Contents Chapter R
Review of Prerequisites
R.1
Sets and the Real Number Line
1
R.2
Models, Algebraic Expressions, and Properties of Real Numbers
8
R.3
Integer Exponents and Scientific Notation
13
R.4
Rational Exponents and Radicals
20
R.5
Polynomials and Multiplication of Radicals
28
Problem Recognition Exercises: Simplifying Algebraic Expressions
35
R.6
Factoring
36
R.7
Rational Expressions and More Operations on Radicals
45
Chapter R Review Exercises
59
Chapter R Test
67
Chapter 1
Equations and Inequalities
1.1
Linear Equations and Rational Equations
71
1.2
Applications and Modeling with Linear Equations
85
1.3
Complex Numbers
96
1.4
Quadratic Equations
102
Problem Recognition Exercises: Simplifying Expressions Versus Solving Equations
118
1.5
Applications of Quadratic Equations
120
1.6
More Equations and Applications
127
1.7
Linear Inequalities and Compound Inequalities
143
1.8
Absolute Value Equations and Inequalities
152
Problem Recognition Exercises: Recognizing and Solving Equations and Inequalities
160
Chapter 1 Review Exercises
164
Chapter 1 Test
180
Chapter 1 Cumulative Review Exercises
186
Chapter 2
Functions and Graphs
2.1
The Rectangular Coordinate System and Graphing Utilities
189
2.2
Circles
202
2.3
Functions and Relations
212
2.4
Linear Equations in Two Variables and Linear Functions
220
2.5
Applications of Linear Equations and Modeling
234
Problem Recognition Exercises: Comparing Graphs of Equations
244
2.6
Transformations of Graphs
246
2.7
Analyzing Graphs of Functions and Piecewise-Defined Functions
258
2.8
Algebra of Functions and Function Composition
271
Chapter 2 Review Exercises
283
Chapter 2 Test
296
Chapter 2 Cumulative Review Exercises
300
Chapter 3
Polynomial and Rational Functions
3.1
Quadratic Functions and Applications
303
3.2
Introduction to Polynomial Functions
318
3.3
Division of Polynomials and the Remainder and Factor Theorems
329
3.4
Zeros of Polynomials
342
3.5
Rational Functions
359
Problem Recognition Exercises: Polynomial and Rational Functions
385
Polynomial and Rational Inequalities
387
Problem Recognition Exercises: Solving Equations and Inequalities
412
Variation
417
Chapter 3 Review Exercises
424
Chapter 3 Test
439
Chapter 3 Cumulative Review Exercises
448
3.6
3.7
Chapter 4
Exponential and Logarithmic Functions
4.1
Inverse Functions
451
4.2
Exponential Functions
460
4.3
Logarithmic Functions
471
Problem Recognition Exercises: Analyzing Functions
482
4.4
Properties of Logarithms
486
4.5
Exponential and Logarithmic Equations
493
4.6
Modeling with Exponential and Logarithmic Functions
511
Chapter 4 Review Exercises
521
Chapter 4 Test
531
Chapter 4 Cumulative Review Exercises
537
Chapter 5
Systems of Equations and Inequalities
5.1
Systems of Linear Equations in Two Variables and Applications
541
5.2
Systems of Linear Equations in Three Variables and Applications
559
5.3
Partial Fraction Decomposition
587
5.4
Systems of Nonlinear Equations in Two Variables
605
5.5
Inequalities and Systems of Inequalities in Two Variables
628
Problem Recognition Exercises: Equations and Inequalities in Two Variables
652
Linear Programming
656
Chapter 5 Review Exercises
668
Chapter 5 Test
685
Chapter 5 Cumulative Review Exercises
697
5.6
Section R.1
Chapter R Review of Prerequisites Section R.1 Sets and the Real Number Line 1. set
2. whole
3. natural
4. integers
5. set-builder
6. roster
7. rational
8. irrational
9. left
10. absolute value
5 , which is the 1 ratio of 5 and 1, so it is an element of . True.
d. 5 can be written as
22. a. 1, 2,3, does not include
b. 0,1, 2,3, does not include
11. a b or b a
1 . 3
False. c. , 2, 1,0,1, 2, does not include
12. base; exponent or power
1 . False. 3
13. square 14. 0; undefined
d.
15. 3 is an element of the set of natural numbers. 16.
1 . False. 3
1 is the ratio of 1 and 3, so it is an 3 element of . True.
23. a. 1, 2,3, does not include 0.25 . False.
2 is an element of the set of rational 5 numbers.
b. 0,1, 2,3, does not include 0.25 . False.
17. 3.1 is not an element of the set of integers. 18. is not an element of the set of rational numbers.
c. , 2, 1,0,1, 2, does not include
19. The set of integers is a proper subset of the set of real numbers.
d. 0.25 is a repeating decimal, so it is an element of . True.
20. The set of rational numbers is a proper subset of the set of real numbers.
24. a. , 2, 1,0,1, 2, does not include . False.
0.25 . False.
21. a. 1, 2,3, does not include 5 . False.
b. cannot be written as the ratio of two integers, so it is not an element of . False.
b. 0,1, 2,3, does not include 5 . False.
c. cannot be represented by a terminating decimal or by a repeating decimal, so it is an element of . True.
c. , 2, 1,0,1, 2, .This includes 5 . True.
d. is an element of , so it is an element of . True.
1
Chapter R Review of Prerequisites 25. The set given can be written in roster form as , 10, 5,0,5,10, . The number 25 is an element of this set. True.
35. a. The first set is not a subset of the second, because the set of whole numbers does not include negative integers. False. b. The first set is a subset of the second, because all whole numbers are also integers. True.
26. The set given can be written in roster form as 1, 2,3,,9 . The number 7 is not an element of this set. False.
n , so all 1 integers are also rational numbers. True.
36. a. Any integer n can be written as
27. The number 0.8 is a repeating decimal, so it is rational. It can be written as 0.88 , and 0.88 0.8 . False.
b. The first set is not a subset of the second, because there all elements in the set of rational numbers that are not integers. False.
45 , so it 100 is rational. The number 0.45 can be written as 0.4545 , and 0.45 0.4545 . True.
28. The number 0.45 can be written as
37. a. The first set is not a subset of the second, because irrational numbers are real numbers that cannot be represented as a ratio of integers, so a number cannot be both rational and irrational. False.
29. The set given can be written in roster form as 0, 2, 4,6, . The number 22 is an element of this set. True.
b. The first set is not a subset of the second, because irrational numbers are real numbers that cannot be represented as a ratio of integers, so a number cannot be both rational and irrational. False.
30. The set given can be written in roster form as , 8, 4, 0, 4,8, . The number 24 is an element of this set. True.
38. a. The first set is a subset of the second, because the real numbers are made up of rational and irrational numbers. True.
31. Irrational numbers are real numbers that cannot be represented as a ratio of integers, so a number cannot be both rational and irrational. False.
b. The first set is a subset of the second, because the real numbers are made up of rational and irrational numbers. True.
n , so all 1 integers are also rational numbers. True.
32. Any integer n can be written as
39. a. 6
b. 6
c. 12 , 6
33. a. The first set is a subset of the second, because all of the elements in the first set are elements of the second set. True.
d. 0.3,0.33, 0.9, 12,
b. The first set is not a subset of the second, because 0 is in the first set but not in the second set. False. 34. a. The first set is a subset of the second, because all of the elements in the first set are elements of the second set. True.
e.
5,
f.
5, 0.3, 0.33, 0.9, 12,
6
40. a. 1
b. The first set is not a subset of the second, because NM and TX are in the first set but not in the second set. False.
c. 0, 4 , 1
2
11 ,6 4
11 , 6, 4 6
b. 0, 1 d. 0, 4, 0.48,1,9.4
Section R.1
41. a 5
42. b 6
5? 9 3 5 5 5? 9 3 3 5 5 3 25 ? 27 False 15 15
43. 3c 9
44. 8d 16
55. 7, ; x | x 7
45. m 4 70
46. n 7 4
56. 2, ; x | x 2
e. f.
2
2
, 13
54.
, 0, 4, 0.48,1, 13,9.4
47. 3.14 ?
57. , 4.1 ; x | x 4.1
?
3.14 3.141592654 True
58. , 2.93 ; x | x 2.93
?
48. 7 7 ?
59. 6, 0 ; x | 6 x 0
?
60. 2, 8 ; x | 2 x 8
7 2.645751311 True
49. 6.7 6.7
6.7 ⱨ 6.7 True
61.
, 6
62.
, 4
63.
7 1 , 6 3
64.
4 7 , 3 4
65.
4,
66.
3,
?
50. 2.1 2.1
2.1ⱨ 2.1 True ?
6.15 6.15
51.
?
6.151515 6.155555 False ?
2.93 2.93
52.
?
2.933333 2.939393 False
53.
9 ? 11 7 8 9 8 ? 11 7 7 8 8 7 72 ? 77 False 56 56
3
67. x | 3 x 7
68. x | 4 x 1
69. x | x 6.7
70. x | x 3.2
3 71. x x 5
7 72. x x 8
Chapter R Review of Prerequisites 73. 6 6 6
74. 4 4 4
88. 8 2 10 10 or 2 8 10 10
75. 0 0
76. 1 1
89. 8 3 8 3 or
3 8 8 3
77.
2 2
2 2 2 2
90. 11 5 11 5 or
78.
6 6
6 6 6 6
91. 6 2 2 6 or 2 6 2 6
79. a. 3 3
92. 3 3 or 3 3
b. 3 3 3
93. a. 42 4 4 16 b. 4 4 4 16
80. a. for m 11, m 11 m 11
2
b. for m 11, m 11 m 11 11 m
c. 42 4 4 16 d.
81. a. for x 2, x 2 x 2
f.
82. a. for t 6, t 6 t 6 t 6
4 is not a real number.
94. a. 92 9 9 81
b. for t 6, t 6 t 6
b. 9 9 9 81 2
z 5 z 5 1 z 5 z 5
c. 92 9 9 81 d.
b. for z 5, z 5 z 5 z 5 1 1 z 5 z 5 z 5 84. a. for x 7,
42
e. 4 2
b. for x 2, x 2 x 2 x 2
83. a. for z 5,
5 11 11 5
93
e. 9 3 f.
7x 7x 1 7x 7x
9 is not a real number.
95. a. 3 8 2 b. 3 8 2
b. for x 7, 7x 7x 7x 1 1 7 x 7 x 7x
c. 3 8 3 8 2
85. 1 6 5 5 or 6 1 5 5 86. 2 9 7 7 or 9 2 5 7
d.
100 10
e.
100 is not a real number.
f. 100 10
87. 3 4 7 7 or 4 3 7 7
96. a. 3 27 3 b. 3 27 3
4
Section R.1
2 106. 5 4 6 2 8 31
c. 3 27 3 27 3
d.
49 7
5 4 6 6 31
e.
49 is not a real number.
5 4 6 36 31
2
5 4 6 5
f. 49 7
5 4 30 5 26 21
3
2 2 2 8 2 97. 3 3 3 3 27
107.
52 32 25 9 16 4
4 4 4 64 4 98. 5 5 5 5 125
108.
62 82 36 64 100 10
99. 0.2 0.2 0.2 0.2 0.2
109.
9 16 3 4 7 49
110.
8 125 2 5 7 343
3
2
4
0.0016 100. 0.1 0.1 0.1 0.1 0.1
3
3
2
3
2
3
3
4
2
2 7 2 2 7 111. 4 4 5 10 2 5 10
0.0001 101.
169 13 25 5
121 11 36 6
102.
4 7 4 10 10
3 4 10 9 4 100
103. 20 12 36 32 2 20 12 36 9 2
20 12 4 2 20 12 2 20 24 4
25
1 2 1 2 1 1 112. 6 6 9 4 3 2
200 4 48 200 192 8
6 12 3 5 18 2
2
1
200 22 48
2
4 9 9 1 100 25
1 104. 200 22 6 4 200 22 12 4 2
2 105. 6 12 3 1 6 18
2
4 1 9 1 6 4 9 9 4 9 4 6 36 36
5 6 36
6 12 3 25 18 6 12 3 7 6 12 21 6 9 3
1 6 5 5 1 36 6 6
5
Chapter R Review of Prerequisites
113. 9 6 3 7 8 25
4000 118. Wh 175 4000 16.1
9 6 4 8 25
4000 175 4016.1
9 6 4 8 25 9 6 4 25
119. V
9 10 5 927
8 2 4 3 5 9 8 2 4 2 9 8 2 4 2 9 8 12 3
124. A bracket is used if an endpoint to an interval is included in the set.
2 8
144 25 3 10 28
125. represents the set of integers. All integers belong to (the set of rational numbers). Therefore, is a subset of . However, 1 contains fractions and decimals, such as 2 and 0.2, that are not integers. Therefore, .
169 13 6 13 13 6 Undefined 0 116.
121 cc
123. A parenthesis is used if an endpoint to an interval is not included in the set.
84 4
122 52 3 10
3
122. All integers can be written as the integer itself over the integer 1.
8 2 6 3
115.
14.44 8
121. All terminating decimals can be written as a decimal fraction—that is, a fraction with a denominator that is a power of 10.
8 24 3 5 9
11 13 4 2
3
1.0 250 120. V 6.0 10.2 L 0.5 293
114. 8 2 4 2 5 5 9
3.8 8 2
2
173.6 lb
9 6 4 25
2
4 92 22 32 5 2 4 9 7 4 12 4 3 12 4
25 4 9 3 3
20 Undefined 0
4000 117. Wh 200 4000 5.5 4000 200 4005.5
126. The statement 1 indicates that the set containing the element 1 is a subset of . However, the statement 1 does not make sense, because the number 1 (not written in set brackets) does not represent a set. A quantity that is not a set cannot be a subset of another set. 127. If n 0 , then n n n n 0
2
128. If n 0 , then n n n n n n 2n
2
129. If n 0 , then n n n n 2n
119.5 lb
6
Section R.1 130. If n 0 , then n n n n n n 0
141.
131. If n 0 , then n n 132. If n 0 , then n n n 133. b 0
134. a 0
135. b 0
136. a 0
6744.25 142.
137. a is negative, b is positive, and c is negative. ab 2 So the sign of 3 can be represented by c
2 , which is positive. 3
37,741.81 143.
138. a is negative, b is positive, and c is negative. a 2c So the sign of 4 can be represented by b
2 , which is negative. 4 0.58
139. a is negative, b is positive, and c is negative. The sum of a and c must be negative. So the
144.
b a c can be represented by a2 3 , which is negative. 2 3
sign of
0.18 140. a is negative, b is positive, and c is negative. The even power of any nonzero number is
145. There are missing parentheses around the denominator.
a b b c can positive. So the sign of b be represented by . Also, 2
4
either a b or b c can equal zero. So the sign of the expression can also be 0 0 or 0 0 . represented by The sign is positive or zero.
7
Chapter R Review of Prerequisites Section R.2 Models, Algebraic Expressions, and Properties of Real Numbers 1. 5, 2 ; x | 5 x 2
b. m 0.04 x 2 3.6 x 49
m 0.04 50 3.6 50 49 2
2. 1, ; x | x 1
m 31 mpg c. m 0.04 x 2 3.6 x 49
2 9 9 2 or 9 2 9 2
3.
m 0.04 75 3.6 75 49 2
4. 6 6 or 6 6 5. True
6. False
7. False
8. True
12 9 9 9.
2
m 4 mpg The model breaks down. The value of 4 mpg is not possible. 26. a. P 1718t 2 82,000t 10,000
P 1718 22 82,000 22 10,000 2
2 3
P 982,000 bacteria
7 1
b. P 1718t 2 82,000t 10,000
12 9 3 2 3 2
12 6 2 3 2
10.
P 1718 36 82,000 36 10,000 2
6 6 12 18 3 6
102 82
16 4
2
P 735,000 bacteria
12 36 2 3
c. P 1718t 2 82,000t 10,000
6 12 54 66 11 6 6
100 64
4 2
2
36
2
2
P 1718 48 82,000 48 10,000 2
P 12,000 bacteria The model breaks down. It is impossible to have 12,000 bacteria.
6 3 4 2
27. a. E 27.45t 704.6
E 27.45 3 704.6
11. dependent
12. breakdown
13. term
14. coefficient
15. commutative
16. 0
17. 1
18. identity
19. x
20. distributive
21. reciprocal
22. associative
23. a b c
24. 0
28. a. W 150 1.2t
25. a. m 0.04 x 3.6 x 49
W 142.8 lb
E $622.25 b. E 27.45t 704.6
E 27.45 15 704.6 E $292.85 c. No. Eventually, the model would produce a negative expenditure. This is not possible.
W 150 1.2 6
2
m 0.04 35 3.6 35 49 2
m 28 mpg
8
Section R.2 43. a. C 159n 0.11159n
b. W 150 1.2t
W 150 1.2 12
C 159n 17.49n C 176.49n
W 135.6 lb c. No. Eventually the model would produce a negative weight or a weight too small to sustain life. This is not possible. 29. Tn Ts 16
30. K C 273.15
31. P W M
32. S P D
33. a. J C 1
b. C J 1
34. a. R F 9
b. F R 9
35. D 0.25P
36. t 0.20c
37. v 2 gh
38. h 8 t
b. C 176.49n
C 176.49 4 C $705.96 44. a. C 149n 0.16 149n 40
C 149n 23.84n 40 C 172.84n 40 b. C 172.84n 40
C 172.84 2 40 C $385.68 45. 12 x 2 y 5 xy 4 9.2 xy 3
12 x 2 y 5 xy 4 9.2 xy 3
39. a. C 0.12k 14.89
Terms: 12x y , xy 4 , 9.2xy 3 ; coefficients: 12, 1 , 9.2 2
b. C 0.12k 14.89
C 0.12 1200 14.89 C $158.89
5
46. 6c 2 d 6.4cd 5 d 6
6c 2 d 6.4cd 5 d 6
40. a. C 3.58n 13.50
Terms: 6c d , 6.4cd 5 , d 6 ; coefficients: 6, 6.4 , 1 2
b. 5000 2000 3000 The family used 3 thousand gal over the 2000-gal base. C 3.58n 13.50
47.
C 3.58 3 13.50 C $24.24 41. a. C 640m 200n 500
5 1 5 m m 5 Term: ; coefficient: 5 m
2 1 2 n n 2 Term: ; coefficient: 2 n
48.
b. C 640m 200n 500
C 640 12 200 3 500 C $8780
49. x y z 1x y z
42. a. C 105n 35 L 40
Term: x y z ; coefficient: 1
b. C 105n 35 L 40
50. m 3 n 1m 3 n
C 105 12 35 2 40
Term: m 3 n ; coefficient: 1
C $1370
51. 7 x x 7
9
Chapter R Review of Prerequisites 52. 9 z z 9
b.
53. 3 w w 3
5 4 1 4 5
The multiplicative inverse is 54. 11 p p 11
7 7 64. a. 0 9 9
1 1 55. y y 3 3 56. z
The additive inverse is
5 5 z 2 2
The multiplicative inverse is
58. c 4 5 c 4 5 c 9
b. 0 ? 1 The multiplicative inverse is undefined.
4 9 4 9 60. p p 1 p p 9 4 9 4
66. a. 1 1 0 The additive inverse is 1 .
61. a. 8 8 0 The additive inverse is 8.
b. 1 1 1 The multiplicative inverse is 1.
1 b. 8 1 8
67. a. 2.1 2.1 0 The additive inverse is 2.1 .
1 The multiplicative inverse is . 8
1 b. 2.1 1 2.1
62. a. 9 9 0 The additive inverse is 9 .
The multiplicative inverse is
1 b. 9 1 9
1 10 or . 2.1 21
68. a. 4.5 4.5 0 The additive inverse is 4.5 .
1 The multiplicative inverse is . 9
1 b. 4.5 1 4.5 The multiplicative inverse is 1 10 2 or . 4.5 45 9
5 5 0 4 4 The additive inverse is
9 . 7
65. a. 0 0 0 The additive inverse is 0.
1 1 5w 5 w 1w w 5 5
63. a.
7 . 9
7 9 b. 1 9 7
57. t 3 9 t 3 9 t 12
59.
4 . 5
5 . 4
69. 14 w3 3w3 w3 14 3 1 w3 16w3
10
Section R.2 70. 12t 5 t 5 6t 5 12 1 6 t 5 5t 5
77. 4 x 1 4 x 4 x
71. 3.9 x3 y 2.2 xy 3 5.1x 3 y 4.7 xy 3
3.9 x y 5.1x y 2.2 xy 4.7 xy 3
3
3
9 x 3 y 6.9 xy 3
3 2
4
3 2
24 x 2 16 x 8
0.004m n 0.01m n 0.005m n 4
4
3 2
80. 6 5 y 2 3 y 4
0.007 m3n 2
81.
82.
3 12 p 5 q 8 p 4 q 2 6 p 3q 4 3 3 3 12 p 5 q 8 p 4 q 2 6 p 3q 4 4 4 9 9 p 5 q 6 p 4 q 2 p 3q 2
8w 16 14w 28 12 22w 84. 3 2 z 4 8 z 9 84
6 z 12 8 z 72 84 14 z 85. 4u 8v 3 7u 2v 2v
4u 8v 21u 6v 2v 25u 16v
2
86. 10 x z 2 8 x 4 z 3 x
12 p 3 18.3 p 2 24 p 6.6 76. 2 5a 4 a 3 0.4a 0.8
10 x z 16 x 8 z 3x 29 x 9 z
87. 12 4 8 2v 5 3w 4v w
2 5a 2 a 2 0.4a 2 0.8 4
12 4 8 2v 15w 20v w
3
12 4 8 22v 15w w
10a 2a 0.8a 1.6 4
83. 2 4w 8 7 2w 4 12
3 4 p 3 6.1 p 3 8 p 3 2.2
75. 3 4 p 3 6.1 p 2 8 p 2.2
2 6 x 2 y 18 yz 2 2 z 3 3 2 2 2 6 x 2 y 18 yz 2 2 z 3 3 3 3 4 4 x 2 y 12 yz 2 z 3 3
1 4 3 4 3 yz y z yz 4 y 4 z 10 4 2 1 4 3 3 yz yz 4 y 4 z y 4 z 10 4 2 1 4 10 3 2 3 yz yz 4 y 4 z y 4 z 10 10 4 2 2 1 4 10 3 6 yz yz 4 y 4 z y 4 z 10 10 4 4 11 4 3 4 yz y z 10 4
3
30 y 2 18 y 24
1 1 2 73. c 7 d cd 7 c 7 d 2cd 7 3 2 5 1 7 2 7 1 c d c d cd 7 2cd 7 3 5 2 5 1 3 2 1 2 c 7 d c 7 d cd 7 2cd 7 5 3 3 5 2 2 5 7 6 7 1 7 4 7 c d c d cd cd 15 15 2 2 1 3 c 7 d cd 7 15 2
6 5 y 2 6 3 y 6 4
0.006m 4 n 0.002m3 n 2
74.
79. 8 3 x 2 2 x 1 8 3 x 2 8 2 x 8 1
72. 0.004m n 0.005m n 0.01m n 0.007 m n 4
1 1 1 78. k 7 1 k 7 k 7 2 2 2
3
3
12 32 88v 60 w w 88v 59w 20
11
Chapter R Review of Prerequisites 88. 6 2 9 z 6 y 8 y z 11
97. W 2 L 3
6 2 9 z 6 y 8 y 8 z 11
98. L 3S 5
6 2 2 y 17 z 11
99. E2009 E2008 489,000
6 4 y 34 z 11 4 y 34 z 5
100. C2011 C2010 0.25
2 89. 2 y 2 13 6 y 2 9 10 9 3
1 1 c h 15 25 1 1 G 240 500 15 25 G 16 20 36 gal
101. G
2 y 4 y 9 9
2 y 2 13 4 y 2 6 10 9 2
2
2 y2 4 y2 9 9 6 y2 3 90. 6 5t 2 12 8t 2 5 11t 2 4
1 1 c h 22 32 1 1 G 220 512 22 32 G 10 16 26 gal
102. G
6 11t 4 11t
6 5t 2 9 6t 2 5 11t 2 2
2
6 11t 2 4 11t 2 10
103. The commutative property of addition indicates that the order in which two quantities are added does not affect the sum. The associative property of addition indicates that the manner in which quantities are grouped under addition does not affect the sum.
b 6 6 3 91. 2a 2 1 2 92.
b 2 4ac
6 2 4 2 4
36 32 4 2 93.
104. For any nonzero number a, the multiplicative 1 inverse is defined as . If a 0 , this would a 1 be which is undefined. 0
x2 x1 2 y2 y1 2
1 2 2 1 4
32 52 9 25 34
2
105.
y y1 3.7 3.1 6.8 2 94. 2 x2 x1 2 1.4 3.4 95.
96.
1 2 1 22 2 r h 7 6 3 3 7 22 6468 49 6 308 21 21
106.
4 3 4 22 3 88 9 792 r 3 27 3 3 7 7 21 7
12
Section R.3 Section R.3 Integer Exponents and Scientific Notation 1.
2.
0
1 1 5 x 2 y 15 2 y 20 5 10 2 1 3 x y 3 y 2 x y 1 5 5 5
2 d. p 1 3 0
3 16. a. 1 7
1 5 3a 2b 6 12a 2b 3 6 2 5 7 a b 2 10a b 9a b 2 3 3 3
b.
30 1 1 1 30 1 7 7 7 7
3 3 3 3 c. w0 w0 1 7 7 7 7
3. a 6 a 6 6 a
0
3 d. w 1 7
4. 2 t 2 t t 2 5. t w 1.93
6. L D 3
7. 1
8. n
9. scientific
10. magnitude
11. m n
12. quotient
17. a. 82
1 1 2 8 64
b. 8 x 2 8
1 8 2 2 x x
c. 8 x
1
2
13. a. 8 1 0
1 64 x 2
1 1 1 d. 82 1 82 1 2 1 8 64 64
b. 80 1 80 11 1 c. 8 x 0 8 x 0 8 1 8
18. a. 7 2
d. 8 x 1 0
14. a. 70 1 b. 7 0 1 70 11 1
1 1 2 7 49
b. 7 y 2 7
1 7 2 2 y y
c. 7 y
1
2
c. 7 y 7 y 7 1 7 0
8x
2
0
d. 7 y 1
7 y
2
1 49 y 2
0
d. 7 2 1 7 2 1
1 1 1 1 2 7 49 49
0
2 15. a. 1 3 b.
19. a.
20 1 1 1 20 1 3 3 3 3
1 1 q2 q2 1 1 1 q 2 q2
b. q 2
2 2 2 2 c. p 0 p 0 1 3 3 3 3
13
1 q2
Chapter R Review of Prerequisites 1 5 p3 2 q2 q
30.
1 2 5q 2 q 3 p3 p
31.
c. 5 p 3 q 2 5 p 3 q 2 5 p 3 d. 5 p 3q 2 5 p 3 q 2 5
20. a.
1 t
4
1 t4 1 t 4 1 1 t4
1 b. t 4 4 t
32.
1 11u 2 c. 11t 4u 2 11 t 4 u 2 11 4 u 2 4 t t d. 11t 4u 2 11 t 4 u 2 11 t 4
1 11t 4 2 u2 u
21. 25 27 257 212 33. 22. 43 48 438 411 23. x 7 x 6 x 2 x 76 2 x11 24. y 3 y 7 y 4 y 37 4 y 8
25. 3c 2 d 7 4c 5 d 3 4 c 25 d 71
34.
12 c 3 d 8
72 1 1 7 24 7 2 2 4 7 7 49 18k 2 p 9 18 k 2 p 9 27 k 5 p 2 27 k 5 p 2 2 2 k 25 p 92 k 3 p 7 3 3 2 1 2 p7 3 p7 3 3 k 3k
10a 3b11 10 a 3 b11 25a 7b3 25 a 7 b3 2 2 a 37 b113 a 4 b8 5 5 8 2 1 2b 4 b8 4 5 a 5a 2m 6 n 4 2 m 6 n 4 6m 2 n 1 6 m 2 n 1 1 1 m 6 2 n 4 1 m 4 n5 3 3 5 1 1 n 4 n5 3 m 3m 4
4 p 7 q 5 4 p 7 q 5 2 2 2 p 7 2 q 5 2 2 2 2p q 2 p q 2 p 5 q 7 2
1 12d 8 12 3 d 8 3 c c
4 4
26. 7 m 3 n 8 3m 5 n 7 3 m 3 5 n 81
35. 42
21 m 8 n 7 1 1 21 21 8 7 8 7 m n mn
36. 23
27.
29.
23
6
2 2
37.
y 3 y 6 y 36 y 3 2 2 y 32 y1 y y2 y y z 8 z12 z 812 z 4 3 3 z 4 3 z1 z z3 z z
5
35
p p
15
2 7
27
p 14
q
42
q 8
38. q 4 28.
3
1 7 2q 7 q 5 p5 p
2
1 p14
1 q8
39. 2cd 2 c d 8c3 d 3 3
65 1 1 658 63 3 8 6 6 216
3
3
3
40. 8mn 8 m n 64m 2 n 2 2
14
2
2
2
Section R.3
7a 7 a 72 a 2 49a 2 7a 41. b b2 b2 b2 b2 2
2
2
36a 6b9 50. 9a 2b 4
2
2
4a 4b13
4 4 p4 q4 pq p4q4 pq 42. 4 256 256 4 4
4
43. 4 x 2 y 3
4 x y 16 x y 16yx 2
2 2
2
3 2
6
4
4
6
3 w z
3 5 2
44. 3w z
3 2
2
5 2
4 x 3 z 5 51. 12 y 2
9 z10 9w z 6 w 6 10
7k 45. 2 n
2
7 kn
5w 46. v 5
7 k n
2 2
2
2
3
4
3 z 2 w1 52. 15 p 4
5 3
3
1 3
3
2
1 4
3
4 a b 2
3
1 5
3
8 26
0
53. 2 y
64 64 1 1
3
1 16a8b 26
3
x y z 3 3
2 3
5 3
27 z15 x9 y 6 3
z w p 2 3
1 3
53 z 6 w3 p 12
1 8 2 43 1 2
13 2
a 8b 26
1 z 2 w1 p 4 5
1 1 v3 53 w15v3 3 15 v3 5 w 125w15 1 47. 8
4 2
33 x 9 y 6 z15
5 w v
5w v
2
1
1 3
1 1 4 n 2 n 49 k 49k 2 5 1 3
4 2
2
a b
2
1
1 x 3 y 2 z 5 3
4
3
4a 62b9 4
4 3
125w3 p12 z 6
6 y z 2 y 6 y z 3
2 8 2
3
2
2 2
8 2
2 62 y 3 y 4 z16 3
1 48. 9
2
1 3
4
0
1 92 34 1 27
2 62 y 7 z16 3
81 81 1 1 16m n 49. 5 2 8m n 2
7
2
2m 25 n7 2
2m 3 n9 2
2
1
2
2
54. 15 z
2
5z w 4
2 y
2
3
7
36 z16 9 z16 7 7 8 y 2y
6 3
w
15 z 2 53 z 4
m n 3 2
2
62 z16
3
6 3
15 53 w18 z 2 z12 2
9 2
15 53 w18 z10 2
m 6 n 18 2
m6
m6 22 n18 4n18
15
53 z10
125 z10 5 z10 152 w18 225w18 9w18
Chapter R Review of Prerequisites 1 1 1 55. 2 3 6
3
3 1 2 1 1 3 2 2 3 6
3
3
3
3 2 1 6 6 6 1 1 1 56. 3 4 2
2 6
3
x 3 4 x 2 61. x 6
1
x12 x 2 x 6
x16
1 3
2
4 3 6 12 12 12
2
1 12
2
y 12 y 3 y 4
y 11
2
1 3
3 2 2
1 6
3
1 58. 3
1 4
2
23 33 63
1 2
2
3 4 2 2
2
2
2
2
1
31 a 2b 4 6 12
3 a b 3
1
3 ab 2
5
v w x 15
10
20
x 32 16 x 32 v 21w28 16 1 v 21w28 2
3 2 3
5
3 2 1
3 6
9
6
1 5
3 2
4
1 4 4
4
16
4 16
143 v 6 w9 x 6 211 v 5 w3 x 2 7
4
v w x 16
4 16
143 21 7 v 15 w2 x8 4
1
3a b 3
1
14v w x 7v w x 64. 21v w x 14 v w x 7 v w x 21 v w x
16 y 4 2 y 4 7 8 x7 x 3
5
4281 x 7 y 4
1 3 60. 2 4 5 4 3a b a b
42 v 2 w6 x 4 5 1 v 15 w10 x 20 4 8 12 8 2 vw x
42 24 1 v 21w28 x 32
1
4 x 3 y 5 8 x13 y 14 2 6 10 1 13 14 4 x y 8 x y
4 5
2
42 v 2 w6 x 4 24 v 8 w12 x8 1
9 16 4 3 1 8 59. 3 5 13 14 4x y x y
1
3 2 4
2
8 27 216 197 2
1
4vw x v w x 63. 2v w x 3
1 57. 2
y11
2
3
1
1
y 2 6 y 3 62. y 4
12 144 3
1
x 16 1 16 x
33 27
4 1 3 1 6 1 4 3 3 4 6 2
5 4 1
143 21 x8
74 v15 w
65. 3 x 5
14
5 4
2
24 x8 v15 w2
3x 5 2 3x 514 2 3x 512
3 a b
66. 2 y 7 z
8
9a b8
16
4
2 y 7 z 13 2 y 7 z 413 9 2 y 7z
Section R.3 67. 6v 7 6v 7 10 9
109
68. 4 x 9 4 x 9 5 11
511
6v 7
c. Move the decimal point 0 places. The exponent on 10 is 0. 2.71 2.71 100
90
4 x 9
55
73. a. Move the decimal point 1 place. The number is between 0 and 1, so use a negative power of 10. 0.86 8.6 101
1 1 1 2 4 22 21 1 1 1 2 4 4 2 1 2 4 8 16 4 4 4 4 4 31 4
69. 22 21 20 21 22
b. Move the decimal point 0 places. The exponent on 10 is 0. 8.6 8.6 100 c. Move the decimal point 1 place. The number is greater than 10, so use a positive power of 10. 86 8.6 101
1 1 1 3 9 32 31 1 1 1 3 9 9 3 1 3 9 27 81 9 9 9 9 9 121 9
70. 32 31 30 31 32
74. a. Move the decimal point 1 place. The number is between 0 and 1, so use a negative power of 10. 0.792 7.92 101 b. Move the decimal point 0 places. The exponent on 10 is 0. 7.92 7.92 100 c. Move the decimal point 1 place. The number is greater than 10, so use a positive power of 10. 79.2 7.92 101
71. a. Move the decimal point 5 places. The number is greater than 10, so use a positive power of 10. 350, 000 3.5 105 b. Move the decimal point 5 places. The number is between 0 and 1, so use a negative power of 10. 0.000 035 3.5 10 5
75. Move the decimal point 10 places. The number is greater than 10, so use a positive power of 10. 29,980,000,000 cm/sec 2.998 1010 cm/sec
c. Move the decimal point 0 places. The exponent on 10 is 0. 3.5 3.5 100
76. Move the decimal point 10 places. The number is greater than 10, so use a positive power of 10. 149,000, 000 km 1.49 108 km
72. a. Move the decimal point 3 places. The number is greater than 10, so use a positive power of 10. 2710 2.71 103
77. Move the decimal point 5 places. The number is between 0 and 1, so use a negative power of 10. 0.000 01 cm 1.0 105 cm
b. Move the decimal point 3 places. The number is between 0 and 1, so use a negative power of 10. 0.002 71 2.71 10 3
78. Move the decimal point 12 places. The number is between 0 and 1, so use a negative power of 10. 0.000 000 000 001 sec 1.0 10 12 sec
17
Chapter R Review of Prerequisites 79. Move the decimal point 0 places. The exponent on 10 is 0. 4.2 L 4.2 100 L
c. 101 is greater than 10. Move the decimal point 1 place to the right. 1.87 101 18.7
80. Move the decimal point 0 places. The exponent on 10 is 0. 8.25 hr 8.25 100 hr
85. 1.67 1021 molecules
81. a. 10 6 is between 0 and 1. Move the decimal point 6 places to the left. 2.61 10 6 0.000 002 61
86. 3.0 1012 bytes 3, 000, 000, 000, 000 bytes
1,670,000,000,000,000,000,000 molecules
87. 7.0 106 m 0.000007 m
b. 106 is greater than 10. Move the decimal point 6 places to the right. 2.61 106 2, 610, 000
88. 4.7 10 7 0.000 000 47
89. 2 10 3 4 108 2 4 103 108
c. 10 1 The decimal point does not move. 2.61 100 2.61 0
8 10
5
90. 3 104 2 101 3 2 104 101
2
82. a. 10 is between 0 and 1. Move the decimal point 2 places to the left. 3.52 102 0.035 2
6 10
b. 102 is greater than 10. Move the decimal point 2 places to the right. 3.52 102 352
91.
8.4 106 8.4 106 2 4 104 2 2.1 10 2.1 10
c. 100 1 The decimal point does not move. 3.52 100 3.52
92.
6.8 1011 6.8 1011 2 108 3.4 103 3.4 103
3
93. 6.2 1011 3 104 6.2 3 1011 104
1
83. a. 10 is between 0 and 1. Move the decimal point 1 place to the left. 6.718 10 1 0.6718
18.6 10
15
1.86 101 1015 1.86 1016
b. 100 1 The decimal point does not move. 6.718 100 6.718
94. 8.1 106 2 105 8.1 2 106 105
c. 101 is greater than 10. Move the decimal point 1 place to the right. 6.718 101 67.18
16.2 1011
1.62 101 1011 1.62 10
12
84. a. 101 is between 0 and 1. Move the decimal point 1 place to the left. 1.87 101 0.187
95.
b. 100 1 The decimal point does not move. 1.87 100 1.87
3.6 1014 3.6 10 14 5 105 5 105 0.72 1019
7.2 101 10 19 7.2 10 20
18
Section R.3
96.
3.68 108 3.68 108 4 102 4 102
101.
0.92 10 10
8 1010 bytes 8 1010 6 songs 4 10 6 bytes 4 10 song 2 104 songs
9.2 101 1010 9.2 1011
6.2 10 4.4 10 5
97.
102.
22
2.2 1017 5 22 6.2 4.4 10 10 10 12.4 10 17 2.2 10
103. 5 L
1.24 101 1010 1.24 10
2.5 1013 red blood cells
3.8 10 4.8 10 2
4
104. 4.3 light-yr
5
2.5 10 4 2 3.8 4.8 10 10 5 2.5 10 7.296 10
25.37 1012 mi
7
2.537 101 1012 mi 2.537 1013 mi
365 days 24 hr 3600 sec 1 yr 1 day 1 hr 31,536,000 sec
105. In the expression 6 x 0 , the exponent 0
applies to x only. In the expression 6 x , the exponent 0 applies to a base of (6x). The first expression simplifies to 6, and the second expression simplifies to 1. 0
3.1536 107 sec 1.0 104 gal 7 b. 3.1536 10 sec 1 sec
1.0 3.1536 104 107 gal
106. Scientific notation is used to represent very large and very small numbers to eliminate the need to write numerous zeros. It is also helpful to keep track of the location of the decimal point when performing calculations.
3.1536 10 gal 11
24 hr 60 min 1 day 1 hr
1440 min 1.44 103 min 65 beats 1.44 103 min b. 1 min
107. Yes; 4 16 0 2
108. Yes; 4 64 0 3
651.44 103 beats
109. No
93.6 103 beats
5.9 1012 mi 1 light-yr
4.3 5.9 1012 mi
99. a. 1 yr
100. a. 1 day
1 106 L 5 106 red blood cells 1L 1 L
25 1012 red blood cells
11
98.
6 1010 bytes 6 1010 videos bytes 5 106 5 106 video 1.2 104 videos
9.36 101 103 beats
110. Yes; 100 1
9.36 104 beats
19
1 1 100
Chapter R Review of Prerequisites 111. F
Gm1m2 d2 6.6726 10 11 5.98 1024 1.901 1027
7.0 10 6.67265.981.901 10 10 10 7.0 10
xm x m8 8 x
123.
x 2 m 7 x 2 m7 m5 x 2 m7m5 x m 2 m 5 x
11 2
11
2
24
27
Gm1m2 112. F d2 6.6726 1011 5.98 1024 8 101 2 6.371 106
6.6726 5.98 8 10 6.3712
11
x
125. x 4 m
3n
4 m3 n
y
126. y 5 m
10 10 10 24
127.
1
2n
5 m2 n
x12 mn y10 mn
x 4 m 3 y 5 n 7 x 4 m3 m7 y 5 n7 3n 2 m 7 3 n 2 x y
6 2
x 4 m3m7 y 5 n73n2 x 3m 4 y 2 n5
7.96 102 N 128.
x 2 n4 y 5n x 2 n4 n1 y 5 n 3n7 n1 3 n 7 x y
113. a. <
b. >
114. a. >
b. <
115. a. >
b. =
129. 6.284 106 6, 284,000
116. a. >
b. =
130. 9.128 104 0.0009128
117. x m x 4 x m 4
yn y n 3 3 y
122.
y 3 n 5 124. 2 n4 y 3n5 2 n4 y 3n52 n 4 y n9 y
11 2
1.55 1018 N
121.
x 2 n4n1 y 5 n3n7 x n5 y 2 n7
118. y n y 7 y n7
131. 2.45 107 0.000000 245
119. x m9 x m2 x m9 m 2 x 2 m7
132. 1.78 1010 17,800,000,000
120. y n9 y n1 y n9 n1 y 2 n8
Section R.4 Rational Exponents and Radicals 1. a. c.
b.
36
36, 0
2.
36, 4, 0 9 , 4, 0 49
d.
36,
e.
37
f.
37, 36,
t 3t 7 t 372 t 2 t2
4. 8 x 5 y 2
1
w4 v2 4 2 w v v 2 w4
3.
8 x 5 y 2 1
y2 8 x5
5. 3 x 5 y 2 2 x 3 y 6 x 53 y 21 6 x 2 y 3
3x 2 6. 3 y
9 , 4, 0 49
20
2
3x 2 y 3
3 x y 9 x1y 2
2
4
6
4
6
Section R.4 7. n 9.
a or a n
m
n
m
8. radicand; index
b. 125
10. n a
c. 1251 3 3 125 5
11. x
12. x
13. n a b
14. rationalizing
15. 4 81 3
16. 3 125 5
4 7 49 8
9 3 121 11
17.
0.09 0.3
19. 21
18.
4
20.
0.16 0.4
121 31. a. 169
12
121 b. 169
1 2
49 32. a. 144
12
49 b. 144
1 2
b. 16 3 4
169 13 121 11
12
144 12 49 7
3
1 163 4
3
1
16 4
3
1
2
3
1 8
2 8
d. 163 4
64 4 26. 3 125 5
3
3
1 1 1 1 3 3 34 4 16 8 2 16
e. 16 is undefined because 4 16 is not a real number. 34
27. a. 251 2 25 5 b. Undefined
f. 16 is undefined because 4 16 is not a real number. 3 4
25 5 5
28. a. 361 2 36 6
34. a. 813 4 4 81 3 27
b. Undefined c. 361 2 36 6 6
b. 813 4
29. a. 271 3 3 27 3 b. 27
144 49
c. 163 4 4 16
24. 4 625 1 4 625 1 5 5
13
12
49 7 144 12
23. 4 81 1 4 81 1 3 3
c. 25
169 121
33. a. 163 4 4 16
22. 4 625 is not a real number
12
121 11 169 13
2 8
81 is not a real number
1 1 25. 3 8 2
3 125 5
13
3
1 813 4
3
1
81
4
3
1
3
3
1 27
c. 813 4 4 81 3 27
3 27 3
d. 813 4
c. 271 3 3 27 3 30. a. 1251 3 3 125 5
3
3
1 1 1 1 3 3 4 813 4 27 3 81
e. 81 is undefined because 4 81 is not a real number. 34
21
Chapter R Review of Prerequisites f. 81 is undefined because 4 81 is not a real number. 3 4
c. 8 z
4 16 2
35. a. 642 3 3 64 b. 64
2
1 23 64
2 3
1
1 2 2 3 4 16 64
41.
4 16 2
c. 642 3 3 64 d. 642 3 e. 64 f. 64
3 64
2 3
36. a. 82 3 3 8 b. 82 3
2
1
64 2
1
64 3
2
2
f. 8
2 3
2
1
4
2
1 16
1
2
2
1 4
2
49.
3w2 3 3 y1 3 23 y 1 3 w
50.
8d 5 7 8c 3 4 57 c 3 4 d
51. 16 x 8 y1 5
34
11
11
4
4
38. a. z 3 10 10 z 3 or 10 z
8 3 4
15 34
3
8 3 4
15 3 4
2 x 24 4 y 3 20 3
1
8 x 6 y 3 20 52. 125a 6b 7 5
13
4
11
x y
34
x y
4
11
16
4 16
b. 6 y 6 y or 6 y c. 6 y 6 y or 6 6 y 4 11
y 7 5 y 4 5 y 7 5 4 5 y11 5 1 5 1 5 y11 51 5 y10 5 y 2 y1 5 y y
1 23 2 2 3 8 2 4 8
13
48.
2
1
a 2 3 a 5 3 a 2 35 3 a 7 3 1 3 1 3 a 7 31 3 a 6 3 a 2 a1 3 a a
2 4
37. a. y 4 11 11 y 4 or 11 y 4 11
15
47.
2
1
3 8
12
44. 11 t 11t1 2
46. 3 m3 n3 m3 n3
2
1 1 1 1 d. 82 3 2 3 2 2 3 8 4 2 8 23
11t 11t
42.
4 16
2 4
e. 8
12
1
c. 82 3 3 8
6 x 6 x
2
8
3
40. 7 z 4 z 4 7
45. 5 a 5 b5 a 5 b5
23
3
10
43. 6 x 6 x1 2
2 4
1 82 3
3
10
3
2
1 1 1 1 2 2 23 3 64 16 4 64
23
3 10
39. 5 a 3 a 3 5
1
8 z or 8 z
b. 8 z 3 10 810 z 3 or 8 10 z
8 y 3 20 x6
125
13
a b 6 13
7 5 1 3
3 125a 61 3b 7 51 3
4
5a 6 3b 7 15 5a 2b 7 15
3
22
5a 2 b 7 15
Section R.4 2m 2 3 n1 5 212 m 2 312 n1 510 53. 3 4 1 2 n 2m n 3 412 210 m1 210 12
63. 4 2 x 5 2 x 5
10
4
212 m8 n 2 n9 210 m5 2 2 m 3 4m 3 7 7 n n
4
3
1 24
12
4
3 z 2 2 3 z 2
65.
w12 w6
67. a.
1 23
3x y 3 x y 54. 3 8 4 3 3 84 3 4 33 y 3x y 3 x 12
64.
66. 4 c32 c8
c 7 c 6 c c6 c c3 c
b. 3 c 7 3 c 6 c 3 c 6 3 c c 2 3 c
34 x 2 y 3 2 3 3 4 2 32 y 3 x x
c. 4 c 7 4 c 4 c3 4 c 4 4 c3 c 4 c3 d. 9 c 7 9 c 7
1
m2 m2 55. m n m n
m 21
12
m 21 2
1
m n m n1 2 m 2
68. a.
m1
1
b. 3 d 11 3 d 9 d 2 3 d 9 3 d 2 d 3 3 d 2
m n m n 12 m n m 1 m m n 1 2 2
c2 c2 56. c d c d
c 2 2
32
12
c. 4 d 11 4 d 8 d 3 4 d 8 4 d 3 d 2 4 d 3 d. 12 d 11 12 d 11
c 2 3 2
69. a.
c d 2 c d 3 2 c 4
c d
2
c 1
c d 1 2
d 11 d 10 d d 10 d d 5 d
24 22 6 22 6 2 6
b. 3 24 3 23 3 3 23 3 3 2 3 3
c3
c d 3 2 12 c d
70. a.
54 32 6 32 6 3 6
b. 3 54 3 33 2 3 33 3 3 3 3 2
c
71. 3 250 x 2 y 6 z11 3 53 2 x 2 y 6 z11
t 2 t for t 0
57. a.
t 2 t for all real numbers
b.
3 53 y 6 z 9 2 x 2 z 2 3 53 y 6 z 9 3 2 x 2 z 2 5 y 2 z3 3 2 x2 z 2
58. a. 4 c 8 c 8 for c 8 4
72. 3 40ab13c17 3 23 5 ab13c17
b. 4 c 8 c 8 for all real numbers 4
3 23 b12 c15 5abc 2 59.
y y 2
61. 3 y 3 y
60.
4
y y 4
3 23 b12 c15 3 5abc 2 2b 4 c 5 3 5abc 2
62. 5 y 5 y
23
Chapter R Review of Prerequisites 73. 4 96 p14 q 7 4 24 6 p14 q 7
3
4 24 p12 q 4 6 p 2 q 3
81.
3
4 24 p12 q 4 4 6 p 2 q 3 2 p3q 4 6 p 2 q3 74. 4 243m19 n10 4 35 m19 n10
83.
4 34 m16 n8 3m3 n 2
3m n 75.
24
3
2m 2 n 7
16m14 n 4
3
5 x5 y x3 x 3 2 4 3 625 x y 125 y 5y
3
2m 2 n 7 n3 n 3 14 4 12 16m n 8m 2m 4
10 14 140 22 35
22 35 2 35
4 34 m16 n8 4 3m3n 2 4
625 x 2 y 4 3
82.
5 x5 y
6 21 126 32 14
84.
3 2
3m n
32 14 3 14
84 y 2 22 21 y 2 3
3
85. 3 xy 2 3 x 2 y 3 xy 2 x 2 y 3 x3 y 3 xy
22 y 2 21 y 2 2
2 y 2 21 y 2
86. 4 a 3b 4 ab3 4 a 3b ab3 4 a 4b 4 ab
2 y 2 21 y 2
87. 3 4 a 3 5 4 a 3 3 5 4 a 3 a 3
2
2
76.
18 w 6 32 2 w 6 3
15 4 a 6 15 4 a 4 a 2
3
15 4 a 4 4 a 2 15a a 2 4
3 w 6 2 w 6 2
2
15a a1 2 15a a
32 w 6 2 w 6 2
88. 7 6 t 5 2 6 t 5 7 2 6 t 5 t 5
3 w 6 2 w 6
14 6 t10 14 6 t 6 t 4 7
77.
78.
p 36 q11 4
79. 4 3
p
7
36 q11 4
p p 6
p p p p 6 6
q10 q 2
q10 q q 5 q 2 2
6
6
3
14 6 t 6 6 t 4 14t t 4 6 14t t 2 3 14t 3 t 2 1 4 89. 3 6a 2b 2 c 3 4a 2 c 2 2 3 2 3 6a 2b 2 c 4a 2 c 2 3 2 3 24a 4b 2 c 3 3 2 3 23 a 3c3 3ab 2 3 2 3 23 a 3c3 3 3ab 2 3 4 ac 3 3ab 2 3
w3 z 5 4 3 w3 z 5 4 3 w3 z 3 z 2 3 8 2 8 2 3 w3 z 3 3 z 2 2wz 3 z 2
c6 d 7 8 3 c6 d 7 8 3 c6 d 6 d 80. 8 3 64 4 64 3
2 3 c 6 d 6 3 d 2c 2 d 2 3 d
24
Section R.4
3 1 90. 3 9m 2 n5 p 3 6m 2 np 4 4 6 1 3 9m 2 n5 p 6m 2 np 4 8 1 3 54m 4 n6 p 5 8 1 3 33 m3n 6 p 3 2mp 2 8 1 3 33 m3n 6 p 3 3 2mp 2 8 3 mn 2 p 3 2mp 2 8
91. 5 x3 y 2 4 x x 3 y
x
25
3 51 4
y y 15
13 4 1 5
95.
x x x x x
x x x x x x1 2
12 12
34 12
y y4 9
13
13
13 9 1 3
y13 27 27 y13
x
14
y 2 5 x12 205 20 y 2 5
97. 3 3 2 y 2 9 3 2 y 2 3 2 y 2 3 9 1 3 2 y 2
5 3 2 y 2
2 13
98. 8 4 3z 3 4 3 z 3 2 4 3z 3 8 1 2 4 3 z 3
9 4 3z 3
1 21 3 1 4 2 3
b
6 12 4 12 3 128 12
b
99.
a10 12b11 12 12 a10b11 93.
6
m 3 m2 m m2 3
m m 16
1 2 3 1 6
53 16
1 7 5 50 18 72 5 3 6 1 2 7 2 5 2 5 2 3 2 6 2 5 3 6 1 7 5 5 2 3 2 6 2 5 3 6
2 7 2 5 2
m5 18 18 m5
100.
8
y
a1 2b1 4 a1 3b 2 3 a
78
13 13
43 13
ab
a
12 12
y y
96. 3 y 3 y 3 y y y y1 3
14
14
x x3 2
7 4 12
x17 20 y 8 20 20 x17 y 8 92. 4 a 2b 3 ab 2 a 2b
7 4 15
y 7 20 20 y 7
x
2 15
x y 35
94. 5 y 4 y 3 y y 3 4
1 7 5 2 2
2 2 1 2 2 2 2 1 2 2 2 1 75 27 12 5 3 3 3 2 3 5 3 3 3 2 3 5 3 2 5 3 2 5 3 2
2 3 2 3 3 2 2 1 3 3
101. 3 x 3 16 xy 4 xy 3 54 xy 5 3 250 x 4 y 4 3 x 3 23 y 3 2 xy xy 3 33 2 xy 5 3 53 x3 y 3 2 xy
6 xy 3 2 xy 3 xy 3 2 xy 25 xy 3 2 xy 28 xy 3 2 xy 102. 8 4 32a 5b 6 5b 4 2a 5b 2 ab 4 162ab 2 8 4 24 a 4b 4 2ab 2 5b 4 a 4 2ab 2 ab 4 34 2ab 2
16ab 4 2ab 2 5ab 4 2ab 2 3ab 4 2ab 2 8ab 4 2ab 2
25
7
Chapter R Review of Prerequisites 103. 12 2 y 5 y 2 y 12 5 y 2 y
111. c a 2 b 2
322 482 1024 2304
104. 8 3w 3w 3w 8 3w 3w 105.
3328 162 13 16 13 in. 58 in.
1 3 8z3 98 z 2 7 1 2 2 3 2 2 z 2z 7 2z 2 7
112. c a 2 b 2 2
36 122 122 182 144 324 2
z 2 z 3 2 z z 3 2 z
106.
468 62 13 6 13 ft 21.6 ft
2 1 2 2 1 2 2 45c 20c 3 3 5c 2 c 5c 3 2 3 2 2 5c c 5c
113. L r 2 h 2
42 102 16 100
2 c 5c 107. s
116 22 29 2 29 in. 10.8 in.
a b c 6 7 9 22 11 2 2 2
114. A r r 2 h 2
6 62 42 6 36 16
A s s a s b s c
6 52 6 22 13
1111 6 11 7 11 9
12 13 m 2 136 m 2
11 5 4 2
115. A r 2
440
3 5 3 5
2 110 2 110 in. 2
2
2
2
a b c 3 5 6 14 7 2 2 2 A s s a s b s c
108. s
2
9 5 45 in.2 116. A r 2
7 7 3 7 5 7 6 7 4 2 1
8 7 2
56 22 14 2 14 cm 2
4 7
109. b c 2 a 2
2
42
7
2
2
16 7 112 cm 2
182 122 324 144 180 62 5 6 5 m
S 117. r 1 C
110. b c 2 a 2
1n
11,500 1 22,990
162 82 256 64 192 82 3 8 3 m
14
1 0.841 0.159 15.9%
26
Section R.4 118. t 0.84 x 3 5 0.84 15
35
4.3 hr
125.
119. a. for n = 1: f 440 2n 12 440 21 12 466.2 Hz for n = 2: f 440 2n 12 440 22 12 493.9 Hz for n = 3: f 440 2n 12 440 23 12 523.3 Hz 126.
12 12
3
62 5 4 3
3
8 9 3
4
11 3 125
81 3 1000 36 25
4
11 5 9 10 6 5
4
16 25 5
12
1.4 4
12
22 2 2 3
1.4 2 2.8 $2.8 million
r 127. Tp 0.7 Ts 273 d
b. P 450 x 2,800,000
450 10,000 2,800, 000
12
273
1.26 106 Tp 0.7 7700 273 4.3 108
4,500,000 2,800, 000 1, 700,000
$1.7 million 121. In each case, add like terms or like radicals by using the distributive property.
12
273
Tp 0.7 7973
1.26 106 273 4.3 108
Tp 0.7 7973
1.26 1 273 4.3 102
1.26 0.1 273 29.1 4.3 Yes; The model gives a mean surface temperature of approximately 29.1C . Tp 0.7 7973
122. The terms within each expression are not like terms or like radicals.
8.0 1012 8.0 1012 4 2.0 10 2.0 104
r 128. Tp 0.7 Ts 273 d
12
8.0 10 2.0 104
12
273
7.0 105 Tp 0.7 5700 273 1.49 108
4.0 108 2.0 104 1.44 1016 1.44 1016 9.0 1010 9.0 1010
255
1.4 0.3 10 1
124.
25 16 9
8 22 2 2 2
120. a. C 1.4 0.3x 1
6 4 16
233
b. for n = 9 f 440 2n 12 440 2 9 12 261.6 Hz
123.
3
1.44 1016 9.0 1010
12
Tp 0.7 5973
7.0 105 273 1.49 108
Tp 0.7 5973
7.0 1 273 1.49 103
273
7.0 1 0.1 273 13.6 1.49 10 Approximately 13.6C . Tp 0.7 5973
0.16 106 0.4 103 4.0 102
27
Chapter R Review of Prerequisites Section R.5 Polynomials and Multiplication of Radicals 13. a b
14. 3 x
15. a. Yes
b. Yes
c. No
d. No
1 4x 6 y 2 11 y 77 x 2 x 3 y
16. a. No
b. Yes
c. No
d. Yes
79 x 8 y
17. 18 x 7 7.2 x 3 4.1 ; Leading coefficient 18 ; Degree 7
1 15 x 6 z 3 24 12 x 5 x 2 z
1. 3 8 4 x
7 x 2 z 24 2. 11 y 7 x
5 5 2
3.
2
5. 3x y
1 4
4.
x x 5
5
3
18. 1.7 y 8 9.1y 5 4.6 y 2 ; Leading coefficient 1.7 ; Degree 8
3
1 3 9 1 3 4 8 4 xy 3 x y x y 2 2 11
1 y; 3 Leading coefficient 1 ; Degree 2
19. y 2
5
81x y 8 81x11 y 5 8
3 2 3
6. 4a b
2
4 20. c 5 c 2 ; Leading coefficient 1; Degree 5 5 2
1 5 3 9 6 1 2 10 ab 4 a b a b 3 3
21. 11
22. 8
23. 8 p 7 4 p 4 2 p 5 2 p 7 6 p 4 p 2
64a11b 4 9
8 p 7 2 p 7 4 p 4 6 p 4 p 2 2 p 5
7. polynomial
8. descending
9. leading
10. coefficient
6 p 7 2 p 4 p 2 2 p 5
7 w5 9w5 3w3 5w3 4w 6 3 2 w5 2w3 4 w 9
25. 0.05c 3b 0.02c 2b 2 0.09cb3 0.03c 3b 0.08c 2b 2 0.1cb3
0.05c b 0.03c b 0.02c b 0.08c b 0.09cb 0.1cb 3
3
2 2
2 2
3
3
0.08c 3b 0.06c 2b 2 0.01cb3
24. 7 w5 3w3 6 9 w5 5w3 4w 3
11. binomial; trinomial 12. perfect
26. 0.004mn5 0.001mn 4 0.05mn3 0.003mn5 0.007 mn 4 0.07 mn3
0.004mn5 0.003mn5 0.001mn 4 0.007 mn 4 0.05mn3 0.07 mn3 0.001mn5 0.008mn 4 0.12mn3
28
Section R.5
5 3 1 1 27. x 2 x 5 2 x 2 x 2 4 8 2 4 1 2 1 2 5 3 x x x x5 2 2 4 2 8 4 1 11 x2 x 6 2 4 8
35. 4u 2 5v 2 2u 2 3v 2
1 7 2 3 28. y 3 y 7 y 3 y 5 7 5 10 10 5
36. 2 z 3 5u 2 7 z 3 u 2
2
2
8u 4 12u 2 v 2 10u 2v 2 15v 4 8u 4 2u 2 v 2 15v 4
5u u
2 z 3 7 z 3 2 z 3 u 2 5u 2 7 z 3 2
2
14 z 6 2u 2 z 3 35u 2 z 3 5u 4 14 z 6 33u 2 z 3 5u 4
1 29. 6a 5b a 2b 2 2a 7b3 3
1 37. 3 y 6 y 2 5 y 4 3 1 3 y y 2 3 y 5 y 3 y 4 3
1 30. 10c 2 d 5 cd 7 5c 3 d 12 2
5v 3v
4u 2 2u 2 4u 2 3v 2 5v 2 2u 2
2 3 1 7 y3 y3 y y 7 5 7 5 10 10 5 1 3 3 y y4 7 10 2
1 6 y 2 6 5 y 6 4 3
7 m 2 2m 4 7 m 2 3m 7 m 2 4
y 3 15 y 2 12 y 2 y 2 30 y 24
31. 7m 2m 3m 4 2
4
y 3 13 y 2 42 y 24
14m6 21m3 28m 2
1 38. 10v 5 v 2 3v 1 5
32. 8 p 5 2 p 2 5 p 1
8 p 5 2 p 2 8 p 5 5 p 8 p 5 1
1 10v v 2 10v 3v 10v 1 5
16 p 40 p 6 8 p 5 7
1 5 v 2 5 3v 51 5
33. 2 x 5 x 4
2 x x 2 x 4 5 x 5 4
2v3 30v 2 10v v 2 15v 5
2 x 2 8 x 5 x 20
2v3 31v 2 25v 5
2 x 3 x 20 2
39. a b a b a b 2
34. w 7 6w 3
a a a b b a b b
w 6 w w 3 7 6w 7 3
a 2 ab ab b 2 a 2 2ab b 2
6w 3w 42 w 21 2
40. a b a b a a a b b a b b
6w2 39w 21
a 2 ab ab b 2 a 2 b 2 41. 4 x 5 4 x 5 4 x 5 16 x 2 25 2
29
2
Chapter R Review of Prerequisites 42. 3 p 2 3 p 2 3 p 2 9 p 2 4 2
2
43. 3w2 7 z 3w2 7 z 3w2
7z 2
52.
2
9 w4 49 z 2
2u 2
44. 9v3 2u 9v3 2u 9v3
2
p 23 p 2 p 22 p 2 p 2 2 p2 22 p 2 p 2 4 p 4 p p 2 p 4 p p 4 2 p 2 2 4 p 24 p3 4 p 2 4 p 2 p 2 8 p 8
81v 6 4u 2
p 3 6 p 2 12 p 8 2
2 1 2 1 2 1 45. c d 3 c d 3 c d 3 5 3 5 3 5 3
2
53. u v w u v w
u v w u 2 2uv v 2 w2
1 2 4 6 c d 25 9
2
54. c d a c d a
4 1 4 1 46. n p 4 n p 4 6 5 6 5 2
c d a c 2 2cd d 2 a 2 2
47. 5m 3 5m 2 5m 3 3 2
2
55. I 45.58 x 460.1
45.58 8 460.1 $824.74 billion
2
56. P 10.86 x 191.5
25m 2 30m 9
10.86 5 191.5 $245.8 billion
48. 7v 2 7v 2 7v 2 2 2
2
2
57. a. I P 45.58 x 460.1 10.86 x 191.5
49v 2 28v 4
2
2
1 2 16 8 1 4 n p n p 4 6 5 36 25
49. 4t 2 3 p 3
2
56.44 x 651.6 b. The polynomial I P represents the total amount of health-related expenditures made by individuals with private insurance.
4t 2 4t 3 p 3 p 2
2 2
2
3 2
3
16t 4 24t 2 p 3 9 p 6 50. 2a 2 11b3
c. I P 56.44 x 651.6
2
2a 2 2 2a 2 11b3 11b3 2
56.44 6 651.6 990.24 $990.24 billion; In the year 2006, a total of $990.24 billion was spent on healthrelated expenses by individuals with private insurance.
2
4a 4 44a 2b3 121b6 51. w 4 w 4 w 4 3
2
w 4 w2 2 w 4 42
w 4 w2 8w 16
58. a. I 45.58 x 460.1
45.58 10 460.1 $915.9 billion
w w2 w 8w w 16 4 w2
b. P 10.86 x 191.5
4 8w 4 16
10.86 10 191.5
w3 8w2 16 w 4 w2 32w 64
$300.1 billion
w3 12 w2 48w 64
30
Section R.5 66. A lw 2 y x z 2 yz xz or xz 2 yz
c. I P 56.44 x 651.6
I P 56.44 10 651.6
67. A lo wo li wi
I P $1216 billion or $1.216 trillion
2 x 3 x 7 x 3 x 2
59. P 2 3x 4 x 3 x 4
2 x x 2 x 7 3 x 3 7
2 5 x 11 10 x 22
x x x 2 3 x 3 2
2 x 2 14 x 3x 21 x 2 2 x 3x 6
60. P 2 y y 1 2 y 3 2 4 y 4 8 y 8 61. A lw
5 y 2 x 3 25 y 2 4 x 2 12 x 9 2
2
68. A lo wo li wi
5 y 6 3 y 1 2 y 4 y 1
25 y 2 4 x 2 12 x 9
5 y 3 y 5 y 1 6 3 y 6 1
62. A lw 6v x 2 6v x 2
6v x 2 36v 2 x 2 4 x 4 2
2
2 y y 2 y 1 4 y 4 1
15 y 2 5 y 18 y 6 2 y 2 2 y 4 y 4 2
13 y 2 17 y 2
63. V lwh x 4 x 4 x 6
69. a. x 1
x 4 x 6
b. x x 1 2 x 1
2
x 8 x 16 x 6
c. x x 1 x 2 x
x x x 6 8 x x 8 x 6 16 x 2
2
d. x 2 x 1 x 2 x 2 2 x 1 2 x 2 2 x 1 2
16 6 x 3 6 x 2 8 x 2 48 x 16 x 96
70. a. x 2
x 3 14 x 2 64 x 96
b. x x 2 2 x 2
64. V lwh 2 x 3 x 1 x 1
c. x x 2 x 2 2 x
2 x 3 x 1
d. x 2 x 2 x 2 x 2 4 x 4 2
2
2 x 3 x 2 2 x 1
x2 x2 4 x 4 4x 4
2 x x 2 2 x 2 x 2 x 1 3 x 2 3 2 x 3 1
71. y 7 2 y 3 2
2 x3 4 x 2 2 x 3x 2 6 x 3
2
y 2 14 y 49 2 y 2 6 y 9
2 x 7 x 8x 3 3
15 y 23 y 6 2 y 6 y 4 2
36v 2 x 2 4 x 4
2
2
x 2 6 x 27
5 y 2 x 3 5 y 2 x 3
2 x 11x 21 x 5 x 6 2
2
y 2 14 y 49 2 y 2 12 y 18
65. A lw a b c ac bc
y 2 26 y 31
31
Chapter R Review of Prerequisites 72. x 4 6 x 1 2
5 2 2 2 5 2 6 3 5 2 4
81. 5 2 2 2 6 3 4
2
x 2 8 x 16 6 x 2 2 x 1
x 2 8 x 16 6 x 2 12 x 6
10 4 30 6 20 2
5 x 2 20 x 10
10 2 30 6 20 2
73. x n 3 x n 7
20 30 6 20 2
x n x n x n 7 3 x n 3 7
5 7 3 7 5 7 2 5 5 7 3
82. 5 7 3 7 2 5 3
x 7 x 3x 21 2n
n
n
x 2 n 4 x n 21
15 49 10 35 15 7
74. y n 4 y n 5
15 7 10 35 15 7
y y y 5 4 y 4 5 n
n
n
n
105 10 35 15 7
y 5 y 4 y 20 2n
n
n
3 6 5 3 3 6 4 2 3 6 6
83. 3 6 5 3 4 2 6
y 2 n y n 20
z 2w z w 2
75. z n wm
n 2
m 2
m n
z 2 n 2 wm z n w 2 m
76. w y n
15 18 12 12 3 36
w 2w y y
m 2
n 2
n
15 32 2 12 22 3 3 6
m 2
m
45 2 24 3 18
w 2 n 2 wn y m y 2 m
5 a 25
77. a n 5 a n 5 a n 78. b n 7 b n 7 b n
2
2
2
4 10 10
56 5 60 2 40
60 x 50
2 y 4 y 2 y 5 3 4 y
85. 2 y 3 4 y 5
80. 2 y 7 2 y 7 2 y 7
28 22 5 12 52 2 40
36 x 2 25 36 x 2 60 x 25
2
2 y 7 2 4 y 2 28 y 49
3 5
4 y 49 4 y 28 y 49 2
28 20 12 50 4 10
2
6 x 52 36 x 2 60 x 25
2
4 10 7 2 4 10 3 5
7 2 b 2 n 49
79. 6 x 5 6 x 5 6 x 5 2
84. 4 10 7 2 3 5 10
2n
2
8 y 2 10 y 12 y 15
28 y 98
8 y 2 y 15
32
Section R.5
5 z 3 z 5 z 6 4 3 z
86. 5 z 4 3 z 6
93. 6 z 5
94. 2v 17
15 z 30 z 12 z 24 2
2
2v 2 2v 17 17 2
2
4 3 6 3 4 3 5 5 2 5 6 3 2 5 5 5
2
2
2
2
2
3
8 15 22
2
2
3
3
3 11 2 11 3 11 9 2 7 2 2 11 7 2 9 2
x 1 5 x 1 5 x 1 5 2
97.
y 2 4 y 2 4 y 2 4 2
66 13 22 126
98.
7 2
2
4 3 7 12 7 5
11 2
99.
4x y 2 y x 2
5u v 6v u 2
2
y 2 4 y 2 2 y 2 4 4 2
100.
2
2
y 2 8 y 2 16 y 8 y 2 18 101.
5 2 x 5 2 x 5 2 x 5 2 x 5 2 x 25 4 x
2
2
25u v 36v u 25u v 36uv 2
2
x 10 x 1 26
2
92. 5u v 6v u 5u v 6v u 2
2
x 1 10 x 1 25
16 x 2 y 4 y 2 x 16 x 2 y 4 xy 2
x 1 5 x 1 2 x 1 5 5
2
25 2 11 50 11 39 91. 4 x y 2 y x 4 x y 2 y x
2
y 2 16 y 14
13 22 60 89. 2 3 7 2 3 7 2 3
2
x 1 25 x 24
6 121 27 22 14 22 63 4
3
4c 6 d 12c 3 d 3 cd 9cd 6
88. 3 11 7 2 2 11 9 2
90. 5 2 11 5 2 11 5 2
2
2
2c d 2 2c d 3d c 3d c
96. 2c 3 d 3d 3 c
72 8 15 50
2
25a 4b 70a 2b 2 ab 49ab 4
24 9 20 15 12 15 10 25
2
5a b 2 5a b 7b a 7b a
95. 5a 2 b 7b 2 a
87. 4 3 2 5 6 3 5 5
2
4v 2 4 17v 17
15 z 18 z 24
2
36 z 2 12 5 z 5
4 6
6z 26z 5 5
2
33
2
2
Chapter R Review of Prerequisites
102.
7 3 z 7 3 z
107. A polynomial consists of a finite number of terms in which the coefficient of each term is a real number, and the variable factor x is raised to an exponent that is a whole number.
7 3 z 7 3 z 7 3 z 49 9 z
2
2
108. In the expression x 7 , the exponent on x is 1 which is a whole number. The expression x 7 is equivalent to x1 2 7 . The exponent on x is 12 , which is not a whole number.
x y x y x y 2 x y x y x y 2
103.
2
2
109. In each case, multiply by using the distributive property.
x y 2 x y x y x y
110. Both expressions involve a product of conjugates a b a b . In each case, the
2 x 2 x2 y 2
a 2 a 2 a 2 2 a 2 a 2 a 2
product is a 2 b 2 .
2
104.
2
111. False
2
a 2 2 a 2 a 2 a 2
112. True
2a 2 a 2 4
113. True
1 bh 2 1 2 5 6 2 5 6 2 2 2 1 2 5 6 2
114. False
105. A
115. a b a b a b 3
a a a b 2ab a 2ab b b a b b a 2 2ab b 2 a b 2
a 3 3a 2b 3ab 2 b3
a b 3 a b 2 a b
a a a b 2ab a 2ab b b a b b a 2 2ab b 2 a b
106. A
2
a 3 a 2b 2a 2b 2ab 2 ab 2 b3
116.
2
2
1 4 5 6 2 1 14 7 m 2 2 1 bh 2 1 4 7 2 4 7 2 2 2 2 1 4 7 2 2
2
2
2
2
2
a 3 a 2b 2a 2b 2ab 2 ab 2 b3 a 3 3a 2b 3ab 2 b3
1 16 7 2 2 1 110 55 ft 2 2
34
Problem Recognition Problem Recognition Exercises: Simplifying Algebraic Expressions
1. a. 641 2 64 8 13
b. 64
d. 2 x 4 y
64 4
d. 64 1
2
42 16
1 64
e. 641 2 64 8 f. 64 g. 64 h. 64
23
3 64
4 16
4
2 3
25a b
2. a. 5ab3
2
b. 5a b3
2
3 64
b.
2
2
2
2
2
1
2
5ab
3 2
3. a. 2 x y
1 x2
1 x15
4
b. 3 x8 3 x 6 x 2 x 2 3 x 2
1 25a 2b6
c. 5 x8 5 x5 x3 x 5 x3
6. a. 3a 4b 2 2a b 2 3a 4b 2 2a b 2
a 5b 2
5a 2 2 5a b 3 b 3
2
b
4b
2x y
2
2
6a 2 3ab 2 8ab 2 4b 4
2
4
3a 2a 3a b 2 4b 2 2a
1 25a 10ab3 b 6 2
1
b. 3a 4b 2 2a b 2
2
2 2 x4 y y
6a 2 5ab 2 4b 4
2
4
2
7. a. a 2 b 2 a 2 b 2 a 2 b 2 a 2 b 2
4x 4x y y2
12
1
2 x4
2
x
x8 x8
d. 9 x8 9 x8
2
8
1 4 x 4 x4 y y 2 8
3
3 2
8
2
x5 x 5 3 x 2 3 x
5. a.
2
5a b
4x y 2
b. 2 x 4 y
c. 2 x 4 y
4
3
1
2
4
2x 2 2x y y
d. x5 x5 3 x 15
25a 10ab b 6
d. 5a b3
1
c. x 5 x 3 x 53 x 2
1 16
5a 2 5a b 3 b 3
2
4 2
2 6
2
c. 5ab3
2x y 4
4. a. x5 x3 x53 x8
64 Undefined
12
1
2
3
c. 642 3 3 64
2a 2
1 4 x8 y 2
b a b
b. a 2 b 2 a 2 b 2 a 2
35
2
2 2
4
4
Chapter R Review of Prerequisites c. a b a b 2
2
a 2 2ab b 2 a 2 2ab b 2 a 2ab b a 2ab b 4ab 2
2
d. a b a b 2
2
c.
2
d.
x y x y x y 2
2
x y
2
a a a 2ab a b 2ab a 2ab 2ab 2ab b b a b 2ab b b 2
2
2
11. a. 3 2 x 3 2 x 3 4 x 2
2
b. 3 2 x 3 2 x 2 3 2 x 2
2
2
2
2
2
12. a. 4 y 3 y y y 14
2
y 3 12 4 12 y 7 12
a 2b 2 2ab3 b 4
12 y 7
a 4 2a 2b 2 b 4
b. 4 y 3 y 4 y 3 y
8. a. for a b, a b a b b a b. for a b, a b a b
13. a. 36 12 6 2 3 6 2
9. a. for x 2, x 2 x 2
x y x y
b.
x2 y 2 x2 y 2
1 1 2 2 4
b. 36 12 6 2 36 2 2 18 2 9
b. for x 2, x 2 x 2 x 2 10. a.
13
y 1 41 3
2
a 4 2a 3b a 2b 2 2a 3b 4a 2b 2 2ab3
2
2
x 2 xy y
2
a 2 2ab b 2 a 2 2ab b 2 2
x y x 2 x y y
c. 36 12 6 2 36 12 3 3 3 1 d. 36 12 6 2 3 3 1
2
62 82 36 64 100 10
14. a.
62 82 6 8 14
b.
Section R.6 Factoring
4x y
2 x 2 2 x y y
1. a. 2 x 4 y 2
2
b. 2 x 4 y 2
8
2
4
25a b
5a 2 5a b b
2. a. 5a 3b 2 4 2
4
2
2 2
2
b. 5a 3 b 2
4 x8 4 x 4 y 2 y 4
6 4
2
3 2
3
2
25a 6 10a 3b 2 b 4
36
2 2
2
Section R.6
3 x 2 x 3 x 5 y 4 y 2 x 4 y 5 y
11. perfect; a b
6 x 2 15 xy 2 8 xy 2 20 y 4
13. 15c5 30c 4 5c3
3. 3x 4 y 2 2 x 5 y 2
2
2
2
12. squares; a b a b
2
5c 3c 6c 1
5c3 3c 2 5c3 6c 5c3 1
6 x 2 7 xy 2 20 y 4
3
6a a 6a 7 k 2k a
4. 6a 2 2k a 2 7 k 2
2
2
3m 4m 5m 1
3m 2 4m 2 3m 2 5m 3m 2 1
6a 42a k 2a k 14k 2
2
6a 40a k 14k 4
2
2
2
2
15. 21a 2b5 14a 3b 4 35a 4b
7 a b 3b 2ab 5a 2
2
2
9 p q 4 p q 2 p 3q 2
9 25 3 5 m5 np m10 n 2 p 2 2 7 4 49
5
2
2
17. 5 z x 6 y 7 x 6 y x 6 y 5 z 7
18. 4t u 8v 3 u 8v u 8v 4t 3
3 4 x 3 6 x 3 9
2 x 4 x 2 x 6 x 2 x 9 2
3k 7 10k 5k 3k 7 5k 2k 5k 1 5k 3k 7 2k 1
19. 10k 2 3k 2 7 5k 3k 2 7
2
2
8 x 12 x 18 x 12 x 18 x 27 2
3
9 p 2 q 5 4 p 2 q 2 9 p 2 q 5 2 p 9 p 2 q 5 3q
2
4
16. 36 p 4 q 7 18 p 3 q 5 27 p 2 q 6
5 3 5 3 6. m5 np m5 np 2 7 2 7
7. 2 x 3 4 x 2 6 x 9
2
2
2
8 x3 27
2
3 y 9 y 3 y 15 y 3 y 25 5 9 y 515 y 5 25
8. 3 y 5 9 y 15 y 25 2
20. 8 j 3 4 j 9 4 j 2 4 j 9
2
4 j 9 8 j 3 4 j 2
2
4 j 9 4 j 2 2 j 4 j 2 1
27 y 3 45 y 2 75 y 45 y 2 75 y 125
4 j 2 4 j 9 2 j 1
27 y 3 125
9. cubes; a b a 2 ab b 2
7 a 2b 3b 4 7 a 2b 2ab3 7 a 2b 5a 2
1 9 1 3 c 4 ab c8 a 2b 2 5 8 25 64
2
2
2
3 1 3 1 5. c 4 ab c 4 ab 5 8 5 8
3
2
14. 12m 4 15m3 3m 2
2
2k 7 k 4
2
10. difference; a b a 2 ab b 2
b. 6 x 12 x 9 3 2 x 4 x 3
21. a. 6 x 2 12 x 9 3 2 x 2 4 x 3
2
37
2
Chapter R Review of Prerequisites
b. 15 y 10 y 25 5 3 y 2 y 5
22. a. 15 y 2 10 y 25 5 3 y 2 2 y 5 2
32. w2 5w 66
w 11 w 6
2
23. 12 x3 y 2 8 x 4 y 3 4 x 2 y
Check: w 11 w 6 w2 6 w 11w 66 w2 5w 66
4 x 2 y 3xy 2 x 2 y 2 1
33. 2t 3 28t 2 80t 2t t 2 14t 40
24. 14a b 21a b 7a b 4 3
3 2
2t
3
2
2t t 14t 40
2a 4 x 9 5 4 x 9
2
4 x 9 2a 5
2t 3 28t 2 80t
26. 6ty 9 y 14t 21 3 y 2t 3 7 2t 3
34. 5u 4 40u 3 35u 2 5u 2 u 2 8u 7
2t 3 3 y 7
5u
27. 12 x 9 x 40 x 30
3 x 4 x 3 10 4 x 3 2
4 x 3 3x 10
28. 30 z 35 z 24 z 28
5 z 6 z 7 4 6 z 7
35. 25 z 6 z 2 14 6 z 2 25 z 14
29. cd 8d 4c 2d cd 4c 2d 8d 2
2
Check: 2 z 7 3z 2 6 z 2 4 z 21z 14
d 4 c 2d
6 z 2 25 z 14
30. 7t 6v tv 42v 7t tv 6v 42v 2
2
t 7 v 6v v 7
36. 8 15m 2 26m 15m 2 26m 8
v 7 t 6v
p
p
z z
2 z 7 3z 2
c d 4 2d d 4
31. p 2 2 p 63
2
5u 4 40u 3 35u 2
2
u u
5u u 8u 7
2
6 z 7 5z 2 4
Check: 5u 2 u 1 u 7 5u 2 u 2 7u 1u 7 2
3
2
5u 2 u 1 u 7
2
2
t t
Check: 2t t 4 t 10 2t t 2 10t 4t 40
25. 8ax 18a 20 x 45
3
2t t 4 t 10
7 a b 2ab 3b 1 3
w w
m m
3m 4 5m 2
Check: 3m 4 5m 2 15m2 6m 20m 8
p 9 p 7
Check: p 9 p 7 p 2 9 p 7 p 63
15m 2 26m 8
p 2 2 p 63
38
Section R.6 37. 7 y 3 z 40 y 2 z 2 12 yz 3
yz 7 y 2 40 yz 12 z 2 yz
43. 4c 4 20c 2 d 3 25d 6
y z y z
yz 7 y 2 z y 6 z Check: yz 7 y 2 z y 6 z
2 3m 7n 7n 3m 7n
38. 11a 3b 18a 2b 2 8ab3
ab 11a 2 18ab 8b 2
a
b
b
2 10u 3v 10u 3v
3
39. t 18t 81 t 2 t 9 9 t 9 2
2
3
2
2
2
2
2
2
2
2
2
2
41. 50 x 3 160 x 2 y 128 xy 2
2 x 25 x 2 80 xy 64 y 2
1 9 z 1 9 z 1 9 z 1 3z 1 9 z 1 3z 1 3z 1
50. 81z 4 1 9 z 2
2 x 5 x 2 5 x 8 y 8 y 2
2
2
2
2
2
2
2
2
2
2
51. y 3 64 y 4 3
2 2 3 y 4 y 2 4 y 3 z 3 z
3 y 4 y 3z
2
4 25 p 4 25 p 4 25 p 4 5 p 2 25 p 4 5 p 2 5 p 2
49. 625 p 4 16 25 p 2
2
40. p 2 8 p 16 p 2 p 4 4 p 4
3 y 16 y 2 24 yz 9 z 2
3 5m 3n 5m 3n 2
42. 48 y 3 72 y 2 z 27 yz 2
3
3
2
2
2 2 3 5m3 3n 2
2
3
48. 75m 6 27 n 4 3 25m6 9n 4
11a b 18a b 8ab
2 2 2 10u 2 3v 3 2
ab 11a 18ab 8b
2 x 5x 8 y
2
47. 200u 4 18v 6 2 100u 4 9v 6
ab 11a 2 22ab 4ab 8b 2
2
2
2
Check: ab 11a 4b a 2b
2 2
4 2
46. 16t 2 49 4t 7 4t 7 4t 7
ab 11a 4b a 2b
3
4 2
4
2
a
2
2
45. 9 w2 64 3w 8 3w 8 3w 8
2
7 y 3 z 40 y 2 z 2 12 yz 3
ab
2
2
yz 7 y 40 yz 12 z
2
44. 9m 4 42m 2 n 4 49n8
3m 2
yz 7 y 2 42 yz 2 yz 12 z 2 2
2
2c 2 2 2c 2 5d 3 5d 3 2c 2 5d 3
3
y 4 y y 4 4 2
2
y 4 y 2 4 y 16
39
2
2
Chapter R Review of Prerequisites 52. u 3 343 u 7 3
58. 4 y 4 10 y 3 36 y 2 90 x
3
2 y 2 y 3 5 y 2 18 y 45
u 7 u u 7 7 2
2
u 7 u 2 7u 49
53. c 4 27c c c3 27
2 y y 2 2 y 5 9 2 y 5
2 y 2 y 5 y 2 9
2 y 2 y 5 y 3 y 3
c c 3 2 2 c c 3 c c 3 3 3
3
c c 3 c 3c 9
54. d 4 8d d d 3 8
2
59. a 2 y 2 10 y 25 a 2 y 2 10 y 25
a 2 y 5 a y 5 a y 5 2
a y 5 a y 5
60. c 2 z 2 8 z 16 c 2 z 2 8 z 16
d d 2 2 2 d d 2 d d 2 2 3
3
d d 2 d 2 2d 4
2
c z 4 c z 4
61. 30 x 3 y 125 x 2 y 120 xy
5 xy 6 x 2 25 x 24 5 xy 3 x 8 2 x 3
2a 5b 2a 2a 5b 5b 2a 5b 4a 10a b 25b 3
2a 2 5b3 2
3
2
3
56. 27 m 64n
c 2 z 4 c z 4 c z 4
55. 8a 6 125b9
12
3
2 2
2
4
6t v 5t 6 2t 5
3 2
3
2 3
62. 60t 4v 78t 3v 180t 2 v 6t 2 v 10t 2 13t 30 2
6
63. Let u x 2 2 .
x 2 3 x 2 28 2
2
9
u 2 3u 28
3m 4n 3m 3m 4n 4n 3m 4n 9m 12m n 16n 3
3m 4 4n3 4
3
4
3
3
4 2
4
8
4 3
57. 30 x 4 70 x 3 120 x 2 280 x
10 x 3x 3 7 x 2 12 x 28
u 4 u 7
3 2
3
x 2 x 9 x 2 x 3 x 3 x2 2 4 x2 2 7
6
2
2
2
64. Let u y 2 2 .
y 2 5 y 2 24 2
2
10 x x 2 3x 7 4 3x 7 10 x 3x 7 x 2 4
2
2
u 2 5u 24
u 8 u 3
y 10 y 1 y 10 y 1 y 1
10 x 3x 7 x 2 x 2
y2 2 8 y2 2 3 2
40
2
2
Section R.6
71. Let u 3n 1 . 2 9m 2 42m 3n 1 49 3n 1
2
65. x 3 12 16
x 12 4 x 12 4 x 16 x 8 x 16 x 2 x 16 x 2 x 2 x 4 2
x 3 12 42 3 3
3
3
3
3m 2 3m 7u 7u 2
3m 7u
3
3
9m 2 42mu 49u 2
3
2
3m 7 3n 1
2
3m 21n 7
2
2
2
2
72. Let u 7 y 1 .
66. y 3 34 49
4 x 2 36 x 7 y 1 81 7 y 1
y 34 7 y 34 7 y 41 y 27 y 41 y 3 y 41 y 3 y 3x 9 2
y 3 34 7 2 3
4 x 2 36 xu 81u 2
3
3
3
3
3
2 x 2 2 x 9u 9u 2
2 x 9u
3
3
2
2
2 x 9 7 y 1
2
2 x 63 y 9
67. x y z 2 x y z x y z
2
2
2
2
73. c 3 2c 5 2
c 3 2c 5 c 3 2c 5
68. a 5 y 2 a 5 y a 5 y 2
69. x y z 3
2
c 3 2c 5 c 3 2c 5
3c 8 c 2 or 3c 8 c 2
3
x y z x y x y z z 2
2
x y z x 2 xy y xz yz z 2
2
2
74. d 6 4d 3 2
2
d 6 4d 3 d 6 4d 3 d 6 4d 3 d 6 4d 3
70. a 5 b3 3
2 a 5 b a 5 a 5 b b 2
a 5 b a 2 10a 25 ab 5b b 2
5d 3 3d 9 =3 5d 3 d 3 or 3 5d 3 d 3
75. p11 64 p8 p 3 64 p11 p 3 64 p8 64 p 3 p8 1 64 p8 1
p 4 p 4 p 16 p 1 p 1 p 4 p 4 p 16 p 1 p 1 p 1 p 4 p 4 p 16 p 1 p 1 p 1 p 1
2 p 3 64 p8 1 p 3 43 p 4 12 2 4 4 p 4 p 4 p 16 p 1 p 1 2 2
2
4
2
4
2
2
4
2
41
2
2
Chapter R Review of Prerequisites 76. t 7 27t 4 t 3 27 t 7 t 3 27t 4 27
80. Let u y 3 .
24 y 7 21y 4 3 y
t 3 t 4 1 27 t 4 1
3 y 8 y 6 7 y3 1
t 3 27 t 4 1 t 3 33 t 12
7 y 1 3 y 8u 7u 1 3 y 8 y
2 2
t 3 t 2 3t 9 t 2 1 t 2 1
3 2
3
2
t 3 t 2 3t 9 t 2 1 t 1 t 1
3 y 8u 1u 1
3 y 8 y 3 1 y 3 1
77. Let u m . m6 26m3 27 3
3 y 2 y 13 y 3 13 3
3 y 2 y 1 4 y 2 2 y 1 y 1 y 2 y 1
m3 26m3 27 2
u 2 26u 27
81. x 2 y 2 x y x y x y 1 x y
u 27 u 1
x y x y 1
m3 27 m3 1 m3 33 m3 13
82. a 2 b 2 a b a b a b 1 a b
m 3 m 2 3m 9 m 1 m 2 m 1
a b a b 1
78. Let u n . n 6 7 n3 8 3
83. a 2 ac 2c 2 c a
a c a 2c 1 a c
7n 8
n3
2
3
a c a 2c 1
u 2 7u 8 u 8 u 1
84. x 2 2 xy 3 y 2 y x
n 2 n 1 n 2 n 2n 4 n 1 n n 1
x y x 3 y 1 x y
n 8 n 1 3
3
3
3
3
2
2
85. 2 x 4 7 x 3 x 2
x 4 2 x 4 4 7 x 3 4 x 2 4
79. Let u x 3 . 16 x 6 z 38 x 3 z 54 z
x y x 3 y 1
3
x 4 2 x 0 7 x1 x 2
2 z 8 x 19 x 27 6
3
x 4 2 7 x x 2 or
19 x 27 2 z 8u 19u 27 2 z 8 x
3 2
3
86. 5t 7 2t 6 t 5
2
2 z 8u 27u 1
t 7 5t 7 7 2t 6 7 t 5 7
3
2 z 2 x 33 x 3 13 3
t 7 5t 0 2t1 t 2
2 z 8 x 27 x 1 3
x2 7 x 2 x4
t 7 5 2t t 2 or
2 x 2 x 3 4 x 2 6 x 9 x 1 x 2 x 1
42
t 2 2t 5 t7
Section R.6 87. y 2 y 3 12 y 4
y 4 y 2 4 y 3 4 12 y 4 4 y
4
89. 2c 7 4 4c 3 4 2c3 4 c 7 4 3 4 2c3 4 3 4
y y 12 y y y y 12 2
1
4
0
y 4 y 4 y 3 or
2
y 4 y 3
2c
34
c 2c
2c
34
c 2
5 y 4 5 2 y1 3 y 0
w5 w35 10w45 9 w55
5 y 4 5 2 y 3
w5 w2 10w1 9 w0 w5 w2 10w 9
w 9 w 1
w5 w 9 w 1 or
92. 7t 4t 1
23
34
93. x 3 x 2
3x 1
4t 1
2 3
53
74
w5
5 x 3x 1 2 3 2 3 1 3x 15 3 2 3 23 0 1 3 x 1 5 x 3x 1 1 3 x 1 3 x 1
23
3 x 1
23
5 x 3x 1 3x 12 3 8 x 1
7t 4t 13 4 3 4 1 4t 1 7 4 3 4 34 0 1 4t 1 7t 4t 1 1 4t 1 4t 1
34
4t 1
34
3x 2
13
7t 4t 1 4t 13 4 11t 1
x 3x 2 2 3 2 3 3x 2 1 3 2 3 2 3 0 1 3x 2 x 3x 2 3x 2 3x 2
2 3
3x 2
2 3
x 3x 2 3x 2 4 x 2 2 2 x 1 2 3 2 3 x 2 2 x 1 or 23 3x 2 94. x 5 x 8
4 5
5 x 8
15
2 3
x 5 x 8 4 5 4 5 5 x 81 5 4 5 4 5 0 1 5 x 8 x 5 x 8 5 x 8 5 x 8
4 5
x 5 x 8 4 5 5 x 8 6 x 8 5 x 8
4 5
2 5 x 8
4 5
0
90. 10 y 9 5 15 y 4 5 5 y 4 5 2 y 9 5 4 5 3 y 4 5 4 5
y4
88. w3 10w4 9 w5
91. 5 x 3x 1
1
3x 4 or
43
2 3 x 4
5 x 8 4 5
Chapter R Review of Prerequisites
z 6 z 6 z 6
95. A x 2 y 2 x y x y
109. z 4 36 z 2 6 z 2 6 2
96. A d 2 4c 2 d 2 2c d 2c d 2c 2
w 7 w 7 w 7
110. w4 49 w2 7 w2 7
97. 2 r 2 2 rh 2 r r h
2
98. P Prt P 1 rt
5 x 5 x 5
111. x 2 2 5 x 5 x 2 2
99.
4 3 4 3 4 R r R3 r 3 3 3 3 4 R r R 2 Rr r 2 3
2
3 c 3 c 3
112. c 2 2 3c 3 c 2 2
100. R 2 r 2 R 2 r 2 R r R r
b. x 1 x 1 x
102. 17 13 17 1317 13 30 4 120
n
103. Expand the square of any binomial. For
114. a. a 5 b5
example: 2c 3 4c 2 12c 9 . 2
a a a a b a a b a ab a b b a b a b b a b b ab b b 4
3
4
2 2
3
3
2 2
4
3
4
a 5 a 4b a 3b 2 a 2b3 ab 4 a 4b a 3b 2 a 2b3 ab 4 b5 a 5 b5
5 4 x .
115. 36 p 2 33 p 12 106. In each case, factor out x to the smallest exponent to which it appears in both terms. That is, 6 x 5 5 x 2 x 2 6 x 3 5 and
107. x 2 5 x 5 x 5
b 2 4ac 33 4 36 12 2
6 x 5 3 5 x 2 3 x 2 3 6 x 5 .
1089 1728 2817 No 116. 24 w2 25w 8
108. y 2 11 y 11 y 11
105. In each case, factor out x to the smallest exponent to which it appears in both terms. That is, 5 x 4 4 x 3 x3 5 x 4 and
5x 4 x x
x n 2 x n 3 1
b. a b a 4 a 3b a 2b 2 ab3 b 4
a b a b a 2 b2 . a b a b a 2 b2 .
4
n 1
a b a 4 a 3b a 2b 2 ab3 b 4
104. After exhausting all possible combinations of factors, no combination results in a 2 b 2 . That is, a b a b a 2 b 2 .
3
113. a. x 5 1 x 1 x 4 x 3 x 2 x 1
2
4
2
2
101. 212 192 21 19 21 19 40 2 80 2
2
b 2 4ac 25 4 24 8 2
625 768 143
No
44
Section R.7 117. 8 x 2 2 x 15
119. 18 y 2 45 y 48
b 2 4ac 2 4 8 15
b 2 4ac 45 4 18 48
2
2
4 480 484 222
2025 3456 5481
Yes
No
118. 6 x 2 7 x 20
120. 54 z 2 39 z 60
b 2 4ac 7 4 6 20
b 2 4ac 39 4 54 60
2
2
1521 12,960 14, 481
49 480 529 232
Yes
No
Section R.7 Rational Expressions and More Operations on Radicals
1. 30 x 3 65 x 2 25 x 5 x 6 x 2 13 x 5
5 x 3 x 1 2 x 5
2. 40 y 3 132 y 2 28 x 4 y 10 y 2 33 y 7
y 2 y 4 y 2 4 y 16
11. 2;1
12. 1
14. rationalizing
3. t 4 t t t 3 1 t t 1 t 2 t 1 4. y 5 64 y 2 y 2 y 3 64
10. zero
13. complex (or compound)
47 2 y 7 5 y 1
9. rational
15.
16.
5. 2 x 3 3 x 2 32 x 48
2 x 3 32 x 3 x 2 48
2 x 3 x 16 2 x 3 x 4 x 4 2 x x 2 16 3 x 2 16 2
17.
x4 : If 7 were substituted for x, then x7 7 7 0 and the denominator would be 0. So, x 7 . y 1 : If 10 were substituted for y, then y 10 10 10 0 and the denominator would be 0. So, y 10 . a a : If 9 or 9 were a 81 a 9 a 9 2
substituted for a, then 9 9 0 or 9 9 0 and the denominator would be 0. So, a 9 , a 9 .
6. 3t 3 t 2 75t 25
3t 3 75t t 2 25
3t 1 t 25 3t 1 t 5 t 5 3t t 2 25 1 t 2 25
18.
2
substituted for t, then 4 4 0 or 4 4 0 and the denominator would be 0. So, t 4 , t 4.
7. 25w4 40 w2u 16u 2
2 5w 4u 4u 5w 4u
5 w2
2
2
2
2
2
19. There are no values for a that make the denominator 0. So there are no restricted values for this expression.
8. 49v 4 42v 2 w 9 w2
7v 2 2 7v 2 3w 3w 7v 2 3w 2
2
t t : If 4 or 4 were t 16 t 4 t 4 2
2
45
Chapter R Review of Prerequisites
21.
22.
1
20. There are no values for t that make the denominator 0. So there are no restricted values for this expression.
3tu 5u 4 v
5
15tu v 28. 3 3t u 3tu t 2 1
u0
6c : If 0 were substituted for a or b, then 7 a 3b 2 the product in the denominator would be 0. So, a 0 , b 0 .
1
29.
5 2 6
10 5 6 15
5 3
5u v ; t 0 , 4
t2
2 6 3
1
11z : If 0 were substituted for x or y, then 8 x5 y the product in the denominator would be 0. So, x 0 , y 0 .
1
3 3 12 4 3 4 3 3 30. 2 8 4 2 1
5 5 and x3 x3 5 5 5 5 , so a and x 3 x 3 x 3 3 x
23.
31.
1
b are equal to the original expression 24.
2 y 2 16 y 2 y y 8 64 y 2 64 y 2
2 2 and ab ab 2 2 2 , so b and c are a b a b a b
2 y y 8
8 y 8 y
2y ; 8 y
1
y 8 , y 8
81 t 2 81 t 2 32. 7t 2 63t 7t t 9
equal to the original expression
1
1
25.
x 3 x 3 x 3 x2 9 ; 2 x 4 x 21 x 3 x 7 x 7
7t t 9 1
1
x 3 , x 7
t 0, t 9
1
y 8 y 8 y 8 y 64 ; y 1 y 7 y 8 y 1 y 8 2
26.
9 t 9 t
2
33.
1
9t ; 7t
4 b a 4b 4a ax xb 2a 2b x a b 2 a b
y 1 , y 8
1
1
3ab 4ac 4ac 12a bc 4 ; a 0, 27. 5 4 b 3ab 3ab b 2
b0
1
4 b a
a b x 2 1
a b, x 2
46
4 ; x2
Section R.7
34.
2 z y 2z 2 y xy xz 3 y 3 z x y z 3 y z
m11n 2 18m9 n5 38. 2 m n 2 9m 2 6mn 15n 2 m11n 2 9m 2 6mn 15n 2 2 m n2 18m9 n5 3 3m 2 2mn 5n 2 m11n 2 18m9 n5 m n m n
1
2 z y
y z x 3 1
y z , x 3
35.
2 ; x3
3a 5b 7 2a 10b 3a 4b7 a 2 a 5b a 5b 12a 4b10 a 5b 3a 4b 7 4 b3
m9 n 2 m 2
m n m n
3 3m 5n m n m9 n 2 18 n3 6
m 3m 5n 6n3 m n 2
2
a 2b3 xy 4 6 y 4
8 x 3 y 6 xy x 3 y 4 24 x 9 y xy 4 x 2 3 8 x 3 y 2 y4 2 x
40.
c 2 d 2 8c 2 4cd 4d 2 37. cd 11 8c 4 d 10 2 2 c d 8c 4 d 10 cd 11 8c 2 4cd 4d 2 8c 4 d 10 c d c d cd 11 4 2c 2 cd d 2
c d c d cd
10
d
a 8b a 8b a b 8b a 2a b a 8b
2 c c d
2c 3d 2c 3d 2 c c 3d 2c 2c 3d c d 2
c
x 3 64 2 x 2 8 x 32 2 41. x 2x 8 16 x x 3 x 3 64 x2 2 x 8 16 x x 3 2 x 2 8 x 32 x 4 x 2 4 x 16 x 4 x 2 x 16 x 2 2 x 2 4 x 16
cd 10 8 c 3 4 2c d c d
2c c d d 2c d
1
3
a b 2a b
2c 2 2cd 4c 2 12cd 9d 2 3c 2 d 2c 3 2c 2 cd 3d 2
2
2a 2b ab 2 a 2 16ab 64b 2 8b 2 ab 2a 2 15ab 8b 2
2
8x 3 y
8
36.
39.
x 4 x 2 4 x 16 x 4 x 2 x 4 x 4 x 2 x 2 4 x 16
47
x2 2x
Chapter R Review of Prerequisites
42.
3 y 2 21y 147 y 3 343 25 y y 3 y 2 12 y 35
49.
m2 6m 9 m 2 6m 9 m3 m3 m3
3 y 2 21 y 147 y 2 12 y 35 25 y y 3 y 3 343
3 y 7 y 49 2
y 25 y
2
m3 m3
y 7 y 5 y 7 y 2 7 y 49 50.
1
y 7 y 5 y 5 y 5 y y 7 y 2 7 y 49
3 y 2 7 y 49
7 n 10 n 2 7 n 10 n2 n5 n5 n5
3 y 5 y 51.
10c 2 21 3 45c 45c 3 10c 2 21 45c 3
46.
47.
4
,
5y 7 y 2 y 5 y 6
,
52.
6
x3 x3 , x 20 x 100 x 10 2
53.
2
3 3 2 2 x 20 x 2 x x 10 LCD = 2 x x 10 48.
12 x 3 35 50 x 4
9 y 11 2 x 9 11 5 2 4 5 2 4 xy 2x y 2 x y y xy 2 x
2
54.
z4 z4 , 4 z 20 z 25 2 z 5 2 2
9 y 22 x 2 x2 y5
2 n3 5 3m 5 2 3 2 4 2 4 3m n m n 3m3 n n3 m n 3m 2n3 15m 3m3 n 4 2n3 15m 3m3 n 4
5 5 12 z 30 z 6 z 2 z 5 2
LCD = 6 z 2 z 5
6 7 6 7 2 4 25 x 10 x 5 x 5 2 x4 6 2 x3 7 5 2 3 5 2 x 4 5 5 x 2x
2 y 5 3 y 6 2 3 4 LCD = y 2 y 5 y 6 4
2 7 2 7 2 3 9c 15c 3 c 5 3c 3 2 5c 2 7 3 2 2 5 3c 3 3 3 c 5c
3t 4 t 2 t 3t 4 2 t 2 3 LCD = t 3t 4 t 2 3
n5
12 12 8 8 , 2 2 3 44. 4 3 4 3 2 3 35b cd 7 5b cd 25b c d 5 b c d LCD = 52 7b 4 c 3d 3 175b 4 c 3 d 3 2t 1
n 5 n 2
n2
7 7 3 3 , 43. 5 4 5 4 2 3 2 6 x yz 2 3x yz 20 xy z 5 2 xy 2 z 3 LCD = 22 3 5 x 5 y 2 z 4 60 x 5 y 2 z 4
45.
m 3 m 3
2
48
Section R.7
55.
1 2 2 x xy x y 2 1 2 x x y x y x y
58.
2
4 t 2 5 t 2 3 t t 2 2 t t 2 t 2 t 2 t t t 2
1 x y 2 x x x y x y x y x y x
x y 2x x x y x y
x y x x y x y
5t 10 4t 2 3t 2 6t t 2 t 2
t 2 t 10 t 2 t 2
x y x x y x y 59.
1 x x y 56.
4 1 2 2 4a b 2a ab 4 1 2a b2a b a 2a b 60.
3w 2w 4 3w 2w 4 1 w 4 4 w w 4 4 w 1
2 x 1 x 6 2 x 1 x 6 1 x 7 7 x x 7 7 x 1 2x 1 x 6 x7 x7 1 x7
4a 2a b a 2a b2a b 2a b a 2a b 2a b
1 1 27 x 1 1 27 x 9 61. 27 x 9 1 1 1 1 27 x 3 9x 3 9x
1 a 2a b 5 2 6 2 y y 1 y
27 x 1 27 x 1 x 1 27 1 9 27 x 1 27 x 1 1 3 1 9x 1 3x 9x 3 1 3x 1 3 3 x 1 3
5 y y 1 2 y 6 y 1
2 y2 5 y y 1 6 y 1 2 2 y y y 1 y 1 y y y 1 2
y 2 y 1
t 2 t 2
3w 2 w 4 w4 w4 1 w4
2
5 t 2 4 t 2 3 t t 2
4 a 1 2a b 2a b2a b a a 2a b 2a b
57.
5 4 3 2 t t2 t
5 y2 5 y 2 y2 6 y 6 7 y2 y 6 2 y 2 y 1 y y 1
49
Chapter R Review of Prerequisites 1 1 8x 1 1 8x 4 62. 8 x 4 1 1 1 1 8x 2 4x 2 4x 8x 1 8x 1 1 8x 1 4 8x 1 8x 1 1 2 1 4x 1 2x 4x 2 1 2x 2 2 x 1
2 1 2a 1 b 1 a b 65. 4 1 4a 2 b 2 a 2 b2 2 1 a 2b 2 a b 4 1 a 2b 2 2 2 a b
a 2b 2 2 a 2 b 2 1 1 b 2 12 a a b 4 a 2b 2 1 1 a2 1 b2 2ab 2 a 2b 4b 2 a 2
1 2
x 5 x 14 6 x x 5 x 14 6 6x 6x 63. 6 1 7 1 7 6x 6 6x 6 6x
ab 2b a
1
u 2v 2 3 u 2v 2 1 2 2 1 v 3uv u v 2 12 u u v 9 u 2v 2 1 9v 2 u 2 2 2 1 u 1 v
x 2 x 7
x7
x2
uv 3v u
3v u 3v u
uv 3v u
3 3 3 3 67. 1 h 1 h 1 h h 3 3 1 h 1 h 1 1 h h
x 2 x 3 3x x 2 x 3 3 3x 3x 64. 3 1 1 1 1 3x 3 3x 3 3x 3x x 3x 2 x 3 3x 1 3 1 3x 1 3x 1 1 3 1 3x x2 2 x 3 x 1
2b a 2b a
3 1 3 1 u 2v 2 u v 3u v 66. u v 2 2 9 1 9 1 9u v 2 u 2 v 2 2 2 2 u v u v 1
6 x x 6 x 5 x 14 1 6 1 6x 6x 1 6x 7 1 6 1 6x x 2 5 x 14 x7
ab 2b a
1 h 3 1 h 3
x 3 x 1 x 1
x3
50
1
1 h 1 1 h h
3 h 3 3 3h 2 hh h 1 h
3 3 or 1 h 1 h
1
Section R.7 4 4 4 1 h 4 4 4 1 h 1 1 h 1 h 1 68. h h 1 h h 1 h 4 1 h 4 1 1 h 1 1 1 h h
4 4 1 h h 1 h
4
71.
y 7
72.
4 4 4h h 1 h
73.
4 4 4 h or 1 h h 1 h 1 h
74.
7 7 x x h 7 7 x h x x h x 69. h x x h h
z
4 3
7 4
75. 3
x x h 7 7 1 x h x x x h h
7 x 7 x h 7 x 7 x 7h hx x h hx x h
7 h 7 7 or x x h h x x h x x h
76. 3
77.
78.
8 8 8 h or x x h h x x h x x h
3
12w2 23 w3
3
50 p 2
3
53 p 3
3
y3
4 3 y2 y
3
12 w2 2w
8wx 2
27a 2b
12 x 1
3
50 p 2 5p
12 4 2 w3 x 2
51
4 3 y2
3 2 3 52 p 2 10 2 3 25 p 5 p 3 5 p 3 52 p 2
4
23 wx 2 4 2w3 x 2
12 4 2w3 x 2 4
2 4 w4 x 4
12 4 2w3 x 2 6 4 2w3 x 2 wx 2wx
79.
7 z z
3 18 3 3 3 2 2 w2 3 12 w 2 w 3 2 w 3 2 2 w2
12 4
4 y y
7 4 z3 7 4 z3 4 4 4 z z 4 z3 z
12 4
7 z
z2
2
7 4 z3
x x h 8 x x h 8 1 x h 1 x x x h h
y 3 y2
3
x x h 8 8 1 x h x x x h h
8 x 8 x h 8 x 8 x 8h hx x h hx x h
4 3 y2
4 y y
z z
3
8 8 x x h 8 8 x h x x h x 70. h x x h h
z
7 z
y y
x x h 7 x x h 7 1 x h 1 x x x h h
y
4 y
12 4 3a 2b3 4
33 a 2b 4 3a 2b3
12 4 3a 2b3 4
34 a 4b 4
12 4 3a 2b3 4 4 3a 2b3 3ab ab 12 x 1 x 1 x 1
43 x 1
x 12
2 3 x 1 2 3x 3 x 1 x 1
Chapter R Review of Prerequisites
80.
50 x2
25 2
x2
5 2 x2
x5
83.
x2 x2
x 5
5 2x 4
5 2x 4 x2 x 2 2
81.
8 15 11
15 11 15 11 8 15 11 8 15 11 15 11 15 11 8 15 11 2 15 11 4 2
82.
2
85.
x 5
2
y 3
84.
y 3
12
2
x 5
6 2 6 2 6 2 6 2 12 6 2 12 6 2 62 6 2 12 6 2 3 6 2 4 12
x 5 x 5 x 5 x 5 x 5 x 5 x 5 2
8 15 11
x 5 x 5
y 3 y 3
y 3 y 3 y 3 y 3 y 3 y 3 y 3 2
2
y 3
2
y 3
2 10 3 5 4 10 2 5
2 10 3 5 2 10 5 2 10 2 10 2 10 5 3 5 2 10 3 5 5 2 2 10 5 2 10 5 2 2 10 5 4 10 2 50 6 50 3 5 40 4 50 15 25 4 5 2 25 20 2 5 5 4 2 5 4 2
2
2
2
2
2 40 5
86.
2
2 35
70
70
70
3 3 6 5 3 2 6 3 3 5 3 3 3 2 6 6 5 3 6 2 6 5 3 2 6 5 3 2 6 5 3 2 6 5 3 2 6 3 3 6
2
2
15 32 6 18 5 18 2 62 45 11 18 12 57 11 32 2 75 24 51 51
57 33 2 3 19 11 2 19 11 2 51 51 17
52
14
Section R.7
87.
7 3x
3x 7 3x 3x x x 3x 3x
88.
89.
4
11y
5 w 7
92. a. R
7 3x 3x 3 10 3x x3 3x 3x
b. R
11y 4 11y 11y y y 11y 11y
4 11y 11y 11 11 y y 11
15 11 y 11y
7 5 7 7 w w 7 7 w
t 2
R1 R2 12 20 240 7.5 R2 R1 20 12 32
b. The concentration of the drug appears to be approaching 0 ng/mL for large values of t.
2 13 2 2 t t t 2 2
350 94. a. T 2 t 3t 10 At 3 hr: 350 350 12.5C T 2 3 3 3 10 28 At 24 hr: 350 350 0.5C T 2 24 3 24 10 658
13 2 2 2 11 2 t2 2t 2t 91. a. S
R1 R2 1 RR 1 2 1 1 R2 R1 R1R2 R1 R2
At 12 hr: 600 12 7200 C 3.9 ng/mL 3 12 125 1853
5 7 7 7 7w w7 2 7 2 7 or 7w 7w 13
1 1 R1 R2
600t 93. a. C 3 t 125 At 1 hr: 600 1 600 C 3 4.8 ng/mL 1 125 126 At 4 hr: 600 4 2400 C 12.7 ng/mL 3 4 125 189
90.
1
2d d d r1 r2
r1r2 2d
d d r1r2 r1 r2
b. The temperature appears to be approaching 0C for large values of t.
2dr1r2 d d r1r2 r1r2 r1 r2
5 2 6 x 1 x 1 x 6 x 1 7 x x 1 x x x 1
95. P
2dr1r2 2 d r1r2 2r r 12 dr2 dr1 d r2 r1 r1 r2
2 400 460 368,000 2r r b. S 1 2 400 460 860 r1 r2 427.9 mph
53
7x 6x 6 x x 1
13 x 6 cm x x 1
Chapter R Review of Prerequisites 6 2 1 2x 1 2x 1 x 1 2 x 1 8 x 2 x 1 x x 2 x 1
96. P
100.
8x 2x 1 x 2 x 1
97. A
10 x 1 yd x 2 x 1
2x in.2 x
98. A
99.
1
y 7 y 5 3 y 5
1 1 or 6 y 7 6 y 7
4t 32 t 2t 1 3t 32 2t 1 1 2m 102. 2m 1 6m 3 m 4
2m 1 2m 1 3 2m 1 m 4 2m 1 2m 1 2m 1 m 4 3 2m 1 2m m 4 2m 1 3 3 m 4 m 4 3
5y 5y 5 5y 5y 2 ft y 5y
2 x3 y x2 6 x 9 5 xy 4 2 2x 6 x y 3xy
4 3 y
1
y 3
5 5y
2 y 5 y 5
t 4 8 t t 8 2t 1 2t 1 t 8
1 bh 2
xy 2 x 2
4 3 y
y 3 1 y 7 y 5 3 y 15
4 t t 8 2t 1 2t 1 t 8
2 1 5 2 2 5 y
t 4 101. 2 t 8 2t 1 2t 17t 8
2 1 1 4 2 2x 2 x 2 2x 2x 2 2x 2x
2 y 2 10 y 25
2
1 bh 2
2 y 2 20 y 50 y 3 2 3 y 15 12 4 y y 12 y 35
x 3 x 3 1
xy x 3
2 x 3
5 xy 4
x2 x 4 4 5 xy 5y
54
2 m 2 8m 6 m 3 3 m 4
2m 2 2m 3 3 m 4
Section R.7
103.
n 2 2n 2 15n 12 2n 5 n 2 2n 2 15n 12 2n 5 n4 n 2 16 n 4 n 4 n 4 n 4 n 4
n 2 n 4 2n2 15n 12 2n 5 n 4 n 4 n 4 n 4 n 4 n 4 n 4
n 2 2n 4n 8 2n 2 15n 12 2n 2 8n 5n 20 n 4 n 4
n 2 16 2 1 n 16 104.
c 2 13c 18 c 1 c 8 c 2 13c 18 c 1 c 3 c 8 c 3 c2 9 c 3 c 3 c 3 c 3 c 3 c 3 c 3 c 3
c 2 13c 18 c 2 3c c 3 c 2 3c 8c 24 c 3 c 3
c2 9 1 c2 9
1 6 1 2 1 a 1 6a 2 a a 105. 1 4a 1 3a 2 1 4 3 a a2 1 6 a 2 1 2 a a 4 3 a 2 1 2 a a
1 12 1 2 1 t 1 12t 2 t t 106. 1 4t 1 3t 2 1 4 3 t t2 1 12 t 2 1 2 t t 4 3 t 2 1 2 t t
a2 a 6 a 2 4a 3
t 2 t 12 t2 4 3
a 2 a 3 a 3 a 1
t 4 t 3 t 4 t 1 t 3 t 1
a2 a 1
107.
34 2 5 3
34 2 5 3
2 5 3 2 5 3 34 2 5 3 2 5 3 34 2 5 3 2
2
20 3
2
55
34 2 5 3 17
4 52 3
Chapter R Review of Prerequisites
108.
13 2 7 2
13 2 7 2
45 9 x 5 x 2 x 3 113. 3 x 3 x 2 25 x 75 45 5 x 2 9 x x 3 3 x 25 x 3x 2 75 5 9 x2 x 9 x2 x x 2 25 3 x 2 25
2 7 2 2 7 2 13 2 7 2 2 7 2 13 2 7 2 2
2
28 2
13 2 7 2
2 7 2
2
114.
42 3
2
3
2
x 3 x 3 x 5 x 3 x 5 x 5 x 3 x 5
98 49 x 2 x 2 x 3 7 x 2 x 3 28 4 x 98 2 x 2 49 x x 3 7 x 2 28 x 3 4 x 2 49 x 2 x 49 x 2 7 x2 4 x x2 4
3
6
2
8 48 8 42 3 6 6 84 3 6 4 2 3
9 x x 5 x 9 x 5 x 25 x 3 x 25 x 3
2
109.
2
2
26
49 x 2 x x 4 7 x
2
10 50 10 5 2 10 5 2 15 15 15 2
110.
5 2 2 15
2
1
2 2
7 x 7 x 2 x x 2 x 2 7 x x 7 x7 or
3
3
111.
14 7x
7x 14 7 x 7x x x 7x 7x 14 7 x 7x 7 7x 7x
112.
6y xy
x2
7x 7 x7 7x x
115.
10 xy 6 y xy 10 xy x x xy xy
6 y xy 10 xy y xy x y 16 y xy xy
x2
t6 t 4 2 1 t t6 t t 4 t 6 t 4 t2 t 2 t t t 6 t t 2 4 t 2 1 t 2 1 t 2 t2 t 2 6t t 2 2t 4t 8 t2 8 8 or t2 t2
16 xy x
56
Section R.7
116.
m4 m2 2 1 m m4 m m 2 m 4 m2 m2 m 2 m m m 4 m m 2 2 m 2 m2 1 m 2 1 m 2
122. a.
15 4 60 6 2 5 10
123.
w3n w z w3n1 w3n z w n 2 wn z 2 wn w2 z 2
124.
2n x 2 n1 x 2 n y x x y x n3 x n y 3 x n x 3 y 3
118. No. The expressions are equal for all real numbers except for x 2 and x 4 . The original expression is undefined for x 2 and x 4 .
, multiply numerator and
x to make a perfect square 1 in the radicand in the denominator. For 3 , x multiply numerator and denominator by
120. Multiplying the denominator by the conjugate of 5 3 will produce a rational number. That is
5 3 5 3 5 3 5 3 2 2
To make a fraction equivalent to the original expression, multiply the numerator by 5 3 as well. 121. a.
127.
14 30 420 20 3 7 21
b. 5 5 5 5 5 2
5
5 3
2
2
x2 y 2 x y m 3 m3 m3 m3
m 3 m 3
2
xn x 2 xy y 2
x y
m3 m3
126.
x n x y x 2 xy y 2
x y x y
x to make a perfect cube in the radicand of the denominator.
xn xn x y
x y x y
2
x y x y
x y x y
125.
denominator by
3
w2 n n w w z w z w z
117. If x y , then the denominator x y will equal zero. Division by zero is undefined.
x
2
wn w2 n w z
m 2 4m m 2 2m 2m 4 m2 4 4 or m2 m2
1
2
93 6
119. For
b. 3 3 3 3 3 3
m 3
5 3 22
2
5 3 22
3 3 2 3 22 2 12 23 36 5 2 5 24 6 2 2 3
5 2 5 2
2
4 16
3
2
25 5 20
57
6
3 4
2
m2 9 m3
Chapter R Review of Prerequisites
128.
7 4
3
7 4 33 4
132.
3 3 4
3
x h x x h x h x h x x h x h x h x
7 4 33
4
34 71 2 33 4 3 24 7 33 4 3
129.
h
3 4
73 3
a3b
h
h
a b a ab b 3
x h x
2 3
3
3
3
2
3
3
2
2
xhx
7 3
ab 3
2
3 14
2
xh x h
x h x 1 xh x
133.
a3b
3 a 2 3 ab 3 b 2
130.
x y 3
x3 y
x y x xy y 3
3
3
3
2
3
3
2
x3 y
3 x 2 3 xy 3 y 2
131.
4 h 2 4 h 2 h 4 h 2 4 h 2 h 4 h 2
4h 2 h
2
134.
2
4h4
h
4 h 2
The expression appears to approach 2 as x approaches infinity.
h
h
The expression appears to approach 6 as x gets close to 3.
4 h 2 1 4h 2
58
Chapter R Review Chapter R Review Exercises 1. a.
9
d.
b. 0, 9
16 4
e. 16 4
c. 0, 8, 9 d. 0, 8,1.45, 9,
2 3
f.
e.
6, 3
f.
2 6, 0, 8,1.45, 9, , 3 3
9.
16 is not a real number. 25 5 4 2
2 2 1 3 4 1 10. 72 72 2 9 8 3 4 8 1 9 72 9 8 8 9
2. a. The first set is a subset of the second, because all of the elements in the first set are elements of the second set. True.
32 9 72 72 72
b. The first set is not a subset of the second, because 2 is in the first set but not in the second set. False. 3. x 4
23 23 72 72
4. 6 y 8
5. a. 11.
3, 7 ; x | 3 x 7
2 12. 5 3 8 2 42 2 5
b.
2.1, ; x | x 2.1
2 5 3 8 2 42 3 5 3 8 2 16 9 5 3 8 2 25 5 3 8 2 5
c.
, 4 ; x | 4 x 6. a. for w 4, w 4 w 4 4 w
5 3 8 10 5 3 2 5 6 1
b. for w 4, w 4 w 4 7. a. 2 5 or
13. a. J E 150
5 2
32 11 3 32 8 24 24 6 36 2 3 2 18 3 2 6 2 4
b. E J 150
14. W 2 L 8
b. 2 5 2 5 5 2
15. a. C 3.6s 50 p 250n b. C 3.6s 50 p 250n
8. a. 162 16 16 256
3.6 2100 50 4 250 2
b. 16 16 16 256 2
7560 200 500 $8260
c. 162 16 16 256
59
Chapter R Review of Prerequisites 2 5x z z 2 13x 2 z 4 5 x z z
b. m 5
16. 13x 2 z 4
d. 8m9 n 2 8 m9 n 2 8 m9
15.2c 2 d 8.7c 2 d 11.1cd 5.4cd 23.9c 2 d 16.5cd
32.
1 18. 8 2 x 2 6 4 x 2 3 13 x 2 2
m 4 m10 m 4 m10 m 6 m 4106 m0 1 6 m
33. 12a 3b 4
12 a b 2
144a 6b8
8 4 x 2 13 x 2 7 x 2 8
20. h 22. i
23. e
24. g
25. a
26. f
81x8 y 5 34. 9 x 6 y 8
x2 x1 y2 y1
2
2
2
2 5 3 6
7 9
2
2
2
9 x y 9 x 2 y 3
2 3
1
9
21 u 5v 2
29. a. 90 1
2
2 2
3 2
x 4 y 6
y6 81x 4
1 4 35. 5 2 3 2 2u v u v
2
49 81 130
1
2 u v 3
2
3 2 1
23 u15v 6 22 u 3v 2
b. 90 1 90 11 1
2u12 v 4
c. 9 x 0 9 x 0 9 1 9
d. 9 x 1 0
2u12 v4
36. a. Move the decimal point 3 places. The number is greater than 10, so use a positive power of 10. 4920 4.92 103
5
30. a.
9 x86 y 58
2
4 2
144b8 a6
2
27. b 28.
3 2
2
8 2 x 2 3 2 x 2 3 13x 2
21. c
1 8m 9 2 n2 n
31. p 8 p12 p 1 p 8121 p 3
17. 15.2c 2 d 11.1cd 8.7c 2 d 5.4cd
19. d
1 2 8n 2 n 9 m9 m
c. 8m 9 n 2 8 m 9 n 2 8
2 Terms: 13x 2 z 4 , , 5 x z ; z coefficients: 13, 2 , 1
1 m5
1 1 m 1 m5 5 1 m 1 m5
60
Chapter R Review
38. 9.2 104 3.0 105 9.23.0 104 105
b. Move the decimal point 3 places. The number is between 0 and 1, so use a negative power of 10. 0.00492 4.92 10 3
27.6 109
2.76 101 109 2.76 10
10
c. Move the decimal point 0 places. The exponent on 10 is 0. 4.92 4.92 100
8.6 10 4.110 8.6 4.1 10 10 39. 3
8
3
2.0 10 6
2.0 10 6
37. a. 101 is between 0 and 1. Move the decimal point 1 place to the left. 9.8 101 0.98
8
17.63 1011
1.763 101 1011
b. 10 1 The decimal point does not move. 9.8 100 9.8 0
1.763 10
12
c. 101 is greater than 10. Move the decimal point 1 place to the right. 9.8 101 98
40. 1 pt
0.47 L 106 L 5 106 red blood cells 2.35 1012 red blood cells 1 pt 1L 1 L
x b. 9 x 9 x or 9 x c. 9 x 9 x or 9 x
41. a. x 2 7 7 x 2 or 27
27
7
7
2
7
b. 10, 0003 4
10 1000 3
3
1 10,0003 4
1
45. a. 10, 0003 4 4 10,000
2
7
2
7
2
2
1
10
42. a. 12 w 12 w1 2
3
b. 12w 12 w
4
3
1 1000
c. 10,0003 4 4 10,000
12
10, 000
3
10 1000 3
43. a. 256 4 4 4
4
4
d. 10,0003 4
b. 4 256 is not a real number. 44. 3
3 8 8 8 3 23 2 2 3 3 3 3 125 125 5 5 125 5
e. 10, 000 4
61
34
1 1 3 34 10,000 4 10, 000
1
10
3
1 1000
is undefined because
10,000 is not a real number.
Chapter R Review of Prerequisites f. 10,000 4
3 4
54. 5 5 a 3 6 5 a 4 5 6 5 a 3 a 4
is undefined because
10,000 is not a real number.
30 5 a 7 30 5 a 5 a 2
p 7 3 p 2 3 p 7 32 3 p 5 3 46. 2 3 p 5 32 3 p 3 3 p p2 3 p2 3 p
47. 9m 4 n 2 3
30 5 a 5 5 a 2 30a 5 a 2
9 m n 12
4 1 2
12
23 12
55. 4 cd 2 3 c 2 d cd 2 c 2 d
3m 2 n1 3
14
3n1 3 m2
c1 4 d 1 2 c 2 3 d 1 3 c1 42 3 d 1 21 3 c 3 128 12b6 124 12
3t 42 3t 4
48. a.
13
c11 12 d 10 12 12 c11d 10
b. 3 3t 4 3t 4 3
56.
c. 4 3t 4 3t 4 4
49. 3 54 xy12 z14 3 33 2 xy12 z14
3 33 y12 z12 2 xz 2
3 33 y12 z12 3 2 xz 2 3 y 4 z 4 3 2 xz 2
1 3 1 125 20 80 5 2 4 1 2 3 2 1 2 5 5 2 5 4 5 5 2 4 1 3 1 5 5 2 5 4 5 5 2 4 5 3 5 5 3 5
57. 2c 3 54c 2 d 3 5cd 3 2c 2 10d 3 250c5
2c 3 33 d 3 2c 2 5cd 3 2c 2
50. 4 32b3c15 d 8 4 24 2 b3c15 d 8
4 24 c12 d 8 2b3c 3
10d 3 53 c 3 2c 2
6cd 3 2c 2 5cd 3 2c 2 50cd 3 2c 2
2 c d 2b c 4
4 12
2c d 3
51.
p13 9
p13 9
24
8
4
3 3
51cd 3 2c 2
3 3
2b c
58. 3 5t 8t 5t 3 8t 5t
p12 p p 6 p 3 3
59. a. Yes b. No
3 xy 5
y3 y 3xy 5 3 52. 3 2 7 2 6 7 2 3 81x y 27 x 3x 81x y 3
53.
c. Yes 60. 2.5 y11 7.61 y 9 y 5 ; Leading coefficient: 2.5; Degree: 11
10 35 350 52 14 52 14 5 14
61. 8
62
Chapter R Review
62. a. R E 14 x 156.67 x 2 8 x 130
c. R E x 2 6 x 26.67
4 6 4 26.67 2
14 x 156.67 x 2 8 x 130
16 24 26.67 66.67 In the year 2010, the profit for cellular and wireless communications industries was approximately $66.67 billion.
x 2 6 x 26.67 b. The polynomial R E represents the profit for cellular and wireless communications industries x years since 2006.
63. 7.2a 2b3 4.1ab 2 3.9b 0.8a 2b3 3.2ab 2 b
7.2a 2b3 4.1ab 2 3.9b 0.8a 2b3 3.2ab 2 b 8a 2b3 7.3ab 2 2.9b
1 64. 8uv 3 u 2 v 2u 3v 4 4
65. 5w3 6 y 2 2w3 y 2
70. 6c 2 5d 3
3
3
6 y2 y2
2
2
3
3 2
2v 1 w2 2
2v 2 2v 1 1 w2 2
10w 7 w y 6 y 3
2 2
71. 2v 1 w 2v 1 w
10w6 5w3 y 2 12w3 y 2 6 y 4 6
2
36c 4 60c 2 d 3 25d 6
5w 2w 5w y 2 6 y 2 2 w3 3
6c 2 6c 5d 5d
2
4v 2 4v 1 w2
4
4 3 2 3 4 3 5 5 4 3 7
72. 4 3 2 3 5 5 7
1 66. 4 p 6 p 2 p 4 2 1 4 p p 2 4 p p 4 p 4 2
8 9 20 15 4 21
1 6 p 2 6 p 6 4 2
24 20 15 4 21
2 p 3 4 p 2 16 p 3 p 2 6 p 24 2
2
2 3 5 3
12 25 30 15 4 15 10 9
1 1 1 1 68. m n3 m n3 3 4 3 4 2
60 26 15 30 30 26 15
2
1 1 1 1 m n 3 m 2 n 6 3 4 9 16
7 2 2 11
74. 7 2 2 11 7 2 2 11 2
69. 5k 3 5k 2 5k 3 3 2
2
6 5 2 5 6 5 5 3 2 3 2 5
67. 9t 4 9t 4 9t 4 81t 2 16 2
73. 6 5 2 3 2 5 5 3
2 p 7 p 22 p 24 3
2
2
49 2 4 11 98 44 54
25k 2 30k 9 63
Chapter R Review of Prerequisites
2c d 2 2c d 5d c 5d c
75. 2c 2 d 5d 2 c 2
2
82. t 12t 2 6 12t 2 t 6 4t 3 3t 2
2
2
83. 8 x 3 40 x 2 y 50 xy 2
2
2 x 4 x 2 20 xy 25 y 2
2
2
2 2 2 x 2 x 2 2 x 5 y 5 y
4c 4 d 20c 2 d 2 cd 25cd 4
x 2 4 x 2 2 x 2 4 4
2x 2x 5 y
2
76.
2
2
25 16a 2516a 25 16a 25 4a 5 16a 25 4a 5 4a 5
84. 256a 4 625 16a 2
2
2
2
2
x 2 8 x 2 16 x 18 8 x 2
2
2
2
77. V lwh
2
x 3 x 2 2 x 5
x 3 2 x 9 x 10
x 3 2 x 2 5 x 4 x 10
3k k 3 3k k 3 k 3k 9
85. 3k 4 81k 3k k 3 27
2
3
2 x 9 x 10 x 6 x 27 x 30 3
2
2
86. c 2 d 3 3
1 bh 2 1 3 2 1 3 2 1 2 2 1 2 3 2 1 2
78. A
2 c 2 d c 2 d c 2 d 2
c 2 d c 2 4c 4 cd 2d d 2
87. 25n 2 m 2 12m 36
1 1 17 2 9 2 1 17 ft 2 2 3
79. 80m 4 n8 48m5 n3 16m 2 n
2
7
3 2
25n 2 m 6
2
5n m 65n m 6
88. Let u x 2 7 .
x 7 8 x 7 15
80. 11 p 2 2 p 1 22 p 2 p 1
25n 2 m 2 12m 36
16m n 5m n 3m n 1 2
3
2
2 x 3 15 x 2 37 x 30
2
2
2
11 p 2 22 p 2 p 1
2
u 2 8u 15 u 5 u 3
11 p 2 p 1 p 2
x 2 x 4 x 2 x 2 x 2
x2 7 5 x2 7 3
81. 15ac 14b 10a 21bc
2
15ac 10a 14b 21bc 5a 3c 2 7b 2 3c
2
5a 3c 2 7b 3c 2 5a 7b 3c 2
64
2
Chapter R Review 89. 2 p 5 4 p 1 2
91. x 4 6 x 2 y 9 y 2 x 2 3 y
2
2 p 5 4 p 1 2 p 5 4 p 1
( x 2 3 y )( x 2 3 y ) x 2 3 y
6 p 4 2 p 6
2 3 p 2 2 p 3
92. 7 w8 5w7 w6
w8 7 w8 8 5w7 8 w6 8
90. Let u m3 . m6 9m3 8 u 2 9u 8
w 7 5w w 8
m 1 m 2 m 1 m 2m 1 m 2 m 2m 4 m 3 1 m3 8 3
3
2
w8 7 w0 5w1 w2
u 1 u 8
3
2
( x 2 3 y )( x 2 3 y 1)
4 3 p 2 p 3
3
( x 3 y ) ( x 3 y ) 1 2
2
7 5 w w2 w8
2
93. 12 x 7 2 4 x 5 2 4 x 5 2 3 x 7 25 2 x 5 25 2 4 x 5 2 3 x1 x 0 4 x 5 2 3 x 1 94. x 2 x 5
95. a.
3 4
2 x 5
14
x 2 x 5 3 4 3 4 2 x 51 4 3 4 3 4 0 1 2 x 5 x 2 x 5 2 x 5 2 x 5
3 4
2 x 5
3 4
x 2 x 5
2 x 5
3 4
3x 5 or
3x 5
2 x 5 3 4
w2 w2 : If 2 or 2 were 2 w 4 w 2 w 2
97.
substituted for w, then 2 2 0 or 2 2 0 and the denominator would be 0. So, w 2 , w 2 .
m 4 , m 3
2 x 5 y x3 y x 2 y 6 2x 5 y x5 y 7 98. x 3 y 6 x 15 y 3 2x 5 y x3 y
b. There are no values for w that make the denominator 0. So there are no restricted values for this expression. 96.
m 4 m 4 m 4 m 2 16 ; 2 m m 12 m 3 m 4 m 3
8c5 d 2 3a 2 24a 2 c 5 d 2 3a 2 ; 16c 6 d 7 2cd 5 8c 5 d 2 2cd 5
99.
c 0, d 0
x2 y6 3
4ac a 2 8ac 16c 2 8a 32c a 2 4ac
a 4c a 4c c 2 8 a 4c a a 4c 4 ac
2
65
Chapter R Review of Prerequisites
100.
5 x 2 25 x 125 x 3 125 16 x x 3 x 2 9 x 20 5 x 2 25 x 125 x 2 9 x 20 16 x x 3 x 3 125
104.
1
x 5 x 4 x 4 x 4 x x 5 x 2 5 x 25 5 x 2 5 x 25
101.
x 1 5 x 1 5 x 1 5 x6 6 x x6 x6 x6 x6 1 x6
1 1 16 x 1 1 16 x 8 105. 16 x 8 1 1 1 1 16 x 4 8x 4 8x
5 5 or x x 4 x x 4
16 x 1 16 x 1 1 8 1 16 x 16 x 1 16 x 1 1 4 1 8x 1 1 2x 2x 1 4 x 2 2 2 x 1 2
7 2 7 2 2 4 20 x 15 x 2 5x 3 5x4 7 3x3 2 4 2 3 3 5 x 4 4 2 5 x 3x
21x 3 8 60 x 4 102.
6 1 2 2 9 x y 3 x xy 6 1 3x y3x y x 3x y 2
103.
5 1 5m 1 n 1 m n 106. 25m 2 n 2 25 1 m2 n2 5 1 m2 n2 m n 25 1 m2n2 2 2 m n
6 x 1 3 x y 3x y3x y x x 3x y 3x y
5 1 m2 n2 m n 25 1 m2n2 2 m2 n2 2 m n 2 2 5mn m n 25n 2 m 2 m2 n2
6 x 3x y x 3 x y3x y 3x y x 3 x y 3x y
1 x 3 x y
4 3 2 2 x x3 x 4 x 3 2 x x 3 3 x2 2 2 x x 3 x 3 x x x x 3
5
107.
4 x 12 3x 2 2 x 2 6 x x 2 x 3
x 2 2 x 12 2 x x 3
108.
66
k
5 4
k
5n m 5n m
5 k
mn 5n m
k k
5 4 k3 4
k 4 k3
5 k
k
2
54 k 3 4
k4
5 k k
54 k3 k
mn 5n m
Chapter R Test 6
109. 4
6
2
4
8x y
2 x y 6 2x y
4
2
x2
3
2 x2 y 4 2 x2 y3 2
6 2x y
3
24 x 4 y 4
3
112.
5y
5
2
5y 3 5y 5y y y 5y 5y
3 5y 5y y 5y
3 5y 5y 5 y 5 5y
3 5y 5 5y 5y 5y
8 5y 5y
2
10 7
15
x 2 x 2 x 4 x 2 x 2 x 4 x 2 x 2
10 7 10 7 10 7 10 7 15 10 7 10 7 15 10 7 2
x 4 x 2
x4
15
2
6 4 2 x2 y3 34 2 x2 y3 xy 2 xy
15
110.
x4
3
4
4
111.
2
4
3
10 7 5 10 7 3
Chapter R Test 1. a. 8
4. for t 5, 5 t 5 t t 5
b. 0, 8
c. 0, 8, 3 5. a.
5 d. 0, 8, , 2.1, 0.4, 3 7 e.
b.
6
f. 0,
6. a.
5 , 8, , 2.1, 0.4, 3 6 7
2. a. M 2 J
b. J
2 2 or 2 2
2 2
2 2 2 2
2 y 11 2 y 22 : y 9 y 22 y 2 y 11 2
If 2 or 11 were substituted for y, then 2 2 0 or 11 11 0 and the denominator would be 0. So, y 2 , y 11 .
1 M 2
3. a. C 12 A 40 L 30
b.
b. C 12 A 40 L 30
12 80 40 2 30 960 80 30 $1070
67
2 y 11 2 y 22 2 y 9 y 22 y 2 y 11 y 2 2
Chapter R Review of Prerequisites
7. 6 4 9 x 2 3x 5 3
14.
6 4 9 x 6 x 10 3 6 4 3x 10 3 6 12 x 40 3
12 x 49 12 x 49 8.
9.
1 2 1 5 a b ab 2 a 2b 4ab 2 3 2 6 1 5 1 a 2b a 2b ab 2 4ab 2 3 6 2 2 1 2 5 2 3 1 6 a b a b ab 2 4ab 2 2 3 6 3 2 6 2 5 3 24 a 2b a 2b ab 2 ab 2 6 6 6 6 3a 2b 21ab 2 6 1 7 a 2b ab 2 2 2 7 2 4 8 2 42 2 4 6
53 ab3 3b 22 5ab 52 b 2 5ab 3b 2 5ab 5b 5ab 6b 5ab b 5ab 16. 3 x 2 y xy x 2 3 y1 3 x1 2 y1 2
x 2 31 2 y1 31 2 x 4 6 3 6 y 2 6 3 6 x7 6 y5 6
1 17. 12n 2 4 n 2 3n 5 2
49 40 16 4 7 7 4 4 10. 2a b
8 15
4 a b 0
3
0
1
13.
15 6
3a b c 2 2
9a 2 2 3a b b 2 c 2 9a 2 6ab b 2 c 2 2
2 1 1 1 19. z p2 z p2 z p2 4 4 4
80k m n 8 k n 10m n 3
18. 3a b c 3a b c
1
3
6n 4 36n3 58n 2 12n 20
10
2 7
6n 4 36n3 60n 2 2n 2 12n 20
a8 15 b
t2 5 t7 5 t 4 t14 1 12. t 41430 t 12 12 3 30 t t t 15
1 4 n 2 4 3n 4 5 2
2 7
1 1 1 11. 1 33 4 2 3 4 1 27 4 32
3
1 12n 2 n 2 12n 2 3n 12n 2 5 2
4a b 4a b 4 a b 6 8
x x1 6 y 5 6 x 6 xy 5
7 2 4 10 42 2 2
2 7 1
125ab3 3b 20ab
15.
3 4 2
34 2 3 4 3 3 8 p q 2 p q 4 8 p 2 q3 2 p3q 4 4 3 4 24 p 4 q 4 p 4 3 2 pq 4 p 4 3 pq 4 p 2
2
2k 5 n 2 3 10m 2 n
68
1 z p4 16
Chapter R Test
5 x 3 x 5 x 2 2 2 3 x 2 2 2
x 9 x 9 4x 4 4x 25. 4 4 x 1 3 1 3 4x 4 4x 4 4x x 9 4x 4x 2 4 4x x 9 1 3 x3 4x 4x 4 4x x 3 x 3 x3 x3
20. 5 x 2 3 x 2 2
15 x 10 2 x 3 2 x 4 15 x 7 2 x 4
x 6z x 2 x 6z 6z 2
21.
2
2
x 12 z x 36 z 2 3x 2 8x 4 22. 3 2 2 x 14 x 49 x 4 x 26 x 14 3x 2 4 x 2 26 x 14 3 x 14 x 2 49 x 8x 4 2 2 2 x 2 13 x 7 3x 4 2 x 1 x x 2 14 x 49
23.
x 3x x x 7 x 7
26.
27.
2 2 x 1 x 7 4 2 x 1 2
x 5 x 5
2
13 10 3
13 10
x5 28.
2
12 y 2 4 y 3 y
6
2
7
t 2
2 7 2 2 7 2 2 2 2t t t t2 t 2 2
1 y 2 4 y
y y 4y
4x2 y xy
6
3 y y y 4 y y y 4 y y 4 y y
3
13 10
12 1 3 24. 3 2 2 y 4y y y 4y
2
2 xy
13 10 13 10 13 10 13 10 6 13 10 13 10 6 13 10 6
2
x5
23 x 3 y 3
2 3 4 x2 y
2
2
2 3 4 x2 y 3
x2 10 x 25 x2 10 x 25 5 x x5 x5 x5 2 x 10 x 25 x5
3 2 3 22 x 2 y 8 8 2 xy 2 3 2 xy 2 3 2 xy 2 3 22 x 2 y
3x 2 x 7
12
3
7 2 2 2 5 2 2t 2t
29. 30 x 3 2 x 2 4 x 2 x 15 x 2 x 2
2
2 x 3x 1 5 x 2
y 2 y 12 y 2 y 4
30. xy 5ay 10ac 2 xc
y 4 y 3 y 3 2 y y 2 y 4
y x 5a 2c 5a x x 5a y 2c
69
Chapter R Review of Prerequisites
3u v 9u 3uv v
33. 27u 3 v 6 3u v 2
31. x 5 2 x 4 81x 162
3
x 5 81x 2 x 4 162
2
x x 81 2 x 81 4
4
x 9 x 3 x 3 x 2
x 4 81 x 2 x 2 9 x 2 9 x 2
w6 4w0 2w1 7 w2
32. c 4a 44a 121
c 2 2a 11
w6 4 2w 7 w2 or
c 4a 44a 121 2
2
4
w6 4 w6 6 2 w5 6 7 w4 6
2
2
2
34. 4w6 2 w5 7 w4
2
2
3
7 w2 2 w 4 w6
2
c 2a 11 c 2a 11 c 2a 11 c 2a 11
35. y 2 y 1
3 4
2 y 1
14
y 2 y 1 3 4 3 4 2 y 11 4 3 4 0 1 3 4 2 y 1 y 2 y 1 2 y 1 2 y 1
3 4
2 y 1
3 4
y 2 y 1 2 y 1 3 4 3 y 1 or
3
5
11 3 8.4 6.0 10 10 12 1013 4.2 105
1.2 101 1013 1.2 1014
2 x 7 x 30 x 4 x 4 2
8.4 10 6.0 10 4.2 10 11
40.
2
2 x 2 12 x 5 x 30 x 2 4 x 4
2 y 13 4
39. 10 7 is between 0 and 1. Move the decimal point 7 places to the left. 8 107 0.000 0008
1 36. V r 2 h 3 1 1 2 5 132 52 25 169 25 3 3 25 25 144 12 100 in.3 3 3
37. A 2 x 5 x 6 x 2
3y 1
2
x 2 11x 34
41. 2.7,
38. Move the decimal point 10 places. The number is greater than 10, so use a positive power of 10. 45,000,000,000 4.5 1010 Move the decimal point 6 places. The number is greater than 10, so use a positive power of 10. 1, 660,000 1.66 106
42. x | 3 x 5
70
Section 1.1
Chapter 1 Equations and Inequalities Section 1.1 Linear Equations and Rational Equations 1. linear
2. first
3. solution
4. set
5. equivalent
6. addition
7. division
8. conditional
48
9. identity
10. contradiction
d. Nonlinear
c. Linear;
e. Linear;
11. rational 12. empty (or null); or
8
13. a. Linear; 2 x 8
4
2 x 8 2 2 x 4
6 x 24 x 4 4 16. 8 y 6 22
1 x8 c. Linear; 2 1 2 x 2 8 2
8 y 16 y 2
2
x 16
17. 4 7 3 4t 1
4 7 12t 3 4 4 12t 0 12t 0t 0
d. Nonlinear e. Linear;
x28 x 2 2 8 2 x 10
10
18. 11 7 2 5 p 2
14. a. Linear; 12 4 x
3
12 4 x 12 4 4 x 4 8 x
15. 6 x 4 20
b. Nonlinear
16
1 x 4 1 4 12 4 x 4 48 x 12
11 7 10 p 4 11 11 10 p 0 10 p 0 p
12 4 x 4 4 3 x
0
b. Nonlinear
71
Chapter 1 Equations and Inequalities 19. 6 v 2 3 9 v 4
25. 2 5 x 6 4 x 3 x 10
10 x 12 4 x 3 x 30
6v 12 3 9 v 4 6v 15 5 v 5v 10 v2 2
10 x 12 4 2 x 30 10 x 12 8 x 120 18 x 132 132 22 x 18 3 22 3
20. 5 u 4 2 11 u 3
5u 20 2 11 u 3 5u 22 14 u 4u 8 u2 2 21.
26. 4 y 3 3 y 2 y 2
4 y 12 3 y 2 y 4 4 y 12 3 3 y 4 4 y 12 9 y 12 5 y 0 y0
2.3 4.5 x 30.2 27.9 4.5 x 6.2 x 6.2
0
22. 9.4 3.5 p 0.4
27.
9.8 3.5 p 2.8 p
2.8 23. 0.05 y 0.02 6000 y 270
1 3 x 2 4 2 3 1 4 x 4 2 4 2 x68
14
0.05 y 120 0.02 y 270 0.03 y 120 270 0.03 y 150 y 5000
28.
5000
1 5 x 1 6 3 5 1 6 x 6 1 6 3 x 10 6
24. 0.06 x 0.04 10, 000 x 520
0.06 x 400 0.04 x 520 0.02 x 120
6000
x 14
16
x 6000
72
x 16
Section 1.1
29.
1 3 2 w w 2 2 4 3 3 1 2 12 w 12 w 2 2 3 4 6w 9 8w 24
32.
7 x 6 3x 7 x 1 42
2 w 33 w
7 x 42 3x 7 x 7 42 10 x 42 7 x 49 3x 91 91 x 3 91 3
33 2
33 2 30.
2 3 7 p p 1 5 10 15 3 2 7 30 p 30 p 1 5 15 10 12 p 9 14 p 30
33.
2 p 21 p
n 3 n 2 n 1 1 4 5 10 n 3 n 2 n 1 20 20 1 4 10 5 5 n 3 4 n 2 2 n 1 20
21 2
5n 15 4n 8 2n 2 20 n 23 2n 18 n 41 n 41 41
21 2
31.
x 6 x x 1 2 3 7 3 x 6 x x 1 21 21 2 3 7 3
y 1 y y 3 1 5 4 2 y 1 y y3 20 20 1 5 2 4
34.
4 y 1 5 y 10 y 3 20 4 y 4 5 y 10 y 30 20 9 y 4 10 y 50
t 2 t7 t4 2 3 5 10 t 2 t 7 t4 30 30 2 3 10 5 10 t 2 6 t 7 3 t 4 60
y 54
10t 20 6t 42 3t 12 60 4t 62 3t 48
y 54
54
110
t 110
1 1 35. a. Vs Vt 50, 040 5560 m3 9 9
73
Chapter 1 Equations and Inequalities
b.
40. a. C 167.95 x 94
1 Vs Vt 9 1 9000 Vt 9 81,000 Vt
C 167.95 9 94 $1605.55 b.
2445.30 167.95 x 94 2351.30 167.95 x 14 x 14 credit-hours
Vt 81,000 m3 36. a. E 0.78d 0.78 500 390 euros b.
41.
E 0.78d 800 0.78d 1026 d d $1026 R 4.25 6 93.9 $119.4 billion
42.
4 x 2005 4 2009 In 2009
43. 2 x 3 4 x 1 1 2 x
38. a. R 17.5 x 138.5
2x 3 4x 4 1 2x 2x 3 2x 5 3 5 Contradiction
R 17.5 2 138.5 $173.5 billion R 17.5 x 138.5 226 17.5 x 138.5
44. 4 3 5n 1 4n 8 16n
87.5 17.5 x 5 x 2005 5 2010 In 2010
12 20n 1 4n 8 16n 20n 13 20n 8 13 8 Contradiction
39. a. T 1.83a 212
T 1.83 4 212 204.68F b.
E 13.7 x 338
475 13.7 x 338 137 13.7 x 10 x 2000 10 2010 In 2010
R 4.25 x 93.9 110.9 4.25 x 93.9 17 4.25 x
b.
E 46.2 x 446.2
862 46.2 x 446.2 415.8 46.2 x 9 x 2000 9 2009 In 2009
37. a. R 4.25 x 93.9
b.
C 167.95 x 94
45. 6 2w 4 w 1 2w 10
T 1.83a 212
6 2w 4 w 4 2w 10 6 2w 2 w 6 00 Identity;
193 1.83a 212 19 1.83a 10.4 a 10.4 103 10, 400 Approximately10,400 ft
74
Section 1.1 46. 5 3x 3 x 1 2
52.
5 3x 3 x 3 2 5 3x 3 x 5 00 Identity;
3x 1 1 x , x 4, x 4 3
1 1 47. x 3 x 1 2 4 1 1 4 x 3 4 x 1 2 4 2 x 12 x 4
1 7 5 2y 2y 2 2y y
2 1 48. y 5 y 4 3 6 2 1 6 y 5 6 y 4 3 6 4 y 30 y 24
17 54.
3y 6 y2
Conditional equation; 2
2 5 2 x 1 x 7 3 x 1 , x 7
t 4 21 t 25
w3 w5 1 4w w w 3 w 5 1 4w 4w 4w w
w 3 4 w 4 w 5 5w 3 4w 20 w 23 23
3 1 1 51. 2 2 2 x 7 x 15 6 x 25 3 1 1 2 x 3 x 5 6 x 5 x 5 2x 3 0 2x 3
x
y 7 10 y 17
1 4 7 3 3t t 1 4 7 3t 3t 3 3t t
25
3 2 5 x5 x4 7 x 5 , x 4 55.
50.
1 7 5 2 2y y
53.
x 8 Conditional equation; 8
49.
6 1 1 2 3x 11x 4 3 x 16 6 1 1 3x 1 x 4 3 x 4 x 4 3x 1 0 2
56.
3 , x 5, x 5 2
x2 x7 1 6x x x 2 x 7 6x 1 6 x 6x x
x 2 6 x 6 x 7 7 x 2 6 x 42 x 44 44
75
Chapter 1 Equations and Inequalities
57.
c 3 3 c3 c3 4 3 c 3 4 c 3 4 c 3 c 3 c 3 4
58.
7 7 d d 7 8 d 7 7 7 d 8 d 7 8 d 7 d 7 8 d 7
4c 12 3 c 3
56 7 d 7 8d
4c 12 3c 9 7c 21
56 7 d 49 8d 15d 105 d 7
c3
; The value 3 does not check.
; The value 7 does not check.
1 3 2 t 1 t 1 1 3 t 1 t 1 t 1
59.
t 1 t 1 2
1 3 t 1 t 1 t 1 t 1 t 1 t 1 3 t2
1 5 2 w 2 w 4 1 5 w 2 w 2 w 2
60.
w 2 w 2
1 5 w 2 w 2 w 2 w 2 w 2 w2 5 w7
7
2 1 11 2 x 5 x 5 x 25 2 1 11 x 5 x 5 x 5 x 5
61.
x 5 x 5
4
2 1 11 x 5 x 5 x5 x5 x 5 x 5
2 x 5 1 x 5 11 2 x 10 x 5 11 x 15 11 x 4
76
Section 1.1 2 1 10 2 c3 c3 c 9 2 1 10 c 3 c 3 c 3 c 3
62.
c 3 c 3
2 1 10 c 3 c 3 c3 c3 c 3 c 3
2 c 3 1 c 3 10
2c 6 c 3 10 c 9 10 c 19
19
5 2 4 2 2 x x 2 x 4 x 3x 2 5 2 4 x 2 x 1 x 2 x 2 x 2 x 1
63.
2
4 x 2 x 2 x 1 x 2 x 1 x 2 x 2 x 2 x 1 5 x 2 2 x 1 4 x 2 5
x 2 x 2 x 1
2
5 x 10 2 x 2 4 x 8 3x 8 4 x 8 16 x
16
4 1 2 2 2 x 2 x 8 x 16 x 6 x 8 4 1 2 x 4 x 2 x 4 x 4 x 4 x 2
64.
2
2 x 4 x 4 x 2 x 4 x 2 x 4 x 4 x 4 x 2 4 x 4 1 x 2 2 x 4
x 4 x 4 x 2
4
1
4 x 16 x 2 2 x 8
22
3x 14 2 x 8 x 22
77
Chapter 1 Equations and Inequalities 5 3m 2 2 m 2 m 2m 8 m 4 5 3m 2 m 2 m 4 m 2 m 4
65.
m 4 m 2
5 3m 2 m 4 m 2 m2 m 4 m 2 m 4
5 m 4 3m 2 m 2 5m 20 3m 2m 4 5m 20 m 4 4m 16 m 4 ; The value 4 does not check.
10 15n 6 2 n 6 n 2n 24 n 4 10 15n 6 n 6 n 6 n 4 n 4
66.
n 6 n 4
10 15n 6 n 6 n 4 n6 n 6 n 4 n 4
10 n 4 15n 6 n 6 10n 40 15n 6n 36 10n 40 9n 36 n 4 ; The value 4 does not check.
5x 1 3 3x 5 x 2 3x 1 2 x 5x 1 3 3x 1 x 2 3x 1 x 2
67.
2
5x 1 3 3x 1 x 2 x 2 3x 1 x 2 3x 1
3x 1 x 2
5 x 1 x 2 3 3x 1 5 x x 2 9 x 3 4 x 2 9 x 3 13x 5 x
5 13
5 13
78
Section 1.1 3x 2 4 2x x 3 2x 3 1 x 3x 2 4 2 x 3 x 1 2 x 3 x 1
68.
2
3x 2 4 2 x 3 x 1 x 1 2 x 3 x 1 2 x 3
2 x 3 x 1
3x 2 x 1 4 2 x 3 3x 2 x 2 8 x 12 x 2 8 x 12 9 x 14 x
14 9
14 9 69. A lw for l
75. 7 x 2 y 8 for y
2 y 7 x 8
A lw w w A A l or l w w
2 y 7 x 8 2 2 7 x 8 7 or y x 4 y 2 2
70. E IR for R
76. 3x 5 y 15 for y
E IR I I E E R or R I I 71.
P a b c for c P a b c or c P a b
72.
W K T for K W T K or K W T
73.
5 y 3x 15 5 y 3x 15 5 5 3 x 15 3 y or y x 3 5 5 77. 5 x 4 y 2 for y 4 y 5 x 2
4 y 5 x 2 4 4 5x 2 5 1 y or y x 4 4 2
s s2 s1 for s1 s s2 s1 s1 s2 s
74.
78. 7 x 2 y 5 for y
t t f ti for ti
2 y 7 x 5 2 y 7 x 5 2 2 7x 5 7 5 y or y x 2 2 2
t t f ti ti t f t
79
Chapter 1 Equations and Inequalities
79.
2 S n 2 d nd 2an
1 1 x y 1 for y 2 3 1 1 6 x y 6 1 2 3 3x 2 y 6 2 y 3 x 6 2 y 3x 6 2 2 3x 6 3 y or y x 3 2 2
2 S n 2 d nd 2an 2n 2n 2 2 S n d nd 2S n 2 d nd a or a 2n 2n 83.
3V r 2 h
1 2 80. x y 2 for y 4 3 2 1 12 x y 12 2 4 3 3x 8 y 24 8 y 3x 24 8 y 3 x 24 8 8 3x 24 3 y or y x 3 8 8
81.
1 V r 2 h for h 3 1 3 V 3 r 2 h 3 3V r 2 h r2 r2 3V 3V h or h 2 2 r r
84.
1 Bh for B 3 1 3 V 3 Bh 3 V
3V Bh
n a d for d 2 n 2 S 2 a d 2 2S n a d 2S na nd 2S na nd 2 S na nd n n 2 S na d n 2S na 2S a d or d n n
3V Bh h h 3V 3V B or B h h
S
85.
6 4 x tx for x 6 x4 t x4 t 6 4t 4t 6 6 x or x 4t 4t
86.
8 3x kx for x 8 x 3 k
n 82. S 2a n 1 d for a 2 n 2 S 2 2a n 1 d 2
x 3 k 8 3 k 3 k 8 8 x or x 3 k 3 k
2S n 2a n 1 d 2S n 2a nd d 2S 2an n 2 d nd
80
Section 1.1 87. 6 x ay bx 5 for x
89.
6 x bx 5 ay
A P (1 rt )
x 6 b 5 ay
A P(1 rt ) 1 rt 1 rt A A P or P 1 rt 1 rt
x 6 b 5 ay 6b 6b 5 ay ay 5 or x x 6b b6
90.
A 1 r C 1 r 1 r C C A or A 1 r 1 r
3x cx d 2 y x 3 c d 2 y x 3 c d 2 y 3c 3c 2y d d 2y or x x 3c c3 2 5 2n 1 3n 4 5 2 2n 1 3n 4 2n 1 3n 4 2n 1 3n 4 5 3n 4 2 2n 1 15n 20 4n 2 19n 18 n
18 19
18 19 92.
C A Ar for A C A 1 r
88. 3 x 2 y cx d for x
91.
A P Prt for P
2 4 5z 3 4z 7 4 2 5 z 3 4 z 7 5 z 3 4 z 7 5z 3 4z 7 4 4 z 7 2 5 z 3 16 z 28 10 z 6 26 z 22 z
22 11 26 13
11 13
81
Chapter 1 Equations and Inequalities
6 12u 8 7u 14 5u 0 u0
93. 5 2 3 5v 3 v 7 8v 6 3 4v 61
5 2 3 5v 3v 21 8v 18 24v 61 5 2 3 8v 21 16v 43
0
5 2 3 8v 21 16v 43
0
5 2 24 8v 16v 43 5 48 16v 16v 43 16v 43 16v 43 32v 0 v0
95.
x 7 x 2 x 2 4 x 13 x 2 2 x 7 x 14 x 2 4 x 13
3
94. 6 4 2 8u 2 u 3 4u 3 2 u 8
96.
m 3 2m 5 2m2 4m 3 2m 2 6m 5m 15 2m 2 4m 3
6 4 2 8u 2u 6 4u 6 3u 8 6 4 2 6u 6 7u 14 6 4 12u 12 7u 14
6 12u 8 7u 14
97.
x 2 5 x 14 x 2 4 x 13 9 x 27 x 3
4
2m 2 m 15 2m 2 4m 3 3m 12 m 4
3 9 2 2 2 c 4c 2c 3c 2c 5c 12 3 9 2 c c 4 c 2c 3 2c 3 c 4 2
3 9 2 c 2c 3 c 4 c 2c 3 c 4 c c 4 c 2c 3 2c 3 c 4 3 2c 3 9 c 4 2c 6c 9 9c 36 2c 3c 45 2c 5c 45 c9 9 98.
4 5 2 2 2 d d 2d 5d 2d 3d 5 4 5 2 d d 1 d 2d 5 2d 5 d 1 2
4 5 2 d 2d 5 d 1 d 2d 5 d 1 d d 1 d 2d 5 2d 5 d 1 4 2d 5 5 d 1 2d 8d 20 5d 5 2d 3d 25 2d d 25 25
82
Section 1.1
99.
1 1 1 1 x x 1 x 3 2 2 6 1 1 1 1 6 x 6 x 1 x 3 2 6 2
101.
t 2 4t 4 t 2 8t 16 12t 12
2 x 3 3 x 1 x
1
2 x 3 3x 3 x 2x 3 2x 3 00
y2 6 y 9 y2 2 y 1 8 y 8 y 1
1 2 2 1 x x 1 x 2 5 5 10 2 1 1 2 10 x 10 x 1 x 2 5 10 5
1
5 x 4 4 x 1 x 5x 4 4 x 4 x 5x 4 5x 4 00
103.
3 5 3a 4 5a 1 3 5 3a 4 5a 1 3a 4 5a 1 3a 4 5a 1 3 5a 1 5 3a 4 15a 3 15a 20
104.
3 20
8 2 8x 3 2x 5 8 2 8 x 3 2 x 5 8 x 3 2 x 5 8x 3 2x 5 8 2 x 5 2 8 x 3 16 x 40 16 x 6
t 1
y 32 y 12
102.
100.
t 2 2 t 4 2
40 6
83
Chapter 1 Equations and Inequalities 40 20 x 1 0.05 x 40 20 x 200 1 0.05 x
109. The value 5 is not defined within the expressions in the equation. Substituting 5 into the equation would result in division by 0.
P
105.
1 0.05 x 200 1 0.05 x
40 20 x 1 0.05 x
110. The equation is an identity. The solution set is all real numbers.
200 10 x 40 20 x
111. The equation cannot be written in the form 3 ax b 0 . The term 3x 1 . Therefore, x 3 the term is not first degree and the x equation is not a first-degree equation.
10 x 160 x 16 yr 180t 2t 10 180t 60 2t 10 v
106.
2t 10 60 2t 10
112. The equation cannot be written in the form ax b 0 . The term 2 x 2 x1 2 . Therefore,
180t 2t 10
the term 2 x is not first degree and the equation is not a first-degree equation.
120t 600 180t 60t 600
113. The equation is a contradiction. There is no real number x to which we add 1 that will equal the same real number x to which we add 2.
t 10 sec 107.
1 1 c h 22 30 1 1 c 165 7 22 30 1 11 c 7 22 2 11 1 22 7 22 c 22 2 154 c 121 A
114. In each case, we can clear fractions by multiplying both sides of the equation by the x 1 LCD. For the equation 1 , the LCD is 3 2 3 1 6, whereas for 1 , the LCD is 2x. x 2 115.
ax 6 4 x 14 a 4 6 4 4 14
c 33 mi
4a 6 16 14
1 1 108. A c h 24 32 1 1 9 60 h 24 32 5 1 9 h 2 32 5 1 32 9 32 h 2 32 288 80 h
4a 6 30 4a 24 a6 116.
ax 3 2 x 9 a 3 3 2 3 9 3a 3 6 9 3a 3 15 3a 18
h 208 mi
a6
84
Section 1.2 a 2 x 5 6 5 x 7
117.
120. Let x 0.826 . 1000 x 826.26 10 x 8.26 990 x 818.00
a 2 16 5 6 5 16 7 a 32 5 6 80 7 27 a 6 87 27 a 81 a3
x
a 2 x 4 12 x 3 2 x
118.
818 409 990 495
121. Let x 0.534 . 1000 x 534.534 x .534 999 x 534.000
a 2 34 4 12 34 3 2 34 a 68 4 408 3 32 72a 408 96 72a 504 a 7
x
534 178 999 333
122. Let x 0.761 . 1000 x 761.761 x .761 999 x 761.000
119. Let x 0.516 . 1000 x 516.16 10 x 5.16 990 x 511.00
x
511 x 990
761 999
Section 1.2 Applications and Modeling with Linear Equations 1. 5 x 2
2. 4 x 10
3. 0.06x
4. 0.6x
5. 40 x
6. 54 x
7. x 1
8. x 2
9. P 2l 2 w
10. 180
11. 90 x
12. 180 x
17. The length of the lot is l 128 2 x . The width of the lot is w 60 2 x . P 2l 2 w 440 2 128 2 x 2 60 2 x 440 256 4 x 120 4 x 440 8 x 376 64 8 x 8 x The width of the easement is 8 ft. 18. The length of the play area is l 78 2 x . The width of the play area is w 36 2 x . P 2l 2w 396 2 78 2 x 2 36 2 x 396 156 4 x 72 4 x 396 8 x 228 168 8 x 21 x The width of the border is 21 ft.
13. I Prt
I 6000 0.075 2 $900 14. 0.08 2 0.16 L 15.
d r
16.
d t
85
Chapter 1 Equations and Inequalities 21. 90 x
19. a. The width of the kitchen is w. The length of the kitchen is l w 4 . P 2l 2w 48 2 w 4 2 w 48 2w 8 2w 48 4w 8 40 4w 10 w l w 4 10 4 14 The kitchen is 14 ft by 10 ft.
23. 180 131 x or 49 x 24. 180 43 x or 137 x 25. The angle between the ladder and the wall is x. The angle between the ladder and the ground is x 22 .
x x 22 90 180
2 x 112 180 2 x 68 x 34 x 22 34 22 56 The angle between the ladder and the ground is 56 and the angle that the ladder makes with the wall is 34 .
b. A lw 0.1lw 1.1lw
A 1.11410 154 ft
22. 180 x
2
c. C 1.06 12154 $1958.88 20. a. The width of the porch is w. The length of the porch is l 2w 2 P 2l 2w
64 2 2w 2 2w
26. The angle that the rope makes with the tree is x. The angle that the rope makes with the ground is 1.5x. x 1.5 x 90 180
64 4w 4 2w 64 6 w 4 60 6 w
2.5 x 90 180 2.5 x 90 x 36 1.5 x 1.5 36 54 The angle that the rope makes with the ground is 54 , and the angle that the rope makes with the tree is 36 .
10 w l 2w 2 2 10 2 22 The porch is 22 ft by 10 ft. b. A lw 0.1lw 1.1lw
A 1.1 2210 242 ft 2 c. C 1.075 5.85 242 $1521.88
27. Let x represent the amount borrowed at 3%. Then, 5000 x is the amount borrowed at 2.5%.
Principal Interest (I = Prt)
3% Interest Loan
2.5% Interest Loan
x
5000 x
x 0.031
5000 x 0.0251
x 0.03 5000 x 0.025 132.50 0.03x 125 0.025 x 132.50 0.005 x 125 132.50 0.005 x 7.50 x 1500 5000 x 5000 1500 3500 Rocco borrowed $1500 at 3% and $3500 at 2.5%.
86
Total
132.50
Section 1.2 28. Let x represent the amount borrowed at 4%. Then, 22,000 x is the amount borrowed at 5.5%. 4% Interest Loan
5.5% Interest Loan
x
22,000 x
x 0.041
22, 000 x 0.0551
Principal Interest (I = Prt)
Total
910
x 0.04 22, 000 x 0.055 910 0.04 x 1210 0.055 x 910 0.015 x 1210 910 0.015 x 300 x 20,000 5000 x 5000 1500 3500 Laura borrowed $20,000 from the bank charging 4% interest, and $2000 from the bank charging 5.5% interest. 29. Let x represent the amount invested in the 3-yr CD. Then, x 2000 is the amount invested in the 18month CD.
Principal Interest (I = Prt)
3-yr CD
18-month (1.5-yr) CD
x
x 2000
x 0.044 3
x 2000 0.031.5
Total
706.50
x 0.044 3 x 2000 0.031.5 706.50 0.132 x 0.045 x 90 706.50 0.177 x 90 706.50 0.177 x 796.50 x 4500 x 2000 4500 2000 2500 Fernando invested $4500 in the 3-yr CD and $2500 in the 18-month CD. 30. Let x represent the amount invested in the 5-yr Treasury note. Then, x 5000 is the amount invested in the 10-yr bond.
Principal Interest (I = Prt)
5-yr Note
10-yr Bond
x
x 5000
x 0.028 5
x 5000 0.03610
Total
5300
x 0.028 5 x 5000 0.03610 5300 0.14 x 0.36 x 1800 5300 0.5 x 1800 5300 0.5 x 3500 x 7000 x 5000 7000 5000 12, 000 Ebony invested $7000 in the Treasury note and $12,000 in the bond.
87
Chapter 1 Equations and Inequalities 31. Let x represent the amount of the 5% solution (in gallons). 5000 gal is the amount of the 10% solution. Therefore, x 5000 is the amount of the resulting 9% solution. 5% Solution
10% Solution
9% Solution
x
5000
x 5000
0.05x
0.1 5000
0.09 x 5000
Amount of Solution Pure Ethanol
0.05 x 0.1 5000 0.09 x 5000 0.05 x 500 0.09 x 450 50 0.04 x 1250 x 1250 gal of E5 should be mixed with the E10. 32. Let x represent the amount of the 10% solution (in cubic centimeters). 60 cc is the amount of the 50% solution. Therefore, x 60 is the amount of the resulting 25% solution. 10% Solution
50% Solution
25% Solution
Amount of Solution
x
60
x 60
Pure Saline
0.1x
0.5 60
0.25 x 60
0.1x 0.5 60 0.25 x 60 0.1x 30 0.25 x 15 15 0.15 x 100 x 100 cc of 10% saline solution should be mixed with the 50% saline solution. 33. Let x represent the amount of the pure sand (in cubic feet). 480 ft2 is the amount of the concrete mix that is 70% sand. Therefore, x 480 is the amount of the resulting 75% sand mixture. 100% Sand
70% Sand
75% Sand
Amount of Mixture
x
480
x 480
Pure Sand
x
0.7 480
0.75 x 480
x 0.7 480 0.75 x 480 x 336 0.75 x 360 0.25 x 24 x 96 96 ft2 of sand should be mixed with the 70% sand mixture.
88
Section 1.2 34. Let x represent the amount of 50% antifreeze solution (in gallons) to be drained (and therefore the amount of 100% antifreeze solution to be added). 4 gal is the amount of the resulting 65% antifreeze solution. Therefore, 4 x is the amount of 50% antifreeze solution that is not drained. 100% Solution
50% Solution
65% Solution
Amount of Solution
x
4 x
4
Pure Antifreeze
x
0.5 4 x
0.65 4
x 0.5 4 x 0.65 4 x 2 0.5 x 2.6 0.5 x 0.6 x 1.2 1.2 gal of 50% antifreeze solution should be drained and replaced with 100% antifreeze. 35. Let x represent the speed of the plane flying to Los Angeles. Then, x 60 is the speed of the plane flying to New York City. Distance
Rate
Time
Los Angeles Flight
3.4x
x
3.4
New York City Flight
2.4 x 60
x 60
2.4
3.4 x 2.4 x 60 2464 3.4 x 2.4 x 144 2464 5.8 x 2320 x 400 x 60 400 60 460 The plane to Los Angeles travels 400 mph and the plane to New York City travels 460 mph. 36. Let x represent the speed of the plane flying to Seattle. Then, x 44 is the speed of the plane flying to Boston. Distance
Rate
Time
Seattle Flight
5.2 x
x
5.2
Boston Flight
2.5 x 44
x 44
37. Let x represent the distance from Darren’s home to his school. Distance
Rate
To School
x
32
To Home
x
48
2.5 x x 1.25 32 48 x x 96 96 1.25 32 48 3x 2 x 120 5 x 120 x 24 The distance is 24 mi.
5.2 x 2.5 x 44 3124 5.2 x 2.5 x 110 3124 7.7 x 3234 x 420 x 44 420 44 376 The plane to Seattle travels 420 mph, and the plane to Boston travels 376 mph.
89
Time x 32 x 48
Chapter 1 Equations and Inequalities 38. Let x represent the distance of the loop. Distance
Rate
Running
x
8
Riding
5x
16
43. Let t represent the time it takes for the 1 runners to cover mile . 4 1 lap 1 lap 1 lap 66 sec 60 sec t sec 1 1 1 660t 660t 66 60 t
Time x 8 5x 16
10t 11t 660
x 5x 1.75 8 16 x 5x 16 16 1.75 8 16 2 x 5 x 28 7 x 28 x4 The loop is 4 mi.
21t 660 220 t 31.4 sec 7 44. Let t represent the time it takes Marta and her daughter to vacuum the house together. 1 job 1 job 1 job 40 min 60 min t min 1 1 1 120t 120t 40 60 t
39. a. S1 45, 000 2250 x b. S 2 48, 000 2000 x c.
3t 2t 120 5t 120 t 24 min
S1 S 2 45,000 2250 x 48,000 2000 x 250 x 3000 x 12 yr
45. Let t represent the time it takes the second pump to fill the pool by itself. 1 job 1 job 1 job 10 hr t hr 6 hr 1 1 1 30t 30t 10 t 6
40. a. S1 25, 000 0.16 x b. S 2 30, 000 0.15 x c.
S1 S 2 25,000 0.16 x 30,000 0.15 x 0.01x 5000 x $500, 000
3t 30 5t 30 2t t 15 hr
41. a. C 7 x 46. Let t represent the time it takes Angelina to mow the lawn by herself. 1 job 1 job 1 job 50 min t min 30 min 1 1 1 150t 150t 50 t 30
b. C 105 7 x 105 x 15 The motorist will save money beginning on the 16th working day. 42. a. C 2.25 x b.
3t 150 5t 150 2t t 75 min or 1 hr 15 min
C 89 2.25 x 89 x 39.6 The commuter will save money on the 40th ride.
90
Section 1.2 x 10 55 10 45 There were 55 Democrat and 45 Republican senators.
47. Let x represent the amount of cement and y represent the amount of gravel. 1 x 2.4 150 2.4 x 150
51. Let x represent the number of deer in the population. 30 5 x 80 5 x 2400
x 62.5 2.4 150 3.6 y 2.4 y 540 y 225 62.5 lb of cement and 225 lb of gravel
x 480 deer 52. Let x represent the number of bass in the lake. 24 4 x 40 4 x 960
48. Let x represent the property tax on a house that is $240,000. 180,000 240, 000 1296 x 180, 000 x 1296 240,000
x 240 bass
1296 240, 000 180,000 x $1728
53. Let x represent the distance from the epicenter to the station.
x
49. Let x represent the patient’s LDL cholesterol level. The HDL cholesterol level is 60 mg/dL, and the total cholesterol is x 60 .
x 60 3.4 60 x 60 60 60 3.4 60 x 60 204 x 144 x 60 144 60 204 LDL is 144 mg/dL and the total cholesterol is 204 mg/dL.
Distance
Rate
P Waves
x
5
S Waves
x
3
Time x 5 x 3
x x 40 3 5 x x 15 15 40 3 5 5 x 3 x 600 2 x 600 x 300 km
50. Let x represent the number of Democrats. Then, x 10 represents the number of Republicans. x 11 x 10 9 x 11 9 x 10 9 x 10 x 10 9 9 x 11 x 10 9 x 11x 110 110 2 x 55 x
54. Let x represent the distance from the epicenter to the station.
91
Distance
Rate
P Waves
x
8
S Waves
x
4.8
Time x 8 x 4.8
Chapter 1 Equations and Inequalities 60. Let x represent the height of the light post.
x x 20 4.8 8 x x 24 20 24 4.8 8
light post x ft 40 ft 6 x 20 40 20 x 6 20 60 20 x 360 x 18 ft
55. Let x represent the price set by the merchant. x 0.25 x 180 0.40 180
0.75 x 180 72 0.75 x 252 x $336
0.90 x 80 28 0.90 x 108 x $120 57. a. C 110 60 x
C 350 110 60 x 350 60 x 240 x 4 hr
24 x 16 x 3 x 36 5 x 36 x 7.2 2 2 x 7.2 4.8 3 3 The pole is 7.2 ft long, and the snow is 4.8 ft deep.
58. a. C 2400 80 x b.
20 ft
61. Let x represent the height of the pole. Then, 1 x is the length of the pole that is in the 8 2 ground, and x is the length of the pole that 3 is in the snow. 2 1 x 1.5 x x 3 8 2 1 x x x 1.5 3 8 2 1 24 x x x 24 1.5 3 8
56. Let x represent the price set by the bookstore. x 0.10 x 80 0.35 80
b.
man
6 ft
5 x 3 x 480 2 x 480 x 240 km
C 5520 2400 80 x 5520 80 x 3120 x 39 hr
5 F 32 9 5 F F 32 9 5 9 F 9 F 32 9 9 F 5F 160 4 F 160 F 40 40C 40F
62. C
59. Let x represent the height of the Washington Monument. 5 x 4 444 4 x 2220
x 555 ft
92
Section 1.2 63. Let x represent the amount of 20% fertilizer solution (in liters) to be drained (and therefore the amount of water to be added). 40 L is the amount of the resulting 15% fertilizer solution. Therefore, 40 x is the amount of 20% fertilizer solution that is not drained. 0% Solution
20% Solution
15% Solution
Amount of Solution
x
40 x
40
Pure fertilizer
0 x
0.20 40 x
0.15 40
0 x 0.20 40 x 0.15 40 8 0.20 x 6 0.20 x 2 x 10 10 L should be drained and replaced by water. 64. Let x represent the amount of water (in liters) to be evaporated. Therefore, 200 x is the amount of the final 25% salt solution. 0% Solution
5% Solution
25% Solution
Amount of Solution
x
200
200 x
Pure salt
0 x
0.05 200
0.25 200 x
0 x 0.05 200 0.25 200 x 10 50 0.25 x 40 0.25 x 160 x 160 mL should be evaporated.
65. Let x represent the measure of one angle. Then, 5 x 18 is the measure of the other angle. x 5 x 18 180
66. Let x represent the measure of one angle. Then, 21x 24 is the measure of the other angle. x 21x 24 90
22 x 24 90 22 x 66 x3 21x 24 21 3 24 63 24 87 The angles measure 87 and 3 .
6 x 18 180 6 x 162 x 27 5 x 18 5 27 18 135 18 153 The angles measure 153 and 27 .
93
Chapter 1 Equations and Inequalities 67. Aliyah had 8000 0.28 8000 8000 2240 5760 to invest. Let x represent the amount she invested
at 11%. Then, 5760 x is the amount she invested at 5%. 11% Investment
5% Investment
x
5760 x
x 0.111
5760 x 0.051
Principal Interest (I = Prt)
Total
453.60
x 0.11 5760 x 0.05 453.60 0.11x 288 0.05 x 453.60 0.06 x 288 453.60 0.06 x 165.60 x 2760 5760 x 5760 2760 3000 Aliyah invested $2760 in the stock returning 11% and $3000 in the stock returning 5%. 68. Let x represent the amount Caitlin invested in the balanced fund. Then, 2x is the amount she invested in the stock fund. Balanced Fund (3.5%)
Stock Fund (17%)
x
2x
x 0.0351
2 x 0.171
Principal Interest (I = Prt)
Total
1125
x 0.035 2 x 0.17 1125 0.035 x 0.34 x 1125 0.375 x 1125 x 3000 2 x 2 3000 6000 Caitlin invested $3000 in the balanced fund and $6000 in the stock fund. 69. Let x represent the length of the shortest side. Then, x 1 is the length of the middle side
70. Let x represent the length of the shortest side. Then, x 9 is the length of the longest side
and x 5 is the length of the longest side.
and x 9 4 x 5 is the length of the
The perimeter is 4x .
middle side. The perimeter is 5x .
P x x 1 x 5
P x x 9 x 5
4 x x x 1 x 5
5 x x x 9 x 5
4 x 3x 6 x6 x 1 6 1 7 x 5 6 5 11 The lengths of the sides are 6 ft, 7 ft, and 11 ft.
5 x 3 x 14 2 x 14 x7 x 5 7 5 12 x 9 7 9 16 The lengths of the sides are 7 cm, 12 cm, and 16 cm.
94
Section 1.2
71.
7 x 8 12.8 8 x 89.6
8 12.8 y 12 12.8 y 96
x 11.2
y 7.5
78. Let x represent the smaller number. Then, x 25 is the larger number.
x 25 1 4 x x 1 x 25 x x4 x x x 25 4 x 1
x = 11.2 ft and y = 7.5 cm 72.
1.2 x 0.96 1.04 0.96 x 1.248
0.5 1.2 y 0.96 1.2 y 0.48
x 1.3
y 0.4
24 3 x 8 x x 25 8 25 33 The numbers are 8 and 33.
x = 1.3 m and y = 0.4 in. 73. No. If x represents the measure of the smallest angle, then the equation x x 2 x 4 180 does not result in an odd integer value for x. Instead the measures of the angles would be even integers.
79. Let x represent the tens digit of the number. Then, 14 x is the ones digit.
10 14 x 1 x 10 x 114 x 18 140 10 x x 10 x 14 x 18 140 9 x 9 x 32 108 18 x
74. No. If x represents the smaller of two consecutive integers, then the equation x x 1 90 does not result in an integer value for x.
6 x 14 x 14 6 8 The original number is 68. 80. Let x represent the tens digit of the number. Then, 9 x is the ones digit.
75. No. If x represents the number of each type of bill, then the solution to the equation 20 x 10 x 5 x 100 is not a whole number.
10 9 x 1 x 10 x 1 9 x 45 90 10 x x 10 x 9 x 45
76. No. If this were true, then the sum of the lengths of the remaining sides would equal the length of the third side, and no triangle can be constructed.
90 9 x 9 x 36 126 18 x 7x 9 x 97 2 The original number is 72.
77. Let x represent the smaller number. Then, x 16 is the larger number. 81.
x 16 2 3 x x 16 2 x x3 x x x
m1 x1 m2 x2 0
30 1.2 20 x2 0 20 x2 36 x2 1.8 m
x 16 3 x 2 14 2 x
82.
m1 x1 m2 x2 0
64 x1 80 2 0
7 x x 16 7 16 23 The numbers are 7 and 23.
64 x1 160 x1 2.5 m
95
Chapter 1 Equations and Inequalities m1 x1 m2 x2 0
83.
m1 x1 m2 x2 0
84.
10 3.2 m2 8 0
m1 10 6 7 0
8m2 32
10m1 42
m2 4 kg
m1 4.2 kg
Section 1.3 Complex Numbers 1. 3x 2 5 x 1 3 x 2 5 x 1 2 x 1
18.
100 i 100 10i
2. 4 y 5 10 y 7 4 y 5 10 y 7
19.
98 i 98 7i 2
20.
63 i 63 3i 7
21.
19 i 19
22.
23 i 23
6 y 2 3. 3a 4 5a 2 15a 6a 20a 8 2
15a 2 14a 8 4. 2c 5 7c 1 14c 2 2c 35c 5
14c 2 33c 5
23. 16 i 16 4i 2
2 1 2 1 2 1 5. m n m n m n 6 5 6 5 6 5 1 4 m2 n2 36 25 2
2 1 2 1 2 1 6. x y x y x y 2 3 2 3 2 3
2
24. 25 i 25 5i
2
25.
4 9 i 4 i 9 2i 3i 6i 2 6 1 6
26.
1 36 i 1 i 36 1i 6i 6i 2 6 1 6
1 2 4 2 x y 4 9
27.
10 5 i 10 i 5 i 2 50 1 52 2 5 2
7. z 2 z 2 2 2 z 22 z 2 4 z 4 2
28.
6 15 i 6 i 15 i 2 90
8. b 7 b 2 2 7 b 7 2 b 2 14b 49
1 32 10 3 10
2
9. 1
10. i b
11. real; imaginary
12. imaginary
13. pure
14. form
15. complex
16. conjugate
17.
29.
6 14 i 6 i 14 i 2 84 1 22 21 2 21
30.
10 15 i 10 i 15 i 2 150 1 52 6 5 6
121 i 121 11i
31.
96
98 2
i 98 i 2
98 49 7 2
Section 1.3
32.
33.
34.
45 5 63 7 80 5
i 45
i 63
i 80
i 5
7
5
45 9 3 5
i
63 i 9 3i 7
i
80 i 16 4i 5
53.
54.
35. Real part: 3; Imaginary part: 7 36. Real part: 2; Imaginary part: 4 37. Real part: 0; Imaginary part: 19
1 ; Imaginary part: 0 4
40. Real part:
4 ; Imaginary part: 0 7
10 125 10 5 5i 5 5 10 5 5i 5 5 2 5i or 2 i 5
55. a. i 20 1
38. Real part: 0; Imaginary part: 40 39. Real part:
18 48 18 4 3i 4 4 18 4 3i 4 4 9 9 3i or i 3 2 2
b. i 29 i 28 i1 1 i1 i c. i 50 i 48 i 2 1 i 2 1 d. i 41 i 44 i 3 1 i 3 i 56. a. i 32 1
41. a. True
b. False
42. a. False
b. True
b. i 47 i 44 i 3 1 i 3 i
43. a. True
b. True
c. i 66 i 64 i 2 1 i 2 1
44. a. False
b. True
d. i 27 i 28 i1 1 i1 i 57. a. i 37 i 36 i1 1 i1 i
45. 5 5 0i 46. 4 4 0i
b. i 37 i 40 i 3 1 i 3 i
47. 4 4 4 2i 8i 0 8i
c. i 82 i80 i 2 1 i 2 1 d. i 82 i 84 i 2 1 i 2 1
48. 2 144 2 12i 24i 0 24i
58. a. i103 i100 i 3 1 i 3 i
49. 2 12 2 2 3i or 2 2i 3
b. i 103 i 104 i1 1 i1 i
50. 6 24 6 2 6 i or 6 2i 6
51.
52.
c. i 52 1 d. i 52 1
8 3i 8 3 4 3 i i 14 14 14 7 14
59. 2 7i 8 3i 2 8 7 3 i
4 5i 4 5 2 5 i i 6 6 6 3 6
10 10i
97
Chapter 1 Equations and Inequalities 60. 6 10i 8 4i 6 8 10 4 i
3 1 7 1 64. i i 5 8 10 6
14 6i
3 7 1 1 i 5 10 8 6
61. 15 21i 18 40i
15 18 21 40 i 3 61i
4 6 7 3 i 10 10 24 24
62. 250 100i 80 25i
250 80 100 25 i 170 75i
1 7 i 10 24
65. 2.3 4i 8.1 2.7i 4.6 6.7i
1 2 5 1 1 5 2 1 63. i i i 2 3 6 12 2 6 3 12
2.3 8.1 4.6 4 2.7 6.7 i
3 5 8 1 i 6 6 12 12
1.2 0i
2 7 1 7 i i 6 12 3 12 66. 0.05 0.03i 0.12 0.08i 0.07 0.05i 0.05 0.12 0.07 0.03 0.08 0.05 i 0.14 0i
67.
73. 3 6i 10 i
1 16 24i 2 3i 8
3 10 3 i 6i 10 6i i 30 3i 60i 6i 2 30 57i 6 1
1 68. 60 30i 10 5i 6
36 57i 74. 2 5i 8 2i
69. 2i 5 i 10i 2i 2 10i 2 1 2 10i
2 8 2 2i 5i 8 5i 2i
70. 4i 6 5i 24i 20i 2
16 4i 40i 10i 2 16 36i 10 1
24i 20 1
26 36i
20 24i 71.
3
75. 3 7i 3 2 3 7i 7i 2
11 7 i 3 11 i 7
2
9 42i 49i 2 9 42i 49 1
i 33 i 2 21
9 42i 49 40 42i
i 33 1 21
76. 10 3i 10 2 10 3i 3i 2
21 i 33 72.
2
2
2
2
100 60i 9i 2 100 60i 9 1
13 5 i 2 13 i 5
100 60i 9 91 60i
i 26 i 2 10 i 26 1 10 10 i 26
98
Section 1.3
77. 3 5 4 5
84. a. 4 5i
3 4 3 i 5 i 5 4 i 5 i 5
b. 4 5i 4 5i 4 5 16 25 41 2
3i 5 4i 5
85. a. 0 8i b. 0 8i 0 8i 0 8 0 64 64
12 3i 5 4i 5 5i 2
2
12 i 5 5 1 17 i 5
78. 2 7 10 7
b. 0 9i 0 9i 0 9 0 81 81 2
2 10 2 i 7 i 7 10 i 7 i 7 20 2i 7 10i 7 7i
2
100 16 116 88. 3 9i 3 9i 3 9 9 81 90 2
90. 5i 5i 5 25 2
11 7i 80. 3 8 3i 6i 2 i 24 9i 12i 6i 2
91.
24 3i 6 1
2 3i 2 3i 2 3 2
2
23 5
18 3i 92. 2
5 7i 5 7i 5 7 2
2
5 7 12
2 2 2 i i 2 2 2 i i 2
2
2
2
4 4i i 2 4 4i i 2
93.
8 2i 2 8 2 1 6
6 2i 6 2i 3 i 3i 3 i 3 i
2
3 2 3 2i 2i 3 2 3 2i 2
2i
2
89. 7i 7i 7 2 49
24 7i 35 1
2
2
2
79. 4 6 2i 5i 3 7i 24 8i 15i 35i 2
82. 3 2i 3 2i
2
87. 10 4i 10 4i 10 4
20 12i 7 7 1 13 12i 7
2
2
86. a. 0 9i
2 i 7 10 i 7
81. 2 i 2 i
2
2
2
18 6i 6i 2i 2
3 1 2
2
18 12i 2 1 9 1
16 12i 16 12 8 6 i i 10 10 10 5 5
2
9 12i 4i 2 9 12i 4i 2
94.
18 8i 2 18 8 1 10
5 i 5 i 4 i 20 5i 4i i 2 4 i 4 i 4 i 4 2 1 2
83. a. 3 6i b. 3 6i 3 6i 3 6 9 36 45 2
2
99
20 9i 1 1 19 9i 19 9 i 16 1 17 17 17
Chapter 1 Equations and Inequalities
95.
8 5i 8 5i 13 2i 13 2i 13 2i 13 2i
100.
104 16i 65i 10i 2
132 22 104 81i 10 1 94 81i 169 4 94 81 i 173 173 96.
101.
1 6 5i
6 5
2
98.
4 3i
4 3 2
2
3i 3i
3i 3i 3i 2 3 1
2 11i
2 11i 11i 11i
2 11i 11i 2
2 11i 2 11 2 11 i 0 i 11 1 11 11
42 4 2 6 16 48
52 4 510 25 200
175 i 175 5i 7 b 2 4ac
105.
6 2 4 2 5 36 40
4 i 4 2i b 2 4ac
106.
42 4 2 4 16 32
16 i 16 4i
107. a.
4 3i 4 3i 4 3i 4 3i
99.
b 2 4ac
6 5i 36 5
1 4 3i
1
1 3i
3i 3i 3i 0 3 3 3
b 2 4ac
104.
6 5i 6 5 i 41 41 41 1
32 i 32 4i 2
6 5i 6 5i
2
3i
103.
1 6 5i
6 5i
11
2
121 16 98 73i 98 73 i 137 137 137
2
102.
110 40i 33i 12i
1
3
1
11 2 4 2 110 73i 12 1
97. 6 5i
1
173
10 3i 10 3i 11 4i 11 4i 11 4i 11 4i
6 6 i 6i 6i 6 6 i 0 i 2 7i 7i i 7i 7 7 7 1
x 2 25 0
5i 2 25 ⱨ 0 25 1 25 ⱨ 0
4 3i 16 3
25 25 ⱨ 0 0ⱨ0
4 3i 4 3 i 19 19 19
b.
x 2 25 0
5i 2 25 ⱨ 0 25 1 25 ⱨ 0
5 5i 5i 5i 5i 5 i 2 13i 13i i 13i 13 1 13 13
25 25 ⱨ 0
5 0 i 13
0ⱨ0
100
Section 1.3 108. a.
x 2 49 0
x 2 6 x 11 = 0
b.
3 i 2 6 3 i 2 11ⱨ 0
7i 2 49 ⱨ 0 49 1 49 ⱨ 0
2
9 6i 3 2i 2 18 6i 2 11ⱨ 0
49 49 ⱨ 0
9 2 1 18 11ⱨ 0
0ⱨ0 b.
9 2 7ⱨ0 0ⱨ0
x 49 0 2
7i 2 49 ⱨ 0 49 1 49 ⱨ 0
111. a bi c di ac adi bci bdi 2
ac ad bc i bd 1
49 49 ⱨ 0
ac bd ad bc i
0ⱨ0
112. a bi a 2 a bi bi 2
x2 4 x 7 = 0
109. a.
2
a 2 2ab i b 2i 2
2 i 3 42 i 3 7 ⱨ 0 2
a 2 2ab i b 2 1
4 4i 3 3i 8 4i 3 7 ⱨ 0 2
a 2 b 2 2ab i
4 3 1 8 7 ⱨ 0 4 3 1ⱨ 0 0ⱨ0
113. The second step does not follow because the multiplication property of radicals can be applied only if the individual radicals are real numbers. Because 9 and 4 are imaginary numbers, the correct logic for simplification would be 9 4 i 9 i 4 i 2 36 1 6 6
x2 4 x 7 = 0
b.
2
2 i 3 42 i 3 7 ⱨ 0 2
4 4i 3 3i 2 8 4i 3 7 ⱨ 0 4 3 1 8 7 ⱨ 0
114. The product a b a b simplifies to
4 3 1ⱨ 0 0ⱨ0
a 2 b 2 . The product a bi a bi
simplifies to a 2 bi , which simplifies to 2
x 6 x 11 = 0 2
110. a.
a2 b2 .
3 i 2 6 3 i 2 11ⱨ 0 2
115. Any real number. For example: 5.
9 6i 3 2i 2 18 6i 2 11ⱨ 0
9 2 1 18 11ⱨ 0
116. Any complex number and its conjugate. For example: 2 5i and 2 5i . In general, for real numbers, a and b, a bi a bi
9 2 7ⱨ0 0ⱨ0
a 2 b 2 , which is a real number. 117. z z a bi a bi a 2 b 2
101
Chapter 1 Equations and Inequalities
b. x 11 x i 11 x i 11
118. z 2 z 2
124. a. x 2 11 x 11 x 11
a bi a bi 2
2
2
2 a 2abi bi a 2 2abi bi 2 2 2 2 2 2 a 2abi b i a 2abi b i 2
2
125.
4ab i 119. a. x 2 9 x 3 x 3 b. x 2 9 x 3i x 3i 126.
120. a. x 100 x 10 x 10 2
b. x 2 100 x 10i x 10i 121. a. x 2 64 x 8 x 8 127.
b. x 2 64 x 8i x 8i 122. a. x 2 25 x 5 x 5 b. x 2 25 x 5i x 5i
b. x 3 x i 3 x i 3
128.
123. a. x 2 3 x 3 x 3 2
Section 1.4 Quadratic Equations 9 27 9 3i 3 9 3i 3 3 3 i 6 6 6 6 2 2
1. 5t 2 7t 6 5t 3 t 2
6.
2. 4 y 2 y 5 4 y 5 y 1
7. quadratic
8. linear
9. a; b
10. 5 x 1 ; x 4
11. k
12. 100
3. x 2 14 x 49 x 7 x 7 x 7 4. z 2 16 z 64 z 8 z 8 z 8
5.
2
2
13. x
8 44 8 2i 11 8 2i 11 11 2 i 4 4 4 4 2
b b 2 4ac 2a
14. b 2 4ac
102
Section 1.4 n 2 5n 24
15.
40 p 2 90 0
19.
n 2 5n 24 0
10 4 p 2 9 0
n 8 n 3 0
10 2 p 3 2 p 3 0
n8 0
or
n 8
n3 0 n3
8, 3
or 2 p 3 0 2 p 3
3 2
p
p y 2 18 7 y
16.
y 7 y 18 0
y 9 y 2 0 y90
or
y 9
20.
y20 y2
32n 2 162 0 2 16n 2 81 0
2 4n 9 4n 9 0
9, 2
4n 9 0
8t t 3 2t 5
n
9 4
n
or
x40
2
9 9 , 4 4
2t 5 4t 1 0 or
4t 1 0
2t 5
4t 1
5 2
t
t
3x 2 12 x
21.
1 4
3x 2 12 x 0 3 x x 4 0
5 1 , 2 4 18.
3x 0 x0
x4
0, 4
6m m 4 m 15 6m 2 24m m 15
z 2 25 z
22.
6m 2 23m 15 0
z 2 25 z 0
6m 5 m 3 0 6m 5 0 or 6m 5 5 m 6 5 , 3 6
4n 9 0 4n 9
8t 24t 2t 5 8t 22t 5 0
or
4n 9
2
2t 5 0
3 2
3 3 , 2 2
2
17.
2p 3 0 2p 3
z z 25 0
m3 0
z0
m 3
or z 25 0 z 25
0, 25 23. x 2 81
x 81 9 9, 9
103
9 4
Chapter 1 Equations and Inequalities 31. w 5 9
24. w2 121
2
w 121 11 11, 11
w5 9
25. 5 y 2 35 0
5 y 2 35 y2 7
8, 2
y 7
7, 7
w 5 9 w 53 w 5 3 or w8
32. c 3 49 2
c 3 49
26. 6v 2 30 0
c 3 49 c 3 7 c 3 7 or c 10 10, 4
6v 2 30 v2 5 v 5
5, 5
4u 2 64 u 2 16
t
u 16 4i 4 i , 4 i
1 17 2 4 t
28. 8 p 2 72 0
1 17 1 i 17 2 4 2 2
1 17 i 2 2 1 17 i 2 2
8 p 72 2
p 2 9 p 9 3i
3i, 3i
2
1 47 34. a 3 9
29. k 2 28 2
a
k 2 28
c 37 c 4
2
17 1 33. t 2 4
27. 4u 2 64 0
k 2 28 2 2 7
2 2 7
w 53 w2
1 47 3 9 1 47 1 i 47 a 3 9 3 3
1 47 i 3 3 47 1 i 3 3
30. z 11 40 2
z 11 40 z 11 40 11 2 10
11 2 10
104
Section 1.4 1 35. x 14 x n x 14 x 14 2 2
2
1 40. v 11v n v 11v 11 2 2
2
x 2 14 x 7
2
x 2 14 x 49 x 7 n 49; x 7
11 v 2 11v 2
2
2
v 2 11v
1 36. y 22 y n y 22 y 22 2 2
121 11 n ; v 4 2
2
2
y 2 22 y 11
y 22 y 121 y 11
1 37. p 2 26 p n p 2 26 p 26 2 p 2 26 p 13
1 38. u 2 4u n u 2 4u 4 2
1 1 n ;m 81 9
2
2
2 2 1 2 42. k 2 k n k 2 k 5 5 2 5
2
2 1 k2 k 5 5
2
1 39. w 3w n w 3w 3 2
1 1 n ;k 25 5
2
2
3 w 2 3w 2
43.
2
9 3 w 2 3w w 4 2
2
2
2 1 1 k2 k k 5 25 5
2
2
9 3 n ;w 4 2
2
2 1 1 m2 m m 9 81 9
2
u 2 4u 4 u 2
2
2
2
u 2 4u 2
2
2
2 1 m2 m 9 9
2
p 2 26 p 169 p 13
n 4; u 2
121 11 v 4 2
1 2 2 2 41. m 2 m n m 2 m 9 9 2 9
2
2
n 169; p 13
2
2
2
n 121; y 11
2
2
2
y 2 22 y 4 0 y 2 22 y 4 2
1 1 y 2 22 y 22 4 22 2 2 2 y 22 y 121 4 121
2
2
y 112 125 y 11 125
11 5 5
105
y 11 5 5
2
2
2
Chapter 1 Equations and Inequalities x 2 14 x 3 0 x 2 14 x 3
44.
2
1 1 x 2 14 x 14 3 14 2 2 x 2 14 x 49 3 49
48. 2
x 72 52
7 2 13
x 7 52 x 7 2 13
5 i 10
t 2 8t 24
45.
2
1 1 t 8t 8 24 8 2 2 2 t 8t 16 24 16 t 4 2 8 t 4 8 t 4 2i 2
2m 2 20m 70 2m 2 20m 70 2 2 2 m 2 10m 35 2 2 1 1 m 2 10m 10 35 10 2 2 2 m 10m 25 35 25 m 5 2 10 m 5 10 m 5 i 10
2
2
2 x x 3 4 x
49.
2x2 6x 4 x 2 x2 7 x 4 2 x2 7 x 4 2 2 2 7 x2 x 2 2
4 2i 2 46.
2
1 7 1 7 7 x x 2 2 2 2 2 2
p 2 24 p 156 2 2 1 1 2 p 24 p 24 156 24 2 2 2 p 24 p 144 156 144 p 122 12 p 12 12 p 12 2i 3
2
x2
2
7 81 x 4 16 x
12 2i 3 47.
7 49 32 49 x 2 16 16 16
4 z 2 24 z 160 4 z 2 24 z 160 4 4 4 z 2 6 z 40 2 2 1 1 2 z 6 z 6 40 6 2 2 z 2 6 z 9 40 9 z 32 31 z 3 31 z 3 i 31
7 9 4 4 16 x 4 4 1 4, 2 x
3 i 31
106
or
7 81 4 16 7 9 x 4 4 7 9 x 4 4 2 1 x 4 2
2
Section 1.4 5c c 2 6 3c
50.
3 14 y 2 2
5c 10c 6 3c 2
14 3 2 2
5c 2 13c 6 5c 2 13c 6 5 5 5 13 6 c2 c 5 5
52.
2
13 1 13 6 1 13 c c 5 5 2 5 2 5
2
2
c2
13 169 120 169 c 5 100 100 100
2 x 2 14 x 5 0 0 2 x 2 14 x 5 2 2 2 2 5 x2 7 x 2 2
5 1 1 x 2 7 x 7 7 2 2 2 49 10 49 x2 7 x 4 4 4
2
289 13 c 10 100 13 289 10 100 13 17 c 10 10 30 4 2 c 3 or c 10 10 5 2 3, 5
2
7 59 x 2 4
c
4 y 2 12 y 5 0
51.
2
7 59 2 4
x
7 59 2 2 7 59 x 2 2
59 7 2 2
4 y 2 12 y 5 0 4 4 4 4 5 y2 3y 4 5 1 1 y 3 y 3 3 4 2 2 9 5 9 y2 3y 4 4 4
x
2
53. False
54. False
55. True
56. True
2
57. x 2 3 x 7 0
a 1, b 3, c 7
2
x
3 7 y 2 2
3 7 y 2 2 y
3 7 2 2 2 2
y
3 14 2 2
b b 2 4ac 2a 3
32 4 1 7 2 1
3 9 28 3 37 2 2 3 37 2
107
2
Chapter 1 Equations and Inequalities 58. x 2 5 x 9 0
t
a 1, b 5, c 9 b b 2 4ac x 2a
5
52 4 1 9 2 1
b b 2 4ac 2a 6
6 36 40 6 4 6 2i 3 i 2 2 2 3 i
5 25 36 5 61 2 2 5 61 2
m m 10 34
62.
m 2 10m 34 0 a 1, b 10, c 34
y 4 y 6 2
59.
m
y2 4 y 6 0
a 1, b 4, c 6 y
b b 2 4ac 2a 4
5c 2 7c 3 0
z 8 z 19
a 5, b 7, c 3
z 8 z 19 0 2
a 1, b 8, c 19
c
b b 2 4ac 2a 8
8 4 119 8 64 76 2 1 2 2
7
72 4 5 3 2 5
7 49 60 7 11 7 i 11 10 10 10 7 i 11 10
8 12 8 2i 3 4 i 3 2 2 4 i 3
64.
t t 6 10
61.
b b 2 4ac 2a
102 4 1 34 2 1
7c 3 5c 2
63.
2
z
10
10 100 136 10 36 2 2 10 6i 5 3i 2 5 3i
4 8 4 2i 2 2 i 2 2 2 2 i 2
60.
b b 2 4ac 2a
4 2 4 1 6 4 16 24 2 1 2
62 4 110 2 1
5d 2 6d 2 6 d 2 5d 2 0
t 6t 10 0 2
a 6, b 5, c 2
a 1, b 6, c 10
108
Section 1.4 b b 2 4ac 2a
d
5
c
52 4 6 2 2 6
5 25 48 5 23 5 i 23 12 12 12 5 i 23 12
6 x 5 x 3 2 x 7 x 5 x 12 6 x 18 x 5 x 15 14 x 2 10 x x 12 6 x 2 13x 15 14 x 2 9 x 12 20 x 2 4 x 3 0 a 20, b 4, c 3
67. 9 x 2 49 0
b b 2 4ac x 2a
a 9, b 0, c 49
4 2 4 20 3 2 20
x
4 16 240 4 256 4 16 40 40 40 4 16 4 16 x or x 40 40 20 1 3 12 x x 40 2 40 10 3 1 , 2 10
0
02 4 9 49 1764 2 9 18
7 42i i 18 3 7 i 3 68. 121x 2 4 0
a 121, b 0, c 4 x
10c 15c 14c 21 2c 30c 35 2
b b 2 4ac 2a
5c 7 2c 3 2c c 15 35
66.
292 4 1214 2 12
2
4
29
29 841 672 24 29 169 29 13 24 24 29 13 29 13 c or c 24 24 16 2 42 7 c c 24 3 24 4 2 7 , 3 4
65.
b b 2 4ac 2a
2
10c 2 c 21 2c 2 30c 35
12c 2 29c 14 0 a 12, b 29, c 14
b b 2 4ac 2a 0
0 2 4 121 4 1936 2 121 242
2 44i i 242 11 2 i 11
109
Chapter 1 Equations and Inequalities
69.
1 2 2 5 x x 2 7 14 2 1 5 14 x 2 14 x 2 14 7
y
7 x 4 5x
b b 2 4ac 2a 20
202 4 4 25 2 4
2
20 400 400 20 0 20 5 8 8 8 2 5 2
7 x 5x 4 0
2
a 7, b 5, c 4 x
b b 2 4ac 2a 5
52 4 7 4 2 7
72.
9n 42n 49 2
5 25 112 5 137 14 14 5 137 14
9n 42n 49 0 2
a 9, b 42, c 49 n
1 2 7 3 x x 3 6 2 1 7 3 6 x 2 6 x 3 2 6
2x 7 9x 2x 9x 7 0 a 2, b 9, c 7 b b 2 4ac x 2a
2 y 4 4 2 y 2 2
9 2 4 2 7 2 2
9 81 56 9 137 4 4 9 137 4
74. Linear 3z 9 0
0.4 y 2 y 2.5 2
42 2 4 9 49 2 9
73. Linear 2y 4 0
71.
42
42 1764 1764 42 0 42 7 18 18 18 3 7 3
2
b b 2 4ac 2a
2
9
100 0.09n 2 100 0.42n 0.49
70.
0.09n 2 0.42n 0.49
3
10 0.4 y 2 10 2 y 2.5 4 y 2 20 y 25 4 y 2 20 y 25 0 a 4, b 20, c 25
110
3z 9 z3
Section 1.4 81. 3x 4 0
75. Quadratic 2 y2 4 y 0
2
3x 4 0 3x 4 4 x 3 4 3
2 y2 4 y 0 2 2 2 y2 2 y 0 y y 2 0 y0
or
y20 y 2
82. 2 x 1 0
0, 2
2
2x 1 0 2 x 1 1 x 2 1 2
76. Quadratic 3z 2 9 z 0
3z 2 9 z 0 3 3 3 z 2 3z 0 z z 3 0 z0
0, 3
or
z 3 0 z3
m 2 4m 2
83.
2
1 1 m 4m 4 2 4 2 2 2 m 4m 4 2 4 2
77. Linear 5 x x 6 5 x 2 27 x 3
m 2 2 2 m2 2
5 x 2 30 x 5 x 2 27 x 3 3x 3 x 1 1
m 2 2
2 2 84.
78. Linear 3 x x 4 3x 2 11x 4
n 2 8n 3 2
1 1 n 2 8n 8 3 8 2 2 2 n 8n 16 3 16
3x 2 12 x 3x 2 11x 4 x 4 x 4 4
n 42 13 n 4 13
4 13
79. Neither 80. Neither
111
n 4 13
2
2
Chapter 1 Equations and Inequalities x2 4x 5x 1 6 3 x2 4 x 5x 6 1 6 3 6
85.
89.
3 2 1 1 x x 5 10 2 1 3 1 10 x 2 x 10 5 2 10
x 2 4 x 10 x 6
6 x2 x 5
x 2 14 x 6
6 x2 x 5 0
2
1 1 x 14 x 14 6 14 2 2 2 x 14 x 49 6 49
x 1 6 x 5 0
2
2
x 1 0 x 1
or
x 72 55 x 7 55
5 1, 6
x 7 55
7 55
90.
m 2m 9 m 3 7 14 2 m 2 2m 9m 3 14 14 2 7 14 2
86.
1 2 11 1 x x 12 24 2 1 2 11 1 24 x x 24 12 2 24
2 x 2 11x 12 2 x 2 11x 12 0
2m 2 4m 9m 21
2 x 3 x 4 0
2m 2 5m 21 0 a 2, b 5, c 21
2 x 3 0 or 2x 3 3 x 2
b b 2 4ac m 2a
5
6x 5 0 6 x 5 5 x 6
52 4 2 21 2 2
x40 x4
3 , 4 2
5 25 168 5 193 4 4 5 193 4
91. x 2 5 x 5 x x 1 4 x 2 1
x2 5x 5x2 5x 4 x2 1 x2 5x x2 5x 1 0 1
87. 2 x 4 x 2 x x 2 8
2 x 8 x2 x2 2 x 8 00
92. p 2 4 p 4 p p 1 3 p 2 2
p2 4 p 4 p2 4 p 3 p2 2
88. 3 y 5 y 2 y y 3 15
p2 4 p p2 4 p 2 02
3 y 15 y 2 y 2 3 y 15 00
112
Section 1.4 93.
2 y 7 y 1 2 y 2 11
97. x 2 5 0
2 y 2 2 y 7 y 7 2 y 2 11
x2 5
2 y 2 9 y 7 2 y 2 11 9 y 18 y 2
x 4 5
5 4
2 94.
98. y 2 11 0
y 2 11
3z 8 z 2 3z 10 2
y 4 11
3z 6 z 8 z 16 3 z 2 10 2
11 4
3z 2 2 z 16 3 z 2 10 2 z 26 z 13 13
99. a. 3x 2 4 x 6 0
b 2 4ac 4 4 3 6 16 72 56 2
b. 56 0 ; there are two imaginary solutions.
95. 7 d 2 5 0
7 d 2 5 5 d2 7
100. a. 5 x 2 2 x 4 0
b 2 4ac 2 4 5 4 4 80 76 2
d
b. 76 0 ; there are two imaginary solutions.
5 5 5 i i 7 7 7
5 7 7 7
i
35 i 7
101. a.
2 w2 8w 3 0
35 i 7
b 2 4ac 8 4 2 3 64 24 40 2
b. 40 0 ; there are two real solutions.
96. 11t 2 3 0 11t 2 3
102. a.
t
6d 2 9d 2 6d 2 9d 2 0
3 t 11 2
2w2 8w 3
6d 2 9d 2 0 2 b 2 4ac 9 4 6 2 81 48 33
3 3 3 i i 11 11 11
3 11
11 11 33 i 11
i
b. 33 0 ;, there are two real solutions.
33 i 11
103. a.
3x x 4 x 4 3 x 2 12 x x 4 3x 2 13x 4 0 2 b 2 4ac 13 4 34 169 48 121
b. 121 0 ; there are two real solutions.
113
Chapter 1 Equations and Inequalities 2 x x 2 x 3
104. a.
109.
2 x2 4 x x 3 2 x2 5x 3 0 2 b 2 4ac 5 4 2 3 25 24 49
2 s gt 2
b. 49 0 ; there are two real solutions.
1.4m 0.1 4.9m 2
105. a.
10 1.4m 0.1 10 4.9m 2
1 2 gt 2 1 2 s 2 gt 2 2 s
2s gt 2 g g 2s 2 t g
14m 1 49m 2
t
49m 2 14m 1 0 b 2 4ac 14 4 491 196 196
2s or t g
2 sg g
2
0
110.
b. The discriminant is 0; there is one real solution.
3.6n 0.4 8.1n 2
106. a.
10 3.6n 0.4 10 8.1n 2
1 2 xh 3 1 3 V 3 x 2 h 3 V
3V x 2 h
3V x 2 h h h 3V x2 h 3V 3Vh x or x h h
36n 4 81n 2 81n 2 36n 4 0 b 2 4ac 36 4 81 4 1296 1296 2
0 b. The discriminant is 0; there is one real solution.
111. a 2 b 2 c 2
107. A r 2
a2 c2 b2
r2
a c2 b2
A A
r
r 108.
112. a 2 b 2 c 2 d 2
2
c2 d 2 a 2 b2 A
or r
A
c d 2 a2 b2
113.
A r h 2
A r 2h h h A r2 h A A h r or r h h
114
L c 2 I 2 Rt L c 2 I 2 Rt 2 c 2 Rt c Rt L I2 c 2 Rt L LRt 1 L I 2 or c Rt c Rt cRt
Section 1.4 114.
I cN 2 r 2 s I cN 2 r 2 s cr 2 s cr 2 s I N2 2 cr s I N cr 2 s N
at 2 2v0t 2s 0 a a, b 2v0 , c 2s
Ics 1 I or N r cs crs
t
kw2 cw r
115.
t
kw2 cw r 0 a k , b c, c r w w
t
b b 2 4ac 2a c
1 s v0t at 2 2 1 2 s 2 v0t at 2 2 2s 2v0t at 2
117.
t
c 2 4 k r 2k
t
c c 4kr w 2k 2
y y
2v0 4v0 2 8as 2a 2v0 2 v0 2 2as 2a v0 v0 2 2as a S 2 rh r 2 h
r
b b 2 4ac 2a m
2v0 2 4 a 2s 2 a
hr 2 2 hr S 0 a h, b 2 h, c S
dy 2 my p 0 a d , b m, c p y
2v0
118.
dy 2 my p
116.
b b 2 4ac 2a
r
m 4 d p 2 d 2
m m 2 4dp 2k
115
b b 2 4ac 2a 2 h
2 h 2 4 h S 2 h
r
2 h 4 2 h 2 4 hS 2 h
r
2 h 2 2 h 2 hS 2 h
r
h 2 h 2 hS h
Chapter 1 Equations and Inequalities
119.
126. 3a 2 2ab b 2 0
1 0 C 1 C LI 2 RI C 0 C LI 2 RI
3a b a b 0 3a b 0 3a b b a 3
CLI 2 CRI 1 0 a CL, b CR, c 1 I I I
b b 2 4ac 2a CR
x 4 x 2 0
127.
CR 4 CL 1 2 CL
x 2x 4x 8 0
2
2
x2 2 x 8 0
CR CR 2 4CL 2CL
x 7 x 1 0
128.
x x 7x 7 0 2
x2 6 x 7 0
A r 2 rs
120.
r 2 sr A 0 a , b s, c A r r r
2 1 x x 0 3 4
129.
b b 2 4ac 2a s
2 1 3 x 4 x 12 0 3 4
s 2 4 A 2
3x 2 4 x 1 0 12 x 2 3 x 8 x 2 0
s 2 s 2 4 A 2
12 x 2 11x 2 0 3 1 x x 0 5 7
130.
121. The right side of the equation is not equal to zero.
3 1 5 x 7 x 35 0 5 7
122. Given ax bx c 0 , where a 0 , the discriminant is b 2 4ac . The discriminant indicates the number and type of solutions to the equation. 2
5 x 3 7 x 1 0 35 x 2 5 x 21x 3 0 35 x 2 26 x 3 0
123. If the discriminant is negative, then the equation has two solutions that are imaginary numbers.
131. x 5 x 5 0
x2 5 0 2
124. To derive the quadratic formula, solve the equation ax 2 bx c 0 for x by completing the square. 125.
x 2 xy 2 y 2 0 x 2 y x y 0 x 2 y 0 or x 2y
or
x2 5 0
132. x 2 x 2 0
x2 2 0 2
x y 0 x y
x2 2 0
116
ab 0 a b
Section 1.4 133. x 2i x 2i 0
3 139. x1 x2 2 5 3 , which is . 1 10 x1 x2 2 5 10 , which is . 1
x 2 2 2 0 x2 4 0 134. x 9i x 9i 0
140. x1 x2 1 2i 1 2i 2 , which is
x 9 0 2
2
135. x 1 2i x 1 2i 0
is
x 1 2i x 1 2i 0 x 1 2i x 1 2i 0 x 12 2 2 0
x2 2 x 5 0
x 2 4 x 4 81 0
142. x
x 2 4 x 85 0 b b 4ac b b 4ac 2a 2a
b b 2 4ac b b 2 4ac 2a 2b b 2a a
2
2
b b 2 4ac b b 2 4ac 138. x1 x2 2a 2a
4a
2
143.
2
4ac c 4a 2
52 4 9 7 2 9
b b 2 4ac 2a 7
7 2 4 11 6 2 11
7 49 264 7 313 22 22 7 313 7 313 or 22 22 or 0.4860 1.1224
5
5 25 252 5 277 18 18 5 277 5 277 or 18 18 or 0.6469 1.2024
x 2 9i x 2 9i 0 x 2 9i x 2 9i 0 x 2 2 9 2 0
b b 2 4ac 2a
136. x 2 9i x 2 9i 0
b 2 b2 4ac 2a 2
2
5 . 1
141. x
x2 2 x 1 4 0
b 2 b 2 4ac
1
x1 x2 1 2i 1 2i 1 2 5 , which
x 81 0
2
2
137. x1 x2
2 .
144.
a
117
Chapter 1 Equations and Inequalities Problem Recognition Exercises: Simplifying Expressions Versus Solving Equations 1. a. Expression 2 x 5 3x 1 6 x 2 2 x 15 x 5
b. Equation 2 x 5 3x 1 0
2x 5 0 2x 5 5 x 2
6 x 2 13x 5
3x 1 0 3 x 1 1 x 3
or
5 1 , 2 3
2. a. Expression 5 1 2 5 1 2 5 x7 1 x3 2 2 x 3 x 7 x 4 x 21 x 3 x 7 x 3 x 7 x 3 x 7 x 7 x 3 x 3 x 7
5 x 7 1 x 3 2 5 x 35 x 3 2 4 x 36 x 3 x 7 x 3 x 7 x 3 x 7
b. Equation
5 1 2 2 x 3 x 7 x 4 x 21
x 3 x 7
5 1 2 x 3 x 7 x3 x7 x 3 x 7
5 x 7 1 x 3 2 5 x 35 x 3 2 4 x 38 2 4 x 36 x 9
9 3. a. Equation
4. a. Equation 5 6 3 2 5 y 2 1 7
2 x 3 8 2
5 6 3 2 5 y 10 1 7
2x 3 8 2 x 3 2 2
5 6 3 5 y 12 1 7
2x 3 2
5 6 15 y 36 1 7
x
3 2 2
5 15 y 43 7
3 2 2
5 15 y 43 7 15 y 38 7 15 y 45 y3
b. Expression
2 x 32 8 2 x 2 2 2 x 3 3 2 8 4 x 2 12 x 9 8
3
4 x 2 12 x 1
118
Problem Recognition b. Expression 5 6 3 2 5 y 2 1
b. Equation 3 x 9 20 x
3 x 27 20 x 4 x 7 7 x 4 7 4
5 6 3 2 5 y 10 1 5 6 3 5 y 12 1 5 6 15 y 36 1 5 15 y 43 5 15 y 43 15 y 38
7. a. Equation 35 12 x 0 x 35 x 12 x x 0 x
5. a. Equation x 2 11x 28 0
x 7 x 4 0
7, 4
x7 0 x7
x40 x4
or
35 12 x x 2 0 x 2 12 x 35 0
x 7 x 5 0
b. Equation x 2 11x 28 0
a 1, b 11, c 28
7, 5
b b 2 4ac x 2a
11
11 121 112 11 233 2 2 11 233 2
x50 x 5
8. a. Equation
x 2 2 x2 3 x2 2 x 2 3 x 2 3 x 2 x 2 3 x 2
6. a. Equation 3 x x 9 20 x
3 x 2 x 2 6
3x 2 27 x 20 x
3x 2 x 4 6 5 x 10 x2 ; The value 2 does not check.
3x 2 28 x 20 0
3x 2 x 10 0 or
or
b. Expression 35 35 x x 12 x 12 x x x x x 2 35 12 x x x
11 2 4 1 28 2 1
3x 2 0 3x 2 2 x 3
x70 x 7
x 10 0 x 10
b. Expression x 2 2 x2 2 x2 3 x2 x2 3 2 5 1 ; for x 2 3 3
2 , 10 3
119
Chapter 1 Equations and Inequalities Section 1.5 Applications of Quadratic Equations 1 bh 2
1. A
16. x 1 4. a b c
3. V lwh 5.
15. x 2
2. A r 2 2
2
2
17. a. x x 2 120
A lw
b.
629 2 x 3 x 6.
x 2 2 x 120 x 2 2 x 120 0
A lw
x 10 x 12 0
1 252 x x 2 4 7.
8.
x 10 0 or x 12 0 x 10 x 12 x 2 12 x 2 10 The integers are 10 and 12 or 10 and 12 .
A r2 88 x
x x 2 120
2
A r2 212 x
18. a. x x 2 35
2
b.
1 bh 2 1 50 x x 8 2
x x 2 35 x 2 2 x 35
9. A
x 2 2 x 35 0
x 5 x 7 0 x5 0
or
x70
x5 x 7 x27 x 2 5 The integers are 5 and 7 or 5 and 7 .
1 10. A bh 2 1 220 x x 4 2
19. a. x 2 x 1 113 2
11.
V lwh 1 640 x 8 x 5
12.
b.
2 x 2 2 x 112 0
V lwh
2 x 2 2 x 112 0 2 2 2 2 2 x x 56 0
a2 b2 c2
x 7 x 8 0
x 2 x 2 2 2 x 2 2 14.
2
x 2 x 2 2 x 1 113
312 x 2 x 1 4 13.
x 2 x 1 113
x7 0 x7
a 2 b2 c2
or
x8 0 x 8
x 1 8 x 1 7 The integers are 7 and 8 or 7 and 8 .
x 2 x 3 2 100 2
120
Section 1.5 V lwh
20. a. x 2 x 1 181 2
1728 x x 12 6
x 2 x 1 181 2
b.
1728 6 x 2 12 x
x x 2 x 1 181 2
2
1728 6 x 72 x
2 x 2 2 x 180 0
6 x 2 72 x 1728 0
2 x 2 2 x 180 0 2 2 2 2 2 x x 90 0
6 x 2 12 x 288 0
x 24 x 12 0
x 9 x 10 0 x9 0
x 24 0 x 10
x 1 10 x 1 9 The integers are 9 and 10 or 9 and 10 .
2000 r 2 2000 r2
504 12 x x 1
r
504 12 x 2 12 x 12 x 2 12 x 504 0
25
24. Let r represent the radius (in miles) of the area in which the earthquake could be felt. A r2
x 7 x 6 0 x7
2000
The radius is approximately 25 yd.
12 x 2 x 42 0 x7 0
x 12 0
23. Let r represent the radius (in yards) of the region watered. A r2
21. Let x represent the width of the cargo space. The length is 12 ft and the height is x 1 ft. V lwh
504 12 x 2 x
or
x 24 x 12 x 12 24 12 36 The dimensions of the cardboard sheet are 24 in. by 36 in.
x 10 0
or
x9
2
or
x60
46,000 r 2 46, 000 r2
x 6
x 1 7 1 6 The dimensions of the cargo space are 6 ft by 7 ft by 12 ft.
r
22. Let x represent the width of the cardboard sheet. Then x 12 in. is the length of the cardboard sheet. The width of the box is x 12 in. The length of the box is
46, 000
121
The earthquake could be felt up to 121 mi from the epicenter. 25. Let x represent the height of the triangle (in feet) and x 3 represent the base of the triangle. 1 lw 2 bh A 2
x 12 12 x in. The height of the box is
6 in.
121
Chapter 1 Equations and Inequalities
20 x 2 x 3 x 348
28. a. Let x represent the distance (in feet) from home plate to second base. a 2 b2 c2
1 2
20 x x 3x 348 2
6 2 6 2 c 2
x 2 17 x 348 0
36 36 c 2
x 29 x 12 0 x 29 0
72 c 2
x 12 0
or
c 72 6 2 in.
x 29 x 12 x 3 12 3 9 The base is 9 ft and the height is 12 ft.
6 2 6 d 2
26. Let x represent the height of the triangle and 3x represent the base of the triangle. 1 lw bh A 2 1 3x x 2 3x x 336 2 3 3x 2 6 x x 2 336 2 6 x 2 12 x 3x 2 672
d 108 6 3 in. 29. a. Let x represent the length (in feet) of the middle leg. a2 b2 c2
x 2 x 2 2 x 2 2 x2 x2 4 x 4 x2 4 x 4 2 x2 4 x 4 x2 4 x 4
x 8 0
9 x 84
x2 8x 0
x8
x x 8 0
84 x 9
x0 x282 6
x 2 8 2 10 The length is 24 ft and the height is 10 ft.
x 2 8 2 10 The lengths of the sides of the lower triangle are 6 ft, 8 ft, and 10 ft.
27. Let x represent the distance (in feet) from home plate to second base. a 2 b2 c2
90 90 x 2
or x 8 0 x8
3x 3 8 24
2
2
108 d 2
9 x 84 x 8 0 or
2
72 36 d 2
9 x 2 12 x 672 0 9 x 84 0
a 2 b2 c2
b.
b. A AT AB
1 1 1 bT hT bB hB bT hT bB hB 2 2 2 1 1 10 4 8 6 40 48 2 2 1 88 44 2 The total area is 44 ft2.
2
8100 8100 x 2 16, 200 x 2 x 16, 200 90 2 127.3 The distance is 90 2 ft or approximately 127.3 ft.
122
Section 1.5 b. 2.91 326 949
30. a. Let x represent the length (in feet) of the shorter leg. a 2 b2 c2
1.94 326 632 Using the rounded values from part (a), the screen is approximately 949 pixels by 632 pixels.
x 2 x 7 2 2 x 1 2 x 2 x 2 14 x 49 4 x 2 4 x 1 2 x 2 14 x 49 4 x 2 4 x 1
32. Let x represent the width (in inches) of the display. Then 1.6x is the length of the display. a2 b2 c2
2 x 2 10 x 48 0 2 x 2 10 x 48 0 2 2 2 2 x 2 5 x 24 0
x 2 1.6 x 2 152 x 2 2.56 x 2 225
x 3 x 8 0 x3 0 x 3 x 7 8 7 15
or
3.56 x 2 225 225 x2 3.56 225 7.95 x 3.56 1.6 x 1.6 7.95 12.72 The length is approximately 12.72 in., and the width is approximately 7.95 in.
x 8 0 x8
2 x 1 2 8 1 17 The lengths of the sides of the triangle on the left are 8 ft, 15 ft, and 17 ft. b. Let x represent the length (in feet) of the unlabeled side. a 2 b2 c2
x 9 15 2
2
33. Let n represent the number of players. 1 N n n 1 2 1 28 n n 1 2 56 n 2 n
2
x 2 81 225 x 2 144 x 144 12 The lengths of the sides of the triangle on the right are 9 ft, 12 ft, and 15 ft.
n 2 n 56 0 n 8 n 7 0 n8 0
31. a. Let x represent the width (in inches) of the cell phone. Then 1.5x is the length of the phone. a2 b2 c2
or
n8 There were 8 players.
x 2 1.5 x 2 3.5 2
34.
x 2 2.25 x 2 12.25 3.25 x 2 12.25 12.25 x2 3.25 12.25 1.94 x 3.25 1.5 x 1.5 1.94 2.91 The length is approximately 2.91 in. and the width is approximately 1.94 in.
n 7
1 n n 1 2 1 171 n n 1 2 342 n 2 n S
n 2 n 342 0 n 18 n 19 0 n 18 0 n 18 The value of n is 18.
123
n7 0
or
n 19 0 n 19
Chapter 1 Equations and Inequalities 35. Let t represent the time(s) at which the population was 600,000. P 1718t 2 82,000t 10,000
600,000 1718t 2 82,000t 10,000 0 1718t 2 82,000t 590, 000 0 1718t 2 82,000t 590, 000 b b 2 4ac 82, 000 2a
t
82, 000 2 4 1718 590, 000 82, 000 2, 669,520,000 2 1718 3436
9 or 39 There were 600,000 organisms approximately 9 hr and 39 hr after the culture was started. 38. a. c 219 x 2 26.7 x 1.64
36. Let x represent the speed of the vehicle in mph. m 0.04 x 2 3.6 x 49
219 0.22 26.7 0.22 1.64 2
219 0.0484 5.874 1.64 10.5996 5.874 1.64 6.4 ng/mL
30 0.04 x 2 3.6 x 49 0 0.04 x 2 3.6 x 79
b. c 219 x 2 26.7 x 1.64 3 219 x 2 26.7 x 1.64 0 219 x 2 26.7 x 1.36
b b 2 4ac x 2a
3.6
3.62 4 0.04 79 2 0.04
x
3.6 0.32 38 or x 52 0.08 The gas mileage will be 30 mpg for speeds of 38 mph and 52 mph.
b b 2 4ac 2a
26.7
26.72 4 219 1.36 2 219
26.7 1904.25 0.16 or 0.04 438 16 %
37. a. d 0.05v 2 2.2v
d 0.05 50 2.2 50 2
1 39. a. s gt 2 v0t s0 2 1 s 32 t 2 16 t 0 2 s 16t 2 16t
0.05 2500 110 125 110 235 ft b.
d 0.05v 2 2.2v 330 0.05v 2 2.2v d 0.05v 2 2.2v 330 v
b. 4 16t 2 16t 16t 2 16t 4 0
b b 4ac 2a 2
2.2
4 4t 2 4t 1 0
2.2 4 0.05 330 2 0.05 2
4 2t 1 0 2t 1 0 2t 1 1 t 2 It would take Michael Jordan 0.5 sec to reach his maximum height of 4 ft. 2
2.2 70.84 62 or 106 0.1 The car can travel at 62 mph and stop in time.
124
Section 1.5 18 30 1.3 or 2.4 9.8 George will catch the bread 1.3 sec after release.
1 40. a. s gt 2 v0t s0 2 1 s 32 t 2 8 5 t 0 16t 2 8 5t 2
t
b. 5 16t 2 8 5t
L L W W L L L 1 1 L L 1 L L L L
43. a.
16t 8 5t 5 0 2
4t 5 0 2
4t 5 0 4t 5
L2 L 1
5 0.56 4 It would take 0.56 sec to reach a height of 5 ft. t
L2 L 1 0 L
1 41. a. s gt 2 v0t s0 2 1 s 32 t 2 75 t 4 16t 2 75t 4 2
L L
b. 80 16t 2 75t 4 0 16t 2 75t 76
t t
b.
b b 4ac 2a 2
75
752 4 16 76 2 16
b b 2 4ac 2a 1
1 5 0.62 or 1.62 2
1.62 l 1 9 l 9 1.62 14.6 ft
44. Since the length of the sides of the square is 18 in., the length of the sides of the right 18 x triangles shown in the figure is . 2 Use the Pythagorean theorem. a2 b2 c 2
75 761 1.5 or 3.2 32 The ball will be at an 80-ft height 1.5 sec and 3.2 sec after being kicked. t
18 x 18 x 2 x 2 2 2
1 42. a. s gt 2 v0t s0 2 1 s 9.8 t 2 18 t 1 4.9t 2 18t 1 2
t t
2
18 x 2 x 2
2
2 18 x 2 2 x 2
b b 2 4ac 2a 18
2
18 x x2 2 2
b. 16 4.9t 18t 10 4.9t 18t 15 0 4.9t 2 18t 15 2
1 2 4 1 1 2 1
324 36 x x 2 2 x 2 x 2 36 x 324 0
18 4 4.915 2 4.9 2
125
Chapter 1 Equations and Inequalities
x
46. Let t represent the time when the ships are 100 nautical miles apart. Then the distance traveled by the first ship is 10t and the distance traveled by the second ship is 15 t 2 .
b b 2 4ac 2a 36
362 4 1 324 2 1
36 2592 7.46 or 43.46 2 The sides are approximately 7.46 in.
a2 b2 c2
10t 2 15 t 2 100 2 2 100t 2 15t 30 10, 000 2
45. a. 4 x 6 y 160 6 y 160 4 x 160 4 x 80 2 x or y y 6 3
100t 2 225t 2 900t 900 10, 000 325t 2 900t 9100 0 13t 2 36t 364 0
b. A lw
t
80 2 x A x 3
80 2 x A x 3
c.
36
36 2 4 13 364 2 13
36 20, 224 6.85 or 4.09 26 85 60 min 6.85 hours 6 hours min 100 1 hr
80 2 x 250 x 3 750 80 x 2 x 2
6 hours 51 min The ships will be 100 nautical miles apart at approximately 6:51 PM.
375 40 x x 2 x 2 40 x 375 0 x 25 x 15 0 x 25 0 or x 25 80 2 x y 3 80 2 25 30 y 10 3 3 or
b b 2 4ac 2a
x 15 0 x 15
80 2 15 50 3 3 Each pen can be 25 yd by 10 yd, or it can 50 yd. be 15 yd by 3 y
126
Section 1.6 Section 1.6 More Equations and Applications 1. x 3 27 x 3 3
3
12.
x 3 x 2 3x 9
98t 3 49t 2 8t 4 0 49t 2 2t 1 4 2t 1 0
49t 4 2t 1 0 2
2. 2 x 3 5 x 2 8 x 20 2 x3 8 x 5 x 2 20
2 x 5 x 4
2x x 4 5 x 4 2
2
7t 2 7t 2 2t 1 0
7t 2 0
2
t
2 x 5 x 2 x 2 3.
13.
2x 5 0 2 x 5 5 x 2
5. 27
23
3
3 2 3
3 23
3
34
2 x 4 x 4 0 2 x 2 x 2 x 4 0 x 2 0 or x 2 0 x2
x 4 x 2i
2i, 2
3 9
4 3 4
14.
5m 4 5 0
5 m 1 m 1 0 5 m 1 m 1 m 1 0 5 m4 1 0
3
n m
8.
9. quadratic; m1 3
10. 4 x 2 1
2 2
m 1 0 or m 1 0 or m 2 1 0 m 1
75 y 3 100 y 2 3 y 4 0 75 y 3 3 y 100 y 2 4 0
x 2 4
2
7. radical
or x 2 4 0
x 2
2
11.
2 2
3 3 27
6. 813 4 34
2t 1 0 1 t 2
2 x 4 32 0
2
x5 0 x 5
and
or
2 x 4 16 0
2 x 15 2 x 15 4. 2 x 2 x 35 x 7 x 5 x7 0 x7
2 7
7t 2 0 2 t 7
2 1 , 7 2
2x 5 2x 5 2 4 x 25 2 x 5 2 x 5 2 x 5 0 and 2x 5 5 x 2
or
i, 1
3 y 25 y 2 1 4 25 y 2 1 0
25 y 1 3 y 4 0 2
5 y 1 5 y 1 3 y 4 0 5 y 1 0 or 5 y 1 0 or 3 y 4 0 1 1 4 y y y 5 5 3 1 4 , 5 3 127
m 1
m 2 1 m 1 m i
Chapter 1 Equations and Inequalities 2 x 4 128 x
15.
2 x 128 x 0 4
2 x x 4 x 4 x 16 0 2 x x3 64 0
2
2 x 0 or x 4 0 or x 0 or x 4 or x
4
42 4 116 2 1
or
n 5
2 y 4 y 18 y 2 2 y 4 5 y 2 18 0 2 y2 9 y2 2 0
2
2 y2 9 0 2 y2 9 9 y2 2
2
10 x 2 0 or x 5 0 or x 2 5 x 25 0 x0 x 5 or
5 4 1 25 2 1 2
5 75 5 5i 3 2 2 5 5i 3 0, 5, 2
3x 5 2 x 2 14 x 2 x 2 x 4 x 2x 8 3x 5 2 x 2 14 x x 2 x 4 x 2 x 4
19.
x 2 x 4
2 x 2 14 x 3x 5 2 4 x x x 2 x 4 x 2 x 4
3x x 4 5 x 2 2 x 2 14 x 3x 2 12 x 5 x 10 2 x 2 14 x x 2 3 x 10 0 x 2 x 5 0 x 2
or
x5
or
9 3 2 2 2 3 2 , i 2 2 y
5 ; The value 2 does not check.
128
n i 5
4
10 x x 5 x 5 x 25 0
n2 5 0 n 2 5
2 y 2 y 2 2 18 y 2
18.
5
4 2 3 3 3 2 3 , i 5 3
10 x5 1250 x 2 5 10 x 1250 x 2 0 10 x 2 x3 125 0
x
2
n
0, 4, 2 2i 3
2
3n 9n 20 2n 2 3n 4 11n 2 20 0 3n 2 4 n 2 5 0 4
3n 2 4 0 3n 2 4 4 n2 3
x 2 4 x 16 0
4 48 4 4i 3 2 2i 3 2 2
16.
3n 2 n 2 3 20 2n 2
17.
y2 2 0 y 2 2 y 2 y i 2
Section 1.6 4c 1 3c 2 3 c 5 c 1 c 2 4c 5 4c 1 3c 2 3 c 5 c 1 c 5 c 1
20.
c 5 c 1
3c 2 3 4c 1 c 5 c 1 c 5 c 1 c 5 c 1
4c c 1 1 c 5 3c 2 3 4c 2 4c c 5 3c 2 3 c 2 3c 2 0
c 1 c 2 0 c 1 21.
or
c 2
2 ; The value 1 does not check.
m 2 1 m3 2m 1 m 2 1 2m 1 m 3 2m 1 m 3 m 3 2m 1 m m 3 1 2m 1 m 3 2 2m 1 m 2 3m 2m 2 6m m 3 4m 2 3m 2 12m 5 0 m
12
122 4 3 5 12 204 12 2 51 6 51 2 3 6 6 3
6 51 3 22.
n 3 2 n3 2n 1 n 3 2 n 3 2n 1 n 3 2n 1 2n 1 n3 n 2n 1 2 n 3 2n 1 3 n 3
2n 2 n 2 2n 2 7 n 3 3n 9 2n 2 n 4n 2 14n 6 3n 9 6n 2 18n 15 0 2n 2 6 n 5 0 n
6
6 2 4 2 5 6 4 6 2i 3 i 2 2 4 4 2
3 i 2
129
Chapter 1 Equations and Inequalities 2
23.
3 5 y y2
25.
5 3 y2 2 y2 2 y y
18 6 m m 3 2 m m 3 m 3 m m 3 18 2m m 3 6m
2 y2 3y 5 2 y2 3y 5 0
2 y 5 y 1 0 2y 5 0 2y 5 5 y 2
18 6 2 m 3m m3 18 6 2 m m 3 m3 2
or
18 2m 2 6m 6m
y 1 0 y 1
2m 2 12m 18 0 m 2 6m 9 0
m 3 2 0
5 ; 1 2
24.
m3 0 m3
; The value 3 does not check.
20 3 2 z z 20 3 z2 7 z2 2 z z 7
26.
7 z 2 20 z 3 7 z 2 20 z 3 0
48 12 3 m m 4 m m 4 m 4 m m 4 48 3m m 4 12m
7 z 1 z 3 0 7z 1 0 7z 1 1 z 7
or
48 12 3 m 4m m4 48 12 3 m m 4 m4 2
z3 0 z 3
48 3m 2 12m 12m 3m 2 24m 48 0
1 ; 3 7
m 2 8m 16 0
m 4 2 0 m4 0 m4
; The value 4 does not check.
130
Section 1.6 27. Let x represent the speed of the boat in still water. Distance (mi)
Rate (mph)
With current
72
x2
Against current
72
x2
Time (hr) 72 x2 72 x2
72 72 9 x2 x2 72 72 x 2 x 2 x 2 x 2 9 x 2 x 2
72 x 2 72 x 2 9 x 2 4
72 x 144 72 x 144 9 x 36 2
288 9 x 2 36 324 9 x 2 36 x 2 x 36 6 or 6 Jesse travels 6 km/hr in still water. 28. Let x represent the speed of the plane in still air. Distance (mi)
Rate (mph)
With wind
800
x 40
Against wind
800
x 40
Time (hr) 800 x 40 800 x 40
800 800 0.5 x 40 x 40 800 800 x 40 x 40 x 40 x 40 0.5 x 40 x 40
800 x 40 800 x 40 0.5 x 2 1600
800 x 32, 000 800 x 32,000 0.5 x 2 800 64,000 0.5 x 2 800 64,800 0.5 x 2 129,600 x 2 x 129, 600 360 or 360 The plane travels 360 mph in still air.
131
Chapter 1 Equations and Inequalities 29. Let x represent the speed at which Jean runs. Then x 8 is the speed at which she rides. Distance (mi)
Rate (mph)
Running
6
x
Riding
24
x8
50 15 1.5 x x 20 100 30 3 x x 20 30 100 x x 20 x x 203 x x 20
Time (hr) 6 x 24 x8
100 x 2000 30 x 3x 2 60 x 130 x 2000 3x 2 60 x
6 24 2.25 x x8 24 96 9 x x8 96 24 x x 8 9 x x 8 x x 8
24 x 8 96 x 9 x 8 x 2
0 3x 2 190 x 2000 0 3x 40 x 50 3 x 40 0 or 3 x 40
24 x 192 96 x 9 x 72 x 120 x 192 9 x 2 72 x 0 9 x 2 48 x 192 0 3 x 2 16 x 64 0 3 x 8 x 8
x
31.
x 8 0 x8
2x 4 6
2 x 4 6 2
2
2 x 4 36
8 3
2 x 40
x 8 8 8 16 Jean runs 8 mph and rides 16 mph.
20
30. Let x represent the speed at which Barbara drives in clear weather. Then x 20 is the speed at which she drives during the thunderstorm. Distance (mi)
Rate (mph)
Clear weather
50
x
Thunderstorm
15
x 20
x 50 0 x 50
40 1 13 3 3 1 2 x 20 13 20 6 3 3 x 20 50 20 30 Barbara drives 30 mph in the thunderstorm and 50 mph in nice weather. x
2
3 x 8 0 or 3x 8
100 x 20 30 x 3 x 2 20 x
32.
x 20
3x 1 11
3x 1 11 2
3 x 1 121 3x 120
Time (hr) 50 x 15 x 20
40
132
x 40
2
Section 1.6 33.
Check: n 10
m 18 2 m
2n 29 3 n
m 18 m 2
m 18 m 2 2
2 10 29 3 ⱨ10
2
49 3 ⱨ10 7 3 ⱨ10 10 ⱨ10 true 10 ; The value 2 does not check.
m 18 m 2 4m 4 0 m 2 5m 14 0 m 2 m 7 m 2 Check: m 2
or
m7
35. 4 3 2 x 5 6 10
m 18 2 m
4 3 2 x 5 4
2 18 2 ⱨ 2
2 x 5 1
3
16 2 ⱨ 2 4 2 ⱨ 2 6 ⱨ 2 false Check: m 7
2 x 5 1 3
3
2 x 5 1 2x 4
m 18 2 m
x2
2
7 18 2 ⱨ 7 25 2 ⱨ 7 5 2ⱨ7 7 ⱨ 7 true 7 ; The value 2 does not check.
36. 3 5 4 x 1 2 8
3 5 4 x 1 6 5
4 x 1 2
4 x 1 2 5
5
34.
2n 29 3 n
2n 29 n 3
2
x
2n 29 n 2 6n 9 0 n 2 n 10 or
31 4
31 4
0 n 2 8n 20 n 2 Check: n 2
5
4 x 1 32 4 x 31
2n 29 n 3 2
3
37. 4 5 y 3 4 2 y 1 0
n 10
4
2n 29 3 n
5y 3 4 2y 1
5 y 3 2 y 1 4
2 2 29 3 ⱨ 2
4
4
5y 3 2y 1
25 3 ⱨ 2 5 3 ⱨ 2 8 ⱨ 2 false
3y 4 y
133
4 3
4
Chapter 1 Equations and Inequalities 4 3 4 5y 3 4 2y 1 0
39.
Check: y
4
8 p 1
8 p 1 p 5 2 p 5 2 p 2 p 5 p 1
p 5 p 1
11 4 11 ⱨ0 3 3 0 ⱨ 0 true
2
0 p2 3 p 4 0 p 1 p 4 p 1
38. 6 y 7 6 4 y 5 0 6
6
6
6
or
p4
Check: p 1
y 7 6 4y 5
y 7 4y 5
8 p p 5 1 6
8 1
1 5 ⱨ1
y 7 4y 5
9 4 ⱨ1
2 3y
Check: p 4
2 y 3
3 2 ⱨ1 true
8 p p 5 1
2 Check: y 3 6 y 7 6 4y 5 0
8 4
4 5 ⱨ1 4 9 ⱨ1 2 3 ⱨ1
2 2 7 6 4 5 ⱨ 0 3 3 6
2
p 5 p2 2 p 1
4 3
6
2
8 p 1 2 p 5 p 5
20 9 4 8 3 ⱨ0 3 3 3 3 4
p5
2
4 4 5 3 4 2 1ⱨ0 3 3 4
8 p p 5 1
1ⱨ1 false 1 ; The value 4 does not check.
2 21 6 8 15 ⱨ0 3 3 3 3
40.
23 6 23 6 ⱨ0 3 3 0 ⱨ 0 true
d 4 6 2 d 1
d 4 6 2d 1
d 4 6 2d 1 2
2 3
2
d 4 6 2d 2 6 2d 1 2 6 2d d 3
2 6 2d d 3 2
2
4 6 2d d 2 6d 9 24 8d d 2 6d 9 0 d 2 2d 15 0 d 3 d 5 or d 3 d 5
134
Section 1.6 Check: d 3 d 4 6 2d 1
k 2 2k 3 2
42.
k 2 2k 3 2 2
3 4 6 2 3 ⱨ 1
k 2 2k 3 4 2k 3 4
1 0 ⱨ 1
4 2k 3 k 9
1ⱨ 1 false
4 2k 3 k 9 2
Check: d 5 d 4 6 2d 1
2
16 2k 3 k 2 18k 81
5 4 6 2 5 ⱨ 1
32k 48 k 2 18k 81
9 16 ⱨ 1
0 k 2 14k 33
3 4 ⱨ 1
0 k 3 k 11
1ⱨ 1 true 5 ; The value 3 does not check.
k 3 Check: k 3
or
k 11
k 2 2k 3 2
3 y 3 2 y
41.
3 y 3 2 y 2
3 2 ⱨ 2 3 3 2
2
1ⱨ 9 2 1ⱨ 3 2
96 y3 y3 2 y
1ⱨ1 true Check: k 11
6 y 3 10 2 y 3 y 3 5 y
3 y 3 5 y 2
k 2 2k 3 2 2
11 2 ⱨ 2 11 3 2
9 y 27 25 10 y y
2
9 ⱨ 25 2
0 y y2
3ⱨ 5 2
2
0 y 2 y 1 y 2
or
43. a.
3 y 3 2 y
34 43
5
43
m 54 3
43
3 1ⱨ 2 2 ⱨ 2 true Check: y 1
b.
m2 3 5
m 5 23 32
3 y 3 2 y
m 53 2
1 3 ⱨ 2 1
5 32
3 4ⱨ 1 3 2 ⱨ1
2,1
m3 4 5
m 5
2 3 ⱨ 2 2 3 1ⱨ 4
3
3 ⱨ 3 true
3,11
y 1
Check: y 2 3
2
1ⱨ1 true
135
32
Chapter 1 Equations and Inequalities n5 6 3
44. a.
n 3 56 65
3
65
n 36 5
65
t2 3
n4 5 3
b.
n4 5
3
54
3
1 125
54
56
t 2
56
21
49. 2v 7 v 3 13
7
t 2 7 6 5 t 2 76 5
3
4 y 3
34
20
y 3
34
5
50. 5u 6
15
3u 1
0
5u 6
3u 1
15 15
y 33 4 5 4 3 y 3 54 3 43
5
2u 7
43
u 7 2
1 8 1 p4 5 16
2 p4 5
47.
p4 5
54
1 16
51. a. P 48t1 5
P 48 2
15
54
5
15
5u 61 5 3u 11 5 5u 6 3u 1
y 54 3 3
5 3
3
v 10
10
76 5 2
46.
0
2v 7 1 3 v 31 3 2v 7 v 3
t 76 5 2
13
2v 71 3 v 31 3
56 65
32
3 1 1 1 t 5 125 25
3
3 t 2
1 25
32
54
n 35 4
45.
1 5 1 t2 3 25
5t 2 3
48.
b.
1 1 1 p 4 2 32 16 5
55%
P 48t1 5 75 48t1 5 25 1 5 t 16
1 32
5
25 15 5 t 16
t 9.3 hr
136
7 2
5
Section 1.6 52. a. h 16 t 4
13
V r 1 C
b.
h 16 14 4
13
h 16 18
13
b.
11, 000 0.15 1 C
42 in.
h 16 t 4
13
11,000 C
60 16 t 4
13
15
0.85 5
3
13 3 15 t 4 4 3
3
15 t 4 49 days 4
55. Let u x 2 2.
x 2 x 2 42 0 2
53. a. v 2 gh 19.6h
2
u 7 u 6 0 u 7 or
v 19.6h
h
h V 54. a. r 1 C
x2 4 x 2
56. Let u y 2 3.
2
y 3 9 y 3 52 0 2
2
26.8 36.6 m 19.6
2
2
u 2 9u 52 0
u 4 u 13 0 u 4
or
u 13
y 3 4
or
y 3 13
y 2 1
or
y 2 16
y i
or
y 4
1n 2
12, 000 r 1 18, 000 2 r 1 3
2
x 2 9 or x 3i or 3i, 2
26.8 19.6 h 26.8 h 19.6 2
u6
x 2 7 or x 2 6 2
26.8 19.6h
26.8 19.6
2
u 2 u 42 0
v 19.6 10 14 m/sec b.
15
11,000 1 5 5 0.85 C 11, 000 5 0.85 C 11, 000 $24,800 C 0.855
15 13 t 4 4
15 t4 4
1n
13
i, 4
13
0.126 or 12.6% per year
137
2
Chapter 1 Equations and Inequalities 59. Let u c1 5 .
3 57. Let u 2 . t
5c 2 5 11c1 5 2 0
2
3 3 2 2 12 t t
5u 2 11u 2 0
5u 1 u 2 0
2
3 3 2 2 12 0 t t
1 5 1 c1 5 5 u
u2 u 2 0
u 3 u 4 0 u 3 3 2 3 t 3 5 t 3 t 5 3 3 , 2 5
or or or or
u4 3 2 4 t 3 2 t 3 t 2
c
15 5
c
c1 5 2
or
c 2
or
c 32
5
1 3125
15 5
5
3d 2 3 d 1 3 4 0 3u 2 u 4 0
3u 4 u 1 0 4 3 4 d1 3 3 u
2
2
5 5 y 3 6 y 3 8 0
d1 3
u 6u 8 0 2
u 2 u 4 0
y 1
or
60. Let u d 1 3 .
5 5 y 3 6 y 3 8
5 3 2 y 5 5 y
u2
1 , 32 3125
5 58. Let u 3. y
u 2
1 5
or
or or or or
3
4 3
d
u 4
64 27
or
u 1
or
d 1 3 1
or
d 1
or
d 1
3
13 3
3
64 , 1 27
5 3 4 y 5 7 y 5 y 7
61.
x2 x2 5 7 x 5x 7 0 Let u x 2 . 4
2
u 2 5u 7 0
5 1, 7
u
138
5
52 4 1 7 5 53 2 1 2
Section 1.6 64. Let u q 1.
5 53 x 2 2
x
3q 2 16q 1 5 0
5 53 2
3u 2 16u 5 0
3u 1 u 5 0
5 53 2 62.
1 3 1 q 1 3 u
x 2 x 2 2 x 2 13
q 3
x 4 2 x 2 x 2 13 x 4 3 x 2 13 0 Let u x 2 .
u
or
q 1 5
or
q
65. a. y 4 y 21
3 2 4 1 13 2 1
4 y 21 y
4 y 21 y 2
3 61 2 3 61 x2 2 x
u 5
1 , 3 5
u 4 3u 2 13 0 3
or
2
16 y 441 42 y y 2 0 441 58 y y 2 0 y 9 y 49
3 61 2
y9 Check: y 9
3 61 2
or
y 49
y 4 y 21
9 4 9 ⱨ 21 9 12 ⱨ 21
63. Let u k 1.
30k 2 23k 1 2 0
21ⱨ 21 true Check: y 49
30u 2 23u 2 0
y 4 y 21
10u 1 3u 2 0
49 4 49 ⱨ 21
1 2 or u u 10 3 1 2 or k 1 k 1 10 3 3 k 10 or k 2 3 , 10 2
49 28 ⱨ 21
9
139
77 ⱨ 21 false
1 5
Chapter 1 Equations and Inequalities b. Let u
Check: w 25 w 3 w 10
y.
y 4 y 21
25 3 25 ⱨ10
y 4 y 21 0
25 15 ⱨ10
u 2 4u 21 0
u 7 u 3 0 u 7
or
u3
y 7
or
y 3
y 7 or y 3 2
2
2
y 49
or
25
b. Let u w.
w 3 w 10
2
w 3 w 10 0
y9
u 2 3u 10 0
9 ; See checks in part (a). 66. a.
10 ⱨ10 true
u 2 u 5 0
w 3 w 10 w 10 3 w
w 10 2 3 w
u5
w 2
or
w5
w 2 or w 5
2
2
or w4 25 ; See checks in part (a).
w2 29 w 100 0
w 4 w 25 0 67.
w 4 or w 25 Check: w 4 w 3 w 10
4 3 4 ⱨ10
1 1 1 f p q 1 1 1 fpq fpq f p q pq fq fp
4 6 ⱨ10 2 ⱨ10 false
pq fp fq p q f fq p
1 1 1 1 R R1 R2 R3 1 1 1 1 RR1 R2 R3 R R1 R2 R3 R1 R2 R3 RR2 R3 RR1R3 RR1 R2 R1 R2 R3 RR2 R3 RR1 R3 RR1 R2 R3 R1R2 RR2 RR1 RR1 R2 R3
or
2
w2 20 w 100 9 w
68.
u 2
RR1 R2 R1 R2 RR2 RR1
140
fq q f
2
w 25
2
Section 1.6 69.
E kT 4
74. 4 x 2 y 2 z
E T4 k E 4 4 4 T k
x2 y 2 z 4 2
x 2 y 2 z 4
2
2
E k
T4
x y z 4 2
2
y 2 z 4 x 2 2
y z 4 x 2 2
4 70. V r3 3 3V r3 4 3V 3 3 3 r 4 r3
71.
PV PV T1T2 1 1 T1T2 2 2 T1 T2 T2 PV 1 1 T1 PV 2 2
3V 4
PV 1 1T2 T1 PV 2 2
kF m kF m a m m ma kF a
V
t t s1v1s2v2 1 s1v1s2 v2 2 s1v1 s2 v2 t1s2 v2 t2 s1v1 t sv v2 2 1 1 t1s2
k P
k P V P P
77.
P
T 2
L g
L T 2 g
PV k
2
2
k V
T2
73. 16 x 2 y 2 z
2
x y z 16
2
x 2 y 2 z 16
2
2
2
4 2 L g
4 2 L g T2 g g
x y z 16 2
2
t1 t 2 s1v1 s2 v2
76.
kF a
m
72.
PV PV 1 1 2 2 T1 T2
75.
gT 2 4 2 L g
x 2 z 16 y 2 2
x z 16 y 2 2
141
4 2 L T2
Chapter 1 Equations and Inequalities
78.
t
2s g
2s t g 2s t2 g
t 2
5
52 4 1 3 2 1
5 37 2 5.5 or 0.5 t 1 5.5 1 6.5 It would take Joan approximately 5.5 hr working alone, and it would take Henry approximately 6.5 hr.
2
g 2s g t2 2 g 2 t2g s 2
84. Let t represent the time it takes Antonio to complete one bathroom. Then t 4 is the time it takes Jeremy to complete one bathroom. 1 job 1 job 1 job t hr t 4 hr 8 hr
79. An equation is in quadratic form if, after a suitable substitution, the equation can be written in the form au 2 bu c 0 , where u is a variable expression.
1 1 1 t t 4 8
80. Given a polynomial equation, set one side equal to zero and factor the other side into linear and quadratic factors. Then set each factor equal to zero and solve for the variable.
1 1 1 8t t 4 8t t 4 t t 4 8 8 t 4 8t t t 4
81. When solving a radical equation, if both sides of the equation are raised to an even power, then the potential solutions must be checked. This is because some or all of the solutions may be extraneous solutions.
8t 32 8t t 2 4t 0 t 2 12t 32 t
82. If m is an even integer, then the mth root of k will be a root with an even index. All positive real numbers have two even-indexed roots: one positive and one negative.
12
122 4 1 32 2 1
12 272 2 14.2 or 2.2 t 4 14.2 4 18.2 It would take Antonio approximately 14.2 hr working alone, and it would take Jeremy approximately 18.2 hr.
83. Let t represent the time it takes Joan to fill 100 orders by herself. Then t 1 is the time it takes Henry to fill 100 orders. 1 job 1 job 1 job t hr t 1 hr 3 hr 1 1 1 t t 1 3
1 1 1 3t t 1 3t t 1 t t 1 3 3 t 1 3t t t 1 3t 3 3t t 2 t 0 t 2 5t 3
142
Section 1.7 86. Let x represent the distance along the shoreline as shown in the figure.
85. Let x represent the distance along the shoreline as shown in the figure. Distance
Rate
Time
Row
4002 x 2
2.5
400 x 2.5
Walk
800 x
5
800 x 5
2
Distance
Rate
Time
Boat
482 x 2
20
482 x 2 20
Car
96 x
60
96 x 60
2
4002 x 2 800 x 300 2.5 5 4002 x 2 800 x 5 5 300 2.5 5
482 x 2 96 x 4 20 60 482 x 2 96 x 60 60 4 20 60
2 4002 x 2 800 x 1500
3 482 x 2 96 x 240
2 4002 x 2 x 700
x 7002 2 4002 x 2
x 2 1400 x 490, 000 4 4002 x 2
3 482 x 2 x 144
3 48 x x 144
2
2
2
3 x 1400 x 150, 000 0
3x 500 x 300 0 or
2
8 x 2 288 x 0
2
500 3
2
x 288 x 20,736 9 x 20,736
2
x
2
x 2 288 x 20,736 9 482 x 2
x 1400 x 490, 000 4 x 640, 000 2
2
8 x x 36 0 x 0 or x 36 The marina is 36 mi up the coast.
x 300
2 Pam can row to a point 166 ft down the 3 beach or to a point 300 ft down the beach to be home in 5 min.
Section 1.7 Linear Inequalities and Compound Inequalities 1. , 5
2. 1,
10. intersection
3. 4,
4. , 5
11. a x b
5 5. x | x 4 6
4 6. x | 3 x 3
7. union, A B
8. inequality
12. union
13. 2 x 5 17
2 x 22 x 11 x | x 11 ; , 11
9. intersection; A B
143
Chapter 1 Equations and Inequalities 19. 5 6 c 4 7
14. 8t 1 17
8t 16 t 2 t | t 2 ; 2,
5 6c 24 7 5 6c 17 12 6c 2 c or c 2 c | c 2 ; , 2
4 15. 3 w 1 3 4 4 w 3 3 w or w 3 w | w 3 ; , 3
16.
20. 14 3 m 7 7
14 3m 21 7 14 3m 14 0 3m 0 m or m 0 m | m 0 ; 0,
5 y2 2 5 10 y 2 4 y or y 4 8
21.
y | y 4 ; 4,
4 x x3 x 2 5 10 4 x x 3 x 10 10 2 10 5 5 4 x 2 x 3 x 20 5 x 2 x 6 x
17. 1.2 0.6a 0.4a 0.5
4 x 26
0.2a 1.7 1.7 a 0.2 a 8.5 a | a 8.5 ; , 8.5
26 4 13 x 2 13 13 x | x ; , 2 2 x
18. 0.7 0.3x 0.9 x 0.4
0.6 x 0.3 0.3 x 0.6 x 0.5 x | x 0.5 ; 0.5,
144
Section 1.7
22.
25. 5 7 x 2 x 6 x 2 9 x
y 3 3y 1 1 4 6 12 y 3 3 y 1 1 12 12 4 12 6
35 5 x 2 x 6 x 2 9 x 35 2
3 y 3 2 3 y 1 1 3 y 9 6 y 2 1
26. 2 3x 1 4 x 2 x 8 5
3 y 8
6 x 2 4 x 2 x 16 5 29
y
8 3
8 8 y | y ; , 3 3
27. 5 3 2 4 x 2 6 2 4 x 3
5 3 2 4 x 8 6 2 4 x 3
5 3 4 x 10 6 2 x 7
23.
5 12 x 30 6 2 x 7
1 5 1 x 4 x 3 x 1 3 6 2 5 1 1 6 x 4 x 3 6 x 1 2 6 3
12 x 25 6 x 5 12 x 25 6 x 30 6 x 5 5 x 6 5 5 x | x ; , 6 6
2 x 4 5 x 3 3 x 6
2 x 8 5 x 15 3 x 6 6 x 17 x
17 6
17 17 x | x ; , 6 6
28. 8 6 10 x 1 2 1 3 2 x 4
8 6 10 x 10 2 1 3 2 x 4
24.
8 10 x 16 2 1 3 x 2
1 4 3 t 6 t 2 t 2 2 3 4 4 1 3 12 t 6 t 2 12 t 2 2 3 4
8 10 x 16 2 1 3 x 6 10 x 8 2 3x 7
10 x 8 6 x 14 4 x 22 11 x 2 11 11 x | x ; , 2 2
6 t 6 16 t 2 9t 24
6t 36 16t 32 9t 24 t 44 t 44 t | t 44 ; , 44
145
Chapter 1 Equations and Inequalities 29. 4 3k 2 k 3 k
34. a. M N y | y 3
4 3k 2k 6 k 4 6 ; ,
b. M N y | y 5 c. M P d. M P y | 3 y 0 e. N P y | y 0 or y 5
30. 2 x 9 6 x 1 4 x
2x 9 6x 6 4x 9 6 ; ,
f. N P 35. A B 1, 0,1, 2, 3, 4, 5 36. A B 0, 1, 2, 3
31. a. A B 0, 3, 4, 6, 8, 9,12 b. A B 0,12
37. M S 1
c. A C 2, 0, 4, 8,12
38. M S 3, 2, 1, 0,1, 2, 3, 4, 5
d. A C 4, 8 e. B C 2, 0, 3, 4, 6, 8, 9,12 f. B C 32. a. X Y 10, 9, 8, 7, 6, 4
39. a. x 4 and 2, 4
x 2
b. x 4 or ,
x 2
40. a. y 2
b. X Y 10, 8
and
y 5
or
y 5
5, 2
c. X Z
10, 9, 8, 7, 6, 5, 4, 3, 2
b. y 2
,
d. X Z e. Y Z 10, 8, 6, 5, 4, 3, 2 f. Y Z 6, 4 33. a. C D b. C D x | 1 x 9 c. C F x | x 9 d. C F x | x 8 e. D F x | x 8 or x 1 f. D F
146
41. a. m 1 6
or
m5 , 5
or
b. m 1 6
and
m5 , 6
and
1 m 2 3 m 6
1 m 2 3 m 6
Section 1.7 42. a. n 6 1
or
n7
or
7,
3 n6 4 n8
and
n7 8,
and
2 y 12 3 y 18
3 n6 4 n8
4 44. a. m 8 5 m 10 2.5,
and
2.08 0.65 y
and
y 3.2
4 b. m 8 5 m 10 10,
, 3.2
45. a. 3 x 2 2 x 8
or
3x 6 2 x 8 2 x 4
or or
x 2
or
4 x 1 2 2 x 4 4 x 4 2 2 x 4 6 x 2 1 x 3
, 2 , 1 3
b. 3 x 2 2 x 8
x 2
and
4 x 1 2 2 x 4 x
and
1 3
46. a. 5 t 4 2 3 t 1 3
or
2t 6 3 t 4 2
5t 20 2 3t 3 3 2t 18
or or
2t 6 3t 12 2 t 8
t9 , 8 9,
or
b. 5 t 4 2 3 t 1 3
2 y 12 3 y 18
or
2.08 0.65 y
or
y 3.2
and
0.85 0.34m
and
m 2.5
or
0.85 0.34m
or
m 2.5
,18
b. n 6 1
43. a.
b.
t9
and and
t 8
2t 6 3 t 4 2
t 8
147
Chapter 1 Equations and Inequalities 47. 2.8 y and y 5 48.
54.
1 z and z 2.4 2
49. 3 2 x 1 9 4 2 x 8
10 5 x 6 2 x
2 x 4 or 4 x 2 4, 2 50.
5 x 2 4 2 5x 2 4 4 2 8 5 x 2 8
4
6 5
6 2, 5
6 3x 9 0 15 3x 9
55. 12.0 x 15.2 g/dL
5 x 3 or 3 x 5 3, 5
56. 18 a 25 yr 57. 90 d 110 ft
5x 4 3 51. 1 2 2 5x 4 6
58. 220 s 410 mph
6 5 x 10
59.
6 x2 5 6 , 2 5
88 92 100 80 90 2.5 x 92 7.5 450 2.5 x 92 7.5 450 2.5 x 690 2.5 x 240 x 96 Marilee needs to score at least 96 on the final exam.
4x 1 5 3 6 4 x 1 15 5 4 x 16
52. 2
60.
5 x4 4 5 4 , 4
36 36.9 37.1 37.4 x 37 5 147.4 x 37 5 147.4 x 185 x 37.6 The child needs a score of at least 37.6.
2 x 1 4 3 6 2 x 1 12
53. 2
61. Let t represent the time it takes for the car to be more than 16 miles ahead of the truck. 50t 40t 16
5 2 x 13
10t 16
5 13 x 2 2 5 13 2 , 2
t 1.6 It will take more than 1.6 hr or 1 hr 36 min.
148
Section 1.7 62. Let t represent the time it takes for a tutor to make over $500 more than a student working in the library. 16.25t 10.75t 500
67.
A P Prt 5000 4000 4000 0.05 t 1000 200t 5t At least 5 yr is required.
5.5t 500 t 91 It would take 91 or more hours.
A P Prt
68.
10,000 P P 0.04 8
63. Let l represent the length of the garden. Then the perimeter of the garden is 2l 200 200 2l 800
10,000 P 0.32 P 10,000 1.32 P
2l 600
7575.76 P or P 7575.76 At least $7575.76 is required.
l 300 The length must be 300 ft or less.
5 F 32 9 5 35 F 32 9 63 F 32
69. C
64. Let x represent the length of the shortest side. Then the lengths of the other two sides are x 1 and x 2 .
x x 1 x 2 24
95 F or F 95 Hypothermia would set in for a core body temperature below 95F .
3x 3 24 3x 21 x7 The shortest side may be 2 ft, 3 ft, 4 ft, 5 ft, 6 ft, or 7 ft.
70.
65. Let x represent the average scores that would produce a nonnegative handicap of 72 or less. 0 0.9 220 x 72
5 F 32 9 5 36.5 F 32 37.5 9 65.7 F 32 67.5 C
97.7 F 99.5 Normal body temperature is between 97.7F and 99.5F , inclusive.
0 220 x 80 220 x 140 220 x 140 or 140 x 220 An average score in league play between 140 and 220, inclusive, would produce a handicap of 72 or less.
71. a. Let s represent the amount of sales. 25,000 0.1s 30, 000 0.08s
0.02s 5000 s 250,000
66. a. 1.1 2015 1.1 300 330 ft 2
b. Job A
b. Let c represent the maximum cost per square foot of tile that Betty can afford. 1.06 6 2015 330c 5600
72. Let x represent the number of nights. 40 1.18 x 169 1.14 x 179
40 199.42 x 204.06 x
1.06 1800 330c 5600
40 4.64 x
1908 349.8c 5600
8.62 x or x 8.62 After 8 nights (9 or more), Hotel B will be less expensive.
349.8c 3692 c 10.55
149
Chapter 1 Equations and Inequalities T 1.83a 212
73.
79. a. 2 x 9 0
1.83a 212 200
2x 9
1.83a 12 a 6.6 Water will boil at temperatures less than 200F at altitudes of 6600 ft or more.
9 2 9 x x 2 x
E 46.2 x 446.2
74.
b. 2 x 9 0
46.2 x 446.2 1000
2x 9
46.2 x 553.8 x 11.987 For the year 2012 and later, the average per capita consumer expenditure for prescription drugs will exceed $1000.
9 2 9 x x 2 x
80. a. 3x 7 0
75. a. x 2 0 x2 x | x 2
3x 7 7 3 7 x x 3 x
b. 2 x 0
2 x or x 2
x | x 2
b. 3x 7 0
76. a. x 6 0
3x 7 7 x 3 7 x x 3
x6 x | x 6 b. 6 x 0 6 x or x 6 x | x 6
81. cd a False
77. a. x 4 0
82. ab c True
x 4 x | x 4
83. If a c , then ad cd . True
b.
84. If a c , then ab bc . False
78. a. x 7 0 x 7 x | x 7
85. , 2 3, 4 1, 3 1, 2 86. , 5 1, 0, 3 0, 3
b.
87. , 2 4, 5, 3 5, 2 88. , 6 10, 8, 12 10, 12
150
Section 1.7 89. 2 x 1 3 or 2 x 4 or x 2 or , 2 0,
2x 3 3 2x 0 x0
96.
1 7 z 3 3 1 10 z 3 30 z or z 30 , 30
90. 4m 2 1 or 4m 2 1 4m 1 or 4m 3 1 3 m or m 4 4 1 3 , , 4 4
97. 1
91. 2 6d 4 2 2 6d 6 2 d 1 6 1 d 1 3 1 3 ,1
98. 0
92.
4 x 3 2 x4 1 3 2 2 x 4 6 2 x 10 2,10
6 x 2 4 x6 0 2 4 0 x68 6 x 14 6,14
8 3k 7 8 15 3k 1 1 5 k 3 1 5, 3
93. 11 6 y 7 18 6 y 3 y
99. The steps are the same with the following exception. If both sides of an inequality are multiplied or divided by a negative real number, then the direction of the inequality sign must be reversed.
and and and
6 y 7 5 6 y 12 y 2
100. The statement 8 x 2 is equivalent to 8 x and x 2 . No real number is greater than 8 and simultaneously less than 2.
3, 2 94. 13 2c 3 10 2c 5 c 5, 4 95.
and and and
101. The statement 3 w 1 is equivalent to w 3 and w 1 . No real number is less than 3 and simultaneously greater than 1 .
2c 3 5 2c 8 c4
102. If the inequalities in a compound inequality are joined by the word “and,” then the solution set to the compound inequality is the intersection of the solution sets of the individual inequalities. If the inequalities are joined by the word “or,” then the solution set to the compound inequality is the union of the solution sets of the individual inequalities.
1 p4 2 1 2 p 2 4 p or p 4 6
, 4 151
Chapter 1 Equations and Inequalities Section 1.8 Absolute Value Equations and Inequalities 1. x 4 or 4 x
14. a. w 2 w 2 or w 2 2, 2
2. x 1 or 1 x
b. w 0
3. 4 2 x 10 4 6 2 x 14
w 0 or w 0 0
3 x 7 x | 3 x 7 ; 3, 7
c. w 2
4. 1 5t 1 1 2 5t 0
15. a. x 3 4
2 t0 5 2 2 t t 0 ; , 0 5 5 or 5m 20 or m 4 m4 m | m 4 or m 4 ; , 4 4,
9. k ; k
10. k ;
11.
12.
x 3 4
x7 7, 1
or
x 1
x3 0
or
x 3 0
x3
or
x3
3
6. 7c 14 or 7c 14 or c 2 c2 c | c 2 or c 2 ; , 2 < 2, 8. u w or u w
or
b. x 3 0
5. 5m 20
7. absolute; k , k
x3 4
c. x 3 7
16. a. m 1 5
m 1 5
or m 1 5
m4 4, 6
or
m 6
b. m 1 0
m 1 0
13. a. p 6
or m 1 0
m 1 or
p 6 or p 6
1
6, 6
c. m 1 1
b. p 0
p 0 or p 0
0 c. p 6
152
m 1
Section 1.8 17. 2 3x 4 7 9
23.
2 3x 4 2
1 1 1 6 4 w 6 2 2 3
3x 4 1 3x 4 1
or
3 x 4 1
3x 5
or
3x 3
5 3
or
x 1
2t 7 5
or
2t 7 5
2t 2
or
2t 12
t 1
or
t 6
x
1 1 1 4 w 2 3 2
24 3w 2 3 24 3w 5 24 3w 5 or 3w 19 or 19 w or 3 19 29 , 3 3
5 ,1 3
18. 4 2t 7 2 22
4 2t 7 20
24 3w 5 3w 29 29 w 3
2t 7 5
1, 6 19.
24.
2
1 7 1 6 2 p 6 2 3 6 12 2 p 7 3
3 c 7 1
12 2 p 10
4 c 7
12 2 p 10 2 p 2 p 1
4 c7 c7 4 c 11 11, 3 20.
or or
c 7 4 c3
1 z 8 or or
z 8 1 z 9
3y 5 y 1
or
2 y 4
or
y 2
or
3 y 5 y 1 4 y 6 3 y 2
3 2, 2
26. 2a 3 a 2
2 8 11y 4 6 11y 4
22.
12 2 p 10 2 p 22 p 11
25. 3 y 5 y 1
4 z 8 3
z 8 1 z 7 7, 9
or or or
1,11
1 z 8
21.
1 7 1 p 3 6 2
6 7 9z 3
2a 3 a 2
or
a5
or
1 5, 3
1 9 z 3
153
2a 3 a 2 3a 1 1 a 3
Chapter 1 Equations and Inequalities
27.
33. a. x 7
1 w 4w 4 1 w 4w 4 w0 0 1 z 3 1 3z z 3 z0 0
x 7 or x 7 7, 7
1 w 4w 4 w0
or or
b. x 7
7 x 7 7, 7 c. x 7
28. 3z
x 7 or x 7 , 7 7,
1 3z z 3 z0
or or
34. a. y 8
y 8 or y 8
8, 8
29. x 4 x 7
x4 x7
or
4 7
or
x 4 x 7
b. y 8
2x 3 x
8 y 8
8, 8
3 2
c. y 8
3 2 30.
y 8 or y 8
, 8 8,
k 3 k 3 k 3 k 3
or
3 3
or
0 31.
35. a. a 9 2 6
k 3 k 3 2k 0
a9 4
k0
a 9 4 or a 5 or 13, 5
2 p 1 1 2 p 2 p 1 1 2 p
or
2 p 1 1 2 p
4p 2
or
00
p
b. a 9 2 6
a9 4 4 a 9 4 13 a 5 13, 5
1 2
c. a 9 2 6
32. 4d 3 3 4d
4d 3 3 4d
or
4d 3 3 4d
8d 6
or
00
d
a 9 4 a 13
a9 4 a 9 4 or a 13 or , 13 13,
3 4
154
a9 4 a 5
Section 1.8 36. a.
b 1 4 1
40. 5 x 1 9 4
b 1 5
5 x 1 5
b 1 5 b4 6, 4
or or
b 1 5 b 6
x 1 1 x 1 1 or x 2 or , 2 0,
b. b 1 4 1
b 1 5
41.
5 b 1 5 6 b 4 6, 4
16 2 p 4 2 p 4 16 16 2 p 4 16 20 2 p 12 10 p 6
b 1 5 b 1 5 b4
10, 6 42.
37. 3 4 x 2 16
18 6 3 z 3 24 3 z 3
3 4 x 18
24 3 z 3
4 x 6
3z 3 24
6 4 x 6 10 x 2 10 x 2 or 2 x 10
24 3z 3 24 27 3z 21 9 z 7 9, 7
2,10
38. 2 7 y 1 17
43.
2 7 y 16
10 5c 4 2 8 5c 4
7 y 8
5c 4 8
8 7 y 8 15 y 1 15 y 1 or 1 y 15
5c 4 8 or 5c 4 8 5c 4 or 5c 12 4 12 c or c 5 5 12 4 , , 5 5
1,15 39. 2 x 3 4 6
2 x 3 10 x3 5 x 3 5 or x 8 or , 8 2,
11 5 2 p 4 16 2 p 4
c. b 1 4 1
b 1 5 or b 6 or , 6 4,
x 11 x0
x35 x2
155
Chapter 1 Equations and Inequalities 44.
15 2d 3 6
49. a. x 9
9 2d 3
2d 3 9 2d 3 9 or 2d 6 or d 3 or , 6 3, 45.
b. x 9
2d 3 9 2d 12 d 6
c. x 9
; ,
y3 2 6 y3 2 2 6 12 y 3 12 15 y 9
50. a. y 2
b. y 2
c. y 2
15, 9 46.
47.
; ,
m4 14 2 m4 14 14 2 28 m 4 28 24 m 32 24, 32
51. a.
14 y 7 14 y 7
b.
14 y 7
1 p 6 0.01 2 1 p 6 0.01 2 1 p 5.99 2 p 11.98
1 p 6 0.01 2 1 or p 6.01 2 p 12.02 or or
c.
18 4 y 7 14 y 7 14 y 7
; , 52. a.
1 q 2 0.05 4 1 q 2 0.05 4 1 q 1.95 4 q 7.8
18 4 y 7 14 y 7
,11.98 12.02, 48.
18 4 y 7
15 2 p 3 13 p 3 13 p 3
or or or
, 7.8 8.2,
1 q 2 0.05 4 1 q 2.05 4 q 8.2
b.
15 2 p 3 13 p 3 13 p 3
156
Section 1.8 c.
15 2 p 3
55. a.
k4 0 k40 or k 4 or 4
13 p 3 13 p 3
; ,
b. k 4 0
53. a. z 0
z 0 or z 0 0
c. k 4 0
k40 k 4 4
c. z 0
d. k 4 0
b. z 0
z0
k 4 0 or k 4 0 k 4 or k 4 k | k 4 or k 4 ;
0
d. z 0
, 4 4,
z | z 0 or z 0 ; , 0 0,
e. k 4 0
e. z 0
; ,
; , 54. a. 2 w 0
2w 0 w0 0
k 4 0 k 4
56. a.
or or
c3 0 c3 0 c3 3
2 w 0 w 0
or or
c 3 0 c3
b. c 3 0
b. 2w 0
c. 2w 0
c. c 3 0
2w 0 w0 0
c3 0 c3 3
d. 2 w 0
d. c 3 0
c 3 0 or c 3 0 c3 c3 or c | c 3 or c 3 ; , 3 3,
2w 0 or 2w 0 w0 w0 or w | w 0 or w 0 ; , 0 0,
e. c 3 0
e. 2w 0
; ,
; ,
157
Chapter 1 Equations and Inequalities 64. y L or L y
57. a. x 4 6 or 4 x 6 b.
x4 6 x 4 6 x 2 2,10
or or
65. a. t 36.5 1.5 or 36.5 t 1.5
x46 x 10
b. t 36.5 1.5
1.5 t 36.5 1.5 35 t 38 35, 38 ; If the refrigerator is set to 36.5F , the actual temperature would be between 35F and 38F , inclusive.
58. a. x 3 8 or 3 x 8 b.
x3 8 x 3 8 x 5 5,11
or or
x38 x 11
66. a. x 16 0.5 or 16 x 0.5 b. x 16 0.5
0.5 x 16 0.5 15.5 x 16.5 15.5,16.5 ; The boxes of cereal vary in weight between 15.5 oz and 16.5 oz, inclusive.
59. a. v 16 0.01 or 16 v 0.01 b. v 16 0.01
0.01 v 16 0.01 15.99 v 16.01 15.99,16.01
67. a. x 0.51 0.03 or 0.51 x 0.03
60. a. t 60 0.2 or 60 t 0.2
b. x 0.51 0.03
0.03 x 0.51 0.03 0.48 x 0.54 0.48, 0.54 ; The candidate is expected to receive between 48% of the vote and 54% of the vote, inclusive.
b. t 60 0.2
0.2 t 60 0.2 59.8 t 60.2 59.8, 60.2 61. a. x 4 1 or 4 x 1 b.
68. a. x 34 3 or 34 x 3
x 4 1 x 4 1 or x3 or , 3 5,
b. x 34 3
x 4 1 x5
3 x 34 3 31 x 37 31, 37 ; The motorist was traveling between 31 mph and 37 mph, inclusive. The motorist should receive a ticket because even at the lower end of the interval, the speed of 31 mph still exceeds the posted speed limit.
62. a. y 10 2 or 10 y 2 b.
y 10 2 y 10 2 y8
or or
y 10 2 y 12
, 8 12, 63. 0 x c or 0 c x
158
Section 1.8 x 500
69. a.
250
1.96
77.
1.96
x 500 5 10
1.96
1.96 5 10 x 500 1.96 5 10
78.
50
1.96
x 60 50
50 x 60 1.96 50
82. Taking the square root of both sides of the
b. No. It appears that the die lands on “1” at a lower rate than expected because 30 outcomes is below the “reasonable” range.
73. 2 z 4
74. 8 p 11
75.
3 x 7 3 2 x 2 7 2 5 x 5 x2 5
76.
2 x6 24 x464 2 x 2 x4 2
x 11 x 5 11 5 x5 6
81. The inequality x 3 0 will be true only for values of x for which x 3 0 (the absolute value will never be less than 0). The solution set is {3}. The inequality x 3 0 is true for all values of x excluding 3. The solution set is x | x 3 or x 3 .
13 x 60 13 47 x 73 47, 73 ; After rolling a fair die 360 times, it is reasonable to expect a “1” to appear between 47 and 73 times, inclusive.
72. 7 y 1
or or or
80. The absolute value of any real numbers is greater than or equal to zero. Therefore, all real numbers have an absolute value greater than 5 .
1.96
71. 3x 1 7
x 1 x 5 1 5 x 5 6
79. The absolute value of any nonzero real number is greater than or equal to zero. Therefore, no real number x has an absolute value of 5 .
1.96
1.96
x 10 x 7 10 7 x73
x5 6
b. Yes, because 560 is above the “reasonable” range.
x 60
or or or
x7 3
30 x 500 30 470 x 530 470, 530 ; In a group of 1000 jurors selected at random, it would be reasonable to have between 470 and 530 women, inclusive.
70. a.
x4 x7 47 x 7 3
equation x 2 4 results in x 2 4 or equivalently x 2 . The solution set for each equation is 2, 2 , indicating that the equations are equivalent. 83. x x 11
x x 11 2 x 11 11 x 2 11 , 2
159
or or
x x 11 0 11
Chapter 1 Equations and Inequalities 84. x x 10
88.
x x 10 0 10
or or
, 5
x x 10 2 x 10 x 5
7 3x 5 13 7 3 x 5 13 12 3 x 18
or or
4 x6
or or
85. 1 x 9
1 x 9
8 2 , < 4, 6 3 3
1 x 9 1 x 9 9 x 1
or
9, 1 < 1, 9
p pˆ z
89.
86. 2 y 11
2 y 11
or or or
11, 2 < 2,11
2 y 11 2 y 11 11 y 2
ˆˆ ˆˆ pq pq p pˆ z n n
pˆ z
ˆˆ ˆˆ pq pq p pˆ z n n
x
87. 5 2 x 1 7
4, 3 < 2, 3
or or or or
5 2 x 1 7 6 2 x 8 3 x 4 4 x 3
ˆˆ pq n
z
90.
5 2x 1 7 4 2x 6 2 x3
7 3x 5 13 2 3x 8 2 8 x 3 3 8 2 x 3 3
x
z n z n
z n
x x
z n z n
Problem Recognition Exercises: Recognizing and Solving Equations and Inequalities 2. a. Absolute value inequality
1. a. Equation in quadratic form and a polynomial equation
b.
b. Let u x 2 5
8 3t 1
x 5 5 x 5 4 0 2
2
2
3t 1 8
u 5u 4 0
3t 1 8 or 3t 7 or 7 t or 3 7 , < 3, 3
2
u 4 u 1 0 u4 x 5 4 2
x2 9
or x 5 1 2
or
x 3 or
3, 6
u 1
or
2 3t 1 6
x2 6 x 6
160
3t 1 8 3t 9 t3
Problem Recognition 3. a. Radical equation
4. a. Absolute value equation
b. 3 2 y 5 4 1
b. 9 3z 7 1 4
2y 5 3
9 3z 7 3
3
2 y 5 3 3
3
3z 7
3
2 y 5 27 2 y 32 y 16
1 3
16 5. a. Rational equation b.
2 5 1 w 3 w 1 2 5 w 3 w 1 w 3 w 11 w 3 w 1 2 w 1 5 w 3 w 3 w 1 2w 2 5w 15 w2 2w 3 7 w 13 w2 2w 3 0 w2 9w 10 w
9
92 4 110 9 41 2 1 2
9 41 2 6. a. Polynomial equation b.
7. a. Compound inequality
48 x 3 80 x 2 3x 5 0
b. 2 m 2 m 5
48 x 3 3 x 80 x 2 5 0
2m 4 m 5 m 9 m 9 9, 3
3x 5 16 x 1 0
3x 16 x 1 5 16 x 1 0 2
2 2
3x 5 4 x 1 4 x 1 0
and
6 m3
and and
3 m m3
8. a. Compound inequality
5 1 1 or x or x 3 4 4 5 1 , 3 4 x
b.
161
6 2c 8
or
2 2c
or
1 c ,12
or
1 c22 3 1 c4 3 c 12
Chapter 1 Equations and Inequalities 9. a. Quadratic equation b.
13. a. Compound inequality
2 p 1 p 5 2 p 40
6 x 7 5 x6 1 7 5 5 x 6 35
b. 1
2 p 2 11 p 5 2 p 40 2 p 2 9 p 35 0
2 p 5 p 7 0 p
5 2
1 x 41 1, 41
p 7
or
5 , 7 2
14. a. Radical equation
5 2 n 5 2n
b. 2 x x 4 7 2 x 2 3 x 5 2 x
5 2 n 5 2n 2
2 x 2 8 x 7 2 x 2 3 x 5 2 x 8 x 7 3 3
2
25 10 2 n 2 n 5 2n
8 x 7 9
10 2 n n 22
8 x 16
10 2 n 22 n 2
x2
2
2
100 2 n 484 44n n 2 200 100n 484 44n n 2
11. a. Linear inequality b.
5 5 2n 2 n
b.
10. a. Linear equation
n 2 144n 284 0
n 142 n 2 0
a 4 3a 1 a 2 4 8 a 4 3a 1 a 8 8 2 4 8
n 142 or n 2 Check: n 142 5 5 2n 2 n
4 a 4 2 3a 1 a
5 ⱨ 5 2 142 2 142
4a 16 6a 2 a
5 ⱨ 289 144
a 18
5 ⱨ17 12
a 18
18,
5 ⱨ 29 false Check: n 2
12. a. Quadratic equation
5 5 2n 2 n
b. 3x 2 11 4
5 ⱨ 5 2 2 2 2 5ⱨ 9 4
3 x 2 7 x2
5ⱨ3 2
7 3
5 ⱨ 5 true 2 ; The value 142 does not check.
7 7 7 3 21 i i i 3 3 3 3 3 21 i 3 x
162
Problem Recognition 15. a. Absolute value equation
u 2
or
u6
4 x 5 3x 2
y 2
or
y 6
b.
4 x 5 3x 2
or
4 x 5 3x 2
x3
or
7x 7
y 2 or y 6 2
y4 Check: y 4
x 1
1, 3
or
2
y 36
y 4 y 12 0
4 4 4 12 ⱨ 0
16. a. Rational equation b.
2
2
4 8 12 ⱨ 0 16 ⱨ 0 false Check: y 36
1 1 2d 0 d 2d 1 2 d 1 1 2d 1 0 d 2d 1 1 1 0 d 1 1 d d 1 1
y 4 y 12 0
36 4 36 12 ⱨ 0 36 24 12 ⱨ 0 0 ⱨ 0 true 36 ; The value 4 does not check. 19. a. Radical equation
17. a. Absolute value inequality
b.
b. x 4 8 3
c 16 23 32
x 4 5
64
x4 5 5 x 4 5
9,1
9 x 1
32
c 64
20. a. Absolute value inequality b. 2 z 14 8 4
2 z 14 4
18. a. Radical equation and an equation quadratic in form b. Let u
c 2 3 16
z 14 2
,
y
y 4 y 12 0 u 2 4u 12 0
u 2 u 6 0
163
Chapter 1 Equations and Inequalities Chapter 1 Review Exercises 1.
5. x 5 2 x 4 3 x 1 5
3 4 2 x 4 2x 7 3 3 4 2 x 2 x 2 2 x 7 3 7 x 2, x 2, x 2 2
x 5 2 x 8 3x 3 5 3x 13 3x 2 13 2
6. 0.2 x 1.6 x 0.8 x 2
2. 8 t 4 7 4 t 3 1 t 6
0.2 x 1.6 x 0.8 x 1.6 0.2 x 1.6 0.2 x 1.6 00
8t 32 7 4 t 3 3t 6 8t 39 4 4t 3 6 8t 39 16t 12 6 8t 39 16t 6 45 24t 45 15 t 24 8
3.
4.
7.
y 2 8 y 16 y 2 6 y 9 7 14 y 7 y 14 1 y 2 1 2
4 2 7 x x2 5 3 10 2 4 7 30 x 30 x 2 5 10 3 24 x 20 21x 60 3x 40 40 x 3 40 3
8.
x3 x4 2 x 5x x3 x 4 5x 2 5 x 5x x x 3 10 x 5 x 20 11x 3 5 x 20 6 x 23 23 x 6 23 6
m 2 m 4 m 1 1 3 4 6 m 2 m 4 m 1 12 1 12 3 6 4 4 m 2 3 m 4 2 m 1 12 4m 8 3m 12 2m 2 12 m 20 2m 10
40
y 4 2 y 3 2
30 m
164
Chapter 1 Review 1 5m 3 2 m 1 m 3m 4 m 4 1 5m 3 m 1 m 1 m 4 m 4
9.
m 1 m 4
1 5m 3 m 1 m 4 m 1 m 1 m 4 m 4
1 m 4 5m 3 m 1 m 4 5m 3m 3 1 m ; The value 1 does not check. 10. 4 x 3 y 6
13.
3 y 4 x 6 y
11.
4 x2 3
t t ta 1 2 2 2ta t1 t2 2ta t1 t2 or t2 2ta t1
1 1 c h 41 36 1 1 11 c 288 41 36 1 11 c 8 41 1 3 c 41 123 c or c 123 mi A
12. 4 x 6 y ax c
4 x ax c 6 y x 4 a c 6 y x
c 6y 6y c or x a4 4a
14. Let x represent the amount invested in the international fund. Then, 12, 000 x is the amount invested in the real estate fund.
Principal Interest (I = Prt)
International Fund
Real Estate Fund
x
12, 000 x
x 0.0821
12, 000 x 0.0151
Total
749.50
x 0.082 12,000 x 0.015 749.50 0.082 x 180 0.015 x 749.50 0.067 x 180 749.50 0.067 x 569.50 x 8500 12, 000 x 12, 000 8500 3500 Shawna invested $8500 in the international fund and $3500 in the real estate fund.
165
Chapter 1 Equations and Inequalities 15. Let x represent the amount invested in the 10-yr Treasury note. Then, x 4000 is the amount invested in the 15-yr bond. 10-yr Note
15-yr Bond
x
x 4000
x 0.03510
x 4000 0.04115
Principal Interest (I = Prt)
Total
10,180
x 0.03510 x 4000 0.04115 10,180 0.35 x 0.615 x 2460 10,180 0.965 x 2460 10,180 0.965 x 7720 x 8000 x 4000 8000 4000 12, 000 Cassandra invested $8000 in the Treasury note and $12,000 in the bond. 16. Let x represent the amount of the 20% acid solution (in cubic centimeters). 100 cc is the amount of the 60% acid solution. Therefore, x 100 is the amount of the resulting 25% acid solution. 20% Solution
60% Solution
25% Solution
Amount of Solution
x
100
x 100
Pure Acid
0.2x
0.6 100
0.25 x 100
0.2 x 0.6 100 0.25 x 100 0.2 x 60 0.25 x 25 35 0.05 x 700 x 700 cc of 20% acid solution should be mixed with the 60% acid solution. 17. Let x represent the amount of the pure sand (in cubic feet). 250 ft2 is the amount of the concrete mix that is 50% sand. Therefore, x 250 is the amount of the resulting 70% sand mixture. 100% Sand
50% Sand
70% Sand
Amount of Mixture
x
250
x 250
Pure Sand
x
0.5 250
0.7 x 250
x 0.5 250 0.7 x 250 x 125 0.7 x 175 0.3 x 50 x 166 166
2 3
2 2 ft of sand should be mixed with the 50% sand mixture. 3
166
Chapter 1 Review 21. a. C 5 x
18. Let x represent the distance from Kevin’s place of work to his home. Distance
Rate
To Work
x
45
To Home
x
30
b. C 80
5 x 80 x 16 The dancer will save money on the 17th dance during a 3-month period.
Time
x 45 x 30
22. Let t represent the time it takes Petra and Dawn to typeset the 150-page manuscript (which is equivalent to three 50-page manuscripts) if they work together. 1 job 1 job 3 jobs 4 days 3.5 days t days
x x 50 45 30 60 x x 50 180 180 45 30 60 4 x 6 x 150 10 x 150 x 15 The distance is 15 mi.
1 1 3 14t 14t 4 3.5 t 3.5t 4t 42 7.5t 42 t 5.6 days
19. Let x represent the speed of the boat traveling north. Then, x 6 is the speed of the boat traveling south. Distance
Rate
Time
Northbound
3x
x
3
Southbound
3 x 6
x6
3
23. Let t represent the time it takes the second pump to drain the pond by itself. 1 job 1 job 1 job 22 hr t hr 10 hr 1 1 1 110t 110t 22 t 10
5t 110 11t 110 6t 55 t 18.3 hr 3
3x 3 x 6 66 3x 3x 18 66 6 x 18 66 6 x 48 x8 x 6 8 6 14 The northbound boat travels 8 mph and the southbound boat travels 14 mph.
24. Let x represent the number of female officers. Then, x 60 represents the number of male officers. x 60 10 x 7 x 60 10 7x 7x x 7
20. a. C A 300 4 x b. C A 360 2 x c.
7 x 60 10 x
C A CB 300 4 x 360 2 x 2 x 60 x 30 If Monique takes 30 classes during the year, the cost for each gym will be the same.
7 x 420 10 x 420 3 x 140 x x 60 140 60 200 There are 140 females and 200 males.
167
Chapter 1 Equations and Inequalities 34. 4 7i 5 i 20 4i 35i 7i 2
25. Let x represent the number of turtles in the pond. 12 3 x 36 3x 432
20 31i 7 1 20 31i 7 27 31i 35. 4 6i 4 2 4 6i 6i 2
x 144 There are approximately 144 turtles in the pond.
2
16 48i 36i 2 16 48i 36 1 16 48i 36 20 48i
26. 169 i 169 13i
28.
16 4 i 16 i 4 4i 2i 8i 2
8 6i 2 2 1 6 6i 2
8 1 8
37. 8 3i 8 3i 8 3 64 9 73 2
29. a. Real part: 3; Imaginary part: 7 b. Real part: 0; Imaginary part: 2
38.
30. a. i 35 i 32 i 3 1 i 3 i
12 13i 3 9 13i 9 13 i 9 1 10 10 10
39. 6 5i
1 6 5i 6 1 5i 6 5i 6 5i
b. i 1 c. i 62 i 60 i 2 1 i 2 1
d. i i i 1 i i 1
e. i 5 i 8 i 3 1 i 3 i
1
6 5i 6 5i 6 5i 62 5 36 5 41 6 5 i 41 41
2 3 1 2 31. i i 3 5 6 5 20 18 5 12 i i 30 30 30 30 20 5 18 12 i 30 30 30 30
15 6 1 1 i i 30 30 2 5
5
40.
7 7i 7i 7i 7 2 i 4i 4i i 4i 4 1 4
41.
3y2 4 y 8 6 y 3y2 2 y 8 0
3 y 4 y 2 0
32. 3i 7 2i 21i 6i 2 21i 6 1 6 21i 33.
3y 4 0 3y 4
11 3 i 5 11 i 3 i 55 i
2
4 3i 4 3i 3 i 12 4i 9i 3i 2 3i 32 12 3 i 3 i
56
1
8 2i 2 4i 2 2i 2
12 i 12 2i 3
16
36. 2 2 4 2 2 i 2 4 i 2
27.
17
2
2
y
15
i 55 1 15
4 , 2 3
15 i 55
168
4 3
or
y20 y 2
Chapter 1 Review 45. x 2 5 x 2 x 4
42. 2v 3 1 6 2
x2 5 x2 4 x 2 x 8
2v 3 2 7
x2 5 x2 2 x 8
2v 3 7
2 x 3
2v 3 7
x
3 7 v 2 3 7 2
3 2 46.
43. 10t 1210 0 2
10t 2 1210 0 10 10 10 2 t 121 0
44.
1 2 2 7 x x 5 3 15 2 1 7 15 x 2 15 x 5 15 3
3x 2 10 7 x 3x 2 7 x 10 0
t 2 121
3x 10 x 1 0
t 121 11i
11i
3 2
3 x 10 0 3x 10
2d d 3 1 4d
x
2d 6d 1 4 d 2
a 2, b 10, c 1
10 3
1 47. x 2 18 x n x 2 18 x 18 2
b b 4ac 2a 2
10
x 1
10 , 1 3
2d 10d 1 0 2
d
x 1 0
or
x 2 18 x 9
10 4 2 1 2 2 2
2
2
x 2 18 x 81 x 9 n 81; x 9
10 100 8 10 108 4 4 10 6 3 5 3 3 4 2 5 3 3 2
48. x 2
2
2
2 2 1 2 x n x2 x 7 7 2 7 x2
2 1 x 7 7
2
2
2 1 1 x x x 7 49 7 2
n
169
1 1 ;x 49 7
2
2
Chapter 1 Equations and Inequalities x 2 10 x 9
49. a.
2 x 2 3x 5 0 2 x 2 3x 5 0 2 2 2 2 3 5 x2 x 2 2
b.
x 2 10 x 9 0
x 1 x 9 0 1, 9
x 1 0 x 1
or
x9 0 x9
2
3 5 1 3 1 3 x x 2 2 2 2 2 2 2
x 2 10 x 9
b.
2
1 1 x 2 10 x 10 9 10 2 2 2 x 10 x 25 9 25
x2
2
2
3 49 x 4 16
x 52 16 x 54 x 1 1, 9
x
x 5 16 x 5 4 or x 5 4 x9
3 7 4 4 10 x 4 5 x 2 5 , 1 2 x
x 2 10 x 9
c.
x 2 10 x 9 0 a 1, b 10, c 9 x
b b 2 4ac 2a 10
x
10 100 36 10 64 10 8 2 2 2 5 4 1 or 9 1, 9
2 x 5 x 1 0 or
x 1
b b 2 4ac 2a 3
32 4 2 5 2 2
3 9 40 3 49 3 7 4 4 4 3 7 3 7 x or x 4 4 4 4 10 4 x x 4 4 5 x x 1 2 5 , 1 2
2 x 2 3x 5 0 2x 5 0 2x 5 5 x 2 5 , 1 2
or
3 49 4 16 3 7 x 4 4 3 7 x 4 4 4 x 4
c. 2 x 2 3x 5 0 a 2, b 3, c 5
102 4 1 9 2 1
50. a.
3 9 40 9 x 2 16 16 16
x 1 0 x 1
170
2
Chapter 1 Review 51. False
s a0t 2 v0t s0
58.
a0t 2 v0t s0 s
52. True
a0t 2 v0t s0 s 0
53. a. 4 x 2 20 x 25 0
a a0 , b v0 , c s0 s
b 2 4ac 20 4 4 25 2
t
400 400 0
v0
v0 4 a0 s0 s t 2 a0 v0 v0 2 4a0 s0 s
b. The discriminant is 0; there is one real solution.
t
2 y 2 5 y 1
54. a.
b b 2 4ac 2a
2a0
2 y 5 y 1 0 2
59. Let x and x 3 represent the width and
b 2 4ac 5 4 21 2
length of the decorative area. Then x 1
25 8
and x 4 are the width and length of the rug. x 1 x 4 108
33 b. 33 0 ; there are two real solutions. 55. a.
5t t 1 4t 11
x 2 4 x x 4 108
5t 2 5t 4t 11
x 2 5 x 104 0
5t t 11 0 2 b 2 4ac 1 4 511
x 8 x 13 0
1 220
x8 x 1 8 1 9
2
x 8 0
219
19.25 x 2 1.5 x 0.5 x 0.75 0 x 2 x 20 0 x 5 x 4
2
y k r x h 2
x 13
19.25 x 0.5 x 1.5
HkRt kRt
57. x h y k r 2 2
x 13 0
60. Let x and x 2 represent the width and length of the finished tablecloth. Then x 0.5 and x 1.5 are the width and length of the cloth. A lw
H kI 2 Rt H kI 2 Rt kRt kRt H I2 kRt H or I I kRt
or
x 4 8 4 12 The rug is 9 ft by 12 ft.
b. 219 0 ; there are two imaginary solutions. 56.
2
2
2
y k r 2 x h
x5 0
x40
x5 x 4 x 0.5 5 0.5 5.5
2
y k r 2 x h
or
2
x 1.5 5 1.5 3.5 The cloth is 3.5 ft by 5.5 ft.
171
Chapter 1 Equations and Inequalities 61. Let x and 1.6x represent the width and length of the screen. a 2 b2 c2
64. a. d 0.048v 2 2.2v
0.048 50 2.2 50 2
0.048 2500 110
x 2 1.6 x 2 502
120 110 230 ft
x 2.56 x 2500 2
2
3.56 x 2 2500
b.
2500 3.56 2500 26.5 x 3.56 1.6 x 1.6 26.5 42.4 The width is 26.5 in. and the length is 42.4 in.
390 0.048v 2 2.2v
x2
d 0.048v 2 2.2v 390 v
x 2 x 2 5.4 x 7.29 49 2 x 5.4 x 41.71 0
b b 2 4ac 2a
5.42 4 2 41.71 2 2
b.
5.4 362.84 4 3.4 or 6.1 x 2.7 3.4 2.7 6.1 The length is 6.1 in. and the width is 3.4 in.
0 8t 2 100t 39 t
1 S n n 1 2 1 4186 n n 1 2 4186 n 2 n
n 91 n 46 0 n 91 The value of n is 91.
or
b b 2 4ac 2a 100
100 2 4 8 39 2 8
100 8752 16 0.4 or 12.1 The mortar will be at an 80-ft height 0.4 sec after launch.
n 2 n 4186 0 n 91 0
s 16t 2 200t 2 80 16t 2 200t 2 40 8t 2 100t 1
63.
2.22 4 0.048 390 2 0.048
1 65. a. s gt 2 v0t s0 2 1 s 32 t 2 200 t 2 2 s 16t 2 200t 2
2
2.2
x 2 x 2.72 72
5.4
b b 2 4ac 2a
2.2 79.72 0.096 70 or 116 The car was traveling 70 mph.
62. Let x and x 2.7 represent the width and length of the screen. a 2 b2 c 2
x
d 0.048v 2 2.2v
n 46 0 n 46
172
Chapter 1 Review 66.
P 0.009t 2 2.05t 182
69.
k 7 3 k 2
0 0.009t 2.05t 118
k 7 3 k 2
2
2
b b 2 4ac t 2a t
2.05
2k 4 3 k
2.05 2 4 0.009 118 2 0.009
2k 2 4 3 k 4k 2 16 3 k
2.05 8.4505 48 or 275 0.018 1950 t 1950 48 1998 The population reached 300 million in 1998.
4k 2 16k 48 0 k 2 4k 12 0
k 2 k 6 0
4 x 20 x 6 x 30 0
2
x 5 4 x 6 0 x 5 2 x 3 0
Check: k 2
4x x 5 6 x 5 0 2
2
2
2 7 3 2 ⱨ 2
x
3 2
k 7 3k 2
6 7 3 6 ⱨ 2
3x x 2 20 x
2
3x 6 x 20 x
2
2
4
k 6
9 1ⱨ2 3 1ⱨ 2 2 ⱨ 2 true Check: k 6
or 2 x 3 0
3 5, 2 68.
or
k 7 3k 2
x 5 or
2
k2
2
x2 5
2
4k 2 48 16k
4 x 3 6 x 2 20 x 30 0 3
2
k 7 3k 4 3k 4
67.
k 7 3 k 2
300 0.009t 2.05t 182 2
2
1 9 ⱨ2 1 3ⱨ 2 2 ⱨ 2 false 2 ; The value 6 does not check.
3x 4 7 x 2 20 0 Let u x 2 . 3u 2 7u 20 0
3u 5 x 4 0 5 3 5 x2 3 u
or
u 4
or x 2 4
x
5 3
x
15 or 3
or
x 4 x 2i
173
Chapter 1 Equations and Inequalities
70.
n 4 1 3n 2 n2 n 4 1 3n 2 n 2 3n 2 n 2 n 2 3n 2 n n 2 1 3n 2 n 2 4 3n 2 n 2 2n 3n 2 4n 4 12n 8 4n 2 18n 12 0 2n 2 9n 6 0 n
9
92 4 2 6 9 129 2 2 4
9 129 4 71. Let u v 1. 11v 2 23v 1 2 0
74.
51 14 x x 6
51 14 x x 6
11u 23u 2 0 2
2
11u 1 u 2 0 1 u 11 1 v 1 11 v 11
u 2
or
0 x 2 2 x 15 0 x 5 x 3
or v 1 2
x 5 or Check: x 5 51 14 x 4 x 2
1 v 2
or
72.
51 70 ⱨ 11
4 x 2x 1 0 Check: x 3
4 x 3 2x 1
4 x 2 x 1 3
3
3
51 14 3 4 ⱨ 3 2 51 42 ⱨ 3
3 3x
9 ⱨ 3
1 x
3 ⱨ 3 false
; The values 5 and 3 do not check.
73. 2 3m 4 3 5
2 3m 4 8
19 ⱨ 11 false
51 14 x 4 x 2
3
4 x 2x 1
1
x3
51 14 5 4 ⱨ 5 2
3
3
2
51 14 x x 2 12 x 36
1 , 11 2 3
51 14 x 4 x 2
3m 4 4
174
Chapter 1 Review p2 3 7
75.
p 7 23 32
1 3 1 d1 3 3 u
32
p 73 2
7 32
d
13 3
p3 4 7
76.
p 7 34 43
7
2 y 5
y 5
54
78.
11
27 8
2
v 2 10v 9 0
v 1 v 9 0 v 1
or
v9
2u 1 1
or
2u 1 9
2u 2
or
2u 2 10
u2 1
or
u2 5
2
y 5 114 5
2
u 1 or
1 10 23 10 w 10 1 10 w2 3
5, 1
w2 3 10 2
81. Let u
w 10 23 32
d
or
2
2
45
d
3 2
2u 1 10 2u 1 9 0
22
y 55 4 11 4 5 y 5 114 5
1 27
or
13 3
80. Let v 2u 2 1. 54
5 114 5
3
1 27 , 27 8
p 74 3
43
77.
or
1 3
d
43
3 2 3 d1 3 2 u
or
2 3 2
w 103
2
u 5
4 1. w 2
4 4 2 1 10 1 0 w w
1 1000
2u 2 10u 0
1 1000
u0
79. Let u d 1 3 . 6d 2 3 7 d 1 3 3 0
4 1 0 w 4 1 w w 4 1, 4
6u 2 7u 3 0
3u 1 2u 3 0
175
2u u 5 0 or u5 4 1 5 or w 4 4 or w or w 1
3
Chapter 1 Equations and Inequalities 4v 10 4 v 5v 25 v 5 5 4v 10 4 v 5 v 5 v 5 5
82.
85.
a1t1 a2t2 v1 v2 at a t v1v2 1 1 v1v2 2 2 v1 v2
4v 10 4 5 v 5 v 5 v 5 5 5 v 5 v 5
v2 a1t1 v1a2t2 va t v2 1 2 2 a1t1
4v 5 10 v 55v 4 4v 50 5v 2 4v 25v 20 5v 2 25v 30 0
86.
v 2 5v 6 0
v 2v 3 0 v 2 or
2, 3 83.
m
v3
p | p 27 ; , 27
1 2a 2 2b 2 2c 2 2
87. 0.6 0.2 x 0.8 x 1.8
2m 2a 2 2b 2 2c 2
2m 2a 2b 2c 2
2
2
2
2 p 14 3 2 18 p 3 27 p or p 27 4
1.2 0.6 x 2 x or x 2 x | x 2 ; 2,
2
4m 2 2a 2 2b 2 2c 2 2m 2 a 2 b 2 c 2 a 2 2m 2 b 2 c 2 88.
a 2m 2 b 2 c 2 84.
1 1 1 a b c 1 1 1 abc abc a b c bc ac ab bc ab ac
4 2 y 3 y 1 2 y 8 4 y 3y 3 2 y 11 y or y 11
y | y 11 ; 11,
b c a ac b
2 y y 1 y 3 4 6 2 y y 1 y 12 12 3 4 6
ac ac or b ca ac
176
Chapter 1 Review
89. 9 5 4 t 1 3 2 5 t 2
9 5 4t 4 3 2 5 t 2
91. a. X Y b. X Y x | 2 x 7
9 4t 9 3 2 t 3
c. X Z x | x 7
9 4t 9 3 2 t 3
d. X Z x | x 3
4t 3 t 1 4t 3t 3 t 3 t | t 3 ; 3,
e. Y Z x | x 3 or x 2 f. Y Z
90. a. A B 10,11,12,13,14,16 b. A B 10,12 c. A C 7, 8, 9,10,11,12,13 d. A C 10,11
92. a. t 2 8
or
t6 , 6 ;
or
b. t 2 8
and
t6
and
, 12 ;
e. B C 7, 8, 9,10,11,12,14,16 f. B C 10 93. a. 2 x 1 4 x 3
2 x 2 4 x 3 3x 3 x 1 , 6< 1, ; b. 2 x 1 4 x 3
x 1
or
5 x 2 3 4 x 1
or or or
5 x 10 3 4 x 1 x 6 x 6
and
5 x 2 3 4 x 1
and
x 6
3x 9 6 4 3x 9 0 6 4 0 3x 9 24 9 3x 33 3 x 11 3,11
95. 0
94. 11 4 x 1 7
10 4 x 8 10 x 2 4 5 2 x 2 5 2, 2
177
1 t 4 3 t 12
1 t 4 3 t 12
Chapter 1 Equations and Inequalities 96. 29, 000 a 31, 000
b. w 2 1 6
w 2 5
97. Let x represent the September rainfall. 8.54 5.79 8.63 x 7.83 4 22.96 x 7.83 4 22.96 x 31.32
7, 3
c. w 2 1 6
w 2 5
x 8.36 More than 8.36 in. is needed.
w 2 5 or w 7 or , 7 3,
98. Let p represent the price of sod. 400 2000t 850
2000t 450 t 0.225 She can afford sod that is $0.225/ft2 or less.
102. a.
3 7x 1 4 1 7 x 1
99. a. x 12 0
x 12 x | x 12
b.
3 7x 1 4 1 7 x 1
,
b. 12 x 0 12 x or x 12 x | x 12
c.
3 7x 1 4 1 7 x 1
100. a. 5 x 7 0
5 x 7 7 x 5 7 x | x 5
103. a. y 5 3 3
y5 0 y5 0 y 5
5
b.
b. y 5 3 3
101. a. w 2 1 6
y5 0
w 2 5 w 2 5 w3 7, 3
5 w 2 5 7 w 3
or or
w 2 5 w 7
c. y 5 3 3
y5 0 y5 0 y 5
5
178
w 2 5 w3
Chapter 1 Review d.
y 5 3 3
106.
y5 0 y50 y 5
or or
1 2 1 x 1 2 3 6
2 1 1 6 x 6 1 2 3 6 3x 4 1 6 3x 4 7 3x 4 7 or 3x 3 or
y50 y 5
, 5 5, e. y 5 3 3
y5 0
, 104. a.
x 1
b.
2
x 1 3x 5 4 x 4 x 1
1 x 2 1 2x x 2 4x x 3x 0 x0 0
107. 2 x
or or or
x 1 x 5 2 x 4 x 2
or or
x 1 1 x 00
5 5 x 1 4 1 5x 1 or or
x0
or
or or or
1 2x x 2 4x x 5x 0 x0
x 2 1 x 1
109. 0.5 x 8 0.01 0.01 0.5 x 8 0.01 7.99 0.5 x 8.01 15.98 x 16.02 15.98,16.02
1 5 x 1 5x 1 1 5x 0
or
108. 4 x 2 10 6 4 x2 4 x 2 1 x 2 1 or x 3 or , 3 1,
x 1 1 x x 1 1 x 2x x x 1 ,
105.
or or or
x 1 x 5 x 1 x 5 1 5
c.
11 1, 3
x 1 3x 5 x 1 3x 5 2 x 6 x 3 3, 1
or
3x 4 7 3x 11 11 x 3
5 x 1 1 5 x 2 2 x 5
110.
2 0, 5
179
9 4 2k 1 13 2k 1 13 2k 1 2k 1 13 13 2k 1 13 12 2k 14 6 k 7 6, 7
Chapter 1 Equations and Inequalities Chapter 1 Test 1.
25 4 5i 2i 10i 2 10 1 10
7. a.
x 2 10 x 25 0
2. a. i i i 1 i i 89
88
1
b 2 4ac 10 4 1 25
1
2
b. i 46 i 44 i 2 1 i 2 1
100 100 0 b. Because the discriminant is 0, there is one real solution.
c. i 35 i 32 i 3 1 i 3 i d. i120 1
8. a.
e. i 11 i 12 i1 1 i1 i
3x 2 10 x 2 0 2 b 2 4ac 10 4 3 2 100 24 76
24 34i 14 1
b. Because 76 0 , there are two real solutions.
24 34i 14 38 34i 2
3
2
9. 3 y 2 5 y 4 2 5 y 6 7 y 3
9 30i 25i 2
3 y 2 5 y 20 2 5 y 42 6 y 3
9 30i 25 1
3 y 2 5 y 22 11 y 39 3 y 10 y 44 11 y 39 13 y 44 11 y 39 2 y 83 83 y 2 83 2
16 30i 5.
3 x x 4 2 x 2 3x 2 12 x 2 x 2
3. 4 7i 6 2i 24 8i 42i 14i 2
4. 3 5i 3 2 3 5i 5i
x 2 25 10 x
4 3i 4 3i 2 5i 2 5i 2 5i 2 5i 8 20i 6i 15i 2 2 2 52 8 26i 15 1 4 25 7 26i 29 7 26 i 29 29
10.
6. a. b 2 4ac 4 4 2 7 2
16 56 40 b. Because 40 0 , there are two imaginary solutions.
2 t 3t 1 2t 5 1 6 4 3 2 t 3t 1 2t 5 12 12 1 6 4 3 2 2 t 3 3t 1 12 4 2t 5 4 2t 9t 3 12 8t 20 7t 7 8t 32 t 25 25
11. 0.4 w 1 0.8 0.1w 0.3 4 w 0.4 w 0.4 0.8 0.1w 1.2 0.3w 0.4w 1.2 0.4 w 1.2 00
180
Chapter 1 Test 11 2 1 2 x x 15 2 x 5 x 3 11 2 1 2 x 5 x 3 2 x 5 x 3
12.
2
11 2 1 2 x 5 x 3 x 3 2 x 5 x 3 2 x 5
2 x 5 x 3
11 2 x 3 1 2 x 5 11 2 x 6 2 x 5 2 x 17 2 x 5
; The value 3 does not check.
12 4 x 3 x
13. 3x 4 2 11 2
6t 2t 1 5 5t
15.
12t 2 11t 5 0
3x 4 2 13
3t 1 4t 5 0
3x 4 13
3t 1 0
3x 4 13 4 13 x 3
2
1 1 y 10 y 10 4 10 2 2 2 y 10 y 25 4 25
16.
2
2
1 3
t
5 4
3x 2 1 x 4 2 2 3x 1 x 4 4 2 4 3x 2 4 x 2
y 52 29
3x 2 4 x 2 0
y 5 29
5 29
4t 5
1 5 , 3 4
y 2 10 y 4
14.
4t 5 0
3t 1 t
4 13 3
or
a 3, b 4, c 2
y 5 29
x
b b 2 4ac 2a 4
4 2 4 3 2 4 16 24 2 3 6
4 8 4 2i 2 2 i 2 6 6 3 2 i 2 3
181
Chapter 1 Equations and Inequalities 12 y 3 24 y 2 3 y 6
17.
19.
12 y 24 y 3 y 6 0 3
12 y 3 3 y 24 y 2 6 0
y 4 y 1 2 4 y 1 0 4 y 1 y 2 0
2 d 7 8 d 2
2 y 1 2 y 1 y 2 0 2y 1
1 2
y
y
4d 28 d 2 16d 64
2 y 1 0 or y 2 0
2 y 1
0 d 2 20d 36 0 d 2 d 18 d 2 or d 18 Check: d 2 2d 1 d 7
y 2
1 2
1 , 2 2
2 2 ⱨ1
18. 2 y 3 4 y 5 13
13
2 y 31 3 4 y 51 3 3
3
2 18 ⱨ1
8 2 y
4 c 72 4 2 c6 c 36 c 72 4 c6 c 6 c 6
c 6 c 6
18 7
36 ⱨ1 25 6 ⱨ1 5 6 ⱨ 4 false ; The values 2 and 18 do not check.
4 y
20.
2 7
4 ⱨ1 9 2 ⱨ1 3 2 ⱨ 2 false Check: d 18 2d 1 d 7
0
2 y 31 3 4 y 51 3 2y 3 4y 5
2
4 d 7 64 16d d 2
2
or
2
2 d 7 8d
2
2y 1 0
2d 1 d 7
2d 1 2 d 7 d 7
3y 4 y2 1 6 4 y2 1 0 2
2d 1 d 7 2
2
c 72 4 c 6 c 6 c6 c 6 c 6
c c 6 4 c 6 c 6 72 c 2 6c 4c 2 144 72 3c 2 6c 72 0 c 2 2c 24 0 c 6 c 4 0
c 6 or c 4 4 ; The value 6 does not check.
182
Chapter 1 Test 21. w4 5 11 0
26.
w4 5 11
w 11 45 54
54
a2 b2 c
a 2 b2
w 11
11
b 2 a 2 c 2
b2 a 2 c2 b a2 c2
2 22. Let u 5 . k
27. 16t 2 v0t 2 0
2
2 2 5 6 5 27 0 k k
16t 2 v0t 2 0
u 2 6u 27 0
a 16, b v0 , c 2
u 3 u 9 0
t
u 3
u9
or
2 5 3 k 2 8 k 2 8k
or
2 5 9 k 2 4 k 2 4k
or
k
or or
1 4 1 1 , 4 2 k
23.
t t
1 2
b b 2 4ac 2a v0
v0 4 16 2 2 16 2
v0 v0 2 128 32
28. a. A B b. A B x | 0 x 2 c. A C x | x 2 d. A C x | x 1
2 x 3 6 4 x3 x3 4
or
x 3 4
x7 1, 7
or
x 1
e. B C x | x 1 or x 0 f. B C 29.
24. 2v 5 2v 1
2v 5 2v 1
or
2v 5 2v 1
5 1
or
4v 4
or
v 1
1 25.
2
a 2 b2 c2
54
54
c 2
4 x 6 3 6 4 x 18 2
10 x 14 10 x 14 or 14 x 10 14,10 30.
aP 4 Pt 2 aP Pt 6 Pa t 6
4 y 24 3 y 18
10,
6 6 P or P at ta
183
or
y 7 2y 3
or
10 y
Chapter 1 Equations and Inequalities 31. 3 x 5 1 4 x 2 6
and
0.3x 1.6 0.2
3 x 15 1 4 x 8 6
and
0.3 x 1.8
0 x
and
x6
6,
2 1 4w 3
32.
b. x 13 4 4
x 13 0
3 4w 3
4w 3 3 4 w 3 3
or
4w 3 3
4w 0
or
4w 6
or
3 w 2
w0
c. x 13 4 4
x 13 0 x 13 0 x 13 13
, 0 , 3 2
d. 33. 8 v 6
x 13 4 4 x 13 0 x 13 0 or x 13 or ,13 13,
8v 6 6 8 v 6 14 v 2 14 v 2
e.
2 v 14
x 13 4 4 x 13 0 x 13 0 x 13 ,
2,14
34. a. 7 x 4 11 2
7 x 4 9
or or
x 13 0 x 13
36. Let x represent the amount of 80% antifreeze solution (in gallons) to be mixed. 2 gal is the amount of the 50% antifreeze solution to be mixed. Therefore, (x + 2) is the amount of the resulting 60% antifreeze solution.
b. 7 x 4 11 2
7 x 4 9
c. 7 x 4 11 2
Amount of Sol. Pure Antifreeze
7 x 4 9
, 35. a. x 13 4 4
80% Sol.
50% Sol.
60% Sol.
x
2
x2
0.8 x
0.5 2
0.6 x 2
0.8 x 0.5 2 0.6 x 2 0.8 x 1 0.6 x 1.2 0.2 x 0.2 x 1 1 gal of 80% antifreeze should be used.
x 13 0 x 13 0
13
x 13 0 x 13
x 13
184
Chapter 1 Test 40. Let x represent the base of the triangular portions and x 7 represent the height.
37. Let x represent the speed of the plane flying to Seattle. Then, x 60 is the speed of the plane flying to New York City.
Seattle Flight New York Flight
Distance
Rate
Time
2.3x
x
2.3
3.3 x 60
x 60
3.3
1 A 2 bh lw 2 1 276 2 x x 7 18 x 7 2 2 276 x 7 x 18 x 126 0 x 2 25 x 150 0 x 30 x 5
2.3 x 3.3 x 60 2662
x 30 or x5 x 7 5 7 12 The base of the triangular portions is 5 ft and the height is 12 ft.
2.3x 3.3x 198 2662 5.6 x 198 2662 5.6 x 2464 x 440 x 60 440 60 500 The plane flying to Seattle flies 440 mph, and the plane flying to New York flies 500 mph.
1 41. a. s gt 2 v0t s0 2 1 32 t 2 60 t 2 16t 2 60t 2 2 b. 52 16t 2 60t 2
38. Let t represent the time it would take the second hose to fill the pool if it worked alone. 1 job 1 job 1 job 3 hr t hr 1.2 hr 1 1 1 6t 6t 3 t 1.2
0 16t 2 60t 50 0 8t 2 30t 25 b b 2 4ac t 2a
2t 6 5t 6 3t t2 The second hose can fill the pool in 2 hr.
30
302 4 8 25 2 8
30 100 30 10 16 16 40 5 20 5 t 2.5 or t 1.25 16 2 16 4 The ball will be at a height of 52 ft at times 1.25 sec and 2.5 sec after being kicked.
39. Let x represent the patient’s LDL cholesterol level. The HDL cholesterol level is 70 mg/dL, and the total cholesterol is x 70 .
x 70 3.8 70 x 70 70 70 3.8 70 x 70 266 x 196 x 70 196 70 266 The LDL level is 196 mg/dL and the total cholesterol is 266 mg/dL.
42. Let s represent the score on the sixth round. 92 88 85 90 89 s 88 6 444 s 88 6 444 s 528
s 84 The golfer would need to score less than 84. 185
Chapter 1 Equations and Inequalities Chapter 1 Cumulative Review Exercises 1 3 5x 1 3 5 x 5 1 3x 5. 5 x 5 2 1 2 1 10 x 5x x 5 x 5
1. 5 x 3 5 x 3
2 2
2
25 x 2 30 x 9 25 x 2 30 x 9
25 x 2 30 x 9 25 x 2 30 x 9
2
2
60 x 3600 x 2 2
6.
2
2. 4 3 2 2 4 3 2 2 4 3 2 2
2
16 3 4 2
3x 2 x 4 3x 4 2 2 4 x 8 x 12 6 x 54 2 3x 2 x 4 6 x 9 3x 4 4 x2 2 x 3
2
2
48 8 40 3.
7 3 7 3 7 3 7 3 2 7 3 7 3 2 7 3 2
2
73
3x 4 x 1 6 x 3 x 3 4 x 3 x 1 3x 4
2
7 3 4 7 3 2
7. 3 81y 5 z 2 w12 3 27 y 3 w12 3 y 2 z 2
3
3x 4 x 1 6 x 3 x 3 3x 4 4 x 3 x 1
3 yw4 3 3 y 2 z 2
2
4.
3 x 3 2
8. a. 4 11 or 11 4 b. 4 11 4 11
6 5 x 2 x2 x2 x 4 x 6 5 x 2 x 2 x 2 x 2
4 x 2 y x 2 xy 4 y
3 9. 4 x 3 32 y 6 4 x 3 8 y 6 4 x3 2 y 2 2
x 2 6 x 2 5 x 2 x 2 x 2 x 2
10.
x
x 2 x 2 6 x 2 5 x 2 x x 2 x 2
6 x 12 5 x 10 x x 2 x 2
2 x 22 x 2 x 2
3 7i 3 7i 2 5i 2 5i 2 5i 2 5i 6 15i 14i 35i 2 4 25 6 29i 35 1 29 29 29i 29 1 i
186
2
2
4
Chapter 1 Cumulative Review 11. Let x represent the amount borrowed at 4%. Then, 8000 x is the amount borrowed at 5%. 4% Interest Loan
5% Interest Loan
x
8000 x
x 0.041
8000 x 0.051
Principal Interest (I = Prt)
Total
380
x 0.04 8000 x 0.05 380 0.04 x 400 0.05 x 380 0.01x 400 380 0.01x 20 x 2000 8000 x 8000 2000 6000 Stephan borrowed $6000 at 5% and $2000 at 4%. 12. 4 x 1 3 6 2
14. Let u
4 x 12 3
2
x x 2 1 5 1 12 0 3 3 2u 2 5u 12 0 2u 3 u 4 0
4x 1 3 4x 1 3 x
1 3 4
3 2 x 3 1 3 2 x 1 3 2 3 x 2 3 , 15 2 u
1 3 4 2 x x 4 2 x 5
13.
2 x2 8x 2 x 5 2 x 2 10 x 5 0 a 2, b 10, c 5 x
b b 2 4ac 2a 10
x 1. 3
102 4 2 5 2 2
15.
u 4
or
x 1 4 3 x 5 3
or or
x 15
or
x4 2 x x4 x2
x 4 x 2
10 100 40 4 10 140 4 10 2 35 5 35 4 2 5 35 2
2
2
x 4 x2 4 x 4 x 2 3x 0 x x 3 0 x0
187
or
x 3
Chapter 1 Equations and Inequalities 18. a. A B
Check: x 0 x4 2 x
b. A B x | 4 x 11
0 4 2 ⱨ 0
c. A C x | x 11
2 2ⱨ0 0 ⱨ 0 true Check: x 3 x4 2 x
d. A C x | x 2 e. B C x | x 2 or x 4 f. B C
3 4 2 ⱨ 3 1 2 ⱨ 3 1ⱨ 3 false
0
19. 2 x 11 1 12
2 x 11 11
16. 5 x 6 4
5 x 2
0,11
5 x 2 5 x 2 x 7 x7
3, 7
or or or
5 x 2 x 3 x3
11 2 x 11 11 0 2 x 22 0 x 11
3 20. y 15 5 y 25
25, x9
17.
72 x 8
x 8 x 9 x 8
72 x 8
x 2 17 x 72 72 x 2 17 x 0 x x 17 0
0,17
x0
or
x 17
188
Section 2.1
Chapter 2 Functions and Graphs Section 2.1 The Rectangular Coordinate System and Graphing Utilities 1. origin 3. d
x x y y2 b. M 1 2 , 1 2 2
2. quadrants
4 2 11 7 , 2 2
x2 x1 2 y2 y1 2
6 18 , 3, 9 2 2
x x y y2 4. M 1 2 , 1 2 2
5. solution
6. 0
7. 0
8. 0; y
12. a. d
x2 x1 2 y2 y1 2
3 1 7 3 2
9.
42 42
16 16 32 4 2 x x y y2 b. M 1 2 , 1 2 2 3 1 7 3 , 2 2 2 10 , 1, 5 2 2
10.
13. a. d
x2 x1 2 y2 y1 2
2 7 5 4 2
9 2 9 2
81 81 162 9 2
11. a. d
x2 x1 2 y2 y1 2
4 2 11 7 2
x x y y2 b. M 1 2 , 1 2 2 2 7 5 4 , 2 2
2
5 1 5 1 , , 2 2 2 2
22 4 2 4 16 20 2 5
189
2
2
Chapter 2 Functions and Graphs 14. a. d
x2 x1 2 y2 y1 2
17. a. d
4 32 1 6 2
72 72
4 5 5 7 2 2 3 5 6 2
2
2
45 72 117
x x y y2 b. M 1 2 , 1 2 2
x x y y2 b. M 1 2 , 1 2 2
4 3 1 6 , 2 2
4 5 5 7 2 2 , 2 2
1 5 1 5 , , 2 2 2 2
5 5 8 2 5 5 , , 4 2 2 2 2
x2 x1 2 y2 y1 2
5.2 2.22 6.4 2.4
32 4 2 9 16 25 5
2
2 7 7 5 3 5 7 4 5 7 80 87 2
6 2 82
2 7 7 5 3 5 , 2 2
x2 x1 2 y2 y1 2
2
x x y y2 b. M 1 2 , 1 2 2
7.4 8.8 , 3.7, 4.4 2 2
31.1 37.1 2 32.7 24.7
2
2
5.2 2.2 6.4 2.4 , 2 2
x2 x1 2 y2 y1 2
18. a. d
x x y y2 b. M 1 2 , 1 2 2
16. a. d
2
2
49 49 98 7 5
15. a. d
x2 x1 2 y2 y1 2
3 7 2 5 3 7 , , 5 2 2 2
2
19. d1
36 64 100 10 x x y y2 b. M 1 2 , 1 2 2
3 1 2 1 32 4 4 2 2
d2
0 3 2 2 12 9 9 3 2
d3
1 02 3 2 1 25 26 2
d12 d 2 2 0 d32
31.1 37.1 32.7 24.7 , 2 2
2 2 3 2 0 26 2
2
2
8 18 0 26
68.2 57.4 , 34.1, 28.7 2 2
26 0 26 True Yes
190
Section 2.1 20. d1
3 1 2 0 2 2 4 4 2 2
d3 1 3 2 2 2
2
16 17 0 1 10 1 False
2
No
16 16 4 2 d12 d32 0 d 2 2 2 2
42 17 0 1
3 3 2 2 0 2 36 4 2 10
d2
2
2
1 3 0 1 2 4 1 3 01 4 4 10 1 Yes
2
8 32 0 40 40 0 40 True Yes 21. d1 5 2 0 4 49 16 65 2
1 3 2 0 4
d3 2 5 4 1 9 9 3 2 2
d d3 0 d 2 2 1
2
2
2 20 4
2
40 4
65 3 2 0 101 2
x3 y 4
24. a.
5 52 1 0 2 100 1 101
d2
x2 y 1
c.
4 2 0 2 10
2
x2 y 1
b.
2
Yes
2
x3 y 4
b.
65 18 0 101
2 3 3 0 4
83 0 101 False
5 30 4
No 22. d1
6 32 2 1 2 81 1 82
d2
3 1 2 1 2 4 9 13
d3
6 12 2 2
8 0 4 False No
2
1 11 3 0 4 10 10
2
49 16 65 d 2 2 d32 0 d12
13 65 0 82 2
2
x3 y 4
c.
1 11 10 3 0 10 4 10 10 1 30 110 40
2
29 110 40
13 65 0 82
40 0 40
78 0 82 False Yes
No
25. x 3 0
x2 y 1
23. a.
x3 x | x 3
2 2 3 0 1 4 30 1 10 1 Yes
191
Chapter 2 Functions and Graphs 26. x 7 0 x 7 x | x 7
32. y x 2
x
3
27. x 10 0
x 10 x | x 10
y
y x2
9
y 3 9
2
4
1
1
0
0
1
1
2
4
3
9
28. x 11 0
x 11 x | x 11
Ordered pair 2
y 2 4 2
y 1 1 2
y 0 0 2
y 1 1 2
y 2 4 2
y 3 9 2
3, 9 2, 4 1,1 0, 0 1,1 2, 4 3, 9
29. 1.5 x 0 x 1.5
x 1.5 x | x 1.5 30. 2.2 x 0
x 2.2 x 2.2 x | x 2.2 33. y x
31. y x x
y
y x
Ordered pair
3
3
y 3
2
2
y 2
1
1
y 1
0
0
y0
1
1
y 1
2
2
y2
3
3
y3
3, 3 2, 2 1, 1 0, 0 1,1 2, 2 3, 3
192
x
y
y
x
Ordered pair
0
0
y 00
1
1
y 1 1
4
2
y 42
9
3
y 93
0, 0 1,1 4, 2 9, 3
Section 2.1 34. y x
36. y
x
y
y x
3
3
y 3 3
2
2
y 2 2
1
1
y 1 1
0
0
y 0 0
1
1
y 1 1
2
2
y 2 2
3
3
y 3 3
1 x
Ordered pair
3, 3 2, 2 1,1 0, 0 1,1 2, 2 3, 3
y
4
1 4
2
1 2
1
1
1 2
x
y
y x
2
8
y 2 8
1
1
0
0
1
1
2
8
Ordered pair 3
y 1 1 3
y 0 0 3
y 1 1 3
y 2 8 3
2, 8 1, 1 0, 0 1,1 2, 8
193
1 4, 4
y
1 1 2 2
1 2, 2
y
1 1 1
1,1
y
1 2 1 2
2
1 , 2 2
y
1 2 1 2
1 , 2 2
Ordered pair
1 2
2
1
1
y
1 1 1
1,1
2
1 2
y
1 1 2 2
1 2, 2
4
1 4
y
1 1 4 4
1 4, 4
35. y x 3 3
1 x 1 1 y 4 4 y
x
Chapter 2 Functions and Graphs 37. y x 2
38. x y 3
y x 2
y 3 x
x
y
y x 2
3
5
y 3 2 5
2
4
y 2 2 4
1
3
y 1 2 3
0
2
y 0 22
1
3
y 1 23
2
4
y 2 24
3
5
y 3 25
Ordered pair
3, 5 2, 4 1, 3 0, 2 1, 3 2, 4 3, 5
39. y 2 x 2 0 y2 x 2 y x2
x
y
y x2
Ordered pairs
2
1
y 2 2 0
1
1
y 1 2 1
2
2
y 2 2 2
2, 0 1,1 , 1, 1 2, 2 , 2, 2
7
3
y 7 2 3
7, 3 , 7, 3
194
x
y
y 3 x
3
0
y 3 3 0
2
1
y 3 2 1
1
2
y 3 1 2
0
3
y 3 0 3
1
2
y 3 1 2
2
1
y 3 2 1
3
0
y 3 3 0
Ordered pair 3, 0
2,1 1, 2 0, 3 1, 2 2,1 3, 0
Section 2.1 40. y 2 x 1 0 y2 x 1 y x 1
41.
x
y
y x 1
Ordered pairs
1
0
y 1 1 0
2
1
y 2 1 1
5
2
y 5 1 2
1, 0 2,1 , 2, 1 5, 2 , 5, 2
10
3
y 10 1 3
10, 3 , 10, 3
Ordered pairs
x y 1 y x 1 y x 1
42.
x
y
y x 1
1
0
y 1 1 0
2
1
y 2 1 1
3
2
y 3 1 2
1, 0 2,1 , 2, 1 3, 2 , 3, 2
4
3
y 4 1 3
4, 3 , 4, 3
5
4
y 5 1 4
5, 4 , 5, 4
x y 3 y x3 y x 3 x
y
y x 3
3
0
y 3 3 0
2
1
y 2 3 1
1
2
y 1 3 2
3, 0 2, 1 , 2, 1 1, 2 , 1, 2
0
3
y 0 3 3
0, 3 , 0, 3
1
4
y 1 3 4
2
5
y 2 3 5
1, 4 , 1, 4 2, 5 , 2, 5
Ordered pairs
195
Chapter 2 Functions and Graphs 45. x-intercepts: 1, 0 , 9, 0 ;
43. y x 1 x
y
y x1
3
2
y 3 1 2
2
1
1
0
0
1
1
2
2
3
3
4
y 2 1 1
y 1 1 0
y 0 1 1
y 1 1 2
y 2 1 3 y 3 1 4
y-intercepts: 0, 3 , 0, 3
Ordered pair 3, 2
46. x-intercepts: 16, 0 , 4, 0 ;
y-intercepts: 0, 8 , 0, 8
2,1 1, 0 0,1 1, 2 2, 3 3, 4
47. x-intercept: 2, 0 ; y-intercept: none 48. x-intercept: none; y-intercept: 0,1 49. x-intercept: 0, 0 ; y-intercept: 0, 0 50. x-intercepts: 0, 0 , 6, 0 ; y-intercept: 0, 0 51. x-intercept: none; y-intercept: none 52. x-intercept: none; y-intercept: none 53. x-intercept: 1, 0 ; y-intercept: 0, 2 54. x-intercept: 5, 0 ; y-intercept: 0, 4
44. y x 2 x
y
y x2
2
4
y 2 2 4
1
3
0
2
1
1
2
0
3
1
4
2
y 1 2 3
y 0 2 2 y 1 2 1
y 2 2 0 y 3 2 1
y 4 2 2
Ordered pair 2, 4
55. Substitute 0 for y: 2 x 4 y 12
Substitute 0 for x: 2 x 4 y 12
2 x 4 0 12
2 0 4 y 12
2 x 12
4 y 12 y3
x 6
1, 3 0, 2 1,1 2, 0 3,1 4, 2
x-intercept: 6, 0 ; y-intercept: 0, 3 56. Substitute 0 for y: 3x 5 y 60
Substitute 0 for x: 3x 5 y 60
3 x 5 0 60
3 0 5 y 60
3 x 60 x 20
5 y 60 y 12
x-intercept: 20, 0 ; y-intercept: 0, 12 57. Substitute 0 for y: x2 y 9
x 2 0 9
Substitute 0 for x: x2 y 9
0 2 y 9
x2 9 y9 x 3 x-intercepts: 3, 0 , 3, 0 ; y-intercept: 0, 9
196
Section 2.1 58. Substitute 0 for y: x 2 y 16
Substitute 0 for x: x 2 y 16
62. Substitute 0 for y: x y2 4
x 2 0 16
02 y 16
x 0 4
0 y 2 4
x 4
4 y2 2 y
x 2 16
Substitute 0 for x: x y2 4
2
y 16
x 4 x-intercepts: 4, 0 , 4, 0 ; y-intercept:
x-intercept: 4, 0 ; y-intercepts:
0,16
0, 2 , 0, 2
59. Substitute 0 for y: y x5 2
63. Substitute 0 for y: x y
0 x 5 2 x5 2 x 5 2
or
x5 2
x3 or Substitute 0 for x: y x5 2
x7
x 0
0 y
x0
0 y
x-intercept: 0, 0 ; y-intercept: 0, 0 64. Substitute 0 for y: x 5y
y 0 5 2 3 x-intercepts: 3, 0 , 7, 0 ; y-intercept: 0, 3
Substitute 0 for x: x 5y
x 5 0
0 5 y
x 0
0 y
x0 x-intercept: 0, 0 ; y-intercept: 0, 0
60. Substitute 0 for y: y x4 3
0 x 4 3 65.
x4 3 x 4 3 or x 7 or Substitute 0 for x: y x4 3
Substitute 0 for x: x y
x 3 2 y 4 2 1 4
9
x 3 y 4 2 36 36 1 9 4
x43 x 1
2
9 x 3 4 y 4 36 Substitute 0 for y: 2
y 0 4 3 1
2
9 x 3 4 y 4 36 2
x-intercepts: 7, 0 , 1, 0 ; y-intercept:
2
9 x 3 4 0 4 36 2
2
0,1
9 x 3 64 36 2
61. Substitute 0 for y: x y2 1
x 0 1 1 2
Substitute 0 for x: x y2 1
9 x 3 28 2
0 y 2 1
x 3 2
1 y2 1 y
28 9
x3
x-intercept: 1, 0 ; y-intercepts:
0, 1 , 0,1
28 9
x 3 Not a real number.
197
28 9
Chapter 2 Functions and Graphs Substitute 0 for x:
67. d AC 4 6 8 10 2
9 x 3 4 y 4 36 2
2
100 4 104 2 26
9 0 3 4 y 4 36 2
2
4 6 2 8 0 2
d BC
81 4 y 4 36 2
4 y 4 45
4 64 68 2 17 Observation tower B is closer.
45 4
68. a. d 1 2 3 3
2
y 4 2
y4
2
45 4
2 1 3 3 , b. M 2 2
Not a real number. x-intercept: none; y-intercept: none
1 0 1 , , 0 2 2 2
x 6 2 y 32 1 16
69. a. l 1 2 3 0 2
4
2
9 9 18 3 2 ft
x 6 2 y 3 2 16 16 1 4 16
w
3 1 2 1 32
4 4 8 2 2 ft
x 62 4 y 3 2 16
b. P 2l 2 w
Substitute 0 for y:
2 5 2 10 2 ft A lw 3 2 2 2 6 2 12 ft
x 6 4 y 3 16 2 x 6 2 4 0 3 16 x 62 36 16 x 6 2 20 2
2
9 36 45 3 5 mi 6.7 mi
45 4
y 4
66.
2
2
2 3 2 2 2 2
70. a. l 5 1 1 4
x 6 20
2
x 6 20
2
2
36 25 61 ft
Not a real number. Substitute 0 for x:
w 1 2 4 3 2
x 62 4 y 32 16 2 2 0 6 4 y 3 16 2 36 4 y 3 16 2 4 y 3 20 y 32 5
2
1 1 2 ft b. P 2l 2 w
61 2 2 2 61 2 ft A lw 61 2 122 ft 2
2
y 3 5 y 3 5 Not a real number. x-intercept: none; y-intercept: none
198
Section 2.1 2 4 1 3 2 4 , 71. C , 1, 2 2 2 2 2
4 2 3 1 d r 2 2 36 4 40 2 10 10 2 2 2 2
75. b
76. e
77. c
78. f
79. d
80. a
2
1975 1995 68.8 72.5 81. M , 2 2 3970 141.3 , 1985, 70.65 2 2 70.65 yr
Center: 1, 2 ; Radius: 10 5 2 3 1 3 2 3 , , ,1 72. C 2 2 2 2 2 2 5 1 3 d r 2 2 49 16 65 2 2 2
3 Center: ,1 ; Radius: 2
10 2810 29 20 82. M , 2 2 2820 49 , 1410, 24.5 2 2 24.5C
2
83. d AB
65 2
d BC
4 2 2 3 2 2 4 1 5
8 4 2 5 32
16 4 20 2 5
7 1 6 2 8 4 , , 4, 2 73. M 2 2 2 2 h
4 0 2 5 16 9 25 5
b
7 12 6 2
2
d AC
36 9 45 3 5 d AB d BC 0 d AC
2
5 2 50 3 5
2
3 5 0 3 5 True Collinear
36 64 100 10 1 1 Area bh 10 5 25 m 2 2 2
84. d AB
2
2
b 7 4 5 4
8 42 3 22 16 1 17
d AC
2 82 1.5 3 2 36 2.25
0.5 17 17 0 1.5 17 2
1.5 17 0 1.5 17 True Collinear
9 81 90 3 10 1 3 10 90 Area 3 10 22.5 m 2 2 2 4
d BC
38.25 2.25 17 1.5 17 d AB d BC 0 d AC
81 9 90 3 10 4 4 4 2 2
4 22 2 1.5 2 4 0.25
4.25 0.25 17 0.5 17
7 4 5 4 11 1 , , 74. M 2 2 2 2 1 11 h 1 2 2 2
2 8 2 2 5 2
199
Chapter 2 Functions and Graphs 85. d AB 1 2 2 8 2
91. d
2
9 36 45 3 5
4 1 3 2 9 25 34
d AC
2 4 8 3
2
2
92. d
2
93. d
3 5 34 0 157 False Not collinear
94. d
2
2
0 32 5 72 1 2
2
1 5 2 5 13
2 92 0 5 1 3 2
2
49 25 16 90 3 10
5 0 2 13 3 2
25 256 281 d AC
2
9 144 9 162 9 2
86. d AB 0 1 3 5 1 4 5 2
2 6 2 3 4 1 1
16 49 4 69
36 121 157 d AB d BC 0 d AC
d BC
2
1 81 9 91
d BC
2
4 52 6 3 1 2 2
95. 2
36 324 360 6 10 d AB d BC 0 d AC 5 281 0 6 10 False Not collinear 87. The points x1 , y1 and x2 , y2 define the endpoints of the hypotenuse d of a right triangle. The lengths of the legs of the triangle are x2 x1 and y2 y1 . Applying the Pythagorean theorem produces 2
96.
2
d 2 x2 x1 y2 y1 , or equivalently
d
x2 x1 2 y2 y1 2 for d 0 .
88. The midpoint formula results in an ordered pair. The x-coordinate of the midpoint is the average of the x-coordinates of the endpoints. The y-coordinate of the midpoint is the average of the y-coordinates of the endpoints.
97. 3 4i
32 42 9 16 25 5
98. 6 8i
6 2 8 2
36 64 100 10
89. To find the x-intercept(s), substitute 0 for y and solve for x. To find the y-intercept(s), substitute 0 for x and solve for y. 90. The graph of the equation represents the set of all solutions to the equation graphed in a rectangular coordinate system.
200
99. 2 7i
2 2 72 4 49 53
100. 1 9i
12 92 1 81 82
Section 2.1 101. The viewing window is part of the Cartesian plane shown in the display screen of a calculator. The boundaries of the window are often denoted by [Xmin, Xmax, Xscl] by [Ymin, Ymax, Yscl].
105.
102. 780 x 42 y 5460 42 y 780 x 5460 780 x 5460 y 42 130 y x 130 7
106.
107.
108.
Window d
103.
109.
110. 104.
201
Chapter 2 Functions and Graphs Section 2.2 Circles
x2 x1 2 y2 y1 2
1. a. d
6 5 7 2 2
5. d 2r 10 2r 5r 5 ft
2
121 25 146 x x y y2 b. M 1 2 , 1 2 2
6. d 2r 2 4.2 8.4 in.
5 6 2 7 1 9 , , 2 2 2 2
2. a. d
x2 x1 2 y2 y1 2
2 4 1 3 2
7. circle; center 9. x h y k r 2 2
10. general 11. x 2 y 7 4 2
2
2 2 2 7 7 2 0 4
x x y y2 b. M 1 2 , 1 2 2
0 00 4 0 0 4 False
4 2 3 1 , 2 2
No
12. x 3 y 5 36 2
6 4 , 3, 2 2 2
2
3 32 5 5 2 0 36 0 0 0 36
Substitute 0 for x: x 2 y 2 16
x 2 0 16
02 y 2 16
x 2 16
y 2 16
x 4
y 4
2
2
2
44 8 2 2
3. Substitute 0 for y: x 2 y 2 16
8. radius
0 0 36 False No
13.
x-intercepts: 4, 0 , 4, 0 ; y-intercepts:
x 1 2 y 3 2 25 4 1 2 7 3 2 0 25 9 16 0 25
0, 4 , 0, 4
25 0 25 True Yes
4. Substitute 0 for y: x 2 y 2 25
Substitute 0 for x: x 2 y 2 25
x 2 0 25
02 y 2 25
x 62 y 12 100 2 62 7 12 0 100
x 2 25
y 2 25
64 36 0 100
x 5
y 5
2
14.
100 0 100 True Yes
x-intercepts: 5, 0 , 5, 0 ; y-intercepts:
0, 5 , 0, 5
15. r 2 81 r 81 9 Center: 4, 2 ; Radius: 9
202
Section 2.2 16. r 2 16
b.
r 16 4 Center: 3,1 ; Radius: 4
17. r 2 6.25
r 6.25 2.5 Center: 0, 2.5 ; Radius: 2.5 18. r 2 2.25 24. a.
r 2.25 1.5 Center: 1.5, 0 ; Radius: 1.5 19. r 2 20
x h2 y k 2 r 2 2 2 2 x 3 y 2 4 x 3 2 y 2 2 16
b.
r 20 2 5 Center: 0, 0 ; Radius: 2 5 20. r 2 28
r 28 2 7 Center: 0, 0 ; Radius: 2 7 81 49 81 9 r 49 7 9 3 3 Center: , ; Radius: 2 4 7
21. r 2
25. a.
x h2 y k 2 r 2 x 4 y 3 2
2
11
2
x 4 2 y 3 2 11 b.
25 22. r 9 25 5 r 9 3 5 1 3 Center: , ; Radius: 7 5 3 2
23. a.
x h2 y k 2 r 2 2 2 2 x 2 y 5 1 x 2 2 y 5 2 1
26. a.
x h2 y k 2 r 2 x 5 y 2 2
2
21
x 5 2 y 22 21
203
2
Chapter 2 Functions and Graphs b.
r
x2 x1 2 y2 y1 2
2 2 2 4 1 2
16 9 25 5
x h2 y k 2 r 2 x 2 2 y 12 5 2 x 2 2 y 12 25 b.
27. a. x h y k r 2 2
2
x 0 2 y 02 2.6 2 x 2 y 2 6.76 b.
x x y y2 30. a. C 1 2 , 1 2 2 7 5 3 1 12 2 , 6,1 , 2 2 2 2
28. a. x h y k r 2 2
2
r
x 0 2 y 02 4.2 2
x2 x1 2 y2 y1 2
7 62 3 1 2 1 4 5 x h2 y k 2 r 2
x 2 y 2 17.64
x 62 y 1 2 5
b.
x 62 y 1 2 5 b.
x x y y2 29. a. C 1 2 , 1 2 2 2 6 4 2 4 2 , , 2,1 2 2 2 2
204
2
Section 2.2 31. a. r
x2 x1 2 y2 y1 2
2 6 2 1 52
33. a. The circle must touch the y-axis at 0, 6 , at one side of a diameter. r
64 36 100 10
4 02 6 62 16 4 x h2 y k 2 r 2 x 4 2 y 6 2 4 2 x 4 2 y 62 16
x h y k r 2 2 2 x 2 y 1 10 x 2 2 y 1 2 100 2
2
x2 x1 2 y2 y1 2
2
b. b.
32. a. r
x2 x1 2 y2 y1 2 3 6 1 5 2
34. a. The circle must touch the x-axis at 2, 0 , at one side of a diameter.
2
r
9 16 25 5
x h2 y k 2 r 2 x 3 2 y 12 5 2 x 3 2 y 12 25
x2 x1 2 y2 y1 2
2 2 4 0 16 4 2
2
x h2 y k 2 r 2 2 2 2 x 2 y 4 4 x 22 y 4 2 16
b. b.
205
Chapter 2 Functions and Graphs 39. x 1 y 5 0 The sum of two squares will equal zero only if each individual term is zero. Therefore, x 1 and y 5 .
35. a. The circle must touch the x-axis and the yaxis at a distance r from the center. Since the center is in quadrant IV, the center must be 5, 5 .
2
x h2 y k 2 r 2 2 x 52 y 5 52 x 52 y 5 2 25
2
1, 5
40. x 3 y 12 0 The sum of two squares will equal zero only if each individual term is zero. Therefore, x 3 and y 12 . 2
b.
2
3, 12
41. x 17 y 1 9 The sum of two squares cannot be negative, so there is no solution. 2
2
42. x 15 y 3 25 The sum of two squares cannot be negative, so there is no solution.
36. a. The circle must touch the x-axis and the yaxis at a distance r from the center. Since the center is in quadrant II, the center must be 3, 3 .
2
2
x h2 y k 2 r 2 2 2 2 x 3 y 3 3 x 3 2 y 3 2 9
43.
x2 y 2 6 x 2 y 6 0
x 6 x y 2 y 6 2
2
2
2
1 2 6 9
b.
1 2 2 1
x 6 x 9 y 2 y 1 6 9 1 2
2
x 32 y 12 4 Center: 3,1 ; Radius: 4 2 44.
x 2 y 2 12 x 14 y 84 0
x 12 x y 14 y 84 2
37.
38.
x h y k r 2 2 x 8 2 y 11 52 x 82 y 11 2 25 2
2
2
2
1 2 12 36
2
1 2 14 49
x 12 x 36 y 14 y 49 84 36 49 2
2
x 6 2 y 7 2 1 Center: 6, 7 ; Radius: 1 1
x h2 y k 2 r 2 2 2 2 x 4 y 16 9 x 42 y 162 81 206
Section 2.2 45.
x 2 y 2 20 y 4 0
x 2 y 2 20 y
49.
4
2
2
2
Center: 0,10 ; Radius: 104 2 26 x 2 y 2 22 x 4 0
50.
x 22 x y 4
x 2 y 2 10 x 22 y 155 0
x 10 x y 22 y 155 2
2
2
2
2
2
2
2
x 5 2 y 11 2 9 Degenerate case:
x 112 y 2 125 Center: 11, 0 ; Radius: 125 5 5
51. 4 x 2 4 y 2 20 y 25 0
47. 10 x 2 10 y 2 80 x 200 y 920 0 x 2 y 2 8 x 20 y 92 0
x2 y 2 5 y
x 8x y 20 y 92 2
2
2
1 2 20 100
2
2
25 25 25 x2 y 2 5 y 4 4 4
x 4 2 y 10 2 24 Center: 4, 10 ; Radius: 24 2 6
2
5 x2 y 0 2
2 x 2 2 y 2 32 x 12 y 90 0
5 Degenerate case (single point): 0, 2
x 2 y 2 16 x 6 y 45 0
x 16 x y 6 y 45 2
2
254
25 1 2 5 4
x 8 x 16 y 20 y 100 92 16 100
2
1 2 16 64
25 0 4
x2 y 2 5 y
2
1 2 8 16
1 2 22 121
x 10 x 25 y 22 y 121 155 25 121
x 22 x 121 y 4 121
2
2
1 2 10 25
2
1 2 22 121
52. 4 x 2 4 y 2 12 x 9 0
1 2 6 9
x 2 y 2 3x
x 16 x 64 y 6 y 9 45 64 9 2
2
x 2 2 y 92 4 Degenerate case:
2
2
2
1 2 18 81
x 4 x 4 y 18 y 81 89 4 81
x 2 y 10 104
48.
2
1 2 4 4
x 2 y 2 20 y 100 4 100
2
x 4 x y 18 y 89 2
1 2 20 100
46.
x 2 y 2 4 x 18 y 89 0
2
x 3x
x 8 2 y 3 2 28 Center: 8, 3 ; Radius: 28 2 7
2 y 2 2
9 1 2 3 4
207
9 0 4 9 4
Chapter 2 Functions and Graphs 57.
2
9 9 9 2 x 3x y 4 4 4 2
3 2 x y 0 2 3 Degenerate case (single point): , 0 2 3 3 y 0 2 4 3 3 y2 y 4 2
x2 y 2 x
53.
x x 2
The approximate location of the earthquake is 8, 7 . 58.
2
9 1 3 2 2 16
2
1 1 2 1 4
1 2 3 9 3 1 9 2 x x y y 4 2 16 4 4 16 2
2
1 3 25 x y 2 4 16 1 3 Center: , ; Radius: 2 4
54.
25 5 16 4
The fire is located at approximately coordinate 2,1 .
2 5 5 x y 0 3 3 9 5 2 5 y y 3 9
x2 y 2 2 2 x x 3
2
59.
36 36 12 y y 2 10 72 12 y y 2 10
2
1 2 1 2 3 9
1 5 25 2 3 36
72 12 y y 2 100 y 2 12 y 28 0
1 2 5 25 5 1 25 2 2 x x y y 3 9 3 36 9 9 36 2
y 2 y 14 0 y 2 or y 14 y 2 and y 14
2
1 5 49 x y 3 6 36 1 5 Center: , ; Radius: 3 6 55. x 4 y 6 1.5 2
2
49 7 36 6
60.
2
56. x 32 y 40 20 2
2
25 8 x x 2 5 25 8 x x 2 25 x2 8x 0 x x 8 0 x 0 or x 8 x 0 and x 8
2
x 32 y 40 400 2
4 x 2 2 1 5 16 8 x x 2 9 5
x 4 2 y 62 2.25 2
2 42 6 y 2 10
2
208
Section 2.2
2 x2 4 x2 6
61.
b.
4 4 x x 2 16 8 x x 2 6 20 12 x 2 x 2 6 20 12 x 2 x 2 36 2 x 2 12 x 16 0 x2 6 x 8 0 x
6
6 2 4 1 8 2 1
6 68 6 2 17 3 17 2 2 3 17, 3 17 and 3 17, 3 17
c.
4 x 2 6 x 4 2
62.
4 x 2 6 x 2 4
d.
16 8 x x 36 12 x x 4 2
2
52 20 x 2 x 2 4 52 20 x 2 x 2 16 2 x 2 20 x 36 0 x 2 10 x 18 0 x
10
102 4 118 2 1
64. a.
10 28 10 2 7 5 7 2 2
5 7, 5 7 and 5 7, 5 7 63. a.
b.
209
Chapter 2 Functions and Graphs c.
d.
d.
66. a.
65. a.
b.
b.
c.
c.
d.
210
Section 2.2 67. A circle is the set of all points in a plane that are equidistant from a fixed point called the center. 68. The center and radius can easily be identified from an equation of a circle written in standard form. 69. The calculator does not connect the pixels at the ends of the semicircles because of limited resolution.
73. 14, 39, 5 by 30, 5, 5
70. The screen is rectangular and is longer horizontally than vertically. If the x and y intervals are the same, then the graph will be elongated horizontally because there is greater distance horizontally over which to spread the graph. 71. 15.1,15.1,1 by 10,10, 1
74. 0.28, 0.18, 0.02 by 0.1, 0.2, 0.02
72. 15.1,15.1,1 by 10,10, 1
211
Chapter 2 Functions and Graphs
Section 2.3 Functions and Relations 1.
x2 x1 2 y2 y1 2
r
2
2 1 4 1 2 2
49 85 85 9 4 4 2
1 Center: , 2 ; Radius: 2
85 2
5. Substitute 0 for y: y x 3 5 x 2 4 x 20
2.
0 x 3 5 x 2 4 x 20 0 x 3 4 x 5 x 2 20
0 x x2 4 5 x2 4
0 x 4 x 5 0 x 2 x 2 x 5 x 2 or x 2 or x 5 Substitute 0 for x: y x 3 5 x 2 4 x 20
x x y y2 3. C 1 2 , 1 2 2
y 0 5 0 4 0 20 20
2 1 7 3 1 10 1 , , , 5 2 2 2 2 2 r
2
3
2
x-intercepts: 2, 0 , 2, 0 , 5, 0 ; y-intercept: 0, 20
x2 x1 2 y2 y1 2 2
1 2 2 7 5 2
6. Substitute 0 for y: y x3 3x 2 x 3 0 x3 3x 2 x 3
9 25 5 4 4 4 2 5 1 Center: , 5 ; Radius: 2 2
0 x3 3x 2 x 3 0 x3 x 3 x 2 3
0 x x2 1 3 x2 1
0 x 1 x 3
x x y y2 4. C 1 2 , 1 2 2
2
0 x 1 x 1 x 3 x 1 or x 1 or x 3 Substitute 0 for x: y x3 3x 2 x 3
3 4 5 1 1 4 1 , , 2 , 2 2 2 2 2
y 0 3 0 0 3 3 3
2
x-intercepts: 1, 0 , 1, 0 , 3, 0 ;
y-intercept: 0, 3
212
Section 2.3 7. relation; domain; y
25. No vertical line intersects the graph in more than one point. This relation is a function.
8. A relation defines y as a function of x if for each value of x in the domain, there is exactly one value of y in the range. 9. does not
10. 2, 4
11. y
12. f x
13. 5
26. There is at least one vertical line that intersects the graph in more than one point. This relation is not a function. 27. No vertical line intersects the graph in more than one point. This relation is a function. 28. There is at least one vertical line that intersects the graph in more than one point. This relation is not a function.
14. 5
15. a. {(Tom Hanks, 5), (Jack Nicholson, 12), (Sean Penn, 5), (Dustin Hoffman, 7)}
29. This mapping defines the set of ordered pairs: 8, 3 , 8, 4 , 8, 5 , 8, 6 . All
b. {Tom Hanks, Jack Nicholson, Sean Penn, Dustin Hoffman} c. {5, 12, 7}
four ordered pairs have the same x value but different y values. This relation is not a function.
d. Yes
16. a. {(Albany, 285), (Denver, 5883), (Miami, 11), (San Francisco, 11)}
30. This mapping defines the set of ordered pairs: 3,11 , 6,13 , 6, 9 . Two
b. {Albany, Denver, Miami, San Francisco} c. {285, 5883, 11}
ordered pairs have the same x value but different y values. This relation is not a function.
d. Yes
17. a. 4, 3 , 2, 3 , 1, 4 , 3, 2 , 3,1 b. 4, 2,1, 3
31. This mapping defines the set of ordered pairs: 1, 4 , 2, 5 , 3, 5 . No two ordered
c. 3, 3, 4, 2,1
pairs have the same x value but different y values. This relation is a function.
d. No 18. a. 2, 2 , 1, 0 , 0, 2 , 1, 0 , 1, 2 b. 2, 1, 0,1
32. This mapping defines the set of ordered pairs: 5, 4 , 6, 4 , 7, 4 , 8, 4 .
c. 2, 0, 2
No two ordered pairs have the same x value but different y values. This relation is a function.
d. No 19. False
20. True 33. x 1 y 5 25 This equation represents the graph of a circle with center 1, 5 and radius 5. This relation is not a function because it fails the vertical line test. 2
21. No vertical line intersects the graph in more than one point. This relation is a function. 22. No vertical line intersects the graph in more than one point. This relation is a function.
2
34. x 3 y 4 1 This equation represents the graph of a circle with center 3, 4 and radius 1. This relation is not a function because it fails the vertical line test.
23. There is at least one vertical line that intersects the graph in more than one point. This relation is not a function.
2
24. No vertical line intersects the graph in more than one point. This relation is a function.
213
2
Chapter 2 Functions and Graphs 35. y x 3
Two or more ordered pairs have the same x value but different y values. This relation is not a function.
38. a. y x y x
No vertical line intersects the graph in more than one point. This relation is a function.
36. y x 4 No vertical line intersects the graph in more than one point. This relation is a function.
b.
x y y x y x
No vertical line intersects the graph in more than one point. This relation is a function.
37. a. y x
b.
x
y
y x
Ordered pairs
0
0
y 0
1
1
y 1
2
2
y 2
3
3
y 3
0, 0 1, 1 2, 2 3, 3
2
Two or more ordered pairs have the same x value but different y values. This relation is not a function.
No vertical line intersects the graph in more than one point. This relation is a function.
39. 4,1
x y2
40. 7, 5
y2 x
41. f x x 2 3x
y x x
y
y x
0
0
y 00
1
1
y 1 1
4
2
y 4 2
9
3
y 9 3
a. f 2 2 3 2 4 6 2
Ordered pairs 0, 0
2
b. f 1 1 3 1 1 3 2 2
1, 1 4, 2 9, 3
c. f 0 0 3 0 0 0 0 2
d. f 1 1 3 1 1 3 4 2
e. f 2 2 3 2 4 6 10 2
214
Section 2.3 42. g x
49. k 5
1 x
5 1
4
Undefined
a. g 2
1 1 2 2
50. f 5 5 3 5 25 15 40
b. g 1
1 1 1
51. k 8
2
52. f 5 5 3 5 25 15 10 2
1 1 c. g 2 2 1 2
53. g t
1 1 d. g 2 2 1 2
e. g 2
1 1 t t
54. f a a 3 a a 2 3a 2
55. k a b
1 1 2 2
a b 1 a b 1
56. h a b 5
43. h x 5 a. h 2 5
57. f a 4 a 4 3 a 4 2
b. h 1 5
c. h 0 5
a 2 8a 16 3a 12
d. h 1 5
a 2 11a 28
e. h 2 5
58. f t 3 t 3 3 t 3 2
44. k x x 1 a. k 2
t 2 6t 9 3t 9 t 2 3t
2 1 1
1 0 Undefined
59. g 0
Undefined b. k 1
1 1 0 0
c. k 0
0 1 1 1
d. k 1
1 1 2
e. k 3
3 1 4 2
45. g 3
8 1 9 3
1 1 3 3
1 1 47. g 3 3 1 3
60. k 10
10 1
9
Undefined
46. h 7 5
61. f 9 7
62. f 1 6
63. f 3 4
64. f 2 3
65. f x 6 when x 1 66. f x 7 when x 9
48. h 7 5
67. f x 3 when x 2
215
Chapter 2 Functions and Graphs 68. f x 4 when x 3
d 2 18 2 36 Joe rides 36 mi in 2 hr.
12
75. Solve h x 0 :
Evaluate h 0 : h x x 8
2 2 d 18 12 3 3 12 mi
h x x 8
r x 250 x
x 8 x-intercepts: 8, 0 , 8, 0 ;
x 8
Evaluate k 0 :
0 x 2
k 0 0 2
k x x 2
x 2
2
x 2 x-intercepts: 2, 0 , 2, 0 ;
C 225 225 0.06 225 0.18 225
y-intercept: 0, 2
225 13.5 40.5 279 If the cost of the food is $225, then the total bill including tax and tip is $279.
77. Solve p x 0 :
Evaluate p 0 :
p x x 2 12
P x 1.075 x 0.40 x
p x x 2 12 p 0 0 12 12
0 x 2 12
P 60 1.075 60 0.40 60
2
x 2 12
1.075 60 24 1.075 84 90.30 If the cost of the book from the publisher is $60, the cost to the student after the bookstore markup and sales tax is $90.30.
0 2x 4
76. Solve k x 0 : k x x 2
C x x 0.06 x 0.18 x
f x 2 x 4
8
y-intercept: 0, 8
r x 250 x
73. Solve f x 0 :
h 0 0 8
0 x 8
r 122 250 122 128 128 mi
72.
g 0 3 0 12
x4 x-intercept: 4, 0 ; y-intercept: 0, 12
r 50 250 50 200 After having driven 50 mi, Frank still has 200 mi remaining.
71.
0 3x 12
g x 3x 12
3 x 12
40 2 b. 40 min is hr. 60 3 d t 18t
b.
Evaluate g 0 :
g x 3x 12
69. a. d t 18t
70. a.
74. Solve g x 0 :
x 12 2 3
x-intercepts: 2 3, 0 , 2 3, 0 ; y-intercept: 0,12
Evaluate f 0 :
78. Solve q x 0 :
f x 2 x 4
q x x 8 2
f 0 2 0 4
0 x2 8
2x 4
4 x2 x-intercept: 2, 0 ; y-intercept: 0, 4
x2 8 x 8 2 2
Evaluate q 0 : q x x2 8
q 0 0 8 8 2
x-intercepts: 2 2, 0 , 2 2, 0 ; y-intercept: 0, 8
216
Section 2.3 79. Solve r x 0 : r x x 8 0 x 8 x 8 0
84. Evaluate B 0 :
Evaluate r 0 : r x x 8
B x 0.12 x 2 0.86 x 1.6
r 0 0 8
B 0 0.12 0 0.86 0 1.6 1.6 2
0,1.6 ; The y-intercept means that for
8
x8 x-intercept: 8, 0 ; y-intercept: 0, 8
80. Solve s x 0 : s x x 3 0 x3 x3 0
x 0 (the year 2002), the number of Botox injection procedures in the United States was 1.6 million.
Evaluate s 0 :
85. Domain: 3, 2, 1, 2, 3 ;
s x x 3
Range: 4, 3, 3, 4, 5
s 0 0 3
86. Domain: 4, 2, 0, 2, 4 ;
3
x 3 x-intercept: 3, 0 ; y-intercept: 0, 3
81. Solve f x 0 :
Evaluate f 0 :
f x x 2
f x x 2
0 x 2 x2
f 0
Range: 2, 0, 2, 4, 6
87. Domain: 3, ; Range: 1, 88. Domain: , 4 ; Range: , 4
0 2
89. Domain: , ; Range: ,
2
x4 x-intercept: 4, 0 ; y-intercept: 0, 2
82. Solve g x 0 :
Evaluate g 0 :
g x x 3
g x x 3
0 x 3
g 0 0 3
x 3
90. Domain: , ; Range: , 91. Domain: , ; Range: 3, 92. Domain: , ; Range: , 2 93. Domain: 5,1 ; Range: 1,1, 3
3
x9 x-intercept: 9, 0 ; y-intercept: 0, 3
94. Domain: 4, 5 ; Range: 2,1, 4 95. x 4 0 x4 , 4 4,
83. Evaluate f 0 :
f x 9.4 x 35.7 f 0 9.4 0 35.7
96. x 2 0
35.7 0, 35.7 ; The y-intercept means that for x 0 (the year 2006), the average amount spent on video games per person in the United States was $35.70.
x2 , 2 2,
217
Chapter 2 Functions and Graphs 106. 6 x 0 x 6 x6 , 6
97. 2t 7 0 2t 7
t
7 2
7 7 , , 2 2
107. There is no restriction on x. ,
98. 3a 4 0 3a 4
a
108. There is no restriction on x. ,
4 3
109. a. f 2 4
4 4 , , 3 3
b. f 3 2 c. f x 1 for x 3 , x 1 , x 1
99. The denominator x 4 is always positive.
,
d. f x 4 for x 2 , x 2
10 e. x-intercepts: 0, 0 and , 0 3
100. The denominator x 2 3 is always positive. ,
f. y-intercept: 0, 0
101. a 15 0 a 15 15,
g. Domain: , h. Range: 4,
102. c 12 0 c 12 12,
110. a. f 2 1 b. f 3 3 c. There are no x-values where f x 1 .
103. 3 x 0 x 3
d. f x 4 for all x on the interval 0, 2
x3 , 3
e. x-intercept: 1, 0 f. y-intercept: 0, 4
104. 6 x 0 x 6 x6 , 6
g. Domain: , h. Range: 4 0, 111. a. f 2 0
105. 3 x 0 x 3 x3 , 3
b. f 3 5 c. f x 1 for x 3 , x 1 , x 1
218
Section 2.3 d. f x 4 for x 4 , x 0
119. f x 3x 2 2
4 e. x-intercepts: 2, 0 and , 0 3
120. f x x 3
f. y-intercept: 0, 4
121. If two points in a set of ordered pairs are aligned vertically in a graph, then they have the same x-coordinate but different ycoordinates. This contradicts the definition of a function. Therefore, the points do not define y as a function of x.
g. Domain: 4, h. Range: 4, 5 112. a. f 2 3
122. To find the x-intercept(s), find the real solutions to the equation f x 0 . To find
b. f 3 2
the y-intercept, evaluate f 0 .
c. f x 1 for x 0 , x 2
123. a. P s 4s
d. f x 4 for x 3 , x 5 e. x-intercept: 1, 0
b. A s s 2
f. y-intercept: 0, 1
P2 P c. A P or A P 4 16
2
g. Domain: , 5
d. P A 4 A
h. Range: 4, 0 3
e. d s s 2 s 2 2 s 2 2
113. r x x 400
d s 2s 2
r x 400 x
d s s 2
114. w x x 200
f. s d s d d 2 2
w x 200 x
2
2 s d d 2 2
115. P x x x x
2 d2 s d 2
P x 3x 116. 2 A x x 180
2 A x 180 x A x
180 x 2
sd
d2 2
sd
d
d 2 g. P d 4 2
117. C x x 90
C x 90 x
P d 2 2d
118. S x x 180
S x 180 x
219
2
or s d
d 2 2
Chapter 2 Functions and Graphs d 2 h. A d 2
2
d f. A d 2 A d
2d 2 A d 4 d2 A d 2
d 2 4
C g. A C 2
124. a. C r 2 r b. A r r 2 c. r d
2
d 2
AC
C 2 4 2
AC
C2 4
h. C A 2
d. d r 2r
2
A
2A C A 2 A C A 2
e. C d d
Section 2.4 Linear Equations in Two Variables and Linear Functions 1. a. No vertical line intersects the graph in more than one point. This relation is a function. b. ,
c. ,
d. 2, 0
e. 0, 3
c. ,
d. 3, 0
e. 0,1
c. 3
d. None
e. 0, 3
x x y y2 5. a. M 1 2 , 1 2 2 3 2 2 5 4 5 , 2 2
2. a. No vertical line intersects the graph in more than one point. This relation is a function. b. ,
b. ,
4 2 3 5 , 2 2 3 5 2 2, 3
3. a. No, a vertical line is not a function. b. 3
c. ,
d. 3, 0
e. None
b. d
x2 x1 2 y2 y1 2
2 3 2 4 5 5 2 2 5 5 2
2
4. a. No vertical line intersects the graph in more than one point. This relation is a function.
8 125 133
220
2
2
Section 2.4 20. 2 x y 4
x x y y2 6. a. M 1 2 , 1 2 2
y 2x 4
6 2 6 3 5 3 , 2 2
x 3 2 1 0
3 6 6 3 3 6 , , 3 3 2 2 2
b. d
x2 x1 2 y2 y1 2
2 6 6 5 3 3 6 4 3 2
2
y 2 0 2 4
2
x-intercept: 2, 0 ; y-intercept: 0, 4
2
21. 2 y 5 x 2
6 48 54 3 6
7. scatter
8. linear
5 y x 1 2
9. vertical
10. horizontal
x
11. True
12. m
y2 y1 x2 x1
1
13. zero
14. undefined
15. slope; intercept
16. mx b
17. m
f x2 f x1 x2 x1
0 2 5 2
18. horizontal
2
0 2
4
22. 3 y 4 x 6
4 y x2 3
4 y 3x 12
x 4
0
2 x-intercept: , 0 ; y-intercept: 0,1 5
19. 3 x 4 y 12 y
y 7 2 1
3 x3 4
x 1
y 0 3 2 3
0 3 2 3
9 2
y 10 3 2
0 2
3 x-intercept: , 0 ; y-intercept: 0, 2 2
x-intercept: 4, 0 ; y-intercept: 0, 3
221
Chapter 2 Functions and Graphs 23. x 6 x 6 6 6 6
26. 3x 2 4 3x 6 x2
y 2 0 2 4
x 2 2 2 2
y 2 0 2 4
x-intercept: 6, 0 ; y-intercept: None x-intercept: 2, 0 ; y-intercept: None
24. y 4 x 2 0 2 4
27. 0.02 x 0.05 y 0.1 2 x 5 y 10 5 y 2 x 10 2 y x2 5
y 4 4 4 4
x 5 0
x-intercept: None; y-intercept: 0, 4 25. 5 y 1 11
1
5 y 10 y2
5
x 2 0 2 4
y 4 2 8 5 0
x-intercept: 5, 0 ; y-intercept: 0, 2
y 2 2 2 2
28. 0.03x 0.07 y 0.21 3x 7 y 21 7 y 3x 21 3 y x3 7
x-intercept: None; y-intercept: 0, 2
x 1
0 1 7
y 24 7 3 18 7 0
x-intercept: 7, 0 ; y-intercept: 0, 3
222
Section 2.4 29. 2 x 3 y 2 y x 3 x 3 0 3 6
y 2 0 2 4
36. m
y2 y1 6 4 10 5 x2 x1 1 9 8 4
37. m
y2 y1 39 52 13 1 22 30 52 x2 x1 4
38. m
y2 y1 30 16 46 1 x2 x1 84 100 184 4
39. m
y2 y1 3.7 4.1 7.8 26 x2 x1 9.5 2.6 6.9 23
40. m
y2 y1 7.9 6.2 1.7 1 x2 x1 5.1 8.5 13.6 8
x-intercept: 0, 0 ; y-intercept: 0, 0
30. 2 x 5 y 2 y x 5 x 5 0
41. m
y 2 0 2 5 2
1 5
42. m
y2 y1 5 1 x2 x1 80 16
33. m
y2 y1 2.5 1 x2 x1 100 40
34. pitch
35. m
43. m
y2 y1 300 3 x2 x1 1000 10
32. m
y2 y1 x2 x1
3 2 3 4 1 1 10 5 10 10 10 4 3 7 7 70
x-intercept: 0, 0 ; y-intercept: 0, 0
31. m
y2 y1 1 6 5 5 20 5 3 10 3 7 x2 x1 7 2 4 4 4 4
44. m
y2 y1 x2 x1 52 5 6 3 6
5 2 6
5 6 2 6 6
y2 y1 5 3 3 3 x2 x1 11 2 11 2 3 11
2 3 11 11 11
2 33 11
45. Use points 1,1 and 0, 4 :
rafter rise 7 rafter run 12
m
y2 y1 1 7 6 3 x2 x1 2 24
4 1 3 y2 y1 3 x2 x1 0 1 1
46. Use points 2, 0 and 0,1 : m
223
y2 y1 1 0 1 x2 x1 0 2 2
30 12
Chapter 2 Functions and Graphs 47. Use points 0, 2 and 3,1 : m
59. a. 2 x 4 y 8
4 y 2 x 8 1 y x2 2 1 m ; y-intercept: 0, 2 2
y2 y1 1 2 1 1 x2 x1 3 0 3 3
48. Use points 0, 3 and 1, 1 : m
y2 y1 1 3 4 4 x2 x1 1 0 1
b.
49. Use points 0, 4 and 2, 4 : m
y2 y1 4 4 0 0 x2 x1 20 2
3 3 50. Use points ,1 and , 3 : 2 2 y y1 3 1 2 m 2 (undefined) x2 x1 3 3 0 2 2
51. Undefined
52. 0
53. 0
54. Undefined
55.
m
60. a. 3x y 6 y 3x 6
m 3 ; y-intercept: 0, 6
b.
y2 y1 x2 x1
4 y2 y1 5 52 5 y2 y1 208 y2 y1 41.6 ft
56.
m
61. a. 3x 2 y 4 2 y 3x 4 3 y x2 2 3 m ; y-intercept: 0, 2 2
y2 y1 x2 x1
5 216 8 x2 x1
5 x2 x1 1728
x2 x1 345.6 m
b.
57. Change in population over change in time 58. Change in distance over change in time which is speed
224
Section 2.4 62. a. 5 x 3 y 6 3 y 5x 6 5 y x2 3 5 m ; y-intercept: 0, 2 3
65. a. 2 y 6 8 2 y 14 y7
m 0 ; y-intercept: 0, 7 b.
b.
66. a. 3 y 9 6 3 y 3 y 1
63. a. 3x 4 y 3 y x 4 3 m ; y-intercept: 0, 0 4
m 0 ; y-intercept: 0, 1 b.
b.
67. a. 0.02 x 0.06 y 0.06 2x 6 y 6 6 y 2 x 6 1 y x 1 3 1 m ; y-intercept: 0,1 3
64. a. 2 x 3 y 2 y x 3 2 m ; y-intercept: 0, 0 3 b.
b.
225
Chapter 2 Functions and Graphs 68. a. 0.03x 0.04 y 0.12 3x 4 y 12 4 y 3x 12
70. a.
3 y x3 4
x y 1 3 4 x y 12 12 1 3 4 4 x 3 y 12 3 y 4 x 12 4 y x4 3 4 m ; y-intercept: 0, 4 3
3 m ; y-intercept: 0, 3 4
b.
b.
69. a.
x y 1 4 7 x y 28 28 1 4 7 7 x 4 y 28 4 y 7 x 28 7 y x7 4 7 m ; y-intercept: 0, 7 4
71. a. Linear
b. Linear
c. Neither
d. Constant
72. a. Linear
b. Neither
c. Constant 73. a. y mx b
1 xb 2 1 9 0 b 2 9b 1 y x9 2 y
b.
b. f x
226
1 x9 2
d. Linear
Section 2.4 74. a.
y mx b 1 y xb 3 1 4 0 b 3 4 b 1 y x4 3
b. f x 75. a.
78. a.
1 x4 3
b. f x
y mx b y 3x b
y 0x b yb 5b y5
6 3 b 3 b y 3 x 3
b. f x 5
b. f x 3x 3
y mx b y 5 x b
80. a.
8 5 2 b 8 10 b 2b y 5 x 2
81. a.
y mx b 2 y xb 3 2 3 5 b 3 10 3 b 3 1 b 3 2 1 y x 3 3
b. f x
y mx b y 0x b yb 3 b y 3
b. f x 3
b. f x 5 x 2 77. a.
3 x4 2
79. a. y mx b
6 3 1 b
76. a.
y mx b 3 y xb 2 3 2 4 b 2 2 6 b 4b 3 y x4 2
y mx b y 1.2 x b 5.1 1.2 3.6 b 5.1 4.32 b 0.78 b y 1.2 x 0.78
b. f x 1.2 x 0.78 82. a.
2 1 x 3 3
y mx b y 2.4 x b 2.8 2.4 1.2 b 2.8 2.88 b 0.08 b y 2.4 x 0.08
b. f x 2.4 x 0.08
227
Chapter 2 Functions and Graphs y mx b
y2 y1 6 2 8 2 0 4 4 x2 x1 y mx b
83. a. m
7 y xb 3 7 3 1 b 3 2 b 3 7 2 y x 3 3
y 2x b 6 2 0 b 6 b y 2x 6
b. f x 2 x 6
7 2 b. f x x 3 3
y y1 3 1 4 1 84. a. m 2 x2 x1 0 8 8 2 y mx b 1 y xb 2 1 3 0 b 2 3 b 1 y x3 2
87. m
88. m
x2 x1 f x2 f x1 x2 x1
4 1 3 3 1 2
2 6 4 1 5 1 4
89. a. Average rate of change
1 b. f x x 3 2
y2 y1 1 3 4 4 x2 x1 47 3 3 y mx b
4 y xb 3 4 1 4 b 3 19 b 3 4 19 y x 3 3
f x2 f x1 x2 x1
8244 6420 1824 $364.80/yr 10 5 5
b. Average rate of change
85. a. m
f x2 f x1 x2 x1
17,452 13,591 3861 $772.20/yr 25 20 5
c. Increasing 90. a. Average rate of change
y2 y1 3 4 7 7 1 2 3 x2 x1 3
f x2 f x1 x2 x1
75 64 11 5.5F/month 53 2
b. Average rate of change
4 19 b. f x x 3 3
86. a. m
f x2 f x1
f x2 f x1 x2 x1
64 79 15 7.5F/month 11 9 2
c. A positive rate of change means that average monthly temperature is increasing. A negative rate of change means that average monthly temperature is decreasing.
228
Section 2.4 f t 0.009t 2 2.10t 182
91. a.
93. f x x 2 3
f 0 0.009 0 2.10 0 182 182
f 0 0 3 3
f 10 0.009 10 2.10 10 182
f 1 1 3 1 3 2
0.9 21 182 203.9 f x2 f x1 Average rate of change x2 x1
f 3 3 3 9 3 6
2
2
2
2
2
f 2 2 3 4 3 1 2
203.9 182 21.9 2.2 million/yr 10 0 10 f t 0.009t 2 2.10t 182
b.
f x2 f x1
a. Average rate of change
f 40 0.009 40 2.10 40 182 2
14.4 84 182 280.4
f 1 f 0 1 0
2 3
1 1 1
1
b. Average rate of change
f 50 0.009 50 2.10 50 182
x2 x1
f x2 f x1 x2 x1
2
22.5 105 182 309.5 f x2 f x1 Average rate of change x2 x1
3 1
6 2 2
c. Average rate of change
309.5 280.4 29.1 2.9 million/yr 50 40 10
c. Increasing f t 0.2t 2 86t 4450
92. a.
f 3 f 1
f 0 f 2 0 2
f x2 f x1 x2 x1
3 1 4 2 2 2
94. g x 2 x 2 2
f 0 0.2 0 86 0 4450 4450
g 0 2 0 2 2
f 10 0.2 10 86 10 4450
g 1 2 1 2 2 2 4
20 860 4450 5290 f x2 f x1 Average rate of change x2 x1
g 3 2 3 2 18 2 20
2
2
2
2
2
g 2 2 2 2 8 2 10 2
5290 4450 840 84 million/yr 10 0 10
a. Average rate of change
f t 0.2t 2 86t 4450
b.
f 20 0.2 20 86 20 4450 2
80 1720 4450 6090
g 1 g 0 1 0
2
180 2580 4450 6850 f x2 f x1 Average rate of change x2 x1
g 3 g 1 3 1
c. Decreasing
229
g 0 g 2 0 2
x2 x1
g x2 g x1 x2 x1
20 4 16 8 2 2
c. Average rate of change
6850 6090 760 76 million/yr 30 20 10
g x2 g x1
42 2 2 1 1
b. Average rate of change
f 30 0.2 30 86 30 4450
8 4 2
g x2 g x1 x2 x1
2 10 8 4 2 2
Chapter 2 Functions and Graphs 95. h x x 3
97. m x x
h 1 1 1
m 0 0 0
h 0 0 0
m 1 1 1
h 1 1 1
m 4 4 2
h 2 2 8
m 9 9 3
3
3
3
3
a. Average rate of change
h 0 h 1 0 1
0 1 1
b. Average rate of change
h 1 h 0 1 0
h 2 h 1 2 1
a. Average rate of change
x2 x1 1 1 1
h x2 h x1 x2 x1
h x2 h x1 x2 x1
8 1 7 7 1 1
3
3
3
a. Average rate of change k 0 k 1 0 1
1
1 2 1
c. Average rate of change
k x2 k x1
2 3
b. Average rate of change
2 1
32 1 5 5
n 10 10 1 9 3
k 2 2 2 8 2 6
94
x2 x1
n 5 5 1 4 2
k 1 1 2 1 2 1
k 2 k 1
m x2 m x1
n 2 2 1 1 1
k 0 0 2 0 2 2
1 0
m 9 m 4
x2 x1
2 1 1 3 3
4 1
m x2 m x1
n 1 1 1 0 0
3
m 4 m 1
x2 x1
98. n x x 1
k 1 1 2 1 2 3
k 1 k 0
1 0
m x2 m x1
1 0 1 1 1 1
c. Average rate of change
96. k x x 3 2
m 1 m 0
b. Average rate of change
1 0 1 1 1 1
c. Average rate of change
h x2 h x1
6 1 1
a. Average rate of change
x2 x1
1 1 1
k x2 k x1
n 2 n 1 2 1
1 1 1 k x2 k x1
n 5 n 2 52
7 7 1
230
n 10 n 5 10 5
n x2 n x1 x2 x1
2 1 1 3 3
c. Average rate of change
x2 x1
x2 x1
1 0 1 1 1 1
b. Average rate of change
x2 x1
n x2 n x1
n x2 n x1
32 1 5 5
x2 x1
Section 2.4 99. a. 1
b. , 1
c. 1,
100. a. 1
b. ,1
c. 1,
101. a. 2
b. , 2
c. 2,
102. a. 1
b. 1,
c. ,1
103. a. 5
b. 5,
c. , 5
110. If C is zero, then the line passes through the origin. 111. The slope and y-intercept are easily determined by inspection of the equation. 112. The average rate of change of f on the interval x1 , x2 is the slope of the secant line passing through the points x1 , f x1 and
x , f x . 2
104. a. 9.5
b. , 9.5 c. 9.5,
105. a. 14
b. 14,
c. ,14
106. a. 1.5
b. ,1.5
c. 1.5,
2
113.
107. a. {8}; The number of men and women in college was approximately the same in 1978.
The x-intercept of the line is 2, 0 . The base of the triangle is 2 units. The y-intercept of the line is 0, 4 . The height of the triangle is 4. 1 1 A bh 2 4 4 units 2 2 2
b. 0, 8 ; The number of women in college was less than the number of men in college from 1970 to 1978. c. 8, 40 ; The number of women in college exceeded the number of men in college from 1978 to 2010.
114.
108. a. {14.7}; The per capita consumption of pork and chicken was roughly the same midyear, 1994. b. 14.7, 30 ; The per capita consumption of pork was less than the per capita consumption of chicken midyear 1994 to 2010.
The x-intercepts of the two lines are 6, 0
and 3, 0 . The base of the triangle is
3 6 9 units . The y-intercept of the two
c. 0,14.7 ; The per capita consumption of pork exceeded the per capita consumption of chicken from 1980 to midyear 1994.
lines is 0, 6 . The height of the triangle is 6. A
109. The line will be slanted if both A and B are nonzero. If A is zero and B is not zero, then the equation can be written in the form y k and the graph is a horizontal line. If B is zero and A is not zero, then the equation can be written in the form x k , and the graph is a vertical line.
231
1 1 bh 9 6 27 units 2 2 2
Chapter 2 Functions and Graphs 119. a. 3.1 2.2 t 1 6.3 1.4t
115.
3.1 2.2t 2.2 6.3 1.4t 2.2t 0.9 1.4t 6.3 3.6t 5.4 t 1.5
1.5
The x-intercepts of the two lines are 4, 0
and 6, 0 . The base of the triangle is
6 4 10 units . The y-intercept of the
two lines is 0, 2 . The height of the triangle is 2. 1 1 A bh 10 2 10 units 2 2 2
b. , 1.5
116. 120. a.
c. 1.5,
11.2 4.6 c 3 1.8c 0.4 c 2 11.2 4.6c 13.8 1.8c 0.4c 0.8 2.8c 2.6 0.4c 0.8 3.2c 1.8
c 0.5625
0.5625
The x-intercepts of the line are 0, 0 and
4, 0 . The diameter of the half circle is 4
units, so the radius is 2 units. 1 1 2 A r 2 2 2 units 2 2 2
117. a. Ax By C By Ax C y
b. , 0.5625
A C x B B
c. 0.5625,
121. a. 2 x 3.8 4.6 7.2
A b. m B
2 x 3.8 11.8 2 x 3.8 11.8 or 2 x 3.8 11.8
C c. 0, B
4, 7.8
A 5 5 118. a. m B 9 9 2 C 6 b. 0, 0, 0, B 9 3
232
2 x 15.6 or
2 x 8
x 7.8 or
x 4
Section 2.4 123. a. 2 4 z 3 14 0
2 4 z 3 14 4z 3 7 4 z 3 49 4 z 52 z 13
13
b. , 4 7.8, c. 4, 7.8 122. a.
3 c. ,13 4
b. 13,
x 1.7 4.95 11.15
124. a. 2 y 1 3 0
x 1.7 6.2
2 y 1 3
x 1.7 6.2 or x 1.7 6.2 x 7.9 or x 4.5 4.5, 7.9
3 2 9 y 1 4 5 y 1.25 4
y 1
1.25
b. 1.25, b. , 4.5 7.9, c. 4.5, 7.9
233
c. 1,1.25
Chapter 2 Functions and Graphs 125.
126.
The lines are not exactly the same. The slopes are different.
The lines are not exactly the same. The y-intercepts are different.
Section 2.5 Applications of Linear Equations and Modeling 1. y mx b 3 y xb 2 3 2 4 b 2 2 6 b 8b 3 y x8 2
5. 2 x 1 11 2 x 10 x5 a. Vertical b. Slope is undefined. c. No y-intercept 6. 3 y 9 6 3 y 15 y 5
2. y mx b 5 y xb 6 5 3 12 b 6 3 10 b 13 b 5 y x 13 6
a. Horizontal b. m 0 7. y y1 m x x1 9. 1 10. R x C x
3. 6 x 3 y y 3 2 y 6x 3 3 y 3x 2
11. y y1 m x x1
y 5 2 x 3 y 5 2 x 3
3 m 3 ; y-intercept: 0, 2
y 5 2 x 6 y 2 x 1
4. 2 x 3 y 1 3y 2x 1 2 1 y x 3 3 2 1 m ; y-intercept: 0, 3 3
12.
y y1 m x x1 y 6 3 x 4 y 6 3 x 12 y 3 x 18
234
c. 0, 5 8. parallel
Section 2.5 13. y y1 m x x1
18. m
2 x 1 3 2 y x 1 3 2 2 y x 3 3
y0
y y1 m x x1
11 x 4 3 11 y 8 x 4 3 11 44 y 8 x 3 3 11 68 y x 3 3 y 8
14. y y1 m x x1
3 x 4 5 3 y x 4 5 3 12 y x 5 5
y0
15.
y2 y1 3 8 11 11 3 x2 x1 7 4 3
19. m
y2 y1 0 8 8 8 x2 x1 5 0 5 5
y y1 m x x1
8 x 5 5 8 y x8 5
y y1 m x x1
y0
y 2.6 1.2 x 3.4 y 2.6 1.2 x 4.08 y 1.2 x 1.48 20. m
16. y y1 m x x1
y2 y1 0 6 6 x2 x1 11 0 11
y 4.1 2.4 x 2.2
y y1 m x x1
y 4.1 2.4 x 5.28 y 2.2 x 1.18
y0
17. m
6 x 11 11 6 y x6 11
y2 y1 1 2 1 1 x2 x1 3 6 9 9
y y1 m x x1
21. m
1 x 6 9 1 2 y2 x 9 3 1 4 y x 9 3
y2 y1 3.7 5.1 1.4 3.5 x2 x1 1.9 2.3 0.4
y y1 m x x1
y2
y 5.1 3.5 x 2.3 y 5.1 3.5 x 8.05 y 3.5 x 2.95 22. m
6 4.8 1.2 y2 y1 1.5 x2 x1 0.8 1.6 0.8
y y1 m x x1
y 6 1.5 x 0.8 y 6 1.5 x 1.2 y 1.5 x 7.2
235
Chapter 2 Functions and Graphs 23.
y y1 m x x1
35. m1 0
y 4 0 x 3
1 1 2 2 0 2 True Perpendicular
y40 y 4
20
24. y y1 m x x1
y 1 0 x 5 y 1 0 y 1
36. m1 0
26. A line with an undefined slope is a vertical line. Every coordinate has the same x-value, 4 so the equation is x . 7
29. a. m
3 11
b. m
1 7 6 6 7
31. a. m 6
b. m
1 1 6 6
32. a. m 10
b. m
1 1 10 10
33. a. m 1
1 b. m 1 1
34. a. m is undefined
5 y 8 x 3 8 3 y x 5 5
1 11 3 3 11
6 7
30. a. m
37. 8 x 5 y 3
28. 0 b. m
1 m2
4 1 0 3 3 4 4 4 0 True 3 3 Perpendicular
25. A line with an undefined slope is a vertical line. Every coordinate has the same x-value, 2 so the equation is x . 3
27. Undefined
1 m2
2x
5 y 1 4
5 y 2x 1 4 8 4 y x 5 5
m1 m2 ; Parallel 38. 2 x 3 y 7
3 y 2 x 7 2 7 y x 3 3 m1 m2 ; Parallel 39. 2 x 6 x3 m undefined Perpendicular
5 y y5 m0
40. 3 y 5
x 1 m undefined
5 3 m0 Perpendicular y
b. m 0
236
4 x 6 y 2 6 y 4 x 2 2 1 y x 3 3
Section 2.5 41. 6 x 7 y 6 y x 7
44. 3 x y 4 y 3x 4; m 3
7 x 3y 0 2
y y1 m x x1
7 3 y x 2 7 y x 6
y 1 3 x 3 y 1 3x 9 y 3x 10 or 3x y 10
m1 m2 m1 0
45. x 5 y 1 5 y x 1 1 1 1 y x ; m 5 5 5 1 1 5 1 m 5 y y1 m x x1
1 m2
6 1 0 7 7 6 6 6 0 False 7 7 Neither 42. 5 y 2 x 2 y x 5
y 4 5 x 6 y 4 5 x 30 y 5 x 26 or 5 x y 26
5 x y0 2 5 y x 2 5 y x 2
46. x 2 y 7 2 y x 7 1 7 1 y x ; m 2 2 2 1 1 2 1 m 2 y y1 m x x1
m1 m2 m1 0
1 m2
2 1 0 5 5 2 2 2 0 False 5 5 Neither
y 4 2 x 5 y 4 2 x 10 y 2 x 14 or 2 x y 14
47. 3x 7 y 5 7 y 3x 5 3 5 3 y x ; m 7 7 7 y y1 m x x1 3 y 8 x 6 7 3 18 y 8 x 7 7 3 38 or y x 7 7
43. 2 x y 6
y 2 x 6; m 2 y y1 m x x1 y 5 2 x 2 y 5 2 x 4 y 2 x 9 or 2 x y 9
237
7 y 3x 38 3 x 7 y 38
Chapter 2 Functions and Graphs 48. 2 x 5 y 4
53. A line that is perpendicular to the y-axis is a horizontal line with slope 0. Every coordinate has the same y-value, so the 3 equation is y . 4
5 y 2x 4 2 4 2 y x ; m 5 5 5 y y1 m x x1 2 x 7 5 2 14 y6 x 5 5 2 44 y x 5 5
54. A line that is perpendicular to the x-axis is a vertical line with undefined slope. Every coordinate has the same x-value, so the 7 equation is x . 9
y 6
or
5 y 2 x 44 55. A line that is parallel to a vertical line is also a vertical line with undefined slope. Every coordinate has the same x-value, so the equation is x 61.5 .
2 x 5 y 44 49. 2 x 4 y y 2 x 4; m 2 1 1 1 0.5 m 2 2 y y1 m x x1
56. A line that is parallel to a horizontal line is also a horizontal line with slope 0. Every coordinate has the same y-value, so the equation is y 0.009 .
y 6.4 0.5 x 2.2 y 6.4 0.5 x 1.1 y 0.5 x 5.3 or
57. a. S x 0.12 x 400 for x 0 b. S x 0.12 8000 400
10 y 5 x 53 5 x 10 y 53
960 400 1360 S 8000 1360 means that the sales person will make $1360 if $8000 in merchandise is sold for the week.
50. 4 x 9 y
y 4 x 9; m 4 1 1 1 0.25 m 4 4 y y1 m x x1
58. a. P t 1.5t 2 for t 0 b. P 1.6 1.5 1.6 2 2.4 2 4.4
y 1.2 0.25 x 3.6 y 1.2 0.25 x 0.9 y 0.25 x 0.3 or
P 1.6 4.4 means that it costs $4.40 to park for 1.6 hr (1 hr 36 min).
20 y 5 x 6 5 x 20 y 6
59. a. W t 120,000 2400 2.7 t for 0 t 4.5
51. A line that is parallel to the x-axis is a horizontal line with slope 0. Every coordinate has the same y-value, so the equation is y 6 .
b. W 2.5 120,000 2400 2.7 2.5
120,000 16,200 103,800 W 2.5 103,800 means that 2.5 hr into the flight, the mass is 103,800 kg.
52. A line that is parallel to the y-axis is a vertical line with undefined slope. Every coordinate has the same x-value, so the equation is x 11 .
60. a. W t 0.25t 6.8 for 0 t 27.2
238
Section 2.5 b. W 20 0.25 20 6.8 5 6.8 1.8
65. a. 730
W 20 1.8 means that after 20 days of drought, the water level will be 1.8 ft.
b. 0, 730 c. 730,
61. a. T x 0.019 x 172 for x 0
66. a. 25,000
b. T 80,000 0.019 80,000 172
b. 0, 25,000
1520 172 1692 T 80,000 1692 means that the property tax is $1692 for a home with a taxable value of 80,000.
c. 25,000, 67. a. C x 0.24 12 x 790 2.88 x 790
62. a. L x 150 x 2400 for 0 x 16
b. R x 6 x
b. L 12 150 12 2400
c. P x R x C x
1800 2400 600 L 12 600 means that after 12 months, Jorge will owe $600.
6 x 2.88 x 790 3.12 x 790 d.
63. a. C x 34.5 x 2275
3.12 x 790 0 3.12 x 790 x 253.2 The business will make a profit if it produces and sells 254 dozen or more cookies.
b. R x 80 x c. P x R x C x
80 x 34.5 x 2275 45.5 x 2275 d.
P x 0
e.
P x 3.12 x 790 P 150 3.12 150 790
45.5 x 2275 0 45.5 x 2275 x 50 items
468 790 322 The business will lose $322. 68. a. C x 36 x 680
64. a. C x 0.4 x 5625
b. R x 60 x
b. R x 1.3x
c. P x R x C x
c. P x R x C x
60 x 36 x 680 24 x 680
1.3x 0.4 x 5625 0.9 x 5625 d.
P x 0
d.
P x 0 24 x 680 0 24 x 680
P x 0 0.9 x 5625 0 0.9 x 5625 x 6250 items
x 28.3 The business will make a profit if 29 or more maintenance calls are made per month.
239
Chapter 2 Functions and Graphs e.
P x 24 x 680
d. y 73.4 x 221
y 73.4 12 221 880.8 221 1102 Approximately 1102
P 42 24 42 680 1008 680 328 The business will make $328. 69. a. m
e. No. There is no guarantee that the linear trend will continue beyond the last observed data point.
y2 y1 41.8 50.8 9 1 x2 x1 19 1 18 2
y y1 m x x1
71. a. m
y 50.8 0.5 x 1
y2 y1 13.0 11.2 1.8 0.18 x2 x1 14 4 10
y y1 m x x1
y 50.8 0.5 x 0.5 y 0.5 x 51.3
y 11.2 0.18 x 4 y 11.2 0.18 x 0.72 y 0.18 x 10.48
b. m 0.5 ; The slope means that alcohol usage among high school students 1 month prior to the CDC survey has dropped by an average rate of 0.5% per yr.
b. m 0.18 means that enrollment in public colleges increased at an average rate of 0.18 million per yr (180,000 per yr).
c. The y-intercept is (0, 51.3) and means that in the year 1990, approximately 51.3% of high school students used alcohol within 1 month prior to the date that the survey was taken.
c. The y-intercept is (0, 10.48) and means that in the year 1990, there were approximately 10,480,000 students enrolled in public colleges. d. y 0.18 x 10.48
d. y 0.5 x 51.3
y 0.18 25 10.48
y 0.5 20 51.3 10 51.3 41.3 Approximately 41.3%
4.5 10.48 14.98 Approximately 14.98 million
e. No. There is no guarantee that the linear trend will continue well beyond the last observed data point.
72. a. m
y2 y1 400 486 86 9.56 x2 x1 13 4 9
y y1 m x x1
y y1 1469 955 514 73.4 70. a. m 2 x2 x1 17 10 7
y 486 9.56 x 4 y 486 9.56 x 38.24 y 9.56 x 524.24 y 9.56 x 524
y y1 m x x1
y 955 73.4 x 10 y 955 73.4 x 734 y 73.4 x 221
b. m 9.56 means that cigarette consumption decreased at an average rate of 9.56 billion per yr.
b. m 73.4 means that the number of transplants increased at an average rate of 73.4 per yr during this time period.
c. The y-intercept is (0, 524) and means that in the year 1990, there were approximately 524 billion cigarettes consumed in the United States.
c. The y-intercept is (0, 221) and means that approximately 221 lung transplants were performed in the United States in the year 1990.
240
Section 2.5 d. y 9.56 x 524
c. m 0.046 means that longevity increases at an average rate of 0.046 yr per 1 day increase in the gestation or incubation period.
y 9.56 25 524 239 524 285 Approximately 285 billion 73. a.
d.
L x 0.0.046 x 6.48 L 80 0.0.046 80 6.48 3.68 6.48 10 Approximately 10 yr
75. Yes; From the graph, the data appear to follow a linear trend.
b. m
76. Yes; From the graph, the data appear to follow a linear trend.
y2 y1 90 60 30 0.125 x2 x1 720 480 240
77. No; From the graph, the data do not appear to follow a linear trend.
y y1 m x x1
y 60 0.125 x 480
78. No; From the graph, the data do not appear to follow a linear trend.
y 60 0.125 x 60 y 0.125 x
79. a. y 0.5 x 52.3
c x 0.125 x
b.
c. m 0.125 means that the amount of cholesterol increases at an average rate of 0.125 mg per calorie of hamburger. d.
c x 0.125 x c 650 0.125 650 81.25 mg
74. a.
c. y 0.5 x 52.3
y 0.5 20 52.3 10 52.3 42.3 Approximately 42.3%, which is close to the result of 41.3% obtained in Exercise 69(d). 80. a. y 67.4 x 333.2 b.
y y1 35 8.5 26.5 0.046 b. m 2 x2 x1 620 44 576 y y1 m x x1
y 8.5 0.046 x 44 y 8.5 0.046 x 2.024 y 0.046 x 6.476 L x 0.0.046 x 6.48
241
Chapter 2 Functions and Graphs c. y 67.4 x 333.2
84. a. y 0.045 x 6.68
y 67.4 12 333.2 808.8 333.2 1142 Approximately 1142, which is close to the result of 1102 obtained in Exercise 70(d).
b.
81. a. y 0.175 x 10.46 b. c. y 0.045 x 6.68
y 0.045 80 6.68 3.6 6.68 10.28 Approximately 10 yr 85. m
c. y 0.175 x 10.46
y 0.175 25 10.46
y2 y1 1 6 5 5 x2 x1 24 2 2
y y1 m x x1
4.375 10.46 14.835 Approximately 14.8 million
5 x 4 2 5 y 6 x 10 2 5 y x4 2 To find the x-intercept, solve y = 0. 5 0 x4 2 5 x4 2 8 8 x , 0 5 5 y 6
82. a. y 9.55 x 527 b.
c. y 9.55 x 527
y 9.55 25 527 238.75 527 288.25 Approximately 288 billion
86. m
y2 y1 7 5 12 2 x2 x1 4 2 6
y y1 m x x1
83. a. y 0.136 x 4.27
y 5 2 x 2
b.
y 5 2 x 4 y 2 x 1 To find the x-intercept, solve y = 0. 0 2 x 1 2 x 1 1 1 x , 0 2 2
c. y 0.136 x 4.27 y 0.136 650 4.27 88.4 4.27 84.13 Approximately 84 mg
242
Section 2.5 87. Use the points 0, 4 and 3,11 . m
91. If the slopes of the two lines are the same and the y-intercepts are different, then the lines are parallel. If the slope of one line is the opposite of the reciprocal of the slope of the other line, then the lines are perpendicular.
y2 y1 11 4 7 x2 x1 3 0 3
y y1 m x x1
7 x 0 3 7 y4 x 3 7 7 y x 4 or f x x 4 3 3 y4
92. The point-slope formula is used to construct an equation of a line if a point on the line and the slope of the line are known. 93. Profit is equal to revenue minus cost. 94. The break-even point is found by determining where revenue equals cost or where profit equals zero.
88. Use the points 0, 7 and 2, 4 .
m
y2 y1 4 7 3 3 x2 x1 2 0 2 2
y y1 m x x1
95. m
3 x 0 2 3 y7 x 2 3 3 y x 7 or g x x 7 2 2 y7
1 1 1 m 1 y y1 m x x1
y 3 1 x 1 y 3 x 1 y x 4
89. Use the points 1, 6 and 3, 2 . m
y2 y1 3 0 3 x2 x1 4 0 4 1 1 4 3 m 3 4 y y1 m x x1
y2 y1 2 6 4 1 x2 x1 3 1 4
96. m
y y1 m x x1 y 6 1 x 1 y 6 x 1
y x 5 or h x x 5
4 x 4 3 4 16 y 3 x 3 3 4 25 y x 3 3 y 3
90. Use the points 2,10 and 5, 18 .
m
y2 y1 3 1 4 1 x2 x1 1 3 4
y2 y1 18 10 28 4 x2 x1 5 2 7
y y1 m x x1
y 10 4 x 2
97. Let c 2.
y 10 4 x 2
c, c 1 2, 2 1 2, 7 3
3
y 10 4 x 8
m 3c 2 3 2 3 4 12 2
y 4 x 2 or h x 4 x 2
243
Chapter 2 Functions and Graphs y y1 m x x1
m
y 7 12 x 2
y2 y1 2 8 10 5 5 1 4 2 x2 x1
y y1 m x x1
y 7 12 x 2
5 x 1 2 5 5 y 8 x 2 2 5 21 y x for 1 x 5 2 2
y 7 12 x 24 y 12 x 17
y 8
98. Let c 2. 1 1 c, 2, c 2 1 1 1 m 2 2 c 4 2
x x y y2 100. M 1 2 , 1 2 2
y y1 m x x1
4 12 1 3 8 4 , , 4, 2 2 2 2 2 y y1 5 2 7 7 m 2 x2 x1 64 2 2
1 1 y x 2 2 4 1 1 1 y x 2 4 2 1 y x 1 4
y y1 m x x1
7 x 4 2 7 y 2 x 14 2 7 y x 16 for 4 x 6 2 y2
x x y y2 99. M 1 2 , 1 2 2 2 4 9 7 2 16 , , 1, 8 2 2 2 2
Problem Recognition Exercises: Comparing Graphs of Equations 1.
3.
2.
4.
244
Problem Recognition 5.
10.
The graphs have the shape of y x with a vertical shift.
6. 11.
7.
The graphs have the shape of y x with a horizontal shift. 12.
8.
The graphs have the shape of y x 2 with a horizontal shift. 13. 9.
The graph of g x x has the shape of the graph of y x but is reflected across the x-axis.
The graphs have the shape of y x with a vertical shift. 2
245
Chapter 2 Functions and Graphs 14.
17.
The graph of g x x has the shape of
The graph of g x x has the shape of
the graph of y x but is reflected across the x-axis.
the graph of y x but is reflected across the y-axis.
15.
18.
The graph of g x 3 x has the shape of
The graphs have the shape of y x 2 but show a vertical shrink or stretch.
the graph of y 3 x but is reflected across the y-axis.
16.
The graphs have the shape of y x but show a vertical shrink or stretch.
Section 2.6 Transformations of Graphs 1. a. m
3 2
b. 0,1
2. a. m 0
c.
c.
246
3 b. 0, 2
Section 2.6 21. The graph of f is the graph of f x x shifted upward 1 unit.
3.
The y-intercepts are different. The graphs differ by a vertical shift. Or we can interpret the difference as a horizontal shift. 22. The graph of g is the graph of f x x shifted upward 2 units.
4.
The centers of the circles are different. The graphs differ by a horizontal shift and by a vertical shift. 5. linear
6. up
7. left
8. right
9. down
10. vertical stretch
11. horizontal shrink
12. horizontal stretch
13. vertical shrink
14. x
15. e
16. f
17. b
18. c
19. a
20. d
23. The graph of k is the graph of f x x 3 shifted downward 2 units.
247
Chapter 2 Functions and Graphs 24. The graph of h is the graph of f x
27. The graph of r is the graph of f x x 2 shifted to the right 4 units.
1 x
shifted downward 2 units.
28. The graph of t is the graph of f x 3 x shifted to the right 2 units.
25. The graph of g is the graph of f x x shifted to the left 5 units.
29. The graph of a is the graph of f x x shifted to the left 1 unit and downward 3 units.
26. The graph of m is the graph of f x x shifted to the left 1 unit.
248
Section 2.6 30. The graph of b is the graph of f x x shifted to the right 2 units and upward 4 units.
33. The graph of m is the graph of f x 3 x stretched vertically by a factor of 4.
34. The graph of n is the graph of f x x stretched vertically by a factor of 3.
31. The graph of c is the graph of f x x 2 shifted to the right 3 units and upward 1 unit.
35. The graph of r is the graph of f x x 2
32. The graph of d is the graph of f x x shifted to the left 4 units and downward 1 unit.
shrunk vertically by a factor of
249
1 . 2
Chapter 2 Functions and Graphs 36. The graph of t is the graph of f x x
shrunk vertically by a factor of
39. The graph of the function is the graph of 1 f x shrunk vertically by a factor of . 3
1 . 3
37. p x 2 x 2 x
40. The graph of the function is the graph of 1 g x shrunk vertically by a factor of . 2
The graph of p is the graph of f x x stretched vertically by a factor of 2.
41. The graph of the function is the graph of f x stretched vertically by a factor of 3.
38. q x 2 x 2 x
The graph of q is the graph of f x x stretched vertically by a factor of
2.
250
Section 2.6 42. The graph of the function is the graph of g x stretched vertically by a factor of 2.
45. The graph of the function is the graph of f x stretched horizontally by a factor of
1 . 3
43. The graph of the function is the graph of f x shrunk horizontally by a factor of 3. 46. The graph of the function is the graph of g x stretched horizontally by a factor of
1 . 2
44. The graph of the function is the graph of g x shrunk horizontally by a factor of 2.
47. The graph of f is the graph of f x
reflected across the x-axis.
251
1 x
Chapter 2 Functions and Graphs 48. The graph of g is the graph of f x x reflected across the x-axis.
51. The graph of p is the graph of f x x 3 reflected across the y-axis.
49. The graph of h is the graph of f x x 3 reflected across the x-axis.
52. The graph of q is the graph of f x 3 x reflected across the y-axis.
50. The graph of k is the graph of f x x reflected across the x-axis.
53. The graph of the function is the graph of f x reflected across the y-axis.
252
Section 2.6 54. The graph of the function is the graph of g x reflected across the y-axis.
57. The graph of the function is the graph of f x reflected across the y-axis.
55. The graph of the function is the graph of f x reflected across the x-axis.
58. The graph of the function is the graph of g x reflected across the y-axis.
56. The graph of the function is the graph of g x reflected across the x-axis.
59. The graph of the function is the graph of f x reflected across the x-axis.
253
Chapter 2 Functions and Graphs 63. The graph of f is the graph of f x x shifted to the left 3 units, stretched vertically by a factor of 2, and shifted downward 1 unit.
60. The graph of the function is the graph of g x reflected across the x-axis.
61. The graph of v is the graph of f x x 2 shifted to the left 2 units, reflected across the x-axis, and shifted upward 1 unit.
64. The graph of g is the graph of f x x shifted to the right 1 unit, stretched vertically by a factor of 2, and shifted upward 3 units.
62. The graph of u is the graph of f x x 2 shifted to the right 1 unit, reflected across the x-axis, and shifted downward 2 units.
65. The graph of p is the graph of f x x shifted to the right 1 unit, shrunk vertically 1 by a factor of , and shifted downward 2 2 units.
254
Section 2.6 66. The graph of q is the graph of f x x shifted to the left 2 units, shrunk vertically 1 by a factor of , and shifted downward 1 3 unit.
69. f x x 3 x 3
The graph of f is the graph of f x x reflected across the y-axis and shifted right 3 units.
70. g x x 4 x 4
67. The graph of r is the graph of f x x reflected across the y-axis, reflected across the x-axis, and shifted upward 1 unit.
The graph of g is the graph of f x x reflected across the y-axis and shifted to the left 4 units.
68. The graph of s is the graph of f x x reflected across the y-axis, reflected across the x-axis, and shifted downward 2 units.
71. n x 2 x 6 2 x 3 4 x 3 2
2
2
The graph of n is the graph of f x x 2 shifted to the left by 3 units and stretched vertically by a factor of 4.
255
Chapter 2 Functions and Graphs 72. m x 2 x 4 2 x 2
75. The graph of the function is the graph of f x shifted to the right 2 units, stretched vertically by a factor of 2, and shifted downward 3 units.
The graph of m is the graph of f x x shifted to the right 2 units and stretched vertically by a factor of 2.
73. The graph of the function is the graph of f x shifted to the right 1 unit, reflected across the x-axis, and shifted upward 2 units.
76. The graph of the function is the graph of f x shifted to the left 2 units, stretched vertically by a factor of 2, and shifted downward 4 units.
74. The graph of the function is the graph of f x shifted to the left 1 unit, reflected across the x-axis, and shifted downward 2 units.
77. The graph of the function is the graph of f x shrunk horizontally by a factor of 2, stretched vertically by a factor of 3, and reflected across the x-axis.
256
Section 2.6 84. As written, g x 2 x is in the form
78. The graph of the function is the graph of f x stretched horizontally by a factor of
g x f ax with a 1 . This indicates a
horizontal shrink. However, g x can also be
1 1 , shrunk vertically by a factor of , and 2 2 reflected across the x-axis.
written as g x 2 x 2 x . This is
written in the form g x af x with a 1 . This represents a vertical stretch. 1 x is in the form 2 h x f ax with 0 a 1 . This indicates
85. As written, h x
a horizontal stretch. However, h x can also 1 x . This is written 2 in the form h x af x with 0 a 1 . This represents a vertical shrink. be written as h x
79. The graph of the function is the graph of f x reflected across the y-axis and shifted downward 2 units.
86. The graph of f is the same as the graph of y y x with a horizontal shift to the right 2 units and a vertical shift downward 3 units. By contrast, the graph of g is the graph of y x with a horizontal shift to the right 3 units and a vertical shift downward 2 units. 87. The graph is f x x 2 shifted to the right 2 units and downward 3 units.
f x x 2 3 2
80. The graph of the function is the graph of f x reflected across the y-axis and shifted upward 3 units.
88. The graph is f x x shifted to the left 3 units and downward 4 units. f x x 3 4 89. The graph is f x
1 shifted to the left 3 x
units. f x
81. c
1 x3
90. The graph is f x x shifted downward 2 units. f x x 2
82. a
83. b
257
Chapter 2 Functions and Graphs 91. The graph is f x x3 reflected across the x-axis and shifted upward 1 unit. f x x3 1
b.
92. The graph is f x x 2 shifted left 2 units and reflected across the x-axis.
f x x 2
2
c. The general shape of y x n is similar to
the graph of y x 2 for even values of n greater than 1.
93. a.
d. The general shape of y x n is similar to
the graph of y x 3 for odd values of n greater than 1.
Section 2.7 Analyzing Graphs of Functions and Piecewise-Defined Functions 1.
3.
2.
4.
5.
f x 3x 2 5 x 1 f x 3 x 5 x 1 2
3x 2 5 x 1
258
Section 2.7 g x 2 x x2 3
6.
This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x by x and y by y .
g x 2 x x 3 2
2 x x2 3
y x 4
7. y
8. x
9. origin
10. y-axis
y x 4 y x 4 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
11. origin 12. x or int(x) or floor(x)
15. x y 4 Replace x by x . x y 4
13. y x 3 Replace x by x . 2
y x 3 2
x y 4 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace y by y .
y x2 3 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y .
x y 4
y x 3
x y 4 This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x by x and y by y .
2
y x2 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x by x and y by y .
x y 4
y x 3
x y 4
y x 3
x y 4 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
2
2
y x2 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
16. x y 2 3 Replace x by x . x y2 3
14. y x 4 Replace x by x . y x 4
x y2 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace y by y .
y x 4 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y .
x y 3 2
y x 4
x y2 3
y x 4
259
Chapter 2 Functions and Graphs This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x by x and y by y .
This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x by x and y by y .
x y 3
x y 4
x y2 3
x y 4 This equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.
2
x y2 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
19. y x 2 x 7 Replace x by x . y x 2 x 7
17. x 2 y 2 3 Replace x by x .
x2 y 2 3
y x 2x 7 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace y by y .
x2 y 2 3 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y .
y x 2x 7 y x 2x 7 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x by x and y by y .
x2 y 3 2
x2 y 2 3 This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x by x and y by y .
y x 2 x 7 y x 2x 7
x2 y 2 3
y x 2x 7 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
x2 y 2 3 This equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.
20. y x 2 6 x 1 Replace x by x .
18. x y 4 Replace x by x . x y 4
y x 6 x 1 2
y x2 6 x 1 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace y by y .
x y 4 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y .
y x2 6 x 1
x y 4
y x2 6 x 1 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.
x y 4
260
Section 2.7 Replace x by x and y by y .
Replace x by x and y by y .
y x 6 x 1
y 4 2 x2
y x2 6 x 1
y 4 2 x2 This equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.
2
y x2 6 x 1 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
1 x3 2 Replace x by x . 1 y x 3 2 1 y x3 2 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace y by y . 1 y x 3 2 1 y x3 2 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x by x and y by y . 1 y x 3 2 1 y x 3 2 1 y x3 2 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
23. y
21. x 2 5 y 2 Replace x by x .
x2 5 y 2 x2 5 y 2 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y . x2 5 y
2
x2 5 y 2 This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x by x and y by y .
x2 5 y 2 x2 5 y 2 This equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 22. y 4 2 x 2 Replace x by x .
y 4 2 x
2
y 4 2 x2 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y .
2 x 1 5 Replace x by x . 2 y x 1 5 2 y x 1 5
24. y
y 4 2 x2 y 4 2 x2 This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis.
261
Chapter 2 Functions and Graphs This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace y by y . 2 y x 1 5 2 y x 1 5 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x by x and y by y . 2 y x 1 5 2 y x 1 5 2 y x 1 5 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
b. Yes c. Even 34. a.
g x x 3 x x8 3 x 8
b. Yes c. Even 35. a.
h x 4 x3 2 x h x 4 x 2 x 4 x 3 2 x 3
b. h x 4 x 3 2 x 4 x 3 2 x c. Yes d. Odd 36. a.
k x 8 x 5 6 x 3 k x 8 x 6 x 8 x 5 6 x 3 5
3
b. k x 8 x 5 6 x 3 8 x 5 6 x 3
25. y-axis symmetry
c. Yes
26. Origin symmetry
d. Odd
27. The function is symmetric to the origin. Therefore, the function is an odd function.
37. a.
2
4 x2 2 x 3
b. m x 4 x 2 2 x 3 4 x 2 2 x 3
29. The function is symmetric to the y-axis. Therefore, the function is an even function.
c. No d. No
30. The function is symmetric to the y-axis. Therefore, the function is an even function.
e. Neither
31. The function is not symmetric with respect to either the y-axis or the origin. Therefore, the function is neither even nor odd.
38. a.
n x 7 x 3x 1 n x 7 x 3 x 1 7 x 3x 1
32. The function is not symmetric with respect to either the y-axis or the origin. Therefore, the function is neither even nor odd.
b. n x 7 x 3 x 1 7 x 3 x 1 c. No
f x 4 x2 3 x
d. No
f x 4 x 3 x 4 x 3 x 2
m x 4 x2 2 x 3 m x 4 x 2 x 3
28. The function is symmetric to the origin. Therefore, the function is an odd function.
33. a.
g x x8 3 x
2
e. Neither
262
Section 2.7 39.
f x 3x6 2 x 2 x
r x r x
r x r x The function is neither even nor odd.
f x 3 x 2 x x 6
f x 3x6 2 x 2 x f x f x The function is even. 40.
45.
q x 16 x 16 x 2 2
p x x 12 x10 5
q x q x
p x x 12 x 5 10
The function is even.
p x x 12 x10 5
46.
p x p x The function is even. 41.
q x 16 x 2
z x 49 x 2 z x 49 x 49 x 2 2
z x z x
k x 13x 3 12 x
The function is even.
k x 13 x 12 x 3
k x 13 x 3 12 x
47.
h x 5 x 5 x
k x 13 x 3 12 x 13 x 3 12 x
h x 5 x 5 x
k x k x
h x h x The function is odd.
The function is odd. 42.
m x 4 x 5 2 x 3 x
48.
m x 4 x 2 x x 5
g x x x
3
g x x
g x x x
3
m x 4 x 2 x x 5
h x 5x
g x g x The function is odd.
m x 4 x5 2 x 3 x 4 x 5 2 x 3 x m x m x The function is odd.
49. a. Use the second rule.
f 3 3 3 9 3 12 2
43.
n x 16 x 3
2
n x 16 x 3
2
n x 16 x 3
2
b. Use the first rule. f 2 3 2 7 6 7 13 c. Use the second rule.
f 1 1 3 1 3 4 2
n x n x n x n x The function is neither even nor odd. 44.
r x 81 x 2
d. Use the third rule. f 4 5 e. Use the third rule. f 5 5
2
r x 81 x 2
2
50. a. Use the first rule. g 3 2 3 3 6 3 9
r x 81 x 2
2
b. Use the third rule. g 3 4
263
Chapter 2 Functions and Graphs c. Use the first rule. g 2 2 2 3 4 3 7
52. a. Use the second rule. t 1.99 2 1.99 3.98 b. Use the first rule. t 0.4 0.4
d. Use the second rule. g 0 5 0 6 0 6 6
c. Use the third rule. t 3 3 3 9
e. Use the third rule. g 4 4
d. Use the first rule. t 1 1
51. a. Use the second rule. h 1.7 1
e. Use the fourth rule. t 3.001 4 3.001 12.004
b. Use the first rule. h 2.5 2 c. Use the second rule. h 0.05 1
53. c
54. a
d. Use the fourth rule. h 2 1
55. d
56. b
e. Use the second rule. h 0 1 57. a.
b.
c.
58. a.
b.
c.
59. a.
b.
c.
264
Section 2.7 60. a.
b.
c.
61.
64.
62.
65.
63.
66.
265
Chapter 2 Functions and Graphs 67.
71. a.
b. y x
68. 72.
f x x
f 3.7 3.7 4 73.
f x x
f 4.2 4.2 5 74.
f x x
f 0.5 0.5 1
69. 75.
f x x
f 0.09 0.09 1 76.
f x x
f 0.5 0.5 0 77.
f x x
f 0.09 0.09 0
70.
78. f x x
f 6 6 6
79.
f x x
f 9 9 9 80.
f x x
f 5 5 5
266
Section 2.7 81.
84.
82.
0.44 0.61 85. a. C x 0.78 0.95
for 0 x 1 for 1 x 2 for 2 x 3 for 3 x 3.5
b.
83.
60 2 x 86. a. L x 32 2.5 x 14 b.
2000 87. S x 2000 0.05 x 40,000
for 2 x 3 for 3 x 3.5
for 0 x 600 63.97 88. C x 63.97 0.40 x 600 for x 600
267
for 0 x 14 for 14 x 19
Chapter 2 Functions and Graphs 89. a. 2,
b. 3, 2
c. 2, 2
90. a. , 0
b. 0,1
c. 1, 4
100. At x 2 , the function has a relative maximum of 3. At x 0 , the function has a relative minimum of 1. At x 2 , the function has a relative maximum of 3.
91. a. ,
101. At x 2 , the function has a relative minimum of 4 . At x 0 , the function has a relative maximum of 0. At x 2 , the function has a relative minimum of 4 .
b. Never decreasing c. Never constant 92. a. ,
102. At x 3 , the function has a relative maximum of 5. At x 0 , the function has a relative minimum of 0. At x 3 , the function has a relative maximum of 5.
b. Never decreasing c. Never constant 93. a. 1,
103. a. (0, 3) and (7, 9); these intervals correspond to the years 2000 to 2003 and 2007 to 2009.
b. ,1
b. (3, 7); this interval corresponds to the years 2003 to 2007.
c. Never constant 94. a. , 1
c. Relative maximum approximately 6%; Relative minimum approximately 4.5%.
b. 1,
104. a. (4, 14) and (24, 27); these intervals correspond to the years 1984 to 1994 and 2004 to 2007.
c. Never constant 95. a. , 2 2,
b. (0, 4) and (14, 24); these intervals correspond to the years 1980 to 1984 and 1994 to 2004.
b. Never decreasing c. 2, 2
c. Relative minima: approximately 166,000 and 132,000; Relative maxima: approximately 195,000.
96. a. Never increasing b. , 0 2,
2 for x 1 105. f x 3 for x 1
c. 0, 2 97. At x 1 , the function has a relative minimum of 3 .
4 for x 2 106. f x 1 for x 2
98. At x 1 , the function has a relative maximum of 2.
x 107. f x 2
99. At x 2 , the function has a relative minimum of 0. At x 0 , the function has a relative maximum of 2. At x 2 , the function has a relative minimum of 0.
for x 2 for x 2
x 2 for x 2 108. f x 4 for x 2
268
Section 2.7 1 109. f x x x
117. If replacing y by y in the equation results in an equivalent equation, then the graph is symmetric with respect to the x-axis. If replacing x by x in the equation results in an equivalent equation, then the graph is symmetric with respect to the y-axis. If replacing both x by x and y by y results in an equivalent equation, then the graph is symmetric with respect to the origin.
for x 0 for x 0
for x 0 x 110. f x x 1 for x 0 111. For t 10 sec, use the first rule. s 10 1.5 10 15 ft/sec For t 20 sec, use the first rule. s 20 1.5 20 30 ft/sec For t 30 sec, use the second rule. 30 30 s 30 2.7 ft/sec 30 19 11 For t 40 sec, use the second rule. 30 30 s 40 1.4 ft/sec 40 19 21
118. If the graph is symmetric with respect to the y-axis, then the function is even. If the graph is symmetric with respect to the origin, then the function is odd. 119. At x 1 , there are two different y values. The relation contains the ordered pairs (1, 2) and (1, 3). 120. The two rules defining f x are stated for x 1 and for x 1 . The function is not defined at x 1 . Therefore, there is a “hole” in the function at the point (1, 3).
112. For t 6 sec, use the first rule. 5 5 2 s 6 6 36 15 mph 12 12 For t 12 sec, use the first rule. 5 2 s 12 12 5 12 60 mph 12 For t 45 sec, use the second rule. s 45 40 mph For t 80 sec, use the third rule. 2 3 s 80 92 80 20 3 2 12 20 21.6 mph 113. Graph b
121. A relative maximum of a function is the greatest function value relative to other points on the function nearby. 122. A function is increasing on an interval I if for all x1 x2 on I, f x1 f x2 . In other words, the function “rises” from left to right over the interval. 123. If the average rate of change is negative and then positive, the function is decreasing and then increasing. Therefore it must have a relative minimum at a.
114. Graph d
c. 4
d. 3
124. If the average rate of change is positive and then negative, the function is increasing and then decreasing. Therefore it must have a relative maximum at a.
e. 3
f. 4
125. a. Concave down
b. Decreasing
b. 1
126. a. Concave up
b. Increasing
127. a. Concave up
b. Decreasing
128. a. Concave down
b. Increasing
115. a. 2
116. a. 5
b. 4
c. 2
d. 6
e. 0
f. 2
269
Chapter 2 Functions and Graphs 129.
132.
The graph does not have a gap at x 2 , however, the limited resolution on the calculator gives the appearance of a gap.
130.
133.
131.
134.
The graph does not have a gap at x 2 , however, the limited resolution on the calculator gives the appearance of a gap.
270
Section 2.8 135.
136.
Relative minimum 7.825
Relative maximum 4.667
Section 2.8 Algebra of Functions and Function Composition 1. 5 2 x 0 2 x 5 5 x 2 5 , 2
6.
f a 4 3 a 4 5 a 4 1 2
3 a 2 8a 16 5a 20 1 3a 24a 48 5a 20 1 2
3a 2 19a 27
2. 1 4 x 0 4 x 1 1 x 4 1 4 ,
7. f x ; g x
9.
f x h f x h
8.
f x ; g x g x
10. f g x
11. f g x x 3 ; Graph d
3. 81 x 2 0
12. f g x x 4 x 4 ; Graph b
x 2 81 x 2 81 x 9 , 9 9, 9 9,
13. f g x x 2 4 x 2 4 ; Graph a 14. f g x x 2 3 ; Graph c
4. 49 x 2 0
15. f g 3 f 3 g 3
x 2 49
2 3 3 4 6 7 13
x 2 49 x 7 , 7 7, 7 7, 5.
f x 3x 2 5 x 1
16. g h 2 g 2 h 2
24
f x 3x 2 5 x 1 f t 2 3 t 2 5 t 2 1
1 1 6 7 23 1
2
17. f g 1 f 1 g 1
3 t 2 4t 4 5t 10 1
2 1 1 4 2 3 6
3t 2 12t 12 5t 10 1 3t 2 17t 21
271
Chapter 2 Functions and Graphs
18. h g 4 h 4 g 4
1 x 0 x 1 x 1 The domain of p is , .
1 1 4 4 8 8 43 1
19. g h 0 g 0 h 0
04
The domain of q is ,1 .
1 1 11 4 03 3 3
The intersection of their domains is , 1 .
20. f h 5 f 5 h 5
2 5
28. r q x r x q x 3x 1 x
1 1 19 10 53 2 2
The domain of r is , . The domain of q is ,1 .
f 8 2 8 16 f 4 21. 8 g 8 8 4 12 3 g
The intersection of their domains is , 1 . q x q 1 x 29. x 2 p x x 3x p
1 1 h 7 h 1 73 4 22. 7 f 7 2 7 14 56 f
The domain of q is ,1 .
The domain of p is , .
g 0 0 4 4 g 23. 0 Undefined f 0 2 0 0 f
The intersection of their domains is , 1 .
x 2 3x 0 x x 3 0
1 1 h 4 4 3 7 h 24. 4 g 4 4 4 0 g
x 0 or x 3 Also exclude the values x 0 and x 3 , which make the denominator zero. The domain is , 3 3, 0 0,1 .
Undefined 25. r p x r x p x
3x x 2 3x
q x 1 x q 30. x r 3 x r x
3x x 2 3x x 2 6 x The domain of r is , .
The domain of q is ,1 .
The domain of r is , .
The domain of p is , .
The intersection of their domains is , 1 . 3 x 0 x0 Also exclude the value x 0 , which makes the denominator zero. The domain is , 0 0,1 .
The intersection of their domains is , .
26. p r x p x r x
27. p q x p x q x x 2 3x 1 x
x 2 3x 3x x 2 3x 3x x 2 6 x The domain of p is , .
The domain of r is , .
The intersection of their domains is , .
272
Section 2.8 p x x 2 3x p 31. x q x q 1 x
r x r 3x 3 2 36. x p x x 3x x 3 p
The domain of p is , .
The domain of q is , .
The domain of q is ,1 .
The domain of p is , .
The intersection of their domains is , 1 . Also exclude the value x 1 , which makes the denominator zero. The domain is ,1 .
The intersection of their domains is , . Also exclude the values x 0 and x 3 , which make the denominator zero. The domain is , 3 3, 0 0, . 37. a. f x h 5 x h 9 5 x 5h 9
r x r 3 x 32. x q x q 1 x The domain of r is , .
b.
The domain of q is ,1 .
The intersection of their domains is , 1 . Also exclude the value x 1 , which makes the denominator zero. The domain is ,1 .
f x h f x 5 x 5h 9 5 x 9 h h 5 x 5h 9 5 x 9 h 5h h 5
38. a. f x h 8 x h 4 8 x 8h 4
33. r q x r x q x
3 x 1 x The domain of r is , .
b.
The domain of q is ,1 .
The intersection of their domains is , 1 . 34. p q x p x q x x 2 3x 1 x The domain of p is , .
f x h f x 8 x 8h 4 8 x 4 h h 8 x 8h 4 8 x 4 h 8h h 8
39. a. f x h x h 4 x h 2
The domain of q is ,1 .
x 2 2 xh h 2 4 x 4h
The intersection of their domains is , 1 . b.
p x x 2 3x x 3 p 35. x r r x 3x 3
f x h f x h 2 x 2 xh h 2 4 x 4h x 2 4 x h 2 2 x 2 xh h 4 x 4h x 2 4 x h 2 2 xh h 4h h 2x h 4
The domain of p is , . The domain of r is , .
The intersection of their domains is , . Also exclude the value x 0 , which makes the denominator zero. The domain is , 0 0, .
273
Chapter 2 Functions and Graphs 40. a. f x h x h 3 x h 2
41.
x 2 xh h 3x 3h 2
b.
2
f x h f x h 2 x 2 xh h 2 3x 3h x 2 3 x h 2 2 x 2 xh h 3x 3h x 2 3 x h 2 2 xh h 3h 2x h 3 h
42.
2
2
f x h f x 2 x h 5 2 x 5 h h 2 x 2h 5 2 x 5 h 2h 2 h
f x h f x 3 x h 8 3x 8 h h 3x 3h 8 3 x 8 h 3h 3 h
f x h f x 5 x h 4 x h 2 5 x 4 x 2 43. h h 2 2 5 x 2 xh h 4 x 4h 2 5 x 2 4 x 2 h 5 x 2 10 xh 5h 2 4 x 4h 2 5 x 2 4 x 2 10 xh 5h 2 4h 10 x 5h 4 h h 2
f x h f x 4 x h 2 x h 6 4 x 2 x 6 44. h h 2 2 4 x 2 xh h 2 x 2h 6 4 x 2 2 x 6 h 2 2 4 x 8 xh 4h 2 x 2h 6 4 x 2 2 x 6 8 xh 4h 2 2h 8 x 4h 2 h h 2
f x h f x x h x3 x 3 3 x 2 h 3xh 2 h3 x3 45. 3x 2 3xh h 2 h h h 3
f x h f x x h 2 x 2 x 3 3x 2 h 3xh 2 h3 x 3 46. 3 x 2 3xh h 2 h h h 3
3
x x h x xh h 1 1 f x h f x x h x x x h x x h x x h x x h 1 47. h h h h h x x h
274
Section 2.8
48.
f x h f x h 1 1 xh2 x2 h x2 xh2 x 2 x h 2 x 2 x h 2 h x2 xh2 h x 2 x h 2 x 2 x h 2 h h 1 x 2 x h 2
49. a.
4 4 4 64 16 48 3
52. h g 2 h
2 2 h 2
2 2 3 4 3 7
53. h f 1 h 13 4 1 h 3
2 3 3 6 3 3
54. g f 3 g 3 4 3 g 27 12 3
g 15 2 15 30
f x h f x 4 x h 4 x h h
55. f g 18 f g 18 f
f
4 11 4 1 b. 4 2 4 1.6569 1 4 1 0.1 4 1 4 1.1 4 1.9524 0.1 0.1 4 1 0.01 4 1 4 1.01 4 1.9950 0.01 0.01 4 1 0.001 4 1 4 1.001 4 1.9995 0.001 0.001
2 18
36 f 6 6 4 6 3
216 24 192 56. f h 1 f h 1 f 2 1 3
f 1 1 4 1 1 4 3 3
57. g f 5 g f 5 g 5 4 5 3
g 125 20 g 105
c. 2
50. a.
2 8 f 16 f 4
51. f g 8 f
2 105 210
f x h f x h
12 12 xh x h
3
3
58. h f 2 h f 2 h 2 4 2
h 8 8 h 0 2 0 3 3
12 12 12 6 b. 2 0.1 2 2.1 2.8571 0.1 0.1 12 12 12 6 2 0.01 2 2.01 2.9851 0.01 0.01 12 12 12 6 2 0.001 2 2.001 2.9985 0.001 0.001 12 12 12 6 2 0.0001 2 2.0001 2.9999 0.0001 0.0001
59. h f 3 h f 3 h 3 4 3
h 27 12 h 15 2 15 3 30 3 27 60. h g 72 h g 72 h
h
144 h 12 2 12 3
24 3 27
c. 3
275
2 72
Chapter 2 Functions and Graphs
61. g f 1 g f 1 g 1 4 1 3
g 1 4 g 3
67. m n x m n x n x 8
x 58 x3
x3 0 x 3 Domain: 3,
2 3 6
Undefined
62. g f 4 g f 4 g 4 4 4 3
g 64 16 g 48
68. n m x n m x m x 5
x8 5
2 48 96
x8 0 x 8 Domain: 8,
Undefined 63. a. f g x f g x
69. q n x q n x
2 g x 4 2 x2 4 2
b. g f x g f x f x
2
2 x 4 4 x 2 16 x 16 2
1 1 x 5 10 x 15
x 15 0 x 15 Domain: ,15 15,
c. No 64. a. k m x k m x 3 m x 1
70. q p x q p x
3 1 3 1 1 x x b. m k x m k x
1 n x 10
1 1 k x 3x 1
1
p x 10
1 x 9 x 10 2
x 9 x 10 0 2
x 10 x 1 0
c. No
x 10 or x 1 Domain: , 1 1,10 10,
65. n p x n p x
p x 5 x2 9 x 5
71. q r x q r x
Domain: ,
66. p n x p n x
1 1 r x 10 2 x 3 10
2 x 3 10 0
n x 9 n x 2
2 x 3 10
x 5 9 x 5 2
2 x 3 10 or 2 x 3 10
x 10 x 25 9 x 45 2
x 19 x 70 Domain: ,
2x 7
or
7 2
or
2 x 13
13 2 13 13 7 7 Domain: , , , 2 2 2 2
2
x
276
x
Section 2.8 72. q m x q m x
1 m x 10
d. T C 4 23.267 4 10.99 104.058 The total cost to purchase 4 boxes of stationery is $104.06.
1 x 8 10
78. a. C x 60 x
x 8 10 0 x 8 10 x 8 100
b. T a a 0.055a 8 1.055a 8 c. T C x T C x 1.055 C x 8
x 92 x8 0 x 8 Domain: 8, 92 92,
1.055 60 x 8 63.3 x 8 d. T C 6 63.3 6 8 387.8 The total cost to purchase 6 tickets is $387.80.
73. n r x n r x r x 5 2 x 3 5
Domain: ,
79. a. r t 80t
c. d r t d r t 7.2 r t
74. r n x r n x
7.2 80t 576t This function represents the distance traveled (in ft) in t minutes.
2 n x 3 2 x 5 3 2 x 10 3 2 x 7 Domain: ,
d. d r 30 576 30 17,280 means that the bicycle will travel 17,280 ft (approximately 3.27 mi) in 30 min.
75. n n x n n x n x 5
x 5 5 x 10 Domain: ,
80. a. C x 2.4 x
76. p p x p p x
p x 9 p x
2
x2 9 x 9 x2 9 x
b. D C
c. D C x D C x
2
b. d r 7.2r
C 0.8
C x 2.4 x 3x 0.8 0.8
d. D C 12 3 12 36 means that 12 croissants cost $36.
x 4 18 x 3 81x 2 9 x 2 81x
81. f x x 2 and g x x 7
x 4 18 x 3 72 x 2 81x Domain: ,
82. f x x 2 and g x x 8
77. a. C x 21.95 x
83. f x 3 x and g x 2 x 1
b. T a a 0.06a 10.99 1.06a 10.99
84. f x 4 x and g x 9 x 5
c. T C x T C x
1.06 C x 10.99
85. f x x and g x 2 x 2 3
1.06 21.95 x 10.99
86. f x x and g x 4 x 2
23.267 x 10.99
277
Chapter 2 Functions and Graphs 87. f x
g. h k 3 h 1 2
5 and g x x 4 x
92. a. m p 1 m 1 p 1 4 2 6
11 88. f x and g x x 3 x
b. p m 4 p 4 m 4
4 2 6
89. a. f g 0 f 0 g 0 3 2 1
m 3 2 m c. 3 Undefined p p 3 0
b. g f 2 g 2 f 2 0 1 1 c. g f 1 g 1 f 1 2 3 6
d. m p 3 m 3 p 3 2 0 0
g 1 1 g 1 d. 1 f 1 2 2 f
e. m p 0 m p 0 m 2 3 f. p m 0 p m 0 p 4 Undefined
e. f g 4 f g 4 f 2 1
g. p m 4 p 2 4
f. g f 0 g f 0 g 3 1 g. g f 4 g 1 2
93. f g 4 f 4 g 4 2 3 1
90. a. f g 0 f 0 g 0 1 3 2
94. g f 0 g 0 f 0 6 3 18
b. g f 1 g 1 f 1 2 1 3
95. g f 2 g f 2 g 4 3
c. g f 2 g 2 f 2 1 2 2
96. f g 0 f g 0 f 6 1
f 3 2 f 2 d. 3 g 3 1 g
97. g g 6 g g 6 g 0 6
e. f g 3 f g 3 f 0 1
98. f f 1 f f 1 f 6 1
f. g f 0 g f 0 g 1 2
99. f g 5 f g 5 f 7 Undefined
g. g f 4 g 3 1
100. g f 0 g f 0 g 3 Undefined
91. a. h k 1 h 1 k 1 2 3 1 b. h k 4 h 4 k 4 1 0 0
101. k p 4 k 4 p 4 6 2 8
k 3 3 k c. 3 Undefined h h 3 0
102. p k 0 p 0 k 0 3 4 7
d. k h 1 k 1 h 1 1 2 3
103. k p 2 k p 2 k 4 6
e. k h 4 k h 4 k 1 3
104. p k 2 p k 2 p 5 0
f. h k 2 h k 2 h 3 0
105. k k 0 k k 0 k 4 6
278
Section 2.8 106. p p 5 p p 5 p 0 3
112. a. A1 x x 5 represents the area of the outer circle in terms of the radius of the inner circle, x. 2
107. T x M W x M x W x
b. A2 x x 2 represents the area of the inner circle based on its radius, x.
0.08 x 5.4 0.03 x 1.1 0.11x 6.5 This function represents the total number of adults (in millions) on probation, on parole, or incarcerated x years after the year 2000.
c. A1 A2 x A1 x A2 x
x 5 x 2 2
108. T x P R x P x R x
x 2 10 x 25 x 2 x 10 x 25 x 2 10 x 25 This function represents the area of the region outside the inner circle and inside the outer circle.
0.6 x 52.9 0.1x 8.4 0.7 x 61.3 This function represents the total school enrollment (in millions) in the United States, x years after 1990.
2
113. a. S1 x x x 4 x 2 4 x
109. a. C x 2.8 x 5000 b. R x 40 x
2
1 1 x 1 x2 1 b. S 2 x r 2 x 2 2 2 2 2 4 8
c. R C x R x C x
c. S1 S 2 x A1 x A2 x 1 x2 4 x x2 8 This function represents the area of the region outside the semicircle, but inside the rectangle.
40 x 2.8 x 5000 40 x 2.8 x 5000 37.2 x 5000 This function represents the profit for selling x CDs. d. R C 2400 37.2 2400 5000
f 114. The domain of x is the intersection of g the domains of f and g with the further restriction to exclude values of x for which g x 0 .
$84,280 110. a. C x 3.5 x 640 b. R x 25 x c. R C x R x C x
115. The domain of f g x is the set of real numbers x in the domain of g such that g(x) is in the domain of f.
25 x 3.5 640 25 x 3.5 x 640 21.5 x 640 This function represents the monthly profit for selling x necklaces.
116. For a positive real number h, the difference quotient represents the average rate of change of a function f between points x, f x and x h, f x h . This is also
d. R C 212 21.5 212 640 $3918
interpreted as the slope of the secant line between the points x, f x and
H L 111. x represents the average of the 2 high and low temperatures for day x.
x h, f x h .
279
Chapter 2 Functions and Graphs
117. a. b.
f x h f x h
xh3 x3 h
xh3 x3 h
xh3 x3 xh3 x3 h xh3 x3 x h 3 x 3
h
h
h c.
118. a. b.
1 x03 x3 f x h f x h
xh3 x3
xh4 x4 h xh4 x4 xh4 x4 h xh4 x4 x h 4 x 4
h
h
x h 4 x 4 h
x04 x4
x h 3 x 3
1
2 x3
xh4 x4 h
1
1
c.
x h 3 x 3
x h 4 x 4
1 xh4 x4
1 2 x4
2 d t h d t 4.84 t h 88 t h 4.84t 88t 119. a. h h 2 2 4.84 t 2th h 88t 88h 4.84t 2 88t h 4.84t 2 9.68th 4.84h 2 88h 4.84t 2 h 2 9.68th 4.84h 88h 9.68t 4.84h 88 h 2
b. 9.68t 4.84h 88 9.68 0 4.84 2 88 9.68 88 78.32 ft/sec c. 9.68t 4.84h 88 9.68 2 4.84 2 88 19.36 9.68 88 58.96 ft/sec d. 9.68t 4.84h 88 9.68 4 4.84 2 88 38.72 9.68 88 39.6 ft/sec e. 9.68t 4.84h 88 9.68 6 4.84 2 88 58.08 9.68 88 20.24 ft/sec
280
Section 2.8
120. a.
2 2 2 2 d t h d t 5 t h 5t 2 5 t 2th h 5t 5t 2 10th 5h 2 5t 2 10th 5h 2 10t 5h h h h h h
b. 10t 5h 10 0 5 2 10 ft/sec c. 10t 5h 10 2 5 2 20 10 30 ft/sec d. 10t 5h 10 4 5 2 40 10 50 ft/sec e. 10t 5h 10 6 5 2 60 10 70 ft/sec 121. a.
N x h N x h
0.28 x h 20.8 x h 194 0.28 x 2 20.8 x 194 2
h 0.28 x 2 xh h 20.8 x 20.8h 194 0.28 x 2 20.8 x 194
2
2
h 0.28 x 0.56 xh 0.28h 20.8h 0.28 x 2 h 2 0.56 xh 0.28h 20.8h 0.56 x 0.28h 20.8 h 2
2
b. 0.56 x 0.28h 20.8 0.56 10 0.28 10 20.8 12.4 The difference quotient of 12.4 means that cigarette production increased at an average rate of 12.4 billion per yr between the years 1950 and 1960. c. 0.56 x 0.28h 20.8 0.56 50 0.28 10 20.8 10 The difference quotient of 10 means that cigarette production decreased at an average rate of 10 billion per yr between the years 1990 and 2000.
A x h A x 0.0092 x h 0.805 x h 21.9 0.0092 x 0.805 x 21.9 122. a. h h 2 2 0.0092 x 2 xh h 0.805 x 0.805h 21.9 0.0092 x 2 0.805 x 21.9 h 0.0092 x 2 0.0184 xh 0.0092h 2 0.805h 0.0092 x 2 h 2 0.0184 xh 0.0092h 0.805h 0.0184 x 0.0092h 0.805 h 2
2
b. 0.0184 x 0.0092h 0.805 0.0184 20 0.0092 5 0.805 0.391 The difference quotient of 0.391 means that the amount of CO2 emitted decreases at an average rate of 0.391 ton per yr for each 1-mpg increase in fuel efficiency for vehicles getting between 20 and 25 mpg. c. 0.0184 x 0.0092h 0.805 0.0184 35 0.0092 5 0.805 0.115 The difference quotient of 0.115 means that the amount of CO2 emitted decreases at an average rate of 0.115 ton per yr for each 1-mpg increase in fuel efficiency for vehicles getting between 35 and 40 mpg.
281
Chapter 2 Functions and Graphs 123.
125. g g x g g x
f f x f f x
1 f x 2
g x 3
1 1 2 x2
Domain of g is 3, . x3 3 0
1 1 2 x 2 1 2 x 4 1 x2 x2 x2 x2 1 2 x 5 2 x 5 x2 Domain of f is , 2 2, . 2 x 5 0
x3 3 x39
126. m m x m m x
m x 4
5 2
x4 40 x4 4 x 4 16
1 h x 6
Domain: 20,
x 20
h h x h h x
x4 4
Domain of g is 4, .
5 5 Domain: , 2 2, , 2 2
124.
Domain: 12,
x 12
2 x 5 x
x3 3
127. f g h x f g h x 2 g h x 1
1
1
2 h x 1 2 3 x
1 6 x6
1 1 1 6 x 36 6 x 6 1 x6 x6 x6 1 x6 6 x 37 6 x 37 x6 Domain of h is , 6 6, . 6 x 37 0 6 x 37
2
2
2 h x 1 2 x 1
128. g f h x g f h x f h x 2
2
3
129. h g f x h g f x 3 g f x
3 f x 3 2 x 1 2
2
130. g h f x g h f x h f x
37 x 6
3 f x
37 37 Domain: , 6 6, , 6 6
2 x 1 2
3
2
131. m x 3 x , n x x 1 , h x 4 x ,
k x x2
132. m x x , n x x 4 , h x 2 x ,
k x x3
282
2
2
Chapter 2 Review Chapter 2 Review Exercises 1. a. d
x2 x1 2 y2 y1 2
4 1 2 8 2
4 x 1 y 18
3. a.
4 3 1 2 0 18
2
4 4 2 0 18
25 100 125 5 5
18 0 18 True
x x y y2 b. M 1 2 , 1 2 2
Yes 4 5 1 2 0 18 4 4 2 0 18
3 6 3 , , 3 2 2 2 2. a. d
14 0 18 False No
x2 x1 2 y2 y1 2
3 3 3 4 6 6 2
3 4. x-intercept: , 0 ; y-intercept: 0, 2 2
2
5. x-intercept: 4, 0 ; y-intercepts:
12 150 162 8 2
0, 4 , 0, 10
x x y y2 b. M 1 2 , 1 2 2
6. x-intercepts: 1, 0 , 7, 0 ; y-intercept: none
3 3 3 6 4 6 , 2 2 4 3 3 6 3 6 2 3, , 2 2 2
7. y x 2 2 x x
y
y x2 2x
2
8
y 2 2 2 8
1
3
0
0
1
1
2
0
3
3
4
8
2
y 1 2 1 3 2
y 0 2 0 0 2
y 1 2 1 1 2
y 2 2 2 0 2
y 3 2 3 3 2
y 4 2 4 8 2
4 x 1 y 18
b.
1 4 8 2 , 2 2
Ordered pair
2, 8 1, 3 0, 0 1, 1 2, 0 3, 3 4, 8
283
Chapter 2 Functions and Graphs 8. d 5 3 2 2 2
x x y y2 13. a. C 1 2 , 1 2 2
2
64 16
7 1 5 3 , 2 2
80 4 5 units
8 2 , 4,1 2 2
9. r 2 4
r 42 Center: 4, 3 ; Radius: 2 10. r 17 2
2
2
11
7 42 5 1 2
x h 2 y k 2 r 2 x 4 2 y 12 5 2 x 4 2 y 12 25
x h2 y k 2 r 2 x 3 y 1
x2 x1 2 y2 y1 2
9 16 25 5
r 17 3 Center: 0, ; Radius: 17 2 11. a.
r
b.
2
x 32 y 12 11 b.
14. a. The circle must touch the x-axis and the yaxis at a distance r from the center. Since the center is in quadrant IV, the center must be 4, 4 . 12. a. x h y k r 2 2
x h2 y k 2 r 2 2 x 42 y 4 4 2 x 42 y 4 2 16
2
x 0 2 y 02 4.2 2 x 2 y 2 17.64 b.
b.
284
Chapter 2 Review 15. a.
x 2 y 2 10 x 2 y 17 0
20. No vertical line intersects the graph in more than one point. This relation is a function.
x 10 x y 2 y 17 2
2
2
21. There is at least one vertical line that intersects the graph in more than one point. This relation is not a function.
2
1 2 10 25
1 2 2 1
x 10 x 25 y 2 y 1 17 25 1 2
2
22. x 2 y 3 4 This equation represents the graph of a circle with center 0, 3 and radius 2. This relation is not a function because it fails the vertical line test. 2
x 5 y 1 9 2
2
b. Center: 5,1 ; Radius:
93
x2 y 2 8 y 3 0
16. a.
x2 y 2 8 y
3
23. x 2 y 3 4
y x2 7
2
1 2 8 16
x 2 y 2 8 y 16 3 16 x y 4 13 2
2
b. Center: 0, 4 ; Radius: 13 17. x 3 y 5 0 The sum of two squares will equal zero only if each individual term is zero. Therefore, x 3 and y 5 . 2
No vertical line intersects the graph in more than one point. This relation is a function.
2
24. f x 2 x 2 4 x
3, 5
18.
a. f 0 2 0 4 0 0 2
b. f 1 2 1 4 1 2 4 6 2
x 2 y 2 6 x 4 y 15 0
x 6 x y 4 y 15 2
c. f 3 2 3 4 3 18 12 6
2
2
1 2 6 9
2
2
1 2 4 4
d. f t 2 t 4 t 2t 2 4t 2
e. f x 4 2 x 4 4 x 4
x 6 x 9 y 4 y 4 15 9 4 2
2
2
x 32 y 22 2
2 x 2 8 x 16 4 x 16
The sum of two squares cannot be negative, so there is no solution.
2 x 16 x 32 4 x 16 2
2 x 2 12 x 16
25. a. f 1 5
19. a. {(Dara Torres, 12), (Carl Lewis, 10), (Bonnie Blair, 6), (Michael Phelps, 16)}
b. f 0 4 c. f x 1 for x 3
b. {Dara Torres, Carl Lewis, Bonnie Blair, Michael Phelps} c. {12, 10, 6, 16}
d. Yes
285
Chapter 2 Functions and Graphs 26.
P x 1.065 x 0.32 x
32.
P 189 1.065 189 0.32 189
x 3
1.065 189 60.48
x 3 and x 3 , 3 3, 3 3,
1.065 248.48 265.70 P 189 265.70 means that a drill that costs $189 from the manufacturer will cost the customer $265.70 after the department store markup and sales tax.
33. There is no restriction on x. , 34. 2 x 0
x 2 x2 , 2
27. Solve p x 0 :
p x x 3 1 0 x 3 1
35. a. f 2 2
x 3 1 x 31 or x 4 or Evaluate p 0 :
x 3 0
b. f 3 1
x 3 1 x2
c. f x 1 for x 1 , x 3 d. f x 4 for x 4
p x x 3 1 p 0 0 3 1 3 1 2
e. x-intercepts: 0, 0 and 2, 0
x-intercepts: 4, 0 , 2, 0 ; y-intercept: 0, 2
f. y-intercept: 0, 0 g. Domain: ,
28. Solve q x 0 :
q x x 2
h. Range: ,1
0 x2
36. f x 2 x 2 4
x2 x4 Evaluate q 0 :
37. 2 x 4 y 8
4 y 2x 8 1 y x2 2
q x x 2 q 0 0 2 2 x-intercept: 4, 0 ; y-intercept: 0, 2
x 4 2 0 2
29. Domain: 4, 2, 0, 2, 3, 5 ;
Range: 3, 0, 2,1
30. Domain: , 4 ; Range: , 3 31. x 5 0 x5 , 5 5,
y 0 1 2 3
x-intercept: 4, 0 ; y-intercept: 0, 2
286
Chapter 2 Review 38. 4 x 5 y
4 2 y2 y1 2 1 42. m 3 3 x2 x1 1 3 4 2
4 y x 5 x 5 0
y 4 0 4 5 4
1 5
43. m
y2 y1 f b f a x2 x1 ba
44. 0 45. Undefined 46. Undefined
x-intercept: 0, 0 ; y-intercept: 0, 0 47. 39. y 2 x 2 0 2 4
48. a. Linear
y 2 2 2 2
b. Neither c. Constant 49.
x-intercept: none; y-intercept: 0, 2 40. 3x 5
x
x 5 3 5 3 5 3 5 3
C represents the change in cost per change t in time.
5 3 y 2
y mx b 2 y xb 3 2 5 1 b 3 2 5 b 3 13 b 3 2 13 2 13 y x ; f x x 3 3 3 3
50. y mx b
y 0x b yb 1 b 4 1 1 y ; f x 4 4
0 2 4
5 x-intercept: , 0 ; y-intercept: none 3
51. m
y y1 4 2 2 1 41. m 2 12 4 16 8 x2 x1 287
f x2 f x1 x2 x1
30 3 4 3 7
Chapter 2 Functions and Graphs
52. a. Average rate of change
m1 0
x2 x1
14,577 10,799 3778 $755.60/yr 10 5 5
b. Average rate of change
f x2 f x1
1 1 0 2 2 1 1 0 True 2 2 Perpendicular
f x2 f x1 x2 x1
35,854 26,561 9293 $1858.60/yr 25 20 5
b. 3 y 1.5 x 5
3 x5 2 1 5 1 y x ; m2 2 3 2 m1 m2 ; Parallel
3y
c. Increasing 53. f x x 3 4
f 0 0 4 4 3
f 2 2 4 8 4 4
c. 4 x 8 y 8
3
8 y 4 x 8
f 4 4 4 64 4 60 3
a. Average rate of change
f 2 f 0 20
f 4 f 2 42
54. a. 1
1 1 y x 1; m2 2 2 m1 m2 1 m1 0 m2
f x2 h x1 x2 x1
4 4 8 4 2 2
b. Average rate of change
f x2 h x1
60 4 2
b. ,1
1 m2
1 1 0 1 2 2 1 0 2 False 2 Neither
x2 x1
56 28 2
c. 1, 57.
2 55. a. 3
y y1 m x x1 y 7 3 x 2 y 7 3 x 2
1 1 3 b. 2 m 2 3
y 7 3x 6 y 3x 1 58. y y1 m x x1
56. 2 x 4 y 3
4 y 2 x 3 y
2 x 0 5 2 y 5 x 5 2 y x5 5 y 5
1 3 1 x ; m1 2 4 2
a. 12 x 6 y 6
6 y 12 x 6 y 2 x 1; m2 2
288
Chapter 2 Review 59. m
b. G 4.5 15 2 4.5 15 9 6
y2 y1 7.1 5.3 1.8 0.9 x2 x1 0.9 1.1 2
G 4.5 6 means that after 4.5 hr of driving, the tank has 6 gal left.
y y1 m x x1
y 5.3 0.9 x 1.1
65. a. C x 1500 35 x
y 5.3 0.9 x 0.99
b. R x 60 x
y 0.9 x 6.29
c. P x R x C x
60. A line with an undefined slope is a vertical line. Every coordinate has the same x-value, so the equation is x 5 .
60 x 1500 35 x 25 x 1500
61. 2 x y 4
25 x 1500 0 25 x 1500
y 2 x 6; m 2
y y1 m x x1
x 60 The studio needs more than 60 private lessons per month to make a profit.
y 6 2 x 2 y 6 2x 4 y 2 x 10
e.
The studio will make $550.
2 2 y x; m 5 5 1 1 5 2 m 2 5 y y1 m x x1
66. a.
5 y 3 x 2 2 5 y 3 x 2 2 5 y 3 x5 2 5 y x2 2
b. m
y2 y1 x2 x1 26,500 22,800 3700 370 12 2 10 y y1 m x x1
y 22,800 370 x 2
63. A line that is perpendicular to the y-axis is a horizontal line with slope 0. Every coordinate has the same y-value, so the equation is y 7 .
G t 15 2t
P x 25 x 1500 P 82 25 82 1500 2050 1500 550
62. 5 y 2 x
64. a. G t 15 t hr
P x 0
d.
y 2 x 6
y 22,800 370 x 740 y 370 x 22,060 E x 370 x 22,060
60 mi 1 gal hr 30 mi
c. The slope is 370 and means that enrollment rose at an average rate of 370 students per year.
289
Chapter 2 Functions and Graphs d. E x 370 x 22,060
70. The graph of g is the graph of f x x shifted upward 1 unit.
E 15 370 15 22,060 5550 22,060 27,610 students 67. a. y 369.6 x 22,111 b.
71. The graph of h is the graph of f x x 2 shifted to the right 4 units. c. y 369.6 15 22,111
5544 22,111 27,655 students 68. The value a provides a vertical stretch or shrink or reflection across the x-axis. The value b provides a horizontal stretch or shrink or reflection across the y-axis. The value c provides a horizontal shift. The value d provides a vertical shift. 72. The graph of k is the graph of f x 3 x shifted to the left 1 unit.
69. The graph of f is the graph of f x x shifted downward 4 units.
290
Chapter 2 Review 76. The graph of v is the graph of f x x reflected across the x-axis and shrunk 1 vertically by a factor of . 2
73. The graph of r is the graph of f x x shifted to the right 3 units and upward 1 unit.
74. The graph of s is the graph of f x x 2 shifted to the left 2 units and downward 3 units.
77. m x x 5 x 5
The graph of m is the graph of f x x reflected across the y-axis and shifted right 5 units.
75. The graph of t is the graph of f x x reflected across the x-axis and stretched vertically by a factor of 2. 78. n x x 4 x 4
The graph of n is the graph of f x x reflected across the y-axis and shifted to the left 4 units.
291
Chapter 2 Functions and Graphs 79. The graph of the function is the graph of f x shrunk horizontally by a factor of 2.
82. The graph of the function is the graph of f x shifted to the right 4 units, reflected across the x-axis, and shifted downward 1 unit.
80. The graph of the function is the graph of f x stretched horizontally by a factor of 2.
83. The graph of the function is the graph of f x shifted to the right 3 units, stretched vertically by a factor of 2, and shifted upward 1 unit.
81. The graph of the function is the graph of f x shifted to the left 1 unit, reflected across the x-axis, and shifted downward 3 units.
84. The graph of the function is the graph of f x shifted to the left 2 units, shrunk
vertically by a factor of downward 3 units.
292
1 , and shifted 2
Chapter 2 Review 85. y x 4 3 Replace x by x .
1 x 1 3 Replace x by x . 1 y x 1 3 1 y x 1 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace y by y . 1 y x 1 3 1 y x 1 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x by x and y by y . 1 y x 1 3 1 y x 1 3 1 y x 1 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
87. y
y x 3 4
y x4 3 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y .
y x4 3 y x4 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x by x and y by y . y x 3 4
y x4 3 y x4 3 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 86. x y y 2 Replace x by x . x y y2
x y y2 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace y by y . x y y
88. x 2 y 2 1 Replace x by x .
x2 y 2 1
2
x y y2 This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x by x and y by y .
x2 y 2 1 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y .
x y y
x2 y 1
2
2
x y y2
x2 y 2 1 This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis.
x y y2 This equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.
293
Chapter 2 Functions and Graphs Replace x by x and y by y . x y 1 This equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.
m x x 2
2
2
m x m x m x m x The function is neither even nor odd.
f x 4 x3 x
95. a. Use the first rule. f 4 4 4 2 16 2 18
f x 4 x x 3
f x 4 x3 x
b. Use the second rule. f 1 1 1 2
f x 4 x x 4 x x
90.
3
3
f x f x
c. Use the third rule. f 3 5
The function is odd.
d. Use the second rule. f 2 2 4 2
g x 3 x
96.
g x 3 x x
g x 3 x x g x g x The function is odd. 91.
p x 4 x2 p x 4 x 4 x2 2
p x p x The function is even. 92.
97.
q x x q x x x q x q x The function is even.
93.
m x x 2 m x x 2
2
89.
94.
x y 1 2
k x x 3
2
k x x 3
98.
2
k x x 3
2
k x k x k x k x The function is neither even nor odd.
294
Chapter 2 Review 99. a.
108. f g 5 f g 5 f 5 2
f x x 1
f 1.5 1.5 1 2.5 3 b.
f 3 3 3 9
f x x 1
109. g f 5 g f 5 g 3 5
f 2 2 1 3 3 c.
g 15 15 2 17
f x x 1
110. a. f g 2 f 2 g 2 3 2 1
f 0.1 0.1 1 0.9 1 d.
b. g f 4 g 4 f 4 2 1 2
f x x 1
f 6.3 6.3 1 5.3 5
g 3 1 g Undefined c. 3 f 3 0 f
100. a. , 3 2, 0
d. f g 4 f 2 3
b. 3, 2 0, 3 c. 3,
e
101. a. 2,
f. g f 5 g f 5 Undefined
b. , 2
111. n m x n x m x
c. Never constant
x 2 4 x 4 x x 2
102. At x 2 , the function has a relative minimum of 2 . At x 4 , the function has a relative minimum of 1 . At x 2 , the function has a relative maximum of 1.
Domain: ,
p x x2 p 112. x 2 n n x x 4 x
103. At x 2 , the function has a relative maximum of 4.
x 4 104. f x 1
for x 3
x20 x2 2 x 4x 0
for x 3
x x 4 0 x 0 and x 4 Domain: 2, 4 4,
105. f h 2 f 2 h 2
3 2
1 1 19 6 2 1 3 3
n x x2 4 x n 113. x p x p x2 x20 x2 Domain: 2,
106. g h 3 g 3 h 3
3 2
g f 4 g f 4 g 1 1
1 1 1 1 3 1 4 4
g 5 5 2 7 g 28 107. 5 1 1 h h 5 5 1 4
114. m p x m x p x 4 x x 2 x20 x2 Domain: 2,
295
Chapter 2 Functions and Graphs 115. q n x q n x
1 1 2 n x 5 x 4 x 5
118.
x 4x 5 0
f x h f x h
x 5 x 1 0 x 5 and x 1 Domain: , 1 1, 5 5,
h 3 x 2 xh h 4 x 4h 9 3 x 2 4 x 9
2
2
116. q p x q p x
1 p x 5
1 x2 5
12 and g x x 5 x 121. a. d t 60t
120. f x
x2 5 x 2 25
b. n d
x 27 Domain: 2, 27 27, 117.
h
2
119. f x x 2 and g x x 4
x20 x2 x2 5 0
f x h f x
h 3x 6 xh 3h 4h 3 x 2 6 xh 3h 2 4h h h 6 x 3h 4 2
3 x h 4 x h 9 3 x 2 4 x 9 2
2
d 28
c. n d t n d t
6 x h 5 6 x 5
d t 28
60t 28
This function represents the number of gallons of gasoline used in t hours.
h 6 x 6h 5 6 x 5 h 6h 6 h
d. n d 7
60 7
15 means that 15 gal 28 of gasoline is used in 7 hr.
Chapter 2 Test x x y y 1. a. C 1 2 , 1 2 2 2
c.
8 2 5 3 , 2 2 6 2 , 3, 1 2 2 b. r
x2 x1 2 y2 y1 2
2 3 2 3 1
x h y k r2 2 2 2 x 3 y 1 41 2 2 x 3 y 1 41 2
2. a. Substitute 0 for y: x y 4
2
2
Substitute 0 for x: x y 4
x 0 4
0 y 4
x 4
y 4
y 4 x-intercept: 4, 0 ; y-intercepts:
25 16 41
0, 4 , 0, 4
b. No
296
Chapter 2 Test 3. a.
x 2 y 2 14 x 10 y 70 0
x 14 x y 10 y 70 2
2
2
2
1 2 14 49
1 2 10 25
x 14 x 49 y 10 y 25 70 49 25 2
2
x 7 2 y 5 2 4 b. Center: 7, 5 ; Radius:
42
4. This mapping defines the set of ordered pairs: 2, 5 , 3, 5 , 4, 5 . No two ordered pairs have the same
x value but different y values. This relation is a function. 5. There is at least one vertical line that intersects the graph in more than one point. This relation is not a function. 6. a. f 1 2 1 7 1 3 2 7 3 12 2
b. f x h 2 x h 7 x h 3 2 x 2 2 xh h 2 7 x 7 h 3 2 x 2 4 xh 2h 2 7 x 7 h 3 2
f x h f x 2 x 4 xh 2h 7 x 7 h 3 2 x 7 x 3 c. h h 2 2 2 x 4 xh 2h 7 x 7 h 3 2 x 2 7 x 3 h 2 4 xh 2h 7 h 4 x 2h 7 h 2
2
d. Substitute 0 for f x :
0 2 x 2 7 x 3 0 2 x2 7 x 3 0 2 x 1 x 3 2x 1 0
or
x3 0
2x 1
or
x3
x
1 2
1 , 0 and 3, 0 2 e. Substitute 0 for x:
f x 2 0 7 0 3 3 2
0, 3
f. 4 x 2h 7 4 1 2 2 7 4 4 7 1
297
2
Chapter 2 Functions and Graphs 7. a. f 0 2
b. f 4 0
d. m
1 4 3 3 4
e. m
3 4
c. f x 2 for x 2 and x 2 d. , 2 0, 2 e. 2, 0 2,
11. x 3 y 4
f. f 0 2 is a relative minimum.
3y x 4
g. f 2 2 and f 2 2 are relative maxima. h. ,
1 4 1 y x ; m 3 3 3 1 1 3 1 m 3 y y1 m x x1
i. , 2
j. f x f x , so the function is even. 8. 3w 7 0
y 6 3 x 2
3w 7
y 6 3 x 2
7 w 3 7 7 , , 3 3
y 6 3x 6 y 3 x 12 12. a. 2
9. 4 c 0
c. x | x 2
c 4 c4 , 4
13.
10. 3x 4 y 8
4 y 3 x 8 3 y x2 4 a. m
3 4
b. 0, 2 14.
c.
298
b. x | x 2
Chapter 2 Test 15.
f x x3 x
18.
f x x x 3
f x x3 x
f x x3 x x3 x f x f x The function is odd. g x x 4 x3 x
19.
g x x x x x 4 x3 x
16.
4
3
g x x 4 x3 x x 4 x3 x g x g x g x g x The function is neither even nor odd. 20. a.
f x x
f 4.27 4.27 4 17. x y 8 Replace x by x . 2
b.
f x x
f 4.27 4.27 5
x2 y 8 x2 y 8 This equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace y by y .
21. f h 6 f 6 h 6
64 65 2 1 1
x2 y 8
22. g h 5 g 5 h 5
x2 y 8
1 55 53 1 0 2 0
This equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x by x and y by y .
x2 y 8 x2 y 8 This equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.
23. h f 1 h f 1
h 1 4 h 3 35 2 Undefined
299
Chapter 2 Functions and Graphs 28. a. f g 3 f 3 g 3 1 2 1
24. f g x f x g x
1 x 4 x 3 x3 0
b. f g 0 f 0 g 0 1 1 1 c. g f 3 g 1 2
x4 x3
d. f g 2 f g 2 f 3 Undefined
x3 Domain: , 3 3,
e. 3, 0 f. , 2
1 g x x 3 g 1 25. x f x x 4 x 3 x 4 f
29. a. m
y y1 m x x1
x 3 x 4 0
y 1833 98 x 0
x 3 and x 4 Domain: , 3 3, 4 4, 1 26. g h x g h x h x 3 x5 0 x5 x5 3 0
y2 y1 2614 1833 781 98 x2 x1 80 8
y 1833 98 x y 98 x 1833
1
b. y 98 x 1833
x5 3
y 98 15 1833 1470 1833 $3303 million
x5 3
30. a. y 92.3 x 1912
x5 9
b. y y 92.3x 1912
x 14 Domain: 5,14 14,
y 92.3 15 1912 1384.5 1912
27. f x x and g x x 7
$3297 million
3
Chapter 2 Cumulative Review Exercises 1. a. f 2 1
2. a.
d. 2,
e. 1,
f. 1,1
x 12 x y 4 y 31 2
b. f x 0 for x 0 and x 3 c. 4,
x 2 y 2 12 x 4 y 31 0 2
2
2
1 2 12 36
1 2 4 4
x 12 x 36 y 4 y 4 31 36 4 2
g. 4, 1
2
x 6 2 y 2 2 9
h. f f 1 f f 1 f 2 1
b. Center: 6, 2 ; Radius:
300
93
Chapter 2 Cumulative Review 3. g f x g f x
8.
1 1 2 f x x 3x
x 2 3x 0 x 2 3x 0 x x 3 0 x 0 and x 3 Domain: , 0 0, 3 3, 4. g h x g x h x
x0 x20 x 2 Domain: 2, 0 0, 5.
1 x2 x
x2 x
9.
f x h f x h
x h 3 x h x 2 3x 2
y2 y1 1 3 4 2 x2 x1 5 2 8 10 y mx b 2 y xb 5 2 1 2 b 5 4 1 b 5 1 b 5 2 1 y x 5 5
10. m
h
x 2 2 xh h 2 3 x 3h x 2 3x
h x 2 xh h 3h x 2 h 2 2 xh h 3h 2 x h 3 h 2
2
6. 2 x h 3 2 0 3 3 0 3 3 0 7. Substitute 0 for f x :
f x x 2 3x 0 x 2 3x
11. x 7 or 7 x
x 2 3x 0 x x 3 0
12. 2 x 3 128 2 x 3 64 2 x 4 2 x 4 x 2 4 x 16
x 0 or x 3 Substitute 0 for x: f x x 2 3x f 0 0 3 0 0 2
x-intercepts: 0, 0 and 3, 0 ;
y-intercept: 0, 0
301
3
3
Chapter 2 Functions and Graphs 13.
3t t 1 2t 6
15. Let u x1 5 . x 2 5 3x1 5 2 0
3t 2 3t 2t 6 3t 2 t 6 0 3t 2 t 6 0
t
1
u 2 3u 2 0 u 1 u 2 0 u 1 or
1 2 4 3 6 2 3
x
1 1 72 1 71 1 71 i 6 6 6 6 71 1 i 6 6
1, 32
14.
17.
15 11
6 15 11
15 11 15 11
x
2 x 32
3a 1 11 11 3a 1 11 12 3a 10 10 4 a 3 10 4, 3
7 4x 2 5
6
u2 15
16. 3a 1 2 9
2 4x 2 4 x 2 2 or 4 x 2 2 4 x 4 or 4x 0 x 1 or x0 0,1
18.
1 x 1
15
6
3 2 x 1 7 2 2 x 6 1 x 3 or 3 x 1 3, 1
15 11 6 15 6 11 3 15 3 11 15 11
4
19. 3c 8c 2 d 3 c 2 50d 3 2d 2c 4 d 3c 22 c 2 d 2 2d c 2 52 d 2 2d 2d
2
c 2d 2 2
6c 2 d 2d 5c 2 d 2d 2c 2 d 2d 9c 2 d 2d 2 1 2 2 2 2 2u 1 w1 uw u w u w 2uw u w uw 2w u 20. 2 2 2 2 2 2 4 1 4u w u w 4w u 2w u 2w u 2w u u 2 w2
302
Section 3.1
Chapter 3 Polynomial and Rational Functions Section 3.1 Quadratic Functions and Applications 1. quadratic
2. axis
3. vertex
4. h, k
5. downward
6. h
7. k
8. k
h. The domain is , .
The range is , 1 .
10. g x x 2 4 x 2 4 2
2
a 1, h 2, k 4
a. Since a 0 , the parabola opens downward.
9. f x x 4 1 2
a 1, h 4, k 1
b. The vertex is h, k 2, 4 .
a. Since a 0 , the parabola opens downward.
c.
2
4 x 2 4 x 2
c. f x x 4 1 2 2 2
2
1 x4 4 1 x x 3 or x 5 The x-intercepts are 3, 0 and 5, 0 .
2
d. g 0 0 2 4 2 4 4 4 0 The y-intercept is 0, 0 . 2
d. f 0 0 4 1 4 1 2
2
4 x2 2 2 x x 4 or x 0 The x-intercepts are 4, 0 and 0, 0 .
0 x 4 1 1 x 4
2
0 x 2 4
b. The vertex is h, k 4,1 .
1 x 4
g x x 2 4
2
e.
2
16 1 15 The y-intercept is 0, 15 . e.
f. The axis of symmetry is the vertical line through the vertex: x 2 . g. The maximum value is 4. h. The domain is , .
f. The axis of symmetry is the vertical line through the vertex: x 4 .
The range is , 4 .
g. The maximum value is 1. 303
Chapter 3 Polynomial and Rational Functions c. k x 2 x 3 2
11. h x 2 x 1 8 2 x 1 8
2
2
2
a 2, h 1, k 8
0 2 x 3 2
a. Since a 0 , the parabola opens upward.
2 2 x 3
b. The vertex is h, k 1, 8 .
1 x 3
c.
2
2
3 1 x x 2 or x 4 The x-intercepts are 2, 0 and 4, 0 .
0 2 x 1 8 2
4 x 1
2
1 x3
h x 2 x 1 8 8 2 x 1
2
2
2
d. k 0 2 0 3 2 2 3 2 2
4 x 1
2
18 2 16 The y-intercept is 0,16 .
1 2 x x 3 or x 1 The x-intercepts are 3, 0 and 1, 0 .
e.
d. h 0 2 0 1 8 2 1 8 2
2
2 8 6 The y-intercept is 0, 6 . e.
f. The axis of symmetry is the vertical line through the vertex: x 3 . g. The minimum value is 2 . h. The domain is , .
The range is 2, .
f. The axis of symmetry is the vertical line through the vertex: x 1 .
13. m x 3 x 1 3 x 1 0 2
2
g. The minimum value is 8 .
a 3, h 1, k 0
h. The domain is , .
a. Since a 0 , the parabola opens upward.
The range is 8, .
b. The vertex is h, k 1, 0 . c. m x 3 x 1
2
a 2, h 3, k 2
0 3 x 1
2
a. Since a 0 , the parabola opens upward.
0 x 1
12. k x 2 x 3 2 2 x 3 2 2
2
b. The vertex is h, k 3, 2 .
2
0 x 1 1 x The x-intercept is 1, 0 .
304
Section 3.1 d. m 0 3 0 1 3 1 3 2
e.
2
The y-intercept is 0, 3 .
e.
f. The axis of symmetry is the vertical line through the vertex: x 2 . g. The minimum value is 0. f. The axis of symmetry is the vertical line through the vertex: x 1 .
h. The domain is , .
The range is 0, .
g. The minimum value is 0. h. The domain is , .
15. p x
The range is 0, .
1 a , h 4, k 1 5
2 1 1 2 14. n x x 2 x 2 0 2 2 1 a , h 2, k 0 2
a. Since a 0 , the parabola opens downward. b. The vertex is h, k 4,1 .
a. Since a 0 , the parabola opens upward. b. The vertex is h, k 2, 0 .
c.
1 x 2 2 2 1 2 0 x 2 2
c. n x
0 x 2
2 1 1 x 4 2 1 x 4 1 5 5
1 x 4 2 1 5 1 2 0 x 4 1 5 1 2 1 x 4 5
p x
5 x 4
2
2
5 x4
0 x2 2 x The x-intercept is 2, 0 .
4 5 x x 4 5
or
x 4 5
The x-intercepts are 4 5, 0 and
1 1 0 2 2 2 2 2 2 2 The y-intercept is 0, 2 .
d. n 0
4 5, 0 .
305
Chapter 3 Polynomial and Rational Functions
The x-intercepts are 1 3, 0 and
1 1 0 4 2 1 4 2 1 5 5 16 5 11 5 5 5 11 The y-intercept is 0, . 5
d. p 0
1 3, 0 .
1 1 0 1 2 1 1 2 1 3 3 1 2 1 3 3 2 The y-intercept is 0, . 3
d. q 0
e.
e.
f. The axis of symmetry is the vertical line through the vertex: x 4 . g. The maximum value is 1. h. The domain is , .
f. The axis of symmetry is the vertical line through the vertex: x 1 .
The range is , 1 .
16. q x
g. The maximum value is 1. h. The domain is , .
1 x 12 1 3
The range is , 1 .
1 a , h 1, k 1 3
17. a. f x x 2 6 x 5
a. Since a 0 , the parabola opens downward.
f x x2 6 x
b. The vertex is h, k 1,1 . c.
f x x 3 4 2
b. The vertex is h, k 3, 4 . c.
f x x2 6 x 5 0 x2 6 x 5
2
0 x 5 x 1
3 x 1
x 5 or x 1 The x-intercepts are 5, 0 and 1, 0 .
1 3 x x 1 3
f x x2 6 x 9 9 5
1 2 q x x 1 1 3 1 2 0 x 1 1 3 1 2 1 x 1 3 3 x 1
2
5 12 6 9
or
x 1 3
d. f 0 0 6 0 5 5 2
The y-intercept is 0, 5 .
306
Section 3.1 f. The axis of symmetry is the vertical line through the vertex: x 4 .
e.
g. The minimum value is 9 . h. The domain is , .
The range is 9, .
19. a. p x 3x 2 12 x 7
p x 3 x 4 x 4 3 4 7 2
h. The domain is , .
p x 3 x 2 19
The range is 4, .
2
b. The vertex is h, k 2, 19 .
18. a. g x x 2 8 x 7
2
1 2 4 4
p x 3 x2 4 x 4 4 7
g. The minimum value is 4 .
g x x 8x
7
2
7
p x 3 x 4 x
f. The axis of symmetry is the vertical line through the vertex: x 3 .
2
c. p x 3x 2 12 x 7
2
1 2 8 16
0 3x 2 12 x 7
g x x 2 8 x 16 16 7
x
g x x 4 9
12
2
12 2 4 3 7 2 3
12 228 12 2 57 6 57 6 6 3 6 57 6 57 x or x 3 3 6 57 The x-intercepts are , 0 and 3
b. The vertex is h, k 4, 9 . c. g x x 2 8 x 7
0 x2 8x 7 0 x 7 x 1 x 7 or x 1 The x-intercepts are 7, 0 and 1, 0 .
6 57 , 0 . 3
d. g 0 0 8 0 7 7 2
d. p 0 3 0 12 0 7 7
The y-intercept is 0, 7 .
2
The y-intercept is 0, 7 .
e. e.
307
Chapter 3 Polynomial and Rational Functions f. The axis of symmetry is the vertical line through the vertex: x 2 .
f. The axis of symmetry is the vertical line through the vertex: x 1 .
g. The minimum value is 19 .
g. The minimum value is 5 .
h. The domain is , .
h. The domain is , .
The range is 19, .
The range is 5, .
20. a. q x 2 x 2 4 x 3
21. a. c x 2 x 2 10 x 4
q x 2 x 2 x 2
4
c x 2 x 2 5 x
2
1 3 2 1 2
2
25 1 2 5 4
q x 2 x 2 x 1 2 1 3 q x 2 x2 2 x 1 1 3
25 25 c x 2 x 2 5 x 4 4 4
2
q x 2 x 1 5 2
b. The vertex is h, k 1, 5 .
25 25 c x 2 x 2 5 x 2 4 4 4
c. q x 2 x 2 4 x 3
5 33 c x 2 x 2 2
2
0 2 x2 4 x 3 x
4
5 33 b. The vertex is h, k , . 2 2
42 4 2 3 2 2
c. c x 2 x 2 10 x 4
4 40 4 2 10 2 10 4 4 2 2 10 2 10 or x x 2 2 2 10 The x-intercepts are , 0 and 2
0 2 x 2 10 x 4 x
10
10 2 4 2 4 2 2
10 132 4 10 2 33 5 33 4 2 5 33 5 33 x or x 2 2 5 33 The x-intercepts are , 0 and 2
2 10 2 , 0 . d. q 0 2 0 4 0 3 3 2
The y-intercept is 0, 3 .
e.
5 33 , 0 . 2 d. c 0 2 0 10 0 4 4 2
The y-intercept is 0, 4 .
308
Section 3.1 9 177 The x-intercepts are , 0 and 6
e.
9 177 , 0 . 6 d. d 0 3 0 9 0 8 8 2
The y-intercept is 0, 8 .
e.
f. The axis of symmetry is the vertical line 5 through the vertex: x . 2
33 . 2
g. The maximum value is
h. The domain is , . 33 The range is , . 2
f. The axis of symmetry is the vertical line 3 through the vertex: x . 2
22. a. d x 3 x 9 x 8 2
d x 3 x 3 x 2
2
9 1 8 3 4 2
g. The maximum value is
9 9 d x 3 x 2 3 x 8 4 4
59 . 4
h. The domain is , .
59 The range is , . 4
9 9 d x 3 x 2 3 x 3 8 4 4 2
23. f x 3x 2 42 x 91
3 59 d x 3 x 2 4
x-coordinate:
3 59 b. The vertex is h, k , . 2 4
b 42 42 7 2a 2 3 6
y-coordinate: f 7 3 7 42 7 91 238 2
The vertex is 7, 238 .
c. d x 3x 2 9 x 8
0 3x 9 x 8 2
x
9
24. g x 4 x 2 64 x 107
9 4 38 2 3 2
x-coordinate:
9 177 9 177 6 6 9 177 9 177 x or x 6 6
b 64 64 8 2a 2 4 8
y-coordinate:
g 8 4 8 64 8 107 149 2
The vertex is 8, 149 .
309
Chapter 3 Polynomial and Rational Functions f. The axis of symmetry is the vertical line through the vertex: x 1 .
1 25. k a a 2 6a 1 3 6 b 6 9 x-coordinate: 2 2a 1 2 3 3 1 2 y-coordinate: k 9 9 6 9 1 28 3 The vertex is 9, 28 .
g. The maximum value is 3 .
h. The domain is , .
The range is , 3 .
28. h x x 2 6 x 10 a. Since a 0 , the parabola opens downward.
1 26. j t t 2 10t 5 4 10 b 10 x-coordinate: 20 1 2a 1 2 4 2 y-coordinate: 2 1 j 20 20 10 20 5 95 4 The vertex is 20, 95 .
b. x-coordinate:
y-coordinate: 2 h 3 3 6 3 10 1 The vertex is 3, 1 .
c. Since the vertex of the parabola is below the x-axis and the parabola opens downward, the parabola cannot cross or touch the x-axis. Therefore, there are no x-intercepts.
27. g x x 2 2 x 4 a. Since a 0 , the parabola opens downward. b. x-coordinate:
b 6 6 3 2a 2 1 2
d. h 0 0 6 0 10 10 2
b 2 2 1 2a 2 1 2
The y-intercept is 0, 10 .
y-coordinate: g 1 1 2 1 4 3
e.
2
The vertex is 1, 3 .
c. Since the vertex of the parabola is below the x-axis and the parabola opens downward, the parabola cannot cross or touch the x-axis. Therefore, there are no x-intercepts. d. g 0 0 2 0 4 4 2
The y-intercept is 0, 4 .
f. The axis of symmetry is the vertical line through the vertex: x 3 .
e.
g. The maximum value is 1 . h. The domain is , .
The range is , 1 .
29. f x 5 x 2 15 x 3 a. Since a 0 , the parabola opens upward.
310
Section 3.1
b. x-coordinate:
30. k x 2 x 2 10 x 5
b 15 15 3 2a 2 5 10 2
a. Since a 0 , the parabola opens upward.
y-coordinate: 2 33 3 3 3 f 5 15 3 2 2 2 4
b. x-coordinate:
y-coordinate: 2 35 5 5 5 k 2 10 5 2 2 2 2 5 35 The vertex is , . 2 2
3 33 The vertex is , . 2 4 c. f x 5 x 2 15 x 3
0 5 x 2 15 x 3 x
15
15 165 10 15 165 x 10
152 4 5 3 2 5
c. k x 2 x 2 10 x 5
0 2 x 2 10 x 5
x
15 165 10 15 165 , 0 and The x-intercepts are 10 or
b 10 10 5 2a 2 2 4 2
x
10
102 4 2 5 2 2
10 140 4 10 2 35 4 5 35 2 5 35 5 35 x or x 2 2 5 35 The x-intercepts are , 0 and 2
15 165 , 0 . 10 d. f 0 5 0 15 0 3 3 2
The y-intercept is 0, 3 .
e.
5 35 , 0 . 2 d. k 0 2 0 10 0 5 5 2
The y-intercept is 0, 5 .
e. f. The axis of symmetry is the vertical line 3 through the vertex: x . 2 g. The minimum value is
33 . 4
h. The domain is , . 33 The range is , . 4
311
Chapter 3 Polynomial and Rational Functions f. The axis of symmetry is the vertical line 5 through the vertex: x . 2 g. The minimum value is
t
42 b 42 4.29 sec 2a 2 4.9 9.8
b. The maximum height is the value of h t at the vertex.
35 . 2
h 4.29 4.9 4.29 42 4.29 3 2
h. The domain is , .
93 m
35 The range is , . 2
35. a. The horizontal distance at which the water will be at its maximum height is the xcoordinate of the vertex. b 0.576 0.576 x 11.1 m 2a 2 0.026 0.052
31. a. The time at which the population will be at a maximum is the t-coordinate of the vertex. b 82, 000 82, 000 t 24 hr 2a 2 1718 3436
b. The maximum height is the value of h x at the vertex. 2 h 11.1 0.026 11.1 0.576 11.1 3
b. The maximum population is the value of P t at the vertex.
P 24 1718 24 82,000 24 10, 000 988, 000
6.2 m
2
c. The distance of the firefighter from the house is the value of x when h x 6 .
32. a. The speed at which the gas mileage will be at a maximum is the x-coordinate of the vertex. b 2.688 2.688 x 48 mph 2a 2 0.028 0.056
6 0.026 x 2 0.576 x 3 0 0.026 x 2 0.576 x 3 x
b. The maximum gas mileage is the value of m x at the vertex.
0.576
0.5762 4 0.026 3 2 0.026
0.576 0.019776 0.052 x 8 or x 14 Since we want the downward branch, we use the second value of x. The firefighter is 14 m from the house.
m x 0.028 48 2.688 48 35.012 29.5 mpg 2
33. a. The time at which the skateboarder will be at his maximum height is the tcoordinate of the vertex. b 5.4 5.4 t 0.55 sec 2a 2 4.9 9.8
36. a. The horizontal distance at which the jumper will be at a maximum height is the x-coordinate of the vertex. b 0.364 0.364 x 3.96 m 2a 2 0.046 0.092
b. The maximum height is the value of h t at the vertex.
b. The maximum height is the value of h x at the vertex. 2 h 3.96 0.046 3.96 0.364 3.96
h 0.55 4.9 0.55 5.4 0.55 3 4.5 m 2
0.72 m
34. a. The time at which the mortar will be at its maximum height is the t-coordinate of the vertex.
312
Section 3.1 b 1 1 0.5 2a 2 1 2 y 1 x 1 0.5 0.5 The numbers are 0.5 and 0.5.
c. The length of the jump is the value of x when h x 0 .
x
0 0.046 x 2 0.364 x 0 x 0.046 x 0.364 x 0 or 0.046 x 0.364 0
39. Let x represent the first number. Let y represent the second number. We know that x y 10
0.046 x 0.364 x 7.9 Since we want the length of the jump, we use the second value of x. The jump is 7.9 m.
y 10 x y x 10 Let P represent the product. P xy
37. Let x represent the first number. Let y represent the second number. We know that x y 24
P x x x 10 x 2 10 x Function P is a quadratic function with a positive leading coefficient. The graph of the parabola opens upward, so the vertex is the minimum point on the function. The xcoordinate of the vertex is the value of x that will minimize the product. b 10 10 5 x 2a 2 1 2 y x 10 5 10 5 The numbers are 5 and 5 .
y 24 x Let P represent the product. P xy P x x 24 x 24 x x 2 x 2 24 x Function P is a quadratic function with a negative leading coefficient. The graph of the parabola opens downward, so the vertex is the maximum point on the function. The xcoordinate of the vertex is the value of x that will maximize the product. b 24 24 x 12 2a 2 1 2 y 24 x 24 12 12 The numbers are 12 and 12.
40. Let x represent the first number. Let y represent the second number. We know that x y 30
y 30 x y x 30 Let P represent the product. P xy
38. Let x represent the first number. Let y represent the second number. We know that x y 1
P x x x 30 x 2 30 x Function P is a quadratic function with a positive leading coefficient. The graph of the parabola opens upward, so the vertex is the minimum point on the function. The xcoordinate of the vertex is the value of x that will minimize the product. b 30 30 15 x 2a 2 1 2 y x 30 15 30 15 The numbers are 15 and 15 .
y 1 x Let P represent the product. P xy P x x 1 x x x 2 x 2 x Function P is a quadratic function with a negative leading coefficient. The graph of the parabola opens downward, so the vertex is the maximum point on the function. The xcoordinate of the vertex is the value of x that will maximize the product.
313
Chapter 3 Polynomial and Rational Functions b. Function V is a quadratic function with a negative leading coefficient. The graph of the parabola opens downward, so the vertex is the maximum point on the function. The x-coordinate of the vertex is the value of x that will maximize the volume. b 240 240 3 x 2a 2 40 80
41. We know that 2 x y 160
y 160 2 x Let A represent the area. A xy A x x 160 2 x 160 x 2 x 2 x 2 160 x Function A is a quadratic function with a negative leading coefficient. The graph of the parabola opens downward, so the vertex is the maximum point on the function. The xcoordinate of the vertex is the value of x that will maximize the area. b 160 160 x 40 2a 2 2 4
The sheet of aluminum should be folded 3 in. from each end. c. V 3 40 3 240 3 360 in.3 2
44. a. The perimeter of the box is 36 in. 2 x 2 y 36 2 y 36 2 x
y 160 2 x 160 2 40 160 80 80 The dimensions are 40 ft by 80 ft.
36 2 x 2 Let A represent the area. A lw y
42. We know that 3x 4 y 120 4 y 120 3 x
A x xy
3 x 4 Let A represent the area. A xy 3 3 3 A x x 30 x 30 x x 2 x 2 30 x 4 4 4 Function A is a quadratic function with a negative leading coefficient. The graph of the parabola opens downward, so the vertex is the maximum point on the function. The xcoordinate of the vertex is the value of x that will maximize the area. 30 b 30 x 20 3 2a 3 2 4 2 3 3 y 30 x 30 20 30 15 15 4 4 The dimensions are 20 ft by 15 ft. y 30
36 2 x x x 18 x 2 18 x x 2 x 2 18 x b. Function A is a quadratic function with a negative leading coefficient. The graph of the parabola opens downward, so the vertex is the maximum point on the function. The x-coordinate of the vertex is the value of x that will maximize the area. b 18 18 9 x 2a 2 1 2
36 2 x 36 2 9 18 9 2 2 2 The dimensions 9 in. by 9 in. should be used. That is, the shadow box should be square. y
c. A 9 9 18 9 81 162 81 in.2 2
43. a. Let V represent the volume. V lwh
V x 20 12 2 x x
20 12 x 2 x 2
240 x 40 x 2 40 x 2 240 x
314
Section 3.1 45. a.
y t 0.000838t 2 0.0812t 0.040 b. From the graph, the time when the population is the greatest is the t-coordinate of the vertex. 0.0812 b 0.0812 t 2a 2 0.000838 0.001676
47. a. x 0 cm . From the table, v 0 195.6 cm/sec .
t
48 hr
d 3000 cm 15.3 sec r 195.6 cm/sec
b. x 9 cm . From the table, v 9 180.0 cm/sec .
c. The maximum population of the bacteria is the y t value at the vertex.
t
y 48 0.000838 48 0.0812 48 2
0.040
d 3000 cm 16.7 sec r 180.0 cm/sec
c.
2g
v x 0.1436 x 2 0.4413x 195.7
46. a.
d. v 5.5 0.1436 5.5 0.4413 5.5 2
m x 0.0235 x 2 2.10 x 15.9
195.7 188.9 cm/sec
b. From the graph, the speed when the gas mileage is the greatest is the x-coordinate of the vertex. 2.10 b 2.10 x 45 mph 2a 2 0.0235 0.047
48. a.
c. The maximum gas mileage is the m x value at the vertex. 2 m 45 0.0235 45 2.10 45 15.9
d s 0.039 x 2 2.53s 2.51
31 mpg
315
Chapter 3 Polynomial and Rational Functions 67. For a parabola opening upward, such as the graph of f x x 2 , the minimum value is the y-coordinate of the vertex. There is no maximum value because the y values of the function become arbitrarily large for large values of x . 68. The vertex 4, 3 is below the x-axis. If the parabola opens downward, then the vertex is the maximum point. If the maximum point is below the x-axis, then no points on the curve will intersect the x-axis.
b. d 62 0.039 62 2.53 62 2.51 2
304 ft c. d 53 0.039 53 2.53 53 2.51 2
241 ft No, the stopping distance required is 242 ft (rounding up), so the car would miss the deer by approximately 3 ft. 49. False
50. True
51. True
52. False
69. No function defined by y f x can have two y-intercepts because the graph would fail the vertical line test. 70. The x-intercepts are the solutions of the equation ax 2 bx c 0 . The discriminant b 2 4ac determines the number and type of solutions to the equation and thus the number of x-intercepts. If b 2 4ac 0 , then the solutions to the equation are imaginary and the function will have no x-intercepts. If b 2 4ac 0 , then the equation has one real solution and the function has one x-intercept. If b 2 4ac 0 , then the equation has two real solutions and the function has two xintercepts.
53. b 2 4ac 12 4 4 9 144 144 0 Discriminant is 0; one x-intercept 2
54. b 2 4ac 20 4 25 4 400 400 0 Discriminant is 0; one x-intercept 2
55. b 2 4ac 5 4 1 8 25 32 57 Discriminant is 57; two x-intercepts 2
56. b 2 4ac 4 4 3 9 16 108 124 Discriminant is 124; two x-intercepts
71. Because a parabola is symmetric with respect to the vertical line through the vertex, the xcoordinate of the vertex must be equidistant from the x-intercepts. Therefore, given y f x , the x-coordinate of the vertex is 4 because 4 is midway between 2 and 6. The ycoordinate of the vertex is f 4 .
2
57. b 2 4ac 6 4 3 11 2
36 132 96 Discriminant is 96 ; no x-intercepts 58. b 2 4ac 5 4 2 10 25 80 55 Discriminant is 55 ; no x-intercepts 2
59. Graph g
60. Graph d
61. Graph h
62. Graph b
63. Graph f
64. Graph e
65. Graph c
66. Graph a
72. Given an equation of a parabola in the form 2 y a f x h k , the orientation is determined by a and the vertex is (h, k). If the vertex is above the x-axis and the parabola opens upward (a > 0), then the graph has no x-intercepts. Likewise, if the vertex is below the x-axis and the parabola opens downward (a < 0), then the graph has no x-intercepts.
316
Section 3.1 73. h, k 2, 3
f 3 9 2 3 12 3 c 9 18 36 c 9 18 c 9 c9
f x a x 2 3 a x 2 3 2
2
2
f 0 5
a 0 2 3 5 2
4a 8 a2
78. Find the x-coordinate of the vertex: b 12 12 2 2a 2 3 6
f x 2 x 2 3 2
f 2 4
74. h, k 3,1
3 2 12 2 c 4 12 24 c 4 12 c 4 c8
f x a x 3 1 a x 3 1 2
2
2
f 0 17
a 0 3 1 17 2
9a 18 a 2
79. Find the x-coordinate of the vertex: b b b b 2a 2 1 2 2
f x 2 x 3 1 2
75. h, k 4, 6
b f 8 2
f x a x 4 6 2
2
b b b 4 8 2 2 b2 b2 4 4 2 b 2 2b 2 16
f 1 3
a 1 4 6 3 9a 3 1 a 3 1 2 f x x 4 6 3 2
b 4 or b 4
76. h, k 2, 5
80. Find the x-coordinate of the vertex: b b b b 2a 2 1 2 2
f x a x 2 5 a x 2 5 2
b 2 16 b 4
2
f 2 13
b f 7 2
a 2 2 5 13 16a 8 1 a 2 1 2 f x x 2 5 2 2
2
b b b 2 7 2 2 b2 b2 9 4 2 b 2 2b 2 36
77. Find the x-coordinate of the vertex: b 12 12 3 2a 2 2 4
b 6 or b 6
317
b 2 36 b 6
Chapter 3 Polynomial and Rational Functions Section 3.2 Introduction to Polynomial Functions 1. f x 6 x 2 18 x 7
x-coordinate:
f x 6 x 2 15 x 36
4.
b 18 18 3 2a 2 6 12 2
0 6 x 2 15 x 36 0 2 x 2 5 x 12 0 2 x 3 x 4
y-coordinate: 2 3 3 3 f 6 18 7 2 2 2
27 13 9 6 27 7 20 4 2 2
2x 3 0
or
x40
2x 3
or
x 4
3 2 3 4 and 2 x
3 13 The vertex is , . 2 2 2. g x 10 x 2 15 x 5
5. 16 x 7 48 x 6 9 x 5 27 x 4
b 15 15 3 x-coordinate: 2a 2 10 20 4
x 4 16 x 2 x 3 9 x 3
2
3 3 3 y-coordinate: g 10 15 5 4 4 4
x 4 x 3 16 x 2 9
3.
x x 3 4 x 3 4 x 3 4
9 45 10 5 16 4
x 4 16 x 3 48 x 2 9 x 27
6. 98 x 6 196 x 5 8 x 4 16 x 3
45 90 40 5 8 8 8 8
2 x 3 49 x 3 98 x 2 4 x 8
3 5 The vertex is , . 4 8
2 x 49 x x 2 4 x 2
f x 4 x 2 18 x 10
2 x 3 x 2 7 x 2 7 x 2
3
2 x 3 x 2 49 x 2 4
0 4 x 2 18 x 10 0 2 x2 9 x 5 0 2 x 1 x 5 2x 1 0
or
x5 0
2 x 1
or
x5
x
2
1 2
1 5 and 2
7. polynomial
8. is
9. is not
10. 2
11. 1
12. False
13. False
14. down; down
15. up; down
16. zeros
17. 3; 4
18. 5; 3
19. 5
20. 4
21. cross 22. touch (without crossing)
318
Section 3.2 23. f has at least one zero on the interval a, b . 24. y
36. Multiply the leading terms from each factor. 3
25. origin
The leading coefficient is positive and the degree is even. The end behavior is up to the left and up to the right.
26. Multiply the leading terms from each factor. 1 4 1 2 x 3x x 4 9 x 2 3 x 6 3 3
37. f x x 3 2 x 2 25 x 50
0 x 3 2 x 2 25 x 50
27. Multiply the leading terms from each factor. 3 4 0.6 x 2 10 x x 0.6 x 2 1000 x3 x 4
5 x 4 x 2 x 5 x 4 x3 2 x 10 x8
0 x 2 x 2 25 x 2
0 x 2 x 2 25
600 x 9
0 x 2 x 5 x 5
28. up to the left and down to the right
x 2, x 5, x 5 Each zero is of multiplicity 1.
29. The leading coefficient is negative and the degree is even. The end behavior is down to the left and down to the right.
38. g x x 3 5 x 2 x 5
0 x3 5 x 2 x 5
30. The leading coefficient is negative and the degree is even. The end behavior is down to the left and down to the right.
0 x 2 x 5 1 x 5
0 x 5 x 2 1
0 x 5 x 1 x 1
31. The leading coefficient is positive and the degree is odd. The end behavior is down to the left and up to the right.
x 5, x 1, x 1 Each zero is of multiplicity 1.
32. The leading coefficient is positive and the degree is odd. The end behavior is down to the left and up to the right.
39. h x 6 x 3 9 x 2 60 x
33. Multiply the leading terms from each factor.
0 3x 2 x 2 3x 20
x 16 x
4 x 2 x x 4 x 4 x 2
4
2
4
0 6 x 3 9 x 2 60 x
The leading coefficient is negative and the degree is odd. The end behavior is up to the left and down to the right.
5 x 0, x , x 4 2 Each zero is of multiplicity 1.
34. Multiply the leading terms from each factor.
2 x 3 x x 2 x 27 x 3
3
x 54 x
0 3x 2 x 5 x 4
7
40. k x 6 x 3 26 x 2 28 x
5
0 6 x 3 26 x 2 28 x
The leading coefficient is negative and the degree is odd. The end behavior is up to the left and down to the right.
0 2 x 3 x 2 13x 14
0 2 x 3 x 7 x 2 7 x 0, x , x 2 3 Each zero is of multiplicity 1.
35. Multiply the leading terms from each factor. 3 2 x 2 x 2 x 2 x 2 x 8 x3 16 x 6
The leading coefficient is positive and the degree is even. The end behavior is up to the left and up to the right.
319
Chapter 3 Polynomial and Rational Functions
0 x 3 5 x 3 5
41. m x x5 10 x 4 25 x 3
0 x 10 x 25 x 5
4
0 x3 x 2 10 x 25 0 x 3 x 5
47. c x x 3 5 x 3 5
3
x 3 5, x 3 5 Each zero is of multiplicity 1.
2
x 0, x 5 The zero 0 has multiplicity 3. The zero 5 has multiplicity 2.
0 x 2 11 x 2 11
48. d x x 2 11 x 2 11
42. n x x 6 4 x 5 4 x 4
0 x 6 4 x5 4 x 4
0 x4 x2 4 x 4 0 x 4 x 2
x 2 11, x 2 11 Each zero is of multiplicity 1.
49. f x 2 x 3 7 x 2 14 x 30
2
f 1 2 1 7 1 14 1 30 3
x 0, x 2 The zero 0 has multiplicity 4. The zero 2 has multiplicity 2.
2 7 14 30 11
f 2 2 2 7 2 14 2 30 3
3
f 3 2 3 7 3 14 3 30 3
0 3 x x 2 x 4 x 0, x 2, x 4 The zero 0 has multiplicity 1. The zero 2 has multiplicity 3. The zero 4 has multiplicity 1. 3
3
2
54 63 42 30 21
f 4 2 4 7 4 14 4 30 3
2
128 112 56 30 10 3 2 f 5 2 5 7 5 14 5 30 250 175 70 30 35
2
a. Since f 1 and f 2 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 1, 2 .
0 2 x 4 x 1 x 2 x 0, x 1, x 2 The zero 0 has multiplicity 4. The zero 1 has multiplicity 3. The zero 2 has multiplicity 2. 3
2
16 28 28 30 10
43. p x 3 x x 2 x 4
44. q x 2 x 4 x 1 x 2
2
2
b. Since f 2 and f 3 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 2, 3 .
0 5 x 3 x 5 2 x 9 x 3 x 3
45. t x 5 x 3x 5 2 x 9 x 3 x 3
c. Since f 3 and f 4 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 3, 4 .
5 9 x , x 3, x 3 3 2 Each zero is of multiplicity 1. x 0, x
0 4 x 5 x 1 3 x 8 x 5 x 5
46. z x 4 x 5 x 1 3 x 8 x 5 x 5
d. Since f 4 and f 5 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 4, 5 .
1 8 x , x 5, x 5 5 3 Each zero is of multiplicity 1. x 0, x
320
Section 3.2 c. Since Y1 2 and Y1 1 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 2, 1 .
50. g x 2 x 3 13 x 2 18 x 5
g 1 2 1 13 1 18 1 5 3
2
2 13 18 5 12
g 2 2 2 13 2 18 2 5 3
2
d. Since Y1 1 and Y1 0 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 1, 0 .
16 52 36 5 5
g 3 2 3 13 3 18 3 5 3
2
54 117 54 5 4
g 4 2 4 13 4 18 4 5 3
2
52. a. Since Y1 4 and Y1 3 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 4, 3 .
128 208 72 5 3
g 5 2 5 13 5 18 5 5 3
2
250 325 90 5 20 a. Since f 1 and f 2 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 1, 2 .
b. Since Y1 3 and Y1 2 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 3, 2 .
b. Since f 2 and f 3 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 2, 3 .
c. Since Y1 2 and Y1 1 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 2, 1 .
c. Since f 3 and f 4 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 3, 4 .
d. Since Y1 1 and Y1 0 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 1, 0 .
d. Since f 4 and f 5 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 4, 5 .
53. f x 4 x 3 8 x 2 25 x 50
f 3 4 3 8 3 25 3 50 3
2
108 72 75 50 55
51. a. Since Y1 4 and Y1 3 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 4, 3 .
f 2 4 2 8 2 25 2 50 3
2
32 32 50 50 36 a. Since f 3 and f 2 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 3, 2 .
b. Since Y1 3 and Y1 2 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 3, 2 .
321
Chapter 3 Polynomial and Rational Functions b. f x 4 x 3 8 x 2 25 x 50
b. The end behavior is down on the left and up on the right. The leading coefficient is positive and the degree is odd.
0 4 x 3 8 x 2 25 x 50 0 4 x 2 x 2 25 x 2
0 x 2 4 x 25 2
c. 4 (odd multiplicity); 1 (odd multiplicity); 3 (odd multiplicity)
0 x 2 2 x 5 2 x 5
58. The graph is smooth and continuous. It can represent a polynomial function.
5 5 x 2, x , x 2 2
a. The function has 3 turning points, so the minimum degree is 4.
5 The zero on the interval 3, 2 is . 2
b. The end behavior is up on the left and up on the right. The leading coefficient is positive and the degree is even.
54. f x 9 x 3 18 x 2 100 x 200
f 4 9 4 18 4 100 4 200 3
2
c. 4 (odd multiplicity); 1 (odd multiplicity); 3 (even multiplicity)
576 288 400 200 264
f 3 9 3 18 3 100 3 200 3
2
59. The graph is smooth and continuous. It can represent a polynomial function.
243 162 300 200 95 a. Since f 4 and f 3 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 4, 3 .
a. The function has 5 turning points, so the minimum degree is 6. b. The end behavior is down on the left and down on the right. The leading coefficient is negative and the degree is even.
b. f x 9 x 3 18 x 2 100 x 200
c. 4 (odd multiplicity); 3 (odd multiplicity); 1 (even multiplicity); 2 7 (odd multiplicity); (odd multiplicity) 2
0 9 x 3 18 x 2 100 x 200 0 9 x 2 x 2 100 x 2
0 x 2 9 x 2 100
0 x 2 3x 10 3 x 10 x 2, x
60. The graph is smooth and continuous. It can represent a polynomial function.
10 10 ,x 3 3
The zero on the interval 4, 3 is
a. The function has 2 turning points, so the minimum degree is 3.
10 . 3
b. The end behavior is up on the left and down on the right. The leading coefficient is negative and the degree is odd.
55. The graph is not smooth. It cannot represent a polynomial function.
c. 4 (odd multiplicity); 1 (odd multiplicity); 2 (odd multiplicity)
56. The graph is not smooth. It cannot represent a polynomial function.
61. The graph is not continuous. It cannot represent a polynomial function.
57. The graph is smooth and continuous. It can represent a polynomial function.
62. The graph is not continuous. It cannot represent a polynomial function.
a. The function has 2 turning points, so the minimum degree is 3.
322
Section 3.2 63. f x x 3 5 x 2
The leading term is x 3 . The end behavior is down to the left and up to the right. 3 2 f 0 0 5 0 0
The y-intercept is 0, 0 .
0 x3 5 x 2 0 x 2 x 5 x 0, x 5 The zeros of the function are 0 (multiplicity 2) and 5 (multiplicity 1). Test for symmetry: 3 2 f x x 5 x x3 5 x 2
1 x 2 x 1 x 3 2 1 1 The leading term is x x x x3 . The 2 2 end behavior is down to the left and up to the right. 1 1 f 0 0 2 0 1 0 3 21 3 2 2 3 The y-intercept is 0, 3 . The zeros of the function are 2 (multiplicity 1), 1 (multiplicity 1), and 3 (multiplicity 1). Test for symmetry: 1 f x x 2 x 1 x 3 2 f x is neither even nor odd.
65. f x
f x is neither even nor odd.
64. g x x 5 2 x 4
The leading term is x 5 . The end behavior is down to the left and up to the right. 5 4 g 0 0 2 0 0 The y-intercept is 0, 0 . 0 x5 2 x 4 0 x 4 x 2 x 0, x 2 The zeros of the function are 0 (multiplicity 4) and 2 (multiplicity 1). Test for symmetry: 5 4 g x x 2 x x5 2 x 4
1 x 1 x 4 x 2 4 1 1 The leading term is x x x x3 . The 4 4 end behavior is down to the left and up to the right. 1 1 h 0 0 10 4 0 2 1 4 2 2 4 4
66. h x
g x is neither even nor odd.
323
Chapter 3 Polynomial and Rational Functions 68. h x x 4 x3 6 x 2
The y-intercept is 0, 2 . The zeros of the function are 1 (multiplicity 1), 4 (multiplicity 1), and 2 (multiplicity 1). Test for symmetry: 1 h x x 1 x 4 x 2 4 h x is neither even nor odd.
The leading term is x 4 . The end behavior is up to the left and up to the right. 4 3 2 h 0 0 0 6 0 0 The y-intercept is 0, 0 . 0 x 4 x3 6 x 2
0 x2 x2 x 6
0 x 2 x 2 x 3 x 0, x 2, x 3 The zeros of the function are 0 (multiplicity 2), 2 (multiplicity 1), and 3 (multiplicity 1). Test for symmetry: 4 3 2 h x x x 6 x x 4 x3 6 x 2 h x is neither even nor odd.
67. k x x 4 2 x 3 8 x 2
The leading term is x 4 . The end behavior is up to the left and up to the right. 4 3 2 k 0 0 2 0 8 0 0 The y-intercept is 0, 0 .
0 x 4 2 x3 8 x 2
0 x2 x2 2 x 8
0 x x 4 x 2 2
x 0, x 4, x 2 The zeros of the function are 0 (multiplicity 2), 4 (multiplicity 1), and 2 (multiplicity 1). Test for symmetry: 4 3 2 k x x 2 x 8 x
69. k x 0.2 x 2 x 4 2
3
The leading term is 0.2 x x 0.2 x5 . The end behavior is down to the left and up to the right. 2 3 k 0 0.2 0 2 0 4 2
x 4 2 x3 8 x 2 k x is neither even nor odd.
0.2 2 4 51.2 2
3
3
The y-intercept is 0, 51.2 . The zeros of the function are 2 (multiplicity 2) and 4 (multiplicity 3). Test for symmetry: 2 3 k x 0.2 x 2 x 4 k x is neither even nor odd.
324
Section 3.2 5 5 x 0, x 1, x , x 3 3 The zeros of the function are 0 (multiplicity 5 2), 1 (multiplicity 1), (multiplicity 1), 3 5 and (multiplicity 1). 3 Test for symmetry: p x 70. m x 0.1 x 3 x 1 2
9 x 9 x 25 x 25 x 5
3
3
2
9 x 5 9 x 4 25 x 3 25 x 2 p x is neither even nor odd.
The leading term is 0.1 x x 0.1x 5 . The end behavior is down to the left and up to the right. 2 3 2 3 m 0 0.1 0 3 0 1 0.1 3 1 0.9 2
4
3
The y-intercept is 0, 0.9 . The zeros of the function are 3 (multiplicity 2) and 1 (multiplicity 3). Test for symmetry: 2 3 m x 0.1 x 3 x 1 m x is neither even nor odd.
72. q x 9 x 5 18 x 4 4 x3 8 x 2
The leading term is x 5 . The end behavior is down to the left and up to the right. 5 4 3 2 q 0 9 0 18 0 4 0 8 0 0 The y-intercept is 0, 0 .
0 9 x 5 18 x 4 4 x 3 8 x 2
0 x 2 9 x 3 18 x 2 4 x 8
0 x 2 9 x 2 x 2 4 x 2
71. p x 9 x 5 9 x 4 25 x3 25 x 2
0 x 2 x 2 9 x 2 4
The leading term is x 5 . The end behavior is down to the left and up to the right. 5 4 3 2 p 0 9 0 9 0 25 0 25 0 0
0 x 2 9 x 3 9 x 2 25 x 25
0 x x 2 3x 2 3x 2 2 2 x 0, x 2, x , x 3 3 The zeros of the function are 0 (multiplicity 2 2), 2 (multiplicity 1), (multiplicity 1), 3 2 and (multiplicity 1). 3
0 x 2 9 x 2 x 1 25 x 1
0 x 2 x 1 9 x 2 25
2
The y-intercept is 0, 0 .
0 9 x 5 9 x 4 25 x 3 25 x 2
0 x x 1 3 x 5 3x 5 2
325
Chapter 3 Polynomial and Rational Functions 74. v x x 4 15 x 2 44
Test for symmetry: q x 9 x 18 x 4 x 8 x 5
4
3
2
The leading term is x 4 . The end behavior is down to the left and down to the right. 4 2 v 0 0 15 0 44 44
9 x 5 18 x 4 4 x3 8 x 2 q x is neither even nor odd.
The y-intercept is 0, 44 . 0 x 4 15 x 2 44 0 x 4 15 x 2 44
0 x 2 4 x 2 11
0 x 2 x 2 x 2 11
x 2, x 2, x 11, x 11 The zeros of the function are 2 (multiplicity 1), 2 (multiplicity 1), 11 (multiplicity 1), and 11 (multiplicity 1). Test for symmetry: 4 2 v x x 15 x 44
73. t x x 4 11x 2 28
The leading term is x 4 . The end behavior is down to the left and down to the right. 4 2 t 0 0 11 0 28 28
x 4 15 x 2 44 v x is even, so it is symmetric with respect to the y-axis.
The y-intercept is 0, 28 . 0 x 4 11x 2 28 0 x 4 11x 2 28
0 x2 4 x2 7
0 x 2 x 2 x 2 7
x 2, x 2, x 7, x 7 The zeros of the function are 2 (multiplicity 1), 2 (multiplicity 1), 7 (multiplicity 1), and 7 (multiplicity 1). Test for symmetry: t x x 11 x 28
75. False
76. True
x 4 11x 2 28 t x is even, so it is symmetric with respect to the y-axis.
77. False
78. False
79. True
80. False
81. False
82. True
83. False
84. True
85. True
86. False
4
2
87. a. The acceleration is increasing over the interval 0,12 68,184 .
326
Section 3.2 92 All polynomial functions y f x are
b. The acceleration is decreasing over the interval 12, 68 184, 200 .
continuous. Therefore, for a b , if f a
and f b have different signs, then at some
c. The graph shows 3 turning points.
point on the interval a, b , f x must be 0. That is, for the function to change sign from positive to negative or from negative to positive, the function must have a y value of 0 somewhere in between.
d. The minimum degree of a polynomial function with 3 turning points is 4. The leading coefficient would be negative since the end behavior of the graph is down on the left and down on the right. e. The acceleration is greatest approximately 184 sec after launch.
93. b
94. d
f. The maximum acceleration is approximately 2.85 G-forces.
95. a
96. c
97. a. f 3 3 3 3 2 2 2
88. a. The number of new AIDS cases increased over the interval 0, 8 14, 20 .
f 4 4 3 4 2 6 2
b. The acceleration is decreasing over the interval 8,14 .
b. By the intermediate value theorem, because f 3 2 and f 4 6 , then f must take on every value between 2 and 6 on the interval 3, 4 .
c. The graph shows 2 turning points. d. The minimum degree of a polynomial function with 2 turning points is 3. The leading coefficient would be positive since the end behavior of the graph is down on the left and up on the right.
c. f x x 2 3 x 2
4 x 2 3x 2 0 x 2 3x 2
e. The number of new AIDS cases is greatest in approximately 1993.
x
f. The maximum number of new cases diagnosed in a single year is approximately 2648.
3
3 2 4 1 2 2 1
3 17 3.56 or 0.56 2 On the interval 3, 4 ,
89. The x-intercepts are the real solutions to the equation f x 0 .
x
3 17 3.56 . 2
98. a. f 4 4 4 4 3 3
90. An x-intercept is a cross point if the corresponding zero of the polynomial has an odd multiplicity. An x-intercept is a touch point if the corresponding zero of the polynomial has an even multiplicity.
2
f 3 3 4 3 3 6 2
b. By the intermediate value theorem, because f 4 3 and f 3 6 , then f must take on every value between 3 and 6 on the interval 4, 3 .
91. A function is continuous if its graph can be drawn without lifting the pencil from the paper.
327
Chapter 3 Polynomial and Rational Functions c. f x x 2 4 x 3
102.
5 x2 4x 3 0 x2 4x 2 0 x2 4x 2 x
4
4 2 4 1 2 2 1
The end behavior is up to the left and up to the right.
4 8 4 2 2 2 2 2 2 0.59 or 3.41 On the interval 4, 3 ,
103. Window b is better.
x 2 2 3.41 . 99.
104. Window b is better.
V t 0.0406t 3 0.154t 2 0.173t 0.0024 100.
105.
T x 0.76 x 3 17.7 x 2 71.2 x 111 101. 106.
The end behavior is down to the left and up to the right.
328
Section 3.3 Section 3.3 Division of Polynomials and the Remainder and Factor Theorems 1. g x x 2 10 x 21
y-coordinate: 1 2 h 6 6 6 6 16 2 18 36 16 2 The vertex is 6, 2 .
a. Since a 0 , the parabola opens upward. b. x-coordinate:
b 10 10 5 2a 2 1 2
y-coordinate: 2 g 5 5 10 5 21 4
c. The axis of symmetry is the vertical line through the vertex: x 6 .
The vertex is 5, 4 .
d. The maximum value is 2.
1 e. h x x 2 6 x 16 2 1 0 x 2 6 x 16 2 0 x 2 12 x 32
c. The axis of symmetry is the vertical line through the vertex: x 5 . d. The minimum value is 4 . e. g x x 2 10 x 21
0 x 8 x 4
0 x 2 10 x 21 0 x 3 x 7
x 8, x 4
x 3, x 7
The x-intercepts are 8, 0 and 4, 0 .
The x-intercepts are 3, 0 and 7, 0 .
1 2 0 6 0 16 16 2 The y-intercept is 0, 16 .
f. h 0
f. g 0 0 10 0 21 21 2
The y-intercept is 0, 21 .
g.
g.
595 3. 42 25, 019
1 2. h x x 2 6 x 16 2
210
a. Since a 0 , the parabola opens downward. b. x-coordinate:
595
29 42
401 378
b 6 6 6 2a 1 1 2 2
239 210 29 Check: 42 595 29 24,990 29 25,019
329
Chapter 3 Polynomial and Rational Functions 2787 4. 23 64,108
c. 2 x 5 3 x 12 65
7 2787 23
46
6 x 2 24 x 15 x 60 65 6 x2 9 x 5
181 161
4x 2 18. a. 3x 4 12 x 2 10 x 3
200
184
12 x 2 16 x
6x 3 6 x 8
7
11
Check: 23 2787 7 64,101 7 64,108
5.
161
168
6 x 5 2 x 4 3x
6.
b. Dividend: 12 x 2 10 x 3 ; Divisor: 3x 4 ; Quotient: 4 x 2 ; Remainder: 11
15 x 3 3x 2 5x
c. 3x 4 4 x 2 11 12 x 2 6 x 16 x 8 11
7. Quotient: x 3 ; Remainder: 0
12 x 2 10 x 3
8. Quotient: x 4 ; Remainder: 0
3x 2 x 4 19. x 4 3x 11x 2 0 x 10 3
9. Dividend: f x ; Divisor: d x ;
3x 3 12 x 2
Quotient: q x ; Remainder: r x
x 0x x2 4 x
10. x 3 2 x 2 x 3 8 2 x3 5 x 2 6 x 1
12. complex
13. True
14. factor; 0
15. True
16. False
6 3x 2 x 4
6 x4
2 x 2 3x 15 20. x 5 2 x 3 7 x 2 0 x 65
2 x 3 10 x 2
3 x 12 17. a. 2 x 5 6 x 2 9 x 5 6 x 2 15 x
4 x 10 4 x 16
2 x3 x 2 3x 6 x 2 3x 9 8
11. f c
2
3x 2 0 x 3x 2 15 x
15 x 65 15 x 75
24 x 5 24 x 60
10
65 2 x 2 3 x 15
b. Dividend: 6 x 2 9 x 5 ; Divisor: 2 x 5 ; Quotient: 3 x 12 ; Remainder: 65
330
10 x5
Section 3.3 4 x 3 20 x 2 13 x 4 21. x 2 4 x 4 12 x 3 27 x 2 30 x 8
4 x 4 8 x3
4 x 3 12 x 2 6 x 18 24. 2 x 6 8 x 4 0 x 3 60 x 2 0 x 108
8 x 4 24 x 3
20 x 3 27 x 2 20 x 3 40 x 2
24 x 3 60 x 2 24 x 3 72 x 2
13x 2 30 x 13 x 2 26 x
4x 8 4 x 8
36 x 108 36 x 108
0
0 4 x 3 12 x 2 6 x 18 x3 4 x 2 5 x 2 25. x 2 5 x 5 4 x 4 0 x 3 18 x 2 20 x 10
3x 3 8 x 2 4 x 16 22. x 3 3x 4 17 x 3 20 x 2 28 x 48
12 x 2 0 x 12 x 2 36 x
4 x 3 20 x 2 13 x 4
3x 4 9 x3
x5 0 x 4 5 x3
8 x 3 20 x 2 8 x 3 24 x 2
4 x 4 5 x 3 18 x 2 4 x 4 0 x 3 20 x 2
4 x 2 28 x 4 x 2 12 x
5 x 3 2 x 2 20 x 5 x 3 0 x 2 25 x
16 x 48 16 x 48
2 x 2 5 x 10 2 x 2 0 x 10
0
5x
3x 3 8 x 2 4 x 16
5x x3 4 x 2 5 x 2 2 x 5
3x3 6 x 2 2 x 4 23. 2 x 4 6 x 4 0 x 3 20 x 2 0 x 16
6 x 4 12 x 3
x3 2 x 2 4 x 6 26. x 3 x 2 x x 0 x 2 8 x 18
2
12 x 3 20 x 2 12 x 3 24 x 2
5
4
3
x5 0 x 4 3x3
4 x2 0 x 4 x2 8x
2 x 4 x 0 x2 2 x 4 0 x 3 6 x 2 4
3
4x 6x 8x 4 x 3 0 x 2 12 x 3
2
8 x 16 8 x 16
6 x 4 x 18 6 x 2 0 x 18 2
0 3x3 6 x 2 2 x 4
4x x3 2 x 2 4 x 6
331
4x x 3 2
Chapter 3 Polynomial and Rational Functions 3x 2 1 27. 2 x 2 x 3 6 x 4 3 x 3 7 x 2 6 x 5
6 x 3x 9 x 4
3
2
31. a. Dividend: 2 x 4 5 x 3 5 x 2 4 x 29 b. Divisor: x 3
c. Quotient: 2 x 3 x 2 2 x 10
2 x2 6 x 5 2 x2 x 3
d. Remainder: 1
32. a. Dividend: x 4 5 x3 2 x 2 x 2
5x 2 3x 2 1
5x 2 2 x2 x 3
b. Divisor: x 2 c. Quotient: x 3 3 x 2 4 x 9
4x2 1 28. 3x x 4 12 x 4 x 13x 2 x 1 2
4
3
d. Remainder: 2
2
12 x 4 4 x 3 16 x 2
33. a. Dividend: x 3 2 x 2 25 x 4
b. Divisor: x 4
3x 2 x 1 3 x 2 x 4 2
c. Quotient: x 2 6 x 1
x5 4 x2 1
d. Remainder: 0
x5 3x x 4
34. a. Dividend: 3x 3 13 x 2 14 x 20
2
b. Divisor: x 5
x 2 3x 9 29. x 3 x 3 0 x 2 0 x 27
x3 3x 2
c. Quotient: 3x 2 2 x 4
3x 2 0 x 3x 2 9 x
d. Remainder: 0 35. 6 4
9 x 27 9 x 27
15
1
24
54
4
9
55
4x 9
55 x6
0 x 2 3x 9
36. 5 6
30
25
6
5
6
6x 5
6 x5
x 2 4 x 16 30. x 4 x 3 0 x 2 0 x 64
x3 4 x 2
4 x2 0 x 4 x 2 16 x
25 19
37. 4 5 17 12 20 12
16 x 64 16 x 64
5
0
5x 3
x 4 x 16 2
332
3
0
Section 3.3 45. a. f x 2 x 4 5 x 3 x 2 7
1 21
38. 3 2
6
21
f 4 2 4 5 4 4 7
2 7
0
2 256 5 64 16 7
4
2x 7 3 8
4
10 20 46
76
5 10 23 38
72
5 x 3 10 x 2 23x 38
72 x2
0
0 2 40. 1 2 5 2 7 7
9
2 7 7 9
14
b. 4 2 5 1 0 8 12 52 2
5
b. 2 3
21
4 12
4
6 8 16 20
42
0
47. a. 1 2
0
b. 3 2
1 49
79
15
2 1 48 2 1 48 127
127 112
1 49 6
0 81 0
79
15
5
0
By the remainder theorem, f 3 0 .
c. 4 2 1 49
1 3 9 27
38
21 84 15
2 7 28
x 4 2 x 3 4 x 2 8 x 16
81
21
By the remainder theorem, f 1 112 .
32
3 9 27
6
The remainder is 38 .
0
2 4 8 16 32 1 2 4 8 16 0
0 0
1
2
3 4 8 10
2 x 4 17 x 3 31x 2 4 x 12 0
2
96 32 24 2 4 38
59
0
4
3 32 2 16 6 4 2 4
42. 2 2 13 3 58 20 24 4 34 62 8 24
0 0
201
5
4 x 4 13x 3 97 x 2 59 x 21
31
3 13 52
g 2 3 2 2 2 6 2 2 4
198 63 41. 3 4 25 58 232 12 39 291 177 63
2 17
208
46. a. g x 3x 5 2 x 4 6 x 2 x 4
2
4 13 97
7
The remainder is 201.
14 2 x 7 x 7 x 9 x 1 3
44. 3 1
2
512 320 16 7 201
39. 2 5
43. 2 1
3
79
15
8 36 52 2 9 13 27
108 123
By the remainder theorem, f 4 123 .
x 3 3x 2 9 x 27
d.
5 2
2 1 49 79 15 5 15 85 15 2 6 34 6 0
5 By the remainder theorem, f 0 . 2
333
Chapter 3 Polynomial and Rational Functions 48. a. 1 3 22 51 3 25
42
8
12
118
d. 1 5 4 15 5 9
76
3 25 76 118
126
5 9
18
By the remainder theorem, g 1 126 .
b. 2 3 22
51 42
6 32 3 16 19
By the remainder theorem, h 1 18 .
8
50. a. 2 2 1 14
38 8 4 0
By the remainder theorem, k 2 9 .
51 42 8 c. 1 3 22 3 19 32 10 32 10
b.
2
By the remainder theorem, g 1 2 . 4 d. 3
27
6
1 2
2 1 14 7 1 0 7 2 0 14 0
1 By the remainder theorem, k 0 . 2
3 22 51 42 8 4 24 36 8 3 18
7
6 16 8 9
4 3
2
By the remainder theorem, g 2 0 .
3 19
6
6
c.
7
1
2
0
4 By the remainder theorem, g 0 . 3
14
7
2 7 14 7
7
2 1 2 7
7
By the remainder theorem, k
49. a. 1 5 4 15 12 5 1 14 1 14
5
d. 2 2
4 2 5
2
By the remainder theorem, h 1 2 . 4 b. 5
0 15
3 5
4
0
15
12
5 3 15 4 3
12
5 4 5 3
7 8 15
51. a. 2 1 3 7 13 10 2 10 6 38 1 5
4 By the remainder theorem, h 0 . 5
c.
10 4
7 0 .
By the remainder theorem, k 2 15 .
5 4 15 12 4 0 12 5
1 14
0
4 3
By the remainder theorem, h
3 19
28
By the remainder theorem, f 2 28 .
Since f 2 0 , 2 is not a zero of f x .
b. 5 1
3 7 5
0
1 2
3 0.
13 10
10 15
10
2
0
3
By the remainder theorem, f 5 0 .
Since f 5 0 , 5 is a zero of f x .
334
Section 3.3 52. a. 2 2
13 10 19
14
4 18
56
74
9 28
37
60
2
55. a. 5i
1 2 5i
14 2 1
7 3
b. 5i
14
2
40
1 20
7
56. a. 3i
57. a. 6 i
36
12i
0
1
11
25
37
By the remainder theorem, g 6 i 0 . Since g 6 i 0 , 6 i is a zero of g x .
b. 6 i
Since q 3 0 , 3 is not a zero of q x .
1
3 1 3 10
0
11
1 5 i
25
37
6 i
0
By the remainder theorem, g 6 i 0 .
10
10
1
6 i 31 i 37
30 10
3 10 30 10
Since g 6 i 0 , 6 i is a zero of g x .
By the remainder theorem, q 10 0 . Since q 10 0 , 10 is a zero of q x .
9
6 i 31 i 37 1 5 i 6 i 0
By the remainder theorem, q 3 8 .
4
Since n 3i 0 , 3i is a zero of n x .
1 30 10 18 9 24 3 8 6 8
0
By the remainder theorem, n 3i 0 .
0
54. a. 3 3
3
1
1 4 3i
33
3 11
12i
3i 9 12i 36
Since p 11 0 , 11 is a zero of
b. 10
36
Since n 3i 0 , 3i is a zero of n x .
By the remainder theorem, p 11 0 . p x .
9
By the remainder theorem, n 3i 0 .
22 33
2 11 22 3 11
4
1 4 3i
b. 3i
1
3i 9 12i 36
p x .
2 3 2 11
50 0
Since m 5i 0 , 5i is a zero of m x .
Since p 2 0 , 2 is not a zero of
3
25 50
By the remainder theorem, m 5i 0 .
By the remainder theorem, p 2 7 .
b. 11 2
0
5i 25 10i 1 2 5i 10i
3 22 33 2
2
1
21 14 2 0
Since g 7 0 , 7 is a zero of g x .
4
10i
Since m 5i 0 , 5i is a zero of m x .
By the remainder theorem, g 7 0 .
53. a. 2 2
50
By the remainder theorem, m 5i 0 .
Since g 2 0 , 2 is not a zero of g x . 13 10 19
25 50
5i 25 10i
By the remainder theorem, g 2 60 .
b. 7 2
2
1
335
Chapter 3 Polynomial and Rational Functions 58. a. 1 5i
2
5
54 26
2 10i 53 5i 2 3 10i
1 5i
1 16
8
26
2
3
13
0
2 3 13
21
61. a. 1 2
By the remainder theorem, f 1 5i 0 .
By the factor theorem, since f 1 0 ,
Since f 1 5i 0 , 1 5i is a zero of
x 1 is not a factor of f x .
f x .
b. 1 5i
b. 2 2 2
5
54 26
2 10i 53 5i 2 3 10i 1 5i
2 1 4 2
8
2 2
0
By the factor theorem, since f 2 2 0 ,
x 2 2 is a factor of f x .
Since f 1 5i 0 , 1 5i is a zero of f x .
62. a. 2 3 1 54 6 10 3 5 44
59. a. 5 1 11 41 61 30 5 30 55 30 1 6 11 6 0
18 88 70
By the factor theorem, since f 2 0 ,
x 2 is not a factor of f x .
By the factor theorem, since f 5 0 , x 5 is a factor of f x .
b. 3 2
1
3
30
54
18
9 2 54 3 2
18
3 1 9 2
390
3 2
0
By the factor theorem, since f 3 2 0 ,
420
By the factor theorem, since f 2 0 ,
x 3 2 is a factor of f x .
x 2 is not a factor of f x .
63. a. 2 5i
60. a. 4 1 10 35 50 24 4 24 44 24 1 6 11 6 0
1
4
29
2 5i 29 1 2 5i
0
Yes, 2 5i is a factor of f x .
By the factor theorem, since g 4 0 ,
b. 2 5i
x 4 is a factor of g x .
b. 1 1 10 35 50 1 11 46 1 11 46 96
16 8
1
4 2 16 2 2
26 0
By the remainder theorem, f 1 5i 0 .
b. 2 1 11 41 61 2 26 134 1 13 67 195
2
1
4
29
2 5i 29 1 2 5i
24
0
Yes, 2 5i is a factor of f x .
96 120
By the factor theorem, since g 1 0 , x 1 is not a factor of g x .
336
Section 3.3
c. x
4
66. a. 2 3 16 5 50 6 20 50
4 2 4 1 29 2 1
3 10 25
4 100 4 10i 2 5i 2 2 The solution set is 2 5i .
f x x 2 3x 2 10 x 25
6
1
b. x 2 3 x 5 x 5 0 5 x 2, x , x 5 3 5 The solution set is 2, , 5 . 3
25
3 4i 25 1 3 4i 0 Yes, 3 4i is a factor of f x .
b. 3 4i
6
1
67. a.
25
3 4i 25 1 3 4i 0 Yes, 3 4i is a factor of f x .
c. x
6
6 4 1 25 2 1
1 f x 4 x 5 x 2 11x 2 4
f x 4 x 1 5 x 1 x 2
6 64 6 8i 3 4i 2 2 The solution set is 3 4i .
b. 4 x 1 5 x 1 x 2 0 1 1 , x , x 2 4 5 1 1 The solution set is , , 2 . 4 5
x
d. The zeros are 3 4i .
2
20 39 3 2 5 11 2 20 44 8 0
2
1 37 36 2
1 4
1 f x x 20 x 2 44 x 8 4
65. a. 1 2
f x x 2 3x 5 x 5
d. The zeros are 2 5i . 64. a. 3 4i
0
1
36
1 36
0
f x x 1 2 x 2 x 36
68. a.
f x x 1 2 x 9 x 4 b. x 1 2 x 9 x 4 0
3 4
8 18 11 15 6 9 15 8 12 20 0
3 f x x 8 x 2 12 x 20 4
9 x 1, x , x 4 2 9 The solution set is 1, , 4 . 2
3 f x 4 x 2 x 2 3x 5 4
f x 4 x 3 2 x 5 x 1
337
Chapter 3 Polynomial and Rational Functions b. 4 x 3 2 x 5 x 1 0
3 2 73. f x x 1 x x 2
3 5 x , x , x 1 4 2 3 5 The solution set is , , 1 . 4 2
5 3 x2 x2 x 2 2 5 3 3 2 x x 2 2 4 5 3 3 2 or f x a x x x 2 2 x4
69. a. 3 9 33 19 3 27 18 3 9 6 1 0
5 3 2 x 4 x3 x 2 2 2
f x x 3 9 x 6 x 1 2
f x x 3 3 x 1
2 x 4 5 x3 3x 2
2
5 3 74. f x x 2 x x 2
b. x 3 3x 1 0 2
x 3, x
1 3
9 x 3 x 2 x 5 2
1 The solution set is 3, . 3
9 3 x 5 x3 2 9 or f x a x 5 x 3 5 x 3 2 x5
33 18 70. a. 2 4 20 8 24 18 4 12 9 0
f x x 2 4 x 2 12 x 9 f x x 2 2 x 3
9 2 x5 x3 5 x3 2
2 x 5 9 x 3 10 x 3
2
x 2 11
75. f x x 2 11 x 2 11
b. x 2 2 x 3 0 2
2
3 x 2, x 2
x 5 2
76. f x x 5 2 x 5 2
71. f x x 2 x 3 x 4
2
77. f x x 2 x 3i x 3i
x 3 x 2 14 x 24
x 2 x 2 9
72. f x x 1 x 6 x 3
2
x 2 50
x 3 x 2 12 x 2 x 2 2 x 24
x 1 x 2 3 x 18
2
x 2 44
3 The solution set is 2, . 2
x 2 x 2 x 12
x 2 x 9 x 18 3
2
78. f x x 4 x 2i x 2i
x 3 3x 2 18 x x 2 3 x 18
x 4 x 2 4
x 3 2 x 2 21x 18
x 4 x 4 x 16 3
338
2
Section 3.3 2 1 79. f x a x x x 4 3 2
84. Direct substitution is easier. q x 5 x 721 2 x 450
q 1 5 1
2 1 6 x x x 4 3 2
f 1 1
100
f x .
6 x 23x 6 x 8 80.
2
b. f 1 1
100
of f x .
2 3 10 x x x 6 5 2
c. g x x99 1
g 1 1 1 1 1 0 99
2 3 5 x 2 x x 6 5 2
5 x 2 2 x 2 15 x 18
Yes, since g 1 0 , x 1 is a factor of g x .
d. g 1 1 1 1 1 2 99
10 x 3 75 x 2 90 x 4 x 2 30 x 36
No, since g 1 0 , x 1 is not a
10 x 71x 60 x 36 3
factor of g x .
2
81. f x x 7 8i x 7 8i
e. Yes
x 7 8i x 7 8i x 7 8i 2
86. third 87. Since x is not a factor, zero is not a zero of the polynomial. False.
82. f x x 5 6i x 5 6i
88. Since x is a factor, zero is a zero of the polynomial. True.
x 5 6i x 5 6i 2
89. 4 4
m 13 5 16 12 28 4 3 7 0
2
x 2 10 x 25 36 x 2 10 x 61
m 28 0
83. Direct substitution is easier. p x 2 x 452 4 x 92 p 1 2 1
452
f. No
2
x 2 14 x 49 64 x 2 14 x 113
x 5 6i
1 11 0
Yes, since f 1 0 , x 1 is a factor
2 3 f x a x x x 6 5 2
5 x 2 2 x 3 x 6
1 11 0
Yes, since f 1 0 , x 1 is a factor of
6 x 3 27 x 2 12 x 4 x 2 18 x 8 3
450
85. a. f x x100 1
3x 2 2 x 1 x 4 3x 2 2 x 9 x 4
2 1
5 1 2 1 7
2 1 3 x 2 x x 4 3 2 2
721
m 28
4 1 2 4 2 92
339
Chapter 3 Polynomial and Rational Functions 95. a. V x x x 1 2 x 1 2 1 x
m 90. 5 3 10 20 22 15 25 25 15 5 5 3 0 3
2 x 3x x 2 x 3
m 15 0
3 1 12 90 2 15 89
b. 6 2
91. 2 4
5 m 2 8 6 2m 12 4 3 m 6 0
0 534 534
The volume is 534 cm3.
2 2m 12 0
96. a. V x x x 1 x 4 2 2 x 5
2m 10 0 2m 10 m 5
x x 2 5 x 4 4 x 5 x 3 5 x 2 4 x 4 x 20 x 3 5 x 2 20
m 6 92. 3 2 7 6 3 3m 9 2 1 m 3 0
b. 10 1 5 0 10 50 1 5 50
6 3m 9 0
20 500 520
The volume is 520 ft3.
3m 3 0 3m 3 m 1
97. The divisor must be of the form x c , where c is a constant. 98. Multiply the quotient times the divisor and add the remainder. The result should be equal to the dividend.
x 2 x 12 r x 3 x4 x4 x 3 x 4 x 3 r x4 x4 r x 3 x 3 x4 r 0 x4 0r
94. 2 1 5 2 1 3
2
2 x3 3x 2 x
m 15
93.
x 2 x 2 3x 1 2 x
99. Compute f c either by direct substitution or by using the remainder theorem. The remainder theorem states that f c is equal to the remainder obtained after dividing f x by x c . 100. Given a polynomial f x , if c is a zero of
8 6 14
f x , then x c is a factor of f x . The
converse is also true. If x c is a factor of f x , then c is a zero of f x .
x2 5x 8 r x 3 x2 x2 14 r x 3 x 3 x4 x4 14 r
340
Section 3.3 104. a. 2 1
4 1 20 20 2 4 10 20 1 2 5 10 0
101. a. 5 1 5 1 5 5 0 5
1
0 1
0
f x x 2 x3 2 x 2 5 x 10
f x x 5 x 2 1
x 2 x 2 x 2 5 x 2
x2 1 0
x 2 x 2 x 2 5
x 2 1
b. x 2 x 5 x 5 0 2
2
b. x 5 x i x i 0
x 2, x 5, x 5
x 5, x i, x i
102. a. 3 1 3 100 300 3 0 300
105. a. The solution appears to be 3.
7 58 24 15 66 24 5 22 8 0
b. 3 5
0
f x x 3 x 2 100
Since the remainder is zero, 3 is a solution.
x 100 0 2
x 2 100
c. f x x 3 5 x 2 22 x 8
x 100 10i f x x 3 x 10i x 10i
x 3 x 4 5 x 2 0
2 5 2 The solution set is 3, 4, . 5 x 3, x 4, x
x 3, x 10i, x 10i The solution set is 3,10i, 10i . 103. a. 1 1
2 2 6 3 1 1 3 3 1 1 3 3 0
106. a. The solution appears to be 5 . b. 5 2
1 41 70 10 55 70 2 11 14 0
f x x 1 x x 3 x 3 2
x 1 x 2 x 1 3 x 1
Since the remainder is zero, 5 is a solution.
x 1 x 1 x 2 3
b. x 1 x 3 x 3 0
c. f x x 5 2 x 2 11x 14
x 1 x 3 x 3 2
x 5 2 x 7 x 2
2
x 1, x 3, x 3
x 3 x 4 5 x 2
b. x 3 x 10i x 10i 0
3
The solution set is 2, 5, 5 .
The solution set is 5, i, i .
0 100
x 2 x 5 x 5
x 1 i f x x 5 x i x i
1
x 5 2 x 7 x 2 0 x 5, x
7 ,x2 2
The solution set is 1, 3, 3 .
7 The solution set is 5, , 2 . 2
341
Chapter 3 Polynomial and Rational Functions Section 3.4 Zeros of Polynomials 1. a. 4,
2 5
10, 4 2
b.
x 4 13 x 2 36 0
6. a.
x 9 x 4 0 2
2
2 c. 4, , 10, 4 2 5
x 3 x 3 x 2 x 2 0
d. 6i, 3 8i
The solution set is 3, 3, 2, 2
x 3, x 3, x 2, x 2
2 e. 4, , 10,, 6i, 3 8i 4 2 5
b. The zeros are 3, 3, 2, 2 .
2. a. x 2 25 x 5 x 5
7. zeros
b. x 2 25 x 5i x 5i
8. Fundamental theorem of algebra
3. f x 3x 4 x 2 4 x 7
2
0 3x 4 x 2 4 x 7
2
3 3
11. Descartes’; f x
7 4 The zeros of the function are 0 (multiplicity 7 4), 2 (multiplicity 3), and (multiplicity 2). 4 x 0, x 2, x
4. f x 5 x x 6 5 x 1
5
0 5 x x 6 5 x 1
5
2 2
12. upper
13. greater than
14. 7 is not the ratio of any of the factors of 6 over the factors of 2.
1 x 0, x 6, x 5 The zeros of the function are 0 (multiplicity 1 1), 6 (multiplicity 2), and (multiplicity 5 5).
15.
Factors of 4 1, 2, 4 1, 2, 4 Factors of 1 1
16.
Factors of 9 1, 3, 9 1, 3, 9 1 Factors of 1
17.
Factors of 6 Factors of 4 1, 2, 3, 6 1, 2, 4 1 3 1 3 1, 2, 3, 6, , , , 2 2 4 4
5. a. f x 2 x3 7 x 2 12 x 31
f 3 2 3 7 3 12 3 31 3
10. a bi
9. n
2
2 27 7 9 36 31
18.
54 63 36 31 4 b. 3 2 7 12 31 6 3 27 2 1 9 4
Factors of 10 Factors of 25 1, 2, 5, 10 1, 5, 25 1 2 1 2 1, 2, 5, 10, , , , 5 5 25 25
By the remainder theorem, f 3 4 .
342
Section 3.4
19.
20.
Factors of 8 Factors of 12 1, 2, 4, 8 1, 2, 3, 4, 6, 12 1 1 2 4 8 1, 2, 4, 8, , , , , , 2 3 3 3 3 1 1 , 6 12
24.
Factors of 6 Factors of 16 1, 2, 3, 6 1, 2, 4, 8, 16 1 3 1 3 1 1, 2, 3, 6, , , , , , 2 2 4 4 8 3 1 3 , , 8 16 16
The remainder is zero. Therefore, 1 is a zero of p x . 2 1
1 7 5 10 2 10 10 2 1 1 5 5 0
The remainder is zero. Therefore, 2 is a zero of q x .
5 are not ratios of any of the factors 3 of 12 over the factors of 2.
21. 7 and
25.
3 are not ratios of any of the factors 2 of 10 over the factors of 4.
22. 3 and
23.
Factors of 2 1, 2 1 1, 2, Factors of 2 1, 2 2 Use synthetic division and the remainder theorem to determine if any of the numbers in the list is a zero of p x . (Successful tests of zeros are shown.) 1 2 1 5 2 2
0 x 5 x 2 x 4 2
x 5 or
1 2
26.
1 5 2 2 1 1 2 2 2 2 4 4 0
zero of p x .
2 2 4 1 4 2 1
2 20 2 2 5 1 5 2 2 The zeros are 5, 1 5 .
2
The remainder is zero. Therefore,
x
2
x
The remainder is zero. Therefore, 1 is a zero of p x .
Factors of 20 1, 2, 4, 5, 10, 20 Factors of 1 1 1, 2, 4, 5, 10, 20 5 1 7 6 20 5 10 20 1 2 4 0 f x x 5 x 2 2 x 4
2 1 4 2 1 4 2 0
2
Factors of 10 1, 2, 5, 10 Factors of 1 1 1, 2, 5, 10 Use synthetic division and the remainder theorem to determine if any of the numbers in the list is a zero of q x . (Successful tests of zeros are shown.) 1 1 1 7 5 10 1 2 5 10 1 2 5 10 0
1 is a 2
343
Factors of 6 1, 2, 3, 6 Factors of 1 1 1, 2, 3, 6 3 1 7 14 6 3 12 6 1 4 2 0
Chapter 3 Polynomial and Rational Functions
0 x 3 x 4 x 2
g x x 3 x 2 4 x 2
29.
2
x 3 or
x
4
4 2 4 1 2 2 1
4 8 42 2 2 2 2 2 The zeros are 3, 2 2 . x
27.
Factors of 7 1, 7 1 7 1, 7, , Factors of 5 1, 5 5 5 1 5 1 35 7 5 1 0 7 5 0 35 0 1 h x x 5 x 2 35 5
1 0 x 5 x 2 35 5 1 x or 5 x 2 35 5 x2 7
m x x 2 3x 3 5 x 2 26 x 8
2
m x x 2 3 x 1 x 4 2
0 x 2 3 x 1 x 4 2
1 x 2, x , x 4 3 The zeros are 2 (multiplicity 2),
30.
1 k x x 7 x 2 21 7
1 0 x 7 x 2 21 7 1 or 7 x 2 21 x 7 x2 3 x 3 1 The zeros are , 3 . 7
m x x 2 3 x 2 11x 4
Factors of 3 1, 3 1 3 1, 3, , Factors of 7 1, 7 7 7 1 7 1 21 3 7 1 0 3 7 0 21 0
Find the zeros of the quotient. 2 3 5 26 8 6 22 8 3 11 4 0
x 7 1 The zeros are , 7 . 5 28.
Factors of 16 Factors of 3 1, 2, 4, 8, 16 1, 3 1, 2, 4, 8, 16, 1 2 4 8 16 , , , , 3 3 3 3 3 2 3 1 36 60 16 6 10 52 16 3 5 26 8 0
1 , 4 . 3
Factors of 45 Factors of 2 1, 3, 5, 9, 15, 45 1, 2 1, 3, 5, 9, 15, 45, 1 3 5 9 15 45 , , , , , 2 2 2 2 2 2 3 2 9 5 57 45 6 9 42 45 2 3 14 15 0
n x x 3 2 x 3 3x 2 14 x 15 Find the zeros of the quotient. 3 2 3 14 15 6 9 15 2 3 5 0
344
Section 3.4
n x x 3 2 x 2 3x 5 2
33. t x x 4 x 2 90
0 x 10 x 9
t x x 2 10 x 2 9
n x x 3 2 x 5 x 1 2
0 x 3 2 x 5 x 1
2
2
x 2 10
5 x 3, x , x 1 2
34. v x x 4 12 x 2 13
Factors of 20 1, 2, 4, 5, 10, 20 31. Factors of 1 1 1, 2, 4, 5, 10, 20 2 1 4 2 20 2 12 20 1 6 10 0
0 x 13 x 1
v x x 2 13 x 2 1 2
2
62 4 110 2 1
35. one
36. 3
37. 7
38. 9
39. a. 2 5i
6 4 6 2i 3 i 2 2 The zeros are 2 , 3 i .
1
7
14 35i
203 0
2 5i 0 14 35i 0 7 0
f x x 2 5i x 2 5i x 2 7
Find the remaining two zeros. x2 7 0
0 x 4 x 4 x 13
x2 7
2
4
28 203
22
Since 2 5i is a zero, 2 5i is also a zero. 2 5i 1 2 5i 7 14 35i
p x x 4 x 2 4 x 13
x
4
1 2 5i
Factors of 52 1, 2, 4, 13, 26, 52 Factors of 1 1 1, 2, 4, 13, 26, 52 4 1 8 29 52 4 16 52 1 4 13 0
or
1
2 5i 29 14 35i
x
x4
x 2 1
or
x 13 or x i The zeros are 13, i .
0 x 2 x 6 x 10
32.
2
x 2 13
q x x 2 x 2 6 x 10
x 2 or x
x 2 9
or
x 10 or x 3i The zeros are 10, 3i .
5 The zeros are 3 (multiplicity 2), , 1 . 2
6
2
x 7 The zeros are 2 5i , 7 .
4 4 113 2 1 2
b. f x
4 36 4 6i x 2 3i 2 2 The zeros are 4, 2 3i .
x 2 5i x 2 5i x 7 x 7
c. The solution set is 2 5i, 7 .
345
Chapter 3 Polynomial and Rational Functions 40. a. 3 i
6
1
3 i 10 15 5i 1 3 i
3x 4 0 3x 4
30 50
5 5
50
15 5i
0
x
Since 3 i is a zero, 3 i is also a zero. 3 i 1 3 i 5 15 5i
The zeros are 4 i ,
3 i 0 15 5i 0 5 0
1
f x x 3 i x 3 i x 2 5
4 c. The solution set is 4 i, . 3
42. a. 5 i 5
x2 5
f x x 5 i x 5 i 5 x 4
Find the remaining zero. 5x 4 0 5x 4
68
16 4i
0
x
Since 4 i is a zero, 4 i is also a zero. 4 i 3 16 3i 16 4i 3
4 c. The solution set is 5 i, . 5
4 37 117 87 193 52 1 9 27 15 52 4 36 108 60 208 0
Find the zeros of the quotient. 3 2i 4 36 108
4 . 5
b. f x x 5 i x 5 i 5 x 4
Find the remaining zero.
1 4
4 5
The zeros are 5 i ,
12 3i 16 4i 4 0
f x x 4 i x 4 i 3 x 4
43. a.
0
25 5i 20 4i 4 0
5
83 68
12 3i 67 4i
20 4i
104
Since 5 i is a zero, 5 i is also a zero. 5 i 5 29 5i 20 4i
c. The solution set is 3 i, 5 .
3 16 3i
170 104
5 29 5i
x 3 i x 3 i x 5 x 5
28
54
25 5i 150 4i
x 5 The zeros are 3 i , 5 .
b. f x
4 . 3
b. f x x 4 i x 4 i 3 x 4
Find the remaining two zeros. x2 5 0
41. a. 4 i 3
4 3
60 208
12 8i 88 24i 108 32i 4 24 8i 20 24i 48 32i Since 3 2i is a zero, 3 2i is also a zero.
346
208 0
Section 3.4 3 2i
24 8i 20 24i 48 32i 48 32i 12 8i 36 24i 4 12 0 16 4
1 f x x 3 2i x 3 2i x 4 x 2 12 x 16 4
f x x 3 2i x 3 2i 4 x 1 x 3 x 4 2
f x x 3 2i x 3 2i 4 x 1 x 1 x 4
0 x 3 2i x 3 2i 4 x 1 x 1 x 4 1 x 3 2i, x 3 2i, x , x 1, x 4 4 1 The zeros are 3 2i, ,1, 4 4
b. f x x 3 2i x 3 2i 4 x 1 x 1 x 4 1 c. The solution set is 3 2i, ,1, 4 . 4
44. a.
5 2
2 5 4 22 50 75 5 0 10 80 75 2 0 4 32 30 0
Find the zeros of the quotient. 1 2i 2 0 4 32 30 2 4i 6 8i 26 12i 30 2 2 4i 10 8i 6 12i 0
Since 1 2i is a zero, 1 2i is also a zero. 1 2i 2 2 4i 10 8i 6 12i 2 4i 4 8i 6 12i 2 4 6 0 5 f x x 1 2i x 1 22i x 2 x 2 4 x 6 2
f x x 1 2i x 1 22i 2 x 5 x 2 x 3 2
f x x 1 2i x 1 22i 2 x 5 x 1 x 3
0 x 1 2i x 1 22i 2 x 5 x 1 x 3 5 x 1 2i, x 1 2i, x , x 1, x 3 2 5 The zeros are 1 2i, , 1, 3 2 b. f x x 1 2i x 1 22i 2 x 5 x 1 x 3 5 c. The solution set is 1 2i, , 1, 3 . 2
347
Chapter 3 Polynomial and Rational Functions 4 45. f x a x 6i x 6i x 5
3 46. f x a x 4i x 4i x 2
4 a x 2 36 x 5
3 a x 2 16 x 2
4 144 a x3 x 2 36 x Let a 5. 5 5
3 a x3 x 2 16 x 24 Let a 2. 2
4 144 5 x3 x 2 36 x 5 5
3 2 x3 x 2 16 x 24 2
5 x3 4 x 2 180 x 144
2 x3 3x 2 32 x 48
47. f x a x 4 x 2
3
a x 4 x 2 x 2
2
a x 4 x 2 x 2 4 x 4
a x 4 x 6 x 12 x 8 a x 6 x 12 x 8 x 4 x 24 x 48 x 32 a x 2 x 12 x 40 x 32 a x 4 x3 4 x 2 4 x 2 x 2 8 x 8 3
2
4
3
2
4
3
2
3
2
f 0 a 0 2 0 12 0 40 0 32 160 32a 160 a 5 4 3 2 f x 5 x 2 x 12 x 40 x 32 4
3
2
f x 5 x 10 x 60 x 200 x 160 4
3
2
48. f x a x 5 x 3 2
2
a x 6 x 9 x 10 x 60 x 90 x 25 x 150 x 225 a x 4 x 26 x 60 x 225 a x 2 10 x 25 x 2 6 x 9 4
3
4
3
2
3
2
2
2
4 3 2 f 0 a 0 4 0 26 0 60 0 225 450 225a 450 a 2 4 3 2 f x 2 x 4 x 26 x 60 x 225
f x 2 x 4 8 x3 52 x 2 120 x 450
348
Section 3.4 52. f x x 3 x 5 10i x 5 10i
2
4 1 49. f x a x x 3 2
x 3 x 5 10i 2
8 16 1 a x2 x x 3 9 2
x x 10 x 125 x 3 x 2 10 x 25 100 3
1 8 4 16 8 a x3 x 2 x 2 x x 2 3 3 9 9
53. f x x 6 2 x 4 4 x 3 2 x 2 5 x 6
3 sign changes in f x . The number of possible positive real zeros is either 3 or 1. 6 4 3 2 f x x 2 x 4 x 2 x
8 3 13 2 4 f 0 a 0 0 0 16 6 9 9 8 a 16 9 a 18 13 4 8 f x 18 x3 x 2 x 6 9 9
5 x 6 f x x6 2 x 4 4 x3 2 x 2 5 x 6 3 sign changes in f x . The number of possible negative real zeros is either 3 or 1.
f x 18 x3 39 x 2 8 x 16
54. g x 3 x 7 4 x 4 6 x 3 5 x 2 6 x 1
2
4 sign changes in g x . The number of possible positive real zeros is either 4, 2, or 0. 7 4 3 2 g x 3 x 4 x 6 x 5 x
5 1 50. f x a x x 6 3 5 25 1 a x2 x x 3 36 3
6 x 1
1 5 5 25 25 a x3 x 2 x 2 x x 3 3 9 36 108
g x 3x 7 4 x 4 6 x 3 5 x 2 6 x 1
4 5 25 a x3 x 2 x 3 36 108
1 sign change in g x . The number of possible negative real zeros is 1.
25 3 4 2 5 f 0 a 0 0 0 25 3 36 108 25 a 25 108 a 108 25 3 4 2 5 f x 108 x x x 3 36 108
55. k x 8 x 7 5 x 6 3 x 4 2 x 3 11x 2 4 x 3 6 sign changes in k x . The number of possible positive real zeros is either 6, 4, 2, or 0. 7 6 4 3 k x 8 x 5 x 3 x 2 x 11 x 4 x 3 2
f x 108 x 144 x 15 x 25 2
k x 8 x 7 5 x 6 3 x 4 2 x 3 11x 2 4 x 3 1 sign change in k x . The number of possible negative real zeros is 1.
51. f x x 4 x 7 4i x 7 4i x 4 x 7 4i 2
2
x x 14 x 65
56. h x 4 x 9 6 x8 5 x 5 2 x 4 3 x 2 x 8
x 4 x 2 14 x 49 16 4
2
x 5 10 x 4 125 x 3
13 4 8 a x3 x 2 x 6 9 9
3
2
5 sign changes in h x . The number of possible positive real zeros is either 5, 3, or 1.
2
x 6 14 x 5 65 x 4
349
Chapter 3 Polynomial and Rational Functions h x 4 x 6 x 5 x 9
8
5
1 1 1 x 6 x 4 x 2 1 1000 100 10 1 6 1 4 1 2 t x x x x 1 1000 100 10 0 sign changes in t x . The number of possible negative real zeros is 0. t x
2 x 3 x x 8 4
2
h x 4 x 9 6 x8 5 x 5 2 x 4 3 x 2 x 8 2 sign changes in h x . The number of possible negative real zeros is 2 or 0.
57. p x 0.11x 4 0.04 x 3 0.31x 2 0.27 x 1.1
61. f x x8 5 x 6 6 x 4 x 3
0 sign changes in p x . The number of possible positive real zeros is 0. 4 3 2 p x 0.11 x 0.04 x 0.31 x
1 sign change. The number of possible positive real zeros is 1. x 5 5 x 3 6 x 1 x 5 5 x 3 6 x 1 0 sign changes. The number of possible negative real zeros is 0. f x has 4 real zeros; 1 positive real zero, no negative real zeros, and the number 0 is a zero of multiplicity 3.
0.27 x 1.1 p x 0.11x 4 0.04 x 3 0.31x 2 0.27 x 1.1 4 sign changes in p x . The number of possible negative real zeros is either 4, 2, or 0.
58. q x 0.6 x 4 0.8 x 3 0.6 x 2 0.1x 0.4
62. f x 5 x8 3x 6 4 x 2 x
4 sign changes in q x . The number of possible positive real zeros is either 4, 2, or 0. 4 3 2 q x 0.6 x 0.8 x 0.6 x
1 sign change. The number of possible positive real zeros is 1. 5 x 3 x 4 x 1 7
5
5 x7 3x5 4 x 1 0 sign changes. The number of possible negative real zeros is 0. f x has 2 real zeros; 1 positive real zero, no negative real zeros, and the number 0 is a zero of multiplicity 1.
q x 0.6 x 0.8 x 0.6 x 0.1x 0.4 3
f x x 5 x 7 3 x 5 4 x 1
0.1 x 0.4 4
f x x3 x5 5 x3 6 x 1
2
0 sign changes in q x . The number of possible negative real zeros is 0. 1 6 1 4 1 2 1 x x x 8 6 3 10 0 sign changes in v x . The number of possible positive real zeros is 0. 1 1 1 1 6 4 2 v x x x x 8 6 3 10 1 1 1 1 v x x6 x4 x2 8 6 3 10 0 sign changes in v x . The number of possible negative real zeros is 0.
59. v x
63. a. 2 1 6 0 5 1 3 2 16 32 74 150 1 8 16 37 75 147 The remainder and all coefficients of the quotient are nonnegative. Therefore, 2 is an upper bound for the real zeros of f x .
b. 2 1
1 6 1 4 1 2 x x x 1 1000 100 10 0 sign changes in t x . The number of possible positive real zeros is 0.
60. t x
1
1
3
2 8 16 42
82
4 8 21 41
79
6
0
5
The signs of the quotient do not alternate.
350
Section 3.4 Therefore, we cannot conclude that 2 is a lower bound for the real zeros of f x . 8 4
7
3
3 33 87
282
1 11 29 94
279
64. a. 3 1
b. 4 6
1
7
3
3 15 57
192
5 19 64
195
8
4
172
6 25
43
102
b. 6 2
40 660 700
11 0 63 50
12 6 36 2 1 6 99
28
594 544
40 3264 3224
The signs of the quotient alternate. Therefore, 6 is a lower bound for the real zeros of f x .
414 442
68. a. 6 3 16 5 90 138 18 12 102 1152 3 2 17 192 1014
28
36 6084 6120
The remainder and all coefficients of the quotient are nonnegative. Therefore, 6 is an upper bound for the real zeros of f x .
83 55
The signs of the quotient alternate. Therefore, 1 is a lower bound for the real zeros of f x .
66. a. 4 6 1 57 24 92 6 23 35
100
The remainder and all coefficients of the quotient are nonnegative. Therefore, 3 is an upper bound for the real zeros of f x .
The remainder and all coefficients of the quotient are nonnegative. Therefore, 6 is an upper bound for the real zeros of f x .
b. 1 8 42 33 8 50 8 50 83
24
67. a. 3 2 11 0 63 50 6 51 153 270 2 17 51 90 220
The signs of the quotient do not alternate. Therefore, we cannot conclude that 3 is a lower bound for the real zeros of f x .
65. a. 6 8 42 33 48 36 8 6 69
70
The signs of the quotient alternate. Therefore, 4 is a lower bound for the real zeros of f x .
The remainder and all coefficients of the quotient are nonnegative. Therefore, 3 is an upper bound for the real zeros of f x .
b. 3 1
1 57
90 138 b. 3 3 16 5 9 75 240 450 3 25 80 150 312
70
36 936 900
The signs of the quotient alternate. Therefore, 3 is a lower bound for the real zeros of f x .
140 210
The remainder and all coefficients of the quotient are nonnegative. Therefore, 4 is an upper bound for the real zeros of f x .
351
69. True
70. False
71. False
72. True
Chapter 3 Polynomial and Rational Functions
73.
Factors of 28 1, 2, 4, 7, 14, 28 1, 2, 4, 8 Factors of 8
2 6 1 57 12
1, 2, 4, 7, 14, 28, 1 1 1 7 7 , , , , 2 4 8 2 4 From Exercise 65, we know that 6 and 1 are upper and lower bounds of f x , respectively. We can restrict the list of possible rational zeros to those on the interval 1, 6 :
0 2 x 7 3x 5 7 5 x ,x 2 3 7 5 The zeros are , , and 2 (each with 2 3 multiplicity 1).
75.
32 40 28 7
0
Find the zeros of the quotient. 0 8 x 2 10 x 7
1 5 , 2 2 From Exercise 67, we know that 3 and 6 are upper and lower bounds of f x , respectively. We can restrict the list of possible rational zeros to those on the interval 6, 3 :
7 1 ,x 4 2
7 1 , , and 4 (each with 4 2 multiplicity 1).
The zeros are
74.
Factors of 40 Factors of 2 1, 2, 4, 5, 8, 10, 20, 40 1, 2 1, 2, 4, 5, 8, 10, 20, 40,
0 4 x 7 2 x 1 x
0
Find the zeros of the quotient. 0 6 x 2 11x 35
1 1 1 7 7 1, 2, 4, , , , , 2 4 8 2 4 4 8 42 33 28 8 10
22 70
11 35
6
70
Factors of 70 Factors of 6 1, 2, 5, 7, 10, 14, 35, 70 1, 2, 3, 6
1 5 1, 2, 4, 5, , 2 2 2 2 11 0 63 50 4 14
1 1, 2, 5, 7, 10, 14, 35, 70, , 2 1 2 5 5 5 7 7 7 , , , , , , , , 3 3 2 3 6 2 3 6 10 14 35 35 35 70 70 , , , , , , 3 3 2 3 6 3 6 From Exercise 66, we know that 4 and 4 are upper and lower bounds of f x , respectively. We can restrict the list of possible rational zeros to those on the interval 4, 4 :
2
28
7 14 35
70 40 20
Find the zeros of the quotient. 4 2 7 14 35 20 8 4 2 1 10
40 20 5 0
Find the zeros of the quotient. 1 2 1 10 5 2 1 0 5 2 0 10 0 Find the zeros of the quotient.
1 1 2 5 5 5 7 1, 2, , , , , , , , 2 3 3 2 3 6 2 7 7 10 , , 3 6 3
352
40 0
Section 3.4 2 x 2 10 0
77.
2 x 2 10 x2 5 x 5
1 1 3 3 9 9 1, 3, 9, , , , , , 2 4 2 4 2 4 3 4 20 13 30 9
1 The zeros are 5, , 2, and 4 (each 2 with multiplicity 1).
76.
12 24 4 8 11
Factors of 36 Factors of 3 1, 2, 3, 4, 6, 9, 12, 18, 36 1, 3
4
1 2 4 , , 3 3 3 From Exercise 68, we know that 6 and 3 are upper and lower bounds of f x , respectively. We can restrict the list of possible rational zeros to those on the interval 3, 6 :
6 20 30 3 10 15
60
3
0 18
4
1
x
0
1 2
The zeros are 3, and
1 (each with 2
multiplicity 2). 36
78.
0
Factors of 4 Factors of 9 1, 2, 4 1, 3, 9 1 1 2 2 4 4 1, 2, 4, , , , , , 3 9 3 9 3 9 2 9 30 13 20 4 18 24 22 4 9 12 11 2 0
18 0
Find the zeros of the quotient. 1 3 1 18 6 3 1 0 6 3
12 3
2 x 12 0
Find the zeros of the quotient. 3 3 10 15 60 18 9 3 54 1 18 6
12
Find the zeros of the quotient. 4 x2 4 x 1 0
120 36 18
33 9 3 0
Find the zeros of the quotient. 3 4 8 11 3
1, 2, 3, 4, 6, 9, 12, 18, 36,
1 2 4 1, 2, 3, 4, , , 3 3 3 2 3 16 5 90 138
Factors of 9 Factors of 4 1, 3, 9 1, 2, 4
Find the zeros of the quotient. 2 9 12 11 2 18 12 2 9 6 1 0
0
Find the zeros of the quotient. 3x 2 18 0
Find the zeros of the quotient. 9 x2 6 x 1 0
3x 2 18 x2 6
3x 12 0
x 6
x
1 The zeros are 6, , 2, and 3 (each with 3 multiplicity 1). 353
1 3
Chapter 3 Polynomial and Rational Functions The zeros are 2, and
81. f x x3 x 2 10 x 34
1 (each with 3
Find the zeros of x 10 x 34 .
multiplicity 2).
79. f x x 2 x 4 2 x 3 11x 2 20 x 10
x
Factors of 10 1, 2, 5, 10 Factors of 1 1 1, 2, 5, 10 1 1 2 11 20 10 1 1 10 10 1 1 10 10 0
102 4 1 34 2 1
10 36 10 6i 5 3i 2 2 The zeros are 0 (with multiplicity 3), and 5 3i (each with multiplicity 1).
82. f x x 4 x 2 12 x 40
Find the zeros of x 2 12 x 40 . x
12
12 2 4 1 40 2 1
12 16 12 4i 6 2i 2 2 The zeros are 0 (with multiplicity 4), and 6 2i (each with multiplicity 1).
Find the zeros of the quotient. x 2 10 0 x 2 10
x i 10 The zeros are 0 (with multiplicity 2), 1 (with multiplicity 2), and i 10 (each with multiplicity 1). 80. f x x 2 x 4 6 x3 12 x 2 18 x 27
10
Find the zeros of the quotient. 1 1 1 10 10 1 0 10 1 0 10 0
2
83. f x x3 3x 2 9 x 13
Factors of 13 1, 13 1, 13 Factors of 1 1 1 1 3 9 13
1 4 13 1 4 13 0
Factors of 27 1, 3, 9, 27 Factors of 1 1 1, 3, 9, 27 3 1 6 12 18 27 3 9 9 27 1 3 3 9 0
Find the zeros of x 2 4 x 13 . x
4
4 2 4 113 2 1
4 36 4 6i 2 3i 2 2 The zeros are 1 and 2 3i (each with multiplicity 1).
Find the zeros of the quotient. 3 1 3 3 9 3 0 9 1 0 3 0
84. f x x3 5 x 2 11x 15
Factors of 15 1, 3, 5, 15 Factors of 1 1 1, 3, 5, 15 3 1 5 11 15 3 6 15 1 2 5 0
Find the zeros of the quotient. x2 3 0 x 2 3 x i 3 The zeros are 0 (with multiplicity 2), 3 (with multiplicity 2), and i 3 (each with multiplicity 1).
354
Section 3.4 Find the zeros of x 2 2 x 5 . x
2
95. If a polynomial has real coefficients, then all imaginary zeros must come in conjugate pairs. This means that if the polynomial has imaginary zeros, there would be an even number of them. A third-degree polynomial has 3 zeros (including multiplicities). Therefore, it would have either 2 or 0 imaginary zeros, leaving room for either 1 or 3 real zeros.
2 2 4 1 5 2 1
2 16 2 4i 1 2i 2 2 The zeros are 3 and 1 2i (each with multiplicity 1).
85. False
86. False
87. False
88. False
89. True
90. False
91. a. True
b. True
c. True
d. True
92. a. x
7
96. The zeros of a second-degree polynomial can be found by applying the quadratic formula. 97. f x has no variation in sign, nor does
f x . By Descartes’ rule of signs, there are no positive or negative real zeros. Furthermore, 0 itself is not a zero of f x
because x is not a factor of f x . Therefore,
7 4 1 5 7 29 2 1 2
there are no real zeros of f x .
2
98. The expression
7 29 7 29 x b. x 2 7 x 5 x 2 2
The term x x has an exponent that is not a positive integer. 99. f x x n 1 has one sign change. The number of possible positive real zeros is 1. Since n is a positive even integer, n f x x 1 x n 1 has one sign change. The number of possible negative real zeros is 1. Therefore, there are 2 possible real roots, and n 2 possible imaginary roots.
93. a. f 2 2 2 7 2 4 8 14 4 2 2
f 3 2 3 7 3 4 18 21 4 1 2
Since f 2 and f 3 have opposite signs, the intermediate value theorem guarantees that f has at least one real zero between 2 and 3. b. x
7
x 3 is not a polynomial. 12
72 4 2 4 7 17 2 2 4
100. f x x n 1 has one sign change. The number of possible positive real zeros is 1. Since n is a positive odd integer,
7 17 2.78 is on the 4 interval [2, 3].
Furthermore,
f x x 1 x n 1 has no sign changes. The number of possible negative real zeros is 0. Therefore, there is 1 possible real root, and n 1 possible imaginary roots. n
94. Let f x x n a n . Then
f a a n a n 0 . By the factor theorem,
since f a 0 , then x a is a factor of xn an .
355
Chapter 3 Polynomial and Rational Functions 1 101. 108 l bh 2
103.
162 x 10 x 7 2 x 5
1 2 108 x 3 x x 2 3 1 108 x3 3 x 2 3 324 x3 3 x 2 0 x3 3 x 2 324 Graph the function f x x3 3x 2 324 on a graphing utility. Use the Zero feature to find positive real roots of the equation.
162 x 10 2 x 2 19 x 35
162 2 x 19 x 35 x 20 x 190 x 350 3
2
2
0 2 x 3 39 x 2 225 x 188 Graph the function f x 2 x3 39 x 2 225 x 188 on a graphing utility. Use the Zero feature to find positive real roots of the equation.
Each dimension was decreased by 1 in.
2 2 x 6 4 3 3 x3 63 9 The triangular front has a base of 6 ft and a height of 4 ft. The length is 9 ft. 102.
5 81 10 x 7 x x 2
104.
240 12 x 8 x 6 x 240 x 12 x 8 x 6
240 x 12 x 2 14 x 48
104 4 3 r r 2h 3 3 104 4 3 r r 2 10 2r 3 3 104 4 3 r 10 r 2 2 r 3 3 3 104 4 3 r 10r 2 2r 3 3 3 52 2r 3 15r 2 3r 3 0 r 3 15r 2 52 Graph the function f r r 3 15r 2 52 on a graphing utility. Use the Zero feature to find positive real roots of the equation.
240 x 3 14 x 2 48 x 12 x 2 168 x 576 0 x 3 26 x 2 216 x 336 Graph the function f x x3 26 x 2 216 x 336 on a graphing utility. Use the Zero feature to find positive real roots of the equation.
Each dimension is decreased by 2 ft. The smaller truck is 10 ft by 6 ft by 4 ft.
105. 6 2 x 4 x 2
3 x 4 x2 3 4x x
3
0 x3 4 x 3
The radius is 2 ft.
356
Section 3.4 Factors of 3 1, 3 1, 3 1 Factors of 1 1 1 0 4 3 1 1 3 1 1 3 0
The width must be positive, so use the positive zero. The width of the rectangle is 5 x , which is 5 2 3 or 3 33 10 3 33 7 33 . 2 2 2 2 2 When x 2 , y x 2 2 4 . 5
Find the zeros of x 2 x 3 . x
1
1 2 4 1 3 1 13 2 1 2
When x
The width must be positive, so use the positive zero. The width of the rectangle is 2x, which is 2 1 13 or 2 1 13 . 2
2
3 33 9 6 33 33 yx 4 2 2
42 6 33 21 3 33 . 4 2 The dimensions are either 3 in. by 4 in. or 7 33 21 3 33 in. by in. 2 2
When x 1 , y 4 x 2 4 1 3 . 2
When x
1 13 , 2
1 13 y 4 x 4 2
2
107. a.
2
1 2 13 13 16 14 2 13 4 4 4 2 2 13 1 1 13 . 4 2 The dimensions are either 2 cm by 3 cm or 1 1 13 1 13 cm by cm. 2 4
f x x 3 x 1 x 2 4
12 5 x 2 x 3
b. Factor x 2 4 . x2 4 0
0 x 3 5 x 2 12 Factors of 12 1, 2, 3, 4, 6, 12 Factors of 1 1 1, 2, 3, 4, 6, 12 2 1 5 0 12 2 6 12 1 3 6 0
x 2 4 x 2i f x x 3 x 1 x 2i x 2i 108. a.
Find the zeros of x 2 3x 6 . x
Factors of 12 1, 2, 3, 4, 6, 12 Factors of 1 1 1, 2, 3, 4, 6, 12 1 1 2 1 8 12 1 3 4 12 1 3 4 12 0 Factor the quotient. 3 1 3 4 12 3 0 12 1 0 4 0
106. 12 5 x x 2
3
3 33 , 2
32 4 1 6 3 33 2 1 2
357
Factors of 8 1, 2, 4, 8 Factors of 1 1 1, 2, 4, 8 2 1 6 9 6 8 2 8 2 8 1 4 1 4 0
Chapter 3 Polynomial and Rational Functions Factor the quotient. 4 1 4 1 4 4 0 4 1 0 1 0
Factor the quotient. 1 1 1 1 1 1 0 1 1 0 1 0
f x x 4 x 2 x 2 1
Factor the quotient. x2 1 0
b. Factor x 2 1 . x2 1 0
x 2 1 x i The fourth roots of 1 are 1, 1 , i, and i .
x 2 1 x i f x x 4 x 2 x i x i 109. a.
112. f x x 6 1
x 1 x x 1 x 1 x x 1
x3 1 x3 1
f x x 2 x 35 4
2
2
0 x 5 x 7
f x x 5 x 7 2
2
2
2
x 5 0 2
Factor x 2 x 1 . x
x 70 2
or
x 5
x 2 7
2
x 5
b. f x x 5 x 5 x 7i x 7i f x x 5 x 5 x 7
x
f x x 4 8 x 2 33
0 x 3 x 11
f x x 2 3 x 2 11 x2 3 0
x 2 11 0
x2 3 x 3
and
x 2 11
x 11i
1 4 11 2 1
1 3 1 i 3 2 2
2
1 i 3 , 2
1 i 3 . 2
113. The number 5 is a real solution to the equation x 2 5 0 and a zero of the polynomial f x x 2 5 . However, by the rational zeros theorem, the only possible rational zeros of f x are 1 and 5 . This
f x x 3 x 3 x 11 b. f x
x 3 x 3 x 11i x 11i
111.
1
The sixth roots of 1 are 1, 1 ,
2
or
1 4 11 2 1
1 3 1 i 3 2 2 2 Factor x x 1 .
x 7i
2
1
2
110. a.
2
means that
Factors of 1 1 1 Factors of 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 0
5 is irrational.
114. a. ax b 0
ax b b x a
358
Section 3.5 116. x 3 9 x 26 m 9, n 26
b b 4ac b. x 2a
2
115.
x 3 3 x 2 m 3, n 2 2
2
n n m x 3 2 3 2
2
3
2
3
3
3 169 27 13 3 169 27 13 3 196 13 3 196 13
2 2 3 2 3 2 2
3
26 26 9 3 2 3 2
2 2 3 3 2 3 2 2
3
n n m 2 3 2
26 26 9 3 2 3 2
n n m 3 2 3 2
3
3
3
2
3
n n m x 3 2 3 2
x 3 3 x 2
3
3 14 13 3 14 13 3 27 3 1 3 1 2
3 1 1 1 3 1 1 1 3 1 3 1 1 1 2
Section 3.5 Rational Functions 1. f x 4 x 4 25 x 2 36
f x 4 x2 9 x2 4
6 x2 6 6 6 h 1 1 2 1 6 6 60 h 1.9 1.9 2 0.1 6 6 600 h 1.99 1.99 2 0.01
3. h x
f x 2 x 3 2 x 3 x 2 x 2 0 2 x 3 2 x 3 x 2 x 2 3 3 x , x , x 2, x 2 2 2 3 3 The zeros are , , 2, 2 . 2 2 2. g x 9 x 4 43 x 2 50
g x 9 x 2 25 x 2 2
4. k x
k 1
2 x2 3 x2 1 2 1 3 2
5 2.5 2
1 1 2 2 10 3 203 2.0099 k 10 102 1 101 2 2 100 3 20003 k 100 2.0001 100 2 1 10001
0 3x 5 3 x 5 x 2 x 2
g x 3x 5 3 x 5 x 2 x 2
5 5 x , x , x 2, x 2 3 3 5 5 The zeros are , , 2, 2 . 3 3
359
2
Chapter 3 Polynomial and Rational Functions 6. q x
5. rational
b.
23. a. 2 c.
d. 2
e. Never increasing
f. , 4 4,
8. x approaches 5 from the left
g. , 4 4,
h. , 2 2,
9. vertical; c
10. nonzero
i. x 4
j. y 2
11. c
12. <; >
24. a. 2
b.
13. left; down
14. one
c.
d. 2
15. denominator
16.
7. x approaches infinity
C x
e. , 3 3, f. Never decreasing
x
g. , 3 3, h. , 2 2,
17. x 5 0
i. x 3
x5 , 5 5,
25. a. 1
b.
c.
d. 1
e. , 3
f. 3,
18. x 3 0 x3 , 3 3, 19.
g. , 3 3, h. 1, i. x 3
4 x 2 3x 1 0
x 1 4 x 1 0
j. y 1 b.
26. a. 4
1 4 1 1 , 1 1, , 4 4 x 1, x
20.
j. y 2
2 x2 5x 7 0
2 x 7 x 1 0
c.
d. 4
e. 1,
f. ,1
g. ,1 1,
h. , 4
i. x 1
j. y 4
27. x 4 0 x4
7 x , x 1 2 7 7 , ,1 1, 2 2
28. x 7 0 x 7
21. x 2 100 0 True for all real x. ,
29.
2 x2 9 x 5 0
x 5 2 x 1 0 x 5, x
22. x 2 49 0 True for all real x. ,
360
1 2
Section 3.5 3x 2 8 x 3 0
30.
b.
x 3 3x 1 0 x 3, x
1 3
37. a. The degree of the numerator is 2. The degree of the denominator is 2. 3 Since n m , the line y or 1 equivalently y 3 is a horizontal asymptote of h.
31. x 2 5 has no real zeros. The function has no vertical asymptotes. 32. x 4 1 has no real zeros. The function has no vertical asymptotes. 33. 2t 2 4t 3 0 t
4
b.
42 4 2 3 4 40 2 2 4
4 2 10 2 10 4 2 2 10 2 10 t ,t 2 2
a
38. a. The degree of the numerator is 2. The degree of the denominator is 2. 4 or Since n m , the line y 1 equivalently y 4 is a horizontal asymptote of r.
4 2 4 3 1 4 28 2 3 6
4 2 7 2 7 6 3
a
2 7 2 7 ,a 3 3
b.
35. a. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of p. b.
3x 2 8 x 5 3 x2 3 3x 2 8 x 5 3x 2 9 8 x 14 7 x 4 7 The graph crosses y 3 at , 3 . 4
34. 3a 2 4a 1 0 4
8 0 x 4x 4 8 0 No solution The graph does not cross y 0 . 2
5 0 x2 2 x 1 5 0 No solution The graph does not cross y 0 .
4 x 2 5 x 1 4 x2 2 4 x 2 5 x 1 4 x 2 8 5 x 7 7 x 5 7 The graph crosses y 4 at , 4 . 5
39. a. The degree of the numerator is 4. The degree of the denominator is 1. Since n m , the function has no horizontal asymptotes.
36. a. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of q.
b. Not applicable
361
Chapter 3 Polynomial and Rational Functions 40. a. The degree of the numerator is 3. The degree of the denominator is 1. Since n m , the function has no horizontal asymptotes.
3x3 2 x 2 7 x 3 3 3x 2 x 7 x x3 x x 44. a. 3 3 5x 1 5x 1 x3 x3 2 7 3 2 x x 1 5 3 x 3
b. Not applicable 41. a. The degree of the numerator is 1. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of t. b.
b. All of the terms approach 0 as x .
2x 4 0 x 7x 4 2x 4 0 2 x 4 x 2 The graph crosses y 0 at 2, 0 .
c. Since the terms from part (b) approach 0 as x , the horizontal asymptote is
2
y
2 x2 3 is in lowest terms x and the denominator is 0 at x 0 . f has a vertical asymptote at x 0 . The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, f has no horizontal asymptote, but does have a slant asymptote. 2 x2 3 2 x2 3 3 2x x x x x The slant asymptote is y 2 x .
x3 0 2 x 3x 5 x3 0 x 3 The graph crosses y 0 at 3, 0 . 2
3x 2 2 is in lowest terms x and the denominator is 0 at x 0 . g has a vertical asymptote at x 0 . The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, g has no horizontal asymptote, but does have a slant asymptote. 3x 2 2 3x 2 2 2 3x x x x x The slant asymptote is y 3x .
46. The expression
2
x 3x 1 3 1 1 2 x 2 3x 1 x 2 x 2 x 2 x x 43. a. 2 2 5 2x 5 2x 5 2 2 2 2 x x x b. All of the terms approach 0 as x . c. Since the terms from part (b) approach 0 as x , the horizontal asymptote is
y
300 3 or y . 50 5
45. The expression
42. a. The degree of the numerator is 1. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of s. b.
2
1 0 0 1 or y . 20 2
3x 2 4 x 5 is in lowest x6 terms and the denominator is 0 at x 6 . h has a vertical asymptote at x 6 .
47. The expression
362
Section 3.5 The quotient is x 5 . The slant asymptote is y x 5 .
The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, h has no horizontal asymptote, but does have a slant asymptote. 6 3 4 5 18 132 3 22 137
50. The expression
lowest terms. x2 7 0
The quotient is 3 x 22 . The slant asymptote is y 3x 22 .
x2 7 x 7 The denominator is 0 at x 7 . q has
2 x 2 3 x 7 is in lowest x3 terms and the denominator is 0 at x 3 . k has a vertical asymptote at x 3 . The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, k has no horizontal asymptote, but does have a slant asymptote. 3 2 3 7 6 9 2 3 2
vertical asymptotes at x 7 and x 7 . The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, q has no horizontal asymptote, but does have a slant asymptote. x3 2 3 2 x 0 x 7 x 3x 2 x 4
48. The expression
x3 0 x 2 7 x
The quotient is x 3 . The slant asymptote is y x 3 . 51. The expression
terms. x2 5 0 x 5 The denominator is 0 at x 5 . p has
vertical asymptotes at x 5 and x 5 . The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, p has no horizontal asymptote, but does have a slant asymptote. x5 2 3 2 x 0x 5 x 5x 4x 1
2
2x 1
x 1 x 2 4
The denominator is 0 at x 1 , x 2 , and x 2 . r has vertical asymptotes at x 1 , x 2 , and x 2 . The degree of the numerator is 1. The degree of the denominator is 3. Since n m , the line y 0 is a horizontal asymptote of r. r has no slant asymptote.
5x x 1 5 x 2 0 x 25
2x 1 is in lowest x3 x 2 4 x 4
terms. 2x 1 2x 1 2 3 2 x x 4 x 4 x x 1 4 x 1
x2 5
5 x 17
x3 5 x 2 4 x 1 is in lowest x2 5
x3 0 x 2 5 x
3x 2 5 x 4 3x 2 0 x 21
The quotient is 2 x 3 . The slant asymptote is y 2 x 3 . 49. The expression
x3 3x 2 2 x 4 is in x2 7
x 26
363
Chapter 3 Polynomial and Rational Functions 3x 4 is in x 2 x 2 9 x 18
52. The expression
The denominator is never 0, so a has no vertical asymptotes. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, a has no horizontal asymptote, but does have a slant asymptote. 3x 2 2 3 2 3x 2 x 1 9 x 0 x 5 x 4
3
lowest terms. 3x 4 3x 4 2 3 2 x 2 x 9 x 18 x x 2 9 x 2
3x 4 x 2 x 2 9
The denominator is 0 at x 2 , x 3 , and x 3 . t has vertical asymptotes at x 2 , x 3 , and x 3 . The degree of the numerator is 1. The degree of the denominator is 3. Since n m , the line y 0 is a horizontal asymptote of t. t has no slant asymptote.
9 x3 6 x 2 3x
6x 8x 4 6 x 2 4 x 2
x
55. The graph of f is the graph of y
42 4 2 3 4 8 2 2 4
4 x3 8 x 2 6 x
56. The graph of g is the graph of y
10 x 2 x 3 10 x 2 20 x 15
shift left 4 units.
21x 12 The quotient is 2 x 5 . The slant asymptote is y 2 x 5 . 54. The expression
9 x3 5 x 4 is in lowest 3x 2 2 x 1
terms. 3x 2 2 x 1 0 x
1 with a x
shift right 3 units.
The denominator is never 0, so f has no vertical asymptotes. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, f has no horizontal asymptote, but does have a slant asymptote. 2x 5 2 x 2 4 x 3 4 x3 2 x 2 7 x 3
2
4x 6 The quotient is 3x 2 . The slant asymptote is y 3x 2 .
4 x3 2 x 2 7 x 3 53. The expression is in 2 x2 4 x 3 lowest terms. 2 x2 4 x 3 0 4
2
22 4 31 2 8 2 3 6 364
1 with a x
Section 3.5 57. The graph of h is the graph of y
1 with a x2
shift upward 2 units.
61. The graph of p is the graph of y
1 x
reflected across the x-axis. 1 58. The graph of k is the graph of y 2 with a x shift downward 3 units.
62. The graph of q is the graph of y
1 x2
reflected across the x-axis. 1 59. The graph of m is the graph of y 2 with a x shift to the left 4 units and a shift downward 3 units.
7 . 2 7 The x-intercepts are 3, 0 and , 0 . 2
63. a. The numerator is 0 at x 3 and x
1 with a x2 shift to the right 1 unit and a shift upward 2 units.
60. The graph of n is the graph of y
b. The denominator is 0 at x 2 and 1 x . f has vertical asymptotes at 4 1 x 2 and x . 4
365
Chapter 3 Polynomial and Rational Functions c. The degree of the numerator is 2. The degree of the denominator is 2. 2 Since n m , the line y or 4 1 equivalently y is a horizontal 2 asymptote of f.
c. The degree of the numerator is 1. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of f. d. The constant term in the numerator is 9 . The constant term in the denominator is 9 . All other terms are 0 when evaluated at x 0 . 9 The y-intercept is 0, 0,1 . 9
d. The constant term in the numerator is 21 . The constant term in the denominator is 2. All other terms are 0 when evaluated at x 0 . 21 The y-intercept is 0, . 2
66. a. The numerator is 0 at x
8 The x-intercept is , 0 . 5
4 64. a. The numerator is 0 at x and x 6 . 3 4 The x-intercepts are , 0 and 6, 0 . 3 b. The denominator is 0 at x
b. The denominator is 0 at x 2 and x 2 . f has vertical asymptotes at x 2 and x 2 .
3 and 2
x 5 . f has vertical asymptotes at x and x 5 .
c. The degree of the numerator is 1. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of f.
3 2
d. The constant term in the numerator is 8 . The constant term in the denominator is 4 . All other terms are 0 when evaluated at x 0 . 8 The y-intercept is 0, 0, 2 . 4
c. The degree of the numerator is 2. The degree of the denominator is 2. 3 Since n m , the line y is a 2 horizontal asymptote of f. d. The constant term in the numerator is 24. The constant term in the denominator is 15 . All other terms are 0 when evaluated at x 0 . 8 24 The y-intercept is 0, 0, . 15 5 65. a. The numerator is 0 at x
8 . 5
1 and x 3 . 5 1 The x-intercepts are , 0 and 3, 0 . 5
67. a. The numerator is 0 at x
b. The denominator is 0 at x 2 . f has a vertical asymptote at x 2 .
9 . 4
c. f x
5 x 1 x 3 5 x 2 14 x 3
x2 x2 The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, f has no horizontal asymptote, but does have a slant asymptote.
9 The x-intercept is , 0 . 4
b. The denominator is 0 at x 3 and x 3 . f has vertical asymptotes at x 3 and x 3 .
366
Section 3.5 2 5
14 10 5 4
3 8 11
69.
The quotient is 5 x 4 . The slant asymptote is y 5 x 4 . d. The constant term in the numerator is 3 . The constant term in the denominator is 2. All other terms are 0 when evaluated at x0. 3 3 The y-intercept is 0, 0, . 2 2
70.
3 and x 2 . 4 3 The x-intercepts are , 0 and 2, 0 . 4
68. a. The numerator is 0 at x
b. The denominator is 0 at x 3 . f has a vertical asymptote at x 3 . c. f x
4 x 3 x 2 4 x 2 11x 6
71.
x3 x3 The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, f has no horizontal asymptote, but does have a slant asymptote. 3 4 11 6 12 3 4 1 9 72.
The quotient is 4 x 1 . The slant asymptote is y 4 x 1 . d. The constant term in the numerator is 6. The constant term in the denominator is 3. All other terms are 0 when evaluated at x0. 6 The y-intercept is 0, 0, 2 . 3
367
Chapter 3 Polynomial and Rational Functions 73. n 0
3 3 3 2 0 7 7 7
3 The y-intercept is 0, . 7 n is never 0. It has no x-intercepts. n is in lowest terms, and 2 x 7 is 0 for 7 x , which is the vertical asymptote. 2 The degree of the numerator is 0. The degree of the denominator is 1. Since n m , the line y 0 is a horizontal asymptote of n. 3 0 2x 7 0 3 No solution n does not cross its horizontal asymptote. 3 3 n x 2 x 7 2 x 7
74. m 0
4 The y-intercept is 0, . 5 m is never 0. It has no x-intercepts. m is in lowest terms, and 2 x 5 is 0 for 5 x , which is the vertical asymptote. 2 The degree of the numerator is 0. The degree of the denominator is 1. Since n m , the line y 0 is a horizontal asymptote of m. 4 0 2x 5 0 4 No solution m does not cross its horizontal asymptote. 4 4 m x 2 x 5 2 x 5
n x n x , n x n x n is neither even nor odd. Interval
7 , 2
7 , 2
Test Point
Comments
4, 3
n(x) is positive. n(x) must approach the horizontal asymptote y 0 from above as x . As x approaches the vertical asymptote 7 x from the 2 left, n x .
2, 1
4 4 4 2 0 5 5 5
m x m x , m x m x m is neither even nor odd. Interval
n(x) is negative. n(x) must approach the horizontal asymptote y 0 from below as x. As x approaches the vertical asymptote 7 x from the 2 right, n x .
5 , 2
368
Test Comments Point m(x) is positive. m(x) must approach the horizontal asymptote y 0 from above as x . 2, 4 As x approaches the vertical asymptote 5 x from the left, 2 m x .
Section 3.5
Interval
5 , 2
Test Point
Comments
3, 4
m(x) is negative. m(x) must approach the horizontal asymptote y 0 from below as x. As x approaches the vertical asymptote 5 x from the right, 2 m x .
Test Point
Comments
, 3
3 5, 8
p(x) is positive. p(x) must approach the horizontal asymptote y 0 from above as x . As x approaches the vertical asymptote x 3 from the left, p x .
3, 0
p(x) is negative. As x approaches the 6 vertical asymptote 2, 5 x 3 from the right, p x .
0, 3
6 2, 5
p(x) is negative. As x approaches the vertical asymptote x 3 from the left, p x .
3 5, 8
p(x) is positive. p(x) must approach the horizontal asymptote y 0 from above as x. As x approaches the vertical asymptote x 3 from the right, p x .
Interval
6
6 2 75. p 0 2 0 9 9 3 2 The y-intercept is 0, . 3 p is never 0. It has no x-intercepts. p is in lowest terms, and x 2 9 is 0 for x 3 and x 3 , which are the vertical asymptotes. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of p. 6 0 2 x 9 0 6 No solution p does not cross its horizontal asymptote. 6 6 p x 2 2 x 9 x 9
3,
p x p x p is even.
369
Chapter 3 Polynomial and Rational Functions 76. q 0
4
0 16 2
4 1 16 4
Interval
1 The y-intercept is 0, . 4 q is never 0. It has no x-intercepts. q is in lowest terms, and x 2 16 is 0 for x 4 and x 4 , which are the vertical asymptotes. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of q. 4 0 2 x 16 0 4 No solution q does not cross its horizontal asymptote. 4 4 q x 2 2 x 16 x 16
4,
Test Point
Comments
1 6, 5
q(x) is positive. q(x) must approach the horizontal asymptote y 0 from above as x. As x approaches the vertical asymptote x 4 from the right, q x .
q x q x q is even.
Test Point
Comments
, 4
1 6, 5
q(x) is positive. q(x) must approach the horizontal asymptote y 0 from above as x . As x approaches the vertical asymptote x 4 from the left, q x .
4, 0
q(x) is negative. As x approaches the 6 vertical asymptote 2, 5 x 4 from the right, q x .
0, 4
q(x) is negative. As x approaches the vertical asymptote x 4 from the left, q x .
Interval
6 2, 5
77. r 0
5 0
0
0 0 6 The y-intercept is 0, 0 . The x-intercept is 0, 0 . 2
r is in lowest terms, and x 2 x 6 x 3 x 2 is 0 for x 3 and x 2 , which are the vertical asymptotes. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of r. 5x 0 2 x x6 5x 0 x0 r crosses its horizontal asymptote at 0, 0 . r x
5 x
5 x x x6
x x 6 r x r x , r x r x 2
r is neither even nor odd.
370
2
Section 3.5
Interval
Test Point
, 2
r(x) is negative. r(x) must approach the horizontal asymptote y 0 5 from below as 3, x . 2 As x approaches the vertical asymptote x 2 from the left, r x .
2, 0
5 1, 4
r(x) is positive. As x approaches the vertical asymptote x 2 from the right, r x .
5 1, 6
r(x) is negative. As x approaches the vertical asymptote x 3 from the left, r x .
5 6, 4
r(x) is positive. r(x) must approach the horizontal asymptote y 0 from above as x. As x approaches the vertical asymptote x 3 from the right, r x .
0, 3
3,
78. t 0
Comments
4 0
0
0 2 0 3 The y-intercept is 0, 0 . The x-intercept is 0, 0 . 2
t is in lowest terms, and x 2 2 x 3 x 3 x 1 is 0 for x 3 and x 1 , which are the vertical asymptotes. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of t. 4x 0 2 x 2x 3 4x 0 x0 t crosses its horizontal asymptote at 0, 0 . t x
4 x
x 2 x 3 t x t x , n x t x 2
4 x x 2x 3 2
t is neither even nor odd. Interval Test Point Comments t(x) is negative. t(x) must approach the horizontal asymptote y 0 from below as , 1 3, 1 x . As x approaches the vertical asymptote x 1 from the left, t x .
1, 0
0, 3
371
3 16 , 4 5
t(x) is positive. As x approaches the vertical asymptote x 1 from the right, t x .
1, 1
t(x) is negative. As x approaches the vertical asymptote x 3 from the left, t x .
Chapter 3 Polynomial and Rational Functions
Interval
3,
79. k 0
5 x 3 5 x 3 2 x 7 2 x 7
Test Point
Comments
k x
k x k x , k x k x k is neither even nor odd.
5 5, 3
t(x) is positive. t(x) must approach the horizontal asymptote y 0 from above as x. As x approaches the vertical asymptote x 3 from the right, t x .
Interval
5 0 3 3 2 0 7 7
3 , 5
8 1, 9
3 7 , 5 2
3, 12
7 , 2
3 The y-intercept is 0, . 7 5x 3 0 5x 3 3 x 5 3 The x-intercept is , 0 . 5 k is in lowest terms, and 2 x 7 is 0 for 7 x , which is the vertical asymptote. 2 The degree of the numerator is 1. The degree of the denominator is 1. 5 Since n m , the line y is a horizontal 2 asymptote of k. 5 5x 3 2 2x 7 10 x 35 10 x 6 35 6 No solution k does not cross its horizontal asymptote.
372
Test Point
4,17
Comments
k(x) is positive. k(x) must approach the horizontal asymptote 5 y from above as 2 x . k(x) is negative. As x approaches the vertical asymptote 7 x from the left, 2 k x . k(x) is positive. k(x) must approach the horizontal asymptote 5 y from above as 2 x. As x approaches the vertical asymptote 7 x from the right, 2 k x .
Section 3.5 80. h 0
4 0 3 3 3 3 0 5 5 5
Test Point
Comments
4,17
h(x) is positive. h(x) must approach the horizontal asymptote 4 y from above as 3 x. As x approaches the vertical asymptote 5 x from the right, 3 h x .
Interval
3 The y-intercept is 0, . 5 4x 3 0
4 x 3 3 x 4
5 , 3
3 The x-intercept is , 0 . 4 h is in lowest terms, and 3x 5 is 0 for 5 x , which is the vertical asymptote. 3 The degree of the numerator is 1. The degree of the denominator is 1. 4 Since n m , the line y is a horizontal 3 asymptote of h. 4 4x 3 3 3x 5 12 x 20 12 x 9 20 9 No solution h does not cross its horizontal asymptote. 4 x 3 4 x 3 h x 3 x 5 3 x 5
81. g 0
Test Point
3 , 4
8 1, 9
3 5 , 4 3
3, 12
2
2 2 1
0 1 The y-intercept is 0, 2 .
h x h x , h x h x h is neither even nor odd. Interval
3 0 5 0 2 2
3x 2 5 x 2 0
3x 1 x 2 0 1 x ,x2 3 1 The x-intercepts are , 0 and 2, 0 . 3
Comments
h(x) is positive. h(x) must approach the horizontal 4 asymptote y 3 from above as x . h(x) is negative. As x approaches the vertical asymptote 5 x from the left, 3 h x .
g is in lowest terms, and x 2 1 is never 0, so there are no vertical asymptotes. The degree of the numerator is 2. The degree of the denominator is 2. 3 Since n m , the line y , or equivalently 1 y 3 , is a horizontal asymptote of g.
373
Chapter 3 Polynomial and Rational Functions 3x 2 5 x 2 x2 1 3x 2 3 3x 2 5 x 2 5 x 5 x 1 g crosses its horizontal asymptote at 1, 3 . 3
g x
3 x 5 x 2 2
3x 2 5 x 2 x2 1
x2 1 g x g x , g x g x
82. c 0
, 1
1 1, 3
Test Point
Comments
0, 2
2,
3,1
2
3 3 1
2 x2 5x 3 0
Since g 2 4 is above the horizontal asymptote y 3 , 2, 4 g(x) must approach the horizontal asymptote from above as x . Plot the point 1 ,1 between 2 1 the horizontal ,1 asymptote and the x2 intercept of 1 , 0 . 3
1 , 2 3
2
0 1 The y-intercept is 0, 3 .
g is neither even nor odd. Interval
2 0 5 0 3
2 x 1 x 3 0 1 x ,x3 2 1 The x-intercepts are , 0 and 3, 0 . 2 c is in lowest terms, and x 2 1 is never 0, so there are no vertical asymptotes. The degree of the numerator is 2. The degree of the denominator is 2. 2 Since n m , the line y , or equivalently 1 y 2 , is a horizontal asymptote of c. 2 x2 5x 3 x2 1 2 x2 2 2 x2 5x 3 5 x 5 x 1 c crosses its horizontal asymptote at 1, 2 . 2
The point 0, 2 is the y-intercept. Since g 3 1 is below the horizontal asymptote y 3 , g(x) must approach the horizontal asymptote from below as x .
c x
2 x 5 x 3 2
x2 1 c x c x , c x c x c is neither even nor odd.
374
2 x2 5x 3 x2 1
Section 3.5
Interval
, 1
1 1, 2
Test Point
x2 2 x 1 0
Comments
x 1 x 1 0
Since c 2 3 is above the horizontal 2, 3 asymptote y 2 , c(x) must approach the horizontal asymptote from above as x . Plot the point 3 6 , between 4 5 3 6 the horizontal , asymptote and the 4 5 x-intercept of 1 , 0 . 2
1 , 3 2
0, 2
3,
6 7, 5
x 1 The x-intercept is 1, 0 . n is in lowest terms, and the denominator is 0 when x 0 , which is the vertical asymptote. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, n has no horizontal asymptote, but does have a slant asymptote. x2 2x 1 x2 2x 1 1 x2 n x x x x x x The quotient is x 2 . The slant asymptote is y x 2 . x2 2 x 1 x 2 2 x 2x x 2x 1 x2
The point 0, 3 is the y-intercept. 6 Since c 7 is 5 below the horizontal asymptote y 2 , c(x) must approach the horizontal asymptote from below as x .
0 1 No solution n does not cross its slant asymptote. 2 x 2 x 1 x2 2 x 1 n x x x n x n x , n x n x
n is neither even nor odd. Select test points from each interval.
Interval
2 0 2 0 1 undefined 83. n 0 0
There is no y-intercept.
375
, 1
Test Point 9 4, 4
Test Point 1 2, 2
1, 0
1 1 , 2 2
1 9 , 4 4
0,
1 25 , 4 4
9 2, 2
Chapter 3 Polynomial and Rational Functions 2 0 4 0 4 84. m 0 undefined 0
There is no y-intercept. x2 4 x 4 0
x 2 x 2 0 x2 The x-intercept is 2, 0 . m is in lowest terms, and the denominator is 0 when x 0 , which is the vertical asymptote. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, m has no horizontal asymptote, but does have a slant asymptote. x2 4 x 4 x2 4 x 4 4 x4 m x x x x x x The quotient is x 4 . The slant asymptote is y x 4 .
85. f 0
10 The y-intercept is 0, . 3 x 2 7 x 10 0
x 5 x 2 0 x 5, x 2 The x-intercepts are 5, 0 and 2, 0 . f is in lowest terms, and x 3 is 0 when x 3 , which is the vertical asymptote. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, f has no horizontal asymptote, but does have a slant asymptote. 3 1 7 10
x2 4 x 4 x4 x 2 2 x 4x x 4x 4 0 4 No solution m does not cross its slant asymptote.
x2 4 x 4 x2 4 x 4 x x m x m x , m x m x m x
3 12 1 4 2
m is neither even nor odd. Select test points from each interval.
Interval , 0
Test Point 4, 9
0, 2
1 9 , 2 2
1,1
2,
1 3, 3
9 8, 2
02 7 0 10 10 3 0 3
The quotient is x 4 . The slant asymptote is y x 4 . x 2 7 x 10 x3 2 2 x 7 x 12 x 7 x 10
Test Point 2, 8
x4
12 10 No solution f does not cross its slant asymptote. 2 x 7 x 10 x 2 7 x 10 f x x 3 x 3 f x f x , f x f x
f is neither even nor odd. Select test points from each interval.
376
Section 3.5 Interval
, 5
Test Point 5 7, 2
5, 3
4, 2
7 9 , 2 2
3, 2
5 5 , 2 2
7 4 , 3 3
2,
1, 2
d does not cross its slant asymptote.
Test Point 4 6, 3
x 2 x 12 x 2 x 12 x 2 x 2 d x d x , d x d x d x
d is neither even nor odd. Select test points from each interval.
9 1, 2
Interval
Test Point
, 3
8, 6
Test Point 4 4, 3
3, 2
3 2, 2
1,12
2, 4
3, 6
7 13 , 2 6
4,
9 6, 2
7, 6
02 0 12 12 6 2 0 2 The y-intercept is 0, 6 .
86. d 0
x 2 x 12 0
x 3 x 4 0 x 3, x 4
The x-intercepts are 3, 0 and 4, 0 . d is in lowest terms, and x 2 is 0 when x 2 , which is the vertical asymptote. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, d has no horizontal asymptote, but does have a slant asymptote. 2 1 1 12 1
2
2
1
10
87. w 0
4 0
2
0 2 4
0
The y-intercept is 0, 0 . The x-intercept is 0, 0 .
w is in lowest terms, and x 2 4 is never 0, so there are no vertical asymptotes. The degree of the numerator is 2. The degree of the denominator is 2. 4 , or Since n m , the line y 1 equivalently y 4 , is a horizontal asymptote of w. 4 x 2 4 2 x 4 2 4 x 16 4 x 2
The quotient is x 1 . The slant asymptote is y x 1 . x 2 x 12 x2 2 2 x x 2 x x 12 x 1
2 12 No solution
16 0 No solution
377
Chapter 3 Polynomial and Rational Functions 3 , or 1 equivalently y 3 , is a horizontal asymptote of u. 3 x 2 3 2 x 1 2 3 x 3 3 x 2
w does not cross its horizontal asymptote. w x
4 x
2
x 4 w x w x 2
Since n m , the line y
4 x x2 4 2
w is even.
Test Point
Interval
, 0 2, 2
0,
88. u 0
2, 2
3 0
Comments
3 0 No solution u does not cross its horizontal asymptote.
Since w 2 2 is above the horizontal asymptote y 4 , w(x) must approach the horizontal asymptote from above as x . Since w 2 2 is above the horizontal asymptote y 4 , w(x) must approach the horizontal asymptote from above as x .
u x
2
x2 1 u x u x
3 x 2 x2 1
u is even.
Interval
Test Point
Comments 3 is 2 above the horizontal asymptote y 3 , u(x) must approach the horizontal asymptote from above as x . 3 Since u 1 is 2 above the horizontal asymptote y 3 , u(x) must approach the horizontal asymptote from above as x .
Since u 1
2
0 2 1
3 x
0
The y-intercept is 0, 0 . The x-intercept is 0, 0 . u is in lowest terms, and x 2 1 is never 0, so there are no vertical asymptotes. The degree of the numerator is 2. The degree of the denominator is 2.
378
, 0
3 1, 2
0,
3 1, 2
Section 3.5 3 2 0 0 4 0 4 4 89. f 0 undefined 0 0 2 3 0
Interval
There is no y-intercept. x3 x 2 4 x 4 0
x 2 x 1 4 x 1 0
x 1 x 2 4 0 x 1 x 2 x 2 0 x 1, x 2, x 2 The x-intercepts are 1, 0 , 2, 0 and
2, 0 .
f is in lowest terms, and x 2 3 x x x 3 is 0 when x 0 and x 3 , which are the vertical asymptotes. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, f has no horizontal asymptote, but does have a slant asymptote. x2 2 3 2 x 3x x x 4 x 4
x3 3x 2
, 3
Test Point 21 8, 2
Test Point
3, 2
5 27 , 2 10
2, 1
4 1 , 3 3
1, 0
1 3 , 2 2
0, 2
1 14 , 3 3
3 1, 2
2,
10 3, 9
15 4, 7
4, 9
2x 4x 2 x 2 6 x 2
3 2 0 3 0 0 3 3 90. g 0 undefined 0 0 2 2 0
2x 4 The quotient is x 2 . The slant asymptote is y x 2 .
There is no y-intercept. x3 3x 2 x 3 0
x3 x 2 4 x 4 x2 x 2 3x 3 2 3 x x 6 x x x2 4 x 4 2 x 4
x 2 x 3 1 x 3 0
x 3 x 2 1 0 x 3 x 1 x 1 0
x2 y x2 22 0
x 3, x 1, x 1
f crosses its slant asymptote at 2, 0 . f x
The x-intercepts are 3, 0 , 1, 0 and
1, 0 .
x x 4 x 4 x 2 3 x 3
2
g is in lowest terms, and x 2 2 x x x 2 is 0 when x 0 and x 2 , which are the vertical asymptotes. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, g has no horizontal asymptote, but does have a slant asymptote.
x3 x 2 4 x 4 x 2 3x f x f x , f x f x f is neither even nor odd. Select test points from each interval.
379
Chapter 3 Polynomial and Rational Functions x5 x 2 x x 3x x 3 2
3
2
x3 2 x 2
5x x 5 x 2 10 x 2
9x 3 The quotient is x 5 . The slant asymptote is y x 5 . x3 3x 2 x 3 x2 2 x 3 2 3 x 3x 10 x x 3 x 2 x 3 x5
91. v 0
y x5
x 3 3 x 2 x 3 x2 2 x
v x
, 3 3, 1
3 2, 8
1, 0
1 3 , 2 2
1 0, 3
1 128 , 5 15
1 ,1 3
1 7 , 2 2
1, 2
3 15 , 2 2
2,
3,16
2 x
4
x2 9 v x v x
x3 3x 2 x 3 x2 2 x g x g x , g x g x g is neither even nor odd. Select test points from each interval. Test Point 48 5, 35
0
v is in lowest terms, and x 2 9 is never 0, so there are no vertical asymptotes. The degree of the numerator is 4. The degree of the denominator is 2. Since n m , the function has no horizontal asymptotes. Since n m 1 , the function has no slant asymptote.
1 16 5 3 3
Interval
0 2 9
The x-intercept is 0, 0 .
1 3
1 16 g crosses its slant asymptote at , . 3 3 g x
4
The y-intercept is 0, 0 .
9 x 3 x
2 0
2 x4 x2 9
v is even. Select test points. 1 The graph passes through 3, 9 , 1, , 5
Test Point 5 4, 8
1 1, , 3, 9 . 5
5,12.8
380
Section 3.5
92. g 0
4 0
d. The average cost would approach $20 per session. This is the same as the fee paid to the gym in the absence of fixed costs.
4
0 2 8
0
The y-intercept is 0, 0 .
94. a. C x 40 x 1200 420 100 200
The x-intercept is 0, 0 .
C x 40 x 1920
g is in lowest terms, and x 2 8 is never 0, so there are no vertical asymptotes. The degree of the numerator is 4. The degree of the denominator is 2. Since n m , the function has no horizontal asymptotes. Since n m 1 , the function has no slant asymptote. 4 x
b. C x
40 x 1920 x
40 20 1920 136 20 40 50 1920 C 50 78.4 50 40 100 1920 C 100 59.2 100 40 200 1920 C 200 49.6 200
c. C 20
4
4 x4 g x x2 8 x 2 8
g x g x g is even. Select test points.
16 The graph passes through 2, , 3
d. C 200 49.6 means that if 200,000 pages are printed, then the average cost per thousand pages is $49.60 (or equivalently $0.0496 per page).
4 4 16 1, , 1, , 2, . 9 9 3
e. The average cost would approach $40 per thousand pages or equivalently $0.04 per page. This is the cost per page in the absence of fixed costs.
59.95 0.24 252 59.95 0.16 C1 366 366 59.95 0.15 C 1 400 400
95. a. C 1 252
93. a. C x 20 x 69.95 39.99
C x 20 x 109.94 b. C x
59.95 0.45 436 400 0.17 252 59.95 0.45 582 400 C1 582 0.24 582 59.95 0.45 700 400 C 1 700 0.28 700
b. C 1 436
20 x 109.94 x
20 5 109.94 41.99 5 20 30 109.94 C 30 23.66 30 20 120 109.94 20.92 C 120 120
c. C 5
c. C 2 x
381
79.95 x
Chapter 3 Polynomial and Rational Functions 79.95 0.32 252 79.95 C 2 400 0.20 252 79.95 C 2 700 0.11 252
d. C 2 252
c.
475 25 5 300 2 475 25 10 C 1 10 362.5 2 475 25 15 C 1 15 425 2
96. a. C 1 5
b. C 2 x
600 x 100 x 600 x 1400 100 x 140,000 1400 x 600 x 140,000 2000 x 70 x 70% of the air pollutants can be removed. C x
80 20 20 ; $20,000 100 20 80 40 53 ; $53,000 C 40 100 40 80 90 C 90 720 ; $720,000 100 90
98. a. C 20
2000 x
2000 400 5 2000 C 2 10 200 10 2000 C 2 15 133.33 15
c. C 2 5
b.
d. The horizontal asymptote is y 0 . If it were possible for a person to have an infinite number of yearly memberships, the average cost per year would drop to $0.
80 x 100 x 80 x 320 100 x 32,000 320 x 80 x 32,000 400 x 80 x 80% of the waste can be removed. C x
99. a. N t P J t
N t 0.091t 3 3.48t 2 15.4t 335 23.0t 159
600 25 200 ; $200,000 97. a. C 25 100 25
N t 0.091t 3 3.48t 2 38.4t 494
N t represents the total number of adults incarcerated in both prisons and jails, t years since 1980.
600 50 600 ; $600,000 b. C 50 100 50 600 75 C 75 1800 ; $1,800,000 100 75 600 90 C 90 5400 ; $5,400,000 100 90
b. R t
J t N
23.0t 159 0.091t 3.48t 2 38.4t 494 R t represents the percentage of incarcerated adults who are in jail, t years since 1980. R t
382
3
Section 3.5 c. R 25
2x 7 x3 Find the quotient. 3 2 7
103. f x
23.0 25 159
0.091 25 3.48 25 38.4 25 494 3
2
0.333 R 25 0.333 means that in the year 2005, 33.3% of incarcerated adults were in jail.
6 2 1 f x 2
1.08t 36.9 V 100. a. P t t N 0.0135t 2 1.09t 73.2 This represents the percentage of people of voting age who voted in U.S. presidential elections in year t.
b. P 60
1 x3
1 with a x shift to the left 3 units and a shift upward 2 units.
The graph of f is the graph of y
1.08 60 36.9
0.0135 60 1.09 60 73.2 2
0.54 This means that in the presidential election in 1992 (60 yr after the year 1932), 54% of the people of voting age actually voted. 101. a. f x
1
x 1 1 f x 3 x 1 2 2
3
5 x 11 x2 Find the quotient. 2 5 11 10 5 1
104. f x
b. Domain: , 1 1, ;
Range: 3,
102. a. f x
1 3 x 4
f x
1 3 x4
f x 5
1 x2
1 with a x shift to the left 2 units and a shift upward 5 units.
The graph of f is the graph of y
b. Domain: , 4 4, ;
Range: , 3 3,
383
Chapter 3 Polynomial and Rational Functions 3 109. Factor of the numerator: x or 2 x 3 2
105. The numerator and denominator share a common factor of x 2 . The value 2 is not in the domain of f. The graph will have a “hole” at x 2 rather than a vertical asymptote.
Factors of the denominator: x 2 x 5 The degree of the numerator is less than the degree of the denominator. 2x 3 f x a x 2 x 5 f 0 3
106. A horizontal asymptote is a horizontal line that the function approaches as x becomes very large—that is, as x approaches infinity or as x approaches negative infinity.
2 0 3 a 3 0 2 0 5 3 3 a 10
107. Factors of the numerator: x 3 x 1
Factor of the denominator: x 2 The degree of the numerator equals the degree of the denominator. x 3 x 1 f x x 2 x a
a 10 2x 3 20 x 30 f x 10 2 x 2 x 5 x 3 x 10
3 4 0 3 0 1 3 0 2 0 a 4 f 0
4 110. Factor of the numerator: x or 3 x 4 3 Factors of the denominator: x 3 x 4 The degree of the numerator is less than the degree of the denominator. 3x 4 f x a x 3 x 4 f 0 1
3 3 2a 4 2a 4 a2 x 3 x 1 x 2 4 x 3 f x x 2 x 2 x 2 4 x 4
3 0 4 a 1 0 3 0 4 4 a 1 12
108. Factors of the numerator: x 4 x 2
Factor of the denominator: x 1 The degree of the numerator equals the degree of the denominator. x 4 x 2 f x x 1 x a
a3 3x 4 9 x 12 f x 3 2 x 3 x 4 x 7 x 12
f 0 8
0 4 0 2 8 0 1 0 a
111. a.
8 8 a a 1 x 4 x 2 x 2 6 x 8 f x x 1 x 1 x 2 2 x 1
384
Problem Recognition b.
c. Vertical asymptotes: 5x2 7 x 6 0
x 2 5 x 3 0 x 2 and x
113. a. f x
3 ; 5
b. On the window 9.4, 9.4,1 by
x 2 3x 2 x 1
6.2, 6.2,1 , the calculator shows a discontinuity or “hole” at 4, 3 . The
x 2 x 1
x2 x 1 No vertical asymptotes
x 1 x 4
x 1 x4 No vertical asymptotes
4 Horizontal asymptote: y 5 112. a. f x
x2 5x 4 x4
graph on the standard viewing window does not show the discontinuity.
b. On the window 4.7, 4.7,1 by
3.1, 3.1,1 , the calculator shows a discontinuity or “hole” at 1,1 . The graph on the standard viewing window does not show the discontinuity.
Problem Recognition Exercises: Polynomial and Rational Functions 1.
Factors of 8 1, 2, 4, 8 Factors of 1 1 1, 2, 4, 8 2 1 3 6 8 2 10 8 1 5 4 0
2.
Factors of 6 1, 2, 3, 6 Factors of 1 1 1, 2, 3, 6 2 1 2 5 6 2 8 6 1 4 3 0
Factor the quotient. x2 5x 4 0
Factor the quotient. x2 4 x 3 0
x 1 x 4 0
x 1 x 3 0
x 1, x 4 The zeros are 2, 1 , and 4 .
x 1, x 3 The zeros are 2 , 1, and 3.
385
Chapter 3 Polynomial and Rational Functions 3. The x-intercepts of q are 2, 0 , 1, 0 , and
10.
3, 0 .
4. The x-intercepts of p are 2, 0 , 1, 0 , and
4, 0 .
Factor the quotient. x2 4x 4 0
5. The x-intercepts of f are 2, 0 , 1, 0 , and
x 2 x 2 0
4, 0 .
x 2 The zeros are 1 and 2 (multiplicity 2).
6. The vertical asymptotes of f occur where q x 0 , at x 2 , x 1 , and x 3 .
11 The x-intercepts of d are 1, 0 and 2, 0 .
7. Since the degree of the numerator is the same as the degree of the denominator, the line 1 y , or equivalently y 1 , is a horizontal 1 asymptote of f.
12 4 5 14 1 281 2 5 10
2, 0 , and
15. Since the degree of the numerator is the same as the degree of the denominator, the line 1 y , or equivalently y 1 , is a horizontal 1 asymptote of g.
1.78,1 and 1.58,1 .
16.
Factors of 8 1, 2, 4, 8 Factors of 1 1 1, 2, 4, 8 4 1 4 2 8 4 0 8 1 0 2 0
x3 4 x 2 2 x 8 1 x3 3x 2 4 x3 4 x 2 2 x 8 x3 3x 2 4
7 x 2 2 x 12 0 x
2
2 2 4 7 12 2 340 2 7 14
2 2 85 1 85 1.17 or 1.46 14 7 The graph crosses the horizontal asymptote 1 85 1 85 at ,1 and ,1 , or 7 7
Factor the quotient. x2 2 0 x2 2 x 2 The zeros are 4,
13. The x-intercepts of g are 4, 0 ,
14. The vertical asymptotes of g occur where d x 0 , at x 1 and x 2 .
1.78 or 1.58 The graph crosses the horizontal asymptote 1 281 1 281 at ,1 and ,1 , or 10 10
9.
2, 0 , and
2, 0 .
5 x 2 x 14 0 1
12. The x-intercepts of c are 4, 0 ,
2, 0 .
x3 3x 2 6 x 8 1 8. 3 x 2 x2 5x 6 x3 3x 2 6 x 8 x3 2 x 2 5 x 6
x
Factors of 4 1, 2, 4 1, 2, 4 1 Factors of 1 1 1 3 0 4 1 4 4 1 4 4 0
1.17,1 and 1.46,1 .
2 , and 2 .
386
Section 3.6 2 x 3 4 x 2 10 x 12 2x 4 21. x 2 11 2 x 3 4 x 2 10 x 12 2 x3 4 x 2 22 x 44 10 x 12 22 x 44 12 x 32 32 8 x 12 3 y 2x 4
17. Graph b has the intercepts and asymptotes of function f. 18. Graph a has the intercepts and asymptotes of function g.
2x 4 19. x 2 0 x 11 2 x3 4 x 2 10 x 12
2 x 3 0 x 2 22 x
4 x 12 x 12 4 x 2 0 x 44
16 12 4 8 y 2 4 3 3 3 3 The graph crosses the slant asymptote at 8 4 , . 3 3
2
12 x 32 a. The quotient is q x 2 x 4 . b. The remainder is r x 12 x 32 .
22. r x 12 x 32
0 12 x 32 12 x 32 32 8 x 12 3 The solution to r x 0 gives the xcoordinate of the point where the graph of f crosses its slant asymptote.
20. The slant asymptote is y 2 x 4 .
Section 3.6 Polynomial and Rational Inequalities 1. 2 x 3 5 x 2 12 x 0
3.
x 2 x 2 5 x 12 0 x 2 x 3 x 4 0 3 x 0, x , x 4 2 3 The solution set is 0, , 4 . 2
1 The solution set is . 2
2. 12 x 3 32 x 2 12 x 0
2x 1 0 x5 2x 1 0 2 x 1 1 x 2
4.
4 x 3 x 2 8 x 3 0 4 x 3x 1 x 3 0 1 x 0, x , x 3 3 1 The solution set is 0, , 3 . 3
4x 3 0 x 1 4x 3 0 4x 3 3 x 4 3 The solution set is . 4
387
Chapter 3 Polynomial and Rational Functions 19. a. ,
5. x 5 0 x5 6. x 1 0
x 1 7. polynomial; 2
8. rational
9. negative
10. , ;
11. a. 4, 1
b. 4, 1
b. ,
c.
d.
20. a.
b.
c. ,
d. ,
21. 5 x 3 x 5 0
x
3 or x 5 5
c. , 4 1,
The boundary points are
d. , 4 1,
3 and 5. 5
Sign of 5 x 3 :
Sign of 5 x 3 x 5 :
Sign of x 5 :
12. a. , 1 3, b. , 1 3, c. 1, 3
d. 1, 3
5
3 5
13. a. , 3 3, b. ,
c.
d. 3 14. a. c. , 4 4,
b. 4 d. ,
15. a. , 2 0, 3
b. , 2 0, 3
c. 2, 0 3,
d. 2, 0 3,
16. a. , 4 0,1
b. , 4 0,1
c. 4, 0 1,
d. 4, 0 1,
17. a. 0, 3 3,
d. , 0 3
18. a. 0,
b. 2 0,
3 , 5 5
b. 5 x 3 x 5 0
3 , 5 5
c. 5 x 3 x 5 0
3 5 , 5
d. 5 x 3 x 5 0
3 , 5, 5
e. 5 x 3 x 5 0
3 , 5, 5
22. 3x 7 x 2 0
x
b. 0,
c. , 0
a. 5 x 3 x 5 0
7 or x 2 3
The boundary points are
7 and 2. 3
Sign of 3 x 7 :
Sign of 3 x 7 x 2 :
Sign of x 2 :
c. , 2 2, 0 d. , 0
388
7 3
2
Section 3.6 a. 3x 7 x 2 0
24.
7 , 2 3
x 9 x 1 0 x 9 or x 1 The boundary points are 9 and 1 .
7 , 2 3
Sign of x 9 : Sign of x 1 :
c. 3x 7 x 2 0
Sign of x 9 x 1 :
7 3 , 2
Sign of x 9 x 1 :
d. 3x 7 x 2 0
b. x 2 10 x 9 0 , 9 1,
e. 3x 7 x 2 0 7 , 2, 3
c. x 2 10 x 9 0 , 9 1,
x 2 x 12 0
d. x 2 10 x 9 0 9, 1
x 2 x 12 0 x 3 x 4 0 x 3 or x 4 The boundary points are 3 and 4. Sign of x 3 :
e. x 2 10 x 9 0 9, 1
25. x 2 12 x 36 0
x 6 2 0
Sign of x 3 x 4 :
Sign of x 4 :
Sign of x 3 x 4 :
9 1
a. x 2 10 x 9 0 9, 1
7 , 2, 3
x 2 10 x 9 0
b. 3x 7 x 2 0
23.
x 2 10 x 9 0
x 6 The boundary point is 6 .
3 4
a. x 2 12 x 36 0 6
a. x 2 x 12 0 3, 4
b. x 2 12 x 36 0 The square of any real number is nonnegative. Therefore, this inequality has no solution.
b. x 2 x 12 0 , 3 4,
c. x 2 x 12 0 , 3 4,
c. x 2 12 x 36 0 The inequality in part (b) is the same as the inequality in part (a) except that equality is included. The expression
d. x 2 x 12 0 3, 4
x 62 0 for x 6 . 6
e. x 2 x 12 0 3, 4
389
Chapter 3 Polynomial and Rational Functions d. x 2 12 x 36 0
27.
The expression x 6 0 for all real 2
3w 2 w 2 w 2 3w 2 w 2 w 4
numbers except where x 6 0 . Therefore, the solution set is all real numbers except 6 . , 6 6,
3w 2 w 4 0 Find the real zeros of the related equation. 3w 2 w 4 0
2
w 1 3w 4 0 w 1 or w
e. x 2 12 x 36 0 The square of any real number is greater than or equal to zero. Therefore, the solution set is all real numbers. ,
4 3
The boundary points are 1 and Sign of w 1 :
Sign of w 1 3w 4 :
Sign of 3w 4 :
26. x 2 14 x 49 0
x 7 0 2
4
1
x7 The boundary point is 7.
4 . 3
3
4 The solution set is 1, . 3
a. x 14 x 49 0 7 2
28.
b. x 2 14 x 49 0 The square of any real number is nonnegative. Therefore, this inequality has no solution.
5 y 2 7 y 3 y 4 5 y 2 7 y 3 y 12 5 y 2 4 y 12 0 Find the real zeros of the related equation. 5 y 2 4 y 12 0
y 2 5 y 6 0
c. x 2 14 x 49 0 The inequality in part (b) is the same as the inequality in part (a) except that equality is included. The expression
y 2 or y
6 5
The boundary points are 2 and
x 7 0 for x 7 . 7 2
Sign of y 2 :
Sign of y 2 5 y 6 :
Sign of 5 y 6 :
d. x 2 14 x 49 0
The expression x 7 0 for all real 2
2
numbers except where x 7 0 . Therefore, the solution set is all real numbers except 7. , 7 7, 2
6 The solution set is 2, . 5
e. x 2 14 x 49 0 The square of any real number is greater than or equal to zero. Therefore, the solution set is all real numbers. ,
390
6 5
6 . 5
Section 3.6 a 2 3a
29.
d 2 6d
30.
a 2 3a 0 Find the real zeros of the related equation. a 2 3a 0
d 2 6d 0 Find the real zeros of the related equation. d 2 6d 0
a a 3 0 a 0 or a 3 The boundary points are 0 and 3.
d d 6 0 d 0 or d 6 The boundary points are 0 and 6.
Sign of a :
Sign of d :
Sign of a a 3 :
Sign of a 3 :
0
Sign of d d 6 :
Sign of d 6 :
3
0
The solution set is , 0 3, .
The solution set is , 0 6, .
10 6 x 5 x 2
31.
5 x 2 6 x 10 0 5 x 2 6 x 10 0 Find the real zeros of the related equation. 5 x 2 6 x 10 0
x
6
6
62 4 5 10 6 236 6 2 59 3 59 2 5 10 10 5
3 59 3 59 or x 5 5 3 59 3 59 and . The boundary points are 5 5 x
3 59 : 5 3 59 Sign of x : 5 3 59 3 59 Sign of x x : 5 5 Sign of x
3 59 3 59 5
3 59 3 59 The solution set is , . 5 5
391
5
Chapter 3 Polynomial and Rational Functions 6 4 x 3x 2
32.
3 x 2 4 x 6 0
3x 2 4 x 6 0 Find the real zeros of the related equation. 3x 2 4 x 6 0 x
4
42 4 3 6 4 88 4 2 22 2 22 2 3 6 6 3
2 22 2 22 or x 3 3 2 22 2 22 and . The boundary points are 3 3 x
2 22 : 3
Sign of x
2 22 : 3 2 22 2 22 Sign of x x : 3 3 Sign of x
2 22 2 22 3
3
2 22 2 22 The solution set is , . 3 3 33. x 4 x 1 x 3 0 Find the real zeros of the related equation. x 4 x 1 x 3 0 x 4 , x 1 , or x 3 The boundary points are 4 , 1, and 3. Sign of x 4 : Sign of x 1 :
Sign of x 3 :
Sign of x 4 x 1 x 3 :
The boundary points are 5 , 2 , and 4. Sign of x 5 :
Sign of x 5 x 2 x 4 :
Sign of x 2 : Sign of x 4 :
5 2
The solution set is 5, 2 4, .
4
35. 5c c 2 4 c 0
4 1 3
2
5c c 2 c 4 0 Find the real zeros of the related equation. 2
The solution set is 4,1 3, .
5c c 2 c 4 0 2
34. x 2 x 5 x 4 0 Find the real zeros of the related equation. x 2 x 5 x 4 0 x 2 , x 5 , or x 4
c c 2 c 4 0 c 0 , c 2 , or c 4 The boundary points are 2 , 0, and 4. 2
392
Section 3.6 Sign of c 2 : 2
Sign of c c 2 c 4 :
Sign of c :
Sign of c 4 : 2
2
0
38. w4 20w2 64 0 Find the real zeros of the related equation. w4 20w2 64 0
w 4 w 16 0 2
w 2 w 2 w 4 w 4 0
4
w 2 , w 2 , w 4 , or w 4 The boundary points are 4 , 2 , 2, and 4.
The solution set is , 2 2, 0 4, .
Sign of w 4 : Sign of w 2 :
36. 6u u 1 3 u 0 2
Sign of w 2 :
6u u 1 u 3 0 Find the real zeros of the related equation. 2 6u u 1 u 3 0 2
Sign of w 4 : Sign of
w 2w 2w 4w 4:
u u 1 u 3 0 u 0 , u 1 , or u 3 The boundary points are 1 , 0, and 3. 2
Sign of u 1 : 2
Sign of u u 1 u 3 :
Sign of u 3 : 2
1 0
x
t 1 t 1 t 3 t 3 :
4 2 2
4
5 , x 2 , or x 2 2
Sign of 2 x 5 : Sign of x 2 :
t 1 , t 1 , t 3 , or t 3 The boundary points are 3 , 1 , 1, and 3.
Sign of
5 The boundary points are , 2 , and 2. 2
t 1 t 1 t 3 t 3 0
Sign of t 3 :
2 x 5 x 2 4 0 2 x 5 x 2 x 2 0
2
Sign of t 1 :
x 2 2 x 5 4 2 x 5 0
3
t 1 t 9 0
Sign of t 1 :
2 x3 5 x 2 8 x 20 0 Find the real zeros of the related equation. 2 x3 5 x 2 8 x 20 0
37. t 4 10t 2 9 0 Find the real zeros of the related equation. t 4 10t 2 9 0
Sign of t 3 :
2 x3 5 x 2 8 x 20
39.
The solution set is , 1 1, 0 3, .
2
The solution set is 4, 2 2, 4 .
Sign of u :
2
Sign of x 2 :
Sign of 2 x 5 x 2 x 2 :
3 1 1
5 2 2
2 5 The solution set is , 2, 2 . 2
3
The solution set is 3, 1 1, 3 .
393
Chapter 3 Polynomial and Rational Functions 3x3 3x 4 x 2 4
40.
42. 4 x 4 4 x3 64 x 2 80 x 0
x x x 16 x 20 0
3x3 4 x 2 3x 4 0 Find the real zeros of the related equation. 3x3 4 x 2 3x 4 0 x 2 3x 4 1 3x 4 0
4 x x3 x 2 16 x 20 0 3
Find the real zeros of the related equation. x x 3 x 2 16 x 20 0
x 1 3x 4 0 2
4 3
The boundary points are 1 , 1, and Sign of x 1 : Sign of x 1 :
Sign of 3 x 4 :
4 . 3
x x 2 x 2 3 x 10 0
x x 2 x 2 x 5 0
Sign of x 1 x 1 3 x 4 : 1 1 4
2
4 The solution set is , 1 1, . 3
Sign of x 2 : 2
x x 5 x 3 x 9 0
2 x x 5 x 3x 9 0 2
3
2
Sign of x x 2 x 5 :
4
5
2
Find the real zeros of the related equation. u 4 5u 2 28u 15 0
u 5u 3 u 5 0 4
2
0
u 5u 28u 15 0
x x 1 x 3 0 x 0 , x 1 , or x 3 The boundary points are 1 , 0, and 3.
Sign of x 3 :
u 4 5u 2 28u 15 0
2
43. 5u 6 28u 5 15u 4 0
x x 1 x 2 6 x 9 0
Sign of x :
The solution set is 2 0, 5 .
Factors of 9 1, 3, 9 1, 3, 9 Factors of 1 1 1 1 5 3 9 1 6 9 1 6 9 0
Sign of x 1 :
2
2
Find the real zeros of the related equation. x x3 5 x 2 3x 9 0
Sign of x 5 :
2
3
Sign of x :
41. 2 x 10 x 6 x 18 x 0 3
x x 2 x 5 0 x 0 , x 2 , or x 5 The boundary points are 2 , 0, and 5.
3
4
Factors of 20 1, 2, 4, 5, 10, 20 Factors of 1 1 1, 2, 4, 5, 10, 20 2 1 1 16 20 2 6 20 1 3 10 0
x 1 x 1 3x 4 0 x 1 , x 1 , or x
2
u 0, u
3 , or u 5 5
The boundary points are 0,
Sign of x x 1 x 3 : 1 0 3 2
The solution set is 1, 0 3 .
394
3 , and 5. 5
Section 3.6
Sign of u 4 :
Sign of 5u 3 :
Sign of u 4 5u 3 u 5 :
Sign of u 5 :
0
Sign of 3 x 1 : 5
Sign of x:
Sign of 2 x 5 : 4
Sign of x 4:
5
3
Sign of
5
x 2 x 5 3 x 1 x 4: 4
3 The solution set is , 5, . 5 5
1
5
0
3
4
2
The solution set is 1 5 5 , 0, , 4 . 3 2 2
4
w 3w 8 w 4 0
w 4 3w 2 8 w 4 0 4
44. 3w 8w 4 w 0 6
5
2
46. 5 x 3 x 2 4 x 1 x 3 0 Find the real zeros of the related equation. 2 3 4 5 x 3x 2 4 x 1 x 3 0 2
Find the real zeros of the related equation. w4 3w2 8w 4 0
w4 3w 2 w 2 0
3
4
Sign of w4 :
x 3x 2 4 x 1 x 3 0 2 1 x 0 , x , x , or x 3 3 4 1 2 The boundary points are , 0, , and 3. 4 3
Sign of 3w 2 :
Sign of 4 x 1 :
2
2 w 0 , w , or w 2 3
The boundary points are 0,
Sign of w 2 :
2 , and 2. 3
2
2
Sign of 3 x 2 :
Sign of x 3 :
1
4
Sign of
2 The solution set is , 2, . 3
5 x 3 x 2 4 x 1 x 3 : 2
45. 6 x 2 x 5 3x 1 x 4 0 5
3
4
0
4
2 3
1 The solution set is , 0 . 4
x 2 x 5 3x 1 x 4 0 Find the real zeros of the related equation. 4 5 x 2 x 5 3x 1 x 4 0 4
2
3
4
Sign of 5 x:
4
0
4
3
5u 3u 5 :
Sign of w
3
5
47. 5 x 3 2 The square of any real number is 2
5 1 , x , or x 4 2 3 1 5 The boundary points are , 0, , and 4. 3 2
x0, x
nonnegative. Therefore, 5 x 3 is nonnegative, and always greater than 0. The solution set is , . 2
395
3
Chapter 3 Polynomial and Rational Functions 48. 4 x 1 6 The square of any real number is 2 nonnegative. Therefore, 4 x 1 is nonnegative, and always greater than 0. The solution set is , . 2
53.
x2 4 x 3 1 0 x2 4 x 4 0
x 2 2 0 The square of any real number is 2 nonnegative. Therefore, x 2 is nonnegative, and always greater than 0. However, the equality is included. The solution set is 2 .
49. 4 x 7 The square of any real number is 2 nonnegative. Therefore, x 7 is nonnegative, and always greater than 0. The solution set is . 2
54. x 2 x 4 1
x2 6 x 8 1 0
50. 1 x 2 The square of any real number is 2
x2 6 x 9 0
x 3 2 0
nonnegative. Therefore, x 2 is nonnegative, and always greater than 0. The solution set is . 2
51.
The square of any real number is
nonnegative. Therefore, x 3 is nonnegative, and always greater than 0. However, the equality is included. The solution set is 3 . 2
16 y 2 24 y 9 16 y 2 24 y 9 0
4 y 32 0 2 The expression 4 y 3 0 for all real
55. a. 2, 3
b. 2, 3
c. , 2 3, d. , 2 3,
numbers except where 4 y 32 0 .Therefore, the solution set is
56. a. , 2 2, b. , 2 2,
3 . 4 3 3 The solution set is , , . 4 4
c. 2, 2
all real numbers except
52.
x 3 x 1 1
57. a. , 2 2,
2 w 5 0 2 The expression 2w 5 0 for all real
b. , 2 2,
c.
d.
58. a.
b.
4w2 20w 25 4w2 20w 25 0
d. 2, 2
2
c. , 3 3, d. , 3 3,
numbers except where 2w 52 0 .Therefore, the solution set is
59.
5 . 2 5 5 The solution set is , , . 2 2
all real numbers except
x2 0 x3 x20 x 2 2 is a boundary point.
396
Section 3.6 The expression is undefined for x 3 . This is also a boundary point. Sign of x 2 :
Sign of x 3 : Sign of
x2 x3
:
c.
60.
x2 0 x3 2, 3
Sign of
c.
x2 x2 0 0 d. x3 x3 , 2 3, , 2 3, 62.
x4 0 x 1 x40
Sign of x 1 : Sign of
x4 x 1
:
4
x4 0 a. x 1 4,1
x 9
:
x4 0 x2 9 0
b
x4 0 x2 9
x4 0 x2 9 ,
d.
x4 0 x2 9 , 0 0,
x2 0 x 2 16 x2 0
Sign of x 2 :
Sign of x 16:
:
2
Sign of
x2 x 2 16
0
1
x4 b 0 x 1 4,1
a.
x4 x4 0 0 c. d. x 1 x 1 , 4 1, , 4 1, 61.
x 2
x0 0 is a boundary point. The denominator is nonzero for all real numbers. Therefore, the only boundary point is 0.
x 4 4 is a boundary point. The expression is undefined for x 1 . This is also a boundary point.
4
0
x2 0 x3 2, 3
Sign of x 4 :
Sign of x 9:
a.
b
2
2 3
a.
Sign of x 4 :
c.
x4 0 x2 9 x4 0
63.
x0 0 is a boundary point. The denominator is nonzero for all real numbers. Therefore, the only boundary point is 0.
397
x2 0 x 2 16 , x2 0 x 2 16 0
b
d.
x2 0 x 2 16 , 0 0, x2 0 x 2 16
5 x 0 x 1 x5 0 x 1 The expression is undefined for x 1 . This is a boundary point. Find the real zeros of the related equation.
Chapter 3 Polynomial and Rational Functions x5 0 x 1 x5 0
The boundary points are 0 and 2.
x5 The boundary points are 1 and 5. Sign of x 1 :
Sign of x 5 :
Sign of
x5 x 1
:
1
66.
5
x2 The boundary points are 6 and 2.
Sign of x 2 : Sign of
x2 x6
:
6
Sign of x 2:
:
0
2
x2 x2
9 3x 0 x2 3x 9 0 x2 x3 0 x2 The expression is undefined for x 0 . This is a boundary point. Find the real zeros of the related equation. x3 0 x2 x3 0 x3 The boundary points are 0 and 3. Sign of x 2 :
Sign of x 3:
:
Sign of
x3 x2
0
3
The solution set is 3, .
2
The solution set is 6, 2 . 65.
The solution set is 2, .
2x 0 x6 x2 0 x6 The expression is undefined for x 6 . This is a boundary point. Find the real zeros of the related equation. x2 0 x6 x20
Sign of x 6 :
Sign of
The solution set is 1, 5 . 64.
Sign of x 2 :
67.
4 2x 0 x2 2x 4 0 x2 x2 0 x2 The expression is undefined for x 0 . This is a boundary point. Find the real zeros of the related equation. x2 0 x2 x20 x2
x2 x 2 0 x3 x 1 x 2 0 x3 The expression is undefined for x 3 . This is a boundary point. Find the real zeros of the related equation. x 1 x 2 0 x3 x 1 x 2 0 x 1 or x 2 The boundary points are 3 , 1 , and 2.
398
Section 3.6 Sign of x 3 : Sign of x 1 :
Sign of x 2 : Sign of
x 1 x 2 : x 3
The expression is undefined for x
is a boundary point. Find the real zeros of the related equation. x6 0 2x 7 x6 0 x6 7 The boundary points are and 6. 2
3 1 2
The solution set is 3, 1 2, . 68.
x2 2 x 8 0 x 1 x 2 x 4 0 x 1 The expression is undefined for x 1 . This is a boundary point. Find the real zeros of the related equation. x 2 x 4 0 x 1 x 2 x 4 0
Sign of 2 x 7 : Sign of x 6 : Sign of
Sign of x 1 :
Sign of x 4 : Sign of
x 2 x 4 : x 1
2
1
6
2
7 The solution set is , 6 . 2
70.
4
The solution set is 2,1 4, . 69.
x 6 : 2 x 7
7
x 2 or x 4 The boundary points are 2 , 1, and 4. Sign of x 2 :
7 . This 2
5 1 2x 7 5 1 0 2x 7 5 2x 7 0 2x 7 2x 7 12 2 x 0 2x 7 2 x 6 0 2x 7 x6 0 2x 7
4 1 3x 8 4 1 0 3x 8 4 3x 8 0 3x 8 3x 8 12 3x 0 3x 8 3 x 4 0 3x 8 x4 0 3x 8 The expression is undefined for x
8 . This 3
is a boundary point. Find the real zeros of the related equation. x4 0 3x 8 x40 x4 8 The boundary points are and 4. 3
399
Chapter 3 Polynomial and Rational Functions Sign of 3 x 8 :
Sign of 7:
:
Sign of x 4 :
x 4 : 3 x 8
Sign of
Sign of
8
3
7 The solution set is , . 3
2x 2 x2
73.
2x 20 x2 2x x2 2 0 x2 x2 4 0 x2 The expression is undefined for x 2 . This is a boundary point. The related equation has no real zeros. Sign of 4:
Sign of x 2 :
:
Sign of
4
x 2
3x 7
7
8 The solution set is , 4 . 3 71.
7
4
3
The solution set is , 2 .
Sign of x 5 :
Sign of x 2 :
3x 1 3x 7 3x 1 0 3x 7 3x 3x 7 0 3x 7 3x 7 7 0 3x 7 The expression is undefined for x
4 x 2 x5 4 x 20 x5 x5 4 x 2 0 x5 x5 3x 6 0 x5 x2 0 x5 The expression is undefined for x 5 . This is a boundary point. Find the real zeros of the related equation. x2 0 x5 x20 x 2 The boundary points are 5 and 2 .
2
72.
Sign of 3 x 7 :
Sign of
x 2 : x 5
5 2
The solution set is 5, 2 . 74. 7 . This 3
is a boundary point. The related equation has no real zeros.
400
3 x 4 x2 3 x 40 x2 x2 3 x 4 0 x2 x2 5 x 5 0 x2 x 1 0 x2
Section 3.6 The expression is undefined for x 2 . This is a boundary point. Find the real zeros of the related equation. x 1 0 x2 x 1 0
The denominator is nonzero for all real numbers. Therefore, the only boundary point is 3.
x 1 The boundary points are 2 and 1 .
Sign of
Sign of x 2 :
The solution set is , 3 .
Sign of x 1 :
x 1 : x 2
Sign of
77.
x2 0 x2 4 Find the real zeros of the related equation. x2 0 x2 4 x20
Sign of x 2 4:
:
Sign of
x2 x2 4
:
x3 x2 1
10 2 x2 x2 10 2 0 x2 x2 8 0 x2 The expression is undefined for x 2 . This is a boundary point. Sign of 10:
Sign of x 2 :
:
10
x 2
2
The solution set is 2, . 78.
2
The solution set is , 2 . 76.
Sign of x 1:
Sign of
x2 2 is a boundary point. The denominator is nonzero for all real numbers. Therefore, the only boundary point is 2.
3
The solution set is 2, 1 .
Sign of x 2:
2
2 1
75.
Sign of x 3:
x3 0 x2 1 Find the real zeros of the related equation. x3 0 x2 1 x3 0
4 1 x3 x3 4 1 0 x3 x3 3 0 x3 The expression is undefined for x 3 . This is a boundary point. Sign of 3:
Sign of x 3 :
:
Sign of
3
x 3
3
The solution set is 3, .
x3 3 is a boundary point.
401
Chapter 3 Polynomial and Rational Functions
79.
4 2 x3 x 4 2 0 x3 x 2 x 3 4x 0 x x 3 x x 3
The boundary points are 0,
Sign of 6 x 1 :
Sign of x:
:
6 x 1
x x 3
Sign of 3 x 2 :
:
2 3 x 2 x x 1
0
3
81.
3 6 4 x 1 x 3 6 x 4 x 1 3 6 0 x 4 x 1 3 x 1 6 x 4 0 x 4 x 1 x 4 x 1 3x 21 0 x 4 x 1
3 1 0
80.
1
2
2 The solution set is 0, 1, . 3
x 1 0 x 1 The boundary points are 3 , 1 , and 0.
Sign of
Sign of
The expression is undefined for x 0 and x 3 . These are boundary points. Find the real zeros of the related equation. 6 x 1 0 x x 3
Sign of x 3 :
Sign of x: Sign of x 1 :
6 x 1 0 x x 3
2 , and 1. 3
The solution set is 3, 1 0, .
3 x 7 0 x 4 x 1
2 4 x 1 x 2 4 0 x 1 x 4 x 1 2x 0 x x 1 x x 1
3 x 7 0 x 4 x 1 The expression is undefined for x 1 and x 4 . These are boundary points. Find the real zeros of the related equation. 3 x 7 0 x 4 x 1
2 3 x 2 0 x x 1
3 x 7 0
The expression is undefined for x 0 and x 1 . These are boundary points. Find the real zeros of the related equation. 2 3 x 2 0 x x 1
x7 The boundary points are 1, 4, and 7. Sign of x 1 : Sign of x 4 : Sign of x 7 :
3x 2 0 x
Sign of
2 3
3 x 7 : x 4 x 1 1
4
7
The solution set is ,1 4, 7 .
402
Section 3.6
82.
5 3 2 x 3 x 5 3 x2 x3 5 3 0 x2 x3 5 x 3 3 x 2 0 x 2 x 3 x 2 x 3
x 2 or x
The boundary points are Sign of 2 x 1 : 2
Sign of x 2 :
Sign of x 4 : 4
Sign of
Sign of x 3 :
Sign of 2 x 9 : Sign of
2x 9
x 2 x 3
4
84.
9 . 2
:
x 3 4 x 1 4 0 x 2 2 x 3 4 x 14 0
9
x 3 or x
2
83.
2
1
4
2
The expression is undefined for x 2 . This is a boundary point. Find the real zeros of the related equation.
3
3 x 4 x 1 4 0 x 2 2 x 3 4 x 1 4 0 x 2 2
2
The solution set is 2, 4 4, .
9 2
Sign of x 2 :
2x 9 0
The boundary points are 2, 3, and
x 2 2 x 1 : x 4
1 , 2, and 4. 2
2
2x 9 0 x 2 x 3 The expression is undefined for x 2 and x 3 . These are boundary points. Find the real zeros of the related equation. 2x 9 0 x 2 x 3 x
1 2
1 4
9 The solution set is 2, 3 , . 2
The boundary points are 2 , 3, and
1 . 4
Sign of x 2 :
2 x 2 x 12 0 x 4 4 x 2 2 x 12 0 x 4 4
Sign of 4 x 1 :
2
4
Sign of x 3 :
x 3 4 x 1 : x 2 4
Sign of
2
2
The expression is undefined for x 4 . This is a boundary point. Find the real zeros of the related equation.
1 4
1 The solution set is 3, . 4
x 2 2 x 12 0 x 4 4 x 2 2 x 1 2 0
1 32 t 2 216 t 0 2 s t 16t 2 216t
85. a. s t
403
3
Chapter 3 Polynomial and Rational Functions
b. t
b 216 216 6.75 2a 2 16 32
1 3 The solution set is , . 4 4 The player will be more than 3 ft in the air between 0.25 sec and 0.75 sec after leaving the ground.
The shell will explode 6.75 sec after launch. c.
16t 2 216t 200 16t 2 216t 200 0 2t 2 27t 25 0 Find the real zeros of the related equation. 2t 2 27t 25 0 t 1 2t 25 0
87. d v 0.06v 2 2v
0.06v 2 2v 250 0.06v 2 2v 250 0 6v 2 200v 25,000 0 3v 2 100v 12,500 0 Find the real zeros of the related equation. 3v 2 100v 12,500 0 3v 250 v 50 0 250 v or v 50 3
25 t 1 or t 2 The boundary points are 1 and
25 . 2
Sign of t 1 :
Sign of t 1 2t 25 :
Sign of 2t 25 :
1
Sign of 3v 250 :
25 The solution set is 1, . 2 The spectators can see the shell between 1 sec and 6.75 sec after launch.
Sign of 3v 250 v 50 :
3
1500t 2 60,000t 450,000 0 t 2 40t 300 0 Find the real zeros of the related equation. t 2 40t 300 0
t 10 t 30 0 t 10 or t 30 Sign of t 10 :
Sign of t 30 :
1 3 and . 4 4
Sign of 4t 1 4t 3 :
250 50
1500t 2 60,000t 10,000 460,000
1 3 or t 4 4
Sign of 4t 3 :
88. P t 1500t 2 60,000t 10,000
16t 2 16t 3 16t 2 16t 3 0 16t 2 16t 3 0 Find the real zeros of the related equation. 16t 2 16t 3 0 4t 1 4t 3 0
Sign of 4t 1 :
250 The solution set is , 50 . 3 The car will stop within 250 ft if the car is traveling less than 50 mph.
1 32 t 2 16 t 0 2 s t 16t 2 16t
The boundary points are
86. a. s t
t
Sign of v 50 :
25 2
b.
1
3
4
4
Sign of t 10 t 30 :
10 30
The solution set is 10, 30 . The population will be greater than 460,000 organisms between 10 hr and 30 hr after the culture is started.
404
Section 3.6 89. a. The degree of the numerator is less than the degree of the denominator. The horizontal asymptote is y 0 and means that the temperature will approach 0C as time increases without bound. b.
91. Let w represent the width. Then 1.2w represents the length. 72 w 1.2 w 96
72 1.2 w2 96 60 w2 80
320 5 x 3x 10 320 5 0 2 x 3 x 10 Find the real zeros of the related equation. 320 5 0 2 x 3x 10 320 5 2 x 3 x 10 5 x 2 15 x 50 320
60 w 80
2
2 15 w 4 5 The width should be between 2 15 ft and 4 5 ft. This is between approximately 7.7 ft and 8.9 ft. 92. Let w represent the width. Then 1.5w represents the length. 480 w 1.5w 720
480 1.5w2 720
5 x 2 15 x 270 0
320 w2 480
x 2 3 x 54 0
320 w 480
x 9 x 6 0
8 5 w 4 30 The width should be between 8 5 yd and
x 9 or x 6 More than 6 hr is required for the temperature to fall below 5C . 90. a. S
4 30 yd. This is between approximately 17.9 yd and 21.9 yd.
2d d d r1 r2
93. 9 x 2 0
x2 9 0 Find the real zeros of the related equation. x2 9 0 x 3 x 3 0 x 3 or x 3 The boundary points are 3 and 3.
2 200 60 200 200 r2 501 400 60 200 4 r2
Sign of x 3 : Sign of x 3 :
400r2 60 4r2 200
Sign of x 3 x 3 : 3 3
400r2 240r2 12,000
3, 3
160r2 12,000 r2 75 The motorist must travel at least 75 mph on the return trip to average at least 60 mph for the round trip.
94. 1 t 2 0
t2 1 0 Find the real zeros of the related equation. t2 1 0
b. No.
t 1 t 1 0 t 1 or t 1 The boundary points are 1 and 1.
405
Chapter 3 Polynomial and Rational Functions Sign of x 1 :
Sign of x 1 x 1 :
Sign of x 1 :
1
1,1
The boundary points are 6 and Sign of x 6 :
Sign of x 6 2 x 3 :
Sign of 2 x 3 :
1
6
95. a 5 0 Find the real zeros of the related equation. a2 5 0
3
2
2
3 The solution set is , 6 , . 2
a 5 a 5 0
98. 4 x 2 7 x 2 0 Find the real zeros of the related equation. 4x2 7 x 2 0
a 5 or a 5 The boundary points are 5 and
5.
Sign of x 5 : Sign of x 5 x 5 :
x 2 4 x 1 0
Sign of x 5 :
, 5 5,
5
x 2 or x
Sign of x 2 :
Sign of x 2 4 x 1 :
1
1 The solution set is , 2 , . 4
99. 2 x 2 9 x 18 0 Find the real zeros of the related equation. 2 x 2 9 x 18 0
7.
Sign of x 7 : Sign of x 7 x 7 : Sign of x 7 :
7
1 . 4
4
a 7 or a 7
2
u 7 u 7 0
Sign of 4 x 1 :
96. u 2 7 0 Find the real zeros of the related equation. u2 7 0
, 7 7,
1 4
The boundary points are 2 and
5
The boundary points are 7 and
3 . 2
x 6 2 x 3 0 x 6 or x
7
3 2
The boundary points are 6 and Sign of x 6 :
Sign of x 6 2 x 3 :
Sign of 2 x 3 :
97. 2 x 9 x 18 0 Find the real zeros of the related equation. 2 x 2 9 x 18 0 2
6
x 6 2 x 3 0
3 . 2
3 2
3 The solution set is , 6 , . 2
3 x 6 or x 2
406
Section 3.6 100. 4 x 2 7 x 2 0 Find the real zeros of the related equation. 4x2 7 x 2 0
x 2 4 x 1 0 x 2 or x
1 4
Sign of x 2 4 x 1 :
Sign of 4 x 1 :
2
Sign of x 2 :
:
2x
x 1
:
1 0
2
3
Sign of b x :
1
Sign of x c :
4
Sign of
3
x a 2 b x x c 3 :
a
b
c
b. b, c c. , a a, b c, 104. a. g x
a x x b2 c x 5
Sign of a x :
Sign of x b :
2
Sign of c x : 5
a x x b : c x 2
Sign of
2 0
5
a
b
c
b. , a c,
The solution set is , 2 0, . 102.
Sign of x a :
x0 The boundary points are 2 and 0.
x2
2
3x 0 x2 The expression is undefined for x 2 . This is a boundary point. Find the real zeros of the related equation. 3x 0 x2 3x 0
Sign of
Sign of 2 x :
103. a. f x x a b x x c
1 The solution set is , 2 , . 4
3x
The solution set is , 1 0, .
Sign of x 2 :
Sign of 3 x :
Sign of
1 The boundary points are 2 and . 4
101.
Sign of x 1 :
c. a, b b, c
2x 0 x 1 The expression is undefined for x 1 . This is a boundary point. Find the real zeros of the related equation. 2x 0 x 1 2x 0 x0 The boundary points are 1 and 0.
105. The solution set to the inequality f x 0 corresponds to the values of x for which the graph of y f x is below the x-axis. 106. The solution set to the inequality f x 0 corresponds to the values of x for which the graph of y f x is on or above the x-axis.
407
Chapter 3 Polynomial and Rational Functions f 2 3
107. Both the numerator and denominator of the rational expression are positive for all real numbers x. Therefore, the expression cannot be negative for any real number.
f x 0
111.
x3 Find the real zeros of the related equation. 2x 6 2 0
f x 0
f x 0
32
f x 0
The solution set is , 32 .
112.
The solution set is 3, 5 . 110.
4 x 60 The radical is real when 4 x0 x 4 x4 Find the real zeros of the related equation. 4 x 6 0 4 x 6 4 x 36 x 32 x 32 The boundary point is 32 . f 45 1 f 21 1
2x 6 2 2x 6 4 2 x 10 x5 The boundary point is 5. 7 15 f 1 f 1 2 2
5
5 The solution set is , 7 . 3
2x 6 2 0 The radical is real when 2x 6 0 2x 6
f x 0
f x 0 7
108. The rational expression is not defined for x 1 . Therefore, x 1 is not included in the solution set. At x 3 , the expression equals zero, indicating that 3 is in the solution set. The solution set is (1, 3]. 109.
f 10 1
3x 5 4 0 The radical is real when 3x 5 0 3x 5 5 x 3 Find the real zeros of the related equation. 3x 5 4 0
5 x 7 0 The radical is real when 5 x 0 x 5 x5 Find the real zeros of the related equation. 5 x 7 0 5 x 7 5 x 49 x 44 x 44 The boundary point is 44 . f 59 1 f 31 1
3x 5 4 3x 5 16 3x 21 x7 The boundary point is 7.
f x 0
44
f x 0
The solution set is , 44 .
408
Section 3.6
113.
1
0
1
0 x3 5 The radical is real when x3 0 x3 The denominator is zero when x3 5 0
114.
x2 4 The radical is real when x20 x2 The denominator is zero when x2 40 x2 4 x 2 16 x 18 The related equation has no zeros. The boundary point is 18. f 11 1 f 27 1 f x 0 f x 0 18
x3 5 x 3 25 x 28 The related equation has no zeros. The boundary point is 28. f 19 1 f 39 1 f x 0 f x 0 28
The solution set is 2,18 .
The solution set is 3, 28 .
115. 3 x 2 6 x 5 5 Rewrite the left inequality f x .
Find the real zeros of the related equation.
x 2 x 4 0
3 x 2 6 x 5
x 2 or x 4
0 x2 6 x 8 x2 6 x 8 0
x 2 x 4 0 Rewrite the right inequality g x .
Find the real zeros of the related equation.
x 6x 5 5
x x 6 0
x 6x 0
x 0 or x 6
2
2
x x 6 0 The boundary points are 0, 2, 4, and 6. f 1 15
f x 0 g 1 7 g x 0
f 1 3
f 3 1
f 5 3
f 1 15
g x 0
g x 0
g x 0
g x 0
f x 0 f x 0 f x 0 f x 0 g 1 5 g 3 9 g 5 5 g 1 7
0 2 The solution set is 0, 2 4, 6 .
4
6
409
Chapter 3 Polynomial and Rational Functions 116. 8 x 2 4 x 3 15 Rewrite the left inequality f x .
Find the real zeros of the related equation.
8 x2 4 x 3
x2 4 x 5 0
0 x2 4 x 5
x 5 x 1 0
x 4x 5 0
x 5 or x 1
Rewrite the right inequality g x .
Find the real zeros of the related equation.
2
x 2 4 x 3 15
x 2 4 x 12 0
x 2 4 x 12 0
x 6 x 2 0 x 6 or x 2
The boundary points are 6 , 5 , 1, and 2.
f 7 16 f 5.5 3.25 f 0 5 f 1.5 3.25 f 3 16 f x 0 f x 0 f x 0 f x 0 f x 0 g 7 9 g 5.5 3.75 g 0 12 g 1.5 3.75 g 3 9 g x 0 g x 0 g x 0 g x 0 g x 0 6 5 1 2 The solution set is 6, 5 1, 2 .
117.
x2 4 5
118.
x 2 1 17
x2 4 5 0
x 2 1 17 0
Find the real zeros of the related equation. x2 4 5 0
Find the real zeros of the related equation. x 2 1 17 0
x2 4 5 x2 4 5
x 2 1 17 or
x 2 4 5
x 2 1 17
x2 9 x 2 1 x 3 The boundary points are 3 and 3. f 4 7 f 0 1 f 4 7 f x 0 f x 0 f x 0 3 3 The solution set is 3, 3 .
or
x 2 1 17
x 2 16 x 2 18 x 4 The boundary points are 4 and 4. f 5 9 f 0 16 f 5 9 f x 0 f x 0 f x 0 4 4 The solution set is 4, 4 .
410
Section 3.6 x 2 18 2
119.
x 2 18 2 0 Find the real zeros of the related equation. x 2 18 2 0
x 2 18 2
x 2 18 2
x 2 20
or
x 2 18 2 x 2 16
x 20
x 4
x 2 5 The boundary points are 2 5 , 4 , 4, and 2 5 . f 5 5 f x 0
f 4.1 0.81
f 0 16
f x 0
2 5
f x 0
4
f 4.1 0.81 f x 0
4
f 5 5
f x 0 2 5
The solution set is , 2 5 4, 4 2 5, . x2 6 3
120.
x2 6 3 0 Find the real zeros of the related equation. x2 6 3 0
x2 6 3
x2 6 3
x2 9
x2 3
x 3
x 3
The boundary points are 3 , 3 ,
3 , and 3.
f 4 7 f x 0
f 2 1 f x 0
f 2 1 f x 0
3
f 0 3 f x 0
3
3
or
x 2 6 3
f 4 7 f x 0 3
The solution set is , 3 3, 3 3, . c. The real zeros are approximately 7.6 , 1.5 , and 1.6.
121. a. 0.552 x 3 4.13x 2 1.84 x 3.5 6.7
0.552 x 4.13 x 1.84 x 10.2 0 3
2
d. , 7.6 1.5,1.6
b.
122. a. 0.24 x 4 1.8 x 3 3.3x 2 2.84 x 1.8 4.5
0.24 x 4 1.8 x 3 3.3x 2 2.84 x 6.3 0
411
Chapter 3 Polynomial and Rational Functions c. The radius should be no more than 1.9 in. to keep the amount of aluminum to at most 90 in.2.
b.
124. a.
c. The real zeros are approximately 5.6 and 0.9. d. , 5.6 0.9, b. 123. a.
b.
c. It is safe to give a second dose 10.8 hr after the first dose.
Problem Recognition Exercises: Solving Equations and Inequalities 1 1 x5 2 2 4 9 1 x7 2 4 18 x 28
1.
2 x2 6 x 5
2.
2 x2 6 x 5 0
x
28 x 18 28, 18
6
6 4 2 5 2 2 2
6 76 4 6 2 19 3 19 4 2 3 19 2
412
Problem Recognition 3.
50 x 3 25 x 2 2 x 1 0
6. 5 y
25 x 2 2 x 1 2 x 1 0
y 5
x
1 2
3 y 3
5, 1
25 x 2 1
or
3 y 4 7
y 1
2 x 1 25 x 2 1 0
2x 1
and
1 25 1 x 5
7. 5t 4 2 7
x2
5t 4 5 5t 4 5
1 1 , 5 2
or
5t 4 5
5t 9
5t 1
9 5
t
t
1 5
1 9 , 5 5
5 x x 3 0 4. 2 x 2
x x 3 0 x2 The expression is undefined for x 2 . This is a boundary point. Find the real zeros of the related equation. 2
8. 3 4 x 5 x 2 3 2 x 2 4 x 1
3 4 x 5 x 6 4 x 2 4 x 1 3 4 x 5 3 x 6 2 x 5
x x 3 0 x2
3 4 x 15 x 30 2 x 10
x x 3 0
3 64 x 120 2 x 10 66 x 133 133 x 66
2
3 4 16 x 30 2 x 10
2
x 0 or x 3 The boundary points are 2 , 0, and 3. Sign of x 2 : Sign of x :
Sign of x 3 : 2
Sign of
x x 3
:
133 66
2
x2
2
0
9.
10 x 2 x 14 29 x 2 100 20 x 2 140 x 29 x 2 100
3
The solution set is , 2 0, .
49 x 2 140 x 100 0
7 x 102 0
5. 4 m 4 5 2 4
77
7 x 10 0 10 x 7
m4 3 m 4 34 m 4 81 m 77
10 7
413
Chapter 3 Polynomial and Rational Functions 5 3y 2 y 2 14 y 2 y 4 y 2 y 2y 8
10.
5 3y 2 y 2 14 y y 4 y 2 y 4 y 2 5 y 2 3 y y 4 2 y 2 14 y y 4 y 2 y 4 y 2 y 4 y 2 5 y 10 3 y 2 12 y 2 y 2 14 y y 2 3 y 10 0 y 2 3 y 10 0
y 5 y 2 0 y 5 or y 2
5 ; The value 2 does not check. x x 14 40
11.
13. x 0.15 x 0.05
x 0.15 x 0.05 or x 0.15 x 0.05
x 14 x 40 0 Find the real zeros of the related equation. x 2 14 x 40 0 2
0.15 0.05
x 0.15 x 0.05 2 x 0.1
x 4 x 10 0 Sign of x 4 :
Sign of x 4 x 10 :
Sign of x 10 :
4
4,10 12.
14.
1 0 x 14 x 40 1 0 x 4 x 10 The expression is undefined for x 4 and x 10 . These are boundary points. The related equation has no real zeros.
Sign of x 10 :
1
:
4,10
x 4 x 10
t 1
t 1 36
10
t 1 0
and
t 1 6 t 37
15. n1 2 7 10
2
Sign of
t 1 5 1
1, 37
Sign of x 4 :
x 0.05
0.05
x 4 or x 10 The boundary points are 4 and 10.
n1 2 3
9
n9
16. 4 x 3 x x 2 x 5 0 2
3
x x 3 x 2 x 5 0 Find the real zeros of the related equation. 2
3
x x 3 x 2 x 5 0 x 0 , x 3 , x 2 , or x 5 The boundary points are 2 , 0, 3, and 5. 2
4 10
414
3
Problem Recognition Sign of x 2 : 2
Sign of x 3:
Sign of x 5 :
Sign of x :
3
Sign of x x 3 x 2 x 5 : 2
3
2
2 0, 3 5,
0
3
15u 2 7u 2 0
5u 1 3u 2 0 1 2 or u 5 3 Back substitute x 1 for u. 1 2 or x 1 x 1 5 3 u
5
x 1 5
17. 2 x 5 x 3 4 x 2 3x
x 5
2 x 5 x 15 4 x 8 3x
3 2
15 8 No solution
21. 8 x 3 10 7
8 x 3 3 Since the absolute value of a number cannot be negative, this inequality has no solution.
7 x 29 3 x 7 x 29 x 3
7 x 29 x 2 6 x 9 x 2 x 20 0
22. 2 x 1
x 2 x 20 0
x 1 16
17
19. x 2 9 5 x 2 9 14 0
Let u x 2 9 .
23.
u 2 5u 14 0
u 2 or u 7 Back substitute x 2 9 for u. x 9 2
or
x2 9 7
x2 7
x 2 16
x 7
x 4
4, 7 1
x 17
x3 3x 2 6 x 8 x3 3x 2 6 x 8 0 Find the real zeros of the related equation. x3 3x 2 6 x 8 0 Factors of 8 1, 2, 4, 8 Factors of 1 1 1, 2, 4, 8 1 1 3 6 8
u 2 u 7 0
2
16
x 1 84 3
x 4 or x 5 5 ; The value 4 does not check. 2
34
x 13 4 8
x 4 x 5 0
x
1
3 5, 2
7 x 15 7 x 8
18.
3 x 1 2
1
1 2 8 1 2 8
0
x 1 x 2 2 x 8 0 x 1 x 2 x 4 0
2
20. 2 7 x 15 x 0
15 x 2 7 x 1 2 0 Let u x 1 .
x 1 , x 2 , or x 4
415
Chapter 3 Polynomial and Rational Functions The boundary points are 2 , 1, and 4. Sign of x 2 :
Sign of x 1 x 2 x 4 :
Sign of x 4 :
2
, 2 1, 4 24.
2
Sign of x 1 :
1
The expression 5 x 7 0 for all real numbers except where 5 x 7 2 0 .Therefore, the solution set is 7 all real numbers except . 5 7 7 , , 5 5
4
3 x 1 x5 3 x 1 0 x5 3 x x5 1 0 x5 x5 2 x 2 0 x5 x 1 0 x5 The expression is undefined for x 5 . This is a boundary point. Find the real zeros of the related equation. x 1 0 x5 x 1 0
27.
11 3 x 3 x 11 3 x 11 or x 14 x 14 , 8 14, 28. 4 x 3 8
Sign of x 1 :
x 1 : x 5
Sign of
29.
5 1
The solution set is 5, 1 . 25. 15 3 x 1 2 x x 18
15 3x 3 2 x x 18
26.
or
1 2 5 x x 1 3 5 6 10 x 12 25 x 30 15 x 42 42 x 15 14 x 5 14 , 5
30. 2 2 x 1 2 8
3x 18 3x 18
2 2 x 1 10
00
,
3 x 11 x 8 x 8
4 x 12 8 4 x 4 x 1 , 4 1,
x 1 The boundary points are 5 and 1 . Sign of x 5 :
2 3 x 9
2x 1 5
25 x 70 x 49 2
25 x 2 70 x 49 0
3, 2
5 x 7 2 0 416
5 2 x 1 5 6 2 x 4 3 x 2
7 x 3 x 4 x 4
Section 3.7 Section 3.7 Variation 1.
9 3 k 4 32 9 32 k 4 3
24 2.
24 k
y
24 24 1
y
24 12 2
y
24 8 3
y
24 6 4
y
24 4 6
b. y is one-half its original value.
7 7 k 10 5 7 5 k 10 7
c. y is one-third its original value. d. decreases 11. C kr
1 k 2
k n
13. C
1 2
3.
24 x
10. a. y
0.24
x
0.24 x 1.68
12. I kA 14. t
15. V khr 2
1.68
e. increases
17. E
ks
19. c
kmn t3
n
k r
16. V klw 18. n
k 2 E2
x 7
49 4.
x 49
0.32
21.
2.56 y
0.32 y 2.56 y 8 y 64
64 23. 5. directly
6. inversely
7. constant; variation
8. jointly
9. a. y 2 x
y 2 1 2
y 2 2 4
y 2 3 6
y 2 4 8
y 2 5 10
25.
b. y is also doubled. c. y is also tripled. d. increases
e. decreases
417
y kx
kuv 20. d 3 T 22.
y kx
20 k 8
42 k 10
20 k 8 5 k 2
42 k 10 21 k 5
k q k 54 18 k 972
k x k 200 50 k 10,000 T
p
24.
y kwv
26. N ktp
40 k 40 0.2
15 k 2 2.5
40 k 8
15 k 5
k5
k 3
Chapter 3 Polynomial and Rational Functions 27. a.
y kx
d.
4 k 10 4 k 10 2 k 5 2 2 y x 5 2 5 5
30. Let S represent the number of people served. Let w represent the weight of the ham. S kw
20 k 8
k x k 4 10 k 40 40 40 y 8 x 5
b. y
28. a.
20 k 8 k 2.5 S 2.5w a. S 2.5w 2.5 10 25 people b. S 2.5w 2.5 15 37 people (rounded down to have more food per person)
y kx 1 24 k 2
c. S 2.5w 2.5 18 45 people
k 48
d.
y 48 x 48 3 144 b.
A 4.5w 135 4.5w 135 w 4.5 w 30 lb
k x k 24 1 2 k 12 12 12 y 4 x 3 y
S 2.5w 30 2.5w 30 w 2.5 w 12 lb
31. Let C represent the average daily cost to rent a car. Let m represent the number of miles driven. k C m k 0.8 100 k 80
29. Let A represent the amount of the medicine. Let w represent the weight of the child. A kw
180 k 40
C
80 m
180 k 40 k 4.5 A 4.5w
a. C
80 80 $0.40 per mile m 200
a. A 4.5w 4.5 50 225 mg
b. C
80 80 $0.27 per mile m 300
c. C
80 80 $0.20 per mile m 400
b. A 4.5w 4.5 60 270 mg c. A 4.5w 4.5 70 315 mg
418
Section 3.7
d.
34. Let p represent the amount of pollution entering the atmosphere. Let n represent the number of people. p kn
80 m 80 0.16 m 0.16m 80 80 m 500 mi 0.16 C
71,000 k 100,000 71,000 k 100,000 k 0.71 d 0.71n 0.71 750,000
32. Let C represent the number of books sold per month. Let p represent the price of a book. k C p k 1500 8 k 12,000
C a. C
532,500 tons 35. a. Let d represent the stopping distance of the car. Let s represent the speed of the car. d ks 2
170 k 50
12,000 p
2
170 k 2500 k 0.068
12,000 12,000 1000 books p 12
d 0.068s 2 0.068 70 333.2 ft 2
12,000 12,000 b. C 800 books p 15 c. C d.
b.
d 0.068s 2 244.8 0.068s 2
12,000 12,000 2000 books p 6
244.8 s2 0.068 3600 s 2
12,000 p 12,000 1200 p 1200 p 12,000 12,000 p $10 1200 C
s 60 mph 36. a. Let A represent the area of a picture projected on a wall. Let d represent the distance from the projector to the wall. A kd 2
36 k 15
33. Let d represent the distance a bicycle travels in 1 min. Let r represent the rpm of the wheels. d kr 440 k 60 440 k 60 22 k 3 22 22 d r 87 638 ft 3 3
2
36 k 225 k 0.16 A 0.16d 2 0.16 25 100 ft 2 2
419
Chapter 3 Polynomial and Rational Functions A 0.16d 2
b.
b.
144 0.16d 2 144 d2 0.16 900 d 2 d 30 ft 37. Let t represent the number of days to complete the job. Let n represent the number of people working on the job. k t n k 12 8 k 96
39. Let I represent the current. Let V represent the voltage. Let R represent the resistance. kV I R k 90 9 10 90 90k
k 1
96 t n a. t b.
I
96 96 6.4 days n 15
96 n 96 8 n 8n 96 n 12 people
0.0005 50k k 0.00001
38. Let y represent the yield of the bond. Let p represent the price of the bond. k y p k 0.05 120 k6
a. y
V 160 32 A 5 R
40. Let R represent the resistance of the wire. Let represent the length of the wire. Let d represent the diameter of the wire. k R 2 d k 50 0.0125 0.2 2
t
y
6 p 6 0.075 p 0.075 p 6 p $80 y
R
0.00001 0.00001 40 0.04 d2 0.1 2
41. Let I represent the amount of interest owed. Let P represent the amount of the principal borrowed. Let t represent the amount of time (in years) that the money is borrowed. I kPt
6 p
480 k 4000 2 480 8000k
6 6 0.06 6% p 100
480 k 8000 k 0.06 I 0.06 6000 4 $1440
420
Section 3.7 45. Let s represent the speed of the canoe. Let represent the length of the canoe. sk
42. Let I represent the amount of interest earned. Let P represent the amount of the principal invested. Let t represent the amount of time (in years) that the money is invested. I kPt
6.2 k 16
750 k 5000 6
6.2
k 16 6.2 k 4 k 1.55
750 30,000k 750 k 30,000 k 0.025
s 1.55 1.55 25 7.75 mph
I 0.025 8000 4 $800
46. Let P represent the period of the pendulum. Let represent the length of the pendulum. Pk
43. Let B represent the BMI of an individual. Let w represent the weight of the individual. Let h represent the weight of the individual. kw B 2 h k 150 21.52 70 2
1.8 k 0.81 1.8
k 0.81 1.8 k 0.9 k2
105,448 150k 105,448 k 150 k 703 B
P 2 2 1 2 sec
703 180
68 2
47. Let C represent the cost to carpet the room. Let represent the length of the room. Let w represent the width of the room. C k w
27.37
3870 k 1015
44. Let S represent the strength of the beam. Let w represent the width of the beam. Let t represent the thickness of the beam. Let represent the length of the beam. kwt 2 S
417
k 6 2
3870 k 150 k 25.8 C k w 25.8 18 24 $11,145.60
2
48. Let C represent the cost to tile the kitchen. Let represent the length of the kitchen. Let w represent the width of the kitchen. C k w
48 20,016 24k 20,016 k 24 k 834 S
1104 k 1012
834 12 4 72
1104 k 120 k 9.2
2
2224 lb
C k w 9.2 2014 $2576
421
Chapter 3 Polynomial and Rational Functions 49. The data appears to show direct variation. Use the first row to find the constant of variation. y kx
k x k 2 4 k 8 y
16 k 5 16 k 5 k 3.2
8 x Check the remaining values in the equation. They check. y
y 3.2 x Check the remaining values in the equation. They check.
53. Make a table of values. x 1 3 y 3 1
50. The data appears to show direct variation. Use the first row to find the constant of variation. y kx
The data appears to show indirect variation. Use the first point to find the constant of variation. k y x k 3 1 k 3
24 k 5 24 k 5 k 4.8 y 4.8 x Check the remaining values in the equation. They check.
3 x Check the other point in the equation. It checks. y
51. The data appears to show indirect variation. Use the first row to find the constant of variation. k y x k 6 2 k 12
54. Make a table of values. x 12 24 36 48 60 y 1 2 3 4 5
The data appears to show direct variation. Use the first point to find the constant of variation. y kx
12 x Check the remaining values in the equation. They check. y
1 k 12 1 12 1 y x 12 Check the other points in the equation. They check. k
52. The data appears to show indirect variation. Use the first row to find the constant of variation.
55. formulas a and c 56. formulas b and c
422
Section 3.7 57. The variable P varies directly as the square of v and inversely as t.
a. T 2 1.66 1019 d 3
58. The variable E varies directly as the square of c and inversely as the square root of b.
T 671 Earth days
T 1.66 1019 1.5 9.3 107
d3
2232 T2 3 1.66 1019 1.66 1019 d 67 million miles
61. I
S 4 r 2 4 20 1600 m 2 2
I
b. The surface area is 4 times as great. 2 Doubling the radius results in 2 times the surface area of the sphere. 1 the 4 intensity at 10 m. This is because the energy from the light is distributed across an area 4 times as great.
62. y
y
k
10d
2
k 1 k 2 100 d 2 100d
k x3 k
3
1 as great. 100
k k 64 3 3 x x 3 4
x 4 y will be 64 times as great.
d. Let I represent the intensity of light from a source. Let d represent the distance of the light from the source. k I 2 d k 200 10 2
63. y
y
k 20,000
kx 2 w4 k 2 x
2 w
y will be
20,000 20,000 I 50 lux d2 202 T kd
k d2
The intensity is
c. The intensity at 20 m should be
60.
T2 1.66 1019
d3
2
400 k 100 k 4
2
3
b. T 2 1.66 10 19 d 3
59. a. Let S represent the surface area of the sphere. Let r represent the radius of the sphere. S kr 2
400 k 10
64. y
3
365 2 k 9.3 107 2 365 1.66 10 19 k 3
y
9.3 10
2
4
2
4
4
1 its original value. 4
kx 5 w2 k 2 x
2 w
1 kx 16w 4 w k 4 x2
5
2
8 kx 4w w
k 32 x 5 2
5
2
y will be 8 times its original value.
7 3
T 2 1.66 1019 d 3
423
Chapter 3 Polynomial and Rational Functions 65. y kxw3
66. y kx 4 w
3 x x y k 3w k 27 w3 9 kxw3 3 3 y will be 9 times its original value.
x4 1 x y k 4 w k kx 4 w 4 w 4 64 256
4
y will be
1 times its original value. 64
Chapter 3 Review Exercises 1. f x x 5 2 x 5 2 2
2
g. The axis of symmetry is the vertical line through the vertex: x 4 .
Vertex: h, k 5, 2
h. The minimum value is 1 .
2. a. f x x 8 x 15
i. The domain is , .
2
15
f x x 8x 2
The range is 1, .
2
1 2 8 16
3. a. f x 2 x 2 4 x 6
f x 2 x 2 2 x
f x x 2 8 x 16 16 15
6
2
1 2 2 1
f x x 4 1 2
b. Since a 0 , the parabola opens upward.
f x 2 x 2 x 1 2 1 6 f x 2 x 2 2 x 1 1 6
c. The vertex is h, k 4, 1 .
2
d. f x x 2 8 x 15
f x 2 x 1 8 2
0 x 2 8 x 15
b. Since a 0 , the parabola opens downward.
0 x 3 x 5 x 3 or x5 The x-intercepts are 3, 0 and 5, 0 .
c. The vertex is h, k 1, 8 . d. f x 2 x 2 4 x 6
e. f 0 0 8 0 15 15 2
0 2 x 2 4 x 6
The y-intercept is 0,15 .
0 x2 2 x 3 0 x 1 x 3
f.
x 1 or x 3 The x-intercepts are 1, 0 and 3, 0 . e. f 0 2 0 4 0 6 6 2
The y-intercept is 0, 6 .
424
Chapter 3 Review Function A is a quadratic function with a negative leading coefficient. The graph of the parabola opens downward, so the vertex is the maximum point on the function. The xcoordinate of the vertex is the value of x that will maximize the area. b 180 180 x 45 2a 2 2 4
f.
y 180 2 x 180 2 45 180 90 90 The dimensions are 45 yd by 90 yd.
g. The axis of symmetry is the vertical line through the vertex: x 1 .
6. a. V p 100 p 1 p 100 p 100 p 2
h. The maximum value is 8.
100 p 2 100 p The value of p at which the variance will be at a maximum is the p-coordinate of the vertex. b 100 100 1 p 2a 2 100 200 2
i. The domain is , .
The range is , 8 .
4. a. f x 2 x 2 12 x 19
f x 2 x2 6 x
19
b. The maximum variance is the value of V p at the vertex.
2
1 2 6 9
2
1 1 1 V 100 100 2 2 2
f x 2 x 6 x 9 2 9 19 f x 2 x 6 x 9 9 19 2 2
f x 2 x 3 1 2
Vertex: h, k 3,1
100 100 25 50 25 4 2
7 a.
b. Since the vertex of the parabola is above the x-axis and the parabola opens upward, the parabola cannot cross or touch the xaxis. Therefore, there are no x-intercepts.
E a 0.476a 2 37.0a 44.6
5. We know that 2 x y 180
b. The age when the yearly expenditure is the greatest is the a-coordinate of the vertex. 37.0 b 37.0 a 39 yr 2a 2 0.476 0.952
y 180 2 x Let A represent the area. A xy A x x 180 2 x 180 x 2 x 2
c. The maximum yearly expenditure is the E a value at the vertex.
x 180 x 2
E 39 0.476 39 37.0 39 44.6 2
$674
425
Chapter 3 Polynomial and Rational Functions 8. a. f x 4 x 3 16 x 2 25 x 100
b. 0 x 4 10 x 2 9
The leading term is 4x 3 . The end behavior is up to the left and down to the right.
x 3, x 3, x 1, x 1 The zeros of the function are 3, 3 , 1, and 1 (each with multiplicity 1).
0 4 x 3 16 x 2 25 x 100 0 4 x 2 x 4 25 x 4
c. The x-intercepts are 3, 0 , 3, 0 ,
1, 0 , and 1, 0 .
0 x 4 2 x 5 2 x 5 x 4, x
0 x 3 x 3 x 1 x 1
b. 0 4 x 3 16 x 2 25 x 100
0 x 4 4 x 2 25
0 x2 9 x2 1
d. f 0 0 10 0 9 9 4
5 5 ,x 2 2
2
The y-intercept is 0, 9 .
The zeros of the function are 4,
5 , and 2
e. f x x 10 x 9 4
2
x 4 10 x 2 9 f x is even.
5 (each with multiplicity 1). 2 f.
5 c. The x-intercepts are 4, 0 , , 0 , and 2
5 , 0 . 2 d. f 0 4 0 16 0 25 0 100 3
2
100 The y-intercept is 0, 100 . e. f x
10. a. f x x 4 3 x 3 3 x 2 11x 6
4 x 16 x 25 x 100 3
2
The leading term is x 4 . The end behavior is up to the left and up to the right.
4 x 16 x 25 x 100 f x is neither even nor odd. 3
2
b. Factor the polynomial to find the zeros. Possible rational zeros: Factors of 6 1, 2, 3 1, 2, 3 Factors of 1 1 1 1 3 3 11 6
f.
1
1 2
5
6
2 5
6
0
f x x 1 x 3 2 x 2 5 x 6
Factor the quotient. Possible rational zeros: Factors of 6 1, 2, 3 Factors of 1
9. a. f x x 4 10 x 2 9
The leading term is x 4 . The end behavior is up to the left and up to the right.
426
Chapter 3 Review 1 1
2 5 6 1 1
6
1 6
0
1
b. 0 x 5 8 x 4 13 x 3
f x x 1 x x 6 2
0 x 3 x 2 8 x 13
2
x
f x x 1 x 2 x 3 2
8
82 4 113 2 1
x 1, x 2, x 3 The zeros of the function are 2, 3 (each with multiplicity 1) and 1 (with multiplicity 2).
8 12 8 2 3 4 3 2 2 x 0, x 4 3, x 4 3 The zeros of the function are 0 (with multiplicity 3), 4 3 and 4 3 (each with multiplicity 1).
c. The x-intercepts are 2, 0 , 3, 0 , and
c. The x-intercepts are 0, 0 , 4 3, 0 ,
x
0 x 1 x 2 x 3 2
1, 0 .
and 4 3, 0 .
d. f 0 0 3 0 3 0 11 0 6 4
3
2
d. f 0 0 8 0 13 0 0 5
6 The y-intercept is 0, 6 .
4
3
The y-intercept is 0, 0 .
e. f x x 8 x 13 x
e. f x
5
x 3 x 3 x 11 x 6 4
3
4
3
x 5 8 x 4 13x 3 f x is neither even nor odd.
2
x 4 3x 3 3 x 2 11x 6 f x is neither even nor odd.
f.
f.
12. f x 2 x 3 5 x 2 6 x 2
f 2 2 2 5 2 6 2 2 3
11. a. f x x 8 x 13x 5
4
3
2
16 20 12 2 22
The leading term is x 5 . The end behavior is down to the left and up to the right.
f 1 2 1 5 1 6 1 2 3
2
2 5 6 2 1
f 0 2 0 5 0 6 0 2 2 3
2
f 1 2 1 5 1 6 1 2 3
2
2 5 6 2 7
427
Chapter 3 Polynomial and Rational Functions f 2 2 2 5 2 6 2 2 3
x3 5 x 4 18. a. 3x 2 3x 4 2 x 3 15 x 2 22 x 8
2
16 20 12 2 14
3x 4 2 x3
a. Since f 2 and f 1 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 2, 1 .
0 15 x 2 22 x 15 x 2 10 x
0 x3 5 x 4 b. Dividend: 3x 4 2 x 3 15 x 2 22 x 8 ; Divisor: 3x 2 ; Quotient: x 3 5 x 4 ; Remainder: 0
c. Since f 0 and f 1 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 0,1 .
19. 2 2
d. Since f 1 and f 2 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 1, 2 . 14. True
15. False
16. True
12 x 8 12 x 8
b. Since f 1 and f 0 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 1, 0 .
13. False
1 5
1
4 8 16 30
50
2 4 8 15 25
49
2 x 4 4 x 3 8 x 2 15 x 25
49 x2
0 0
2 20. 3 1 3 1 7 3 18 51 174 1 6 17 58
2 x 3x 9 17. a. x 2 x 3 2 x 4 x 3 0 x 2 4 x 1
176
2
2 x 4 2 x 3 6 x 2
x 3 6 x 2 17 x 58
21. 2 3
3x3 6 x 2 4 x 3x3 3x 2 9 x
9 x 2 13x 1 9 x 2 9 x 27
176 x3
4
1
6 12 28
64
3 6 14 32
65
0
2
By the remainder theorem, f 2 65 .
22 x 28
22.
22 x 28 2 x 3 x 9 2 x x3 2
5
1
2
4
10 5
5 5 2 5 10 5 1 2 5
b. Dividend: 2 x 4 x 3 4 x 1 ; Divisor: x 2 x 3 ; Quotient: 2 x 2 3 x 9 ; Remainder: 22 x 28
1 2 5
By the remainder theorem, f
428
5
5 0 .
5 0
Chapter 3 Review 23. a. 2 3 13 2 6 38
52 80 3 19 40 132
40 264 224
26. a. 2 1 4 2
1 6
By the factor theorem, since f 2 0 ,
By the factor theorem, since f 2 0 ,
x 2 is not a factor of f x .
b.
2 3
46 12 34
x 2 is not a factor of f x .
b. 2 5 2
3 13 2 52 40 2 10 8 40 3 15 12 60 0
4 46
1
25 2
46
1 2 5 2
0
2 By the remainder theorem, f 0 . 3
By the factor theorem, since
2 2 Since f 0 , is a zero of f x . 3 3
of f x .
27.
24. a. 5 3
13 2 52 40 15 10 60 40 3 2 12 8 0
2 15 67 26 8 3 10 38 8 15 57 12 0 2 f x a x 15 x 2 57 x 12 3
By the remainder theorem, f 5 0 .
Since f 5 0 , 5 is a zero of f x .
2 3 x 5 x 2 19 x 4 3
b.
3x 2 5 x 1 x 4
13 2 52 40 6i 12 26i 52 20i 40 3 13 6i 10 26i 0 20i
2i 3
28. f x x 1 x 3 2 x 3 2
x 1 x 2 18 x3 x 2 18 x 18
By the remainder theorem, f 2i 0 . Since f 2i 0 , 2i is a
zero of f x .
1 1 29. f x a x x x 3 4 2
25. a. 4 1
4 9 36 4 0 36 1 0 9 0
1 1 8 x x x 3 4 2 1 1 4 x 2 x x 3 4 2
By the factor theorem, since f 4 0 ,
4 x 1 2 x 1 x 3
x 4 is a factor of f x .
b. 3i
f 2 5 2 0 , x 2 5 2 is a factor
4 x 1 2 x 2 5 x 3
1
4 9 36 3i 9 12i 36 1 4 3i 12i 0
8 x 20 x 12 x 2 x 2 5 x 3 3
2
8 x 3 22 x 2 7 x 3
By the factor theorem, since f 3i 0 ,
x 3i is a factor of f x .
429
Chapter 3 Polynomial and Rational Functions 30. a. f x is a fifth-degree polynomial. It has 5 zeros. b.
Factor the quotient. 2 1 2 2 4 2 0 4 1 0 2 0
Factors of 40 Factors of 2 1, 2, 4, 5, 8, 10, 20, 40 1, 2 1, 2, 4, 5, 8, 10, 20, 40, 1 5 . 2 2
The rational zero is 2 (with multiplicity 2). d. Find the remaining two zeros. x2 2 0
x2 2 x 2 The zeros are 2 (with multiplicity 2), 2.
c. 1 2 7 2
9 18 4 40 9 18 36 40 2 9 18 36 40 0
32. Since 2 7i is a zero, then 2 7i is also a zero. The total number of zeros is 6, so the minimum degree of f x is 6.
Factor the quotient. 2 2 9 18 36 40 4 10 16 40 2 5 8 20 0
33. a. 3 i
Factor the quotient. 5 2 5 8 20 2 5 0 20 2 0 8 0 The rational zeros are
Since 3 i is a zero, 3 i is also a zero. 3 i 1 3 i 5 15 5i 3 i 0 15 5i 1 0 5 0
5 , 2, 1 . 2
f x x 3 i x 3 i x 2 5
d. Find the remaining two zeros. 2 x2 8 0
Find the remaining two zeros. x2 5 0
2 x 2 8 x 2 4 x 2i
x2 5 x 5 The zeros are 3 i , 5 .
5 The zeros are , 2, 1 , 2i, 2i . 2
b. f x
31. a. f x is a fourth-degree polynomial. It has 4 zeros. b.
6 5 30 50 3 i 10 15 5i 50 1 3 i 5 15 5i 0 1
x 3 i x 3 i x 5 x 5
Factors of 8 1, 2, 4, 8 Factors of 1 1 1, 2, 4, 8
c. The solution set is 3 i, 5 .
c. 2 1
4 2 8 8 2 4 4 8 1 2 2 4 0
430
Chapter 3 Review 34. f x x 2 x 2 3i x 2 3i
38. a. 2 1 3 0 2 2 2 4
f x x 2 x 2 3i x 2 3i
x 2 x 2 3i 2
1 1 2 2
2
The remainder and 3 of the coefficients of the quotient are negative. Therefore, 2 is not an upper bound for the real zeros of f x .
x x 4 x 13 x2 x2 4 x 4 9 2
2
x 4 4 x3 13x 2
2 b. 2 1 3 0 2 10 20
5 35. f x a x 2i x 2i x 3
1 5 10 18
5 a x2 4 x 3
39. a. 5 1 4 2 5 5 1 1 7
5 20 3 x3 x 2 4 x 3 3 3 x 3 5 x 2 12 x 20
1 35 36
The remainder and all coefficients of the quotient are nonnegative. Therefore, 5 is an upper bound for the real zeros of f x .
36. g x 3x 4 x 2 x 5 x 4 6
3 36 33
The signs of the quotient alternate. Therefore, 2 is a lower bound for the real zeros of f x .
5 20 a x 3 x 2 4 x Let a 3. 3 3
7
3 4 7
2
4 sign changes in g x . The number of possible positive real zeros is either 4, 2, or 0. 7 6 2 g x 3 x 4 x 2 x
b. 2 1 4
2 2 12 1 6 14
5 x 4 g x 3x7 4 x6 2 x 2 5 x 4
1 28 27
The signs of the quotient alternate. Therefore, 2 is a lower bound for the real zeros of f x .
1 sign change in g x . The number of possible negative real zeros is 1. 1 2 37. n x x 6 x 4 x3 4 x 2 3 3 7 0 sign changes in n x . The number of possible positive real zeros is 0. 1 2 6 4 3 2 n x x x x 4 x 3 3 7 1 2 n x x 6 x 4 x3 4 x 2 3 3 7 2 sign changes in n x . The number of possible negative real zeros is either 2 or 0.
40. a. 3
b.
c.
d. 3
e. , 2 2, f. Never decreasing g. , 2 2, h. , 3 3, i. x 2 41.
2 x 2 x 15 0
2 x 5 x 3 0 5 x , x 3 2
431
j. y 3
Chapter 3 Polynomial and Rational Functions 2x 1 x 0x 3 2x x 6x 7
42. x 2 3 is never zero. There are no vertical asymptotes.
2
3
2 x3 0 x 2 6 x
43. a. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of r.
x 0x 7 x2 0x 3 2
4
3 b. 2 0 x 2x 1 3 0 No solution The graph does not cross y 0 .
The quotient is 2 x 1 . The slant asymptote is y 2 x 1 . 47. The expression
44. a. The degree of the numerator is 2. The degree of the denominator is 2. 2 Since n m , the line y , or 1 equivalently y 2 , is a horizontal asymptote of q.
4 x 2 5 is in lowest 3x 2 14 x 5
terms. 3 x 2 14 x 5 0
3x 1 x 5 0 1 x ,x5 3
2 x 2 3x 4 2 b. x2 1 2 x 2 3 x 4 2 x 2 2
n has vertical asymptotes at x
1 and 3
x5. The degree of the numerator is 2. The degree of the denominator is 2. 4 or Since n m , the line y 3 4 equivalently y is a horizontal 3 asymptote of n.
3 x 4 2 3x 6 x2 The graph crosses y 2 at 2, 2 . 45. a. The degree of the numerator is 3. The degree of the denominator is 1. Since n m , k has no horizontal asymptotes.
1 with a x shift to the right 4 units and a shift upward 2 units.
48. The graph of f is the graph of y
b. Not applicable 46. The expression
2
2 x3 x 2 6 x 7 is in x2 3
lowest terms. x2 3 0 x2 3 x 3 The denominator is 0 at x 3 . m has vertical asymptotes at x 3 and x 3 . The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, m has no horizontal asymptote, but does have a slant asymptote.
432
Chapter 3 Review 2 0 49. k 0 0 02 0 12 The y-intercept is 0, 0 . The x-intercept is 0, 0 .
Interval
0, 4
k is in lowest terms, and x 2 x 12 x 3 x 4 is 0 for x 3 and x 4 , which are the vertical asymptotes. The degree of the numerator is 2. The degree of the denominator is 2. 1 Since n m , the line y , or equivalently 1 y 1 , is a horizontal asymptote of k.
4,
Test Point
Comments
3 3, 2
k(x) is negative. As x approaches the vertical asymptote x 4 from the left, k x .
6, 2
k(x) is positive. As x approaches the vertical asymptote x 4 from the right, k x . k(x) is greater than 1. k(x) must approach the horizontal asymptote y 1 from above as x .
x2 x 2 x 12 x 2 x 12 x 2 1
x 12 k crosses its horizontal asymptote at 12,1 . x2 x2 x 2 x 12 x 2 x 12 k x k x , k x k x
k x
k is neither even nor odd. Interval
, 12
12, 3
3, 0
Test Point Comments k(x) is less than 1. k(x) must approach the 98 14, horizontal 99 asymptote y 1 from below as x . k(x) is positive. As x approaches the vertical 4, 2 asymptote x 3 from the left, k x .
2 2, 3
2 0 6 0 9 undefined 50. m 0 0
There is no y-intercept. x2 6 x 9 0
x 3 x 3 0 x 3 The x-intercept is 3, 0 . m is in lowest terms, and the denominator is 0 when x 0 , which is the vertical asymptote. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, m has no horizontal asymptote, but does have a slant asymptote. x2 6 x 9 x2 6 x 9 9 m x x6 x x x x x The quotient is x 6 .
k(x) is negative. As x approaches the vertical asymptote x 3 from the right, k x .
433
Chapter 3 Polynomial and Rational Functions The slant asymptote is y x 6 .
x 6 q x q x
x 6x 9 x 2 2 x 6x x 6x 9 x6
12
q x
2
2
12 x 6 2
q is even.
0 9 No solution m does not cross its slant asymptote.
Test Point
Interval
x 6 x 9 x2 6 x 9 m x x x m x m x , m x m x 2
, 3
Test Point 3 6, 2
Test Point 1 4, 4
3, 0
1 2, 2
1, 4
0,
1,16
3, 12
, 0
0,
12
12 51. q 0 2 2 0 6 6
The y-intercept is 0, 2 . q is never 0. It has no x-intercepts nor vertical asymptotes. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of q. 12 0 2 x 6 0 12 No solution q does not cross its horizontal asymptote.
Comments
q(x) is positive. q(x) must approach the 4 3, horizontal asymptote 5 y 0 from above as x . q(x) is positive. q(x) must approach the 6 2, horizontal asymptote 5 y 0 from above as x.
m is neither even nor odd. Select test points from each interval. Interval
52. a. P 1
1 90 78% 0.16 1 1
P 4
4 90 57% 0.16 4 1
P 6
6 90 49% 0.16 6 1
1 6.25 0.16 P t will approach 6.25%.
b. Horizontal asymptote:
53. a. , 4 c. 4, 1 1, 54. a. 1, 2 c. ,1 2,
434
b. , 4 1 d. 4, b. 1, 2 d. ,1 2,
Chapter 3 Review x 2 7 x 10 0
55.
x 5 x 2 0 x 5 or x 2 The boundary points are 5 and 2 .
t 3t 18 0 Find the real zeros of the related equation. t 2 3t 18 0
Sign of x 5 :
2
t 3 t 6 0
Sign of x 5 x 2 :
t 3 or t 6 The boundary points are 3 and 6.
Sign of x 2 :
Sign of t 3 :
5 2
6
Find the real zeros of the related equation. w3 w2 9w 9 0
Sign of x 5 :
w2 w 1 9 w 1 0
w 9 w 1 0 2
w 3 w 3 w 1 0
w 3 , w 3 , or w 1 The boundary points are 3 , 1 , and 3. Sign of w 3 :
Sign of w 3 w 3 w 1 :
Sign of w 1 :
Sign of w 3 :
3 1
The solution set is 3, 1 3, .
1 5
c.
w 3 w 3 w 1 0
x 1 0 x5 x 1 0 x 1 1 is a boundary point. The expression is undefined for x 5 . This is also a boundary point.
x 1 0 a. x5 1, 5
Sign of t 3 t 6 :
2
e. x 7 x 10 0 , 5 2,
w 9 w 1 0
2
:
w2 w 1 9 w 1 0
d. x 2 7 x 10 0 , 5 2,
x5
58. w3 w2 9w 9 0 w3 w2 9w 9 0
c. x 2 7 x 10 0 5, 2
Sign of
The solution set is , 3 6,
b. x 7 x 10 0 5, 2
x 1
3
2
Sign of x 1 :
Sign of t 6 :
a. x 2 7 x 10 0 5, 2
56.
t t 3 18
57.
59. x 2 2 x 4 3
x 1 b 0 x5 1, 5
x2 2x 1 0
x 1 2 0 The square of any real number is
x 1 x 1 0 0 d. x5 x5 , 1 5, , 1 5,
nonnegative. Therefore, x 1 is nonnegative, and always greater than 0. However, the equality is included. The solution set is 1 . 2
435
3
Chapter 3 Polynomial and Rational Functions 62. 4 x 5 0
60. 6 x 4 3 x 4 x 2 0 2
4
3
The expression 4 x 5 0 for all real
6 x 4 3 x 4 x 2 0 Find the real zeros of the related equation. 2 3 6 x 4 3 x 4 x 2 0 2
4
3
numbers except where 4 x 5 0 . Therefore, the solution set is all real numbers 5 except . 4 5 5 , , 4 4 4
x 4 3 x 4 x 2 0 4 x 0 , x , or x 2 3 2
3
The boundary points are 2 , 0, and
4 . 3
Sign of x 2 :
Sign of 6 x 4 :
Sign of 6 x 4 3 x 4 x 2 :
3
Sign of 3 x 4 : 2
2
3
2
0
63.
4 3
The solution set is 2, . z 3 3z 2 10 z 24
61.
z 3 3 z 2 10 z 24 0 Find the real zeros of the related equation. z 3 3 z 2 10 z 24 0 Factors of 24 Factors of 1 1, 2, 3, 4, 6, 8, 12, 24 1 1, 2, 3, 4, 6, 8, 12, 24 2 1 3 10 24 2
x3 The boundary points are 0 and 3.
64.
x 2 , x 3 , or x 4 The boundary points are 3 , 2, and 4.
Sign of x 2 x 3 x 4 :
Sign of x 2 : Sign of x 4 :
3
The solution set is 3, 2 4, .
2
Sign of x 3:
:
x3 x2
3
The solution set is , 0 0, 3 .
0
0
x 2 x 2 x 12 0 x 2 x 3 x 4 0 Sign of x 3 :
Sign of x 2 :
Sign of
2 24
1 1 12
6 2x 0 x2 2x 6 0 x2 x3 0 x2 The expression is undefined for x 0 . This is a boundary point. Find the real zeros of the related equation. x3 0 x2 x3 0
4
436
8 1 3x 4 8 1 0 3x 4 8 3x 4 0 3x 4 3x 4 12 3 x 0 3x 4 3 x 4 0 3x 4 x4 0 3x 4
Chapter 3 Review The expression is undefined for x
Sign of x :
4 . This 3
is a boundary point. Find the real zeros of the related equation. x4 0 3x 4 x40
Sign of 6 5 x 4 :
Sign of x 2 : Sign of
6 x 1
x x 3
:
0
x4
Sign of 3 x 4 :
Sign of x 4 :
x 4 : 3 x 4
4
4 The solution set is , 0 , 2 . 5
4 and 4. 3
66.
4
3
1 x 3x 52 0 x 3 4 x 1 3x 52 0 x 3 4 The expression is undefined for x 3 . This is a boundary point. Find the real zeros of the related equation.
4 The solution set is , 4, . 3
x 1 3x 52 0 x 3 4 x 1 3x 52 0
3 2 65. x2 x 3 2 0 x2 x 2 x 2 3x 0 x x 2 x x 2
x 1 or x
5 3
5 The boundary points are , 1, and 3. 3
5x 4 0 x x 2 The expression is undefined for x 0 and x 2 . These are boundary points. Find the real zeros of the related equation. 5x 4 0 x x 2
Sign of 3 x 5 : 2
Sign of x 1 :
Sign of x 3 : 4
x 1 3x 5 : x 3
Sign of
4
4 5
The boundary points are 0,
2
5x 4 0 x
2
5
The boundary points are
Sign of
4
5 3
1
The solution set is 1, 3 3, . 4 , and 2. 5
80 40 15 x x 120 15 x C x x
67. a. C x
437
3
Chapter 3 Polynomial and Rational Functions
b.
120 15 x 16 x 120 15 x 16 0 x 120 15 x 16 x 0 x 120 x 0 x x 120 0 x The expression is undefined for x 0 . This is a boundary point. Find the real zeros of the related equation x 120 0 x x 120 0
Sign of 2t 1 :
Sign of x : Sign of
x 120 x
:
Sign of 2t 1 t 2 :
2
2
1 The solution set is , 2 or 2 0.5 t 2 sec. The ball will be more than 18 ft high 0.5 sec and 2 sec after it is thrown.
69. m kw
71. y
72.
70. x
Q kp t
73.
d 1.8
132 k 33 k4
k p2
kx z t3
132 k 11 9
kc x2 k 3
2 2
7.2 3k k 2.4
74. Let w represent the weight of the ball. Let r represent the radius of the ball. w kr 3
3.24 k 3
1 68. a. s t 32 t 2 40 t 2 2 s t 16t 2 40t 2
3
3.24 k 27 k 0.12
b. 16t 2 40t 2 18
w 0.12r 3 0.12 5 15 lb 3
16t 2 40t 16 0 2t 2 5t 2 0 Find the real zeros of the related equation. 2t 2 5t 2 0
75.
2t 1 t 2 0 1 or t 2 2
The boundary points are
1
0 120 The solution set is , 0 120, . The trainer must have more than 120 sessions with his clients for his average cost to drop below $16 per session.
t
Sign of t 2 :
x 120 The boundary points are 0 and 120. Sign of x 120 :
1 and 2. 2
438
k L k 6.25 1.6 k 10 10 10 5 lb F L 2 F
Chapter 3 Test 76. Let P represent power in a circuit. Let I represent the current in the circuit. Let R represent the resistance in the circuit. P kIR 2
144 k 4 6
77. F
F
2
144 k 144 k 1
kq1q2 d2 k 2q1 2q2 d 2
2
4kq1q2 d2 4 kq q 16 12 2 d The force will be 16 times as great.
P IR 2 310 300 W 2
Chapter 3 Test 1. a. f x 2 x 2 12 x 16
f x 2 x2 6 x
f.
16
2
1 2 6 9
f x 2 x 6 x 9 2 9 16 f x 2 x 2 6 x 9 9 16 2
f x 2 x 3 2 2
b. Since a 0 , the parabola opens upward.
g. The axis of symmetry is the vertical line through the vertex: x 3 .
c. The vertex is h, k 3, 2 .
h. The minimum value is 2 .
d. f x 2 x 2 12 x 16
i. The domain is , .
The range is 2, .
0 2 x 2 12 x 16 0 x2 6 x 8
2. a. f x 2 x 4 5 x3 17 x 2 41x 21 The leading coefficient is positive and the degree is even. The end behavior is up to the left and up to the right.
0 x 2 x 4 x 2 or x 4 The x-intercepts are 2, 0 and 4, 0 .
b. Possible rational zeros: Factors of 21 Factors of 2 1, 3, 7, 21 1, 2
e. f 0 2 0 12 0 16 16 2
The y-intercept is 0,16 .
1 3 7 21 1, 3, 7, 21, , , , 2 2 2 2
439
Chapter 3 Polynomial and Rational Functions 41 21 c. 1 2 5 17 2 3 20 21 2 3 20
21
1 2 3 20
21
3. a. Multiply the leading terms from each factor.
0.25 x 3 x x 0.25 x 9 2
0
b. The leading coefficient is negative and the degree is odd. The end behavior is up to the left and down to the right.
2 1 21 1 21 0
2
2
f x x 1 2 x 7 x 3
x 1 2 x 7 x 3 0 7 , x 3 2 The zero 1 has multiplicity 2. The zeros 7 and 3 have multiplicity 1. 2 x 1, x
b. 0 x 4 5 x 2 36
0 x 2 x 2 x 3i x 3i x 2, x 2, x 3i, x 3i The zeros are 2, 2 , 3i, and 3i .
2
21 f. f x 2 x 5 x 17 x
0 x2 4 x2 9
e. f 0 2 0 5 0 17 0 41 0 21 3
4
4. a. The leading term has degree 4. There are 4 zeros.
7 d. The x-intercepts are , 0 , 3, 0 , and 2 1, 0 .
4
0 0.25 x 3 x 2 x 1
x 0, x 2, x 1 The zero 0 has multiplicity 3. The zero 2 has multiplicity 2. The zero 1 has multiplicity 4.
2
3
4
2
2
4
c. f x 0.25 x 3 x 2 x 1 2
f x x 1 2 x x 21 2
4
c. The x-intercepts are 2,0 and 2, 0 . 2
d. f x x 5 x 36 4
41 x 21
2
x 4 5 x 2 36 f x is even.
2 x 4 5 x 3 17 x 2 41x 21 f x is neither even nor odd.
5. f x x3 5 x 2 2 x 5
g.
f 2 2 5 2 2 2 5 3
2
8 20 4 5 27
f 1 1 5 1 2 1 5 3
2
1 5 2 5 3
f 0 0 5 0 2 0 5 5 3
2
f 1 1 5 1 2 1 5 3
2
1 5 2 5 3
f 2 2 5 2 2 2 5 3
2
8 20 4 5 3
440
Chapter 3 Test a. Since f 2 and f 1 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 2, 1 .
80 51 9 b. 1 5 47 5 42 38 89 5 42
c. Since f 1 0 , x 1 is not a factor of
f x .
d. 3 5
47 80 51 9 15 96 48 9 5 32 16 3 0
c. Since f 0 and f 1 have the same sign, the intermediate value theorem does not guarantee that the function has at least one zero on the interval 0,1 .
Since f 3 0 , x 3 is a factor of f x .
d. Since f 1 and f 2 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 1, 2 .
e. 2 5
47 80 51 9 10 74 12 126 5 37 6 63 117
By the remainder theorem, f 2 117 .
2 x2 2 x 4 6. a. x 2 3x 1 2 x 4 4 x3 0 x 2 x 5
2 x 6 x 2 x 3
2
8. a. 2i
2 x3 2 x 2 x 2 x3 6 x 2 2 x
4x x 5 4 x 2 12 x 4
11x 9
f x x 2i x 2i x 2 8 x 17
11x 9 2x 2x 4 2 x 3x 1 2
Find the remaining two zeros. x 2 8 x 17 0
b. Dividend: 2 x 4 4 x 3 x 5 ; Divisor: x 2 3x 1 ; Quotient: 2 x 2 2 x 4 ; Remainder: 11x 9 7. a.
8 21 32 68 2i 4 16i 32 34i 68 1 8 2i 17 16i 34i 0 1
Since 2i is a zero, 2i is also a zero. 2i 1 8 2i 17 16i 34i 2i 16i 34i 1 8 17 0
2
80
Since f 1 0 , 1 is a zero of f x .
b. Since f 1 and f 0 have opposite signs, the intermediate value theorem guarantees that the function has at least one zero on the interval 1, 0 .
4
38 89
x
8
82 4 117 2 1
8 4 8 2i 4i 2 2 The zeros are 2i , 4 i .
3 5 47 80 51 9 5 3 30 66 9 5 50 110 15 0
b. f x
x 2i x 2i x 4 i x 4 i
3 3 Since f 0 , is a zero of f x . 5 5
c. The solution set is 2i, 4 i .
441
Chapter 3 Polynomial and Rational Functions 9. a. f x is a fourth-degree polynomial. It has 4 zeros.
g. Factor 3x 2 6 . 3x 2 6 0
3x 2 6
Factors of 12 1, 2, 3, 4, 6, 12 b. Factors of 3 1, 3 1, 2, 3, 4, 6, 12, 1 2 4 , , 3 3 3 7 12 14 6 26 28 3 13 14 14
c. 2 3
x2 2 x 2 The zeros are
2 , 3 , 2 . 3
h.
12 28 40
The remainder and all coefficients of the quotient are nonnegative. Therefore, 2 is an upper bound for the real zeros of f x . d. 4 3
7 12 14 12 20 32 3 5 8 46
12 184 196
1 2 10. f x a x x x 4 Let a 15. 5 3
The signs of the quotient alternate. Therefore, 4 is a lower bound for the real zeros of f x .
1 2 15 x x x 4 5 3 5 x 1 3x 2 x 4
e. We can restrict the list of possible rational zeros to those on the interval 4, 2 :
5 x 1 3x 2 10 x 8
1 2 4 1, 2, 3, , , 3 3 3 From part (c), the value 2 itself is not a zero of f x . Likewise, from part (d), the value 4 itself is not a zero. Therefore, 2 and 4 are also eliminated from the list of possible rational zeros.
15 x 3 50 x 2 40 x 3x 2 10 x 8 15 x 3 53x 2 30 x 8 11. f x 6 x 7 4 x5 2 x 4 3 x 2 1
3 sign changes in f x . The number of possible positive real zeros is either 3 or 1. 7 5 4 f x 6 x 4 x 2 x
f. 3 3
7 12 14 12 9 6 18 12 3 2 6 4 0
3 x 1 2
f x 6 x 7 4 x5 2 x 4 3x 2 1
2 sign changes in f x . The number of possible negative real zeros is either 2 or 0.
Find the zeros of the quotient. 2 3 2 6 4 3 2 0 4 3 0 6 0 The rational zeros are
2 and 3 . 3
442
Chapter 3 Test 2 x 2 3x 5 12. The expression is in lowest x7 terms and the denominator is 0 at x 7 . r has a vertical asymptote at x 7 . The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, r has no horizontal asymptote, but does have a slant asymptote. 7 2 3 5 14 77 2 11 82
1 with a x2 shift upward 3 units and a reflection across the x-axis.
15. The graph of m is the graph of y
The quotient is 2 x 11 . The slant asymptote is y 2 x 11 . 16. h 0
3x 1 13. The expression is in lowest terms. 4 x2 1 3x 1 3 x 1 2 4 x 1 2 x 1 2 x 1
0 4 2
4 1 4
The y-intercept is 0,1 . h is never 0. It has no x-intercepts. h is in lowest terms, and x 2 4 is 0 for x 2 and x 2 , which are the vertical asymptotes. The degree of the numerator is 0. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of q. 4 0 2 x 4 0 4 No solution h does not cross its horizontal asymptote. 4 4 h x 2 2 x 4 x 4
1 1 , and x . 2 2 1 p has vertical asymptotes at x , and 2 1 x . 2 The degree of the numerator is 1. The degree of the denominator is 2. Since n m , the line y 0 is a horizontal asymptote of p. p has no slant asymptote.
The denominator is 0 at x
14. The expression
4
5x2 2 x 1 is in lowest 3x 2 4
h x h x h is even.
terms. The denominator 3x 2 4 is never zero. k has no vertical asymptotes. The degree of the numerator is 2. The degree of the denominator is 2. 5 Since n m , the line y is a horizontal 3 asymptote of k.
443
Interval
Test Point
, 2
h(x) is negative. h(x) must approach the horizontal asymptote y 0 from below as 1 4, x . 3 As x approaches the vertical asymptote x 2 from the left, h x .
Comments
Chapter 3 Polynomial and Rational Functions
Interval
2, 0
0, 2
2,
Test Point
Comments
4 1, 3
h(x) is positive. As x approaches the vertical asymptote x 2 from the right, h x .
4 1, 3
h(x) is positive. As x approaches the vertical asymptote x 2 from the left, h x .
1 4, 3
h(x) is negative. h(x) must approach the horizontal asymptote y 0 from below as x. As x approaches the vertical asymptote x 2 from the right, h x .
x2 2 x 1 x2 2 x 1 1 k x x2 x x x x x The quotient is x 2 . The slant asymptote is y x 2 . x2 2 x 1 x x2 2 x x2 2 x 1 0 1 No solution k does not cross its slant asymptote. x2
2 x 2 x 1 x2 2 x 1 k x x x k x k x , k x k x
k is neither even nor odd. Select test points from each interval. Interval
, 1
Test Point 9 4, 4
Test Point 1 2, 2
1, 0
1 1 , 2 2
1 9 , 4 4
0,
1 25 , 4 4
9 2, 2
2 0 2 0 1 undefined 17. k 0 0
There is no y-intercept. x2 2 x 1 0
18.
x 1 x 1 0
c 2 c 20 c 2 c 20 0 Find the real zeros of the related equation. c 2 c 20 0
x 1 The x-intercept is 1, 0 . k is in lowest terms, and the denominator is 0 when x 0 , which is the vertical asymptote. The degree of the numerator is exactly one greater than the degree of the denominator. Therefore, k has no horizontal asymptote, but does have a slant asymptote.
c 4 c 5 0 c 4 or c 5 The boundary points are 4 and 5.
444
Chapter 3 Test Sign of c 4 :
Sign of c 4 c 5 :
4
5
Sign of c 5 :
The solution set is 4, 5 .
21. 9 x 2 42 x 49 0
3x 7 2 0 2 The expression 3x 7 0 for all real 2 numbers except where 3x 7 0 . Therefore, the solution set is all real numbers 7 except . 3 7 7 , , 3 3
y 3 13 y 12
19.
y 3 13 y 12 0 Find the real zeros of the related equation. y 3 13 y 12 0 Factors of 12 1, 2, 3, 4, 6, 12 Factors of 1 1 1, 2, 3, 4, 6, 12 1 1 0 13 12 1 1 12 1 1 12 0
22.
x 1 x 2 x 12 0 x 1 x 4 x 3 0 x 1 , x 4 , or x 3 The boundary points are 4 , 1, and 3. Sign of x 4 :
Sign of x 1 x 4 x 3 :
Sign of x 1 :
Sign of x 3 :
4
The solution set is 4,1 3, .
1
Sign of x 3 :
Sign of x 2 :
Sign of
3
x x 4 x 1 0 Find the real zeros of the related equation. 2 3 x x 4 x 1 0 x 0 , x 4 , or x 1 The boundary points are 1 , 0, and 4. 3
Sign of x x 4 x 1 :
Sign of x:
Sign of x 4 : 2
2
3
1 0
:
2
4 0 x 9 4 0 2 x 9
23.
3
Sign of x 1 :
x2
The solution set is , 3 2, .
3
2
x3
3
20. 2 x x 4 x 1 0 2
x3 0 2 x x3 0 x2 The expression is undefined for x 2 . This is a boundary point. Find the real zeros of the related equation. x3 0 x2 x3 0 x 3 The boundary points are 3 and 2.
2
4
x 3 x 3
0
The expression is undefined for x 3 and x 3 . These are boundary points. The related equation has no real zeros.
4
The solution set is , 1 0, .
445
Chapter 3 Polynomial and Rational Functions Sign of x 3 :
Sign of x 3 :
4
:
Sign of
x 3 x 3
3
3
The solution set is 3, 3 . 24.
26.
27. Let A represent the surface area of a cube. Let s represent the length of an edge. A ks 2
4 3 x 1 x 4 3 0 x 1 x 3 x 1 4x 0 x x 1 x x 1 7x 3 0 x x 1 The expression is undefined for x 0 and x 1 . These are boundary points. Find the real zeros of the related equation. 7x 3 0 x x 1
24 k 2
A 6s 2 6 7 294 ft 2 2
28. Let w represent the weight of a body. Let d represent the distance from the center of the Earth. k w 2 d k 180 40002
Sign of 7 x 3 :
:
Sign of x 1 : Sign of
7x 3 x x 1
0
3
k 2,880,000,000 2,880,000,000 2,880,000,000 178 lb w d2 4020 2
3 , and 1. 7
Sign of x:
2
24 k 4 k6
7x 3 0 3 x 7 The boundary points are 0,
ky x z k 6 4 7.2 7 50.4 12k k 4.2 w
29. Let P represent the pressure on the wall. Let A represent the area of the wall. Let v represent the velocity of the wind. P kAv 2
P kA 3v kA 9v 2 9 kAv 2
1
2
7
The pressure is 9 times as great.
3 The solution set is , 0 ,1 . 7
2000 1 1000 11 2000 5 P 5 1667 5 1 2000 10 P 10 1818 10 1 There will be 1000 rabbits after 1 yr, 1667 rabbits after 5 yr, and 1818 rabbits after 10 yr.
30. a. P 1
25. E kv 2
446
Chapter 3 Test 2000 2000 1 The rabbit population will approach 2000 as t increases.
4.9t 2 98t 200
d.
b. Horizontal asymptote:
4.9t 2 98t 200 0 4.9t 2 98t 200 0 Find the real zeros of the related equation. 4.9t 2 98t 200 0
31. a. y 20 140.3 means that with 20,000 plants per acre, the yield will be 140.3 bushels per acre; y 30 172 means that with 30,000 plants per acre, the yield will be 172 bushels per acre; y 60 143.5 means that with 60,000 plants per acre, the yield will be 143.5 bushels per acre.
t
98
98 2 4 4.9 200 2 4.9
98 5684 2.3 or 17.7 9.8 4.9t 2 98t 200 0
t 2.3 t 17.7 0
b. The number of plants per acre that will maximize the yield is the n-coordinate of the vertex. 8.32 b 8.32 n 2a 2 0.103 0.206
The boundary points are 2.3 and 17.7. Sign of t 2.3 :
Sign of t 2.3 t 17.7 :
Sign of t 17.7 :
40.4 thousand plants 40,400 plants
2.3 17.7
The solution set is 2.3,17.7 . The rocket will be more than 200 m high between 2.3 sec and 17.7 sec after launch.
c. The maximum yield is the value of y n at the vertex. 2 y 40.4 0.103 40.4 8.32n 15.1
33 a.
183 bushels 1 9.8 t 2 98 t 0 2 s t 4.9t 2 98t
32. a. s t
b. t
n a 0.0011a 2 0.027 a 2.46
98 b 98 10 2a 2 4.9 9.8
b. The age when the number of yearly visits is the least is the a-coordinate of the vertex. b 0.027 0.027 a 12 yr 2a 2 0.0011 0.0022
The rocket will reach maximum height 10 sec after launch. c.
s t 4.9t 2 98t s 10 4.9 10 98 10 2
c. The minimum number of yearly visits is the n a value at the vertex.
490 980 490 The rocket’s reach maximum height is 490 m.
n 12 0.001112 0.027 12 2.46 2
2.3 visits per year
447
Chapter 3 Polynomial and Rational Functions Chapter 3 Cumulative Review Exercises 5 c. The x-intercepts are , 0 and 1, 0 . 2
1. a. x 2 16 0
x 2 16 x 4, x 4
d. f 0 2 0 0 8 0 5 5 3
2
The y-intercept is 0, 5 .
b. Since the degree of the numerator is greater than the degree of the denominator by more than 1, the horizontal asymptote 2 is y 2 . 1
e.
2. f x x 2 x 3 2i x 3 2i
f x x 2 x 3 2i x 3 2i 2 f x x 2 x 3 4
f x x 2 x 6 x 13 f x x 2 x 2 6 x 9 4 2
4.
f x x3 6 x 2 13x 2 x 2 12 x 26 f x x3 8 x 2 25 x 26 3. f x 2 x3 x 2 8 x 5 a. The leading term is 2x 3 . The end behavior is down to the left and up to the right. b.
5.
Factors of 5 1, 5 Factors of 2 1, 2
3 2i 3 2i 4 i 4i 4i 4i 12 3i 8i 2 16 1 10 11i 17 10 11 i 17 17 x 2 y 2 8 x 14 y 56 0
x 8x y 14 y 56 2
2
2
1 2 8 16
1 5 1, 5, , 2 2 1 2 1 8 5 3 5 2 2 3 5 0
2
1 2 14 49
x 8 x 16 y 14 y 49 56 16 49 2
2
x 4 2 y 7 2 9 Center: 4, 7 ; Radius: 9 3
Find the zeros of the quotient. 2 x 2 3x 5 0
2 x 5 x 1 0 5 x , x 1 2 5 (multiplicity 1) and 1 2 (multiplicity 2). The zeros are
448
Chapter 3 Cumulative Review 10. a. f x 2 x 2 6 x 1
y y1 3 8 5 5 6. m 2 x2 x1 24 2 2
2 x2 6 x 1 0 6 6 4 21 x 2 2
y y1 m x x1
5 x 4 2 5 y 8 x 10 2 5 y x2 2
y 8
6 28 4 62 7 4 3 7 2
7. x y 2 9
3 7 , 0 and The x-intercepts are 2
x 0 9 2
x 9 The x-intercept is 9, 0 .
3 7 2 , 0 .
x y 9 2
0 y2 9
b. f 0 2 0 6 0 1 1 2
y 9 y 3 2
The y-intercept is 0,1 .
The y-intercepts are 0, 3 and 0, 3 .
c. x
8.
9.
b 6 6 3 2a 2 2 4 2 2
3 3 3 f 2 6 1 2 2 2 9 9 1 2 7 2 3 7 The vertex is , . 2 2
y 5 x y 25 x 5 x y y
v t mt
11. 125 x 6 y 9 5 x 2
v0t m t 2 m v0t t 2
2
449
3
3 3
3
4
2
3
6
Chapter 3 Polynomial and Rational Functions 3
4 x 3 y 5 4 y 6 12. 2 12 z x
18. c 2 5c 9 c c 3
12
c 2 5c 9 c 2 3c 8c 9 9 c 8 9 , 8
4 3 x 9 y15 41 2 y 3 x 6 z6 y15 2 x 6 3 9 6 3 4 x z y
y12 32 x 3 z 6
19.
13. 3 250 z 5 xy 21 3 125 z 3 y 21 2 z 2 x 5 zy 7 3 2 z 2 x
49 x 2 14 x 1 0 x
7 x 1 2 0
x The expression is undefined for x 0 . This is a boundary point. Find the real zeros of the related equation. 7 x 12 0 x
1 1 1 1 2 2 3x 2 x3 14. 3 x x 3x x 2 2 1 3 1 3 3x x 3 2 2 3 x 3 x 1 x3 x 3 x 3 x 3
7 x 1 2 0 x
1 1 15. 5 x 3 4 2 20 x 12 2
The boundary points are
32 x 10 10 x 32 10, 32
Sign of 7 x 1 : 2
Sign of x : Sign of
16. x 3 4 10
x3 6
3, 9
0, 20.
5,1
x
:
2
or
2 x 1 x 4 2x 1 x 4 3x 3 x 1
5
450
1 0 7
4 x 3 x 12 0
17. 2 x 1 x 4
x 5
7 x 1
1 and 0. 7
6 x 3 6 3 x 9
2x 1 x 4
1 7
4 x 3 x 12 4 x 3 x 12 3 x 15 x5
Section 4.1
Chapter 4 Exponential and Logarithmic Functions Section 4.1 Inverse Functions 1. 2,1 , 3, 2 , 4, 3 2. one-to-one 3. y x
4. is not
5. is
6. =
7. x; x
8. f 1
9. b, a
10. is not; is
11. Yes
12. Yes
13. No
14. No
15. Yes
16. Yes
17. No
18. No
26. The graph does not define y as a one-to-one function of x because a horizontal line intersects the graph in more than one point. 27. The relation does not define y as a function of x, because it fails the vertical line test. If the relation is not a function, it is not a oneto-one function. 28. The relation does not define y as a function of x, because it fails the vertical line test. If the relation is not a function, it is not a oneto-one function. 29. Assume that f a f b . 5a 4 5b 4 5a 5b ab Since f a f b implies that a b , then f is one-to-one.
19. The graph does not define y as a one-to-one function of x because a horizontal line intersects the graph in more than one point.
30. Assume that h a h b . 3a 2 3b 2 3a 3b ab Since h a h b implies that a b , then f is one-to-one.
20. The graph does not define y as a one-to-one function of x because a horizontal line intersects the graph in more than one point. 21. The graph does define y as a one-to-one function of x because no horizontal line intersects the graph in more than one point.
31. Assume that g a g b .
22. The graph does define y as a one-to-one function of x because no horizontal line intersects the graph in more than one point.
a 3 8 b3 8 a 3 b3
23. The graph does not define y as a one-to-one function of x because a horizontal line intersects the graph in more than one point.
a 3 3 b3 ab Since g a g b implies that a b , then f is one-to-one. 3
24. The graph does not define y as a one-to-one function of x because a horizontal line intersects the graph in more than one point. 25. The graph does not define y as a one-to-one function of x because a horizontal line intersects the graph in more than one point. 451
Chapter 4 Exponential and Logarithmic Functions 32. Assume that k a k b .
x 3 38. h k x h k x h 7
a 3 27 b3 27
x 3 7 3 x33 x 7
a 3 b3 a 3 3 b3 ab Since k a k b implies that a b , then f is one-to-one. 3
k h x k h x k 7 x 3 7 x 3 3 7 x x
7 7 Yes, k is the inverse of h.
33. No; For example the points 1, 3 and
2 x 39. m n x m n x m 6
1, 3 have the same y values but different x values. That is, m a m b 3 , but
2 x 6 2 2 x 2 6
a b.
x4 No, since m n x x , n is not an inverse of m.
34. No; For example the points 3,10 and
3,10 have the same y values but different x values. That is, n a n b 10 , but
3 x 40. p q x p q x p 4
a b.
35. No; For example the points 2, 3 and
3 x 4 3 3 x 3 4
4, 3 have the same y values but different x values. That is, p a p b 3 , but
x6 No, since p q x x , q is not an inverse of p.
a b.
36. No; For example the points 0, 3 and 6, 3 have the same y values but different x values. That is, q a q b 3 , but a b .
4 x 4 41. t v x t v x t x x 4 1 x
x 4 37. f g x f g x f 5
x 4 5 4 x44 x 5
g
4 4 4 x 1 v t x v t x v x 1 4 x 1
f x g f x g 5 x 4
5 x 4 4 5 x x
5 Yes, g is the inverse of f.
4x 4x x x 4 x 4
4 4 x 1 4 4 x 4 4 x 4 4 4 x Yes, v is the inverse of t.
5
452
Section 4.1
g
6 2x 42. w z x w z x w x
6 6x 6 2x 6 2 x 2 x 2 x
60,000 2500 x 60,000
2500 2500 x x 2500 Yes, g is the inverse of f.
6x x 6
b. The value g(x) represents the number of months since January based on the monthly amount in sales, x.
6 z w x z w x z x 2 6 6 2 x 2 6 x 2 12 6 6 x2
45. a. Assume that f a f b . 2a 3 2b 3 2a 2b ab Since f a f b implies that a b , then f is one-to-one.
6 x 12 12 6 x x 6 6 Yes, z is the inverse of w.
b.
x 2000 43. a. f g x f g x f 150
f x 2x 3 y 2x 3 x 2y 3 x 3 2y x3 y 2 x3 f 1 x 2
x 2000 2000 150 150 2000 x 2000 x g f x g f x g 2000 150 x
f x g f x g 60,000 2500 x
2000 150 x 2000
150 150 x x 150 Yes, g is the inverse of f.
c.
b. The value g(x) represents the number of years since the year 2010 based on the number of applicants to the freshman class, x. x 60,000 44. a. f g x f g x f 2500
46. a. Assume that f a f b . 4a 4 4b 4 4a 4b ab Since f a f b implies that a b , then f is one-to-one.
x 60,000 60,000 2500 2500 60,000 x 60,000 x
453
Chapter 4 Exponential and Logarithmic Functions f x 4x 4
b.
49.
y 4x 4 x 4y 4 x 4 4y x4 y 4 x4 f 1 x 4
h x 3 x 5 y 3 x5 x 3 y 5 x3 y 5 x3 5 y h 1 x x 3 5
50.
c.
k x 3 x 8 y 3 x8 x 3 y8 x3 y 8 x3 8 y k 1 x x 3 8
51.
f
48.
y 4 x3 2
4 x f x 9 4 x y 9 4 y x 9 9x 4 y y 4 9x
47.
1
m x 4 x3 2
x 4 y3 2 x 2 4 y3 x2 y3 4 x2 3 y 4 m 1 x 3
x 4 9 x 8 x 3 8 x y 3 8 y x 3 3x 8 y y 8 3x
g x
52.
x2 4
n x 2 x3 5
y 2 x3 5 x 2 y3 5 x 5 2 y3 x5 y3 2 x5 3 y 2
g 1 x 8 3 x
n 1 x 3
454
x5 2
Section 4.1
53.
5 x2 5 y x2 5 x y2
c x
56.
x y 2 5 xy 2 x 5 xy 5 2 x 5 2x y x 5 2x c 1 x x
54.
2 x3 2 y x3 2 x y 3
x5 x 1 x5 y x 1 y 5 x y 1 x y 1 y 5 xy x y 5 xy y x 5 y x 1 x 5 x5 y x 1 x5 v 1 x x 1 v x
s x
3 x a f x c
57.
b 3 x a y c b y a 3 c x b y a 3 xc b 3 b x c y a
x y 3 2 xy 3 x 2 xy 3 x 2 3x 2 y x 3 x 2 s 1 x x
55.
3 3
b x c y a
b x c a y f 1 x 3 b x c a
x4 t x x2 x4 y x2 y4 x y2
58.
g x b x a c 3
y b x a c 3
x b y a c 3
x c b y a xc 3 y a b x c 3 ya b xc 3 a y b xc g 1 x 3 a b 3
x y 2 y 4 xy 2 x y 4 xy y 2 x 4 y x 1 2 x 4 2x 4 x 1 2x 4 t 1 x x 1 y
455
Chapter 4 Exponential and Logarithmic Functions 59. a.
e.
f x x2 1 y x2 1 x y2 1 x 1 y2 x 1 y f 1 x x 1
f. c. , 0
b. Yes d. 3, f x x2 3
e.
y x2 3 x y2 3 x 3 y2
g. 1,
x3 y f 1 x x 3
h. , 0
61. a.
f.
c. 1,
b. Yes g. 3,
h. , 0
d. 0,
60. a.
e.
f x x 1 y x 1 x
y 1
x y 1 2
x 1 y 2
f 1 x x 2 1; x 0
b. Yes
f. The range of f is 0, . Therefore, the
c. , 0
domain of f 1 must be 0, .
d. 1, 456
Section 4.1 h. 0,
g.
i. 2,
63. Domain: 0, 4 ; Range: 0, 64. Domain: 2, ; Range: 3, 5 f x x 3
65.
y x 3
h. 0,
x y 3
i. 1,
x3 y y x3
62. a.
y x 3
or
y 3 x f
1
x 3 x; x 3
f x x 3
66.
y x 3 x y 3 x3 y
c. 2,
b. Yes
y x3
y x 3
d. 0, e.
y x 3
or
f
f x x 2 y x2 x
1
x x 3; x 3
67. subtracts; x 6
68. divides;
y2
x y2 2
69. f 1 x
x 2 y 2
f 1 x x 2 2; x 0
x4 7
70. f 1 x 3 x 11
71. f 1 x 3 x 20
f. The range of f is 0, . Therefore, the domain of f 1 must be 0, .
73. f 1 x
x 2
x 1 8
72. f 1 x x 10 74. p 1 x
x 10 2
g. 75. q 1 x x 1 4 76. m 1 x 5
457
x 33 4
3
Chapter 4 Exponential and Logarithmic Functions 82. a. f 20 5 , so f 1 5 20 .
77.
b. f 2.2 9.45 , so f 1 9.45 2.2 . c. f 8 8 , so f 1 8 8 . 83. True
84. False
85. False
86. True
87. a. T x 6.33x
78.
b.
T x 6.33 x y 6.33 x x 6.33 y x y 6.33 T 1 x
x 6.33
c. T 1 x represents the mass of a mammal based on the amount of air inhaled per breath, x.
79.
170 27 means that a 6.33 mammal that inhales 170 mL of air per breath during normal respiration is approximately 27 kg (this is approximately 60 lb—the size of a Labrador retriever).
d. T 1 170
88. a. d t 555t
80.
b.
d t 555t y 555t t 555 y t y 555 d 1 t
81. a. f 12 32 , so f 1 32 12 .
t 555
c. d 1 t represents the amount of time (in hr) at which the aircraft has been at its cruising altitude based on t hours traveling at that altitude.
b. f 0.5 2.5 , so f 1 2.5 0.5 . c. f 10 26 , so f 1 26 10 .
458
Section 4.1 4 91. V r r 3 3 4 V r3 3 3V r3 4 3V 3 r 4
2553 4.6 means that if 555 2553 mi is traveled at cruising altitude, then the plane has been at its cruising altitude for 4.6 hr.
d. d 1 2553
89. a. T x 24 x 108 b.
T x 24 x 108 y 24 x 108 x 24 y 108 x 108 24 y x 108 y 24 x 108 T 1 x 24
r V 3
3V represents the radius of a 4 sphere as a function of its volume. r V 3
92.
c. T 1 x represents the taxable value of a home (in $1000) based on x dollars of property tax paid on the home. 2988 108 120 means that 24 if a homeowner is charged $2998 in property taxes, then the taxable value of the home is $120,000.
d. T 1 2988
9 F C C 32 5 9 F C 32 5 9 F 32 C 5 5 F 32 C 9 5 C F F 32 9 5 C F F 32 represents the 9 temperature in degrees Celsius as a function of the corresponding temperature in degrees Fahrenheit.
90. a. L x 2.5 0.015x b.
3V 4
L x 2.5 0.015 x y 2.5 0.015 x x 2.5 0.015 y x 2.5 0.015 y 2.5 x y 0.015 2.5 x L1 x 0.015
93. The domain and range of a function and its inverse are reversed. 94. A one-to-one function y f x is a function in which no two ordered pairs have the same y-coordinate but different xcoordinates.
c. L1 x represents the number of days since January 1 based on the water level, x.
95. If a horizontal line intersects the graph of a function in more than one point, then the function has at least two ordered pairs with the same y-coordinate but different xcoordinates. This conflicts with the definition of a one-to-one function.
2.5 1.9 40 means that given 0.015 a water level of 1.9 ft, the time elapsed since January 1 is 40 days.
d. L1 1.9
459
Chapter 4 Exponential and Logarithmic Functions 96. The graph of f x x 2 k is a parabola opening upward or downward. The function is not one-to-one. However, if the domain is restricted to x values strictly to one side or the other of the axis of symmetry, then the resulting function is one-to-one and the inverse function can be found.
100. Let f be a periodic function and let p 0 be the period. Then f x p f x for all x in the domain of f. Since p 0 , then x p x , so f is not one-to-one. 101. a.
97. a. 23 8 , so f 8 3 . b. 25 32 , so f 32 5 . c. 21 2 , so f 2 1 . d. 23
The graphs of f and g appear to be symmetric with respect to the line y x . This suggests that f and g are inverses.
1 1 1 , so f 3 . 3 8 2 8
98. a. 33 27 , so f 27 3 .
b.
b. 34 81 , so f 81 4 . c. 31 3 , so f 3 1 . d. 32
1 1 1 , so f 2 . 2 9 3 9
c.
99. Let f be an increasing function. Then for every value a and b in the domain of f such that a b we have f a f b . Now if u v , then either u v or v u . Then either f u f v or f v f u . In
The expressions Y1(Y2) and Y2(Y1) represent the composition of functions f g x and g f x . In each case, Y1(Y2) and Y2(Y1) equal the value of x. This suggests that f and g are inverses.
either case, f u f v , and f is one-toone.
Section 4.2 Exponential Functions 2. No; For example the points 2, 4 and
1. y is a linear function of x, so the relation is a function, which we can write as f x 3x 5 .
2, 4 have the same y values but different x values. That is, f a f b 4 , but
Assume that f a f b . 3a 5 3b 5 3a 3b ab Since f a f b implies that a b , then f is one-to-one.
a b.
460
Section 4.2
3.
6.
6 x2 6 y x2 6 x y2
f x
x y 2 6 xy 2 x 6 xy 6 2 x 6 2x y x 6 2x f 1 x x
4.
2 x5 2 y x5 2 x y 5
g x
x y 5 2 xy 5 x 2 xy 5 x 2 5x 2 y x 5 x 2 g 1 x x
7. b x
8. is not; is
9. increasing
10. decreasing
11. ,
12. 0,
13. 0,1
14. y 0
15. is not
16. e
17. e; natural
18. continuously
19. a. 51 0.2
b. 54.8 2264.9364
c. 5 2 9.7385 20. a. 72 0.0204 c. 7 11 635.1453
5.
1 21. a. 4
3
1 c. 4
3
1 22. a. 6
3
1 c. 6
23. a and d
461
d. 5 156.9925 b. 75.9 96,845.2749 d. 7e 198.2507 1.4
64
1 b. 4
0.5 e
0.0906
1 d. 4 1 b. 6
1.4
216
1 0.2817 d. 6
0.5
0.5
0.1436
0.1520
0.0814
24. b and d
0.0599
Chapter 4 Exponential and Logarithmic Functions 25.
29.
Domain: , ; Range: 0,
Domain: , ; Range: 0,
26.
30.
Domain: , ; Range: 0,
Domain: , ; Range: 0,
27.
31.
Domain: , ; Range: 0,
Domain: , ; Range: 0,
28.
32.
Domain: , ; Range: 0,
Domain: , ; Range: 0,
462
Section 4.2 33. f x 3x 2
36. n x 4 x3
Since k 2 , shift the graph of y 3x up 2 units.
Since h 3 , shift the graph of y 4 x right 3 units.
Domain: , ; Range: 2,
Domain: , ; Range: 0,
34. g x 4x 3 4x 3
37. p x 3x4 1 3x4 1 Since h 4 and k 1 , shift the graph of y 3x right 4 units and down 1 unit.
Since k 3 , shift the graph of y 4 x down 3 units.
Domain: , ; Range: 3,
Domain: , ; Range: 1,
35. m x 3x2 3x 2
38. q x 4x4 2 4x 4 2 Since h 4 and k 2 , shift the graph of y 4 x left 4 units and up 2 units.
Since h 2 , shift the graph of y 3x left 2 units.
Domain: , ; Range: 0,
Domain: , ; Range: 2,
463
Chapter 4 Exponential and Logarithmic Functions 39. k x 3x 1 3x
42. v x 4 x
Since a 0 , reflect the graph of y 3x across the x-axis.
1 1 4x 4
Domain: , ; Range: 0,
Domain: , ; Range: , 0
x 1
40. h x 4 1 4 x
x
1 y left 1 unit and down 3 units. 3
Domain: , ; Range: , 0 1 1 3x 3
x 1
1 1 43. f x 3 3 3 3 Since h 1 and k 3 , shift the graph of
x
Since a 0 , reflect the graph of y 4 x across the x-axis.
41. t x 3 x
x
Domain: , ; Range: 3,
x
x 2
1 44. g x 1 4 Since h 2 and k 1 , shift the graph of x
1 y right 2 units and up 1 unit. 4
Domain: , ; Range: 0,
Domain: , ; Range: 1, 464
Section 4.2 x
49. f x e x4
x
1 1 45. k x 2 1 2 3 3 Since a 0 and k 2 , reflect the graph of
Since h 4 , shift the graph of y e x right 4 units.
x
1 y across the x-axis and shift it up 2 3 units.
Domain: , ; Range: 0, 50. g x e x2
Domain: , ; Range: , 2 x
Since h 2 , shift the graph of y e x right 2 units.
x
1 1 46. h x 2 1 2 4 4 Since a 0 and k 2 , reflect the graph of x
1 y across the x-axis and shift it down 4 2 units.
Domain: , ; Range: 0, 51. h x e x 2 Since k 2 , shift the graph of y e x up 2 units. Domain: , ; Range: , 2 47. a. e4 54.5982
b. e3.2 0.0408
c. e 13 36.8020
d. e 23.1407
48. a. e3 0.0498 c. e 7 14.0940
b. e6.8 897.8473 d. ee 15.1543
Domain: , ; Range: 2,
465
Chapter 4 Exponential and Logarithmic Functions 52. k x e x 1 e x 1
r 55. A P 1 n
Since k 1 , shift the graph of y e x down 1 unit.
nt
0.04 A 10,000 1 n
n5
a. For annual compounding, n 1 . 0.04 A 10,000 1 1
15
12,166.53
b. For quarterly compounding, n 4 . 0.04 A 10,000 1 4
Domain: , ; Range: 1,
45
12,201.90
c. For monthly compounding, n 12 .
53. m x e x 3 1 e x 3 Since a 0 and k 3 , reflect the graph of y e x across the x-axis and shift it down 3 units.
0.04 A 10,000 1 12
125
12,209.97
d. For daily compounding, n 365 . 0.04 A 10,000 1 365
3655
12,213.89
e. For continuous compounding, use A Pert . A Pert 10,000e 0.045 12,214.03 r 56. A P 1 n
Domain: , ; Range: , 3
nt
0.035 A 8000 1 n
54. n x e x 4 1 e x 4 Since a 0 and k 4 , reflect the graph of y e x across the x-axis and shift it up 4 units.
n20
a. For annual compounding, n 1 . 0.035 A 8000 1 1
120
15,918.31
b. For quarterly compounding, n 4 . 0.035 A 8000 1 4
420
16,061.05
c. For monthly compounding, n 12 . 0.035 A 8000 1 12
1220
16,093.62
d. For daily compounding, n 365 . Domain: , ; Range: , 4
0.035 A 8000 1 365
466
36520
16,109.48
Section 4.2 62. I Prt 35,000 0.048 20 $33,600
e. For continuous compounding, use A Pert . A Pert 8000e 0.03520 16,110.02
I Pe rt P 35,000e 0.036 20 35,000 71,905.16 35,000 $36,905.16 3.6% compounded continuously for 20 yr results in more interest.
57. A Pert 20,000e10r a. A 20,000e10 0.03 $26,997.18 b. A 20,000e10 0.04 $29,836.49
1 63. A t 10 2
c. A 20,000e10 0.055 $34,665.06 58. A Pert 6000e12r
t 28.9
28.9 28.9
1
57.8 28.9
2
1 1 a. A 28.9 10 10 5 2 2 After 28.9 yr, the amount of 90Sr remaining is 5 g . After one half-life, the amount of substance has been halved.
a. A 6000e12 0.01 $6764.98 b. A 6000e12 0.02 $7627.49 c. A 6000e12 0.45 $10,296.04
1 1 10 2.5 b. A 57.8 10 2 2 After 57.8 yr, the amount of 90Sr remaining is 2.5 g . After two half-lives, the amount of substance has been halved, twice.
59. a. I Prt 10,000 0.055 4 $2200 b. I Pe rt P 10,000e 0.05 4 10,000 12,214.03 10,000 $2214.03
100 28.9
1 0.909 c. A 100 10 2 After 100 yr, the amount of 90Sr remaining is approximately 0.909 g .
c. 5.5% simple interest results in less interest. 60. a. I Prt 15,000 0.067 5 $5025
1 64. A t 0.1 2
b. I Pe rt P 15,000e 0.064 5 15,000 20,656.92 15,000 $5656.92
t 138.4
1 a. A 138.4 0.1 2
138.4 138.4
1 0.1 2
1
0.05 After 138.4 yr, the amount of 210Po remaining is 0.05 mg. After one half-life, the amount of substance has been halved.
c. 6.7% simple interest results in less interest. 61. I Prt 25,000 0.052 30 $39,000
1 b. A 276.8 0.1 2
I Pe rt P 25,000e 0.038 30 25,000 78,169.21 25,000 $53,169.21 3.8% compounded continuously for 30 yr results in more interest.
276.8 138.4
1 0.1 2
2
0.025 After 276.8 yr, the amount of 210Po remaining is 0.025 mg. After one half-life, the amount of substance has been halved.
467
Chapter 4 Exponential and Logarithmic Functions d. P 15 34 1.00804 38
500 138.4
15
1 c. A 500 0.1 0.008 2 After 500 yr, the amount of 90Sr remaining is approximately 0.008 mg.
65. P t 310 1.0097
P 25 34 1.00804 42 25
P 200 34 1.00804
200
169
67. P a 760e0.13a
t
a. P 0 760e0.13 0 760e0 760 mmHg
a. The base is 1.0097 1 , so the graph is an increasing exponential function.
b. P 8.848 760e0.138.848 241 mmHg
b. P 0 310 1.0097 310 In the year 2010, the U.S. population was approximately 310 million. This is the initial population in 2010. 0
68. A t 100e0.0318t a. A15 100e0.031815 $161.12 In the year 2025, approximately $161.12 will be needed to buy something that cost $100 in 2010.
c. P 10 310 1.0097 341 In the year 2020, the U.S. population will be approximately 341 million if this trend continues. 10
b. A 22 100e0.0318 22 $201.29
P 20 310 1.0097 376 20
d.
69. a. T t 78 350 78 e 0.046t
P 30 310 1.0097 414 30
T t 78 272e 0.046t
e. P 200 310 1.0097 2137 In the year 2210 the U.S population will be approximately 2.137 billion. The model cannot continue indefinitely because the population will become too large to be sustained from the available resources. 200
66. P t 34 1.00804
b. T 10 78 272e0.04610 250F c. T 60 78 272e0.046 60 95.2F Yes; after 60 min, the cake will be approximately 95.2F . 70. a. T t 60 122 60 e 0.00351t T t 60 62e 0.00351t
t
a. The base is 1.00804 1 , so the graph is an increasing exponential function.
b. T 12 60 62e0.0035112 119F c. T 24 60 62e0.00351 24 117F Yes; the temperature after 24 hr will be approximately 117F .
b. P 0 34 1.00804 34 In the year 2010, the Canadian population was approximately 34 million. This is the initial population in 2010. 0
71. a. V 5 120 0.8 39 The resale value is $39,000. 5
c. P 5 34 1.00804 35 In the year 2015, the Canadian population will be approximately 35 million if this trend continues. 5
b. V 1 120 0.8 96 The depreciation after 1 yr is 120,000 96,000 24,000 . 1
C h 18 36 22 h 76h C 800 76 800 60,800
468
Section 4.2 It costs the farmer $60,800 $24,000 $84,800 to run the tractor for 800 hr during the first year.
e. x is between 3 and 4. 76. a. 1 c. 3
72. a. V 5 10,000 0.9 $6561 4
d. x is between 1 and 2.
b. V 8 10,000 0.9 $4300 8
1 73. a. P x 4
b. 2
e. x is between 2 and 3.
x
77. a. See part (d). b. Yes
2
c. Domain: , ; Range: 0,
3
d.
1 1 P 2 0.0625 4 16 1 1 P 3 0.015625 4 64 4
1 1 P 4 0.003906 4 256 5
1 1 P 5 0.000977 4 512
b. The probability decreases. e. Domain: 0, ; Range: ,
10
1 c. P 10 9.54 107 4 It is very unlikely.
1 74. a. P x 2
f. f 1 1 0 ; f 1 2 1 ; f 1 4 2 78. a. See part (d).
x
b. Yes c. Domain: , ; Range: 0,
2
1 1 P 2 0.25 2 4
d.
3
1 1 P 3 0.125 2 8 4
1 1 P 4 0.0625 2 16 10
1 1 P 10 0.000977 2 512
b. It is unlikely. 75. a. 2
e. Domain: 0, ; Range: ,
b. 3
1 f. g 1 1 0 ; g 1 3 1 ; g 1 1 3
c. 4 d. x is between 2 and 3.
469
Chapter 4 Exponential and Logarithmic Functions 79. a.
b. 0
c. 80. a. c.
d.
b.
b. 0
c.
d.
81. The range of an exponential function is the set of positive real numbers; that is, 2 x is nonnegative for all values of x in the domain. d. 82. The base is variable and the exponent is constant rather than the other way around. 83. f x 2 x
f b f a 0.25 1 0.75 0.375 ba 20 2
f b f a 0.0625 0.25 ba 42 0.1875 2 0.0938 f b f a 0.015625 0.0625 ba 64 0.046875 2 0.0234
a.
f b f a 1 0.25 0.75 0.375 ba 0 2 2
b.
f b f a 4 1 3 1.5 ba 20 2
a.
c.
f b f a 16 4 12 6 ba 42 2
f b f a 1 9 8 4 ba 0 2 2
b.
d.
f b f a 64 16 48 24 ba 64 2
f b f a 0.1111 1 0.8888 ba 20 2 0.4444
1 86. f x 3
c.
84. f x 3x a.
f b f a 1 0.1111 0.8888 ba 0 2 2 0.4444
b.
f b f a 9 1 8 4 ba 20 2
c.
f b f a 81 9 72 36 ba 42 2
d.
a.
f b f a 0.0123 0.1111 ba 42 0.0988 2 0.0494 f b f a 0.0014 0.0123 ba 64 0.0109 2 0.0055
87. 3x 2e x 6 xe x 0
f b f a 729 81 648 d. 324 ba 64 2
1 85. f x 2
x
3xe x x 2 0 xe x x 2 0
x
x 0 or e x 0 or x 2 e x 0 has no solution because e raised to a power is always greater than 0, so the real solutions are 0, 2 .
f b f a 1 4 3 1.5 ba 0 2 2
470
Section 4.3 x 2e x e x 0
88.
2
e x e x e x e x 93. 2 2
ex x2 1 0 e x 1 x 1 0
e x 0 or x 1 or x 1 e x 0 has no solution because e raised to a power is always greater than 0, so the real solutions are 1,1 .
e
89. a. e x eh e xh
b. e x
2
2
2x
d. e x e x e x x e0 1
e1
2x
91. e x e x
2x
2x
x
x
e 2e e 2
2x
0
e 2e e 2
2x
0
1 2x e 2 x e 2 x e e 2 x 2 2
x h f x h f h 2 x h 2 x 2 2 1 h h h
97.
e4 x 2e2 x 1 e2 x
2 x
e2 x 2 e 2 x
The graphs of Y2 and Y3 are close approximations of Y1 = ex near x = 0.
e 2e 1 e2 x 4x
2x
Section 4.3 Logarithmic Functions 1.
2 x
e2 x 2 e 2 x
92. e x e x
96.
e 1 e e 1 e 1
b. e e e
2x
x h f x h f x e x h e x e e 1 95. h h h
1
90. a. e xh e x e x eh e x e x eh 1 2x
1 2 e x e x e x e x 4
e e xh eh
4x
e x e x e x e x 94. 2 2 2
x
e. e 2 x e2 x
1 2x e 2 e 2 x e 2 x 2 e 2 x 4 1 4 1 4
x
c.
2
f x 3 x 4
2.
g x 2 x 7 y 2x 7 x 2y 7 x 7 2y x7 y 2 x7 g 1 x 2
y 3 x4 x 3 y4 x3 y 4 x3 4 y f 1 x x 3 4
471
Chapter 4 Exponential and Logarithmic Functions 3. 3 4 81
4. 2 3 8
1 36
6. 4 3
5. 6 2
1 30. 2
1 64
32 log1 2 32 5
31. 109 1,000,000,000 log1,000,000,000 9
7. logarithmic
32. 101 10 log10 1
8. logarithm; base; argument 9. exponential
10. logarithmic
11. common; natural
12. 10; e
13. 0; 0
14. 1; 1
33. a7 b log a b 7 34. M 3 N log M N 3 35. log 3 9 y 3 y 9 32 y2
15. x; x 16. increasing; decreasing 17. x 0 ; vertical
5
36. log 2 16 y 2 y 16 24 y4
18. left; downward
19. log8 64 2 82 64
37. log 5 5 y
20. log9 81 2 92 81
5 y 5 51 y 1
1 1 4 104 21. log 10,000 10,000
38. log 6 6 y 6 y 6 61 y 1
1 1 6 106 22. log 1,000,000 1,000,000
39. log100,000,000 y
23. log4 1 0 40 1
10 y 100,000,000 108 y8
24. log8 1 0 80 1
40. log10,000,000 y
25. log a b c ac b
10 y 10,000,000 107 y7
26. log x M N x N M 27. 53 125 log5 125 3
1 41. log 2 y 16
28. 25 32 log 2 32 5 1 29. 5
1 1 4 24 16 2 y 4
2y
3
125 log1 5 125 3
472
Section 4.3 49. log1 7 49 y
1 42. log 3 y 9
y
1 1 49 7 7
1 1 32 9 32 y 2
3y
2
y 2
50. log1 4 16 y
1 43. log y 10
y
1 1 16 4 4
1 1 1 101 10 10 y 1
10 y
2
y 2
1 51. log1 2 y 32
1 y 44. log 10,000
y
1 1 1 2 32 2 y5
1 1 4 104 10,000 10 y 4
10 y
5
1 52. log1 6 y 36
45. ln e y 6
y
1 1 1 6 36 6 y2
e y e6 y6
46. ln e10 y
2
53. log 0.00001 y
e y e10 y 10
10 y 0.00001 105 y 5
1 47. ln 3 y e 1 e y 3 e 3 e y 3
54. log 0.0001 y 10 y 0.0001 104 y 4
55. a. 104 46,832 105 , so 4 log 46,832 5 . log 46,832 4.6705
1 48. ln 8 y e
b. 106 1,247,310 107 , so 6 log1,247,310 7 . log1,247,310 6.0960
1 e 8 e8 y 8
ey
c. 101 0.24 100 , so 1 log 0.24 0 . log 0.24 0.6198
473
Chapter 4 Exponential and Logarithmic Functions 58. a. f 1860 ln1860 7.5283
d. 106 0.0000032 105 , so 6 log 0.0000032 5 . log 0.0000032 5.4949
b. f 0.0694 ln 0.0694 2.6679
e. 105 5.6 105 106 , so 5 log 5.6 105 6 .
c. f
d. f 2 ln 2 1.8379
log 5.6 10 5.7482 5
3
3
f. f 8.5 10 ln 8.5 10 37.0039
e. f 1.3 1012 ln 1.3 1012 27.8934
2
e. 10 5.1 10 10 , so 3 log 5.1 103 2 .
17
log 5.1 103 2.2924 56. a. 105 293,416 106 , so 5 log 293,416 6 . log 293,416 5.4675 b. 102 897 103 , so 2 log897 3 . log897 2.9528 c. 102 0.038 101 , so 2 log 0.038 1 . log 0.038 1.4202
log 8.2 102 1.0862 57. a. f 94 ln 94 4.5433 b. f 0.182 ln 0.182 1.7037 c. f
155 ln 155 2.5217
d. f 4 ln 4 2.5310
f. f 7.1 10 ln 7.1 10 7.2502
e. f 3.9 109 ln 3.9 109 22.0842 4
62. log d d 1
63. 5log5 x y x y
64. 4log4 ac a c
65. ln eab a b
66. ln e x 1 x2 1
67. log 5 1 0
68. log 1 0
0 1 2
f. 102 8.2 102 101 , so 2 log 8.2 102 1 .
61. logc c 1
1
log 9.1 108 8.9590
60. log6 67 7
2
59. log 4 411 11
x
e. 108 9.1 108 109 , so 8 log 9.1 108 9 .
17
2
69. Interchange the x- and y-coordinates of the ordered pairs from its inverse function.
d. 104 0.00061 103 , so 4 log 0.00061 3 . log 0.00061 3.2147
87 ln 87 2.2330
4
474
y 3x 1 9 1 3 1 3 9
x 1 9 1 3 1 3 9
y log 3 x 2 1 0 1 2
Section 4.3 70. Interchange the x- and y-coordinates of the ordered pairs from its inverse function. x 2 1 0 1 2
y 5x 1 25 1 5 1 5 25
72. Interchange the x- and y-coordinates of the ordered pairs from its inverse function.
y log 5 x
x 1 25 1 5 1 5 25
x
2
2 1 0
1 0 1 2
1 2
71. Interchange the x- and y-coordinates of the ordered pairs from its inverse function. x 2 1 0 1 2
1 y 3 9 3 1 1 3 1 9
x
x 9 3 1 1 3 1 9
1 y 5
x
25 5 1 1 5 1 25
x 25 5 1 1 5 1 25
y log1 5 x 2 1 0 1 2
73. Interchange the x- and y-coordinates of the ordered pairs from its inverse function.
y log1 3 x 2 1 0
y ex 1 2 2 0.14 e 1 1 0.37 e 0 1 1 e 2.72 2 e 2 7.39 x
1 2
475
x 0.14 0.37 1 2.72 7.39
y ln x 2 1 0 1 2
Chapter 4 Exponential and Logarithmic Functions 76. a. The graph of y log 2 x 3 is the graph
74. Interchange the x- and y-coordinates of the ordered pairs from its inverse function. x 2 1 0 1 2
y 1 100 1 10 1 10 100
x
x 1 100 1 10 1 10 100
of y log 2 x shifted to the left 3 units.
y log x 2 1 0 1 2
b. Domain: 3, ; Range: , c. Vertical asymptote: x 3 77. a. The graph of y 2 log3 x is the graph of y log3 x shifted up 2 units.
75. a. The graph of y log3 x 2 is the graph of y log3 x shifted to the left 2 units.
b. Domain: 0, ; Range: , c. Vertical asymptote: x 0 78. a. The graph of y 3 log 2 x is the graph of y log 2 x shifted up 3 units. b. Domain: 2, ; Range: , c. Vertical asymptote: x 2
b. Domain: 0, ; Range: , c. Vertical asymptote: x 0 476
Section 4.3 b. Domain: 0, ; Range: ,
79. a. The graph of y log3 x 1 3 is the graph of y log3 x shifted right 1 unit and down 3 units.
c. Vertical asymptote: x 0 82. a. The graph of y log 2 x is the graph of y log 2 x reflected across the x-axis.
b. Domain: 1, ; Range: , c. Vertical asymptote: x 1
b. Domain: 0, ; Range: ,
80. a. The graph of y log 2 x 2 1 is the
c. Vertical asymptote: x 0
graph of y log 2 x shifted right 2 units and down 1 unit.
b. Domain: 2, ; Range: , c. Vertical asymptote: x 2 81. a. The graph of y log3 x is the graph of y log3 x reflected across the x-axis.
477
83. f x log 8 x 8 x 0 x 8 x8
Domain: , 8
84. g x log 3 x 3 x 0 x 3 x3
Domain: , 3
85. h x log 2 6 x 7 6x 7 0 6 x 7 7 x 6
7 Domain: , 6
86. k x log3 5x 6 5x 6 0 5 x 6 6 x 5
6 Domain: , 5
Chapter 4 Exponential and Logarithmic Functions
87. m x ln x 2 14
91. r x log3 4 x
x 2 14 0
4 x2 0
x 2 14 The square of any real number is positive, so this condition is always true. Domain: ,
This is true for all real x, except when 4 x 0 , or 4 x . Domain: , 4 4,
92. s x log5 3 x
88. n x ln x 11 2
x 11 0
This is true for all real x, except when 3 x 0, 3 x. Domain: , 3 3,
x 2 11 The square of any real number is positive, so this condition is always true. Domain: ,
I 106.9 I 0 93. a. M log log I0 I 0
log106.9 6.9
x x 12 0 Find the real zeros of the related equation. x 2 x 12 0 2
I 103.2 I 0 b. M log log I0 I 0
x 3 x 4 0
log103.2 3.2
x 3 or x 4 The boundary points are 3 and 4. Sign of x 3 :
Sign of x 3 x 4 :
Sign of x 4 :
3
Domain: , 3 4,
90. q x log x 2 10 x 9
c.
4
log105.4 5.4
x 10 x 9 0 Find the real zeros of the related equation. x 2 10 x 9 0
I 105.8 I 0 b. M log log I0 I 0
x 9 x 1 0
log105.8 5.8
x 9 or x 1 The boundary points are 9 and 1 .
Sign of x 9 x 1 :
Sign of x 1 :
106.9 I 0 106.93.2 103.7 5012 3.2 10 I 0 The Loma Prieta earthquake was approximately 5012 times more intense than an earthquake with magnitude 3.2.
I 105.4 I 0 94. a. M log log I0 I 0
2
Sign of x 9 :
2
3 x2 0
2
89. p x log x2 x 12
2
c.
9 1
Domain: , 9 1,
478
105.8 I 0 105.85.4 100.4 2.5 5.4 10 I 0 The earthquake was approximately 2.5 times more intense than originally thought.
Section 4.3 100. a. d log 2 d
1015 I 0 95. a. L 10log 10log1015 I 0 10 15 150 dB
1 log 2 d 1 log 2 d d 21 2 mm
109 I 0 10log109 b. L 10log I
b. d log 2 d
0
10 9 90 dB
5 log 2 d 5 log 2 d
1015 I 0 c. 10159 106 1,000,000 9 10 I 0 The sound of a jet plane taking off is 1,000,000 times more intense than the noise from city traffic.
d 25 32 mm
101. a. log3 x 1 4 34 x 1 b. 34 x 1 81 x 1 80 x The solution set is 80 .
1010 I 0 10log1010 96. a. L 10log I 0 10 10 100 dB
c.
107 I 0 10log107 b. L 10log I
log 3 x 1 4
log 3 80 1 0 4
0
log 3 810 4
10 7 70 dB
40 4
10
c.
10 I 0 10107 103 1000 7 10 I 0 The sound of a motorcycle is 1000 times more intense than a vacuum cleaner.
102. a. log2 x 5 4 24 x 5 b. 24 x 5 16 x 5 21 x The solution set is 21 .
b. pH log 1 10 2 2
97. a. pH log 5.0 103 2.3 2
c. log 2 x 5 4
c. Lemon juice is more acidic.
log 2 21 5 0 4
12.7 b. pH log 4.1 10 9.4
98. a. pH log 2.0 10
13
log 2 16 0 4 40 4
10
103. a. log 4 7 x 6 3 43 7 x 6
c. Bleach is more basic. 99.
b. 43 7 x 6 64 7 x 6 70 7 x 10 x The solution set is 10 .
d log 2 d
128 log 2 128 7 8 log 2 8 3 4 log 2 4 2 1 log 2 1 0 0.5 log 2 0.5 1 1 479
Chapter 4 Exponential and Logarithmic Functions c.
log 4 7 x 6 3
111. a. log4 64 log4 4 3 1 2
log 4 7 10 6 0 3
64 b. log 4 log 4 16 2 4
log 4 70 6 0 3 log 4 64 0 3
c. They are the same.
30 3
112. a. log100,000 log100 5 2 3
104. a. log5 9 x 11 2 52 9 x 11
100,000 b. log log1000 3 100
b. 52 9 x 11 25 9 x 11 36 9 x 4 x The solution set is 4 . c.
c. They are the same. 113. a. log 2 25 5 b. 5 log 2 2 5 1 5
log 5 9 x 11 2
c. They are the same.
log 5 9 4 11 0 2
114. a. log7 76 6
log 5 36 11 0 2
b. 6 log7 7 6 1 6
log 5 25 0 2
c. They are the same.
20 2
105. log3 log 4 64 log3 3 1
115. a.
1 106. log 2 log1 2 log 2 2 1 4
ln 2 r ln 2 t 0.035 19.8 yr 0.035 t r
ln 2 17.3 yr 0.04 ln 2 t 0.06 11.6 yr 0.06 ln 2 t 0.08 8.7 yr 0.08
b. t 0.04
1 1 107. log16 log81 3 log16 4 2
1 1 108. log 4 log16 4 log 4 2 2
Ar log 1 12 P 116. a. n r log 1 12
109. a. log2 2 log 2 4 1 2 3 b. log 2 2 4 log 2 8 3 c. They are the same.
3000 0.06 log 1 12 200 0.06 log 1 12
110. a. log3 3 log3 27 1 3 4 b. log3 3 27 log3 81 4 c. They are the same.
480
log 0.925 16 months log 1.005
Section 4.3 Ar log 1 12 P b. n r log 1 12
d.
x 1 121. t x log 4 x 3 x 1 0 x3 The boundary points are 1 and 3.
128,000 0.04 log 1 12 611.09 360 months 0.04 log 1 12
Sign of x 1 :
23
Sign of x 3 :
34
Sign of
23.7797 b. log 6.626 10 33.1787
117. a. log 6.022 10
b. log 1.602 10 18.7953 19
Sign of x 4 :
inclusive.
c.
f b f a 1.3010 1 0.0301 ba 20 10
d.
f b f a 1.4771 1.3010 0.0176 ba 30 20
120. a.
Sign of
f b f a 0 0.3010 0.6020 ba 1 0.5 f b f a 1 0 1 0.1111 ba 10 1 9
3
x2 x4
:
2
4
Domain: , 2 4, 123. s x ln
x 5 1
x 5 1 0 x 5 1 x 5 1 x 4
f b f a 0 0.6931 1.3862 ba 1 0.5
124. v x ln
Domain: 4,
x 8 1
x 8 1 0
f b f a 2.3026 0 b. 0.2558 ba 10 1
c.
Sign of x 2 :
c. Given a number a 10n , then log a 10n is between n and n + 1,
b.
:
x 2 122. r x log5 x 4 x2 0 x4 The boundary points are 2 and 4.
118. a. log 2.9979 108 8.4768
119. a.
x3
Domain: ,1 3,
inclusive.
x 1
1
c. Given a number a 10n , then log a 10n is between n and n + 1,
f b f a 3.4012 2.9957 0.0406 ba 30 20
x 8 1 x 8 1 x9
f b f a 2.9957 2.3026 0.0693 ba 20 10
481
Domain: 9,
Chapter 4 Exponential and Logarithmic Functions 1 125. c x log x 6 x6 0 x60 x6
128.
Domain: 6,
1 126. d x log x 8 x8 0 x8 0 x 8
It appears that the functions are inverses. 129.
Domain: 8,
127. a. The graphs are the same. 130.
The graphs match closely on the interval 0, 2 b. ln1.5 0.4010 The graphs are the same.
Problem Recognition Exercises: Analyzing Functions 1. f x 3 a. ,
h. Graph E b. 3
2. g x 2 x 3 a. ,
c. f x 3 03 No x-intercept
b. ,
c. g x 2 x 3 0 2x 3 3 2x 3 x 2 3 x-intercept: , 0 2
d. f x 3 f 0 3
y-intercept: 0, 3 e. No asymptotes f. The slope is 0, so the function is never increasing.
d. g x 2 x 3 g 0 2 0 3 3
g. The slope is 0, so the function is never deceasing.
y-intercept: 0, 3 482
Problem Recognition c. h x 3 x 2
e. No asymptotes f. The slope is positive, so the function is always increasing. ,
0 3 x2 0 x2 2 x x-intercept: 2, 0
g. The slope is positive, so the function is never deceasing. h. Graph G
d. h x 3 x 2 h 0 3 0 2 3 2
3. d x x 3 4 1 x 3 4 The graph of this function is the graph of y x 2 shifted right 3 units and down 4 units. 2
2
y-intercept: 0, 3 2 e. No asymptotes
b. 4,
a. , 2
5. k x
0 x 3 4 2
2 x 1
a. Undefined for x 1 . ,1 1,
2
b. The function is never 0. , 0 0,
2 x 3 x 1 or x 5 x-intercepts: 1, 0 and 5, 0
2 x 1 2 0 x 1 02 No x-intercept
c. k x
d. d x x 3 4 2
d 0 0 3 4 9 4 5 2
y-intercept: 0, 5 e. No asymptotes
2 x 1 2 k 0 2 0 1 y-intercept: 0, 2
d. k x
f. Vertex: 3, 4 Since a is positive, the function is increasing to the right of the vertex. 3,
e. Vertical asymptote: x 1 The degree of the numerator is 0. The degree of the denominator is 1. Horizontal asymptote: y 0
g. Since a is positive, the function is decreasing to the left of the vertex. , 3 h. Graph N
f. Never increasing
4. h x x 2 The graph of this function is the graph of y 3 x shifted right 2 units. 3
a. ,
f. ,
g. Never decreasing h. Graph B
c. d x x 3 4 4 x 3
g. ,1 1,
h. Graph L 6. z x
b. ,
3x x2
a. Undefined for x 2 . , 2 2,
483
Chapter 4 Exponential and Logarithmic Functions
b.
3x x2 3x 6 3x 6 0 No solution The function does not cross its horizontal asymptote, so it is never 3. , 3 3,
4 d. p x 3
3x x2 3x 0 x2 0 3x 0 x x-intercept: 0, 0
f. ,
3
0
4 p 0 1 3
y-intercept: 0,1 e. Horizontal asymptote: y 0
c. z x
g. Never decreasing h. Graph M 8. q x x2 6 x 9 a. , b. Vertex: b 6 6 x 3 2a 2 1 2
3x d. z x x2 3 0 z 0 0 02 y-intercept: 0, 0
y 3 6 3 9 9 18 9 0 2
, 0
c. q x x 2 6 x 9
e. Vertical asymptote: x 2 The degree of the numerator is 1. The degree of the denominator is 1. 3 Horizontal asymptote: y 3 1
0 x2 6 x 9
0 x2 6 x 9 0 x 3
f. , 2 2,
d. q x x 2 6 x 9
x
q 0 0 6 0 9 9 2
y-intercept: 0, 9
a. ,
e. No asymptotes
b. 0,
f. , 3
4 c. p x 3
2
x 3 x-intercept: 3, 0
g. Never decreasing h. Graph A 4 7. p x 3
x
x
g. 3,
h. Graph C 9. m x x 4 1 The graph of this function is the graph of y x shifted right 4 units and down 1 unit.
x
4 0 Not possible 3 No x-intercept
a. ,
484
b. 1,
Problem Recognition c. m x x 4 1
c. r x 3 x
0 x 4 1
0 3 x 0 3 x x3 x-intercept: 3, 0
1 x 4 x 4 1 or x 4 1 x5 x3 x-intercepts: 3, 0 and 5, 0
d. r x 3 x
d. m x x 4 1
r 0 3 0 3
m 0 0 4 1 4 1 3
y-intercept: 0, 3
y-intercept: 0, 3 e. No asymptotes
f. 4,
g. , 4
h. Graph I
f. Never increasing
g. , 3
h. Graph F
a. 3,
0 x3 0 x3 3 x x-intercept: 3, 0
0 x 3 x 3 x 3 or x 3 x-intercepts: 3, 0 and 3, 0
d. s x x 3 s 0 0 3 3 No real solution No y-intercept
d. n x x 3 n 0 0 3 3
y-intercept: 0, 3
e. No asymptotes
e. No asymptotes
f. , 0
g. 0,
h. Graph D
b. 0,
c. s x x 3
b. , 3
c. n x x 3
f. 3,
g. Never decreasing h. Graph K 13. t x e x 2 The graph of this function is the graph of f x e x shifted up 2 units.
11. r x 3 x x 3
a. ,
The graph of this function is the graph of f x x reflected across the y-axis and shifted right 3 units. a. , 3
e. No asymptotes
12. s x x 3 The graph of this function is the graph of f x x shifted right 3 units.
10. n x x 3 The graph of this function is the graph of y x reflected across the x-axis and shifted up 3 units. a. ,
b. 2,
c. t x e x 2 0 ex 2
b. 0,
2 e x No solution No x-intercept
485
Chapter 4 Exponential and Logarithmic Functions c. v x ln x 2
d. t x e x 2
0 ln x 2
t 0 e 2 1 2 3 0
y-intercept: 0, 3
e0 x 2 1 x 2 1 x x-intercept: 1, 0
e. Horizontal asymptotes: y 2 f. ,
g. Never decreasing
d. v x ln x 2
h. Graph H
v 0 ln 0 2 ln 2
14. v x ln x 2 The graph of this function is the graph of f x ln x shifted left 2 units.
y-intercept: 0, ln 2
e. Vertical asymptotes: x 2
a. 2,
f. 2,
b. ,
h. Graph J
g. Never decreasing
Section 4.4 Properties of Logarithms 1. a. log2 4 log2 8 2 3 5
12. quotient; logb x logb y
b. log 2 4 8 log 2 32 5 13. p logb x
14.
log a x log a b
15. 10; e
16.
log x ln x ; log 5 ln 5
2. a. log5 5 log5 25 1 2 3 b. log5 5 25 log5 125 3 3. a. log4 256 log4 16 4 2 2
17. log5 125z log5 125 log5 z 3 log5 z
256 log 4 16 2 b. log 4 16
18. log7 49k log7 49 log7 k 2 log7 k 19. log 8cd log8 log c log d
4. a. log1,000,000 log10 6 1 5
20. log 24vw log 24 log v log w
1,000,000 b. log log100,000 5 10
5. a. log3 81 4
b. log3 34 4
6. a. log8 64 2
b. log8 82 2
7. 1
8. 0
9. x
10. x
21. log 2 x y z log 2 x y log 2 z 22. log3 a b c log3 a b log3 c p 23. log12 log12 p log12 q q m 24. log9 log9 m log9 n n
11. logb x logb y 486
Section 4.4 a4 38. log8 9 log8 a 4 log8 b9 log8 c b c
e 25. ln ln e ln 5 1 ln 5 5
4log8 a 9log8 b log8 c
x 26. ln ln x ln e ln x 1 e
10 12 39. log log10 log a 2 b 2 2 2 a b 1 1 log a 2 b 2 2
m2 n log m2 n log100 27. log 100
log m n 2 2
d 2 1 12 2 40. log log d 1 log10, 000 10, 000
1000 28. log 2 log1000 log c 2 1 c 1
1 log d 2 1 4 2 1 4 log d 2 1 2
3 log c 2 1
29. log 2 x 3 4log 2 x 3 4
30. log 8t 3 2log 8t 3
3 xy 13 41. ln 2 ln xy ln w ln z 2 wz
2
3 31. log 6 7 x3 log 6 x3 7 log 6 x 7
1 ln x ln y ln w 2ln z 3 1 1 ln x ln y ln w 2ln z 3 3
3 32. log8 4 x3 log8 x3 4 log8 x 4
4 pq 14 42. ln 3 ln pq ln t 3 ln m tm
33. ln 2kt kt ln 2 34. ln 0.5 rt ln 0.5 rt
1 ln p ln q 3ln t ln m 4 1 1 ln p ln q 3ln t ln m 4 4
1 35. log 7 mn 2 log 7 7 log 7 m log 7 n 2 7 1 log 7 m 2log 7 n
a2 4 a2 4 43. ln 4 ln e3 e3
1 36. log 4 t 3v log 4 16 log 4 t 3 log 4 v 16
14
1 a2 4 ln 4 e3
1 ln a 2 4 ln e3 4 1 1 ln a 2 4 3 4 4 1 3 ln a 2 4 4 4
2 3log 4 t log 4 v
p5 log log 6 p 5 log 6 q log 6 t 3 37. 6 3 qt 5log 6 p log 6 q 3log 6 t
487
Chapter 4 Exponential and Logarithmic Functions e2 e2 44. ln 2 ln 2 c 5 c 5
15
5
48. log 2 4 y 2 log 2 y 2 1 12 log 2 y 2 4 1 12 log 2 y log 2 2 4 1 1 log 2 y log 2 2 4 2
1 e2 ln 2 5 c 5
12 14
1 ln e 2 ln c 2 5 5 1 1 2 ln c 2 5 5 5 2 1 ln c 2 5 5 5
1 1 log 2 y 1 4 2 1 1 log 2 y 4 8
2 x x2 3 8 45. log 4 3x 8 log 2 x x 2 3 log 4 3 x
49. log15 3 log15 5 log15 3 5 log15 15 1
log 2 log x log x 2 3 log 4 3 x 1 log 2 log x 8log x 2 3 log 4 3 x 2 8
12
50. log12 8 log12 18 log12 8 18 log12 144 2 98 51. log 7 98 log 7 2 log 7 log 7 49 2 2
5 y 4 x 1 7 46. log 3 2 7 x
144 log 6 36 2 52. log 6 144 log 6 4 log 6 4
7 log 5 y 4 x 1 log 3 2 7 x 7 13 log 5 log y log 4 x 1 log 2 7 x 1 log 5 log y 7 log 4 x 1 log 2 7 x 3
53. log150 log 3 log 5 log150 log 3 log 5 150 log log10 1 3 5
54. log 3 693 log 3 33 log 3 7
12 47. log5 3 x 5 log5 x 5 1 12 log 5 x 5 3 1 12 log 5 x log 5 5 3 13
log 3 693 log 3 33 log 3 7 693 log 3 log 3 3 1 33 7
55. 2log 2 x log 2 t log 2 x 2 log 2 t log 2 x 2t
1 1 log 5 x log 5 5 3 2
56. 5log 4 y log 4 w log4 y5 log4 w log4 y 5 w
1 1 log 5 x 1 3 2 1 1 log 5 x 3 6
57. 4 log8 m 3log8 n 2 log 8 p log8 m 4 log8 n3 log8 p 2
log8 m 4 log8 n3 log8 p 2 m4 log8 3 2 n p
488
Section 4.4 58. 8log 3 x 2 log 3 z 7 log 3 y log 3 x8 log 3 z 2 log 3 y 7
log 3 x8 log 3 z 2 log 3 y 7
63.
1 1 ln x 1 ln x 1 2 2 1 ln x 1 ln x 1 2 x 1 ln x 1
12
60.
ln
64.
13
x 2 1 ln x 1
ln 3
1 log 2 w log 2 w2 100 log 2 w 10 4 w1 4 w2 100 log 2 w 10 w1 4 w 10 w 10 log 2 w 10 14 log 2 w w 10 log 2 4 w w 10
x 1 x 1
1 1 ln x 2 1 ln x 1 3 3 1 ln x 2 1 ln x 1 3
x8 log 3 2 7 z y
59.
1 log 4 p log 4 q 2 16 log 4 q 4 3 p1 3 q 2 16 log 4 q4 p1 3 q 4 q 4 log 4 q4 13 log 4 p q 4 log 4 3 p q 4
x2 1 x 1
65.
1 2 61. 6log x log y log z 3 3 1 6log x log y 2log z 3 13 1 log x 6 log yz 2 log x 6 log yz 2 3 x6 x6 log log 13 2 yz 2 3 yz
1 6ln x 2 ln x ln x 2 2 1 6 ln x 2 ln x ln x 2 2 6 6 1 x x 2 1 x 2 ln ln 2 x2 2 x x 2 6 ln x x 23 ln x
1 3 62. 15log c log d log k 4 4 1 15log c log d 3log k 4 14 1 log c15 log dk 3 log c15 log dk 3 4 c15 c15 log log 14 4 3 dk 3 dk
66.
12
x 23 ln 1 2 x
1 12ln x 5 ln x ln x3 3 1 12 ln x 5 ln x ln x 3 3 12 12 1 x x 5 1 x 5 ln ln 2 3 x3 3 x
x 512 ln 2 x 4 x 5 ln 3 2 x
489
13
x 5 4 ln 2 3 x
Chapter 4 Exponential and Logarithmic Functions
77. log b 100 log b 2 5 2log b 2 5
67. log 8 y 2 7 y log y 1
2
2 log b 2 log b 5
log 8 y 2 7 y y 1
2 0.356 0.827 2.366
y 8 y 7 log log 8 y 7 y
78. log b 225 log b 3 5 2log b 3 5 2
2 log b 3 log b 5
68. log 9t 3 5t log t 1
2 0.565 0.827 2.784
t 9t 5 log 9t 5 log t log 9t 3 5t t 1
79. a. 8 15 16
2
2
23 15 24 3 log 2 15 4 Between 3 and 4
69. logb 15 log b 3 5 log b 3 log b 5 70. logb 10 log b 2 5 log b 2 log b 5
c. 23.9069 15
0.356 0.827 1.183
80. a. 9 15 27
71. logb 81 logb 3 4logb 3 4 0.565 2.26
32 15 23 2 log 3 15 3 Between 2 and 3
4
72. logb 125 log b 53 3log b 5 3 0.827
b. log3 15
2.481
0.356 2 0.827 2.01
log15 2.4650 log 3
c. 32.4650 15
73. logb 50 logb 2 52 log b 2 2log b 5
81. a.
1 3 5 50 3 51 0 log 5 3 1 Between 0 and 1
74. logb 12 log b 3 2 log b 3 2log b 2 2
log15 3.9069 log 2
b. log 2 15
0.565 0.827 1.392
0.565 2 0.356 1.277
b. log5 3
15 3 5 75. log b log b 2 2
log 3 0.6826 log 5
c. 50.6826 3
log b 3 log b 5 log b 2
82. a. 1 5 8
0.565 0.827 0.356 1.036
80 5 81 0 log8 5 1 Between 0 and 1
6 2 3 76. log b log b 5 5
b. log8 5
log b 2 log b 3 log b 5 0.356 0.565 0.827 0.094
log 5 0.7740 log8
c. 80.7740 5
490
Section 4.4 83. a. 0.25 0.3 0.5 22 0.3 21 2 log 2 0.3 1 Between 2 and 1 b. log 2 0.3
1 log e 1 ln10 0 ln e ln10 ln10 ln10 0 ln e ln10 0 ln10
90. ln10 0
log 0.3 1.7370 log 2
c. 21.7370 0.3 84. a. 0.125 0.2 0.25 23 0.2 22 3 log 2 0.2 2 Between 3 and 2 b. log 2 0.2 c. 2
2.3219
1 1 91. False; log5 125 log 5 125
(The left side is 3 and the right side is
log 0.2 2.3219 log 2
log 4.68 107
(The left side is 2 and the right side is
log 2 25.4800 25.4800 2 46,800,000
1 93. log 4 0 log 4 p p
log 2.54 10
86. log 2 2.54 1010
10
log 4 p 0 log 4 p 1
log 4 p 0 log 4 p
log 2 34.5641 234.5641 2.54 1010
87. log 4 5.68 106
log 5.68 106 log 4
True
1 94. log8 0 log8 w w log8 w 0 log8 w 1
8.7128
log8 w 0 log8 w
48.7128 5.68 106
88. log 4 9.84 105
1 .) 3
1 1 92. False; log 6 36 log 6 36
0.2
85. log 2 4.68 107
True
log 9.84 105
True
95. False; log 10 10 log10 log10 (The left side is 2 and the right side is 1.)
log 4 6.6555 46.6555 9.84 105
100 log100 96. False; log 10 log10 (The left side is 1 and the right side is 2.)
1 ln10 ln e 1 0 ln10 ln10 1 1 0 ln10 ln10
97. log 2 7 y log 2 10 log 2 7 y
89. log e 0
log 2 7 y 1 0 log 2 7 y log 2 7 y 0 log 2 7 y
True
491
True
1 .) 2
Chapter 4 Exponential and Logarithmic Functions 98. log 4 3d log 4 10 log 4 3d
c c2 x2 104. ln c c2 x2
log 4 3d 1 0 log 4 3d log 4 3d 0 log 4 3d
True
c c2 x2 c c2 x2 ln c c2 x2 c c2 x2
99. The given statement log5 5 log5 25 is not defined because the arguments to the logarithmic expressions are not positive real numbers.
b.
2
105. log 2 5 log5 9
1h
log 5 log 9 log 9 log 2 9 log 2 log 5 log 2
implies that b M x and b N y . Then x bM N b M N . Writing the expression y b x b M N in logarithmic form, we have y x logb M N , or equivalently, y
x logb logb x logb y as desired. y
b b 2 4ac b b 2 4ac 103. log log 2a 2a
108. Let y logb x , which implies that b y x . Raising both sides to the p power yields
b x or b x . Converting back to y p
b b 2 4ac b b 2 4ac log 2a 2a
2
107. Let M logb x and N logb y , which
1 x x2 1 ln x x2 1 x x2 1
2
log11 log 4 log 3 log11 log 4 log 3 4 log 3
1 ln 2 x x 1
2
106. log 3 11 log11 4
1
x x2 1 ln x x 2 1 ln 2 2 x x 1
ln x 2 ln c c x 2 ln x
ln x h ln x 1 ln x h ln x h h
102. ln x x 2 1 ln x x 2 1
ln c c 2 x 2
ln x h ln x h
1 x h x h ln ln x h x
100. The change-of-base formula enables us to write a logarithm of one base in terms of a ratio of logarithms of a different base. This is important among other reasons because it enables us to approximate the value of a logarithm of a base other than 10 or e. 101. a.
2 2 2 2 2 2 c c x c c x ln 2 ln 2 2 x2 c c x
p
py
p
logarithmic form, we have logb x p py , but since y logb x , we have logb x p p logb x as desired.
b 2 b 2 4ac 4ac log 2 log 2 4a 4a c log log c log a a
492
Section 4.5 109.
113. a.
The graphs are the same. 110. b.
12 1 log x 2 log x 2 log x 2 log x 2
114.
111. The graphs differ by a vertical shift. This can be shown algebraically by using the product property of logarithms. For example, ln 2 x ln 2 ln x . Therefore, the graph of y ln x is shifted upward by ln 2 units. 112.
Section 4.5 Exponential and Logarithmic Equations 6. 7 x log 3 2log 5 3log11 2log 5 3log11 x 7 log 3
1. 400 e0.2t ln 400 0.2t 2. 8721 100.003k log8721 0.003k
2log 5 3log11 7 log 3
3. log11,000 4w 6 104 w6 11,000 4. ln850 3z 2 e3 z2 850 5. 14 x ln 7 2ln 3 7 ln 2 2ln 3 7 ln 2 x 14ln 7
2ln 3 7 ln 2 14ln 7
493
7. exponential
8. x; y
9. x; y
10. logarithmic
Chapter 4 Exponential and Logarithmic Functions 11. 3x 81
18.
3x 34
4
x4
1 49
7 2 x3 7 2
2 x 25
5
x 13. 3 5 5t
51 3 5t
14.
1 4
1 4
82 x5 32 x6
2
25
3 2 x 5
x 6
3 3w 31 2 3w
x 15 1 2
20.
3 y 1
2
3 x 4
1
y 1
21.
2 3t 5
103
3t
106t 10 1093t 6t 10 9 3t 9t 19
52 z 2 54 2z 2 4 2z 2
1
z 1
t
19 9
19 9
c 5
113c1 111
22. 100,0002 w1 10,0004w
10
5 2 w1
c 5
104
4 w
1010 w5 10164 w 10w 5 16 4 w 14w 11
113c1 11 c5 3c 1 c 5 4c 4 c 1
14
1003t 5 10003t
10
52 z 2 625
17. 11
2 x 1
14 x
3 y 3
1 11
32
33 x12 34 x2 3x 12 4 x 2
4
3 y 1 4
3 c 1
15
27 x4 92 x1
3
23 y 1 16 2
16.
x 1
26 x15 25 x30 6 x 15 5 x 30
1 w 2 15.
19.
1 3
1 t 3
x 1
7 2 x3 7 2 x2 2 x 3 2 x 2 4x 1
12. 2 x 32
x5
7
2 x 3
1
w
494
11 14
11 14
Section 4.5 23.
27. 103 4 x 8100 120,000
6t 87 log 6t log 87
103 4 x 128,100
t log 6 log 87
log103 4 x log128,100 3 4 x log128,100 4 x log128,100 3 log128,100 3 x 4 x 0.5269 log128,100 3 ; x 0.5269 4
log 87 log 6 t 2.4925 log 87 ; t 2.4925 log 6 t
24.
2 z 70 log 2 z log 70 z log 2 log 70 log 70 z log 2 z 6.1293 log 70 ; z 6.1293 log 2
25.
28. 1058 x 4200 84,000
1058 x 79,800 log1058 x log 79,800 5 8 x log 79,800 8 x log 79,800 5 log 79,800 5 8 x 0.0122 log 79,800 5 ; x 0.0122 8 x
1024 19 x 4 1020 19 x log1020 log19 x log1020 x log19
29. 21,000 63,000e 0.2t 1 e 0.2t 3 1 ln ln e 0.2t 3
log1020 log19 x 2.3528 log1020 ; x 2.3528 log19 x
26.
1 ln 0.2t 3
801 23 y 6
1 ln 3 ln1 ln 3 0 ln 3 t 0.2 0.2 0.2 ln 3 ln 3 t 5ln 3 5.4931 1 0.2 5 ln 3 or 5ln 3 ; t 5.4931 0.2
795 23 y log 795 log 23 y log 795 y log 23 log 795 y log 23 y 2.1299 log 795 ; y 2.1299 log 23
495
Chapter 4 Exponential and Logarithmic Functions 30.
80 320e 0.5t 1 e 0.5t 4 1 ln ln e 0.5t 4
1 6 x ln 2 3x 4 ln 7 ln 2 6 x ln 2 3 x ln 7 4 ln 7 6 x ln 2 3x ln 7 4 ln 7 ln 2
1 ln 0.5t 4
x 6 ln 2 3ln 7 4 ln 7 ln 2 ln 2 4 ln 7 6 ln 2 3ln 7 x 0.7093
x
1 ln 4 ln1 ln 4 0 ln 4 t 0.5 0.5 0.5 ln 4 ln 4 2ln 4 2.7726 t 1 0.5 2 ln 4 or 2ln 4 ; t 2.7726 0.5 31.
216 x 73 x 4 ln 216 x ln 73 x 4
33.
ln 2 4 ln 7 ; x 0.7093 6 ln 2 3ln 7 1118 x 92 x3 ln1118 x ln 92 x3
34.
1 8 x ln11 2 x 3 ln 9 ln11 8 x ln11 2 x ln 9 3ln 9
36 x5 52 x ln 36 x5 ln 52 x
8 x ln11 2 x ln 9 3ln 9 ln11 x 8ln11 2 ln 9 3ln 9 ln11
6 x 5 ln 3 2 x ln 5
ln11 3ln 9 8ln11 2 ln 9 x 0.1779
6 x ln 3 5ln 3 2 x ln 5 6 x ln 3 2 x ln 5 5ln 3
x
x 6ln 3 2ln 5 5ln 3
ln11 3ln 9 ; x 0.1779 8ln11 2 ln 9
5ln 3 x 2ln 5 6ln 3 x 1.6286 5ln 3 ; x 1.6286 2ln 5 6 ln 3 32.
4 x 1
35.
e2 x 9e x 22 0
e 9 e 22 0 x 2
x
Let u e x . u 2 9u 22 0
7 3 4 x 1 ln 7 ln 35 x 5x
u 11 u 2 0
4 x 1 ln 7 5 x ln 3 4 x ln 7 ln 7 5 x ln 3 4 x ln 7 5 x ln 3 ln 7
u 11
or
u 2
e x 11
or
e x 2 No solution
ln e x ln11
x 4 ln 7 5ln 3 ln 7
x ln11 x 2.3979 ln11 ; x 2.3979
ln 7 x 4 ln 7 5ln 3 x 0.8495 ln 7 ; x 0.8495 4 ln 7 5ln 3
496
Section 4.5 40. a. log 4 64 0 3 log 4 64 63
e 2 x 6e x 16 0
36.
e 6 e 16 0 x 2
3 0 3 log 4 1
x
30 3 Yes, the value is a solution.
Let u e x . u 2 6u 16 0
b. log 4 1 0 3 log 4 1 63
u 8 u 2 0 u8
or
u 2
ex 8
or
e x 2 No solution
log 4 1 0 3 log 4 64 No, the value is not a solution.
c. log 4 32 0 3 log 4 32 63
ln e x ln 8
log 4 32 0 3 log 4 31 No, the value is not a solution.
x ln 8 2.0794
ln 8 ; x 2.0794
41. log 4 3w 11 log 4 3 w
e 2 x 9e x
37.
e 9e 0 e e 9 0 x 2 x
3w 11 3 w 4w 8
x
x
w 2
e x 0 or e x 9
42. log 7 12 t log 7 t 6
12 t t 6 6 2t
e 2 x 7e x
38.
e 7e 0 e e 7 0 x 2 x
2
x
3t
x
3
43. log x 2 7 x log18
e x 0 or e x 7
x 2 7 x 18 x 2 7 x 18 0
39. a. log 2 16 31 0 5 log 2 16
x 9 x 2 0
log 2 5 0 5 4
x 9 or x 2
log 2 5 0 1
9, 2
44. log p 2 6 p log 7
No, the value is not a solution. b. log 2 32 31 0 5 log 2 32
p2 6 p 7 p2 6 p 7 0
log 2 10 5 5
p 7 p 1 0
00 0 Yes, the value is a solution.
p 7 or p 1
c. log 2 16 1 0 5 log 2 1
log 2 17 0 5 log 2 1 No, the value is not a solution.
497
7, 1
Chapter 4 Exponential and Logarithmic Functions 45. 6 log 5 4 p 3 18
51. 2 ln 4 3t 1 7
4 p 3 53
ln 4 3t 3
4 p 3 125
4 3t e3
2 ln 4 3t 6
log 5 4 p 3 3
4 p 128 p 32
3t e3 4
32
t
46. 5log 6 7 w 1 10
4 e3 ; t 5.3618 3
log 6 7 w 1 2
7 w 1 62
52. 4 ln 6 5t 2 22
7 w 1 36
4 ln 6 5t 20
7 w 35 w5
ln 6 5t 5
5
6 5t e5
47. log8 3 y 5 10 12
5t e5 6
log8 3 y 5 2
t
3 y 5 82
6 e5 28.4826 5
6 e5 ; t 28.4826 5
3 y 5 64 3 y 69 y 23
4 e3 5.3618 3
23
53.
log 2 w 3 log 2 w 2
48. log 3 7 5 z 14 17
log 2 w log 2 w 2 3
7 5 z 33
w w 2 23
7 5 z 27
w2 2 w 8
log 2 w w 2 3
log 3 7 5 z 3
w2 2 w 8 0 w 2 w 4 0 w 2 or w 4 Check: log 2 w 3 log 2 w 2
5 z 20 z 4
4
49. log p 17 4.1
p 17 104.1
log 2 2 3 0 log 2 2 2
p 104.1 17 12,572.2541
log 2 2 3 0 log 2 4 1 3 0 2 2 0 2 log 2 w 3 log 2 w 2
10 17 ; p 12,572.2541 4.1
50. log q 6 3.5
q 6 103.5
log 2 4 3 0 log 2 4 2
q 103.5 6 q 3168.2777
log 2 4 3 0 log 2 2 undefined undefined 2 ; The value 4 does not check.
10 6 ; q 3168.2777 3.5
498
Section 4.5 54. log 3 y log 3 y 6 3
57.
log 3 y y 6 3
log 5 z log 5 z 20 3
y 2 6 y 27 0
z z 20 53
log 5 z z 20 3
y y 6 33
y 9 y 3 0
z 2 20 z 125
y 9 or y 3 Check: log 3 y log 3 y 6 3
z 2 20 z 125 0
z 5 z 25 0 z 5 or z 25 Check: log 5 z 3 log 5 z 20
log 3 9 log 3 9 6 0 3 log 3 9 log 3 3 0 3
log 5 5 0 3 log 5 5 20
undefined undefined log 3 y log 3 y 6 3
log 5 5 0 3 log 5 25
log 3 3 log 3 3 6 0 3
undefined undefined log 5 z 3 log 5 z 20
log 3 3 log 3 9 0 3
log 5 25 0 3 log 5 25 20
1 20 3 3 ; The value 9 does not check.
2 0 3 log 5 5
20 3 1 20 2 25 ; The value 5 does not check.
log 6 7 x 2 1 log 6 x 5
55.
log 6 7 x 2 log 6 x 5 1 7x 2 1 log 6 x 5
58.
32
log 2 x x 6 4
x x 6 2 4
x 2 6 x 16 0
x 32
x 2 x 8 0
log 4 5 x 13 1 log 4 x 2
x 2 or x 8 Check: log 2 x 4 log 2 x 6
log 2 2 0 4 log 2 2 6
log 4 5 x 13 log 4 x 2 1
log 2 2 0 4 log 2 8
5 x 13 1 log 4 x 2
undefined undefined log 2 x 4 log 2 x 6
5 x 13 41 x2 5 x 13 4 x 8
5
log 2 x 4 log 2 x 6
log 2 x log 2 x 6 4
7x 2 61 x5 7x 2 6 x5 7 x 2 6 x 30
56.
log 5 z 3 log 5 z 20
log 2 8 0 4 log 2 8 6 3 0 4 log 2 2
x5
30 4 1 30 3 8 ; The value 2 does not check.
499
Chapter 4 Exponential and Logarithmic Functions 59.
ln x ln x 4 ln 3x 10
Check: ln x ln x 3 ln 5 x 7
x x 4 ln 0 3x 10 x x 4 e0 3x 10 x2 4 x 1 3x 10 x 2 4 x 3 x 10
ln1 ln 1 3 0 ln 5 1 7
ln x ln x 4 ln 3 x 10 0
ln1 ln 2 0 ln 2
undefined undefined ln x ln x 3 ln 5 x 7 ln 7 ln 7 3 0 ln 5 7 7 ln 7 ln 4 0 ln 28 ln 28 0 ln 28
7 ; The value 1 does not check.
x 7 x 10 0 2
x 2 x 5 0
61.
x 2 or x 5 Check: ln x ln x 4 ln 3x 10
log x log x 7 log x 15 0
x x 7 log 0 x 15 x x 7 100 x 15 x2 7 x 1 x 15 x 2 7 x x 15
ln 2 ln 2 4 0 ln 3 2 10 ln 2 ln 2 0 ln 4
undefined undefined ln x ln x 4 ln 3x 10 ln 5 ln 5 4 0 ln 3 5 10 ln 5 ln10 ln 5
x 2 8 x 15 0
ln 5 0 ln 5 5 ; The value 2 does not check. 60.
log x log x 7 log x 15
x 3 x 5 0 x 3 or x 5 Check: log x log x 7 log x 15
ln x ln x 3 ln 5 x 7
ln x ln x 3 ln 5 x 7 0
log 3 log 3 7 0 log 3 15
x x 3 ln 0 5x 7 x x 3 e0 5x 7 x 2 3x 1 5x 7 x 2 3x 5 x 7
log 3 log 4 0 log 12 undefined undefined log x log x 7 log x 15 log 5 log 5 7 0 log 5 15 log 5 log 2 0 log 10 undefined undefined ; The values 3 and 5 do not check.
x2 8x 7 0
x 1 x 7 0 x 1 or x 7
500
Section 4.5 log x log x 10 log x 18
62.
nt
r 65. A P 1 n Interest is A P $1000 .
log x log x 10 log x 18 0
x x 10 log 0 x 18 x x 10 100 x 18 x 2 10 x 1 x 18 x 2 10 x x 18
0.022 4000 1 12
x 2 11x 18 0
x 2 x 9 0 x 2 or x 9 Check: log x log x 10 log x 18
12 t
4000 1000
0.022 4000 1 12
12 t
0.022 1 12
12 t
0.022 ln 1 12
12 t
undefined undefined log x log x 10 log x 18 log 9 log 9 10 0 log 9 18
nt
r 66. A P 1 n Interest is A P $2000 .
undefined undefined ; The values 2 and 9 do not check.
0.025 10,000 1 12
rt
3 10,000 10,000e0.055t 0.055t
ln 3 0.055t ln 3 t 0.055 t 20 yr 64.
5 ln 4
10 yr 2 months
log 9 log 1 0 log 9
3 e
5 4
5 ln 4 t 0.022 12 ln 1 12 t 10.15
log 2 log 8 0 log 16
A Pe
0.022 5 12t ln 1 ln 4 12
log 2 log 2 10 0 log 2 18
63.
5000
A Pe rt
12 t
10,000 2000
0.025 10,000 1 12
12 t
0.025 1 12
12 t
0.025 ln 1 12
12 t
12,000
6 5
6 ln 5
0.025 6 12t ln 1 ln 5 12
1,000,000 80,000e0.06t
6 ln 5 t 0.025 12ln 1 12
12.5 e0.06t ln12.5 0.06t ln12.5 t 0.06 t 42 yr
t 7.3 7 yr 4 months
501
Chapter 4 Exponential and Logarithmic Functions 67. a. R 100 2 b.
t 4.2
100 2
30 100 2
2 4.2
72 mCi
69.
0.5 e kx
t 4.2
ln 0.5 ln e kx
3 t 4.2 2 10 3 t 4.2 ln ln 2 10
ln 0.5 kx ln 0.5 k ln 0.5 ln 0.5 14.1 m Ocean: x k 0.0491 ln 0.5 ln 0.5 8.7 m Tahoe: x k 0.0799 ln 0.5 ln 0.5 Erie: x 3.5 m k 0.1980 x
t ln 2 3 ln 10 4.2 3 4.2 ln 10 t 7.3 ln 2 7.3 2 5.3 days 68. a.
C 80 2
t 6
60 80 2
t 6
70.
ln 0.2 ln e kx ln 0.2 kx ln 0.2 k ln 0.2 ln 0.2 Ocean: x 32.8 m k 0.0491 ln 0.2 ln 0.2 20.1 m Tahoe: x k 0.0799 ln 0.2 ln 0.2 8.1 m Erie: x k 0.1980 x
t ln 2 3 ln 4 6 3 6ln 4 2.5 hr t ln 2 C 80 2
t 6
30 80 2
t 6
P e kx 0.2 e kx
3 t 6 2 4 t 6 3 ln ln 2 4
b.
P e kx
71.
P e kx 0.01 e kx
3 t 6 2 8 t 6 3 ln ln 2 8
ln 0.01 ln e kx ln 0.01 kx ln 0.01 k ln 0.01 ln 0.01 Ocean: x 93.8 m k 0.0491 x
t ln 2 3 ln 8 6 3 6ln 8 8.5 hr t ln 2
ln 0.01 ln 0.01 57.6 m k 0.0799 ln 0.01 ln 0.01 23.3 m Erie: x k 0.1980
72. Tahoe: x
502
Section 4.5 73.
T 50 1550e 0.05t 100 50 1550e 0.05t
b.
L I log 12 10 10 L log I log1012 10 L log I 12log10 10 L log I 12 10
50 1550e 0.05t 1 e 0.05t 31 1 ln ln e 0.05t 31 1 ln 0.05t 31 1 ln 31 t 69 or 1 hr 9 min 0.05 74.
L
I 1010 30
I 1010
T 70 255e 0.017 t 110 70 255e 0.017 t
71
76. I 1010
40 255e 0.017 t
12
12
12
10312 10 9 W/m 2
107.112 10 4.9
1.26 10 5 W/m 2 90
8 e 0.017 t 51 8 ln ln e 0.017 t 51
12
I 1010 10912 10 3 1 10 3 W/m 2 1.26 105 I 1 103 W/m 2 77. a. L D 8.8 5.1log D
8 ln 0.017t 51
L 6 8.8 5.1log 6 12.8 b. L D 8.8 5.1log D
8 ln 51 109 or 1 hr 49 min t 0.017
15.5 8.8 5.1log D 6.7 5.1log D 6.7 log D 5.1 D 106.7 5.1 21 in.
I 75. a. L 10 log I 0
3.4 10 8 L 10 log 1012
I L 10 log 12 10
78. a. L D 8.8 5.1log D
L 4 8.8 5.1log 4 11.9
10 log 3, 4 104 45.3 dB
b. L D 8.8 5.1log D
17 8.8 5.1log D 8.2 5.1log D 8.2 log D 5.1 D 108.2 5.1 41 in.
503
Chapter 4 Exponential and Logarithmic Functions 79. a.
83. For 0 t 72 : C t 0.088 0.89ln t 2
pH log H 8.5 log H
1.5 0.088 0.89ln t 2
8.5 log H
1.588 0.89 ln t 2
10 8.5 H
1.588 ln t 2 0.89 e1.588 0.89 t 2
H 3.16 10 9 mol/L b.
pH log H
t e1.588 0.89 2 4 min For t 72 : C t 4.64e 0.003t
2.3 log H 2.3 log H 102.3 H
H 5.01 10 80. a.
3
1.5 4.64e 0.003t 1.5 e 0.003t 4.64 1.5 ln 0.003t 4.64
mol/L
pH log H 6.2 log H 6.2 log H 10
6.2
H
1.5 ln 4.64 t 376 min 0.003 4 min t 376 min
H 6.31 10 7 mol/L b.
pH log H
84. a.
8.4 log H
D t 91 160 ln t 1 D 15 91 160 ln 15 1
8.4 log H
91 160 ln16 535
10 8.4 H
b.
H 3.98 10 9 mol/L
D t 91 160 ln t 1 600 91 160 ln t 1
81. S t 94 18ln t 1
509 160 ln t 1
65 94 18ln t 1
509 ln t 1 160 e509 160 t 1
29 18ln t 1 29 ln t 1 18 e 29 18 t 1
t e509 160 1 23 days
t e 29 18 1 4 months
85.
f x 2x 7 y 2x 7
82. S x 5 7 ln x 1
x 2y 7
19.1 5 7 ln x 1
x 7 2y
14.1 7 ln x 1 14.1 ln x 1 7 e14.1 7 x 1
log 2 x 7 y f 1 x log 2 x 7
x e14.1 7 1 6.50 or $650 504
Section 4.5 f x 5x 6
86.
91.
y 5x 6
y log x 7 9
x5 6
x log y 7 9
y
x65
x 9 log y 7
y
log 5 x 6 y
10 x9 y 7
f 1 x log 5 x 6
10 x9 7 y f 1 x 10 x9 7
f x ln x 5
87.
y ln x 5
92.
x log y 11 8
e y5 x
x 8 log y 11
ex 5 y f 88.
x e 5
10 x 8 y 11
x
10 x 8 11 y
f x ln x 7
f 1 x 10 x 8 11
y ln x 7 x ln y 7
93. 5 x 3 122
ex y 7
5 x 125
ex 7 y
5 x 53
f 1 x e x 7 89.
x 3 x 3 or x 3
f x 10 x 3 1 y 10 x 3 1
94. 11 x 9 130
x 10 y 3 1
11 x 121
x 1 10 y 3
3, 3
11 x 112
log x 1 y 3
x 2
log x 1 3 y
x 2 or x 2
f 1 x log x 1 3 90.
f x log x 11 8 y log x 11 8
x ln y 5
1
f x log x 7 9
f x 10
x2
4
y 10
x2
4
x 10
y2
4
2, 2
95. log x 2log 3 2
x log 2 2 3 x log 2 9
x 4 10 y 2
x 102 9 x 900
log x 4 y 2 log x 4 2 y f 1 x log x 4 2
505
900
Chapter 4 Exponential and Logarithmic Functions 96. log y 3log 5 3
x2 6x 6x
102.
x log 3 3 5
x2 6x 6x 0
x 3 log 125
6 x 1 x 1 0 x
6 x 0 or x 1 or x 1
x 103 125 x 125,000 97.
1,1
undefined
125,000
103. log 3 log 3 x 0
6 x 2 36
log 3 x 30
6 x 2 62
log 3 x 1
x2 2 2
x 31
x2 4
x3
2
2
4, 4
x 4 98.
6x x2 1 0
104. log 5 log 5 x 1
8 y 7 64
log 5 x 51
8 y 7 82
log 5 x 5
2
2
x 55
y2 7 2
3, 3
x 3
105. 3 ln x 12 0
3 ln x 12
99. log 9 x 4 log 9 6
ln x 4
x4 6
ln x 4
x 4 6 or x 4 6 x2
x 10
xe
10, 2
100. log8 3 x log8 5
3 x 5 or 3 x 5 x 2 x 8 x 2
x8
ln x 4 x e 4 1 x 4 e
106. 7 ln x 14 0
2, 8
7 ln x 14 ln x 2
x 2 e x 9e x
ln x 2
x 2 e x 9e x 0
x e2
ex x2 9 0 e x x 3 x 3 0 e x 0 or x 3 or x 3 undefined
or
4
4 1 e , 4 e
3 x 5
3125
x 3125
y2 9
101.
3
2 1 e , 2 e
3, 3 506
or
ln x 2 x e 2 1 x 2 e
Section 4.5 107. log 3 x log 3 2 x 6
Let u e x . 2u 2 9u 4 0
1 log 3 4 2
x log 3 log 3 2 2 x 6
2u 1 u 4 0
x 2 2x 6 x 4 x 12 3x 12 x 4
or
1 2 1 ex 2 u
1 log 3 4 2 1 log 3 4 log 3 2 4 6 0 log 3 4 2 1 log 3 4 log 3 2 0 log 3 4 2 undefined undefined ; The value 4 does not check. log 3 x log 3 2 x 6
1 3 1 ex 3 u
or
u7 ex 7
111. If two exponential expressions of the same base are equal, then their exponents must be equal. 112. If two logarithmic expressions of the same base are equal, then their arguments must be equal. 113. Take a logarithm of any base b on each side of the equation. Then apply the power property of logarithms to write the product of x and the log b 4 . Finally divide both sides by log b 4 .
6e 3e 4 0 2 e 9e 4 0 x
x
1 x ln x ln 7 3 1 ln , ln 7 ; x 1.0986, x 1.9459 3
x 2
x
3u 1 u 7 0
2e x e x 3 3e x 4 x
x
Let u e x . 3u 2 22u 7 0
1 log 5 x log 5 x 1 log 5 8 3 1 log 5 2 log 5 2 1 0 log 5 8 3 1 log 5 2 log 5 1 0 log 5 8 3 undefined undefined ; The value 2 does not check.
x
2
x 2
Check:
2 e
18e 4e 7 0 3 e 22e 7 0
3 ex
x 2 x 1 x 2x 2 x 2 x 2
3e x e x 6 4e x 7
110.
1 108. log 5 x log 5 x 1 log 5 8 3 x log 5 2 log 3 x 1
x 2
ex 4
1 x ln x ln 4 2 1 ln , ln 4 ; x 0.6931, x 1.3863 2
Check:
109.
u4
507
Chapter 4 Exponential and Logarithmic Functions 114. Use the product property of logarithms to write a single logarithm on the left side of the equation. Then write the logarithmic equation in its equivalent exponential form. The resulting equation is linear which is solved by applying the addition and multiplication properties of equality.
10 13 10 4 3 10 x 13 10 x 12
u 1 u 4 0 u 1
10
118.
12 10 x 13 0
Let u 10 . u 2 12u 13 0 x
or
10 x 1
or
u 2 ln x 2
x log13
x e 2
119.
e 8 e 9 0 x
Let u e . u 2 8u 9 0
x 10
u9 ex 9
x ln 1
x ln 9
or
log x 2 x 102
0
x 1 1,100
u 1 u 9 0 e 1
x e 1
log x 2 log x 2
log x 0
x
or
1 e2
log x 2 2 log x 0 log x log x 2 0
x
u 1 ln x 1
1 1 2 , ; x 0.1353, x 0.3679 e e
e x e x 9e x 8 e x
u 1
ln x 2 ln x3 2 ln x 2 3 ln x 2 0
10 x 13
e x 9e x 4 2 e x 9e x 8
x 2
u 2 u 1 0
undefined log13 ; x 1.1139
xe x e4 e, e 4 ; x 2.7183, x 54.5982
u 13
x log 1
116.
ln x 4
Let u ln x . u 2 3u 2 0
u 1 u 13 0 u 1
u4
or
ln x 1
10 x 10 x 13 10 x 12 10 x x 2
ln x 2 ln x5 4 ln x 2 5 ln x 4 0 Let u ln x . u 2 5u 4 0
x
x
115.
117.
120.
x 100
log x 2 log x3 log x 2 3log x 0 log x log x 3 0
undefined ln 9 ; x 2.1972
log x 0 x 10 x 1 1,1000
508
or 0
log x 3 x 103 x 1000
1 e
Section 4.5 log w 4 log w 12 0
121.
124.
e 6e 4 0
log w 4 log w 12 0
x 2
2
u 2 4u 12 0
u 6 u 2 0 or
u
u2
log w 6
log w 2
undefined
log w 4
x
Let u e x . u 2 6u 4 0
Let u log w .
u 6
e 2 x 6e x 4 0
6
6 2 4 1 4 2 1
6 20 6 2 5 3 5 2 2 u 3 5 or u 3 5
w 104 10,000
10,000
ex 3 5
x ln 3 5 ln x 3 ln x 10 0
122.
x ln 3 5
x 1.6556
x 0.2693
ln x 3 ln x 10 0
ln 3 5 , ln 3 5 ; x 1.6556,
Let u ln x . u 2 3u 10 0
x 0.2693
2
125. log 5 6c 5 log 5 c 1
u 5 u 2 0 u 5
or
ln x 6 undefined
u2
log 5
ln x 2 ln x 4
e ; x 54.5981
2
6c 2 5c 25 0
2c 5 3c 5 0 5 5 or c 2 3 Check: log 5 6c 5 log 5 c 1 c
82 4 1 6 2 1
5 5 log 5 6 5 log 5 0 1 2 2
8 40 8 2 10 4 10 2 2 u 4 10 or u 4 10
e 4 10 x
6c 2 5c 52
8e x 6 0
8
2
Let u e x . u 2 8u 6 0 u
6c 5c 1
1 log 5 6c 2 5c 1 2 log 5 6c 2 5c 2
x e4 54.5981
e 2 x 8e x 6 0 ex
6c 5 c 1
log 5
4
123.
ex 3 5
5 log 5 10 log 5 0 1 2 undefined undefined
e 4 10 x
x ln 4 10 ln 4 10 , ln 4 10 ; x 1.9688, x ln 4 10
x 0.1771
509
Chapter 4 Exponential and Logarithmic Functions log 5 6c 5 log 5 c 1 5 5 01 log 5 6 5 log 5 3 3 5 01 3 5 log 5 15 0 1 3
log 5 15 log 5
1.4408, 2.8584
log 5 5 0 1 10 1 5 5 ; The value does not check. 2 3
128.
126. log 3 x 8 log 3 x 1
log 3
x 8 x 1
log 3
x 8x 1 2
1 log 3 x 2 8 x 1 2 log 3 x 2 8 x 2 x 2 8 x 32 x2 8x 9 0 x 1 x 9 0 x 1 or x 9 Check: log 3 x 8 log 3 x 1
1.5818, 0.7417 129.
log 3 1 8 log 3 1 0 1 log 3 9 log 3 1 0 1 undefined undefined log 3 x 8 log 3 x 1
2.0960
log 3 9 8 log 3 9 0 1 log 3 1 log 3 3 0 1 10 1 9 ; The value 1 does not check.
130.
127.
2.0087
510
Section 4.6 Section 4.6 Modeling with Exponential and Logarithmic Functions 2. f x log b x
1. a and e only
14.
N e 0.025t N0
3. 102 x 4 80,600 log80,600 2 x 4
N ln 0.025t N0
4. ln x 4 6 e6 x 4 5.
f x e x2 ye
N ln N0 t 0.025
x2
x e y 2 ln x y 2
15.
f 1 x 2 ln x
2 ln x y
M 8.8 5.1log D M 8.8 5.1log D M 8.8 log D 5.1 10 M 8.8 5.1 D
6. g x 4 ln x y 4 ln x x 4 ln y x ln y 4 ex 4 y
16. log E 12.2 1.44 M
log E 1.44 M 12.2
g 1 x e x 4
7. 5 3x 0 3x 5 5 x 3 8.
N N 0e 0.025t
E 101.44 M 12.2 17.
pH log H
5 , 3
x2 0 x 0 or x 0 x 0 or x 0
10 pH H 18.
, 0 0,
I L 10 log I0 I L log 10 I0
10. y ae ln b x
9. growth; decay
pH log H
10 L 10 11. logistic 13.
12. c
Q Q0 e Q e kt Q0
I I0
I 0 10 L 10 I
kt
19.
A P 1 r
t
A t 1 r P t A ln ln 1 r P
Q ln kt Q 0
Q ln Q0 k t
A ln t ln 1 r P
511
Chapter 4 Exponential and Logarithmic Functions
20.
A ln P t ln 1 r
5 e0.044t 3 5 ln 0.044t 3
A Pe rt A e rt P A ln rt P
5 ln 3 t 11.6 yr 0.044
A 50,000e rt
24. a.
194,809.67 50,000e 20 r
A ln P r t
194,809.67 e 20 r 50,000 194,809.67 ln 20r 50,000
k E 21. ln A RT
194,809.67 ln 50,000 r 0.68 or 6.8% 20
k e E RT A k Ae E RT
A 50,000e rt
b.
1 P 22. ln A k 14.7
250,000 50,000e0.068t 250,000 e0.068t 50,000
P kA ln 14.7
5 e0.068t
P e kA 14.7 P 14.7e kA 23. a.
ln 5 0.068t ln 5 t 23.7 yr 0.068
A 12,000e rt 14,309.26 12,000e
P 806.07 Pe0.032 3
14,309.26 e4 r 12,000
P Pe0.096 806.07
P 1 e0.096 806.07
14,309.26 ln 4r 12,000
P
14,309.26 ln 12,000 r 0.44 or 4.4% 4 b.
A Pert
25. a.
4r
b.
806.07 $8000 1 e0.096
A Pert 10,000 8000e0.032t 10,000 e0.032t 8000 5 e0.032t 4 5 ln 0.032t 4
A 12,000e rt 20,000 12,000e0.044t 20,000 e0.044t 12,000
512
Section 4.6 5 ln 4 7 yr t 0.032
Taiwan: P t P0 ekt P t 22.9ekt 23.7 22.9e k 10 23.7 e k 10 22.9 23.7 10k ln 22.9
A Pe rt
26. a.
P 193.03 Pe0.021 2 P Pe0.042 193.03
P 1 e0.042 193.03 193.03 1 e0.042 P $4500
23.7 ln 22.9 0.00343 k 10 P t 22.9e0.00343t
P
b.
A Pert 6000 4500e0.021t 6000 e0.021t 4500 4 e0.021t 3 4 ln 0.021t 3
b. Australia: P 20 19e0.01735 20 26.9 million Taiwan: P 20 22.9e0.00343 20 24.5 million c. The population growth rate for Australia is greater. d. Australia: 30 19e0.01735t
4 ln 3 t 13.7 yr 0.021
30 e0.01735t 19 30 ln 0.01735t 19
27. a. Australia: P t P0 ekt
30 ln 19 t 26.3 years 0.01735 In the year 2026 Taiwan: 30 22.9e0.00343t 30 e0.00343t 22.9 30 ln 0.00343t 22.9
P t 19ekt 22.6 19e k 10 22.6 e k 10 19 22.6 10k ln 19 22.6 ln 19 k 0.01735 10 P t 19e0.01735t
30 ln 22.9 t 78.7 years 0.00343 In the year 2078
513
Chapter 4 Exponential and Logarithmic Functions 28. a. Switzerland: P t P0 e kt
Israel: 10 6.7e0.01391t 10 e0.01391t 6.7 10 ln 0.01391t 6.7
P t 7.3ekt 7.8 7.3e k 10 7.8 e k 10 7.3 7.8 10k ln 7.3
10 ln 6.7 t 28.8 years 0.01391 In the year 2028
7.8 ln 7.3 0.00662 k 10 P t 7.3e0.00662t Israel: P t P0 e kt P t 6.7e
29. a. Costa Rica: t P t 4.3 1.0135
P t 4.3e ln1.0135t P t 4.3e0.01341t In 2000, the population is 4.3 million. Norway: t P t 4.6 1.0062
kt
7.7 6.7e k 10 7.7 e k 10 6.7 7.7 10k ln 6.7
P t 4.6e ln1.0062t P t 4.6e0.00618t In 2000, the population is 4.6 million.
7.7 ln 6.7 k 0.01391 10 P t 6.7e0.01391t
b. Costa Rica: 5 4.3e0.01341t
5 e0.01341t 4.3 5 ln 0.01341t 4.3
b. Switzerland: P 20 7.3e0.00662 20 8.3 million Israel: P 20 6.7e0.01391 20 8.8 million
5 ln 4.3 11.2 years t 0.01341 In the year 2011 Norway: 5 4.6e0.00618t 5 e0.00618t 4.6 5 ln 0.00618t 4.6
c. The population growth rate is greater for Israel. d. Switzerland: 10 7.3e0.00662t
10 e0.00662t 7.3 10 ln 0.00662t 7.3
5 ln 4.6 13.5 years t 0.00618 In the year 2013
10 ln 7.3 t 47.5 years 0.00662 In the year 2047 514
Section 4.6 c. The population growth rate for Costa Rica is greater.
Q t Q0 e 0.000121t
32.
0.294Q0 Q0 e 0.000121t 0.294 e 0.000121t ln 0.294 0.000121t ln 0.294 t 10,117 yr 0.000121
30. a. Haiti: t P t 8.5 1.0158
P t 8.5e ln1.0158t P t 8.5e0.01568 In 2000, the population is 8.5 million. Sweden: t P t 9.0 1.0048
33. a.
0.5Q0 Q0 e k 87.7 0.5 e k 87.7 ln 0.5 87.7 k ln 0.5 k 0.0079 87.7 Q t 2e 0.0079t
P t 9.0e ln1.0048t P t 9.0e0.00479t In 2000, the population is 9.0 million. b. Haiti: 10.5 8.5e0.01568
b. Q t 2e 0.0079t
10.5 e0.01568t 8.5 10.5 ln 0.01568t 8.5
1.6 2e 0.0079t 0.8 e 0.0079t ln 0.8 0.0079t ln 0.8 t 28 yr 0.0079
10.5 ln 8.5 t 13.5 years 0.01568 In the year 2013 Sweden: 10.5 9.0e0.00479t 10.5 e0.00479t 9.0 10.5 ln 0.00479t 9.0
34. a.
Q t Q0 e kt 0.5Q0 Q0 e k 6 0.5 e k 6 ln 0.5 6k ln 0.5 k 0.1155 6 Q t 10e 0.1155t
10.5 ln 9.0 t 32.2 years 0.00479 In the year 2032
b. Q t 10e 0.1155t
3 10e 0.1155t 0.3 e 0.1155t ln 0.3 0.1155t ln 0.3 t 10.4 hr 0.1155
c. The population growth rate for Haiti is greater. 31.
Q t Q0 e kt
Q t Q0 e 0.000121t
35. a. Q t Q0 e kt 1 Q0 Q0 e k 174 3 1 e k 174 3
0.78Q0 Q0 e 0.000121t 0.78 e 0.000121t ln 0.78 0.000121t ln 0.78 t 2053 yr 0.000121 515
Chapter 4 Exponential and Logarithmic Functions 1 ln 174k 3
b.
1 ln 3 k 0.0063 174 Q t 300e 0.0063t b.
Q t Q0 e 0.0063t 0.5Q0 Q0 e 0.0063t 0.5 e 0.0063t ln 0.5 0.0063t ln 0.5 t 110 min 0.0063
36. a.
t 6
1 P 0 2,000,000 2
06
1 P 6 2,000,000 2
66
1 P 12 2,000,000 2
12 6
1 P 60 2,000,000 2
60 6
1,000,000 500,000 1953
P 0 2,000,000e 0.1155t 2,000,000 P 6 2,000,000e 0.1155t 1,000,000
Q t Q0 e kt
P 12 2,000,000e 0.1155t 500,000 P 60 2,000,000e 0.1155t 1956
0.2 e k 90 ln 0.2 90k
1 38. a. Q t 2
ln 0.2 k 0.01788 90 Q t Q0 e 0.01788t
t 1620
1 t 2
Q t Q0 e 0.01788t
1 1620
0.5Q0 Q0 e 0.01788t
eln1 2t
0.5 e 0.01788t ln 0.5 0.01788t ln 0.5 t 38.8 days 0.01788
e e 0.000428t
1 Q t 2
t 1620
1 Q 0 2
0 1620
1 1620
ln 1 2 /1620t
b.
1 37. a. P t 2,000,000 2
2,000,000
P t 2,000,000e 0.1155t
0.2Q0 Q0 e k 90
b.
1 P t 2,000,000 2
t 6
t 16
1 2,000,000 2
2,000,000 eln1 2t
16
ln 1 2 /6t 2,000,000 e 0.1155t 2,000,000e
1
1 Q 1620 2
1620 1620
1 Q 3240 2
3240 1620
0.5 0.25
Q t e 0.000428t Q 0 e 0.000428 0 1 Q 1620 e 0.0004281620 0.5 Q 3240 e 0.000428 3240 0.25 516
Section 4.6 39. a. P t
P 0
725 1 8.295e 0.0165t 725 0.0165 0
40. a. P t
725 1 8.295e0
P 0
1 8.295e 725 78 1 8.295 On January 1, 1900, the U.S. population was approximately 78 million. b.
P t P 120
P t P 150
725 1 8.295e 0.0165t 725
b.
c.
P t P 140
725 1 8.295e 0.0165t 725 500 1 8.295e 0.0165t 500 1 8.295e 0.0165t 725
d.
55.1 1 9.6e 0.02515t 55.1
55.1 1 9.6e 0.02515t 55.1
55.1 1 9.6e 0.02515t 55.1 45 1 9.6e 0.02515t 45 1 9.6e 0.02515t 55.1 P t
1 8.295e 0.0165t 1.45
55.1 45 55.1 1 9.6e 0.02515t 45 55.1 1 0.02515t 45 e 9.6 55.1 1 0.02515t ln 45 9.6
1 9.6e 0.02515t
0.0165t
0.45 0.45 e 0.0165t 8.295 0.45 0.0165t ln 8.295 0.45 ln 8.295 t 176.6 0.0165
In the year 2076
55.1 1 ln 45 9.6 t 149.3 0.02515
8.295 e0.0165t approaches , so the value of the expression approaches 0.
e. As t , The denominator of
f. Limiting value: P t
55.1 1 9.6e0
1 9.6e 0.02515140 42.9 million
P t
8.295e
1 9.6e 0.02515115 36.0 million
725 1 8.295e 0.0165t 725
P t P 115
1 8.295e 0.0165150 427 million
d.
0.02515 0
1 9.6e 55.1 5.2 1 9.6 On January 1, 1900, the Canadian population was approximately 5.2 million.
1 8.295e 0.0165120 338 million
c.
55.1 1 9.6e 0.02515t 55.1
725 725 million 1 0
In the year 2049
517
Chapter 4 Exponential and Logarithmic Functions e. As t , The denominator of
9.6
b.
0.02515t
e approaches , so the value of the expression approaches 0.
f. Limiting value: P t 41. a. N t
N 0
55.1 55.1 million 1 0
72 70 72 1 9e 0.36t 70 72 1 e 0.36t 70 9 72 0.36t ln 70 1 9
2.4 1 15e 0.72t 2.4
2.4
1 15e 0.72 6 2,000,000
c.
1 9e 0.36t
1 15e 0.72 0 2.4 0.15 1 15 150,000 were infected initially. b. N 6
72 1 9e 0.36t 72 70 1 9e 0.36t 70 1 9e 0.36t 72 S t
2
72 1 ln 70 9 t 16 weeks 0.36
2.4 1 15e 0.72t 2.4 1 1 15e 0.72t 1 15e 0.72t 2.4 N t
c. Limiting value: 72 S t 72 thousand $72,000 1 0
15e 0.72t 1.4 1.4 e 0.72t 15 1.4 0.72t ln 15 1.4 ln 15 t 3.3 months 0.72
43. exponential
44. linear
45. logarithmic
46. logistic
47. a. exponential
d. Limiting value: 2.4 N t 2.4 million 2,400,000 1 0 42. a. S t
S 3
72 1 9e 0.36t 72
1 9e 0.36 3 $18,000
b. y 2.3 1.12
18
518
x
Section 4.6 b. y 5.08 15ln x
48. a. exponential
53. a. logistic
b. y 51.6 1.3
x
b. y
49. a. linear
18 1 496e 1.1x
54. a. logistic
b. y 2.28 x 4.08 50. a. linear b. y
12 1 219e 1.07 x
55. a. y 1.4663 1.096
t
b. y 1.4663 1.096
t
1.4663e ln1.096t 1.4663e0.09167 t
b. y 5.47 x 641
c. y 1.4663e0.09167 912 29,000 million Approximately 29 billion; This is unreasonable because this exceeds the total world population. A logistic model would probably fit the long-term trend better.
51. a. logarithmic
b. y 20.7 9.72ln x 52. a. logarithmic
d. y
889 1 576e 0.0986t
e. y
889 1 576e 0.0986 912
56. a. y 11,988 1.035
t
b. y 11,988 1.035
t
877 million
11,988e ln1.035t 11,988e0.0344t c. y 11,988e0.0344 7 $15,300
519
Chapter 4 Exponential and Logarithmic Functions 57. a. H t 4.86 6.35ln t b.
61. A visual representation of the data can be helpful in determining the type of equation or function that best models the data.
25 4.86 6.35ln t 20.14 6.35ln t
62. The average rate of change is constant for a linear function for all intervals [a, b] in the domain of the function. For an exponential function defined by f x b x b 1 , the average rate of change increases for increasing values of x.
20.14 ln t 6.35 t e 20.14 6.35 24 yr c. No, the tree will eventually die. 58. a. N t 19.7 10.9 ln t
63. An exponential growth model has unbounded growth, whereas a logistic growth model imposes a limiting value on the dependent variable. That is, a logistic growth model has an upper bound restricting the amount of growth.
b. N 7 19.7 10.9 ln 7 41 books c. No. The trend cannot continue indefinitely. At some point, book sales will begin to decrease as most reading enthusiasts have read the book.
64. The corresponding expression with base e is given by e ln bt .
59. a. y mt b 2920t 29,200 b. y V0bt
y 29,200 0.8
Ar 12 P 12 t r 1 1 12
t
65. a.
c. y 2920t 29,200
y 2920 5 29,200 $14,600
12 t r Ar P 1 1 12 12
y 2920 10 29,200 $0 d. y 29,200 0.8
t
y 29,200 0.8 $9568 5
y 29,200 0.8 $3135 10
60. a. y mt b y 6750t 54,000
b. y V0bt y 54,000 0.7
t
12 t
r P 1 12
12 t
r 1 12
12 t
r ln 1 12
12 t
Ar 12
Ar P 12
1
Ar 12 P
Ar ln 1 12 P
r Ar 12t ln 1 ln 1 12 12 P
c. y 6750t 54,000
y 6750 4 54,000 $27,600
Ar ln 1 12 P t r 12ln 1 12
y 6750 8 54,000 $0 d. y 54,000 0.7
r P P 1 12
t
y 54,000 0.7 $12,965 4
y 54,000 0.7 $3113 8
520
Chapter 4 Review b. This represents the amount of time (in yr) required to completely pay off a loan of A dollars at interest rate r, by paying P dollars per month.
P0 N P P0 N P0 e kt
66. a.
P P0 N P0 e kt P0 N PP0 P N P0 e
kt
P0 N
P N P0 e
kt
P0 N PP0
P N P0 e kt P0 N P e kt e kt
P0 N P P N P0 P N P0 P0 N P
P N P0 kt ln P0 N P P N P0 1 t ln k P0 N P
N 3 N 2 0.87916 0.52702 32 1 0.35214
c.
N 4 N 3 1.3029 0.87916 43 1 0.42374
d.
N 5 N 4 1.7023 1.3029 54 1 0.39940
e.
N 6 N 5 2.0008 1.7023 65 1 0.29850
f.
N 7 N 6 2.1876 2.0008 76 1 0.18680
g. The value 0.23791 means that the number of additional computers infected with the virus between month 1 and month 2 increased at a rate of 237,910 per month. h. The rate of change increases up through approximately month 4 and then begins to decrease. This can be visualized in the graph of the function.
b. This represents the amount of time required for the population to reach a value P, given an initial population P0 and a limiting value N. 67. a.
b.
N 2 N 1 0.52702 0.28911 2 1 1 0.23791
Chapter 4 Review Exercises 3. Assume that f a f b .
1. The relation does not define y as a one-toone function of x because a horizontal line intersects the graph in more than one point.
a 3 1 b3 1 a 3 b3
2. Yes. Each x value in the domain is associated with only one y value in the range, and each y value in the range is associated with only one x value in the domain.
3
a 3 3 b3
ab Since f a f b implies that a b , then f is one-to-one.
521
Chapter 4 Exponential and Logarithmic Functions x y 7 2
4. No; For example the points 3, 8 and
3, 8 have the same y values but different x values. That is, f a f b 8 , but
xy 7 x 2 xy 7 x 2 2 7x y x 2 y 7 x 2 f 1 x 7 x
ab. x 3 5. f g x f g x f 4 x 3 4 3 x 33 x 4
g f x g f x g 4 x 3 4 x 3 3 4 x x
4 Yes, g is the inverse of f.
9. a.
4
6. m n x m n x m x 1
3
3 x 1 1 3
3 x3 3x 2 3x 1 1
7.
d. 3, e.
f x 2 x3 5
x y2 3
x 2 y3 5
x 3 y2
x 5 2 y3 x5 y3 2 x5 3 y 2
8.
f x x2 3 y x2 3
y 2 x3 5
f 1 x 3
c. , 0
b. Yes
3 x3 3x 2 3x No, since m n x x , n is not an inverse of m.
x3 y f 1 x x 3 f.
x5 2
2 x7 2 y x7 2 x y7
f x
g. 3,
522
h. , 0
Chapter 4 Review 10. a.
c. f 1 22,176
22,176 4.2 mi 5280
12. b and c 13. a.
c. 1,
b. Yes d. 0, e.
g x x 1 y x 1
b. Domain: , c. Range: 0,
x
d. y 0
y 1
x y 1 2
14. a.
x 1 y 2
g 1 x x 2 1; x 0 f.
b. Domain: , c. Range: 0, d. y 0 g. 0, 11. a.
h. 1,
15. a. Since a 0 and k 1 , reflect the graph of y 3x across the x-axis and shift the graph up 1 unit.
f x 5280 x y 5280 x x 5280 y x y 5280 f 1 x
x 5280
b. f 1 x represents the conversion from x
feet to f 1 x miles.
b. Domain: , c. Range: 1, d. y 1
523
Chapter 4 Exponential and Logarithmic Functions
16. a. Since h 3 and k 4 , shift the graph of y 2 x left 3 units and down 4 units.
21. log b x 2 y 2 4 b 4 x 2 y 2 22. ln x c d ec d x 23. 106 1,000,000 log1,000,000 6 24. 81 3
1 1 1 log8 2 2 3
25. log 3 81 y
3 y 81 34 y4
b. Domain: , c. Range: 4, d. y 4
26. log100,000 y
17. The graph of y e is an increasing function.
10 y 100,000 105 y5
x
nt
r 0.05 18. a. A P 1 24,000 1 n 12
1210
1 27. log 2 y 64
$39,528.23
1 1 6 2 6 64 2 y 6
2y
b. A Pe rt 24,000e 0.045 30 $92,578.21 19. a. I Prt 12,000 0.072 4 $3456
28. log1 4 16 y
b. I Pe P rt
y
1 1 2 16 4 4 4
12,000e 0.065 4 12,000
y 2
15,563.16 12,000 $3563.16
29. log11 1 y
c. 7.2% simple interest results in less interest. 20. R t 128 2
11y 1 y0
t 4.2
a. R 6 128 2
6 4.2
30. log 5 5 y
48 mCi
5 y 5 51 y 1
b. R 4.2 128 2 128 2 64 After 4.2 days, (1 biological half-life), the radioactivity level is 64 mCi (half the original amount). 4.2 4.2
1
31. 4log 4 7 7 32. ln e11 11
c. After each half-life the radioactivity level is halved. After 2 half-lives it will be 32 mCi, and after 3 half-lives it will be 16 mCi.
524
2
Chapter 4 Review 33. f x log x 4 x40
38. a. The graph of y log 2 x 3 is the graph
of y log 2 x shifted to the right 3 units.
x4 Domain: 4, 34. g x ln 3 2 x
3 2x 0 2 x 3 3 2 3 Domain: , 2 x
35. h x log 2 x 2 4
b. Domain: 3, c. Range: ,
d. Vertical asymptote: x 3
x2 4 0
39. a. The graph of y 2 ln x is the graph of y ln x shifted up 2 units.
x 2 4 The square of any real number is positive, so this condition is always true. Domain: ,
36. k x log 2 x 2 4
x2 4 0 Find the real zeros of the related equation. x2 4 0
x 2 x 2 0 x 2 or x 2 The boundary points are 2 and 2. Sign of x 2 :
Sign of x 2 x 2 :
Sign of x 2 :
2
Domain: , 2 2, 37. m x log 2 x 4
b. Domain: 0, c. Range: , d. Vertical asymptote: x 0
2
40. a. pH log 5.0 10 9 8.3 b. Since the pH is greater than 7, baking soda is alkaline.
2
x 4 2 0
41. a. pH log 3.16 10 5 4.5
This is true for all real x, except: x40
b. Since the pH is less than 7, tomatoes are acidic.
x4 Domain: , 4 4,
525
42. 0
43. 1
44. p
45. x
Chapter 4 Exponential and Logarithmic Functions 46. log b x log b y
47. log b x log b y
55.
48. p log b x
1 1 ln x 2 9 ln x 3 4 4
100 12 49. log log100 log c 2 10 2 c 10 1 2 log c 2 10 2
14
x2 9 1 2 ln x 9 ln x 3 ln 4 x3
x 3 x 3 ln 4 x 3
ln 4
x3
56. log b 8 log b 23 3log b 2 3 0.289 0.867
1 50. log 2 a 2b log 2 8 log 2 a 2 log 2 b 8
57. log b 45 log b 32 5 2 log b 3 log b 5
2 0.458 0.671 1.587
3 2 log 2 a log 2 b 3 ab 2 13 ln ab 2 ln c ln d 5 51. ln 5 cd
1 58. log b log b 32 2 log b 3 9
2 0.458 0.916
1 ln a 2 ln b ln c 5ln d 3 1 2 ln a ln b ln c 5ln d 3 3
log 596 3.2839 log 7 73.2839 596
59. log 7 596
x 2 2 x 15 52. log 1 x
log 0.982 0.0131 log 4 40.0131 0.982
60. log 4 0.982
log x 2 2 x 1 log 1 x 5
log x 2 5log 2 x 1 log 1 x
12
61.
42 y 7 64 4 2 y 7 43
1 2 log x 5log 2 x 1 log 1 x 2
2y 7 3 2 y 10
1 53. 4 log 5 y 3log 5 x log 5 z 2 4 3 log 5 y log 5 x log 5 z1 2
5
y 5
y z log 5 3 x
62. 1000
4
10
2 x 1
3 2 x 1
250 2 54. log 250 log 2 log 5 log 5
1 100 102
x4
x 4
106 x3 102 x8 6 x 3 2 x 8
log100 2
8x 5
x
526
5 8
5 8
Chapter 4 Review 63.
67. 400e 2t 2.989 2.989 e 2t 400 2.989 ln e 2t ln 400
7 x 51 log 7 x log 51
x log 7 log 51 log 51 2.0206 x log 7
2.989 2t ln 400
log 51 ; x 2.0206 log 7 64.
2.989 ln 400 2.4483 t 2 2.989 ln 400 ; t 2.4483 2
516 11w 21 537 11w log 537 log11w log 537 w log11
w
log 537 2.6215 log11
68. 2 101.2t 58
log 537 ; w 2.6215 log11 65.
101.2t 29 log101.2t log 29
32 x1 43 x 2 x 1
ln 3
ln 4
1.2t log 29 3x
2 x 1 ln 3 3x ln 4
log 29 ; t 1.2187 1.2
2 x ln 3 ln 3 3x ln 4 2 x ln 3 3x ln 4 ln 3
x 2 ln 3 3ln 4 ln 3 x
e 3 e 40 0
ln 3 0.5600 3ln 4 2 ln 3
x 2
ln 2
ln 7
x
Let u e x . u 2 3u 40 0
u 8 u 5 0
2c 3 7 2 c 5 c3
e2 x 3e x 40 0
69.
ln 3 ; x 0.5600 3ln 4 2 ln 3
66.
log 29 1.2187 1.2
t
2 c5
c 3 ln 2 2c 5 ln 7
u8
or
u 5
ex 8
or
e x 5 No solution
ln e x ln 8
c ln 2 3ln 2 2c ln 7 5ln 7 c ln 2 2c ln 7 5ln 7 3ln 2
x ln 8 2.0794
ln11 ; x 2.3979
c ln 2 2 ln 7 5ln 7 3ln 2
e2 x 10e x
70.
5ln 7 3ln 2 2.3917 c ln 2 2 ln 7 5ln 7 3ln 2 ; c 2.3917 ln 2 2 ln 7
e 10 e 0 e e 10 0 x 2 x
527
x
x
e x 0 or e x 10
Chapter 4 Exponential and Logarithmic Functions 71. log 5 4 p 7 log 5 2 p
75. 3ln n 8 6.3
ln n 8 2.1
4p 7 2 p 5 p 5
n 8 e 2.1
p 1
1
72. log 2 m 2 10m log 2 11
76.
4 log 2 x log 2 x 6
m 10m 11
log 2 x log 2 x 6 4
m 10m 11 0
log 2 x x 6 4
2
2
x 11 x 1 0
x x 6 24
x 11 or x 1 11,1
x 2 6 x 16 x 2 6 x 16 0
x 8 x 2 0
73. 2 log 6 4 8 y 6 10
x 8 or x 2 Check: 4 log 2 x log 2 x 6
2 log 6 4 8 y 4
log 6 4 8 y 2 4 8 y 62
4 log 2 8 0 log 2 8 6
4 8 y 36
4 log 2 8 0 log 2 2
8 y 32 y 4
undefined undefined 4 log 2 x log 2 x 6
4 74.
n e 2.1 8 16.1662 e2.1 8 ; n 16.1662
4 log 2 2 0 log 2 2 6 4 10 log 2 8
5 4 log 3 2 5 x 1
3 0 3 2 ; The value 8 does not check.
4 4 log 3 2 5 x
1 log 3 2 5 x 31 2 5 x
77.
log 6 3x 2 log 6 x 4 1
log 6 3 x 2 log 6 x 4 1
1 2 5x 3 5 5 x 3 1 x 3 1 3
3x 2 1 log 6 x 4 3x 2 61 x4 3x 2 6 x4 3 x 2 6 x 24
3x 22 22 x 3
528
Chapter 4 Review Check:
80.
log 6 3 x 2 log 6 x 4 1
22 22 log 6 3 2 0 log 6 4 1 3 3 10 log 6 20 0 log 6 3 undefined undefined 22 ; The value does not check. 3
log x 2 log x 2 35 log x 2 2 log x 35 0 Let u log x .
u 2 2u 35 0
u 7 u 5 0 u7 log x 7 x 10
ln x ln x 2 ln x 6 0
or
u 5 log x 5 x 105 1 x 100,000
1 10,000, 000, 100,000
x x 2 ln 0 x6 x x 2 e0 x6 x2 2 x 1 x6 x2 2 x x 6
f x 4x
81.
y 4x x 4y log 4 x log 4 4 y log 4 x y
f 1 x log 4 x
x x6 0 2
x 3 x 2 0 x 3 or x 2 Check: ln x ln x 2 ln x 6
g x log x 5 1
82.
y log x 5 1 x log y 5 1
ln 3 ln 3 2 0 ln 3 6
x 1 log y 5
ln 3 ln 1 0 ln 3
10 x1 y 5
undefined undefined ln x ln x 2 ln x 6
10 x1 5 y g 1 x 10 x1 5
ln 2 ln 2 2 0 ln 2 6 ln 2 ln 4 0 ln 8
83. a.
ln 8 0 ln 8 2 ; The value 3 does not check.
P e 0.5 x 0.5 e 0.5 x ln 0.5 ln e 0.5 x
79. log 5 log 2 x 1
ln 0.5 0.5 x ln 0.5 1.39 m 0.5 Long Island Sound is murky.
log 2 x 51
x
log 2 x 5
32
or 7
x 10,000,000 or
ln x ln x 2 ln x 6
78.
or
x 25 x 32
529
Chapter 4 Exponential and Logarithmic Functions P e 0.5 x
b.
0.01 e
5 ln 3 t 9.8 yr 0.052
0.5 x
ln 0.01 ln e 0.5 x ln 0.01 0.5 x x
87. a. Since the exponent on e is negative, the population is decreasing.
ln 0.01 9.2 m 0.5
80 85.5e 0.00208t 80 e 0.00208t 85.5 80 ln 0.00208t 85.5
84. log B 1.7 2.3M
log B 2.3M 1.7 B 102.3 M 1.7 T T f T0 e kt
85.
80 ln 85.5 t 32 yr 0.00208
T0 e kt T f T T0 e kt T T f e kt
T Tf
88. a. P t 16.9 1.00836
T0
P t 16.9e0.00833t
0
1 T Tf t ln k T0
b.
P t 16.9e0.00833t 20 16.9e0.00833t 20 e0.00833t 16.9 20 ln 0.00833t 16.9
A 18,000e rt 23,344.74 18,000e5 r 23,344.74 e5 r 18,000
20 ln 16.9 t 20 yr 0.00833
23,344.74 ln 5r 18,000 23,344.74 ln 18,000 r 0.52 or 5.2% 5 b.
t
P t 16.9e ln1.00836t
T Tf kt ln T
86. a.
P 85.5e 0.00208t
b.
89.
Q t Q0 e 0.000121t 0.712Q0 Q0 e 0.000121t 0.712 e 0.000121t ln 0.712 0.000121t ln 0.712 t 2800 yr 0.000121
A 18,000e rt 30,000 18,000e0.052t 30,000 e0.052t 18,000
90. a. P t
5 e0.052t 3 5 ln 0.052t 3
3000 1 2e 0.37 t 3000
3000 3000 1000 1 2e0 1 2 1 2e The lake was initially stocked with 1000 bass. P 0
530
0.37 0
Chapter 4 Test b. P t
P 2 c. P t
P 4 d.
3000 1 2e 0.37 t 3000 1 2e 0.37 2 3000 1 2e 0.37 t 3000 1 2e 0.37 4
e. As t , The denominator of
e approaches , so the value of the expression approaches 0.
1535
f. Limiting value: P t
2061
91. a. Y1 2.38 1.5
3000 3000 bass 1 0
x
b.
3000 1 2e 0.37 t 3000 2800 1 2e 0.37 t 2800 1 2e 0.37 t 3000 15 1 2e 0.37 t 14 1 0.37 t 2e 14 1 0.37 t e 28 1 0.37t ln 28 1 ln 28 9 yr t 0.37 P t
2 0.37 t
92. a. y 6698 350 ln x b. y 6698 350 ln 26 5558 hospitals
Chapter 4 Test 1. a.
f x 4 x3 1
b.
f f x f f x 1
y 4 x3 1
3
x 1 x 1 f 3 43 1 4 4
x 4 y3 1 x 1 4 y3 x 1 y3 4 x 1 3 y 4 f 1 x 3
1
x 1 4 1 x 1 1 x 4
f f x f f x f 4 x 1 4 x 1 1 1
1
1
x 1 4
3
3
3
4
3
531
3
4x 3 x3 x 4
Chapter 4 Exponential and Logarithmic Functions 5. f x log x
2. a. f is a one-to-one function of x because no horizontal line intersects the graph in more than one point.
a. Domain: 0, ; Range: ,
b.
b.
f x log x y log x x log y 10 x 10log y 10 x y f 1 x 10 x
c. Domain: , ; Range: 0, x3 x4 x3 y x4 y3 x y4
6. f x 3x 1
f x
3.
a. Domain: , ; Range: 1, b.
y 3x 1
x y 4 y 3
x 3y 1 x 1 3y
xy 4 x y 3 xy y 4 x 3
log 3 x 1 log 3 3 y
y x 1 4 x 3
log 3 x 1 y
4x 3 x 1 4x 3 f 1 x x 1
f 1 x log 3 x 1
y
c. Domain: 1, ; Range: , x
1 7. a. f x 2 3
4. f x x 2 1, x 0 a. Domain: , 0 ; Range: , 1
1 Since k 2 , shift the graph of y 3 up 2 units.
f x x2 1
b.
f x 3x 1
y x2 1 x y2 1 y2 x 1 y 1 x f
1
x 1 x
c. Domain: , 1 ; Range: , 0 b. Domain: ,
532
x
Chapter 4 Test c. Range: 2, d. Horizontal asymptote: y 2 8. a. g x 2 x 4
Since h 4 , shift the graph of y 2 x right 4 units.
b. Domain: 1, c. Range: , d. Vertical asymptote: x 1 11. ln x y a e a x y b. Domain: ,
12. 104 x3 y log y 4 x 3
c. Range: 0, 13. log 9
d. Horizontal asymptote: y 0
1 y 81 1 1 9y 811 92 9 2 81 y 2
9. a. The graph of y ln x is the graph of y ln x reflected across the x-axis.
14. log 6 216 y
6 y 216 63 y3 15. ln e8 y
e y e8 y8 b. Domain: 0,
16. log10 4 y
c. Range: ,
10 y 10 4 y 4
d. Vertical asymptote: x 0 10. a. The graph of y log 2 x 1 3 is the
17.
graph of y log 2 x shifted left 1 unit and down 3 units.
10
log a 2 b 2
y
log y
log 10
log a 2 b2
log a 2 b 2 log y a2 b2 y
533
Chapter 4 Exponential and Logarithmic Functions 18. log1 2 1 y
2 23. 6 log 2 a 4 log 2 b log 2 c 3 6 4 log 2 a log 2 b log 2 c 2 3
y
1 1 2
a6 3 c2 a 6c 2 3 log 2 4 log 2 4 b b
y0 19. f x log 7 2 x
7 2x 0
24.
2 x 7
7 2 7 Domain: , 2 x
x 3 x 4 ln x4
12
20. g x log 4 x 2 25
x 25 0 Find the real zeros of the related equation. x 2 25 0
1 26. log b log b 53 3log b 5 125
x 5 or x 5 The boundary points are 5 and 5.
Sign of x 5 x 5 :
Sign of x 5 :
5
Domain: , 5 5,
3 0.671 2.013 27.
5
5y 1 2y 6 3 y 7 y
1 5ln x 2 ln y ln w ln z 3
28.
a2 b2 2 2 12 4 log10 22. log log a b 4 10
7 3
5 x3 53 log 5 x3 log 53
x 3 log 5 log 53
y 3
7 3
25 y 1 22 y 6
25 y 1 4 y 3 25 y 1 22
x5 y 2 21. ln 3 ln x 5 ln y 2 ln w ln z1 3 w z
3 0.289 2 0.458 1.783
x 5 x 5 0 Sign of x 5 :
ln x 3
25. log b 72 log b 23 32 3log b 2 2 log b 3
2
1 1 ln x 2 x 12 ln x 4 2 2 1 ln x 2 x 12 ln x 4 2
x log 5 3log 5 log 53
x log 5 log 53 3log 5
1 log a 2 b 2 4 2
log 53 3log 5 log 5 log 53 x 3 0.5331 log 5 x
log 53 3 ; x 0.5331 log 5 534
Chapter 4 Test 2c7 32 c3
29.
c7
34.
log x log x 1 log12
log x log x 1 log12 0
2 c3
ln 2 ln 3 c 7 ln 2 2c 3 ln 3 c ln 2 7 ln 2 2c ln 3 3ln 3 c ln 2 2c ln 3 3ln 3 7 ln 2 c ln 2 2 ln 3 3ln 3 7 ln 2 3ln 3 7 ln 2 1.0346 x ln 2 2 ln 3 3ln 3 7 ln 2 ; x 1.0346 ln 2 2 ln 3
x x 1 log 0 12 x x 1 100 12 x2 x 1 12 x 2 x 12 x 2 x 12 0
30. 7e 4 x 14
x 3 x 4 0
e 2
x 3 or x 4 Check: log x log x 1 log12
4x
ln e 4 x ln 2 4 x ln 2 ln 2 x 0.1733 4 ln 2 ; x 0.1733 4
log 3 log 4 1 0 log12 log 3 log 5 0 log12 undefined undefined log x log x 1 log12 log 4 log 4 1 0 log12
e 2 x 7e x 8 0
31.
ex
2
log 4 log 3 0 log12
7 ex 8 0
log12 0 log12
4 ; The value 3 does not check.
Let u e . u 2 7u 8 0 u 1 u 8 0 u 1 or x
e 1 x
35.
log 4 x log 4 x 30 3
u 8 e 8 No solution
log 4 x x 30 3
x
or
ln e ln1 x ln1 0 x
x x 30 43
0
32. log 5 3 x log 5 x 1 3 x x 1 2 x 2 x 1
x 2 30 x 64 x 2 30 x 64 0
x 32 x 2 0 x 32 or x 2 Check: 3 log 4 x log 4 x 30
1
3 log 4 32 0 log 4 32 30
33. 5ln x 2 1 16
3 log 4 32 0 log 4 2
ln x 2 3
x 2 e3
undefined
x e 2 18.0855 e 2 ; x 18.0855 3
3
3 log 4 x log 4 x 30
535
undefined
Chapter 4 Exponential and Logarithmic Functions 3 log 4 x log 4 x 30
39. a.
3 log 4 2 0 log 4 2 30 3 0 log 4 32 log 4 2
A Pe rt 13,566.25 10,000e r 5 13,566.25 e5 r 10,000
3 0 log 4 32 log 4 2
13,566.25 5r ln 10,000
3 0 log 4 64 3 0 3 2 ; The value 32 does not check.
13,566.25 ln 10,000 r 5 0.061 or 6.1%
log 3 log x 3 log 4 x 5
36.
log 3 log x 3 log 4 x 5 0
3 x 3 log 0 4x 5 3x 9 100 4x 5 3x 9 1 4x 5 3x 9 4 x 5
5 e0.061t ln 5 0.061t t
ln 5 26.4 yr 0.061
40. a. P t 10,000 2
t 5
10,000 2t S 92 k ln t 1
15
10,000e
ln 2 5t
S 92 ln t 1 k e 92 S k t 1
10,000e0.1386t b. 5,000,000 10,000e0.1386t
t e 92 S k 1 r A P 1 n A r 1 P n
15
10,000 e ln 2t
S 92 k ln t 1
38.
A Pe rt 50,000 10,000e0.061t
4 x
4 37.
b.
500 e0.1386t ln 500 0.1386t ln 500 t 45 hr 0.1386
nt
nt
A r ln ln 1 P n
1200 1 2e 0.12t 1200 1200 1200 P 0 400 0.12)0 1 2e 1 2e 0 1 2 400 deer were present when the park service began tracking the herd.
41. a. P t nt
A r ln nt ln 1 P n A ln P t r n ln 1 n
b. P t
P 4
536
1200 1 2e 0.12t 1200 1 2e 0.12 4
536 deer
Chapter 4 Cumulative Review c. P t
P 8 d.
1200 1 2e 0.12t 1200 1 2e 0.128
e. As t , the denominator of
e approaches , so the value of the expression approaches 0.
680 deer
f. Limiting value: P t
1200 1 2e 0.12t 1200 900 1 2e 0.12t 0.12 t 1200 900 1 2e P t
2 0.12 t
1200 1200 deer 1 0
42. a. y 88.6 7.475ln x
b. y 88.6 7.475ln12 70
4 3 1 2e 0.12t 3 1 e 0.12t 6
1 2e 0.12t
1 0.12t ln 6 1 ln 6 t 15 yr 0.12t
Chapter 4 Cumulative Review Exercises 1.
3.0 10 8.2 10 20 10 2.0 10 4.
3x 1 6 x 2 x 2 3 x 1 6 x 2 2 x 2 x 1 x 2 2 x 2 x 1 3x 6 2 x 3 x 2 x 2
9
1.23 10
5.
5 3
2 x2
3
3
5 2 x2
3
5 3 2x 7 2x 7 2 2 x 7 2 or 2x 5 5 x 2 5 9 , , 2 2
53 4x 53 4x 3 2x 4x 8 x3
4x
a b a ab b 1
3. a 3 b3 a b a b a 2 ab b 2 a b 2
5
2 2x 7
3 2.
3
7
2
537
2x 7 2 2x 9 9 x 2
10
Chapter 4 Exponential and Logarithmic Functions 6.
3 x x 1 x 6
Find the real zeros of the related equation. x 5 x 2 x 2 0 x 5 , x 2 , or x 2 The boundary points are 5 , 2 , and 2.
3x 2 3x x 6 3x 2 4 x 6 0 x
4
4 2 4 3 6 2 3
Sign of x 5 :
Sign of x 2 :
4 88 4 2 22 2 22 6 6 3 2 22 3
Sign of x 2 :
7.
Sign of x 5 x 2 x 2 : 5 2 2
The solution set is 5, 2 2, . 10. 5 x 1 3 4 x
t 3 t 3
5x 1 3 4x
or
5x 1 3 4 x
9x 4 4 x 9 4 , 2 9
or
5 x 1 3 4 x
2
t 3 t 2 6t 9 0 t 7t 6 2
0 t 1 t 6 t 1 or t 6 Check: t 3 4 t 1 1 3 40 11
t 3 4 t 1
2 2 m 3
33
2
Let u x 9 . u 2 2u 35 0 2
6 3 40 6 1
u 5 u 7 0 u 5
x 9 7
x 4 x 2 2, 4
x 2 16 x 4
2
m 1
34 m6 33m3 4m 6 3m 3 m9 9
12.
u7
or
x 9 5 2
92 m3 27 m1
3
x 2
11. x 2 9 2 x 2 9 35 0
4 40 2 9 40 7 2 40 2 3 40 7 60 2 70 7 6 ; The value 1 does not check.
9.
t 3 4 t 1 t 3 t 3
8.
2
log 2 3 x 1 log 2 x 1 3
log 2 3x 1 log 2 x 1 3
3x 1 log 2 3 x 1 3x 1 3 2 x 1 3x 1 8 x 1 8 x 8 3x 1 5 x 9 9 x 5
x 3 5 x 2 4 x 20 0 x 2 x 5 4 x 5 0
x 5 x 2 4 0 x 5 x 2 4 0 x 5 x 2 x 2 0 538
Chapter 4 Cumulative Review 15. a. f x x 2 16 x 55 Since a 0 , the graph of the function opens upward.
Check: log 2 3 x 1 log 2 x 1 3 9 9 log 2 3 1 0 log 2 1 3 5 5 32 4 log 2 0 log 2 3 5 5 undefined undefined 9 ; The value does not check. 5 13.
b. x
y 8 16 8 55 9 2
Vertex: 8, 9
c. The parabola opens upward, so the vertex is the minimum point: 8, 9 .
x4 0 x2 The expression is undefined for x 2 . This is a boundary point. Find the real zeros of the related equation. x4 0 x2 x40
d. The minimum value is 9 . e. 0 x 2 16 x 55
0 x 5 x 11 x 5 or x 11 x-intercepts: 5, 0 and 11, 0 f. f 0 0 16 0 55 55 2
x4 The boundary points are 2 and 4. Sign of x 2 :
Sign of x 4 :
x 4 : x 2
Sign of
b 16 8 2a 2 1
y-intercept: 0, 55
g.
2 4
The solution set is 2, 4 . 14.
Factors of 9 1, 3, 9 1, 3, 9 Factors of 1 1 1 1 10 10 10 9
h. The axis of symmetry is x 8 . i. ,
1 9 1 9 1 9 1 9 0
f x x 1 x 3 9 x 2 x 9
j. 9,
16. f x 1.5 x 2 x 2 x 1 The leading term is 2 3 1.5 x x x 1.5 x 6 . The end behavior is down to the left and down to the right.
Find the zeros of the quotient. 9 1 9 1 9 9 0 9 1 0 1 0
3
f x x 1 x 9 x 2 1
f 0 1.5 0 0 2 0 1 0 2
3
The y-intercept is 0, 0 .
x 1, x 9, x i The zeros are 1 , 9 , i .
539
Chapter 4 Exponential and Logarithmic Functions 18. a. f x 2 x 2 3 2 x 2 3
The zeros of the function are 0 (multiplicity 2), 2 (multiplicity 3), and 1 (multiplicity 1). Test for symmetry: 2 3 f x 1.5 x x 2 x 1
This is the graph of y 2 x shifted left 2 units and down 3 units.
1.5 x 2 x 2 x 1 3
f x is neither even nor odd.
b. Asymptote: y 3 c. , . d. 3,
3x 6 x2 f is in lowest terms, and x 2 is 0 for x 2 , which is the vertical asymptote.
17. a. f x
40 50 19. log 40 log 50 log 2 log 2 2000 log 2
b. The degree of the numerator is 1. The degree of the denominator is 1. 3 Since n m , the line y 3 is a 1 horizontal asymptote of f.
log1000 3 20.
f x 3 x 4 1 y 3 x 4 1 x 3 y 4 1
c.
x 1 3 y 4
x 13 y 4 x 13 4 y 3 f 1 x x 1 4
540
Section 5.1
Chapter 5 Systems of Equations and Inequalities Section 5.1 Systems of Linear Equations in Two Variables and Applications 2. solution
7 x 3 y 23
3. substitution; addition 4. parallel
7 7 1 3 ⱨ 23 6
1. system
5. inconsistent
7 7 ⱨ 23 2 21 ⱨ 23 false 2
6. dependent
3 x 5 y 7
7. a.
3 1 5 2 ⱨ 7 3 10 ⱨ 7 7 ⱨ 7 true x 4 y 7
No 11 2 6 3 ⱨ 4
1 4 2 ⱨ 7
22 18 ⱨ 4
1 8 ⱨ 7 7 ⱨ 7 true
4 ⱨ 4 true 7 x 3 y 23
7 2 3 3 ⱨ 23
Yes 3x 5 y 7
b.
11x 6 y 4
b.
14 9 ⱨ 23
2 3 5 1 ⱨ 7 3
23 ⱨ 23 true Yes
2 5 ⱨ 7 7 ⱨ 7 true x 4 y 7
9. a.
2 4 1 ⱨ 7 3 2 4 ⱨ 7 3 14 ⱨ 7 false 3
3 x5 2 3 2 ⱨ 2 5 2 2 ⱨ 3 5 2 ⱨ 2 true 6 x 4 y 20 y
6 2 4 2 ⱨ 20 12 8 ⱨ 20 20 ⱨ 20 true
No 8. a.
11x 6 y 4
Yes
7 111 6 ⱨ 4 6
b.
11 7 ⱨ 4 4 ⱨ 4 true
541
3 x5 2 3 11 ⱨ 4 5 2 11ⱨ 6 5 11ⱨ 11 true y
Chapter 5 Systems of Equations and Inequalities 6 x 4 y 20
15. x 3 y 5
6 4 4 11 ⱨ 20
x 3 y 5
24 44 ⱨ 20 20 ⱨ 20 true
3x 2 y 18 3 3 y 5 2 y 18
Yes
9 y 15 2 y 18
1 10. a. y x 2 5 1 1 ⱨ 5 2 5 1ⱨ 1 2 1ⱨ1 true 2 x 10 y 10
11 y 33 y3 x 3 y 5 3 3 5 9 5 4
4, 3
16. 2 x y 2
2 5 10 1 ⱨ10
y 2 x 2
10 10 ⱨ10
5x 3 y 9 5 x 3 2 x 2 9
20 ⱨ10 false No
5x 6 x 6 9 x 3 x 3 y 2 x 2 2 3 2 6 2 8
1 b. y x 2 5 1 4 ⱨ 10 2 5 4ⱨ2 2 4 ⱨ 4 true 2 x 10 y 10
3, 8
17. 3 y 7 2
3y 9 y3
2 10 10 4 ⱨ10 20 40 ⱨ10 20 ⱨ10 false
2x 7 y 1 2 x 7 3 1
No
2 x 21 1 2 x 20
11. Intersecting lines, one solution.
x 10
12. Parallel lines, no solution. The system is inconsistent.
10, 3
18. 6 5 x 1
13. Same line, infinitely many solutions. The equations are dependent.
5 x 5 x 1
14. Intersecting lines, one solution.
3x 2 y 11 3 1 2 y 11 3 2 y 11 8 2y 4 y
542
1, 4
Section 5.1 19. 2 x y 2 y
2 21. 3 x 7 y 1 6 x 14 y 2 6 x 5 y 17 6 x 5 y 17 19 y 19 y 1 3x 7 y 1
2x 2 y 2 y 2x 2 3y 2 3y x 2
3x 7 1 1
4x 1 2 5 y
3x 7 1 3x 6
2 3y 4 1 2 5y 2
2, 1
x 2
2 2 3y 1 2 5 y 4 6 y 1 2 5y
2 10 x 4 y 4 22. 5 x 2 y 2 3 x 4 y 30 3x 4 y 30 13x 26 x 2 5 x 2 y 2
y 1 y 1 2 3 y 2 3 1 2 3 1 2 2 2 2 1 ,1 2 x
5 2 2 y 2 10 2 y 2 2 y 12
20. 5 x y 9 2 y
2, 6
y6
5x 5 y 9 2 y 5x 9 3 y 9 3y x 5
23.
11x 5 4 y
2 x 2 y 22 y
11x 4 y 5
6 y 2 10 7 x
2 x 4 y 22 y 2 x 5 y 22
5 11x 4 y 5 55 x 20 y 25 4 2 x 5 y 22 8 x 20 y 88 63x 63 x 1
9 3y 6 y 2 10 7 5 30 y 10 50 7 9 3 y
11x 5 4 y
30 y 10 50 63 21y 9 y 3 1 y 3 1 9 3 3 9 1 9 3y x x 2 5 5 5 1 2, 3
111 5 4 y 11 5 4 y 16 4 y 4 y
1, 4
24. 3 x y y 14
2x 2 7 y
3x 3 y y 14 3 x 2 y 14
2 x 7 y 2
2 3 x 2 y 14 6 x 4 y 28 3 2 x 7 y 2 6 x 21y 6 17 y 34 y 2
543
Chapter 5 Systems of Equations and Inequalities 3x 6 y 5
2x 2 7 y 2 x 2 7 2
2 3x 6 5 45
2 x 2 14 2 x 12
4 5 15 45 x 4 75 45 x 79
3x
x6
6, 2
25. 0.6 x 0.1y 0.4
6x y 4
2 x 0.7 y 0.3
20 x 7 y 3
x
7 6 x y 4 42 x 7 y 28 20 x 7 y 3 20 x 7 y 3 62 x 31 0 1 0 x 0 2 0 6x y 4
12 2x 9 2 35
3 y 4
26. 0.25 x 0.04 y 0.24
0.15 x 0.12 y 0.12
1 ,1 2
108 2 35 70 x 108 70 70 x 178 2x
25 x 4 y 24 15 x 12 y 12
x
3 25 x 4 y 24 75 x 12 y 72 15 x 12 y 12 15 x 12 y 12 60 x 60 x 1
29. 3x 4 y 6
89 35
89 12 , 35 35 9 x 12 y 4 9 x 12 y 4
25 x 4 y 24
3 3x 4 y 6 9 x 12 y 18 9 x 12 y 4 9 x 12 y 4 0 14
25 1 4 y 24 25 4 y 24 4 y 1 1 y 4
79 2 , 45 45
2 28. 3 x 4 y 9 6 x 8 y 18 3 2 x 9 y 2 6 x 27 y 6 35 y 12 0 12 0 y 0 35 0 2x 9 y 2
1 6 y 4 2
y 1
79 45
; The system is inconsistent.
1 1, 4
30. 4 x 8 y 2
3 27. 2 x 11 y 4 6 x 33 y 12 2 3 x 6 y 5 6 x 12 y 10 45 y 2 0 2 0 y 0 45 0
2x 8 4 y 2x 4 y 8
4 x 8 y 2 4 x 8 y 2 2 2 x 4 y 8 4 x 8 y 16 0 18
; The system is inconsistent.
544
Section 5.1 34. 3 x 3 y 2 y
31. 3x y 6
y 3x 6 x
3x 9 y 2 y 3 x 11y
1 y2 3
x
1 3x 6 2 3 xx2 2 22 x, y | 3x y 6 ; The equations are x
2x 5 5 7 y 2 x 7 y 11 2 y 7 y 3
dependent.
22 y 7 y 3 22 y 21y
32. 2 x y 8
y 2 x 8
43 y 0
y 2x 8
y0 11 11 x y 0 0 3 3
1 x y4 2 x
11 y 3
1 2 x 8 4 2 xx4 4
35. 5 y 800 x
x 5 y 800
44 x, y | 2 x y 8 ; The equations are
3x 10 y 1900 3 5 y 800 10 y 1900
dependent.
15 y 2400 10 y 1900 5 y 500
33. 2 x 4 4 5 y
y 100
2x 5 y x
0, 0
x 5 y 800 5 100 800
5 y 2
500 800 300 300, 100
2 4 x y 7 y 2 4 x y 7 y
36. 4 x 1800 y
4x 4 y 7 y
y 4 x 1800
4x 3y
2 x 7 y 2400
5 4 y 3 y 2
2 x 7 4 x 1800 2400
10 y 3 y
2 x 28 x 12,600 2400
7y 0
30 x 15,000
y0 5 5 x y 0 0 2 2
x 500 y 4 x 1800 4 500 1800
0, 0
2000 1800 200 500, 200
545
Chapter 5 Systems of Equations and Inequalities 37. x
2 x 1 3 1 y x2 6
3 5 y 2 2 3 5 x y 2 2
39. y
5 2x y y x 8 10 x 5 y y x 8 11x 4 y 8 5 3 11 y 4 y 8 2 2 33 55 y 4 y 8 2 2 33 y 55 8 y 16 41 y 71 71 y 41 3 5 3 71 5 213 5 x y 2 2 2 41 2 82 2 213 205 8 4 82 82 82 41 4 71 , 41 41 38. x
2 1 x 1 x 2 3 6 4 x 6 x 12 3x 18 x6 1 1 y x 2 6 2 1 2 3 6 6 6, 3 1 40. y x 7 4 3 y x 17 2 1 3 x 7 x 17 4 2 2 x 56 12 x 136 10 x 80 x8 1 1 y x 7 8 7 2 7 5 4 4 8, 5
5 3 y 4 2 x
5 3 y 4 2
3 2 x y 2 x 6x 3y 2 x 7x 3y 2 3 5 7 y 3y 2 4 2 7 5 y 6 12 y 8 35 y 42 12 y 8 47 y 34 34 y 47 5 3 5 34 3 170 3 x y 4 2 4 47 2 188 2 170 282 112 28 188 188 188 47 28 34 , 47 47
41. 4 x 2 6 y 3
4x 8 6 y 3
1 3 1 x y 4 8 2 2 x 3 y 4
4 x 6 y 11 4 x 6 y 11 4 x 6 y 11 2 2 x 3 y 4 4 x 6 y 8 0 19
; The system is inconsistent.
546
Section 5.1
42.
1 1 1 x y 14 7 2 x 2y 7
46. y 0.18 x 0.129 y 0.15 x 0.1275
2 x 2 y 3 20 2 x 4 y 17
0.18 x 0.129 0.15 x 0.1275 0.03x 0.0015 x 0.05 y 0.18 x 0.129 0.18 0.05 0.129 0.009 0.129 0.12 0.05, 0.12
2
2 x 4 y 14 x 2 y 7 2 x 4 y 17 2 x 4 y 17 0 3
; The system is inconsistent. 43.
y 1 2 4x y 2 4x 2 y 2x
47.
0.04 x 0.01 y 0.02
x 2 y 1 6 8 2 x 2 4 y 4 48 x 4 y 50
x 2 y 1 12 2 4 2 x 2 y 1 48 2 x 4 y 1 48 y 2 x 53 y 2 4 y 50 53 y 8 y 100 53 9 y 153 y 17 x 4 y 50 4 17 50 68 50 18
4x y 2 4 x 4 x 2 2 4x 4x 2 2 22 x, y | 4 x y 2 ; The equations are dependent. y 3 5 5 5x y 3
44. x
18, 17
y 5 x 3 0.05 x 0.01y 0.03
48.
5x y 3 5 x 5 x 3 3 5x 5x 3 3 33 x, y | 5 x y 3 ; The equations are
x 1 y 2 1 2 10 5 x 1 y 2 10 5 x 5 y 2 10 5 x 17 y x 1 y 2 21 6 2 x 1 3 y 2 126 x 1 3 y 6 126 x 3 y 131 y 5 x 17 y 5 3 y 131 17 y 15 y 655 17 16 y 672 y 42
dependent. 45. y 2.4 x 1.54
y 3.5 x 7.9 2.4 x 1.54 3.5 x 7.9 5.9 x 9.44 x 1.6 y 2.4 x 1.54 2.4 1.6 1.54 3.84 1.54 2.3 1.6, 2.3
x 3 y 131 3 42 131 126 131 5
5, 42 547
Chapter 5 Systems of Equations and Inequalities 49. Let x represent the amount of 36% alcohol solution. Let y represent the amount of 20% alcohol solution.
Amount of mixture Pure alcohol
36% Solution
20% Solution
30% Solution
x
y
40
0.36x
0.20y
0.30 40 12
20 x y 40 x y 40 20 x 20 y 800 100 0.36 x 0.20 y 12 36 x 20 y 1200 36 x 20 y 1200 16 x 400 x 25 x y 40
25 y 40 y 15
25 L of 36% solution and 15 L of 20% solution should be mixed.
50. Let x represent the amount of 30% saline solution. Let y represent the amount of 10% saline solution.
Amount of mixture Pure saline
30% Solution
10% Solution
12% Solution
x
y
200
0.30x
0.10y
0.12 200 24
x y 200 x y 200 x 100 10 0.30 x 0.10 y 24 3 30 x 10 y 2400 x
y 200 y 240 2 x 40 x 20
x y 200
20 y 200 y 180 The pharmacist should mix 20 mL of the 30% solution and 180 mL of the 10% solution. 51. Let x represent the amount of 100% antifreeze solution. Let y represent the amount of 36% antifreeze solution. 100% Solution
36% Solution
50% Solution
Amount of mixture
x
y
16
Pure antifreeze
x
0.36y
0.50 16 8
100 x y 16 x y 16 100 x 100 y 1600 100 100 x 36 y 800 x 0.36 y 8 100 x 36 y 800 64 y 800 y 12.5 x y 16 x 12.5 16 x 3.5 3.5 L should be replaced.
548
Section 5.1 52. a. Let x represent the amount of 0% vinegar solution (water). Let y represent the amount of 100% vinegar solution. 0% Solution
100% Solution
40% Solution
Amount of mixture
x
y
10
Pure vinegar
0
y
0.40 10 4
y4 x y 10 x 4 10 x6
6 cups of water and 4 cups of vinegar should be mixed.
b. Let x represent the amount of 100% vinegar solution. Let y represent the amount of 40% vinegar solution. 100% Solution
40% Solution
60% Solution
Amount of mixture
x
y
10
Pure antifreeze
x
0.40y
0.60 10 6
100 x y 10 x y 10 100 x 100 y 1000 100 x 0.40 y 6 100 x 40 y 600 100 x 40 y 600 60 y 400 0 0 20 0 0 y 0 0 3 0 0 x y 10 20 x 10 3 2 1 10 x He should mix 6 cups of the 40% solution and 3 cups of pure vinegar. 3 3 3
53. Let x represent the grams of fat in Cherry ice cream. Let y represent the grams of fat in Mint Chocolate Chunk ice cream. Monique’s sundae: 2 x y 43
54. Let x represent the milligrams of sodium in 1 oz of chips. Let y represent the milligrams of sodium in cup of soda Bryan’s intake: 3x 2 y 700 Jadyn’s intake: x 3 y 350 x 3 y 350 3 x 2 y 700 3 3 y 350 2 y 700 9 y 1050 2 y 700 7 y 350 y 50
y 2 x 43 Tara’s sundae: x 2 y 47 x 2 2 x 43 47 x 4 x 86 47 3 x 39 x 13 y 2 x 43 2 13 43 26 43 17 Cherry has 13 g of fat and Mint Chocolate Chunk has 17 g of fat.
x 3 y 350 3 50 350 150 350 200 1 oz of chips contains 200 mg of sodium and 1 cup of soda contains 50 mg of sodium.
549
Chapter 5 Systems of Equations and Inequalities 55. Let x represent the amount borrowed at 4.6%. Let y represent the amount borrowed at 6.2%.
Principal Interest (I = Prt)
4.6% interest
6.2% interest
Total
x
y
5000
x 0.046 3 0.138 x
y 0.062 3 0.186 y
762
138 x y 5000 x y 5000 138 x 138 y 690,000 1000 0.138 x 0.186 y 762 138 x 186 y 762,000 138 x 186 y 762,000 48 y 72,000 y 1500 x y 5000 x 1500 5000 x 3500 She borrowed $3500 at 4.6% and $1500 at 6.2%.
56. Let x represent the amount borrowed from the school at 4.5% interest. Let y represent the amount borrowed from the government at 6% interest.
Principal Interest (I = Prt)
4.5% interest
6% interest
Total
x
y
100,000
x 0.045 4 0.18 x
y 0.06 4 0.24 y
19,200
18 x y 100,000 x y 100,000 18 x 18 y 1,800,000 100 0.18 x 0.24 y 19,200 18 x 24 y 1,920,000 18 x 24 y 1,920,000 6 y 120,000 y 20,000 x y 100,000 x 20,000 100,000 x 80,000 He borrowed $80,000 from the school and $20,000 through the government loan.
57. Let x represent one employee’s weekly salary. Let y represent the other employee’s weekly salary. x y 1350 2 x y 2700 x y 300 y 300 y 2700
58. Let x represent the electrician’s hourly wage. Let y represent the plumber’s hourly wage. x y 33 2 x y 66
x y 66 8 x 5 y 438 8 y 66 5 y 438
2 y 2400
8 y 528 5 y 438
y 1200 x y 300 1200 300 1500 One makes $1200 and the other makes $1500.
3 y 90 y 30 x y 66 30 66 36 The electrician makes $36/hr and the plumber makes $30/hr. 550
Section 5.1 2 x 2 y 44
59. Let x represent Josie’s walking speed. Let y represent the speed of the sidewalk.
With sidewalk Against sidewalk
Distance (ft)
Rate (ft/sec)
Time (sec)
200
x y
40
90
x y
30
2 18 2 y 44 36 2 y 44 2y 8 y4 The current is 4 mph and the boat travels 18 mph in still water.
d rt
61. Let x represent the speed of one runner Let y represent the speed of the other runner.
200 x y 40 40 x 40 y 90 x y 30 30 x 30 y 4 40 x 40 y 200 10 x 10 y 50 3 30 x 30 y 90 10 x 10 y 30 20 x 80 x 4
Opposite direction Same direction
40 x 40 y 200
Distance (m)
Rate (m/sec)
Time (sec)
390
x y
30
390
x y
130
d rt
40 4 40 y 200
390 x y 30 30 x 30 y
160 40 y 200 40 y 40 y 1 The sidewalk moves at 1 ft/sec and Josie walks 4 ft/sec on nonmoving ground.
390 x y 130 130 x 130 y 30 30 x 30 y 390 x y 13 130 130 x 130 y 390 x y 3 2 x 16 x 8
30 x 30 y 390
60. Let x represent the speed of the boat in still water. Let y represent the speed of the current.
With current Against current
Distance (mi)
Rate (mi/hr)
Time (hr)
44
x y
2
56
x y
4
30 8 30 y 390 240 30 y 390 30 y 150 y5 The speeds are 8 m/sec and 5 m/sec. 62. Let x represent the speed of one particle. Let y represent the speed of the other particle.
d rt 44 x y 2 2 x 2 y 56 x y 4 4 x 4 y
Opposite direction Same direction
2 x 2 y 44 4 x 4 y 88 4 x 4 y 56 4 x 4 y 56 8 x 144 x 18 2
Distance (ft)
Rate (ft/sec)
Time (sec)
390
x y
30
390
x y
130
d rt 280 x y 10 10 x 10 y 280 x y 70 70 x 70 y
551
Chapter 5 Systems of Equations and Inequalities 10 10 x 10 y 280 x y 28 70 70 x 70 y 280 x y 4 2 x 32 x 16
b. m
y y1 m x x1 y 188 67 x 2
10 x 10 y 280
y 188 67 x 134
10 16 10 y 280
y 67 x 54
160 10 y 280 10 y 120 y 12 The particles travel 16 ft/sec and 12 ft/sec. 63. a. m
c. 1.25 x 54.3 0.32 x 64.5
1.57 x 10.2 x 6.5 y 0.32 x 64.5 0.32 6.5 64.5 62.4 {(18, 1260)}; The solution indicates that in the year 2018, the yearly number of visitors from Brazil and from Australia will equal 1,260,000 each if these trends continue.
y2 y1 66.8 56.8 10 1.25 10 2 8 x2 x1
y y1 m x x1
y 56.8 1.25 x 2 y 56.8 1.25 x 2.5 y 1.25 x 54.3
65. a. C x 52 x 480
y y1 61.3 64.5 3.2 0.32 b. m 2 10 0 10 x2 x1
b. R x 100 x
y y1 m x x1
c.
y 64.5 0.32 x 0
C x R x
52 x 480 100 x 48 x 480 x 10 offices
y 64.5 0.32 x y 0.32 x 64.5
d. 28 10 ; The company will make money.
c. 1.25 x 54.3 0.32 x 64.5
1.57 x 10.2 x 6.5 y 0.32 x 64.5
66. a. C x 0.75 x 100 b. R x 2 x
0.32 6.5 64.5 62.4 {(6.5, 62.4)}; The solution indicates that midyear in 2006, the per capita consumption of beef and chicken was approximately equal at 62.4 lb each. 64. a. m
y2 y1 724 188 536 67 10 2 8 x2 x1
c.
C x R x
0.75 x 100 2 x 1.25 x 100 x 80 products
y2 y1 666 342 324 54 7 1 6 x2 x1
d. 60 80 ; The vendor will lose money.
y y1 m x x1 y 342 54 x 1 y 342 54 x 54
y 54 x 288
552
Section 5.1 y x 2 0 2 2 0, 2 Area of small triangle: 1 1 A bh 2 2 2 2 2 Area of large triangle: 1 1 A bh 2 4 4 2 2 Total area: 2 4 6 square units
67. a.
b. Find the intersection of the lines. 1 2x x 5 2 4 x x 10 5 x 10 x2 y 2 x 2 2 4 2, 4 The slopes are negative reciprocals, so the lines are perpendicular and the triangle is a right triangle. Length of short leg:
69. a. m
y y1 m x x1
4 x 3 5 4 12 y 1 x 5 5 4 7 y x 5 5
y 1
b. m
d 22 42 4 16 20 2 5 Length of long leg: d
10 2 0 4 82 4 2
2
y2 y1 7 1 8 4 x2 x1 7 3 10 5
y2 y1 1 5 4 1 x2 x1 4 0 4
y y1 m x x1
2
y 5 1 x 0
64 16 80 4 5 1 A bh 2 1 4 5 2 5 20 square units 2
y 5 x y x 5
c.
68. a.
4 7 x x 5 5 5 4 x 7 5 x 25 9 x 18 x2 y x 5 2 5 3 The centroid is 2, 3 .
70. a. m
y2 y1 6 2 8 x2 x1 3 2 5
y y1 m x x1
8 x 2 5 8 16 y2 x 5 5 8 6 y x 5 5
y 2
b. Find the intersection of the lines. 1 x2 x2 2 3 `x 0 2 x0
553
Chapter 5 Systems of Equations and Inequalities b. m
73. y 60 x 28 180
y2 y1 4 0 4 4 x2 x1 0 1 1
y x 148 x 28 x y 180
y y1 m x x1
y 2 x 208 y x 148 2 x 208 x 148
y 0 4 x 1 y 4 x 4 c.
8 6 x 4 x 4 5 5 8 x 6 20 x 20
x 60 x 60 y 2 x 208 2 60 208
28 x 14
120 208 88 x 60, y 88
1 x 2
74. x y 46 x y 82 180
1 y 4 x 4 4 4 2 4 2 2
y 46 y 82 180
1 The centroid is , 2 . 2
2 y 144 y 72 x y 46 72 46 26 x 26, y 72
71. Let x represent one angle. Let y represent the other angle. x y 90 x 2y 6 x y 90
75. For example: x y a
2 y 6 y 90
3 5 a 2a System: x y 2
3 y 96 y 32
x 2 y 6 2 32 6 64 6 58 The angles are 32 and 58 .
2x y b 2 3 5 b 1 b
2 x y 1
72. Let x represent one angle. Let y represent the other angle. x y 180 x 5 y 12 x y 180
76. For example: x y a
3x 4 y b
4 3 a
3 4 4 3 b
1 a
12 12 b
System: x y 1
5 y 12 y 180
0b
3x 4 y 0
6 y 168 y 28
77.
x 5 y 12 5 28 12 140 12 152 The angles are 28 and 152 .
Cx 5 y 13
2 x Dy 5
C 4 5 1 13
2 4 D 1 5
4C 5 13
8 D 5
4C 8
D3
C2
554
Section 5.1 3 x Ay 3
Bx y 12
3 5 A 2 3
B 5 2 12
15 2 A 3 2 A 12
5B 2 12 5B 10
78.
f x mx b
f 12 8
m 3 b 3
m 12 b 8
b 3m 3
v2 u 2v 5 2 2 5 4 1 1 x 1 1 x 1 u 1
u
1 b 3 3 3
1 3
g 2 1
g 4 10
m 2 b 1
m 4 b 10
b 2m 1
b 2m 1 3 b 2 1 2
3 2
1 1 and v . x y u 2v 1 u 2v 1
83. Let x represent Marta’s bicycling speed. Let y represent Marta’s running speed. d Use d rt t . r 12 4 4 x y 3 21 3 5 x y 3 1 1 Let u and v . x y 4 1 4 12u 4v 3u v 3 3 5 3 5 7u v 21u 3v 3 9 0 2 4u 0 9 0 1 u 18 0
b 4m 10
2m 1 4m 10 6m 9
b 3 1 4
81. Let u
u 4v 7 u 4v 7 u 4v 7
2v 1 4v 7 6v 6
v 1 u 2v 1 2 1 1 2 1 3 1 x 1 1 x u 3
u
1 y 1 1 y v 2 v
1 1, 2
b 1 3 4 g x mx b
m
2v 4
b 3m 3
g x mx b
80.
6v 15 4v 11
b 12m 8
3m 3 12m 8 15m 5
u 2v 5 u 2v 5
3 2v 5 4v 11
f x mx b
f 3 3
m
3u 4v 11
B2
A6 79.
1 1 and v . x y 3u 4v 11
82. Let u
1 y 1 1 y 1 v 1 v
1 , 1 3 555
Chapter 5 Systems of Equations and Inequalities 1 1 v x y 1 1 1 1 x 2 y 8 u 1 v 1 2 8 Sheila swims 2 mph and runs 8 mph.
4 3 4 1 12 4v 18 3 2 4 4v 3 3 2 4v 3 1 v 6 1 1 u v x y 1 1 1 1 x 18 y 6 1 u v 1 18 6 Marta bicycles 18 mph and runs 6 mph.
u
12u 4v
85. Let x represent city miles. Let y represent highway miles. mi mi Use mpg gal . gal mpg x y 254 x y 254 x y 14 16 22 y x 176 176 14 16 22 11x 8 y 2464 11 y 254 8 y 2464 11 y 2794 8 y 2464 3 y 330 y 110 x y 254 110 254 144 The truck was driven 144 mi in the city and 110 mi on the highway.
84. Let x represent Shelia’s swimming speed. Let y represent Shelia’s running speed. d Use d rt t . r 1 6 5 x y 4 2 8 2 x y 1 1 Let u and v . x y 5 5 u 6v u 6v 4 4 2 2u 8v 2 u 4v 1 1 0 2v 4 0 1 0 v 8 0 5 u 6v 4 1 5 u 6 8 4 3 5 u 4 4 2 1 u 4 2
86. Let x represent city miles. Let y represent highway miles. mi mi Use mpg gal . gal mpg x y 420 x y 420 x y 26 12 18 y x 36 36 26 12 18 3 x 2 y 936 3 y 420 2 y 936 3 y 1260 2 y 936 y 324 y 324 x y 420 324 420 96 The driver drove 96 mi in the city and 324 mi on the highway.
556
Section 5.1 87. p 0.025 x
93.
p 0.04 x 104 a. 0.025 x 0.04 x 104
92 b 2 c 2 c 2 b 2 81 c b c b 81 Positive factors of 81: 1, 3, 9, 27, 81 c b 1 c b 81 c b 3 c b 81 c b 1 c b 27 2c 82 2c 82 2c 30 c 41 c 41 c 15 c b 81 c b 1 c b 27
0.065 x 104 x 1600 p 0.025 x 0.025 1600 40
1600, 40 b. $40
41 b 81
c. 1600 tickets
b 40
88. p 0.002 x
c b 27 c b 3 2c 30 c 15
p 0.005 x 70 a. 0.002 x 0.005 x 70
0.007 x 70 x 10,000
cb3 15 b 3
p 0.002 x 0.002 10,000 20
b 12
10,000, 20
41 b 1 b 40
15 b 27 b 12
c b 9 c b 9 2c 18 c 9 cb9 9b 9 b0
9,12,15 and 9, 40, 41
b. $20 94.
c. 10,000 cookbooks 89. If the system represents two intersecting lines, then the lines intersect in exactly one point. The solution set consists of the ordered pair representing that point. If the lines in the system are parallel, then the lines do not intersect and the system has no solution. If the equations in a system of linear equations represent the same line, then the solution set is the set of points on the line. 90. If the system of equations reduces to an identity such as 0 0 , then the equations are dependent. 91. If the system of equations reduces to a contradiction such as 0 1 , then the system has no solution and is said to be inconsistent. 92. The solutions to the system are ordered pairs generated by the equation y x 2 . In each ordered pair, y depends on x (y is 2 more than x). Likewise, x depends on y (x is 2 less than y).
a 2 122 c 2 c 2 a 2 144 c a c a 144 Positive factors of 144: 1, 2, 3, 4, 6, 8, 9,12,16,18, 24, 36, 48, 72,144 c a 1 c a 144 c a 3 c a 144 c a 1 c a 48 2c 145 2c 145 2c 51 145 145 51 c c c 2 2 2 c a 48 c a 3 2c 51 51 c 2
c a 9 c a 16 2c 25 25 c 2
c a 16 c a 9 2c 25 25 c 2
c a 2 c a 72 2c 74 c 37
c a 4 c a 36 2c 40 c 20
c a 72
c a 72 c a 2 2c 74 c 37 ca 2
37 a 72
37 a 2
20 a 36
a 35 557
a 35
c a 36 a 16
Chapter 5 Systems of Equations and Inequalities c a 36 c a 4 2c 40 c 20 ca 4
c a 6 c a 24 2c 30 c 15 c a 24
c a 24 c a 6 2c 30 c 15 ca6
20 a 4 a 16
15 a 24 a9
15 a 6 a 9
c a 8 c a 18 2c 26 c 13 c a 18
c a 18 c a 8 2c 26 c 13 ca8
c a 12 c a 12 2c 24 c 12 c a 12
13 a 18 a5
13 a 8 a 5
12 a 12 a0
F2
100 1 3
100
1 3
96.
2.470, 2.253 97.
2.017, 0.015 98.
0.529, 13.806
1 3
1 3 1 3
100 1 3
50
3
3 2 F1 F2 2 2 3 2 F1 F2 0 2 2 1 2 F1 F2 50 2 2 3 2 F1 F2 0 2 2 1 2 F1 F2 50 2 2 1 3 F1 50 2 100 F1 1 3 F1
2 3
3 F1 2
F1 2 2 6 F1 F1 2 2 2 6 50 3 1 6 25 2 25 2 3 3 lb 44.8 lb
35,12, 37 , 16,12, 20 , 9,12,15 , and 5,12,13 95.
2
100 1 3
99.
2
3 1 lb 36.6 lb
2 3 F2 F1 2 2
1.028, 15.772 558
3 1
Section 5.2 Section 5.2 Systems of Linear Equations in Three Variables and Applications 1. x 5 y 12 1 1 1 x y 2 3 3 3x 2 2 y
5. linear 6. triple 7. plane
3 5 y 12 2 2 y 15 y 36 2 2 y
8. triple; intersection
17 y 34
9. For example: Let x = 0 and y = 0. 2 x 4 y 6 z 12
y 2 x 5 y 12 5 2 12 10 12 2
2 0 4 0 6 z 12
2, 2
6 z 12 z 2 Let x = 0 and z = 0. 2 x 4 y 6 z 12
2. y 2 x 6 0.03 x 0.07 y 0.02
3x 7 y 2
0, 0, 2
2 0 4 y 6 0 12
3x 7 2 x 6 2
4 y 12 y3 Let y = 0 and z = 0. 2 x 4 y 6 z 12
3x 14 x 42 2 11x 44 x 4 y 2 x 6 2 4 6 8 6 2
0, 3, 0
2 x 4 0 6 0 12
4, 2
2 x 12 x6
3 3. y x 1 4 3x 4 y 4
6, 0, 0
10. For example: Let x = 0 and y = 0. 3 x 5 y z 15
3 3x 4 x 1 4 4
3 0 5 0 z 15
3x 3x 4 4 44 3 x, y | y x 1 4
z 15 Let x = 0 and z = 0. 3x 5 y z 15
0, 0,15
3 0 5 y 0 15 5 y 15 y 3 Let y = 0 and z = 0. 3 x 5 y z 15
4. y 2 x 5 1 1 x y 1 5 10 2 x y 10
0, 3, 0
3x 5 0 0 15
2 x 2 x 5 10
3x 15 x5
2 x 2 x 5 10 5 10 ; inconsistent system 559
5, 0, 0
Chapter 5 Systems of Equations and Inequalities x 3y 7z 7
11. a.
2 x 3 y z 12
b.
2 1 3 4 2 ⱨ 12
2 3 3 7 0 ⱨ 7 2 9 ⱨ 7 7 ⱨ 7 true 2 x 4 y z 16
2 12 2 ⱨ 12 12 ⱨ 12 true x y 2z 9
2 2 4 3 0 ⱨ16
1 4 2 2 ⱨ 9
4 12 ⱨ16
1 4 4ⱨ9
16 ⱨ16 true 3x 5 y 6 z 9
9 ⱨ 9 true 3 x 2 y z 7
3 2 5 3 6 0 ⱨ 9
3 1 2 4 2 ⱨ 7
6 15 ⱨ 9
3 8 2 ⱨ 7
9 ⱨ 9 true
7 ⱨ 7 true
Yes
Yes x 3y 7z 7
b.
x y z 2
13. a.
2 3 4 7 1 ⱨ 7
2 0 0 ⱨ 2
2 12 7 ⱨ 7 7 ⱨ 7 true 2 x 4 y z 16
2 ⱨ 2 true x 2y z 2
2 2 0 0 ⱨ 2
2 2 4 4 1 ⱨ16
2 ⱨ 2 true 3x 5 y z 6
4 16 1ⱨ16 13 ⱨ16 false
3 2 5 0 0 ⱨ 6
No
6 ⱨ 6 true 2 x 3 y z 12
12. a.
Yes
2 2 3 5 1 ⱨ 12
x y z 2
b.
1 2 1 ⱨ 2
4 15 1ⱨ 12 12 ⱨ 12 true x y 2z 9
1 2 1ⱨ 2 2 ⱨ 2 true x 2y z 2
2 5 2 1 ⱨ 9
1 2 2 1 ⱨ 2
2 5 2ⱨ9 9 ⱨ 9 true 3x 2 y z 7
1 4 1ⱨ 2 2 ⱨ 2 true 3x 5 y z 6
3 2 2 5 1 ⱨ 7
3 1 5 2 1 ⱨ 6
6 10 1ⱨ 7
3 10 1ⱨ 6 6 ⱨ 6 true
5 ⱨ 7 false No Yes
560
Section 5.2 14. a.
b.
x y z 3
3x 4 y z 1
1 2 6 ⱨ 3
3 1 4 2 6 ⱨ1
1 2 6 ⱨ 3 3 ⱨ 3 true
3 8 6 ⱨ1 5 ⱨ1 false
x y z 3
3x 4 y z 1
3 1 5 ⱨ 3
3 3 4 1 5 ⱨ1
3 1 5 ⱨ 3 3 ⱨ 3 true
9 4 5 ⱨ1 0 ⱨ1 false
No
No
x 2 y z 9 15. A B 3x 4 y 5 z 9 C 2 x 3 y z 12 Eliminate z from equations A and C. A x 2 y z 9 C 2 x 3 y z 12 x y 3 D Eliminate z from equations B and C. B 3x 4 y 5 z 9 3x 4 y 5 z 9 Multiply by 5 C 2 x 3 y z 12 10 x 15 y 5 z 60 7 x 19 y 69 E Solve the systems of equations D and E. Multiply by 7 D x y 3 7 x 7 y 21 E 7 x 19 y 69 7 x 19 y 69 12 y 48 y 4 Back substitute. A x 2 y z 9 D x y 3 x 4 3
1 2 4 z 9
x 1 12 x 1
1 8 z 9 z 2
16. A 2 x y z 6 B x 5 y z 10 C 3 x y 3 z 12 Eliminate z from equations A and B. A 2x y z 6 B x 5 y z 10 x 4y 16 D
561
1, 4, 2
Chapter 5 Systems of Equations and Inequalities Eliminate z from equations A and C. Multiply by 3 A 2 x y z 6 6 x 3 y 3 z 18 C 3x y 3 z 12 3x y 3 z 12 9x 2 y 30 E
Solve the systems of equations D and E. x 4 y 16 x 4 y 16 D Multiply by 2 E 9 x 2 y 30 18 x 4 y 60 19 x 76 x 4 Back substitute. A 2x y z 6 D x 4 y 16
4 4 y 16
2 4 3 z 6
4 y 12
83 z 6
y 3
z 1
17. A 4 x 3 y 2 z 5
B 2 x y y z 6
4, 3, 1 4 x 3 y 2 z 5
2x 2 y y z 6
2 x y z 6
C 6 x y z x 5 y 8 6x 6 y z x 5 y 8
5 x y z 8
Eliminate y and z from equations B and C. B 2 x y z 6 C 5 x y z 8 7x 14 x 2 Eliminate x from equations A and B. A
4 x 3 y 2 z 5
2 x y z 6
B
4 2 3 y 2 z 5
2 2 y z 6
8 3 y 2 z 5
4 y z 6
3 y 2 z 3 D y z 2 E Solve the systems of equations D and E. 3 y 2 z 3 D 3 y 2 z 3 Multiply by 2 E y z 2 2 y 2 z 4 y 1 y 1 Back substitute. E y z 2
1 z 2 z 3
2, 1, 3
562
Section 5.2 18. A 3x 5 y z 13
B x y z x 3
x y z x 3
3x 5 y z 13
2 x y z 3
C 5 x y 3 y 3z 4 5 x 5 y 3 y 3 z 4 5 x 2 y 3z 4 Eliminate z from equations A and B. A 3x 5 y z 13 B 2 x y z 3 10 D x 4y Eliminate z from equations B and C. Multiply by 3 B 2 x y z 3 6 x 3 y 3 z 9 C 5 x 2 y 3 z 4 5 x 2 y 3 z 4 x 5y 13 E
Solve the systems of equations D and E. x 4 y 10 D E x 5 y 13 y 3 Back substitute. B 2 x y z 3 D x 4 y 10 x 4 3 10
2 2 3 z 3
x 12 10 x 2
4 3 z 3 z4
2, 3, 4
5z 2 19. A 2 x B 3y 7z 9 C 5 x 9 y 22 Eliminate y from equations B and C. Multiply by 3 B 3 y 7 z 9 9 y 21z 27 C 5 x 9 y 22 5 x 9 y 22 5 x 21z 5 D
Solve the systems of equations A and D. Multiply by 5 10 x 25 z 10 A 2 x 5 z 2 Multiply by 2 D 5 x 21z 5 10 x 42 z 10 67 z 0 z 0
Back substitute. A 2x 5z 2
B
3y 7z 9
2x 5 0 2
3 y 7 0 9
2x 2
3y 9
x 1
y3
1, 3, 0 563
Chapter 5 Systems of Equations and Inequalities 20. A 3x 2 y 8 B 5 y 6z 2 C 7x 11z 33 Eliminate y from equations A and B.
A 3x 2 y 8 15 x 10 y 40 Multiply by 2 B 5 y 6 z 2 10 y 12 z 4 15 x 12 z 36 D Multiply by 5
Solve the systems of equations C and D. Multiply by 15 105 x 165 z 495 C 7 x 11z 33 Multiply by 7 D 15 x 12 z 36 105 x 84 z 252 81z 243 z 3
Back substitute. B 5 y 6z 2
A
3 x 2 y 8
5 y 6 3 2
3x 2 4 8
5 y 18 2 5 y 20 y4
3x 8 8 3x 0 x0
0, 4, 3
21. A 4 x 3 y 0 B 3 y z 1 z 12 C 4x Eliminate x from equations A and C. A 4 x 3 y 0 C 4x z 12 3 y z 12 D
Eliminate y and z from equations B and D. B 3 y z 1 D 3 y z 12 0 11 The system of equations reduces to a contradiction. There is no solution. 22. A 4 x y 2 z 1 B 3x 5 y z 2 C 9 x 15 y 3z 0 Eliminate z from equations A and B.
A 4x y 2z 1 4x y 2z 1 Multiply by 2 B 3 x 5 y z 2 6 x 10 y 2 z 4 10 x 9 y 3 D
564
Section 5.2 Eliminate z from equations B and C. Multiply by 3 B 3x 5 y z 2 9 x 15 y 3z 6 C 9 x 15 y 3z 0 9 x 15 y 3z 0 0 6
The system of equations reduces to a contradiction. There is no solution. 23. A 2 x 3 y 6 z 1
2 x 3 y 6 z 1
B 6 y 12 z 10 x 9 10 x 6 y 12 z 9 C 3z 6 y 3x 1 3 x 6 y 3 z 1 Eliminate y from equations B and C. B 10 x 6 y 12 z 9 C 3 x 6 y 3 z 1 13 x 9z 8 D Eliminate y and z from equations A and B. Multiply by 2 A 2 x 3 y 6 z 1 4 x 6 y 12 z 2 B 10 x 6 y 12 z 9 10 x 6 y 12 z 9 7 14 x 0 0 1 0 x 0 0 2 0 A 2 x 3 y 6 z 1 B 3 x 6 y 3 z 1
1 2 3 y 6 z 1 2
1 3 6 y 3 z 1 2
1 3 y 6 z 1
3 12 y 6 z 2
3 y 6 z 2 D Solve the systems of equations D and E.
12 y 6 z 5 E
Multiply by 1 D 3 y 6 z 2 3y 6z 2 12 y 6 z 5 E 12 y 6 z 5 9 y 3 0 0 1 0 y 0 0 3 0 Back substitute. A 2 x 3 y 6 z 1
1 1 2 3 6 z 1 2 3 1 1 6 z 1 6 z 1 z
1 6
1 1 1 , , 2 3 6
565
Chapter 5 Systems of Equations and Inequalities 24. A 5 x 2 y 3z 3
5 x 2 y 3 z 3
B 4 y 1 10 x 5 z 10 x 4 y 5 z 1 C 2z 5 x 6 y 5 x 6 y 2 z 0 Eliminate x from equations A and C. A 5 x 2 y 3 z 3 C 5 x 6 y 2 z 0 4 y 5 z 3 D Eliminate x from equations B and C. B 10 x 4 y 5 z 1 10 x 4 y 5 z 1 Multiply by 2 C 5 x 6 y 2 z 0 10 x 12 y 4 z 0 16 y 9 z 1 E Solve the systems of equations D and E. Multiply by 4 D 4 y 5 z 3 16 y 20 z 12 E 16 y 9 z 1 16 y 9 z 1 11z 11 z 1 Back substitute. D 4 y 5 z 3 A 5 x 2 y 3z 3 4 y 5 1 3 4 y 5 3
1 5 x 2 3 1 3 2
4y 2
5 x 1 3 3
y
1 2
5x 1 x
1 5
1 1 , , 1 5 2
25. A x 2 y 4 z 3 B y 3z 5 C x 2 z 7 Eliminate x from equations A and C. Multiply by 1 A x 2 y 4 z 3 x 2 y 4 z 3 C x x 2 z 7 2 z 7 2 y 6 z 10 D
Pair up equations B and D to solve for y and z. Multiply by 2 y 3z 5 2 y 6 z 10 B D 2 y 6 z 10 2 y 6 z 10 0 0
The system reduces to the identity 0 0 . It has infinitely many solutions.
566
Section 5.2 26. A 3 x 2 y 5 z 6 B 3y z 4 C 3 x 17 y 26 Eliminate x from equations A and C. Multiply by 1 A 3x 2 y 5 z 6 3 x 2 y 5 z 6 C 3x 17 y 3 x 17 y 26 26 15 y 5 z 20 D
Pair up equations B and D to solve for y and z. Multiply by 5 15 y 5 z 20 B 3 y z 4 D 15 y 5 z 20 15 y 5 z 20 0 0
The system reduces to the identity 0 0 . It has infinitely many solutions. 27. A 0.2 x 0.1y 0.6 z
2x y 6z 0
B 0.004 x 0.005 y 0.001z 0 4 x 5 y z 0 C 30 x 50 z 20 y 3x 2 y 5 z 0 Eliminate y from equations A and C. Multiply by 2 A 2 x y 6 z 0 4 x 2 y 12 z 0 C 3x 2 y 5 z 0 3x 2 y 5 z 0 7x 7z 0 D
Eliminate y from equations A and B. Multiply by 5 A 2 x y 6 z 0 10 x 5 y 30 z 0 B 4x 5 y z 0 4x 5 y z 0 30 z 0 14 x 7 x 15 z 0 E
Solve the systems of equations D and E. Multiply by 1 D 7 x 7 z 0 7 x 7 z 0 E 7 x 15 z 0 7 x 15 z 0 8z 0 z 0 Back substitute. A 2x y 6z 0 D 7x 7z 0 7 x 7 0 0
2 0 y 6 0 0
7x 0 0
0 y00
7x 0
y 0
x0
y0
567
0, 0, 0
Chapter 5 Systems of Equations and Inequalities 28. A 0.3x 0.5 y 1.2 z
3x 5 y 12 z 0
B 0.05 x 0.1 y 0.04 z 5 x 10 y 4 z 0 C 100 x 300 y 700 z x 3y 7z 0 Eliminate x from equations A and C. A 3x 5 y 12 z 0 3x 5 y 12 z 0 Multiply by 3 C x 3 y 7 z 0 3x 9 y 21z 0 4 y 9z 0 D Eliminate y from equations B and C. B 5 x 10 y 4 z 0 5 x 10 y 4 z 0 Multiply by 5 C x 3 y 7 z 0 5 x 15 y 35 z 0 25 y 39 z 0 E Solve the systems of equations D and E. Multiply by 25 100 y 225 z 0 D 4 y 9 z 0 Multiply by 4 E 25 y 39 z 0 100 x 156 z 0 69 z 0 z 0
Back substitute. D 4 y 9z 0
C
x 3y 7z 0
4 y 9 0 0
x 3 0 7 0 0
4y 0 0
x00 0
y0
x0
1 1 1 7 x y z x 3y 4z 7 12 4 3 12 1 1 1 17 B x y z x 5 y 2 z 17 10 2 5 10 1 1 C x y z 3 2 x y 4 z 12 2 4 Eliminate x from equations A and B. A x 3y 4z 7 B x 5 y 2 z 17 8 y 2 z 10 D
29. A
Eliminate x and z from equations B and C. Multiply by 2 2 x 10 y 4 z 34 B x 5 y 2 z 17 C 2 x y 4 z 12 2 x y 4 z 12 22 11 y y 2
568
0, 0, 0
Section 5.2 Back substitute. D 8y 2 z 10
x 3y 4z 7
A
8 2 2 z 10
x 3 2 4 3 7
16 2 z 10
x 6 12 7
1, 2, 3
x 1
2z 6 z3 7 1 y z4 2x 7 y z 8 2 2 3 1 B x y z 1 3x 4 y 2 z 4 4 2 1 2 3 C x y z 1 x 4 y 3 z 10 10 5 10 Eliminate x from equations A and C.
30. A x
A 2x 7 y z 8 2x 7 y z 8 Multiply by 2 C x 4 y 3z 10 2 x 8 y 6 z 20 15 y 7 z 12 D Eliminate x from equations B and C. B 3x 4 y 2 z 4 3x 4 y 2 z 4 Multiply by 3 C x 4 y 3z 10 3x 12 y 9 z 30 16 y 11z 34 E Solve the systems of equations D and E. Multiply by 11 165 y 77 z 132 D 15 y 7 z 12 Multiply by 7 E 16 y 11z 34 112 y 77 z 238 53 y 106 y 2
Back substitute. D 15y 7 z 12
C
x 4 y 3 z 10
15 2 7 z 12
x 4 2 3 6 10
30 7 z 12 7 z 42
x 8 18 10 x0
0, 2, 6
z 6 31. A 3x 2 y 5 z 12 B 3 y 8 z 8 C 10 z 20
Solve C for z. 10 z 20 z2
6, 8, 2 569
Solve B for y. 3 y 8 z 8
Solve A for x. 3x 2 y 5 z 12
3 y 8 2 8
3 x 2 8 5 2 12
y 16 8 3 y 24 y 8
3 x 16 10 12 3x 18 x6
Chapter 5 Systems of Equations and Inequalities 32. A 4 x 6 y z 4 B 2 y 4 z 24 C 6 z 48
Solve C for z. 6 z 48 z8
3, 4, 8
Solve B for y. 2 y 4 z 24
Solve A for x. 4 x 6 y z 4
2 y 4 8 24
4 x 6 4 8 4
2 y 32 24 2 y 8 y 4
4 x 24 8 4 4 x 12 x 3
x 2 y 4 z 1 8 2 x 4 3 y 12 z 1 48 2 x 3 y z 55 3 2 6 x 2 z 1 B 8 2 x 4 3 z 3 48 2 x 3z 49 3 2 y 4 z 1 1 C 3 y 12 2 z 2 12 3y 2z 2 4 6 Eliminate x from equations A and B. A 2 x 3 y z 55 B 2 x 3 z 49 3 y 4 z 104 D
33. A
Solve the systems of equations C and D. Multiply by 1 C 3 y 2 z 2 3 y 2 z 2 3 y 4 z 104 D 3 y 4 z 104 6 z 102 z 17 Back substitute. 2 x 3z 49 D 3y 4 z 104 B 3y 4 17 104
2 x 3 17 49
3y 68 104 3 y 36 y 12
2 x 51 49 2 x 2 x 1
1,12,17
x 1 y 2 z 2 13 12 x 12 28 y 56 21z 42 1092 12 x 28 y 21z 1118 7 3 4 y2 z2 B 8 y 16 9 z 18 216 8 y 9 z 250 3 9 8 x 1 z 2 C 3 2 x 2 7 z 14 42 2 x 7 z 30 7 2 Eliminate x from equations A and C.
34. A
A 12 x 28 y 21z 1118 12 x 28 y 21z 1118 Multiply by 6 C 2x 7 z 30 12 x 42 z 180 28 y 21z 938 D
570
Section 5.2 Solve the systems of equations B and D. Multiply by 7 56 y 63z 1750 B 8 y 9 z 250 Multiply by 2 D 28 y 21z 938 56 y 42 z 1876 21z 126 6 z
Back substitute. B 8y 9 z 250
C
2 x 7 z 30
8y 9 6 250
2 x 7 6 30
8y 54 250 8 y 304 y 38
2 x 42 30 2 x 12 x 6
35. A 3 x y 6 4 z y
B 4 6 y 5z
6, 38, 6
3x 3 y 6 4 z y 3x 2 y 4 z 6 6 y 5z 4
C 3x 4 y z 0 3x 4 y z 0 Eliminate x from equations A and C. A 3x 2 y 4 z 6 C 3x 4 y z 0 6 y 5z 6 D
6 y 5z 4
3x 4 y z 0
Solve the systems of equations B and D. Multiply by 1 6 y 5 z 4 B 6 y 5 z 4 D 6 y 5z 6 6 y 5z 6 0 2 The system of equations reduces to a contradiction. There is no solution.
36. A 4 x y 8 z y
B 3 3x 4 z
4x 4 y 8 z y 3x 4 z 3
C x 3 y 3z 1 x 3 y 3z 1 Eliminate y from equations A and C. A 4x 3y z 8 C x 3 y 3z 1 4z 9 D 3x
4x 3y z 8 3x 4 z 3
x 3 y 3z 1
Solve the systems of equations B and D. Multiply by 1 3x 4 z 3 B 3x 4 z 3 D 3y 4z 9 3y 4z 9 0 6 The system of equations reduces to a contradiction. There is no solution.
571
Chapter 5 Systems of Equations and Inequalities 37. Let x represent the amount invested in the large cap stock fund. Let y represent the amount invested in the real estate fund. Let z represent the amount invested in the bond fund. A x y z 8000 x y z 8000
B x 2y
x 2y 0
C 0.062 x 0.135 y 0.044 z 66 62 x 135 y 44 z 66,000 Eliminate x from equations A and B and from equations B and C. A x y z 8000 62 B 62 x 124 y 0 1 B x 2 y 0 C 62 x 135 y 44 z 66,000 3 y z 8000 D 11y 44 z 66,000 y 4 z 6000 E Solve the systems of equations D and E. D 3y z 8000 3 E 3 y 12 z 18,000 13z 26,000 z 2000 Back substitute. D 3 y z 8000 B x 2y 0 3 y 2000 8000 x 2 2000 0 3 y 6000 x 4000 0 y 2000 x 4000 He invested $4000 in the large cap fund, $2000 in the real estate fund, and $2000 in the bond fund. 38. Let x represent the amount put in the money market account. Let y represent the amount invested in the stock. Let z represent the amount invested in the mutual fund. A x y z 120,000 x y z 120,000
B y z 10,000
y z 10,000
C 0.022 x 0.06 y 0.02 z 2820 22 x 60 y 20 z 2,820,000 Eliminate z from equations A and B and from equations B and C. 20 B 20 y 20 z 200,000 A x y z 120,000 B y z 10,000 C 22 x 60 y 20 z 2,820,000 2,620,000 E x 2y 130,000 D 22 x 40 y Solve the systems of equations D and E. 20 D 20 x 40 y 2,600,000 E 22 x 40 y 2,620,000 2x 20,000 x 10,000
572
Section 5.2 Back substitute. y z 10,000 B D x 2 y 130,000 60,000 z 10,000 10,000 2 y 130,000 z 50, 000 2 y 120,000 z 50,000 y 60,000 $10,000 was put in the money market account. $60,000 was invested in the stock, and $50,000 was invested in the mutual fund.
39. Let x represent the number of free-throws. Let y represent the number of 2-point shots. Let z represent the number of 3-point shots. A x 2 y 3z 26 x 2 y 3z 26
B y z4
40. Let x represent the rate of saw A. Let y represent the rate of saw B. Let z represent the rate of saw C. A 6x 6 y 6 z 7200 x y z 1200
yz4
B 9.6 x 9.6 y 7200
C x yz x yz 0 Eliminate y from equations A and B and from equations B and C. A x 2 y 3z 26 2 B 2 y 2 z 8 x 5 z 18 D B C
x y 750
C 9 y 9 z 7200 y z 800 Eliminate x and y from equations A and B. A x y z 1200 1 B x y 750 z 450 Back substitute. C 9 y 9 z 7200
y z4 x y z0 2z 4 E x
9 y 9 450 7200 9 y 4050 7200 9 y 3150 y 350
Solve the systems of equations D and E. 1 D x 5 z 18 x 2z 4 E 7 z 14 z 2 Back substitute. D x 5 z 18
x y 750 x 350 750 x 400 Saw A cuts 400 ft/hr, saw B cuts 350 ft/hr, and saw C cuts 450 ft/hr. B
x 5 2 18 x 10 18 x8 B yz4 y2 4 y6 He made eight free-throws, six 2-point shots, and two 3-point shots.
573
Chapter 5 Systems of Equations and Inequalities 41. Let x represent the percentage of nitrogen. Let y represent the percentage of phosphorous. Let z represent the percentage of potassium. A z 2y 2y z 0
B x yz
Eliminate z from equations A and D. A 2y z 0 D y z 21 3y 21 y 7 Back substitute. A 2y z 0
x yz0
C x y z 42 x y z 42 Eliminate y and z from equations B and C. B x y z 0 C x y z 42 2x 42 x 21 Back substitute. C x y z 42 21 y z 42
2 7 z 0 14 z 0 z 14 The proper N-P-K label is 21-7-14, which is choice b.
y z 21 D
42. Let x represent the number of tickets sold for seats in section A. Let y represent the number of tickets sold for seats in section B Let z represent the number of tickets sold for seats in section C. A x y z 4000 x y z 4000
B 50x 30 y 20 z 120,000 5x 3 y 2 z 12,000 C y x z 1000 x y z 1000 Eliminate x and z from equations A and C. x y z 4000 A C x y z 1000 5000 2y y 2500 Back substitute. B 5 x 3 y 2 z 12,000 A x y z 4000 5 x 3 2500 2 z 12,000
x 2500 z 4000
5 x 7500 2 z 12,000
x z 1500 D Eliminate z from equations D and E. 2 D 2 x 2 z 3000 E 5 x 2 z 4500 1500 3x x 500
5 x 2 z 4500 E Back substitute. A
x y z 4000 500 2500 z 4000 3000 z 4000
z 1000 500 tickets in section A, 2500 tickets in section B, and 1000 tickets in section C were sold.
574
Section 5.2 43. Let x represent the length of the shortest side. Let y represent the length of the middle side. Let z represent the length of the longest side. A x y z 55 x y z 55
B x z7
Eliminate z from equations C and D. C w l 12 1 D 4 w l 52 5w 40 w 8 Back substitute. C w l 12 A wh 2
x z 7
C y x z 19 x y z 19 Eliminate x and z from equations A and C. A x y z 55 C x y z 19 36 2y y 18 Back substitute. A x y z 55
8h 2 h 6
45. Let x represent the measure of the largest angle. Let y represent the measure of the middle angle. Let z represent the measure of the smallest angle. A x y z 180 x y z 180
x z 37 D Eliminate z from equations B and D. B x z 7 D x z 37 30 2x x 15 Back substitute. A x y z 55
B x y z 100 x y z 100 2 y 2 y 3z 0 3 Eliminate x from equations A and B. x y z 180 A 1 B x y z 100 2 y 2 z 80 y z 40 D Eliminate z from equations C and D. C 2 y 3z 0 3 D 3 y 3 z 120 120 5y y 24 Back substitute. C 2 y 3z 0 A x y z 180 C z
15 18 z 55 33 z 55 z 22 The sides are 15 in., 18 in., and 22 in. 44. Let w represent the width of the package. Let h represent the height of the package. Let l represent the length of the package. A 2 w 2h l 48 2 w 2h l 48
l 20
h6 The dimensions are w = 8 in., h = 6 in., and l = 20 in.
x 18 z 55
B w h2
8 l 12
wh 2
2 24 3 z 0
C l w 12 w l 12 Eliminate h from equations A and B. A 2 w 2h l 48 2 B 2 w 2h 4 4w l 52 D
48 3 z 0
x 24 16 180 x 40 180
3 z 48
x 140
z 16 The angles are 16 , 24 , and 140 . 46. Let x represent the measure of the largest angle. Let y represent the measure of the middle angle. Let z represent the measure of the smallest angle.
575
Chapter 5 Systems of Equations and Inequalities A x y z 180 x y z 180 B x y z 18
Eliminate z from equations C and D. C y 2z 0 2 D 2 y 2 z 162 162 3y y 54 Back substitute. C y 2z 0 A x y z 180
x y z 18
1 y y 2z 0 2 Eliminate x from equations A and B. x y z 180 A 1 B x y z 18 2 y 2 z 162 y z 81 D C z
10 0 10 5 3 1 2 The slopes are 5 and 1.
47. a. m
m
54 2 z 0 2 z 54
z 27 x 99 The angles are 27 , 54 , and 99 .
15 10 5 1 2 3 5
y ax 2 bx c
b.
Substitute 1, 0 : 0 a 1 b 1 c 2 Substitute 3,10 : 10 a 3 b 3 c 2 Substitute 2,15 : 15 a 2 b 2 c 2
A a bc 0 B 9a 3b c 10 C 4a 2b c 15
Eliminate b from equations A and B and from equations A and C. 3 A 3a 3b 3c 0 2 A 2a 2b 2c 0 B 9a 3b c 10 C 4a 2b c 15 6a 2c 10 D 6a 3c 15 E Solve the system of equations D and E. D 6a 2c 10 1 E 6a 3c 15 5c 5 c 1
D
6a 2c 10 6a 2 1 10 6a 2 10 6a 12 a2
Back substitute. A
abc 0 2 b 1 0 b 3
x 54 27 180 x 81 180
y 2 x 2 3x 1
c.
576
Section 5.2 6 9 15 5 1 2 3 The slopes are 5 and 2.
m
48. a. m
3 9 12 2 4 2 6
y ax 2 bx c
b.
Substitute 2, 9 : 9 a 2 b 2 c 2 Substitute 1, 6 : 6 a 1 b 1 c 2 Substitute 4, 3 : 3 a 4 b 4 c 2
A 4a 2b c 9 B a b c 6 C 16a 4b c 3
Eliminate b from equations A and B and from equations B and C. A 4a 2b c 9 4 B 4a 4b 4c 24 2 B 2a 2b 2c 12 C 16a 4b c 3 6a 3c 3 D 12a 3c 21 E Solve the system of equations D and E. D 6a 3c 3 D 6a 3c 3 E 12a 3c 21 6 1 3c 3 18 18a 6 3c 3 a 1 3c 9
Back substitute. B
a b c 6 1 b 3 6 1 b 3 6 b4
c 3 y x 4x 3 2
c.
y ax 2 bx c
49.
Substitute 0, 6 : 6 a 0 b 0 c 2 Substitute 2, 6 : 6 a 2 b 2 c 2 Substitute 1, 9 : 9 a 1 b 1 c 2
A c 6 B 4a 2b c 6 C a bc 9
Eliminate c from equations B and C. B 4a 2b c 6 C a bc 9 4a 2b 6 6
ab6 9
4a 2b 12
a b 3 E
2a b 6 D Solve the system of equations D and E. D 2a b 6 E a b 3 3 3a a 1
E
a b 3 1 b 3 b 4 b 4
y x 4x 6 2
577
Chapter 5 Systems of Equations and Inequalities y ax 2 bx c
50.
4 a 0 b 0 c Substitute 0, 4 : 2 6 a 2 b 2 c Substitute 2, 6 : 2 Substitute 3, 31 : 31 a 3 b 3 c
A c 4 B 4a 2b c 6 C 9a 3b c 31
2
Eliminate c from equations B and C. B 4a 2b c 6 C 9a 3b c 31 4a 2b 4 6
9a 3b 4 31
4a 2b 2
9a 3b 27
2a b 1 D Solve the system of equations D and E. D D 2a b 1
3a b 9 E 2a b 1
2 2 b 1
E 3a b 9 10 5a a 2
4 b 1 b3
y 2 x 2 3 x 4 y ax 2 bx c
51. a.
Substitute 0, 6 : 6 a 0 b 0 c 2 Substitute 4, 7 : 7 a 4 b 4 c 2 Substitute 9, 9 : 9 a 9 b 9 c
A c 6 B 16a 4b c 7 C 81a 9b c 9
Eliminate c from equations B and C. B 16a 4b c 7
C 81a 9b c 9
2
81a 9b 6 9
16a 4b 6 7
81a 9b 3
16a 4b 1 D
27 a 3b 1 E Solve the system of equations D and E.
3D 48a 12b 3 4 E 108a 12b 4 60a 1 0 0 1 a 0 0 60
y
b. y
D
16a 4b 1 1 16 4b 1 60 4 4b 1 15 4 60b 15 11 b 60
1 2 11 x x6 60 60
1 2 11 1 11 144 132 2 x x 6 12 12 6 6 10.6% 60 60 60 60 60 60
578
Section 5.2 y ax 2 bx c
52. a.
Substitute 0, 3 : 3 a 0 b 0 c 2 Substitute 3, 4 : 4 a 3 b 3 c 2 Substitute 10, 8 : 8 a 10 b 10 c 2
A c 3 B 9a 3b c 4 C 100a 10b c 8
Eliminate c from equations B and C. C 100a 10b c 8 B 9a 3b c 4 100a 10b 3 8 9a 3b 3 4 100a 10b 5 9a 3b 1 D 20a 2b 1 E Solve the system of equations D and E. 2 D 18a 6b 2 3 E 60a 6b 3 42a 1 0 0 1 a 0 0 42
y b. y
53.
D
9a 3b 1 1 9 3b 1 42 3 3b 1 14 3 42b 14 11 b 42
1 2 11 x x3 42 42 1 2 11 1 11 225 165 2 x x 3 15 15 3 3 12 billion metric tons 42 42 42 42 42 42
1 2 at v0t s0 2 1 1 2 a v0 s0 30 Substitute 1, 30 : 30 a 1 v0 1 s0 A 2 2 1 2 Substitute 2, 54 : 54 a 2 v0 2 s0 B 2a 2v0 s0 54 2 1 2 9 Substitute 3, 82 : 82 a 3 v0 3 s0 C a 3v0 s0 82 2 2 Eliminate v0 from equations A and B and from equations A and C. 3 2 A a 2v0 2s0 60 3 A a 3v0 3s0 90 2 9 B 2a 2v0 s0 54 C a 3v0 s0 82 2 a s0 6 D 3a 2s0 8 s t
Solve the system of equations D and E. 2 D 2a 2s0 12 E 3a 2s0 8 a 4
D a s0 6 4 s0 6 s0 10
579
D
Chapter 5 Systems of Equations and Inequalities Back substitute. 1 A a v0 s0 30 2 1 4 v0 10 30 2 2 v0 10 30 v0 18 54.
a 4, v0 18, s0 10
1 2 at v0t s0 2 1 1 2 Substitute 1, 7 : 7 a 1 v0 1 s0 A a v0 s0 7 2 2 1 2 Substitute 2, 12 : 12 a 2 v0 2 s0 B 2a 2v0 s0 12 2 1 2 9 Substitute 3, 37 : 37 a 3 v0 3 s0 C a 3v0 s0 37 2 2 Eliminate v0 from equations A and B and from equations A and C. 3 2 A a 2v0 2s0 14 3 A a 3v0 3s0 21 2 9 B 2a 2v0 s0 12 C a 3v0 s0 37 2 a s0 26 D 3a 2s0 58 s t
D
Solve the system of equations D and E. 2 D 2a 2s0 52 E 3a 2s0 58 a 6 Back substitute. 1 a v0 s0 7 A 2 1 6 v0 20 7 2 3 v0 20 7 v0 10 55. a.
D a s0 26 6 s0 26 s0 20
a 6, v0 10, s0 20
y ax1 bx2 c Substitute 28, 0.5, 225 : 225 a 28 b 0.5 c Substitute 25, 0.8, 207 : 207 a 25 b 0.8 c Substitute 18, 0.4,154 : 154 a 18 b 0.4 c
A 28a 0.5b c 225 B 25a 0.8b c 207 C 18a 0.4b c 154
Eliminate c from equations A and B and from equations A and C. 1 A 28a 0.5b c 225 1 A 28a 0.5b c 225 B 25a 0.8b c 207 B 18a 0.4b c 154 3a 0.3b 18 10a 0.1b 71 E a 0.1b 6 D
580
Section 5.2 Solve the system of equations D and E. D a 0.1b 6
1 D a 0.1b 6 E 10a 0.1b 71 11a 77 a 7
7 0.1b 6 0.1b 1 b 10
Back substitute. A 28a 0.5b c 225 28 7 0.5 10 c 225 196 5 c 225
y 7 x1 10 x2 24
c 24 b. y 7 x1 10 x2 24 y 7 20 10 0.4 24 140 4 24 168 $168,000 56. a.
y ax1 bx2 c Substitute 3500, 6, 20 : 20 a 3500 b 6 c Substitute 3200, 4, 26 : 26 a 3200 b 4 c Substitute 4100, 8,18 : 18 a 4100 b 8 c
A 3500a 6b c 20 B 3200a 4b c 26 C 4100a 8b c 18
Eliminate c from equations A and B and from equations A and C. 1 A 3500a 1 A 3500a 6b c 20 6b c 20 B 3200a 4b c 26 B 4100a 8b c 18 300a 2b 6 600a 2b 2 150a b 3 D 300a b 1 D Solve the system of equations D and E. D 150a b 3 E 300a b 1 150a 2 1 a 75
D 150a b 3 1 150 b 3 75 2 b 3 b 5
Back substitute. A 3500a 6b c 20 1 3500 6 5 c 20 75 140 30 c 20 3 140 90 3c 60 3c 10 10 c 3
y
1 10 x1 5 x2 75 3
581
Chapter 5 Systems of Equations and Inequalities 1 10 x1 5 x2 75 3 1 10 152 10 y 3800 5 6 30 24 mpg 75 3 3 3
b. y
57. The set of all ordered pairs that are solutions to a linear equation in three variables forms a plane in space.
Solve the system of equations F and G. 2 F 46c 26d 104 13 G 39c 26d 104 7c 0 0 c Back substitute. G 3c 2d 8
58. The planes do not all intersect in a common point in space. 59. Pair up two equations in the system and eliminate a variable. Choose a different pair of two equations from the system and eliminate the same variable. The result should be a system of two linear equations in two variables. Solve this system using either the substitution or addition method. Then back substitute to find the third variable.
3 0 2d 8
E
2d 8 d4 b 7c 5d 21 b 7 0 5 4 21 b 20 21 b 1
60. Substitute the ordered triple into each individual equation in the system and verify that it is a solution to each equation.
A 2a b c d 7 2a 1 0 4 7 2a 4 a2 2, 1, 0, 4
61. A 2a b c d 7 B 3b 2c 2d 11 C a 3c 3d 14 D 4a 2b 5c 6 Eliminate a from equations A and C. A 2a b c d 7 2 C 2a 6c 6d 28 b 7c 5d 21 E
62. A 3a 4b 2c d 8 B 2a 3b 2d 7 C 5b 3c 4d 4 7 D a b 2c Eliminate a from equations B and D. B 2a 3b 2d 7 2 D 2a 2b 4c 14 5b 4c 2d 7 E
Eliminate b from equations B and E. B 3b 2c 2d 11 3 D 3b 21c 15d 63 23c 13d 52 F Eliminate a and b from equations A and D. 2 A 4a 2b 2c 2d 14 6 D 4a 2b 5c 3c 2d 8 G
Eliminate b from equations C and E. C 5b 3c 4d 4 1 E 5b 4c 2d 7 c 2d 3 F
582
Section 5.2 Eliminate a from equations A and D. A 3a 4b 2c d 8 21 3 D 3a 3b 6c b 4c d 13 G
Eliminate y from equations A and C. A 6 x 8 y 3 z 24 4 C 4 x 8 y 8 z 96 11z 72 E 10 x
Eliminate b from equations C and G. C 5b 3c 4d 4 5 G 5b 20c 5d 65 23c 9d 69 H
Solve the systems of equations D and E. 10 D 10 x 10 z 80 E 10 x 11z 72 z 8 Back substitute. D x z 8 x 8 8 x 16 B 2x 2 y z 0
Solve the system of equations F and H. 23 F 23c 46d 69 H 23c 9d 69 55d 0 0 d Back substitute. F c 2d 3
2 16 2 y 8 0 32 2 y 8 0 2 y 24 y 12 x u 3 y v 1 16 u 3 12 v 1 13 u 11 v 13,11,10
c 2 0 3 C
c3 5b 3c 4d 4 5b 3 3 4 0 4
5b 9 4 5b 5 b 1 A 3a 4b 2c d 8 3a 4 1 2 3 0 8 3a 4 6 8 3a 6 a2 2,1, 3, 0
z w2 8 w2 10 w
64. Let x u 1, y v 1, and z w 3 . x y z A 11 2 x 2 y 3z 132 6 6 4 x y z B 7 4 x 6 y 3z 84 3 2 4 x y z C 20 3x y 3z 120 2 6 2 Eliminate z from equations A and B. A 2 x 2 y 3z 132 1 B 4 x 6 y 3z 84 2 x 8 y 48 x 4 y 24 D Eliminate z from equations A and C. A 2 x 2 y 3z 132 1 C 3x y 3z 120 x 3y 12 E
63. Let x u 3, y v 1, and z w 2 . x y z A 1 6 x 8 y 3z 24 4 3 8 x y z B 0 2x 2 y z 0 2 2 4 x y z C 6 x 2 y 2 z 24 4 2 2 Eliminate y from equations B and C. B 2x 2 y z 0 C x 2 y 2 z 24 3x 3 z 24 x z 8 D
583
Chapter 5 Systems of Equations and Inequalities Solve the systems of equations D and E. D x 4 y 24 1 E x 3 y 12 y 12 Back substitute. D x 4 y 24
2 x 2 y 3z 132
A
2 24 2 12 3z 132
x 4 12 24 x 48 24 x 24 x 24
48 24 3z 132 3z 60 z 20 x u 1 y v 1 24 u 1 12 v 1
z w3 20 w 3
23 u
17 w
23,13,17
x 2 y 2 Ax By C 0
65. a.
Substitute 2, 2 :
2 2 2 2 A 2 B 2 C 0 Substitute 6, 0 : 6 2 0 2 A 6 B 0 C 0 2 2 Substitute 7, 3 : 7 3 A 7 B 3 C 0
A 2 A 2 B C 8 B 6A
Solve the system of equations B and D. B 6 A C 36 1 D 4 A C 28 2A 8 A 4
4 A C 28
D
4 4 C 28 16 C 28 C 12
Back substitute. A 2 A 2 B C 8 2 4 2 B 12 8 8 2 B 12 8 2 B 12 B6
x 2 y 2 4 x 6 y 12 0
x 2 y 2 4 x 6 y 12 0 y2 6 y 12 x2 4 x 2 2 x 4 x 4 y 6 y 9 12 4 9
x 22 y 3 2 25 x 22 y 3 2 52
Center: 2, 3 ; Radius: 5 584
C 36
C 7 A 3B C 58
Eliminate B from equations A and C. 3 A 6 A 6 B 3C 24 2 C 14 A 6 B 2C 116 20 A 5C 140 4 A C 28 D
b.
13 v
Section 5.2 x 2 y 2 Ax By C 0
66. a.
Substitute 1,12 : 1 12 A 1 B 12 C 0 2
Substitute 5,10 : Substitute 9, 2 :
2
5 10 A 5 B 10 C 0 9 2 2 2 A 9 B 2 C 0 2
2
A A 12 B C 145 B 5 A 10 B C 125 C 9 A 2 B C 85
Eliminate A from equations A and B. 5 A 5 A 60 B 5C 725 B 5 A 10 B C 125 70 B 6C 850 D Eliminate A from equations A and C. 9 A 9 A 108 B 9C 1305 C 9 A 2 B C 85 110 B 10C 1390 11B C 139 E Solve the system of equations D and E. D 70 B 6C 850 6 E 66 B 6C 834 4B 16 4 B
11B C 139
E
11 4 C 139 44 C 139 C 95
Back substitute. A A 12 B C 145 A 12 4 95 145 A 48 95 145 A 2 A2 b.
x 2 y 2 2 x 4 y 95 0
x 2 y 2 2 x 4 y 95 0 x2 2 x y2 4 y 95 2 2 x 2 x 1 y 4 y 4 95 1 4
x 12 y 2 2 100 x 12 y 2 2 102 67.
Substitute 3,1, 2 :
Substitute 1,1, 0 :
Center: 1, 2 ; Radius: 10
c1 x c2 y c3 z 1
c1 3 c2 1 c3 2 1
c1 1 c2 1 c3 0 1
Substitute 1, 3, 2 : c1 1 c2 3 c3 2 1 Eliminate c3 from equations A and C. A 3c1 c2 2c3 1 C c1 3c2 2c3 1 4c1 2c2 2 2c1 c2 1 D 585
A 3c1 c2 2c3 1 B c1 c2 C
1
c1 3c2 2c3 1
Chapter 5 Systems of Equations and Inequalities Solve the system of equations B and D. B c1 c2 1
B c1 c2 1 D 2c1 c2 1 c1 2
2 c2 1 c2 3
Back substitute. A 3c1 c2 2c3 1
3 2 3 2c3 1 6 3 2c3 1 2c3 8 c3 4
2x 3y 4z 1 68.
Substitute 5, 3, 4 :
c1 x c2 y c3 z 1
c1 5 c2 3 c3 4 1
A 5c1 3c2 4c3 1
Substitute 1, 0,1 : c1 1 c2 0 c3 1 1
B c1
Substitute 1, 2, 5 : c1 1 c2 2 c3 5 1
c3 1
C c1 2c2 5c3 1
Eliminate c2 from equations A and C. 2 A 10c1 6c2 8c3 2 3 C 3c1 6c2 15c3 3 13c1 7c3 1 D Solve the system of equations B and D. 7 B 7c1 7c3 7 D 13c1 7c3 1 6c1 6 c1 1 Back substitute. A 5c1 3c2 4c3 1
B c1 c3 1 1 c3 1 c3 2
5 11 3c2 4 2 1 5 3c2 8 1
3c2 12 c2 4
x 4 y 2z 1
69. A 2 A B 2C 11
B 5 A 2 B 7C 26 C 3 A 8B 4C 5 Eliminate B from equations A and B and equations A and C. 2 A 4 A 2 B 4C 22 8 A 16 A 8 B 16C 88 B 5 A 2 B 7C 26 C 3 A 8 B 4C 5 9A 3C 48 19 A 12C 83 E 3 A C 16 D 586
Section 5.2 Solve the system of equations D and E. 12 D 36 A 12C 192 E 19 A 12C 83 55 A 275 5 A
3 A C 16
D
3 5 C 16 15 C 16 C 1
Back substitute. A 2 A B 2C 11 2 5 B 2 1 11 10 B 2 11 B3 70. A
B
A 3B 3C
A 5, B 3, C 1
3
A 10 B 11C 37
C 12 A 3B 4C 82 Eliminate A from equations A and B and equations A and C. 1 A A 3B 3C 3 12 A 12 A 36 B 36C 36 B A 10 B 11C 37 C 12 A 3B 4C 82 13B 8C 34 D 39 B 32C 46 E Solve the system of equations D and E. D 13B 8C 34 13B 8 34 13B 26 B 2
3 D 39 B 24C 102 E 39 B 32C 46 56C 56 1 C Back substitute. A A 3B 3C 3 A 3 2 3 1 3 A63 3 A6
A 6, B 2, C 1
Section 5.3 Partial Fraction Decomposition 1. a. 3x 2 y 6 2 y 3x 6 3 y x3 2 b.
4 y 6 x 12 3 y x3 2
3 c. x, y y x 3 2 2. a. 2 y 7 4 x 2 y 4 x 7 7 y 2 x 2
3 3 x3 x3 2 2 3 3 Identity Infinitely many solutions
587
6 x 3 y 10 3 y 6 x 10 10 y 2 x 3
Chapter 5 Systems of Equations and Inequalities A 3A 7C 18 3 C 3 A 12 B 3 12 B 7C 21 D
7 10 2 x 2 3 7 10 Contradiction 2 3 No solution
b. 2 x
Solve the system of equations B and D. 12 B 12 B 60C 180 D 12 B 7C 21 67C 201 C 3
c. 3. A 3 A 5 B C 19 3 A 5B C 19 B B 5C 2 A 1 2 A B 5C 1
B B 5C 15
C 7 A 3B C 13 7 A 3B C 13 Eliminate C from equations A and B and equations A and C. 5 A 15 A 25B 5C 95 B 2 A B 5C 1 17 A 24 B 96 D
B 5 3 15 B 15 15 B0 Back substitute. A 3 A 7C 18 3 A 7 3 18
1 A 3 A 5 B C 19 C 7 A 3B C 13 4 A 8B 32 E
3 A 21 18 3 A 3 A 1
Solve the system of equations D and E. D 17 A 24 B 96 3 E 12 A 24 B 96 29 A 0 A 0
3x 4 5. x 2 2 x 1 3x 3 2 x 2 x 5
3x3 6 x 2 3x
4x 4x 5 4 x 2 8 x 4
4 0 8 B 32
4x 1
8 B 32 B4 Back substitute. A 3 A 5 B C 19
3x 4
4x 1 x 2x 1 2
2x 5 6. x 6 x 9 2 x 17 x 54 x 68
3 0 5 4 C 19
2
3
2
0 20 C 19
2 x 3 12 x 2 18 x
C 1
4. A 3 A 7C 18 3 A B B 5C 15
2
E 4 A 8 B 32
0, 4,1
1, 0, 3
5 x 36 x 68 5 x 2 30 x 45 2
6 x 23
7C 18 B 5C 15
2x 5
1 C A 4 B 1 A 4 B Eliminate A from equations A and C.
6 x 23 x 6x 9 2
7. fraction decomposition 8. constant; A
588
9. linear; Ax B
Section 5.3 10. When the degree of the numerator is greater than or equal to the degree of the denominator, apply long division first. x 37 A B 11. x 4 2 x 3 x 4 2 x 3 12.
20 x 4 A B x 5 3x 1 x 5 3x 1
13.
8 x 10 8 x 10 A B 2 x 2 x x x 2 x x 2
14.
y 12 y 12 A B 2 y 3 y y y 3 y y 3
15.
6w 7 6w 7 A B w2 w 6 w 2 w 3 w 2 w 3
16.
10t 11 6w 7 A B 2 t 5t 6 t 6 t 1 t 6 t 1
17.
x 2 26 x 100 x 2 26 x 100 x 2 26 x 100 A B C 2 5 x3 10 x 2 25 x x x 2 10 x 25 x x x x 5 x 5 2
18.
B C 3x 2 2 x 8 3 x 2 2 x 8 3x 2 2 x 8 A 2 3 2 2 x 4x 4x x x 4x 4 x x 2 x 2 2 x x 2
19.
13x 2 2 x 45 13 x 2 2 x 45 A Bx C 2 x3 18 x 2 x x2 9 2 x x2 9
20.
17 x 2 7 x 18 17 x 2 7 x 18 A Bx C 7 x3 42 x 7 x x2 6 7 x x2 6
21.
2 x3 x 2 13x 5 2 x3 x 2 13x 5 Ax B Cx D 2 2 2 x 4 10 x 2 25 x 5 x2 5 x2 5
22.
3x3 4 x 2 11x 12 3x3 4 x 2 11x 12 Ax B Cx D 2 2 2 x4 6 x2 9 x 3 x2 3 x2 3
23.
5x2 4 x 8
x 4 x x 4 2
A Bx C 2 x4 x x4
589
Chapter 5 Systems of Equations and Inequalities
24.
25.
26.
x 2 15 x 6
x 6 x 2 x 6 2
2 x5 3x3 4 x 2 5
x x 2 x 2 x 7 3
2
A Bx C 2 x 6 x 2x 6
2
A B C D Ex F Gx H 2 2 3 2 x x 2 x 2 x 2 x 2x 7 x 2x 7 2
A B C Dx E Fx G 2 2 2 x 3 2 x 9 2 x 9 x 1 x2 1
6 x 4 5 x3 2 x 2 5
x 3 2 x 9 x 1 2
2
2
x 37 A B x 4 2 x 3 x 4 2 x 3
27.
x 37 B A x 4 2 x 3 x 4 2x 3 x 4 2 x 3
x 4 2 x 3
x 37 A 2 x 3 B x 4 x 37 2 Ax 3 A Bx 4 B
x 37 2 A B x 3 A 4 B 1 2 A B 1
Multiply by 4 1 1 2 A B 8 A 4 B 4 3 A 4 B 37 2 37 3 A 4 B 11A 33 A 3
2 3 B 1 6 B 1 B 7
x 37 3 7 x 4 2 x 3 x 4 2 x 3 20 x 4 A B x 5 3x 1 x 5 3x 1
28.
20 x 4 B A x 5 3x 1 x 5 3 x 1 x 5 3 x 1
x 5 3x 1
20 x 4 A 3x 1 B x 5 20 x 4 3 Ax A Bx 5 B
20 x 4 3 A B x A 5 B Multiply by 5 15 A 5B 100 1 20 3 A B 2 4 A 5 B A 5B 4 96 16 A A 6
20 x 4 6 2 x 5 3x 1 x 5 3x 1
590
2 A 5 B 4 6 5 B 4 5 B 10 B2
Section 5.3
29.
8 x 10 8 x 10 A B 2 x 2 x x x 2 x x 2
30.
8 x 10 B A x x 2 x x 2 x x 2 x x 2 8 x 10 A x 2 Bx
y 12 A B y y 3 y y 3 y y 3 y y 3 y 12 A y 3 By
8 x 10 Ax 2 A Bx
y 12 Ay 3 A By
8 x 10 A B x 2 A
y 12 A B y 3 A
A B 8 5 B 8 B3
2 A 10 A5
3 A 12 A 4
y 12 4 5 2 y 3y y y3
8 x 10 5 3 2 x 2x x x 2
31.
y 12 y 12 A B 2 y 3 y y y 3 y y 3
6w 7 6w 7 A B w w 6 w 2 w 3 w 2 w 3 2
6w 7 B A w 2 w 3 w 2 w 3 w 2 w 3
w 2 w 3
6w 7 A w 3 B w 2 6w 7 Aw 3 A Bw 2 B
6w 7 A B w 3 A 2 B Multiply by 2 6 A B 2 A 2 B 12 1 2 7 3 A 2 B 3 A 2 B 7 5 5A A 1
1 A B 6 1 B 6 B5
6w 7 1 5 w w6 w2 w3 2
32.
10t 11 6w 7 A B 2 t 5t 6 t 6 t 1 t 6 t 1 10t 11 B A t 6 t 1 t 6 t 1 t 6 t 1
t 6 t 1
10t 11 A t 1 B t 6 10t 11 At A Bt 6 B
10t 11 A B t A 6 B
591
A B 1 4 B 1 B5
Chapter 5 Systems of Equations and Inequalities 1 10 A B 2 11 A 6 B 21 7B 3 B 10t 11 7 3 2 t 5t 6 t 6 t 1 33.
1 A B 10 A 3 10 A 7
x 2 26 x 100 x 2 26 x 100 x 2 26 x 100 A B C 2 3 2 2 x 10 x 25 x x x 10 x 25 x x 5 x 5 2 x x 5
x 2 26 x 100 A B C 2 2 x x 5 x x 5 2 2 x x 5 x x 5 x 5 x 2 26 x 100 A x 5 Bx x 5 Cx 2
x 2 26 x 100 A x 2 10 x 25 Bx 2 5 Bx Cx x 26 x 100 Ax 10 Ax 25 A Bx 2 5 Bx Cx 2
2
x 2 26 x 100 A B x 2 10 A 5B C x 25 A
25 A 100 A 4
A B 1 4 B 1 B 3
10 A 5 B C 26 10 4 5 3 C 26 40 15 C 26 C 1
x 2 26 x 100 4 3 1 3 2 x 10 x 25 x x x 5 x 5 2 3x 2 2 x 8 3 x 2 2 x 8 3x 2 2 x 8 A B C 34. 3 2 2 2 x 4x 4x x x 4x 4 x x 2 x 2 2 x x 2
3 x 2 2 x 8 A B C 2 2 x x 2 x x 2 2 2 x x 2 x x 2 x 2 3 x 2 2 x 8 A x 2 Bx x 2 Cx 2
3 x 2 2 x 8 A x 2 4 x 4 Bx 2 2 Bx Cx 3 x 2 x 8 Ax 4 Ax 4 A Bx 2 2 Bx Cx 2
2
3 x 2 2 x 8 A B x 2 4 A 2 B C x 4 A 4A 8 A 2
4 A 2B C 2
A B 3 2 B 3
4 2 2 5 C 2
B 5
8 10 C 2 C4
3x 2 x 8 2 5 4 3 2 x 4 x 4 x x x 2 x 2 2 2
592
Section 5.3 13x 2 2 x 45 13 x 2 2 x 45 A Bx C 35. 2 x 3 18 x 2 x x2 9 2 x x2 9
13x 2 2 x 45 A Bx C 2 x x2 9 2 2x x 9 2 2x x 9 2 x x 9
2
13 x 2 2 x 45 A x 2 9 Bx C 2 x
9 A 45 A5
13 x 2 x 45 Ax 9 A 2 Bx 2 2Cx 2
2
A 2 B 13 5 2 B 13 2B 8 B4
2C 2
A 7 B 17
7C 7 C 1
C 1
13 x 2 2 x 45 A 2 B x 2 2C x 9 A 13x 2 2 x 45 5 4 x 1 2 x 3 18 x 2 x x2 9 17 x 2 7 x 18 17 x 2 7 x 18 A Bx C 36. 7 x 3 42 x 7 x x2 6 7 x x2 6
17 x 2 7 x 18 A Bx C 7 x x2 6 2 7x x 6 2 7x x 6 7 x x 6
2
17 x 2 7 x 18 A x 2 6 Bx C 7 x
6 A 18 A3
B2
17 x 7 x 18 Ax 6 A 7 Bx 2 7Cx 2
3 7 B 17 7 B 14
2
17 x 2 7 x 18 A 7 B x 2 7C x 6 A 17 x 2 7 x 18 3 2 x 1 7 x 3 42 x 7 x x2 6 x3 37. x 0 x 7 x x 3x 13 x 28 x 28 3
2
4
3
2
x 4 0 x3 7 x 2
3x 6 x 28 x 3x 3 0 x 2 21x 3
2
6 x 7 x 28 2
x 3x 13x 28 x 28 6 x 2 7 x 28 x 3 x3 7 x x3 7 x 6 x 2 7 x 28 6 x 2 7 x 28 A Bx C 2 x3 7 x x x 7 x x2 7 4
3
2
7 A 28
A B 6
A 4
4 B 6 B2
x 4 3x 3 13x 2 28 x 28 4 2x 7 x 3 2 3 x 7x x x 7
593
C 7
Chapter 5 Systems of Equations and Inequalities x4 38. x 0 x 2 x x 4 x 11x 13 x 12 3
2
4
3
2
x 4 0 x3 2 x 2
4 x 9 x 13x 4 x 3 0 x 2 8 x 3
2
9 x 5 x 12 2
x 4 x 11x 13 x 12 9 x 2 5 x 12 x 4 x3 2 x x3 2 x 9 x 2 5 x 12 9 x 2 5 x 12 A Bx C 2 x3 2 x x x 2 x x2 2 4
3
2
9 x 2 5 x 12 A Bx C x x2 2 2 x x2 2 2 x x 2 x x 2
2 A 12 A6
9 x 5 x 12 A x 2 Bx C x 2
2
2
2
9 x 2 5 x 12 A B x 2 C x 2 A x 4 4 x 3 11x 2 13 x 12 6 3x 5 x4 2 3 x 2x x x 2 39.
2 x 3 x 2 13x 5 2 x 3 x 2 13 x 5 Ax B Cx D 2 2 2 x 4 10 x 2 25 x 5 x2 5 x2 5
B Cx D x 5 2 x x 13x 5 x 5 Ax x 5 x 5 x 5 2 x x 13x 5 Ax B x 5 Cx D 2
2
3
2
2
2
2
2
3
2
2
2
2
2
2 x 3 x 2 13x 5 Ax 3 5 Ax Bx 2 5 B Cx D 2 x 3 x 2 13x 5 Ax 3 Bx 2 5 A C x 5 B D A 2
B 1
5 A C 13
5B D 5
5 2 C 13
5 1 D 5
10 C 13
5 D 5
C3
D0
2 x 3 x 2 13x 5 2 x 1 3x 2 4 2 x 10 x 25 x 5 x2 5 2
6 B 9 B3
9 x 5 x 12 Ax 2 A Bx Cx 2
A B 9
594
C 5
Section 5.3
40.
3x 3 4 x 2 11x 12 3x 3 4 x 2 11x 12 Ax B Cx D 2 2 2 x4 6 x2 9 x 3 x2 3 x2 3
3 2 Ax B Cx D 3x 4 x 2 11x 12 2 x 3 x 3 2 2 2 2 2 x 3 x x 3 3
2
2
3 x 3 4 x 2 11x 12 Ax B x 2 3 Cx D 3 x 3 4 x 2 11x 12 Ax 3 3 Ax Bx 2 3B Cx D 3 x 3 4 x 2 11x 12 Ax 3 Bx 2 3 A C x 3B D B 4
A3
3 A C 11
3B D 12
3 3 C 11
3 4 D 12
9 C 11
12 D 12 D0
C2 3x 3 4 x 2 11x 12 3x 4 2x 2 4 2 x 6x 9 x 3 x2 3 2
41.
5x2 4 x 8
x 4 x x 4 2
A Bx C 2 x4 x x4
5x2 4 x 8
Bx C A x 4 x 2 x 4 2 2 x 4 x x 4 x 4 x x 4
x 4 x 2 x 4
5 x 2 4 x 8 A x 2 x 4 Bx C x 4 5 x 2 4 x 8 Ax 2 Ax 4 A Bx 2 4 Bx Cx 4C 5 x 2 4 x 8 A B x 2 A 4 B C x 4 A 4C 1 5 A B 2 4 A 4 B C 3 8 4A 4C
1 2
Multiply by 4 20 5 A B 4 A 4B 4 A 4 B C A 4 B C 4 5A C 16 4
4 A 4C 8 3 4 A 4C 8 Multiply by 4 4 5 A C 16 20 A 4 B 64 24 A 72 A 3
1 A B 5
3
4 A 4C 8
3 B 5
4 3 4C 8
B2
12 4C 8 4C 4 C 1
5x2 4 x 8
3 2x 1 2 2 x 4 x x 4 x 4 x x 4
595
Chapter 5 Systems of Equations and Inequalities
42.
x 2 15 x 6
x 6 x 2 x 6 2
A Bx C 2 x 6 x 2x 6
x 2 15 x 6 Bx C A 2 x 6 x 2 2 x 6 x x x 2 6 2 6 2 x 6 x 2x 6 x 6 x 2 x 6
x 2 15 x 6 A x 2 2 x 6 Bx C x 6 x 15 x 6 Ax 2 Ax 6 A Bx 2 6 Bx Cx 6C 2
2
x 2 15 x 6 A B x 2 2 A 6 B C x 6 A 6C 1 1 A B 2 15 2 A 6 B C 3 6 6 A 6C Multiply by 6 6 6 A 6 B 1 1 A B 2 15 2 A 6 B C 2 A 6 B C 15 4 A C 9 4
3 4
6 A 6C 6 6 A 6C 6 Multiply by 6 4 A C 9 24 A 6C 54 30 A 60 A 2
1 A B 1
3
6 A 6C 6 A C 1 2 C 1
2 B 1 B3
C 1 x 2 15 x 6
2 3x 1 2 2 x 6 x 2 x 6 x 6 x 2 x 6
43.
4 x 3 4 x 2 11x 7 4 x 3 4 x 2 11x 7 Ax B Cx D 2 2 x4 5x2 6 x 2 x 3 x2 2 x2 3
4 x 3 4 x 2 11x 7 Ax B Cx D x2 2 x2 3 2 x2 2 x2 3 2 2 2 x 3 x 2 x 2 x 3
4 x 3 4 x 2 11x 7 Ax B x 2 3 Cx D x 2 2
4 x 4 x 11x 7 Ax 3 Ax Bx 3B Cx 2Cx Dx 2 2 D 3
2
3
2
3
4 x 3 4 x 2 11x 7 A C x 3 B D x 2 3 A 2C x 3B 2 D AC 4 A 4C
3 A 2C 11
AC 4
3 4 C 2C 11
A 1 4
12 3C 2C 11
A3
C 1 596
Section 5.3 B D 4 B 4 D
3B 2 D 7
B D 4
3 4 D 2 D 7
B 5 4
12 3D 2 D 7
B 1
D 5 D 5 4 x 4 x 11x 7 3x 1 x 5 2 x4 5x2 6 x 2 x2 3 3
44.
2
3x 3 4 x 2 6 x 7 3x 3 4 x 2 6 x 7 Ax B Cx D 2 2 x4 5x2 4 x 4 x 1 x2 4 x2 1
3x3 4 x 2 6 x 7 Ax B Cx D x2 4 x2 1 2 x2 4 x2 1 2 2 2 x 1 x 4 x 4 x 1
3x 3 4 x 2 6 x 7 Ax B x 2 1 Cx D x 2 4
3x 3 4 x 2 6 x 7 Ax 3 Ax Bx 2 B Cx3 4Cx Dx 2 4 D 3x 3 4 x 2 6 x 7 A C x 3 B D x 2 A 4C x B 4 D
AC 3 A 3C
A 4C 6 3 C 4C 6 3C 3 C 1
AC 3 A 1 3 A2
B D 4 B 4 D
B 4 D 7 4 D 4 D 7 3D 3 D 1 3 2 3x 4 x 6 x 7 2 x 3 x 1 2 x4 5x2 4 x 4 x2 1
B D 4 B 1 4 B 3
2x 5 45. x 2 3 x 10 2 x3 11x 2 4 x 24
2 x 3 6 x 2 20 x
5 x 16 x 24 5 x 2 15 x 50 2
x 26 2 x 11x 4 x 24 x 26 2x 5 2 2 x 3 x 10 x 3 x 10 x 26 x 26 A B 2 x 3x 10 x 2 x 5 x 2 x 5 3
2
597
Chapter 5 Systems of Equations and Inequalities
x 26 B A x 2 x 5 x 2 x 5 x 2 x 5
x 2 x 5
x 26 A x 5 B x 2 x 26 Ax 5 A Bx 2 B
x 26 A B x 5 A 2 B A B 1
5 A 2 B 26 5 1 B 2 B 26
A 1 B
5 5B 2 B 26 7 B 21 B 3
A B 1 A31 A 4
2 x 3 11x 2 4 x 24 4 3 2x 5 2 x 3 x 10 x2 x5 3x 2 46. x 3 x 4 3 x 11x x 10 2
3
2
3x 3 9 x 2 12 x
2 x 13 x 10 2 x2 6 x 8 2
7 x 18 3x 11x x 10 7 x 18 3x 2 2 2 x 3x 4 x 3x 4 7 x 18 7 x 18 A B 2 x 3x 4 x 1 x 4 x 1 x 4 3
2
7 x 18 B A x 1 x 4 x 1 x 4 x 1 x 4
x 1 x 4
7 x 18 A x 4 B x 1 7 x 18 Ax 4 A Bx B
7 x 18 A B x 4 A B A B 7 A7B
4 A B 18 4 7 B B 18 28 4 B B 18 5 B 10 B2
3x 3 11x 2 x 10 5 2 3x 2 2 x 3x 4 x 1 x 4
598
A B 7 A 2 7 A5
Section 5.3 3x 4 47. x 2 x 1 3x 2 x x 5 2
3
2x 5 48. x 6 x 9 2 x 17 x 54 x 68 2
2
3x3 6 x 2 3x
3
2 x 3 12 x 2 18 x
4x 4x 5 4 x 2 8 x 4
5 x 36 x 68 5 x 2 30 x 45
4x 1
2 x 17 x 54 x 68 6 x 23 2x 5 2 2 x 6x 9 x 6x 9 A B 6 x 23 6 x 23 2 2 x 6 x 9 x 3 x 3 x 3 2 3
2
x 32
4 x 1 Ax A B
6 x 23 Ax 3 A B
4 x 1 A x A B
6 x 23 A x 3 A B A6
A B 1 4 B 1 B 5
49. a.
18 B 23 B 5
2
2 x 3 17 x 2 54 x 68 x2 6 x 9 6 5 2x 5 x 3 x 3 2
Factors of 45 1, 3, 5, 9, 15, 45 Factors of 1 1 1, 3, 5, 9, 15, 45 5 1 1 21 45 5 30 45 1 6 9 0
3 A B 23 3 6 B 23
3x 2 x x 5 4 5 3x 4 2 x 2x 1 x 1 x 1 2 3
2
6 x 23 A B 2 x 3 2 2 x 3 x 3 x 3 6 x 23 A x 3 B
4 x 1 A B 2 x 1 2 2 x 1 x 1 x 1 4 x 1 A x 1 B
x 12
A 4
6 x 23
3x 2 x x 5 4x 1 3x 4 2 2 x 2x 1 x 2x 1 4x 1 4x 1 A B 2 2 x 2 x 1 x 1 x 1 x 1 2 3
2
2
2
x 3 x 2 21x 45 x 2 6 x 9 x 5 x 3 x 5 2
599
Chapter 5 Systems of Equations and Inequalities 3 x 2 35 x 70
b.
x 3 x 5 2
A B C 2 x 3 x 3 x5
3 x 2 35 x 70 A B C 2 x 3 x 5 2 2 x 5 x 3 x 5 x 3 x 3
x 3 2 x 5
3x 2 35 x 70 A x 3 x 5 B x 5 C x 3
2
3x 2 35 x 70 A x 2 2 x 15 B x 5 C x 2 6 x 9
3x 2 35 x 70 Ax 2 2 Ax 15 A Bx 5 B Cx 2 6Cx 9C 3x 2 35 x 70 A C x 2 2 A B 6C x 15 A 5 B 9C 5 2 3
C 3 A 1 2 2 A B 6C 35 3 15 A 5B 9C 70
10 A 5 B 30C 175 15 A 5 B 9C 70 25 A 39C 245 4
Solve the system of equations 1 and 4. 1 A C 3 25 1 25 A 25C 75 A 5 3 4 25 A 39C 245 64C 320 A 2 C 5
2
2A B 6C 35 2 2 B 6 5 35 4 B 30 35 B 1
3x 35 x 70 2 1 5 2 3 2 x x 21x 45 x 3 x 3 x5 2
50. a.
Factors of 4 1, 2, 4 1, 2, 4 Factors of 1 1 4 1 2 7 4 4 1 2
8 4 1
0
x 3 2 x 2 7 x 4 x 2 2 x 1 x 4 x 1 x 4 b.
10 x 2 17 x 17
x 1 x 4 2
2
A B C 2 x 1 x 1 x4
10 x 2 17 x 17 A B C 2 2 x 1 x 4 x 1 x 4 2 2 x 4 x 1 x 4 x 1 x 1 10 x 2 17 x 17 A x 1 x 4 B x 4 C x 1
2
10 x 2 17 x 17 A x 2 3x 4 B x 4 C x 2 2 x 1
10 x 17 x 17 Ax 3 Ax 4 A Bx 4 B Cx 2Cx C 2
2
2
10 x 2 17 x 17 A C x 2 3 A B 2C x 4 A 4 B C
600
Section 5.3 4 2 3
C 10 A 1 2 3 A B 2C 17 3 4 A 4 B C 17
12 A 4 B 8C 68 4 A 4 B C 17 16 A 9C 85 4
Solve the system of equations 1 and 4. 1 A C 10 9 1 9 A 9C 90 7 C 10 4 16 A 9C 85 25 A 175 C3 A 7
3A B 2C 17
2
3 7 B 2 3 17 21 B 6 17 B2
10 x 2 17 x 17 7 2 3 2 3 2 x 2 x 7 x 4 x 1 x 1 x4 51. a.
Factors of 8 1, 2, 4, 8 1, 2, 4, 8 Factors of 1 1 2 1 6 12 8 2 8 8 1 4 4 0
x 3 6 x 2 12 x 8 x 2 4 x 2 x 2 x 2 x 2 x 2 b.
3x 2 8 x 5
x 2
3
2
3
A B C 2 x 2 x 2 x 2 3
3x 2 8 x 5 A B C 3 3 x 2 x 2 3 2 3 x 2 x 2 x 2 x 2 3 x 2 8 x 5 A x 2 B x 2 C 2
3 x 2 8 x 5 A x 2 4 x 4 B x 2 C 3 x 8 x 5 Ax 4 Ax 4 A Bx 2 B C 2
2
3 x 2 8 x 5 A x 2 4 A B x 4 A 2 B C 3 1 A 8 2 4A B 3 4 A 2B C 5
4A B 8
4A 2 B C 5
4 3 B 8
4 3 2 4 C 5
12 B 8
12 8 C 5
B 4
C 1
3x 2 8 x 5 3 4 1 2 3 2 x 6 x 12 x 8 x 2 x 2 x 23
601
Chapter 5 Systems of Equations and Inequalities
52. a.
Factors of 27 1, 3, 9, 27 1, 3, 9, 27 Factors of 1 1 3 1 9 27 27 3 18 27 1 6 9 0
x 3 9 x 2 27 x 27 x 2 6 x 9 x 3 x 3 x 3 x 3 2 x 2 17 x 37
b.
x 3
3
2
3
A B C 2 x 3 x 3 x 33
2 x 2 17 x 37 A B C 3 x 3 3 2 3 x 3 x 3 x 3 x 3
x 33
2 x 2 17 x 37 A x 3 B x 3 C 2
2 x 2 17 x 37 A x 2 6 x 9 B x 3 C 2 x 2 17 x 37 Ax 2 6 Ax 9 A Bx 3B C 2 x 2 17 x 37 A x 2 6 A B x 9 A 3B C A 2 1 17 2 6 A B 3 9 A 3B C 37
6 A B 17
9A 3B C 37
6 2 B 17
9 2 3 5 C 37
12 B 17
18 15 C 37
B 5
C4
2 x 2 17 x 37 2 5 4 2 3 2 x 9 x 27 x 27 x 3 x 3 x 33
53. Partial fraction decomposition is a procedure in which a rational expression is written as a sum of two or more simpler rational expressions.
55. A proper rational expression is a rational expression in which the degree of the numerator is less than the degree of the denominator.
54. For a repeated linear factor that occurs three times, one fraction must be given with a denominator of ax b , one fraction must
56. Given an improper rational expression, use long division to divide the numerator by the denominator. Then perform partial fraction decomposition to the expression represented by the remainder/divisor.
have denominator ax b , and one 2
denominator must have ax b . The numerators in each fraction must be constants, which we might call A, B, and C. 3
602
Section 5.3 2 A B n n 2 n n 2
57. a.
3 A B n n 3 n n 3
58. a.
2 B A n n 2 n n 2 n n 2 n n 2 2 A n 2 Bn
3 B A n n 3 n n 3 n n 3 n n 3 3 A n 3 Bn
2 An 2 A Bn
3 An 3 A Bn 3 A B n 3 A A B 0 3A 3 1 B 0 A 1 B 1 3 1 1 1 1 n n 3 n n 3 n n 3
2 A B n 2 A A B 0 1 B 0 B 1 2 1 1 1 1 n n 2 n n 2 n n 2
2A 2 A 1
b.
2 2 2 2 2 1 3 2 4 3 5 4 6 5 7
b.
1 1 1 1 1 1 1 4 2 5 3 6
1 1 1 1 1 1 1 3 2 4 3 5
1 1 1 1 4 7 5 8
1 1 1 1 4 6 5 7 c. As n ,
3 3 3 3 3 1 4 2 5 3 6 4 7 5 8
c. As n ,
1 0. n2
1 0. n3
1 1 1 1 1 1 d. 1 4 2 5 3 6
1 1 1 1 1 1 1 1 d. 1 3 2 4 3 5 4 6
1 1 1 1 4 7 5 8
1 1 5 7
1 1 1 1 1 1 1 1 2 3 4 4 5 5
1 1 1 1 1 1 1 2 3 3 4 4
1 1 6 6
1 1 5 5
3 3 0 0 0 2 2
603
11 11 0 0 0 6 6
Chapter 5 Systems of Equations and Inequalities 1 A B x a bx x a bx
59.
1 B A x a bx x a bx x a bx x a bx 1 A a bx Bx
aA 1 1 A a
1 aA bAx Bx
B
1 bA B x aA 1 b 1 1 b a a x a bx x a bx ax a a bx 1 1 2 a x a x a x
60.
bA B 0 1 b B0 a
2
1 A B 2 a x ax a x 1 B A a x a x a x a x a x a x a x a x 1 A a x B a x 2
1 aA Ax aB Bx 1 A B x aA aB aA aB 1 aB aB 1 2aB 1 1 B A 2a 1 1 1 1 1 2 a 2 a 2 2 a x a x a x 2a a x 2a a x A B 0 A B
61. Let u e x . 5e x 7 5u 7 5u 7 A B 2 2x x e 3e 2 u 3u 2 u 1 u 2 u 1 u 2
5u 7 B A u 1 u 2 u 1 u 2 u 1 u 2
u 1 u 2
5u 7 A u 2 B u 1 5u 7 Au 2 A Bu B
5u 7 A B u 2 A B
604
b a
Section 5.4 A B 5 A3 5 A 2
2A B 7
A B 5 A B 5
2 B 5 B 7 2 B 10 B 7 B 3 B3
5e x 7 A B 2 3 x x x 2x e 3e 2 u 1 u 2 e 1 e 2 62. Let u e x . 3e x 22 3u 22 3u 22 A B 2 x 2x e 3e 4 u 3u 4 u 1 u 4 u 1 u 4
3u 22 B A u 1 u 4 u 1 u 4 u 1 u 4
u 1 u 4
3u 22 A u 4 B u 1 3u 22 Au 4 A Bu B
3u 22 A B u 4 A B 4 A B 22
A B 3 A B 3
4 B 3 B 22 4 B 12 B 22 5B 10 B2
A B 3 A 2 3 A 5
3e x 22 A B 5 2 x x 2x x e 3e 4 u 1 u 4 e 1 e 4
Section 5.4 Systems of Nonlinear Equations in Two Variables 1. y x 2 1 x 2 1 The graph of the equation is the graph of y x 2 shifted to the right 2 units and downward 1 unit. 2
2
2. y 3x 2 Slope: 3 , y-intercept: 2
605
Chapter 5 Systems of Equations and Inequalities 3.
x 3 2 y 22 16 2 x 3 2 y 2 42 Center: 3, 2 , radius: 4
3 6. 5 x 4 y 3 15 x 12 y 9 5 3 x 7 y 17 15 x 35 y 85 47 y 94 y 2 5 x 4 y 3
5 x 4 2 3 5 x 8 3 5x 5
1, 2
x 1 7. nonlinear 8. ordered; intersection 4. y x 2 3 x 0 3 The graph of the equation is the graph of y x 2 reflected across the x-axis and shifted downward 3 units. 2
9. a. A y x 2 2
B 2x y 2 y 2x 2
b. B
2 x x2 2 2
2x 3y 3 2 4 y 7 3 y 3 8 y 14 3 y 3 11y 11 y 1
2 x x2 2 2
5. x 4 y 7
2x 3y 3
2x y 2
x2 2 x 0
x 4y 7
x x 2 0 x 0 or x 2
x 4 1 7 x 4 7 x3
A y x 2 2 0 2 2 2
The solution is 0, 2 .
3, 1
A y x 2 2 2 2 4 2 2 2
The solution is 2, 2 .
2, 2 , 0, 2
606
Section 5.4 B y x 1 3 1 3 1 4
10. a. A y x 2 3
The solution is 3, 4 .
B y 2x 0 y 2x
B y x 1 4 1 4 1 3
The solution is 4, 3 .
3, 4 , 4, 3
12. a. A x 2 y 2 25 x 2 y 2 52
B 3y 4x
y
4 x 3
y 2x 0
b. B
x 3 2x 0 2
x2 2 x 3 0
x 3 x 1 0 x 3 or x 1 A y x 2 3 3 3 9 3 6 2
The solution is 3, 6 . 2
2
4 x x 25 3
The solution is 1, 2 .
2
1, 2 , 3, 6
9 2 16 2 x x 25 9 9 25 2 x 25 9 x2 9 x 3 4 4 B y x 3 4 3 3 The solution is 3, 4 .
11. a. A x 2 y 2 25 x 2 y 2 52
B x y 1
x 2 y 2 25
b. A
A y x 2 3 1 3 1 3 2
y x 1
4 4 x 3 4 3 3 The solution is 3, 4 . B y
b. A
3, 4 , 3, 4
x 2 y 2 25 x 2 x 1 25 2
x 2 x 2 2 x 1 25 2 x 2 2 x 24 0 x 2 x 12 0 x 3 x 4 0 x 3 or x 4 607
Chapter 5 Systems of Equations and Inequalities 13. a. A y x
B x 2 y 2 20 x 2 y 2 2 5
B y x 2 4 2= 6 Not real B y x 2 3 2= 1 1
2
The solution is 3,1 .
3,1
15. a. A
x 2 2 y 2 9 2 2 x 2 y 0 32
B y 2x 4
x 2 y 2 20
b. B
x2
x 20 2
x 2 x 20 x 2 x 20 0 x 5 x 4 0 x 5 or x 4 A y x 5 Not real
b. A
A y x 42 The solution is 4, 2 .
x 2 2 y 2 9 x 2 2 2 x 4 2 9
x 2 4 x 4 4 x 2 16 x 16 9
4, 2
5 x 2 12 x 11 0 b 2 4ac 12 4 511 76 There are no real solutions. 2
14. a. A x 2 y 2 10 x 2 y 2
10
2
B y x2
16. a. A x 2 y 3 4 2
x 0 2 y 3 2 22 B y x 4
x 2 y 2 10
b. A
x2
x 2 10 2
x 2 x 2 10 x 2 x 12 0 x 4 x 3 0 x 4 or x 3 608
Section 5.4 x 2 y 3 4
18. a. A y 3 x
2
b. A
x 2 x 4 3 4
B yx
2
x2 x 7 4 2
x 2 x 2 14 x 49 4 2 x 2 14 x 45 0
b 2 4ac 14 4 2 45 164 There are no real solutions. 2
17. a. A y x 3
y3 x
b. A
B yx
x 3 x x3 x
x3 x 0
x x2 x 0 x x 1 x 1 0 x 0 or x 1 or x 1 B y x0 The solution is 0, 0 .
b. A y x
3
B y x 1
xx
3
The solution is 1, 1 .
0 x x 3
B y x 1
The solution is 1,1 .
0 x x 1 2
0 x x 1 x 1
1, 1 , 0, 0 , 1,1
x 0 or x 1 or x 1 B yx0 The solution is 0, 0 . B y x 1 The solution is 1, 1 . B y x 1 The solution is 1,1 .
1, 1 , 0, 0 , 1,1
609
Chapter 5 Systems of Equations and Inequalities 2 x 2 3 y 2 11 19. A 2 x 2 3 y 2 11 Multiply by 2. 2 x 2 8 y 2 16 B x 2 4 y 2 8 5 y 2 5 y2 1 y 1 y 1: B x 2 4 y 2 8
y 1: B x 2 4 y 2 8
x 2 4 1 8
x 2 4 1 8
x2 4 8
x2 4 8
x2 4 x 2
x2 4 x 2
2
2
2,1 , 2, 1 , 2, 1 , 2, 1 2 2 6 x 2 2 y 2 42 20. A 3 x y 21 4 x 2 2 y 2 2 B 4 x 2 2 y 2 2 10 x 2 40 2 x 4 x 2 Multiply by 2.
x 2 : B 4x 2 2 y 2 2
x 2 : B 4x 2 2 y 2 2
4 2 2 y 2 2
4 2 2 y 2 2
16 2 y 2 2
16 2 y 2 2
2 y 2 18
2 y 2 18
y2 9 y 3
y2 9 y 3
2
2
2, 3 , 2, 3 , 2, 3 , 2, 3 Multiply by 3. 21. A 3x 2 3xy 60 x 2 xy 20 B 2 x 2 3 xy 44 2 x 2 3xy 44 x2 16 x 4
x 4: A
x 2 xy 20
x 4 : A
x 2 xy 20
42 4 y 20
42 4 y 20
16 4 y 20
16 4 y 20
4 y 4 y 1
4y 4 y 1
4, 1 , 4,1
610
Section 5.4 22. A 4 xy 3 y 2 9 4 xy 3 y 2 9 Multiply by 2. B 2 xy y 2 5 4 xy 2 y 2 10 y2 1 y 1
y 1: B
2 xy y 2 5
y 1: B
2 xy y 2 5
2 x 1 1 5
2 x 1 1 5
2 x 1 5 2 x 6 x 3
2 x 1 5 2 x 6 x3
2
3,1 , 3, 1
2
Multiply by 3. 2 2 15 x 2 6 y 2 3 23. A 5 x 2 y 1 2 2 Multiply by 2. B 2 x 2 3 y 2 4 4 x 6 y 8 2 11 11x x2 1 x 1
x 1: B
2 x 2 3 y 2 4
x 1: B
2 x 2 3 y 2 4
2 1 3 y 2 4
2 1 3 y 2 4
2 3 y 2 4
2 3 y 2 4
3 y 2 6
3 y 2 6
y2 2
y2 2
y 2
y 2
2
2
1, 2 , 1, 2 , 1, 2 , 1, 2 2 2 18 x 2 15 y 2 114 24. A 6 x 5 y 38 2 2 Multiply by 5. B 7 x 2 3 y 2 9 35 x 15 y 45 159 53 x 2 2 x 3 x 3 Multiply by 3.
x 3: B
7 x2 3 y 2 9
7 x2 3 y 2 9
3 3y 9
7 3 3y2 9
21 3 y 2 9
21 3 y 2 9
2
7
x 3: B
2
2
3 y 2 12
3 y 2 12
y2 4 y 2
y2 4 y 2
3, 2 , 3, 2 , 3, 2 , 3, 2 611
Chapter 5 Systems of Equations and Inequalities 2 2 Multiply by 9. 25. A x 2 y 2 1 9 x 9 y 2 2 2 2 9x 4 y B 9 x 4 y 36 13 y 2
9 36 27 27 y2 13 27 y i 13
The values for y are all imaginary numbers. 2 2 Multiply by 16. 26. A 4 x 2 y 2 4 64 x 16 y 2 2 9 x 2 16 y 2 B 9 x 16 y 144 73 x 2
64 144 80 80 x2 73 80 x i 73
The values for x are all imaginary numbers.
27. A x 2 4 xy 4 y 2 1
B x y 4
x 2 6 xy 9 y 2 0
A
x 2 6 x x 2 9 x 2 0
y x 4
2
x 2 4 xy 4 y 2 1
x 2 6 x 2 12 x 9 x 2 4 x 4 0
x 2 4 x x 4 4 x 4 1
5 x 12 x 9 x 36 x 36 0
A
2
2
2
4 x 2 24 x 36 0
x 2 4 x 2 16 x 4 x 2 8 x 16 1
x2 6 x 9 0
5 x 2 16 x 4 x 2 32 x 64 1
x 3 2 0
9 x 2 48 x 63 0
x3
3 x 2 16 x 21 0
B y x 2 3 2 1
3x 7 x 3 0
3,1
7 or x 3 3 7 12 5 7 B y x 4 4 3 3 3 3
x
29. A y x 2 4 x 5
B y 4x 5
B y x 4 3 4 1 7 5 3,1 , , 3 3
4 x 5 x2 4 x 5 0 x2 x0 B y 4 x 5 4 0 5 5
28. A x 2 6 xy 9 y 2 0
B x y2
y x2 4 x 5
A
y x2
0, 5 612
Section 5.4 30. A y x 2 6 x 9
Check: 1, 1
B y 2 x 5
1 x 1 1ⱨ 1 1ⱨ1 false 1,1
A
A y
y x2 6 x 9 2 x 5 x 2 6 x 9 0 x2 4 x 4
0 x 2 x2 B y 2 x 5 2 2 5 1 2
33. A x 2 y 4 25 2
2,1
B y x2 9
31. A y x 1 B y x
x 2 y 4 25 2
A
2
x x 5 25 2
x 2 x 2 9 4 25 2
2
2
x 2 x 4 10 x 2 25 25 x4 9 x2 0
y x2 1 x2 x x3 1 x 1 1 1 B y 1 x 1 1,1 A
x2 x2 9 0 x x 3 x 3 0 x 0 or x 3 or x 3 2
B y x 2 9 0 9 9 2
B y x 2 9 3 9 0 2
B y x 2 9 3 9 0 2
32. A y
1 x
0, 9 , 3, 0 , 3, 0
B y x 34. A
1 x 1 x x 1 x 2 x x3 1 x 1 B y x 1 1
A
y
B x y2 A
1 x 1 1ⱨ 1 1ⱨ1 true
x 102 y 2 100
y 10 y 100 2
2
2
y 4 20 y 2 100 y 2 100 y 4 19 y 2 0
y 2 y 2 19 0 y 0 or y 19 or y 19 B x y 2 0 0
Check: 1,1 A y
x 102 y 2 100
2
19 B x y 19 19 0, 0 , 19, 19 , 19, 19 B x y 2 19
B y x 1ⱨ 1 1ⱨ1 true
2
613
2
2
Chapter 5 Systems of Equations and Inequalities 37. Let x represent one number. Let y represent the other number. A x y 12 y x 12
1 1 and v 2 . 2 x y 4 3 A 2 2 23 4u 3v 23 x y 5 1 B 2 2 14 5u v 14 v 5u 14 x y
35. Let u
A
B xy 35 x x 12 35
4u 3v 23
x 2 12 x 35 0
4u 3 5u 14 23
x 2 12 x 35 0
4u 15u 42 23 19u 19 u 1 B v 5u 14 5 1 14 9 u
x 5 x 7 0 x 5 or x 7 A y x 12 5 12 7 A y x 12 7 12 5 The numbers are 5 and 7.
1 1 1 1 , so x 2 u 1 x
38. Let x represent one number. Let y represent the other number. A x y 9 y x 9
1 1 1 1 v 2 , so y v 9 3 y 1 1 1 1 1, , 1, , 1, , 1, 3 3 3 3
B xy 36 B
x 2 9 x 36 0 x 2 9 x 36 0
x 12 x 3 0 x 12 or x 3 A y x 9 12 9 3
3u v 13 3u 5u 5 13 2u 8 u4 B v 5u 5 5 4 5 25
A y x 9 3 9 12 The numbers are 3 and 12.
A
1 1 1 1 , so x 2 x u 4 2
v
1 1 1 1 , so y 2 v 25 5 y
xy 36 x x 9 36
1 1 36. Let u 2 and v 2 . x y 3 1 A 2 2 13 3u v 13 x y 5 1 B 2 2 5 5u v 5 v 5u 5 x y
u
xy 35
B
39. Let x represent one number. Let y represent the other number. A x 2 y 2 29 B x 2 y 2 21 2 x2 50 2 x 25 x 5
1 1 1 1 1 1 1 1 , , , , , , , 2 5 2 5 2 5 2 5
614
Section 5.4 Both numbers must be positive. x 5 : A x 2 y 2 29
42. Let x represent one number. Let y represent the other number. A x y 4 y x 4
52 y 2 29
B x 2 y 2 64
25 y 2 29 y2 4 y 2 The numbers are 5 and 2.
x 2 y 2 64
B
x 2 x 4 64 2
x 2 x 2 8 x 16 64 x x 8 x 16 64 8 x 80 x 10 A y x 4 10 4 6 The numbers are 10 and 6 .
40. Let x represent one number. Let y represent the other number. A x 2 y 2 145 B x 2 y 2 17 2 x2 162 2 x 81 x 9
2
2
43. Let x represent one number. Let y represent the other number. x 3 4 3y 4x y x A y 4 3
Both numbers must be negative. x 9 : A x 2 y 2 145
9 2 y 2 145
B x 2 y 2 225
81 y 145 2
2
4 x x 225 3 2
16 2 x 225 9 25 2 x 225 9 x 2 81 x 9 4 4 A y x 9 12 3 3 4 4 A y x 9 12 3 3 The numbers are 9 and 12 or 9 and 12 . x2
41. Let x represent one number. Let y represent the other number. A x y 2 x y2
B x 2 y 2 44 B
x 2 y 2 225
B
y 2 64 y 8 The numbers are 9 and 8 .
x 2 y 2 44
y 2 2 y 2 44 y 2 4 y 4 y 2 44 4 y 40 y 10 A x y 2 10 2 12 The numbers are 12 and 10.
44. Let x represent one number. Let y represent the other number. x 5 12 5 y 12 x y x A y 12 5
B x 2 y 2 676
615
Chapter 5 Systems of Equations and Inequalities x 2 y 2 676
B
A
y x 28 12 28 16
A y x 28 16 28 12 The rectangle is 16 cm by 12 cm.
2
12 x 2 x 676 5 144 2 x 676 25 169 2 x 676 25 x 2 100 x 10 12 12 A y x 10 24 5 5 12 12 A y x 10 24 5 5 The numbers are 10 and 24 or 10 and 24 . x2
47. Let x represent the length. Let y represent the width. A xy 240
B 2 x 2 y 6 58 2 x 2 y 64 x y 32 y x 32 xy 240
A
x x 32 240 x 32 x 240 0 2
45. Let x represent one dimension. Let y represent the other dimension. A 2 x 2 y 36 x y 18 y x 18
x 2 32 x 240 0
x 12 x 20 0
B xy 80
A xy 80
B
A y x 32 20 32 12 The floor is 20 ft by 12 ft.
x x 18 80 x 2 18 x 80 0
48. Let x represent the length. Let y represent the width. A xy 320
x 2 18 x 80 0
x 8 x 10 0 A
x 12 or x 20 y x 32 12 32 20
x 8 or x 10 y x 18 8 18 10
B 2 x 2 y 72 x y 36 y x 36 xy 320
A
A y x 18 10 18 8 The rectangle is 10 m by 8 m.
x x 36 320 x 36 x 320 0 2
46. Let x represent one dimension. Let y represent the other dimension. A 2 x 2 y 56 x y 28 y x 28
x 2 36 x 320 0
x 16 x 20 0
B xy 192 B
A xy 192
x 16 or x 20 y x 36 16 36 20
A y x 36 20 36 16 The sign is 20 ft by 16 ft.
x x 28 192 x 2 28 x 192 0 x 2 28 x 192 0
x 12 x 16 0 x 12 or x 16
616
Section 5.4 49. Let x represent the length. Let y represent the 288 48 ft 2 . width. The floor area is 6 48 y A xy 48 x
51. Let x represent the length. Let y represent the width. 288 A 16xy 4608 y x B 2 16 x 2 16 y xy 1440 32 x 32 y xy 1440
B x 2 y 2 102 x 2 y 2 100
32 x 32 y xy 1440 288 288 x 1440 32 x 32 x x 9216 288 1440 32 x x 9216 1152 0 32 x x 32 x 2 1152 x 9216 0 x 2 36 x 288 0 x 24 x 12 0 x 24 or x 12 288 288 A y 12 x 24 288 288 A y 24 12 x The aquarium is 24 in. by 12 in. by 16 in.
x 2 y 2 100
B
B
2
48 x 2 100 x 2304 100 x2 x 4 2304 100 x 2
x2
x 4 100 x 2 2304 0
x 36 x 64 0 2
2
x 6 or x 8 48 48 A y 8 6 x 48 48 A y 6 x 8 The truck is 6 ft by 6 ft by 8 ft. 50. Let x represent the length. Let y represent the width. 108 y A xy 108 x
52. Let x represent the length. Let y represent the width. 80 A 3xy 240 y x
B x 2 y 2 152 x 2 y 2 225
B 2 3x 2 3 y 2 xy 268
x 2 y 2 225
B
6 x 6 y 2 xy 268 3x 3 y xy 134
2
108 x2 225 x B
11,664 225 x2 x 4 11,664 225 x 2
x2
80 80 3x 3 x 134 x x 240 80 134 x 240 54 0 3x x 3x 2 54 x 240 0
x 4 225 x 2 11,664 0
3x
x 81 x 144 0 2
3 x 3 y xy 134
2
x 9 or x 12 108 108 A y 12 x 9 108 108 A y 9 x 12 The window is 9 yd by 12 yd.
x 2 18 x 80 0
x 8 x 10 0 x 8 or x 10 617
Chapter 5 Systems of Equations and Inequalities 80 80 10 x 8 80 80 A y 8 x 10 The box is 8 m by 10 m by 3 m. A
y
55. A y
B y
2
2
x 2 x 11 65 2
x 0 or x 128 3
2 x 2 22 x 56 0 x 2 11x 28 0
3 3 x 0 0 3 3 3 3 B y x 128 3 128 3 3 The ball will hit the ground at the point B y
x 4 x 7 0
x 4 or x 7 y x 11 4 11 7
221.7, 128 .
54. Let x represent the length of one side. Let y represent the length of the other side. A x y 11 y x 11
x2 3x 36 3 B y x 3
56. A y
73 x y 73 2
2
2
x 2 y 2 73
B
x x 11 73 2
A
2
x 2 x 2 22 x 121 73 2 x 2 22 x 48 0 x 2 11x 24 0
x 3 x 8 0 A
128 3, 128 or approximately
A y x 11 7 11 4 The legs are 4 ft and 7 ft.
B x2 y 2
x x 128 3 0
x 2 x 2 22 x 121 65
A
x2 3 x 192 3
x2 3 3 x x 3 192 3 x2 2 3 x0 192 3 x 2 128 3x 0
x 2 y 2 65
B
y
65 x y 65 2
3 x 3
A
53. Let x represent the length of one side. Let y represent the length of the other side. A x y 11 y x 11
B x2 y 2
x2 3 x 192 3
x 3 or x 8 y x 11 3 11 8
x2 3x 36 3 x2 x 3x 3 36 2 x 4 3 x0 36 3 x 2 48 3x 0 y
x x 48 3 0 x 0 or x 48 3
A y x 11 8 11 3 The legs are 3 in. and 8 in.
3 3 x 0 0 3 3 3 3 B y x 48 3 48 3 3 B y
618
Section 5.4 b. Q0 44.1e 4 k 44.1e 4 0.3786 201 MBq
The ball will hit the ground at the point
48 3, 48 or approximately 83.1, 48 .
c.
Q t 201e 0.3786t
57. A system of linear equations contains only linear equations, whereas a nonlinear system has one or more equations that are nonlinear.
Q 12 201e 0.378612 2.1 After 12 hr, the radioactivity level is 2.1 MBq.
58. The addition method is an efficient technique to eliminate like terms from two or more equations.
61. a. A 60,000 P0 e7 k P0 60,000e 7 k
B 80,000 P0 e12 k P0 80,000e 12 k
59. a. A t A0e kt
60,000e 7 k 80,000e 12 k
0.69 A0e 3t b.
e 7 k 80,000 e 12 k 60,000
A t A0e kt 0.655 A0e 4t
e5 k
c. A 0.69 A0 e 3t A0 0.69e3k
4 5k ln 3
B 0.655 A0 e 4t A0 0.655e 4 k
1 4 k ln 5 3 k 0.058
0.69e3k 0.655e 4 k 0.69 e 4 k 0.655 e3k 0.69 ek 0.655 0.69 ln k 0.655 0.052 k
b. P0 60,000e 7 k 60,000e 7 0.058 40,000 The original population is 40,000. c.
15 e0.058t 2 15 ln 0.058t 2
A t 0.81e 0.052t A 12 0.81e 0.05212 0.43 g/dL
60. a. A 44.1 Q0 e
4 k
Q0 44.1e
P t 40,000e0.058t 300,000 40,000e0.058t
d. A0 0.69e3k 0.69e3 0.052 0.81 g/dL e.
4 3
15 ln 2 t 0.058 35 t The population will reach 300,000 approximately 35 hr after the culture is started.
4k
B 30.2 Q0e 5 k Q0 30.2e5 k
44.1e 4 k 30.2e5 k 44.1 e5 k 30.2 e 4 k 44.1 ek 30.2 44.1 k ln 30.2 0.3786 k
619
Chapter 5 Systems of Equations and Inequalities b. P 7328.70e 2 k 7328.70e 2 0.06 6500 The original principal is 6500.
62. a. A 7328.70 Pe 2 k P 7328.70e 2 k
B 8774.10 Pe5 k P 8774.10e 5 k
c.
7328.70e 2 k 8774.10e 5 k
A t 6500e0.06t 15,000 6500e0.06t
e 2 k 8774.10 e 5 k 7328.70 8774.10 e3 k 7328.70 8774.10 3k ln 7328.70
30 e0.06t 13 30 ln 0.06t 13 30 ln 13 t 0.06 14 t The investment will be worth $15,000 in approximately 14 yr.
1 8774.10 k ln 3 7328.70 k 0.06 6%
63. A y 2 x1 log 2 y x 1 1 log 2 y x
B 1 log 2 y x The equations are the same. The system has infinitely many solutions. 64. A x 2 y 2 0 x 2 y 2 x y
B x y The equations are the same. The system has infinitely many solutions. log x 2 log y 5 65. A log x 2 log y 5 Multiply by 2. B 2 log x log y 0 4 log x 2 log y 0 5log x 5 log x 1 x 10 A
log x 2 log y 5 log10 2 log y 5 1 2 log y 5 2 log y 4 log y 2
10,100
y 102 100
620
Section 5.4 log 2 x 3log 2 y 6 66. A log 2 x 3log 2 y 6 Multiply by 1. B log 2 x log 2 y 2 log 2 x log 2 y 2 4 log 2 y 4 log 2 y 1 y 2 A log 2 x 3log 2 y 6 log 2 x 3log 2 2 6 log 2 x 3 6 log 2 x 3 x 23 8
8, 2
67. A 2 x 2 y 6 2 y 2 x 6
68. A 3x 9 y 18 3x 32 y 18
B 4 x 2 y 14 22 x 2 y 14
B 3x 3 y 30 3x 3 y 30
B
22 x 2 y 14
3x 32 y 18
A
22 x 2 x 6 14
3 y 30 32 y 18
22 x 2 x 6 14
32 y 3 y 12 0
22 x 2 x 20 0 Let u 2 x u 2 u 20 0
32 y 3 y 12 0 Let u 3 y u 2 u 12 0
u 4 u 3 0
u 5 u 4 0 u 5 or u 4
u 4 or u 3
2 x 5 or 2 x 4
3 y 4 or 3 y 3
x2 A 2 2 6
B 3 3 30
x
y 1 x
y
y
22 2 y 6
3x 31 30
4 2y 6
3x 3 30
2y 2
3x 27
y 1
x3
2,1
3,1
621
Chapter 5 Systems of Equations and Inequalities 69. A
B x 2 y 4 29
x 1 2 y 12 5
2
x 2 29 y 4
x2 2 x 1 y 2 2 y 1 5 x2 2 x y 2 2 y 3 0 x 29 y 4 : 2
x 29 y 4
2
x2 2 x y 2 2 y 3 0
A
2
29 y 4
2
2 29 y 4 y 2 y 3 0 2
2
2
29 y 4 2 29 y 4 y 2 2 y 3 0 2
2
29 y 2 8 y 16 2 29 y 4 y 2 2 y 3 0 2
2 29 y 4 6 y 10 2
29 y 4 3 y 5 2
29 y 4 3 y 5 2
2
29 y 2 8 y 16 9 y 2 30 y 25 10 y 2 22 y 12 0 5 y 2 11y 6 0
5 y 6 y 1 0 y x 29 y 4 : 2
x2 2 x y 2 2 y 3 0
A
6 or y 1 5
29 y 4
2
2 29 y 4 y 2 y 3 0 2
2
2
29 y 4 2 29 y 4 y 2 2 y 3 0 2
2
29 y 2 8 y 16 2 29 y 4 y 2 2 y 3 0 2
2 29 y 4 6 y 10 2
29 y 4 3 y 5 2
29 y 4 3 y 5 2
2
29 y 2 8 y 16 9 y 2 30 y 25 10 y 2 22 y 12 0 5 y 2 11 y 6 0
5 y 6 y 1 0 y
622
6 or y 1 5
Section 5.4 6 y : 5
B x 2 y 4 29 2
B x 2 y 4 29
y 1:
2
x 2 1 4 29
2
2
6 x 2 4 29 5
x 2 5 29 2
2
26 x 2 29 5
x 2 25 29
676 725 25 25 49 x2 25 7 x 5
x 2
x2
Check: 2,1
Check: 2,1
A
x2 4
x 1 2 y 1 2 5 2 12 1 1 2 ⱨ 5
B x 2 y 4 29
1 4ⱨ5
4 25 ⱨ 29
2
2 2 1 4 2 ⱨ 29
5 ⱨ 5 true
A
29 ⱨ 29 true
x 1 y 1 5 2 1 2 1 12 ⱨ 5 2
2
9 4ⱨ5 13ⱨ 5 false 7 6 Check: , 5 5
A
x 1 y 12 5 2
2
2
7 6 1 1 ⱨ 5 5 5 4 121 ⱨ5 25 25 125 ⱨ5 25 5 ⱨ 5 true
7 6 Check: , 5 5
A
x 12 y 12 5 2
2
7 6 1 1 ⱨ 5 5 5 144 121 ⱨ5 25 25 265 ⱨ5 25 53 ⱨ 5 false 5
7 6 2,1 , , 5 5 623
B
x 2 y 4 29 2
2
2
7 6 4 ⱨ 29 5 5 49 676 ⱨ 29 25 25 725 ⱨ 29 25 29 ⱨ 29 true
Chapter 5 Systems of Equations and Inequalities 70. A
x 32 y 22 4
B
x 12 y 2 8 y 2 8 x 1
x2 6 x 9 y 2 4 y 4 4 x2 6 x y 2 4 y 9 0 y 8 x 1 : 2
2
y 8 x 1
2
x2 6 x y 2 4 y 9 0
A
4 8 x 1 9 0 x 6 x 8 x 1 4 8 x 1 9 0 x 6 x 8 x 2 x 1 4 8 x 1 9 0
x2 6 x
8 x 1
2
2
2
2
2
2
2
2
2
8 x 16 4
8 x 1 2
2 x 4 8 x 1
2
2 x 4 2 8 x 1 2 4 x 2 16 x 16 8 x 2 2 x 1 5 x 2 14 x 9 0
5 x 9 x 1 0 x y 8 x 1 : 2
9 or x 1 5
x2 6 x y 2 4 y 9 0
A
4 8 x 1 9 0 x 6 x 8 x 1 4 8 x 1 9 0 x 6 x 8 x 2 x 1 4 8 x 1 9 0
x 2 6 x 8 x 1
2
2
2
2
2
2
2
2
2
8 x 16 4
8 x 1
2 x 4 8 x 1
2
2
2 x 4 2 8 x 1 2 4 x 2 16 x 16 8 x 2 2 x 1 5 x 2 14 x 9 0
5 x 9 x 1 0 x
624
9 or x 1 5
Section 5.4 9 x : B 5
x 1 2 y 2 8
x 1: B
x 12 y 2 8 1 12 y 2 8 22 y 2 8
2
9 2 1 y 8 5 2
14 2 y 8 5
4 y2 8 y2 4 y 2
196 200 y2 25 25 4 y2 25 2 y 5 9 2 Check: , 5 5
A
x 3 2 y 22 4 2
B
2
9 2 1 ⱨ 8 5 5
2
14 2 ⱨ 8 5 5
9 2 3 2 ⱨ 4 5 5 2
6 8 ⱨ 4 5 5 36 64 ⱨ4 25 25 100 ⱨ4 25 4 ⱨ 4 true 9 2 Check: , 5 5
A
x 12 y 2 8 2
2
2
2
196 4 ⱨ8 25 25 200 ⱨ8 25 8 ⱨ 8 true
x 3 2 y 2 2 4 2
2
9 2 3 2 ⱨ 4 5 5 2
2
6 12 ⱨ 4 5 5
Check: 1, 2
A
36 144 ⱨ4 25 25 180 ⱨ 4 false 25 x 32 y 22 4
1 32 2 22 ⱨ 4 2 2 0 2 ⱨ 4 4 ⱨ 4 true
625
B
x 1 2 y 2 8 1 1 2 2 2 ⱨ8 22 4 ⱨ8 8 ⱨ 8 true
Chapter 5 Systems of Equations and Inequalities Check: 1, 2
A
x 32 y 22 4 1 3 2 2 22 ⱨ 4 2 2 42 ⱨ 4 4 16 ⱨ 4 20 ⱨ 4 false
9 2 1, 2 , , 5 5
71. A 2 2 x
x
B 6 2 y
y
1
3
C
B 2 2 y
y
C 2x y 9
x 2 y 2 10
C
2
2
2
1
2
9
2
10
2
1
2
2x 2 y 2 9 2
1 3 10
2
2 1 2 9 8
10
2
1
2 9
10
9 9
10 2 10
2 9 2 9
2 1 1 A 2 2 x
B 6 2 y
2 1 1 A 8 4 x
B 2 2 y
2 2 1 x
6 2 1 y
8 4 1 x
2 2 1 y
2 2x
6 2y
8 4x
2 2y
1 x
3 y
2 x
1 y
2
1:
x
2
C x y 10 2
72. A 8 4 x
1: A 2 2 x
1:
B 6 2 y
1: A 8 4 x
B 2 2 y
2 2 1 x
6 2 1 y
8 4 1 x
2 2 1 y
2 2 x 1 x
6 2 y 3 y
8 4 x 2 x
2 2 y 1 y
1, 3,1 , 1, 3, 1
2,1,1 , 2, 1, 1
626
Section 5.4 74. a. The center of the circle is 0, 0 . Find the equation of the line that passes through 0, 0 and 4, 5 .
73. a. A x 2 y 2 25
x 4 y 2 25 2
B
2
The circles appear to intersect at 0, 5 and 4, 3 .
m
Check: 0, 5
y y1 m x x1
x 2 y 2 25
A
5 x 0 4 5 y x 4 Solve the system A x2 y 2 9 5 B y x 4 A x2 y 2 9
y0
0 5 ⱨ 25 2
2
0 25 ⱨ 25 25 ⱨ 25 true 2 x 4 y 22 25
B
0 4 2 5 2 2 ⱨ 25 Check: 4, 3
16 9 ⱨ 25 25 ⱨ 25 true
2
5 x 2 x 9 4
x y 25 2
A
y2 y1 5 0 5 x2 x1 4 0 4
2
4 32 ⱨ 25 2
x2
16 9 ⱨ 25 25 ⱨ 25 true 2 2 B x 4 y 2 25
4 42 3 2 2 25
0, 5 , 4, 3
0 25 ⱨ 25 25 ⱨ 25 true
12 41 41 41 41 The point on the circle closest to 4, 5 will have the positive x-coordinate. 5 5 12 41 15 41 B y x 4 4 41 41
b. Find the equation of the line that passes through 0, 5 and 4, 3 .
m
25 2 x 9 16 41 2 x 9 16 144 x2 41 144 x 41
y2 y1 3 5 8 2 40 4 x2 x1
y y1 m x x1
12
41
12 41 15 41 41 , 41
y 51 2 x 0 y 5 2x y 2x 5 The domain of the line segment is 0 x4.
b. The point on the circle furthest from 4, 5 will have the negative x-coordinate.
B y
5 5 12 41 15 41 x 4 4 41 41
12 41 15 41 41 , 41 627
Chapter 5 Systems of Equations and Inequalities 75. 2.359, 5.584
78. { 4.054, 3.946 , 4.054, 3.946 ,
2.237, 5.196 , 2.237, 5.196}
76. 3.805,1.336
79.
80.
77. {1.538, 6.135 , 1.538, 6.135 ,
3.693, 5.135 , 3.693, 5.135}
Section 5.5 Inequalities and Systems of Inequalities in Two Variables 1. 3x y 4
2. x 2 y 2
y 3x 4
2 y x 2 1 y x 1 2
628
Section 5.5 3. y x 2 4
7. linear
8. left; vertical
9. above; horizontal
10. solid
11. II
12
0, 0 ; 2; inside
3 x 4 y 12
13. a.
3 1 4 3 12 ? ?
3 12 12 ?
9 12 Yes 4. y x 2 4
b.
3x 4 y 12 3 5 4 1 12 ? ?
15 4 12 ?
19 12 No c.
3 x 4 y 12 3 4 4 0 12 ? ?
12 0 12 ?
12 12 No 5. 3x y 4
2x 3y 6
14. a.
2 3 3 3 6
y 3x 4 x 2y 2
? ?
6 9 6
x 2 3x 4 2
?
3 6 No
x 6x 8 2 5 x 10
2x 3y 6
b.
2 5 3 1 6 ?
x 2 y 3x 4 3 2 4 6 4 2
?
10 3 6
2, 2 6.
?
7 6 Yes c.
y x2 4
2x 3y 6 2 0 3 2 6 ?
x2 4 x2 4
?
06 6
2 x 8 2
?
6 6 No
x2 4 x 2
15. a.
y x 2 4 2 4 4 4 0 2
y x 3
2
30 3 3 ?
y x 2 4 2 4 4 4 0
30 6
2
?
2, 0 , 2, 0
?
2
30 36 No
629
2
Chapter 5 Systems of Equations and Inequalities b. y x 3
4 1 3 ?
4 2 ?
2
2
2
?
4 4 Yes c. y x 3
5 5 3 ?
5 2 ?
2
2
2
5 4 Yes
b. The bounding line would be drawn as a dashed line.
y x3 1
c. The bounding line would be dashed and the graph would be shaded strictly below the line.
?
16. a.
2 1 1 ?
3
?
2 1 1
18. a. 2 x 5 y 10 2 x 5 y 10 related
?
2 2 Yes
equation
x-intercept: 2 x 5 0 10
b. y x3 1
6 2 1 ?
3
5 y 10 2 x 10 y2 x5 The inequality is strict, so draw the line as a dashed line. Test (0, 0): 2 x 5 y 10
?
6 8 1 ?
6 7 Yes c.
y x3 1 50 4 1 ?
y-intercept: 2 0 5 y 10
2 0 5 0 10
3
?
?
50 64 1
?
0 10 false The points on the side of the line not containing (0, 0) are solutions.
?
50 65 No 17. a. 4 x 5 y 20 4 x 5 y 20 related equation
x-intercept: 4 x 5 0 20
y-intercept: 4 0 5 y 20
4 x 20 5 y 20 x5 y 4 The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): 4 x 5 y 20 4 0 5 0 20 ?
b. The bounding line would be drawn as a solid line.
?
0 0 true The points on the side of the line containing (0, 0) are solutions.
c. The bounding line would be dashed and the graph would be shaded strictly below the line.
630
Section 5.5 19. 2 x 5 y 5 2x 5 y 5 related equation
x-intercept: 2 x 5 0 5
y-intercept: 2 0 5 y 5
2x 5 5y 5 5 y 1 x 2 The inequality is strict, so draw the line as a dashed line. Test (0, 0): 2x 5 y 5
21. 30 x 20 y 600 30 x 20 y 600 related
2 0 5 0 5 ?
equation
?
30 x 20 y 600 30 x 20 y 600 3 x 2 y 60 y-intercept: x-intercept: 3x 2 0 60 3 0 2 y 60
0 5 false The points on the side of the line not containing (0, 0) are solutions.
2 y 60 3 x 60 y 30 x 20 The inequality symbol allows for equality, so draw the line as a solid line. 30 x 20 y 600 Test (0, 0): 30 0 20 0 600 ? ?
0 600 false The points on the side of the line not containing (0, 0) are solutions.
20. 5 x 4 y 8 5 x 4 y 8 related equation
x-intercept: 5 x 4 0 8
y-intercept: 5 0 4 y 8
5 x 8
4y 8 y2
8 5 The inequality symbol allows for equality, so draw the line as a solid line. 5 x 4 y 8 Test (0, 0): x
5 0 4 0 8 ? ?
0 8 true The points on the side of the line containing (0, 0) are solutions.
631
Chapter 5 Systems of Equations and Inequalities 22. 400 x 100 y 8000
400 x 100 y 8000 related equation
400 x 100 y 8000 400 x 100 y 8000 4 x y 80 y-intercept: x-intercept: 4 x 0 80 4 0 y 80
y 80 4 x 80 x 20 The inequality is strict, so draw the line as a dashed line. Test (0, 0): 400 x 100 y 8000
24. 3x 2 y 3x 2 y related equation
3x 2 y 3 y x0 2 The inequality is strict, so draw the line as a dashed line. Test (1, 1): 3x 2 y
400 0 100 0 8000 ? ?
0 8000 true The points on the side of the line containing (0, 0) are solutions.
3 1 2 1 ? ?
3 2 true The points on the side of the line containing (1, 1) are solutions.
23. 5 x 6 y 5x 6 y related equation
5x 6 y 5 y x0 6 The inequality symbol allows for equality, so draw the line as a solid line. Test (1, 1): 5x 6 y
25. 3 2 x y y 3
3 2 x y y 3 related equation
3 2 x y y 3
5 1 6 1 ?
2x 2 y y y 2 x 0 The inequality is strict, so draw the line as a dashed line. Test (1, 1): 3 2 x y y 3
?
5 6 true The points on the side of the line containing (1, 1) are solutions.
3 2 1 1 1 3 ? ?
3 4 4 ?
7 4 true 632
Section 5.5 The points on the side of the line containing (1, 1) are solutions.
28. y 5 The related equation y 5 is a horizontal line. The inequality y 5 represents all points below or on the line y 5 . The inequality symbol allows for equality, so draw the line as a solid line.
26. 4 3 x y 2 y 4 4 3 x y 2 y 4 related equation
4 3 x y 2 y 4 3x 3 y 2 y
y 3x 0 The inequality is strict, so draw the line as a dashed line. Test (1, 1): 4 3 x y 2 y 4 4 3 1 1 2 1 4 ? ?
4 2 true The points on the side of the line containing (1, 1) are solutions.
1 y45 2 1 y 1 2 y 2 The related equation y 2 is a horizontal line. The inequality y 2 represents all points above or on the line y 2 . The inequality symbol allows for equality, so draw the line as a solid line.
29.
27. x 6 The related equation x 6 is a vertical line. The inequality x 6 represents all points strictly to the left of the line x 6 . The inequality is strict, so draw the line as a dashed line.
633
Chapter 5 Systems of Equations and Inequalities c. The shaded region would contain points on the circle (solid curve) and points outside the circle.
1 30. x 2 4 3 1 x2 3 x 6 The related equation x 6 is a vertical line. The inequality x 6 represents all points strictly to the right of the line x 6 . The inequality is strict, so draw the line as a dashed line.
32. a. y x 2 1 y x2 1 related equation
The equation represents a parabola opening upward with vertex 0, 1 . The inequality symbol allows for equality, so draw the parabola as a solid curve. Test (0, 0): y x 2 1 0 0 1 ?
2
?
0 1 true The test point satisfies the inequality. The solution set consists of the points above or on the parabola.
31. a. x 2 y 2 4 x2 y 2 4 related equation
The equation represents a circle centered at 0, 0 with radius 2. The inequality is strict, so draw the circle as a dashed curve. Test (0, 0): x2 y 2 4
b. The region on and below the parabola would be shaded.
0 2 0 2 4 ?
c. The region strictly above the parabola would be shaded (the parabola would be drawn as a dashed curve).
?
0 4 true The test point inside the circle satisfies the inequality. The solution set consists of the points strictly inside the circle.
y x2 33. y x 2 related equation
The equation represents a parabola opening downward with vertex 0, 0 . The inequality is strict, so draw the parabola as a dashed curve. y x2 Test (0, 1): 1 0 ? ?
2
1 0 false The test point does not satisfy the inequality. The solution set consists of the points below the parabola.
b. The region outside the circle would be shaded.
634
Section 5.5
y x 1 2 36. y x 1 2 related
34. x 2 y 2 16 x 2 y 2 16 related
2
equation
equation
The equation represents a circle centered at 0, 0 with radius 4. The inequality symbol allows for equality, so draw the circle as a solid curve. Test (0, 0): x 2 y 2 16
The equation represents a parabola opening downward with vertex 1, 2 . The inequality symbol allows for equality, so draw the parabola as a solid curve. 2 Test (0, 0): y x 1 2
02 0 2 16 ?
0 0 1 2 ?
?
0 16 false The test point does not satisfy the inequality. The solution set consists of the points outside or on the circle.
2
2
?
0 3 true The test point satisfies the inequality. The solution set consists of the points above or on the parabola.
35. y x 2 1 y x 2 1 related 2
37. x 3 x 3 related
equation
equation
The equation represents a parabola opening upward with vertex 2,1 . The inequality symbol allows for equality, so draw the parabola as a solid curve. 2 Test (0, 0): y x 2 1
The related equation x 3 represents two vertical lines, x 3 and x 3 . The inequality symbol allows for equality, so draw the lines as solid lines. Test (0, 0): x 3
0 0 2 1 ?
2
2
?
0 3
?
0 5 true The test point satisfies the inequality. The solution set consists of the points below or on the parabola.
?
0 3 true The test point satisfies the inequality. The solution set consists of all points between or on the lines x 3 and x 3 .
635
Chapter 5 Systems of Equations and Inequalities The test point does not satisfy the inequality. The solution set consists of all points strictly outside the lines y 1 and y 1 .
38. y 2 y 2 related equation
The related equation y 2 represents two horizontal lines, y 2 and y 2 . The inequality symbol allows for equality, so draw the lines as solid lines. Test (0, 0): y 2
40. x 1 3 x 1 3 related equation
x 1 3 x 2
?
0 2
The related equation x 2 represents two vertical lines, x 2 and x 2 . The inequality is strict, so draw the lines as dashed lines. Test (0, 0): x 1 3
?
0 2 true The test point satisfies the inequality. The solution set consists of all points between or on the lines y 2 and y 2 .
?
0 1 3 ?
1 3 false The test point does not satisfy the inequality. The solution set consists of all points outside the lines x 2 and x 2 .
39. 2 y 2 2 y 2 related equation
2 y 2 y 1 The related equation y 1 represents two horizontal lines, y 1 and y 1 . The inequality is strict, so draw the lines as dashed lines. Test (0, 0): 2 y 2
41. y x y x related equation
The equation represents the parent square root function. The inequality symbol allows for equality, so draw the function as a solid curve.
?
20 2 ?
0 2 false
636
Section 5.5 Test (1, 2):
43. a. y 2 x 3
y x ?
2 1
1 2 2 3
?
1 7 Yes
? ?
2 1 true The test point satisfies the inequality. Since the domain of the function is 0, , the yaxis is also a boundary of the solution set. The solution set consists of the points that are above or on the curve, and to the right of or on the y-axis.
b. x y 1 ?
2 1 1 ?
3 1 No c. No. Since the point 2,1 is not a solution to the second inequality, it is not a solution to the system of inequalities. 44. a. y x 5 ?
2 3 5 ?
2 2 No b.
3x y 11 3 3 2 11 ? ?
11 11 Yes 42. y x 1 y x 1 related
c. No. Since the point 3, 2 is not a solution to the first inequality, it is not a solution to the system of inequalities.
equation
The equation represents the parent square root function, shifted to the right 1 unit. The inequality is strict, so draw the function as a dashed curve. Test (1, 2): y x 1
45. a. x y 4
1 2 0 1
?
1 1 true
0 1 4 1 4 true
?
2 11
y 2x 1
?
? ?
Yes
?
2 0 false The test point does not satisfy the inequality. Since the domain of the function is 1, , the line x 1 is also a boundary of the solution set. The solution set consists of the points that are strictly below the curve, and to the right of or on the line x 1 .
b. x y 4 ?
3 1 4 ?
4 4 false No c. x y 4
0 2 2 1
?
0 5 true
20 4 2 4 true Yes d. x y 4 ?
1 4 4 ?
5 4 false No
637
y 2x 1
?
? ?
Chapter 5 Systems of Equations and Inequalities y x2 3
46. a.
Find the point of intersection. 1 x 4 2 x 1 2 1 x 2 x 5 2 x 4 x 10
1 2 3 ?
2
?
1 1 false No y x2 3
x 2y 2
2 0 3
0 2 2 2
b.
? ?
?
2 3 true Yes
4 2 true x 2y 2
c. y x 2 3
0 2 1 2
1 0 3 ?
5 x 10 x2 1 1 y x 4 2 4 3 2 2 Graph 2, 3 as an open dot. It is not a solution to either inequality.
?
2
?
2
?
?
2 2 true
1 3 true Yes y x2 3
d.
6 3 3 ?
2
?
6 6 false No 1 1 x 4 y x4 related 2 2 equation The inequality is strict, so draw the line as a dashed line. 1 Test (0, 0): y x4 2 ? 1 0 0 4 2
47. y
1 1 x 2 y x2 related 3 3 equation The inequality symbol allows for equality, so draw the line as a solid line. 1 Test (0, 0): y x2 3 ? 1 0 0 2 3
48. y
?
0 4 false The points on the side of the line not containing (0, 0) are solutions. y 2 x 1 y 2 x 1 related
?
0 2 true The points on the side of the line containing (0, 0) are solutions. y x 4 y x4 related
equation
The inequality is strict, so draw the line as a dashed line. y 2 x 1 Test (0, 0):
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): y x4
0 2 0 1 ? ?
0 1 false The points on the side of the line not containing (0, 0) are solutions.
?
0 0 4 ?
0 4 false The points on the side of the line not containing (0, 0) are solutions.
638
Section 5.5 Find the point of intersection. 1 x2 x4 3 1 x x2 3 x 3x 6
3x 4 y 4
Test (0, 0):
3 0 4 0 4 ? ?
0 4 false The points on the side of the line not containing (0, 0) are solutions. Find the point of intersection. 3 2 x 5 y 5 6 x 15 y 15 2 3 x 4 y 4 6 x 8 y 8 23 y 23 y 1 2x 5 y 5
2 x 6 x3 1 1 y x 2 3 2 1 3 3 Graph 3, 1 as a closed dot. It is a solution to both inequalities.
2 x 5 1 5 2x 0 x0 Graph 0,1 as a closed dot. It is a solution to both inequalities.
49. 2 x 5 y 5 2x 5 y 5 related equation
x-intercept: 2 x 5 0 5
y-intercept: 2 0 5 y 5
2x 5
5y 5 50. 4 x 3 y 3 4x 3y 3 related
5 y 1 2 The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): 2x 5 y 5 x
equation
2 0 5 0 5 ?
4y 4
5y 5
4 0 3 0 3 ?
equation
3x 4
4x 3
3 y 1 4 The inequality is strict, so draw the line as a dashed line. Test (0, 0): 4x 3y 3
0 5 true The points on the side of the line containing (0, 0) are solutions. 3 x 4 y 4 3 x 4 y 4 related y-intercept: 3 0 4 y 4
y-intercept: 2 0 5 y 5
x
?
x-intercept: 3 x 4 0 4
x-intercept: 4 x 3 0 3
?
0 3 false The points on the side of the line not containing (0, 0) are solutions. x 4 y 4 x 4 y 4 related
4 y 1 3 The inequality symbol allows for equality, so draw the line as a solid line. x
equation
639
Chapter 5 Systems of Equations and Inequalities x-intercept: x 4 0 4
y-intercept: 0 4 y 4
x 4
4 y 4
x 2 y 2 16 x 2 y 2 16 related equation
The equation represents a circle centered at 0, 0 with radius 4. The inequality symbol allows for equality, so draw the circle as a solid curve. x 2 y 2 16 Test (0, 0):
y 1 The inequality is strict, so draw the line as a dashed line. x 4 y 4 Test (0, 0):
02 02 16 ?
0 4 0 4 ?
?
0 16 true The test point satisfies the inequality. The solution set consists of the points inside or on the circle. The graph shows that the circles do not intersect.
?
0 4 false The points on the side of the line not containing (0, 0) are solutions. Find the point of intersection. 4x 3y 3 4x 3y 3 4 x 4 y 4 4 x 16 y 16 19 y 19 y 1 x 4 y 4
x 4 1 4 x0 Graph 0, 1 as an open dot. It is not a solution to either inequality.
52. x 2 y 2 1 x2 y 2 1 related equation
The equation represents a circle centered at 0, 0 with radius 1. The inequality symbol allows for equality, so draw the circle as a solid curve. x2 y 2 1 Test (0, 0):
0 2 0 2 1 ?
51. x 2 y 2 9 x2 y 2 9 related
?
0 1 false The test point does not satisfy the inequality. The solution set consists of the points outside or on the circle. x 2 y 2 25 x 2 y 2 25 related
equation
The equation represents a circle centered at 0, 0 with radius 3. The inequality symbol allows for equality, so draw the circle as a solid curve. Test (0, 0): x2 y 2 9
equation
The equation represents a circle centered at 0, 0 with radius 5. The inequality is strict, so draw the circle as a dashed curve. x 2 y 2 25 Test (0, 0):
0 0 9 2
2 ?
?
0 9 false The test point does not satisfy the inequality. The solution set consists of the points outside or on the circle.
02 0 2 25 ? ?
0 25 true
640
Section 5.5 54. y 2 x 4 y 2x 4 related
The test point satisfies the inequality. The solution set consists of the points inside the circle. The graph shows that the circles do not intersect.
equation
The inequality is strict, so draw the line as a dashed line. Test (0, 0): y 2x 4 0 2 0 4 ? ?
0 4 false The points on the side of the line containing (0, 0) are solutions. 2 x y 2 y 2x 2 related equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): 2 x y 2 2 0 0 2 ?
53. y 3x 3 y 3x 3 related
?
0 2 false The points on the side of the line containing (0, 0) are not solutions. Find the point of intersection. 2x 4 2x 2 4 2 The lines do not intersect; they are parallel. The solutions to the first inequality are below the line, and the solutions to the second inequality are above the line. The solution sets do not overlap. The solution set of the system is the empty set, .
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): y 3x 3 0 3 0 3 ? ?
0 3 false The points on the side of the line not containing (0, 0) are solutions. 3 x y 1 y 3x 1 related equation
The inequality is strict, so draw the line as a dashed line. Test (0, 0): 3 x y 1
55. x 3 x 3 related
3 0 0 1 ?
equation
The related equation x 3 represents two vertical lines, x 3 and x 3 . The inequality is strict, so draw the lines as dashed lines. Test (0, 0): x 3
?
0 1 true The points on the side of the line containing (0, 0) are solutions. Find the point of intersection. 3x 3 3x 1 31 The lines do not intersect; they are parallel. The solutions to the first inequality are above the line, and the solutions to the second inequality are below the line. The solution sets do not overlap. The solution set of the system is the empty set, .
?
0 3 ?
0 3 true The test point satisfies the inequality. The solution set consists of all points between the lines x 3 and x 3 . y 3 y 3 related equation
The related equation y 3 represents two horizontal lines, y 3 and y 3 . The inequality is strict, so draw the lines as dashed lines. 641
Chapter 5 Systems of Equations and Inequalities Test (0, 0):
The lines intersect at 2, 2 , 2, 2 ,
y 3 ?
2, 2 , and 2, 2 . These points are
?
solutions to the inequalities, so graph them with closed dots.
0 3 0 3 true The test point satisfies the inequality. The solution set consists of all points between the lines y 3 and y 3 . The lines intersect at 3, 3 , 3, 3 , 3, 3 ,
and 3, 3 . These points are not solutions to the strict inequalities, so graph them with open dots.
57. y x 2 2 y x2 2 related equation
The equation represents a parabola opening upward with vertex 0, 2 . The inequality symbol allows for equality, so draw the parabola as a solid curve. Test (0, 0): y x2 2
x 2 56. x 2 related
0 0 2 ?
equation
2
?
0 2 true The test point satisfies the inequality. The solution set consists of the points above or on the parabola. y x yx related
The related equation x 2 represents two vertical lines, x 2 and x 2 . The inequality symbol allows for equality, so draw the lines as solid lines. Test (0, 0): x 2
equation
?
0 2
The inequality is strict, so draw the line as a dashed line. yx Test (1, 2):
?
0 2 false The test point does not satisfy the inequality. The solution set consists of all points outside or on the lines x 2 and x 2 . y 2 y 2 related
?
2 1 true The points on the side of the line containing (1, 2) are solutions. y4 The related equation y 4 is a horizontal line. The inequality y 4 represents all points below or on the line y 4 . The inequality symbol allows for equality, so draw the line as a solid line. Find the points of intersection between y x 2 2 and y x :
equation
The related equation y 2 represents two horizontal lines, y 2 and y 2 . The inequality symbol allows for equality, so draw the lines as solid lines. y 2 Test (0, 0): ?
0 2 ?
0 2 false The test point does not satisfy the inequality. The solution set consists of all points outside or on the lines y 2 and y 2 .
x2 2 x x2 x 2 0
x 1 x 2 0 642
Section 5.5 x 1 or x 2 y x 1 yx2
Test (0, 0):
The solutions are 1, 1 and 2, 2 . They do not satisfy the strict inequalities, so plot them as open dots. Find the points of intersection between y x 2 2 and y 4 :
0 5 true The points on the side of the line containing (0, 0) are solutions. y 1 The related equation y 1 is a horizontal line. The inequality y 1 represents all points above the line y 1 . The inequality is strict, so draw the line as a dashed line. Find the points of intersection between y x 2 7 and y x 5 :
0 0 5 ? ?
x2 2 4 x2 6 0 x2 6 x 6 y4 The solutions are
y x 5
x2 7 x 5
6, 4 and 6, 4 .
x2 x 2 0
x2 x 2 0
They satisfy both inequalities, so plot them as closed dots.
x 1 x 2 0 x 1 or x 2 y x 5 1 5 6 y x 5 2 5 3 The solutions are 1, 6 and 2, 3 . They satisfy the inequalities, so plot them as closed dots. Find the points of intersection between y x 2 7 and y 1 : x2 7 1
58. y x 2 7 y x2 7 related
x 2 6
equation
x2 6
The equation represents a parabola opening downward with vertex 0, 7 . The inequality symbol allows for equality, so draw the parabola as a solid curve. Test (0, 0): y x2 7
y 1
The solutions are
6,1 and 6,1 .
They do not satisfy the inequalities, so plot them as open dots.
0 0 7 ?
x 6
2
?
0 7 true The test point satisfies the inequality. The solution set consists of the points below or on the parabola. y x 5 y x 5 related equation
The inequality symbol allows for equality, so draw the line as a solid line.
643
Chapter 5 Systems of Equations and Inequalities 59. x 2 y 2 100 x 2 y 2 100 related
4 4 x 6 8 3 3 The solutions are 6, 8 and 6, 8 . They do not satisfy the inequalities, so plot them as open dots. Find the points of intersection between x 2 y 2 100 and x 8 :
y
equation
The equation represents a circle centered at 0, 0 with radius 10. The inequality symbol allows for equality, so draw the circle as a solid curve. Test (0, 0): x 2 y 2 100
02 02 100 ?
x 2 y 2 100
?
8 2 y 2 100
0 100 true The test point satisfies the inequality. The solution set consists of the points inside or on the circle. 4 4 y x y x related 3 3 equation The inequality is strict, so draw the line as a dashed line. 4 y x Test (1, 0): 3 ? 4 0 1 3 ? 4 0 true 3 The points on the side of the line containing (1, 0) are solutions. x8 The related equation x 8 is a vertical line. The inequality x 8 represents all points above or on the line x 8 . The inequality symbol allows for equality, so draw the line as a solid line. Find the points of intersection between 4 x 2 y 2 100 and y x : 3 x 2 y 2 100
y 2 36 y 6 x8 The solutions are 8, 6 and 8, 6 . They satisfy the inequalities, so plot them as closed dots.
60. x 2 y 2 100 x 2 y 2 100 related equation
The equation represents a circle centered at 0, 0 with radius 10. The inequality is strict, so draw the circle as a dashed curve. Test (0, 0): x 2 y 2 100
02 02 100 ? ?
2
0 100 true The test point satisfies the inequality. The solution set consists of the points inside the circle. y x yx related
4 x 2 x 100 3 16 2 x 100 9 25 2 x 100 9 x 2 36 x 6 4 4 y x 6 8 3 3
x2
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (1, 0): yx ?
0 1 false The points on the side of the line not containing (1, 0) are solutions. 644
Section 5.5 y 1 The related equation y 1 is a horizontal line. The inequality y 1 represents all points above or on the line y 1 . The inequality symbol allows for equality, so draw the line as a solid line. Find the points of intersection between x 2 y 2 100 and y x : x 2 y 2 100 x 2 x 2 100
61. y e x y ex related
2 x 100 2
equation
The equation represents the parent natural exponential function. The inequality is strict, so draw the function as a dashed curve. Test (0, 0): y ex
x 2 50 x 5 2 y x5 2
?
0 e0
y x 5 2 The solutions are 5 2, 5 2 and 5 2, 5 2 . They do
?
0 1 true The test point satisfies the inequality. The solution set consists of the points below the function. y 1 y 1 related
not satisfy the inequalities, so plot them as open dots. Find the points of intersection between x 2 y 2 100 and y 1 :
equation
The related equation y 1 is a horizontal line. The inequality y 1 represents all points above the line y 1 . The inequality is strict, so draw the line as a dashed line. x2 The related equation x 2 is a vertical line. The inequality x 2 represents all points to the left of the line x 2 . The inequality is strict, so draw the line as a dashed line. Find the point of intersection between y e x and y 1 :
x y 100 2
2
x 2 1 100 2
x 2 99 x8
x 3 11
The solutions are 3 11,1 and 3 11, 1 . They do not satisfy the inequalities, so plot them as open dots. Find the point of intersection between y x and y 1 : yx y 1 x 1 The solution is 1,1 . It satisfies the inequalities, so plot it as a closed dot.
y ex 1 ex x0 The solution is 0,1 . It does not satisfy the inequalities, so plot it as an open dot. Find the points of intersection between y e x and x 2 : y ex y e 2 7.4 The solution is 2, 7.4 . It does not satisfy the inequalities, so plot it as an open dot.
645
Chapter 5 Systems of Equations and Inequalities The point of intersection between y 1 and
Find the point of intersection between y
x 2 is 2,1 . It does not satisfy the inequalities, so plot it as an open dot.
and y x : 2 y x 2 x x 2 x 2
2 x
x 2 The solutions are
2, 2 and 2, 2 . The do not
satisfy the inequalities, so plot them as open dots. Find the points of intersection between 2 y and y 0 : x 2 y x 2 0 no solution x Find the points of intersection between y x and y 0 : yx x0 y0 The solution is 0, 0 . It does not satisfy the inequalities, so plot it as an open dot.
2 2 y related x equation x The equation represents a reciprocal function. The inequality symbol allows for equality, so draw the function as a dashed curve. 2 1 y Test 2, : 2 x 1 ? 2 2 2 1 ? 1 true 2 The test point satisfies the inequality. The solution set consists of the points below or on the function. y 0 y0 related
62. y
equation
The related equation y 0 is a horizontal line. The inequality y 0 represents all points above the line y 0 . The inequality is strict, so draw the line as a dashed line. y x yx related equation
The inequality is strict, so draw the line as a dashed line. yx Test (2, 1): ?
63. x 2 y 3 9
1 2 true The points on the side of the line containing (2, 1) are solutions.
2
2
x 2 y 3 9 related 2
2
equation
The equation represents a circle centered at 2, 3 with radius 3. The inequality symbol allows for equality, so draw the circle as a solid curve.
646
Section 5.5
x 22 y 32 9 ? 0 22 0 32 9
Test (0, 0):
The test point satisfies the inequality. The solution set consists of the points inside the circle. 2 x y 4 y 2x 4 related
?
49 9
equation
?
13 9 false The test point does not satisfy the inequality. The solution set consists of the points inside or on the circle. x y 2 y x2 related
The inequality is strict, so draw the line as a dashed line. 2 x y 4 Test (0, 0): 2 0 0 4 ? ?
2 4 false The points on the side of the line not containing (0, 0) are solutions. Find the points of intersection.
equation
The inequality is strict, so draw the line as a dashed line. Test (0, 0): x y2 ?
00 2 ?
0 2 false The points on the side of the line not containing (0, 0) are solutions. Find the points of intersection.
The line does not intersect the circle. The solutions to the first inequality are inside the circle, and the solutions to the second inequality are above the line and the circle. The solution sets do not overlap. The solution set of the system is the empty set, .
The line does not intersect the circle. The solutions to the first inequality are inside the circle, and the solutions to the second inequality are below the line and the circle. The solution sets do not overlap. The solution set of the system is the empty set, . 64. x 4 y 1 25 2
2
x 4 y 1 25 related 2
2
equation
The equation represents a circle centered at 4, 1 with radius 5. The inequality is strict, so draw the circle as a dashed curve. Test (0, 0):
65. x 6
66. y 7
67. y 2
68. x
69. x y 18
70. x y 4
71. a. x y 9
b. x 3
c. y 4
d. x 0
1 2
e. y 0 f. The inequalities x 0 and y 0 together represent the set of points in the first quadrant, including the bounding axes. x y 9 x y 9 related
x 4 2 y 1 2 25 ? 0 42 0 1 2 25
equation
?
16 1 25 ?
17 25 true 647
Chapter 5 Systems of Equations and Inequalities Test (0, 0):
x y9 ?
00 9 ?
0 9 true The inequality x y 9 represents the set of points on and below the line x y 9 . In addition, the inequality x 3 represents the set of points on or to the right of the line x 3 , and the inequality y 4 represents the set of points on or below the line y 4 . 73. a. x y 60,000
b. y 2 x d. y 0
c. x 0
e. The inequalities x 0 and y 0 together represent the set of points in the first quadrant, including the bounding axes. x y 60,000 x y 60,000 related equation
Test (0, 0):
x y 60,000 ?
0 0 60,000 ?
72. a. x 6
0 60,000 true The inequality x y 60,000 represents the set of points on and below the line x y 60,000 . y 2 x y 2x related
b. y 10
c. x y 20
d. x 0
e. y 0
equation
f. The inequalities x 0 and y 0 together represent the set of points in the first quadrant, including the bounding axes. x y 20 x y 20 related
Test (5000, 20,000): y 2x 20,000 2 5000 ? ?
20,000 10,000 true The inequality y 2 x represents the set of points on and above the line y 2 x .
equation
Test (0, 0):
x y 20 ?
0 0 20 ?
0 20 true The inequality x y 20 represents the set of points on and below the line x y 20 . In addition, the inequality x 6 represents the set of points on or to the right of the line x 6 , and the inequality y 10 represents the set of points on or below the line y 10 .
648
Section 5.5 74. a. x y 2000
b. x 3 y
c. x 0
d. y 0
77. Find the equations of the three sides of the triangle. 2 4 6 1 Line 1: m 3 3 6
e. The inequalities x 0 and y 0 together represent the set of points in the first quadrant, including the bounding axes. x y 2000 x y 2000 related
y 2 1 x 3 y 2 x3 y x 1
equation
Test (0, 0):
x y 2000
Line 2: m
?
0 0 2000 ?
0 2000 true The inequality x y 2000 represents the set of points on and below the line x y 2000 . x 3 y x 3y related
1 y 4 x 5 4 1 5 y4 x 4 4 1 11 y x 4 4 4 4 8 4 Line 3: m 3 5 2
equation
Test (1000, 100):
42 2 1 4 5 3 8
x 3y 1000 3 100 ? ?
1000 300 true The inequality x 3 y represents the set of points on and below the line x 3 y .
y 4 4 x 3 y 4 4 x 12 y 4 x 16 Graph the equations to determine the inequalities.
Line 1 is on the right. Line 2 is on the top. Line 3 is on the left. 1 11 y x 1 , y x , y 4 x 16 4 4
75. These points are the Quadrant I part of a circle with vertex at the origin and radius 3: x2 y 2 9 , x 0 , y 0 76. These points are the Quadrant II part of a circle with vertex at the origin and radius 4: x 2 y 2 16 , x 0 , y 0
78. Find the equations of the three sides of the triangle. 1 4 5 Line 1: m 1 1 4 5
y 1 1 x 1 y 1 x 1 yx
649
Chapter 5 Systems of Equations and Inequalities Line 2: m
1 1 2 1 5 1 4 2
b. The distance from the tower to the resident is
d
1 x 1 2 1 1 y 1 x 2 2 1 3 y x 2 2 4 1 3 1 Line 3: m 4 5 9 3 1 y 1 x 5 3 1 5 y 1 x 3 3 1 8 y x 3 3 Graph the equations to determine the inequalities. y 1
4 52 52 81 25 106
No, the resident is 106 10.3 mi from the tower. 81. If the inequality is strict—that is, posed with < or >—then the bounding line or curve should be dashed. 82. After graphing the bounding line or curve to the solution set, select a test point from a given region of the plane. If the ordered pair is a solution to the inequality, then all other points in that region are also solutions. 83. Find the solution set to each individual inequality in the system. Then to find the solution set for the system of inequalities, take the intersection of the solution sets to the individual inequalities. 84. The solution set is a unit square (a square with sides 1 unit in length) in the first quadrant with one vertex at the origin. 85. x y x y related equation
Graph the equivalent form: x y and x y . The inequality symbol allows for equality, so draw the lines as solid lines. Test points in each region. Test (1, 0): x y
Line 1 is on the left. Line 2 is on the right. Line 3 is on the bottom. 1 3 1 8 y x, y x , y x 2 2 3 3 79. a. This is a circle with vertex 12, 9 and radius 16.
?
1 0
x 122 y 92 256
?
1 0 true The test point satisfies the inequality. The points on the side of the lines containing (1, 0) are solutions. Test (0, 1): x y
b. The distance from Hawthorne to the point of the earthquake is
d 92 122 81 144 225 15 km Yes, the earthquake could be felt at Hawthorne, which is 15 km from the earthquake.
?
01 ?
0 1 false The test point does not satisfy the inequality. The points on the side of the lines not containing (0, 1) are solutions.
80. a. This is a circle with vertex 4, 5 and radius 8.
x 42 y 52 64
650
Section 5.5 Test 1, 0 :
x y
1 1 Test , : 4 4
x y 1 1 1 ? 1 4 4 1 ? 1 true 2 The test point satisfies the inequality. The points on the side of the line containing 1 1 , are solutions. 4 4
?
1 0 ?
1 0 true The test point satisfies the inequality. The points on the side of the lines containing 1, 0 are solutions. Test 0, 1
x y ?
0 1 ?
0 1 false The test point does not satisfy the inequality. The points on the side of the lines not containing 0, 1 are solutions.
1 1 Test , : 4 4
x y 1 1 1 ? 1 4 4 1 ? 1 true 2 The test point satisfies the inequality. The points on the side of the line containing 1 1 , are solutions. 4 4 1 1 Test , : 4 4 x y 1 1 1 ? 1 4 4 1 ? 1 true 2 The test point satisfies the inequality. The points on the side of the line containing 1 1 , are solutions. 4 4
x y 1 86. x y 1 related equation
Graph the equivalent forms: x y 1 if x 0 and y 0 x y 1 if x 0 and y 0 x y 1 if x 0 and y 0 x y 1 if x 0 and y 0 The inequality symbol allows for equality, so draw the lines as solid lines. Test points in each form. 1 1 x y 1 Test , : 4 4 1 1 ? 1 4 4 1 ? 1 true 2 The test point satisfies the inequality. The points on the side of the line containing 1 1 , are solutions. 4 4
651
Chapter 5 Systems of Equations and Inequalities 87.
88.
Problem Recognition Exercises: Equations and Inequalities in Two Variables The points on the side of the line not containing (0, 0) are solutions. 2 x y 0 y 2x related
1. a.
equation
The inequality is strict, so draw the line as a dashed line. Test (1, 1): 2 x y 0 2 1 1 0 ? ?
1 0 true The points on the side of the line containing (1, 1) are solutions. From part (c), the point of intersection is (1, 2). Graph (1, 2) as an open dot. It is not a solution to either inequality.
b.
2 x y 0
c.
2 x 3 x 5 0 2 x 3 x 5 0 5 x 5
e. y 3 x 5 y 3x 5 related
x 1 y 3x 5 3 1 5 3 5 2
equation
The inequality is strict, so draw the line as a dashed line. Test (0, 0): y 3 x 5
1, 2
d. y 3 x 5 y 3x 5 related
0 3 0 5 ?
equation
The inequality is strict, so draw the line as a dashed line. Test (0, 0): y 3 x 5
?
0 2 true The points on the side of the line containing (0, 0) are solutions. 2 x y 0 y 2x related
0 3 0 5 ?
equation
?
0 2 false 652
Problem Recognition d. y 2 x 3 y 2x 3 related
The inequality is strict, so draw the line as a dashed line. 2 x y 0 Test (1, 1):
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): y 2 x 3
2 1 1 0 ? ?
1 0 false The points on the side of the line not containing (1, 1) are solutions. From part (c), the point of intersection is (1, 2). Graph (1, 2) as an open dot. It is not a solution to either inequality.
0 2 0 3 ? ?
0 3 true The points on the side of the line containing (0, 0) are solutions. 4 x 2 y 2 4 x 2 y 2 related equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): 4 x 2 y 2 4 0 2 0 2 ? ?
0 2 true The points on the side of the line containing (0, 0) are solutions. From part (c), the lines do not intersect. 2. a.
e. y 2 x 3 y 2x 3 related
b.
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): y 2 x 3 0 2 0 3 ? ?
0 3 false The points on the side of the line not containing (0, 0) are solutions. 4 x 2 y 2 4 x 2 y 2 related
4 x 2 y 2
c.
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): 4 x 2 y 2
4 x 2 2 x 3 2 4 x 4 x 6 2
6 2
4 0 2 0 2 ? ?
0 2 false
653
Chapter 5 Systems of Equations and Inequalities The inequality symbol allows for equality, so draw the parabola as a solid curve. 1 Test (1, 2): y x 2 2 ? 1 2 2 1 2 ? 1 2 true 2 The points on the side of the parabola containing (1, 2) are solutions. From part (a), the point of intersection is (0, 0). Graph (0, 0) as a closed dot. It is a solution to both inequalities.
The points on the side of the line not containing (0, 0) are solutions. From part (c), the lines do not intersect; they are parallel. The solutions to the first inequality are below the line, and the solutions to the second inequality are above the line. The solution sets do not overlap. The solution set of the system is the empty set, . 3. a. y x 2 1 y x2 2
1 2 x 2 1 x2 x2 2 y
c. y x 2 y x2 related equation
The inequality symbol allows for equality, so draw parabola as a solid curve. Test (1, 2): y x 2
1 2 x 0 2 x2 0 x0
2 1 ?
y x 0 0 2
2
?
2 1 true The points on the side of the parabola containing (1, 2) are solutions. 1 1 y x 2 y x2 related 2 2 equation The inequality symbol allows for equality, so draw the parabola as a solid curve. 1 Test (1, 2): y x 2 2 ? 1 2 2 1 2 ? 1 2 false 2 The points on the side of the parabola not containing (1, 2) are solutions.
0, 0
b. y x 2 y x2 related equation
The inequality symbol allows for equality, so draw the parabola as a solid curve. Test (1, 2): y x 2 2 1 ?
2
2
?
2 1 false The points on the side of the parabola not containing (1, 2) are solutions. 1 1 y x 2 y x2 related 2 2 equation
654
Problem Recognition The solution sets do not overlap, except for the point of intersection. From part (a), the point of intersection is (0, 0). Graph (0, 0) as a closed dot. It is a solution to both inequalities.
Test (0, 0):
x y 1 ?
0 0 1 ?
0 1 false The points on the side of the line not containing (0, 0) are solutions. y x 3 y x 3 related 2
2
equation
The inequality symbol allows for equality, so draw the parabola as a solid curve. Test (0, 0):
y x 3
2
0 0 3
2
? ?
0 9 false The points on the side of the parabola not containing (0, 0) are solutions. From part (a), the points of intersection are (2, 1) and (5, 4). Graph (2, 1) and (5, 4) as closed dots. They are solutions to both inequalities.
4. a. x y 1 y x 3
2
x y 1 x x 3 1 2
x x2 6 x 9 1
c. x y 1 x y 1 related
x 2 7 x 10 0
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): x y 1
x 2 7 x 10 0
x 2 x 5 0 x 2 or x 5
?
0 0 1
y x 3 2 3 1 1 2
2
2
?
0 1 true The points on the side of the line containing (0, 0) are solutions.
y x 3 5 3 2 4 2
2
2
2,1 , 5, 4
y x 3 y x 3 related 2
b. x y 1 x y 1 related
2
equation
The inequality symbol allows for equality, so draw the parabola as a solid curve.
equation
The inequality symbol allows for equality, so draw the line as a solid line.
655
Chapter 5 Systems of Equations and Inequalities Test (0, 0):
y x 3
2
0 0 3
2
? ?
0 9 true The points on the side of the parabola containing (0, 0) are solutions. From part (a), the points of intersection are (2, 1) and (5, 4). Graph (2, 1) and (5, 4) as closed dots. They are solutions to both inequalities.
Section 5.6 Linear Programming 1. x y 70
2. x y 10
3. 3x 4 y 60
4. 5 x 2 y 54
5. y 3x
6. x
7. 60 x 80
8. 10 x 20
9. linear
10. objective
11. feasible
12. vertices
13. z 0.80 x 1.10 y
14. z 10 x 8 y
15. z 0.62 x 0.50 y
16. z 82 x 55 y
18. a. z 1.8 x 2.2 y
at 0, 0 z 1.8 0 2.2 0 0
at 0, 27 z 1.8 0 2.2 27 59.4
at 18, 27 z 1.8 18 2.2 27
2 y 3
32.4 59.4 91.8 at 27, 9 z 1.8 27 2.2 9
48.6 19.8 68.4 at 27, 0 z 1.8 27 2.2 0 48.6 The values x 18 and y 27 produce the maximum value of the objective function. b. Maximum value: 91.8 19. a. z 1000 x 900 y
at 5, 45 z 1000 5 900 45
17. a. z 3x 2 y
at 0, 0 z 3 0 2 0 0
5000 40,500 45,500
at 45, 45 z 1000 45 900 45
at 0, 9 z 3 0 2 9 18
45,000 40,500 85,500
at 4, 8 z 3 4 2 8 12 16 28
at 45, 10 z 1000 45 900 10
at 7, 5 z 3 7 2 5 21 10 31
45,000 9000 54,000
at 9, 0 z 3 9 2 0 27 The values x 7 and y 5 produce the maximum value of the objective function.
at 25,15 z 1000 25 900 15
25,000 13,500 38,500
at 10, 30 z 1000 10 900 30
b. Maximum value: 31
10,000 27,000 37,000 The values x 10 and y 30 produce the minimum value of the objective function. b. Minimum value: 37,000
656
Section 5.6 at 20, 40 z 250 20 150 40
20. a. z 6 x 9 y
at 8,18 z 6 8 9 18
5000 6000 11,000 at 60, 0 z 250 60 150 0 15,000 The values x 60 and y 0 produce the maximum value of the objective function.
48 162 210 at 18, 8 z 6 18 9 8 108 72 180 at 14, 2 z 6 14 9 2
c. Maximum value: 15,000
84 18 102 at 2,14 z 6 2 9 14
22. a. Bounding lines: x0, y 0, 2 x y 40 , x 2 y 50 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
12 126 138 The values x 14 and y 2 produce the minimum value of the objective function.
b. Minimum value: 102
x 2 y 50 0 2 y 50 y 25
21. a. Bounding lines: x 0 , y 0 , x y 60 , y 2 x Find the points of intersection between pairs of bounding lines. y 2x
y 2 0 y0 x y 60 x 2 x 60 3x 60 x 20 y 2 x 2 20 40 x y 60 x 0 60 x 60
0, 25
2 x y 40 2 2 y 50 y 40 4 y 100 y 40 3 y 60 y 20 x 2 y 50
0, 0
x 2 20 50
20, 40
x 40 50 x 10 2 x y 40 2 x 0 40 2 x 40 x 20
60, 0
b. z 250 x 150 y
at 0, 0 z 250 0 150 0 0
657
10, 20
20, 0
Chapter 5 Systems of Equations and Inequalities b. z 9.2 x 8.1y
b. z 3x 2 y
at 0, 0 z 9.2 0 8.1 0 0
at 0, 50 z 3 0 2 50 100
at 10, 20 z 9.2 10 8.1 20
30 40 70 at 20, 0 z 3 20 2 0 60 The values x 20 and y 0 produce the minimum value of the objective function.
at 10, 20 z 3 10 2 20
at 0, 25 z 9.2 0 8.1 25 202.5
92 162 254 at 20, 0 z 9.2 20 8.1 0 184 The values x 10 and y 20 produce the maximum value of the objective function.
c. Minimum value: 60 24. a. Bounding lines: x0, y 0, 4 x 3 y 60 , 2 x 3 y 36 Find the points of intersection between pairs of bounding lines. 4 x 3 y 60 4 0 3 y 60 3 y 60 y 20 0, 20
c. Maximum value: 254 23. a. Bounding lines: x0, y 0, 3x y 50 , 2 x y 40 Find the points of intersection between pairs of bounding lines. 3x y 50
3 0 y 50
0, 50
y 50
4 x 3 y 60 4 x 2 x 36 60 2 x 24 x 12
3x y 50 3x 2 x 40 50 x 10 2 x y 40 2 10 y 40 y 20 2 x y 40 2 x 0 40 2 x 40 x 20
10, 20
20, 0
658
2 x 3 y 36 2 12 3 y 36 24 3 y 36 3 y 12 y4
12, 4
2 x 3 y 36 2 x 3 0 36 2 x 36 x 18
18, 0
Section 5.6 b. z 4.5 x 6 y
b. z 150 x 90 y
at 0, 20 z 4.5 0 6 20 120
at 0, 0 z 150 0 90 0 0
54 24 78 at 18, 0 z 4.5 18 6 0 81 The values x 12 and y 4 produce the minimum value of the objective function.
at 8, 40 z 150 8 90 40
at 12, 4 z 4.5 12 6 4
at 0, 40 z 150 0 90 40 3600
1200 3600 4800 at 36,12 z 150 36 90 12 5400 1080 6480 at 36, 0 z 150 36 90 0 5400 The values x 36 and y 12 produce the maximum value of the objective function.
c. Minimum value: 78 25. a. Bounding lines: x 0 , y 0 , y 40 x 36 , x y 48 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
x 0, y 40
c. Maximum value: 6480 26. a. Bounding lines: x 0 , y 0 , x 10 y 8 , x y 12 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
0, 40
x y 48 x 40 48 x8
8, 40
x 0, y 8
x y 48 36 y 48
0, 8
x y 12
y 12
36,12
x 36, y 0
36, 0
x 8 12 x4
4, 8
x y 12 10 y 12
659
y2
10, 2
x 10, y 0
10, 0
Chapter 5 Systems of Equations and Inequalities b. z 50 x 70 y
a. z 100 x 120 y
at 0, 0 z 50 0 70 0 0
at 0, 0 z 100 0 120 0 0
at 0, 8 z 50 0 70 8 560
at 0, 9 z 100 0 120 9 1080
at 4, 8 z 50 4 70 8
at 4, 9 z 100 4 120 9
200 560 760 at 10, 2 z 50 10 70 2
400 1080 1480 at 8, 6 z 100 8 120 6
500 140 640 at 10, 0 z 50 10 70 0 500 The values x 4 and y 8 produce the maximum value of the objective function.
800 720 1520 at 11, 0 z 100 11 120 0 1100
The ordered pair 8, 6 produces the maximum value of the objective function, 1520.
c. Maximum value: 760
b. z 100 x 140 y
27. Bounding lines: x 0 , y 0 , 3x 4 y 48 , 2 x y 22 , y 9 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
at 0, 0 z 100 0 140 0 0
at 0, 9 z 100 0 140 9 1260
at 4, 9 z 100 4 140 9
400 1260 1660 at 8, 6 z 100 8 140 6
0, 9
x 0, y 9
800 840 1640 at 11, 0 z 100 11 140 0 1100
3x 4 y 48
The ordered pair 4, 9 produces the maximum value of the objective function, 1660.
3x 4 9 48 3x 36 48 3x 12
4, 9
x4
28. Bounding lines: x 0 , y 0 , x y 20 , x 2 y 36 , x 14 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0 x 2 y 36
3 x 4 y 48 3x 4 2 x 22 48 3x 8 x 88 48 5 x 40 x8
0 2 y 36
2 x y 22 16 y 22 y6
x y 20
2 y 36 y 20
8, 6
y 16
2 x y 22
y 16
2 x 0 22 2 x 22 x 11
0,18
y 18
2 8 y 22
x y 20 x 16 20
11, 0
x4
660
4,16
Section 5.6 x y 20 14 y 20 y6
14, 6
x 14, y 0
14, 0
d. Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
0, 90
x 0, y 90 240 x 320 y 48,000
a. z 12 x 15 y
240 x 320 90 48,000
at 0, 0 z 12 0 15 0 0
240 x 28,800 48,000
at 0,18 z 12 0 15 18 270
240 x 19,200
at 4,16 z 12 4 15 16
x 80
48 240 288 at 14, 6 z 12 14 15 6
80, 90
240 x 320 y 48,000 240 120 320 y 48,000
168 90 258 at 14, 0 z 12 14 15 0 168
28,800 320 y 48,000
The ordered pair 4,16 produces the maximum value of the objective function, 288.
320 y 19,200 y 60
120, 0
x 120, y 0
b. z 15 x 12 y
at 0, 0 z 15 0 12 0 0
120, 60
e. z 160 x 240 y
at 0, 0 z 160 0 240 0 0
at 0,18 z 15 0 12 18 216
at 4,16 z 15 4 12 16
at 0, 90 z 160 0 240 90 21,600
60 192 252 at 14, 6 z 15 14 12 6
12,800 21,600 34,400
at 80, 90 z 160 80 240 90
at 120, 60 z 160 120 240 60
210 72 282 at 14, 0 z 15 14 12 0 210
19,200 14,400 33,600
The ordered pair 14, 6 produces the maximum value of the objective function, 282.
at 120, 0 z 160 120 240 0 19,200
f. The greatest profit is realized when 80 kitchen tables and 90 dining room tables are produced.
29. a. Profit: z 160 x 240 y b. x 0 , y 0 , x 120 , y 90 , 240 x 320 y 48,000
g. The maximum profit is $34,400. 30. a. Income: z 24 x 20 y
c.
b. x 0 , y 0 , x y 18 , y 4 , y 2 x
661
Chapter 5 Systems of Equations and Inequalities 31. a. Let x represent the number of large trees. Let y represent the number of small trees. Profit: z 35 x 30 y Constraints: x 0 , y 0 , x y 400 , 120 x 80 y 43,200 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
c.
x y 400 0 y 400 y 400
d. Find the points of intersection between pairs of bounding lines. x 0, y 4 0, 4
120 x 80 y 43,200 120 x 80 x 400 43,200 120 x 80 x 32,000 43,200 40 x 11,200 x 280 x y 400 280 y 400 y 120 280,120
x y 18 0 y 18
0,18
y 18 x y 18 x 2 x 18 3 x 18 x6 y 2 x 2 6 12
y 2x 4 2x
120 x 80 y 43,200 120 x 80 0 43,200 120 x 43,200 x 360
6,12
360, 0
z 35 x 30 y
2, 4
2 x
0, 400
at 0, 0 z 35 0 30 0 0
at 0, 400 z 35 0 30 400 12,000
e. z 24 x 20 y
at 0, 4 z 24 0 20 4 80
at 280,120 z 35 280 30 120
at 0,18 z 24 0 20 18 360
9800 3600 13,400
at 6,12 z 24 6 20 12
at 360, 0 z 35 360 30 0 12,600 280 large trees and 120 small trees would maximize profit.
144 240 384 at 2, 4 z 24 2 20 4
48 80 128
b. The maximum profit is $13,400. c. z 50 x 30 y
f. Josh should tutor 6 hr of chemistry and 12 hr of math to maximize his income.
at 0, 0 z 50 0 30 0 0
at 0, 400 z 50 0 30 400 12,000
g. Josh’s maximum income is $384 per week.
at 280,120 z 50 280 30 120
h. Josh will maximize his income when he tutors the maximum number of hours that he can work into his schedule which is 18 hr.
14,000 3600 17,600
at 360, 0 z 50 360 30 0 18,000 In this case, the nursery should have 360 large trees and no small trees. 662
Section 5.6 at 26, 0 z 150 26 120 0 3900 In this case, the store should have 12 standard bikes and 18 deluxe bikes.
32. a. Let x represent the number of deluxe exercise bikes. Let y represent the number of standard exercise bikes. Profit: z 180 x 120 y Constraints: x 0 , y 0 , x y 30 , 540 x 420 y 14,040 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
x y 30 0 y 30 y 30
33. a. Let x represent the number of trips by the large truck. Let y represent the number of trips by the small truck. Cost: z 150 x 120 y Constraints: x 0 , y 0 , 3 24 x 18 y 288 ; x y 4 Find the points of intersection between pairs of bounding lines. 24 x 18 y 288
0, 30
540 x 420 y 14,040
24 0 18 y 288
540 x 420 x 30 14,040
18 y 288 y 16
540 x 420 x 12,600 14,040 120 x 1440 x 12 x y 30 12 y 30 y 18 12,18
24 x 18 y 288 3 24 y 18 y 288 4 18 y 18 y 288 36 y 288 y8 3 3 x y 8 6 4 4
540 x 420 y 14,040 540 x 420 0 14,040 540 x 14,040 x 26
0,16
26, 0
6, 8
z 150 x 120 y
z 180 x 120 y
at 0,16 z 150 0 120 16 1920
at 0, 0 z 180 0 120 0 0
at 6, 8 z 150 6 120 8
at 0, 30 z 180 0 120 30 3600
900 960 1860 The company should make 8 trips with the small truck and 6 trips with the large truck.
at 12,18 z 180 12 120 18
2160 2160 4320 at 26, 0 z 180 26 120 0 4680 26 deluxe bikes and 0 standard model bikes would maximize profit.
b. The minimum cost is $1860. 34. a. Let x represent the number of full-time employees. Let y represent the number of part-time employees. Cost: z 40 20 x 25 12 y
b. The maximum profit is $4680. c. z 150 x 120 y
at 0, 0 z 150 0 120 0 0
at 0, 30 z 150 0 120 30 3600
z 800 x 300 y Constraints: x 0 , y 0 , 40 x 25 y 3600 ; x 1.25 y
at 12,18 z 150 12 120 18
1800 2160 3960
663
Chapter 5 Systems of Equations and Inequalities 0.4 x 0.6 y 540
Find the points of intersection between pairs of bounding lines. 40 x 25 y 3600
0.4 1.2 y 1200 0.6 y 540 0.48 y 0.6 y 540 1.08 y 540 y 500 x 1.2 y 1200
40 x 25 0 3600 40 x 3600 x 90
90, 0
x 1.2 500 1200
40 x 25 y 3600 40 1.25 y 25 y 3600
x 600 1200 x 600
50 y 25 y 3600 75 y 3600 y 48 x 1.25 y 1.25 48 60
z 90 x 120 y at 0, 0 z 90 0 120 0 0
60, 48
at 0, 900 z 90 0 120 900
z 800 x 300 y
108,000
at 90, 0 z 800 90 300 0 72,000
at 1200, 0 z 90 1200 120 0
at 60, 48 z 800 60 300 48
108,000
at 600, 500 z 90 600 120 500
48,000 14, 400 62,400 The store should have 60 full-time employees and 48 part-time employees.
54,000 60,000 114,000 The manufacturer should produce 600 grill A units and 500 grill B units to maximize profit.
b. The minimum labor cost for the week under these constraints is $62,400. 35. a. Let x represent the number of grill A units produced. Let y represent the number of grill B units produced. Profit: z 90 x 120 y Constraints: x 0 , y 0 , x 1.2 y 1200 ; 0.4 x 0.6 y 540 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
b. The maximum profit is $114,000. c. z 110 x 120 y
at 0, 0 z 110 0 120 0 0
at 0, 900 z 110 0 120 900 108,000 at 1200, 0 z 110 1200 120 0 132,000
at 600, 500 z 110 600 120 500
0.4 x 0.6 y 540
66,000 60,000 126,000 In this case, the manufacturer should produce 1200 grill A units and 0 grill B units.
0.4 0 0.6 y 540 0.6 y 540 y 900
0, 900
x 1.2 y 1200 x 1.2 0 1200 x 1200
600, 500
1200, 0
664
Section 5.6 36. a. Let x represent the number of model A units produced. Let y represent the number of model B units produced. Profit: z 150 x 200 y Constraints: x 0 , y 0 , 2 x 3 y 1200 ; 1.2 x 1.5 y 660 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0 2 x 3 y 1200
2 0 3 y 1200 3 y 1200 y 400
1.2 x 1.5 y 660 1.2 x 1.5 0 660
0, 400
x 550
2 x 3 y 1200 2 x 3 y 1200 Multiply by 2. 1.2 x 1.5 y 660 2.4 x 3 y 1320 0.4 x 120 x 300
550, 0
2 x 3 y 1200 2 300 3 y 1200 600 3 y 1200 3 y 600 y 200
300, 200
1.2 x 1.5 y 660 1.2 x 1.5 0 660 1.2 x 660 x 550
550, 0
z 150 x 200 y at 0, 0 z 150 0 200 0 0
at 0, 400 z 150 0 200 400 80,000 at 550, 0 z 150 550 200 0 82,500
at 300, 200 z 150 300 200 200 45,000 40,000 85,000 The manufacturer should produce 300 model A units and 200 model B units. b. The maximum profit is $85,000. c. z 180 x 200 y
at 0, 0 z 180 0 200 0 0
at 0, 400 z 180 0 200 400 80,000 at 550, 0 z 180 550 200 0 99,000
at 300, 200 z 180 300 200 200 54,000 40,000 94,000 In this case, the manufacturer should produce 550 model A units and no model B units.
665
Chapter 5 Systems of Equations and Inequalities z 120 x 100 y
37. a. Let x represent the number of acres of corn planted. Let y represent the number of acres of soybeans planted. Profit: z 120 x 100 y Constraints: x 0 , y 0 , x y 1200 , 180 x 120 y 198,000 ; 80 x 100 y 110,000 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
at 0, 0 z 120 0 100 0 0
at 0,1100 z 120 0 100 1100 110,000
at 1100, 0 z 120 1100 100 0 132,000
at 500, 700 z 120 500 100 700 60,000 70,000 130,000 at 900, 300 z 120 900 100 300
80 x 100 y 110,000
108,000 30,000 138,000 The farmer should plant 900 acres of corn and 300 acres of soybeans.
80 0 100 y 110,000 100 y 110,000 y 1100
0,1100
180 x 120 y 198,000
b. The maximum profit is $138,000.
180 x 120 0 198,000 180 x 198,000 x 1100
c. z 100 x 120 y
at 0, 0 z 100 0 120 0 0
1100, 0
at 0,1100 z 100 0 120 1100 132,000
80 x 100 y 110,000
at 1100, 0 z 100 1100 120 0
80 x 100 x 1200 110,000
110,000
80 x 100 x 120,000 110,000 20 x 10,000 x 500 x y 1200 500 y 1200 y 700 500, 700 180 x 120 y 198,000
at 500, 700 z 100 500 120 700 50,000 84,000 134,000 at 900, 300 z 100 900 120 300 90,000 36,000 126,000 In this case, 500 acres of corn and 700 acres of soybeans should be planted.
180 x 120 x 1200 198,000 180 x 120 x 144,000 198,000 60 x 54,000 x 900 x y 1200 900 y 1200 y 300 900, 300
666
Section 5.6 z 42 x 35 y
38. a. Let x represent the number of acres of wheat planted. Let y represent the number of acres of rye planted. Profit: z 42 x 35 y Constraints: x 0 , y 0 , x y 800 , 90 x 120 y 90,000 ; 50 x 40 y 36,000 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
at 0, 0 z 42 0 35 0 0
at 0, 750 z 42 0 35 750 26,250
at 720, 0 z 42 720 35 0 30,240 at 200, 600 z 42 200 35 600
8400 21,000 29,400 at 400, 400 z 42 400 35 400 16,800 14,000 30,800 The farmer should plant 400 acres of wheat and 400 acres of rye.
90 x 120 y 90,000 90 0 120 y 90,000 120 y 90,000 y 750
b. The maximum profit is $30,800. c. z 40 x 45 y
0, 750
at 0, 0 z 40 0 45 0 0
50 x 40 y 36,000
at 0, 750 z 40 0 45 750 33,750
50 x 40 0 36,000 50 x 36,000 x 720
at 720, 0 z 40 720 45 0 28,800 at 200, 600 z 40 200 45 600
720, 0
8000 27,000 35,000 at 400, 400 z 40 400 45 400
90 x 120 y 90,000 90 x 120 x 800 90,000
16,000 18,000 34,000 In this case, 200 acres of wheat and 600 acres of rye should be planted.
90 x 120 x 96,000 90,000 30 x 6000 x 200 x y 800 200 y 800 y 600 200, 600
39. Linear programming is a technique that enables us to maximize or minimize a function under specific constraints. 40. An objective function is a function that is to be optimized, such as a function representing cost or profit.
50 x 40 y 36,000 50 x 40 x 800 36,000 50 x 40 x 32,000 36,000 10 x 4000 x 400 x y 800 400 y 800 y 400 400, 400
41. The feasible region for a linear programming application is found by first identifying the constraints on the relevant variables. Then the regions defined by the individual constraints are graphed. The intersection of the constraints define the feasible region. 42. The vertices are found by finding the points of intersection of the bounding lines representing the constraints.
667
Chapter 5 Systems of Equations and Inequalities Chapter 5 Review Exercises 2x 3y 0
5. 5 x y 19 2 y
0.2 x 0.7 y 1.7
2 2 1 3 ⱨ 0 3
5 x 5 y 19 2 y
2 x 7 y 17
1. a.
5 x 3 y 19
2 2ⱨ0 0 ⱨ 0 true 5 x 6 y 1
2 5 x 3 y 19 10 x 6 y 38 5 2 x 7 y 17 10 x 35 y 85 41 y 123 y 3 5 x 3 y 19
2 5 1 6 ⱨ 1 3 5 4 ⱨ 1 1ⱨ 1 true
5 x 3 3 19 5 x 9 19 5 x 10
Yes 2x 3y 0
b.
x2
2 6 3 4 ⱨ 0
2 6. 9 x 2 y 4 18 x 4 y 8 2x 4 y 7 2x 4 y 7 20 x 15 0 0 3 x 0 0 4 2x 4 y 7
12 12 ⱨ 0 0 ⱨ 0 true 5 x 6 y 1 5 6 6 4 ⱨ 1 30 24 ⱨ 1 6 ⱨ 1 false
3 2 4y 7 4
No 2. Parallel lines, no solution. The system is inconsistent.
3 4y 7 2 3 8 y 14 8 y 11
3. Intersecting lines, one solution. 4.
4x y 7 4x y 7 2 2 x 5 y 19 4 x 10 y 38 9 y 45 y 5 4x y 7
y
7.
4x 5 7 4 x 12 x3
2, 3
3, 5
11 8
1 1 x y 1 10 2 x 5 y 10 x 5 y 10
3 11 , 4 8 2 x 10 y 6 2 x 10 y 6
2 x 10 y 6 2 5 y 10 10 y 6 10 y 20 10 y 6 20 6 ; The system is inconsistent.
668
Chapter 5 Review
8.
x y 20 8 y 20 y 12 8 L of the 40% mixture and 12 L of the 10% mixture should be used.
3 x 4 4 y x x y
4 y 4x x 4 y 3x 3 y x 4 3 x, y y x 4
11. Let x represent the speed of the plane in still air. Let y represent the speed of the wind. Rate (mi/hr)
Time (hr)
960
x y
2
960
x y
8 3
With wind Against wind
9. Let x represent the milligrams of calcium in 1 cup of milk. Let y represent the milligrams of calcium in 1 cup of cooked spinach. Day 1: 3x y 1140
y 3x 1140 3 Day 2: 2 x y 960 2 3 2 x 3 x 1140 960 2 4 x 3 3 x 1140 1920
d rt 960 x y 2 2 x 2 y 8 8 8 960 x y x y 3 3 3 2880 8 x 8 y 4 2 x 2 y 960 8 x 8 y 3840 8 x 8 y 2880 8 x 8 y 2880 16 x 6720 x 420
4 x 9 x 3420 1920 5 x 1500 x 300 y 3 x 1140 3 300 1140
2 x 2 y 960
900 1140 240 Milk has 300 mg per cup and spinach has 240 mg per cup.
2 420 2 y 960 840 2 y 960 2 y 120
10. Let x represent the amount of 40% alcohol solution. Let y represent the amount of 10% alcohol solution.
Amount of mixture Pure alcohol
Distance (mi)
y 60 The speed of the plane in still air is 420 mph and the speed of the wind is 60 mph.
40% Sol.
10% Sol.
22% Solution
12. a. C x 100 x 1200
x
y
20
b. R x 250 x
0.40x
0.10y
0.22 20 4.4
c.
C x R x 100 x 1200 250 x 150 x 1200
x y 20 x y 20 10 0.40 x 0.10 y 4.4 4 x y 44 3 x 24 x 8
x8 If eight excursions are run per month, then the captain will break even.
669
Chapter 5 Systems of Equations and Inequalities d. R x C x 250 x 100 x 1200
Solve the systems of equations B and D. 10 B 30 y 70 z 140 3 D 30 y 27 z 54 97 z 194 z 2
250 x 100 x 1200 150 x 1200 150 18 1200 2700 1200 1500 The captain will earn $1500.
D
10 y 9 2 18
13. A 3a 4b 2c 17 B 2a 3b c 1 C 4a b 3c 7 Eliminate c from equations A and B. and from equations B and C. A 3a 4b 2c 17 2 B 4a 6b 2c 2 a 10b 19 D
10 y 18 18 10 y 0 y0 Back substitute. A 6 x 5 y 24 6 x 5 0 24 6 x 24 x4
3 B 6a 9b 3c 3 C 4a b 3c 7 10 10a 10b a b 1 E Solve the systems of equations D and E. E a b 1 D a 10b 19 b 1 9b 18 b 2 Back substitute. A 3a 4b 2c 17 E
a
c 3
4, 0, 2
15. A x 2y z 5 B x y z 1 C 4 x 7 y 2 z 16 Eliminate z from equations A and B and from equations B and C. A x 2y z 5 B x y z1 2x 3y 6 D
a 2 1 a 1
2 B 2x 2 y 2z 2 C 4 x 7 y 2 z 16 6x 9 y 18 2x 3y 6 E Solve the systems of equations D and E. D 2x 3y 6 1 E 2 x 3 y 6 0 0
3 1 4 2 2c 17 3 8 2c 17 2c 6
10 y 9 z 18
1, 2, 3
14. A 6 x 5 y 24 B 3 y 7 z 14 C 4x 3z 10 Eliminate x from equations A and C. 2 A 12 x 10 y 48 3 C 12 x 9 z 30 10 y 9 z 18 10 y 9 z 18 D
The system reduces to the identity 0 0 . It has infinitely many solutions.
670
Chapter 5 Review 16. A u v 2 w 1 B 2v 5 w 2 C 3u 5v w 1
Eliminate u from equations A and C. 3 A 3u 3v 6 w 3 C 3u 5v w 1 2v 5w 2 D Solve the systems of equations B and D. B 2v 5 w 2 1 D 2v 5w 2 0 4 The system of equations reduces to a contradiction. There is no solution. 17. Let x represent the number of seats in Section A. Let y represent the number of seats in Section B Let z represent the number of seats in Section C. x y z 12,000 A x y z 12,000
B z x y 0 x y z C 90 x 65 y 40 z 655,000 18 x 13 y 8 z 131,000 Eliminate x and y from equations A and B. x y z 12,000 A 1 B x y z 0 2 z 12,000 z 6000 Back substitute. A x y z 12,000
18 x 13 y 8 z 131,000
B
18 x 13 y 8 6000 131,000
x y 6000 12,000
18 x 13 y 48,000 131,000
x y 6000 D Solve the system of equations D and E. 13 D 13 x 13 y 78,000
18 x 13 y 83,000 E Back substitute. A
x y z 12,000
18 x 13 y 83,000 1000 y 6000 12,000 5x 5000 y 5000 1000 x There are 1000 seats in Section A, 5000 in Section B, and 6000 in Section C. E
671
Chapter 5 Systems of Equations and Inequalities 18. Let x represent the amount put in the savings account. Let y represent the amount invested in the bond fund. Let z represent the amount invested in the stock fund. x y A x y z 20,000
z 20,000
B y 2z y 2z 0 C 0.015 x 0.045 y 0.062 z 942 15 x 45 y 62 z 942,000 Eliminate x from equations A and C. 15 A 15 x 15 y 15 z 300,000 C 15 x 45 y 62 z 942,000 30 y 47 z 642,000 D Solve the systems of equations B and D. 30 B 30 y 60 z 0 D 30 y 47 z 642,000 107 z 642,000 z 6000
y 2z 0
D
y 2 6000 0 y 12,000 0 y 12,000
Back substitute. x y z 20,000 A x 12,000 6000 20,000 x 2000 She put $2000 in savings, and invested $12,000 in the bond fund and $6000 in the stock fund. 19.
y ax 2 bx c Substitute 1, 4 : 4 a 1 b 1 c 2 Substitute 1, 6 : 6 a 1 b 1 c 2 Substitute 3, 8 : 8 a 3 b 3 c
A a b c 4 B a bc 6 C 9a 3b c 8
2
Eliminate b from equations A and B and from equations A and C. A a b c 4 3 A 3a 3b 3c 12 B a b c 6 C 9a 3b c 8 2a 2c 2 12a 4c 4 a c 1 D 3a c 1 E Solve the system of equations D and E. D a c 1 1 E 3a c 1 2a 2 a 1 Back substitute. A a b c 4 1 b 2 4 b 5 b5
D
a c 1 1 c 1 c2
y x2 5x 2
672
Chapter 5 Review y ax 2 bx c
20.
Substitute 1, 2 : 2 a 1 b 1 c 2 Substitute 2,1 : 1 a 2 b 2 c 2 Substitute 3,10 : 10 a 3 b 3 c 2
A a b c 2 B 4a 2b c 1 C 9a 3b c 10
Eliminate c from equations A and B and from equations A and C. A a b c 2 A a b c 2 1 B 4a 2b c 1 1 C 9a 3b c 10 3a b 3 8a 2b 12 3a b 3 D 4a b 6 E Solve the system of equations D and E. D 3a b 3 1 E 4a b 6 a 3 a 3
3a b 3
D
3 3 b 3 9b 3 b 6
Back substitute. A a b c 2 3 6 c 2 c 1 y 3x 2 6 x 1 21.
x 11 A B x 2 x 1 x 2 x 1
22.
5 x 22 5 x 22 A B 2 x 8 x 16 x 4 x 4 x 4 2
23.
7 x 2 19 x 15 7 x 2 19 x 15 A B C 2 3 2 2 2 x 3x x 2 x 3 x x 2x 3
24.
2 x 2 x 10 2 x 2 x 10 A Bx C 2 x3 5 x x x 5 x x2 5
25.
2 x3 x 2 8 x 16 2 x3 x 2 8 x 16 Ax B Cx D 2 2 x4 5x2 4 x 1 x 4 x2 1 x2 4
26.
2
4 x 4 3x 2 2 x 5
x 2 x 5 x 2 2 3
2
A B C D Ex F Gx H 2 2 3 2 x 2 x 5 2 x 5 2 x 5 x 2 x2 2
27. Use long division to divide the numerator by the denominator. Then perform partial fraction decomposition on the expression (remainder/divisor).
673
Chapter 5 Systems of Equations and Inequalities x 11 A B x 2 x 1 x 2 x 1
28.
x 11 B A x 2 x 1 x 2 x 1 x 2 x 1
x 2 x 1
x 11 A x 1 B x 2 x 11 Ax A Bx 2 B
x 11 A B x A 2 B A B 1 A B 1
B 1 2 B 11
3 x 11 4 x 2 x 1 x 2 x 1 29.
A B 1 A 4 1 A3
A 2 B 11 B 1 2 B 11 3B 12 B 4
5 x 22 5 x 22 A B 2 x 8 x 16 x 4 x 4 x 4 2 2
5 x 22 A B 2 x 4 2 2 x 4 x 4 x 4 5 x 22 A x 4 B
x 4 2
5 x 22 Ax 4 A B 5 x 22 A x 4 A B A5
4 A B 22 4 5 B 22
5 x 22 5 2 x 8 x 16 x 4 x 4 2
B2
2
674
Chapter 5 Review x2 30. 2 x 3x 2 x 7 x 13 x 19 x 15 3
2
4
3
2 x 4 3x3
2
4 x 13 x 2 4 x3 6 x 2 3
7 x 19 x 15 2
2 x 7 x 13 x 19 x 15 7 x 2 19 x 15 x 2 2 x3 3x 2 2 x3 3x 2 7 x 2 19 x 15 7 x 2 19 x 15 A B C 2 3 2 2 2 x 3x x 2 x 3 x x 2x 3 4
3
2
7 x 2 19 x 15 C A B 2 x 2 2 x 3 2 x 2 x 3 2 2 x 3 x x x 2 x 3 7 x 2 19 x 15 Ax 2 x 3 B 2 x 3 Cx 2 7 x 2 19 x 15 2 Ax 2 3 Ax 2 Bx 3B Cx 2 7 x 2 19 x 15 2 A C x 2 3 A 2 B x 3B 3B 15 B5
3 A 2 B 19
2A C 7
3 A 2 5 19
2 3 C 7
3A 9
6C 7
A3
C 1
2 x 7 x 13x 19 x 15 3 5 1 x2 2 3 2 x x 2 x 3x 2x 3 4
31.
3
2
2 x 2 x 10 2 x 2 x 10 A Bx C 2 x3 5 x x x 5 x x2 5
2 x 2 x 10 A Bx C x x2 5 2 x x2 5 2 x x 5 x x 5
2 x 2 x 10 A x 2 5 Bx C x 2 x x 10 Ax 5 A Bx 2 Cx 2
2
2 x 2 x 10 A B x 2 C x 5 A 5 A 10 A 2
C 1
A B 2 2 B 2 B4
2 x 2 x 10 2 4 x 1 x3 5 x x x2 5
675
Chapter 5 Systems of Equations and Inequalities 2 x 3 x 2 8 x 16 2 x 3 x 2 8 x 16 Ax B Cx D 2 2 32. x4 5x2 4 x 1 x 4 x2 1 x2 4
2 x 3 x 2 8 x 16 Ax B Cx D x2 1 x2 4 2 2 x2 1 x2 4 2 2 x 4 x 1 x 1 x 4
2 x 3 x 2 8 x 16 Ax B x 2 4 Cx D x 2 1
2 x x 8 x 16 Ax 4 Ax Bx 4 B Cx Cx Dx 2 D 3
2
3
2
3
2 x 3 x 2 8 x 16 A C x 3 B D x 2 4 A C x 4 B D AC 2 A 2C
4A C 8
AC 2
4 2 C C 8
A0 2
8 4C C 8
A2
3C 0 C0 B D 1 B 1 D
4 B D 16
B D 1
4 1 D D 16
B 4 1
4 4 D D 16 3D 12
B 5
D4 2 x 3 x 2 8 x 16 2 x 5 4 2 2 4 2 x 5x 4 x 1 x 4
33. a. A y x 2 1 y x 2 1
b. B
x y 3
x x 2 1 3
B x y 3 y x 3
x x 2 1 3 x2 x 2 0 x2 x 2 0
x 1 x 2 0 x 1 or x 2 A y x 2 1 1 1 1 1 2 2
The solution is 1, 2 .
A y x 2 1 2 1 4 1 5 2
The solution is 2, 5 .
1, 2 , 2, 5
676
Chapter 5 Review 34. a. A y x 1
x2 y 2 5
b. B
B x2 y 2 5 x2 y 2
5
2
x2
x 1 5 2
x2 x 1 5 x2 x 6 0
x 3 x 2 0 x 3 or x 2 A y x 1 3 1 4 Not real A y x 1 2 1 1 The solution is 2,1 .
2,1
2 2 6 x 2 2 y 2 8 35. A 3x y 4 x 2 2 y 2 36 B x 2 2 y 2 36 28 7 x2 x2 4 x 2 Multiply by 2.
x 2: A
3x 2 y 2 4
x 2 : A
3x 2 y 2 4
3 2 y 2 4
3 2 y 2 4
12 y 2 4
12 y 2 4
y 2 16
y 2 16
y 2 16
y 2 16
y 4
y 4
2
2
2, 4 , 2, 4 , 2, 4 , 2, 4 Multiply by 3. 2 6 x 2 3xy 72 36. A 2 x xy 24 2 x 2 3xy 9 B x 3xy 9 63 7 x2 2 x 9 x 3
x 3: B
x 2 3xy 9
x 3: B
x 2 3 xy 9
32 3 3 y 9
32 3 3 y 9
9 9 y 9
9 9 y 9
9 y 18
9 y 18
y 2
y2
3, 2 , 3, 2
677
Chapter 5 Systems of Equations and Inequalities 37. A y
39. Let x represent one number. Let y represent the other number. x 4 3 A 4 y 3x y x y 3 4
8 x
B y x
8 x 8 x x 64 x 2 x x 3 64 x4 B y x 4 2 A
B x 2 y 2 100
y
B
2
3 x 2 100 4 9 2 x 100 16 25 2 x 100 16 x 2 64 x 8 3 3 A y x 8 6 4 4 3 3 A y x 8 6 4 4 The numbers are 8 and 6 or 8 and 6 . x2
Check: 4, 2
8 x 8 2ⱨ 4 2 ⱨ 2 true Check: 4, 2 A y
x 2 y 2 100
B y x 2ⱨ 4 2 ⱨ 2 true
8 x 8 2 ⱨ 4 2 ⱨ 2 false 4, 2
40. Let x represent the length of one side. Let y represent the length of the other side. A x y 12 y x 12
A y
B x2 y 2 B
74 x y 74 2
2
x 2 y 2 74 x 2 x 12 74 2
38. Let x represent one number. Let y represent the other number. A x 2 y 2 97 B x 2 y 2 65 2 x2 162 x 2 81 x 9
x 2 x 2 24 x 144 74 2 x 2 24 x 70 0 x 2 12 x 35 0
x 5 x 7 0 A
Both numbers must be negative. x 9 : A x 2 y 2 97
x 5 or x 7 y x 12 5 12 7
A y x 12 7 12 5 The legs are 5 ft and 7 ft.
9 2 y 2 97 81 y 2 97 y 2 16 y 4 The numbers are 9 and 4 .
678
2
Chapter 5 Review 41. Let x represent the length. Let y represent the width. A xy 288
3 0 4 0 8 ? ?
0 8 false The points on the side of the line not containing (0, 0) are solutions.
B 2 x 2 y 72 x y 36 y x 36 xy 288
A
3x 4 y 8
Test (0, 0):
x x 36 288 x 36 x 288 0 2
x 2 36 x 288 0
x 12 x 24 0 A
x 12 or x 24 y x 36 12 36 24
A y x 36 24 36 12 The sign is 12 ft by 24 ft. y x 4 43. a. y x 4 related 2
42. a. 3x 4 y 8 3x 4 y 8 related
equation
equation
The equation represents a parabola opening upward with vertex 4, 0 .The inequality is strict, so draw the parabola as a dashed curve. 2 Test (0, 0): y x 4
y-intercept: 3 0 4 y 8
x-intercept: 3 x 4 0 8
2
4y 8 3x 8 y2 8 x 3 The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): 3x 4 y 8
0 0 4 ?
2
?
0 16 true The points on the side of the parabola containing (0, 0) are solutions.
3 0 4 0 8 ? ?
0 8 true The points on the side of the line containing (0, 0) are solutions.
y x 4 b. y x 4 related 2
2
equation
The equation represents a parabola opening upward with vertex 4, 0 .The inequality symbol allows for equality, so draw the parabola as a solid curve.
b. 3x 4 y 8 3x 4 y 8 related equation
The inequality is strict, so draw the line as a dashed line.
679
Chapter 5 Systems of Equations and Inequalities Test (0, 0):
y x 4
2
0 0 4
2
?
45. x 3.5 The related equation x 3.5 is a vertical line. The inequality x 3.5 represents all points to the left or on the line x 3.5 . The inequality symbol allows for equality, so draw the line as a solid line.
?
0 16 false The points on the side of the parabola not containing (0, 0) are solutions.
44. 5 x y 8 x 15 5 x y 8 x 15 related
3 y 1 4 2 3 y3 2 y 2 The related equation y 2 is a horizontal line. The inequality y 2 represents all points strictly above the line y 2 . The inequality is strict, so draw the line as a dashed line.
46.
equation
5 x y 8 x 15 5 x 5 y 8 x 15 5 y 3 x 15 3 y x3 5 The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): 5 x y 8 x 15 5 0 0 8 0 15 ? ?
0 15 false The points on the side of the line not containing (0, 0) are solutions.
680
Chapter 5 Review 47. x 2 y 2 4 x 2 y 2 4 related 2
0 2 1 4 ?
equation
The equation represents a circle centered at 0, 2 with radius 2. The inequality is strict, so draw the circle as a dashed curve. 2 Test (0, 1) : x 2 y 2 4
?
2 4 true 3x 4 y 6 3 0 4 1 6 ? ?
4 6 false No, the ordered pair is not a solution to the system.
0 1 2 4 2
x 2y 4
49. a.
2
2 ?
?
1 4 true The test point inside the circle satisfies the inequality. The solution set consists of the points strictly inside the circle.
x 2y 4
b.
1 2 4 4 ? ?
7 4 true 3x 4 y 6 3 1 4 4 6 ? ?
19 6 true Yes, the ordered pair is a solution to the system. 1 1 x 1 y x 1 related 2 2 equation The inequality is strict, so draw the line as a dashed line. 1 Test (0, 0): y x 1 2 ? 1 0 0 1 2
50. y 48. y 2 y 2 related equation
The related equation y 2 represents two horizontal lines, y 2 and y 2 . The inequality is strict, so draw the lines as dashed lines. Test (0, 0): y 2
?
0 1 false The points on the side of the line not containing (0, 0) are solutions. 3x 2 y 4 3x 2 y 4 related
?
0 2 ?
0 2 false The test point does not satisfy the inequality. The solution set consists of all points strictly outside the lines y 2 and y 2 .
equation
y-intercept: x-intercept: 3 0 2 y 4 3 x 2 0 4 3x 4 2y 4 4 y2 x 3 The inequality is strict, so draw the line as a dashed line. Test (0, 0): 3x 2 y 4 3 0 2 0 4 ? ?
0 4 true The points on the side of the line containing (0, 0) are solutions.
681
Chapter 5 Systems of Equations and Inequalities Find the point of intersection. 3x 2 y 4 1 3x 2 x 1 4 2 3x x 2 4 4x 2 1 x 2 1 1 1 5 y x 1 1 2 2 2 4 1 5 Graph , as an open dot. It is not a 2 4 solution to either inequality.
The test point does not satisfy the inequality. The solution set consists of the points outside or on the circle.
y x2 3 52. y x 2 3 related equation
The equation represents a parabola opening upward with vertex 0, 3 . The inequality symbol allows for equality, so draw the parabola as a solid curve. Test (0, 0): y x2 3 0 0 3 ?
2
?
0 3 true The test point satisfies the inequality. The solution set consists of the points above or on the parabola. y 1 The related equation y 1 is a horizontal line. The inequality y 1 represents all points above the line y 1 . The inequality is strict, so draw the line as a dashed line. x y 3 x y 3 related
51. x 2 y 2 9 x2 y 2 9 related equation
The equation represents a circle centered at 0, 0 with radius 3. The inequality symbol allows for equality, so draw the circle as a solid curve. Test (0, 0): x2 y 2 9
0 2 0 2 9 ?
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): x y3
?
0 9 true The test point satisfies the inequality. The solution set consists of the points inside or on the circle.
?
00 3
x 12 y 2 4 x 1 2 y 2 4 related
?
0 3 true The points on the side of the line containing (0, 0) are solutions. Find the points of intersection between y x 2 3 and x y 3 : x y 3 x x2 3 3 x2 x 6 0 x 3 x 2 0
equation
The equation represents a circle centered at 1, 0 with radius 2. The inequality symbol allows for equality, so draw the circle as a solid curve. Test (0, 0): x 1 2 y 2 4
0 1 2 02 4 ? ?
1 4 false 682
Chapter 5 Review x 3 or x 2
The equation represents a parabola opening downward with vertex 0,1 . The inequality is strict, so draw the parabola as a dashed curve. y x2 1 Test (0, 0):
y x 3 3 3 9 3 6 2
2
y x 2 3 2 3 4 3 1 2
The solutions are 3, 6 and 2,1 . Find the points of intersection between y x 2 3 and y 1 :
0 0 1 ?
2
?
0 1 false The test point does not satisfy the inequality. The solution set consists of the points below the curve. The solution sets do not overlap. The solution set of the system is the empty set, .
y x2 3 1 x2 3 4 x2 2 x y 1 The solutions are 2,1 and 2,1 .
The point 2,1 is a point of intersection of all three equations. Test it in y 1 : y 1
54. a. x y 12
b. y 2 x
c. x 0
d. y 0
e. The inequalities x 0 and y 0 together represent the set of points in the first quadrant, including the bounding axes. x y 12 x y 12 related
?
1 1 false The point 2,1 does not satisfy y 1 , so
plot it as an open dot. Plot 3, 6 as a
closed dot, and plot 2,1 as an open dot.
equation
Test (0, 0):
x y 12 ?
0 0 12 ?
0 12 true The inequality x y 12 represents the set of points on and below the line x y 12 . y 2 x y 2x related equation
Test (1, 4):
y 2x 4 2 1 ?
y ex 53. y e x related
?
4 2 true The inequality y 2 x represents the set of points on and above the line y 2 x .
equation
The equation represents the parent natural exponential function. The inequality is strict, so draw the function as a dashed curve. Test (0, 0): y ex ?
0 e0 ?
0 1 false The test point does not satisfy the inequality. The solution set consists of the points above the curve. y x 2 1 y x2 1 related equation
683
Chapter 5 Systems of Equations and Inequalities 55. z 24 x 20 y 56. a. z 36 x 50 y
at 0, 0 z 36 0 50 0
at 0,18 z 36 0 50 18 900
at 10,14 z 36 10 50 14 at 16,10
360 700 1060 z 36 16 50 10
576 500 1076 at 18, 0 z 36 18 50 0 648 The values x 16 and y 10 produce the maximum value of the objective function.
b. z 55 x 40 y
at 4,10 z 55 4 40 10 220 400 620 at 0,18 z 55 0 40 18 720
b. Maximum value: 1076
at 12, 0 z 55 12 40 0 660 The values x 4 and y 10 produce the minimum value of the objective function.
57. a. Bounding lines: x0, y 0, 2 x y 18 , 5 x 4 y 60 Find the points of intersection between pairs of bounding lines. 5 x 4 y 60
c. Minimum value: 620 58. a. Let x represent the number of scoops of brand A powder. Let y represent the number of acres of brand B powder. Fiber: z 10 x 8 y Constraints: x 0 , y 0 , x y 180 , 3x 2 y 480 ; 3x 4 y 696 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0
5 x 4 2 x 18 60 5 x 8 x 72 60 3 x 12 2 x y 18
x4
2 4 y 18 8 y 18 y 10
4,10
3x 4 y 696
2 x y 18 2 0 y 18 y 18
3 0 4 y 696 4 y 696
0,18
y 174
5 x 4 y 60
3x 2 y 480
5 x 4 0 60 5 x 60 x 12
0,174
3x 2 x 180 480 3x 2 x 360 480
12, 0
x y 180 120 y 180 y 60
684
x 120
120, 60
Chapter 5 Test at 24, 156 z 10 24 8 156
3 x 4 y 696 3x 4 x 180 696
240 1248 1488 at 160, 0 z 10 160 8 0 1600 She should use 120 scoops of brand A and 60 scoops of brand B.
3 x 4 x 720 696 x 24 x 24
x y 180
b. The maximum amount of fiber is 1680 g.
24 y 180
c. z 8 x 10 y
24, 156
y 156
at 0, 0 z 8 0 10 0 0
at 0,174 z 8 0 10 174 1740
3 x 2 y 480 3x 2 0 480 3 x 480 x 160
at 120, 60 z 8 120 10 60
960 600 1560 at 24, 156 z 8 24 10 156
160, 0
192 1560 1752 at 160, 0 z 8 160 10 0 1280 In this case, 24 scoops of brand A and 156 scoops of brand B should be used.
z 10 x 8 y at 0, 0 z 10 0 8 0 0
at 0,174 z 10 0 8 174 1392 at 120, 60 z 10 120 8 60
1200 480 1680
Chapter 5 Test 1. a.
x 5 y 3
2 x 3 y z 5
2. a.
7 5 2 ⱨ 3
2 0 3 1 2 ⱨ 5
7 10 ⱨ 3
0 3 2 ⱨ 5
3 ⱨ 3 true y 2 x 12
5 ⱨ 5 true 5 x y 3 z 18
2 ⱨ 2 7 12
5 0 1 3 2 ⱨ 18 0 1 6 ⱨ 18
2 ⱨ14 12
7 ⱨ 18 false
2 ⱨ 2 true Yes b.
No
x 5 y 3
b.
3 5 0 ⱨ 3
2 x 3 y z 5 2 3 3 0 1 ⱨ 5
3 ⱨ 3 true y 2 x 12
6 0 1ⱨ 5 5 ⱨ 5 true 5 x y 3 z 18
0 ⱨ 2 3 12
5 3 0 3 1 ⱨ 18
0 ⱨ 6 12
15 0 3 ⱨ 18
0 ⱨ 18 false No
18 ⱨ 18 true
685
Chapter 5 Systems of Equations and Inequalities x 2 y 5z 8
5. 0.2 x 0.35 y 2.5
20 x 35 y 250
3 2 0 5 1 ⱨ 8
0.16 x 0.5 y 5.8
16 x 50 y 580
3 0 5ⱨ8
5 20 x 35 y 250 4x 7 y 50 4 16 x 50 y 580 4 12.5 x y 145 19.5 y 195 y 10
8 ⱨ 8 true Yes 3. a.
2x 4 y 9 ?
20 x 35 y 250
?
20 x 35 10 250
2 6 4 1 9 16 9 true 3 x y 4
20 x 350 250
3 6 1 4 ?
20 x 100
?
x5
19 4 true Yes, the ordered pair is a solution to the system. b.
5,10
2 3 y 5 10 2 3 x y 5 10
6. x
2x 4 y 9 2 1 4 4 9 ? ?
3 2 10 x 4 y 3
14 9 true 3x y 4
5x 2 y
3 1 4 4 ?
3 2 10 y 4 y 3 5 10
?
1 4 false No, the ordered pair is not a solution to the system.
4 y 3 4 y3 33 x, y |10 x 4 y 3
4. x 5 4 y
3x 7 y 4 3 5 4 y 7 y 4 15 12 y 7 y 4 19 y 19 y 1
7. 7 x y 3 5 y
4 3x y 2 x
7x 7 y 3 5y 7x 2 y 3
12 x 4 y 2 x 14 x 4 y 0
2 7 x 2 y 3 14 x 4 y 6 14 x 4 y 0 14 x 4 y 0 0 6
x 5 4 y 5 4 1 5 4 1
1,1
; The system is inconsistent.
686
Chapter 5 Test 8. A a 6b 3c 14 B 2a b 2c 8 C 3a 2b c 8 Eliminate a from equations A and B. and from equations A and C. 2 A 2a 12b 6c 28 B 2a b 2c 8 11b 8c 20 D
3 A C
B
3 5 2 z 9 15 2 z 9 2 z 6 z3 Back substitute. A x 4 z 10 x 4 3 10
3a 18b 9c 42 3a 2b c 8 20b 10c 50 2b c 5 E
x 12 10 x 2
2b c 5 2 4 c 5
Solve the systems of equations B and D. B x 3y 2 D x 3 y 4 0 2 The system of equations reduces to a contradiction. There is no solution.
8 c 5 c3 Back substitute. A a 6b 3c 14 a 6 4 3 3 14
2, 5, 3
10. A 2 x y z 3 2 B x 3y C x 2 y z 7 Eliminate z from equations A and C. 1 A 2 x y z 3 C x 2 y z 7 x 3y 4 D
Solve the systems of equations D and E. D 11b 8c 20 8 E 16b 8c 40 20 5b b 4
E
3y 2z 9
1, 4, 3
a 24 9 14 a 1
11. A
x 42 y 2 25
B x y 3 x y 3
9. A x 4 z 10 B 3y 2z 9 C 2x 5 y 21 Eliminate x from equations A and C. 2 A 2 x 8 z 20 C 2x 5 y 21 5 y 8z 1 D
A
x 4 2 y 2 25 y 3 4 2 y 2 25 y 1 2 y 2 25 y 2 2 y 1 y 2 25 2 y 2 2 y 24 0 y 2 y 12 0
y 3 y 4 0
Solve the systems of equations B and D. 4 B 12 y 8 z 36 D 5 y 8z 1 7 y 35 y 5
y 3 or y 4 B x y 3 3 3 0 B x y3 43 7
0, 3 , 7, 4 687
Chapter 5 Systems of Equations and Inequalities 2 2 10 x 2 2 y 2 28 12. A 5 x y 14 x 2 2 y 2 17 B x 2 2 y 2 17 11 11x 2 2 x 1 x 1 Multiply by 2.
x 1: A
5x 2 y 2 14
x 1: A
5x 2 y 2 14
5 1 y 2 14
5 1 y 2 14
5 y 2 14
5 y 2 14
y2 9 y 3
y2 9 y 3
2
2
1, 3 , 1, 3 , 1, 3 , 1, 3
Multiply by 3. 2 6 xy 3 y 2 72 13. A 2 xy y 24 2 Multiply by 2. B 3 xy 2 y 2 38 6 xy 4 y 76 y2 4 y 2
y 2: A
2 xy y 2 24
y 2 : A
2 xy y 2 24
2 x 2 2 24
2 x 2 2 24
4 x 4 24 4 x 20 x 5
4 x 4 24 4 x 20 x5
2
2
5, 2 , 5, 2
1 1 and v . x3 y 1 1 2 A 7 u 2v 7 u 2v 7 x 3 y 1 3 1 B 7 3u v 7 x 3 y 1
14. Let u
B
3u v 7 3 2v 7 v 7
6v 21 v 7 7v 28 v4 1 1 u , so 1 x3 x3 x 31 x 2
A u 2v 7 2 4 7 1
v
1 , so y 1
1 y 1 4y 4 1 4y 5 4
y
688
5 4
5 2, 4
Chapter 5 Test 15. Let x represent the number of pounds of 100% peanuts. Let y represent the number of pounds of 45% peanuts. 100% Peanuts
45% Peanuts
56% Peanuts
Amount of mixture
x
y
20
Peanuts
x
0.45y
0.56 20 11.2
100 100 x 100 y 2000 x y 20 x y 20 100 100 x 45 y 1120 x 0.45 y 11.2 100 x 45 y 1120 55 y 880 16 y x y 20 x 16 20 x4
The manager should mix 4 lb of peanuts with 16 lb of the 45% mixture. 16. Let x represent the speed of one runner. Let y represent the speed of the other runner. Distance (m) Opposite direction
400
Rate (m/sec) x y
Same direction
400
x y
d rt
400 x y 40 10 x y
x y 10 x y 2 2 x 12 x 6
Time (sec)
40 200
400 x y 200 2 x y
x y 10 6 y 10 y4
One runner runs 6 m / sec and the other runs 4 m/sec. 17. Let x represent the amount invested in the first stock. Let y represent the amount invested in the second stock. Let z represent the amount invested in the third stock. x y A x y z 15,000
z 15,000
B 0.08 x 0.032 y 0.058 z 274 80 x 32 y 58 z 274,000 C y z 2000 2000 y z Eliminate x from equations A and B. 80 A 80 x 80 y 80 z 1,200,000 B 80 x 32 y 58 z 274,000 112 y 138 z 1,474,000 D
689
Chapter 5 Systems of Equations and Inequalities Solve the systems of equations C and D. 138 C 138 y 138 z 276,000 D 112 y 138 z 1,474,000 250 y 1,750,000 y 7000
D
y z 2000 7000 z 2000 z 5000 z 5000
Back substitute. A x y z 15,000 x 7000 5000 15,000 x 3000 Dylan invested $3000 in the risky stock, $7000 in the second stock, and $5000 in the third stock. 18. Let x represent one number. Let y represent the other number. A x y 3 x y 3
B x 2 y 2 33 x 2 y 2 33
B
A
x y3 43 7
y 3 y 2 33 2
y 2 6 y 9 y 2 33 6 y 24 y4 The numbers are 7 and 4. 19. Let x represent the length. Let y represent the width. A xy 1452
B 2 x 2 y 154 x y 77 y x 77 xy 1452
A
x x 77 1452 x 77 x 1452 0 2
x 2 77 x 1452 0
x 44 x 33 0 x 44 or x 33 A
y x 77 44 77 33
A
y x 77 33 77 44
The screen is 44 in. by 33 in.
690
Chapter 5 Test y ax 2 bx c
20.
Substitute 1, 1 : 1 a 1 b 1 c 2 Substitute 2, 1 : 1 a 2 b 2 c 2 Substitute 1, 7 : 7 a 1 b 1 c
A a b c 1 B 4a 2b c 1 C a bc 7
2
Eliminate c from equations A and B and from equations A and C. a b c 1 A A a b c 1 1 B 4a 2b c 1 1 C a b c 7 3a b 2 2b 8 b 4 3a b 2 D Back substitute. D 3a b 2
A
a b c 1 2 4 c 1
3a 4 2 3a 6
c 1
a2 y 2x 4x 1 2
21.
22.
23.
15 x 15 15 x 15 A B 3x 2 x 2 x 1 3 x 2 x 1 3 x 2 5 x 6 3x5 4 x3 x 3
x 3 x 3 x 2 5 x 1
2
A B C D Ex F Gx H 2 3 2 3 x x x x 3 x 5x 1 x2 5x 1 2
12 x 29 12 x 29 A B 2 x 2 11x 15 x 3 2 x 5 x 3 2 x 5
12 x 29 B A x 3 2 x 5 5 3 2 5 3 2 x x x x
x 3 2 x 5
12 x 29 A 2 x 5 B x 3 12 x 29 2 Ax 5 A Bx 3B
12 x 29 2 A B x 5 A 3B 2 A B 12 B 2 A 12
5 A 3B 29
2 A B 12
5 A 3 2 A 12 29
2 7 B 12
5 A 6 A 36 29 A 7
14 B 12
12 x 29 7 2 2 x 2 11x 15 x 3 2 x 5
A 7
691
B2
Chapter 5 Systems of Equations and Inequalities
24.
6x 8 6x 8 A B 2 x 4 x 4 x 2 x 2 x 2 2 2
6x 8 A B 2 x 2 2 2 x 2 x 2 x 2
x 2 2
6 x 8 A x 2 B 6 x 8 Ax 2 A B
6 x 8 A x 2 A B
A6
2A B 8 2 6 B 8 12 B 8
6x 8 6 4 2 x 4 x 4 x 2 x 2 2
B 4
x2 25. x 4 x x 6 x 4 x 20 x 32 3
2
4
3
x 4 4 x3
2
2 x 4 x2 2 x 3 8 x 2 3
4 x 20 x 32 2
x 4 6 x 3 4 x 2 20 x 32 4 x 2 20 x 32 x 2 x3 4 x 2 x3 4 x 2 2 2 4 x 20 x 32 4 x 20 x 32 A B C 2 3 2 2 x 4x x x 4 x x x4 4 x 2 20 x 32 C A B 2 x x 4 x x 4 2 2 x x 4 x 4 x x 4 x 2 20 x 32 Ax x 4 B x 4 Cx 2 2
4 x 2 20 x 32 Ax 2 4 Ax Bx 4 B Cx 2 4 x 2 20 x 32 A C x 2 4 A B x 4 B 4 B 32 B8
4 A B 20
A C 4
4 A 8 20 4 A 12
3 C 4 C 1
A 3 3 8 1 x 4 6 x 3 4 x 2 20 x 32 x2 2 3 2 x 4x x x x4
692
Chapter 5 Test x 2 2 x 21 x 2 2 x 21 A Bx C 2 26. x3 7 x x x 7 x x2 7
x 2 2 x 21 A Bx C x x2 7 2 x x2 7 2 x x 7 x x 7
x 2 2 x 21 A x 2 7 Bx C x x 2 x 21 Ax 7 A Bx 2 Cx 2
2
x 2 2 x 21 A B x 2 C x 7 A A B 1
C 2
7 A 21 A 3
3 B 1 B4
x 2 2 x 21 3 4 x 2 x3 7 x x x2 7 27.
7 x 3 4 x 2 63x 15 7 x 3 4 x 2 63 x 15 Ax B Cx D 2 2 x 4 11x 2 18 x 2 x 9 x2 2 x2 9
B Cx D x 2 x 9 7 x x 4x2 x63x 9 15 x 2 x 9 Ax x 2 x 9 7 x 4 x 63 x 15 Ax B x 9 Cx D x 2 2
2
3
2
2
3
2
2
2
2
2
2
2
2
7 x 3 4 x 2 63 x 15 Ax 3 9 Ax Bx 2 9 B Cx3 2Cx Dx 2 2 D 7 x 3 4 x 2 63 x 15 A C x 3 B D x 2 9 A 2C x 9 B 2 D AC 7 A 7C
9 A 2C 63
AC 7
9 7 C 2C 63
A0 7
63 9C 2C 63
A7
7C 0 C0 BD4 B 4D
9 B 2 D 15
BD4
9 4 D 2 D 15
B3 4
36 9 D 2 D 15
B 1
7 D 21 D3 7 x 3 4 x 2 63x 15 7 x 1 3 2 2 4 2 x 11x 18 x 2 x 9
693
Chapter 5 Systems of Equations and Inequalities 28. 2 x y 6 y 2 x y 6 y related
30. x 4 x 4 related
equation
equation
2 x y 6 y 2x 2 y 6 y 3 y 2 x 6 2 y x2 3 The inequality is strict, so draw the line as a dashed line. Test (0, 0): 2 x y 6 y
The related equation x 4 represents two vertical lines, x 4 and x 4 . The inequality is strict, so draw the lines as dashed lines. Test (0, 0): x 4 ?
0 4 true The test point satisfies the inequality. The solution set consists of all points inside the lines x 4 and x 4 .
2 0 0 6 0 ? ?
0 6 false The points on the side of the line not containing (0, 0) are solutions.
31. x y 4 x y 4 or y x 4 related equation
The inequality symbol allows for equality, so draw the line as a solid line. x y4 Test (0, 0):
29. x 3 y 2 9 x 3 y 2 9 related 2
2
?
00 4
equation
?
0 4 true The points on the side of the line containing (0, 0) are solutions. 2 x y 2 2 x y 2 or y 2 x 2 related
The equation represents a circle centered at 3, 0 with radius 3. The inequality symbol allows for equality, so draw the circle as a solid curve. Test (1, 1): x 32 y 2 9
equation
The inequality is strict, so draw the line as a dashed line. 2 x y 2 Test (0, 0):
1 3 2 1 2 9 ? ?
17 9 true The test point satisfies the inequality. The solution set consists of the points outside or on the circle.
2 0 0 2 ? ?
0 2 true The points on the side of the line containing (0, 0) are solutions. Find the point of intersection. x 4 2x 2 3x 2 2 x 3 2 10 y x 4 4 3 3 694
Chapter 5 Test 2 10 Graph , as an open dot. It is not a 3 3 solution to first inequality.
Find the points of intersection between y x 2 5 and x y 3 : x y 3
x x2 5 3 x2 x 5 3 x2 x 2 0
x 1 x 2 0 x 1 or x 2 y x 2 5 1 5 1 5 4 2
y x 2 5 2 5 4 5 1 2
The solutions are 1, 4 and 2,1 . Find the points of intersection between y x 2 5 and y 1 :
y x 5 32. y x 5 related 2
2
equation
The equation represents a parabola opening downward with vertex 0, 5 . The inequality symbol allows for equality, so draw the parabola as a solid curve. Test (0, 0): y x 2 5
y x2 5 1 x2 5 x2 4 x 2 y 1
?
0 02 5
The solutions are 2,1 and 2,1 .
?
0 5 true The test point satisfies the inequality. The solution set consists of the points below or on the parabola. y 1 The related equation y 1 is a horizontal line. The inequality y 1 represents all points above the line y 1 . The inequality is strict, so draw the line as a dashed line. x y 3 x y 3 related
The point 2,1 is a point of intersection of all three equations. Test it in y 1 : y 1 ?
1 1 false The point 2,1 does not satisfy y 1 , so
plot it as an open dot. Plot 1, 4 as a closed dot, and plot 2,1 as an open dot.
equation
The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): x y 3 ?
00 3 ?
0 3 true The points on the side of the line containing (0, 0) are solutions.
695
Chapter 5 Systems of Equations and Inequalities 33. z 2.4 x 0.55 y
b. z 600 x 850 y
at 0, 0 z 600 0 850 0 0
34. a. z 4 x 5 y
at 12, 36 z 600 12 850 36
at 2, 8 z 4 2 5 8 8 40 48
7200 30,600 37,800
at 6,14 z 4 6 5 14 24 70 94
at 48, 0 z 600 48 850 0 28,800 The values x 12 and y 36 produce the maximum value of the objective function. c. Maximum value: 37,800
at 12, 6 z 4 12 5 6 48 30 78
at 10, 2 z 4 10 5 2 40 10 50 The values x 2 and y 8 produce the minimum value of the objective function. b. Minimum value: 48
36. a. Let x represent the number of scoops of the whey powder. Let y represent the number of acres of the soy powder. Protein: z 20 x 18 y Constraints: x 0 , y 0 , x y 60 , 3x 2 y 150 ; 3 x 4 y 216 Find the points of intersection between pairs of bounding lines. x 0, y 0 0, 0 3x 4 y 216
35. a. Bounding lines: x 0 , y 0 , x y 48 , y 3x Find the points of intersection between pairs of bounding lines. y 3x y 3 0 y0 0, 0
x y 48 x 3x 48 4 x 48 x 12 y 3x 3 12 36 x y 48 x 0 48 x 48
3 0 4 y 216 4 y 216 y 54
12, 36
0, 54
3x 2 y 150 3x 2 x 60 150
48, 0
3 x 2 x 120 150 x 30 x y 60 30 y 60 y 30
30, 30
3x 4 y 216 3x 4 x 60 216 3x 4 x 240 216 x 24 x 24 x y 60 24 y 60 y 36
696
24, 36
Chapter 5 Cumulative Review 3x 2 y 150
b. The maximum protein content is 1140 g.
3x 2 0 150
c. z 18 x 20 y
3 x 150 x 50
at 0, 0 z 18 0 20 0 0
50, 0
at 0, 54 z 18 0 20 54 1080
at 30, 30 z 18 30 20 30
z 20 x 18 y at 0, 0 z 20 0 18 0 0
540 600 1140 at 24, 36 z 18 24 20 36
at 0, 54 z 20 0 18 54 972
at 30, 30 z 20 30 18 30
432 720
600 540
1152 at 50, 0 z 18 50 20 0 900 In this case, 24 scoops of whey protein should be mixed with 36 scoops of soy protein.
1140 at 24, 36 z 20 24 18 36 480 648 1128 at 50, 0 z 20 50 18 0 1000 30 scoops of each type of protein powder should be mixed to maximize protein content.
Chapter 5 Cumulative Review Exercises 2 x x 2 5 x 7
1.
2 x2 4 x 5x 7 0
u 5 u 12 0
12 4 2 7 1 57 2 2 4
u 5 or
1 57 4 2.
4, i
2t 8 t 4 2t 8 t 4 2t 8 t 2 8t 16 0 t 2 6t 8 0 t 4 t 2
2, 4
Let u x 4 . u 2 7u 60 0
2x x 7 0
1
2
2
2
x
3. x 2 4 7 x 2 4 60 0
t 4 or t 2
697
u 12
x 2 4 5
x 2 4 12
x 2 1 x i
x 2 16 x 4
Chapter 5 Systems of Equations and Inequalities log 2 x 4 log 2 x 1 3
4.
The expression is undefined for x 2 . This is a boundary point. The related equation has no real zeros.
log 2 x 4 log 2 x 1 3 x 4 3 log 2 x 1
x4 23 x 1 x4 8 x 1 x 4 8 x 1
Sign of x 2 :
:
Sign of
6
x 2
2
The solution set is , 2 .
x 4 8x 8 12 7 x 12 x 7
7. 3 x 2 1 8
3 x2 9 x2 3 x 2 3 or x 2 3 x 5 or x 1 , 5 1,
Check: log 2 x 4 log 2 x 1 3 12 12 log 2 4 ⱨ log 2 1 3 7 7 40 5 log 2 3 ⱨ log 2 3 7 7
8.
5 x 17 5 x 17 A B 2 2 x 6 x 9 x 3 x 3 x 3 2 5 x 17 A B 2 x 3 2 2 x 3 x 3 x 3 5 x 17 A x 3 B
x 32
undefined
undefined 12 ; The value does not check. 7 2 x 1 5. 50e 2000
5 x 17 Ax 3 A B
e 2 x1 40 2 x 1 ln 40 2 x 1 ln 40 1 ln 40 x 2 1 ln 40 2 6.
Sign of 6:
A 5
5 x 17 A x 3 A B 3 A B 17 3 5 B 17 15 B 17
5 x 17 5 2 2 x 6 x 9 x 3 x 3 2
x4 1 x2 x4 1 0 x2 x4 x2 1 0 x2 x2 x4 x2 0 x2 6 0 x2
B2
9. a. f g x f g x
2 g x 3 g x 2
2 5 x 1 3 5 x 1 2
2 25 x 2 10 x 1 15 x 3 50 x 20 x 2 15 x 3 2
50 x 2 5 x 1
698
Chapter 5 Cumulative Review b. g f x g f x
14.
5 2 x 2 3x 1 10 x 2 15 x 1 10.
f x 3 x 2
f x x 1 x3 x 2 9 x 9
y 3 x2
Find the zeros of the quotient. 1 1 1 9 9 1 0 9 1 0 9 0
x 3 y2 x3 y 2 y x3 2 f 1 x x3 2 11. log 5 256
Factors of 9 1, 3, 9 1, 3, 9 Factors of 1 1 1 1 2 10 18 9 1 1 9 9 1 1 9 9 0
0 x 1 x 9
f x x 1 x 2 9 2
2
log 256 3.4454 log 5
2
x 1, x 3i The zeros are 3i (multiplicity 1), 3i (multiplicity 1),and 1 (multiplicity 1.
1 12. f x x 3 4 x 2 2 2 1 3 1 11 2 f 1 1 4 1 2 4 2 2 2 2 1 3 2 f 3 3 4 3 2 2 27 49 36 2 2 2 f x2 f x1 Average rate of change x2 x1
15. a. The graph of f x 2 x 1 3 is the
graph of f x x shifted to the right 1 unit, stretched vertically by a factor of 2, and shifted downward 3 units.
49 11 38 f 3 f 1 19 2 2 2 3 1 2 2 2 13. x 3 y 6
3y x 6 1 y x2 3 1 The slope is . The slope of a 3 1 1 perpendicular line is 3 . 3 y y1 m x x1
b. Domain: , c. Range: 3,
y 1 3 x 2 y 1 3x 6 y 3x 7
699
Chapter 5 Systems of Equations and Inequalities 16. a. The graph of f x ln x 3 is the
Test for symmetry: 2 2 f x x x 1 x 2
graph of f x ln x shifted to the right 3 units.
x 2 x 1 x 2
2
f x is neither even nor odd.
b. Domain: 3, b. Domain: ,
c. Range: , 17. a. f x x 2 x 1 x 2
c. Range: ,
2
The leading term is x 2 x x x5 . The end behavior is up to the left and down to the right. 2 2 f 0 0 0 1 0 2 0
18. 3x 5 y 1
2
3 3x 5 x 4 1 5 3x 3 x 20 1 0 21 contradiction
The y-intercept is 0, 0 . The zeros of the function are 0 (multiplicity 2), 1 multiplicity 1, and 2 (multiplicity 2).
19. A 5a 2b 3c 10 B 3a b 2c 7 a 4b 4c 3 C Eliminate b from equations A and B.
A 5a 2b 3c 10 5a 2b 3c 10 Multiply by 2 B 3a b 2c 7 6a 2b 4c 14 a c 4 D Eliminate b from equations B and C. Multiply by 4 B 3a b 2c 7 12a 4b 8c 28 C a 4b 4c 3 a 4b 4c 3 13a 4c 25 E
700
Chapter 5 Cumulative Review Solve the systems of equations D and E. Multiply by 4 4a 4c 16 D a c 4 E 13a 4c 25 13a 4c 25 9a 9 a 1 Back substitute. C a 4b 4c 3 D a c 4 1 4b 4 3 3
1 c 4 c 3 c3
4b 8 b2
1, 2, 3 Multiply by 2. 2 2 4 x 2 6 y 2 20 20. A 2 x 3 y 10 2 2 Multiply by 3. B 5 x 2 2 y 2 13 15 x 6 y 39 2 19 x 19 x2 1 x 1
x 1: A
2 x 2 3 y 2 10
x 1: A
2 x 2 3 y 2 10
2 1 3 y 2 10
2 1 3 y 2 10
2 3 y 2 10
2 3 y 2 10
3 y 2 12
3 y 2 12
y2 4 y 2
y2 4 y 2
2
2
1, 2 , 1, 2 , 1, 2 , 1, 2
21. 3x y 1 3x y 1 related
The points on the side of the line containing (0, 0) are solutions. x 2 y 4 x 2y 4 related
equation
3x y 1 y 3 x 1 y 3x 1 The inequality symbol allows for equality, so draw the line as a solid line. Test (0, 0): 3x y 1
equation
x-intercept: x 2 0 4
y-intercept: 0 2y 4 2y 4 x4 y2 The inequality is strict, so draw the line as a dashed line.
3 0 0 1 ? ?
0 1 true 701
Chapter 5 Systems of Equations and Inequalities Test (0, 0):
x 2y 4 23.
0 2 0 4 ?
10 5x
4 5x 10 5x 4 5x 5 x x 5 5x 5x
?
0 4 true The points on the side of the line containing (0, 0) are solutions. Find the point of intersection. x 2y 4
10 5 x 20 5 x 5x 5x 30 5 x 5x 6 5x x
x 2 3 x 1 4 x 6x 2 4 7x 6 6 x 7 18 7 11 6 y 3x 1 3 1 7 7 7 7 6 11 Graph , as an open dot. It is not a 7 7 solution to the second inequality.
A t Pe rt
24. a.
A t 8000e rt A 5 8000e5 r 10,907.40 8000e5 r 10,907.40 e5 r 8000 10,907.40 5r ln 8000 10,907.40 ln 8000 r 0.062 5 A t 8000e0.062t b. 16,000 8000e0.062t
22.
2 e0.062t ln 2 0.062t ln 2 t 11.2 yr 0.062
f x h f x h 2 x h 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 2 xh h 5h h 2 x h 5
25.
y kxz 2 36 k 10 3
2
36 90k 36 2 k 90 5 2 2 384 2 y xz 2 12 4 76.8 5 5 5
702