
1518 Copyright © 2025 Pearson Education, Inc.
Section 13.3: Geometric Sequences; Geometric Series 80. 1, 1, 2, 3, 5, 8, ... There is no common difference and there is no common ratio. Therefore the sequence is neither arithmetic nor geometric. 81.
87. a.
a10 2(0.9)10 1 2(0.9)9 0.775 feet
3 n/2
r
b. Find n when an 1 :
n 1 3 2
2(0.9)n 1 1
n 1 n 3 2 2 31/ 2
n 3 2
0.9 n 1 0.5 (n 1) log 0.9 log 0.5 log 0.5 n 1 log 0.9 log 0.5 n 1 7.58 log 0.9
The ratio of consecutive terms is constant. Therefore the sequence is geometric. 50
S50 3
k/2
3
1/ 2
k 1
1 31/ 2
50
1 31/ 2
2.004706374 1012
82.
On the 8th swing the arc is less than 1 foot.
(1)
c.
n
(1) n 1
(1) n 1 n 1 (1) n The ratio of consecutive terms is constant. Therefore the sequence is geometric. 50 1 (1)50 0 S50 (1) k (1) 1 (1) k 1 r
Find the 10th term of the geometric sequence: a1 2, r 0.9, n 10
20 1 0.9
88. a.
an 24(0.8) n 1 24 0.8
1
0.8n
30 0.8 ft n
c.
Find n when an 0.5 : 24(0.8) n 1 0.5
0.8 n 1 0.020833 (n 1) log 0.8 log 0.020833 log 0.020833 n 1 log 0.8 log 0.020833 n 1 18.35 log 0.8
42000 1.03 $47, 271.37 4
86. This is a geometric series with a1 $15, 000, r 0.85, n 6 . Find the 6th term: 6 1
Find the 3rd term of the geometric sequence: a1 24, r 0.8, n 3
b. The height after the n th bounce is:
85. This is a geometric series with a1 $42, 000, r 1.03, n 5 . Find the 5th term:
a6 15000 0.85
15.88 feet
a3 24(0.8)31 24(0.8) 2 15.36 feet
84. Find the common ratio of the terms and solve the system of equations: x x2 r; r x 1 x x2 x x2 x 2 x2 x 2 x x 1
5 1
15
d. Find the infinite sum of the geometric series: 2 2 S 20 feet 1 0.9 0.1
83. Find the common ratio of the terms and solve the system of equations: x2 x3 r; r x x2 x2 x3 x 2 4 x 4 x 2 3x x 4 x x2
a5 42000 1.03
Find the sum of the first 15 swings: 1 0.9 15 1 (0.9)15 S15 2 2 0.1 1 0.9
On the 19th bounce the height is less than 0.5 feet.
15000 0.85 $6655.58 5
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Chapter 13: Sequences; Induction; the Binomial Theorem d. Find the infinite sum of the geometric series: 24 24 S 120 feet on the upward 1 0.8 0.2 bounce. For the downward motion of the ball: 30 30 S 150 feet 1 0.8 0.2 The total distance the ball travels is 120 + 150 = 270 feet. 89. This is an ordinary annuity with P $100 and n 12 30 360 payment periods. The 0.08 0.0067 . Thus, interest rate per period is 12 1 .08 360 1 $149, 035.94 A 100 12.08 12
90. This is an ordinary annuity with P $400 and n 12 3 36 payment periods. The interest 0.06 . Thus, 12 0.06 36 1 1 12 $15, 734.44 A 400 0.06 12
rate per period is
91. This is an ordinary annuity with P $500 and n 4 20 80 payment periods. The interest 0.05 0.0125 . Thus, 4 1 0.012580 1 $68, 059.40 A 500 0.0125
rate per period is
92. This is an ordinary annuity with P $1000 and n 2 15 30 payment periods. The interest 0.07 0.035 . Thus, 2 1 0.03530 1 $51, 622.68 A 1000 0.035
rate per period is
0.035 . Thus, 12 1 0.035 120 1 12 50, 000 P 0.035 12 0.035 12 $348.60 P 50, 000 1 0.035 120 1 12
interest rate per period is
94. This is an ordinary annuity with A $185, 000 and n 12 18 216 payment periods. The 0.0475 . Thus, 12 0.0475 216 1 1 12 185, 000 P 0.0475 12 0.0475 12 $543.48 P 185, 000 0.0475 216 1 1 12
interest rate per period is
95. This is a geometric sequence with a1 1, r 2, n 64 . Find the sum of the geometric series: 1 264 1 264 S64 1 264 1 1 2 1 1.845 1019 grains
96. This is an infinite geometric series with a1 1 , r 1 . 4 4 Find the sum of the infinite geometric series: 1 1 1 S 4 4 1 3 3 1 4 4 1 of the square is eventually shaded. 3
97. The common ratio, r 0.90 1 . The sum is: 1 1 S 10 . 1 0.9 0.10 The multiplier is 10.
93. This is an ordinary annuity with A $50, 000 and n 12 10 120 payment periods. The 1520 Copyright © 2025 Pearson Education, Inc.
Section 13.3: Geometric Sequences; Geometric Series 98. The common ratio, r 0.95 1 . The sum is: 1 1 S 20 . 1 0.95 0.05 The multiplier is 20.
On the 111th day or December 20, 2014, the amount will be less than $0.01. Find the sum of the geometric series: 1 0.9 111 1 rn S111 a1 1000 1 0.9 1 r 1 0.9 111 $9999.92 1000 0.1
99. This is an infinite geometric series with 1.03 a 4, and r . 1.09 4 $72.67 . Find the sum: Price 1.03 1 1.09
102. First, determine the number of seats in the section: n Sn 2a1 (n 1)d 2 14 2 2 13 2 210 seats 2 Now, find the sum of a geometric sequence with a1 0.01 and r 1.05 and n 210.
100. This is an infinite geometric series with 1.04 a1 2.5, and r . 1.11 2.5 Find the sum: Price $39.64 . 1.04 1 1.11
1 1.05 210 S210 0.01 $5633.36 1 1.05
101. Given: a1 1000, r 0.9 Find n when an 0.01 : 1000(0.9) n 1 0.01
0.9 n 1 0.00001 (n 1) log 0.9 log 0.00001 log 0.00001 n 1 log 0.9 log 0.00001 n 1 110.27 log 0.9 (cont on next column) ________________________________________________________________________________________________ y x r, z x r 2 103. x y z x x r x r 2 103 x(1 r r 2 ) 103 x 2 y 2 z 2 x 2 x 2 r 2 x 2 r 4 6901 x 2 (1 r 2 r 4 ) 6901
x(1 r r ) 103 2
2
2
x(1 2r 3r 2 2r 3 r 4 ) 10, 609 x 2 (1 r 2 r 4 ) x 2 (2r 2r 2 2r 3 ) 10, 609 x 2 (1 r 2 r 4 ) 2 xr x(1 r r 2 ) 10, 609 6901 2 xr 103 10, 609 206 xr 3708 xr 18 y
1521
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Chapter 13: Sequences; Induction; the Binomial Theorem
104. a.
The total area is the sum of the areas from each iteration: A A1 A2 A3 1 1 A1 2[1 triangle]; A2 2 3 [3 triangles]; Now each additional iteration adds four times as many 9 9 1 triangles as the previous iteration, each of which has an area the area of a triangle in the previous iteration. 9 2
3
1 1 So, A3 2 12 [12 triangles],A4 2 48 [48 triangles], and so on. The total area of the Koch 9 9 2
3
1 1 1 snowflake, in m2, is given by A 2 2 3 2 12 2 48 . 9 9 9
b. Splitting up the first term and regrouping factors, the series can be rewritten as 2
A
term a1 A
3
1 3 34 34 34 1 . The terms after form an infinite geometric series with the first 2 2 29 29 29 2 3 4 27 3 and common ratio r , which sums to S 1 4 . Therefore, the total area is 9 1 9 10 2
1 27 32 3.2 m 2 . 2 10 10
105. Both options are geometric sequences: Option A: a1 $40, 000; r 1.06; n 5 a5 40, 000(1.06)5 1 40, 000(1.06) 4 $50, 499 1 1.06 5 S5 40000 $225, 484 1 1.06
Option B: a1 $44, 000; r 1.03; n 5 a5 44, 000(1.03)5 1 44, 000(1.03) 4 $49,522 1 1.035 S5 44000 $233, 602 1 1.03 Option A provides more money in the 5th year, while Option B provides the greatest total amount of money over the 5 year period. 106. Find the sum of each sequence: A: Arithmetic series with: a1 $1000, d 1, n 1000 Find the sum of the arithmetic series: 1000 S1000 1000 1 500(1001) $500,500 2 B: This is a geometric sequence with a 1 1, r 2, n 19 . Find the sum of the geometric series: 1 219 1 219 219 1 $524, 287 S19 1 1 2 1
107. Option 1: Total Salary $2, 000, 000(7) $100, 000(7) $14, 700, 000
Option 2: Geometric series with: a1 $2, 000, 000, r 1.045, n 7 Find the sum of the geometric series: 1 1.045 7 $16, 038,304 S 2, 000, 000 1 1.045 Option 3: Arithmetic series with: a1 $2, 000, 000, d $95, 000, n 7 Find the sum of the arithmetic series: 7 S7 2(2, 000, 000) (7 1)(95, 000) 2 $15,995, 000 Option 2 provides the most money; Option 1 provides the least money. 108. The amount paid each day forms a geometric sequence with a1 0.01 and r 2 . 1 r 22 1 222 0.01 41,943.03 1 r 1 2 The total payment would be $41,943.03 if you worked all 22 days.
S22 a1
1522 Copyright © 2025 Pearson Education, Inc.
Section 13.3: Geometric Sequences; Geometric Series a22 a1 r 22 1 0.01 2 20,971.52
116.
21
The payment on the 22nd day is $20,971.52. Answers will vary. With this payment plan, the bulk of the payment is at the end so missing even one day can dramatically reduce the overall payment. Notice that with one sick day you would lose the amount paid on the 22nd day which is about half the total payment for the 22 days.
3 1 0 0 6 0 2 2 6 0 2 6 3 1 0 1 2 4 2 4 1 4 1 2 3(4 6) 1(0 24) 0 3(10) ( 24) 30 24 54
109. Yes, a sequence can be both arithmetic and geometric. For example, the constant sequence 3,3,3,3,..... can be viewed as an arithmetic sequence with a1 3 and d 0. Alternatively, the same sequence can be viewed as a geometric sequence with a1 3 and r 1.
117.
110. Answers will vary. 111. Answers will vary.
Angle A would be 180 – 7 = 173⁰. And thus angle B = 2⁰. Now use the Law of Sines to solve sin 2 sin 5 6 x for X: 6sin 5 x 14.98 sin 2 We also have angle C = 90 – 7 = 83⁰. So, using the sine function we have: Y sin 7 14.98 Y 14.98sin 7 1.825 Adding Liv’s height of 5.5 gives 1.825 5.5 7.3 ft
112. Answers will vary. Both increase (or decrease) exponentially, but the domain of a geometric sequence is the set of natural numbers while the domain of an exponential function is the set of all real numbers. 113. log 7 62 114. u
log 62 2.121 log 7
8i 15 j 8i 15 j v 8i 15 j v 82 ( 15) 2 8i 15 j 8 15 i j 17 17 17
115. Hyperbola: Vertices: (–2, 0), (2, 0); Focus: (4, 0); Center: (0, 0); Transverse axis is the x-axis; a 2; c 4 . Find b: b 2 c 2 a 2 16 4 12
118.
4 a(0 4) 2 (0 2)(0 1) 4 a(4) 2 (2)(1) 4 32 1 a 8 1 f ( x) ( x 4) 2 ( x 2)( x 1) 8
b 12 2 3
Write the equation:
f ( x) a( x 4) 2 ( x 2)( x 1)
x2 y 2 1 4 12
119.
16t 2 3t (16 3) t 1 16t 2 3t 13 (16t 13)(t 1) t 1 t 1 16t 13
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Copyright © 2025 Pearson Education, Inc.
Chapter 13: Sequences; Induction; the Binomial Theorem
120. Solve for t in the first equation and substitute into the second. x t 5 t x 5
3. I:
n 1: 1 2 3 and
1 k (k 5) , then 2 3 4 5 (k 2) [(k 1) 2]
II: If 3 4 5 (k 2)
y t
3 4 5 (k 2) (k 3)
y x5
121. The new equation would be: g ( x) 7 x 5 . 122. x 4 29 x 2 100 ( x 2 4)( x 2 25) ( x 2)( x 2)( x 5)( x 5)
1 k (k 5) (k 3) 2 1 5 k2 k k 3 2 2 1 2 7 k k 3 2 2 1 2 k 7k 6 2 1 (k 1)(k 6) 2 1 (k 1) k 1 5 2 Conditions I and II are satisfied; the statement is true.
Section 13.4 1. I:
n 1: 2 1 2 and 1(1 1) 2
II: If 2 4 6 2k k (k 1) , then 2 4 6 2k 2(k 1) 2 4 6 2k 2(k 1) k (k 1) 2(k 1) (k 1)(k 2)
4. I:
n 1: 2 1 1 3 and 1(1 2) 3
II: If 3 5 7 (2k 1) k (k 2) , then 3 5 7 (2k 1) [2(k 1) 1]
k 1 k 1 1
3 5 7 (2k 1) (2k 3)
Conditions I and II are satisfied; the statement is true.
k (k 2) (2k 3)
n 1: 4 1 3 1 and 1(2 1 1) 1
k 2 2k 2k 3
II: If 1 5 9 (4k 3) k (2k 1) , then 1 5 9 (4k 3) (4(k 1) 3)
k 2 4k 3 (k 1)(k 3)
2. I:
1 1(1 5) 3 2
(k 1) k 1 2
1 5 9 (4k 3) 4k 4 3
Conditions I and II are satisfied; the statement is true.
k (2k 1) 4k 1 2
2k k 4k 1 2k 2 3k 1 (k 1)(2k 1)
1 1(3 1 1) 2 2 1 II: If 2 5 8 (3k 1) k (3k 1) , 2 then
5. I:
k 1 2 k 1 1
Conditions I and II are satisfied; the statement is true.
n 1: 3 1 1 2 and
1524 Copyright © 2025 Pearson Education, Inc.
Section 13.4: Mathematical Induction 2 5 8 (3k 1) [3(k 1) 1]
8. I:
2 5 8 (3k 1) (3k 2) 1 3 1 k (3k 1) (3k 2) k 2 k 3k 2 2 2 2 3 2 7 1 k k 2 3k 2 7 k 4 2 2 2 1 (k 1)(3k 4) 2 1 (k 1) 3 k 1 1 2
1 3 32 3k 1 3k 1 k 1 1 (3 1) 3k 3k 3k 2 2 2 3 k 1 1 3 3 3k 1 2 2 2 1 k 1 3 1 2
9. I:
2
1 2 2 2
k 1
1 2 2 2 2
1 1 1 4k 1 4k 4k 4k 3 3 3 4 k 1 1 4 4 4k 1 3 3 3 1 k 1 4 1 3 Conditions I and II are satisfied; the statement is true.
10. I:
n 1: 511 1 and
2 1 , then
k
1 1 5 1 1 4
II: If 1 5 52 5k 1
2k 11 2
k
k 1
1 4 42 4k 1 4k
n 1: 211 1 and 21 1 1
II: If 1 2 2 2
1 4 42 4k 1 4k 11
1 3 1 k (3k 1) (3k 1) k 2 k 3k 1 2 2 2 3 2 5 1 2 k k 1 3k 5k 2 2 2 2 1 (k 1)(3k 2) 2 1 (k 1) 3 k 1 1 2 Conditions I and II are satisfied; the statement is true.
k 1
1 n 1: 411 1 and 41 1 1 3
2
1 II: If 1 4 42 4k 1 4k 1 , then 3
1 4 7 (3k 2) (3k 1)
7. I:
Conditions I and II are satisfied; the statement is true.
1 1(3 1 1) 1 2 1 II: If 1 4 7 (3k 2) k (3k 1) , 2 then 1 4 7 (3k 2) [3(k 1) 2] n 1: 3 1 2 1 and
1 k (3 1) , then 2
1 3 32 3k 1 3k 11
Conditions I and II are satisfied; the statement is true. 6. I:
1 1 (3 1) 1 2
II: If 1 3 32 3k 1
n 1: 311 1 and
1 k 5 1 , then 4
1 5 52 5k 1 5k 11 1 5 52 5k 1 5k
2k 1 2k 2 2k 1 2k 1 1 Conditions I and II are satisfied; the statement is true.
1 k 1 1 5 1 5k 5k 5k 4 4 4 5 1 1 5k 5 5 k 1 4 4 4 1 k 1 5 1 4
Conditions I and II are satisfied; the statement is true. 1525
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Chapter 13: Sequences; Induction; the Binomial Theorem
11. I:
n 1:
II: If
1 1 1 1 and 1(1 1) 2 11 2
1 1 1 1 k , then 1 2 2 3 3 4 k (k 1) k 1
1 1 1 1 1 1 1 1 1 1 1 2 2 3 3 4 k (k 1) (k 1)(k 1 1) 1 2 2 3 3 4 k (k 1) (k 1)(k 2)
k k k 2 1 1 k 1 (k 1)(k 2) k 1 k 2 (k 1)(k 2)
k 2 2k 1 k 1 (k 1)(k 1) k 1 (k 1)(k 2) (k 1)(k 2) k 2 k 1 1
Conditions I and II are satisfied; the statement is true. 12. I:
n 1:
II: If
1 1 1 1 and (2 1 1)(2 1 1) 3 2 1 1 3
1 1 1 1 k , then 1 3 3 5 5 7 (2k 1)(2k 1) 2k 1
1 1 1 1 1 1 3 3 5 5 7 (2k 1)(2k 1) (2(k 1) 1)(2(k 1) 1) 1 1 1 1 1 (2k 1)(2k 1) (2k 1)(2k 3) 1 3 3 5 5 7 k k 1 2k 3 1 2k 1 (2k 1)(2k 3) 2k 1 2k 3 (2k 1)(2k 3)
2k 2 3k 1 (k 1)(2k 1) k 1 k 1 (2k 1)(2k 3) (2k 1)(2k 3) 2k 3 2 k 1 1
Conditions I and II are satisfied; the statement is true. 13. I:
n 1: 12 1 and
1 1(1 1)(2 1 1) 1 6
II: If 12 22 32 k 2
1 k (k 1)(2k 1) , then 6
1 k (k 1)(2k 1) (k 1) 2 6 1 7 1 1 1 1 (k 1) k (2k 1) k 1 (k 1) k 2 k k 1 (k 1) k 2 k 1 (k 1) 2k 2 7k 6 6 6 6 3 3 6 1 (k 1)(k 2)(2k 3) 6 1 (k 1) k 1 1 2 k 1 1 6
12 22 32 k 2 (k 1) 2 12 22 32 k 2 (k 1) 2
Conditions I and II are satisfied; the statement is true.
1526 Copyright © 2025 Pearson Education, Inc.
Section 13.4: Mathematical Induction
14. I:
n 1: 13 1 and
1 2 1 (1 1) 2 1 4
II: If 13 23 33 k 3
1 2 k (k 1) 2 , then 4
13 23 33 k 3 (k 1)3 13 23 33 k 3 (k 1)3
1 2 k (k 1) 2 (k 1)3 4
1 1 (k 1) 2 k 2 k 1 (k 1) 2 k 2 4k 4 4 4 1 (k 1) 2 (k 2) 2 4 1 (k 1) 2 ((k 1) 1) 2 4
Conditions I and II are satisfied; the statement is true. 15. I:
n 1: 5 1 4 and
1 1(9 1) 4 2
II: If 4 3 2 (5 k )
1 k (9 k ) , then 2
4 3 2 (5 k ) 5 (k 1) 4 3 2 (5 k ) (4 k ) 9 1 1 7 1 k k 2 4 k k 2 k 4 k 2 7k 8 2 2 2 2 2 1 1 1 (k 1)(k 8) (k 1)(8 k ) (k 1) 9 (k 1) 2 2 2 Conditions I and II are satisfied; the statement is true.
16. I:
1 n 1: (1 1) 2 and 1(1 3) 2 2
1 II: If 2 3 4 (k 1) k (k 3) , then 2 2 3 4 (k 1) (k 1) 1 2 3 4 (k 1) (k 2) 1 1 3 1 5 k (k 3) (k 2) k 2 k k 2 k 2 k 2 2 2 2 2 2 1 2 1 k 5k 4 (k 1)(k 4) 2 2 1 (k 1)((k 1) 3) 2
Conditions I and II are satisfied; the statement is true. 17. I:
1 n 1: 1(1 1) 2 and 1(1 1)(1 2) 2 3
1527
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1 k (9 k ) (4 k ) 2
Chapter 13: Sequences; Induction; the Binomial Theorem
1 II: If 1 2 2 3 3 4 k (k 1) k (k 1)(k 2) , then 3 1 2 2 3 3 4 k (k 1) (k 1)(k 1 1) 1 2 2 3 3 4 k (k 1) (k 1)(k 2) 1 1 k (k 1)(k 2) (k 1)(k 2) (k 1)(k 2) k 1 3 3 1 (k 1)(k 2)(k 3) 3 1 (k 1)((k 1) 1)((k 1) 2) 3
Conditions I and II are satisfied; the statement is true. 18. I:
1 n 1: (2 1 1)(2 1) 2 and 1(1 1)(4 1 1) 2 3
1 II: If 1 2 3 4 5 6 (2k 1)(2k ) k (k 1)(4k 1) , then 3 1 2 3 4 5 6 (2k 1)(2k ) (2(k 1) 1)(2(k 1)) 1 2 3 4 5 6 (2k 1)(2k ) (2k 1)(k 1) 2 1 1 k (k 1)(4k 1) 2(k 1)(2k 1) (k 1) k (4k 1) 2(2k 1) 3 3
1 4 1 (k 1) k 2 k 4k 2 (k 1) 4k 2 k 12k 6 3 3 3 1 1 (k 1) 4k 2 11k 6 (k 1)(k 2)(4k 3) 3 3 1 (k 1)((k 1) 1)(4(k 1) 1) 3
Conditions I and II are satisfied; the statement is true.
19. I:
n 1: 12 1 2 is divisible by 2
II: If k 2 k is divisible by 2 , then (k 1) 2 (k 1) k 2 2k 1 k 1 (k 2 k ) (2k 2)
Since k 2 k is divisible by 2 and 2k 2 is divisible by 2, then (k 1) 2 (k 1) is divisible by 2. Conditions I and II are satisfied; the statement is true.
20. I:
n 1: 13 2 1 3 is divisible by 3
II: If k 3 2k is divisible by 3 , then (k 1)3 2(k 1) k 3 3k 2 3k 1 2k 2 (k 3 2k ) (3k 2 3k 3)
Since k 3 2k is divisible by 3 and 3k 2 3k 3 is divisible by 3, then (k 1)3 2(k 1) is divisible by 3. Conditions I and II are satisfied; the statement is true.
1528 Copyright © 2025 Pearson Education, Inc.
Section 13.4: Mathematical Induction
21. I:
n 1: 12 1 2 2 is divisible by 2
25. I:
II: If k 2 k 2 is divisible by 2 , then
II:
a k 1 b k 1 a a k b b k
(k 2 k 2) (2k )
a ak a bk a bk b bk
Since k 2 k 2 is divisible by 2 and 2k is divisible by 2, then (k 1) 2 (k 1) 2 is divisible by 2. Conditions I and II are satisfied; the statement is true.
Since a b is a factor of a k b k and a b is a factor of a b , then a b is a factor of a k 1 b k 1 . Conditions I and II are satisfied; the statement is true.
n 1: 1(1 1)(1 2) 6 is divisible by 6
26. I:
II:
2 k 1 1
a
2( k 1) 1
2 k 1 1
b
.
2( k 1) 1
a 2k 3 b2k 3
a 2 a 2 k 1 b 2 k 1 b 2 k 1 (a 2 b 2 )
Since a b is a factor of a 2 k 1 b 2 k 1 and a b is a factor of a 2 b 2 a b a b ,
n 1: If x 1 then x1 x 1.
then a b is a factor of a 2 k 3 b 2 k 3 . Conditions I and II are satisfied; the statement is true.
II: Assume, for some natural number k, that if x 1 , then x k 1 . Then x k 1 1, for x 1, x k 1 x k x 1 x x 1
27. I:
n 1 : 1 a 1 a 1 1 a 1
II: Assume that there is an integer k for which the inequality holds. We need to show that if
k
( x 1)
1 a 1 ka then k 1 1 a 1 k 1 a . k
Conditions I and II are satisfied; the statement is true. n 1: If 0 x 1 then 0 x1 1.
1 a
II: Assume, for some natural number k, that if 0 x 1 , then 0 x k 1 . Then, for 0 x 1,
k 1
1 a 1 a k
1 ka 1 a 1 ka 2 a ka 1 k 1 a ka 2
k
x x 1 x x 1
k 1
b
a 2 a 2 k 1 a 2 b 2 k 1 a 2 b 2 k 1 b 2 b 2 k 1
is divisible by 6. Conditions I and II are satisfied; the statement is true.
Thus, 0 x
If a b is a factor of a 2 k 1 b 2 k 1 , show that a b is a factor of a
k k 1 k 2 3 k 1 k 2
0 x
a3 b3 a b a 2 ab b 2
Thus, k 1 k 2 k 3
k 1
n 1: a b is a factor of a 211 b 211 a 3 b3 .
3 k 1 k 2 is divisible by 6.
24. I:
a a k b k b k ( a b)
II: If k (k 1)(k 2) is divisible by 6 , then (k 1)(k 1 1)(k 1 2) (k 1)(k 2)(k 3) k (k 1)(k 2) 3(k 1)(k 2). Now, k (k 1)( k 2) is divisible by 6; and since either k 1 or k 2 is even,
23. I:
If a b is a factor of a k b k , show that a b is a factor of a k 1 b k 1 .
(k 1) 2 (k 1) 2 k 2 2k 1 k 1 2
22. I:
n 1: a b is a factor of a1 b1 a b.
1.
1 k 1 a
Conditions I and II are satisfied; the statement is true.
Conditions I and II are satisfied, the statement is true. 1529
Copyright © 2025 Pearson Education, Inc.
Chapter 13: Sequences; Induction; the Binomial Theorem a ( a d ) ( a 2d ) a ( k 1) d a kd
28. n 1:
1 1 41 41 is a prime number.
a ( a d ) ( a 2d ) a ( k 1) d ( a kd )
n 41:
k (k 1) (a kd ) 2 k (k 1) (k 1)a d k 2
2
2
2
41 41 41 41 is not a prime number. 29. II: If 2 4 6 2k k 2 k 2 , then 2 4 6 2k 2(k 1) 2 4 6 2k 2k 2 k 2 k 2 2k 2 (k 2 2k 1) (k 1) 2 (k 1)2 (k 1) 2
I: 30. I:
n 1: 2 1 2 and 12 1 2 4 2 n 1: a r
11
1 r1 a and a a 1 r
1 rk II: If a a r a r 2 a r k 1 a , 1 r then a a r a r 2 a r k 1 a r k 11 a a r a r 2 a r k 1 a r k 1 rk k a a r 1 r a (1 r k ) a r k (1 r ) 1 r k a a r a r k a r k 1 1 r 1 r k 1 a 1 r Conditions I and II are satisfied; the statement is true.
31. I:
ka d
k 2 k 2k (k 1)a d 2 k2 k (k 1)a d 2 (k 1)k (k 1)a d 2
k 1 k 1 1 k 1 a d 2 Conditions I and II are satisfied; the statement is true. n 4 : The number of diagonals of a 1 quadrilateral is 4(4 3) 2 . 2 II: Assume that for any integer k, the number of diagonals of a convex polygon with k sides 1 (k vertices) is k (k 3) . A convex 2 polygon with k 1 sides ( k 1 vertices) consists of a convex polygon with k sides (k vertices) plus a triangle, for a total of ( k 1 ) vertices. The diagonals of this k 1 -sided convex polygon consist of the diagonals of the k-sided polygon plus k 1 additional diagonals. For example, consider the following diagrams.
32. I:
n 1: a (1 1)d a and 1 a d
1(1 1) a 2
II: If a (a d ) (a 2d ) a (k 1)d ka d
then
k (k 1) 2
k = 5 sides
1530 Copyright © 2025 Pearson Education, Inc.
Section 13.4: Mathematical Induction
Conditions I and II are satisfied; the statement is true. 33. I: n 3 : (3 2) 180 180 which is the sum of the angles of a triangle.
II: Assume that for any integer k, the sum of the angles of a convex polygon with k sides is (k 2) 180 . A convex polygon with k 1 sides consists of a convex polygon with k sides plus a triangle. Thus, the sum of the angles is (k 2) 180 180 ((k 1) 2) 180.
k + 1 = 6 sides k 1 = 4 new diagonals
Thus, we have the equation: 1 1 3 k (k 3) (k 1) k 2 k k 1 2 2 2 1 2 1 k k 1 2 2 1 k2 k 2 2 1 (k 1)(k 2) 2 1 (k 1)((k 1) 3) 2
Conditions I and II are satisfied; the statement is true.
5 8 4 1 1 8 1 5 8 34. I: n 1: (Condition I holds) 2 3 2 1 1 4 1 2 3 5 8 4k 1 8k II: If n 1: , 1 4k 2 3 2k 5 8 then 2 3
k 1
k
5 8 5 8 4k 1 8k 5 8 1 4k 2 3 2 3 2 3 2k 5 4k 1 2(8k ) 8(4k 1) 3 8k 4k 5 8k 8 5 2k 2(1 4k ) 8(2k ) 3 1 4k 2k 2 4k 3 4k 4 1 8k 8 4(k 1) 1 8(k 1) 2k 2 1 4k 4 2(k 1) 1 4(k 1)
(Condition II holds) 35.
a.
c3 2(3) 1 7
c.
I:
c1 21 1 1 .
c4 2(7) 1 15
b.
n 1: one fold results in 1 crease and
II: If ck 2k 1 , then
c1 1 21 1
ck 1 2ck 1 2 2k 1 1
c2 3 22 1 3
c3 7 2 1 4
c4 15 2 1
d.
So cn 2n 1 .
2k 1 2 1 2k 1 1 Each fold doubles the thickness so the stack thickness will be
1531
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Chapter 13: Sequences; Induction; the Binomial Theorem
225 0.02 mm 671, 088.64 mm (or about 671 meters). 36. Answers will vary.
F1 F2 F3
37. log 2 x 5 4
22 F1 23 F2
4
2
x5
16
x5
22 F1 i 22 F1 j
23 F2 i 12 F2 j 500 j
256 x 5 x 251
4 x 3 y 7 38. 2 x 5 y 16 Multiply each side of the second equation by –2 and add to eliminate x: 4 x 3 y 7 4 x 10 y 32 13 y 39 y 3 Substitute and solve for x:
2 2
the result into the second equation to solve the system: 2
F2 2
F1
2 3
2 2
F1 12
500 0
2 2
2 2 3
3 2
2 3
F1
F1
F 500 1
F1 448.3 lb F2
4x 2
2 (448.3) 366.0 lb 3
The tension in the left cable is about 448.3 kg and the tension in the right cable is about 366.0 kg.
1 x 2
The solution is x
1 1 , y 3 or , 3 . 2 2
39. Let F1 = the tension on the left cable, F2 = the tension on the right cable, and F3 = the force of the weight of the box. F1 F1 cos 135º i sin 135º j
i j 2 2
2 2
F2 F2 cos 30º i sin 30º j F2
F1 12 F2 500 j 0
Set the i and j components equal to zero and solve: 2 F 3 F 0 2 1 2 2 2 1 2 F1 2 F2 500 0 Solve the first equation for F2 and substitute
4 x 3( 3) 7 4 x 9 7
F1
i
3 2
i 12 j
é 3 -1ù ú é 1 2 -1ù ê ú ⋅ ê 1 0ú 40. AB = ê ê ú ê0 1 4úû ê ë ú 2 2 ë û é 1(3) + 2(1) -1(-2) 1(-1) + 2(0) -1(2)ù ú =ê ê0(3) + 1(1) + 4(-2) 0(-1) + 1(0) + 4(2)ú ë û é 7 -3ù ú =ê ê- 7 8úû ë 3x
A B + x + 2 x -1 3 x = A( x -1) + B( x + 2)
41. ( x + 2)( x -1)
F3 500 j For equilibrium, the sum of the force vectors must be zero.
=
Letting x = 1: 3 = A(1-1) + B (1 + 2) 3 = 3B B =1
1532 Copyright © 2025 Pearson Education, Inc.
Section 13.5: The Binomial Theorem
Letting x = -2: -6 = A(-2 -1) + B (-2 + 2)
n n! n(n 1)! n 1 n 1! ( 1)! 1 (n 1)!
-6 = -3 A A= 2
n n! 3. False; j j ! n j !
3x 2 1 = + ( x + 2)( x -1) x + 2 x -1
4. Binomial Theorem
42. Using the L.O.C.: 2
2
2
b = a + c - 2ac cos B
5 5! 5 4 3 2 1 5 4 5. 10 3 3! 2! 3 2 1 2 1 2 1
92 = 42 + 10.22 - 2(4)(10.2)cos B 81 = 120.04 - 81.6cos B -39.14 = -81.6cos B æ -39.14 ö÷ = 61.4 B = cos-1 çç çè -81.6 ÷÷ø
43.
7 7! 7 6 5 4 3 2 1 7 6 5 35 6. 3 3! 4! 3 2 1 4 3 2 1 3 2 1
7 7! 7 6 5 4 3 2 1 7 6 7. 21 5 5! 2! 5 4 3 2 1 2 1 2 1
e3 x-7 = 4 ln e3 x-7 = ln 4
(3x - 7)ln e = ln 4
9 9! 9 8 7 6 5 4 3 2 1 9 8 8. 36 7 7! 2! 7 6 5 4 3 2 1 2 1 2 1
3 x - 7 = ln 4 3 x = ln 4 + 7 ln 4 + 7 x= » 2.795 3
50 50! 50 49! 50 50 9. 1 49 49!1! 49! 1
1- 5 1 - cos 8 44. tan = = 2 1 + cos 1+ 5 8
100 100! 100 99 98! 100 99 10. 4950 98! 2 1 2 1 98 98! 2!
3
=
8 = 3 = 3 13 13 13 13 13 8
=
39 13
1000 1000! 1 11. 1 1000 1000! 0! 1 1000 1000! 1 1 12. 0 0!1000! 1
45. 0 ( x 1) 2 (3 x 2 ) ( x 1)( x 2 x 1)(2 x 2)
55 55! 13. 1.8664 1015 23 23! 32!
( x 1) 2 [(3 x 2 ) 2( x 2 x 1)] ( x 1) 2 [5 x 2 2 x 2]
60 60! 14. 4.1918 1015 20 20! 40!
The only real solution is x = 1.
47 47! 1.4834 1013 15. 25 25! 22!
Section 13.5
37 37! 16. 1.7673 1010 19 19!18!
1. Pascal Triangle n n! n! 1 2. 0 0! (n 0)! 1 n ! 1533
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Chapter 13: Sequences; Induction; the Binomial Theorem
5 5 5 5 5 5 17. ( x 1)5 x5 x 4 x3 x 2 x1 x 0 x5 5 x 4 10 x3 10 x 2 5 x 1 0 1 2 3 4 5 5 5 5 5 5 5 18. ( x 1)5 x5 (1) x 4 (1) 2 x3 (1)3 x 2 (1) 4 x1 (1)5 x 0 0 1 2 3 4 5 x5 5 x 4 10 x3 10 x 2 5 x 1 6 6 6 6 6 6 6 19. ( x 2)6 x 6 x5 ( 2) x 4 ( 2) 2 x3 ( 2)3 x 2 ( 2) 4 x( 2)5 x0 ( 2)6 0 1 2 3 4 5 6 x 6 6 x5 ( 2) 15 x 4 4 20 x3 ( 8) 15 x 2 16 6 x (32) 64 x 6 12 x5 60 x 4 160 x3 240 x 2 192 x 64 5 5 5 5 5 5 20. ( x 3)5 x5 x 4 (3) x3 (3) 2 x 2 (3)3 x1 (3) 4 x0 (3)5 0 1 2 3 4 5 x5 5 x 4 (3) 10 x3 9 10 x 2 (27) 5 x 81 243 x5 15 x 4 90 x3 270 x 2 405 x 243 4 4 4 4 4 21. (3x 1) 4 (3 x) 4 (3 x)3 (3 x) 2 (3 x) 0 1 2 3 4 81x 4 4 27 x3 6 9 x 2 4 3 x 1 81x 4 108 x3 54 x 2 12 x 1 5 5 5 5 5 5 22. (2 x 3)5 (2 x)5 (2 x) 4 3 (2 x)3 32 (2 x) 2 33 2 x 34 35 0 1 2 3 4 5 32 x5 5 16 x 4 3 10 8 x3 9 10 4 x 2 27 5 2 x 81 243 32 x5 240 x 4 720 x3 1080 x 2 810 x 243
23.
x y 0 x 1 x y 2 x y 3 x y 4 x y 5 y 2
5
2 5
5
2 5
2 4
5
2
2 3
5
2 2
2 2
5
2 3
5
2 4
2
2 5
x10 5 x8 y 2 10 x 6 y 4 10 x 4 y 6 5 x 2 y8 y10
24.
x y 0 x 1 x y 2 x y 3 x y 4 x y 2
6
2 6
6
2 6
2 5
6
2
2 4
6
2 2
2 3
6
2 3
5 6 6 6 x2 y 2 y 2 5 6
x12 6 x10 y 2 15 x8 y 4 20 x 6 y 6 15 x 4 y8 6 x 2 y10 y12
25.
x 2 0 x 1 x 2 2 x 2 3 x 2 6
6
6 4
6
6
5
1
6
4
x 2 5 x 2 6 2 2
4
6
5
6
6
2
3
6
x3 6 2 x5/ 2 15 2 x 2 20 2 2 x3/ 2 15 4 x 6 4 2 x1/ 2 8 x3 6 2 x5/ 2 30 x 2 40 2 x3/ 2 60 x 24 2 x1/ 2 8
1534 Copyright © 2025 Pearson Education, Inc.
3
2 2
2 4
Section 13.5: The Binomial Theorem
26.
x 3 0 x 1 x 3 2 x 3 3 x 3 4 3 4
4
4
4
3
1
4
2
2
4
3
4
4
x 2 4 3x3/ 2 6 3 x 4 3 3 x1/ 2 9 x 2 4 3x3/ 2 18 x 12 3x1/ 2 9
27.
5 0
5 1
5 2
5 3
5 4
5 5
ax by 5 ax 5 ax 4 by ax 3 by 2 ax 2 by 3 ax by 4 by 5 a 5 x5 5a 4 x 4by 10a3 x3b 2 y 2 10a 2 x 2b3 y 3 5axb 4 y 4 b5 y 5
28.
4 0
4 1
4 2
4 3
4 4
ax by 4 ax 4 ax 3 (by ) ax 2 by 2 ax by 3 by 4 a 4 x 4 4a 3 x3by 6a 2 x 2 b 2 y 2 4axb3 y 3 b 4 y 4
29. n 10, j 4, x x, a 3
33. n 9, j 2, x 2 x, a 3 9 9! 7 2 128 x7 (9) (2 x) 3 2 2! 7! 9 8 128 x7 9 2 1 41, 472 x 7
10 6 4 10! 10 9 8 7 81x 6 81x 6 x 3 4 4! 6! 4 3 2 1 17, 010 x 6
The coefficient of x 6 is 17, 010. 30. n 10, j 7, x x, a 3
The coefficient of x 7 is 41,472.
10 3 10! 7 2187 x3 x (3) 7 7! 3! 10 9 8 2187 x3 3 2 1 262, 440 x3
34. n 9, j 7, x 2 x, a 3 9 9! 2 7 4 x 2 ( 2187) (2 x) (3) 7! 2! 7 9 8 4 x 2 2187 2 1 314,928 x 2
The coefficient of x3 is 262,440.
31. n 12, j 5, x 2 x, a 1
The coefficient of x 2 is 314,928.
12 12! 7 5 128 x 7 (1) (2 x) (1) 5 5! 7! 12 11 10 9 8 (128) x 7 5 4 3 2 1
35. n 7, j 4, x x, a 3 7 3 4 7! 7 65 81x3 81x3 2835 x3 x 3 4! 3! 3 2 1 4
101,376 x 7
36. n 7, j 2, x x, a 3
The coefficient of x 7 is 101,376.
7 5 7! 76 2 9 x5 9 x5 189 x5 x (3) 2! 5! 2 1 2
32. n 12, j 9, x 2 x, a 1 12 12! 3 9 8 x3 (1) (2 x) (1) 9 9! 3! 12 11 10 8 x3 3 2 1 1760 x3 The coefficient of x3 is 1760.
37. n 9, j 2, x 3 x, a 2 9 9! 7 2 2187 x 7 4 (3x) ( 2) 2 2! 7! 9 8 8748 x 7 314,928 x7 2 1
1535 Copyright © 2025 Pearson Education, Inc.
Chapter 13: Sequences; Induction; the Binomial Theorem 38. n 8, j 5, x 3 x, a 2
41. The x 4 term in
8 8! 3 5 27 x3 32 (3x) ( 2) 5 5! 3! 87 6 864 x3 48,384 x3 3 2 1
j 12 12 12 2 12 j 1 24 3 j x j j x x j 0 j 0 occurs when: 24 3 j 0
24 3 j j 8 The coefficient is 12 12! 12 11 10 9 8 8! 4! 4 3 2 1 495
42. The x 2 term in 8
9
9
occurs when: 4 j 2
j
j 0
occurs when: 93j 0
j 2 j2 The coefficient is 8 2 8! 87 9 9 252 3 6! 2! 2 1 2
9 3j j 3 The coefficient is 9 9! 9 8 7 3 84 1 3! 6! 3 2 1 3
43. (1.001)5 1 103
j
8 8 j 4 j 3 3 x x j 0 j
8 j
j 0
9 9 1 j 93 j 1 x 2 x j 0 j
9 j
8
j x
40. The x 0 term in
j x
0 1 1 1 10 2 1 10 3 1 10 5
5
5
5
4
3
3
10 10 10 10 j 2 j 10 2 j x j 2 x x j 0 j 0 j occurs when: 3 10 j 4 2 3 j 6 2 j4 The coefficient is 10 10! 10 9 8 7 4 16 16 3360 2 6! 4! 4 3 2 1 4
39. The x 0 term in 12
j
10
5
3
3 2
5
2
3 3
1 5(0.001) 10(0.000001) 10(0.000000001) 1 0.005 0.000010 0.000000010 1.00501 (correct to 5 decimal places) 6 6 6 6 2 6 3 44. (0.998)6 1 0.002 16 15 ( 0.002) 14 0.002 13 0.002 0 1 2 3 1 6 0.002 15 0.000004 20 0.000000008 ... 1 0.012 0.00006 0.00000016 0.98805 (correct to 5 decimal places)
1536
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Section 13.5: The Binomial Theorem n n 1 ! n n! n! n 45. n 1 n n n n 1 !( 1 )! 1 !(1)! n 1! n n! n! n! n! 1 n n !( n n )! n ! 0! n ! 1 n!
46. 12! 479, 001, 600 4.790016 108 20! 2.432902008 1018 25! 1.551121004 1025 12
1 12 12! 2 12 1 479, 013,972.4 e 12 12 1 20
1 20 18 20! 2 20 1 2.43292403 10 e 12 20 1 25
1 25 25 25! 2 25 1 1.551129917 10 e 12 25 1
47. 2n (1 1) n n n n n 1n 1n 1 1 1n 2 12 1n n 1n 0 1 2 n n n n 0 1 n
n n n n 48. Show that (1) n 0 0 1 2 n n 0 (1 1) n n n n 1n 1n 1 (1) 1n 2 (1) 2 1n n (1) n 0 1 2 n n n n n (1) n 0 1 2 n 5 4 3 2 2 3 4 5 5 5 1 5 1 3 5 1 3 5 1 3 5 1 3 5 3 1 3 49. (1)5 1 0 4 1 4 4 2 4 4 3 4 4 4 4 4 5 4 4 4
50. a. The number of hexagons that can be formed using 8 points is the third entry (or 7th , by symmetry) in the row for n = 8 of the Pascal Triangle, 28. b. The number of triangles that can be formed using 10 points is the fourth entry (or 8th by symmetry) in the row for n = 10 of the Pascal Triangle, 120. c. The number of dodecagons that can be formed using 20 points is the 9th entry (or 13th by symmetry) in the row for n = 20 of the Pascal Triangle, 125,970.
1537
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Chapter 13: Sequences; Induction; the Binomial Theorem
51.
f ( x) (1 x 2 ) (1 x 2 ) 2 (1 x 2 )10 . This is a geometric series with a1 1 x 2 , r 1 x 2 , and n 10.
Therefore f ( x)
(1 x 2 ) 1 (1 x 2 )10 1 (1 x 2 )
(1 x 2 ) (1 x 2 )11 x2
. Since the denominator is x 2 , the coefficient of
x 4 in f ( x) will be the coefficient of x6 in the numerator which is the coefficient of x6 in the expansion of k 11 11 11 k (1 x 2 )11 1 ( x 2 ) . In general, the terms of this expansion are given by x 2 . To 1 11 k 3 11 113 11 x 2 x6 . The coefficient is obtain the term with x6 we need k 3 which gives 1 11 3 8
11 11! 165 . 8 8!3! 8
52. Using the binomial theorem the general coefficients of the expansion of a (b c) 2 are given by 8 j 5 4 2 2 8 j a (b c) . To obtain the term containing a b c , we need j 5 which gives j 8 5 8 5 6 k 6 k 6 2 3 6 a (b c) a (b c) . The terms in the expansion of (b c) are given by b c . To obtain 5 5 k 6 the term containing a5b 4 c 2 , we need k 4 which gives b 4 c 2 . Combining these results, the term containing 4 8 8 6 8 6 a5b 4 c 2 in the expansion of a (b c) 2 is a 5 b 4 c 2 a 5b 4 c 2 . The coefficient is 5 4 5 4
8 6 8! 6! 840 . 5 4 5!3! 4!2!
6 x 5 x 1
53.
x
ln 6 ln 5
x y z 0 55. 2 x y 3 z 1 4 x 2 y z 12
x 1
x ln 6 ( x 1) ln 5 x ln 6 x ln 5 ln 5
Add the first equation and the second equation to eliminate y: x y z 0 2 x y 3z 1 3x 2 z 1 Multiply each side of the first equation by 2 and add to the third equation to eliminate y: 2x 2 y 2z 0 4x 2 y z 12
x ln 6 x ln 5 ln 5 x(ln 6 ln 5) ln 5 ln 5 x 8.827 ln 6 ln 5 The solution set is 8.827
54. a. v w (2)(3) (3)( 2) 0
b. c.
cos 1 (0) 90 The vectors are orthogonal.
6 x 3z
12
2x z 4 Multiple the second derived equation by 2 and add the two results to eliminate z:
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Section 13.5: The Binomial Theorem 3 x 2 z 1
(4, 2).
4x 2z 8 7x 7 x 1
Substituting and solving for the other variables: 1 y ( 2) 0 2(1) z 4 1 y 2 0 z 2 y 3 z 2 y3 The solution is x 1, y 3, z 2 or (1,3, 2) .
57. g ( f ( x)) x 2 6 2 x 2 4 ( x 2)( x 2)
The domain is , 2 2,
x0 y0 56. x y 6 2 x y 10 Graph x 0; y 0 . Shaded region is the first quadrant. Graph the line x y 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 0 6 is true, shade the side of the line containing (0, 0). Graph the line 2 x y 10 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) 0 10 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:
58.
5 3 x 2x C 3 5 5 (3)3 2(3) C 3 5 45 6 C C 46 y
59. sin 2 sin 2 tan 2 sin 2 (1 tan 2 ) sin 2 sec 2 sin 2
1 cos 2
tan 2
60.
The x-axis and y-axis intersect at (0, 0). The intersection of x y 6 and the y-axis is (0, 6). The intersection of 2 x y 10 and the x-axis is (5, 0). To find the intersection of 2 x y 10 and x y 6 , solve the system:
1 1 2 1 23 3 x ( x3 1) x 3 x 3 (3 x 2 ) ( x 1) x(9 x 2 ) 3 3 ( x3 1) 2 ( x3 1) 2
( x3 1) 9 x3
x y 6 2 x y 10 Solve the first equation for x: x 6 y .
2
3x 3 ( x3 1) 2
Substitute and solve: 2(6 y ) y 10 12 2 y y 10 12 y 10 y 2 y2 x 6 (2) 4 The point of intersection is (4, 2). The four corner points are (0, 0), (0, 6), (5, 0), and
61.
1 8 x3 2
3 x 3 ( x3 1) 2
f ( x) is in lowest terms and is undefined at x 3 and x 1 . Therefore the vertical asymptotes are x 3 and x 1 .
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Chapter 13: Sequences; Induction; the Binomial Theorem
62.
(2) 2 1 4 1 2(2) 5 4 5 5 5 1
f ( x)
The point on the graph is (2,5) .
Chapter 13 Review Exercises 1. a1 (1)1
1 3 4 23 5 33 6 43 7 53 8 , a2 (1) 2 , a3 (1)3 , a4 (1) 4 , a5 (1)5 1 2 3 22 4 3 2 5 42 6 52 7
21 2 22 4 23 8 24 16 25 32 2. c1 2 2, c2 2 1, c3 2 , c4 2 1, c5 2 1 4 9 16 25 1 2 3 4 5
3. a1 3, a2
2 2 4 2 4 8 2 8 16 3 2, a3 2 , a4 , a5 3 3 3 3 3 9 3 9 27
4. a1 2, a2 2 2 0, a3 2 0 2, a4 2 2 0, a5 2 0 2 5.
4
(4k 2) 4 1 2 4 2 2 4 3 2 4 4 2 6 10 14 18 48 k 1
10. 0, 4, 8, 12, ... Arithmetic d 4 0 4 n n Sn 2(0) (n 1)4 4(n 1) 2n(n 1) 2 2
1 1 1 1 13 k 1 1 6. 1 1 2 3 4 13 k 1 k
7.
an n 5 Arithmetic
3 3 3 3 11. 3, , , , , ... Geometric 2 4 8 16 3 3 1 1 2 r 3 2 3 2 1 n 1 n 1 1 1 n 2 2 Sn 3 3 6 1 1 2 1 1 2 2
d (n 1 5) (n 5) n 6 n 5 1 n n Sn 6 n 5 (n 11) 2 2
8.
cn 2n3 Examine the terms of the sequence: 2, 16, 54, 128, 250, ... There is no common difference; there is no common ratio; neither.
9.
sn 23n Geometric r
23( n 1)
23n 3
12. Neither. There is no common difference or common ratio.
23n 33n 23 8
23 n 23 n 1 8n 1 8n 8 n Sn 8 8 8 1 1 8 7 7
13.
30
30
30
k 1
k 1
k 1
(k 2 2) k 2 2 30 30 1 2 30 1 2(30) 9515 6
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Chapter 13 Review Exercises
14.
40
40
k 1
k 1
Subtract the second equation from the first equation and solve for d. 13d 65 d 5 a1 31 6(5) 31 30 1
40
( 2k 8) 2k 8 40
k 1 40
k 1
k 1
2 k 8 40(1 40) 2 40(8) 2 1640 320 1320
an a1 n 1 d
1 n 1 5 1 5n 5 5n 4 General formula:
1 7 1 7 1 1 7 1 1 3 3 1 15. 1 2 3 3 3 k 1 1 3 3 1 1 1 2 2187 1 2186 1093 0.49977 2 2187 2187 k
10
16.
2 k 1
k
21. a10 a1 9d 0 a18 a1 17 d 8 ; Solve the system of equations: a1 9d 0
a1 17d 8 Subtract the second equation from the first equation and solve for d. 8d 8
1 2 10 2 1 ( 2) 2 1 1024 2 1023 3 3 682
d 1 a1 9(1) 9
an a1 n 1 d 9 n 11
9 n 1 n 10 General formula:
17. Arithmetic a1 3, d 4, an a1 (n 1)d a9 3 (9 1)4 3 8(4) 3 32 35
1 3 Since r 1, the series converges.
1 , n 11; an a1r n 1 10 111
1 a11 1 10
an n 10
22. a1 3, r
18. Geometric a1 1, r
an 5n 4
Sn
10
1 10
1 10, 000, 000, 000
a1 3 3 9 1 r 1 2 2 1 3 3
1 2 Since r 1 , the series converges.
23. a1 2, r
19. Arithmetic a1 2, d 2, n 9, an a1 (n 1)d
Sn
a9 2 (9 1) 2 2 8 2 9 2 12.7279
a1 2 2 4 1 r 1 3 3 1 2 2
1 3 , r 2 2 Since r 1 , the series diverges.
24. a1
20. a7 a1 6d 31 a20 a1 19d 96 ; Solve the system of equations: a1 6d 31 a1 19d 96
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Chapter 13: Sequences; Induction; the Binomial Theorem
then
1 2 Since r 1 , the series converges.
25. a1 4, r
Sn
26. I:
12 42 7 2 (3k 2) 2 3(k 1) 2 12 42 7 2 (3k 2) 2 (3k 1) 2
a1 4 4 8 1 r 1 1 1 2 2 n 1: 3 1 3 and
3 1 (1 1) 3 2
3k (k 1) , then 2 3 6 9 3k 3(k 1)
1 (k 1) 6k 2 9k 2 2 1 6k 3 6k 2 9k 2 9k 2k 2 2 1 2 6k k 1 9k k 1 2 k 1 2 1 (k 1) 6k 2 12k 6 3k 3 1 2
3k (k 1) 3(k 1) 2 3k 3(k 1) ((k 1) 1) (k 1) 3 2 2
1 (k 1) 6(k 2 2k 1) 3(k 1) 1 2 1 (k 1) 6(k 1) 2 3(k 1) 1 2 Conditions I and II are satisfied; the statement is true.
Conditions I and II are satisfied; the statement is true. n 1: 2 311 2 and 31 1 2
II: If 2 6 18 2 3k 1 3k 1 , then 2 6 18 2 3k 1 2 3k 11 2 6 18 2 3k 1 2 3k 3k 1 2 3k 3 3k 1 3k 1 1 Conditions I and II are satisfied; the statement is true. n 1:
(3 1 2) 2 1 and
1 1(6 12 3 1 1) 1 2
II: If 12 42 (3k 2) 2
3 6 9 3k 3(k 1)
28. I:
1 k 6k 2 3k 1 (3k 1) 2 2 1 6k 3 3k 2 k 18k 2 12k 2 2 1 3 6k 15k 2 11k 2 2
II: If 3 6 9 3k
27. I:
2
1 k 6k 2 3k 1 , 2
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Chapter 13 Review Exercises
5 5! 5 4 3 2 1 5 4 10 29. 2 2! 3! 2 1 3 2 1 2 1 5 5 5 5 5 5 30. ( x 2)5 x5 x 4 2 x3 22 x 2 23 x1 24 25 0 1 2 3 4 5 x5 5 2 x 4 10 4 x3 10 8 x 2 5 16 x 1 32 x5 10 x 4 40 x3 80 x 2 80 x 32 4 4 4 4 4 31. (3x 4) 4 (3x) 4 (3x)3 ( 4) (3 x) 2 ( 4) 2 (3 x)( 4)3 ( 4) 4 0 1 2 3 4 81x 4 4 27 x3 ( 4) 6 9 x 2 16 4 3 x( 64) 1 256 81x 4 432 x3 864 x 2 768 x 256
32. n 9, j 2, x x, a 2
a.
After striking the ground the third time, the 3
3 135 8.44 feet . height is 20 16 4
9 7 2 9! 9 8 4 x7 4 x 7 144 x 7 x 2 2 2! 7! 2 1 The coefficient of x 7 is 144.
b. After striking the ground the n th time, the n
3 height is 20 feet . 4
33. n 7, j 5, x 2 x, a 1 7 7! 76 2 5 4 x 2 (1) 4 x 2 84 x 2 (2 x) 1 5 5! 2! 2 1
c.
The coefficient of x 2 is 84.
If the height is less than 6 inches or 0.5 feet, then: 3 0.5 20 4
34. This is an arithmetic sequence with a1 80, d 3, n 25 a. b.
3 0.025 4
a25 80 (25 1)(3) 80 72 8 bricks
n
n
3 log 0.025 n log 4 log 0.025 12.82 n 3 log 4 The height is less than 6 inches after the 13th strike. d. Since this is a geometric sequence with r 1 , the distance is the sum of the two
25 (80 8) 25(44) 1100 bricks 2 1100 bricks are needed to build the steps. S25
35. This is an arithmetic sequence with a1 30, d 1, an 15 15 30 (n 1)(1) 15 n 1 16 n n 16 16 S16 (30 15) 8(45) 360 tiles 2 360 tiles are required to make the trapezoid.
infinite geometric series - the distances going down plus the distances going up. Distance going down: 20 20 Sdown 80 feet. 3 1 1 4 4 Distance going up:
36. This is a geometric sequence with 3 a1 20, r . 4 1543
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Chapter 13: Sequences; Induction; the Binomial Theorem 2. a1 4; an 3an 1 2
15 15 60 feet. 3 1 1 4 4 The total distance traveled is 140 feet.
Sup
a2 3a1 2 3 4 2 14 a3 3a2 2 3 14 2 44 a4 3a3 2 3 44 2 134
a5 3a4 2 3 134 2 404 The first five terms of the sequence are 4, 14, 44, 134, and 404.
37. This is an ordinary annuity with P $350 and n 12 20 240 payment periods. The 0.065 . Thus, 12 0.065 240 1 1 12 $171, 647.33 A 350 0.065 12
interest rate per period is
3.
3
1 k 1
k 1 k 1 2
k
11 1 1
2 1 2 1 31 3 1 2 1 2 1 2 1 2 3 22 33 44 1 1 1 1 4 9 3 4 61 2 4 9 36
1
38. This is a geometric sequence with a1 50, 000, r 1.04, n 5 . Find the fifth term of the sequence: a5 50, 000(1.04)51 50, 000(1.04) 4 58, 492.93 Her salary in the fifth year will be $58,492.93.
4
4.
2 k
3 k k 1
2
3
1
2 2
1
3
2 3
2
3
2 4
3
3
4
2 4 8 16 1 2 3 4 3 9 27 81 130 680 10 81 81
Chapter 13 Test 1. an
n2 1 n8
2 3 4 11 5. ... 5 6 7 14 Notice that the signs of each term alternate, with the first term being negative. This implies that the general term will include a power of 1 . Also note that the numerator is always 1 more than the term number and the denominator is 4 more than the term number. Thus, each term is in k k 1 the form 1 . The last numerator is 11 k 4 which indicates that there are 10 terms. 2 3 4 11 10 k k 1 ... 1 5 6 7 14 k 1 k 4
12 1 0 0 1 8 9 22 1 3 a2 2 8 10 32 1 8 a3 3 8 11 42 1 15 5 a4 4 8 12 4 52 1 24 a5 5 8 13 a1
The first five terms of the sequence are 0,
3 , 10
6. 6,12,36,144,... 12 6 6 and 36 12 24 The difference between consecutive terms is not constant. Therefore, the sequence is not arithmetic.
8 5 24 , , and . 13 11 4
12 2 and 36 3 6 12
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Chapter 13 Test
The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric.
n 9. an 7 2
n n 1 an an 1 7 7 2 2
1 7. an 4n 2 1 4n 1 4n 1 4 an 2 n 1 2 4 an 1 1 4 1 4n 1 2
n n 1 7 7 2 2 1 2 The difference between consecutive terms is constant. Therefore, the sequence is arithmetic 1 with common difference d and first term 2 1 13 a1 7 . 2 2 The sum of the first n terms of the sequence is given by n Sn a1 an 2 n 13 n 7 2 2 2
2
Since the ratio of consecutive terms is constant, the sequence is geometric with common ratio 1 r 4 and first term a1 41 2 . 2 The sum of the first n terms of the sequence is given by 1 rn Sn a1 1 r 1 4n 2 1 4 2 1 4n 3
8. 2, 10, 18, 26,... 10 2 8 , 18 10 8 , 26 18 8
The difference between consecutive terms is constant. Therefore, the sequence is arithmetic with common difference d 8 and first term a1 2 .
n 27 n 2 2 2
n 27 n 4
8 10. 25,10, 4, ,... 5
an a1 n 1 d
8
10 2 4 2 5 8 1 2 , , 25 5 10 5 4 5 4 5 The ratio of consecutive terms is constant. Therefore, the sequence is geometric with common ratio r 52 and first term a1 25 . The sum of the first n terms of the sequence is given by n 2 n 2 1 1 5 5 1 rn Sn a1 25 25 2 3 1 r 1 5 5 n n 5 2 125 2 25 1 1 3 5 3 5
2 n 1 8 2 8n 8 6 8n The sum of the first n terms of the sequence is given by n Sn a an 2 n 2 6 8n 2 n 4 8n 2 n 2 4n
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Chapter 13: Sequences; Induction; the Binomial Theorem
11. an
2n 3 2n 1
an an 1
2n 3 an 2n 3 2n 1 2n 3 2n 1 2n 1 an 1 2 n 1 3 2n 1 2n 5 2n 1 2n 5 2 n 1 1
2n 3 2 n 1 3 2n 3 2n 5 2n 1 2 n 1 1 2n 1 2n 1
The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric.
2n 3 2n 1 2n 5 2n 1 2n 1 2n 1
4 n 8n 3 4 n 8n 5 2
2
12. For this geometric series we have r
4n 2 1
and a1 256 . Since r
8
4n 2 1 The difference of consecutive terms is not constant. Therefore, the sequence is not arithmetic.
13.
64 1 256 4
1 1 1 , the series 4 4
converges and we get a 256 256 1024 5 S 1 1 1 r 1 5
4
4
3m 2 5 0 3m 5 1 3m 4 2 2 3m 3 2 2 3 3m 2 2 3 4 3m 2 4 5 2 5 5
5
5
5
4
3
5
5
5
2
243m 5 81m 2 10 27 m 4 10 9m 8 5 3m 16 32 243m5 810m 4 1080m3 720m 2 240m 32
14. First we show that the statement holds for n 1 . 1 1 1 1 1 2 1 1 1 1 The equality is true for n 1 so Condition I holds. Next we assume that 1 1 1 ... 1 n 1 is 1 2 3 n true for some k, and we determine whether the formula then holds for k 1 . We assume that 1 1 1 1 1 1 1 2 1 3 ... 1 k k 1 . 1 1 1 1 1 Now we need to show that 1 1 1 ... 1 1 k 1 1 k 2 . 1 2 3 k k 1 We do this as follows: 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 3 ... 1 k 1 k 1 1 1 1 2 1 3 ... 1 k 1 k 1 1 k 1 1 (using the induction assumption) 1 k 1 k 1 1 k 1 k 11 k 1 k2 Condition II also holds. Thus, formula holds true for all natural numbers.
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Chapter 13 Cumulative Review 15. The yearly values of the car form a geometric sequence with first term a1 31, 000 and common ratio r 0.85 (which represents a 15% loss in value). an 31, 000 0.85
2. a.
n 1
The nth term of the sequence represents the value of the car at the beginning of the nth year. Since we want to know the value after 10 years, we are looking for the 11th term of the sequence. That is, the value of the car at the beginning of the 11th year.
b. solving 3 x 2 y 2 100 3 x 2 3 y 2 300 2 3x 2 y 0 y 3x
a11 a1 r111 31, 000 0.85 6,103.11 10
After 10 years, the car will be worth $6,103.11.
3 y 2 y 300
16. The weights for each set form an arithmetic sequence with first term a1 100 and common difference d 30 . If we imagine the weightlifter only performed one repetition per set, the total weight lifted in 5 sets would be the sum of the first five terms of the sequence. an a1 n 1 d
3 y 2 y 300 3 y 2 y 300 0
y
S5 5 100 220 5 320 800 2
Since he performs 10 repetitions in each set, we multiply the sum by 10 to obtain the total weight lifted. 10 800 8000
1 3601 6
1 3601 1 3601 x2 x 18 18
The weightlifter will have lifted a total of 8000 pounds after 5 sets.
1 3601 0 18 Therefore, the system has solutions undefined since
Chapter 13 Cumulative Review 1.
2 3
1 3601 1 3601 x2 x 18 18 or 1 3601 1 3601 y 3x 2 6 6
n a an 2 2
1 12 4 3 300
Substitute and solve for x: 1 3601 1 3601 y 3x 2 6 6
a5 100 5 1 30 100 4 30 220 Sn
graphing x 2 y 2 100 and y 3x 2 .
x2 9
x
1 3601 1 3601 and ,y 18 6
x
1 3601 1 3601 . ,y 18 6
1 3601 1 3601 , , 18 6 1 3601 1 3601 , 18 6
x 2 9 or x 2 9 x 3 or x 3i
The solution set is 3, 3, 3i,3i
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Chapter 13: Sequences; Induction; the Binomial Theorem
The graphs intersect at the points 1 3601 1 3601 1.81,9.84 , 18 6
c.
a.
g 2 2 2 1 5 3 5
15 5 52 3 f g 2 f g 2 f 5 5 f 5
1 3601 1 3601 1.81,9.84 , 18 6
b.
3. 2e x 5 5 ex 2
3 4
12 6 42 2 g 6 2 6 1 13 f 4
c.
f g x f g x 3 2 x 1 2 x 1 2
5 The solution set is ln . 2
6x 3 2x 1
d. To determine the domain of the composition f g x , we start with the domain of g
4. slope m 5 ; Since the x-intercept is 2, we
know the point 2, 0 is on the graph of the line
and exclude any values in the domain of g that make the composition undefined. g x is defined for all real numbers and
and is a solution to the equation y 5 x b . y 5x b 0 5 2 b
f g x is defined for all real numbers
0 10 b
1 . Therefore, the domain of the 2 1 composite f g x is x | x . 2
except x
10 b Therefore, the equation of the line with slope 5 and x-intercept 2 is y 5 x 10 .
5. Given a circle with center (–1, 2) and containing the point (3, 5), we first use the distance formula to determine the radius.
e.
g f x 2
3x 1 x 2 6x 1 x2 6x x 2 x2 7x 2 x2
3 1 5 2 2 2
42 32 16 9 25 5 Therefore, the equation of the circle is given by
x 1 y 2 2 52 2
f.
x 12 y 2 2 52
To determine the domain of the composition g f x , we start with the domain of f and exclude any values in the domain of f that make the composition undefined. f x is defined for all real numbers except
x 2 2 x 1 y 2 4 y 4 25 x 2 y 2 2 x 4 y 20 0
6.
g f 4 g f 4 g 6 13
5 ln e x ln 2 5 x ln 0.916 2
r
x 2 and g f x is defined for all real
3x f x , g x 2x 1 x2
numbers except x 2 . Therefore, the domain of the composite g f x is
x | x 2 . 1548
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Chapter 13 Cumulative Review
g.
g x 2x 1
9. Center point (0, 4); passing through the pole (0,4) implies that the radius = 4 using rectangular coordinates:
y 2x 1 x 2 y 1 x 1 2 y
x h 2 y k 2 r 2 x 0 2 y 4 2 42
x 1 y 2
h.
x 2 y 2 8 y 16 16
g 1 x
x 1 2 The domain of g 1 x is the set of all real
x2 y 2 8 y 0 converting to polar coordinates: r 2 8r sin 0
numbers.
r 2 8r sin r 8sin
3x x2 3x y x2 3y x y2
f x
10. 2sin 2 x sin x 3 0, 0 x 2 2sin x 3 sin x 1 0 3 , which is impossible 2 3 sin x 1 0 sin x 1 x 2 3 Solution set . 2 2sin x 3 0 sin x
x y 2 3 y xy 2 x 3 y xy 3 y 2 x y x 3 2 x 2x x3 2 x f 1 x x 3
11. cos 1 0.5
y
We are finding the angle , , whose cosine equals 0.5 . cos 0.5 2 2 cos 1 0.5 3 3
The domain of f 1 x is x | x 3 . 7. Center: (0, 0); Focus: (0, 3); Vertex: (0, 4); Major axis is the y-axis; a 4; c 3 .
1 12. sin , is in Quadrant II 4
Find b: b 2 a 2 c 2 16 9 7 b 7 Write the equation using rectangular coordinates: x2 y 2 1 7 16
a.
is in Quadrant II cos 0 1 cos 1 sin 2 1 4
8. The focus is 1,3 and the vertex is 1, 2 .
Both lie on the vertical line x 1 . We have a = 1 since the distance from the vertex to the focus is 1 unit, and since 1, 3 is above
1
1, 2 , the parabola opens up. The equation of the parabola is:
b.
x h 2 4a y k
x 1 4 1 y 2 2
2
1 15 15 16 16 4
1 sin 4 1 4 tan cos 15 4 15 4
x 12 4 y 2 1549
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1 15
15 15
15 15
Chapter 13: Sequences; Induction; the Binomial Theorem 4. p1 (1.0043) p0 1127467 c.
p1 (1.0043)(324459463) 1127467
sin(2 ) 2sin cos
p1 326,982,106 The population is predicted to be 326,982,106 in 2018.
1 15 2 4 4 15 8
d.
5. Actual population in 2018: 326,766,748. The formula’s prediction was higher but fairly close. 6. Birth rate: 42.4 per 1000 population (0.0424) Death rate: 9.9 per 1000 population (0.0099) Population for 2017: 42,862,958 I = net immigration = 150000 r 0.0424 0.0099 0.0325 pn (1 0.0325) pn 1 150000
cos(2 ) cos 2 sin 2 2
15 1 2 4 4 15 1 14 7 16 16 16 8
e.
pn (1.0325) pn 1 150000 p0 42,862,958
2 4 2 2
p1 (1.0325) p0 150000
is in Quadrant I sin 0 2 2
p1 (1.0325)(42,862,958) 150000 p1 44,106, 004 The population is predicted to be 44,106,004 in 2018. Actual population in 2018: 44,270,563. The formula’s prediction was lower but fairly close.
15 1 1 cos 4 sin 2 2 2 4 15 4 2
4 15 8
7. Answers will vary. This appears to support the article. The growth rate for the U.S. is much smaller than the growth rate for Uganda.
4 15
8. It could be but one must consider trends in each of the pieces of data to find if the growth rate is increasing or decreasing over time. The same thing must be examined with respect to the net immigration.
2 2
Chapter 13 Projects Project II
Project I – Internet-based Project
1. 2, 4, 8, 16, 32, 64
Answers will vary based on the year that is used. Data used in these solutions will be from 2017.
2. length n 2n levels This is a geometric sequence: an 2n Recursive expression: an 2an 1 , a0 1
1. I = net immigration = 1,127,467 Population for 2017 = 325,700,000 2. r = 0.0125 – 0.0082 = 0.0043
3. 256 2n
3. pn (1 0.0043) pn 1 1127467
28 2n n8
pn (1.0043) pn 1 1127467 p0 324459463
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Chapter 13 Projects Project IV
Project III 1. Qst 3 2 Pt 1 , Qdt 18 3Pt
1. 1, 2, 4, 7, 11, 16, 22, 29
P0 2, b 2, d 3, c 18
2. It is not arithmetic because there is no common difference. It is not geometric because there is no common ration.
a 3 a 3 3 18 2 Pt 1 21 2 Pt 1 3 3 2 Pt 7 Pt 1 , P0 2 3
3. Scatter diagram
Pt
2.
2
6 2
4. y 2.5 x 2.5 The graph does not pass through any of the points. y6 12.5
Pt
y7 15 y8 17.5
Pt 1
y1 0 y2 2.5
y3 5
17 3. P1 3 Qs1 3 2(2) Qs1 1 P2
y4 7.5 y5 10
17 Qd 1 18 3 3 Qd 1 1
5
( yr yi ) i
i 1
(0 1) (2.5 2) (5 4) (7.5 7) (10 11) 1 This is the sum of the errors.
29 9
17 29 Qs 2 3 2 Qd 2 18 3 3 9 25 25 Qs 2 Qd 2 3 3 The market (supply and demand) are getting closer to being the same.
5. y 0.5 x 2 0.5 x 1 The graph passes through all of the points. y6 16 y7 22 y8 29
4. The equilibrium price is 4.20.
y1 1
5. It takes 17 time periods.
y2 2 y3 4
6. Qd 17 18 3(4.20) 5.40
y4 7
Qs17 3 2(4.20) 5.40 The equilibrium quantity is 5.4.
y5 11 5
( yr yi ) 0 i 1
i
The sum of the errors is zero. 1551
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Chapter 13: Sequences; Induction; the Binomial Theorem 6. When trying to obtain the cubic and quartic polynomials of best fit, the cubic and quartic terms have coefficient zero and the polynomial of best fit is given as the quadratic in part e. For the exponential function of best fit, y (0.59)(1.83) x . y6 22.2 y7 40.6 y8 74.2 The sum of these errors becomes quite large. This error shows that the function does not fit the data very well as x gets larger.
7. The quadratic function is best. 8. The data does not appear to be either logarithmic or sinusoidal in shape, so it does not make sense to try to fit one of those functions to the data.
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Chapter 14 Counting and Probability 13. n( A B ) 50, n( A B ) 10, n( B ) 20 n( A B ) n( A) n( B ) n( A B ) 50 n( A) 20 10 40 n( A)
Section 14.1 1. union 2. intersection 3. True; the union of two sets includes those elements that are in one or both of the sets. The intersection consists of the elements that are in both sets. Thus, the intersection is a subset of the union. 4. True; every element in the universal set is either in the set A or the complement of A.
14. n( A B ) 60, n( A B ) 40, n( A) n( B) n( A B ) n( A) n( B ) n( A B ) 60 n( A) n( A) 40 100 2n( A) n( A) 50
5. subset;
15. From the figure: n( A) 15 3 5 2 25
6. finite
16. From the figure: n( B ) 10 3 5 2 20
7. True
17. From the figure: n( A or B ) n( A B) 15 2 5 3 10 2 37
8. b 9. , a , b , c , d , a, b , a, c , a, d ,
18. From the figure: n( A and B ) n( A B ) 3 5 8
b, c , b, d , c, d , a, b, c , a, b, d , a, c, d , b, c, d , a, b, c, d
19. From the figure: n( A but not C ) n( A) n( A C ) 25 7 18
10. , a , b , c , d , e , a, b , a, c ,
a, d , a, e , b, c , b, d , b, e , c, d , c, e , d , e , a, b, c , a, b, d , a, b, e , a, c, d , a, c, e , a, d , e , b, c, d , b, c, e , b, d , e , c, d , e , a, b, c, d , a, b, c, e , a, b, d , e , a, c, d , e , b, c, d , e , a, b, c, d , e
20. From the figure: n( A) 10 2 15 4 31 21. From the figure: n( A and B and C ) n( A B C ) 5 22. From the figure: n( A or B or C ) n( A B C ) 15 3 5 2 10 2 15 52
11. n( A) 15, n( B ) 20, n( A B) 10 n( A B) n( A) n( B) n( A B)
23. There are 5 choices of shirts and 3 choices of ties; there are (5)(3) = 15 different arrangements.
15 20 10 25
24. There are 5 choices of blouses and 8 choices of skirts; there are (5)(8) = 40 different outfits.
12. n( A) 30, n( B ) 40, n( A B) 45 n( A B ) n( A) n( B ) n( A B) 45 30 40 n( A B ) n( A B ) 30 40 45 25
25. There are 9 choices for the first digit, and 10 choices for each of the other three digits. Thus, there are (9)(10)(10)(10) = 9000 possible fourdigit numbers. 1553
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Chapter 14: Counting and Probability
There are 8 different kinds of blood: A-Rh+, B-Rh+, AB-Rh+, O-Rh+, A-Rh–, B-Rh–, AB-Rh–, O-Rh–
26. There are 8 choices for the first digit, and 10 choices for each of the other four digits. Thus, there are (8)(10)(10)(10)(10) = 80,000 possible five-digit numbers.
31. a. n( widowed or divorced ) n( widowed ) n(divorced ) 3.7 11.0 14.7 There were 14.7 million men 15 years old and older who were widowed or divorced.
27. Let A those who will purchase a major appliance and B those who will buy a car
n(U ) 500, n( A) 200, n( B ) 150, n( A B ) 25 n( A B ) n( A) n( B ) n( A B )
b.
200 150 25 325 n(purchase neither) n U n A B 500 325 175 n(purchase only a car) n B n A B 150 25 125
32. a. n(divorced or separated ) n(divorced ) n(separated ) 14.6 2.6 17.2 There were 17.2 million women 15 years old and older who were divorced or separated.
28. Let A those who will attend Summer Session I and B those who will attend Summer Session II n( A) 200, n( B) 150, n( A B ) 75,
n A B 275
b.
n( A B ) n( A) n( B ) n( A B ) 200 150 75 275
n(U ) n( A B) n A B 275 275 550
550 students participated in the survey. 29. Construct a Venn diagram: 15
15
15
5 15
n(married, widowed or divorced) n(married ) n( widowed ) n(divorced) 64.0 11.5 14.6 90.1 There were 91.6 million women 15 years old and older who were married, widowed, or divorced.
33. There are 8 choices for the DOW stocks, 15 choices for the NASDAQ stocks, and 4 choices for the global stocks. Thus, there are (8)(15)(4) = 480 different portfolios.
AT&T
IBM
n(married, divorced or separated) n(married ) n(divorced ) n(separated) 63.8 11.0 2.0 76.8 There were 76.8 million males 15 years old and older who were married, divorced, or separated.
10 10
34 – 35. Answers will vary.
15
36. The graph is a circle with center at (2, 1) and radius 3.
GE
(a) 15
(b) 15
(c) 15
(d) 25
(e) 40 30. Construct a Venn diagram:
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Section 14.1: Counting
The solutions are: 8,3 and 3, 2 .
37. a 2, b 2 c 3 a b c 2bc cos A 2
2
2
b 2 c 2 a 2 22 32 22 3 cos A 2bc 2(2)(3) 4
42. (2 x 7)(3x 2 5 x 4) 6 x 10 x 8 x 21x 35 x 28 3
3 A cos 1 41.4º 4
cos B
43. There is a common ratio between the terms:
a c b 2 3 32 3 2ac 2(2)(3) 4 2
2
2
2
2
C 180º A B 180º 41.4º 41.4º 97.2º f ( x) ( x 2)( x 2 3x 10)
44.
1 2 2 1 ( x 2) 3 ( x 2) 3 3 3 1 2 1 ( x 2) 3 2( x 2) 3 1 3 1 1 2( x 2) 3 1 2 3 3( x 2)
f ( x)
0 ( x 2)( x 5)( x 2) ( x 2) 0 ( x 5) 0 ( x 2) 0 x2 x5 The zeros are 2,5, 2 .
x 2
39. log 3 x log 3 2 2
a.
log 3 (2 x) 2
1
1 2 1 x2 8 15 x 8 b. f ( x) is undefined when the denominator is zero. 1
( x 2) 3
x3 72 x x3 72 x 0 x( x 2 72) 0 x 2 72
f ( x) 0 when the numerator is zero. 2( x 2) 3 1 0
32 2 x 1 1 2x x 9 18 1 The solution set is . 18
40.
12
5 0.6 . Because r < 1, the series 4 converges. The first term is 4. So the sum is a 4 4 10 . S 1 r 1 0.6 0.4 r
3 B cos 1 41.4º 4
38.
2
6 x3 31x 2 43 x 28
b 2 a 2 c 2 2ac cos B 2
2
2
( x 2) 3 0 ( x 2) 0 x2
or x 0
x 72 6 2
The solution set is 6 2 . 41. Solve the first equation for x and substitute into the second equation. x y 5 x y5 y 5 y 2 1 y y6 0 ( y 3)( y 2) 0 2
y 3 or y 2 For y 3 : x 3 5 8 For y 2 : x 2 5 3
45. Find the partial fraction decomposition: 3x 2 15 x 5 A B C x x 1 ( x 1) 2 x( x 1) 2
Multiplying both sides by x( x 1) 2 , we obtain: 3x 2 15 x 5 A( x 1) 2 Bx( x 1) Cx Let x 1 , then 3(1) 2 15(1) 5 A(1 1) 2 B(1)(1 1) C (1) 7 1C C 7
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Chapter 14: Counting and Probability
Let x 0 , then 3(0) 2 15(0) 5 A(0 1) 2 B (0)(0 1) C (0) 5 A 0 B 0C 5 A Let x 1 , then 3(1) 2 15(1) 5 5(1 1) 2 B (1)(1 1) 7(1) 23 20 2 B 7 2 B 4 B 2 2 3x 15 x 5 5 2 7 2 x x 1 ( x 1) 2 ( x 2)( x 1)
Section 14.2 1. 1; 1 2. b 3. permutation
11. P (7, 0)
7! 7! 1 (7 0)! 7!
12. P (9, 0)
9! 9! 1 (9 0)! 9!
13. P (8, 4)
8! 8! 8 7 6 5 4! 1680 (8 4)! 4! 4!
14. P (8, 3)
8! 8! 8 7 6 5! 336 (8 3)! 5! 5!
15. C (8, 2)
8! 8! 8 7 6! 28 (8 2)! 2! 6! 2! 6! 2 1
16. C (8, 6)
8! 8! 8 7 6! 28 (8 6)! 6! 2! 6! 6! 2 1
17. C (7, 4)
7! 7! 7 6 5 4! 35 (7 4)! 4! 3! 4! 4! 3 2 1
18. C (6, 2)
6! 6! 6 5 4! 15 (6 2)! 2! 4! 2! 4! 2 1
19. C (15, 15)
4. combination
18! 18! 18 17! 18 (18 1)!1! 17!1! 17! 1
5. P (n, r )
n! (n r )!
20. C (18, 1)
6. C (n, r )
n! (n r )! r !
21. C (26, 13)
6! 6! 6 5 4! 30 7. P (6, 2) (6 2)! 4! 4!
22. C (18, 9)
26! 26! 10, 400, 600 (26 13)!13! 13!13!
18! 18! 48, 620 (18 9)! 9! 9! 9!
23. {abc, abd , abe, acb, acd , ace, adb, adc, ade, aeb, aec, aed , bac, bad , bae, bca, bcd , bce, bda, bdc, bde, bea, bec, bed , cab, cad , cae, cba, cbd , cbe, cda, cdb, cde, cea, ceb, ced , dab, dac, dae, dba, dbc, dbe, dca, dcb, dce, dea, deb, dec, eab, eac, ead , eba, ebc, ebd , eca, ecb, ecd , eda, edb, edc}
7! 7! 7 6 5! 42 8. P (7, 2) (7 2)! 5! 5!
9. P (4, 4)
15! 15! 15! 1 (15 15)!15! 0!15! 15! 1
4! 4! 4 3 2 1 24 (4 4)! 0! 1
8! 8! (8 8)! 0! 8 7 6 5 4 3 2 1 40,320 1
10. P (8, 8)
P (5,3)
5! 5! 5 4 3 2! 60 (5 3)! 2! 2!
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Section 14.2: Permutations and Combinations
24. {ab, ac, ad , ae, ba, bc, bd , be, ca, cb, cd , ce, da, db, dc, de, ea, eb, ec, ed } P (5, 2)
5! 5! 5 4 3! 20 (5 2)! 3! 3!
30. {123, 124, 125, 126, 134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256, 345,346, 356, 456} C (6,3)
6! 6 5 4 3! 20 (6 3)! 3! 3 2 1 3!
25. {123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241, 243, 312, 314, 321, 324, 341, 342, 412, 413, 421, 423, 431, 432}
31. There are 4 choices for the first letter in the code and 4 choices for the second letter in the code; there are (4)(4) = 16 possible two-letter codes.
4! 4! 4 3 2 1 24 (4 3)! 1! 1
32. There are 5 choices for the first letter in the code and 5 choices for the second letter in the code; there are (5)(5) = 25 possible two-letter codes.
P (4,3)
26. {123, 124, 125, 126, 132, 134, 135, 136, 142, 143, 145, 146, 152, 153, 154, 156, 162, 163, 164, 165, 213, 214, 215, 216, 231, 234, 235, 236, 241, 243, 245, 246, 251, 253, 254, 256, 261, 263, 264, 265, 312, 314, 315, 316, 321, 324, 325, 326, 341, 342, 345, 346, 351, 352, 354, 356, 361, 362, 364, 365, 412, 413, 415, 416, 421, 423, 425, 426, 431, 432, 435, 436, 451, 452, 453, 456, 461, 462, 463, 465, 512, 513, 514, 516, 521, 523, 524, 526, 531, 532, 534, 536, 541, 542, 543, 546, 561, 562,563, 564, 612, 613, 614, 615, 621, 623, 624, 625, 631, 632, 634, 635, 641, 642, 643, 645, 651, 652, 653, 654} P (6,3)
6! 6! 6 5 4 3! 120 (6 3)! 3! 3!
27. {abc, abd , abe, acd , ace, ade, bcd , bce, bde, cde} C (5,3)
5! 5 4 3! 10 (5 3)! 3! 2 1 3!
28. {ab, ac, ad , ae, bc, bd , be, cd , ce, de} C (5, 2)
5! 5 4 3! 10 (5 2)! 2! 3! 2 1
29. {123, 124, 134, 234} C (4,3)
4! 4 3! 4 (4 3)! 3! 1! 3!
33. There are two choices for each of three positions; there are (2)(2)(2) = 8 possible threedigit numbers. 34. There are ten choices for each of three positions; there are (10)(10)(10) = 1000 possible three-digit numbers. (Note this is if we allow numbers with initial zeros such as 012.) 35. To line up the four people, there are 4 choices for the first position, 3 choices for the second position, 2 choices for the third position, and 1 choice for the fourth position. Thus there are (4)(3)(2)(1) = 24 possible ways four people can be lined up. 36. To stack the five boxes, there are 5 choices for the first position, 4 choices for the second position, 3 choices for the third position, 2 choices for the fourth position, and 1 choice for the fifth position. Thus, there are (5)(4)(3)(2)(1) = 120 possible ways five boxes can be stacked. 37. Since no letter can be repeated, there are 5 choices for the first letter, 4 choices for the second letter, and 3 choices for the third letter. Thus, there are (5)(4)(3) = 60 possible threeletter codes. 38. Since no letter can be repeated, there are 6 choices for the first letter, 5 choices for the second letter, 4 choices for the third letter, and 3 choices for the fourth letter. Thus, there are (6)(5)(4)(3) = 360 possible three-letter codes. 39. There are 26 possible one-letter names. There are (26)(26) = 676 possible two-letter names. There are (26)(26)(26) = 17,576 possible threeletter names. Thus, there are 26 + 676 + 17,576 = 18,278 possible companies that can be listed on the New York Stock Exchange.
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Chapter 14: Counting and Probability 40. There are (26)(26)(26)(26) = 456,976 possible four-letter names. There are 26)(26)(26)(26)(26) = 11,881,376 possible five-letter names. Thus, there are 456,976 + 11,881,376 = 12,338,352 possible companies that can be listed on the NASDAQ.
47. The 1st person can have any of 365 days, the 2nd person can have any of the remaining 364 days. Thus, there are (365)(364) = 132,860 possible ways two people can have different birthdays. 48. The first person can have any of 365 days, the second person can have any of the remaining 364 days, the third person can have any of the remaining 363 days, the fourth person can have any of the remaining 362 days, and the fifth person can have any of the remaining 361 days. Thus, there are (365)(364)(363)(362)(361) = 6,302,555,018,760 possible ways five people can have different birthdays.
41. A committee of 4 from a total of 7 students is given by: 7! 7! 7 6 5 4! C (7, 4) 35 (7 4)! 4! 3! 4! 3 2 1 4! 35 committees are possible. 42. A committee of 3 from a total of 8 professors is given by: 8! 8! 8 7 6 5! C (8,3) 56 (8 3)! 3! 5! 3! 3 2 1 5! 56 committees are possible.
49. Choosing 2 boys from the 4 boys can be done C(4,2) ways. Choosing 3 girls from the 8 girls can be done in C(8,3) ways. Thus, there are a total of: 4! 8! (4 2)! 2! (8 3)! 3! 4! 8! 2! 2! 5! 3! 4 3 2! 8 7 6 5! 2 1 2! 5! 3!
C (4,2) C (8,3)
43. There are 2 possible answers for each question. Therefore, there are 210 1024 different possible arrangements of the answers. 44. There are 4 possible answers for each question. Therefore, there are 45 1024 different possible arrangements of the answers.
6 56 336
50. The committee is made up of 2 of 4 administrators, 3 of 8 faculty members, and 5 of 20 students. The number of possible committees is:
45. There are 5 choices for the first position, 4 choices for the second position, 3 choices for the third position, 2 choices for the fourth position, and 1 choice for the fifth position. Thus, there are (5)(4)(3)(2)(1) = 120 possible arrangements of the books. 46. a.
There are 26 choices for each of the first two positions, and 10 choices for each of the next four positions. Thus, there are (26)(26)(10)(10)(10)(10) = 6,760,000 possible license plates.
b.
There are 26 choices for each of the first two positions, 10 choices for the first digit, 9 choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit. Thus, there are (26)(26)(10)(9)(8)(7) = 3,407,040 possible license plates.
c.
C (4,2) C (8,3) C (20,5) 4! 8! 20! (4 2)! 2! (8 3)! 3! (20 5)! 5! 4! 8! 20! 2! 2! 5! 3! 15! 5! 4! 8 7 6 5! 20 19 18 17 16 15! 2 1 2 1 5 4! 3! 15! 5!
5,209,344 possible committees
51. This is a permutation with repetition. There are 9! 90, 720 different words. 2! 2!
There are 26 choices for the first letter, 25 choices for the second letter, 10 choices for the first digit, 9 choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit. Thus, there are (26)(25)(10)(9)(8)(7) = 3,276,000 possible license plates.
52. This is a permutation with repetition. There are 11! 4,989, 600 different words. 2! 2! 2! 53. a.
C (7, 2) C (3,1) 21 3 63
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Section 14.2: Permutations and Combinations
b.
C (7,3) C (3, 0) 35 1 35
c.
C (3,3) C (7, 0) 1 1 1
54. a. b. c.
C (15,5) C (10, 0) 3003 1 3003 C (15,3) C (10, 2) 455 45 20, 475 C (15, 4) C (10,1) C (15,5) C (10, 0) 1365 10 3003 1 13, 650 3003 16, 653
55. There are C(100, 22) ways to form the first committee. There are 78 senators left, so there are C(78, 13) ways to form the second committee. There are C(65, 10) ways to form the third committee. There are C(55, 5) ways to form the fourth committee. There are C(50, 16) ways to form the fifth committee. There are C(34, 17) ways to form the sixth committee. There are C(17, 17) ways to form the seventh committee. The total number of committees C (100, 22) C (78,13) C (65,10) C (55,5) C (50,16) C (34,17) C (17,17) 1.157 1076
56. The team is made up of 5 of 10 linemen, 3 of 10 linebackers, and 3 of 5 safeties. The number of possible teams is: C (10,5) C (10,3) C (5,3) 252 120 10 302, 400 There are 302,400 possible defensive teams. 57. There are 9 choices for the first position, 8 choices for the second position, 7 for the third position, etc. There are 9 8 7 6 5 4 3 2 1 9! 362,880 possible batting orders.
58. There are 8 choices for the first position, 7 choices for the second position, 6 for the third position, etc. and 1 choice for the last position. There are 8 7 6 5 4 3 2 1 1 8! 1 40,320 possible batting orders. 59. The team must have 1 pitcher and 8 position players (non-pitchers). For pitcher, choose 1 player from a group of 4 players, i.e., C(4, 1). For position players, choose 8 players from a group of 11 players, i.e., C(11, 8). Thus, the number different teams possible is C (4,1) C (11,8) 4 165 660. 60. Consider the ways that the American League can win. Then multiply by 2 to get the total for both leagues. To win the World Series, the last game must be won. There is 1 way to win in four games. To win in 5 games, three of the first four must be won, so there are C(4, 3) = 4 ways to win in 5 games. To win in 6 games, three of the first five must be won, so there are C(5, 3) = 10 ways to win in 6 games. To win in 7 games, three of the first six must be won, so there are C(6, 3) = 20 ways to win in 7 games. Therefore, there are 1 + 4 + 10 + 20 = 35 ways the American League can win the World Series. There are also 35 ways the National League can win the World Series. There are a total of 70 different sequences possible. 61. Choose 2 players from a group of 6 players. Thus, there are C (6, 2) 15 different teams possible. 62. Choose 1 of 2 centers, 2 of 3 guards, and 2 of 7 forwards. There are C (2,1) C (3, 2) C (7, 2) 2 3 21 126 different teams possible. 63. a.
b.
If numbers can be repeated, there are (50)(50)(50) = 125,000 different lock combinations. If no number can be repeated, then there are 50 49 48 117, 600 different lock combinations. Answers will vary. Typical combination locks require two full clockwise rotations to the first number, followed by a full counter-clockwise rotation past the first number to the second number, followed by a clockwise rotation to the third number (not past the second). This is not clear from the given directions. Perhaps a better name for a combination lock would be a permutation lock since the order in which the numbers are entered matters.
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Chapter 14: Counting and Probability 64. For each possible number of characters for the password, 8 through 12, find the total number of arrangements of the 36 letters and digits, and then subtract both the number of arrangements consisting only of the 26 letters (because the password must have at least one digit) and the number os arrangements consisting only of the 10 digits (because the password must have at least one letter). Finally, add the results to obtain the total number of passwords: 368 268 108 369 269 109 3610 2610 1010 3611 2611 1011 3612 2612 1012 8-char passwords
9-char passwords
10-char passwords
11-char passwords
12-char passwords
18
=4.774516364 10
70. sin 75 sin(45 30)
65 – 66. Answers will vary.
sin 45 cos30 cos 45 sin 30
67. A permutation is an ordered arrangement of objects while with a combination order does not matter. For example, the number of ways the 11 teams in the Big Ten can come in first, second, and third would be a permutation problem. The number of ways to pick 6 numbers in the Illinois State lottery is a combination problem because the order in which the numbers are selected is irrelevant.
2 3 2 1 2 2 2 2 6 2 6 2 4 4 4 cos15 cos(45 30) cos 45 cos30 sin 45 sin 30
2 3 2 1 2 2 2 2
68. First solve for s r 5 4 5 4 1 A r 2 2 1 2 5 (4) 4 2
6 2 4 4
Also, cos15
10 ft 2
6 2 4
1 cos30 2 1 2
3 2
2 3 4
2 3 2
71. an a1r n 1
69. ( g f )( x) g (2 x 1)
a5 5( 2)5 1
(2 x 1) 2 (2 x 1) 2
5( 2)4 80
4x2 4x 1 2 x 1 2 4x2 2x 2
5 5 5 2 5 3 5 4 5 5 72. (2 x 3)5 ( x)5 ( x) 4 2 y ( x)3 2 y ( x) 2 2 y x 2 y 2 y 0 1 2 3 4 5 x5 5 x 4 2 y 10 x3 4 y 2 10 x 2 8 y 3 5 x 16 y 4 32 y 5 x5 10 x 4 y 40 x3 y 2 80 x 2 y 3 80 xy 4 32 y 5
3x 4 y 5 73. 5 x 2 y 17 Multiply each side of the second equation by 2, and add the equations to eliminate y:
3x 4 y 5 10 x 4 y 34 13x
39
x3 Substitute and solve for y:
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Section 14.3: Probability 3 3 4 y 5
77.
9 4y 5
6x
3
2
10( x 3) 5
( x 3) 5
4 y 4
y 1 The solution of the system is x 3, y 1 or
using ordered pairs 3, 1 .
6 6 14 5
3 1 4 2 2
2
y 1 3 x 3 3
5 6 The polar form of z 3 i is 5 5 i sin z r cos i sin 2 cos 6 6
5 i 6
76. Find the partial fraction decomposition: 5 x 2 3x 14 Ax B Cx D 2 ( x 2 2) 2 x 2 ( x 2 2) 2
Multiplying both sides by ( x 2 2) 2 , we obtain: 5 x 2 3x 14 ( Ax B)( x 2 2) Cx D 5 x 2 3x 14 Ax3 Bx 2 2 Ax 2 B Cx D 5 x 2 3x 14 Ax3 Bx 2 (2 A C ) x 2 B D
A0;
B 5;
2A C 3
2 B D 14
2(0) C 3
2(5) D 14
C 3
D4
5 x 2 3x 14 ( x 2 2) 2
10( x 3) 2
( x 3) 5 ( x 3) 5 6 x 10 x 30 2
( x 3) 5 16 x 30 2
Section 14.3 1. equally likely 2. complement
75. r x 2 y 2
2e
2
( x 3) 5
0 2 4 2 0 1 74. BC 3 1 3 1 5 0 4(0) 2(3) 0(5) 4( 2) 2(1) 0(0) 1(0) 3(3) 1(5) 1(2) 3(1) 1(0)
tan
6x
5
x2 2
3x 4 ( x 2 2) 2
3. False; probability may equal 0. In such cases, the corresponding event will never happen. 4. True; in a valid probability model, all probabilities are between 0 and 1, and the sum of the probabilities is 1. 5. Probabilities must be between 0 and 1, inclusive. Thus, 0, 0.01, 0.35, and 1 could be probabilities. 6. Probabilities must be between 0 and 1, inclusive.
Thus,
1 3 2 , , , and 0 could be probabilities. 2 4 3
7. All the probabilities are between 0 and 1. The sum of the probabilities is 0.2 + 0.3 + 0.1 + 0.4 = 1. This is a probability model. 8. All the probabilities are between 0 and 1. The sum of the probabilities is 0.4 + 0.3 + 0.1 + 0.2 = 1. This is a probability model. 9. All the probabilities are between 0 and 1. The sum of the probabilities is 0.3 + 0.2 + 0.1 + 0.3 = 0.9. This is not a probability model. 10. One probability is not between 0 and 1. This is not a probability model. 11. a. The sample space is: S HH , HT , TH , TT . b. Each outcome is equally likely to occur; so
1561 Copyright © 2025 Pearson Education, Inc.
Chapter 14: Counting and Probability
P( E )
1 . The probability of probability of each is 12
n( E ) . The probabilities are: n( S ) 1
1
1
1
4
4
4
4
getting a 2 or 4 followed by a Red is 1 1 1 P(2 Red) P(4 Red) . 12 12 6
P ( HH ) , P ( HT ) , P (TH ) , P (TT ) .
12. a. The sample space is: S HH , HT , TH , TT . b. Each outcome is equally likely to occur; so P( E )
18. The sample space is: S = {Forward Yellow, Forward Red, Forward Green, Backward Yellow, Backward Red, Backward Green} There are 6 equally likely events and the
n( E ) . The probabilities are: n( S ) 1
1
1
1
4
4
4
4
P ( HH ) , P ( HT ) , P (TH ) , P (TT )
probability of each is
getting Forward followed by Yellow or Green is:
13. a. The sample space of tossing two fair coins and a fair die is: S {HH 1, HH 2, HH 3, HH 4, HH 5, HH 6, HT 1, HT 2, HT 3, HT 4, HT 5, HT 6, TH 1, TH 2, TH 3, TH 4, TH 5, TH 6, TT 1, TT 2, TT 3, TT 4, TT 5, TT 6} b. There are 24 equally likely outcomes and the
probability of each is
P (Forward Yellow) P ( Forward Green)
1 . 24
1 . 24
P(1 Red Backward) P(1 Green Backward)
15. a. The sample space for tossing three fair coins is: S {HHH , HHT , HTH , HTT , THH , THT , TTH , TTT } b. There are 8 equally likely outcomes and the
probability of each is
1 1 1 24 24 12
20. The sample space is: S = {Yellow 1 Forward, Yellow 1 Backward, Red 1 Forward, Red 1 Backward, Green 1 Forward, Green 1 Backward, Yellow 2 Forward, Yellow 2 Backward, Red 2 Forward, Red 2 Backward, Green 2 Forward, Green 2 Backward, Yellow 3 Forward, Yellow 3 Backward, Red 3 Forward, Red 3 Backward, Green 3 Forward, Green 3 Backward, Yellow 4 Forward, Yellow 4 Backward, Red 4 Forward, Red 4 Backward, Green 4 Forward, Green 4 Backward} There are 24 equally likely events and the 1 . probability of each is 24
1 . 8
16. a. The sample space for tossing one fair coin three times is: S {HHH , HHT , HTH , HTT , THH , THT , TTH , TTT } . b. There are 8 equally likely outcomes and the
probability of each is
1 1 1 6 6 3
19. The sample space is: S = {1 Yellow Forward, 1 Yellow Backward, 1 Red Forward, 1 Red Backward, 1 Green Forward, 1 Green Backward, 2 Yellow Forward, 2 Yellow Backward, 2 Red Forward, 2 Red Backward, 2 Green Forward, 2 Green Backward, 3 Yellow Forward, 3 Yellow Backward, 3 Red Forward, 3 Red Backward, 3 Green Forward, 3 Green Backward, 4 Yellow Forward, 4 Yellow Backward, 4 Red Forward, 4 Red Backward, 4 Green Forward, 4 Green Backward} There are 24 equally likely events and the probability of each is 1 . The probability of 24 getting a 1, followed by a Red or Green, followed by a Backward is:
14. a. The sample space of tossing a fair coin, a fair die, and a fair coin is: S {H 1H , H 2 H , H 3H , H 4 H , H 5 H , H 6 H , H 1T , H 2T , H 3T , H 4T , H 5T , H 6T , T 1H , T 2 H , T 3H , T 4 H , T 5H , T 6 H , T 1T , T 2T , T 3T , T 4T , T 5T , T 6T } b. There are 24 equally likely outcomes and the
probability of each is
1 . The probability of 6
1 . 8
17. The sample space is: S = {1 Yellow, 1 Red, 1 Green, 2 Yellow, 2 Red, 2 Green, 3 Yellow, 3 Red, 3 Green, 4 Yellow, 4 Red, 4 Green} There are 12 equally likely events and the 1562
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Section 14.3: Probability
The probability of getting a Yellow, followed by a 2 or 4, followed by a Forward is P(Yellow 2 Forward) P(Yellow 4 Forward)
1 1 1 24 24 12
21. The sample space is: S = {1 1 Yellow, 1 1 Red, 1 1 Green, 1 2 Yellow, 1 2 Red, 1 2 Green, 1 3 Yellow, 1 3 Red, 1 3 Green, 1 4 Yellow, 1 4 Red, 1 4 Green, 2 1 Yellow, 2 1 Red, 2 1 Green, 2 2 Yellow, 2 2 Red, 2 2 Green, 2 3 Yellow, 2 3 Red, 2 3 Green, 2 4 Yellow, 2 4 Red, 2 4 Green, 3 1 Yellow, 3 1 Red, 3 1 Green, 3 2 Yellow, 3 2 Red, 3 2 Green, 3 3 Yellow, 3 3 Red, 3 3 Green, 3 4 Yellow, 3 4 Red, 3 4 Green, 4 1 Yellow, 4 1 Red, 4 1 Green, 4 2 Yellow, 4 2 Red, 4 2 Green, 4 3 Yellow, 4 3 Red, 4 3 Green, 4 4 Yellow, 4 4 Red, 4 4 Green} There are 48 equally likely events and the 1 . The probability of probability of each is 48 getting a 2, followed by a 2 or 4, followed by a Red or Green is P( 2 2 Red) P (2 4 Red) P (2 2 Green) P (2 4 Green) 1 1 1 1 1 48 48 48 48 12 22. The sample space is: S = {Forward 11, Forward 12, Forward 13, Forward 14, Forward 21, Forward 22, Forward 23, Forward 24, Forward 31, Forward 32, Forward 33, Forward 34, Forward 41, Forward 42, Forward 43, Forward 44, Backward 11, Backward 12, Backward 13, Backward 14, Backward 21, Backward 22, Backward 23, Backward 24, Backward 31, Backward 32, Backward 33, Backward 34, Backward 41, Backward 42, Backward 43, Backward 44} There are 32 equally likely events and the 1 . The probability of probability of each is 32 getting a Forward, followed by a 1 or 3, followed by a 2 or 4 is P (Fwd 12) P (Fwd 14) P( Fwd 32) P (Fwd 34) 1 1 1 1 1 32 32 32 32 8 23. A, B, C, F 24. A (equally likely outcomes)
25. B 26. F 27. Let P (tails) x, then P (heads) 4 x x 4x 1 5x 1 1 x 5 1 4 P (tails) , P(heads) 5 5 28. Let P (heads) x, then P(tails) 2 x x 2x 1 3x 1 1 x 3 1 2 P (heads) , P (tails) 3 3 29. P (2) P (4) P (6) x P (1) P (3) P (5) 2 x P (1) P (2) P (3) P(4) P(5) P (6) 1 2x x 2x x 2x x 1 9x 1 1 x 9 1 P (2) P (4) P (6) 9 2 P (1) P (3) P(5) 9 30. P (1) P (2) P (3) P (4) P (5) x; P (6) 0 P (1) P (2) P (3) P(4) P(5) P (6) 1 x x x x x 0 1 5x 1 1 x 5 1 P(1) P(2) P(3) P(4) P(5) ; P(6) 0 5 31. P ( E )
n( E ) n{1, 2,3} 3 n( S ) 10 10
32. P ( F )
n( F ) n{3, 5, 9, 10} 4 2 n( S ) 10 10 5
1563 Copyright © 2025 Pearson Education, Inc.
Chapter 14: Counting and Probability
33. P ( E )
n( E ) n{2, 4, 6,8,10} 5 1 n( S ) 10 10 2
34. P ( F )
n( F ) n{1, 3, 5, 7, 9} 5 1 n( S ) 10 10 2
n(sum of two dice is 3) n( S ) n{1,2 or 2,1} 2 1 n( S ) 36 18
43. P (sum of two dice is 3)
n(sum of two dice is 12) n( S ) n{6, 6} 1 n( S ) 36
44. P (sum of two dice is 12)
n(white) 5 5 1 35. P (white) n( S ) 5 10 8 7 30 6
36. P (black)
n(black) 7 7 n( S ) 5 10 8 7 30
45. P ( A B ) P ( A) P( B) P( A B) 0.25 0.45 0.15 0.55
37. The sample space is: S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG} n(3 boys) 1 P (3 boys) 8 n( S )
46. P ( A B ) P ( A) P( B) P( A B) 0.25 0.45 0.6 0.10 47. P ( A B ) P ( A) P ( B ) 0.25 0.45 0.70
38. The sample space is: S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG} n(3 girls) 1 P (3 girls) 8 n( S )
48. P ( A B ) 0 49. P ( A B ) P ( A) P( B) P( A B) 0.85 0.60 P( B ) 0.05 P ( B ) 0.85 0.60 0.05 0.30
39. The sample space is: S = {BBBB, BBBG, BBGB, BGBB, GBBB, BBGG, BGBG, GBBG, BGGB, GBGB, GGBB, BGGG, GBGG, GGBG, GGGB, GGGG} n(1 girl, 3 boys) 4 1 P (1 girl, 3 boys) 16 4 n( S )
50. P ( A B ) P ( A) P( B) P( A B) 0.65 P ( A) 0.30 0.15 P ( A) 0.65 0.30 0.15 0.50 51. P (not licensed) 1 P (licensed) 1 0.074 0.926 or 92.6%
40. The sample space is: S = {BBBB, BBBG, BBGB, BGBB, GBBB, BBGG, BGBG, GBBG, BGGB, GBGB, GGBB, BGGG, GBGG, GGBG, GGGB, GGGG} n(2 girls, 2 boys) 6 3 P (2 girls, 2 boys) 16 8 n( S )
52. P (does not own a pet) 1 P (owns a pet) 1 0.67 0.33
n(sum of two dice is 7) n( S ) n{1,6 or 2,5 or 3,4 or 4,3 or 5,2 or 6,1} 6 1 n( S ) 36 6
41. P (sum of two dice is 7)
53. P (does not use YouTube) 1 P (uses YouTube) 1 0.81 0.19
n(sum of two dice is 11) n( S ) n{5,6 or 6,5} 2 1 n( S ) 36 18
54. P (not drawing Flush) 1 P(drawing Flush) 1 0.001965 0.998035 or 99.8035%
42. P (sum of two dice is 11)
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Section 14.3: Probability
55.
P(never makes online purchases) 1 P (makes online purchases) 1 0.79 0.21
56.
P( not Samoas/Caramel deLites) 1 P (Samoas/Caramel deLites) 1 0.19 0.81
64. There are 40 households out of 100 with an income between $25,000 and $74,999. n( E ) n(25000 to 74999) 40 2 P( E ) n( S ) n(total households) 100 5
57. P (white or green) P (white) P(green) n(white) n(green) n( S ) 98 17 9 8 3 20 58. P (white or orange) P (white) P (orange) n(white) n(orange) n( S ) 93 12 3 9 8 3 20 5 59. P (not white) 1 P (white) n(white) 1 n( S ) 9 11 1 20 20
65. There are 45 households out of 100 with an income of less than $50,000. n( E ) n(less than 50,000) 45 9 P( E ) n( S ) n(total households) 100 20 66. There are 55 households out of 100 with an income of $50,000 or more. n( E ) n($50,000 or more) 55 11 P( E ) n( S ) n(total households) 100 20 67. a.
P (1 or 2) P (1) P (2) 0.24 0.33 0.57
b.
P (1 or more) 1 P none 1 0.05 0.95
c.
P (3 or fewer) 1 P 4 or more 1 0.17 0.83
60. P (not green) 1 P(green) n(green) 1 n( S ) 8 12 3 1 20 20 5
d.
P (3 or more) P(3) P (4 or more) 0.21 0.17 0.38
e.
P (fewer than 2) P (0) P (1) 0.05 0.24 0.29
f.
P (fewer than 1) P (0) 0.05
g.
P(1, 2, or 3) P (1) P (2) P (3) 0.24 0.33 0.21 0.78
61. P (strike or one) P (strike) P (one) n(strike) n(one) n( S ) 3 1 4 1 8 8 2 62. P (100 or 30) P (100) P(30) n(100) n(30) n( S ) 11 2 1 20 20 10
63. There are 38 households out of 100 with an income of $75,000 or more. n( E ) n(75, 000 or more) 38 19 P( E ) n( S ) n(total households) 100 50
h.
P(2 or more) P(2) P(3) P(4 or more) 0.33 0.21 0.17 0.71
68. a.
P (at most 2) P (0) P (1) P (2) 0.10 0.15 0.20 0.45
b.
P (at least 2) P(2) P(3) P(4 or more) 0.20 0.24 0.31 0.75
c.
P (at least 1) 1 P(0) 1 0.10 0.90
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Chapter 14: Counting and Probability
69. a.
73. The number of different selections of 6 numbers is the number of ways we can choose 5 white balls and 1 red ball, where the order of the white balls is not important. This requires the use of the Multiplication Principle and the combination formula. Thus, the total number of distinct ways to pick the 6 numbers is given by n white balls n red ball
P (freshman or female) P (freshman) P (female) P (freshman and female)
n(freshman) n(female) n(freshman and female)
n( S )
18 15 8 25 33 33
b.
P (sophomore or male)
C 52,5 C 10,1
P (sophomore) P ( male) P (sophomore and male)
70. a.
n(sophomore) n( male) n(sophomore and male)
15 18 8 25 33 33
b.
71.
72.
n( women) n( under 40) n(women and under 40) n( S ) 452 13
7 13
P (men or over 40) P (men) P (over 40) P(men and over 40) n(men) n(over 40) n(men and over 40) n( S )
52! 10! 5! 47! 1! 9!
52 51 50 49 48 10 9! 5 4 3 2 1 9! 52 51 50 49 48 10 5 4 3 2 25,989, 600 Since each possible combination is equally likely, the probability of winning on a $1 play is 1 P win on $1 play 25,989, 600 0.0000000385
P ( women) P ( under 40) P ( women and under 40)
52! 10! 5! 52 5 ! 1! 10 1!
n( S )
P (women or under 40)
986 13
74. The outcome “exactly two dice have the same reading” can happen 6 ways; exactly two 1’s, or exactly two 2’s, or exactly two 3’s or exactly two 4’s or exactly two 5’s or exactly two 6’s. So, we find the porabaility or each of these six outcomes and combine their results. Now, when three dice are towwec, “exactly two 1’s” can occur in 3 C2 3 ways. So
11 13
P(at least 2 with same birthday) 1 P (none with same birthday) n(different birthdays) 1 n( S ) 365 364 363 362 361 360 354 1 36512 1 0.833 0.167
2
1 5 P (exactly two 1's) 3 C2 6 6 2
1 5 5 3 6 6 72
P(at least 2 with same birthday) 1 P (none with same birthday) n(different birthdays) 1 n( S ) 365 364 363 362 361 360 331 1 36535 1 0.186 0.814
Similarly, P(exactly two 2's) 5 , 72 5 , P (exactly two 4's) 72 5 P(exactly two 5's) , and 72 5 . So, P (exactly two 6's) 72 P (exactly two 3's)
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5 , 72
Section 14.3: Probability
P (exactly two dice have the same reading 6
5 5 72 12
é3 1 2 1ùú ê ê 2 -2 5 5úú ê ê ú 3 2 - 9ú êë 1 û é 1 3 2 -9 ù ê ú ê 2 -2 5 5 ú ( R1 « R3 ) ê ú ê3 1 2 1 ú ë û é1 - 9ù 3 2 ê ú ê æ ö 1 23 ú çç R2 = - 1 r1 ÷÷ ê 0 - ú 1 ç ê ú è 8 8ú 8 ø÷ ê ê 0 -8 -4 28úû ë é 19 3ù ê1 0 - ú ê 8 8ú ê ú æ R3 = 8r2 + r3 ö÷ ê 1 23 ú ÷ ê0 1 - ú çç çè R1 = -3r2 + r1 ÷÷ø ê 8 8ú ê ú 5ú ê 0 0 -5 ê ú êë úû é 19 3ù ê1 0 - ú ê 8 8ú ê ú æ ö 1 23 ú ê çç R3 = - 1 r3 ÷÷ ê0 1 - ú çè ê 8 8ú 5 ø÷ ê ú 1 -1ú ê0 0 ê ú êë úû æ ö 1 é1 0 0 2ùú ê ççç R2 = r3 + r2 ÷÷÷ 8 ç ÷÷ êê 0 1 0 -3úú ççç ÷÷÷ 19 ê ú ÷ ç êë 0 0 1 -1úû çè R1 = - 8 r3 + r2 ÷ø The solution is x 2, y 3, z 1 or
75. 2; left; 3; down 2 76. x 6 cos 3 3 2 y 6sin 3 3 3
The ordered pair is 3,3 3
77. log5 ( x 3) 2 52 x 3 25 x 3 x 22
The solution set is 22 . ïìï3 x + y + 2 z = 1 ï 78. ïí2 x - 2 y + 5 z = 5 ïï ïïî x + 3 y + 2 z = -9
Write the augmented matrix:
(2, 3, 1) .
79. 7 -6 3 -8
0 5 =7
6 -4 2
0 5 -4 2
+6
-8 5 6 2
+3
-8
0
6 -4
= 7(0 + 20) + 6(-16 - 30) + 3(32 - 0) = 7(20) + 6(-46) + 3(32) = 140 - 276 + 96 = -40
80.
108 147 363 36 3 49 3 121 3 6 3 7 3 11 3 10 3
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Chapter 14: Counting and Probability
obtain: 7 x 2 5 x 30 A( x 2 2 x 4) ( Bx C )( x 2)
81. Let t1 be the time to his friend’s house and t2 be the time back to this own house. Then, 60t1 40t2
Let x 2 , then
7(2) 2 5(2) 30 A 2 2 2(2) 4 B (2) C (2 2) 48 12 A
2 t1 t2 3 The average speed is:
A4
Let x 0 , then 0 0 30 4 0 0 4 ( B (0) C )(0 2)
2 60 t2 40t2 r1t1 r2 t2 60t1 40t2 3 2 t1 t2 t1 t2 t2 t2 3 40t2 40t2 80t2 5 5 t2 t2 3 3 3 80 48 mph 5
30 16 2C 14 C C 7
Let x 1 , then 7 5 30 4 1 2 4 ( B 7)(1) 32 28 B 7 3 B B3 7 x 5 x 30 4 3x 7 x 2 x2 2 x 4 x3 8 2
82. This sequence is an arithmetic sequence where a = 5 and d = 7. Thus the 85 term would be a85 5 (85 1)(7) 5 84(7) 593
83.
Chapter 14 Review Exercises 1. , Dave , Joanne , Erica , Dave, Joanne ,
Dave, Erica , Joanne, Erica , Dave, Joanne,Erica 2. n( A) 8, n( B) 12, n( A B ) 3 n( A B ) n( A) n( B ) n( A B ) 8 12 3 17
The area of the semicircle would be: 1 (4) 2 8 2 The area of the triangle would be: 1 1 bh (8)(3) 12 . The total area is: 2 2 8 12 sq units
3. n( A) 12, n( A B ) 30, n( A B) 6 n( A B ) n( A) n( B ) n( A B ) 30 12 n( B ) 6 n( B ) 30 12 6 24 4. From the figure: n( A) 20 2 6 1 29
84. Find the partial fraction decomposition: 7 x 2 5 x 30 2x 4 x3 8 ( x 2)( x 2 2 x 4) A Bx C 2 x 2 x 2x 4
5. From the figure: n( A or B ) 20 2 6 1 5 0 34 6. From the figure: n( A and C ) n( A C ) 1 6 7
Multiplying both sides by ( x 2)( x 2 2 x 4) , we 1568
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Chapter 14 Review Exercises 7. From the figure: n(not in B) 20 1 4 20 45 8. From the figure: n(neither in A nor in C ) n( A C ) 20 5 25 9. From the figure: n(in B but not in C ) 2 5 7 10. P (8,3)
8! 8! 8 7 6 5! 336 (8 3)! 5! 5!
11. C (8,3)
8! 8! 8 7 6 5! 56 (8 3)! 3! 5! 3! 5! 3 2 1
20. Since there are repeated colors: 10! 10 9 8 7 6 5 4 3 2 1 12, 600 4! 3! 2! 1! 4 3 2 1 3 2 1 2 1 1 different vertical arrangements. 21. a.
b.
C (9, 4) C (5,3) C (2, 2) 126 10 1 1260 committees can be formed.
22. a.
365 364 363 348 8.634628387 1045
b.
P (no one has same birthday) 365 364 363 348 0.6531 365 18
c.
P (at least 2 have same birthday)
12. There are 2 choices of material, 3 choices of color, and 10 choices of size. The complete assortment would have: 2 3 10 60 suits. 13. There are two possible outcomes for each game or 2 2 2 2 2 2 2 27 128 outcomes for 7 games.
C (9, 4) C (9,3) C (9, 2) 126 84 36 381, 024 committees can be formed.
1 P (no one has same birthday) 1 0.6531 0.3469 P (unemployed) 0.038
14. Since order is significant, this is a permutation. 9! 9! 9 8 7 6 5! P (9, 4) 3024 (9 4)! 5! 5! ways to seat 4 people in 9 seats.
23. a.
15. Choose 4 runners –order is significant: 8! 8! 8 7 6 5 4! P (8, 4) 1680 (8 4)! 4! 4! ways a squad can be chosen.
24. P ($1 bill)=
16. Choose 2 teams from 14–order is not significant: 14! 14! 14 13 12! C (14, 2) 91 (14 2)! 2! 12! 2! 12! 2 1 ways to choose 2 teams. 17. There are 8 10 10 10 10 10 2 1, 600, 000 possible phone numbers. 18. There are 24 9 10 10 10 216, 000 possible license plates. 19. There are two choices for each digit, so there are 28 256 different numbers. (Note this allows numbers with initial zeros, such as 011.)
b.
P (not unemployed) 1 P(unemployed) 1 0.038 0.962 n($1 bill) 4 9 n( S )
25. Let S be all possible selections, so n( S ) 100 . Let D be a card that is divisible by 5, so n( D) 20. Let PN be a card that is 1 or a prime number, so n( PN ) 26 . n( D) 20 1 0.2 n( S ) 100 5 n( PN ) 26 13 0.26 P ( PN ) n( S ) 100 50 P( D)
26. Let S be all possible selections, let T be a car that needs a tune-up, and let B be a car that needs a brake job. a.
P Tune-up or Brake job
P T B P T P B P T B 0.6 0.1 0.02 0.68
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Chapter 14: Counting and Probability
b.
P Tune-up but not Brake job
7. C 11,5
P Tune-up P Tune-up and Brake job
11 10 9 8 7 6! 5 4 3 2 1 6! 11 10 9 8 7 5 4 3 2 1 462
P T P T B
0.6 0.02 0.58
c.
11! 11! 5!11 5 ! 5!6!
P Neither Tune-up nor Brake job
1 P (Tune-up or Brake job)
8. Since the order in which the colors are selected doesn’t matter, this is a combination problem. We have n 21 colors and we wish to select r 6 of them. 21! 21! C 21, 6 6! 21 6 ! 6!15!
1 P (T ) P( B ) P(T B) 1 0.6 0.1 0.02 0.32
21 20 19 18 17 16 15! 6!15! 21 20 19 18 17 16 6 5 4 3 2 1 54, 264 There are 54,264 ways to choose 6 colors from the 21 available colors.
Chapter 14 Test 1. From the figure: n physics 4 2 7 9 22 2. From the figure: n biology or chemistry or physics
9. Because the letters are not distinct and order matters, we use the permutation formula for nondistinct objects. We have four different letters, two of which are repeated (E four times and D two times). n! 8! n1 !n2 !n3 !n4 ! 4!2!1!1!
22 8 2 4 9 7 15 67 Therefore, n none of the three 70 67 3
8 7 6 5 4! 4! 2 1 87 65 2 4765
3. From the figure: n only biology and chemistry
n biol. and chem. n biol. and chem. and phys.
8 2 2 8
840 There are 840 distinct arrangements of the letters in the word REDEEMED.
4. From the figure: n physics or chemistry 4 2 7 9 15 8 45
10. Since the order of the horses matters and all the horses are distinct, we use the permutation formula for distinct objects. 8! 8! 8 7 6! P 8, 2 8 7 56 8 2 ! 6! 6!
5. 7! 7 6 5 4 3 2 1 5040 6. P 10, 6
10! 10! 10 6 ! 4!
There are 56 different exacta bets for an 8-horse race.
10 9 8 7 6 5 4! 4! 10 9 8 7 6 5
11. We are choosing 3 letters from 26 distinct letters and 4 digits from 10 distinct digits. The letters
151, 200
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Chapter 14 Cumulative Review
and numbers are placed in order following the
format LLL DDDD with
repetitions being allowed. Using the Multiplication Principle, we get 26 26 23 10 10 10 10 155, 480, 000 Note that there are only 23 possibilities for the third letter. There are 155,480,000 possible license plates using the new format.
Thus, the probability of winning is: n( E ) n(winning) P( E ) n( S ) n(total possible outcomes) 1 0.000033069 30, 240
12. Let A = Kiersten accepted at USC, and B = Kiersten accepted at FSU. Then, we get P A 0.60 , P B 0.70 , and P A B 0.35 .
a. Here we need to use the Addition Rule. P A B P A P B P A B
16. The number of elements in the sample space can be obtained by using the Multiplication Principle: 6 6 6 6 6 7, 776 Consider the rolls as a sequence of 5 slots. The number of ways to position 2 fours in 5 slots is C 5, 2 . The remaining three slots can be filled
with any of the five remaining numbers from the die. Repetitions are allowed so this can be done in 5 5 5 125 different ways. Therefore, the total number of ways to get exactly 2 fours is 5! 5 4 125 C 5, 2 125 125 1250 2! 3! 2 The probability of getting exactly 2 fours on 5 rolls of a die is given by 1250 625 P exactly 2 fours 0.1608 . 7776 3888
0.60 0.70 0.35 0.95 Kiersten has a 95% chance of being admitted to at least one of the universities.
b. Here we need the Complement of an event.
P B 1 P B 1 0.70 0.30
Kiersten has a 30% chance of not being admitted to FSU. 13. a.
b.
Since the bottle is chosen at random, all bottles are equally likely to be selected. Thus, 5 5 1 P Coke 0.25 8 5 4 3 20 4 There is a 25% chance that the selected bottle contains Coke. 83 11 0.55 8 5 4 3 20 There is a 55% chance that the selected bottle contains either Pepsi or IBC.
P Pepsi IBC
Chapter 14 Cumulative Review 1.
3x 2 2 x 1 0
x
14. Since the ages cover all possibilities and the age groups are mutually exclusive, the sum of all the probabilities must equal 1. 0.03 0.23 0.29 0.25 0.01 0.81 1 0.81 0.19 The given probabilities sum to 0.81. This means the missing probability (for 18-20) must be 0.19. 15. The sample space for picking 5 out of 10 numbers in a particular order contains 10! 10! P (10,5) 30, 240 possible (10 5)! 5! outcomes. One of these is the desired outcome.
3 x 2 2 x 1
b b 2 4ac 2a 2
2 2 4 31 2 3
2 4 12 2 8 6 6 2 2 2i 1 2i 6 3 2 1 2 1 The solution set is i, i . 3 3 3 3
2.
f ( x) x 2 4 x 5
a 1, b 4, c 5. Since a 1 0, the graph
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Chapter 14: Counting and Probability
is concave up. The x-coordinate of the vertex is b 4 x 2 . 2a 2(1) The y-coordinate of the vertex is 2 b f f 2 2 4 2 5 . 2 a 4 8 5 9 Thus, the vertex is 2, 9 . The axis of
5.
f x 5 x 4 9 x3 7 x 2 31x 6
Step 1: Step 2:
Step 3:
f ( x) has at most 4 real zeros. Possible rational zeros: p 1, 2, 3, 6; q 1, 5; p 1 2 3 6 1, , 2, , 3, , 6, q 5 5 5 5 Using the Bounds on Zeros Theorem:
f ( x) 5 x 4 1.8 x3 1.4 x 2 6.2 x 1.2
symmetry is the line x 2 .The discriminant is: b 2 4ac (4) 2 4(1)(5) 16 20 36 0 . So the graph has two x-intercepts. The x-intercepts are found by solving: x2 4x 5 0 ( x 5)( x 1) 0 x 5 or x 1 The x-intercepts are 5 and 1. The y-intercept is f (0) (0) 2 4 (0) 5 5 .
a3 1.8, a2 1.4, a1 6.2, a0 1.2
Max 1, 1.2 6.2
1.4 1.8
Max 1, 10.6 10.6 1 Max 1.2 ,
6.2 , 1.4 , 1.8
1 6.2 7.2 The smaller of the two numbers is 7.2. Thus, every zero of f lies between –7.2 and 7.2. Graphing using the bounds: (Second graph has a better window.)
3. y 2 x 1 4 2
Using the graph of y x 2 , horizontally shift to the left 1 unit, vertically stretch by a factor of 2, and vertically shift down 4 units.
Step 4: From the graph we see that there are x-intercepts at 0.2 and 3. Using synthetic division with 3: 3 5 9 7 31 6 15 18 33 6 5 6 11 2 0 Since the remainder is 0, x 3 is a factor. The
4.
x 4 0.01 0.01 x 4 0.01 0.01 4 x 0.01 4 3.99 x 4.01 The solution set is x 3.99 x 4.01 or
other factor is the quotient: 5 x3 6 x 2 11x 2 . Using synthetic division with 2 on the quotient: 0.2 5 6 11 2 1 1 2 5
5
10
0
Since the remainder is 0, x 0.2 x 0.2 is a
3.99, 4.01
factor. The other factor is the quotient:
5 x 2 5 x 10 5 x 2 x 2 .
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Chapter 14 Cumulative Review
Factoring, f ( x) 5( x 2 x 2) x 3 x 0.2 The real zeros are 3 and 0.2. The complex zeros come from solving x 2 x 2 0. x
2 b b 2 4ac 1 1 4 1 2 2a 2 1
1 1 8 1 7 2 2 1 7i 2 Therefore, over the set of complex numbers, f x 5 x 4 9 x3 7 x 2 31x 6 has zeros
1 7 1 7 1 i, i , , 3 . 2 2 5 2 2
9. Multiply each side of the first equation by –3 and add to the second equation to eliminate x; multiply each side of the first equation by 2 and add to the third equation to eliminate x: x 2 y z 15 3x y 3 z 8 2 x 4 y z 27 3x 6 y 3 z 45 3x y 3z 8 7 y 6 z 53 2 x 2 y z 15 2 x 4 y 2 z 30
2 x 4 y z 27 2 x 4 y z 27 z 3 Substituting and solving for the other variables: z 3 7 y 6 3 53
6. g ( x) 3x 1 5 x
Using the graph of y 3 , shift the graph horizontally 1 unit to the right, then shift the graph vertically 5 units upward. Domain: All real numbers or (, ) Range: { y | y 5} or (5, ) Horizontal Asymptote: y 5
7 y 35 y 5 z 3, y 5 x 2(5) 3 15 x 10 3 15 x 2 The solution is x 2, y 5, z 3 or (2, 5, 3) .
10. 3, 1, 5, 9, ... is an arithmetic sequence with a 3, d 4 . Using an a (n 1)d , a33 3 (33 1) 4 3 32 4 3 128 125
To compute the sum of the first 20 terms, we use 20 S 20 a a20 . 2 a20 3 (20 1) 4
7. log 3 (9) log3 (32 ) 2 8. log 2 (3x 2) log 2 x 4
3 19 4 3 76 73
log 2 x(3 x 2) 4
x(3x 2) 24 3x 2 2 x 16 2
3 x 2 x 16 0 (3x 8)( x 2) 0 8 or x 2 3 Since x 2 makes the original logarithms x
8 3
undefined, the solution set is .
Therefore, 20 S20 a a20 2 20 3 73 2 10 70 700.
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Chapter 14: Counting and Probability
Chapter 14 Projects
11. y 3sin 2 x 3sin 2 x 2 Amplitude: A 3 3
Project I 1. Table 3 is a probability model since the total of the probabilities is 1.
2 2 Phase Shift: 2 2 T
Period:
Cash Prize Jackpot $1,000,000 $10, 000 $500 $200 $10 $4 $2 $0 Total
12. Use the Law of Cosines: a 2 b 2 c 2 2bc cos A a 2 52 92 2 5 9 cos 40o
2. There are 70 numbers to choose for the first 5 ‘white’ numbers and 25 numbers to choose for the last ‘gold’ number. Thus there are 70 C5 possibilities for the ‘white’ numbers and 25 C1 possibilities for the ‘gold’ number. Multiply these together to find the total possibilities.
o
106 90 cos 40 a 6.09 b 2 a 2 c 2 2ac cos B cos B
a 2 c 2 b 2 6.092 92 52 93.0881 2ac 2 6.09 9 109.62
70 C5 25 C1 (12103014)(25)
93.0881 o B cos 1 31.9 109.62
302,575,350
So the probability of winning is: 1 P win 302,575,350 0.00000000330
C 180o A B 180o 40o 31.9o 108.1o
Area of the triangle =
Probability 0.00000000330 0.00000007932 0.00000107411 0.00002577851 0.00006874270 0.00309316646 0.01123595506 0.02702702703 0.95854817351 1.00000000000
3. To calculate the expected value, multiply each numeric outcome by its corresponding probability and then add these products.
1 5 9 sin 40o 14.46 2
square units.
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Chapter 14 Projects
Cash Prize
Probability
prize prob
Cash Prize
Probability
prize prob
$40,000,000
0.00000000330
0.13200000000
JP
0.00000000330
J (0.00000000330)
$1,000,000
0.00000007932
0.07932000000
$1,000,000
0.00000007932
0.07932000000
$10, 000
0.00000107411 0.01074110000
$10, 000
0.00000107411
0.01074110000
$500
0.00002577851 0.01288925500
$500
0.00002577851
0.01288925500
$200
0.00006874270
0.01374854000
$200
0.00006874270
0.01374854000
$10
0.00309316646
0.03093166460
$10
0.00309316646
0.03093166460
$4
0.01123595506
0.04494382020
$4
0.01123595506
0.04494382020
$2
0.02702702703
0.05405405410
$2
0.02702702703
0.05405405410
$0
0.95854817351 0.00000000000
$0
0.95854817351
0.00000000000
expect value
0.37862843390
So the expected cash prize is 0.38. 4. The expected financial result from purchasing one ticket is $0.38 $2.00 $1.62 . Therefore, your expected profit from one ticket is $1.62 . 5. Use the same procedure replacing the Jackpot with $100,000,000. Cash Prize
Probability
prize prob
$250,000,000
0.00000000330
0.82500000000
$1,000,000
0.00000007932
0.07932000000
$10, 000
0.00000107411 0.01074110000
$500
0.00002577851 0.01288925500
$200
0.00006874270
0.01374854000
$10
0.00309316646
0.03093166460
$4
0.01123595506
0.04494382020
$2
0.02702702703
0.05405405410
$0
0.95854817351 0.00000000000
expect value
1.07162843400
The expected financial result from purchasing one ticket is $1.07 $2.00 $0.93 . Therefore, your expected profit from one ticket is $0.93 . 6. We need to solve the expected value equation for the Jackpot amount that would make the expected value equal to $1.00. Thus:
expect value
2.00000000000
J (0.00000000330) 0.07932 0.0107411 0.012889255 0.01374854 0.0309316646 0.0449438202 0.05405405410 2 J (0.00000000330) 0.2466283898 2 J (0.00000000330) 1.75337161 J $531,324, 730
7. Answers will vary. Project II 1. 0 bit errors: 1011
1 bit errors: 0011 1111 1001 1010 2 bit errors: 0111 0001 0010 1101 1110 1000 3 bit errors: 0110 0101 0000 1100 4 bit errors: 0100 4
2 16 2. P symbol received correctly 81 3
3. # of received symbols with 2 bit errors: C (8, 2) 28 1575 Copyright © 2025 Pearson Education, Inc.
Chapter 14: Counting and Probability Project V
8
256 2 P received correctly 6561 3 P received incorrectly 1 P received correctly 6305 6561
One simulation might be: Woman Woman told you has about Boy-Boy
4. Let k = # of errors, n = 8 = length of symbol. Probability of k errors : n k nk P (n, k ) p 1 p k k
Boy-Girl Younger boy Girl-Boy
8 k
3
8 1 2 8 1 2 1 3 3 3 3 3
probability that she has 2 boys is
Man has
5
Girl-Boy Older boy
probabilities are the same.
0.499923 To find the probability that an error occurred but is not detected, we need to assume that an even number of errors occurred: P (error occured, but not detected) P (8, 2) P (8, 4) P(8, 6) P (8,8) 6
4
1 2 1 2
Thus the probability he has two boys is
0.156464 0.272992 0.068044 0.002423
2
1 1 1 . 4 4 2
Man told Probability you about
Boy-Boy Older boy
5 3 7 1 8 1 2 8 1 2 5 3 3 7 3 3
8 1 2 8 1 2 2 3 3 4 3 3
Older boy
We leave out the combinations where she would have to tell you about a girl. Thus, the
P (8,1) P (8,3) P (8,5) P (8, 7) 7
1 4 1 4 1 4 1 4
Older boy
Boy-Boy Younger boy
8 1 2 P (8, k ) k 3 3 Since this parity code only detects odd numbers of errors, P (error detected) 1
Probability
4
6 2 8 0 8 1 2 8 1 2 6 3 3 8 3 3
0.273402 0.170364 0.016985 0.000151 0.460951
Project III
Answers will vary. Project IV e. Answers will vary, depending on the L2 generated by the calculator. f. The data accumulates around y = 0.5.
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1 . The 2
Appendix Graphing Utilities Section 1 1.
1, 4 ; Quadrant II
2. (3, 4); Quadrant I 3. (3, 1); Quadrant I 4.
6, 4 ; Quadrant III
5. X min 6 X max 6 X scl 2 Y min 4 Y max 4 Y scl 2
6. X min 3 X max 3 X scl 1 Y min 2 Y max 2 Y scl 1 7. X min 6 X max 6 X scl 2 Y min 1 Y max 3 Y scl 1 8. X min 9 X max 9 X scl 3 Y min 12 Y max 4 Y scl 4
9. X min 3 X max 9 X scl 1 Y min 2 Y max 10 Y scl 2 10. X min 22 X max 10 X scl 2 Y min 4 Y max 8 Y scl 1
In Problems 11 – 16, answers will vary. One possible setting is given. 11.
X min 11 X max 5 X scl 1 Y min 3 Y max 6 Y scl 1
12.
X min 3 X max 7 X scl 1 Y min 4 Y max 9 Y scl 1
13.
X min 30 X max 50 X scl 10 Y min 90 Y max 50 Y scl 10
1577 Copyright © 2025 Pearson Education, Inc.
Appendix: Graphing Utilities
14.
X min 90 X max 30
2.
a.
X scl 10 Y min 50 Y max 70 Y scl 10
15.
X min 10 X max 110
b.
X scl 10 Y min 10 Y max 160 Y scl 10
16.
X min 20 X max 110 X scl 10 Y min 10 Y max 60 Y scl 10
3.
a.
b.
Section 2 1.
a.
4.
a.
b.
b.
1578 Copyright © 2025 Pearson Education, Inc.
Section 2: Using a Graphing Utility to Graph Equations
5.
a.
8.
b.
b.
6.
a.
9.
a.
b.
a.
b.
b.
7.
a.
10.
a.
b.
1579 Copyright © 2025 Pearson Education, Inc.
Appendix: Graphing Utilities
11.
14.
a.
b.
b.
12.
15.
a.
b.
13.
a.
a.
b.
a.
16.
b.
a.
b.
1580 Copyright © 2025 Pearson Education, Inc.
Section 2: Using a Graphing Utility to Graph Equations are 3,1 , 2, 0 , and 0, 2 .
17. 21.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 1 , 2, 0 , and 1,1 .
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 4 , 2, 2 , and 1, 0 .
18. 22.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 5 , 2, 4 , and 1, 3 .
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 8 , 2, 6 , and 1, 4 .
19. 23.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3,5 , 2, 4 , and 1,3 . 20.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3,8 , 2, 6 , and 1, 4 .
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph 1581 Copyright © 2025 Pearson Education, Inc.
Appendix: Graphing Utilities
27.
24.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 7 , 2, 2 , and 1,1 .
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 4 , 2, 2 , and 1, 0 . 28.
25.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 11 , 2, 6 , and 1, 3 .
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3,11 , 2, 6 , and 1,3 . 29.
26.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 7.5 , 2, 6 , and 1, 4.5 .
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 7 , 2, 2 , and 1, 1 .
1582 Copyright © 2025 Pearson Education, Inc.
Section 3: Using a Graphing Utility to Locate Intercepts and Check for Symmetry
32.
30.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3,1.5 , 2, 0 , and 1, 1.5 .
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 7.5 , 2, 6 , and 1, 4.5 .
31.
Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are 3, 1.5 , 2, 0 , and 1,1.5 .
Section 3 1.
The smaller x-intercept is roughly 4.65 . 3.
The smaller x-intercept is roughly 3.41 . 2.
The smaller x-intercept is roughly 1.71 .
1583 Copyright © 2025 Pearson Education, Inc.
Appendix: Graphing Utilities
8.
4.
The smaller x-intercept is roughly 1.43 .
The positive x-intercept is 2.00. 9.
5.
The smaller x-intercept is roughly 0.28 .
The positive x-intercept is 4.50. 10.
6.
The smaller x-intercept is roughly 0.22 .
The positive x-intercept is 1.70. 11.
7.
The positive x-intercepts are 1.00 and 23.00.
The positive x-intercept is 3.00.
1584 Copyright © 2025 Pearson Education, Inc.
Section 5: Square Screens
6. Answers will vary. X max X min 10 6 16
12.
We want a ratio of 8:5, so the difference between Ymax and Ymin should be 10. In order to see the
point 4,8 , the Ymax value must be greater than 8. We might choose Ymax 10 , which means 10 Ymin 10 , or Ymin 0 . Since our range of Y values is more than 10, we might consider using a scale of 2. Thus, Ymin 0 , Ymax 10 , and Yscl 1 will make the point 4,8 visible and
The positive x-intercept is 1.07
have a square screen.
Section 5 Problems 1-4 assume that a ratio of 3:2 is required for a square screen, as with a TI-84 Plus. 1.
X max X min 6 6 12 3 Ymax Ymin 2 2 4
for a ratio of 3:1, resulting in a screen that is not square. 2.
X max X min 5 5 10 5 Ymax Ymin 4 4 8 4
for a ratio of 5:4, resulting in a screen that is not square. 3.
4.
X max X min 16 0 16 8 Ymax Ymin 8 2 10 5 for a ratio of 8:5, resulting in a square screen.
X max X min 14 10 24 8 Ymax Ymin 8 7 15 5 for a ratio of 8:5, resulting in a square screen.
5. Answers will vary. X max X min 12 4 16
We want a ratio of 8:5, so the difference between Ymax and Ymin should be 10. In order to see the
point 4,8 , the Ymax value must be greater than 8. We might choose Ymax 10 , which means 10 Ymin 10 , or Ymin 0 . Since we are on the order of 10, we would use a scale of 1. Thus, Ymin 0 , Ymax 10 , and Yscl 1 will make the point 4,8 visible and have a square screen.
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