Corrections: p. 1, 3, 5, 36 NOTE THAT THESE CORRECTIONS SHOULD BE MADE IN CA-8 ALSO.
CHAPTER P
PREREQUISITES
P.1
Modeling the Real World with Algebra 1
P.2 P.3
Real Numbers 2 Integer Exponents and Scientific Notation 7
P.4
Rational Exponents and Radicals 12
P.5
Algebraic Expressions 16
P.6
Factoring 19
P.7
Rational Expressions 24
P.8
Solving Basic Equations 31
P.9
Modeling with Equations 36 Chapter P Review 42 Chapter P Test 48
¥
FOCUS ON MODELING: Making Optimal Decisions 51
1
P
PREREQUISITES
P.1
MODELING THE REAL WORLD WITH ALGEBRA
1. Using this model, we find that 15 cars have W 4 15 60 wheels. To find the number of cars that have a total of W .If the cars in a parking lot have a total of 124 wheels, we find that there are W wheels, we write W 4X X 4 X 124 4 31 cars in the lot.
2. If each gallon of gas costs $350, then x gallons of gas costs $35x. Thus, C 35x. We find that 12 gallons of gas would cost C 35 12 $42. 3. If x $120 and T 006x, then T 006 120 72. The sales tax is $720. 4. If x 62,000 and T 0005x, then T 0005 62,000 310. The wage tax is $310. 5. If 70, t 35, and d t, then d 70 35 245. The car has traveled 245 miles. 6. V r 2 h 32 5 45 1414 in3 240 N 30 miles/gallon G 8 175 175 G 7 gallons (b) 25 G 25
8. (a) T 70 0003h 70 0003 1500 655 F
7. (a) M
(b) 64 70 0003h 0003h 6 h 2000 ft
10. (a) P 006s 3 006 123 1037 hp
9. (a) V 95S 95 4 km3 38 km3 (b) 19 km3 95S S 2 km3 11. (a)
(b) 75 006s 3 s 3 125 so s 5 knots
Depth (ft)
Pressure (lb/in2 )
0
045 0 147 147
10
045 10 147 192
20
045 20 147 237
30
Insert solution for part (c)
(b) We know that P 30 and we want to find d, so we solve the equation 30 147 045d 153 045d
153 340. Thus, if the pressure is 30 lb/in2 , the depth 045 is 34 ft. d
045 30 147 282
40
045 40 147 327
50
045 50 147 372
60
045 60 147 417 (b) We solve the equation 40x 120,000
12. (a) Population
Water use (gal)
0
0
1000
40 1000 40,000
2000 3000 4000 5000
x
120,000 3000. Thus, the population is about 3000. 40
40 2000 80,000
40 3000 120,000 40 4000 160,000 40 5000 200,000
13. The number N of cents in q quarters is N 25q. ab 14. The average A of two numbers, a and b, is A . 2
1
2
CHAPTER P Prerequisites
15. The cost C of purchasing x gallons of gas at $350 a gallon is C 35x.
16. The amount T of a 15% tip on a restaurant bill of x dollars is T 015x. 17. The distance d in miles that a car travels in t hours at 60 mi/h is d 60t. d 18. The speed r of a boat that travels d miles in 3 hours is r . 3 19. (a) $12 3 $1 $12 $3 $15 (b) The cost C, in dollars, of a pizza with n toppings is C 12 n.
(c) Using the model C 12 n with C 16, we get 16 12 n n 4. So the pizza has four toppings.
20. (a) 3 30 280 010 90 28 $118 daily days cost miles (b) The cost is , so C 30n 01m. rental rented per mile driven (c) We have C 140 and n 3. Substituting, we get 140 30 3 01m 140 90 01m 50 01m m 500. So the rental was driven 500 miles.
21. (a) (i) For an allelectric car, the energy cost of driving x miles is Ce 004x.
(ii) For an average gasoline powered car, the energy cost of driving x miles is C g 012x.
(b) (i) The cost of driving 10,000 miles with an allelectric car is Ce 004 10,000 $400.
(ii) The cost of driving 10,000 miles with a gasoline powered car is C g 012 10,000 $1200.
22. (a) If the width is 20, then the length is 40, so the volume is 20 20 40 16,000 in3 .
(b) In terms of width, V x x 2x 2x 3 . 4a 3b 2c d 4a 3b 2c 1d 0 f . 23. (a) The GPA is abcd f abcd f (b) Using a 2 3 6, b 4, c 3 3 9, and d f 0 in the formula from part (a), we find the GPA to be 54 463429 284. 649 19
P.2
THE REAL NUMBERS
1. (a) The natural numbers are 1 2 3 .
(b) The numbers 3 2 1 0 are integers but not natural numbers. p 5 , 1729 . (c) Any irreducible fraction with q 1 is rational but is not an integer. Examples: 32 , 12 23 q p (d) Any number which cannot be expressed as a ratio of two integers is irrational. Examples are 2, 3, , and e. q
2. (a) ab ba; Commutative Property of Multiplication
(b) a b c a b c; Associative Property of Addition (c) a b c ab ac; Distributive Property
3. (a) In setbuilder notation: x 3 x 5 (b) In interval notation: 3 5
(c) As a graph: _3
5
4. The symbol x stands for the absolute value of the number x. If x is not 0, then the sign of x is always positive.
5. The distance between a and b on the real line is d a b b a. So the distance between 5 and 2 is 2 5 7. 6. (a) If a b, then any interval between a and b (whether or not it contains either endpoint) contains infinitely many ba numbers—including, for example a n for every positive n. (If an interval extends to infinity in either or both 2 directions, then it obviously contains infinitely many numbers.)
SECTION P.2 The Real Numbers
3
(b) No, because 5 6 does not include 5. 7. (a) No: a b b a b a in general.
(b) No; by the Distributive Property, 2 a 5 2a 2 5 2a 10 2a 10.
8. (a) Yes, absolute values (such as the distance between two different numbers) are always positive. (b) Yes, b a a b.
9 3, 10 (b) Integers: 2, 100 2 50, 9 3, 10 (c) Rational numbers: 45 92 , 13 , 16666 53 , 2, 100 2 , 9 3, 10 (d) Irrational numbers: 2, 314
10. (a) Natural numbers: 2,
9. (a) Natural number: 100 (b) Integers: 0, 100, 8
(c) Rational numbers: 15, 0, 52 , 271, 314, 100, 8 (d) Irrational numbers: 7,
11. Commutative Property of addition
12. Commutative Property of multiplication
13. Associative Property of addition
14. Distributive Property
15. Distributive Property
16. Distributive Property
17. Commutative Property of multiplication
18. Distributive Property
19. x 3 3 x
20. 7 3x 7 3 x
21. 4 A B 4A 4B
22. 5x 5y 5 x y
23. 2 x y 2x 2y
24. a b 5 5a 5b 26. 43 6y 43 6 y 8y
25. 5 2x y 5 2 x y 10x y 27. 52 2x 4y 52 2x 52 4y 5x 10y
28. 3a b c 2d 3ab 3ac 6ad
15 29 29. (a) 23 57 14 21 21 21
15 1 30. (a) 25 38 16 40 40 40
31. (a) 23 6 32 23 6 23 32 4 1 3 1 5 4 13 1 13 (b) 3 14 1 45 12 4 4 5 5 4 5 20
2 32. (a) 2 3 2 32 23 12 3 13 93 13 83 2
33. (a) 2 3 6 and 2 72 7, so 3 72
34. (a) 3 23 2 and 3 067 201, so 23 067
5 3 10 9 1 (b) 12 8 24 24 24
(b) 6 7
wrong problem
(c) 35 72
15 4 25 (b) 32 58 16 36 24 24 24 24 2
3 2 1 2 1 2 1 45 9 (b) 15 23 51 21 51 21 10 10 12 3 3 10 15 10 5 10 5
(b) 23 067
(c) 06 06
35. (a) False
36. (a) False:
(b) True
(b) False
37. (a) True
(b) False
38. (a) True
3 173205 17325.
(b) True
4
CHAPTER P Prerequisites
39. (a) x 0
(b) t 4
40. (a) y 0
(d) 5 x 13
(c) a (e) 3 p 5
(b) z 3
(c) b 8
(d) 0 17
(e) y 2
41. (a) A B 1 2 3 4 5 6 7 8
42. (a) B C 2 4 6 7 8 9 10
(b) A B 2 4 6
(b) B C 8
43. (a) A C 1 2 3 4 5 6 7 8 9 10 (b) A C 7
44. (a) A B C 1 2 3 4 5 6 7 8 9 10 (b) A B C 46. (a) A C x 1 x 5
45. (a) B C x x 5
(b) A B x 2 x 4
(b) B C x 1 x 4
48. 2 8] x 2 x 8
47. 3 0 x 3 x 0 _3
0
2
50. 6 12 x 6 x 12
49. [2 8 x 2 x 8 2
8
_6
51. [2 x x 2
1
53. x 1 x 1]
54. 1 x 2 x [1 2] 1
1
55. 2 x 1 x 2 1]
_5
1
58. 5 x 2 x 5 2
_1
_5
(b) 3 5]
(c) 3
60. (a) [0 2
2
(b) 2 0]
(c) 0]
62. 2 0 1 1 0
61. 2 0 1 1 2 1 _2
2
56. x 5 x [5
57. x 1 x 1
59. (a) [3 5]
1
_ _2
52. 1 x x 1
2
_2
8
1
_1
0
SECTION P.2 The Real Numbers
63. [4 6] [0 8 [0 6] 0
64. [4 6] [0 8 [4 8 6
65. 4 4 _4
_4
8
66. 6] 2 10 2 6] 4
67. (a) 50 50 (b) 13 13 69. (a) 6 4 6 4 2 2 1 1 1 (b) 1 1
71. (a) 2 6 12 12 (b) 13 15 5 5
73. 2 3 5 5 75. (a) 17 2 15
(b) 21 3 21 3 24 24 3 12 55 67 67 (c) 10 11 8 40 40 40 40
2
6
68. (a) 2 8 6 6 (b) 8 2 8 2 6 6 70. (a) 2 12 2 12 10 10 (b) 1 1 1 1 1 1 1 0 1 1 1 72. (a) 6 24 4 4 5 (b) 712 127 5 1 1
74. 25 15 4 4 7 1 49 5 54 18 18 76. (a) 15 21 105 105 105 35 35 (b) 38 57 38 57 19 19.
(c) 26 18 26 18 08 08.
77. (a) Let x 0777 . So 10x 77777 x 07777 9x 7. Thus, x 79 .
13 (b) Let x 02888 . So 100x 288888 10x 28888 90x 26. Thus, x 26 90 45 . 19 (c) Let x 0575757 . So 100x 575757 x 05757 99x 57. Thus, x 57 99 33 .
78. (a) Let x 52323 . So 100x 5232323 1x 52323 99x 518. Thus, x 518 99 .
62 (b) Let x 13777 . So 100x 1377777 10x 137777 90x 124. Thus, x 124 90 45 .
1057 (c) Let x 213535 . So 1000x 21353535 10x 213535 990x 2114. Thus, x 2114 990 495 . 79. 3, so 3 3. 80. 2 1, so 1 2 2 1.
81. a b, so a b a b b a.
82. a b a b a b b a 2b
83. (a) a is negative because a is positive.
(b) bc is positive because the product of two negative numbers is positive. (c) a ba b is positive because it is the sum of two positive numbers.
wrong problem
(d) ab ac is negative: each summand is the product of a positive number and a negative number, and the sum of two negative numbers is negative. 84. (a) b is positive because b is negative.
(b) a bc is positive because it is the sum of two positive numbers.
(c) c a c a is negative because c and a are both negative. (d) ab2 is positive because both a and b2 are positive.
85. Distributive Property
5
6
CHAPTER P Prerequisites
86. (a) When L 60, x 8, and y 6, we have L 2 x y 60 2 8 6 60 28 88. Because 88 108 the post office will accept this package. When L 48, x 24, and y 24, we have L 2 x y 48 2 24 24 48 96 144, and since 144 108, the post office will not accept this package. (b) If x y 9, then L 2 9 9 108 L 36 108 L 72. So the length can be as long as 72 in. 6 ft.
m1 m2 m1 m m1n2 m2n1 and y be rational numbers. Then x y 2 , n1 n2 n1 n2 n1 n2 m m n m 2 n1 m m m m m , and x y 1 2 1 2 . This shows that the sum, difference, and product xy 1 2 1 2 n1 n2 n1 n2 n1 n2 n1n2 of two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarily irrational; for example, 2 2 2, which is rational. Also, the sum of two irrational numbers is not necessarily irrational; for example, 2 2 0 which is rational.
87. Let x
88. 12
2 is irrational. If it were rational, then by Exercise 6(a), the sum 12 2 12 2 would be rational, but
this is not the case. Similarly, 12 2 is irrational. (a) Following the hint, suppose that r t q, a rational number. Then by Exercise 6(a), the sum of the two rational numbers r t and r is rational. But r t r t, which we know to be irrational. This is a contradiction, and hence our original premise—that r t is rational—was false. a (b) r is a nonzero rational number, so r for some nonzero integers a and b. Let us assume that rt q, a rational b a c bc c , implying number. Then by definition, q for some integers c and d. But then r t q t , whence t d b d ad that t is rational. Once again we have arrived at a contradiction, and we conclude that the product of a rational number and an irrational number is irrational.
89. x 1 x
1
2
10
100
1000
1
1 2
1 10
1 100
1 1000
As x gets large, the fraction 1x gets small. Mathematically, we say that 1x goes to zero. x
1
05
01
001
0001
1 x
1
1 05 2
1 01 10
1 001 100
1 0001 1000
As x gets small, the fraction 1x gets large. Mathematically, we say that 1x goes to infinity. 90. We can construct the number 2 on the number line by Ï2
transferring the length of the hypotenuse of a right triangle with legs of length 1 and 1. Similarly, to locate 5, we construct a right triangle with legs of length 1 and 2. By the Pythagorean Theorem, the length of the hypotenuse is 12 22 5. Then transfer the
length of the hypotenuse to the number line.
_1
0
1 1 Ï5
_1
0
1
Ï2
2
3
2 Ï5
3
1
The square root of any rational number can be located on a number line in this fashion. The circle in the second figure in the text has circumference , so if we roll it along a number line one full rotation, we have found on the number line. Similarly, any rational multiple of can be found this way.
SECTION P.3 Integer Exponents and Scientific Notation
7
abab a b a b a. 2 2 On the other hand, if b a, then max a b b and a b a b b a. In this case, abba a b a b b. 2 2 If a b, then a b 0 and the result is trivial. a b b a a b a b a. (b) If a b, then min a b a and a b b a. In this case 2 2 a b a b Similarly, if b a, then b; and if a b, the result is trivial. 2 92. Answers will vary.
91. (a) Suppose that a b, so max a b a and a b a b. Then
93. (a) Subtraction is not commutative. For example, 5 1 1 5. (b) Division is not commutative. For example, 5 1 1 5.
(c) Putting on your socks and putting on your shoes are not commutative. If you put on your socks first, then your shoes, the result is not the same as if you proceed the other way around. (d) Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result. (e) Washing laundry and drying it are not commutative. 94. (a) If x 2 and y 3, then x y 2 3 5 5 and x y 2 3 5. If x 2 and y 3, then x y 5 5 and x y 5. If x 2 and y 3, then x y 2 3 1 and x y 5. In each case, x y x y and the Triangle Inequality is satisfied. (b) Case 0: If either x or y is 0, the result is equality, trivially. xy Case 1: If x and y have the same sign, then x y x y
if x and y are positive x y. if x and y are negative
Case 2: If x and y have opposite signs, then suppose without loss of generality that x 0 and y 0. Then x y x y x y.
P.3
INTEGER EXPONENTS AND SCIENTIFIC NOTATION
1. Using exponential notation we can write the product 5 5 5 5 5 5 as 56 .
2. Yes, there is a difference: 54 5 5 5 5 625, while 54 5 5 5 5 625. 3. In the expression 34 , the number 3 is called the base and the number 4 is called the exponent.
4. (a) When we multiply two powers with the same base, we add the exponents. So 34 35 39 . 35 (b) When we divide two powers with the same base, we subtract the exponents. So 2 33 . 3 2 5. When we raise a power to a new power, we multiply the exponents. So 34 38 .
6. (a) 21
1 2
(b) 23
1 8
(c)
1 1 2 2
1 (d) 3 23 8 2
7. To move a number raised to a power from numerator to denominator or from denominator to numerator change the sign of 1 1 a 3 1 6a 2 the exponent. So a 2 2 , 2 b2 , 2 3 2 , and 3 6a 2 b3 . a b b a b b
8. Scientists express very large or very small numbers using scientific notation. In scientific notation, 8,300,000 is 83 106 and 00000327 is 327 105 .
8
CHAPTER P Prerequisites
9. (a) No,
2 2 9 2 3 . 3 2 4
(b) No, 53 125 and 53 53 125.
3 10. (a) No, x 2 x 23 x 6 .
3 3 (b) No, 2x 4 23 x 4 8x 12 .
11. (a) 26 26 64
12. (a) 53 125 (b) 53 53 125 2 (c) 52 25 4
(b) 26 64 2 27 33 (c) 15 33 25 52 0 1 5 21 13. (a) 3 2
14. (a) 23 20 23 1 8 1 1 (b) 23 20 3 1 8 2 53 125 3 3 (c) 5 27 33
23 1 1 3 8 30 2 2 2 9 3 2 (c) 3 2 4
(b)
15. (a) 53 5 531 625
16. (a) 38 35 385 313
107 1074 1000 104 4 (c) 35 354 320
(b) 54 52 542 25 3 (c) 22 223 64 710 7108 72 49 78 62 (b) 2 622 64 1296 6
(b)
83 83 1 831 82 64 8 8 81 81 1 (b) 1 811 82 8 64 8 1 3 1 31 4 5 (c) 5 5 5 625
17. (a)
(c) 41 42 412 43
18. (a)
1 64
19. (a) t 5 t 2 t 52 t 7 2 2 (b) 4z 3 42 z 3 16z 32 16z 6
20. (a) a 4 a 6 a 46 a 10 3 3 8 (b) 2b3 23 b3 8b33 9 b 2 (c) 2y 10 y 11 2y 1011 2y 1 y
1 21. (a) x 5 x 3 x 53 x 2 2 x
1 22. (a) y 2 y 5 y 25 y 3 3 y
(c) x 3 x 5 x 35 x 2
(b) 2 4 5 245 1 (c)
1
x 16 x 1610 x 6 x 10
a 9 a 2 a 921 a 6 a 3 3 3 (b) a 2 a 4 a 24 a 6 a 63 a 18
23. (a)
x 3 5x 6 x 3 5x 9 5x 6 3 2 8 2 25. (a) 3x 3 y 2 2y 3 3 2x 3 y 2 y 3 6x 3 y 5 (c)
1 (b) z 5 z 3 z 4 z 534 z 2 2 z y7 y0 1 (c) 10 y 7010 y 3 3 y y z2 z4 24. (a) 3 1 z 2431 z 4 z z 4 4 4 (b) 2a 3 a 2 24 a 32 16 a 5 16a 20 3 2z 3 33 z 23 2z 3 54z 9 (c) 3z 2
SECTION P.3 Integer Exponents and Scientific Notation
2 z3 (b) 52 z 2
52 z2
2
9
2 52 2 z 3 254 z3 2 z z2
4n 2 8m 2 n 4 12 n 2 8 12 m 2 n 42 2 m 3 27a 14 a 2 b1 33 a 43 b23 a 2 b1 33 a 122 b61 (b) 3a 4 b2 b7
26. (a)
7 x 2 y 1 25 y 1 x x y x 5 3 a 33 a9 a3 (b) 3 6 2 2b 8b 23 b2
27. (a)
y 2 z 3 y 1 2 3 3 y 1 y z yz 2 2 2 y8 x 3 y 2 33 y 22 6 y 4 (b) x x x 3 y 2 x 12 3 5 19 a 3 b2 a2 2533 b523 c33 a b 29. (a) a b c3 c9 2 u 1 2 10 1233 2223 (b) 3 u u 11 u 3 2 28. (a)
2 22 432 522 232 x 10 2x 3 y 2 x 4 z2 x 30. (a) y z 4 5 3 4 4y z yz 3 rs 2 332 s 2322 r 9 s 2 (b) 2 r r 3 s 2
31. (a)
8a 3 b4 4a 8 4a 35 b45 9 5 5 2a b b
32. (a)
5x y 2 5x 11 y 23 5x 2 y x 1 y 3
33. (a) 69,300,000 693 107
3 125 y 513 x 23 y 3 6 3 5x 2 x y 3 9 2a 1 b 3 a 1323 b333 a (b) 2 a 2 b3 8b12 (b)
34. (a) 129,540,000 12954 108
(b) 7,200,000,000,000 72 1012
(b) 7,259,000,000 7259 109
(c) 0000028536 28536 105
(c) 0000 000 001 4 14 109
(d) 00001213 1213 104
(d) 00007029 7029 104
35. (a) 319 105 319,000
36. (a) 71 1014 710,000,000,000,000
(b) 2721 108 272,100,000
(b) 6 1012 6,000,000,000,000
(c) 2670 108 0000 000 026 70
(c) 855 103 000855
(d) 9999 109 0000 000 009 999
(d) 6257 1010 0000 000 000 625 7
37. (a) 5,900,000,000,000 mi 59 1012 mi
10
CHAPTER P Prerequisites
(b) 0000 000 000 000 4 cm 4 1013 cm
(c) 33 billion billion molecules 33 109 109 33 1019 molecules
38. (a) 93,000,000 mi 93 107 mi
(b) 0000 000 000 000 000 000 000 053 g 53 1023 g
(c) 5,970,000,000,000,000,000,000,000 kg 597 1024 kg 39. 72 109 1806 1012 72 1806 109 1012 130 1021 13 1020 40. 1062 1024 861 1019 1062 861 1024 1019 914 1043
1295643 1295643 109 109176 01429 1019 1429 1019 41. 3610 2511 3610 1017 2511 106 731 10 16341 1028 731 16341 1028 731 16341 101289 63 1038 42. 9 00000000019 19 19 10 5 2 1582 10 162 10 162 1582 00000162 001582 105283 0074 1012 43. 594621 58 594621000 00058 594621 108 58 103 74 1014
9 3542 106 8774796 35429 1054 105448 319 104 10102 319 10106 44. 12 12 48 27510376710 4 505 10 505 10 45. 1050 1010 1050 , whereas 10101 10100 10100 10 1 9 10100 1050 . So 1010 is closer to 1050 than 10100 is to 10101 .
46. (a) b5 is negative since a negative number raised to an odd power is negative.
(b) b10 is positive since a negative number raised to an even power is positive. (c) ab2 c3 we have positive negative2 negative3 positive positive negative which is negative.
(d) Since b a is negative, b a3 negative3 which is negative. (e) Since b a is negative, b a4 negative4 which is positive. (f)
a 3 c3 negative positive3 negative3 positive negative which is negative. 6 6 positive positive positive b c negative6 negative6
47. Since one light year is 59 1012 miles, Centauri is about 43 59 1012 254 1013 , or 25,400,000,000,000 miles away. 93 107 mi t st s 500 s 8 13 min. s 186 000 103 liters 3 14 2 133 1021 liters 49. Volume Average depth Area 37 10 m 36 10 m m3
48. 93 107 mi 186 000
50. Each person’s share is equal to
2670 1013 National debt 80555 104 $80,555. Population 33145 108
51. First, we estimate the total mass of the stars in the observable universe: Number of stars Total star mass Number of galaxies Mass of typical star 15 2 177 1012 1011 1030 Galaxy 531 1053 kg
SECTION P.3 Integer Exponents and Scientific Notation
11
Thus, the number of hydrogen atoms in the observable universe is 531 1053 Total star mass 318 1080 Mass of a single hydrogen atom 167 1027
52. (a)
Weight
Height
A
295 lb
4232
obese
B
105 lb
5 ft 10 in. 70 in.
1695
underweight
C
220 lb
5 ft 6 in. 66 in.
overweight
110 lb
6 ft 4 in. 76 in.
2678
D
5 ft 2 in. 62 in.
2012
normal
(b) Answers will vary. 53. I 15,000 10037512n 1
BMI 703
W H2
Patient
Year
Total interest
1
$68910
2
140985
3
216372
4
295222
5
377694
Result
54. Since 106 103 103 it would take 1000 days 274 years to spend the million dollars.
Since 109 103 106 it would take 106 1,000,000 days 273972 years to spend the billion dollars. 5 18 185 25 32 (b) 206 056 20 056 106 1,000,000 55. (a) 5 9 9 m factors
am a a a 56. (a) n . Because m n, we can cancel n factors of a from numerator and denominator and are left with a a a a n factors
m n factors of a in the numerator. Thus, n factors
(b)
57. (a)
a n b
n factors
a a a a a a an n b b b b b b b
a n b
am a mn . an
n factors
1 1 bn n n n n a b a ab
a n 1a n 1 bm (b) m n bm n m b 1b a a
12
CHAPTER P Prerequisites
P.4
RATIONAL EXPONENTS AND RADICALS
1. (a) Using exponential notation we can write 3 5 as 513 . (b) Using radicals we can write 512 as 5. 2 12 2 5212 5 and 5 512 5122 5. (c) No. 52 52 3 12 2. 412 23 8; 43 6412 8 3. Because the denominator is of the form
a, we multiply numerator and denominator by a: 1 1 3 33 . 3
3
3
4. 513 523 51 5
4a 2 2a. 6. No. For example, if a 2, then a 2 4 8 2 2, but a 2 0. 7. (a) 144 122 12 8. (a) 49 72 7 5 3 (b) 5 32 25 2 (b) 3 125 53 5 3 3 3 1 4 1 4 4 1 1 (c) 27 13 13 (c) 5. No. If a is negative, then
81
3
3
9. (a) 3 3 16 3 3 8 2 3 3 8 3 2 6 3 2 18 18 2 2 (b) 81 9 3 81 3 3 3 3 27 93 (c) 4 2 2 4 11. (a) 3 15 45 9 5 3 5 48 48 (b) 16 4 3 3 3 (c) 3 24 3 18 3 24 18 63 2 6 3 2 72 72 93 13. (a) 8 8 (b) 3 9 3 24 3 9 24 3 216 6 1 4 1 1 1 1 1 4 4 (c) 4 32 8 32 8 256 4 4 4 15. (a) x x 4 4 (b) 16x 8 24 x 8 2x 2
3 3 274 33 3 3 3 (b) 4 64y 6 4 24 y 4 4y 2 2 y 4 4y 2 3 3 19. (a) 3 8x 9 y 3 23 x 3 y 3 2x 3 y (b) 4 8x 6 y 2 4 2x 2 y 2 4 8x 6 y 2 2x 2 y 2 4 16x 8 y 4 4 4 24 x 2 y 4 2x 2 y 17. (a)
10. (a) 2 3 81 2 3 27 3 2 3 27 3 3 6 3 3 18 92 3 2 (b) 5 5 25 2 3 12 43 (c) 49 7 49 12. (a) 10 32 320 64 5 8 5 54 54 (b) 3 6 6 3 (c) 3 15 3 75 3 15 75 53 9 5 3 9 3 27 3 3 3 9 3 3 9 14. (a) 3 4 2 4 2 8 2 5 5 1 5 5 1 (b) 4 128 4 128 32 2 3 1 12 3 1 3 12 (c) 3 96 8 2 96 15 5 16. (a) x 10 x 10 x2 13 (b) 3 x 3 y 6 x 3 y 6 x y2 4 4 4 18. (a) x 2 z 5 z 4 x 2 z z x 2 z, z 0 2 (b) x 6 y 6 x 3 y 3 x3 y3 2 3 2 2 20. (a) 16x 4 y 6 z 2 42 x 2 y z 4x 2 y3 z 19 3 3 512x 9 512x 9 2x (b)
SECTION P.4 Rational Exponents and Radicals
32 18 16 2 9 2 16 2 9 2 4 2 3 2 7 2 (b) 75 48 25 3 16 3 25 3 16 3 5 3 4 3 9 3 22. (a) 125 45 25 5 9 5 5 5 3 3 8 5 (b) 3 54 3 16 3 27 2 3 8 2 3 27 3 2 3 8 3 2 3 2 23. (a) 9a 3 a 9 a 2 a a 3a a a 3a 1 a x (b) 16x x 5 16 x x 4 x x 2 4 3 3 3 24. (a) x 4 3 8x x 3 x 23 x x 2 3 x (b) 4 18rt 3 5 32r 3 t 5 4 32 2 r t 2 t 5 42 2 r 2 r t 4 t 4 3 t 2rt 5 4 r t 2 2rt 20rt 2 12t 2rt 25. (a) 36x 2 36x 4 62 x 2 1 x 2 6x 1 x 2 (b) 81x 2 81y 2 92 x 2 y 2 9 x 2 y 2 26. (a) 27a 3 63a 2 32 a 2 3a 7 3a 3a 7 (b) 25t 2 100t 2 125t 2 5 5t, t 0 28. 5 6 615 27. 10 1012 15 1 1 1 5 29. 735 73 73 30. 652 52 12 5 6 65 6 21. (a)
1 1 31. 12 512 5 5 1 1 1 33. 534 34 4 3 4 5 125 5
1 1 32. 13 713 3 7 7 1 34. y 15 y 32 y3 1 1 36. 34 x 34 4 3 x x
1 1 35. 23 x 23 3 2 x x 37. (a) 1614 4 16 2
(b) 813 3 8 2
38. (a) 2713 3 27 3
(b) 813 3 8 2
39. (a) 3225
2 5 32 22 4
(b)
2 3 125 52 25 32 3 3 125 25 5 25 (b) 64 8 512 64
40. (a) 12523
1 1 1 (c) 2743 4 4 81 3 3 27
41. (a) 523 513 52313 51 5
335 33525 315 5 3 325 3 3 (c) 3 4 413 4133 41 4 (b)
42. (a) 327 3127 327127 32 9
12 3 1 9 4 9 4 2 49
1 1 (c) 912 3 9 13 1 1 1 (c) 3 8 2 8 3 34 4 8 16 16 (c) 4 81 27 81
13
14
CHAPTER P Prerequisites
723 1 72353 71 7 753 10 10 1 1 (c) 5 6 615 61510 62 2 36 6 43. When x 3, y 4, z 1 we have x 2 y 2 32 42 9 16 25 5. 4 44. When x 3, y 4, z 1 we have 4 x 3 14y 2z 4 33 14 4 2 1 4 27 56 2 4 81 34 3. (b)
45. When x 3, y 4, z 1 we have
23 23 23 113 9x23 2y23 z 23 9 323 2 423 123 33 32 22 1 9 4 1 14.
1 . 46. When x 3, y 4, z 1 we have x y2z 3 42 1 122 144 48. (a) 4b12 8b14 412 8b1214 16b34 47. (a) x 34 x 54 x 3454 x 2 2 (b) y 23 y 43 y 2343 y 2 5a 12 32 5a 34212 45a 2 (b) 3a 34
43 23 432313 53 13 6 (b) 3x 12 y 13 36 x 126 y 136 36 729x 3 y 2
49. (a)
51. (a)
23 8a 6 b32 823 a 623 b3223 4a 4 b
35
23 1 50. (a) 8y 3 823 y 323 2 4y 13 1 4 6 413 613 (b) u u 43 2 u 32 8b9 (b) 4a 4 b6 432 a 432 b632 8a 6 b9 6 a
x3 x 535 y 1335 15 y 13 12 16u 6 23 (b) u 3 2 u 313 213 1612 u 612 2312 4u 23 u 3 13 4u 4
52. (a)
x 5 y 13
12 x 1 32 x 142 x 112 9x 12 x 12 9 4 53. (a) 4 2 412 y y y y2 y2 y 4 2 44323 34 134 22 132 43 423 42 213 313 (b) z 1 z 12 z 14 z 122 81z 4 z 81z 5 81z 5 3 3 y 23 y 2 x 1 x 1 x 1 y 23 2 54. (a) 3 1 3 2 x 1 2 x y x y x 1 y 2 2 2 13 13 2 13 2 y3 23 3 4 3 23 3 6 z x y6 16y 6 z 43 x 1 y 2 8y 8 x y 4y z (b) 2 2 13 12 4 43 x x 8z 2z x 813 z 4 x 12 12 12 55. (a) x 3 x 3 x 312 x 32 56. (a) x 5 x 5 x 512 x 52 15 14 5 4 (b) x 6 x 6 x 615 x 65 (b) x 6 x 6 x 614 x 32 4 57. (a) 6 y 5 3 y 2 y 56 y 23 y 5623 y 32 58. (a) b3 b b3412 b54 3 2 (b) 5 3 x 2 4 x 5 2x 1314 10x 712 (b) 2 a a 2a 1223 2a 76 6 60. (a) 5 x 3 y 2 10 x 4 y 16 x 35410 y 251610 x y 2 59. (a) 4st 3 s 3 t 2 412 s 1236 t 3226 2st 116 4 7 3 x 8x 2 7434 (b) x x (b) 813 x 2312 2x 16 4 3 x x
3x 14 y
2
SECTION P.4 Rational Exponents and Radicals
15
13 12 61. (a) 3 y y y 112 y 3213 y 12 s 54 62. (a) s s 3 s 132 2 4 3 3y 3u 18u 5 9u 2 3 27y 3 54x y (b) (b) 5 3 3 3 2 x 2x y x 2u 12 1 6 6 3 1 12 12 3 64. (a) 4 3 63. (a) 6 3 6 6 6 3 3 3 2 15 3 3 2 6 12 43 5 (b) (b) 2 2 2 2 5 5 5 5 9 234 948 9 8 8 513 835 14 34 (c) (c) 23 13 4 3 2 2 5 2 2 5 5 2 5 1 5x 5x s s 3t 3st 1 65. (a) 66. (a) 5x 3t 3t 3t 3t 5x 5x 5x 3 2 3 x x 5 5x b a a a b2 (b) (b) 6 3 3 5 5 5 5 b b b2 b2 5 2 5 2 25 25 1 x 1 x 1 c 1 c (c) 5 3 35 25 (c) 35 35 25 x x x x c c c c 1 1 1 1 5 5 1 2 2 67. (a) 4 4 4 (b) 4 64 4 40 8 16 4 4 43 40 4 x 32 x 324 x6 68. (a) x 6 y2 y 12 y 124 y 2 12 5 u 5u (b) 25u 2 4 2512 u 212 412 5 u 2 2 . Since u 0, this is equivalent to 2 . 4 2 12 24 y 69. (a) y y y y 12 4 (b) 81 8 z 8 8112 8412 z 8412 9 z. Since 0 and z 0, this is equivalent to 9z. 70. (a)
(b)
x 32
y 12
4
4 3 3 z 6
12
4
y3
12 z 212
x 2
x 32
y 12
412
4
x 2 x 6 x 2 x4 2 3 3 y y y y
32 2 z z
71. (a) Since 12 13 , 212 213 . 12 13 12 13 (b) 12 212 and 12 213 . Since 12 13 , we have 12 12 .
112 112 343112 ; 413 4412 44 256112 . So 714 413 . 72. (a) We find a common root: 714 7312 73 16 16 (b) We find a common root: 3 5 513 526 52 2516 ; 3 312 336 33 2716 . So 3 5 3. 1 mile 73. First convert 1135 feet to miles. This gives 1135 ft 1135 0215 mi. Thus the distance you can see is given 5280 feet by D 2r h h 2 2 3960 0215 02152 17028 413 miles. 74. (a) Using f 04 and substituting d 65, we obtain s 30 f d 30 04 65 28 mi/h. (b) Using f 05 and substituting s 50, we find d. This gives s 30 f d 50 30 05 d 50 15d 2500 15d d 500 3 167 feet.
16
CHAPTER P Prerequisites
75. Since 1 day 86,400 s, 36525 days 31,557,600 s. Substituting, we obtain 13 23 667 1011 199 1030 d 315576 107 15 1011 m 15 108 km. 2 4 7523 005012 17707 ft/s. 24123 0040 (b) Since the volume of the flow is V A, the canal discharge is 17707 75 13280 ft3 s.
76. (a) Substituting the given values we get V 1486 77. (a)
n
1
2
5
10
100
21n
211 2
212 1414
215 1149
2110 1072
21100 1007
So when n gets large, 21n decreases to 1. (b) n 1n 1 2
1 11 1 2
2 05
12 1 2
0707
1n So when n gets large, 12 increases to 1.
P.5
5 15 1 2
0871
10 110 1 0933 2
100 1100 1 0993 2
ALGEBRAIC EXPRESSIONS
1. (a) 2x 3 12 x 3 is a polynomial. (The constant term is not an integer, but all exponents are integers.) (b) x 2 12 3 x x 2 12 3x 12 is not a polynomial because the exponent 12 is not an integer. (c)
1
x 2 4x 7
is not a polynomial. (It is the reciprocal of the polynomial x 2 4x 7.)
(d) x 5 7x 2 x 100 is a polynomial. 3 (e) 8x 6 5x 3 7x 3 is not a polynomial. (It is the cube root of the polynomial 8x 6 5x 3 7x 3.) 4 2 (f) 3x 5x 15x is a polynomial. (Some coefficients are not integers, but all exponents are integers.)
2. To add polynomials we add like terms. So 3x 2 2x 4 8x 2 x 1 3 8 x 2 2 1 x 4 1 11x 2 x 5.
3. To subtract polynomials we subtract like terms. So 2x 3 9x 2 x 10 x 3 x 2 6x 8 2 1 x 3 9 1 x 2 1 6 x 10 8 x 3 8x 2 5x 2.
4. We use FOIL to multiply two polynomials:x 2 x 3 x x x 3 2 x 2 3 x 2 5x 6. 5. The Special Product Formula for the “square of a sum” is A B2 A2 2AB B 2 . So 2x 32 2x2 2 2x 3 32 4x 2 12x 9.
6. The Special Product Formula for the “product of the sum and difference of terms” is A B A B A2 B 2 . So 6 x 6 x 62 x 2 36 x 2 .
7. (a) No; x 52 x 2 2 5x 25 x 2 25. (b) Yes; x a2 x 2 2xa a 2 .
8. (a) Yes; by a Special Product Formula, x 5 x 5 x 2 25. (b) No, x a x a x 2 a 2 , by a Special Product Formula.
9. Type: binomial. Terms: 5x 3 and 6. Degree: 3.
10. Type: trinomial. Terms: 2x 2 , 5x, and 3. Degree: 2.
SECTION P.5 Algebraic Expressions
11. Type: monomial. Term: 8. Degree: 0.
12. Type: monomial. Terms: 12 x 7 . Degree: 7. 13. Type: quadrinomial. Terms: x, x 2 , x 3 , and x 4 . Degree: 4. 14. Type: binomial. Terms: 2x and 3. Degree: 1. 15. 12x 7 5x 12 12x 7 5x 12 7x 5
16. 5 3x 2x 8 x 3 17. 2x 2 3x 1 3x 2 5x 4 2x 2 3x 1 3x 2 5x 4 x 2 2x 3 18. 3x 2 x 1 2x 2 3x 5 3x 2 x 1 2x 2 3x 5 x 2 4x 6 19. 5x 3 4x 2 3x x 2 7x 2 5x 3 4x 2 3x x 2 7x 2 5x 3 3x 2 10x 2 20. 5x 2 3 1 4x 3x 2 5x 2 3 1 4x 3x 2 8x 2 4x 4 21. 3 x 1 4 x 2 3x 3 4x 8 7x 5
22. 8 2x 5 7 x 9 16x 40 7x 63 9x 103 23. 4 x 2 3x 5 3 x 2 2x 1 4x 2 12x 20 3x 2 6x 3 x 2 6x 17 24. 5x 3 3x 2 2x 2 3 x 4x 3 5x 3 3x 2 2x 2 3 2 x 2 4x 3 13x 3 3x 2 6
25. 4 5 42 20
26. 10x 1 2x 10x 20x 2 20x 2 10x 27. y 3 y 2 y y 3 y 2 y 3 y y 5 y 4 28. 4z 2 z 2 4z 3 8z 29. x 3 x 2 3x 2x x 4 3x 2 x 5 3x 4 2x 5 6x 3 x 5 3x 4 6x 3 30. 4x 1 x 3 3x 3 x 3 x 4x 4x 4 3x 6 3x 4 3x 6 7x 4 4x 31. 4x x 2 2 x 2 4x 2x 2 x 1 4x 2 8x 2x 2 8x 2x 3 2x 2 2x 3 32. 6x 3 x 2 1 2x 2 3x 2 2 2x 4 6x 5 6x 3 4x 6x 3 4x 8 6x 5 12x 3 8 33. x 4 x 10 x 2 10x 4x 4 10 x 2 14x 40 34. y 1 y 7 y 2 7y y 1 7 y 2 8y 7
35. z 1 4z 5 4z 2 5z 4z 1 5 4z 2 z 5
36. 3x 2 x 5 3x 2 3 5x 2x 2 5 3x 2 13x 10
37. 3t 2 7t 4 21t 2 12t 14t 8 21t 2 26t 8 38. 4s 1 2s 5 8s 2 18s 5 39. 3x 5 2x 1 6x 2 10x 3x 5 6x 2 7x 5 40. 7y 3 2y 1 14y 2 13y 3 41. x 3y 2x y 2x 2 5x y 3y 2
42. 4x 5y 3x y 12x 2 19x y 5y 2
43. 3u 4 4u 3 12u 2 25u 12 2
44. 2r s r 2s 2r 2 3rs 2s 2
45. 4x 32 16x 2 24x 9
46. 2 7y2 49y 2 28y 4
47. 1 3z2 9z 2 6z 1
48. 2u 32 4u 2 12u 9
49. y 3x2 y 2 6x y 9x 2
50. 5x y2 25x 2 10x y y 2
17
18
CHAPTER P Prerequisites
51. 2x 3y2 4x 2 12x y 9y 2 52. r 2s2 r 2 4rs 4s 2 2 2 2 53. 5 y 3 52 y 3 2 5 y 3 y 6 10y 3 25 54. x 4 3 x 8 6x 4 9 55. 7 7 2 49
56. 5 y 5 y 25 y 2
57. 3x 4 3x 4 3x2 42 9x 2 16 58. 2y 5 2y 5 4y 2 25 x 2 x 2 x 4 60. y 2 y 2 y 2 59. 61. y 23 y 3 3y 2 2 3y 22 23 y 3 6y 2 12y 8 62. x 33 x 3 9x 2 27x 27
63. 2 5u3 125u 3 150u 2 60u 8
64. 3x 23 27x 3 54x 2 36x 8 65. x 2 x 2 2x 3 x 3 2x 2 3x 2x 2 4x 6 x 3 4x 2 7x 6 66. x 1 2x 2 x 1 2x 3 x 2 1 67. 2x 5 x 2 x 1 2x 3 2x 2 2x 5x 2 5x 5 2x 3 7x 2 7x 5 68. 1 2x x 2 3x 1 2x 3 5x 2 x 1 2 x x x x x x x xx 71. y 13 y 23 y 53 y 1323 y 1353 y 2 y 69.
73. 75.
x 12 y 12 x 2 a2
2
x 2 xy y
x 2 a2 x 4 a4
x 1 x x 2 x 72. x 14 2x 34 x 14 2x x 1 2 1 74. u u2 u u 76. x 12 y 12 x 12 y 12 x y 78. h2 1 1 h2 1 1 h2 70. x 32
ab a b a b2 2 79. x 1 x 2 x 1 x 2 x 12 x 2 x 2 2x 1 x 4 x 4 x 2 2x 1 80. x 2 x 2 x 2 x 2 x 4 3x 2 4
77.
81. 2x y 3 2x y 3 2x y2 32 4x 2 4x y y 2 9
82. x y z x y z x 2 y 2 z 2 2yz 83. (a) 12 a b2 a 2 b2 12 a 2 2ab b2 a 2 b2 12 2ab ab. 2 2 a 2 b2 a 2 b2 (b) a 2 b2 a 2 b2 a 2 b2 a 2 b2 a 2 b2 a 2 b2 a 2 b2 a 2 b2 2b2 2a 2 4a 2 b2 84. LHS a 2 b2 c2 d 2 a 2 c2 a 2 d 2 b2 c2 b2 d 2 .
RHS ac bd2 ad bc2 a 2 c2 2abcd b2 d 2 a 2 d 2 2abcd b2 c2 a 2 c2 a 2 d 2 b2 c2 b2 d 2 . So LHS RHS, that is, a 2 b2 c2 d 2 ac bd2 ad bc2 .
85. (a) The height of the box is x, its width is 6 2x, and its length is 10 2x. Since Volume height width length, we have V x 6 2x 10 2x. (b) V x 60 32x 4x 2 60x 32x 2 4x 3 , degree 3.
SECTION P.6 Factoring
19
(c) When x 1, the volume is V 60 1 32 12 4 13 32, and when x 2, the volume is V 60 2 32 22 4 23 24.
86. (a) The width is the width of the lot minus the setbacks of 10 feet each. Thus width x 20 and length y 20. Since Area width length, we get A x 20 y 20. (b) A x 20 y 20 x y 20x 20y 400
(c) For the 100 400 lot, the building envelope has A 100 20 400 20 80 380 30,400. For the 200 200, lot the building envelope has A 200 20 200 20 180 180 32,400. The 200 200 lot has a larger building envelope. 87. (a) A 4000 1 r3 2000 1 3r 3r 2 r 3 2000 6000r 6000r 2 2000r 3 , degree 3. (b) Remember that % means divide by 100, so 2% 002. Interest rate r
2%
35%
5%
6%
10%
Amount A
$424483
$443487
$463050
$476406
$532400
88. (a) When x 1, x 52 1 52 36 and x 2 25 12 25 26. (b) x 52 x 2 10x 25
89. (a) The degree of the product is the sum of the degrees of the original polynomials. (b) The degree of the sum could be lower than either of the degrees of the original polynomials, but is at most the largest of the degrees of the original polynomials. (c) Product: 2x 3 x 3 2x 3 x 7 4x 6 2x 4 14x 3 2x 4 x 2 7x 6x 3 3x 21 4x 6 4x 4 20x 3 x 2 10x 21 Sum: 2x 3 x 3 2x 3 x 7 4.
P.6
FACTORING
1. (a) The polynomial 2x 3 3x 2 10x has three terms: 2x 3 , 3x 2 , and 10x. (b) The factor x is common to each term, so 2x 3 3x 2 10x x 2x 2 3x 10 .
2. The greatest common factor in the expression 18x 3 30x is 6x, and the expression factors as 6x 3x 2 5 .
3. To factor the trinomial x 2 8x 12 we look for two integers whose product is 12 and whose sum is 8. These integers are 6 and 2, so the trinomial factors as x 6 x 2. 4. The Special Factoring Formula for the “difference of squares” is A2 B 2 A B A B. So 49x 2 9 7x 3 7x 3.
5. The Special Factoring Formula for a “perfect square” is A2 2AB B 2 A B2 . So x 2 10x 25 x 52 .
6. The greatest common factor in the expression 4 x 12 x x 12 is x 12 , and the expression factors as 4 x 12 x x 12 x 12 4 x.
7. 10x 15 5 2x 3 9. 4x 4 x 2 x 2 4x 2 1 x 2 2x 1 2x 1 11. 4x 3 y 2 6x y 3 8x 2 y 4 2x y 2 2x 2 3y 4x y 2
13. y y 6 9 y 6 y 6 y 9
8. 12 3y 3 4 y
10. 3x 4 6x 3 x 2 x 2 3x 2 6x 1 12. 7x 4 y 2 14x y 3 21x y 4 7x y 2 x 3 2y 3y 2
14. z 22 5 z 2 z 2 [z 2 5] z 2 z 3
20
CHAPTER P Prerequisites
15. z 2 11z 18 z 9 z 2
16. x 2 4x 5 x 5 x 1
17. 10x 2 19x 6 5x 2 2x 3
18. 6y 2 11y 21 y 3 6y 7
19. 3x 2 16x 5 3x 1 x 5
20. 5x 2 7x 6 5x 3 x 2
21. 3x 22 8 3x 2 12 [3x 2 2] [3x 2 6] 3x 4 3x 8
22. 2 a b2 5 a b 3 [a b 3] [2 a b 1] a b 3 2a 2b 1 23. x 2 400 x 20 x 20
24. 64 y 2 8 y 8 y
25. 36a 2 49 6a 7 6a 7
26. 4z 2 81 2z 9 2z 9
27. x 2 49y 2 x 7y x 7y
28. 4u 2 25 2 2u 5 2u 5
29. z 52 92 z 5 3 z 5 3
30. y 2 x 12 y x 1 y x 1
31. x 2 12x 36 x 2 2 6x 62 x 62
32. 16 8y y 2 y 2 2 4y 42 y 42
33. 100 20z z 2 z 2 2 10z 102 z 102
34. 2 14 49 2 2 7 72 72
35. 25 30t 9t 2 3t2 2 3t 5 52 3t 52
36. 16z 2 24z 9 4z2 2 4z 3 32 4z 32
37. 4u 2 36u 81 2 2u2 2 2u 9 92 2u 92
38. 4y 2 20yz 25z 2 2y2 2 2y 5z 5z2 2y 5z2 39. x 3 125 x 3 53 x 5 x 2 5x 25 40. y 3 64 y 3 43 y 4 y 2 4y 16 41. 83 27 23 33 2 3 42 6 9 42. 1 27a 3 13 3a3 1 3a 1 3a 9a 2 43. 27x 3 y 3 3x3 y 3 3x y 3x2 3x y y 2 3x y 9x 2 3x y y 2 44. 8x 3 y 3 2x3 y 3 2x y 4x 2 2x y y 2 3 2 45. a 3 b6 a 3 b2 a b2 a 2 ab2 b2 a b2 a 2 ab2 b4 46. 64x 3 y 3 27 4x y3 33 4x y 3 16x 2 y 2 12x y 9 47. x 3 4x 2 x 4 x 2 x 4 1 x 4 x 4 x 2 1 48. 3x 3 x 2 6x 2 x 2 3x 1 2 3x 1 3x 1 x 2 2 49. 5x 3 x 2 5x 1 x 2 5x 1 5x 1 x 2 1 5x 1 50. 18x 3 9x 2 2x 1 9x 2 2x 1 2x 1 9x 2 1 2x 1 51. x 3 x 2 x 1 x 2 x 1 1 x 1 x 1 x 2 1 52. x 5 x 4 x 1 x 4 x 1 1 x 1 x 1 x 4 1
53. Start by factoring out the power of x with the smallest exponent, that is, x 23 . Thus, x 23 3x 53 x 23 1 3x
54. Start by factoring out the power of x with the smallest exponent, that is, x 14 . Thus, x 34 5x 14 x 14 x 5
55. Start by factoring out the power of x with the smallest exponent, that is, x 32 . Thus, x 32 x 12 x 12 x 32 1 x x 2 x 32 x 2 x 1
SECTION P.6 Factoring
56. Start by factoring out the power of x with the smallest exponent, that is, x 13 . Thus, x 53 x 23 2x 13 x 13 x 2 x 2 12 57. Start by factoring out the power of x 2 1 with the smallest exponent, that is, x 2 1 .
12 12 12 12 x2 3 x2 1 2 x2 1 x2 3 2 x2 1 x2 1 . Thus, x 2 1 x2 1 2x 1 58. x 12 x 112 x 12 x 112 x 12 x 112 [x 1 x] x 12 x 112 2x 1 x x 1 3 2 2 3 2 60. 12x 3x 3x 4 x 59. 2x 12x 2x 1 6x 61. 15x 3 10x 2 5x 2 3x 2
62. 6x yz 3xz 3xz 2y 1
63. x 2 2x 8 x 4 x 2
64. x 2 14x 48 x 8 x 6
65. y 2 4y 21 y 7 y 3
66. t 2 10t 24 t 4 t 6
67. 2x 2 5x 3 2x 3 x 1 69. 9x 2 36x 45 9 x 2 4x 5 9 x 5 x 1
68. 2x 2 7x 4 2x 1 x 4
71. 12x 2 x 20 4x 5 3x 4
72. 15 26t 8t 2 4t 3 2t 5
73. x 2 121 x 11 x 11
74. 4t 2 900 2t 30 2t 30
75. 49 4y 2 7 2y 7 2y
76. 4t 2 9s 2 2t 3s 2t 3s
77. t 2 6t 9 t 32
78. x 2 10x 25 x 52
70. 8x 2 10x 3 4x 3 2x 1
79. y 2 10yz 25z 2 y 5z2 81. 1 8t 3 1 2t 1 2t 4t 2
80. r 2 6rs 9s 2 r 3s2 82. 27x 3 1000 3x 10 9x 2 30x 100 83. 8x 3 125 2x3 53 2x 5 2x2 2x 5 52 2x 5 4x 2 10x 25 3 2 x 2 4 x 2 42 x 2 4 x 4 4x 2 16 84. x 6 64 x 6 26 x 2 43 x 2 4 85. x 3 2x 2 x x x 2 2x 1 x x 12 86. 3x 3 27x 3x x 2 9 3x x 3 x 3 87. x 6 x 5 42x 4 x 4 x 2 x 42 x 4 x 7 x 6 88. 3t 4 2t 3 5t 2 t 2 3t 2 2t 5 t 2 t 1 3t 5 89. x 4 y 3 x 2 y 5 x 2 y 3 x 2 y 2 x 2 y 3 x y x y
90. 18y 3 x 2 2x y 4 2x y 3 9x y 3 91. 27x 6 y 3 3x 2 y 3 3x 2 y 9x 4 3x 2 y y 2 92. a 3 64b6 a 4b2 a 2 4ab2 16b4 93. y 3 5y 2 9y 45 y y 2 9 5 y 2 9 y 5 y 3 y 3 94. y 3 y 2 y 1 y 2 y 1 y 1 y 2 1 y 1 y 1 y 1 y 1 y 12 y 1
21
22
CHAPTER P Prerequisites
95. 3x 3 x 2 12x 4 3x x 2 4 x 2 4 3x 1 x 2 4 3x 1 x 2 x 2 96. 9x 3 18x 2 x 2 9x 2 x 2 x 2 9x 2 1 x 2 3x 1 3x 1 x 2
97. a b2 a b2 [a b a b] [a b a b] 2b 2a 4ab 1 2 1 1 1 1 2 1 98. 1 1 1 1 1 1 x x x x x x 1 1 1 1 2 4 1 1 1 1 2 x x x x x x 99. x 2 x 2 1 9 x 2 1 x 2 1 x 2 9 x 1 x 1 x 3 x 3 100. a 2 1 b 22 4 a 2 1 a 2 1 b 22 4 a 1 a 1 [b 2 2] [b 2 2] a 1 a 1 b b 4
101. Start by factoring out the power of x with the smallest exponent, that is, x 32 . So 1 x2 x 32 2x 12 x 12 x 32 1 2x x 2 x 32 1 x2 . x 32 102. x 172 x 132 x 132 x 12 1 x 132 [x 1 1] [x 1 1] x 132 x 2 x
103. x 1 x 22 x 12 x 2 x 1 x 2 [x 2 x 1] 3 x 1 x 2 104. y 4 y 23 y 5 y 24 y 4 y 23 1 y y 2 y 4 y 23 y 2 2y 1 y 4 y 23 y 12
105. Start by factoring y 2 7y 10, and then substitute a 2 1 for y. This gives 2 a 2 1 7 a 2 1 10 a 2 1 2 a 2 1 5 a 2 1 a 2 4 a 1 a 1 a 2 a 2 2 a 2 2a 1 a 2 2a 3 a 2 2a 1 106. a 2 2a 2 a 2 2a 3 a 2 2a 3
a 1 a 3 a 12 2 2 107. x 2 3 x 13 4x 12 x 13 x 13 x 2 3 4x 12 x 2 3 4x 1 x 13 x 2 4x 4 x 2 4x 2 x 13 x 2 3 4x 1 x 13 x 22 x 2 4x 2 108. x 12 x 2 6 x 1 x 2 9 x 2 x 2 x 12 6 x 1 9
x 2 [x 1 3]2 x 2 x 22 4 5 4 109. 5 x 2 4 2x x 24 x 2 4 4 x 23 2 x 2 4 x 23 5 x x 2 x 2 4 2 4 4 2 x 2 4 x 23 5x 2 10x 2x 2 8 2 x 2 4 x 23 7x 2 10x 8 110. 3 2x 12 2 x 312 2x 13 12 x 312 2x 12 x 312 6 x 3 2x 1 12 2x 12 x 312 6x 18 x 12 2x 12 x 312 7x 35 2 111.
1 2 13 43 43 43 1 x2 3 3 x 3 x2 3 x 2 3 23 x 2 x 2 3 23 x 2 x 2 3 x2 3 43 3 x2 3
112. 12 x 12 3x 412 32 x 12 3x 412 12 x 12 3x 412 [3x 4 3x] 12 x 12 3x 412 4 2x 12 3x 412
SECTION P.6 Factoring
23
2 113. 4a 2 c2 c2 b2 a 2 2ac2 c2 b2 a 2 2ac c2 b2 a 2 2ac c2 b2 a 2 (difference of squares) 2ac c2 b2 a 2 2ac c2 b2 a 2 c2 2ac a 2 b2 (regrouping) b2 c2 2ac a 2 b2 c a2 c a2 b2 (perfect squares)
[b c a] [b c a] [c a b] [c a b] (each factor is a difference of squares) b c a b c a c a b c a b a b c a b c a b c a b c
114. (a) Mowed portion field habitat
(b) Using the difference of squares, we get b2 b 2x2 [b b 2x] [b b x] 2x 2b 2x 4x b x.
115. The volume of the shell is the difference between the volumes of the outside cylinder (with radius R) and the inside cylinder R r (with radius r). Thus V R 2 h r 2 h R 2 r 2 h R r R r h 2 h R r. The 2 R r R r and 2 is the average circumference (length of the rectangular box), h is the height, and average radius is 2 2 R r R r is the thickness of the rectangular box. Thus V R 2 h r 2 h 2 h R r 2 average radius 2 height thickness R
length
rl
h
h
thickness
116. (a) 5282 5272 528 527 528 527 1 1055 1055
(b) 10202 10102 1020 1010 1020 1010 10 2030 20,300
(c) 501 499 500 1 500 1 5002 12 250,000 1 249,999
(d) 1002 998 1000 2 1000 2 10002 22 1,000,000 4 999,996 117. (a)
A2 A 1
A 1
A 1
A 1
A2 A A2
1
A 1
A2 A 1
A3 A2 A
A3
A3 A2 A 1
1
A 1
A3 A2 A 1
A 4 A3 A2 A A4
1 (b) Based on the pattern in part (a), we suspect that A5 1 A 1 A4 A3 A2 A 1 . Check:
A 4 A 3 A2 A 1
A 1
A4 A3 A2 A 1
A 5 A 4 A 3 A2 A
A5 1 The general pattern is An 1 A 1 An1 An2 A2 A 1 , where n is a positive integer.
24
CHAPTER P Prerequisites
118. (a) A B A2 AB B 2 A3 A2 B AB 2 A2 B AB 2 B 3 A3 B 3 (b) A B A2 AB B 2 A3 A2 B AB 2 A2 B AB 2 B 3 A3 B 3
P.7
RATIONAL EXPRESSIONS
P x , where P and Q are polynomials. 1. A rational expression has the form Q x 3x (a) 2 is a rational expression. x 1 x 1 (b) is not a rational expression. A rational expression must be a polynomial divided by a polynomial, and the 2x 3 numerator of the expression is x 1, which is not a polynomial.
x3 x xx 2 1 is a rational expression. x 3 x 3 2. To simplify a rational expression we cancel factors that are common to the numerator and denominator. So, the expression x 1x 2 x 1 simplifies to . x 3x 2 x 3 3. To multiply two rational expressions we multiply their numerators together and multiply their denominators together. So 2 x 2x 2x is the same as . 2 x 1 x 3 x 1 x 3 x 4x 3 (c)
4. (a)
2 x 1 has three terms. x x 1 x 12
(b) The least common denominator of all the terms is x x 12 . (c)
2 x 2x x 1 1 x x x 12 x 12 2x x 1 x 2 2 2 2 x x 1 x 1 x 1 x x 1 x x 12 x 1
x 2 2x 1 2x 2 2x x 2
5. (a) Yes. Cancelling x 1, we have
x x 12
x x 1 x 12
x . x 1
(b) No; x 52 x 2 10x 25 x 2 25, so x 5
2x 2 1
x x 12
x 2 10x 25
x 2 25.
3 a a 3a 1 . 3 3 3 3 (b) No. We cannot “separate” the denominator in this way; only the numerator, as in part (a). (See also Exercise 101.)
6. (a) Yes,
7. The domain of 4x 2 10x 3 is all real numbers.
8. The domain of x 4 x 3 9x is all real numbers.
9. Since x 3 0 we have x 3. Domain: x x 3
10. Since 3t 6 0 we have t 2. Domain: t t 2
11. Since x 3 0, x 3. Domain; x x 3
12. Since x 1 0, x 1. Domain; x x 1
13. x 2 x 2 x 1 x 2 0 x 1 or 2, so the domain is x x 1 2. x x 14. 2 . x 2 x 2 0 x 2, so the domain is x x 2. x 2 x 2 x 4 x 2 15. is defined for x x 2 x 3, so its domain is x x 2. x 3 x 2 16. 2 is defined for x x 2 x 3, so its domain is x x 2 x 3. x 9
SECTION P.7 Rational Expressions
17. 19. 21. 22.
x 5 x 5 x 5 x 5 12 x 5 2x 10 2 x 5 x 2 1 x 2 2 x 2 2 2 x x x 4
x 1 x 2 7x 8 x 8 x 1 x 2 x 8 x 2 x 2 10x 16
18. 20.
x x2 2
x2 2 x 3 2x 2 x x 1 x 1 x x
x2 x 2 x 2 x 2 x 1 2 x 1 1 1 x x x 1
x 2 x 12 x 4 x 4 x 3 x 2 x 2 x 3 x 2 5x 6
y y 1 y y2 y y1 y 1 y 1 y2 1 y y 2 5y 24 y8 y y 8 y 3 y 3 5y 2 24y 24. y y 3 y 3 y3 y 3 9y y y2 9 2x 6 x 2x 3 2 x 2x 3 x 2 x 2x 3 2x x 6x 25. 2x 3 2x 3 x 2 2x 3 x 2 2x 2 7x 6 23.
26. 27. 28. 29.
x 1 1 x x 1 1 x2 1 x 1 x 3 2 2 x 1 x2 x 1 x 1 x x 1 x 1 x x 1 4x
x2 4
4x 1 x 2 x 2 16x 4 x 2 x 2 x 2 16x
x 5 x 2 25 x 4 x 5 x 5 x 4 x 4 x 4 x 4 x 5 x 2 16 x 5
x 3 x 2 2x 15 x 5 x 5 x 3 x 5 x 2 x 2 x 5 x 5 x 2 x 2 25
x 1 x 1 1x x 2 2x 3 3 x x 3 x 1 x 3 x 3 x 1 x 1 1x x 3 x 1 x 2 2x 3 3 x 2t 3 2t 3 1 2t 3 2t 3 31. 2 2 2 t 9 2t 3 2t 3 t 9 4t 2 9 t 9 30.
32.
33. 34. 35. 36.
3y 2 9y y2 9 3y y 3 y 3 y 3 3 2 2 3 2y 3 y 9y 2y 3y 9 y y 9 y 3 2y 3
x 3 2x 2 8x x 2 2x 24 x 4 x x 2 x 4 x 4 x 6 x 4 x 6 x 2 x x 4 x 4 x 2 8x 12 x 3 16x
xy 2x 2 x y y 2 x 2 2x y y 2 2x y x y x y2 2 2 2 2 x 2y x y x 2y 2x y x y x x y 2y 2x 3x y y
x 3 x 2 7x 12 x 3 2x 2 7x 15 x 3 x 5 x 5 2x 3 2 2 2 2x 3 2x 3 x 3 x 4 2x 3 x 4 4x 9 2x 7x 15 4x 9 x 2 7x 12 2x 1
2x 2 x 15
2x 1 6x 2 x 2 x 3 1 x 3 x 3 2x 5 2x 1 3x 2 2x 5 3x 2
x3 x 2 2x 1 x 3 x 1 x 1 x3 x 1 37. x 2 x 1 x x 1 x x 1 x x 2 2x 1
2x 2 3x 2 x 2 x2 x 2 x 2 2x 1 x 1 x 2 2x 2 3x 2 x2 1 38. x 1 x 1 x 1 x 2 2x 1 x2 1 2x 2 5x 2 2x 2 5x 2 x2 x 2
25
26
CHAPTER P Prerequisites
x 1 x xy z y z yz y x z xz x x 40. yz z 1 y y 39.
1 x 3 1 x 4 x 3 x 3 x 3 x 3 3x 2 3x 2 2 x 1 3x 2 2x 2 x 4 42. 2 x 1 x 1 x 1 x 1 x 1 1 2 x 3 2 x 5 x 3 2x 10 3x 7 43. x 5 x 3 x 5 x 3 x 5 x 3 x 5 x 3 x 5 x 3 41. 1
44. 45. 46. 47. 48.
1 1 x 1 x 1 x 1x 1 2x x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1
3 1 3 x 2 1 x 1 3x 6 x 1 2x 5 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2
3 x x 6 x 3 x 4 x 2 6x 3x 12 x 2 3x 12 x 4 x 6 x 4 x 6 x 4 x 6 x 4 x 6 x 4 x 6
5 3 5 2x 3 3 10x 15 3 10x 18 2 5x 9 2 2 2 2 2x 3 2x 32 2x 3 2x 3 2x 3 2x 3 2x 32 x
x 12
49. u 1
x 2 2 x 1 3x 2 x 2x 2 2 2 x 1 1 1 x x x 1 x 1 x 12
u u u 2 2u 1 u u 2 3u 1 u 1 u 1 u1 u1 u1 u1 u1
3 2b2 3ab 4a 2 2b2 3ab 4a 2 4 2 2 2 2 2 2 2 2 2 ab b a a b a b a b a 2 b2 x 1 1 1 1 1 x 2x 1 2 51. 2 2 2 2 2 x x 1 x x x x x x 1 x x 1 x x 1 50.
1 1 x2 x 1 x2 x 1 1 2 3 3 3 3 x x x x x x x3 1 1 2 2 2 x 4 1 53. x 3 x 2 7x 12 x 3 x 3 x 4 x 3 x 4 x 3 x 4 2x 8 1 2x 7 x 3 x 4 x 3 x 4 52.
54.
55. 56.
57. 58.
x
x2 4
x x 1 1 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 2x 2 2 x 1 x 2 x 2 x 2 x 2
1 1 1 1 x 3 1 x 2 x 3 x2 9 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3
x 2 x 2 x 1 x 2 x 1 x 4 x 2 x 2 x 2 5x 4 2 x 2 x 2 4x 2x 4 x 2 6x 4 x x 4 x 1 x 2 x 4 x 1 x 2 x 4 x 1 x 2 x 4 x 1 x 2 x 4 3 4 3 4 2 x 1 3x 4 2x 2 3x 4 5x 6 2 2 x x 1 x2 x x x 1 x x 1 x x 1 x x 1 x x 1 x x 1 x x 1 1 1 1 z2 x2 y2 x 2 y2 z2 x 3 y3 z x y3 z3 x 3 yz 3 x 3 y3 z3 x 3 y3 z3 x 3 y3 z3 x 3 y3 z3
SECTION P.7 Rational Expressions
59.
1 1 x x 1 x 1 x 2 2x 1 1 x 2 x 1 x 2 x 12 x 3 x 12 x 3 x 12 x 3 x 12 x 3 x 12 x 3 x 12
60.
27
x 12
x 3 x 12
1 3 x
2 2 1 3 3 1 2 x 1 x 12 x 1 x 12 x 1 x 1 x 1 2 x 1 3 x 1 x 1 x 1 x 1 x 12 x 1 x 12 x 1 x 12
x2 1
2x 2 3x 3 x 2 1 2x 2 3x 3 x2 x 4 x 1 x 12 x 1 x 12 x 1 x 12 x 1 x 12 x 1 x 12
x 1 1x x 1 61. 1 1 1 2x x x 2 x 2 1 1x
y 1 2y y 2 62. 3 3 3y y y 1 y 1 1 2y
1 1 1 2 x 1 x 3 x 2 1 x 2 x 2 63. 1 1 x 1 x 2 1 1 x 2 1 x 2 x 2 1 c 1 c11 c 64. 1 c11 c2 1 c1 1
1 1 1 1 x 1 x 3 2 x 1 x 3 x 1 x 1 x 3 65. x 1 x 3 x 1 x 1 x 3 x 1 x 1 x 1 x 3 x 1 x 1 x 3 2 x 1 x 3 x 3 x 2 x 2 2x 3 x 2 2x 8 5 3 1 2 4 x x x x 66. x 4 x 1 x 3 x 4 x 3 x 1 x 4 x 3 x 1 x 4 x 3 x 1 x x xy x x 2 y 1 x2 y x2 y y 67. y y x y 2 y 2 y 2 x 1 y xy y x x x
y y xy x y y x2 2 y y2 x x x 68. x x x y2 x 2 x x y2 y x y y y y x
28
CHAPTER P Prerequisites
y x x 2 y2 x 2 y2 x 2 y2 xy y x xy x y. An alternative method is to multiply the 69. 2 2 2 2 1 1 x y 1 y x y x x2 y2 x 2 y2 numerator and denominator by the common denominator of both the numerator and denominator, in this case x 2 y 2 : y x y x 2 y2 x y x 2 2 3 3 x y xy x y y x y x 2 x y. 1 1 1 1 x 2 y2 y x2 y2 x 2 x2 y2 x2 y2 x x 2 y2 y xy x y2 y x y2 x 3 x y2 x y2 x3 70. x x 2 y x x y x y x x 2 y2 x 2 y2 x 2 y2 2 2 x y x y2 y x y x 1 x2 1 y2 y2 x 2 xy yx y x y x x y x2 y2 x 2 y2 x 2 y2 71. 1 y x 2 y2 2 y 2 y x 1 1 y x xy x y 1 x x xy xy x y 1 1 x 2 y2 yx y2 x 2 x 2 y 2 y x y x x2 y2 . 2 Alternatively, 1 1 2 2 2 1 1 x y y x xy x y x y xy x y x y 1 1 u 1 72. 1 1 1u 1u u1 1 u 1 x 1x 1 x 73. 1 1 1 x 1 x 1 1x 1 x x 2x 1 2x 3 x 1 1x 1 1 1 74. 1 1 x 2 x 2 x 2 1 x 1 1 1x 1 1 1 1 x h 1 x 1 x 1 x h 75. h h 1 x 1 x h 1 x 1 x h 76. In calculus it is necessary to eliminate the h in the denominator, and we do this by rationalizing the numerator: 1 1 x x h 1 x x h x x h x x h . h h x x h x x h h x x h x x h x x h x x h 1 1 2 x 2 2xh h 2 2 2 2 x 2 x x h 2x h x h x 77. h hx 2 x h2 hx 2 x h2 x 2 x h2 x 2 y 2
x 2 h 2 2xh 3x 3h x 2 3x h h 2x 3 x h2 3 x h x 2 3x 2x h 3 h h h 2 x2 x2 1 1 x2 1 x 79. 1 1 2 2 2 2 1x 1x 1x 1x 1 x2 1 x2 1 2 2x 3 1 1 1 80. 1 x 3 3 1 x6 3 1 x 6 12 x 6 12 6 6 6 4x 4x 16x 16x 16x 2 1 1 x3 3 x 3 3 4x 4x 78.
SECTION P.7 Rational Expressions
81.
82.
2 x 3 x 53 3 x 52 x 32 x 56
4x 3 1 x3 3 1 x2 1 x 4 1 x6
x 3 x 52 [2 x 5 3 x 3] x 56
x 3 1 x2 [4 1 x 3x] 1 x6
x 3 19 x x 54
x 3 4 x 1 x4
x 2 2 1 x12 x 1 x12 1 x12 [2 1 x x] 1x 1x 1 x32 12 12 12 1 x2 1 x2 1 x2 x2 x2 1 x2 1 84. 32 2 2 1x 1x 1 x2 83.
85.
3 1 x13 x 1 x23 1 x23
1 x23 [3 1 x x] 1 x23
2x 3
1 x43 7 3x12 7 3x 32 x 7 32 x 7 3x12 32 x 7 3x12 86. 7 3x 7 3x 7 3x32 1 3 10 3 10 87. 9 10 10 3 3 10 3 10 3 10 2 5 3 3 63 5 88. 6 3 5 45 2 5 2 5 2 5
2 5 3 5 3 5 3 89. 53 5 3 5 3 5 3 x 1 x 1 1 1 90. x 1 x 1 x 1 x 1 3 y y y 3y y 3 y y y 91. 3 y 3y 3 y 3 y 3 y 2 x y x y x y 2 x y 2 x y 2 x y 2 x 2 y 92. xy x y x y x y 2 5 2 5 2 5 45 1 93. 5 5 2 5 5 2 5 5 2 5 3 5 3 5 3 5 35 2 1 94. 2 2 3 5 3 5 2 3 5 2 3 5 r 2 r 2 r 2 r 2 95. 5 5 r 2 5 r 2 x x h x x h x x h x x h 96. h x x h h x x h x x h h x x h x x h h 1 h x x h x x h x x h x x h x2 1 x x2 1 x x2 1 x2 1 97. x 2 1 x 2 2 2 1 x 1x x 1x x 1x x 1x 1 x 1 x x 1 x 98. x 1 x 1 x 1 x x 1 x x 1 x 2
2
29
30
CHAPTER P Prerequisites
99. (a) R
1 1 R R R1 R2 1 2 1 1 1 1 R1 R2 R2 R1 R1 R2 R1 R2
(b) Substituting R1 10 ohms and R2 20 ohms gives R 100. (a) The average cost A
200 10 20 67 ohms. 30 20 10
Cost 500 6x 001x 2 . number of shirts x
(b) x
10
20
50
100
200
500
1000
Average cost
$5610
$3120
$1650
$1200
$1050
$1200
$1650
101. x x2 9 x 3
280
290
295
299
2999
3
3001
301
305
310
320
580
590
595
599
5999
?
6001
601
605
610
620
From the table, we see that the expression
x2 9 approaches 6 as x approaches 3. We simplify the expression: x 3
x2 9 x 3 x 3 x 3, x 3. Clearly as x approaches 3, x 3 approaches 6. This explains the result in the x 3 x 3 table. 2 2 102. No, squaring changes its value by a factor of . x x 103. Answers will vary. Algebraic Error 1 1 1 a b ab
a b2 a 2 b2 a 2 b2 a b ab b a 1 a ab b am a mn an 104. (a)
Counterexample 1 1 1 2 2 22
1 32 12 32 52 122 5 12 26 6 2 1 1 11 5 3 352 32
5 a a 5a 1 , so the statement is true. 5 5 5 5
(b) This statement is false. For example, take x 5 and y 2. Then LHS RHS
5 5 x , and 2 . y 2 2
(c) This statement is false. For example, take x 0 and y 1. Then LHS RHS
1 1 1 1 , and 0 . 1y 11 2 2
x 1 51 6 2, while y1 21 3 0 x 0, while xy 01
(d) This statement is false. For example, take x 1 and y 1. Then LHS 2 2a 2 RHS 1, and 2 1. 2b 2
a b
2
1 2, while 1
SECTION P.8 Solving Basic Equations
31
a a a 1 1 a 1 a 1 . (e) This statement is true: b b b b b (f) This statement is true:
1 x x2 1 1 x x2 1 x. x x x x x
105. (a) x x
1 x
1
3
1 2
9 10
99 100
999 1000
9999 10,000
2
3333
25
2011
20001
2000001
200000001
It appears that the smallest possible value of x
1 is 2. x
(b) Because x 0, we can multiply both sides by x and preserve the inequality: x
1 1 2x x 2x x x
x 2 1 2x x 2 2x 1 0 x 12 0. The last statement is true for all x 0, and because each step is 1 reversible, we have shown that x 2 for all x 0. x
P.8
SOLVING BASIC EQUATIONS
1. Substituting x 3 in the equation 4x 2 10 makes the equation true, so the number 3 is a solution of the equation.
2. Subtracting 4 from both sides of the given equation, 3x 4 10, we obtain 3x 4 4 10 4 3x 6. Multiplying by 13 , we have 13 3x 13 6 x 2, so the solution is x 2.
x 2x 10 is equivalent to 52 x 10 0, so it is a linear equation. 2 2 2 (b) 2x 1 is not linear because it contains the term , a multiple of the reciprocal of the variable. x x (c) x 7 5 3x 4x 2 0, so it is linear.
3. (a)
4. (a) x x 1 6 x 2 x 6 is not linear because it contains the square of the variable. (b) x 2 x is not linear because it contains the square root of x 2.
(c) 3x 2 2x 1 0 is not linear because it contains a multiple of the square of the variable.
5. (a) This is true: If a b, then a x b x.
(b) This is false, because the number could be zero. However, it is true that multiplying each side of an equation by a nonzero number always gives an equivalent equation.
(c) This is false. For example, 5 5 is false, but 52 52 is true.
6. To solve the equation x 3 125 we take the cube root of each side. So the solution is x 3 125 5.
7. (a) When x 2, LHS 4 2 7 8 7 1 and RHS 9 2 3 18 3 21. Since LHS RHS, x 2 is not a solution. (b) When x 2, LHS 4 2 7 8 7 15 and RHS 9 2 3 18 3 15. Since LHS RHS, x 2 is a solution.
8. (a) When x 1, LHS 2 5 1 2 5 7 and RHS 8 1 7. Since LHS RHS, x 1 is a solution. (b) When x 1, LHS 2 5 1 2 5 3 and RHS 8 1 9. Since LHS RHS, x 1 is not a solution.
9. (a) When x 2, LHS 1 [2 3 2] 1 [2 1] 1 1 0 and RHS 4 2 6 2 8 8 0. Since LHS RHS, x 2 is a solution. (b) When x 4 LHS 1 [2 3 4] 1 [2 1] 1 3 2 and RHS 4 4 6 4 16 10 6. Since LHS RHS, x 4 is not a solution.
32
CHAPTER P Prerequisites
1 1 12 12 12 1 and RHS 1. Since LHS RHS, x 2 is a solution. 24 2 1 is not defined, so x 4 is not a solution. (b) When x 4 the expression 44
10. (a) When x 2, LHS 12
11. (a) When x 1, LHS 2 113 3 2 1 3 2 3 5. Since LHS 1, x 1 is not a solution. (b) When x 8 LHS 2 813 3 2 2 3 4 3 1 RHS. So x 8 is a solution.
432 23 8 4 and RHS 4 8 4. Since LHS RHS, x 4 is a solution. 46 2 2 32 23 832 292 (b) When x 8, LHS 272 and RHS 8 8 0. Since LHS RHS, x 8 is not a 86 2 2 solution. 0a a a 13. (a) When x 0, LHS RHS. So x 0 is a solution. 0b b b ba ba is not defined, so x b is not a solution. (b) When x b, LHS bb 0 2 b b b2 b b2 b2 b 14. (a) When x , LHS b 14 b2 0 RHS. So x is a solution. 2 2 2 4 2 4 2 2 2 1 1 b 1 1 1 14 b2 2 1 , so x is not a solution. b (b) When x , LHS b b b 4 b b 12. (a) When x 4, LHS
15. 5x 6 14 5x 20 x 4
16. 3x 4 7 3x 3 x 1
17. 7 2x 15 2x 8 x 4
18. 4x 95 1 4x 96 x 24
19. 12 x 7 3 12 x 4 x 8
20. 2 13 x 4 13 x 6 x 18
21. 3x 3 5x 3 0 8x x 0
22. 2x 3 5 2x 4x 2 x 12
23. 7x 1 4 2x 9x 3 x 13
24. 1 x x 4 3 2x x 32
25. x 3 4x 3 5x x 35
26. 2x 3 7 3x 5x 4 x 45
27. x3 1 53 x 7 x 3 5x 21 4x 24 x 6
3 x 3 4x 10 3x 30 x 40 28. 25 x 1 10
29. 2 1 x 3 1 2x 5 2 2x 3 6x 5 2 2x 8 6x 6 8x x 34
30. 5 x 3 9 2 x 2 1 5x 15 9 2x 4 1 5x 24 2x 3 7x 21 x 3 31. 4 y 12 y 6 5 y 4y 2 y 30 6y 3y 2 30 6y 9y 32 y 32 9
32. r 2 [1 3 2r 4] 61 r 2 1 6r 12 61 r 2 6r 11 61 r 12r 22 61 13r 39 r 3 33. x 13 x 12 x 5 0 6x 2x 3x 30 0 (multiply both sides by 6) x 30
34. 23 y 12 y 3 y 21 11
y1 8y 6 y 3 3 y 1 8y 6y 18 3y 3 14y 18 3y 3 11y 21 4
x 1 x 1 6x 8x 2x x 1 24x 7x 1 24x 1 17x x 17 2 4 x 1 1 5x 18x 15x 2 x 1 1 3x 2x 1 x 1 36. 3x 2 3 6 35. 2x
37. x 1 x 2 x 2 x 3 x 2 x 2 x 2 5x 6 x 2 5x 6 6x 8 x 43
SECTION P.8 Solving Basic Equations
33
38. x x 1 x 32 x 2 x x 2 6x 9 x 6x 9 5x 9 x 95
39. x 1 4x 5 2x 32 4x 2 x 5 4x 2 12x 9 x 5 12x 9 13x 14 x 14 13
40. t 42 t 42 32 t 2 8t 16 t 2 8t 16 32 16t 32 t 2 1 4 41. 1 3 4 3x (multiply both sides by the LCD, 3x) 1 3x x 13 x 3x 6 2 42. 5 4 2 5x 6 4x 4 9x 49 x x x 4 2x 1 5 2x 1 4 x 2 10x 5 4x 8 6x 13 x 13 43. 6 x 2 5 2x 7 23 2x 7 3 2 2x 4 (cross multiply) 6x 21 4x 8 2x 29 x 29 44. 2 2x 4 3 2 2 t 1 3 t 6 [multiply both sides by the LCD, t 1 t 6] 2t 2 3t 18 20 t 45. t 6 t 1 5 6 6 x 4 5 x 3 6x 24 5x 15 x 39 46. x 3 x 4 3 1 47. 1 3 6 3x 3 2 [multiply both sides by 6 x 1] 18 3x 3 2 3x 15 2 x 1 2 3x 3 3x 13 x 13 3 48.
49.
5 12x 5 2 12x 5 x 2x 6x 3 5 6x 3 12x 2 5x 12x 2 6x 30x 15 6x 3 x 12x 2 5x 12x 2 24x 15 19x 15 x 15 19
1 1 1 10 10 z 1 5 z 1 2 z 1 10 10z [multiply both sides by 10z z 1] z 2z 5z z1 3 z 3 z 1 100z 3z 3 100z 3 97z 97
1 4 15 0 3 t 4 3 t 15 0 3 t 12 4t 15 0 3t 30 0 3t 30 3 t 3 t 9 t2 t 10 x 1 51. 2 x 2 2x 4 2 [multiply both sides by 2 x 2] x 4x 8 2 3x 6 x 2. 2x 4 x 2 But substituting x 2 into the original equation does not work, since we cannot divide by 0. Thus there is no solution. 5 2 1 x 3 5 2 x 3 x 2 2x 6 x 4 52. x 3 x2 9 x 3 3 1 6x 12 53. 2 3 x x 4 6x 12 (multiply both sides by x x 4] 3x 7x 16 4x 16 x 4 x x 4x x 4. But substituting x 4 into the original equation does not work, since we cannot divide by 0. Thus, there is no solution. 1 2 1 54. 2x 1 2 x 1 1 1. This is an identity for x 0 and x 12 , so the solutions are 2 x 2x 1 2x x all real numbers except 0 and 12 . 50.
55. x 2 25 x 5
56. 3x 2 48 x 2 16 x 4 57. 5x 2 15 x 2 3 x 3 58. x 2 1000 x 1000 10 10
59. 8x 2 64 0 x 2 8 0 x 2 8 x 8 2 2 60. 5x 2 125 0 5 x 2 25 0 x 2 25 x 5 61. x 2 16 0 x 2 16 which has no real solution.
34
CHAPTER P Prerequisites
62. 6x 2 100 0 6x 2 100 x 2 50 3 , which has no real solution. 2 63. x 3 5 x 3 5 x 3 5 4 7 2 64. 3x 4 7 3x 4 7 3x 4 7 x 3 65. x 3 27 x 2713 3
66. x 5 32 0 x 5 32 x 3215 2 67. 0 x 4 16 x 2 4 x 2 4 x 2 4 x 2 x 2 x 2 4 0 has no real solution. If x 2 0, then x 2. If x 2 0, then x 2. The solutions are 2. 16 27 2716 3 27 6 6 x 16 68. 64x 27 x 64 64 2 64
69. x 4 64 0 x 4 64 which has no real solution.
70. x 13 8 0 x 13 8 x 1 813 2 x 1. 14 8114 x 2 3. So x 2 3, then x 1. If 71. x 24 81 0 x 24 81 x 24 x 2 3, then x 5. The solutions are 5 and 1.
72. x 14 16 0 x 14 16, which has no real solution.
73. 3 x 33 375 x 33 125 x 3 12513 5 x 3 5 8 74. 4 x 25 1 x 25 14 x 2 5 14 x 2 5 14 75. 3 x 5 x 53 125 3 3 14 76. x 43 16 0 x 43 16 24 x 43 24 212 x 4 212 x 212 23 8 15 23 8 77. 2x 53 64 0 2x 53 64 x 53 32 x 3235 25
32 32 62 x 63 216 78. 6x 23 216 0 6x 23 216 x 23 36 62 x 23
944 313 302 161 80. 836 095x 997 095x 161 x 169 095 582 506 81. 215x 463 x 119 115x 582 x 119 195 059 82. 395 x 232x 200 195 332x x 332
79. 302x 148 1092 302x 944 x
4497 4366 103 84. 214 x 406 227 011x 214x 86684 227 011x 225x 109584 x 48704 487 026x 194 85. 176 026x 194 176 303 244x 026x 194 533 429x 455x 727 303 244x 727 160 x 455 173x 320 86. 151 173x 151 212 x 173x 320 151x 022x 320 x 1455 212 x 022 12 d 12 M 88. d r T H T 87. r M r rH 83. 316 x 463 419 x 724 316x 1463 419x 3034 4497 103x x
89. P V n RT R
PV nT
90. F G
mM Fr 2 m 2 GM r
SECTION P.8 Solving Basic Equations
91. P 2l 2 2 P 2l
35
P 2l 2
1 1 1 R1 R2 R R2 R R1 (multiply both sides by the LCD, R R1 R2 ). Thus R1 R2 R R1 R R2 R R1 R2 R R2 . R1 R2 R R R2 R1 R2 R 3V 3V r 93. V 13 r 2 h r 2 h h mM mM mM 2 r G 94. F G 2 r G F F r 3V 3V r 3 95. V 43 r 3 r 3 4 4 2 2 2 2 2 2 96. a b c b c a b c2 a 2 i 2 i i 2 A i A A A 1 1 i 100 100 97. A P 1 1 100 P 100 100 P 100 P P 98. a 2 x a 1 a 1 x a 2 x a 1 x a 1 a 2 a 1 x a 1 a 2 a 1 x a 1 92.
a 1 x 2 a a1 ax b 2d b 99. 2 ax b 2 cx d ax b 2cx 2d ax 2cx 2d b a 2c x 2d b x cx d a 2c a1 a1 b1 100. a a 1 a a 1 b b 1 a 2 a a 2 a b2 b 2a b2 b b b a a 12 b2 b
8 25 0032 250 25 000055. So the beam shrinks 10,000 10,000 000055 12025 0007 m, so when it dries it will be 12025 0007 12018 m long. 0032 25 5 0032 25 75 0032 (b) Substituting S 000050 we get 000050 10,000 75 234375. So the water content should be 234375 kg/m3 . 0032 3150 840. So the toy manufacturer can 102. Substituting C 3600 we get 3600 450 375x 3150 375x x 375 manufacture 840 toy trucks. 101. (a) The shrinkage factor when 250 is S
103. (a) Solving for when P 10,000 we get 10,000 156 3 3 64102 86 km/h.
(b) Solving for when P 50,000 we get 50,000 156 3 3 320513 147 km/h.
104. Substituting F 300 we get 300 03x 34 1000 103 x 34 x 14 10 x 104 10,000 lb. 105. (a) C 12 1000 22 1000 981 1 200,000 211,810 (b) 12 2 gh P C P C 12 2 gh.
If the height increases to 5 m, then P 211,810 12 1000 22 1000 981 5 160,760 Pa. If h 1 m and 4 ms, then P 211,810 12 1000 42 1000 981 1 194,000 Pa.
106. (a) 3 0 k 5 k 0 k 1 k 5 k 1 2k 6 k 3
(b) 3 1 k 5 k 1 k 1 3 k 5 k k 1 k 2 1 k 3
(c) 3 2 k 5 k 2 k 1 6 k 5 2k k 1 k 1 k 1. x 2 is a solution for every value of k. That is, x 2 is a solution to every member of this family of equations.
36
CHAPTER P Prerequisites
107. When we multiplied by x, we introduced x 0 as a solution. When we divided by x 1, we are really dividing by 0, since x 1 x 1 0.
P.9
MODELING WITH EQUATIONS
1. An equation modeling a realworld situation can be used to help us understand a realworld problem using mathematical methods. We translate realworld ideas into the language of algebra to construct our model, and translate our mathematical results back into realworld ideas in order to interpret our findings. 2. In the formula I Pr t for simple interest, P stands for principal, r for interest rate, and t for time (in years). 3. (a) A square of side x has area A x 2 .
(b) A rectangle of length l and width has area A l. (c) A circle of radius r has area A r 2 .
5 16 ounces 4. Balsamic vinegar contains 5% acetic acid, so a 32 ounce bottle of balsamic vinegar contains 32 5% 32 100
of acetic acid.
1 1 wall . x hours x d rt d d rt d 6. Solving d rt for r, we find r . Solving d rt for t, we find t . t t t r r r 7. If n is the first integer, then n 1 is the middle integer, and n 2 is the third integer. So the sum of the three consecutive integers is n n 1 n 2 3n 3. 5. A painter paints a wall in x hours, so the fraction of the wall she paints in one hour is
8. If n is the middle integer, then n 1 is the first integer, and n 1 is the third integer. So the sum of the three consecutive integers is n 1 n n 1 3n. 9. If n is the first even integer, then n 2 is the second even integer and n 4 is the third. So the sum of three consecutive even integers is n n 2 n 4 3n 6.
10. If n is the first integer, then the next integer is n 1. The sum of their squares is n 2 n 12 n 2 n 2 2n 1 2n 2 2n 1.
11. If s is the third test score, then since the other test scores are 78 and 82, the average of the three test scores is 160 s 78 82 s . 3 3 12. If q is the fourth quiz score, then since the other quiz scores are 8, 8, and 8, the average of the four quiz scores is 888q 24 q . 4 4 13. If x dollars are invested at 2 12 % simple interest, then the first year you will receive 0025x dollars in interest. 14. If n is the number of months the apartment is rented, and each month the rent is $945, then the total rent paid is 945n. 15. Since is the width of the rectangle, the length is three times the width, or 4. Then area length width 4 42 ft2 .
four
16. Since is the width of the rectangle, the length is 6. The perimeter is 2 length 2 width 2 6 2 4 12 d distance . 17. If d is the given distance, in miles, and distance rate time, we have time rate 55 1h 34 s mi. 18. Since distance rate time we have distance s 45 min 60 min 19. If x is the quantity of pure water added, the mixture will contain 25 oz of salt and 3 x gallons of water. Thus the 25 concentration is . 3x
SECTION P.9 Modeling with Equations
37
20. If p is the number of pennies in the purse, then the number of nickels is 2p, the number of dimes is 4 2 p, and the number of quarters is 2 p 4 2 p 4 p 4. Thus the value (in cents) of the change in the purse is 1 p 5 2 p 10 4 2 p 25 4 p 4 p 10 p 40 20 p 100 p 100 131 p 140. 21. If d is the number of days and m the number of miles, then the cost of a rental is C 65d 040m. In this case, d 3 and 88 C 283, so we solve for m: 283 65 3 040m 283 195 04m 04m 88 m 220. Thus, the 04 truck was driven 220 miles. 100 if m 250 22. The cost of speaking for m minutes on this plan is C In this case m 250 100 025 m 250 if m 250
and C 12050, so we solve 100 025 m 250 1205 025 m 250 205 m 250 4 205 82 m 250 82 332. Therefore, the tourist used 332 minutes of talk in the month of June. 23. If x is the student’s score on their final exam, then because the final counts twice as much as each midterm, the average 82 75 71 2x 228 2x 114 x 114 x score is . For the average to be 80%, we must have 80% 08 3 100 200 500 250 250 114 x 250 08 200 x 86. So the student scored 86% on their final exam. 24. Six students scored 100 and three students scored 60. Let x be the average score of the remaining 25 6 3 16 students. 6 100 3 60 16x 084 780 16x 084 2500 2100 Because the overall average is 84% 084, we have 25 100 16x 1320 x 1320 16 825. Thus, the remaining 16 students’ average score was 825%.
25. Let m be the amount invested at 2 12 %. Then 12,000 m is the amount invested at 3%. Since the total interest is equal to the interest earned at 25% plus the interest earned at 3%, we have 318 0025m 003 12,000 m 318 0025m 360 003m 0005m 42 m
42 8400. Thus, 0005
$8400 is invested at 2 12 % and 12,000 8400 $3600 is invested at 3%.
26. Let m be the amount invested at 5%. Then 8000 m is the total amount invested. Thus 4% of the total investment interest earned at 3 12 % interest earned at 5% 40 4000. Thus, $4000 must be So 004 8000 m 0035 8000 005m 320 004m 280 005m m 001 invested at 5%. 2625 0075 or 75%. 27. Using the formula I Prt and solving for r, we get 26250 3500 r 1 r 3500 28. If $3000 is invested at an interest rate a%, then $5000 is invested at a 12 %, so, remembering that a is expressed as a 1
a 2 a 1 5000 1 30a 50a 25 80a 25. Since the total interest 100 100 is $265, we have 265 80a 25 80a 240 a 3. Thus, $3000 is invested at 3% interest. 29. Let x be the monthly salary. Since annual salary 12 monthly salary Christmas bonus, we have 180,100 12x 7300 172,800 12x x 14,400. The monthly salary is $14,400. 30. Let s be the assistant’s annual salary. Then the foreman’s annual salary is 115s. Their total income is the sum of their salaries, so s 115s 113,305 215s 113,305 s 52,700. Thus, the assistant’s annual salary is $52,700. 31. Let x be the overtime hours worked. Since gross pay regular salary overtime pay, we obtain the equation 16650 814 1850 35 1850 15 x 814 64750 2775x 16650 2775x x 6. Thus, the lab 2775 technician worked 6 hours of overtime. 32. Let x be the number of hours worked by the plumber. Then the cabinet maker works for 9x hours. The total labor charge is percentage, the total interest is I 3000
the sum of their charges, so 2610 150x 80 9x 2610 150x 720x x 2610 870 3. Thus, the plumber works for 3 hours and the cabinet maker works for 3 9 27 hours.
38
CHAPTER P Prerequisites
33. All ages are in terms of the daughter’s age 7 years ago. Let y be age of the daughter 7 years ago. Then 11y is the age of the movie star 7 years ago. Today, the daughter is y 7, and the movie star is 11y 7. But the movie star is also 4 times his daughter’s age today. So 4 y 7 11y 7 4y 28 11y 7 21 7y y 3. Thus, today the movie star is 11 3 7 40 years old. 34. Let h be number of home runs Babe Ruth hit. Then h 41 is the number of home runs that Hank Aaron hit. So 1469 h h 41 1428 2h h 714. Thus Babe Ruth hit 714 home runs. 35. Let n be the number of nickels. Then there are also n dimes and n quarters. The total value of the coins in the purse is the sum of the values of nickels, dimes, and quarters, so 280 005n 010n 025n 28 04 p p 28 04 7. So the
purse contains 7 nickels, 7 dimes, and 7 quarters. 36. Let q be the number of quarters. Then 2q is the number of dimes, and 2q 5 is the number of nickels. Thus 300 value of nickels value of dimes value of quarters, so 300 005 2q 5 010 2q 025q 300 010q 025 020q 025q 275 055q q 275 055 5. Thus you have 5 quarters, 2 5 10 dimes, and 2 5 5 15 nickels.
37. Let l be the length of the garden. Since area width length, we obtain the equation 1125 25l l 1125 25 45 ft. So
the garden is 45 feet long. 38. Let be the width of the pasture. Then the length of the pasture is 3. Since area length width we have 132,300 3 32 2 1323,300 44,100 210. Thus the width of the pasture is 210 feet.
39. Let be the width of the building lot. Then the length of the building lot is 5. Since a halfacre is 12 43,560 21,780 and area is length times width, we have 21,780 5 52 2 4,356 66. Thus the width of the building lot is 66 feet and the length of the building lot is 5 66 330 feet.
40. Let x be the length of a side of the square plot. As shown in the figure,
x
area of the plot area of the building area of the parking lot. Thus,
x 2 60 40 12,000 2,400 12,000 14,400 x 120. So the plot of
x
land measures 120 feet by 120 feet.
60 40
41. The shaded area is the sum of the area of a square and the area of a triangle. So A y 2 12 y y 32 y 2 . We are given that the area is 120 in2 , so 120 32 y 2 y 2 80 y 80 4 5. y is positive, so y 4 5 894 in. x
42. First we write a formula for the area of the figure in terms of x. Region A has dimensions 10 cm and x cm and region B has dimensions 6 cm and x cm. So the shaded region has area 10 x 6 x 16x cm2 . We are given that this is equal to 144 cm2 , so 144 16x x 144 16 9 cm.
10 cm
A
6 cm B
x
43. Let x be the width of the strip. Then the length of the mat is 20 2x, and the width of the mat is 15 2x. Now the perimeter is twice the length plus twice the width, so 102 2 20 2x 2 15 2x 102 40 4x 30 4x 102 70 8x 32 8x x 4. Thus the strip of mat is 4 inches wide. 44. Let x be the width of the strip. Then the width of the poster is 100 2x and its length is 140 2x. The perimeter of the printed area is 2 100 2 140 480, and the perimeter of the poster is 2 100 2x 2 140 2x. Now we use the
fact that the perimeter of the poster is 1 12 times the perimeter of the printed area: 2 100 2x 2 140 2x 32 480
480 8x 720 8x 240 x 30. The blank strip is thus 30 cm wide.
45. Let x be the length of the person’s shadow, in meters. Using similar triangles, x 5. Thus the person’s shadow is 5 meters long.
x 10 x 20 2x 6x 4x 20 6 2
SECTION P.9 Modeling with Equations
39
46. Let x be the height of the tall tree. Here we use the property that corresponding sides in similar triangles are proportional. The base of the similar triangles starts at x 5 150 15 25 25 x 5 15 150 25x 125 2250 25x 2375 x 95. Thus the
x-5
eye level of the woodcutter, 5 feet. Thus we obtain the proportion
5
tree is 95 feet tall.
25
15
125
47. Let x be the amount (in mL) of 60% acid solution to be used. Then 300 x mL of 30% solution would have to be used to yield a total of 300 mL of solution. 60% acid
30% acid
Mixture
mL
x
300
Rate (% acid)
060
300 x
Value
060x
030
050
030 300 x
050 300
60 200. 03 So 200 mL of 60% acid solution must be mixed with 100 mL of 30% solution to get 300 mL of 50% acid solution.
Thus the total amount of pure acid used is 060x 030 300 x 050 300 03x 90 150 x
48. The amount of pure acid in the original solution is 300 50% 150. Let x be the number of mL of pure acid added. Then 150 x 60% 06 the final volume of solution is 300 x. Because its concentration is to be 60%, we must have 300 x 30 150 x 06 300 x 150 x 180 06x 04x 30 x 75. Thus, 75 mL of pure acid must be 04 added.
49. Let x be the number of grams of silver added. The weight of the rings is 5 18 g 90 g. 5 rings
Pure silver
Mixture
Grams
90
x
Rate (% gold)
090
0
90 x
090 90
0x
Value
075
075 90 x
So 090 90 0x 075 90 x 81 675 075x 075x 135 x 135 075 18. Thus 18 grams of silver
must be added to get the required mixture.
50. Let x be the number of liters of water to be boiled off. The result will contain 6 x liters. Original
Water
Final
Liters
6
Concentration
120
x
6x
Amount
120 6
0
200 6 x
0
200
So 120 6 0 200 6 x 720 1200 200x 200x 480 x 24. Thus 24 liters need to be boiled off.
40
CHAPTER P Prerequisites
51. Let x be the number of liters of coolant removed and replaced by water. 60% antifreeze
60% antifreeze (removed)
Water
Mixture
Liters
36
x
x
36
Rate (% antifreeze)
060
060
0
050
060 36
060x
0x
050 36
Value
036 06. Thus 06 liters 06
So 060 36 060x 0x 050 36 216 06x 18 06x 036 x must be removed and replaced by water.
52. Let x be the number of gallons of 2% bleach removed from the tank. This is also the number of gallons of pure bleach added to make the 5% mixture. Original 2%
Pure bleach
5% mixture
100 x
x
100
002
1
005
002 100 x
1x
005 100
Gallons Concentration Bleach
So 002 100 x x 005 100 2 002x x 5 098x 3 x 306. Thus 306 gallons need to removed and replaced with pure bleach. 53. Let c be the concentration of fruit juice in the cheaper brand. The new mixture consists of 650 mL of the original fruit punch and 100 mL of the cheaper fruit punch. Original Fruit Punch
Cheaper Fruit Punch
mL
650
100
750
Concentration
050
c
048
050 650
100c
048 750
Juice
Mixture
So 050 650 100c 048 750 325 100c 360 100c 35 c 035. Thus the cheaper brand is only 35% fruit juice. 54. Let x be the number of ounces of $300oz tea Then 80 x is the number of ounces of $275oz tea. $300 tea
$275 tea
Ounces
x
Rate (cost per ounce)
300
80 x
Value
300x
Mixture 80
275
290
275 80 x
290 80
So 300x 275 80 x 290 80 300x 220 275x 232 025x 12 x 48. The mixture uses 48 ounces of $300oz tea and 80 48 32 ounces of $275oz tea.
1 of the car per 55. Let t be the time in minutes it would take to wash the car if the friends worked together. Friend 1 washes 25 1 of the car per minute. The sum of these fractions is equal to the fraction of the job they minute, while Friend 2 washes 35
1 1 875t 875t 875t 1 875 35t 25t 60t t 875 60 t 25 35 t 25 35 t 14 35 60 minutes, or 14 minutes 35 seconds can do working together, so we have
56. Let t be the time, in minutes, it takes the landscaper to mow the lawn. Since the assistant is half as fast, it would take them 1 1 10t 10t 10t 1 t 10 5 15. Thus, it would the 2t minutes to mow the lawn alone. Thus, 10 t 2t 10 t 2t assistant 2 15 30 minutes to mow the lawn alone.
SECTION P.9 Modeling with Equations
41
57. Let t be the number of hours it would take your friend to paint a house alone. Then working together, it takes 23 t hours. Because it takes you 7 hours, we have
3 14t 1 1 1 14t 14t 14t 2 2 2 2t 14 21 t 72 . Thus, it 7 t 7 t t t t 3
3
would take your friend 35 h to paint a house alone. 58. Let h be the time, in hours, to fill the swimming pool using the smaller hose alone. Since the larger hose takes 20% less 1 1 1 16 08 16 08h 128 16 08h time, it takes 08h to fill the pool alone. Thus 16 16 h 08h h 288 08 36. Thus, the smaller hose takes 36 hours to fill the pool alone, and the larger hose takes 08 36 288 hours. 59. Let t be the time in hours that the commuter spent on the train. Then 11 2 t is the time in hours that commuter spent on the bus. We construct a table: Rate By train By bus
Time
40
t
60
11 t 2
Distance
40t
60 11 2 t
The total distance traveled is the sum of the distances traveled by bus and by train, so 300 40t 60 11 2 t
300 40t 330 60t 30 20t t 30 20 15 hours. So the time spent on the train is 55 15 4 hours.
60. Let r be the speed of the slower cyclist, in mi/h. Then the speed of the faster cyclist is 2r. Rate
Time
Distance
Slower cyclist
r
2
2r
Faster cyclist
2r
2
4r
When they meet, they will have traveled a combined total of 90 miles, so 2r 4r 90 6r 90 r 15. The speed of the slower cyclist is 15 mi/h, while the speed of the faster cyclist is 2 15 30 mi/h. 61. Let r be the speed of the plane from Montreal to Los Angeles. Then r 020r 120r is the speed of the plane from Los Angeles to Montreal. Rate Montreal to L.A.
r
L.A. to Montreal
12r
Time 2500 r 2500 12r
Distance 2500 2500
2500 55 2500 2500 2500 r 12r 6 r 12r 33,000 55 12r 2500 6 12 2500 6 66r 18,000 15,000 66r 33,000 r 66 500. Thus the plane flew The total time is the sum of the times each way, so 9 16
at a speed of 500 mi/h on the trip from Montreal to Los Angeles.
ft 22 62. Let x be the speed of the car in mi/h. Since a mile contains 5280 ft and an hour contains 3600 s, 1 mi/h 5280 3600 s 15 ft/s. 220 220 The truck is traveling at 50 22 15 3 ft/s. So in 6 seconds, the truck travels 6 3 440 feet. Thus the back end
of the car must travel the length of the car, the length of the truck, and the 440 feet in 6 seconds, so its speed must be 1430440 242 ft/s. Converting to mi/h, we have that the speed of the car is 242 15 55 mi/h. 6 3 3 22
63. Let x be the distance from the fulcrum to where the 125pound friend sits. Then substituting the known values into the formula given, we have 100 8 125x 800 125x x 64. So the 125pound friend should sit 64 feet from the fulcrum.
42
CHAPTER P Prerequisites
64. Let be the largest weight that can be hung. In this exercise, the edge of the building acts as the fulcrum, so the 240 lb man is sitting 25 feet from the fulcrum. Then substituting the known values into the formula given in Exercise 63, we have 240 25 5 6000 5 1200. Therefore, 1200 pounds is the largest weight that can be hung. l
65. Let l be the length of the lot in feet. Then the length of the diagonal is l 10. We apply the Pythagorean Theorem with the hypotenuse as the diagonal. So
l 2 502 l 102 l 2 2500 l 2 20l 100 20l 2400 l 120.
50
l+10
Thus the length of the lot is 120 feet.
66. Let r be the radius of the running track. The running track consists of two semicircles and two straight sections 110 yards long, so we get the equation 2r 220 440 2r 220 r 110 3503. Thus the radius of the semicircle is about 35 yards. 67. Let h be the height in feet of the structure. The structure is composed of a right cylinder with radius 10 and height 23 h and a cone with base radius 10 and height 13 h. Using the formulas for the volume of a cylinder and that of a cone, we obtain the 100 equation 1400 102 23 h 13 102 13 h 1400 200 3 h 9 h 126 6h h (multiply both sides
by
9 ) 126 7h h 18. Thus the height of the structure is 18 feet. 100
68. Let h be the height of the break, in feet. Then the portion of the bamboo above the break is 10 h. Applying the Pythagorean Theorem, we obtain
h 2 32 10 h2 h 2 9 100 20h h 2 91 20h h 91 20 455. Thus the break is 455 ft above the ground.
10-h
h
3
69. Answers will vary.
CHAPTER P REVIEW 1. (a) Since there are initially 250 tablets and the patient takes 2 tablets per day, the number of tablets T that are left in the bottle after x days is T 250 2x. (b) After 30 days, there are 250 2 30 190 tablets left.
(c) We set T 0 and solve: T 250 2x 0 250 2x x 125. The patient will run out after 125 days.
2. (a) The total cost is $11 per calzone plus the $7 delivery charge, so C 11x 7. (b) Four calzones would cost 11 4 7 $51.
(c) We solve C 11x 7 40 11x 33 x 3. You can order three calzones.
3. (a) 16 is rational. It is an integer, and more precisely, a natural number.
(b) 16 is rational. It is an integer, but because it is negative, it is not a natural number. 16 4 is rational. It is an integer, and more precisely, a natural number. (d) 2 is irrational. (c)
(e) 83 is rational, but is neither a natural number nor an integer. (f) 82 4 is rational. It is an integer, but because it is negative, it is not a natural number.
4. (a) 5 is rational. It is an integer, but not a natural number.
(b) 25 6 is rational, but is neither an integer nor a natural number. (c) 25 5 is rational, a natural number, and an integer. (d) 3 is irrational.
CHAPTER P 3 (e) 24 16 2 is rational, but is neither a natural number nor an integer.
(f) 1020 is rational, a natural number, and an integer. 5. Commutative Property for addition.
6. Commutative Property for multiplication.
7. Distributive Property.
8. Distributive Property.
5 4 9 3 5 2 6 3 6 6 6 2 5 4 1 5 2 (b) 6 3 6 6 6 15 12 33 9 15 12 11. (a) 8 5 85 21 2 15 5 55 25 15 12 (b) 8 5 8 12 84 32
11 21 22 1 7 10 15 30 30 30 11 21 22 43 7 (b) 10 15 30 30 30 30 35 55 25 30 12 12. (a) 7 35 7 12 12 2 30 12 6 12 72 30 12 (b) 7 35 7 35 77 49
9. (a)
10. (a)
13. x [2 6 2 x 6
14. x 0 10] 0 x 10
_2
6
15. x 4] x 4
0
16. x [2 2 x 4
17. x 5 x [5
_2
18. x 3 x 3
5
_3
20. 0 x 12 x 0 12
19. 1 x 5 x 1 5] _1
5
0
21. (a) A B 1 0 12 1 2 3 4
22. (a) C D 1 2]
(b) A B 1
(b) C D 0 1]
23. (a) A C 1 2 (b) B D 12 1
24. (a) A D 0 1 (b) B C 12 1
25. 1 4 1 4 3 3 27. 212 812 2 8 16 4 29. 21613 31.
1 1 16 3 13 216 216
242 242 2 121 11 2
33. (a) 5 3 2 2
(b) 5 3 8 8
10
1 2
26. 5 10 4 5 10 4 5 6 1
9 8 1 28. 23 32 18 19 72 72 72 23 30. 6423 43 42 16
32.
2 50 100 10
34. (a) 4 0 4 4 (b) 4 4 8 8
Review
43
44
CHAPTER P Prerequisites
35. (a) 3 7 713 5 (b) 74 745 6
x 5 x 56 9 12 9 (b) x x x 92
37. (a)
3 7 5 573 3 3 (b) 4 5 514 534 12 38. (a) y 3 y 3 y 32 2 2 (b) 8 y y 18 y 14 36. (a)
3 2 4 a3 b b3 a 6 a 6 b2 b12 a 66 b212 b14 a2 3 2 x 1 y 2 27x 3 y 6 4 x 2 y 2 27 4 x 32 y 62 12x y 8 (b) 3x y 2 3 9 9
39. (a)
x 4 3x2 x 4 9x 2 9x 423 9x 3 3 x x3 6 r 2 s 43 r 12 s 8 (b) r 122 s 86 r 10 s 2 r 13 s r 2s6 2 3 x 3 y y4 3 x 6 y4 y2 3 x 6 y6 x 2 y2 41. (a) (b) 8 z 10 8 z 10 4 z5 40. (a)
8r 12 s 3 4r 52 4r 122 s 34 4r 52 s 7 2 4 s7 2r s 2 4 6 a 2 b4 c6 ab2 c3 2 a 26 b48 c6 4a 4 b12 c6 4a c (b) 2 2a 3 b4 22 a 6 b8 b12
42. (a)
43. 78,250,000,000 7825 1010
44. 208 108 0000 000 020 8 293 106 1582 1014 000000293 1582 1014 293 1582 ab 1061412 45. 12 12 c 28064 28064 10 28064 10 165 1032 46. 80
times 60 minutes 24 hours 365 days 90 years 38 109 times minute hour day year
47. x 2 5x 14 x 7 x 2 48. 12x 2 10x 8 2 6x 2 5x 4 2 3x 4 2x 1 49. 2x 2 y 6x y 2 2x y x 3y
50. 4y 2 z 3 10y 3 z 12y 5 z 2 2y 2 z 2z 2 5y 6y 3 z
51. 3x 2 2x 1 3x 1 x 1 2 52. x 4 x 2 2 x 2 x 2 2 x 2 2 x 2 1 x 2 2 x 1 x 1
53. 4t 2 13t 12 4t 3 t 4 2 54. x 4 2x 2 1 x 2 1 [x 1 x 1]2 x 12 x 12 55. 16 4t 2 4 4 t 2 4 2 t 2 t 4 t 2 t 2 56. 2y 6 32y 2 2y 2 y 4 16 2y 2 y 2 4 y 2 4 2y 2 y 2 4 y 2 y 2 57. x 6 1 x 3 1 x 3 1 x 1 x 2 x 1 x 1 x 2 x 1
CHAPTER P
Review
58. 16a 4 b2 2ab5 2ab2 8a 3 b3 2ab2 2a b 4a 2 2ab b2 59. x 3 27 x 3 x 2 3x 9 60. 3y 3 81x 3 3 y 3 27x 3 3 y 3x y 2 3x y 9x 2 61. 3x 12 2x 12 5x 32 x 12 5x 3 2x 2 3x x 12 5x 3 x 1 62. 7x 32 8x 12 x 12 x 32 x 2 8x 2 7 x 32 x 1 x 7 63. 5x 3 15x 2 x 3 5x 2 x 3 x 3 5x 2 1 x 3 64. 3 32 4 12 2 3 4 3 2 4 3 3 2 2 65. a b2 3 a b 10 a b 5 a b 2
66. 3x 22 3x 2 6 [3x 2 3] [3x 2 2] 3x 1 3x 4
67. 2y 7 2y 7 4y 2 49
68. 1 x 2 x 3 x 3 x 2 x x 2 9 x 2 2 x x 2 9 x 2 7 x 69. x 2 x 2 x x 22 x 3 2x 2 x x 2 4x 4 x 3 2x 2 x 3 4x 2 4x 2x 3 6x 2 4x x x 2 2x 3 x 3 2x 2 3x 70. x 2 2x 3 x x 71. x x 1 2 x 1 x x 2 x 1 2x x x 2x x 2x 32 x x 12
72. 2x 13 2x3 3 2x2 1 3 2x 12 13 8x 3 12x 2 6x 1 x 2 2x 3 x 3 x 3 x 1 2 2x 3 x 1 2x 3 2x 5x 3 2 t 1 t t 1 t2 t 1 t3 1 74. 2 t 1 t 1 t 1 t 1 73.
75. 76.
x 2 4x 5 x 2 12x 36 x 6 x 1 x 6 x 5 x 1 x 62 2 2 x 5 x 1 x 5 x 5 x 5 x 6 x 1 x 25 x 7x 6
x 2 2x 15 x 2 x 12 x 1 x 5 x 3 x 1 x 1 x 4 x 5 x 1 x 4 x 3 x 2 6x 5 x2 1
77. x 78.
x x 1 1 x2 x 1 1 x 1 x 1 x 1 x 1
1 x x2 1 x x 1 x2 1 x2 x x 1 2 2 2 2 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 2 1
1 3 2 x 22 x x 2 3x 2 2 x x 2 x 2 x x 22 x x 22 x x 22 2 x 2 4x 4 x 2 2x 3x 2x 2 8x 8 x 2 2x 3x 3x 2 7x 8 x x 22 x x 22 x x 22 1 1 1 2 1 2 80. x 2 x2 4 x2 x 2 x 2 x 2 x 2 x 2 x 1 x 1 2 x 2 x 2 x 1 x 2 x 1 x 2 x 2 x 1 x 2 x 2 x 1 x 2 79.
x 2 2x 5 x 2 x 2 x 1 2x 4 x 2 x 1 x 2 x 2 x 1 x 2
45
46
CHAPTER P Prerequisites
x 2 1 1 2 2x 2x 2 x 1 1 x 2 1 1 81. x x 2 x 2 2x x 2 2x x 2 2x 1 1 1 1 1 x 1 x x 1 x x 1 x 1 x 82. x 1 1 1 1 x x 1 2x 1 x 1 x x x 1 x x 1 x 44 x x 2 x 2 x 2 83. x 4 x 2 x 2 x 2 h 1 x h x x h x x h x x h x 84. h h x h x h x h x h x h x x h x 1 11 11 1 85. 11 11 11 11 3 6 6 3 3 6 86. 6 2 6 6 6 10 10 10 2 10 21 87. 10 10 2 2 1 21 21 21 14 3 2 42 14 2 14 42 14 2 88. 62 2 2 7 3 2 3 2 3 2 32 2 x x 2 x 2 x 2x x x x 89. 2 x 4 2 x 2 x 2 x 22 x x 4 x 4 x 2 x 2 x 2 90. x 4 x 2 x 2 x 2
x 5 is defined whenever x 10 0 x 10, so its domain is x x 10. x 10 2x 92. 2 is defined whenever x 2 9 0 x 2 9 x 3, so its domain is x x 3 and x 3. x 9 x is defined whenever x 0 (so that x is defined) and x 2 3x 4 x 1 x 4 0 x 1 and 93. 2 x 3x 4 x 4. Thus, its domain is x x 0 and x 4. x 3 is defined whenever x 3 0 x 3 and x 2 4x 4 x 22 0 x 2. Thus, its domain is 94. 2 x 4x 4 x x 3. 91.
95. This statement is false. For example, take x 1 and y 1. Then LHS x y3 1 13 23 8, while RHS x 3 y 3 13 13 1 1 2, and 8 2. 1a 1 a 1 a 1 a 1 . 96. This statement is true for a 1: 1a 1a 1 a 1 a 1 a 1 a
97. This statement is true:
12 y 12 12 y 1. y y y y
98. This statement is false. For example, take a 1 and b 1. Then LHS 3 a b 3 1 1 3 2, while RHS 3 a 3 b 3 1 3 1 1 1 2, and 3 2 2. 99. This statement is false. For example, take a 1. Then LHS a 2 12 1 1, which does not equal a 1. The true statement is a 2 a.
CHAPTER P
100. This statement is false. For example, take x 1. Then LHS and
1 5 . 5 4
Review
47
1 1 1 1 1 1 5 1 , while RHS , x 4 14 5 x 4 1 4 4
101. 3x 12 24 3x 12 x 4
102. 5x 7 42 5x 49 x 49 5
103. 7x 6 4x 9 3x 15 x 5
104. 8 2x 14 x 3x 6 x 2
105. 13 x 12 2 2x 3 12 2x 15 x 15 2 6 3 106. 23 x 35 15 2x 10x 9 3 30x 40x 6 x 40 20
107. 2 x 3 4 x 5 8 5x 2x 6 4x 20 8 5x 2x 26 8 5x 3x 18 x 6 5 x 5 2x 5 3 x 5 2 2x 5 5 3x 15 4x 10 5 x 30 x 30 108. 2 3 6 x 1 2x 1 109. x 1 2x 1 2x 1 x 1 2x 2 3x 1 2x 2 3x 1 6x 0 x 0 x 1 2x 1 x 1 7 110. 3 x 3 x 2 1 x 3x 6 1 2x 7 x x 2 x 2 2 3x 3x x x 1 2 x 1 x 2 x x 1 x x 2 x 2 x 2 0. Since this 111. x 1 3x 6 3 x 2 x 2 last equation is never true, there is no real solution to the original equation. 112. x 22 x 42 x 22 x 42 0 [x 2 x 4] [x 2 x 4] 0 [x 2 x 4] [x 2 x 4] 6 2x 2 0 2x 2 0 x 1. 113. x 2 144 x 12
7 114. 4x 2 49 x 2 49 4 x 2
115. x 3 27 0 x 3 27 x 3.
116. 6x 4 15 0 6x 4 15 x 4 52 . Since x 4 must be nonnegative, there is no real solution.
117. x 13 64 x 1 4 x 1 4 5. 118. x 22 2 0 x 22 2 x 2 2 x 2 2. 119. 3 x 3 x 33 27. 2 120. x 23 4 0 x 13 4 x 13 2 x 8.
121. 4x 34 500 0 4x 34 500 x 34 125 x 12543 54 625. 122. x 215 2 x 2 25 32 x 2 32 34. xy 2A x y x 2A y. 123. A 2 V xz . 124. V x y yz xz V y x z xz V xz y x z y x z 1 1 1 1 1 11 11 125. Multiply through by t: J tJ 1 t , J 0. t 2t 3t 2 3 6 6J q q q q q q 126. F k 1 2 2 r 2 k 1 2 r k 1 2 . (In realworld applications, r represents distance, so we would take the F F r positive root.) 127. Let x be the number of pounds of raisins. Then the number of pounds of nuts is 50 x. Raisins
Nuts
Mixture
Pounds
x
50
Rate (cost per pound)
320
50 x 240
272
So 320x 240 50 x 272 50 320x 120 240x 136 08x 16 x 20. Thus the mixture uses 20 pounds of raisins and 50 20 30 pounds of nuts.
48
CHAPTER P Prerequisites
128. Let t be the number of hours that the district supervisor drives. Then the store manager drives for t 14 hours. Supervisor Manager
Rate
Time
Distance
45
t
40
t 14
45t 40 t 14
When they pass each other, they will have traveled a total of 160 miles. So 45t 40 t 14 160 45t 40t 10 160 85t 170 t 2. Since the supervisor leaves at 2:00 P. M . and travels for 2 hours, they pass each other at 4:00 P. M .
129. Let x be the amount invested in the account earning 15% interest. Then the amount invested in the account earning 25% is 7000 x. 15% Account
25% Account
Total
Amount invested
x
7000
Interest earned
0015x
7000 x
0025 7000 x
12025
From the table, we see that 0015x 0025 7000 x 12025 0015x 175 0025x 12025 5475 001x x 5475. Thus, $5475 is invested in the account earning 15% interest and $1525 is invested in the account earning 25% interest. 130. The amount of interest currently earned is 6000 003 $180 per year. If a total of $300 is desired, another $120 in interest must be earned at a rate of 125% per year. If the additional amount invested is x, we have 00125x 120 x 9600. Thus, an additional $9600 must be invested at 125% simple interest to earn a total of $300 interest per year. 131. Let t be the time it would take the interior decorator to paint a living room if they work alone. It would take the assistant 1 2t hours alone, and it would take the apprentice 3t hours alone. Thus, the decorator does of the job per hour, the assistant t 1 1 1 1 1 of the job per hour, and the apprentice does of the job per hour. So 1 6 3 2 6t does 2t 3t t 2t 3t 6t 11 t 11 6 . Thus, it would take the decorator 1 hour 50 minutes to paint the living room alone.
132. Let be width of the pool. Then the length of the pool is 2, and its volume is 8 2 8464 162 8464 2 529 23. Since 0, we reject the negative value. The pool is 23 feet wide, 2 23 46 feet long, and 8 feet deep.
CHAPTER P TEST 1. (a) The cost is C 9 15x.
(b) There are four extra toppings, so x 4 and C 9 15 4 $15.
(c) We have C 195 9 15x 15x 105 x 7. Thus, a $1950 pizza has 7 toppings.
2. (a) 5 is rational. It is an integer, and more precisely, a natural number. (b) 5 is irrational. (c) 93 3 is rational, and it is an integer.
(d) 1,000,000 is rational, and it is an integer.
3. (a) A B 0 1 5 (b) A B 2 0 12 1 3 5 7 4. (a)
_4
2
[4 2
0
3
[0 3]
CHAPTER P
(b) 0
(c) 2 4 6 6 5. (a)
2
_4
[4 2 [0 3] [0 2
_5
3
3
[4 2 [0 3] [4 3]
2
5 3] (b) x 3 x 3]; 1 x 4 x [1 4
2
6. (a) 34 81
1 1 (c) 34 4 81 3 2 32 9 2 2 (e) 3 4 2 8 32 9 4 3 (g) 16 216 24 7. (a) 186,000,000,000 186 1011
(b) 34 81
375 (d) 72 37572 33 27 3 5 2 32 1 (f) 4 2 16 34 (h) 1634 24 23 18
(b) 0000 000 396 5 3965 107
a 3 b2 a2 b ab3 (b) 200 32 10 2 4 2 6 2 2 (c) 2x 12 y 2 3x 14 y 1 2 32 x 12214 y 221 18x 2 (d) 3a 3 b3 4ab2 3a 3 b3 42 a 2 b4 48a 5 b7 2 (e) 9x 2 y 4 3x y 2 3 x y 2 12 12 12 y2 4x 9 y 3 4x 8 y4 (f) 4 7 4 8 xy y 4x 2x
8. (a)
9. (a) z 4z 3 2z 3 2z 4z 2 3z 6z 4z 2 3z
(b) x 3 4x 5 4x 2 5x 12x 15 4x 2 7x 15 2 2 a b a b a b ab (c) (d) 2x 32 2x2 2 2x 3 32 4x 2 12x 9
(e) x 23 x3 3 x2 2 3 x 22 23 x 3 6x 2 12x 8 (f) x 2 x 3 x 3 x 2 x 2 9 x 4 9x 2
10. (a) 4x 2 25 2x 5 2x 5
(b) 2x 2 5x 12 2x 3 x 4
(c) x 3 3x 2 4x 12 x 2 x 3 4 x 3 x 3 x 2 4 x 3 x 2 x 2 (d) x 4 27x x x 3 27 x x 3 x 2 3x 9 (e) 2x 32 8x 12 10x 12 2x 12 x 2 4x 5 2x 12 x 5 x 1
Test
49
50
CHAPTER P Prerequisites
(f) x 4 y 2 9x 2 y 2 x 2 y 2 x 2 9 x 2 y 2 x 3 x 3 11. (a) (b)
2 4 3 3 3 1 3 3 1 2 2 3
x 1 2x 2 x 1 x 3 2x 1 x 1 x 3 2x 1 x 3 x 3 x 3 2x 1 x2 9
x2 x2 x 1 x 1 x 1 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 2 2 x x x 2 x 2 1 x 2 x 2 x 2 x 2 x 2 y x x y y2 x 2 x y y x y x y x x y x y xy (d) x y 1 1 xy 1 1 xy xy xy y x y x 3 2 6 6 6 632 12. (a) 332 3 3 2 3 2 3 2 4 2 2 2 10 52 10 10 52 (b) 50 2 10 5 2 2 10 5 4 52 52 52 1 1 x 1 x 1 (c) 1x 1 x 1 x 1 x (c)
x2
x2 4
13. (a) 4x 3 2x 7 4x 2x 7 3 2x 10 x 5. 3 (b) 8x 3 125 8x 3 3 125 2x 5 x 52 . 32 6432 x 83 512. (c) x 23 64 0 x 23 64 x 23
x 3 x x 2x 1 x 3 2x 5 2x 2 x 2x 2 5x 6x 15 x x 15 2x 15 2x 5 2x 1 x 15 2 . (e) 3 x 12 18 0 3 x 12 18 x 12 6 x 1 6 x 6 1 E E 14. E mc2 c2 c . (We take the positive root because c represents the speed of light, which is positive.) m m 15. Let t be the time (in hours) it took the trucker to drive from Amity to Belleville. Then 44 t is the time it took the trucker to drive from Belleville to Amity. Since the distance from Amity to Belleville equals the distance from Belleville to Amity, we have 50t 60 44 t 50t 264 60t 110t 264 t 24 hours. Thus the distance is 50 24 120 mi. (d)
Making Optimal Decisions
51
FOCUS ON MODELING Making Optimal Decisions
1. (a) The total cost is
cost of printer
maintenance cost
number of months
printing cost
number of months
. Each
month the printing cost is 8000 003 240. Thus we get C1 5800 25n 240n 5800 265n. rental number printing number . Each month the printing cost (b) In this case the cost is cost of months cost of months
(c)
is 8000 006 480. Thus we get C2 95n 480n 575n. Years
n
Purchase
Rental
1
12
8,980
6,900
2
24
12,160
13,800
3
36
15,340
20,700
4
48
18,520
27,600
5
60
21,700
34,500
6
72
24,880
41,400
(d) The cost is the same when C1 C2 are equal. So 5800 265n 575n 5800 310n n 1871 months.
daily
daily
2. (a) The cost of Plan 1 is 3
cost
The cost of Plan 2 is 3
cost
cost per mile
number of miles
3 65 015x 195 015x.
3 90 270.
(b) When x 400, Plan 1 costs 195 015 400 $255 and Plan 2 costs $270, so Plan 1 is cheaper. When x 800, Plan 1 costs 195 015 800 $315 and Plan 2 costs $270, so Plan 2 is cheaper.
(c) The cost is the same when 195 015x 270 015 75x x 500. So both plans cost $270 when you drive 500 miles.
setup
cost per
price per
number
3. (a) The total cost is (b) The revenue is
cost
tire
tire
of tires
number of tires
. So C 8000 22x.
. So R 49x.
(c) Profit Revenue Cost. So P R C 49x 8000 22x 27x 8000.
(d) Break even is when profit is zero. Thus 27x 8000 0 27x 8000 x 2963. So they need to sell at least 297 tires to break even.
52
FOCUS ON MODELING
4. (a) Option 1: In this option the width is constant at 100. Let x be the increase in length. Then the additional area is increase 100x. The cost is the sum of the costs of moving the old fence, and of installing the width in length
new one. The cost of moving is $6 100 $600 and the cost of installation is 2 10 x 20x, so the total cost is C 600 . Substituting in the area C 20x 600. Solving for x, we get C 20x 600 20x C 600 x 20 C 600 we have A1 100 5 C 600 5C 3,000. 20 Option 2: In this option the length is constant at 180. Let y be the increase in the width. Then the additional area is increase 180y. The cost of moving the old fence is 6 180 $1080 and the cost of installing the new length in width
(b)
one is 2 10 y 20x, so the total cost is C 20y 1080. Solving for y, we get C 20y 1080 20y C 1080 C 1080 C 1080 . Substituting in the area we have A2 180 9 C 1080 9C 9,720. y 20 20 Cost C
Area gain A1 from Option 1
Area gain A2 from Option 2
$1100
2,500 ft2
180 ft2
$1200
3,000 ft2
1,080 ft2
$1500
4,500 ft2
3,780 ft2
$2000
7,000 ft2
8,280 ft2
$2500
9,500 ft2
12,780 ft2
$3000
12,000 ft2
17,280 ft2
(c) If the farmer has only $1200, Option 1 gives the greatest gain. If the farmer has only $2000, Option 2 gives the greatest gain. 5. (a) Design 1 is a square and the perimeter of a square is four times the length of a side. 24 4x, so each side is x 6 feet long. Thus the area is 62 36 ft2 .
12 . Thus, the area is Design 2 is a circle with perimeter 2r and area r 2 . Thus we must solve 2r 24 r 2 12 144 458 ft2 . Design 2 gives the largest area.
(b) In Design 1, the cost is $3 times the perimeter p, so 120 3 p and the perimeter is 40 feet. By part (a), each side is then 40 10 feet long. So the area is 102 100 ft2 . 4
In Design 2, the cost is $4 times the perimeter p. Because the perimeter is 2r, we get 120 4 2r so 2 120 15 225 15 r . The area is r 2 716 ft2 . Design 1 gives the largest area. 8
6. (a) Plan 1: Tomatoes every year. Profit acres Revenue cost 100 1600 300 130,000. Then for n years the profit is P1 130,000n. (b) Plan 2: Soybeans followed by tomatoes. The profit for two years is Profit acres soybean tomato 100 1200 1600 280,000. Remember that no fertilizer is revenue revenue needed in this plan. Then for 2k years, the profit is P2 280,000k.
(c) When n 10, P1 130,000 10 1,300,000. Since 2k 10 when k 5, P2 280,000 5 1,400,000. So Plan B is more profitable.
Making Optimal Decisions
53
7. (a) Data (GB) 1 15 2 25 3 35 4
Plan A
Plan B $25
25 5 200 $35
Plan C $40
40 5 150 $4750
$60 60 5 100 $65
25 10 200 $45
40 10 150 $55
60 10 100 $70
25 20 200 $65
40 20 150 $70
60 20 100 $80
25 15 200 $55 25 25 200 $75 25 30 200 $85
40 15 150 $6250 40 25 150 $7750
40 30 150 $85
60 15 100 $75 60 25 100 $85 60 30 100 $90
(b) For Plan A: CA 25 2 10x 10 20x 5. For Plan B: CB 40 15 10x 10 15x 25. For Plan C: CC 60 1 10x 10 10x 50. Note that these equations are valid only for x 1.
(c) If 22 GB are used, Plan A costs 25 12 2 $49, Plan B costs 40 12 15 $58, and Plan C costs 60 12 1 $72. If 37 GB are used, Plan A costs 25 27 2 $79, Plan B costs 40 27 15 $8050, and Plan C costs 60 27 1 $87. If 49 GB are used, Plan A costs 25 39 2 $103, Plan B costs 40 39 15 $9850, and Plan C costs 60 39 1 $99.
(d) (i) We set CA CB 20x 5 15x 25 5x 20 x 4. Plans A and B cost the same when 4 GB are used.
(ii) We set CA CC 20x 5 10x 50 10x 45 x 45. Plans A and C cost the same when 45 GB are used.
(iii) We set CB CC 15x 25 10x 50 5x 25 x 5. Plans B and C cost the same when 5 GB are used. 8. (a) In this plan, Company A gets $32 million and Company B gets $32 million. Company A’s investment is $14 million, so they make a profit of 32 14 $18 million. Company B’s investment is $26 million, so they make a profit of 32 26 $06 million. So Company A makes three times the profit that Company B does, which is not fair.
(b) The original investment is 14 26 $4 million. So after giving the original investment back, they then share the profit of $24 million. So each gets an additional $12 million. So Company A gets a total of 14 12 $26 million and Company B gets 26 12 $38 million. So even though Company B invests more, they make the same profit as Company A, which is not fair.
(c) The original investment is $4 million, so Company A gets 14 4 64 $224 million and Company B gets 26 64 $416 million. This seems the fairest. 4
Corrections: p. 2 NOTE THAT THESE CORRECTIONS SHOULD BE MADE IN CA-8 ALSO.
CHAPTER 1
EQUATIONS AND GRAPHS
1.1 1.2
The Coordinate Plane 1 Graphs of Equations in Two Variables 8
1.3 1.4 1.5
Circles 19 Lines 24 Solving Quadratic Equations 34
1.6
Complex Numbers 43
1.7
Solving Other Types of Equations 46
1.8
Solving Inequalities 54
1.9
Solving Absolute Value Equations and Inequalities 73
1.10
Solving Equations and Inequalities Graphically 76
1.11
Modeling Variation 86 Chapter 1 Review 90 Chapter 1 Test 109
¥
FOCUS ON MODELING: Fitting Lines to Data 113
1
1
EQUATIONS AND GRAPHS
1.1
THE COORDINATE PLANE
1. (a) The point that is 3 units to the right of the yaxis and 5 units below the xaxis has coordinates 3 5.
(b) The point 2 7 is 2 units to the right of the yaxis and 7 units above the xaxis, so it is closer to the yaxis.
2. If x is positive and y is negative, then the point x y is in Quadrant IV. 3. The distance between the points a b and c d is c a2 d b2 . So the distance between 1 2 and 7 10 is 7 12 10 22 62 82 36 64 100 10. 4. The point midway between a b and c d is 8 12 1 7 2 10 4 6. 2 2 2 2
ac bd . So the point midway between 1 2 and 7 10 is 2 2
5. The points have coordinates A 5 1, B 1 2, C 2 6, D 6 2, E 4 1, F 2 0, G 1 3, and H 2 2. 6. A and B lie in Quadrant I and E and G lie in Quadrant III. y
7.
5l (0, 5)
(_1, 0)
y
8. (_2.5, 3.5)
(21 , 23) 0
x
5
(_5, 0)
(_1, _2)
9. x y x 2
5l
10. x y y 2
y 5l
0
0
(2, 0)
x 5 (2.6, _1.3)
y 5l
5
x
0
5
x
1
2
≤
CHAPTER 1 Equations and Graphs
11. x y y 1
12. x y x 3
y 5l
0
13. x y 2 x 4
5l
5
x
0
14. x y 0 y 2
y 5l
0
15. x y x 1 and y 3
5
x
x
5
x
5
x
y
0
16. x y x 2 and y 1
5l
y 5l
5
x
17. x y 1 x 1 and y 4
0
18. x y 1 x 1 and 2 y 2 y
y
5l
5l
0
5
5l
y
0
y
5
x
0
5
x
SECTION 1.1 The Coordinate Plane
19. x y x y 0
20. x y x y 0
y
y
5l
5l
0
5
x
0
5
x
21. The two points are 0 2 and 3 0. (a) d 3 02 0 22 32 22 9 4 13 30 02 (b) Midpoint: 32 1 2 2 22. The two points are 2 1 and 2 2. (a) d 2 22 1 22 42 32 16 9 25 5 2 2 1 2 0 12 (b) Midpoint: 2 2 23. The two points are 3 3 and 5 3. (a) d 3 52 3 32 82 62 64 36 100 10 3 5 3 3 1 0 (b) Midpoint: 2 2 24. The two points are 2 3 and 4 1. (a) d 2 42 3 12 62 22 36 4 50 2 10 2 4 3 1 1 2 (b) Midpoint: 2 2 25. (a)
26. (a)
y
y (_2, 5)
(6, 16) 1l
(10, 0) 2
(0, 8) 2l 1
(b) d 0 62 8 162 62 82 100 10 0 6 8 16 (c) Midpoint: 3 12 2 2
x
2 102 5 02 122 52 169 13 2 10 5 0 (c) Midpoint: 4 52 2 2
(b) d
x
3
4
CHAPTER 1 Equations and Graphs y
27. (a)
y
28. (a)
5l
(_4,5)
(_1, 1)
0
5
1
(3,_2)
x
(_6, _3)
3 42 2 52 72 72 49 49 98 7 2 4 3 5 2 (c) Midpoint: 12 32 2 2
(b) d
y
29. (a)
1l
x
1 62 1 32 52 42 41 6 1 3 1 (c) Midpoint: 72 1 2 2
(b) d
y
30. (a)
(_6, 2) 1l
1l 1
(6, _2)
(5, 0)
1
x
x
(0, _6)
(b) d 6 62 2 22 122 42 144 16 160 4 10 6 6 2 2 0 0 (c) Midpoint: 2 2
1 52 3 32 42 4. d A C 1 12 3 32 62 6. So
31. d A B
the area is 4 6 24.
y
A
B
(b) d 0 52 6 02 52 62 25 36 61 0 5 6 0 52 3 (c) Midpoint: 2 2 32. The area of a parallelogram is its base times its height. Since two sides are parallel to the xaxis, we use the length of one of these as the base. Thus, the base is d A B 1 52 2 22 42 4. The height is the change in the y coordinates, thus, the height is 6 2 4. So the area of the parallelogram is base height 4 4 16. y
C
1l 1 C
D
x D
1l
A 1
B x
SECTION 1.1 The Coordinate Plane
33. From the graph, the quadrilateral ABC D has a pair of parallel sides, so ABC D is a trapezoid. The area is b1 b2 h. From the graph we see that 2 b1 d A B 1 52 0 02 42 4; b2 d C D 4 22 3 32 22 2; and h is the difference in ycoordinates is 3 0 3. Thus 42 the area of the trapezoid is 3 9. 2
34. The point S must be located at 0 4. To find the area, we find the length of one side and square it. This gives d Q R 5 02 1 62 52 52 25 25 50 2 50 50. So the area is y
y
D 1l A
1
C
R B
5
Q
P
1l 1
x
x
6 02 7 02 62 72 36 49 85. d 0 B 5 02 8 02 52 82 25 64 89.
35. d 0 A
Thus point A 6 7 is closer to the origin. 36. d E C 6 22 3 12 42 22 16 4 20. d E D 3 22 0 12 52 12 25 1 26.
Thus point C is closer to point E. 37. d P R 1 32 1 12 42 22 16 4 20 2 5. d Q R 1 12 1 32 0 42 16 4. Thus point Q 1 3 is closer to point R.
38. (a) The distance from 7 3 to the origin is 7 02 3 02 72 32 49 9 58. The distance from 3 7 to the origin is 3 02 7 02 32 72 9 49 58. So the points are the same distance from the origin.
(b) The distance from a b to the origin is a 02 b 02 a 2 b2 . The distance from b a to the origin is b 02 a 02 b2 a 2 a 2 b2 . So the points are the same distance from the origin.
39. Since we do not know which pair are isosceles, we find the length of all three sides. d A B 3 02 1 22 32 32 9 9 18 3 2. d C B 3 42 1 32 12 42 1 16 17. d A C 0 42 2 32 42 12 16 1 17. So sides AC and C B have the same length.
6
CHAPTER 1 Equations and Graphs
40. Since the side AB is parallel to the xaxis, we use this as the base in the formula area 12 base height. The height is the change in the ycoordinates. Thus, the base is 2 4 6 and the height is 4 1 3. So the area is 12 6 3 9.
41. (a) Here we have A 2 2, B 3 1, and C 3 3. So d A B 3 22 1 22 12 32 1 9 10; d C B 3 32 1 32 62 22 36 4 40 2 10; d A C 3 22 3 22 52 52 25 25 50 5 2.
Since [d A B]2 [d C B]2 [d A C]2 , we conclude that the triangle is a right triangle. (b) The area of the triangle is 12 d C B d A B 12 10 2 10 10.
52 42 25 16 41; 11 62 3 72 16 25 41; d A C 2 62 2 72 42 52 d B C 2 112 2 32 92 12 81 1 82.
42. d A B
Since [d A B]2 [d A C]2 [d B C]2 , we conclude that the triangle is a right triangle. The area is 41 1 41 41 2 . 2
43. We show that all sides are the same length (its a rhombus) and then show that the diagonals are equal. Here we have A 2 9, B 4 6, C 1 0, and D 5 3. So d A B 4 22 6 92 62 32 36 9 45; d B C 1 42 0 62 32 62 9 36 45; d C D 5 12 3 02 62 32 36 9 45; d D A 2 52 9 32 32 62 9 36 45. So the points form a rhombus. Also d A C 1 22 0 92 32 92 9 81 90 3 10, and d B D 5 42 3 62 92 32 81 9 90 3 10. Since the diagonals are equal, the rhombus is a square.
42 82 16 64 80 4 5. 3 12 11 32 22 42 4 16 20 2 5. d B C 5 32 15 112 d A C 5 12 15 32 62 122 36 144 180 6 5. So d A B d B C d A C,
44. d A B
and the points are collinear.
45. Let P 0 y be such a point. Setting the distances equal we get 0 52 y 52 0 12 y 12 25 y 2 10y 25 1 y 2 2y 1 y 2 10y 50 y 2 2y 2 12y 48 y 4. Thus, the point is P 0 4. Check: 0 52 4 52 52 12 25 1 26; 0 12 4 12 12 52 25 1 26.
7
SECTION 1.1 The Coordinate Plane
13 06 2 2
2 3. So the length of the median CC is d C C 18 02 2 2 92 1 . So the length of the median B B 2 8 3 2 37. The midpoint of AC is B 2 2 9 3 2 1 62 109 . The midpoint of BC is A 3 8 6 2 11 4 . So the length is d B B 2 2 2 2 2 11 1 2 4 02 145 . of the median A A is d A A 2 2
46. The midpoint of AB is C
47. As indicated by Example 3, we must find a point S x1 y1 such that the midpoints
y
of P R and of QS are the same. Thus x1 1 y1 1 4 1 2 4 . Setting the xcoordinates equal, 2 2 2 2
x 1 4 1 1 4 1 x1 1 x1 2Setting the we get 2 2 y 1 2 4 1 2 4 y1 1 y1 3. ycoordinates equal, we get 2 2 Thus S 2 3. 48. We solve the equation 6 8
y
C D
A
1
P
x S
2 7 1 7 2 2 41 24 52 3 . of B D is 2 2
(b) The midpoint of AC is
52 3 , the midpoint
(c) Since the they have the same midpoint, we conclude that the
B
1l
1
2x 2x to find the x coordinate of B. This gives 6 12 2 x x 10. Likewise, 2 2
3y 16 3 y y 13. Thus, B 10 13. 2
49. (a)
1l
R
Q
x
diagonals bisect each other.
a b a0 b0 . Thus, 2 2 2 2 2 a 2 b b2 a2 a 2 b2 d C M 0 0 ; 2 2 4 4 2 2 a a 2 b 2 2 b b2 a2 a 2 b2 d A M a 0 ; 2 2 2 2 4 4 2 2 2 b a a 2 b 2 b2 a2 a 2 b2 d B M 0 b . 2 2 2 2 4 4 2
50. We have M
51. (a) The point 5 3 is shifted to 5 3 3 2 8 5. (b) The point a b is shifted to a 3 b 2.
(c) Let x y be the point that is shifted to 3 4. Then x 3 y 2 3 4. Setting the xcoordinates equal, we get x 3 3 x 0. Setting the ycoordinates equal, we get y 2 4 y 2So the point is 0 2.
8
CHAPTER 1 Equations and Graphs
(d) A 5 1, so A 5 3 1 2 2 1; B 3 2, so B 3 3 2 2 0 4; and C 2 1, so C 2 3 1 2 5 3.
52. (a) The point 3 7 is reflected to the point 3 7.
(b) The point a b is reflected to the point a b. (c) Since the point a b is the reflection of a b, the point 4 1 is the reflection of 4 1.
(d) A 3 3, so A 3 3; B 6 1, so B 6 1; and C 1 4, so C 1 4. 53. (a) d A B 32 42 25 5. (b) We want the distances from C 4 2 to D 11 26. The walking distance is 4 11 2 26 7 24 31 blocks. Straightline distance is 4 112 2 262 72 242 625 25 blocks.
(c) The two points are on the same avenue or the same street. 3 27 7 17 15 12, which is at the intersection of 15th Street and 12th Avenue. 54. (a) The midpoint is at 2 2 (b) They each must walk 15 3 12 7 12 5 17 blocks.
55. The midpoint of the line segment is 66 45. The pressure experienced by an ocean diver at a depth of 66 feet is 45 lb/in2 .
56. Let the area of the triangle be A. As hinted, we draw a rectangle
y
5 1 4 units high, so its area is A R 6 4 24. The areas of the three
right triangles inside the rectangle but outside the original triangle are A1 12 7 1 5 2 9, A2 12 4 1 2 1 32 , and
A3 12 7 4 5 1 6. Since the area of the rectangle is the sum of
(7, 5)
(1, 5)
circumscribing the triangle. This rectangle is 7 1 6 units long and
AÁ (1, 2)
A
1
Aª (1, 1)
0
1
(4, 1)
A£ (7, 1) x
the areas of the right triangles and the original triangle, we have A A R A1 A2 A3 24 9 32 6 15 2 .
57. Following the hint, we consider an equilateral triangle with side length 1 anywhere in the plane. Each of its vertices is either red or blue, and each is 1 unit away from the other two. So at least two have the same color and are exactly 1 unit apart.
1.2
GRAPHS OF EQUATIONS IN TWO VARIABLES
1. If the point 2 3 is on the graph of an equation in x and y, then the equation is satisfied when we replace x by 2 and y by 3. ?
?
We check whether 2 3 2 1 6 3. This is false, so the point 2 3 is not on the graph of the equation 2y x 1.
To complete the table, we express y in terms of x: 2y x 1 y 12 x 1 12 x 12 . x
y
2
12
1
0
x y
0
1 2
1
1
2
3 2
2 12 1 0 0 12 1 1 2 32
y 1
0
1
x
2. (a) The xcoordinates of the points where the graph of an equation intersects the xaxis are called xintercepts. Similarly, the ycoordinates of the points where the graph of an equation intersects the xaxis are called yintercepts.
SECTION 1.2 Graphs of Equations in Two Variables
9
(b) To find the xintercept(s) of the graph of an equation we set y equal to 0 in the equation and solve for x: 2 0 x 1 x 1, so the xintercept of 2y x 1 is 1.
(c) To find the yintercept(s) of the graph of an equation we set x equal to 0 in the equation and solve for y: 2y 0 1 y 12 , so the yintercept of 2y x 1 is 12 .
3. (a) If a graph is symmetric with respect to the xaxis and a b is on the graph, then a b is also on the graph. (b) If a graph is symmetric with respect to the yaxis and a b is on the graph, then a b is also on the graph. (c) If a graph is symmetric about the origin and a b is on the graph, then a b is also on the graph. 4. (a) The xintercepts are 5 and 3, and the yintercepts are 2. (b) The graph is symmetric about the xaxis. ?
?
?
?
5. 0 5: 3 0 5 5 0 0 5 5 0. Yes. 2 1: 3 2 1 5 0 6 1 5 0. No. ?
?
2 1: 3 2 1 5 0 6 1 5 0. Yes. So 0 5 and 2 1 lie points on the graph of this equation. ? ? 6. 4 3: 3 1 2 4 3 9. Yes. ? ? 1 0: 0 1 2 1 0 1. No. ? ? 0 1: 1 1 2 0 1 1. Yes. So 4 3 and 0 1 lie on the graph of the equation. ?
?
7. 1 1: 5 1 2 1 7 5 2 7. Yes. ?
?
?
?
2 1: 5 2 2 1 7 10 2 7. No. 1 6: 5 1 2 6 7 5 12 7. Yes. So 1 1 and 1 6 lie on the graph of the equation.
? ? 8. 1 1: 1 12 1 1 1 2 1. No. ? ? 1 12 : 12 12 1 1 12 2 1. Yes. ? ? 1 12 : 12 12 1 1 12 2 1. Yes. So both 1 12 and 1 12 lie on the graph of this equation. ?
?
?
?
?
?
9. 0 2: 02 0 2 22 4 0 0 4 4. Yes. 1 2: 12 1 2 22 4 1 2 4 4. No.
2 2: 22 2 2 22 4 4 4 4 4. Yes. So 0 2 and 2 2 lie on the graph of this equation. ?
?
10. 0 1: 02 12 1 0 0 1 1 0. Yes. 2 2 ? ? 1 1 : 1 1 1 0 12 12 1 0. Yes. 2 2 2 2 2 2 ? ? 3 1 3 12 1 0 34 14 1 0. Yes. 2 2 : 2 So 0 1, 1 1 , and 23 12 all lie on the graph of this equation. 2
2
10
CHAPTER 1 Equations and Graphs
11. y 3x x
y
3
9
12. y 5x
y 5l
6
2 1
3
0
0
1
3
2
0
5
x
y
3
15
2
10
1
5
1
5
2
10
3
15
0
6
3
9
13. y 2x 3 y
3
4
2
1
5l
y
5
1
3
1
1
3
1
0
3
1
5
1
2
7
3
3
9
5
0
5
x
0
15. 4x 5y 40 y 8 45 x y
2
96
0
8
2
64
4
48
6
32
8
16
10
0
12
16
5
x
5
x
y 5l
0
4 5 7 9
16. 2x 3y 12 y 23 x 4
y
y
x
y
3
6
1
47
0
5l
1 3
0
0
0
x
1
x
5l
14. y x 4
y
x
y
x
5
x
6
4 33 2
0
9
2
12
4
5
0
_5
5
x
11
SECTION 1.2 Graphs of Equations in Two Variables
17. y x 2 3
y
x
y
3
6
2
1
1
2
0 1 2
18. y 3 x 2
5l
y
3
6
2
1
1
2
0
3
1
2
1
2
6
3
1
3
0
x
1
2
3
19. y x 2 1 x
y
3
8
2
3
5l
0
0
1
5
x
x
y
3
15
2
5
1
1
0
0
1
2
3
2
y
5
4
3 1
0 1
y 5l
0 0
5
x
y 5l
0
5
x
5
15
x
y
5
4
3
x
1
1
22. y x 1
2
1
3
3
8
x
0
6
1
21. y x 1
5l
20. y 2x 2 3
y
0
1
3
y
x
y 5l
2
1
0
0
1
0
1
2
3
2
3
4
5
4
5
6
0
5
x
12
CHAPTER 1 Equations and Graphs
23. y x 3
24. x y 3 . Since x y 3 is solved for x in terms of y, we
insert values for y and find the corresponding values of x
y
x
y
3
27
in the table below.
5l
x
y
27
3
8
2
1
1
1
1
8
2
27
3
8
2 1
1
0
0
1
1
2
0
1
x
0
8
3
27
25. x y 4 . Since x y 3 is solved for x in terms of y, we
insert values for y and find the corresponding values of x in the table below. x
y
81
3
16
y
2
1
1
1
0
0
0
1
1
16
2
81
3
5
x
2
10
1
8
0 1 2 3 4
6 4 2
0 2
0
0
y
3
80
2
15
1
0
0
1
1
0
2
15
2 0
1
x
80
y 0 0 2x 6 x 3, so the xintercept is 3, and
y
xaxis symmetry: 2x y 6, which is not the same as
1l
x 0 y 2 0 6 6, so the yintercept is 6.
2x y 6, so the graph is not symmetric with respect to the xaxis. yaxis symmetry: 2x y 6, which is not the same as
2x y 6, so the graph is not symmetric with respect to the yaxis.
Origin symmetry: 2x y 6, which is not the same as
2x y 6, so the graph is not symmetric with respect to the origin.
x
5
y
x
27. (a) Solve for y: 2x y 6 y 2x 6. y
5l
26. y 1 x 4
3
x
y
0
1
x
SECTION 1.2 Graphs of Equations in Two Variables
(b) y 2 x 12 x
y
2
18
1
8
0
2
1
0
2
2
3
8
13
y
y 0 0 2 x 12 x 1, so the xintercept is 1,
and x 0 y 2 0 12 2, so the yintercept is 2.
xaxis symmetry: y 2 x 12 , which is not the same as
y 2 x 12 , so the graph is not symmetric with respect to
the xaxis.
yaxis symmetry: y 2 x 12 , which is not the same as
1
y 2 x 12 , so the graph is not symmetric with respect to
1
the yaxis.
x
Origin symmetry: y 2 x 12 y 2 x 12 ,
which is not the same as y 2 x 12 , so the graph is not
symmetric with respect to the origin.
28. (a) Solve for y: x 4y 8 y 14 x 2. x
y
2
52
0 2 4 8
2
32
1
0
12
1
y 0 0 14 x 2 x 8, so the xintercept is 8, and
y
xaxis symmetry: y 2x 6, which is not the same as
1l
x 0 y 14 0 2 2, so the yintercept is 2.
y 2x 6, so the graph is not symmetric with respect to the
xaxis. yaxis symmetry: y 2 x 6 y 2x 6, which is
0
x
1
not the same as y 2x 6, so the graph is not symmetric with respect to the yaxis.
Origin symmetry: y 2 x 6 y 2x 6, which is not the same as y 2x 6, so the graph is not symmetric
with respect to the origin. (b) y x 2 4 x
y
3
5
2
0
1
3
0
4
1
3
2
0
3
5
y 0 0 x 2 4 x 2, so the xintercepts are
y
xaxis symmetry: y x 2 4, which is not the same as
1l
2, and x 0 y 02 4 4, so the yintercept is 4. y x 2 4, so the graph is not symmetric with respect to
the xaxis.
yaxis symmetry: y x2 4 x 2 4, so the graph
is symmetric with respect to the yaxis.
Origin symmetry: y x 2 4, which is not the same as y x 2 4, so the graph is not symmetric with respect to
the origin.
0
1
x
14
CHAPTER 1 Equations and Graphs
29. (a) y
x 2
x
y
0
2
1
3 1 2 1 3
2 3 4
3 6
6
1
9
5
y 0 0 x 2, so there is no xintercept, and x 0 y 0 2 2, so the yintercept is 2. xaxis symmetry: y x 2, which is not the same as y x 2, so the graph is not symmetric with respect to
y
the xaxis.
yaxis symmetry: y x 2, which is not the same as y x 2, so the graph is not symmetric with respect to
the yaxis.
x 2 y x 2, which is not the same as y x 2, so the graph is not Origin symmetry: y
1l 0
x
1
symmetric with respect to the origin. (b) y x x
y
5
5
3
3
1
1
1
1
0 3 5
30. (a) y
0
x 0 y 0 0, so the yintercept is 0.
same as y x, so the graph is not symmetric with
yaxis symmetry: y x x, so the graph is
respect to the origin.
4
0
5 6
1 2
8
2
13
3
5
x
symmetric with respect to the yaxis.
5
y
0
respect to the xaxis.
Origin symmetry: y x y x, which is not the
x
5l
xaxis symmetry: y x y x, which is not the
3
x 4
y
y 0 x 0 x 0, so the xintercept is 0, and
same as y x, so the graph is not symmetric with
y 00
x 4 x 4, so the xintercept is 4, and x 0 y 0 4, so there is no yintercept. xaxis symmetry: y x 4, which is not the same as y x 4, so the graph is not symmetric with respect to
the xaxis.
yaxis symmetry: y x 4, which is not the same as y x 4, so the graph is not symmetric with respect to
the yaxis.
Origin symmetry: y x 4 y x 4, which is not the same as y x 4, so the graph is not
symmetric with respect to the origin.
y
1l
0
1
x
SECTION 1.2 Graphs of Equations in Two Variables
(b) x y. Here y is not a function of x. y
x
3
3 2
2
31. (a) y
y 0 x 0 0, so the xintercept is 0, and x 0
x y, so the graph is symmetric with respect to the xaxis.
0
0
graph is not symmetric with respect to the yaxis.
1
1
2
2
Origin symmetry: x y y, which is not the same,
3
3
y
2
0 3
1
0 1
2 3
2
0
5l
xaxis symmetry: x y y, which is the same as
1
x
y
y 0 y 0, so the yintercept is 0.
1
4 x2
15
yaxis symmetry: x y, which is not the same, so the
0
5
x
so the graph is not symmetric with respect to the origin.
4 x 2 0 x 2, so the xintercepts are 2, and x 0 y 4 02 2, so the yintercept is 2. xaxis symmetry: y 4 x 2 , which is not the same as y 4 x 2 , so the graph is not symmetric with respect to y 0
y
2l
0
the xaxis.
yaxis symmetry: y 4 x2 4 x 2 , so the graph
2
x
is symmetric with respect to the yaxis. Origin symmetry: y 4 x2 4 x 2 y 4 x 2 , which is not the same as y 4 x 2 , so
the graph is not symmetric with respect to the origin. (b) y x 3 4x y
x
3
15
2
0
1
3
0
0
1
3
2
0
3
15
y 0 0 x 3 4x x x 2 4 x x 2 x 2, so
y
the xintercepts are 0 and 2, and x 0 y 03 4 0 0, so the yintercept is 0.
xaxis symmetry: y x 3 4x y x 3 4x, which
is not the same as y x 3 4x, so the graph is not symmetric with respect to the xaxis.
yaxis symmetry: y x3 4 x x 3 4x, which
is not the same as x x 3 4x, so the graph is not symmetric
with respect to the yaxis.
Origin symmetry: y x3 4 x y x 3 4x, so
the graph is symmetric with respect to the origin.
10l 0
1
x
16
CHAPTER 1 Equations and Graphs
32. (a) y 4 x 2 x
y
2
0 3
1
0 1 2
y 0 4 x 2 0 x 2, so the xintercepts are 2, and x 0 y 4 2, so the yintercept is 2. xaxis symmetry: y 4 x 2 y 4 x 2 , which is not the same as y 4 x 2 , so the graph is not
2 3
symmetric with respect to the xaxis. yaxis symmetry: y 4 x2 4 x 2 , so the
0
y
2l
0
2
x
graph is symmetric with respect to the yaxis. Origin symmetry: y 4 x2 y 4 x 2 , which is not the same as y 4 x 2 , so the graph is not symmetric with respect to the origin.
(b) x y 3 y
3
y 0 x 03 0, so the xintercept is 0, and x 0
y
x
8 4
2 34
1
1
0 1
x.
0
4
1 3 4
8
2
y
0 y 3 y 0, so the yintercept is 0.
xaxis symmetry: x y3 y 3 , which is not the same
as x y 3 , so the graph is not symmetric with respect to the xaxis.
1l 0
yaxis symmetry: x y 3 x y 3 , which is not the
10
x
same as x y 3 , so the graph is not symmetric with respect to the yaxis.
Origin symmetry: x y3 x y 3 , so the graph is
symmetric with respect to the origin.
33. (a) To find xintercepts, set y 0. This gives 0 x 6 x 6, so the xintercept is 6. To find yintercepts, set x 0. This gives y 0 6 6, so the yintercept is 6. (b) To find xintercepts, set y 0. This gives 0 x 2 5 x 2 5 x 5, so the xintercepts are 5. To find yintercepts, set x 0. This gives y 02 5 5, so the yintercept is 5.
34. (a) To find xintercepts, set y 0. This gives 4x 2 25 02 100 x 2 25 x 5, so the xintercepts are 5. To find yintercepts, set x 0. This gives 4 02 25y 2 100 y 2 4 y 2, so the yintercepts are 2.
(b) To find xintercepts, set y 0. This gives x 2 x 0 3 0 1 x 2 1 x 1, so the xintercepts are 1. To find yintercepts, set x 0. This gives 02 0y 3y 1 y 13 , so the yintercept is 13 .
35. (a) To find xintercepts, set y 0. This gives 9x 2 4 02 36 9x 2 36 x 2 4 x 2, so the xintercepts are 2. To find yintercepts, set x 0. This gives 9 02 4y 2 36 y 2 9, so there is no yintercept.
(b) To find xintercepts, set y 0. This gives 0 2x 0 4x 1 x 14 , so the xintercept is 14 . To find yintercepts,
set x 0. This gives y 2 0 y 4 0 1, so the yintercept is 1. 36. (a) To find xintercepts, set y 0. This gives 0 x 2 16 x 2 16 x 4, so the xintercepts are 4. To find yintercepts, set x 0. This gives y 02 16, so there is no yintercept. (b) To find xintercepts, set y 0. This gives 0 64 x 3 x 3 64 x 4, so the xintercept is 4. To find yintercepts, set x 0. This gives y 64 03 y 8, so the yintercept is 8.
SECTION 1.2 Graphs of Equations in Two Variables
17
37. To find xintercepts, set y 0. This gives 0 4x x 2 0 x 4 x 0 x or x 4, so the xintercepts are 0 and 4. To find yintercepts, set x 0. This gives y 4 0 02 y 0, so the yintercept is 0.
x 2 02 x2 1 1 x 2 9 x 3, so the xintercepts are 3 and 3. 9 4 9 y2 02 y 2 1 1 y 2 4 x 2, so the yintercepts are 2 and 2. To find yintercepts, set x 0. This gives 9 4 4 39. To find xintercepts, set y 0. This gives x 4 02 x 0 16 x 4 16 x 2. So the xintercepts are 2 and 2.
38. To find xintercepts, set y 0. This gives
To find yintercepts, set x 0. This gives 04 y 2 0 y 16 y 2 16 y 4. So the yintercepts are 4 and 4.
40. To find xintercepts, set y 0. This gives x 2 03 x 2 02 64 x 2 64 x 8. So the xintercepts are 8 and 8. To find yintercepts, set x 0. This gives 02 y 3 02 y 2 64 y 3 64 y 4. So the yintercept is 4.
41. xaxis symmetry: y x 4 x 2 y x 4 x 2 , which is not the same as y x 4 x 2 , so the graph is not symmetric with respect to the xaxis. yaxis symmetry: y x4 x2 x 4 x 2 , so the graph is symmetric with respect to the yaxis.
Origin symmetry: y x4 x2 y x 4 x 2 , which is not the same as y x 4 x 2 , so the graph is not symmetric with respect to the origin. 42. xaxis symmetry: x y4 y2 y 4 y 2 , so the graph is symmetric with respect to the xaxis.
yaxis symmetry: x y 4 y 2 , which is not the same as x y 4 y 2 , so the graph is not symmetric with respect to the yaxis.
Origin symmetry: x y4 y2 x y 4 y 2 , which is not the same as x y 4 y 2 , so the graph is not symmetric with respect to the origin. 43. xaxis symmetry: y x 3 10x y x 3 10x, which is not the same as y x 3 10x, so the graph is not symmetric with respect to the xaxis. yaxis symmetry: y x3 10 x y x 3 10x, which is not the same as y x 3 10x, so the graph is not symmetric with respect to the yaxis. Origin symmetry: y x3 10 x y x 3 10x y x 3 10x, so the graph is symmetric with respect to the origin. 44. xaxis symmetry: y x 2 x y x 2 x, which is not the same as y x 2 x, so the graph is not symmetric with respect to the xaxis. yaxis symmetry: y x2 x y x 2 x, so the graph is symmetric with respect to the yaxis. Note that x x.
Origin symmetry: y x2 x y x 2 x y x 2 x, which is not the same as y x 2 x, so the graph is not symmetric with respect to the origin. 45. xaxis symmetry: x 2 y4 x 4 y2 2 x 2 y 4 x 4 y 2 2, so the graph is symmetric with respect to the xaxis. yaxis symmetry: x2 y 4 x4 y 2 2 x 2 y 4 x 4 y 2 2, so the graph is symmetric with respect to the yaxis.
Origin symmetry: x2 y4 x4 y2 2 x 2 y 4 x 4 y 2 2, so the graph is symmetric with respect to the origin. (Note that if a graph is symmetric with respect to each coordinate axis, it is symmetric with respect to the origin. The converse is not true, as shown in the next exercise.) 46. xaxis symmetry: x 3 y x y3 1 x 3 y x y 3 1 x 3 y x y 3 1 1, so the graph is not symmetric with respect to the xaxis. yaxis symmetry: x3 y x y 3 1 x 3 y x y 3 1 x 3 y x y 3 1 1, so the graph is not symmetric with respect to the yaxis. Origin symmetry: x3 y x y3 1 x 3 y x y 3 1, so the graph is symmetric with respect to the origin.
18
CHAPTER 1 Equations and Graphs
47. Symmetric with respect to the yaxis.
48. Symmetric with respect to the xaxis.
y
y
x
0
0
49. Symmetric with respect to the origin.
x
50. Symmetric with respect to the origin. y
y
x
0
0
51. (a) Symmetric about the xaxis:
(b) Symmetric about the yaxis:
x
(c) Symmetric about the origin:
y
y
y
1
1
1
x
1
52. (a) Symmetric about the xaxis.
x
1
(b) Symmetric about the yaxis.
(c) Symmetric about the origin.
y
y
y
1
1
1
1
x
0
1
x
1
x
1
x
53. (a) From the graph, it appears that I 10 900 lm, I 30 100 lm, and I 40 55 lm. (b) It appears that I x 100 lm for 10 cm x 30 cm.
54. (a) From the graph, it appears that the closest the satellite gets to the center of the moon is approximately 5 Mm and farthest approximately 70 Mm. x 3252 Setting y 0 in the equation (which obviously corresponds to the desired distances), we have 1 1420 x 3252 1420 x 325 1420 377 x 52 or 702. Thus, the smallest and largest distances are 52 Mm and 702 Mm.
SECTION 1.3 Circles
19
102 100 x 3252 x 3252 1 1 x 3252 1420 113 (b) Setting y 10, we have 213 1420 213 1420 213 x 325 75333 x 505 or 5995. The distances of these points from the center of the moon are d1 5052 102 112 Mm and d2 59952 102 608 Mm.
55. True. If a b is on the graph, then by symmetry about the xaxis, the point a b is on the graph. Then by symmetry about the yaxis, the point a b is on the graph. Thus, the graph is symmetric with respect to the origin. 56. False. For example, the graph of y x 3 is symmetric with respect to the origin, but not with respect to either axis.
1.3
CIRCLES
1. The equation of a circle with center C h k and radius r is x h2 y k2 r 2 . So the equation of the circle with
center 1 3 and radius 2 is x 12 y 32 4. The equation of the circle in the figure is [x 1]2 y 22 32 x 12 y 22 9.
9 3. The graph of 2 2 x 5 y 8 16 is a circle with center 5 8 5 8 and radius 16 4. 2 3. To complete the square of the expression x 2 bx, we add b2 14 b2 . To complete the square of x 2 12x, we add 1 122 36 to get x 2 12x 36 x 62 . 4
2. The graph of the equation x 2 y 2 9 is a circle with center 0 0 and radius
4. To determine if the graph of x 2 2x y 2 4y 31 is a circle, we complete the squares of the xterms and the yterms: x 2 2x 1 y 2 4y 4 31 1 4 x 12 y 22 36. So the graph of this equation is a circle with center 1 2 and radius 6.
5. x 2 y 2 9 has center 0 0 and radius 3.
6. x 2 y 2 5 has center 0 0 and radius
y
5.
y
1l
1l 1
1
x
7. x 22 y 2 9 has center 2 0 and radius 3. y
x
8. x 2 y 12 4 has center 0 1 and radius 2. y
1l 1l 0
0 1
x
1
x
20
CHAPTER 1 Equations and Graphs
9. x 32 y 42 25 has center 3 4 and radius 5. 10. x 12 y 22 36 has center 1 2 and radius 6.
y
y
1l 1
x
1l 1
x
11. Using h 3, k 1, and r 2, we get x 32 y 12 22 x 32 y 12 4. 12. Using h 2, k 5, and r 3, we get x 22 y 52 32 x 22 y 52 9.
13. The equation of a circle centered at the origin is x 2 y 2 r 2 . Using the point 4 7 we solve for r 2 . This gives 42 72 r 2 16 49 65 r 2 . Thus, the equation of the circle is x 2 y 2 65.
14. Using h 1 and k 5, we get x 12 y 52 r 2 x 12 y 52 r 2 . Next, using the point 4 6, we solve for r 2 . This gives 4 12 6 52 r 2 130 r 2 . Thus, an equation of the circle is
x 12 y 52 130.
1 5 1 9 2 5. The radius is one half the 15. The center is at the midpoint of the line segment, which is 2 2 diameter, so r 12 1 52 1 92 12 36 64 12 100 5. Thus, the equation of the circle is x 22 y 52 52 or x 22 y 52 25.
1 7 3 5 16. The center is at the midpoint of the line segment, which is 3 1. The radius is one half the 2 2 diameter, so r 12 1 72 3 52 4 2. Thus, an equation of the circle is x 32 y 12 32.
17. Since the circle is tangent to the xaxis, it must contain the point 7 0, so the radius is the change in the ycoordinates. That is, r 3 0 3. So the equation of the circle is x 72 y 32 32 , which is x 72 y 32 9.
18. Since the circle with r 5 lies in the first quadrant and is tangent to both the xaxis and the yaxis, the center of the circle is at 5 5. Therefore, the equation of the circle is x 52 y 52 25.
19. From the figure, the center of the circle is 2 2. The radius is distance from the center to 0 2, so r 2. Thus, an equation of the circle is x 22 y 22 22 x 22 y 22 4.
20. From the figure, the center of the circle is 1 1. The radius is the distance from the center to the point 2 0. Thus, r 1 22 1 02 9 1 10, and an equation of the circle is x 12 y 12 10. 21. From the figure, the center of the circle is 0 12 . The radius is the distance from the center to 0 1, so r 12 . Thus, an 2 2 2 equation of the circle is x 02 y 12 12 x 2 y 12 14 . 22. From the figure, the center of the circle is 12 0 . The radius is the distance from the center to 1 0, so r 12 . Thus, an 2 2 2 equation of the circle is x 12 y 02 12 x 12 y 2 14 . 2 23. To complete the square, we add 82 16: x 2 8x 16 x 42 .
SECTION 1.3 Circles
21
2 24. To complete the square, we add 12 36: y 2 12y 36 x 62 . 2
2 2 94 : y 2 3x 94 y 32 . 25. To complete the square, we add 3 2 26. To complete the square, we add
2 3 2
34 : x 2
2 3x 34 x 23 .
2 2 2 2 27. Completing the square gives x 2 y 2 4x 6y 12 0 x 2 4x 42 y 2 6y 62 12 42 62
x 2 4x 4 y 2 6y 9 12 4 9 x 22 y 32 1. Thus, the circle has center 2 3 and radius 1.
2 2 28. Completing the square gives x 2 y 2 8x 5 0 x 2 8x 82 y 2 5 82 x 2 8x 16 y 2 5 16 x 42 y 2 11. Thus, the circle has center 4 0 and radius 11. 2 2 2 2 y 2 12 y 12 18 12 12 29. Completing the square gives x 2 y 2 12 x 12 y 18 x 2 12 x 12 2 2 2 2 2 2 1 y2 1 y 1 1 1 1 2 1 x 1 x 2 12 x 16 y 14 14 . Thus, the circle has center 2 16 8 16 16 8 4 4 1 1 and radius 1 . 4 4 2 1 0 x 2 1 x 12 2 y 2 2y 2 2 1 12 2 2 2 30. Completing the square gives x 2 y 2 12 x 2y 16 2 2 2 16 2 2 2 x 14 y 12 1. Thus, the circle has center 14 1 and radius 1. 2 2 9. 31. Completing the square gives 2x 2 2y 2 3x 0 x 2 y 2 32 x 0 x 34 y 2 34 16 Thus, the circle has center 34 0 and radius 34 .
2 1 37 . 32. Completing the square gives 3x 2 3y 2 6x y 0 x 2 y 2 2x 13 y 0 x 12 y 16 1 36 36 37 1 Thus, the circle has center 1 6 and radius 6 .
33. x 2 8x y 2 6y 25 0 x 42 y 32 25 16 9 x 42 y 32 25 25 0. The graph is the point 4 3. 2 34. x 2 x y 2 10y 25 0 x 12 y 52 14 . The graph is a circle with center 12 5 and radius 12 .
35. x 2 6x y 2 2y 8 0 x 32 y 12 8 9 1 x 32 y 12 18. The graph is a circle with center 3 1 and radius 3 2. 2 2 36. x 2 x y 2 4y 9 0 x 12 y 22 9 14 4 x 12 y 22 19 4 . The graph is empty.
2 7 2 7 . The graph is empty. 2 37. x 2 8x y 2 7y 30 0 x 42 y 72 30 16 49 4 x 4 y 2 4
38. x 2 12x y 2 4y 40 0 x 62 y 22 40 36 4 x 62 y 22 0. The graph is the point 6 2.
22
CHAPTER 1 Equations and Graphs
39. x 32 y 22 4. Setting y 0, we have x 32 4 4 x 3 0 x 3, so the
xintercept is 3. Setting x 0, we have 9 y 22 4
y 22 5, so there is no yintercept.
40. x 2 6x y 2 6y 9 0 x 32 y 32 9.
Setting y 0, we have x 32 9 9 x 3, so the xintercept is 3. Setting x 0, we have 9 y 32 9
y 3, so the yintercept is 3. y
y
1l
1l 1
1
x
41. x 32 y 12 10. Setting y 0, we have
x 32 9 x 0 or x 6, so the xintercepts are 0
and 6. Setting x 0, we have y 12 1 y 0 or y 2, so the yintercepts are 0 and 2.
x
42. x 2 2x y 2 7y 8 0 2 x 12 y 72 85 4 . Setting y 0 in the original
equation, we have x 2 2x 8 x 2 x 4 0, so the xintercepts are 2 and 4. Setting x 0, we have
y
y 2 7y 8 y 1 y 8 0, so the yintercepts
are 1 and 8.
y
1l 1
x
1l 1
43. x y x 2 y 2 1 . This is the set of points on or inside the circle x 2 y 2 1. y
1l
x
44. x y x 2 y 2 4 . This is the set of points outside the circle x 2 y 2 4.
y
2l
1
x
2
x
SECTION 1.3 Circles
45. Completing the square gives x 2 y 2 4y 12 0 4 2 4 2 12 x 2 y 2 4y 2 2
23
46. x y x 2 y 2 9 and y x . This is the top
x 2 y 22 16. Thus, the center is 0 2, and the
radius is 4. So the circle x 2 y 2 4, with center 0 0
quarter of the interior of the circle of radius 3. Thus, the area is 14 32 94 . y
and radius 2 sits completely inside the larger circle. Thus,
the area is 42 22 16 4 12.
3l 3
47. We find the distance between the transmitter at 10 5 and the person at 50 60: d1
x
50 102 60 52 68 mi.
This is less than 75 mi, so this person would be able to receive the signal. For the person at 50 60, d2 50 102 60 52 885 mi, so this person would not be able to receive the signal.
y
48. By definition, if the epicenter is 30 mi from A 10 50, then it lies on the circle with radius 30 centered at A, with equation
50
x 102 y 502 302 . Similarly, it lies on the circle with radius 50
centered at B 30 10, with equation x 302 y 102 502 , and the circle with radius 40 centered at C 50 20, with equation
A E
B
C 50
x 502 y 202 402 .
Graphing these three circles, we see that the epicenter lies at E 10 20. Check: 10 102 20 502 302 , 10 302 20 102 502 , and 10 502 20 202 402 .
49. (a) x 22 y 12 9 has center 2 1 and radius 3.
x 62 y 42 16 has center 6 4 and radius 4. The distance between centers is 2 62 1 42 42 32 16 9 25 5.
Because 5 3 4, these circles intersect.
(b) x 2 y 22 4 has center 0 2 and radius 2.
x 52 y 142 9 has center 5 14 and radius 3. The distance between centers is 0 52 2 142 52 122 25 144 169 13.
Because 13 2 3, these circles do not intersect.
(c) x 32 y 12 1 has center 3 1 and radius 1.
x 22 y 22 25 has center 2 2 and radius 5. The distance between centers is 3 22 1 22 12 32 1 9 10. Since 10 1 5, these circles intersect.
x
24
CHAPTER 1 Equations and Graphs
50. If the distance d between the centers of the circles is greater than the sum r1 r2 of their radii, then the circles do not intersect, as shown in the first diagram. If d r1 r2 , then the circles intersect at a single point, as shown in the second diagram. If d r1 r2 , then the circles intersect at two points, as shown in the third diagram.
rª CÁ rÁ
Cª
CÁ rÁ
d
rª
Cª
rÁ
d
rª
Cª
CÁ
d
Case 1 d r1 r2
Case 2 d r1 r2
51. Completing the square gives x 2 y 2 ax by c 0 x 2 ax
x
a 2 2
y
b 2
2
c
a 2 b2 4
a 2 2
Case 3 d r1 r2 y 2 by
2 a 2 b 2 b c 2 2 2
.
a 2 b2 a 2 b2 0, this is the equation of a circle; if c 0, it is the equation of a single point, and if 4 4 a 2 b2 0, its graph is empty. c 4 a b a 2 b2 If the equation represents a circle, its center is and its radius is c 12 a 2 b2 4ac. 2 2 4 If c
1.4
LINES
1. We find the “steepness” or slope of a line passing through two points by dividing the difference in the ycoordinates of these 51 2. points by the difference in the xcoordinates. So the line passing through the points 0 1 and 2 5 has slope 20 2. (a) The line with equation y 3x 2 has slope 3. (b) Any line parallel to this line has slope 3.
(c) Any line perpendicular to this line has slope 13 . 3. The pointslope form of the equation of the line with slope 3 passing through the point 1 2 is y 2 3 x 1. 4. For the linear equation 2x 3y 12 0, the xintercept is 6 and the yintercept is 4. The equation in slopeintercept form is y 23 x 4. The slope of the graph of this equation is 23 .
5. The slope of a horizontal line is 0. The equation of the horizontal line passing through 2 3 is y 3. 6. The slope of a vertical line is undefined. The equation of the vertical line passing through 2 3 is x 2. 7. (a) Yes, the graph of y 3 is a horizontal line 3 units below the xaxis.
(b) Yes, the graph of x 3 is a vertical line 3 units to the left of the yaxis.
(c) No, a line perpendicular to a horizontal line is vertical and has undefined slope. (d) Yes, a line perpendicular to a vertical line is horizontal and has slope 0.
SECTION 1.4 Lines y
8.
25
Yes, the graphs of y 3 and x 3 are perpendicular lines.
5l x=_3
0
x
5 y=_3
y y1 1 1 1 0 10. m 2 x2 x1 30 3 3 y2 y1 3 3 2 1 12. m x2 x1 3 5 8 8 y2 y1 2 1 3 1 14. m x2 x1 2 4 6 3 2 2 y2 y1 16. m 0 x2 x1 63 y y1 20 2. 17. For 1 , we find two points, 1 2 and 0 0 that lie on the line. Thus the slope of 1 is m 2 x2 x1 1 0 y y1 32 For 2 , we find two points 0 2 and 2 3. Thus, the slope of 2 is m 2 12 . For 3 we find the points x2 x1 20 y y1 1 2 3. For 4 , we find the points 2 1 and 2 2 and 3 1. Thus, the slope of 3 is m 2 x 2 x1 32 1 1 y y1 2 1 2 2. Thus, the slope of 4 is m 2 . x2 x1 2 2 4 4
2 02 y y1 2 9. m 2 x2 x1 0 1 1 y y1 5 3 2 11. m 2 x2 x1 3 3 6 y2 y1 44 13. m 0 x2 x1 05 y y1 53 2 15. m 2 1 x2 x1 68 2
18. (a)
y
m=2
m=1
(b)
y
m=3
1 m=_ 2
1 m=_ 2
m=0
1 m=_ 3
1l
1l 1
1
x
x 1 m=_ _ 3
m=_1
04 1. Since the yintercept is 4, 40 the equation of the line is y mx b 1x 4. So y x 4, or x y 4 0. 04 20. We find two points on the graph, 0 4 and 2 0. So the slope is m 2. Since the yintercept is 4, the 2 0 equation of the line is y mx b 2x 4, so y 2x 4 2x y 4 0. 0 3 21. We choose the two intercepts as points, 0 3 and 2 0. So the slope is m 32 . Since the yintercept is 3, 20 the equation of the line is y mx b 32 x 3, or 3x 2y 6 0. 19. First we find two points 0 4 and 4 0 that lie on the line. So the slope is m
22. We choose the two intercepts, 0 4 and 3 0. So the slope is m equation of the line is y mx b 43 x 4 4x 3y 12 0.
0 4 43 . Since the yintercept is 4, the 3 0
26
CHAPTER 1 Equations and Graphs
23. Using y mx b, we have y 3x 2 or 3x y 2 0. 24. Using y mx b, we have y 25 x 4 2x 5y 20 0.
25. Using the equation y y1 m x x1 , we get y 1 3 x 4 y 1 3x 12 3x y 11 0.
26. Using the equation y y1 m x x1 , we get y 4 1 x 2 y 4 x 2 x y 2 0.
27. Using the equation y y1 m x x1 , we get y 7 23 x 1 3y 21 2x 2 2x 3y 19 2x 3y 19 0.
28. Using the equation y y1 m x x1 , we get y 5 72 x 3 2y 10 7x 21 7x 2y 31 0.
5 61 y y1 5. Substituting into y y1 m x x1 , we get 29. First we find the slope, which is m 2 x2 x1 12 1 y 6 5 x 1 y 6 5x 5 5x y 11 0. 3 2 y y1 55 1. Substituting into y y1 m x x1 , we get 30. First we find the slope, which is m 2 x2 x1 4 1 y 3 1 x 4 y 3 x 4 x y 1 0. y y1 6 1 5 31. We are given two points, 2 5 and 5 1. Thus, the slope is m 2 2. Substituting into x2 x1 52 3 y y1 m x x1 , we get y 5 2 x 2 y 2x 9 or 2x y 9 0. 77 y y1 0. Substituting into 32. We are given two points, 1 7 and 4 7. Thus, the slope is m 2 x2 x1 41 y y1 m x x1 , we get y 7 0 x 1 y 7 or y 7 0. 3 3 0 y y1 3. Using the yintercept, 33. We are given two points, 1 0 and 0 3. Thus, the slope is m 2 x2 x1 01 1 we have y 3x 3 y 3x 3 or 3x y 3 0. y y1 60 68 34 . Using the yintercept 34. We are given two points, 8 0 and 0 6. Thus, the slope is m 2 x2 x1 0 8 we have y 34 x 6 3x 4y 24 0.
35. Since the equation of a line with slope 0 passing through a b is y b, the equation of this line is y 3.
36. Since the equation of a line with undefined slope passing through a b is x a, the equation of this line is x 3.
37. Since the equation of a line with undefined slope passing through a b is x a, the equation of this line is x 2. 38. Since the equation of a line with slope 0 passing through a b is y b, the equation of this line is y 1.
39. Any line parallel to y 2x 8 has slope 2. The desired line passes through 1 4, so substituting into y y1 m x x1 , we get y 4 2 [x 1] y 2x 6 or 2x y 6 0. 1 2. The desired line passes through 3 2, so substituting 40. Any line perpendicular to y 12 x 7 has slope 12 into y y1 m x x1 , we get y 2 2 [x 3] y 2x 8 or 2x y 8 0. 41. Since the equation of a horizontal line passing through a b is y b, the equation of the horizontal line passing through 4 5 is y 5. 42. Any line parallel to the yaxis has undefined slope and an equation of the form x a. Since the graph of the line passes through the point 4 5, the equation of the line is x 4. 43. Since 3x 2y 4 2y 3x 4 y 32 x 2, the slope of this line is 32 . Thus, the line we seek is given by y 4 32 x 3 y 4 32 x 92 y 32 x 17 2 3x 2y 17 0.
44. Since 2x 3y 4 0 3y 2x 4 y 23 x 43 , the slope of this line is m 23 . Substituting m 23 and b 6 into the slopeintercept formula, the line we seek is given by y 23 x 6 2x 3y 18 0.
45. Any line parallel to x 5 has undefined slope and an equation of the form x a. Thus, an equation of the line is x 1.
46. Any line perpendicular to y 1 has undefined slope and an equation of the form x a. Since the graph of the line passes through the point 2 6, an equation of the line is x 2.
SECTION 1.4 Lines
27
47. First find the slope of 3x 4y 7 0 4y 3x 7 y 34 x 74 . Thus, the slope of any line perpendicular 1 to 3x 4y 7 0 is m 34 43 . Since it passes through 2 1, the equation of the line we seek is
y 1 43 x 2 3y 3 4x 8 4x 3y 11 0.
48. First find the slope of the line 4x 8y 1. This gives 4x 8y 1 8y 4x 1 y 12 x 18 . So the slope of the 1 2. The equation of the line we seek is y 23 2 x 12 line that is perpendicular to 4x 8y 1 is m 12 y 23 2x 1 6x 3y 1 0.
49. First find the slope of the line passing through 2 5 and 2 1. This gives m of the line we seek is y 7 1 x 1 x y 6 0. 50. First find the slope of the line passing through 3 5 and 6 2: m
4 15 1, and so the equation 2 2 4
25 3 1 , and so the slope of the line 6 3 9 3
1 3. Thus an equation of the line we seek is y 10 3 x 4 3x y 22 0. that is perpendicular is m 13
51. (a)
52. (a)
y
(_2, 1)
y
1l 1
x 1l 1
(b) y 1 32 x 2 2y 2 3 x 2
b=_1 b=_3
8
b=_6
4 _4
_2
0
2x y 7 0.
54.
_4
b=6
_8
b=3 b=1
b=0
y 2x b, b 0, 1, 3, 6. They have the same
slope, so they are parallel.
_2
0 _4
4
3
m= 2
4
_4 2
x
(b) y 1 2 x 4 y 1 2x 8
2y 2 3x 6 3x 2y 8 0.
53.
(4, _1)
_8
2
4
m= 34 m= 41
m=0 1 m=_ 4 3
m=_ 4 3
m=_ 2
y mx 3, m 0, 14 , 34 , 32 . Each of the lines
contains the point 0 3 because the point 0 3
satisfies each equation y mx 3. Since 0 3 is on the yaxis, they all have the same yintercept.
28
CHAPTER 1 Equations and Graphs 3
m= 2
55. 4 2 _2
0 _2
m=6 y
56. m= 34
4
6
8
m=0
m=1
10
m= 41 2
m=2 1
m= 2
1
m=_ 4
_12
_8
_4
x m=0 1 4 m=_ 2
0
m=_1
3
m=_ 4
_4
_10 m=_6
3
m=_ 2
y m x 3, m 0, 14 , 34 , 32 . Each of the lines
m=_2
y 2 m x 3, m 0, 12 , 1, 2, 6. Each of the
contains the point 3 0 because the point 3 0 satisfies
lines contains the point 3 2 because the point 3 2
each equation y m x 3. Since 3 0 is on the xaxis,
satisfies each equation y 2 m x 3.
we could also say that they all have the same xintercept.
57. y x 4. So the slope is 1 and the yintercept is 4. y
58. y 12 x 1. So the slope is 12 and the yintercept is 1.
y
1 1
x
1 1
59. 2x y 7 y 2x 7. So the slope is 2 and the yintercept is 7.
y
x
60. 2x 5y 0 5y 2x y 25 x. So the slope is 2 and the yintercept is 0. 5
y
1l 1 1 1
x
x
SECTION 1.4 Lines
61. 4x 5y 10 5y 4x 10 y 45 x 2. So
62. 3x 4y 12 4y 3x 12 y 34 x 3. So
y
y
the slope is 45 and the yintercept is 2.
29
the slope is 34 and the yintercept is 3.
1l
1l 1
1
x
x
63. y 4 can also be expressed as y 0x 4. So the slope is 64. x 5 cannot be expressed in the form y mx b. So 0 and the yintercept is 4.
the slope is undefined, and there is no yintercept. This is a vertical line.
y
y
1l 1
x
1l 1
x
65. x 3 cannot be expressed in the form y mx b. So the 66. y 2 can also be expressed as y 0x 2. So the slope slope is undefined, and there is no yintercept. This is a vertical line.
is 0 and the yintercept is 2. y
y
1l 1
1l 1
x
x
30
CHAPTER 1 Equations and Graphs
67. 3x 2y 6 0. To find xintercepts, we set y 0 and solve for x: 3x 2 0 6 0 3x 6 x 2, so the xintercept is 2. To find yintercepts, we set x 0 and solve for y: 3 0 2y 6 0 2y 6 y 3, so the yintercept is 3. y
68. 6x 7y 42 0. To find xintercepts, we set y 0 and solve for x: 6x 7 0 42 0 6x 42 x 7, so the xintercept is 7. To find yintercepts, we set x 0 and solve for y: 6 0 7y 42 0 7y 42 y 6, so the yintercept is 6. y
1
2 1
2
x
69. 12 x 13 y 1 0. To find xintercepts, we set y 0 and
solve for x: 12 x 13 0 1 0 12 x 1 x 2,
x
70. 13 x 15 y 2 0. To find xintercepts, we set y 0 and solve for x: 13 x 15 0 2 0 13 x 2 x 6, so
so the xintercept is 2. To find yintercepts, we set x 0 and solve for y:
the xintercept is 6. To find yintercepts, we set x 0 and solve for y:
yintercept is 3.
yintercept is 10.
1 0 1 y 1 0 1 y 1 y 3, so the 2 3 3 y
1 0 1 y 2 0 1 y 2 y 10, so the 3 5 5
1
y
2 1
2
x
71. y 6x 4. To find xintercepts, we set y 0 and solve for x: 0 6x 4 6x 4 x 23 , so the xintercept is 23 . To find yintercepts, we set x 0 and solve for y: y 6 0 4 4, so the yintercept is 4. y
x
72. y 4x 10. To find xintercepts, we set y 0 and
solve for x: 0 4x 10 4x 10 x 52 , so the xintercept is 52 .
To find yintercepts, we set x 0 and solve for y: y 4 0 10 10, so the yintercept is 10. y
2 1
1 1
x
x
31
SECTION 1.4 Lines
73. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y 2x 3 has slope 2. The line with equation 2y 4x 5 0 2y 4x 5 y 2x 52 also has slope 2, and so the lines are parallel.
74. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y 12 x 4 has slope 12 .
1 , and so the lines are neither The line with equation 2x 4y 1 4y 2x 1 y 12 x 14 has slope 12 12
parallel nor perpendicular.
75. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 2x 5y 8
5y 2x 8 y 25 x 85 has slope 25 . The line with equation 10x 4y 1 4y 10x 1 y 52 x 14 has 1 , and so the lines are perpendicular. slope 52 25
76. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 15x 9y 2
9y 15x 2 y 53 x 29 has slope 53 . The line with equation 3y 5x 5 3y 5x 5 y 53 x 53 has slope 5 , and so the lines are parallel. 3
77. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 7x 3y 2
3y 7x 2 y 73 x 23 has slope 73 . The line with equation 9y 21x 1 9y 21x 1 y 73 19 has 1 , and so the lines are neither parallel nor perpendicular. slope 73 73
78. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 6y 2x 5
6y 2x 5 y 13 x 56 has slope 13 . The line with equation 2y 6x 1 2y 6x 1 y 3x 12 has 1 , and so the lines are perpendicular. slope 3 13
79. We first plot the points to find the pairs of points that determine each side. Next we find the slopes of opposite sides. The slope of AB is slope of DC is
y
41 3 1 , and the 71 6 2
C
10 7 3 1 . Since these slope are equal, these two sides 5 1 6 2
are parallel. The slope of AD is
D
6 71 3, and the slope of BC is 1 1 2
B
6 10 4 3. Since these slope are equal, these two sides are parallel. 57 2 Hence ABC D is a parallelogram.
1l
A
x
1
80. We first plot the points to determine the perpendicular sides. Next find the slopes of 4 2 3 1 , and the slope of AC is the sides. The slope of AB is 3 3 6 3
y
C
9 3 8 1 . Since 9 3 6 2
slope of AB slope of AC 23 32 1 the sides are perpendicular, and ABC is a right triangle.
B
1l
A
1
x
32
CHAPTER 1 Equations and Graphs
81. We first plot the points to find the pairs of points that determine each side. Next we 2 1 31 and the find the slopes of opposite sides. The slope of AB is 11 1 10 5 2 1 68 . Since these slope are equal, these two sides slope of DC is 0 10 10 5 61 5 are parallel. Slope of AD is 5, and the slope of BC is 01 1 5 38 5. Since these slope are equal, these two sides are parallel. 11 10 1
y
C D
1l
B A
1
x
Since slope of AB slope of AD 15 5 1, the first two sides are each perpendicular to the second two sides. So the sides form a rectangle.
82. (a) The slope of the line passing through 1 1 and 3 9 is
8 91 4. The slope of the line passing through 1 1 31 2
21 1 20 4. Since the slopes are equal, the points are collinear. 61 5 4 73 2. The slope of the line passing through (b) The slope of the line passing through 1 3 and 1 7 is 1 1 2 12 15 3 . Since the slopes are not equal, the points are not collinear. 1 3 and 4 15 is 4 1 5 and 6 21 is
83. We need the slope and the midpoint of the line AB. The midpoint of AB is
17 42 2 2
4 1, and the slope of
2 4 6 1 1 1. The slope of the perpendicular bisector will have slope 1. Using the 71 6 m 1 pointslope form, the equation of the perpendicular bisector is y 1 1 x 4 or x y 3 0. AB is m
84. We find the intercepts (the length of the sides). When x 0, we have 2y 3 0 6 0 2y 6 y 3, and when y 0, we have 2 0 3x 6 0 3x 6 x 2. Thus, the area of the triangle is 12 3 2 3.
85. (a) We start with the two points a 0 and 0 b. The slope of the line that contains them is
b b0 . So the equation 0a a
b of the line containing them is y x b (using the slopeintercept form). Dividing by b (since b 0) gives a y x x y 1 1. b a a b x y (b) Setting a 6 and b 8, we get 1 4x 3y 24 4x 3y 24 0. 6 8
86. (a) The line tangent at 3 4 will be perpendicular to the line passing through the points 0 0 and 3 4. The slope of 4 0 4 1 3 this line is . Thus, the slope of the tangent line will be . Then the equation of the tangent 30 3 4 43 line is y 4 34 x 3 4 y 4 3 x 3 3x 4y 25 0.
(b) Since diametrically opposite points on the circle have parallel tangent lines, the other point is 3 4.
SECTION 1.4 Lines
33
87. Using the diagram provided, we calculate the coordinates of the midpoints: m 1 has coordinates ca 0b 0a 0b 12 a 12 b and m 2 has coordinates 12 a c 12 b . 2 2 2 2
1b 1b 2 0, and so it is parallel to the third side (a segment of the The line joining the midpoints thus has slope 1 2 c 12 a a 2 xaxis). 1 a c 1 a 2 1 b 1 b 2 1 c2 1 c, half the length of the third The line joining the midpoints has length 2 2 2 2 4 2
side of the triangle; and so we have verified both conclusions of the theorem.
88. (a) The slope represents the increase in the average surface temperature in C per year. The T intercept is the average surface temperature in 1950, or 15 C. (b) In 2050, t 2050 1950 100, so T 002100 15 17 degrees Celsius.
89. (a) The slope is 00417D 00417 200 834. It represents the increase in dosage for each oneyear increase in the child’s age. (b) When a 0, c 834 0 1 834 mg. 90. (a)
(b) The slope, 4, represents the decline in number of spaces sold for
y
each $1 increase in rent. The yintercept is the number of spaces at the flea market, 200, and the xintercept is the cost per space when the
200l
manager rents no spaces, $50. 100l
20
91. (a)
40
60
80 100
x
(b) The slope is the cost per toaster oven, $6. The yintercept, $3000, is
y
the monthly fixed cost—the cost that is incurred no matter how many toaster ovens are produced.
10,000l
5000l
500
1000
20
10
x
(b) Substituting a for both F and C, we have
92. (a) C F
30 22
4
14
0
10
20
30
32
50
68
86
a 95 a 32 45 a 32
a 40 . Thus both scales agree at
40 .
t t1 80 70 10 5 93. (a) Using n in place of x and t in place of y, we find that the slope is 2 . So the linear n2 n1 168 120 48 24 5 n 168 t 80 5 n 35 t 5 n 45. equation is t 80 24 24 24
5 150 45 7625 F 76 F. (b) When n 150, the temperature is approximately given by t 24
34
CHAPTER 1 Equations and Graphs
94. (a) Using t in place of x and V in place of y, we find the slope of the line
(b)
y
using the points 0 4000 and 4 200. Thus, the slope is
4000l
3800 200 4000 950. Using the V intercept, the m 40 4 linear equation is V 950t 4000.
3000l 2000l
(c) The slope represents a decrease in the value of the computer of $950 each year. The V intercept is the original price of the computer.
1000l
(d) When t 3, the value of the computer is given by
1
V 950 3 4000 1150.
95. (a) We are given
434 change in pressure 0434. Using P for 10 feet change in depth 10
pressure and d for depth, and using the point P 15 when d 0, we have P 15 0434 d 0 P 0434d 15.
(c) The slope represents the increase in pressure per foot of descent. The yintercept represents the pressure at the surface. (d) When P 100, then 100 0434d 15 0434d 85 d 1959 ft. Thus the pressure is 100 lb/in3 at a depth of
approximately 196 ft.
(b)
2
3
4
x
y 60l 50l 40l 30l 20l 10l 10
20
30
40
50
60 x
96. Slope is the rate of change of one variable per unit change in another variable. So if the slope is positive, then the temperature is rising. Likewise, if the slope is negative then the temperature is decreasing. If the slope is 0, then the temperature is not changing. 97. We label the three points A, B, and C. If the slope of the line segment AB is equal to the slope of the line segment BC, then the points A, B, and C are collinear. Using the distance formula, we find the distance between A and B, between B and C, and between A and C. If the sum of the two smaller distances equals the largest distance, the points A, B, and C are collinear. Another method: Find an equation for the line through A and B. Then check if C satisfies the equation. If so, the points are collinear.
1.5
SOLVING QUADRATIC EQUATIONS
1. (a) The Quadratic Formula states that x
b
b2 4ac . 2a
(b) In the equation 12 x 2 x 4 0, a 12 , b 1, and c 4. So, the solution of the equation is 1 12 4 12 4 13 x 2 or 4. 1 1 2 2
2. (a) To solve the equation x 2 4x 5 0 by factoring, we write x 2 4x 5 x 5 x 1 0 and use the ZeroProduct Property to get x 5 or x 1. 2 (b) To solve by completing the square, we add 5 to both sides to get x 2 4x 5, and then add 42 to both sides to get x 2 4x 4 5 4 x 22 9 x 2 3 x 5 or x 1.
SECTION 1.5 Solving Quadratic Equations
35
(c) To solve using the Quadratic Formula, we substitute a 1, b 4, and c 5, obtaining 4 42 4 1 5 4 36 2 3 x 5 or x 1. x 2 1 2 3. For the quadratic equation ax 2 bx c 0 the discriminant is D b2 4ac. If D 0, the equation has two real solutions; if D 0, the equation has one real solution; and if D 0, the equation has no real solution. 4. There are many possibilities. For example, x 2 1 has two solutions, x 2 0 has one solution, and x 2 1 has no solution.
5. x 2 13x 30 0 x 15 x 2 0 x 15 0 or x 2 0. Thus, x 15 or x 2. 6. x 2 13x 30 0 x 10 x 3 0 x 10 0 or x 3 0. Thus, x 3 or x 10.
7. x 2 x 6 x 2 x 6 0 x 2 x 3 0 x 2 0 or x 3 0. Thus, x 2 or x 3.
8. x 2 4x 21 x 2 4x 21 0 x 3 x 7 0 x 3 0 or x 7 0. Thus, x 3 or x 7.
9. 5x 2 9x 2 0 5x 1 x 2 0 5x 1 0 or x 2 0. Thus, x 15 or x 2.
10. 6x 2 x 12 0 3x 4 2x 3 0 3x 4 0 or 2x 3 0. Thus, x 43 or x 32 .
11. 2s 2 5s 3 2s 2 5s 3 0 2s 1 s 3 0 2s 1 0 or s 3 0. Thus, s 12 or s 3.
12. 4y 2 9y 28 4y 2 9y 28 0 4y 7 y 4 0 4y 7 0 or y 4 0. Thus, y 74 or y 4.
13. 12z 2 44z 45 12z 2 44z 45 0 6z 5 2z 9 0 6z 5 0 or 2z 9 0. Thus, z 56 or z 92 .
14. 42 4 3 42 4 3 0 2 1 2 3 0 2 1 0 or 2 3 0. If 2 1 0, then 12 ; if 2 3 0, then 32 .
15. x 2 5 x 100 x 2 5x 500 x 2 5x 500 0 x 25 x 20 0 x 25 0 or x 20 0. Thus, x 25 or x 20.
16. 6x x 1 21 x 6x 2 6x 21 x 6x 2 5x 21 0 2x 3 3x 7 0 2x 3 0 or 3x 7 0. If 2x 3 0, then x 32 ; if 3x 7 0, then x 73 .
17. x 2 10x 2 0 x 2 10x 2 x 2 10x 25 2 25 x 52 23 x 5 23 x 5 23 18. x 2 6x 2 0 x 2 6x 2 x 2 6x 9 2 9 x 32 11 x 3 11 x 3 11. 19. x 2 6x 11 0 x 2 6x 11 x 2 6x 9 11 9 x 32 20 x 3 2 5 x 3 2 5. 2 3 3 20. x 2 3x 74 0 x 2 3x 74 x 2 3x 94 74 94 x 32 16 4 4 x 2 2 x 2 2 x 12 or x 72 .
2 3 x 2 x 14 34 14 x 12 1 x 12 1 x 12 1. So 4 x 12 1 32 or x 12 1 12 . 25 x 5 2 21 x 5 21 21 22. x 2 5x 1 0 x 2 5x 1 x 2 5x 25 1 4 4 2 4 2 4 2 21. x 2 x 34 0 x 2 x
x 52 221 . 23. x 2 22x 21 0 x 2 22x 21 x 2 22x 112 21 112 21 121 x 112 100 x 11 10 x 11 10. Thus, x 1 or x 21.
24. x 2 18x 19 x 2 18x 92 19 92 19 81 x 92 100 x 9 10 x 9 10, so x 1 or x 19. 25. 5x 2 10x 1 5 x 2 2x 1 5 x 2 2x 1 1 5 x 12 65 x 1 65 x 1 530 26. 2x 2 16x 5 0 x 2 8x 52 0 x 2 8x 52 x 2 8x 16 52 16 x 42 27 2 3 6 x 4 27 2 x 4 2 .
36
CHAPTER 1 Equations and Graphs
49 7 2 17 27. 2x 2 7x 4 0 x 2 72 x 2 0 x 2 72 x 2 x 2 72 x 49 16 2 16 x 4 16 17 7 x 74 17 16 x 4 4 . 25 x 5 2 153 x 5 153 28. 4x 2 5x 8 0 x 2 54 x 2 0 x 2 54 x 2 x 2 54 x 25 2 64 64 8 64 8 64
x 58 3 817 .
29. x 2 8x 12 0 x 2 x 6 0 x 2 or x 6.
30. x 2 3x 18 0 x 3 x 6 0 x 3 or x 6.
31. x 2 8x 20 0 x 10 x 2 0 x 10 or x 2.
32. 10x 2 9x 7 0 5x 7 2x 1 0 x 75 or x 12 .
33. 2x 2 x 3 0 x 1 2x 3 0 x 1 0 or 2x 3 0. If x 1 0, then x 1; if 2x 3 0, then x 32 .
34. 3x 2 7x 4 0 3x 4 x 1 0 3x 4 0 or x 1 0. Thus, x 43 or x 1.
35. 3x 2 6x 5 0 x 2 2x 53 0 x 2 2x 53 x 2 2x 1 53 1 x 12 83 x 1 83
x 1 2 3 6 .
36. x 2 6x 1 0 2 6 62 4 1 1 b b 4ac 6 36 4 6 32 64 2 x 3 2 2. 2a 2 1 2 2 2 37. x 2 43 x 49 0 9x 2 12x 4 0 3x 22 0 x 23 . b b2 4ac 6 62 4 4 1 6 52 3 13 1 2 2 . 38. 2x 3x 2 0 4x 6x 1 0 x 2a 2 4 8 4 39. 4x 2 16x 9 4x 2 16x 9 0 2x 1 2x 9 0 2x 1 0 or 2x 9 0. If 2x 1 0, then x 12 ; if 2x 9 0, then x 92 .
40. 0 x 2 4x 1 0 2 4 42 4 1 1 b b 4ac 4 16 4 4 12 42 3 x 2 3. 2a 2 1 2 2 2 41. 4x x 2 1 x 2 4x 1 0 4 42 4 1 1 b b2 4ac 4 16 4 4 12 42 3 x 2 3 2a 2 1 2 2 2 2 5 52 4 1 3 5 25 12 5 13 b b 4ac . 42. 3 5z z 2 0 z 2a 2 1 2 2 2 4ac 2 22 4 7 4 2 4 112 b b 2 2 , so 43. 7x 2x 4 7x 2x 4 0 x 2a 2 7 4 there is no real solution. 3 9 4 1 5 3 29 b b2 4ac 44. z z 3 5 z 2 3z 5 0 z . 2a 2 2 2 4ac 2 22 4 3 2 2 4 24 2 20 b b 45. 3x 2 2x 2 0 x . Since the 2a 2 3 6 6 discriminant is less than 0, the equation has no real solution.
SECTION 1.5 Solving Quadratic Equations
7 72 4 5 5
b2 4ac 2a 2 5 Since the discriminant is less than 0, the equation has no real solution.
b 46. 5x 2 7x 5 x
7
37
7 51 49 100 . 10 10
47. x 2 0011x 0064 0 0011 00112 4 1 0064 0011 0000121 0256 0011 0506 x . 2 1 2 2 0011 0506 0011 0506 Thus, x 0259 or x 0248. 2 2 48. x 2 2450x 1500 0 2450 24502 4 1 1500 2450 60025 6 2450 00025 2450 0050 x . Thus, 2 1 2 2 2 2450 0050 2450 0050 x 1250 or x 1200. 2 2 49. x 2 2450x 1501 0 2450 24502 4 1 1501 2450 60025 6004 2450 00015 x . 2 1 2 2 Thus, there is no real solution. 50. x 2 1800x 0810 0 1800 18002 4 1 0810 1800 324 324 1800 0 x 0900. Thus the only 2 1 2 2 solution is x 0900. 51. h
1 gt 2 t 1 gt 2 t h 0 0 2 2
0. Using the Quadratic Formula, 0 0 2 4 12 g h 0 02 2gh t . g 2 12 g
n n 1 2S n 2 n n 2 n 2S 0. 2 1 12 4 1 2S 1 1 8S n . 2 1 2
52. S
Using the Quadratic Formula,
53. A 2x 2 4xh 2x 2 4xh A 0. Using the Quadratic Formula, 4h 4 4h 2 2A 4h 4h2 4 2 A 4h 16h 2 8A 4h 2 4h 2 2A x 2 2 4 4 4 2 2h 4h 2 2A 2h 4h 2 2A 4 2 54. A 2r 2 2r h 2r 2 2r h A 0. Using the Quadratic Formula, 2h 2h2 4 2 A 2h 4 2 h 2 8 A h 2 h 2 2 A . r 2 2 4 2
38
CHAPTER 1 Equations and Graphs
1 1 1 c s b c s a s a s b cs bc cs ac s 2 as bs ab sa s b c s 2 a b 2c s ab ac bc 0. Using the Quadratic Formula, a b 2c a b 2c2 4 1 ab ac bc s 2 1 a b 2c a 2 b2 4c2 2ab 4ac 4bc 4ab 4ac 4bc 2 2 2 2 a b 2c a b 4c 2ab 2 2 4 2 1 1 4 2 r 2 1 r r 2 1 r 2 r 1 r 2r 2 4 1 r r r 2 2r 2 4 4r 56. r 1r r 1r r r 5 52 4 1 4 5 25 16 5 41 . r 2 5r 4 0. Using the Quadratic Formula, r 2 1 2 2 55.
57. D b2 4ac 62 4 1 1 32. Since D is positive, this equation has two real solutions.
58. x 2 6x 9 x 2 6x 9, so D b2 4ac 62 4 1 9 36 36 0. Since D 0, this equation has one real solution.
59. D b2 4ac 2202 4 1 121 484 484 0. Since D 0, this equation has one real solution.
60. D b2 4ac 2212 4 1 121 48841 484 00441. Since D 0, this equation has two real solutions. 61. D b2 4ac 52 4 4 13 8 25 26 1. Since D is negative, this equation has no real solution.
62. D b2 4ac r2 4 1 s r 2 4s. Since D is positive, this equation has two real solutions. 1 63. a 2 x 2 2ax 1 0 ax 12 0 ax 1 0. So ax 1 0 then ax 1 x . a 64. ax 2 2a 1 x a 1 0 [ax a 1] x 1 0 ax a 1 0 or x 1 0. If ax a 1 0, a1 ; if x 1 0, then x 1. then x a 65. We want to find the values of k that make the discriminant 0. Thus k 2 4 4 25 0 k 2 400 k 20
66. We want to find the values of k that make the discriminant 0. Thus D 362 4 k k 0 4k 2 362 2k 36 k 18.
67. Let n be one number. Then the other number must be 55 nsince n 55 n 55. Because
the product is 684, we have n 55 n 684 55n n 2 684 n 2 55n 684 0
55 552 41684 5517 72 36 or n 55 30252736 552 289 5517 21 2 2 . So n 2 2 5517 38 n 2 2 19. In either case, the two numbers are 19 and 36. 68. Let n be one even number. Then the next even number is n 2. Thus we get the equation n 2 n 22 1252
n 2 n 2 4n 4 1252 0 2n 2 4n 1248 2 n 2 2n 624 2 n 24 n 26. So n 24 or n 26. Thus the consecutive even integers are 24 and 26 or 26 and 24.
69. Let be the width of the garden in feet. Then the length is 30. Thus 2800 30 2 30 2800 0 70 40 0. If 70 0, then 70, which is impossible. Therefore 40 0, and so 40. The garden is 40 feet wide and 70 feet long. 70. Let be the width of the bedroom. Then its length is 5. Since area is length times width, we have
234 5 2 5 2 5 234 0 18 13 0 18 or 13. Since the width must be positive, the width is 13 feet.
SECTION 1.5 Solving Quadratic Equations
39
71. Let be the width of the garden in feet. We use the perimeter to express the length l of the garden in terms of width. Since the perimeter is twice the width plus twice the length, we have 200 2 2l 2l 200 2 l 100 . Using the formula for area, we have 2400 100 100 2 2 100 2400 0 40 60 0. So 40 0 40, or 60 0 60. If 40, then l 100 40 60. And if 60, then l 100 60 40. So the length is 60 feet and the width is 40 feet.
72. Let be the width of the lot in feet. Then the length is 8. Using the Pythagorean Theorem, we have
2 82 2322 2 2 16 64 53824 22 16 53670 0 2 8 26880 0 168 160 0. Since widths must be positive, the width is 160 feet and the length is 168 feet.
73. First we write a formula for the area of the figure in terms of x. Region A has dimensions 14 in. and x in. and region B has dimensions 13 x in. and x in. So
the area of the figure is 14 x [13 x x] 14x 13x x 2 x 2 27x. We
are given that this is equal to 160 in2 , so 160 x 2 27x x 2 27x 160 0 x 32 x 5 x 32 or x 5. But x must be positive, so x 5 in.
x A
14 in. 13 in. x
B
74. The shaded area is the sum of the area of a rectangle and the area of a triangle. So A y 1 12 y y 12 y 2 y. We
are given that the area is 1200 cm2 , so 1200 12 y 2 y y 2 2y 2400 0 y 50 y 48 0. y is positive, so y 48 cm.
2 2 75. Let h be the height the ladder reaches (in feet). Using the Pythagorean Theorem we have 7 12 h 2 19 12 2 2 2 2 15 225 1296 h 2 39 h 2 39 15 1521 2 4 4 2 4 4 4 324. So h 324 18 feet. 76. Let h be the height of the flagpole, in feet. Then the length of each guy wire is h 5. Since the distance between the points where the wires are fixed to the ground is equal to one guy wire, the triangle is equilateral, and the flagpole is the perpendicular bisector of the base. Thus from the Pythagorean Theorem, we get 1 h 5 2 h 2 h 52 h 2 10h 25 4h 2 4h 2 40h 100 h 2 30h 75 0 2
30 302 4175 3 . Since h 3020 3 0, we reject it. Thus 30 900300 30 2 1200 3020 h 2 2 2 21 3 15 10 3 3232 ft 32 ft 4 in. the height is h 3020 2
77. Let t be the time in hours that it takes you to wash all the windows. Then it takes your roommate t 32 hours to wash all the windows, and the sum of the fractions of the job you can do individually per hour equals the fraction 1 1 1 4 9 9 of the job you can do together. Since 1 hour 48 minutes 1 48 60 1 5 5 , we have t t 3 2
5
2 1 59 9 2t 3 2 9t 5t 2t 3 18t 27 18t 10t 2 15t 10t 2 21t 27 0 t 2t 3 21 212 4 10 27 21 441 1080 21 39 21 39 9 t . So t 2 10 20 20 20 10 21 39 or t 3. Since t 0 is impossible, you can wash the windows alone in 3 hours, and it takes your roommate 20 3 32 4 12 hours.
40
CHAPTER 1 Equations and Graphs
78. Let t be the time, in hours, it takes your manager to deliver all the flyers alone. Then it takes the assistant t 1 1 1 1 1 hours to deliver all the flyers alone, and it takes the group 04t hours to do it together. Thus 4 t t 1 04t 1 1 1 4t 04t 04t 10 t t 1 4 t 1 4t 10 t 1 04t 1 t 4 4 t t 1 t 1 t 2 t 4t 4 4t 10t 10 t 2 t 6 0 t 3 t 2 0. So t 3 or t 2. Since t 2 is impossible, it takes your manager 3 hours to deliver all the flyers alone.
79. Let x be the rate, in mi/h, at which the salesman drove between Ajax and Barrington. Cities
Distance
Rate
Ajax Barrington
120
x
Barrington Collins
150
x 10
Time 120 x 150 x 10
distance to fill in the “Time” column of the table. Since the second part of the trip rate 1 hour) more than the first, we can use the time column to get the equation 120 1 150 took 6 minutes (or 10 x 10 x 10 120 10 x 10 x x 10 150 10x 1200x 12,000 x 2 10x 1500x x 2 290x 12,000 0 We have used the equation time
290 2902 4112,000 290 84,10048,000 290 36,100 290190 145 95. Hence, the salesman x 2 2 2 2
drove either 50 mi/h or 240 mi/h between Ajax and Barrington. (The first choice seems more likely!)
80. Let x be the rate, in mi/h, at which the trucker drove from Tortula to Cactus. Cities
Distance
Rate
Tortula Cactus
250
x
Cactus Dry Junction
360
x 10
Time 250 x 360 x 10
distance to fill in the time column of the table. We are given that the sum of rate 360 250 11 250 x 10 360x the times is 11 hours. Thus we get the equation x x 10 11x x 10 250x 2500 360x 11x 2 110x 11x 2 500x 2500 0 500 5002 4 11 2500 500 250,000 110,000 500 360,000 500 600 . Hence, x 2 11 22 22 22 the trucker drove either 454 mi/h (impossible) or 50 mi/h between Tortula and Cactus. We have used time
SECTION 1.5 Solving Quadratic Equations
41
81. Let r be the rowing rate in km/h of the crew in still water. Then their rate upstream was r 3 km/h, and their rate downstream was r 3 km/h. Distance
Rate
Upstream
6
r 3
Downstream
6
r 3
Time 6 r 3 6 r 3
Since the time to row upstream plus the time to row downstream was 2 hours 40 minutes 83 hour, we get the equation
6 8 6 6 3 r 3 6 3 r 3 8 r 3 r 3 18r 54 18r 54 8r 2 72 r 3 r 3 3 0 8r 2 36r 72 4 2r 2 9r 18 4 2r 3 r 6. Since 2r 3 0 r 32 is impossible, the solution is
r 6 0 r 6. So the rate of the rowing crew in still water is 6 km/h.
82. Let r be the speed of the southbound boat. Then r 3 is the speed of the eastbound boat. In two hours the southbound boat has traveled 2r miles and the eastbound boat has traveled 2 r 3 2r 6 miles. Since they are traveling is directions with are 90 apart, we can use the Pythagorean Theorem to get 2r 2 2r 62 302 4r 2 4r 2 24r 36 900 8r 2 24r 864 0 8 r 2 3r 108 0 8 r 12 r 9 0. So r 12 or r 9. Since speed is positive, the speed of the southbound boat is 9 mi/h.
83. Let x be the length of one side of the cardboard, so we start with a piece of cardboard x by x. When 4 inches are removed from each side, the base of the box is x 8 by x 8. Since the volume is 100 in3 , we get 4 x 82 100
x 2 16x 64 25 x 2 16x 39 0 x 3 x 13 0So x 3 or x 13. But x 3 is not possible, since then the length of the base would be 3 8 5 and all lengths must be positive. Thus x 13, and the piece of cardboard is 13 inches by 13 inches. 84. Let r be the radius of the can. Now using the formula V r 2 h with V 40 cm3 and h 10, we solve for r. Thus 40 r 2 10 4 r 2 r 2. Since r represents radius, r 0. Thus r 2, and the diameter is 4 cm.
85. Using h 0 288, we solve 0 16t 2 288, for t 0. So 0 16t 2 288 16t 2 288 t 2 18 t 18 3 2. Thus it takes 3 2 424 seconds for the ball the hit the ground.
86. (a) Using h 0 96, half the distance is 48, so we solve the equation 48 16t 2 96 48 16t 2 3 t 2 t 3. Since t 0, it takes 3 1732 s. (b) The ball hits the ground when h 0, so we solve the equation 0 16t 2 96 16t 2 96 t 2 6 t 6. Since t 0, it takes 6 2449 s. 87. We are given o 40 ft/s.
(a) Setting h 24, we have 24 16t 2 40t 16t 2 40t 24 0 8 2t 2 5t 3 0 8 2t 3 t 1 0
t 1 or t 1 12 Therefore, the ball reaches 24 feet in 1 second (ascending) and again after 1 12 seconds (descending).
(b) Setting h 48, we have 48 16t 2 40t 16t 2 40t 48 0 2t 2 5t 6 0 5 23 5 25 48 . However, since the discriminant D 0, there are no real solutions, and hence the t 4 4 ball never reaches a height of 48 feet.
(c) The greatest height h is reached only once. So h 16t 2 40t 16t 2 40t h 0 has only one solution. Thus D 402 4 16 h 0 1600 64h 0 h 25. So the greatest height reached by the ball is 25 feet.
(d) Setting h 25, we have 25 16t 2 40t 16t 2 40t 25 0 4t 52 0 t 1 14 . Thus the ball reaches the highest point of its path after 1 14 seconds.
42
CHAPTER 1 Equations and Graphs
(e) Setting h 0 (ground level), we have 0 16t 2 40t 2t 2 5t 0 t 2t 5 0 t 0 (start) or t 2 12 . So the ball hits the ground in 2 12 s.
88. If the maximum height is 100 feet, then the discriminant of the equation 16t 2 0 t 100 0 must equal zero. So
0 b2 4ac 0 2 4 16 100 02 6400 0 80. Since 0 80 does not make sense, we must have 0 80 ft/s.
89. Let x be the distance from the center of the earth to the dead spot (in thousands of miles). Now setting F 0, we have 0012K K 0012K K 2 K 239 x2 0012K x 2 57121 478x x 2 0012x 2 0 2 x x 239 x2 239 x2
0988x 2 478x 57121 0. Using the Quadratic Formula, we obtain 478 4782 4 0988 57,121 478 228,484 225,742192 x 2 0988 1976 478 52362 478 2741808 241903 26499. 1976 1976 So either x 241903 26499 268 or x 241903 26499 215. Since 268,000 is greater than the distance from the earth to the moon, we reject the first root; thus x 215,000 miles.
90. Let y be the circumference of the circle, so 360 y is the perimeter of the square. Use the circumference to find the 2 radius, r, in terms of y: y 2r r y 2. Thus the area of the circle is y 2 y 2 4. Now if the 2 perimeter of the square is 360 y, the length of each side is 14 360 y and the area of the square is 14 360 y . 2 Setting these areas equal, we obtain y 2 4 14 360 y y 2 14 360 y 2y 360 y 2 y 360 . Therefore, y 360 2 1691. Thus one wire is 1691 in. long and the other is 1909 in. long. 91. Let x equal the original length of the reed in cubits. Then x 1 is the piece that fits 60 times along the length of the field, that is, the length is 60 x 1. The width is 30x. Then converting cubits to ninda, we have
2 2 375 60 x 1 30x 12 25 2 x x 1 30 x x x x 30 0 x 6 x 5 0. So x 6 or 12 x 5. Since x must be positive, the original length of the reed is 6 cubits.
92. (a) x 2 9x 20 0 x 4 x 5 0 x 4 or x 5. The product of the solutions is 4 5 20, the constant term in the original equation; and their sum is 4 5 9, the negative of the coefficient of x in the original equation. b b2 4c b b2 4c 2 and r2 . (b) In general, the equation x bx c 0 has solutions r1 2 2 2 1 b b2 4c b b2 4c 14 b2 b2 4c c and b2 4c Thus, r1r2 b2 2 2 4 2b b b2 4c b b2 4c b. r1 r2 2 2 2 b b 2 b 93. (a) We make the substitution x u : x 2 bx c 0 u c 0 b u 2 2 2 b2 b2 b2 bu c 0 u2 c 0. u 2 bu 4 2 4 49 2 (b) Here b 5 and c 6, so the substitution x u 52 results in the equation u 2 25 4 6 0u 4 u 72 x 72 52 6 or x 72 52 1.
SECTION 1.6 Complex Numbers
1.6
43
COMPLEX NUMBERS
1. The imaginary number i has the property that i 2 1.
2. For the complex number 3 4i the real part is 3 and the imaginary part is 4.
3. (a) The complex conjugate of 3 4i is 3 4i 3 4i. (b) 3 4i 3 4i 32 42 25
4. If 3 4i is a solution of a quadratic equation with real coefficients, then 3 4i 3 4i is also a solution of the equation. 5. Yes, every real number a is a complex number of the form a 0i.
6. Yes. For any complex number z, z z a bi a bi a bi a bi 2a, which is a real number. 7. 3 8i: real part 3, imaginary part 8.
8. 5 i 5 i: real part 5, imaginary part 1.
9.
10.
2 5i 23 53 i: real part 23 , imaginary part 53 . 3
4 7i 2 72 i: real part 2, imaginary part 72 . 2
13. 23 i: real part 0, imaginary part 23 . 15. 3 4 3 2i: real part 3, imaginary part 2.
12. 12 : real part 12 , imaginary part 0. 14. 3i: real part 0, imaginary part 3. 16. 2 5 2 i 5: real part 2, imaginary part 5.
17. 3 2i 5i 3 2 5 i 3 7i
18. 3i 2 3i 2 [3 3] i 2 6i
19. 5 3i 4 7i 5 4 3 7 i 1 10i
20. 3 4i2 5i 3 2[4 5] i 59i 22. 3 2i 5 13 i 3 5 2 13 i 2 73 i
11. 3: real part 3, imaginary part 0.
21. 6 6i 9 i 6 9 6 1 i 3 5i 23. 7 12 i 5 32 i 7 5 12 32 i 2 2i
24. 4 i 2 5i 4 i 2 5i 4 2 1 5 i 6 6i
25. 12 8i 7 4i 12 8i 7 4i 12 7 8 4 i 19 4i 26. 6i 4 i 6i 4 i 4 6 1 i 4 7i
27. 4 1 2i 4 8i
28. 2 3 4i 6 8i
29. 3 4i 2 5i 6 15i 8i 20i 2 6 7i 20 1 26 7i
30. 5 i 6 2i 30 10i 6i 2i 2 30 16i 2 1 28 16i
31. 6 5i 2 3i 12 18i 10i 15i 2 12 15 18 10 i 27 8i 32. 2 i 3 7i 6 14i 3i 7i 2 6 7 14 3 i 1 17i
33. 3 2i 3 2i 32 2i2 13
34. 10 i 10 i 102 i 2 101
35. 3 2i2 9 12i 4 1 5 12i
36. 10 i2 100 20i 1 99 20i 1 i i i 1 i 37. 2 i i i 1 i 1 1 1i 1i 1i 1 1 1i 38. 2 2i 1i 1i 1i 11 2 1 i2 1 3i 5 5i 1 3i 1 2i 39. 1 i 1 2i 5 1 2i 1 2i 5 5i 2i 2 i 1 3i 12 12 i 40. 1 3i 10 1 3i 1 3i
44
CHAPTER 1 Equations and Graphs
10i 1 2i 10i 20i 2 5 4 2i 10i 20 10i 4 2i 1 2i 1 2i 1 2i 14 5 1 4i 2 1 2 3i 2 3i 2 3i 1 2 3i 2 3i 13 42. 2 3i1 13 2 2 3i 2 3i 2 3i 4 9 13 4 9i
41.
43.
4 6i 4 6i 3i 12i 18i 2 18 12 18 12i i 2 43 i 3i 3i 3i 9 9 9 9i 2
3 5i 3 5i 15i 45i 75i 2 75 45 75 45i i 13 15 i 15i 15i 15i 225 225 225 225i 2 1 1 1i 1 1i 1i 1 i 1i 1i 1 i 45. 1i 1i 1i 1i 1i 1i 2 2 1 i2 1 i2 44.
3 i 6i 2i 2 5 5i 2 i 10 5i 10i 5i 2 10 5 5 10 i 1 2i 3 i 2i 2i 2i 2i 5 4 i2 15 5i 15 5 5 5i 3 i 5 5 48. i 10 i 2 15 1 47. i 3 i 2 i i 2 49. 3i5 35 i 2 i 243 12 i 243i 50. 2i4 24 i 4 16 1 16 250 250 51. i 1000 i 4 1250 1 52. i 1002 i 4 i 2 1i 2 1 46.
54. 8 4 1 2i 2 1 55. 4 9 2i 3i 6 56. 2 32 16 1 4i 57. 2 1 3 3 6 2 3i 3i 3 6 3 3 2 3 i 58. 3 4 6 8 3 2i 6 2 2i 18 2 6i 2 6i 4 2i 2 3 2 4 2 2 6 2 6 i 2 4 6i 2 1 2i 2 2 2i 2 8 2 59. 1 2 1 2i 1 2i 2 2 2i 6i 36 2i 2i 60. 2i 2 1 2i 2 9 2i 3i 2i 2i
53.
25 25 1 5i
61. x 2 25 0 x 2 25 x 5i
62. 2x 2 5 0 x 2 52 x 210 i 6 36 4 13 3 12 16 3 2i 63. x 2 6x 13 0 x 2 Or: x 2 6x 13 0 x 2 6x 9 13 9 x 32 4 x 3 2i x 3 2i 2 22 4 1 2 2 4 8 2 4 2 2i 2 64. x 2x 2 0 x 1 i 2 1 2 2 2 2 36 2 4 4 2 5 65. 2x 2 2x 5 0 x 12 32 i 2 2 4 12 122 4 8 5 12 16 2 34 14 i 66. 8x 12x 5 0 x 2 8 16 1 12 4 1 1 1 1 4 1 3 1 3i 2 12 23 i 67. x x 1 0 x 2 1 2 2 2
SECTION 1.6 Complex Numbers
68. x 2 3x 3 0 x
3 32 4 1 3 2 1 4 42 4 9 4
3
45
3 3 3 3i 9 12 32 23 i 2 2 2
128 29 4 9 2 i 2 9 18 6 2 20 2 4 4 1 6 70. t 2 0 t 2 2t 6 0 t 1 5i t 2 2 71. 6x 2 12x 7 0 69. 9x 2 4x 4 0 x
4
12 122 467 144168 6i 12 2 6i 1 6 i 12 12 1212 24 122 x 12 12 12 6 26
72. x 2 12 x 1 0 2 1 12 4 1 1 12 14 4 12 15 1 1 15i 2 4 x 2 2 14 415 i 2 1 2 2 2 73. z 3 4i 5 2i 3 4i 5 2i 8 2i 74. z 3 4i 5 2i 8 2i 8 2i
75. z z 3 4i 3 4i 32 42 25
76. z 3 4i 5 2i 15 6i 20i 8i 2 23 14i
77. LHS z a bi c di a bi c di a c b d i a c b d i. RHS z a bi c di a c b d i a c b d i. Since LHS RHS, this proves the statement.
78. LHS z a bi c di ac adi bci bdi 2 ac bd ad bc i ac bd ad bc i.
RHS z a bi c di a bi c di ac adi bci bdi 2 ac bd ad bc i. Since LHS RHS, this proves the statement. 2 79. LHS z2 a bi a bi2 a 2 2abi b2 i 2 a 2 b2 2abi. RHS z 2 a bi2 a 2 2abi b2 i 2 a 2 b2 2abi a 2 b2 2abi. Since LHS RHS, this proves the statement.
80. z a bi a bi a bi z.
81. z z a bi a bi a bi a bi 2a, which is a real number.
82. z z a bi a bi a bi a bi a bi a bi 2bi, which is a pure imaginary number.
83. z z a bi a bi a bi a bi a 2 b2 i 2 a 2 b2 , which is a real number.
84. Suppose z z. Then we have a bi a bi a bi a bi 0 2bi b 0, so z is real. Now if z is real, then z a 0i(where a is real). Since z a 0i, we have z z. b b2 4ac 85. Using the Quadratic Formula, the solutions to the equation are x . Since both solutions are nonreal, we 2a b 4ac b2 have b2 4ac 0 4ac b2 0, so the solutions are x i, where 4ac b2 is a real number. 2a 2a Thus the solutions are complex conjugates of each other. 86. i i, i 5 i 4 i i, i 9 i 8 i i;
i 3 i, i 7 i 4 i 3 i, i 11 i 8 i 3 i;
i 2 1, i 6 i 4 i 2 1, i 10 i 8 i 2 1; i 4 1, i 8 i 4 i 4 1, i 12 i 8 i 4 1.
Because i 4 1, we have i n i r , where r is the remainder when n is divided by 4, that is, n 4 k r , where k is an
integer and 0 r 4. Since 4446 4 1111 2, we must have i 4446 i 2 1.
46
CHAPTER 1 Equations and Graphs
1.7
SOLVING OTHER TYPES OF EQUATIONS
Note: In cases where both sides of an equation are squared, the implication symbol is sometimes used loosely. For example, 2 x x 1 “” x x 12 is valid only for positive x. In these cases, inadmissible solutions are identified later in the
solution.
1. (a) To solve the equation x 3 4x 2 0 we factor the lefthand side: x 2 x 4 0, as above. (b) The solutions of the equation x 2 x 4 0 are x 0 and x 4. 2. (a) Isolating the radical in 2x x 0, we obtain 2x x. 2 2x x2 2x x 2 . (b) Now square both sides:
(c) Solving the resulting quadratic equation, we find 2x x 2 x 2 2x x x 2 0, so the solutions are x 0 and x 2. (d) We substitute these possible solutions into the original equation: 2 0 0 0, so x 0 is a solution, but 2 2 2 4 0, so x 2 is not a solution. The only real solution is x 0.
3. The equation x 12 5 x 1 6 0 is of quadratic type. To solve the equation we set W x 1. The resulting
quadratic equation is W 2 5W 6 0. Solving this equation we get W 3 W 2 0 W 2 or W 3. Thus the solution to the original equation is x 1 2 or x 1 3 x 1 or x 2. You can verify that these are both solutions to the original equation.
4. The equation x 6 7x 3 8 0 is of quadratic type. To solve the equation we set W x 3 . The resulting quadratic equation is W 2 7W 8 0. Solving this equation we get W 2 7W 8 W 8 W 1 W 8 or W 1. Thus the solution to the given equation is x 3 8 2 or x 3 1 1. You can verify that these are both solutions to the original equation.
5. x 2 x 0 x x 1 0 x 0 or x 1 0. Thus, the two real solutions are 0 and 1.
6. 3x 3 6x 2 0 3x 2 x 2 0 x 0 or x 2 0. Thus, the two real solutions are 0 and 2. 7. x 3 25x x 3 25x 0 x x 2 25 0 x x 5 x 5 0 x 0 or x 5 0 or x 5 0. The three real solutions are 5, 0, and 5.
8. x 5 5x 3 x 5 5x 3 0 x 3 x 2 5 0 x 0 or x 2 5 0. The solutions are 0 and 5. 9. x 5 3x 2 0 x 2 x 3 3 0 x 0 or x 3 3 0. The solutions are 0 and 3 3. 10. 6x 5 24x 0 6x x 4 4 0 6x x 2 2 x 2 2 0. Thus, x 0, or x 2 2 0 (which has no solution), or x 2 2 0. The solutions are 0 and 2. 11. 0 4z 5 10z 2 2z 2 2z 3 5 . If 2z 2 0, then z 0. If 2z 3 5 0, then 2z 3 5 z 3 52 . The solutions are 0 and 3 52 . 3 2 2 . The solutions are 0 12. 0 125t 10 2t 7 t 7 125t 3 2 . If t 7 0, then t 0. If 125t 3 2 0, then t 3 125 5 3
and 52 .
13. 0 x 5 8x 2 x 2 x 3 8 x 2 x 2 x 2 2x 4 x 2 0, x 2 0, or x 2 2x 4 0. If x 2 0, then x 0; if x 2 0, then x 2, and x 2 2x 4 0 has no real solution. Thus the solutions are x 0 and x 2.
SECTION 1.7 Solving Other Types of Equations
47
14. 0 x 4 64x x x 3 64 x 0 or x 3 64 0. If x 3 64 0, then x 3 64 x 4. The solutions are 0 and 4.
15. 0 x 3 5x 2 6x x x 2 5x 6 x x 2 x 3 x 0, x 2 0, or x 3 0. Thus x 0, or x 2, or x 3. The solutions are x 0, x 2, and x 3. 16. 0 x 4 x 3 6x 2 x 2 x 2 x 6 x 2 x 3 x 2. Thus either x 2 0, so x 0,or x 3, or x 2. The solutions are 0, 3, and 2.
17. 0 x 4 4x 3 2x 2 x 2 x 2 4x 2 . So either x 2 0 x 0, or using the Quadratic Formula on x 2 4x 2 0, 42 412 8 42 2 2 2. The solutions are 0, 2 2, and 4 2 168 4 we have x 4 21 2 2 2 2. 18. 0 y 5 8y 4 4y 3 y 3 y 2 8y 4 . If y 3 0, then y 0. If y 2 8y 4 0, then using the Quadratic Formula, we 8 82 4 1 4 8 48 have y 4 2 3. Thus, the three solutions are 0, 4 2 3, and 4 2 3. 2 1 2 19. 3x 54 3x 53 0. Let y 3x 5. The equation becomes y 4 y 3 0 y y 3 1 y y 1 y 2 y 1 0. If y 0, then 3x 5 0 x 53 . If y 1 0, then 3x 5 1 0 x 43 . If y 2 y 1 0, then 3x 52 3x 5 1 0 9x 2 33x 31 0. The discriminant is b2 4ac 332 4 9 31 27 0, so this case gives no real solution. The solutions are x 53 and x 43 .
20. x 54 16 x 52 0. Let y x 5. The equation becomes y 4 16y 2 y 2 y 4 y 4 0. If y 2 0, then x 5 0 and x 5. If y 4 0, then x 5 4 0 and x 1. If y 4 0, then x 5 4 0 and x 9. Thus, the solutions are 9, 5, and 1. 21. 0 x 3 5x 2 2x 10 x 2 x 5 2 x 5 x 5 x 2 2 . If x 5 0, then x 5. If x 2 2 0, then x 2 2 x 2. The solutions are 5 and 2. 22. 0 2x 3 x 2 18x 9 x 2 2x 1 9 2x 1 2x 1 x 2 9 2x 1 x 3 x 3. The solutions are 12 , 3, and 3.
23. x 3 x 2 x 1 x 2 1 0 x 3 2x 2 x 2 x 2 x 2 x 2 x 2 x 2 1 . Since x 2 1 0 has no real solution, the only solution comes from x 2 0 x 2.
24. 7x 3 x 1 x 3 3x 2 x 0 6x 3 3x 2 2x 1 3x 2 2x 1 2x 1 2x 1 3x 2 1 2x 1 0 or 3x 2 1 0. If 2x 1 0, then x 12 . If 3x 2 1 0, then 3x 2 1 x 2 13 x 13 . The solutions are 12 and 13 .
x2 50 x 2 50 x 100 50x 5000 x 2 50x 5000 0 x 100 x 50 0 x 100 0 x 100 or x 50 0. Thus x 100 or x 50. The solutions are 100 and 50. 1 2 26. 0 x 2 2 x 1 0 x 2 2x 2 0 x 1 x2 2 22 4 1 2 2 48 2 4 x . Since the radicand is negative, there is no real solution. 2 1 2 2 2 y 1 y 1 y 2 y 2 1 2 y 2 3y 2 2y 2 2 y 2 3y 0 y y 3 0 27. 2 y2 y 1 y 0 or 3. 25.
48
CHAPTER 1 Equations and Graphs
3 3 1 3 1 1 3 32 2 1 2 3 22 1 0 1 1 2 1 0 1 or 12 . 1 1 1 1 54 4 x 1 x 2 4 x 1 x 2 54 29. x 1 x 2 x 1 x 2 28.
4 x 2 4 x 1 5 x 1 x 2 4x 8 4x 4 5x 2 5x 10 5x 2 3x 14 0
5x 7 x 2 0. If 5x 7 0, then x 75 ; if x 2 0, then x 2. The solutions are 75 and 2.
x 5 5 28 2 x 2 x 5 5 x 2 28 x 2 7x 10 5x 10 28 x 2 2x 8 0 x 2 x 2 x 4 x 2 x 4 0 x 2 0 or x 4 0 x 2 or x 4. However, x 2 is inadmissible since we can’t divide by 0 in the original equation, so the only solution is 4. x x2 x 2x x x2 2 5x x 2 2 5x 3x 4 x 2 2 15x 2 20x 0 14x 2 20x 2 5x 31. x 3x 4 3 4x 3 4x 20 202 4 14 2 20 400 112 20 512 20 16 2 5 4 2 . The x 2 14 28 28 28 7 5 4 2 . solutions are 7 3 1 3 1x x 4 4 x 3x 1 2x 2 4x 2x 2 7x 1 0. Using the Quadratic 32. x 2 x x 2 x x 2 4x 2 4x 7 72 4 2 1 7 57 7 57 . Both are admissible, so the solutions are . Formula, we find x 2 2 4 4 2 4x 3 25 4x 3 4x 28 x 7 33. 5 4x 3 52 34. 8x 1 3 8x 1 9 8x 10 x 54 35. 2x 1 3x 5 2x 1 3x 5 x 4 36. 3 x x 2 1 3 x x 2 1 x 2 x 2 0 x 1 x 2 0 x 1 or x 2. 37. 2x 1 1 x 2x 1 x 1 2x 1 x 12 2x 1 x 2 2x 1 0 x 2 4x x x 4. Potential solutions are x 0 and x 4 x 4. These are only potential solutions since squaring is not a reversible ? operation. We must check each potential solution in the original equation. Checking x 0: 2 0 1 1 0 ? ? ? ? 1 1 0 is false. Checking x 4: 2 4 1 1 4, 9 1 4, 3 1 4 is true. Thus, the only solution is x 4. 38. 2x x 1 8 x 1 8 2x x 1 8 2x2 x 1 64 32x 4x 2 30.
0 4x 2 33x 63 4x 21 x 3. Potential solutions are x 21 4 and x 3. Substituting each of these solutions
into the original equation, we see that x 3 is a solution, but x 21 4 is not. Thus 3 is the only solution. 2 x 1 x 2 6x 9 x 1 x 2 7x 10 0 39. x x 1 3 x 3 x 1 x 32 x 2 x 5 0. Potential solutions are x 2 and x 5. We must check each potential solution in the original equation. Checking x 2: 2 2 1 3, which is false, so x 2 is not a solution. Checking x 5: 5 5 1 3 5 2 3, which is true, so x 5 is the only solution. 2 40. 3 x 2 1 x 3 x 1 x 3 x 1 x2 3 x x 2 2x 1 x 2 3x 2 0. Using 3 17 3 32 4 1 2 . Substituting the Quadratic Formula to find the potential solutions, we have x 2 1 2
each of these solutions into the original equation, we see that x 32 17 is a solution, but x 32 17 is not. Thus
x 32 17 is the only solution.
SECTION 1.7 Solving Other Types of Equations
49
41. x 4 4x 2 3 0. Let y x 2 . Then the equation becomes y 2 4y 3 0 y 1 y 3 0, so y 1 or y 3. If y 1, then x 2 1 x 1, and if y 3, then x 2 3 x 3.
42. 2x 4 4x 2 1 0. The LHS is the sum of two nonnegative numbers and a positive number, so 2x 4 4x 2 1 1 0. This equation has no real solution. 43. 0 x 6 26x 3 27 x 3 27 x 3 1 . If x 3 27 0 x 3 27, so x 3. If x 3 1 0 x 3 1, so x 1. The solutions are 3 and 1.
44. x 8 15x 4 16 0 x 8 15x 4 16 x 4 16 x 4 1 . If x 4 16 0, then x 4 16 which is impossible (for real numbers). If x 4 1 0 x 4 1, so x 1. The solutions are 1 and 1.
45. 0 x 52 3 x 5 10 [x 5 5] [x 5 2] x x 7 x 0 or x 7. The solutions are 0 and 7. x 1 x 1 2 x 1 46. Let . Then 0 3 becomes 0 2 4 3 1 3. Now if 1 0, 4 x x x x 1 x 1 x 1 x 1 then 10 1 x 1 x x 12 , and if 3 0, then 3 0 3 x x x x 1 1 1 x 1 3x x 4 . The solutions are 2 and 4 . 2 1 1 1 . Then 8 0 becomes 2 2 8 0 4 2 0. So 4 0 2 47. Let x 1 x 1 x 1 1 4, and 2 0 2. When 4, we have 4 1 4x 4 3 4x x 34 . When x 1 1 2 1 2x 2 3 2x x 32 . Solutions are 34 and 32 . 2, we have x 1 2 x x 4x 48. Let . Then 4 becomes 2 4 4 0 2 4 4 22 . Now if x 2 x 2 x 2 x x 20 2 x 2x 4 x 4. The solution is 4. 2 0, then x 2 x 2 49. Let u x 23 . Then 0 x 43 5x 23 6 becomes u 2 5u 6 0 u 3 u 2 0 u 3 0 or u 2 0. If u 3 0, then x 23 3 0 x 23 3 x 332 3 3. If u 2 0, then x 23 2 0 x 23 2 x 232 2 2. The solutions are 3 3 and 2 2. 50. 4 x 112 5 x 132 x 152 0 x 1 4 5 x 1 x 12 0 x 1 4 5x 5 x 2 2x 1 0 x 1 x 2 3x 0 x 1 x x 3 0 x 1 or x 0 or x 3. The solutions are 1, 0, and 3.
51. Let u x 4; then 0 2 x 473 x 443 x 413 2u 73 u 43 u 13 u 13 2u 1 u 1. So
u x 4 0 x 4, or 2u 1 2 x 4 1 2x 7 0 2x 7 x 72 , or u 1 x 4 1 x 5 0
x 5. The solutions are 4, 72 ,and 5.
52. x 32 10x 12 25x 12 0 x 12 x 2 10x 25 0 x 12 x 52 0. Now x 12 0, so the only solution is x 5.
53. x 12 x 12 6x 32 0 x 32 x 2 x 6 0 x 32 x 2 x 3 0. Now x 12 0, and furthermore
the original equation cannot have a negative solution. Thus, the only solution is x 3. 54. Let u x. Then 0 x 5 x 6 becomes u 2 5u 6 u 3 u 2 0. If u 3 0, then x 3 0 x 3 x 9. If u 2 0, then x 2 0 x 2 x 4. The solutions are 9 and 4. 55. 2x 12 14 x 3 4 2x 12 4 14 x 3 8x 2 x 12 7x 14 x 2
50
CHAPTER 1 Equations and Graphs
56. x 2 7x 12 0 x 4 x 3 0 x 4 0 or x 3 0. Thus, x 4 or x 3. 57.
4a a 4 a a 3a 2 4 a 3a 2 2a 3a 2 a 4 0 a 1 3a 4 0. Thus, a 1 or 3a 2 a 43 .
3x 5 8 3x 5 8 x 5 3x 5 8x 40 5x 45 x 9 x 5 1 1 2 60 3 10 15 1 35 165 x 33 59. 3 7 5 3 2 58.
60. 6x x 1 21 x 6x 2 6x 21 x 6x 2 5x 21 0 2x 3 3x 7 0 2x 3 0 or 3x 7 0.
If 2x 3 0, then x 32 ; if 3x 7 0, then x 73 . 61. x 9 3x 0 x 9 3x x 2 9 3x x 2 3x 9 0 3 32 4 1 9 3 9 36 3 3 5 3 3 5 b b2 4ac . However, is not a x 2a 2 1 2 2 2 3 3 5 3 3 5 93 is the sum of two negative nonzero numbers and hence is nonzero. solution because 2 2 3 3 5 Thus, the only solution is x . 2 x 5 62. 3x 12 3x 6 x 5 (multiply both sides by 3) 2x 1 x 12 3 4 63. Let u x. Then 0 x 3 4 x 4 u 2 3u 4 u 4 u 1. So u 4 4 x 4 0 4 x 4 x 44 256, or u 1 4 x 1 0 4 x 1. However, 4 x is the positive fourth root, so this cannot equal 1. The only solution is 256. 64. x 6 x 3 6 0 x 3 2 x 3 3 0 x 3 2 or x 3 3 x 3 2 or 3 3 65. z 12 8 z 1 15 0 [z 1 5] [z 1 3] 0 z 6 z 4 0 z 4 or 6
66. Let u x 16 . (We choose the exponent 16 because the LCD of 2, 3, and 6 is 6.) Then x 12 3x 13 3x 16 9 x 36 3x 26 3x 16 9 u 3 3u 2 3u 9 0 u 3 3u 2 3u 9 u 2 u 3 3 u 3 u 3 u 2 3 . So u 3 0 or u 2 3 0. If u 3 0, then x 16 3 0 x 16 3 x 36 729. If u 2 3 0, then x 13 3 0 x 13 3 x 33 27. The solutions are 729 and 27.
67. 2t 52 2t 32 8 4t 2 20t 25 4t 2 12t 9 8 32t 8 t 14 2 x x x 6 0. Let , as in Example 6. Then 2 7 6 1 6 0 1 68. 7 x 1 x 1 x 1 x x 6 or 6 either 1, which has no solution; or 6 x . The solution is 65 . x 1 x 1 5 1 4 4 69. 3 2 0 1 4x 4x 2 0 1 2x2 0 1 2x 0 2x 1 x 12 . The solution is 12 . x x x x4 we get, 0 1 4x 2 x 4 . Substituting u x 2 , we get 0 1 4u u 2 , and 4 4 42 411 4 2164 42 12 422 3 2 3. Substituting using the Quadratic Formula, we get u 21 back, we have x 2 2 3, and since 2 3 and 2 3 are both positive we have x 2 3 or x 2 3. Thus the solutions are 2 3, 2 3, 2 3, and 2 3.
70. 0 4x 4 16x 2 4. Multiplying by
SECTION 1.7 Solving Other Types of Equations
71.
x 5 x 5. Squaring both sides, we get
x 5 x 25
51
x 5 25 x. Squaring both sides again, we
get x 5 25 x2 x 5 625 50x x 2 0 x 2 51x 620 x 20 x 31. Potential solutions are x 20 and x 31. We must check each potential solution in the original equation. Checking x 20: 20 5 20 5 25 20 5 5 20 5, which is true, and hence x 20 is a solution. Checking x 31: 36 31 5 37 5, which is false, and hence x 31 is not a 31 5 31 5 solution. The only real solution is x 20.
72.
3 4x 2 4x x 4x 2 4x x 3 0 x 3 4x 2 4x x x 2 4x 4 x x 22 . So x 0 or x 2. The
solutions are 0 and 2.
73. x 2 x 3 x 332 0 x 2 x 3 x 332 0 x 3 x 2 x 3 0 x 3 x 2 x 3 .
If x 312 0, then x 3 0 x 3. If x 2 x 3 0, then using the Quadratic Formula x 12 13 . The
solutions are 3 and 12 13 .
74. Let u
2 2 11 x 2 . By definition of u we require it to be nonnegative. Now 11 x 2 1 u 1. 2 u 11 x
Multiplying both sides by u we obtain u 2 2 u 0 u 2 u 2 u 2 u 1. So u 2 or u 1. But since u must be nonnegative, we only have u 2 11 x 2 2 11 x 2 4 x 2 7 x 7. The solutions are 7. 75. x x 2 2. Squaring both sides, we get x x 2 4 x 2 4 x. Squaring both sides again, we get x 2 4 x2 16 8x x 2 0 x 2 9x 14 0 x 7 x 2. If x 7 0, then x 7. If x 2 0, then x 2. So x 2 is a solution but x 7 is not, since it does not satisfy the original equation.
76.
1 x 2x 1 5 x. We square both sides to get 1 x 2x 1 5 x 2 x 2x 1 4 x 16 8 x x 2x 1 16 8 x. Again, squaring both sides, we obtain 2 2x 1 16 8 x 256 256 x 64x 62x 255 256 x. We could continue squaring both sides until we found possible solutions; however, consider the last equation. Since we are working with real numbers, for x to be defined, we must have x 0. Then 62x 255 0 while 256 x 0, so there is no solution.
77. 0 x 4 5ax 2 4a 2 a x 2 4a x 2 . Since a is positive, a x 2 0 x 2 a x a. Again, since a is positive, 4a x 2 0 x 2 4a x 2 a. Thus the four solutions are a and 2 a. b 78. 0 a 3 x 3 b3 ax b a 2 x 2 abx b2 . So ax b 0 ax b x or a ab ab2 4 a 2 b2 ab 3a 2 b2 b x , but this gives no real solution. Thus, the solution is x . 2 2 a 2 a 2a 79.
x a x a 2 x 6. Squaring both sides, we have x a x a 2 x 6 2x 2 x a x a 2x 12 2 x a x a 12 x a 2 x a x a x a 6. Squaring both sides again we have x a x a 36 x 2 a 2 36 x 2 a 2 36 x a 2 36. Checking these answers, we see that x a 2 36 is not a solution (for example, try substituting a 8), but x a 2 36 is a solution.
52
CHAPTER 1 Equations and Graphs
80. Let x 16 . Then x 13 2 and x 12 3 , and so
0 3 a2 b ab 2 a b a 2 b a 3 x b 6 x a . So 6 x a 0 a 6 x x a 6 is one solution. Setting the first factor equal to zero, we have 3 x b 0 3 x b x b3 . However, the original equation includes the term b 6 x, and we cannot take the sixth root of a negative number, so this is not a solution. The only solution is x a 6 .
81. The volume is 180 ft3 , so x x 4 x 9 180 x 3 5x 2 36x 180 x 3 5x 2 36x 180 0 x 2 x 5 36 x 5 0 x 5 x 2 36 0 x 5 x 6 x 6 0 x 6 is the only positive solution. So the box is 2 feet by 6 feet by 15 feet.
82. Let r be the radius of the larger sphere, in millimeters. Equating the volumes, we have 43 r 3 43 23 33 43 r 3 23 33 44 r 3 99 r 3 99 463. Therefore, the radius of the larger sphere is about 463 mm. 83. Let r be the radius of the tank, in feet. The volume of the spherical tank is 43 r 3 and is also 750 01337 100275. So 4 r 3 100275 r 3 23938 r 288 feet. 3
84. Let x be the length of the hypotenuse of the triangle, in feet. Then one of the other sides has length x 7 feet, and since the perimeter is 392 feet, the remaining side
must have length 392 x x 7 399 2x. From the Pythagorean Theorem,
x-7
x
we get x 72 399 2x2 x 2 4x 2 1610x 159250 0. Using the
Quadratic Formula, we get 2 44159250 x 1610 161024 16108 44100 1610210 , and so x 2275 or x 175. But if x 2275, then the 8
side of length x 7 combined with the hypotenuse already exceeds the perimeter of 392 feet, and so we must have x 175. Thus the other sides have length 175 7 168 and 399 2 175 49. The lot has sides of length 49 feet, 168 feet, and 175 feet.
900 . After 5 people 85. Let x be the number of people originally intended to take the trip. Then originally, the cost of the trip is x 900 900 4500 cancel, there are now x 5 people, each paying 2. Thus 900 x 5 2 900 900 2x 10 x x x 4500 0 2x 10 0 2x 2 10x 4500 2x 100 x 45. Thus either 2x 100 0, so x 50, or x x 45 0, x 45. Since the number of people on the trip must be positive, originally 50 people intended to take the trip. 120,000 . If one person joins the group, then there would n 120,000 120,000 be n 1 members in the group, and each person would pay 6000. So n 1 6000 120,000 n n n n 120,000 6000 n 1 120,000 20 n n 1 20n n 2 19n 20 20n 6000 n 6000
86. Let n be the number of people in the group, so each person now pays
0 n 2 n 20 n 4 n 5. Thus n 4 or n 5. Since n must be positive, there are now 4 friends in the group.
87. Let x be the height of the pile in feet. Then the diameter is 3x and the radius is 32 x feet. Since the volume of the cone is 4000 3x 2 3x 3 4000 3 3 1000 ft , we have x 1000 1000 x x 3 752 feet. 3 2 4 3 3
SECTION 1.7 Solving Other Types of Equations
53
88. Let h be the height of the screens in inches. The width of the smaller screen is h 7 inches, and the width of the bigger screen is 18h inches. The diagonal measure of the smaller screen is h 2 h 72 , and the diagonal measure of the larger screen is h 2 18h2 424h 2 206h. Thus h 2 h 72 3 206h h 2 h 72 206h 3. Squaring both sides gives h 2 h 2 14h 49 424h 2 1236h 9 0 224h 2 2636h 40. Applying
2636 26362 422440 10532496 2636 3245 . So the Quadratic Formula, we obtain h 2636 448 2224 448
h
2636 3245 1313. Thus, the screens are approximately 131 inches high. 448
89. Let x be the length, in miles, of the abandoned road to be used. Then the length of the abandoned road not used is 40 x, and the length of the new road is 102 40 x2 miles, by the Pythagorean Theorem. Since the cost of the road is cost per mile number of miles, we have 100,000x 200,000 x 2 80x 1700 6,800,000 2 x 2 80x 1700 68 xSquaring both sides, we get 4x 2 320x 6800 4624 136x x 2
1 18488 x 136 3x 2 184x 2176 0 x 184 3385626112 6 6 3 or x 16. Since 45 3 is longer than the existing road, 16 miles of the abandoned road should be used. A completely new road would have length 102 402 (let x 0) and would cost 1700 200,000 83 million dollars. So no, it would not be cheaper.
90. Let x be the distance, in feet, that you go on the boardwalk before veering off onto the sand. The distance along the boardwalk from where you started to the point on the boardwalk closest to the umbrella is 7502 2102 720 ft. Thus the distance you walk on the sand is 720 x2 2102 518,400 1440x x 2 44,100 x 2 1440x 562,500. Along boardwalk Across sand
Distance
Rate
x
4
x 2 1440x 562,500
2
Time x 4
x 2 1440x 562,500 2
Since 4 minutes 45 seconds 285 seconds, we equate the time it takes to walk along the boardwalk and across the sand x x 2 1440x 562,500 1140 x 2 x 2 1440x 562,500. Squaring both to the total time to get 285 4 2 sides, we get 1140 x2 4 x 2 1440x 562,500 1,299,600 2280x x 2 4x 2 5760x 2,250,000 0 3x 2 3480x 950,400 3 x 2 1160x 316,800 3 x 720 x 440. So x 720 0 x 720, and x 440 0 x 440. Checking x 720, the distance across the sand is
210 210 feet. So 720 4 2 180 105 285 seconds. Checking x 440, the distance across the sand is 350 720 4402 2102 350 feet. So 440 4 2 110 175 285 seconds. Since both solutions are less than or equal
to 720 feet, we have two solutions: Either you walk 440 feet down the boardwalk and then head towards the umbrella, or you walk 720 feet down the boardwalk and then head towards the umbrella.
91. Let d be the distance from the lens to the object. Then the distance from the lens to the image is d 4. So substituting 1 1 1 . Now we multiply by the F 48, x d, and y d 4, and then solving for x, we have 48 d d 4 LCD, 48d d 4, to get d d 4 48 d 4 48d d 2 4d 96d 192 0 d 2 136d 192 136 104 . So d 16 or d 12. Since d 4 must also be positive, the object is 12 cm from the lens. d 2
54
CHAPTER 1 Equations and Graphs
d d 1 2 1 2 1 3 0 . Letting d, we have 3 14 1090 92. Since the total time is 3 s, we have 3 1090 4 4 1090 545 591054 . Since 0, we have d 1151, so d 13256. The well 22 545 6540 0 4 is 1326 ft deep. 93. (a) Method 1: Let u x, so u 2 x. Thus x x 2 0 becomes u 2 u 2 0 u 2 u 1 0. So u 2 or u 1. If u 2, then x 2 x 4. If u 1, then x 1 x 1. So the possible solutions are 4 and 1. Checking x 4 we have 4 4 2 4 2 2 0. Checking x 1 we have 1 1 2 1 1 2 0. The only solution is 4. Method 2: x x 2 0 x 2 x x 2 4x 4 x x 2 5x 4 0 x 4 x 1 0. So the possible solutions are 4 and 1. Checking will result in the same solution. 1 1 12 10 (b) Method 1: Let u , so u 2 1 0 becomes 12u 2 10u 1 0. Using . Thus 2 2 x 3 x 3 3 3 x x 102 4121 52 102 13 5 13 . If u 5 13 , then the quadratic formula, we have u 10 212 10 24 24 12 12 12 5 13 5 13 1 12 513 x 3 5 13. So x 2 13. 12 5 13 5 13 x 3 12 12 5 13 5 13 1 5 13 5 13 12 x 3 , then 5 13. So If u 12 12 5 13 5 13 x 3 12 x 2 13. The solutions are 2 13. 10 12 1 0 x 32 Method 2: Multiplying by the LCD, x 32 , we get x 32 x 3 x 32 12 10 x 3 x 32 0 12 10x 30 x 2 6x 9 0 x 2 4x 9 0. Using the Quadratic 2 4 52 42 13 2 13. The solutions are 2 13. Formula, we have u 4 4 419 2 2 22
1.8
SOLVING INEQUALITIES
1. (a) If x 5, then x 3 5 3 x 3 2. (b) If x 5, then 3 x 3 5 3x 15.
(c) If x 2, then 3 x 3 2 3x 6.
(d) If x 2, then x 2.
2. To solve the nonlinear inequality
x 1 0 we x 2
first observe that the numbers 1 and 2 are zeros of the numerator and denominator. These numbers divide the real line into the three intervals 1, 1 2, and 2 . The endpoint 1 satisfies the inequality, because
Sign of x 1 Sign of x 2
Sign of x 1 x 2
1
1 2
2
21 1 1 0 0, but 2 fails to satisfy the inequality because is not 1 2 22
defined. Thus, referring to the table, we see that the solution of the inequality is [1 2.
3. (a) No. For example, if x 2, then x x 1 2 1 2 0, but x 0. (b) No. For example, if x 2, then x x 1 2 3 6 5, but x 5.
SECTION 1.8 Solving Inequalities
4. (a) To solve 3x 7, start by dividing both sides of the inequality by 3.
(b) To solve 5x 2 1, start by adding 2 to both sides of the inequality.
5. x
2 3x 13
6. x
5 17 13 ; no 1 5 13 ; no 0 2 3 5 6
1 5 3
1 2x 5x
11 25; yes
5
2 0; no
1
3 5; yes
2 3 5 6
13 10 3 ; no 23 25 6 ; no
0
0 13 ; no 1 1 ; yes 2 3 1 13 ; yes 47 13 ; yes 7 13 ; yes 13 13 ; yes
1 0; yes
1 5
1 5; no
347 1118; no
3
5 The elements 56 , 1, 5, 3, and 5 satisfy the inequality.
5 15; no
5
9 25; no
The elements 5, 1, and 0 satisfy the inequality. 8.
7. x 5
1 2x 4 7
1 14 7; no
1
1 6 7; no
2 3 5 6
1 83 7; no 1 73 7; no
0
1 5 3 5
x
x
1 4 7; no
5 1 0 2 3 5 6
1 5 3 5
1
2 4 2; no
2 3 5 6
2 73 2; no
0
1 2 7; no
1 5
1 2 7; yes
3
1 047 7; no
1 6 7; yes
1 1 x 2 15 12 ; yes 1 12 ; yes
1 is undefined; no 0 3 1 ; no 2 2 6 1 ; no 5 2 1 12 ; no 045 12 ; yes 1 1 ; yes 3 2 1 1 ; yes 5 2
The elements 5, 1, 5, 3, and 5 satisfy the inequality.
2 8 2; no
5
2 3 2; no
2 13 6 2; no 2 2 2; no
2 076 2; yes
2 0 2; yes
5 2 2 2; yes The elements 5, 3, and 5 satisfy the inequality.
The elements 3 and 5 satisfy the inequality. 9.
2 3 x 2
10. x 5
x2 2 4 27 4; no
1
3 4; yes
0
2 4; yes
2 3 5 6
22 4; yes 9 97 4; yes 36
1 5
3 4; yes
3
11 4; no
5
27 4; no
7 4; no
The elements 1, 0, 23 , 56 , and 1 satisfy the inequality.
55
56
CHAPTER 1 Equations and Graphs
11. 2x 7 x 72 . Interval: 72 Graph:
12. 4x 10 x 52 . Interval: 52 Graph:
7 2
13. 2x 5 3 2x 8 x 4 Interval: 4 . Graph:
14. 3x 11 5 3x 6 x 2 Interval: 2. Graph:
4
15. 7 x 5 x 2 x 2
Interval: [7 . Graph:
2
17. 2x 1 0 2x 1 x 12 1
__ 2
12 __ 11
4 1 22. 25 x 1 15 2x 12 5 x 5 x 3
Interval: 13 . Graph:
3
23. 13 x 2 16 x 1 16 x 3 x 18 18. Graph:
1
__ 3
24. 23 12 x 16 x (multiply both sides by 6)
Interval:
4 3x 1 6x 3 9x 13 x Interval: 13 . Graph:
_18
25. 4 3x 1 8x 4 3x 1 8x 5x 5 x 1
Interval: 1]. Graph:
_1
27. 2 x 5 4 3 x 1
1 _ 3
26. 2 7x 3 12x 16 14x 6 12x 16 2x 22 x 11
Interval: 11]. Graph:
11
28. 8 x 3 12 5 x 15 _3
_1
29. 1 2x 5 7 4 2x 12 2 x 6 2
2
Interval: 12 11 . Graph:
_3
21. x 32 12 x 12 x 32 x 3
Interval: 2 6. Graph:
Interval: 2 . Graph:
20. 5 3x 8x 7 11x 12 x 12 11
19. 4x 7 8 9x 5x 15 x 3.
Interval: [3 1. Graph:
7
18. 0 4x 8 0 x 2 x 2
Interval: 12 . Graph:
Interval: 3 . Graph:
_2
16. 5 3x 16 3x 21 x 7
Interval: 2]. Graph:
Interval: 3 . Graph:
_ 52
6
Interval: [5 15]. Graph:
_5
15
30. 1 3x 4 16 3 3x 12 1 x 4 Interval: 1 4]. Graph:
_1
4
57
SECTION 1.8 Solving Inequalities
31. 2 8 2x 1 10 2x 9 5 x 92 92 x 5
Interval: 92 5 . Graph:
33.
9 _ 2
32. 3 3x 7 12 10 3x 13 2 13 10 3 x 6
13 Interval: 10 3 6 . Graph:
5
2x 13 2 1 2 2x 13 8 (multiply each 6 12 3
21 Interval: 15 2 2 . Graph:
15 __ 2
13
_ _6_
4 3x 1 1 (multiply each expression by 20) 2 5 4 10 4 4 3x 5 10 16 12x 5
34.
21 expression by 12) 15 2x 21 15 2 x 2.
10
_ _3_
11 11 13 26 12x 11 13 6 x 12 12 x 6 .
21 __ 2
13 . Graph: Interval: 11 12 6
11 __ 12
13 __ 6
35. x 2 x 3 0. The expression on the left of the inequality changes sign where x 2 and where x 3. Thus we must check the intervals in the following table. Interval Sign of x 2
2
2 3
3
Sign of x 3
Sign of x 2 x 3
From the table, the solution set is x 2 x 3. Interval: 2 3. Graph:
_2
3
36. x 5 x 4 0. The expression on the left of the inequality changes sign when x 5 and x 4. Thus we must check the intervals in the following table. Interval Sign of x 5
4
4 5
5
Sign of x 4
Sign of x 5 x 4
From the table, the solution set is x x 4 or 5 x. Interval: 4] [5 . Graph:
_4
5
37. x 2x 7 0. The expression on the left of the inequality changes sign where x 0 and where x 72 . Thus we must check the intervals in the following table. Interval Sign of x Sign of 2x 7
Sign of x 2x 7
72
72 0
0
From the table, the solution set is x x 72 or 0 x . Interval: 72 [0 . Graph:
7
__ 2
0
58
CHAPTER 1 Equations and Graphs
38. x 2 3x 0. The expression on the left of the inequality changes sign when x 0 and x 23 . Thus we must check the intervals in the following table. Interval
0
Sign of x
Sign of 2 3x
Sign of x 2 3x
0 23
From the table, the solution set is x x 0 or 23 x . Interval: 0] 23 .
2 3
Graph:
0
2 _ 3
39. x 2 3x 18 0 x 3 x 6 0. The expression on the left of the inequality changes sign where x 6 and where x 3. Thus we must check the intervals in the following table. Interval Sign of x 3 Sign of x 6
Sign of x 3 x 6
3
3 6
6
From the table, the solution set is x 3 x 6. Interval: [3 6]. Graph:
_3
6
40. x 2 8x 7 0 x 7 x 1 0. The expression on the left of the inequality changes sign where x 7 and where x 1. Thus we must check the intervals in the following table. Interval Sign of x 7 Sign of x 1
Sign of x 7 x 1
1
1 7
7
From the table, the solution set is x x 1 or x 7. Interval: 1 7 . Graph:
1
7
41. 3x 2 5x 2 3x 2 5x 2 0 x 2 3x 1 0. The expression on the left of the inequality changes sign where x 2 and where x 13 . Thus we must check the intervals in the following table. Interval Sign of x 2
Sign of 3x 1
Sign of x 2 3x 1
2
2 13
1 3
From the table, the solution set is x x 2 or x 13 . Interval: 2] 13 . Graph:
_2
1 _ 3
42. x 2 x 2 x 2 x 2 0 x 1 x 2 0. The expression on the left of the inequality changes sign when x 1 and x 2. Thus we must check the intervals in the following table. Interval Sign of x 1 Sign of x 2
Sign of x 1 x 2
1
1 2
2
From the table, the solution set is x 1 x 2. Interval: 1 2. Graph:
_1
2
SECTION 1.8 Solving Inequalities
59
43. 3x 2 3x 2x 2 4 x 2 3x 4 0 x 1 x 4 0. The expression on the left of the inequality changes sign where x 1 and where x 4. Thus we must check the intervals in the following table. Interval Sign of x 1 Sign of x 4
Sign of x 1 x 4
1
1 4
4
From the table, the solution set is x 1 x 4. Interval: 1 4. Graph:
_1
4
44. 5x 2 3x 3x 2 2 2x 2 3x 2 0 2x 1 x 2 0. The expression on the left of the inequality changes sign when x 12 and x 2. Thus we must check the intervals in the following table. Interval Sign of 2x 1 Sign of x 2
Sign of 2x 1 x 2
2
2 12
1 2
From the table, the solution set is x x 2 or 12 x . Interval: 2] 12 . Graph:
_2
1 _ 2
45. x 2 3 x 6 x 2 3x 18 0 x 3 x 6 0. The expression on the left of the inequality changes sign where x 6 and where x 3. Thus we must check the intervals in the following table. Interval Sign of x 3 Sign of x 6
Sign of x 3 x 6
3
3 6
6
From the table, the solution set is x x 3 or 6 x. Interval: 3 6 . Graph:
_3
6
46. x 2 2x 3 x 2 2x 3 0 x 3 x 1 0. The expression on the left of the inequality changes sign when x 3 and x 1. Thus we must check the intervals in the following table. Interval Sign of x 3 Sign of x 1
Sign of x 3 x 1
3
3 1
1
From the table, the solution set is x x 3 or 1 x. Interval: 3 1 . Graph:
_3
1
47. x 2 4 x 2 4 0 x 2 x 2 0. The expression on the left of the inequality changes sign where x 2 and where x 2. Thus we must check the intervals in the following table. Interval Sign of x 2 Sign of x 2
Sign of x 2 x 2
2
2 2
2
From the table, the solution set is x 2 x 2. Interval: 2 2. Graph:
_2
2
60
CHAPTER 1 Equations and Graphs
48. x 2 9 x 2 9 0 x 3 x 3 0. The expression on the left of the inequality changes sign when x 3 and x 3. Thus we must check the intervals in the following table. Interval Sign of x 3
3
3 3
3
Sign of x 3
Sign of x 3 x 3
From the table, the solution set is x x 3 or 3 x. Interval: 3] [3 . Graph:
_3
3
49. x 2 x 1 x 3 0. The expression on the left of the inequality changes sign when x 2, x 1, and x 3. Thus we must check the intervals in the following table. Interval Sign of x 2 Sign of x 1 Sign of x 3
Sign of x 2 x 1 x 3
2
2 1
1 3
3
From the table, the solution set is x x 2 or 1 x 3. Graph:
_2
1
Interval: 2] [1 3].
3
50. x 5 x 2 x 1 0. The expression on the left of the inequality changes sign when x 5, x 2, and x 1. Thus we must check the intervals in the following table. Interval Sign of x 5 Sign of x 2 Sign of x 1
Sign of x 5 x 2 x 1
1
1 2
2 5
5
From the table, the solution set is x 1 x 2 or 5 x. Interval: 1 25 . Graph:
_1
2
5
51. x 4 x 22 0. Note that x 22 0 for all x 2, so the expression on the left of the original inequality changes sign only when x 4. We check the intervals in the following table. Interval Sign of x 4
Sign of x 22
Sign of x 4 x 22
2
2 4
4
From the table, the solution set is x x 2 and x 4. We exclude the
endpoint 2 since the original expression cannot be 0. Interval: 2 2 4. Graph:
_2
4
SECTION 1.8 Solving Inequalities
61
52. x 4 x 22 0. Note that x 22 0 for all x 2, so the expression on the left of the original inequality changes sign only at x 4. We check the intervals in the following table. Interval Sign of x 4
2
2 4
4
Sign of x 22
Sign of x 32 x 1
From the table, the solution set is x x 4.
(The point 2 is already excluded.) Interval: 4 . Graph:
4
53. x 32 x 2 x 5 0. The lefthand side is 0 when x 5, 3, or 2. We check the intervals in the following table. Interval Sign of x 32
5
5 3
3 2
2
_3
2
Sign of x 5 Sign of x 2
Sign of x 22 x 3 x 1
When x 3, the lefthand side is equal to 0 and the inequality is satisfied. Thus, the solution set is x x 5, x 3, or x 2. Interval: 5] 3 [2 . Graph:
_5
54. 4x 2 x 2 9 0 4x 2 x 3 x 3 0. The expression on the left of the inequality changes sign when x 3 and x 0. Thus we must check the intervals in the following table. Interval Sign of 4x 2 Sign of x 3
Sign of x 3 Sign of 4x 2 x 2 9
3
3 0
0 3
3
From the table, the solution set is x 3 x 3. (The endpoint 0 is included since the original expression is allowed to be 0.) Interval: [3 3]. Graph:
_3
3
55. x 3 4x 0 x x 2 4 0 x x 2 x 2 0. The expression on the left of the inequality changes sign where x 0, x 2 and where x 4. Thus we must check the intervals in the following table. Interval Sign of x Sign of x 2 Sign of x 2
Sign of x x 2 x 2
2
2 0
0 2
2
From the table, the solution set is x 2 x 0 or x 2. Interval: 2 02 . Graph:
_2
0
2
62
CHAPTER 1 Equations and Graphs
56. 9x x 3 0 x 3 9x x x 2 9 x x 3 x 3. The expression on the left of the inequality changes sign when x 3, x 0, and x 3. Thus we must check the intervals in the following table. Interval Sign of x 3 Sign of x
Sign of x 3
Sign of x x 3 x 3
3
3 0
0 3
3
From the table, the solution set is x 3 x 0 or 3 x. Interval: [3 0][3 . Graph:
_3
0
3
57. x 4 x 2 x 4 x 2 0 x 2 x 2 1 0 x 2 x 1 x 1 0. The expression on the left of the inequality changes sign where x 0, where x 1, and where x 1. Thus we must check the intervals in the following table. Interval Sign of x 2 Sign of x 1 Sign of x 1
Sign of x 2 x 1 x 1
1
1 0
0 1
1
From the table, the solution set is x x 1 or 1 x. Interval: 1 1 . Graph:
_1
1
58. x 5 x 2 x 5 x 2 0 x 2 x 3 1 0 x 2 x 1 x 2 x 1 0. The expression on the left of the inequality 1 12 4 1 1 1 3 2 changes sign when x 0 and x 1. But the solution of x x 1 0 are x . 2 1 2 Since these are not real solutions. The expression x 2 x 1 does not changes signs, so we must check the intervals in the following table. Interval Sign of x 2 Sign of x 1
Sign of x 2 x 1 Sign of x 2 x 1 x 2 x 1
0
0 1
1
From the table, the solution set is x 1 x. Interval: 1 . Graph:
1
SECTION 1.8 Solving Inequalities
59.
x 3 0. The expression on the left of the inequality changes sign where x 1 and where x 3. Thus we must check x 1 the intervals in the following table. Interval Sign of x 1
Sign of x 3 x 3 Sign of x 1
60.
1
1 3
3
Sign of 2x 6
cannot equal 0 we must have x 1. Interval: 1 [3 . Graph:
3
3 2
2
Sign of x 2 2x 6 Sign of x 2
_1
3
From the table, the solution set is x 3 x 2. Interval: 3 2. Graph:
_3
2
x x 5 x 2 4x 10 2x 5 x 5 50 0 0 0. The expression on the left x 2 x 2 x 2 x 2 x 2 x 2 of the inequality changes sign when x 2 and x 52 . Thus we must check the intervals in the following table. Interval
2
Sign of 2x 5
Sign of x 2 2x 5 Sign of x 2
62.
From the table, the solution set is x x 1 or x 3. Since the denominator
2x 6 0. The expression on the left of the inequality changes sign when x 3 and x 2. Thus we must check the x 2 intervals in the following table. Interval
61.
63
2 52
5 2
From the table, the solution set is x 2 x 52 . Interval: 2 52 . Graph:
2
5 _ 2
x 4 x 4 5 2x 1 9x 9 x 4 5 50 0 0. The expression on the left of the inequality 2x 1 2x 1 2x 1 2x 1 changes sign when x 1 and x 12 . Thus we must check the intervals in the following table. Interval Sign of 9x 9
Sign of 2x 1 9x 9 Sign of 2x 1
1 12
12
1
From the table, the solution set is x x 1 or 12 x . Interval: 1 12 . Graph:
_1
1
__ 2
64
63.
CHAPTER 1 Equations and Graphs
2x 1 2x 1 3 x 5 x 16 2x 1 3 3 0 0 0. The expression on the left of the inequality x 5 x 5 x 5 x 5 x 5 changes sign where x 16 and where x 5. Thus we must check the intervals in the following table. Interval
5
5 16
16
Sign of x 16
Sign of x 5 x 16 Sign of x 5
64.
From the table, the solution set is x x 5 or x 16. Since the denominator cannot equal 0, we must have x 5. Interval: 5 [16 . Graph:
5
16
3x 3x 3x 2x 3x 1 1 0 0 0. The expression on the left of the inequality changes 3x 3x 3x 3x 3x sign when x 0 and x 3. Thus we must check the intervals in the following table. Since the denominator cannot equal 0, we must Interval Sign of 3 x
Sign of 2x 2x Sign of 3x
65.
0
0 3
3
have x 3. The solution set is x 0 x 3. Interval: [0 3. Graph:
0
3
4 4 xx 4 x2 4 2 x 2 x x x 0 0 0 0. The expression on the left of the x x x x x x inequality changes sign where x 0, where x 2, and where x 2. Thus we must check the intervals in the following table. Interval Sign of 2 x Sign of x
Sign of 2 x 2 x 2 x Sign of x
2
2 0
0 2
2
From the table, the solution set is x 2 x 0 or 2 x. Interval: 2 02 . Graph:
_2
0
2
SECTION 1.8 Solving Inequalities
66.
65
x x 3x x 1 2x 3x 2 x 2 3x x 3x 3x 0 0 0 0. The expression on x 1 x 1 x 1 x 1 x 1 x 1 the left of the inequality changes sign whenx 0, x 23 , and x 1. Thus we must check the intervals in the following table. Interval Sign of x
Sign of x 1 2 x 2 x Sign of x
Graph:
67. 1
_1
2
__ 3
23 0
0
Sign of 2 3x
From the table, the solution set is
1 23
1
x x 1 or 23 x 0 . Interval: 1 23 0 .
0
2 2 2 x x 1 2x 2 x 1 x 2 x 2x 2x 2 2 1 0 0 0 x 1 x x 1 x x x 1 x x 1 x x 1 x x 1
x2 x 2 x 2 x 1 0 0. The expression on the left of the inequality changes sign where x 2, where x x 1 x x 1 x 1, where x 0, and where x 1. Thus we must check the intervals in the following table. Interval Sign of x 2 Sign of x 1 Sign of x
Sign of x 1 x 2 x 1 Sign of x x 1
2
2 1
1 0
0 1
1
Since x 1 and x 0 yield undefined expressions, we cannot include them in the solution. From the table, the solution set is x 2 x 1 or 0 x 1. Interval: [2 1 0 1]. Graph:
_2
_1
0
1
66
68.
CHAPTER 1 Equations and Graphs
4 3 4 3x 4 x 1 x x 1 3x 4x 4 x 2 x 3 1 1 0 0 0 x 1 x x 1 x x x 1 x x 1 x x 1 x x 1 4 x2 2 x 2 x 0 0 The expression on the left of the inequality changes sign when x 2, x 2, x x 1 x x 1 x 0, and x 1. Thus we must check the intervals in the following table. Interval Sign of 2 x Sign of 2 x Sign of x
Sign of x 1 2 x 2 x Sign of x x 1
2
2 0
0 1
1 2
2
Since x 0 and x 1 give undefined expressions, we cannot include them in the solution. From the table, the solution set is x 2 x 0 or 1 x 2. Interval: [2 0 1 2]. Graph:
69.
_2
0
1
2
x 1 x 2 x 1 x 2 x 2 x 2 x 1 x 3 0 0 x 3 x 2 x 3 x 2 x 3 x 2 x 2 x 3 x 2 4 x 2 2x 3 2x 1 0 0. The expression on the left of the inequality x 3 x 2 x 3 x 2 changes sign where x 12 , where x 3, and where x 2. Thus we must check the intervals in the following table. Interval Sign of 2x 1 Sign of x 3
Sign of x 2 2x 1 Sign of x 3 x 2 From the table, the solution set is
Graph:
_3
1
__ 2
2
3 12
12 2
2
3
x 3 x 12 or 2 x . Interval: 3 12 2 .
SECTION 1.8 Solving Inequalities
70.
67
1 x 2 1 x 1 x 2x 1 2x 3 0 0 0 0. The x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 expression on the left of the inequality changes sign when x 32 , x 1, and x 2. Thus we must check the intervals in the following table. Interval Sign of 2x 3 Sign of x 1
Sign of x 2 2x 3 Sign of x 1 x 2
2 32
32 1
1
2
From the table, the solution set is x x 2 or 32 x 1 The points x 2 and x 1 are
excluded from the solution because the expression is undefined at those values. Interval: 2 32 1 . Graph:
_2
3
__ 2
_1
71. 0 5 2x 2x 5 x 52 . Interval: 52 . Graph:
72.
5 _ 2
6 6 6 6x 6 x 1 x x 1 6 1 10 0 x 1 x x 1 x x x 1 x x 1 x x 1 x 2 x 6 6x 6x 6 x 2 x x 3 x 2 0 0 0. The x x 1 x x 1 x x 1 expression on the left of the inequality changes sign where x 3, where x 2, where x 0, and where x 1. Thus we must check the intervals in the following table. Interval Sign of x 3 Sign of x 2 Sign of x
Sign of x 1 x 3 x 2 Sign of x x 1
2
2 0
0 1
1 3
3
From the table, the solution set is x 2 x 0 or 1 x 3. The points x 0 and x 1 are excluded from the solution set because they make the denominator zero. Interval: [2 0 1 3]. Graph:
_2
0
1
3
68
CHAPTER 1 Equations and Graphs
73. 16x x 3 0 x 3 16x x x 2 16 x x 4 x 4. The expression on the left of the inequality changes sign when x 4, x 0, and x 4. Thus we must check the intervals in the following table. Interval Sign of x 4
4
4 0
0 4
4
Sign of x
Sign of x 4
Sign of x x 4 x 4
From the table, the solution set is x 4 x 0 or 4 x. Interval: [4 0] [4 . Graph:
74. 5 3x 4 14 9 3x 18 3 x 6. Interval: [3 6]. Graph:
3
_4
0
4
6
75. x 2 5x 6 0 x 3 x 2 0. The expression on the left of the inequality changes sign when x 3 and x 2. Thus we must check the intervals in the following table. Interval Sign of x 3
3
3 2
2
Sign of x 2
Sign of x 3 x 2 76.
From the table, the solution set is x x 3 or 2 x.
Interval: 3 2 . Graph:
_3
_2
x x 3 x 1 2x 3 x 3 3 0 0 0. The expression on the left of the inequality x 1 x 1 x 1 x 1 x 1 changes sign when x 32 and x 1. Thus we must check the intervals in the following table. Interval Sign of 2x 3
32
32 1
1
From the table, the solution set is x 32 x 1 . Interval: 32 1 .
Graph:
Sign of x 1 2x 3 Sign of x 1
3
_1
__ 2
77. x 32 x 1 0. Note that x 32 0 for all x 3, so the expression on the left of the original inequality changes sign only when x 1. We check the intervals in the following table. Interval Sign of x 32 Sign of x 1
Sign of x 32 x 1
3
3 1
1
From the table, the solution set is x x 1. (The endpoint 3 is already excluded.) Interval: 1 . Graph:
_1
SECTION 1.8 Solving Inequalities
78. 1 3 2x 5 2 2x 8 1 x 4. Interval: [1 4].
79. For
Graph:
1
4
x 2 9 to be defined as a real number we must have x 2 9 0 x 3 x 3 0. The expression in the
inequality changes sign at x 3 and x 3. Interval Sign of x 3 Sign of x 3
Sign of x 3 x 3
3
3 3
3
Thus x 3 or x 3.
80. For
x 2 5x 50 to be defined as a real number we must have x 2 5x 50 0 x 5 x 10 0. The
expression on the left of the inequality changes sign when x 5 and x 10. Thus we must check the intervals in the following table. Interval Sign of x 5
Sign of x 10
Sign of x 5 x 10
5
5 10
10
Thus x 5 or x 10.
81. For
12 1 to be defined as a real number we must have x 2 3x 10 0 x 2 x 5 0. The x 2 3x 10
expression in the last inequality changes sign at x 2 and x 5. Interval Sign of x 2 Sign of x 5
Sign of x 2 x 5
2
2 5
5
Thus x 2 or x 5, and the solution set is 2 5 .
69
70
CHAPTER 1 Equations and Graphs
82. For 4
1x 1x to be defined as a real number we must have 0The expression on the left of the inequality changes 2x 2x
sign when x 1 and x 2. Thus we must check the intervals in the following table. Interval Sign of 1 x
Sign of 2 x 1x Sign of 2x
2
2 1
1
Thus 2 x 1 and the solution set is 2 1]. Note that x 2 has been excluded from the solution set because the expression is undefined at that value.
83. a bx c bc (where a, b, c 0) bx c x
bc 1 bc bx cx a a b
c c c c bc c x a a b a b
a b c cb ca . ab ab
84. We have a bx c 2a, where a, b, c 0 a c bx 2a c
ac 2a c x . b b
85. Inserting the relationship C 59 F 32, we have 20 C 30 20 59 F 32 30 36 F 32 54 68 F 86. 86. Inserting the relationship F 95 C 32, we have 50 F 95 50 95 C 32 95 18 95 C 63 10 C 35.
87. Let x be the average number of miles driven per day. Each day the cost of Plan A is 95 040x, and the cost of Plan B is 135. Plan B saves money when 135 95 040x 40 04x 100 x. So Plan B saves money when you average more than 100 miles a day.
88. Let m be the number of minutes of international calls placed per month. Then under Plan A, the cost will be 25 005m, and under Plan B, the cost will be 5012m. To determine when Plan B is advantageous, we must solve 25005m 5012m 20 007m 2857 m. So Plan B is advantageous if a person places fewer than 286 minutes of international calls per month.
89. We need to solve 6400 035m 2200 7100 for m. So 6400 035m 2200 7100 4200 035m 4900 12,000 m 14,000. She plans on driving between 12,000 and 14,000 miles.
SECTION 1.8 Solving Inequalities
71
h , where T is the temperature in C, and h is the height in meters. 100 (b) Solving the expression in part (a) for h, we get h 100 20 T . So 0 h 5000 0 100 20 T 5000 0 20 T 50 20 T 30 20 T 30. Thus the range of temperature is from 20 C down to 30 C.
90. (a) T 20
91. (a) Let x be the number of $3 increases. Then the number of seats sold is 120 x. So P 200 3x
3x P 200 x 13 P 200. Substituting for x we have that the number of seats sold is
120 x 120 13 P 200 13 P 560 3 .
(b) 90 13 P 560 3 115 270 360 P 200 345 270 P 560 345 290 P 215
290 P 215. Putting this into standard order, we have 215 P 290. So the ticket prices are between $215 and $290.
92. If the customer buys x pounds of coffee at $650 per pound, then his cost c will be 650x. Thus x
c . Since the 65
scale’s accuracy is 003 lb, and the scale shows 3 lb, we have 3 003 x 3 003 297
c 303 65
650 297 c 650 303 19305 c 19695. Since the customer paid $1950, he could have been over or undercharged by as much as 195 cents. 93. 00004
4,000,000 001. Since d 2 0 and d 0, we can multiply each expression by d 2 to obtain d2
00004d 2 4,000,000 001d 2 . Solving 00004d 2 4,000,000, we have d 2 10,000,000,000 d 100,000; solving 4,000,000 001d 2 , we have 400,000,000 d 2 20,000 d. Putting these together, we have 20,000 d 100,000. 94.
600,000 500 600,000 500 x 2 300 (Note that x 2 300 300 0, so we can multiply both sides by the 2 x 300
denominator and not worry that we might be multiplying both sides by a negative number or by zero.) 1200 x 2 300 0 x 2 900 0 x 30 x 30. The expression in the inequality changes sign at x 30 and x 30. However, since x represents distance, we must have x 0. Interval Sign of x 30 Sign of x 30
Sign of x 30 x 30
0 30
30
So x 30 and you must stand at least 30 meters from the center of the fire. 95. (a) We substitute d 1320 ft and t 10 s into the given formula and find the value of a that results in a quartermile time 2 of exactly 10 s: 1320 12 a 102 1320 50a a 1320 50 264 fts . Thus, the quartermile time will be less
than 10 s if a 264 fts2 .
2 1320 . Taking the positive root, we find 32 that the quartermile time for a (downward) quartermile under Earth’s gravity is 21320 91 s. 32
(b) We substitute a g 32 fts2 and solve for t: 1320 12 32 t 2 t 2
72
CHAPTER 1 Equations and Graphs
96. Solve 30 10 09 001 2 for 10 75. We have 30 10 09 001 2 001 2 09 20 0 01 4 01 5 0. The possible endpoints are 01 4 0 01 4 40 and 01 5 0 01 5 50. Interval
10 40
40 50
50 75
Sign of 01 4 Sign of 01 5
Sign of 01 4 01 5 Thus he must drive between 40 and 50 mi/h. 97. 240
2 1 2 240 0 1 3 80 0. The expression in the inequality changes sign at 20 20 20
60 and 80. However, since represents the speed, we must have 0. Interval
0 60
60
1 3 Sign of 20
Sign of 80 1 3 80 Sign of 20
So Kerry must drive between 0 and 60 mi/h.
98. Solve 2400 20x 2000 8x 00025x 2 2400 20x 2000 8x 00025x 2 00025x 2 12x 4400 0 00025x 1 x 4400 0. The expression on the left of the inequality changes sign whenx 400 and x 4400.
Since the manufacturer can only sell positive units, we check the intervals in the following table. Interval Sign of 00025x 1 Sign of x 4400
Sign of 00025x 1 x 4400
0 400
400 4400
4400
So the manufacturer must sell between 400 and 4400 units to enjoy a profit of at least $2400. 99. Let x be the length of the garden and its width. Using the fact that the perimeter is 120 ft, we must have 2x 2 120 60 x. Now since the area must be at least 800 ft2 , we have 800 x 60 x 800 60x x 2 x 2 60x 800 0 x 20 x 40 0. The expression in the inequality changes sign at x 20 and x 40. However, since x represents length, we must have x 0. Interval Sign of x 20 Sign of x 40
Sign of x 20 x 40 The length of the garden should be between 20 and 40 feet.
0 20
20 40
40
SECTION 1.9 Solving Absolute Value Equations and Inequalities
100. Case 1: a b 0
73
We have a a a b, since a 0, and b a b b, since b 0. So a 2 a b b2 , that
is a b 0 a 2 b2 . Continuing, we have a a 2 a b2 , since a 0 and b2 a b2 b, since b2 0. So a 3 ab2 b3 . Thus a b 0 a 3 b3 . So a b 0 a n bn , if n is even, and a n b, if n is odd. Case 2: 0 a b
We have a a a b, since a 0, and b a b b, since b 0. So a 2 a b b2 . Thus 0
a b a 2 b2 . Likewise, a 2 a a 2 b and b a 2 b b2 , thus a 3 b3 . So 0 a b a n bn , for all positive integers n. Case 3: a 0 b
If n is odd, then a n bn , because a n is negative and bn is positive. If n is even, then we could have
either a n bn or a n bn . For example, 1 2 and 12 22 , but 3 2 and 32 22 . 101. The rule we want to apply here is “a b ac bc if c 0 and a b ac bc if c 0 ”. Thus we cannot simply multiply by x, since we don’t yet know if x is positive or negative, so in solving 1
3 , we must consider two cases. x
Case 1: x 0
Multiplying both sides by x, we have x 3. Together with our initial condition, we have 0 x 3.
Case 2: x 0
Multiplying both sides by x, we have x 3. But x 0 and x 3 have no elements in common, so this
gives no additional solution. Hence, the only solutions are 0 x 3. 102.
c a c ad ad a , so by Rule 3, d d c. Adding a to both sides, we have a c a. Rewriting the lefthand b d b d b b side as
ad ab a b d a ac and dividing both sides by b d gives . b b b b bd
Similarly, a c
c b d ac c cb c , so . d d bd d
103. (a) Because x is nonnegative, x y x 2 x y, and because y is nonnegative, x y x y y 2 . Thus, by transitivity, x 2 y2.
xy x y2 xy . Expanding, this becomes 4x y x y2 x 2 y 2 2x y 2 4 x 2 y 2 2x y 0 x y2 0. This is true for any x and y, so the original inequality is true for all nonnegative x and y.
(b) By part (a),
1.9
xy
SOLVING ABSOLUTE VALUE EQUATIONS AND INEQUALITIES
1. The equation x 3 has the two solutions 3 and 3.
2. (a) The solution of the inequality x 3 is the interval [3 3].
(b) The solution of the inequality x 3 is a union of two intervals 3] [3 .
3. (a) The set of all points on the real line whose distance from zero is less than 3 can be described by the absolute value inequality x 3.
(b) The set of all points on the real line whose distance from zero is greater than 3 can be described by the absolute value inequality x 3.
4. (a) 2x 1 5 is equivalent to the two equations 2x 1 5 and 2x 1 5. (b) 3x 2 8 is equivalent to 8 3x 2 8.
74
CHAPTER 1 Equations and Graphs
5. 5x 20 5x 20 x 4.
6. 3x 10 3x 10 x 10 3.
7. 5 x 3 28 5 x 25 x 5 x 5.
8. 12 x 7 2 12 x 9 x 18 x 18.
9. x 3 2 is equivalent to x 3 2 x 3 2 x 1 or x 5.
10. 2x 3 7 is equivalent to either 2x 3 7 2x 10 x 5; or 2x 3 7 2x 4 x 2. The two solutions are x 5 and x 2. 11. x 4 05 is equivalent to x 4 05 x 4 05 x 45 or x 35.
12. x 4 3. Since absolute value is always nonnegative, there is no solution.
13. 2x 3 11 is equivalent to either 2x 3 11 2x 14 x 7; or 2x 3 11 2x 8 x 4. The two solutions are x 7 and x 4.
14. 2 x 11 is equivalent to either 2 x 11 x 9; or 2 x 11 x 13. The two solutions are x 9 and x 13.
15. 4 3x 6 1 3x 6 3 3x 6 3, which is equivalent to either 3x 6 3 3x 3 x 1; or 3x 6 3 3x 9 x 3. The two solutions are x 1 and x 3.
16. 5 2x 6 14 5 2x 8 which is equivalent to either 5 2x 8 2x 3 x 32 ; or 5 2x 8 3 13 2x 13 x 13 2 . The two solutions are x 2 and x 2 .
17. 3 x 5 6 15 3 x 5 9 x 5 3, which is equivalent to either x 5 3 x 2; or x 5 3 x 8. The two solutions are x 2 and x 8.
18. 20 2x 4 15 2x 4 5. Since the absolute value is always nonnegative, there is no solution. 19. 8 5 13 x 56 33 5 13 x 56 25 13 x 56 5, which is equivalent to either 13 x 56 5 13 x 35 6
1 5 1 25 25 25 35 x 35 2 ; or 3 x 6 5 3 x 6 x 2 . The two solutions are x 2 and x 2 . 3 9 20. 35 x 2 12 4 35 x 2 92 which is equivalent to either 35 x 2 92 35 x 52 x 25 6 ; or 5 x 2 2 3 x 13 x 65 . The two solutions are x 25 and x 65 . 5 2 6 6 6
21. x 1 3x 2, which is equivalent to either x 1 3x 2 2x 3 x 32 ; or x 1 3x 2 x 1 3x 2 4x 1 x 14 . The two solutions are x 32 and x 14 .
22. x 3 2x 1 is equivalent to either x 3 2x 1 x 2 x 2; or x 3 2x 1 x 3 2x 1 3x 4 x 43 . The two solutions are x 2 and x 43 .
23. x 5 5 x 5. Interval: [5 5].
24. 2x 20 20 2x 20 10 x 10. Interval: [10 10].
25. 2x 7 is equivalent to 2x 7 x 72 ; or 2x 7 x 72 . Interval: 72 72 .
26. 12 x 1 x 2 is equivalent to x 2 or x 2. Interval: 2] [2 .
27. x 4 10 is equivalent to 10 x 4 10 6 x 14. Interval: [6 14].
28. x 3 9 is equivalent to x 3 9 x 6; or x 3 9 x 12. Interval: 6 12 . 29. x 1 1 is equivalent to x 1 1 x 0; or x 1 1 x 2. Interval: 2] [0 .
30. x 4 0 is equivalent to x 4 0 x 4 0 x 4. The only solution is x 4.
31. 2x 1 3 is equivalent to 2x 1 3 2x 4 x 2; or 2x 1 3 2x 2 x 1. Interval: 2] [1 .
32. 3x 2 7 is equivalent to 3x 2 7 3x 5 x 53 ; or 3x 2 7 3x 9 x 3. Interval: 53 3 .
SECTION 1.9 Solving Absolute Value Equations and Inequalities
75
33. 2x 3 04 04 2x 3 04 26 2x 34 13 x 17. Interval: [13 17]. 34. 5x 2 6 6 5x 2 6 4 5x 8 45 x 85 . Interval: 45 85 . x 2 2 2 x 2 2 6 x 2 6 4 x 8. Interval: 4 8. 35. 3 3 x 1 4 1 x 1 4 1 x 1 4 x 1 8 which is equivalent to either x 1 8 x 7; or 36. 2 2 2 x 1 8 x 9. Interval: 9] [7 .
37. x 6 0001 0001 x 6 0001 6001 x 5999. Interval: 6001 5999. 38. x a d d x a d a d x a d. Interval: a d a d.
39. 4 x 2 3 13 4 x 2 16 x 2 4 4 x 2 4 6 x 2. Interval: 6 2.
40. 3 2x 4 1 2x 4 2 2x 4 2 which is equivalent to either 2x 4 2 2x 2 x 1; or 2x 4 2 2x 6 x 3. Interval: 3] [1 .
41. 8 2x 1 6 2x 1 2 2x 1 2 2 2x 1 2 1 2x 3 12 x 32 . Interval: 12 32 .
42. 7 x 2 5 4 7 x 2 1 x 2 17 . Since the absolute value is always nonnegative, the inequality is true for all real numbers. In interval notation, we have . 1 43. 12 4x 13 56 4x 13 53 , which is equivalent to either 4x 13 53 4x 43 x 13 ; or 4x 53 3
4x 2 x 12 . Interval: 12 13 . 44. 2 12 x 3 3 51 2 12 x 3 48 12 x 3 24 24 12 x 3 24 27 12 x 21 54 x 42. Interval: [54 42].
45. 1 x 4. If x 0, then this is equivalent to 1 x 4. If x 0, then this is equivalent to 1 x 4 1 x 4 4 x 1. Interval: [4 1] [1 4].
46. 0 x 5 12 . For x 5, this is equivalent to 12 x 5 12 92 x 11 2 . Since x 5 is excluded, the solution is 9 5 5 11 . 2 2
47.
48.
1 13 2 1 2 x 7 (x 7) x 7 12 12 x 7 12 15 2 x 2 and x 7. x 7 13 . Interval: 15 7 7 2 2 1 5 15 2x 3, since 2x 3 0, provided 2x 3 0 x 32 . Now for x 32 , we have 2x 3
1 2x 3 is equivalent to either 1 2x 3 16 2x 8 x; or 2x 3 1 2x 14 x 7 . 5 5 5 5 5 5 5 7 8 Interval: 5 5 .
49. x 3
50. x 2
51. x 7 5
52. x 2 4
53. x 2
54. x 1
55. x 3
56. x 4
57. (a) Let x be the thickness of the laminate. Then x 0020 0003.
(b) x 0020 0003 0003 x 0020 0003 0017 x 0023. h 682 2 2 h 682 2 58 h 682 58 624 h 740. Thus 95% of the adult males are 58. 29 29 between 624 in and 740 in.
76
CHAPTER 1 Equations and Graphs
59. x 1 is the distance between x and 1; x 3 is the distance between x and 3. So x 1 x 3 represents those points closer to 1 than to 3, and the solution is x 2, since 2 is the point halfway between 1 and 3. If a b, then the ab solution to x a x b is x . 2
1.10 SOLVING EQUATIONS AND INEQUALITIES GRAPHICALLY 1. The solutions of the equation x 2 2x 3 0 are the xintercepts of the graph of y x 2 2x 3. 2. The solutions of the inequality x 2 2x 3 0 are the xcoordinates of the points on the graph of y x 2 2x 3 that lie above the xaxis. 3. (a) From the graph, it appears that the graph of y x 4 3x 3 x 2 3x has xintercepts 1, 0, 1, and 3, so the solutions to the equation x 4 3x 3 x 2 3x 0 are x 1, x 0, x 1, and x 3.
(b) From the graph, we see that where 1 x 0 or 1 x 3, the graph lies below the xaxis. Thus, the inequality x 4 3x 3 x 2 3x 0 is satisfied for x 1 x 0 or 1 x 3 [1 0] [1 3].
4. (a) The graphs of y 5x x 2 and y 4 intersect at x 1 and at x 4, so the equation 5x x 2 4 has solutions x 1 and x 4. (b) The graph of y 5x x 2 lies strictly above the graph of y 4 when 1 x 4, so the inequality 5x x 2 4 is satisfied for those values of x, that is, for x 1 x 4 1 4.
5. (a)
(b) xintercepts 0, 1; yintercept 0. y x 3 x 2 x 2 x 1, so y 0
1
x 0 or x 1 and x 0 y 0.
(c) The graph is not symmetric with respect to either axis or the origin. 2
1
1
2
1
y x 3 x 2 ; [2 2] by [1 1] (b) xintercepts 0, 2; yintercept 0. y x 4 2x 3 x 3 x 2, so y 0
6. (a)
x 0 or x 2; x 0 y 0.
2
(c) The graph is not symmetric with respect to either axis or the origin. 2
1
1
2
3
2
y x 4 2x 3 ; [2 3] by [3 3]
SECTION 1.10 Solving Equations and Inequalities Graphically
7. (a)
(b) No xintercept; yintercept 2. y 0 has no solution; x 0
1 4
77
2
2
y
4
2 2. 1
(c) The graph is symmetric with respect to the yaxis:
1
2 3
2 x2 1
2 . 2 x 1
2 y 2 ; [5 5] by [3 1] x 1 (b) xintercepts 1, yintercept 1. y 0 3 x 1; x 0 y 1 02 1.
8. (a) 2
4
2
2
(c) The graph is symmetric with respect to the yaxis: 3 3 1 x2 1 x 2 .
4
2 4
y
3 1 x2 0 1 x2 0
3 1 x 2 ; [5 5] by [5 3]
9. Although the graphs of y 3x 2 6x 12 and 7 x 2 appear to intersect in the viewing y 7 12 rectangle [4 4] by [1 3], there is no point of intersection. You can verify this by zooming in.
10. Although the graphs of y
y 15 41 3x appear to intersect in the viewing
rectangle [8 8] by [1 8], there is no point of intersection. You can verify this by zooming in. 8
3
6
2
4
1 4
2
49 x 2 and
2 2
1
4
8 6 4 2
2 4 6 8
11. The graphs of y 6 4x x 2 and y 3x 18 appear to 12. The graphs of y x 3 4x and y x 5 appear to have have two points of intersection in the viewing rectangle
one point of intersection in the viewing rectangle [4 4]
[6 2] by [5 20]. You can verify that x 4 and
by [15 15]. The solution is x 2627.
x 3 are exact solutions.
10
20 4
10
2
2 10
6
4
2
2
4
78
CHAPTER 1 Equations and Graphs
13. Algebraically: 3x 2 5x 4 6 2x x 3.
Graphically: We graph the two equations y1 3x 2 and y2 5x 4 in the viewing rectangle [1 4] by [1 13].
Zooming in, we see that the solution is x 3.
14. Algebraically: 23 x 11 1 x 53 x 10 x 6.
Graphically: We graph the two equations y1 23 x 11
and y2 1 x in the viewing rectangle [12 2] by
[2 8]. Zooming in, we see that the solution is x 6.
10 5
5
1
1
2
3
4 12 10 8 6 4 2
15. Algebraically:
1 2 7 2x x 2x
5. 4 1 14x x 14
2 1 x 2x
2x 7
1 2 x 2x and y2 7 in the viewing rectangle [2 2] by [2 8]. Zooming in, we see that the solution is x 036.
Graphically: We graph the two equations y1
5
2
6 5 4 16. Algebraically: x 2 2x 2x 4 6 5 4 2x x 2 2x x 2 x 2 2x 2x 4 2x 4 x 2 6 x 5 8x 6x 12 5x 12 3x 4 x. Graphically: We graph the two equations 6 4 5 and y2 in the viewing y1 x 2 2x 2x 4 rectangle [5 5] by [10 10]. Zooming in, we see that there is only one solution at x 4. 10
2
1
1
2 4
2
2
4
10
17. Algebraically: 4x 2 8 0 x 2 2 x 2.
Graphically: We graph the equation y1 4x 2 8 and determine where this curve intersects the xaxis. We use the viewing rectangle [2 2] by [4 4]. Zooming in, we see that solutions are x 141 and x 141. 4
18. Algebraically: x 3 10x 2 0 x 2 x 10 0 x 0 or x 10. Graphically: We graph the equation y x 3 10x 2 and determine where this curve intersects the xaxis. We use the viewing rectangle [12 2] by [5 5]. We see that the solutions are x 0 and x 10. 4
2
2
2
1
2 4
1
2 12 10 8 6 4 2 2 4
2
SECTION 1.10 Solving Equations and Inequalities Graphically
19. Algebraically: x 2 9 0 x 2 9, which has no real solution. Graphically: We graph the equation y x 2 9 and see that this curve does not intersect the xaxis. We use the viewing rectangle [5 5] by [5 30]. 30 20 10
79
20. Algebraically: x 2 3 2x x 2 2x 3 0 2 22 4 1 3 2 8 . x 2 1 2 1 Because the discriminant is negative, there is no real solution. Graphically: We graph the two equations y1 x 2 3 and y2 2x in the viewing rectangle [4 6] by [6 12], and see that the two curves do not intersect. 10
4
2
2
4
5 4
2
2
4
6
5
4 21. Algebraically: 81x 4 256 x 4 256 81 x 3 .
Graphically: We graph the two equations y1 81x 4 and
y2 256 in the viewing rectangle [2 2] by [250 260].
Zooming in, we see that solutions are x 133. 260
22. Algebraically: 2x 5 243 0 2x 5 243 x 5 243 2 5 5 243 3 x 2 2 16. Graphically: We graph the equation y 2x 5 243 and
determine where this curve intersects the xaxis. We use the viewing rectangle [5 10] by [5 5]. Zooming in, we see that the solution is x 261.
255
4 2 2
1
0
1
2 5
5
2
10
4
23. Algebraically: x 54 80 0 x 54 80 x 5 4 80 2 4 5 x 5 2 4 5.
Graphically: We graph the equation y1 x 54 80
and determine where this curve intersects the xaxis. We use the viewing rectangle [1 9] by [5 5]. Zooming in, we see that solutions are x 201 and x 799.
24. Algebraically: 3 x 53 72 x 53 72 3 24 x 5 3 24 x 5 3 24.
Graphically: We graph the two equations y1 3 x 53 and y2 72 in the viewing rectangle [3 1] by [65 78]. Zooming in, we see that the solution is x 212
4 75
2 2 4
2
4
6
70
8 3
2
1
65
80
CHAPTER 1 Equations and Graphs
25. We graph y x 2 11x 30 in the viewing rectangle
26. We graph y x 2 075x 0125 in the viewing
[2 8] by [01 01]. The solutions appear to be exactly
rectangle [2 2] by [01 01]. The solutions are
x 5 and x 6. [In fact
x 025 and x 050.
x 2 11x 30 x 5 x 6.]
0.1
0.1
2
0.0
4
6
1
8
1
2
0.1
0.1
27. We graph y x 3 6x 2 11x 6 in the viewing
28. Since 16x 3 16x 2 x 1 16x 3 16x 2 x 1 0, we graph y 16x 3 16x 2 x 1 in the viewing
rectangle [1 4] by [01 01]. The solutions are x 100, x 200,and x 300.
rectangle [2 2] by [01 01]. The solutions are: x 100, x 025, and x 025.
0.1
0.1 1
1
2
3
4 2
0.1
1
1
2
0.1
29. We first graph y x
x 1 in the viewing rectangle [1 5] by [01 01] and
find that the solution is near 16. Zooming in, we see that solutions is x 162.
0.1
1
1
2
3
4
0.1
1 x2 1 x 1 x 2 0Since x is only defined
30. 1
x
for x 0, we start with the viewing rectangle [1 5] by [1 1]. In this rectangle, there
appears to be an exact solution at x 0 and
another solution between x 2 and x 25. We then use the viewing rectangle [23 235] by [001 001], and isolate the second solution as x 2314. Thus the solutions are x 0 and x 231.
1
1 1
0.01
1
2
3
4
5
0.00
0.01
2.32
2.34
5
SECTION 1.10 Solving Equations and Inequalities Graphically
31. We graph y x 13 x in the viewing rectangle [3 3] by [1 1]. The solutions
1
are x 1, x 0, and x 1, as can be verified by substitution.
3
2
1
1
2
3
1
32. Since x 12 is defined only for x 0, we start by
1
graphing y x 12 x 13 x in the viewing
0.01
rectangle [1 5] by [1 1] We see a solution at x 0 and another one between x 3 and x 35. We then use the viewing rectangle
[33 34] by [001 001], and isolate the second
1
1
2
3
4
0.00
5
1
3.35
3.40
0.01
solution as x 331. Thus, the solutions are x 0 and x 331. 2x 1 and y 3x 5 in the viewing rectangle [0 9] by [0 5] and see that the only solution to the equation 2x 1 3x 5 is x 4, which
33. We graph y
can be verified by substitution.
4 2 0
0
2
3 x and y x 2 1 in the viewing rectangle [4 4] by [0 4] and see that the equation 3 x x 2 1 has solutions x 1 and x 2,
34. We graph y
4
6
8
4 3
which can be verified by substitution.
2 1 4
2x 1 1 and y x in the viewing rectangle [1 6] by [0 6] and see that the only solution to the equation 2x 1 1 x is x 4, which can
35. We graph y
be verified by substitution.
2
0
2
4
6 4 2 0
4
6
x 1 and y 8 in the viewing rectangle [2 6] by [2 12] and see that the only solution to the equation 2x x 1 8 is x 3,
36. We graph y 2x
2
10
which can be verified by substitution.
5
2
2
4
6
81
82
CHAPTER 1 Equations and Graphs
37. x 3 2x 2 x 1 0, so we start by graphing
100
the function y x 3 2x 2 x 1 in the viewing
1
rectangle [10 10] by [100 100]. There
appear to be two solutions, one near x 0 and
10
5
another one between x 2 and x 3. We then use the viewing rectangle [1 5] by [1 1] and
5
10
1
100
1
2
3
4
5
1
zoom in on the only solution, x 255. 38. x 4 8x 2 2 0. We start by graphing the
10
function y x 4 8x 2 2 in the viewing
1
rectangle [10 10] by [10 10]. There appear to be four solutions between x 3 and x 3. We then use the viewing rectangle [5 5] by
10
5
5
[1 1], and zoom to find the four solutions x 278, x 051, x 051, and x 278. 39. x x 1 x 2 16 x
x x 1 x 2 16 x 0. We start by graphing
the function y x x 1 x 2 16 x in the
4
viewing rectangle [5 5] by [10 10]. There
4
2
10
1
10
1
2
appear to be three solutions. We then use the
10
2
4
2
1
10
viewing rectangle [25 25] by [1 1] and
2
4
1
2
1
zoom into the solutions at x 205, x 000, and x 105. 40. x 4 16 x 3 . We start by graphing the functions y1 x 4 and y2 16 x 3 in the viewing rectangle [10 10] by [5 40]. There appears to be two solutions, one near x 2 and another one near x 2. We then use the viewing
rectangle [24 22] by [27 29], and zoom in to find the solution at x 231. We then use the viewing rectangle [17 18] by [95 105], and zoom in to find the solution at x 179. 40
29
30 28
20 10 10
5
5
10
2.4
2.3
27 2.2
10.4 10.2 10.0 9.8 9.6 1.70
1.75
41. We graph y x 2 and y 3x 10 in the viewing rectangle [4 7] by [5 30].
1.80
30
The solution to the inequality is [2 5].
20 10 4
2
2
4
6
SECTION 1.10 Solving Equations and Inequalities Graphically
42. Since 05x 2 0875x 025 05x 2 0875x 025 0, we graph
83
4
y 05x 2 0875x 025 in the viewing rectangle [3 1] by [5 5]. Thus the
2
solution to the inequality is [2 025].
3
2
1
1
2 4
43. Since x 3 11x 6x 2 6 x 3 6x 2 11x 6 0, we graph
4
y x 3 6x 2 11x 6 in the viewing rectangle [0 5] by [5 5]. The solution
2
set is 100] [200 300].
0
2
2
4
4
44. Since 16x 3 24x 2 9x 1
0.01
4
16x 3 24x 2 9x 1 0, we graph
2
y 16x 3 24x 2 9x 1 in the viewing
rectangle [3 1] by [5 5]. From this rectangle,
3
2
1
we see that x 1 is an xintercept, but it is
2
1
1.0
0.00
0.5
4
unclear what is occurring between x 05 and
0.01
x 0. We then use the viewing rectangle [1 0] by [001 001]. It shows y 0 at x 025. Thus in interval notation,
the solution is 1 025 025 .
45. Since x 13 x x 13 x 0, we graph y x 13 x
in the viewing rectangle [3 3] by [1 1]. From this, we
05x 2 1 2 x 05x 2 1 2 x 0, we graph y 05x 2 1 2 x in the viewing rectangle
46. Since
find that the solution set is 1 0 1 .
[1 1] by [1 1]. We locate the xintercepts at
and y2 x in the same viewing rectangle, and see that
x 0535. Thus in interval notation, the solution is approximately 0535] [0535 .
Another Method: As in Example 7, we graph y1 x 13 x 13 x for 1 x 0 and for 1 x. 1
1
2 1
1.0 2
2 1
2
1 2
2
0.5
0.5 1
1.0
84
CHAPTER 1 Equations and Graphs
47. Since x 12 x 12 x 12 x 12 0, we graph y x 12 x 12 in the viewing
rectangle [2 2] by [5 5]. The solution set is 0.
48. Since x 12 x 3 x 12 x 3 0, we graph
y x 12 x 3 in the viewing rectangle [4 4] by
[1 1]. The xintercept is close to x 2. Using a trace
function, we obtain x 2148. Thus the solution is [2148 .
4 2
1 2
1
2
1
2
4
4
2
2
4
1
49. We graph the equations y 3x 2 3x and y 2x 2 4 in the viewing rectangle
[2 6] by [10 50]. We see that the two curves intersect at x 1 and at x 4,
40
the inequality 3x 2 3x 2x 2 4 has solution set 1 4.
20
and that the first curve is lower than the second for 1 x 4 . Thus, we see that
2
2
50. We graph the equations y 5x 2 3x and y 3x 2 2 in the viewing rectangle
[3 2] by [5 25]. We see that the two curves intersect at x 2 and at x 12 ,
6
20
which can be verified by substitution. The first curve is higher than the second for
x 2 and for x 12 , so the solution set of the inequality 5x 2 3x 3x 2 2 is 2] 12 .
4
10
3
2
1
1
2
51. We graph the equation y x 32 x 2 x 5 in the viewing rectangle
[7 3] by [120 20] and see that the inequality x 32 x 2 x 5 0 has
the solution set 5] 3 [2 .
6
4
2
2
50 100
52. We graph the equation y 4x 2 x 2 9 in the viewing rectangle [5 5] by [100 100] and see that the inequality 4x 2 x 2 9 0 has the solution set [3 3].
100
4
2 100
2
4
SECTION 1.10 Solving Equations and Inequalities Graphically
53. To solve 5 3x 8x 20 by drawing the graph of a single equation, we isolate
85
1
all terms on the lefthand side: 5 3x 8x 20 5 3x 8x 20 8x 20 8x 20 11x 25 0 or 11x 25 0. We graph y 11x 25, and see that the solution is x 227, as in Example 2.
1
1
2
3
1
54. Graphing y x 3 6x 2 9x and y
x in the viewing rectangle [001 002]
0.2
by [005 02], we see that x 0 and x 001 are solutions of the equation x 3 6x 2 9x x.
0.1
0.01
0.01
0.02
(c) We graph the equations y 15,000 and
55. (a) We graph the equation y 10x 05x 2 0001x 3 5000 in the viewing
y 10x 05x 2 0001x 3 5000 in the viewing
rectangle [0 450] by [5000 20000].
rectangle [250 450] by [11000 17000]. We use a zoom or trace function on a graphing calculator, and find that
20000
the company’s profits are greater than $15,000 for 279 x 400.
10000
16000
0
100 200 300 400
14000
(b) From the graph it appears that
12000
0 10x 005x 2 0001x 3 5000 for
300
400
100 x 500, and so 101 cooktops must be produced
to begin to make a profit. 56. (a)
(b) Using a zoom or trace function, we find that y 10 for x 667. We x 2 could estimate this since if x 100, then 5280 000036. So for x 2 15x. Solving 15x 10 we x 100 we have 15x 5280
15 10 5 0
0
50
100
get 15 100 or x 100 15 667 mi.
57. Answers will vary. 58. Calculators perform operations in the following order: exponents are applied before division and division is applied before addition. Therefore, Y_1=x^1/3 is interpreted as y interpreted as y
x x1 , which is the equation of a line. Likewise, Y_2=x/x+4 is 3 3
x 4 1 4 5. Instead, enter the following: Y_1=x^(1/3), Y_2=x/(x+4). x
86
CHAPTER 1 Equations and Graphs
1.11 MODELING VARIATION 1. If the quantities x and y are related by the equation y 5x then we say that y is directly proportional to x, and the constant of proportionality is 5. 5 2. If the quantities x and y are related by the equation y then we say that y is inversely proportional to x, and the constant x of proportionality is 5. x 3. If the quantities x, y, and z are related by the equation z 5 then we say that z is directly proportional to x and inversely y proportional to y. 4. Because z is jointly proportional to x and y, we must have z kx y. Substituting the given values, we get 10 k 4 5 20k k 12 . Thus, x, y, and z are related by the equation z 12 x y.
5. (a) In the equation y 3x, y is directly proportional to x. (b) In the equation y 3x 1, y is not proportional to x.
3 , y is not proportional to x. x 1 3 (b) In the equation y , y is inversely proportional to x. x
6. (a) In the equation y
7. T kx, where k is constant.
8. P k, where k is constant.
9.
10. kmn, where k is constant.
k , where k is constant. z ks 11. y , where k is constant. t
12. P
k , where k is constant. T
13. z k y, where k is constant.
14. A
kx 2 , where k is constant. t3
15. V klh, where k is constant.
16. S kr 2 2 , where k is constant.
k P2t 2 , where k is constant. 18. A k x y, where k is constant. 3 b 19. Since y is directly proportional to x, y kx. Since y 32 when x 8, we have 32 k 8 k 4. So y 4x. k k 24 20. is inversely proportional to t, so . Since 3 when t 8, we have 3 k 24, so . t 8 t k k 75 21. A varies inversely as r, so A . Since A 15 when r 5, we have 15 k 75. So A . r 5 r 17. R
22. P is directly proportional to T , so P kT . Since P 60 when T 72, we have 60 k 72 k 56 . So P 56 T . 23. Since A is directly proportional to x and inversely proportional to t, A have 42
k 7 18x k 18. Therefore, A . 3 t
kx . Since A 42 when x 7 and t 3, we t
24. S kpq. Since S 350 when p 7 and q 20, we have 350 k 7 20 350 140k k 52 . So S 52 pq.
k k k 216. 25. Since W is inversely proportional to the square of r, W 2 . Since W 24 when r 3, we have 24 r 32 216 So W 2 . r 10 15 xy xy k 10. So t 10 . 26. t k . Since t 125 when x 10, y 15, and r 12, we have 125 k r 12 r
SECTION 1.11 Modeling Variation
87
27. Since C is jointly proportional to l, , and h, we have C klh. Since C 128 when l h 2, we have 128 k 2 2 2 128 8k k 16. Therefore, C 16lh. 2 28. H kl 2 2 . Since H 36 when l 2 and 13 , we have 36 k 22 13 36 49 k k 81. So H 81l 2 2 .
k k 275 k 29. R . Since R 25 when x 121, 25 k 275. Thus, R . 11 x x 121 a 2 2 abc abc . Since M 128 when a d and b c 2, we have 128 k 4k k 32. So M 32 . 30. M k d a d x3 31. (a) z k 2 y (b) If we replace x with 3x and y with 2y, then z k
x3 27 , so z changes by a factor of 27 k 4. 4 y2 2y2
3x3
x2 32. (a) z k 4 y
x2 9 9. k 4 , so z changes by a factor of 16 (b) If we replace x with 3x and y with 2y, then z k 16 y 2y4 3x2
33. (a) z kx 3 y 5
(b) If we replace x with 3x and y with 2y, then z k 3x3 2y5 864kx 3 y 5 , so z changes by a factor of 864.
k 34. (a) z 2 3 x y
(b) If we replace x with 3x and y with 2y, then z
k 3x2 2y3
1 k 1 . , so z changes by a factor of 72 72 x 2 y 3
35. (a) The force F needed is F kx.
(b) Since F 30 N when x 9 cm and the spring’s natural length is 5 cm, we have 30 k 9 5 k 75 Ncm. (c) From part (b), we have F 75x. Substituting x 11 5 6 into F 75x gives F 75 6 45 N.
36. (a) C kpm
(b) Since C 60,000 when p 120 and m 4000, we get 60,000 k 120 4000 k 18 . So C 18 pm.
(c) Substituting p 92 and m 5000, we get C 18 92 5000 $57,500.
37. (a) P ks 3 .
(b) Since P 96 when s 20, we get 96 k 203 k 0012 W(mih)3 . So P 0012s 3 .
(c) Substituting x 30, we get P 0012 303 324 watts.
38. Let the amount of soluble CO2 be x and the temperature of the water be T . Because these quantities are inversely k k proportional, x . We are given that when T 273 K, x 3 g, so 3 k 819. Thus, if T 298, then T 273 819 275 g. x 298 39. D ks 2 . Since D 150 when s 40, we have 150 k 402 , so k 009375. Thus, D 009375s 2 . If D 200, then 200 009375s 2 s 2 21333, so s 46 mi/h (for safety reasons we round down). 40. L ks 2 A. Since L 1700 when s 50 and A 500, we have 1700 k 502 500 k 000136. Thus L 000136s 2 A. When A 600 and s 40 we get the lift is L 000136 402 600 13056 lb.
41. F k As 2 . Since F 220 when A 40 and s 5. Solving for k we have 220 k 40 52 220 1000k k 022. Now when A 28 and F 175 we get 175 0220 28 s 2 284090 s 2 so s 284090 533 mi/h.
88
CHAPTER 1 Equations and Graphs
42. (a) T 2 kd 3
3 (b) Substituting T 365 and d 93 106 , we get 3652 k 93 106 k 166 1019 . 3 (c) T 2 166 1019 279 109 360 109 T 600 104 . Hence the period of Neptune is 6.00104 days 164 years.
43. (a) P
kT . V
(b) Substituting P 332, T 400, and V 100, we get 332 P
83T . V
(c) Substituting T 500 and V 80, we have P about 519 kPa.
k 400 k 83. Thus k 83 and the equation is 100
83 500 51875 kPa. Hence the pressure of the sample of gas is 80
s 2 r (b) For the first car we have 1 1600 and s1 60 and for the second car we have 2 2500. Since the forces are equal
44. (a) F k
we have k
2500 s22 16 602 1600 602 k s22 , so s2 48 mi/h. r r 25
k 45. (a) The loudness L is inversely proportional to the square of the distance d, so L 2 . d k (b) Substituting d 10 and L 70, we have 70 2 k 7000. 10 1 k k , so the loudness is changed by a factor of 14 . (c) Substituting 2d for d, we have L 4 d2 2d2 k k , so the loudness is changed by a factor of 4. (d) Substituting 12 d for d, we have L 2 4 d2 1d 2
46. (a) The power P is jointly proportional to the area A and the cube of the velocity , so P k A 3 . (b) Substituting 2 for and 12 A for A, we have P k 12 A 23 4k A 3 , so the power is changed by a factor of 4. 3 (c) Substituting 12 for and 3A for A, we have P k 3A 12 38 Ak 3 , so the power is changed by a factor of 38 . 47. (a) R
kL d2
k 12 7 0002916. k 2400 00052 7 3 4375 (c) Substituting L 3 and d 0008, we have R 137 ohms. 2400 00082 32 (b) Since R 140 when L 12 and d 0005, we get 140
(d) If we substitute 2d for d and 3L for L, then R
k 3L 2d2
3 kL , so the resistance is changed by a factor of 34 . 4 d2
48. Let S be the final size of the cabbage, in pounds, let N be the amount of nutrients it receives, in ounces, and let c be the N number of other cabbages around it. Then S k . When N 20 and c 12, we have S 30, so substituting, we have c N 20 30 k 12 k 18. Thus S 18 . When N 10 and c 5, the final size is S 18 10 5 36 lb. c
SECTION 1.11 Modeling Variation
89
ES k60004 6000 4 204 160,000. So the sun 300 EE k3004 produces 160,000 times the radiation energy per unit area than the Earth.
49. (a) For the sun, E S k60004 and for earth E E k3004 . Thus
(b) The surface area of the sun is 4 435,0002 and the surface area of the Earth is 4 3,9602 . So the sun has 4 435,0002 435,000 2 times the surface area of the Earth. Thus the total radiation emitted by the sun is 3,960 4 3,9602 435,000 2 160,000 1,930,670,340 times the total radiation emitted by the Earth. 3,960 50. (a) Let T and l be the period and the length of the pendulum, respectively. Then T k l.
T2 T2 2T 2 (b) T k l T 2 k 2 l l 2 . If the period is doubled, the new length is 4 2 4l. So we would 2 k k k quadruple the length l to double the period T .
51. (a) Since f is inversely proportional to L, we have f (b) If we replace L by 2L we have
k , where k is a positive constant. L
k k 12 12 f . So the frequency of the vibration is cut in half. 2L L
52. (a) Since r is jointly proportional to x and P x, we have r kx P x, where k is a positive constant.
(b) When 10 people are infected the rate is r k10 5000 10 49,900k. When 1000 people are infected the rate is r k 1000 5000 1000 4,000,000k. So the rate is much higher when 1000 people are infected. Comparing 1000 people infected 4,000,000k these rates, we find that 80. So the infection rate when 1000 people are infected 10 people infected 49,900k is about 80 times as large as when 10 people are infected. (c) When the entire population is infected the rate is r k 5000 5000 5000 0. This makes sense since there are no more people who can be infected.
02 2105 . 1002 Now, because the range is R 500 km for these values of e and , we can use the first equation to calculate the value of C: C 500 C 100, so a full charge for this vehicle is 100 kWh. 02 C Substituting e k 2 into the first given equation, we have R 2 . k 100 100 296 km; and at 80 kmh, its range is R 781 km. So at 130 kmh, its range is R 2 5 2 10 130 2 105 802
53. We substitute e 02 kWhkm and 100 km/h into the second given equation: 02 k 1002 k
54. (a) We are given that the mass flow rate is m A0 0 . Because m is constant, we can equate the flow rates for the wider A and narrower sections of pipe: A0 0 A 0 0 . The velocity of the fluid is indeed inversely proportional A to the crosssectional area of the pipe. r02 0 A0 0 06 m2 5 ms 45 ms. Thus, A r 2 02 m2 the fluid flows through the constricted section of pipe at 45 m/s.
(b) We substitute the given values into our expression for :
L 25 1026 14 . 55. Using B k 2 with k 0080, L 25 1026 , and d 24 1019 , we have B 0080 2 347 10 d 24 1019 The star’s apparent brightness is about 347 1014 Wm2 .
90
CHAPTER 1 Equations and Graphs
L L L 2 56. First, we solve B k 2 for d: d k d k because d is positive. Substituting k 0080, L 58 1030 , and B B d 58 1030 16 B 82 10 , we find d 0080 238 1022 , so the star is approximately 238 1022 m from earth. 82 1016 57. Examples include radioactive decay and exponential growth in biology.
CHAPTER 1 REVIEW 1. (a)
y
Q
(b) The distance from P to Q is d P Q 5 22 12 02 49 144 193 3 5 2 12 0 (c) The midpoint is 6 . 2 2 2
1 1
(d) The line has slope m
P
x
12 0 12 7 , and has 5 2
equation y 0 12 7 x 2
24 y 12 7 x 7 12x 7y 24 0. y
Q
(e) The radius of this circle was found in part (b). It is r d P Q 193. So an equation is 2 193 x 22 y 2 193. x 22 y 02 Q
y
2
1 1
2. (a)
x
1
Q
x
P
P
y 1
P 2
P
x
(b) The distance from P to Q is d P Q 2 72 11 12 25 100 125 5 5 2 7 11 1 9 (c) The midpoint is 6 . 2 2 2
CHAPTER 1
10 11 1 2, 27 5 and its equation is y 11 2 x 2
(d) The line has slope m
y 11 2x 4 y 2x 15. y
Review
(e) The radius of this circle was found in part (b). It is r d P, Q 5 5. So an equation is 2 x 72 y 12 5 5 x 72 y 12 125.
1 1
y
x
P
2 2
x
P
Q
Q
y
3. (a)
4
P
x
4 Q
(d) The line has slope m
(c) The midpoint is
16 8 2 14 6 4 10 5
and equation y 2 85 x 6
8 38 y 2 85 x 48 5 y 5x 5 . y
6 4 2 14 2 2
x
4
1 6.
(e) The radius of this circle was found in part (b). It is r d P Q 2 89. So an equation is 2 [x 6]2 y 22 2 89 x 62 y 22 356.
4
P
(b) The distance from P to Q is d P Q 6 42 [2 14]2 100 256 356 2 89
P
Q
y
4 4
x
Q
y
4. (a)
2 Q
2
P
x
(b) The distance from P to Q is d P Q [5 3]2 [2 6]2 64 16 80 4 5. (c) The midpoint is
5 3 2 6 2 2
1 4.
91
92
CHAPTER 1 Equations and Graphs
(d) The line has slope m
2 6 48 12 , and 5 3
has equation y 2 12 x 5 y 2 12 x 52 y 12 x 92 .
(e) The radius of this circle was found in part (b). It is r d P Q 4 5. So an equation is 2 2 x 52 y 2 4 5 x 52 y 22 80.
y
y
2 2
Q
P
x
2 Q
5. x y x 4 or y 2
2
x
P
6. x y x 1 and y 4
y
y
1
1 1
x
1
x
7. d A C 74 and 4 12 4 32 4 12 4 32 d B C 5 12 3 32 5 12 3 32 72. Therefore, B is closer to C. 8. The circle with center 3 4 and radius
2 6 has equation x 32 y 42 6 x 32 y 42 6.
9. The center is C 5 1, and the point P 0 0 is on the circle. The radius of the circle is r d P C 0 52 0 12 = 0 52 0 12 26. Thus, the equation of the circle is x 52 y 12 26.
1 11 21 38 , and the radius is 12 of the distance from P to Q, or 10. The midpoint of segment P Q is 2 2 2 2 r 12 d P, Q 12 2 12 3 82 12 2 12 3 82 r 12 34. Thus the equation is 2 2 x 12 y 11 17 2 2 .
CHAPTER 1
11. (a) x 2 y 2 8x 2y 13 0 x 2 8x y 2 2y 13 x 2 8x 16 y 2 2y 1 13 16 1
Review
93
(b) The circle has center 4 1 and radius 2.
y
x 42 y 12 4, an equation of a circle.
1 x
1
12. (a) 2x 2 2y 2 2x 8y 12 x 2 x y 2 4y 14 x 2 x 14 y 2 4y 4 14 14 4
x 12
2
(b) The circle has center 12 2
and radius 3 2 2 . y
y 22 92 , an equation of a circle.
1l 1
x
13. (a) x 2 y 2 72 12x x 2 12x y 2 72 x 2 12x 36 y 2 72 36 x 62 y 2 36. Since the left side of this equation must be greater than or equal to zero, this equation has no graph.
14. (a) x 2 y 2 6x 10y 34 0 x 2 6x y 2 10y 34 x 2 6x 9 y 2 10y 25 34 9 25 x 32 y 52 0, an equation of a point.
(b) This is the equation of the point 3 5. y
1l 1
x
94
CHAPTER 1 Equations and Graphs
15. y 2 3x x
y
2
8
0
2 3
17.
16. 2x y 1 0 y 2x 1
y
2 0
1
x
18. y
x
y
2
14 7
2
19. y 16 x 2 x
y
3
7
1
15
0
16
1
15
3
7
y
0
0
1
1
2
4
3
9
3
12
0
1 1 1
x
1
x
1
x
y
x
y
4
5
0
0
4
5
20. 8x y 2 0 y 2 8x
y
y
y x 0 5x 4y 0 4 5
x
x
y
8
8
2
4
0
0
1
y
1
2 1
x
y x
2
2
0
2
21. x
y 0
1
y x 1 y 72 x 7 2 7
0
x
y
22. y 1 x 2
y
0
1
x
1 1 2
1
0 1
x
1
y
23
1
0
1
x
CHAPTER 1
Review
95
23. x 16 y 2 (a) xaxis symmetry: replacing y by y gives x 16 y2 16 y 2 , which is the same as the original equation, so the graph is symmetric with respect to the xaxis. yaxis symmetry: replacing x by x gives x 16 y 2 x y 2 16, which is not the same as the original equation, so the graph is not symmetric with respect to the yaxis. Origin symmetry: replacing x by x and y by y gives x 16 y2 x 16 y 2 , which is not the same as the original equation, so the graph is not symmetric with respect to the origin. (b) To find xintercepts, we set y 0 and solve for x: x 16 02 16, so the xintercept is 16.
To find yintercepts, we set x 0 and solve for y: 0 16 y 2 y 4, so the yintercept are 4 and 4.
24. x 2 4y 2 9 (a) xaxis symmetry: replacing y by y gives x 2 4 y2 9 x 2 4y 2 9, which is the same as the original equation, so the graph is symmetric with respect to the xaxis. yaxis symmetry: replacing x by x gives x2 4 y2 9 x 2 4y 2 9, which is the same as the original equation, so the graph is symmetric with respect to the yaxis. Origin symmetry: replacing x by x and y by y gives x2 4 y2 9 x 2 4y 2 9, the same as the original equation, so the graph is symmetric with respect to the origin. (b) To find xintercepts, we set y 0 and solve for x: x 2 4 02 9 x 3, so the xintercepts are 3 and 3. To find yintercepts, we set x 0 and solve for y: 02 4y 2 9 y 32 , so the yintercepts are 32 and 32 .
25. x 2 9y 9 (a) xaxis symmetry: replacing y by y gives x 2 9 y 9 x 2 9y 9, so the graph is not symmetric with respect to the xaxis. yaxis symmetry: replacing x by x gives x2 9y 9 x 2 9y 9, so the graph is symmetric with respect to the yaxis. Origin symmetry: replacing x by x and y by y gives x2 9 y 9 x 2 9y 9, so the graph is not symmetric with respect to the origin. (b) To find xintercepts, we set y 0 and solve for x: x 2 9 0 9 x 3, so the xintercepts are 3 and 3. To find yintercepts, we set x 0 and solve for y: 02 9y 9 y 1, so the yintercept is 1.
26. x 12 y 2 4 (a) xaxis symmetry: replacing y by y gives x 12 y2 4 x 12 y 2 4, so the graph is symmetric with respect to the xaxis. yaxis symmetry: replacing x by x gives x 12 y 2 4 1 x2 y 2 4, so the graph is not symmetric with respect to the yaxis. Origin symmetry: replacing x by x and y by y gives x 12 y2 4 1 x2 y 2 4, so the graph is not symmetric with respect to the origin. (b) To find xintercepts, we set y 0 and solve for x: x 12 02 4 x 12 4 x 3 or 1, so the xintercepts are 3 and 1. To find yintercepts, we set x 0 and solve for y: 0 12 y 2 4 y 2 3 y 3, so the yintercepts are 3 and 3.
96
CHAPTER 1 Equations and Graphs
27. 9x 2 16y 2 144 (a) xaxis symmetry: replacing y by y gives 9x 2 16 y2 144 9x 2 16y 2 144, so the graph is symmetric with respect to the xaxis. yaxis symmetry: replacing x by x gives 9 x2 16y 2 144 9x 2 16y 2 144, so the graph is symmetric with respect to the yaxis. Origin symmetry: replacing x by x and y by y gives 9 x2 16 y2 144 9x 2 16y 2 144, so the graph is symmetric with respect to the origin. (b) To find xintercepts, we set y 0 and solve for x: 9x 2 16 02 144 9x 2 144 x 4, so the xintercepts are 4 and 4. To find yintercepts, we set x 0 and solve for y: 9 02 16y 2 144 16y 2 144, so there is no yintercept. 4 x 4 (a) xaxis symmetry: replacing y by y gives y , which is different from the original equation, so the graph is not x symmetric with respect to the xaxis. 4 , which is different from the original equation, so the graph is not yaxis symmetry: replacing x by x gives y x symmetric with respect to the yaxis. 4 4 Origin symmetry: replacing x by x and y by y gives y y , so the graph is symmetric with respect x x to the origin. 4 (b) To find xintercepts, we set y 0 and solve for x: 0 has no solution, so there is no xintercept. x To find yintercepts, we set x 0 and solve for y. But we cannot substitute x 0, so there is no yintercept.
28. y
29. x 2 4x y y 2 1 (a) xaxis symmetry: replacing y by y gives x 2 4x y y2 1, which is different from the original equation, so the graph is not symmetric with respect to the xaxis. yaxis symmetry: replacing x by x gives x2 4 x y y 2 1, which is different from the original equation, so the graph is not symmetric with respect to the yaxis. Origin symmetry: replacing x by x and y by y gives x2 4 x y y2 1 x 2 4x y y 2 1, so the graph is symmetric with respect to the origin. (b) To find xintercepts, we set y 0 and solve for x: x 2 4x 0 02 1 x 2 1 x 1, so the xintercepts are 1 and 1. To find yintercepts, we set x 0 and solve for y: 02 4 0 y y 2 1 y 2 1 y 1, so the yintercepts are 1 and 1. 30. x 3 x y 2 5 (a) xaxis symmetry: replacing y by y gives x 3 x y2 5 x 3 x y 2 5, so the graph is symmetric with respect to the xaxis. yaxis symmetry: replacing x by x gives x3 x y 2 5, which is different from the original equation, so the graph is not symmetric with respect to the yaxis. Origin symmetry: replacing x by x and y by y gives x3 x y2 5, which is different from the original equation, so the graph is not symmetric with respect to the origin. (b) To find xintercepts, we set y 0 and solve for x: x 3 x 02 5 x 3 5 x 3 5, so the xintercept is 3 5. To find yintercepts, we set x 0 and solve for y: 03 0y 2 5 has no solution, so there is no yintercept.
CHAPTER 1
31. (a) We graph y x 2 6x in the viewing rectangle [10 10] by [10 10].
32. (a) We graph y
97
Review
5 x in the viewing rectangle
[10 6] by [1 5].
10
4 2
10
5
5
10 10
10
(b) From the graph, we see that the xintercepts are 0
5
5
(b) From the graph, we see that the xintercept is 5 and
and 6 and the yintercept is 0.
the yintercept is approximately 224.
33. (a) We graph y x 3 4x 2 5x in the viewing rectangle [4 8] by [30 20]. 20
x2 x2 y2 1 y2 1 34. (a) We graph 4 4 x2 in the viewing rectangle [3 3] by y 1 4 [2 2].
4
2
2
4
6
8
2 1
20 3
(b) From the graph, we see that the xintercepts are 1,
2
1 1
1
2
3
2
0, and 5 and the yintercept is 0.
(b) From the graph, we see that the xintercepts are 2 and 2 and the yintercepts are 1 and 1.
35. (a) The line that has slope 2 and yintercept 6 has the slopeintercept equation
(c)
y
y 2x 6. (b) An equation of the line in general form is 2x y 6 0.
1 1
x
98
CHAPTER 1 Equations and Graphs
36. (a) The line that has slope 12 and passes through the point 6 3 has
(c)
y
equation y 3 12 x 6 y 3 12 x 6 y 12 x.
(b) 12 x 3 y 3 x 6 2y 6 x 2y 0.
1
37. (a) The line that passes through the points 3 2 and 1 4 has slope m
(c)
1
x
1
x
1
x
1
x
y
3 4 2 , so, using the second point for convenience, an 1 3 2
equation of the line is y 4 32 x 1 y 32 x 52 .
1
(b) y 32 x 52 2y 3x 5 3x 2y 5 0.
38. (a) The line that has xintercept 4 and yintercept 12 passes through the points
(c)
y
12 0 3 and the equation is 4 0 and 0 12, so m 04 y 0 3 x 4 y 3x 12. (b) y 3x 12 3x y 12 0. 2
39. (a) The vertical line that passes through the point 3 2 has equation x 3.
(c)
y
(b) x 3 x 3 0. 1
CHAPTER 1
40. (a) The horizontal line with yintercept 5 has equation y 5.
(c)
99
Review
y
(b) y 5 y 5 0. 1 1
41. (a) The line containing 2 4 and 4 4 has slope
(c)
x
y
4 4 8 m 4, and the line passing through the origin with 42 2 this slope has equation y 4x. 1
(b) y 4x 4x y 0.
42. (a) 2x 5y 10 5y 2x 10 y 25 x 2, so the given line has slope (c)
1
x
1
x
y
m 25 . Thus, an equation of the line passing through 1 1 parallel to this line is y 1 25 x 1 y 25 x 35 .
1
(b) y 25 x 35 5y 2x 3 2x 5y 3 0.
43. (a) The line y 12 x 10 has slope 12 , so a line perpendicular to this one has
(c)
y
1 2. In particular, the line passing through the origin slope 12
perpendicular to the given line has equation y 2x. (b) y 2x 2x y 0.
1 1
x
100
CHAPTER 1 Equations and Graphs
44. (a) The line with equation x 4y 7 0 4y x 7 y 14 x 74 has (c)
y
1 4. Since the slope 14 , so any line perpendicular to it has slope 14
desired line passes through 2 3, it has equation y 3 4 x 2 y 4x 11.
(b) y 4x 11 4x y 11 0. 1 1
x
45. The line with equation y 13 x 1 has slope 13 . The line with equation 9y 3x 3 0 9y 3x 3 y 13 x 13 also has slope 13 , so the lines are parallel.
46. The line with equation 5x 8y 3 8y 5x 3 y 58 x 38 has slope 58 . The line with equation 10y 16x 1 1 has slope 8 1 , so the lines are perpendicular. 10y 16x 1 y 85 x 10 5 58
47. (a) The slope, 03, represents the increase in length of the spring for each unit increase in weight . The sintercept is the resting or natural length of the spring. (b) When 5, s 03 5 25 15 25 40 inches.
48. (a) We use the information to find two points, 0 60000 and 3 70500. Then the slope is 70,500 60,000 10,500 m 3,500. So S 3,500t 60,000. 30 3 (b) The slope represents the accountant’s annual increase in salary, $3500, and the Sintercept represents the accountant’s initial salary, $60,000. (c) When t 12, the accountant’s salary will be S 3500 12 60,000 42,000 60,000 $102,000.
49. x 2 9x 14 0 x 7 x 2 0 x 7 or x 2.
50. x 2 24x 144 0 x 122 0 x 12 0 x 12.
51. 2x 2 x 1 2x 2 x 1 0 2x 1 x 1 0. So either 2x 1 0 2x 1 x 12 ; or x 1 0 x 1. 52. x 1 x 2 3 x 1 x 2 3 x 2 x 2 0 x 1 x 2 0 x 1 or x 2. However, x 1 fails to satisfy the original equation because 12 3 2, which is undefined. Thus, the only solution is x 2. 53. 0 4x 3 25x x 4x 2 25 x 2x 5 2x 5 0. So either x 0; or 2x 5 0 2x 5 x 52 ; or 2x 5 0 2x 5 x 52 .
54. x 3 2x 2 5x 10 0 x 2 x 2 5 x 2 0 x 2 x 2 5 0 x 2 or x 5.
55. 3x 2 4x 1 0 2 2 7 4 42 431 b b2 4ac 4 1612 42 7 7 4 28 6 2 . x 2a 6 6 6 3 23
3 32 419 2 4ac b b 3 2936 3 227 , which are not real numbers. 56. x 2 3x 9 0 x 2a 21
There is no real solution. 1 2 57. 3 x 1 2 x 3 x x 1 x 1 2x 3x 2 3x 0 3x 2 6x 1 x x 1 2 3 6 6 62 431 b b2 4ac 6 3612 6 24 62 6 3 6 . x 2a 6 6 6 6 3 23
CHAPTER 1
58.
Review
101
1 8 x 2 x x 2 x 2 8 x 2 2x x 2 8 x 2 3x 10 0 x 2 x 5 0 x 2 x 2 x 4 x 2 or x 5. However, since x 2 makes the expression undefined, we reject this solution. Hence the only solution is x 5.
59. x 4 8x 2 9 0 x 2 9 x 2 1 0 x 3 x 3 x 2 1 0 x 3 0 x 3, or x 3 0 x 3, however x 2 1 0 has no real solution. The solutions are x 3.
60. x 4 x 32. Let u x. Then u 2 4u 32 u 2 4u 32 0 u 8 u 4 0 So either u 8 0 or u 4 0. If u 8 0, then u 8 x 8 x 64. If u 4 0, then u 4 x 4, which has no real solution. So the only solution is x 64. 61. x 32 3 x 3 4 0 [x 3 4] [x 3 1] 0 x 7 x 2 0 x 2 or x 7. 62. 9x 12 6x 12 x 32 0 x 12 9 6x x 2 0 x 12 x 32 0 x 3.
63. x 7 4 x 7 4 x 7 4, so x 11 or x 3.
64. 2x 5 9 is equivalent to 2x 5 9 2x 5 9 x
59 . So x 2 or x 7. 2
65. (a) 2 3i 1 4i 2 1 3 4 i 3 i
(b) 2 i 3 2i 6 4i 3i 2i 2 6 i 2 8 i
66. (a) 3 6i 6 4i 3 6 [6 4] i 3 2i (b) 4i 2 12 i 8i 2i 2 8i 2 2 8i 4 2i 4 2i 2 i 8 8i 2i 2 6 8i 8 8i 2 65 85 i 2i 2i 2i 41 5 4 i2 (b) 1 1 1 1 1 i 1 i 1 i i i 2 1 1 2
67. (a)
2 3i 1 5i 2 5i 3i 2 2 3i 1 i 12 52 i 2 1i 2 1 i 1 i 1i (b) 10 40 i 10 2i 10 20i 2 20
68. (a)
69. x 2 16 0 x 2 16 x 4i 70. x 2 12 x 12 2 3i b 71. x 2 6x 10 0 x 72. 2x 2 3x 2 0 x
6 62 4 1 10 6 36 40 b2 4ac 3 i 2a 2 1 2
3
32 4 2 2 2 2
3
3 7 7 i 4 4 4
73. x 4 256 0 x 2 16 x 2 16 0 x 4 or x 4i 74. x 3 2x 2 4x 8 0 x 2 x 2 4 x 2 or x 2i
102
CHAPTER 1 Equations and Graphs
75. Let r be the athlete’s running speed, in mi/h. Then they cycle at r 8 mi/h. Rate
Cycle
r 8
Run
r
Time
Distance
4 r 8 25 r
4 25
25 4 1. Multiplying by 2r r 8, we r 8 r get 4 2r 25 2 r 8 2r r 8 8r 5r 40 2r 2 16r 0 2r 2 3r 40 Since the total time of the workout is 1 hour, we have
3 32 4240 3 9320 3 329 . Since r 0, we reject the negative value. The athlete’s running r 22 4 4 3 329 378 mi/h. speed is r 4 x2 1500 20x x 2 x 2 20x 1500 0 x 30 x 50 0. So 76. Substituting 75 for d, we have 75 x
20 x 30 or x 50. The speed of the car was 30 mi/h.
77. Let x be the length of one side in cm. Then 28 x is the length of the other side. Using the Pythagorean Theorem, we have x 2 28 x2 202 x 2 784 56x x 2 400 2x 2 56x 384 0 2 x 2 28x 192 0
2 x 12 x 16 0. So x 12 or x 16. If x 12, then the other side is 28 12 16. Similarly, if x 16, then the other side is 12. The sides are 12 cm and 16 cm. 80 and the total amount of fencing material is 78. Let l be length of each garden plot. The width of each plot is then l 480 80 4 l 6 88. Thus 4l 88 4l 2 480 88l 4l 2 88l 480 0 4 l 2 22l 120 0 l l 4 l 10 l 12 0. So l 10 or l 12. If l 10 ft, then the width of each plot is 80 10 8 ft. If l 12 ft, then the
width of each plot is 80 12 667 ft. Both solutions are possible.
80. 12 x 7x 12 8x 32 x. Interval: 32
79. 3x 2 11 3x 9 x 3. Interval: 3 . Graph:
-3
Graph:
81. 3 x 2x 7 10 3x 10 3 x Interval: 10 3 Graph:
3 2
82. 7 3x 1 1 6 3x 0 2 x 0 Interval: [2 0]. Graph:
10 3
_2
0
83. x 2 7x 8 0 x 8 x 1 0. The expression on the left of the inequality changes sign where x 8 and where x 1. Thus we must check the intervals in the following table. Interval Sign of x 8 Sign of x 1
Sign of x 8 x 1
1
1 8
8
Interval: 1 8 Graph:
_1
8
CHAPTER 1
Review
103
84. x 2 1 x 2 1 0 x 1 x 1 0. The expression on the left of the inequality changes sign when x 1 and x 1. Thus we must check the intervals in the following table. Interval: [1 1]
Interval Sign of x 1
1
1 1
1
Sign of x 1
Sign of x 1 x 1
85.
Graph:
_1
1
2x 5 2x 5 x 1 x 4 2x 5 1 1 0 0 0. The expression on the left of the inequality x 1 x 1 x 1 x 1 x 1 changes sign where x 1 and where x 4. Thus we must check the intervals in the following table. Interval Sign of x 4
4 1
1
defined at this value. Thus the solution is [4 1.
Graph:
Sign of x 1 Sign of
We exclude x 1, since the expression is not
4
x 4 x 1
_4
_1
86. 2x 2 x 3 2x 2 x 3 0 2x 3 x 1 0. The expression on the left of the inequality changes sign when 1 and 32 . Thus we must check the intervals in the following table. Interval
1
Sign of 2x 3
Sign of x 1
Sign of 2x 3 x 1
87.
1 32
3 2
Interval: 1] 32 Graph:
3 2
_1
x 4 x 4 0 0. The expression on the left of the inequality changes sign where x 2, where x 2, 2 2 x 2 x x 4 and where x 4. Thus we must check the intervals in the following table. Interval
Sign of x 4 Sign of x 2
2
2 2
2 4
4
Sign of x 2 x 4 Sign of x 2 x 2 Since the expression is not defined when x 2we exclude these values and the solution is 2 2 4]. Graph:
_2
2
4
104
88.
CHAPTER 1 Equations and Graphs
5 5 5 0 0 2 0. The 0 2 x 1 x 2 x 2 x x 1 4 x 1 x 1 x 4 expression on the left of the inequality changes sign when 2 1and 2. Thus we must check the intervals in the following table. 5
x 3 x 2 4x 4
Interval Sign of x 1
2
2 1
1 2
2
Sign of x 2 Sign of x 2
5 Sign of x 1 x 2 x 2 Interval: 2 1 2 Graph:
_2
1
2
89. x 5 3 3 x 5 3 2 x 8. Interval: [2 8] Graph:
2
90. x 4 002 002 x 4 002 398 x 402 Interval: 398 402
8
Graph:
3.98
4.02
91. 2x 1 1 is equivalent to 2x 1 1 or 2x 1 1. Case 1: 2x 1 1 2x 0 x 0. Case 2: 2x 1 1 2x 2 x 1. Interval: 1] [0 . Graph:
_1
0
92. x 1 is the distance between x and 1 on the number line, and x 3 is the distance between x and 3. We want those points that are closer to 1 than to 3. Since 2 is midway between 1 and 3, we get x 2 as the solution. Graph: 2
93. (a) For
24 x 3x 2 to define a real number, we must have 24 x 3x 2 0 8 3x 3 x 0. The expression
on the left of the inequality changes sign where 8 3x 0 3x 8 x 83 ; or where x 3. Thus we must
check the intervals in the following table. Interval Sign of 8 3x Sign of 3 x
Sign of 8 3x 3 x
3
3 83
8 3
Interval: 3 83 . Graph:
_3
8 3
CHAPTER 1
(b) For 4
1 x x4
Review
105
to define a real number we must have x x 4 0 x 1 x 3 0 x 1 x 1 x x 2 0.
The expression on the left of the inequality changes sign where x 0; or where x 1; or where 1 x x 2 0 12 411 x 1 21 1 214 which is imaginary. We check the intervals in the following table. Interval: 0 1. Interval
0
0 1
Sign of x
1
Sign of 1 x
Sign of 1 x x 2
Sign of x 1 x 1 x x 2
6 9 6 9 r3 3 r 3 . Thus r 94. We have 8 43 r 3 12
Graph: 0
3
1
6 3 9 .
95. From the graph, we see that the graphs of y x 2 4x and y x 6 intersect at x 1 and x 6, so these are the solutions of the equation x 2 4x x 6.
96. From the graph, we see that the graph of y x 2 4x crosses the xaxis at x 0 and x 4, so these are the solutions of the equation x 2 4x 0.
97. From the graph, we see that the graph of y x 2 4x lies below the graph of y x 6 for 1 x 6, so the inequality x 2 4x x 6 is satisfied on the interval [1 6].
98. From the graph, we see that the graph of y x 2 4x lies above the graph of y x 6 for x 1 and 6 x , so the inequality x 2 4x x 6 is satisfied on the intervals 1] and [6 .
99. From the graph, we see that the graph of y x 2 4x lies above the xaxis for x 0 and for x 4, so the inequality x 2 4x 0 is satisfied on the intervals 0] and [4 .
100. From the graph, we see that the graph of y x 2 4x lies below the xaxis for 0 x 4, so the inequality x 2 4x 0 is satisfied on the interval [0 4]. 101. x 2 4x 2x 7. We graph the equations y1 x 2 4x 102.
x 4 x 2 5. We graph the equations y1
x 4
and y2 2x 7 in the viewing rectangle [10 10] by
and y2 x 2 5 in the viewing rectangle [4 5] by
solutions x 1 and x 7.
solutions x 250 and x 276.
[5 25]. Using a zoom or trace function, we get the
10
5
[0 10]. Using a zoom or trace function, we get the
20
10
10
5
5
10
4
2
0
2
4
106
CHAPTER 1 Equations and Graphs
103. x 4 9x 2 x 9. We graph the equations y1 x 4 9x 2 104. x 3 5 2. We graph the equations and y2 x 9 in the viewing rectangle [5 5] by
y1 x 3 5 and y2 2 in the viewing rectangle
[25 10]. Using a zoom or trace function, we get the
[20 20] by [0 10]. Using Zoom and/or Trace, we get the
solutions x 272, x 115, x 100, and x 287.
solutions x 10, x 6, x 0, and x 4.
10 4
10
2 10
2
5
4
20
20
105. x 2 12 4x. We graph the equations y1 x 2 and
y2 12 4x in the viewing rectangle [8 4] by [0 40].
10
0
10
20
106. x 3 4x 2 5x 2. We graph the equations
y1 x 3 4x 2 5x and y2 2 in the viewing rectangle
Using a zoom or trace function, we find the points of
[10 10] by [5 5]. We find that the point of intersection
intersection are at x 6 and x 2. Since we want
is at x 507. Since we want x 3 4x 2 5x 2, the solution is the interval 507 .
x 2 12 4x, the solution is the union of intervals
6 2 .
4
40
2
30 20
10
5
10
2
5
10
4 8
6
4
2
0
2
4
108. x 2 16 10 0. We graph the equation y1 x 4 4x 2 and y2 12 x 1 in the viewing rectangle y x 2 16 10 in the viewing rectangle [10 10] by [5 5] by [5 5]. We find the points of intersection are [10 10]. Using a zoom or trace function, we find that the at x 185, x 060, x 045, and x 200. Since xintercepts are x 510 and x 245. Since we we want x 4 4x 2 12 x 1, the solution is want x 2 16 10 0, the solution is approximately 185 060 045 200. 510] [245 245] [510 .
107. x 4 4x 2 12 x 1. We graph the equations
4
10
2 4
2
2 4
2
4 10
5
5 10
10
CHAPTER 1
Review
107
109. Here the center is at 0 0, and the circle passes through the point 5 12, so the radius is r 5 02 12 02 25 144 169 13. The equation of the circle is x 2 y 2 132
x 2 y 2 169. The line shown is the tangent that passes through the point 5 12, so it is perpendicular to the line 12 12 0 . The slope of the line we seek is through the points 0 0 and 5 12. This line has slope m 1 5 0 5 5 1 1 5 x 5 y 12 5 x 25 m2 . Thus, an equation of the tangent line is y 12 12 12 12 m1 125 12 5 x 169 5x 12y 169 0. y 12 12
110. Because the circle is tangent to the xaxis at the point 5 0 and tangent to the yaxis at the point 0 5, the center is at 5 5 and the radius is 5. Thus an equation is x 52 y 52 52 x 52 y 52 25. The slope of 4 4 51 , so an equation of the line we seek is the line passing through the points 8 1 and 5 5 is m 58 3 3 y 1 43 x 8 4x 3y 35 0.
111. Since M varies directly as z we have M kz. Substituting M 120 when z 15, we find 120 k 15 k 8. Therefore, M 8z. k k 112. Since z is inversely proportional to y, we have z . Substituting z 12 when y 16, we find 12 k 192. y 16 192 Therefore z . y k 113. (a) The intensity I varies inversely as the square of the distance d, so I 2 . d k (b) Substituting I 1000 when d 8, we get 1000 k 64,000. 82 64,000 64,000 (c) From parts (a) and (b), we have I . Substituting d 20, we get I 160 candles. d2 202 114. Let f be the frequency of the string and l be the length of the string. Since the frequency is inversely proportional to the k 5280 k k 5280. Therefore f . For length, we have f . Substituting l 12 when k 440, we find 440 l 12 l 5280 l 5280 f 660, we must have 660 660 8. So the string needs to be shortened to 8 inches. l 115. Let be the terminal velocity of the parachutist in mi/h and be his weight in pounds. Since the terminal velocity is directly proportional to the square root of the weight, we have k . Substituting 9 when 160, we solve 9 for k. This gives 9 k 160 k 0712. Thus 0712 . When 240, the terminal velocity is 160 0712 240 11 mi/h. 116. Let r be the maximum range of the baseball and be the velocity of the baseball. Since the maximum range is directly proportional to the square of the velocity, we have r l 2 . Substituting 60 and r 242, we find 242 k 602
k 00672. If 70, then we have a maximum range of r 00672 702 3294 feet. k 117. The speed is inversely proportional to the square root of the density d, so . In fresh water with density d k 3 d1 1 gcm , the speed is 1 1480 ms, so 1480 k 1480. Thus, in seawater with density 10273 gcm3 , we 1 1480 have 1460 ms. 10273 118. (a) Because 122 , for fixed D the angular distance is directly proportional to the wavelength . Thus, the angular D distance is smaller for shorter wavelengths.
108
CHAPTER 1 Equations and Graphs
(b) For fixed , if we substitute D1 2D in the formula 122 then the angular distance becomes D 1 1 122 122 . Thus, for a fixed wavelength, if the diameter of the mirror is doubled, 1 122 D1 2D 2 D 2 the angular distance is halved. c c 1 z2 c 1 z2 c 119. We solve the first equation for : 1 z c c c 1 z2 1 . c 1 z2 1 z2 c 1 z2 1 c 1 z2 1 1 z2 1 3 105 1 22 1 8 5 24 105 kms. 3 10 Substituting z 2 and c 3 105 , we find 2 10 1 2 1 , so substituting from above and H0 208 Mlykms, we have By Hubble’s Law, H0 D D H0 24 105 11,538 megalightyears. 208 120. (a) The first compression wave is traveling at about 6 kms, so it reaches the seismograph station 120 km away in about D
120 20 seconds. 6
The first transverse wave is traveling at about 4 kms, so it reaches the seismograph station 120 km away in about 120 30 seconds. 4
Thus, the time difference is about 10 seconds. (b) If the epicenter is d km from the seismograph station, then the time difference x between the detection of compression d d d and transverse waves is x seconds. Thus, d 12x. This agrees with our result in part (a). 4 6 12 121. (a) y 2 x 3 is the graph of the absolute value function y x, stretched vertically by a factor of 2 and shifted downward 3 units. It has graph III (b) 2y 3x 2 y 32 x 1 represents a line with slope 32 , xintercept 23 , and yintercept 1. It has graph V.
(c) y x 4 has y 0 0 and y 0 elsewhere. It is an even function, and has graph II.
(d) x 12 y 12 9 is the equation of a circle with center 1 1 and radius 3. It has graph IV.
(e) 6y x 3 y 16 x 12 represents a line with slope 16 , xintercept 3, and yintercept 12 . It has graph I.
(f) y
6x is an odd function with y 0 0 and horizontal asymptote the xaxis. It has graph VII. 1 x4
(g) x y 3 is an odd function whose graph contains the points 1 1 and 1 1. It has graph VIII.
(h) x 2 2x y 2 4y 1 0 x 12 y 22 1 1 4 4 is the equation of a circle with center 1 2 and radius 2. It has graph VI.
CHAPTER 1
Test
109
CHAPTER 1 TEST 1. (a)
y
There are several ways to determine the coordinates of S. The diagonals of a
S
P
square have equal length and are perpendicular. The diagonal P R is horizontal and has length is 6 units, so the diagonal QS is vertical and also has length 6.
R
Thus, the coordinates of S are 3 6. (b) The length of P Q is 0 32 3 02 18 3 2. So the area of 2 P Q RS is 3 2 18.
1 1
Q
x
y
2. (a)
(b) The xintercepts occur when y 0, so 0 4 x 2 x 2 4 x 2. The yintercept occurs when x 0, so y 4.
(c) xaxis symmetry: y 4 x 2 y x 2 4, which is not the same as the original equation, so the graph is not symmetric with respect to the xaxis.
1 x
1
yaxis symmetry: y 4 x2 4 x 2 , which is the same as the original
equation, so the graph is symmetric with respect to the yaxis.
Origin symmetry: y 4 x2 y 4 x 2 , which is not the same
as the original equation, so the graph is not symmetric with respect to the origin. 3. (a)
(b) The distance between P and Q is d P Q 3 52 1 62 64 25 89. 3 5 1 6 (c) The midpoint is 1 72 . 2 2 (d) The center of the circle is the midpoint, 1 72 , and the length of the radius is 1 89 . Thus the equation of the circle whose diameter is P Q is 2 2 2 2 x 12 y 72 12 89 x 12 y 72 89 4.
y Q
P
1 1
4.
x
(a) x 2 y 2 5 has center 0 0 and radius 5.
(b) x 12 y 32 4 22 has
(c) x 2 y 2 10x 16 0
y
y
center 5 0 and radius 3.
center 1 3 and radius 2.
x 52 y 2 16 25 9 has y
1
1 1
x
1
x
1 1
x
110
CHAPTER 1 Equations and Graphs
5. (a) x 4 y 2 . To test for symmetry with respect to the xaxis, we replace y
y
with y: x 4 y2 x 4 y 2 , so the graph is symmetric with
respect to the xaxis.
To test for symmetry with respect to the yaxis, we replace x with x:
1
x 4 y 2 is different from the original equation, so the graph is not
1
x
1
x
symmetric with respect to the yaxis.
For symmetry with respect to the origin, we replace x with x and y with
y: x 4 y2 x 4 y 2 , which is different from the
original equation, so the graph is not symmetric with respect to the origin. To find xintercepts, we set y 0 and solve for x: x 4 02 4, so the xintercept is 4.
To find yintercepts, we set x 0 and solve for y:: 0 4 y 2 y 2 4 y 2, so the yintercepts are 2 and 2.
(b) y x 2. To test for symmetry with respect to the xaxis, we replace y
y
with y: y x 2 is different from the original equation, so the
graph is not symmetric with respect to the xaxis.
To test for symmetry with respect to the yaxis, we replace x with x:
1
y x 2 x 2 is different from the original equation, so the
graph is not symmetric with respect to the yaxis.
To test for symmetry with respect to the origin, we replace x with x and
y with y: y x 2 y x 2, which is different from the
original equation, so the graph is not symmetric with respect to the origin. To find xintercepts, we set y 0 and solve for x: 0 x 2 x 2 0 x 2, so the xintercept is 2.
To find yintercepts, we set x 0 and solve for y: y 0 2 2 2, so the yintercept is 2.
6. (a) To find the xintercept, we set y 0 and solve for x: 3x 5 0 15
(b)
y
3x 15 x 5, so the xintercept is 5.
To find the yintercept, we set x 0 and solve for y: 3 0 5y 15 5y 15 y 3, so the yintercept is 3.
(c) 3x 5y 15 5y 3x 15 y 35 x 3.
1 1
x
(d) From part (c), the slope is 35 .
(e) The slope of any line perpendicular to the given line is the negative 1 5. reciprocal of its slope, that is, 35 3
7. (a) The line through 6 7 and 1 3 has slope 2x y 5 0.
3 7 2, so an equation is y 7 2 x 6 y 2x 5 16
(b) 3x y 10 0 y 3x 10, so the slope of the line we seek is 3. Using the pointslope form, y 6 3 x 3 y 6 3x 9 3x y 3 0. x y (c) Using the intercept form we get 1 2x 3y 12 2x 3y 12 0. 6 4
CHAPTER 1
8. (a) When x 100 we have T 008 100 4 8 4 4, so the
(b)
Test
111
T
temperature at one meter is 4 C.
(c) The slope represents the raise in temperature as the depth increase.
5
The T intercept is the surface temperature of the soil and the xintercept represents the depth of the “frost line”, where the soil 20
below is not frozen.
40
60
80
100 120 x
_5
9. (a) x 5 14 12 x 2x 10 28 x 3x 18 x 6 (b)
2x 2x 1 2x x 2x 1 x 1 (x 1, x 0) 2x 2 2x 2 x 1 0 x 1 x 1 x 1 x
(c) x 2 x 12 0 x 4 x 3 0. So x 4 or x 3. 4 42 4 2 1 4 16 8 4 8 4 2 2 2 2 . (d) 2x 2 4x 1 0 x 2 2 4 4 4 2 (e) 3 x 5 2 3 x 5 4 1 x 5. (Note that this is impossible, so there can be no solution.) Squaring both sides again, we get 1 x 5 x 4. But this does not satisfy the original equation, so there is no solution. (You must always check your final answers if you have squared both sides when solving an equation, since extraneous answers may be introduced, as here.) (f) x 4 3x 2 2 0 x 2 1 x 2 2 0. So x 2 1 0 x 1 or x 2 2 0 x 2. Thus the solutions are x 1, x 1, x 2, and x 2. 10 10 10 2 (g) 3 x 4 10 0 3 x 4 10 x 4 10 3 x 4 3 x 4 3 . So x 4 3 3 or 22 2 22 x 4 10 3 3 . Thus the solutions are x 3 and x 3 .
10. (a) 3 2i 4 3i 3 4 2i 3i 7 i
(b) 3 2i 4 3i 3 4 2i 3i 1 5i
(c) 3 2i 4 3i 3 4 3 3i 2i 4 2i 3i 12 9i 8i 6i 2 12 i 6 1 18 i
3 2i 4 3i 12 17i 6i 2 6 17 12 17i 6 3 2i i 4 3i 4 3i 4 3i 16 9 25 25 16 9i 2 24 124 1 (e) i 48 i 2 2 (f) 2 2 8 2 2 8 2 2 2 8 2 4 2i 4i 2 6 2i (d)
4 11. Using the Quadratic Formula, 2x 2 4x 3 0 x
4 8 42 4 2 3 1 22 i. 2 2 4
12. Let be the width of the parcel of land. Then 70 is the length of the parcel of land. Then 2 702 1302
2 2 140 4900 16,900 22 140 12,000 0 2 70 6000 0 50 120 0. So 50 or 120. Since 0, the width is 50 ft and the length is 70 120 ft.
13. (a) 4 5 3x 17 9 3x 12 3 x 4. Expressing in standard form we have: 4 x 3. Interval: [4 3. Graph:
_4
3
112
CHAPTER 1 Equations and Graphs
(b) x x 1 x 2 0. The expression on the left of the inequality changes sign when x 0, x 1, and x 2. Thus we must check the intervals in the following table. Interval Sign of x Sign of x 1 Sign of x 2
Sign of x x 1 x 2
2
2 0
0 1
1
From the table, the solution set is x 2 x 0 or 1 x. Interval: 2 0 1 . Graph:
_2
0
1
(c) x 4 3 is equivalent to 3 x 4 3 1 x 7. Interval: 1 7. Graph: (d)
1
7
3 2x x 2 x 3 2x 2x x 2 3 2x 3 2x x x 0 0 0 2x 2x 2x 2x 2x x 2 4x 3 x 1 x 3 0. The expression on the left of the inequality changes sign at x 1, x 2, and x 2 x 2 x 3. Thus we must check the intervals in the following table. Interval
1
1 2
2 3
3
Sign of x 1 Sign of x 3
Sign of x 2 x 1 x 3 Sign of x 2 Intervals: 1 2 3
Graph:
1
2
3
14. 5 59 F 32 10 9 F 32 18 41 F 50. Thus the medicine is to be stored at a temperature between 41 F and 50 F.
15. For 6x x 2 to be defined as a real number 6x x 2 0 x 6 x 0. The expression on the left of the inequality changes sign when x 0 and x 6. Thus we must check the intervals in the following table. Interval
Sign of x Sign of 6 x
0
0 6
6
Sign of x 6 x From the table, we see that 6x x 2 is defined when 0 x 6.
Fitting Lines to Data
16. (a) x 3 9x 1 0. We graph the equation
113
(b) x 2 1 x 1. We graph the equations
y x 3 9x 1 in the viewing rectangle [5 5]
y1 x 2 1 and y2 x 1 in the viewing
by [10 10]. We find that the points of
rectangle [5 5] by [5 10]. We find that the
intersection occur at x 294, 011, 305.
points of intersection occur at x 1 and x 2. Since we want x 2 1 x 1, the solution is
10
the interval [1 2].
10 4
2
2
4 5
10 4
2
2
4
5
17. (a) M k
h 2 L
4 62
h 2 (b) Substituting 4, h 6, L 12, and M 4800, we have 4800 k k 400. Thus M 400 . 12 L 3 102 12,000. So the beam can support 12,000 pounds. (c) Now if L 10, 3, and h 10, then M 400 10
FOCUS ON MODELING Fitting Lines to Data
1. (a) Using a graphing calculator, we obtain the regression line y 18807x 8265.
we get y 18807 58 8265 1917 cm.
y
180
160
140 40
(b) Using x 58 in the equation y 18807x 8265,
50
Femur length (cm)
x
114
FOCUS ON MODELING
2. (a) Using a graphing calculator,
(b) For a per capita GDP of $80,000, the model predicts
we obtain the regression line y 017565x 18265.
y
carbon emissions of y 017565 80 18265 159 tons per capita. For $32,000, it predicts
14 12 10 8 6 4 2
y 017565 32 18265 74 tons per capita. (c) A linear model is reasonable for these data. One limitation is that there are no data points beyond percapita income above $51,000, so the model cannot reliably make predictions for GDPs much larger than $51,000.
0
10
60 x
20 30 40 50 GDP per capita ($000)
3. (a) Using a graphing calculator, we obtain the regression line y 6451x 01523.
(b) Using x 18 in the equation y 6451x 01523, we get y 6451 18 01523 116 years.
y 100
80 60 40 20 0
2
4
6
8
10
12
14
16
18
20 x
Diameter (in.)
4. (a) Using a graphing calculator, we obtain the regression
y 4857x 22097, we get y 265 chirps per
line y 4857x 22097.
minute.
y
200
100
0
50
60
(b) Using x 100 F in the equation
70
80
Temperature (°F)
90
x
Fitting Lines to Data
5. (a) Using a graphing calculator, we obtain the regression line y 013198x 72514
115
(b) Using x 25 in the regression line equation, we get y 013198 25 72514 395 million km2 .
y
This is roughly 10% less than the actual figure of
8
44 million km2 . (c) Despite fluctuations over brief periods, the model
6
seems fairly accurate. If external circumstances 4
change (reduced or increased CO2 emissions, for example), it may become less reliable. It is unlikely
2
to be accurate far into the future.
0
10
x
20
Years since 1994 6. (a) Using a graphing calculator, we obtain the regression line y 0168x 1989.
(b) Using the regression line equation y 0168x 1989, we get y 813% when
x 70%.
y
20
10
0
20
40
60
80
100 x
Flow rate (%) 7. (a) Using a graphing calculator, we obtain
(b) The correlation coefficient is r 098, so linear model is appropriate for x between 80 dB and
y 39018x 4197.
104 dB.
y
(c) Substituting x 94 into the regression equation, we
100
get y 39018 94 4197 53. So the intelligibility is about 53%.
50
0
80
90
100
110 x
Noise level (dB) 8. Students should find a fairly strong correlation between shoe size and height. 9. Results will depend on student surveys in each class.
Corrections: 19, 20, 33, 42, 70, 77, 79, 88, 101 NOTE THAT THESE CORRECTIONS SHOULD BE MADE IN ALL THREE BOOKS.
CHAPTER 2
FUNCTIONS
2.1 2.2
Functions 1 Graphs of Functions 10
2.3
Getting Information from the Graph of a Function 22
2.4
Average Rate of Change of a Function 33
2.5 2.6 2.7
Linear Functions and Models 38 Transformations of Functions 44 Combining Functions 59
2.8
One-to-One Functions and Their Inverses 69 Chapter 2 Review 81 Chapter 2 Test 95
¥
FOCUS ON MODELING: Modeling with Functions 99
1
2
FUNCTIONS
2.1
FUNCTIONS
1. If f x x 3 1, then (a) the value of f at x 1 is f 1 13 1 0. (b) the value of f at x 2 is f 2 23 1 9.
(c) the net change in the value of f between x 1 and x 2 is f 2 f 1 9 0 9.
2. For a function f , the set of all possible inputs is called the domain of f , and the set of all possible outputs is called the range of f . x 5 have 5 in their domain because they are defined when x 5. However, 3. (a) f x x 2 3x and g x x h x x 10 is undefined when x 5 because 5 10 5, so 5 is not in the domain of h. 0 55 (b) f 5 52 3 5 25 15 10 and g 5 0. 5 5 4. (a) Verbal: “Subtract 4, then square and add 3.” (b) Numerical: x
f x
0
19
2
7
4
3
6
7
5. A function f is a rule that assigns to each element x in a set A exactly one element called f x in a set B. Diagram (a) does not represent a function because the input value 5 in set A has two output values in set B. Diagram (b) does represent a function because each number in set A has exactly one output value in set B. 6. (a) From the table, f 1 4 and f 2 4.
(b) Yes, a function can have the same output for two different inputs. (However, no input can have more than one possible output.)
7. Yes, it is possible that f 1 f 2 5. [For instance, let f x 5 for all x.]
8. No, it is not possible to have f 1 5 and f 1 6. A function assigns each value of x in its domain exactly one value of f x. 9. Multiplying x by 3 gives 3x, then subtracting 5 gives f x 3x 5.
10. Adding 2 gives x 2, then multiplying by 5 gives f x 5 x 2.
11. Squaring gives x 2 , adding 1 gives x 2 1, then taking the square root gives f x
x 2 1. x 1 . 12. Adding 1 gives x 1, taking the square root gives x 1, then dividing by 6 gives f x 6 13. 5x 1: Multiply by 5, then add 1. 14. 4 x 2 2 : Square, subtract 2, then multiply by 4. x 4 : Take the square root, subtract 4, then divide by 3. 15. 3
1
2
CHAPTER 2 Functions
16.
x2 9 : Square, then add 9, then take the square root, then divide by 2. 2
17. Machine diagram for f x
x 1.
18. Machine diagram for f x
1
subtract 1, then take square root
0
2
subtract 1, then take square root
1
5
subtract 1, then take square root
2
19. f x 2 x 12
3 . x 2
3
subtract 2, take reciprocal, multiply by 3
3
_1
subtract 2, take reciprocal, multiply by 3
_1
1
subtract 2, take reciprocal, multiply by 3
_3
20. g x 2x 3
x
f x
x
g x
1
2 1 12 8
3
2 3 3 3
0 1 2 3
2 12 2
2 1 12 0 2 2 12 2 2 3 12 8
2
2 2 3 1
1
2 1 3 5
0 3
2 0 3 3 2 3 3 9
21. f x 3x 2 1; f 2 3 22 1 12 1 13; f 2 3 22 1 12 1 13; f 0 3 02 1 1; 2 2 5 3 5 1 15 1 16. f 13 3 13 1 3 19 1 43 ; f
22. f x 4x x 3 ; f 2 4 2 23 8 8 0; f 0 4 0 03 0; f 2 4 2 23 8 8 0; 3 f 1 4 1 13 3; f 12 4 12 12 2 18 15 8 .
1 2 3 10 1 12 1 1 a a1 1x ; g 2 ; g 0 ; g 2 ; g a ; 5 5 5 5 5 5 5 5 5 1 x2 1 a 2 3a g x2 ; g a 2 . 5 5 5 x 3 1 3 2 03 3 13 a3 ; h 1 ; h 0 ; h 1 1; h a ; 24. h x 2 2 2 2 2 2 2 a2 2 3 x 1 2 a2 1 x 2 3 h x 2 ;h a 2 . 2 2 2 2 23. g x
25. f x x 2 2x; f 0 02 2 0 0; f 3 32 2 3 9 6 15; f 3 32 2 3 9 6 3; 2 1 2 1 1 1 2 . 2 f a a 2 2 a a 2 2a; f x x2 2 x x 2 2x; f a a a a a 1 1 1 1 2; h 2 2 1 5 ; h 1 1 1 1 2 5 ; ; h 1 1 1 2 2 2 1 t 2 2 2 2 1 1 1 1 1 ;h h x 1 x 1 x. 1 x 1 x x x x
26. h t t
SECTION 2.1 Functions
3
1 1 1 1 2 1 1 1 1 1x 1 1 2 2 ; ; g 2 ; g 1 , which is undefined; g 3 1 1x 1 2 3 3 1 1 2 3 1 2 2 1 x2 1 2 x2 1 a 1a 1 a 1 1a1 2a 2 g a ; g a 1 ;g x 1 . 1 a 1a 1 a 1 1a1 a 1 x2 1 x2 t 2 2 2 22 02 a2 28. g t ; g 2 0; g 2 , which is undefined; g 0 1; g a ; t 2 2 2 22 02 a2 a2 2 2 a3 a2 a12 . 2 ; g a 1 g a2 2 2 a 1 2 a 1 a 22 a 4 29. k x 3x 2 x 1; k 1 3 12 1 1 5; k 0 3 02 0 1 1; k 2 3 22 2 1 11; 2 k 5 3 5 5 1 16 5; k a 1 3 a 12 a 1 1 3a 2 6a 3 a 1 1 3a 2 7a 5; 2 k x 2 3 x 2 x 2 1 3x 4 x 2 1. 27. g x
30. k x x 4 x 3 ; k 2 24 23 16 8 24; k 1 14 13 1 1 2; 4 3 a 4 a 3 a a3 a 4 3a 3 2 a4 ; k a a2 a2 a8 a6 ; k 1 14 13 0; k 3 3 3 81 27 81 4 3 1 1 1 1 1t 1 k 4 3 4 . t t t t t t 31. f x 2 x 1; f 2 2 2 1 2 3 6; f 0 2 0 1 2 1 2; f 12 2 12 1 2 12 1; f 2 2 2 1 2 1 2; f x 1 2 x 1 1 2 x; f x 2 2 2 x 2 2 1 2 x 2 1 2x 2 2 (since x 2 1 0 ). 2 1 x 2 1 ; f 2 1; f 1 1; f x is not defined at x 0; x 2 2 1 1 x 2 1x 5 5 x x2 1 . 1; f x 2 2 2 1 since x 2 0, x 0; f f 5 x 5 5 x 1x x x
32. f x
33. Since 5 5, we have f 5 3 5 1 14. Since 0 5, we have f 0 3 0 1 1. Since 13 5, we have f 13 3 13 1 2. Since 5 5, we have f 5 52 1 24. Since 6 5, we have f 6 62 1 35. 34. Since 3 2, we have f 3 5. Since 0 2, we have f 0 5. Since 2 2, we have f 2 5. Since 3 2, we have f 3 2 3 3 3. Since 5 2, we have f 5 2 5 3 7.
35. Since 4 1, we have f 4 42 2 4 16 8 8. Since 32 1, we have 2 f 32 32 2 32 94 3 34 . Since 1 1, we have f 1 12 2 1 1 2 1. Since
1 0 1, we have f 0 0. Since 25 1, we have f 25 1. 36. Since 5 0, we have f 5 3 5 15. Since 0 0 2, we have f 0 0 1 1. Since 0 1 2, we have f 1 1 1 2. Since 0 2 2, we have f 2 2 1 3. Since 5 2, we have f 5 5 22 9.
37. f x 2 x 22 1 x 2 4x 4 1 x 2 4x 5; f x f 2 x 2 1 22 1 x 2 1 4 1 x 2 6. 38. f 2x 3 2x 1 6x 1; 2 f x 2 3x 1 6x 2. 2 39. f x 2 x 2 4; f x [x 4]2 x 2 8x 16. x x f x 6x 18 3 2x 6 6 18 2x 18; 2x 6 40. f 3 3 3 3 3 41. f x 3x 2, so f 1 3 1 2 1 and f 5 3 5 2 13. Thus, the net change is f 5 f 1 13 1 12. 42. f x 4 5x, so f 3 4 5 3 11 and f 5 4 5 5 21. Thus, the net change is f 5 f 3 21 11 10.
4
CHAPTER 2 Functions
43. g t 1 t 2 , so g 2 1 22 1 4 3 and g 5 1 52 24. Thus, the net change is g 5 g 2 24 3 21. 44. h t t 2 5, so h 3 32 5 14 and h 6 62 5 41. Thus, the net change is h 6h 3 4114 27. 45. f a 3 a; f a h 3 a h 3 a h;
h f a h f a 3 a h 3 a 1. h h h
46. f a a 2 4a; f a h a h2 4 a h a 2 2ah h 2 4a 4h a 2 4a 2ah 4h h 2 ; a 2 2ah h 2 4a 4h a 2 4a f a h f a 2ah h 2 4h 2a 4 h h h h 47. f a 5; f a h 5;
55 f a h f a 0. h h
1 1 ; f a h ; a1 ah1 a1 ah1 1 1 f a h f a a 1 a h 1 a 1 a h 1 ah1 a1 h h h h 1 a 1 a h 1 . h a 1 a h 1
48. f a
ah a ; f a h ; a1 ah1 a a h 1 a h a 1 a ah f a h f a a h 1 a 1 a h 1 a 1 a h 1 a 1 h h h a h a 1 a a h 1 2 a a ah h a 2 ah a a h 1 a 1 h h a h 1 a 1 1 a h 1 a 1
49. f a
ah1 a1 ; f a h ; a ah a h 1 a a 1 a h ah1 a1 a 2 ah a a 2 ah a h f a h f a a a h a h a h h h ah a h h 1 ah a h a a h
50. f a
51. f a 3 5a 4a 2 ;
f a h 3 5 a h 4 a h2 3 5a 5h 4 a 2 2ah h 2
3 5a 5h 4a 2 8ah 4h 2 ; 2 8ah 4h 2 3 5a 4a 2 3 5a 5h 4a f a h f a h h 3 5a 5h 4a 2 8ah 4h 2 3 5a 4a 2 5h 8ah 4h 2 h h h 5 8a 4h 5 8a 4h. h
SECTION 2.1 Functions
5
52. f a a 3 ; f a h a h3 a 3 3a 2 h 3ah 2 h 3 ; a 3 3a 2 h 3ah 2 h 3 a 3 f a h f a 3a 2 h 3ah 2 h 3 h h h 2 2 h 3a 3ah h 3a 2 3ah h 2 . h 53. f x 3x. Since there is no restriction, the domain is all real numbers, . Since every real number y is three times the real number 13 y, the range is all real numbers .
54. f x 5x 2 4. Since there is no restriction, the domain is all real numbers, . Since 5x 2 0 for all x, 5x 2 4 4 for all x, so the range is [4 .
55. f x x 3. Since there is no restriction, the domain is all real numbers, . Since x 0 for all x, x 3 3 for all x, so the range is [3 . 56. f x 2 x 1. We must have x 1 0 x 1, so the domain is [1 . Since x 1 0 for all x, 2 x 1 2, so the range is [2 .
57. f x 3x, 2 x 6. The domain is [2 6], f 2 3 2 6, and f 6 3 6 18, so the range is [6 18].
58. f x 5x 2 4, 0 x 2. The domain is [0 2], f 0 5 02 4 4, and f 2 5 22 4 24, so the range is [4 24]. 2 59. f x . Since the denominator cannot equal 0, we have 3 x 0 x 3. Thus the domain is x x 3. In 3x interval notation, the domain is 3 3 . x . Since the denominator cannot equal 0, we have 4 x 0 x 4. Thus the domain is x x 4. In 60. f x 4x interval notation, the domain is 4 4 . x 2 . Since the denominator cannot equal 0, we have x 2 1 0 x 2 1 x 1. Thus the domain is 61. f x 2 x 1 x x 1. In interval notation, the domain is 1 1 1 1 . x4 62. f x 2 . Since the denominator cannot equal 0, x 2 x 6 0 x 3 x 2 0 x 3 or x 2. x x 6 In interval notation, the domain is 3 3 2 2 . 63. f t 2 t. We must have 2 t 0 t 2. Thus, the domain is 2]. 64. g t t 2 9. The argument of the square root is positive for all t, so the domain is . 65. f t 3 2t 5. Since the odd root is defined for all real numbers, the domain is the set of real numbers, . 66. g x 7 3x. For the square root to be defined, we must have 7 3x 0 7 3x 73 x. Thus the domain is 73 . 67. f t t 2 25. Since the square root is defined as a real number only for nonnegative numbers, we require that
t 2 25 0 t 5. So the domain is t t 5 or t 5. In interval notation, the domain is 5] [5 . 68. g t 36 t 2 . We must have 36 t 2 0 t 6. Thus the domain is [6 6]. 2x 69. g x . We require 2 x 0, and the denominator cannot equal 0. Now 2 x 0 x 2, and 3 x 0 3x x 3. Thus the domain is x x 2 and x 3, which can be expressed in interval notation as [2 3 3 . x 70. g x 2 . We must have x 0 for the numerator and 2x 2 x 1 0 for the denominator. So 2x 2 x 1 0 2x x 1 2x 1 x 1 0 2x 1 0 or x 1 0 x 12 or x 1. Thus the domain is 0 12 12 .
6
CHAPTER 2 Functions
71. g x a table:
4
x 2 6x. Since the input to an even root must be nonnegative, we have x 2 6x 0 x x 6 0. We make
Sign of x Sign of x 6
Sign of x x 6
0
0 6
6
Thus the domain is 0] [6 . 72. g x x 2 2x 8. We must have x 2 2x 8 0 x 4 x 2 0. We make a table: Sign of x 4 Sign of x 2
Sign of x 4 x 2
2
2 4
4
Thus the domain is 2] [4 . 4 . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 73. f x 2x 2 x 0 x 2. Thus the domain is 2. 3x . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 74. f x x 2 x 2 0 x 2. Thus the domain is 2 .
x 12 . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 75. f x 2x 1 2x 1 0 x 12 . Thus the domain is 12 .
x 76. f x . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 4 9 x2 9 x 2 0 3 x 3 x 0. We make a table: Interval Sign of 3 x Sign of 3 x Thus the domain is 3 3.
Sign of x 4 x 2
3
3 3
3
77. To evaluate f x, square the input and add 1 to the result. (a) f x x 2 1
(c)
y
(b) x
f x
2
5
1
2
0
1
1
2
2
5
1 1
x
SECTION 2.1 Functions
78. To evaluate f x, add 2 to the input and square the result. (a) f x x 22
y
(c)
(b) x
g x
5
9 1
3
0
2
1
1
1
1
1
9
x
79. Let T x be the amount of sales tax charged in Lemon County on a purchase of x dollars. To find the tax, take 8% of the purchase price. (a) T x 008x
(c)
y
(b) x
T x
2
016
4
032
6
048
8
064
1l
1
x
80. Let V d be the volume of a sphere of diameter d. To find the volume, take the cube of the diameter, then multiply by and divide by 6. 3 (a) V d d 3 6 6d
(c)
y
(b) x
f x
2
4 42 3 32 335 3
4 6 8
81. f x
36 113
256 268 3
10l 1
x
1 if x is rational
The domain of f is all real numbers, since every real number is either rational or
1
The domain of f is all real numbers, since every real number is either rational or
5 if x is irrational
irrational; and the range of f is 1 5. 82. f x
if x is rational 5x if x is irrational
irrational. If x is irrational, then 5x is also irrational [see Exercise P.2.88(b).] Also, if x is irrational, then f x5 x, so every irrational number is in the range of f . Thus, the range of f is x x 1 or x is irrational.
7
8
CHAPTER 2 Functions
0 2 50 and V 20 50 1 20 2 0. 83. (a) V 0 50 1 20 20
(c)
(b) V 0 50 represents the volume of the full tank at time t 0, and
V 20 0 represents the volume of the empty tank twenty minutes
later.
(d) The net change in V as t changes from 0 minutes to 20 minutes is V 20 V 0 0 50 50 gallons.
x
V x
0
50
5
28125
10
125
15
3125
20
0
84. (a) S 2 4 22 16 5027, S 3 4 32 36 11310.
(b) S 2 represents the surface area of a sphere of radius 2, and S 3 represents the surface area of a sphere of radius 3. 05c2 075c2 866 m, L 075c 10 1 661 m, and 85. (a) L 05c 10 1 2 c c2 09c2 436 m. L 09c 10 1 c2 (b) It will appear to get shorter. a 4 567 108 W W T , where a 567 108 3004 1462 2 , , so L 300 4 2 m m K 567 108 W 567 108 W L 350 3504 2708 2 , and L 1000 10004 18,050 2 . m m (b) The radiance increases dramatically (to the fourth power) as temperature increases.
86. (a) L T
(c) The sun’s radiance is approximately L 5778 87. (a) R 1
R 10
13 7 104
1 4 104 13 7 1004
R 100
1 4 1004
567 108 W 57784 201 107 2 . m
20 2 mm, 5
(b)
166 mm, and
13 7 10004
148 mm. 1 4 10004
(c) The net change in R as x changes from 10 to 100 is R 100 R 10 148 166 018 mm. 88. (a) 01 18500 025 012 4440, 04 18500 025 042 1665.
(b) They tell us that the blood flows much faster (about 275 times faster) 01 cm from the center than 01 cm from the edge.
(d) The net change in V as r changes from 01 cm to 05 cm is V 05 V 01 0 4440 4440 cms. 89. (a) D 01 2 3960 01 012 79201 281 miles D 02 2 3960 02 022 158404 398 miles
x
R x
1
2
10
166
100
148
200
144
500
141
1000
139
r
r
0
4625
01
4440
02
3885
03
2960
04
1665
05
0
(c)
SECTION 2.1 Functions
9
(b) 1135 feet 1135 miles 0215 miles. D 2 3960 0215 02152 1702846 413 miles 0215 5280 (c) D 7 2 3960 7 72 55489 2356 miles
(d) The net change in D as h changes from 1135 ft (or 0215 mi) to 7 mi is D 7 D 0215 2356413 1943 miles.
90. (a) From the table, P 1960 13 million, P 1980 27 million, and P 2020 72 million.
(b) The net change in population from 1960 to 1980 was P 1980 P 1960 27 13 14 million and the net change from 1980 to 2020 was P 2020 P 1980 72 27 45 million.
91. (a) Since 0 5000 10,000 we have T 5000 0. Since 10,000 12,000 20,000 we have T 12,000 008 12,000 10,000 160. Since 20,000 25,000 we have T 25,000 800 015 25,000 20,000 1550. (b) There is no tax on $5000, a tax of $160 on $12,000 income, and a tax of $1550 on $25,000.
92. (a) C 25 25 9 $34; C 45 45 9 $54; C 50 $50; and C 65 $65. (b) The answers represent the total prices of the goods purchased, including shipping. 114x if 0 x 2 93. (a) T x 228 99 x 2 if x 2
(b) T 2 114 2 $228; T 3 228 99 3 2 $327; and T 5 228 99 5 2 $525.
(c) The answers represent the total costs of the lodgings. 15 40 x if 0 x 40 94. (a) F x 0 if 40 x 65 15 x 65 if x 65
(b) F 30 15 40 10 15 10 $150; F 50 $0; and F 75 15 75 65 15 10 $150. (c) The answers represent fines for violating speed limits on the freeway.
T (¡F) 400
96.
95. We assume the grass grows linearly. h (in.) 2
300 200 100
1
0
97.
0
W
W
W
T (¡F)
W
1
t (h)
t (d)
98.
V ($000) 20
50
10
0
noon
midnight Time of day
0
10
20
30
t (yr)
If the car becomes a collectible antique, the value begins to rise again. 99. Answers will vary. 100. Answers will vary. 101. Answers will vary.
10
CHAPTER 2 Functions
2.2
GRAPHS OF FUNCTIONS
1. To graph the function f we plot the points x f x in a coordinate plane. To graph f x x 2 2, we plot the points x x 2 2 . So, the point 3 32 2 3 7 is
on the graph of f . The height of the graph of f above the
x
f x
x y
2
2
2 2
1
1
1 1
0
xaxis when x 3 is 7.
0
1
x
1 1
1
2
1
0 2
2
1
y
2
2 2
2. If f 4 10 then the point 4 10 is on the graph of f . 3. If the point 3 7 is on the graph of f , then f 3 7.
4. (a) f x x 2 is a power function with an even exponent. It has graph IV. (b) f x x 3 is a power function with an odd exponent. It has graph II. (c) f x x is a root function. It has graph I. (d) f x x is an absolute value function. It has graph III.
5. Because the input 1 has two different outputs (1 and 2), this set of ordered pairs does not define y as a function of x. 6. For any positive x, there are two possible values of y, since x 4y 2 14 x y 2 y 12 x. Thus, the equation does not define y as a function of x.
7. Because the input x 10 corresponds to outputs y 10 and y 15, this table does not define y as a function of x. 8. This graph fails the Vertical Line Test, and so does not define y as a function of x. 9.
10.
y
x
f x x 2
6
4
2
4
2
1
6
0
4
1
2
2
0 2
2
0
x
1
0
2
2
4
4
6
3
6
8
4
11. x 3 2
0
f x x 3, 3 x 3
1
x
12.
y
x
6 5 3
1
2
2
1
3
0
0 1
1 1
x
2 3
y
f x 4 2x 8
1 x
1
4 x 3 , 2 0x 5
f x
15 1 05 0
4
05
5
1
y
1 1
x
11
SECTION 2.2 Graphs of Functions
13.
14.
y
x
f x x 2
4
16
5
3
9
4
2
4
1
1
0
x
23 14 7
3
1 x
1
0
2
2
1
1
0
15.
g x x 2 6x 9
x
16.
y
16
1
1
4
2
1
3
0
4
1
5
4
7
16
17. x 3 2 1
r x 3x 4
10
0
x
1
3
0
0
1
3
2
48
3
243
g x x 3 8
3
19
1
16
5
4
4
1
3
0
2
1
1
4
48
x
2
7
243
0 7
0
8
1
9
2
16
3
35
x
r x 20 x 4
3
61
0
x
1
20.
4 1
x
19
0
20
1
19
2
4
3
61
x
g x x 23
5
27
4
8
3
1
2
x
y 0
1
x
_10
y 20 10
4
1
100
1
16
2
y
19.
g x x 32
18.
y
1
2
x
1
y
f x x 2 2
0
1
1
0
8
1
27
_1
0
x
1
_10
y
0
4 1
x
12
CHAPTER 2 Functions
21.
3
27 1
1
0
0
1
1
8
0
f x 2
y
x
4
0
9
1
_1
2
_2
3
1
12 14
0
1 C t 2 t
2
4
6
1 4
5
3
12
4
21
5
32
6
x
C t
32
4 16
1 _1 0
1
1
1
1
2
2
1 4
x
H x 2x
0
y
27.
2 1
0
2
4 2 0
1 1
x
y
2 0
2
t
2 1
12 13
y
x
8
x
4
1 t 1
28.
10 6
1 _4 0
2
1
12
t
x 4
1
2
4
3
1 2
3
1
y
1 2
16
4
0
1
1 4 1 2
5
f x
26.
y
x
10
3
3
x
8
0 _1
2
24.
1
0
2
1
x
4
x
0
1
2
1
25.
0
27
2
25
1
8
1
16
1
3
23. x
x
10
1
2
8
2
27
3
27
0 _1
y
k x 3 x
x 1
2
8
22.
y
k x 3 x
x
H x x 2 4
1
3
0
2
1
1
2
0
3
1
4
2
2 0
2
x
13
SECTION 2.2 Graphs of Functions
29. x
x
0
5 0
0
1
2
2
4
5
10
10 4
2 1 1
31.
x
1
2
0
0
1
0
3
0
x
f x
3
1
2
1
1
1
32.
y
f x 2x 2 12
5
8
2
0
2
1
0
2
2
5
8
1 1
33. f x 8x x 2
(a) [5 5] by [5 5]
x
y
G x x x
5
0
2
x
30.
y
G x x x
1 x
1
y
x x
1
0
undefined
1
1
2
1
3
1
x
1
(b) [10 10] by [10 10] 10
4 2 4
2
2
2
4
10
5
4
5
10
5
10
10
(c) [2 10] by [5 20]
(d) [10 10] by [100 100]
20
100
10 10 2
2
4
6
8
10
The viewing rectangle in part (c) produces the most appropriate graph of the equation.
5 100
14
CHAPTER 2 Functions
34. f x x 2 4x 32
(b) [10 10] by [10 10]
(a) [3 3] by [5 5]
10
4 2 3
2
1 2
1
2
3
10
5
5
4
10
10
(c) [7 7] by [30 5]
(d) [6 10] by [40 5] 5
6 4 2 10
2
4
5
10
2
4
6 10 20
20
30
30
40
The viewing rectangle in part (d) produces the most appropriate graph of the equation. 35. f x 3x 3 9x 20
(b) [5 5] by [10 20]
(a) [2 2] by [10 10] 10
20 10
2
1
1
2 4
2 10
10
(c) [5 5] by [20 20]
(d) [3 3] by [40 20] 20
20
10 3 4
2 10
2
4
2
1 20
20
1
2
3
40
The viewing rectangle in part (d) produces the most appropriate graph of the equation. 36. f x x 4 10x 2 5x
(b) [5 5] by [10 10]
(a) [1 1] by [10 10] 10
1.0
0.5 10
10
0.5
1.0
4
2 10
2
4
SECTION 2.2 Graphs of Functions
(c) [5 5] by [40 20]
(d) [10 10] by [20 20] 20
20 10
4
2 20
2
4 10
5
40
10
5
10
20
The viewing rectangle in part (c) produces the most appropriate graph of the equation. 1 2x if x 0 if x 1 38. f x 37. f x x 1 if x 1 2 if x 0 y
y
1
2 x
1
39. f x
x
if x 0
40. f x
x 1 if x 0
y
x
1
2x 3 if x 1 3x
if x 1 y
1
1 x
1
1
3 x if x 1 41. f x 2x 2 if x 1
x
4 x 2 if x 1 42. f x x 5 if x 1
y
y
4 2 0 _2
1 1
x
_4 _6
1
x
15
16
CHAPTER 2 Functions
43. f x
0 if x 2
3 if x 2
x 2 if x 1 44. f x 1 if x 1
y
y
1
1 x
1
if x 2 4 2 45. f x x if 2 x 2 x 6 if x 2
1 if x 1 46. f x x if 1 x 1 1 if x 1
y
y
1
1
x 2 6x 12 if x 2 6x
x
1
x
1
47. f x
x
1
if x 2
20 10
2
2x x 2 if x 1 48. f x x 13 if x 1
0
2
4
6
The first graph shows the output of a typical graphing device. However, the actual graph
of this function is also shown, and its difference from the graphing device’s version should be noted. y 1
2 3 2 1 2
2 if x 2 49. f x x if 2 x 2 2 if x 2
1
1
2
3
if x 1 1 50. f x 1 x if 1 x 2 2 if x 2
x
SECTION 2.2 Graphs of Functions
17
51. The curves in parts (a) and (c) are graphs of a function of x, by the Vertical Line Test. 52. The curves in parts (b) and (c) are graphs of functions of x, by the Vertical Line Test. 53. Solving for y in terms of x gives x 2 3y 7 3y x 2 7 y 13 x 2 7 . This defines y as a function of x.
54. Solving for y in terms of x gives 10x y 5 y 10x 5. This defines y as a function of x. 55. Solving for y in terms of x gives y 3 x 5 y 3 x 5 y 3 x 5. This defines y as a function of x. 3 56. Solving for y in terms of x gives x 2 y 13 1 y 13 x 2 1 y x 2 1 . This defines y as a function of x.
57. Solving for y in terms of x gives x y 2 y x. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 58. Solving for y in terms of x gives x 2 y 12 4 y 12 4 x 2 y 1 4 x 2 y 1 4 x 2 . The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 59. Solving for y in terms of x gives 2x 4y 2 3 4y 2 2x 3 y 12 2x 3. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 60. Solving for y in terms of x gives 2x 2 4y 2 3 4y 2 2x 2 3 y 12 2x 2 3. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.
61. Solving for y in terms of x using the Quadratic Formula gives 2x y 5y 2 4 5y 2 2x y 4 0 2x 2x2 4 5 4 2x 4x 2 80 x x 2 20 . The last equation gives two values of y for a y 2 5 10 5 given value of x. Thus, this equation does not define y as a function of x. 62. Solving for y in terms of x gives y x 5 y x 5 y x 52 . This defines y as a function of x. 63. Solving for y in terms of x gives 2 x y 0 y 2 x. This defines y as a function of x.
64. Solving for y in terms of x gives 2x y 0 y 2x. Since a a, the last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 65. Solving for y in terms of x gives x y 3 y 3 x. This defines y as a function of x. 66. Solving for y in terms of x gives x y 4 y 4 x. The last equation gives two values of y for any positive value of x. Thus, this equation does not define y as a function of x. 67.
y
68.
3
2
10
1 0
y 20
1
2
3
4
5
6 x
0
1
2
3
4
5 x
The input value 1 has two output values, so the relation
Each input value has a single output value, so this relation
does not define y as a function of x. Its domain is
defines y as a function of x. Its domain is 1 2 3 4 5
0 1 4 5 6 and its range is 1 2 3.
and its range is 5 10 15 20.
69. y 13 x y 1 3 x y 3 x 1. This defines y as a function of x since every real number x has one and only one cube root. 70. The set x y yx does not define y as a function of x. For example, 2 4 and 2 6 both belong to the set.
18
CHAPTER 2 Functions
71. (a) f x x 2 c, for c 0, 2, 4, and 6. c=4
10 8
c=2
(b) f x x 2 c, for c 0, 2, 4 , and 6. c=_2
6
4
0 2
c=_6
4
c=0
2 2
c=0
8
6
4
c=_4
10
c=6
2
2
4
4
0 2
2
4
4
6
6
8
8
10
10
2
4
(c) The graphs in part (a) are obtained by shifting the graph of f x x 2 upward c units, c 0. The graphs in part (b) are obtained by shifting the graph of f x x 2 downward c units.
72. (a) f x x c2 , for c 0, 1, 2, and 3.
(b) f x x c2 , for c 0, 1, 2, and 3.
c=0 c=1
10 8
c=_1 c=0
c=2
c=_2
c=3
c=_3
10 8
6
6
4 2 4
0 2
2
2
4
4 2 4
0 2
2
4
4
6
6
8
8
10
10
2
4
(c) The graphs in part (a) are obtained by shifting the graph of y x 2 to the right 1, 2, and 3 units, while the graphs in part (b) are obtained by shifting the graph of y x 2 to the left 1, 2, and 3 units.
73. (a) f x cx 2 , for c 1, 12 , 2, and 4. c=1
c=4
10
(b) f x cx 2 , for c 1, 1, 12 , and 2.
c=2
10
8 4
8
c=1
1 c=_ 2
6
6 4 2
2 4
2
0 2
2
4
4
2
4
4
6
6
2
4 1
c=__ 2
8
8 10
0 2
c=_4
10
c=_2
(c) As c increases, the graph of f x cx 2 is stretched vertically. As c decreases, the graph of f is flattened. When c 0, the graph is reflected about the xaxis.
SECTION 2.2 Graphs of Functions
74. (a) f x x 1n , for n 2, 4, and 6.
(b) f x x 1n , for n 3, 5, and 7. 2
3
n=2
2
n=3 n=5 n=7
1
n=4 n=6
1
0
1
1
2
19
3
3
2
1
0
1
2
3
1
4
2
1
(c) Graphs of even roots are similar to y
x, graphs of odd roots are similar to y 3 x. As n increases, the graph of
y x 1n becomes steeper near x 0 and flatter for x 1.
75. The slope of the line segment joining the points 2 1 and 4 6 is m
6 1 76 . Using the pointslope form, 4 2
we have y 1 76 x 2 y 76 x 73 1 y 76 x 43 . Thus the function is f x 76 x 43 for 2 x 4.
76. The slope of the line containing the points 3 2 and 6 3 is m of the line, we have y 3 59 x 6 y 59 x 10 3 3
3 x 6.
2 3 5 59 . Using the pointslope equation 3 6 9
5 x 13 . Thus the function is f x 59 x 13 , for 9
77. First solve the circle for y: x 2 y 2 9 y 2 9 x 2 y 9 x 2 . Since we seek the top half of the circle, we choose y 9 x 2 . So the function is f x 9 x 2 , 3 x 3. 78. First solve the circle for y: x 2 y 2 9 y 2 9 x 2 y 9 x 2 . Since we seek the bottom half of the circle, we choose y 9 x 2 . So the function is f x 9 x 2 , 3 x 3. 05 for 10 r 100. As the balloon r2 is inflated, the skin gets thinner, as we would expect.
79. We graph T r
80. We graph P 141 3 for 1 10. As wind speed increases, so does power output, as expected. P
T
20,000
0.004 0.003 10,000
0.002 0.001 0
50
r
0
label horizontal axis "v"
5
10
Match label on graph V(t)
20
Match label on graph:
CHAPTER 2 Functions
t 2 81. y 50 1 , 0 t 20 20
82. y
L(T)
567 108
V
20000
20
10000
0
20 t
10
0
0
1000 T
500
Match label on graph
84. y 18,500 025 r 2 , 0 r 05
2 83. L 10 1 , 0 300,000 300,0002
L
v
v(r)
10
4000
5
0
T 4 , 0 T 1000
L
40
0
2000
0
1e+5
2e+5
0
3e+5 v
0.0
0.1
0.2
0.3
0.4
0.5
r
300
400
500
600
x
P
85. 060 084 P x 108 132
1.40
if 0 x 1 if 1 x 2 if 2 x 3
if 3 x 35
1.20 1.00 0.80 0.60 0.40 0.20 0
86. (a) E x
1000 012x
if 0 x 300 4600 017 x 300 if 300 x
1
2
4
3
x
(b) E
100 80 60 40 20 0
100
200
SECTION 2.2 Graphs of Functions
21
87. The graph of x y 2 is not the graph of a function because both 1 1 and 1 1 satisfy the equation x y 2 . The graph of x y 3 is the graph of a function because x y 3 x 13 y. If n is even, then both 1 1 and 1 1 satisfies the equation x y n , so the graph of x y n is not the graph of a function. When n is odd, y x 1n is defined for all real
numbers, and since y x 1n x y n , the graph of x y n is the graph of a function.
88. (a) The graphs of f x x 2 x 6 and g x x 2 x 6 are shown in the viewing rectangle [10 10] by [10 10].
10
10
10
5
5
10
10
5
5
10
10
10
For those values of x where f x 0 , the graphs of f and g coincide, and for those values of x where f x 0, the graph of g is obtained from that of f by reflecting the part below the xaxis about the xaxis. (b) The graphs of f x x 4 6x 2 and g x x 4 6x 2 are shown in the viewing rectangle [5 5] by [10 15]. 10
4
2
10
2
4
4
10
2
2
4
10
For those values of x where f x 0 , the graphs of f and g coincide, and for those values of x where f x 0, the graph of g is obtained from that of f by reflecting the part below the xaxis above the xaxis. (c) In general, if g x f x, then for those values of x where f x 0, the graphs of f and g coincide, and for those values of x where f x 0, the graph of g is obtained from that of f by reflecting the part below the xaxis above the xaxis. y
y
x
y f x
x
y g x
89. (a) The relation “x is the sibling of y” is not a function, in general, because a person x might have more than one sibling. (b) The relation “x is the birth mother of y” is not a function, in general, because x might have given birth to more than one child. (c) The relation “x is a student in your school and y is their ID number” is a function because every student x has a unique ID number. (d) The relation “x is the age of a student and y is their shoe size” is not a function because two students might well be the same age and have different shoe sizes. For example, it is possible that 21 10 and 21 8 are both in the relation.
22
CHAPTER 2 Functions
90. The relation in Exercise 5 is 1 1 1 2 2 3 3 4. Because each yvalue has one corresponding xvalue, the relation defines x as a function of y. The relation in Exercise 6 is defined by x 4y 2 . Because every value of y has one corresponding xvalue, the relation defines x as a function of y. The relation in Exercise 7 is 8 11 10 15 2 11 10 10. Because the yvalue 11 has two corresponding xvalues (8 and 2), the relation does not define x as a function of y. There exist horizontal lines that intersect the graph in Exercise 8 more than once. Thus, some yvalues have multiple corresponding xvalues, and so the relation does not define x as a function of y.
2.3
GETTING INFORMATION FROM THE GRAPH OF A FUNCTION
1. To find a function value f a from the graph of f we find the height of the graph above the xaxis at x a. From the graph of f we see that f 3 4 and f 5 6. The net change in f between x 3 and x 5 is f 5 f 3 6 4 2. 2. The domain of the function f is all the xvalues of the points on the graph, and the range is all the corresponding yvalues. From the graph of f we see that the domain of f is the interval [1 7] and the range of f is the interval [0 7]. 3. (a) If f is increasing on an interval, then the yvalues of the points on the graph rise as the xvalues increase. From the graph of f we see that f is increasing on the intervals 1 2 and 4 5. (b) If f is decreasing on an interval, then yvalues of the points on the graph fall as the xvalues increase. From the graph of f we see that f is decreasing on the intervals 2 4 and 5 7. 4. (a) A function value f a is a local maximum value of f if f a is the largest value of f on some interval containing a. From the graph of f we see that there are two local maximum values of f : one maximum is 7, and it occurs when x 2; the other maximum is 6, and it occurs when x 5.
(b) A function value f a is a local minimum value of f if f a is the smallest value of f on some interval containing a. From the graph of f we see that there is one local minimum value of f . The minimum value is 2, and it occurs when x 4.
5. The solutions of the equation f x 0 are the xintercepts of the graph of f . The solution of the inequality f x 0 is the set of xvalues at which the graph of f is on or above the xaxis. From the graph of f we find that the solutions of the equation f x 0 are x 1 and x 7, and the solution of the inequality f x 0 is the interval [1 7]. 6. (a) To solve the equation 2x 1 x 4 graphically we graph the
y
functions f x 2x 1 and g x x 4 on the same set of axes and determine the values of x at which the graphs of f and g intersect. From the graph, we see that the solution is x 1.
1 0
1
x
(b) To solve the inequality 2x 1 x 4 graphically we graph the functions f x 2x 1 and g x x 4 on the same set of axes and find the values of x at which the graph of g is higher than the graph of f . From the graphs in part (a) we see that the solution of the inequality is 1. 7. (a) h 2 1, h 0 1, h 2 3, and h 3 4. (b) Domain: [3 4]. Range: [1 4].
(c) h 3 3, h 2 3, and h 4 3, so h x 3 when x 3, x 2, or x 4.
(d) The graph of h lies below or on the horizontal line y 3 when 3 x 2 or x 4, so h x 3 for those values of x. (e) The net change in h between x 3 and x 3 is h 3 h 3 4 3 1.
8. (a) g 4 3, g 2 2, g 0 2, g 2 1, and g 4 0. (b) Domain: [4 4]. Range: [2 3].
SECTION 2.3 Getting Information from the Graph of a Function
(c) g 4 3. [Note that g 2 1 not 3.]
(d) It appears that g x 0 for 1 x 18 and for x 4; that is, for x 1 x 18 4.
(e) g 1 0 and g 2 1, so the net change between x 1 and x 2 is 1 0 1. 9. (a) f 0 3 15 g 0, so f 0 is larger.
(b) f 1 25 1 g 1, so f 1 is larger.
(c) f x g x for x 2 and x 2.
(d) f x g x for 4 x 2 and 2 x 4; that is, on the intervals [4 2] and [2 4]. (e) f x g x for 2 x 2; that is, on the interval 2 2.
10. (a) The graph of g is higher than the graph of f at x 6, so g 6 is larger.
(b) The graph of f is higher than the graph of g at x 3, so f 3 is larger.
(c) The graphs of f and g intersect at x 2, x 5, and x 7, so f x g x for these values of x. (d) f x g x for 1 x 2 and approximately 5 x 7; that is, on [1 2] and [5 7].
(e) f x g x for 2 x 5 and approximately 7 x 8; that is, on 2 5 and 7 8].
11. From the graph, the domain of f is 3 3] and its range is [2 3]. 12. From the graph, the domain of f is 3 3 and its range is 3 2]. 13. From the graph, the domain of f is [3 3] and its range is 3 2 3. 14. From the graph, the domain of f is 3 3] and its range is [3 3]. y
15. (a)
y
16. (a)
2 0
x
1
2 0
1
(b) Domain: ; Range: 17. (a)
y
x
(b) Domain: ; Range: 18. (a)
y
1 0
1
x
(b) Domain: ; Range: [3
4
0
1
x
(b) Domain: ; Range: [2
23
24
CHAPTER 2 Functions
19. (a)
20. (a)
y
y
1 0
1 0
1
x
1
x
(b) Domain: 1 4; Range: 4 2 (b) Domain: [2 5]; Range: [4 3]
21. (a)
22. (a)
y
y
10 0 2 0
x
1
(b) Domain: [3 3]; Range: [1 8]
(b) Domain: [3 3]; Range: [28 26]
23. (a)
24. (a)
6
4
10
2
5
1
2
4
6
1
(b) Domain: [1 ; Range: [0
26. (a)
2 6 4 2 2
2
2
(b) Domain: ; Range: [1
25. (a)
x
1
2 2
4 6
(b) Domain: [6 6]; Range: [6 0]
4
6
1 6 4 2 1
2
4
(b) Domain: ; Range: 0 2]
6
SECTION 2.3 Getting Information from the Graph of a Function
27. (a)
28. (a)
10
1 6 4 2 1
2
4
6
4
2
2
4
10
(b) Domain: ; Range: [1 1]
(b) Domain: ; Range: [9
y
29.
y
30.
y=8x-9
y=4x-5 1l y=5-x
1l 1
1
x y=3-4x
x
(a) From the graph, we see that 4x 5 5 x when
(a) From the graph, we see that 3 4x 8x 9 when
(b) From the graph, we see that 4x 5 5 x when
(b) From the graph, we see that 3 4x 8x 9 when
x 1.
x 2.
x 1.
x 2.
y
31.
y
32.
1l
1
x y=_x@l
y=x@l y=3-4x 1l
y=2-xl 1
x
(a) From the graph, we see that x 2 2 x when
(a) From the graph, we see that x 2 3 4x when
(b) From the graph, we see that x 2 2 x when
(b) From the graph, we see that x 2 3 4x when
x 2 or x 1.
2 x 1.
x 1 or x 3.
1 x 3.
25
26
CHAPTER 2 Functions
34.
33.
20
10
10 4
2
2
10 20
10
30
20
(a) We graph y x 3 3x 2 (black) and
y x 2 3x 7 (gray). From the graph, we see
4
6
(a) We graph y 5x 2 x 3 (black) and
y x 2 3x 4 (gray). From the graph, we see
that the graphs intersect at x 432, x 112,
that the graphs intersect at x 058, x 129, and
and x 144.
(b) From the graph, we see that
2
x 529.
(b) From the graph, we see that
x 3 3x 2 x 2 3x 7 on approximately
5x 2 x 3 x 2 3x 4 on approximately
[432 112] and [144 .
[058 129] and [529 .
(a) We graph y 16x 3 16x 2 (black) and y x 1 (gray). From the
35.
graph, we see that the graphs intersect at x 1, x 025, and
2
x 025.
2
1
(b) From the graph, we see that 16x 3 16x 2 x 1 on [1 025]
1
and [025 .
2
36.
(a) We graph y 1
4
x (black) and y
x 2 1 (gray). From the
graph, we see that the solutions are x 0 and x 231. (b) From the graph, we see that 1 x x 2 1 on approximately
3 2
0 231.
1 1
0
1
2
3
4
5
37. (a) The domain is [1 4] and the range is [1 3]. (b) The function is increasing on 1 1 and 2 4 and decreasing on 1 2. 38. (a) The domain is [2 3] and the range is [2 3]. (b) The function is increasing on 0 1 and decreasing on 2 0 and 1 3. 39. (a) The domain is [3 3] and the range is [2 2]. (b) The function is increasing on 2 1 and 1 2 and decreasing on 3 2, 1 1, and 2 3. 40. (a) The domain is [2 2] and the range is [2 2]. (b) The function is increasing on 1 1 and decreasing on 2 1 and 1 2.
SECTION 2.3 Getting Information from the Graph of a Function
41. (a) f x x 2 5x is graphed in the viewing rectangle [2 7] by [10 10].
27
42. (a) f x x 3 4x is graphed in the viewing rectangle [10 10] by [10 10].
10
10
2
2
4
6
10
5
10
5
10
10
(b) The domain is and the range is [625 .
(c) The function is increasing on 25 . It is decreasing on 25.
43. (a) f x 2x 3 3x 2 12x is graphed in the viewing rectangle [3 5] by [25 20].
(b) The domain and range are . (c) The function is increasing on 115 and 115 . It is decreasing on 115 115.
44. (a) f x x 4 16x 2 is graphed in the viewing rectangle [10 10] by [70 10].
20 10
10 2
5
5
10
20 2
10
4
40 60
20
(b) The domain and range are .
(b) The domain is and the range is [64 .
(c) The function is increasing on 1 and 2 .
(c) The function is increasing on 283 0 and
It is decreasing on 1 2.
283 . It is decreasing on 283 and 0 283.
45. (a) f x x 3 2x 2 x 2 is graphed in the viewing rectangle [5 5] by [3 3].
46. (a) f x x 4 4x 3 2x 2 4x 3 is graphed in the viewing rectangle [3 5] by [5 5]. 4
2
2 4
2
2
4
2
2
2
2
4
4
(b) The domain and range are .
(b) The domain is and the range is [4 .
(c) The function is increasing on 155 and
(c) The function is increasing on 04 1 and 24 .
022 . It is decreasing on 155 022.
It is decreasing on 04 and 1 24.
28
CHAPTER 2 Functions
47. (a) f x x 25 is graphed in the viewing rectangle [10 10] by [5 5].
48. (a) f x 4 x 23 is graphed in the viewing rectangle [10 10] by [10 10].
10
4 2 10
5
5
2
10
10
5
5
10
4
10
(b) The domain is and the range is [0 .
(b) The domain is and the range is 4].
(c) The function is increasing on 0 . It is decreasing
(c) The function is increasing on 0. It is
on 0.
49. (a) f x 2
x 3 is graphed in the viewing
rectangle [4 10] by [0 6].
decreasing on 0 . 50. (a) f x 25 x 2 is graphed in the viewing rectangle [6 6] by [1 6].
6
6
4
4
2
2
4 2 0
2
4
6
6
8 10
4
2
2
4
6
(b) The domain is [3 and the range is [2 .
(b) The domain is [5 5] and the range is [0 5].
(c) The function is increasing on 3 .
(c) The function is increasing on 5 0 and decreasing on 0 5.
51. (a) Local maxima: 3 at x 1 and 4 at x 3. Local minimum: 3 at x 1.
(b) The function is increasing on 1 and 1 3 and decreasing on 1 1 and 3 .
52. (a) Local maximum: 2 at x 0. Local minima: 1 at x 2 and 0 at x 2.
(b) The function is increasing on 2 0 and 2 and decreasing on 2 and 0 2.
53. (a) Local maximum: 3 at x 0. Local minima: 1 at x 2 and 1 at x 1.
(b) The function is increasing on 2 0 and 1 and decreasing on 2 and 0 1.
54. (a) Local maxima: 3 at x 2 and 2 at x 1. Local minima: 0 at x 1 and 1 at x 2.
(b) The function is increasing on 2, 1 1, and 2 and decreasing on 2 1 and 1 2.
55. (a) In the first graph, we see that f x x 3 x has a local minimum and a local maximum. Smaller x and yranges show that f x has a local maximum of about 038 when x 058 and a local minimum of about 038 when x 058. 5
0.5
0.50 0.3
0.4 5
0.4
5 5
0.60
0.55
0.3 0.50
0.5
(b) The function is increasing on 058 and 058 and decreasing on 058 058.
0.55
0.60
SECTION 2.3 Getting Information from the Graph of a Function
29
56. (a) In the first graph, we see that f x 3 x x 2 x 3 has a local minimum and a local maximum. Smaller x and yranges show that f x has a local maximum of about 400 when x 100 and a local minimum of about 281 when x 033. 4 2
2
0
2
0.40
0.35
2.9
4.1
2.8
4.0
2.7 0.30
3.9
0.9
1.0
1.1
(b) The function is increasing on 033 100 and decreasing on 033 and 100 .
57. (a) In the first graph, we see that g x x 4 2x 3 11x 2 has two local minimums and a local maximum. The local maximum is g x 0 when x 0. Smaller x and yranges show that local minima are g x 1361 when x 171 and g x 7332 when x 321. 5
5
1.75
1.70
1.65 13.4
73.0
3.1
50
13.6
73.5
100
13.8
74.0
3.2
3.3
(b) The function is increasing on 171 0 and 321 and decreasing on 171 and 0 321.
58. (a) In the first graph, we see that g x x 5 8x 3 20x has two local minimums and two local maximums. The local maximums are g x 787 when x 193 and g x 1302 when x 104. Smaller x and yranges show that local minimums are g x 1302 when x 104 and g x 787 when x 193. Notice that since g x is odd, the local maxima and minima are related. 20
2.0
1.8 7.8
13.1 13.0
5
7.9
5
12.9 20
8.0 1.2
1.0
1.0
1.2
7.90 12.8 7.85 13.0 13.2
7.80 1.90
1.95
2.00
(b) The function is increasing on 193, 104 104, and 193 and decreasing on 193 104 and 104 193.
30
CHAPTER 2 Functions
59. (a) In the first graph, we see that U x x 6 x has only a local maximum. Smaller x and yranges show that U x has a local maximum of about 566 when x 400. 10
5.70
5
5.65 5.60 5
3.9
4.0
4.1
(b) The function is increasing on 400 and decreasing on 400 600. 60. (a) In the first viewing rectangle below, we see that U x x x x 2 has only a local maximum. Smaller x and yranges show that U x has a local maximum of about 032 when x 075. 1.0
0.40
0.5
0.35
0.0
0.0
0.5
0.30
1.0
0.7
0.8
0.9
(b) The function is increasing on 0 075 and decreasing on 075 1. 1 x2 has a local minimum and a local maximum. Smaller x and yranges x3 show that V x has a local maximum of about 038 when x 173 and a local minimum of about 038 when x 173.
61. (a) In the first graph, we see that V x
2
0.40
1.6
1.7
1.8
0.30 0.35 5
0.35
5 1.8
2
1.7
0.30 1.6
0.40
(b) The function is increasing on 173 and 173 and decreasing on 173 0 and 0 173.
1 has only a local maximum. Smaller x and x2 x 1 yranges show that V x has a local maximum of about 133 when x 050.
62. (a) In the first viewing rectangle below, we see that V x 2
1.40 1.35
5
5
0.6
0.5
1.30 0.4
(b) The function is increasing on 050 and decreasing on 050 .
63. (a) At 5 A . M . the graph shows that the power consumption is about 11 gigawatts. Since t 22 represents 10 P. M ., the graph shows that the power consumption at 10 P. M . is about 14 gigawatts. (b) The power consumption is lowest between 3 A . M . and 4:30 A . M . The power consumption is highest at about 7 P. M .
31
SECTION 2.3 Getting Information from the Graph of a Function
(c) The net change in power consumption from 5 A . M . to 10 P. M . is P 22 P 5 14 11 3 gigawatts. 64. (a) The first noticeable movements occurred at time t 5 seconds. (b) It seemed to end at time t 30 seconds.
(c) Maximum intensity was reached at t 17 seconds.
65. (a) This person appears to be gaining weight steadily until the age of 21 when this person’s weight gain slows down. The person continues to gain weight until the age of 30, at which point this person experiences a sudden weight loss. Weight gain resumes around the age of 32, and the person dies at about age 68. Thus, the person’s weight W is increasing on 0 30 and 32 68 and decreasing on 30 32. (b) The sudden weight loss could be due to a number of reasons, among them major illness, a weight loss program, etc. (c) The net change in the person’s weight from age 10 to age 20 is W 20 W 10 150 50 100 lb. 66. (a) Measuring in hours since midnight, the representative’s distance from home D is increasing on 8 9, 10 12, and 15 17, constant on 9 10, 12 13, and 17 18, and decreasing on 13 15 and 18 19. (b) The representative travels away from home and stops to make a sales call between 9 A . M . and 10 A . M ., and then travels further from home for a sales call between 12 noon and 1 P. M . Next, the representative travels along a route that veers closer to home before going further away again. A final sales call is made between 5 P. M . and 6 P. M ., then the representative returns home. (c) The net change in the distance D from noon to 1 P. M . is D 1 P. M . D noon 0. 67. (a) The function W is increasing on 0 150 and 300 365 and decreasing on 150 300. (b) W has a local maximum at x 150 and a local minimum at x 300.
(c) The net change in the depth W from 100 days to 300 days is W 300 W 100 25 75 50 ft.
68. (a) The function P is increasing on 0 25 and decreasing on 25 50. (b) The maximum population was 50,000, and it was attained at x 25 years, which represents the year 1995. (c) The net change in the population P from 1990 to 2010 is P 40 P 20 40 40 0.
69. Runner A won the race. All runners finished the race. Runner B fell, but got up and finished the race.
70. (a)
a (m/s²)
71. (a)
2
E 400 300
1
200 100
0
2000
4000 x (km)
(b) As the distance x increases, the acceleration a due to gravity decreases. The rate of decrease is rapid at first, and slows as the distance increases.
0
50
100 150 200 250 300
(b) As the temperature T increases, the energy E increases. The rate of increase gets larger as the temperature increases.
T
32
CHAPTER 2 Functions
72. In the first graph, we see the general location of the minimum of V 99987 006426T 00085043T 2 00000679T 3 is around T 4. In the second graph, we isolate the minimum, and from this graph, we see that the minimum volume of 1 kg of water occurs at T 396 C. 1005
999.76
1000
999.75
995
0
999.74
20
3.5
4.0
4.5
10 . In the second graph, we isolate the 5 minimum, and from this graph, we see that energy is minimized when 75 mi/h.
73. In the first graph, we see the general location of the minimum of E 273 3 10000
4700 4650
5000 6
8
10
4600
7.4
7.5
7.6
74. In the first graph, we see the general location of the maximum of r 32 1 r r 2 is around r 07 cm. In the second graph, we isolate the maximum, and from this graph we see that at the maximum velocity is approximately 047 when r 067 cm. 1.0
0.50 0.48
0.5
0.46 0.0
0.6
0.8
1.0
75. (a) f x is always increasing, and f x 0 for all x. y
0
0.60
0.65
0.70
(b) f x is always decreasing, and f x 0 for all x. y
x
0
x
SECTION 2.4 Average Rate of Change of a Function
(c) f x is always increasing, and f x 0 for all x.
33
(d) f x is always decreasing, and f x 0 for all x.
y
y
x 0
0
x
76. To find the fixed points of a function algebraically, set f x x and solve. To find the fixed points graphically, find the points of intersection with the graphs of y f x and y x. (a) 5x x 2 x x 2 4x 0 x x 4 0 x 0 or 4. make bold
(b) x 3 x 1 x x 3 1 0 x 1. (c)
(d)
y
y
1 0
2.4
1 1
0
x
1
x
From the graph, the fixed points of the function are
From the graph, we see that this function has no
4, 0, and 3.
fixed point.
AVERAGE RATE OF CHANGE OF A FUNCTION
100 miles 50 mi/h. 2 hours f b f a 2. The average rate of change of a function f between x a and x b is average rate of change . ba
1. If you travel 100 miles in two hours then your average speed for the trip is average speed
3. The average rate of change of the function f x x 2 between x 1 and x 5 is f 1 average rate of change f 5 51
25 1 24 52 12 6. 4 4 4
4. (a) The average rate of change of a function f between x a and x b is the slope of the secant line between a f a and b f b. (b) The average rate of change of the linear function f x 3x 5 between any two points is 3.
5. (a) Yes, the average rate of change of a function between x a and x b is the slope of the secant line through a f a f b f a and b f b; that is, . ba (b) Yes, the average rate of change of a linear function y mx b is the same (namely m) for all intervals.
6. (a) No, the average rate of change of an increasing function is positive over any interval.
34
CHAPTER 2 Functions
(b) No, just because the average rate of change of a function between x a and x b is negative, it does not follow
that the function is decreasing on that interval. For example, f x x 2 has negative average rate of change between x 2 and x 1, but f is increasing for 0 x 1.
7. (a) The net change is f 4 f 1 5 3 2.
(b) We use the points 1 3 and 4 5, so the average rate of change is 8. (a) The net change is f 5 f 1 2 4 2. (b) We use the points 1 4 and 5 2, so the average rate of change is 9. (a) The net change is f 5 f 0 2 6 4. (b) We use the points 0 6 and 5 2, so the average rate of change is 10. (a) The net change is f 5 f 1 4 0 4.
53 2 . 41 3 24 2 1 . 51 4 2 4 26 . 50 5
(b) We use the points 1 0 and 5 4, so the average rate of change is 11. (a) The net change is f 7 f 4 [5 7 3] [5 4 3] 15. f 7 f 4 15 (b) The average rate of change is 5. 74 3
40 4 2 . 5 1 6 3
12. (a) The net change is s 5 s 1 [4 2 5] [4 2 1] 8. 8 s 5 s 1 2. (b) The average rate of change is 51 4 13. (a) The net change is g 10 g 6 2 12 10 2 12 6 8. (b) The average rate of change is
8 1 g 10 g 6 . 10 6 16 2
14. (a) The net change is h 5 h 2
9 3 5 4 3 2 4 . 5 5 5 9
h 5 h 2 3 5 . 52 3 5 15. (a) The net change is f 3 f 1 3 32 3 3 12 1 30 4 26. (b) The average rate of change is
26 f 3 f 1 13. 31 2 16. (a) The net change is f 0 f 3 3 0 02 3 3 32 0 18 18. (b) The average rate of change is
f 0 f 3 18 6. 0 3 3 17. (a) The net change is f 10 f 0 103 4 102 03 4 02 600 0 600. (b) The average rate of change is
f 10 f 0 600 60. 10 0 10 18. (a) The net change is g 2 g 2 24 23 22 24 23 22 12 28 16. (b) The average rate of change is
(b) The average rate of change is
19. The difference quotient is
g 2 g 2 16 4. 2 2 4
4a 2 8ah 4h 2 4a 2 4 a h2 4a 2 f a h f a 8a 4h. h h a h a
SECTION 2.4 Average Rate of Change of a Function
35
20. The difference quotient is
3 10 a h2 3 10a 2 10a 2 20ah 10h 2 10a 2 f a h f a 20a 10h. h h a h a 1 1 f a h f a a a h 1 a h a 21. The difference quotient is . h ah a h a a h a h a 22. The difference quotient is 2 2 f a h f a 2 a 1 2 a h 1 2h 2 ah1 a1 . h h a 1 a h 1 h a 1 a h 1 a h a a 1 a h 1 23. The difference quotient is f a h f a h 1 ah a ah a a h a . h a h a ah a h ah a h ah a ah a
24. The difference quotient is f a h f a a h a
2 22 a ah2
h
2a 2 2 a h2 a 2 a h2 h
2a 2 2 a 2 2ah h 2 ha 2 a h2
25. (a) The average rate of change is 1b 3 1a 3 1b 3 1a 3 1 b a 1 f b f a 2 2 2 2 2 . ba ba ba ba 2
2h 2a h ha 2 a h2
2 2a h
a 2 a h2
.
(b) The slope of the line f x 12 x 3 is 12 , which is also the average rate of change.
26. (a) The average rate of change is g b g a 4 b a 4b 2 4a 2 4. ba ba ba (b) The slope of the line g x 4x 2 is 4, which is also the average rate of change.
27. The function f has a greater average rate of change between x 0 and x 1. The function g has a greater average rate of change between x 1 and x 2. The functions f and g have the same average rate of change between x 0 and x 15. 28. The average rate of change of f is constant, that of g increases, and that of h decreases. 29. (a) The average rate of change is that the depth decreased.
50 75 25 1 W 200 W 100 ft/day. The negative sign indicates 200 100 200 100 100 4
(b) Answers will vary. One interval on which the average rate of change is 0 is [200 350]. 50 40 10 P 25 P 20 2. Thus, the average rate of change over this time 25 20 25 20 5 period is 2000 people per year.
30. (a) The average rate of change is
(b) On average, the population grew at a rate of 2000 per year between 1990 and 1995. (c) There are many, including 1990–2010 and 1970–2016 (approximately). 6375 4869 3765 persons/yr. 2012 2008 4921 6288 34175 persons/yr. (b) The average rate of change of population between 2014 and 2018 is 2018 2014 (c) The population was increasing from 2002 to 2012.
31. (a) The average rate of change of population between 2008 and 2012 is
(d) The population was decreasing from 2012 to 2020. 800 400 400 100 476 m/s. 152 68 84 21 400 1,600 1,200 268 m/s. (b) The average speed is 412 263 149
32. (a) The average speed is
36
CHAPTER 2 Functions
(c) Lap
Length of time to run lap
Average speed of lap
1
32
625 m/s
2
36
556 m/s
3
40
500 m/s
4
44
455 m/s
5
51
392 m/s
6
60
333 m/s
7
72
278 m/s
8
77
260 m/s
The runner is slowing down throughout the run.
3629 1146 2483 cakesyr. 2020 2010 638 1042 404 cakesyr. (b) The average rate of change between 2011 and 2012 was 2012 2011 1145 638 507 cakesyr. (c) The average rate of change between 2012 and 2013 was 2013 2012 (d) Year Snack cakes sold Change in sales from previous year
33. (a) The average rate of change of sales between 2010 and 2020 was
2010
1146
—
2011
1042
2012
638
104
2013
1145
2014
1738
593
2015
1804
66
2016
1121
2017
1987
683
2018
2533
546
2019
2983
450
2020
3629
646
404
507
866
Sales increased most quickly between 2016 and 2017, and decreased most quickly between 2015 and 2016.
SECTION 2.4 Average Rate of Change of a Function
37
34. Year
Number of books
2000
530
2001
590
2002
650
2006
890
2010
1130
2012
1250
2015
1430
2017
1550
2018
1610
2019
1670
2020
1730
35. The average rate of change of the temperature of the soup over the first 20 minutes is 119 200 81 T 20 T 0 405 F/min. Over the next 20 minutes, it is 20 0 20 0 20 T 40 T 20 89 119 30 15 F/min. The first 20 minutes had a higher average rate of change of 40 20 40 20 20 temperature (in absolute value).
36. (a) (i) Between 1860 and 1890, the average rate of change was about 83,300 farms per year. (ii) Between 1950 and 1980, the average rate of change was 100,000 farms per year.
45 20 y 1890 y 1860 00833, a gain of 1890 1860 30 25 55 y 1980 y 1950 01, a loss of about 1980 1950 30
(b) From the graph, it appears that the steepest rate of decline was during the period from 1950 to 1960.
d 10 d 0 100 10 ms. 10 0 10 (b) Skier A gets a great start, but slows at the end of the race. Skier B maintains a steady pace. Runner C is slow at the beginning, but accelerates down the hill.
37. (a) For all three skiers, the average rate of change is
38. (a) Skater B won the race, because he travels 500 meters before Skater A. A 10 A 0 200 0 (b) Skater A’s average speed during the first 10 seconds is 20 ms. 10 0 10 B 10 B 0 100 0 Skater B’s average speed during the first 10 seconds is 10 ms. 10 0 10 500 400 A 40 A 25 7 ms. (c) Skater A’s average speed during his last 15 seconds is 40 25 15 B 35 B 20 500 200 Skater B’s average speed during his last 15 seconds is 20 ms. 35 20 15
38
CHAPTER 2 Functions
39. t a
t b
3
35
3
31
3
301
3
3001
3
30001
Average Speed
f b f a ba
16 352 16 32 104 35 3
16 312 16 32 976 31 3
16 3012 16 32 9616 301 3
16 30012 16 32 96016 3001 3
16 300012 16 32 960016 30001 3
From the table it appears that the average speed approaches 96 fts as the time intervals get smaller and smaller. It seems reasonable to say that the speed of the object is 96 fts at the instant t 3.
2.5
LINEAR FUNCTIONS AND MODELS
1. If f is a function with constant rate of change, then (a) f is a linear function of the form f x ax b. (b) The graph of f is a line.
2. If f x 5x 7, then (a) The rate of change of f is 5.
(b) The graph of f is a line with slope 5 and yintercept 7.
3. From the graph, we see that y 2 50 and y 0 20, so the slope of the graph is 50 20 y 2 y 0 15 galmin. m 20 2
4. From Exercise 3, we see that the pool is being filled at the rate of 15 gallons per minute. 5. If a linear function has positive rate of change, its graph slopes upward.
6. f x 3 is a linear function because it is of the form f x ax b, with a 0 and b 3. Its slope (and hence its rate of change) is 0. 7. f x 5 2x 2x 5 is linear with a 2 and b 5. 9. f x
20 x 15 x 4 is linear with a 15 and 5
8. f x 12 x 4 12 x 2 is linear with a 12 and b 2.
10. f x
4 2x 4 2 is not linear. x x
b 4. 11. f x x 2 3x 3x 2 2x is not linear. 13. f x
x 1 is not linear.
12. f x 3 6 5x 15x 18 is linear with a 15 and b 18.
14. f x 2x 52 4x 2 20x 25 is not linear.
SECTION 2.5 Linear Functions and Models y
15. x
f x 2x 3 7
2
5
1
3
1
1
1
0
2
1
0
3
x
1
3
The slope of the graph of f x 2x 3 is 2. y
16. x
g x 3x 1
2
7
1
4
0 1
1
1
0
2
2
5
3
8
1
x
The slope of the graph of g x 3x 1 is 3. 17.
r t 23 t 2
t 1
267
0
2
1
133
2
067
3
0
4
067
y
2 0
2
t
The slope of the graph of r t 23 t 2 is 23 . 18.
h t 12 34 t
t 2 1
2
125
0
05
1
025
2 3
y
1 175
The slope of the graph of h t 12 34 t is 34 .
1 0
1
t
39
40
CHAPTER 2 Functions y
19. (a)
y
20. (a)
1 _1
1 0
0
1
z
x
1
_10
(b) The graph of f x 2x 6 has slope 2.
(b) The graph of g z 3z 9 has slope 3.
(c) f x 2x 6 has rate of change 2.
(c) g z 3z 9 has rate of change 3.
y
21. (a)
y
22. (a)
1
1 0
x
1
(b) The graph of f x 2 3x has slope 3. (c) f x 2 3x has rate of change 3.
z
1
(b) The graph of g z z 3 z 3 has slope 1. (c) g z z 3 has rate of change 1.
y
23. (a)
0
24. (a)
y
1 1 0
(b) The graph of h t 15 .
1
t
5 2t 15 t 12 has slope 10
(c) h t 15 t 12 has rate of change 15 .
0
1
w
(b) The graph of s 05 2 has slope 05. (c) s 05 2 has rate of change 05.
SECTION 2.5 Linear Functions and Models y
25. (a)
26. (a)
41
y 1 0
x
1
1 _1
0 _1
t
1
(b) The graph of f t 32 t 2 has slope 32 . (c) f t 32 t 2 has rate of change 32 .
(b) The graph of g x 54 x 10 has slope 54 . (c) g x 54 x 10 has rate of change 54 .
27. The linear function f with rate of change 5 and initial value 10 has equation f x 5x 10.
28. The linear function f with rate of change 3 and initial value 1 has equation f x 3x 1. 29. The linear function f with slope 12 and yintercept 3 has equation f x 12 x 3.
30. The linear function f with slope 45 and yintercept 2 has equation f x 45 x 2.
31. (a) From the table, we see that for every increase of 2 in the value of x, f x increases by 3. Thus, the rate of change of f is 32 . (b) When x 0, f x 7, so b 7. From part (a), a 32 , and so f x 32 x 7.
32. (a) From the table, we see that f 3 11 and f 0 2. Thus, when x increases by 3, f x decreases by 9, and so the rate of change of f is 3. (b) When x 0, f x 2, so b 2. From part (a), a 3, and so f x 3x 2.
33. (a) From the graph, we see that f 0 3 and f 1 4, so the rate of change of f is (b) From part (a), a 1, and f 0 b 3, so f x x 3. 34. (a) From the graph, we see that f 0 4 and f 2 0, so the rate of change of f is (b) From part (a), a 2, and f 0 b 4, so f x 2x 4. 35. (a) From the graph, we see that f 0 2 and f 4 0, so the rate of change of f is (b) From part (a), a 12 , and f 0 b 2, so f x 12 x 2.
43 1. 10 04 2. 20 02 1 . 40 2
36. (a) From the graph, we see that f 0 1 and f 2 0, so the rate of change of f is (b) From part (a), a 12 , and f 0 b 1, so f x 12 x 1. 37.
f
Increasing the value of a makes the graph of f steeper. In other words, it
a=2
increases the rate of change of f .
a=1 a= 21
1 01
1 0 1 . 20 2
t
42
CHAPTER 2 Functions
38.
f
Increasing the value of b moves the graph of f upward, but does not affect
b=2
the rate of change of f . b=1 b= 21
1
0
39. (a)
t
1
T 38
(b) The slope of T x 150x 32 is the value of a, 150. (c) The amount of trash is changing at a rate equal to the slope of the graph,
36
150 thousand tons per year.
34
The graph is incorrect (it does not match the function--for instance T(10)=1532). The revsied graph should use 0 < x < 3 and 0 < T < 500 labeling tics every 100 on the Taxis.
32
0
40. (a)
10
20
30
x
f 900
(b) The slope of the graph of f x 200 32x is 32.
700
(c) Ore is being produced at a rate equal to the slope of the graph, 32 thousand tons per year.
800 600 500 400 300 200 100
0
5
10
15
20
25 x
41. (a) Let V t at b represent the volume of hydrogen. The balloon is being filled at the rate of 05 ft3 s, so a 05, and initially it contains 2 ft3 , so b 2. Thus, V t 05t 2.
(b) We solve V t 15 05t 2 15 05t 13 t 26. Thus, it takes 26 seconds to fill the balloon. 42. (a) Let V t at b represent the volume of water. The pool is being filled at the rate of 10 galmin, so a 10, and initially it contains 300 gal, so b 300. Thus, V t 10t 300. (b) We solve V t 1300 10t 300 1300 10t 1000 t 100. Thus, it takes 100 minutes to fill the pool.
1 . The ramp 43. (a) Let H x ax b represent the height of the ramp. The maximum rise is 1 inch per 12 inches, so a 12 1 x. starts on the ground, so b 0. Thus, H x 12
1 150 125. Thus, the ramp reaches a height of 125 inches. (b) We find H 150 12
1200 0075, or 75%. 15,000 500 The road biker descends 500 vertical feet over 10,000 feet, so the grade of the road is 005, or 5%. 10,000
44. The mountain biker descends 1200 vertical feet over 15,000 feet, so the grade of the road is
43
SECTION 2.5 Linear Functions and Models
45. (a) From the graph, we see that the slope of the engineer’s trip is steeper than that of the manager. Thus, the engineer traveled faster. 70 7 (b) The points 0 0 and 6 7 are on the engineer’s graph, so their speed is miles per minute or 60 6 7 70 mih. 60 6 16 10 The points 0 10 and 6 16 are on the manager’s graph, so their speed is 60 60 mih. 60 1 hmin 1 mi/min and the engineer’s speed is (c) t is measured in minutes, so the manager’s speed is 60 mih 60
1 hmin 7 mi/min. Thus, the manager’s distance is modeled by f t 1 t 0 10 t 10 and the 70 mih 60 6
engineer’s distance is modeled by g t 76 t 0 0 76 t.
46. (a) Let d t represent the distance traveled. When t 0, d 0, and when
(b)
40 0 t 50, d 40. Thus, the slope of the graph is 08. The 50 0 yintercept is 0, so d t 08t.
(c) The speed of the bus is equal to the slope of the graph of d, that is, 08 mimin or 08 60 48 mih.
d 110 100 90 80 70 60 50 40 30 20 10 0
20
40
60
80 100 120 t
6 , so if we take 0 0 as the starting point, the 47. Let x be the horizontal distance and y the elevation. The slope is 100
6 x. We have descended 1000 ft, so we substitute y 1000 and solve for x: 1000 6 x elevation is y 100 100 1 16,667 316 mi. x 16,667 ft. Converting to miles, the horizontal distance is 5280
48. (a) Taking x 0 to correspond to the year 1980, we have a 20 and b 024, so D x 20 024x.
(b)
D 30
(c) The rate of sedimentation is equal to the slope of the graph, 024 cmyr or 24 mmyr.
20 10
0
10
20
30
40
50
x
49. The atmospheric pressure is 100 kPa at sea level (x 0 km), and it decreases by 12 kPa for each kilometer increase in elevation, so a linear model is f x 100 12x. At the peak of Mt. Rainier, the atmospheric pressure is estimated to be f 44 100 12 44 472 kPa. 50. The boiling point of water is 100 C at 100 kPa and it decreases by 375 C for each 10 kPa drop in pressure. In 375 this case the linear model g x has g 100 100 and its graph has slope 0375, so an equation is 10 g x 100 0375 x 100 g x 0375 x 625. We estimate the boiling point of water at 88 kPa to be g 88 0375 88 625 955 C.
44
CHAPTER 2 Functions
51. (a) Let C x ax b be the cost of driving x miles. In May, driving
(b)
480 miles cost $380, and in June, driving 800 miles cost $460. Thus, the
C 600
points 480 380 and 800 460 are on the graph, so the slope is
500
a
1 460 380 . We use the point 480 380 to find the value of b: 800 480 4
380 14 480 b b 260. Thus, C x 14 x 260.
(c) The rate at which the cost increases is equal to the slope of the line, that is 1 . So the cost increases by $025 for every additional mile driven. 4
400 300 200 100 200 400 600 800 1000 1200 1400 x
0
The slope of the graph of C x 14 x 260 is the value of a, 14 . 52. (a) Let C x ax b be the cost of producing x chairs in one day. The first (b) day, it cost $2200 to produce 100 chairs, and the other day it cost $4800 to produce 300 chairs.. Thus, the points 100 2200 and 300 4800 are on 4800 2200 13. We use the point 300 100 100 2200 to find the value of b: 2200 13 100 b b 900. Thus,
the graph, so the slope is a C x 13x 900.
(c) The rate at which the factory’s cost increases is equal to the slope of the line, that is $13chair.
C 10,000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0
100 200 300 400 500 600 x
The slope of the graph of C x 13x 900 is the value of a, 13. f x2 f x1 ax ax1 ax2 b ax1 b 2 . x2 x1 x2 x1 x 2 x1 a x2 x1 ax ax1 a. (b) Factoring the numerator and cancelling, the average rate of change is 2 x2 x1 x2 x1
53. (a) By definition, the average rate of change between x1 and x2 is
f x f a c. x a (b) Multiplying the equation in part (a) by x a, we obtain f x f a c x a. Rearranging and adding f a to both sides, we have f x cx f a ca, as desired. Because this equation is of the form f x Ax B with constants A c and B f a ca, it represents a linear function with slope c and yintercept f a ca.
54. (a) The rate of change between any two points is c. In particular, between a and x, the rate of change is
2.6
TRANSFORMATIONS OF FUNCTIONS
1. (a) The graph of y f x 3 is obtained from the graph of y f x by shifting upward 3 units. (b) The graph of y f x 3 is obtained from the graph of y f x by shifting left 3 units.
2. (a) The graph of y f x 3 is obtained from the graph of y f x by shifting downward 3 units. (b) The graph of y f x 3 is obtained from the graph of y f x by shifting right 3 units.
3. (a) The graph of y f x is obtained from the graph of y f x by reflecting about the xaxis.
(b) The graph of y f x is obtained from the graph of y f x by reflecting about the yaxis.
SECTION 2.6 Transformations of Functions
45
4. (a) The graph of f x 2 is obtained from that of y f x by shifting upward 2 units, so it has graph II.
(b) The graph of f x 3 is obtained from that of y f x by shifting to the left 3 units, so it has graph I.
(c) The graph of f x 2 is obtained from that of y f x by shifting to the right 2 units, so it has graph III.
(d) The graph of f x 4 is obtained from that of y f x by shifting downward 4 units, so it has graph IV. 5. If f is an even function, then f x f x and the graph of f is symmetric with respect to the yaxis. 6. If f is an odd function, then f x f x and the graph of f is symmetric with respect to the origin. 7. (a) The graph of y f x 11 can be obtained by shifting the graph of y f x upward 11 units. (b) The graph of y f x 8 can be obtained by shifting the graph of y f x to the left 8 units.
8. (a) The graph of y f x 7 can be obtained by shifting the graph of y f x to the right 7 units.
(b) The graph of y f x 10 can be obtained by shifting the graph of y f x downward 10 units.
9. (a) The graph of y 14 f x can be obtained by reflecting the graph of y f x about the yaxis, then shrinking the resulting graph vertically by a factor of 14 .
(b) The graph of y 5 f x can be obtained by reflecting the graph of y f x about the xaxis, then stretching the resulting graph vertically by a factor of 5. 10. (a) The graph of y 6 f x can be obtained by reflecting the graph of y f x about the xaxis, then stretching the resulting graph vertically by a factor of 6. (b) The graph of y 23 f x can be obtained by reflecting the graph of y f x about the yaxis, then shrinking the resulting graph vertically by a factor of 23 .
11. (a) The graph of y f x 1 5 can be obtained by shifting the graph of y f x to the right 1 unit and downward 5 units. (b) The graph of y f x 2 4 can be obtained by shifting the graph of y f x to the left 2 units and downward 4 units. 12. (a) The graph of y f x 4 6 can be obtained by shifting the graph of y f x to the right 4 units and upward 6 units. (b) The graph of y f x 2 9 can be obtained by shifting the graph of y f x to the left 2 units and upward 9 units. 13. (a) The graph of y 5 f x can be obtained by reflecting the graph of y f x about the yaxis, then shifting the resulting graph upward 5 units. (b) The graph of y 3 12 f x 2 can be obtained by shifting the graph of y f x to the left 2 units, then shrinking vertically by a factor of 12 , then reflecting about the xaxis, then shifting upward 3 units.
14. (a) The graph of y 10 f x 1 can be obtained by shifting the graph of y f x to the left 1 unit, then reflecting about the xaxis, then shifting upward 10 units. (b) The graph of y 4 f x 5 8 can be obtained by reflecting the graph of y f x about the yaxis, shifting to the right 5 units, stretching vertically by a factor of 4, and shifting down 8 units. 15. (a) The graph of y 2 f 5x can be obtained by shrinking the graph of y f x horizontally by a factor of 15 ,
reflecting about the xaxis, and shifting up 2 units. (b) The graph of y 1 f 12 x 1 can be obtained by shifting the graph of y f x to the left 1 unit, stretching
horizontally by a factor of 2, then shifting upward 1 unit. 16. (a) The graph of y f 13 x 2 can be obtained by stretching the graph of y f x horizontally by a factor of 3 and shifting downward 2 units.
(b) The graph of y f 2 x 3 1 can be obtained by shifting the graph of y f x to the right 3 units, shrinking horizontally by a factor of 12 , then shifting downward 1 unit.
46
CHAPTER 2 Functions
17. (a) The graph of g x x 22 is obtained by shifting the graph of f x to the left 2 units. (b) The graph of g x x 2 2 is obtained by shifting the graph of f x upward 2 units.
18. (a) The graph of g x x 43 is obtained by shifting the graph of f x to the right 4 units. (b) The graph of g x x 3 4 is obtained by shifting the graph of f x downward 4 units.
19. (a) The graph of g x x 2 2 is obtained by shifting the graph of f x to the left 2 units and downward 2 units.
(b) The graph of g x g x x 2 2 is obtained from by shifting the graph of f x to the right 2 units and upward 2 units.
20. (a) The graph of g x x 1 is obtained by reflecting the graph of f x about the xaxis, then shifting the resulting graph upward 1 unit. (b) The graph of g x x 1 is obtained by reflecting the graph of f x about the yaxis, then shifting the resulting graph upward 1 unit. 21. (a)
y
y
(b) y=x@
y=x@
1 0
1
x
y=2(x+3)@
y=x@-4
2 0
(c)
y
1
x
y
(d)
y=x@
y=x@
1
1
0
0
1
x
1
x
y=1-x@ y=(x+1)@-3
22. (a)
y
(b)
y
y=Ïx+1
y=Ïx y=Ïx-2
1 1
x
y=Ïx
1 1
x
SECTION 2.6 Transformations of Functions y
(c)
y
(d)
y=Ïx+2+2
y=Ïx
1
y=Ïx
1
1
47
1
x
x y=_Ïx+1
23. The graph of y x 1 is obtained from that of y x by shifting to the left 1 unit, so it has graph II. Its range is [0 . 24. y x 1 is obtained from that of y x by shifting to the right 1 unit, so it has graph IV. Its range is [0 . 25. The graph of y x 1 is obtained from that of y x by shifting downward 1 unit, so it has graph I. Its range is [1 . 26. The graph of y x is obtained from that of y x by reflecting about the xaxis, so it has graph III. Its range is 0]. 27. f x x 2 5. Shift the graph of y x 2 downward 5 units.
28. f x x 2 2. Shift the graph of y x 2 upward 2 units. y
y
y=x@+2 y=x@
y=x@ 1 0
1
x
1
0
29. f x 3 units.
x 3. Shift the graph of y y
x upward
x
1
y=x@-5
30. f x x 5. Shift the graph of y x downward 5 units.
y
y=Ïx+3
y=| x |
1 1
y=Ïx
1 1
x
x y=| x |-5
48
CHAPTER 2 Functions
31. f x x 52 . Shift the graph of y x 2 to the right 5 units.
32. f x x 12 . Shift the graph of y x 2 to the left 1 unit.
y
y y=x@ y=(x-5)@
y=x@
y=(x+1)@
5 1
x
1
33. f x x 2. Shift the graph of y x to the left 2 units.
y
1
34. f x 4 units.
x 4. Shift the graph of y
x to the right
y
y=| x+2 |
y=Ïx
y=| x |
y=Ïx-4
1
1 1
x
1
x
35. f x x 3 . Reflect the graph of y x 3 about the xaxis.
y
x
36. f x x. Reflect the graph of y x about the xaxis.
y
y=| x |
y=x#
2 1
1
x y=_x#
1
x y=_| x |
SECTION 2.6 Transformations of Functions
37. f x 4 x. Reflect the graph of y 4 x about the yaxis.
49
38. f x 3 x. Reflect the graph of y 3 x about the yaxis.
y
y
2 y=Îx 4 y=Ï_x
4 y=Ïx
2
1
x y=Î_x
10
x
39. f x 5x 2 . Stretch the graph of y x 2 vertically by a factor of 5.
y
y=5x@
40. f x 13 x. Shrink the graph of y x vertically by a factor of 13 .
y
y=|x| y=x@ 1
1
2 1
x
41. f x 15 x. Reflect the graph of y x about the yaxis, then shrink vertically by a factor of 15 .
0
y=3|x| x
1
42. f x 2 x. Reflect the graph of y x about the yaxis, then stretch vertically by a factor of 2. y
y
y=2Ï_x y=Ïx
1
y=Ïx
1 1 1 y=_5 Ïx
x
1
x
50
CHAPTER 2 Functions
43. f x x 4 2. Shift the graph of y x to the right 44. f x x 12 3. Shift the graph of y x 2 to the left 4 units, then shift upward 2 units.
1 unit, then shift downward 3 units. y
y
y=(x+1)@-3 y=|x|
y=x@ 1
y=|x-4|+2
0
1 0
1
x
x
1
45. f x 2 x 4 3. Shift the graph of y x to the left 4 units, stretch vertically by a factor of 2, reflect about the xaxis, then shift upward 3 units.
46. f x 1 12 x 2. Shift the graph of y x to the right 2 units, shrink vertically by a factor of 12 , reflect
about the xaxis, then shift upward 1 unit.
y
y y=|x|
y=Ïx
1
1 x
1
x
1
y=_2Ïx+4+3
1
y=1- 2 |x-2|
47. f x 12 x 22 3. Shift the graph of y x 2 to the left 2 units, shrink vertically by a factor of 12 , then shift
downward 3 units.
48. f x 2 x 1 3. Shift the graph of y x to the right 1 unit, stretch vertically by a factor of 2, then shift upward 3 units. y
y y=x@
y=2Ïx-1+3
1
y=Ïx 1
1
y= 2 (x+2)@-3
x
1 1
x
51
SECTION 2.6 Transformations of Functions
49. f x 12 x 4 3. Shrink the graph of y x
vertically by a factor of 12 , then shift the result to the left
50. f x 3 2 x 12 . Stretch the graph of y x 2 vertically by a factor of 2, reflect the result about the
xaxis, then shift the result to the right 1 unit and upward
4 units and downward 3 units.
3 units.
y
y y=x@
y=Ïx 1
1 4
1
x 1
x
y=3-2(x-1)@
y=2Ïx+4-3
51. y f x 10. When f x x 2 , y x 2 10.
52. y f x 4. When f x
53. y f x 3. When f x x 4 , y x 34 .
54. y f x 8. When f x x 3 , y x 83 .
55. y f x 2 5. When f x x, y x 2 5.
56. y f x 4 3. When f x x,
57. y f x 1. When f x 4 x, y 4 x 1.
58. y f x 2. When f x x 2 , y x 22 .
59. y 2 f x 3 2. When f x x 2 ,
60. y 12 f x 1 3. When f x x,
61. g x f x 2 x 22 x 2 4x 4
62. g x f x 3 x 3 3
63. g x f x 1 2 x 1 2 65. g x f x 2 x 2
64. g x 2 f x 2 x 66. g x f x 2 1 x 22 1 x 2 4x 3
67. (a) y f x 4 is graph #3.
68. (a) y 13 f x is graph #2.
y 2 x 32 2.
x, y
y x 4 3.
y 12 x 1 3.
(b) y f x 3 is graph #1.
(b) y f x 4 is graph #3.
(c) y 2 f x 6 is graph #2.
(c) y f x 4 3 is graph #1.
(d) y f 2x is graph #4.
(d) y f x is graph #4.
69. (a) y f x 2
(b) y f x 2
y
1
(c) y 2 f x
y
1 1
x
x 4.
y
1 1
x
1
x
52
CHAPTER 2 Functions
(d) y f x 3
(e) y f x
y
y
1 x
(b) y f x
y
1
1
1
x
(c) y f x 2
y
1 1
(d) y f x 2
y
1 1
70. (a) y f x 1
(f) y 12 f x 1
(e) y f x
y
1
(f) y 2 f x
y
71. (a) y f 2x
x
(b) y f
y
1x 2
y=f(2x)
y=f(x)
x
1
x
y
y
1 1
x
1 1
x
1
y
x
1 1
x
1 1
x
1
1 x) y=f(_ 2
1 1
y=f(x)
x
SECTION 2.6 Transformations of Functions
(c) y 2 f 2x
(d) y 2 f
y
1x 2
y y=2f(2x)
1
y=f(3x)
1x 3
y
y=f(x)
1 y=f(x)
1 y=f( _ 3 x)
x
1
3
(d) y 1 f
y
y=f(3(x+1))
x 1 x) y=_2f(_ 2
(b) y f
y
(c) y f 3 x 1
1
x
y=f(x)
72. (a) y f 3x
y=f(x)
1
1
1
1x 3
y=f(x)
1 y=1-f( _ 3 x)
1
y=f(x)
73. y [[2x]]
74. y 14 x
y
1
8
(d)
4
8
6
4
2
4
6
x
2
x
y
x
For part (b), shift the graph in (a) to the left 5 units; for part (c), shift the graph in (a) to the left 5 units, and stretch it vertically by a factor of 2; for part (d), shift
(c)
the graph in (a) to the left 5 units, stretch it vertically by a factor of 2, and then shift
(b)
it upward 4 units.
(a) 2 0
3
1 1
75.
x
y
x
1
53
8
54
CHAPTER 2 Functions
76. (a)
8
6
4
2
(b)
(c)
6
For (b), reflect the graph in (a) about the xaxis; for (c), stretch the graph in (a)
4
vertically by a factor of 3 and reflect about the xaxis; for (d), shift the graph in (a)
2
to the right 5 units, stretch it vertically by a factor of 3, and reflect it about the 2
4
6
8
2
4
(c)
(d)
For part (b), shrink the graph in (a) vertically by a factor of 13 ; for part (c), shrink
(a) (b)
the graph in (a) vertically by a factor of 13 , and reflect it about the xaxis; for
2
4
2
2 2
important to obtain the graphs in part (c) and (d).
4 6
77.
xaxis. The order in which each operation is applied to the graph in (a) is not
4
(c)
6 (d)
part (d), shift the graph in (a) to the right 4 units, shrink vertically by a factor of 13 , and then reflect it about the xaxis.
4
78. (b)
6
4
4
For (b), shift the graph in (a) to the left 3 units; for (c), shift the graph in (a) to the
2 (a)
left 3 units and shrink it vertically by a factor of 12 ; for (d), shift the graph in (a) to the left 3 units, shrink it vertically by a factor of 12 , and then shift it downward
(c)
2
4
2
6
3 units. The order in which each operation is applied to the graph in (a) is not important to sketch (c), while it is important in (d).
(d) 4
(b) y f 2x 2 2x 2x2 4x 4x 2
79. (a) y f x 2x x 2
4
2
1x 2
(c) y f x 14 x 2
2 2 12 x 12 x
4
4
4
2
2
2
2
2
4
4
2
4
2
2
4
4
2
4
2
2
4
4
The graph in part (b) is obtained by horizontally shrinking the graph in part (a) by a factor of 12 (so the graph is half as wide). The graph in part (c) is obtained by horizontally stretching the graph in part (a) by a factor of 2 (so the graph is twice as wide). 80. (a) y f x 2x x 2 (b) y f x 2 x x2 (c) y f x 2 x x2 2x x 2 2x x 2
4
2
4
4
4
2
2
2
2 4
2
4
4
2
2 4
2
4
4
2
2 4
2
4
SECTION 2.6 Transformations of Functions
(d) y f 2x 2 2x 2x2 2x x 2 4x 4x 2
(e) y f 12 x x 14 x 2
4
2 2 12 x 12 x
4
2 4
2
55
2 2
2
4
4
4
2
2
2
4
4
The graph in part (b) is obtained by reflecting the graph in part (a) about the yaxis. The graph in part (c) is obtained by rotating the graph in part (a) through 180 about the origin [or by reflecting the graph in part (a) first about the xaxis and then about the yaxis]. The graph in part (d) is obtained by reflecting the graph in part (a) about the yaxis and then horizontally shrinking the graph by a factor of 12 (so the graph is half as wide). The graph in part (e) is obtained by reflecting the graph in part (a) about the yaxis and then horizontally stretching the graph by a factor of 2 (so the graph is twice as wide).
81. f x x 4 . f x x4 x 4 f x. Thus f x
82. f x x 3 . f x x3 x 3 f x. Thus
y
y
is even.
f x is odd.
2 1
x
2 1
x
83. f x x 2 x. f x x2 x x 2 x. Thus 84. f x x 4 4x 2 . f x f x. Also, f x f x, so f x is
neither odd nor even.
y
f x x4 4 x2 x 4 4x 2 f x. Thus f x is even.
y
1 1
x
1 1
x
56
CHAPTER 2 Functions
85. f x x 3 x.
86. f x 3x 3 2x 2 1.
f x 3 x3 2 x2 1 3x 3 2x 2 1.
f x x3 x x 3 x x 3 x f x.
Thus f x f x. Also f x f x, so f x is neither odd nor even.
Thus f x is odd.
y
y
1
1
87. f x 1 3 x. f x 1 3 x 1 3 x. Thus f x f x. Also f x f x, so f x is
neither odd nor even.
x
1
x
1
88. f x x 1x. f x x 1 x x 1x x 1x f x.
y
Thus f x is odd.
y
1 x
1
1 x
1
89. (a) Even
(b) Odd y
y
1
1 1
x
1
x
57
SECTION 2.6 Transformations of Functions
90. (a) Even
(b) Odd y
y
1
1 1
1
x
x
91. (a) Since f x x 2 4 0, for 2 x 2, the graph of y g x is found by sketching the graph of y f x for x 2 and x 2, then reflecting about the xaxis the part of the graph of y f x for 2 x 2. y
(b)
y
1
1 1
x
g x 4x x 2 y
92. (a)
x
1
h x x 3 y
(b)
y=f(|x|) y=|f(x)| 0 y=f(x)
0
x
x
y=f(x)
93. (a) The bungee jumper drops to a height of 200 feet, bounces up and down, then settles at 350 feet.
(b)
y (ft) 500
(c) To obtain the graph of H from that of h, we shift downward 100 feet. Thus, H t h t 100.
0
4
t (s)
58
CHAPTER 2 Functions
(b)
94. (a) The swimmer swims two and a half laps, slowing down with each
80
successive lap. In the first 30 seconds they swim 50 meters, so their
60
average speed is 50 30 167 ms.
40 20
(c) Here the swimmer swims 60 meters in 30 seconds, so the average speed is 60 2 ms. 30
0
100
200
t
This graph is obtained by stretching the original graph vertically by a factor of 12.
95. (a) The trip to the park corresponds to the first piece of the graph. The class travels 800 feet in 10 minutes, so their average speed is 800 10 80 ftmin. The second (horizontal) piece of the graph stretches from t 10 to t 30, so the class spends 20 minutes at the park. The park is 800 feet from the school.
(b)
(c)
y 600
y 1000 800
400
600 400
200 0
200 20
40
0
60 t
20
40
60 t
The new graph is obtained by shrinking the original
This graph is obtained by shifting the original graph
graph vertically by a factor of 050. The new average
to the right 10 minutes. The class leaves ten minutes
speed is 40 ftmin, and the new park is 400 ft from
later than it did in the original scenario.
the school.
96. (a) For z 11, the graph of g x f 1 z x f 12x is obtained from that of f x by shrinking horizontally by a 1. factor of 12
(b) g x f 12x, so in this case 12x 480 x 40. The dip occurs at 40 nm at the source. (c) The observed dip at 410 nm corresponds to a wavelength of 410 12 34 nm at the source. Similarly,
the observed dips at 430, 480, and 660 nm
480 correspond to dips at 430 12 36, 12 40, and 660 55 nm at the source. 12
g (lm) 2 1 0
30
40
50
60
x (nm)
97. To obtain the graph of g x x 22 5 from that of f x x 22 , we shift to the right 4 units and upward 5 units.
98. To obtain the graph of g x from that of f x, we reflect the graph about the yaxis, then reflect about the xaxis, then shift upward 6 units.
SECTION 2.7 Combining Functions
99.
y
y
1
y
1 1
x
59
1 1
x
1
x
f x [[x]]
g x [[2x]] h x [[3x]] 1 The graph of k x [[nx]] is that of a step function with steps units wide, and the graph of k x n1 x is that of a n step function with steps n units wide. 100. f even implies f x f x; g even implies g x g x; f odd implies f x f x; and g odd implies g x g x. If f and g are both even, then f g x f x g x f x g x f g x and f g is even. If f and g are both odd, then f g x f x g x f x g x f g x and f g is odd. If f is odd and g is even, then f g x f x g x f x g x, which is neither odd nor even.
101. f even implies f x f x; g even implies g x g x; f odd implies f x f x; and g odd implies g x g x. If f and g are both even, then f g x f x g x f x g x f g x. Thus f g is even. If f and g are both odd, then f g x f x g x f x g x f x g x f g x. Thus f g is even If f if odd and g is even, then f g x f x g x f x g x f x g x f g x. Thus f g is odd.
102. f x x n is even when n is an even integer and f x x n is odd when n is an odd integer. These names were chosen because polynomials with only terms with odd powers are odd functions, and polynomials with only terms with even powers are even functions. 103. Any function that is periodic with period 1 has this property. So if we draw any function on the interval [0 1 and repeat its graph on every other interval [n n 1, n an integer, we will have such a function. One example is f x x [[x]].
2.7
COMBINING FUNCTIONS
1. From the graphs of f and g in the figure, we find f g 2 f 2 g 2 3 5 8, f 2 3 f . 2 f g 2 f 2 g 2 3 5 2, f g 2 f 2 g 2 3 5 15, and g g 2 5
2. By definition, f g x f g x. So, if g 2 5 and f 5 12, then f g 2 f g 2 f 5 12.
3. If the rule of the function f is “add one” and the rule of the function g is “multiply by 2” then the rule of f g is “multiply by 2, then add one” and the rule of g f is “add one, then multiply by 2.” 4. We can express the functions in Exercise 3 algebraically as f x x 1, g x 2x, f g x 2x 1, and g f x 2 x 1.
5. (a) The function f g x is defined for all values of x that are in the domains of both f and g. (b) The function f g x is defined for all values of x that are in the domains of both f and g.
(c) The function f g x is defined for all values of x that are in the domains of both f and g, and g x is not equal to 0. 6. The composition f g x is defined for all values of x for which x is in the domain of g and g x is in the domain of f .
60
CHAPTER 2 Functions
7. Both f x 3x and g x 1 x have domain , and the intersection of their domains is . f g x 3x 1 x 2x 1 with domain ; f g x 3x 1 x 4x 1 with domain ; 3x f with domain 1 1 . x f g x 3x 1 x 3x 3x 2 with domain ; and g 1x 8. Both f x 3 2x and g x 2x 1 have domain , and the intersection of their domains is . f g x 3 2x 2x 1 4 with domain ; f g x 3 2x 2x 1 2 4x with domain 3 2x f with domain x ; f g x 3 2x 2x 1 4x 2 4x 3 with domain ; and g 2x 1 12 12 . 9. Both f x x 3 x 2 and g x x 2 have domain , and the intersection of their domains is .
f g x x 3 2x 2 with domain ; f g x x 3 with domain ; f g x x 5 x 4 with x3 x2 f x 1 with domain 0 0 . domain ; and x g x2
10. Both f x x 2 1 and g x x 2 2 have domain , and the intersection of their domains is .
f g x 2x 2 3 with domain ; f g x 1 with domain ; x2 1 f with domain . f g x x 2 1 x 2 2 x 4 3x 2 2 with domain ; and x 2 g x 2
11. Both f x 5 x and g x x 2 3x have domain , and the intersection of their domains is . f g x 5 x x 2 3x x 2 4x 5 with domain ; f g x 5 x x 2 3x x 2 2x 5 with domain ; f g x 5 x x 2 3x x 3 8x 2 15x with domain ; and 5x f 5x with domain 0 0 3 3 . x 2 g x x 3 x 3x 12. f x x 2 2x has domain ; g x 3x 2 1 has domain ; and the intersection of their domains is . f g x x 2 2x 3x 2 1 4x 2 2x 1 with domain ; f g x x 2 2x 3x 2 1 2x 2 2x 1 with domain ; f g x x 2 2x 3x 2 1 3x 4 6x 3 x 2 2x with domain ; and x 2 2x f , 3x 2 1 0 x 33 with domain x x 33 . x 2 g 3x 1 13. f x [3 5].
25 x 2 has domain [5 5]; g x x 3 has domain [3 ; and the intersection of their domains is
f g x 25 x 2 x 3 with domain [3 5]; f g x 25 x 2 x 3 with domain [3 5]; 25 x 2 f 2 with domain 3 5]. x f g x 25 x x 3 with domain [3 5]; and g x 3
61
SECTION 2.7 Combining Functions
14. f x 16 x 2 has domain [4 4]; g x x 2 1 has domain 1] [1 ; and the intersection of their domains is [4 1] [1 4]. f g x 16 x 2 x 2 1 with domain [4 1] [1 4]; f g x 16 x 2 x 2 1 with domain 16 x 2 f 2 2 [4 1] [1 4]; f g x with 16 x x 1 with domain [4 1] [1 4]; and x g x2 1 domain [4 1 1 4]. 1 3 15. f x has domain x x 1; g x has domain x x 2; the intersection of their domains is x 1 x 2 x x 1 2 or, in interval notation, 1 1 2 2 . 1 3 4x 1 x 2 3 x 1 with domain 1 1 2 2 ; f g x x 1 x 2 x 1 x 2 x 1 x 2 3 2x 5 1 x 2 3 x 1 with domain 1 1 2 2 ; f g x x 1 x 2 x 1 x 2 x 1 x 2 3 3 1 with domain 1 1 2 2 ; and f g x x 1 x 2 x 1 x 2 1 f x 1 x 2 with domain 1 1 2 2 . x 3 g 3 x 1 x 2 3 x 16. f x has domain x x 3; g x has domain x x 3; and the intersection of their domains is x 3 x 3 x x 3 or, in interval notation, 3 3 3 3 .
3 x x2 9 x x 2 6x 9 3 2 with domain x x 3; f g x x 3 x 3 x 3 x 3 x 9 x2 9 3 x 3x with domain x x 3; f g x 2 with domain x x 3; and x 3 x 3 x 9 3 3 x 3 3x 9 f x 2 with domain x x 0 3. x x 3 g x x 3 x 3x x 3 17. f x x 3 x. The domain of x is [0 , and the domain of 3 x is 3]. Thus, the domain of f is 3] [0 [0 3]. 1x 1x . The domain of x 4 is [4 , and the domain of is 0 0 1]. Thus, the 18. f x x 4 x x domain of f is [4 0 0 1] [4 0 0 1]. 1 19. h x x 314 . Since the fourth root is an even root and the denominator cannot equal 0, x 3 0 x 314 x 3 So the domain is 3 . 1 x 3 . The domain of x 3 is [3 , and the domain of is x 1. Since x 1 corresponds to the 20. k x x 1 x 1 union of intervals 1 1 , the domain is [3 1 1 [3 1 1 . f g x
21.
22.
y
y
23.
f
g 0
f+g 0
4
f+g
f
g
2
f
x
f+g
g
x 4
2
0
2
4
62
CHAPTER 2 Functions
24. 5
25.
4 3
4
2
0
2
26.
3
f
2
4
f+g f
2
2
1
20
g 1
g
20 f
1
f+g
40
f+g
g
40
4
2
00
In Solutions 27–32, f x 4x 5 and g x x 2 2. 27. (a) f g 1 f 12 2 f 3 4 3 5 17 (b) g f 1 g 4 1 5 g 9 92 2 83
28. (a) f f 0 f 4 0 5 f 5 4 5 5 25 (b) g g 1 g 12 2 g 3 32 2 11 29. (a) f g 2 f g 2 f 22 2 f 6 4 6 5 29 (b) g f 1 g f 1 g 4 1 5 g 1 12 2 3
30. (a) f f 1 f f 1 f 4 1 5 f 9 4 9 5 41 (b) g g 0 g g 0 g 02 2 g 2 22 2 6 31. (a) f g x f g x f x 2 2 4 x 2 2 5 4x 2 13
(b) g f x g f x g 4x 5 4x 52 2 16x 2 40x 27
32. (a) f f x f f x f 4x 5 4 4x 5 5 16x 25 2 (b) g g x g g x g x 2 2 x 2 2 2 x 4 4x 2 6
33. f g 2 f 5 4
34. f 0 0, so g f 0 g 0 3.
35. g f 4 g f 4 g 2 5
36. g 0 3, so f g 0 f 3 0.
37. g g 2 g g 2 g 1 4
38. f 4 2, so f f 4 f 2 2.
39. From the table, g 2 5 and f 5 6, so f g 2 6.
40. From the table, f 2 3 and g 3 6, so g f 2 6.
41. From the table, f 1 2 and f 2 3, so f f 1 3. 42. From the table, g 2 5 and g 5 1, so g g 2 1.
43. From the table, g 6 4 and f 4 1, so f g 6 1.
44. From the table, f 2 3 and g 3 6, so g f 2 6.
45. From the table, f 5 6 and f 6 3, so f f 5 3. 46. From the table, g 3 6 and g 6 4, so g g 3 4.
47. f x 2x 3 has domain ; g x 4x 1 has domain . f g x f 4x 1 2 4x 1 3 8x 1 with domain . g f x g 2x 3 4 2x 3 1 8x 11 with domain . f f x f 2x 3 2 2x 3 3 4x 9 with domain . g g x g 4x 1 4 4x 1 1 16x 5 with domain .
2
4
SECTION 2.7 Combining Functions
x 48. f x 6x 5 has domain . g x has domain . 2 x x 6 5 3x 5 with domain . f g x f 2 2 6x 5 3x 52 with domain . g f x g 6x 5 2 f f x f 6x 5 6 6x 5 5 36x 35 with domain . x x x 2 with domain . g g x g 2 2 4 49. f x x 2 has domain ; g x x 1 has domain .
f g x f x 1 x 12 x 2 2x 1 with domain . g f x g x 2 x 2 1 x 2 1 with domain . 2 f f x f x 2 x 2 x 4 with domain . g g x g x 1 x 1 1 x 2 with domain .
1 50. f x 3 x has domain . g x 3 has domain x x 0 0 0 . x 1 1 1 3 3 with domain x x 0 0 0 . f g x f x x3 x 1 1 g f x g 3 x 3 with domain x x 0 0 0 . 3 x x f f x f 3 x 3 3 x 9 x with domain . 1 1 x 3 with domain x x 0 0 0 . g g x g x3 1x 3 1 51. f x x 2 1 has domain . g x has domain 0 . x 1 1 1 2 1 1 with domain 0 . f g x f x x x 1 with domain . g f x g x 2 1 x2 1 2 f f x f x 2 1 x 2 1 1 x 4 2x 2 2 with domain . 1 1 4 x with domain 0 . g g x g x 1 x 52. f x x 4 has domain . g x x 4 has domain . f g x f x 4 x 4 4 with domain . g f x g x 4 x 4 4 x with domain . f f x f x 4 x 4 4 x 8 with domain . g g x g x 4 x 4 4 x 4 4 (x 4 4 is always positive). The domain is .
63
64
CHAPTER 2 Functions
x has domain x x 1; g x 2x 1 has domain x 1 2x 1 2x 1 with domain x x 0 0 0 . f g x f 2x 1 2x 2x 1 1 2x x x 2 1 1 with domain x x 1 1 1 g f x g x 1 x 1 x 1 x x x x 1 x x x 1 . f f x is defined whenever both f x and f f x f x 1 x 1 x x 1 2x 1 1 x 1 f f x are defined; that is, whenever x 1 and 2x 1 0 x 12 , which is 1 1 12 12 .
53. f x
g g x g 2x 1 2 2x 1 1 4x 2 1 4x 3 with domain . x 1 has domain x x 1; g x has domain x x 0. x 1 x 1 1 1 1 1 x . f g x is defined whenever both g x and f g x are f g x f x x 1 1 x 1 1
54. f x
x
x
defined, so the domain is x x 1 0. 1 x 1 x x . g f x is defined whenever both f x and g f x are defined, so the g f x g x 1 x x1
domain is x x 1 0. x x x x xx1 . f f x is defined whenever both f x and f f x f x x 1 1 2x 1 x 1 x1 1 x1 f f x are defined, so the domain is x x 1 12 . 1 1 1 x. g g x is defined whenever both g x and g g x are defined, so the domain is g g x g x x
x x 0.
2 x has domain x x 0; g x has domain x x 2. x x 2 2x 4 2 x x . f g x is defined whenever both g x and f g x are defined; that f g x f x 2 x x 2 is, whenever x 0 and x 2. So the domain is x x 0 2. 2 1 2 2 x . g f x is defined whenever both f x and g f x are defined; g f x g 2 x 2 2x 1x 2 x that is, whenever x 0 and x 1. So the domain is x x 0 1. 2 2 x. f f x is defined whenever both f x and f f x are defined; that is, whenever f f x f 2 x x x 0. So the domain is x x 0. x x x x xx 2 . g g x is defined whenever both g x and g g x g x 2 x 2 2 3x 4 x 2 x 2 g g x are defined; that is whenever x 2 and x 43 . So the domain is x x 2 43 .
55. f x
SECTION 2.7 Combining Functions
65
1 1 has domain x x 1; g x 2 has domain . x 1 x 1 1 1 x2 1 1 1 with domain x x 0. f g x f 1 x2 1 1 x2 1 x2 1 x2 1 1 x 2 2x 1 1 x 12 2 with domain x x 1. g f x g 2 2 x 1 x 2x 2 1 x 1 1 1 x 1 1 1x x 1 1 , with domain x x 1 2. f f x f 1 x 1 1 x 1 x 2 1 x 1 2 21 x x 4 2x 2 1 1 1 with domain . g g x g 2 2 4 2 x 1 x 2x 2 2 1 1 x2 1 1 x2 1
56. f x
1 57. f x has domain x x 0; g x x 2 4x has domain . x 1 . f g x is defined whenever 0 x 2 4x x x 4. The product of two f g x f x 2 4x x 2 4x numbers is positive either when both numbers are negative or when both numbers are positive. So the domain of f g is x x 0 and x 4 x x 0 and x 4 which is 0 4 . 1 4 1 2 1 1 . g f x is defined whenever both f x and g f x are 4 g f x g x x x x x defined, that is, whenever x 0. So the domain of g f is 0 . 1 1 x 14 . f f x is defined whenever both f x and f f x are defined, that is, f f x f x 1 x whenever x 0. So the domain of f f is 0 . 2 g g x g x 2 4x x 2 4x 4 x 2 4x x 4 8x 3 16x 2 4x 2 16x x 4 8x 3 12x 2 16x
with domain .
58. f x x 2 has domain . g x x 3 has domain [3 . 2 x 3 x 3 x 3 with domain [3 . f g x f g f x g x 2 x 2 3. For the domain we must have x 2 3 x 3 or x 3. Thus the domain is 3 3 . 2 f f x f x 2 x 2 x 4 with domain . x 3 3. For the domain we must have x 3 3 x 3 9 x 12, so the g g x g x 3 domain is [12 .
66
CHAPTER 2 Functions
x has domain [0 ; g x 3 x has domain . f g x f 3 x 1 3 x 1 6 x. For 6 x to be defined we must have x 0, so the domain of f g is [0 . g f x g 1 x 3 1 x with domain [0 . f f x f 1 x 1 1 x. As in Example 4, for x to be defined we must have x 0, and for 1 x to be defined we must have 1 x 0 x 1 x 1. Thus, to satisfy both conditions we must have 0 x 1, so the domain of f f is [0 1]. g g x g 3 x 3 3 x 9 x with domain . 60. f x x 2 1 has domain 1] [1 ; g x 1 x has domain 1]. 2 1x 1 x 1 1 x 1 x with domain 0]. f g x f x 2 1 1 x 2 1. For x 2 1 to be defined we must have x 2 1 0 x 2 1 x 1 g f x g or x 1, and for 1 x 2 1 to be defined we must have 1 x 2 1 0 x 2 1 1 x 2 1 1 x 2 2 x 2 2 x 2. Thus, the domain of g f is 2 1 1 2 . 2 2 x 1 x 2 1 1 x 2 2. For x 2 2 to be defined we must have x 2 2 0 f f x f x 2 x 2 or x 2. These values also satisfy x 2 1 0, so the domain of f f is 2 2 . g g x g 1 x 1 1 x. For 1 x to be defined we must have 1 x 0 x 1, and for 1 1 x to be defined we must have 1 1 x 0 1 x 1 1 x 1 x 0. Thus, the domain of g g is [0 1]. 61. f g h x f g h x f g x 1 f x 1 x 11 3 62. g h x g x 2 2 x 2 2 x 6 6x 4 12x 2 8. 1 . f g h x f x 6 6x 4 12x 2 8 6 x 6x 4 12x 2 8 4 x 5 x 5 1 63. f g h x f g h x f g x f 3 3 3 x x x 3 . f g h x f . 64. g h x g x 3 3 3 x 1 x 1 x 1 59. f x 1
For Exercises 65–78, many answers are possible.
65. F x x 95 . Let f x x 5 and g x x 9, then F x f g x. 66. F x x 1. If f x x 1 and g x x, then F x f g x.
x x2 and g x x 2 , then F x f g x. . Let f x 67. F x 2 x 4 x 4 1 1 68. F x . If f x and g x x 3, then F x f g x. x 3 x 3 69. F x 1 x . Let f x x and g x 1 x 3 , then F x f g x. 70. F x 1 x. If f x 1 x and g x x, then F x f g x. 71. F x 1 x 3 1. If f x 1 x and g x x 3 1, then F x f g x. 1 1 and g x x 2 x 1, then F x f g x. 72. F x . If f x 3 3 2 x x x 1
1 1 73. F x 2 . Let f x , g x x 1, and h x x 2 , then F x f g h x. x x 1
SECTION 2.7 Combining Functions
67
74. F x 3 x 1. If g x x 1 and h x x, then g h x x 1, and if f x 3 x, then F x f g h x. 9 75. F x 4 3 x . Let f x x 9 , g x 4 x, and h x 3 x. Then F x f g h x. 2 2 76. F x 2 . If g x 3 x and h x x, then g h x 3 x, and if f x x 2 , then 3 x F x f g h x. 3 x x 77. F x . If h x x, g x , and f x x 3 , then F x f g h x. x 1 x 1 1 1 1 . If h x x 2 1, g x 1 , and f x , then F x f g h x. 78. F x 1 x x 1 x2 1 79. Yes. If f x m 1 x b1 and g x m 2 x b2 , then f g x f m 2 x b2 m 1 m 2 x b2 b1 m 1 m 2 x m 1 b2 b1, which is a linear function, because it is of the form y mx b. The slope is m 1 m 2 . 80. (a) From the graphs we can see that f g 1 f 1 g 1 2 1 3. The only graph that satisfies this condition is graph VI. (b) The graphs of f and g intersect at x 23 and f x g x for x & 23, so f g x 0 for x & 23, corresponding to graph I. (c) Because f and g are both negative for x 1, have opposite signs for 1 x 0, and are both positive for x 0, the graph of f g is negative only on 1 0, corresponding to graph III. (d) g 2 x is everywhere nonnegative, corresponding to graph II. (e) Because f 1 0, the graph of g f has a vertical asymptote at x 1, corresponding to graph V.
(f) f x x 1, so the graph of f g is the graph of g translated upward 1 unit, corresponding to graph IV. 81. The price per sticker is 015 0000002x and the number sold is x, so the revenue is R x 015 0000002x x 015x 0000002x 2 .
82. As found in Exercise 81, the revenue is R x 015x 0000002x 2 , and the cost is 0095x 00000005x 2 , so the profit is P x 015x 0000002x 2 0095x 00000005x 2 0055x 00000015x 2 .
83. (a) Because the ripple travels at a speed of 60 cm/s, the distance traveled in t seconds is the radius, so g t 60t. (b) The area of a circle is r 2 , so f r r 2 .
(c) f g t g t2 60t2 3600t 2 cm2 . This function represents the area of the ripple as a function of time.
84. (a) Let f t be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 2 cm/s, the radius is r f t 2t after t seconds. (b) The volume of the balloon can be written as g r 43 r 3 .
3 (c) g f t 43 2t3 32 3 t . g f represents the volume as a function of time.
85. Let r be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 2 cm/s, the radius is r 2t after t seconds. Therefore, the surface area of the balloon can be written as S 4r 2 4 2t2 4 4t 2 16t 2 .
86. (a) From Exercises 2.5.49–50, f x 100 12x and g x 0375 x 625. Thus, h x g f x g 100 12x 0375 100 12x 625 100 45x. The inputs of h are the inputs of f (elevations), and its outputs are the outputs of g (boiling points). The function h models the boiling point of water as a function of elevation. (b) Using the formula for h, we estimate the boiling point of water at the peak of Mt. Rainier to be h 44 100 45 44 802 C.
68
CHAPTER 2 Functions
(c) We solve h x 91 100 45x 91 x 2. Water boils at 91 C approximately 2 km above sea level. 87. (a) f x 080x
(b) g x x 50
(c) f g x f x 50 080 x 50 080x 40. f g represents applying the $50 coupon, then the 20% discount. g f x g 080x 080x 50. g f represents applying the 20% discount, then the $50 coupon. So applying the 20% discount, then the $50 coupon gives the lower price.
88. (a) f x 090x
(b) g x x 100
(c) f g x f x 100 090 x 100 090x 90. f g represents applying the $100 coupon, then the 10% discount. g f x g 090x 090x 100. g f represents applying the 10% discount, then the $100 coupon. So applying the 10% discount, then the $100 coupon gives the lower price.
89. Let t be the time since the plane flew over the radar station. (a) Let s be the distance in miles between the plane and the radar station, and let d be the horizontal distance that the plane has flown. Using the Pythagorean theorem, s f d 1 d 2 . (b) Since distance rate time, we have d g t 350t. (c) s t f g t f 350t 1 350t2 1 122,500t 2 .
90. A x 105x A A x A A x A 105x 105 105x 1052 x. A A A x A A A x A 1052 x 105 1052 x 1053 x. A A A A x A A A A x A 1053 x 105 1053 x 1054 x. A represents the amount in
the account after 1 year; A A represents the amount in the account after 2 years; A A A represents the amount in the account after 3 years; and A A A A represents the amount in the account after 4 years. We can see that if we compose n copies of A, we get 105n x.
91. g x 2x 1 and h x 4x 2 4x 7.
Method 1: Notice that 2x 12 4x 2 4x 1. We see that adding 6 to this quantity gives
2x 12 6 4x 2 4x 1 6 4x 2 4x 7, which is h x. So let f x x 2 6, and we have
f g x 2x 12 6 h x. Method 2: Since g x is linear and h x is a second degree polynomial, f x must be a second degree polynomial, that is, f x ax 2 bx c for some a, b, and c. Thus f g x f 2x 1 a 2x 12 b 2x 1 c
4ax 2 4ax a 2bx b c 4ax 2 4a 2b x a b c 4x 2 4x 7. Comparing this with f g x, we
have 4a 4 (the x 2 coefficients), 4a 2b 4 (the x coefficients), and a b c 7 (the constant terms) a 1 and
2a b 2 and a b c 7 a 1, b 0 c 6. Thus f x x 2 6.
f x 3x 5 and h x 3x 2 3x 2. Note since f x is linear and h x is quadratic, g x must also be quadratic. We can then use trial and error to find g x.
Another method is the following: We wish to find g so that f g x h x. Thus f g x 3x 2 3x 2 3 g x 5 3x 2 3x 2 3 g x 3x 2 3x 3 g x x 2 x 1.
SECTION 2.8 One-to-One Functions and Their Inverses
69
92. If g x is even, then h x f g x f g x h x. So yes, h is always an even function.
If g x is odd, then h is not necessarily an odd function. For example, if we let f x x 1 and g x x 3 , g is an odd function, but h x f g x f x 3 x 3 1 is not an odd function.
If g x is odd and f is also odd, then h x f g x f g x f g x f g x f g x h x. So in this case, h is also an odd function. If g x is odd and f is even, then h x f g x f g x f g x f g x f g x h x, so in this case, h is an even function.
93.
y
y
y
y
y
1l
1l
1l
1l
1l
1
0
x
0
1
x
0
1
x
0
x
1
0
1
y u x y u x 3 y xu x y u x 1 u x 2 y 2u x In parts (a)–(d), many answers are possible. Note that for any function f x, the graph of y u x f x is 0 for x 0 and identical to the graph of f x for x 0. (a) This is the graph of y u x 2.
(b) This is the graph of y 3u x 1.
(c) This is similar to the fourth graph above. One possible equation is y 3u x 2 3u x 4.
(d) This is the fifth graph above shifted one unit to the right. One possible equation is y x 1 u x 1.
94. f 0 x
1 , f 1 x f 0 f 0 x 1x
1
1 1 1x
1x 1x 1
x 1 , x
1 1 x x, and f 3 x f 0 f 2 x f 0 x. x 1 x x 1 1x 1 x This cycle continues, so f 3n f 0 for any natural number n. Since 1000 3 333 1, we conclude that x 1 . f 1000 x f 1 x x f 2 x f 0 f 1 x
2.8
ONE-TO-ONE FUNCTIONS AND THEIR INVERSES
1. A function f is onetoone if different inputs produce different outputs. You can tell from the graph that a function is onetoone by using the Horizontal Line Test. 2. (a) For a function to have an inverse, it must be onetoone. f x x 2 is not onetoone, so it does not have an inverse. However g x x 3 is onetoone, so it has an inverse. (b) The inverse of g x x 3 is g1 x 3 x.
3. (a) Proceeding backward through the description of f , we can describe f 1 as follows: “Take the third root, subtract 5, then divide by 3.” 3 x 5 . (b) f x 3x 53 and f 1 x 3 4. Yes, the graph of f is onetoone, so f has an inverse. Because f 4 1, f 1 1 4, and because f 5 3, f 1 3 5.
x
70
CHAPTER 2 Functions
5. If the point 3 4 is on the graph of f , then the point 4 3 is on the graph of f 1 . [This is another way of saying that f 3 4 f 1 4 3.] 6. (a) This is false in general. For instance, if f x x, then f 1 x x, but
1 1 f 1 x. f x x
(b) This is true, by definition. 7. By the Horizontal Line Test, f is not onetoone.
8. By the Horizontal Line Test, f is onetoone.
9. By the Horizontal Line Test, f is onetoone.
10. By the Horizontal Line Test, f is not onetoone.
11. By the Horizontal Line Test, f is not onetoone.
12. By the Horizontal Line Test, f is onetoone.
13. f x 2x 4. If x1 x2 , then 2x1 2x2 and 2x1 4 2x2 4. So f is a onetoone function.
14. f x 3x 2. If x1 x2 , then 3x1 3x2 and 3x1 2 3x2 2. So f is a onetoone function. 15. g x x. If x1 x2 , then x1 x2 because two different numbers cannot have the same square root. Therefore, g is a onetoone function. 16. g x x. Because every number and its negative have the same absolute value (for example, 1 1 1), g is not a onetoone function. 17. h x x 2 2x. Because h 0 0 and h 2 2 2 2 0 we have h 0 h 2. So f is not a onetoone function. 18. h x x 3 8. If x1 x2 , then x13 x23 and x13 8 x23 8. So f is a onetoone function.
19. f x x 4 5. Every nonzero number and its negative have the same fourth power. For example, 14 1 14 , so f 1 f 1. Thus f is not a onetoone function.
20. f x x 4 5, 0 x 2. If x1 x2 , then x14 x24 because two different positive numbers cannot have the same fourth power. Thus, x14 5 x24 5. So f is a onetoone function.
21. r t t 6 3, 0 t 5. If t1 t2 , then t16 t26 because two different positive numbers cannot have the same sixth power. Thus, t16 3 t26 3. So r is a onetoone function.
22. r t t 4 1. Every nonzero number and its negative have the same fourth power. For example, 14 1 14 , so r 1 r 1. Thus r is not a onetoone function. 1 1 1 23. f x 2 . Every nonzero number and its negative have the same square. For example, 1 , so 2 x 1 12 f 1 f 1. Thus f is not a onetoone function. 1 1 1 24. f x . If x1 x2 , then . So f is a onetoone function. x x1 x2 25. (a) f 5 9. Since f is onetoone, f 1 9 5. (b) f 1 10 0. Since f is onetoone, f 0 10.
26. (a) f 3 6. Since f is onetoone, f 1 6 3. (b) f 1 12 8. Since f is onetoone, f 8 12.
27. f x 5 2x. Since f is onetoone and f 1 5 2 1 3, then f 1 3 1. (Find 1 by solving the equation 5 2x 3.)
28. To find g1 5, we find the x value such that g x 5; that is, we solve the equation g x x 2 4x 5. Now
x 2 4x 5 x 2 4x 5 0 x 1 x 5 0 x 1 or x 5. Since the domain of g is [2 , x 1 is
the only value where g x 5. Therefore, g1 5 1. 29. (a) Because f 6 2, f 1 2 6.
(b) Because f 2 5, f 1 5 2.
(c) Because f 0 6, f 1 6 0.
30. (a) Because g 4 2, g1 2 4.
(b) Because g 7 5, g1 5 7.
(c) Because g 8 6, g1 6 8.
In #30, replace "g" with "f"
SECTION 2.8 One-to-One Functions and Their Inverses
31. From the table, f 4 5, so f 1 5 4. 33. f 1 f 1 1
71
32. From the table, f 5 0, so f 1 0 5. 34. f f 1 6 6
35. From the table, f 6 1, so f 1 1 6. Also, f 2 6, so f 1 6 2. Thus, f 1 f 1 1 f 1 6 2.
36. From the table, f 5 0, so f 1 0 5. Also, f 4 5, so f 1 5 4. Thus, f 1 f 1 0 f 1 5 4. 37. f g x f 4 x 5 14 [4 x 5 5] x 5 5 x for all x. g f x g 14 x 5 4 14 x 5 5 x for all x. Thus f and g are inverses of each other.
3x 3 1 x for all x. 38. f g x f 3x 3 x 3x g f x g 1 3 1 3 x for all x. Thus f and g are inverses of each other. 3 3 39. f g x f 32 x 9 23 32 x 9 6 x for all x. g f x g 23 x 6 32 23 x 6 9 x for all x. Thus f and g are inverses of each other.
x 7 7 x for all x. 4 4x 7 7 g f x g 4x 7 x for all x. Thus f and g are inverses of each other. 4 1 1 41. f g x f x for all x 0. Since f x g x, we also have g f x x for all x 0. Thus f and x 1x g are inverses of each other. 5 42. f g x f 5 x 5 x x for all x. 5 g f x g x 5 x 5 x for all x. Thus f and g are inverses of each other. 40. f g x f
x 7 4
4
2 x 9 x 9 9 x 9 9 x for all x 9. g f x g x 2 9 x 2 9 9 x 2 x for all x 0. Thus f and g are inverses of each other.
43. f g x f
3 44. f g x f x 113 x 113 1 x 1 1 x for all x. 3 g f x g x 3 1 x 113 1 x 1 1 x for all x. Thus f and g are inverses of each other.
1 1 x for all x 0. 1 1 x 1 1 x 1 1 1 x 1 1 x for all x 1. Thus f and g are inverses of each other. g f x g 1 x 1 x 1 2 4 x2 4 4 x 2 4 4 x 2 x 2 x, for all 0 x 2. (Note that the last 46. f g x f
45. f g x f
equality is possible since x 0.) 2 2 g f x g 4x 4 4 x 2 4 4 x 2 x 2 x, for all 0 x 2. (Again, the last equality is possible since x 0.) Thus f and g are inverses of each other.
72
CHAPTER 2 Functions
2x2 2
4x 2x 2 2 x 1 x for all x 1. 2x 2 2 x 1 4 2 x1 2 x2 2 2 x 2 2 x 2 4x x 2 x2 x for all x 2. Thus f and g are inverses of x2 g f x g x 2 x 2 1 x 2 4 1
47. f g x f
2x 2 x 1
x1 2x2
x2
each other.
54x 5 5 4x 5 1 3x 19x 13x x for all x 13 . 54x 3 5 4x 4 1 3x 19 3 13x 4 5 4 x5 19x 5 3x 4 4 x 5 x 5 3x4 for all x 43 . Thus f and g are inverses g f x g x5 3x 4 3x 4 3 5 19 x 13
48. f g x f
5 4x 1 3x
3x4
of each other.
49. f x 3x 15. y 3x 15 3x y 15 x 13 y 5. So f 1 x 13 x 5. Check: f f 1 x 3 13 x 5 15 x and f 1 f x 13 3x 15 5 x 5 5 x.
50. f x 8 3x. y 8 3x 3x 8 y x 83 13 y. So f 1 x 13 x 83 . Check: f f 1 x 8 3 13 x 83 x and f 1 f x 13 8 3x 83 83 x 83 x.
51. f x 34 x 12. y 34 x 12 34 x y 12 x 43 y 16. So f 1 x 43 x 16. Check: f f 1 x 34 43 x 16 12 x and f 1 f x 43 34 x 12 16 x 16 16 x.
3x 3x .y 10y 3 x x 3 10y. So f 1 x 10x 3. 10 10 3 10x 3 3x x and f 1 f x 10 3 3 x 3 x. Check: f f 1 x 10 10 53. f x 5 4x 3 . y 5 4x 3 4x 3 5 y x 3 14 5 y x 3 14 5 y. So f 1 x 3 14 5 x. 3 3 3 1 1 x and f 1 f x 3 14 5 5 4x 3 x 3 x. Check: f f x 5 4 4 5 x 52. f x
54. f x 3x 3 8. y 3x 3 8 3x 3 y 8 x 3 13 y 83 x 3 13 y 83 . So f 1 x 3 13 x 8. 3 3 Check: f f 1 x 3 3 13 x 8 8 x and f 1 f x 3 13 3x 3 8 8 x 3 x. 1 1 1 1 1 .y x 2 x 2. So f 1 x 2. x 2 x 2 y y x 1 1 x and f 1 f x 1 2 x 2 2 x. Check: f f 1 x 1 2 2 x2 x
55. f x
x 2 x 2 .y y x 2 x 2 x y 2y x 2 x y x 2 2y x y 1 2 y 1 x 2 x 2 2 y 1 2 x 1 x . So f 1 x . y1 x 1 2 x 1 2 4x 2 x 1 2 x 1 x 1 1 x and Check: f f x 2 x 1 2 x 1 2 x 1 4 2 x 1 x2 2 x2 1 2 [x 2 x 2] 2 2x x. f 1 f x x2 1 4 x 2 x 2
56. f x
x2
SECTION 2.8 One-to-One Functions and Their Inverses
73
x 2y x 2x .y y 2 x x x 1 y 2y x . So f 1 x . 2x 2x y 1 x 1 2x x 2 2x 2x 2x 2x Check: f f 1 x x 1 x and f 1 f x x. 2x x 2 x 1 2x x 2 x 2 2 2x 1 x 1
57. f x
4x 5y 5x 4x .y y x 5 4x x 4 y 5y x . So f 1 x . x 5 x 5 4 y 4x 5x 4 20x 4 x Check: f f 1 x x and 5x 5x 5 4 x 5 4x 4x 5 x5 5 4x 20x f 1 f x x. 4 x 5 4x 20 4 4x
58. f x
x5
2x 5 2x 5 .y y x 7 2x 5 x y 7y 2x 5 x y 2x 7y 5 x y 2 7y 5 x 7 x 7 7y 5 7x 5 x . So f 1 x . y2 x 2 7x 5 5 2 2 7x 5 5 x 2 x 2 Check: f f 1 x x and 7x 5 7x 5 7 x 2 7 x 2 2x5 7 x7 5 7 2x 5 5 x 7 19x f 1 f x x. 2x5 2 2x 5 2 x 7 19
59. f x
x7
4x 2 4x 2 .y y 3x 1 4x 2 3x y y 4x 2 4x 3x y y 2 x 4 3y y 2 3x 1 3x 1 y2 x 2 x . So f 1 x . 4 3y 4 3x x 2 2 4 4 x 2 2 4 3x 4 3x Check: f f 1 x x and x 2 3 x 2 4 3x 1 3 4 3x 4x2 2 4x 2 2 3x 1 10x 3x1 f 1 f x x. 4x2 4 1 3 2 10 3x 4x 43
60. f x
3x1
2x 3 2x 3 .y y 1 5x 2x 3 y 5x y 2x 3 2x 5x y y 3 x 2 5y y 3 1 5x 1 5x y3 x 3 x . So f 1 x . 5y 2 5x 2 2x3 3 2x 3 3 1 5x 17x 15x Check: f f 1 x x and f 1 f x x. 5 2x 3 2 1 5x 17 5 2x3 2
61. f x
15x
74
CHAPTER 2 Functions
3 4x y3 3 4x .y y 8x 1 3 4x 8x y y 3 4x 4x 2y 1 y 3 x . 8x 1 8x 1 4 2y 1 x 3 So f 1 x . 4 2x 1 34x 3 3 4x 3 8x 1 20x 8x1 Check: f f 1 x x and f 1 f x x. 34x 8 4x 4 1 20 3 8x 1 4 2
62. f x
8x1
x3 1
x3 1
x 3 3y 1 x 3 3y 1. So f 1 x 3 3x 1. 3 3 3 x3 1 3x 1 1 3 3 1 1 x and f f x 3 1 x 3 1 1 x. Check: f f x 3 3
63. f x
3
.y
7 7 64. f x x 5 6 . y x 5 6 7 y x 5 6 x 5 7 y 6 x 5 7 y 6. Thus, f 1 x 5 7 x 6. 7 5 7 5 7 7 x 6 6 x 66 x and Check: f f 1 x 7 5 7 5 5 f 1 f x x 6 6 x 5 6 6 x. 65. f x 2 3 x. y 2 3 x y 2 3 x x y 23 . Thus, f 1 x x 23 . 3 3 3 Check: f f 1 x 2 x 23 x and f 1 f x 2 3 x 2 3 x x. 66. f x 3 6x 5. y 3 6x 5 y 3 6x 5 x 16 y 3 5 . Thus, f 1 x 16 x 3 5 . 3 3 x 5 5 x and 3 6 16 x 3 5 5 Check: f f 1 x 3 f 1 f x 16 3 6x 5 5 16 6x 5 5 x. 67. f x x 32 1. y x 32 1 y 1 x 32 x y 123 . Thus, f 1 x x 123 . 32 23 23 1 x and f 1 f x x 32 1 1 x 32 x. Check: f f 1 x x 123
68. f x x 235 . y x 235 y 53 x 2 x y 53 2. Thus, f 1 x x 53 2. 35 53 x and f 1 f x x 235 2 x 2 2 x. Check: f f 1 x x 53 2 2
69. (a), (b) f x 3x 6
70. (a), (b) f x 16 x 2 , x 0
y
y
f f Ð!
1 1
f
x
f Ð!
2 2
(c) f x 3x 6. y 3x 6 3x y 6 x 13 y 6. So f 1 x 13 x 6.
x
(c) f x 16 x 2 , x 0. y 16 x 2 x 2 16 y x 16 y. So f 1 x 16 x, x 16. (Note: x 0 f x 16 x 2 16.)
SECTION 2.8 One-to-One Functions and Their Inverses
71. (a), (b) f x x 3 1
72. (a), (b) f x
y
x 1
y
f f Ð!
f Ð!
1 x
1
f 1 x
1
(c) f x x 3 1 y x 3 1 x 3 y 1 x 3 y 1. So f 1 x 3 x 1.
73. (a), (b) f x 3
y
(c) f x
x 1, x 1. y
f 1 x x 2 1, x 0.
x 1, x 1
f f Ð!
f Ð!
1
(c) f x 3 x 1, x 1. y 3 x 1 y 3 x 1 y 32 x 1 x y 32 1 and y 3. So
x
1
x
1
(c) f x 2 x 1, x 1. y 2 x 1 y 2 x 1 y 22 x 1 x y 22 1 and y 2. So f 1 x x 22 1, x 2.
f 1 x x 32 1, x 3.
75. f x x 3 x. Using a graphing device and the
Horizontal Line Test, we see that f is not a onetoone function. For example, f 0 0 f 1.
2
y
f 1
x 1, y 0
y 2 x 1 x y 2 1 and y 0. So
74. (a), (b) f x 2
x 1
2
76. f x x 3 x. Using a graphing device and the
Horizontal Line Test, we see that f is a onetoone function.
2
2
75
76
CHAPTER 2 Functions
x 12 . Using a graphing device and the x 6 Horizontal Line Test, we see that f is a onetoone
77. f x
function.
78. f x
x 3 4x 1. Using a graphing device and the
Horizontal Line Test, we see that f is not a onetoone function. For example, f 0 1 f 2.
10 2 20
20 2
10
79. f x x x 6. Using a graphing device and the Horizontal Line Test, we see that f is not a onetoone function. For example f 0 6 f 2.
0
2
80. f x x x. Using a graphing device and the
Horizontal Line Test, we see that f is a onetoone function. 20
10
10
5
10
5 20
10
82. (a) y f x 2 12 x 12 x 2 y x 4 2y.
81. (a) y f x 2 x x y 2. So f 1 x x 2.
So f 1 x 4 2x.
(b)
(b)
10
5
5
5
10
10
5 10
83. (a) y g x
x 3, y 0 x 3 y 2 , y 0
x y 2 3, y 0. So g1 x x 2 3, x 0.
(b)
84. (a) y g x x 2 1, x 0 x 2 y 1, x 0 x y 1. So g1 x x 1. (b)
10
5
10
5
5
10 5
10
f
SECTION 2.8 One-to-One Functions and Their Inverses
77
4 y (since x 0, we take the
85. If we restrict the domain of f x to [0 , then y 4 x 2 x 2 4 y x positive square root). So f 1 x 4 x.
If we restrict the domain of f x to 0], then y 4 x 2 x 2 4 y x 4 y (since x 0, we take the negative square root). So f 1 x 4 x. g
86. If we restrict the domain of g x to [1 , then y x 12 x 1 root) x 1 y. So g 1 x 1 x.
y (since x 1 we take the positive square
If we restrict the domain of g x to 1], then y x 12 x 1 y (since x 1 we take the negative square root) x 1 y. So g 1 x 1 x. h
87. If we restrict the domain of h x to [2 , then y x 22 x 2 root) x 2 y. So h 1 x 2 x.
y (since x 2, we take the positive square
If we restrict the domain of h x to 2], then y x 22 x 2 y (since x 2, we take the negative square root) x 2 y. So h 1 x 2 x.
88. k x x 3 k
x 3 if x 3 0 x 3 x 3
if x 3 0 x 3
If we restrict the domain of k x to [3 , then y x 3 x 3 y. So k 1 x 3 x.
If we restrict the domain of k x to 3], then y x 3 y x 3 x 3 y. So k 1 x 3 x. y
89. f Ð!
90. y
2 f
2
f
x 1
f Ð! 1
x
91. f x x 2 9, x 0. The range is [9 . y x 2 9 y 9 x 2 x x 9.
y 9. Thus, f 1 x
92. f x x 2 2x 1 x 12 , x 1. The range is [0 . y x 12 x x 0.
y 1. Thus, f 1 x
x 9, x 1,
1 1 1 1 , x 0. . Thus, f 1 x 93. f x 4 , x 0. The range is 0 . y 4 x 4 4 y x x x 1 1 94. f x 2 , x 0. The range is 0 1]. y 2 x2 y 1 y x x 1 x 1 0 x 1. 95. f x
x, 0 x 9. The range is [0 3]. y
1y . Thus, f 1 x y
1x , x
x x y 2 . Thus, the inverse function is f 1 x x 2 , 0 x 3.
96. f x x 2 6x, x 3.The range is [9 . y x 2 6x x 2 6x 9 y 9 x 32 y 9 x y 9 3. Thus, the inverse function is f 1 x x 9 3, x 9.
78
CHAPTER 2 Functions
97. (a)
(b) Yes, the graph is unchanged upon reflection about
y
the line y x. (c) y
1 0
98. (a)
1
1 1 1 x , so f 1 x . x y x
x
(b) Yes, the graph is unchanged upon reflection about
y
the line y x.
x 3 y x 1 x 3 x 1 y3 x y 1 y 3 x . Thus, y1
(c) y 1 0
1
x
f 1 x
x 3 . x 1
99. (a) The price of a pizza with no toppings (corresponding to the yintercept) is $16, and the cost of each additional topping (the rate of change of cost with respect to number of toppings) is $150. Thus, f n 16 15n. (b) p f n 16 15n p 16 15n n 23 p 16. Thus, n f 1 p 23 p 16. This function represents the number of toppings on a pizza that costs x dollars.
(c) f 1 25 23 25 16 23 9 6. Thus, a $25 pizza has 6 toppings. 100. (a) f x 500 80x.
p 500 1 p 500 . So x f 1 p . f 80 80 represents the number of hours the investigator spends on a case for x dollars. 720 1220 500 (c) f 1 1220 9. If the investigator charges $1220, they spent 9 hours investigating the case. 80 80 V t 2 t 2 t 2 V t 101. (a) V f t 100 1 1 , 0 t 40. V 100 1 1 40 40 100 40 40 100 t V 1 t 40 4 V . Since t 40, we must have t f 1 V 40 4 V . f 1 represents time that 40 10 has elapsed since the tank started to leak. (b) f 1 15 40 4 15 245 minutes. In 245 minutes the tank has drained to just 15 gallons of water. 102. (a) g r 18,500 025 r 2 . 18,500 025 r 2 4625 18,500r 2 18,500r 2 4625 4625 4625 4625 1 2 1 r r . Since r represents a distance, r 0, so g . g 18,500 18,500 18,500 represents the radial distance from the center of the vein at which the blood has velocity . 4625 30 (b) g 1 30 0498 cm. The velocity is 30 cms at a distance of 0498 cm from the center of the artery 18,500 or vein. (b) p f x 500 80x. p 500 80x 80x p 500 x
SECTION 2.8 One-to-One Functions and Their Inverses
79
103. (a) D f p 3 p 150. D 3 p 150 3 p 150 D p 50 13 D. So f 1 D 50 13 D. f 1 D represents the price that is associated with demand D.
(b) f 1 30 50 13 30 40. So when the demand is 30 units, the price per unit is $40. g
104. (a) F g C 95 C 32. F 95 C 32 95 C F 32 C 59 F 32. So g1 F 59 F 32. g1 F represents the Celsius temperature that corresponds to the Fahrenheit temperature of F.
(b) F 1 86 59 86 32 59 54 30. So 86 Fahrenheit is the same as 30 Celsius. 105. (a) U f x 079x
(b) U 079x x U079 1265823U , so f 1 U 1265823U . f 1 U represents the value of U US dollars in Canadian dollars.
(c) f 1 12,250 1265823 12,250 15,50633. So $12,250 in US currency is worth $15,50633 in Canadian currency. 01x, if 0 x 20,000 106. (a) f x 2000 02 x 20,000 if x 20,000 (b) We find the inverse of each piece of the function f :
f 1 x 01x. T 01x x 10T . So f 11 T 10T .
f 2 x 2000 02 x 20,000 02x 2000. T 02x 2000 02x T 2000 x 5T 10,000. So f 21 T 5T 10,000. Since f 0 0 and f 20,000 2000 we have f 1 T taxpayer’s income.
10T ,
if 0 T 2000
5T 10,000 if T 2000
This represents the
(c) f 1 10,000 5 10,000 10,000 60,000. The required income is 60,000. 107. (a) f x 085x.
(b) g x x 1000.
(c) H x f g x f x 1000 085 x 1000 085x 850.
(d) P H x 085x 850. P 085x 850 085x P 850 x 1176P 1000. So
H 1 P 1176P 1000. The function H 1 represents the original sticker price for a given discounted price P.
(e) H 1 13,000 1176 13,000 1000 16,288. So the original price of the car is $16,288 when the discounted price ($1000 rebate, then 15% off) is $13,000. 108. f x mx b. Notice that f x1 f x2 mx1 b mx2 b mx1 mx2 . We can conclude that x1 x2 if and only if m 0. Therefore f is onetoone if and only if m 0. If m 0, f x mx b y mx b mx y b x
yb x b . So, f 1 x . m m
2x 1 is “multiply by 2, add 1, and then divide by 5 ”. So the reverse is “multiply by 5, subtract 1, and then 5 5x 1 2 1 5x 1 1 5x 5x 1 5x 1 2 . Check: f f 1 x f x divide by 2 ” or f 1 x 2 2 5 5 5 2x 1 5 1 2x 1 2x 1 1 2x 5 and f 1 f x f 1 x. 5 2 2 2
109. (a) f x
80
CHAPTER 2 Functions
1 1 3 is “take the negative reciprocal and add 3 ”. Since the reverse of “take the negative x x reciprocal” is “take the negative reciprocal ”, f 1 x is “subtract 3 and take the negative reciprocal ”, that is, 1 x 3 1 1 1 1 . Check: f f x f 3 3 x 3 x and f x 3 1 1 x 3 x 3 1 x 3 1 1 1 x x. f 1 f x f 1 3 1 1 1 x 1 3 3 x x (c) f x x 3 2 is “cube, add 2, and then take the square root”. So the reverse is “square, subtract 2, then take 3 the cube root ” or f 1 x x 2 2. Domain for f x is 3 2 ; domain for f 1 x is [0 . Check: 3 3 2 3 2 1 x 2 x 2 2 x 2 2 2 x 2 x (on the appropriate domain) and f f x f 2 3 3 3 1 1 3 f f x f x 2 x 3 2 2 x 3 2 2 x 3 x (on the appropriate domain). (b) f x 3
(d) f x 2x 53 is “double, subtract 5, and then cube”. So the reverse is “take the cube root, add 3 x 5 Domain for both f x and f 1 x is . Check: 5, and divide by 2” or f 1 x 2 3 3 3 3 3 x 5 x 5 3 f f 1 x f 2 5 3 x 5 5 3 x x 3 x and 2 2 x 2x 53 5 2x 2x 5 5 1 1 3 f x. f x f 2x 5 2 2 2 In a function like f x 3x 2, the variable occurs only once and it easy to see how to reverse the operations step by step. But in f x x 3 2x 6, you apply two different operations to the variable x (cubing and multiplying by 2) and then add 6, so it is not possible to reverse the operations step by step.
110. f I x f x; therefore f I f . I f x f x; therefore I f f .
By definition, f f 1 x x I x; therefore f f 1 I . Similarly, f 1 f x x I x; therefore f 1 f I .
111. (a) We find g 1 x: y 2x 1 2x y 1 x 12 y 1. So g1 x 12 x 1. Thus 2 f x h g1 x h 12 x 1 4 12 x 1 4 12 x 1 7 x 2 2x 1 2x 2 7 x 2 6.
(b) f g h f 1 f g f 1 h I g f 1 h g f 1 h. Note that we compose with f 1 on the left
on each side of the equation. We find f 1 : y 3x 5 3x y 5 x 13 y 5. So f 1 x 13 x 5. Thus g x f 1 h x f 1 3x 2 3x 2 13 3x 2 3x 2 5 13 3x 2 3x 3 x 2 x 1. 112. f g g 1 f 1 x f g g1 f 1 x f g g1 f 1 x f f 1 x x and g 1 f 1 f g x g 1 f 1 f g x g 1 f 1 f g x g1 g x x.
CHAPTER 2
Review
81
CHAPTER 2 REVIEW 1. “Square, then subtract 5” can be represented by the function f x x 2 5. x 2. “Divide by 2, then add 9” can be represented by the function g x 9. 2 3. f x 3 x 10: “Add 10, then multiply by 3.” 4. f x 6x 10: “Multiply by 6, then subtract 10, then take the square root.”
5. g x x 2 4x
6. h x 3x 2 2x 5
x
g x
x
1
5
2
3
1
4
0
0
1
3
2 3
4 3
0 1 2
h x
5
0
11
7. C x 5000 30x 0001x 2 (a) C 1000 5000 30 1000 0001 10002 $34,000 and
C 10,000 5000 30 10,000 0001 10,0002 $205,000.
(b) From part (a), we see that the total cost of printing 1000 copies of the book is $34,000 and the total cost of printing 10,000 copies is $205,000. (c) C 0 5000 30 0 0001 02 $5000. This represents the fixed costs associated with getting the print run ready.
(d) The net change in C as x changes from 1000 to 10,000 is C 10,000 C 1000 205,000 34,000 $171,000, and 171,000 C 10,000 C 1000 $19copy. the average rate of change is 10,000 1000 9000 8. E x 400 003x (a) E 2000 400 003 2000 $460 and E 15 000 400 003 15,000 $850.
(b) From part (a), we see that if the salesperson sells $2000 worth of goods, they make $460, and if they sell $15,000 worth of goods, they make $850. (c) E 0 400 003 0 $400 is the salesperson’s base weekly salary.
(d) The net change in E as x changes from 2000 to 15,000 is E 15,000 E 2000 850 460 $390, and the average 390 E 15,000 E 2000 $003 per dollar. rate of change is 15,000 2000 13,000 (e) Because the value of goods sold x is multiplied by 003 or 3%, we see that the salesperson earns a percentage of 3% on the goods that they sell. 9. f x x 2 4x 6; f 0 02 4 0 6 6; f 2 22 4 2 6 2;
f 2 22 4 2 6 18; f a a2 4 a 6 a 2 4a 6; f a a2 4 a 6 a 2 4a 6;
f x 1 x 12 4 x 16 x 2 2x 14x 46 x 2 2x 3; f 2x 2x2 4 2x6 4x 2 8x 6. 10. f x 4 3x 6; f 5 4 15 6 1; f 9 4 27 6 4 21; f a 2 4 3a 6 6 4 3a; f x 4 3 x 6 4 3x 6; f x 2 4 3x 2 6.
11. f x x 2 8 f a a 2 8, f a h a h2 8 a 2 h 2 2ah 8, and 2 h 2 2ah 8 a 2 8 a 2ah h 2 f a h f a 2a h. h h h
82
CHAPTER 2 Functions
1 1 1 f a , f a h , and x 2 a2 ah2 1 1 h 1 f a h f a a 2 a h 2 ah2 a2 . h h h a 2 a h 2 h a 2 a h 2 a 2 a 2 h
12. f x
13. By the Vertical Line Test, figures (b) and (c) are graphs of functions. By the Horizontal Line Test, figure (c) is the graph of a onetoone function. 14. (a) f 2 1 and f 2 2.
(b) The net change in f from 2 to 2 is f 2 f 2 2 1 3, and the average rate of change is 3 f 2 f 2 . 2 2 4 (c) The domain of f is [4 5] and the range of f is [4 4]. (d) f is increasing on 4 2 and 1 4; f is decreasing on 2 1 and 4 5. (e) f has local maximum values of 1 (at x 2) and 4 (at x 4).
(f) f is not a onetoone, for example, f 2 1 f 0. There are many more examples. 15. Domain: We must have x 5 0 x 5. In interval notation, the domain is [5 . Range: For x in the domain of f , we have x 5 x 5 0 x 5 0 f x 0. So the range is [0 .
1 has domain x x 2. For x 2, f takes on all negative real numbers, and for x 2, it takes on all x 2 positive real numbers. Thus, the range of f is 0 0 .
16. f x
17. f x 7x 15. The domain is all real numbers, .
2x 1 . Then 2x 1 0 x 12 . So the domain of f is x x 12 . 2x 1 19. f x x 2 4. x 2 4 0 for all x, so the domain is all real numbers, . 18. f x
20. f x 3x
21. f x
2 . The domain of f is the set of x where x 1 0 x 1. So the domain is 1 . x 1
1 1 1 . The denominators cannot equal 0, therefore the domain is x x 0 1 2. x x 1 x 2
2x 2 5x 3 2x 2 5x 3 . The domain of g is the set of all x where the denominator is not 0. So the 2 2x 1 x 3 2x 5x 3 domain is x 2x 1 0 and x 3 0 x x 12 and x 3 .
22. g x
23. h x
4 x x 2 1. We require the expression inside the radicals be nonnegative. So 4 x 0 4 x; also
x 2 1 0 x 1 x 1 0. We make a table: Interval Sign of x 1 Sign of x 1
Sign of x 1 x 1
1
1 1
1
Thus the domain is 4] 1] [1 1] [1 4]. 3 2x 1 . Since the roots are both odd, the domain is the set of all x where the denominator is not 0. Now 24. f x 3 2x 2 3 2x 2 0 3 2x 2 2x 8 x 4. Thus the domain of f is x x 4.
CHAPTER 2
25. f x 2 34 x
26. f x 3 1 2x, 2 x 2 y
y
1
2
x
1
27. f x 3x 2 4
Review
1
28. f x 12 x 2 8
y
x
y 1 1
1 1
29. f x
x
x
x 5
30. f x
y
3 x 1 y
1 1 1
x
1
x
83
84
CHAPTER 2 Functions
31. f x 13 x 32 2
32. f x 2 x 4 3
y
y
1 1
x
1
x
1 1
33. f x 4 x 2
x
34. f x 12 x 12 2
y
y
1
1 1
x
36. f x 3 x
35. f x 12 x 3 y
1
2 0
37. f x 5 x
y
1
0
x
38. f x 3 x 2
y
1
x
2
y
1 1
x
1
x
CHAPTER 2
1 39. f x 2 x
40. f x y
Review
85
1 x 13
y
1 0
1
x 5 0
1
x
x if x 0 42. f x x 2 if 0 x 2 1 if x 2
1 x if x 0 41. f x 1 if x 0
y
y
1 1
x
1 1
x
43. x y 2 14 y 2 14 x y 14 x, so the original equation does not define y as a function of x.
44. 3x
y 8
y 3x 8 y 3x 82 , so the original equation defines y as a function of x.
13 45. x 3 y 3 27 y 3 x 3 27 y x 3 27 , so the original equation defines y as a function of x (since the cube root function is onetoone).
46. 2x y 4 16 y 4 2x 16 y 4 2x 16, so the original equation does not define y as a function of x.
86
CHAPTER 2 Functions
47. (a) The ordered pairs in this relation are 3 3, 2 0, 0 1, 2 3, and 3 3.
(b) The ordered pairs in this relation are 3 3, 2 1, 0 2, 2 5, and 3 3. y
y
1 x
1
1 x
1
Each input in the domain has exactly one output in the range, so this relation defines y as a function of
The input 2 has two different outputs (1 and 5), so
x. The domain is 3 2 0 2 3 and the range is
this relation does not define y as a function of x. The
3 1 0 3.
domain is 3 0 2 3 and the range is 2 1 3 5.
48. (a) This relation defines y as a function of x because each value x in the domain has exactly one corresponding value of y. (b) This relation does not define y as a function of x because the xvalue 68 corresponds to two different yvalues. (c) This relation does not define y as a function of x because every integer xvalue corresponds to an infinite number of different integer yvalues. (d) This relation does not define y as a function of x because every positive number x has two fourth roots.
49. We graph f x
9 x 2 in the viewing rectangle
[4 4] by [1 4].
50. We graph f x
[5 5] by [1 6].
4
x 2 3 in the viewing rectangle
5
2
4
2
2
4
(a) From the graph, the domain of f is [3 3] and the range of f is [0 3]. (b) f x 0 at x 3. (c) f x 1 on approximately 283 283.
5
5
(a) From the graph, the domain of f is 173] [173 and the range of f is [0 .
(b) f x 0 at x 173. (c) f x 1 on 2 and 2 .
CHAPTER 2
51. We graph f x
[5 5] by [1 5].
x 3 4x 1 in the viewing rectangle
Review
87
52. We graph f x x 4 x 3 x 2 3x 6 in the viewing rectangle [3 4] by [20 100]. 100
4
50
2 5
5
(a) From the graph, the domain of f is approximately [211 025] [186 and the range of f is
[0 .
(b) f x 0 at x 211, x 025, and x 186. (c) f x 1 on 2 0 and 2 .
2
2
4
(a) From the graph, the domain of f is and the range of f is approximately [710 .
(b) f x 0 at x 149 and x 127. (c) f x 1 on approximately 154 and 138 .
53. We graph f x 2x 2 4x 5 in the viewing rectangle [2 4] by [1 10].
54. We graph f x 1 x x 2 in the viewing rectangle [4 3] by [10 2].
10 4
2
5
5
2
2
4
(a) The local minimum value is f 1 3. There is no local maximum.
(b) f is decreasing on 1 and increasing on 1 . 55. We graph f x 33 16x 25x 3 in the viewing rectangle [2 2] by [10 10].
10
(a) The local maximum value is f 12 54 . There is
no local minimum. (b) f is increasing on 12 and decreasing on 12 .
56. We graph f x x 3 4x 2 in the viewing rectangle [2 5] by [15 10].
10
2
10
2 10
(a) The local minimum value is approximately f 046 281 and the local maximum value is
approximately f 046 379.
(b) f is decreasing on approximately 046 and 046 and increasing on approximately 046 046.
2
2
2
4
10
(a) The local maximum value is f 0 0 and the local
minimum value is approximately f 267 948.
(b) f is decreasing on approximately 0 267 and increasing on 0 and approximately 267 .
88
CHAPTER 2 Functions
57. We graph f x x 23 6 x13 in the viewing rectangle [5 10] by [8 8].
58. We graph f x x 4 16 in the viewing rectangle [3 3] by [5 30].
5
20
5
5
10
5 2
(a) The local maximum value is f 4 317 and the local minimum value is f 0 0.
(b) f is decreasing on 0 and 4 and increasing on 0 4.
2
(a) The local maximum value is f 0 16 and the local minimum values are f 2 0.
(b) f is decreasing on 2 and 0 2 and increasing on 2 0 and 2 .
f 8 f 4 4 1. 84 4 35 7 g 30 g 10 . 60. The net change is g 30 g 10 30 5 35 and the average rate of change is 30 10 20 4 4 f 2 f 1 . 61. The net change is f 2 f 1 6 2 4 and the average rate of change is 2 1 3 Replace "f" with "g" f 3 f 1 6 62. The net change is f 3 f 1 1 5 6 and the average rate of change is 3. 31 2 63. The net change is f 4 f 1 42 2 4 12 2 1 8 1 9 and the average rate of change is 59. The net change is f 8 f 4 8 12 4 and the average rate of change is
9 f 4 f 1 3. 41 3
64. The net change is g a h g a a h 12 a 12 2ah 2h h 2 and the average rate of change is 2ah 2h h 2 g a h g a 2a 2 h. aha h
65. f x 2 3x2 9x 2 12x 4 is not linear. It cannot be expressed in the form f x ax b with constant a and b. 2x 10 66. g x 2 5 5 x 2 5 is linear with a 2 5 5 and b 2 5. 5 67. (a)
68. (a)
y
1 0
1
x
y
1 0
(b) The slope of the graph is the value of a in the equation f x ax b 3x 2; that is, 3. (c) The rate of change is the slope of the graph, 3.
1
x
(b) The slope of the graph is the value of a in the equation f x ax b 12 x 3; that is, 12 .
(c) The rate of change is the slope of the graph, 12 .
69. The linear function with rate of change 2 and initial value 3 has a 2 and b 3, so f x 2x 3.
CHAPTER 2
Review
89
70. The linear function whose graph has slope 12 and yintercept 1 has a 12 and b 1, so f x 12 x 1. 71. Between x 0 and x 1, the rate of change is f x 2x 3. 72. Between x 0 and x 2, the rate of change is is f x 14 x 6.
f 1 f 0 53 2. At x 0, f x 3. Thus, an equation is 10 1
55 6 f 2 f 0 14 . At x 0, f x 6. Thus, an equation 20 2
73. The points 0 4 and 8 0 lie on the graph, so the rate of change is y 12 x 4.
04 1 . At x 0, y 4. Thus, an equation is 80 2
74. The points 0 4 and 2 0 lie on the graph, so the rate of change is is y 2x 4.
0 4 2. At x 0, y 4. Thus, an equation 20
75. f x 12 x 6 (a) The average rate of change of f between x 0 and x 2 is 1 2 6 1 0 6 f 2 f 0 5 6 1 2 2 , and the average rate of change of f between x 15 20 2 2 2 and x 50 is 1 50 6 1 15 6 19 32 1 f 50 f 15 2 2 . 50 15 35 35 2 (b) The rates of change are the same.
(c) Yes, f is a linear function with rate of change 12 . 76. f x 8 3x
f 2 f 0 [8 3 2] [8 3 0] 28 3, 20 2 2 and the average rate of change of f between x 15 and x 50 is [8 3 50] [8 3 15] f 50 f 15 142 37 3. 50 15 35 35 (b) The rates of change are the same. (a) The average rate of change of f between x 0 and x 2 is
(c) Yes, f is a linear function with rate of change 3.
77. f x x 12
(a) The average rate of change of f between x 0 and x 2 is average rate of change of f between x 15 and x 50 is
f 2 f 0 2 12 0 12 0, and the 20 20
f 50 f 15 2401 196 50 12 15 12 .50 12 63. 50 15 35 35 (b) The rates of change are not the same. (c) No, f is not a linear function. 1 78. f x x 3
1
1
f 2 f 0 1 23 03 , and the average 20 2 15 1 1 1 f 50 f 15 503 153 . rate of change of f between x 15 and x 50 is 50 15 35 954 (b) The rates of change are not the same.
(a) The average rate of change of f between x 0 and x 2 is
(c) No, f is not a linear function.
90
CHAPTER 2 Functions
79. (a) (i) y f x 8. Shift the graph of f x upward 8 units. (ii) g x x 3 8
(b) (i) y f x 8. Shift the graph of f x to the left 8 units. (ii) g x x 83
(c) (i) y 1 2 f x. Stretch the graph of f x vertically by a factor of 2, then shift it upward 1 unit. (ii) g x 1 2x 3
(d) (i) y f x 2 2. Shift the graph of f x to the right 2 units, then downward 2 units. (ii) g x x 23 2
(e) (i) y f x. Reflect the graph of f x about the yaxis. (ii) g x x3 x 3
(f) (i) y f x. Reflect the graph of f x first about the yaxis, then reflect about the xaxis. (ii) g x x3 x 3
(g) (i) y f x. Reflect the graph of f x about the xaxis. (ii) g x x 3
(h) (i) y f 1 x. Reflect the graph of f x about the line y x. (ii) g x 3 x 80. (a) y f x 2
(b) y f x
y
1
(c) y 3 f x
y
1 1
(d) y 12 f x 1
(e) y f 1 x
y
1 1
x
1
x
(f) y f x
y
1
1 1
x
y
x
y
1 1
x
1
x
81. (a) f x 2x 5 3x 2 2. f x 2 x5 3 x2 2 2x 5 3x 2 2. Since f x f x, f is not even. f x 2x 5 3x 2 2. Since f x f x, f is not odd. (b) f x x 3 x 7 . f x x3 x7 x 3 x 7 f x, hence f is odd.
(c) f x
1 x2 1 x2 1 x2 . f f x. Since f x f x, f is even. x 1 x2 1 x2 1 x2
CHAPTER 2
Review
91
1 1 1 1 . f x . f x . Since f x f x , f is not even, and since x 2 2x x 2 x 2 f x f x, f is not odd.
(d) f x
82. (a) This function is odd. (b) This function is neither even nor odd. (c) This function is even. (d) This function is neither even nor odd. 2. 83. (a) The average rate of change is the same between any two points if the function is linear, as is graph 5 is increasing on 2 and decreasing on 2 . (b) Graph 4 has domain [1 . (c) Graph
3 has f x f x for all x in its domain, so it is even. (d) Graph 1 is decreasing on . (e) Graph 3 has two local minima. (f) Graph
1, 2 , and 4 all pass the Horizontal Line Test and thus have inverse functions. (g) Graphs
84. h t 16t 2 48t 32 16 t 2 3t 32 16 t 2 3t 94 32 36 2 16 t 2 3t 94 68 16 t 32 68 The stone reaches a maximum height of 68 feet.
85. (a) If the rate is doubled, the volume is given by Vc 2V t 16t. (b) The volume is given by Vb V t 4 08 t 4, t 4.
(c) In this case, the volume is given by Va V t 50 08t 5.
3V 3V r V 3 86. (a) Solving V 43 r 3 for r, we have r 3 4 4 3 08t 3t 3 . This function models the radius of the balloon as a function (b) r V t r V t r 08t 3 4 5 of t. 3 50 30 (c) r V 50 3 3 212 ft 5 2 12 3960 12 3960 3960 h 3960 h 3960 1 . Thus, 87. (a) 144 3960 h 144 3960 h 12 1 x 3960 1 . This represents the astronaut’s height above the surface of the earth as a function of their x weight. 12 1 (b) 64 3960 1 1980. This means that if the astronaut weighs 64 lb, then they are 1980 mi above the 64 surface of the earth.
88. The higher the value of k, the higher the maximum yield. (In fact, it appears that the maximum yield is 12 k.) A nitrogen level of 1 ppm seems to deliver the maximum yield, regardless of the value of k. 5x . The maximum yield is Y 1 52 tons per acre. For k 5, Y x 1 x2
92
CHAPTER 2 Functions
89. f x x 2, g x x 2
90. f x x 2 1, g x 3 x 2 5
5
4
2
2
2
2
91. f x x 2 3x 2 and g x 4 3x.
(a) f g x x 2 3x 2 4 3x x 2 6x 6 (b) f g x x 2 3x 2 4 3x x 2 2 (c) f g x x 2 3x 2 4 3x 4x 2 12x 8 3x 3 9x 2 6x 3x 3 13x 2 18x 8 x 2 3x 2 f , x 43 (d) x g 4 3x
(e) f g x f 4 3x 4 3x2 3 4 3x 2 16 24x 9x 2 12 9x 2 9x 2 15x 6 (f) g f x g x 2 3x 2 4 3 x 2 3x 2 3x 2 9x 2
92. f x 1 x 2 and g x x 1. (Remember that the proper domains must apply.) 2 x 1 1 x 1 1x 1 x (a) f g x f (b) g f x g 1 x 2 1 x 2 1 x 2 x (c) f g 2 f g 2 f 2 1 f 1 1 12 2. (d) f f 2 f f 2 f 1 22 f 5 1 52 26.
(e) f g f x f g f x f x 1 x2 1 x 2 . Note that g f x x by part (b). (f) g f g x g f g x g x x 1. Note that f g x x by part (a).
x 1 and g x x x 2 . f g x f x x 2 x x 2 1, and the domain is [0 1]. 2 g f x g x 1 x 1 x 1 x 1 x 2 x 1 x x, and the domain is [0 . x 1 x 1 1, and the domain is [0 . f f x f 2 g g x g x x 2 x x 2 x x 2 x x 2 x 2 2x 3 x 4 x 4 2x 3 2x 2 x, and the
93. f x
domain is .
CHAPTER 2
Review
93
2 , has domain x x 4. x, has domain x x 0. g x x 4 2 2 . f g x is defined whenever both g x and f g x are defined; that is, f g x f x 4 x 4 2 2 whenever x 4 and 0. Now 0 x 4 0 x 4. So the domain of f g is 4 . x 4 x 4 2 . g f x is defined whenever both f x and g f x are defined; that is, whenever g f x g x x 4 x 0 and x 4 0. Now x 4 0 x 16. So the domain of g f is [0 16 16 . x x x 14 . f f x is defined whenever both f x and f f x are defined; that is, f f x f whenever x 0. So the domain of f f is [0 . 2 x 4 2 x 4 2 g g x is defined whenever both g x and g g x g 2 x 4 2 4 x 4 9 2x 4 x 4 g g x are defined; that is, whenever x 4 and 9 2x 0. Now 9 2x 0 2x 9 x 92 . So the domain of g g is x x 92 4 . 95. f x 1 x, g x 1 x 2 and h x 1 x. 2 f 1 12 x x f g h x f g h x f g 1 x f 1 1 x 2 f x 2 x 1 x 2 x 1 2 x x 1 x 1 x
94. f x
96. If h x
x and g x 1 x , then g h x g
1 f g h x f 1 x T x. 1 x
1 x 1 x. If f x , then x
97. f x 3 x 3 . If x1 x2 , then x13 x23 (unequal numbers have unequal cubes), and therefore 3 x13 3 x23 . Thus f is a onetoone function. 98. g x 2 2x x 2 x 2 2x 1 1 x 12 1. Since g 0 2 g 2 , as is true for all pairs of numbers
equidistant from 1, g is not a onetoone function. 1 99. h x 4 . Since the fourth powers of a number and its negative are equal, h is not onetoone. For example, x 1 1 1 and h 1 1, so h 1 h 1. h 1 14 14 100. r x 2 x 3. If x1 x2 , then x1 3 x2 3, so x1 3 x2 3 and 2 x1 3 2 x2 3. Thus r is onetoone. 101. p x 33 16x 25x 3 . Using a graphing device and 102. q x 33 16x 25x 3 . Using a graphing device and the Horizontal Line Test, we see that p is not a onetoone
the Horizontal Line Test, we see that q is a onetoone
function.
function. 10
5
10
5 10
5
5 10
103. f x 3x 2 y 3x 2 3x y 2 x 13 y 2. So f 1 x 13 x 2.
94
CHAPTER 2 Functions
2x 1 2x 1 y 2x 1 3y 2x 3y 1 x 12 3y 1 So f 1 x 12 3x 1. 3 3 105. f x x 13 y x 13 x 1 3 y x 3 y 1. So f 1 x 3 x 1. 106. f x 1 5 x 2. y 1 5 x 2 y 1 5 x 2 x 2 y 15 x 2 y 15 . So
104. f x
f 1 x 2 x 15 .
107. The graph passes the Horizontal Line Test, so f has an inverse. Because f 1 0, f 1 0 1, and because f 3 4, f 1 4 3.
108. The graph fails the Horizontal Line Test, so f does not have an inverse. 109. (a), (b) f x x 2 4, x 0
110.
y
f
(a) If x1 x2 , then 4 x1 4 x2 , and so 1 4 x1 1 4 x2 . Therefore, f is a onetoone function.
f Ð!
(b), (c)
y
1 1
f Ð!
x
f 1
(c) f x x 2 4, x 0 y x 2 4, y 4 x 2 y 4 x y 4. So f 1 x x 4, x 4.
1
x
(d) f x 1 4 x. y 1 4 x 4 x y 1 x y 14 . So f 1 x x 14 ,
x 1. Note that the domain of f is [0 , so y 1 4 x 1. Hence, the domain of f 1 is
[1 .
111. (a) y x 2 is the graph of the absolute value function y x, shifted downward 2 units. It has Graph VI, and defines y as a function of x. (b) y x 12 2 is the graph of y x 2 , shifted to the left 1 unit and downward 2 units. It has Graph IV, and defines y as a function of x. (c) x 22 y 12 4 is the graph of a circle centered at 2 1 with radius 2. It has Graph V, and does not define y as a function of x. (d) y x 23 is the graph of y x 3 , shifted to the right 2 units. It has Graph I, and defines y as a function of x.
(e) x y 2 3 is the graph of y x 2 , reflected about the line y x and shifted to the left 3 units. It has Graph VIII, and does not define y as a function of x. 1 (f) y is undefined at x 0 and has Graph III. It defines y as a function of x. x (g) y 3 x 32 3 is the graph of y x 2 , shifted to the left 3 units, reflected about the xaxis, stretched vertically by a factor of 3, and shifted 3 units upward. It has Graph VII, and defines y as a function of x. (h) y 2 x is the graph of of y x, reflected about the yaxis and shifted to the right 2 units. It has Graph II, and defines y as a function of x.
CHAPTER 2
95
Test
CHAPTER 2 TEST
1. By the Vertical Line Test, figures (a) and (b) are graphs of functions. By the Horizontal Line Test, only figure (a) is the graph of a onetoone function.
0 2 2 a2 a2 2. (a) f 0 0; f 2 ; f a 2 . 01 21 3 a21 a3 x (b) f x . Our restrictions are that the input to the radical is nonnegative and that the denominator must not be 0. x 1 Thus, x 0 and x 1 0 x 1. (The second restriction is made irrelevant by the first.) In interval notation, the domain is [0 . 10 2 f 10 f 2 3 10 11 2 10 1 2 1 (c) The average rate of change is . 10 2 10 2 264
3. (a) “Subtract 2, then cube the result” can be expressed
(c)
y
algebraically as f x x 23 . (b) x
f x
1
27
0 1 2 3
2
1 x
8 1
0 1
4 8 (d) We know that f has an inverse because it passes the Horizontal Line Test. A verbal description for f 1 is, “Take the cube root, then add 2.” (e) y x 23 3 y x 2 x 3 y 2. Thus, a formula for f 1 is f 1 x 3 x 2.
4. (a) f 3 2 and f 2 3.
(b) The net change between x 3 and x 2 is f 2 f 3 3 2 5. The average rate of change over this 5 f 2 f 3 1. interval is 2 3 5 (c) f has domain [5 5] and range [4 4]. (d) f is increasing on 5 4 and 1 3 and decreasing on 4 1 and 3 5.
(e) f has a local minimum value of 4 at x 1 and local maximum values of 1 at x 4 and 4 at x 3. (f) No, f fails the Horizontal Line Test and so is not onetoone.
96
CHAPTER 2 Functions
5. R x 500x 2 3000x
(a) R 2 500 22 3000 2 $4000 represents their total
(b) We can see from the graph that the
sales revenue when their price is $2 per bar and
revenue increases until the price reaches
R 4 500 42 3000 4 $4000 represents their total
$3, then decreases. R
sales revenue when their price is $4 per bar
5000
(c) The maximum revenue is $4500, and it is achieved at a price
4000
of x $3.
3000 2000 1000 0
1
2
3
4
x
5
6. The net change is f 2 h f 2 2 h2 2 2 h 22 2 2 4 h 2 4h 4 2h 0 2h h 2 and the average rate of change is
2h h 2 f 2 h f 2 2 h. 2h2 h
7. (a) f x x 52 x 2 10x 25 is not linear because it cannot be
(b)
y
expressed in the form f x ax b for constants a and b. g x 1 5x is linear.
y=g(x)
y=f(x)
(c) g x has rate of change 5. 10 0
8. (a) f x x 2
1
x
(b) g x x 42 1. To obtain the graph of g, shift the graph of f to the left 4 units and
y
downward 1 unit. y
y=x@
2 0
1
x
y=x@
2 y=(x+4)@-1
0
1
x
9. (a) The graph of y f x 3 2 can be obtained by shifting the graph of f x to the right 3 units, then upward 2 units. (b) If f x x, then y f x 3 2 x 3 2.
10. (a) The graph of y f x can be obtained by reflecting the graph of f x about the yaxis. (b) If f x x, then y f x x.
CHAPTER 2
11. (a) f 2 1 2 1 2 3 (since 2 1 ).
97
Test
y
(b)
f 1 1 1 0 (since 1 1 ).
1 x
1
12. f x x 2 x 1; g x x 3.
(a) f g x f x g x x 2 x 1 x 3 x 2 2x 2 (b) f g x f x g x x 2 x 1 x 3 x 2 4
(c) f g x f g x f x 3 x 32 x 3 1 x 2 6x 9 x 3 1 x 2 5x 7 (d) g f x g f x g x 2 x 1 x 2 x 1 3 x 2 x 2
(e) f g 2 f 1 12 1 1 1. [We have used the fact that g 2 2 3 1.] (f) g f 2 g 7 7 3 4. [We have used the fact that f 2 22 2 1 7.]
(g) g g g x g g g x g g x 3 g x 6 x 6 3 x 9. [We have used the fact that g x 3 x 3 3 x 6.]
13. (a) f x x 3 1 is onetoone because each real number has a unique cube.
(b) g x x 1 is not onetoone because, for example, g 2 g 0 1.
1 1 1 x for all x 0, and g f x 2 x 2 2 x for all x 2. Thus, by 1 1 1 2 2 x x 2 x the Inverse Function Property, f and g are inverse functions.
14. f g x
x 3 5y 3 x 3 . y 2x 5 y x 3 x 2y 1 5x 3 x . Thus, 2x 5 2x 5 2y 1 5x 3 f 1 x . 2x 1 (b) f x 3 x, x 3 and f 1 x 3 x 2 , 16. (a) f x 3 x, x 3 y 3 x
15. f x
y 2 3 x x 3 y 2 . Thus f 1 x 3 x 2 , x 0.
x 0
y
f 1 x
1
f Ð!
17. The domain of f is [0 6], and the range of f is [1 7]. 18. The graph passes through the points 0 1 and 4 3, so f 0 1 and f 4 3.
98
CHAPTER 2 Functions
19. The graph of f x 2 can be obtained by shifting the graph of f x to the right
y
2 units. The graph of f x 2 can be obtained by shifting the graph of f x
y=f(x)+2
upward 2 units.
y=f(x-2) 1
f 1
x
20. The net change of f between x 2 and x 6 is f 6 f 2 7 2 5 and the average rate of change is 5 f 6 f 2 . 62 4 21. Because f 0 1, f 1 1 0. Because f 4 3, f 1 3 4. 22. y
f fÐ! 1
1
x
23. (a) f x 3x 4 14x 2 5x 3. The graph is shown in the viewing rectangle [10 10] by [30 10].
10
10 20
(b) No, by the Horizontal Line Test. (c) The local maximum is approximately 255 when x 018, as shown in the first viewing rectangle [015 025] by [26 25]. One local minimum is approximately 2718 when x 161, as shown in the second viewing rectangle [165 155] by [275 27]. The other local minimum is approximately 1193 when x 143, as shown is the viewing rectangle [14 15] by [12 119]. 0.15 2.50 2.55 2.60
0.20
0.25
1.65
1.60
1.55 27.0 27.2 27.4
1.40 11.90
1.45
1.50
11.95 12.00
(d) Using the graph in part (a) and the local minimum, 2718, found in part (c), we see that the range is [2718 .
(e) Using the information from part (c) and the graph in part (a), f x is increasing on the intervals 161 018 and 143 and decreasing on the intervals 161 and 018 143.
Modeling with Functions
99
FOCUS ON MODELING Modeling with Functions 1. Let be the width of the building lot. Then the length of the lot is 3. So the area of the building lot is A 32 , 0. 2. Let be the width of the poster. Then the length of the poster is 10. So the area of the poster is A 10 2 10.
3. Let be the width of the base of the rectangle. Then the height of the rectangle is 12 . Thus the volume of the box is given by the function V 12 3 , 0. 4. Let r be the radius of the cylinder. Then the height of the cylinder is 4r. Since for a cylinder V r 2 h, the volume of the cylinder is given by the function V r r 2 4r 4r 3 .
5. Let P be the perimeter of the rectangle and y be the length of the other side. Since P 2x 2y and the perimeter is 20, we have 2x 2y 20 x y 10 y 10 x. Since area is A x y, substituting gives A x x 10 x 10x x 2 , and since A must be positive, the domain is 0 x 10.
6. Let A be the area and y be the length of the other side. Then A x y 16 y P 2x 2
32 16 2x , where x 0. x x
7.
16 . Substituting into P 2x 2y gives x
Let h be the height of an altitude of the equilateral triangle whose side has length x, x
x
h 1 _x 2
as shown in the diagram. Thus the area is given by A 12 xh. By the Pythagorean 2 Theorem, h 2 12 x x 2 h 2 14 x 2 x 2 h 2 34 x 2 h 23 x.
Substituting into the area of a triangle, we get A x 12 xh 12 x 23 x 43 x 2 , x 0.
8. Let d represent the length of any side of a cube. Then the surface area is S 6d 2 , and the volume is V d 3 d 3 V . 2 Substituting for d gives S V 6 3 V 6V 23 , V 0. A r 9. We solve for r in the formula for the area of a circle. This gives A r 2 r 2 r A
A , A 0.
A , so the model is
C 10. Let r be the radius of a circle. Then the area is A r 2 , and the circumference is C 2r r . Substituting for r 2 2 C2 C , C 0. gives A C 2 4 60 . x2 The surface area, S, of the box is the sum of the area of the 4 sides and the area of the base and top. Thus 240 240 60 2x 2 S 4xh 2x 2 4x 2x 2 , so the model is S x 2x 2 , x 0. 2 x x x
11. Let h be the height of the box in feet. The volume of the box is V 60. Then x 2 h 60 h
12. By similar triangles,
12 5d 5 5 L d 12L 5d 7L L . The model is L d 57 d. L L d 7
100
FOCUS ON MODELING
13.
Let d1 be the distance traveled south by the first ship and d2 be the distance
dª
traveled east by the second ship. The first ship travels south for t hours at 5 mi/h, so dÁ
d1 15t and, similarly, d2 20t. Since the ships are traveling at right angles to
D
each other, we can apply the Pythagorean Theorem to get D t d12 d22 15t2 20t2 225t 2 400t 2 25t.
14. Let x be one of the numbers. Then the other number is 60 x, so the product is given by the function P x x 60 x 60x x 2 .
Let b be the length of the base, l be the length of the equal sides, and h be the
15. l
h
b
l
height in centimeters. Since the perimeter is 8, 2l b 8 2l 8 b 2 l 12 8 b. By the Pythagorean Theorem, h 2 12 b l 2 h l 2 14 b2 . Therefore the area of the triangle is b1 1 2 2 A 12 b h 12 b l 2 14 b2 4 8 b 4 b 2 b b b 64 16b b2 b2 64 16b 4 4 b b 4 b 4 4 4 so the model is A b b 4 b, 0 b 4.
16. Let x be the length of the shorter leg of the right triangle. Then the length of the other triangle is 2x. Since it is a right triangle, the length of the hypotenuse is x 2 2x2 5x 2 5 x (since x 0 ). Thus the perimeter of the triangle is P x x 2x 5 x 3 5 x.
2 2 17. Let be the length of the rectangle. By the Pythagorean Theorem, 12 h 2 102 h 2 102 4 2 4 100 h 2 2 100 h 2 (since 0 ). Therefore, the area of the rectangle is A h 2h 100 h 2 , so the model is A h 2h 100 h 2 , 0 h 10.
18. Using the formula for the volume of a cone, V 13 r 2 h, we substitute V 100 and solve for h. Thus 100 13 r 2 h h r
300 . r 2
Modeling with Functions
(b) Let x be one number: then 19 x is the other number, and so the product, p, is
19. (a) We complete the table. First number
Second number
Product
1 2 3 4 5 6 7 8 9 10 11
18 17 16 15 14 13 12 11 10 9 8
18 34 48 60 70 78 84 88 90 90 88
101
p x x 19 x 19x x 2 . (c) p x 19x x 2 x 2 19x 2 2 19 x 2 19x 19 2 2 x 952 9025
So the product is maximized when the numbers are both 95.
From the table we conclude that the numbers is still increasing, the numbers whose product is a maximum should both be 95. 20. Let the positive numbers be x and y. Since their sum is 100, we have x y 100 y 100 x. We wish to minimize
the sum of squares, which is S x 2 y 2 x 2 100 x2 . So S x x 2 100 x2 x 2 10,000 200x x 2 2x 2 200x 10,000 2 x 2 100x 10,000 2 x 2 100x 2500 10,000 5000 2 x 502 5000.
Thus the minimum sum of squares occurs when x 50. Then y 100 50 50. Therefore both numbers are 50.
21. (a) Let x be the width of the field (in feet) and l be the length of the field (in feet). Since the farmer has 2400 ft of fencing we must have 2x l 2400. 2000
200
400
200
Area=2000(200)=400,000 1000 700
1000
1000
700 Area=400(1000)=400,000
Area=1000(700)=700,000
Width
Length
Area
200
2000
400,000
300
1800
540,000
400
1600
640,000
500
1400
700,000
600
1200
720,000
700
1000
700,000
800
800
640,000
It appears that the field of largest area is about 600 ft 1200 ft.
(b) Let x be the width of the field (in feet) and l be the length of the field (in feet). Since the farmer has 2400 ft of fencing we must have 2x l 2400 l 2400 2x. The area of the fencedin field is given by A x l x 2400 2x x 2x 2 2400x 2 x 2 1200x . (c) The area is A x 2 x 2 1200x 6002 2 6002 2 x 6002 720,000. So the maximum area occurs when x 600 ft and l 2400 2 600 1200 ft.
rancher
22. (a) Let be the width of the rectangular area (in feet) and l be the length of the field (in feet). Since the farmer has 750 feet of fencing, we must have 5 2l 750 2l 750 5 l 52 150 . Thus the total area of the four pens is A l 52 150 52 2 150 .
102
FOCUS ON MODELING
(b) We complete the square to get A 52 2 150 52 2 150 752 52 752 52 752 140625. Therefore, the largest possible total area of the four pens is 14,0625 ft2 .
23. (a) Let x be the length of the fence along the road. If the area is 1200, we have 1200 x width, so the width of the garden 1200 7200 1200 . Then the cost of the fence is given by the function C x 5 x 3 x 2 8x . is x x x (b) We graph the function y C x in the viewing
(c) We graph the function y C x and y 600 in
cost is minimized when x 30 ft. Then the
From this we get that the cost is at most $600
rectangle [0 75] [0 800]. From this we get the width is 1200 30 40 ft. So the length is 30 ft and
the width is 40 ft.
the viewing rectangle [10 65] [450 650].
when 15 x 60. So the range of lengths he can fence along the road is 15 feet to 60 feet. 600
500
500 0
0
50
20
40
24. (a) Let x be the length of wire in cm that is bent into a square. So 10 x is the length of wire in 10 x x and , and the area cm that is bent into the second square. The width of each square is 4 4 x 2 x2 100 20x x 2 10 x 2 and . Thus the sum of the areas is of each square is 4 16 4 16
x2 100 20x x 2 100 20x 2x 2 18 x 2 54 x 25 4 . 16 16 16 (b) We complete the square. 1 x 2 10x 25 1 x 2 10x 25 25 25 1 x 52 25 . A x 18 x 2 54 x 25 4 8 4 8 8 8 8 4 25 2 So the minimum area is 8 cm when each piece is 5 cm long. A x
25. (a) Let h be the height in feet of the straight portion of the window. The circumference of the semicircle is C 12 x. Since the perimeter of the window is 30 feet, we have x 2h 12 x 30.
Solving for h, we get 2h 30 x 12 x h 15 12 x 14 x. The area of the window is 2 A x xh 12 12 x x 15 12 x 14 x 18 x 2 15x 12 x 2 18 x 2 . 120 (b) A x 15x 18 4 x 2 18 4 x 2 x 4 2 2 60 450 120 450 60 1 2 x 18 4 x 8 4 x 4 4 4 4 4 The area is maximized when x
60 840, and hence h 15 12 840 14 840 420. 4
60
Modeling with Functions
26. (a) The height of the box is x, the width of
103
(b) We graph the function y V x in the viewing rectangle
the box is 12 2x, and the length of the
[0 6] [200 270].
box is 20 2x. Therefore, the volume of the box is
250
V x x 12 2x 20 2x 4x 3 64x 2 240x, 0 x 6
200
(c) From the graph, the volume of the box
0
5
From the calculator we get that the volume of the box is
with the largest volume is 262682 in3
greater than 200 in3 for 1174 x 3898 (accurate to 3
when x 2427.
decimal places).
27. (a) Let x be the length of one side of the base and let h be the height of the box in feet. Since the volume of 12 the box is V x 2 h 12, we have x 2 h 12 h 2 . The surface area, A, of the box is sum of the x area of the four sides and the area of the base. Thus the surface area of the box is given by the formula 12 48 A x 4xh x 2 4x x2 x 2 , x 0. x x2 (b) The function y A x is shown in the first viewing rectangle below. In the second viewing rectangle, we isolate the minimum, and we see that the amount of material is minimized when x (the length and width) is 288 ft. Then the 12 height is h 2 144 ft. x 26
50
25 0
0
2
24
4
3.0
28. Let A, B, C, and D be the vertices of a rectangle with base AB on the xaxis and its other two vertices C and D above the xaxis and lying on the parabola y 8 x 2 . Let C have the coordinates x y, x 0. By symmetry, the coordinates of D must be x y. So the width of the rectangle is 2x, and the length is y 8 x 2 . Thus the area of the rectangle is A x length width 2x 8 x 2 16x 2x 3 . The graphs of A x below show that the area is maximized when
x 163. Hence the maximum area occurs when the width is 326 and the length is 533. y
y=8-x@
D
A
C
x
B
x
20
18
10
17
0
0
2
4
16
1.5
2.0
104
FOCUS ON MODELING
29. (a) Let x be the width of the pen and l be the length in meters. We use the area to establish a relationship between 100 . So the amount of fencing used is x and l. Since the area is 100 m2 , we have l x 100 l x 2 200 2x 100 F 2l 2x 2 2x . x x (b) Using a graphing device, we first graph F in the viewing rectangle [0 40] by [0 100], and locate the approximate location of the minimum value. In the second viewing rectangle, [8 12] by [39 41], we see that the minimum value of F occurs when x 10. Therefore the pen should be a square with side 10 m. 100
41
50
40
0
0
20
40
39
8
10
12
30. (a) The distance from B to R is d1 0 32 x 02 x 2 9 and the distance from R to A is d2 6 02 6 x2 x 62 36, so the total distance is d x d1 x d2 x x 2 9 x 62 36. (b) Using a graphing device, we find that the distance is minimized when x 2. 11.0 10
10.8 10.6
0
0
5
31. (a) Let t1 represent the time, in hours, spent walking, and let t2 represent the time spent rowing. Since the distance walked is x and the walking speed is 5 mi/h, the time spent walking is t1 15 x. By the
Pythagorean Theorem, the distance rowed is d 22 7 x2 x 2 14x 53, and so the time spent rowing is t2 12 x 2 14x 53. Thus the total time is T x 12 x 2 14x 53 15 x.
1.8
2.0
2.2
(b) We graph y T x. Using the zoom function, we see that T is minimized when x 613. You should land at a point 613 miles from point B. 4 2 0
0
2
4
6
Modeling with Functions
32. (a) Let x be the distance from point B to C, in miles. Then the distance from A to C is flying from A to C then C to D is f x 14 x 2 25 10 12 x.
105
x 2 25, and the energy used in
(b) Using a graphing device, we find that the energy expenditure is minimized when the distance from B to C is about 51 miles. 200
169.1
100
169.0
0
0
5
10
168.9
5.0
5.1
5.2
33. (a) Using the Pythagorean Theorem, we have that the height of the upper triangles is 25 x 2 and the height of the lower triangles is 144 x 2 . So the area of the each of the upper triangles is 12 x 25 x 2 , and the area of the each of the lower triangles is 12 x 144 x 2 . Since there are two upper triangles and two lower triangles, we get that the total area 25 x 2 144 x 2 . is A x 2 12 x 25 x 2 2 12 x 144 x 2 x (b) The function y A x x 25 x 2 144 x 2 is shown in the first viewing rectangle below. In the second viewing rectangle, we isolate the maximum, and we see that the area of the kite is maximized when x 4615. So the length of the horizontal crosspiece must be 2 4615 923. The length of the vertical crosspiece is 52 46152 122 46152 1300. 100
60.1
50
60.0
0
0
2
4
59.9 4.60
4.62
4.64
Corrections: 1, 12, 13, 16, 45, 85, 90, 91, 94, 95, 96, 97, 99, 113, 116, 120, 122, 123, 124, 130, 134, 138
CHAPTER 3
NOTE THAT THESE CORRECTIONS SHOULD BE MADE IN ALL THREE BOOKS.
POLYNOMIAL AND RATIONAL FUNCTIONS
3.1
Quadratic Functions and Models 1
3.2
Polynomial Functions and Their Graphs 11
3.3
Dividing Polynomials 28
3.4
Real Zeros of Polynomials 37
3.5
Complex Zeros and the Fundamental Theorem of Algebra 70
3.6 3.7
Rational Functions 80 Polynomial and Rational Inequalities 101 Chapter 3 Review 115 Chapter 3 Test 135
¥
FOCUS ON MODELING: Fitting Polynomial Curves to Data 139
1
3
POLYNOMIAL AND RATIONAL FUNCTIONS
3.1
QUADRATIC FUNCTIONS AND MODELS
1. To put the quadratic function f x ax 2 bx c in vertex form we complete the square. 2. The quadratic function f x a x h2 k is in vertex form. (a) The graph of f is a parabola with vertex h k. (b) If a 0 the graph of f opens upward. In this case f h k is the minimum value of f .
(c) If a 0 the graph of f opens downward. In this case f h k is the maximum value of f .
3. The graph of f x 3 x 22 6 is a parabola that opens upward, with its vertex at 2 6, and f 2 6 is the minimum value of f . 4. The graph of f x 3 x 22 6 is a parabola that opens downward, with its vertex at 2 6, and f 2 6 is the maximum value of f . 6. f x 12 x 2 2x 6
5. f x x 2 6x 5 (a) The vertex is 3 4, the xintercepts are 1 and 5, and
(a) The vertex is 2 8, the xintercepts are 6 and 2,
it appears that the yintercept is approximately 5.
and the yintercept is 6.
(b) Maximum value of f : 4
(b) Maximum value of f : 8
(c) Domain , range: 4]
(c) Domain: , range: 8]
2x^2 - 4x -1
7. f x x 2 6x 5
8. f x 3x 2 6x 1
(a) The vertex is 1 3, the xintercepts are
(a) The vertex is 1 4, the xintercepts are
(b) Minimum value of f : 3
(b) Minimum value of f : 4
(c) Domain: , range: [3
(c) Domain: , range: [4
2 6 02 and 22, and the yintercept is 1. 2
32 3 22 and 02, and the yintercept is 1. 3
9. (a) f x x 2 4x 9 x 22 4 9 x 22 5
(c)
y
(b) The vertex is at 2 5. xintercepts: y 0 0 x 22 5 x 22 5. This has
no real solution, so there is no xintercept.
yintercept: x 0 y 0 22 5 9. The yintercept is 9. (d) Domain: , range: [5 1 0
1
x
1
2
CHAPTER 3 Polynomial and Rational Functions
10. (a) f x x 2 6x 8 x 32 9 8 x 32 1
y
(c)
(b) The vertex is at 3 1.
xintercepts: y 0 0 x 32 1 x 3 1 x 4
or 2. The xintercepts are 4 and 2.
yintercept: x 0 y 0 32 1 8. The yintercept is 8.
(d) Domain: , range: [1 .
11. (a) f x x 2 6x x 2 6x x 2 6x 9 9 x 32 9
1 1x
(c)
(b) The vertex is at 3 9.
y
1
xintercepts: y 0 0 x 2 6x x x 6. So x 0 or x 6.
x
1
The xintercepts are 0 and 6.
yintercept: x 0 y 0. The yintercept is 0. (d) Domain: , range: [9
12. (a) f x x 2 8x x 2 8x 16 16 x 42 16
(c)
y
(b) The vertex is at 4 16.
xintercepts: y 0 0 x 2 8x x x 8. So x 0 or
x 8. The xintercepts are 0 and 8.
3
yintercept: x 0 y 0. The yintercept is 0.
1
x
1
x
(d) Domain: , range: [16
13. (a) f x 3x 2 6x 3 x 2 2x 3 x 2 2x 1 3
(c)
y
3 x 12 3
(b) The vertex is at 1 3.
xintercepts: y 0 0 3 x 12 3 x 12 1
x 2 or 0. The xintercepts are 2 and 0.
yintercept: x 0 y 3 02 6 0 0. The yintercept is 0.
(d) Domain: , range: [3
1
3
SECTION 3.1 Quadratic Functions and Models
14. (a) f x x 2 10x x 2 10x x 2 10x 25 25 x 52 25
(c)
y
(b) The vertex is at 5 25. 10
xintercepts: y 0 0 x 52 25 x 52 25
x 0 or 10. The xintercepts are 0 and 10.
x
1
yintercept: x 0 y 02 10 0 0. The yintercept is 0.
(d) Domain: , range: 25]
15. (a) f x x 2 4x 3 x 2 4x 4 4 3 x 22 1
(c)
y
(b) The vertex is at 2 1.
xintercepts: y 0 0 x 2 4x 3 x 1 x 3. So x 1 or x 3. The xintercepts are 1 and 3.
1
yintercept: x 0 y 3. The yintercept is 3.
1
x
1
x
(d) Domain: , range: [1
16. (a) f x x 2 2x 2 x 2 2x 1 1 2 x 12 1
(c)
y
(b) The vertex is at 1 1. xintercepts: y 0 x 12 1 0 x 12 1. Since
this last equation has no real solution, there is no xintercept.
1
yintercept: x 0 y 2. The yintercept is 2. (d) Domain: , range: [1
17. (a) f x x 2 10x 15 x 2 10x 25 25 15
(c)
y
x 52 10
(b) The vertex is at 5 10. xintercepts: y 0 0 x 52 10 x 52 10
x 5 10 x 5 10. yintercept: x 0 y 15. The yintercept is 15. (d) Domain: , range: 10]
2 1
x
4
CHAPTER 3 Polynomial and Rational Functions
18. (a) f x x 2 12x 11 x 2 12x 36 36 11
(c)
y
x 62 25
(b) The vertex is at 6 25. xintercepts: y 0 0 x 62 25 x 62 25
x 1 or 11. yintercept: x 0 y 11. The yintercept is 11. (d) Domain: , range: 25]
19. (a) f x 3x 2 6x 7 3 x 2 2x 1 3 7 3 x 12 4
2
(c)
1
x
y
(b) The vertex is at 1 4.
xintercepts: y 0 0 3 x 12 4 3 x 12 4.
Since this equation has no real solution, there is no xintercept. yintercept: x 0 y 7. The yintercept is 7. (d) Domain: , range: [4
2 x
1
20. (a) f x 3x 2 6x 2 3 x 2 2x 1 3 2
(c)
y
3 x 12 1
(b) The vertex is at 1 1. xintercepts: y 0 0 3 x 12 1 0 x 12 13 x 1 13 x 1 13 . The xintercepts are 1 13 and 1 13 .
2 x
1
yintercept: x 0 y 2. The yintercept is 2.
(d) Domain: , range: 1] 21. (a) f x 05x 2 6x 16 12 x 2 12x 36 18 16
(c)
y
12 x 62 2
(b) The vertex is at 6 2.
xintercepts: y 0 0 12 x 62 2 x 62 4
x 8 or 4. yintercept: x 0 y 16. The yintercept is 16. (d) Domain: , range: [2
2 1
x
5
SECTION 3.1 Quadratic Functions and Models
22. (a) f x 2x 2 12x 10 2 x 2 6x 9 18 10 2 x 32 8
(c)
y
(b) The vertex is at 3 8. The xintercepts are 5 and 1 and the yintercept is 10.
2 0
(d) Domain: , range: [8
23. (a) f x 4x 2 12x 1 4 x 2 3 1 2 2 4 x 32 9 1 4 x 32 10 (b) The vertex is at 32 10 .
x
y
(c)
10
2 2 xintercepts: y 0 0 4 x 32 10 x 32 52 x 32 52 x 32 52 32 210 . The xintercepts are
32
1
5
1
10 10 3 2 and 2 2 .
_3
_2
x
_1
yintercept: x 0 y 4 02 12 0 1 1. The yintercept
is 1.
(d) Domain: , range: 10]
24. (a) f x 3x 2 2x 2 3 x 2 23 x 2 2 2 3 x 13 13 2 3 x 13 73 (b) The vertex is at 13 73 .
xintercepts: y 0 0 3 x 13
2
73 x 13
(c)
2
10
79
x 13 37 x 13 37 . The xintercepts are 13 37 and
13 37 .
yintercept: x 0 y 3 02 2 0 2 2. The yintercept
is 2.
(d) Domain: , range: 73
y 20
1
x
6
CHAPTER 3 Polynomial and Rational Functions
25. (a) f x x 2 2x 1 x 2 2x 1 x 2 2x 1 1 1 x 12 2 (b)
y
26. (a) f x x 2 8x 8 x 2 8x 16 8 16 x 42 8
(b)
y
2
1 1
(c) The minimum value is f 1 2. 27. (a) f x 4x 2 8x 1 4 x 2 2x 1 4 x 2 2x 1 4 1
(c) The minimum value is f 4 8. 28. (a) f x 2x 2 12x 14 2 x 2 6x 14 2 x 32 18 14
2 x 32 4
4 x 12 5
(b)
y
x
1
x
(b)
y
1
x
1
(c) The minimum value is f 1 5. 29. (a) f x x 2 3x 3 x 2 3x 3 x 2 3x 94 3 94 2 x 32 21 4 (b)
x
1
1
y
(c) The minimum value is f 3 4. 30. (a) f x 1 6x x 2 x 2 6x 1 x 2 6x 9 1 9 x 32 10
(b)
y
2
1 1
(c) The maximum value is f 32 21 4.
x
1
(c) The maximum value is f 3 10.
x
SECTION 3.1 Quadratic Functions and Models
31. (a) f x 3x 2 12x 13 3 x 2 4x 13 3 x 2 4x 4 13 12 3 x 22 1
(b)
32. (a) f x 2x 2 12x 20 2 x 2 6x 20 2 x 2 6x 9 18 20 2 x 32 2
y
(b)
y
2 x
1
2 1
(c) The minimum value is f 2 1.
(c) The minimum value is f 3 2. 34. (a) f x 3 4x 4x 2 4 x 2 x 3 4 x 2 x 14 3 1 2 4 x 12 4
33. (a) f x 1 x x 2 x 2 x 1 x 2 x 14 1 14 2 x 12 54 (b)
x
(b)
y
y
1
1 1
(c) The maximum value is f 12 54 .
x
1
x
(c) The maximum value is f 12 4.
35. For f x 7x 2 14x 5 we have a 7, b 14, and c 5. Because a 0, the maximum value is b f 14 f 7 12 14 1 5 2. f 2a 1 27
36. For f x 6x 2 48x 1 we have a 6, b 48, and c 1. Because a 0, the minimum value is b f 48 f 4 6 42 48 4 1 95. f 2a 26
37. For f t 4t 2 40t 110 we have a 4, b 40, and c 110. Because a 0, the minimum value is b f 40 f 5 4 52 40 5 110 10. f 2a 24
38. For g x 5x 2 60x 200 we have a 5, b 60, and c 200. Because a 0, the maximum value is b g 60 2 g 2a 25 g 6 5 6 60 6 200 20.
39. For f s s 2 12s 16 we have a 1, b 12, and c 16. Because a 0, the minimum value is b f 12 f 06 062 12 06 16 1564. f 2a 21
7
8
CHAPTER 3 Polynomial and Rational Functions
40. For g x 100x 2 1500x we have a 100, b 1500, and c 0. Because a 0, the minimum value is b g 1500 g 15 100 15 2 1500 15 5625. g 2a 2100 2 2 2 41. For h x 12 x 2 2x 6 we have a 12 , b 2, and c 6. Because a 0, the minimum value is b h 2 1 2 h 2a 212 h 2 2 2 2 2 6 8
x2 2x 7 13 x 2 2x 7 we have a 13 , b 2, and c 7. Because a 0, the maximum value is 42. For f x 3 b f 2 1 2 f 2a 213 f 3 3 3 2 3 7 10. 43. For f x 3 x 12 x 2 12 x 2 x 3 we have a 12 , b 1, and c 3. Because a 0, the maximum value is b f 1 1 7 2 f 2a 212 f 1 2 1 1 3 2 .
44. For g x 2x x 4 7 2x 2 8x 7 we have a 2, b 8, and c 7. Because a 0, the minimum value is b g 8 g 2 2 22 8 2 7 1. g 2a 22 45. (a) The graph of f x x 2 179x 321 is shown. The minimum value is f 090 401. 1.0
0.9
0.8 3.9 4.0
(b) f x x 2 179x 321 has a 1, b 179, and c 321. Because a 0, the minimum value is b f 179 f 0895 f 2a 21 08952 179 0895 321 4011025
4.1
46. (a) The graph of f x 1 x
2 2x is
shown. The maximum value is f 035 118. 1.180
(b) f x 1 x 2 x 2 2x 2 x 1 has a 2, b 1, and c 1. Because a 0, the maximum value is b 1 f 42 f 2a f 2 2
2 2 42 42 1 82 1
1.175 1.170 0.30
0.35
0.40
1176777
47. The vertex is 2 3, so the parabola has equation y a x 22 3. Substituting the point 3 1, we have 1 a 3 22 3 a 4, so f x 4 x 22 3.
48. The vertex is 1 5, so the parabola has equation y a x 12 5. Substituting the point 3 7, we have
7 a 3 12 5 a 3, so f x 3 x 12 5. 49. Substituting t x 2 , we have f t 3 4t t 2 t 2 4t 4 4 3 t 22 7. Thus, the maximum value is 7, when t 2 (or x 2). 50. Substituting t x 3 , we have f t 2 16t 4t 2 4 t 2 4t 4 16 2 4 t 22 14. Thus, the minimum value is 14, when t 2 (or x 3 2).
SECTION 3.1 Quadratic Functions and Models
9
2 2 2 16 54 16 t 54 25. Thus the maximum 51. y f t 40t 16t 2 16 t 2 52 16 t 2 52 t 54 height attained by the ball is f 54 25 feet. 32 x 2 x 5 2 x 2 25 x 5 2 x 2 25 x 25 2 2 25 2 5 52. (a) We complete the square: y 400 25 2 25 2 4 25 4 2 x 25 2 65 , so the maximum height of the ball is 65 8125 ft. y 25 4 8 8
2 x 2 x 5. Using the Quadratic (b) The ball hits the ground when its vertical displacement y is 0, that is, when 0 25 2 5 1 1 4 25 25 5 65 . Taking the positive root, we find that x 163 ft. Formula, we find x 4 4 25 53. R x 80x 04x 2 04 x 2 200x 04 x 2 200x 10,000 4,000 04 x 1002 4,000. So
revenue is maximized at $4,000 when 100 units are sold. 54. P x 0001x 2 3x 1800 0001 x 2 3000x 1800 0001 x 2 3000x 2 250 000 1800
2250 0001 x 15002 450. The maximum profit is $450, and occurs when 1500 cans are sold. 1 n 2 1 n 2 60n 1 n 2 60n 900 10 1 n 302 10. Since the maximum of 55. E n 23 n 90 90 90 90 the function occurs when n 30, the viewer should watch the commercial 30 times for maximum effectiveness. 56. C t 006t 00002t 2 00002 t 2 300t 00002 t 2 300t 22,500 45 00002 t 1502 45. The maximum concentration of 45 mg/L occurs after 150 minutes.
57. A n n 900 9n 9n 2 900n is a quadratic function with a 9 and b 900, so by the formula, the maximum 900 b 50 trees, and because a 0, this gives a maximum value. or minimum value occurs at n 2a 2 9 58. A n 700 n 10 001n 001n 2 10n 001 700 n 7000 001n 2 3n 7000. This is a quadratic b 3 function with a 001 and b 3, so the maximum (a 0) occurs at x 150. Since n 150 2a 2 001 is the number of additional vines that should be planted, the total number of vines that maximizes grape production is 700 150 850 vines. 59. The area of the fencedin field is given by A x 2400 2x x 2x 2 2400x. Thus, by the formula in this section, 2400 b 600. The maximum area occurs when x 600 feet the maximum or minimum value occurs at x 2a 2 2 and l 2400 2 600 1200 feet. 60. The total area of the four pens is A 52 150 52 2 375. Thus, by the formula, the maximum or minimum value occurs at
375 b 75. Therefore, the largest possible total area of the four pens is 2a 2 5 2
A 75 52 752 375 75 14,0625 square feet.
61. A x 15x 18 4 x 2 , so by the formula, the maximum area occurs when x and h 15 12 840 14 840 42 ft.
15 b 84 ft 2a 2 1 4
62. A x 18 x 2 54 x 25 4 , so by the formula, the maximum or minimum area occurs where x 25 2 minimum area is 18 52 54 5 25 4 8 cm when each piece is 5 cm long.
8
5 b 4 5. The 2a 2 1 8
10
CHAPTER 3 Polynomial and Rational Functions
63. (a) The area of the corral is A x x 1200 x 1200x x 2 x 2 1200x.
(b) A is a quadratic function with a 1 and b 1200, so by the formula, it has a maximum or minimum at 1200 b 600, and because a 0, this gives a maximum value. The desired dimensions are 600 ft by x 2a 2 1 600 ft.
64. (a) The dimensions of the gutter are x inches and 30 x x 30 2x inches, so the cross sectional area is A x 30 2x 30x 2x 2 .
(b) Since A is a quadratic function with a 2 and b 30, the maximum occurs at x
b 30 75 inches. 2a 2 2
(c) The maximum cross section is A 75 2 752 30 75 1125 in2 .
65. (a) To model the revenue, we need to find the total attendance. Let x be the ticket price. Then the amount by which the ticket price is lowered is 10 x, and we are given that for every dollar it is lowered, the attendance increases by 3000; that is, the increase in attendance is 3000 10 x. Thus, the attendance is 27,000 3000 10 x, and since each spectator pays $x, the revenue is R x x [27,000 3000 10 x] 3000x 2 57,000x.
(b) Since R is a quadratic function with a 3000 and b 57,000, the maximum occurs at 57,000 b 95; that is, when admission is $950. x 2a 2 3000 (c) We solve R x 0 for x: 3000x 2 57,000x 0 3000x x 19 0 x 0 or x 19. Thus, if admission is $19, nobody will attend and no revenue will be generated.
66. (a) Let x be the price per feeder. Then the amount by which the price is increased is x 10, and we are given that for every dollar increase, sales decrease by 2; that is, the change in sales is 2 x 10, so the total number sold is 20 2 x 10 40 2x. The profit per feeder is equal to the sale price minus the cost, that is, x 6. Multiplying the number of feeders sold by the profit per feeder sold, we find the profit to be P x 40 2x x 6 2x 2 52x 240.
(b) Using the formula, profit is maximized when x
52 b 13; that is, when the society charges $13 per 2a 2 2
feeder. The maximum weekly profit is P 13 2 132 52 13 240 $98.
y
67. Because f x x m x n 0 when x m or x n, those are its xintercepts. By symmetry, we expect that the vertex is halfway between mn . We obtain the graph shown at right. these values; that is, at x 2
Expanding, we see that f x x 2 m n x mn, a quadratic
function with a 1 and b m n. Because a 0, the minimum value occurs at x
m n b , the xvalue of the vertex, as expected. 2a 2
y=(x-a)(x-b)
a
0
a+b 2
b
x
SECTION 3.2 Polynomial Functions and Their Graphs
11
68. The revenue is R x 23,500x 1000x 2 and the attendance is A x 9500 1000 14 x, so at a ticket
price of $1175, the revenue is R 1175 23,500 1175 1000 11752 $138,060 and the attendance is A 1175 9500 1000 14 1175 11,750. The arena’s capacity is 15,000, so we set A x 15,000 and solve for x: 15,000 9500 1000 14 x 1000 14 x 5500 14 x 55 x $850, the ticket price at which the arena will be filled to capacity. The
revenue in that case is R 85 23,500 85 1000 852 $127,500, which is less than the maximum revenue of $138,060. The model reflects the fact that lower prices generally correspond to more attendance, and vice versa; and revenue is the product of price and attendance. In general, quadratic models are appropriate for modeling profit and revenue due to this fact.
69. The point Q x y lies on the line y 2x 3, so it can be written as Q x 2x 3. The distance from P 3 2 to Q is thus g x [2x 3 2]2 x 32 2x 12 x 32 5x 2 10x 10. Note that in general, the minimum value of g x f x occurs at the value of x at which f x is minimized. Let f x 5x 2 10x 10. We find the vertex form of f : f x 5 x 2 2x 1 5 10 5 x 12 5. Thus, f (and g) have minimum values at x 1. The point closest to P on the line y 2x 3 is therefore 1 2 1 3 1 1.
3.2
POLYNOMIAL FUNCTIONS AND THEIR GRAPHS
1. Graph I cannot be that of a polynomial because it is not smooth (it has a cusp.) Graph II could be that of a polynomial function, because it is smooth and continuous. Graph III could not be that of a polynomial function because it has a break. Graph IV could not be that of a polynomial function because it is not smooth.
2. (a) y x 3 8x 2 2x 15 has odd degree and a positive leading coefficient, so y as x and y as x .
(b) y 2x 4 12x 100 has even degree and a negative leading coefficient, so y as x and y as x .
3. (a) If c is a zero of the polynomial P, then P c 0.
(b) If c is a zero of the polynomial P, then x c is a factor of P x.
(c) If c is a zero of the polynomial P, then c is an xintercept of the graph of P.
4. (a) This is impossible. If P has degree n 3, it has at most n 1 2 local extrema.
(b) This is possible. For example, y x 3 has degree 3 and no local maxima or minima.
(c) This is possible. For example, P x x 4 has one local maximum and no local minima.
Q(x)
12
CHAPTER 3 Polynomial and Rational Functions
5. (a) P x 12 x 2 2
(b) P x 2 x 32 y
y
1 1
(_2, 0)
x
(2, 0)
(0, 18) 10
(0, _2)
1
Domain: ; range: [2 (c) P x 4x 2 1
R(x)
y
x
(3, 0)
Domain: ; range: [0 (d) P x x 22 (_2, 0)
S(x)
y 1 0
1
x
(0, _4)
2 (0, 1)
1
x
Domain: ; range: 0]
Domain: ; range: [1
Q(x)
6. (a) P x x 4 1
(_1, 0)
(b) P x x 14
y (0, 1)
(1, 0)
y
x
2
(0, 1)
(_1, 0)
Domain: ; range: 1]
Domain: ; range: [0
1x
R(x)
SECTION 3.2 Polynomial Functions and Their Graphs
(c) P x 6x 4 6
S(x)
(d) P x 19 x 34
y
y (0, 9)
1
(_1, 0)
(1, 0)
0
x 1 0
(0, _6)
x
(3, 0)
Domain: ; range: [0
Domain: ; range: [6
7. (a) P x x 3 8
1
(b) Q x x 3 27
y
y
1 1
(2, 0)
x (0, 27)
(0, _8)
4
(3, 0) 1
Domain: ; range: (c) R x x 23
(_2, 0)
x
Domain: ; range: (d) S x 12 x 13 4
y
y
1 1
x
(0, _72) (0, _8)
Domain: ; range:
(_1, 0)
1 1
Domain: ; range:
x
13
14
CHAPTER 3 Polynomial and Rational Functions
8. (a) P x x 35
(b) Q x 2 x 35 64
y
y (0, 422)
(0, 243) (_1, 0)
50
(_3, 0)
100 1 x
1 x
Domain: ; range:
Domain: ; range:
(c) R x 12 x 25
(d) S x 12 x 25 16
y
y (0, 16) (0, 32)
2
(2, 0) 1
x
4 1
Domain: ; range:
(4, 0)
x
Domain: ; range:
9. (a) P x x x 2 4 x 3 4x has odd degree and a positive leading coefficient, so y as x and y as x .
(b) This corresponds to graph III. 10. (a) Q x x 2 x 2 4 x 4 4x 2 has even degree and a negative leading coefficient, so y as x and y as x .
(b) This corresponds to graph I. 11. (a) R x x 5 5x 3 4x has odd degree and a negative leading coefficient, so y as x and y as x . (b) This corresponds to graph V.
12. (a) S x 12 x 6 2x 4 has even degree and a positive leading coefficient, so y as x and y as x . (b) This corresponds to graph II.
13. (a) T x x 4 2x 3 has even degree and a positive leading coefficient, so y as x and y as x . (b) This corresponds to graph VI.
14. (a) U x x 3 2x 2 has odd degree and a negative leading coefficient, so y as x and y as x . (b) This corresponds to graph IV.
SECTION 3.2 Polynomial Functions and Their Graphs
15. P x x 2 x 5
16. P x 4 x x 1 y
y
(0, 4) (_1, 0) 2 (_2, 0)
2
(4, 0)
1
x
(5, 0) x
1
(0, _10)
17. P x x x 2 x 3
18. P x x 3 x 1 x 4
y
y
1 (_3, 0)
(2, 0)
1 (0, 0)
(12, 0) x
(_3, 0)
4
(1, 0) 2
19. P x 2x 1 x 1 x 3
y
(0, 12)
(0, 3) 1
(_1, 0)
1
(_2, 0)
21. P x x 2 x 1 x 2 x 3 y
(0, 12) 0 (_1, 0)
(2, 0) 1
(3, 0) x
( _23 , 0) 1
(3, 0)
x
22. P x x x 1 x 1 2 x y
(_1, 0)
(_2, 0)
4
(1/2, 0) x
5
x
20. P x x 3 x 2 3x 2
y
(_3, 0)
(4, 0)
1 0
(0, 0) 1 (1, 0)
(2, 0)
x
15
16
CHAPTER 3 Polynomial and Rational Functions
24. P x 15 x x 52
23. P x 2x x 22 y
y
1 (0, 0)
2 (0, 0)
(2, 0)
1
x
1
x
(5, 0)
2
25. P x x 1 x 12 2x 3
26. P x x 23 x 1 x 32 y
y
(_2, 0) (_1, 0) 2 0 (_2, 0)
(0, 72)
20
(1, 0)
0
(3, 0)
x
2
(3/2, 0) 1
x
(0, _6)
1 x 22 x 32 27. P x 12
28. P x x 12 x 23
y
y
(0, 8) 2 (_2, 0)
(0, 3)
1 (1, 0)
x
1 (_2, 0)
1
(3, 0)
x
30. P x 19 x 2 x 32 x 32
29. P x x x 22 x 23 y
10 (_2, 0)
y
1 (0, 0)
5 (2, 0)
x
(_3, 0)
(0, 0)
1
(3, 0) x
SECTION 3.2 Polynomial Functions and Their Graphs
31. P x x 3 x 2 6x x x 2 x 3
32. P x x 3 2x 2 8x x x 2 x 4
y
2 (0, 0)
(_2, 0)
1
y
4
(_4, 0) (3, 0)
x
33. P x x 3 x 2 12x x x 3 x 4 y
(0, 0)
1
(2, 0)
x
34. P x 2x 3 x 2 x x 2x 2 x 1 x 2x 1 x 1 y
4 (0, 0) 1
(_3, 0)
(4, 0)
1 x
( _21 , 0)
(0, 0)
1
(_1, 0)
35. P x x 4 3x 3 2x 2 x 2 x 1 x 2 y
1
36. P x x 5 9x 3 x 3 x 3 x 3 y
(_3, 0)
10 (0, 0)
(1, 0) (0, 0)
1
(2, 0)
x
x
(3, 0) 1
x
17
18
CHAPTER 3 Polynomial and Rational Functions
37. P x x 3 x 2 x 1 x 1 x 12 y
38. P x x 3 3x 2 4x 12 x 3 x 2 x 2 y
(_2, 0)
2
(_3, 0) (_1, 0)
1
(2, 0) 1
(1, 0)
x
x
1 (0, _1)
(0, _12)
2 40. P x 18 2x 4 3x 3 16x 24 2 18 x 22 2x 32 x 2 2x 4
39. P x 2x 3 x 2 18x 9 x 3 2x 1 x 3 y
y
40 (0, 9)
(_3, 0)
( _21 , 0)
1
(3, 0)
x (0, 72) 20 1
(_ _32 , 0)
41. P x x 4 2x 3 8x 16 x 22 x 2 2x 4
y
(_2, 0)
10 (2, 0) 1 (0, _16)
(0, 16) 1 (2, 0)
x
42. P x x 4 2x 3 8x 16 x 2 x 2 x 2 2x 4
y
10
(2, 0)
x
x
SECTION 3.2 Polynomial Functions and Their Graphs
43. P x x 4 3x 2 4 x 2 x 2 x 2 1 y
19
2 44. P x x 6 2x 3 1 x 3 1 2 x 12 x 2 x 1 y
1
(_2, 0)
(2, 0)
x
1 (0, _4)
(0, 1)
1
1 (1, 0)
x
45. P x 3x 3 x 2 5x 1; Q x 3x 3 . Since P has odd degree and positive leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, we see that the graphs of P and Q have different intercepts. 100
10
50 3
2
1
P 1
5
Q
50
2
3
1.5 1
0.5 5
100
10
P
Q
0.5
1
1.5
46. P x 18 x 3 14 x 2 12x; Q x 18 x 3 . Since P has odd degree and negative leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 60 40
1000 P
Q
Q
500 6
30 20 10 500 1000
10
20
30
4
P
20 2
0 20 40 60 80
2
4
6
20
CHAPTER 3 Polynomial and Rational Functions
47. P x x 4 7x 2 5x 5; Q x x 4 . Since P has even degree and positive leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and we see that they have different intercepts. 600
10
500 Q
400
P
P
300
Q 3
200
2
1
6
4
1
2
3
10
100 2 0 100
0
2
4
6
20
48. P x x 5 2x 2 x; Q x x 5 . Since P has odd degree and negative leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different. 3
200 Q
P
100
3
2
1 0 100
Q 1
2
P 1
3
2 1 0 1
1
2
200
3
49. P x x 11 9x 9 ; Q x x 11 . Since P has odd degree and positive leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look like they have the same end behavior. On a small viewing rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 600000 400000
Q
P
2
4
100
Q
6
4
2 0 200000
6
1
0
1
50
P
400000
P
50
200000
Q
100
600000
50. P x 2x 2 x 12 ; Q x x 12 . Since P has even degree and negative leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different. 200 3
2
1 0 200
1
2
2
3
400 600 P Q
800
P
1
0
Q
2
1000 1200
4
1
SECTION 3.2 Polynomial Functions and Their Graphs
51. (a) xintercepts are 0 and 4, yintercept is 0.
52. (a) xintercepts are 0 and 45, yintercept is 0.
(b) Local maximum at 2 4, no local minimum.
(b) Local maximum at 0 0, local minimum at 3 3.
(c) Domain , range 4].
(c) Domain , range .
53. (a) xintercepts are 2 and 1, yintercept is 1. (b) Local maximum at 1 0, local minimum at 1 2.
54. (a) xintercepts are 0 and 4, yintercept is 0. (b) No local maximum, local minimum at 3 3. (c) Domain , range [3 .
(c) Domain , range . 55. y 10x x 2 , [2 12] by [15 30]
No local minimum. Local maximum at 5 25. Domain: , range: 25].
56. y x 3 3x 2 , [2 5] by [10 10]
Local minimum at 2 4. Local maximum at 0 0. Domain: , range: .
30
10
20 10 2 2 10
2
4
6
2
4
8 10 12 10
57. y x 3 12x 9, [5 5] by [30 30]
Local maximum at 2 25. Local minimum at 2 7. Domain: , range: .
58. y 2x 3 3x 2 12x 32, [5 5] by [60 30] Local minimum at 2 52. Local maximum at 1 25.
Domain: , range: .
20
20 4
2
2
4
4
20
2 20
2
4
40 60
59. y x 5 9x 3 , [4 4] by [50 50]
Local minimum at approximately 232 4517, local maximum at approximately 232 4517.
60. y x 4 18x 2 32, [5 5] by [100 100]
Local minima at 3 49 and 3 49. Local maximum at 0 32.
Domain: , range: [49 .
Domain: , range: .
100
40 20 4
2
20 40
2
4
4
2 100
2
4
21
22
CHAPTER 3 Polynomial and Rational Functions
61. y 3x 5 5x 3 3, [3 3] by [5 10]
62. y x 5 5x 2 6, [3 3] by [5 10]
Domain: , range: .
Domain: , range: .
Local maximum at 1 5. Local minimum at 1 1.
Local minimum at 126 124. Local maximum at 0 6.
10
10
5 3
2
1
1
2
3
3
2
1
1
2
3
5
63. y 2x 2 3x 5 has one local maximum at 075 613.
64. y x 3 12x has no local maximum or minimum. 20
10
10
5 4
2
4 2
2 10
2
4
20
4
5
65. y x 3 x 2 x has one local maximum at 033 019 66. y 6x 3 3x 1 has no local maximum or minimum. and one local minimum at 100 100.
20 10
2 3 3
2
1
1
2
3
2
1 10
67. y x 4 5x 2 4 has one local maximum at 0 4 and two local minima at 158 225 and 158 225. 10
1 5
3
68. y 12x 5 375x 4 7x 3 15x 2 18x has two local
maxima at 050 465 and 297 1210 and two local minima at 140 2744 and 140 254. 20
5 2
2
20
2
3
1
1
2
3
4
2
2 20
SECTION 3.2 Polynomial Functions and Their Graphs
69. y x 25 32 has no maximum or minimum.
23
3 70. y x 2 2 has one local minimum at 0 8.
60
10
40 20
3
2
2
4
2
6
1
1
2
3
10
71. y x 8 3x 4 x has one local maximum at 044 033 and two local minima at 109 115 and 112 336.
72. y 13 x 7 17x 2 7 has one local maximum at 0 7 and one local minimum at 171 2846. 20
4 2 3
2
1 2
3 1
2
2
3
1
1
2
3
20
4
73. y cx 3 ; c 1, 2, 5, 12 . Increasing the value of c stretches 74. P x x c4 ; c 1, 0, 1, 2. Increasing the value of c shifts the graph to the right.
the graph vertically.
100
c=5
c=0
c=1
c=_1 4
2
2
c=2
1
4
0
2
1 c=_ 2
2
1
100
c=2
c=1
2
75. P x x 4 c; c 1, 0, 1, and 2. Increasing the value of c moves the graph up.
76. P x x 3 cx; c 2, 0, 2, 4. Increasing the value of c makes the “bumps” in the graph flatter.
c=2 c=1 c=0 c=_1
6 4
10
c=0
c=_2
c=2
c=_4
2 2
1
0
0
1
2 10
2
24
CHAPTER 3 Polynomial and Rational Functions
77. P x x 4 cx; c 0, 1, 8, and 27. Increasing the value 78. P x x c ; c 1, 3, 5, 7. The larger c gets, the flatter the of c causes a deeper dip in the graph, in the fourth
graph is near the origin, and the steeper it is away from the
quadrant, and moves the positive xintercept to the right.
origin.
c=1 c=0
4
10
20
2
0
c=7
c=5
c=3
c=27
c=1 2 c=8
4
2
1
0
1
2
20
10 40
79. (a)
y
y=x#-2x@-x+2 y=_x@+5x+2
5 1
x
(b) The two graphs appear to intersect at 3 points. (c) x 3 2x 2 x 2 x 2 5x 2 x 3 x 2 6x 0 x x 2 x 6 0 x x 3 x 2 0. Then either x 0, x 3, or x 2. If x 0, then y 2; if x 3 then
y 8 if x 2, then y 12. Hence the points where the
two graphs intersect are 0 2, 3 8, and 2 12.
80. Graph 1 belongs to y x 4 . Graph 2 belongs to y x 2 . Graph 3 belongs to y x 6 . Graph 4 belongs to y x 3 . Graph 5 belongs to y x 5 .
81. (a) Let P x be a polynomial containing only odd powers of x. Then each term of P x can be written as C x 2n1 , for some constant C and integer n. Since C x2n1 C x 2n1 , each term of P x is an odd function. Thus by part (a), P x is an odd function. (b) Let P x be a polynomial containing only even powers of x. Then each term of P x can be written as C x 2n , for some constant C and integer n. Since C x2n C x 2n , each term of P x is an even function. Thus by part (b), P x is an even function. (c) Since P x contains both even and odd powers of x, we can write it in the form P x R x Q x, where R x contains all the evenpowered terms in P x and Q x contains all the oddpowered terms. By part (d), Q x is an odd function, and by part (e), R x is an even function. Thus, since neither Q x nor R x are constantly 0 (by assumption), by part (c), P x R x Q x is neither even nor odd. (d) P x x 5 6x 3 x 2 2x 5 x 5 6x 3 2x x 2 5 PO x PE x where PO x x 5 6x 3 2x and PE x x 2 5. Since PO x contains only odd powers of x, it is an odd function, and since PE x contains only even powers of x, it is an even function.
SECTION 3.2 Polynomial Functions and Their Graphs
82. (a) From the graph, P x x 3 4x x x 2 x 2 has three xintercepts, one local maximum, and one local minimum.
25
(b) From the graph, Q x x 3 4x x x 2 4 has one xintercept and no local maximum or minimum.
10
5
10
5
5
5
10
10
(c) For the xintercepts of P x x 3 ax, we solve x 3 ax 0. Then we have x x 2 a 0 x 0 or x 2 a. If x 2 a, then x a. So P has 3 xintercepts. Since P x x x 2 a x x a x a , by part (c) of problem 67, P has 2 local extrema. For the xintercepts of Q x x 3 ax, we solve x 3 ax 0. Then we have x x 2 a 0 x 0 or x 2 a. The equation x 2 a has no real solutions because a 0. So Q has 1
xintercept. We now show that Q is always increasing and hence has no extrema. Ifx1 x2 , then ax1 ax2 (because
a 0) and x13 x23 . So we have x13 ax1 x23 ax2 , and hence Q x1 Q x2 . Thus Q is increasing, that is, its graph always rises, and so it has no local extrema. 83. (a) P x x 1 x 3 x 4. Local maximum at 18 21.
Local minimum at 36 06.
(b) Since Q x P x 5, each point on the graph of Q has ycoordinate 5 units more than the corresponding point on the graph of P. Thus Q has a local maximum at 18 71 and a local minimum at 35 44.
10 10 5 5
10
84. (a) P x x 2 x 4 x 5 has one local maximum and one local minimum.
(b) Since P a P b 0, and P x 0 for a x b (see the table
below), the graph of P must first rise and then fall on the interval a b,
and so P must have at least one local maximum between a and b. Using similar reasoning, the fact that P b P c 0 and P x 0 for
10
b x c shows that P must have at least one local minimum between b and c. Thus P has at least two local extrema.
5 10
Interval Sign of x a Sign of x b Sign of x c
Sign of x a x b x c
a a b b c c
85. Since the polynomial shown has five distinct zeros, it has at least five factors, and so its degree is greater than or equal to 5. (Alternately, we can use the theorem on Local Extrema of Polynomials: Because the function has four local extrema, its degree is at least 5.)
26
CHAPTER 3 Polynomial and Rational Functions c=2 c=1
y 2
86. If c 0, the polynomial P x cx 4 2x 2 has only a maximum at x 0. For c 0, P x has a
c=3
c=21
1
maximum at x 0 and minima at q, where q
decreases as c increases. Because the graph of P is
_1
_2
fundamentally different for positive and negative values of c, c 0 is called a transitional value.
0
1
_1
x
2
_2 _3 _4
87. P x 8x 03x 2 00013x 3 372 4000
c=_2 c=_1
c=0
c=41
(a) For the firm to break even, P x 0. From the graph, we see that P x 0 when x 252. But unless the firm also manufactures
metablenders, it cannot produce fractions of a blender, so it must produce at least 26 blenders a year.
2000
(b) No, the profit does not increase indefinitely. The largest profit is 0
0
approximately $327622, which occurs when the firm produces
200
166 blenders per year.
88. P t 120t 04t 4 1000
(a) A maximum population of approximately 1380 is attained after 422 months. (b) The rabbit population disappears after approximately 842 months.
1000 500 0
0
5
10
89. (a) The length of the bottom is 40 2x, the width of the bottom is 20 2x, and the height is x, so the volume of the box is V x 20 2x 40 2x 4x 3 120x 2 800x.
(c) Using the domain from part (b), we graph V in the viewing rectangle [0 10] by [0 1600]. The maximum volume is V 15396 when x 423.
(b) Since the height and width must be positive, we must have x 0 and 20 2x 0, and so the domain of V is 0 x 10.
1000
0
0
5
10
SECTION 3.2 Polynomial Functions and Their Graphs
90. (a) Let h be the height of the box. Then the total length of all
(c) Using the domain from part (b), we graph V in
12 edges is 8x 4h 144 in. Thus, 8x 4h 144
the viewing rectangle [0 18] by [0 2000]. The
2x h 36 h 36 2x. The volume of the box is equal to area of baseheight x 2 36 2x 2x 3 36x 2 .
27
maximum volume is V 1728 in3 when x 12 in. 2000
Therefore, the volume of the box is
1000
V x 2x 3 36x 2 2x 2 18 x. (b) Since the length of the base is x, we must have x 0. Likewise, the height must be positive so 36 2x 0
0
0
10
x 18. Putting these together, we get that the domain of V
is 0 x 18. 91.
The graph of y x 100 is close to the xaxis for x 1, but passes through the
y (_1, 1)
points 1 1 and 1 1. The graph of y x 101 behaves similarly except that the
(1, 1)
1
yvalues are negative for negative values of x, and it passes through 1 1
y=x@
y=x$
instead of 1 1. x
1
y=x%
y=x#
(1, _1)
92. No, it not possible for a thirddegree polynomial to have exactly one local extremum. The end behavior of such a polynomial is the same as that of y kx 3 , and for this function, the values of y go off in opposite directions as x and x . But for a function with just one extremum, the values of y go in the same direction (either both positive or both negative) on both sides of the extremum. Neither is it possible for any polynomial to have two local maxima and no local minimum. All polynomials are continuous and defined everywhere, so any two local maxima must have a local minimum between them. An example of a polynomial with six local extrema is P x x 1 x 2 x 3 x 4 x 5 x 6 x 7. 93.
y
1
Recall that if f has a fixed point at x, then the graph of f intersects the (1, 1)
f£
functions f i with domain [0 1] and range contained in [0 1]. It quickly
fÁ
becomes clear that all such functions have a fixed point.
y=x fª 0 (0, 0)
line y x. We draw the line y x and sketch a few graphs of continuous
1
x
28
CHAPTER 3 Polynomial and Rational Functions
3.3
DIVIDING POLYNOMIALS
1. If we divide the polynomial P by the factor x c, and we obtain the equation P x x c Q x R x, then we say that x c is the divisor, Q x is the quotient, and R x is the remainder.
2. (a) If we divide the polynomial P x by the factor x c, and we obtain a remainder of 0, then we know that c is a zero of P. (b) If we divide the polynomial P x by the factor x c, and we obtain a remainder of k, then we know that P c k.
3.
3
3 3
4.
4
6
9
9
3
13
P x D x
x 3
5. 3x 2
3x 3
4x 2
12x 3 12x 3
8x
16x 2 8x 2
24x 2
x 1 x
24x 2 16x
15x 1 15x 10 11
Thus, the quotient is 4x 2 8x 5 and the remainder is 11, and
12x 3 16x 2 x 1 P x D x 3x 2 11 2 4x 8x 5 3x 2
10
0
7
0
7
2x 2 10x 7 7 P x 2x . D x x 5 x 5
2x 2
6.
5
10
Thus, the quotient is 2x and the remainder is 7, and
13 . x 3
2 2
Thus, the quotient is 3x 3 and the remainder is 13, and 3x 2 6x 4
5
5x 1
45
10x 3 2x 2 4x 1 10x 3 2x 2
4x 1 4x 45 9 5
Thus, the quotient is 2x 2 45 and the remainder is 95 , and 9
10x 3 2x 2 4x 1 2 4 P x 5 2x 5 . D x 5x 1 5x 1
SECTION 3.3 Dividing Polynomials
2x 2 x
7. x2 4
1
8.
2x 4 x 3 9x 2 2x 4
x 2 3x 1
8x 2
x 3 x 2 x 3
2x 3 6x 2 17x
2x 5 0x 4
x2
x 3 2x 2
6x 4 18x 3 6x 2
17x 3 8x 2
43x 2 14x 5
43x 2 129x 43
Thus, the quotient is 2x 2 x 1 and the remainder is P x D x
2x 4 x 3 9x 2 x2 4
3x
17x 3 51x 2 17x
4
4x 4
4x 4, and
3x 5
2x 5 6x 4 2x 3
4x
x 2 4x
43
x 3 2x 2
6x 4
4x 4 2x 2 x 1 2 . x 4
115x 48
Thus, the quotient is 2x 3 6x 2 17x 43 and the remainder is 115x 48 and
2x 5 x 3 2x 2 3x 5 P x D x x 2 3x 1
115x 48 2x 3 6x 2 17x 43 2 x 3x 1
9. Long division:
x 5
10. Long division: 3x 2
10x
3x 3
15x 2
3x 3
50
5x 2
0x
5
x 2
10x 2 0x
5x 3 10x 2 10x 23 5x 4
5x 4 10x 3
10x 2 3x 2
10x 3 10x 2
10x 2 50x
50x 50x
10x 3 20x 2
10x 2 3x
5
10x 2 20x
250
23x 2
245
23x 46
Thus, the quotient is 3x 2 10x 50 and the remainder is
245, so
P x 3x 3 5x 2 5 x 5 3x 2 10x 50 245.
Synthetic division:
5
3 3
5
0
5
15
50
250
10
50
245
Thus, the quotient is 3x 2 10x 50 and the remainder is
245, as above.
29
48
P x 5x 4 10x 2 3x 2 x 2 5x 3 10x 2 10x 23 48
Synthetic division: 2
5 5
0
10
3
2
10
20
20
46
10
10
23
48
Thus, the quotient is 5x 3 10x 2 10x 23 and the
remainder is 48, as above.
30
CHAPTER 3 Polynomial and Rational Functions
x2
11.
1
12.
2x 3 3x 2 2x
2x 3
2x 1
2x 3 3x 2
2x 2
4x 3
x 4
4x 3 2x 2
7x 9
2x 2 7x
2x
2x 2 x
2x 3
8x 9
3
8x 4
Thus, the quotient is x 2 1 and the remainder is 3, and P x 2x 3 3x 2 2x 2x 3 x 2 1 3.
13. 2x 2 1
4x 2 2x 1
5
Thus
P x 4x 3 7x 9 2x 1 2x 2 x 4 5.
14.
8x 4 4x 3 6x 2 8x 4
3x 2 3x 1
4x 2
4x 3 2x 2 4x 3
2x 2 2x 2
9x 3 6x 2
3x 2
27x 5 9x 4 0x 3 3x 2 0x 3 27x 5 27x 4 9x 3
18x 4 9x 3 3x 2 18x 4 18x 3 6x 2
2x
9x 3 3x 2
2x
9x 3 9x 2 3x
1
6x 2 3x 3
2x 1
6x 2 6x 2
Thus, the quotient is 4x 2 2x 1 and the remainder is 2x 1, and
P x 27x 5 9x 4 3x 2 3 3x 2 3x 1 9x 3 6x 2 3x 2 3x 5
2x 2 1 4x 2 2x 1 2x 1
15. x 2
3x 5
Thus
P x 8x 4 4x 3 6x 2
x 1
16.
x 2 3x 7
x 2 2x
x 7 x 2
5
Thus, the quotient is x 1 and the remainder is 5.
x 3
x2
x 2
x 3 2x 2 x 3 3x 2
x 1
x 2 x x 2 3x
2x 1 2x 6 5
Thus, the quotient is x 2 x 2 and the remainder is 5.
SECTION 3.3 Dividing Polynomials
17. 3x 1
3x 2
2x 2 2x 1
18.
x
9x 3 6x 2 x 1
8x 3 2x 2 2x 3
4x 3
9x 3 3x 2
31
8x 3 6x 2
3x 2 x
8x 2 2x
3x 2 x
8x 2 6x
1
4x 3 4x 3
Thus, the quotient is 3x 2 x and the remainder is 1.
0
Thus, the quotient is 2x 2 2x 1 and the remainder is 0. 19.
4x x2 x 1
2
20.
4x 3 2x 2
4x 3 4x 2 4x
x 2 3x 2
3
2x 2 4x
x2 x 5
x 4 4x 3 0x 2
x 3
x 4 3x 3 2x 2
x 3 2x 2
2x 2 2x 2
x
x 3 3x 2
2x
5x 2
6x 5
5x 2
Thus, the quotient is 4x 2 and the remainder is 6x 5.
x 3
15x 10
14x 13
Thus, the quotient is x 2 x 5 and the remainder is 14x 13.
21. 2x 2 0x 5
3x
1
22.
6x 3 2x 2 22x 0 6x 3
15x
2x 2
2x 2
3 3x 2 7x
7x 0
9x 2
x 5
9x 2 21x
20x 5
5
Thus, the quotient is 3 and the remainder is 20x 5.
7x 5
Thus, the quotient is 3x 1 and the remainder is 7x 5. x4
23. x2 1
24.
1
x 6 0x 5 x 4 0x 3 x 2 0x 1
x6
x4
0
x2 x2
1
4x 2 6x 8
1 x3 x2 5 x 7 2 2 4 2x 5 7x 4 0x 3 0x 2 0x 13
2x 5 3x 4 4x 3
4x 4 4x 3 0x 2 4x 4 6x 3 8x 2
1
10x 3 8x 2 0x
0
10x 3 15x 2 20x
Thus, the quotient is x 4 1 and the remainder is 0.
7x 2 20x 13
7x 2 21 2 x 14
19 x 1 2
Thus, the quotient is 12 x 3 x 2 52 x 74 and the
remainder is 19 2 x 1.
32
CHAPTER 3 Polynomial and Rational Functions
25. The synthetic division table for this problem takes the following form.
26. The synthetic division table for this problem takes the following form.
3
2
3
5
2
6
3
1
6
1
Thus, the quotient is 2x 1 and the remainder is 6. 27. The synthetic division table for this problem takes the
1
1
4
1
2
2
1
6
Thus, the quotient is x 2 and the remainder is 6. 28. The synthetic division table for this problem takes the following form.
following form. 3
1
3
1
0
3
2
2
0 8
2
2
4 4
8
3
16 13
Thus, the quotient is 3x 2 and the remainder is 2.
Thus, the quotient is 4x 8 and the remainder is 13.
29. The synthetic division table for this problem takes the
30. Since x 5 x 5, the synthetic division table for this problem takes the following form.
following form. 2
3 3
2
1
6
8
4
9
18 13
5
Thus, the quotient is 3x 2 4x 9 and the remainder is 13. 31. Since x 4 x 4 and
x 3 10x 13 x 3 0x 2 10x 13, the synthetic
division table for this problem takes the following form. 1
4
0
16
1
11
Thus, the quotient is x 2 4x 6 and the remainder is 11. 33. Since x 5 3x 3 6 x 5 0x 4 3x 3 0x 2 0x 6, the synthetic division table for this problem takes the following form. 1
1
0
3
0
0
1
1
4
4
1
4
4
4
6
4
2
Thus, the quotient is x 4 x 3 4x 2 4x 4 and the remainder is 2.
10
25
25
25
5
5
35
following form. 3
1
0
1
1
3
6
2
6
10
0
18
24
8
24
Thus, the quotient is x 3 2x 2 6x 8 and the remainder is 24.
34. The synthetic division table for this problem takes the following form. 3
1
30
32. Since x 4 x 3 10x x 4 x 3 0x 2 10x 0, the synthetic division table for this problem takes the
24
6
4
20
Thus, the quotient is 5x 2 5x 5 and the remainder is 35.
13
10
4
5
5
5
1
9
3
1
6
27 18 9
27
27 0
Thus, the quotient is x 2 6x 9 and the remainder is 0.
SECTION 3.3 Dividing Polynomials
35. The synthetic division table for this problem takes the following form.
36. The synthetic division table for this problem takes the following form.
1 2
2
3 1
2
2
1
2
0
0
1
4
23
37. Since x 3 27 x 3 0x 2 0x 27, the synthetic
division table for this problem takes the following form. 3
1 1
0
0
3
9
3
9
0
4
3
3
0
5
6
12
6
3
12
5
5
4
8
4
16 16
0
8
is 0.
0
2
0 0
40
2 2
9
1
1
5
10
6
Therefore, by the Remainder Theorem, P 12 6. 1
1 1 2
1
5
2
3
3
2
Therefore, by the Remainder Theorem, P 1 2. 44. P x 2x 3 21x 2 9x 200, c 11
2 0
1 2
1
11
7
2
0 2
7
45. P x 5x 4 30x 3 40x 2 36x 14, c 7 30
2 2
1
Therefore, by the Remainder Theorem, P 2 7.
5
0
Thus, the quotient is x 3 2x 2 4x 8 and the remainder
6
10
43. P x x 3 2x 2 7, c 2
1
0
42. P x x 3 x 2 x 5, c 1
Therefore, by the Remainder Theorem, P 2 12.
1
1
0
17
6
7
2
2
1 1
41. P x x 3 3x 2 7x 6, c 2
1
4
following form.
0
1
2
1
1 29 7 9
38. The synthetic division table for this problem takes the
27
Therefore, by the Remainder Theorem, P 2 17.
2
6
1 23 1 3
40. P x 2x 2 9x 1, c 12
2
1
4
5
remainder is 79 .
2
39. P x x 4 x 2 5, c 2 1
10
Thus, the quotient is 6x 3 6x 2 x 13 and the
27
Thus, the quotient is x 2 3x 9 and the remainder is 0.
2
6 6
Thus, the quotient is 2x 2 4x and the remainder is 1.
7
33
36
14 497
35
35
35
5
5
71
483
Therefore, by the Remainder Theorem, P 7 483.
21
9
22
11
1
20
200
220 20
Therefore, by the Remainder Theorem, P 11 20. 46. P x 6x 5 10x 3 x 1, c 2 2
6 6
0
10
0
1
1
12
24
68
136
274
12
34
68
137
273
Therefore, by the Remainder Theorem, P 2 273.
34
CHAPTER 3 Polynomial and Rational Functions
47. P x x 7 3x 2 1 c3
3
48. P x 2x 6 7x 5 40x 4 7x 2 10x 112, c 3
x 7 0x 6 0x 5 0x 4 0x 3 3x 2 0x 1 1 1
0
0
0
0
0
3
9
27
81
243
720
2160
3
9
27
81
240
720
2159
3
3
2
40
0
6
39
3
13
2
1
7
1
3
7
9
2
10
112
6
12
4
100
Therefore, by the Remainder Theorem, P 3 100.
Therefore by the Remainder Theorem, P 3 2159.
49. P x 3x 3 4x 2 2x 1, c 23 2 3
3
4 2
3
6
2
4 2
50. P x x 3 x 1, c 14 1 4
1 4 3 7 3
01
1 1
2
3
021
21
0279
279
8279
52. (a) P x 6x 7 40x 6 16x 5 200x 4 60x 3 69x 2 13x 139, c 7 6 6
40
16
42
14
2
30
1
1
1 16 15 16
1 15 64
49 64
8
01 Therefore, by the Remainder Theorem, P 01 8279.
7
0 1 4 1 4
Therefore, by the Remainder Theorem, P 14 49 64 .
Therefore, by the Remainder Theorem, P 23 73 . 51. P x x 3 2x 2 3x 8, c 01
1
200
60
69
13
70
7
10
10
1
20
210
70
139
140 1
Therefore, by the Remainder Theorem, P 7 1.
(b) P 7 6 77 40 76 16 75 200 74 60 73 69 72 13 7 139 6 823,543 40 117,649 16 16,807 200 2401 60 343 69 49 13 7 139 1
which agrees with the value obtained by synthetic division, but requires more work.
53. P x x 3 3x 2 3x 1, c 1 1
1
3
1
1
2
3 2 1
54. P x x 3 2x 2 3x 10, c 2 1
2
1
1
0
Since the remainder is 0, x 1 is a factor.
2 2
1
4
3
10
5
0
8
10
Since the remainder is 0, x 2 is a factor.
SECTION 3.3 Dividing Polynomials
55. P x 2x 3 7x 2 6x 5, c 12 1 2
2 2
7
6
1
4
8
10
35
56. P x x 4 3x 3 16x 2 27x 63, c 3, 3 3
5
1
3
16
3
5
1
0
18
6
63
27
6
2
63
0
21
Since the remainder is 0, x 3 is a factor. We next show that x 3 is also a factor by using synthetic division on
Since the remainder is 0, x 12 is a factor.
the quotient of the above synthetic division, x 3 6x 2 2x 21. 3
1 1
6
2
3
9
3
21 21
0
7
Since the remainder is 0, x 3 is a factor.
57. P x x 3 3x 2 18x 40, c 4 4
1 1
3
18 28
4
10
7
58. P x x 3 5x 2 2x 10, c 5
1
1 3
1
2
11 6
5
15 15
0
Since the remainder is 0, we know that 3 is a zero and P x x 3 x 2 2x 5 . Now x 2 2x 5 0 2 22 4 1 5 1 6. Hence, when x 2 1 the zeros are 3 and 1 6.
5
1
10
2
5
0
Hence, the zeros are 4, 2, and 5.
3
1
40
Since the remainder is 0, we know that 4 is a zero and P x x 4 x 2 7x 10 x 4 x 2 x 5.
59. x 3 x 2 11x 15, c 3
5
40
0
0
10
0
2
Since the remainder is 0, we know that 5 is a zero and P x x 5 x 2 2 x 5 x 2 x 2 . Hence, the zeros are 5 and 2.
60. P x 3x 4 x 3 21x 2 11x 6, c 2, 13 2
3 3
1
21
6
14
7
7
6
11 14
6
3
0
Since the remainder is 0, we know that 2 is a zero and P x x 2 3x 3 7x 2 7x 3 . 1 3
3
7
1
3
6
7
3
2
3
9
0
Since the remainder is 0, we know that 13 is a zero and P x x 2 x 13 3x 2 6x 9 3 x 2 x 13 x 1 x 3.
Hence, the zeros are 2, 13 , 1, and 3.
36
CHAPTER 3 Polynomial and Rational Functions
61. P x 3x 4 8x 3 14x 2 31x 6, c 2, 3 3
2
8 6
3
31
6
28
28
14
3
6
14
14
3
14
3
2
1
0
Since the remainder is 0, we know that 2 is a zero and P x x 2 3x 3 14x 2 14x 3 . 3
62. P x 2x 4 13x 3 7x 2 37x 15, c 1, 3
14
3
9
15
3
5
1
3x 2 5x 1 0 when 5 52 4 3 1 5 37 x , and 2 3 6 5 37 . hence, the zeros are 2, 3, and 6
2
2
7
37
15
15
22
15
22
15
15
0
Since the remainder is 0, we know that 1 is a zero and P x x 1 2x 3 15x 2 22x 15 . 3
0
Since the remainder is 0, we know that 3 is a zero and P x x 2 x 3 3x 2 5x 1 . Now
13
2
15
22
15
2
6
27
15
9
5
0
The remainder is 0, so 3 is a zero and P x x 1 x 3 2x 2 9x 5
x 1 x 3 2x 1 x 5.
Hence, the zeros are 1, 3, 12 , and 5.
63. Since the zeros are x 1, x 1, and x 3, the factors are x 1, x 1, and x 3. Thus P x x 1 x 1 x 3 x 3 3x 2 x 3.
64. Since the zeros are x 2, x 0, x 2, and x 4, the factors are x 2, x, x 2, and x 4.
Thus P x c x 2 x x 2 x 4. If we let c 1, then P x x 4 4x 3 4x 2 16x.
65. Since the zeros are x 1, x 1, x 3, and x 5, the factors are x 1, x 1, x 3, and x 5. Thus P x x 1 x 1 x 3 x 5 x 4 8x 3 14x 2 8x 15.
66. Since the zeros are x 2, x 1, x 0, x 1, and x 2, the factors are x 2, x 1, x, x 1, and x 2. Thus P x c x 2 x 1 x x 1 x 2. If we let c 1, then P x x 5 5x 3 4x.
67. Since the zeros of the polynomial are 2, 0, 1, and 3, it follows that P x C x 2 x x 1 x 3 C x 4 2C x 3
5C x 2 6C x. Since the coefficient of x 3 is to be 4, 2C 4, so C 2. Therefore, P x 2x 4 4x 3 10x 2 12x is the polynomial. 68. Since the zeros of the polynomial are 1, 0, 2, and 12 , it follows that P x C x 1 x x 2 x 12 C x 4
3 C x 3 3 C x 2 C x. Since the coefficient of x 3 is to be 3, 3 C 3, so C 2. Therefore, P x 2x 4 3x 3 3x 2 2x 2 2 2
is the polynomial.
2 and integer coefficients, the fourth zero must be 2, otherwise the constant term would be irrational. Thus, P x C x 1 x 1 x 2 x 2 C x 4 3C x 2 2C. Requiring
69. Since the polynomial degree 4 and zeros 1, 1, and
that the constant term be 6 gives C 3, so P x 3x 4 9x 2 6. 70. Since the polynomial degree 5 and zeros 2, 1, 2, and 5 and integer coefficients, the fourth zero must be 5, otherwise the constant term would be irrational. Thus, P x C x 2 x 1 x 2 x 5 x 5 C x 5 C x 4 9C x 3 9C x 2 20C x 20C. Requiring that the constant term be 40 gives C 2, so P x 2x 5 2x 4 18x 3 18x 2 40x 40.
SECTION 3.4 Real Zeros of Polynomials
37
71. The yintercept is 2 and the zeros of the polynomial are 1, 1, and 2. It follows that P x C x 1 x 1 x 2 C x 3 2x 2 x 2 . Since P 0 2 we have 2 C 03 2 02 0 2 2 2C C 1 and P x x 1 x 1 x 2 x 3 2x 2 x 2.
72. The yintercept is 4 and the zeros of the polynomial are 1 and 2 with 2 being degree two. It follows that P x C x 1 x 22 C x 3 3x 2 4 . Since P 0 4 we have 4 C 03 3 02 4 4 4C C 1 and P x x 3 3x 2 4.
73. The yintercept is 4 and the zeros of the polynomial are 2 and 1 both being degree two. It follows that P x C x 22 x 12 C x 4 2x 3 3x 2 4x 4 . Since P 0 4 we have 4 C 04 2 03 3 02 4 0 4 4 4C C 1. Thus P x x 22 x 12 x 4 2x 3 3x 2 4x 4.
74. The yintercept is 2 and the zeros of the polynomial are 2, 1, and 1 with 1 being degree two. It follows that P x C x 2 x 1 x 12 C x 4 x 3 3x 2 x 2 . Since P 0 2 we have 4 C 04 03 3 02 0 2 2 2C so C 1 and P x x 4 x 3 3x 2 x 2.
75. (a) By the Remainder Theorem, the remainder when P x 6x 1000 17x 562 12x 26 is divided by x 1 is P 1 6 11000 17 1562 12 1 26 6 17 12 26 3.
(b) If x 1 is a factor of Q x x 567 3x 400 x 9 1, then Q 1 must equal 0. By the Remainder Theorem,
Q 1 1567 3 1400 19 1 1 3 1 1 0, so x 1 is a factor. 76. R x x 5 2x 4 3x 3 2x 2 3x 4 x 4 2x 3 3x 2 2x 3 x 4 x 3 2x 2 3x 2 x 3 x 4 x 2 2x 3 x 2 x 3 x 4 [x 2 x 3] x 2 x 3 x 4
So to calculate R 3, we start with 3, then subtract 2, multiply by 3, add 3, multiply by 3, subtract 2, multiply by 3, add 3, multiply by 3, and add 4, to get 157.
3.4
REAL ZEROS OF POLYNOMIALS
1. If the polynomial function P x an x n an1 x n1 a1 x a0 has integer coefficients, then the only numbers that p could possibly be rational zeros of P are all of the form , where p is a factor of the constant coefficient a0 and q is a factor q of the leading coefficient an . The possible rational zeros of P x 6x 3 5x 2 19x 10 are 1, 12 , 13 , 16 , 2, 23 , 5, 52 , 53 , 56 , 10, and 10 3.
2. Using Descartes’ Rule of Signs, we can tell that the polynomial P x x 5 3x 4 2x 3 x 2 8x 8 has 1, 3 or 5 positive real zeros and no negative real zero. 3. This is true. If c is a real zero of the polynomial P, then P x x c Q x, and any other zero of P x is also a zero of Q x P x x c. 4. False. Consider the polynomial P x x 1 x 2 x 4 x 3 x 2 10x 8. An upper bound is 3, but 3 is not a lower bound.
5. P x x 3 4x 6. The constant term is 6, with factors 1, 2, 3, 6; and since the leading term is 1, these are the possible rational zeros of P. 6. Q x x 5 3x 2 5x 10 has possible rational zeros 1, 2, 5, 10.
38
CHAPTER 3 Polynomial and Rational Functions
7. R x 3x 4 2x 3 8x 2 9. The constant term is 9, with factors 1, 3, 9; and the leading term is 3, with factors 1, 3. Thus, the possible rational zeros of R are 11 , 31 , 91 , 13 , 33 , 93 . Eliminating duplicates, they are 1, 3, 9, 13 .
8. S x 5x 6 3x 4 20x 2 15. The constant term is 15, with factors 1, 3, 5, 15; and the leading term is 5, with 1 3 5 15 factors 1, 5. Thus, the possible rational zeros of S are 11 , 31 , 51 , 15 1 , 5 , 5 , 5 , 5 . Eliminating duplicates,
they are 1, 3, 5, 15, 15 , 35 .
9. T x 6x 5 8x 3 5. The constant term is 5, with factors 1, 5; and the leading term is 6, with factors 1, 2, 3, 6. Thus, the possible rational zeros of T are 11 , 51 , 12 , 52 , 13 , 53 , 16 , 56 ; that is, 1, 5, 12 , 52 , 13 , 53 , 16 , 56 .
10. U x 12x 5 6x 3 2x 8. The constant term is 8, with factors 1, 2, 4, 8; and the leading term is 12, with
factors 1, 2, 3, 4, 6, 12. Thus, the possible rational zeros of U are 11 , 21 , 41 , 81 , 12 , 22 , 42 , 82 , 13 ,
1 , 2 , 4 , 8 . Eliminating duplicates, they are 1, 2, 4, 23 , 43 , 83 , 14 , 24 , 44 , 84 , 16 , 26 , 46 , 86 , 12 12 12 12 1. 8, 12 , 13 , 14 , 23 , 43 , 83 , 16 , 12
11. (a) P x 5x 3 x 2 5x 1 has possible rational zeros 1, 15 . (b) From the graph, the actual zeros are 1, 15 , and 1.
12. (a) P x 3x 3 4x 2 x 2 has possible rational zeros 1, 2, 13 , 23 . (b) From the graph, the actual zeros are 1 and 23 .
13. (a) P x 2x 4 9x 3 9x 2 x 3 has possible rational zeros 1, 3, 12 , 32 . (b) From the graph, the actual zeros are 12 , 1, and 3.
14. (a) P x 4x 4 x 3 4x 1 has possible rational zeros 1, 12 , 14 . (b) From the graph, the actual zeros are 14 and 1.
15. P x x 3 5x 2 8x 12. The possible rational zeros are 1, 2, 3, 4, 6, 12. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 3 5x 2 8x 12 has 1 variation in sign and hence P has 1 negative real zero. 1
1
5
1
8
12
4
12
1 4 12 0 x 1 is a zero. P x x 3 5x 2 8x 12 x 1 x 2 4x 12 x 1 x 6 x 2. Therefore, the zeros are 2, 1, and 6.
16. P x x 3 4x 2 19x 14. The possible rational zeros are 1, 7, 14. P x has 1 variation in sign and hence 1 positive real zero. P x x 3 4x 2 19x 14 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
4 1
1
19
14
3
22
22
36
1
1
4 1
19 5
14 14
x 1 is not a zero. 1 5 14 0 x 1 is a zero. P x x 3 4x 2 19x 14 x 1 x 2 5x 14 x 2 x 1 x 7. Thus, the zeros are 2, 1, and 7. 3
SECTION 3.4 Real Zeros of Polynomials
39
17. P x x 3 5x 2 3x 9. The possible rational zeros are 1, 3, 9. P x has 2 variation in sign and hence 0 or 2 positive real zero.P x x 3 5x 2 3x 9 has 1 variation in sign and hence P has 1 negative real zero. 1
1
3
9
4
1
5
1
1
1
1
5 1
9
9
x 1 is not a zero. 1 6 9 0 x 1 is a zero. P x x 3 5x 2 3x 9 x 1 x 2 6x 9 x 1 x 32 . Therefore, the zeros are 1 and 3. 4
1
8
3
6
18. P x x 3 3x 2. The possible rational zeros are 1, 2. P x has 1 variation in sign and hence 1 positive real zero.P x x 3 3x 2 has 2 variations in sign and P has hence 0 or 2 negative real zeros. 1
1
0
3
1 1
1
1
1
2
2
0
3
2
2
2
4
2
x 1 is not a zero. 1 2 1 0 x 2 is a zero. P x x 3 3x 2 x 2 x 2 2x 1 x 2 x 12 . Therefore, the zeros are 2 and 1. 2
4
19. P x x 3 6x 2 12x 8. The possible rational zeros are 1, 2, 4, 8. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x x 3 6x 2 12x 8 has no variations in sign and hence P has no negative real zero. 1
1
6
1
1
12
5 7
1
2
8
12
6
7
2
8 8
8
x 1 is not a zero. 1 4 4 0 x 2 is a zero. P x x 3 6x 2 12x 8 x 2 x 2 4x 4 x 23 . Therefore, the only zero is x 2. 5
1
20. P x x 3 12x 2 48x 64. The possible rational zeros are 1, 2, 4, 8, 16, 32, 64. P x has 0 variations in sign and hence no positive real zero. P x x 3 12x 2 48x 64 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1
1
1
12
48
64
1
11
37
11
37
27
2 x 1 is not a zero. 4
1 1
1
12
48
64
4
32
64
8
16
1
0
12
48
64
2
20
56
10
28
6
x 2 is not a zero.
x 4 is a zero.
P x x 3 12x 2 48x 64 x 4 x 2 8x 16 x 43 . Therefore, the only zero is x 4. 21. P x x 3 27x 54. The possible rational zeros are 1, 2, 3, 6, 9, 18, 27. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 3 27x 30 has 1 variation in sign and hence P has 1 negative real zero. 1
1
0
1 1
1
27
54
1
26
26
80
2
1
0
2 x 1 is not a zero. 3
1
0 3
1
9
2
4
23
54
46
8
x 2 is not a zero.
54 54
18 0 x 3 is a zero. P x x 3 27x 54 x 3 x 2 3x 18 x 3 x 6 x 3. Therefore, the zeros are 6 and 3.
3
27
1
27
40
CHAPTER 3 Polynomial and Rational Functions
22. P x x 3 5x 2 9x 45. The possible rational zeros are 1, 3, 9, 15, 45. P x has 1 variation in sign and
hence 1 positive real zero. P x x 3 5x 2 9x 45 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
5
9
1 1
6
6
2
45 3
3
48
1
5 2
x 1 is not a zero. 3
1
5
9
3 1
1
7
9
45
5
35
14
10 x 2 is not a zero.
45
24
45
8
15 0 x 3 is a zero. P x x 3 5x 2 9x 45 x 3 x 2 8x 15 x 3 x 5 x 3. Therefore, the zeros are 5, 3, and 3.
23. P x x 3 3x 2 x 3. The possible rational zeros are 1, 3. P x has 1 variation in sign and hence 1 positive real zero. P x x 3 3x 2 x 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
1
3
1
1
2
2
3 3
3 0 x 1 is a zero. So P x x 3 3x 2 x 3 x 1 x 2 2x 3 x 1 x 3 x 1. Therefore, the zeros are 1, 3, and 1.
24. P x x 3 4x 2 11x 30. The possible rational zeros are 1, 2, 3, 5, 10, 15, 30. P x has 2 variations
in sign and hence 0 or 2 positive real zeros. P x x 3 4x 2 11x 30 has 1 variation in sign and hence P has 1 negative real zero. 1
1
4 1
1
11
30
3
14
14
16
1
2
4
2
11
30
4
30
1 2 15 0 x 2 is a zero. So P x x 3 4x 2 11x 30 x 2 x 2 2x 15 x 2 x 5 x 3. Thus, the zeros are 3, 2, and 5. 3
25. Method 1: P x x 4 5x 2 4The possible rational zeros are 1, 2, 4. P x has 1 variation in sign and hence 1 positive real zero. P x x 4 5x 2 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
0
1
0
4
4
4
5
1
1 1 4 4 0 x 1 is a zero. Thus P x x 4 5x 2 4 x 1 x 3 x 2 4x 4 . Continuing with the quotient we have: 1
1
1
1
4 0
4 4
1 0 4 0 x 1 is a zero. P x x 4 5x 2 4 x 1 x 1 x 2 4 x 1 x 1 x 2 x 2. Therefore, the zeros are 1, 2.
Method 2: Substituting u x 2 , the polynomial becomes P u u 2 5u 4, which factors: u 2 5u 4 u 1 u 4 x 2 1 x 2 4 , so either x 2 1 or x 2 4. If x 2 1, then x 1; if x 2 4, then x 2. Therefore, the zeros are 1 and 2.
SECTION 3.4 Real Zeros of Polynomials
41
26. P x x 4 2x 3 3x 2 8x 4. Using synthetic division, we see that x 1 is a factor of P x: 1
1
8
1
3 1
4
1
4
2
1
4 4
4
0
x 1 is a zero.
We continue by factoring the quotient, and we see that x 1 is again a factor: 1
1
1
1
1
4
4
0
4
0
4 0 x 1 is a zero. P x x 4 2x 3 3x 2 8x 4 x 1 x 1 x 2 4 x 12 x 2 x 2
Therefore, the zeros are 1 and 2.
27. P x x 4 6x 3 7x 2 6x 8. The possible rational zeros are 1, 2, 4, 8. P x has 1 variation in sign and hence 1 positive real zero. P x x 4 6x 3 7x 2 6x 8 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1
1
6
1
7
1
7
7
14
6
8
8
0
14
8
x 1 is a zero
and there are no other positive zeros. Thus P x x 4 6x 3 7x 2 6x 8 x 1 x 3 7x 2 14x 8 . Continuing
by factoring the quotient, we have:
1
1
1
7
14
8
1
6
8
6
8
0
x 1 is a zero. So P x x 4 6x 3 7x 2 6x 8 x 1 x 1 x 2 6x 8 x 1 x 1 x 2 x 4. Therefore, the
zeros are 4, 2, and 1.
28. P x x 4 x 3 23x 2 3x 90. The possible rational zeros are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x x 4 x 3 23x 2 3x 90 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
1
3
90
0
0
23
26
23
26
1
1
1
1
1 3
23
3
21
12
1
2
27
1
42
90
21
45
1
21
45
2
64
45
3
90
2
1
1 5
1
5
23
30
0
x 2 is a zero.
45
1 6 9 0 x 5 is a zero. 4 9 72 P x x 2 x 5 x 2 6x 9 x 2 x 5 x 32 . Therefore, the zeros are 3, 2, and 5. 1
42
CHAPTER 3 Polynomial and Rational Functions
29. P x 9x 4 82x 2 9 has possible rational zeros 1, 3, 9, 13 , 19 . Since P x has 2 variations in sign, there are 0 or 2 positive real zeros, and since P x 9x 4 9x 2 36 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
9
0
9 9
9
0
9
9
73
73
73
73
64
82
9
3
0
9
0
9
81
3
9
1
3
82
27 27
0
x 3 is a zero
Because P is even, we know that x 3 is also a zero. Thus, P x 9x 4 82x 2 9 x 2 9 9x 2 1 x 3 x 3 3x 1 3x 1, and the zeros are 3 and 13 . Note: Since P x contains only even powers of x, factoring by substitution also works. Let x 2 u; then P u 9u 2 82u 9 u 9 9u 1 x 2 9 9x 2 1 , which gives the same result.
30. P x 6x 4 23x 3 13x 2 32x 16. The possible rational zeros are 1, 2, 4, 8, 16, 12 , 13 , 23 , 43 , 83 , 16 3 , 16 . Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x 6x 4 23x 3 13x 2 32x 16 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
6 6
32
16
6
13 17
30
2
17
30
23
4
6
2
18
23
13
6
1
23
12
2
24
6 6
32
16
4
36
16
9
4
0
11
13
32
16
22
70
76
35
38
60
x 4 is a zero
P x 6x 4 23x 3 13x 2 32x 16 x 4 6x 3 x 2 9x 4 . We continue: 12
6
1
9
4
1
3
4
2 8 0 x 12 is a zero Thus, P x x 4 2x 1 3x 2 x 4 x 1 2x 1 3x 4 x 4 with zeros 1, 12 , 43 , and 4. 6
31. P x 6x 4 7x 3 9x 2 7x 3. The possible rational zeros are 1, 2, 3, 12 , 13 , 16 , 32 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x 6x 4 7x 3 9x 2 7x 3 has 2 variations in sign and hence P has
0 or 2 negative real zeros.
1 6
7 9 7 6
13
3
4 3
6 13
4 3 0 x 1 is a zero. Thus P x 6x 4 7x 3 9x 2 7x 3 x 1 6x 3 13x 2 4x 3 . Continuing by factoring the quotient, we have 1
6
13
4
6
7
3 3
6 7 3 0 x 1 is a zero. Thus P x x 1 x 1 6x 2 7x 3 x 1 x 1 2x 3 3x 1 with zeros 32 , 1, 13 , and 1.
SECTION 3.4 Real Zeros of Polynomials
43
32. P x 6x 3 37x 2 5x 6. The possible rational zeros are 1, 2, 3, 6, 12 , 13 , 16 , 23 , 32 . Since P x has 1 variation in sign, P has 1 positive real zero. Since P x 6x 3 37x 2 5x 6 has 2 variations in sign, P has 0 or 2
negative real zeros. 1
6 2
37
5
2
39
39
44
1 2
6
6
37
5
3
20
40
25
44 38
x 1 is an upper bound. 1 3
6
37
5
2
13
6
6 25 2 13 2
x 12 is an upper bound.
6
6
6 39 18 0 x 13 is a zero. P x 3x 1 2x 2 13x 6 3x 1 2x 1 x 6. Therefore, the zeros are 6, 12 , and 13 . 33. Factoring by grouping can be applied to this exercise.
4x 3 4x 2 x 1 4x 2 x 1 x 1 x 1 4x 2 1 x 1 2x 1 2x 1. Therefore, the zeros are 1 and 12 .
34. We use factoring by grouping: P x 2x 3 3x 2 2x 3 2x x 2 1 3 x 2 1 x 2 1 2x 3 x 1 x 1 2x 3. Therefore, the zeros are 32 and 1. 35. P x 4x 3 7x 3. The possible rational zeros are 1, 3, 12 , 32 , 14 , 34 . Since P x has 2 variations in sign, there are 0 or 2 positive zeros. Since P x 4x 3 7x 3 has 1 variation in sign, there is 1 negative zero. 1 2
4
0 2
3
7
1
3
2 6 0 x 12 is a zero. P x x 12 4x 2 2x 6 2x 1 2x 2 x 3 2x 1 x 1 2x 3 0. Thus, the zeros are 32 , 4
1 , and 1. 2
1 . Since P x has 2 36. P x 12x 3 25x 2 x 2. The possible rational zeros are 1, 2, 12 , 13 , 23 , 14 , 16 , 12
variations in sign, P has 0 or 2 positive real zeros, and since P x 12x 3 25x 2 x 2 has 1 variation in sign, P has 1 negative real zero. 1
12
25 12
12
1
2
13
12
2
12
25
24
1
2
2
2
10 12 1 1 0 x 2 is a zero. P x 12x 3 25x 2 x 2 x 2 12x 2 x 1 4x 1 3x 1 x 2. Therefore, the zeros are 14 , 13 ,
and 2.
13
12
44
CHAPTER 3 Polynomial and Rational Functions
37. P x 24x 3 10x 2 13x 6. The possible rational zeros are 1, 2, 3, 6, 12 , 32 , 13 , 23 , 14 , 34 , 16 , 18 , 1 , 1 . P x has 1 variation in sign and hence 1 positive real zero. P x 24x 3 10x 2 13x 6 has 2 38 , 12 24
variations in sign, so P has 0 or 2 negative real zeros. 1 24
10 13 6 24
24 14 3 24
2 24
14 1
1 7 x 1 is not a zero.
10 13
24 38 6 24
6
72 186 519
24 62 173 525 x 3 is not a zero. 12
24
10
6
76 126
63 132 x 2 is not a zero.
10 13
6
144 804 4746
24 134 791 4752 x 6 is not a zero.
13
6
1
12
10 13 48
6
2 12 0 x 12 is a zero. Thus, P x 24x 3 10x 2 13x 6 2x 1 12x 2 x 6 3x 2 2x 1 4x 3 has zeros 23 , 12 , and 34 . 24
1 , 3 , 3 . P x has 2 38. P x 12x 3 20x 2 x 3. The possible rational zeros are 1, 2, 3, 12 , 13 , 14 , 16 , 12 2 4
variations in sign and hence 0 or 2 positive real zeros. P x 12x 3 20x 2 x 3 has 1 variation in sign and hence P has 1 negative real zero. 1
3
12
1
3
12
12
8
7
8
7
4
12
20
1
3
36
48
147
16
49
150
20
12
2 x 1 is not a zero.
12 1 2
1
3
24
8
18
4
9
21
20
1
x 3 is not a zero. 12 14 Thus, P x 12x 3 20x 2 x 3 x 12 12x 2 14x 6 . Continuing:
6
12
12
20
6
3 2
12
14
6
4
0
18
12
6 x 32 is a zero.
Thus, P x 2x 1 2x 3 3x 1 has zeros 12 , 32 , and 13 .
7
x 2 is not a zero. 3 3 0
x 12 is a zero.
SECTION 3.4 Real Zeros of Polynomials
45
39. P x 6x 4 13x 3 32x 2 45x 18. The possible rational zeros are 1, 2, 3, 6, 9, 18, 12 , 13 , 16 , 23 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x 6x 4 13x 3 32x 2 45x 18 has 2
variations in sign and hence P has 0 or 2 negative real zeros. 1
6
13
6
19
32
6
19
45
18
13
58
2
6
13 12
13 58 40 x 1 is not a zero. P x x 2 6x 3 25x 2 18x 9 . Continuing: 6
3
25
18
18
21
6
32
45
18
9
50
25
18
36
18
0
x 2 is a zero.
9
9
6 7 3 0 x 3 is a zero. P x x 2 x 3 3x 2 8x 3 x 2 x 3 3x 1 2x 3, so the zeros are 3, 32 , 13 , and 2.
40. P x 2x 4 11x 3 11x 2 15x 9. The possible rational zeros are 1, 3, 9, 12 , 32 , 92 . Since P x has 1
variation in sign, P has 1 or 3 positive real zeros. Since P x 2x 4 11x 3 11x 2 15x 9 has 3 variations in sign, P has 1 or 3 negative real zeros.
1
2
11
11
2
13
15
24
9
9
2 13 24 9 0 x 1 is a zero. P x x 1 2x 3 13x 2 24x 9 . We continue by factoring the quotient: 3
2
13
24
9
6
57
243
12
2
13
24
9
1
6
9
252 x 2 is an upper bound. 2 12 18 0 x 12 is a root. P x x 1 2x 1 x 2 6x 9 x 1 2x 1 x 32 . Therefore, the zeros are 3, 12 , and 1. 2
19
81
zero.
46
CHAPTER 3 Polynomial and Rational Functions
41. P x x 5 3x 4 9x 3 31x 2 36. The possible rational zeros are 1, 2, 3, 4, 6, 8, 9, 12, 18. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 5 3x 4 9x 3 31x 2 36 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1
1
3
9
1
4
1
4
5
36
31
0
36
5
36
36
36 36 0 x 1 is a zero. So P x x 5 3x 4 9x 3 31x 2 36 x 1 x 4 4x 3 5x 2 36x 36 . Continuing by factoring the quotient, 5
we have:
1
1
4 1
1
5
1
0
0
36
36
72
3
1
5
4 3
1
7
2
36
21
1
4 2
1 36
48
5
12
6
7
36
14
22
36 44 80
36
36
16
12 0 x 3 is a zero. So P x x 1 x 3 x 3 7x 2 16x 12 . Since we have 2 positive zeros, there are no more positive zeros, so
we continue by factoring the quotient with possible negative zeros. 1
1 1
7
16
12
1
6
10
10
2
1
1
7
16
12
2
10
12
0 x 2 is a zero. Then P x x 1 x 3 x 2 x 2 5x 6 x 1 x 3 x 22 x 3. Thus, the zeros are 1, 3, 2,
and 3.
6
2
5
6
SECTION 3.4 Real Zeros of Polynomials
47
42. P x x 5 4x 4 3x 3 22x 2 4x 24 has possible rational zeros 1, 2, 3, 4, 6, 8, 12, 24. Since P x has 3 variations in sign, there are 1 or 3 positive real zeros. Since P x x 5 4x 4 3x 3 22x 2 4x 24 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
1
4
1
1
3
22
3
6
4
2
7
1
0
2
P x x 22 x 3 7x 6 2
1
0
7
2
1
1
4
14
2
7
8
12
0
14
12
7
6
0
4
24
12
0
16
8
24 x 2 is a zero.
x 2 is a zero again.
1
3
6
4
3
22
2
4
12
3 6 16 12 12 P x x 2 x 4 2x 3 7x 2 8x 12 2
1
2
24
16
0
7
3
6
9
6
6
1 3 2 0 x 3 is a zero. 2 3 12 P x x 22 x 3 x 2 3x 2 x 22 x 3 x 1 x 2 0. Therefore, the zeros are 1, 2, and 3. 1
43. P x 3x 5 14x 4 14x 3 36x 2 43x 10 has possible rational zeros 1, 2, 5, 10, 13 , 23 , 53 , 10 3 . Since P x has 2 variations in sign, there are 0 or 2 positive real zeros. Since P x 3x 5 14x 4 14x 3 36x 2 43x 10
has 3 variations in sign, P has 1 or 3 negative real zeros. 1
3
14 3
3
14
36
43
10
11
25
11
54
2
3
8
3
2
6
30
24
5
4
68
184
14 6
11 25 11 54 64 P x x 2 3x 4 8x 3 30x 2 24x 5 2
3 3
8
5
3
14
36
43
10
16
60
48
10
30
24
5
8 15
30 35
24
25
0
x 2 is a zero.
5
5
34 92 189 3 7 5 1 0 x 5 is a zero. P x x 2 x 5 3x 3 7x 2 5x 1 . Since 3x 3 7x 2 5x 1 has no variation in sign, there are no more
positive zeros.
1
3
7
5
1
3
4
1
3 4 1 0 x 1 is a zero. P x x 2 x 5 x 1 3x 2 4x 1 x 2 x 5 x 1 x 1 3x 1. Therefore, the zeros are 1,
13 , 2, and 5.
48
CHAPTER 3 Polynomial and Rational Functions
44. P x 2x 6 3x 5 13x 4 29x 3 27x 2 32x 12 has possible rational zeros 1, 2, 3, 4,
6, 12, 12 , 32 . Since P x has 5 variations in sign, there are 1 or 3 or 5 positive real zeros. Since
P x 2x 6 3x 5 13x 4 29x 3 27x 2 32x 12 has 3 variations in sign, P has 1 or 3 negative real zeros. 1
2
3
13
29
2
1
14
1
14
2
2
2
13
1
11
1
11
4
2
29
3
2
15
15
32
14
2
4
10
13
7
13
26
20
20
8
12
12
0 x 2 is a zero. P x x 2 2x 5 x 4 11x 3 7x 2 13x 6 . We continue with the quotient: 2
7
12
12
12
27
22
32
27
6
6
10
2
6
2 5 1 5 3 0 x 2 is a zero again. P x x 22 2x 4 5x 3 x 2 5x 3 . We continue with the quotient, first noting 2 is no longer a possible rational
solution:
3
2
5 6
2
11
5
1
22
42
21
47
3
94 91
x 3 is an upper bound.
We know that there is at least 1 more positive zero. 1 2
2
5 1
1
2
5 1
3
3
2 6 2 6 0 x 12 is a zero. P x x 22 x 12 2x 3 6x 2 2x 6 . We can factor 2x 3 6x 2 2x 6 by grouping; 2x 3 6x 2 2x 6 2x 3 6x 2 2x 6 2x 6 x 2 1 . So P x 2 x 22 x 12 x 3 x 2 1 .
Since x 2 1 has no real zeros, the zeros of P are 3, 2, and 12 .
45. P x 3x 3 5x 2 2x 4. The possible rational zeros are 1, 2, 4, 13 , 23 , 43 . P x has 1 variation in sign and hence 1 positive real zero. P x 3x 3 5x 2 2x 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1 3 5 2 4 5 2 4 1 3 3
3
2
0
3
2
4
2 0 4 3 2 4 0 x 1 is a zero. So P x x 1 3x 2 2x 4 . Using the Quadratic Formula on the second factor, we have 2 434 13 13 . Therefore, the zeros of P are 1 and 13 13 . x 2 223
SECTION 3.4 Real Zeros of Polynomials
49
46. P x 3x 4 5x 3 16x 2 7x 15. The possible rational zeros are 1, 3, 5, 15, 13 , 53 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x 3x 4 5x 3 16x 2 7x 15 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
3
3
3
5
16
2 18 11 4 So P x x 3 3x 3 4x 2 4x 5 . Continuing:
3
4
5
3
16
7
15
2
18
11
3
1
9
3
4
4
3
1
7
15
12
12
15
4
5
0
x 3 is a zero.
5 5
3 1 5 0 x 1 is a zero. Thus, P x x 3 3x 3 4x 2 4x 5 x 1 x 3 3x 2 x 5 . Using the Quadratic Formula on the last 2 435 factor, we have x 1 123 16 61 . Therefore, the zeros of P are 1, 3, and 16 61 .
47. P x x 4 6x 3 4x 2 15x 4. The possible rational zeros are 1, 2, 4. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 4 6x 3 4x 2 15x 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1 1 6 4 15 4 4 15 4 2 1 6 1
1
5
5
1 4
14
1 14
1
18
6
4
4
15
4
8
16
4
8
4
4
2 4 1 0 x 4 is a zero. So P x x 4 x 3 2x 2 4x 1 . Continuing by factoring the quotient, we have: 4
1
2
4
1
2
1
1
2
4
8
4
1
1
16
15
x 4 is an upper bound.
1 1
2
8
7
4
1
3
1
3
1
0
14
18
1
x 1 is a zero.
So P x x 4 x 1 x 2 3x 1 . Using the Quadratic Formula on the third factor, we have: 3 32 411 3 13 . Therefore, the zeros are 4, 1, and 3 13 . x 2 2 21
50
CHAPTER 3 Polynomial and Rational Functions
48. P x x 4 2x 3 2x 2 3x 2. The possible rational zeros are 1, 2. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 4 2x 3 2x 2 3x 2 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
2
1
1
3
2
3 1
2
3
1
2
2
0
3
1 2
P x x 1 x 3 3x 2 x 2 . Continuing with the quotient: 1 1 3 1 2 1 4
1 1
5
2 1
1
1 2
1 4 5
x 1 is a zero. 3
1 2
2 2
2
3 x 1 is an 1 2 1 1 1 1 1 0 x 2 is a zero. upper bound. So P x x 1 x 2 x 2 x 1 . Using the Quadratic Formula on the third factor, we have 1 12 4 1 1 1 5 5. x . Therefore, the zeros are 1, 2, and 1 2 2 1 2
49. P x x 4 7x 3 14x 2 3x 9. The possible rational zeros are 1, 3, 9. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x x 4 7x 3 14x 2 3x 4 has 1 variation in sign and hence P has 1 negative real zero. 1
1
7
1
1
14
6
3
8
9
1
3
14
7
5
3
3
12
6
2
3
3
3
3
1
1
9 9
6 8 5 4 1 4 2 3 0 x 3 is a zero. So P x x 3 x 3 4x 2 2x 3 . Since the constant term of the second term is 3, 9 are no longer possible zeros.
Continuing by factoring the quotient, we have:
3
1 1
4
0
x 3 is a zero again.
So P x x 32 x 2 x 1 . Using the Quadratic Formula on the second factor, we have: 1 12 411 1 5 . Therefore, the zeros are 3 and 1 5 . x 2 2 21
SECTION 3.4 Real Zeros of Polynomials
51
50. P x x 5 4x 4 x 3 10x 2 2x 4. The possible rational zeros are 1, 2, 4. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x x 5 4x 4 x 3 10x 2 2x 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
4 1
1
1
10
2
3
4
6
2
4
8 1
3
is 2, 4 are no longer possible zeros. 1
2
1
0
2
0
2
1
10
2
4
10
0
0
5
4
4
2
0
x 2 is a zero.
Since the constant term of the second factor
Continuing by factoring the quotient, we have:
0
2
10
20
1
1
5
4
2
4 6 8 4 So P x x 2 x 4 2x 3 5x 2 2 . 2
1
1
1
2 1
5 3
0
2
2
2
5 10 18 1 3 2 2 0 x 1 is a zero. So P x x 2 x 1 x 3 3x 2 2x 2 . Continuing by factoring the quotient, we have:
1
3
1
2
4
2
2
4 2 0 x 1 is a zero again. So P x x 2 x 12 x 2 4x 2 . Using the Quadratic Formula on the second factor, we have:
4 42 412 4 8 42 2 2 2. Therefore, the zeros are 1, 2, and 2 2. x 21 2 2
51. P x 4x 3 6x 2 1. The possible rational zeros are 1, 12 , 14 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x 4x 3 6x 2 1 has 1 variation in sign and hence P has 1 negative real zero. 1
4
6 4
So P x
0
1
2
2
1 2
4
6
2
0
1
2
1
4 2 2 1 4 4 2 0 x 12 is a zero. x 12 4x 2 4x 2 . Using the Quadratic Formula on the second factor, we have:
4 42 442 4 48 44 3 1 3 . Therefore, the zeros are 1 and 1 3 . x 8 8 2 2 2 24
52
CHAPTER 3 Polynomial and Rational Functions
52. P x 3x 3 5x 2 8x 2. The possible rational zeros are 1, 2, 13 , 23 . P x has 1 variation in sign and hence 1 positive real zero. P x 3x 3 5x 2 8x 2 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1 3
3
5 3
8
2
3
2
10
2
10
12
3
5
8
2
43
1
3
2
28 3
5
8
2
0
2
5
8 2
3
1
12
6
14
3
5
6
2 3
28 9
4
3
46 9
3
8
3
3
8 13
2
0
3
5
8
3
2
3
10
26 3
3
5
8
2
3
11
6
22
14
28 30
2
2
1
2
20 3
2
Thus we have tried all the positive rational zeros, so we try the negative zeros. 1
8
2
2
3 6 6 0 x 13 is a zero. So P x x 13 3x 2 6x 6 3 x 13 x 2 2x 2 . Using the Quadratic Formula on the second factor, we 2 22 412 22 12 222 3 1 3. Therefore, the zeros are 13 and 1 3. have: x 21
53. P x 2x 4 15x 3 17x 2 3x 1. The possible rational zeros are 1, 12 . P x has 1 variation in sign and hence 1 positive real zero. P x 2x 4 15x 3 17x 2 3x 1 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1 2
2 2 12
15
17
3
1
8
16
25
25 2 31 2
2
15
17
3
1
7
5
1 31 4 27 4
x 12 is an upper bound. 1
1
14 10 2 0 x 12 is a zero. So P x x 12 2x 3 14x 2 10x 2 2 x 12 x 3 7x 2 5x 1 . 2
1
1
7
5
1
6
1 1
1 6 1 0 x 1 is a zero. So P x x 12 2x 3 14x 2 10x 2 2 x 12 x 1 x 2 6x 1 Using the Quadratic Formula on the
6 62 411 10 3 10. Therefore, the zeros are 1, 1 , third factor, we have x 62 40 62 21 2 2
and 3
10.
53
SECTION 3.4 Real Zeros of Polynomials
54. P x 4x 5 18x 4 6x 3 91x 2 60x 9. The possible rational zeros are 1, 3, 9, 12 , 32 , 92 , 14 , 34 , 94 .
P x has 4 variations in sign and hence 0 or 2 or 4 positive real zeros. P x 4x 5 18x 4 6x 3 91x 2 60x 9 has 1 variation in sign and hence P has 1 negative real zero. 1
4
18
4
14
4
6
91
14
20
9
60
71
3
4
6
91
18
72
18
1
12
9
60 57
9
20 71 11 10 4 6 24 19 3 0 x 3 is a zero. So P x x 3 4x 4 6x 3 24x 2 19x 3 . Continuing by factoring the quotient, we have: 4
3
6
24
12
18
19
3
18
3
4 6 6 1 0 x 3 is a zero again. So P x x 32 4x 3 6x 2 6x 1 . Continuing by factoring the quotient, we have: 3
4
6
12
1 2
1
6
54
144
4
6
2
6
4
1
1
48 1445 x 3 is an upper bound. 4 8 2 0 x 12 is a zero. So P x x 32 x 12 4x 2 8x 2 2 x 32 x 12 2x 2 4x 1 . Using the Quadratic Formula on 2 421 the second factor, we have x 4 422 44 24 1 26 . Therefore, the zeros are 12 , 3, and 1 26 . 4
18
55. (a) P x x 3 3x 2 x 3 has possible rational zeros 1, 3. 1
1
3
1
4
3
4
3
1
3
1
0
(b)
y
1
x 1 is a zero.
So P x x 1 x 2 4x 3 x 1 x 3 x 1.
1
x
1
x
The real zeros of P are 3, 1, and 1.
56. (a) P x x 3 3x 2 6x 8 has possible rational zeros 1, 2, 4,
(b)
y
8. 1
1
3 1
1 2
1
4
8
2
10
3
6
8
4
0
5
10
2
2
4
2 1
6
8 x 2 is a zero.
So P x x 2 x 2 5x 4 x 2 x 4 x 1. The real zeros of P are 4, 1, and 2.
54
CHAPTER 3 Polynomial and Rational Functions
57. (a) P x 2x 3 7x 2 4x 4 has possible rational zeros 1, 2, 4,
(b)
y
12 .
1 2 7
4
4
2 2 7
2 5 1
4
4
4 6 4
2 5 1 3
2 3 2 0 x 2 is a zero. So P x x 2 2x 2 3x 2 . Continuing: 2
2
3
2
1
0
4
2
1 x
1
2 x 2 is a zero again.
Thus P x x 22 2x 1. The real zeros of P are 2 and 12 . 58. (a) P x 3x 3 17x 2 21x 9 has possible rational zeros 1, 3,
(b)
y 2
9, 13 , 23 . 3
1
1
17
21
3
20
3
20
41
1 3
3
17 1
6
3
18
27
x
9
41 32
21
x 1 is an upper bound. 9
9 x 13 is a zero.
0
So P x x 13 3x 2 18x 27 3 x 13 x 2 6x 9 3 x 13 x 32 . The real zeros of P are 3 and 13 .
59. (a) P x x 4 5x 3 6x 2 4x 8 has possible rational zeros 1,
(b)
y
2, 4, 8. 1
1
5
1
2
6
4
4
2
8
6
1
4
2
6
2
1
5
6
4
6
0
8
2
1
3
0
2 1
x
8
4
0
x 2 is a zero.
So P x x 2 x 3 3x 2 4 and the possible rational zeros are restricted to 1, 2, 4. 2
1
3
2
0
4
2
4
1 1 2 0 x 2 is a zero again. P x x 22 x 2 x 2 x 22 x 2 x 1 x 23 x 1. So the real zeros of P are 1 and 2.
55
SECTION 3.4 Real Zeros of Polynomials
60. (a) P x x 4 10x 2 8x 8 has possible rational zeros 1, 2,
(b)
y
4, 8. 1 1
0 10 1 1
1 1 4 1
0
8 8
9 17
9 17
10
9 8
0 10
2 1
8 8
2 4 12 40
6 20 32
1 2
5
8
1
4 16 24 64
6 16 72 x 4 is an upper bound.
1 4 1
x
1
0
10
8
1
1
9
8
1
2
1
0
10
8
2
4
12
1 9 1 7 1 2 So P x x 2 x 3 2x 2 6x 4 . Continuing, we have: 1
2
1
2
6
2
4
8
0
x 2 is a zero.
4
4
8
4
6
8
x 2 is a zero again. P x x 22 x 2 4x 2 . Using the Quadratic Formula on the second factor, we have 2 412 8 42 2 2 2. So the real zeros of P are 2 and 2 2. 4 x 4 421 2 2 1
0
2
61. (a) P x x 5 x 4 5x 3 x 2 8x 4 has possible rational zeros
(b)
y
1, 2, 4. 1
1
1
1
2
1
1
2
1
1
8
4
0
5
4
4
5
4
5
1
0
5
1
8
4
6
10
4
2
1
4
8 1 1
3 5 2 0 x 2 is a zero. So P x x 2 x 4 x 3 3x 2 5x 2 , and the possible rational zeros are restricted to 1, 2. 2
1
1
1
1
3
5
2
3
3
1
0
1
3
3
1
1
2
1
2
6
6
2
x 2 is a zero again. So P x x 22 x 3 3x 2 3x 1 , and the possible rational zeros are restricted to 1.
1
2 1 0 x 1 is a zero. So P x x 22 x 1 x 2 2x 1 x 22 x 13 ., and the real zeros of P are 1 and 2.
x
56
CHAPTER 3 Polynomial and Rational Functions
62. (a) P x x 5 2x 4 8x 3 16x 2 16x 32 has possible rational
(b)
y
zeros 1, 2, 4, 8, 16. 1
1
2
8
1 1 2
1
2 2
1
4
8
3
3
5
0
0
16
32
5
21
5
21
5
16
32
32
32
16
8
16
10 1
27
x
x 2 is a zero. So P x x 2 x 4 4x 3 16x 16 . Continuing:
16
0
16
2
1 1
4
0
2
12
6
12
16
24
16
16
8
0 x 2 is a zero again. Thus, P x x 22 x 3 6x 2 12x 8 x 22 x 23 , and the real zeros of P are 2.
63. P x x 3 x 2 x 3. Since P x has 1 variation in sign, P has 1 positive real zero. Since P x x 3 x 2 x 3 has 2 variations in sign, P has 2 or 0 negative real zeros. Thus, P has 1 or 3 real zeros. 64. P x 2x 3 x 2 4x 7. Since P x has 3 variations in signs, P has 3 or 1 positive real zeros. Since P x 2x 3 x 2 4x 7 has no variation in sign, there is no negative real zero. Thus, P has 1 or 3 real zeros. 65. P x 2x 6 5x 4 x 3 5x 1. Since P x has 1 variation in sign, P has 1 positive real zero. Since P x 2x 6 5x 4 x 3 5x 1 has 1 variation in sign, P has 1 negative real zero. Therefore, P has 2 real zeros. 66. P x x 4 x 3 x 2 x 12. Since P x has no variations in sign, P has no positive real zero. Since P x x 4 x 3 x 2 x 12 has 4 variations in sign, P has 4, 2, or 0 negative real zeros. Therefore, P x has 0, 2, or 4 real zeros. 67. P x x 5 4x 3 x 2 6x. Since P x has 2 variations in sign, P has 2 or 0 positive real zeros. Since P x x 5 4x 3 x 2 6x has no variation in sign, P has no negative real zero. Therefore, P has a total of 1 or 3 real zeros (since x 0 is a zero, but is neither positive nor negative). 68. P x x 8 x 5 x 4 x 3 x 2 x 1. Since P x has 6 variations in sign, the polynomial has 6, 4, 2, or 0 positive real zeros. Since P x has no variation in sign, the polynomial has no negative real zeros. Therefore, P has 6, 4, 2, or 0 real zeros. 69. P x 2x 3 5x 2 x 2; a 3, b 1 3
2 2
1
2 2
5
1
6
3
1
4
5
1
2
7
7
8
2 12 14
alternating signs lower bound.
2
8 6
Therefore a 3 and b 1 are lower and upper bounds.
all nonnegative upper bound.
SECTION 3.4 Real Zeros of Polynomials
70. P x x 4 2x 3 9x 2 2x 8; a 3, b 5 1
3
2
1 5
1
2
8
18
48
9 15
3 5
6
2
9
2
8
15
30
160
3
6
32
168
5
1
56
16
Alternating signs lower bound.
All nonnegative upper bound.
Therefore a 3 and b 5 are lower and upper bounds.
71. P x 8x 3 10x 2 39x 9; a 3, b 2 8
3
10
8 2
42
14
3
10
39
8
16 8
9
39
24
26
57
9
0
alternating signs lower bound.
9
52
26
13
35
all nonnegative upper bound.
Therefore a 3 and b 2 are lower and upper bounds. Note that x 3 is also a zero.
72. P x 3x 4 17x 3 24x 2 9x 1; a 0, b 6 0
3
3
24
0
0
17
9
1
0
0
24
9
1
17
24
18
6
9
180
1026
1
30
171
1027
3 6
17
3
Alternating signs lower bound. 1 All nonnegative upper bound.
Therefore a 0 and b 6 are lower and upper bounds. Note that because P x alternates in sign, by Descartes’ Rule of Signs, 0 is automatically a lower bound.
73. P x x 4 2x 3 3x 2 5x 1; a 2, b 1 2
1 1
1
1 1
2
3
5
2
0
6
0
3
2
3
1
3
6
3
6
11
1 2
1
1 5
Alternating signs lower bound.
1
11 10
Therefore a 2 and b 1 are lower and upper bounds.
All nonnegative upper bound.
58
CHAPTER 3 Polynomial and Rational Functions
74. P x x 4 3x 3 4x 2 2x 7; a 4, b 2 1
4
3
1 2
4
2
7
0
2
1
4
4 1
1
3 2
1
0
8
4
2
7
6
10
13
10
5
12
20 All nonnegative upper bound.
Therefore a 4 and b 2 are lower and upper bounds.
75. P x 2x 4 6x 3 x 2 2x 3; a 1, b 3 2
1
3
2
8
2
2 3
1
6
11
9
8
9
11
14
6
1
2
3
3
3
1
6
2 2
6
0
0
1
Alternating signs lower bound.
All nonnegative upper bound.
Therefore a 1 and b 3 are lower and upper bounds.
76. P x 3x 4 5x 3 2x 2 x 1; a 1, b 2 3
1
5
8
3
3 2
2 6
5
2 2
0
1
0
1
3
6
3
1
1 5
6
8
Alternating signs lower bound.
4
5 1
Alternating signs lower bound.
1
2 1
Therefore a 1 and b 2 are lower and upper bounds.
All nonnegative upper bound.
77. P x x 3 3x 2 4 and use the Upper and Lower Bounds Theorem: 1
1
1 3
1
3 1 4
4
4
4
0
0
4
3
0
0
0
0
4
3
1
0
4
alternating signs lower bound.
all nonnegative upper bound.
Therefore 1 is a lower bound (and a zero) and 3 is an upper bound. (There are many possible solutions.)
78. P x 2x 3 3x 2 8x 12 and using the Upper and Lower Bounds Theorem: 2
2
3 4
2 3
8 14
12
12
7
6
2
3
8
12
9
3
3
1
15
6
2 (There are many possible solutions.)
0
Alternating signs x 2 is a lower bound (and a zero).
All nonnegative x 3 is an upper bound.
SECTION 3.4 Real Zeros of Polynomials
79. P x x 4 2x 3 x 2 9x 2. 1
1
1
2
1
1
2
9
1
0
9
1
0
9
7
3
1
2
1
3
3
1
4
1 1
1
1
2
1
9
2
12
9
3
11
2
1
2
0
0
1
2
9
2
14
7
12
all positive upper bound.
2 13
4
4
3
1 1
9
3
1
2
15
13
alternating signs lower bound.
Therefore 1 is a lower bound and 3 is an upper bound. (There are many possible solutions.)
80. Set P x x 5 x 4 1. 1
1 1
1
1 1
0
0
0
1
1
0
0
0
0
0
0
0
0
1
0
0
0
1
2
2
2
2
1
1 1 2
2
(There are many possible solutions.)
81. P x 2x 4 3x 3 4x 2 3x 2. 1
2
2
2
3 2
P x x 1 2x 3 5x 2 x 2
1
2
Alternating signs x 1 is a lower bound.
1
4
5
5
2
All nonnegative x 1 is an upper bound.
1
2
3
1
2 0
2
5
1
2
3
x 1 is a zero.
2 2
2 3 2 0 x 1 is a zero. P x x 1 x 1 2x 2 3x 2 x 1 x 1 2x 1 x 2. Therefore, the zeros are 2, 12 , 1.
59
60
CHAPTER 3 Polynomial and Rational Functions
82. P x 2x 4 15x 3 31x 2 20x 4. The possible rational zeros are 1, 2, 4, 12 . Since all of the coefficients are positive, there are no positive zeros. Since P x 2x 4 15x 3 31x 2 20x 4 has 4 variations in sign, there are 0, 2, or 4 negative real zeros. 1 2 15
31
20
4
2 2 15
2 13 18 2
2 13
18
2
2
9
2 11
2
2
4 2 11
4 14 10 7
20
4
4 22 18 4
P x x 2 2x 3 11x 2 9x 2 : 2 2 11
31
3
2
12 2 11
2
8 12 12
2
5 12
9
9
0 x 2 is a zero.
9
2
1 5 2
3 14
2 10
4
0 x 12 is a zero.
P x x 2 2x 1 x 2 5x 2 . Now if x 2 5x 2 0, then x 5 25412 52 17 . Thus, the zeros 2
are 2, 12 , and 52 17 .
83. Method 1: P x 4x 4 21x 2 5 has 2 variations in sign, so by Descartes’ rule of signs there are either 2 or 0 positive zeros. If we replace x with x, the function does not change, so there are either 2 or 0 negative zeros. Possible rational zeros are 1, 12 , 14 , 5, 52 , 54 By inspection, 1 and 5 are not zeros, so we must look for noninteger solutions: 1 2
4
0 2
4
2
21
0
5
1
10
5
20
10
0
x 12 is a zero.
P x x 12 4x 3 2x 2 20x 10 , continuing with the quotient, we have: 12
4 4
P x x 12
2 2
0
20
10
20
0
0
10
x 12 is a zero.
1 2 4x 20 0. If 4x 2 20 0, then x 5. Thus the zeros are 12 5. x 2
Method 2: Substituting u x 2 , the equation becomes 4u 2 21u 5 0, which factors: 4u 2 21u 5 4u 1 u 5 4x 2 1 x 2 5 . Then either we have x 2 5, so that x 5, or we have x 2 14 , so that x 14 12 . Thus the zeros are 12 5.
SECTION 3.4 Real Zeros of Polynomials
61
84. P x 6x 4 7x 3 8x 2 5x x 6x 3 7x 2 8x 5 . So x 0 is a zero. Continuing with the quotient,
1 Q x 6x 3 7x 2 8x 5. The possible rational zeros are 1, 5, 12 , 52 , , 53 , 16 , 56 . Since Q x has 2 3
variations in sign, there are 0 or 2 positive real zeros. Since Q x 6x 4 7x 3 8x 2 5x has 1 variation in sign, there is 1 negative real zero. 6
1
6
5
6
8 1
9
1
9
4
7
1 2
6
5
7
8
5
30
115
535
6
23
107
540
6
7
5
3
8
6
2
5
4
10
0
All positive upper bound.
x 12 is a zero.
P x x 2x 1 3x 2 2x 5 x 2x 1 3x 5 x 1. Therefore, the zeros are 0, 1, 12 and 53 .
85. P x x 5 7x 4 9x 3 23x 2 50x 24. The possible rational zeros are 1, 2, 3, 4, 6, 8, 12, 24. P x has 4 variations in sign and hence 0, 2, or 4 positive real zeros. P x x 5 7x 4 9x 3 23x 2 50x 24 has 1 variation in sign, and hence P has 1 negative real zero. 1
1
7 1
1
6
9
23
6
3
3
26
50
26
24
24 24
0
x 1 is a zero.
P x x 1 x 4 6x 3 3x 2 26x 24 ; continuing with the quotient, we try 1 again. 1
1 1
3
26
1
5
2
5
2
24
1
5
2
24
6
16
6
24
24 0
x 1 is a zero again.
P x x 12 x 3 5x 2 2x 24 ; continuing with the quotient, we start by trying 1 again. 1
1 1
2
24
1
4
6
4
6
5
18
2
2
1
3
8
8
3
1 1
2
24
3
6
24
2
8
5
0
x 3 is a zero.
P x x 12 x 3 x 2 2x 8 x 12 x 3 x 4 x 2. Therefore, the zeros are 2, 1, 3 4.
62
CHAPTER 3 Polynomial and Rational Functions
1 86. P x 8x 5 14x 4 22x 3 57x 2 35x 6. The possible rational zeros are 1, 2, 3, 6, , 2 1 32 , 14 , 34 , , 38 . Since P x has 4 variations in sign, there are 0, 2, or 4 positive real zeros. Since 8 P x 8x 5 14x 4 22x 3 57x 2 35x 6 has 1 variation in sign, there is 1 negative real zero. 1
8
14
8
22
22
57
22
0
0
57
8
35 57
6
2
92
92
98
8
30
8
14
8
30
22
57
38
19
60
16
6
35 38
76
6
3
0
x 2 is a zero.
P x x 2 8x 4 30x 3 38x 2 19x 3 . All the other real zeros are positive. 1
38
8
8
16
22
16
22
8
16
22 8
8 Since f
1 2
14
2
14
3
0
3
P x x 2 x 1 8x 3 22x 2 16x 3 . 1
3
19
x 1 is a zero.
1 2
3
8
2
22 4
8
1
16
3 7 2 1 2
9
7
18
0 f 1, there must be a zero between 12 and 1. We try 34 : 3 4
8
6
8
16
22
12 4
16
3
3 0
x 34 is a zero.
P x x 2 x 1 4x 3 2x 2 4x 1 . Now, 2x 2 4x 1 0 when x 4 16421 22 2 . Thus, the 22
zeros are 1, 34 , 2, and 22 2 .
87. P x x 3 x 2. The only possible rational zeros of P x are 1 and 2. 1
1
0 1
1
1
1
2
0
2
1
2
1
0
0 2
1
2
1
2
3
4
4
6
1
1 1
0 1 1
1
2
0
2
1
0
Since the row that contains 1 alternates between nonnegative and nonpositive, 1 is a lower bound and there is no need to try 2. Therefore, P x does not have any rational zeros.
SECTION 3.4 Real Zeros of Polynomials
63
88. P x 2x 4 x 3 x 2. The only possible rational zeros of P x are 1, 2, 12 . 1 2
2
1
2 2
1
1
2 1 2 5 2
1
0
0
0
0
1
1 2
2
0
3
0
1
2
3
3
2
3
2 12
All nonnegative x 12 is an upper bound.
4
Alternating signs x 1 is a lower bound. 2 2
Therefore, there is no rational zero.
1 1 2
0
1
2
1
12 1 2
14 7 4
1
89. P x 3x 3 x 2 6x 12 has possible rational zeros 1, 2, 3, 4, 6, 12, 13 , 23 , 43 . 3 1
3
1 2
2
3
5
1
3
4
2
3
7
12
6
3 1 3 2 3 4 3 13 23 43
8
4 4
20
2
14
8
4
all positive x 2 is an upper bound alternating signs x 2 is a lower bound
Therefore, there is no rational zero.
3
1 0
6 6
16 3
3
1
3
3
3
2
16 3
3
4
3 3
5
2
2 3
12 10 76 9 28 3 124 9 44 3 100 9
90. P x x 50 5x 25 x 2 1. The only possible rational zeros of P x are 1. Since P 1 150 5 125 12 1 4 and P 1 150 5 125 12 1 6, P x does not have a rational zero.
91. P x x 3 3x 2 4x 12, [4 4] by [15 15]. The
possible rational zeros are 1, 2, 3, 4, 6, 12. By
observing the graph of P, the rational zeros are x 2, 2, 3.
92. P x x 4 5x 2 4, [4 4] by [30 30].
The possible rational solutions are 1, 2, 4.
By observing the graph of the equation, the solutions of the given equation are x 1, 2.
10
4
2
20
2 10
4
4
2
2 20
4
64
CHAPTER 3 Polynomial and Rational Functions
93. P x 2x 4 5x 3 14x 2 5x 12, [2 5] by
94. P x 3x 3 8x 2 5x 2, [3 3] by [10 10]
[40 40]. The possible rational zeros are 1, 2, 3,
The possible rational solutions are 1, 2, 13 , 23 . By
4, 6, 12, 12 , 32 . By observing the graph of P, the
observing the graph of the equation, the only real solution of the given equation is x 2.
zeros are 32 , 1, 1, 4.
10
40 20 2
2
20
2
4
2 10
40
95. x 4 x 4 0. Possible rational solutions are 1, 2, 4. 1
1 1
1
0
0
1
1
1
1
1
0
1
1
1
1
4
0
4
1
0 1
1
0
0
2
4
1
2
1
0
1
2
0
1
1
2
4
1
2
2
2
2
2
1
4
4
7
10
0
1
4
8
4
14
18
8
4
x 2 is an upper bound.
14
9
x 2 is a lower bound.
Therefore, we graph the function P x x 4 x 4 in the viewing rectangle [2 2] by [5 20] and see there are two
solutions. In the viewing rectangle [13 125] by [01 01], we find the solution x 128. In the viewing rectangle
[1516] by [01 01], we find the solution x 153. Thus the solutions are x 128, 153. 20 10 1.30 2
1
1
1.28
1.26
2
0.1
0.1
0.0
0.0
0.1
0.1
1.55
1.60
96. 2x 3 8x 2 9x 9 0. Possible rational solutions are 1, 3, 9, 12 , 32 , 92 . 1
2
8
2
2
6
9 6 3
9
3
2
3
6
6
2
9
8
6
2
3
9 9 0
x 3 is a zero.
We graph P x 2x 3 8x 2 9x 9 in the viewing rectangle [4 6] by
[40 40]. It appears that the equation has no other real solution. We can factor 2x 3 8x 2 9x 9 x 3 2x 2 2x 3 . Since the quotient is a quadratic expression, we can use the Quadratic Formula to locate the other possible 2 423 , which are not real. So the only solution is x 3. solutions: x 2 222
40 20 4
2 20 40
2
4
6
SECTION 3.4 Real Zeros of Polynomials
65
97. 400x 4 400x 3 1096x 2 588x 909 0. 1 4 4 1096 588 4
4 8 2 4
909
8 296 884
296 884
025
4 1096 588
909
8
8
4 4
2 4
8
4 12 3 4
592 008
004
296
4 1096 588
901
4
24 2608 4040
1304
4 1096 12 8
909
202 4949 x 2 is an upper bound. 588
24 3912
1304
909 135
45 14409 x 3 is a lower bound.
Therefore, we graph the function P x 400x 4 400x 3 1096x 2 588x 909 in the viewing rectangle [3 2] by [10 40]. There appear to be two solutions. In the viewing rectangle [16 14] by [01 01], we find the solution x 150. In the viewing rectangle [08 12] by [0 1], we see that the graph comes close but does not go through the xaxis. Thus there is no solution here. Therefore, the only solution is x 150. 40
0.1
20 1.6 3
2
1
1
1.0 0.5
0.0
1.5
0.0
2
0.1
0.8
0.9
1.0
1.1
1.2
98. x 5 2x 4 096x 3 5x 2 10x 48 0. Since all the coefficients are positive, there is no positive solution. So x 0 is an upper bound. 2 1 1
2 096 2
5
10
48
0 192 616 768
0 096
308
3 1
384 288
2 096 3
5
10
48
3 1188 2064 9192
1 1 396
688 3064 8712 x 3 is a lower bound.
Therefore, we graph P x x 5 2x 4 096x 3 5x 2 10x 48 in the viewing rectangle [3 0] by [10 5] and see that there are three possible solutions. In the viewing rectangle [175 17] by [01 01], we find the solution x 171. In the viewing rectangle [125 115] by [01 01], we find the solution x 120. In the viewing
rectangle [085 075] by [01 01], we find the solution x 080. So the solutions are x 171, 120, 080. 0.1 2
0.1
0.1
0 1.74 10
1.72
0.0 0.1
1.25
1.20
0.0 0.1
0.85
0.80
0.0 0.1
66
CHAPTER 3 Polynomial and Rational Functions
99. Let r be the radius of the silo. The volume of the hemispherical roof is 12 43 r 3 23 r 3 . The volume of the cylindrical section is r 2 30 30r 2 . Because the total volume of the silo is 15,000 ft3 , we get the following equation: 2 r 3 30r 2 15000 2 r 3 30r 2 15000 0 r 3 45r 2 22500 0. Using a graphing device, we 3 3
first graph the polynomial in the viewing rectangle [0 15] by [10000 10000]. The solution, r 1128 ft., is shown in the viewing rectangle [112 114] by [1 1]. 1
10000
0
11.3
0
11.4
1
10
10000
100. The volume of the box is V 1500 x 20 2x 40 2x 4x 3 120x 2 800x 4x 3 120x 2 800x 1500 4 x 3 30x 2 200x 375 0. Clearly, we must have 20 2x 0, and so 0 x 10. 5
1
30 5
200
375
125
375
1 25 75 0 x 5 is a zero. x 3 30x 2 200x 375 x 5 x 2 25x 75 0. Using the Quadratic Formula, we find the other zeros:
252 325 2552 13 . Since 2552 13 10, the two answers are: x height 5 cm, x 25 6254175 2
width 20 2 5 10 cm, and length 40 2 5 30 cm; and x height 2552 13 349 cm, width 20 25 5 13 5 13 5 1303 cm, and length 40 25 5 13 15 5 13 3303 cm. 101. Let r be the radius of the cone and cylinder and let h be the height of the cone. Since the height and diameter are equal, we get h 2r. So the volume of the
cylinder is V1 r 2 cylinder height 20r 2 , and the volume of the cone is V2 13 r 2 h 13 r 2 2r 23 r 3 . Since the total volume is
500 , it follows 3
500 r 3 30r 2 250 0. By Descartes’ Rule of that 23 r 3 20r 2 3 Signs, there is 1 positive zero. Since r is between 276 and 2765 (see the table), the radius should be 276 m (correct to two decimals).
r 1 2 3 27 276 277
r 3 30r 2 250 219 122
47
1162 233
144
2765
144
28
715
102. (a) Let x be the length, in ft, of each side of the base and let h be the height. The volume of the box is V 2 2 hx 2 , and so hx 2 2 2. The length of the diagonal on the base is x 2 x 2 2x 2 , and hence the length of the diagonal between opposite corners is 2x 2 h 2 x 1. Squaring both sides of the equation, we have 2x 2 h 2 x 2 2x 1 h 2 x 2 2x 1 h x 2 2x 1. Therefore, 2 2 hx 2 x 2 2x 1 x 2 x 2 2x 1 x 4 8 x 6 2x 5 x 4 8 0.
SECTION 3.4 Real Zeros of Polynomials
67
(b) We graph y x 6 2x 5 x 4 8 in the viewing rectangle [0 5] by [10 10], and we see that there are two solutions. In the second viewing rectangle, [14 15] by [1 1], we see the solution x 145. The third viewing rectangle, [225 235] by [1 1], shows the solution x 231. 10
1
0
2
1
0
4
10
1.45
0
1.50
1
2.30
2.35
1
If x width length 145 ft, then height x 2 2x 1 134 ft, and if x width length 231 ft, then height x 2 2x 1 053 ft.
103. Let b be the width of the base, and let l be the length of the box. Then the length plus girth is l 4b 108, and the volume
is V lb2 2200. Solving the first equation for l and substituting this value into the second equation yields l 108 4b V 108 4b b2 2200 4b3 108b2 2200 0 4 b3 27b2 550 0. Now P b b3 27b2 550
has two variations in sign, so there are 0 or 2 positive real zeros. We also observe that since l 0 b 27, so b 27 is an upper bound. Thus the possible positive rational real zeros are 1 2 3 10 11 22 25. 1
1
27
0
550
1
1
26
26
26
26
524
1
27
5
5
2
0
550
110
550
1
27
0
550
1
2
50
100
25
50
450
1 22 110 0 b 5 is a zero. P b b 5 b2 22b 110 . The other zeros are b 22 48441110 222 924 2230397 . The positive 2 2
answer from this factor is b 2620Thus we have two possible solutions, b 5 or b 2620. If b 5, then
l 108 4 5 88; if b 2620, then l 108 4 2620 320. Thus the length of the box is either 88 in. or 320 in.
104. Given that x is the length of a side of the rectangle, we have that the length of the diagonal is x 10, and the length of the other side of the rectangle is x 102 x 2 . Hence x x 102 x 2 5000 x 2 20x 100 25,000,000 2x 3 10x 2 2,500,000 0 x 3 5x 2 1,250,000 0. The first viewing rectangle, [0 120] by [100 500],
shows there is one solution. The second viewing rectangle, [106 1061] by [01 01], shows the solution is x 10608.
Therefore, the dimensions of the rectangle are 47 ft by 106 ft.
0.1
400 200 0
0.0 50
100
106.05
106.10
0.1
105. (a) An odddegree polynomial must have a real zero. The end behavior of such a polynomial requires that the graph of the polynomial heads off in opposite directions as x and x . Thus the graph must cross the xaxis. (b) There are many possibilities one of which is P x x 4 1.
68
CHAPTER 3 Polynomial and Rational Functions
(c) P x x x 2 x 2 x 3 2x. (d) P x x 2 x 2 x 3 x 3 x 4 5x 2 6. If a polynomial with integer coefficients has no real zeros, then the polynomial must have even degree.
a for x we have 3 a 3 a 2 a x 3 ax 2 bx c u c a u b u 3 3 3 a3 a2 ab a2 2a 3 2 2 u au u a u u bu c 3 27 3 9 3
106. (a) Substituting u
a3 a3 ab a2 2a 2 u au 2 u bu c 3 27 3 9 3 2a 2 a3 a3 ab a2 3 2 b u c u a a u 3 3 27 9 3 a2 2a 3 ab 3 u b u c 3 27 3 u 3 au 2
(b) x 3 6x 2 9x 4 0. Setting a 6, b 9, and c 4, we have: u 3
62 9 3
u 16 18 4 u 3 3u 2.
2 3 3 3 2 2 2 22 3 33 3 1 3 1 1 1 2 107. (a) Using the cubic formula, x 2 4 27 2 4 27 2
1
0
2
3 4
2
2
1 2 1 0 So x 2 x 2 2x 1 x 2 x 12 0 x 2, 1. Using the methods from this section, we have
1
1
0
1
3
1
2
2
1 1 2 0 So x 3 3x 2 x 1 x 2 x 2 x 12 x 2 0 x 2, 1.
Since this factors easily, the factoring method was easier.
(b) Using the cubic formula, 2 3 3 3 4 3 33 4 42 4 x 2 4 27 2 4 27 3 3 3 3 2 4 1 2 4 1 2 5 2 5
SECTION 3.4 Real Zeros of Polynomials
Using a graphing calculator, we see that P x x 3 3x 4 has one zero. Using methods from this section, P x has possible rational zeros 1, 2, 4. 1
1
1
0
3
4
1
1
4
1
4
4
1
1
0
3
4
1
1
4
69
1 is an upper bound. 1 1 4 0 x 1 is a zero. P x x 3 3x 4 x 1 x 2 x 4 . Using the Quadratic Formula we have
1 12 414 1 215 which is not a real number. Since it is not easy to see that x 2
3 3 2 5 2 5 1, we see that the factoring method was much easier.
108. (a) Since z b, we have z b 0. Since all the coefficients of Q x are nonnegative, and since z 0, we have Q z 0 (being a sum of positive terms). Thus, P z z b Q z r 0, since the sum of a positive number and a nonnegative number. (b) In part (a), we showed that if b satisfies the conditions of the first part of the Upper and Lower Bounds Theorem and z b, then P z 0. This means that no real zero of P can be larger than b, so b is an upper bound for the real zeros. (c) Suppose b is a negative lower bound for the real zeros of P x. Then clearly b is an upper bound for P1 x P x. Thus, as in Part (a), we can write P1 x x b Q x r, where r 0 and the coefficients of Q are all nonnegative, and P x P1 x x b Q x r x b [Q x] r . Since the coefficients of Q x are all nonnegative, the coefficients of Q x will be alternately nonpositive and nonnegative, which proves the second part of the Upper and Lower Bounds Theorem. 109. P x x 5 x 4 x 3 5x 2 12x 6 has possible rational zeros 1, 2, 3, 6. Since P x has 1 variation in sign,
there is 1 positive real zero. Since P x x 5 x 4 x 3 5x 2 12x 6 has 4 variations in sign, there are 0, 2, or 4 negative real zeros. 1
1
1
1
1
0
1
0
1
5
12
6
1
6
18
6
18
24
2
1
6 15
1 2
1
3 1 1 1 5 12 6 3
1
1
1
1
5 2
3
12
6
6
36
18
42
1 1 1 1 5 12 6
30 54
1
2
2 1
5 10 18 48 3 is an upper bound. 1 2 1 6 P x x 1 x 4 2x 3 x 2 6x 6 , continuing with the quotient we have 1
2
1 1
2 1 3
1
3 4
6 4 10
6
6
6
0 x 1 is a zero.
6 10
4
1 is a lower bound.
Therefore, there is 1 rational zero, namely 1. Since there are 1, 3 or 5 real zeros, and we found 1 rational zero, there must
be 0, 2 or 4 irrational zeros. However, since 1 zero must be positive, there must be at least one irrational zero. Therefore, there is exactly 1 rational zero and there are either 2 or 4 irrational zeros.
70
CHAPTER 3 Polynomial and Rational Functions
3.5
COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA
1. The polynomial P x 5x 2 x 43 x 7 has degree 6. It has zeros 0, 4, and 7. The zero 0 has multiplicity 2, and the zero 4 has multiplicity 3. 2. (a) If a is a zero of the polynomial P, then x a must be a factor of P x.
(b) If a is a zero of multiplicity m of the polynomial P, then x am must be a factor of P x when we factor P completely.
3. A polynomial of degree n 1 has exactly n zeros, if a zero of multiplicity m is counted m times.
4. If the polynomial function P has real coefficients and if a bi is a zero of P, then a bi is also a zero of P. So if 3 i is a zero of P, then 3 i is also a zero of P.
5. P x x 4 1 (a) True, P has degree 4, so by the Zeros Theorem it has four (not necessarily distinct) complex zeros. (b) True, by the Complete Factorization Theorem, this is true.
(c) False, the fourth power of any real number is nonnegative, so P x 1 for all real x and P has no real zeros.
6. P x x 3 x
(a) False. P x x x 2 1 , and x 2 1 has no real zeros. (b) True, P 0 0.
(c) False, P x cannot be factored into linear factors with real coefficients because x 2 1 has no real zeros. 7. (a) x 4 4x 2 0 x 2 x 2 4 0. So x 0 or x 2 4 0. If x 2 4 0 then x 2 4 x 2i. Therefore, the solutions are x 0 and 2i.
(b) To get the complete factorization, we factor the remaining quadratic factor P x x 2 x 4 x 2 x 2i x 2i. 8. (a) x 5 9x 3 0 x 3 x 2 9 0. So x 0 or x 2 9 0. If x 2 9 0 then x 3i. Therefore, the zeros of P are x 0, 3i.
(b) Since 3i and 3i are the zeros from x 2 9 0, x 3i and x 3i are the factors of x 2 9. Thus the complete factorization is P x x 3 x 2 9 x 3 x 3i x 3i. 9. (a) x 3 2x 2 2x 0 x x 2 2x 2 0. So x 0 or x 2 2x 2 0. If x 2 2x 2 0 then 2 22 412 2 4 22i x 2 2 2 1 i. Therefore, the solutions are x 0, 1 i.
(b) Since 1 i and 1 i are zeros, x 1 i x 1 i and x 1 i x 1 i are the factors of x 2 2x 2. Thus the complete factorization is P x x x 2 2x 2 x x 1 i x 1 i. 10. (a) x 3 x 2 x 0 x x 2 x 1 0. So x 0 or x 2 x 1 0. If x 2 x 1 0 then
1 12 411 12 3 12 i 23 . Therefore, the zeros of P are x 0, 12 i 23 . x 21 (b) The zeros of x 2 x 1 0 are 12 i 23 and 12 i 23 , so factoring we get x 2 x x 12 i 23 x 12 i 23 x 12 i 23 . Thus the complete factorization is 1 x 12 i 23 P x x x 2 x 1 x x 12 i 23 x 12 i 23 .
2 11. (a) x 4 2x 2 1 0 x 2 1 0 x 2 1 0 x 2 1 x i. Therefore the zeros of P are x i.
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
71
(b) Since i and i are zeros, x i and x i are the factors of x 2 1. Thus the complete factorization is 2 P x x 2 1 [x i x i]2 x i2 x i2 .
12. (a) x 4 x 2 2 0 x 2 2 x 2 1 0. So x 2 2 0 or x 2 1 0. If x 2 2 0 then x 2 2 x 2. And if x 2 1 0 then x 2 1 x i. Therefore, the zeros of P are x 2, i. (b) To get the complete factorization, we factor the quadratic factors to get P x x 2 2 x 2 1 x 2 x 2 x i x i.
13. (a) x 4 16 0 0 x 2 4 x 2 4 x 2 x 2 x 2 4 . So x 2 or x 2 4 0. If x 2 4 0 then x 2 4 x 2i. Therefore the zeros of P are x 2, 2i.
(b) Since i and i are zeros, x i and x i are the factors of x 2 1. Thus the complete factorization is P x x 2 x 2 x 2 4 x 2 x 2 x 2i x 2i. 2 14. (a) x 4 6x 2 9 0 x 2 3 0 x 2 3. So x i 3 are the only zeros of P (each of multiplicity 2). (b) To get the complete factorization, we factor the quadratic factor to get 2 2 2 2 x i 3 . x i 3 P x x 2 3 x i 3 x i 3
15. (a) x 3 8 0 x 2 x 2 2x 4 0. So x 2 or x 2 2x 4 0. If x 2 2x 4 0 then
2 22 414 2 212 22i2 3 1i 3. Therefore, the zeros of P are x 2, 1 i 3. x 2
(b) Since 1 i 3 and 1 i 3 are the zeros from the x 2 2x 4 0, x 1 i 3 and x 1 i 3 are the factors of x 2 2x 4. Thus the complete factorization is P x x 2 x 2 2x 4 x 2 x 1 i 3 x 1i 3 x 2 x 1 i 3 x 1 i 3
16. (a) x 3 8 0 x 2 x 2 2x 4 0. So x 2 or x 2 2x 4 0. If x 2 2x 4 0 then
2 22 414 3 1i 3. Therefore, the zeros of P are x 2, 1 i 3. 22 12 22i x 2 2
(b) Since 1 i 3 and 1 i 3 are the zeros from x 2 2x 4 0, x 1 i 3 and x 1 i 3 are the factors of x 2 2x 4. Thus the complete factorization is P x x 2 x 2 2x 4 x 2 x 1 i 3 x 1 i 3 x 2 x 1 i 3 x 1 i 3
17. (a) x 6 1 0 0 x 3 1 x 3 1 x 1 x 2 x 1 x 1 x 2 x 1 . Clearly, x 1 are solutions.
If x 2 x 1 0, then x 1 1411 12 3 12 23 so x 12 i 23 . And if x 2 x 1 0, then 2
1 2 3 12 x 1 1411 2
3 1 i 3 . Therefore, the zeros of P are x 1, 1 i 3 , 1 i 3 . 2 2 2 2 2 2 2
72
CHAPTER 3 Polynomial and Rational Functions
(b) The zeros of x 2 x 1 0 are 12 i 23 and 12 i 23 , so x 2 x 1 factors as x 12 i 23 x 12 i 23 x 12 i 23 . Similarly,since x 12 i 23
the zeros of x 2 x 1 0 are 12 i 23 and 12 i 23 , so x 2 x 1 factors as 1 x 12 i 23 x 12 i 23 x 12 i 23 . Thus the complete i 23 x 2 factorization is P x x 1 x 2 x 1 x 1 x 2 x 1
x 1 x 1 x 12 i 23 x 12 i 23 x 12 i 23 x 12 i 23
18. (a) x 6 7x 3 8 0 0 x 3 8 x 3 1 x 2 x 2 2x 4 x 1 x 2 x 1 . Clearly, x 1 and x 2 are solutions. If x 2 2x 4 0, then x 2 4414 22 12 22 2 23 so x 1 i 3. If 2
1 2 3 12 23 12 i 23 . Therefore, the zeros of P are x 1, 2, x 2 x 1 0, then x 1 1411 2 1 i 3, 12 i 23 . (b) From Exercise 10, x 2 2x 4 x 1 i 3 x 1 i 3 and from Exercise 11, x 2 x 1 x 12 i 23 x 12 i 23 . Thus the complete factorization is P x x 2 x 2 2x 4 x 1 x 2 x 1
x 2 x 1 x 1 i 3 x 1 i 3 x 12 i 23 x 12 i 23
19. P x x 4 16x 2 x 2 x 2 16 x 2 x 4i x 4i. The zeros of P are 0 (multiplicity 2) and 4i (multiplicity 1). 20. P x 9x 6 16x 4 x 4 9x 2 16 x 4 3x 4i 3x 4i. The zeros of P are 0 (multiplicity 4) and 43 i (multiplicity 1).
21. Q x x 6 2x 5 2x 4 x 4 x 2 2x 2 . Using the Quadratic Formula to solve x 2 2x 2 0, we 22 412 1 i. The zeros of Q are 0 (multiplicity 4) and 1 i (multiplicity 1), and obtain x 2 21
Q x x 4 x 1 i x 1 i. 22. Q x x 5 x 4 x 3 x 3 x 2 x 1 . Using the Quadratic Formula to solve x 2 x 1 0, we obtain 12 411 12 i 23 . The zeros of Q are 0 (multiplicity 3) and 12 i 23 (multiplicity 1); and x 1 21 Q x x 3 x 12 i 23 x 12 i 23 . 23. P x x 3 4x x x 2 4 x x 2i x 2i. The zeros of P are 0, 2i, and 2i, all of multiplicity 1. 24. P x x 3 x 2 x x x 2 x 1 . Using the Quadratic Formula, we have
1 12 411 1 2 3 12 i 23 . The zeros of P are 0, 12 i 23 , and 12 i 23 , all of multiplicity 1; and x 21 P x x x 12 i 23 x 12 i 23 .
25. Q x x 4 1 x 2 1 x 2 1 x 1 x 1 x 2 1 x 1 x 1 x i x i. The zeros of Q are 1, 1, i, and i, all of multiplicity 1.
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
73
26. Q x x 4 625 x 2 25 x 2 25 x 5 x 5 x 2 25 x 5 x 5 x 5i x 5i. The zeros
of Q are 5, 5, 5i, and 5i, all of multiplicity 1. 27. P x 16x 4 81 4x 2 9 4x 2 9 2x 3 2x 3 2x 3i 2x 3i. The zeros of P are 32 , 32 , 32 i, and 32 i, all of multiplicity 1.
28. P x x 3 64 x 4 x 2 4x 16 .
Using the Quadratic Formula, we have
4 48 44i 3 2 2i 3. The zeros of P are 4, 2 2i 3, and 2 2i 3, all of x 4 164116 2 2 2
multiplicity 1; and P x x 4 x 2 2i 3 x 2 2i 3 . 29. P x x 3 x 2 9x 9 x 2 x 1 9 x 1 x 1 x 2 9 x 1 x 3i x 3i. The zeros of P are 1, 3i, and 3i, all of multiplicity 1. 30. P x x 6 729 x 3 27 x 3 27 x 3 x 2 3x 9 x 3 x 2 3x 9 . Using the
27 2 . Using the 3 227 32 27 32 3 2 3 i Quadratic Formula for x 2 3x 9, we obtain x 3 9419 2 2 The zeros of P are 3, 3, 32 3 2 3 i, 32 3 2 3 i, 32 3 2 3 i, and 32 3 2 3 i, all of multiplicity 1; and P x x 3 x 3 x 32 3 2 3 i x 32 3 2 3 i x 32 3 2 3 i x 32 3 2 3 i .
Quadratic Formula for x 2 3x 9, we obtain x 3 9419 32 27 32 2
2 31. P x x 6 10x 4 25x 2 x 2 x 4 10x 2 25 x 2 x 2 5 . The zeros of P are 0 (multiplicity 2) and 5i 2 2 x 5i . (multiplicity 2); and P x x 2 x 5i 2 32. P x x 5 18x 3 81x x x 4 18x 2 81 x x 2 9 . The zeros of P are 0 (multiplicity 1) and 3i
(multiplicity 2); and P x x x 3i2 x 3i2 . 33. P x x 4 3x 2 4 x 2 1 x 2 4 x 1 x 1 x 2i x 2i. The zeros of P are 1, 1, 2i, and 2i (all of multiplicity 1).
34. P x x 5 7x 3 x 3 x 2 7 x 3 x i 7 x i 7 . The zeros of P are 0 (multiplicity 3), i 7, and i 7, both of multiplicity 1.
2 2 2 x i 3 . The zeros of P are 0 35. P x x 5 6x 3 9x x x 4 6x 2 9 x x 2 3 x x i 3 (multiplicity 1), i 3 (multiplicity 2), and i 3 (multiplicity 2). 2 2 36. P x x 6 16x 3 64 x 3 8 x 22 x 2 2x 4 . Using the Quadratic Formula, on x 2 2x 4 we
2 22 414 2 212 22i2 3 1 i 3. The zeros of P are 2, 1 i 3, and 1 i 3, all of have x 2 2 2 multiplicity 2; and P x x 22 x 1 i 3 x 1i 3 .
37. Since 1 i and 1 i are conjugates, the factorization of the polynomial must be P x a x [1 i] x [1 i] a x 2 2x 2 . If we let a 1, we get P x x 2 2x 2.
38. Since 1 i 2 and 1 i 2 are conjugates, the factorization of the polynomial must be x 1 i 2 c x 2 2x 3 . If we let c 1, we get P x x 2 2x 3. P x c x 1 i 2
74
CHAPTER 3 Polynomial and Rational Functions
39. Since 2i and 2i are conjugates, the factorization of the polynomial must be Q x b x 3 x 2i x 2i] b x 3 x 2 4 b x 3 3x 2 4x 12 . If we let b 1, we get Q x x 3 3x 2 4x 12.
40. Since i is a zero, by the Conjugate Roots Theorem, i is also a zero. So the factorization of the polynomial must be Q x b x 0 x i x i bx x 2 1 b x 3 x . If we let b 1, we get Q x x 3 x.
41. Since i is a zero, by the Conjugate Roots Theorem, i is also a zero. So the factorization of the polynomial must be P x a x 2 x i x i a x 3 2x 2 x 2 . If we let a 1, we get P x x 3 2x 2 x 2.
42. Since 1 i is a zero, by the Conjugate Roots Theorem, 1 i is also a zero. So the factorization of the polynomial must be Q x a x 3 x [1 i] x [1 i] a x 3 x 2 2x 2 a x 3 x 2 4x 6 . If we let a 1, we get Q x x 3 x 2 4x 6.
43. Since the zeros are 1 2i and 1 (with multiplicity 2), by the Conjugate Roots Theorem, the other zero is 1 2i. So a factorization is R x c x [1 2i] x [1 2i] x 12 c [x 1] 2i [x 1] 2i x 12 c [x 1]2 [2i]2 x 2 2x 1 c x 2 2x 1 4 x 2 2x 1 c x 2 2x 5 x 2 2x 1 c x 4 2x 3 x 2 2x 3 4x 2 2x 5x 2 10x 5 c x 4 4x 3 10x 2 12x 5 If we let c 1 we get R x x 4 4x 3 10x 2 12x 5.
44. Since S x has zeros 2i and 3i, by the Conjugate Roots Theorem, the other zeros of S x are 2i and 3i. So a factorization of S x is S x C x 2i x 2i x 3i x 3i C x 2 4i 2 x 2 9i 2 C x 2 4 x 2 9 C x 4 13x 2 36 If we let C 1, we get S x x 4 13x 2 36.
45. Since the zeros are i and 1 i, by the Conjugate Roots Theorem, the other zeros are i and 1 i. So a factorization is T x C x i x i x [1 i] x [1 i] C x 2 i 2 [x 1] i [x 1] i C x 2 1 x 2 2x 1 i 2 C x 2 1 x 2 2x 2 C x 4 2x 3 2x 2 x 2 2x 2 C x 4 2x 3 3x 2 2x 2 C x 4 2C x 3 3C x 2 2C x 2C Since the constant coefficient is 12, it follows that 2C 12 C 6, and so T x 6 x 4 2x 3 3x 2 2x 2 6x 4 12x 3 18x 2 12x 12.
46. Since U x has zeros 12 , 1 (with multiplicity two), and i, by the Conjugate Roots Theorem, the other zero is i. So a factorization of U x is U x c x 12 x 12 x i x i 12 c 2x 1 x 2 2x 1 x 2 1 12 c 2x 5 3x 4 2x 3 2x 2 1 Since the leading coefficient is 4, we have 4 12 c 2 c. Thus we have U x 12 4 2x 5 3x 4 2x 3 2x 2 1 4x 5 6x 4 4x 3 4x 2 2.
47. P x x 3 2x 4 x 2 x 2 2x 2 , so x 2 or x 2 2x 2 0. Using the Quadratic Formula, we have 22 412 1 i. Thus the zeros are 2 and 1 i. x 2 21
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
48. P x x 3 7x 2 16x 10. We start by trying the possible rational factors of the polynomial: 1
1
7 1
1
6
16
10 10
6
10
0
x 1 is a zero.
So P x x 1 x 2 6x 10 . Using the Quadratic Formula on the second factor, we have 3 i. Thus the zeros are 1 and 3 i. x 6 364110 2 49. P x x 3 2x 2 2x 1. By inspection, P 1 1 2 2 1 0, and hence x 1 is a zero. 1
1
2
1
1
1
2
1 1
1
1
0
Thus P x x 1 x 2 x 1 . So x 1 or x 2 x 1 0.
Using the Quadratic Formula, we have x 1 1411 1i2 3 . Hence, the zeros are 1 and 1i2 3 . 2 50. P x x 3 7x 2 18x 18 has possible rational zeros 1, 2, 3, 6, 9, 18. Since all of the coefficients are positive, there are no positive real zeros. 1 1
7 18
18
2 1
1 6 12
7
18
18
3 1
2 10 16
7
18
18
3 12 18
1 5 8 2 1 4 6 0 x 3 is a zero. 6 12 6 So P x x 3 x 2 4x 6 . Using the Quadratic Formula on the second factor, we have 2 2 i 2. Thus the zeros are 3, 2 i 2. x 4 16416 42 8 42i 2 2 1
51. P x x 3 3x 2 3x 2.
2
1
3 2
1
1
3
2 2
2
1
0
Thus P x x 2 x 2 x 1 . So x 2 or x 2 x 1 0 Using the Quadratic Formula we have x 1 1411 1i2 3 . Hence, the zeros are 2, and 1i2 3 . 2 52. P x x 3 x 6 has possible zeros 1, 2, 3. 1
1
0
1
1
1
6
0
2
1
0 2
1
4
6
6
1 2 3 0 x 2 is a zero. 0 6 2 4413 2 1 i 2. Thus the 22i P x x 2 x 2 2x 3 . Now x 2 2x 3 has zeros x 2 2 zeros are 2, 1 i 2. 1
1
75
76
CHAPTER 3 Polynomial and Rational Functions
53. P x 2x 3 7x 2 12x 9 has possible rational zeros 1, 3, 9, 12 , 32 , 92 . Since all coefficients are positive, there are no positive real zeros. 2
1
2
7
12
9
2
5
7
5
7
There is a zero between 1 and 2.
32
2
2
2
2
2
7
12
9
3
6
9
7
12
9
4
6
12
3
6
3
2 4 6 0 x 32 is a zero. P x x 32 2x 2 4x 6 2 x 32 x 2 2x 3 . Now x 2 2x 3 has zeros 22 2 x 2 4431 1 i 2. Hence, the zeros are 32 and 1 i 2. 2 2
54. Using synthetic division, we see that x 3 is a factor of the polynomial: 1
2
9
8
2
2
3
9
9
8
3
6
6
9 9
6
2 2 3 0 x 3 is a zero. So P x 2x 3 8x 2 9x 9 x 3 2x 2 2x 3 . Using the Quadratic Formula, we find the other two solutions: 2 4 4 3 2 2 20 x 12 25 i. Thus the zeros are 3, 12 25 i. 2 2 4 2
3
6
6
55. P x x 4 x 3 7x 2 9x 18. Since P x has one change in sign, we are guaranteed a positive zero, and since P x x 4 x 3 7x 2 9x 18, there are 1 or 3 negative zeros. 1
1
1
7
9
1
2
9
18
18
1 2 9 18 0 Therefore, P x x 1 x 3 2x 2 9x 18 . Continuing with the quotient, we try negative zeros.
1
1
2
9
18
1
1
8
2
1
2
1
8
10
0
9
18
0
18
0 P x x 1 x 2 x 2 9 x 1 x 2 x 3i x 3i. Therefore,the zeros are 1, 2, and 3i. 1
1
2
9
56. P x x 4 2x 3 2x 2 2x 3 has possible zeros 1, 3. 1
1
2
1
2
2
3
1
3
5
3
1
2 3
2 3
2 3
3 3
1 1 1 1 0 x 3 is a zero. 5 8 P x x 3 x 3 x 2 x 1 . If we factor the second factor by grouping, we get x 3 x 2 x 1 x 2 x 1 1 x 1 x 1 x 2 1 . So we have P x x 3 x 1 x 2 1 x 3 x 1 x i x i. Thus the zeros are 3, 1, i, and i. 1
1
3
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
77
57. We see a pattern and use it to factor by grouping. This gives P x x 5 x 4 7x 3 7x 2 12x 12 x 4 x 1 7x 2 x 1 12 x 1 x 1 x 4 7x 2 12 x 1 x 2 3 x 2 4 x 1 x i 3 x i 3 x 2i x 2i
Therefore,the zeros are 1, i 3, and 2i.
58. P x x 5 x 3 8x 2 8 x 3 x 2 1 8 x 2 1 x 2 1 x 3 8 x 2 1 x 2 x 2 2x 4
(factoring a sum of cubes). So x 2, or x 2 1 0. If x 2 1 0, then x 2 1 x i. If x 2 2x 4 0, then 12 x 2 4414 1 1 i 3. Thus, the zeros are 2, i, 1 i 3. 2 2
59. P x x 4 6x 3 13x 2 24x 36 has possible rational zeros 1, 2, 3, 4, 6, 9, 12, 18. P x has 4 variations in sign and P x has no variation in sign. 1 1 6 13 24 1 5
1 5
8 16
Continuing:
36
2 1 6 13 24
8 16
2 8
20
1 4
36
3 1 6 13 24
10 28
5 14
3 9
8
1 3
36
12 36
4 12
0 x 3 is a zero.
3 1 3 4 12 3 0
1
12
0 4
0 x 3 is a zero. P x x 32 x 2 4 x 32 x 2i x 2i. Therefore,the zeros are 3 (multiplicity 2) and 2i.
60. P x x 4 x 2 2x 2 has possible rational zeros 1, 2. 1 1 0 1 2 2 1
1 1
1 0 2
0 1 2
1 1 1 0
1 0 2
1
1 1
2
2
1 2 2
0 2 4 1 is an upper bound. 1 1 0 2 0 1 2 2 0 P x x 12 x 2 2x 2 . Using the Quadratic Formula on x 2 2x 2, we have x 2 248 22i 2 1 i.
Thus, the zeros of P x are 1 (multiplicity 2) and 1 i.
61. P x 4x 4 4x 3 5x 2 4x 1 has possible rational zeros 1, 12 , 14 . Since there is no variation in sign, all real zeros (if there are any) are negative. 1
4
4
5
4
1
4
0
5
1
12
4
12
4 0 5 1 2 P x x 12 4x 3 2x 2 4x 2 . Continuing:
4
4 4
2
4
2
2
0
2
0
4
0
4
5
4
1
2
1
2
1
2
4
2
0
x 12 is a zero again.
2 P x x 12 4x 2 4 . Thus, the zeros of P x are 12 (multiplicity 2) and i.
x 12 is a zero.
78
CHAPTER 3 Polynomial and Rational Functions
62. P x 4x 4 2x 3 2x 2 3x 1 has possible rational zeros 1, 12 , 14 . P has one variation in sign, so P has one positive real zero. 4
1
2
2
3
1
4 6 4 P x x 1 4x 3 6x 2 4x 1 . Continuing:
1
0
12
4
4
4
1
6
4
1
4
2
2
6
4
1 1 is a zero. 6
4
1
2
2
1
2 2 3 4 4 2 0 12 is a zero. P x x 1 x 12 4x 2 4x 2 . Using the Quadratic Formula on 4x 2 4x 2, we find 4
x 4 81632 12 12 i. Thus, P has zeros 1, 12 , 12 12 i.
63. P x x 5 3x 4 12x 3 28x 2 27x 9 has possible rational zeros 1, 3, 9. P x has 4 variations in sign and P x has 1 variation in sign.
1
1
12
3 1
1 1
1
10
2
1
9
10
2
10
18
9
18
9 9
0
x 1 is a zero.
1
1
9
18
1
1
2
27
28
9
1 1
9
0
9
9
1 9 9 0 x 1 is a zero. 1 0 9 0 x 1 is a zero. P x x 13 x 2 9 x 13 x 3i x 3i. Therefore,the zeros are 1 (multiplicity 3) and 3i
64. P x x 5 2x 4 2x 3 4x 2 x 2 has possible rational zeros 1, 2. 1
1
1
1
2
2
1
1
2
3
2
4
1
2
1
2 2
2
4
0
4
1 0
2
2
1 1 3 2 4 1 0 2 0 1 0 x 2 is a zero. 2 P x x 2 x 4 2x 2 1 x 2 x 2 1 x 2 x i2 x i2 . Thus, the zeros of P x are 2, i. 65. (a) P x x 3 5x 2 4x 20 x 2 x 5 4 x 5 x 5 x 2 4 (b) P x x 5 x 2i x 2i
66. (a) P x x 3 2x 4
1
1
0 1
1
1
2
1
1
4
2
1 5
P x x 3 2x 4 x 2 x 2 2x 2
(b) P x x 2 x 1 i x 1 i 67. (a) P x x 4 8x 2 9 x 2 1 x 2 9 x 1 x 1 x 2 9 (b) P x x 1 x 1 x 3i x 3i
1
0 2
1
2
2
4
2
0
4
4
79
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
2 68. (a) P x x 4 8x 2 16 x 2 4
(b) P x x 2i2 x 2i2 69. (a) P x x 6 64 x 3 8 x 3 8 x 2 x 2 2x 4 x 2 x 2 2x 4 (b) P x x 2 x 2 x 1 i 3 x 1 i 3 x 1 i 3 x 1 i 3 70. (a) P x x 5 16x x x 4 16 x x 2 4 x 2 4 x x 2 x 2 x 2 4 (b) P x x x 2 x 2 x 2i x 2i 71. (a) x 4 2x 3 11x 2 12x x x 3 2x 2 11x 12 0. We first find the bounds for our viewing rectangle. 1 5
1
4
1
11 4
32
6
13
50
3
5
12
2
20
x 5 is an upper bound.
40
x 4 is a lower bound.
We graph P x x 4 2x 3 11x 2 12x in the viewing rectangle [4 5] by [50 10] and see that it has 4 real solutions. Since this matches the degree of P x, P x has no nonreal solution. (b) x 4 2x 3 11x 2 12x 5 0. We use the same bounds for our viewing rectangle, [4 5] by [50 10],
(c) x 4 2x 3 11x 2 12x 40 0. We graph
T x x 4 2x 3 11x 2 12x 40 in the viewing
and see that R x x 4 2x 3 11x 2 12x 5 has
rectangle [4 5] by [10 50], and see that T has no
2 real solutions. Since the degree of R x is 4, R x
real solution. Since the degree of T is 4, T must have
must have 2 nonreal solutions.
4 nonreal solutions.
5 20
40 20
40
5
72. (a) 2x 4i 1 2x 1 4i x 12 2i. (b) x 2 i x 0 x x i 0 x 0, i.
(c) x 2 2i x 1 0 x i2 0 x i.
(d) i x 2 2x i 0. Using the Quadratic Formula, we get 2 22 4ii 2 1 2 1 2 i 1 2 i. x 22i 8 22 i 2i 2i
73. (a) P x x 2 1 i x 2 2i. So P 2i 2i2 1 i 2i 2 2i 4 2i 2 2 2i 0, and P 1 i 1 i2 1 i 1 i 2 2i 1 2i 1 1 1 2 2i 0.
Therefore, 2i and 1 i are solutions of the equation x 2 1 i x 2 2i 0. However, P 2i 2i2 1 i 2i 2 2i 4 2i 2 2 2i 4 4i, and
P 1 i 1 i2 1 i 1 i 2 2i 2 2i. Since, P 2i 0 and P 1 i 0, 2i and 1 i are not solutions. (b) This does not violate the Conjugate Roots Theorem because the coefficients of the polynomial P x are not all real.
80
CHAPTER 3 Polynomial and Rational Functions
74. (a) Because i and 1 i are zeros, i and 1 i are also zeros. Thus,
P x C x i x i x [1 i] x [1 i] C x 2 1 x 2 2x 2 C x 4 2x 3 2x 2 x 2 2x 2 C x 4 2x 3 3x 2 2x 2
Because C 1, the polynomial is P x x 4 2x 3 3x 2 2x 2.
(b) Because i and 1 i are zeros,
P x C x i x [i 1] C x 2 xi x xi 1 i C x 2 1 2i x 1 i
Because C 1, the polynomial is P x x 2 1 2i x 1 i.
75. Because P has real coefficients, the imaginary zeros come in pairs: a bi (by the Conjugate Roots Theorem), where b 0. Thus there must be an even number of nonreal zeros. Since P is of odd degree, it has an odd number of zeros (counting multiplicity). It follows that P has at least one real zero. 76. P x 2x 4 3x 3 6x 2 12x 8 50
From the graph, we see that P has real zeros 12 and 2. Thus P x 2x 1 x 2 x 2 4 has complex zeros x 2i, and so P x 2x 1 x 2 x 2i x 2i.
2
2
4
50
3.6
RATIONAL FUNCTIONS
1. If the rational function y r x has the vertical asymptote x 2, then as x 2 , either y or y . 2. If the rational function y r x has the horizontal asymptote y 2, then y 2 as x . 3. The function r x
x 1 x 2 has xintercepts 1 and 2. x 2 x 3
4. The function r has yintercept 13 .
5. The function r has vertical asymptotes x 2 and x 3.
6. The function r has horizontal asymptote y 1.
7. The graph of s x r x 1 can be obtained by shifting the graph of r x to the right 1 unit, so its vertical asymptote is also shifted to the right 1 unit (to x 3). The horizontal asymptote is unchanged at y 4.
8. The graph of t x r x 5 can be obtained by shifting the graph of r x downward 5 units, so its horizontal asymptote is also shifted downward 5 units (to y 1). Its vertical asymptote is unchanged at x 2. x2 x x x 1 x for x 1. 2 x 2 x 1 2x 4 x 1 2x 4 (a) True, r has vertical asymptote x 2.
9. r x
(b) False. r does not have vertical asymptote x 1. It has a “hole” at 1 16 , because r is not defined at x 1.
(c) False, r has horizontal asymptote y 12 but not horizontal asymptote y 1. (d) True, r has horizontal asymptote y 12 .
SECTION 3.6 Rational Functions
x2 x 10. True, the graph of a rational function may cross a horizontal asymptote. For example, r x 2 crosses its x x 1 horizontal asymptote y 1 at the point 12 1 . x 11. r x x 2 (a)
x
r x
x
r x
x
r x
x
r x
15
3
25
5
10
125
0833
21
21
50
1042
10
201
201
100
1020
2001
2001
1000
1002
19 199 1999
19 199 1999
0962
50 100
0980
1000
0998
(b) r x as x 2 and r x as x 2 .
(c) r has horizontal asymptote y 1. 4x 1 12. r x x 2 (a) x
r x
x
r x
x
r x
x
r x
15
14
25
22
10
5125
325
21
94
50
4188
10
201
904
100
4092
2001
9004
1000
4009
19 199 1999
86 896 8996
50 100
3827 3912
1000
3991
(b) r x as x 2 and r x as x 2 .
(c) r has horizontal asymptote y 4. 3x 10 13. r x x 22 (a) x 15
r x
x
22
19 199 1999
430 40,300 4,003,000
r x
25
10
21 201 2001
370 39,700 3,997,000
x
r x
x
r x
10
03125
10
02778
50
00608
100
00302
1000
50
00592
00030
100
00298
1000
00030
x
r x
x
r x
10
209
(b) r x as x 2 and r x as x 2 . (c) r has horizontal asymptote y 0.
14. r x (a)
3x 2 1
x 22 x
r x
x
r x
15
31
25
79
10
4703
19
1183
21
1423
50
3256
199
128,803
201
131,203
100
3124
1999
12,988,003
2001
13,012,003
1000
3012
(b) r x as x 2 and r x as x 2 .
(c) r has horizontal asymptote y 3.
50 100 1000
2774 2884 2988
81
82
CHAPTER 3 Polynomial and Rational Functions
In the solutions to Exercises 15–22, let f x
1 . x
4 1 15. r x 4 4 f x 2. From this form we see that the graph x 2 x 2
y
of r is obtained from the graph of f by shifting 2 units to the right and stretching vertically by a factor of 4. Thus r has vertical asymptote x 2 and horizontal
asymptote y 0. The domain of r is 2 2 and its range is
4
1
x
0 0 .
9 9 f x 3. From this form we see that the graph of r is obtained x 3 from the graph of f by shifting 3 units to the left and stretching vertically by a
y
16. r x
factor of 9. Thus r has vertical asymptote x 3 and horizontal asymptote y 0.
10
The domain of r is 3 3 and its range is 0 0 .
1x
2 1 2 2 f x 1. From this form we see that the 17. s x x 1 x 1
y
graph of s is obtained from the graph of f by shifting 1 unit to the left, stretching vertically by a factor of 2, and reflecting about the xaxis. Thus s has vertical
asymptote x 1 and horizontal asymptote y 0. The domain of s is
2
1 1 and its range is 0 0 .
18. s x
3 1 3 3 f x 4. From this form we see that the x 4 x 4
1
x
y
graph of s is obtained from the graph of f by shifting 4 units to the right, stretching vertically by a factor of 3, and then reflecting about the xaxis. Thus s has vertical asymptote x 4 and horizontal asymptote y 0. The domain of s is 4 4 and its range is 0 0 .
1 1
x
83
SECTION 3.6 Rational Functions
1 2x 3 2 f x 2 2 (see the long x 2 x 2 division at right). From this form we see that the graph of t is
19. t x
y
2 x 2
obtained from the graph of f by shifting 2 units to the right
2x 3 2x 2 1
and 2 units vertically. Thus t has vertical asymptote x 2
1
and horizontal asymptote y 2. The domain of t is 9 3x 3 1 3 39 x 2 x 2 x 2 9 f x 2 3
20. t x
x
1
2 2 and its range is 2 2 .
y
3 x 2
3x 3
From this form we see that the graph of t is obtained from the
3x 6
graph of f by shifting 2 units to the left, stretching vertically
9
5
x
1
by a factor of 9, reflecting about the xaxis, and then shifting 3 units vertically. Thus t has vertical asymptote x 2 and horizontal asymptote y 3. The domain of t is
2 2 and its range is 3 3 .
1 x 2 1 f x 3 1 (see the long x 3 x 3 division at right). From this form we see that the graph of r is
21. r x
y
1 x 3
x 2
obtained from the graph of f by shifting 3 units to the left,
x 3
reflect about the xaxis, and then shifting vertically 1 unit.
1
Thus r has vertical asymptote x 3 and horizontal
1 1
x
asymptote y 1. The domain of r is 3 3 and its range is 1 1 .
1 2x 9 2 2 x 4 x 4 f x 4 2
22. t x
1 x 4
y
2 x 4
2x 9
From this form we see that the graph of t is obtained from the
2x 8
graph of f by shifting 4 units to the right, reflecting about the
1
xaxis, and then shifting 2 units vertically. Thus t has vertical
1 1
x
asymptote x 4 and horizontal asymptote y 2. The domain of r is 4 4 and its range is 2 2 .
x 1 . When x 0, we have r 0 14 , so the yintercept is 14 . The numerator is 0 when x 1, so the x 4 xintercept is 1.
23. r x
3x . When x 0, we have s 0 0, so the yintercept is 0. The numerator is zero when 3x 0 or x 0, so x 5 the xintercept is 0.
24. s x
25. t x
x2 x 2 2 . When x 0, we have t 0 13 , so the yintercept is 13 . The numerator is 0 when x 6 6
x 2 x 2 x 2 x 1 0 or when x 2 or x 1, so the xintercepts are 2 and 1.
84
CHAPTER 3 Polynomial and Rational Functions
2 2 12 , so the yintercept is 12 . The numerator is never zero, so . When x 0, we have r 0 26. r x 2 4 x 3x 4 there is no xintercept. 27. r x
x2 9 . Since 0 is not in the domain of r x, there is no yintercept. The numerator is 0 when x2
x 2 9 x 3 x 3 0 or when x 3, so the xintercepts are 3.
x3 8 . When x 0, we have r 0 84 2, so the yintercept is 2. The xintercept occurs when x 3 8 0 28. r x 2 x 4 x 2 x 2 2x 4 0 x 2 or x 1 i 3, which has only one real solution, so the xintercept is 2.
29. From the graph, the xintercept is 3, the yintercept is 3, the vertical asymptote is x 2, and the horizontal asymptote is y 2.
30. From the graph, the xintercept is 0, the yintercept is 0, the horizontal asymptote is y 0, and the vertical asymptotes are x 1 and x 2.
31. From the graph, the xintercepts are 1 and 1, the yintercept is about 14 , the vertical asymptotes are x 2 and x 2, and the horizontal asymptote is y 1.
32. From the graph, the xintercepts are 2, the yintercept is 6, the horizontal asymptote is y 2, and there are no vertical asymptotes 5 has a vertical asymptote where x 2 0 x 2, and y 0 is a horizontal asymptote because the degree x 2 of the denominator is greater than that of the numerator.
33. r x
2x 3 34. r x 2 has are vertical asymptotes where x 2 1 0 x 1 or x 1, and y 0 is a horizontal asymptote x 1 because the degree of the denominator is greater than that of the numerator. 3x 10 has no vertical asymptote since x 2 5 0 for all x. There is a horizontal asymptote at y 0 because the x2 5 degree of the denominator is greater than that of the numerator.
35. r x
2x 3 x 2 has vertical asymptotes where x 4 16 x 2 4 x 2 4 0 x 2, and y 0 is a horizontal 4 x 16 asymptote because the degree of the denominator is greater than that of the numerator.
36. r x
37. s x
10x 3 7 has vertical asymptotes where x 3 x x x 2 1 x x 1 x 1 0 x 1, 0, or 1; and 3 x x
horizontal asymptote y 10 1 10. 38. s x 39. r x
18x 2 9 has no vertical asymptote since 9x 2 1 0 for all x. It has horizontal asymptote y 18 9 2. 9x 2 1
x 1 2x 3 has vertical asymptotes where x 2 4x 7 0 x 74 or x 2, and horizontal x 2 4x 7
asymptote y 40. r x
1 12 . 14 2
x 3 x 2 has vertical asymptotes where 5x 1 2x 3 0 x 15 or x 32 , and horizontal 5x 1 2x 3
asymptote y
1 11 . 52 10
85
SECTION 3.6 Rational Functions
41. r x
6x 3 2 6x 3 2 . Because the quadratic in the denominator has no real zero, r has vertical 2x 3 5x 2 6x x 2x 2 5x 6
asymptote x 0 and horizontal asymptote y 62 3.
5x 3 5x 3 5x 2 42. r x 3 . Because the denominator has no real zero, r has no vertical x 2x 2 5x x x 2 2x 5 x 2 2x 5 asymptote. r has horizontal asymptote y 51 5.
t(x)
x2 2 . A vertical asymptote occurs when x 1 0 x 1. There is no horizontal asymptote because the degree x 1 of the numerator is greater than the degree of the denominator.
43. y
x 3 3x 2 x 2 x 3 . Because the degree of the numerator is greater than the degree of the denominator, x 2 x 2 x2 4 the function has no horizontal asymptote. Two vertical asymptotes occur at x 2 and x 2. By using long division, we 4x 12 so y x 3 is a slant asymptote. see that r x x 3 2 x 4 2x 2 y . When x 0, y 2, so the yintercept is 2. When y 0, 45. r x x 1 2x 2 0 x 1, so the xintercept is 1. Since the degree of the numerator
44. r x
and denominator are the same, the horizontal asymptote is y 21 2. There is a
vertical asymptote at x 1. As x 1 , y
2x 2 , and as x 2 , x 2
1
3 x 13 1 3x . When x 0, y 14 , so the yintercept is 14 . When 46. r x 2x 4 2 x 2
y
y 0, we have x 13 0 x 13 , so the xintercept is 13 . There is a vertical
1
asymptote where x 2 0 x 2. Because the degree of the denominator
1
3 3 . The and the numerator are the same, the horizontal asymptote is y 2 2 3 domain is x x 2 and the range is y y 2 .
47. r x
3x 2 12x 13 x 2 4x 4
3 x 2 4x 4 1 x 2 4x 4
3
1 x 22
13 y 13 4 , so the yintercept is 4 . There is no xintercept since
. When x 0,
1
x 22
x
1
2x 2 y . The domain is x x 1 and the range is y y 2. x 2
x
y
is positive
on its domain. There is a vertical asymptote at x 2. The horizontal asymptote is y 3. The domain is x x 2 and the range is y y 3.
1 1
x
86
CHAPTER 3 Polynomial and Rational Functions
2 x 2 4x 4 1 2x 2 8x 9 1 48. r x . When 2 2 2 x 4x 4 x 4x 4 x 22 x 0, y 94 , so the yintercept is 94 . There is no xintercept since
1
x 22
y
is
1 0
positive on its domain. There is a vertical asymptote at x 2. The horizontal
1
x
asymptote is y 2. The domain is x x 2 and the range is y y 2.
x 2 8x 16 2 x 2 8x 18 2 49. r x 2 . When 1 2 x 8x 16 x 8x 16 x 42 x 0, y 98 , and so the yintercept is 98 . There is no xintercept since 2
x 42
is positive on its domain. There is a vertical asymptote at x 4. The
y
1 0
1
x
horizontal asymptote is y 1. The domain is x x 4 and the range is y y 1.
1 2x 2 4x 2 2 1 x 2 2x 3 1 2 50. r x 2 . is true. When 2 2 2x 4x 2 2x 4x 2 x 12
y
x 0, we have y 32 , so the yintercept is 2. There is no xintercept since 1
x 12
is positive on its domain. There is a vertical asymptote at x 1. The
horizontal asymptote is y 12 . The domain is x x 1 and the range is y y 12 .
51. s x
4x 8 8 . When x 0, y 2, so the yintercept is 2. x 4 x 1 4 1
1 0
1
x
y
When y 0, 4x 8 0 x 2, so the xintercept is 2. The vertical asymptotes are x 1 and x 4, and because the degree of the numerator is less than the
degree of the denominator, the horizontal asymptote is y 0. The domain is x x 1 4 and the range is .
1 1
x
87
SECTION 3.6 Rational Functions
9 . When x 0, y 94 , so the yintercept is 94 . Since the 52. s x 2 x 5x 4 numerator is never zero, there is no xintercept. The vertical asymptotes occur when x 2 5x 4 x 1 x 4 0 x 1 or x 4, and because the
degree of the numerator is less than the degree of the denominator, the horizontal
y
2 x
1
asymptote is y 0. The domain is x x 1 4 and the range is 4] 0 .
9x 18 9 x 2 53. s x 2 . When x 0, y 9, so the yintercept is x 1 x 2 x x 2 9. When y 0, we have x 2 0 x 2, so the xintercept is 2. There are
y
vertical asymptotes where x 1 x 2 0 x 1 or x 2. Because the degree of the denominator is greater than the degree of the numerator, the
4
horizontal asymptote is y 0. The domain is x x 2 1 and the range is
1
x
1
x
1] [9 .
54. s x
x 2 2 , so the yintercept is 23 When . When x 0, y 3 x 3 x 1
y
y 0, we have x 2 0 x 2, so the xintercept is 2. A vertical
asymptote occurs when x 3 x 1 0 x 3 and x 1. Because the degree of the denominator is greater than the degree of the numerator, the
1
horizontal asymptote is y 0. The domain is x x 3 1 and the range is .
55. r x
x 1 x 2 . When x 0, y 23 , so the yintercept is 23 . When x 1 x 3
y
y 0, x 1 x 2 0 x 2, 1, so, the xintercepts are 2 and 1. The
vertical asymptotes are x 1 and x 3, and because the degree of the
numerator and denominator are the same the horizontal asymptote is y 11 1.
The domain is x x 1 3 and the range is .
1 1
x
88
CHAPTER 3 Polynomial and Rational Functions
2x 2 10x 12 2 1 6 2 x 1 x 6 . When x 0, y 2, x 2 x 3 2 3 x2 x 6 so the yintercept is 2. When y 0, 2 x 1 x 6 0 x 6, 1, so the
y
56. r x
xintercepts are 6 and 1. Vertical asymptotes occur when x 2 x 3 0
x 3 or x 2. Because the degree of the numerator and denominator are the
2
same the horizontal asymptote is y 21 2. The domain is x x 3 2 and
x
1
the range is .
x 2 2x 8 x 2 x 4 . The function has vertical asyptotes 2 x x 2 x 2x x 2 and x 0. Since x cannot equal 0, there is no yintercept. The
y
57. r x
xintercepts are 4 and 2. Because the degree of the denominator and numerator
are the same, the horizontal asymptote is y 11 1. The domain is
4
x x 2 0 and the range is 1 [9 .
3x 2 6
3 x2 2
. We cannot have x 0 or y 0 so, there is no x 58. r x 2 x x 4 x 4x or yintercept. There are vertical asymptotes at x 0 and x 4. Because the
1
y
degree of the numerator and denominator are the same, the horizontal asymptote is.y 31 3. The domain is x x 0 4 and the range is 3] 32 .
59. s x
x 2 2x 1 x 12 2 . Since x 0 is not in the domain of s x, 3 2 x 3x x x 3
there is no yintercept. The xintercept occurs when y 0
x 2 2x 1 x 12 0 x 1, so the xintercept is 1. Vertical asymptotes
occur when x 0, 3. Since the degree of the numerator is less than the degree of
the denominator, the horizontal asymptote is y 0. The domain is x x 0 3 and the range is .
x
2 2
x
1
x
y 1
89
SECTION 3.6 Rational Functions
x2 x 6 x 3 x 2 . The xintercept occurs when y 0 x x 3 x 2 3x x 3 x 2 0 x 2, 3, so the xintercepts are 2 and 3. There is no
y
60. y
yintercept because y is undefined when x 0. The vertical asymptotes are x 0
and x 3. Because the degree of the numerator and denominator are the same,
2
the horizontal asymptotes is y 11 1. The domain is x x 3 0 and the
x
1
range is .
x 2 2x 1 x 12 61. r x 2 x 2x 1 x 12
x 1 2 . When x 0, y 1, so the x 1
y
yintercept is 1. When y 0, x 1, so the xintercept is 1. A vertical asymptote
occurs at x 1 0 x 1. Because the degree of the numerator and
denominator are the same the horizontal asymptote is y 11 1. The domain is x x 1 and the range is y y 0.
1 x
1
9x 2 9x 2 . When x 0, we have y 0, so the 4 x 2 x 1 4x 2 4x 8 graph passes through the origin. Vertical asymptotes occur at x 2 and x 1.
62. r x
y
Because the degree of the denominator and numerator are the same, the horizontal asymptote is y 94 . The domain is x x 2 1 and the range is
1
63. r x
5 x 12 5x 2 10x 5 . When x 0, y 59 , so the yintercept is x 2 6x 9 x 32
x
1
0] [2 .
y
5 . Since r 1 0, the xintercept is 1. The vertical asymptote is x 3. 9
Because the degree of the denominator and numerator are the same, the horizontal asymptote occurs at y 51 5. The domain is x x 3 and the range is
y y 0.
2 2
x
90
CHAPTER 3 Polynomial and Rational Functions
x3 x2 x 2 x 1 64. r x 3 3 . When x 0, we have y 0, so the x 3x 2 x 3x 2
y
yintercept is 0. When y 0, we have x 2 x 1 0, so the xintercepts are 0
and 1. Vertical asymptotes occur when x 3 3x 2 0. Since x 3 3x 2 0
when x 2, we can factor x 2 x 12 0, so the vertical asymptotes occur
1
at x 2 and x 1. Because the degree of the denominator and numerator are
the same, the horizontal asymptote is y 11 1. The domain is x x 1 2
x
1
and the range is .
65. r x
x 2 4x 5 x 5 x 5 x 1 for x 1. When x 0, y 52 , 2 x 2 2 1 x x x x 2
y
so the yintercept is 52 . When y 0, x 5, so the xintercept is 5. Vertical asymptotes occur when x 2 0 x 2. Because the degree of the
(1, 2)
denominator and numerator are the same, the horizontal asymptote is y 1. The
1
domain is x x 2 1 and the range is y y 1 2.
0
1
x
Move
66. r x
x 2 3x 10 x 2 x 2 x 5 for x 1 x 3 x 5 x 1 x 3 x 5 x 1 x 3
y
x 5. When x 0, we have y 23 , so the yintercept is 23 . When y 0, we
have x 2, so the xintercept is 2. Vertical asymptotes occur when
1
x 1 x 3 0 x 1 or 3, so the vertical asymptotes occur at x 1 and x 3. Because the degree of the denominator is greater than that of the
numerator, the horizontal asymptote is y 0. The domain is x x 5 1 3
0
(_5, _ 327 )
1
x
and the range is .
x 2 2x 3 x 3 x 1 x 3 for x 1. When x 0, x 1 x 1 y 3, so the yintercept is 3. When y 0, x 3, so the xintercept is 3.
y
67. r x
There are no asymptotes. The domain is x x 1 and the range is
1
y y 4.
0 (_1, _4)
1
x
91
SECTION 3.6 Rational Functions
x x 1 x 3 x 3 2x 2 3x x x 1 for x 3. When x 0, x 3 x 3 we have y 0, so the yintercept is 0. When y 0, x 1 or 0, so the
y
68. r x
(3, 12)
xintercepts are 1 and 0. There are no asymptotes. The domain is x x 3 and the range is y y 14 .
1 0
x 3 5x 2 3x 9 . We use synthetic division to check whether the x 1 denominator divides the numerator:
y
69. r x
1
1
5 1
3
9
6
9
1 6 9 0 x 2 6x 9 x 1 Thus, r x x 2 6x 9 x 32 for x 1. x 1 When x 0, y 9, so the yintercept is 9. When y 0, x 3, so the xintercept
x
1
(_1, 16)
2 0
x
1
is 3. There are no asymptotes. The domain is x x 1 and the range is
y y 0.
x 1 x 2 4x 5 x 1 x 5 for x 5. x 0 is 70. r x 3 x x 2 x 5 x x 2 x 7x 2 10x not in the domain of r , so there is no yintercept. When y 0, x 1, so the
y
xintercept is 1. There are vertical asymptotes at x 2 and x 0. Because the
degree of the denominator is less than the degree of the numerator, y 0 is a
Move 1
horizontal asymptote. The domain is x x 5 2 0 and the range is y y 0134 or y 1866.
71. r x
x2 . When x 0, y 0, so the graph passes through the origin. There x 2
0
(_5, _ 327 )
1
x
y
is a vertical asymptote when x 2 0 x 2, with y as x 2 , and y as x 2 . Because the degree of the numerator is greater than the
degree of the denominator, there is no horizontal asymptote. Using long division, we see that y x 2
4 , so y x 2 is a slant asymptote. x 2
2 2
x
92
CHAPTER 3 Polynomial and Rational Functions
x x 2 x 2 2x . When x 0, we have y 0, so the graph passes x 1 x 1 through the origin. Also, when y 0, we have x 0 or 2, so the xintercepts are
y
72. r x
2 and 0. The vertical asymptote is x 1. There is no horizontal asymptote, and the line y x 3 is a slant asymptote because by long division, we have y x 3
2 . x 1
2 x
1
x 2 2x 8 x 4 x 2 . The vertical asymptote is x 0, thus, x x there is no yintercept. If y 0, then x 4 x 2 0 x 2, 4, so the
y
73. r x
xintercepts are 2 and 4. Because the degree of the numerator is greater than the
degree of the denominator, there are no horizontal asymptotes. By using long
2
8 division, we see that y x 2 , so y x 2 is a slant asymptote. x
3x x 2 x 3 x . When x 0, we have y 0, so the graph passes 2x 2 2 x 1 through the origin. Also, when y 0, we have x 0 or x 3, so the xintercepts are 0 and 3. The vertical asymptote is x 1. There is no horizontal asymptote, and the line y 12 x 1 is a slant asymptote because by long division we have
1
x
1
1 y 12 x 1 . x 1
x 2 5x 4 x 4 x 1 . When x 0, y 43 , so the yintercept x 3 x 3
x
y
74. r x
75. r x
2
y
is 43 . When y 0, x 4 x 1 0 x 4, 1, so the two xintercepts are 4 and 1. A vertical asymptote occurs when x 3, with y as x 3 , and y as x 3 . Using long division, we see that 28 , so y x 8 is a slant asymptote. y x 8 x 3
5 2
x
93
SECTION 3.6 Rational Functions
x3 4 x3 4 . When x 0, we have 2x 1 x 1 2x 2 x 1 04 4, so the yintercept is 4. Since x 3 4 0 x 3 4, y 001 the xintercept is x 3 4. There are vertical asymptotes where
y
76. r x
1 1
2x 1 x 1 0 x 12 or x 1. Since the degree of the numerator is
x
greater than the degree of the denominator, there is no horizontal asymptote. By long division, we have y 12 x 14 slant asymptote.
77. r x
3 x 15 4 4 , so the line y 1 x 1 is a 2 4 2x 2 x 1
x3 x2 x 2 x 1 . When x 0, y 0, so the graph passes x 2 x 2 x2 4
y
through the origin. Moreover, when y 0, we have x 2 x 1 0 x 0, 1, so the xintercepts are 0 and 1. Vertical asymptotes occur when x 2; as
x 2 , y and as x 2 , y . Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal 4x 4 , so y x 1 is a asymptote. Using long division, we see that y x 1 2 x 4 slant asymptote.
78. r x
2x 3 2x x2 1
2x x 2 1
x 1 x 1
2
50 5
5 20 20
1
x
2
2x 2 6x 6 , g x 2x. f has vertical asymptote x 3. x 3
10
x
y
. When x 0, we have y 0, so the graph
passes through the origin. Also, note that x 2 1 0, for all real x, so the only xintercept is 0. There are two vertical asymptotes at x 1 and x 1. There is no horizontal asymptote, and the line y 2x is a slant asymptote because by long 4x . division, we have y 2x 2 x 1
79. f x
1
20 50
94
CHAPTER 3 Polynomial and Rational Functions
80. f x
x 3 6x 2 5 , g x x 4. f has vertical asymptotes x 0 and x 2. x 2 2x 20
20
5
5
20
20
20
81. f x
20
x 3 2x 2 16 , g x x 2 . f has vertical asymptote x 2. x 2 50
50
10
10
5 50
82. f x
x 4 2x 3 2x x 12
4
, g x 1 x 2 . f has vertical asymptote x 1.
2
2
4
4
2
2
5
5
10
10
4
2x 2 5x has vertical asymptote x 15, xintercepts 0 and 25, yintercept 0, local maximum 39 104, 2x 3 12 . From the graph, we see that the end and local minimum 09 06. Using long division, we get f x x 4 2x 3 behavior of f x is like the end behavior of g x x 4.
83. f x
f
2x 3
x 4
2x 2 5x
50
20
2x 2 3x 8x
10
8x 12
12
10 20
20
20 50
SECTION 3.6 Rational Functions
95
x 4 3x 3 x 2 3x 3 has vertical asymptotes are x 0, x 3, xintercept 082, and no yintercept. The x 2 3x local minima are 080 263 and 338 1476. The local maximum is 256 488. By using long division, we see that 3 f x x 2 1 2 . From the second graph, we see that the end behavior of f x is the same as the end behavior x 3x f 2 of g x x 1.
84. f x
x2
x 2 3x
1
20
x 4 3x 3 x 2 3x 3
50
x 4 3x 3
0x 3 x 2 3x
5
x 2 3x
5
5 50
20
3
5
x5 85. f x 3 has vertical asymptote x 1, xintercept 0, yintercept 0, and local minimum 14 31. f x 1 x2 Thus y x 2 3 . From the graph we see that the end behavior of f x is like the end behavior of g x x 2 . x 1 x2 x3 1
10
x5
10
x5 x2 x2
5
5
5
5
10
10
Graph of f 86. f x
f
x4
Graph of f and g
has vertical asymptotes x 141, xintercept 0, and yintercept 0. The local maximum is 0 0. The
x2 2
4 . From the second graph, local minima are 2 8 and 2 8. By using long division, we see that f x x 2 2 2 x 2 2 we see that the end behavior of f x is the same as the end behavior of g x x 2. x2
x2 2
2
20
x 4 0x 3 0x 2 0x 0
x4
50
2x 2 2x 2 2x 2
r(x)
4
5
5
5
4
5
x 4 3x 3 6 has vertical asymptote x 3, xintercepts 16 and 27, yintercept 2, local maxima 04 18 x 3 6 and 24 38, and local minima 06 23 and 34 543. Thus y x 3 . From the graphs, we see that the end x 3 behavior of f x is like the end behavior of g x x 3 .
87. f x
r
x3
x 3
x 4 3x 3 6
x 4 3x 3
6
100
5
100
5 100
5
5 100
96
CHAPTER 3 Polynomial and Rational Functions
4 x2 x4 x 4 x 2 4 has vertical asymptotes x 1, xintercepts 16, and yintercept 4. The local x 1 x 1 x2 1 maximum is 0 4 and there is no local minimum. 6 Thus y x 2 2 . From the graphs, we see that the end behavior of f x is like the end behavior of g x x 2 . x 1 x 2
88. r x
x2 1
89.
20
x 4 0x 3 x 2 0x 4 x 4
x2
0
2
4
20
5
5
5
20
5 20
90.
c=4 c=3 c=2
5
c=1 c=5
5
c=4
5
10
c=3
c=2
5
2
cx has the same basic shape for The graph of r x 2 x 1 all values of c. Increasing the value of c stretches the graph vertically.
x 1 has y 1 as a horizontal x c asymptote and x c as a vertical asymptote for all values of c. The location of the vertical asymptote changes as c changes.
The graph of r x
r
SECTION 3.6 Rational Functions
92.
91.
c=1 c=2 c=3 c=4
c=4
5
97
20 4 9 16
c=3
10
c=2 c=1
2
2
20
5
40
cx 2 has the same basic shape, x 1 local maximum r 0 0, vertical asymptotes x 1, and horizontal asymptote y c for all values of c. The horizontal asymptote moves upward as c increases. The graph of r x
x2 c has the same basic shape and x 5 vertical asymptote x 5 for all values of c. The local maximum and minimum values get closer together as c increases.
The graph of r x
93. (a) The total cost of producing x purses is C x, so the average cost per purse is (b)
y 100
158, the average cost per purse decreases, presumably due to For x
158, the average cost steadily increases, economies of scale. For x
80
perhaps because of supply issues or overtime costs. The local
60
minimum of approximately A 158 545 tells us that the lowest
40
average cost is approximately $5450 per purse, and is achieved when
20 0
Total cost C x . Number of purses x
158 purses are produced. 100 200 300 400 500 600 700 x
3000 3000t 3000 . So as t , we have t 1 t 1 p t 3000.
94. (a)
(b) p t 2000 0
0
20
40
Please don't use this symbol. I don't think students don't know what it means.
98
CHAPTER 3 Polynomial and Rational Functions
5t 95. c t 2 t 1
(a) The highest concentration of drug is 250 mg/L, and it is reached 1 hour after the drug is administered. (b) The concentration of the drug in the bloodstream goes to 0.
2
0
(c) From the first viewing rectangle, we see that an approximate solution
0
10
5t is near t 15. Thus we graph y 2 and y 03 in the t 1 viewing rectangle [14 18] by [0 05]. So it takes about 1661 hours for the concentration to drop below 03 mg/L.
20
0.4 0.2 0.0
14
16
64 106 2 96. Substituting for R and g, we have h . The vertical 2 98 64 106 2
2e+7
asymptote is 11,000, and it represents the escape velocity from the earth’s
1e+7
gravitational pull: 11,000 m/s 1900 mi/h.
0
97. P P0
s0 s0
P 440
332 332
18
0
10000
4000
If the speed of the train approaches the speed of sound, the pitch of the whistle becomes very loud. This would be experienced as a “sonic boom”—an effect
2000
seldom heard from trains. 0
98. (a)
1 1 1 1 1 1 xF xF 1 y . Using x y F y F x y xF xF
55x . Since y 0, we use the viewing F 55, we get y x 55 rectangle [0 1000] by [0 250]. (b) y approaches 55 millimeters. (c) y approaches .
0
200
400
200 100 0
0
500
1000
99
SECTION 3.6 Rational Functions
99. (a) The initial amount of salt is 4 lb and salt is being poured in at 5 lb per C (lb/gal) minute. Thus, at time t there is S t 4 5t pounds of salt in the
0.1
tank. Similarly, there is W t 100 50t gallons of water in the tank, so the concentration at time t is C t
4 5t . 100 50t
0
10
20
t (minutes)
4 5 10 009 lbgal. The tank contains 1000 gal of 100 50 10 water when W t 100 50t 1000 50t 900 t 18 minutes, at which time the concentration is 4 5 18 C 18 0094 lbgal. 100 50 18 (c) As more salt and water is added, the concentration in the tank approaches the ratio of added salt to added water, that is, 4 5t 5 01 lbgal. We can also see this from the graph. As t becomes large, the concentration 50 C t 100 50t C t approaches the horizontal asymptote y 01, regardless of the initial concentration.
(b) After 10 minutes, the concentration is C 10
2x 1 . Vertical asymptote x 3 and horizontal asymptote y 2: r x . x 3 x 3 x 4 Vertical asymptotes x 1 and x 1, horizontal asymptote 0, and xintercept 4: q x . Of course, x 1 x 1
100. Vertical asymptote x 3: p x
other answers are possible.
zeros
zeros
x 6 10
130 49x 2
x2 8 4 has no xintercept since the numerator has no real roots. Likewise, x 4 8x 2 15 x 8x 2 15 r x has no vertical asymptotes, since the denominator has no real roots. Since the degree of the numerator is two greater
101. r x
than the degree of the denominator, r x has no horizontal or slant asymptotes. From the division below, we see that the graph of r is close to that of y x 2 8 for large x. x 4 8x 2 15
x2
8
x 6 0x 5 0x 4 0x 3 0x 2 0x 10 x6
8x 4
15x 2
8x 4
64x 2
8x 4
15x 2
10
49x 2
130
1 1 f x 2. From this form we see that 102. (a) Let f x 2 . Then r x x x 22
120
y
the graph of r is obtained from the graph of f by shifting 2 units to the right. Thus
r has vertical asymptote x 2 and horizontal asymptote y 0.
1 1
x
100
CHAPTER 3 Polynomial and Rational Functions
2x 2 4x 5 3 3 f x 1 2. From this form we see that the graph of s is obtained from 2 x 2 2x 1 x 12 the graph of f by shifting 1 unit to the left, stretching vertically by a factor of 3, and shifting 2 units vertically. Thus r has vertical asymptote x 1 and horizontal asymptote y 2.
(b) s x
y
2 x 2 2x 1
2x 2
2x 2
4x 4x
5 2 3 1 x
1
2 3x 2 12x 14 (c) Using long division, we see that p x 2 3 2 which cannot be graphed by transforming x 4x 4 x 4x 4 1 f x 2 . Using long division on q we have: x x 2 4x 4
3
3x 2
3x 2
12x 3x 2
0x 12x 12x
2 12
3
x 2 4x 4
3x 2 3x 2
14
12x 12x
0 12 12
12 So q x 2 12 f x 2 3. From this form we see that the graph of q is obtained 3 x 4x 4 x 22 from the graph of f by shifting 2 units to the right, stretching vertically by a factor of 12, and then shifting 3 units vertically down. Thus the vertical asymptote is x 2 and the horizontal asymptote is y 3. We show y p x just 1 to verify that we cannot obtain p x from y 2 . x
y 1 1
y
1
1
y q x
x
y p x
x
SECTION 3.7 Polynomial and Rational Inequalities
3.7
101
POLYNOMIAL AND RATIONAL INEQUALITIES
1. To solve a polynomial inequality, we factor the polynomial into irreducible factors and find all the real zeros of the polynomial. Then we find the intervals determined by the real zeros and use test points in each interval to find the sign of the polynomial on that interval. Sign of x
2
2 0
0 1
1
x 2 x 1
P x x x 2 x 1
From the table, we see that P x 0 on the intervals [2 0] and [1 .
2. To solve a rational inequality, we factor the numerator and the denominator into irreducible factors. The cut points are the real zeros of the numerator and the real zeros denominator. Then we find the intervals determined by the cut points, and we use test points to find the sign of the rational function on each interval. Sign of x 2 x 1 x 3 x 4
4
4 2
2 1
1 3
3
x 2 x 1 r x x 3 x 4 From the table, we see that r x 0 on the intervals 4, [2 1], and 3 .
3. The inequality x 3 x 5 2x 5 0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 3, 5, and 52 are 5, 5 52 , 52 3 , and 3 . We make a sign diagram: 5 52 52 3 Sign of 5 3 x 3 x 5
2x 5
P x x 3 x 5 2x 5
None of the endpoints satisfies the inequality. The solution is 5 52 3 .
4. The inequality x 1 x 2 x 3 x 4 0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 1, 2, 3, and 4 are 4, 4 2, 2 1, 1 3, and 3 . We make a sign diagram: Sign of x 1 x 2 x 3 x 4
P x x 1 x 2 x 3 x 4
4
4 2
2 1
1 3
3
All of the endpoints satisfy the inequality. The solution is 4] [2 1] [3 .
102
CHAPTER 3 Polynomial and Rational Functions
5. The inequality x 52 x 3 x 1 0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 5, 3, and 1 are 5, 5 3, 3 1, and 1 . We make a sign diagram: Sign of
x 52
5
5 3
3 1
1
x 3 x 1
P x x 52 x 3 x 1
None of the endpoints satisfies the inequality. The solution is 5 5 3 1 .
6. The inequality 2x 74 x 13 x 1 0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 72 , 1, and 1 are 1, 1 1, 1 72 , and 72 . We make a sign diagram: 7 1 72 Sign of 1 1 1 2 2x 74 x 13 x 1
P x 2x 74 x 13 x 1
All of the endpoints satisfy the inequality [note that P 72 0]. The solution is [1 1] 72 .
7. We start by moving all terms to one side and factoring: x 3 4x 2 4x 16 x 3 4x 2 4x 16 x 4 x 2 x 2 0. The intervals determined by the zeros 4, 2, and 2 are 4, 4 2, 2 2, and 2 . We make a sign diagram: Sign of x 4 x 2 x 2
P x x 4 x 2 x 2
4
4 2
2 2
2
All of the endpoints satisfy the inequality. The solution is [4 2] [2 .
8. We start by moving all terms to one side and factoring: 2x 3 18x x 2 9 2x 3 x 2 18x 9 x 3 2x 1 x 3 0. The intervals determined by the zeros 3, 12 , and 3 are 3, 3 12 , 12 3 , and 3 . We make a sign diagram: Sign of x 3
2x 1 x 3
P x x 3 2x 1 x 3
3
3 12
13 2
3
None of the endpoints satisfies the inequality. The solution is 3 12 3 .
9. We start by moving all terms to one side and factoring: 2x 3 x 2 918x 2x 3 x 2 18x 9 2x 1 x 2 9 0. Note that x 2 9 0 for all x, so the sign of P x 2x 1 x 2 9 is negative where 2x 1 is negative and positive where 2x 1 is positive. The endpoint x 12 does not satisfy the inequality, so the solution is 12 .
SECTION 3.7 Polynomial and Rational Inequalities
103
10. We start by moving all terms to one side and factoring: x 4 3x 3 x 3 x 4 3x 3 x 3 0. The possible rational zeros of P x x 4 3x 3 x 3 are 1 and 3.
1 1 3 1 3 1
4
3
1 4 3 0 Thus, P x x 1 x 3 4x 2 4x 3 x 1 x 3 x 2 x 1 . The last factor is positive everywhere, so
we test the intervals 3, 3 1, and 1 . Sign of x 3 x 1
3
3 1
1
P x x 1 x 3 x 2 x 1
Neither of the endpoints satisfies the inequality. The solution is 3 1.
11. All the terms are on the left size. We factor, using the substitution t x 2 : x 4 7x 2 18 0 t 2 7t 18 t 2 t 9 0 x 2 2 x 2 9 0 x 2 2 x 3 x 3 0. The first factor is positive everywhere, so we test 3, 3 3, and 3 : Sign of x 3 x 3
P x x 2 2 x 3 x 3
3
3 3
3
Neither of the endpoints satisfies the inequality. The solution is 3 3.
12. All the terms are on the left size. We factor, using the substitution t x 2 : 4x 4 25x 2 36 0 4t 2 25t 36 4t 9 t 4 0 4x 2 9 x 2 4 2x 3 2x 3 x 2 x 2 0. The zeros are 32 and 2, so we test 2, 2 32 , 32 32 , 32 2 , and 2 : 32 2 32 32 32 Sign of 2 2 2 x 2
x 32 x 32 x 2 P x
All of the endpoints satisfy the inequality. 2 32 32 2 .
104
CHAPTER 3 Polynomial and Rational Functions
13. All the terms are on the left size. To factor, note that the possible rational zeros of P x x 3 x 2 17x 15 are 1, 3, 5, 15. 1 1 1 17 1
15
2 15
1 2 15 0 Thus, P x x 1 x 2 2x 15 x 5 x 1 x 3. The zeros are 5, 1, and 3, so we test 5,
5 1, 1 3, and 3 :
Sign of x 5
5
5 1
1 3
3
x 1 x 3 P x
All of the endpoints satisfy the inequality P x 0, so the solution is [5 1] [3 .
14. All the terms are on the left size. To factor, note that the possible rational zeros of P x x 4 3x 3 3x 2 3x 4 are 1, 2, 4. 1 1 3 3 3 4 1
4 1
4
1 4
1 4
0
Thus, P x x 1 x 3 4x 2 x 4 x 1 x 4 x 2 1 . The last factor is positive everywhere, so we test 4, 4 1, and 1 :
Sign of x 4
4
4 1
1
x 1 P x
None of the endpoints satisfies P x 0, so the solution is 4 1.
3 3 3 15. We start by moving all terms to one side and factoring: x 1 x 2 7 1 x 2 x 7 1 x 2 0
P x x 7 1 x3 1 x3 0. The intervals determined by the zeros 1, 1, and 7 are 1, 1 1, 1 7, and 7 . We make a sign diagram: Sign of x 7
1 x3 1 x3 P x
1
1 1
1 7
7
None of the endpoints satisfies the inequality. The solution is 1 1 7.
SECTION 3.7 Polynomial and Rational Inequalities
105
16. We start by moving all terms to one side and factoring: x 2 7 6x 1 6x 3 7x 2 1 0. The possible rational zeros of P x 6x 3 7x 2 1 are 1, 12 , 13 , 16 .
1 6
7 0 1
6 1
1
6 1 1 0 Thus, P x x 1 6x 2 x 1 1 x 2x 1 3x 1. The intervals determined by the zeros 13 , 12 , and 1 13 , 13 12 , 12 1 , and 1 . We make a sign diagram: 11 13 13 12 Sign of 1 2 3x 1 2x 1 1x
P x
All of the endpoints satisfy the inequality P x 0. The solution is 13 12 [1 .
x 1 0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring x 10 needed, we find the intervals determined by the cut points 1 and 10. These are 1, 1 10, and 10 . We make a sign diagram:
17. r x
Sign of
1
1 10
x 1
10
x 10
r x
The cut point 1 does not satisfy the inequality, and the cut point 10 is not in the domain of r. Thus, the solution is 1 10.
3x 7 0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring x 2 needed, we find the intervals determined by the cut points 73 and 2. These are 2, 2 73 , and 73 . We
18. r x
make a sign diagram:
Sign of 3x 7 x 2
r x
2
2 73
7 3
The cut point 73 satisfies equality, but the cut point 2 is not in the domain of r. Thus, the solution is 2 73 .
106
CHAPTER 3 Polynomial and Rational Functions
x 3 x 3 2x 5 x 3 1 1 0 0 2x 5 2x 5 2x 5 x 8 r x 0. The intervals determined by the cut points are 8, 8 52 , and 52 . 2x 5 8 52 52 Sign of 8
19. We start by moving all terms to one side and simplifying:
x 8 2x 5
r x
The cut point 8 satisfies equality, but 52 is not in the domain of r. Thus, the solution is 8 52 .
20. We start by moving all terms to one side and simplifying:
x 4 x 4 x 4 4 x 5 4 4 0 0 x 5 x 5 x 5
3 x 8 3x 24 0 r x 0. The intervals determined by the cut points are 5, 5 8, and 8 . x 5 x 5 Sign of
5
5 8
x 8
8
x 5
r x
The cut point 8 satisfies equality, but 5 is not in the domain of r . Thus, the solution is 5 [8 .
5x 7 5x 7 4x 10 5x 7 1 1 0 0 4x 10 4x 10 4x 10 x 3 r x 0. The intervals determined by the cut points are 52 , 52 3 , and 3 . 2 2x 5 52 52 3 Sign of 3
21. We start by moving all terms to one side and simplifying:
x 3
2x 5
r x
The cut point 3 satisfies equality, but 52 is not in the domain of r. Thus, the solution is 52 3 .
22. We start by moving all terms to one side and simplifying:
4x 6 4x 6 4x 6 2 x 7 2 2 0 0 x 7 x 7 x 7
2 x 10 2x 20 0 0. The intervals determined by the cut points are 7, 7 10, and 10 . x 7 x 7 Sign of x 10 x 7
r x
7
7 10
10
The cut point 10 fails to satisfy the strict inequality and 7 is not in the domain of r, so the solution is 7 10 .
SECTION 3.7 Polynomial and Rational Inequalities
23. r x
2x 5
x 2 2x 35 2x 5
0.
107
Since all nonzero terms are already on one side, we factor:
2x 5 . Thus, the cut points are 7, 52 , and 5. The intervals determined by these x 7 x 5 points are 7, 7 52 , 52 5 , and 5 . We make a sign diagram: 7 52 52 5 Sign of 7 5
r x
x 2 2x 35
2x 5 x 7 x 5
r x
The cut point 52 satisfies equality, but the cut points 7 and 5 are not in the domain of r. Thus, the solution is 7 52 5 .
4x 2 25 4x 2 25 2x 5 2x 5 0. Since all nonzero terms are already on one side, we factor: r . x x 3 x 3 x2 9 x2 9 Thus, the cut points are 52 and 3. The intervals determined by these points are 3, 3 52 , 52 52 , 52 3 ,
24. r x
and 3 . We make a sign diagram: Sign of 2x 5
3 52
52 52
3
2x 5 x 3 x 3
r x
53 2
3
The cut points 52 satisfy equality, but the cut points 3 are not in the domain of r. Thus, the solution is 3 52 52 3 .
x2 x2 25. r x 2 0. Since all nonzero terms are already on one side, we factor: r x . The cut x 5 x 2 x 3x 10 points are 5, 0, and 2. We make a sign diagram: Sign of x 5
5
5 0
0 2
2
x 2
r x
The cut points 5 and 2 are not in the domain of r, so the solution is 5 2.
x 3 x 3 0. Since all nonzero terms are already on one side, we factor: r x . The cut points are 26. r x 2 x 6x 9 x 32 3. We make a sign diagram: Sign of
x 32 x 3
r x
3
3 3
3
The cut point 3 is not in the domain of r, and the cut point 3 satisfies the inequality, so the solution is 3 3 3].
108
CHAPTER 3 Polynomial and Rational Functions
27. We start by moving all terms to one side and factoring: r x
x 2 3 2 x 1 x2 3 2 0 x 1 x 1
x 2 2x 1 x 12 0. The cut points are 1. We make a sign diagram: x 1 x 1 Sign of
x 12 x 1
r x
1
1 1
1
The cut point 1 is not in the domain of r and the cut point 1 does not satisfy the original equation, so the solution is 1 1 1 . 4x 3 x 2 1 x 2 4x 4 4x 3 1 0 0 28. We start by moving all terms to one side and factoring: 2 2 x 1 x 1 x2 1 x 22 0, which is true for all x. The solution is . r x 2 x 1 x 2 2x 3 x 3 x 1 0. The intervals determined by the cut points are 3, 3 23 , 29. r x 2 3x 2 x 3 3x 7x 6 2 3 1 , 1 3, and 3 . 3 23 23 1 Sign of 3 1 3 3 x 3
3x 2 x 1 x 3
r x
None of the cut points satisfies the strict inequality, so the solution is 3 23 1 3 .
x 1 x 1 0. The second factor in the denominator is positive for all x, so the intervals 30. r x 3 x 1 x 1 x 2 x 1 determined by the cut points are 1, 1 1, and 1 . Sign of x 1 x 1
r x
1
1 1
1
4
4 3
3
The cut point 1 satisfies equality, but 1 is not in the domain of r. Thus, the solution is 1 [1 . x x 2 6x 9 3 x 2 6x 9 x 3 x 32 x 3 3x 2 9x 27 0. The second factor in the 31. r x x 4 x 4 x 4 numerator is positive for all x, so the intervals determined by the cut points are 4, 4 3, and 3 . Sign of x 4 x 3
r x
The cut point 3 satisfies equality, but 4 is not in the domain of r. Thus, the solution is 4 3].
SECTION 3.7 Polynomial and Rational Inequalities
109
x 2 16 x 4 x 4 x 4 x 4 0. The last factor in the denominator is positive for 32. r x 4 2 x 4 x2 4 x 16 x 2 x 2 x 2 4 all x, so the intervals determined by the cut points are 4, 4 2, 2 2, 2 4, and 4 . Sign of x 4
4
4 2
2 2
2 4
4
x 2 x 2 x 4
r x
None of the cut points satisfy the strict inequality. Thus, the solution is 4 2 2 4.
x 12 0. The numerator is nonnegative for all x, but note that x 1 fails to satisfy the strict inequality. x 1 x 2 The intervals determined by the cut points are 2, 2 1, and 1 .
33. r x
Sign of
x 12
2
2 1
1
x 2 x 1
r x
(except at x 1)
The cut points 2 and 1 are not in the domain of r, so the solution is 2 1 1 1 .
34. r x 1 .
x 2 2x 1
x 3 3x 2 3x 1
x 12
x 13
0. The intervals determined by the cut points are 1, 1 1, and
Sign of x 13
1
1 1
1
x 12
r x
The cut point 1 is not in the domain of r and the cut point 1 satisfies the inequality. Thus, the solution is 1] 1. 5 x 5 x 4 4 0 2 x 1 2 x 1 x 2 7x 18 x x 1 5 2 4 2 x 1 x 2 x 9 0 0 0. The intervals determined by the cut 2 x 1 2 x 1 2 x 1 points are 2, 2 1, 1 9, and 9 .
35. We start by moving all terms to one side and factoring:
Sign of x 2 x 1 x 9
r x
2
2 1
1 9
9
The cut point 1 is inadmissible in the original inequality, so the solution is [2 1 [9 .
110
CHAPTER 3 Polynomial and Rational Functions
x 1 x 2 x 1 x 2 0 36. We start by moving all terms to one side and factoring: x 3 x 2 x 3 x 2 2 x 12 x 2 x 2 x 1 x 3 0 r x 0. The intervals determined by the cut points are x 3 x 2 x 3 x 2 3, 3 12 , 12 2 , and 2 . 3 12 12 2 Sign of 3 2 x 3
The cut points fail to satisfy the strict inequality, so the solution is 3 12 2 .
x 12 x 2
r x (note negative sign)
6 6 6 6 1 1 0 x 1 x x 1 x x2 x 6 6x 6 x 1 x x 1 x 2 x 3 0 0 r x 0. The intervals determined by the x x 1 x x 1 x x 1 cut points are 2, 2 0, 0 1, 1 3, and 3 .
37. We start by moving all terms to one side and factoring:
Sign of x 2 x
x 1 x 3
r x (note negative sign)
2
2 0
0 1
1 3
3
The cut points 0 and 1 are inadmissible in the original inequality, so the solution is [2 0 1 3]. 1 2x 1 2 x 3 x 2 x x 2 1 1 2x x 2 x 1 x 3 x 1 2x x 3 0 0 x 3 x 2 x 2 x 1 x 2 x 1 x 3 3x 1 0. The intervals determined by the cut points are 2, 2 13 , 13 1 , 1 3, and x 2 x 1 x 3 3 . 2 13 13 1 Sign of 2 1 3 3
38. We start by moving all terms to one side and factoring:
x 2
3x 1 x 1 x 3
r x
The cut point 13 satisfies equality, but 2, 1, and 3 are not in the domain of r, so the solution is 2 13 1 3 .
SECTION 3.7 Polynomial and Rational Inequalities
111
39. We start by moving all terms to one side and factoring: 1 1 1 x 22 x 2 2 x 1 x 2 x 2 x 1 x 22 x 22 r x
x 22 x 2 x 1 x 1 x 22
and 2. We make a sign diagram:
x 2 2x 1
x 1 x 22
Sign of x 1
x 1
x 12
x 1 x 22
2 1
1
r x
x 1
x 22
2
x 22
x 1 x 22
0. The cut points are 1
Neither cut point satisfies the original inequality, so the solution is 2 2 1.
1 2 1 1 2 1 0 x x 1 x 2 x x 1 x 2 x 2 3x 2 x 2 2x 2x 2 2x x 1 x 2 x x 2 2x x 1 0 0 x x 1 x 2 x x 1 x 2 3x 2 r x 0. The intervals determined by the cut points are 2, 2 1, 1 23 , 23 0 , x x 1 x 2 and 0 . 1 23 23 0 Sign of 2 2 1 0
40. We start by moving all terms to one side and simplifying:
x 2 x 1
3x 2 x
r x
None of the cut points satisfies the strict inequality. Thus, the solution is 2 1 23 0 .
41. The graph of f lies above that of g where f x g x; that is, where x 2 3x 10 x 2 3x 10 0 x 2 x 5 0. We make a sign diagram: Sign of x 2 x 5
x 2 x 5
2
2 5
5
Thus, the graph of f lies above the graph of g on 2 and 5 . 42. The graph of f lies above that of g where f x g x; that is, where
1 . We make a sign diagram: x x 1
Sign of x
1 1 1 1 x 1 x 0 0 x x 1 x x 1 x x 1
0
0 1
1
x 1 1 x x 1 Thus, the graph of f lies above the graph of g on 0 1.
112
CHAPTER 3 Polynomial and Rational Functions
43. The graph of f lies above that of g where f x g x; that is, where 4x r x
2x 1 2x 1 0. We make a sign diagram: x 12 12 0 Sign of 2x 1
1 4x 2 1 1 4x 0 0 x x x
0 12
Thus, the graph of f lies above the graph of g on 12 0 and 12 .
x
2x 1
r x
1 2
x3 x2 2 2 2 0 44. The graph of f lies above that of g where f x g x; that is, where x 2 x x 2 x 0 x x x x 1 x 2 2x 2 0. The second factor in the numerator is positive for all x. We make a sign diagram: r x x Sign of
0
0 1
1
x x 1
r x
Thus, the graph of f lies above the graph of g on 0 and 1 . 45. f x 6 x x 2 is defined where 6 x x 2 x 2 x 3 0. We make a sign diagram: Sign of x 2
2
2 3
3
x 3
x 2 x 3
Thus, the domain of f is [2 3]. 5x 5x is defined where 0 and 5 x 0. We make a sign diagram: 46. g x 5x 5x Sign of 5x
5
5 5
5
1
1 1
1
5x 5x 5x The cut point 5 is permissible, and so the domain of g is [5 5. 4 47. h x x 4 1 is defined where x 4 1 x 1 x 1 x 2 1 0. The last factor is positive for all x. We make a sign diagram:
Sign of x 1 x 1
x2 1
x 1 x 1 x 2 1
Thus, the domain of h is 1] [1 .
SECTION 3.7 Polynomial and Rational Inequalities
48. f x
1
x 4 5x 2 4 make a sign diagram:
is defined where x 4 5x 2 4 x 2 4 x 2 1 x 2 x 2 x 1 x 1 0. We Sign of x 2 x 1 x 1 x 2
x 4 5x 2 4
2
2 1
1 1
1 2
2
Thus, the domain of h is 2 1 1 2 . 49.
50.
20
20 10
10 4
2
2
4
4
2
10
2
10 20
From the graph, we see that x 3 2x 2 5x 6 0 on
From the graph, we see that 2x 3 x 2 8x 4 0 on 2] 12 2 .
[2 1] [3 .
51.
52.
2
1
4
4
2
2 1
2
2
2
4
1 2
2
3
4
From the graph, we see that x 4 4x 3 8x 0 on
approximately 137 037 1.
approximately 124 0 2 324 .
54.
53.
20
40
10
20 1.6
1
1
4
From the graph, we see that 2x 3 3x 1 0 on
0
1
2
From the graph, we see that 5x 4 8x 3 on 0 85 . 55.
113
2
1
1
2
10
From the graph, we see that x 5 x 3 x 2 6x on
approximately [131 0] [151 .
1 x2 x 4 x x 1 1 x2 1 x2 4 x x 1 4 x x 1 0 0 x x x
3x 2 2x 1 1 2x x 2 4x 2 4x 1 x 3x 1 0 0 r x 0. The domain of r is 0 , and x x x both 3x 1 and x are positive there. 1 x 0 for x 1, so the solution is 0 1].
114
CHAPTER 3 Polynomial and Rational Functions
7x 8 56. 23 x 13 x 212 12 x 23 x 212 0 16 x 212 x 13 [4 x 2 3x] 0 r x 0. 3 6 x x 2 Note that the domain of r is 2 . We make a sign diagram with cut points 87 and 0: 2 87 87 0 Sign of 0 7x 8 3 x x 2
Neither cut point is a solution. The solution is 87 0 .
r x
57. We want to solve P x x a x b x c x d 0, where a b c d. We make a sign diagram with cut points a, b, c, and d: Sign of
a
a b
b c
c d
x a
d
x b x c
x d P x
Each cut points satisfies equality, so the solution is a] [b c] [d .
[x a] x b x 2 a b x ab x a x b 0. Note that x c x c x c 0 a c, so c a 0. We make a sign diagram with cut points c, a, and b:
58. Factoring the numerator, we have r x Sign of
x c x b
x a P x
c
c a
a b
b
The cut points a and b satisfy equality, but c is not in the domain of P. Thus, the solution is c [a b]. 2500 275 x 2 2 2500 2500 300 2 275 0 0 59. We want to solve the inequality T x 300 25 2 x 2 x 2 x2 2 25 11x 2 78 0. Because x represents distance, it is positive, and the denominator is positive for all x. Thus, x2 2 the inequality holds where 11x 2 78 x 78 11 266. The temperature is below 300 C at distances greater than 266 meters from the center of the fire.
2 175 0 2 25 4375 0. Using the Quadratic Formula, we find 60. We want to solve d 175 25 25 252 4 1 4375 25 18,125 25 18,125 . Because is positive, we have 548, so 2 1 2 2 Kerry can travel at up to 548 mih.
CHAPTER 3
61.
Review
88x x 2 and y 40 in the viewing rectangle 17 17 20
60
We graph N x
40
[0 100] by [0 60], and see that N x 40 for approximately 95 x 423.
20
Thus, cars can travel at between 95 and 423 mih.
0
0
50
115
100
62. (a) F1 is inversely proportional to the square of the distance between the balls with 5 3 constant of proportionality 3, so F1 2 . F2 is inversely proportional to the x 0 1 cube of the distance with constant of proportionality 1, so F2 3 . Thus, 1 2 3 x 5 1 3 F x F1 F2 3 2 . x x 1 3 (b) F x 0 3 2 0 x 2 3x 3 x 13 . F x 0 on 0 13 and F x 0 on 13 . From the graph x x in part (a), the greatest repulsive force occurs when x 12 .
(c) As the distance gets very small, the attractive force increases without bound, and as the distance gets very large, the repulsive force diminishes, becoming arbitrarily close to 0.
CHAPTER 3 REVIEW
1. (a) f x x 2 6x 2 x 2 6x 2 x 2 6x 9 2 9
2. (a) f x 2x 2 8x 4 2 x 2 4x 4 2 x 2 4x 4 4 8 2 x 22 4
x 32 7
(b)
(b)
y
y
1 _1
0 x
1 0
1
x
116
g(x)
CHAPTER 3 Polynomial and Rational Functions
3. (a) f x 1 10x x 2 x 2 10x 1 x 2 10x 25 1 25 x 52 26
(b)
4. (a) f x 2x 2 12x 2 x 2 6x 2 x 2 6x 9 18 2 x 32 18
(b)
y
y
10 0
1 x
5 0
1
x
2 5. f x x 2 3x 1 x 2 3x 1 x 2 3x 94 1 94 x 32 54 has the maximum value 54 when x 32 .
6. f x 3x 2 18x 5 3 x 2 6x 5 3 x 2 6x 9 5 27 3 x 32 22 has the minimum value 22 when x 3.
7. We write the height function in vertex form: h t 16t 2 48t 32 16 t 2 3t 32 16 t 2 3t 94 2 32 36 16 t 32 68. The stone reaches a maximum height of 68 ft. 8. We write the profit function in vertex form:
P x 1500 12x 0004x 2 0004 x 2 3000x 1500 0004 x 2 3000x 15002 1500 0004 15002 0004 x 15002 7500
Thus, the maximum profit of $7500 is achieved when 1500 units are sold.
9. P x x 3 64
10. P x 2x 3 16 2 x 2 x 2 2x 4
y
y
5
(0, 64)
(0, _16)
10 1
The domain and range are .
(2, 0) 1
(4, 0)
x
The domain and range are .
x
CHAPTER 3
11. P x 2 x 14 32
117
12. P x 81 x 34 y
y
5
25
(1, 0) x
(_3, 0)
Review
(0, 0)
(6, 0) 1
x
(0, _30)
The domain is and the range is [32 .
13. P x 32 x 15
The domain is and the range is 81].
14. P x 3 x 25 96
y
y
(0, 31) 10
(_1, 0)
1
x
20 (0, 0)
1
The domain and range are .
The domain and range are .
15. (a) P x x 3 x 1 x 5 x 3 7x 2 7x15 has odd degree and a positive leading coefficient, so
16. (a) P x x 5 x 2 9 x 2
x 4 3x 3 19x 2 27x 90
y as x and y as x . (b)
x
has even degree and a negative leading coefficient, so
y
y as x and y as x . (b)
y 20
10 1
1 x
x
118
CHAPTER 3 Polynomial and Rational Functions
17. (a) P x x 12 x 4 x 22 x 5 2x 4 11x 3 8x 2 20x 16
18. (a) P x x 2 x 2 4 x 2 9 x 6 13x 4 36x 2 has even degree and a positive leading coefficient, so y as x and y as x .
has odd degree and a negative leading coefficient, so y as x and y as x .
(b)
y
y
(b)
100
20 x
1 _5
5
x
_100
19. (a) P x x 3 x 22 . The zeros of P are 0 and 2,
20. (a) P x x x 13 x 12 . The zeros of P are 1,
(b) We sketch the graph using the guidelines from
(b) We sketch the graph using the guidelines from
with multiplicities 3 and 2, respectively.
Section 3.2.
0, and 1, with multiplicities 3, 1, and 2, respectively.
Section 3.2. y
y
1
1
x
0.1 1
x
21. y x 2 8x, [4 12] by [50 30]. xintercepts: 0 and 22. P x x 3 4x 1. xintercepts: 21, 03, and 19. 8. yintercept: 0. No local minimum. Local maximum
yintercept: 1. Local maximum 12 41. Local
4 16. End behavior: y as x .
minimum 12 21. End behavior: y as
20
x and y as x . 4
4 2 20 40
2 4 6 8 10 12
2 3
2
1 2 4
1
2
3
CHAPTER 3
23. P x 2x 3 6x 2 2, [2 5] by [6 8].
xintercepts: 05, 07, and 29. yintercept: 2. Local
Review
119
24. y x 4 4x 3 , [5 5] by [30 30]. xintercepts: 4 and 0. yintercept: 0. Local minimum at 3 27. No local
maximum is 2 6. Local minimum is 0 2. End
maximum. End behavior: y as x .
behavior: y as x and y as x .
20
5 4
2
2
2
2
4
20
4
5
25. P x 3x 4 4x 3 10x 1. xintercepts: 01 and 21. 26. P x x 5 x 4 7x 3 x 2 6x 3. xintercepts: yintercept: 1. Local minimum is 14 145. There is
30, 13, and 19. yintercept: 3. Local maxima are
no local maximum. End behavior: y as x .
24 332 and 05 50. Local minima are 06 06 and 16 16. End behavior: y as x
20
and y as x .
10 1
1
10
2
40
3
20
20 4
2
2 20
27. (a) Use the Pythagorean Theorem and solving for y 2 we have, x 2 y 2 102 y 2 100 x 2 . Substituting we get S 138x 100 x 2 1380x 138x 3 .
28. (a) The area of the four sides is 2x 2 2x y 1200
600 x 2 2x y 1200 2x 2 y . Substituting x 600 x 2 600x x 3 . we get V x 2 y x 2 x
(b) Domain is [0 10]. (c)
(b) 4000
5000
2000 0
0
5
0
10
(d) The strongest beam has width 58 inches. 29.
x 2 5x 2 3
1 1
5
6
2
4
Using synthetic division, we see that Q x x 2 and R x 4.
20
3x 2 x 5 x 2
2
3
10
(c) V is maximized when x 1414, y 2828. 30.
x 3
0
2
3 3
1 6 5
5 10
5
Using synthetic division, we see that Q x 3x 5 and R x 5.
120
31.
CHAPTER 3 Polynomial and Rational Functions
2x 3 x 2 3x 4 x 5
32. 2
5
3
1
2
Using synthetic division, we see that
294
1
34.
1
0
25
5
1
2
7
85
415
8 17
5
7
49
329
7
47
325
2x 4 3x 3 12 x 4 4
422
83
4
Q x x 2 7x 47 and R x 325.
x 4 8x 2 2x 7 x 5 5
2
Using synthetic division, we see that
Q x 2x 2 11x 58 and R x 294. 33.
0
1
290
58
11
7
4
55
10
x 3 2x 4 x 7
2 2
3
0
0
8
20
80
5
20
80
12 320
308
Using synthetic division, we see that
Using synthetic division, we see that
Q x x 3 5x 2 17x 83 and R x 422.
Q x 2x 3 5x 2 20x 80 and R x 308.
35. [AU: Erroneously numbered 31 in ms]
36.
2x 3 x 2 8x 15 x 2 2x 1 x 2 2x 1
2x
x 4 2x 2 7x x2 x 3 x2 x 3
3
x2
x
x4
x 3 3x 3
2x 3 x 2 8x 15
2x 3 4x 2 2x
x 3 5x 2 7x 4x 2 4x 0
3x 2 6x 3
4x 2 4x 12
12
37. P x 2x 3 9x 2 7x 13; find P 5. 5
2
9
10
2
1
Therefore, P 5 3.
7
5
2
13 10
3
0
x 3 x 2 3x
3x 2 6x 15
Therefore, Q x 2x 3, and R x 12.
4
x 4 0x 3 2x 2 7x
12
Therefore, Q x x 2 x 4, and R x 12. 38. Q x x 4 4x 3 7x 2 10x 15; find Q 3 3
1 1
4
7
10
15
3
3
12
6
1
4
2
21
By the Remainder Theorem, we have Q 3 21.
39. The remainder when dividing P x x 500 6x 101 x 2 2x 4 by x 1 is
P 1 1500 6 1201 12 2 1 4 8.
40. The remainder when dividing P x x 101 x 4 2 by x 1 is P 1 1101 14 2 0. Q(x) Q
CHAPTER 3
41. 12 is a zero of P x 2x 4 x 3 5x 2 10x 4 if P 12 0. 1 2
2
1
5
1 2 Since P
1 2
1
2
4
10
2 8
4
121
42. x 4 is a factor of
P x x 5 4x 4 7x 3 23x 2 23x 12 if P 4 0. 4
4
0
1 1
0, 12 is a zero of the polynomial.
Review
4 4
0
7 0
7
23
12
20
12
23 28
5
3
0
Since P 4 0, x 4 is a factor of the polynomial.
43. (a) P x x 5 6x 3 x 2 2x 18 has possible rational zeros 1, 2, 3, 6, 9, 18.
(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x x 5 6x 3 x 2 2x 18 has 3 variations in sign, there are 1 or 3 negative real zeros.
44. (a) P x 6x 4 3x 3 x 2 3x 4 has possible rational zeros 1, 2, 4, 12 , 13 , 23 , 43 , 16 .
(b) Since P x has no variations in sign, there are no positive real zeros. Since P x 6x 4 3x 3 x 2 3x 4 has 4 variations in sign, there are 0, 2, or 4 negative real zeros.
45. (a) P x 3x 7 x 5 5x 4 x 3 8 has possible rational zeros 1, 2, 4, 8, 13 , 23 , 43 , 83 .
(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x 3x 7 x 5 5x 4 x 3 8 has 3 variations in sign, there are 1 or 3 negative real zeros.
46. (a) P x 6x 10 2x 8 5x 3 2x 2 12 has possible rational zeros 1, 2, 3, 4, 6, 12, 12 , 13 , 16 , 23 , 32 , 43 .
(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x 6x 10 2x 8 5x 3 2x 2 12 has 2 variations in sign, there are 0 or 2 negative real zeros.
47. (a) P x x 3 16x x x 2 16
48. (a) P x x 3 3x 2 4x
x x 1 x 4
x x 4 x 4
has zeros 0, 1, and 4 (all of multiplicity 1).
has zeros 4, 0, 4 (all of multiplicity 1). (b)
(b)
y
y
2
5 1
x
1
x
122
CHAPTER 3 Polynomial and Rational Functions
49. (a) P x x 4 x 3 2x 2 x 2 x 2 x 2
50. (a) P x x 4 5x 2 4 x 2 4 x 2 1 x 2 x 2 x 1 x 1
x 2 x 2 x 1
Thus, the zeros are 1, 1, 2, 2 (all of multiplicity
The zeros are 0 (multiplicity 2), 2 (multiplicity 1),
1).
and 1 (multiplicity 1).
(b)
(b)
y
y
5
1
x
1
x
1
51. (a) P x x 4 2x 3 7x 2 8x 12. The possible rational zeros are 1, 2, 3, 4, 6, 12. P has 2 variations in sign, so it has either 2 or 0 positive real zeros. 1
1
2
8
1
7
1
1
8
1
8
2 1 0 6
6
0
12
2
1
2
7
2
0
0
2
12
12
0
14
12
7
6
P x x 4 2x 3 7x 2 8x 12 x 2 x 3 7x 6 . Continuing: 2
zero
4
4
1 2 2 10
3 1 0 7 6 3
9
6
1 3
2
0
8
(b)
zero.
0
x 2 is a root.
y
so x 3 is a root and
P x x 2 x 3 x 2 3x 2
5
x 2 x 3 x 1 x 2
Therefore the real roots are 2, 1, 2, and 3 (all of multiplicity 1).
x
1
x
zeros
52. (a) P x x 4 2x 3 2x 2 8x 8 x 2 x 2 2x 2 4 x 2 2x 2 x 2 4 x 2 2x 2 x 2 x 2 x 2 2x 2
(b)
y
5
The quadratic is irreducible, so the real roots are 2 (each of multiplicity 1).
1
zeros
zeros
CHAPTER 3
Review
123
53. (a) P x 2x 4 x 3 2x 2 3x 2. The possible rational roots are 1, 2, 12 . P has one variation in sign, and hence 1 positive real root. P x has 3 variations in sign and hence either 3 or 1 negative real roots. 1
zero.
2 2
1
2
2
3
3
2
5
zeros.
2
3
5 2 0 x 1 is a zero. P x 2x 4 x 3 2x 2 3x 2 x 1 2x 3 3x 2 5x 2 . (b)
Continuing: 1 2 2
3
2
2 2
2 1 4
12 2 2
5
1
4 2
3
5
3 5
2
4 2 14
2 1 7 12
2
1 1 2 2
4
y
5
0 x 12 is a zero.
1
x
P x x 1 x 12 2x 2 2x 4 . The quadratic is irreducible, so the real zeros are 1 and 12 (each of
multiplicity 1).
54. (a) P x 9x 5 21x 4 10x 3 6x 2 3x 1. The possible rational zeros are 1, 13 , 19 . P has 3 variations in sign, hence 3 or 1 positive real roots. P x has 2 variations in sign, hence 2 or 0 real negative roots. 1 9 21
zeros.
10
6 3 1
9 12 2
4
1
zeros.
9 12 2 4 1 0 x 1 is a zero. P x x 1 9x 4 12x 3 2x 2 4x 1 . Continuing: 1 9 12 2
4
1
9 3 5 1
9 3 5 1 P x x 12 9x 3 3x 2 5x 1 .
0 x 1 is a zero again. (b)
y
Continuing:
1 9 3 5 1 9
9
6
6
1
4
1
0 x 1 is a zero yet again. P x x 13 9x 2 6x 1 x 13 3x 12
So the real zeros of P are 1 (multiplicity 3) and 13 (multiplicity 2).
1
x
124
CHAPTER 3 Polynomial and Rational Functions
55. Because it has degree 3 and zeros 12 , 2, and 3, we can write P x C x 12 x 2 x 3
C x 3 92 C x 2 72 C x 3C. In order that the constant coefficient be 12, we must have 3C 12 C 4, so P x 4x 3 18x 2 14x 12.
56. Because it has degree 4 and zeros 4, 4, and 3i, 3i must also be a zero. Thus, P x C x 42 x 2 9 C x 4 8C x 3 25C x 2 72C x 144C.
The coefficient of x 2 is 25 25C, so C 1 and hence P x x 4 8x 3 25x 2 72x 144.
57. No, there is no polynomial of degree 4 with integer coefficients that has zeros i, 2i, 3i and 4i. Since the imaginary zeros of polynomial equations with real coefficients come in complex conjugate pairs, there would have to be 8 zeros, which is impossible for a polynomial of degree 4. 58. P x 3x 4 5x 2 2 3x 2 2 x 2 1 . Since 3x 2 2 0 and x 2 1 0 have no real zeros, it follows that 3x 4 5x 2 2 has no real zeros.
solutions,
59. P x x 3 x 2 x 1 has possible rational zeros 1. 1
1
1
1
1
0
1
0
1
1
1
0
So P x x 1 x 2 1 . Therefore, the zeros are 1 and i.
x 1 is a zero.
60. P x x 3 8 x 2 x 2 2x 4 0, so 2 is a zero. Using the Quadratic Formula, the other zeros are 2 22 4 1 4 1 12 x 1 3i. 2 2 61. P x x 3 3x 2 13x 15 has possible rational zeros 1, 3, 5, 15. 1
1
3
1
1
13
15
2
15
2 15 0 x 1 is a zero. So P x x 3 3x 2 13x 15 x 1 x 2 2x 15 x 1 x 5 x 3. Therefore, the zeros are 3, 1, and 5.
62. P x 2x 3 5x 2 6x 9 has possible rational zeros 1, 3, 9, 12 , 32 , 92 . Since there is one variation in sign, there is a positive real zero. 1 2 5 6 9 2
7
1
3 2
5 6 9
6 33 81
3 2 5 6 9 2
3 12
9
2 11 27 72 x 3 is an upper bound 2 8 6 0 x 32 is a zero. So P x 2x 3 5x 2 6x 9 2x 3 x 2 4x 3 2x 3 x 3 x 1. Therefore, the zeros are 3, 1 2 7
and 32 .
1 8
CHAPTER 3
Review
125
63. P x x 4 6x 3 17x 2 28x 20 has possible rational zeros 1, 2, 4, 5, 10, 20. Since all of the coefficients are positive, there are no positive real zeros. 1
1
6
17
28
20
1
5
12
16
2
1
17
28
20
2
8
18
20
9 10 0 x 2 is a zero. P x x 4 6x 3 17x 2 28x 20 x 2 x 3 4x 2 9x 10 . Continuing with the quotient, we have 1
5
12
16
1
6
4
2
1
4
2
1
2
9
10
4
10
5
4
0 x 2 is a zero. Thus P x x 4 6x 3 17x 2 28x 20 x 22 x 2 2x 5 . Now x 2 2x 5 0 when
x 2 4451 24i 1 2i. Thus, the zeros are 2 (multiplicity 2) and 1 2i. 2 2
64. P x x 4 7x 3 9x 2 17x 20 has possible rational zeros 1, 2, 4, 5, 10, 20. 1 1 7
1
9 17 20 2 1 7 8
17
0
9 17 20
2 18
54
1 9 27
37
1 1
74
7
9 17 20
1 6
3
20
54 x 2 is an 1 6 3 20 0 x 1 is a zero. upper bound. So P x x 4 7x 3 9x 2 17x 20 x 1 x 3 6x 2 3x 20 . Continuing with the quotient, we have 1 8 17
0 20
1 1
6
3 20
1 5
2
2 1
3 20
4 1
10
2 8
5 2 18
4 5 10
6
3 20
4 8
20
1 2 5 0 x 4 is a zero. So P x x 4 7x 3 9x 2 17x 20 x 1 x 4 x 2 2x 5 . Now using the Quadratic Formula on 2 2 6 2 4 4 1 5 1 6. Thus, the zeros are 4, 1, and 1 6. x 2 2x 5 we have: x 2 2 1
1
6
65. P x x 5 3x 4 x 3 11x 2 12x 4 has possible rational zeros 1, 2, 4. 1 1 3 1 11 12 1 2 3
4
8 4
1 2 3
8 4 0 x 1 is a zero. P x x 5 3x 4 x 3 11x 2 12x 4 x 1 x 4 2x 3 3x 2 8x 4 . Continuing with the quotient, we have 1 1 2 3
8 4
1 1 4
4
1 1 4 4 0 x 1 is a zero. x 5 3x 4 x 3 11x 2 12x 4 x 12 x 3 x 2 4x 4 x 13 x 2 4 x 13 x 2 x 2
Therefore, the zeros are 1 (multiplicity 3), 2, and 2. 66. P x x 4 81 x 2 9 x 2 9 x 3 x 3 x 2 9 x 3 x 3 x 3i x 3i. Thus, the zeros are 3, 3i.
126
CHAPTER 3 Polynomial and Rational Functions
67. P x x 6 64 x 3 8 x 3 8 x 2 x 2 2x 4 x 2 x 2 2x 4 . Now using the Quadratic
Formula to find the zeros of x 2 2x 4, we have 3 1 i 3, and using the Quadratic Formula to find the zeros of x 2 2x 4, we have x 2 4441 22i 2 2 22i2 3 1 i 3. Therefore, the zeros are 2, 2, 1 i 3, and 1 i 3. x 2 4441 2
1 . 68. P x 18x 3 3x 2 4x 1 has possible rational zeros 1, 12 , 13 , 16 , 19 , 18
1 18
1 18 2
3 4 1
18 21 17
3 4 1
9
6
1
18 21 17 16 x 1 is an upper bound. 18 12 2 0 x 12 is a zero. So P x 18x 3 3x 2 4x 1 2x 1 9x 2 6x 1 2x 1 3x 12 . Thus the zeros are 12 and 13
(multiplicity 2).
69. P x 6x 4 18x 3 6x 2 30x 36 6 x 4 3x 3 x 2 5x 6 has possible rational zeros 1, 2, 3, 6. 1 6 18
6 30
6 12
6 12
36
6 36
6 36
0 x 1 is a zero. 6 x 1 x 3 2x 2 x 6 .
So P x 6x 4 18x 3 6x 2 30x 36 x 1 6x 3 12x 2 6x 36 Continuing with the quotient we have 1 1 2 1 6
2 1 2 1 6
1 1 2
2
1
3 1 2 1 6
0 2
3
6
1 2 0 x 3 is a zero. So P x 6x 4 18x 3 6x 2 30x 36 6 x 1 x 3 x 2 x 2 . Now x 2 x 2 0 when 1 1 2 8
0 1 8
3
1
7 7 x 1 1412 1i , and so the zeros are 1, 3, and 1i . 2 2 2
70. P x x 4 15x 2 54 x 2 9 x 2 6 . If x 2 9, then x 3i. If x 2 6, then x i 6. Therefore, the zeros are 3i and i 6. 71. 2x 2 5x 3 2x 2 5x 3 0. The solutions are x 05, 3.
2
4
72. Let P x x 3 x 2 14x 24. The solutions to P x 0 are x 3, 2, and 4. 5
5 20
5 40
CHAPTER 3
73. x 4 3x 3 3x 2 9x 2 0 has solutions x 024, 424.
Review
127
74. x 5 x 3 x 5 x 3 0. We graph
P x x 5 x 3. The only real solution is 134. 10
5
2
2
50 10
75. P x x 3 2x 4
1
1
0 1
1
1
2
1
1
4 1 5
2
1
0 2
1
2
4
2
0
4
2
4
P x x 3 2x 4 x 2 x 2 2x 2 . Since x 2 2x 2 0 has no real solution, the only real zero of P is
x 2.
76. P x x 4 3x 2 4 x 2 1 x 2 4 x 1 x 1 x 2 4 . Since x 2 4 0 has no real solution, the only real zeros of P are x 1 and x 1.
3 . The vertical asymptote is x 4. Because the x 4 denominator has higher degree than the numerator, the horizontal
77. (a) r x
y
asymptote is y 0. When x 0, y 34 , so the yintercept is 34 .
There is no xintercept because the numerator is never 0. The domain
1
of r is 4 4 and its range is 0 0 . 3 1 1 3 3 f x 4, so we (b) If f x , then r x x x 4 x 4
1
x
obtain the graph of r by shifting the graph of f to the left 4 units and stretching vertically by a factor of 3.
78. (a) r x
1 . The vertical asymptote is x 5 and the horizontal x 5
y
asymptote is y 0. When x 0, y 15 , so the yintercept is 15 .
There is no xintercept. The domain of r is 5 5 and its range is 0 0 .
1 1 1 f x 5. (b) If f x , then r x x x 5 x 5
Thus, we obtain the graph of r by shifting the graph of f to the right 5 units and reflecting in the xaxis.
1 1
x
128
CHAPTER 3 Polynomial and Rational Functions
3x 4 . The vertical asymptote is x 1. Because the x 1 denominator has the same degree as the numerator, the horizontal
79. (a) r x
y
asymptote is y 31 3. When x 0, y 4, so the yintercept is 4. When y 0, 3x 4 0 x 43 , so the xintercept is 43 . The
domain of r is 1 1 and its range is 3 3 .
1 , then x 3 x 1 1 1 r x 3 3 f x 1. Thus, we x 1 x 1 obtain the graph of r by shifting the graph of f to the right 1 unit,
(b) If f x
1 x
1
reflecting in the xaxis, and shifting upward 3 units. 80. (a) r x
2x 5 . The vertical asymptote is x 2 and the horizontal x 2
y
asymptote is y 2. When x 0, y 52 , so the yintercept is 52 .
When y 0, x 52 , so the xintercept is 52 . The domain of r is 2 2 and its range is 2 2 .
1
1 , then x 2 x 2 1 1 r x 2 2 f x 2. Thus, we x 2 x 2 obtain the graph of r by shifting the graph of f to the left 2 units and
(b) If f x
1x
_1
upward 2 units. 12 3x 12 . When x 0, we have r 0 12, so the yintercept is x 1 1 12. Since y 0, when 3x 12 0 x 4, the xintercept is 4. The vertical
81. r x
y
asymptote is x 1. Because the denominator has the same degree as the
numerator, the horizontal asymptote is y 31 3. The domain of r is
5
1 1 and its range is 3 3 .
82. r x
1 x 22
1 1 . When x 0, we have r 0 2 , so the yintercept is 14 . 4 2
1
x
y
Since the numerator is 1, y never equals zero and there is no xintercept. There is a vertical asymptote at x 2. The horizontal asymptote is y 0 because the
degree of the denominator is greater than the degree of the numerator. The domain of r is 2 2 and its range is 0 . 1 1
x
CHAPTER 3
x 2 x 2 1 83. r x 2 . When x 0, we have r 0 2 8 4 , x 2 x 4 x 2x 8
y
xintercept is 2. There are vertical asymptotes at x 2 and x 4. The domain
1
129
Review
so the yintercept is 14 . When y 0, we have x 2 0 x 2, so the of r is 2 2 4 4 and its range is .
84. r x
x
1
x 3 27 27 . When x 0, we have r 0 27 4 , so the yintercept is y 4 . x 4
y
When y 0, we have x 3 27 0 x 3 27 x 3. Thus the xintercept
is x 3. The vertical asymptote is x 4. Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. By long division, we have r x x 2 4x 16
37 . So the end behavior of y is x 4
20
like the end behavior of g x x 2 4x 16. The domain of r is
x
2
4 4 and its range is .
85. r x
x2 9 x 3 x 3 . When x 0, we have r 0 9 1 , so the 2x 2 1 2x 2 1
y 1
yintercept is 9. When y 0, we have x 2 9 0 x 3 so the xintercepts
1
x
are 3 and 3. Since 2x 2 1 0, the denominator is never zero so there are no
vertical asymptotes. The horizontal asymptote is at y 12 because the degree of
the denominator and numerator are the same. The domain of r is and its range is 9 12 .
86. r x
2x 2 6x 7 7 . When x 0, we have r 0 7 4 4 , so the yintercept x 4
y
is y 74 . We use the Quadratic Formula to find the xintercepts:
6 62 427 6 92 3 23 . Thus the xintercepts are x 4 2 22
x 39 and x 09. The vertical asymptote is x 4. Because the degree of the
numerator is greater than the degree of the denominator, there is no horizontal
1 asymptote. By long division, we have r x 2x 2 , so the slant x 4 asymptote is r x 2x 2. The domain of r is 4 4 and its range is approximately 717] [1283 .
2 1
x
130
CHAPTER 3 Polynomial and Rational Functions
x 2 5x 14 x 7 x 2 x 7 for x 2. The xintercept is x 2 x 2 7 and the yintercept is 7, there are no asymptotes, the domain is x x 2, and
87. r x
y
(2, 9)
the range is y y 9.
1 0
x x 2 3x 10
x x 5 x 2 x 3 3x 2 10x x x 5 x 2 x 2 x 2 for x 2. The xintercepts are 0 and 5 and the yintercept is 0, there are no asymptotes, the domain is x x 2, and the range is y y 25 4 .
88. r x
1
x
y (_2, 14)
2 0
x 2 3x 18 x 6 x 6 x 3 for x 3. The xintercept is 89. r x 2 x 5 x 5 x 3 x 8x 15
6 and the yintercept is 65 , the vertical asymptote is x 5, the horizontal asymptote is y 1, the domain is x x 3 5, and the range is y y 1 92 .
1
x
y
x-5
2 2
x 2 2x 15 x 3 x 3 x 5 90. r x 3 for 2 x 1 x 2 x 5 x 1 x 2 x 4x 7x 10
y
x 5. The xintercept is 3 and the yintercept is 32 , the vertical asymptotes are
x 1 and x 2, the horizontal asymptote is y 0, the domain is x x 5 1 2, and the range is y y 19 or y 1 (You can use a graphing calculator to find the range.)
x 3 . From the graph we see that the xintercept is 3, the yintercept is 2x 6 05, there is a vertical asymptote at x 3 and a horizontal asymptote at
91. r x
y 05, and there is no local extremum.
x
(3, _ 92 )
1
(_5, _ 27 )
0
1
x
5
10 8 6 4 2 5
2
4
CHAPTER 3
2x 7 . From the graph we see that the xintercept is 35, the yintercept 92. r x 2 x 9 is 078, there is a horizontal asymptote at y 0 and no vertical asymptote, the
20
local minimum is 111 090, and the local maximum is 811 012.
Review
131
20
1
x3 8 . From the graph we see that the xintercept is 2, the 93. r x 2 x x 2 yintercept is 4, there are vertical asymptotes at x 1 and x 2, there is no
10
horizontal asymptote, the local maximum is 0425 3599, and the local
5
minimum is 4216 7175. By using long division, we see that
10 x f x x 1 2 , so f has a slant asymptote of y x 1. x x 2
5 10
2x 3 x 2 . From the graph we see that the xintercepts are 0 and 12 , the x 1 yintercept is 0, there is a vertical asymptote at x 1, the local maximum is
94. r x
50
0 0, and the local minima are 157 1790 and 032 003. Using long division, we see that r x 2x 2 3x 3
3 . So the end x 1
5
behavior of r is the same as the end behavior of g x 2x 2 3x 3.
5
95. 2x 2 x 3 2x 2 x 3 0 P x x 1 2x 3 0. The cut points occur where x 1 0 and where 2x 3 0; that is, at x 1 and x 32 . We make a sign diagram: 1 32 Sign of 1 x 1
2x 3 P x
3 2
Both endpoints satisfy the inequality. The solution is 1] 32 . 96. x 3 3x 2 4x 12 0 x 2 x 3 4 x 3 0 P x x 2 x 2 x 3 0. The cut points are 2, 2, and 3. We make a sign diagram: Sign of x 2 x 2 x 3 P x
2
2 2
2 3
3
All endpoints satisfy the inequality. The solution is 2] [2 3].
132
CHAPTER 3 Polynomial and Rational Functions
97. x 4 7x 2 18 0 x 2 9 x 2 2 0 P x x 3 x 3 x 2 2 0. The last factor is positive for all x. We make a sign diagram:
Sign of x 3
3
3 3
3
x 3 P x
Neither endpoint satisfies the strict inequality. The solution is 3 3.
98. x 8 17x 4 16 0 x 4 16 x 4 1 0 x 2 4 x 2 4 x 2 1 x 2 1 0 P x x 2 4 x 2 1 x 4 x 4 x 1 x 1 0. The first two factors are positive for all x. We make a sign diagram:
Sign of x 4
4
4 1
1 1
1 4
4
x 1 x 1 x 4 P x
None of the endpoints satisfies the strict inequality. The solution is 2 1 1 2 .
99.
5 5 5 0 r x 0. We make a sign diagram: 0 2 x 1 x 2 x 2 x 3 x 2 4x 4 x x 4 x2 4 Sign of x 2 x 1 x 2
r x
2
2 1
1 2
2
None of the endpoints is in the domain of r . The solution is 2 1 2.
100.
2 3x 1 2 3 3x 1 2 x 2 9x 3 2x 4 7x 1 3x 1 0 0 0 r x 0. We x 2 3 x 2 3 3 x 2 3 x 2 3 x 2 make a sign diagram: 1 2 17 Sign of 2 7 x 2
7x 1
r x
The cut point 17 satisfies equality, but 2 is not in the domain of r. The solution is 2 17 .
CHAPTER 3
101.
Review
133
2 3 1 2 3 x x 3 2x x 2 3 x 2 x 3 1 0 0 x 2 x 3 x x 2 x 3 x x x 2 x 3 2 9 2x 4x 18 0 r x 0. We make a sign diagram: x x 2 x 3 x x 2 x 3 9 2 92 Sign of 3 3 0 0 2 2 x 3 x
x 2
9 2x
r x
The cut point 92 satisfies equality, but the other cut points are not in the domain of r . The solution is 3 0 2 92 . 102.
3 4 1 3 4 x x 3 3x x 2 4 x 2 x 3 1 0 0 x 2 x 3 x x 2 x 3 x x x 2 x 3 7x 24 r x 0. We make a sign diagram: x x 2 x 3 24 24 Sign of 2 0 0 3 3 7 7 2 7x 24 x 2 x
x 3
r x
24 The cut point 24 7 satisfies equality, but the other cut points are not in the domain of r. The solution is 7 2 0 3.
103. f x
24 x 3x 2 x 3 8 3x is defined where r x x 3 8 3x 0. We make a sign diagram: 8 3 83 Sign of 3 3 x 3
8 3x
r x
Both cut points satisfy equality, so the domain of f is 3 83 . 104. g x 4
1 1 is defined where r x x 1 x x 2 x 1 0. 4 4 x x4 x 1 x3 x 1 x x 2 x 1 1
(Equality is excluded because the denominator cannot be 0.) The last factor of r x is positive for all x. We make a sign diagram: Sign of x 1x
x2 x 1
r x
0
0 1
1
Neither cut point satisfies the strict inequality, so the domain of g is 0 1.
134
CHAPTER 3 Polynomial and Rational Functions
105.
106.
40
20
20 2 3
2
1
1
2
1
3
1
2
3
>
20
From the graph, we see that x 4 x 3 5x 2 4x 5 on
From the graph, we see that x 5 4x 4 7x 3 12x 2 0
approximately [074 195].
on approximately 106 017 191 .
107. (a) We use synthetic division to show that 1 is a zero of P x 2x 4 5x 3 x 4: 2
1
2
5
0
1
4
2
3
3
4
3 3 4 0 Thus, 1 is a zero and P x x 1 2x 3 3x 2 3x 4 .
(b) P x has no change of sign, and hence no positive real zeros. But P x x 1 Q x, so Q cannot have a positive real zero either. 108. We want to find the solutions of x 4 x 2 24x 6x 3 20 P x x 4 6x 3 x 2 24x 20 0. The possible rational zeros of P are 1, 2, 4, 5, 10, 20. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x x 4 6x 3 x 2 24x 20 has 1 variation in sign, and so P has 1 negative real zeros. 1
1
6
1
1
1
24
5
4
2
1
20
5
2
x 1 is a zero. 1 3 P x x 1 x 2 x 2 3x 10 x 1 x 2 x 2 x 5, and 5
4
20
20 0
4
20
6
20
10
0
y
x 2 is a zero.
so the zeros of P (and hence the xcoordinates of the points of intersection) are 2, 1, 2, and 5. From the original equation, the coordinates of the points of intersection are 2 28, 1 26, 2 68, 5 770.
100 1
109. (a) y
x
x2 4 has vertical asymptotes at x 0 and x 2, so it has graph VII. x x2 4
1 x 22 x 23 has xintercepts 2 and yintercept 1 0 22 0 23 2, so it has graph V. (b) y 16 16
(c) y x 1 x 22 has xintercepts 1 and 2 and yintercept 4, so it has graph III.
(d) y 2x x 2 x 2 x is a parabola that opens downward with xintercepts 0 and 2 and yintercept 0. It has graph I. (e) y x 2 2x 3 x 1 has xintercepts 32 , 0, and 1, so it has graph IV.
(f) x 2y y 2 y 2 y is a parabola that opens leftward, so it has graph VIII
CHAPTER 3
135
Test
x2 x2 has vertical asymptotes x 1 and horizontal asymptote y 11 1, so it has x 1 x 1 x2 1 graph VI. (h) y 9 x 2 is a semicircle with domain [3 3], so it has graph II.
(g) y
CHAPTER 3 TEST y
1. f x x 2 x 6 x2 x 6 x 2 x 14 6 14 2 x 12 25 4
2. g x 2x 2 6x 3 is a quadratic
function with a 2 and b 6, so it has a maximum or minimum value where b 6 3 x . Because 2a 2 2 2
1 1
x
a 0, this gives the minimum value 2 g 32 2 32 6 32 3 32 .
Domain: , range: 25 . 4
3. (a) We write the function in vertex form: h x 10x 001x 2 001 x 2 1000x 001 x 2 1000x 5002 001 5002 001 x 5002 2500
Thus, the cannonball reaches a maximum height of 2500 feet. (b) By the symmetry of the parabola, we see that the cannonball’s height will be 0 again (and thus it will splash into the water) when x 1000 ft. y
4. f x x 23 27 has yintercept y 23 27 19 and xintercept where x 23 27 x 1.
10
(0, 19) (1, 0)
x
1
5. (a)
2
1
0 2
1
4
2
4 0
2
5
0
4
2
9
(b) 2x 2 1
x 3 2x 2
2x 5 4x 4 x 3 2x 5
Therefore, the quotient is
4x 4
Q x x 3 2x 2 2, and the remainder is
4x 4
x3
1 2 x 2 0x
x2 2x 2 x2
R x 9.
7
x2
7
12 15 2
Therefore, the quotient is Q x x 3 2x 2 12 and the remainder is R x 15 2 .
136
CHAPTER 3 Polynomial and Rational Functions
6. (a) Possible rational zeros are: 1, 3, 12 , 32 . (b)
1
2
5
4 7
2
y
3 3
2
2 7 3 0 x 1 is a zero. P x x 1 2x 2 7x 3 x 1 2x 1 x 3 2 x 1 x 12 x 3
1
x
(c) The zeros of P are x 1, 3, 12 .
7. P x x 3 x 2 4x 6. Possible rational zeros are: 1, 2, 3, 6. 1 1 1 4 1
0
1
6
2 1 1 4 2
4
6
2
3 1 1 4 6 3
4
6
6
0 4 10 1 1 2 10 1 2 2 0 x 3 is a zero. So P x x 3 x 2 2x 2 . Using the Quadratic Formula on the second factor, we have 2 4 2 2 1 2 22 4 1 2 1 i. So zeros of P x are 3, 1 i, and 1 i. x 2 1 2 2 8. P x x 4 2x 3 5x 2 8x 4. The possible rational zeros of P are: 1, 2, and 4. Since there are four changes in sign, P has 4, 2, or 0 positive real zeros. 1
1
5
2
1
4
8
4
1
4
1 1 4 4 0 So P x x 1 x 3 x 2 4x 4 . Factoring the second factor by grouping, we have P x x 1 x 2 x 1 4 x 1 x 1 x 2 4 x 1 x 12 x 2i x 2i.
9. Since 3i is a zero of P x, 3i is also a zero of P x. And since 1 is a zero of multiplicity 2, P x x 12 x 3i x 3i x 2 2x 1 x 2 9 x 4 2x 3 10x 2 18x 9. 10. P x 2x 4 7x 3 x 2 18x 3. (a) Since P x has 4 variations in sign, P x can have 4, 2, or 0 positive real zeros. Since P x 2x 4 7x 3 x 2 18x 3 has no variations in sign, there are no negative real zeros. (b)
4
2 2
7
1
8
4
1
5
18
3
20
8
2
11
Since the last row contains no negative entry, 4 is an upper bound for the real zeros of P x. 1
2 2
7 1 9
1 9 10
18 10 28
3 28 31
Since the last row alternates in sign, 1 is a lower bound for the real zeros of P x.
CHAPTER 3
(c) Using the upper and lower limit from part (b), we graph P x in the viewing rectangle [1 4] by
Test
137
(d) Local minimum 282 7031.
[1 1]. The two real zeros are 017 and 393.
50
1 2
2
4
50 2
4
1
11. (a) P 1 13 2 12 1 4, Q 1 13 2 12 4 1 8 3, R 1 13 4 1 5 0,
S 1 13 2 12 4 1 8 5, and T 1 13 9 12 27 1 27 8. Thus, only R has value 0 at x 1. (b) P x x 3 2x 2 x x x 2 2x 1 x x 12 , Q x x 3 2x 2 4x 8 x 2 x 2 4 x 2 x 2 4 x 2 x 2 x 22 , R x x 3 4x 5 x 1 x x 2 5 , S x x 3 2x 2 4x 8 x 2 x 24 x 2 x 2 x 2 4 , and T x x 3 9x 2 27x 27 x 33 , so only the polynomial P has remainder 0 when divided by x 1.
(c) From part (b), only Q x has x 2 as a factor.
(d) From part (b), only T x has 3 as a zero of multiplicity 3. (e) A polynomial with 2 and 2i as zeros must have x 2 and x 2 4 as factors. From part (b), this describes only S x. x2 x 6 2x 1 x 3 27 x 3 9x x 3 6x 2 9x , u x . 12. r x 2 , s x 2 , t x , and x 2 x 2 x 3 x x 2 x 4 x 25 (a) r has horizontal asymptote y 0 because the degree of the denominator is greater than the degree of the numerator. u has horizontal asymptote y 11 1 because the degrees of the numerator and the denominator are the same.
(b) The degree of the numerator of s x is one more than the degree of the denominator, so s has a slant asymptote. x x 2 6x 9 x x 32 (c) The denominator of s x is never 0, so s has no vertical asymptote. x x x 3 x 3 x 3 for x 3, so has no vertical asymptote.
(d) From part (c), has a “hole” at 3 0. 2x 1 2x 1 , so r has vertical asymptotes at x 1 (e) r x 2 x 1 x 2 x x 2
y
and x 2. y 0 is a horizontal asymptote because the degree of the numerator is less than the degree of the denominator.
(f) u x
x2 x 6 x 3 x 2 . When x 0, we have 2 x 5 x 5 x 25
6 6 , so the yintercept is y 6 . When y 0, we have x 3 or u x 25 25 25
x 2, so the xintercepts are 3 and 2. The vertical asymptotes are x 5 and
x 5. The horizontal asymptote occurs at y 11 1 because the degree of the
denominator and numerator are the same.
1 2
x
138
CHAPTER 3 Polynomial and Rational Functions
x 2 2x 5
(g) x 2
0 x 3 2x 2 2x 2 9x 2x 2 4x 5x 0 5x 10 10
Thus P x x 2 2x 5 and t x 13. x
y
x 3 0x 2 9x
10 2
x
x 3 9x have the same end behavior. x 2
6x 6 x x 2x 5 2x 2 4x 6 6x x 1 3 x 0 x 0 0 r x. We 2x 5 2x 5 2x 5 2x 5 x5 2
make a sign diagram:
Sign of x 1
3x
x 52
r x
1
1 52
53 2
3
The cut point 52 is excluded so that the denominator is not 0. The solution is 1] 52 3 .
1 is defined where 4 2x x 2 0. Using the Quadratic Formula to solve x 2 2x 4 0, we 14. f x 4 2x x 2 2 22 4 1 4 1 5. The radicand is positive between these two roots, so the domain of have x 2 1 f is 1 5 1 5 .
15. (a)
P x x 4 4x 3 8x. From the graph, the xintercepts are approximately
10
124, 0, 2, and 324, P has local maximum P 1 5, and P has local minima P 073 P 273 4.
2
2
4
(b) From part (a), P x 0 on approximately 124] [0 2] [324 .
solutions,
139
Fitting Polynomial Curves to Data
FOCUS ON MODELING Fitting Polynomial Curves to Data
1. (a) Using a graphing calculator, we obtain the quadratic
2. (a) Using a graphing calculator, we obtain the quadratic
polynomial
polynomial
y 0275428x 2 197485x 2735523 (where
y 02783333x 2 184655x 166732 (where
miles are measured in thousands). (b)
y
plants/acre are measured in thousands). y
(b)
100
120
80 60
80
40
40
20 0
25
30
35
40
45
50 x
0
20
30
40
60 x
50
Density (thousand plants/acre)
Pressure (lb/in@)
(c) Moving the cursor along the path of the polynomial,
(c) Moving the cursor along the path of the polynomial, we find that yield when 37,000 plants are planted per
we find that 3585 lb/in2 gives the longest tire life.
3. (a) Using a graphing calculator, we obtain the cubic polynomial
acre is about 135 bushels/acre.
4. (a)
y 60 50
y 000203709x 3 0104522x 2
1966206x 145576.
(b)
10
y
40 30 20 10
20
0 10
1
2
3
x
Time (s) A quadratic model seems appropriate. 35 x
(b) Using a graphing calculator, we obtain the quadratic
(c) Moving the cursor along the path of the polynomial,
(c) Moving the cursor along the path of the polynomial,
0
5
10
15
20
25
30
Seconds
polynomial y 160x 2 518429x 420714.
we find that the subjects could name about
we find that the ball is 20 ft. above the ground 03
43 vegetables in 40 seconds.
seconds and 29 seconds after it is thrown upward.
(d) Moving the cursor along the path of the polynomial,
(d) Again, moving the cursor along the path of the
we find that the subjects could name 5 vegetables in
polynomial, we find that the maximum height is
about 20 seconds.
462 ft.
140
FOCUS ON MODELING
5. (a) Using a graphing calculator, we obtain the quadratic polynomial y 00120536x 2 0490357x 496571. (c) Moving the cursor along the path of the polynomial, we find that the tank should drain in 190 minutes.
(b)
y 6 5 4 3 2 1 0
10
Time (min)
20 x
COMMENTS: 1,2,9,11,15,36,39,45,51,54,55
CHAPTER 4
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
4.1
Exponential Functions 1
4.2
The Natural Exponential Function 9
4.3
Logarithmic Functions 16
4.4
Laws of Logarithms 24
4.5
Exponential and Logarithmic Equations 28
4.6
Modeling with Exponential Functions 36
4.7
Logarithmic Scales 42 Chapter 4 Review 44 Chapter 4 Test 52
¥
FOCUS ON MODELING: Fitting Exponential and Power Curves to Data 55
1
4
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
4.1
EXPONENTIAL FUNCTIONS
1 , f 0 50 1, f 2 52 25, 1. The function f x 5x is an exponential function with base 5; f 2 52 25
and f 6 56 15,625.
2. (a) f x 2x is an exponential function with base 2. It has graph III. x (b) f x 2x 12 is an exponential function with base 12 . It has graph I.
(c) The graph of f x 2x is obtained from the graph of 2x by reflecting about the xaxis. Thus, it has graph II.
(d) The graph of f x 2x is obtained from the graph of 2x by reflecting about the x and yaxes. Thus, it has graph IV. 3. (a) To obtain the graph of g x 2x 1 we start with the graph of f x 2x and shift it downward 1 unit.
(b) To obtain the graph of h x 2x1 we start with the graph of f x 2x and shift it to the right 1 unit. nt for compound interest, the letters P, r, n, and t stand for principal, interest rate per 4. In the formula A t P 1 nr year, number of times interest is compounded per year, and number of years, respectively, and A t stands for the amount accumulated after t years. So if $100 is invested at an interest rate of 6% compounded quarterly, then the amount after 4 2 2 years is 100 1 006 11265. 4 x 5. The exponential function f x 12 has the horizontal asymptote y 0. This means that as x , we have x 1 0. 2 x 6. The exponential function f x 12 3 has the horizontal asymptote y 3. This means that as x , we have x 1 3 3. 2 5 22195, f 2 0063, f 03 1516. 7. f x 4x ; f 12 412 4 2, f 8. f x 3x1 ; f 12 0577, f 25 5196, f 1 0111, f 14 0439. x1 ; g 12 0192, g 2 0070, g 35 15588, g 14 1552. 9. g x 13
3x 10. f x 43 ; f 12 0650, f 6 8281, f 3 0075, f 43 3160.
11.
f x 6x
y
12.
f x 01x
x
y
x
y
2
1 36 1 6
2
100
1
1
0 1
6
2
36
1 1
x
1
10
0
1
1
1 10 1 100
2
Replace "f" with "g" (5x) y
1 1
x
1
2
13.
x
y
g x 23 x
y
2
9 4 3 2
1
0
1
1
2 3 4 9
2 15.
Replace with "g"
CHAPTER 4 Exponential and Logarithmic Functions
y
14. h x 4x x
y
2
1 16 1 4
1
1
x
1
1
1
4
2
16
2 1
x 16. h x 4 58
y
h x 5 22x
0
x
y
x
y
2
1775
3
1638
1
2308
0
30
1
39
2
507
3 4
2
64
0
4
1
25
6591
2
156
8568
3
098
x
1
y
1024
0
1
1 x
1
x 18. f x 8x and g x 18 8x
17. f x 4x and g x 4x y
y
f=g g
f
2
1
x
1
x
1
x 20. f x 34 and g x 15x .
19. f x 4x and g x 7x . y
y
g
f
f
g
2
5 1
x
Something is wrong here. Both graphs should pass through the point (0, 1).
x
0
1
x
SECTION 4.1 Exponential Functions
3
21. From the graph, f 2 a 2 9, so a 3. Thus f x 3x .
22. From the graph, f 1 a 1 15 , so a 5. Thus f x 5x . 1 , so a 1 . Thus f x 1 x . 23. From the graph, f 2 a 2 16 4 4
x 24. From the graph, f 3 a 3 8, so a 12 . Thus f x 12 .
25. The graph of f x 5x1 is obtained from that of y 5x by shifting 1 unit to the left, so it has graph II. 26. The graph of f x 5x 1 is obtained from that of y 5x by shifting 1 unit upward, so it has graph I. x 27. f x 3x 1. The graph of f is obtained by shifting the 28. f x 13 2. The graph of f is obtained by shifting x graph of y 3x upward 1 unit. The yintercept is the graph of y 13 downward 2 units. y f 0 30 1 2. Domain: . 0 Range: 1 . Horizontal asymptote: y 1. yintercept: y f 0 13 2 1. y
Domain: . Range: 2 . Horizontal symptote: y 2.
y
(1, 4)
(0, 2) 1
x
1
1 x
1
29. The graph of g x 5x is obtained by reflecting the graph of y 5x about the xaxis.
yintercept: y g 0 50 1. Domain: .
Range: 0. Horizontal asymptote: y 0. y 1
(0, _1)
30. The graph of g x 5x is obtained by reflecting the graph of y 5x about the yaxis.
yintercept: y g 0 50 1. Domain: .
Range: 0 . Horizontal asymptote: y 0. y
x
1
(1, _5) 2
(0, 1) 1
x
4
CHAPTER 4 Exponential and Logarithmic Functions
31. h x 3x2 . The graph of h is obtained by shifting the graph of y 3x to the right 2 units.
yintercept: y h 0 302 19 . Domain: .
Range: 0 . Horizontal asymptote: y 0. y
1
32. h x 10x1 . The graph of h is obtained by shifting the graph of y 10x to the left 1 unit.
yintercept: y h 0 1001 10.
Domain: . Range: 0 . Horizontal asymptote: y 0.
y
(2, 1)
(0, 1/9)
20
x
1
(0, 10) 1
33. y 2x 3. The graph is obtained by reflecting the
graph of y 2x about the xaxis and then shifting upward 3 units. yintercept: y 20 3 4.
Domain: . Range: 3 . Horizontal asymptote: y 3.
y
x
34. y 2x 3. The graph is obtained by reflecting the
graph of y 2x about the xaxis and shifting upward 3 units. yintercept: y 20 3 2.
Domain: . Range: 3. Horizontal asymptote: y 3.
y
(0, 2)
(1, 1)
1 (0, 4)
1
x
1 x
1
35. y 10x 1. The graph is obtained by reflecting the
graph of y 10x about the xaxis and shifting downward 1 unit. yintercept: y 100 1 2.
Domain: . Range: 1. Horizontal asymptote: y 1.
y 1
x 36. y 12 4. The graph is obtained by shifting the x graph of y 12 downward 4 units.
0 yintercept: y 12 4 3. Domain: .
Range: 4 . Horizontal asymptote: y 4. 1
y
x
(0, _2) 1 1 (0, _3)
x
SECTION 4.1 Exponential Functions
5
37. h x 2x4 1. The graph of h is obtained by shifting
38. y 10x1 5. The graph of y is obtained by shifting the
yintercept: y 204 1 17 16 . Domain: .
yintercept: y 1001 5 5. Domain: .
the graph of y 2x to the right 4 units and upward 1 unit.
Range: 1 . Horizontal asymptote: y 1.
graph of y 10x to the left 1 unit and downward 5 units. Range: 5 . Horizontal asymptote: y 5. y
y
5
2
(4, 2)
(0, 17/16)
1 4
1
x
x
x 39. g x 1 3x 3x 1. The graph of g is obtained 40. y 3 15 5x 3. The graph of y is obtained by reflecting the graph of y 3x about the x and yaxes
and then shifting upward 1 unit.
by reflecting the graph of y 5x about the x and yaxes
yintercept: y 30 1 0. Domain: .
yintercept: y 50 3 2. Domain: .
and then shifting upward 3 unit.
Range: 1. Horizontal asymptote: y 1. y
Range: 3. Horizontal asymptote: y 3. y
1 (_1, _2)
41. (a)
1
(0, 2)
1
x
x
1
42. (a)
y
y
g f
1
1 1
x
(b) Since g x 3 2x 3 f x and f x 0, the
height of the graph of g x is always three times the height of the graph of f x 2x , so the graph of g
is steeper than the graph of f .
1
x
x2 (b) f x 9x2 32 32x2 3x g x. So f x g x, and the graphs are the same.
6
CHAPTER 4 Exponential and Logarithmic Functions
43.
f x x 3
x 0
y
g x 3x
0
1
1
1
3
2
8
9
3
27
27
4
64
81
6
216
729
8
512
6561
10
1000
59,049
x
g(x)=3
f(x)=x#
20 0
x
1
From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f . 44.
f x x 4
x 0
y
g x 4x
0
1
1
1
4
2
16
16
3
81
64
4
256
256
6
1296
4096
8
4096
65,536
10
10,000
1,048,576
g(x)=4
x
f(x)=x$
100 0
x
1
From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f . 45. (a) From the graphs below, we see that the graph of f ultimately increases much more quickly than the of g. graph (i) [0 5] by [0 20] (ii) [0 25] by 0 107 (iii) [0 50] by 0 108 20
10,000,000
f
8,000,000 6,000,000
10 g
f
1
2
4,000,000
3
4
80,000,000
f
g
40,000,000 20,000,000
0
5
100,000,000
60,000,000
2,000,000
0
g
10
20
0
20
40
(b) From the graphs in parts (a)(i) and (a)(ii), we see that the approximate solutions are x 12 and x 224. 46. (a) (i) [4 4] by [0 20]
(iii) [0 20] by 0 105
(ii) [0 10] by [0 5000]
20 g
f
4000
f
g
100,000
80,000
10
60,000
2000
f
g
40,000 20,000 4
3
2
1 0
1
2
3
4
0
2
4
6
8
10
0
10
(b) From the graphs in parts (i) and (ii), we see that the solutions of 3x x 4 are x 080, x 152 and x 717.
20
SECTION 4.1 Exponential Functions
47.
c=2 c=1
5
48.
10
c=0.5
4
2
1
c=0.25
2
0 1
1
2
3
0
2
The larger the value of c, the more rapidly the graph of shifted horizontally 1 unit. This is because of our choice of c; each c in this exercise is of the form 2k . So f x 2k 2x 2xk . 2
2
4
The larger the value of c, the more rapidly the graph of x f x 2cx increases. In general, f x 2cx 2c ; x so, for example, f x 22x 22 4x .
f x c2x increases. Also notice that the graphs are just
49. y 10xx
c=0.5
4
1 3
c=1
6
c=0.25
2
c=2
8
3 c=4
c=4
50. y x2x 2 1 4 4
2
0
2
(a) From the graph, we see that the function is increasing
on 144 and decreasing on 144.
(b) From the graph, we see that the range is
(b) From the graph, we see that the range is
approximately 0 178].
approximately [053 .
10xh 10x f x h f x 51. f x 10x , so 10x
52. f x 3x1 , so
2
(a) From the graph, we see that the function is increasing
on 050 and decreasing on 050 .
h
2
4
h
10h 1 . h
3xh1 3x1 f x h f x 3x1 h h
3h 1 . h
53. (a) After 1 hour, there are 1500 2 3000 bacteria. After 2 hours, there are 1500 2 2 6000 bacteria. After 3 hours, there are 1500 2 2 2 12,000 bacteria. We see that after t hours, there are N t 1500 2t bacteria. (b) After 24 hours, there are N 24 1500 224 25,165,824,000 bacteria.
54. (a) Because the population doubles every year, there are N t 320 2t mice after t years. (b) After 8 years, there are approximately N 8 320 28 81,920 mice.
7
8
CHAPTER 4 Exponential and Logarithmic Functions
55. Using the formula A t P 1 ik with P 5000,
004 per month, and k 12 number 12 of years, we fill in the table:
i 4% per year
56. Using the formula A t P 1 ik with P 5000, rate per year i per month, and k 12 5 60 months, 12 we fill in the table:
Amount
Rate per year
Amount
1
$520371
1%
$525625
2
$541571
2%
$552539
3
$563636
3%
$580808
4
$586599
4%
$610498
5
$610498
5%
$641679
6
$635371
6%
$674425
Time (years)
2t 57. P 8000, r 00625, and n 2. So A t 8000 1 00625 8000 1031252t . 2
(a) A 5 8000 10312510 10,88252, and so the value of the investment is $10,88252.
(b) A 10 8000 10312520 14,80366, and so the value of the investment is $14,80366. (c) A 15 8000 10312530 20,13765, and so the value of the investment is $20,13765.
365t . 58. P 3500, r 0035, and n 365. So A t 3500 1 0035 365 365 2 (a) A 2 3500 1 0035 375377, and so the value of the investment is $375377. 365 365 3 388747, and so the value of the investment is $388747. (b) A 3 3500 1 0035 365 365 6 431783, and so the value of the investment is $431783. (c) A 6 3500 1 0035 365
4t . 59. P 1200, r 00275, and n 4. So A t 1200 1 00275 4 4 123334, and so the value of the investment is $123334. (a) A 1 1200 1 00275 4 8 126761, and so the value of the investment is $126761. (b) A 2 1200 1 00275 4 40 157835, and so the value of the investment is $157835. (c) A 10 1200 1 00275 4 4t . 60. P 14,000, r 00525, and n 4. So A t 14000 1 00525 4 16 17,24791, and so the amount due is $17,24791. (a) A 4 14000 1 00525 4 24 19,14436, and so the amount due is $19,14436. (b) A 6 14000 1 00525 4 32 21,24932, and so the amount due is $21,24932. (c) A 8 14000 1 00525 4
23 P 10456 10000 13023P P 767896. 61. We must solve for P in the equation 10000 P 1 009 2 Thus, the present value is $7,67896.
125 62. We must solve for P in the equation 100000 P 1 008 P 10066760 100000 14898P 12 P $67,12104.
9
SECTION 4.2 The Natural Exponential Function
r n 008 12 63. rAPY 1 1. Here r 008 and n 12, so rAPY 1 1 1006666712 1 0083000. n 12 Thus, the annual percentage yield is about 83%. r n 0055 4 1. Here r 0055 and n 4, so rAPY 1 1 0056145. Thus, the annual 64. rAPY 1 n 4 percentage yield is about 561%. 65. Let T n be the thickness of the paper after n folds. The sheet is 0001 in thick and each fold doubles the thickness, 1 ft so T n 0001 2n . After 50 folds, the folded paper is T 50 0001 250 1126 1012 in 12 in 1 mi 17,771,460 mi—roughly onefifth the distance from the earth to the sun! 5280 ft 66. Since f 40 240 1,099,511,627,776, it would take a sheet of paper 4 inches by 1,099,511,627,776 inches. Since there are 12 inches in a foot and 5,280 feet in a mile, 1,099,511,627,776 inches 174 million miles. So the dimensions of the sheet of paper required are 4 inches by about 174 million miles.
4.2
THE NATURAL EXPONENTIAL FUNCTION
1. The function f x e x is called the natural exponential function. The number e is approximately equal to 271828. 2. In the formula A t Pert for continuously compound interest, the letters P, r, and t stand for principal, interest rate per year, and number of years respectively, and A t stands for amount accumulated after t years. So, if $100 is invested at an interest rate of 6% compounded continuously, then the amount after 2 years is A 2 100 e0062 $11275. 3. h x e x ; h 1 2718, h 23141, h 3 0050, h
2 4113
4. h x e3x ; h 13 0368, h 15 0011, h 1 20086, h 12,391648 5.
f x 15e x
y
6. g x 4e13x
x
y
x
y
2
020
3
1087
1 05
055
2
091
0
15
05
247
1
408
2
1108
2 0
1
x
Replace with "f" y
779
1
558
0
4
1
287
2
205
3
147
2 0
1
x
10
CHAPTER 4 Exponential and Logarithmic Functions
7. f x e x 3. The graph of f is obtained from the graph of y e x by shifting it upward 3 units. yintercept: y f 0 e0 3 4. Domain: .
Range: 3 . Horizontal asymptote: y 3. y
8. f x ex 1. The graph of f is obtained from the graph of y e x by reflecting it about the yaxis and shifting it upward 1 unit. yintercept:
y f 0 e0 1 2. Domain: .
Range: 1 . Horizontal asymptote: y 1. y
1 1
x
1 x
1
9. g x ex 3. The graph of g is obtained from the graph of y e x by reflecting it about the yaxis and
shifting it downward 3 units. yintercept:
y g 0 e0 3 2. Domain: .
Range: 3 . Horizontal asymptote: y 3.
10. h x e x 4. The graph of h is obtained by shifting the graph of y e x downward 4 units. yintercept:
y h 0 e0 4 3. Domain: .
Range: 4 . Horizontal asymptote: y 4. y
y
1 1
1
x
x
1
11. f x 3 e x . The graph of f is obtained from the graph 12. y 2 ex . The graph is obtained by reflecting the of y e x by reflecting it about the xaxis, then shifting
graph of y e x about the yaxis, then reflecting about the
Domain: . Range: 3. Horizontal
y 2 e0 1. Domain: . Range: 2.
upward 3 unit. yintercept: y f 0 3 e0 2. asymptote: y 3.
y
xaxis, then shifting upward 2 units. yintercept:
Horizontal asymptote: y 2. y
1
1 1
x
1
x
SECTION 4.2 The Natural Exponential Function
13. y 4 3ex . The graph is obtained by reflecting that of y e x about the yaxis, then stretching vertically by a factor of 3, then reflecting about the xaxis, then shifting upward 4 units. yintercept: y 4 3e0 1.
Domain: . Range: 4. Horizontal asymptote: y 4.
11
14. f x 3e x 1. The graph of f is obtained by stretching that of y e x vertically by a factor of 3, then shifting 1 unit upward. yintercept: y f 0 3e0 1 4. Domain: . Range: 1 . Horizontal asymptote: y 1. y
y
1 x
1
1
f(x)
1
x
y
f
y
15. y e x2 . The graph of y e x2 is obtained from the graph of y e x by shifting it to the right 2 units.
= f(0)
16. f x e x3 4. The graph of f x e x3 4 is obtained by shifting the graph of y e x to the right 3 units, and then upward 4 units. yintercept: yintercept: y e02 e2 014. Domain: . Range: 0 . Horizontal asymptote: y 0. y f 0 e03 4 e3 4 405. Domain: . Range: 4 . Horizontal y asymptote: y 4. y
1
(3, 5)
(2, 1) x
1
1 x
1
17. h x e x1 3. The graph of h is obtained from the graph of y e x by shifting it to the left 1 unit and downward 3 units. yintercept: y h 0 e01 3 e1 3 028. Domain: . Range: 3 . Horizontal asymptote: y 3. y
18. g x e x1 2. The graph of g is obtained by shifting the graph of y e x to the right 1 unit, reflecting it about the xaxis, then shifting it downward 2 units. yintercept: y g 0 e01 2 e1 2 237. Domain: . Range: 2. Horizontal asymptote: y 2. y
1 1 1 (0, e-3) (_1, _2)
x (1, _3)
1
x
12
CHAPTER 4 Exponential and Logarithmic Functions 2
2
19. f x 5ex 2
20. f x 10ex5
y
y 6
10
4 5
2 4
2
2
4
No xintercept. yintercept: 5. Horizontal asymptote: y 0. Local maximum: f 0 5. No local minimum.
21. l x
2
x
2
4
6
8
x
No xintercept. yintercept: 10e25 139 1010 .
Horizontal asymptote: y 0. Local
maximum: f 5 10. No local minimum.
20 1 3e15x
22. l x
100 1 4e05x
y 20
y 100
10
50
2
2
4
10
x
x
No xintercept. yintercept: 5. Horizontal
No xintercept. yintercept: 20. Horizontal
asymptotes: y 0 and y 20. No local maximum or
symptotes: y 0 and y 100. No local maximum or
minimum.
minimum.
23. s x 04xe15x
24. s x 25xe35x
y 0.2
y 0.4 0.2
1
1
2
3
0.2
4
5
x
0.5
0.4
0.2
0.6
0.4
0.5
1.0
1.5
xintercept: 0. yintercept: 0. Horizontal
xintercept: 0. yintercept: 0. Horizontal
asymptote: y 0. Local maximum:
asymptote: y 0. Local maximum:
approximately s 067 0098. No local minimum.
x
approximately s 029 026. No local minimum.
Answers to Exercises 25–28 will vary. 2
2.0
25. F x 2ex10 can be expressed as f g x, where g x x 102 and f x 2e x . 1 26. F x e x ex can be expressed as f g x, where g x e x and f x x . x 27. F x 1 e x can be expressed as f g x, where g x 1 e x and f x x.
SECTION 4.2 The Natural Exponential Function
13
3 28. F x 3 e x can be expressed as f g x, where g x 3 e x and f x x 3 . 29. (a)
30. (a)
y
y 1
1 y=_ 2 e¨
y=sinh x
y=cosh x 1 y=_ 2 eШ
1
1 y=_ 2 e¨
1
x
1
ex e x ex ex 2 2 e x ex cosh x 2
ex e x ex ex 2 2 ex e x e x ex sinh x 2 2
(b) cosh x
31. (a)
a=1
a=1.5
10 a=0.5
a=2
8
4
0
32.
4
0
2 2
(b) sinh x
2
6
4
x
1
y=_ _ 2 eШ
2
4
a xa e (b) As a increases the curve y exa 2 flattens out and the y intercept increases. 33. g x x x . Notice that g x is only defined for x 0. The graph of g x is shown in the viewing rectangle
[0 15] by [0 15]. From the graph, we see that there is a local minimum of about 069 when x 037.
0
20
40
x From the graph, we see that y 1 1x approaches e as x get large.
34. g x e x e2x . The graph of g x is shown in the
viewing rectangle [1 2] by [1 6]. From the graph, we see that there is a local minimum of about 189 when x 023. 5
1
0
0
1
1
1
2
35. D t 50e02t . So when t 3 we have D 3 50e023 274 milligrams. 36. m t 13e0015t (a) m 0 13 kg.
(b) m 45 13e001545 13e0675 6619 kg. Thus the mass of the radioactive substance after 45 days is about 66 kg.
14
CHAPTER 4 Exponential and Logarithmic Functions
37. (a)
(b) From the graph, it appears that the maximum
s 0.2
concentration is reached at t 083 h, or approximately 50 minutes.
0.1
(c) From the graph, it appears that s t decreases to 001 at t 49 h. 1
2
3
4
5
6
t
The drug concentration in the bloodstream increases rapidly until it reaches a peak level, then decreases slowly. 38. (a)
(b) From the graph, the average height appears to be
g 0.004
185 cm.
0.002 0.000
160
180
200
220
x
39. t 180 1 e02t
(a) 0 180 1 e0 180 1 1 0. (b) 5 180 1 e025 180 0632 11376 ft/s. So the
(c)
y 200
velocity after 5 s is about 1138 ft/s. 10 180 1 e0210 180 08646647 1556 ft/s. So
100
the velocity after 10 s is about 1556 ft/s.
(d) The terminal velocity is 180 ft/s. 40. (a) Q 5 15 1 e0045 15 01813 27345. Thus
x
5
(c)
approximately 27 lb of salt are in the barrel after 5 minutes. (b) Q 10 15 1 e00410 15 03297 4946Thus
Q
20
10
approximately 49 lb of salt are in the barrel after 10 minutes. 0
(d) The amount of salt approaches 15 lb. This is to be expected, since
0
100
200
t
50 gal 03 lb/gal 15 lb. 41. n t
5000 1 39e004t
(a) n 0
5000 125 1 39e0040
(c) From the graph, we see that n t approaches 5000 as t gets large.
(b)
n
6000 4000 2000 0
0
100
200
t
SECTION 4.2 The Natural Exponential Function
42. D t
15
54 54 . So D 20 1600 ft. 001t 1 29e 1 29e00120
43. Using the formula A t Pert with P 7000 and r 3% 003, we fill in the table:
44. Using the formula A t Pert with P 7000 and t 10 years, we fill in the table:
Time (years)
Amount
Rate per year
Amount
1
$721318
1%
$773620
2
$743286
2%
$854982
3
$765922
3%
$944901
4
$789248
4%
$10,44277
5
$813284
5%
$11,54105
6
$838052
6%
$12,75483
45. We use the formula A t Pert with P 2000 and r 35% 0035. (a) A 2 2000e0035 2 $214502
(b) A 42 2000e0035 4 $230055 (c) A 12 2000e0035 12 $304392
46. We use the formula A t Pert with P 3500 and r 625% 00625.
(a) A 3 3500e00625 3 $422181 (b) A 6 3500e00625 6 $509247 (c) A 9 3500e00625 9 $614269
47. (a) Using the formula A t P 1 ik with P 600, i 25% per year 0025, and k 10, we calculate A 10 600 102510 $76805.
0025 20 $76922. semiannually and k 10 2 20, so A 600 1 (b) Here i 0025 10 2 2 0025 40 $76982. quarterly and k 10 4 40, so A 600 1 (c) Here i 25% per year 0025 10 4 4 (d) Using the formula A t Pert with P 600, r 25% 0025, and t 10, we have A 10 600e0025 10 $77042.
48. We use the formula A t Pert with P 8000 and t 12.
(a) If r 2% 002, then A 12 8000e002 12 $10,16999.
(b) If r 3% 003, then A 12 8000e003 12 $11,46664.
(c) If r 45% 0045, then A 12 8000e0045 12 $13,72805. (d) If r 7% 007, then A 12 8000e007 12 $18,53094.
2 49. Investment 1: After 1 year, a $100 investment grows to A 1 100 1 0025 10252. 2 4 10227. Investment 2: After 1 year, a $100 investment grows to A 1 100 1 00225 4 Investment 3: After 1 year, a $100 investment grows to A 1 100e002 10202. We see that Investment 1 yields the highest return.
2 10519. 50. Investment 1: After 1 year, a $100 investment grows to A 1 100 1 005125 2 Investment 2: After 1 year, a $100 investment grows to A 1 100e005 10512. We see that Investment 1 yields a higher return.
12 12
16
CHAPTER 4 Exponential and Logarithmic Functions
51. (a) A t Pert 5000e009t (b) 20000
0
0
10
20
(c) A t 25,000 when t 1788 years.
52. We know that e x is increasing and ex is positive, so f x e x ex is increasing on 0 and its minimum value on that interval is f 0 e0 e0 2. But f is even, so because it is increasing on 0 , it is decreasing on 0. Therefore, its absolute minimum value is indeed f 0 2.
4.3
LOGARITHMIC FUNCTIONS
1. log x is the exponent to which the base 10 must be raised in order to get x. x
103
102
101
100
101
102
103
1012
log x
3
2
1
0
1
2
3
12
2. The function f x log9 x is the logarithm function with base 9. So f 9 log9 91 1, f 1 log9 90 0, f 19 log9 91 1, f 81 log9 92 2, and f 3 log9 912 12 . 3. (a) 53 125, so log5 125 3.
(b) log5 25 2, so 52 25.
4. (a) f x log2 x is a logarithmic function with base 2. It has graph III.
(b) The graph of f x log2 x is obtained from that of y log2 x by reflecting about the yaxis. It has graph II.
(c) f x log2 x is obtained from that of y log2 x by reflecting about the xaxis. It has graph I.
(d) f x log2 x is obtained from that of y log2 x by reflecting about the x and yaxes. It has graph IV.
5. The natural logarithmic function f x ln x has the vertical asymptote x 0.
6. The natural logarithmic function f x ln x 1 has the vertical asymptote x 1. 7.
Logarithmic form
Exponential form
log8 8 1
81 8
log4 64 3
log8 4 23
823 4
log4 8 32
log8 81 1
81 18
log8 64 2
log8 512 3
1 2 log8 64
9. (a) 34 81 0 (b) 13 1
13. (a) 3x 5 3 (b) 16 2y
82 64 83 512
1 82 64
10. (a) 51 15 1 (b) 15 5 2 1 z 14. (a) 10 (b) 102t 3
8.
Logarithmic form
Exponential form 43 64
log4 2 12
412 2
1 2 log4 16 log4 21 12 1 5 log4 32 2
1 42 16
11. (a) 813 2 (b) 102 001 15. (a) e2y 10 (b) e2 3x 1
432 8
412 12
1 452 32 1 12. (a) 53 125
(b) 823 4 16. (a) e3 x 2 (b) e1 2x 3
SECTION 4.3 Logarithmic Functions
17. (a) log10 10,000 4 1 2 (b) log5 25 21. (a) log4 70 x
18. (a) log6 36 2 1 1 (b) log10 10
22. (a) log3 10 2x
(b) log12 3
(b) log10 01 4x
25. (a) log2 2 1 (b) log5 1 log5 50 0 1 (c) log12 2 log12 12 1
27. (a) log6 36 log6 62 2 (b) log9 81 log9 92 2
(c) log7 710 10 1 log 33 3 29. (a) log3 27 3 3 3 (b) log13 27 log13 13 (c) log7 7 log7 712 12 31. (a) 3log3 5 5
19. (a) log8 81 1 (b) log2 18 3
23. (a) ln 2 x
2 (c) log12 025 log12 12 2
28. (a) log2 32 log2 25 5 (b) log5 513 13 (c) log6 1 log6 60 0
30. (a) log5 125 log5 53 3
(b) log49 7 log49 4912 12 (c) log9
12 3 log9 312 log9 912 14
32. (a) eln 3 3
(c) 10log 13 13
(b) log10 0001 x 10x 0001 x 3 37. (a) ln x 3 x e3 (b) ln e2 x x 2 ln e 2 1 x 4x 1 x 3 39. (a) log4 64 64 3 (b) log12 x 3 12 x x 18 41. (a) log2 12 x 2x 12 x 1
1 (b) log10 x 3 103 x x 1000
43. (a) logx 16 4 x 4 16 x 2
(b) logx 8 32 x 32 8 x 823 4
(b) ln t 05x
(b) log4 64 log4 43 3
(c) eln 10 10
35. (a) log6 x 2 x 62 36
24. (a) ln 05 x 1
26. (a) log3 37 7
1 (b) eln1
(b) ln e4 4 1 ln e1 1 (c) ln e
(b) log12 8 3
(b) ln y 3
(b) 5log5 27 27
33. (a) log8 025 log8 823 23
20. (a) log4 0125 32
12 2 log4 212 log4 412 14 1 (b) log4 12 log4 21 log4 412 12 3 (c) log4 8 log4 23 log4 412 log4 432 32
34. (a) log4
0 36. (a) log13 x 0 x 13 1 (b) log4 1 x 4x 1 x 0
38. (a) ln x 1 x e1 1e (b) ln 1e x x ln e1 ln e 1 40. (a) log9 13 x 9x 13 x 12 (b) log9 x 05 x 912 3
42. (a) logx 1000 3 x 3 1000 x 10 (b) logx 25 2 x 2 25 x 5 44. (a) logx 6 12 x 12 6 x 36 (b) logx 3 13 x 13 3 x 27
17
18
CHAPTER 4 Exponential and Logarithmic Functions
46. (a) log 50 16990 (b) log 2 01505 (c) log 3 2 06276
45. (a) log 2 03010 (b) log 352 15465 (c) log 23 01761
48. (a) ln 27 32958
47. (a) ln 5 16094
49.
(b) ln 739 20001
(b) ln 253 32308 (c) ln 1 3 10051
(c) ln 546 40000
y
x 1 33 1 32 1 3
f x
y
50.
2
x
1
1
_3
1 43 1 42 1 4
1
0
_4
3
1
32
_5
2
3
0
2
_1
1
2
3
4
5
6
x
_2
f x log3 x
x 1 103 1 102 1 10
f x 6
_4
4
1
42
_5
2
2
3
1
1
0
2
1
_1
0
_4
10
2
102
_5
4
x
g x
3
4
5
6
x
4 8
1
2 3
f x 2 log x y
54.
2
x
f x
1
1 4 1 2
4
0
0
0
4
1
1
1
1
_2
2
_3
f x
2
_1
1
2
3
4
5
_4
10
2
102
_5
3
1 4
_3
1
x
2
_2
1
6
8
2
3
4
5
6
x
_2
1
52.
1
1
_1
1 16 1 8 1 2
_3
1 103 1 102 1 10
0
2
x
2
53.
1
3
2
0
4
2
y 3
y
51.
g x
0 _1
1
2
3
4
5
6
7
8 x
7
8 x
_2 _3 _4
f x log12 x y 1 0
3
_1
2
_2
1
_3
1
_4
0
g x log4 x
1
2
3
4
5
6
_5
g x log x 1
f x log2 x 2
55. Since the point 5 1 is on the graph, we have 1 loga 5 a 1 5. Thus the function is y log5 x. 56. Since the point 12 1 is on the graph, we have 1 loga 12 a 1 12 a 2. Thus the function is y log2 x.
SECTION 4.3 Logarithmic Functions
19
57. Since the point 3 12 is on the graph, we have 12 loga 3 a 12 3 a 9. Thus the function is y log9 x.
58. Since the point 9 2 is on the graph, we have 2 loga 9 a 2 9 a 13 . Thus the function is y log13 x. 59. The graph of f x 2 ln x is obtained from that of y ln x by shifting it upward 2 units, as in graph I.
60. The graph of f x ln x 2 is obtained from that of y ln x by shifting it to the right 2 units, as in graph II. 61. The graph of y log4 x is obtained from that of y 4x
62. The graph of y log3 x is obtained from that of y 3x
y
y
by reflecting it in the line y x.
by reflecting it in the line y x.
y=4¨
y=3¨
y=x
y=log£ x
y=log¢ x
1
1 x
1
63. The graph of g x log5 x is obtained from that of y log5 x by reflecting it about the yaxis.
Domain: 0. Range: . Vertical asymptote: x 0.
y=x
1
x
64. The graph of f x log10 x is obtained from that of y log10 x by reflecting it about the xaxis.
Domain: 0 . Range: . Vertical asymptote: x 0.
y
y
1
1 1
x
1
x
65. The graph of f x log2 x 4 is obtained from that of 66. The graph of g x ln x 2 is obtained from that of
y ln x by shifting to the left 2 units. Domain: 2 .
y log2 x by shifting it to the right 4 units.
Range: . Vertical asymptote: x 2.
Domain: 4 . Range: . Vertical asymptote: x 4.
y
y
1 1
1 1
x
x
20
CHAPTER 4 Exponential and Logarithmic Functions
67. The graph of h x ln x 5 is obtained from that of
y ln x by shifting to the left 5 units. Domain: 5 .
Range: . Vertical asymptote: x 5. y
68. The graph of f x 2 log13 x is obtained from that of y log13 x by reflecting about the xaxis, then shifting
upward 2 units. Domain: 0 . Range: . Vertical asymptote: x 0. y
1 1
x (1, 2) 1 x
1
69. The graph of y 2 log3 x is obtained from that of
y log3 x by shifting upward 2 units. Domain: 0 .
Range: . Vertical asymptote: x 0. y
70. The graph of y 1 log10 x is obtained from that of y log10 x by reflecting it about the xaxis, and then
shifting it upward 1 unit. Domain: 0 .
Range: . Vertical asymptote: x 0. y
(1, 2) 1
(1, 1)
1 1
x
x
1
71. The graph of y log3 x 1 2 is obtained from that of 72. The graph of y 1 ln x is obtained from that of y log3 x by shifting to the right 1 unit and then
downward 2 units. Domain: 1 . Range: . Vertical asymptote: x 1. 1
y
y ln x by reflecting it about the yaxis and then shifting
it upward 1 unit. Domain: 0. Range: . Vertical asymptote: x 0.
y
x
1
(2, _2)
(_1, 1) 1 1
x
SECTION 4.3 Logarithmic Functions
73. The graph of y ln x is obtained from that of y ln x
by reflecting the part of the graph for 0 x 1 about the
74. Note that y ln x
ln x
if x 0 ln x if x 0
21
The graph
of y ln x is obtained by combining the graph of
xaxis. Domain: 0 . Range: [0 .
y ln x and its reflection about the yaxis.
Vertical asymptote: x 0.
Domain: 0 0 . Range: . Vertical
y
asymptote: x 0.
y 1 x
1 1 x
1
75. f x log10 x 3. We require that x 3 0 x 3, so the domain is 3 . 76. f x log5 8 2x. Then we must have 8 2x 0 8 2x 4 x, and so the domain is 4. 77. g x log3 x 2 1 . We require that x 2 1 0 x 2 1 x 1 or x 1, so the domain is 1 1 . 78. g x ln x x 2 . Then we must have x x 2 0
x 1 x 0. Using the methods from Chapter 1 with the
endpoints 0 and 1, we get the table at right. Thus the domain is 0 1.
Interval
0
0 1
Sign of x
1
Sign of 1 x
Sign of x 1 x
79. h x ln x ln 2 x. We require that x 0 and 2 x 0 x 0 and x 2 0 x 2, so the domain is 0 2. 80. h x x 2 log5 10 x . Then we must have x 2 0 and 10 x 0 x 2 and 10 x 2 x 10. So the domain is [2 10. 81. y log10 1 x 2 has domain 1 1, vertical
asymptotes x 1 and x 1, and local maximum y 0 at x 0.
82. y ln x 2 x ln x x 1 has domain
0 1 , vertical asymptotes x 0 and x 1, and no local maximum or minimum. 10
2
2 10 2
10 10
22
CHAPTER 4 Exponential and Logarithmic Functions
83. y x ln x has domain 0 , vertical asymptote x 0, and no local maximum or minimum.
84. y x ln x2 has domain 0 , no vertical asymptote, local minimum y 0 at x 1, and local maximum y 054 at x 014. 10
2 5 5 0
ln x has domain 0 , vertical asymptote x 0, x horizontal asymptote y 0, and local maximum y 037
85. y
at x 272.
2
4
86. y x log10 x 10 has domain 10 , vertical
asymptote x 10, and local minimum y 362 at x 587.
5 0
10
20 10
10
2 5
Answers to Exercises 87–90 will vary. 87. F x ln x 2 1 can be expressed as f g x, where f x ln x and g x x 2 1.
88. F x ln x3 can be expressed as f g x, where f x x 3 and g x ln x. 89. F x 1 ln x can be expressed as f g x, where f x ln x and g x 1 x. 90. F x 5 log x can be expressed as f g x, where f x 5 x and g x log x.
91. f x 2x and g x x 1 both have domain , so f g x f g x 2gx 2x1 with domain and g f x g f x 2x 1 with domain . 2
92. f x 3x and g x x 2 1 both have domain , so f g x f g x 3x 1 with domain 2 and g f x g f x 3x 1 32x 1 with domain .
93. f x log2 x has domain 0 and g x x 2 has domain , so f g x f g x log2 x 2 with domain 2 and g f x g f x log2 x 2 with domain 0 . 94. f x log x has domain 0 and g x x 2 has domain , so f g x f g x log x 2 with
domain 0 0 and g f x g f x log x2 with domain 0 . 6 95. The graph of g x x grows faster than the graph of 96. (a) f x ln x.
4
6 4 2
0
g
10
g
2
f
0
20
30
f
10
20
30
(b) From the graph, we see that the solution to the equation x 1 ln 1 x is x 1350.
SECTION 4.3 Logarithmic Functions
97. (a)
98. (a)
c=4 c=3 c=2 c=1
2
c=4
6
c=3
4
c=2
1 2
0
23
10
20
30
0
(b) Notice that f x log cx log c log x, so
c=1 10
20
30
(b) As c increases, the graph of f x c log x
as c increases, the graph of f x log cx is
stretches vertically by a factor of c.
shifted upward log c units.
99. (a) f x log2 log10 x . Since the domain of log2 x is the positive real numbers, we have: log10 x 0 x 100 1. Thus the domain of f x is 1 . y x (b) y log2 log10 x 2 y log10 x 102 x. Thus f 1 x 102 .
100. (a) f x ln ln ln x. We must have ln ln x 0 ln x 1 x e. So the domain of f is e . y
(b) y ln ln ln x e y ln ln x ee ln x ee 2x 2x .y y y2x 2x x 12 1 2x y y 2x y2x 2x 1 y 2x 1 y y . Thus x log2 1y x f 1 x log2 . 1x
101. (a) f x
ey
ex
x. Thus the inverse function is f 1 x ee . (b)
x 0. Solving this using the methods from 1x Chapter 1, we start with the endpoints, 0 and 1. Interval
0
0 1
Sign of x
1
Sign of 1 x x Sign of 1x
Thus the domain of f 1 x is 0 1. I 2500 ln 07 89169 moles/liter. 102. Using I 07I0 we have C 2500 ln I0 D 8267 ln 073 2602 years. 103. Using D 073D0 we have A 8267 ln D0
104. Substituting N 1,000,000 we get t 3 105. When r 6% we have t t
ln 2 87 years. 008
log N50 log 20,000 3 4286 hours. log 2 log 2
ln 2 ln 2 116 years. When r 7% we have t 99 years. And when r 8% we have 006 007
106. Using k
025 and substituting C 09C0 we have C 025 ln 1 09 025 ln 01 058 hours. t 025 ln 1 C0
log 2AW log 2 1005 log 40 532. Using A 100 and log 2 log 2 log 2 log 2 10010 log 20 523 log 2AW 432. So the smaller icon is 123 times W 10 we find the ID to be log 2 log 2 log 2 432 harder.
107. Using A 100 and W 5 we find the ID to be
24
CHAPTER 4 Exponential and Logarithmic Functions
108. (a) Since 2 feet 24 inches, the height of the graph is 224 1677216 inches. Now, since there are 12 inches per foot and ,216 5280 feet per mile, there are 12 5280 63,360 inches per mile. So the height of the graph is 1,677 63360 2648, or
about 265 miles. (b) Since log2 224 24, we must be about 224 inches 265 miles to the right of the origin before the height of the
graph of y log2 x reaches 24 inches or 2 feet. 109. log log 10100 log 100 2 log log log 10googol log log googol log log 10100 log 100 2
110. Notice that loga x is increasing for a 1. So we have log4 17 log4 16 log4 42 2. Also, we have log5 24 log5 25 log5 52 2. Thus, log5 24 2 log4 17.
111. The numbers between 1000 and 9999 (inclusive) each have 4 digits, while log 1000 3 and log 10,000 4. Since log x 3 for all integers x where 1000 x 10,000, the number of digits is log x 1. Likewise, if x is an integer where 10n1 x 10n , then x has n digits and log x n 1. Since log x n 1 n log x 1, the number of digits in x is log x 1. 112. We express each side in exponential form: log1a x y 1a y x a y x. Using the definition of the logarithm on the last equation, we have loga x y y loga x.
Thus, we have shown that log1a x loga x, and so the graph of
g x log1a x can be obtained by reflecting the graph of f x loga x
about the xaxis.
y 4 3 2
f(x)=log4 x
1 0 _1
1
2
3
_2
4 g(x)=log1/4 x
_3 _4
4.4
LAWS OF LOGARITHMS
1. The logarithm of a product of two numbers is the same as the sum of the logarithms of these numbers. So log5 25 125 log5 25 log5 125 2 3 5.
2. The logarithm of a quotient of two numbers is the same as the difference of the logarithms of these numbers. So 25 log 25 log 125 2 3 1. log5 125 5 5
3. The logarithm of a number raised to a power is the same as the power times the logarithm of the number. So log5 2510 10 log5 25 10 2 20. x2 y log x 2 y log z log x 2 log y log z 2 log x log y log z 4. log z x2 y 2 2 . 5. 2 log x log y log z log x log y log z log x y log z log z
6. (a) To express log7 12 in terms of common logarithms, we use the Change of Base Formula to write log 12 1079 log7 12 1277. log 7 0845 2485 ln 12 (b) Yes, the result is the same: log7 12 1277. ln 7 1946
5 x
SECTION 4.4 Laws of Logarithms
7. (a) False. log A B log A log B. (b) True. log AB log A log B.
A log A log B. B log A log A log B. (b) False. log B
8. (a) True. log
9. log 50 log 200 log 50 200 4
10. log6 9 log6 24 log6 9 24 3
60 2 11. log2 60 log2 15 log2 15
12. log3 135 log3 45 log3 135 45 1
13. 14 log3 81 14 4 1 15. log5 5 log5 512 12
14. 13 log3 27 13 3 1 16. log5 1 log5 532 32 125 2 6 17. log2 6 log2 15 log2 20 log2 15 log2 20 log2 5 20 log2 8 log2 23 3 100 log3 19 log3 32 2 18. log3 100 log3 18 log3 50 log3 18 50 100 19. log4 16100 log4 42 log4 4200 200 33 log2 299 99 20. log2 833 log2 23 21. log log 1010,000 log 10,000 log 10 log 10,000 1 log 10,000 log 104 4 log 10 4 200 ln e200 ln e ln e200 200 ln e 200 22. ln ln ee 23. log3 8x log3 8 log3 x
24. log6 7r log6 7 log6 r
25. log3 2x y log3 2 log3 x log3 y
26. log5 4st log5 4 log5 s log5 t 28. log t 5 log t 52 52 log t 27. ln a 3 3 ln a 29. log3 x yz log3 x yz12 12 log3 x log3 y log3 z 30. log5 x y6 6 log5 x y 6 log5 x log5 y 32. ln x y 2 ln x 12 ln y 2 12 ln x 2 ln y 31. ln a 3 b2 ln a 3 ln b2 3 ln a 2 ln b
4a y log2 4a log2 b 2 log2 a log2 b log6 y log6 6z log6 y log6 z 1 34. log6 b 6z a 3 b2 35. log8 log8 a 3 log8 b2 log8 c 3 log8 a 2 log8 b log8 c c 3x 4 y 2 log7 3 4 log7 x 2 log7 y log7 2 3 log7 z 36. log7 2z 3 3x 5 12 52 log3 x log3 y 37. log3 y
33. log2
y3 38. log 3 log y 12 log 2 12 log x 2x x 3 y4 log x 3 y 4 log z 6 3 log x 4 log y 6 log z 39. log 6 z x2 loga x 2 loga yz 3 2 loga x loga y 3 loga z 2 loga x loga y 3 loga z 40. loga 3 yz
25
26
CHAPTER 4 Exponential and Logarithmic Functions
x 4 2 12 ln x 4 2 3 42. log x 2 4 13 log x 2 4
41. ln
x z x z 12 log 12 log x z 12 log y y y 4x 2 44. ln 2 ln 4 2 ln x ln x 2 3 x 3 2 2 3 x y 13 ln x 2 y 2 13 ln x y 45. ln xy 43. log
x2 2 ln x 12 ln x 1 x 1 2 x2 4 x2 4 1 log 1 log x 2 4 log x 2 1 3 7 47. log x 2 2 2 2 x2 1 x3 7 x2 1 x3 7 12 log x 2 4 log x 2 1 2 log x 3 7 48. log x y z 12 log x y z 12 log x log y z 12 log x 12 log y z 12 log x 12 log y 12 log z 12 log x 14 log y 18 log z 49. log4 6 2 log4 7 log4 6 log4 72 log4 6 72 log4 294 46. ln
50. 12 log2 5 2 log2 7 log2
5 log2 49 log2 495
51. 2 log x 3 log x 1 log x 2 log x 13 log 52. 3 ln 2 2 ln x 12 ln x 4 ln 23 ln x 2 ln
x2 x 13
8x 2 x 4 ln x 4
53. log x 1 log x 1 3 log x log x 1 x 1 log x 3 log
x2 1 x3
x 3 1 54. ln x 3 ln x 2 9 ln 2 ln x 3 x 9 x 2 x 2 log5 55. 12 log5 x 2 log5 x 2 4 log5 x 12 log5 2 x 4 x x 3 4x ac2 a 4 c8 56. 4 log3 a 3 log3 b 2 log3 c 4 log3 a log3 b3 log3 c2 4 log3 3 log3 12 b b 2 4 x 3 13 log x 2 12 log 57. 13 log x 23 12 log x 4 log x 2 x 6 2 2 x x 6 12 x4 x2 x 2 x 2 x2 log x 2 log log x 2 log log log 2 x 3 x 3 x 2 x 3 x 2 [x 3 x 2]
bd c 58. loga b c loga d r loga s loga bd c loga s r loga r s log 10 log 7 1430677 60. log14 7 0737350 59. log5 10 log 5 log 14 log 4 log 30 0630930 62. log5 30 2113283 61. log9 4 log 9 log 5
SECTION 4.4 Laws of Logarithms
27
log 261 log 532 0493008 64. log6 532 3503061 log 7 log 6 log 125 log 25 65. log4 125 3482892 66. log12 25 0368743 log 4 log 12 ln x 1 1 loge x 67. log3 x ln x. The graph of y ln x is loge 3 ln 3 ln 3 ln 3 2
63. log7 261
shown in the viewing rectangle [1 4] by [3 2]. 2
4
2
68. Note that logc x
1 ln x (by the change of base formula). So ln c
a=2 a=e a=5 a=10
2
the graph of y logc x is obtained from the graph of y ln x by
1 depending ln c on whether ln c 1 or ln c 1. All of the graphs pass through 1 0 either shrinking or stretching vertically by a factor of
0
2
4
2
because logc 1 0 for all c.
1 ln e ln 10 ln 10 log b log c log d log d 70. loga b logb c logc d log a log b log c log a 1 1 1 1 log abc 1 1 log a log b log c 71. log x log x log x loga x logb x logc x log x log x
69. log e
log a
72. loga n x
log b
log c
1 log x 1 log x log x loga x log a n n log a n log a n
73. (a) log P log c k log W log P log c log W k log P log
c Wk
P
1 log x logabc
1 logabc x
c . Wk
8000 8000 1866 and when W 10 we have P 21 64. 221 10 80 P0 . Substituting P0 80, t 24, and c 03 we have P 305. So the 74. From Example 4(a), P t 1c 24 103 student should get a score of 30. (b) Using k 21 and c 8000, when W 2 we have P
75. (a) M 25 log BB0 25 log B 25logB0 .
(b) Suppose B1 and B2 are the brightness of two stars such that B1 B2 and let M1 and M2 be their respective magnitudes. Since log is an increasing function, we have log B1 log B2 . Then log B1 log B2 log B1 log B0 log B2 log B0 log B1 B0 log B2 B0 25 log B1 B0 25 log B2 B0 M1 M2 . Thus the brighter star has less magnitudes.
(c) Let B1 be the brightness of the star Albiero. Then 100B1 is the brightness of Betelgeuse, and its magnitude is M 25log 100B1 B0 25 log 100 log B1 B0 25 2 log B1 B0 5 25 log B1 B0 5 magnitude of Albiero
76. (a) log S log c k log A log S log c log Ak log S log c Ak S c Ak .
(b) If A 2A0 when k 3 we get S c 2A0 3 c 23 A30 8 c A30 . Thus doubling the area increases the number of species eightfold.
28
CHAPTER 4 Exponential and Logarithmic Functions
77. (a) False; log xy log x log y log x log y.
(b) False; log2 x log2 y log2 xy log2 x y. (c) True; the equation is an identity: log5 ab2 log5 a log5 b2 log5 a 2 log5 b. (d) True; the equation is an identity: log 2z z log 2.
(e) False; log P log Q log P Q log P log Q. (f) False; log a log b log ab log a log b. x (g) False; x log2 7 log2 7x log2 7 .
(h) True; the equation is an identity. loga a a a loga a a 1 a.
(i) False; log x y log x log y. For example, 0 log 3 2 log 3 log 2. (j) True; the equation is an identity: ln 1A ln A1 1 ln A ln A.
78. The error is on the first line: log 01 0, so 2 log 01 log 01.
79. Let f x x 2 . Then f 2x 2x2 4x 2 4 f x. Now the graph of f 2x is the same as the graph of f shrunk
horizontally by a factor of 12 , whereas the graph of 4 f x is the same as the graph of f x stretched vertically by a factor of 4. Let g x e x . Then g x 2 e x2 e2 e x e2 g x. This shows that a horizontal shift of 2 units to the right is the same as a vertical stretch by a factor of e2 .
Let h x ln x. Then h 2x ln 2x ln 2 ln x ln 2 h x. This shows that a horizontal shrinking by a factor of 12 is the same as a vertical shift upward by ln 2. 80. If the identity is true, then ln x x 2 1 ln x x 2 1 0 ln x x 2 1 x x 2 1 0 ln x 2 x 2 1 0 ln 1 0. The last statement is true, and all steps are reversible, so the original identity is true. 81. Note that x ln y x ln Cekx x kx ln C, which is on the line Y k X ln C. 82. We show that the point ln x ln y ln x ln ax n satisfies the equation Y n X ln a. Substituting, we get Y n X ln a ln ax n n ln x ln a ln ax n ln ax n , which is true.
4.5
EXPONENTIAL AND LOGARITHMIC EQUATIONS
1. (a) First we isolate e x to get the equivalent equation e x 25.
(b) Next, we take the natural logarithm of each side to get the equivalent equation x ln 25.
(c) Now we use a calculator to find x 3219.
2. (a) First we combine the logarithms to get the equivalent equation log 3 x 2 log x.
(b) Next, we write each side in exponential form to get the equivalent equation 3 x 2 x. (c) Now we find x 3.
3. Because the function 5x is onetoone, 5x1 625 5x1 54 x 1 4 x 5. 2
4. Because the function e x is onetoone, e x e9 x 2 9 x 3.
5. Because the function 5x is onetoone, 52x3 1 52x3 50 2x 3 0 x 32 . 1 2x 3 1 x 1. 6. Because the function 10x is onetoone, 102x3 10
7. Because the function 7x is onetoone, 72x3 765x 2x 3 6 5x 3x 9 x 3. 8. Because the function e x is onetoone, e12x e3x5 1 2x 3x 5 5x 6 x 65 . 2
2
9. Because the function 6x is onetoone, 6x 1 61x x 2 1 1 x 2 2x 2 2 x 1.
SECTION 4.5 Exponential and Logarithmic Equations 2
2
10. Because the function 10x is onetoone, 102x 3 109x 2x 2 3 9 x 2 3x 2 12 x 2.
11. (a) e x 16 ln e x ln 16 x ln e ln 16 x ln 16 4 ln 2 (b) x 2772589
12. (a) e2x 5 ln e2x ln 5 2x ln e ln 5 2x ln 5 x 12 ln 5 (b) x 0804719
13. (a) 10x 6 log 10x log 6 x log 10 log 6 x log 6 (b) x 0778151
14. (a) 105x 24 log 105x log 24 5x log 10 log 24 x 15 log 24 (b) x 0276042
15. (a) 3x5 4 ln 3x5 ln 4 x 5 ln 3 ln 4 x 5
ln 4 ln 4 x 5 ln 3 ln 3
(b) x 3738140
16. (a) e23x 11 ln e23x ln 11 2 3x ln 11 x 13 ln 11 2 (b) x 0132632
17. (a) 3 61x 15 61x 5 ln 61x ln 5 1 x ln 6 ln 5 x 1 (b) x 0101756
ln 5 ln 6
ln 8 1 18. (a) 5 432x 8 432x 85 3 2x ln 4 ln 85 3 2x 5 x ln 4 2 (b) x 1330482 15 19. (a) 200 1024t 1500 1024t 15 2 4t ln 102 ln 2 t
3
ln 85 ln 4
ln 15 2 4 ln 102
(b) t 25437319 12 20. (a) 25 101512t 60 101512t 12 5 12t ln 1015 ln 5 t
ln 12 5 12 ln 1015
(b) t 4900103
21. (a) 3 e5t 12 5 t ln 4 t 5 ln 4 (b) t 3613706
22. (a) 5
2t3 1 2
24
(b) t 0368483
2t3 1 2
ln 24 5 t 1 1 24 24 5 2t 3 ln 2 ln 5 2t 3 2 ln 12
3
ln 24 5 ln 12
10 ln 03 x ln 2 ln 03 x 10 ln 2 (b) x 17369656
23. (a) 2x10 03
24. (a) 2x50 06
50 ln 06 x ln 2 ln 06 x 50 ln 2
(b) x 36848280 25. (a) 4 1 105x 9 1 105x 94 105x 54 5x log 10 log 54 x 15 log 54
(b) x 0019382 ln 45 ln 45 x 1 26. (a) 2 5 3x1 100 5 3x1 50 3x1 45 x 1 ln 3 ln 45 x 1 ln 3 ln 3 (b) x 2464974
29
30
CHAPTER 4 Exponential and Logarithmic Functions
27. (a) 8 e14x 20 e14x 12 1 4x ln 12 x (b) x 0371227 28. (a) 1 e4x1 20 e4x1 19 4x 1 ln 19 x
1 ln 12 4 ln 19 1 4
(b) x 0486110 29. (a) 4x 212x 50 22x 212x 50 22x 1 2 50 22x 50 3 2x ln 2 ln 50 ln 3 x
ln 50 3 ln 4
(b) x 2029447
30. (a) 125x 53x1 200 53x 53x1 200 53x 1 5 200 53x 100 3 3x log 5 log 100 log 3 2 log 3 3 log 5 (b) x 0726249 x
31. (a) 5t 103t2 t log 5 3t 2 t log 5 3 2 t (b) t 0869176 32. (a) et 31t t 1 t ln 3 t 1 ln 3 ln 3 t (b) t 0523495 33. (a) 5x3 3x1 (b) x 1954364
2 log 5 3
ln 3 ln 3 1
3 ln 3 x ln 5 x 1 ln 3 x ln 5 3 x 1 ln 3 x ln 5 3 ln 3 3 ln 3 x 3 ln 5 3 ln 3
34. (a) 32x1 24x1 2x 1 ln 3 4x 1 ln 2 x 2 ln 3 4 ln 2 ln 3 ln 2 x (b) x 3114131
ln 6 ln 9 ln 16
50 4 50 4 4ex 46 4ex 115 ex ln 115 x x ln 115 1 ex (b) x 2442347
35. (a)
10 2 10 2 2ex 8 2ex 4 ex ln 4 x x ln 4 1 ex (b) x 1386294 37. e2x 5e x 6 0 e x 6 e x 1 0 e x 6 or e x 1. e x 6 has no solution because e x 0 for all x. 36. (a)
Thus, the only solution occurs where e x 1 x 0. 38. e2x 3e x 10 0 e x 5 e x 2 0 e x 5 or e x 2. The former equation has no solution because e x 0 for all x, so the only solution is x ln 2 06931. 39. e4x 4e2x 21 0 e2x 7 e2x 3 0 e2x 7 or e2x 3. Now e2x 7 has no solution because
e2x 0 for all x. But we can solve e2x 3 2x ln 3 x 12 ln 3 05493. So the only solution is x 05493. 40. 34x 32x 6 0 32x 3 32x 2 0 32x 3 or 32x 2. The latter equation has no solution, so we solve
32x 3 2x 1 x 12 . 41. 2x 10 2x 3 0 2x 22x 10 3 2x 0 2x 2x 5 2x 2 0. The first two factors are positive
everywhere, so we solve 2x 2 0 x 1. 42. e x 15ex 8 0 ex e2x 15 8e x 0 ex e x 5 e x 3 0. The first factor is positive everywhere, so the solutions occur where e x 5 or e x 3; that is, x ln 3 10986 or x ln 5 16094.
SECTION 4.5 Exponential and Logarithmic Equations
31
43. x 2 2x 2x 0 2x x 2 1 0 2x 0 (never) or x 2 1 0. If x 2 1 0, then x 2 1 x 1. So the only
solutions are x 1. 44. x 2 10x x10x 2 10x x 2 10x x10x 2 10x 0 10x x 2 x 2 0 10x 0(never) or x 2 x 2 0. If x 2 x 2 0, then x 2 x 1 0 x 2, 1. So the only solutions are x 2, 1.
45. 4x 3 e3x 3x 4 e3x 0 x 3 e3x 4 3x 0 x 0 or e3x 0 (never) or 4 3x 0. If 4 3x 0, then
3x 4 x 43 . So the solutions are x 0 and x 43 . 46. x 2 e x xe x e x 0 e x x 2 x 1 0 e x 0 (impossible) or x 2 x 1 0. If x 2 x 1 0, then
5 . So the solutions are x 1 5 . x 1 2 2
47. log x 2 log x 3 log 4x log x 2 x 3 log 4x x 2 x 3 4x x 2 5x 6 0 x 6 x 1 0 x 1 or x 6. However, x 1 does not satisfy the original equation because log x is defined only for x 0. Thus, the only solution is x 6.
48. log5 x 2 log5 x 5 log5 6x log5 x 2 x 5 log5 6x x 2 3x 10 6x x 2 9x 10 0 x 10 x 1 0 x 1 or x 10. However, only x 10 satisfies the original equation, and so that is the only solution. 49. 2 log x log 2 log 3x 4 log x 2 log 6x 8 x 2 6x 8 x 2 6x 8 0 x 4 x 2 0 x 4 or x 2. Thus the solutions are x 4 and x 2. ln x 2 2x 1 x 2 x 2 2x 1 0 x 12 0 x 1. 50. ln x 12 ln 2 2 ln x ln 2 x 12
Thus the only solution is x 1. 51. log2 3 log2 x log2 5 log2 x 2 log2 3x log2 5x 10 3x 5x 10 2x 10 x 5 52. log4 x 2 log4 3 log4 5 log4 2x 3 log4 3 x 2 log4 5 2x 3 3x 6 10x 15 7x 21 x 3 53. log x 9 x 109 1,000,000,000
54. log x 3 4 x 3 104 x 9997 55. ln 4 x 1 4 x e x 4 e 12817 56. ln 3x 1 0 3x 1 1 x 0
57. log3 5 x 1 5 x 31 x 14 3 1 58. log12 3x 2 1 3x 2 12 3x 0 x 0
59. 4 log 3 x 3 log 3 x 1 3 x 10 x 7 60. log2 x 2 x 2 2 x 2 x 2 22 4 x 2 x 6 0 x 3 x 2 0 x 3 or x 2. Thus, the solutions are x 3 and x 2. 61. log2 x log2 x 3 2 log2 [x x 3] 2 x 2 3x 22 x 2 3x 4 0 x 4 x 1 x 1 or x 4. Since log 1 3 log 4 is undefined, the only solution is x 4. 62. log x log x 3 1 log [x x 3] 1 x 2 3x 10 x 2 3x 10 0 x 2 x 5 0 x 2 or x 5. Since log 2 is undefined, the only solution is x 5.
63. log9 x 5 log9 x 3 1 log9 [x 5 x 3] 1 x 5 x 3 91 x 2 2x 24 0 x 6 x 4 0 x 6 or4. However, x 4 is inadmissible, so x 6 is the only solution. 64. ln x 1 ln x 1 0 ln x 1 x 1 0 x 1 x 1 1 x 2 2 0 x 2. However, x 2 fails to satisfy the original equation, so x 2 is the only solution. 2 x 1 1 x 1 1 65. log15 x 1 log15 x 1 2 log15 2 25 x 1 x 1 24x 26 x 1 x 1 5 25 x 13 12 .
32
CHAPTER 4 Exponential and Logarithmic Functions
66. log3 x 15 log3 x 1 2 log3 x 15 9x 9 8x 24 x 3
x 15 x 15 x 15 2 32 9 x 15 9 x 1 x 1 x 1 x 1
67. ln x 3 x ln x x 3 0. Let f x ln x x 3. We need to solve the equation f x 0. From the graph of f , we get x 221.
68. log x x 2 2 log x x 2 2 0. Let
f x log x x 2 2. We need to solve the equation
f x 0. From the graph of f , we get x 001 or
x 147.
2 0
2
4
0
1
2
2
69. x 3 x log10 x 1 x 3 x log10 x 1 0.
Let f x x 3 x log10 x 1. We need to solve the
equation f x 0. From the graph of f , we get x 0 or x 114.
70. x ln 4 x 2 x ln 4 x 2 0. Let f x x ln 4 x 2 . We need to solve the equation f x 0. From the graph of f , we get x 196 or
x 106.
2
1
1
2
2
2
2
71. e x x e x x 0. Let f x e x x. We need to
solve the equation f x 0. From the graph of f , we get x 057.
72. 2x x 1 2x x 1 0. Let
f x 2x x 1. We need to solve the equation
f x 0. From the graph of f , we get x 138.
2
1.0
0.5
2
0 1 2
1 2
2
SECTION 4.5 Exponential and Logarithmic Equations
73. 4x
x 4x
x 0. Let f x 4x
2
33
2
74. e x 2 x 3 x e x 2 x 3 x 0. Let
x.
2
We need to solve the equation f x 0. From the graph
f x e x 2 x 3 x. We need to solve the equation
of f , we get x 036.
f x 0. From the graph of f , we get x 089 or
x 071.
2
2 0
1
2 2
2
2 2
75. log x 2 log 9 x 1 log [x 2 9 x] 1 log x 2 11x 18 1 x 2 11x 18 101 0 x 2 11x 28 0 x 7 x 4. Also, since the domain of a logarithm is positive we must have
0 x 2 11x 18 0 x 2 9 x. Using the methods from Chapter 1 with the endpoints 2, 4, 7, 9 for the intervals, we make the following table: Interval
2
2 4
4 7
7 9
9
Sign of x 7 Sign of x 4 Sign of x 2 Sign of 9 x
Sign of x 7 x 4 Sign of x 2 9 x
Thus the solution is 2 4 7 9.
76. 3 log2 x 4 23 x 24 8 x 16.
77. 2 10x 5 log 2 x log 5 03010 x 06990. Hence the solution to the inequality is approximately the interval 03010 06990. 78. x 2 e x 2e x 0 e x x 2 2 0 e x x 2 x 2 0. We use the methods of Chapter 1 with the endpoints 2 and 2, noting that e x 0 for all x. We make a table: Interval Sign of e x Sign of x 2 Sign of x 2 Sign of e x x 2 x 2
Thus 2 x 2.
2
2 2
2
ln y 79. To find the inverse of f x 22x , we set y f x and solve for x. y 22x ln y ln 22x 2x ln 2 x . 2 ln 2 ln x ln x Interchange x and y: y . Thus, f 1 x . 2 ln 2 2 ln 2
34
CHAPTER 4 Exponential and Logarithmic Functions
80. To find the inverse of f x 3x1 , we set y f x and solve for x. y 3x1 ln y ln 3x1 x 1 ln 3 x 1
ln y ln y ln x ln x x 1. Interchange x and y: y 1. Thus, f 1 x 1. ln 3 ln 3 ln 3 ln 3
81. To find the inverse of f x log2 x 1, we set y f x and solve for x. y log2 x 1 2 y 2log2 x1 x 1 x 2 y 1. Interchange x and y: y 2x 1. Thus, f 1 x 2x 1.
82. To find the inverse of f x log 3x, we set y f x and solve for x. y log 3x 10 y 3x x 10x 10x . Thus, f 1 x . 3 3 1 log 22 log5 x log 1 83. 22 log5 x 16 2 2 16
10 y . Interchange 3
x and y: y
2 4 log5 x 12 x 512 1 04472 5 log5 x
84. log2 log3 x 4 log3 x 24 16 x 316 43,046,721
1. 85. log x4 log x3 0 log x3 log x 1 log x 0 or log x 1 x 1 or x 10
log x 3 log x log 3 log x 3 log 3x x 3 3x 2x 3 x 32 86. log x3 3 log x log x3 3 log x 0 log x log x2 3 log x 0 or log x2 3 0. Now log x 0 x 1. Also log x2 3 0 log x2 3 log x 3 x 10 3 , so x 10 3 539574 or
x 10 3 00185. Thus the solutions to the equation are x 1, x 10 3 539574 and x 10 3 00185. 00225 45 87. (a) A 5 5000 1 5000 100562520 559360. Thus the amount after 5 years is $559360. 4 00225 4t 5000 10056254t 2 10056254t log 2 4t log 1005625 (b) 10000 5000 1 4 log 2 t 3089 years. Thus the investment will double in about 3089 years, or about 30 years 10 months. 4 log 1005625 88. (a) A 2 6500e00454 $778191
0045t ln 16 0045t t (b) 8000 6500e0045t 16 e 13 13
1 ln 16 13 461. So the investment will reach 0045
$8000 in about 461 years, or about 4 years 8 months. 4t 89. 8000 5000 1 0035 5000 1008754t 16 1008754t log 16 4t log 100875 4 t
log 16 1349 years. The investment will increase to $8000 in approximately 13 years and 6 months. 4 log 100875
2t 90. 5000 3000 1 0065 53 103252t log 53 2t log 10325 t 2
log 53 2 log 10325
799. So it takes about
8 years to grow to $5000.
ln 2 815 years. Thus the investment will double in about 815 years. 0085 r 8 r r r 24 8 143577 8 143577 1 143577 1 1 92. 143577 1000 1 2 2 2 2 r 2 8 143577 1 00925. Thus the rate was about 925%. 91. 2 e0085t ln 2 0085t t
ln 3 126277. So only 5 grams remain after 93. 15e0087t 5 e0087t 13 0087t ln 13 ln 3 t 0087 approximately 13 days.
SECTION 4.5 Exponential and Logarithmic Equations
35
94. We want to solve for t in the equation 80 e02t 1 70 (when motion is downwards, the velocity is negative). Then ln 18 7 1 1 80 e02t 1 70 e02t 1 8 e02t 8 02t ln 8 t 104 seconds. Thus the 02 velocity is 70 ft/s after about 10 seconds. 10 7337, so there are approximately 7337 fish after 3 years. 1 4e083 10 08t 1 e08t 025 08t ln 025 5 1 4e08t 10 (b) We solve for t. 5 2 4e 1 4e08t ln 025 t 173. So the population will reach 5000 fish in about 1 year and 9 months. 08 k 96. (a) The length y is inversely proportional to 2x , so y x k 2x . To find k, we substitute 200 k 28 2 k 200 28 51200. Thus, we have y 51200 2x .
95. (a) P 3
(b) y 51200 25 1600 basepairs
5 2000 5 x ln 2 ln 5 x ln 128 47 cm (c) 2000 51200 2x 2x 51200 128 128 ln 2 h P P 97. (a) ln ehk P P0 ehk . Substituting k 7 and P0 100 we get P 100eh7 . P0 k P0
(b) When h 4 we have P 100e47 5647 kPa. T 20 T 20 98. (a) ln 011t e011t T 20 200e011t T 20 200e011t 200 200
(b) When t 20 we have T 20 200e01120 20 200e22 422 F. 13t5 13 I 1 e13t5 e13t5 1 13 I 13 t ln 1 13 I 99. (a) I 60 13 1 e 60 60 5 60 5 ln 1 13 I . t 13 60 5 ln 1 13 2 0218 seconds. (b) Substituting I 2, we have t 13 60
100. (a) P M Cekt Cekt M P ekt 1 MP MP t ln kt ln C k C
MP C
(c)
(b) P t 20 14e0024t . Substituting M 20, C 14, k 0024, 1 MP and P 12 into t ln , we have k C 1 20 12 t ln 2332. So it takes about 23 months. 0024 14
20 10 0
0
100
200
101. Since 91 9, 92 81, and 93 729, the solution of 9x 20 must be between 1 and 2 (because 20 is between 9 and 81), whereas the solution to 9x 100 must be between 2 and 3 (because 100 is between 81 and 729). 1 102. Notice that log x 1 log x log x 1, so 1 log x 101 for all x 0. So log x x 1 log x 5 has no solution, and x 1 log x k has a solution only when k 10.
10
This is verified by the graph of f x x 1 log x .
0
0
5
10
36
CHAPTER 4 Exponential and Logarithmic Functions
103. (a) x 1logx1 100 x 1 log x 1logx1 log 100 x 1 2 log x 1 log x 1 log 100 log x 1 log x 1 log x 1 2 0 log x 1 2 log x 1 1 0. Thus either log x 1 2 x 101 or log x 1 1 x 11 10 . (b) log2 x log4 x log8 x 11 log2 x log2 x log2 3 x 11 log2 x x 3 x 11 log2 x 116 11 6 11 6 log2 x 11 log2 x 6 x 2 64
2 ln 3 or 2x 1, which (c) 4x 2x1 3 2x 2 2x 3 0 2x 3 2x 1 0 either 2x 3 x ln 2 ln 3 has no real solution. So x is the only real solution. ln 2
4.6
MODELING WITH EXPONENTIAL FUNCTIONS
1. (a) Here n 0 10 and a 15 hours, so n t 10 2t15 10 22t3 .
(b) After 35 hours, there will be n 35 10 22353 106 108 bacteria. 2t 3 ln 1000 (c) n t 10 22t3 10,000 22t3 1000 ln 22t3 ln 1000 ln 2 ln 1000 t 149, 3 2 ln 2 so the bacteria count will reach 10,000 in about 149 hours.
2. (a) Here n 0 25 and a 5 hours, so n t 25 2t5 .
(b) After 18 hours, there will be n 18 25 2185 303 bacteria. t (c) n t 25 2t5 1,000,000 2t5 40,000 ln 2t5 ln 40,000 ln 2 ln 40,000 5 5 ln 40,000 764, so the bacteria count will reach 1,000,000 in about 764 hours. t ln 2
3. (a) A model for the squirrel population is n t n 0 2t6 . We are given
(c)
n
1,000,000
that n 30 100,000, so n 0 2306 100,000
100,000 3125. Initially, there were approximately 25 3125 squirrels.
800,000
(b) In 10 years, we will have t 40, so the population will be
200,000
n0
600,000 400,000
n 40 3125 2406 317,480 squirrels.
0
10
20
30
50 t (years)
40
Strange notation 4. (a) A model for the bird population is n t n 0 2t10 . We are given that n 25 13,000, so n 0 22510 13,000
13,000 n 0 52 2298. Initially, there were approximately 2300 birds. 2
(b) In 5 years, we will have t 30, so the population will be
approximately n 30 2298 23010 18,384 18,400 birds.
(c)
n (’000) 60 50 40 30 20 10 0
10
20
30
40 t
37
SECTION 4.6 Modeling with Exponential Functions
5. (a) Taking t 0 in the year 2005 and measuring n in thousands, we have
(d)
n 0 128 and r 012. Therefore, an exponential model is
n 80 60
n t 128e012t .
40
(b) In 2010, t 5, so the population was approximately n 5 128e0125 233, or 23,300 beavers.
20
(c) Solving n t 50, we get 128e012t 50 ln 128e012t ln 50 ln 128 012t ln 50
0
5
10
15 t
5
10
15
1 ln 50 ln 128 114, so the beaver population will reach t 012
50,000 after about 114 years.
6. (a) Here r 006 and n 0 450 (thousand). Thus, the population at
(d)
n 400
time t modeled by n t 450e006t , where t 0 corresponds to
300
the year 2010.
200
(b) t 2025 2010 15. Then we have
100
n 15 450e00615 183, and we estimate the prairie dog
population in 2025 to be approximately 183,000.
0
(c) Solving n t 300,000, we get 450e006t 300 e006t 23
t
006t ln 23 t 68 years.
7. n t n 0 ert ; n 0 110 million, t 2036 2011 25.
(a) r 003; n 25 110,000,000e00325 110,000,000e075 232,870,000. Thus at a 3% growth rate, the projected population will be approximately 233 million people by the year 2036.
(b) r 002; n 25 110,000,000e00225 110,000,000e050 181,359,340. Thus at a 2% growth rate, the projected population will be approximately 181 million people by the year 2036. 8. (a) In this case, a model for the bacteria population is n t n 0 ert 22e012t , so after 24 hours the population is approximately n 24 22e01224 392 bacteria.
(b) In this case, a model is n t n 0 ert 22e005t , so after 24 hours the population is approximately n 24 22e00524 73 bacteria.
9. (a) The doubling time is 18 years and the initial population is 112,000, so a model is n t 112,000 2t18 .
(c)
n (million) 2
(b) We need to find the relative growth rate r. Since the population is 2 112,000 224,000 when t 18, we have 224,000 112,000e18r 2 e18r ln 2 18r r ln182 00385. Thus, a model is
1
0
20
40
60
n t 112,000e00385t . (d) Using the model in part (a), we solve the equation n t 112,000 2t18 500,000 2t18 125 28 t 125 ln 2t18 ln 125 28 18 ln 2 ln 28 t
population to reach 500,000.
18 ln 125 28 ln 2
3885. Therefore, it takes about 3885 years for the
t
38
CHAPTER 4 Exponential and Logarithmic Functions
10. (a) The doubling time is 25 years and the initial population is 350,000, so
(c)
a model is n t 350,000 2t25 .
n (million) 2
(b) r ln252 00277, so a model is n t 350,000e00277t .
1
(d) We solve the equation n t 350,000 2t25 2,000,000 t25 ln 40 t ln 2 ln 40 2t25 40 7 ln 2 7 7 25
0
25 ln 40 7 6286. Therefore, it takes about 629 years for the t ln 2 population to reach 2,000,000.
20
40
60
t
11. (a) The deer population in 2010 was 20,000. (b) Using the model n t 20,000ert and the point 4 31000, we have 31,000 20,000e4r 155 e4r 4r ln 155 r 14 ln 155 01096. Thus n t 20,000e01096t
(c) n 8 20,000e010968 48,218, so the projected deer population in 2018 is about 48,000. ln 5 1468. Thus, it takes about 147 years (d) 100,000 20,000e01096t 5 e01096t 01096t ln 5 t 01096 for the deer population to reach 100,000. 12. (a) From the graph, we see that the initial bullfrog population was 100. (b) We use a model of the form n t n 0 ert with n 0 100. Because we know that the population was 225 at t 2, we 04055t . solve n 2 225 100e2r 225 r 12 ln 225 100 04055. Thus, a model is n t 100e
(c) The estimated population after 15 years is n 15 100e0405515 43,800 frogs.
(d) The population will reach 75,000 when n t 100e04055t 75,000 e04055t 750 t will take about 163 years for the population to reach 75,000.
ln 750 1632. So it 04055
13. (a) Using the formula n t n 0 ert with n 0 8600 and n 1 10000, we solve for r, giving 10000 n 1 8600er 50 r 01508t . 50 43 e r ln 43 01508. Thus n t 8600e (b) n 2 8600e015082 11627. Thus the number of bacteria after two hours is about 11,600. ln 2 (c) 17200 8600e01508t 2 e01508t 01508t ln 2 t 4596. Thus the number of bacteria will 01508 double in about 46 hours.
14. (a) Using n t n 0 ert with n 2 400 and n 6 25,600, we have n 0 e2r 400 and n 0 e6r 25,600. Dividing the
n e6r 25,600 second equation by the first gives 0 2r 64 e4r 64 4r ln 64 r 14 ln 64 104. Thus the 400 n0e relative rate of growth is about 104%.
(b) Since r 14 ln 64 12 ln 8, we have from part (a) n t n 0 e
1 2 ln 8 t
. Since n 2 400, we have 400 n 0 eln 8
400 n 0 ln 8 400 8 50. So the initial size of the culture was 50. e (c) Substituting n 0 50 and r 104, we have n t n 0 ert 50e104t .
(d) n 45 50e10445 50e468 53885, so the size after 45 hours is approximately 5400. ln 1000 (e) n t 50,000 50e104t e104t 1000 104t ln 1000 t 664. Hence the population will 104 reach 50,000 after roughly 6 hours 40 minutes.
SECTION 4.6 Modeling with Exponential Functions
39
15. (a) Using n t n 0 ert with n 0 49 and n 19 44, we have n 0 49 and n 0 e19r 44 49e19t 44 e19r 44 49 1 44 000566t , we solve 35 49e000566t 19r ln 44 49 r 19 ln 49 000566. Using the model n t 49e
000566t ln 35 49 t 594 years, so the population is projected to decline to 35 million in the year 2059.
(b) We solve n t 12 n 0 n 0 e000566t 000566t ln 12 t 1225 years.
10 t 239 years, so the 16. (a) A model is given by n t 78e00104t . We solve 10 78e00104t 00104t ln 78
population of the world is projected to reach 10 billion in late 2043.
(b) 2n 0 n 0 e00104t 2 e00104t ln 2 00104t t 666. So at the current growth rate, it will take approximately 666 years for the population to double. 17. (a) Setting t 0 in 1950 and substituting M 11, r 00189, and A
M n0 11 25 34 into the logistic growth model n0 25
11 M , we have n t . Solving 1 Aert 1 34e00189t 11 1 34e00189t 10 n t 10, we obtain 10 1 34e00189t
(b)
n (bn) 10
n t
0
50
100
1 , t 1 ln 1 1866. Thus, the world’s e00189t 34 00189 34
population is projected to reach 10 billion in the year 2136.
18. (a) From Exercise 5, n 0 128 and r 012. Therefore, with M 80 and A
M n0 80 128 5 25, a logistic model is n0 128
n t
150
200 t
Strange notation n (’000) 160 120
80 M . 1 Aert 1 525e012t
80
(b) We see from the graph that the models diverge dramatically as time goes on. The exponential model increases without bound, while the logistic
40 0
10
20
30
5
10
15
40 t
model approaches the carrying capacity. M with n 0 8, M 6000, r 057, and 1 Aert M n0 6000 6000 8 A 749, we have n t . n0 8 1 749e057t
19. (a) Using n t
(b) From the graph, the number of infections reaches 5900 on the 18th or 19th day.
n 6000 5000 4000 3000 2000 1000 0
M with n 0 18, M 10, r 02, and 1 Aert 4 10 10 18 M n0 , we have n t . A n0 18 9 1 4 e02t
20. (a) Using n t
9
(b) The fish population declines rapidly at first, then more gradually, approaching 10,000 as t .
20 t
Strange notation
n (’000) 20 15 10 5 0
5
10
15
20 t
21. (a) Because the halflife is 1600 years and the sample weighs 22 mg initially, a suitable model is m t 22 2t1600 .
40
CHAPTER 4 Exponential and Logarithmic Functions
(b) From the formula for radioactive decay, we have m t m 0 ert , where m 0 22 and r
ln 2 ln 2 0000433. h 1600
Thus, the amount after t years is given by m t 22e0000433t .
(c) m 4000 22e00004334000 389, so the amount after 4000 years is about 4 mg.
9 e0000433t (d) We have to solve for t in the equation 18 22 e0000433t . This gives 18 22e0000433t 11 9 ln 11 9 t 0000433t ln 11 4634, so it takes about 463 years. 0000433
22. (a) Because the halflife is 30 years and the sample weighs 10 g initially, a suitable model is m t 10 2t30 . ln 2 ln 2 (b) Using m t m 0 ert with m 0 10 and h 30, we have r 00231. Thus m t 10e00231t . h 30 (c) m 80 10e0023180 16 grams. ln 5 (d) 2 10e00231t 15 e00231t ln 15 00231t t 70 years. 00231 ln 2 23. By the formula in the text, m t m 0 ert where r , so m t 50e[ln 229]t . We need to solve for t in the h ln 2 32 t 29 ln 32 1867, so it t ln equation 32 50e[ln 229]t . This gives e[ln 229]t 32 50 50 50 29 ln 2 takes about 18 years and 8 months. 24. From the formula for radioactive decay, we have m t m 0 ert , where r
ln 2 . Since h 30, we have h
ln 2 00231 and m t m 0 e00231t . In this exercise we have to solve for t in the equation 005m 0 m 0 e00231t 30 ln 005 1297. So it will take about 130 s. e00231t 005 00231t ln 005 t 00231 ln 2 25. By the formula for radioactive decay, we have m t m 0 ert , where r , in other words m t m 0 e[ln 2 h ]t . In h ln 2 48 this exercise we have to solve for h in the equation 200 250e[ln 2 h ]48 08 e[ln 2 h ]48 ln 08 h ln 2 48 1491 hours. So the halflife is approximately 149 hours. h ln 08 ln 2 . In other words, m t m 0 e[ln 2 h ]t . 26. From the formula for radioactive decay, we have m t m 0 ert , where r h (a) Using m 3 058m 0 , we have to solve for h in the equation 058m 0 m 3 m 0 e[ln 2 h ]3 . 3 ln 2 3 ln 2 ln 058 h 38 days. Thus the Then 058m 0 m 0 e[3 ln 2 h ] e[3 ln 2 h ] 058 h ln 058 halflife of radon222 is about 38 days. (b) Here we have to solve for t in the equation 02m m e[ln 2382]t . So we have 02m m e[ln 2382]t r
0
0
0
0
382 ln 02 ln 2 t ln 02 t 887. So it takes roughly 9 days for a sample of 02 e[ln 2382]t
382 Radon222 to decay to 20% of its original mass.
ln 2
27. By the formula in the text, m t m 0 e[ln 2 h ]t , so we have 065 1 e[ln 25730]t ln 065 t
5730 ln 065 3561. Thus the artifact is about 3560 years old. ln 2
ln 2 t 5730
ln 2 ln 2 . Since h 5730, r 0000121 h 5730 and m t m 0 e0000121t . We need to solve for t in the equation 059m 0 m 0 e0000121t e0000121t 059 ln 059 43606. So the mummy was buried about 4360 years ago. 0000121t ln 059 t 0000121
28. From the formula for radioactive decay, we have m t m 0 ert where r
SECTION 4.6 Modeling with Exponential Functions
29. We use the radioactive decay model m t m 0 ert with m 0 005 mol and r
41
ln 2 . Solving m t 004 for t, we get 45
ln 2 t ln 45 t 145. Thus, the age of the sample is approximately 145 billion years. 45 ln 2 30. We use the radioactive decay model m t m 0 ert with m 0 00005 00015 0002 mol and r . Solving 07 ln 2 t ln 14 t 14. According to this model, the age m t 00005 for t, we get 00005 0002e[ln 207]t 07 of the sample is approximately 14 billion years. This is concordant with the age estimate given by the 238 Uto206 Pb decay model. ln 2 31. We use the radioactive decay model m t m 0 ert with m t 08m 0 and r and solve for t: 08 e[ln 2432]t 432 ln 2 t ln 08 t 1391 years. (Note that the absolute quantity of 141 Am is irrelevant.) 432 004 005e[ln 245]t
32. (a) We use Newton’s Law of Cooling: T t Ts D0 ekt with k 01947, Ts 60, and D0 986 60 386 . So T t 60 386e01947t .
12 01947t ln (b) Solve T t 72. So 72 60 386e01947t 386e01947t 12 e01947t 1 12 ln 600, and the time of death was about 6 hours ago. t 01947 386
386
12 386
33. (a) T 0 65 145e0050 65 145 210 F.
(b) T 10 65 145e00510 1529. Thus the temperature after 10 minutes is about 153 F.
(c) 100 65 145e005t 35 145e005t 02414 e005t ln 02414 005t t
ln 02414 284. 005
Thus the temperature will be 100 F in about 28 minutes. kt 34. Using Newton’s Law of Cooling, T t Ts D0 ekt with Ts 75 and D0 185 75 110. So T t 75 110e . 15 (a) Since T 30 150, we have T 30 75 110e30k 150 110e30k 75 e30k 15 22 30k ln 22 1 ln 15 . Thus we have T 45 75 110e4530 ln1522 1369, and so the temperature of the turkey k 30 22
after 45 minutes is about 137 F.
25 5 (b) The temperature will be 100 F when 75 110et30 ln1522 100 et30 ln1522 22 110 5 ln 22 t 5 t 30 1161. So the temperature will be 100 F after about 2 hours. ln ln 15 22 22 30 ln 15 22
35. We use Newton’s Law of Cooling: T t Ts D0 ekt , with Ts 20 and
D0 100 20 80. So T t 20 80ekt . Since T 15 75, we have 11 20 80e15k 75 80e15k 55 e15k 11 16 15k ln 16 1 ln 11 . Thus T 25 20 80e2515ln1116 628, and so the k 15 16 temperature after another 10 min is 63 C. The function
T t 20 80e115ln1116t is shown in the viewing rectangle [0 30] by
[50 100].
100 80 60 0
20
42
CHAPTER 4 Exponential and Logarithmic Functions
4.7
LOGARITHMIC SCALES
1. (a) pH log H log 50 103 23 (b) pH log H log 32 104 35 (c) pH log H log 50 109 83 2. pH log H log 31 108 75 and the substance is basic.
3. (a) pH log H 30 H 103 M (b) pH log H 65 H 1065 32 107 M 4. (a) pH log H 46 H 1046 M 25 105 M (b) pH log H 73 H 1073 M 50 108 M 5. 40 107 H 16 105 log 40 107 log H log 16 105 log 40 107 pH log 16 105 64 pH 48. Therefore the range of pH readings for cheese is approximately 48 to 64.
6. 158 104 10pH 158 103 log 158 104 log 10pH log 158 103
log 158 4 pH log 158 3 28 pH 38. The pH ranges from 28 to 38. 7. (a) For the California red wine, we have pH log H log H 32 H 1032 63 104 M. For the Italian white wine, pH log H log H 29 H 1029 13 103 M. (b) The California red wine has lower hydrogen ion concentration. 8. (a) pH log H log H 55 H 1055 32 106 M. (b) The saliva was more acidic when the patient was sick.
(c) As pH increases, hydrogen ion concentration decreases. So as the patient gets better, the pH of their saliva will increase and its hydrogen ion concentration will decrease. 3125 I with S 104 and I 3125, so M log 4 55. S 10 I I (b) M log 10 M I S 10 M . We have M 48 and S 104 , so I 104 1048 63. S S
9. (a) M log
10. (a) M log
I 721 with S 104 and I 721, so M log10 4 59. S 10
(b) I 104 1058 631.
11. Let I0 be the intensity of the smaller earthquake and I1 the intensity of the larger earthquake. Then I1 20I0 . 20I0 I I log 20 log I0 log S. Then Notice that M0 log 0 log I0 log S and M1 log 1 log S S S M1 M0 log 20 log I0 log S log I0 log S log 20 13. Therefore the magnitude is 13 times larger.
I 12. Let the subscript S represent the San Francisco earthquake and J the Japan earthquake. Then we have M S log S 83 S 83 I I 10 J S 49 I J S 1049 . So 49 1034 25119, and so the San Francisco I S S 1083 and M J log S IJ 10 earthquake was 2500 times more intense than the Japan earthquake.
SECTION 4.7 Logarithmic Scales
13. Let the subscript J represent the Japan earthquake and S represent the San Francisco earthquake. Then M J log
43
IJ 91 S
I I S 1091 I J S 1091 and M S log S 83 I S S 1083 . So J 1008 63, and hence the Japan S IS S 1083 earthquake was about six times more intense than the San Francisco earthquake. 14. Let the subscript N represent the Northridge, California earthquake and K the Kobe, Japan earthquake. Then I I I 1072 M N log N 68 I N S 1068 and M K log K 72 I K S 1072 . So K 68 1004 251, and S S IN 10 so the Kobe, Japan earthquake was 25 times more intense than the Northridge, California earthquake. 20 105 I 10 log 10 log 2 107 10 log 2 log 107 10 log 2 7 73. Therefore the I0 10 1012 intensity level was 73 dB.
15. 10 log
32 102 I 10 log 10 log 32 1010 105 dB. I0 10 1012 I I 70 10 log log I 12 7 log I 5, so the intensity was 105 wattsm2 . 17. 10 log I0 10 1012 I 18. 98 10 log 12 log I 1012 98 log I 98 log 1012 22 I 1022 63 103 . So the 10 intensity was 63 103 wattsm2 . 16. 10 log
I 31 105 10 log 10 log 31 107 75 dB. 12 I0 10 10 I I (b) Here 90 dB, so 90 10 log 12 9 log 12 9 log I log 1012 log I 12 9 10 10 I 103 Wm2 .
19. (a) The intensity is 31 105 Wm2 , so 10 log
(c) The ratio of the intensities is
103 Ie 323. Is 31 105
IM 20. Let the subscript M represent the power mower and C the rock concert. Then 106 10 log 1012 IC log IC 1012 120 log I M 1012 106 I M 1012 10106 . Also 120 10 log 12 10
IC 1012 106 1014 2512, and so the ratio of intensity is roughly 25. IM 10 k I1 k k k 2 log d1 10 log and I1 2 1 10 log 2 10 log 20 log d1 . Similarly, 21. (a) 1 10 log I0 I I d1 d1 I0 0 0 k 20 log d2 . Substituting the expression for 1 gives 2 10 log I0 k d 20 log d1 20 log d1 20 log d2 1 20 log d1 20 log d2 1 20 log 1 . 2 10 log I0 d2 d1 2 120 20 log 02 106, and so the (b) 1 120, d1 2, and d2 10. Then 2 1 20 log 120 20 log 10 d2 intensity level at 10 m is approximately 106 dB. IC 1012 10120 . So
44
CHAPTER 4 Exponential and Logarithmic Functions
CHAPTER 4 REVIEW 2 9739, f 25 55902 7 18775, f 55 135765 2. f x 3 2x ; f 22 0653, f
1. f x 5x ; f 15 0089, f
3. g x 4e x2 ; g 07 0269, g 1 1472, g 12527 3 26888, g 36 174098 4. g x 74 e x1 ; g 2 0644, g 5. f x 3x 1. Domain , range 1 , horizontal asymptote y 1.
x 6. f x 12 5. Domain , range 5 , horizontal asymptote y 5.
y
y
1 1
(0, 2) 1
x
(0, _4) x
1
7. g x 4x1 . Domain , range 0 , horizontal asymptote y 0. y
8. g x 2x1 . Domain , range 0, horizontal asymptote y 0. y
1
1
(0, _1/2)
x
1 (0, 1/4) x
1
9. h x e x2 1. Domain , range 1, 10. h x 3ex 1. Domain , range 1 , horizontal asymptote y 1.
horizontal asymptote y 1. y
y 1 1
x
(0, 4) (0, _1-e@)
1 1
x
CHAPTER 4
11. f x log3 x 2. Domain 2 , range , vertical asymptote x 2.
Review
12. f x log4 x 2. Domain 2 ,
range , vertical asymptote x 2. y
y
1 x
1
1 (0, _1)
x
1
(0, - 1/2)
13. f x log13 x 1. Domain 0 , range , vertical asymptote x 0. y
14. f x 1 log12 x 2. Domain 2 , range , vertical asymptote x 2. y
(0, 2) 1 1
x
1 x
1
15. g x log2 x 2. Domain 0,
16. g x log3 x 3 2. Domain 3 ,
range , vertical asymptote x 0.
range , vertical asymptote x 3.
y
y
1 1
x 1 2
x
45
46
CHAPTER 4 Exponential and Logarithmic Functions
17. g x 2 ln x. Domain 0 , range , vertical asymptote x 0.
18. g x ln x 2 .
Domain x x 0 0 0 ,
y
range , vertical asymptote x 0. y
1 x
1
1 1
x
2
19. f x 10x log 1 2x. Since log u is defined only for u 0, we require 1 2x 0 2x 1 x 12 , and so the domain is 12 . 20. g x log 2 x x 2 . We must have 2 x x 2 0 (since log y is defined only for y 0) x 2 x 2 0 x 2 x 1 0. The endpoints of the intervals are 2 and 1. Interval Sign of x 2 Sign of x 1
Sign of x 2 x 1
1
1 2
2
2
2 2
2
Thus the domain is 1 2. 21. h x ln x 2 4 . We must have x 2 4 0 ( since ln y is defined only for y 0) x 2 4 0 x 2 x 2 0. The endpoints of the intervals are 2 and 2. Interval Sign of x 2 Sign of x 2
Sign of x 2 x 2
Thus the domain is 2 2 .
22. k x ln x. We must have x 0. So x 0 x 0 or x 0. Since x 0 x 0, the domain is x 0 or x 0 which is equivalent to x 0. In interval notation, 0 0 . 23. log2 1024 10 210 1024
24. log6 37 x 6x 37
25. log x y 10 y x
26. ln c 17 e17 c
27. 26 64 log2 64 6
28. 4912 17 log49 17 12
29. 10x 74 log10 74 x log 74 x 31. log2 128 log2 27 7
30. ek m ln m k 32. log8 1 log8 80 0
33. 10log 45 45 35. ln e6 6
34. log 0000001 log 106 6 36. log4 8 log4 432 32
CHAPTER 4 1 log 33 3 37. log3 27 3
39. log5
5 log5 512 12
41. log 25 log 4 log 25 4 log 102 2
38. 2log2 13 13 2 40. e2 ln 7 eln 7 72 49 42. log3 243 log3 352 52
Review
23 3 43. log2 1623 log2 24 log2 292 92 44. log5 250 log5 2 log5 250 2 log5 125 log5 5 3 45. log8 6 log8 3 log8 2 log8 63 2 log8 4 log8 823 23 46. log10 log10 10100 log10 100 log10 102 2 47. log AB 2 C 3 log A 2 log B 3 log C 48. log2 x x 2 1 log2 x log2 x 2 1 log2 x 12 log2 x 2 1 1 x2 1 x2 1 1 ln x 2 1 ln x 2 1 1 ln x 1 x 1 ln x 2 1 ln 49. ln 2 2 2 x2 1 x2 1 12 ln x 1 ln x 1 ln x 2 1 4x 3 log 4x 3 log y 2 x 15 log 4 3 log x 2 log y 5 log x 1 50. log 5 2 y x 1 x 2 1 5x3/2 log5 x 2 1 5x3/2 log5 x x 2 1 2 log5 x 32 log5 1 5x 12 log5 x 3 x 51. log5 x3 x 2 log5 x 32 log5 1 5x 12 log5 x log5 x 2 1 2 log5 x 32 log5 1 5x 12 log5 x log5 x 1 log5 x 1 3 4 x 12 13 ln x 4 12 ln x 16 12 ln x 3 52. ln x 16 x 3 53. log 6 4 log 2 log 6 log 24 log 6 24 log 96 54. log x log x 2 y 3 log y log x x 2 y y 3 log x 3 y 4 2 x y3/2 55. 32 log2 x y 2 log2 x 2 y 2 log2 x y3/2 log2 x 2 y 2 log2 2 x 2 y2
2 x 1 56. log5 2 log5 x 1 13 log5 3x 7 log5 [2 x 1] log5 3x 713 log5 3 3x 7 x2 4 57. log x 2 log x 2 12 log x 2 4 log [x 2 x 2] log x 2 4 log x2 4 5 5 ln x 4 x 2 4x 58. 12 ln x 4 5 ln x 2 4x 12 ln x 4 x 2 4x
59. 26x3 8 26x3 23 6x 3 3 x 1 1x 1x 2 60. 13 19 13 13 1 x 2 x 1
61. 53x2 2 ln 53x2 ln 2 3x 2 ln 5 ln 2 3x 2
ln 2 2 ln 2 x ln 5 0523 ln 5 3
62. 1043x 5 log 1043x log 5 4 3x log 5 x 13 4 log 5 1100
47
48
CHAPTER 4 Exponential and Logarithmic Functions
63. 25x1 34x ln 25x1 ln 34x 5x 1 ln 2 4 x ln 3 x 5 ln 2 ln 3 4 ln 3 ln 2 x
ln 81 4 ln 3 ln 2 2 0811 5 ln 2 ln 3 5 ln 2 ln 3
64. 102x5 e3x1 ln 102x5 ln e3x1 25 x ln 10 3x 1 x 3 25 ln 10 1 x 2
1
5 ln 10 3
5 0481 2 ln 10 15
65. x 3 34x x 2 34x 6x 34x 34x x x 2 x 6 0 34x 0 or x 0 or x 2 x 6 x 2 x 3 0. The first of these equations has no solution, so the solutions are x 0, x 2, and x 3.
2 2 66. e2x 6e x 9 0 e x 6e x 9 0 e x 3 0 e x 3 0 ln e x ln 3 x ln 3 110
67. log x log x 1 log 12 log x x 1 log 12 x x 1 12 x 2 x 12 0 x 3 x 4 0 x 4 or 3. Since log 4 is undefined, the only solution is x 3. 68. ln x 2 ln 3 ln 5x 7 ln 3 x 2 ln 5x 7 3 x 2 5x 7 2x 1 x 12 , but since ln x 2 is undefined for x 12 , there is no solution.
69. log2 1 x 4 1 x 24 x 1 16 15
70. ln 2x 3 1 0 ln 2x 3 1 2x 3 e1 x 12 1e 3 168
71. log3 x 8 log3 x 2 log3 x x 8 2 x x 8 9 x 2 8x 9 0 x 9 x 1 0 x 1 or 9. We reject 1 because it does not satisfy the original equation, so the only solution is x 9. 72. log12 x 5 log12 x 2 1 log12
x 5 1 x 5 1 2 x 5 x 2 x 12 x 2 x 2 2
3 log 063 2x log 5 log 063 x 0430618 3 2 log 5 log 7 2602452 74. 23x5 7 3x 5 log 2 log 7 x 13 5 log 2
73. 52x /3 063
75. 52x1 34x1 2x 1 log 5 4x 1 log 3 2x log 5 log 5 4x log 3 log 3 log 3 log 5 2303600 x 2 log 5 4 log 3 log 3 log 5 x 4 log 3 2 log 5 1 ln 10000 0614023 76. e15k 10000 15k ln 10000 k 15
77. y e xx2 . Vertical asymptote x 2, horizontal asymptote y 272, no maximum or minimum. 10
78. y 10x 5x . No vertical asymptote, horizontal
asymptote y 0, local minimum of about 013 at x 052.
2 5 1 20
0
20 2
2
CHAPTER 4
79. y log x 3 x . Vertical asymptotes x 1, x 0,
x 1, no horizontal asymptote, local maximum of about
041 when x 058.
Review
49
80. y 2x 2 ln x. Vertical asymptote x 0, no horizontal asymptote, local minimum of about 119 at x 050. 100 50
1
1
2
0
0
5
10
2
81. 3 log x 6 2x. We graph y 3 log x and y 6 2x in 82. 4 x 2 e2x . From the graphs, we see that the solutions the same viewing rectangle. The solution occurs where the two graphs intersect. From the graphs, we see that the
are x 064 and x 2.
5
solution is x 242. 10 2 5
2
10
10
83. ln x x 2We graph the function f x ln x x 2, and we see that the graph lies above the xaxis for
016 x 315. So the approximate solution of the given inequality is 016 x 315.
84. e x 4x 2 e x 4x 2 0. We graph the function
f x e x 4x 2 , and we see that the graph lies below the
xaxis for 041 071 431.
2
5 10
0
5
2
85. f x e x 3ex 4x. We graph the function f x,
and we see that the function is increasing on 0 and 110 and that it is decreasing on 0 110.
2
86. The line has xintercept at x e0 1. When x ea ,
y ln ea a. Therefore, using the pointslope equation,
a a0 we have y 0 a x 1 y a x 1. e 1 e 1
2 5
87. log4 15
log 15 1953445 log 4
log 3 4 0147839 88. log7 34 log 7
50
CHAPTER 4 Exponential and Logarithmic Functions
89. log9 028
log 028 0579352 log 9
90. log100 250
log 250 1198970 log 100
91. Notice that log4 258 log4 256 log4 44 4 and so log4 258 4. Also log5 620 log5 625 log5 54 4 and so log5 620 4. Then log4 258 4 log5 620 and so log4 258 is larger. x x 92. f x 23 . Then y 23 log2 y 3x log3 log2 y x, and so the inverse function is f 1 x log3 log2 x . Since log3 y is defined only when y 0, we have log2 x 0 x 1. Therefore the domain is 1 , and the range is . r nt . 93. P 12,000, r 010, and t 3. Then A P 1 n 23
(a) For n 2, A 12,000 1 010 12,000 1056 $16,08115. 2 123 $16,17818. (b) For n 12, A 12,000 1 010 12 3653 $16,19764. (c) For n 365, A 12,000 1 010 365
(d) For n , A Pert 12,000e0103 $16,19831.
94. P 5000, r 0085, and n 2. 215 (a) For t 15, A 5000 1 0085 5000 104253 $566498. 2
7 (b) We want to find t such that A 7000. Then A 5000 104252t 7000 104252t 7000 5000 5 log 75 log 75 7 t 404, and so the investment will amount to $7000 2t log 10425 log 5 2t log 10425 2 log 10425 after approximately 4 years.
(c) In this case, we solve n t n 0 ert for t when n 0 5000, r 0085, and n t 7000: 7000 5000e0085t 7 7 e0085t ln 7 0085t t ln 5 396, so the investment will grow to $7000 in just under 4 years. 5 5 0085
r nt with P 100,000, r 0052, n 365, and A 100,000 10,000 110,000, 95. We use the formula A P 1 n 365t 365t and solve for t: 110,000 100,000 1 0052 11 1 0052 log 11 365t log 1 0052 365 365 365 t
log 11 1833. The account will accumulate $10,000 in interest in approximately 18 years. 365 log 1 0052 365
96. We solve n t n 0 ert for n t 2n 0 and r 0045: 2n 0 n 0 e0045t ln 2 0045t t retirement savings plan will double in about 154 years.
ln 2 15403. The 0045
00425 365 97. After one year, a principal P will grow to the amount A P 1 P 104341. The formula for simple 365 interest is A P 1 r. Comparing, we see that 1 r 104341, so r 004341. Thus the annual percentage yield is 4341%. 0032 12 98. A P 1 P 103247 P 1 r r 3247% 12 99. (a) Using the model n t n 0 ert , with n 0 30 and r 015, we have the formula n t 30e015t . (b) n 4 30e0154 55.
015t 015t ln 50 t (c) 500 30e015t 50 3 e 3 reach 500 in about 19 years.
1 ln 50 3 1876. So the stray cat population will 015
CHAPTER 4
51
Review
100. Using the model n t n 0 ert , with n 0 10000 and n 1 25000, we have 25000 n 1 10000er1 er 52 r ln 52 0916. So n t 10000e0916t .
(a) Here we must solve the equation n t 20000 for t. So n t 10000e0916t 20000 e0916t 2 ln 2 0756. Thus the doubling period is about 45 minutes. 0916t ln 2 t 0916 (b) n 3 10000e09163 156250, so the population after 3 hours is about 156,250.
101. (a) From the formula for radioactive decay, r
ln 2 and n t 150 et ln 275,380 or, equivalently, 75,380
n t 150 2t75,380 .
(b) n 1000 150 2100075380 14863 mg (c) We solve n t 50 13 2t75380 t 75,380 119,474 years.
ln 3 119,474. Thus, only 50 mg remains after approximately ln 2
ln 2 . So m t m 0 e[ln 2 h ]t . h (a) Using m 8 033m 0 , we solve for h. We have 033m 0 m 8 m 0 eln 2 h 033 e8 ln 2 h 8 ln 2 8 ln 2 ln 033 h 5002. So the halflife of this element is roughly 5 days. h ln 033 (b) m 12 m e[ln 25]12 019m , so about 19% of the original mass remains.
102. From the formula for radioactive decay, we have m t m 0 ert , where r
0
0
103. From the formula for radioactive decay, we have m t m 0 ert , where r
ln 2 ln 2 . Since h 4, we have r 0173 h 4
and m t m 0 e0173t . (a) Using m 20 0375, we solve for m 0 . We have 0375 m 20 m 0 e017320 003125m 0 0375 0375 12. So the initial mass of the sample was about 12 g. m0 003125 (b) m t 12e0173t or, equivalently, m t 12 2t4 . (c) m 3 12e01733 7135. So there are about 71 g remaining after 3 days.
(d) Here we solve m t 015 for t: 015 12e0173t 00125 e0173t 0173t ln 00125 ln 00125 t 253. So it will take about 25 days until only 15% of the substance remains. 0173
104. (a) Using n 0 1500 and n 5 3200 in the formula n t n 0 ert , we have 3200 n 5 1500e5r e5r 32 15 32 1 32 01515t . 5r ln 15 r 5 ln 15 01515. Thus n t 1500 e
(b) We have t 2031 2020 11 so n 11 1500e0151511 7940. Thus in 2031 the bird population should be about 7940.
105. (a) The doubling time is a 15 hr, so the relative growth rate is 2 r ln 15 0462.
carrying capacity
(b) Since the initial population is 1400, the model we want has the form 1400 n t , where r 0462 and 1 Aert M n0 1400 100 A 13. Thus, a model is n0 100 n t
1400 . 1 13e0462t
(c)
n 1400 1000 700
0
10
It appears to take about 555 hr for the yeast to reach 700 colonies.
t
52
CHAPTER 4 Exponential and Logarithmic Functions
106. We use Newton’s Law of Cooling: T t Ts D0 ekt with k 00341, Ts 60 and D0 190 60 130.
3 So 90 T t 60 130e00341t 90 60 130e00341t 130e00341t 30 e00341t 13 3 t ln 313 430, so the engine cools to 90 F in about 43 minutes. 00341t ln 13 00341 8 107. H 13 10 M. Then pH log H log 13 108 79, and so fresh egg whites are basic. 108. pH 19 log H . Then H 1019 126 102 M.
109. Let I0 be the intensity of the smaller earthquake and I1 be the intensity of the larger earthquake. Then I1 35I0 . Since I I M log , we have M0 log 0 65 and S S 35I0 I I1 log log 35 log 0 log 35 M0 log 35 65 804. So the magnitude on the M1 log S S S Richter scale of the larger earthquake is approximately 80. I I 110. Let the subscript J represent the jackhammer and W the whispering: J 132 10 log J log J 132 I0 I0 I I I 10132 J 10132 . Similarly W 1028 . So J 10104 251 1010 , and so the ratio of intensities is I0 I0 IW 1028
251 1010 .
111. (a) y 2x has graph VI because it grows without bound as x increases and has asymptote y 0. (b) y ln x has graph VIII because it has domain 0 and is decreasing.
(c) 2x 3y 6 has graph V because it is linear with xintercept 3 and yintercept 2. 1 (d) y 1 3 has graph III because it has vertical asymptote x 0 and horizontal asymptote y 1. x (e) y log2 x has graph II because it has domain 0 and is increasing. 2
(f) y 4e x 4 has graph VII because it has range 0 and is even.
(g) y 2 2x x 2 has graph IV because it is a parabola that opens downward.
(h) y 10xex has graph I because it has the characteristic shape of a surge function.
CHAPTER 4 TEST 1. (a)
y
(b) (0, 2)
1 0
(1, 0)
y 1 1
2
x
f x 3 3x has domain , range 3,
g x log3 x 3 has domain 3 , range
, and vertical asymptote x 3. 2. (a) f t ln 2t 3 is defined where 2t 3 0 2t 3 t 32 , so its domain is 32 . (b) g x log x 2 1 is defined where x 2 1 0 x 1, so its domain is 1 1 . and horizontal asymptote y 3.
x
CHAPTER 4
3. (a) 62x 25 log6 62x log6 25 2x log6 25
Test
(b) ln A 3 eln A e3 A e3
4. (a) 10log 36 36
(b) ln e3 3 12 log3 332 32 (c) log3 27 log3 33 3 (d) log2 80 log2 10 log2 80 10 log2 8 log2 2 3
(e) log8 4 log8 823 23
(f) log6 4 log6 9 log6 4 9 log6 62 2
5. (a) log
x y3 z2
log x log y 3 log z 2 log x 3 log y 2 log z
1 x x 12 x ln 12 ln x 12 ln y ln (b) ln y y 2 y x2 1 1 x2 1 (c) log log 3 12 log x 2 1 3 log x log x 1 3 2 x x 1 x x 1 12 log x 2 1 32 log x 12 log x 1
6. (a) log a 2 log b log a log b2 log ab2
x 2 25 x 5 x 5 ln ln x 5 (b) ln x 2 25 ln x 5 ln x 5 x 5
(c) log3 x 2 log3 x 1 3 log3 y log3 x log3 x 12 log3 y 3 log3
x y3 x 12
7. (a) 34x 3100 4x 100 x 25 2
(b) e3x2 e x 3x 2 x 2 x 2 3x 2 0 x 1 x 2 0 x 1 or x 2 3x1 3x1 6 23 65 3x 1 ln 23 ln 65 3x ln 23 ln 65 ln 23 ln 65 32 ln 95 (c) 5 23 x
ln 95
3 ln 23
0483
(d) 10x3 62x log 10x3 log 62x x 3 2x log 6 x 2 log 6 1 3 x
3 539 2 log 6 1
8. (a) log 2x 3 2x 103 x 12 1000 500
(b) log x 1 log 2 log 5x log 2 x 1 log 5x 2x 2 5x 3x 2 x 23
(c) 5 ln 3 x 4 ln 3 x 45 3 x e45 x 3 e45 0774 (d) log4 x 3 log4 x 1 2 log4
x 3 x 3 2 42 16 x 3 16x 16 x 19 15 x 1 x 1
9. Using the Change of Base Formula, we have log12 27
log 27 1326. log 12
53
54
CHAPTER 4 Exponential and Logarithmic Functions
10. (a) From the formula for population growth, we have 8000 1000er1
n (d)
y
8 er r ln 8 207944. Thus n t 1000e207944t .
(b) n 15 1000e20794415 22,600 (c) 15,000 1000e207944t 15 e207944t ln 15 207944t ln 15 13. Thus the population will reach 15,000 after t 207944 approximately 13 hours. 12t 11. (a) A t 12,000 1 0056 , where t is in years. 12
50,000 1
x
t
365t 0056 3653 0056 . So A 3 12,000 1 $14,19506. (b) A t 12,000 1 365 365 (c) A t 12,000e0056t . So 20,000 12,000e0056t 5 3e0056t ln 5 ln 3e0056t ln 5 ln 3 0056t 1 ln 5 ln 3 912. Thus, the amount will grow to $20,000 in approximately 912 years. t 0056
12. (a) The initial mass is m 0 3 and the halflife is h 10, so using the formula m t m 0 2t h , we have m t 3 2t10 .
(b) Using the radioactive decay model with m 0 3 and r lnh2 ln102 , we have m t 3e[ln 210]t 3e00693t . (c) After 1 minute 60 seconds, the amount remaining is m 60 3e0069360 0047 g. 106 106 106 00693t 6 00693t 00693t ln e 00693t ln 10 e ln (d) We solve 3e 3 3 3 1 106 ln 215, so there is 1 g of 91 Kr remaining after about 215 seconds, or 36 minutes. t 00693 3 IJ IJ 64 1064 13. Let the subscripts J and P represent the two earthquakes. Then we have M J log S S I I I 1064 S 1064 S I J . Similarly, M P log P 31 1031 P 1031 S I P . So J 31 1033 19953, S S IP 10 S and so the Japan earthquake was about 1995 times more intense than the Pennsylvania earthquake.
Fitting Exponential and Power Curves to Data
55
FOCUS ON MODELING Fitting Exponential and Power Curves to Data 1. (a)
(b) Let t be the time (in hours) and y be amount of
y 5
iodine131 (in grams). Using a graphing device, we obtain the exponential model y abt , where
4.8
a 479246 and b 099642.
4.6
(c) To find the halflife of iodine131, we must find the time when the sample has decayed to half its original
4.4
mass. Setting y 240 g, we get
4.2
240 479246 099642t
4 3.8 0
10
20
30
40
ln 240 ln 479246 t ln 099642 ln 240 ln 479246 1928 h, or approximately t ln 099642 8 days.
50 x
Time (h)
t 2. (a)
y
(b) We let t represent the time (in seconds) and y the
5
distance fallen (in meters). Using a graphing device, we obtain the power model: y 49622t 20027 .
4
(c) When t 3 the model predicts that y 44792 m.
3 2 1
0
0.2
0.4
0.6
0.8
1
x
Time (s)
3. (a) Let A be the area of the cave and S the number of species of bat. Using a graphing device, we obtain the power function model S 014A064 . From the graph,
we see that the model fits the data reasonably well. (b) According to the model, there are
S 014 205064 4 species of bat living in the El
Sapo cave.
t
S
7 6 5 4 3 2 1
0
100
200
300
400
Area (m@ )
500 A
56
FOCUS ON MODELING
4. (a)
ln L
ln L
2
2
1
0
1
2
4
6
8
10
12
14
16
18
20 w
_2
0
_1
1
2 ln w
Let denote weight and L wingspan. From semilog and loglog plots of the data, it appears that a power function is more appropriate. Using a graphing device, we calculate the power function model L 3056238039520 . L
100
50
0
2
4
6
8
10
12
14
16
18
20 w
(b) According to our model from part (a), the wingspan required for a 300lb bird to fly is approximately L 300 3056238 300039520 291 in. This is almost three times the actual wingspan of an ostrich.
5. (a) Using a graphing device, we find the model ln I 57267458 002999x
I e57267458002999x 3069687e002999x . The “murkiness” constant is k 003.
(b) At the bottom of the twilight zone, the light intensity is approximately
I 1000 307e0031000 29 1011 3 1012 , so these species can thrive in the twilight zone. 6. (a) Using a graphing device, we find the model c , where a 4910976596, y 1 aebx b 04981144989, and c 500855793. The model appears to fit the data very well. (b) From the model, the carrying capacity is c 500.
y 500 400 300 200 100
0
2 4 6 8 10 12 14 16 18 x
COMMENTS: 8, 37
CHAPTER 5
TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH
5.1
Angle Measure 1
5.2
Trigonometry of Right Triangles 6
5.3
Trigonometric Functions of Angles 11
5.4
Inverse Trigonometric Functions and Right Triangles 15
5.5 5.6
The Law of Sines 20 The Law of Cosines 25 Chapter 5 Review 30 Chapter 5 Test 35
¥
FOCUS ON MODELING: Surveying 37
1
5
TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH
5.1
ANGLE MEASURE y
1. (a) The radian measure of an angle is the length of the arc that subtends the angle in a circle of radius 1.
400¡
. (b) To convert degrees to radians we multiply by 180
2¹/3
(c) To convert radians to degrees we multiply by 180 . (d) An angle is in standard position if it is drawn in the x yplane with its
210¡
0
x
_¹/4
vertex at the origin and its initial side on the positive xaxis. 2. (a) If a central angle is drawn in a circle of radius r, the length of the arc subtended by is s r . (b) The area of the sector with central angle is A 12 r 2 .
3. (a) The angular speed of the point is
. t
s (b) The linear speed of the point is . t (c) The linear speed and the angular speed ome are related by the equation r.
4. No, if the common angular speed is , Object A has linear speed 2, while Object B has angular speed 5. Object B has greater linear speed. rad rad 0349 rad 5. 20 20 180 9
rad 2 0698 rad 6. 40 40 180 9
rad 3 rad 0942 rad 7. 54 54 180 10
rad 5 rad 1309 rad 8. 75 75 180 12
rad rad 0785 rad 9. 45 45 180 4
rad rad 0524 rad 10. 30 30 180 6
rad 5 rad 1745 rad 11. 100 100 180 9
rad 10 rad 3491 rad 12. 200 200 180 9
rad 50 rad 17453 rad 13. 1000 1000 180 9
rad 20 rad 62832 rad 14. 3600 3600 180
rad 7 rad 1222 rad 15. 70 70 180 18
rad 5 rad 2618 rad 16. 150 150 180 6
17. 76 76 180 210
18. 43 43 180 240
19. 56 56 180 150
20. 32 32 180 270
540 21. 3 3 180 1719
360 22. 2 2 180 1146
630 23. 35 35 180 2005
324 24. 18 18 180 1031
180 18 25. 10 10
5 180 50 26. 518 18
2 180 24 27. 215 15
13 180 28. 13 12 12 195
29. 50 is coterminal with 50 360 410 , 50 720 770 , 50 360 310 , and 50 720 670 . (Other answers are possible.) 30. 135 is coterminal with 135 360 495 , 135 720 855 , 135 360 225 , and 135 720 585 . (Other answers are possible.) 1
2
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
31. 34 is coterminal with 34 2 114 , 34 4 194 , 34 2 54 , and 34 4 134 . (Other answers are possible.) 11 13 32. 116 is coterminal with 116 2 236 , 116 4 356 , 116 2 6 , and 6 4 6 . (Other answers
are possible.)
7 15 9 17 33. 4 is coterminal with 4 2 4 , 4 4 4 , 4 2 4 , and 4 4 4 . (Other answers are possible.)
34. 45 is coterminal with 45 360 315 , 45 720 675 , 45 360 405 , and 45 720 765 . (Other answers are possible.) 35. Since 430 70 360 , the angles are coterminal.
36. Since 330 30 360 , the angles are coterminal.
37. Since 176 56 126 2; the angles are coterminal.
38. Since 323 113 213 7 is not a multiple of 2, the angles are not coterminal.
39. Since 875 155 720 2 360 , the angles are coterminal.
40. Since 340 50 290 is not a multiple of 360 , the angles are not coterminal.
41. Since 400 360 40 , the angles 400 and 40 are coterminal. 42. Since 375 360 15 , the angles 375 and 15 are coterminal.
43. Since 780 2 360 60 , the angles 780 and 60 are coterminal.
44. Since 100 260 360 is a multiple of 360 , the angles 100 and 260 are coterminal. 45. Since 800 3 360 280 , the angles 800 and 280 are coterminal.
46. Since 1270 190 1080 3 360 is a multiple of 360 , the angles 1270 and 190 are coterminal.
47. Since 196 2 76 , the angles 196 and 76 are coterminal.
5 48. Since 53 2 3 , the angles 3 and 3 are coterminal.
49. Since 25 12 2 , the angles 25 and are coterminal.
50. Since 10 2 3717, the angles 10 and 10 2 are coterminal. 17 51. Since 174 2 2 4 , the angles 4 and 4 are coterminal.
52. Since 512 32 24 12 2, the angles 512 and 32 are coterminal.
53. Using the formula s r, s 56 9 152 .
7 35 54. Using the formula s r, the length of the arc is s 140 5 5 122. 180 9 9 10 180 s 2 rad 2 1146 55. r 5 8 s 56. Solving for r, we have r , so the radius of the circle is r 4. 2 57. Solving for s, we have s r , so the length of the arc is 4 2 8 cm. 8 838 m. 58. Solving for s, we have s r 12 40 180 3 14 14 180 s rad 891 . 59. Solving for , we have r 9 9 15 5 5 180 s rad 955 . 60. Solving for , we have r 9 3 3 s 15 18 61. r 573 m 56 s 20 72 62. r 2292 cm 50 180
SECTION 5.1 Angle Measure 32 4 128 4468 63. (a) A 12 r 2 12 82 80 180 9 9
(b) A 12 r 2 12 102 05 25
2A 2 12 , so the radius is 586. 64. (a) A 12 r 2 r 07 2A 2 12 (b) r 303 150 180 65. 34 rad and r 8 m, so A 12 r 2 12 82 34 24 754 m2 . 66. 145 145
29 and r 6 ft, so A 1 r 2 1 62 29 29 456 ft2 . 36 2 2 36 2 180
67. A 90 m2 and 160 160
8 1 8 2 rad. Thus, A 12 r 2 90 r r 180 9 2 9
, so A 1 r 2 20 5 r 2 r 68. A 20 m2 and 512 2 12
2 20
2 90
9 80 m. 8
4 6 12 55 m. 5
69. r 80 mi and A 1600 mi2 , so A 12 r 2 1600 12 802 12 rad. 70. The area of the circle is r 2 600 m2 , so r A 12 r 2
1 600 900 3 2865 m2 . 2
600 .
Thus, the area of the sector is
71. Referring to the figure, we have AC 3 1 4, BC 1 2 3, and AB 2 3 5. Since
2
AB 2 AC 2 BC 2 , then by the Pythagorean Theorem, the
triangle is a right triangle. Therefore, 2 and 2 A 12 r 2 12 12 2 4 ft .
3 A
72. The triangle is equilateral, so 1 3 rad. To find 2 , we use the formula 2 1 2 3 1 0047 rad, or approximately 27 .
3
B 2
1 1 C
1 s 1 rad. Thus, r 1
73. (a) Between 1:00 P. M . and 1:45 P. M ., the minute hand traverses threequarters of a complete revolution, or 3 2 3 rad. 4 2
The hour hand moves threequarters of the way from 12 to 1, which is itself onetwelfth of a revolution. So the hour 1 2 rad. hand traverses 34 12 8
(b) Between 1:00 P. M . and 6:45 P. M ., the minute hand traverses five complete revolutions plus threequarters of a revolution; that is, 5 2 34 2 232 rad. The hour hand moves through fivetwelfths of a revolution, plus threequarters of the way from 6 to 7; that is, 5 3 1 23 12 2 4 12 2 24 rad.
3
4
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
74. The area of the sector is A B 12 12 3 6 . The height h of the 12
equilateral triangle is B 12
3 2
B 1
1 2 3 , and so its area is 2 2
1/2 1/2
h
A
1
43 . Thus, the area of the region in question is
3 A 6 B 6 4 .
75. The circumference of each wheel is d 28 in. If the wheels revolve 10,000 times, the distance traveled is 1 ft 1 mi 10,000 28 in. 1388 mi. 12 in. 5280 ft 76. Since the diameter is 30 in., we have r 15 in. In one revolution, the arc length (distance traveled) is s r 2 1 rev 67227 rev. 15 30 in. The total distance traveled is 1 mi 5280 ft/mi 12 in/ft 63,360 in. 63,360 in. 30 in Therefore the car wheel will make approximately 672 revolutions. rad rad. 77. We find the measure of the angle in degrees and then convert to radians. 405 255 15 and 15 180 12 3960 330 1036725 and so the distance between the two cities is Then using the formula s r , we have s 12 roughly 1037 mi.
rad rad. Then using the formula s r , the length of the arc is 78. 35 30 5 5 180 3
s 3 3960 110 345575. So the distance between the two cities is roughly 346 mi.
1 of its orbit which is 2 rad. Then s r 2 93,000,000 1,600,9113, so the 79. In one day, the earth travels 365 365 365
distance traveled is approximately 16 million miles.
80. Since the sun is so far away, we can assume that the rays of the sun are parallel when striking the earth. Thus, the angle 500 s 180 500 formed at the center of the earth is also 72 . So r 3980 mi, and the circumference 72 180 72 2 180 500 25,000 mi. 72 1 1 rad rad. Then s r 3960 1152, and so a 81. The central angle is 1 minute 60 60 180 10,800 10,800 is c 2r
nautical mile is approximately 1152 mi. 219,900 ft2 . 82. (a) The area is A 12 r 2 12 3002 280 180
(b) The nozzle 300 ft from the center has three times as much area to cover as the nozzle 100 ft from the center, so it should spray 3 40 120 gallons per minute.
83. The area is equal to the area of the large sector (with radius 34 in.) minus the area of the small sector (with radius 14 in.) 1131 in.2 . Thus, A 12 r12 12 r22 12 342 142 135 180
84. The area available to the cow is shown in the diagram. Its area is the sum of four quartercircles: A 14 1002 502 402 302 3750
11,781 ft2
50 ft 30 ft
20 ft 50 ft
100 ft
60 ft 50 ft
40 ft
SECTION 5.1 Angle Measure
5
45 2 rad 90 rad/min. 1 min 45 2 16 1440 in./min 45239 in./min. (b) The linear speed is 1 6000 2 rad 12,000 rad/min. 86. (a) The angular speed is 1 min
85. (a) The angular speed is
(b) The linear speed is 87.
5 6000 2 12 250 ft/s 2618 ft/s. 60 3
8 2 2 32 6702 ft/s. 15 15
600 2 11 12 ft 1 mi 60 min 125 mi/h 393 mi/h. 1 min 5280 ft 1 hr 1 2 3960 1 day 89. 23 h 56 min 4 s 239344 hr. So the linear speed is 103957 mi/h. 1 day 239344 hr
88.
50 mi/h 1h 5280 ft linear speed 2200 rad/min. radius 2 ft 60 min 1 mi angular speed 2200 rad/min (b) The rate of revolution is 350 rev/min. 2 2 2 100 2 020 m 209 m/s. 91. 60 s 3 40 2 4 linear speed of pedal 160 rad/min. 92. (a) The angular speed is radius of wheel sprocket 2 (b) The linear speed of the bicycle is angular speed radius 160 rad/min 13 in 2080 in/min 62 mi/h. 90. (a) The radius is 2 ft, so the angular speed is
93. (a) The circumference of the opening is the length of the arc subtended by the angle on the flat piece of paper, that is, C s r 6 53 10 314 cm.
C 10 5 cm. 2 2 (c) By the Pythagorean Theorem, h 2 62 52 11, so h 11 33 cm. (d) The volume of a cone is V 13 r 2 h. In this case V 13 52 11 868 cm3 . (b) Solving for r, we find r
94. (a) With an arbitrary angle , the circumference of the opening is 92 3 C 2 2 , h 6 r 36 2 , and C 6, r 2 92 9 2 9 V 13 r 2 h 2 36 2 2 2 42 2 . 3
(b)
100 50 0
0
2
4
6
(c) The volume seems to be maximized for 513 rad, or about 293 .
95. Refer to Exercise 74. The equilateral triangle has area 43 and each of the three edge regions has area 1 12 3 3 , so the total area of the Reuleaux triangle is 3 3 3 3 . Its perimeter is 2 3 4 6 4 4 6 4 2 2 . 3 3 96. Answers will vary—although of course everybody prefers radians.
6
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
5.2 1. (a)
TRIGONOMETRY OF RIGHT TRIANGLES (b) sin
opposite
adjacent opposite opposite , cos , and tan . hypotenuse hypotenuse adjacent
adjacent
¬
hypotenuse
2. The trigonometric ratios do not depend on the size of the triangle because all right triangles with angle are similar. 1 1 1 3. The reciprocal identities state that csc , sec , and cot . sin cos tan 4. (a) x r cos and y r sin . (b) If r 6 and 30 then x 6 cos 30 3 3 and y 6 sin 30 3.
5. sin 45 , cos 35 , tan 43 , csc 54 , sec 53 , cot 34
7 , cos 24 , tan 7 , csc 25 , sec 25 , cot 24 6. sin 25 7 7 25 24 24 9 2 7. The remaining side is obtained by the Pythagorean Theorem: 41 402 81 9. Then sin 40 41 , cos 41 , 41 41 9 tan 40 9 , csc 40 , sec 9 , cot 40
8. The hypotenuse is obtained by the Pythagorean Theorem: 17 17 8 tan 15 8 , csc 15 , sec 8 , cot 15
8 82 152 289 17. Then sin 15 17 , cos 17 ,
9. The remaining side is obtained by the Pythagorean Theorem:
32 22 13. Then sin 2
13
2 1313 ,
cos 3 3 1313 , tan 23 , csc 213 , sec 313 , cot 32 13 10. The remaining side is obtained by the Pythagorean Theorem: 82 72 15. Then sin 78 , cos 815 ,
tan 7 7 1515 , csc 87 , sec 8 8 1515 , cot 715 15 15 2 2 11. c 5 3 34
(a) sin cos 3 3 3434 34 12. b 72 42 33 (a) sin cos 47
(b) tan cot 35
(b) tan cot 4
33
(c) sec csc 534 (c) sec csc 7
13. (a) sin 22 037461
(b) cot 38 2 1 041421
14. (a) cos 45 080902
(b) csc 48 134563
15. (a) sec 1 185082
(b) tan 51 123490
16. (a) csc 10 575877 x 17. Since sin 30 , we have x 25 sin 30 25 12 25 2. 25 12 12 12 18. Since sin 45 , we have x 1 12 2. x sin 45
(b) sin 35 035078
2 x , we have x 13 sin 60 13 23 132 3 . 19. Since sin 60
13
33
SECTION 5.2 Trigonometry of Right Triangles
20. Since tan 30
7
4 4 4 , we have x 1 4 3. x tan 30 3
12 12 , we have x 1651658. 21. Since tan 36 x tan 36 25 25 , we have x 3130339. 22. Since sin 53 x sin 53 y x cos x 28 cos , and sin y 28 sin . 23. 28 28 4 4 x tan x 4 tan , and cos y 4 sec . 24. 4 y cos 26. cos 12 25. tan 56 . Then the third side is x 52 62 61. 13 . The third side is y 132 122 25 5. The other five ratios are The other five ratios are sin 5 61 , cos 6 61 , 61 61 61 csc 5 , sec 6 , and cot 65 . Ï61
61
5 , tan 5 , csc 13 , sec 13 , and sin 13 12 5 12
cot 12 5.
13
5
5
¬
¬
12
6
27. cot 1. Then the third side is r
12 12 2.
The other five ratios are sin 1 22 , 2 cos 1 22 , tan 1, csc 2, and 2 sec 2.
Ï2
28. tan
3. The third side is r 12 3 2. The other
five ratios are sin 23 , cos 12 , csc 2 , 3
sec 2, and cot 1 . 3
¬
2
1
Ï3
1
¬ 1
29. csc 11 6 . The third side is x
112 62 85. The 30. cot 53 . The third side is x 52 32 34. The
6 , cos 85 , other five ratios are sin 11 11
other five ratios are sin 3 3434 , cos 5 3434 ,
tan 6 8585 , sec 118585 , and cot 685 .
tan 35 , csc 334 and sec 534 .
11
Ï34
6
¬
¬ Ï85
1 3 1 3 31. sin cos 6 6 2 2 2
32. sin 30 csc 30 sin 30
1 1 sin 30
33. sin 30 cos 60 sin 60 cos 30 12 12 23 23 14 34 1
5
3
8
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
2 3 12 34 14 1 2 2 2 35. cos 30 2 sin 30 2 23 12 34 14 12 2 2 2 3 1 1 1 2 1 36. sin 31 18 3 1 18 3 2 3 1 3 cos 4 sin 4 cos 3 2 2 2 2 2 2 18 4 2 3 14 2 3 2 2 2 3 2 2 1 12 14 2 12 37. cos 4 sin 6 2 2 2 4 2 2 2 3 3 2 38. sin 12 2 94 2 3 tan 6 csc 4 2 3 2 39. This is an isosceles right triangle, so the other leg has length 16 tan 45 16, the hypotenuse has length 34. sin 60 2 cos 60 2
2
16 16 2 2263, and the other angle is 90 45 45 . sin 45
40. The other leg has length 100 tan 75 2679, the hypotenuse has length 90 75 15 . 41. The other leg has length 35 tan 52 4480, the hypotenuse has length 90 52 38 .
100 10352, and the other angle is sin 75 35 5685, and the other angle is cos 52
42. The adjacent leg has length 1000 cos 68 37461, the opposite leg has length 1000 sin 68 92718, and the other angle is 90 68 22 .
43. The adjacent leg has length 335 cos 8 3095, the opposite leg has length 335 sin 8 1282, and the other angle is 3 . 2
8
8
44. The opposite leg has length 723 tan 6 4174, the hypotenuse has length . 2
6
723 8348, and the other angle is cos 6
3
45. The adjacent leg has length
106 106 14590, the hypotenuse has length 18034, and the other angle is tan sin 5 5
3 . 2
5
10
46. The adjacent leg has length 425 cos 38 16264, the opposite leg has length 425 sin 38 39265, and the other angle is 3 . 2
8
comma
8
1 045 cos 2 089, tan 1 , csc 224, sec 224 112, cot 200. 47. sin 224 224 2 2
sin 40 064, cos 40 077, tan 40
48.
064 083, csc 40 156, 077
sec 40 131, cot 40 120. 40¡
49. x
100 100 2309 tan 60 tan 30
50. Let d be the length of the base of the 60 triangle. Then tan 60 dx
85 85 x d 981. tan 30 tan 30
51. Let h be the length of the shared side. Then sin 60
85 85 85 d 49075, and so tan 30 d tan 60 dx
50 h 50 h h 57735 sin 65 x 637 h sin 60 x sin 65
9
SECTION 5.2 Trigonometry of Right Triangles
52. Let h be the hypotenuse of the top triangle. Then sin 30 x
10 h 58. tan 60 tan 60
53.
From the diagram, sin
10 ¬
¹ -¬ 2
y
5 5 h h 10, and so tan 60 h sin 30 x
y x and tan , so x y sin 10 sin tan . y 10
x
¬
54. sin
b 1 d a a sin , tan b tan , cos c sec , cos d cos 1 1 c 1
55. Let h be the height, in feet, of the Empire State Building. Then tan 11
h h 5280 tan 11 1026 ft. 5280
56. (a) Let r be the distance, in feet, between the plane and the Gateway Arch. Therefore, sin 22
35,000 r
35,000 93,431 ft. sin 22 (b) Let x be the distance, also in feet, between a point on the ground directly below the plane and the Gateway Arch. Then 35,000 35,000 x 86,628 ft. tan 22 x tan 22 r
57. (a) Let h be the distance, in miles, that the beam has diverged. Then tan 05
h 240,000
h 240,000 tan 05 2100 mi.
(b) Since the deflection is about 2100 mi whereas the radius of the moon is about 1000 mi, the beam will not strike the moon. 58. Let x be the distance, in feet, of the ship from the base of the lighthouse. Then tan 23
200 200 x 471 ft. x tan 23
59. Let h represent the height, in feet, that the ladder reaches on the building. Then sin 72
h h 20 sin 72 19 ft. 20
60. Let h be the height, in feet, of the communication tower. Then sin 65
h h 600 sin 65 544 ft. 600
61. Let h be the height, in feet, of the kite above the ground. Then sin 50
h h 450 sin 50 345 ft. 450
62. 18¡ x 14¡
hÁ hª
Let h 1 be the height of the flagpole above elevation and let h 2 be the height below, h as shown in the figure. So tan 18 1 h 1 x tan 18 . Similarly, x h 2 x tan 14 . Since the flagpole is 60 feet tall, we have h 1 h 2 60, so x tan 18 tan 14 60 x
60 1045 ft. tan 18 tan 14
63. Let h 1 be the height of the window in feet and h 2 be the height from the window to the top of the tower. Then tan 25
h1 325
h2 h 2 325 tan 39 263 ft. Therefore, the height of the window 325 is approximately 152 ft and the height of the tower is approximately 152 263 415 ft.
h 1 325 tan 25 152 ft. Also, tan 39
10
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
64.
52¡ 52¡
car, and d2 be the distance, in feet, between the same point and the other car. Then
5150
dÁ
d2
Let d1 be the distance, in feet, between a point directly below the plane and one
35¡
dª
35¡
tan 52
5150 5150 5150 d1 402362 ft, and tan 35 d1 tan 52 d2
5150 735496 ft. So the distance between the two cars is about d1 d2 402362 735496 11,379 ft. tan 35
65. Let d1 be the distance, in feet, between a point directly below the plane and one car, and d2 be the distance, in feet, between d d the same point and the other car. Then tan 52 1 d1 5150 tan 52 65917 ft. Also, tan 38 2 5150 5150 d2 5150 tan 38 40236 ft. So in this case, the distance between the two cars is about 2570 ft. 66. Let x be the distance, in feet, between a point directly below the balloon and the first mile post. Let h be the height, h h in feet, of the balloon. Then tan 22 and tan 20 . So h x tan 22 x 5280 tan 20 x x 5280 5280 tan 20 x 47,977 ft. Therefore h 47,9769 tan 22 19,384 ft 37 mi. tan 22 tan 20 67. Let x be the horizontal distance, in feet, between a point on the ground directly below the top of the mountain and h the point on the plain closest to the mountain. Let h be the height, in feet, of the mountain. Then tan 35 x h 1000 tan 32 and tan 32 . So h x tan 35 x 1000 tan 32 x 82942. Thus x 1000 tan 35 tan 32 h 82942 tan 35 5808 ft.
68.
Since the angle of elevation from the observer is 45 , the distance from the observer is h, as shown in the figure. Thus, the length of the leg in the smaller right h 75¡ h
ã600ã
h 600 h tan 75 h 600 h 600 tan 75 600 tan 75 h 1 tan 75 h 473 m. 1 tan 75 triangle is 600 h. Then tan 75
600-h
69. Let d be the distance, in miles, from the earth to the sun.
Then sec 8985
d 240,000
d 240,000 sec 8985 917 million miles.
70. (a) s r rs 6155 3960 15543 rad 8905
(b) Let d represent the distance, in miles, from the center of the earth to the moon. Since cos d
3960 , we have d
3960 3960 239,9615. So the distance AC is 239,9615 3960 236,000 mi. cos cos 8905
r 71. Let r represent the radius, in miles, of the earth. Then sin 60276 r600 r 600 sin 60276 r sin 60276 600 sin 60276 r 1 sin 60276 r 600 1sin 60276 3960099. So the earth’s radius is about 3960 mi.
,000 , we have 72. Let d represent the distance, in miles, from the earth to Alpha Centauri. Since sin 0000211 93,000 d 93,000,000 13 d sin 0000211 25,253,590,022,410. So the distance from the earth to Alpha Centauri is about 253 10 mi.
73. Let d be the distance, in AU, between Venus and the sun. Then sin 463
d d, so d sin 463 0723 AU. 1
74. If two triangles are similar, then their corresponding angles are equal and their corresponding sides are proportional. That is, if triangle ABC is similar to triangle A B C then AB r A B , AC r A C , and BC r B C . Thus when we express any trigonometric ratio of these lengths as a fraction, the factor r cancels out.
SECTION 5.3 Trigonometric Functions of Angles 75. From the diagram, we see that tan 60 x1 x cot 60 33 . Thus, the perimeter of ABC is 3 1 33 3 3.
B
1 A
5.3
11
60¡ x
P
C
1
TRIGONOMETRIC FUNCTIONS OF ANGLES
1. If the angle is in standard position and P x y is a point on the terminal side of , and r is the distance from the origin to x y y P, then sin , cos , and tan . r r x 2. The sign of a trigonometric function of depends on the quadrant in which the terminal angle of lies. For example, if is in quadrant II, sin is positive. In quadrant III, cos is negative. In quadrant IV, sin is negative. 3. (a) If is in standard position, then the reference angle is the acute angle formed by the terminal side of and the xaxis. So the reference angle for 100 is 80 and that for 190 is 10 .
(b) If is any angle, the value of a trigonometric function of is the same, except possibly for sign, as the value of the trigonometric function of . So sin 100 sin 80 and sin 190 sin 10 .
4. The area A of a triangle with sides of lengths a and b and with included angle is given by the formula A 12 ab sin . So the area of the triangle with sides 4 and 7 and included angle 30 is 12 4 7 sin 30 7.
5. (a) The reference angle for 135 is 180 135 45 .
6. (a) The reference angle for 155 is 180 155 25 .
(b) The reference angle for 195 is 195 180 15 .
(b) The reference angle for 280 is 360 280 80 .
(c) The reference angle for 300 is 360 300 60 .
(c) The reference angle for 390 is 390 360 30 .
7. (a) The reference angle for 250 is 250 180 70 . (b) The reference angle for 485 is 540 485 55 . (c) The reference angle for 100 is
8. (a) The reference angle for 99 is 180 99 81 . (b) The reference angle for 199 is 199 180 19 .
(c) The reference angle for 359 is 360 359 1 .
180 100 80 .
is 7 3 . 9. (a) The reference angle for 710 10 10
(b) The reference angle for 98 is 98 8.
(c) The reference angle for 103 is 103 3 3.
11. (a) The reference angle for 57 is 57 27 .
10. (a) The reference angle for 56 is 56 6.
(b) The reference angle for 109 is 109 9.
(c) The reference angle for 237 is 237 3 27 .
12. (a) The reference angle for 23 is 23 2 03.
(b) The reference angle for 14 is 14 04.
(b) The reference angle for 23 is 23 084.
(c) The reference angle for 14 is 14 because 14 2.
(c) The reference angle for 10 is 10 10 0.
13. cos 150 cos 30 23
14. sin 210 sin 30 12
15. tan 315 tan 45 1
16. cot 60 cot 60 tan160 33
17. csc 300 csc 60 sin160 2 3 3
19. sin 225 sin 45 22
18. sec 330 sec 30 cos130 2 3 3 20. csc 135 csc 45 sin145 2
12
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
22. tan 120 tan 60 3
21. sec 420 sec 60 cos160 2 23. cot 570 cot 30 tan130 3
24. sin 810 sin 90 1
25. sin 32 sin 2 1 27. tan 43 tan 3 3 29. csc 56 csc 6 2 31. sec 173 sec 3
1 26. cos 43 cos 3 2 3 28. cos 116 cos 6 2
2 3 30. sec 76 sec 6 3
1 2 cos 3
32. csc 54 csc 4
1 2 sin 4
1 33. cot 4 cot 4 tan 1 4
2 1 34. cos 74 cos 4 2 2
35. tan 52 tan 2 which is undefined.
1 36. sin 116 sin 6 2
37. Since sin 0 and cos 0, is in quadrant III. 38. Since both tan and sin are negative, is in quadrant IV. 39. sec 0 cos 0. Also tan 0 in quadrant IV.
sin 0 sin 0 (since cos 0). Since sin 0 and cos 0, is cos
40. Since csc 0 sin 0 and cos 0, is in quadrant II. 41. tan 24 12
42. sin 23
3 42
35
43. cos 23
3 32
22
2 2 2 2 45. sec 2 33 313 46. csc 1 33 310 47. cos 35 . Then y 52 32 42 4, since is in quadrant IV. Thus, sin 45 , tan 43 , csc 54 , 44. cot 23
sec 53 , and cot 34 . 48. tan 34 , so r 32 42 5. Because is in quadrant III, x 4 and y 3, and so sin 35 , cos 45 , csc 53 , sec 54 , and cot 43 . 49. cot 2, so tan 12 and r 22 12 5. Thus, because is in quadrant IV, sin 1 55 , 5 cos 2 5 5 , tan 12 , csc 5, and sec 25 . 50. csc 2. Then sin 12 and x 22 12 3. Because is in quadrant I, sin 12 , cos 23 , tan 1 33 , sec 2 2 3 3 , and cot 3. 3 3 51. sin 23 , so x 32 22 5. Because is in quadrant III, cos 35 , tan 2 5 5 , csc 32 ,
sec 3 5 5 , and cot 25 . 130 , cos 7 130 , tan 9 , 52. cot 79 , so r 72 92 130. Because is in quadrant II, sin 9 130 7 130 csc
130 , and sec 130 . 7 9
53. sin 14 0, so because cos 0, is in quadrant I and x csc 4, sec 4 1515 , and cot 15.
42 12 15. Thus, cos 415 , tan 1515 ,
54. cos 13 0, so because tan 0, is in quadrant IV and y 32 12 2 2. Thus, sin 2 3 2 , tan 2 2, csc 3 4 2 , sec 3, and cot 42 .
SECTION 5.3 Trigonometric Functions of Angles
13
32 12 10. Because sin 0 and tan 0, is in quadrant III and we have sin 3 1010 , cos 1010 , csc 310 , sec 10, and cot 13 . 7 2 2 56. csc 12 7 , so sin 12 and because cos 0 and csc 0, is in quadrant III. Thus, x 12 7 95 55. tan 3 31 , so r
7 , cos 95 , tan 7 95 , sec 12 95 , and cot 95 . and sin 12 7 12 95 95
57. csc 4, so sin 14 and because tan 0 and csc 0, is in quadrant IV. Thus, x 42 12 15 and sin 14 , cos 415 , tan 1515 , sec 4 1515 , and cot 15. 3 , so because sec 0 and sin 0, is in quadrant I and x 102 32 91. Thus, cos 91 , 58. sin 10 10
10 91 91 tan 3 9191 , csc 10 3 , sec 91 , and cot 3 . sin 1 cos2 2 59. Since sin is negative in quadrant III, sin 1 cos and we have tan . cos cos 1 sin2 cos because cos 0 in quadrant II. 60. cot sin sin 61. cos2 sin2 1 cos 1 sin2 because cos 0 in quadrant IV.
1 1 because all trigonometric functions are positive in quadrant I. cos 1 sin2 63. sec2 1 tan2 sec 1 tan2 because sec 0 in quadrant II. 64. csc2 1 cot2 csc 1 cot2 because csc 0 in quadrant III. 3 2 65. If 3 , then sin 2 sin 3 2 and 2 sin 2 sin 3 3. 2 sin 2 3 and sin 2 sin 2 0890. 66. If , then sin 3 3 4 9 62. sec
67. The area is 12 7 9 sin 72 300.
68. The area is 12 10 22 sin 10 191. 69. The area is 12 102 sin 60 25 3 433.
70. The area is 12 132 sin 60 1694 3 732.
71. Let the length of the other side be x. Then A 16 12 5 x sin 36 x 32 5 csc 36 1089 in. 48 96 4 6 98 cm. 72. Let the lengths of the equal sides be x. Then A 12 x 2 sin 24 12 x 2 sin 56 x 5 sin 6
4 . For the triangle defined by the two sides, 73. For the sector defined by the two sides, A1 12 r 2 12 22 120 180 3 1 1 A2 2 ab sin 2 2 2 sin 120 2 sin 60 3. Thus the area of the region is A1 A2 43 3 246.
74. The area of the entire circle is r 2 122 144, the area of the sector is 12 r 2 12 122 3 24, 1 1 2 and the area of the triangle is 2 ab sin 2 12 sin 3 36 3, so the area of the shaded region is A1 A2 A3 144 24 36 3 120 36 3 4393. 75. The height of the equilateral triangle is 22 12 3, and so its ¹/3 ¹/3 1 area is 12 2 3 3. The area of each sector within the triangle is ¹/3 1 12 , and so the area of the shaded area is 3 . 1 2 3 6 2
14
CHAPTER 5 Trigonometric Functions: Right Triangle Approach A
76. We use the area formula A 12 ab sin .
a
3 2 2 4 For P Q R, A1 12 a 2 sin 3 4 a 1 a 3 ; for APC, A2 12 a 2a sin 23 23 a 2 23 4 2.
P 2¹/3 a
3
B
Thus, the area of ABC is A A1 3A2 7.
Q
¹/3 a
2a R C
77. (a) tan
5280 ft h , so h tan 1 mile 5280 tan ft. 1 mile 1 mile
(b)
20
60
80
85
h
1922
9145
29,944
60,351
78. (a) Let the depth of the water be h. Then the crosssectional area of the gutter is a trapezoid whose height is h 10 sin . The bases are 10 and 10 2 10 cos 10 20 cos . Thus, the area is b b2 10 10 20 cos A 1 h 10 sin 100 sin 100 sin cos . 2 2 (b)
(c)
200
130.0 129.5
100 0
0
129.0 1.00
1
1.05
1.10
From the graph, the largest area is achieved when 1047 rad 60 . 79. (a) From the figure in the text, we express depth and width in terms of .
(b)
depth width Since sin and cos , we have depth 20 sin 20 20 and width 20 cos . Thus, the crosssection area of the beam is A depth width 20 cos 20 sin 400 cos sin . (c) The beam with the largest crosssectional area is the square beam, 10 2 by 10 2 (about 1414 by 1414).
200
0
0
1
80. Using depth 20 sin and width 20 cos , from Exercise 79, we have strength k width depth2 . Thus S k 20 cos 20 sin 2 8000k cos sin2 .
81. (a) On Earth, the range is R
02 sin 2 g
122 sin 3 32
9 3 3897 ft and the height is 4
122 sin2 2 sin2 6 9 05625 ft. H 0 2g
2 32
16
2 2 122 sin 3 23982 ft and H 12 sin 6 3462 ft (b) On the moon, R
82. Substituting, t
52
2000 5 10 158 s. 16 sin 30
2 52
SECTION 5.4 Inverse Trigonometric Functions and Right Triangles
83. (a) W 302 038 cot 065 csc
15
10
(b) From the graph, it appears that W has its minimum value at about 0946 542 .
5 0
84. (a) We label the lengths L 1 and L 2 as shown in the figure. Since sin
¬ 6 Lª
9 6 and cos , we L1 L2
have L 1 9 csc and L 2 6 sec . Thus
40
0
LÁ ¬
(d) The minimum value of L is the shortest
2
20
¬
L L 1 L 2 9 csc 6 sec .
(c) The minimum value of L is 2107.
(b)
0
0.0
0.5
1.0
1.5
9
distance the pipe must pass through.
85. sin 4 00697564737 but sin 4 07568024953. Your partner has found sin 4 instead of sin 4 radians.
O P O P adj O P. Since QS is tangent to the circle at R, O R Q is a right triangle. Then O R hyp 1 R Q O Q opp hyp tan R Q and sec O Q. Since SO Q is a right angle S O Q is a right O R O R adj adj O S S R hyp adj O S and cot S R. Summarizing, we have triangle and O S R . Then csc O R O R opp opp sin P R, cos O P, tan R Q, sec O Q, csc O S, and cot S R. 1 1 1 tan2 1 sec2 87. (a) sin2 cos2 1 sin2 cos2 2 cos cos2 1 1 (b) sin2 cos2 1 sin2 cos2 1 1 cot2 csc2 sin2 sin2
86. cos
88. Let x represent the desired real number, in radians. We wish to find
the smallest nonzero solution of the equation sin x sin 180x . (Since
y 1
180x y=sin ¹
x is in radians, 180x is in degrees.) Now if a is in degrees with
0 a 180 and sin a b, then either a b or 180 a b. The graph shows that the first positive solution occurs when
y=sin x 0
2
x
_1 180x 180x 180, so we have 180 x. 180 180 3088. You can verify that the value of sin is the same whether Solving this equation, we get x 180 180 your calculator is set to radian mode or degree mode.
0
5.4
INVERSE TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES
1. For a function to have an inverse, it must be onetoone. To define the inverse sine function we restrict the domain of the sine function to the interval 2 2 . 2. (a) The function sin1 has domain [1 1] and range 2 2 .
16
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
(b) The function cos1 has domain [1 1] and range [0 ]. (c) The function tan1 has domain and range 2 2 .
8 sin1 4 3. (a) sin1 10 5
6 cos1 3 (b) cos1 10 5
5 , we let cos1 5 . So 4. To find tan cos1 13 13
5 . We then complete the right triangle in cos 13
the figure and use the triangle to find that 5 12 . tan cos1 13 5
(c) tan1 68 tan1 43
13 ¬
12
5
3 3
5. (a) sin1 1 2
(b) cos1 0 2
(c) tan1
6. (a) sin1 0 0
(b) cos1 1
(c) tan1 0 0
7. (a) sin1 22 4 8. (a) sin1 23 3
9. sin1 030 0305 17458
(b) cos1 22 34 (b) cos1 12 23
11. cos1 13 1231 70529
(c) tan1 1 4 (c) tan1 3 3
10. cos1 02 1772 101537 12. sin1 56 0985 56443
13. tan1 3 1249 71565
14. tan1 4 132582 75964
15. cos1 3 is undefined.
16. sin1 2 is undefined.
6 3 , so sin1 3 369 . 17. sin 10 5 5
7 , so tan1 7 213 . 18. tan 18 18
9 , so tan1 9 347 . 19. tan 13 13
3 1 3 254 . 20. sin 30 7 70 7 , so sin
21. sin 47 , so sin1 47 348 .
22. cos 89 , so cos1 98 273 .
23. tan 34 , so tan1 43 369 .
25. tan 32 , so tan1 32 563 .
24. tan 24 12 , so 180 tan1 12 1534
26. tan 3, so 180 tan1 3 2516 .
27. We use sin1 to find one solution in the interval 90 90 . sin 23 sin1 23 418 . Another solution with between 0 and 180 is obtained by taking the supplement of the angle: 180 418 1382 . So the solutions of the equation with between 0 and 180 are approximately 418 and 1382 . 28. One solution is given by cos1 43 414 . This is the only solution, because cos x is onetoone on 0 180 . 29. One solution is given by cos1 25 1136 . This is the only solution, because cos x is onetoone on 0 180 .
30. tan1 20 871 , so the only solution in 0 180 is approximately 180 871 929 . 31. tan1 5 787 . This is the only solution on 0 180 . 32. One solution is sin1 45 531 . Another solution on 0 180 is approximately 180 531 1269 .
SECTION 5.4 Inverse Trigonometric Functions and Right Triangles
33. To find cos sin1 45 , first let sin1 54 . Then is the number in the interval 2 2 whose sine is 45 . We draw a right triangle with as one of its acute
5
angles, with opposite side 4 and hypotenuse 5. The remaining leg of the triangle is
¬
found by the Pythagorean Theorem to be 3. From the figure we get cos sin1 54 sin 35 .
17
4
3
Another method: By the cancellation properties of inverse functions, sin sin1 54 is exactly 45 . To find cos sin1 45 , we
first write the cosine function in terms of the sine function. Let u sin1 45 . Since 0 u 2 , cos u is positive, and since 2 9 3 cos2 u sin2 u 1, we can write cos u 1 sin2 u 1 sin2 sin1 45 1 45 1 16 25 25 5 . Therefore, cos sin1 54 35 . 34. To find cos tan1 34 , we draw a right triangle with angle , opposite side 4, and adjacent side 3. From the figure we see that cos tan1 34 cos 35 .
12 , we draw a right triangle with 35. To find sec sin1 13
angle , opposite side 12, and hypotenuse 13. From the 12 sec 13 . figure we see that sec sin1 13 5 13
¬
5
5 ¬
4
3
7 7 , we draw a csc cos1 25 36. To find csc cos1 25 right triangle with angle , adjacent side 7, and hypotenuse 25. From the figure we see that 7 csc 25 csc cos1 25 24 . 25
¬
12
7
24
1 12 , we draw sin tan 37. To find sin tan1 12 5 5 a right triangle with angle , opposite side 12, and adjacent side 5. From the figure we see that sin 12 sin tan1 12 5 13 . 13 12
¬
38. To find sec tan1 23 sec tan1 32 , we draw a
right triangle with angle , opposite side 2, and adjacent side 3. From the figure we see that sec tan1 23 sec 313 .
5
Ï13 ¬
3
2
18
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
39. To find sin sec1 4 sin sec1 4 , we draw a right triangle with angle , adjacent side 1, and
hypotenuse 4. From the figure we see that sin sec1 4 sin 415 . 4
¬
40. To find csc cot1 43 , we draw a right triangle with angle
, adjacent side 3, and opposite side 4. From the figure we see that csc cot1 34 csc 54 . 5
1
Ï15
¬
4
3
41. We want to find cos sin1 x . Let sin1 x, so sin x. We sketch a right
1
triangle with an acute angle , opposite side x, and hypotenuse 1. By the Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have cos sin1 x cos 1 x 2 .
x
¬ Ï1-x@
Another method: Let u sin1 x. We need to find cos u in terms of x. To do so, we write cosine in terms of sine. Note 1 x. Now cos u 1 sin2 u is positive because u lies in the interval . that 2 u 2 because u sin 2 2 Substituting u sin1 x and using the cancellation property sin sin1 x x gives cos sin1 x 1 x 2 .
42. We want to find cot cos1 x . Let cos1 x, so cos x. We sketch a right triangle with an acute angle , adjacent side x, and hypotenuse 1. By the Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have x cot cos1 x cot . 1 x2
1 ¬
x
43. We want to find sec cos1 x . Let cos1 x, so cos x. We sketch a right triangle with an acute angle , opposite side x, and hypotenuse 1. By the Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have 1 sec cos1 x sec . x
1 ¬
Ïx@+1 ¬
45. We want to find sec tan1 x . Let tan1 x, so tan x. We sketch a right triangle with an acute angle , opposite side x, and adjacent side 1. By the Pythagorean Theorem, the hypotenuse is x 2 1. From the figure we have sec tan1 x sec x 2 1.
Ï1-x@
x
44. We want to find cos tan1 x . Let tan1 x, so tan x. We sketch a right triangle with an acute angle , opposite side x, and adjacent side 1. By the Pythagorean Theorem, the hypotenuse is x 2 1. From the figure we have 1 cos tan1 x sin . 2 x 1
Ï1-x@
1
Ïx@+1 ¬
x
1
x
SECTION 5.4 Inverse Trigonometric Functions and Right Triangles
46. We want to find tan sin1 x . Let sin1 x, so sin x. We sketch a right triangle with an acute angle , opposite side x, and hypotenuse 1. By the Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have x . tan sin1 x tan 1 x2
1
19
x
¬ Ï1-x@
47. Let represent the angle of elevation of the ladder. Let h represent the height, in feet, that the ladder reaches on the 6 03 cos1 03 1266 rad 725 . By the Pythagorean Theorem, h 2 62 202 building. Then cos 20 h 400 36 364 19 ft. 96 08 tan1 08 0675 387 . 48. Let be the angle of elevation of the sun. Then tan 120
49. (a) Solving tan h2 for h, we have h 2 tan .
(b) Solving tan h2 for we have tan1 h2.
50. (a) Solving tan 50s. Solving for , we have tan1 50s.
5 1 5 682 . (b) Set s 20 ft to get tan 50 20 2 . Solving for , we have tan 2
51. (a) Solving sin h680 for we have sin1 h680. (b) Set h 500 to get sin1 500 680 0826 rad 473 .
52. The tourist, the base of the low side of the tower, and the top of the low side form a right triangle in which tan 39 56 tan1 39 56 40 .
3960 . Solving for , we get 53. (a) Since the radius of the earth is 3960 miles, we have the relationship cos h3960 3960 . cos1 h3960
(b) The arc length is s radius included angle 3960 2 7920. 3960 (c) s 7920 cos1 h3960 3960 (d) When h 100 we have s 7920 cos1 1003960 7920 cos1 3960 4060 17615 miles. 3960 3960 3960 2450 1 (e) When s 2450 we have 2450 7920 cos1 h3960 2450 7920 cos h3960 h3960 cos 7920 2450 h 3960 3960 sec 2450 7920 h 3960 sec 7920 3960 1973 mi.
1 1 1 sin sin1 08102 541 7 tan 10 2 3 1 tan 10 1 1 1 1 (b) For n 2, sin 483 . For n 3, sin 322 . For n 4, 5 tan 15 7 tan 15 1 1 245 . n 0 and n 1 are outside of the domain for 15 , because 3732 sin1 9 tan 15 tan 15 1 and 1244, neither of which is in the domain of sin1 . 3 tan 15
54. (a) sin1
55. We have sin k sin , where 594 and k 133. Substituting, sin 594 133 sin sin 594 06472. Using a calculator, we find that sin1 06472 403 , so sin 133 4 2 4 403 2 594 424 .
20
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
1 56. Let sec1 x. Then sec x, as shown in the figure. Then cos , so x 1 1 cos1 . Thus, sec1 x cos1 , x 1. x x In particular, sec1 2 cos1 12 3.
1 Let csc1 x. Then csc x, as shown in the figure. Then sin , so x 1 1 . Thus, csc1 x sin1 , x 1. sin1 x x In particular, csc1 3 sin1 31 0340.
1 Let cot1 x. Then cot x, as shown in the figure. Then tan , so x 1 1 1 1 1 tan . Thus, cot x tan , x 1. x x In particular, cot1 4 tan1 41 0245.
5.5
x ¬ 1
x
1
º
1
x
THE LAW OF SINES
sin B sin C sin A . a b c 2. (a) The Law of Sines can be used to solve triangles in cases ASA or SSA.
1. In triangle ABC with sides a, b, and c the Law of Sines states that
(b) The Law of Sines can give ambiguous solutions in case SSA. sin 110 6 sin 110 sin 40 , so x 88. 3. For ABC we have 6 x sin 40 sin sin 50 4. For P Q R we have , so sin1 65 sin 50 668 . 6 5 376 sin 57 5. C 180 984 246 57 . x 3188. sin 984 17 sin 1144 254. 6. C 180 375 281 1144 . x sin 375 267 sin 52 7. C 180 52 70 58 . x 248. sin 58 563 sin 67 8. sin 0646. Then sin1 0646 403 . 802 36 sin 120 9. sin C 0693 C sin1 0693 439 . 45 185 sin 50 10. C 180 102 28 50 . x 1449. sin 102 65 sin 46 65 sin 20 11. C 180 46 20 114 . Then a 512 and b 243. sin 114 sin 114 2 sin 100 2 sin 30 12. B 180 30 100 50 . Then c 257 and a 131. sin 50 sin 50 12 sin 44 13. B 68 , so A 180 68 68 44 and a 899. sin 68 34 sin 80 14. sin B 0515, so B sin1 0515 310 . Then C 180 80 31 690 and 65 65 sin 69 c 62. sin 80
SECTION 5.5 The Law of Sines
15. C 180 50 68 62 . Then 230 sin 68 230 sin 50 1995 and b 2415. a sin 62 sin 62
16. C 180 110 23 47 . Then 50 sin 110 50 sin 23 267 and b 642. a sin 47 sin 47 C
C
50¡
A
68¡ 230
23¡
A
110¡ 50
B
B
17. B 180 30 65 85 . Then 10 sin 30 10 sin 65 a 50 and c 91. sin 85 sin 85
18. C 180 95 22 63 . Then 420 sin 95 420 sin 63 b 11169 and c 9990. sin 22 sin 22 C
C 10
65¡
30¡
A
420 95¡
22¡
A
B
19. A 180 51 29 100 . Then 44 sin 51 44 sin 100 894 and c 705. a sin 29 sin 29
C
10¡
100¡
51¡
A
44 29¡
B
20. A 180 100 10 70 . Then 115 sin 10 115 sin 70 1097 and b 203. a sin 100 sin 100
C
A
21
115
B
B
15 sin 110 0503 B sin1 0503 302 . Then 28 28 sin 398 C 180 110 302 398 , and so c 191. Thus B 302 , C 398 , and c 191. sin 110 40 sin 37 0802 C1 sin1 0822 534 or C2 180 534 1266 . 22. sin C 30 30 sin 896 If C1 534 , then B1 180 37 534 896 and b1 498. sin 37 30 sin 164 141. If C2 1266 , then B2 180 37 1266 164 and b2 sin 37 Thus, one triangle has B1 896 , C1 534 , and b1 498; the other has B2 164 , C2 1266 , and b2 141.
21. Since A 90 there is only one triangle. sin B
23. A 125 is the largest angle, but since side a is not the longest side, there can be no such triangle.
45 sin 38 0660 B1 sin1 0660 413 or B2 180 413 1387 . 42 42 sin 1007 67. If B1 413 , then A1 180 38 413 1007 and a1 sin 38 42 sin 33 If B2 1387 , then A2 180 38 1387 33 and a2 39. sin 38 Thus, one triangle has A1 1007 , B1 413 , and a1 67; the other has A2 33 , B2 1387 , and a2 39.
24. sin B
22
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
30 sin 25 0507 C1 sin1 0507 3047 or C2 180 3047 14953 . 25 25 sin 12453 4873. If C1 3047 , then A1 180 25 3047 12453 and a1 sin 25 25 sin 547 If C2 14953 , then A2 180 25 14953 547 and a2 564. sin 25 Thus, one triangle has A1 125 , C1 30 , and a1 49; the other has A2 5 , C2 150 , and a2 56. 100 sin 30 26. sin B 23 B1 sin1 32 418 or B2 180 418 1382 . 75 75 sin 1082 1425. If B1 418 , then C1 180 30 418 1082 and c1 sin 30 75 sin 118 If B2 1382 , then C2 180 30 1382 118 and c2 307. sin 30 Thus, one triangle has B1 418 , C1 1082 , and c1 1425; the other has B2 1382 , C2 118 , and c2 307. 100 sin 50 27. sin B 1532. Since sin 1 for all , there can be no such angle B, and thus no such triangle. 50 80 sin 135 28. sin B 0566 B1 sin1 0566 344 or B2 180 344 1456 . 100 100 sin 106 If B1 344 , then C 180 135 344 106 and c 259. sin 135 If B2 180 344 1456 , then A B2 135 1456 180 , so there is no such triangle. Thus, the only possible triangle is B 344 , C 106 , and c 259. 26 sin 29 29. sin A 0840 A sin1 0840 572 or A 180 572 1228 . Only the second solution 15 15 sin 281 satisfies A 90 , so B 180 29 1228 282 and b 146. sin 29 82 sin 58 0953, so C sin1 0953 724 or C 180 724 1076 . Only the first solution 30. sin C 73 73 sin 496 satisfies C 90 , so A 180 58 724 496 and a 656. sin 58 sin 45 sin 31. Here B 45 , AC 22 32 13, and B D 5. Thus, by the Law of Sines, 5 13 5 sin 45 787 . sin1 13 sin 45 sin 3 2 sin 45 1 2 2 716 . sin 32. Here A 45 , BC 1 3 10, and AB 3 2, so 10 10 3 2 25. sin C
sin B 28 sin 30 sin 30 sin B 07, so 20 28 20 B sin1 07 44427 . Since BC D is isosceles, B B DC 44427 . Thus, BC D 180 2 B 91146 911 .
33. (a) From ABC and the Law of Sines we get
(b) From ABC we get BC A 180 A B 180 30 44427 105573 . Hence DC A BC A BC D 105573 91146 144 . 12 sin 25 525. 34. By symmetry, DC B 25 , so A 180 25 50 105 . Then by the Law of Sines, AD sin 105 35. (a) Let a be the distance from satellite to the tracking station A in miles. Then the subtended angle at the satellite is 50 sin 842 C 180 93 842 28 , and so a 1018 mi. sin 28 (b) Let d be the distance above the ground in miles. Then d 10183 sin 87 1017 mi.
SECTION 5.5 The Law of Sines
36. (a) Let x be the distance from the plane to point A. Then x 5
sin 48 377 mi. sin 100
(b) Let h be the height of the plane. Then sin 32
200 sin 52 219 ft. sin 46
sin 48 x sin 48 AB sin 180 32 48 sin 100
h h 377 sin 32 200 mi. x
37. C 180 82 52 46 , so by the Law of Sines, have AC
23
AC AB AB sin 52 AC , so substituting we sin 52 sin 46 sin 46
312 sin 486 0444 ABC sin1 0444 264 , and so BC A 180 486 264 105 . 527 527 sin 105 6785 ft. Then the distance between A and B is AB sin 486
38. sin ABC
39.
We draw a diagram. A is the position of the tourist and C is the top of the tower.
C
B 90 397 8603 and so C 180 301 8603 6387 .
Thus, by the Law of Sines, the length of the tower is
3.97¡
B
30.1¡
100
40.
A
BC
The situation is illustrated in the diagram.
C 165
180
67¡ B
100 sin 301 559 m. sin 6387
A
D
AC 165 sin 67 1519 ft, so using the Pythagorean Theorem we can calculate B A 1652 15192 644 ft and AD 1802 15192 966 ft. Thus the anchor points are B A AD 644 966 161 ft apart.
41. The angle subtended by the top of the tree and the sun’s rays is A 180 90 52 38 . Thus the height of the tree 215 sin 30 is h 175 ft. sin 38 42.
C
x
Œ
Let x be the length of the wire, as shown in the figure. Since 12 , other angles in ABC are 90 58 148 , and 180 12 148 20 .
Thus,
100 sin 148 x x 100 155 m. sin 148 sin 20 sin 20
º B 100
A 58¡
43. Call the balloon’s position R. Then in P Q R, we see that P 62 32 30 , and Q 180 71 32 141 . Q R P Q sin 30 Therefore, R 180 30 141 9 . So by the Law of Sines, Q R 60 192 m. sin 30 sin 9 sin 9
24
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
44.
Label the diagram as shown, and let the hill’s angle of elevation be . Then
B º
applying the Law of Sines to ABC,
30
sin 4 sin 8 055669 sin1 055669 338 . But from AB D,
C
B AD B 8 90 , so 90 8 338 482 .
120
8¡
A
sin sin 8 120 30
Œ
D
45. Let d be the distance from the earth to Venus, and let be the angle formed by sun, Venus, and earth. By the Law sin 394 sin 0878, so either sin1 0878 614 or 180 sin1 0878 1186 . of Sines, 1 0723 0723 d d 1119 AU; in the second case, In the first case, sin 180 394 614 sin 394 d 0723 d 0427 AU. sin 180 394 1186 sin 394 sin 60 b sin 60 sin B or sin B . Similarly, applying the Law of b c c sin B sin 120 r sin 120 r b Sines to BC D gives or sin B . Since sin 120 sin 60 , we have r cd cd c cd c sin D sin 60 b sin 60 b (). Similarly, from ADC and the Law of Sines we have or sin D , and r cd b d d a sin 120 b sin 60 a sin 120 b d from B DC we have sin D . Thus, . Combining this with cd d cd a cd b c d cd b b ab r a b 1. Solving for r, we find 1 (), we get r a cd cd cd r a a b ab ab . r ab 43 12 cm. (b) r 43 (c) If a b, then r is infinite, and so the face is a flat disk.
46. (a) Applying the Law of Sines to ABC, we get
47. By the area formula from Section 5.3, the area of ABC is A 12 ab sin C. Because we are given a and the three sin A a sin B sin B b . Thus, angles, we need to find b in terms of these quantities. By the Law of Sines, b a sin A a 2 sin B sin C a sin B A 12 ab sin C 12 a sin C . sin A 2 sin A 48. By the area formula from Section 5.3,
1 ab sin C Area of ABC sin C 12 , because a and b are the same for both Area of A B C sin C ab sin C 2
triangles. 49.
C b
a
B
C b
B
A
a b: One solution
B
a
C b
a» B
b a b sin A: Two solutions
A
C b
a B
a b sin A: One solution
A
a b sin A: No solution
SECTION 5.6 The Law of Cosines
25
A 30 , b 100, sin A 1 . If a b 100 then there is one triangle. If 100 a 100 sin 30 50, then there are 2
two possible triangles. If a 50, then there is one (right) triangle. And if a 50, then no triangle is possible.
5.6
THE LAW OF COSINES
1. For triangle ABC with sides a, b, and c the Law of Cosines states c2 a 2 b2 2ab cos C.
2. The Law of Cosines is required to solve triangles in cases SSS and SAS. 3. For ABC we have x 2 32 42 2 3 4 cos 35 , so x 25 24 cos 35 23.
13 62 52 32 13 299 . cos1 15 265 15 5. x 2 212 422 2 21 42 cos 39 441 1764 1764 cos 39 834115 and so x 834115 289. 6. x 2 152 182 2 15 18 cos 108 225 324 540 cos 108 715869 and so x 715869 268. 7. x 2 252 252 2 25 25 cos 140 625 625 1250 cos 140 2207556 and so x 2207556 47. 8. x 2 22 82 2 2 8 cos 88 4 64 32 cos 88 66883 and so x 66883 82.
4. For P Q R we have 32 62 52 2 6 5 cos , so cos
9. 37832 68012 42152 2 6801 4215 cos . Then cos cos1 0867 2989 . 10. 15462 6012 12252 2 601 1225 cos . Then cos cos1 0359 1110 .
37832 68012 42152 0867 2 6801 4215
15462 6012 12252 0359 2 601 1225
11. x 2 242 302 2 24 30 cos 30 576 900 1440 cos 30 228923 and so x
228923 151.
156 202 102 122 065 cos1 065 13054 . 2 10 12 240 13. c2 102 182 2 10 18 cos 120 100 324 360 cos 120 604 and so c 604 24576. Then 18 sin 120 sin A 0634295 A sin1 0634295 394 , and B 180 120 394 206 . 24576 12. 202 102 122 2 10 12 cos . Then cos
14. 122 402 442 2 40 44 cos B cos B
122 402 442 0964 B cos1 0964 155 . Then 2 40 44
40 sin 155 0891 A sin1 0891 630 , and so C 180 155 630 1015 . 12 4 sin 53 15. c2 32 42 2 3 4 cos 53 9 16 24 cos 53 10556 c 10556 32. Then sin B 0983 325 B sin1 0983 795 and A 180 53 795 475 . sin A
16. a 2 602 302 2 60 30 cos 70 3600 900 3600 cos 70 326873 30 sin 70 0493 a 326873 572. Then sin C 572 C sin1 0493 295 , and B 180 70 295 805 . 2
2
2
25 22 0644 A cos1 0644 499 . Then 17. 202 252 222 2 25 22 cos A cos A 20 22522
sin B 25 sin20499 0956 B sin1 0956 729 , and so C 180 499 729 572 .
2
2
2
12 16 078125 18. 102 122 162 2 12 16 cos A cos A 10 21216 A cos1 078125 386 . Then sin B 12 sin 386 0749 10
B sin1 0749 485 , and so C 180 386 485 929 .
26
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
sin 40 0833 C sin1 0833 564 or C 180 564 1236 . 19. sin C 162125 1 2
sin 836 1932. If C1 564 , then A1 180 40 564 836 and a1 125sin 40
sin 164 550. If C2 1236 , then A2 180 40 1236 164 and a2 125sin 40 Thus, one triangle has A 836 , C 564 , and a 1932; the other has A 164 , C 1236 , and a 550.
52 1024. Since sin 1 for all , there is no such A, and hence there is no such triangle. 20. sin A 65 sin 50
55 1065. Since sin 1 for all , there is no such B, and hence there is no such triangle. 21. sin B 65 sin 50
sin 61 1094 and c 735 sin 83 1241. 22. A 180 61 83 36 . Then b 735 sin 36 sin 36
sin 35 20. 23. B 180 35 85 60 . Then x 3sin 60
24. x 2 102 182 2 10 18 cos 40 100 324 360 cos 40 148224 and so x
sin 30 25. x 50 sin 100 254
2
2
148224 122.
2
10 11 205 0932 cos1 0932 213 . 26. 42 102 112 2 10 11 cos . Then cos 4 21011 220
27. b2 1102 1382 2 110 138 cos 38 12,100 19,044 30,360 cos 38 72200 and so b 850. Therefore, 2
2
85 128 using the Law of Cosines again, we have cos 1102110138
2
8915 .
40 0803 sin1 0803 535 or 180 535 1265 , but 535 doesn’t fit the 28. sin 10 sin 8
picture, so 1265 .
29. x 2 382 482 2 38 48 cos 30 1444 2304 3648 cos 30 588739 and so x 243. sin 98 11808. 30. A 180 98 25 57 . Then x 1000 sin 57
31. The semiperimeter is s 91215 18, so by Heron’s Formula the area is 2 A 18 18 9 18 12 18 15 2916 54. 32. The semiperimeter is s 122 52 , so by Heron’s Formula the area is 2 15 A 52 52 1 52 2 52 2 15 16 4 0968.
12, so by Heron’s Formula the area is 33. The semiperimeter is s 789 2 A 12 12 7 12 8 12 9 720 12 5 268.
34. The semiperimeter is s 11100101 106, so by Heron’s Formula the area is 2 A 106 106 11 106 100 106 101 302,100 10 3021 5496.
35. The semiperimeter is s 346 13 2 2 , so by Heron’s Formula the area is 13 3 13 4 13 6 455 455 533. A 13 2 2 2 2 16 4
6, so by Heron’s Formula the area of 36. The smaller triangles have the same area. The semiperimeter of each is s 255 2 each is A 6 6 2 6 5 6 5 24 2 6, so the shaded area is 4 6 980.
37. We draw a diagonal connecting the vertices adjacent to the 100 angle. This forms two triangles. Consider the triangle with sides of length 5 and 6 containing the 100 angle. The area of this triangle is A1 12 5 6 sin 100 1477. To use
Heron’s Formula to find the area of the second triangle, we need to find the length of the diagonal using the Law of Cosines: c2 a 2 b2 2ab cos C 52 62 2 5 6 cos 100 71419 c 845. Thus the second triangle has semiperimeter 8 7 845 117255 and area A2 117255 117255 8 117255 7 117255 845 2600. The area s 2 of the quadrilateral is the sum of the areas of the two triangles: A A1 A2 1477 2600 4077.
SECTION 5.6 The Law of Cosines
27
38. We draw a line segment with length x bisecting the 60 angle to create two triangles. By the Law of Cosines, 32 42 x 2 2 4x cos 30 x 2 4 3x 7 0. Using the Quadratic Formula, we find x 2 3 5. The minus sign provides the correct length of about 123 (the other solution is about 57, which corresponds to a convex quadrilateral 342 3 5 with the same side lengths), so the semiperimeter of each triangle is s and the total area of the figure 2 is A 2 s s 3 s 4 s 2 3 5 246. 39.
Label the centers of the circles A, B, and C, as in the figure. By the Law of
C
6
5
6
B 5
4
4 A
Cosines, cos A
92 102 112 AB 2 AC 2 BC 2 13 A 7053 . 2 AB AC 2 9 10
Now, by the Law of Sines,
sin 7053 sin B sin C . So 11 AC AB
1 085710 5899 and sin B 10 11 sin 7053 085710 B sin
9 sin 7053 077139 C sin1 077139 5048 . The area of sin C 11
ABC is 12 AB AC sin A 12 9 10 sin 7053 42426.
2 7053 9848. Similarly, the areas of sectors B and C 4 360 360 are S B 12870 and SC 15859. Thus, the area enclosed between the circles is A ABC S A S B SC
The area of sector A is given by S A R 2
A 42426 9848 12870 15859 385 cm2 .
40. By the Law of Cosines we have a 2 62 42 2 6 4 cos 45 52 24 2, b2 62 x 2 2 6 x cos 30 x 2 6 3x36, and c2 x 2 42 2 x 4 cos 30 45 x 2 2 2 6 x16. By the Pythagorean Theorem, a 2 b2 c2 , so we have 52 24 2 x 2 6 3x 36 x 2 2 2 6 x 16 12 3 2 26 12 2 3 3x 18 2 6 x 8 36 12 2 3 3 2 6 x, so x . 3 3 2 6
41. Let c be the distance across the lake, in miles. Then c2 2822 3562 2 282 356 cos 403 5313 c 230 mi. D
42. Suppose ABC D is a parallelogram with AB DC 5, AD BC 3, and A 50 (see the figure). Since opposite angles are equal in a
parallelogram, it follows that C 50 , and
C
3 50¡l
A B 5 260 130 . 2 By the Law of Cosines, AC 2 32 52 2 3 5 cos 130 AC 9 25 30 cos 130 73. Similarly, B D 32 52 2 3 5 cos 50 38.
B D 360 100 260 . Thus, B D
43. In half an hour, the faster car travels 25 miles while the slower car travels 15 miles. The distance between them is given by the Law of Cosines: d 2 252 152 2 25 15 cos 65 d 252 152 2 25 15 cos 65 5 25 9 30 cos 65 231 mi.
44. Let x be the car’s distance from its original position. Since the car travels
x
at a constant speed of 40 miles per hour, it must have traveled 40 miles east, and then 20 miles northeast (which is 45 east of “due north”). From the diagram, we see that 135 , so x 202 402 2 20 40 cos 135 10 4 16 16 cos 135 560 mi.
º 45¡ 40
20
28
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
45. The airplane travels a distance of 625 15 9375 miles in its original
direction and 625 2 1250 miles in the new direction. Since it makes a course correction of 10 to the right, the included angle is
10¡
937.5
1250
d
180 10 170 . From the figure, we use the Law of Cosines to get
the expression d 2 93752 12502 2 9375 1250 cos 170 4,749,54942, so d 2179 miles. Thus, the airplane’s distance from its original position is approximately 2179 miles.
46. Let d be the distance between the two boats in miles. After one hour, the boats have traveled distances of 30 miles and 26 miles. Also, the angle subtended by their directions is 180 50 70 60 . Then d 2 302 262 2 30 26 cos 60 796 d 796 282. Thus the distance between the two boats is about 28 miles. 47. (a) The angle subtended at Egg Island is 100 . Thus using the Law of
Forrest Island
Cosines, the distance from Forrest Island to the boat’s home port is x 2 302 502 2 30 50 cos 100
900 2500 3000 cos 100 3920945 and so x 3920945 6262 miles.
10¡
(b) Let be the angle shown in the figure. Using the Law of Sines, sin
50 sin 100
07863 sin1 07863 518 . Then
50
x
80¡ Egg Island
20¡
6262 90 20 518 182 . Thus the bearing to the boat’s home port
¬
30
70¡ Home Port
is S 182 E.
48. (a) In 30 minutes the plane flies 100 miles due east, so using the Law of
B
Cosines we have x 2 1002 3002 2 100 300 cos 40
300
1002 1 9 6 cos 40 1002 5404. Thus, x
1002 5404 2325, and so the plane is 2325 miles from
its destination.
(b) Using the Law of Sines, sin
A
50¡ 40¡
x ¬
100
300 sin 40 0829 2325
sin1 0829 56 . However, since 90 , the angle we seek is
180 56 124 . Hence the bearing is 124 90 34 , that is, N 34 E.
49. The largest angle is the one opposite the longest side; call this angle . Then by the Law of Cosines, 442 362 222 2 36 22 cos cos
362 222 442 009848 cos1 009848 957 . 2 36 22
50. Let be the angle formed by the cables. The two tugboats and the barge form a triangle: the side opposite has a length of 120 ft and the other two sides have lengths of 212 and 230 ft. Therefore, 1202 2122 2302 2 212 230 cos cos
2122 2302 1202 cos 08557 cos1 08557 31 . 2 212 230
51. Let d be the distance between the kites. Then d 2 3802 4202 2 380 420 cos 30 d 3802 4202 2 380 420 cos 30 211 ft.
SECTION 5.6 The Law of Cosines
29
52. Let x be the length of the wire and the angle opposite x, as shown in the figure. Since the mountain is inclined 32 , we must have
125
x
180 90 32 122 . Thus, x 552 1252 2 55 125 cos 122 161 ft.
¬ 55 32¡
53. Solution 1: From the figure, we see that 106 and sin 74
3400 b
3400 3537. Thus, x 2 8002 35372 2 800 3537 cos 106 sin 74 x 8002 35372 2 800 3537 cos 106 x 3835 ft.
b
Solution 2: Notice that tan 74
3400 a
a
Pythagorean Theorem, x 2 a 8002 34002 . So x 9749 8002 34002 3835 ft.
x
3400 9749. By the tan 74
b
3400
74¡
800
54. Let the person be at point A, the first landmark (at 62 ) be at point B, and the other landmark be at point C. We 1150 1150 1150 AB 2450. Similarly, cos 54 want to find the length BC. Now, cos 62 AB cos 62 AC 1150 1956. Therefore, by the Law of Cosines, BC 2 AB 2 AC 2 2 AB AC cos 43 AC cos 54 BC 24502 19562 2 2450 1956 cos 43 BC 1679. Thus, the two landmarks are roughly 1679 feet apart.
112 148 190 abc 225. Thus, 55. By Heron’s formula, A s s a s b s c, where s 2 2 A 225 225 112 225 148 225 190 82777 ft2 . Since the land value is $20 per square foot, the value of the lot is approximately 82777 20 $165,554.
56. Having found a 132 using the Law of Cosines, we use the Law of Sines to find B:
sin 465 sin B 105 132
105 sin 465 0577. Now there are two angles B between 0 and 180 which have sin B 0577, namely 132 B 352 and B 1448 . But we must choose B , since otherwise A B 180 . 1 2 1 sin 465 180 sin 465 sin C sin C 0989, so either C 815 Using the Law of Sines again, 180 132 132 or C 985 . In this case we must choose C 985 so that the sum of the angles in the triangle is A B C 465 352 985 180 . (The fact that the angles do not sum to exactly 180 , and the discrepancies between these results and those of Example 3, are due to roundoff error.) The method in this exercise is slightly easier computationally, but the method in Example 3 is more reliable. sin B
57. In any ABC, the Law of Cosines gives a 2 b2 c2 2bccos A, b2 a 2 c2 2accos B, and c2 a 2 b2 2abcos C. Adding the second and third equations gives b2 a 2 c2 2ac cos B
c2 a 2 b2 2ab cos C
b2 c2 2a 2 b2 c2 2a c cos B b cos C
30
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
Thus 2a 2 2a c cos B b cos C 0, and so 2a a c cos B b cos C 0. Since a 0 we must have a c cos B b cos C 0 a b cos C c cos B . The other laws follow from the symmetry of a, b, and c.
CHAPTER 5 REVIEW 7 183 rad 2. (a) 105 105 180 12
052 rad 1. (a) 30 30 180 6
5 262 rad (b) 150 150 180 6
2 126 rad (b) 72 72 180 5
9 707 rad (c) 405 405 180 4
035 rad (c) 20 20 180 9
5 393 rad (d) 225 225 180 4
7 550 rad (d) 315 315 180 4
3. (a) 56 rad 56 180 150
4. (a) 53 rad 53 180 300
180 (b) 9 rad 9 20
(b) 109 rad 109 180 200
(c) 43 rad 43 180 240
900 (c) 5 rad 5 180 2865
720 (d) 4 rad 4 180 2292
(d) 113 rad 113 180 660
5. r 10 m, 25 rad. Then s r 10 25 4 126 m. s 7 28 rad 1604 6. s 7 cm, r 25 cm. Then 25 r 5 rad. Then r s 25 18 90 286 ft. 7. s 25 ft, 50 50 180 18 5
13 rad. Then r s 13 18 18 m. 8. s 13 m, 130 130 180 18 13
9. Since the diameter is 28 in, r 14 in. In one revolution, the arc length (distance traveled) is s r 2 14 28 in. The total distance traveled is 60 mi/h 05 h 30 mi 30 mi 5280 ft/mi 12 in./ft 1,900,800 in. The number of 1 rev revolution is 1,900,800 in 216087 rev. Therefore the car wheel will make approximately 21,609 revolutions. 28 in. s 2450 35448 and so the angle is 10. r 3960 miles, s 2450 miles. Then 0619 rad 0619 180 r 3960 approximately 354 . 11. r 5 m, 2 rad. Then A 12 r 2 12 52 2 25 m2 . 18,151 ft2 12. A 12 r 2 12 2002 52 180
2A 2 125 250 625 04 rad 229 r2 252 2A 100 14. A 50 m2 and 116 rad. Thus, r 11 42 m. 13. A 125 ft2 , r 25 ft. Then
6
150 2 rad 300 rad/min 9425 rad/min. The linear speed is 15. The angular speed is 1 min 150 2 8 2400 in./min 75398 in./min 6283 ftmin. 1 rad 7000 rad/min 21,9911 rad/min. 16. (a) The angular speed of the engine is e 35002 1 min
e 7000rad/min 09 (b) To find the angular speed of the wheels, we calculate g
77778 rad/min 24,4346 rad/min.
(c) The speed of the car is the angular speed of the wheels times their radius: 77778 rad 11 in 60 min 1 mile 2545 mi/h. min 1 hr 63,360 in.
CHAPTER 5
Review
31
17. r 52 72 74. Then sin 5 , cos 7 , tan 57 , csc 574 , sec 774 , and cot 75 . 74 74
18. x 19.
3 , cos 91 , tan 3 , csc 10 , sec 10 , and cot 91 . 102 32 91. Then sin 10 10 3 3 91
x cos 40 5
20. cos 35 21.
y 5 sin 40 321.
2 x cos235 244, and tan 35 2y y 2 tan 35 140. x
1 sin 20 x
22. cos 30
x 5 cos 40 383, and 5y sin 40
91
x sin120 292, and xy cos 20
2924 311. y cosx20 09397
x x 4 cos 30 346, and sin 30 xy y x sin 30 346 05 173. 4
23. A 90 20 70 , a 3 cos 20 2819, and b 3 sin 20 1026. A
24. C 90 60 30 , cos 60 20a
a 20 cos 60 40, and tan 60 b20
C
b
b 20 tan 60 3464. C
70¡ 3
20¡
a b B A
25. c
7 028379 163 , 252 72 24, A sin1 25
and C sin1 24 25 12870 737 . 25 A
27. tan
c
26. b
C
30¡
a 60¡ 20
B
5 03948 226 , 122 52 13, A sin1 13
and C sin1 12 13 11760 674 . b
7 B
A
12
C 5 B
1 1 1 1 a cot , sin b csc a tan b sin
28. Let h be the height of the tower in meters. Then tan 2881
h h 1000 tan 2881 550 m. 1000
One side of the hexagon together with radial line segments through its endpoints
29. x
8
forms a triangle with two sides of length 8 m and subtended angle 60 . Let x be the
60¡
length of one such side (in meters). By the Law of Cosines,
8
x 2 82 82 2 8 8 cos 60 64
hexagon is 6x 6 8 48 m.
x 8. Thus the perimeter of the
32
CHAPTER 5 Trigonometric Functions: Right Triangle Approach y
30. As the crankshaft moves in its circular pattern, point Q is
y
determined by the angle , namely it has coordinates Q 2 cos , 2 sin . We split the triangle into two right triangles O Q R and P Q R, as shown in the figure. Let h be the height of the piston. We consider two cases, 0 180 and 180 360 .
If 0 180 , then h is the sum of O R and R P. Using the Pythagorean Theorem, we find R P 82 2 cos 2 , while
8
h
h
8 R ¬ O 2
Q(2 cos ¬, 2 sin ¬) x
Q(2 cos ¬, 2 sin ¬)
O R is the ycoordinate of the point Q, 2 sin . Thus h 64 4 cos2 2 sin .
¬
O 2 R
x
If 180 360 , then h is the difference between R P and R O. Again, R P 64 4 cos2 and O R is the ycoordinate of the point Q, 2 sin . Thus h 64 4 cos2 2 sin . Since sin 0 for 180 360 , this also reduces to h 64 4 cos2 2 sin . Since we get the same result in both cases, the height of the piston in terms of is h 64 4 cos2 2 sin .
r , 0518 r r 236,900sin 0259 2 r AB 236,900 sin 0259 r 1 sin 0259 236,900 sin 0259 r 1076 and so the radius of the moon is 1 sin 0259 roughly 1076 miles.
31. Let r represent the radius, in miles, of the moon. Then sin
32. Let d1 represent the horizontal distance from a point directly below the plane to the closer ship in feet, and d2 represent the 35,000 35,000 35,000 d1 , and similarly tan 40 horizontal distance to the other ship in feet. Then tan 52 d1 tan 52 d2 35,000 35,000 35,000 d2 . So the distance between the two ships is d2 d1 14,400 ft. tan 40 tan 40 tan 52 34. csc 94 csc 33. sin 315 sin 45 1 22 4 2 2
35. tan 135 tan 45 1 1 3 37. cot 223 cot 23 cot 3 3 3
39. cos 585 cos 225 cos 45 1 22 2
2 3 2 41. csc 83 csc 23 csc 3 3 3
3 36. cos 56 cos 6 2
38. sin 405 sin 45 1 22 2 40. sec 223 sec 43 sec 3 2
2 3 42. sec 136 sec 6 3
43. cot 390 cot 30 cot 30 3 44. tan 234 tan 34 tan 4 1 5 12 13 13 45. r 52 122 169 13. Then sin 12 13 , cos 13 , tan 5 , csc 12 , sec 5 , and 5 . cot 12
46. If is in standard position, then the terminal point of on the unit circle is simply cos sin . Since the terminal point is given as 23 12 , sin 12 . 47. y 3x 1 0 y 3x 1, so the slope of the line is m 3. Then tan m 3 60 . 48. 4y 2x 1 0 y 12 x 14 . The slope of the line is m 12 . Then tan m 12 and r 12 22 5. So sin 1 , cos 2 , tan 12 , csc 5, sec 25 , and cot 2. 5
5
CHAPTER 5
Review
33
sin 1 cos2 2 . 49. Since sin is positive in quadrant II, sin 1 cos and we have tan cos cos 50. sec
1 1 (because cos 0 in quadrant III). cos 1 sin2
51. tan2
sin2 sin2 2 cos 1 sin2
52. csc2 cos2
1 sin2
cos2
1 sin2 sin2
1 sin2
1
53. tan 37 , sec 43 . Then cos 34 and sin tan cos 47 , csc 4 4 7 7 , and cot 3 3 7 7 . 7 7 41 9 40 54. sec 41 40 , csc 9 . Then sin 41 , cos 41 , tan
9 sin 41 9 , and cot 40 . 40 40 9 cos 41
55. sin 35 . Since cos 0, is in quadrant II. Thus, x 52 32 16 4 and so cos 45 , tan 34 , csc 53 , sec 54 , cot 43 .
5 56. sec 13 5 and tan 0. Then cos 13 , and must be in quadrant III sin 0. Therefore, 25 12 , tan sin 12 , csc 13 , and cot 5 . sin 1 cos2 1 169 13 5 12 12 cos
4 cos 45 2 since 5 5 sin cos 0 in quadrant II. But tan 12 sin 12 cos 12 2 1 . Therefore, 5 5 cos 5 1 2 1 sin cos 5 .
57. tan 12 . sec2 1 tan2 1 14 54 cos2
5
5
5
1 1 1 sin 1 sin 1 sin 2 3 2 1 3. 58. sin 2 for in quadrant I. Then tan sec cos cos cos 2 1 sin2 1 12 59. By the Pythagorean Theorem, sin2 cos2 1 for any angle .
3 5 10 5 5 60. cos 23 and 2 . Then 6 2 6 3 . So sin 2 sin 3 sin 3 2 .
61. sin1 23 3
62. tan1 33 6
63. Let u sin1 25 and so sin u 25 . Then from the triangle, 64. Let u cos1 83 then cos u 38 . From the triangle, we tan sin1 25 tan u 2 . have sin cos1 38 sin u 855 . 21
5
2
u Ï21
8 u
Ï55
3
34
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
65. Let tan1 x
tan x. Then from the x . triangle, we have sin tan1 x sin 1 x2
66. Let sin1 x. Then sin x. From the triangle, we 1 have csc sin1 x csc . x
Ï1+x@
1
¬
x
¬
1
Ï1-x@
x x tan1 2 2 10 sin 30 69. B 180 30 80 70 , and so by the Law of Sines, x 532. sin 70 2 sin 45 70. x 146 sin 105 67. cos
x x cos1 3 3
x
68. tan
71. x 2 1002 2102 2 100 210 cos 40 21,926133 x 14807 20 sin 60 0247 B sin1 0247 1433 . Then C 180 60 1433 10567 , and so 72. sin B 70 70 sin 10567 7782. x sin 60 73. x 2 22 82 2 2 8 cos 120 84 x 84 917 6 sin 3121 4 sin 110 33. 0626 B 3879 . Then C 180 110 3879 3121 , and so x 74. sin B 6 sin 110 sin 25 23 sin 25 sin 23 sin 25 sin sin1 541 or 75. By the Law of Sines, 23 12 12 12 180 541 1259 . Because 1259 does not fit the diagram, we must have 541 . sin 80 4 sin 80 sin 4 sin 80 1 sin sin 520 . 76. By the Law of Sines, 4 5 5 5 77. By the Law of Cosines, 1202 1002 852 2 100 85 cos , so cos cos1 016618 804 .
1202 1002 852 016618. Thus, 2 100 85
sin 10 5 sin 10 sin A sin A 02894, so A sin1 02894 168 5 3 3 or A 180 168 1632 . Therefore, 180 10 168 1532 or 180 10 1632 68 .
78. We first use the Law of Sines to find A:
79. After 2 hours the ships have traveled distances d1 40 mi and d2 56 mi. The subtended angle is 180 32 42 106 . Let d be the distance between the two ships in miles. Then by the Law of Cosines, d 2 402 562 2 40 56 cos 106 5970855 d 773 miles.
80. Let h represent the height of the building in feet, and x the horizontal distance from the building to point B. Then h h h tan 241 and tan 302 x h cot 302 . Substituting for x gives tan 241 x 600 x h cot 302 600 600 tan 241 h tan 241 h cot 302 600 h 1160 ft. 1 tan 241 cot 302 81. Let d be the distance, in miles, between the points A and B . Then by the Law of Cosines, d 2 322 562 2 32 56 cos 42 14966 d 39 mi. 120 sin 689 12008 miles. Let d be the shortest distance, in miles, to 82. C 180 423 689 688 . Then b sin 688 the shore. Then d b sin A 12008 sin 423 808 miles.
CHAPTER 5
Test
83. A 12 ab sin 12 8 14 sin 35 3212
568 abc 84. By Heron’s Formula, A s s a s b s c, where s 95. Thus, 2 2 A 95 95 5 95 6 95 8 1498. 85. (a) csc 2, so sin 22 . The sine function is negative for terminal points below the xaxis, and the reference angle in this case is 4 , so this corresponds to graph VII.
(b) sin 1 22 has reference angle 4 and its terminal point lies above the xaxis, so this corresponds to graph III. 2 (c) sin1 35 sin 35 , which corresponds to graph I.
(d) tan 43 corresponds to graph VI.
(e) tan 32 corresponds to graph V, since tangent is negative in quadrant III. (f) sin 2 corresponds to VIII, since 5
4 y 2 . r 5 22 42
2 x 1 , which lies in the (g) cos1 1 cos 1 corresponds to graph II since 5 5 r 5 2 2 2 4 domain of the inverse cosine function.
(h) sin 35 corresponds to graph IV since that terminal point has
3 y 3 . 2 2 r 5 3 4
CHAPTER 5 TEST 11 rad. 135 135 3 rad. 1. 330 330 180 6 180 4
180 234 2. 43 rad 43 180 240 . 13 rad 13 745
3. (a) The angular speed is
120 2 rad 240 rad/min 75398 rad/min. 1 min
(b) The linear speed is 120 2 16 3840 ft/min 1 12,0637 ft/min
4. (a) sin 405 sin 45 1 22 2
(b) tan 150 tan 30 1 33 3 (c) sec 53 sec 3 2
(d) csc 52 csc 2 1
137 mi/h
2 13 3 2 2 26 6 13 5. r 32 22 13. Then tan sin . 3 39 13 3 13 b a a 24 sin . Also, cos b 24 cos . 6. sin 24 24 7. cos 13 and is in quadrant III, so r 3, x 1, and y 32 12 2 2. Then
1 23 43 2. 3 2 csc 1 4 2 2 2 2 tan 1 sin 1 5 13 13 1 5 , tan 5 . Then sec tan . 8. sin 13 12 cos cos sin sin 12 5 12 2 2 2 2 9. sec 1 tan tan sec 1. Thus, tan sec 1 since tan 0 in quadrant II.
tan cot csc tan
35
36
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
h h 6 tan 73 196 ft. 6 x x 11. (a) tan tan1 4 4 3 3 (b) cos cos1 x x 10. tan 73
9 so tan u 9 . From the triangle, r 12. Let u tan1 40 40 40 9 1 cos tan 40 cos u 41 .
92 402 41. So
41 u
9
40
13. By the Law of Cosines, x 2 102 122 2 10 12 cos 48 8409 x 91. 14. C 180 52 69 59 . Then by the Law of Sines, x
230 sin 69 2505. sin 59
h x h h 50 tan 20 and tan 28 x h 50 tan 28 50 50 x 50 tan 28 h 50 tan 28 50 tan 20 84. 15 sin 108 16. Let A and X be the other angles in the triangle. Then sin A 0509 A 3063 . Then 28 28 sin 4137 X 180 108 3063 4137 , and so x 195. sin 108 82 62 92 01979, so cos1 01979 786 . 17. By the Law of Cosines, 92 82 62 2 8 6 cos cos 2 8 6 15. Let h be the height of the shorter altitude. Then tan 20
18. We find the length of the third side x using the Law of Cosines: x 2 52 72 2 5 7 cos 75 5588 x 7475. sin 75 5 sin 75 sin sin 06461, so sin1 06461 402 . Therefore, by the Law of Sines, 5 7475 7475 50 72 . A triangle 1 r r sin 1 102 sin 72 . Thus, the area of the 19. (a) A sector 12 r 2 12 102 72 180 180 2 2 72 shaded region is A shaded A sector A triangle 50 180 sin 72 153 m2 .
(b) The shaded region is bounded by two pieces: one piece is part of the triangle, the other is part of the circle. The first part has length l 102 102 2 10 10 cos 72 10 2 2 cos 72 . The second has length 4. Thus, the perimeter of the shaded region is p l s 102 2 cos 72 4 243 m. s 10 72 180
20. (a) If is the angle opposite the longest side, then by the Law of Cosines cos
92 132 202 06410. Therefore, 2 9 20
cos1 06410 1299 .
(b) From part (a), 1299 , so the area of the triangle is A 12 9 13 sin 1299 449 units2 . Another way to find 9 13 20 abc 21. Thus, the area is to use Heron’s Formula: A s s a s b s c, where s 2 2 A 21 21 20 21 13 21 9 2016 449 units2 . 21. Label the figure as shown. Now 85 75 10 , so by the Law of Sines, 100 x sin 75 sin 10
x 100
sin 75 h . Now sin 85 sin 10 x
sin 75 sin 85 554 ft. h x sin 85 100 sin 10
º
x 75¡ 100
85¡
h
Surveying
37
FOCUS ON MODELING Surveying 1. Let x be the distance between the church and City Hall. To apply the Law of Sines to the triangle with vertices at City Hall, the church, and the first bridge, we first need the measure of the angle at the first bridge, which is 180 25 30 125 . x 086 086 sin 125 Then x 14089. So the distance between the church and City Hall is about sin 125 sin 30 sin 30 141 miles. 2. To find the distance z between the fire hall and the school, we use the distance found in the text between the bank and the cliff. To find z we first need to find the length of the edges labeled x and y.
bank 50¡
In the banksecond bridgecliff triangle, the third angle is
1.55
155 sin 50 180 50 60 70 , so x 126. sin 70 In the second bridgeschoolcliff triangle, the third angle is
cliff
126 sin 45 109. sin 55 Finally, in the schoolfire hallcliff triangle, the third angle is
80¡
180 80 55 45 , so y
109 sin 80 131. sin 55 Thus, the fire hall and the school are about 131 miles apart.
180 45 80 55 , so z
fire hall
55¡
z
70¡
x
60¡
second bridge
45¡
80¡ y 45¡
55¡ school
equal
3. First notice that D BC 180 20 95 65 and D AC 180 60 45 75 . AC 20 20 sin 45 AC 146 . From BC D we get From AC D we get sin 45 sin 75 sin 75 BC 20 20 sin 95 BC 220. By applying the Law of Cosines to ABC we get sin 95 sin 65 sin 65 AB2 AC2 BC2 2 AC BC cos 40 1462 2202 2146220cos 40 205, so AB 205 143 m. Therefore, the distance between A and B is approximately 143 m. 4. Let h represent the height in meters of the cliff, and d the horizontal distance to the cliff. The third horizontal 200 sin 516 h angle is 180 694 516 59 and so d 182857. Then tan 331 sin 59 d h d tan 331 182857 tan 331 1192 m. AB BC sin sin sin d sin d sin sin BC AB . Thus, h BC sin . sin sin cos sin cos cos sin cos sin d tan tan Dividing numerator and denominator by cos cos , we have h . tan tan 800 tan 25 tan 29 d tan tan 2350 ft (b) h tan tan tan 29 tan 25
5. (a) In ABC, B 180 , so C 180 180 . By the Law of Sines,
R R h cos Rh R h R 1 cos h cos R . With cos 1 cos sin 2 141 cos 0021 h 141 km and 0021 rad, we have R 6390 km. 1 cos 0021
6. By the Law of Sines,
38
FOCUS ON MODELING
7. Let the surveyor be at point A, the first landmark (with angle of depression 42 ) be at point B, and the other landmark 2430 2430 2430 36316. Similarly, sin 39 be at point C. We want to find BC. Now sin 42 AB AB AC sin 42 2430 38613. Therefore, by the Law of Cosines, BC2 AB2 AC2 2 AB AC cos 68 AC sin 39 BC 363162 386132 2 36316 38613 cos 68 4194. Thus, the two landmarks are approximately 4194 ft apart. 8. We start by labeling the edges and calculating the remaining angles, as shown in the first figure. Using the Law of Sines, a 150 150 sin 29 b 150 150 sin 91 we find the following: a 8397, b 17318, sin 29 sin 60 sin 60 sin 91 sin 60 sin 60 17318 17318 sin 32 d 17318 17318 sin 61 c c 9190, e 15167, sin 32 sin 87 sin 87 sin 61 sin 87 sin 87 e 15167 15167 sin 41 f 15167 15167 sin 88 e 12804, f 19504, sin 41 sin 51 sin 51 sin 88 sin 51 sin 51 g 19504 19504 sin 50 h 19504 19504 sin 38 g 14950, and h 12015. Note that sin 50 sin 92 sin 92 sin 38 sin 92 sin 92 we used two decimal places throughout our calculations. Our results are shown (to one decimal place) in the second figure.
29¡ 150
c
61¡
87¡ 88¡
91¡
60¡ a
32¡ 41¡
128.0
51¡ 38¡
d
b
91.9
e
g
f 50¡ h
92¡
150
173.2
84
151.7
195.0
149.5
120.2
9. Answers will vary. Measurements from the Great Trigonometric Survey were used to calculate the height of Mount Everest to be exactly 29,000 ft, but in order to make it clear that the figure was considered accurate to within a foot, the height was published as 29,002 ft. The accepted figure today is 29,029 ft.
CORRECTIONS: p. 20,21,26,36,56,57
CHAPTER 6
TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH
6.1 6.2
The Unit Circle 1 Trigonometric Functions of Real Numbers 6
6.3
Trigonometric Graphs 11
6.4
More Trigonometric Graphs 26
6.5
Inverse Trigonometric Functions and Their Graphs 37
6.6
Modeling Harmonic Motion 40 Chapter 6 Review 46 Chapter 6 Test 54
¥
FOCUS ON MODELING: Fitting Sinusoidal Curves to Data 56
1
6
TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH
6.1
THE UNIT CIRCLE
1. (a) The unit circle is the circle centered at 0 0 with radius 1. (b) The equation of the unit circle is x 2 y 2 1.
(c) (i) Since 12 02 1, the point is P 1 0.
(ii) P 0 1
(iii) P 1 0
(iv) P 0 1
2. (a) If we mark off a distance t along the unit circle, starting at 1 0 and moving in a counterclockwise direction, we arrive at the terminal point determined by t. (b) The terminal points determined by 2 , , 2 , 2 are 0 1, 1 0, 0 1, and 1 0, respectively.
3. If the terminal point determined by t is P, then the terminal point determined by t 2 is P. The terminal point for t 3 3 3 1 7 1 and so the terminal point for t . is 2 2 3 is 2 2 2 2 and the y The terminal point determined by t is 4. 4 2 2 ¹ Ï2 Ï2 ; , 5¹ ; _Ï3 ,1 4 l2l l2l 6 l2l 2 terminal point determined by t 54 is 22 22 . 3 1 The terminal point determined by t 1 x 0 6 is 2 2 and the Ï3 1 _¹6 ; l2l , _ 2 terminal point determined by t 56 is 23 12 .
(
5¹ ; 4
)
(_Ï2l2l , _Ï2l2l)
(
)
(
)
In general, if the terminal point determined by t is P a b, then by symmetry, the terminal point determined by t is P a b.
2 2 9 16 1, P 3 4 lies on the unit circle. 5. Since 35 45 25 25 5 5 2 7 2 576 49 1, P 24 7 lies on the unit circle. 25 6. Since 24 25 625 25 25 25 2 2 9 7 1, P 3 7 lies on the unit circle. 7. Since 34 47 16 16 4 4 2 2 24 1, P 5 2 6 lies on the unit circle. 8. Since 57 2 7 6 25 7 7 49 49 2 2 9. Since 35 23 59 49 1, P 35 23 lies on the unit circle. 2 2 11 5 25 10. Since 611 56 11 36 36 1, P 6 6 lies on the unit circle. 2 9 y 2 16 y 4 . Since P x y is in quadrant III, y is negative, so the point is 11. 35 y 2 1 y 2 1 25 25 5 3 4 P 5 5 . 7 2 1 x 2 1 49 x 2 576 x 24 . Since P is in quadrant IV, x is positive, so the point is 12. x 2 25 625 625 25 24 7 P 25 25 . 2 13. x 2 13 1 x 2 1 19 x 2 89 x 2 3 2 . Since P is in quadrant II, x is negative, so the point is P 2 3 2 13 .
1
2
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
14.
2
4 y 2 21 y 21 . Since P is in quadrant I, y is positive, so the point is y 2 1 y 2 1 25 25 5 21 2 P 5 5 . 2 5
2 4 x 2 45 x 3 5 . Since P x y is in quadrant IV, x is positive, so the point is 15. x 2 27 1 x 2 1 49 7 49 P 3 7 5 27 . 2 16. 23 y 2 1 y 2 1 49 y 2 59 y 35 . Since P is in quadrant II, y is positive, so the point is P 23 35 .
17.
5 2 y 2 1 y 2 1 25 y 2 144 y 12 . Since its ycoordinate is negative, the point is P 5 12 . 13 169 169 13 13 13
2 9 x 2 16 x 4 . Since its xcoordinate is positive, the point is P 4 3 . 18. x 2 35 1 x 2 1 25 25 5 5 5 2 19. x 2 23 1 x 2 1 49 x 2 59 x 35 . Since its xcoordinate is negative, the point is P 35 23 . 2 5 x 2 20 x 2 5 . Since its xcoordinate is positive, the point is 1 x 2 1 25 20. x 2 55 25 5 2 5 5 P 5 5 .
2 21. 32 y 2 1 y 2 1 29 y 2 79 y 37 . Since P lies below the xaxis, its ycoordinate is negative, so the point is P 32 37 .
2 4 y 2 21 y 21 . Since P lies above the xaxis, its ycoordinate is positive, so 22. 25 y 2 1 y 2 1 25 25 5 21 2 the point is P 5 5 .
24.
23.
t
Terminal Point
0
1 0 2 2 2 2
4
2 3 4
t
Terminal Point
t
Terminal Point
1 0 22 22
0
1 0 3 1 2 2 1 3 2 2
5 4 3 2 7 4
0 1
22 22
2
1 0
0 1
2 2 2 2
1 0
6
3
2 2 3 5 6
25. t 0, so t 5 corresponds to P x y 1 0. y
P(_1, 0) t=5¹
Q(1, 0)
t=0 x
t
Terminal Point
1 0 23 12 12 23
7 6 4 3 3 2 5 3 11 6
0 1 12 23 23 12
2
1 0
0 1
1 3 2 2 31 2 2
1 0
26. t 0, so t 3 corresponds to P x y 1 0. y
P(_1, 0)
t=_3¹
Q(1, 0)
t=0
x
SECTION 6.1 The Unit Circle
27. t 0, so t 4 corresponds to P x y 1 0. y
28. t 0, so t 6 corresponds to P x y 1 0. y
t=_4¹ t=0 P(1, 0) x
t=6¹ t=0 P(1, 0) x
3 29. t 2 , so t 2 corresponds to P x y 0 1. y ¹ Q(0, 1) t= 2
5 30. t 2 , so t 2 corresponds to P x y 0 1. y P(0, 1)
5¹
t= 2
¹
t= 2 x
3¹
t= 2
x
P(0, _1)
5 31. t 2 , so t 2 corresponds to P x y 0 1. y ¹ Q(0, 1) t= 2
3 32. t 2 , so t 2 corresponds to P x y 0 1. y
3¹
¹
t=_ 2 t= 2
P(0, 1) Q(0, 1)
x
x
5¹ P(0, _1)
t=_ 2
3 1 5 33. t 6 and t 6 corresponds to P x y 2 2 . y
P(_ 2 , 2 )
34. t 3 and t 3 corresponds to P x y 12 23 . y
Ï3 1
5¹
t= 6
x
Q ( 21 , Ï3 2)
¹
t= 3
x ¹ t=_ 3
P ( 21 , _ Ï3 ) 2
3
4
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
3 35. t 4 and t 4 corresponds to P x y 22 22 . y
5 36. t 4 and t 4 corresponds to P x y 22 22 .
¹ 4
x
P (_ 2 , _ 2 ) Ï2
Ï2
x
5¹
t= 4
P(_ Ï2 , _ Ï2 ) 2 2
3¹
t=_ 4
5 corresponds to P x y 1 3 . 37. t and t 3 3 2 2 y
y
Q(Ï2 , Ï2 ) 2 2
5¹
t=_ 3
P ( 21 , Ï3 2)
¹
t= 3
7 38. t 6 and t 6 corresponds to P x y 23 12 . y
P(_ 2 , 2 ) Ï3 1
x
x
7¹
t=_ 6
2 corresponds to P x y 1 3 . 40. t and t 3 3 2 2
7 39. t 4 and t 4 corresponds to P x y 22 22 .
y
P (_ 21 , Ï3 2)
y
2¹
t= 3
7¹
t= 4
x x P( Ï2 , _ Ï2 ) 2 2
7 41. t 6 and t 6 corresponds to P x y 23 12 .
7 42. t 4 and t 4 corresponds to P x y 22 22 .
y
y
7¹
t= 6
Q( 2 , 2) Ï3 1
¹ 6
¹ 4
x
P(_ 2 , _2 ) Ï3
P( Ï2 , Ï2 2 2)
1
7¹
t=_ 4
x
SECTION 6.1 The Unit Circle
43. (a) t 43 3
44. (a) t 9 9 0 (b) t 54 4
(b) t 2 53 3 (c) t 76 6
(c) t 256 4 6
(d) t 35 036
(d) t 4 086
45. (a) t 57 27
46. (a) t 115 2 5
(c) t 3 0142
(c) t 2 6 0283
(d) t 2 5 1283
(d) t 7 2 0717
(b) t 79 29
(b) t 97 27
47. (a) t 34 4 2 2 (b) P 2 2
48. (a) t 54 4 2 2 (b) P 2 2
51. (a) t 2 116 6 3 1 (b) P 2 2
52. (a) t 6 (b) P 23 12
49. (a) t 56 6 3 1 (b) P 2 2
50. (a) t 43 3 3 1 (b) P 2 2
53. (a) t 134 3 4 2 2 (b) P 2 2
54. (a) t 136 2 6 3 1 (b) P 2 2
57. (a) t 4 113 3 3 1 (b) P 2 2
58. (a) t 316 5 6 3 1 (b) P 2 2 60. (a) t 10 414 4 2 2 (b) P 2 2
55. (a) t 7 416 6 3 1 (b) P 2 2
56. (a) t 174 4 4 2 2 (b) P 2 2
59. (a) t 163 5 3 3 1 (b) P 2 2
61. t 1 05 08
62. t 25
63. t 11 05 09 65. Let Q x y 35 45 be the terminal point determined by t. y
¹+t
Q( 35 , 45) ¹-t
2¹+t
t x _t
64. t 42
P 08 06
P 05 09 (a) t determines the point P x y 35 45 . (b) t determines the point P x y 35 45 . (c) t determines the point P x y 35 45 . (d) 2 t determines the point P x y 35 45 .
5
6
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
66. Let Q x y 34 47 be the terminal point determined by t. y ¹-t 4¹+t t-¹
Q( 34 , Ï7 4) t _t
x
(a) t determines the point P x y 34 47 . (b) 4 t determines the point P x y 34 47 . (c) t determines the point P x y 34 47 . (d) t determines the point P x y 34 47 .
67. The distances P Q and P R are equal because they both subtend arcs of length 3 . Since P x y is a point on the unit circle, x 2 y 2 1. Now d P Q x x2 y y2 2y and d R S x 02 y 12 x 2 y 2 2y 1 2 2y (using the fact that x 2 y 2 1). Setting these equal gives 2y 2 2y 4y 2 2 2y 4y 2 2y 2 0 2 2y 1 y 1 0. So y 1 or y 12 . Since 2 P is in quadrant I, y 12 is the only viable solution. Again using x 2 y 2 1 we have x 2 12 1 x 2 34 x 23 . Again, since P is in quadrant I the coordinates must be 23 12 . 68. P is the reflection of Q about the line y x. Since Q is the point Q 23 12 it follows that P is the point P 12 23 .
6.2
TRIGONOMETRIC FUNCTIONS OF REAL NUMBERS
1. If Px y is the terminal point on the unit circle determined by t, then sin t y, cos t x, and tan t yx.
2. If Px y is on the unit circle, then x 2 y 2 1. So for all t we have sin2 t cos2 t 1. So we can write cosine in terms of sine as cos t 1 sin2 t and we can write sine in terms of cosine as sin t 1 cos2 t.
3. Because t3 is in quadrant IV, sin t3 0. Also, sin t2 sin t1 because on the interval 0 t , the value of sin t is larger the closer t is to 2 . Thus, sin t3 sin t1 sin t2 . 4. Because t2 is in quadrant II, cos t2 0. Also, cos t1 cos t3 because on the interval 2 t 2 , the value of cos t is
larger the closer t is to 0. Thus, cos t2 cos t3 cos t1 .
6.
5. t 0
cos t 1
sin t 0
t 0
2 2
2 2
0
1
22
2 2
1
0
4
2 3 4
5 4 3 2 7 4
22
22
0 2 2
1
22
2
1
0
6 3
2 2 3 5 6
cos t
sin t
1
3 2 1 2
t
cos t
sin t
0
1
0
1 2 3 2
0
1
12
3 2 1 2
7 6 4 3 3 2 5 3 11 6
1
0
2
23
23 12
0
12
23 1
1 2 3 2
23
1
0
12
SECTION 6.2 Trigonometric Functions of Real Numbers 7. (a) sin 23 23
(b) cos 174 22
(c) tan 176 33
11. (a) cos 34 22 (b) cos 54 22 (c) cos 74 22
1 15. (a) cos 3 2 (b) sec 3 2 3 (c) sin 3 2
19. (a) csc 76 2 23 (b) sec 6 3 (c) cot 56 3
23. (a) sin 13 0
(b) cos 14 1 (c) tan 15 0
8. (a) sin 56 12
(b) cos 56 23 (c) tan 143 3
12. (a) sin 34 22 (b) sin 54 22 (c) sin 74 22
16. (a) tan 4 1 (b) csc 4 2 (c) cot 4 1 20. (a) sec 34 2 (b) cos 23 12 (c) tan 76 33
9. (a) sin 134 22
(b) cos 34 22
(c) tan 76 33
13. (a) sin 73 23 (b) csc 73 2 3 3 (c) cot 73 33 3 17. (a) cos 6 2
2 3 (b) csc 3 3 3 (c) tan 6 3
21. (a) sin 43 23
10. (a) sin 113 23 (b) cos 116 23
(c) tan 74 1
14. (a) csc 54 2 (b) sec 54 2
(c) tan 54 1 2 18. (a) sin 4 2 (b) sec 4 2 (c) cot 6 3
25. t 0 sin t 0, cos t 1, tan t 0, sec t 1, csc t and cot t are undefined. 26. t 2 sin t 1, cos t 0, csc t 1, cot t 0, tan t and sec t are undefined. 27. t
28. t 32
sin t 0, cos t 1, tan t 0, sec t 1, csc t and cot t are undefined.
sin t 1, cos t 0, csc t 1, cot t 0, tan t and sec t are undefined.
3 2 2 5 3. 9 3 4 29. 45 35 16 25 25 1. So sin t 5 , cos t 5 , and tan t 4 4 5
30.
2 3 5
22. (a) csc 23 2 3 3 (b) sec 53 2 (b) sec 116 2 3 3 3 10 1 (c) cot (c) cos 3 3 3 2 25 24. (a) sin 2 sin 2 24 1 (b) cos 252 cos 2 24 0 (c) cot 252 cot 2 24 0
4 2 9 16 1. So sin t 4 , cos t 3 , and tan t 5 4 . 45 25 25 5 5 3 3 5
1 2 3 . 3 23 2 2 23 3 3 1 1 3 1 4 4 1. So sin t 2 , cos t 2 , and tan t 3. 32. 2 2 1 2 13 2 2 13 7 13 . 13 6 33. 67 713 36 49 49 1. So sin t 7 , cos t 7 , and tan t 6 6 7 9 2 2 9 1600 81 1. So sin t 9 , cos t 40 , and tan t 41 9 . 34. 40 41 41 1681 1681 41 41 40 40 41 12 5 2 12 2 25 144 1. So sin t 12 , cos t 5 , and tan t 13 12 . 35. 13 13 169 169 13 13 5 5 13 2 5 2 2 5 20 1. So sin t 2 5 , cos t 5 , and tan t 5 2. 36. 55 2 5 5 25 25 5 5 5 5 2 2 31. 23 12 34 14 1. So sin t 12 , cos t 23 , and tan t
7
8
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
21 2 2 21 400 441 1. So sin t 21 , cos t 20 , and tan t 29 21 . 37. 20 29 29 841 841 29 29 20 20 29
38.
24 25
2
7 7 2 576 49 1. So sin t 7 , cos t 24 , and tan t 25 7 . 25 625 625 25 25 24 24
39. (a) 08 (b) 084147 43. (a) 10 (b) 102964
25
40. (a) 07 (b) 069671 44. (a) 36 (b) 360210
41. (a) 09 (b) 093204 45. (a) 06 (b) 057482
42. (a) 03 (b) 028366 46. (a) 09 (b) 088345
47. sin t cos t. Since sin t is positive in quadrant IV and cos t is negative in quadrant IV, their product is negative. 48. sin t tan t is positive in quadrant IV because both sin t and tan t are negative in that quadrant. 1 tan t sin t tan t sin t tan t tan t sin t tan2 t sin t. Since tan2 t is always positive and sin t is negative in 49. cot t cot t quadrant III, the expression is negative in quadrant III. 1 1, provided cos t 0. 50. cos t sec t is positive in any quadrant, since cos t sec t cos t cos t 51. Quadrant II
52. Quadrant III
53. Quadrant II 54. Quadrant II 55. Because cosine is negative in quadrant III, cos t 1 sin2 t in that quadrant. 56. Because sine is negative in quadrant IV, sin t 1 cos2 t in that quadrant. 57. Because sine is positive in quadrant II, sin t 1 cos2 t in that quadrant. 1 cos2 t in that quadrant. 58. Because tangent is negative and cosine is positive in quadrant IV, tan t cos t 1 cos2 t 59. Because tangent and cosine are both negative in quadrant II, tan t in that quadrant. cos t sin t in that quadrant. 60. Because tangent is negative and sine is positive in quadrant II, tan t 1 sin2 t 61. Because tangent is negative in quadrant IV, tan t sec2 t 1 in that quadrant. 62. Because secant is positive in quadrant IV, sec t 1 tan2 t in that quadrant. 63. Because cosecant is positive in quadrant II, csc t 1 cot2 t in that quadrant. 1 sec2 t 1 in that 64. Because sine and secant are both negative in quadrant III, sin t 1 cos2 t 1 sec t sec2 t quadrant. sin2 t sin2 t . (Because all trigonometric functions in this expression are squared, the quadrant is 2 cos t 1 sin2 t immaterial.) 1 1 1 cos2 t 1. (Because all trigonometric functions in this expression are squared, the 66. sec2 t sin2 t 2 cos t cos2 t quadrant is immaterial.) 67. sin t 45 and the terminal point of t is in quadrant IV, so the terminal point determined by t is P x 45 . Since P is 2 9 3 on the unit circle, x 2 45 1. Solving for x gives x 1 16 25 25 5 . Since the terminal point is in quadrant IV, x 35 . Thus the terminal point is P 35 45 . Thus, cos t 35 , tan t 43 , csc t 54 , sec t 53 , 65. tan2 t
cot t 34 .
SECTION 6.2 Trigonometric Functions of Real Numbers
9
7 and the terminal point of t lies in quadrant III, so the terminal point determined by t is P 7 y . Since 68. cos t 25 25 2 7 49 576 24 . Since the terminal y 2 1. Solving for y gives x 1 625 P is on the unit circle, 25 625 25 7 24 24 24 25 point is in quadrant III, y 24 25 . Thus the terminal point is P 25 25 . Thus, sin t 25 , tan t 7 , csc t 24 , 7 sec t 25 7 , cot t 24 .
69. sec t 3 and the terminal point of t lies in quadrant IV. Thus, cos t 13 and the terminal point determined by t is P 13 y . 2 Since P is on the unit circle, 13 y 2 1. Solving for y gives y 1 19 89 2 3 2 . Since the terminal point is in quadrant IV, y 2 3 2 . Thus the terminal point is P 13 2 3 2 . Therefore, sin t 2 3 2 , cos t 13 , 3 3 2 , cot t 1 2. tan t 2 2, csc t 4 4 2 2
2 2
2 1 1 17 . 70. tan t 14 and the terminal point of t lies in quadrant III. Since sec2 t tan2 t 1 we have sec2 t 14 1 16 16 17 17 Thus sec t 17 16 4 . Since sec t 0 in quadrant III we have sec t 4 , so 1 4 1 4 17 1 1717 . . Since tan t cos t sin t we have sin t 14 4 cos t 17 17 sec t 17 17 417 Thus, the terminal point determined by t is P 4 1717 1717 . Therefore, sin t 1717 , cos t 4 1717 , csc t 17,
sec t 417 , cot t 4.
12 2 1 169 , so because 2 2 2 71. tan t 12 5 and sin t 0, so t is in quadrant II. Since sec t tan t 1 we have sec t 5 25 1 13 5 5 secant is negative in quadrant II, sec t 169 25 5 . Thus, cos t sec t 13 and we have P 13 y . Since 5 12 . Thus, the terminal point determined by t is P 5 12 , and so 13 tan t cos t sin t we have sin t 12 5 13 13 13 5 13 13 5 sin t 12 13 , cos t 13 , csc t 12 , sec t 5 , cot t 12 .
72. csc t 5 and cos t 0, so t is in quadrant II. Thus, sin t 15 and the terminal point determined by t is P x 15 . Since 2 1 2 6 . Since the terminal point is in P is on the unit circle, x 2 15 1. Solving for x gives x 1 25 5 2 6 2 6 1 1 1 6, quadrant II, x 5 and the terminal point is P 5 5 . Thus, sin t 5 , cos t 2 5 6 , tan t 12 2 6 5 6 5 sec t 12 , cot t 2 6. 2 6
73. sin t 14 , sec t 0, so t is in quadrant III. So the terminal point determined by t is P x 14 . Since P is on the unit 2 1 15 15 . Since the terminal point is in quadrant III, circle, x 2 14 1. Solving for x gives x 1 16 16 4 15 . Thus, the terminal point determined by t is P 415 14 , and so cos t 415 , tan t 1 1515 , x 15 4 csc t 4, sec t 4 4 1515 , cot t 15. 15
10
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
74. tan t 4 and the terminal point of t lies in quadrant II. Since sec2 t tan2 t 1 we have sec2 t 42 1 161 17. 1 17 1 . Since tan t cos t sin t Thus sec t 17. Since sec t 0, we have sec t 17and cos t 17 sec t 17 we have sin t 4 1717 4 1717 . Thus, the terminal point determined by t is P 1717 4 1717 . Thus, sin t 4 1717 , cos t 1717 , csc t 417 , sec t 17, cot t 14 . For Exercises 75–82, many answers are possible. 75. If f x x 2 and g x cos x, then F x f g x cos x2 cos2 x. 76. If f x e x and g x sin x, then F x f g x esin x . 77. If f x x and g x 1 tan x, then F x f g x 1 tan x. 78. If f x
x sin x and g x sin x, then F x f g x . 1x 1 sin x
2 79. If f x e x , g x x 2 , and h x sin x, then F x f g h x f sin2 x esin x . 80. If f x sin x, g x
x, and h x ln x, then F x f g h x f
ln x sin ln x .
81. If f x ln x, g x x 2 , and h x cos x, then F x f g h x f cos2 x ln cos2 x . 82. If f x sin x, g x
x , and h x e x , then F x f g h x f 1x
ex 1 ex
sin
ex . 1 ex
83. f x x2 sin x x 2 sin x f x, so f is odd. 84. f x x2 cos 2 x x 2 cos 2x f x, so f is even. 85. f x sin x cos x sin x cos x f x, so f is odd. 86. f x sin x cos x sin x cos x, which is neither f x nor f x, so f is neither even nor odd. 87. f x x cos x x cos x f x, so f is even. 88. f x x sin3 x x [sin x]3 x sin x3 x sin3 x f x, so f is even. 89. f x x3 cos x x 3 cos x, which is neither f x nor f x, so f is neither even nor odd. 90. f x cos sin x cos sin x cos sin x f x, so f is even. 91. t
0
025
y t 4 283
050 075 100 125 0
92. (a) B 6 80 7 sin 2 87 mmHG
(b) B 105 80 7 sin 105 12 827 mmHG
283 4 283
(c) B 12 80 7 sin 80 mmHG
7 3 (d) B 20 80 7 sin 20 12 80 2 739 mmHG
93. (a) I 01 08e03 sin 1 0499 A (b) I 05 08e15 sin 5 0171 A
94. t
0
1
2
4
6
8
12
H t 175 1504 100 386 100 1503 588
SECTION 6.3 Trigonometric Graphs
95. We see from the diagram that every summand in
y
2 100 101 102 200 sin sin sin sin sin sin 100 100 100 100 100 100 has a corresponding term whose terminal point is diametrically opposed to its own on the unit circle. But for each of these pairs of terms, n n sin n 100 n sin sin sin 0, and so the 100 100 100 100 pair sums to 0.
11
k¹
t=100
1
k¹ sin 100
sin (k+100)¹ 100
1
0
x
(k+100)¹ 100
t=
Algebraically: 2 100 101 102 200 sin sin sin sin sin sin 100 100 100 100 100 100 2 2 sin sin sin [sin x sin 2x] sin 100 100 100 100 0 0 0 0 96. Notice that if P t x y, then P t x y. Thus, (a) sin t y and sin t y. Therefore, sin t sin t.
(b) cos t x and cos t x. Therefore, cos t cos t. y y sin t sin t tan t. (c) tan t cos t x x cos t 97. To prove that AO B C DO, first note that O B O D 1 and O AB OC D 2 . Now C O D AO B 2
C O D AO B AB O. Since we know two angles and one side to be 2
equal, the triangles are (SAA) congruent. Thus AB OC and O A C D, so if B has coordinates x y, then D has coordinates y x. Therefore, (a) sin t 2 x cos t (b) cos t 2 y, and sin t y. Therefore, cos t 2 sin t. x cos t x (c) tan t 2 y y sin t cot t
6.3
TRIGONOMETRIC GRAPHS
1. If a function f is periodic with period p, then f t p f t for every t. The trigonometric functions y sin x and cos x are periodic, with period 2 and amplitude 1. y
y
1
1
2¹ ¹
_1
2¹ x
¹
x
_1
2. To obtain the graph of y 5 sin x, we start with the graph of y sin x, then shift it 5 units upward. To obtain the graph of y cos x, we start with the graph of y cos x, then reflect it in the xaxis.
12
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
3. (a) The sine and cosine curves y a sin kx and y a cos kx, k 0, have amplitude a and period 2k. The sine
(b)
y 3
curve y 3 sin 2x has amplitude 3 3 and period
0
22 ; an appropriate interval on which to graph one period is [0 ].
¹ 4
¹ 2
3¹ 4
_3
(b) y
4. (a) The sine curve y a sin k x b has amplitude a,
period 2k, and horizontal shift b. The sine curve 2 y 4 sin 3 x 6 has amplitude 4 4, period 3 ,
4 2
and horizontal shift 6 ; an appropriate interval on which to 5 . graph one period is 6 6
5. f x 2 sin x has domain and range [1 3]. y
0
_2
¹ 6
¹ 3
¹ 2
2¹
y 1
2¹
x
x _2
7. f x sin x has domain and range [1 1]. y
8. f x 2 cos x has domain and range [1 3]. y
1 ¹
x
2¹
1 ¹
9. f x 2 sin x has domain and range [3 1]. y
¹
2¹
x
5¹ 6
6. f x 2 cos x has domain and range [3 1].
¹
¹
2¹ 3
_4
2
_1
x
¹
2¹
x
10. f x 1 cos x has domain and range [2 0]. y 1
¹
2¹
x
x
SECTION 6.3 Trigonometric Graphs
11. g x 3 cos x has domain and range [3 3]. y
2
12. g x 2 sin x has domain and range [2 2]. y
2 ¹
2¹
x
13. g x 12 sin x has domain and range 12 12 . y
1
¹
2¹
x
14. g x 23 cos x has domain and range 23 23 . y
1 ¹
2¹
x
x
15. g x 3 3 cos x has domain and range [0 6]. y
2
¹
2¹
16. g x 4 2 sin x has domain and range [2 6]. y
2 ¹
2¹
x
17. h x cos x has domain and range [0 1]. y
1
¹
2¹
x
18. h x sin x has domain and range [0 1]. y
1 ¹
2¹
x
¹
2¹
x
13
14
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
19. y cos 2x has amplitude 1 and period . An appropriate interval on
which to sketch one complete period is [0 ]. To find the key points on
y 1
this interval we divide the interval into four subintervals, each of length 2 1 . Now, we start at the left endpoint of the interval (x 0) 2 4 4
and successively add 4 to obtain the xcoordinates of the key points. We 3 0, and y 1. calculate y 0 1, y 4 0, y 2 1, y 4 20. y sin 2x has amplitude 1 and period . An appropriate interval on
which to sketch one complete period is [0 ]. To find the key points on
¹ 2
¹ x
¹ 2
¹ x
¹ 4
¹ x 2
_1
y 1
this interval we divide the interval into four subintervals, each of length 2 1 . Now, we start at the left endpoint of the interval (x 0) 2 4 4
and successively add 4 to obtain the xcoordinates of the key points. We 3 1, and y 0. calculate y 0 0, y 4 1, y 2 0, y 4 21. y cos 4x has amplitude 1 and period 2 . An appropriate interval on which to sketch one complete period is 0 2 . To find the key points on
_1
y 1
this interval we divide the interval into four subintervals, each of length 2 1 . Now, we start at the left endpoint of the interval (x 0) 4 4 8
and successively add 8 to obtain the xcoordinates of the key points. We 3 0, and 0, y 1, y calculate y 0 1, y 8 4 8 y 2 1. 22. y sin x has amplitude 1 and period 2. An appropriate interval on
which to sketch one complete period is [0 2]. To find the key points on
_1
y 1
this interval we divide the interval into four subintervals, each of length 2 1 1 . Now, we start at the left endpoint of the interval (x 0) and 4 2 successively add 12 to obtain the xcoordinates of the key points. We
calculate y 0 0, y 12 1, y 1 0, y 32 1, and y 2 0.
23. y 3 sin 2x has amplitude 3 and period 1. An appropriate interval on which to sketch one complete period is [0 1]. To find the key points on
2 1
x
1 2
1 x
_1
y 3
this interval we divide the interval into four subintervals, each of length 2 1 1 . Now, we start at the left endpoint of the interval (x 0) and 2 4 4 successively add 14 to obtain the xcoordinates of the key points. We
calculate y 0 0, y 14 3, y 12 0, y 34 3, and y 1 0.
_3
SECTION 6.3 Trigonometric Graphs
24. y 2 cos 8x has amplitude 2 and period 4 . An appropriate interval on which to sketch one complete period is 0 4 . To find the key points on
y 2
this interval we divide the interval into four subintervals, each of length
¹ 8
1 . Now, we start at the left endpoint of the interval (x 0) and 4
4
16
to obtain the xcoordinates of the key points. We successively add 16 3 0, y calculate y 0 2, y 16 8 2, y 16 0, and y 4 2.
25. y 10 sin 12 x has amplitude 10 and period 4. An appropriate interval on which to sketch one complete period is [0 4]. To find the key points on
_2
y 10
this interval we divide the interval into four subintervals, each of length
4¹
2 1 12 4 . Now, we start at the left endpoint of the interval (x 0) and
successively add to obtain the xcoordinates of the key points. We calculate y 0 0, y 10, y 2 0, y 3 10, and
¹ x 4
2¹
x
4¹
8¹ x
3¹
6¹ x
¹ 2
¹ x
_10
y 4 0.
26. y 5 cos 14 x has amplitude 5 and period 8. An appropriate interval on
which to sketch one complete period is [0 8]. To find the key points on
y 5
this interval we divide the interval into four subintervals, each of length 2 1 14 4 2. Now, we start at the left endpoint of the interval (x 0)
and successively add 2 to obtain the xcoordinates of the key points. We calculate y 0 5, y 2 0, y 4 5, y 6 0, and
_5
y 8 5.
27. y 13 cos 13 x has amplitude 13 and period 6. An appropriate interval
on which to sketch one complete period is [0 6]. To find the key points
y 1/3
on this interval we divide the interval into four subintervals, each of length 2 1 3 13 4 2 . Now, we start at the left endpoint of the interval (x 0)
and successively add 32 to obtain the xcoordinates of the key points. We calculate y 0 13 , y 32 0, y 3 13 , y 92 0, and
_1/3
y 6 13 .
28. y 4 sin 2x has amplitude 4 and period . An appropriate interval on which to sketch one complete period is [0 2]. To find the key points on
y 4
this interval we divide the interval into four subintervals, each of length 2 1 . Now, we start at the left endpoint of the interval (x 0) 2 4 4
and successively add 4 to obtain the xcoordinates of the key points. We 3 4, and y 0. calculate y 0 0, y 4 4, y 2 0, y 4
_4
15
16
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
29. y 2 sin 8x has amplitude 2 and period 14 . An appropriate interval on which to sketch one complete period is 0 14 . To find the key points on
y 2
this interval we divide the interval into four subintervals, each of length
2 1 1 . Now, we start at the left endpoint of the interval (x 0) 8 4 16 1 to obtain the xcoordinates of the key points. We and successively add 16
1 2, y 1 0, y 3 2, and calculate y 0 0, y 16 8 16 y 14 0.
30. y 3 sin 4x has amplitude 3 and period 12 . An appropriate interval on which to sketch one complete period is 0 12 . To find the key points on
1 8
1 x 4
1 4
1 x 2
¹ 3
2¹ 3
_2
y 3
this interval we divide the interval into four subintervals, each of length
2 1 1 . Now, we start at the left endpoint of the interval (x 0) and 4 4 8 successively add 18 to obtain the xcoordinates of the key points. We
_3
calculate y 0 0, y 18 3, y 14 0, y 38 3, and y 12 0. 31. y 2 sin 3x has amplitude 2 and period 23 . An appropriate interval on which to sketch one complete period is 0 23 . To find the key points on
y 2
this interval we divide the interval into four subintervals, each of length
x
2 1 . Now, we start at the left endpoint of the interval (x 0) 3 4 6
and successively add 6 to obtain the xcoordinates of the key points. We calculate y 0 0, y 6 2, y 3 0, y 2 2, and y 23 0. 32. y 4 cos 6x has amplitude 4 and period 3 . An appropriate interval on which to sketch one complete period is 0 3 . To find the key points on
_2
y 4
this interval we divide the interval into four subintervals, each of length
2 1 . Now, we start at the left endpoint of the interval (x 0) 6 4 12
to obtain the xcoordinates of the key points. We and successively add 12 calculate y 0 4, y 12 0, y 6 4, y 4 0, and y 3 4.
¹ 6
¹ x 3
1
2 x
_4
33. y 1 12 cos x has amplitude 12 and period 2. An appropriate interval
y
this interval we divide the interval into four subintervals, each of length
1
on which to sketch one complete period is [0 2]. To find the key points on
2 1 1 . Now, we start at the left endpoint of the interval (x 0) and 4 2 successively add 12 to obtain the xcoordinates of the key points. We
calculate y 0 32 , y 12 1, y 1 12 , y 32 1, and y 2 32 .
17
SECTION 6.3 Trigonometric Graphs y
34. y 2 cos 4x has amplitude 1 and period 12 . An appropriate interval on which to sketch one complete period is 0 12 . To find the key points
0 _1
on this interval we divide the interval into four subintervals, each of length
2 1 1 . Now, we start at the left endpoint of the interval (x 0) and 4 4 8 successively add 18 to obtain the xcoordinates of the key points. We
calculate y 0 1, y y 12 1.
1 8
2, y
1 4
3, y
3 8
1 x 2
1 4
_2 _3
2, and
35. y cos x 2 has amplitude 1, period 2, and horizontal shift 2 . An 5 appropriate interval on which to sketch one complete period is 2 2 . To find
y 1
the key points on this interval we divide the interval into four subintervals, each of
¹
length 24 2 . Now, we start at the left endpoint of the interval (x 2 ) and
successively add 2 to obtain the xcoordinates of the key points. We calculate 3 1, y 2 0, and y 5 1. y 2 1, y 0, y 2 2
36. y 2 sin x 3 has amplitude 2, period 2, and horizontal shift 3 . An 7 appropriate interval on which to sketch one complete period is 3 3 . To find
y 2 4¹ 3
the key points on this interval we divide the interval into four subintervals, each of
¹ 3
length 24 2 . Now, we start at the left endpoint of the interval (x 3 ) and
successively add 3 to obtain the xcoordinates of the key points. We calculate y 3 0, y 56 2, y 43 0, y 116 2, and y 73 0.
37. y 2 sin x 6 has amplitude 2, period 2, and horizontal shift 6 . An 13 appropriate interval on which to sketch one complete period is 6 6 . To find
¹ 6
successively add 2 to obtain the xcoordinates of the key points. We calculate y 4 3, y 4 0, y 34 3, y 54 0, and y 74 3.
7¹ 6
¹
successively add 2 to obtain the xcoordinates of the key points. We calculate y 6 0, y 23 2, y 76 0, y 53 2, and y 136 0.
length 24 2 . Now, we start at the left endpoint of the interval (x 4 ) and
x
2¹
y
length 24 2 . Now, we start at the left endpoint of the interval (x 6 ) and
the key points on this interval we divide the interval into four subintervals, each of
¹
7¹ 3
2
the key points on this interval we divide the interval into four subintervals, each of
38. y 3 cos x 4 has amplitude 3, period 2, and horizontal shift 4 . An 7 . To find appropriate interval on which to sketch one complete period is 4 4
x
2¹
13¹ 6
2¹
x
y 3 _
¹ 4
3¹ 4
7¹ 4
¹
2¹
x
18
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
39. y 4 sin x 12 has amplitude 4, period 2, and horizontal shift 12 . An appropriate interval on which to sketch one complete period is 12 52 . To find the
y 4 2
key points on this interval we divide the interval into four subintervals, each of
length 2 14 12 . Now, we start at the left endpoint of the interval (x 12 ) and
successively add 12 to obtain the xcoordinates of the key points. We calculate y 12 0, y 1 4, y 32 0, y 2 4, and y 52 0.
0
x
2
1
_2 _4
40. y 2 cos x 14 has amplitude 2, period 2, and horizontal shift 14 . An appropriate interval on which to sketch one complete period is 14 74 . To find
y 1
the key points on this interval we divide the interval into four subintervals, each of
0 _1
length 2 14 12 . Now, we start at the left endpoint of the interval (x 14 ) and successively add 12 to obtain the xcoordinates of the key points. We calculate
x
1
y 14 2, y 14 0, y 34 2, y 54 0, and y 74 2.
41. y 2 cos 4 x 4 has amplitude 2, period 2 , and horizontal shift 4 . An 3 appropriate interval on which to sketch one complete period is 4 4 . To find
the key points on this interval we divide the interval into four subintervals, each of length 24 14 8 . Now, we start at the left endpoint of the interval (x 4 ) and
successively add 8 to obtain the xcoordinates of the key points. We calculate 3 0, y 2, y 5 0, and y 3 2. y 4 2, y 8 2 8 4
2 42. y 3 cos 3 x 3 has amplitude 3, period 3 , and horizontal shift 3 . An appropriate interval on which to sketch one complete period is 3 3 . To find
y
2 1 0 _1
¹ 4
y 2 _¹ 3
and successively add 6 to obtain the xcoordinates of the key points. We calculate y 3 3, y 6 0, y 0 3, y 6 0, and y 3 3.
43. y cos 2x cos 2 x 2 has amplitude 1, period , and horizontal shift 2 . An appropriate interval on which to sketch one complete period is 2 2 .
x
3¹ 4
_2
the key points on this interval we divide the interval into four subintervals, each of length 23 14 6 . Now, we start at the left endpoint of the interval (x 3 )
¹ 2
0 _2
¹ 3
x
y 1
To find the key points on this interval we divide the interval into four subintervals, each of length 22 14 4 . Now, we start at the left endpoint of the interval
(x 2 ) and successively add 4 to obtain the xcoordinates of the key points. We calculate y 2 1, y 4 0, y 0 1, y 4 0, and y 2 1.
¹
_2
0
¹ x 2
19
SECTION 6.3 Trigonometric Graphs
2 44. y sin 3x 2 sin 3 x 6 has amplitude 1, period 3 , and horizontal
shift 6 . An appropriate interval on which to sketch one complete period is 5 . To find the key points on this interval we divide the interval into four 6 6
subintervals, each of length 23 14 6 . Now, we start at the left endpoint of the
interval (x 6 ) and successively add 6 to obtain the xcoordinates of the key 2 1, and points. We calculate y 6 0, y 3 1, y 2 0, y 3 y 56 0.
2 45. y 2 sin 23 x 6 2 sin 3 x 4 has amplitude 2, period 3, and
horizontal shift 4 . An appropriate interval on which to sketch one complete period 13 is 4 4 . To find the key points on this interval we divide the interval into four
y 1 5¹ 6
0
¹ 6
x
¹ 2
_1
y 2 7¹ 4
¹ 4
2 1 3 . Now, we start at the left endpoint of subintervals, each of length 23 4 4
13¹ 4
x
3 the interval (x 4 ) and successively add 4 to obtain the xcoordinates of the 7 0, y 5 2, and key points. We calculate y 4 0, y 2, y 4 2 y 134 0.
2 46. y 5 cos 3x 4 5 cos 3 x 12 has amplitude 5, period 3 , and
y 5
. An appropriate interval on which to sketch one complete horizontal shift 12 3 . To find the key points on this interval we divide the interval period is 12 4
¹ 12
into four subintervals, each of length 23 14 6 . Now, we start at the left
5¹ 12
3¹ 4
x
¹ 3
x
) and successively add to obtain the endpoint of the interval (x 12 6 xcoordinates of the key points. We calculate y 12 5, y 4 0, 5, y 7 0, and y 3 5. y 512 12 4
2 47. y 2 2 cos 3 x 3 has amplitude 2, period 3 , and horizontal shift 3 . An appropriate interval on which to sketch one complete period is 3 3 . To find
y
the key points on this interval we divide the interval into four subintervals, each of
2
length 23 14 6 . Now, we start at the left endpoint of the interval (x 3 )
and successively add 6 to obtain the xcoordinates of the key points. We calculate y 3 0, y 6 2, y 0 4, y 6 2, and y 3 0.
¹
_3
0
20
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
48. y 3 sin 2 x 18 has amplitude 1, period 1, and horizontal shift 18 . An appropriate interval on which to sketch one complete period is 18 78 . To find
y
the key points on this interval we divide the interval into four subintervals, each of 1 1 1 length 22 4 4 . Now, we start at the left endpoint of the interval (x 8 )
and successively add 14 to obtain the xcoordinates of the key points. We calculate y 18 3, y 18 4, y 38 3, y 58 2, and y 78 3.
1 1 1 has amplitude 1 , period 1, 49. y 12 12 cos 2x cos 2 x 3 2 2 6 2
and horizontal shift 16 . An appropriate interval on which to sketch one complete period is 16 76 . To find the key points on this interval we divide the interval into
1 1 four subintervals, each of length 22 4 4 . Now, we start at the left endpoint of
the interval (x 16 ) and successively add 14 to obtain the xcoordinates of the key 5 1 , y 2 1, y 11 1 , and points. We calculate y 16 0, y 12 2 3 12 2 7 y 6 0.
1 _81
0
7 1 x 8
3 8
y
1
0
1 6
2 50. y 1 cos 3x 2 1 cos 3 x 6 has amplitude 1, period 3 , and
2 3
1
7 6
x
y 2
horizontal shift 6 . An appropriate interval on which to sketch one complete period is 6 2 . To find the key points on this interval we divide the interval
1
into four subintervals, each of length 23 14 6 . Now, we start at the left
endpoint of the interval (x 6 ) and successively add 6 to obtain the xcoordinates of the key points. We calculate y 6 2, y 0 1, y 6 0, y 3 1, and y 2 2.
¹
_6
¹ 6
¹ 2
x
51. This function has amplitude a 4 and period 2k 2. As a sine curve its horizontal shift is bs 0, and as a cosine curve it is bc 2 . Equations are y a sin k x bs 4 sin x and y a cos k x bc 4 cos x 2 . 52. This curve has amplitude a 2 and period 2k . As a cosine curve its horizontal shift is bc 0, and as a sine curve it is bs 4 . Equations are y a cos k x bc 2 cos 2x and y a sin k x bs 2 sin 2 x 4 .
53. This curve has amplitude a 32 and period 2k 23 . As a cosine curve its horizontal shift is bc 0, and as a sine curve it 3 3 is bs 6 . Equations are y a cos k x bc 2 cos 3x and y a sin k x bs 2 sin 3 x 6 .
54. This curve has amplitude a 3 and period 2k 4. As a sine curve its horizontal shift is bs 0, and as a cosine curve it is bc . Equations are y 3 sin 12 x and y 3 cos 12 x .
pi/6
, and as a cosine 55. This curve has amplitude a 12 and period 2k . As a sine curve its horizontal shift is bs 712 - pi/6 1 7 1 curve it is bc 3 . Equations are y 2 sin 2 x 12 and y 2 cos 2 x 3 .
56. This curve has amplitude a 2 and period 2k . As a sine curve its horizontal shift is bs 0, and as a cosine curve it is bc 4 . Equations are y 1 2 sin 2x and y 1 2 cos 2 x 4 .
-
Since -7pi/12 is the horizontal shift for the reflected sine curve, lets give these two curves as alternative solutions and instead use -pi/12 for the horizontal shift of the sine curve and pi/6 for the horizontal shift of the cosine curve.
+
SECTION 6.3 Trigonometric Graphs
21
57. This curve has amplitude a 1 and period 2k . As a sine curve its horizontal shift is bs 2 , and as a cosine curve it 3 is bc 34 . Equations are y 1 sin 2 x 2 and y 1 cos 2 x 4 .
3/4
58. This curve has amplitude a 4 and period 2k 2. As a sine curve its horizontal shift is bs 14 , and as a cosine curve it is Since -pi/4 is the horizontal - 3/4 bc 14 . Equations are y 4 4 sin x 14 and y 4 4 cos x 14 .
+
59. f x cos 100x, [01 01] by [15 15]
60. f x 3 sin 120x, [01 01] by [4 4]
1
0.1
0.1
0.1
0.1
1
61. f x sin
x , [250 250] by [15 15] 40
62. f x cos
1
1
200
x , [0 500] by [1 1] 80
0
200 1
200
1
63. y tan 25x, [02 02] by [3 3]
64. y csc 40x, [01 01] by [10 10] 10
2
0.2
400
0.2
0.1
2
0.1 10
65. y sin2 20x, [05 05] by [02 12]
66. y
cos 10x, [04 04] by [01 15]
1.0 1 0.5
0.5
0.5
0.4
0.2
0.0
0.2
0.4
shift for the reflected cossine curve, lets give this as an alternative solution and instead use 3/4 for the horizontal shift of the cosine curve.
22
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
67. f x x, g x sin x
68. f x sin x, g x sin 2x
6
g 6
f+g
2 4
2
2
f
4
0 2
2
4
6
6
f
g 1
4
2
4
f+g
0
2
4
6
1
6
2
69. f x sin 3x, g x cos 12 x
_5
70. f x 05 sin 5x, g x cos 2x.
2
f+g
2
1
g
1
0 _1 _2
5
f+g f
_2
2 _1
f
g
_2
71. y x 2 sin x is a sine curve that lies between the graphs of y x 2 and y x 2 .
72. y x cos x is a cosine curve that lies between the graphs of y x and y x.
200
5
10
5
10
5 5
200
73. y
x sin 5x is a sine curve that lies between the graphs of y x and y x. 2
cos 2x is a cosine curve that lies between the 1 x2 1 1 and y . graphs of y 1 x2 1 x2
74. y
1 2 2
4 5
5 1
SECTION 6.3 Trigonometric Graphs
23
75. y cos 3x cos 21x is a cosine curve that lies between
76. y sin 2x sin 10x is a sine curve that lies between the
1
1
the graphs of y cos 3x and y cos 3x.
0.5
graphs of y sin 2x and y sin 2x.
0.5
0.5
1
0.5 1
77. y sin x sin 2x. The period is 2, so we graph the
function over one period, . Maximum value 176 when x 094 2n, minimum value 176 when x 094 2n, n any integer.
78. y x 2 sin x, 0 x 2. Maximum value 697 when x 524, minimum value 068 when x 105. 5
2 0 2
2
2
4
6
2
79. y 2 sin x sin2 x. The period is 2, so we graph the
function over one period, . Maximum value 300 when x 157 2n, minimum value 100 when x 157 2n, n any integer.
cos x . The period is 2, so we graph the function 2 sin x over one period. Maximum value 058 when
80. y
x 576 2n (exact value x 116 2n); Minimum
value 058 when x 367 2n (exact value x 76 2n) for any integer n.
2
1 2
2 0
2
2
4
6
1
81. cos x 04, x [0 ]. The solution is x 116.
82. tan x 2, x [0 ]. The solution is x 111.
1 2 0
1
2
3 0
1
0
1
2
3
24
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
83. csc x 3, x [0 ]. The solutions are x 034, 280.
84. cos x x, x [0 ]. The solution is x 074.
4
2
2 0
85. f x
0 0
1
2
1
3
2
3
1 cos x x
(a) Since f x is odd.
1 cos x 1 cos x f x, the function x x
(c)
1
(b) The function is undefined at x 0, so the xintercepts occur when
20
1 cos x 0, x 0 cos x 1, x 0 x 2, 4, 6,
20 1
(d) As x , f x 0. (e) As x 0, f x 0.
86. f x
sin 4x 2x
(a) f x
sin 4x sin 4x f x, so the function is even. 2 x 2x
(b) The xintercepts occur when sin 4x 0, x 0 4x n x 14 n, n any nonzero integer.
(c) 2
1
(d) As x , f x 0.
2
(e) As x 0, f x 2.
2
2 20 seconds. 10 (b) Since h 0 3 and h 10 3, the wave height is 3 3 6 feet.
87. (a) The period of the wave is
88. (a) The period of the vibration is
1 2 second. 880 440
1 of a second, there are 440 vibrations (b) Since each vibration takes 440 per second.
(c)
v 1
0
_1
0.002 0.004 0.006 0.008
t
SECTION 6.3 Trigonometric Graphs
89. (a) The period of p is
1 2 minute. 160 80
(c)
25
p 160
(b) Since each period represents a heart beat, there are 80 heart beats per
140
minute.
120 115
(d) The maximum (or systolic) is 115 25 140 and the minimum (or
100 90 80
diastolic) is 115 25 90. The read would be 14090 mmHg, which is higher than normal.
90. (a) The period of R Leonis is
0
2 312 days. 156
(c)
2 t (s)
1
b 10
(b) The maximum brightness is 79 21 10; the minimum brightness
8 6
is 79 21 58.
4 2 0
x are the points of intersection of the graphs of 91. The solutions to sin x 100
200
400
600
t
1
x . We first find the number of positive solutions. Since y sin x and y 100
x 1 sin x 1, the yvalues of the intersection points must satisfy 100
20
x 100. We can use a graphing device to see that the graphs intersect at
exactly two points on each of the intervals [0 ], [2 3], and [4 5]. This
20 1
pattern holds throughout [0 100], so because 31 100 32, the interval [30 31] is the rightmost interval with two solutions. Thus, there are 32 nonnegative solutions, including x 0, and so by symmetry, there are 31 negative solutions. Therefore, the given equation has 63 solutions on [100 100]. 92. (a) y sin x . This graph looks like a sine function (b) y sin x 2 . This graph looks like a graph of sin x which which has been stretched horizontally (stretched
more for larger values of x). It is defined only for x 0, so it is neither even nor odd.
has been shrunk for x 1 (shrunk more for larger values of x) and stretched for x 1. It is an even function, whereas
sin x is odd.
1 0 1
93. (a) y
1 200
400
5
5 1
1 is periodic with period 2, and it is neither even nor odd. y 0 12 , and it has graph III. 2 sin x
(b) y esin 2x is periodic with period , and it is and neither even nor odd. y 0 e0 1, and it has graph I. sin x (c) y is aperiodic and odd. It has graph IV. 1 x 1 is aperiodic and even. It has graph II. (d) y cos 1 x2
26
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
6.4
MORE TRIGONOMETRIC GRAPHS
1. The trigonometric function y tan x has period and asymptotes x n, n an integer. 2
2. The trigonometric function y csc x has period 2 and asymptotes x n, n an integer. y
y 10
5
¹/2 x
_¹/2
0
_¹
_10
¹ x
_5
3 3. f x tan x 4 corresponds to Graph II. f is undefined at x 4 and x 4 , and Graph II has the shape of a graph of a tangent function. 3 4. f x sec 2x corresponds to Graph III. f is undefined at x 4 and x 4 , and Graph III has the shape of a graph of a secant function.
5. f x cot 4x corresponds to Graph VI.
6. f x tan x corresponds to Graph I.
7. f x 2 sec x corresponds to Graph IV.
8. f x 1 csc x corresponds to Graph V.
9. y 4 tan x has period .
_2¹
3
10. y 3 tan x has period .
y 10
0
_¹
_2¹ ¹
_¹
x
_10
0
¹
x
¹
x
_10
11. y 32 tan x has period .
12. y 34 tan x has period . y 5
_2¹
y 10
y 4 2
_¹
0
¹
x
_2¹
0
_¹
_2 _5
_4
SECTION 6.4 More Trigonometric Graphs
13. y cot x has period .
27
14. y 2 cot x has period .
y
y 5
4
_2¹
0
_¹
_¹
x
¹
x
¹
_5
15. y 2 csc x has period 2.
16. y 12 csc x has period 2.
y
y
5
1
_¹
¹
17. y 3 sec x has period 2.
x
_¹
y
y
5
_¹
¹
x
_¹
x
19. y tan 3x has period 3 . An appropriate interval is 6 6 . The graph has xintercepts where 3x n x 3 n, n an integer; and vertical asymptotes where
3x 2 n x 6 3 n, n an integer. y
¹
20. y tan 4x has period 4 . An appropriate interval is 8 8 . The graph has xintercepts where 4x n x 2 n, n an integer; and vertical asymptotes where
4x 2 n x 8 4 n, n an integer. y
4
4
2 0
_¹/2
x
18. y 3 sec x has period 2.
5
_¹
¹
2
¹ ¹/2
x
_¹
0
_¹/2
_2
_2
_4
_4
¹ ¹/2
x
28
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
21. y 5 tan x has period 1. An appropriate interval is 12 12 . The graph has xintercepts where x n x n, n an integer; and vertical asymptotes where
x n x 1 n, n an integer. 2
22. y 3 tan 4x has period 14 . An appropriate interval is 18 18 . The graph has xintercepts where 4x n x 14 n, n an integer; and vertical asymptotes where
1 1 4x 2 n x 8 4 n, n an integer.
2
y 20
y 10
10 _3
_2
0
_1
5 1
_1
3 x
2
0
_1/2
_10
_5
_20
_10
23. y 2 cot 3x has period 13 . An appropriate interval is 0 13 . The graph has xintercepts where 3x 2 n x 16 13 n, n an integer; and vertical asymptotes
where 3x n x 13 n, n an integer.
_2/3
_1/3
x 14 12 n, n an integer; and vertical asymptotes
where 2x n x 12 n, n an integer. y 10
5
5 1/3
1 x
2/3
_1
_5
_10
_10
4
is [2 2]. The graph has xintercepts where 4 x n x 4n, n an integer; and vertical asymptotes where x n x 2 2n, n an integer. 2
y
1 x
1/2
26. y cot 2 x has period 2. An appropriate 2
interval is [0 2]. The graph has xintercepts where
x n x 1 2n, n an integer; and vertical 2
2
asymptotes where 2 x n x 2n, n an integer. y
2 _2
0
_1/2
_5
25. y tan 4 x has period 4. An appropriate interval
4
24. y 3 cot 2x has period 12 . An appropriate interval is 0 12 . The graph has xintercepts where 2x 2 n
y 10
0
1 x
1/2
2 2
x
_1
1
x
SECTION 6.4 More Trigonometric Graphs
29
27. y 2 tan 3x has period 13 . An appropriate interval is 28. y 2 tan 2 x has period 2. An appropriate 2 16 16 . The graph has xintercepts where 3x n interval is [1 1]. The graph has xintercepts where x 13 n, n an integer; and vertical asymptotes where
3x n x 1 1 n, n an integer. 2
6
3
y
x n x 2n, n an integer; and vertical 2
asymptotes where 2 x 2 n x 1 2n, n an
integer.
y
4 _0.5
4 0.5
x
_1
1
x
2 29. y csc 4x has period 24 2 . An appropriate interval is 30. y 5 csc 3x has period 3 . An appropriate interval is 4 4 . The graph has vertical asymptotes where 3 3 . The graph has vertical asymptotes where 4x n x 4 n, n an integer; maxima where 3x n x 3 n, n an integer; maxima where
4x 32 2n x 38 2 n, n an integer; and
minima where 4x 2 2n x 8 2 n, n an
integer.
2 3x 32 2n x 2 3 n, n an integer; and
2 minima where 3x 2 2n x 6 3 n, n an
integer.
y
y
2
10
_¹ 2
¹ 2
x
31. y sec 2x has period 22 . An appropriate interval is [0 ]. The graph has vertical asymptotes where 2x n
x 2 n, n an integer; maxima where 2x 2n
x 2 n, n an integer; and minima where 2x 2n x n, n an integer.
_¹
x
32. y 12 sec 4x has period 12 . An appropriate interval is 0 12 . The graph has vertical asymptotes where 4x n x 14 n, n an integer; maxima where 4x 2n x 14 12 n, n an integer; and
minima where 4x 2n x 12 n, n an integer.
y
y 3
2 _¹
¹
2 ¹
x
1 _1
_1/2
0 1/8 _1 _2 _3
1/2
1 x
30
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
33. y 5 csc 32 x has period 322 43 . An appropriate interval is 0 43 . The graph has vertical asymptotes
where 32 x n x 23 n, n an integer; maxima where
34. y 5 sec 2x has period 22 1. An appropriate
interval is [0 1]. The graph has vertical asymptotes where 2x n x 12 n, n an integer; maxima where
2x 2n x 12 n, n an integer; and minima
3 x 3 2n x 1 4 n, n an integer; and 2 2 3 3 minima where 2 x 2 2n x 13 43 n, n an
where 2x 2n x n, n an integer.
y
10
y
integer.
_0.5
10 _1
1
0.5
xintercepts where x 4 n x 4 n, n an
integer; and vertical asymptotes where x 4 2 n
x 4 n, n an integer. y 4
36. y tan x 4 has period and horizontal shift 4 . An 3 appropriate interval is 4 4 . The graph has xintercepts where x 4 n x 4 n, n an
integer; and vertical asymptotes where x 4 2 n
x 4 n, n an integer.
y 4
2 _¹
2
0 ¹/4
_2
¹
2¹ x
_2¹
_¹
x
_4
xintercepts where x 4 2 n x 4 n, n
an integer; and vertical asymptotes where x 4 n
38. y 2 cot x 3 has period and horizontal shift 3 . 4 An appropriate interval is 3 3 . The graph has
5 xintercepts where x 3 2 n x 6 n, n
an integer; and vertical asymptotes where x 3 n x 3 n, n an integer.
y 4
4
2 _¹
2¹
_2
37. y cot x 4 has period and horizontal shift 4 . 3 An appropriate interval is 4 4 . The graph has
_2¹
¹ 3¹/4
0
_4
x 4 n, n an integer.
x
x
35. y tan x 4 has period and horizontal shift 4 . 5 An appropriate interval is 4 4 . The graph has
_2¹
1
0
¹ 3¹/4
y
2 2¹ x
_2¹
_¹
0 ¹/3
_2
_2
_4
_4
¹
2¹
x
SECTION 6.4 More Trigonometric Graphs
31
39. y csc x 4 has period 2 and horizontal shift 4 . 9 An appropriate interval is 4 4 . The graph has
40. y sec x 4 has period 2 and horizontal shift 4 . 7 An appropriate interval is 4 4 . The graph has
3 n an integer; maxima where x 4 2 2n
n an integer; maxima where x 4 2n
3 x 4 2 2n x 4 2n, n an integer.
x 4 2n x 4 2n, n an integer.
vertical asymptotes where x 4 n x 4 n,
x 74 2n, n an integer; and minima where
_2¹
_¹
vertical asymptotes where x 4 n x 4 n,
x 34 2n, n an integer; and minima where
y 4
y
2
2
0 ¹/4
_2
2¹ x
¹
_¹
¹
x
_4
41. y 12 sec x 6 has period 2 and horizontal shift 6 . 13 An appropriate interval is 6 6 . The graph has
42. y 3 csc x 2 has period 2 and horizontal shift 3 2 . An appropriate interval is 2 2 . The graph has
n an integer; maxima where x 6 2n
3 n an integer; maxima where x 2 2 2n
vertical asymptotes where x 6 n x 6 n,
x 76 2n, n an integer; and minima where x 6 2n x 6 2n, n an integer.
vertical asymptotes where x 2 n x 2 n,
x 2n, n an integer; and minima where
x 2 2 2n x 2n, n an integer. y
y
6
1 ¹
2¹
x
_¹
¹
x
Maybe use the same labeling as done in the answer shown here. 32
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
43. y tan 2 x 3 has period 2 and horizontal shift 3 . 7 . The graph has An appropriate interval is 12 12 xintercepts where 2 x 3 n x 3 2 n, n an integer; and vertical asymptotes where 7 2 x 3 2 n x 12 n, n an integer. y 4 2 0 _2
5¹/6 ¹/3 ¹/2
44. y cot 2x 4 cot 2 x 8 has period 2 and 5 . horizontal shift . An appropriate interval is 8 8 8 The graph has xintercepts where 2 x 8 2 n x 38 2 n, n an integer; and vertical asymptotes where 2 x 8 n x 8 2 n, n an integer. y 4 2
x
The label pi/2 looks like it is for the asymptote. Please use different labeling.
_4
45. y 5 cot 3x 2 5 cot 3 x 6 has period 3 and horizontal shift 6 . An appropriate interval is 6 6 . The graph has xintercepts where 3 x 6 2 n x 3 n, n an integer; and vertical asymptotes where 3 x 6 n x 6 3 n, n an integer. y 10
_¹/2
_¹
_¹/2
¹
x
_4
46. y 4 tan 4x 2 4 tan 4 x 2 has period 4 and 3 5 horizontal shift 2 . An appropriate interval is 8 8 . The graph has xintercepts where 4 x 2 n
x 2 4 n, n an integer; and vertical asymptotes where 5 4 x 2 2 n x 8 4 n, n an integer. y 10 5
0 ¹/6 ¹/2
¹ x
0
_5
x
¹/2
_5
_10
_10
47. y cot 2x 2 cot 2 x 4 has period 2 and 3 . horizontal shift . An appropriate interval is 4 4 4 The graph has xintercepts where 2 x 4 2 n x 2 2 n, n an integer; and vertical asymptotes where 2 x 4 n x 4 2 n, n an integer. y
48. y 12 tan x 12 tan x 1 has period 1 and horizontal shift 1. An appropriate interval is 12 32 . The graph has xintercepts where x 1 n x n, n an integer; and vertical asymptotes where
1 x 1 2 n x 2 n, n an integer. y 1
2 _¹
¹/2
_2
5 _¹
0 ¹/8
¹ x
0
1/2
1
3/2
x
SECTION 6.4 More Trigonometric Graphs
1 49. y 2 csc x 3 2 csc x 3 has period
2 2 and horizontal shift 1 . An appropriate interval is 3
1 7 . The graph has vertical asymptotes where 3 3
x 13 n x 3 n, n an integer; maxima
where x 13 32 2n x 16 2n, n an
integer; and minima where x 13 2 2n x 56 2n, n an integer. y
33
1 2 50. y 3 sec 14 x 6 3 sec 4 x 3 has period 8
and horizontal shift 23 . An appropriate interval is 2 26 . The graph has vertical asymptotes where 3 3 1 4
x 23 n x 23 4n, n an integer;
maxima where 14 x 23 2n
x 143 8n, n an integer; and minima where 1 x 2 2n x 2 8n, n an integer. 4 3 3 y 10
4
5 1
2
x
_ 4¹ 3
_5
0
8¹ 3
x
20¹ 3
_10
51. y sec 2 x 52. y csc 2 x 4 has period and horizontal shift 4 . 2 has period and horizontal shift 2 . 5 An appropriate interval is An appropriate interval is 2 2 . The graph has 4 4 . The graph has vertical vertical asymptotes where 2 x n, n an 2 n asymptotes where 2 x n x 4 4 2 x 2 2 n, n an integer; maxima where integer; maxima where 2 x 4 2n 3 2 x 2 2 2n x 4 n, n an integer; x 34 n, n an integer; and minima where and minima where 2 x n, n an integer. 2 2 2n 2 x 2n x 4 4 x 4 n, n an integer.
y 5
0
y 5
¹/2
¹
x _¹
0
_5 _5
¹
x
34
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
2 53. y 5 sec 3x 2 5 sec 3 x 6 has period 3 and 5 . . An appropriate interval is horizontal shift 6 6 6 The graph has vertical asymptotes where 3 x 6 n x 6 3 n, n an integer; maxima where
2 3 x 6 2n x 2 3 n, n an integer;
2 and minima where 3 x 6 2n x 6 3 n, n
an integer.
54. y 12 sec 2x 12 sec 2 x 12 has period
2 1 and horizontal shift 1 . An appropriate interval is 2 2
1 3 . The graph has vertical asymptotes where 2 2
2 x 12 n x 12 12 n, n an integer; maxima
where 2 x 12 2n x n, n an integer;
and minima where 2 x 12 2n x 12 n, n an integer.
y
y 10 _¹/2
¹/2
1
x
0
2 55. y tan 23 x 6 tan 3 x 4 has period
0.5
1
1.5
x
56. y tan 12 x 4 has period 12 2 and horizontal
is 2 . The graph has xintercepts where
5 3 . The shift . An appropriate interval is 4 4 4 graph has xintercepts where 12 x 4 n
vertical asymptotes where 23 x 4 2 n
3 where 12 x 4 2 n x 4 2n, n an
3 23 2 and horizontal shift 4 . An appropriate interval 2 x n x 3 n, n an integer; and 3 4 4 2
x 32 n, n an integer.
_¹
x 4 2n, n an integer; and vertical asymptotes
integer.
y
y
2
2 ¹
2¹
x
5¹
_ 4
¹
3¹ 4
7¹ 4
2¹ x
SECTION 6.4 More Trigonometric Graphs
57. y 3 sec x 12 has period 2 2 and horizontal shift 12 . An appropriate interval is 12 32 . The graph has vertical asymptotes where x 12 n x 12 n, n an integer; maxima where
x 12 2n x 12 2n, n an integer; and minima where x 12 2n x 12 2n, n an
2 58. y sec 3x 2 sec 3 x 6 has period 3 and horizontal shift 6 . An appropriate interval is 6 2 . The graph has vertical asymptotes where 3 x 6 n
x 6 3 n, n an integer; maxima where 2 3 x 6 2n x 6 3 n, n an integer; 2 and minima where 3 x 6 2n x 6 3 n,
n an integer.
y
integer.
y
2 6
_¹/2
1
1
¹/2
and horizontal shift 6 . An appropriate interval is 5 . The graph has xintercepts where 12 12 2 x 6 n x 6 2 n, n an integer; and vertical asymptotes where 2 x 6 2 n n, n an integer. x 512 2
60. y 2 cot 3x 3 2 cot 3 x 1 has period 13 and horizontal shift 1. An appropriate interval is 1 23 . The graph has xintercepts where
5 1 3 x 1 2 n x 6 3 n, n an integer;
and vertical asymptotes where 3 x 1 n x 1 13 n, n an integer.
y
¹/6
y 4 2
4 _¹/3
x
x
59. y 2 tan 2x 3 2 tan 2 x 6 has period 2
2¹/3 x
0
_1
x
_2 _4
61. y tan 30x, [03 03] by [3 3]
62. y csc 50x, [03 03] by [6 6] 5
2
0.2
0.2 2
35
0.2
0.2 5
36
tan 20π x
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
63. y
tan 20x, [05 05] by [02 3]
64. y sec2 10x, [1 1] by [02 5]
0.15
4
Replace "0.5" with "0.15" (4 times)
2
2
0.5
0.0
1
0.5
65. (a) d t 3 tan t, so d 015 153,
(b)
d 025 300, and d 045 1894.
d
0
S 8 346, and S 1175 9154.
1
10
(c) d as t 12 .
t , so S 2 1039, S 6 0, 66. (a) S t 6 cot 12
0
(b)
0.1
0.2
0.3
0.4
0.5 t
S 5
(c) From the graph, it appears that S t 6 at
approximately t 3 and t 9, corresponding to 9 A . M . and 3 P. M .
0
6
t
(d) As t 12 , the sun approaches the horizon and the person’s shadow grows longer and longer.
67. (a) If f is periodic with period p, then by the definition of a period, f x p f x for all x in the domain of f . 1 1 1 for all f x 0. Thus, is also periodic with period p. Therefore, f x p f x f 1 also has period 2. Similarly, since cos x has (b) Since sin x has period 2, it follows from part (a) that csc x sin x 1 also has period 2. period 2, we conclude sec x cos x 68. If f and g are periodic with period p, then f x p f x for all x and g x p g x for all x. Thus f x p f x for all x [unless g x is undefined, in which case both are undefined.] But consider f x sin x g x p g x f x sin x (period 2) and g x cos x (period 2). Their quotient is tan x, whose period is . g x cos x 69. (a) The graph of y cot x is the same as the graph of y tan x shifted 2 units to the right. (b) The graph of y csc x is the same as the graph of y sec x shifted 2 units to the right.
SECTION 6.5 Inverse Trigonometric Functions and Their Graphs
6.5
37
INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR GRAPHS
1. (a) To define the inverse sine function we restrict the domain of sine to the interval 2 2 . On this interval the
sine function is onetoone and its inverse function sin1 is defined by sin1 x y sin y x. For example, 1 sin1 21 6 because sin 6 2 .
(b) To define the inverse cosine function we restrict the domain of cosine to the interval [0 ]. On this interval the cosine function is onetoone and its inverse function cos1 is defined by cos1 x y cos y x. For example, 1 cos1 12 3 because cos 3 2 .
1 sin 2. (a) The cancellation property sin1 sin x x is valid for x in the interval 2 2 . By this property, sin 4 4 . and sin1 sin 3 3 (b) If x is not in the interval in part (a), then the cancellation property does not apply. For example, sin1 sin 56 sin1 12 6. 3. (a) sin1 1 2 because sin 2 1 and 2 lies in 2 2 . 3 (b) sin1 23 3 because sin 3 2 and 3 lies in 2 2 . (c) sin1 2 is undefined because there is no real number x such that sin x 2.
4. (a) sin1 1 2
(b) sin1 22 4
5. (a) cos1 1 6. (a) cos1 22 4
(b) cos1 21 3
7. (a) tan1 1 4 8. (a) tan1 0 0 9. (a) cos1 12 23
10. (a) cos1 0 2
(b) cos1 1 0 3 3 1 3 (b) tan 3 2 1 2 4 (b) sin (b) tan1
(b) sin1 0 0
11. sin1 23 072973 13. cos1 37 201371
(c) sin1 2 is undefined. (c) cos1 23 56 (c) cos1 22 34
(c) tan1 33 6 3 1 3 (c) tan 6
(c) tan1 1 4 (c) sin1 12 6
12. sin1 89 109491 14. cos1 49 111024
15. cos1 092761 275876
16. sin1 013844 013889
17. tan1 10 147113
18. tan1 26 153235
19. tan1 123456 088998
20. cos1 123456 is undefined because 123456 1.
21. sin1 025713 026005 23. sin sin1 14 14 25. tan tan1 5 5 27. sin sin1 32 is undefined because 32 1. 29. cos cos1 15 15
22. tan1 025713 025168 24. cos cos1 32 23 26. sin sin1 5 is undefined because 5 1. 28. tan tan1 32 32 30. sin sin1 34 34
38
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
4 31. sin1 sin 4 33. sin1 sin 34 4 5 1 cos 6 56 35. cos 56 37. cos1 cos 76
4 32. cos1 cos 4 34 34. cos1 cos 34 36. sin1 sin 56 6 7 38. sin1 sin 6 6
4 39. tan1 tan 40. tan1 tan 4 3 3 2 11 1 1 tan 3 3 sin 4 4 41. tan 42. sin 43. sin cos1 21 23 44. tan sin1 23 3 46. tan cos1 22 tan 34 1 45. cos sin1 1 cos 2 0 23 2 3 1 3 47. sec tan1 33 sec 48. csc tan csc 6 3 3 3 3 2 1 1 49. csc cot 1 csc 4 2 50. sin sec 2 sin 3 2 51. We want to find sec tan1 x . Let u tan1 x. Using the Pythagorean identity, we have sec u 1 tan2 u. Since u lies in the interval 2 2 and sec u is positive on that interval, we can use the cancellation property to write sec u 1 tan2 u 1 x 2 , and so sec tan1 x 1 x 2 . 52. We want to find cos sin1 x . Let u sin1 x. Using the Pythagorean identity, we have cos u 1 sin2 u. Since u lies in the interval 2 2 and cos u is positive on that interval, we can use the cancellation property to write cos u 1 sin2 u 1 x 2 , and so cos sin1 x 1 x 2 . 53. We want to find tan sin1 x . Let u sin1 x. Using the Reciprocal and Pythagorean identities, we have 2u 1 1 sin x2 . If x 0, then u 0 and 1 1 tan u sec2 u 1 2 2 2 cos u 1 x2 1 sin u 1 sin u x . tan u 0, and similarly if x 0 then tan u 0, so tan sin1 x 1 x2 54. We want to find sin sec1 x . Let u sec1 x. Using the Pythagorean identity, we have sec2 u 1 1 2 . Since sin u, u, and x have the same sign, sin u 1 cos u 1 2 sec u sec2 u x2 1 . sin sec1 x x 55. If f x e x and g x arcsin x, then F x f g x earcsin x . 2 56. If f x x 2 and g x tan1 x, then F x f g x tan1 x . 57. If f x sin1 x and g x 1x, then F x f g x sin1 1x. 1 1 and g x tan1 x, then F x 58. If f x . 1x 1 tan1 x
2
59. If f x e x , g x arcsin x, and h x x 2 , then F x f g h x earcsin x . 60. If f x tan1 x, g x x, and h x x 2 1, then F x f g h x tan1 x 2 1. 2 61. If f x tan1 x, g x e x , and h x 1 x 2 , then then F x f g h x tan1 e1x .
39
SECTION 6.5 Inverse Trigonometric Functions and Their Graphs
62. If f x ln x, g x arctan x, and h x x 4 , then F x f g h x ln arctan x 4 . 63. (a) The arctangent function is defined everywhere, as is x 2 , so the domain of f x tan1 x 2 is .
2 1
(b) Because x 2 is even, f x is itself even. (See Exercise 2.7.92.) Also,
x 2 0 for all x and tan1 x 0 for x 0, so f x 0 everywhere.
2 Because tan1 x has a horizontal asymptote at y 2 and x increases
10
10
without bound, f x has the same asymptote.
64. (a) For f x sin1 x 2 to be defined we must have 1 x 2 1
2
1 x 1, so the domain of f x is [1 1]. (b) Because x 2 is even, f x sin1 x 2 is even. (See Exercise 2.7.92.)
1
Also, x 2 0 for all x and sin1 x 0 for x 0, so f x 0 on its
domain.
1
1
65. (a) The cosine function has domain and its range is [1 1], which lies within the domain of arcsin. Thus, f x sin1 cos x has domain . (b) Because cos x is even, f x sin1 cos x is even. It is also periodic x if x 0 2 2 with period . Note that f x x if 0 x 2
1
10
10 1
2
66. (a) The arctangent function is defined everywhere and has range 2 2 , so tan1 x is never 0 and thus f x 2
1
4
tan1 x has domain . 2
2
2 is the yintercept. Because tan1 x approaches as x (b) f 0 2
1 1 . As 2 2
becomes large, f x has a horizontal asymptote at y
4
2
2
4
x gains negative magnitude, the denominator of f x approaches 0 and so f x grows without bound in the negative xdirection. 67. (a)
From the graph of y sin1 x cos1 x, it appears that y 157. We suspect
that the actual value is 2.
2
1
(b) To show that sin1 x cos1 x 2 , start with the identity sin a 2 cos a and take arcsin of both sides to obtain 1
1 cos a. Now let a cos1 x. Then a 2 sin 1 cos cos1 x cos1 x sin sin1 x sin1 x, so 2
sin1 x cos1 x 2.
40
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
68. (a)
From the graph of y tan1 x tan1 1x , it appears that y 157 for x 0
2
and y 157 for x 0. We suspect that the actual values are 2.
2
(b) To show that tan1 x tan1 x1 2 for x 0, we start with the identity tan a cot 2 a and take arccot of both sides to obtain
2
cot1 tan a a. Now use the identity cot1 x tan1 x1 to write 2 1 1 , we have tan1 tan1 a x 2 a. Substituting a tan 1 tan1 1 tan1 1 tan1 1 tan1 1 x x 2 1x 2
2
tan tan
1x
. For the case x 0, simply note that tan1 x tan1 x, so for positive x, tan1 x tan1 x1 2 1 tan1 x tan1 x1 tan1 x tan1 x 2.
69. The domain of f x sin sin1 x is the same as that of sin1 x, [1 1], and the graph of f is the same as that of y x on [1 1].
The domain of g x sin1 sin x is the same as that of sin x, , because for all x, the value of sin x lies within
the domain of sin1 x. g x sin1 sin x x for 2 x 2 . Because the graph of y sin x is symmetric about the
3 line x 2 , we can obtain the part of the graph of g for 2 x 2 by reflecting the graph of y x about this vertical line. The graph of g is periodic with period 2. y
y
¹/2
1
y=g(x)
y=f(x) _1
1
x
_3¹/2
_¹
_¹/2
_1
6.6
¹/2
¹
3¹/2 x
_¹/2
MODELING HARMONIC MOTION
1. (a) Because y 0 at time t 0, y a sin t is an appropriate model.
(b) Because y a at time t 0, y a cos t is an appropriate model.
2. (a) Because y 0 at time t 0, y aect sin t is an appropriate model.
(b) Because y a at time t 0, y aect cos t is an appropriate model.
2 , and the phase 3. (a) For an object in harmonic motion modeled by y A sin kt b the amplitude is A, the period is k b is b. To find the horizontal shift, we factor k to get y A sin k t . From this form of the equation we see that the k b horizontal shift is . k (b) For an object in harmonic motion modeled by y 5 sin 4t the amplitude is 5, the period is 2 , the phase is , and the horizontal shift is 4.
4. Objects A and B are in harmonic motion modeled by y 3 sin 2t and y 3 sin 2t 2 . The phase of A is and the phase of B is 2 . The phase difference is 2 , so the objects are moving out of phase.
SECTION 6.6 Modeling Harmonic Motion
5. y 2 sin 3t
1 (a) Amplitude 2, period 23 , frequency period 23 .
(b)
6. y 3 cos 12 t 2 4, frequency (a) Amplitude 3, period 12 1
y
1
period 4 .
2
(b) ¹ 3
2¹ 3
y 3
x 4¹ x
2¹
7. y cos 03t
20 , frequency 3 . (a) Amplitude 1, period 203 3 20
(b)
8. y 24 sin 36t
5 , frequency 9 . (a) Amplitude 24, period 236 9 5
(b)
y
y 2.4
1
5¹ 18
5¹ 9
x
20¹ x 3
10¹ 3
_1
3t 025 cos 9. y 025 cos 15t 3 2 3 025 cos 32 t 29
2 4 , frequency 3 . (a) Amplitude 025, period 32 3 4
(b)
10. y 32 sin 02t 14 32 sin 02 t 7 10, frequency 1 . (a) Amplitude 32 , period 202 10
(b)
y
y
1.5
0.25 -7 2¹ 9
8¹ 9
14¹ 9
x
5¹-7
x
10¹-7
_1.5
_0.25
11. y 5 cos 23 t 34 5 cos 23 t 98
12. y 16 sin t 18
2 3, frequency 1 . (a) Amplitude 5, period 23 3
(b)
y
(a) Amplitude 16, period 2, frequency 21 .
(b)
y 1.6
5
¹
3¹
1.8+2 9
_8
3¹ 9 2 _8
3¹ _ 9 8
x
1.8
13. The amplitude is a 10 cm, the period is 2k 3 s, and f 0 0, so f t 10 sin 23 t.
1.8+2 1.8+¹
x
1.8+2¹
41
42
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
14. The amplitude is 24 ft, the period is 2k 2 min, and f 0 0, so f t 24 sin t. 5 Hz, and f 0 0, so f t 6 sin 10t. 15. The amplitude is 6 in., the frequency is 2k
16. The amplitude is 12 m, the frequency is 2k 05 Hz, and f 0 0, so f t 12 sin t. 17. The amplitude is 60 ft, the period is 2k 05 min, and f 0 60, so f t 60 cos 4t. 18. The amplitude is 35 cm, the period is 2k 8 s, and f 0 35, so f t 35 cos 4 t. 19. The amplitude is 24 m, the frequency is 2k 750 Hz, and f 0 24, so f t 24 cos 1500t. 20. The amplitude is 625 in., the frequency is 2k 60 Hz, and f 0 625, so f t 625 cos 120t. 21. (a) k 2, c 15, and f 3 6, so we have y 2e15t cos 6t.
(b)
have y 100e005t cos 2 t. y 100
5
y 075e3t cos 23 t. y 0.75
x
25. (a) k 7, c 10, and p 6 12, so we have y 7e10t sin 12t.
¹/5
x
26. (a) k 1, c 1, and p 1 2, so we have y et sin 2t.
(b)
y
x
24. (a) k 075, c 3, and p 3 23 , so we have (b)
4
y 15
x
23. (a) k 100, c 005, and p 4 2 , so we
(b)
have y 15e025t cos 12t.
(b)
y 2
1
(b)
22. (a) k 15, c 025, and f 06 12, so we
y 0.5
1 ¹/30
x
1
x
SECTION 6.6 Modeling Harmonic Motion
27. (a) k 03, c 02, and f 20 40, so we have y 03e02t sin 40t.
28. (a) k 12, c 001, and f 8 16, so we
y
(b)
have y 12e001t sin 16t.
(b)
y 10
0.2 0.2
0.4
0.6
0.8
x
1
x
2
29. y 5 sin 2t 2 5 sin 2 t 4 has amplitude 5, period , phase 2 , and horizontal shift 4 . 30. y 10 sin t 3 has amplitude 10, period 2, phase 3 , and horizontal shift 3 .
2 31. y 100 sin 5t 100 sin 5 t 5 has amplitude 100, period 5 , phase , and horizontal shift 5 . 1 2 2 32. y 50 sin 12 t 5 50 sin 2 t 5 has amplitude 50, period 4, phase 5 , and horizontal shift 5 .
33. y 20 sin 2 t 4 20 sin 2t 2 has amplitude 20, period , phase 2 , and horizontal shift 4 . 8 sin 4t has amplitude 8, period , phase , and horizontal shift . 34. y 8 sin 4 t 12 3 2 3 12 5 35. y1 10 sin 3t 2 ; y2 10 sin 3t 2 5 (a) y1 has phase 2 and y2 has phase 2 .
y 10
(d)
5 (b) The phase difference is 2 2 2.
yÁ, yª
5
(c) Because the phase difference is a multiple of 2, the curves are in
_¹
phase.
0
¹ t
_5 _10
36. y1 15 sin 2t 3 ; y2 15 sin 2t 6
(a) y1 has phase 3 and y2 has phase 6 .
y
(d)
10
(b) The phase difference is 3 6 6.
_¹
(c) Because the phase difference is not a multiple of 2, the curves are out of phase.
yª
yÁ
0
¹ t
_10
80 sin 5t ; y 80 sin 5t 37. y1 80 sin 5 t 10 2 2 3 (a) y1 has phase 2 and y2 has phase 3 .
y 80
(d)
(b) The phase difference is 2 3 6.
(c) Because the phase difference is not a multiple of 2, the curves are out of phase.
_¹
0
_80
yª
yÁ
¹ t
43
44
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
3 20 sin 2t 3 20 sin ; y 38. y1 20 sin 2 t 20 sin 2 t 2t 2 2 2
y 20
(d)
(a) y1 has phase and y2 has phase 3.
(b) The phase difference is 3 2.
yÁ, yª
(c) Because the phase difference is a multiple of 2, the curves are in
0
_¹
phase.
¹ t
_20
39. p t 115 25 sin 160t
2 1 00125, (a) Amplitude 25, period 160 80
frequency
1 80. period
(c) The period decreases and the frequency increases.
P
(b)
140 120 100 80 0
40. y 02 cos 20t 8
(a) The frequency is 20 2 10 cycles/min.
(c) Since y 02 cos 20t 8 02 1 8 82 and when t 0, y 82, the maximum displacement is 82 m.
(b)
0.01
0.02
t
y 8.4 8.2 8 7.8
0
0.2
0.4
t
41. The graph resembles a sine wave with an amplitude of 5, a period of 25 , and no phase shift. Therefore, a 5,
2 2 5
5, and a formula is d t 5 sin 5t. 42. From the graph we see that the amplitude is 6 feet and the period is 12 hours. Also, the sine curve is shifted 6 hours (exactly half its period) to the right, which is equivalent to reflecting the curve about the taxis. Thus, the equation is t 6 sin t. y 6 sin 212 6
43. Since the mass travels from its highest point (compressed spring) to its lowest point in 12 s, it completes half a period in 12 s. 2 12 2. Also, a 5. So y 5 cos 2t. So, 12 one period 12 s 12 2 1 2. So y 2 cos 2t. 44. a 2, 45. Since the Ferris wheel has a radius of 10 m and the bottom of the wheel is 1 m above the ground, the minimum height is 2 , and so y 11 10 sin t , where t is in 1 m and the maximum height is 21 m. Then a 10 and 20 s 10 10 seconds. 46. Let f t represent the measure of the angle at time t. The amplitude of this motion is 10. Since the period is 2s, we have 2 2 . Thus f t 10 sin t. 47. a 02, 2 10, b 38 5 . Then y 02 sin 5 t 38.
SECTION 6.6 Modeling Harmonic Motion
48. (a) m 10 g, k 3, a 5 cm. Then f t 5 cos
45
3 t . 10
(b) The function f t a cos kmt describes an object oscillating in simple harmonic motion, so by comparing it with the general equation y a cos t we see that km. This means the frequency is 1 km k . f 2 2 2 m k (c) If the mass is increased, then the denominator of increases, so overall the frequency decreases. If the frequency m has decreased, then by definition the oscillations are slower. k increases. Thus, overall the frequency increases. If (d) If a stiffer spring is used, k is larger and so the numerator of m the frequency has increased, then by definition the oscillations are faster. 49. The amplitude is 12 100 80 10 mmHG, the period is 24 hours, and the phase shift is 8 hours, so f t 10 sin 12 t 8 90.
50. The signal consists of the seven phases corresponding to the digits 01, 11, 10, 01, 00, 00, and 10. Thus, the encoded string of digits is 01111001000010. 51. (a) The maximum voltage is the amplitude, that is, Vmax a 45 V.
(b) From the graph we see that 4 cycles are completed every 01 seconds, or equivalently, 40 cycles are completed every second, so f 40. f 40. (c) The number of revolutions per second of the armature is the frequency, that is, 2 (d) a 45, f 40 80. Then V t 45 cos 80t. 2 1130 52. (a) As the car approaches, the perceived frequency is f 500 1130110 5539 Hz. As it moves away, the perceived 1130 frequency is f 500 1130110 4556 Hz. 1130 1130 (b) The frequency is , so 1000 1130110 11078 (approaching) or 9113 (receding). 500 1130110 2 Thus, models are y A sin 11078t and A sin 9113t. 53. k 1, c 09, and
1 . Since f 0 0, f t e09t sin t. 2 2
2 4. 2 (a) Since f 0 6, f t 6e28t cos 4t.
54. k 6, c 28, and
1 (b) The amplitude of the vibration is 05 when 6e28t 05 e28t 12 1 ln 12 089 s. t 12 ln 12 28
55.
1 28t ln 12
kect 4 ectct3 4 e3c 4 3c ln 4 c 13 ln 4 046. kect3
56. (a)
3 kec0 5 e2c 5 2c ln 5 c 12 ln 5 080. 06 kec2
(b) f t 3e08t cos t. If the frequency is 165 cycles per second, then f t 3e08t cos 330t.
165 330. Thus, 2
46
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
57. (a) For fan A, the amplitude is 1, the frequency is 100, and the phase is 0, so y sin 200t.
100 200 and an equation is 2
For fan B, the amplitude is 1, the frequency is 100, and the phase is 34 , so again 200 and an equation is y sin 200t 34 .
(b) The phase difference is 34 , so the fans are out of phase. If fan A were rotated 34 counterclockwise, the phase difference would become 0 and the fans would be in phase. 58. (a) E I 50 sin 120t, so the voltage phase is 0. E I I 50 sin 120t 54 , so the voltage phase is 54 .
(b) The phase difference is 54 , so the generators are out of phase. If the armature in the second generator were rotated 34 , then the phase difference would become 54 34 2, and the generators would produce voltage in phase.
1 59. From left to right: at t 6 , sin t 2 and the values are increasing; at t 2 , sin t 1 and the values reach a maximum;
at t 56 , sin t 12 and the values are decreasing; at t , sin t 0 and the values are decreasing; at t 76 , sin t 12 and the values are decreasing; at t 32 , sin t 1 and the values reach a minimum, and at t 116 , sin t 12 and the
values are increasing. 60. From left to right: new moon, waxing crescent moon, waxing gibbous moon, full moon, waning gibbous moon, third quarter moon, waning crescent moon, new moon. Tides and werewolf sightings are in phase with the lunar cycle; paying rent and cellphone bills are out of phase.
CHAPTER 6 REVIEW 2 2 1. (a) Since 23 12 34 14 1, the point P 23 12 lies on the unit circle.
(b) sin t 12 , cos t 23 , tan t
1 2 3 . 3 23
2 2 9 1 1, the point P 3 4 lies on the unit circle. 2. (a) Since 35 45 25 2 5 5 4 (b) sin t 45 , cos t 35 , tan t 35 43 . 5
3. t 23
4. t 53
(a) t 23 3 3 1 (b) P 2 2
(c) sin t 23 , cos t 12 , tan t 3, csc t 2 3 3 ,
sec t 2, and cot t 33 .
5. t 114
(a) t 3 114 4 2 2 (b) P 2 2
(c) sin t 22 , cos t 22 , tan t 1, csc t 2, sec t 2, and cot t 1.
(a) t 2 53 3 3 1 (b) P 2 2
(c) sin t 23 , cos t 12 , tan t 3,
csc t 2 3 3 , sec t 2, and cot t 33 . 6. t 76 (a) t 76 6 3 1 (b) P 2 2
(c) sin t 12 , cos t 23 , tan t 33 , csc t 2, sec t 2 3 3 , and cot t 3.
CHAPTER 6 2 7. (a) sin 34 sin 4 2
Review
47
8. (a) tan 3 3 (b) tan 3 3
2 (b) cos 34 cos 4 2
10. (a) cos 5 080902 (b) cos 5 080902
9. (a) sin 11 089121 (b) cos 11 045360 11. (a) cos 92 cos 2 0
12. (a) sin 7 043388
(b) sec 92 is undefined
(b) csc 7 230476
13. (a) tan 52 is undefined
14. (a) sin 2 0
(b) cot 52 cot 2 0
(b) csc 2is undefined.
15. (a) tan 56 33
1 16. (a) cos 3 2
(b) cot 56 3
1 (b) sin 6 2
sin t tan t cos t sin t sin t 17. cos t cos t cos2 t 1 sin2 t
sin2 t 1 cos2 t 1 1 2 cos t cos t cos t cos3 t sin t sin t sin t (because t is in quadrant IV, cos t is positive). 19. tan t cos t 1 sin2 t 1 sin2 t 1 1 1 (because t is in quadrant II, cos t is negative). 20. sec t 2 cos t 1 sin t 1 sin2 t 18. tan2 t sec t tan2 t
5
5 , cos t 12 . Then tan t 13 5 , csc t 13 , sec t 13 , and cot t 12 . 21. sin t 13 13 12 5 12 5 12 13
22. sin t 12 , cos t 0. Since sin t is negative and cos t is positive, t is in quadrant IV. Thus, t determines the terminal point 2 3 3 P 23 12 , and cos t 23 , tan t , csc t 2, sec t , cot t 3. 3 3 1 cos t 5 1 23. cot t 2 , csc t 2 . Since csc t , we know sin t 2 2 5 5 . Now cot t , so 5 sin t sin t 1 1 1 5 5. cos t sin t cot t 2 5 5 12 55 , and tan t 2 while sec t 5 1 5 cos t 2 5
24. cos t 35 , tan t 0. Since cos t and tan t are both negative, t is in quadrant II. Thus, t determines the terminal point 5 P 35 , 45 , and sin t 45 , tan t 43 , csc t 54 , sec t , cot t 34 . 3 2 1 sin t 1 1 sin t 4 cos t. Thus, cos2 t sin2 t 1 cos2 t 14 cos t 1 25. tan t 4 , so cot t 4 and cos t 4
17 17 2 cos2 t 16 17 sec t 16 . Because cosine and secant are negative in Quadrant III, we have sec t 4 , and thus,
sec t cot t 4 417 .
8 , so csc t 17 and cos2 t 1 sin2 t 1 8 2 225 sec2 t 289 . Because cosine and secant 26. sin t 17 8 17 289 225 17 17 119 are positive in Quadrant IV, sec t 17 15 , and so csc t sec t 8 15 120 .
27. cos t 35 , so because the terminal point is in Quadrant I, sin t 45 . Thus, tan t 43 and sec t 53 , so tan t sec t 43 53 3.
48
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
28. sin2 t cos2 t 1 for any value of t.
29. y 10 cos 12 x
2
(a) This function has amplitude 10, period 1 4, 2
and horizontal shift 0. (b)
30. y 4 sin 2x
(a) This function has amplitude 4, period 22 1, and horizontal shift 0.
(b)
y
y 4 10 2¹
4¹
0.5
x
31. y sin 12 x 2 4, and (a) This function has amplitude 1, period 12
(a) This function has amplitude 2, period 2, and horizontal shift 4.
(b)
y
y
2
1
2¹ 2¹
4¹
¹/4
x
33. y 3 sin 2x 2 3 sin 2 x 1
(a) This function has amplitude 3, period 22 , and horizontal shift 1.
(b)
x
32. y 2 sin x 4
horizontal shift 0. (b)
1
x
¹
34. y cos 2 x 2
(a) This function has amplitude 1, period 22 , and horizontal shift 2.
y
(b)
y
3 x 1-¹/2
1
1+¹/2
1 ¹/2
¹
x
CHAPTER 6
1 35. y cos 2 x 6 cos 2 x 3
(a) This function has amplitude 1, period 2 2 4, and horizontal shift 13 . y
(b)
Review
49
36. y 10 sin 2x 2 10 sin 2 x 4
(a) This function has amplitude 10, period 22 , and horizontal shift 4.
(b)
y
1
_13/3
11/3 x
_1/3
10 x ¹/2
¹
37. From the graph we see that the amplitude is 5 and the period is 2 . Considered as a sine function, there is no horizontal shift, and considered as a cosine function, the horizontal shift is 8 . Therefore, the function’s equation can be written as y 5 sin 4x or y 5 cos 4 x 8 . 38. From the graph we see that the amplitude is 2 and the period is 4 (since 14 of the period has been completed at 1 2. Thus, 2 4 k . Considered as a sine function, there is no horizontal shift, and considered as a cosine function, the k 2
horizontal shift is 1. Therefore, the function’s equation can be written as y 2 sin 2 x or y 2 cos 2 x 1.
39. From the graph we see that the amplitude is 12 and the period is 1. Considered as a sine function, the horizontal shift is 1 . Therefore, the function’s equation can be written as 13 , and considered as a cosine function, the horizontal shift is 12 1 . y 12 sin 2 x 13 or y 12 cos 2 x 12
40. From the graph we see that the amplitude is 4 and the period is 43 , so 2k 43 k 32 . Considered as a sine function, the horizontal shift is 3 , and considered as a cosine function, the horizontal shift is 0. Therefore, the function’s equation 3 can be written as y 4 sin 32 x 3 or y 4 cos 2 x.
41. y 3 tan x has period .
42. y tan x has period 1.
y
y
10
_¹
¹
2
x _1
1
x
50
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
1 44. y sec 12 x 2 sec 2 x has period
43. y 2 cot x 2 has period . y
2 4. 1 2
4
y
x
¹
2 _2¹
45. y 4 csc 2x 4 csc 2 x 2 has period 2 . 2
2¹
x
5¹/6
x
46. y tan x 6 has period . y
y
2 5
_¹
_¹/6 ¹
x
1 47. y tan 12 x 8 tan 2 x 4 has period 1 2. 2
1 48. y 4 sec 4x has period 24 2. y
y
10 2 ¹/4
5¹/4
x
_0.5 _0.25
0.25
x
0.5
49. sin1 1 50. cos1 12 23 2 51. sin1 sin 136 52. tan cos1 12 tan 6 3 3 100 sin 8t has amplitude 100, period , phase , and horizontal shift . 53. y 100 sin 8 t 16 2 4 2 16 3 has amplitude 80, period 2 , phase 3 , and horizontal shift . 54. y 80 sin 3 t 80 sin 3t 2 2 3 2 2 3 5 55. y1 25 sin 3 t 2 25 sin 3t 2 ; y2 10 sin 3t 2 (a) y1 has phase 32 and y2 has phase 52 .
y
(d)
20
(b) The phase difference is 32 52 . (c) Because the phase difference is not a multiple of 2, the curves are out of phase.
_¹
0 _20
yÁ
yª ¹ x
CHAPTER 6
56. y1 50 sin 10t 2 ; y2 50 sin 10 t 20 50 sin 10t 2 (a) y1 has phase 2 and y2 has phase 2 .
y
(d)
yÁ, yª
40
(b) The phase difference is 0. 2
2
(c) Because the phase difference is a multiple of 2, the curves are in
0
_¹/2
phase.
¹/2 x
_40
58. (a) y sin cos x
57. (a) y cos x
1
1 5 5
5 1
5
(b) This function has period .
(b) This function has period 2.
(c) This function is even.
(c) This function is even.
59. (a) y cos 201x
60. (a) y 1 2cos x 4 1 2
50
50 1
10
10
(b) This function is not periodic.
(b) This function has period 2.
(c) This function is neither even nor odd.
(c) This function is even.
61. (a) y x cos 3x
62. (a) y
x sin 3x (x 0)
5
0 5
Review
5
5
5
(b) This function is not periodic.
(b) This function is not periodic.
(c) This function is even.
(c) This function is neither even nor odd.
10
51
52
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
63. y x sin x is a sine function whose graph lies between those of y x and y x.
64. y 2x cos 4x is a cosine function whose graph lies between the graphs of y 2x and y 2x .
10
5
10
10
2
10
2 5
65. y x sin 4x is the sum of the two functions y x and y sin 4x.
66. y sin2 x cos2 x is the sum of the two functions
y sin2 x and y cos2 x. Note that sin2 x cos2 x 1
for all x. 2
1 2
2 2 5
67. We graph y f x cos x sin 2x in the viewing
rectangle [0 2] [2 2] and see that the function has local maxima of approximately f 063 176 and
5
68. We graph y f x cos x sin2 x in the viewing
rectangle [0 2] [2 2] and see that the function has local maxima of approximately
f 414 037 and local minima of approximately
f 105 f 524 125 and local minima of
periodic with period 2.
periodic with period 2.
f 251 176 and f 528 037. The function is
f 0 f 2 1 and f 1. The function is 2
2
0
2
4
0
6
69. We want to find solutions to sin x 03 in the interval [0 2], so we plot the functions y sin x and
y 03 and look for their intersection. We see that the
graphs intersect at x 0305 and at x 2837.
6
70. We want to find solutions to cos 3x x in the interval [0 ], so we plot the functions y cos 3x and y x and
look for their intersection. We see that the graphs intersect at x 0390. 1
1
1
4
2
2
0
2
2
4
6
0
1
1
2
3
CHAPTER 6
71. y1 cos sin x, y2 sin cos x (a)
1
Review
53
(b) y1 has period , while y2 has period 2.
yÁ
(c) sin cos x cos sin x for all x. 10
0
10
1
yª
72. The amplitude is a 50 cm. The frequency is 8 Hz, so 8 2 16. Since the mass is at its maximum displacement when t 0, the motion follows a cosine curve. So a function describing the motion of P is f t 50 cos 16t. 73. The amplitude is 12 100 50 cm, the frequency is 4 Hz, so 4 2 8. Since the mass is at its lowest point when t 0, a function describing the distance of the mass from its rest position is f t 50 cos 8t.
74. (a) The initial amplitude is 16 cm and the frequency is 14 Hz, so
(b)
a function describing the motion is y 16e072t cos 28t.
20
(c) When t 10 s, y 16e72 cos 28 00119 cm.
0
2
20
75. (a) y
x x has vertical asymptotes at x 2 and yintercept 0, so it has graph VII. 2 x 2 x 4 x2
2 (b) y 4 cos 2 x 1 has amplitude 4 and period 2 4, so it has graph I.
(c) y tan 4x has period 4 4, yintercept 0, and vertical asymptotes where 4x 2 n x 4n 2, n an integer, so it has graph V.
(d) y x 3 is a polynomial of odd degree and increases without bound in the positive direction, so it has graph III.
(e) y x cos 4x is bounded by y x since cos 4x 1 for all x. It fluctuates between x and x on every interval of length 2 , so it has graph VI.
(f) y 1 3 sin 2x has amplitude 3, period 2 2 4, and yintercept 1. Its graph can be obtained by stretching that of y sin x horizontally by a factor of 2 , then stretching vertically by a factor of 3, then shifting 1 unit upward. This corresponds to graph IV.
x (g) y sec 4x has period 2 4 8, yintercept 1, and vertical asymptotes where 4 2 n 4n 2, so it has
graph II.
4 tan1 x has horizontal asymptotes at y 4 2, so it has graph VIII. (h) y 2
54
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
CHAPTER 6 TEST
1. Since P x y lies on the unit circle, x 2 y 2 1 y 1 quadrant. Therefore y is negative
y 56 .
2 11 6
5 25 36 6 . But P x y lies in the fourth
2 9 , and so x 3 . From the diagram, 2. Since P is on the unit circle, x 2 y 2 1 x 2 1 y 2 . Thus, x 2 1 45 25 5 3 3 4 x is clearly negative, so x 5 . Therefore, P is the point 5 5 . (a) sin t 45
(b) cos t 35
4
(c) tan t 53 43 5
(d) sec t 53 13 22 4 3 (d) csc 2 1
3. (a) sin 76 05 (c) tan 53 3 4. tan t
(b) cos
sin t sin t . But t is in quadrant II cos t 1 sin2 t
Thus, tan t
sin t . 1 sin2 t
8 , t in quadrant III 5. cos t 17
quadrant III) 1 6. y 5 cos 4x
cos t is negative, so we choose the negative square root.
tan t cot t csc t 1
1 1 2 . 1 15 15 64 1 289 17
(a) This function has amplitude 5, period 24 2, phase 0, and horizontal shift 0.
(b)
1 (since t is in 1 cos2 t
1 7. y 2 sin 12 x 6 sin 2 x 3
2 (a) This function has amplitude 2, period 1 4, 2
phase 6 , and horizontal shift 3 .
y
(b)
5 0
¹/2 x
¹/4
y 2 ¹/3
9. y tan 2 x 4 has period 2 .
8. y csc 2x has period 22 . y
y
2 _¹/2
10. (a) tan1 1 4
13¹/3 x
7¹/3
2 ¹/2
_¹/2 x
(b) cos1 23 56
¹/2 x
(c) tan1 tan 3 0
1 (d) cos tan1 3 cos 3 2
CHAPTER 6 2 11. From the graph, we see that the amplitude is 2 and the phase shift is 3 . Also, the period is , so k k 2 2. Thus, the function is y 2 sin 2 x 3 . 12. y1 30 sin 6t 2 ; y2 30 sin 6t 3
(a) y1 has phase 2 and y2 has phase 3 .
y 30
(d)
(b) The phase difference is 2 3 6.
(c) Because the phase difference is not a multiple of 2, the curves are out of phase.
yª
55
yÁ
0
_¹/2
Test
¹/2 x
_30
13. y (a)
cos x 1 x2
(b) The function is even. 1.0
(c) The function has a minimum value of approximately 011 when x 254 and a maximum value of 1 when x 0.
0.5
10
10
14. The amplitude is 12 10 5 cm and the frequency is 2 Hz. Assuming that the mass is at its rest position and moving upward when t 0, a function describing the distance of the mass from its rest position is f t 5 sin 4t.
15. (a) The initial amplitude is 16 in. and the frequency is 12 Hz, so a function describing the motion is y 16e01t cos 24t.
(b)
20
0
20
0.5
1.0
56
FOCUS ON MODELING
FOCUS ON MODELING Fitting Sinusoidal Curves to Data Replace "x" with "t" in all graphs in #1-5 1. (a) Using the method of Example 1, we find the vertical shift b 12 maximum value minimum value 12 21 21 0, the
y 2
amplitude
1
a 12 maximum value minimum value 12 21 21 21, the
0
period
_1
2 2 6 0 12 (so 05236), and the phase shift c 0. Thus, our model is y 21 cos 6 t.
2
4
6
8
10
12
14 x
2
4
6
8
10
12
14 x
_2
From the graph, we see that the curve fits the data quite well. (b) Using the SinReg command on the TI84, we find
y 2048714222 sin 05030795477t 1551856108 00089616507.
This model is equivalent to y 205 cos 050t 155 2 001 205 cos 050t 002 001.
This is the same as the function in part (a), correct to one decimal place.
y 2 1 0 _1 _2
2. (a) Using the method of Example 1, we find the vertical shift b 12 maximum value minimum value 12 063 010 0265,
y 0.6
the amplitude
0.4
a 12 maximum value minimum value 12 063 010 0365,
0.2
2 2 55 25 6 (so 1047), and the phase shift the period c 05. Thus, our model is y 0365 cos 1047 t 05 0265. From the graph, we see that the curve fits the data reasonably well.
(b) Using the SinReg command on the TI84, we find y 0327038879 sin 1021164911t 2118186963 02896017397.
This model is equivalent to 033 cos 102t 212 2 029 033 cos 102t 055 029.
This is the same as the function in part (a), correct to one decimal place.
0
2
4
6
x
2
4
6
x
_0.2 y 0.6 0.4 0.2 0 _0.2
57
Fitting Sinusoidal Curves to Data
3. (a) Let t be the time since midnight. We find a function of the form y a sin t c b. a 12 374 366 04. The period is 24 and
026. b 1 374 366 37. Because the maximum so 224 2
y
37
value occurs at t 16, we get c 16. Thus the function is y 04 cos 026 t 16 37.
36
(b) Using the SinReg command on the TI84 we obtain the function y a sin bt c d, where a 04, b 026, c 262, and
d 370. Thus we get the model y 04 sin 026t 262 370. 4. (a) Let t be the time in years. We find a function of the form y a sin t c b, where y is the owl population.
a 12 80 20 30. The period is 2 9 3 12 and so
052. b 1 80 20 50. Because the values start at the 212 2
middle we have c 0. Thus the function is y 30 sin 052t 50.
(b) Using the SinReg command on the TI84 we find that for the function y a sin bt c d, where a 258, b 052, c 002, and
d 506. Thus we get the model y 258 sin 052t 002 506. 5. (a) Let t be the time since 1985. We find a function of the form y a sin t c b. a 12 63 22 205. The period is
0
10
x
20
Time y 80 60 40 20
0
2
4
6
8
10
10
12
12 x
Year y 60
052. b 1 63 22 425. The 2 15 9 12 and so 220 2
40
y 205 sin 052 t 6 425.
20
average value occurs in the ninth year, so c 6. Thus, our model is
(b) Using the SinReg command on the TI84, we find that for the function y a sin bt c d, where a 178, b 052, c 311, and
d 424. Thus we get the model y 178 sin 052t 311 424.
0
2
4
6
8
Year since 1985
14
x
CORRECTIONS: p. 8,14,25,40
CHAPTER 7
ANALYTIC TRIGONOMETRY
7.1
Trigonometric Identities 1
7.2 7.3
Addition and Subtraction Formulas 9 Double-Angle, Half-Angle, and Product-Sum Formulas 17
7.4
Basic Trigonometric Equations 27
7.5
More Trigonometric Equations 31 Chapter 7 Review 38 Chapter 7 Test 45
¥
FOCUS ON MODELING: Traveling and Standing Waves 46
1
7
ANALYTIC TRIGONOMETRY
7.1
TRIGONOMETRIC IDENTITIES
1. An equation is called an identity if it is valid for all values of the variable. The equation 2x x x is an algebraic identity
and the equation sin2 x cos2 x 1 is a trigonometric identity. 2. The fact that cos x has the same value as cos x for any x can be expressed as the identity cos x cos x. sin t sin t 3. cos t tan t cos t cos t 1 cot t 4. cos t csc t cos t sin t 1 tan 5. sin sec sin cos sin 1 1 6. tan csc sec cos sin cos 1 sin2 x 1 cos2 x sin2 x 7. tan2 x sec2 x 1 2 2 2 cos x cos x cos x cos2 x 1 sec x sin x cos x tan x 8. 1 csc x cos x sin x 1 2 9. sin 2 y sec y cos2 y sec y cos2 y cos y cos y sin 2 u cos u 10. tan 2 u sin u cos u sin u sin u sin u cos u 2 sin2 u cos2 u 1 cos u cos u csc u sin u sin u sin u sin2 12. cos2 1 tan2 cos2 1 cos2 sin2 1 cos2
11. sin u cot u cos u sin u
1 cos 1 cos2 sin2 sin sec cos cos tan 13. sin sin sin cos sin cos cos cos cot cos cos sin 14. sec 2 2 1 csc sin cos 1 sin sin sin 1 sin x sin x sec x cos x 1 15. cos x tan x sin x 1 cos x cos x sec x cos x sin x 16. cos x cot x cos x sin x sin t sin t tan t tan t cos t 1 17. sin t tan t tan t cos t
1
2
CHAPTER 7 Analytic Trigonometry
18.
cos A 1 cot A 1 cot A sin A sin A sin A cot A sin A sin A sin A cos A csc A sin A
2 3 2 19. sin3 2 x sin x cos x cos x 1 cos x cos x cos x
20. sin4 cos4 cos2 sin2 cos2 sin2 cos2 cos2 sin2 21.
tan2 x sin2 x sec2 x 1 cos2 x sin2 x. 2 2 sec x sec x cos2 x 1 sec2 x 1 1 1 cos2 x sin2 x Another method: 2 sec x sec2 x
cos2 x 1 2 2 sec x cos x cos x 1 cos x sin x sin x 22. cos x sin x tan x sin x sin x cos x 23.
1 cos y cos y 1 cos y 1 cos y 1 cos y cos y cos y 1 1 1 sec y 1 cos y 1 1 cos y cos y
1 cos y 1 sin y sin y 1 sin y sin y sin2 y 2 24. sin y 1 1 csc y sin y 1 1 sin y 1 sin y 25.
cos u 1 sin u2 cos2 u 1 2 sin u sin2 u cos2 u 1 2 sin u 1 1 sin u cos u 1 sin u cos u 1 sin u cos u 1 sin u cos u 1 sin u 2 2 sin u 2 1 sin u 2 2 sec u cos u 1 sin u cos u 1 sin u cos u
26. Note that because sin2 t 1 cos2 t 1 cos t 1 cos t, we can write sin t 1 cos t 1 cos t csc t cot t. 1 cos t sin t sin t sin t 27.
cos x cos x sec x tan x sec x tan x
28. Because tan A tan A, 29.
30.
1 cos t sin t . Thus, 1 cos t sin t
cos x 1 sin2 x cos2 x 1 sin x 1 sin x 1 sin x sin x 1 1 sin x 1 sin x 1 sin x cos x cos x
cot A 1 cot A 1 tan A 1 tan A cot A 1 tan A 1 tan A tan A tan A 1 tan A
1 1 sin 1 sin 1 2 2 sec2 1 sin 1 sin cos2 1 sin2 1 1 tan2 x 1 1 tan2 x 1 sec2 x 2 tan2 x 1 1 1 1 2 2 2 2 2 sec x sec x sec x sec x sec x sec2 x 1 1 11 cos2 x sec2 x sec2 x
SECTION 7.1 Trigonometric Identities
cos x 31. (a) sec x sin x
sin y tan y 1 cos2 y sin2 y cos y 32. (a) 1 csc y cos y cos y sin y
1 sin2 x cos2 x sin x sin x
cos x
1 sin x cos x sin2 x 1 csc x sin x sin x sin x
(b) We graph each side of the equation and see that the cos x and y csc x sin x are graphs of y sec x sin x identical, confirming that the equation is an identity.
cos2 y 1 sec y cos y cos y cos y
(b) We graph each side of the equation and see that the tan x graphs of y and y sec x cos x are csc x identical, confirming that the equation is an identity.
1
-5
3
1 5
-5
-1
5 -1
sin sin 1 tan cos sec 34. sin sin cos sin
cos cos 33. cos cos cos2 1 sec cos 1 cos u sec u cos u cot u cot u 35. tan u cos u 1 2 37. cos2 2 y csc y sin y sin y sin y
36.
cot x sec x cos x 1 sin x 1 csc x sin x cos x
sin x 2 cos x sin x 38. sin x cos x 2 , so tan x 2 sin x 2 2 cos x
2
39. sin x cos x2 sin2 x 2 sin x cos x cos2 x 1 2 sin x cos x
cos sin2 cos2 1 sin sec csc cos sin cos sin cos sin 41. cos x sin x cos x sin x cos x sin x 40. tan cot
42. cot cos sin
cos2 sin2 1 cos cos sin csc sin sin sin
1 1 cos A sec A 1 1 cos A cos A 43. 1 sec A 1 cos A 1 cos A 1 cos A cos2 1 sin2 1 44. sin csc sin sin sin sin 45. 1 cos 1 cos 1 cos2 sin2
1 csc2
sin x cos x cos2 x sin2 x 1 sec x csc x 1 1 47. sec2 y 1 tan2 y 2 cos2 y 1 sin y
46.
48. csc x sin x
1 sin2 x cos2 x 1 sin x cos x cot x sin x sin x sin x
49. tan x cot x2 tan2 x 2 tan x cot x cot2 x tan2 x 2 cot2 x tan2 x 1 cot2 x 1 sec2 x csc2 x
4
CHAPTER 7 Analytic Trigonometry
50. sin2 y cos2 y tan2 y sin2 y cos2 y tan2 y 1 tan2 y sec2 y 2 2 51. 1 sin2 t cos2 t 4 sin2 t cos2 t 2 cos2 t 4 sin2 t cos2 t 4 cos2 t cos2 t sin2 t 4 cos2 t 52.
2 sin x cos x
sin x cos x2 1
2 sin x cos x
sin2 x cos2 x 2 sin x cos x 1 2 sin x cos x 1 2 sin x cos x 1 1
2 sin x cos x 2 sin x cos x sin2 x cos2 x 1
sin2 x 1 cos2 x csc x sin x sin x sin x cos2 t 1 2 t cos2 t 2 t csc2 t 1 cot2 t cos2 t 54. cot2 t cos2 t cos 1 cos sin2 t sin2 t 1 1 cos t cos t cos t 1 cos2 t sec t cos t cos t cos t sin2 t 55. 1 1 sec t cos t 1 cos t cos t cos x cos x 1 cos2 x 56. cot x csc x cos x 1 cot x cos x cot x csc x cos x csc x sin x sin x sin x sin x cos2 x 1 sin2 x sin x sin x sin x 57. cos2 x sin2 x cos2 x 1 cos2 x 2 cos2 x 1 58. 2 cos2 x 1 2 1 sin2 x 1 2 2 sin2 x 1 1 2 sin2 x 2 2 59. sin4 cos4 sin2 cos2 sin2 cos2 sin2 cos2 sin2 cos2 cos2 x sin2 x cos2 x 1 60. 1 cos2 x 1 cot2 x sin2 x 1 sin2 x 53. csc x cos2 x sin x
sin2 t 2 sin t cos t cos2 t sin2 t cos2 t 2 sin t cos t 1 sin t cos t2 2 2 sec t csc t sin t cos t sin t cos t sin t cos t sin t cos t sin t cos t 1 1 sin t cos t 1 1 62. sec t csc t tan t cot t 2 sec2 t csc2 t cos t sin t cos t sin t cos2 t sin t
61.
sin2 u sin2 u 1 1 2 2 2 2 1 tan2 u 1 cos u cos2 u cos u cos u sin u 63. 2 2 2 2 2 2 2 1 tan u cos u sin u cos u sin2 u sin u sin u cos u 1 1 2 2 cos u cos u 1 sec2 x 1 sec2 x 1 64. 1 cos2 x 1 1 tan2 x sec2 x sec2 x 1 1 1 1 sec x csc x cos x sin x cos x sin x sin x cos x sin x cos x sin x cos x 65. cos x cos x sin x cos x sin x sin x tan x cot x sin2 x cos2 x cos x sin x cos x sin x sin x cos x sin x cos x sin x cos x cos x sin x cos x sin x 66. sin x cos x 1 1 sin x cos x sec x csc x sin x cos x cos x sin x cos x sin x sin x 1 cos x 1 cos x sin x sin x 1 2 cos x cos2 x sin2 x 1 cos x 67. sin x 1 cos x sin x 1 cos x 1 cos x sin x sin x 1 cos x 2 2 cos x 2 1 cos x 2 csc x sin x 1 cos x sin x 1 cos x
SECTION 7.1 Trigonometric Identities
68.
csc2 y cot2 y csc2 y cot2 y sin2 y 1 cos2 y sin2 y 2 2 2 sec y sec y tan2 y sin y 1 Another method:
csc2 y cot2 y sec2 y
sin2 y
cos2 y sin2 y
cos2 y
cos2 y sin2 y
1 cos2 y
1 cos2 y sin2 y
cos2 y cos2 y
sin2 u sin2 u cos2 u sin2 u 2 u tan2 u sin2 u 1 cos cos2 u cos2 u cos2 u 4 4 2 2 2 70. sec x tan x sec x tan x sec x tan2 x 1 sec2 x tan2 x sec2 x tan2 x
69. tan2 u sin2 u
sin x 1 1 tan x cos x cos x cos x sin x 71. sin x cos x 1 tan x cos x sin x 1 cos x sin2 1 1 sin cos2 cos sin csc sin sin 72. cos cos sin cos cos cot cos sin 1 1 sin cos sin sin 1 sec x tan x sec x tan x 2 sec x 2 sec x 1 2 sec x 73. 2 2 sec x tan x sec x tan x 1 sec x tan x sec x tan x sec x tan x 74.
cos2 t tan2 t 1
1 sin2 t 1 sec2 t tan2 t cos2 t sin2 t sin2 t sin2 t 1 sin x 1 sin x 4 sin x 4 sin x 1 sin x 1 sin x2 1 sin x2 4 tan x sec x 4 75. 1 sin x 1 sin x cos x cos x 1 sin x 1 sin x cos2 x 1 sin2 x sin x sin x cos y cos x sin y sin y tan x tan y cos x cos y cos x cos y cos x 76. cos y cos x sin y sin x cos y cot x cot y sin x sin y sin x sin y sin x sin y sin x cos y cos x sin y sin x sin y tan x tan y cos x cos y cos x sin y sin x cos y cos x cos y sin x cos x sin2 x sin x cos x cos2 x sin3 x cos3 x 77. sin2 sin x cos x cos2 x 1 sin x cos x sin x cos x sin x cos x sin y csc y sin2 y sin y csc y csc2 y sin3 y csc3 y sin2 y csc2 y 1 78. sin y csc y sin y csc y 79.
1 cos 1 cos 1 cos2 sin2 sin 1 cos sin sin 1 cos sin 1 cos sin 1 cos 1 cos
sin2 t tan2 t
1
sin x 1 sin x 1 sin x 1 sin2 x 1 cos2 x 2 sin x 1 sin x 1 sin x 1 sin x 1 sin x 12 1 sin sin cos tan 81. 1 sin cos sin cos 1 tan cos sin A sin A 1 cos A sin A 1 cos A sin A 1 cos A cos A 82. cot A cot A cot A 1 cos A 1 cos A 1 cos A sin A 1 cos2 A sin2 A 1 cos A cos A 1 csc A sin A sin A sin A sin A sec x sec x tan x sec x sec x tan x sec x sec x tan x sec x sec x sec x tan x 83. 2 2 sec x tan x sec x tan x sec x tan x 1 sec x tan x
80.
5
6
CHAPTER 7 Analytic Trigonometry
sec2 tan2 1 sec tan sec tan sec tan sec tan sin x cos x sin2 x cos2 x sin x cos x sin x cos x sin x cos x2 sin x cos x2 85. 2 2 sin x cos x sin x cos x sin x cos x sin x cos x sin x cos x sin x cos x sin x cos x2 tan sin tan sin tan sin tan sin tan sin tan sin tan sin 86. tan sin tan sin tan sin tan2 sin2 sin2 sec2 1 84. sec tan sec tan
tan sin tan sin
tan sin tan sin tan2 sin2 1 sin x 1 sin x 1 2 sin x sin2 x 1 2 sin x sin2 x 1 sin x 1 2 sin x sin2 x 87. 2 2 2 2 1 sin x 1 sin x 1 sin x cos x cos x cos x cos2 x 1 sin x
sec2 x 2 sec x tan x tan2 x sec x tan x2
88.
1 sin x 1 sin x 1 sin x 1 sin x2 1 sin x2 1 sin x 1 sin x 1 sin x cos2 x 1 sin2 x
1 sin x 2 tan x sec x2 cos x
cos x 1 cos x 1 cos x 1 cos x 1 cos2 x sin2 x 1 sin x sin x sin x sin x 1 cos x sin x 1 cos x sin x 1 cos x 1 1 1 1 1 cos x csc x cot x 1 cos x sin x sin x sin x 1 sin u 1 sin u sin u tan u sin u sec u 1 cos u cos u 90. 1 sin u sec u 1 sin u tan u sin u 1 sin u cos u cos u sin sin x sin 91. x sin ; then tan (since cos 0 for 0 2 ). 2 2 2 cos cos 1x 1 sin 92. x tan ; then 1 x 2 1 tan2 1 sec2 1 sec2 sec (since sec 0 for 0 2 ). 93. x sec ; then x 2 1 sec2 1 tan2 1 1 tan2 tan (since tan 0 for 0 2) 89. csc x cot x
94. x 2 tan ; then 1 1 1 1 2 2 2 4 tan 2 sec2 x 4x 4 tan2 4 1 tan2 2 tan 2 4 2 tan 2 1 1 cos2 cos 18 cot2 cos 2 8 8 tan sec sin2 2 2 2 95. x 3 sin ; then 9 x 9 3 sin 9 9 sin 9 1 sin2 3 cos2 3 cos cos 0 for 0 ). 2
96. x 5 sec ; then
x 2 25 x
5 sec 2 25 5 sec
tan 0 for 0 2 ).
25 sec2 1 5 sec
(since
sin 5 tan2 tan cos sin (since 1 5 sec sec cos
97. f x cos2 x sin2 x, g x 1 2 sin2 x. From the graph, f x g x this appears to be an identity. Proof:
1
f x cos2 x sin2 x cos2 x sin2 x 2 sin2 x 1 2 sin2 x g x. Since
f x g x for all x, this is an identity.
-5
5 -1
SECTION 7.1 Trigonometric Identities
98. f x tan x 1 sin x, g x
sin x cos x . From the graph, f x g x does 1 sin x
not appear to be an identity. In order to show this, let x 4 . Then, 21 . However, f 4 1 1 1 2
2
7
1
-5
1 1 1 1 2 2 2 . Since f g , this is not an g 4 2 4 4 1 21 2 1 2 2
5 -1
identity.
99. f x sin x cos x2 , g x 1. From the graph, f x g x does not
appear to be an identity. In order to show this, we can set x 4 . Then we have 2 2 2 1 2 2 2 1 g f 4 1 4 . Since 2 2 2 f 4 g 4 , this is not an identity.
2 1
-5
100. f x cos4 x sin4 x, g x 2 cos2 x 1. From the graph, f x g x appears to be an identity. In order to prove this, simplify the expression f x: f x cos4 x sin4 x cos2 x sin2 x cos2 x sin2 x cos2 x sin2 x 1 2 cos2 x cos2 x sin2 x 2 cos2 x cos2 x sin2 x 2 cos2 x 1 g x
5
1
-5
5 -1
Since f x g x for all x, this is an identity.
101. tan x cot x2 tan2 x 2 tan x cot x cot2 x tan2 x 2 cot2 x tan2 x 1 cot2 x 1 sec2 x csc2 x 102.
1 cos x sin x 1 cos2 x sin2 x 2 cos x sin x cos x sin x 1 cos x sin x 1 cos x sin x 1 cos x sin x 1 cos x sin x 1 cos x sin x 1 cos2 x sin2 x 2 cos x 2 2 cos x 2 sin x 2 sin x cos x 1 sin x 1 cos x sin x 1 cos x 2 cos x 1 cos x cos x 2 cos x 2 cos x
sin cos 1 1 103. sin tan cos cot sin cos sin 1 cos 1 cos sin cos sin 1 1 sin 1 cos 1 sin 1 cos 1 cos sin
104. sin6 cos6 sin2 cos2 sin4 sin2 cos2 cos4 sin2 1 cos2 sin2 cos2 cos2 1 sin2 sin2 cos2 3 sin2 cos2 1 3 sin2 cos2
105.
sin2 y tan2 y
cos2 y cot2 y
sin2 y cos2 y
sin2 y sin2 y cos2 y sin2 y 2 y cos2 y 1 sin 2 2 sin6 y sin2 y cos y cos y tan6 y 2 2 2 2 2 cos6 y sin y cos y cos y cos y cos2 y sin2 y 1 cos y sin2 y
sin2 y
8
106.
CHAPTER 7 Analytic Trigonometry
1 sin x cos x2
1 sin2 x 2 sin x cos x cos2 x
cos2 x
sec2 x
1
1 cos2 x tan2 x 2 tan x 1 sin x 1 2 cos x cos2 x sin2 x
1 tan x2
sin x ln sin x ln sin x ln cos x ln sin x 107. ln tan x sin x ln tan x ln sin x ln cos x 1 2 ln sin x ln sec x 2 ln sin x ln cos x 108. ln tan x ln cot x ln tan x cot x ln 1 0 2
2
2
2
2
2
2
2
109. esin x etan x e1cos x esec x1 e1cos xsec x1 esec x e cos x
110. e x2 lnsin x e x e2 lnsin x e x elnsin x elnsin x e x sin2 x
111. LHS R cos sin 2 R sin sin 2 R cos 2 R sin 2 cos2 sin2 R cos 2 R sin 2 R cos 2 RHS
112. (a) x y2 x 2 2x y y 2 is an identity.
(b) x 2 y 2 1 is not an identity. For example, 12 12 1. (c) x y z x y xz is the Distributive Law, an identity. 2
2
(d) This is an identity: LHS esin xcos x e1 RHS. 2
(e) xeln x x x 2 x 3 is an identity, because it is true for all values of x for which both sides of the equation are defined (x 0). (f) sin t cos t 1 is not an identity. For example, sin 4 cos 4 2. (g) x 2 tan2 x 0 is not an identity.
(h) t 2 cos2 t t cos t t cos t (difference of squares).
113. (a) Choose x 2 . Then sin 2x sin 0, whereas 2 sin x 2 sin 2 2. 2 2 2 2 2 2 4 1. (b) Choose 4 . Then sec csc
replace "θ" with "x" in part (b) (3 times)
1 1 2 . (c) Choose x 4 and y 4 . Then sin x y sin 2 1, whereas sin x sin y sin 4 sin 4 2
2
2
Since these are not equal, the equation is not an identity. 1 1 1 (d) Choose x 1 whereas 4 . Then sin x cos x sin cos 1 2 1 4 4 2 2 csc x sec x csc 4 sec 4 2 2. Since these are not equal, the equation is not an identity.
114. No. All this proves is that f x g x for x in the range of the viewing rectangle. It does not prove that these functions
x4 x6 x2 and g x cos x. In the first viewing rectangle 2 24 720 the graphs of these two functions appear identical. However, when the domain is expanded in the second viewing rectangle, you can see that these two functions are not identical. are equal for all values of x. For example, let f x 1
1
-2
2 -1
it doesnt really prove anything if the graphs "appear" to be the same
-10
10 -5 -10
shows
SECTION 7.2 Addition and Subtraction Formulas
9
115. Answers will vary. 116. Label a the side opposite , b the side opposite u, and c the hypotenuse. Since u 2 , we must have u 2 a 2 u. Next we express all six trigonometric function for each angle: cos u c sin cos u sin 2 u , b a sin u bc cos sin u cos 2 u , tan u a cot tan u cot 2 u , cot u b tan c c cot u tan 2 u , sec u a csc sec u csc 2 u , and csc u b sec csc u sec 2 u .
117. From the diagram, we see that sin a 23 , cos b 23 , tan c 23 , and
c
cot d 23 . So since a b 2 and c d 2 ,
a
sin1 23 cos1 23 tan1 32 cot1 32 abcd 2 2 .
3
3
d
b 2
7.2
2
ADDITION AND SUBTRACTION FORMULAS
1. If we know the values of the sine and cosine of x and y we can find the value of sin x y using the addition formula for sine, sin x y sin x cos y cos x sin y. 2. If we know the values of the sine and cosine of x and y we can find the value of cos x y using the subtraction formula for cosine, cos x y cos x cos y sin x sin y.
6 2 4 2 3 2 6 2 1 4. sin 15 sin 45 30 sin 45 cos 30 cos 45 sin 30 2 2 2 2 4 6 5. cos 105 cos 60 45 cos 60 cos 45 sin 60 sin 45 12 22 23 22 2 4 2 6. cos 195 cos 15 cos 45 30 cos 45 cos 30 sin 45 sin 30 22 23 22 12 6 4 3 tan 30 1 3 3 tan 45 3 7. tan 15 tan 45 30 2 3 3 1 tan 45 tan 30 3 3 11 3 3 1 tan 45 tan 30 33 3 8. tan 165 tan 15 tan 45 30 32 3 1 tan 45 tan 30 3 3 11 3 2 1 2 3 6 2 7 9. sin 19 12 sin 12 sin 4 3 sin 4 cos 3 cos 4 sin 3 2 2 2 2 4 2 3 2 2 6 17 5 1 10. cos 12 cos 12 cos 4 6 cos 4 cos 6 sin 4 sin 6 2 2 2 2 4
3. sin 75 sin 45 30 sin 45 cos 30 cos 45 sin 30 22 23 22 12
tan tan 3 tan 4 1 3 3 2 tan 12 11. tan 12 3 4 1 tan 1 3 3 tan 4 sin 5 sin sin cos cos sin 2 3 2 1 6 2 12. sin 512 12 4 6 4 6 4 6 2 2 2 2 4 3 2 2 6 2 1 13. cos 11 12 cos 12 cos 3 4 cos 3 cos 4 sin 3 sin 4 2 2 2 2 4 3 tan 5 tan tan 4 tan 6 1 3 3 3 2 3 14. tan 712 12 4 6 1 tan 3 3 4 tan 6 11 3 3
15. cos 23 cos 67 sin 23 sin 67 cos 23 67 cos 90 0
16. sin 35 cos 25 cos 35 sin 25 sin 35 25 sin 60 23
10
CHAPTER 7 Analytic Trigonometry
3 sin sin 3 sin 1 17. sin 34 cos cos 4 4 4 4 4 2 18.
tan 35 tan 15
3 tan 2 3 tan 5 15 3
1 tan 35 tan 15
tan 55 tan 10 tan 55 10 tan 45 1 1 tan 55 tan 10 cos sin 5 sin cos 5 cos 3 20. cos 518 9 18 9 18 9 6 2 sin 2 u sin 2 cos u cos 2 sin u 1 cos u 0 sin u cos u 21. tan 2 u cos u cos cos u sin sin u 0 cos u 1 sin u sin u cot u 2 2 2 cos 2 cos u sin 2 sin u cos 2 u 0 cos u 1 sin u sin u 22. cot 2 u sin u sin cos u cos sin u 1 cos u 0 sin u cos u tan u 2 2 2 19.
23. sec 2 u
24. csc 2 u
1 1 1 1 cos u sin sin u 0 cos u 1 sin u sin u csc u cos cos u 2 2 2 1 1 1 1 cos u cos sin u 1 cos u 0 sin u cos u sec u sin sin u 2 2 2
25. sin x 2 sin x cos 2 cos x sin 2 0 sin x 1 cos x cos x 26. cos x 2 cos x cos 2 sin x sin 2 0 cos x 1 sin x sin x 27. sin x sin x cos cos x sin 1 sin x 0 cos x sin x
28. cos x cos x cos sin x sin 1 cos x 0 sin x cos x tan x tan tan x 0 29. tan x tan x 1 tan x tan 1 tan x 0 cos x cos cos x sin x 2 2 sin x sin 2 30. cot x 2 sin x sin x cos cos x sin cos x tan x 2 2 2 cos x cos sin x 1 cos x 0 sin x cos x and x sin 31. LHS sin 2 2 2 cos x cos sin x 1 cos x 0 sin x cos x. Therefore, LHS RHS. x sin RHS sin 2 2 2 1 32. cos x 3 sin x 6 2 cos x 23 sin x 23 sin x 12 cos x 0 tan x tan 3 tan x 3 33. tan x 3 1 tan x tan 1 3 tan x 3 tan x tan 4 tan x 1 tan x 1 34. tan x 4 1 tan x tan 1 tan x 1 tan x 1 4
35. sin x y sin x y sin x cos y cos x sin y sin x cos y cos x sin y 2 cos x sin y
36. cos x y cos x y cos x cos y sin x sin y cos x cos y sin x sin y 2 cos x cos y 1 1 1 1 tan x tan y cot x cot y 1 1 cot x cot y cot x cot y 37. cot x y 1 1 tan x y tan x tan y cot x cot y cot y cot x cot x cot y 1 1 1 1 tan x tan y cot x cot y 1 1 cot x cot y cot x cot y 38. cot x y 1 1 tan x y tan x tan y cot x cot y cot x cot y cot x cot y sin y sin x cos y cos x sin y sin x y sin x 39. tan x tan y cos x cos y cos x cos y cos x cos y sin x sin y cos x cos y sin x sin y cos x y 40. 1 tan x tan y 1 cos x cos y cos x cos y cos x cos y
SECTION 7.2 Addition and Subtraction Formulas
41.
sin x y tan x tan y sin x cos y cos x sin y tan x tan y cos x cos y 1 tan x tan y cos x cos y sin x sin y cos x y 1 tan x tan y cos x cos y
42.
sin x y sin x y sin x cos y cos x sin y sin x cos y cos x sin y 2 cos x sin y tan y cos x y cos x y cos x cos y sin x sin y cos x cos y sin x sin y 2 cos x cos y
11
43. cos x y cos x y cos x cos y sin x sin y cos x cos y sin x sin y cos2 x cos2 y sin2 x sin2 y cos2 x 1 sin2 y 1 cos2 x sin2 y cos2 x sin2 y cos2 x sin2 y cos2 x sin2 y cos2 x sin2 y
44. cos x y cos y sin x y sin y cos x cos y sin x sin y cos y sin x cos y cos x sin y sin y cos2 y cos x sin2 y cos x cos2 y sin2 y cos x cos x 45. sin x y z sin x y z sin x y cos z cos x y sin z
cos z sin x cos y cos x sin y sin z cos x cos y sin x sin y
sin x cos y cos z cos x sin y cos z cos x cos y sin z sin x sin y sin z 46. The addition formula for the tangent function can be written as tan A tan B tan A B 1 tan A tan B. Also note that tan A tan A. Using these facts, we get tan x y tan y z tan z x tan x y y z 1 tan x y tan y z tan z x tan x z 1 tan x y tan y z tan z x tan x z tan z x tan x y tan y z tan x z
tan x z tan x z tan x y tan y z tan x z 0 tan x y tan y z tan x z tan x y tan y z tan z x
47. We want to write cossin1 x tan1 y in terms of x and y only. We let
1
sin1 x and tan1 y and sketch triangles with angles and such that sin x and tan y. From the triangles, we have cos 1 x 2 , 1
x
Ï1+y@ ú
¬
Ï1-x@
y
y
1
cos , and sin . tan y sin x 1 y2 1 y2 From the subtraction formula for cosine we have 1 y 1 x2 x y 1 1 2 x cos sin x tan y cos cos cos sin sin 1 x 2 2 1y 1y 1 y2
x 48. Let sin1 x and cos1 y. From the triangles, tan and 1 x2 1 y2 , so using the addition formula for tangent, we have tan y x 1 y2 y 1 x2 tan sin1 x cos1 y tan 1 y2 x 1 y 1 x2 x y 1 x 2 1 y2 y 1 x 2 x 1 y2
1
x
1
Ï1-y@
¬
ú
sin x
cos y
Ï1-x@
y
12
CHAPTER 7 Analytic Trigonometry
1 49. Let tan1 x and tan1 y. From the triangles, cos , 1 x2 x 1 y sin , cos , and sin , so using the 1 y2 1 y2 1 x2
Ï1+x@ ¬
subtraction formula for sine, we have sin tan1 x tan1 y sin sin cos cos sin
Ï1+y@
x
ú
1
y
1
tan y
tan x
x 1 1 y xy 2 2 2 2 1y 1y 1x 1x 1 x 2 1 y2
50. Let sin1 x and cos1 y. From the triangles, cos 1 x 2 and sin 1 y 2 , so using the addition formula for sine, we have sin sin1 x cos1 y sin sin cos cos sin x y 1 x 2 1 y2 x y 1 x 2 1 y2
1
1
x
Ï1-y@
¬
ú
sin x
cos y
Ï1-x@
y
1 1 , so the addition formula for sine gives 51. We know that cos1 21 3 and tan 4 sin cos cos sin 3 2 1 2 6 2 . sin cos1 21 tan1 1 sin 3 4 3 4 3 4 2 2 2 2 4 1 3 , so the addition formula for cosine gives 52. We know that sin1 23 3 and cot 6 cos cos sin sin 1 3 3 1 0. cos sin1 23 cot1 3 cos 3 6 3 6 3 6 2 2 2 2
53. We sketch triangles such that sin1 43 and cos1 31 . From the triangles, we have tan 3 and tan 2 2, so the subtraction formula for 7
4
tangent gives 3 2 2 tan tan 7 tan sin1 34 cos1 13 1 tan tan 1 3 2 2 7 3 2 14 76 2
3
3
¬
ú
Ï7
sin 34
2Ï2
1
cos 13
54. We sketch triangles such that cos1 23 and tan1 12 . From the
triangles, we have sin 35 , cos 2 , and sin 1 , so the addition 5 5 formula for sine gives sin cos1 32 tan1 12 sin cos cos sin 35 2 23 1 2 10 2 5 23 15 3 5
5
5
3
¬
Ï5
Ï5 ú
2
cos 23
2
tan 12
1
13
SECTION 7.2 Addition and Subtraction Formulas
55. As in Example 7, we sketch the angles and in standard position
y
(_3, 4)
3
with terminal sides in the appropriate quadrants and find the remaining sides using the Pythagorean Theorem. To find sin ,
5
4
we use the subtraction formula for sine and the triangles we have
1 (_3, _1)
¬
sketched: sin sin cos cos sin 3 12 3 4 3 1 5 5 10 10 5 10 3 10 10
sin 45
3
have cos cos cos sin sin 3 2 4 6 4 5 5 5 3 5 3 15
4
2 Ï5
(_3, _4)
(2, _Ï5)
tan 43 y
cos 23 (12, 5)
13 ¬
x
3
5
(_2Ï5, Ï5)
x
12
y
y
2Ï2
x
(_1, _2Ï2)
3
x
cos 2 5 5
¬
15
ú
2Ï5
5 sin 13
1
y 5
Ï5
5
have
2 30 1 15 2 2
ú
x
58. Using the addition formula for tangent and the triangles shown, we 2 2 1 tan tan 15 tan 1 tan tan 1 2 2 1
y
y ¬
56. Using the addition formula for cosine and the triangles shown, we
sin sin cos cos sin 5 2 5 12 2 5 5 13 5 13 5 65
x
Ï10
tan 13
x
3
57. Using the addition formula for sine and the triangles shown, we have
y ú
(_Ï15, 1) 1
4 Ï15
ú x
sin 14
cos 13 2 3 12 4 2. Thus, sin 12 and cos 2 3 56 , so 59. k A2 B 2 3 sin x cos x k sin x 2 sin x 56 . 60. k A2 B 2 12 12 2 and satisfies sin 1 , cos 1 4 . Thus, 2 2 sin x cos x k sin x 2 sin x 4 . 5 1 and cos 5 1 7 , so 61. k A2 B 2 52 52 50 5 2. Thus, sin 4 5 2 2 5 2 2 7 7 5 sin 2x cos 2x k sin 2x 5 2 sin 2x 4 5 2 sin 2 x 8 . 2 62. k A2 B 2 32 3 3 36 6 and satisfies sin 3 6 3 23 , cos 36 12 3 . Thus, 1 3 sin x 3 3 cos x k sin x 6 sin x 3 6 sin x 3 .
14
Try to have the graph match the one in the answers CHAPTER 7 Analytic Trigonometry
graph is incorrect for #63
63. (a) g x cos 2x 3 sin 2x 2 3 4 2, and satisfies k 12
(b) This is a sine curve with amplitude 2, period , and . phase shift 12 y
sin 12 , cos 23 6 . Thus, we can
2
write
g x k sin 2x 2 sin 2x 6 2 sin 2 x 12 .
1 _2¹
0
_¹
¹
2¹ x
_1 _2
64. (a) f x sin x cos x k
12 12 2, and
satisfies sin cos 1 4 . Thus, 2
(b) This is a sine curve with amplitude
2, period 2,
and phase shift 4. y
we can write
2
f x k sin x 2 sin x 4 .
1 _2¹
_¹
0
¹
2¹ x
_1 _2
65. f x cos x. Now
f x h f x cos x h cos x cos x cos h sin x sin h cos x h h h 1 cos h sin h cos x 1 cos h sin h sin x cos x sin x h h h
66. g x sin x. Now
sin x h sin x sin x cos h cos x sin h sin x g x h g x h h h sin h 1 cos h sin h cos x sin x 1 cos h cos x sin x h h h
2 67. (a) y sin2 x 4 sin x 4 . From the graph we see that the value of y seems to always be equal to 1.
1
-5
2 sin x cos cos x sin 2 2 (b) y sin2 x 4 sin x 4 sin x cos 4 cos x sin 4 4 4 2 2 1 sin x cos x 1 sin x cos x 12 sin x cos x2 sin x cos x2 2 2 1 2 2 sin x 2 sin x cos x cos2 x sin2 x 2 sin x cos x cos2 x 12 [1 2 sin x cos x 1 2 sin x cos x] 12 2 1
5
SECTION 7.2 Addition and Subtraction Formulas
68. (a) y 12 [cos x cos x ]. The graph of y appears to be the same as
15
1
that of cos x.
-5
5 -1
(b) y 12 [cos x cos x ]
12 [cos x cos sin x sin cos x cos sin x sin ]
12 [ cos x 0 cos x 0] 12 2 cos x cos x 69. If 2 , then 2 . Now if we let y x , then sin x cos x 2 sin y cos y 2 sin y sin y 0.
Therefore, sin x cos x 0.
70. Let A and B be the two angles shown in the diagram. Then 180 A B, 90 A, and 90 B. Subtracting the second and third equation from the first, we get
6
4
Œ
A
180 90 90 A B A B 4
71. tan1
xy 1 xy
72. cot u
B
º
4
. Then
tan tan
3
3
8 9 tan tan 17 6 4 4 3 12 112 2 17 12 6 . 1 tan tan 1 1
tan1
6
tan u tan 1 tan u tan
4
2
tan1 tan u u tan1 x tan1 y
x 1x 1 cot cot u 1 0, so tan1 x tan1 cot cot u x 1x
1 u cot1 cot u cot1 0 . x 2
y y and tan . Thus, m tan . x x tan 2 tan 1 (b) tan tan 2 1 . From part (a), we have m 1 tan 1 and m 2 tan 2 . Then by 1 tan 2 tan 1 m m1 . substitution, tan 2 1 m1m2
73. (a) By definition, m
(c) Let be the unknown angle as in part (b). Since m 1 13 and m 2 12 , 1
tan
1
1
2
6
m2 m1 2 3 67 17 tan1 17 0142 rad 81 . 1 m1m2 1 1 1 3
(d) From part (b), we have cot have 0
1 m1m2 . If the two lines are perpendicular then 90 and so cot 0. Thus we m2 m1
1 m1m2 0 1 m 1 m 2 m 1 m 2 1 m 2 1m 1 . Thus m 2 is the negative reciprocal of m 1 . m2 m1
74. Clearly C 4 . Now tan A B
1
1
tan A tan B 3 1 2 1 1. Thus A B 4 , so A B C 4 4 2 . 1 tan A tan B 1 3
2
16
CHAPTER 7 Analytic Trigonometry
75. (a) y f 1 t f 2 t 5 sin t 5 cos t 5 sin t cos t 10
-5
5
A2 B 2 52 52 5 2. Therefore, 1 B 5 sin and 2 2 5 2 2 A B A 1 cos . Thus, 4. 2 A2 B 2
(b) k
-10
76. (a) f t C sin t C sin t C sin t C sin t cos cos t sin C 1 cos sin t C sin cos t A sin t B cos t
where A C 1 cos and B C sin . (b) In this case, f t 10 1 cos 3 sin t 10 sin 3 cos t 15 sin t 5 3 cos t. Thus 2 3 and sin 5 3 1 , so . Therefore, k 152 5 3 10 3 and has cos 15 2 2 6 10 3 10 3 f t 10 3 sin t 6 . 77. sin s t cos 2 s t cos 2 s t cos 2 s cos t sin 2 s sin t sin s cos t cos s sin t. The last equality comes from again applying the cofunction identities. sin s cos t cos s sin t sin s t 78. tan s t cos s t cos s cos t sin s sin t sin t 1 sin s sin s cos t cos s sin t cos s cos t cos t tan s tan t cos s sin s sin t 1 cos s cos t sin s sin t 1 tan s tan t 1 cos s cos t cos s cos t 79. Suppose such a triangle exists. We translate it (preserving its supposed equilaterality) so that one of its vertices lies at the origin and its other vertices become P1 m 1 n 1 and P2 m 2 n 2 . n n Then O P1 has slope 1 and O P2 has slope 2 . The angle between these sides is 3 , so from Exercise 73(b), m1 m2 n n2 1 m 1 n2 m 2 n1 m2 m1 3 tan n 1 n 2 m m n n —a rational number. But 3 is irrational, so such a triangle cannot exist. 3 1 2 1 2 1 m1 m2 Note: The triangle pictured in the text has side lengths 17, 17, and 3 2 and approximate interior angles 5904 , 5904 , and 6193 . 80. Using the addition formula for tangent twice, we get tan A tan B tan C tan A tan B 1 tan A tan B tan C tan A B tan C 1 tan A tan B . tan A B C tan A tan B 1 tan A B tan C 1 tan A tan B tan tan B tan C tan C 1 1 tan A tan B But A B C 180 and tan 180 0, so the numerator of the last expression must be 0. Thus, tan A tan B 1 tan A tan B tan C 0 tan A tan B tan C tan A tan B tan C 0 tan A tan B tan C tan A tan B tan C.
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
7.3
17
DOUBLE-ANGLE, HALF-ANGLE, AND PRODUCT-SUM FORMULAS
1. If we know the values of sin x and cos x, we can find the value of sin 2x using the DoubleAngle Formula for Sine: sin 2x 2 sin x cos x.
x x 2. If we know the value of cos x and the quadrant in which lies, we can find the value of sin using the HalfAngle Formula 2 2 x 1 cos x . for Sine: sin 2 2 5 , x in quadrant I cos x 12 and tan x 5 . Thus, sin 2x 2 sin x cos x 2 5 12 120 , 3. sin x 13 13 12 13 13 169 120 2 2 sin 2x 169 5 119 120 169 120 cos 2x cos2 x sin2 x 12 13 14425 13 169 169 , and tan 2x cos 2x 119 169 119 119 . 169
4. tan x 43 . Then sin x 45 and cos x 35 (x is in quadrant II). Thus, sin 2x 2 sin x cos x 2 45 35 24 25 , 24 2 2 7 , and tan 2x sin 2x 25 24 25 24 . cos 2x cos2 x sin2 x 35 45 916 7 25 25 25 7 cos 2x 7 25
5. cos x 45 . Then sin x 35 (csc x 0) and tan x 34 . Thus, sin 2x 2 sin x cos x 2 35 45 24 25 , 24 2 2 7 , and tan 2x sin 2x 25 24 25 24 . cos 2x cos2 x sin2 x 45 35 169 7 25 25 25 7 7 cos 2x 25
6. csc x 4. Then sin x 14 , cos x 415 , and tan x 1 (tan x 0). Thus, 15 2 2 7 sin 2x 2 sin x cos x 2 14 415 815 , cos 2x cos2 x sin2 x 415 14 151 16 8 , and 15 sin 2x tan 2x 78 815 87 715 . cos 2x 8
7. sin x 35 . Then, cos x 45 and tan x 34 (x is in quadrant III). Thus, sin 2x 2 sin x cos x 2 35 45 24 25 , 24 2 2 sin 2x 25 7 24 25 24 cos 2x cos2 x sin2 x 45 35 169 25 25 , and tan 2x cos 2x 7 25 7 7 . 25
8. sec x 2. Then cos x 12 , sin x 23 , and tan x 3 (x is in quadrant IV). Thus, 2 1 3 , cos 2x cos2 x sin2 x 1 2 3 12 , and sin 2x 2 sin x cos x 2 23 2 2 2 2 23 sin 2x tan 2x 3. 1 cos 2x 2
9. tan x 13 and cos x 0, so sin x 0. Thus, sin x 1 and cos x 3 . Thus, 10 10 6 3 , cos 2x cos2 x sin2 x 3 2 1 2 8 4 , 3 10 sin 2x 2 sin x cos x 2 1 5 10 5 10
3
10
sin 2x 45 35 54 34 . and tan 2x cos 2x 5
10
10
18
CHAPTER 7 Analytic Trigonometry
10. cot x 23 .
Then tan x 32 , sin x 3 (sin x 0), and cos x 2 . Thus, 13 13 2 2 5 2 x sin2 x 2 2 3 12 , cos 2x cos 49 sin 2x 2 sin x cos x 2 3 13 13 13 , and
13 13 12 sin 2x 12 135 12 13 tan 2x 13 5 5. cos 2x 13 2 1 cos 2x 2 14 12 cos 2x 14 cos2 2x 11. sin4 x sin2 x 2
13
13
1 cos 4x 14 12 cos 2x 14 14 12 cos 2x 18 18 cos 4x 38 12 cos 2x 18 cos 4x 2 12 34 cos 2x 14 cos 4x 2 1 cos 2x 2 4 2 14 12 cos 2x 14 cos2 2x 12. cos x cos x 2 1 cos 4x 14 12 cos 2x 14 14 12 cos 2x 18 18 cos 4x 38 12 cos 2x 18 cos 4x 2 12 34 cos 2x 14 cos 4x
13. We use the result of Example 4 to get 1 1 cos 2x cos 4x cos 2x cos 4x. cos2 x sin4 x sin2 x cos2 x sin2 x 18 18 cos 4x 12 12 cos 2x 16
14. Using Example 4, we have 1 1 cos 4x cos 2x cos 2x cos 4x. cos4 x sin2 x cos2 x cos2 x sin2 x 12 1 cos 2x 18 1 cos 4x 16 2 15. Since sin4 x cos4 x sin2 x cos2 x we can use the result of Example 4 to get sin4 x cos4 x
1 1 cos 4x 2 1 1 cos 4x 1 cos2 4x 8 8 64 32 64
1 1 cos 4x 1 1 1 cos 8x 1 1 cos 4x 1 1 cos 8x 64 32 64 2 64 32 128 128 3 1 1 1 3 1 128 32 cos 4x 128 cos 8x 32 4 cos 4x 4 cos 8x
16. Using the result of Exercise 12, we have 2 cos6 x cos2 x cos2 x 38 12 cos 2x 18 cos 4x 12 12 cos 2x
3 1 cos 2x 1 cos 4x 3 cos 2x 1 cos2 2x 1 cos 2x cos 4x 16 4 16 16 4 16
Because 14 cos2 2x
1 cos 4x 1 1 cos 4x , the last expression is equal to 4 2 8 8
3 1 1 3 1 1 1 16 4 cos 2x 16 cos 4x 16 cos 2x 8 8 cos 4x 16 cos 2x cos 4x 5 7 cos 2x 3 cos 4x 1 cos 2x cos 4x 1 5 7 cos 2x 3 cos 4x cos 2x cos 4x 16 16 16 16 16 17. sin 15 12 1 cos 30 12 1 23 14 2 3 12 2 3 3
1 2 1 cos 30 18. tan 15 2 3 1 sin 30 2
1 22 1 cos 45 21 19. tan 225 2 sin 45 2 20. sin 75 12 1 cos 150 12 1 23 14 2 3 12 2 3
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
19
1 1 21. cos 165 2 1 cos 330 2 1 cos 30 12 1 23 12 2 3
22. cos 1125 12 1 cos 225 12 1 cos 45 12 1 22 12 2 2 1 cos 54 1 22 5 21 23. tan 8 5 sin 4 22
24. cos 38 cos 12 34
1 cos 34 2
1 22 2
root because 38 is in quadrant I, so cos 38 0. 1 1 cos 1 1 3 1 2 3 25. cos 12 2 6 2 2 2
2 2 12 2 2. Note that we have chosen the positive 4
1 cos 56 1 23 5 2 3 26. tan 12 5 1 sin 6 2
1 1 cos 9 1 1 2 1 2 2. We have chosen the negative root because 9 is in 2 4 2 2 2 8
27. sin 98
quadrant III, so sin 98 0. 11 1 11 12 1 23 12 2 3. We have chosen the positive root because 11 28. sin 12 2 1 cos 6 12 is in quadrant II, so sin 11 12 0.
29. (a) 2 sin 16 cos 16 sin 32 (b) 2 sin 4 cos 4 sin 8 31. (a) cos2 21 sin2 21 cos 42 (b) cos2 9 sin2 9 cos 18 sin 8 8 tan 4 tan 1 cos 8 2 4 1 cos 4 tan tan 2 (b) sin 4 2
33. (a)
35. sin x x sin x cos x cos x sin x 2 sin x cos x
2 tan 5 tan 10 1 tan2 5 2 tan 5 tan 10 (b) 1 tan2 5 32. (a) cos2 sin2 cos 2 2 (b) 2 sin cos sin 2 2 1 cos 30 34. (a) sin 15 2 1 cos 8 sin 4 (b) 2 tan x tan x 2 tan x 36. tan x x 1 tan x tan x 1 tan2 x
30. (a)
37. sin x 35 . Since x is in quadrant I, cos x 45 and x2 is also in quadrant I. Thus, sin x2 12 1 cos x 12 1 45 1 1010 , cos x2 12 1 cos x 12 1 45 3
3 1010 , and
10 10 x sin 2 1 310 13 . tan x2 10 cos x2 38. cos x 45 . Since x is in quadrant III, sin x 35 and tan x 34 . Also, since 180 x 270 ,
90 x2 135 and so x2 is in quadrant II. Thus, sin x2 12 1 cos x 12 1 45 3 3 1010 , 10 x sin 2 10 3. 3 1 cos x2 12 1 cos x 12 1 45 1 1010 , and tan x2 10 10 cos x2
20
CHAPTER 7 Analytic Trigonometry
2 2 . Since 90 x 180 , we have 39. csc x 3. Then, sin x 13 and since x is in quadrant II, cos x 3 x x x 1 45 2 90 and so 2 is in quadrant I. Thus, sin 2 2 1 cos x 12 1 2 3 2 16 3 2 2 , cos x2
1 1 cos x 2
x 1 1 2 2 1 3 2 2 , and tan x sin 2 322 3 2 2. x 2 3 6 2 32 2 cos 2
2 2 2 and cos x 2 , since x is in quadrant I. Also, since 0 x 90 , 0 x2 45 and so x2 is also in quadrant I. Thus, sin x2 12 1 cos x 12 1 22 12 2 2, 1 22 1 cos x 2 x 1 1 1 x cos 2 2 1 cos x 2 1 2 2 2 2, and tan 2 2 1. 2 sin x 2 41. sec x 32 . Then cos x 23 and since x is in quadrant IV, sin x 35 . Since 270 x 360 , we have x x x 1 135 2 180 and so 2 is in quadrant II. Thus, sin 2 2 1 cos x 12 1 23 1 66 , 6
40. tan x 1. Then sin x
sin x2 30 cos x2 12 1 cos x 12 1 23 5 , and tan x2 1 6 1 55 . x 6 6 5 5 6 cos 2
42. cot x 5.
Then, cos x 5
26
and sin x 1 (csc x 0). 2
Since cot x 0 and
csc x 0, it follows that x is in quadrant III. Thus 180 x 270 and so 90 x2 135 . 26 , x x 1 1 1 5 Thus 2 is in quadrant II. sin 2 12 265 2 1 cos x 2 13 26
1 5 26 26 , and tan x 1 cos x cos x2 12 1 cos x 12 1 5 12 265 5 26. 13 2 1 26 sin x 2
7 and find that sin 24 . Thus, using 43. We sketch a triangle with cos1 25 25
the double-angle formula for sine, 7 sin 2 2 sin cos 2 24 7 336 . sin 2 cos1 25 25 25 625
12 44. We sketch a triangle with tan1 12 5 and find that sin 13 . Thus, using a
double-angle formula for cosine, 2 2 1 2 12 cos 2 1 2 sin 119 cos 2 tan1 12 5 13 169 .
25
¬ 7
24
7 cos 25 ¬
13 12
tan 12 5
45. Rewriting the given expression and using a double-angle formula for cosine, we have 1 1 1 8 sec 2 sin1 14 . 2 7 cos 2 sin1 14 1 2 sin2 sin1 41 1 2 14
46. We sketch a triangle with cos1 23 and find that sin 35 . Thus, using a half-angle formula for tangent, 5 sin 5 3 2 . tan 12 cos1 23 tan 12 1 cos 5 1 3
3
Ï5
cos 23
¬ 2
5
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
47. To write sin 2 tan1 x as an algebraic expression in x, we let tan1 x
Ï1+x@
x and sketch a suitable triangle. We see that sin and 1 x2 1 cos , so using the double-angle formula for sine, we have 1 x2 x 1 2x sin 2 tan1 x sin 2 2 sin cos 2 . 2 2 1 x2 1x 1x
¬
x
1
48. To write tan 2 cos1 x as an algebraic expression in x, we let cos1 x 1 1 x2 and sketch a suitable triangle. We see that tan , so using the ¬ x x double-angle formula for tangent, we have 1 x2 2 2 tan 2 1 x2 2x 1 x 2 x 1 . tan 2 cos x tan 2 2 1 tan2 2x 2 1 1 x2 1 x2 x 1 1 x2 x
Ï1-x@
1 cos cos1 x 1x 1 1 49. Using the half-angle formula for sine, we have sin 2 cos x . Because cos1 2 2 1 cos1 x is positive. Thus, sin 1 cos1 x 1 x . and so sin has range [0 ], 12 cos1 x lies in 0 2 2 2 2
50. Using a double-angle formula for cosine, we have cos 2 sin1 x 1 2 sin2 sin1 x 1 2x 2 .
y
51. To evaluate cos 2, we first sketch the angle in standard position with 4
terminal side in quadrant III. Using a double-angle formula for cosine, we have 2 7. cos 2 1 2 sin2 1 2 35 25
3
x
5
(_4, _3)
52. To evaluate sin 2, we first sketch the angle in standard position with
y ¬
terminal side in quadrant IV and find the remaining side using the Pythagorean Theorem. Using the half-angle formula for sine, we have 1 12 1 cos 13 26 . Because 2 lies in sin 26 2 2 2
¬
12 13
x
5
(12, _5)
quadrant II, where sine is positive, we take the positive value. Thus, sin 2 2626 . 53. To evaluate sin 2, we first sketch the angle in standard position with terminal side in quadrant II and find the remaining side using the Pythagorean Theorem. Using the double-angle formula for sine, we have sin 2 2 sin cos 2 17 4 7 3 8493 .
1
(_4Ï3, 1)
y 7 4Ï3
¬ x
21
22
CHAPTER 7 Analytic Trigonometry
54. To evaluate tan 2, we first sketch the angle in standard position with terminal
y
(3, 4)
side in quadrant I and find the remaining side using the Pythagorean Theorem. Using the double-angle formula for tangent, we have 2 tan tan 2 1 tan2
2 43
24 2 . 7 4
1 3
5 ¬
3
4 x
55. sin 5x cos 4x 12 [sin 5x 4x sin 5x 4x] 12 sin 9x sin x
56. sin 2x sin 3x 12 [cos 2x 3x cos 2x 3x] 12 [cos x cos 5x] 12 cos x cos 5x
57. cos x sin 4x 12 [sin x 4x sin x 4x] 12 [sin 5x sin 3x] 12 sin 5x sin 3x 58. cos 5x cos 3x 12 [cos 5x 3x cos 5x 3x] 12 cos 8x cos 2x
59. 3 cos 4x cos 7x 3 12 [cos 4x 7x cos 4x 7x] 32 [cos 11x cos 3x] 32 cos 11x cos 3x 3x sin x sin 60. 11 sin x2 cos x4 11 12 sin x2 x4 sin x2 x4 11 2 4 4 7x 5x 7x 5x cos 2 sin 6x cos x 61. sin 7x sin 5x 2 sin 2 2 5x 4x 5x 4x 9x x 62. sin 5x sin 4x 2 cos sin 2 cos sin 2 2 2 2 4x 6x 4x 6x sin 2 sin 5x sin x 2 sin 5x sin x 63. cos 4x cos 6x 2 sin 2 2 11x 7x 9x 2x 9x 2x 64. cos 9x cos 2x 2 cos cos 2 cos cos 2 2 2 2 2x 7x 9x 5x 9x 5x 2x 7x sin 2 cos sin 2 cos sin 65. sin 2x sin 7x 2 cos 2 2 2 2 2 2 x 7x 7x x 3x 4x 3x 4x 66. sin 3x sin 4x 2 sin cos 2 sin cos 2 sin cos 2 2 2 2 2 2 67. 2 sin 525 sin 975 2 12 cos 525 975 cos 525 975 cos 45 cos 150 cos 45 cos 150 22 23 12 2 3 2 3 68. 3 cos 375 cos 75 32 cos 45 cos 30 32 22 23 34 21 69. cos 375 sin 75 12 sin 45 sin 30 12 22 12 14 75 15 75 15 cos 2 sin 45 cos 30 2 22 23 26 70. sin 75 sin 15 2 sin 2 2 255 195 255 195 sin 2 sin 225 sin 30 2 22 12 22 71. cos 255 cos 195 2 sin 2 2 2 3 6 5 1 5 1 5 2 cos 72. cos 12 cos 12 2 cos 2 12 12 cos 2 12 12 4 cos 6 2 cos 4 cos 6 2 2 2 2 73. cos2 5x sin2 5x cos 2 5x cos 10x 74. sin 8x sin 2 4x 2 sin 4x cos 4x
75. sin x cos x2 sin2 x 2 sin x cos x cos2 x 1 2 sin x cos x 1 sin 2x 76. cos4 x sin4 x cos2 x sin2 x cos2 x sin2 x cos2 x sin2 x cos 2x 2 tan x 2 tan x sin x 2 cos2 x 2 sin x cos x sin 2x cos x 1 tan2 x sec2 x 1 1 2 sin2 x 1 cos 2x 2 sin2 x 78. tan x sin 2x 2 sin x cos x 2 sin x cos x
77.
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
79. tan
x 2 x
cos x tan
x 2
1 cos x cos x sin x
1 cos x sin x
23
1 cos x cos x cos2 x sin2 x sin x sin x sin x
1 cos x 1 2 cos x csc x 2 sin x sin x sin x sin 4x 2 sin 2x cos 2x 2 2 sin x cos x cos 2x 81. 4 cos x cos 2x sin x sin x sin x 1 1 sin 2x 1 2 sin x cos x 1 1 12 csc x sec x 82. sin 2x 2 sin x cos x 2 sin x cos x 80. tan
83.
84.
cos2 x sin2 x cos x sin x 1 cos x cos 2x cos x sin x cos x sin x 1 2 sin x cos x cos x sin x 1 cos x cos x sin x cos x sin x cos2 x sin2 x 2 sin x cos x 1 tan x 1 tan x sin 2x 2 sin x cos x tan x 1 cos 2x 1 2 cos2 x 1
1 1 tan2 x 2 tan x 2 tan x 1 tan2 x 2 86. sin4 x cos4 x sin2 x cos2 x 2 sin2 x cos2 x 1 12 sin2 2x 1 12 1 cos2 2x 12 1 cos2 2x 85. cot 2x
1 tan 2x
2 tan x tan x 2 tan x tan x 1 tan2 x 2 tan 2x tan x 1 tan x 87. tan 3x tan 2x x 2 tan x 1 tan 2x tan x 1 tan2 x 2 tan x tan x tan x 1 2 1 tan x 3 tan x tan3 x 1 3 tan2 x 2x 1 sin 2x 1 2 cos 2 sin 2x sin 4x 2 sin 2x 2 sin 2x cos 2x sin 2x 1 cos 2x 88. 3x x 3x x 2 cos x 2 cos 3x 2 cos 2x cos x 2 cos 2x cos x cos 2 2 cos 2 2
sin 2x cos x sin 2x cos2 x tan 2x cos x cos 2x cos x cos 2x 2 sin 3x cos 2x sin 3x sin x sin 5x tan 3x 89. cos x cos 5x 2 cos 3x cos 2x cos 3x
90. 91.
sin 3x sin 7x 2 sin 5x cos 2x cos 2x cot 2x cos 3x cos 7x 2 sin 5x sin 2x sin 2x sin 10x 2 sin 5x cos 5x cos 5x sin 9x sin x 2 sin 5x cos 4x cos 4x
sin x sin 3x sin 5x sin x sin 5x sin 3x 2 sin 3x cos 2x sin 3x sin 3x 2 cos 2x 1 tan 3x cos x cos 3x cos 5x cos x cos 5x cos 3x 2 cos 3x cos 2x cos 3x cos 3x 2 cos 2x 1 xy xy xy cos 2 sin sin sin x sin y xy 2 2 2 93. tan xy xy xy cos x cos y 2 cos 2 cos cos 2 2 2 xyxy 2x xyxy 2y 2 cos 2 cos sin sin sin y sin x y sin x y 2 2 2 2 tan y 94. xyxy 2x x yx y 2y cos x y cos x y cos y 2 cos 2 cos cos cos 2 2 2 2 92.
24
CHAPTER 7 Analytic Trigonometry
x 2 4
. Then 2 x 1 cos x 1 cos 2y 2 1 sin x 1 sin x tan2 tan2 y 2 4 1 cos 2y 1 sin x 1 sin x 1 cos x 2
95. Let y
2y
x
tan2 x 1 cot2 x 1 2 sin2 2x sec2 x csc2 x 96. 1 cos 4x 2 tan2 x cot2 x 1 1 2 sin2 2x 2 4 cos2 x sin2 x sec2 x csc2 x 8 sin2 x cos2 x 8 97. sin 130 sin 110 2 cos
130 110 130 110 sin 2 cos 120 sin 10 2 12 sin 10 sin 10 2 2
100 200 100 200 sin 2 sin 150 sin 50 98. cos 100 cos 200 2 sin 2 2 2 12 sin 50 sin 50
45 15 45 15 cos 2 sin 30 cos 15 2 12 cos 15 2 2 cos 15 sin 90 15 sin 75 (applying the cofunction identity)
99. sin 45 sin 15 2 sin
87 33 87 33 cos 2 cos 60 cos 27 2 2 2 12 cos 27 cos 27 sin 90 27 sin 63
100. cos 87 cos 33 2 cos
101.
sin x sin 2x sin 3x sin 4x sin 5x sin x sin 5x sin 2x sin 4x sin 3x cos x cos 2x cos 3x cos 4x cos 5x cos x cos 5x cos 2x cos 4x cos 3x sin 3x 2 cos 2x 2 cos x 1 2 sin 3x cos 2x 2 sin 3x cos x sin 3x tan 3x 2 cos 3x cos 2x 2 cos 3x cos x cos 3x cos 3x 2 cos 2x 2 cos x 1
102. n 1: sin 21 x 2 sin x cos x 21 sin x cos 20 x n 2: sin 22 x sin 4x 2 sin 2x cos 2x 2 2 sin x cos x cos 2x 4 sin x cos x cos 2x 22 sin x cos x cos 21 x n 3: sin 23 x sin 8x 2 sin 4x cos 4x 2 4 sin x cos x cos 2x cos 4x 8 sin x cos x cos 2x cos 4x 23 sin x cos x cos 2x cos 22 x In general, for n 0 we have sin 2n x sin 2 2n1 x 2 sin 2n1 x cos 2n1 x 2 2n1 sin x cos x cos 2x cos 4x cos 8x cos 2n2 x cos 2n1 x 2n sin x cos x cos 2x cos 4x cos 8x cos 2n2 x cos 2n1 x
103. With u sin1 x for 0 x 1, we can write cos1 1 2x 2 cos1 1 2 sin2 u cos1 cos 2u 2u 2 sin1 x. 104. With u tan1
x2 1 1 1 1 cos 2u 2u 2 tan1 1 , we can write cos cos x x x2 1
25
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
105. (a) f x
sin 3x cos 3x sin x cos x sin 3x x sin 3x cos x cos 3x sin x sin x cos x sin x cos x 2 sin x cos x sin 2x 2 sin x cos x sin x cos x for all x for which the function is defined.
cos 3x sin 3x sin x cos x
(b) f x
2 1
-5
5
The function appears to have a constant value of 2 wherever it is defined. 106. (a) f x cos 2x 2 sin2 x
(b) f x cos 2x 2 sin2 x cos2 x sin2 x 2 sin2 x cos2 x sin2 x 1
1
-5
5
The function appears to have a constant value of 1. 107. (a) y sin 6x sin 7x
(b) By a sum-to-product formula,
2
-5
5 -2
y sin 6x sin 7x 6x 7x 6x 7x 2 sin cos 2 2 13 1 2 sin 2 x cos 2 x
(c) We graph y sin 6x sin 7x, y 2 cos 12 x , and y 2 cos 12 x . 2
1 2 sin 13 2 x cos 2 x
-5
5 -2
The graph of y f x lies
between the other two graphs. 1 3 108. From Example 2, we have cos 3x 4 cos3 x 3 cos x. If 3x 3 , then cos 3 2 4 cos x 3 cos x
1 8 cos3 x 6 cos x 8 cos3 x 6 cos x 1 0. Substituting y cos x gives 8y 3 6y 1 0. 2 109. (a) cos 4x cos 2x 2x 2 cos2 2x 1 2 2 cos2 x 1 1 8 cos4 x 8 cos2 x 1. Thus the desired polynomial is P t 8t 4 8t 2 1.
(b) cos 5x cos 4x x cos 4x cos x sin 4x sin x cos x 8 cos4 x 8 cos2 x 1 2 sin 2x cos 2x sin x 8 cos5 x 8 cos3 x cos x 4 sin x cos x 2 cos2 x 1 sin x [from part (a)] 8 cos5 x 8 cos3 x cos x 4 cos x 2 cos2 x 1 sin2 x 8 cos5 x 8 cos3 x cos x 4 cos x 2 cos2 x 1 1 cos2 x 8 cos5 x 8 cos3 x cos x 8 cos5 x 12 cos3 x 4 cos x 16 cos5 x 20 cos3 x 5 cos x
Thus, the desired polynomial is P t 16t 5 20t 3 5t.
Replace "P" with "Q"
26
CHAPTER 7 Analytic Trigonometry
110. Let c1 and c2 be the lengths of the segments shown in the figure. By the Law of
C b b sin 2x c or c . Also by the x sin 2x sin B sin B b a x s c1 s and Law of Sines applied to BC D and AC D we have sin B sin x B A cÁ cª D s c2 s sin x s sin x . So c1 and c2 . Since c c1 c2 , we have sin A sin x sin B sin A s sin x s sin x 1 b a b sin 2x 1 s sin x . By applying the Law of Sines to ABC, sin B sin B sin A sin B sin A sin B sin A b b sin 2x 1 b sin x b 1 . Substituting we have s sin x s 1 sin A a sin B sin B sin B a sin B sin B a b b ab 2ab cos x b 2b sin x cos x sin x 1 2b cos x s 1 s s . b sin 2x s sin x 1 a a a a ab
Sines applied to ABC, we have
111. Using
a
product-to-sum
formula, RHS 4 sin A sin B sin C 4 sin A 12 [cos B C cos B C] 2 sin A cos B C 2 sin A cos B C.
Using another product-to-sum formula, this is equal to 2 12 [sin A B C sin A B C] 2 12 [sin A B C sin A B C]
sin A B C sin A B C sin A B C sin A B C Now A B C , so A B C 2C, A B C 2B, and A B C 2A . Thus our expression simplifies to sin A B C sin A B C sin A B C sin A B C sin 2C sin 2B 0 sin 2A sin 2C sin 2B sin 2A LHS
112. (a) The length of the base of the inscribed rectangle twice the length of the adjacent side which is 2 5 cos and the length of the opposite side is 5 sin . Thus the area of the rectangle is modeled by A 2 5 cos 5 sin 25 2 sin cos 25 sin 2.
(b) The function y sin u is maximized when u 2 . So 2 2 4 . Thus the maximum cross-sectional area is A 4 25 sin 2 4 25 cm2 . 5 2 354 cm. (c) The length of the base is 2 5 cos 5 2 707 cm and the width of the rectangle is 5 sin 4 4 2
113. (a) In both logs the length of the adjacent side is 20 cos and the length of the opposite side is 20 sin . Thus the cross-sectional area of the beam is modeled by A 20 cos 20 sin 400 sin cos 200 2 sin cos 200 sin 2.
(b) The function y sin u is maximized when u 2 . So 2 2 4 . Thus the maximum cross-sectional area is A 4 200 sin 2 4 200.
114. We first label the figure as shown. Because the sheet of paper is folded over, E AC C AB . Thus BC A AC E 90 . It follows that EC D 180 BC A AC E 180 90 90 2. Also,
from the figure we see that BC L sin and C E L sin , so
A ¬ L
E
DC EC cos 2 L sin cos 2. Thus 6 D B DC
C B L sin cos 2 L sin L sin 1 cos 2 L sin 2 cos2 . So L
3 6 . 2 sin cos2 sin cos2
D
C
B
27
SECTION 7.4 Basic Trigonometric Equations
115. (a) y f 1 t f 2 t cos 11t cos 13t
(c) We graph y cos 11t cos 13t, y 2 cos t, and y 2 cos t.
2
2 -5
5 -5
-2
5 -2
(b) Using the identity
y cos cos y 2 cos 2
The graph of f lies between the graphs
y cos , we have 2 11t 13t 11t 13t f t cos 11t cos 13t 2 cos cos 2 2 2 cos 12t cos t 2 cos 12t cos t
116. (a) f 1 770 Hz and f 2 1209 Hz, so
of y 2 cos t and y 2 cos t. Thus,
the loudness of the sound varies between y 2 cos t.
(c) 2
y sin 2 770t sin 2 1209t sin 1540t sin 2418t
0
0.005
(b) Using a sum-to-product formula, we have 1540t 2418t 1540t 2418t cos 2 2 2 sin 1979t cos 439t
-2
y 2 sin
117. We find the area of ABC in two different ways. First, let AB be the base and C D
C
be the height. Since B OC 2 we see that C D sin 2. So the area is
1 base height 1 2 sin 2 sin 2. 2 2 On the other hand, in ABC we see that AC B is a right angle. So BC 2 sin
and AC 2 cos , and the area is
1
A
¬
1
O
D
B
1 base height 1 2 sin 2 cos 2 sin cos . 2 2
Equating the two expressions for the area of ABC, we get sin 2 2 sin cos .
7.4
BASIC TRIGONOMETRIC EQUATIONS
1. Because the trigonometric functions are periodic, if a basic trigonometric equation has one solution, it has infinitely many solutions. 2. The basic equation sin x 2 has no solution (because the sine function has range [1 1]), whereas the basic equation sin x 03 has infinitely many solutions.
3. We can find some of the solutions of sin x 03 graphically by graphing y sin x and y 03. The solutions shown are x 97, x 60, x 34, x 03, x 28, x 66, and x 91.
4. (a) To find one solution of sin x 03 in the interval [0 2, we take sin1 to get x sin1 03 030 The other solution in this interval is x sin1 03 284.
(b) To find all solutions, we add multiples of 2 to the solutions in [0 2. The solutions are x 030 2k and x 284 2k.
28
CHAPTER 7 Analytic Trigonometry
5. Because sine has period 2, we first find the solutions in the interval
y 1
[0 2. From the unit circle shown, we see that sin 23 in quadrants I 2 and II, so the solutions are 3 and 3 . We get all solutions of the
5¹
¬= 3
equation by adding integer multiples of 2 to these solutions: 2 3 2k and 3 2k for any integer k.
y=Ï3/2
_1
0
¹
¬= 3
1
x
6. The sine function is negative in quadrants III and IV, so solutions of sin 22 on the interval [0 2 are 54 and
74 . Adding integer multiples of 2 to these solutions gives all solutions: 54 2k, 74 2k for any integer k.
7. The cosine function is negative in quadrants II and III, so the solution of cos 1 on the interval [0 2 is . Adding integer multiples of 2 to this solution gives all solutions: 2k 2k 1 for any integer k.
8. The cosine function is positive in quadrants I and IV, so the solutions of cos 23 on the interval [0 2 are 6 and 11 116 . Adding integer multiples of 2 to these solutions gives all solutions: 6 2k, 6 2k for any integer k.
9. The cosine function is positive in quadrants I and IV, so the solutions of cos 14 on the interval [0 2 are
cos1 41 132 and 2 cos1 41 497. Adding integer multiples of 2 to these solutions gives all solutions:
132 2k, 497 2k for any integer k.
10. The sine function is negative in quadrants III and IV, so the solutions of sin 03 on the interval [0 2 are
sin1 03 345 and 2 sin1 03 598. Adding integer multiples of 2 to these solutions gives all solutions, 345 2k and 598 2k for any integer k.
11. The sine function is negative in quadrants III and IV, so the solutions of sin 045 on the interval [0 2 are
sin1 045 361 and 2 sin1 045 582. Adding integer multiples of 2 to these solutions gives all solutions, 361 2k, 582 2k for any integer k.
12. The cosine function is positive in quadrants I and IV, so the solutions of cos 032 on the interval [0 2 are
cos1 032 125 and 2 cos1 032 504. Adding integer multiples of 2 to these solutions gives all solutions, 125 2k, 504 2k for any integer k. 13. We first find one solution by taking tan1 of each side of the equation: tan1 3 3 . By definition, this is the only solution in the interval 2 2 . Since tangent has period , we get all solutions of the equation by adding integer multiples of : 3 k for any integer k.
14. One solution of tan 1 is tan1 1 4 . Adding integer multiples of to this solution gives all solutions: 4 k for any integer k.
15. One solution of tan 5 is tan1 5 137. Adding integer multiples of to this solution gives all solutions: 137 k for any integer k. 16. One solution of tan 13 is tan1 13 032. Adding integer multiples of to this solution gives all solutions: 032 k for any integer k.
17. One solution of cos 23 is cos1 23 56 and another is 2 56 76 . All solutions are
56 2k, 76 2k for any integer k. Specific solutions include 56 2 76 , 76 2 56 , 56 , 76 , 56 2 176 , and 76 2 196 .
5 18. One solution of cos 12 is cos1 12 3 and another is 2 3 3 . All solutions are 3 2k and 5 7 11 53 2k for any integer k. Specific solutions include 53 , 3 , 3 , 3 , 3 , and 3 .
SECTION 7.4 Basic Trigonometric Equations
29
3 19. One solution of sin 22 is sin1 22 4 and another is 4 4 . All solutions are 4 2k and 3 9 11 34 2k for any integer k. Specific solutions include 74 , 54 , 4 , 4 , 4 , and 4 . 20. One solution of sin 23 is sin1 23 43 and another is 2 sin1 23 53 . All solutions are 4 5 10 11 43 2k and 53 2k for any integer k. Specific solutions include 23 , 3 , 3 , 3 , 3 , and 3 . 21. One solution of cos 028 is cos1 028 129 and another is 2 cos1 028 500. All solutions are
129 2k and 500 2k for any integer k. Specific solutions include 500, 129, 129, 500, 757, and 1128.
22. One solution of tan 25 is tan1 25 119. All solutions are 119 k for any integer k. Specific solutions include 509, 195, 119, 433, 747, and 1061.
23. One solution of tan 10 is tan1 10 147. All solutions are 147 k for any integer k. Specific solutions include 775, 461, 147, 167, 481, and 795.
24. One solution of sin 09 is sin1 09 426 and another is 2 sin1 09 516. All solutions are 426 2k and 516 2k for any integer k. Specific solutions include 202, 112, 426, 516, 1054, and 1144. 25. cos 1 0 cos 1. In the interval [0 2 the only solution is Thus the solutions are 2k 1 for any integer k. 26. sin 1 0 sin 1. In the interval [0 2 the only solution is 32 . Therefore, the solutions are
32 2k for any integer k. 11 27. 2 cos 3 0 cos 23 . The solutions in the interval [0 2 are 6 , 6 . Thus the solutions are 11 6 2k, 6 2k for any integer k.
28. 2 sin 1 0 sin 12 . The solutions in the interval [0 2 are 76 , 116 . Thus the solutions are 76 2k, 116 2k for any integer k.
29. 3 cos 1 0 cos 13 . The solutions in the interval [0 2 are cos1 13 123 and 2 cos1 13 505. Thus the solutions are 123 2k, 505 2k for any integer k.
3 . The solutions in the interval [0 2 are sin1 3 345 and 30. 10 sin 3 0 sin 10 10
3 598. Thus the solutions are 345 2k, 598 2k for any integer k. 2 sin1 10 31. 3 tan2 1 0 tan2 13 tan 33 . The solutions in the interval 2 2 are 6 , so all solutions are
6 k, 6 k for any integer k.
32. cot 1 0 cot 1. The solution in the interval 0 is 34 . Thus, the solutions are 34 k for any integer k. 3 5 7 33. 2 cos2 1 0 cos2 12 cos 1 4 , 4 , 4 , 4 in [0 2. Thus, the solutions are 4 k, 2
3 k for any integer k. 4
2 4 5 34. 4 sin2 3 0 sin2 34 sin 23 3 , 3 , 3 , 3 in [0 2. Thus, the solutions are 3 k, 2 k for any integer k. 3
35. 3 sin2 1 0 sin 33 sin1 33 062 in 2 2 . Thus, the solutions are 062 k for any integer k. 36. tan2 9 0 tan 3 tan1 3 125 in 2 2 . Thus, the solutions are 125 k for any integer k.
30
CHAPTER 7 Analytic Trigonometry
3 5 7 37. sec2 2 0 sec2 2 sec 2. In the interval [0 2 the solutions are 4 , 4 , 4 , 4 . Thus, the solutions are 2k 1 4 for any integer k.
5 7 11 38. csc2 4 0 csc2 4 csc 2. In the interval [0 2 the solutions are 6 , 6 , 6 , 6 . So the
5 solutions are 6 k, 6 k for any integer k. 39. tan2 4 2 cos 1 0 tan2 4 or 2 cos 1.
If tan2 4 0, then tan2 4 tan 2 tan1 2 111 or tan1 2 111 in 2 2 . If 2 cos 1, then cos 12 cos1 12 23 or 43 on [0 2, so 23 2k or 43 2k for any
integer k.
Thus, the original equation has solutions 111 k, 111 k, 23 2k, and 43 2k for any integer k. 1 . The first equation has solution tan1 2 111 40. tan 2 16 sin2 1 0 tan 2 or sin2 16 on 2 2 . The second equation is equivalent to sin 14 , which has solutions sin1 41 025, sin1 41 289, sin1 41 339, and 2 sin1 41 603 on [0 2.
Thus, the original equation has solutions 111 k, 025 2k, 289 2k, 339 2k, and 603 2k for any integer k. 5 41. 4 cos2 4 cos 1 0 2 cos 12 0 2 cos 1 0 cos 12 3 2k, 3 2k for any
integer k.
42. 2 sin2 sin 1 0 2 sin 1 sin 1 0 2 sin 1 0 or sin 1 0. Since 2 sin 1 0 7 2 sin 12 76 , 116 in [0 2 and sin 1 2 in [0 2. Thus the solutions are 6 2k,
11 2k, 2k for any integer k. 6 2 2 43. tan tan 6 0 tan 3 tan 2 0 tan 2 or tan 3 tan1 2 111 or
tan1 3 125 in 2 2 . Thus, the solutions are 111 k, 125 k for any integer k. 44. 3 cos4 5 cos2 2 cos2 1 3 cos2 2 0 cos 1 cos 1 3 cos2 2 0 cos 1
or cos 36 . The left-hand side of the original equation has period , and the solutions on [0 are 0 or
cos1 36 062. Thus, the solutions are k, 062 k for any integer k.
45. 2 cos2 7 cos 3 0 2 cos 1 cos 3 0 cos 12 or cos 3 (which is inadmissible) 3,
5 . Therefore, the solutions are 2k, 5 2k for any integer k. 3 3 3 2 46. sin sin 2 0 sin 2 sin 1 0 sin 2 (inadmissible) or sin 1. Thus, the solutions are
32 2k for any integer k.
47. cos2 cos 6 0 cos 2 cos x 3 0 cos x 2 or cos x 3, neither of which has a solution. Thus, the original equation has no solution. 48. 2 sin2 5 sin 12 0 sin x 4 2 sin x 3 0 sin x 4 or sin x 32 , neither of which has a solution. Thus, the original equation has no solution. 49. sin2 2 sin 3 sin2 2 sin 3 0 sin 3 sin 1 0 sin 3 0 or sin 1 0. Since
sin 1 for all , there is no solution for sin 3 0. Hence sin 1 0 sin 1 32 2k for any integer k.
50. 3 tan3 tan 3 tan3 tan 0 tan 3 tan2 1 0 tan 0 or 3 tan2 1 0. Now tan 0
5 k and 3 tan2 1 0 tan2 13 tan 1 6 k, 6 k. Thus the solutions are k, 3
k, 5 k for any integer k. 6
6
SECTION 7.5 More Trigonometric Equations
31
7 11 51. cos 2 sin 1 0 cos 0 or sin 12 2 k, 6 2k, 6 2k for any integer k. 52. sec 2 cos 2 0 sec 0 or 2 cos 2 0. Since sec 1, sec 0 has no solution. Thus 7 7 2 cos 2 0 2 cos 2 cos 22 4 4 in [0 2. Thus 4 2k, 4 2k for any
integer k.
53. cos sin 2 cos 0 cos sin 2 0 cos 0 or sin 2 0. Since sin 1 for all , there is no 3 solution for sin 2 0. Hence, cos 0 2 2k, 2 2k 2 k for any integer k.
54. tan sin sin 0 sin tan 1 0 sin 0 or tan 1 0. Now sin 0 when k and tan 1 0 tan 1 34 k. Thus, the solutions are k, 34 k for any integer k.
55. 3 tan sin 2 tan 0 tan 3 sin 2 0 tan 0 or sin 23 . tan 0 has solution 0 on 2 2 and sin 23 has solutions sin1 23 073 and sin1 32 241 on [0 2, so the original equation has solutions k, 073 2k, 241 2k for any integer k.
3 56. 4 cos sin 3 cos 0 cos 4 sin 3 0 cos 0 or sin 34 . cos 0 has solutions 2 , 2 on
[0 2, while sin 34 has solutions sin1 43 399 and 2 sin1 43 544 on [0 2. Thus, the original equation has solutions 2 k, 399 2k, 544 2k for any integer k.
sin 70 sin 70 07065 133 sin 2 57. We substitute 1 70 and 1 133 into Snell’s Law to get 2 sin 2 133 2 4495 . sin 1 1 0658, so we substitute 2 90 into Snell’s Law to get 0658 58. The index of refraction from glass to air is 152 sin 90 sin 1 0658 1 411 .
59. (a) F 12 1 cos 0 cos 1 0
(b) F 12 1 cos 025 1 cos 05 cos 05 60 or 360 60 300 (c) F 12 1 cos 05 1 cos 1 cos 0 90 or 270
(d) F 12 1 cos 1 1 cos 2 cos 1 180
60. Statement A is true: every identity is an equation. However, Statement B is false: not every equation is an identity. The difference between an identity and an equation is that an identity is true for all values in the domain, whereas an equation may be true only for certain values in the domain and false for others. For example, x 0 is an equation but not an identity, because it is true for only one value of x.
7.5
MORE TRIGONOMETRIC EQUATIONS
1. Using a Pythagorean identity, we calculate sin x sin2 x cos2 x 1 sin x 1 1 sin x 0, whose solutions are x k for any integer k.
2. Using a double-angle formula, we we see that the equation sin x sin 2x 0 is equivalent to the equation sin x 2 sin x cos x 0. Factoring the left-hand side as sin x 1 2 cos x, we see that solving this equation is equivalent to solving the two basic equations sin x 0 and 1 2 cos x 0. 3. 2 cos2 sin 1 2 1 sin2 sin 1 0 2 sin2 sin 1 0 2 sin2 sin 1 0. From Exercise 7.4.42, the solutions are 76 2k, 116 2k, 2 2k for any integer k.
4. sin2 4 2 cos2 sin2 cos2 cos2 4 1 cos2 4 cos2 3 Since cos 1 for all , it follows that the original equation has no solution.
32
CHAPTER 7 Analytic Trigonometry
5. tan2 2 sec 2 sec2 1 2 sec 2 sec2 2 sec 3 0 sec 3 sec 1 0 sec 3 or sec 1. If sec 3, then cos 13 , which has solutions cos1 31 123 and 2 cos1 31 505
on [0 2. If sec 1, then cos 1, which has solution on [0 2. Thus, solutions are 2k, 123 2k, 505 2k for any integer k.
6. csc2 cot 3 1 cot2 cot 3 cot2 cot 2 0 cot 2 cot 1 0 cot 2 or cot 1. If cot 2, then tan 12 , which has solution tan1 12 046 on 2 2 , and if cot 1, then tan 1, which has solution 4 on 2 2 . Thus, solutions are 4 k, 046 k for any integer k. 7. sin 2 sin 0 2 sin cos sin 0 sin 2 cos 1 0 sin 0 or cos 12 . The first equation has 5 solutions 0, on [0 2, and the second has solutions 3 , 3 on [0 2. Thus, solutions are k, 3 2k, 5 2k for any integer k. 3
8. 3 sin 2 2 sin 0 3 2 sin cos 2 sin 0 2 sin 3 cos 1 0 sin 0 or cos 13 . The first equation has solutions 0, on [0 2, and the second has solutions cos1 13 123 and
2 cos1 13 505 on [0 2. Thus, solutions are k, 123 2k, 505 2k for any integer k. 5 9. 3 cos 2 2 cos2 0 3 2 cos2 1 2 cos2 0 4 cos2 3 cos 23 6 , 6 on [0 5 (the left-hand side of the given equation has period ). Thus, solutions are 6 k, 6 k for any integer k.
10. cos 2 cos2 12 2 cos2 1 cos2 12 cos2 12 cos 22 . Thus, the solutions are 4 k,
3 k for any integer k. 4 11. 2 sin2 cos 1 2 1 cos2 cos 1 0 2 cos2 cos 1 0 2 cos 1 cos 1 0 5 2 cos 1 0 or cos 1 0 cos 12 or cos 1 3 2k, 3 2k, 2k 1 for any integer k.
12. tan 3 cot 0
sin 3 cos sin2 3 cos2 sin2 cos2 4 cos2 0 0 0 cos sin cos sin cos sin
1 4 cos2 2 0 1 4 cos2 0 4 cos2 1 cos 12 3 k, 3 k for any integer k. cos sin 13. sin 1 cos sin cos 1. Squaring both sides, we have sin2 cos2 2 sin cos 1 sin 2 0,
3 which has solutions 0, 2 , , 2 in [0 2. Checking in the original equation, we see that only 2 and are
valid. (The extraneous solutions were introduced by squaring both sides.) Thus, the solutions are 2k 1 , 2 2k for any integer k.
14. Square both sides of cos sin 1 to get cos2 sin2 2 sin cos 1 sin 2 0, which has solutions 0, , , 3 on [0 2. Checking in the original equation, we see that only 0 and 3 are valid. Thus, the solutions 2
2
2
are 2k, 32 2k for any integer k.
1 sin 1 sin cos 1. Squaring both sides, we have sin2 cos2 2 sin cos 1 cos cos 3 sin 2 0, which has solutions 0, 2 , , 2 on [0 2. Checking in the original equation, we see that only 0
15. tan 1 sec
is valid. Thus, the solutions are 2k for any integer k. 16. 2 tan sec2 4 2 tan 1 tan2 4 tan2 2 tan 3 0 tan 3 tan 1 0 tan 3 or tan 1. The first equation has the solution tan1 3 125 on 2 2 , and the second has solution 4 on 2 2 . Thus, the solutions are 125 k, 4 k for any integer k.
5 2 5 2 17. (a) 2 cos 3 1 cos 3 12 3 3 , 3 for 3 in [0 2. Thus, solutions are 9 3 k, 9 3 k for any
integer k.
SECTION 7.5 More Trigonometric Equations
33
5 7 11 13 17 (b) We take k 0, 1, 2 in the expressions in part (a) to obtain the solutions 9 , 9 , 9 , 9 , 9 , 9 in [0 2. 5 5 18. (a) 2 sin 2 1 sin 2 12 2 6 , 6 for 2 in [0 . Thus, solutions are 12 k, 12 k for any integer k.
, 5 , 13 , 17 in [0 2. (b) Take k 0, 1 in the expressions in part (a) to obtain the solutions 12 12 12 12
2 19. (a) 2 cos 2 1 0 cos 2 12 2 23 2k, 43 2k 3 k, 3 k for any integer k. 2 4 5 (b) The solutions in [0 2 are 3, 3 , 3 , 3 .
2 k, 11 2 k for any integer k. 20. (a) 2 sin 3 1 0 2 sin 3 1 sin 3 12 , which has solutions 718 3 18 3
, 11 , 19 , 23 , 31 , 35 . (b) The solutions in [0 2 are 718 18 18 18 18 18 1 5 1 k for any integer k. 21. (a) 3 tan 3 1 0 tan 3 3 6 k 518 3 3 5 11 23 29 35 (b) The solutions in [0 2 are 18 , 18 , 17 18 , 18 , 18 , 18 .
5 1 5 1 22. (a) sec 4 2 0 sec 4 2 4 3 2k, 3 2k 12 2 k, 12 2 k for any integer k.
, 5 , 7 , 11 , 13 , 17 , 19 , 23 . (b) The solutions in [0 2 are 12 12 12 12 12 12 12 12
23. (a) cos 2 1 0 cos 2 1 2 2k 4k for any integer k.
(b) The only solution in [0 2 is 0. 24. (a) tan 4 3 0 tan 4 3 4 23 k 83 4k for any integer k.
(b) There is no solution in [0 2. 25. (a) 2 sin 3 3 0 2 sin 3 3 sin 3 23 3 43 2k, 53 2k 4 6k, 5 6k for any integer k.
(b) There is no solution in [0 2. 26. (a) sec 2 cos 2 cos2 2 1 cos 2 1 2 k 2k for any integer k. (b) The only solution in [0 2 is 0.
1 27. (a) sin 2 3 cos 2 tan 2 3 12 tan1 3 062 on 4 4 . Thus, solutions are 062 2 k for any integer k.
(b) The solutions in [0 2 are 062, 219, 376, 533.
1 5 sin 3 1 5 sin2 3 sin 3 55 , which has solutions 28. (a) csc 3 5 sin 3 sin 3 5 5 089, 1 sin1 5 120, and 1 1 1 1 015, 3 sin 5 3 3 sin 5 3 3 5 23 13 sin1 55 194 on 0 23 . Thus, solutions are 015 13 k, 089 13 k for any integer k.
(b) The solutions in [0 2 are 015, 089, 120, 194, 224, 298, 329, 403, 434, 508, 539, 613.
29. (a) 1 2 sin cos 2 1 2 sin 1 2 sin2 2 sin2 2 sin 0 2 sin sin 1 0 sin 0 or sin 1 0, , or 2 in [0 2. Thus, the solutions are k, 2 2k for any integer k. (b) The solutions in [0 2 are 0, 2 , .
30. (a) tan 3 1 sec 3 tan 3 12 sec2 3 tan2 3 2 tan 3 1 sec2 3 sec2 3 2 tan 3 sec2 3 2 tan 3 0 3 k for any integer k. Because squaring both sides is an operation that can introduce extraneous solutions, we must check each of the possible solutions in the original equation, and we see that only 23 k are valid solutions.
(b) The solutions in [0 2 are 0, 23 , 43 .
34
CHAPTER 7 Analytic Trigonometry
31. (a) 3 tan3 3 tan2 tan 1 0 tan 1 3 tan2 1 0 tan 1 or 3 tan2 1 tan 1 or 5 tan 1 6 k, 4 k, 6 k for any integer k. 3
5 7 5 11 (b) The solutions in [0 2 are 6, 4, 6 , 6 , 4 , 6 .
32. (a) 4 sin cos 2 sin 2 cos 1 0 2 sin 1 2 cos 1 0 2 sin 1 0 or 2 cos 1 0 5 2 4 sin 12 or cos 12 6 2k, 6 2k, 3 2k, 3 2k for any integer k.
5 2 4 (b) The solutions in [0 2 are 6, 6 , 3 , 3 .
33. (a) 2 sin tan tan 1 2 sin 2 sin tan tan 2 sin 1 0 2 sin 1 tan 1 0
5 3 2 sin 1 0 or tan 1 0 sin 12 or tan 1 6 2k, 6 2k, 4 k for any integer k.
3 5 7 (b) The solutions in [0 2 are 6, 4 , 6 , 4 .
cos sin 1 sin cos2 cos sin sin . cos cos sin sin cos2 Multiplying both sides by the common denominator cos2 sin gives sin2 cos4 sin2 cos2 sin2 cos4 1 cos2 cos2 sin2 cos4 cos2 cos4 sin2 cos2 sin2 1 sin2
34. (a) sec tan cos cot sin
3 2 sin2 1 sin 1 4 k, 4 k for any integer k. Since we multiplied the original 2
equation by cos2 sin (which could be zero) we must check to see if we have introduced extraneous solutions. However, each of the values of does indeed satisfy the original equation. 3 5 7 (b) The solutions in [0 2 are 4, 4 , 4 , 4 .
35. (a)
(b) f x 3 cos x 1; g x cos x 1. f x g x when
4
3 cos x 1 cos x 1 2 cos x 2 cos x 1 x 2k 2k 1 . The points of intersection are
2
2k 1 2 for any integer k.
-6
-4
-2
2
4
6
-2
The points of intersection are approximately 314 2. 36. (a)
(b) f x sin 2x 1; gx 2 sin 2x 1. f x g x when
sin 2x 1 2 sin 2x 1 sin 2x 0 x 12 k. The points of intersection are 12 k 1 for any integer k.
2
-6
-4
-2
2
4
6
The points of intersection are approximately 628 1, 471 1, 314 1, 157 1, and 0 1.
SECTION 7.5 More Trigonometric Equations
37. (a)
35
3. f x g x when tan x 3 x 3 k. The intersection points are 3 k 3 for any integer k.
(b) f x tan x; g x
10
-1
1 -10
The point of intersection is approximately 105 173. 38. (a)
(b) f x sin x 1; g x cos x. f x g x when sin x 1 cos x sin x 12 cos2 x sin2 x 2 sin x 1 cos2 x
-6
-4
-2
2
4
6
sin2 x 2 sin x 1 cos2 x 0 2 sin2 x 2 sin x 0
2 sin x sin x 1 0 sin x 0 or sin x 1 x k, 2 2k. However, x k is not a solution when k is even. (The extraneous
-2
solutions were introduced by squaring both sides.) So the solutions are
The points of intersection are approximately 471 0, 314 1, 157 0, and 314 1.
x 2k 1 , 2 2k, and the intersection points are 2k 1, 2 2k 0 for any integer k.
3 5 7 9 11 13 15 39. cos cos 3 sin sin 3 0 cos 3 0 cos 4 0 4 2 , 2 , 2 , 2 , 2 , 2 , 2 , 2 in 3 5 7 9 11 13 15 [0 8 8 , 8 , 8 , 8 , 8 , 8 , 8 , 8 in [0 2.
5 40. cos cos 2 sin sin 2 12 cos 2 12 cos 12 cos 12 3 , 3 in [0 2.
2 41. sin 2 cos cos 2 sin 23 sin 2 23 sin 23 3 , 3 in [0 2.
42. sin 3 cos cos 3 sin 0 sin 3 0 sin 2 0 2 0, , 2, 3, 4 in [0 4 0, 2 , , 3 in [0 2. 2
7 43. sin 2 cos 0 2 sin cos cos 0 cos 2 sin 1 0 cos 0 or sin 12 2, 6 , 3 , 11 in [0 2. 2 6
sin sin 0 sin 0 sin sin 1 cos 0 (and cos 1 ) 2 1 cos 3 sin cos 0 sin 0 or cos 0 0, 2 , 2 in [0 2 ( is inadmissible).
44. tan
45. cos 2 cos 2 2 cos2 1 cos 2 0 2 cos2 cos 3 0 2 cos 3 cos 1 0
2 cos 3 0 or cos 1 0 cos 32 (which is impossible) or cos 1 0 in [0 2. cos sin cos sin 8 sin cos sin cos 8 sin cos sin cos 46. tan cot 4 sin 2 cos sin cos sin sin2 cos2 8 sin2 cos2 1 2 2 sin cos 2 sin 22 12 sin 2 1 . Therefore, 2 4 k
2 3 k 3 k 3 5 7 or 2 4 k 8 2 or 8 2 . Thus on the interval [0 2 the solutions are 8, 8 , 8 , 8 , 9 , 11 , 13 , 15 . 8 8 8 8
47. cos 2 cos2 0 2 cos2 1 cos2 0 cos2 1 k for any integer k. On [0 2, the solutions are 0, .
36
CHAPTER 7 Analytic Trigonometry
48. 2 sin2 2 cos 2 2 sin2 2 1 2 sin2 4 sin2 3 sin 23 23 k, 43 k for any 2 4 5 integer k. On [0 2, the solutions are 3, 3 , 3 , 3 .
49. cos 2 cos 4 0 cos 2 2 cos2 2 1 0 cos 2 1 2 cos 2 1 0. The first factor has zeros at
2 4 5 2 0, and the second has zeros at 3 , 3 , 3 , 3 . Thus, solutions of the original equation are are 0, 3 , 3 ,
, 43 , 53 in [0 2.
50. sin 3 sin 6 0 sin 3 2 sin 3 cos 3 0 sin 3 1 2 cos 3 0. The first factor has zeros at 0, 3,
2 , , 4 , 5 and the second has zeros at , 5 , 7 , 11 , 13 , 17 , so these are all solutions of the original equation 3 3 3 9 9 9 9 9 9
in [0 2. 1 cos . Squaring both sides, we have 51. cos sin 2 sin 2 cos sin 2 2 cos2 sin2 2 sin cos 1 cos 1 2 sin cos 1 cos either 2 sin 1 or cos 0 6, , 5 , 3 in [0 2. Of these, only and 3 satisfy the original equation. 2
6
2
6
2
52. Square both sides of sin cos 12 to get sin2 cos2 2 sin cos 14 1 sin 2 14 sin 2 34 .
1 1 3 115, 1 sin1 3 357, and Thus, the possible solutions in [0 2 are 12 sin1 34 042, 2 2 sin 4 2 4 3 1 sin1 3 429. We find that only valid solutions to the original equation on [0 2 are 115, 357. 2 2 4
53. sin sin 3 0 2 sin 2 cos 0 2 sin 2 cos 0 sin 2 0 or cos 0 2 k or k 2 12 k for any integer k.
54. cos 5 cos 7 0 2 sin 6 sin 0 sin 6 sin 0 sin 6 0 or sin 0 6 k or k 16 k for any integer k.
55. cos 4 cos 2 cos 2 cos 3 cos cos cos 2 cos 3 1 0 cos 0 or cos 3 12 2 or
5 7 11 13 17 2 5 2 3 3 2k, 3 2k, 3 2k, 3 2k, 3 2k, 3 2k 2 k, 9 3 k, 9 3 k for
any integer k.
56. sin 5 sin 3 cos 4 2 cos 4 sin cos 4 cos 4 2 sin 1 0 cos 4 0 or sin 12 5 1 5 4 2 k or 6 2k, 6 2k 8 4 k, 6 2k, 6 2k for any integer k.
58. cos x
57. sin 2x x
x 3
1
-1
1 1
-1
The three solutions are x 0 and x 095.
-4
-2
2 -1
The three solutions are x 117, 266, and 294.
SECTION 7.5 More Trigonometric Equations
59. 2sin x x
60. sin x x 3 1
2 -1 2
1 -1
4
The only solution is x 192. 61.
37
The three solutions are x 0 and x 093.
cos x x2 1 x2
62. cos x 12 e x ex 1
1
-1
-1
1
The two solutions are x 071.
1
The only solution is x 0.
x 2x tan u tan 63. With u tan1 x and tan1 2x, we have u 4 tan u 1 1 tan u tan 1 1 x 2x 1 3 17 3 32 4 2 1 2 2 x 2x 1 2x 2x 3x 1 0 x . Because tan x is not one-to-one, 2 2 4 we must check both roots, and find that only
173 is a solution to the original equation. 4
64. With u sin1 x and cos1 x, we have 2u cos 2u 1 cos 2u cos sin 2u sin 1 1 2 sin2 u cos 2 sin u cos u sin 1. Referring to the diagram, this becomes 1 2x 2 x 2x 1 x 2 1 x 2 1 x 2x 3 2x 2x 3 1 x 1.
22002 sin 2 5000 151250 sin 2 32 sin 2 003308 2 189442 or 2 180 189442 17810558 . If 2 189442 , then 094721 , and if 2 17810558 , then 8905279 .
65. We substitute 0 2200 and R 5000 and solve for . So 5000
66. Since 4e3t 0, we have 0 4e3t sin 2t 0 sin 2t 2t 0, , 2, t 0, 12 , 1, 32 , . 67. (a) 10 12 283 sin 23 t 80 283 sin 23 t 80 2 sin 23 t 80 070671. Now
sin 070671 and 078484. If 23 t 80 078484 t 80 456 t 344. Now in the interval [0 2, we have 078484 392644 and 2 078484 549834. If 23 t 80 392644
t 80 2281 t 3081. And if 23 t 80 549834 t 80 3194 t 3994
3994 365 344. So according to this model, there should be 10 hours of sunshine on the 34th day (February 3) and on the 308th day (November 4). (b) Since L t 12 283 sin 23 t 80 10 for t [34 308], the number of days with more than 10 hours of daylight is 308 34 1 275 days.
38
CHAPTER 7 Analytic Trigonometry
. The part of the belt touching the larger 2 2 pulley has length 2 R R and similarly the part
68. (a) First note that
touching the smaller belt has length r. To calculate a and b, we write cot
a b a R cot and b r cot , so 2 R r 2 2
R
a Œ
¬l
Œ
b
R
r r
the length of the straight parts of the belt is 2a 2b 2 R r cot . Thus, the total length of the belt is 2 L and so 2 cot L R r 2 R r cot R r 2 cot 2 2 2 R r L . 2 cot 2 R r L 2778 (b) We plot 2 cot and 45113 in the same 2 R r 242 121 10 viewing rectangle. The solution is 1047 rad 60 . 5 0
0
2
69. sin cos x is a function of a function, that is, a composition of trigonometric
y
functions (see Section 2.7). Most of the other equations involve sums, products,
1
differences, or quotients of trigonometric functions. sin cos x 0 cos x 0 or cos x . However, since cos x 1, the only
solution is cos x 0 x 2 k. The graph of f x sin cos x is shown.
_2¹
0
_¹
_1
CHAPTER 7 REVIEW
sin cos 1. sin cot tan sin sin cos
cos
sin2 cos2 sin2 1 sec cos cos cos
2. sec 1 sec 1 sec2 1 tan2 3. cos2 x csc x csc x 1 sin2 x csc x csc x csc x sin2 x csc x csc x sin2 x 4.
1
1 sin2 x
1 sec2 x 1 tan2 x cos2 x
5.
cos2 x tan2 x
6.
1 1 cos x 1 cos2 x sin2 x 1 sec x 1 1 cos x 1 cos x sec x sec x 1 cos x 1 cos x 1 cos x
7.
sin2 x
cos2 x 1 sin x
1 sin x sin x
cos2 x sin2 x
tan2 x sin2 x
cot2 x
1 cot2 x sec2 x cos2 x
cos x cos x cos x sin x 1 1 sec x tan x 1 sin x cos x cos x cos x
8. 1 tan x 1 cot x 1 cot x tan x tan x cot x 2 cot x tan x 2 2
1 cos2 x sin2 x 2 2 sec x csc x cos x sin x cos x sin x
sin x cos x sin x cos x
¹
x
CHAPTER 7
39
sin2 x cos2 x sin2 x 1 cos2 x sin2 x 2 2 cos x 2 sin2 x cos2 x sin x 1 2 10. tan x cot x sec x csc x2 csc2 x sec2 x cos x sin x cos x sin x cos x sin x 9. sin2 x cot2 x cos2 x tan2 x sin2 x
cos2 x
Review
cos2 x
2 sin x cos x 2 sin x cos x 2 sin x sin 2x tan x 1 cos 2x 2 cos x 1 2 cos2 x 1 2 cos2 x cos x cos y sin x sin y cos x cos y sin x sin y cos y sin x cos x y cot y tan x 12. cos x sin y cos x sin y cos x sin y cos x sin y sin y cos x x 1 cos x 13. csc x tan csc x csc x csc x cot x cot x 2 sin x x sin x 1 cos x 1 cos x 1 1 14. 1 tan x tan 1 1 1 1 sec x 2 cos x sin x cos x cos x cos x cos 2x 2 sin x cos x 2 cos2 x 1 1 sin 2x 2 cos x 2 cos x sec x 15. sin x cos x sin x cos x cos x tan x tan 1 tan x 4 16. tan x 4 1 tan x tan 1 tan x 4
11.
1 1 1 1 cos x sec x 1 1 cos x x cos x cos x 17. tan 1 1 sin x sec x cos x sin x 2 sin x sin x cos x cos x
18. cos x cos y2 sin x sin y2 cos2 x 2 cos x cos y cos2 y sin2 x 2 sin x sin y sin2 y cos2 x sin2 x sin2 y cos2 y 2 cos x cos y sin x sin y 2 2 cos x y x x x 2 x x x x x x x x 19. cos sin 1 sin x cos2 2 sin cos sin2 sin2 cos2 2 sin cos 1 sin 2 2 2 2 2 2 2 2 2 2 2 2 3x 7x 3x 7x 2 sin sin 2 sin 5x sin 2x sin 2x cos 3x cos 7x 2 sin 5x sin 2x 2 2 tan 2x 20. 3x 7x 3x 7x sin 3x sin 7x 2 sin 5x cos 2x 2 sin 5x cos 2x cos 2x cos 2 sin 2 2 x y x y x y x y 2 sin cos 2 sin x cos y sin x sin x y sin x y 2 2 tan x 21. x y x y x y x y cos x y cos x y 2 cos x cos y cos x cos 2 cos 2 2 1 22. sin x y sin x y 2 cos x y x y cos x y x y 12 cos 2y cos 2x 12 1 2 sin2 y 1 2 sin2 x 12 2 sin2 x 2 sin2 y sin2 x sin2 y
2 23. (a) f x 1 cos x2 sin x2 , g x sin x
5 -1
prove this, expand f x and simplify, using the double-angle formula for sine:
1
-5
(b) The graphs suggest that f x g x is an identity. To
2 f x 1 cos x2 sin x2 1 cos2 x2 2 cos x2 sin x2 sin2 2x 1 2 cos x2 sin x2 cos2 x2 sin2 2x 1 sin x 1 sin x g x
40
CHAPTER 7 Analytic Trigonometry
24. (a) f x sin x cos x, g x 2
sin2 x cos2 x
-5
5
(b) The graphs suggest that f x g x in general. For example, choose x 6 and evaluate the functions: 1 3 f 6 2 2 12 3 , whereas 1 3 g 6 4 4 1 1, so f x g x.
-2
25. (a) f x tan x tan x2 , g x
1 cos x
(b) The graphs suggest that f x g x in general. For example, choose x 3 and evaluate: f 3 tan 3
5
-5
tan 6
5
1 3 1 1, whereas g 3 1 2, so 3
2
f x g x.
-5
26. (a) f x 1 8 sin2 x 8 sin4 x, g x cos 4x 1
-5
5
(b) The graphs suggest that f x g x is an identity. To
show this, expand g x by using double-angle identities: g x cos 4x cos 2 2x 1 2 sin2 2x 1 2 2 sin x cos x2 1 2 4 sin2 x cos2 x 1 8 sin2 x 1 sin2 x
-1
1 8 sin2 x 8 sin4 x f x
27. (a) f x 2 sin2 3x cos 6x
(b) The graph suggests that f x 1 for all x. To prove this, we use the double angle formula to note that
1
-5
5
cos 6x cos 2 3x 1 2 sin2 3x, so f x 2 sin2 3x 1 2 sin2 3x 1.
-1
x 28. (a) f x sin x cot , g x cos x 2
(b) Proof: f x sin x cot x2 sin x
2
-5
Conjecture: f x g x 1
5
2 sin x2 cos x2
cos x2 sin x2
cos x2 sin x2
Now subtract and add 1, so f x 2 x cos2 1 1 cos x 1 g x 1. 2
Graph for #28 looks kinda jagged
2 cos2 x2
CHAPTER 7
41
Review
29. 4 sin 3 0 4 sin 3 sin 34 sin1 34 08481 or sin1 34 22935. 30. 5 cos 3 0 5 cos 3 cos 35 cos1 35 22143 or 2 cos1 35 40689.
31. cos x sin x sin x 0 sin x cos x 1 0 sin x 0 or cos x 1 x 0, or x 0. Therefore, the solutions are x 0 and . 5 32. sin x 2 sin2 x 0 sin x 1 2 sin x 0 sin x 0 or sin x 12 x 0, or x 6 , 6 . Therefore, the 5 solutions in [0 2 are x 0, 6 , 6 , .
33. 2 sin2 x 5 sin x 2 0 2 sin x 1 sin x 2 0 sin x 12 or sin x 2 (which is inadmissible) x 6, 5 . Thus, the solutions in [0 2 are x and 5 . 6 6 6
34. sin x cos x tan x 1 sin x cos x cos2 x sin x cos x sin x cos x sin x cos2 x cos x 0 sin x cos x 1 cos x cos x 1 0 sin x cos x cos x 1 0 sin x cos x or cos x 1 tan x 1 or 5 5 cos x 1 x 4 , 4 or x 0. Therefore, the solutions in [0 2 are x 0, 4 , 4 .
35. 2 cos2 x 7 cos x 3 0 2 cos x 1 cos x 3 0 cos x 12 or cos x 3 (which is inadmissible) x 3, 5 . Therefore, the solutions in [0 2 are x , 5 . 3 3 3
36. 4 sin2 x 2 cos2 x 3 2 sin2 x 2 sin2 x cos2 x 3 0 2 sin2 x 2 3 0 2 sin2 x 1 sin x 1 . 2
3 5 7 So the solutions in [0 2 are x 4, 4 , 4 , 4 .
37. Note that x is not a solution because the denominator is zero. 4 cos x 2 cos x 12 x 23 , 43 in [0 2.
1 cos x 3 1 cos x 3 3 cos x 1 cos x
38. sin x cos 2x sin x 1 2 sin2 x 2 sin2 x sin x 1 0 2 sin x 1 sin x 1 0 sin x 12 or
5 3 3 5 sin x 1 x 6 , 6 or x 2 . Thus, the solutions in [0 2 are x 6 , 2 , 6 . 39. Factor by grouping: tan3 x tan2 x 3 tan x 3 0 tan x 1 tan2 x 3 0 tan x 1 or tan x 3 2 4 5 2 3 4 5 7 x 34 , 74 or x 3 , 3 , 3 , 3 . Therefore, the solutions in [0 2 are x 3 , 3 , 4 , 3 , 3 , 4 . 40. cos 2x csc2 x 2 cos 2x cos 2x csc2 x 2 cos 2x 0 cos 2x csc2 x 2 0 cos 2x 0 or csc2 x 2
cos 2x 0or sin2 x 12 cos 2x 0 or sin x 1 . 2
3 5 7 3 5 7 For cos 2x 0, the solutions in [0 4 are 2x 2 , 2 , 2 , 2 the solutions in [0 2 are x 4 , 4 , 4 , 4 . 3 5 7 For sin x 1 , the solutions in [0 2 are x 4, 4 , 4 , 4 . 2
3 5 7 Thus, the solutions of the equation in [0 2 are x 4, 4 , 4 , 4 .
1 cos x 1 4 sin x cos x 1cos x 4 sin2 x cos x 1 4 sin2 x cos x cos x 0 41. tan 12 x 2 sin 2x csc x sin x sin x 3 5 7 11 cos x 4 sin2 x 1 0 cos x 0 or sin x 12 x 2 , 2 or x 6 , 6 , 6 , 6 . Thus, the solutions in 5 7 3 11 [0 2 are x 6, 2, 6 , 6 , 2 , 6 .
42. cos 3xcos 2xcos x 0 cos 2x cos xsin 2x sin xcos 2xcos x 0 cos 2x cos xcos 2xsin 2x sin xcos x 0 cos 2x cos x 1 2 sin2 x cos x cos x 0 cos 2x cos x 1 cos x 1 2 sin2 x 0 cos 2x cos x 1 cos x cos 2x 0 cos 2x cos x 1 cos x 0 cos 2x 2 cos x 1 0 cos 2x 0 or
3 5 7 2 4 cos x 12 2x 2 , 2 , 2 , 2 (in [0 4) or x 3 , 3 (in [0 2). Thus, the solutions in [0 2 are x 4 , 2 , 3 , 5 , 4 , 7 . 3 4 4 3 4
42
CHAPTER 7 Analytic Trigonometry 1 sin x 3 sin x 1 3 cos x 3 cos x sin x 1 23 cos x 12 sin x 12 3 cos x cos x sin x 1 cos x 1 x , 5 x , 3 . However, x 3 is cos cos x sin 6 6 2 6 2 6 3 3 6 2 2
43. tan x sec x
inadmissible because sec 32 is undefined. Thus, the only solution in [0 2 is x 6.
sin x 0 2 cos2 x 3 sin x 0 (cos x 0) 2 1 sin2 x 3 sin x 0 cos x 2 sin2 x 3 sin x 2 0 2 sin x 1 sin x 2 0 sin x 12 or sin x 2 (which has no solution) x 6,
44. 2 cos x 3 tan x 0 2 cos x 3 5 . 6
45. We graph f x cos x and g x x 2 1 in the viewing 46. We graph f x esin x and g x x in the viewing rectangle [0 65] by [2 2]. The two functions intersect
rectangle [0 65] by [1 3]. The two functions intersect
at only one point, x 118.
at only one point, x 222.
2 2 0
2
4
6
0
-2
2
4
6
4002 sin2 sin2 08 sin 08944 634 64 4002 sin2 2500 sin2 2500. Therefore it is impossible for the projectile to reach a height of 3000 ft. (b) 64 (c) The function M 2500 sin2 is maximized when sin2 1, so 90 . The projectile will travel the highest when it is shot straight up. k 48. Since e02t 0 we have f t e02t sin 4t 0 sin 4t 0 4t k t where k is any integer. Thus 4 the shock absorber is at equilibrium position every quarter second. 1 cos 30 2 3 2 3 49. Since 15 is in quadrant I, cos 15 . 2 4 2
47. (a) 2000
2 , which is Another method: cos 15 cos 45 30 cos 45 cos 30 sin 45 sin 30 22 23 22 12 6 4 equal to 12 2 3. 3 5 1 1 cos 2 3 2 3 2 6 is in quadrant I, sin 5 . 50. Since 512 12 2 2 4 2 sin sin cos cos sin 1 3 1 1 31 6 2 , which is Another method: sin 512 4 6 4 6 4 6 2 2 4 2 2 2 2 1 equal to 2 2 3. 1
51. tan 8
1 1 cos 2 4 1 1 2 21 1 2 sin 4 2
cos sin 2 sin 1 52. 2 sin 12 12 12 6 2
53. sin 5 cos 40 cos 5 sin 40 sin 5 40 sin 45 1 22 2 tan 66 tan 6 54. tan 66 6 tan 60 3 1 tan 66 tan 6
CHAPTER 7
43
Review
2 cos 2 cos 1 2 55. cos2 sin 8 8 8 4 2 2
3 sin sin cos cos sin sin sin 1 2 56. 12 cos 12 2 12 6 12 6 12 6 12 4 2 2 57. We use a product-to-sum formula: cos 375 cos 75 12 cos 45 cos 30 12 22 23 14 2 3 .
1 1 cos 45 675 225 675 225 cos 2 cos 45 cos 225 2 2 2 2 2 1 cos 45 1 1 22 2
58. cos 675 cos 225 2 cos
2
In the solutions to Exercises 59–64, x and y are in quadrant I, so we know that sec x 32 cos x 23 , so sin x 35 and
1 2. tan x 25 . Also, csc y 3 sin y 13 , and so cos y 2 3 2 , and tan y 4 2 2 59. sin x y sin x cos y cos x sin y 35 2 3 2 23 13 29 1 10 .
60. cos x y cos x cos y sin x sin y 23 2 3 2 35 13 19 4 2 5 .
5 2 5 2 2 2 5 2 8 10 tan x tan y 8 2 5 61. tan x y 23 2 4 2 4 5 5 2 2 1 tan x tan y 8 8 10 8 10 1 1 2
4
2
4
62. sin 2x 2 sin x cos x 2 35 23 4 9 5 .
y 63. cos 2
1 cos y 2
1 2 2 3 2
2 2
1 3 1 cos y y 64. tan 1 2 sin y 3
32 2 (since cosine is positive in quadrant I) 6
32 2 32 2 1
65. We sketch a triangle such that cos1 73 . We see that tan 2 310 , and the double-angle formula for tangent gives 2 tan tan 2 1 tan2
7
4 10 2 2 310 12 10 3 . 2 31 1 40 9 1 2 310
¬
2Ï10
3
cos 37
5 . From the 66. We sketch triangles such that tan1 43 and cos1 13
¬
triangles, we see that sin 35 , cos 45 , and sin 12 13 , so the addition
5
sin sin cos cos sin
3
12
tan 34
5 cos 13
formula for sine gives
5 4 12 63 35 13 5 13 65
67. The double-angle formula for tangent gives tan 2 tan1 x
2 tan tan1 x 2x . 1 tan2 tan1 x 1 x2
4
13
ú
5
44
CHAPTER 7 Analytic Trigonometry
68. Let sin1 x and cos1 y. From the triangles, cos 1 x 2 and sin 1 y 2 , so using the addition formula for cosine, we have cos cos cos sin sin 1 x 2 y x 1 y 2 y 1 x 2 x 1 y2 10 10 1 tan 69. (a) tan x x 10 , for x 0. Since the road sign can first be seen when 2 , (b) tan1 x 10 10 x we have 2 tan1 2864 ft. Thus, the sign can first be x tan 2
1
1
x
ú
sin x
cos y
Ï1-x@
y
2 1 0
20
seen at a height of 2864 ft.
40
70. (a) Let be the angle formed by the top of the tower, the car, and the base of the
40
building, and let be the angle formed by base of the tower, the car, and the base 420 420 of the building, as shown in the diagram. Then tan tan1 x x 380 380 and tan tan1 . Thus, x x 420 380 tan1 tan1 . x x (b) We graph x and find that is maximized when x 400.
Ï1-y@
¬
380
¬
º x
0.055 0.050 0.045 300
71. (a) y sin 2 x cos x corresponds to graph VII. (b) y 4 sin x cos x 2 sin 2x corresponds to graph III. (c) y 1 cos 2x 1 1 2 sin2 x 2 sin2 x corresponds to graph VI. 1 1 2 sin2 x 1 cos 2x tan2 x corresponds to graph II. (d) y 1 cos 2x 1 2 cos2 x 1 (e) y 12 sin x 23 cos x cos 3 sin x sin 3 cos x sin x 3 corresponds to graph IV. (f) y 1 tan2 x sec2 x corresponds to graph VIII.
(g) y cos2 x sin2 x cos 2x corresponds to graph I. sin x tan 12 x corresponds to graph V. (h) y 1 cos x
400
500
CHAPTER 7
Test
45
CHAPTER 7 TEST sin2 cos2 1 sin sin cos sec cos cos cos cos sin x 1 cos x tan x tan x 1 cos x tan x 1 cos x 1 cos x 1 cos x 2. csc x 1 sec x 2 2 1 cos x 1 cos x 1 cos x sin x cos x 1 cos x sin x 2 tan x 2 tan x 2 sin x 3. cos2 x 2 sin x cos x sin 2x cos x 1 tan2 x sec2 x x 1 cos x sin x 1 cos x 4. sin x tan 2 sin x 1. tan sin cos
5. 2 sin2 3x 1 cos 2 3x 1 cos 6x
6. cos 4x 1 2 sin2 2x 1 2 2 sin x cos x2 1 8 sin2 x 1 sin2 x 1 8 sin2 x 8 sin4 x 2 x x 2 1 cos x 1 cos x 1 cos x 1 cos x 1 cos2 x 7. sin cos 2 1 sin x 2 2 2 2 2 2 4 x x x x 2 x x x Another method: sin cos 2 sin cos cos2 1 sin 2 1 sin x sin2 2 2 2 2 2 2 2 sin x 2 sin 2 sin 2 sin sin sin 8. tan (because 2 2 2 cos cos 2 cos 4 x2 4 4 sin 2 1 sin 4 2 sin cos 0 for 2 2)
9. (a) sin 8 cos 22 cos 8 sin 22 sin 8 22 sin 30 12
(b) sin 75 sin 45 30 sin 45 cos 30 cos 45 sin 30 22 23 22 12 14 6 2 1 3 1 cos 150 2 3 2 3 2 Another method: Since 75 is in quadrant I, sin 75 , 2 2 4 2 6 2 . which is equal to 14 1 cos 1 23 2 3 6 12 2 3 (c) sin 12 2 2 4
is in quadrant I, Another method: Since 12 sin sin cos cos sin 3 2 1 2 1 6 2 , which is equal to sin 12 3 4 3 4 3 4 2 2 2 2 4 1 2 3. 2
52 102 5 . 10. From the figures, we have cos cos cos sin sin 2 35 1 23 2 15 5 5 3 5
11. sin 3x cos 5x 12 [sin 3x 5x sin 3x 5x] 12 sin 8x sin 2x 7x 3x 2x 5x 2x 5x 12. sin 2x sin 5x 2 cos sin 2 cos sin 2 2 2 2
1 35
1 3 53 2. 4 5 4 4 5 5 14. 3 sin 1 0 3 sin 1 sin 13 sin1 13 034 or sin1 31 280 on [0 2.
13. sin 45 . Since is in quadrant III, cos 35 . Then tan
1 cos 2 sin
46
FOCUS ON MODELING
15. 2 cos 1 sin 1 0 cos 12 or sin 1. The first equation has solutions 3 105 and 53 524 on [0 2, while the second has the solution 2 157.
16. 2 cos2 5 cos 2 0 2 cos 1 cos 2 0 cos 12 or cos 2 (which is impossible). So in the interval [0 2, the solutions are 23 209, 43 419.
3 17. sin 2 cos 0 2 sin cos cos 0 cos 2 sin 1 0 cos 0 or sin 12 2 , 2 or 5 5 3 6 , 6 . Therefore, the solutions in [0 2 are 6 052, 2 157, 6 262, 2 471.
18. 5 cos 2 2 cos 2 25 2 cos1 04 1159279. The solutions in [0 4 are 2 1159279, 2 1159279,
2 1159279, 4 1159279 2 1159279, 5123906, 7442465, 11407091 057964, 256195, 372123, 570355 in [0 2. 19. 2 cos2 x cos 2x 0 2 cos2 x 2 cos2 x 1 0 cos x 12 . The solutions in [0 4 are x 3 105, x 23 209, x 43 419, and x 53 524. x 1 1 cos x 20. 2 tan csc x 0 2 0 cos x 12 . The solutions in [0 4 are x 3 105 and 2 sin x sin x x 53 524.
9 so tan u 9 . From the triangle, cos u 40 , so using a 21. Let u tan1 40 40 41 2 1 1519 double-angle formula for cosine, cos 2u 2 cos2 u 1 2 40 41 1681 .
41 u
22. We sketch triangles such that cos1 x and tan1 y. From the y 1 triangles, we have sin 1 x 2 , sin , and cos , 1 y2 1 y2
1 ¬
sin sin cos cos sin 1 y 1 x2 x 2 1y 1 y2
40
Ï1-x@
Ï1+y@ ú
x
so the addition formula for sine gives
9
cos x
y
1
tan y
1 x2 x y 1 y2
FOCUS ON MODELING Traveling and Standing Waves 1. (a) Substituting x 0, we get y 0 t 5 sin 4 0 8 t 5 sin 8 t 5 sin 8 t. (b)
y
6
t=0
3.2
6.4
4 2 _2
x
4¹ 2¹
6¹
_4 _6
t=1.6
4.8
(c) We express the function in the standard form y x t A sin k x t: y x t 5 sin 4x 8 t 5 sin 4 x 32 t . . Comparing this to the standard form, we see that this is a traveling wave with velocity 32
2. (a) y 02 sin 1047x 0524t 02 sin 1047x 0524t 2 02 sin 1047x cos 0524t 04 sin 1047x cos 0524t. m 3m. So the nodes are at 3, 6, 9, 12, . The nodes occur when 1047x m x 1047
Traveling and Standing Waves
(b)
47
y t=0
0.4
t=1
0.2
t=2 1
_0.2
2
3
4
t=3l t=4
5
x
6
t=5
_0.4
t=6
Note that when t 3, cos 0524 3 0001, so y x 3 00004. Thus, the graph of y x 3 cannot be distinguished from the x-axis in the diagram. Yes, this is a standing wave. 068. Since 6, we have 3. From the graph, we see that the amplitude is A 27 and the period is 92, so k 292 6 410, so the equation we seek is y x t 27 sin 068x 410t. k 292
4. (a) We are given A 5, period 23 , and 05. Since the period is 23 , we have k 223 3. Thus, expressing the (b)
function in standard form, we have y x t 5 sin 3 x 05t 5 sin 3x 15t. y
t=0
1
2
4 2
_2
x
¹ ¹/2
3¹/2
_4 t=0.5
1.5
5. From the graphs, we see that the amplitude is A 06. The nodes occur at x 0, 1, 2, 3. Since sin x 0 when x k (k any integer), we have . Then since the frequency is 2, we get 20 2 40. Thus, an equation for this model is f x t 06 sin x cos 40t. 6. From the graph, we see that the amplitude is A 7. Now sin x 0 when x k (k an integer). So for k 1, we must 1 have 2 2. Then since the period is 4, we have 2 4 2 . Thus, an equation for this model is f x t 7 sin 2x cos 12 t.
7. (a) The first standing wave has 1, the second has 2, the third has 3, and the fourth has 4.
(b) is equal to the number of nodes minus 1. The first string has two nodes and 1; the second string has three nodes and 2, and so forth. Thus, the next two values of would be 5 and 6, as sketched below.
(c) Since the frequency is 2, we have 440 2 880.
(d) The first standing wave has equation y sin x cos 880t, the second has equation y sin 2x cos 880t, the third has equation y sin 3x cos 880t, and the fourth has equation y sin 4x cos 880t.
8. (a) The nodes of the tube occur when cos 12 x 0 and 0 x 377. So 12 x 2k 1 2 x 2k 1 . Thus, the nodes are at x , 3, 5, 7, 9, and 11. We stop there since 13 377. Note that the endpoints of the tube (x 0 and x 377) are not nodes. (b) In the function y A cos x cos t, the frequency is 2. In this case, the frequency is 50 2 25 Hz.
CORRECTIONS: p. 27,34,35,39,42
CHAPTER 8
POLAR COORDINATES, PARAMETRIC EQUATIONS, AND VECTORS
8.1 8.2
Polar Coordinates 1 Graphs of Polar Equations 5
8.3
Polar Form of Complex Numbers; De Moivre’s Theorem 12
8.4
Plane Curves and Parametric Equations 24
8.5 8.6
Vectors 35 The Dot Product 41 Chapter 8 Review 45 Chapter 8 Test 53
¥
FOCUS ON MODELING: The Path of a Projectile 56
1
8
POLAR COORDINATES, PARAMETRIC EQUATIONS, AND VECTORS
8.1
POLAR COORDINATES
1. We can describe the location of a point in the plane using different coordinate systems. The point P shown in the figure has rectangular coordinates 1 1 and polar coordinates 2 4 . 2. (a) If a point P in the plane has polar coordinates r then it has rectangular coordinates x y where x r cos and y r sin . y (b) If P has rectangular coordinates x y then it has polar coordinates r where r 2 x 2 y 2 and tan . x 7 correspond to the point and 2 3 1 in Cartesian coordinates. 3. Yes; both 2 6 6 4. No; adding a multiple of 2 to gives the same point, as does adding an odd multiple of and reversing the sign of r .
5.
6.
(2, ¹2 )
O
(1, 0)
7.
(3, _ ¹4 )
¹ 2
O
O
8.
(
5¹ 4, _ 6
)
¹ 4
O
9. 5¹
(_2, 4¹ 3) 4¹ 3
_ 6
10. 7¹ 3
O
O
(_3, 7¹ 3 ) Answers to Exercises 11–16 will vary. 5 3 11. 3 2 has polar coordinates 3 2 or 3 2
(3, ¹2 )
O
(2, 3¹ 4 )1
¹ 2
5 13. 1 76 has polar coordinates 1 6 or 1 6 . 7¹ 6
1
O
12. 2 34 has polar coordinates 2 114 or 2 74 .
(_1, 7¹ 6)
3¹ 4
O
2 5 14. 2 3 has polar coordinates 2 3 or 2 3 .
(_2, _ ¹3 ) O
1
¹ 3
1
2
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
15. 5 0 has polar coordinates 5 or 5 2. (_5, 0)
16. 3 1 has polar coordinates 3 1 2 or 3 1 . (3, 1)
O 1 O
17. Q has coordinates 4 34 .
18. R has coordinates 4 34 4 54 . 3 20. P has coordinates 4 134 4 54 4 19. Q has coordinates 4 4 4 4 . 4 . 23 4 4 4 3 . . 22. Q has coordinates 4 21. P has coordinates 4 234 4 4 4 4 4 5 4 . 4 7 . 24. S has coordinates 4 103 23. P has coordinates 4 101 4 4 4 4 4 25. P 3 3 in rectangular coordinates, so r 2 x 2 y 2 32 32 18 and we can take r 3 2. 3 y 1, so since P is in quadrant 2 we take 34 . Thus, polar coordinates for P are 3 2 34 . tan x 3 26. Q 0 3 in rectangular coordinates, so r 3 and 32 . Polar coordinates for Q are 3 32 . 27. Here r 5 and 23 , so x r cos 5 cos 23 52 and y r sin 5 sin 23 5 2 3 . R has rectangular coordinates 52 5 2 3 . 28. r 2 and 56 , so S has rectangular coordinates r cos r sin 2 cos 56 2 sin 56 3 1 . 29. r 3 2 . So x r cos 3 cos 2 3 0 0 and y r sin 3 sin 2 3 1 3. Thus, the rectangular coordinates are 0 3. 30. r 6 23 . So x r cos 6 cos 23 6 12 3 and y r sin 6 sin 23 6 23 3 3. Thus, the rectangular coordinates are 3 3 3 . 2 2 cos 2 1 1, and 31. r 4 . So x r cos 4 2 1 y r sin 2 sin 4 2 1. Thus, the rectangular coordinates are 1 1. 2 32. r 1 52 . So x r cos 1 cos 52 1 0 0 and y r sin 1 sin 52 1 1 1. Thus, the rectangular coordinates are 0 1.
33. r 5 5. So x r cos 5 cos 5 5, and y r sin 5 sin 5 0. Thus, the rectangular coordinates are 5 0. 34. r 0 13. So x y 0 0 because r 0. 3 3 35. r 3 6 . So x r cos 3 cos 6 2 and y r sin 3 sin 6 2 . Thus, the rectangular coordinates are 32 23 . 36. r 2 2 34 . So x r cos 2 2 cos 34 2 and y r sin 2 2 sin 34 2. Thus, the rectangular coordinates are 2 2.
y 1 1, so, since 37. x y 1 1. Since r 2 x 2 y 2 , we have r 2 12 12 2, so r 2. Now tan x 1 the point is in the second quadrant, 34 . Thus, polar coordinates are 2 34 .
SECTION 8.1 Polar Coordinates
38. x y
3
2 3 3 3 . Since r 2 x 2 y 2 , we have r 2 3 3 32 36, so r 6. Now
y 1 3 , so, since the point is in the fourth quadrant, 116 . Thus, polar coordinates are 6 116 . 3 3 x 3 2 2 y 39. x y 8 8 . Since r 2 x 2 y 2 , we have r 2 8 8 16, so r 4. Now tan 8 1, so, 8 x . since the point is in the first quadrant, . Thus, polar coordinates are 4 4 4 2 2 2 2 2 2 40. x y 6 2 . Since r x y , we have r 6 2 8, so r 2 2. Now 2 y tan 1 , so, since the point is in the third quadrant, 76 . Thus, polar coordinates are 2 2 76 . 3 x 6 y 2 41. x y 3 4. Since r x 2 y 2 , we have r 2 32 42 25, so r 5. Now tan 43 , so, since the point is in x the first quadrant, tan1 34 . Thus, polar coordinates are 5 tan1 43 . tan
2 y 2, and 5. Now, tan x 1 since the point is in the fourth quadrant, 2 tan1 2 (since we need 0 2). Thus, polar coordinates are 5 2 tan1 2 . y 43. x y 6 0. r 2 62 36, so r 6. Now tan 0, so since the point is on the negative xaxis, . x Thus, polar coordinates are 6 . 44. x y 0 3 . r 3 and since the point is on the negative yaxis, 32 . Thus, polar coordinates are 3 32 . 42. x y 1 2. Since r 2 x 2 y 2 , we have r 2 12 22 2, so r
45. x y r cos r sin tan 1, and so 4.
46. x 2 y 2 9. By substitution, r cos 2 r sin 2 9 r 2 cos2 sin2 9 r 2 9 r 3.
47. x y 2 . We substitute and then solve for r: r cos r sin 2 r 2 sin2 cos r sin2 cos cot csc . r sin2 5 5 csc . 48. y 5. By substitution, r sin 5 r sin 4 4 sec . 49. x 4. We substitute and then solve for r: r cos 4 r cos 50. x 2 y 2 1. By substitution, r cos 2 r sin 2 1 r 2 cos2 sin2 1 r 2 cos 2 1 r2
1 sec 2. cos 2
51. x 2 y 2 y. By substitution, r cos 2 r sin 2 r sin r 2 cos2 sin2 r sin r sin . 32 6x y r 3 6 r cos r sin r 6 sin cos 3 sin 2. 52. x 2 y 2
53. r 7. But r 2 x 2 y 2 , so x 2 y 2 r 2 49. Hence, the equivalent equation in rectangular coordinates is x 2 y 2 49. 54. r 3 x 2 y 2 r 2 9, so an equivalent equation in rectangular coordinates is x 2 y 2 9.
55. 2 cos 0, so an equivalent equation in rectangular coordinates is x 0. y 56. tan 0 0 y 0. x 57. r cos 6. But x r cos , and so x 6 is an equivalent rectangular equation. 2 r sin 2. But r sin y, so y 2 is an equivalent rectangular equation. 58. r 2 csc r sin
4
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
59. r 4 sin r 2 4r sin . Thus, x 2 y 2 4y is an equivalent rectangular equation. Completing the square, it can be written as x 2 y 22 4.
60. r 6 cos r 2 6r cos . By substitution, x 2 y 2 6x x 2 6x 9 y 2 9 x 32 y 2 9.
61. r 1 cos . If we multiply both sides of this equation by r we get r 2 r r cos . Thus r 2 r cos r, and squaring 2 2 both sides gives r 2 r cos r 2 , or x 2 y 2 x x 2 y 2 in rectangular coordinates.
62. r 3 1 sin 3 3 sin r 2 3r 3r sin r 2 3r sin 3r. Squaring both sides gives 2 2 r 2 3r sin 9r 2 , or x 2 y 2 3y 9 x 2 y 2 in rectangular coordinates.
63. r 1 2 sin . If we multiply both sides of this equation by r we get r 2 r 2r sin . Thus r 2 2r sin r, and 2 2 squaring both sides gives r 2 2r sin r 2 , or x 2 y 2 2y x 2 y 2 in rectangular coordinates. 64. r 2 cos r 2 2r r cos r 2 r cos 2r 2 x 2 y2 x 4 x 2 y2 1 sin cos y x 1.
65. r
2 2 r 2 r cos 2r2 r 2 r cos 4r 2
r sin cos 1 r sin r cos 1, and since r cos x and r sin y, we get
1 r 1 sin 1 r r sin 1. Thus r 1 r sin , and squaring both sides gives 1 sin r 2 1 r sin 2 x 2 y 2 1 y2 1 2y y 2 x 2 2y 1 0.
66. r
4 r 1 2 sin 4 r 2r sin 4. Thus r 4 2r sin . Squaring both sides, we get 1 2 sin r 2 4 2r sin 2 . Substituting, x 2 y 2 4 2y2 x 2 y 2 16 16y 4y 2 x 2 3y 2 16y 16 0.
67. r
2 r 1 cos 2 r r cos 2 r 2 r cos . Squaring both sides, we get r 2 2 r cos 2 . 1 cos Substituting, x 2 y 2 2 x2 4 4x x 2 y 2 4x 4. y y 69. r 2 tan . Substituting r 2 x 2 y 2 and tan , we get x 2 y 2 . x x 2 70. r 2 sin 2 2 sin cos r 4 2r 2 sin cos 2 r cos r sin . By substitution, x 2 y 2 2x y 68. r
x 4 2x 2 y 2 y 4 2x y 0.
y 2 2 2 2 71. sec 2 cos 12 3 tan 3 x 3 y 3x y 3x y 3x 0. y 72. cos 2 1 means that 2 0 0 tan 0 0 y 0 x 73. (a) In rectangular coordinates, the points r1 1 and r2 2 are x1 y1 r1 cos 1 r1 sin 1 and x2 y2 r2 cos 2 r2 sin 2 . Then, the distance between the points is D x1 x2 2 y1 y2 2 r1 cos 1 r2 cos 2 2 r1 sin 1 r2 sin 2 2 r12 cos2 1 sin2 1 r22 cos2 2 sin2 2 2r1r2 cos 1 cos 2 sin 1 sin 2
r12 r22 2r1r2 cos 2 1
SECTION 8.2 Graphs of Polar Equations
5
(b) The distance between the points 3 34 and 1 76 is D
340 32 12 2 3 1 cos 76 34 9 1 6 cos 512
(c) In rectangular coordinates, 3 34 corresponds to 3 cos 34 3 sin 34 3 2 2 3 2 2 and 1 76 corresponds 2 2 3 7 7 1 3 2 2 23 3 2 2 12 340, to cos 6 sin 6 2 2 , so the distance formula gives D
as above.
74. (a) Because streets are laid out in a grid, rectangular coordinates are more appropriate when giving driving directions to a friend. (b) Descartes’ famous declaration cogito ergo sum does not apply to pigeons, so polar coordinates are probably more useful in this case.
8.2
GRAPHS OF POLAR EQUATIONS
1. To plot points in polar coordinates we use a grid consisting of circles centered at the pole and rays emanating from the pole. 2. (a) To graph a polar equation r f we plot all the points r that satisfy the equation.
(b) The graph of the polar equation r 3 is a circle with radius 3 centered at the pole. The graph of the polar equation 4 is a line passing through the pole with slope 1.
3
O
3. VI
4. III
5. II
O
6. IV
7. I
8. V
9. Polar axis: 2 sin 2 sin r, so the graph is not symmetric about the polar axis. Pole: 2 sin 2 sin cos cos sin 2 sin 2 sin r, so the graph is not symmetric about the pole. Line 2 : 2 sin 2 sin cos cos sin 2 sin r, so the graph is symmetric about 2 . 10. Polar axis: 4 8 cos 4 8 cos r, so the graph is symmetric about the polar axis. Pole: 4 8 cos 4 8 cos cos sin sin 4 8 cos r, so the graph is not symmetric about the pole. Line 2 : 4 8 cos 4 8 cos cos sin sin 4 8 cos r, so the graph is not symmetric about 2.
11. Polar axis: 3 sec 3 sec r, so the graph is symmetric about the polar axis. 3 1 3 Pole: 3 sec 3 sec r , so the graph is not symmetric cos cos cos sin sin cos about the pole. 1 3 3 Line 2 : 3 sec cos cos cos sin sin cos 3 sec r, so the graph is not symmetric about 2.
6
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
12. Polar axis: 5 cos csc 5 cos csc r, so the graph is not symmetric about the polar axis. 1 Pole: 5 cos csc 5 cos sin 1 1 5 cos cos sin sin 5 cos 5 cos csc r , sin cos cos sin sin so the graph is symmetric about the pole. 1 Line 2 : 5 cos csc 5 cos sin 1 1 5 cos cos sin sin 5 cos 5 cos csc r , sin cos cos sin sin so the graph is not symmetric about 2 . 4 4 r, so the graph is not symmetric about the polar axis. 3 2 sin 3 2 sin 4 4 4 4 Pole: r, so the graph is not 3 2 sin 3 2 sin cos cos sin 3 2 sin 3 2 sin symmetric about the pole. 4 4 4 Line 2 : 3 2 sin 3 2 sin cos cos sin 3 2 sin r, so the graph is symmetric about 2.
13. Polar axis:
5 5 r , so the graph is symmetric about the polar axis. 1 3 cos 1 3 cos 5 5 5 5 Pole: r, so the graph is not 1 3 cos 1 3 cos cos sin sin 1 3 cos 1 3 cos symmetric about the pole. 5 5 5 Line 2 : 1 3 cos 1 3 cos cos sin sin 1 3 cos r, so the graph is not symmetric about 2.
14. Polar axis:
15. Polar axis: 4 cos 2 4 cos 2 r 2 , so the graph is symmetric about the polar axis. Pole: r2 r 2 , so the graph is symmetric about the pole.
2 Line 2 : 4 cos 2 4 cos 2 2 4 cos 2 4 cos 2 r , so the graph is symmetric about 2 .
16. Polar axis: 9 sin 9 sin r 2 , so the graph is not symmetric about the polar axis. Pole: r2 r 2 , so the graph is symmetric about the pole.
2 Line 2 : 9 sin 9 sin cos cos sin 9 sin r , so the graph is symmetric about 2 .
17. r 2 r 2 4 x 2 y 2 4 is an equation of a circle with radius 2 centered at the origin.
18. r 1 r 2 1 x 2 y 2 1 is an equation of a circle with radius 1 centered at the origin.
(2, ¹2 ) (2, ¹)
O
(1, ¹2 ) (2, 0)
1
(2, 3¹ 2)
(1, ¹)
(1, 0)
O
1
(1, 3¹ 2 )
SECTION 8.2 Graphs of Polar Equations
19. 2 cos 0 x 0 is an equation of a vertical line.
O
1
O
21. r 6 sin r 2 6r sin x 2 y 2 6y
x 2 y 32 9, a circle of radius 3 centered at 0 3.
(6, ¹2 )
1
22. r cos r 2 r cos x 2 y 2 x 2 x 12 y 2 14 , a circle of radius 12 centered at 10 . 2
(0, ¹2 ) O
(1, 0)
O
1
23. r 2 cos . Circle.
1
24. r 3 sin . Circle.
(3, ¹2 ) (_2, 0)
O
1
O
25. r 2 2 cos . Cardioid.
(
¹ 2, 2
)
26. r 1 sin . Cardioid.
1
(2, ¹2 )
(4, ¹) 1
(2, 3¹ 2 )
7
y 20. 56 tan 33 33 y 33 x, a x 3 line with slope 3 passing through the origin.
(1, ¹)
(0, 3¹ 2 ) (1, 0) 1
8
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
27. r 3 1 sin . Cardioid.
28. r cos 1. Cardioid.
(_3, 0)
(_1, 3¹ 2 )
(_3, ¹) 1
(_2, ¹)
(0, 0)
1
(_1, ¹2 ) (_6, ¹2 ) 29. r sin 2
30. r 2 cos 3
O
O
1
31. r cos 5
1
32. r sin 4 1
1
_1
_1
1
O
1
_1
_1
33. r 2 sin 5
O
34. r 3 cos 4
(2, ¹2 )
(_3, 3¹ 2 )
(_3, 0)
O
(_3, ¹)
O
(_3, ¹2 )
35. r
3 2 sin
(Ï3, ¹)
(Ï3, 0)
_1 O
1
_1
(Ï3-2, ¹2 )
36. r 2 sin
(3, ¹2 )
_2 _3
(2, ¹)
(Ï3+2, 3¹ 2)
O
1
(1, 3¹ 2 )
(2, 0)
SECTION 8.2 Graphs of Polar Equations
37. r
3 cos
2
(Ï3-1, ¹) _2
_1
1
38. r 1 2 cos (Ï3, ¹2 ) (Ï3+1, 0)
O _1 _2
(1, ¹2 )
1
(3, ¹) (_1, 0)
2
O
(Ï3, 3¹ 2)
39. r 2 2 2 cos
40. r 3 6 sin
1
(1, 3¹ 2 )
(9, ¹2 )
(2, ¹2 ) (2-2Ï2, 0) (2+2Ï2, ¹) O
(_3, 3¹ 2 ) (2, 3¹ 2 )
41. r 2 cos 2
(3, ¹)
(3, 0)
O
42. r 2 4 sin 2
O
O
1
43. r , 0
1
44. r 1, 0
( ¹2 , ¹2) (¹, ¹)
O
10 1 , 2¹) ( 2¹
0.4
( ¹1 , ¹)
45. r 2 sec
2 3¹ , 2) ( 3¹
46. r sin tan
2 O
1
(3, 0)
O
1
9
10
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
47. r cos
, [0 4] 2
48. r sin
8 , [0 10] 5
1
1
1
1
1
1
49. r 1 2 sin
1 1
, [0 4] 2
50. r
1 08 sin2 , [0 2] 1
2
1 2
1 1
51. r 1 sin n. There are n loops. 2
2
2
2
2
2
n1 52. r 1 c sin 2 2
2
2 2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
n2
n3
n4
n5
2
2
2
2
2
2 2
2
2 2
2
2 2
2
2 2
c 03 c 06 c1 c 15 c2 3 As c increases, the graph becomes “pinched” along the line 4 , eventually growing more “leaves” with that line as their axis. 1 1 , and so on. is IV, since the graph must contain the points 0 0 53. The graph of r sin 2 2 2 1 54. The graph of r is I, since as increase, r decreases ( 0). So this is a spiral into the origin. 5 7 5 7 55. The graph of r sin is III, since for 2 2 2 the values of r are also 2 2 2 . Thus the graph must cross the vertical axis at an infinite number of points. 56. The graph of r 1 3 cos 3 is II, since when 0 r 1 3 4 and when r 1 3 2, so there should be two intercepts on the positiveaxis.
SECTION 8.2 Graphs of Polar Equations
57.
x 2 y2
3
3 4x 2 y 2 r 2 4 r cos 2 r sin 2
r 6 4r 4 cos2 sin2 r 2 4 cos2 sin2 r 2 cos sin sin 2. The equation is r sin 2, a rose.
58.
x 2 y2
3
O
1
2 2 3 x 2 y 2 r 2 r cos 2 r sin 2
2 2 r 6 r 2 cos2 r 2 sin2 r 6 r 2 cos2 sin2
2 r 6 r 4 cos2 sin2 r 2 cos 22 r cos 2. The equations
1
r cos 2 and r cos 2 have the same graph, a rose.
59.
2 x 2 y 2 r 2 r cos 2 r sin 2 r 4 r 2 cos2 r 2 sin2 r 4 r 2 cos2 sin2
x 2 y2
2
r 2 cos2 sin2 cos 2. The graph is r 2 cos 2, a leminiscate.
O
1
2 2 60. x 2 y 2 x 2 y 2 x r 2 r 2 r cos r 2 [r r cos ]2 r 2 r 2 r cos 2 1 r cos 2 1 r cos r cos 1.
Both r cos 1 and r cos 1 give the same graph. To see this, replace
1
with : cos 1 cos 1 cos 1. This is a cardioid.
61. (a) r a cos b sin r 2 ar cos br sin
(b) r 2 sin 2 cos has center 1 1 and radius 12 22 22 2.
x 2 y 2 ax by x 2 ax y 2 by 0
x 2 ax 14 a 2 y 2 by 14 b2 14 a 2 14 b2 2 2 x 12 a y 12 b 14 a 2 b2 . Thus, in rectangular coordinates the center is 12 a 12 b and the radius is 12 a 2 b2 .
62. (a)
(b) r tan sec y x 2.
5
2
2
(2, ¹2 )
(0, 3¹ 4 )
sin cos
1 cos
yr xr2
O
1
(2, 0)
ry y 2 2 1 x x
11
12
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
64. (a)
63. (a) 5000
5000
5000
5000
5000
5000
5000
5000
At 0, the satellite is at the “rightmost” point in
The satellite starts out at the same point, 5625 0,
its orbit, 5625 0. As increases, it travels
but its orbit decays. It narrowly misses the earth at its
counterclockwise. Note that it is moving fastest
first perigee (closest approach to the earth), and
when .
crashes during its second orbit.
(b) The satellite is closest to earth when . Its height above the earth’s surface at this point is
(b) The satellite crashes when 837 rad 480 .
22,500 4 cos 3960 45003960 540 mi. 65. The graphs of r 1 sin 6 and r 1 sin 3 have the same shape as r 1 sin , rotated through angles of 6 and
2
, respectively. Similarly, the graph of r f is the 3
1
graph of r f rotated by the angle .
2
2 1
66. The circle r 2 (in polar coordinates) has rectangular coordinate equation x 2 y 2 4. The polar coordinate
form is simpler. The graph of the equation r sin 2 is a fourleafed rose. Multiplying both sides by r 2 , we get 32 r 3 r 2 2 cos sin 2 r cos r sin which is x 2 y 2 2x y in rectangular form. The polar form is definitely simpler.
67. y 2 r sin 2 r 2 csc . The rectangular coordinate system gives the simpler equation here. It is easier to study lines in rectangular coordinates.
8.3
POLAR FORM OF COMPLEX NUMBERS; DE MOIVRE’S THEOREM
1. A complex number z a bi has two parts: a is the real part and b is the imaginary part. To graph a bi we graph the ordered pair a b in the complex plane. 2. (a) The modulus of z is r a 2 b2 and an argument of z is an angle satisfying tan ba.
(b) We can express z in polar form as z r cos i sin where r is the modulus of z and is the argument of z. 3. (a) The complex number z 1 i in polar form is z 2 cos 34 i sin 34 . (b) The complex number z 2 cos 6 i sin 6 in rectangular form is z 3 i. (c) The complex number z can be expressed in rectangular form as 1 i or in polar form as 2 cos 4 i sin 4 .
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
4. A nonzero complex number has n different nth roots. The number 16 has four
Im
fourth roots. These roots are 2, 2, 2i, and 2i. In the complex plane, these
2i
roots all lie on a circle of radius 2.
_2
2
0 _2i
5. 4i
02 42 4
6. 3i
Im
09 3
Im
4i i 0
i 1
0
Re
1
Re
_3i
7. 2
402
8. 6 6
Im
Im
i _2
9. 5 2i
0
i 1
0
Re
52 22 29
10. 7 3i
Im
6
1
Re
49 9 58 Im
5+2i i 0
i 1
Re
0
1
Re 7-3i
11. 3 i 3 1 2
12. 1 33 i 1 13 43 2 3 3
Im
Im
0
i
Ï3+i
i 1
0
Re _1- Ï3 i 3
1
Re
Re
13
14
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
3 4i 9 16 13. 25 25 1 5
2 i 2 14. 12 12 1 2
Im
Im
i
3+4i 5
0
1
Im 2z
0
0
Re
15. z 1 i, 2z 2 2i, z 1 i, 12 z 12 12 i
i
i
_Ï2+iÏ2 2
16. z 1
1
Re
3i, 2z 2 2 3i, z 1 3i,
1z 1 3i 2 2 2
Im
z
2z
1z 2
1
Re
z
_z
i
1z 2
0
1
Re
_z
17. z 8 2i, z 8 2i
18. z 5 6i, z 5 6i
Im
z
z
i 0
Im
1
i 0
Re
1
Re
z
z
19. z 1 2 i, z 2 2 i, z 1 z 2 2 i 2 i 4, z 1 z 2 2 i 2 i 4 i 2 5
20. z 1 1 i, z 2 2 3i,
z 1 z 2 1 i 2 3i 1 2i,
z 1 z 2 1 i 2 3i 2 3i 2i 3i 2 1 5i
Im
Im
i 0
zÁzª
zª 1
zÁzª zÁ+zª
zÁ
Re
zÁ i 0
1
zÁ+zª zª
Re
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
21. z a bi a 0 b 0
22. z a bi a 1 b 1
Im
Im
i
i 0
23. z z 3
1
0
Re
24. z z 1
Im
25. z z 2
1
0
Re
26. z 2 z 5
Im
1
Re
Im
i
0
1
0
Re
28. z a bi a b
Im
i 0
Re
i
i
27. z a bi a b 2
1
Im
i 0
15
1
Re
1
Re
Im
i 1
0
Re
29. For 1 i, tan 11 1 with in quadrant I 4 , and r
12 12 2. Hence, 1 i 2 cos 4 i sin 4 .
7 30. For 1 i, tan 1 1 1 with in quadrant IV 4 , and r 1 i 2 cos 74 i sin 74 .
2 1 with in quadrant II 3 , and r 31. For 2 2i, tan 2 4 2 2i 2 2 cos 34 i sin 34 .
12 12 2. Hence,
22 22 2 2. Hence,
16
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
32. For 2 2i, tan 2 1 with in quadrant III 54 , and r 2 5 2 2i 2 cos 4 i sin 54 .
3 with in quadrant III 7 , and r 33. For 3 i, tan 1 3 6 3 7 7 3 i 2 cos 6 i sin 6 .
2 2 2 2 2. Hence, 2 3 12 2. Hence,
34. For 5 5 3i, tan 553 3 with in quadrant II 23 , and r 5 5 3i 10 cos 23 i sin 23 .
2 52 5 3 10. Hence,
2 3 with in quadrant IV 11 , and r 35. For 2 3 2i, tan 2 2 3 22 4. Hence, 3 6 2 3 11 2 3 2i 4 cos 6 i sin 116 . 2 36. For 3 3 3i, tan 3 3 3 3 with in quadrant I , and r 32 3 3 6. Hence, 3 3 3 3i 6 cos 3 i sin 3 . 37. For 2i, tan is undefined, 2 , and r 2. Hence, 2i 2 cos 2 i sin 2 . 38. For 5i, tan is undefined, 32 , and r 5. Hence, 5i 5 cos 32 i sin 32 . 39. For 3, tan 0, , and r 3. Hence, 3 3 cos i sin . 40. For 2, tan 0, 0, and r 2. Hence, 2 2 cos 0 i sin 0.
41. For 6 2i, tan 2 33 with in quadrant II 56 , and r 6 6 2i 2 2 cos 56 i sin 56 .
2 2 6 2 2 2. Hence,
42. For 5 15i, tan 15 3 with in quadrant III 43 , and r 5 Hence, 5 15i 2 5 cos 43 i sin 43 .
2 2 5 15 2 5.
43. For 4 3i, tan 34 with in quadrant I tan1 34 06435, and r 42 32 5. Hence, 4 3i 5 cos tan1 43 i sin tan1 34 . 44. For 3 2i, tan 23 with in quadrant I tan1 23 05880, and r 32 22 13. Hence, 3 2i 13 cos tan1 23 i sin tan1 23 . 2 3 with in quadrant IV 11 , and r 45. For 4 4 3 42 8. 3 i 4 3 4i, tan 4 3 6 4 3 11 3 i 8 cos 6 i sin 116 . Hence, 4 46. For i 2 6i 6 2i, tan 2 33 with in quadrant I tan1 33 6 , and 6 2 2 6 2 2 2. Hence, i 2 6i 2 2 cos r 6 i sin 6 . 3 1 with in quadrant II 3 , and r 32 32 3 2. Hence, 47. For 3 1 i 3 3i, tan 3 4 3 1 i 3 2 cos 34 i sin 34 .
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
2 1 with in quadrant II 3 , and r 22 22 2 2. Hence, 48. For 2i 1 i 2 2i, tan 2 4 2i 1 i 2 2 cos 34 i sin 34 .
17
6 cos i sin 49. z 1 3 cos 3 i sin 3 , z 2 2 cos 6 i sin 6 , z 1 z 2 3 2 cos 3 6 i sin 3 6 2 2 i sin 3 cos i sin z 1 z 2 32 cos 3 6 3 6 2 6 6 3 cos 54 i sin 54 , z 2 2 cos i sin , z1 z2 3 2 cos 54 i sin 54 2 3 cos 4 i sin 4 , z 1 z 2 23 cos 54 i sin 54 23 cos 4 i sin 4
50. z 1
2 cos 53 i sin 53 , z 2 2 2 cos 32 i sin 32 , 4 cos 76 i sin 76 , 2 2 2 cos 53 32 i sin 53 32 z1 z2 z 1 z 2 2 cos 53 32 i sin 53 32 12 cos 6 i sin 6
51. z 1
2 2
, z z cos 3 i sin 3 cos 13 i sin 13 , i sin 52. z 1 cos 34 i sin 34 , z 2 cos 1 2 3 3 4 3 4 3 12 12 3 3 5 5 z 1 z 2 cos 4 3 i sin 4 3 cos 12 i sin 12 53. z 1 4 cos 120 i sin 120 , z 2 2 cos 30 i sin 30 , z 1 z 2 4 2 cos 120 30 i sin 120 30 8 cos 150 i sin 150 , z 1 z 2 42 cos 120 30 i sin 120 30 2 cos 90 i sin 90
2 cos 75 i sin 75 , z 2 3 2 cos 60 i sin 60 , z 1 z 2 2 3 2 cos 75 60 i sin 75 60 6 cos 135 i sin 135 , z 1 z 2 2 cos 75 60 i sin 75 60 13 cos 15 i sin 15
54. z 1
3 2
55. z 1 4 cos 200 i sin 200 , z 2 25 cos 150 i sin 150 , z 1 z 2 4 25 cos 200 150 i sin 200 150 100 cos 350 i sin 350 , 4 cos 200 150 i sin 200 150 4 cos 50 i sin 50 z 1 z 2 25 25
56. z 1 45 cos 25 i sin 25 , z 2 15 cos 155 i sin 155 , 4 cos 180 i sin 180 , z 1 z 2 45 15 cos 25 155 i sin 25 155 25 z 1 z 2 45 15 cos 25 155 i sin 25 155 4 cos 130 i sin 130 4 cos 130 i sin 130 3 i, so tan 1 1 with 1 in quadrant I 1 6 , and r1 3 1 2. 3 z 2 1 3i, so tan 2 3 with 2 in quadrant I 2 3 , and r1 1 3 2. Hence, z 1 2 cos 6 i sin 6 and z 2 2 cos 3 i sin 3 . 4 cos i sin , Thus, z 1 z 2 2 2 cos 6 3 i sin 6 3 2 2 i sin cos i sin , and 1z 1 cos i sin . z 1 z 2 22 cos 1 6 3 6 3 6 6 2 6 6
57. z 1
18
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
2 2i, so tan 1 1 with 1 in quadrant IV 1 74 , and r1 2 2 2. z 2 1 i, so tan 2 1 with 2 in quadrant IV 2 74 , and r2 1 1 2. Hence, z 1 2 cos 74 i sin 74 and z 2 2 cos 74 i sin 74 . Thus, z 1 z 2 2 2 cos 74 74 i sin 74 74 2 2 cos 72 i sin 72 2 2 cos 32 i sin 32 , z 1 z 2 2 cos 74 74 i sin 74 74 2 cos 0 i sin 0, and 2 1z 1 12 cos 74 i sin 74 12 cos 4 i sin 4 .
58. z 1
1 with 1 in quadrant IV 1 11 , and r1 12 4 4. 59. z 1 2 3 2i, so tan 1 2 6 2 3 3 z 2 1 i, so tan 2 1 with 2 in quadrant II 2 34 , and r2 1 1 2. Hence, z 1 4 cos 116 i sin 116 and z 2 2 cos 34 i sin 34 . i sin 7 , Thus, z 1 z 2 4 2 cos 116 34 i sin 116 34 4 2 cos 712 12 4 13 z 1 z 2 cos 116 34 i sin 116 34 2 2 cos 13 12 i sin 12 , and 2 1 14 cos 1z 1 4 cos 116 i sin 116 6 i sin 6 . 60. z 1 2i, so 1 32 , and r1 2. z 2 3 3 3i, so tan 2 3 with 2 in quadrant IV 2 43 , and r2 9 27 6. Hence, z 1 2 cos 32 i sin 32 and z 2 6 cos 43 i sin 43 . Thus, z 1 z 2 6 2 cos 56 i sin 56 , 2 6 cos 32 43 i sin 32 43 z 1 z 2 62 cos 32 43 i sin 32 43 62 cos 6 i sin 6 , and 1z 1 1 cos 32 i sin 32 22 cos 2 i sin 2 . 2
61. z 1 5 5i, so tan 1 55 1with 1 in quadrant I 1 4 , and r1
25 25 5 2.
z 2 4, so 2 0, and r2 4. Hence, z 1 5 2 cos 4 i sin 4 and z 2 4 cos 0 i sin 0. i sin , z z 5 2 cos i sin , and 0 i sin 0 20 2 cos Thus, z 1 z 2 5 2 4 cos 1 2 4 4 4 4 4 4 4 1 cos i sin 2 cos i sin . 1z 1 4 4 10 4 4 5 2
3 with 1 in quadrant III 1 11 , and r1 48 16 8. 62. z 1 4 3 4i, so tan 1 4 3 6 4 3
, and r2 8. z 2 8i, so 2 2 Hence, z 1 8 cos 116 i sin 116 and z 2 8 cos 2 i sin 2 . 11 64 cos Thus, z 1 z 2 8 8 cos 116 2 i sin 6 2 3 i sin 3 , 11 z 1 z 2 88 cos 116 cos 43 i sin 43 , and 1z 1 18 cos 2 i sin 6 2 6 i sin 6 .
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
19
63. z 1 20, so 1 , and r1 20. z 2 3 i, so tan 2 1 with 2 in quadrant I 2 6 , and r2 3 1 2. 3 Hence, z 1 20 cos i sin and z 2 2 cos 6 i sin 6 . 40 cos 7 i sin 7 , Thus, z 1 z 2 20 2 cos 6 i sin 6 6 6 10 cos 5 i sin 5 , and z 1 z 2 20 2 cos 6 i sin 6 6 6 1 [cos i sin ] 1 cos i sin . 1z 1 20 20
64. z 1 3 4i, so tan 1 43 with 1 in quadrant I 1 tan1 43 , and r1 9 16 5. z 2 2 2i, so tan 2 1 with 2 in quadrant IV 2 74 , and r2 4 4 2 2. Hence, z 1 5 cos tan1 43 i sin tan1 43 5 cos 0927 i sin 0927 and z 2 2 2 cos 74 i sin 74 . Thus, z 1 z 2 5 2 2 cos tan1 43 74 i sin tan1 34 74 10 2 cos 0142 i sin 0142, 5 cos tan1 34 74 i sin tan1 34 74 5 4 2 cos tan1 43 74 i sin tan1 43 74 z 1 z 2 22
5 4 2 [cos 457 i sin 457], and 1z 1 15 cos tan1 43 i sin tan1 43 15 [cos 0927 i sin 0927].
6 65. 3 i 2 cos 56 i sin 56 , so 3 i 26 cos 306 i sin 306 64 cos 5 i sin 5 64. 66. 1 i
10 cos 704 i sin 704 32 cos 32 i sin 32 32i. 2 cos 74 i sin 74 , so 1 i10 2
5 67. 2 2i 2 cos 54 i sin 54 , so 2 2i 25 cos 254 i sin 254 16 2 16 2i.
7 7 68. 1 i 2 cos cos 74 i sin 74 8 8i. 2 4 i sin 4 , so 1 i 1 1 1 and tan 1 . Thus 2 2 i cos i sin . Therefore, 69. r 2 2 4 2 2 4 4 12 2 2 cos 12 2 2 i 4 i sin 12 4 cos 3 i sin 3 1. 1 with in quadrant IV 11 . Thus 3 i 2 cos 11 i sin 11 , so 70. r 3 1 2 and tan 6 6 6 3 10 10 3 i sin 110 1 cos i sin 1 1 1 cos 110 3i 12 6 6 1024 3 3 1024 2 2 i 2048 1 3i . 71. r 4 4 4 2 and tan 1 with in quadrant IV 74 . Thus 2 2i 2 2 cos 74 i sin 74 , so 8 2 2i8 2 2 cos 14 i sin 14 4096 1 0i 4096. 72. r 14 34 1 and tan 3 with in quadrant III 43 . Thus 12 23 i cos 43 i sin 43 , so 15 12 23 i cos 603 i sin 603 cos 20 i sin 20 1. 73. r 1 1 2 and tan 1 with in quadrant III 54 . Thus 1 i 2 cos 54 i sin 54 , so 7 cos 354 i sin 354 8 2 cos 34 i sin 34 8 2 1 i 1 8 1 i. 2 1 i7 2
2
74. r 9 3 2 3 and tan 33 with in quadrant I 6 . Thus 3 3i 2 3 cos 6 i sin 6 , so 4 3 3i 144 cos 23 i sin 23 144 12 23 i 72 1 3i .
20
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
2 1 . Thus 2 3 2i 4 cos i sin , so 75. r 12 4 4 and tan 6 6 6 2 3 3 5 5 1 1 2 3 2i cos 56 i sin 56 1024 23 12 i 2048 3i 14
1 1 2 and tan 1 with in quadrant IV 74 . Thus 1 i 2 cos 74 i sin 74 , so 1 cos 56 i sin 56 1 cos 14 i sin 14 1 . 1 i8 16 4 4 16 16
76. r
4 1 . Thus 77. r 48 16 8 and tan 6 4 3 3 4 3 4i 8 cos 6 i sin 6 . So, 12 6 2k 6 2k i sin for k 0, 1. 8 cos 4 3 4i 2 2 i sin and Thus the two roots are 0 2 2 cos 12 12 13 13 1 2 2 cos 12 i sin 12 .
Im
i 0 wÁ
4 1 . Thus 78. r 48 16 8 and tan 6 4 3 3 4 3 4i 8 cos 6 i sin 6 . So 13 6 2k 6 2k i sin for k 0, 1, 2. 2 cos 4 3 4i 3 3 i sin , Thus the three roots are 0 2 cos 18 18 13 13 i sin 25 . 1 2 cos 18 i sin 8 , and 2 2 cos 25 18 18 79. 81i 81 cos 32 i sin 32 . Thus, 32 2k 32 2k i sin for k 0, 1, 81i14 8114 cos 4 4 2, 3. The four roots are 0 3 cos 38 i sin 38 , 1 3 cos 78 i sin 78 , 2 3 cos 118 i sin 118 , and 3 3 cos 158 i sin 158 . 2k 2k cos i sin for k 0, 1, 2, 3, 5 5 4. Thus the five roots are 0 2 cos 0 i sin 0, 1 2 cos 25 i sin 25 , 2 2 cos 45 i sin 45 , 3 2 cos 65 i sin 65 , and 4 2 cos 85 i sin 85 .
wü 1
Re
Im wÁ i 0
wü 1
Re
wª
Im
wÁ
wü
i 0
1
Re w£
wª
80. 32 32 cos 0 i sin 0. Thus, 3215
Im wÁ wª
i 0
w£
wü 1
w¢
Re
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
2k 2k i sin , for k 0, 1, 2, 3, 4, 5, 6, 8 8 7. So the eight roots are 0 cos 0 i sin 0 1,
81. 1 cos 0 i sin 0. Thus, 118 cos
2 2 1 cos 4 i sin 4 2 i 2 , 2 cos 2 i sin 2 i, 3 cos 34 i sin 34 22 i 22 , 4 cos i sin 1, 5 cos 54 i sin 54 22 i 22 , 6 cos 32 i sin 32 i, and 7 cos 74 i sin 74 22 i 22 .
Im wª
w£
wÁ wü
0
w¢
1
Re
w¦
w°
2 cos 4 i sin 4 . So 13 4 2k 4 2k cos i sin for k 0, 1, 2 1 i13 3 3 i sin , 2. Thus the three roots are 0 216 cos 12 12 i sin 9 , and 216 cos 17 i sin 17 . 1 216 cos 912 2 12 12 12
i
w§
82. 1 i
Im wÁ
i wü 0
1
Re
wª
13 cos 83. i cos 2 i sin 2 , so i
2 2k 3
i sin
2 2k 3
for
Im
3 1 k 0, 1, 2. Thus the three roots are 0 cos 6 i sin 6 2 2 i,
i
wÁ
wü
1 cos 56 i sin 56 23 12 i, and 2 cos 32 i sin 32 i.
0
1
Re
wª
15 cos 84. i cos 2 i sin 2 , so i
2 2k 5
i sin
2 2k 5
i sin , k 0, 1, 2, 3, 4. Thus the five roots are 0 cos 10 10
for
Im i
9 9 13 13 1 cos 2 i sin 2 , 2 cos 10 i sin 10 , 3 cos 10 i sin 10 , and
wü
wª 0
17 4 cos 17 10 i sin 10 .
2k 4
2k i sin 4
2 2 k 0, 1, 2, 3. So the four roots are 0 cos 4 i sin 4 2 i 2 ,
for
1 cos 34 i sin 34 22 i 22 , 2 cos 54 i sin 54 22 i 22 , and 3 cos 74 i sin 74 22 i 22 .
1
Re
1
Re
w¢
w£
85. 1 cos i sin . Then 114 cos
wÁ
Im wÁ
i
0 wª
wü
w£
21
22
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors 162 1 3 32 and tan 1616 3 3 with in quadrant III 43 . Thus 16 16 3i 32 cos 43 i sin 43 . So 15 43 2k 43 2k 16 16 3i i sin for 3215 cos 5 5 i sin 4 , k 0, 1, 2, 3, 4. The five roots are 0 2 cos 415 15 i sin 16 , 1 2 cos 23 i sin 23 , 2 2 cos 16 15 15 22 22 28 3 2 cos 15 i sin 15 , and 4 2 cos 15 i sin 28 15 .
86. r
Im wÁ
wü i
wª
0
1
Re w¢
w£
87. z 4 1 0 z 114 z 22 22 i, z 22 22 i (from Exercise 85) , then, z i 18 cos 2 2k i sin 2 2k for i sin 88. z 8 i 0 x i 18 . Since i cos 2 2 8 8
i sin , cos 5 i sin 5 , cos 9 i sin 9 , k 0, 1, 2, 3, 4, 5, 6, 7. Thus there are eight solutions: z cos 16 16 16 16 16 16
13 17 17 21 21 25 25 29 29 cos 13 16 i sin 16 , cos 16 i sin 16 , cos 16 i sin 16 , cos 16 i sin 16 , and cos 16 i sin 16 . 13 89. z 3 4 3 4i 0 z 4 3 4i . Since 4 3 4i 8 cos 6 i sin 6 , 13 6 2k 6 2k 13 cos i sin , for k 0, 1, 2. Thus the three roots are 8 4 3 4i 3 3 i sin , z 2 cos 13 i sin 13 , and z 2 cos 25 i sin 25 . z 2 cos 18 18 18 8 18 18
90. z 6 1 0 z 116 . Since 1 cos 0 i sin 0 z 116 cos 2k6 i sin 2k6 for k 0, 1, 2, 3, 4, 5. Thus there are
six solutions: z 1, 12 23 i, 12 23 i.
2 cos 54 i sin 54 , 91. z 3 1 i z 1 i13 . Since 1 i 54 2k 54 2k i sin for k 0, 1, 2. Thus the three solutions to this z 1 i13 216 cos 3 3 i sin 5 , 216 cos 13 i sin 13 , and 216 cos 21 i sin 21 . equation are z 216 cos 512 12 12 12 12 12
92. z 3 1 0 z 113 . Since 1 cos 0 i sin 0 z 113 cos
2k 2k i sin for k 0, 1, 2. Thus the three 3 3
solutions to this equation are z cos 0 i sin 0, cos 23 i sin 23 , and cos 43 i sin 43 or z 1, 12 23 i, 12 23 i. i i2 4 1 1 i 5 1 5 2 i 93. z i z 1 0 z 2 1 2 2 i 9 i i 2 4 1 2 2i or i 94. z 2 i z 2 0 z 2 2 2i 2i2 4 1 2 2i 4 95. z 2 2i z 2 0 z i 1 2 2 96. z 2 1 i z i z i z 1 0, so z 1 or i. 97. 1
1 cos 0 i sin 0, so by De Moivre’s Theorem, its n roots are 2k 2k 0 2k 0 2k i sin cos i sin for k 0 1 2 n 1. So z 0 1, z k 11n cos n n n n 2 4 2 4 i sin , z 2 cos i sin 2 , and so on. z 1 cos n n n n
23
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
n 98. From Exercise 97, k 1 for 0 k n 1. Multiplying both sides by s n and noting that s n z, we have n n s n k s n sk z for 0 k n 1, showing that 1 s s2 sn1 are nth roots of z. 99. Let a bi and z c di.
(a) zz c di c di c2 d 2 i 2 c2 d 2 z2 .
(b) z a bi c di ac bd ad bc i, so z ac bd2 ad bc2 ac2 2abcd bd2 ad2 2abcd bc2 ac2 ad2 bc2 bd2 and z a 2 b2 c2 d 2 a 2 b2 c2 d 2 ac2 ad2 bc2 bd2 z. (c)
(d)
1 c di c di 1 c d 2 2 2 i, so 2 2 z c di c di c d c d c d2 2 2 1 d2 1 1 c2 1 c d . 2 2 z 2 d2 2 2 z 2 2 2 2 c2 d 2 c2 d 2 c c d c d c d
a bi c di ac bd bc ad a bi c di 2 2 i, so z c di c di c2 d 2 c d2 c d2 ac bd 2 bc ad 2 ac2 bd2 2abcd bc2 ad2 2abcd 2 z c2 d 2 c2 d 2 c2 d 2 a 2 b2 c2 d 2 2 2 ac bd bc ad a 2 b2 2 2 2 2 z 2 2 2 2 c d c d c d
100. The cube roots of 1 are 0 1, 1 cos 23 i sin 23 12 23 i, and 2 cos 43 i sin 43 12 23 i, so their sum is 0 1 2 1 12 23 i 12 23 i 0.
The fourth roots of 1 are 0 1, 1 i, 2 1, and 3 i, so their sum is 0 1 2 3 1 i 1 i 0. The fifth roots of 1 are 0 1, 1 cos 25 i sin 25 , 2 cos 45 i sin 45 , 3 cos 65 i sin 65 , and
4 cos 85 i sin 85 , so their sum is 1 cos 25 i sin 25 cos 45 i sin 45 cos 65 i sin 65 cos 85 i sin 85 5 5 1 1 1 2 cos 25 2 cos 65 (most terms cancel) 1 2 0 4 4 4 4
3 3 1 2 2 1 2 The sixth roots of 1 are 0 1, 1 cos 3 i sin 3 2 2 i, cos 3 i sin 3 2 2 i,
3 1, 4 cos 43 i sin 43 12 23 i, and 5 cos 53 i sin 53 12 23 i, so their sum is 1 1 12 23 i 23 i 1 12 23 i 12 23 i 0. 2
2 2 2 2 3 3 2 3 The eight roots of 1 are 0 1, 1 cos 4 i sin 4 2 2 i, i, cos 4 i sin 4 2 2 i,
4 1, 5 cos 54 i sin 54 22 22 i, 6 i, and 7 cos 74 i sin 74 22 22 i, so their sum is 1 22 22 i i 22 22 1 22 22 i i 22 22 i 0. It seems that the sum of any set of nth roots is 0. To prove this, factor n 1 1 1 1 1 2 3 n1 . Since this is 0 and 1 0, we must have
1 1 2 3 n1 0.
24
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
101. The cube roots of 1 are 0 1, 1 cos 23 i sin 23 , and 2 cos 43 i sin 43 , so their product is 0 1 2 1 cos 23 i sin 23 cos 43 i sin 43 cos 2 i sin 2 1. The fourth roots of 1 are 0 1, 1 i, 2 1, and 3 i, so their product is 0 1 2 3 1 i 1 i i 2 1.
The fifth roots of 1 are 0 1, 1 cos 25 i sin 25 , 2 cos 45 i sin 45 , 3 cos 65 i sin 65 , and 4 cos 85 i sin 85 , so their product is 1 cos 25 i sin 25 cos 45 i sin 45 cos 65 i sin 65 cos 85 i sin 85 cos 4 i sin 4 1.
3 2 2 1 2 3 The sixth roots of 1 are 0 1, 1 cos 3 i sin 3 , cos 3 i sin 3 2 2 i, 1,
4 cos 43 i sin 43 12 23 i, and 5 cos 53 i sin 53 12 23 i, so their product is cos 2 i sin 2 1 cos 4 i sin 4 5 i sin 5 cos 5 i sin 5 1. 1 cos cos i sin 3 3 3 3 3 3 3 3 3 3 2 3 The eight roots of 1 are 0 1, 1 cos 4 i sin 4 , i, cos 4 i sin 4 ,
4 1, 5 cos 54 i sin 54 , 6 i, 7 cos 74 i sin 74 , so their product is 3 3 5 5 7 7 i2 1 cos 4 i sin 4 i cos 4 i sin 4 1 cos 4 i sin 4 i cos 4 i sin 4
cos 2 i sin 2 1. The product of the nth roots of 1 is 1 if n is even and 1 if n is odd. m m 1 . The proof requires the fact that the sum of the first m integers is 2 2 2 2k 2k Let cos i sin . Then k cos i sin for k 0 1 2 n 1. The argument of the n n n n product of the n roots of unity can be found by adding the arguments of each k . So the argument of the product is 2 n 2 2 n 1 2 2 1 2 2 2 3 [0 1 2 3 n 2 n 1]. 0 n n n n n n 2 n 1 n n 1 . Thus the product of the n roots of Since this is the sum of the first n 1 integers, this sum is n 2 unity is cos n 1 i sin n 1 1 if n is even and 1 if n is odd.
102.
z1 r cos 1 i sin 1 cos 2 i sin 2 r cos 1 cos 2 i 2 sin 1 cos 2 i sin 1 cos 2 i sin 2 cos 1 1 1 z2 r2 cos 2 i sin 2 cos 2 i sin 2 r2 cos2 2 i 2 sin2 2 r 1 cos 1 2 i sin 1 2 r2
103. Let 1 i and z a bi. Then z 1 i a bi a b a b i, 2 12 12 2, and z2 a 2 b2 . Then by Exercise 99(b), z2 a b2 a b2 2 z2 2 a 2 b2 . Thus, for any given a and b, we have c a b and d a b such that c2 d 2 2 a 2 b2 .
8.4
PLANE CURVES AND PARAMETRIC EQUATIONS
1. (a) The parametric equations x f t and y g t give the coordinates of a point x y f t g t for appropriate values of t. The variable t is called a parameter. (b) When t 0 the object is at 0 02 0 0 and when t 1 the object is at 1 12 1 1. (c) If we eliminate the parameter in part (b) we get the equation y x 2 . We see from this equation that the path of the moving object is a parabola.
SECTION 8.4 Plane Curves and Parametric Equations
25
2. (a) It is true that the same curve can be described by parametric equations in many different ways. For instance, the parabola y x 2 can be represented by x t, y t 2 or by x t, y t 2 or by x t, y t.
(b) The parametric equations x 2t, y 2t2 model the position of a
y
moving object at time t. When t 0 the object is at 0 0, and when t 1 the object is at 2 4.
t=1 [Ex. 2(b)]
(c) If we eliminate the parameter we get the equation y x 2 , which is the same equation as in Exercise 1(b). So the objects in Exercises 1(b) and
1
2(b) move along the same path, but traverse the path differently.
t=1 [Ex. 1(b)]
0
1
2
t=0 [1(b) and 2(b)]
3. (a) x 2t, y t 6
4. (a) x 6t 4, y 3t, t 0
y
y
1 1
1 2
(b) Since x 2t, t
3 x
x
x
x x and so y 6 2 2
x 2y 12 0.
y (b) Since y 3t, t and so 3 y 4 2y 4 x 2y 4 0, y 0. x 6 3
2 6. (a) x 2t 1, y t 12
5. (a) x t 2 , y t 2, 2 t 4 y
y
1
1 4
x
(b) Since y t 2 t y 2, we have x t 2
x y 22 , and since 2 t 4, we have
4 x 16.
1
x
2 2 (b) Since y t 12 , 4y 4 t 12 2t 12 ,
and since x 2t 1, we have x 2 2t 12 4y
y 14 x 2 .
26
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
7. (a) x
t, y 1 t t 0 y
8. (a) x t 2 , y t 4 1 y
1
x
20 10
_10
x
1
(b) Since x
t, we have x 2 t, and so y 1 x 2
with x 0.
9. (a) x
1 , y t 1 t
(b) Since x t 2 , we have x 2 t 4 and so y x 2 1, x 0.
10. (a) x t 1, y y
t t 1
1
(b) Since x
1 1 1 we have t and so y 1. t x x
11. (a) x 4t 2 , y 8t 3
1
x
1
y
x
1
(b) Since x t 1, we have t x 1, so y
x 1 . x
12. (a) x t, y 1 t
y
y
1 1
x 1 x
1
(b) Since y 8t 3 have y 2 x 3 .
3 y 2 64t 6 4t 2 x 3 , we
(b) Since x t, we have y 1 x, where x 0.
14. (a) x 2 cos t, y 3 sin t, 0 t 2
13. (a) x 2 sin t, y 2 cos t, 0 t y
y
1
1 1
x
(b) x 2 2 sin t2 4 sin2 t and y 2 4 cos2 t. Hence, x 2 y 2 4 sin2 t 4 cos2 t 4 x 2 y 2 4,
where x 0.
1
x
(b) We have cos t x2 and sin t y3, so
x22 y32 cos2 t sin2 t 1 1 x 2 1 y 2 1. 4 9
SECTION 8.4 Plane Curves and Parametric Equations
15. (a) x sin2 t, y sin4 t
16. (a) x sin2 t, y cos t
y
27
y 1
1
1
x
1
(b) Since x sin2 t we have x 2 sin4 t and so y x 2 . But since 0 sin2 t 1 we only get the part of this parabola for which 0 x 1.
17. (a) x cos t, y cos 2t
x
(b) Since y cos t, we have y 2 cos2 t and so
x y 2 sin2 t cos2 t 1 or x 1 y 2 (actually y 1 x), where 0 x 1.
18. (a) x cos 2t, y sin 2t
y 1
y
This curve is goimg counterclockwise
1
1
x
(b) Since x cos t we have x 2 cos2 t, so
2x 2 1 2 cos2 t 1 cos 2t y. Hence, the
rectangular equation is y 2x 2 1, 1 x 1.
19. (a) x sec t, y tan t, 0 t 2 x 1 and y 0.
1
x
(b) x 2 cos2 2t and y 2 sin2 2t. Then
x 2 y 2 cos2 2t sin2 2t 1 x 2 y 2 1.
20. (a) x cot t, y csc t, 0 t so y 1. y
y
1 1
1 1
x
x
(b) x 2 sec2 t, y 2 tan2 t, and
y 2 1 tan2 t 1 sec2 t x 2 . Therefore, y 2 1 x 2 x 2 y 2 1, x 1, y 0.
(b) x 2 cot2 t, y 2 csc2 t, and so
x 2 1 cot2 t 1 csc2 t y 2 . Therefore,
y 2 x 2 1, with y 1. This is the top half of the
hyperbola y 2 x 2 1.
28
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
21. (a) x tan t, y cot t, 0 t 2 x 0.
22. (a) x et , y et , so y 0.
y
y
1
1 x
1
x
1
(b) x y tan t cot t 1, so y 1x for x 0. 23. (a) x e2t , y et , so y 0.
(b) x y et et 1, so y 1x, x 0. 24. (a) x sec t, y tan2 t, 0 t 2 , so y 0 and
y
x 1.
2
y
1 _1 0
1
2
3
4
5
7 x
6
2
2 (b) x e2t et y 2 , y 0.
x
1
(b) x 2 sec2 t 1 tan2 t, so x 2 1 y, x 1, y 0.
25. (a) x cos2 t, y sin2 t, so 0 x 1 and 0 y 1. y
26. (a) x cos3 t, y sin3 t, 0 t 2 y
1
1
1
x
(b) x y cos2 t sin2 t 1. Hence, the equation is x y 1 for 0 x 1 and 0 y 1.
1
x
(b) x 23 cos2 t and y 23 sin2 t, and so x 23 y 23 1.
27. x 3 cos t, y 3 sin t. The radius of the circle is 3, the position at time 0 is x 0 y 0 3 cos 0 3 sin 0 3 0 and the orientation is counterclockwise (because x is decreasing and y is increasing initially). x y 3 0 again when t 2, so it takes 2 units of time to complete one revolution.
28. x 2 sin t, y 2 cos t. The radius is 2, the position at time 0 is 0 2, the orientation is clockwise (because x is increasing and y is decreasing initially), and it takes 2 units of time to complete one revolution.
29. x sin 2t, y cos 2t. The radius of the circle is 1, the position at time 0 is x 0 y 0 sin 0 cos 0 0 1 and the orientation is clockwise (because x is increasing and y is decreasing initially). x y 0 1 again when t , so it takes units of time to complete one revolution. 30. x 4 cos 3t, y 4 sin 3t. The radius is 4, the position at t 0 is 4 0, the orientation is counterclockwise, and it takes 2 units of time to complete one revolution. 3
Answers to Exercises 31–36 will vary.
SECTION 8.4 Plane Curves and Parametric Equations
29
31. The radius is 5 and it takes 4 seconds to complete a clockwise revolution, so if the position at t 0 is 0 5, then parametric equations are x 5 sin 12 t, y 5 cos 12 t.
32. The radius is 1 and it takes 2 seconds to complete a counterclockwise revolution, so if the position at t 0 is 1 0, parametric equations are x cos t, y sin t. 33. Since the line passes through the point 4 1 and has slope 12 , parametric equations for the line are x 4 t, y 1 12 t.
34. Since the line passes through the points 6 7 and 7 8, its slope is x 6 t, y 7 t.
87 1. Thus, parametric equations for the line are 76
35. Since cos2 t sin2 t 1, we have a 2 cos2 t a 2 sin2 t a 2 . If we let x a cos t and y a sin t, then x 2 y 2 a 2 . Hence, parametric equations for the circle are x a cos t, y a sin t.
a 2 cos2 t b2 sin2 t x2 y2 1. If we let x a cos t and y b sin t, then 2 2 1. 2 2 a b a b Hence, parametric equations for the ellipse are x a cos t, y b sin t. x 37. x 0 cos t, y 0 sin t 16t 2 . From the equation for x, t . Substituting into the equation for y gives 0 cos 2 x x 16x 2 . Thus the equation is of the form y c1 x c2 x 2 , 16 y 0 sin x tan 2 0 cos 0 cos 0 cos2
36. Since cos2 t sin2 t 1, we have
where c1 and c2 are constants, so its graph is a parabola. 38. 0 2048 fts and 30 . (a) The projectile will hit the ground when y 0. Since y 0 sin t 16t 2 , we have 0 2048 sin 30 t 16t 2 0 1024t 16t 2 64 seconds.
16t t 64 0 t 0 or t 64. Hence, the projectile will hit the ground after
(b) After 64 seconds, the projectile will hit the ground at x 2048 cos 30 64 65,536 3 113,5117 ft 215 miles. (c) The maximum height attained by the projectile is the maximum value of y. Since y 1024t 16t 2 16 t 2 64t 16 t 2 64t 1024 16,384 16 t 322 16,384, y is maximized when t 32, and therefore the maximum height is 16,384 ft 31 miles. 40. x 2 sin t, y cos 4t
39. x sin t, y 2 cos 3t
2 1 1
1
2
1
2
2
41. x 3 sin 5t, y 5 cos 3t
42. x sin 4t, y cos 3t 5
1
1 5
1 1
30
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
43. x sin cos t, y cos t 32 , 0 t 2
44. x 2 cos t cos 2t, y 2 sin t sin 2t 2
1
1
2
1
2
1
2
45. (a) r 212 , 0 4 x 2t12 cos t, y 2t12 sin t
46. (a) r sin 2 cos x sin t 2 cos t cos t, y sin t 2 cos t sin t
(b)
(b)
2
2
2
2 2
47. (a) r (b)
1
4 cos t 4 sin t 4 x ,y 2 cos 2 cos t 2 cos t
48. (a) r 2sin x 2sin t cos t, y 2sin t sin t (b)
2
2
2
2 2
2
4 2
2
2 49. x t 3 2t, y t 2 t is Graph III, since y t 2 t t 2 t 14 14 t 12 14 , and so y 14 on this curve, while x is unbounded.
50. x sin 3t, y sin 4t is Graph IV, since when t 0 and t the curve passes through 0 0. Thus this curve must pass through 0 0 twice.
51. x t sin 2t, y t sin 3t is Graph II, since the values of x and y oscillate about their values on the line x t, y t y x. 52. x sin t sin t, y cos t cos t is Graph I, since this curve does not pass through the point 0 0.
31
SECTION 8.4 Plane Curves and Parametric Equations
53. (a) It is apparent that x O Q and
(b) The curve is graphed with a 3 and b 2.
y Q P ST . From the diagram,
x O Q a cos and y ST b sin .
2
Thus, parametric equations are x a cos and y b sin .
4
2
y
a
b O
2
4
2 S ¬ T
(c) To eliminate we rearrange: sin yb
P Q
sin2 yb2 and cos xa
x
cos2 xa2 . Adding the two equations:
sin2 cos2 1 x 2 a 2 y 2 b2 . As
indicated in part (b), the curve is an ellipse. 54. (a) A has coordinates a cos a sin . Since O A is perpendicular to AB, 2
O AB is a right triangle and B has coordinates a sec 0. It follows that P has coordinates a sec b sin . Thus, the parametric equations are x a sec , y b sin .
10
10
(b) The curve is graphed at right with a 3, b 2, and . 55. (a) If we modify Figure 8 so that PC b, then by the
2
(b) 5
same reasoning as in Example 6, we see that x OT P Q a b sin and y T C C Q a b cos .
We graph the case where a 3 and b 2.
20
20
56. (a) From the figure, we see that x OT P Q a b sin and y T C C Q a b cos . When a 1 and b 2, the parametric equations become x 2 sin , y 1 2 cos . The curve is graphed for 0 4. (b)
y
C(a¬,a)
b
P(x, y) O
¬
a¬
10
Q
T
10
x
2 x2 y 2 y . Since sec . Also, y b sec sec b a2 b2 2 2 2 2 x y y x tan2 sec2 1, we have 2 2 1 2 2 1, which is the equation of a hyperbola. a b b a
57. x a tan tan
x a
tan2
32
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
58. Substituting the given values for x and y into the equation we derived in Exercise 57, we get 2 2 b t 1 a t t 1 t 1. Thus the points on this curve satisfy the equation, which is that of a hyperbola. b2 a2 x2 y2 However, this hyperbola is only the part of 2 2 1 for which x 0 and y 0. b a
59. x t cos t, y t sin t, t 0 t
x
y
t
0
0
0
5 4 3 2 7 4
2 8
2 8
0
4
2 3 4
61. x
2 3 2 8
38 2
2
x
y
y
y
1
58 2 58 2 0 32 7 2 7 2 8 8
2
60. x sin t, y sin 2t
x
1
0
0
3t 3t 2 ,y , t 1 3 1t 1 t3 t
09
x
y
t
x
y
t
996 897
2
067 133
11
075 389 292
25 045 113
05
171 086
3
032 096
0
4
018 074
05
133 067
5
012 060
1
15
6
008 050
15
103 154
0
0
x
1
1
15
x
y
t
997 1097
x
y
4
019 076 012 060
125 393
492
45 015 067
15
284
5
171
6
128
7
104
8
2 25 3 35
189 086 051 035 025
087
y 1
1
x
008 050 006 043 005 038
As t 1 we have x and y . As t 1 we have x and y . As t we have x 0 and y 0 . As t we have x 0 and y 0 .
62. x cot t, y 2 sin2 t, 0 t
y
2
1
1
x
SECTION 8.4 Plane Curves and Parametric Equations y
63. (a) We first note that the center of circle C (the small circle) has coordinates [a b] cos [a b] sin . Now the arc P Q has the same length as the arc
a ab a , and so . Thus the b b b xcoordinate of P is the xcoordinate of the center of circle C plus ab , and the ycoordinate of P is the b cos b cos b ab ycoordinate of the center of circle C minus b sin b sin . b ab So x a b cos b cos and b ab y a b sin b sin . b P Q, so b a
(b) If a 4b, b
33
¬ ú
Q
P»
¬
a , and x 34 a cos 14 a cos 3, y 34 a sin 14 a sin 3. 4
(a, 0) P x
y
a
From Example 2 in Section 7.3, cos 3 4 cos3 3 cos . Similarly, one can prove that sin 3 3 sin 4 sin3 . Substituting, we get x 34 a cos 14 a 4 cos3 3 cos a cos3 y 34 a sin 14 a 3 sin 4 sin3 a sin3 . Thus,
a x
x 23 y 23 a 23 cos2 a 23 sin2 a 23 , so x 23 y 23 a 23 .
64. The coordinates of C are [a b] cos [a b] sin . Let be OC P as shown in the figure. Then the arcs traced a out by rolling the circle along the outside are equal, that is, b a . P is displaced from C by amounts b equal to the legs of the right triangle C PT , where C P is the hypotenuse. Since OC Q is a right triangle, it follows that a ab OC Q . Thus, . So 2 2 b 2 b 2 ab ab a b cos b sin x a b cos b sin b 2 2 b ab a b cos b cos b and ab ab a b sin b cos y a b sin b cos b 2 2 b ab a b sin b sin b Therefore, parametric equations for the epicycloid are ab ab x a b cos b cos and y a b sin b sin . b b
y
C a ¬
ú
Œ
T
P Q
x
34
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
65. A polar equation for the circle is r 2a sin . Thus the coordinates of Q are x r cos 2a sin cos and
y r sin 2a sin2 . The coordinates of R are x 2a cot and y 2a. Since P is the midpoint of Q R, we use the midpoint formula to get x a sin cos cot and y a 1 sin2 .
66. (a) C 2a cot 2a, so the xcoordinate of P is x 2a cot . Let B 0 2a. Then O AB is a right angle and O B A , so O A 2a sin and A 2a sin cos 2a sin2 . Thus, the ycoordinate of P is y 2a sin2 .
5
Exercise asks for a = 3
(b) The curve is graphed with a 1.
10
10
ay cos 67. We use the equation for y from Example 6 and solve for . Thus for 0 , y a 1 cos a ay ay ay cos1 . Substituting into the equation for x, we get x a cos1 sin cos1 . a a a 2ay y 2 2ay y 2 ay ay ay 2 However, sin cos1 1 . Thus, x a cos1 , a a a a a 2ay y 2 x 2ay y 2 x ay y 1 cos 1 cos and we have a a a a 2ay y 2 x . y a 1 cos a 68. (a)
2 2
5
2 2
2 2
2
2
5
5
5
5
5 5
5
R 05 R1 R3 (b) The graph with R 5 seems to most closely resemble the profile of the engine housing.
R5
69. (a) In the figure, since O Q and QT are perpendicular and OT and T D are perpendicular, the angles formed by their intersections are equal, that is, DT Q. Now the coordinates of T are cos sin . Since T D is the length of the string that has been unwound from the circle, it must also have arc length , so T D . Thus the xdisplacement from T to D is sin while the ydisplacement from T to D is cos . So the coordinates of D are x cos sin and y sin cos . (b) y 10
T 1 ¬
D Q
1
x
20
20 10
70. (a) It takes 2 units of time. The parametric equations of the particle that moves twice as fast around the circle are x sin 2t, y cos 2t.
35
SECTION 8.5 Vectors
(b) From the first table, we see that the particle travels counterclockwise. If we want the particle to travel in a clockwise direction, then we want the second table to apply. Possible parametric equations for clockwise traversal of the unit circle are x sin t, y cos t. t
x sin t
0
0
1
2
0
3 2
1
2
y
y cos t
t
x
y
1
0
0
1
0
1
0
t=¹
equation.
2
x
0 3¹ t= 2
1
71. C: x t, y t 2 ; D: x t, y t, t 0 (a) For C, x t, y t 2 y x 2 . For D, x t, y t y x 2 . For F, x et
t=0
1
1
0
For E, x sin t
¹
1 t= 2
E: x sin t, y 1 cos2 t
0
3 2
1
2
0
1
0 1
F: x et , y e2t
In #71, replace "e" with "3" (4 times)
x 2 sin2 t 1 cos2 t y and so y x 2 .
x 2 e2t y and so y x 2 . Therefore, the points on all four curves satisfy the same rectangular
(b) Curve C is the entire parabola y x 2 . Curve D is the right half of the parabola because t 0 and so x 0. Curve E is the portion of the parabola for 1 x 1. Curve F is the portion of the parabola where x 0, since et 0 for all t. y
y
y
y
1
1
1
1
0
1
0
x
C
8.5
1
D
x
0
E
1
x
0
1
x
F
VECTORS
1. (a) The vector u has initial point A and terminal point B.
(b) The vector u has initial point 2 1 and terminal point 4 3. In component form we write u 2 2
v
u
2u
u+v
and v 3 6. Then 2u 4 4 and u v 1 8.
u
2. (a) The length of a vector w a1 a 2 is w u 22 22 8 2 2.
a12 a22 , so the length of the vector u in Figure II is
(b) If we know the length w and direction of a vector w then we can express the vector in component form as w w cos w sin .
36
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
3. 2u 2 2 3 4 6
4. v 3 4 3 4
y
y
2u v
u 1
v
u 1 x
1
5. u v 2 3 3 4 2 3 3 4 1 7 y
1
_v
x
6. u v 2 3 3 4 2 3 3 4 5 1 y
v u+v _v
u 1 x
1
7. v 2u 3 4 2 2 3 3 2 2 4 2 3
u 1 1
u-v
x
8. 2u v 2 2 3 3 4 2 2 3 2 3 4 1 10
7 2 y
y
v 2u+v _2u
v
1
2u
x
1
1
v-2u
1
x
In Solutions 9–18, v represents the vector with initial point P and terminal point Q.
9. P 2 1, Q 5 4. v 5 2 4 1 3 3
10. P 2 1, Q 3 4. v 3 2 4 1 5 3
11. P 1 2, Q 4 1. v 4 1 1 2 3 1
12. P 3 1, Q 1 2. v 1 3 2 1 2 3
13. P 1 3, Q 4 5. v 4 1 5 3 3 2
14. P 2 5, Q 3 1. v 3 2 1 5 1 4
15. P 5 3, Q 1 0. v 1 5 0 3 4 3
16. P 1 3, Q 6 1.
17. P 1 1, Q 1 1. v 1 1 1 1 0 2
18. P 8 6, Q 1 1. v 1 8 1 6 7 5
v 6 1 1 3 5 4
SECTION 8.5 Vectors y
19.
(6, 7)
y
20.
(3, 5)
u
u (4, 3)
(4, 3)
1
1 x
1
The terminal point is 4 2 3 4 6 7. y
21.
The terminal point is 4 1 3 2 3 5. y
22.
(4, 3)
u
(4, 3)
u
1
(_4, 2)
1
(8, 0) x
1
y
x
1
The terminal point is 4 4 3 3 8 0. 23.
x
1
The terminal point is 4 8 3 1 4 2. y
24. (_3, 5)
(2, 3)
u u 1
u
(_3, 5) u
1
x
1
u
x
u
1
25.
u
(2, 3)
y
26.
(_3, 5)
y (2, 3)
(2, 3)
u u
(_3, 5)
u
1 1
1 1
x u
27. u 2 3 2i 3j
28. u 1 0 i
29. u 0 2 2j
30. u 4 5 4i 5j
u
x
37
38
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
31. u 1 4, v 1 2. 2u 2 1 4 2 8; 3v 3 1 2 3 6; u v 1 4 1 2 0 6; 3u 4v 3 12 4 8 7 4 32. u 2 5 v 2 8. 2u 2 2 5 4 10; 3v 3 2 8 6 24; u v 2 5 2 8 0 3; 3u 4v 6 15 8 32 14 47 33. u 0 1, v 2 0. 2u 20 1 0 2; 3v 32 0 6 0; uv 0 12 0 2 1; 3u 4v 0 3 8 0 8 3 34. u i, v 2j. 2u 2i; 3v 3 2j 6j; u v i 2j; 3u 4v 3i 8j 35. u 2i j, v j. 2u 2 2i j 4i 2j; 3v 3j; u v 2i j j 2i; 3u 4v 3 2i j 4j 6i 7j 36. u i j, v i j. 2u 2i 2j 3v 3i 3j u v i j i j 2i; 3u 4v 3 i j 4 i j i 7j 37. u 3i j v 2i 3j. Then u 32 12 10; v 22 32 13; 2u 6i 2j; 2 1 3 2 2 2u 6 2 2 10; 2 v i 2 j; 12 v 12 32 12 13; u v 5i 2j; u v 52 22 29; u v 3i j 2i 3j i 4j; u v 12 42 17; u v 10 13 38. u 2i 3j v i 2j. Then u 4 9 13; v 1 4 5; 2u 4i 6j; 2u 16 36 2 13; 1 v 1 i j; 1 v 1 1 1 5; u v i j; u v 1 1 2; u v 3i 5j; 2 2 2 4 2 u v 9 25 34; u v 13 5 39. u 10 1, v 2 2. Then u 102 12 101; v 22 22 2 2; 2u 20 2; 2u 202 22 404 2 101; 12 v 1 1; 12 v 12 12 2; u v 8 3; u v 82 32 73; u v 12 1; u v 122 12 145; u v 101 2 2 40. u 6 6, v 2 1. Then u 36 36 6 2; v 4 1 5; 2u 12 12; 2u 144 144 12 2; 12 v 1 12 ; 12 v 1 14 12 5; u v 8 5; u v 64 25 89; u v 4 7; u v 16 49 65; u v 6 2 5 In Solutions 41–46, x represents the horizontal component and y the vertical component. 41. v 10, direction 60 . x 10 cos 60 5 and y 10 sin 60 5 3. Thus, v xi yj 5i 5 3j. 42. v 20, direction 150 . x 20 cos 150 10 3 and y 20 sin 150 10. Thus, v xi yj 10 3i 10j. 43. v 1, direction 225 . x cos 225 1 and y sin 225 1 . Thus,
2 2 2 2 1 1 v xi yj i j 2 i 2 j. 2 2 44. v 800, direction 125 . x 800 cos 125 45886 and y 800 sin 125 65532. Thus, v xi yj 800 cos 125 i 800 sin 125 j 45886i 65532j.
45. v 4, direction 10 . x 4 cos 10 394 and y 4 sin 10 069. Thus, v xi yj 4 cos 10 i 4 sin 10 j 394i 069j. 46. v 3, direction 300 . x 3 cos 300 23 and y 3 sin 300 32 . Thus, v xi yj 23 i 32 j. 47. v 3 4. The magnitude is v 32 42 5. The direction is , where tan 43 tan1 43 5313 . 48. v 22 22 . The magnitude is v 12 12 1. The direction is , where tan 1 with in quadrant III 180 tan1 1 225 .
5 with in 49. v 12 5. The magnitude is v 122 52 169 13. The direction is , where tan 12 5 15738 . quadrant II tan1 12
SECTION 8.5 Vectors
50. v 40 9. The magnitude is v 9 1268 . tan1 40
39
9 with in quadrant I 1600 81 41. The direction is , where tan 40
2 3 2. The direction is , where tan 3 with in quadrant I 51. v i 3j. The magnitude is v 12 tan1 3 60 . 52. v i j. The magnitude is v 1 1 2. The direction is , where tan 1 with in quadrant I tan1 1 45 .
3 53. v 30, direction 30 . x 30 cos 30 30 2598, y 30 sin 30 15. So the horizontal component of 2 force is 15 3 lb and the vertical component is 15 lb. 54. v 500, direction 70 . x 500 cos 70 17101, y 500 sin 70 46985. So the east component of the velocity is 17101 mi/h and the north component is 46985 mi/h. 55. The flow of the river can be represented by the vector v 3j and the swimmer can be represented by the vector u 2i. Therefore the true velocity is u v 2i 3j. 56. If the current is 12 mi/h, it can be represented by the vector v 12j. If the swimmer heads in a direction north of east, his velocity relative to the water is u 2 cos i 2 sin j and so his velocity relative to land is u v 2 cos i 2 sin 12 j. We want the ycomponent of his velocity to be 0, so we calculate 2 sin 12 0 sin 35 sin1 53 369 . Therefore, he should swim 369 north of east, or N 531 E.
57. The ocean currents can be represented by the vector v 3i and the salmon can be represented by the vector u 5 2 2 i 5 2 2 j. Therefore u v 5 2 2 3 i 5 2 2 j represents the true velocity of the fish. 58. The direction of the plane is N 45 W, so its velocity relative to the air is u ux u y , where ux 585 cos 135 4137 and u y 585 sin 135 4137. The wind direction is W, so its velocity is w wx 0 40 0. Thus, the actual flight path is v u w 4137 40 4137 4537 4137, the airplane’s true speed is v 45372 41372 614 mih, and its true direction is 180 tan1 4137 4537 1376 or approximately N 476 W.
= <0, 40>
59. (a) The velocity of the wind is 40j.
In #71, the exercise asks for component form
(b) The velocity of the jet relative to the air is 425i.
= <425, 0>
(c) The true velocity of the jet is v 425i 40j 425 40. 40 54 is (d) The true speed of the jet is v 4252 402 427 mi/h, and the true direction is tan1 425 N 846 E.
55 3 60. (a) The velocity of the wind is w 55 cos 60 55 sin 60 55 2 2 . (b) The velocity of the jet is w 765 cos 45 765 sin 45 7652 2 7652 2 . 765 2 55 3 765 2 56844 58857. (c) The true velocity of the jet is v w 55 2 2 2 2 (d) The true speed of the jet is w v 568442 588572 818 mih. The direction of the vector w v is tan1 58857 56844 46 . Thus the true direction of the jet is approximately N 44 E.
61. Let v be the velocity vector of the jet and let be the direction of this vector. Thus v 765 cos i 765 sin j. If w 55 3 j. is the velocity vector of the wind then the true course of the jet is v w 765 cos 55 i 765 sin 2 2 To achieve a course of true north, the eastwest component (the i component) of the jet’s velocity vector must be 0. That 275 cos1 275 921 . Thus the pilot should head his plane in the is 765 cos 55 0 cos 2 765 765
direction N 21 W.
40
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
u
62. The speed of the jet is 300 mi/h, so its velocity relative to the air is v 300 cos i 300 sin j. The wind has velocity w 30j, so
w
¬
v
the true course of the jet is given by
u v w 300 cos i 300 sin 30 j. We want the ycomponent of the jet’s velocity to be 0, so we solve
1 574 . Therefore, the jet should head in the direction 18574 (or S 8426 W). 300 sin 30 0 sin 10
63. (a) The velocity of the river is represented by the vector r 10 0 10i.
(b) Since the boater direction is 60 from the shore at 20 mih, the velocity of the boat is represented by the vector b 20 cos 60 20 sin 60 10 10 3 10i 10 3j 10 1732. (c) w r b 10 10 0 10 3 20i 10 3j 20 1732 (d) The true speed of the boat is w 202 17322 265 mih, and the true direction is 409 N 491 E. tan1 1732 20
64. Let w be the velocity vector of the water, let v be the velocity vector of the boat, and let be the direction of v. Then v 20 cos i 20 sin j and w 10i. The true course of the boat is v w 20 cos 10 i 20 sin j. To achieve a course of true north, the eastwest component of the boat’s velocity vector must be 0. Thus, 20 cos 10 0 10 12 cos1 12 120 . Thus the boater should head the boat in the direction N 30 W. cos 20 65. (a) Let b bx b y represent the velocity of the boat relative to the water. Then b 24 cos 18 24 sin 18 24 cos 18 i 24 sin 18 j 228 74. (b) Let w x y represent the velocity of the water. Then w 0 where is the speed of the water. So the true velocity of the boat is b w 24 cos 18 24 sin 18 24 cos 18 i 24 sin 18 j. For the direction to be due east, we must have 24 sin 18 0 742 mih. Therefore, the true speed of the water is 74 mi/h. Since b w 24 cos 18 0 24 cos 18 i, the true speed of the boat is b w 24 cos 18 228 mih. 66. Let represent the velocity of the sailor and l the velocity of the ocean liner. Then w 2 0, and l 0 25, and so r 2 0 0 25. Hence, relative to the water, the sailor’s speed is r 4 625 2508 mih, and their direction 25 9457 or approximately N 457 W. is tan1 2
67. F1 2 5 and F2 3 8. (a) F1 F2 2 3 5 8 5 3
(b) The additional force required is F3 0 0 5 3 5 3.
68. F1 3 7, F2 4 2, and F3 7 9. (a) F1 F2 F3 3 4 7 7 2 9 0 0 (b) No additional force is required.
69. F1 4i j, F2 3i 7j, F3 8i 3j, and F4 i j. (a) F1 F2 F3 F4 4 3 8 1 i 1 7 3 1 j 0i 4j 4j (b) The additional force required is F5 0i 0j 0i 4j 4j.
70. F1 i j, F2 i j, and F3 2i j. (a) F1 F2 F3 1 1 2 i 1 1 1 j 0i j j
(b) The additional force required is F4 0i 0j 0i j j. 71. F1 10 cos 60 10 sin 60 5 5 3 , F2 8 cos 30 8 sin 30 4 3 4 , and F3 6 cos 20 6 sin 20 5638 2052. (a) F1 F2 F3 5 4 3 5638 5 3 4 2052 757 1061.
(b) The additional force required is F4 0 0 757 1061 757 1061.
SECTION 8.6 The Dot Product
41
72. F1 3 1, F2 1 2, F3 2 1, and F4 0 4. (a) F1 F2 F3 F4 3 1 2 0 1 2 1 4 2 4 (b) The additional force required is F5 0 0 2 4 2 4.
73. From the figure we see that T1 T1 cos 50 i T1 sin 50 j and T2 T2 cos 30 i T2 sin 30 j. Since T1 T2 100j we get T1 cos 50 T2 cos 30 0 and T1 sin 50 T2 sin 30 100. From the first cos 50 cos 50 sin 30 equation, T2 T1 , and substituting into the second equation gives T1 sin 50 T1 100 cos 30 cos 30 T1 sin 50 cos 30 cos 50 sin 30 100 cos 30 T1 sin 50 30 100 cos 30 cos 30 T1 100 879385. sin 80 cos 30 Similarly, solving for T1 in the first equation gives T1 T2 and substituting gives cos 50 cos 30 sin 50 T2 T2 sin 30 100 T2 cos 30 sin 50 cos 50 sin 30 100 cos 50 cos 50 100 cos 50 T2 652704. Thus, T1 879416 cos 50 i 879416 sin 50 j 565i 674j and sin 80 T2 652704 cos 30 i 652704 sin 30 j 565i 326j. 74. From the figure we see that T1 T1 cos 223 i T1 sin 223 j and T2 T2 cos 415 i T2 sin 415 j. Since T1 T2 18 278j, we get T1 cos 223 T2 cos 415 0 and T1 sin 223 T2 sin 415 18 278. From the cos 223 first equation, T2 T1 , and substituting into the second equation gives cos 415 cos 223 sin 415 T1 sin 223 T1 18 278 cos 415 T1 sin 223 cos 415 cos 223 sin 415 18 278 cos 415 cos 415 T1 sin 223 415 18,278 cos 415 T1 18,278 15 257. sin 638 cos 415 and substituting gives Similarly, solving for T1 in the first equation gives T1 T2 cos 223 cos 415 sin 223 T2 T2 sin 415 18,278 T2 sin 223 cos 415 sin 415 cos 223 18,278 cos 223 cos 223 18,278 cos 223 T2 18,847. sin 638 Thus, T1 15,257 cos 223 i 15,257 sin 223 j 14,116i 5,789j and T2 18,847 cos 415 i 18,847 sin 415 j 14,116i 12,488j. 75. When we add two vectors, the resultant vector can be found by first placing the initial point of the second vector at the terminal point of the first vector. The resultant vector can then found by using the new terminal point of the second vector and the initial point of the first vector. When these n vectors are placed head to tail in the plane so that they form a polygon, the initial point and the terminal point are the same. Thus the sum of the n vectors is the zero vector.
8.6
THE DOT PRODUCT
1. The dot product of u a1 a2 and v b1 b2 is defined by u v a1 a2 b1 b2 . The dot product of two vectors is a real number, or scalar, not a vector.
42
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
uv . So if u v 0, the vectors are perpendicular. u v To find the angle between the vectors u and v in the figure, we first find 4 3 3 2 6 13 12 6 , and so cos1 6 6513 109 , rounded to the cos 4 3 3 2 65 42 32 32 22
2. The angle satisfies cos
nearest degree.
u 3. (a) The component of u along v is the scalar u cos and can be uv expressed in terms of the dot product as . v projv u ¬ v uv compv u v. (b) The projection of u onto v is the vector projv u v2 4. The work done by a force F in moving an object along a vector D is W F D. 5. (a) u v 2 0 1 1 2 0 2 6. (a) u v 1 3 3 1 3 3 0 uv uv 2 1 45 (b) cos 0 90 (b) cos 2 2 2 u v u v 7. (a) u v 1 0 1 13 1 0 1 replace "13" 8. (a) u v 6 6 1 1 6 6 12 with "3" uv uv 1 60 (b) cos 12 1 180 (b) cos 12 6 2 2 u v u v
9. (a) u v 3 2 1 2 3 4 1 uv 1 97 (b) cos 13 5 u v 11. (a) u v 0 5 1 3 0 5 3 5 3
10. (a) u v 3 4 4 3 12 12 24 uv 24 163 55 (b) cos u v
13. (a) u v i 3j 4i j 4 3 1 uv (b) cos 1 86 10 17 u v
14. (a) u v 3i 4j 2i j 6 4 10 uv (b) cos 10 5 5 u v 153 cos1 10
uv (b) cos 5523 23 30 u v
15. (a) u v 1 3 1 3 1 3 2
uv 1 2 22 2 120 u v 17. u v 12 12 0 vectors are orthogonal (b) cos
12. (a) u v 1 1 1 1 1 1 0 uv (b) cos 0 0 90 2 2 u v
5 5
16. (a) u v 6 8 3 4 18 32 50 uv 50 1 0 105 (b) cos u v
18. u v 0 4 4 0 0 0 0 vectors are orthogonal
replace "orthogonal" with "perpendicular"
19. u v 8 12 4 0 vectors are not orthogonal
20. u v 2 0 0 7 0 0 0 vectors are orthogonal 21. u v 24 24 0 vectors are orthogonal
22. u v 4 1 0 3 4 0 4 vectors are not orthogonal
#17-22 use "perpendicular"
23. u v u w 2 1 1 3 2 1 3 4 2 3 6 4 9
24. u v w 2 1 [1 3 3 4] 2 1 4 1 8 1 9
25. u v u v [2 1 1 3] [2 1 1 3] 3 2 1 4 3 8 5
26. u v u w 2 1 1 3 2 1 3 4 1 10 10 12 u v 4 6 3 4 12 24 27. x v 3 4 5 5
SECTION 8.6 The Dot Product
28. x
1 1 3 5 u v 3 5 2 2 2 2 2 2 1 1 v 2 1 1 2
2
2
43
2
u v 7i 24j j 0 24 24 v j 1 uv 28 56 7i 8i 6j 56 0 30. x v 8i 6j 10 5 64 36 2 4 1 1 uv 1 1 1 1. v 31. (a) u1 projv u 12 12 v2 (b) u2 u u1 2 4 1 1 3 3. We resolve the vector u as u1 u2 , where u1 1 1 and u2 3 3. 7 4 2 1 uv 2 1 2 2 1 4 2 32. (a) u1 projv u v 22 12 v2 (b) u2 u u1 7 4 4 2 3 6. We resolve the vector u as u1 u2 , where u1 4 2 and u2 3 6. 1 2 1 3 uv 1 3 12 1 3 12 32 v 33. (a) u1 projv u 2 2 v 12 3 1 3 (b) u2 u u1 1 2 2 2 32 12 . We resolve the vector u as u1 u2 , where u1 12 32 and u2 32 12 . 11 3 3 2 uv 3 2 3 3 2 9 6 34. (a) u1 projv u v v2 32 22 (b) u2 u u1 11 3 9 6 2 3. We resolve the vector u as u1 u2 , where u1 9 6 and u2 2 3. 2 9 3 4 uv 6 3 4 18 24 3 35. (a) u1 projv u v 4 5 5 5 v2 33 42 24 28 21 18 24 (b) u2 u u1 2 9 18 5 5 5 5 . We resolve the vector u as u1 u2 , where u1 5 5 and 21 . u2 28 5 5 1 1 2 1 uv 1 2 1 2 1 2 36. (a) u1 projv u v 1 5 5 5 v2 22 12 (b) u2 u u1 1 1 25 15 35 65 . We resolve the vector u as u1 u2 , where u1 25 15 and u2 35 65 .
29. x
37. W F d 4 5 3 8 28
38. W F d 400 50 201 0 80 400
39. W F d 10 3 4 5 25
40. W F d 4 20 5 15 280
41. Let u u1 u2 and v v1 v2 . Then u v u 1 u 2 1 2 u 1 1 u 2 2 1 u 1 2 u 2 1 2 u 1 u 2 v u 42. Let u u1 u2 and v v1 v2 . Then
cu v c u 1 u 2 1 2 cu 1 cu 2 1 2 cu 1 1 cu 2 2 c u 1 1 u 2 2 c u v u 1 c 1 u 2 c 2 u 1 u 2 c 1 c 2 u cv
43. Let u u 1 u 2 , v 1 2 , and w 1 2 . Then u v w u 1 u 2 1 2 1 2 u 1 1 u 2 2 1 2
u 1 1 1 1 u 2 2 2 2 u 1 1 u 2 2 1 1 2 2 u 1 u 2 1 2 1 2 1 2 u w v w
44. Let u u 1 u 2 and v 1 2 . Then u v u v u 1 u 2 1 2 u 1 u 2 1 2 u 1 1 u 2 2 u 1 1 u 2 2 2 2 u 21 u 22 12 22 u2 v2 u 21 12 u 22 22 u 21 u 22 12 22
44
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
uv
v. Then v2 uv uv uv uv uv projv u u projv u v u v u v v v v2 v2 v2 v2 v2
45. We use the definition that projv u
u v2 v2
u v2 v4
v2
u v2 v2
u v2
Thus u and u projv u are orthogonal. uv u v v v u v v2 46. v projv u v v u v. v2 v2 v2
v2
0
47. W F d 4 7 4 0 16 ftlb
48. The displacement of the object is D 11 13 2 5 9 8. Hence, the work done is W F D 2 8 9 8 18 64 82 ftlb.
49. The distance vector is D 200 0 and the force vector is F 50 cos 30 50 sin 30 . Hence, the work done is W F D 200 0 50 cos 30 50 sin 30 200 50 cos 30 8660 ftlb.
50. W F d, and in this problem F 0 2500, d 500 cos 12 500 sin 12 Thus,
W 0 2500 500 cos 12 500 sin 12 0 2500 1040 260,000 ftlb. This is the work done by gravity; the
car does (positive) work in overcoming the force of gravity. So the work done by the car is 260,000 ftlb. 51. (a) Since the force parallel to the driveway is 490 w sin 10 w is about 2822 lb.
490 28218, and thus the weight of the car sin 10
(b) The force exerted against the driveway is 28218 cos 10 2779 lb.
52. Since the weight of the car is 2755 lb, the force exerted perpendicular to the earth is 2755 lb. Resolving this into a force u perpendicular to the driveway gives u 2766 cos 65 1164 lb. Thus, a force of about 1164 lb is required.
53. Since the force required parallel to the plane is 80 lb and the weight of the package is 200 lb, it follows that 80 200 sin , 80 2358 , and so the angle of inclination is where is the angle of inclination of the plane. Then sin1 200 approximately 236 .
54. Let R represent the force exerted by the rope and d the force causing the cart to roll down the ramp. Gravity acting on the cart exerts a force w of 40 lb directly downward. So the magnitude of the part of that force causing the cart to roll down the ramp is d 40 sin 15 . The angle between R and d is 45 , so the magnitude of the force holding the cart up is R cos 45 . Equating these two, we have R cos 45 40 sin 15 , so R
40 sin 15 1464 lb. cos 45
55. (a) 2 0 4 2 8, so Q 0 2 lies on L. 2 2 4 1 4 4 8, so R 2 1 lies on L. (b) u Q P 0 2 3 4 3 2. v Q R 0 2 2 1 2 1. 3 2 2 1 uv 2 1 v w projv u 2 2 v 2 2 1 8 85 2 1 16 5 5
(c) From the graph, we can see that u w is
orthogonal to v (and thus to L). Thus, the
distance from P to L is u w. y
L
P u
Q 0
w v
u-w R x
CHAPTER 8
CHAPTER 8 REVIEW
1. (a)
O
(8, _ 3¹ 4)
¹/6
3 (b) x 12 cos 12 6 2 6 3,
1 y 12 sin 6 12 2 6. Thus, the rectangular coordinates of P are 6 3 6 .
3. (a)
2¹ 3
O
O
(b) x 3 cos 74 3 22 3 2 2 , y 3 sin 74 3 22 3 2 2 . Thus, the rectangular coordinates of P are 3 2 2 3 2 2 .
5. (a)
(_Ï3, 2¹ 3 ) (b) x 3 cos 23 3 12 23 , y 3 sin 23 3 23 32 . Thus, the rectangular coordinates of P are 23 32 .
6. (a)
y
y P
P
1
2 2
(b) r
_ 3¹ 4
(b) x 8 cos 34 8 22 4 2, y 8 sin 34 8 22 4 2. Thus, the rectangular coordinates of P are 4 2 4 2 .
4. (a)
7¹ 4
(_3, 7¹ 4 )
O
2. (a)
(12, ¹6)
Review
x
82 82 128 8 2 and tan1 88 .
Since P is in quadrant I, 4 . Polar coordinates for P are 8 2 4 . (c) 8 2 54
1 x
(b) r
2 2 2 6 8 2 2 and
tan1
6 tan1 3 . Since x is 3 2
negative and y is positive, 23 . Polar coordinates for P are 2 2 23 . (c) 2 2 53
45
46
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
7. (a)
8. (a)
y
y
2 2
1
x
x
1
P
P
2 2 6 2 6 2 144 12 and (b) r
tan1 62 4 . Since P is in quadrant III, 6 2 54 . Polar coordinates for P are 12 54 . (c) 12 4
(b) r
42 42 32 4 2 and
7 tan1 4 4 . Since P is in quadrant IV, 4 . Polar coordinates for P are 4 2 74 . (c) 4 2 34
9. (a) x y 4 r cos r sin 4
10. (a) x y 1 r cos r sin 1
r cos sin 4 r
r 2 cos sin 1 r 2
4 cos sin
(b) The rectangular equation is easier to graph. y
2 1 or cos sin sin 2
r 2 2 csc 2. (b) The rectangular equation is easier to graph. y
1
1
11. (a) x 2 y 2 4x 4y r 2 4r cos 4r sin r 2 r 4 cos 4 sin r 4 cos 4 sin
(b) The polar equation is easier to graph.
x
1
x
1
2 2 12. (a) x 2 y 2 2x y r 2 2 r cos r sin r 4 2r 2 cos sin r 2 2 cos sin
r 2 sin 2
(b) The polar equation is easier to graph.
O
1
O
1
CHAPTER 8
13. (a)
(3, ¹2 )
14. (a)
(3, ¹2 )
Review
(6, 0)
2
(3, 3¹ 2 )
O
(b) r 3 3 cos r 2 3r 3r cos , which gives x 2 y 2 3 x 2 y 2 3x x 2 3x y 2 3 x 2 y 2 . Squaring both sides 2 gives x 2 3x y 2 9 x 2 y 2 .
15. (a)
1
(b) r 3 sin r 2 3r sin , so x 2 y 2 3y 2 x 2 y 2 3y 94 94 x 2 y 32 94 .
16. (a)
O
O
2
(b) r 2 sin 2 r 2 2 sin cos
r 3 4r 2 sin cos 32 4 r sin r cos and so, since r2 x r cos and y r sin , we get 3 x 2 y 2 16x 2 y 2 .
2
(b) r 4 cos 3 4 cos 2 cos sin 2 sin 4 cos cos2 sin2 2 sin2 cos r 4 cos3 3 sin2 cos r 4 4r 3 cos3 3 sin2 cos , which gives
x 2 y2
2
4x 3 12x y 2 .
18. (a)
17. (a) (1, ¹)
O
1
(1, 0)
1 1 2 cos 2 cos sin2 r 2 cos2 sin2 1
(b) r 2 sec 2
r 2 cos2 r 2 sin2 1
r cos 2 r sin 2 1 x 2 y 2 1.
O
1
(b) r 2 4 sin 2 8 sin cos 2 r 4 8r 2 sin cos , so x 2 y 2 8x y.
47
48
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
19. (a)
(b) r sin cos r 2 r sin r cos , so x 2 y 2 y x 2 2 x 2 x 14 y 2 y 14 12 x 12 y 12 12 .
(1, ¹2) (1, 0)
O
1
(b) r
20. (a)
(
¹ 2, 2
(4, ¹)
)
2 x 2 y 2 4 x. Squaring both sides gives 4 x 2 y 2 4 x2 16 8x x 2
1 ( 4, 0)
O
4 2r r cos 4, so 2 x 2 y 2 x 4 2 cos
3
3x 2 8x 4y 2 16.
(2, 3¹ 2 )
21. r cos 3, [0 3].
22. r sin 94, [0 8]
1
1
1
1
1
1
1
1
24. r sin 2 2, [0 2]. This graph is bounded.
23. r 1 4 cos 3, [0 6]. 5
2
5
2
2 2
5
CHAPTER 8
25. (a)
26. (a)
Im
Review
49
Im 2 2
4+4i 1 1
(b) 4 4i has r
_10i
Re
16 16 4 2, and
(b) 10i has r 10 and 32 . (c) 10i 10 cos 32 i sin 32
tan1 44 4 (in quadrant I). (c) 4 4i 4 2 cos 4 i sin 4
27. (a)
Re
28. (a)
Im
Im 1+Ï3 i 1
5+3i 1 1
1
Re
(b) 5 3i. Then r 25 9 34, and
tan1 53 . (c) 5 3i 34 cos tan1 53 i sin tan1 53
29. (a)
Im
_1+i
(b) 1 i has r
30. (a)
Im
5
1
_1
Re
(b) 1 3i has r 1 3 2, and tan1 3 3 (since is in quadrant I). (c) 1 3i 2 cos 3 i sin 3
_20
5
Re
Re
1 with 1 1 2 and tan 1
in quadrant II 34 . (c) 1 i 2 cos 34 i sin 34
(b) 20 has r 20 and . (c) 20 20 cos i sin
1 3 2 and tan 1 3 3 with in quadrant III 53 . Therefore, 1 3i 2 cos 53 i sin 53 , and so 4 1 3i 24 cos 203 i sin 203 16 cos 23 i sin 23 16 12 i 23 8 1 i 3 . 32. 1 i has r 11 2 and 2 cos 4 . Thus, 1 i 4 i sin 4 , and so 8 2 cos 2 i sin 2 16 1 i 0 16. 1 i8 33. 3 i has r 3 1 2 and tan 1 with in quadrant I 6 . Therefore, 3 i 2 cos 6 i sin 6 ,
31. 1
3i has r
3
and so 4 1 cos 2 i sin 2 1 1 i 3 3i 24 cos 46 i sin 46 16 3 3 16 2 2 1 1 i 3 1 1 i 3 32 32
50
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
32 1 3 i cos i sin . Thus 34. 12 23 i has r 14 34 1 and tan1 12 (since is in quadrant I). So 3 2 2 3 3 20 1 3i 120 cos 203 i sin 203 cos 23 i sin 23 12 23 i 12 1 i 3 2 2 35. 16i has r 16 and 32 . Thus, 16i 16 cos 32 i sin 32 and so 3 4k 3 4k i sin for k 0, 1. Thus 16i12 1612 cos 4 4 4 1 i 1 2 2 1 i and the roots are 0 4 cos 34 i sin 34 2 2 1 4 cos 74 i sin 74 4 1 i 1 2 2 1 i. 2
2
36. 4 4 3i has r 16 48 8 and tan1 4 4 3 3 . Therefore, 4 4 3i 8 cos 3 i sin 3 . 13 6k 6k Thus 4 4 3i i sin for k 0, 1, 2. Thus the three roots are 3 8 cos 9 9 7 7 13 13 0 2 cos 9 i sin 9 , 1 2 cos 9 i sin 9 , and 2 2 cos 9 i sin 9 .
2k 2k 37. 1 cos 0 i sin 0. Then 116 1 cos i sin
for k 0, 1, 2, 3, 4, 5. Thus the six roots are 3 1 2 2 1 i 3 , 0 1 cos 0 i sin 0 1, 1 1 cos 3 i sin 3 2 i 2 , 2 1 cos 3 i sin 3 2 2 3 1 cos i sin 1, 4 1 cos 43 i sin 43 12 i 23 , and 5 1 cos 53 i sin 53 12 i 23 .
18 cos 38. i cos 2 i sin 2 . Then i
6
2 2k 8
6
i sin
2 2k 8
cos
4k 16
i sin
4k 16
for
i sin , cos 5 i sin 5 , cos 9 i sin 9 , k 0, 1, 2, 3, 4, 5, 6, 7. Thus the eight roots are 0 cos 16 1 2 16 16 16 16 16
13 17 17 21 21 25 25 3 cos 13 16 i sin 16 , 4 cos 16 i sin 16 , 5 cos 16 i sin 6 , 6 cos 16 i sin 16 , and
29 7 cos 29 16 i sin 16 .
39. (a)
40. (a)
y
y
1 1 x
1 x
1
(b) x 1 t 2 , y 1 t t y 1. Substituting for t gives x 1 y 12 x 2y y 2 in
rectangular coordinates.
(b) x t 2 1, y t 2 1
t 2 y 1.
Substituting for t 2 gives x y 1 1 y x 2 where x 1 and y 1.
CHAPTER 8
41. (a)
42. (a)
y
Review
51
y
1 1
x
1
1
(b) x 1 cos t cos t x 1, and y 1 sin t sin t 1 y. Since cos2 t sin2 t 1, it follows that x 12 1 y2 1
x 12 y 12 1. Since t is restricted by 0t 2 , 1 cos 0 x 1 cos 2
1 x 2, and similarly, 0 y 1. (This is the lower right quarter of the circle.) 43. x cos 2t, y sin 3t
x
1 2 1 (b) x 2 x 2, y 2 . Substituting for t t t 1 gives y 2 x 22 . Since t is restricted by t 1 1 0 t 2, we have , so x 52 and y 12 . t 2 The rectangular coordinate equation is y 2 x 22 , x 52 . 44. x sin t cos 2t, y cos t sin 3t
1
1
1
1 1
1
1 1
45. The coordinates of Q are x cos and y sin . The coordinates of R are x 1 and y tan . Hence, the midpoint P 1 cos sin tan 1 cos sin tan , so parametric equations for the curve are x and y . is 2 2 2 2 46. Q has coordinates a cos a sin and O R a cos Q P P has coordinates 2a cos a sin . Thus, parametric equations are x 2a cos y a sin , 0 2. 47. u 2 3, v 8 1. u 22 32 13, u v 2 8 3 1 6 4,
u v 2 8 3 1 10 2, 2u 2 2 2 3 4 6, and 3u 2v 3 2 2 8 3 3 2 1 22 7. 48. u 2i j, v i 2j. u 22 12 5, u v 2 1 i 1 2 j 3i j, u v 2 1 i 1 2 j i 3j, 2u 4i 2j, and 3u 2v 3 2i j 2 i 2j 4i 7j. 49. The vector with initial point P 0 3 and terminal point Q 3 1 is 3 0 1 3 3 4.
50. If the vector 5i 8j is placed in the plane with its initial point at P 5 6, its terminal point is 5 5 6 8 10 2. 2 2 3 2 51. u 2 2 3 has length 2 2 3 4. Its direction is given by tan 3 with in quadrant II, 2 so tan1 3 120 . 52. v 2i 5j has length 22 52 29. Its direction is given by tan 52 with in quadrant IV, so 2 tan1 52 2918 . 53. u u cos u sin 20 cos 60 20 sin 60 10 10 3 .
54. u u cos u sin 135 cos 125 135 sin 125 135 cos 125 135 sin 125 77 111
52
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
55. (a) The force exerted by the first tugboat can be expressed in component form as u 20 104 cos 40 20 104 sin 40 15321 12856, and that of the second tugboat is 34 104 cos 15 34 104 sin 15 32841 8800. Therefore, the resultant force is w u v 15321 32841 12856 8800 48162 4056. (b) The magnitude of the resultant force is 481622 40562 48,332 lb. Its direction is given by 4056 0084, so tan1 0084 48 or N 852 E. tan 48,162
56. (a) The velocity of the airplane is the sum of its velocity relative to the air, which is u 600 cos 30 600 sin 30 300 3 300 , and the wind velocity v 50 cos 120 50 sin 120 25 25 3 . Thus, its velocity is w 300 3 25 300 25 3 . 2 2 300 3 25 300 25 3 602 mi/h, and its direction is given by (b) The true speed of the airplane is w 300 25 3 tan 0694, so tan1 0694 348 , or N 552 E. 300 3 25 57. u 4 3, v 9 8. u 42 32 5, u u 42 32 25, and u v 4 9 3 8 60. 58. u 5 12, v 10 4. u 52 122 13, u u 52 122 169, and u v 5 10 12 4 2. 59. u 2i 2j, v i j. u 22 22 2 2, u u 22 22 8, and u v 2 1 2 1 0. 60. u 10j, v 5i 3j. u 102 10, u u 102 100, and u v 10 3 30. 61. u v 4 2 3 6 4 3 2 6 0, so the vectors are perpendicular.
62. u v 5 3 2 6 5 2 3 6 8 0, so the vectors are not perpendicular. The angle between them is given by uv 340 8 340 775 . cos , so cos1 85 u v 85 2 52 32 2 62
63. u v 2i j i 3j 2 1 1 3 5, so the vectors are not perpendicular. The angle between them is given by uv 2 5 cos , so cos1 22 45 . u v 2 22 12 12 32
64. u v i j i j 1 1 1 1 0, so the vectors are perpendicular.
3 6 1 1 uv 17 37 65. (a) u 3 1, v 6 1. The component of u along v is . v 37 62 12 uv 17 6 1 102 17 v (b) projv u 37 37 37 v2 17 and u u proj u 3 1 102 17 9 54 . (c) u1 projv u 102 2 v 37 37 37 37 37 37 1 4 2 9 14 97 uv . 66. (a) u i 2j, v 4i 9j. The component of u along v is v 97 42 92 uv 56 126 (b) projv u v 14 97 4i 9j 97 i 97 j v2 126 56 126 153 68 (c) u1 projv u 56 97 i 97 j and u2 u projv u i 2j 97 i 97 j 97 i 97 j. 67. The distance vector is D 1 7 1 1 6 2, so the work done is W F D 2i 9j 6 2 12 18 6 ftlb.
CHAPTER 8
Test
53
68. Here W F D F D cos , where is the angle between F and D. Thus, we have 3800 250 20 cos 3800 405 . cos1 5000
CHAPTER 8 TEST 1. (a) x 8 cos 54 8 22 4 2, y 8 sin 54 8 22 4 2. So the point has rectangular coordinates 4 2 4 2 . (b) P 6 2 3 in rectangular coordinates. So tan 263 and the reference angle is 6 . Since P is in 2 quadrant II, we have 56 . Next, r 2 62 2 3 36 12 48, so r 4 3. Thus, polar coordinates for the point are 4 3 56 or 4 3 116 . (b) r 8 cos
2. (a)
x 2 8x y 2 0
O
x 42 y 2 16
1
r 2 8r cos
x 2 y 2 8x
x 2 8x 16 y 2 16
The curve is a circle.
(9, ¹2)
3.
The curve is a limaçon.
(_3,3¹ 2) (3, ¹)
4. (a)
O
2 (3, 0)
1+Ï3 i 1
1
3i has r 1 3 2 and tan1 3 3 . So, in trigonometric form, 1 3i 2 cos 3 i sin 3 . (c) z 1 3i 2 cos 3 i sin 3 z 9 29 cos 93 i sin 93 512 cos 3 i sin 3
(b) 1
Im
Re
512 1 i 0 512
i sin 7 and z 2 cos 5 i sin 5 . 5. z 1 4 cos 712 2 12 12 12 7 5 7 5 i sin 8 cos i sin 8 and Then z 1 z 2 4 2 cos 12 12 7 5 7 5 3 1 i 3 i. 2 i sin z 1 z 2 42 cos i sin 2 cos 6 6 2 2 12 12
54
CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors
6. 27i has r 27 and 2 , so 27i 27 cos 2 i sin 2 . Thus, 2k 2k 3 2 2 13 i sin 27 cos 27i 3 3 4k 4k i sin 3 cos 6 6
Im wÁ
wü 1 0
for k 0, 1, 2. Thus, the three roots are 3 1 i 3 3 i , 3 i sin 0 3 cos 6 6 2 2 2 1 3 cos 56 i sin 56 3 23 12 i 32 3 i , and 2 3 cos 96 i sin 96 3i.
7. (a) x 3 sin t 3, y 2 cos t, 0 t . From the
work of part (b), we see that this is the halfellipse y
(b) x 3 sin t 3 x 3 3 sin t
y 2 cos t
x 3 sin t. 3
x 32 sin2 t. Similarly, 9
y cos t, and squaring both sides gives 2
y2 cos2 t. Since sin2 t cos2 t 1, it follows that 4
1 x
1
Re
wª
Squaring both sides gives
shown.
1
y2 x 32 1. Since 0 t , sin t 0, so 9 4 3 sin t 0 3 sin t 3 3, and so x 3. Thus the curve consists of only the right half of the ellipse.
8. We start at the point 3 5. Because the line has slope 2, for every 1 unit we move to the right, we must move up 2 units. Therefore, parametric equations are x 3 t, y 5 2t.
9. (a) x 3 sin 2t, y 3 cos 2t. The radius is 3 and the position at time t 0 is 3 sin 2 0 3 cos 2 0 0 3. Initially x is increasing and y is decreasing, so the motion is clockwise. At time t the object is back at 0 3, so it takes units of time to complete one revolution. (b) If the speed is doubled, the time to complete one revolution is halved, to 2 . Parametric equations modeling this motion are x 3 sin 4t, y 3 cos 4t. (c) x 2 y 2 3 sin 2t2 3 cos 2t2 9 sin2 2t cos2 2t 9, so an equation in rectangular coordinates is x 2 y 2 9.
(d) In polar coordinates, an equation is r 3. 10. (a)
y
(b) u 3 3 i [9 1] j 6i 10j (c) u 62 102 2 34
(_3, 9)
u
1 1
(3, _1)
x
11. (a) u 3v 1 3 3 6 2 1 3 6 3 3 2 19 3
CHAPTER 8
Test
55
(b) u v 1 3 6 2 5 5 52 52 5 2
(c) u v 1 3 6 2 1 6 3 2 0
(d) Because u v 0, u and v are perpendicular. y
12. (a)
(b) The length of u is u
(_4Ï3, 4)
4 3 with in quadrant II, so 3 4 3 180 tan1 33 150 .
tan
u
1 1
x
2 4 3 42 8. Its direction is given by
13. (a) The river’s current can be represented by the vector u 8 0 and the motorboat’s velocity relative to the water by the vector v 12 cos 60 12 sin 60 6 6 3 . Thus, the true velocity is w u v 14 6 3 . 2 6 13 0742, so (b) The true speed is w 142 6 3 174 mi/h. The direction is given by tan 14 tan1 0742 366 or N 534 E. uv 338 3 5 2 1 338 cos1 2 45 . , so cos1 26 14. (a) cos 2 u v 26 32 22 52 12 uv 26 13 (b) The component of u along v is . v 2 26 uv 5 1 v 13 (c) projv u 26 5i j 2 i 2 j v2 15. The work is W F d 3i 5j 7 2 i 13 2 j 3i 5j 5i 15j 3 5 5 15 90 ftlb.
56
FOCUS ON MODELING
FOCUS ON MODELING The Path of a Projectile x . Substituting this value for t into the equation for y, we get 0 cos 2 x g x x 2 . This shows y 0 sin t 12 gt 2 y 0 sin 12 g y tan x 2 0 cos 0 cos 2 0 cos2
1. From x 0 cos t, we get t
that y is a quadratic function of x, so its graph is a parabola as long as 90 . When 90 , the path of the projectile is a straight line up (then down). 2. (a) Applying the given values, we get x 0 cos t 15t and y 4 0 sin t 12 gt 2 4 15 3t 16t 2 as the parametric equations for the path of the baseball.
y 10
0
10
x
20
(b) The baseball will hit the ground when y 0 4 2598t 16t 2 . Using the Quadratic Formula, 2598 25982 4164 t 177 seconds (since t must be positive). So the baseball travels 32 x 15 177 265 ft (horizontally) before hitting the ground after 177 s.
3. (a) We find the time at which the projectile hits the ground using the equation t 2 0 gsin , where 0 1000 and g 16. Since the rocket is fired 5 from the vertical axis and is measured from the horizontal, we have
sin 85 6226, so the rocket is in the air for 6226 seconds. 90 5 85 . Then t 200032
(b) Substituting the given values into y 0 sin t 12 gt 2 , we get
(d)
y 1000 sin 85 t 16t 2 996t 16t 2 . Then
y 15,000
y f t at 2 bt c, where a 16, b 996, and c 0. So y is a
10,000
b 996 31125, and the maximum value is t 2a 216
5000
quadratic function whose maximum value is attained at
f 31125 996 31125 16 311252 15,500 (see Section 3.1 for a
guide to finding the maximum value of a quadratic function). Thus, the
0
2000
4000
6000 x
rocket reaches a maximum height of 15,500 feet. (c) We use the equation x 0 cos t to find the horizontal distance traveled after t seconds. Since the rocket hits the ground after
6226 seconds, substituting into the equation for horizontal distance gives x 1000 cos 85 6226 5426. Thus, the rocket travels a horizontal
distance of 5426 feet.
2 0 sin , so the projectile travels g
y (m)
2 2 sin x 0 cos 0 0 sin 2 meters. Substituting 0 330 ms,
3000
4. (a) The projectile hits the ground when t g
g
3302 sin 2 sin 2 08999. x 10 km, and g 98 m/s2 , we get 10000
98 So 2 6415 3208 or 2 11585 5793 .
4000
2000 1000 0
4000
8000
x (m)
The Path of a Projectile
57
(b) The projectile fired at 3208 will hit the target in 357 seconds, while the projectile fired at 5793 reaches the target in 571 seconds. 5. We use the equation of the parabola from Exercise 1 and find its vertex: 2 02 sin cos x g g 2 2 x y 2 x y tan x 2 g 2 0 cos2 2 0 cos2 2 2 sin cos x 2 sin cos 2 02 sin cos 2 g g 0 0 2 y 2 2 x g g g 2 0 cos2 2 0 cos2 2 02 sin cos 02 sin cos 02 sin2 02 sin2 g y 2 x . Thus the vertex is at , so the maximum g 2g g 2g 2 cos2 0
02 sin2
. 2g 6. Since the horizontal component of the projectile’s velocity has been reduced by , the parametric equations become height is
x 0 cos t, y 0 sin t 12 gt 2 . 7. In Exercise 6 we derived the equations x 0 cos t,
y
y 0 sin t 12 gt 2 . We plot the graphs for the given values of
10
0 , , and in the figure to the right. The projectile will be blown
8
¬=60¡
backwards if the horizontal component of its velocity is less than the
6
speed of the wind, that is, 32 cos 24 cos 34 414 .
4 ¬=45¡
¬=40¡ ¬=30¡
2
_14 _12 _10 _8 _6 _4 _2 0
The optimal firing angle appears to be between 15 and 30 . We
y
graph the trajectory for 20 , 23 , and 25 . The solution
3
2
4 x
¬=15¡ ¬=5¡
¬=25¡
appears to be close to 23 .
¬=23¡
2
¬=20¡ 1
0
8. (a) Here 0 200 and 60 , so parametric equations are x 0 cos t 200 cos 60 t 100t,
y 0 sin t 12 gt 2 200 sin 60 t 12 98 t 2 100 3t 49t 2 . (b) Here vt 0 cos i 0 sin gt j 100i 100 3 98t j. At its highest point, the vertical component is 0, so the velocity is 100i. It lands when y 100 3t 49t 2 0 t 353 s, at which point its 2 v speed is 353 1002 100 3 98 353 200 ms. Notice
that the speed of the rocket when it hits the ground is the same as its initial speed.
(c)
1
2
3
y 1200 800
x
vÁö
1400 1000
4
vÁü v§ v£ü
600 400 200 0
1000
2000
3000
x
CORRECTIONS: p. 8,9,25,44,48,49,56
CHAPTER 9
SYSTEMS OF EQUATIONS AND INEQUALITIES
9.1
Systems of Linear Equations in Two Variables 1
9.2
Systems of Linear Equations in Several Variables 9
9.3 9.4
Partial Fractions 16 Systems of Nonlinear Equations 27
9.5
Systems of Inequalities 35 Chapter 9 Review 47 Chapter 9 Test 55
¥
FOCUS ON MODELING: Linear Programming 58
1
9
SYSTEMS OF EQUATIONS AND INEQUALITIES
9.1
SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
1. The given system is a system of two equations in the two variables x and y. To check if 5 1 is a solution of this system, we check if x 5 and y 1 satisfy each equation in the system. The only solution of the given system is 2 1. 2. A system of equations in two variables can be solved by the substitution method, the elimination method, or the graphical method. 2x 3y 7 Using the substitution method with we solve the second equation for y in terms of x: 5x y 9 5x y 9 5x y 9 y 5x 9. Substituting this into the first equation gives 2x 3 5x 9 7 2x 15x 27 7 17x 34 x 2. Substituting this into the first equation then gives 2 2 3y 7 3y 3 y 1. Using the elimination method, we multiply the second equation by 3 to get 15x 3y 27, then add this to the first equation:
4
2x 3y 15x 3y 7 27 17x 34 x 2. Substituting, we find that y 1, as above.
Using the graphical method, we graph 2x 3y 7 y 73 23 x and
5x y 9 y 5x 9 on the same screen and see that the lines intersect at
2 -4
-2
-2
2
4
-4
x y 2 1.
3. A system of two linear equations in two variables can have one solution, no solution, or infinitely many solutions. 4. For the given system, the graph of the first equation is the same as the graph of the second equation, so the system has x t where t is any real number. Some of the infinitely many solutions. We express these solutions by writing y 1t solutions of this system are 1 0, 3 4, and 5 4.
5.
x y 2
2x 4y 16
Solving the first equation for x, we get x 2 y, and substituting this into the second equation gives
2 2 y 4y 16 4 2y 4y 16 6y 12 y 2. Substituting for y we get x 2 2 4. Thus, the solution is 4 2. 6.
x y 8
5x 4y 35
Solving the first equation for y, we get y x 8, and substituting this into the second equation gives
5x 4 x 8 35 5x 4x 32 35 x 3. Substituting for x we get y 3 8 5. Thus, the solution is 3 5. 7.
x 3y 11
3x 5y 17
Solving the first equation for x, we get x 3y 11, and substituting this into the second equation gives
3 3y 11 5y 17 9y 33 5y 17 4y 16 y 4. Substituting for y we get x 3 4 11 1. Thus, the solution is 1 4. 1
2
CHAPTER 9 Systems of Equations and Inequalities
8.
2x y 7
Solving the first equation for y, we get y 7 2x, and substituting this into the second equation gives
x 2y 2
x 2 7 2x 2 x 14 4x 2 3x 12 x 4. Substituting for x we get y 7 2x 7 2 4 1. Thus, the solution is 4 1. 9.
2x 3y 7
Solving the second equation for x, we get x 5y, and substituting into the first equation gives
x 5y 0
2 5y 3y 7 7y 7 y 1. Thus, x 5 1 5, and the solution is 5 1. 10.
4x y 5
Solving the first equation for y gives y 5 4x, and substituting into the second equation, we get
5x 2y 4
5x 2 5 4x 4 5x 10 8x 4 3x 6 x 2. Then y 5 4 2 3, and the solution is 2 3. 11.
3x 2y 13
6x 5y
28
Multiplying the first equation by 2 gives the system
6x 4y 26
6x 5y
28
Adding, we get y 2,
and substituting into the first equation in the original system gives 3x 2 2 13 3x 9 x 3. The solution is 3 2. 12.
2x 5y 18
Multiplying the first equation by 3 and the second by 2 gives the 3x 4y 19 6x 15y 54 system Subtracting the equations gives 23y 92 y 4. Substituting this value into the first 6x 8y 38
equation in the original system gives 2x 5 4 18 2x 2 x 1. Thus, the solution is 1 4. 13.
2x y 1
x 2y 8
By inspection of the graph, it appears that 2 3 is the solution to the system. We check this in both
equations to verify that it is a solution. 2 2 3 4 3 1 and 2 2 3 2 6 8. Since both equations are satisfied, the solution is 2 3. 14.
xy2
2x y 5
By inspection of the graph, it appears that 3 1 is the solution to the system. We check this in both
equations to verify that it is a solution. 3 1 2 and 2 3 1 6 1 5. Since both equations are satisfied, the solution is 3 1. 15.
xy4
16.
2x y 2
2x y 4
3x y 6
The solution is x 2, y 0.
The solution is x 2, y 2. y
y
1 1
x
1 1
x
SECTION 9.1 Systems of Linear Equations in Two Variables
17.
2x 3y 12
18.
x 32 y 4
2x 6y 0
3x 9y 18
The lines are parallel, so there is no intersection and hence
The lines are parallel, so there is no intersection and hence
no solution.
no solution. y
y
1 1
19.
x 1 y 5 2
2x
12x 15y 18
2x 52 y 3
y
1
5x 3y 12
x
There are infinitely many solutions. The lines are the same.
1
5x 3y 18
1
20.
y 10 y
1
x
There are infinitely many solutions.
21.
3
x
1 1
x
Adding the two equations gives 10x 30 x 3. Substituting for x in the first equation gives
5 3 3y 18 3y 3 y 1. Hence, the solution is 3 1. 2x y 10 22. Solving the second equation for x gives x 3y 16, and substituting into the first equation, we x 3y 16
have 2 3y 16 y 10 6y 32 y 10 7y 42 y 6. Substituting, we have x 3 6 16 2, and so the solution is 2 6. 2x 3y 9 23. Adding the two equations gives 6x 18 x 3. Substituting for x in the second equation gives 4x 3y 9 4 3 3y 9 12 3y 9 3y 3 x 1. Hence, the solution is 3 1. 3x 2y 0 24. Adding the two equations gives 2x 8 x 4. Substituting for x in the second equation gives x 2y 8
3 4 2y 0 12 2y 0 y 6. Hence, the solution is 4 6. x 3y 5 25. Solving the first equation for x gives x 3y 5. Substituting for x in the second equation gives 2x y 3
2 3y 5 y 3 6y 10 y 3 7y 7 y 1. Then x 3 1 5 2. Hence, the solution is 2 1. x y 7 Adding 3 times the first equation to the second equation gives 5x 20 x 4. So 26. 2x 3y 1 4 y 7 y 3, and the solution is 4 3.
4
CHAPTER 9 Systems of Equations and Inequalities
27. x y 2 y x 2. Substituting for y into 4x 3y 3 gives 4x 3 x 2 3 4x 3x 6 3 x 3, and so y 3 2 5. Hence, the solution is 3 5. 28. 9x y 6 y 9x 6. Substituting for y into 4x 3y 28 gives 4x 3 9x 6 28 23x 46 x 2, and so y 9 2 6 12. Thus, the solution is 2 12. 29. x 2y 7 x 72y. Substituting for x into 5x y 2 gives 5 7 2y y 2 3510y y 2 11y 33 y 3, and so x 7 2 3 1. Hence, the solution is 1 3. 30. 4x 12y 0 x 3y. Substituting for x into 12x 4y 160 gives 12 3y 4y 160 40y 160 y 4, and so x 3 4 12. Therefore, the solution is 12 4. 31. 13 x 16 y 1 2x y 6 y 2x 6. Substituting for y into 23 x 16 y 3 gives 23 x 16 2x 6 3
4x 2x 6 18 2x 12 x 6, and so y 2 6 6 6. Hence, the solution is 6 6. 32. 34 x 12 y 5 y 10 32 x. Substituting for y into 14 x 32 y 1 gives 14 x 32 10 32 x 1 x 60 9x 4 8x 64 x 8, and so y 10 32 8 2. Hence, the solution is 8 2.
33. 12 x 13 y 2 x 23 y 4 x 4 23 y. Substituting for x into 15 x 23 y 8 gives 15 4 23 y 23 y 8
4 2 y 10 y 8 12 2y 10y 120 y 9, and so x 4 2 9 10. Hence, the solution is 10 9. 5 15 15 3
34. 02x 02y 18 x y9. Substituting for x into 03x 05y 33 gives 03 y 905y 33 02y 06 y 3, and so x 3 9 6. Hence, the solution is 6 3. 3x 2y 8 3x 2y 8 Multiplying the second equation by 3 gives the system Subtracting the second 35. x 2y 0 3x 6y 0
equation from the first gives 8y 8 y 1. Substituting into the first equation we get 3x 2 1 8 3x 6 x 2. Thus, the solution is 2 1. 4x 2y 16 36. Adding the first equation to 4 times the second equation gives 22y 264 y 12, so x 5y 70
4x 2 12 16 x 10, and the solution is 10 12. x 4y 8 37. Adding 3 times the first equation to the second equation gives 0 22, which is never true. Thus, 3x 12y 2 the system has no solution. 3x 5y 2 Adding 3 times the first equation to the second equation gives 0 12, which is false. Therefore, 38. 9x 15y 6
there is no solution to this system. 2x 6y 10 39. Adding 3 times the first equation to 2 times the second equation gives 0 0. Writing the equation 3x 9y 15 in slope-intercept form, we have 2x 6y 10 6y 2x 10 y 13 x 53 , so the solutions are all pairs of the form t 13 t 53 where t is a real number. 2x 3y 8 Adding 7 times the first equation to 1 times the second equation gives 0 59, which is false. 40. 14x 21y 3 Therefore, there is no solution to this system. 6x 4y 12 41. Adding 3 times the first equation to 2 times the second equation gives 0 0. Writing the equation in 9x 6y 18
slope-intercept form, we have 6x 4y 12 4y 6x 12 y 32 x 3, so the solutions are all pairs of the form t 32 t 3 where t is a real number.
SECTION 9.1 Systems of Linear Equations in Two Variables
42.
25x 75y 100
5
1 times the first equation to 1 times the second equation gives 0 0, which is always Adding 25 10
10x 30y 40
true. We now put the equation in slope-intercept form. We have x 3y 4 3y x 4 y 13 x 43 , so the solutions are all pairs of the form t 13 t 43 where t is a real number. 43.
8s 3t 3
Adding 2 times the first equation to 3 times the second equation gives s 3, so
5s 2t 1
8 3 3t 3 24 3t 3 t 7. Thus, the solution is 3 7. 44.
u 30 5
3u 80
Adding 3 times the first equation to the second equation gives 10 10 1, so
5
u 30 1 5 u 25. Thus, the solution is u 25 1. 1x 3y 3 2 5 45. 5 x 2y 10 3
Adding 10 times the first equation to 3 times the second equation gives 0 0. Writing the equation
in slope-intercept form, we have 12 x 35 y 3 35 y 12 x 3 y 56 x 5, so the solutions are all pairs of the form t 56 t 5 where t is a real number.
46.
3x 1y
1 3 2 2x 1 y 1 2 2 2
Adding 6 times the first equation to 4 times the second equation gives x 5 x 5. So
9 5 2y 3 y 21. Thus, the solution is 5 21. 47.
04x 12y 14 12x
Adding 30 times the first equation to 1 times the second equation gives 41y 410 y 10, so
5y 10
12x 5 10 10 12x 60 x 5. Thus, the solution is 5 10. 48.
26x 10y 4
06x 12y
Adding 3 times the first equation to 25 times the second equation gives 63x 63 x 1, so
3
26 1 10y 4 10y 30 y 3. Thus, the solution is 1 3. 1x 1y 2 3 4 49. 8x 6y 10
Adding 24 times the first equation to the second equation gives 0 58, which is never true. Thus,
the system has no solution.
1 x 1y 4 10 2 50. 2x 10y 80
Adding 20 times the first equation to the second equation gives 0 0, which is always true. We
now put the equation in slope-intercept form. We have 2x 10y 80 10y 2x 80 y 15 x 8, so the solutions are all pairs of the form t 15 t 8 where t is a real number.
6
CHAPTER 9 Systems of Equations and Inequalities
51.
021x 317y 951
52.
235x 117y 589
The solution is approximately 387 274.
1872x 1491y 1233
621x 1292y 1782
The solution is approximately 071 172.
5
-5
5
5
-5
5
-5
53.
-5
2371x 6552y 13,591
54.
9815x 992y 618,555
The solution is approximately 6100 2000.
435x 912y
0
132x 455y 994
The solution is approximately 285 136. 5
20 -5 0
50
5 -5
1 , a 1. So 55. Subtracting the first equation from the second, we get ay y 1 y a 1 1 y a1 1 1 1 1 1 x 0x . Thus, the solution is . a1 1a a1 a1 a1 a a , a b. So x 1 56. Subtracting the first equation from a times the second, we get a b y a y ab ab b a b . Hence, the solution is . x ba ba ab ab 1 57. Subtracting b times the first equation from a times the second, we get a 2 b2 y a b y 2 , 2 ab a b a 1 1 b 1 1 ax x . Thus, the solution is . a 2 b2 0. So ax ab ab ab ab ab
1 b 58. Subtracting a times the first equation from the second, we get b b a y 1 y . So ax 0 b b a b b a 1 1 1 1 1 . Hence, the solution is 2 . x 2 a b a a b a b b a a ab b ab x y 34 Adding these two equations gives 2x 44 x 22. So 59. Let the two numbers be x and y. Then x y 10 22 y 34 y 12. Therefore, the two numbers are 22 and 12. x 3y 0 x y 2 x y 60. Let x be the larger number and y be the other number. This gives x 2y 6 x 6 2y two equations gives y 6, so x 6 2 6 18. Therefore, the two numbers are 18 and 6.
Adding these
SECTION 9.1 Systems of Linear Equations in Two Variables
61. Let d be the number of dimes and q be the number of quarters. This gives
d
q 14
7
Subtracting the
010d 025q 275
first equation from 10 times the second gives 15q 135 q 9. So d 9 14 d 5. Thus, the number of dimes is 5 and the number of quarters is 9. 62. Let c be the number of children and a be the number of adults. This gives
c
a 2200
150c 400a 5050
Subtracting 3 times
the first equation from 2 times the second gives 5a 3500 a 700, so c 700 2200 c 1500. Therefore, the number of children admitted was 1500 and the number of adults was 700. r p 185 63. Let r be the amount of regular gas sold and p the amount of premium gas sold. Then 360r 400 p 690
Subtracting the second equation from 4 times the first equation gives 4r 360r 4 185 690 04r 50 r 125. Substituting this value of r into the original first equation gives 125 p 185 p 60. Thus, 125 gallons of regular gas and 60 gallons of premium were sold.
64. Let s be the number of boxes of regular strawberries sold and d the number of deluxe strawberries sold. Then s d 135 Subtracting 7 times the first equation from the second equation gives 10d 7d 1110 7 135 7s 10d 1110
3d 165 d 55. Substituting this value of d into the original first equation gives s 55 135 s 80. Thus, 80 boxes of standard strawberries and 55 boxes of deluxe strawberries were sold. 2x 2y 180 65. Let x be the speed of the plane in still air and y be the speed of the wind. This gives 12x 12y 180 Subtracting 6 times the first equation from 10 times the second gives 24x 2880 x 120, so 2 120 2y 180 2y 60 y 30. Therefore, the speed of the plane is 120 mi/h and the wind speed is 30 mi/h.
66. Let x be speed of the boat in still water and y be speed of the river flow.
Downriver:
x
y 20
Upriver: 25x 25y 20
Adding 5 times the first equation to 2 times the second gives 10x 140 x 14, so 14 y 20 y 6. Therefore, the boat’s speed relative to the water is 14 mi/h and the current in the river flows at 6 mi/h. 67. Let a and b be the number of grams of food A and food B. Then
012a 020b 100a
32
50b 22,000
Subtracting 250 times the first equation from the second, we get 70a 14,000 a 200, so 012 200 020b 32 020b 8 b 40. Thus, 200 grams of food A and 40 grams of food B is fed to the rats each day. 68. Let x be the number of pounds of Kenyan coffee and y be the number of pounds of Sri Lankan coffee. This gives 700x 1120y 2310 Adding the first equation and 7 times the second gives 420y 210 y 05, so x y 3 x 05 3 x 25. Thus, 25 pounds of Kenyan coffee and 05 pounds of Sri Lankan coffee should be mixed.
69. Let x and y be the sulfuric acid concentrations in the first and second containers. 300x 600y 900 015 Subtracting the first equation from 3 times the second gives 900y 90 y 010, so 100x 500y 600 0125 100x 500 010 75 x 025. Thus, the concentrations of sulfuric acid are 25% in the first container and 10% in the second.
8
CHAPTER 9 Systems of Equations and Inequalities
70. Let x and y be the number of milliliters of the two brine solutions. Quantity: x y 1000 Subtracting the first equation from 20 times the second gives 3y 1800 Concentrations: 005x 020y 014 y 600, so x 600 1000 x 400. Therefore, 400 milliliters of the 5% solution and 600 milliliters of the 20% solution should be mixed. Total invested: x y 20,000 71. Let x be the amount invested at 5% and y the amount invested at 8%. Interest earned: 005x 008y 1180
Subtracting 5 times the first equation from 100 times the second gives 3y 18,000 y 6,000, so x 6,000 20,000 x 14,000. Thus, $14,000 is invested at 5% and $6,000 at 8%. 72. Let x be the amount invested at 6% and y the amount invested at 10%. The ratio of the amounts invested gives x 2y. Then the interest earned is 006x 010y 3520 6x 10y 352,000. Substituting gives 6 2y 10y 352,000 22y 352,000 y 16,000. Then x 2 16,000 32,000. Thus, $32,000 is invested at 6% and $16,000 at 10%. 73. Let x be the length of time the truck travels and y be the length of time the SUV travels. Then (since 15 min 025 hr), y x 025, so x y 025. Multiplying by 40, we get 40x 40y 10. Comparing the distances, we get 40x 40y 10 60x 40y 35, or 60x 40y 35. This gives the system Adding, we get 20x 45 x 225, 60x 40y 35
76
so y 225 025 25. Thus, the truck travels for 2 14 hours and the SUV travels for 2 12 hours.
74. Let x be the weight of the gold in the crown and y the weight of the silver. Writing equations representing the weight and x y 235 volume of the crown, we have Adding the first equation to 193 times the second, we have x y 14 193 105
88 y 193 105 y 235 14 193 105 y 352 y 42. Thus, x 235 42 193, and so the crown contains 193 g
of gold and 42 g of silver.
75. Let x be the tens digit and y be the ones digit of the number.
xy7
Adding 9 times the first
10y x 27 10x y
74
equation to the second gives 18x 36 x 2, so 2 y 7 y 5. Thus, the number is 25. 76. First let us find the intersection point of the two lines. The y-coordinate of the intersection point is the height of the triangle. y 2x 4 We have Adding 2 times the first equation to the second gives 3y 12, so the triangle has height 4. y 4x 20
78
length 5 2 3. Therefore, the area of the triangle is A 12 bh 12 3 4 6. 77. n 5, so nk1 xk 1 2 3 5 7 18, nk1 yk 3 5 6 6 9 29, n k1 xk yk 1 3 2 5 3 6 5 6 7 9 124, and n 2 2 2 2 2 2 k1 xk 1 2 3 5 7 88. Thus we get the system 18a 5b 29 Subtracting 18 times the first equation from 5 times the 88a 18b 124
Furthermore, y 2x 4 intersects the x-axis at x 2, and y 4x 20 intersects the x-axis at x 5. Thus the base has
second, we get 116a 98 a 0845. Then
b 15 [18 0845 29] 2758. So the regression line is y 0845x 2758.
Please give solution to #77, as done in CA-8 solutions to Exercise 5.1.77
Note that no corrections are required for CA-8 or PMFC-8 for the corresponding exercise solutions.
y 10 8 6 4 2 0
1
2 3 4 5 6 7 x
SECTION 9.2 Systems of Linear Equations in Several Variables
9.2
9
SYSTEMS OF LINEAR EQUATIONS IN SEVERAL VARIABLES
1. If we add 2 times the first equation to the second equation, the second equation becomes x 3z 1. 2. To eliminate x from the third equation, we add 3 times the first equation to the third equation. The third equation becomes 4y 5z 4. 3. The equation 6x 3y 12 z 0 is linear. 4. The equation x 2 y 2 z 2 4 is not linear, since it contains squares of variables. x y 3y z 5 is not a linear system, since the first equation contains a product of variables. In fact 5. The system x y 2 5z 0 move to the left, as in the 2x yz 3
both the second and the third equation are not linear. x 2y 3z 10 6. The system 2x 5y is linear. 2 y 2z 4
x 2y z 5 7. yz 2 z 4
statement of the exercise
Substituting z 4 into the second equation gives y 4 2 y 2. Substituting z 4 and
y 2 into the first equation gives x 2 2 4 5 x 3. Thus, the solution is 3 2 4. 3x 3y z 0 Substituting z 3 into the second equation gives y 4 3 10 y 2. Substituting z 3 8. y 4z 10 z 3 and y 2 into the first equation gives 3x 3 2 3 0 x 3. Thus, the solution is 3 2 3. x 2y z 7 9. Solving we get 2z 6 z 3. Substituting z 3 into the second equation gives y 3z 9 2z 6
y 3 3 9 y 0. Substituting z 3 and y 0 into the first equation gives x 2 0 3 7 x 4. Thus, the solution is 4 0 3. x 2y 3z 10 Solving we get 3z 12 z 4. Substituting z 4 into the second equation gives 10. 2y z 2 3z 12
2y 4 2 y 3. Substituting z 4 and y 3 into the first equation gives x 2 3 3 4 10 x 4. Thus, the solution is 4 3 4. 2x y 6z 5 Solving we get 2z 1 z 12 . Substituting z 12 into the second equation gives 11. y 4z 0 2z 1 y 4 12 0 y 2. Substituting z 12 and y 2 into the first equation gives 2x 2 6 12 5 x 5. Thus, the solution is 5 2 12 .
10
12.
CHAPTER 9 Systems of Equations and Inequalities
4x
2y
3z 10
z 6
1z 2
4
Solving we get 12 z 4 z 8 . Substituting z 8 into the second equation gives
2y 8 6 2y 2 y 1. Substituting z 8 into the first equation gives 4x 3 8 10 4x 14 x 72 . Thus, the solution is 72 1 8 .
3x y z 4 Add the third equation to the second equation: y z 1 x 2y z 1 3x y z 4 Or, add the first equation to three times the second equation: 4y 7z 4 x 2y z 1
3x y z 4 13. x y 2z 0 x 2y z 1
3 5x 2y 3z 14. 10x 3y z 20 x 3y z 8
3 5x 2y 3z Add twice the first equation to the second equation: y 5z 14 x 3y z 8 5x 2y 3z 3 Or, add 10 times the third equation to the second equation: 27y 11z 60 x 3y z 8
2x y 3z 5 15. 2x 3y z 13 6x 5y z 7
2x y 3z 5 Add 3 times the first equation to the third equation: 2x 3y z 13 8y 8z 8 5 2x y 3z Or, add 3 times the second equation to the third equation: 2x 3y z 13 14y 4z 32
x 3y 2z 1 Add 2 times the first equation to 3 times the third equation: y z 1 2x z 1 x 3y 2z 1 Or, add 2 times the second equation to the third equation: y z 1 3z 3
x 3y 2z 1 16. y z 1 2y z 1
x 2y z 3 x 2y z 3 17. y z 4 y z 4 x 2y 3z 9 4z 12
Eq. 1 Eq. 3
So z 3 and y 3 4 y 1. Thus, x 2 1 3 3 x 2. So the solution is 2 1 3. x 5y 3z 4 x 5y 3z 4 x 5y 3z 4 18. 3y 5z 11 3y 5z 11 3y 5z 11 x 2y z 11 3y 4z 7 z 4 Eq. 2 Eq. 3 1 Eq. 1 Eq. 3 So 3y 5 4 11 y 3. Then x 5 3 3 4 4 x 1. So the solution is 1 3 4.
SECTION 9.2 Systems of Linear Equations in Several Variables
x 3y 2z 5 x 3y 2z 5 19. 3x 2y z 8 7y 7z 7 y 3z 1 y 3z 1
x 3y 2z 5 Eq. 2 3 Eq. 1 7y 7z 7 28z 0
11
Eq. 2 7 Eq. 3
So z 0 and 7y 0 7 y 1. Then x 3 1 2 0 5 x 2. So the solution is 2 1 0. x 2y 3z 10 x 2y 3z 10 x 2y 3z 10 20. 3y z 7 3y z 7 3y z 7 x y z 3y 4z 17 Eq. 2 1 Eq. 3 5z 10 1 Eq. 1Eq. 3 7 So z 2 and 3y 2 7 y 3. Then x 2 3 3 2 10 x 2. So the solution is 2 3 2. x y z 4 x y z 4 x y z 4 21. 2y 2z 6 x 3y 3z 10 y 3z 5 1 Eq. 1 Eq. 2 Eq. 3 2x y z 3 y 3z 5 2 Eq. 1 1 Eq. 3 2y 2z 6 Eq. 2 x y z 4 y 3z 5 4z 4 2 Eq. 2 Eq. 3
So z 1 and y 3 1 8 y 2. Then x 2 1 4 x 1. So the solution is 1 2 1. x y z 0 x y z 0 x y z 0 22. x 2y 5z 3 y 2z 1 13 Eq. 1 13 Eq. 2 y 2z 1 3x y 6 5z 10 4 Eq. 2 Eq. 3 4y 3z 6 3 Eq. 1 Eq. 3 So z 2 and y 2 2 1 y 3. Then x 3 2 0 x 1. So the solution is 1 3 2. 4z 1 4z 1 4z 1 x x x 23. 2x y 6z 4 y 2z 2 2 Eq. 1 Eq. 2 y 2z 2 2x 3y 2z 8 3y 6z 6 2 Eq. 1 Eq. 3 12z 12 3 Eq. 2 Eq. 3
So z 1 and y 2 1 2 y 0. Then x 4 1 1 x 5. So the solution is 5 0 1. x y 2z 2 x y 2z 2 2 x y 2z 24. y 6z 11 3 Eq. 1 Eq. 3 3x y 5z 8 y 6z 11 4y z 2x y 2z 7 23z 46 4 Eq. 2 Eq. 2 2 2 Eq. 1 Eq. 2
So z 2 and y 6 2 11 y 1. Then x 1 2 2 x 1. So the solution is 1 1 2. 2x 4y z 2 2x 4y z 2 x 2y 3z 4 25. 5z 10 x 2y 3z 4 Eq. 1 2 Eq. 2 14y 5z 4 Eq. 2 Eq. 3 14y 5z 4 3 Eq. 1 2 Eq. 3 3x y z 1 5z 10 Eq. 2 Eq. 3
So z 2 and 14y 5 2 4 y 1. Then x 2 1 3 2 4 x 0. So the solution is 0 1 2. 2x y z 8 2x y z 8 2x y z 8 26. 3y z 2 Eq. 1 2 Eq. 2 x y z 3 3y z 2 2x 4z 18 8z 32 Eq. 2 3 Eq. 3 y 3z 10 Eq. 1 Eq. 3
So z 4 and 3y 4 2 y 2. Then 2x 2 4 8 x 1. So the solution is 1 2 4. Eq. 2 2y 4z 1 2x y 2z 1 27. 2x y 2z 1 2y 4z 1 Eq. 1 4x 2y 0 4z 2 2Eq. 2 Eq. 3 So z 12 and 2y 4 12 1 y 12 . Then 2x 12 2 12 1 x 14 . So the solution is 14 12 12 .
12
CHAPTER 9 Systems of Equations and Inequalities
Eq. 2 2 z 2 6x 2y z 6x 2y 28. y z 1 Eq. 1 6x 2y z 2 y z 1 x y 3z 2 21z 14 4 Eq. 2 Eq. 3 4y 17z 10 Eq. 2 6 Eq. 3 So z 23 and y 23 1 y 13 . Then 6x 2 13 23 2 x 13 . So the solution is 13 13 23 . x 3y z 2 x 3y z 2 29. Since 0 7 is false, this system is inconsistent. 3x 4y 2z 1 13y 5z 5 3 Eq. 1 Eq. 2 2x 6y 2z 3 0 7 2 Eq. 1 Eq. 3 y z 1
2z 0 Eq. 2 2z 0 x 2y 5z 4 x x 30. x 2z 0 x 2y 5z 4 Eq. 1 2y 3z 4 4x 2y 11z 2 4x 2y 11z 2 2y 3z 2 2z 0 x Since 0 6 is false, this system is inconsistent. 2y 3z 4 0 6 Eq. 2 Eq. 3
Eq. 1 Eq. 2
4 Eq. 1 Eq. 3
3 Eq. 2 3 x 2y x 2y 2x 3y z 1 31. y z 5 Eq. 2 2 Eq. 1 x 2y 3 2x 3y z 1 Eq. 1 x 3y z 4 x 3y z 4 yz 1 Eq. 3 Eq. 1 3 x 2y Since 0 4 is false, this system is inconsistent. y z 5 0 4 Eq. 2 Eq. 3 5 x 2y 3z x 2y 3z 5 32. 2x y z 5 5y 5z 5 4x 3y 7z 5 5y 5z 5
5 x 2y 3z 5y 5z 5 Eq. 2 2 Eq. 1 0 10 2 Eq. 2 Eq. 3
Since 0 10 is false, this system is inconsistent. x y z0 x y z 0 x y z 0 33. x 2y 3z 3 y 2z 3 y 2z 3 Eq. 2 Eq. 1 2x 3y 4z 3 00 y 2z 3 2 Eq. 1 Eq. 3
Eq. 3 Eq. 2
Eq. 2 Eq. 3
So z t and y 2t 3 y 2t 3. Then x 2t 3 t 0 x t 3. So the solutions are t 3 2t 3 t, where t is any real number. x 2y z 3 x 2y z 3 x 2y z 3 34. 2x 5y 6z 7 y 4z 1 y 4z 1 Eq. 2 2 Eq. 1 2x 3y 2z 5 0 0 Eq. 2 Eq. 3 y 4z 1 Eq. 3 2 Eq. 1 So z t and we have y 4t 1 y 4t 1. Substituting into the first equation, we have x 2 4t 1 t 3 x 7t 1. So the solutions are 7t 1 4t 1 t, where t is any real number. x 3y 2z 0 x 3y 2z 0 x 3y 2z 0 35. 2x 4z 4 6y 8z 4 6y 8z 4 Eq. 2 2 Eq. 1 4x 6y 4 0 0 Eq. 2 Eq. 3 6y 8z 4 Eq. 3 4 Eq. 1 So z t and 6y 8t 4 6y 8t 4 y 43 t 23 . Then x 3 43 t 23 2t 0 x 2t 2. So the solutions are 2t 2 43 t 23 t , where t is any real number.
SECTION 9.2 Systems of Linear Equations in Several Variables
2x 4y z 3 x 2y 2z 0 36. x 2y 4z 6 2x 4y z 3 x 2y 2z 0 x 2y 4z 6 x 2y 2z 0 3z 3 0 0 2 Eq. 2 Eq. 3
Eq. 3 Eq. 1 Eq. 2
x 2y 2z 0 3z 3 6z 6
2 Eq. 1 Eq. 2
13
Eq. 3 Eq. 1
So z 1 and y t, and substituting into the first equation we have x 2t 2 1 0 x 2t 2. Thus, the solutions are 2t 2 t 1, where t is any real number. x z 2 6 x z 2 6 y 2z y 2z 3 3 37. x 2y z 2 2y 2z 2 8 Eq. 3 Eq. 1 2x y 3z 2 0 y z 6 12 Eq. 4 2 Eq. 1 x z 2 6 x z 2 6 y 2z 3 y 2z 3 1 Eq. 3 2z 2 2 z 1 Eq. 3 2 Eq. 2 2 3z 6 9 Eq. 4 Eq. 2 6 12 3 Eq. 3 2 Eq. 4
So 2 and z 2 1 z 1. Then y 2 1 3 y 1 and x 1 2 2 6 x 1. Thus, the solution is 1 1 1 2. x y z 0 x y z 0 x y z 0 x y 2z 2 0 z 0 Eq. 2 Eq. 1 y z 1 Eq. 4 38. 2x 2y 3z 4 1 z 2 1 Eq. 3 2 Eq. 1 z 0 Eq. 2 2x 3y 4z 5 2 y z 1 Eq. 4 Eq. 3 z 2 1 Eq. 3 x y z 0 yz 1 z 0 1 Eq. 4 Eq. 3 So 1 and z 1 0 z 1. Then y 1 1 1 y 1 and x 1 1 1 0 x 1. So the solution is 1 1 1 1.
39. Let x be the amount invested at 4%, y the amount invested at 5%, and z the amount invested at 6%. We set x y z 100,000 Total money: up a model and get the following equations: Annual income: 004x 005y 006z 0051 100,000 Equal amounts: xy x y z 100,000 x y z 100,000 4x 5y 6z 510,000 y 2z 110,000 Eq. 2 4 Eq. 1 x y 2y z 100,000 0 Eq. 3 Eq. 1 x y z 100,000 y 2z 110,000 3z 120,000 2 Eq. 2 Eq. 3
So z 40,000 and y 2 40,000 110,000 y 30,000. Since x y, x 30,000. the financial planner should invest $30,000 in short-term bonds, $30,000 in intermediate-term bonds, and $40,000 in long-term bonds.
14
CHAPTER 9 Systems of Equations and Inequalities
40. Let x be the amount invested at 3%, y the amount invested at 5 12 %, and z the amount invested at 9%. We x y z 50,000 Total investment: set up a model and get the following equations: Annual income: 003x 0055y 009z 2540 Twice as much: x 2z x y z 50,000 x y z 50,000 6x 11y 18z 508,000 200 Eq. 2 5x 7z 42,000 Eq. 2 11 Eq. 1 x x 2z 0 2z 0 x y z 50,000 5x 7z 42,000 3z 42,000 Eq. 2 5 Eq. 3
So z 14,000 and x 2z 28,000. Since x y z 50,000, we have y 50,000 14,000 28,000 8000. Thus, $28,000 should be invested in the least risky account, $8,000 in the intermediate account, and 14,000 in the highest-yielding account.
41. Let x, y, and z be the number of acres of land planted with corn, wheat, and soybeans. We set up a model and x y z 1200 Total acres: Substituting 2x for y, we get get the following equations: Market demand: 2x y Total cost: 45x 60y 50z 63,750 x 2x z 1200 z 1200 z 1200 3x 3x 2x y 2x y 0 2x y 0 45x 60 2x 50z 63,750 15x 165x 3750 Eq. 3 50 Eq. 1 50z 63,750
So 15x 3,750 x 250 and y 2 250 500. Substituting into the original equation, we have 250 500 z 1200 z 450. Thus the farmer should plant 250 acres of corn, 500 acres of wheat, and 450 acres of soybeans.
42. Let a, b, and c be the number of gallons of Regular, Performance Plus, and Premium gas sold. The information provided a b c 6500 b c 6500 a gives the following system: 300a 320b 330c 20,050 02b 03c 550 Eq. 2 3 Eq. 1 a 3c 0 b 4c 6500 Eq. 1 Eq. 3 b c 6500 a 02b 03c 550 5c 7500 10 Eq. 2 2 Eq. 3
Thus, c 1500, so 02b 03 1500 550 b 500 and a 500 1500 6500 a 4500. The gas station sold 4500 gallons of Regular, 500 gallons of Performance Plus, and 1500 gallons of Premium gas.
43. Let a, b, and c be the number of ounces of Type A, Type B, and Type C pellets used. The requirements for the different 2a 3b c 9 2a 3b c 9 vitamins gives the following system: 3a b 3c 14 Equations 2 7b 3c 1 2 Eq. 2 3 Eq. 1 8a 5b 7c 32 7b 3c 4 Eq. 3 4 Eq. 1 and 3 are inconsistent, so there is no solution.
SECTION 9.2 Systems of Linear Equations in Several Variables
15
44. Let a, b, and c represent the number of servings of toast, cottage cheese, and fruit, respectively. Then 2c 6 2a the given dietary requirements give rise to the following system: a 5b 11 100a 120b 60c 460 2a 2c 6 c 6 2a 10b 2c 16 10b 2c 16 2 Eq. 2 Eq. 1 16c 32 Eq. 3 12 Eq. 2 120b 40c 160 Eq. 3 50 Eq. 1
Thus, c 2, so 10b 2 2 16 10b 20 b 2 and 2a 2 2 6 a 1. The patient should eat one serving of toast and two each of cottage cheese and fruit.
45. Let a, b, and c represent the number of Midnight Mango, Tropical Torrent, and Pineapple Power smoothies sold. The 8a 6b 2c 820 8a 6b 2c 820 given information leads to the system 3a 5b 8c 690 22b 58c 3060 8 Eq. 2 3 Eq. 1 3a 3b 4c 450 2b 4c 240 Eq. 2 Eq. 3 8a 6b 2c 820 22b 58c 3060 14c 420 Eq. 2 11 Eq. 3
Thus, c 30, so 22b 58 30 3060 22b 1320 b 60 and 8a 6 60 2 30 820 a 50. Thus, The Juice Company sold 50 Midnight Mango, 60 Tropical Torrent, and 30 Pineapple Power smoothies on that particular day.
46. Let a, b, and c represent the number of days required at each plant. The given information leads 8a 10b 14c 110 8a 10b 14c 110 to the system 16a 12b 10c 150 8b 18c 70 2 Eq. 1 1 Eq. 2 10a 18b 6c 114 84b 2c 162 5 Eq. 2 8 Eq. 3 8a 10b 14c 110 8b 18c 70 382c 1146 21 Eq. 2 2 Eq. 3
Thus, c 3, so 8b 18 3 70 8b 16 b 2 and 8a 10 2 14 3 110 a 6. Thus, Factory A should be scheduled for 6 days, Factory B for 2 days, and Factory C for 3 days.
47. Let a, b, and c be the number of shares of Stock A, Stock B, and Stock C in the investor’s portfolio. Since the total value remains unchanged, we get the following system: 10a 25b 29c 74,000 10a 25b 29c 74,000 12a 20b 32c 74,000 50b 14c 74,000 6 Eq. 1 5 Eq. 2 16a 15b 32c 74,000 125b 72c 222,000 8 Eq. 1 5 Eq. 3 10a 25b 29c 74,000
50b 14c
74,000
74c 74,000
5 Eq. 2 2 Eq. 3
So c 1,000. Back-substituting we have 50b 14 1000 74,000 50b 60,000 b 1,200 . And finally 10a 25 1200 29 1000 74,000 10a 30,000 29,000 74,000 10a 15,000 a 1,500. Thus the portfolio consists of 1,500 shares of Stock A, 1,200 shares of Stock B, and 1,000 shares of Stock C.
16
CHAPTER 9 Systems of Equations and Inequalities
I1 I2 I3 0 48. 16I1 8I2 4 24I2 16I3 4 16 Eq. 1 Eq. 2 8I2 4I3 5 8I2 4I3 5 I2 I3 0 I1 24I2 16I3 4 28I3 19 Eq. 2 3 Eq. 3 19 2 2 19 11 So I3 19 28 068 and 24I2 16 28 4 I2 7 029. Then I1 7 28 0 I1 28 039.
I 1 I2 I3 0
x x1 y0 y1 z z1 49. (a) We begin by substituting 0 , , and 0 into the left-hand side of the first equation: 2 2 2 x0 x1 y0 y1 z z1 b1 c1 0 12 a1 x0 b1 y0 c1 z 0 a1 x1 b1 y1 c1 z 1 a1 2 2 2 12 d1 d1 d1
Thus the given ordered triple satisfies the first equation. We can show that it satisfies the second and the third in exactly the same way. Thus it is a solution of the system.
(b) We have shown in part (a) that if the system has two different solutions, we can find a third one by averaging the two solutions. But then we can find a fourth and a fifth solution by averaging the new one with each of the previous two. Then we can find four more by repeating this process with these new solutions, and so on. Clearly this process can continue indefinitely, so there are infinitely many solutions.
9.3
PARTIAL FRACTIONS
1. (iii): r x 2. (ii): r x 3. 5. 7. 8.
4 x x 22
B C A x x 2 x 22
Bx C A 2x 8 2 x 1 x 4 x 1 x 2 4
B A 1 x 1 x 2 x 1 x 2 x 2 3x 5
x 22 x 4
B C A x 2 x 22 x 4
Bx C x2 A 2 x 3 x 4 x 3 x 2 4
x A B x 4. 2 x 1 x 4 x 1 x 4 x 3x 4 1 B 1 C D A 6. 4 3 2 3 x x 1 x x3 x x 1 x x
B Cx D 1 1 A 1 2 2 x 1 x 1 x 1 x2 1 x4 1 x 1 x 1 x 1 x 2 1
Ax B Cx D x 3 4x 2 2 2 9. 2 x 1 x2 2 x2 1 x 2
10.
B A Cx D x4 x2 1 Ex F 2 x 2 2 2 2 2 x x 4 x2 4 x x 4
B C x3 x 1 A D Ex F Gx H 2 2 x 2x 5 2 x 2x 5 2x 52 2x 53 x 2 2x 5 x 2x 53 x 2 2x 5 12. Since x 3 1 x 2 1 x 1 x 2 x 1 x 1 x 1 x 12 x 1 x 2 x 1 , we have 11.
B Dx E C 1 A 1 2 . x 1 x 12 x 1 x3 1 x2 1 x x 1 x 12 x 1 x 2 x 1
17
SECTION 9.3 Partial Fractions
13.
B 2 A . Multiplying by x 1 x 1, we get 2 A x 1 B x 1 x 1 x 1 x 1 x 1 AB 0 Adding we get 2A 2 A 1. Now A B 0 B A, so 2 Ax A Bx B. Thus AB 2 B 1. Thus, the required partial fraction decomposition is
14.
B A 2x . Multiplying by x 1 x 1, we get 2x A x 1 B x 1 x 1 x 1 x 1 x 1 AB 2 2x Ax A Bx B. Thus Adding we 2A 2 A 1. Since A B 0 B A, B 1. The AB 0 required partial fraction decomposition is
15.
2 1 1 . x 1 x 1 x 1 x 1
1 2x 1 . x 1 x 1 x 1 x 1
5 B A . Multiplying by x 1 x 4, we get 5 A x 4 B x 1 x 1 x 4 x 1 x 4 AB 0 5 Ax 4A Bx B. Thus Now A B 0 B A, so substituting,we get 4A A 5 4A B 5 5A 5 A 1and B 1. The required partial fraction decomposition is
16.
A B x 6 . x x 3 x x 3
Multiplying by x x 3we get x 6 A x 3 x B AB1 x 6 Ax 3A Bx A B x 3A. Thus Now 3A 6 A 2, and 2 B 1 B 1. 3A 6 The required partial fraction decomposition is
17.
1 1 5 . x 1 x 4 x 1 x 4
x 6 2 1 . x x 3 x x 3
B 12 A 12 . Multiplying by x 3 x 3, we get 12 A x 3 B x 3 x 3 x 3 x 3 x 3 x2 9 A B 0 AB 0 12 Ax 3A Bx 3B. Thus Adding, we get 2A 4 A 2. So 2 B 0 3A 3B 12 AB 4 2 2 12 . B 2. The required partial fraction decomposition is 2 x 3 x 3 x 9
18.
x 12 A B x 12 . Multiplying by x x 4, we get x x 4 x x 4 x 2 4x x 12 A x 4 Bx Ax 4A Bx A B x 4A. Thus we must solve the system 2 3 x 12 . gives 4A 12 A 3, and 3 B 1 B 2. Thus 2 x x 4 x 4x
19.
AB 4A
4 B 4 A . Multiplying by x 2 4, we get x 2 x 2 x 2 x 2 x2 4 A B 0 AB 0 4 A x 2 B x 2 A B x 2A 2B, and so 2A 2B 4 AB 2 1 1 4 . A 1, and B 1. Therefore, 2 x 2 x 2 x 4
1
This
12
Adding we get 2A 2
18
20.
CHAPTER 9 Systems of Equations and Inequalities
2x 1
x2 x 2 and so
B 2x 1 A . Thus, 2x 1 A x 1 B x 2 A B x A 2B, x 2 x 1 x 2 x 1
A B 2
A 2B 1
Adding the two equations, we get 3B 3 B 1. Thus A 1 2 A 1. Therefore,
2x 1 1 1 . x 2 x 1 x2 x 2 21.
x 14 B x 14 A . Hence, x 14 A x 2 B x 4 A B x 2A 4B, x 4 x 2 x 4 x 2 x 2 2x 8 A B 1 2A 2B 2 and so Adding, we get 3A 9 A 3. So 3 B 1 B 2. 2A 4B 14 A 2B 7 2 3 x 14 . Therefore, 2 x 4 x 2 x 2x 8
22.
23.
A B 8x 3 8x 3 . Hence, 8x 3 A 2x 1 Bx 2A B x A, giving x 2x 1 x 2x 1 2x 2 x 2A B 8 2 3 8x 3 . So A 3 A 3, and 2 3 B 8 B 2. Therefore, 2 x 2x 1 2x x A 3 B x A x . Hence, 4x 3 2x 1 4x 3 2x 1 8x 2 10x 3 x A 2x 1 B 4x 3 2A 4B x A 3B, and so Adding, we get 2B 1 B 12 , and A 32 . Therefore,
24.
2A 4B 1
A 3B 0
3 2
2A 4B 1
2A 6B 0
1 x 2 . 4x 3 2x 1 8x 2 10x 3
A B C 7x 3 7x 3 . Hence, x x 3 x 1 x x 3 x 1 x 3 2x 2 3x 7x 3 A x 3 x 1 Bx x 1 C x x 3 A x 2 2x 3 B x 2 x C x 2 3x Thus
A B C x 2 2A B 3C x 3A 0
Coefficients of x 2
2A B 3C 7 3A 3
Coefficients of x
AB C
1 B C 0
2 B 3C 7
Constant terms
So 3A 3 A 1. Substituting, the system reduces to
Adding these two equations, we get 3 4C 7 C 1. Thus 1 B 1 0 B 2.
7x 3 2 1 1 Therefore, 3 . x x 3 x 1 x 2x 2 3x 25.
B C A 9x 2 9x 6 9x 2 9x 6 . Thus, x 2 x 2 2x 1 x 2 x 2 2x 1 2x 3 x 2 8x 4
9x 2 9x 6 A x 2 2x 1 B x 2 2x 1 C x 2 x 2 A 2x 2 3x 2 B 2x 2 5x 2 C x 2 4 2A 2B C x 2 3A 5B x 2A 2B 4C
SECTION 9.3 Partial Fractions
19
Coefficients of x 2 2A 2B C 9 2A 2B C 9 This leads to the system 16B 3C 45 3A 5B 9 Coefficients of x 2A 2B 4C 6 4B 3C 15 Constant terms 9 2A 2B C Hence, 15C 15 C 1; 16B 3 45 B 3; and 2A 6 1 9 A 2. 16B 3C 45 15C 15 Therefore,
26.
9x 2 9x 6
2x 3 x 2 8x 4
3 1 2 . x 2 x 2 2x 1
B C 3x 2 3x 27 A 3x 2 3x 27 . Thus, 2 x 2 2x 3 x 3 x 2 2x 3 x 3 x 2 2x 3x 9
3x 2 3x 27 A 2x 3 x 3 B x 2 x 3 C x 2 2x 3 A 2x 2 3x 9 B x 2 5x 6 C 2x 2 x 6
2A B 2C 3 So 3A 5B C 3 9A 6B 6C 27
2A B 2C x 2 3A 5B C x 9A 6B 6C Coefficients of x 2 2A B 2C 3 2A B 2C 3 7B 4C 3 7B 6C 3 Coefficients of x 21B 6C 27 24C 24 Constant terms
Hence, 24C 24 C 1; then 7B 4 3 B 1; and 2A 1 2 3 A 3. Therefore,
27.
1 1 3 3x 2 3x 27 . 2 x 2 2x 3 x 3 x 2 2x 3x 9
x2 1 B x2 1 C A . Hence, 2 x x 1 x3 x2 x 2 x 1 x
x 2 1 Ax x 1 B x 1 C x 2 A C x 2 A B x B, and so B 1; A 1 0 A 1; and
1 x2 1 1 2 2 . 1 C 1 C 2. Therefore, 3 2 x x 1 x x x 28.
3x 2 5x 13 B C A 3x 2 5x 13 . Thus, 2 2 3x 2 x 2 3x 2 x 4x 4 3x 2 x 2 x 22
3x 2 5x 13 A x 22 B 3x 2 x 2 C 3x 2 A x 2 4x 4 B 3x 2 4x 4 C 3x 2
A 3B x 2 4A 4B 3C x 4A 4B 2C A 3B 3 Coefficients of x 2 3 A 3B This leads to the following system: 4A 4B 3C 5 Coefficients of x 8B 3C 17 4A 4B 2C 13 Constant terms 8B 5C 8 3 A 3B 109 Hence, 8C 9 C 98 ; 8B 3 98 8B 27 8B 3C 17 8 17 B 64 ; and 8C 9 135 109 9 2 64 64 8 327 3 A 135 . Therefore, 3x 5x 13 A 3 109 A . 64 64 64 3x 2 x 2 x 22 3x 2 x 22
20
29.
CHAPTER 9 Systems of Equations and Inequalities
B A . Hence, 2x A 2x 3 B 2Ax 3A B. So 2A 2 2x 3 2x 32 2x 3 1 A 1; and 3 1 B 0 B 3. Therefore, 2 . 2x 3 2x 32 4x 12x 9 2x
4x 2 12x 9
2x
2x 32
B A . Hence, x 4 A 2x 5 B 2Ax 5A B, and so 30. 2 2x 5 2x 5 2x 52 x 4
A 12 and 5 12 B 4 B 32 . Therefore,
31.
x 4
1
3
2A
1
5A B 4
2 2 . 2x 5 2x 52 2x 52
B 4x 2 x 2 C D A 4x 2 x 2 . Hence, 3 2 3 4 3 x x 2 x 2x x x 2 x x 4x 2 x 2 Ax 2 x 2 Bx x 2 C x 2 Dx 3 A D x 3 2A B x 2 2B C x 2C So 2C 2 C 1; 2B 1 1 B 0; 2A 0 4 A 2; and 2 D 0 D 2. Therefore, 1 2 2 4x 2 x 2 . 3 4 3 x x 2 x 2x x
32.
x 3 2x 2 4x 3 B A C D 2 3 4 . Hence, x 3 2x 2 4x 3 Ax 3 Bx 2 C x D. Thus A 1; B 2; 4 x x x x x C 4; and D 3. Therefore,
33.
10x 2 27x 14 x 13 x 2
2 1 4 3 x 3 2x 2 4x 3 2 3 4. x x4 x x x
B C D A . Thus, x 2 x 1 x 12 x 13
10x 2 27x 14 A x 13 B x 2 x 12 C x 2 x 1 D x 2
A x 3 3x 2 3x 1 B x 2 x 2 2x 1 C x 2 x 2 D x 2 A x 3 3x 2 3x 1 B x 3 3x 2 C x 2 x 2 D x 2
A B x 3 3A C x 2 3A 3B C D x A 2B 2C 2D
SECTION 9.3 Partial Fractions
21
which leads to the system A B 0 3A C 10 3A 3B C D 27 A 2B 2C 2D 14
A B 0 2 3B C 10 Coefficients of x 3B 2C D 17 Coefficients of x 3B 5C 7D 15 Constant terms Coefficients of x 3
A B 0 A B 0 3B C 10 3B C 10 3C D 7 3C D 7 3C 8D 2 9D 9
Hence, 9D 9 D 1, 3C 1 7 C 2, 3B 2 10 B 4, and A 4 0 A 4. Therefore, 10x 2 27x 14 x 13 x 2
34.
4 1 4 2 . x 2 x 1 x 12 x 13
2x 2 5x 1 B C 2x 2 5x 1 2x 2 5x 1 D A . Thus, x 1 x 1 x 12 x 4 2x 3 2x 1 x 1 x 3 x 2 x 1 x 13 x 1 x 13 2x 2 5x 1 A x 13 B x 1 x 12 C x 1 x 1 D x 1 A x 3 3x 2 3x 1 B x 1 x 2 2x 1 C x 2 1 D x 1 A x 3 3x 2 3x 1 B x 3 x 2 x 1 C x 2 1 D x 1
A B x 3 3A B C x 2 3A B D x A B C D AB 3A B C
0
A B 0 2B C 2 Coefficients of x 2 2B C D 3 Coefficients of x 2B 3C 4D 2 Constant terms Coefficients of x 3
2 3A B D 5 A B C D 1 A B 0 A B 0 2B C 2 2B C 2 2C D 1 2C D 1 2C 5D 5 6D 6 which leads to the system
Hence, 6D 6 D 1, 2C 1 1 C 0,
2x 2 5x 1 1 1 1 2B 0 2 B 1, and A 1 0 A 1. Therefore, 4 . x 1 x 1 x 13 x 2x 3 2x 1
22
35.
CHAPTER 9 Systems of Equations and Inequalities
3x 3 22x 2 53x 41 x 22 x 32
B D A C . Thus, x 2 x 22 x 3 x 32
3x 3 22x 2 53x 41 A x 2 x 32 B x 32 C x 22 x 3 D x 22 A x 3 8x 2 21x 18 B x 2 6x 9
C x 3 7x 2 16x 12 D x 2 4x 4
A C x 3 8A B 7C D x 2
21A 6B 16C 4D x 18A 9B 12C 4D A C 3 A C 3 Coefficients of x 3 8A B 7C D 22 2 B C D 2 Coefficients of x so we must solve the system 6B 5C 4D 10 21A 6B 16C 4D 53 Coefficients of x 9B 6C 4D 13 18A 9B 12C 4D 41 Constant terms A C 3 A C 3 B C D 2 B C D 2 Hence, D 1, C 2 2 C 0, B 0 1 2 C 2D 2 C 2D 2 3C 5D 5 D 1 B 1, and A 0 3 A 3. Therefore,
36.
3x 3 22x 2 53x 41 x 22 x 32
1 3 1 . x 2 x 22 x 32
B D 3x 2 12x 20 A C 3x 2 12x 20 3x 2 12x 20 . Thus, 2 2 x 22 2 x 2 x 2 2 x 4 8x 2 16 2 2 22 x x x x 4
3x 2 12x 20 A x 2 x 22 B x 22 C x 22 x 2 D x 22 A x 3 2x 2 4x 8 B x 2 4x 4 C x 3 2x 2 4x 8 D x 2 4x 4
A C x 3 2A B 2C D x 2 4A 4B 4C 4D x 8A 4B 8C 4D A C 0 Coefficients of x 3 A C 0 2A B 2C D 2 B 4C D 3 3 Coefficients of x which leads to the system 4A 4B 4C 4D 12 6B 8C 2D 6 Coefficients of x 4B 16C 12D 4 8A 4B 8C 4D 20 Constant terms A C 0 A C 0 B 4C D 3 B 4C D 3 Hence, 48D 48 D 1, 16C 8D 24 16C 8D 24 32C 32D 0 48D 48 16C 8 24 C 1; B 4 1 3 B 2, and A 1 0 A 1. Therefore, 3x 2 12x 20 2 1 1 1 . x 2 x 22 x 2 x 22 x 4 8x 2 16
37.
Bx C x 3 A x 3 2 2 . Hence, x 3 A x 2 3 Bx 2 C x A B x 2 C x 3A. So 3 x x 3x x x 3 x 3 x 3 x 1 1 3A 3 A 1; C 1; and 1 B 0 B 1. Therefore, 3 2 . x x 3x x 3
SECTION 9.3 Partial Fractions
38.
23
3x 2 2x 8 3x 2 2x 8 Ax B C . Thus, 2 x 1 x 2 2 x 1 x 3 x 2 2x 2 x 2 3x 2 2x 8 Ax B x 1 C x 2 2 A C x 2 A B x B 2C, which leads to the system 2 C3 Coefficients of x 2 A C 3 Coefficients of x A A B 2 Coefficients of x B C1 Coefficients of x B 2C 8 Constant terms 2C 6 Constant terms Hence, 2C 6 C 3, B 3 1 B 2; and A 3 3 A 0. Therefore, 3x 2 2x 8 2 3 . 2 x 3 x 2 2x 2 x 2 x 1
Ax B Cx D 2x 3 7x 5 2 . Thus, 39. 2 2 2 x x 2 x 1 x x 2 x 1 2x 3 7x 5 Ax B x 2 1 C x D x 2 x 2
Ax 3 Ax Bx 2 B C x 3 C x 2 2C x Dx 2 Dx 2D
A C x 3 B C D x 2 A 2C D x B 2D
We must solve the system A C 2 B C D0 A 2C D 7 B 2D 5
Coefficients of x 3 Coefficients of x 2 Coefficients of x Constant terms
A C 2 BC D 0 0 C D 5 C D 5 2D 10 C D 5
A C BC D
2
Hence, 2D 10 D 5, C 5 5 C 0, B 0 5 0 B 5, and A 0 2 A 2. Therefore,
40.
2x 5 5 2x 3 7x 5 . x2 x 2 x2 1 x2 x 2 x2 1
x2 x 1 Ax B Cx D x2 x 1 2 . Thus, 2x 4 3x 2 1 2x 2 1 x 1 2x 2 1 x 2 1 x 2 x 1 Ax B x 2 1 C x D 2x 2 1 Ax 3 Ax Bx 2 B 2C x 3 2Dx 2 C x D A 2C x 3 B 2D x 2 A C x B D
which leads to the system A 2C 0 B 2D 1 C 1 A B D1
A 2C 0 2 B 2D 1 Coefficients of x C 1 Coefficients of x D 0 Constant terms Coefficients of x 3
Hence, D 0, C 1, B 0 1 B 1, and A 2 0 A 2. Therefore,
x2 x 1 2x 1 x 2 2 . 4 2 2x 3x 1 2x 1 x 1
24
41.
CHAPTER 9 Systems of Equations and Inequalities
Bx C x4 x3 x2 x 1 A Dx E 2 2 2 . Hence, x 2 x 1 x2 1 x x 1
2 x 4 x 3 x 2 x 1 A x 2 1 Bx C x x 2 1 x Dx E A x 4 2x 2 1 Bx 2 C x x 2 1 Dx 2 E x A x 4 2x 2 1 Bx 4 Bx 2 C x 3 C x Dx 2 E x A B x 4 C x 3 2A B D x 2 C E x A
So A 1, 1 B 1 B 0; C 1; 2 0 D 1 D 1; and 1 E 1 E 2. Therefore, x4 x3 x2 x 1 1 1 x 2 2 2 2 . x x 1 x2 1 x x2 1
42.
Cx D Ax B 2x 2 x 8 2 2 2 . Thus, 2 x 4 x 4 x2 4 2x 2 x 8 Ax B x 2 4 C x D Ax 3 4Ax Bx 2 4B C x D Ax 3 Bx 2 4A C x 4B D
2x 2 x 8 2 1 and so A 0, B 2, 0 C 1 C 1, and 8 D 8 D 0. Therefore, 2 2 2 . x 4 x2 4 x2 4
43. We must first get a proper rational function. Using long division, we find that 2 2x 2 x 5 x 5 2x 4 x 3 x 5 2 2 2x x 5 x 2 A Bx C . Hence, x x x 2 x 3 2x 2 x 2 x 3 2x 2 x 2 x2 1 x 2 x 2 1 2x 2 x 5 A x 2 1 Bx C x 2 Ax 2 A Bx 2 C x 2Bx 2C
A B x 2 C 2B x A 2C
Equating coefficients, we get the system 2 Coefficients of x 2 2 2 A B A B A B 2B C 1 Coefficients of x 2B C 1 2B C 1 A 2C 5 Constant terms B 2C 3 5C 5
Therefore, 5C 5 C 1, 2B 1 1 B 1, and A 1 2 A 3, so x 1 3 x 5 2x 4 x 3 x 5 2 x2 . 3 2 x 2 x 1 x 2x x 2
This is the solution to the wrong problem (it was changed in CA-8 and AT-5). The correct solution is given in CA-8 Exercise #5.3.44. [Note that in PMFC-8 the corresponding problem was not changed and the solution is correct.] 44.
SECTION 9.3 Partial Fractions
25
x 5 3x 4 3x 3 4x 2 4x 12 x 5 3x 4 3x 3 4x 2 4x 12 . We use long division to get a proper rational x 4 4x 3 6x 2 8x 8 x 22 x 2 2
function:
x 4 4x 3 6x 2 8x 8
x
1
x 5 3x 4 3x 3 4x 2 4x 12
x 5 4x 4 6x 3 8x 2 8x
x 4 3x 3 4x 2 4x 12
x 4 4x 3 6x 2 8x 8 x 3 2x 2 4x 4
B x 5 3x 4 3x 3 4x 2 4x 12 Cx D x 3 2x 2 4x 4 A , so 2 x 1 x 1 2 2 2 2 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 2x 2 4x 4 A x 2 x 2 2 B x 2 2 C x D x 22
Thus,
A x 3 2x 2 2x 4 B x 2 2 C x D x 2 4x 4
Ax 3 2Ax 2 2Ax 4A Bx 2 2B C x 3 4C x 2 4C x Dx 2 4Dx 4D
A C x 3 2A B 4C D x 2 2A 4C 4D x 4A 2B 4D A C 1 Coefficients of x 3 2A B 4C D 2 Coefficients of x 2 which leads to the system 2A 4C 4D 4 Coefficients of x 4A 2B 4D 4 Constant terms A C 1 A C 1 B 2C D 0 2 Eq. 1 Eq. 2 B 2C D 0 2C 4D 2 B 3D 2 Eq. 2 Eq. 3 Eq. 2 Eq. 3 2B 8C 4D 12 2 Eq. 3 Eq. 4 8C 2D 8 2 Eq. 3 Eq. 4 A C 1 B 2C D 0 Then D 0; 2C 2 C 1; B 2 0 B 2; and A 1 1 2C 4D 2 18D 0 4 Eq. 3 Eq. 4 A 0. Therefore,
45.
x x 5 3x 4 3x 3 4x 2 4x 12 2 2 . x 1 2 2 2 x 2 x 2 x 2 x 2
A B ax b . Hence, ax b A x 1 B x 1 A B x A B. x 1 x 1 x2 1 AB a ab . So Adding, we get 2A a b A 2 AB b Substituting, we get B a A
ab ab ab ab 2a . Therefore, A and B . 2 2 2 2 2
26
CHAPTER 9 Systems of Equations and Inequalities
ax 3 bx 2 Ax B Cx D 46. 2 2 2 . Hence, 2 x 1 x 1 x2 1 ax 3 bx 2 Ax B x 2 1 C x D Ax 3 Ax Bx 2 B C x D Ax 3 Bx 2 A C x B D
and so A a, B b, a C 0 C a, and b D 0 D b. Therefore, A a, B b, C a, and D b.
1 x is already a partial fraction decomposition. The denominator in the first term is a 47. (a) The expression 2 x 1 x 1 quadratic which cannot be factored and the degree of the numerator is less than 2. The denominator of the second term is linear and the numerator is a constant. x (b) The term can be decomposed further, since the numerator and denominator both have linear factors. x 12 B A x . Hence, x A x 1 B Ax A B. So A 1, B 1, and 2 x 1 x 1 x 12 1 1 x . x 1 x 12 x 12 (c) The expression
1 2 is already a partial fraction decomposition, since each numerator is constant. x 1 x 12
x 2 2 is already a partial fraction decomposition, since the denominator is the square of a quadratic x2 1 which cannot be factored, and the degree of the numerator is less than 2.
(d) The expression
48. Combining the terms, we have 1 x 2 2x 1 2 x2 1 2 1 x 1 1 1 x 1 x 12 x 1 x 12 x 1 x 12 x 1 x 12 x 1
2x 2 2 x 1 x 2 2x 1 x 12 x 1
3x 2 x
x 12 x 1
3x 2 x
A B C , and so x 1 x 12 x 1 x 12 x 1 3x 2 x A x 1 x 1 B x 1 C x 12 A x 2 1 B x 1 C x 2 2x 1
Now to find the partial fraction decomposition, we have
Ax 2 A Bx B C x 2 2C x C A C x 2 B 2C x A B C 2 C 3 A C 3 Coefficients of x A which result in the system B 2C 1 B 2C 1 Coefficients of x A B C 0 B 2C 3 Eq. 1 Eq. 3 Constant terms
A
C
3
4C
4
B 2C 1
so 4C 4 C 1, B 2 1 1 B 1, and A 1 3 A 2. 1 Eq. 2 Eq. 3
Therefore, we get back the same expression:
3x 2 x
x 12 x 1
1 1 2 . x 1 x 12 x 1
SECTION 9.4 Systems of Nonlinear Equations
9.4
27
SYSTEMS OF NONLINEAR EQUATIONS
1. The solutions of the system are the points of intersection of the two graphs, namely 2 2 and 4 8. 2. For 2 2: 2y x 2 2 2 22 0 and y x 2 2 4, so 2 2 is a solution.
For 4 8: 2y x 2 2 8 42 0 and y x 8 4 4, so 4 8 is a solution. y x2 Substituting y x 2 into the second equation gives x 2 x 12 3. y x 12
0 x 2 x 12 x 4 x 3 x 4 or x 3. So since y x 2 , the solutions are 3 9 and 4 16. x 2 y 2 25 Substituting for y in the first equation gives x 2 2x2 25 5x 2 25 x 2 5 x 5 4. y 2x 5 2 5 and When x 5 then y 2 5, and when x 5 then y 2 5 2 5. Thus the solutions are 5 2 5 .
5.
x 2 y2 8 x y0
Solving the second equation for y gives y x, and substituting this into the first equation gives
x 2 x2 8 2x 2 8 x 2. So since y x, the solutions are 2 2 and 2 2. x2 y 9 Solving the first equation for y, we get y 9 x 2 . Substituting this into the second equation gives 6. x y 3 0 x 9 x 2 3 0 x 2 x 6 0 x 3 x 2 0 x 3 or x 2. If x 3, then y 9 32 0,
and if x 2, then y 9 22 5. Thus the solutions are 3 0 and 2 5. x y2 0 7. Solving the first equation for x gives x y 2 , and substituting this into the second equation gives 2x 5y 2 75 2 y 2 5y 2 75 3y 2 75 y 2 25 y 5. So since x y 2 , the solutions are 25 5 and 25 5. 8.
x2 y 1
Solving the first equation for y, we get y x 2 1. Substituting this into the second equation gives 2x 2 3y 17 2x 2 3 x 2 1 17 2x 2 3x 2 3 17 5x 2 20 x 2 4 x 2. If x 2, then y 22 1 3,
and if x 2, then y 22 1 3. Thus the solutions are 2 3 and 2 3. x 2 2y 1 9. Subtracting the first equation from the second equation gives 7y 28 y 4. Substituting y 4 into x 2 5y 29
the first equation of the original system gives x 2 2 4 1 x 2 9 x 3. The solutions are 3 4 and 3 4. 3x 2 4y 17 10. Multiplying the first equation by 2 and the second by 3 gives the system 2x 2 5y 2 6x 2 8y 34 Adding we get 7y 28 y 4. Substituting this value into the second equation gives 6x 2 15y 6 2x 2 5 4 2 2x 2 22 x 2 11 x 11. Thus the solutions are 11 4 and 11 4 .
28
11.
CHAPTER 9 Systems of Equations and Inequalities
3x 2 y 2 11
Multiplying the first equation by 4 gives the system
x 2 4y 2 8
12x 2 4y 2 44 x 2 4y 2 8
Adding the equations
gives 13x 2 52 x 2. Substituting into the first equation we get 3 4 y 2 11 y 1. Thus, the solutions are 2 1, 2 1, 2 1, and 2 1.
12.
2x 2 4y 13
Multiplying the second equation by 2 gives the system
x 2 y 2 72
2x 2 4y 13
2x 2 2y 2 7
Subtracting the
equations gives 4y 2y 2 6 y 2 2y 3 0 y 3 y 1 0 y 3, y 1. If y 3, then 5 2 2x 2 4 3 13 x 2 25 . If y 1, then 2x 2 4 1 13 x 2 92 x 3 2 2 . Hence, the x 2 2 5 2 3 2 solutions are 2 3 and 2 1 . 13.
x y2 3 0
14.
x 2 y2 1
Adding the two equations gives 2x 2 x 1 0. Using the Quadratic Formula we have 2x 2 y 2 4 0 1 1 4 2 1 1 9 1 3 1 3 1 3 x . So x 1 or x 12 . Substituting x 1 2 2 4 4 4 4 into the first equation gives 1 y 2 3 0 y 2 2 y 2. Substituting x 12 into the first equation gives 1 y 2 3 0 y 2 7 y 7 . Thus the solutions are 1 2 and 1 7 . 2 2 2 2 2
2x 2 y 2 x 3
Subtracting the first equation from the second equation gives x 2 x 2
x 2 x 2 0 x 2 x 1 0 x 2, x 1. Solving the first equation for y 2 we have y 2 x 2 1. When x 1, y 2 12 1 0 so y 0 and when x 2, y 2 22 1 3 so y 3. Thus, the solutions are 1 0, 2 3 , and 2 3 . 15.
x2 y
8
x 2y 6
By inspection of the graph, it appears that 2 4 is a solution, but is difficult to get accurate values
for the other point. Multiplying the first equation by 2 gives the system
2x 2 2y 16 x 2y 6
Adding the equations gives
2x 2 x 10 2x 2 x 10 0 2x 5 x 2 0. So x 52 or x 2. If x 52 , then 52 2y 6 2y 72 y 74 , and if x 2, then 2 2y 6 2y 8 y 4. Hence, the solutions are 52 74 and
2 4.
16.
x y 2 4
x y
2
By inspection of the graph, it appears that 0 2 and 5 3 are solutions to the system. We check
each point in both equations to verify that it is a solution. For 0 2: 0 22 4 and 0 2 2. For 5 3: 5 32 5 9 4 and 5 3 2. Thus, the solutions are 0 2 and 5 3.
SECTION 9.4 Systems of Nonlinear Equations
17.
x2
y0
x 3 2x y 0
29
By inspection of the graph, it appears that 2 4, 0 0, and 1 1 are solutions to the system.
We check each point in both equations to verify that it is a solution. For 2 4: 22 4 4 4 0 and 23 2 2 4 8 4 4 0.
For 0 0: 02 0 0 and 03 2 0 0 0.
For 1 1: 12 1 1 1 0 and 13 2 1 1 1 2 1 0. Thus, the solutions are 2 4, 0 0, and 1 1. x 2 y 2 4x 18. By inspection of the graph, it appears that 0 0 is a solution, but is difficult to get accurate values for x y2
the other points. Substituting for y 2 we have x 2 x 4x x 2 3x 0 x x 3 0. So x 0 or x 3. If x 0, then y 2 0 so y 0. And is x 3 then y 2 3 so y 3. Hence, the solutions are 0 0, 3 3 , and 3 3 . y x 2 4x Subtracting the second equation from the first equation gives x 2 4x 4x 16 19. y 4x 16
x 2 8x 16 0 x 42 0 x 4. Substituting this value for x into either of the original equations gives y 0. Therefore, the solution is 4 0. x y2 0 x y2 20. Substituting for y in the Solving the first equation for x and the second equation for y gives 2 yx 0 y x2 first equation gives x x 4 x x 3 1 0 x 0, x 1. Thus, the solutions are 0 0 and 1 1. x 2y 2 21. Now x 2y 2 x 2y 2. Substituting for x gives y 2 x 2 2x 4 y 2 x 2 2x 4
y 2 2y 22 2 2y 2 4 y 2 4y 2 8y 4 4y 4 4 y 2 4y 4 0 y 22 0 y 2. Since x 2y 2, we have x 2 2 2 2. Thus, the solution is 2 2. y 4 x2 22. Setting the two equations equal, we get 4 x 2 x 2 4 2x 2 8 x 2. Therefore, the solutions y x2 4 are 2 0 and 2 0. xy 4 23. Now x y 4 x 4 y. Substituting for x gives x y 12 4 y y 12 y 2 4y 12 0 x y 12 y 6 y 2 0 y 6, y 2. Since x 4 y, the solutions are 2 6 and 6 2. x y 24 24 Since x 0 is not a solution, from the first equation we get y 24. . Substituting into the 2 2 x 2x y 4 0 2 24 2x 4 4x 2 576 x 4 2x 2 288 0 x 2 18 x 2 16 0. second equation, we get 2x 2 4 x
Since x 2 18 cannot be 0 if x is real, we have x 2 16 0 x 4. When x 4, we have y 24 4 6 and when 24 6. Thus the solutions are 4 6 and 4 6. x 4, we have y 4 x 2 y 16 16 16 . Substituting for x 2 gives 4y 16 0 4y 2 16y 16 0 25. Now x 2 y 16 x 2 y y x 2 4y 16 0 y 2 4y 4 0 y 22 0 y 2. Therefore, x 2 has no solution.
16 8, which has no real solution, and so the system 2
30
26.
27.
CHAPTER 9 Systems of Equations and Inequalities
x
x 2 y2 9
y 0 Solving the first equation for x, we get x y. Substituting for x gives y 2 4x 2 12 2 y 2 4 y 12 y 2 4y 12 0 y 6 y 2 0 y 6, y 2. Since x 2 is not a real solution, the only solution is 6 6 . x 2 y2 1
Adding the equations gives 2x 2 10 x 2 5 x 5. Now x 5 y 2 9 5 4
y 2, and so the solutions are
28.
x 2 2y 2 2
5 2 , 5 2 , 5 2 , and 5 2 .
Multiplying the first equation by 2 gives the system
2x 2 3y 15
equations gives 4y 2 3y 11 4y 2 3y 11 0 y
3
2x 2 4y 2 4
2x 2 3y 15
Subtracting the two
9 4 4 11 which is not a real number. 2 4
Therefore, there are no real solutions.
29.
2x 2 8y 3 19
4x 2 16y 3 34
Multiplying the first equation by 2 gives the system
4x 2 16y 3 38 4x 2 16y 3 34
Adding the two
equations gives 8x 2 72 x 3, and then substituting into the first equation we have 2 9 8y 3 19 y 3 18 y 12 . Therefore, the solutions are 3 12 and 3 12 . 30.
x 4 y 3 17
3x 4 5y 3 53
Multiplying the first equation by 3 gives the system
3x 4 3y 3 51 3x 4 5y 3 53
Subtracting the equations
gives 2y 3 2 y 3 1 y 1, and then x 4 1 15 x 2. Therefore, the solutions are 2 1 and 2 1.
2 3 1 x y 31. 4 7 1 x y
1 1 If we let u and ,the system is equivalent to x y
equation by 4 gives the system
4u 6 2
4u 7 1
2u 3 1
4u 7 1
Multiplying the first
Adding the equations gives 3, and then substituting into the first
equation gives 2u 9 1 u 5. Thus, the solution is 15 13 .
4 6 7 2 4 2 x y 32. 2 1 2 4 0 x y
4u 6 7 1 1 2 , and multiplying the If we let u 2 and 4 , the system is equivalent to u 2 0 x y
second equation by 3, gives
4u 6 7 2
3u 6 0
Adding the equations gives 7u 72 u 12 , and 14 . Therefore,
1 1 2 2 , 2 2 , 2 2 , x 2 2 x 2, and y 4 4 y 2. Thus, the solutions are u and 2 2 .
SECTION 9.4 Systems of Nonlinear Equations
33.
y x 2 8x
34.
y 2x 16
The solutions are 8 0 and 2 20.
y x 2 4x
2x y 2
y x 2 4x
y 2x 2
The solutions are approximately 035 130 and 565 930.
20
10
-10
10 -20
35.
x 2 y 2 25
x 3y 2
5
y 25 x 2
36.
y 13 x 23
The solutions are 451 217 and 491 097.
x 2 y 2 17
x 2 2x y 2 13
y 17 x 2 y 13 2x x 2
The solutions are approximately 2 361.
5
5
-5
5
-5
-5
5 -5
2 y2 x y 18 2x 2 1 37. 9 18 y x 2 6x 2 y x 2 6x 2
38.
x 2 y2 3
y x 2 2x 8
y x2 3
y x 2 2x 8
The solutions are approximately 222 140,
The solutions are 123 387 and 035 421.
188 072, 345 299, and 465 431.
5
5
-5
5 -5
5
-5 -5
39.
4 32 x 4 y 2 x 2 2x y 0 y x 2 2x x 4 16y 4 32
The solutions are 230 070 and 048 119.
40.
y e x ex y 5 x2
119 359 and 119 359. 5
2
-2
The solution are approximately
2 -2
-2
2
31
32
CHAPTER 9 Systems of Equations and Inequalities
log x log y 3 2 41. 2 log x log y 0
Adding the two equations gives 3 log x 32 log x 12 x
10. Substituting into the
second equation we get 2 log 1012 log y 0 log 10 log y 0 log y 1 y 10. Thus, the solution is 10 10 .
42.
2x 2 y 10
4x 4 y 68
2x 2 y 10
22x 22y 68
If we let u 2x and 2 y , the system becomes
u 10
u 2 2 68
Solving the first equation for u, and substituting this into the second equation gives u 10 u 10 , so 10 2 2 68 100 20 2 2 68 2 10 16 0 8 2 0 2 or 8. If 2, then u 8, and so y 1 and x 3. If 8, then u 2, and so y 3 and x 1. Thus, the solutions are 1 3 and 3 1. xy 3 43. Solving the first equation for x gives x 3 y and using the hint, x 3 y 3 387 3 x y 3 387 x y x 2 x y y 2 387. Next, substituting for x, we get 3 3 y2 y 3 y y 2 387 9 6y y 2 3y y 2 y 2 129 3y 2 9y 9 129 y 8 y 5 0 y 8 or y 5. If y 8, then
x 3 8 5, and if y 5, then x 3 5 8. Thus the solutions are 5 8 and 8 5. x2 xy 1 44. Adding the equations gives x 2 x yx y y 2 4 x 2 2x y y 2 4 x y2 4 x y 2. x y y2 3 If x y 2, then from the first equation we get x x y 1 x 2 1 x 12 , and so y 2 12 32 . If x y 2, then from the first equation we get x x y 1 x 2 1 x 12 , and so y 2 12 32 . Thus the solutions are 12 32 and 12 32 . 45. Let and l be the lengths of the sides, in cm. Then we have the system
l 180
2l 2 54
We solve the second equation
for giving, 27 l, and substitute into the first equation to get l 27 l 180 l 2 27l 180 0
l 15 l 12 0 l 15 or l 12. If l 15, then 27 15 12, and if l 12, then 27 12 15. Therefore, the dimensions of the rectangle are 12 cm by 15 cm.
46. Let b be the length of the base of the triangle, in feet, and h be the height of the triangle, 1 bh 84 168 2 . By substitution, The first equation gives b in feet. Then b2 h 2 252 625 h
168 2 h 2 625 h 4 625h 2 1682 0 h 2 49 h 2 576 0 h 7 or h 24. Thus, the lengths of h
the other two sides are 7 ft and 24 ft.
47. Let l and be the length and width, respectively, of the rectangle. Then, the system of equations is 2l 2 70 Solving the first equation for l, we have l 35 , and substituting into the second gives l 2 2 25 l 2 2 25 l 2 2 625 35 2 2 625 1225 70 2 2 625 22 70 600 0 15 20 0 15 or 20. So the dimensions of the rectangle are 15 and 20.
SECTION 9.4 Systems of Nonlinear Equations
33
48. Let be the width and l be the length of the rectangle, in inches. From the figure, the diagonals of the rectangle are l 160 1602 160 simply diameters of the circle. Then, . By substitution, 2 l 2 400 2 2 2 l l l 20 400 l 4 400l 2 1602 0 l 2 80 l 2 320 0 l 80 4 5 or l 320 8 5. Therefore, the dimensions of the rectangle are 4 5 in. and 8 5 in..
49. At the points where the rocket path and the hillside meet, we have
y 1x 2
y x 2 401x
Substituting for y in the second
801 0 x 0, x 801 . When x 0, the rocket has x 0 x x equation gives 12 x x 2 401x x 2 801 2 2 2 801 1 801 801 801 not left the pad. When x 2 , then y 2 2 4 . So the rocket lands at the point 801 2 4 . The distance from
the base of the hill is
801 2 2
801 2 44777 meters. 4
50. Let x be the circumference and y be length of the stove pipe. Using the circumference we can determine x 2 1 2 x . Thus the volume is x y. So the system is given by y the radius, 2r x r 2 2 4 x y 1200 1 1 1 2 x y x x y x 1200 600 Substituting for x y in the second equation gives 1 4 4 4 x 2 y 600 4 1200 600 600 1200 . Thus the dimensions of the sheet metal are 2 63 in and 1910 in. x 2. So y x 2
34
CHAPTER 9 Systems of Equations and Inequalities
51. The point P is at an intersection of the circle of radius 26 centered at A 22 32
y
40 and the circle of radius 20 centered at B 28 20. We have the system A B x 222 y 322 262 20 2 2 2 x 28 y 20 20 0 _20 20 40 x x 2 44x 484 y 2 64y 1024 676 2 2 x 56x 784 y 40y 400 400 _20 x 2 44x y 2 64y 832 Subtracting the two equations, we get 12x 24y 48 x 2y 4, x 2 56x y 2 40y 784
which is the equation of a line. Solving for x, we have x 2y 4. Substituting into the first equation gives
2y 42 44 2y 4 y 2 64y 832 4y 2 16y1688y176 y 2 64y 832 5y 2 168y192 832 2 451024 5y 2 168y 1024 0. Using the Quadratic Formula, we have y 168 16825 16810 7744 16888 10
y 8 or y 2560. Since the y-coordinate of the point P must be less than that of point A, we have y 8. Then x 2 8 4 12. So the coordinates of P are 12 8.
To solve graphically, we must solve each equation for y. This gives x 222 y 322 262 y 322 262 x 222 y 32 676 x 222 y 32 676 x 222 . We use the function y 32 676 x 222 because the intersection we at interested in is below the point A. Likewise, solving the second equation for y, we would get the function y 20 400 x 282 . In a three-dimensional situation, you would need a minimum of three satellites, since a point on the earth can be uniquely specified as the intersection of three spheres centered at the satellites. 52. The graphs of y x 2 and y x k for various values of k are shown. If we solve y x2 we get x 2 x k 0. Using the Quadratic Formula, the system y x k 1 1 4k . So there is no solution if 1 4k is undefined, that we have x 2 is, if 1 4k 0 k 14 . There is exactly one solution if 1 4k 0 k 14 , and there are two solutions if 1 4k 0 k 14 .
y k=2 1
k=_ 4 1
1
x
k=_3
SECTION 9.5 Systems of Inequalities
9.5
35
SYSTEMS OF INEQUALITIES
1. If the point 2 3 is a solution of an inequality in x and y, then the inequality is satisfied when we replace x by 2 and y by 3. Because 4 2 2 3 8 6 2 1, the point 2 3 is a solution of the inequality 4x 2y 1.
2. To graph an inequality we first graph the corresponding equation. So to
y
graph y x 1, we first graph the equation y x 1. To decide which
side is the graph of the inequality we use test points. Test Point
Inequality y x 1 ?
0 0
Part of graph
201
Not part of graph
x
1
y=x+1
001X ?
0 2
1
Conclusion
3. If the point 2 3 is a solution of a system of inequalities in x and y, then each inequality is satisfied when we replace x by 2 and y by 3. Because 2 2 4 3 16 17 and 6 2 5 3 27 29, the point 2 3 is a solution of the given system.
4. (a)
xy0 xy2
(b)
xy0 xy2
y
(c)
y
1 x
(d)
xy0 xy2 y
x-y=0
x+y=2 1
xy0 xy2 y
x-y=0
1
x-y=0
1
x+y=2 1
5.
x
x-y=0
1
x+y=2 1
x
x+y=2 1
6. Test Point Inequality x 5y 3 ?
Conclusion
1 2 1 5 2 3 X
Solution
1 2
1 5 2 3 X
Solution
1 2
1 5 2 3
Not a solution
8 5 1 3
Not a solution
8 1
?
? ?
Test Point Inequality 3x 2y 2 2 1 1 3
?
3 2 2 1 2 X ?
3 1 2 3 2 ?
Conclusion Solution Not a solution
1 3
3 1 2 3 2 X
Solution
0 1
3 0 2 1 2 X
Solution
?
x
36
CHAPTER 9 Systems of Equations and Inequalities
7. Test Point System
8. 3x 2y 5 2x y 3
? 3 0 2 0 5X
0 0
? 2 0 0 3 ? 3 1 2 2 5X
1 2
? 2 1 2 3X ? 3 1 2 1 5 X
1 1
? 2 1 1 3X ? 3 3 2 1 5
3 1
?
2 3 1 3 X
9. y 2x. The test point 1 0 satisfies the
Conclusion
Test Point System
Not a solution
0 0
Solution
1 3
Solution
3 0
Not a solution
1 2
10. y 3x. The test point
x 2y 4
4x 3y 11 ?
0 2 0 4
? 4 0 3 0 11 ? 1 2 3 4 X ? 4 1 3 3 11 X ? 3 2 0 4 ? 4 3 3 0 11 ? 1 2 2 4 X ? 4 1 3 2 11
11. y 2. The test point
1 0 satisfies the
Conclusion Not a solution
Solution
Not a solution
Not a solution
12. x 1. The test point
0 3 satisfies the
2 0 satisfies the
inequality.
inequality.
inequality.
inequality.
y
y
y
y
y=3x 1
1
x
1
x
1
1
x=_1
y=2 1
1
x
1
x
y=_2x
13. x 2. The test point
14. y 1. The test point
0 0 satisfies the
0 2 satisfies the
inequality.
inequality.
inequality.
inequality.
y
y
y
y
x=2
1 1
x
1
15. y x 3. The test
point 0 0 satisfies the
point 0 0 satisfies the
y=x-3
y=1 1
16. y 1 x. The test
1 x
1
x
1 1
x y=1-x
SECTION 9.5 Systems of Inequalities
17. 2x y 4. The test
point 0 0 satisfies the
point 0 0 satisfies the
inequality.
1
inequality. y
y
2 2
x
10
x
_x@+y=5
3x-y-9=0
21. x 2 y 2 9. The test point 0 4 satisfies the inequality.
y=x@+1
1
x
1
y
point 0 2 satisfies the
inequality. y
1
20. y x 2 1. The test
point 0 6 satisfies the
inequality.
y
2x-y=_4
19. x 2 y 5. The test
18. 3x y 9 0. The test
37
1
x
22. x 2 y 22 4. The test point 0 1 satisfies the inequality.
y
x@+y@=9
x@+(y-2)@=4
1 x
1
1 x
1
23. 3x 2y 18
24. 4x 3y 9
4 -2 0
2
4
6
8
8 4
10 12
-4
-4
-8
-2 0
-12
25. 5x 2y 8
2
4
6
8
-4
26. 5x 3y 15
8
4
4 0 -2
0 -4
2
4
-8
6
5
-4 -8
27. The boundary is a solid curve, so we have the inequality y 12 x 1. We take the test point 0 2 and verify that it satisfies the inequality: 2 12 0 1.
28. The boundary is a solid curve, so we have the inequality y x 2 2. We take the test point 0 0 and verify that it satisfies the inequality: 0 02 2.
29. The boundary is a broken curve, so we have the inequality x 2 y 2 4. We take the test point 0 4 and verify that it satisfies the inequality: 02 42 4.
30. The boundary is a solid curve, so we have the inequality y x 3 4x. We take the test point 1 1 and verify that it satisfies the inequality: 1 13 4 1.
38
31.
CHAPTER 9 Systems of Equations and Inequalities
xy4 yx
The vertices occur where
xy4 yx
y
Substituting, we have
y=x 1
2x 4 x 2. Since y x, the vertex is 2 2, and the solution set is not
32.
2x 3y 12
2x 3y 12
3x y 21
The vertices occur where
x
1
bounded. The test point 0 1 satisfies each inequality.
x+y=4
2x 3y 12 3x y 21
y
2x+3y=12 1
and adding the two equations gives 11x 75 x 75 11 .
1
9x 3y 63 132150 18 y 6 . Therefore, the Then 2 75 11 3y 12 3y 11 11 11 6 vertex is 75 11 11 , and the solution set is not bounded. The test point 0 5
x
3x-y=21
satisfies each inequality.
33.
y 1x 2 4
y 2x 5
The vertex occurs where
y 1x 2
y
4
Substituting for y
y 2x 5
y=41 x+2
gives 14 x 2 2x 5 74 x 7 x 4, so y 3. Hence, the vertex is 4 3,
1
34.
xy0
4 y 2x
The vertices occur where
y x
4 y 2x
y
set is not bounded. The test point 3 0 satisfies each inequality.
One vertex occurs where
y 2x 8
y 12 x 5
4+y=2x 2 x-y=0
x
2
y
Substituting for
y gives 2x 8 12 x 5 32 x 3 x 2, so y 2 2 8 4. y 2x 8 Hence, this vertex is 2 4. Another vertex occurs where y0 2x 8 0 x 4; this vertex is 4 0. Another occurs where y 12 x 5 y 5; this gives the vertex 0 5. The origin is another x 0 vertex, and the solution set is bounded. The test point 1 1 satisfies each inequality.
y=2x-5
Substituting for y
gives 4 x 2x x 4, so y 4. Hence, the vertex is 4 4, and the solution
y 2x 8 35. y 12 x 5 x 0, y 0
x
1
and the solution is not bounded. The test point 0 1 satisfies each inequality.
y=_2x+8 1
y=_ 2 x+5 x
1 1
SECTION 9.5 Systems of Inequalities
4x 3y 18 36. 2x y 8 x 0, y 0
One vertex occurs where
4x 3y 18 2x y 8
y
Subtracting twice
2x+y=8
the second equation from the first gives y 2, so 2x 2 8 x 3 and the vertex is 3 2. Other vertices occur at the x-intercept of 2x y 8 and the
4x+3y=18
1
x
1
y-intercept of 4x 3y 18; these are 4 0 and 0 6, respectively. The origin is
another vertex, and the solution set is bounded. The test point 1 1 satisfies each inequality.
37.
x 0
y 0 3x 5y 15 3x 2y 9
From the graph, the points 3 0, 0 3 and 0 0 are vertices,
y 3x+2y=9
and the fourth vertex occurs where the lines 3x 5y 15 and 3x 2y 9
intersect. Subtracting these two equations gives 3y 6 y 2, and so x 53 . Thus, the fourth vertex is 53 2 , and the solution set is bounded. The test point
3x+5y=15
1
x
1
1 1 satisfies each inequality.
38.
y
x 2 y 12
2x 4y 8
From the graph, the vertices occur at 2 1 and 28 12.
39.
y 9 x2
x 0, y 0
y=12
6
2x-4y=8 x
6
The solution set is not bounded. The test point 6 0 satisfies each inequality.
x=2
y
From the graph, the vertices occur at 0 0, 3 0, and 0 9.
y=9-x@
The solution set is bounded. The test point 1 1 satisfies each inequality. y=0
1 x
1 x=0
2 y x 40. y 4 x 0
From the graph, the vertices occur at 0 0, 0 2, and 2 4. The
y x=0 y=4
solution set is bounded. The test point 1 3 satisfies each inequality. y=x@
1 1
x
39
40
41.
CHAPTER 9 Systems of Equations and Inequalities
y 9 x2 y x 3
The vertices occur where
y 9 x2
y x 3
Substituting for y
y=9-x@
y y=x+3
gives 9 x 2 x 3 x 2 x 6 0 x 2 x 3 0 x 3, x 2. Therefore, the vertices are 3 0 and 2 5, and the solution set is
1
42.
y
y x2
xy6
The vertices occur where
y x2
xy6
x
1
bounded. The test point 0 4 satisfies each inequality.
Substituting for y y=x@
gives x 2 x 6 x 2 x 6 0 x 3 x 2 0 x 3, x 2. Since y x 2 , the vertices are 3 9 and 2 4, and the solution set is not
43.
x 2 y2 4 xy0
The vertices occur where
x 2 y2 4 xy0
x+y=6
2
bounded. The test point 0 7 satisfies each inequality.
1 y
x-y=0
Since x y 0 1
x y, substituting for x gives y 2 y 2 4 y 2 2 y 2, and x 2. Therefore, the vertices are 2 2 and 2 2 , and the
44.
x@+y@=4
y
x 0 y 0
x y 10 2 x y2 9
x
1
solution set is bounded. The test point 1 0 satisfies each inequality.
x
From the graph, the vertices are 0 3, 0 10, 3 0, and
10 0. The solution set is bounded. The test point 4 1 satisfies each inequality.
x+y=10
x=0
1
x
y=0
1
x@+y@=9
45.
x2 y 0
2x 2 y 12 2x 2 2y 0
2x 2 y 12
The vertices occur where
x2 y 0
2x 2 y 12
y
x@-y=0
Subtracting the equations gives 3y 12 y 4, and
x 2. Thus, the vertices are 2 4 and 2 4, and the solution set is bounded.
46.
2x 2 y 4 x2 y 8
The vertices occur where
x
1
2x 2 y 4 x2 y 8
y
Adding the
equations gives 2x 2 x 2 12 x 2 4 x 2. Therefore, the vertices are 2 4 and 2 4. The solution set is not bounded. The test point 0 6 satisfies each inequality.
2x@+y=12
1
The test point 0 1 satisfies each inequality.
2 x@-y=8
2x@+y=4
(_2, _4)
1
x (2, _4)
41
SECTION 9.5 Systems of Inequalities
47.
x 2 y2 9
The vertices occur where
2x y 2 1
x 2 y2 9
Subtracting the
2x y 2 1
equations gives x 2 2x 8 x 2 2x 8 x 2 x 4 0. Therefore,
y
2x+y@=1
1
the vertices are 2 5 and 2 5 . The solution set is bounded. The test
point 0 0 satisfies each inequality.
48.
x 2 y2 4
x 2 y2 4
Subtracting the
x 2 2y 1
equations gives y 2 2y 3 y 2 2y 3 y 3 y 1 0. However, y 3 is extraneous, and the vertices are 3 1 and 3 1 . The solution
y x@-2y=1
(_Ï3, 1)
(Ï3, 1)
1
We find the vertices of the region by solving pairs of the
corresponding equations:
3x y 0 xy2
x 2y 14
x y 2
3x y
0
2x y 2
x 2y 14
3y 12
x
1
set is bounded. The test point 0 2 satisfies each inequality. x 2y 14 49. 3x y 0 x y 2
x
1
(_2, _Ï5)
The vertices occur where
x 2 2y 1
x@+y@=9
(_2, Ï5)
x@+y@=4
x+2y=14
y
3x-y=0
y 4 and x 6.
x 1 and y 3. Therefore, the vertices
1 x
1 x-y=2
are 6 4 and 1 3, and the solution set is not bounded. The test point 0 4 satisfies each inequality.
50.
y x 6
3x 2y 12
x 2y 2
To find where the line y x 6 intersects the lines
y 3x+2y=12
3x 2y 12 and x 2y 2, we substitute for y: 3x 2 x 6 12 x 0 and y 6; x 2 x 6 2 x 14 and y 8. Next, adding the equations 3x 2y 12 and x 2y 2 gives 4x 14 x 72 , so these lines intersect at the point 72 34 . Since the vertex 14 8 is not part of the solution set, the vertices are 0 6 and 72 34 , and the solution set is not bounded. The test
y=x+6
2 2
x
x-2y=2
point 4 2 satisfies each inequality.
51.
x 0, y 0
x 5, x y 7
y
The points of intersection are 0 7, 0 0, 7 0, 5 2,
and 5 0. However, the point 7 0 is not in the solution set. Therefore, the
x=5 x=0
vertices are 0 7, 0 0, 5 0, and 5 2, and the solution set is bounded. The test point 1 1 satisfies each inequality.
x+y=7
1 y=0
1
x
42
52.
CHAPTER 9 Systems of Equations and Inequalities
x 0, y 0
y 4, 2x y 8
y
2x+y=8
The points of intersection are 0 8, 0 4, 4 0, 2 4,
and 3 2. However, the point 0 8 is not in the solution set. Therefore, the
y=4 x=0 1
vertices are 0 4, 2 4, 4 0, and 0 0. The solution set is bounded. The test point 1 1 satisfies each inequality.
53.
y x 1
1
We find the vertices of the region by solving pairs of the
x 2y 12 x 1 0
corresponding equations. Using x 1 and substituting for x in the line y x 1 gives the point 1 0. Substituting for x in the line x 2y 12 gives y x 1 13 the point 1 2 . x y 1 and y 1 2y 12 x 2y 12 10 13 3y 13 y 13 3 and x 3 . So the vertices are 1 0, 1 2 , and 10 13 , and none of these vertices is in the solution set. The solution set is 3 3
y
x
y=0
x+2y=12
x+1=0 y=x+1
1
x
1
bounded. The test point 0 2 satisfies each inequality.
x y 12 54. y 12 x 6 3x y 6
y
Graphing these inequalities, we see that there are no points 3x+y=6
that satisfy all three, and hence the solution set is empty.
x+y=12 2
y=21 x-6
55.
x 2 y2 8
x 2, y 0
The intersection points are 2 2, 2 0, and 2 2 0 .
x@+y@=8
x
2 1 2
y
x=2
1
However, since 2 2 is not part of the solution set, the vertices are 2 2, 2 0, and 2 2 0 . The solution set is bounded. The test point 21 1 satisfies each
y=0
1
x
inequality.
2 x y 0 56. xy6 xy6
Adding the equations x y 6 and x y 6 yields 2x 12
x 6. So these curves intersect at the point 6 0. To find where x 2 y 0
and x y 6 intersect, we solve the first for y, giving y x 2 , and then substitute
x+y=6
y
x@-y=0
2 1
x-y=6
x
into the second equation to get x x 2 6 x 2 x 6 0 x 3 x 2 0 x 3 or x 2. When x 3, we have
y 9, and when x 2, we have y 4, so the points of intersection are 3 9 and 2 4. Substituting y x 2 into the
equation x y 6 gives x x 2 6, which has no solution. Thus the vertices (which are not in the solution set) are 6 0, 3 9, and 2 4. The solution set is not bounded. The test point 1 0 satisfies each inequality.
SECTION 9.5 Systems of Inequalities
57.
x 2 y2 9
x y 0, x 0
Substituting x 0 into the equations x 2 y 2 9 and
43
x+y=0 y
1
x y 0 gives the vertices 0 3 and 0 0. To find the points of intersection
x=0
for the equations x 2 y 2 9 and x y 0, we solve for x y and substitute
x 1 3 2 2 2 into the first equation. This gives y y 9 y 2 . The points x@+y@=9 0 3and 3 2 2 3 2 2 lie away from the solution set, so the vertices are 0 0, 0 3, and 3 2 2 3 2 2 . Note that the vertices are not solutions in this case. The solution set is bounded. The test point
0 1 satisfies each inequality.
58.
y x3
y 2x 4 x y 0
y
The curves y x 3 and x y 0 intersect when
y=2x+4
x 3 x 0 x 0 y 0. The lines x y 0 and y 2x 4 intersect
y=x#
1
when x 2x 4 3x 4 x 43 y 43 . To find where y x 3 and y 2x 4 intersect, we substitute for y and get x 3 2x 4
x
1 x+y=0
x 3 2x 4 0 x 2 x 2 2x 2 0 x 2 (the other factor has no
real solution). When x 2 we have y 8Thus, the vertices of the region are 0 0, 43 43 , and 2 8. The solution set is bounded. The test point 0 2 satisfies each inequality.
x 2y 14 59. 3x y 0 x y2
The lines x 2y 14 and 3x y 0 intersect when
14 x 6x x 2 and y 6. The lines 3x y 0 and x y 2 intersect where 3x 2 x x 1 and y 3. The lines x 2y 14 and x y 2 intersect where 14 2y 2 y y 4 and x 6. Thus, the vertices of the region are 1 3, 2 6, and 6 4. The solution set is bounded. The test point 1 0 satisfies each inequality.
x 2y 14 60. 3x y 0 x y2
y (2, 6) 2 (_1, _3)
2
(6, 4)
x
y
The vertices are 1 3, 2 6, and 6 4. In this case the
(2, 6)
solution set is not bounded. The test point 3 0 satisfies each inequality. 2 (_1, _3)
2
(6, 4)
x
44
CHAPTER 9 Systems of Equations and Inequalities
x y 12 61. y 12 x 6 y 2x 6
The lines x y 12 and y 12 x 6 intersect where
y (2, 10)
12 x 12 x 6 x 12 and y 0. The lines y 12 x 6 and y 2x 6 intersect where 12 x 6 2x 6 x 8 and y 10. The lines x y 12
and y 2x 6 intersect where 12 x 2x 6 x 2 and y 10. The solution set is not bounded. The test point 0 8 satisfies each inequality.
62.
y x 1
x 2y 12
x 1 0
The lines y x 1 and x 2y 12 intersect where
2 2
(12, 0)
(_8, _10)
(_1, 132 )
y
( 103 , 133 )
10 y 1 12 2y y 13 3 and x 3 . The lines y x 1 and x 1 0
intersect at 1 0, and the lines x 2y 12 and x 1 0 intersect where
12 2y 1 y 13 2 and x 1. Thus, the vertices of the region are 10 13 . The solution set is bounded. The test point , and 1 0, 1 13 2 3 3
(_1, 0)
3x y 5 30x 10y 50 x 2y 5 10x 20y 50 63. x 6y 9 10x 60y 90 x 0, y 0 x 0 y 0
1
y (0, 5)
(1, 2)
The vertices are 1 5,
point 4 3 satisfies each inequality. y x 3 65. y 2x 6 Using a graphing device, we find the region shown. The y 8 vertices are 3 0, 1 8, and 11 8.
x y 12 66. 2x y 24 x y 6
1 y
17 7 , and 8 1. The solution set is bounded. The test 4 4
Using the graphing device, we find the region shown. The
(3, 1)
1
The vertices are 0 5, 1 2, 3 1, and 9 0. The solution set is not bounded. The test point 1 3 satisfies each inequality.
Add dots for the graphs of #62-64. [see #61 graph] x
1
0 2 satisfies each inequality.
x y6 4x 7y 39 64. x 5y 13 x 0 y 0
x
1
(9, 0)
x
(1, 5)
( 174 , 74 )
(8, 1) x
1
10 8 6 4 2 0 -2 -4
10
10
vertices are 3 9, 6 12, and 12 0. 0
10
SECTION 9.5 Systems of Inequalities
67.
y 6x x 2
xy4
10 8 6 4 2
Using a graphing device, we find the region shown. The
vertices are 06 34 and 64 24.
-4 -2 -2 0 2 -4
68.
y x3
2x y 0
45
4
6
8 10
2
4
12 10 8 6 4 2
Using a graphing device, we find the region shown. The
y 2x 6
vertices are 0 0, 22 103 and 15 3. -4
-2
0 -2
y
69. (a) Let x and y be the numbers of acres of potatoes and corn, respectively. x y 500 90x 50y 40,000 A system describing the possibilities is 30x 80y 30,000 x 0 y 0
(0, 375) (200, 300) (375, 125)
100
x ( 4000 9 , 0)
100
(b) Because the point 300 180 lies in the feasible region, the farmer can plant 300 acres of potatoes and 180 acres of corn. (c) Because the point 150 325 lies outside the feasible region, the farmer cannot plant this combination of crops.
70. (a) Let x and y be the numbers of acres of cauliflower and cabbage, respectively. x y 300 70x 35y 17,500 A system describing the possibilities is 25x 55y 12,000 x 0 y 0
y
(0, 2400 11 ) (150, 150)
(200, 100)
50 50
(250, 0)
(b) Because the point 155 115 lies in the feasible region, the farmer can plant 155 acres of cauliflower and 115 acres of cabbage. (c) Because the point 115 175 lies outside the feasible region, the farmer cannot plant this combination of crops. 71. Let x be the number of fiction books published in a year and y the number of nonfiction books. Then the following system of inequalities holds: x 0, y 0 From the graph, we see that the vertices are 50 50, 80 20 x y 100 y 20, x y and 20 20.
y
120 100 80 60 40 20
x+y=100 x=y y=20 20 40 60 80 100
x
x
46
CHAPTER 9 Systems of Equations and Inequalities
72. Let x be the number of chairs made and y the number of tables. Then the following 2x 3y 12 system of inequalities holds: 2x y 8 The intersection points are 0 4, x 0, y 0
y 2x+y=8
2x+3y=12
1
0 8, 6 0, 4 0, 0 0, and 3 2. Since the points 0 8 and 6 0 are not in
73. Let x be the number of Standard Blend packages and y be the number of Deluxe
y
200
Blend packages. Since there are 16 ounces per pound, we get the following system
3 x+38 y=90 4
of inequalities:
x
1
the solution set, the vertices are 0 4, 0 0, 4 0, and 3 2.
1 x+58 y=80 4
100
x 0
y 0 1 x 5 y 80 4 8 3 3 4 x 8 y 90
200 x
100
From the graph, we see that the vertices are 0 0, 120 0, 70 100 and 0 128. 74. Let x be the amount of fish and y the amount of beef (in ounces) in each can. Then
y
the following system of inequalities holds: 12x 6y 60
12x+6y=60
3x 9y 45 x 0, y 0
3x+9y=45
2
x
2
From the graph, the vertices are 15 0, 3 4, and 0 10.
75. x 2y 4, x y 1, x 3y 9, x 3. Method 1: We shade the solution to each inequality with lines perpendicular to the boundary. As you can see, as the number of inequalities in the system increases, it gets harder to locate the region where all of the shaded parts overlap. Method 2: Here, if a region is shaded then it fails to satisfy at least one inequality. As a result, the region that is left unshaded satisfies each inequality, and is the solution to the system of inequalities. In this case, this method makes it easier to identify the solution set. To finish, we find the vertices of the solution set. The line x 3 intersects the line x 2y 4 at 3 12 and the line
x 3y 9 at 3 2. To find where the lines x y 1 and x 2y 4 intersect, we add the two equations, which gives
3y 5 y 53 , and x 23 . To find where the lines x y 1 and x 3y 9 intersect, we add the two equations, 5 3 1 2 5 3 5 which gives 4y 10 y 10 4 2 , and x 2 . The vertices are 3 2 , 3 2, 3 3 , and 2 2 , and the solution set is bounded.
_x+y=1
y
1
x+3y=9 x
1 x=3
Method 1
Method 2
Solution Set
x+2y=4
CHAPTER 9
Review
CHAPTER 9 REVIEW 1.
3x y 5 2x y 5
y
Adding, we get 5x 10 x 2. So 2 2 y 5 y 1. 1
Thus, the solution is 2 1.
2.
y 2x 6
y x 3
3.
2x 7y 28
2x 7y 28
6x 8y 15 6x 8y 15 4. 3 x 2y 4 6x 8y 16
y
Since these equations represent the
2 2
x
1
x
1
x
y
Adding gives 0 1 which is
false. Hence, there is no solution. The lines are parallel.
2x y 1 5. x 3y 10 3x 4y 15
x
1
same line, any point on this line will satisfy the system. Thus the solution are t 27 t 4 , where t is any real number.
2
1
y
y 27 x 4
x
Subtracting the second equation from the first, we get
0 3x 3 x 1. So y 1 3 4. Thus, the solution is 1 4.
2x 7y 28
1
1
y
Solving the first equation for y, we get y 2x 1.
Substituting into the second equation gives x 3 2x 1 10 5x 7 7 12 x 75 . So y 75 1 12 5 Checking the point 5 5 in the third ? 21 48 equation we have 3 75 4 12 5 15 but 5 5 15. Thus, there is no
solution, and the lines do not intersect at one point.
1
47
48
CHAPTER 9 Systems of Equations and Inequalities
2x 5y 9 6. x 3y 1 7x 2y 14
y
Adding the first equation to twice the second equation gives
11y 11 y 1. Substituting back into the second equation, we get
1
x 3 1 1 x 4. Checking point 4 1 in the third equation gives
7 4 2 1 26 14. Thus there is no solution, and the lines do not intersect at
7.
y x 2 2x y 6x
x
1
one point.
Substituting for y gives 6 x x 2 2x x 2 x 6 0. Factoring, we have x 2 x 3 0.
Thus x 2 or 3. If x 2, then y 8, and if x 3, then y 3. Thus the solutions are 3 3 and 2 8.
8.
x 2 y2 8
y x 2
Substituting for y in the first equation gives x 2 x 22 8
2x 2 4x 4 0 2 x 2 2x 2 0. Using the Quadratic Formula, we have 2 2 3 2 4 8 1 3. If x 1 3. then y 1 3 2 1 3, and if x 1 3, x 2 2 then y 1 3 2 1 3. Thus, the solutions are 1 3 1 3 and 1 3 1 3 . 4 3x 6 y 9. 8 x 4 y
Adding twice the first equation to the second gives 7x 16 x 16 7 . So
16 8 4 7 y
16 14 16y 56 28y 12y 56 y 14 3 . Thus, the solution is 7 3 .
10.
x 2 y 2 10
x 2 2y 2 7y 0
Subtracting the first equation from the second gives y 2 7y 10
y 2 7y 10 0 y 2 y 5 0 y 2, y 5. If y 2, then x 2 4 10 x 2 6 x 6, and if y 5, then x 2 25 10 x 2 15, which leads to no real solution. Thus the solutions are 6 2 and 6 2 .
32x y 0 43 11. 7x 12y 341 y 7x 341 12 The solution is approximately 2141 1593. 032x 043y
12.
12x 32y
660
7137x 3931y 20,000
y 6 x 1102 3
y 7137 x 20,000 3931
3931
The solution is approximately 6104 10573.
Graphs missing in #11,12
Note that in CA-8 and PMFC-8 the corresponding graphs are fine.
CHAPTER 9
13.
x y 2 10
1 y 12 x 22
y x 10
y 22 x 12
The solutions are 1194 139 and 1207 144.
14.
y 5x x
y x5 5
Review
49
The solutions are approximately
145 135, 1 6, and 151 1293.
graphs missing in #13,14 15.
x 2y z
8
x
4x z 9 2x y z 8
2y z 8 8 x 2y z 8y 3z 23 8y 3z 23 3y z 8 z 5
Therefore, z 5, 8y 3 5 23
y 1, and x 2 1 5 8 x 1. Hence, the solution is 1 1 5. x y 3z 4 x y 3z 4 x y 3z 4 16. 4x 2y z 11 6y 13z 27 6y 13z 27 5x y z 16 3z 9 6y 16z 36
Therefore, z 3,
6y 13 3 27 y 2, and x 2 3 3 4 x 3. Hence, the solution is 3 2 3. x y 2z 6 x y 2z 6 x y 2z 6 17. 2x Therefore, 3z 6 z 2, 2y 2 0 5z 12 2y z 0 2y z 0 x 2y 3z 9 4y z 6 3z 6 y 1, and x 1 2 2 6 x 1. Hence, the solution is 1 1 2. x 2y 3z 1 x 2y 3z 1 x 2y 3z 1 18. y 4z 1 y 4z 1 x 3y z 0 2x 28z 0 6y 4z 6 6z 6
x 3, and so the solution is 3 1 0. x 2y 3z 1 x 2y 3z 1 x 2y 3z 1 19. 2x y z 3 3y 5z 1 3y 5z 1 2x 7y 11z 2 6y 10z 1 0 1
Thus, z 0, y 1, and x 2 1 3 0 1
which is impossible. Therefore, the
system has no solution. x y z 2 x y z 2 x y z 2 2x 3z 5 2y 5z 2 1 2y 5z 2 1 20. 4 9 3y z 3 7 13z 12 11 x 2y x y 2z 3 5 z 2 3 z 2 3 x y z 2 2y 5z 2 1 Therefore, 14 28 2, 13z12 2 11 z 1; 2y5 12 2 1 13z 12 11 14 28 2y 1 1 y 0; and x 1 2 2 x 1. So the solution is 1 0 1 2.
50
21.
CHAPTER 9 Systems of Equations and Inequalities
x 3y
z4
4x y 15z 5
x 3y z
4
y z 1
Thus, the system has infinitely many solutions given by z t,
y t 1 y 1 t, and x 3 1 t t 4 x 1 4t. Therefore, the solutions are 1 4t 1 t t, where t is any real number. 2x 3y 4z 3 2x 3y 4z 3 3 2x 3y 4z 22. 4x 5y 9z 13 y z 7 y z 7 2x 7z 0 3y 3z 3 0 24
Since this last equation is impossible, the system is inconsistent and has no solution. x z 2 x z 2 x z 2 2x y y 2z 4 8 y 2z 4 8 2 12 23. 3y z 4 5z 13 20 3y z 4 x yz z 4 20 y 8 10 x z 2 y 2z 4 8 Therefore, 33 120 40 20 z 60 , 5z 13 40 11 11 11 , 5z 13 20 33 120 40 8 y 48 ; and x 60 40 2 x 2 . Hence, the solution is y 2 60 4 11 11 11 11 11 11 2 48 60 40 . 11 11 11 11 x 4y z 8 x 4y z 8 x 4y z 8 24. 2x 6y z 9 2y 3z 7 2y 3z 7 x 6y 4z 15 6y 9z 21 00
Thus, the system has infinitely many solutions. Letting z t, we find 2y 3t 7 y 72 32 t, and x 4 72 32 t t 8 x 6 5t. Therefore, the solutions are 6 5t 72 32 t t , where t is any real number.
25. Let the age of the younger child be x, so the older child’s age is x 4. Then because the sum of their ages is 22, we have x x 4 22 2x 4 22 2x 18 x 9. Thus, the younger child is 9 years old and the older child is 9 4 13 years old. 26. Let x be the amount in the account yielding 6% and y the amount in the one yielding 7%. The system is y 2x Substituting gives 006x 007 2x 600 02x 600 x 3000, so 006x 007y 600 y 2 3000 6000. Hence, $3000 is invested at 6% and $6000 is invested at 7%.
27. Let the time spent driving be x and the time spent flying be y. The total travel time is 6 hours, so x y 6. The distance driven is 70x and the distance flown is 550y. The total distance traveled is 2700 miles, so 70x 550y 2700. x y 6 We have the system Subtracting 70 times the first equation from the second, we have 70x 550y 2700 550y 70y 2700 70 6 480y 2280 y 19 4 . The traveler drove for 4 hours 15 minutes and flew for 1 hour 45 minutes.
CHAPTER 9
Review
51
28. Let the volume of the 15% solution be x mL and the volume of the 25% solution be y mL. Then x y 500, and the volume of acid in the mixture is 015x 025y 022 500 110. x y 500 Thus, we solve the system Subtracting 015 times the first equation from the second, we have 015x 025y 110 025y 015y 110 015 500 01y 35 y 350. Thus, x 500 350 150, so the chemist should mix 150 mL of the 15% solution and 500 150 350 mL of the 25% solution.
29. Let n, d, and q be the numbers of nickels, dimes, and quarters in the piggy bank. We get the following system: 50 n d q Since 10d 25n, we have d 52 n, so substituting into the first equation we get 5n 10d 25z 560 10d 5 5n
n 52 n q 50 72 n q 50 q 50 72 n. Now substituting this into the second equation we have 115 115 5n 10 52 n 25 50 72 n 560 5n 25n 1250 175 2 n 560 1250 2 n 560 2 n 650
n 12. Then d 52 12 30 and q 50 n d 50 12 30 8. Thus the piggy bank contains 12 nickels, 30 dimes, and 8 quarters.
30. Let c, s, and p be the number of each kind of salmon caught. The given information leads to the system c s p 25 c s p 25 c s p 25 c s p 25 c s p3 c s p 3 2s 2 p 22 2s 2 p 22 p 4, so c 2s 0 4 p 16 3s p 25 c 2s 2s 2 4 22 s 7 and c 7 4 25 c 14. The fisherman caught 14 coho, 7 sockeye, and 4 pink salmon.
31.
3x 1 B A 3x 1 .Thus, 3x 1 A x 3 B x 5 x A B 3A 5B, x 5 x 3 x 5 x 3 x 2 2x 15 AB 3 3A 3B 9 and so Adding, we have 8B 8 B 1, and A 2. Hence, 3A 5B 1 3A 5B 1 3x 1 1 2 . x 5 x 3 x 2 2x 15
32.
8
x 3 4x
8
x x2 4
A B C 8 . Then 8 A x 2 4 Bx x 2 x x 2 x 2 x x 2 x 2
C x x 2 x 2 A B C x 2B 2C 4A. Thus, 4A 8 A 2, 2B 2C 0 B C, and 1 1 8 2 . 2 2B 0 B 1, so C 1. Therefore, 3 x x 2 x 2 x 4x 33.
2x 4
x x 12
B C A . Then 2x 4 A x 12 Bx x 1 C x Ax 2 2Ax A Bx 2 x x 1 x 12
Bx C x x 2 A B x 2A B C A. So A 4, 4 B 0 B 4, and 8 4 C 2 C 2. 4 2 4 2x 4 . Therefore, x x 1 x 12 x x 12 34.
x 6 Ax B C x 6 . Thus, 2 2 x 2 x 4 x 2 x 3 2x 2 4x 8 x 4 x 6 Ax B x 2 C x 2 4 Ax 2 2Ax Bx 2B C x 2 4C x 2 A C x 2A B 2B 4C
52
CHAPTER 9 Systems of Equations and Inequalities
and we get the system
A 2A B
C0
1
2B 4C 6
A
C0
B 2C 1
2B 4C 6
A
C0
B 2C 1
8C 8
x 6 x 1 1 Thus, 8C 8 C 1, B 2 1 B 1, and A 1 0 A 1. So 3 . 2 x 2x 2 4x 8 x 4 x 2
35.
2x 1 2x 1 A Bx C 2 1 Bx C x Ax 2 A Bx 2 C x A B x 2 . Then 2x 1 A x x x3 x x x2 1 x2 1 x 2 1 2x 1 2 . C x A. So A 1, C 2, and A B 0 gives us B 1. Thus 3 x x x x 1
B Cx D A 5x 2 3x 10 2 . 36. Since x 4 x 2 2 x 2 1 x 2 2 x 1 x 1 x 2 2 , we have 4 2 x 1 x 1 x x 2 x 2 Thus 5x 2 3x 10 x 1 x 2 2 A x 1 x 2 2 B x 2 1 C x D Ax 3 Ax 2 2Ax 2A Bx 3 Bx 2 2Bx 2B C x 3 Dx 2 C x D
A B C x 3 A B D x 2 2A 2B C x 2A 2B D A BC 0 Coefficients of x 3 A B C 0 A B 2 D 5 Coefficients of x 2B C D 5 This leads to the system 2A 2B C 3 Coefficients of x 3C 3 2A 2B D 10 Constant terms 4B C D 13 A B C 0 2B C D 5 Thus C 1, 3 1 3D 3 D 0, 2B 1 0 5 B 3, and C 1 3C 3D 3 A 3 1 0 A 2. Thus,
37.
3 x 2 5x 2 3x 10 2 . x 1 x 1 x4 x2 2 x 2
Cx D Ax B 3x 2 x 6 2 2 . Thus x 2 22 x 2 x2 2 3x 2 x 6
x 2 2 Ax B C x D Ax 3 Bx 2 2Ax 2B C x D
x 3 A x 2 B x 2A C 2B D A 0 A 0 B 3 B 3 This leads to the system C 1 2A C 1 D 0 2B D6
Thus
3x 2 x 6 x 3 2 2 . 2 2 2 x 2 x 2 x 2
CHAPTER 9
Review
53
Bx C x2 x 1 A Dx E 2 2 . Thus x xx 2 12 x 1 x2 1 2 x 2 x 1 x 2 1 A x x 2 1 Bx C x Dx E Ax 4 2Ax 2 A Bx 4 C x 3 Bx 2 C x Dx 2 E x 38.
x 4 A B x 3 C x 2 2A B D x C E A AB 0 C 0 Thus A 1, B 1, C 0, D 0, and E 1, so This leads to the system 2A B D 1 C E 1 A 1 x2 x 1 x 1 1 2 2 . x 2 xx 2 12 x 1 x 1
39.
2x 3y 7 x 2y 0
By inspection of the graph, it appears that 2 1 is the solution to the system. We check this in both
equations to verify that it is the solution. 2 2 3 1 4 3 7 and 2 2 1 2 2 0. Since both equations are satisfied, the solution is indeed 2 1.
40.
3x y 8
y x 2 5x
By inspection of the graph, it appears that 2 14 and 4 4 are solutions to the system.
We check each possible solution in both equations to verify that it is a solution: 3 2 14 6 14 8 and
22 5 2 4 10 14; also 3 4 4 12 4 8 and 42 5 4 16 20 4. Since both points satisfy both equations, the solutions are 2 14 and 4 4. 41.
x2 y 2
x 2 3x y 0
By inspection of the graph, it appears that 2 2 is a solution to the system, but is difficult
to get accurate values for the other point. Adding the equations, we get 2x 2 3x 2 2x 2 3x 2 0 2 2x 1 x 2 0. So 2x 1 0 x 12 or x 2. If x 12 , then 12 y 2 y 74 . If x 2, then 22 y 2 y 2. Thus, the solutions are 12 74 and 2 2. 42.
x y 2 x 2 y 2 4y 4
By inspection of the graph, it appears that 2 0 and 2 4 are solutions to the system. We
check each possible solution in both equations to verify that it is a solution: 2 0 2 and 22 02 4 0 4;
also 2 4 2 and 22 42 4 4 4 16 16 4. Since both points satisfy both equations, the solutions are 2 0 and 2 4.
43. The boundary is a solid curve, so we have the inequality x y 2 4. We take the test point 0 0 and verify that it satisfies the inequality: 0 02 4.
44. The boundary is a solid curve, so we have the inequality x 2 y 2 8. We take the test point 0 3 and verify that it satisfies the inequality: 02 32 8.
54
CHAPTER 9 Systems of Equations and Inequalities
46. y x 2 3
45. 3x y 6
y
47. x 2 y 2 9
48. x y 2 4
y
y
y
1
1 1 1
49.
x
50.
y 1x 1 3
0
x y 2 51. y x 2 x 3
y x 1
x 2 y2 1 y
y
x
0
1
x
53.
xy0
2
y 4
1 1
x 2 y2 9
y 2x 52. y 2x y 1x 2
y
1
1
4
x
y
The vertices occur where y x. By substitution, 1
x 2 x 2 9 x 3 , and so y 3 . Therefore, the vertices are 2 2 3 3 and 3 3 and the solution set is bounded. The test point 2
2
x
1
x
x
1
y x 2 3x
1
1
1
2
x
1
2
1 1 satisfies each inequality.
y x2 4 54. y 20
y
The vertices occur where y x 2 4 and y 2. By
substitution, x 2 4 20 x 2 16 x 4 and y 20. Thus, the vertices
(_4, 20)
(4, 20)
are 4 20 and the solution set is bounded. The test point 0 10 satisfies each inequality.
2 2
x 0, y 0 55. x 2y 12 y x 4
The intersection points are 4 0, 0 4,
y
4 16 , 0 6, 3 3
0 0, and 12 0. Since the points 4 0 and 0 6 are not in the solution set, the vertices are 0 4, 43 16 3 , 12 0, and 0 0. The solution set is bounded. The test point 1 1 satisfies each inequality.
x
y=x+4
x+2y=12 1 1
x
x
CHAPTER 9
56.
x4
x y 24
x 2y 12
y
point 20 4. The lines x 4 and x 2y 12 intersect at the point 4 4, but intersect at the point 4 20. Hence, the vertices are 4 20 and 20 4. The solution set is not bounded. The test point 12 12 satisfies each inequality.
55
x=4
The lines x y 24 and x 2y 12 intersect at the
this vertex does not satisfy the other inequality. The lines x y 24 and x 4
Test
x+y=24 1 1
x=2y+12 x
x y z a x y z a ab ac ab ac ,z , and x a 57. x yz b 2z a b Thus, y 2 2 2 2 x yz c 2y ac bc bc ac ab . The solution is . x 2 2 2 2 ax by cz a b c ax by cz a b c 58. bx by cz a b c 0 bx by cz Subtracting the second equation from c c cx cy cz c x yz 1 the first gives a b x a b x 1Subtracting the third equation from the second, b c x b c y 0 y x 1. So 1 1 z 1 z 1, and the solution is 1 1 1.
59. Solving the second equation for y, we have y kx. Substituting for yin the first equation gives us 12 x kx 12 1 k x 12 x . Substituting for y in the third equation gives us kx x 2k k1 2k 12 2k . These points of intersection are the same when the x-values are equal. Thus, k 1 x 2k x k 1 k1 k 1 12 k 1 2k k 1 12k 12 2k 2 2k 0 2k 2 10k 12 2 k 2 5k 6 2 k 3 k 2. Hence, k 2 or k 3.
60. The system will have infinitely many solutions when the system has solutions other than 0 0 0. So we solve the system with x 0, y 0, and z 0: kx y z0 kx y z 0 kx y z 0 2y k 3 z 0 x 2y kz 0 2y k 3 z 0 x 3z 0 y 3k 1 z 0 [k 3 2 3k 1] z 0 Since z 0, we must have k 3 2 3k 1 5k 1 0 k 15 .
CHAPTER 9 TEST 1. (a) The system is linear. x 3y 7 (b) Multiplying the first equation by 5 and then adding gives 13y 39 y 3. So 5x 2y 4 x 3 3 7 x 2. Thus, the solution is 2 3.
2. (a) The system is nonlinear.
56
CHAPTER 9 Systems of Equations and Inequalities
(b)
6x y 2 10 3x y 5
6x y 2 10 y 2 2y 0
Thus y 2 2y y y 2 0, so either y 0 or y 2. If y 0, then
3x 5 x 53 and if y 2, then 3x 2 5 3x 3 x 1. Thus the solutions are 53 0 and 1 2.
3. (a) The system is nonlinear. x 2 y 2 100 Substituting y 3x into the first equation gives x 2 3x2 100 x 2 10 x 10. (b) y 3x If x 10, then y 3 10, and if x 10, then y 3 10. We can verify that 10 3 10 and 10 3 10 are valid solutions to the first equation in the given system.
graph missing in #4
4.
x 2y 1
Note that in CA-8 and PMFC-8 the corresponding graph is not missing
y x 3 2x 2
The solutions are approximately 055 078, 043 029, and 212 056.
x 2y z 3 x 2y z 3 5. (a) x 3y 2z 3 y z 0 2x 3y z 8 y 3z 2
x 2y z 3 y z 0 Eq. 1 Eq. 2 2z 2 2 Eq. 1 Eq. 3
z 1, Eq. 2 Eq. 3
so y 1 0 y 1 and z 2 1 1 3 x 2. Thus, the solution is 2 1 1.
(b) The system is neither inconsistent nor dependent.
x y 9z 3 x y 9z 3 6. (a) x 4z 7 y 13z 10 3x y z 5 2y 26z 14
x y 9z 3 y 13z 10 Eq. 1 Eq. 2 0 6 3 Eq. 1 Eq. 3
2 Eq. 2 Eq. 3
The last equation cannot be satisfied, so the system has no solution.
(b) The system is inconsistent.
2x y z 0 7y 9z 2 3 Eq. 1 2 Eq. 2 0 0 Eq. 2 Eq. 3 Eq. 1 2 Eq. 3 Letting z t, we have 7y 9t 2 y 27 97 t, so 2x 27 97 t t 0 x 17 17 t. The solutions are 1 1 t 2 9 t t where t is any real number. 7 7 7 7
2x y z 0 2x y z 0 7. (a) 3x 2y 3z 1 7y 9z 2 x 4y 5z 1 7y 9z 2
(b) The system is dependent.
x y 2z 8 x y 2z 8 8. (a) 2x y 20 3y 4z 4 2x 2y 5z 15 z 1
2 Eq. 1 Eq. 2 2 Eq. 1 Eq. 3
x 0 2 1 8 x 10. Thus, the solution is 10 0 1.
(b) The system is neither inconsistent nor dependent.
z 1, so 3y 4 1 4 y 0 and
CHAPTER 9
Test
57
9. Let be the speed of the wind and a the speed of the airplane in still air, in kilometers per hour. Then the speed of the of the plane flying against the wind is a and the speed of the plane flying with the wind is a . Using distance rate time, 600 25 a 240 a Adding the two equations, we get 600 2a a 300. we get the system 50 300 360 a a 60
So 360 300 60Thus the speed of the airplane in still air is 300 km/h and the speed of the wind is 60 km/h.
10. Let x, y, and z represent the price in dollars for coffee, juice, and donuts respectively. Then the system of equations is 2x y 2z 625 Friend A 2x y 2z 625 2x y 2z 625 x 3z 375 Friend B y 4z 125 y 4z 125 3x y 4z 925 Friend C y 5z 200 z 075
Thus, z 075, y 4 075 125 y 175, and 2x 175 2 075 625 x 15. Thus coffee costs $150, juice costs $175, and donuts cost $075.
11. 3x 4y 6 x@+y²5
y
12. x 2 y 3 y=2x+5
1
1 1
2x y 8 13. x y 2 x 2y 4
x
y
these two equations gives 3x 6 x 2, and so y 8 2 2 4 . Thus, the
third vertex is 2 4. x2 y 5
x
1
From the graph, the points 4 0 and 0 2 are vertices. The
third vertex occurs where the lines 2x y 8 and x y 2 intersect. Adding
14.
y
y 2x 5
1 x
1
Substituting y 2x 5 into the first equation gives
x@+5²5
y
y=2x+5
x 2 2x 5 5 x 2 2x 0 x x 2 0 x 0 or x 2. If
x 0, then y 5 2 0 5, and if x 2, then y 5 2 2 1. Thus, the
vertices are 0 5 and 2 1.
1 1
15.
4x 1
x 12 x 2
B C A . Thus, x 1 x 12 x 2
x
4x 1 A x 1 x 2 B x 2 C x 12 A x 2 x 2 B x 2 C x 2 2x 1 A C x 2 A B 2C x 2A 2B C
58
FOCUS ON MODELING
which leads to the system of equations A C 0 A C 0 C 0 A A B 2C 4 B 3C 4 B 3C 4 2A 2B C 1 2B 3C 1 9C 9
Therefore, 9C 9 C 1, B 3 1 4 B 1, and A 1 0 A 1. Therefore, 4x 1
x 12 x 2
16.
1 1 1 . x 1 x 12 x 2
2x 3 Bx C 2x 3 A 2 . Then x x 3 3x x x 3 x2 3 2x 3 A x 2 3 Bx C x Ax 2 3A Bx 2 C x A B x 2 C x 3A.
x 2 1 2x 3 2 . So 3A 3 A 1, C 2 and A B 0 gives us B 1. Thus 3 x x x x 3
FOCUS ON MODELING Linear Programming
2.
1. Vertex 0 2 0 5 4 0
M 200 x y
200 0 2 198 200 0 5 195 200 4 0 196
Thus, the maximum value is 198 and the minimum value is 195.
x 0, y 0 3. 2x y 10 2x 4y 28
N 12 x 14 y 40
Vertex 1 0 1 1 2 2 2 2 4 0
1 2
1 1 1 0 40 405 4 2 1 1 1 40 40375 2 4 2 1 2 1 2 40 415 2 4 1 4 1 0 40 42 2 4
Thus, the maximum value is 42 and the minimum value is 40375.
The objective function is P 140 x 3y. From the graph,
y
the vertices are 0 0, 5 0, 2 6, and 0 7. Vertex 0 0 5 0 2 6 0 7
2x+y=10
P 140 x 3y
140 0 3 0 140 140 5 3 0 135 140 2 3 6 156 140 0 3 7 161
Thus the maximum value is 161, and the minimum value is 135.
2x+4y=28 1 1
x
Linear Programming
x 0, y 0 x 10, y 20 4. xy5 x 2y 18
The objective function is Q 70x 82y. From the graph,
y
x+2y=18
the vertices are at 0 9, 0 5, 5 0, 10 0, and 10 4. Note that the restriction y 20 is irrelevant, superseded by x 2y 18 and x 0. Vertex 0 9 0 5 5 0 10 0 10 4
59
Q 70x 82y
x+y=5
1
70 0 82 9 738
x
1
70 0 82 5 410
x=10
70 5 82 0 350
70 10 82 0 700
70 10 82 4 1028
Thus, the maximum value of Q is 1028 and the minimum value is 350.
5. Let t be the number of tables made daily and c be the number of chairs made daily. Then the data given can be summarized by the following table: Tables t
Chairs c
Available time
2h
3h
108 h
Finishing
1h
1 h 2
20 h
Profit
$35
$20
Carpentry
2t 3c 108 Thus we wish to maximize the total profit P 35t 20c subject to the constraints t 12 c 20 t 0, c 0 From the graph, the vertices occur at 0 0, 20 0, 0 36, and 3 34.
y
Vertex 0 0 20 0 0 36 3 34
P 35t 20c
35 0 20 0
0
1
t+2 c=20
35 20 20 0 700
35 0 20 36 720 35 3 20 34 785
2t+3c=108 10 10
Hence, 3 tables and 34 chairs should be produced daily for a maximum profit of $785.
x
60
FOCUS ON MODELING
6. Let c be the number of colonial homes built and r the number of ranch homes built. Since there are 100 lots available, c r 100. From the capital restriction, we get 30,000c 40,000r 3,600,000, or 3c 4r 360. Thus, we wish to c 0, r 0 maximize the profit P 4000c 8000r subject to the constraints c r 100 3c 4r 360 From the graph, the vertices occur at 0 0, 100 0, 40 60, and 0 90.
y
Vertex 0 0 100 0 40 60 0 90
P 4000c 8000r
4000 0 8000 0
c+r=100
0
4000 100 8000 0 400,000
4000 40 8000 60 640,000
3c+4r=360
20
4000 0 8000 90 720,000
x
20
Therefore, the contractor should build 90 ranch style houses for a maximum profit of $720,000. Note that ten of the lots will be left vacant.
7. Let x be the number of crates of oranges and y the number of crates of grapefruit. Then the data given can be summarized by the following table: Oranges
Grapefruit
Available
Volume
4 ft3
6 ft3
300 ft3
Weight
80 lb
100 lb
5600 lb
Profit
$250
$400
In addition, x y. Thus we wish to maximize the total profit P 25x 4y subject to the constraints x 0, y 0, x y 4x 6y 300 80x 100y 5600 From the graph, the vertices occur at 0 0, 30 30, 45 20, and 70 0.
y
Vertex 0 0 30 30 45 20 70 0
P 25x 4y
25 0 4 0
80x+100y=5600
0
25 30 4 30 195
25 45 4 20 1925 25 70 4 0 175
x=y
4x+6y=300 10 10
x
Thus, the truck should carry 30 crates of oranges and 30 crates of grapefruit for a maximum profit of $195.
Linear Programming
8. Let x be the daily production of standard calculators and y the daily production of x 100, y 80 scientific calculators. Then the inequalities x 200, y 170 describe the x y 200
y
180 120 90
100 100 100 170 200 170 200 80
C 5x 7y
x+y=200
30
30 60 90 120 150 180 x
(b) Maximize the objective function P 2x 5y: Vertex
5 100 7 100 1200
100 100
5 200 7 170 2190
200 170
5 120 7 80 1160
120 80
5 100 7 170 1690
100 170
5 200 7 80 1560
200 80
120 80
y=80
60
200 170, 200 80, and 120 80.
Vertex
x=200
150
constraints. From the graph, the vertices occur at 100 100, 100 170,
(a) Minimize the objective function C 5x 7y:
x=100 y=170
61
P 2x 5y
2 100 5 100 300 2 100 5 170 650 2 200 5 170 450 2 200 5 80
0
2 120 5 80 160
So to minimize cost, they should produce
So to maximize profit, they should produce
120 standard and 80 scientific calculators.
100 standard and 170 scientific calculators.
9. Let x be the number of television sets shipped from Long Beach to Santa Monica and y the number of television sets shipped from Long Beach to El Toro. Thus, 15 x sets must be shipped to Santa Monica from Pasadena and 19 y sets
to El Toro from Pasadena. Thus, x 0, y 0, 15 x 0, 19 y 0, x y 24, and 15 x 19 y 18. x 0, y 0 x 15, y 19 Simplifying, we get the constraints x y 24 x y 16 The objective function is the cost C 5x 6y 4 15 x 55 19 y x 05y 1645, which we wish to
minimize. From the graph, the vertices occur at 0 16, 0 19, 5 19, 15 9, and 15 1. Vertex 0 16 0 19 5 19 15 9 15 1
C x 05y 1645
y
x=15
0 05 16 1645 1725 0 05 19 1645 174 5 05 19 1645 179 15 05 9 1645 184 15 05 1 1645 180
10 x+y=16 10
x+y=24 x
The minimum cost is $17250 and occurs when x 0 and y 16. Hence, 16 TVs should be shipped from Long Beach to
El Toro, 15 from Pasadena to Santa Monica, and 3 from Pasadena to El Toro.
62
FOCUS ON MODELING
10. Let x be the number of sheets shipped from the east side warehouse to customer A and y be the number of sheets shipped from the east side warehouse to customer B. Then 50 x sheets must be shipped to customer A from the west side
warehouse and 70 y sheets must be shipped to customer B from the west side warehouse. Thus, we obtain the constraints x 0, y 0 x 0, y 0 x 50, y 70 x 50, y 70 x y 80 x y 80 x y 75 50 x 70 y 45
The objective function is the cost C 05x 06y 04 50 x 055 70 y 01x 005y 585, which we wish to minimize. From the graph, the vertices occur at 5 70, 10 70, 50 30, and 50 25. y
Vertex 5 70 10 70 50 30 50 25
x=70
C 01x 005y 585
01 5 005 70 585 625
y=70
01 10 005 70 585 63
x+y=75
01 50 005 30 585 65
01 50 005 25 585 6475
x+y=80
10
x
10
Therefore, the minimum cost is $6250 and occurs when x 5 and y 70. So 5 sheets should be shipped from the east side warehouse to customer A, 70 sheets from the east side warehouse to customer B, and 45 sheets from the west side warehouse to customer A.
11. Let x be the number of bags of standard mixtures and y be the number of bags of deluxe mixtures. Then the data can be summarized by the following table: Standard
Deluxe
Available
Cashews
100 g
150 g
15 kg
Peanuts
200 g
50 g
20 kg
Selling price
$195
$220
Thus the total revenue, which we want to maximize, is given by R 195x 225yWe have the constraints x 0, y 0, x y x 0, y 0, x y 01x 015y 15 10x 15y 1500 02x 005y 20 20x 5y 2000
From the graph, the vertices occur at 0 0, 60 60, 90 40, and 100 0.
y
Vertex 0 0 60 60 90 40 100 0
R 195x 225y
196 0 225 0 0
10x+15y=1500
195 60 225 60 252
195 90 225 40 2655 195 100 225 0 195
x=y 20x+5y=2000
10 10
x
Hence, the confectioner should pack 90 bags of standard and 40 bags of deluxe mixture for a maximum revenue of $26550.
Linear Programming
63
12. Let x be the quantity of type I food and y the quantity of type II food, in ounces. Then the data can be summarized by the following table: Type I
Type II
Required
Fat
8g
12 g
24 g
Carbohydrate
12 g
12 g
36 g
Protein
2g
1g
4g
Cost
$020
$030
x 0, y 0, Also, the total amount of food must be no more than 5 oz. Thus, the constraints are x y 5, 8x 12y 24 12x 12y 36, 2x y 4
The objective function is the cost C 02x 03y, which we wish to minimize. From the graph, the vertices occur at 1 2, 0 4, 0 5, 5 0, and 3 0. Vertex 1 2 0 4 0 5 5 0 3 0
y
C 02x 03y
x+y=5
02 1 03 2 08 02 0 03 4 12 02 0 03 5 15
2x+y=4 12x+12y=36
02 5 03 0 10 02 3 03 0 06
8x+12y=24 1 x
1
Hence, the rabbits should be fed 3 oz of type I food and no type II food, for a minimum cost of $060.
13. Let x be the amount in municipal bonds and y the amount in bank certificates, both in dollars. Then 12000 x y is the amount in high-risk bonds. So our constraints can be stated as x 0, y 0, x 3y x 0, y 0, x 3y x y 12,000 12,000 x y 0 12,000 x y 2000 x y 10,000
From the graph, the vertices occur at 7500 2500, 10000 0, 12000 0, and 9000 3000. The objective function is P 007x 008y 012 12000 x y 1440 005x 004y, which we wish to maximize. y
Vertex 7500 2500 10000 0 12000 0 9000 3000
P 1440 005x 004y
x+y=12,000
1440 005 7500 004 2500 965
1440 005 10,000 004 0 940 1440 005 12,000 004 0 840
1440 005 9000 004 3000 870
x+y=10,000
x=3y
2000 2000
x
Hence, the advisor should invest $7500 in municipal bonds, $2500 in bank certificates, and the remaining $2000 in high-risk bonds for a maximum yield of $965.
64
FOCUS ON MODELING
14. The only change that needs to be made to the constraints in Exercise 13 is that the 2000 in the last inequality becomes 3000. Then we have x 0, y 0, x 3y x 0, y 0, x 3y x y 12,000 12,000 x y 0 12,000 x y 3000 x y 9000
From the graph, the vertices occur at 6750 2250, 9000 0, 12000 0, and 9000 3000. The objective function is Y 007x 008y 012 12000 x y 1440 005x 004y, which we wish to maximize. y
Vertex 6750 2250
Y 1440 005x 004y
1440 005 9000 004 0 990
9000 0 12000 0 9000 3000
x+y=12,000
1440 005 6750 004 2250 10125 1440 005 12,000 004 0 840
1440 005 9000 004 3000 870
x=3y
x+y=9000 2000
x
2000
Hence, she should invest $6750 in municipal bonds, $2250 in bank certificates, and $3000 in high-risk bonds for a maximum yield of $101250, which is an increase of $4750, over her yield in Exercise 13.
15. Let g be the number of games published and e be the number of educational programs published. Then the number of utility programs published is 36 g e. Hence we wish to maximize profit, P 5000g 8000e 6000 36 g e 216,000 1000g 2000e, subject to the constraints g 4, e 0 g 4, e 0 36 g e 0 g e 36 36 g e 2e g 3e 36.
From the graph, the vertices are at 4 32 3 , 4 32, and 36 0. The objective function is P 216,000 1000g 2000e. e
Vertex 4 32 3
P 216,000 1000g 2000e 216,000 1000 4 2000 32 3 233,33333
36 0
216,000 1000 36 2000 0 180,000
4 32
216,000 1000 4 2000 32 276,000
g=4 g+e=36 g+3e=36
10 10
g
So, they should publish 4 games, 32 educational programs, and no utility program for a maximum profit of $276,000 annually.
Linear Programming
4y 3x 5 4x y 15 x 3y 7
16.
The minimum value of P for which the line P x y intersects the feasible region is P 3, and the maximum value is P 10. These are the extreme values of P on the feasible region because the range of P on that region is
[3 10]. The minimum value occurs at 1 2 and the maximum value occurs at 5 5.
P=5 y 5 P=4 4 P=3 3 P=2 2 P=1 1 0
P=11 P=10 P=9 P=8 P=7 P=6 1
2
3
4
5
x
65
CORRECTIONS p. 19
CHAPTER 10
MATRICES AND DETERMINANTS
10.1
Matrices and Systems of Linear Equations 1
10.2
The Algebra of Matrices 12
10.3
Inverses of Matrices and Matrix Equations 20
10.4
Determinants and Cramer’s Rule 29 Chapter 10 Review 43 Chapter 10 Test 53
¥
FOCUS ON MODELING: Computer Graphics 56
1
10 MATRICES AND DETERMINANTS 10.1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS 1. A system of linear equations with infinitely many solutions is called dependent. A system of linear equations with no solution is called inconsistent. 1 1 1 1 x y z 1 2. x 2z 3 1 0 2 3 2y z 3 0 2 1 3
3. (a) The leading variables are x and y. (b) The system is dependent.
(c) The solution of the system is x 3 t y 5 2t z t. x 2 4. (a) y1 0 1 0 1 z3 0 0 1 3
1 0 0 2
1 0 1 2
The solution is 2 1 3.
x z2 The solution is z t, y 1 t, x 2 t. (b) 0 1 1 1 yz1 0 0 0 0 1 0 0 2 2 x This system is inconsistent and has no solution. (c) y 1 0 1 0 1 03 0 0 0 3
5. 3 2
3
6. 2 4
1 1 2
11. 2 1 1
7. 2 1
0 1
0 1 3
13. (a) Yes, this matrix is in rowechelon form. (b) Yes, this matrix is in reduced rowechelon form. x 3 (c) y 5
8. 3 1
12.
1 0
9. 1 3
0
1 1
3 2
1 1
3
10. 2 2
7 3
14. (a) Yes, this matrix is in rowechelon form. (b) No, this matrix not in reduced rowechelon form. The entry above the leading 1 in the second row is not 0. x 3y 3 (c) y 5
1
2
CHAPTER 10 Matrices and Determinants
16. (a) Yes, the matrix is in rowechelon form.
15. (a) Yes, this matrix is in rowechelon form.
(b) Yes, the matrix is in reduced rowechelon form. 7z 0 x (c) y 3z 0 01
(b) No, this matrix is not in reduced rowechelon form, since the leading 1 in the second row does not have a zero above it. x 2y 8z 0 (c) y 3z 2 00 17. (a) No, this matrix is not in rowechelon form, since the row of zeros is not at the bottom.
18. (a) Yes, the matrix is in rowechelon form. (b) Yes, the matrix is in reduced rowechelon form. 1 x (c) y 2 z3
(b) No, this matrix is not in reduced rowechelon form. 0 x (c) 00 y 5z 1 19. (a) Yes, this matrix is in rowechelon form.
20. (a) No, this matrix is not in rowechelon form, since the fourth column has the leading 1 of two rows.
(b) Yes, this matrix is in reduced rowechelon form. x 3y 0 z 2 0 (c) 01 00
(b) No, this matrix is not in reduced rowechelon form. x 3y 0 y 4 0 (c) u2 0
Notice that this system has no solution.
21.
1 3
1
2
1
1
4
5
2 3
3
1
3
8
2
1 3
5
6 5 1
7
23. 2
3R1 R2 R2
1 2 1 1
22. 10 3
0
3
1 3
24. 0 0
1
1 20
1
1 13
2 1
1 1
2 1
1
2R2 R3 R3
0
1
2
4
7
0
4
1 2 1 1
2R1 R2 R2
3R1 R3 R3
1
5 2 3
3
1 3
8
0 1 5 14
2
1 3
1
2 3 0 8
5
1 13 8 8
2 1
0 0
1 3
1
1 1
0 3
3
3
SECTION 10.1 Matrices and Systems of Linear Equations
x 2y 4z 3 25. (a) y 2z 7 z2
x y 3z 8 26. (a) y 3z 5 z 1
(b) y 2 2 7 y 3, so x 2 3 4 2 3
(b) y 3 1 5 y 2, so x 2 3 1 8
x 1. The solution is 1 3 2.
x 3. The solution is 3 2 1.
x 2y 3z 7 y 2z 5 27. (a) z 2 5 3
x 2z 2 5 y 3z 1 28. (a) z 0 1
(b) z 2 3 5 z 1, so y 2 1 5
(b) z 1 0 z 1, so y 3 1 1
y 3 and x 2 3 3 1 3 7 x 7. The
y 2 and x 2 1 2 1 5 x 5. The
solution is 5 2 1 1.
solution is 7 3 1 3.
1 2 1 1
29. 0
1 2 5
1
R3 R1 R3
1 3 8
1 2 1 1
0 0
1 2 5
1 1
6 3
0 0
R3 3R2 R3
3 2 7
1 2
1
1
1
5 . Thus, 4z 8
2
0 4 8
z 2; y 2 2 5 y 1; and x 2 1 2 1 x 1. Therefore, the solution is 1 1 2.
1 1 6 3
30. 1 1 3 3
R2 R1 R2
1 2 4 7
13 R2
0 0 3 0 1 2 4 7
R3 R1 R3
1 1
6 3
0 0 1 0 0 1 2 4
1 1
6 3
0 1 2 4 . 0 0 1 0
R3 R2
Thus, z 0; y 2z y 0 y 4; and x y 6z 3 x 4 0 3 x 1. Therefore, the solution is 1 4 0.
31.
1 2 1
3
1 2
1
4
2 3 1 5
1 2
1
1
0 7 1 1 0 7 3 3
R2 3R1 R2 R3 2R1 R3
R3 R2 R3
1 2
0 0
1
1
1 .
7 1 0
2 2
Thus, 2z 2 z 1; 7y 1 1 y 0; and x 2 0 1 1 x 2. Therefore, the solution is 2 0 1.
1
0
1
7
5
3 1
2
1 1 2
0
32. 2 1 2 5
1
0
1
7
0 1 4 19 0 3 6 33
R2 2R1 R2 R3 5R1 R3
R2 R2 R3 3R2 R3
1 0
1
7
0 1 4 19 . 0 0 18 90
Thus, 18z 90 z 5; y 4 5 19 y 1; and x 5 7 x 2. The solution is 2 1 5.
33. 2 3
3
1
5
1 8 16
R1 15 R2 R1 6R2 R3 R2
1 30 R3
R2 2R1 R2 R3 3R1 R3
1 0 1 1 0 30 0 90 0 0 1 2
1 1 2
0 0
5
5
0
5
4 2 16
R1 R3 R1
1 30 R2
4R2 5R3 R3
1 1 2
0 0
5
0
5 5
0 30 60
1 0 0 1 0 1 0 3 . Therefore, the solution is 1 3 2. 0 0 1 2
4
CHAPTER 10 Matrices and Determinants
1
34. 1
0 2
1 3
1 3
R2 R1 R2 2R1 R3 R3
1 0 1 3
0 2 0 0 0 3 4 4
1 2 R2
2R3 3R2 R3
2 3 2 2 1 0 0 2 0 1 0 0 . The solution is 2 0 1. 0 0 1 1 1 2 1 9 1 2 1 9 R2 2R1 R2 0 4 1 20 35. 2 0 1 2 R3 3R1 R3 0 1 5 5 3 5 2 22
1 0 1 3
0 1 0 0 0 0 8 8
4R3 R2 R3
1
R1 18 R3 R1
2 1
1R 8 3
9
0 4 1 20 . 0 0 19 0
Thus, 19x3 0 x3 0; 4x2 20 x2 5; and x1 2 5 9 x1 1. Therefore, the solution is 1 5 0. 2 1 0 7 2 1 0 7 2 1 0 7 R2 R1 R2 R3 7R2 R3 0 2 1 1 2 0 2 1 1 . Then, 9x3 9 36. 2 1 1 6 2 R3 3R1 R3 3 2 4 11 0 7 8 1 0 0 9 9 x3 1; 2x2 1 1 x2 1; and 2x1 1 7 x1 3. Therefore, the solution is x1 x2 x3 3 1 1. 2 3 1 13 2 3 1 13 2 3 1 13 2R2 R1 R2 3 13R2 R3 0 1 11 25 R 0 1 11 37. 25 1 2 5 6 2R3 . 5R1 R3 5 1 1 49 0 13 3 33 0 0 146 292
Thus, 146z 292 z 2; y 11 2 25 y 3; and 2x 3 3 2 13 x 10. Therefore, the solution is 10 3 2. 10 10 20 60 10 10 20 60 10 10 20 60 2R2 3R1 R2 3 7R2 R3 0 10 120 230 R 38. 15 20 30 25 2R3 0 10 120 230 . R1 R3 5 30 10 45 0 70 40 150 0 0 880 1760
Thus, 880z 1760 z 2; 10y 120 2 230 y 1; and 10x 10 40 60 x 1. Therefore, the solution is 1 1 2. 1 2 1 3 1 2 1 3 1 0 11 11 R2 3R1 R2 1 2R2 R1 0 1 5 4 R 0 1 39. 5 4 3 7 2 5 R3 . The third row of the 2R1 R3 R3 R2 R3 2 3 7 4 0 1 5 2 0 0 0 6
matrix states 0 6, which is impossible. Hence, the system is inconsistent, and there is no solution. 1 2 3 4 1 2 3 4 3 0 1 3 R2 3R1 R2 R1 R2 R1 0 6 10 15 3 0 6 10 15 . 40. 3 0 1 3 R3 R1 R3 R3 R2 R3 1 4 7 2 0 6 10 2 0 0 0 13
The system is inconsistent, and there is no solution. 2 3 9 5 1 0 3 2 1 0 3 2 R R R2 2R1 R2 1 R2 1 2 3 0 3 15 9 41. 1 0 3 2 2 3 9 5 R3 3R1 R3 3 1 4 3 3 1 4 3 0 1 5 3 1 0 3 2 1 0 3 2 3 R2 R3 0 1 5 3 R 0 1 5 3 . Therefore, this system has infinitely many solutions, given by x 3t 2 0 1 5 3 0 0 0 0 x 2 3t, and y 5t 3 y 3 5t. Hence, the solutions are 2 3t 3 5t t, where t is any real number.
5
SECTION 10.1 Matrices and Systems of Linear Equations
1 42. 2
5
2
3
1
6 11
3 16
R2 2R1 R2 R3 3R1 R3
20 26
1 2 5 3 0 2 1 7 0 10 5 35
R3 5R2 R3
1 2
5 3 2 1 7 .
0
0
0
0 0
The system is dependent; there are infinitely many solutions, given by 2y t 7 2y t 7 y 12 t 72 ; and x 2 12 t 72 5t 3 x t 7 5t 3 x 4t 10. The solutions are 4t 10 12 t 72 t , where t is any real number.
1 1
3
3
43. 4 8 32 24 2 3 11
3
3
14 R2
0 4 20 12 0 1 5 2
R2 4R1 R2 R3 2R1 R3
4
1 1
1 1
0 0
R3 R2 R3
3
3
1 5 3 . The third row of 0
0 5
the matrix states 0 5, which is impossible. Hence, the system is inconsistent, and there is no solution.
44.
6 2 12
2
1 3 3
1
1 R1 2 10
2
2
6
1 3 1
6
1 3 1
R2 R1 R2 1 3 2 10 0
1
3 2
6
R3 R1 R3
0
0 1
6
1 3 1 6
R 3R R 3 2 3 4 0
0 3 12
0
0 1 4 . 0 0 0
The system is dependent; there are infinitely many solutions, given by z 4, and x 3y z 6, so x 3t 4 6 x 3t 2. The solutions are 3t 2 t 4, where t is any real number.
1
4 2 3
45. 2 1 8
5 12 5 11 30
R2 2R1 R2 R3 8R1 R3
1
4 2 3
0 9 9 18 0 27 27 54
R3 3R2 R3
1
4 2 3
0 9 0 0
9 18 . 0 0
Therefore, this system has infinitely many solutions, given by 9y 9t 18 y 2 t, and x 4 2 t 2t 3 x 5 2t. Hence, the solutions are 5 2t 2 t t, where t is any real number.
3
2 3 10
46. 1 1 1 5 1
4 1 20
R1 R2
1 1 1 5
3 1
2 3 10 4 1 20
R2 3R1 R2 R3 R1 R3
1 1 1 5
0 0
5 5
0 25 0 25
R3 R2 R3
1 1 1 5 0 5 0 25 . The system is dependent; there are infinitely many solutions, given by 5s 25 s 5, and 0 0 0 0
r 5 t 5 r t. Hence, the solutions are t 5 t, where t is any real number.
2
1 2 12
3
3 3 2
1 47. 1 2
1 6 18
R1 R2 R1
1 12 1
6
2 1 2 12 3 32 3 18
R2 2R1 R2 R3 3R1 R3
1 12 1 6
0 0 0 0
0 0 . 0 0
Therefore, this system has infinitely many solutions, given by x 12 s t 6 x 6 12 s t. Hence, the solutions are 6 12 s t s t , where s and t are any real numbers.
0 1 5 7 R R 2 1 48. 3 2 0 12 3 0 10 80
3 2 0 12 3 R1 R3 0 1 5 7 R 3 0 10 80
Therefore, the system is inconsistent and has no solution.
3 2 0 12 3 2R2 R3 0 1 5 7 R 0 2 10 68
3 2 0 12 0 1 5 7 . 0 0 0 82
6
CHAPTER 10 Matrices and Determinants
4 3
49. 2
1 3 4
1 1
1 8
2
1 1
2
3
1 1
2
3
R 2 2R1 R2 2 2 1 3 4 R 0 1 1 2 R3 4R1 R3 4 3 1 8 0 1 7 20 1 1 2 3 R3 R2 R3 . Therefore, 6z 18 z 3; y 3 2 0 1 1 2 0 0 6 18 R1 R3
3
1 1 2 3 0 1 1 2 0 1 7 20
y 1; and x 1 2 3 3 x 2. Hence, the solution is 2 1 3.
2 3
5
14
1
1 1
3
1
1 1 3
R2 4R1 R2 R1 R3 50. 4 1 2 17 4 1 2 17 0 5 1 1
1
3 1
1 3
R1
3
2 3
5
14
R3 2R1 R3
1
1 1 3
0
0
R R R 3 2 3 2 5 0 5
0 5
7 20
2 5 . 5 25
Thus 5z 25 z 5; 5y 2 5 5 5y 15 y 3; and x 3 5 3 x 1. Hence, the solution is 1 3 5.
51. 1
2 4
0
2
0 0
3R2 R1 R2 R3 2R2 R3
1 11 1
3 1
1
3
9
1 7
0 0
R2 R3 R3
1 7 7
Therefore, the system is inconsistent and there is no solution.
52.
1 3
2
5
2 3 2 2 0
1
4
7
1 3
2
5
0 3 6 12 0 3 6 12
R2 2R1 R2 R3 R1 R3
3 1
1 3 R2
R2 R3 R3
1
3
9 .
1 7 0
0 16
1 3
2
5
0
0
0
0 0
1 2 4 . Therefore,
the system is dependent. Let z t. Then y 2t 4 y 4 2t and x 3 4 2t 2t 5 x 3 4 2t 2t 5 4t 7. The solutions are 4t 7 4 2t t, where t is any real number.
1
2 3
53. 2 4 6 3
5
10
7 2 13
1 2
3 5
0 0 12 0 1 7
R2 2R1 R2 R3 3R1 R3
0
2
1 2
3 5
0 1 7 0 0 12
R2 R3
2 . Therefore, 0
12z 0 z 0; y 7 0 2 y 2; and x 2 2 3 0 5 x 9. Hence, the solution is 9 2 0.
3 1 0 2
54. 4 3 1 4
3R2 4R1 R2 2R3 R2 R3
2 5 1 0
3
1 0
2
0 13 3 20 0 13 3 4
R2 R3 R3
Therefore, the system is inconsistent and there is no solution.
1 1
55. 1 1
0
6 1
8
5
3 14 4
R2 R1 R2 R3 R1 R3
1 1
0 0
1
6 5
8
3
4 20 12
3
1 0
2
1 1
6
8
0
0
0
0 13 3 20 . 0 0 0 16
R3 4R2 R3
0 0
1 5 3 . Therefore,
the system is dependent. Let z t. Then y 5t 3 y 35t and x y 6t 8 x 83 5t6t 5t. The solutions are 5 t 3 5t t, where t is any real number.
7
SECTION 10.1 Matrices and Systems of Linear Equations
56.
3 1
4 2 1
2 1 1 7
3 2 1
1 3 2 1 0 2 1 1 0 10 7 11
1 3 2 1 1 3 2 1 R2 1 R3 2 4R1 R2 4 2 1 7 R 4 0 10 7 11 R3 3R1 R3 3 1 2 1 0 8 4 4 1 3 2 1 R3 5R2 R3 0 2 1 1 . Thus 2z 6 z 3; 2y 3 1 2y 2 0 0 2 6 R1 R3
R1
y 1; and x 3 1 2 3 1 x 2. Hence, the solution is 2 1 3. 1 2 1 3 3 1 2 1 3 3 1 2 1 3 3 R2 3R1 R2 3 4 1 1 9 0 2 4 8 18 3 4 1 1 9 R1 R3 R1 R3 57. 0 3 0 4 3 1 1 1 1 0 1 1 1 1 0 R4 2R1 R4 2 1 4 2 3 0 5 6 8 9 2 1 4 2 3 1 2 1 3 3 1 2 1 3 3 0 1 2 4 9 R 3R R 0 1 2 4 1 9 323 3R4 2R3 R4 2 R2 0 3 0 4 3 R4 5R2 R4 0 0 6 8 24 0 5 6 8 9 0 0 4 12 36 1 2 1 3 3 0 1 2 4 9 . Therefore, 20 60 3; 6z 24 24 z 0. Then y 12 9 y 3 and 0 0 6 8 24 0 0 0 20 60
x 6 9 3 x 0. Hence, the solution is 0 3 0 3. 1 1 1 1 6 1 1 1 1 6 R2 R4 R2 R2 2R1 R2 2 0 1 3 0 2 3 1 4 8 R3 R4 R3 R3 R1 R3 58. 1 1 1 0 4 10 0 2 1 5 16 R4 3R1 R4 2 R4 3 5 1 1 20 0 2 2 2 2 1 1 1 1 6 1 1 1 1 6 0 1 1 1 0 1 1 1 1 1 R4 R2 5 R 3 R R 4 3 4 . R3 R4 0 0 5 1 2 0 0 5 1 2 0 0 3 7 14 0 0 0 32 64
1 1 1 1
0 0 0 0 0 1
5 3
1
2 7 14 1 1 1
Thus 32 64 2; 5z 2 2 5z 0 z 0; y 0 2 1 y 3; and x 3 0 2 6 x 1. Hence the solution is 1 3 0 2. 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 0 3 1 2 2 0 3 1 2 2 0 3 1 2 2 R3 R1 R3 R R2 R4 4 59. 1 1 0 3 2 R4 3R1 R4 0 0 2 4 4 0 0 2 4 4 0 3 7 1 1 0 0 6 3 3 3 0 1 2 5 1 1 2 1 2 0 3 1 2 2 R4 3R3 R4 . Therefore, 9 9 1; 2z 4 1 4 z 0. Then 0 0 2 4 4 0 0 0 9 9 3y 0 2 1 2 y 0and x 0 2 0 1 2 x 1. Hence, the solution is 1 0 0 1.
6
8
CHAPTER 10 Matrices and Determinants
1 3 2
1
2
1 2 0 2 10 60. 0 0 1 5 15 3
0 2
1
R4 14R3 R4
1 3
0 0
R2 R1 R2 R4 3R1 R4
2
1 2
1 2 3 8 0 1 5 15 9 4 2 3
R4 9R2 R4
1 3
0 0
2
1 2
1 2 3 8 0 1 5 15 0 14 25 75
3 0 0 1 3 2 1 2 0 1 2 3 8 . Thus, 45 135 3; z 5 3 15 z 0; 0 0 1 5 15 0 0 0 45 135
y 2 0 3 3 8 y 1; and x 3 1 2 0 3 2 x 2. Hence, the solution is 2 1 0 3.
1 1
61. 3
0 1 0
0 1 2 0
1 4
R2 3R1 R2 R3 R1 R3
1 2 0
1 1
0
1 0
0 3 1 1 0 0 3 1 1 0
R3 R2 R3
1 1
0
1 0
0
0
0 0
0 0
3 1 1 0 .
Therefore, the system has infinitely many solutions, given by 3y s t 0 y 13 s t and x 13 s t t 0 x 13 s 2t. So the solutions are 13 s 2t 13 s t s t , where s and t are any real numbers.
2 1 2 1 5 1 1 4 1 3 1 1 4 1 3 R1 R2 2 2R1 R2 2 1 2 1 5 R 0 1 10 1 11 62. 1 1 4 1 3 R1 R3 3R1 R3 3 2 1 0 0 3 2 1 0 0 0 1 11 3 9 1 1 4 1 3 R3 R2 R3 0 1 10 1 11 . Thus, the system has infinitely many solutions, given by z 2t 2 0 0 1 2 2
z 2 2t; y 10 2 2t t 11 y 31 19t; and x 31 19t 4 2 2t t 3 x 20 12t. Hence, the solutions are 20 12t 31 19t 2 2t t, where t is any real number.
1
0
1 1
4
0 1 1 0 4 63. 1 2 3 1 12 2
R3 R1 R3
R4 2R1 R4
1
0
1 1
4
0 1 1 0 4 0 2 2 0 8
R3 2R2 R3
1 0
1 1
4
0 1 1 0 4 0 0 0 0 0
0 2 5 1 0 0 4 3 9 0 0 4 3 9 1 0 1 1 4 0 1 1 0 4 R3 R4 . Therefore, 4z 3t 9 4z 9 3t z 94 34 t. Then we have 0 0 4 3 9 0 0 0 0 0 3 9 3 7 7 y 94 34 t 4 y 7 4 4 t and x 4 4 t t 4 x 4 4 t. Hence, the solutions are 7 7 t 7 3 t 9 3 t t , where t is any real number. 4 4 4 4 4 4
SECTION 10.1 Matrices and Systems of Linear Equations
9
0 0 1 1 2 0 1 0 0 2 6 R2 R3 R2 1 2 0 3 12 0 2 0 5 18 0 R R R4 R3 R4 1 3 1 R2 R1 R2 1 0 0 2 0 1 1 2 2 0 0 4 12 6 0 R 3 2 2 0 2 5 6 0 0 2 9 18 0 0 2 9 18 1 0 0 2 6 1 0 0 2 6 0 1 1 2 0 1 1 2 0 0 R2 2R3 R2 R R R 4 3 4 . R2 R3 0 0 2 9 18 0 0 2 9 18 0 0 2 9 18 0 0 0 0 0 Thus, the system has infinitely many solutions given by 2z 9t 18 z 92 t 9; y 92 t 9 2t 0 y 52 t 9; and x 2t 6 x 2t 6. So the solutions are 2t 6 52 t 9 92 t 9 t , where t is any real number.
64.
0 1 1 2
3 2
0 1
075 375 295 40875
65. Using a graphing device to find the reduced rowechelon form of 095 875 0 is x 125, y 025, z 075.
3375 , we find that the solution 125 015 275 36625
131 272 371 139534
66. Using a graphing device to find the reduced rowechelon form of 021 0 0
solution is x 371, y 172, z 381.
0
373
134322 , we find that the
234 456 213984
42 31
6 67. Using a graphing device to find the reduced rowechelon form of 35 solution is x 12, y 34, z 52, 13.
0 42
04
45 , we find that the 0 67 32 3488 31 48 52 766 0
49 27 52
0
9
0 145
0 27 0 43 1187 68. Using a graphing device to find the reduced rowechelon form of , we find that the solution 0 31 42 0 721 73 54 0 0 1327 is x 13, y 07, z 12, 32.
69. Let x, y, z represent the number of VitaMax, Vitron, and VitaPlus pills taken daily. The matrix representation for the system of equations is 1 5 10 15 50 1 2 3 10 1 2 3 10 1 2 3 10 5 R1 1 2 2 3R1 R2 3 R2 R3 15 20 0 50 5R 3 4 0 10 R 0 2 9 20 R 1 0 2 9 20 . R3 2R1 R3 5 R3 10 10 10 50 2 2 2 10 0 2 4 10 0 0 5 10 Thus, 5z 10 z 2; 2y 18 20 y 1; and x 2 6 10 x 2. Hence, the patient should take 2 VitaMax, 1 Vitron, and 2 VitaPlus pills daily.
10
CHAPTER 10 Matrices and Determinants
70. Let x be the quantity, in mL, of 10% acid, y the quantity of 20% acid, and z the quantity of 40% acid. Then 1 1 1 100 10R2 01x 02y 04z 18 R2 R1 R2 Writing this equation in matrix form, we get x y z 100 01 02 04 18 R 3 R1 R3 x 4z 0 1 0 4 0
1
1
1
100
0 1 3 80 0 1 5 100
1 1 1 100
0 1 3 0 0 2
R3 R2 R3 R3
80 . Thus 2z 20 z 10; y 30 80 y 50; and 20
x 50 10 100 x 40. So the chemist should mix together 40 mL of 10% acid, 50 mL of 20% acid, and 10 mL of 40% acid.
71. Let x, y, and z represent the distance, in miles, of the run, swim, and cycle parts of the race respectively. Then, since distance time , we get the following equations from the three contestants’ race times: speed y x z 10 4 20 25 2x 5y z 50 x y z 3 4x 5y 2z 90 which has the following matrix representation: 75 6 15 x y z 175 8x 40y 3z 210
15
2
3
5 1
50
4 5 2 90 8 40 3 210
40
R2 2R1 R2 R3 4R1 R3
2
5
1
50
0 5 0 10 0 20 1 10
2
5
1
50
0 5 0 10 . 0 0 1 30
R3 4R2 R3
Thus, z 30 z 30; 5y 10 y 2; and 2x 10 30 50 x 5. So the race consists of a 5mile run, a 2mile swim, and a 30mile cycle.
72. Let a, b, and c be the number of students in classrooms A, B, and C, respectively, where a, b, c 0. Then, the a b c 100 a b c 100 1a 1b By substitution, a 52 a 32 a 100 a 20; system of equations is b 52 a 2 5 c 3b 3a 1b 1c 5 3 5 2 5 3 b 2 20 50; and c 2 20 30. So there are 20 students in classroom A, 50 students in classroom B, and
30 students in classroom C.
73. Let t be the number of tables produced, c the number of chairs, and a the number of armoires. Then, the system of equations 1 t 2c 2a 600 2 t c a 300 1 3 and a matrix representation is is 2 t 2 c a 400 t 3c 2a 800 2t 3c 4a 1180 t 3 c 2a 590
2
1 2 2
600
1 3 2 800 2 3 4 1180
R2 R1 R2 R3 2R1 R3
1
2 2 600
0 1 0 200 0 1 0 20
R3 R2 R3
1 2 2 600
0 1 0 200 . 0 0 0 180
The third row states 0 180, which is impossible, and so the system is inconsistent. Therefore, it is impossible to use all of the available laborhours.
11
SECTION 10.1 Matrices and Systems of Linear Equations
74. The number of cars entering each intersection must equal the number of cars leaving that intersection. This leads 200 180 x z x 70 20 to the following equations: Simplifying and writing this in matrix form, we get 200 y 30 y z 400 200 1 0 1 0 380 1 0 1 0 380 1 0 1 0 380 1 0 0 1 50 R R R 0 0 1 1 430 0 1 0 1 170 2 1 2 R R 2 3 . 0 1 0 1 170 R4 R3 R4 0 1 0 1 170 0 0 1 1 430 0 1 1 0 600 0 0 1 1 430 0 0 1 1 430
Therefore, z t 430 z 430 t; y t 170 y 170 t; and x 430 t 380 x t 50. Since x, y, z, 0, it follows that 50 t 430, and so the solutions are t 50 170 t 430 t t, where 50 t 430.
75. Line containing the points 0 0 and 1 12: Using the general form of a line, y ax b, we substitute for x and y and solve for a and b. The point 0 0 gives 0 a 0 b b 0; the point 1 12 gives 12 a 1 b a 12. Since a 12 and b 0, the equation of the line is y 12x. Quadratic containing the points 0 0, 1 12, and 3 6: Using the general form of a quadratic, y ax 2 bx c, we substitute for x and y and solve for a, b, and c. The point 0 0 gives 0 a 02 b 0 c c 0; the point 1 12
gives 12 a 12 b 1 c a b 12; the point 3 6 gives 6 a 32 b 3 c 9a 3b 6. Subtracting the third equation from 3 times the third gives 6a 30 a 5. So a b 12 b 12 a b 17. Since a 5, b 17, and c 0, the equation of the quadratic is y 5x 2 17x.
Cubic containing the points 0 0, 1 12, 2 40, and 3 6: Using the general form of a cubic, y ax 3 bx 2 cx d,
we substitute for x and y and solve for a, b, c, and d. The point 0 0 gives 0 a 03 b 02 c 0 d d 0; the point the point 1 12 gives 12 a 13 b 12 c 1 d a b c d 12; the point 2 40 gives
40 a 23 b 22 c 2 d 8a 4b 2c d 40; the point 3 6 gives 6 a 33 b 32 c 3 d a b c 12 27a 9b 3c d 6. Since d 0, the system reduces to 8a 4b 2c 40 which has representation 27a 9b 3c 6
1 1 1 12
1
1
1
12
12 R2
1 1 1 12
1 1
1 12
R 3R R 2 8R1 R2 3 2 3 8 4 2 40 R 0 4 6 56 0 2 3 28 2 3 28 0 2 . R3 1 27R1 R3 6 R3 0 18 24 318 0 3 4 53 0 0 1 22 27 9 3 6
So c 22 and backsubstituting we have 2b 3 22 28 b 47 and a 47 22 0 a 13. So the
cubic is y 13x 3 47x 2 22x. Fourthdegree polynomial containing the points 0 0, 1 12, 2 40, 3 6, and 1 14: Using the general form of a
fourthdegree polynomial, y ax 4 bx 3 cx 2 dx e, we substitute for x and y and solve for a, b, c, d, and e. The point
0 0 gives 0 a 04 b 03 c 02 d 0e e 0; the point 1 12 gives 12 a 14 b 13 c 12 d 1e;
the point 2 40 gives 40 a 24 b 23 c 22 d 2e; the point 3 6 gives 6 a 34 b 33 c 32 d 3e; the point 1 14 gives 14 a 14 b 13 c 12 d 1 e.
12
CHAPTER 10 Matrices and Determinants
Because the first equation is e 0, we eliminate e from the other equations to get a b c d 12 1 1 1 1 12 1 1 1 1 12 1 16a 8b 4c 2d 40 16 8 4 2 40 R2 16R1 R2 0 8 12 14 152 2 R4 R2 R3 81R1 R3 R2 R3 R R R 6 6 4 81a 27b 9c 3d 1 4 81 27 9 3 0 54 72 78 966 R3 R4 a b c d 14 1 1 1 1 14 0 2 0 2 26 1 1 1 1 12 1 1 1 1 12 1 1 1 1 12 0 0 1 1 0 1 13 0 1 13 0 1 13 3 8R2 R3 0 1 R R 6 R R 434 . 0 8 12 14 152 R4 54R2 R4 0 0 12 6 48 0 0 12 6 48 0 54 72 78 966 0 0 72 24 264 0 0 0 12 24
So d 2. Then 12c 6 2 48 c 3and b 2 13 b 11.
Finally, a 11 3 2 12 a 4. So the fourthdegree polynomial
40
containing these points is y 4x 4 11x 3 3x 2 2x.
20
2
2
4
20
10.2 THE ALGEBRA OF MATRICES 1. We can add (or subtract) two matrices only if they have the same dimension. 2. (a) We can multiply two matrices only if the number of columns in the first matrix is the same as the number of rows in the second matrix. (b) If A is a 3 3 matrix and B is a 4 3 matrix, then (ii) B A and (iii) A A are defined, but (i) AB and (iv) B B are not.
3. (i) A A and (ii) 2A exist for all matrices A, but (iii) A A is not defined when A is not square.
4. The entry in the first row, second column is a12 3 3 1 2 2 1 9; the entry in the second row, third column is a23 1 2 2 1 0 0 0; the entry in the third row, first column is a31 1 1 3 3 2 2 4, and 3 1 2 1 3 2 4 9 7 so on: 1 2 0 3 2 1 7 7 0 1 3 2 2 1 0 4 5 5 5. The matrices have different dimensions, so they cannot be equal. 6. Because 14 025, ln 1 0, 2 4, and 3 62 , the corresponding entries are equal, so the matrices are equal.
7. All corresponding entries must be equal, so a 5 and b 3.
8. All corresponding entries must be equal, so a 5 and b 7. 2 6 1 3 1 3 0 1 1 2 1 1 2 0 2 9. 10. 5 3 6 2 1 5 1 1 0 1 3 2 0 2 2 1 1 0 1 1 1 2 3 6 11. 3 12. 2 1 0 1 2 1 is undefined because these 4 1 12 3 0 1 1 3 1 1 0 3 0 matrices have incompatible dimensions.
SECTION 10.2 The Algebra of Matrices
2 6 1 2 3 6 is undefined because these 13. 1 3 2 4 2 0 matrices have incompatible dimensions.
1 2
3
5
2
1
2 1 2
14. 6 3 4
2 3
1 2 6 7 6 0
1 2
3 2
13
7
5 1 1 1 4 2 2 1 7 10 7 1 7 1 2 1 2 5 4 6 1 2 1 1 12 1 17. 2X A B X 2 B A . 2 2 1 2 3 7 2 4 1 3
15.
1 2
16. 0
18. 3X B C 3X C B. but C is a 3 2 matrix and B is a 2 2 matrix, so C B is impossible. Thus, there is no solution. 19. 2 B X D. Since B is a 2 2 matrix, B X is defined only when X is a 2 2 matrix, so 2 B X is a 2 2 matrix. But D is a 3 2 matrix. Thus, there is no solution. 20. 5 X C D X C 15 D 10 20 2 3 2 4 2 3 4 7 1 X 15 D C 30 20 1 0 6 4 1 0 7 4 . 5 10 0 0 2 2 0 0 2 2 2
21. 15 X D C X D 5C 2 3 10 20 10 15 10 20 0 5 X 5C D 5 1 0 30 20 5 0 30 20 25 20 . 0 2 10 0 0 10 10 0 10 10 22. 2A B 3X 2A B 3X
X 13 2A B 1 8 12 2 5 1 6 7 2 73 1 4 6 2 5 2 1 1 3 3 3 1 1 1 3 3 7 2 6 3 7
In Solutions 23–36, the matrices A, B, C, D, E, F, G, and H are defined as follows: 0 2 5 2 52 3 12 5 A C B 1 1 3 0 2 3 0 7 1 1 0 0 5 3 10 E F G 2 0 1 0 6 1 0 0 0 0 1 5 2 2 0 3 12 5 2 52 5 2 5 23. (a) B C 1 1 3 0 2 3 1 1 0
(b) B F is undefined because B 2 3 and F 3 3 don’t have the same dimensions. 0 3 12 5 1 3 5 2 52 24. (a) C B 0 2 3 1 1 3 1 3 6
D
3
3
7 3
3
1
H
2 1
14
CHAPTER 10 Matrices and Determinants
(b) 2C 6B 2
0
2 5
25. (a) 5A 5
0
10 25
2 52
7
0
2 3
6
0
35
1 5 14 8 30 2
3
1 1 3
6 10 24
(b) C 5A is undefined because C 2 3 and A 2 2 don’t have the same dimensions. 2 52 0 13 72 15 3 12 5 2 26. (a) 3B 2C 3 1 1 3 0 2 3 3 1 3
(b) 2H D is undefined because 2H 2 2 and D 1 2 don’t have the same dimensions.
27. (a) AD is undefined because A 2 2 and D 1 2 have incompatible dimensions. 2 5 14 14 (b) D A 7 3 0 7 28. (a) D H
7 3
3
1
2 1
27 4
(b) H D is undefined because H and D have incompatible dimensions. 4 7 2 5 3 1 29. (a) AH 14 7 0 7 2 1 3 1 2 5 6 8 (b) H A 2 1 0 7 4 17
30. (a) BC is undefined because B 2 3 and C 2 3 have incompatible dimensions. 1 0 0 1 3 12 5 3 2 5 0 1 0 (b) B F 1 1 3 1 1 3 0 0 1
31. (a) G F
5 3 10 1
1 0 0
0 0 1 0
5 3 10
6
1
0 0 1 5 5 3 10 1 1 (b) G E 6 1 0 2 8 5 2 2 0 1
2
6
2
5
2
32. (a) B 2 is undefined because B 2 3 is not square. 1 0 0 1 0 0 1 0 0 (b) F 2 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1
33. (a) A2
2 5
0
7
2 5
0
7
4 45
0
49
0 2
SECTION 10.2 The Algebra of Matrices
(b) A3
2 5
0
34. (a) D A B
7
2 5
0
7 3
7
2 5
2 5
0
7
3
4 45
0
49
2 5 0
7
1 5 3 2 14 14
8 335
0
343
15
1 5 2 28 21 28
1 1 3 1 6 5 2 5 3 12 5 7 3 28 21 28 (b) D AB 7 3 1 1 3 7 7 21 0 7 1 1 1 1 6 5 2 5 3 2 5 13 2 2 35. (a) AB E 1 1 3 7 7 21 0 7 7 0 0 0
7
1 1 3
(b) AH E is undefined because the dimensions of AH 2 2 and E 3 1 are incompatible. 5 2 5 0 2 52 3 12 5 7 3 38 11 35 36. (a) D BDC D B C 7 3 1 1 3 0 2 3 1 1 0 (b) B F F E is undefined because the dimensions 2 3 3 3 2 3 and 3 3 3 1 3 1 are incompatible.
In Solutions 37–42, the matrices A, B, and C are defined as follows: 12 01 03 11 24 02 02 01 C B A 0 05 09 01 04 11 21 21 05 21 07 03 05 156 562 37. AB 128 088 38. B has 2 columns and A has 3 rows, so B A is undefined. 109 097 035 003 033 019 029 39. BC 40. C B 055 105 105 027 325 241 431 446 06 094 004 41. B and C have different dimensions, so B C is undefined. 42. A2 112 192 01 041 095 131 x 2 2y 2 x 2y 2 2 . Thus we must solve the system So x 2 and 2y 2 y 1. Since 43. 4 2x 4 6 2x 6y 6 6y these values for x and y also satisfy the last two equations, the solution is x 2, y 1.
3x 6 x y 6 9 x y 3x 3y 3y 9 . Since 3 , we must solve the system 44. 3 y x 9 6 y x 3y 3x 3x 6 3y 9
x 2 and 3y 9 y 3. Thus, the solution is x 2, y 3.
So 3x 6
16
CHAPTER 10 Matrices and Determinants
45. 2
x
y
2 4
x
y
2x
2y
. Since 2 , Thus we must solve the xy xy 2 6 xy xy 2 x y 2 x y 2x 2 2y 4 So x 1 and y 2. Since these values for x and y also satisfy the last two equations, the system 2 x y 2 2 x y 6
solution is x 1, y 2. x y y x 4 4 x y yx 4 4 . Thus we must solve the system 46. y x x y 6 6 x y x y 6 6 xy 4 y x 4 Adding the first equation to the last equation gives 2x 10 x 5 and 5 y 6 y 1. So the x y 6 xy 6 solution is x 5, y 1. 2x 5y 7 2 5 x 7 . 47. written as a matrix equation is 3x 2y 4 3 2 y 4 6x y z 12 48. 2x z 7 y 2z 4
written as a matrix equation is 2
3
x y z2 4x 2y z 2 50. x y 5z 2 x y z 2
4
1
x
12
1 y 7 .
0
0
3x1 2x2 x3 x4 0 49. x1 x3 5 3x x x 4 2
6 1
1 2
4
z
x1
2
0 x2 written as a matrix equation is 5 . 1 0 1 0 x3 4 0 3 1 1 x4
written as a matrix equation is
1 1
3 2 1
1
1
x 4 2 1 2 y . 2 1 1 5 z 1 1 1 2 1
0 . ABC is undefined because the dimensions of A (2 4) 1 7 9 2 , and C 1 2 12 4 0 2 3 3 21 27 6 1 7 9 2 . B AC is undefined because and B (1 4) are not compatible. AC B 2 2 14 18 4
51. A
1 0 6 1
, B
the dimensions of B (1 4) and A (2 4) are not compatible. BC A is undefined because the dimensions of C (4 1) and A (2 4) are not compatible. C AB is undefined because the dimensions of C (4 1) and A (2 4) are not compatible. C B A is undefined because the dimensions of B (1 4) and A (2 4) are not compatible.
SECTION 10.2 The Algebra of Matrices
52. (a) Let A
a b c d
and B
e f g h
. Then A B
a b
a b
A2
AB
c d
c d
cg d h
cg d h
a 2 bc ab bd
e2 f g e f f h
ae bg a f bh
ae c f be d f
a 2 bc ae bg ae c f e2 f g ab bd e f f h a f bh be d f
c g a e d h c g
a b
e f
g h
A2 AB B A B 2
c d
, and
a e b f b f d h
a e2 b f c g
cg d h
ae b f ae b f A B2
ae b f
ac cd bc d 2
ce dg c f dh
c g b f d h2
; B 2
; B A
e f g h e f g h
e f g h a b c d
;
eg gh f g h 2
ag ch bg dh
ac cd eg gh ce dg ag ch bc d 2 f g h 2 c f dh bg dh a 2 2ae e2 b c g f c g
;
. Then
a b f e b f b d h f d h
c b f g b f d 2 2dh h 2 a e b f b f d h a e2 b f c g A B2 c g a e d h c g c g b f d h2 c a e g a e d c g h c g
17
(b) No. From part (a), A B2 A B A B A2 AB B A B 2 A2 2AB B 2 unless AB B A which is not true in general, as we saw in Example 5. 5 075 010 0 4 53. (a) AB 025 070 070 20 22 7 0 020 030 10
(b) Five members of the group have no postsecondary education, 22 have 1 to 4 years, and seven have more than 4 years. 075 020 005 80 96 54. (a) AB 060 030 010 170 103 040 030 030 40 95 (b) 96 students slept less than 4 hours, 103 slept 4 to 7 hours, and 95 slept more than 7 hours. 50 20 15 350 35375 55. (a) AB 40 75 20 575 65625 35 60 100 425 89250 (b) The total revenue for Monday is the 1 1th entry of the product matrix, $35375.
(c) The total revenue is the sum of the three entries in the product matrix, $190250. 4000 1000 3500 $4690 $1690 $13,210 56. (a) B A $090 $080 $110 400 300 200 700 500 9000
18
CHAPTER 10 Matrices and Determinants
(b) The entries in the product matrix represent the total food sales in Santa Monica, Long Beach, and Anaheim, respectively. 12 10 0 $1000 $500 $32,000 $18,000 57. (a) AB 4 4 20 $2000 $1200 $42,000 $26,800 8 9 12 $1500 $1000 $44,000 $26,800 (b) The daily profit in January from the Biloxi plant is the 2 1 matrix entry, namely $42,000.
(c) The total daily profit from all three plants in February was $18,000 $26,800 $26,800 $71,600. 2000 2500 3000 1500 58. (a) AB 6 10 14 28 105,000 58,000 2500 1000 1000 500
(b) That day they canned 105,000 ounces of tomato sauce and 58,000 ounces of tomato paste. 120 50 60 010 9700 59. (a) AC 40 25 30 050 4650 Ashton’s stand sold $97 worth of produce on Saturday, Bryn’s 60 30 20 100 4100
stand sold $4650 worth, and Cimeron’s stand sold $41 worth. 7000 100 60 30 010 (b) BC 35 20 20 050 3350 Ashton’s stand sold $70 worth of produce on Sunday, Bryn’s stand 4850 60 25 30 100 sold $3350 worth, and Cimeron’s stand sold $4850 worth. 120 50 60 100 60 30 220 110 90 (c) A B 40 25 30 35 20 20 75 45 50 60 30 20 60 25 30 120 55 50
This represents the melons, squash, and
tomatoes they sold during the weekend. 16700 120 50 60 100 60 30 010 220 110 90 010 (d) A B C 40 25 30 35 20 20 050 75 45 50 050 8000 100 8950 60 30 20 60 25 30 100 120 55 50 During the weekend, Ashton’s stand sold $167 worth, Bryn’s stand sold $80 worth, and Cimeron’s stand sold 9700 7000 16700 $8950 worth of produce. Notice that A B C AC BC 4650 3350 8000 . 4100 4850 8950
60. Suppose A is n m and B is i j. If the product AB is defined, then m i. If the product B A is defined, then j n. Thus if both products are defined and if A is n m, then B must be m n. 1 1 1 2 1 3 1 n 2 3 n A A A 61. (a) A 0 1 0 1 0 1 0 1 n1 2n1 2 2 4 4 8 8 2 1 1 A3 A4 An A2 (b) A 2n1 2n1 2 2 4 4 8 8 1 1
SECTION 10.2 The Algebra of Matrices
62. Let A
a b c d
.
For the first matrix, we have A2
a b c d
19
insert exponent (2 times)
a 2 bc 4 b a d 0 4 0 a 2 bc b a d a b a 2 bc ab bd . So A2 2 2 c a d 0 ac cd bc d c a d bc d 0 9 c d bc d 2 9
If a d 0, then a d, so 4 a 2 bc d2 bc d 2 bc 9, which is a contradiction. Thus a d 0. Since b a d 0 and c a d 0, we must have b 0 and c 0. So the first equation becomes a 2 4 a 2, and the
fourth equation becomes d 2 9 d 3. 4 0 2 0 2 0 2 0 2 0 are A1 , A2 , A3 , and A4 . Thus the square roots of 0 9 0 3 0 3 0 3 0 3 a 2 bc 1 b a d 5 1 5 For the second matrix, we have A2 Since a d 0 and c a d 0, we must have c a d 0 0 9 bc d 2 9 a2 1 a 1 c 0. The equations then simplify into the system b a d 5 b a d 5 d2 9 d 3 We consider the four possible values of a and d. If a 1 and d 3, then b a d 5 b 4 5 b 54 . If a 1
and d 3, then b a d 5 b 2 5 b 52 . If a 1 and d 3, then b a d 5 b 2 5 1 5 are b 52 . If a 1 and d 3, then b a d 5 b 4 5 b 54 . Thus, the square roots of 0 9 1 54 1 52 1 52 1 54 , A2 , A3 , and A4 . A1 0 3 0 3 0 3 0 3
20
CHAPTER 10 Matrices and Determinants
10.3 INVERSES OF MATRICES AND MATRIX EQUATIONS
1. (a) The matrix I
1 0 0 1
is called an identity matrix.
(b) If A is a 2 2 matrix, then AI A and I A A.
(c) If A and B are 2 2 matrices with AB I then B is the inverse of A.
2. (a)
5 3 3 2
x
y
4 3
d b 2 3 2 3 1 1 . ad bc c a 5 2 3 3 3 5 3 5 2 3 4 1 . (c) The solution of the matrix equation is X A1 B 3 5 3 3
(b) The inverse of A is A1
(d) The solution of the system is x 1, y 3. 4 1 2 1 ; B . 3. A 7 2 7 4 1 0 4 1 2 1 1 0 2 1 4 1 . and B A AB 0 1 7 2 7 4 0 1 7 4 7 2
4. A
7 3 2 . ; B 2
2 3
2 1 7 3 2 3 1 0 2 2 . BA 2 1 4 7 0 1
5. A
4 7
1 1
3 1 4
8 3
0 ; B 2
AB
4
2 3
1
4 7
1 1 . AB
3 2
4
9 10 8
6. A 1 1 6 ; B 12
8 3
4
1
1
1
2 1
3 1 4
1 3 1 0 1 1 3 2 8 3 4 1 3 1 1 0 0 B A 2 1 1 1 4 0 0 1 0 1 0 1 1 3 2 0 0 1
7 3 2 1 0 and 2
3 2
4
0 2
2
0 1
1 0 0
1 1 0 1 0 and
0
9 10 8
14 11 . AB 1 1 6 12 1 1 2 1 12 12 2 1 12 12 2 2 9 10 8 3 2 4 1 0 0 B A 12 14 11 1 1 6 0 1 0 . 1 1 1 2 2 1 12 0 0 1 2 2
0 0 1
1 0 0
14 11 0 1 0 and 1 1 0 0 1 2 2
21
SECTION 10.3 Inverses of Matrices and Matrix Equations
7 4
5 7 3 4
1
2
5
2 4 7 4 1 2 1 0 1 2 1 . Then, A A1 3 3 7 7 14 12 3 7 3 2 3 2 2 2 0 1 2 2 7 4 1 2 1 0 . and A1 A 3 7 3 2 2 2 0 1 1 3 2 8. B 0 2 2 We begin with a 3 6 matrix whose left half is B and whose right half is I3 . 2 1 0 1 3 2 1 0 0 1 3 2 1 0 0 1 3 2 1 0 0 2R 5R R 3 2R1 R3 3 2 3 0 2 2 0 1 0 0 2 2 0 1 0 R 0 2 2 0 1 0 0 5 4 2 0 1 0 0 2 4 5 2 2 1 0 0 0 1 0 1 3 2 1 0 0 1 0 1 1 32 1 R1 3R2 R1 R R R 2 R2 1 3 1 0 1 1 0 1 0 1 0 2 2 1 0 2 1 R2 R3 R2 2 R3 5 5 0 0 1 2 2 1 0 0 1 2 2 1 1 1 1 1 1 1 1 3 2 1 0 0 1 0 0 1 1 1 0 1 0 2 2 1 . Then B 1 2 2 1 ; B 1 B 2 2 1 0 2 2 0 1 0 ; 2 52 1 2 52 1 2 1 0 0 0 1 0 0 1 2 52 1 1 3 2 1 1 1 1 0 0 and B B 1 0 2 2 2 2 1 0 1 0 . 2 1 0 2 52 1 0 0 1 1 1 2 and verify that A1 A A A1 I . 9. Using a graphing device, we find A1 3 2 2 2 1 1 0 1 1 10. Using a graphing device, we find B 1 33 31 3 and verify that B B B B I3 . 13 12 1 1 3 2 9 2 9 2 1 11. 27 26 13 3 13 9 13 3 7. A
12. 13.
14.
15.
5 13 7
4
8 5 6 3
8
4
A1
4 7 4 7 4 7 1 20 21 3 5 3 5 3 5
1
1
1
13 5 13 5 1 26 25 5 2 5 2
5 4 53 43 1 35 32 8 7 83 73
4 3 1 , which is not defined, and so there is no inverse. 24 24 8 6
22
CHAPTER 10 Matrices and Determinants
1 1 3
1 16. 2
17.
5 4
04 12
03
06
1 2 53
1
4 2 3 1 0 0
4 13 5
1 2
12 1 15
3 2
1 2 06 12 1 024 036 03 04 1 2 2
18. 3 3 2 0 1 0
4
2
3
3
1 0 0
4 0
4
0 0
4
1 4 R1
R1 R3 R1 0 6 1 3 4 0 0 6 1 3 4 0 3R3 R2 R3 1 0 1 0 0 1 0 2 1 1 0 4 0 0 2 6 4 12 1 0 1 0 0 1 1 0 0 3 2 5 1 0 0 3 2 5 1 1 R3 R1 0 6 1 3 4 0 R 6 R2 0 6 0 6 6 6 0 1 0 1 1 1 . R2 R3 R2 0 0 1 3 2 6 0 0 1 3 2 6 0 0 1 3 2 6 3 2 5 Therefore, the inverse matrix is 1 1 1 . 3 2 6
2 4
4R2 3R1 R2 4R3 R1 R3
1 1 0 0
2 4
1
1 0 0
6 0
5
0
1 1 0
1 2 R3
1 4 0
3R3 2R2 R3 0 6 1 1 2 0 0 6 1 1 2 0 3R1 2R2 R1 0 4 1 1 0 2 0 0 1 5 4 6 1 4 0 0 0 1 6 0 0 24 24 30 1 0 0 4 4 5 1 R1 5R3 R1 6 R1 0 6 0 0 1 0 1 1 1 . 6 6 6 1 R2 R3 R2 6 R2 , R3 0 0 1 5 4 6 0 0 1 5 4 6 4 4 5 Therefore, the inverse matrix is 1 1 1 . 5 4 6
19. 1 1 1 0 1 0
5
7 4 1 0 0
2R2 R1 R2 2R3 R1 R3
1
0 1 1 0 1
1
2 3R1 R2 3 1 3 0 1 0 R 0 1 0 R3 5R1 R3 5 7 4 1 0 0 0 7 1 6 7 5 0 0 1 1 0 1 1 0 1 1 0 0 26 7 25 1 R3 R1 R3 7R2 R3 0 1 0 3 1 3 R 0 1 0 3 1 3 R . 2 , R3 0 0 1 27 7 26 0 0 1 27 7 26 26 7 25 Therefore, the inverse matrix is 3 1 3 . 27 7 26
20. 3 1 3 0 1 0
1 2 3 1 0 0 21. 4 5 1 0 1 0 1 1 10 0 0 1
R3 R1 R3 R1 R3
R2 4R1 R2 R3 R1 R3
1 2 3 1 0 0 0 3 13 4 1 0 0 3 13 1 0 1
R3 R2 R3
Since the left half of the last row consists entirely of zeros, there is no inverse matrix.
1
3 1 3
6 0 5
1 2 3 1 0 0 0 3 13 4 1 0 . 0 0 0 3 1 1
SECTION 10.3 Inverses of Matrices and Matrix Equations
0 2 2 1 0 0
3 2 0 1 0 0
2 1 0 1 0 0 22. 1 1 4 0 1 0
23
1 1 4 0 1 0 1 1 4 0 1 0 R2 2R1 R2 R1 R2 R1 2 1 0 1 0 0 R3 2R1 R3 0 1 8 1 2 0 R3 R2 R3 2 1 2 0 0 1 2 1 2 0 0 1 0 1 6 0 2 1 1 0 4 1 1 0 1 0 0 1 1 2 1 0 0 1 1 2 R1 2R3 R1 R2 0 1 8 1 2 0 0 1 0 3 2 4 0 1 0 3 2 4 1 R2 4R3 R2 . 2 R3 1 1 0 2 0 0 2 1 0 1 0 0 2 1 0 1 0 0 1 2 1 1 2 Therefore, the inverse matrix is 3 2 4 . 12 0 12 R1 R2
1 2 3 0 0 1
1
1 2
3 0 0
1
1
0 0
1 0 1
2 3R1 R2 3 1 3 0 1 0 R 0 7 6 0 1 3 0 2 2 1 0 0 0 2 2 1 0 0 1 2 3 0 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 R1 R3 R1 3 2R2 R3 0 1 0 3 1 3 R 2 R3 0 1 0 3 1 3 R2 3R3 R2 0 2 2 1 0 0 0 0 2 7 2 6 1 0 1 1 0 1 1 0 0 92 1 4 1 R3 R1 0 1 0 3 1 3 R 3 1 3 0 1 0 . Therefore, the inverse matrix is 7 7 0 0 1 1 3 0 0 1 2 1 3 2 9 1 4 2 3 1 3 . 7 1 3 2
23. 3
24. 5
1 3 0 1 0
R1 R3
R R R 1 3 1 1 1 0 1 0 5
0 0 1 0 1 1 1 0 1
R2 5R1 R2 0 0 R3 2R1 R3
1 1 5 1
R 2R R 3 2 3 5
2 2 0 0 0 1 0 2 0 2 0 3 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 2 R3 0 1 1 5 1 5 0 1 0 1 0 3 . Therefore, the inverse matrix is 1 0 3 . R2 R 2 2 R 3 2 13 0 0 1 6 1 13 6 1 0 0 2 12 2 13 2 2
2 2 0 0 0 1
24
CHAPTER 10 Matrices and Determinants
1 2 0 3 1 0 0 0
0 1 1 1 0 1 0 0 25. 0 1 0 1 0 0 1 0
R3 R2 R3 R4 R1 R4
1 2 0 2 0 0 0 1
1 2 0 3 1 0
0
0
0 0 0 1
1 2
0
0 1 1 0 0 1 0 0
3
1
1
0
0 0 0
1 0 0 0 0 1 1 0 0 1 1 0 0 1
1 0 2 1 1 2
0
R3 R4
0
0 1 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 2 0 1 0 0 0 0 0 2 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 R1 R1 R4 . 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 0 1 1 0 0 1 0 0 0 1 1 0 0 1 0 0 2 1 1 0 1 1 Therefore, the inverse matrix is . 0 1 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 0 0 1 1
R1 2R2 R1 R2 R3 R2
0 1 0 0
0 1 0
0
1
R1 2R3 R1 R2 R4 R2
1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 R R R 0 1 0 1 0 1 0 0 R R R 0 1 0 1 0 1 0 0 414 424 26. . 1 1 1 0 0 0 1 0 R3 R1 R3 0 1 0 0 1 0 1 0 R3 R2 R3 0 0 0 1 1 1 1 0 1 1 1 1 0 0 0 1
Therefore, there is no inverse matrix.
27.
29.
3
2
0 1
1
0 2
3 0
2 3
1
1
0 4 1 2 2 2 0 1
1 0 0 0
0 0 0 7
2 1 3
3 1 2
2
0 1 0 1 2 2 2 1 2 1 1 1 0
1 1 72 6 1 1 0 2 6 1 0 0 3
1
0 2 0 0 33. 0 0 4 0
4 3 3
1 1 3 1 3
0 1
1 7 3 31. 0 2 1 0 0 3
1
0 2
1 4 1
3
0 1 0 1 1 0 0 1
1 0 0 0
28.
30.
0 0 0
5 5
2
1
0
1
0 1 2 3
0 1
1
0 1 0 0 2 0 0 1 0 4 0 0 0 17
3
2 3 1 0 0 0
1
2 5 0 0 32. 4 2 3 0
34.
1 0 0 2
1
2 25
1
1 1 0 1 1 0
3 1
5 1 2 1
1
0
0 0 3
2 5 15
1 3 25 25 2 1 5 5 1 5 35
1 16 16 1 1 0 0 4 2 1 1 3 3 3 5 6
13 6
2 56
1
0
0 0
1 0
0
1 6
1 2 0 0 5 5 16 2 1 0 15 3 15 1 2 1 37 15 15 3
1
0
0 1 1 0 1
0 0
0 0 13 1 2
SECTION 10.3 Inverses of Matrices and Matrix Equations
In Solutions 35–38, the matrices A and B are defined as follows: 1 0 2 A 0 2 1 4 2 1
14
3 3 4 4 7 23 3 35. A1 B 16 16 16 7 1 5 8 8 8
7 3 4
22 2 37. B AB 1 7 7 50 7
39.
3x 2y 1
40.
5x 7y 9
26 7
2 1 2
B
0
3
1
0
1 2
0
0
1
25 7
13 7
22 7
1 5 2 36. AB 1 7 7 7
16 7 37 7
4 8 13 7 7 7 1 8 17 6 38. B AB 7 7 7 31 9 10 7 7 7
3 2
x
1
5 7
x
9
2
5
x
2
7
4
x
0
1
x
is equivalent to the matrix equation
25
. 13 9 y 3 x 1 9 2 3 . Therefore, x 3 and y 4. Using the inverse from Exercise 11, y 13 3 3 4
13x 9y 3
is equivalent to the matrix equation
. 3 4 y 6 x 4 7 9 6 . Therefore, x 6 and y 3. Using the inverse from Exercise 12, y 3 5 6 3 3x 4y 6
2x 5y 2 41. 5x 13y 20
. 5 13 y 20 x 13 5 2 126 . Therefore, x 126 and y 50. Using the inverse from Exercise 13, y 5 2 20 50
42.
7x 4y
0
8x 5y 100
is equivalent to the matrix equation
is equivalent to the matrix equation
Using the inverse from Exercise 14,
2x 4y z 7 43. x y z 0 x 4y 2
x
y
53
83
. 8 5 y 100 43 0 400 3 . Therefore, x 400 and y 700 . 3 3 73 700 100 3
2 4
is equivalent to the matrix equation 1 1 1 y
1 4 0 z x 4 4 5 7 38 Using the inverse from Exercise 19, 9 y 1 1 1 0 . z 5 4 6 2 47
Therefore, x 38, y 9, and z 47.
7
0 .
2
26
CHAPTER 10 Matrices and Determinants
5x 7y 4z 1 44. 3x y 3z 1 6x 7y 5z 1
5
7 4
x
1
is equivalent to the matrix equation 3 1 3 y 1 . 6
7 5 z 1 x 26 7 25 1 8 3 1 1 1 . Using the inverse from Exercise 20, y 3 z 27 7 26 1 8
Therefore, x 8, y 1, and z 8. 2y 2z 12 0 2 2 x 12 45. 3x y 3z 2 is equivalent to the matrix equation 3 1 3 y 2 . x 2y 3z 8 1 2 3 z 8
x
Using the inverse from Exercise 23, y
92 1
z
3
7 2
4
12
1 3 2 1 3
8
20
10 . 16
Therefore, x 20, y 10, and z 16. x 2y 3 0 1 2 0 3 x 0 0 1 1 1 y 1 yz 1 is equivalent to the matrix equation 46. . 0 1 0 1 z 2 y 2 x 2y 2 3 1 2 0 2 3 1 0 0 2 1 0 x y 1 0 1 1 1 5 Using the inverse from Exercise 25, . z 0 1 1 0 2 1 3 1 0 0 1 3 Therefore, x 1, y 5, z 1, and 3.
47. Using a calculator, we get the result 3 2 1.
48. Using a calculator, we get the result 1 2 3.
49. Using a calculator, we get the result 3 2 2.
50. Using a calculator, we get the result 6 12 24.
51. Using a calculator, we get the result 8 1 0 3.
52. Using a calculator, we get the result 8 4 2 1. 53. This has the form M X C, so M 1 M X M 1 C and M 1 M X M 1 M X X. 1 3 2 3 2 3 2 1 . Since X M 1 C, we get Now M 1 98 4 3 4 3 4 3 x y z 3 2 1 0 1 7 2 3 . u 4 3 2 1 3 10 3 5 39 92 1 4 3 6 39 2 54. Using the inverse matrix from Exercise 23, we see that 3 1 3 6 12 15 30 . 7 33 1 3 0 0 33 2 2 39 39 x u 2 Hence, y 15 30 . 33 33 z 2
SECTION 10.3 Inverses of Matrices and Matrix Equations
1
a a 1 1 a a 1 1 1 55. 2a 1 1 a 2 a 2 2a 2 a a a a a a 1 a 0 0 0 1 0 0 0 1 0 0 0 1a 0 0 0 R 1 a 1 0 b 0 0 0 1 0 0 0 1 0 0 0 1b 0 0 b R2 56. . 1 0 0 c 0 0 0 1 0 0 0 1 0 0 R3 0 1c 0 c 1 R d 4 0 0 0 d 0 0 0 1 0 0 0 1 0 0 0 1d 1a 0 0 0 a 0 0 0 0 1b 0 0 b 0 0 0 Thus the matrix . has inverse 0 0 0 c 0 0 1c 0 0 0 0 1d 0 0 0 d 1 2 x 1 1x x 2 x x 2 x 1 1 . 57. 2x 2 x 2 x 2 x 2 x 2 x x2 1x 2x 2
a a
The inverse does not exist when x 0. 1 x e2x e3x e2x e e x e2x 1 1 . The inverse exists for all x. 58. 2 e4x e4x e2x e x e2x e3x e2x e3x 0 1 0 0 1 ex 1 ex 0 1 0 0 R ex R R 12 e2x R2 2 1 2 x e2x 0 0 1 0 0 2e2x 0 e x 1 0 59. e 1 2 R3 0 0 2 0 0 1 0 0 2 0 0 1 1 1 ex 1 0 0 0 1 ex 0 1 0 0 2 2 x 1 x 1 e2x 0 . 1 e R2 R1 0 1 0 1 ex 1 e2x 0 R 0 1 0 2 e 2 2 2 1 1 0 0 1 0 0 0 0 1 0 0 2 2 1 1 ex 0 2 2 1 ex 1 e2x 0 . The inverse exists for all x. Therefore, the inverse matrix is 2 2 1 0 0 2 1 1 x 1 1 1 x 1 1 1 1 x 1 x 1 x2 x 1 x2 60. . x 1 x 1 x 2 1 x x x x x x x x 1 x 1 x x The inverse exists for x 0, 1. 3 1 3 1 0 0 4 2 4 0 1 0 1 1 1 1 1 0 R3 R1 R3 1 R2 R1 3 1 3 1 0 0 R 61. (a) 4 2 4 0 1 0 3 1 3 1 0 0 R1 R2 3 2 4 0 0 1 0 1 1 1 0 1 0 1 1 1 0 1 R1 12 R2 R1 1 1 1 1 1 0 1 0 1 1 12 0 R R R 12 R2 R2 3R1 R2 3 0 1 3 1 0 2 0 4 3 0 0 1 0 2 2 1 R3 2 R2 R3 0 0 1 1 32 1 0 1 1 1 0 1 0 1 1 1 0 0 0 1 1 3 3 0 1 0 2 0 0 . Therefore, the inverse of the matrix is 2 . 2 2 3 3 0 0 1 1 2 1 2 1 1
27
28
CHAPTER 10 Matrices and Determinants
A
0
(b) B 2 C
1 1
10
1
0 14 1 .
3 2 1 32
1
13
2
Therefore, the rats should be fed 1 oz of food A, 1 oz of food B, and 2 oz of food C. A 0 1 1 9 2 3 (c) 0 12 0 B 2 . 2 3 C 1 2 1 10 1 Therefore, the rats should be fed 2 oz of food A, no food B, and 1 oz of food C. 2 7 A 0 1 1 3 (d) 0 4 2 . B 2 2 1 32 1 11 7 C Since A 0, there is no such combination of foods.
3 1 4 1 0 0 1 1 2 1 1 0 3 1 4 1 0 0 R R R2 R1 R2 1 1 2 1 1 0 12 3 1 4 1 0 0 62. 4 2 6 0 1 0 R3 R1 R3 0 1 1 1 0 1 0 1 1 1 0 1 3 2 5 0 0 1 1 1 2 1 1 0 1 1 2 1 1 0 1 0 2 2 4 3 0 R3 2 R2 R3 0 2 2 4 3 0 . 0 1 1 1 0 1 0 0 0 1 32 1
R2 3R1 R2
Since the inverse matrix does not exist, it would not be possible to use matrix inversion in the solutions of parts (b), (c), and (d).
9x 11y 8z 740 63. (a) 13x 15y 16z 1204 8x 7y 14z 828
9 11 8 1 0 0 (c) 13 15 16 0 1 0
9 11
x
740
(b) 13 15 16 y 1204 8
8
7 14
z
828
9 11 8 1 0 0 R3 31R2 R3 0 8 40 13 9 0 20 8 7 14 0 0 1 0 25 62 8 0 9 9 11 8 1 0 0 9 11 8 1 0 0 1 R R 252 R2 2 3 0 8 40 13 9 0 0 252 0 243 279 180 63 R3 2R2 R3 0 252 0 243 279 180 0 8 40 13 9 0 9 11 8 1 0 0 9 11 8 1 0 0 1 R 11R 8R R R3 27 5 27 5 31 31 1 2 3 1 0 1 2520 7 0 28 0 1 0 28 28 7 28 1 25 0 0 2520 1305 1125 360 0 0 1 29 56 56 7 63 63 7 7 9 1 9 0 0 1 0 0 4 4 4 4 1 31 5 0 1 0 27 0 1 0 27 31 5 . 9 R1 7 7 28 28 28 28 25 1 25 1 0 0 1 29 0 0 1 29 7 7 56 56 56 56 9R2 13R1 R2 9R3 8R1 R3
SECTION 10.4 Determinants and Cramer’s Rule
29
7 7 1 4 4 27 31 5 . (You could also use a calculator to find the inverse.) Therefore, Thus, the inverse of the matrix is 28 28 7 25 1 29 7 56 56 7 7 1 x 740 16 4 4 y 27 31 5 1204 28 . The salesperson earns $16 on a standard phone, $28 on a deluxe 28 28 7 25 1 29 z 828 36 7 56 56
phone, and $36 on a super deluxe phone.
64. No. Consider the following counterexample: A
0 1 0 0
and B
0 2 0 0
. Then, AB O, but neither A O nor
0 1 . Then, A2 O, but A O. B O. There are infinitely many matrices for which A2 O. One example is A 0 0
65. To show that B 1 A1 is the inverse of AB, we need to show that the inverse property holds. Using the associative property of matrices, we calculate AB B 1 A1 A B 1 B A1 AIn A1 A A1 In and B 1 A1 AB B 1 A1 A B B 1 In B B 1 B In . This shows that the inverse property of matrices holds for AB and B 1 A1 , and so the two matrices are inverses of one another.
10.4 DETERMINANTS AND CRAMER’S RULE 1. True. det A is defined only for a square matrix A. 2. True. det A is a number not a matrix. 3. True. If det A 0 then A is not invertible. 2 1 2 4 1 3 11 4. (a) 3 4 1 0 2 (b) 3 2 1 1 [2 4 1 3] 0 [3 4 1 0] 2 [3 3 2 0] 8 3 2 9 7 0 3 4 2 0 has determinant D 2 3 0 0 6. 5. The matrix 0 3 0 1 has determinant D 0 0 1 2 2. 6. The matrix 2 0 3 1 2 has determinant D 3 2 1 1 0. 7. The matrix 2 3 2 1 3 02 04 has determinant D 02 08 04 04 0. 8. The matrix 04 08 4 5 has determinant D 4 1 5 0 4. 9. The matrix 0 1
30
CHAPTER 10 Matrices and Determinants
10. The matrix
1
2
3 2
has determinant D 2 2 1 3 1.
2 5 does not have a determinant because it is not square. 3 12. The matrix does not have a determinant because it is not square. 0 11. The matrix
1 13. The matrix 2
14. The matrix
1 8 has determinant D 1 1 1 1 1 1 1 . 2 2 8 4 8 8 1 12
22 14
05
10
has determinant D 22 10 05 14 22 07 29. 1 0 12
In Solutions 15–20, A 3 5 2 . 0 0 4
15. M11 5 4 0 2 20, A11 12 M11 20
16. M33 1 5 3 0 5, A33 16 M33 5
17. M12 3 4 0 2 12, A12 13 M12 12
18. M13 3 0 0 5 0, A13 14 M13 0
19. M23 1 0 0 0 0, A23 15 M23 0 20. M32 1 2 3 12 72 , A32 15 M32 72 2 1 0 2 4 2 6 4 4. Since M 0, the . Therefore, expanding by the first column, M 2 21. M 0 2 4 1 3 0 1 3
matrix has an inverse. 1 2 5 22. M 2 3 2 . Therefore, 3 5 3 2 2 3 2 2 3 2 9 10 2 6 6 5 10 9 19 24 5 0, and so 5 M 1 3 3 5 3 3 5
the matrix does not have an inverse. 30 0 20 23. M 0 10 20 . Therefore, expanding by the first row, 40 0 10 0 10 10 20 20 30 100 0 20 0 400 3000 8000 5000, and so M 1 exists. M 30 0 0 10 40
24. M
2
1 2
2 32 2 4 1 1 4 2 8 3 1 5 4, and the M 1 . Therefore, 4 0 2 2 1 2 2 4 2 2 1
2 32 12
matrix has an inverse.
SECTION 10.4 Determinants and Cramer’s Rule
31
1 3 7 1 3 3 7 2 6 14 16 0. Since 8 . Therefore, expanding by the second row, M 2 25. M 2 0 8 0 2 2 2 0 2 2 M 0, the matrix does not have an inverse. 0 1 0 2 4 6 4 2. Since M 0, the matrix . Therefore, expanding by the first row, M 1 26. M 2 6 4 1 3 1 0 3
has an inverse. 1 3 3 0 0 2 0 1 . Therefore, expanding by the third row, 27. M 1 0 0 2 1 6 4 1 1 3 3 3 3 0 1 3 3 3 3 3 6 6 4 4, and so M 1 exists. 4 1 M 1 2 0 1 2 0 2 0 1 1 4 2 0 6 4 1 6 4 6 4 1
1 2 2 2 0 2 1 2 2 2 3 4 0 4 6 16 2 2 92, 2 . Therefore, M 1 4 0 4 2 3 4 4 6 28. M 3 4 4 4 0 1 6 0 0 1 0 1 6 0 1 0 2 0 1
2 0 2
and so M 1 exists. 10 20 31 1 2 1 30. 10 11 45 1080. The matrix has an inverse. 29. 2 2 1 6. The matrix has an inverse. 20 40 50 1 2 2 7 1 3 2 5 1 10 2 3 9 11 5 2 18 18 13 31. 0. The matrix has no inverse. 12. The matrix has an inverse. 32. 2 6 3 30 4 24 0 31 5 15 10 39 1 10 2 10 3 5 10 2 4 3 2 10 2 2 26 8 6 24 1 3 34. 33. 8. The matrix has an inverse. 0. The matrix has no inverse. 6 20 15 3 27 9 16 45 8 12 20 36 12 9 6 1 0 0 4 6 0 0 4 6 2 1 1 3 2 1 1 3 , by replacing R3 with R3 R2 . Then, expanding by the third row, 35. M 2 1 2 3 0 0 1 0 3 0 1 7 3 0 1 7 0 0 6 2 1 6 2 0 3 1 18. M 1 2 1 3 6 3 0 3 0 7
32
CHAPTER 10 Matrices and Determinants
36. M
2
3 1 7
6 2 3 . 7 7 0 5 3 12 4 0 4
2 0 1 7 2 3 1 7 4 0 2 3 4 6 2 3 , M Then 7 0 5 7 7 0 5 7 3 0 4 0 3 12 4 0
by replacing C2 with C2 3C3 . So expanding about the second column, 2 1 7 2 1 4 2 7 7 22 3 5 1183. 3 M 7 4 2 3 7 7 3 4 3 4 3 4 0 1 2 3 4 5 1 2 3 4 1 2 3 0 2 4 6 8 0 2 4 6 1 2 60 2 120. 37. M 0 0 3 6 9 , so M 5 5 4 0 2 4 20 3 0 0 3 6 0 2 0 0 0 4 8 0 0 3 0 0 0 4 0 0 0 0 5 2 1 6 4 2 1 6 4 0 1 6 4 7 2 2 5 11 2 2 5 7 2 2 5 38. M . Then, M , by replacing C1 with C1 2C2 . So 4 2 10 8 0 2 10 8 4 2 10 8 6 1 1 4 8 1 1 4 6 1 1 4 1 6 0 1 6 0 1 6 4 1 6 4 1 6 1 6 104 M 11 2 10 8 8 2 2 5 11 2 10 0 8 2 2 13 88 2 10 2 10 2 10 0 1 1 8 2 10 8 1 1 4 88 2 104 2 32 4 1 0 39. B 2 1 1 4 0 3 4 1 4 0 1 0 6 12 4 2 1 1 (a) B 2 4 0 4 3 0 3 4 1 4 1 4 6 2 3 (b) B 1 2 1 4 0 (c) Yes, as expected, the results agree.
40. If we expand along the first row of each submatrix, we see that the determinant is 210 1024. 2x y 9 2 9 9 1 2 1 25. 41. Then D 10, and D y 5, Dx x 2y 8 8 2 1 2 1 8 Dy Dx 10 25 Hence, x 2, y 5, and so the solution is 2 5. D D 5 5 6x 12y 33 6 33 33 12 6 12 9, and D y 6, Dx 12. 42. Then D 4x 7y 20 20 7 4 7 4 20 Dy Dx 3 9 12 Hence, x and y 2, and so the solution is 32 2 . D D 6 2 6
SECTION 10.4 Determinants and Cramer’s Rule
33
1 3 3 6 1 6 8. 43. Then, D 12, and D y 20, Dx 3x 2y 1 3 1 1 2 3 2 Dy Dx 8 12 Hence, x 04,and so the solution is 06 04. 20 06, y D D 20 1 1 1 1 1x 1y 1 1 1 1 1 2 3 2 3 3 2 1. 44. Then, D 1 6 , Dx 3 3 , and D y 1 1 1 3 1x 1 y 3 4 6 2 4 6 2 6 4 2 1 Dy Dx 1 Hence, x 31 2, y 1 6, and so the solution is 2 6. D D x 6y 3
6
6
04 04 04 12 04 12 08. 32, and D y 08, Dx 45. Then, D 12x 16y 32 12 32 32 16 12 16 Dy Dx 32 08 Hence, x 4, y 1,and so the solution is 4 1. D D 08 08 10x 17y 21 10 21 21 17 10 17 30. 12, and D y 30, Dx 46. Then, D 20x 31y 39 20 39 39 31 20 31 Dx 2 , y D y 30 1, and so the solution is 2 1 . Hence, x 12 30 5 5 D D 30 x y 2z 0 47. Then expanding by the second row, 3x z 11 x 2y 0 04x 12y 04
0 1 2 1 1 2 1 1 1 2 1 2 12 1 11, Dx 11 0 1 11 44, 1 D 3 0 1 3 1 2 2 0 2 0 0 2 0 1 2 0 1 1 0 1 0 2 1 1 1 2 D y 3 11 1 11 11. 22, and Dz 3 0 11 11 1 2 1 0 1 2 0 1 0 0
22 11 Therefore, x 44 11 4, y 11 2, z 11 1, and so the solution is 4 2 1.
5 3 1 5 3 0 4 28 426 398, 6 Then D 0 4 6 1 7 10 7 10 7 10 0 5 6 3 1 6 1 22 4 0 22 6 3 272 126 398, D y 0 22 6 1 6 Dx 22 4 6 1 13 10 13 10 7 13 13 10 0 7 13 0 5 3 6 5 3 5 5 6 6 428 1562 1990. 6 154 642 796, and Dz 0 4 22 4 22 7 10 7 13 7 13 7 10 13
6 5x 3y z 48. 4y 6z 22 7x 10y 13
796 1990 Therefore, x 398 398 1, y 398 2, and z 398 5, and so the solution is 1 2 5.
34
CHAPTER 10 Matrices and Determinants
2x1 3x2 5x3 1 49. x 1 x 2 x3 2 2x x 8 2
3
Then, expanding by the third row, 2 3 5 2 5 2 3 6 1 7, D 1 1 1 2 1 1 1 1 0 2 1
1 3 5 2 1 2 1 1 1 3 30 20 7, Dx 2 1 1 5 3 1 8 2 8 1 2 1 8 2 1
2 1 5 2 5 2 1 Dx 1 2 1 8 24 3 21, and 2 1 1 1 2 0 8 1
2 3 1 2 3 2 1 Dx 1 1 2 2 6 8 14. 8 3 1 1 1 2 0 2 8
21 14 Thus, x1 7 7 1, x2 7 3, x 3 7 2, and so the solution is 1 3 2.
c 2 2a 50. a 2b c 9 3a 5b 2c 22
2 0 1 1 2 2 1 18 1 19, 1 Then D 1 2 1 2 3 5 5 2 3 5 2 2 0 1 9 2 2 1 18 1 19, 1 Da 9 2 1 2 22 5 5 2 22 5 2
2 2 1 1 9 1 1 9 1 80 10 5 95, and 1 2 Db 1 9 1 2 3 22 3 2 22 2 3 22 2
2 0 2 1 2 2 9 2 2 0. 2 Dc 1 2 9 2 3 5 5 22 3 5 22
Hence, a 1, b 5, and c 0, and so the solution is 1 5 0.
SECTION 10.4 Determinants and Cramer’s Rule
1x 1y 1z 7 3 5 2 10 10x 6y 15z 21 2 2 3 11 51. 3 x 5 y 2 z 10 20x 12y 45z 33 Then 4 9 5x 4y 5z 9 x 5y z 5 10 6 15 20 12 20 45 12 45 15 6 2400 1950 300 750, D 20 12 45 10 5 4 5 5 4 5 5 4 5 21 6 15 33 12 33 45 12 45 5040 1440 3600 0, 15 6 Dx 33 12 45 21 9 4 9 5 4 5 9 4 5 10 21 15 20 33 20 45 33 45 D y 20 33 45 10 2400 6825 5175 750, and 15 21 5 9 5 5 9 5 5 9 5 10 6 21 20 12 20 33 12 33 2400 2070 420 750. 21 6 Dz 20 12 33 10 5 4 5 9 4 9 5 4 9
Therefore, x 0, y 1, z 1, and so the solution is 0 1 1. 5 2 1 0 2x y 5 3 0 3 24 35 11, 1 52. 5x Then D 5 0 3 2 3z 19 0 7 4 7 0 4 7 4y 7z 17 5 1 0 19 3 0 3 60 82 22, 1 Dx 19 0 3 5 17 7 4 7 17 4 7 2 5 0 5 3 19 3 D y 5 19 3 2 164 175 11, and 5 0 7 17 7 0 17 7 2 1 5 2 1 2 5 52 85 33. Thus, x 2, y 1, and z 3, and so the 17 Dz 5 0 19 4 5 0 5 19 0 4 17
solution is 2 1 3. 0 3 5 3y 5z 4 2 0 2 1 12 70 58, 5 53. 2x Then D 2 0 1 3 z 10 4 7 4 0 4x 7y 4 7 0 0 0 4 5 4 3 5 4 5 4 5 216, and 378, D y 2 10 1 4 Dx 10 0 1 7 10 1 10 1 4 0 0 0 7 0 0 3 4 0 4 3 4 120 56 176. 7 Dz 2 0 10 4 2 10 0 10 4 7 0 108 88 189 108 88 Thus, x 189 29 , y 29 , and z 29 , and so the solution is 29 29 29 .
35
36
CHAPTER 10 Matrices and Determinants
4 2x 5y 54. x y z8 3x 5z 0
2 5 0 1 1 1 1 10 40 50, 5 Then D 1 1 1 2 3 5 0 5 3 0 5
2 4 0 4 5 0 1 1 8 1 4 5 80 32 48, and 4 220, D y 1 8 1 2 Dx 8 1 1 5 3 5 0 5 8 1 3 0 5 0 0 5
2 5 4 5 4 132. Thus, x 22 , y 24 , and z 66 , and so the solution is 22 24 66 . Dz 1 1 8 3 5 25 25 5 25 25 1 8 3 0 0
55.
x y z0 2z 0 x
y z
0
Then
2z 1 1 1 1 1 1 1 1 2 0 1 2 0 0 1 1 1 2 1 0 1 1 0 2 1 1 D 1 0 1 0 1 2 0 1 2 0 1 1 0 2 1 1 0 1 0 2 0 1 2 0 1 2 0 1 0 2 0 2 0 1 1 1 1 2 1 4, 0 1 1 1 1 1 1 0 0 0 1 1 1 2, Dx 1 0 0 1 1 1 0 1 1 0 1 1 1 1 0 1 0 2 0
1 0 1 1 1 1 1 2 0 0 1 1 1 0 1 1 2 1 1, Dy 1 2 0 1 1 2 0 0 1 0 1 0 1 0 0 1 0 1 1 2 0
1 1 0 1 1 1 1 2 0 0 1 1 1 0 1 1 1 2 1 1, and Dz 2 1 1 2 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 1 0
1 1 1 0 1 1 1 2 0 0 0 1 1 2 2 4. Hence, we have x Dx 2 1 , D 2 1 2 0 0 0 1 1 0 D 4 2 1 1 0 1 1 1 0 2 1 y
Dy D
Dz D 1 1 1 1 4 ,z , and 1, and the solution is 12 14 14 1 . D D 4 4 4 4 4
SECTION 10.4 Determinants and Cramer’s Rule
xy1 1 1 0 0 0 1 0 1 1 0 yz 2 0 1 1 0 1 1 1 0 1 1 1 2, 56. Then D 1 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 1 z 3 1 0 1 0 0 1 x 4 1 0 0 1 1 1 0 0 2 1 0 1 1 0 2 1 1 0 3 1 1 1 1 1 1 3 2, Dx 1 0 1 1 1 3 1 1 1 2 3 0 1 1 4 1 0 1 0 1 4 0 1 0 0 1 4 0 0 1 1 1 0 0 0 1 0 2 1 0 0 2 1 0 1 0 3 1 1 1 3 1 4, Dy 1 2 1 1 1 0 1 1 3 1 1 0 3 1 1 1 1 4 1 0 1 1 0 1 4 0 1 1 4 0 1 1 1 1 0 1 1 0 1 2 0 0 1 2 0 1 1 3 1 1 1 0, 1 Dz 1 0 3 1 1 1 2 0 1 0 0 3 1 1 2 4 1 0 3 1 0 4 1 1 0 4 1 1 1 0 1 1 0 1 1 1 2 0 1 1 2 0 1 1 2 1 3 4 2 6. 1 1 D 1 0 1 3 1 1 1 2 0 0 1 3 1 3 1 3 0 4 0 1 3 0 0 4 1 0 0 4 Dy Dx Dz D 2 4 0 6 Hence, the solution is x 1, y 2, z 0, and 3. D D D D 2 2 2 2
0 0 1 6 2 1 1 1 48 6 1 42 21 57. Area 6 2 1 2 2 2 2 3 8 3 8 1
y
12 [5 2 6 10] 12 [3 16] 12 19 19 2
(6, 2)
1 (0, 0)
1 0 1 3 5 5 1 1 1 58. Area 3 5 1 1 1 2 2 2 2 2 1 2 2 1
(3, 8)
x
1
y (_2, 2)
(3, 5)
1 (1, 0)
x
37
38
CHAPTER 10 Matrices and Determinants
1 3 1 2 9 2 1 9 1 1 1 1 3 59. Area 2 9 1 1 2 2 5 6 5 1 6 1 5 6 1
y (_1, 3)
12 [1 9 6 3 2 5 1 12 45]
2 1
12 [15 3 3 57] 12 63 63 2
2 5 1 7 2 7 1 2 1 1 1 1 5 60. Area 7 2 1 2 2 2 3 4 3 1 4 1 3 4 1
(2, 9)
x (5, _6)
(_2, 5)
12 [2 2 4 5 7 3 28 6] 12 [2 6 5 4 34] 12 12 20 34 12 66 33
y (7, 2)
1
x
1 (3, _4)
a 0 0 0 0 b 0 0 0 c 0 0 0 b 0 0 0 0 c 0 0 d 0 abcde 61. 0 0 c 0 0 a ab 0 d 0 abc 0 0 d 0 0 e 0 0 0 d 0 0 0 e 0 0 0 e 0 0 0 0 e a a a a a a a a a a a a 0 a a a a 0 a a a a a 2 3 a5 62. 0 0 a a a a a 0 a a a 0 0 a a 0 a 0 0 0 a a 0 0 a 0 0 0 a 0 0 0 0 a
x 12 13 x 12 0 x 2 x x 1 0 x 0, 1, or 2 63. 0 x 1 23 0 x 2 0 x 1 0 0 x 2
x 1 1 1 x 1 x 1 1 x 0 x x 2 1 x x 2 2x 1 0 x 12 0 x 1 64. 1 1 x x 1 x x x x 1 x 1 x
1 0 x 2 x 1 1 0 x 0 1 x 2 0 x 2 1 x 1 65. x 2 1 0 0 1 x 0 0 1 x 0 1
a b x a a x a a b 1 [ax x x a] a x b bx 66. x x b x 1 x x x b x 0 1 1
ax x 2 ax ax ab bx x 2 ax bx ab x a x b 0 x a or x b
SECTION 10.4 Determinants and Cramer’s Rule
39
1 x x2 y y2 x x2 x x2 2 1 1 yz 2 y 2 z xz 2 x 2 z x y 2 x y 2 67. 1 y y 1 2 2 2 z z z z y y 1 z z2
yz 2 y 2 z x z 2 x 2 z x y 2 x y 2 x yz x yz x yz xz 2 y 2 z yz 2 x 2 y x 2 z zy 2 x yz z x y xz y 2 yz x x y xz y 2 yz z x x y x z y 2 yz z x x y z y y z z x x y y z
68.
x 2y 6z 5
3x 6y 5z 8 2x 6y 9z 7 (a) If x 1, y 0, and z 1, then x 2y 6z 12 06 1 5, 3x 6y 5z 3 16 05 1 8, and 2x 6y 9z 2 1 6 0 9 1 7. Therefore, x 1, y 0, z 1 is a solution of the system. 1 0 6 1 2 6 1 2 6 . Then, M 3 6 5 3 0 5 (replacing C2 with C2 2C1 ), so (b) M 3 6 5 2 2 9 2 6 9 2 6 9 1 6 2 5 18 46. M 2 3 5 1 2 6 x 5 (c) We can write the system as a matrix equation: 3 6 5 y 8 or M X B. Since M 0, M has 2 6 9 z 7 an inverse. If we multiply both sides of the matrix equation by M 1 , then we get a unique solution for X, given by X M 1 B. Thus, the equation has no other solution.
(d) Yes, since M 0.
a b 1 1 1 1 69. (a) If three points lie on a line then the area of the “triangle” they determine is 0, that is, 12 Q a2 b2 1 0 2 a3 b3 1 Q 0. If the points are not collinear, then the point form a triangle, and the area of the triangle determined by these
points is nonzero. If Q 0, then 12 Q 12 0 0, so the “triangle” has no area, and the points are collinear. 6 4 1 y 2 10 6 4 6 4 (6, 13) (b) (i) 2 10 1 6 13 6 13 2 10 6 13 1 (2, 10) 26 60 78 24 60 8 34 104 68 0
Thus, these points are collinear.
(_6, 4) 1 1
x
40
CHAPTER 10 Matrices and Determinants
5 10 1 2 6 5 10 5 10 (ii) 2 6 1 15 2 15 2 2 6 15 2 1
y (_5, 10) (2, 6)
4 90 10 150 30 20 94 140 50 4
These points are not collinear. Note that this is difficult to determine from the diagram.
1 2
(15, _2)
x
x y 1 70. (a) Let M x1 y1 1 . Then, expanding by the third column, x2 y2 1 x1 y1 x y x y x1 y2 x2 y1 x y2 x2 y x y1 x1 y M x2 y2 x2 y2 x1 y1 x1 y2 x2 y1 x y2 x2 y x y1 x1 y x2 y x1 y x y2 x y1 x1 y2 x2 y1
x2 x1 y y2 y1 x x1 y2 x2 y1
So M 0 x2 x1 y y2 y1 x x1 y2 x2 y1 0 x2 x1 y y2 y1 x x1 y2 x2 y1 y y1 x y y1 y x x1 x 1 2 1 2 x2 x1 y y2 y1 x x1 y2 x1 y1 x1 y1 x2 y1 y 2 x2 x1 x2 x1 x2 x1 y y1 y y1 y 2 x x1 y1 y y1 2 x x1 , which is the “twopoint” form of the equation for the line x2 x1 x2 x1 passing through the points x1 y1 and x2 y2 . (b) Using the result of part (a), the line has equation x y 1 20 50 x y x y 0 20 50 1 0 10 25 10 25 20 50 10 25 1
500 500 25x 10y 50x 20y 0 25x 30y 1000 0 5x 6y 200 0.
71. (a) Let x, y, and z be the weights (in pounds) of apples, peaches, and pears, respectively. x y z 18 We get the model 075x 090y 060z 1380 075x 090y 060z 180
SECTION 10.4 Determinants and Cramer’s Rule
41
1 1 1 075 090 075 060 090 060 0 090 135 045, 1 1 (b) D 075 090 060 1 075 090 075 060 090 060 075 090 060 18 1 1 1380 090 1380 060 090 060 0 72 108 36, 1 1 Dx 1380 090 060 18 180 090 180 060 090 060 180 090 060 1 18 1 075 1380 075 060 1380 060 1 18 D y 075 1380 060 1 075 180 075 060 180 060 075 180 060 72 162 117 27, and 1 1 18 075 090 075 1380 090 1380 18 1 Dz 075 090 1380 1 075 090 075 180 090 180 075 090 180
108 117 243 18. Dy Dx Dz 36 27 18 So x 8; y 6; and z 4. D D D 045 045 045 Thus, the customer bought 8 pounds of apples, 6 pounds of peaches, and 4 pounds of pears.
72. (a) Using the points 10 25, 15 3375, and 40 40,we substitute for x and y and get the system 100a 10b c 25 225a 15b c 3375 1600a 40b c 40 100 10 1 225 15 100 10 100 10 1 1 (b) D 225 15 1 1 1600 40 225 15 1600 40 1600 40 1
9000 24,000 4000 16,000 1500 2250 15,000 12,000 750 3750, 25 10 1 25 25 10 3375 15 10 1 1 Da 3375 15 1 1 40 3375 15 40 40 40 40 40 1
1350 600 1000 400 375 3375 750 600 375 1875, 100 25 1 100 25 100 25 225 3375 1 1 Db 225 3375 1 1 225 3375 1600 40 1600 40 1600 40 1
9000 54,000 4000 40,000 3375 5625 45,000 36,000 2250 11,250, and
42
CHAPTER 10 Matrices and Determinants
100 10 25 100 10 100 10 225 15 40 3375 Dc 225 15 3375 25 225 15 1600 40 1600 40 1600 40 40 25 9,000 24,000 3375 4,000 16,000 40 1,500 2,250
25 15,000 3375 12,000 40 750 375,000 405,000 30,000 0.
Thus, a
Db Dc Da 1875 11,250 0 005, b 3, and c 0. The model is D D D 3750 3,750 3,750
y 005x 2 3x.
73. Using the determinant formula for the area of a triangle, we have 1000 2000 1 1000 2000 1000 2000 5000 4000 1 1 1 1 Area 5000 4000 1 1 2 2 5000 4000 2000 6000 2000 6000 2000 6000 1 12 22,000,000 2,000,000 6,000,000 12 14,000,000 7,000,000
Thus, the area is 7,000,000 ft2 .
74. (a) The coordinates of the vertices of the surrounding rectangle are a1 b1 , a2 b1 , a2 b3 , and a1 b3 . The area of the surrounding rectangle is given by a2 a1 b3 b1 a2 b3 a1 b1 a2 b1 a1 b3 a1 b1 a2 b3 a1 b3 a2 b1 . (b) The area of the three blue triangles are as follows:
Area of a1 b1 a2 b1 a2 b2 : 12 a2 a1 b2 b1 12 a2 b2 a1 b1 a2 b1 a1 b2 Area of a2 b2 a2 b3 a3 b3 : 12 a2 a3 b3 b2 12 a2 b3 a3 b2 a2 b2 a3 b3 Area of a1 b a1 b3 a3 b3 : 12 a3 a1 b3 b1 12 a3 b3 a1 b1 a3 b1 a1 b3 . Thus the sum of the areas of the blue triangles, B, is
B 12 a2 b2 a1 b1 a2 b1 a1 b2 12 a2 b3 a3 b2 a2 b2 a3 b3 12 a3 b3 a1 b1 a3 b1 a1 b3 12 a1 b1 a1 b1 a2 b2 a2 b3 a3 b2 a3 b3 12 a1 b2 a1 b3 a2 b1 a2 b2 a3 b1 a3 b3 a1 b1 12 a2 b3 a3 b2 12 a1 b2 a1 b3 a2 b1 a3 b1
So the area of the red triangle A is the area of the rectangle minus the sum of the areas of the blue triangles, that is, A a1 b1 a2 b3 a1 b3 a2 b1 a1 b1 12 a2 b3 a3 b2 12 a1 b2 a1 b3 a2 b1 a3 b1 a1 b1 a2 b3 a1 b3 a2 b1 a1 b1 12 a2 b3 a3 b2 12 a1 b2 a1 b3 a2 b1 a3 b1
12 a1 b2 a2 b3 a3 b1 12 a1 b3 a2 b1 a3 b2 a b 1 1 1 (c) We first find Q a2 b2 1 by expanding about the third column. a3 b3 1 a1 b1 a1 b1 a2 b2 1 1 a2 b3 a3 b2 a1 b3 a3 b1 a1 b2 a2 b1 Q 1 a3 b3 a2 b2 a3 b3 a1 b2 a2 b3 a3 b1 a1 b3 a2 b1 a3 b2
So 12 Q 12 a1 b2 a2 b3 a3 b1 12 a1 b3 a2 b1 a3 b2 , the area of the red triangle. Since 12 Q is not always positive, the area is 12 Q.
75. (a) If A is a matrix with a row or column consisting entirely of zeros, then if we expand the determinant by this row or column, we get A 0 A1 j 0 A2 j 0 Anj 0.
CHAPTER 10
Review
43
(b) Use the principle that if matrix B is a square matrix obtained from A by adding a multiple of one row to another, or a multiple of one column to another, then A B. If we let B be the matrix obtained by subtracting the two rows (or columns) that are the same, then matrix B will have a row or column that consists entirely of zeros. So B 0 A 0.
(c) Again use the principle that if matrix B is a square matrix obtained from A by adding a multiple of one row to another, or a multiple of one column to another, then A B. If we let B be the matrix obtained by subtracting the proper multiple of the row (or column) from the other similar row (or column), then matrix B will have a row or column that consists entirely of zeros. So B 0 A 0.
76. Gaussian elimination is superior, since it takes much longer to evaluate six 5 5 determinants than it does to perform one fiveequation Gaussian elimination. 77. Let B A1 . Then, using the given formula, det A A1 det A det A1 det I det A det A1 1 det A det A1 det A1
1 . det A
CHAPTER 10 REVIEW 1. (a) 2 3
2. (a) 2 3
(b) Yes, this matrix is in rowechelon form.
(b) Yes, this matrix is in rowechelon form.
(c) No, this matrix is not in reduced rowechelon form, since the leading 1 in the second row does not have a
(c) Yes, this matrix is in reduced rowechelon form. x 6 (d) y 0
0 above it. x 2y 5 (d) y 3 3. (a) 3 4 (b) Yes, this matrix is in rowechelon form. (c) Yes, this matrix is in reduced rowechelon form. 8z 0 x (d) y 5z 1 0 0
4. (a) 3 4 (b) No, this matrix is not in rowechelon form, since the leading 1 in the second row is not to the left of the one above it. (c) Since this matrix is not in rowechelon form, it is not in reduced rowechelon form. x 3y 6z 2 (d) 2x y 5 z0
44
CHAPTER 10 Matrices and Determinants
6. (a) 4 4
5. (a) 3 4 (b) No, this matrix is not in rowechelon form. The leading 1 in the second row is not to the left of the
above it.
one above it. (c) No, this matrix is not in reduced rowechelon form. y 3z 4 (d) x y 7 x 2y z 2
1
2 2
6
(b) No, this matrix is not in rowechelon form. The leading 1 in the fourth row is not to the left of the one
1 1 0 1
R R 1 2 7. 1 1 0 1 1 2 1 3 7 2
2 2 1 3
(c) No, this matrix is not in reduced rowechelon form. x 8y 6z 4 y 3z 5 (d) 2z 7 x y z 0
1 1 0 1
R2 R1 R2 0 6 R3 2R1 R3 7 0
3 2 3 3
1 1 0 1
R R R 3 2 3 7 0 9 0
3 2 0 1
7 . 2
Thus, z 2, 3y 2 2 7 3y 3 y 1, and x 1 1 x 0, and so the solution is 0 1 2. 1 1 1 2 1 1 1 2 1 1 1 2 R R R R R R 2 1 2 3 2 3 8. 1 1 3 6 0 2 2 4 0 2 2 4 . Thus, z 1; 2y 2 1 4 0 2 3 5 0 2 3 5 0 0 1 1
y 1; and x 1 1 2 x 2, and so the solution is 2 1 1. 1 2 3 2 1 2 3 2 R2 2R1 R2 3 R2 R3 0 3 5 6 R 9. 2 1 1 2 R3 2R1 R3 0 3 5 5 2 7 11 9
1 2
0 0
3 2
3 5 0
0
6 . 1
The last row corresponds to the equation 0 1, which is always false. Thus, there is no solution. 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2 1 R2 R1 R2 3 R2 R3 0 2 2 4 R 0 2 2 4 0 1 1 2 2 R2 10. 1 1 3 6 R3 3R1 R3 3 1 5 10 0 2 2 4 0 0 0 0 0 0 0 0 1 0 2 4 R1 R2 R1 0 1 1 2 . Let z t. Then y t 2 y 2 t and x 2t 4 x 4 2t, and so the 0 0 0 0
solutions are 4 2t 2 t t, where t is any real number. 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 R2 R1 R2 1 1 4 1 1 0 2 5 2 1 0 1 4 5 6 R3 R1 R3 R3 R2 R3 11. 1 2 0 4 7 0 3 1 3 7 0 3 1 3 7 R4 2R1 R4 2 2 3 4 3 0 0 1 2 3 0 0 1 2 3 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 6 6 R3 R4 0 1 4 5 6 R4 13R3 R4 0 1 4 5 R3 3R2 R3 0 1 4 5 . 0 0 13 12 11 0 0 0 0 1 2 3 1 2 3 0 0 1 2 3 0 0 13 12 11 0 0 0 14 28 Therefore, 14 28 2, z 2 2 3 z 1, y 4 1 5 2 6 y 0, and x 0 1 2 0 x 1. So the solution is 1 0 1 2.
CHAPTER 10
1 0 3 0 1 0 1 0 4 5 12. 0 2 1 1 0
1 0 0 1 0 2
R4 2R1 R4
3
0 1
5 0 6
0 4 1
1
1 0 0 1 0 0
R3 2R2 R3 R4 R2 R4
3
0
45
5 9 10 0 1
0 4 1
1
Review
2 1 5 4 4 0 1 1 4 0 0 1 1 0 3 0 1 1 0 3 0 1 0 1 0 4 R3 5 R4 3R3 R4 0 1 0 4 5 R3 R4 1 0 0 1 0 1 9 R4 0 0 1 0 1 0 0 1 9 10 0 0 0 9 9 1 0 3 0 1 1 0 0 0 2 0 1 0 4 5 0 1 0 0 1 R1 3R3 R1 . Therefore, the solution is 2 1 1 1. R2 4R4 R2 0 0 1 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1
1 1 3 2
13. 2
1 1 2
1 1
0
1 1
3
2
1 1
3
2
1 R3 R2 R3 0 3 5 2 0 3 5 2 3 R2 3 0 4 4 0 3 5 2 0 0 0 0 4 4 1 1 3 2 1 0 3 3 5 2 . The system is dependent, so let z t: y 5 t 2 1 R2 R1 0 1 5 2 R 0 1 3 3 3 3 3 3 0 0 0 0 0 0 0 0 y 53 t 23 and x 43 t 43 x 43 t 43 . So the solution is 43 t 43 53 t 23 t , where t is any real number.
14. 1
1
2
R2 2R1 R2 R3 3R1 R3
1
3
1 1
0
1
1 1 0 1
1 3 R2 R3 0 2 2 2 R 2 R4 0 2 2 2 0 2 2 2 0 0 0 0 1 0 1 2 0 1 1 1 . Since the system is dependent, let z t. Then y t 1 0 0 0 0
R2 R1 R2 R3 R1 R3
1 3 2 1 1 1 0 1 1 R2 R1 0 1 1 1 R 0 0 0 0
y t 1 and x t 2 x t 2. So, the solution is t 2 t 1 t where t is any real number.
1 1
1 1 0
1 1
1 1 0
1 2 R4
1 1
1 1 0
2 3R1 R2 1 R2 R1 R R 0 2 4 2 2 0 1 2 1 1 3 1 1 1 2 1 0 1 0 1 . Since the system is dependent, Let z s and t. Then y 2s t 1 y 2s t 1 and 0 1 2 1 1
15.
x s 1 x s 1. So the solution is s 1 2s t 1 s t, where s and t are any real numbers.
1 1 3 16. 2 1 6 1 2 9
R2 2R1 R2 R3 R1 R3
1 1 3 0 3 0 0 1 6
R2 3R3 R2
1 1 3 0 0 18 . Since the second row corresponds 0 1 6
to the equation 0 18, which is always false, this system has no solution.
46
CHAPTER 10 Matrices and Determinants
1 1
17. 3
1
1 0
2 1 6
R2 3R1 R2 R3 R1 R3
4 3 3
1 1
0 0
1 0
5 4 6 5 4 3
1 1
1 0
0
0 3
0 0
R3 R2 R3
5 4 6 . The last row of this
matrix corresponds to the equation 0 3, which is always false. Hence there is no solution. 1 2 3 2 1 2 3 2 1 2 3 2 R2 2R1 R2 3 R2 R3 0 5 11 3 R 18. 0 5 11 3 . Since the third 2 1 5 1 R3 4R1 R3 0 5 11 2 0 0 0 1 4 3 1 6
row corresponds to the equation 0 1, which is always false, this system has no solution. 1 0 0 1 1 1 0 0 1 1 1 1 1 1 2 R2 R1 R2 1 1 1 1 0 0 1 1 2 1 1 1 1 1 0 R1 12 R3 R3 R1 R3 19. 0 1 1 2 1 2 0 0 2 2 1 1 1 1 2 R4 2R1 R4 2 4 4 2 6 0 4 4 4 4 2 4 4 2 6 1 0 0 1 1 1 0 0 1 1 R1 R3 R1 0 1 1 2 1 0 1 1 2 1 14 R3 R3 R2 R3 R2 2R3 R2 R4 4R2 R4 0 0 0 4 0 121 R4 0 0 0 0 0 R4 R3 R4 0 0 0 12 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 1 . This system is dependent. Let z t, so y t 1 y t 1; x 1 x 1. So the 0 0 0 1 0 0 0 0 0 0 solution is 1 t 1 t 0, where t is any real number. 1 1 2 3 0 1 1 2 3 0 R 3R R 3 1 3 20. 0 1 1 1 1 0 1 1 1 1 3 2 7 10 2 0 1 1 1 2
R3 R2 R3
1 1 2 3 0 0 1 1 1 1 . Since the 0 0 0 0 1
third row corresponds to the equation 0 1, which is always false, this system has no solution.
21. A 3 3 and B 2 3 have different dimensions, so they are not equal. 25 1 5 1 5 e0 B, so A and B are equal. 22. A 0 21 log 1 12 0 12 In Solutions 23–34, the matrices A, B, C, D, E, F, and G are defined as follows: A
1
2 0 1 4
D 0 1 2
0
B
E
2 1
12
1
1 2 4 2 1 0
4 0 2
F 1 1 0
C
7 5 0
23. A B is not defined because the matrix dimensions 1 3 and 2 3 are not compatible.
3
G
1 2
2 32 2 1 5
CHAPTER 10
24. C D
1 2
3
1
1
4
2 32 0 1
2 1
25. 2C 3D 2
1 2
3
2
0
4
12 1 2
4
5 2
1
Review
1 6
3 12
4 18
2 32 3 0 1 4 3 0 3 4 2 6 0 4 2 2 0 2 1
0 2
26. 5B 2C is not defined because the matrix dimensions 2 3 and 3 2 are not compatible. 27. G A 5 2 0 1 10 0 5
28. AG is undefined because the matrix dimensions 1 3 and 1 1 are not compatible. 1 3 1 3 4 2 11 2 2 2 7 1 2 4 1 2 4 10 1 11 2 3 2 29. BC 30. C B 2 32 8 2 2 9 2 1 0 2 1 0 1 2 4 3 8 2 1 2 1 1 3 4 0 2 4 0 2 2 14 2 1 2 4 30 22 2 2 3 3 3 1 1 0 31. B F 32. FC 1 1 0 2 2 2 2 1 0 9 1 4 27 57 7 5 0 7 5 0 2 1 2 2 1 3 3 7 1 11 1 4 2 2 2 2 2 1 2 1 15 3 1 3 33. C D E 2 2 0 1 1 2 2 1 4 2 1 1 2 2 2 1 2 0 0 1 12 1 4 0 2 1 6 1 4 4 0 2 0 2 12 12 34. F 2C D 4 2 1 1 0 4 3 0 1 1 1 0 4 4 7 5 0 4 2 2 0 7 5 0 6 2 20 34 27 0 21 6 42 24 35. AB 2 36. A2 B 20 5 13 3 37 22 5 22 7 3 42 27 19 14 26 8 32 4 4 7 1 11 7 38. B AB 1 37. A1 B A 3 3 2 1 4 3 35 13 18 80 7 16 3 3 2 39. AB 12 42.
1 1 A 3
45. AB
40. B A 12
2 5 2
6
2 1 3
3 52 1 1
32
46. AB 2 2 1 1 0
1 1
2
1
43. A1 B A 4
1 0 0 1
5 2
and B A
1 0 0
3 52
1 1
2 5
6
2
32
2 0 1 0 and B A 1
1 1 1
0 0 1
1 41. A1 3 44. A1 B A 4
2
5 2 2 1 3
1
1 0 0 1
.
1 0 0
2 2 2 1 0 1 0 .
1 1 1
0
1 1
0 0 1
47
48
CHAPTER 10 Matrices and Determinants
In Solutions 47–52, A
2 1 3 2
, B
1 2 2
4
, and C
0 1 3
.
2 4 0
47. A 3X B 3X B A X 13 B A. Thus, X 13
1 2 4
2
2 1 3 2
48. 12 X 2B A X 2B 2A X 2A 2B 2 A B. 2 1 1 2 3 1 6 2 2 . X 2 3 2 2 4 1 6 2 12
1 3 . 1 3 5
2
Thus,
49. 2 X A 3B X A 32 B X A 32 B. Thus, 3 3 7 2 1 2 2 1 2 1 2 . 3 2 X 2 3 6 0 8 2 4 3 2 3 2 50. 2X C 5A 2X 5A C, but the difference 5A C is not defined because the dimensions of 5A and C are not the same. 1 AX X A1 C. Now 51. AX C A
A1
2 1 0 1 3 2 2 6 1 2 1 2 1 1 . . Thus, X A C 4 3 3 2 3 2 2 4 0 4 5 9 3 2
52. AX B A1 AX X A1 B. From Exercise 65, 2 1 2 1 1 2 4 8 . Thus, X A1 B . A1 3 2 3 2 2 4 7 14
1 4 2 9
2
2
53. D 54. D
55. D
. Then D 1 9 2 4 1, and so D 1
1 3
2
1
.
. Then D 2 3 1 2 8, and so D 1 1 8
4 12 2
9 4
6
1
3 1 4 . 8 1 1 2 8 4
3 2
. Then D 4 6 2 12 0, and so D has no inverse.
1 2 1 2 2 2 6 4 2 0, and so D has no inverse. 4 . Then D 2 56. D 1 1 2 0 2 3 2 0 3 2 2 4 0
CHAPTER 10
Review
49
3 0 1 3 0 2 3 4 12 9 1. So D 1 exists. 1 . Then, D 1 57. D 2 3 0 2 3 4 2 4 2 1 1 3 1 1 1 0 1 3 1 1 1 0 3 0 1 1 0 0 R2 2R1 R2 R2 1 R2 R1 0 9 2 2 3 0 3 2 3 0 0 1 0 R 2 3 0 0 1 0 2R3 R3 4R1 R3 4 2 1 0 0 1 0 14 3 4 4 1 4 2 1 0 0 1 1 3 1 1 1 0 1 3 1 1 1 0 1 3 1 1 1 0 R3 R2 R3 9R2 R3 3 R2 R3 0 27 6 6 9 0 R 0 27 6 6 9 0 1 R3 0 1 0 2 1 2 R1 3R2 R1 3 0 28 6 8 8 2 0 1 0 2 1 2 0 9 2 2 3 0 3 2 3 1 0 1 5 4 6 1 0 0 3 2 3 1 2 R3 0 1 0 . Thus, D 1 2 1 2 2 1 2 0 1 0 2 1 2 . R1 R3 R1 8 6 9 0 0 2 16 12 18 0 0 1 8 6 9 1 2 3 1 0 0 1 2 3 2 4 2 5 4 5 1 58. D 2 4 5 . Then, D 1 5 6 2 2 6 3 2 5 1 4 6 1. So D exists. 2 4 5 0 1 0 2 5 6 0 0 1 2 5 6 1 2 3 1 0 0 1 2 3 1 0 0 R R R 3R R R2 2R1 R2 2 3 1 3 1 0 0 1 2 1 0 0 1 0 2 0 1 R3 2R1 R3 0 1 0 2 0 1 0 0 1 2 1 0 1 3 2 1 2 0 5 3 0 1 0 0 1 3 2 1 1 2R2 R1 0 1 0 2 0 1 R 0 1 0 2 0 1 . Thus, D 2 0 1 . 2 1 0 0 0 1 2 1 0 0 0 1 2 1 0 1 0 0 1 1 0 0 1 1 0 0 0 1 2 R2 2 0 2 0 2 0 2 0 2 0 2 0 1 0 0 1 3 3 R3 1 59. D . Thus, D 0 3 3 2 3 24 and D exists. 0 0 3 3 1 0 4 0 0 3 3 0 0 1 0 4 R4 0 0 4 0 0 0 4 0 0 0 4 0 0 0 1 1 0 0 14 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 14 R1 R4 R1 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 1 0 1 R2 R4 R2 2 2 4 2 4 1 . Therefore, D . 0 0 1 1 0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 1 R3 R4 R3 3 3 4 3 4 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 14 4 4 1 0 1 0 0 1 0 1 60. D . Thus, 1 1 1 2 1 2 1 2 1 0 1 0 1 1 1 2 1 1 1 2 1 1 0 1 0 1 0. D 1 1 2 1 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 2 Hence, D 1 does not exist.
50
CHAPTER 10 Matrices and Determinants
2 5 2 5 1 , and so . If we let A , then A1 61. 24 25 5 12 5 2 y 5 12 17 5 2 x 2 5 10 65 . Therefore, the solution is 65 154. y 5 12 17 154 6 5 x 1 6 5 . , then 62. If we let A 8 7 y 1 8 7 7 5 7 5 1 7 5 7 5 x 6 1 2 , and so 2 2 . 2 1 A1 2 42 40 8 6 4 3 4 3 1 8 6 y 7
12 5
x
10
12 5
Therefore, the solution is 6 7. 1 2 1 5 2 1 5 1 0 0 1 2 2 0 1 0 2 1 5 x 3 R R 1 2 1 63. 1 2 2 y 4 . Let A 1 2 2 . Then 1 2 2 0 1 0 2 1 5 1 0 0 1 1 0 3 1 0 3 0 0 1 1 0 3 0 0 1 1 0 3 z 6 1 2 2 0 1 0 1 2 2 0 1 0 R 2R R R1 2R2 R1 R2 2R1 R2 2 3 2 0 3 1 1 2 0 0 1 1 1 0 2 R3 R1 R3 R3 R3 2R2 0 2 1 0 1 1 0 2 1 0 1 1 1 0 4 2 1 4 1 0 4 2 1 4 1 0 0 6 3 8 R1 4R3 R1 R3 0 1 1 1 0 2 0 1 0 1 1 1 . 0 1 1 1 0 2 R2 R3 R2 0 0 1 2 1 3 0 0 1 2 1 3 0 0 1 2 1 3 1 1 6 3 8 x 6 3 8 12 3 1 1 Hence, A1 1 1 1 and y 1 1 1 4 12 , and so the solution is 1 1 2 1 3 z 2 1 3 6 12 1 1 1 . 12 12 12 2 0 3 x 5 2 0 3 2 0 3 1 0 0 R R 1 2 64. 1 1 6 y 0 . Let A 1 1 6 . Then 1 1 6 0 1 0 3 1 1 z 5 3 1 1 3 1 1 0 0 1 1 1 6 0 1 0 1 1 6 0 1 0 1 1 6 0 1 0 2 2R1 R2 2 2R3 R2 2 0 3 1 0 0 R 0 2 9 1 2 0 R 0 2 9 1 2 0 R3 3R1 R3 3 1 1 0 0 1 0 4 17 0 3 1 0 0 1 2 1 1 1 1 0 12 5 6 1 1 0 12 5 6 1 R1 9R3 R1 1 R2 R1 0 2 0 17 7 9 2 R2 0 1 0 17 7 9 R 2 2 2 R2 6R3 2R2 0 0 1 2 1 1 0 0 1 2 1 1 7 3 3 7 3 3 x 10 1 0 0 72 32 32 2 2 2 2 2 2 5 0 1 0 17 7 9 . Hence, A1 17 7 9 and y 17 7 9 0 20 , and 2 2 2 2 2 2 2 2 2 z 5 0 0 1 2 1 1 2 1 1 2 1 1 5 so the solution is 10 20 5.
65. (a) The i jth entry of A represents how many pounds of vegetable j were sold on day i, and the ith entry of B represents the price of vegetable i.
CHAPTER 10
Review
51
150 685 100 . The jth entry of AB represents the total revenue on day j. (b) AB 14 12 16 410 050 25 16 30
66. (a) We are given that x y 18, and since the total is $600, 20x 50y 600. 20 50 x 600 (b) 1 1 y 18 1 5 1 50 1 3 , so 30 (c) A1 1 2 20 1 50 1 1 20 30 3 1 1 5 18 5 20 30 10 600 600 3 3 . The customer 30 X A1 B 30 1 2 1 600 2 18 20 12 8 18 30 3 30 3 received 10 $20 bills and 8 $50 bills.
2 13 13 7 2 7 60 78 18. 208 210 2, and D y 32 42 10, Dx 67. D 6 30 30 16 6 16 2 1 and y 18 9 , and so the solution is 1 9 . Therefore, x 10 5 10 5 5 5 140 11 12 11 1260 220 1480, and 108 77 185, Dx 68. D 9 9 20 7 12 140 Dy 240 980 740. Therefore, x 1480 8 and y 740 4, and so the solution is 8 4. 185 185 7 20 2 1 5 2 1 1 7 195 39 156, 3 69. D 1 7 0 5 1 7 5 4 5 4 3 0 1 5 0 1 9 7 495 27 522, 3 Dx 9 7 0 5 9 7 9 4 9 4 3 2 0 5 2 0 1 9 D y 1 9 0 5 180 54 126, and 3 1 9 5 9 5 9 3 2 1 0 2 1 2 1 117 117 234. 9 Dz 1 7 9 9 1 7 5 4 5 4 9
522 87 , y 126 21 , and z 234 3 , and so the solution is 87 21 3 . Therefore, x 156 26 156 26 156 2 26 26 2
52
CHAPTER 10 Matrices and Determinants
3 4 1 3 1 1 4 52 11 41, 1 70. D 1 0 4 4 1 4 2 5 2 1 5 10 4 1 10 1 20 4 880 20 860, 1 Dx 20 0 4 4 20 4 30 5 30 1 5 3 10 1 10 1 10 1 20 4 660 80 40 540, and D y 1 20 4 3 2 1 20 4 30 5 30 5 2 30 5 3 4 10 3 10 1 20 40 50 10. 1 Dz 1 0 20 4 1 20 2 30 2 1 30 860 540 10 860 540 10 Therefore, x 860 41 41 , y 41 , and z 41 , and so the solution is 41 41 41 . 1 3 1 3 1 1 3 1 3 1 1 1 4 8 10 11. 71. The area is 3 1 1 2 2 2 2 2 2 2 3 1 2 2 1 5 2 1 1 5 5 2 5 2 1 1 1 21 3 27 51 . 72. The area is 1 5 1 2 2 2 2 4 1 4 1 1 5 4 1 1
73. Let x be the amount invested in Bank A, y the amount invested in Bank B, and z the amount invested in Bank C. x y z 60,000 y z 60,000 x We get the following system: 002x 0025y 003z 1575 2x 25y 3z 157,500 which 2x 2x 2z y y 2z 0
1
1 1
60,000
has matrix representation 2 25 3 157,500
1 1
1 60,000
0 1 0 40,000 0 05 1 37,500
2 1 2
R1 R2 R1 R3 05R2 R3
0
0 1 0 40,000 0 0 1 17,500
1
1
1
60,000
R2 1 R3 0 05 1 3 37,500 0 3 0 120,000 1 0 1 2500 R1 R3 R1 0 1 0 40,000 . Thus, $2500 is 0 0 1 17,500
R2 2R1 R2 R3 2R1 R3
1 0 1 20,000
invested in Bank A, $40,000 in Bank B, and $17,500 in Bank C. 74. Let x, y, and z be the weights (in pounds) of haddock, sea bass, and red snapper. Our system has the matrix 1 1 1 560 1 1 1 560 1 R2 R2 R3 R2 225 representation 0 225 0 765 375 225 600 1725 R3 375R1 R3 375R2 225R3 R3 0 375 225 1140 375 0 600 960 1 1 1 560 1 0 1 220 1 0 0 160 R1 R2 R1 1 R3 R1 0 1 0 0 1 0 340 R 340 0 1 0 340 . Thus, he caught 160 lb 1 R 50625 3 0 0 50625 30375 0 0 1 60 0 0 1 60 of haddock, 340 lb of sea bass, and 60 lb of red snapper.
CHAPTER 10
Test
53
CHAPTER 10 TEST
1 8 0
0
0 0 0
0
1. 0 1 7 10 is in rowechelon form, but not reduced rowechelon form because the 1 in the second row does not have a 0 above it. 0 0 0 4 0 0 2 5 2. is in neither rowechelon nor reduced rowechelon form. 0 1 2 7 1 0 3 0 1 0 0 is in reduced rowechelon form. 3. 0 0 1
1 0 0
3
0 0 1
3 2
4. 0 1 0 2 is in reduced rowechelon form.
x y 2z 0 5. 2x 4y 5z 5 2y 3z 5
1 1
has the matrix representation 2 4
0
5 5
R2 R1 R2
1 1
2
0
0 2 1 5 0 2 3 5
2 3 5 1 1 2 0 1 1 2 0 1 R3 R2 R3 5 0 2 1 5 2 R3 0 2 1 5 . Thus z 0, 2y 0 5 y 2 , and 0 0 2 0 0 0 1 0 5 5 5 x 2 2 0 0 x 2 . Thus, the solution is 2 52 0 .
2x 3y z 3 6. x 2y 2z 1 4x y 5z 4 R2 2R1 R2 R3 4R1 R3
0
2
2 3 1
has the matrix representation 1
1 2 2 1 0 7 3 5 0 7 3 8
R3 R2 R3
4
3
2 2 1 1 5
R1 R2
4
1 2 2 1 0 7 3 5 . 0 0 0 3
Since the last row corresponds to the equation 0 3, this system has no solution. x 2y 3 1 2 0 3 R R R 3 1 3 7. has the matrix representation 3y z 2 0 3 1 2 x 2y z 2 1 2 1 2 3R3 4R2 R3
1 2 0 3 0 3 1 2 0 0 7 7
1 7 R3
1 2 0 3 0 3 1 2 . 0 0 1 1
1
2 2 1
2 3 1 4 1 5
1
2 0
3 4
3
0 3 1 2 0 4 1 5
Thus z 1, 3y 1 2 y 1, and x 2 1 3 x 1. Hence, the solution is 1 1 1.
54
CHAPTER 10 Matrices and Determinants
x 3y z 0 8. 3x 4y 2z 1 x 2y 1
has the matrix representation
1 3 1 0 0 5 1 1 0 0 0 0
R3 R2 R3
1 3 1
0
3 4 2 1
15 R2
1
3 1
0
0 5 1 1 0 5 1 1 1 0 25 35 1 . 0 1 1 5 5 0 0 0 0
R2 3R1 R2 R3 R1 R3
1 2 0 1 1 3 1 0 1 3R2 R1 0 1 1 1 R 5 5 0 0 0 0
Since this system is dependent, let z t. Then y 15 t 15 y 15 t 15 and x 25 t 35 x 25 t 35 . Thus, the solution is 25 t 35 15 t 15 t .
In Solutions 9–16, A
2 3 2 4
2 4
1 0 4
, B 1 1 , and C 1 1 2 . 3 0
0 1 3
9. A B is undefined because A is 2 2 and B is 3 2, so they have incompatible dimensions. 10. AB is undefined because A is 2 2 and B is 3 2, so they have incompatible dimensions. 12 22 6 12 6 10 2 4 2 4 2 3 3 1 1 0 1 3 3 3 2 11. B A 3B 1 1 2 4 3 0 6 9 9 0 3 9 3 0
1 0 4
2 3
2 4
14
4
36 58
2 3 2 3 12. C B A 1 1 2 1 1 2 4 3 3 2 4 0 3 0 1 3 3 0 8 1 18 28
13. A
2 4
A1
1 4 3 2 32 8 6 2 2 1 1
14. B 1 does not exist because B is not a square matrix.
15. det B is not defined because B is not a square matrix. 1 0 4 1 1 1 2 1 4 3 4 16. det C 1 1 2 1 0 1 1 3 0 1 3 17. (a) The system
4x 3y 10
3x 2y 30
is equivalent to the matrix equation
4 3
3 2
x y
10 30
.
4 3 2 3 x 2 3 10 70 4 2 3 3 1. So D 1 and . (b) We have D 3 4 y 3 4 30 90 3 2 Therefore, x 70 and y 90.
CHAPTER 10
Test
55
1 4 0 1 4 1 1 0 1 1 2. Since A 0, A does not have an inverse, and 0, B 0 2 0 2 18. A 0 2 0 2 3 1 1 1 3 0 1 1 0 1
since B 0, B does have an inverse.
1 0 0 1 2 0
0 2 0 0 1 0 0 0 1 3 6 1
z 14 2x 19. 3x y 5z 0 4x 2y 3z 2
1 2 R2
1 4 0 1 0 0
0 2 0 0 1 0
1
4 0 1 0 0
0 2 0 0 1 0 0 12 1 3 0 1
R3 3R1 R3
3 0 1 0 0 1
1 0 0 1 2 0
1 2 0
R1 2R2 R1 R3 6R2 R3
0 1 0 0 1 0 . Therefore, B 1 0 1 0 . 2 2 0 0 1 3 6 1 3 6 1
2 0 1 3 1 1 5 26 10 36, 1 Then D 3 1 5 2 4 2 2 3 4 2 3
14 0 1 0 1 1 5 182 2 180, 1 Dx 0 1 5 14 4 2 2 3 2 2 3
2 14 1 2 14 14 1 D y 3 0 5 3 120 300 180, and 5 4 2 2 3 4 2 3 2 0 14 3 1 1 0 4 140 144. Dz 3 1 0 2 14 2 2 2 2 4 2 2
180 144 Therefore, x 180 36 5, y 36 5, z 36 4, and so the solution is 5 5 4.
20. Let x and y represent the number of pounds of almonds and walnuts respectively. Then the problem is modeled by the system
of equations
x
y 3
475x 345y 1191
1 1 345 475 13, , so det D Then D 475 345 475 345
1
1
3 1 3 1 1191 1425 234. Then det Dx 1035 1191 156, and det D y 475 1191 1191 345 x
D y 234 Dx 156 12 and y 18, so the customer bought 12 pounds of almonds and 18 pounds D D 13 13
of walnuts.
56
FOCUS ON MODELING
FOCUS ON MODELING Computer Graphics
1. The data matrix D
0 1 1 0
represents the gray
0 0 1 1
square.
Reflection using T
TD
y
2
1
0
0 1
1
0
0 1
:
0 1 1 0
0 1
0 0 1 1
y
1
1
0
0 0 1 1
2
0
1
1
x
2
_1
1
0
2
x
1 1
2
x
_1
Expansion with c 2 using T
2 0
: 0 1 2 0 0 1 1 0 0 2 2 0 TD 0 1 0 0 1 1 0 0 1 1 y
Shearing with c 1 using T
: 0 1 1 1 0 1 1 0 0 1 2 1 TD 0 1 0 0 1 1 0 0 1 1 y
2
2
1
1
0
1
0
x
2
_1
_1
2. The data matrix D
0 1 1 0 0 0 1 1
square.
represents the
Reflection in yaxis using T1
T1 D
y
2
1
0
0 1
1
0
1
0 1 1 0 0 0 1 1 y
1 0
0 1
3 1
2
2
x
_1
1
_1
0
1
x
:
0 1 1 0
0
0
1 1
57
Computer Graphics
Expansion in ydirection with c 3 using T2
T2 D
1 0 0 3
0 1 1 0 0 0 1 1 y
3. (a) T
1 15
0 1 1 0 0 0 3 3
0 3
:
Shear in ydirection with c 1 using T3
y
3
2
2
1
1
x
1
1 0
: 1 1 1 0 0 1 1 0 0 1 1 0 T3 D 1 1 0 0 1 1 0 1 2 1
3
0
_1
1 0
_1
0
x
1
is a rightward shear in the xdirection. 0 1 1 1 15 1 15 1 (b) T 1 0 1 0 1
(c) T 1 is a leftward shear in the xdirection.
(d) The result is the original matrix. Algebraically, T 1 T D T 1 T D I D D where I is the 2 2 identity 1 15 1 15 0 1 1 0 1 15 0 1 25 15 0 1 1 0 matrix: 0 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1
4. (a) T
3 0 0 1
xdirection.
(b) S
is an expansion by a factor of 3 in the
1 0 0 2
ydirection.
is an expansion by a factor of 2 in the
y
y
2
2
1
1
0
1
2
3
x
0
_1
3 0
0 3 3 0
(c) T S D
0 1
1 0
0 2
0 1 1 0 0 0 1 1
3 0 0 1
0 1 1 0 0 0 2 2
This corresponds to expansion by a factor of 3 in the xdirection and a factor of 2 in the ydirection.
x
3
_1
0 0 2 2
2
1
y
2 1
0 _1
1
2
3
x
58
FOCUS ON MODELING
(d) W T S
3 0
1 0
3 0
0 2 3 0 0 1 1 0 0 3 3 0 . It is the same as T S D. (e) W D 0 2 0 0 1 1 0 0 2 2 0 1 1 4 4 1 1 6 6 0 0 5. (a) D 0 0 4 4 5 5 7 7 8 8 0 075 0 , (b) T 0 1 075 0 0 1 1 4 4 1 1 6 6 0 0 0 075 075 3 3 075 075 45 45 0 0 TD 0 0 4 4 5 5 7 7 8 8 0 0 1 0 0 4 4 5 5 7 7 8 8 0 1 025 , (c) S 0 1 1 025 0 1 1 4 4 1 1 6 6 0 0 0 1 2 5 525 225 275 775 8 2 0 SD 0 1 0 0 4 4 5 5 7 7 8 8 0 0 0 4 4 5 5 7 7 8 8 0 0 1 2 1 0 0 y represents the 6. (a) The data matrix D 4 0 0 2 4 4 0 0 1
0 2
figure at right.
3
2 1
(b) T
1
1
, 0 1 1 1 0 1 2 1 0 0 TD 0 1 0 0 2 4 4 0 0 1 4 5 4 0 0 0 2 4 4 0
The transformation is a reflection about the xaxis and a shear in the xdirection. 1 1 1 0 1 1 (c) T 0 1 0 1 0 1
0
1
2
3
4
5
x
1
2
3
4
5
x
y 0 _1 _2 _3 _4
CORRECTIONS: p. 3,25,31,39,42,59,50,53,67
CHAPTER 11
CONIC SECTIONS
11.1 11.2
Parabolas 1 Ellipses 5
11.3
Hyperbolas 14
11.4 11.5 11.6
Shifted Conics 22 Rotation of Axes 34 Polar Equations of Conics 46 Chapter 11 Review 54 Chapter 11 Test 69
¥
FOCUS ON MODELING: Conics in Architecture 72
1
11 CONIC SECTIONS 11.1 PARABOLAS 1. A parabola is the set of all points in the plane equidistant from a fixed point called the focus and a fixed line called the directrix of the parabola. 2. The graph of the equation x 2 4 py is a parabola with focus F 0 p, directrix y p, and vertical axis. So the graph of x 2 12y is a parabola with focus F 0 3 and directrix y 3.
3. The graph of the equation y 2 4 px is a parabola with focus F p 0, directrix x p, and horizontal axis. So the graph of y 2 12x is a parabola with focus F 3 0 and directrix x 3.
4. (a) From top to bottom: focus 0 3, vertex 0 0, directrix y 3. (b) From left to right: directrix x 3, vertex 0 0, focus 3 0.
5. y 2 2x is Graph III, which opens to the right and is not as wide as the graph for Exercise 9.
6. y 2 14 x is Graph V, the only graph that opens to the left.
7. x 2 6y is Graph II, which opens downward and is narrower than the graph for Exercise 10. 8. 2x 2 y x 2 12 y is Graph I, the only graph that opens upward.
9. y 2 8x 0 y 2 8x is Graph VI, which opens to the right and is wider than the graph for Exercise 5.
10. 12y x 2 0 x 2 12y is Graph IV, which opens downward and is wider than the graph for Exercise 7. 11. (a) x 2 16y, so 4 p 16 p 4. The focus is 0 4, the directrix is y 4, and the focal diameter is 16.
12. (a) x 2 8y, so 4 p 8 p 2. The focus is
0 2, the directrix is y 2, and the focal diameter is 8.
y
(b)
(b)
1
2 x
2
13. (a) y 2 4x, so 4 p 4 p 1. The focus is
1 0, the directrix is x 1, and the focal diameter is 4.
(b)
y
the directrix is x 6, and the focal diameter is 24. y
2 1
1
x
14. (a) y 2 24x, so 4 p 24 p 6. The focus is 6 0, (b)
y
1
1
x
x
1
2
CHAPTER 11 Conic Sections
1 y 2 y 2 16x, so 4 p 16 p 4. The 15. (a) x 16
focus is 4 0, the directrix is x 4, and the focal diameter is 16.
16. (a) y 12 x 2 x 2 2y, so 4 p 2 p 12 . The focus is 0 12 , the directrix is y 12 , and the focal diameter is 2.
y
(b)
y
(b) 2
1
x
1
x
1
17. (a) y 2x 2 x 2 12 y, so 4 p 12 p 18 . The focus is 0 18 , the directrix is y 18 , and the focal diameter is 12 .
(b)
1 y 2 y 2 12x, so 4 p 12 18. (a) x 12
p 3. The focus is 3 0, the directrix is x 3,
and the focal diameter is 12. y
(b)
y 1
x
2
19. (a) 5y x 2 , so 4 p 5 p 54 . The focus is 0 54 , the directrix is y 54 , and the focal diameter is 5.
(b)
x
1
_1
20. (a) 9x y 2 , so 4 p 9 p 94 . The focus is 94 0 , the directrix is x 94 , and the focal diameter is 9.
(b)
y
y
2 1 1
21. (a) x 2 12y 0 x 2 12y, so 4 p 12
p 3. The focus is 0 3, the directrix is y 3,
and the focal diameter is 12. (b)
x
1
x
22. (a) x 15 y 2 0 y 2 5x, so 4 p 5 p 54 . The focus is 54 0 , the directrix is x 54 , and the focal diameter is 5.
y
(b) 1 2
y
x 1 1
x
SECTION 11.1 Parabolas
23. (a) 5x 3y 2 0 y 2 53 x. Then 4 p 53 5 . The focus is 5 0 , the directrix is p 12 12 5 , and the focal diameter is 5 . x 12 3 y
(b)
2
x= 5 12 1
24. (a) 8x 2 12y 0 x 2 32 y. Then 4 p 32 p 38 . The focus is 0 38 , the directrix is y 38 , and the focal diameter is 32 . y
(b)
1
25. x 2 20y
x
1
x
26. x 2 8y
5
-5
5 -2
-10
10
27. y 2 13 x
28. 8y 2 x 2
1
5
-2
10
-2
-1
29. 4x y 2 0
30. x 2y 2 0 2
In #29, ndicate y-axis scale. -2
-1
1
2
4
-2
31. Since the focus is 0 3, p 3 4 p 12. Hence, the standard equation of the parabola is x 2 12y. 32. Since the focus is 0 18 , p 18 4 p 12 . Hence, the standard equation of the parabola is x 2 12 y. 33. Since the focus is 8 0, p 8 4 p 32. Hence, the standard equation of the parabola is y 2 32x.
34. Since the focus is 5 0, p 5 4 p 20. Hence, the standard equation of the parabola is y 2 20x. 35. Since the focus is 0 34 , p 34 4 p 3. Hence, the standard equation of the parabola is x 2 3y. 1 0 , p 1 4 p 1 . Hence, the standard equation of the parabola is y 2 1 x. 36. Since the focus is 12 12 3 3 37. Since the directrix is x 2, p 2 4 p 8. Hence, the standard equation of the parabola is y 2 8x.
3
4
CHAPTER 11 Conic Sections
38. Since the directrix is y 14 , p 14 4 p 1. Hence, the standard equation of the parabola is x 2 y.
1 , p 1 4 p 2 . Hence, the standard equation of the parabola is x 2 2 y. 39. Since the directrix is y 10 10 5 5
40. Since the directrix is x 18 , p 18 4 p 12 . Hence, the standard equation of the parabola is y 2 12 x.
1 , p 1 4 p 1 . Hence, the standard equation of the parabola is y 2 1 x. 41. Since the directrix is x 20 20 5 5
42. Since the directrix is y 5, p 5 4 p 20. Hence, the standard equation of the parabola is x 2 20y.
43. The focus is on the positive x-axis, so the parabola opens horizontally with 2 p 2 4 p 4. Hence, the standard equation of the parabola is y 2 4x.
44. The focus is on the negative y-axis, so the parabola opens vertically with 2 p 6 4 p 12. Thus, the standard equation of the parabola is x 2 12y.
45. The parabola opens downward with focus 10 units from 0 0, so p 10 4 p 40 and the standard equation of the parabola is x 2 40y.
46. Since the parabola opens upward with focus 5 units from the vertex, the focus is 5 0. So p 5 4 p 20. Thus, the standard equation of the parabola is x 2 20y.
47. The directrix has y-intercept 6, and so p 6 4 p 24. Therefore, the standard equation of the parabola is x 2 24y.
48. Since the focal diameter is 8 and the focus is on the negative y-axis, 4 p 8. So the standard equation is x 2 8y.
49. p 6 4 p 24. Since the parabola opens upward, its standard equation is x 2 24y.
50. The directrix is x 2, and so p 2 4 p 8. Since the parabola opens to the left, its standard equation is y 2 8x.
51. p 4 4 p 16. Since the parabola opens to the left, its standard equation is y 2 16x.
52. p 3 4 p 12. Since the parabola opens downward, its standard equation is x 2 12y.
53. The focal diameter is 4 p 32 32 3. Since the parabola opens to the left, its standard equation is y 2 3x.
54. The focal diameter is 4 p 2 5 10. Since the parabola opens upward, its standard equation is x 2 10y.
55. The equation of the parabola has the form y 2 4 px. Since the parabola passes through the point 4 2, 22 4 p 4 4 p 1, and so its standard equation is y 2 x.
56. Since the directrix is x p, we have p2 16, so p 4, and the standard equation is y 2 4 px or y 2 16x. 57. The area of the shaded region is width height 4 p p 8, and so p2 2 p 2 (because the parabola opens downward). Therefore, its standard equation is x 2 4 py 4 2y x 2 4 2y. 58. The focus is 0 p. Since the line has slope 12 , an equation of the line is y 12 x p. Therefore, the point where the
line intersects the parabola has y-coordinate 12 2 p p 1. The parabola’s equation is of the form x 2 4 py, so 1 5 2 2 (since p 0). Hence, the standard equation of the parabola is 2 4 p p 1 p p 1 0 p 2 x2 2 5 1 y.
59. (a) A parabola with directrix y p has equation x 2 4 py. If the directrix is y 12 , then p 12 , so an equation is x 2 4 12 y x 2 2y.
If the directrix is y 1, then p 1, so an equation is x 2 4 1 y x 2 4y. If the directrix is y 4, then p 4, so an equation is
x 2 4 4 y x 2 16y. If the directrix is y 8, then p 8, so
the standard equation is x 2 4 8 y x 2 32y.
(b)
-4
-2
1
0 -1 -2
2
4 x@=_32y x@=_16y
-3 -4 -5
x@=_4y x@=_2y
As the directrix moves further from the vertex, the parabolas get flatter.
SECTION 11.2 Ellipses
60. (a) If the focal diameter of a parabola is 4 p, it has standard equation
(b)
10
x 2 4 py. If the focal diameter is 4 p 1, the standard equation is
2y=x@
8 4y=x@
6
x 2 y. If the focal diameter is 4 p 2, the standard equation is x 2 2y.
4
If the focal diameter is 4 p 4, the standard equation is x 2 4y. If the
focal diameter is 4 p 8, the standard equation is x 2 8y.
y=x@
5
8y=x@
2 -4
-2
0
2
4
As the focal diameter increases, the parabolas get flatter.
61. (a) Since the focal diameter is 12 cm, 4 p 12. Hence, the parabola has standard equation y 2 12x.
(b) At a point 20 cm horizontally from the vertex, the parabola passes through the point 20 y, and hence from part (a), y 2 12 20 y 2 240 y 4 15. Thus, C D 8 15 31 cm.
62. The equation of the parabola has the form x 2 4 py. From the diagram, the parabola passes through the point 10 1, and so 102 4 p 1 4 p 100 and p 25. Therefore, the receiver is 25 ft from the vertex.
63. With the vertex at the origin, the top of one tower will be at the point 300 150. Inserting this point into the equation x 2 4 py gives 3002 4 p 150 90000 600 p p 150. So the standard equation of the parabolic part of the
cables is x 2 4 150 y x 2 600y.
64. The equation of the parabola has the form x 2 4 py. From the diagram, the parabola passes through the point 100 379, and so 1002 4 p 379 1516 p 10000 and p 65963. Therefore, the receiver is about 65963 inches 55 feet from the vertex.
65. Many answers are possible: satellite dish TV antennas, sound surveillance equipment, solar collectors for hot water heating or electricity generation, bridge pillars, etc. 66. Yes. If a cone intersects a plane that is parallel to a line on the cone, the resulting curve is a parabola, as shown in the text.
11.2 ELLIPSES 1. An ellipse is the set of all points in the plane for which the sum of the distances from two fixed points F1 and F2 is constant. The points F1 and F2 are called the foci of the ellipse. x2 y2 2. The graph of the equation 2 2 1 with a b 0 is an ellipse with horizontal major axis, vertices a 0 and a 0 a b x2 y2 and foci c 0, where c a 2 b2 . So the graph of 2 2 1 is an ellipse with vertices 5 0 and 5 0 and foci 5 4 3 0 and 3 0. x2 y2 3. The graph of the equation 2 2 1 with a b 0 is an ellipse with vertical major axis, vertices 0 a and 0 a b a x2 y2 and foci 0 c, where c a 2 b2 . So the graph of 2 2 1 is an ellipse with vertices 0 5 and 0 5 and foci 4 5 0 3 and 0 3. 4. (a) From left to right: vertex 5 0, focus 3 0, focus 3 0, vertex 5 0.
(b) From top to bottom: vertex 0 5, focus 0 3, focus 0 3, vertex 0 5.
y2 x2 1 is Graph II. The major axis is horizontal and the vertices are 4 0. 16 4 y2 6. x 2 1 is Graph IV. The major axis is vertical and the vertices are 0 3. 9 7. 4x 2 y 2 4 x 2 14 y 2 1 is Graph I. The major axis is vertical and the vertices are 0 2. 5.
6
CHAPTER 11 Conic Sections 1 x 2 1 y 2 1 is Graph III. The major axis is horizontal and the vertices are 5 0. 8. 16x 2 25y 2 400 25 16
9.
y2 x2 1. 25 9
(c)
(a) This ellipse has a 5, b 3, and so c2 a 2 b2 16 c 4. The vertices c are 5 0, the foci are 4 0, and the eccentricity is e 45 08. a
y
1 1
x
1
x
2
x
(b) The length of the major axis is 2a 10, and the length of the minor axis is 2b 6.
10.
y2 x2 1 16 25
(c)
(a) This ellipse has a 5, b 4, and so c2 25 16 9 c 3. The vertices
y
1
are 0 5, the foci are 0 3, and the eccentricity is e ac 35 06.
(b) The length of the major axis is 2a 10, and the length of the minor axis is 2b 8.
11.
y2 x2 1 36 81
(c)
(a) This ellipse has a 9, b 6, and so c2 81 36 45 c 3 5. The vertices are 0 9, the foci are 0 3 5 , and the eccentricity is
y
2
e ac 35 .
(b) The length of the major axis is 2a 18 and the length of the minor axis is 2b 12.
12.
x2 y2 1 4
(c)
(a) This ellipse has a 2, b 1, and so c2 4 1 3 c 3. The vertices are 2 0, the foci are 3 0 , and the eccentricity is e ac 23 .
y
1 1
x
(b) The length of the major axis is 2a 4 and the length of the minor axis is 2b 2.
13.
y2 x2 1 49 25 (a) This ellipse has a 7, b 5, and so c2 49 25 24 c 2 6. The vertices are 7 0, the foci are 2 6 0 , and the eccentricity is e ac 2 7 6 .
(b) The length of the major axis is 2a 14 and the length of the minor axis is 2b 10.
(c)
y
2 2
x
7
SECTION 11.2 Ellipses
14.
y2 x2 1 9 64
(c)
(a) This ellipse has a 8, b 3, and so c2 64 9 55 c 55. The vertices are 0 8, the foci are 0 55 , and the eccentricity is
y
2 x
2
e ac 855 .
(b) The length of the major axis is 2a 16 and the length of the minor axis is 2b 6.
15. 9x 2 4y 2 36
y2 x2 1 4 9
(c)
(a) This ellipse has a 3, b 2, and so c2 9 4 5 c 5. The vertices are 0 3, the foci are 0 5 , and the eccentricity is e ac 35 .
y
1 1
x
(b) The length of the major axis is 2a 6, and the length of the minor axis is 2b 4.
16. 4x 2 25y 2 100
y2 x2 1 25 4
(c)
(a) This ellipse has a 5, b 2, and so c2 25 4 21 c 21. The vertices are 5 0, the foci are 21 0 , and the eccentricity is
y
1 1
x
1
x
1
x
e ac 521 .
(b) The length of the major axis is 2a 10, and the length of the minor axis is 2b 4.
17. x 2 4y 2 16
y2 x2 1 16 4
(c)
(a) This ellipse has a 4, b 2, and so c2 16 4 12 c 2 3.The vertices are 4 0, the foci are 2 3 0 , and the eccentricity is
y
1
e ac 2 4 3 23 .
(b) The length of the major axis is 2a 8, and the length of the minor axis is 2b 4.
18. 4x 2 y 2 16
y2 x2 1 4 16
(a) This ellipse has a 4, b 2, and so c2 16 4 12 c 2 3. The vertices are 0 4, the foci are 0 2 3 , and the eccentricity is e ac 2 4 3 23 .
(b) The length of the major axis is 2a 8, and the length of the minor axis is 2b 4.
(c)
y
1
8
CHAPTER 11 Conic Sections
19. 16x 2 25y 2 1600
y2 x2 1 100 64
(c)
(a) This ellipse has a 10, b 8, and so c2 100 64 36 c 6. The
y
2
vertices are 10 0, the foci are 6 0, and the eccentricity is e ac 35 .
x
2
(b) The length of the major axis is 2a 20 and the length of the minor axis is 2b 16.
y2 x2 1 49 2 (a) This ellipse has a 7, b 2, and so c2 49 2 47 c 47. The vertices are 7 0, the foci are 47 0 , and the eccentricity is
20. 2x 2 49y 2 98
(c)
y
1 x
1
e ac 747 .
(b) The length of the major axis is 2a 14 and the length of the minor axis is 2b 2 2. y2 x2 1 3 9 (a) This ellipse has a 3, b 3, and so c2 9 3 6 c 6. The vertices are 0 3, the foci are 0 6 , and the eccentricity is
21. 3x 2 y 2 9
(c)
y
1 1
x
1
x
1
x
e ac 36 .
(b) The length of the major axis is 2a 6 and the length of the minor axis is 2b 2 3. y2 x2 1 9 3 (a) This ellipse has a 3, b 3, and so c2 9 3 6 c 6. The vertices are 3 0, the foci are 6 0 , and the eccentricity is
22. x 2 3y 2 9
(c)
y
1
e ac 36 .
(b) The length of the major axis is 2a 6 and the length of the minor axis is 2b 2 3. y2 x2 1 2 4 (a) This ellipse has a 2, b 2, and so c2 4 2 2 c 2. The vertices are 0 2, the foci are 0 2 , and the eccentricity is
23. 2x 2 y 2 4
e ac 22 .
(b) The length of the major axis is 2a 4 and the length of the minor axis is 2b 2 2.
(c)
y
1
9
SECTION 11.2 Ellipses
y2 x2 1 4 3 (a) This ellipse has a 2, b 3, and so c2 4 3 c 1. The vertices are
24. 3x 2 4y 2 12
(c)
y
1
2 0, the foci are 1 0, and the eccentricity is e ac 12 .
x
1
(b) The length of the major axis is 2a 4 and the length of the minor axis is 2b 2 3. 25. x 2 4y 2 1
y2 x2 1 1 1
(c)
y 1
4
(a) This ellipse has a 1, b 12 , and so c2 1 14 34 c 23 . The vertices are 1 0, the foci are 23 0 , and the eccentricity is 3 e ac 32 1 2 .
1
x
(b) The length of the major axis is 2a 2, and the length of the minor axis is 2b 1.
26. 9x 2 4y 2 1
y2 x2 1 19 14
(c)
5 c 5 . The vertices are (a) This ellipse has a 12 , and so c2 14 19 36 6 56 5 1 0 2 , the foci are 0 6 , and the eccentricity is e ac 12 35 .
y 0.5
0.5 x
(b) The length of the major axis is 2a 1, and the length of the minor axis is 2b 23 .
27. x 2 4 2y 2 x 2 2y 2 4
y2 x2 1 4 2
(c)
2, and so c2 4 2 2 c 2. The vertices are 2 0, the foci are 2 0 , and the eccentricity is
(a) This ellipse has a 2, b
y
1 x
1
e ac 22 .
(b) The length of the major axis is 2a 4, and the length of the minor axis is 2b 2 2. x2 y2 1 28. y 2 1 2x 2 2x 2 y 2 1 1 1
2 2 (a) This ellipse has a 1, b 2 , and so c2 1 12 12 c 22 . The vertices are 0 1, the foci are 0 22 , and the eccentricity is e ac 11 2 22 .
(b) The length of the major axis is 2a 2, and the length of the minor axis is 2b 2.
(c)
y 1
1
x
10
CHAPTER 11 Conic Sections
29. This ellipse has a horizontal major axis with a 5 and b 4, so its standard equation is
x2
y2 x 2 y2 1. 1 25 16 52 42
y2 y2 x2 x2 1. 30. This ellipse has a vertical major axis with a 5 and b 2. Thus, its standard equation is 2 2 1 4 25 2 5 31. This ellipse has a vertical major axis with c 2 and b 2. So a 2 c2 b2 22 22 8 a 2 2. So its standard equation is
x2
y2 y2 x2 1. 1 2 4 8 22 2 2
32. This ellipse has a vertical major axis with a 4 and c 3. So c2 a 2 b2 9 16 b2 b2 7. Thus, its standard equation is
x2 y2 y2 x2 2 1 1. 7 7 16 4
x2 y2 1. Substituting the 162 b2 64 36 1 36 1 1 36 3 b2 4 36 48. Thus, the standard point 8 6 into the equation, we get 256 4 4 3 b2 b2 b2 2 2 y x 1. equation of the ellipse is 256 48
33. This ellipse has a horizontal major axis with a 16, so its standard equation has the form
x2 y2 34. This ellipse has a vertical major axis with b 2, so its standard equation has the form 2 2 1. Substituting the 2 a 4 16 4 4 1 3 4 4 1 . Thus, the standard point 1 2 into the equation, we get 4 2 1 2 1 2 a 2 4 4 3 3 a a a y2 x2 3y 2 x2 1 1. equation of the ellipse is 4 163 4 16 35.
y2 y2 x2 4x 2 x2 1 1 y 2 20 25 20 20 25 5 4x 2 . y 20 5
y2 y2 1 1 x 2 y 2 12 12x 2 12 12 y 12 12x 2 .
36. x 2
4 2
5 -2 -5
-1
5
-2
1
2
-4
-5
37. 6x 2 y 2 36 y 2 36 6x 2 y 36 6x 2 . 5
-10
x2 38. x 2 2y 2 8 2y 2 8 x 2 y 2 4 2 x2 y 4 . 2
10
2
-5 -2
2 -2
SECTION 11.2 Ellipses
11
39. The foci are 4 0, and the vertices are 5 0. Thus, c 4 and a 5, and so b2 25 16 9. Therefore, the standard equation of the ellipse is
y2 x2 1. 25 9
40. The foci are 0 3 and the vertices are 0 5. Thus, c 3 and a 5, and so c2 a 2 b2 9 25 b2 b2 25 9 16. Therefore, the standard equation of the ellipse is
y2 x2 1. 16 25
41. The foci are 1 0 and the vertices are 2 0. Thus, c 1 and a 2, so c2 a 2 b2 1 4 b2 b2 4 1 3. Therefore, the standard equation of the ellipse is
y2 x2 1. 4 3
42. The foci are 0 2 and the vertices are 0 3. Thus, c 2 and a 3, so c2 a 2 b2 4 9 b2 b2 9 4 5. Therefore, the standard equation of the ellipse is
y2 x2 1. 5 9
43. The foci are 0 10 and the vertices are 0 7. Thus, c 10 and a 7, so c2 a 2 b2 10 49 b2 b2 49 10 39. Therefore, the standard equation of the ellipse is
y2 x2 1. 39 49
44. The foci are 15 0 and the vertices are 6 0. Thus, c 15 and a 6, so c2 a 2 b2 15 36 b2 b2 36 15 21. Therefore, the standard equation of the ellipse is
y2 x2 1. 36 21
45. The length of the major axis is 2a 4 a 2, the length of the minor axis is 2b 2 b 1, and the foci are on the y-axis. Therefore, the standard equation of the ellipse is x 2
y2 1. 4
46. The length of the major axis is 2a 6 a 3, the length of the minor axis is 2b 4 b 2, and the foci are on the x-axis. Therefore, the standard equation of the ellipse is
y2 x2 1. 9 4
47. The foci are 0 2, and the length of the minor axis is 2b 6 b 3. Thus, a 2 4 9 13. Since the foci are on the y-axis, the standard equation is
x2 y2 1. 9 13
48. The foci are 5 0, and the length of the major axis is 2a 12 a 6. Thus, c2 a 2 b2 25 36 b2 b2 36 25 11. Since the foci are on the x-axis, the standard equation is
y2 x2 1. 36 11
49. The endpoints of the major axis are 10 0 a 10, and the distance between the foci is 2c 6 c 3. Therefore, b2 100 9 91, and so the standard equation of the ellipse is
x2 y2 1. 100 91
50. Since the endpoints of the minor axis are 0 3, we have b 3. The distance between the foci is 2c 8, so c 4. Thus, a 2 b2 c2 9 16 25, and the standard equation of the ellipse is
x2 y2 1. 25 9
51. The length of the major axis is 10, so 2a 10 a 5, and the foci are on the x-axis, so the form of the equation is 2 5 x2 y2 4 5 4 4 22 2 1. Since the ellipse passes through 2 1 2 1 2 5 2 , we have 25 25 25 5 b b b b y2 x2 2 1. b 5, and so the standard equation is 25 5
12
CHAPTER 11 Conic Sections
52. The length of the minor axis is 10, so 2b 10 b 5, and the foci are on the y-axis, so the form of the equation is 2 2 5 40 40 5 40 4 x 2 y2 2 1. Since the ellipse passes through 2 1 2 5 40 , we have 1 2 25 a 25 25 a 5 a a 2 2 y x 1. a 2 50, and so the standard equation is 25 50 2 6 and 53. The eccentricity is 13 , so e 13 , and the foci are 0 2, so c 2. Thus, e ac a ce 13
b2 a 2 c2 36 4 32. The major axis lies on the y-axis, so the standard equation is 54. The eccentricity is e 075 34 and the foci are 15 0
y2 x2 1. 32 36
3 0 , so c 3 . Thus, a c 32 2 and e 2 2 34
b2 a 2 c2 4 94 74 . The major axis lies on the x-axis, so the standard equation is
y2 x2 7 1. 4
4 55. Since the length of the major axis is 2a 4, we have a 2. The eccentricity is 23 ac 2c , so c 3. Then y2 b2 a 2 c2 4 3 1, and since the foci are on the y-axis, the standard equation of the ellipse is x 2 1.
4
56. The eccentricity is e 35 and the major axis has length 2a 12, so a 6. Thus, c ae 2 5 and y2 x2 1. b2 a 2 c2 36 20 16. The foci are on the x-axis, so the standard equation of the ellipse is
36
4x 2 y 2 4 57. 4x 2 9y 2 36
16
y
Subtracting the first equation from the second gives
(0, 2) 1
8y 2 32 y 2 4 y 2. Substituting y 2 in the first equation gives
(0, _2)
2 y2 x 1 9x 2 16y 2 144 144x 2 256y 2 2304 16 9 58. 16x 2 9y 2 144 144x 2 81y 2 1296 y2 x2 1 9 16
Adding gives 175y 2 1008 y 12 5 . Substituting for y gives 2 1296 12 144 9x 2 144 2304 9x 2 16 12 5 25 25 x 5 , and so the 12 four points of intersection are 12 5 5 .
100x 2 25y 2 100 59. y2 1 x 2 9
Dividing the first equation by 100 gives x 2
y
(_ 125 , 125)
( 125 , 125)
1
x
1
(
12 12 _5 ,_5
)
(
12 12 5,_5
y
y2 1. 4
y2 y2 0 Subtracting this equation from the second equation gives 9 4 1 1 y 2 0 y 0. Substituting y 0 in the second equation gives 9 4 x 2 02 1 x 1, and so the points of intersection are 1 0.
x
1
4x 2 22 4 x 0, and so the points of intersection are 0 2.
(_1, 0)
1
(1, 0) 1
x
)
SECTION 11.2 Ellipses
60.
25x 2 144y 2 3600
144x 2 25y 2 3600
3600x 2 20,736y 2 518,400 3600x 2
625y 2 90,000
61. (a) The ellipse x 2 4y 2 16
y
Subtracting
3600 the second equation from the first, we have 20,111y 2 428,400, so y 2 169 2 60 2 y 60 3600 x 60 13 . Substituting for y gives 25x 144 13 13 , and so 60 the four points of intersection are 60 13 13 .
13
(_ 6013 , 6013 )
10
( 6013 , 6013 ) 10 x
(
_ 60 , _ 60 13 13
)
(
60 , _ 60 13 13
)
y2 x2 1 has a 4 and b 2. Thus, an equation of the ancillary circle is 16 4
x 2 y 2 4.
(b) If s t is a point on the ancillary circle, then s 2 t 2 4 4s 2 4t 2 16 2s2 4 t2 16, which implies that 2s t is a point on the ellipse. 1 k=4 100 x 2 . For the top 62. (a) x 2 ky 2 100 ky 2 100 x 2 y k 2 1 half, we graph y 100 x 2 for k 4, 10, 25, and 50. k=10 k 1 (b) This family of ellipses have common major axes and vertices, and the eccentricity increases as k increases. 63.
k=50
-10-8 -6 -4 -2 0
k=25
2 4 6 8 10
y2 x2 1 is an ellipse for k 0. Then a 2 4 k, b2 k, and so c2 4 k k 4 c 2. Therefore, all of k 4k the ellipses’ foci are 0 2 regardless of the value of k.
64. The foci are c 0, where c2 a 2 b2 . The endpoints of one latus rectum are the points c k, and the length is 2k. b2 a 2 c2 k2 k2 c2 a 2 c2 c2 k2 . Since Substituting this point into the equation, we get 2 2 1 2 1 2 a b b a a2 a2 2b2 b4 b2 b2 b2 a 2 c2 , the last equation becomes k 2 2 k . Thus, the length of the latus rectum is 2k 2 . a a a a 65. Using the perihelion, a c 147,000,000, while using the aphelion, a c 153,000,000. Adding, we have 2 2 2a 300,000,000 a 150,000,000. So b2 a 2 c2 150 106 3 106 22,4911012 224911016 .
y2 x2 1. 16 22500 10 22491 1016 c 66. Using the eccentricity, e 025 c 025a. Using the length of the minor axis, 2b 10,000,000,000 a 18 b 5 109 . Since a 2 c2 b2 , a 2 025a2 25 1018 15 a 2 25 1018 a 2 80 3 10 16 5 5 9 9 9 5 109 . Since the Sun is at one focus of the ellipse, a 80 3 10 4 3 10 . Then c 025 4 3 10 3 the distance from Pluto to the Sun at perihelion is a c 4 53 109 53 109 3 53 109 387 109 km; the distance from Pluto to the Sun at aphelion is a c 4 53 109 53 109 5 53 109 645 109 km. Thus, an equation of the orbit is
67. Using the perilune, a c 1075 68 1143, and using the apolune, a c 1075 195 1270. Adding, we get
2a 2413 a 12065. So c 1270 12065 c 635. Therefore, b2 120652 6352 1,451,610. Since
a 2 1,455,642, an equation of Apollo 11’s orbit is
y2 x2 1. 1,455,642 1,451,610
14
CHAPTER 11 Conic Sections
68. Placing the origin at the center of the sheet of plywood and letting the x-axis be the long central axis, we have 2a 8, so that a 4, and 2b 4, so that b 2. So c2 a 2 b2 42 22 12 c 2 3 346. So the tacks should be located 2 346 692 feet apart and the string should be 2a 8 feet long. 69. From the diagram, a 40 and b 20, and so an equation of the ellipse whose top half is the window is
x2 y2 1. 1600 400
252 h2 Since the ellipse passes through the point 25 h, by substituting, we have 1 625 4y 2 1600 1600 400 5 39 975 1561 in. Therefore, the window is approximately 156 inches high at the specified point. y 2 2 70. Have each friend hold one end of the string on the blackboard. These fixed points will be the foci. Then, keeping the string taut with the chalk, draw the ellipse. 71. We start with the flashlight perpendicular to the wall; this shape is a circle. As the angle of elevation increases, the shape of the light changes to an ellipse. When the flashlight is angled so that the outer edge of the light cone is parallel to the wall, the shape of the light is a parabola. Finally, as the angle of elevation increases further, the shape of the light is hyperbolic. 72. The shape drawn on the paper is almost, but not quite, an ellipse. For example, when the bottle has radius 1 unit and the compass legs are set 1 unit apart, then it can be shown that an equation of the resulting curve is 1 y 2 2 cos x. The graph of this curve differs very slightly from the ellipse with the same major and minor axis. This example shows that in mathematics, things are not always as they appear to be.
11.3 HYPERBOLAS 1. A hyperbola is the set of all points in the plane for which the difference of the distances from two fixed point F1 and F2 is constant. The points F1 and F2 are called the foci of the hyperbola. x2 y2 2. The graph of the equation 2 2 1 with a 0, b 0 is a hyperbola with horizontal transverse axis, vertices a 0 a b x2 y2 and a 0 and foci c 0, where c a 2 b2 . So the graph of 2 2 1 is a hyperbola with vertices 4 0 and 4 3 4 0 and foci 5 0 and 5 0. y2 x2 3. The graph of the equation 2 2 1 with a 0, b 0 is a hyperbola with vertical transverse axis, vertices 0 a a b y2 x2 and 0 a and foci 0 c, where c a 2 b2 . So the graph of 2 2 1 is a hyperbola with vertices 0 4 and 4 3 0 4 and foci 0 5 and 0 5. 4. (a) From left to right: focus 5 0, vertex 4 0, asymptote y 34 x, asymptote y 34 x, vertex 4 0, focus 5 0.
(b) From top to bottom: focus 0 5, vertex 0 4, asymptote y 43 x, asymptote y 43 x, vertex 0 4, focus 0 5.
5.
x2 y 2 1 is Graph III, which opens horizontally and has vertices at 2 0. 4
6. y 2
x2 1 is Graph IV, which opens vertically and has vertices at 0 1. 9
1 x 2 1 is Graph II, which opens vertically and has vertices at 0 3. 7. 16y 2 x 2 144 19 y 2 144 1 x 2 1 y 2 1 is Graph I, which opens horizontally and has vertices at 5 0. 8. 9x 2 25y 2 225 25 9
15
SECTION 11.3 Hyperbolas
9.
y2 x2 1 4 16
(c)
(a) The hyperbola has a 2, b 4, and c2 16 4 c 2 5. The vertices are 2 0, the foci are 2 5 0 , and the asymptotes are y 42 x
y
1 1
x
1
x
1
x
2
x
1
x
y 2x.
(b) The transverse axis has length 2a 4. 10.
x2 y2 1 9 16
(c)
(a) The hyperbola has a 3, b 4, and c2 9 16 25 c 5. The vertices
y
1
are 0 3, the foci are 0 5, and the asymptotes are y 34 x.
(b) The transverse axis has length 2a 6.
11.
x2 y2 1 36 4
(c)
(a) The hyperbola has a 6, b 2, and c2 36 4 40 c 2 10. The vertices are 0 6, the foci are 0 2 10 , and the asymptotes are y 3x.
y
2
(b) The transverse axis has length 2a 12.
12.
y2 x2 1 9 64
(c)
(a) The hyperbola has a 3, b 8, and c2 9 64 73 c 73. The vertices are 3 0, the foci are 73 0 , and the asymptotes are y 83 x.
y
5
(b) The transverse axis has length 2a 6.
13.
x2 y2 1 1 25
(c)
y 1
(a) The hyperbola has a 1, b 5, and c2 1 25 26 c 26. The vertices are 0 1, the foci are 0 26 , and the asymptotes are y 15 x.
(b) The transverse axis has length 2a 2.
14.
y2 x2 1 2 1 (a) The hyperbola has a 2, b 1, and c2 2 1 3 c 3. The vertices are 2 0 , the foci are 3 0 , and the asymptotes are y 22 x.
(b) The transverse axis has length 2a 2 2.
(c)
y
1 1
x
16
CHAPTER 11 Conic Sections
15. x 2 y 2 1
(c)
(a) The hyperbola has a 1, b 1, and c2 1 1 2 c 2. The vertices are 1 0, the foci are 2 0 , and the asymptotes are y x.
y
1 1
x
(b) The transverse axis has length 2a 2.
16.
y2 x2 1 16 12
(c)
(a) The hyperbola has a 4, b 2 3, and c2 16 12 28 c 28 2 7. The vertices are 4 0, the foci are 2 7 0 , and the
y
2 2
x
1
x
2
x
2
x
1
x
asymptotes are y 23 x.
(b) The transverse axis has length 2a 8. 17. 9x 2 4y 2 36
y2 x2 1 4 9
(c)
(a) The hyperbola has a 2, b 3, and c2 4 9 13 c 13. The vertices are 2 0, the foci are 13 0 , and the asymptotes are y 32 x.
y
1
(b) The transverse axis has length 2a 4. 18. 25y 2 9x 2 225
x2 y2 1 9 25
(c)
(a) The hyperbola has a 3, b 5, and c2 25 9 34 c 34. The vertices are 0 3, the foci are 0 34 , and the asymptotes are y 35 x.
y
2
(b) The transverse axis has length 2a 6. 19. 4y 2 9x 2 144
x2 y2 1 36 16
(c)
(a) The hyperbola has a 6, b 4, and c2 a 2 b2 52 c 2 13. The vertices are 0 6, the foci are 0 2 13 , and the asymptotes are
y
2
y 32 x.
(b) The transverse axis has length 2a 12. 20. y 2 25x 2 100
x2 y2 1 100 4
(a) The hyperbola has a 10, b 2, and c2 100 4 104 c 2 26. The vertices are 0 10, the foci are 0 2 26 , and the asymptotes are y 5x.
(b) The transverse axis has length 2a 20.
(c)
y
5
17
SECTION 11.3 Hyperbolas
y2 x2 1 8 2 (a) The hyperbola has a 2 2, b 2, and c2 8 2 10 c 10. The vertices are 2 2 0 , the foci are 10 0 , and the asymptotes are
21. x 2 4y 2 8 0
(c)
y
1 1
x
1
x
1
x
1
x
y 2 x 12 x. 8
(b) The transverse axis has length 2a 4 2. x2 y2 1 3 9 (a) The hyperbola has a 3, b 3, and c2 3 9 12 c 2 3. The vertices are 0 3 , the foci are 0 2 3 , and the asymptotes are
22. 3y 2 x 2 9 0
(c)
y
1
y 33 x.
(b) The transverse axis has length 2a 2 3.
23. x 2 y 2 4 0 y 2 x 2 4
x2 y2 1 4 4
(c)
(a) The hyperbola has a 2, b 2, and c2 4 4 8 2 2. The vertices are 0 2, the foci are 0 2 2 , and the asymptotes are y x.
y
1
(b) The transverse axis has length 2a 4.
x2 y2 1 4 12 (a) The hyperbola has a 2, b 2 3, and c2 4 12 16 c 4. The
24. x 2 3y 2 12 0
(c)
y
1
vertices are 0 2, the foci are 0 4, and the asymptotes are
2 x 3 x. y 3 2 3
(b) The transverse axis has length 2a 4. y2 25. 4y 2 x 2 1 1 x 2 1 4
(a) The hyperbola has a 12 , b 1, and c2 14 1 54 c 25 . The vertices are 0 12 , the foci are 0 25 , and the asymptotes are 1 y 12 1 x 2 x.
(b) The transverse axis has length 2a 1.
(c)
y 1
1
x
18
CHAPTER 11 Conic Sections
26. 9x 2 16y 2 1
y2 x2 1 19 116
y
(c)
1
1 25 c 5 . The (a) The hyperbola has a 13 , b 14 , and c2 19 16 144 12 1 5 vertices are 3 0 , the foci are 12 0 , and the asymptotes are
1
x
3 y 14 13 x 4 x.
(b) The transverse axis has length 2a 23 . 27. From the graph, the foci are 4 0, and the vertices are 2 0, so c 4 and a 2. Thus, b2 16 4 12, and since the vertices are on the x-axis, the standard equation of the hyperbola is
x2 y2 1. 4 12
28. From the graph, the foci are 0 13 and the vertices are 0 12, so c 13 and a 12. Then
b2 c2 a 2 169 144 25, and since the vertices are on the y-axis, the standard equation of the hyperbola is x2 y2 1. 144 25
29. From the graph, the vertices are 0 4, the foci are on the y-axis, and the hyperbola passes through the point 3 5. So
y2 x2 9 25 32 52 2 1. Substituting the point 3 5, we have 2 1 1 2 16 16 16 b b b 9 y2 9 x2 2 2 b 16. Thus, the standard equation of the hyperbola is 1. 16 16 16 b
the equation is of the form
x2 y2 30. The vertices are 2 3 0 , so a 2 3, so the standard equation of the hyperbola is of the form 2 2 1. b 2 3 Substituting the point 4 4 into the equation, we get
standard equation of the hyperbola is
16 16 16 16 16 4 1 2 b2 48. Thus, the 1 2 12 b2 12 12 b b
y2 x2 1. 12 48
a 3 31. From the graph, the vertices are 0 3, so a 3. Since the asymptotes are y 3x x, we have 3 b 1. b b y2 x2 y2 Since the vertices are on the x-axis, the standard equation is 2 2 1 x 2 1. 9 3 1 b 1 3 b 32. The vertices are 3 0, so a 3. Since the asymptotes are y 12 x x, we have b . Since the a 3 2 2 4y 2 y2 x2 x2 1. 1 vertices are on the x-axis, the standard equation is 2 9 9 3 322 33. x 2 2y 2 8 2y 2 x 2 8 y 2 12 x 2 4 y 12 x 2 4
34. 3y 2 4x 2 24 3y 2 4x 2 24 y 2 43 x 2 8 y 43 x 2 8 10
5
-5
5 -5
-10
10 -10
SECTION 11.3 Hyperbolas
35.
x2 y2 x2 y2 x2 1 1 y2 2 2 6 2 6 3 x2 2 y 3 5
-5
36.
19
y2 x2 1 16x 2 25y 2 1600 100 64 2 25y 2 16x 2 1600 y 2 16 25 x 64 2 y 16 25 x 64
20
5
-20
20
-5 -20
37. The foci are 5 0 and the vertices are 3 0, so c 5 and a 3. Then b2 25 9 16, and since the vertices are on the x-axis, the standard equation of the hyperbola is
y2 x2 1. 9 16
38. The foci are 0 10 and the vertices are 0 8, so c 10 and a 8. Then b2 c2 a 2 100 64 36, and since the vertices are on the y-axis, the standard equation of the hyperbola is
x2 y2 1. 64 36
39. The foci are 0 2 and the vertices are 0 1, so c 2 and a 1. Then b2 4 1 3, and since the vertices are on the y-axis, the standard equation is y 2
x2 1. 3
40. The foci are 6 0 and the vertices are 2 0, so c 6 and a 2. Then b2 c2 a 2 36 4 32, and since the vertices are on the x-axis, the standard equation is
y2 x2 1. 4 32
b b 41. The vertices are 1 0 and the asymptotes are y 5x, so a 1. The asymptotes are y x, so 5 b 5. a 1 2 y 1. Therefore, the standard equation of the hyperbola is x 2 25 6 a 42. The vertices are 0 6, so a 6. The asymptotes are y 13 x x 13 b 18. Since the vertices are on b b y2 x2 the y-axis, the standard equation of the hyperbola is 1. 36 324 43. The vertices are 0 6, so a 6. Since the vertices are on the y-axis, the standard equation of the hyperbola has the form y2 81 25 45 x2 25 2 1. Since the hyperbola passes through the point 5 9, we have 2 1 2 b2 20. 36 36 36 b b b x2 y2 1. Thus, the standard equation is 36 20 44. The vertices are 2 0, so a 2. Since the vertices are on the x-axis, the hyperbola has a standard equation of the form y2 30 30 9 5 x2 2 1. Since the hyperbola passes through the point 3 30 , we have 2 1 2 b2 24. 4 4 4 b b b x2 y2 Thus, the standard equation is 1. 4 24 45. The asymptotes of the hyperbola are y x, so b a. Since the hyperbola passes through the point 5 3, its foci are x2 y2 25 9 on the x-axis, and its standard equation has the form, 2 2 1, so it follows that 2 2 1 a 2 16 b2 . a a a a y2 x2 1. Therefore, the standard equation of the hyperbola is 16 16
20
CHAPTER 11 Conic Sections
46. The asymptotes of the hyperbola are y x, so b a. Since the hyperbola passes through the point 1 2, its foci are on
y2 x2 4 1 the y-axis, and its standard equation has the form 2 2 1, so it follows that 2 2 1 a 2 3. Therefore, the a a a a x2 y2 1. standard equation of the hyperbola is 3 3
x2 42 12 y2 47. The foci are 0 3, so c 3 and an equation is 2 2 1. The hyperbola passes through 1 4, so 2 2 1 a b a b 16b2 a 2 a 2 b2 16b2 9 b2 9 b2 b2 b4 8b2 9 0 b2 1 b2 9 0, Thus, b2 1, a 2 8, and the standard equation is
y2 x 2 1. 8
x2 y2 1. The hyperbola passes through 4 18 , 48. The foci are 10 0 , so c 10 and an equation is 2 b a 18 16 2 2 2 2 2 so 2 2 1 16b 18a a b 16b 18 10 b2 10 b2 b2 b4 24b2 180 0 a b y2 x2 b2 30 b2 6 0. Thus, b2 6, a 2 4, and the standard equation is 1. 4 6 49. The foci are 5 0, and the length of the transverse axis is 6, so c 5 and 2a 6 a 3. Thus, b2 25 9 16, and the standard equation is
x2 y2 1. 9 16
50. The foci are 0 1, and the length of the transverse axis is 1, so c 1 and 2a 1 a 12 . Then b2 c2 a 2 1 14 34 , and since the foci are on the y-axis, the standard equation is
x2 y2 4x 2 1 4y 2 1. 14 34 3
y2 x2 1 has a 5 and b 5. Thus, the asymptotes are y x, and their 5 5 slopes are m 1 1 and m 2 1. Since m 1 m 2 1, the asymptotes are perpendicular.
51. (a) The hyperbola x 2 y 2 5
(b) Since the asymptotes are perpendicular, they must have slopes 1, so a b. Therefore, c2 2a 2 a 2
y2 c2 x2 since the vertices are on the x-axis, the hyperbola’s standard equation is 1 1 1 x 2 y 2 . 2 2 2 2c 2c
y2 x2 y2 x2 52. The hyperbolas 2 2 1 and 2 2 1 are conjugate to each other. a b a b (a) x 2 4y 2 16 0
y2 x2 1 and 4y 2 x 2 16 0 16 4
y2 x2 1. So the hyperbolas are conjugate to each other. 16 4 (b) They both have the same asymptotes, y 12 x. (c) The two general conjugate hyperbolas both have asymptotes b y x. a
y
1 1
x
c2 , and 2
SECTION 11.3 Hyperbolas
21
53. x c2 y 2 x c2 y 2 2a. Let us consider the positive case only. Then x c2 y 2 2a x c2 y 2 , and squaring both sides gives x 2 2cx c2 y 2 4a 2 4a x c2 y 2 x 2 2cx c2 y 2 4a x c2 y 2 4cx 4a 2 . Dividing by 4 and squaring both sides gives a 2 x 2 2cx c2 y 2 c2 x 2 2a 2 cx a 4 a 2 x 2 2a 2 cx a 2 c2 a 2 y 2 c2 x 2 2a 2 cx a 4 a 2 x 2 a 2 c2 a 2 y 2 c2 x 2 a 4 . Rearranging the order, we have c2 x 2 a 2 x 2 a 2 y 2 a 2 c2 a 4 c2 a 2 x 2 a 2 y 2 a 2 c2 a 2 . The negative case gives the same result.
54. (a) The hyperbola F2 5 0.
x2 y2 1 has a 3, b 4, so c2 9 16 25, and c 5. Therefore, the foci are F1 5 0 and 9 16
25 2569 25 16 (b) Substituting P 5 16 3 into an equation of the hyperbola, we get 9 16 9 9 1, so P lies on the hyperbola. (c) d P F1 5 52 163 02 16 , and d F P 5 52 163 02 13 900 256 34 2 3 3 .
16 (d) d P F2 d P F1 34 3 3 6 2 3 2a.
55. (a) From the equation, we have a 2 k and b2 16 k. Thus, c2 a 2 b2 k 16 k 16 c 4. Thus the foci of the family of hyperbolas are 0 4. x2 kx 2 y2 x2 2 (b) 1 y k 1 y k . For k=12 k 16 k 16 k 16 k 8 k=8 6 2 kx , k 1, 4, 8, 12. As k the top branch, we graph y k k=4 4 16 k increases, the asymptotes get steeper and the vertices move further apart.
-10
2
k=1
0
10
56. d A B 500 2c c 250. (a) Since t 2640 s and 980 fts, we have d d P A d P B t 980 fts 2640 s 2,587,200 ft 490 mi.
(b) c 250, 2a 490 a 245 and the foci are on the y-axis. Then b2 2502 2452 2475. Hence, the standard equation of the hyperbola is
x2 y2 1. 60,025 2475
2502 x2 (c) Since P is due east of A, c 250 is the y-coordinate of P. Therefore, P is at x 250, and so 1 2475 2452 2502 x 2 2475 1 10205. Then x 101, and so P is approximately 101 miles from A. 2452 57. Since the asymptotes are perpendicular, a b. Also, since the sun is a focus and the closest distance is 2 109 , it follows 2 109 and that c a 2 109 . Now c2 a 2 b2 2a 2 , and so c 2a. Thus, 2a a 2 109 a 21 4 1018 x2 y2 a 2 b2 1 23 1019 . Therefore, the standard equation of the hyperbola is 23 1019 23 1019 32 2 x 2 y 2 23 1019 .
22
CHAPTER 11 Conic Sections
58. (a) These equally spaced concentric circles can be used as a kind of measure where we count the number of rings. In the case of the red dots, the sum of the number of wave crests from each center is a constant, in this case 17. As you move out one wave crest from the left stone you move in one wave crest from the right stone. Therefore this satisfies the geometric definition of an ellipse. (b) Similarly, in the case of the blue dots, the difference of the number of wave crests from each center is a constant. As you move out one wave crest from the left center you also move out one wave crest from the right stone. Therefore this satisfies the geometric definition of a hyperbola. 59. Some possible answers are: as cross-sections of nuclear power plant cooling towers, or as reflectors for camouflaging the location of secret installations. 60. The wall is parallel to the axis of the cone of light coming from the top of the shade, so the intersection of the wall and the cone of light is a hyperbola. In the case of the flashlight, hold it parallel to the ground to form a hyperbola.
11.4 SHIFTED CONICS 1. (a) If we replace x by x 3 the graph of the equation is shifted to the right by 3 units. If we replace x by x 3 the graph is shifted to the left by 3 units. (b) If we replace y by y 1 the graph of the equation is shifted upward by 1 unit. If we replace y by y 1 the graph is shifted downward by 1 unit. 2. x 2 12y, from top to bottom: focus 0 3, vertex 0 0, directrix y 3. x 32 12 y 1, from top to bottom: focus 3 4, vertex 3 1, directrix y 2. 3.
4.
5.
x2 y2 x 32 y 12 2 1, from left to right: vertex 5 0, focus 3 0, focus 3 0, vertex 5 0. 1, from 2 2 5 4 5 42 left to right: vertex 2 1, focus 0 1, focus 6 1, vertex 8 1. y2 x2 1, from left to right: focus 5 0, vertex 4 0, asymptote y 34 x, asymptote y 34 x, vertex 4 0, 42 32 x 32 y 12 focus 5 0. 1, from left to right: focus 2 1, vertex 1 1, asymptote y 34 x 13 4 , 2 4 32 asymptote y 34 x 54 , vertex 7 1, focus 8 1. x 22 y 12 1 9 4 x2 y2 1 by shifting it 2 units to 9 4 the right and 1 unit upward. So a 3, b 2, and c 9 4 5. The
(a) This ellipse is obtained from the ellipse
center is 2 1, the vertices are 2 3 1 1 1 and 5 1, and the foci are 2 5 1 .
(b) The length of the major axis is 2a 6 and the length of the minor axis is 2b 4.
(c)
y
1 1
x
23
SECTION 11.4 Shifted Conics
6.
x 32 y 32 1 16
(c) x2
1 x
1
y 2 1 by shifting to the right 3 units and downward 3 units. So a 4, b 1, and c 16 1 15. The center is 3 3, the vertices are 3 4 3 1 3 and 7 3, and the foci are 3 15 3 .
(a) This ellipse is obtained from the ellipse
y
16
(b) The length of the major axis is 2a 8 and the length of the minor axis is 2b 2.
7.
x2 y 52 1 9 25
(c)
y
1 x
1
x2 y2 1 by shifting it 5 units (a) This ellipse is obtained from the ellipse 9 25 downward. So a 5, b 3, and c 25 9 4. The center is 0 5, the vertices are 0 5 5 0 10 and 0 0, and the foci are 0 5 4 0 9 and 0 1.
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 6.
8. x 2
y 22 1 4
(c)
y2 (a) This ellipse is obtained from the ellipse x 2 1 by shifting it 2 units 4 downward. So a 2, b 1, and c 4 1 3. The center is 0 2, the
y
1 x
1
vertices are 0 2 2 0 4 and 0 0, and the foci are 0 2 3 0 2 3 and 0 2 3 .
(b) The length of the major axis is 2a 4 and the length of the minor axis is 2b 2.
9.
y 12 x 52 1 16 4 y2 x2 1 by shifting it 5 units to 16 4 the left and 1 units upward. So a 4, b 2, and c 16 4 2 3. The
(c)
y
(a) This ellipse is obtained from the ellipse
center is 5 1, the vertices are 5 4 1 9 1 and 1 1, and the foci are 5 2 3 1 5 2 3 1 and 5 2 3 1 .
(b) The length of the major axis is 2a 8 and the length of the minor axis is 2b 4.
1 1
x
24
CHAPTER 11 Conic Sections
10.
y 1 x 12 1 36 64
y
(c)
y2 x2 1 by shifting it 1 unit to the 36 64 left and 1 unit downward. So a 8, b 6, and c 64 36 2 7. The center is 1 1, the vertices are 1 1 8 1 9 and 1 7, and the foci are 1 1 2 7 1 1 2 7 and 1 2 7 1 .
(a) This ellipse is obtained from the ellipse
2 2
x
(b) The length of the major axis is 2a 16 and the length of the minor axis is 2b 12.
11. 4x 2 25y 2 50y 75 4x 2 25 y 12 25 75
x2 y 12 1 25 4
y
(c)
y2 x2 1 by shifting it 1 unit 25 4 upward. So a 5, b 2, and c 25 4 21. The center is 0 1, the
(a) This ellipse is obtained from the ellipse
1 1
vertices are 5 1 5 1 and 5 1, and the foci are 21 1 21 1 and 21 1 .
x
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 4.
12. 9x 2 54x y 2 2y 46 0 9 x 32 81 y 12 1 46 0
(c)
x 32 y 12 1 4 36
y
1 2
y2 x2 1 by shifting it 3 units to 4 36 the right and 1 unit downward. So a 6, b 2, and c 36 4 4 2. The
(a) This ellipse is obtained from the ellipse
x
center is 3 1, the vertices are 3 1 6 3 7 and 3 5, and the foci are 3 1 4 2 3 1 4 2 and 3 4 2 1 .
(b) The length of the major axis is 2a 12 and the length of the minor axis is 2b 4.
13. x 32 8 y 1
(a) This parabola is obtained from the parabola x 2 8y by shifting it 3 units to the right and 1 unit down. So 4 p 8 p 2. The vertex is 3 1, the focus is 3 1 2 3 1, and the directrix is y 1 2 3.
(b)
y
1 1
x
25
SECTION 11.4 Shifted Conics
14. y 12 16 x 3
y
(b)
(a) This parabola is obtained from the parabola y 2 16x by shifting it 3 units to
2
the right and 1 unit down. So 4 p 16 p 4. The vertex is 3 1, the
15. y 52 6x 12 6 x 2
x
1
focus is 3 4 1 7 1, and the directrix is x 3 4 1.
y
(b)
1 1
(a) This parabola is obtained from the parabola y 2 6x by shifting to the right
x
2 units and down 5 units. So 4 p 6 p 32 . The vertex is 2 5, the focus is 2 32 5 12 5 , and the directrix is x 2 32 72 .
16. y 2 16x 8 16 x 12
"1" seems to be in the wrong location.
y
(b)
(a) This parabola is obtained from the parabola y 2 16x by shifting to the right 12 unit. So 4 p 16 p 4. The vertex is 12 0 , the focus is 1 4 0 9 0 , and the directrix is x 1 4 7 . 2 2 2 2
2 x
1
The directrix is at x = -7/2
17. 2 x 12 y x 12 12 y
(b)
y
(a) This parabola is obtained from the parabola x 2 12 y by shifting it 1 unit to the right. So 4 p 12 p 18 . The vertex is 1 0, the focus is 1 18 , and the directrix is y 18 .
1 x
1
2 2 18. 4 x 12 y x 12 14 y
y
(b)
1
(a) This parabola is obtained from the parabola x 2 14 y by shifting it 12 unit to 1 . The vertex is 1 0 , the focus is the left, so 4 p 14 p 16 2 1 1 1 1 1 1 . 2 0 16 2 16 , and the directrix is y 0 16 16
19. y 2 6y 12x 33 0 y 32 9 12x 33 0 y 32 12 x 2
1
(b)
x
y
(a) This parabola is obtained from the parabola y 2 12x by shifting it 2 units to the right and 3 units upward. So 4 p 12 p 3. The vertex is 2 3, the focus is 2 3 3 5 3, and the directrix is x 2 3 1.
2 1
x
In #16, something went wrong with the tick spacing.
26
CHAPTER 11 Conic Sections
20. x 2 2x 20y 41 0 x 12 1 20y 41 0 x 12 20 y 2
y
(b)
(a) This parabola is obtained from the parabola x 2 20y by shifting it 1 unit to the left and 2 units upward, so 4 p 20 p 5. The vertex is 1 2, the focus
1
is 1 2 5 1 7, and the directrix is y 2 5 3.
21.
y 32 x 12 1 9 16
1
x
1
x
10
20
1
x
y
(b)
x2 y2 1 by shifting it 9 16 1 unit to the left and 3 units up. So a 3, b 4, and c 9 16 5. The
(a) This hyperbola is obtained from the hyperbola
1
center is 1 3, the vertices are 1 3 3 4 3 and 2 3, the foci are 1 5 3 6 3 and 4 3, and the asymptotes are
y 3 43 x 1 y 43 x 1 3 y 43 x 13 3 and y 43 x 53 .
22. x 82 y 62 1
(b)
y
(a) This hyperbola is obtained from the hyperbola x 2 y 2 1 by shifting to the right 8 units and downward 6 units. So a 1, b 1, and c
1 1 2.
The center is 8 6, the vertices are 8 1 6 7 6 and 9 6, the foci are 8 2 6 , and the asymptotes are y 6 x 8
_10
_20
y x 14 and y x 2.
23. y 2
x 12 1 4
(b)
x2 (a) This hyperbola is obtained from the hyperbola y 2 1 by shifting it 1 unit 4 to the left. So a 1, b 2, and c 1 4 5. The center is 1 0, the
y
1
vertices are 1 1 1 1 and 1 1, the foci are 1 5 1 5 and 1 5 , and the asymptotes are y 12 x 1 y 12 x 12 and y 12 x 12 .
24.
y 12 x 32 1 25 y2 x 2 1 by shifting to the 25 left 3 units and upward 1 unit. So a 5, b 1, and c 25 1 26. The
(a) This hyperbola is obtained from the hyperbola
center is 3 1, the vertices are 3 1 5 3 4 and 3 6, the foci are 3 1 26 , and the asymptotes are y 1 5 x 3 y 5x 16 and y 5x 14.
(b)
y
5 1
x
x
27
SECTION 11.4 Shifted Conics
25.
y 12 x 12 1 9 4
y
(b)
x 2 y2 1 by shifting it 1 unit 9 4 to the left and 1 unit downward, so a 3, b 2, and c 9 4 13. The center is 1 1, the vertices are 1 3 1 4 1 and 2 1, the foci are 1 13 1 1 13 1 and 13 1 1 , and the
1
(a) This hyperbola is obtained from the hyperbola
1
x
5
x
asymptotes are y 1 23 x 1 y 23 x 13 and y 23 x 53 .
26.
x2 y 22 1 36 64
(b)
x2 y2 1 by shifting it (a) This hyperbola is obtained from the hyperbola 36 64 2 unit downward. So a 6, b 8, and c 36 64 10. The center is 0 2, the vertices are 0 2 6 0 8 and 0 4, the foci are
y 5
0 2 10 0 12 and 0 8, and the asymptotes are y 2 34 x y 34 x 2 and y 34 x 2.
27. 36x 2 72x 4y 2 32y 116 0 36 x 12 36 4 y 42 64 116 0
y
(b) x 12 y 42 1 36 4
y2 x 2 1 by shifting it 1 unit 36 4 to the left and 4 units upward. So a 6, b 2, and c 36 4 2 10.
2
(a) This hyperbola is obtained from the hyperbola
1
x
The center is 1 4, the vertices are 1 4 6 1 2 and 1 10, the foci are 1 4 2 10 1 2 10 4 and 1 4 2 10 , and
the asymptotes are y 4 3 x 1 y 3x 7 and y 3x 1. 28. 25x 2 9y 2 54y 306 25x 2 9 y 32 81 306
x 2 y 32 1 9 25
x2 y2 1 by shifting it 9 25 3 units downward. So a 3, b 5, and c 9 25 34. The center is 0 3, the vertices are 3 3 3 3 and 3 3, the foci are 34 3 , and the asymptotes are 34 3 34 3 and
(a) This hyperbola is obtained from the hyperbola
(b)
y
2 1
x
y 3 53 x y 53 x 3 and y 53 x 3.
29. This is a parabola that opens down with its vertex at 0 4, so its equation is of the form x 2 a y 4. Since 1 0 is a point on this parabola, we have 12 a 0 4 1 4a a 14 . Thus, an equation is x 2 14 y 4.
30. This is a parabola that opens to the right with its vertex at 6 0, so its equation is of the form y 2 4 p x 6, with
p 0. Since the distance from the vertex to the directrix is p 6 12 6, an equation is y 2 4 6 x 6
y 2 24 x 6
28
CHAPTER 11 Conic Sections
31. This is an ellipse with the major axis parallel to the x-axis, with one vertex at 0 0, the other vertex at 10 0, and one focus at 8 0. The center is at 010 2 0 5 0, a 5, and c 3 (the distance from one focus to the center). So b2 a 2 c2 25 9 16. Thus, an equation is
y2 x 52 1. 25 16
32. This is an ellipse with the major axis parallel to the y-axis. From the graph the center is 2 3, with a 3 and b 2. Thus, an equation is
x 22 y 32 1. 4 9
33. This is a hyperbola with center 0 1 and vertices 0 0 and 0 2. Since a is the distance form the center to a vertex, a we have a 1. The slope of the given asymptote is 1, so 1 b 1. Thus, an equation of the hyperbola is b 2 2 y 1 x 1. 34. From the graph, the vertices are 2 0 and 6 0. The center is the midpoint between the vertices, so the center is 26 00 4 0. Since a is the distance from the center to a vertex, a 2. Since the vertices are on the x-axis, the 2 2 y2 x 42 42 0 42 2 1. Since the point 0 4 lies on the hyperbola, we have 2 1 4 4 b b 2 2 2 16 16 4 y 3y 2 16 16 4 x x 2 1 3 2 b2 . Thus, an equation of the hyperbola is 16 1 1. 4 3 4 4 16 b b
equation is of the form
3
35. The ellipse with center C 2 3, vertices V1 8 3 and V2 12 3, and foci F1 4 3 and F2 8 3 has
x 22 y 32 1. The distance between the vertices a2 b2 is 2a 12 8 20, so a 10. Also, the distance from the center to each focus is c 2 4 6, so a horizontal major axis, so its equation has the form
b2 a 2 c2 100 36 64. Thus, an equation is
y 32 x 22 1. 100 64
36. The ellipse with vertices V1 1 4 and V2 1 6 and foci F1 1 3 and F2 1 5 is centered midway between the vertices; that is, it has C 1 12 4 6 1 1. The distance between the vertices is 2a 6 4 10, so a 5.
Also, the distance from the center to each focus is c 1 3 4, so b2 a 2 c2 25 16 9. Thus, an equation is x 12 y 12 1. 9 25
37. The hyperbola with center C 1 4, vertices V1 1 3 and V2 1 11, and foci F1 1 5 and F2 1 13 has x 12 y 42 1. The distance between the vertices 2 a b2 is 2a 11 3 14, so a 7. Also, the distance from the center to each focus is c 4 5 9, so a vertical transverse axis, so its equation has the form
b2 c2 a 2 81 49 32. Thus, an equation is
x 12 y 42 1. 49 32
38. The hyperbola with vertices V1 1 1 and V2 5 1, and foci F1 4 1 and F2 8 1 is centered midway between the vertices; that is, it has C 12 1 5 1 2 1. It has a horizontal transverse axis, so its equation has the form
x 22 y 12 1. The distance between the vertices is 2a 5 1 6, so a 3. Also, the distance from the 2 a b2 x 22 y 12 1. center to each focus is c 2 4 6, so b2 c2 a 2 36 9 27. Thus, an equation is 9 27
39. The parabola with vertex V 3 5 and directrix y 2 has an equation of the form x 32 4 p y 5. The distance from the vertex to the directrix is p 5 2 3, so an equation is x 32 12 y 5.
SECTION 11.4 Shifted Conics
29
40. The parabola with focus F 1 3 and directrix x 3 has vertex midway between the focus and the directrix; that is, it has V 12 3 1 3 2 3. The distance from the vertex to the directrix is p 2 3 1. (Since the directrix is to the right of the focus, p is negative.) Thus, an equation is y 32 4 1 x 2 4 x 2.
41. The hyperbola with foci F1 1 5 and F2 1 5 that passes through the point 1 4 is centered midway between the foci; y2 x 12 that is, it has C 1 12 5 5 1 0. It has a vertical transverse axis, so its equation has the form 2 1. a b2 The point 1 4 lies on the transverse axis, so it is a vertex and we have a 4 0 4. Also, the distance from the center to each focus is c 5 0 5, so b2 c2 a 2 25 16 9. Thus, an equation is
y2 x 12 1. 16 9
42. The hyperbola with foci F1 2 2 and F2 4 2 that passes through the point 3 2 is centered midway between the foci; that is, it has C 12 2 4 2 1 2. It has a horizontal transverse axis, so its equation has the form
y 22 x 12 1. The point 3 2 lies on the transverse axis, so it is a vertex and we have a 3 1 2. a2 b2 Also, the distance from the center to each focus is c 4 1 3, so b2 c2 a 2 9 4 5. Thus, an equation is x 12 y 22 1. 4 5
43. The ellipse with foci F1 1 4 and F2 5 4 that passes through the point 3 1 is centered midway between the foci; y 42 x 32 that is, it has C 12 1 5 4 3 4, and so its equation has the form 1. The distance from 2 a b2 the center to each focus is c 3 1 2, so c2 a 2 b2 a 2 b2 4. Substituting the point x y 3 1 into the 3 32 1 42 1 b2 25. From above, we have a 2 b2 4 29. Thus, an 2 a b2 y 42 x 32 1. equation of the ellipse is 29 25 equation of the ellipse, we have
44. The ellipse with foci F1 3 4 and F2 3 4 and x-intercepts 0 and 6 is centered midway between the foci; that is, it has x 32 y 2 C 3 12 4 4 3 0, and so its equation has the form 2 1. Because 0 is an x-intercept, we substitute b2 a 02 32 2 1 b 3. We also have c 4 0 4, so a 2 b2 c2 25 and an 0 0 into this equation, obtaining b2 a y2 x 32 1. equation is 9 25 45. The parabola that passes through the point 6 1, with vertex V 1 2 and horizontal axis of symmetry has an equation of 1 . the form y 22 4 p x 1. Substituting the point 6 1 into this equation, we have 1 22 4 p 6 1 p 28
Thus, an equation is y 22 17 x 1.
46. Because 6 2 is lower than V 4 1, the parabola opens downward, so an equation is x 42 4 p y 1.
Substituting the point 6 2, we have 6 42 4 p 2 1 p 1, so an equation is x 42 4 y 1.
47. y 2 4 x 2y y 2 8y 4x y 2 8y 16 4x 16
y
y 42 4 x 4. This is a parabola with 4 p 4 p 1. The vertex is
4 4, the focus is 4 1 4 3 4, and the directrix is x 4 1 5. 1 1
x
30
CHAPTER 11 Conic Sections
48. 9x 2 36x 4y 2 0 9 x 2 4x 4 36 4y 2 0
y
y2 x 22 1. This is an ellipse with a 3, 4 9 b 2, and c 9 4 5. The center is 2 0, the foci are 2 5 , the
9 x 22 4y 2 36
1 x
1
vertices are 2 3, the length of the major axis is 2a 6, and the length of the minor axis is 2b 4.
49. x 2 5y 2 2x 20y 44 x 2 2x 1 5 y 2 4y 4 44 1 20
y
x 12 y 22 x 12 5 y 22 25 1. This is a hyperbola 25 5 with a 5, b 5, and c 25 5 30. The center is 1 2, the foci are 1 30 2 , the vertices are 1 5 2 4 2 and 6 2, and the asymptotes
1 x
2
are y 2 55 x 1 y 55 x 1 2 y 55 x 2 55 and
y 55 x 2 55 . y
50. x 2 6x 12y 9 0 x 2 6x 9 12y x 32 12y. This is a
parabola with 4 p 12 p 3. The vertex is 3 0, the focus is 3 3,
_1
and the directrix is y 3.
51. 4x 2 25y 2 24x 250y 561 0 4 x 2 6x 9 25 y 2 10y 25 561 36 625 4 x 32 25 y 52 100
x 32
1
x
y 1 x
1
y 52
1. This is an ellipse 25 4 with a 5, b 2, and c 25 4 21. The center is 3 5, the foci are 3 21 5 , the vertices are 3 5 5 2 5 and 8 5, the length of the major axis is 2a 10, and the length of the minor axis is 2b 4.
52. 2x 2 y 2 2y 1 2x 2 y 2 2y 1 1 1 2x 2 y 12 2
y 12 1. This is an ellipse with a 2, b 1, and c 2 1 1. 2 The center is 0 1, the foci are 0 1 1 0 0 and 0 2, the vertices are 0 1 2 , the length of the major axis is 2a 2 2, and the length of the minor x2
axis is 2b 2.
y
1 1
x
SECTION 11.4 Shifted Conics
53. 16x 2 9y 2 96x 288 0 16 x 2 6x 9y 2 288 0 16 x 2 6x 9 9y 2 144 288 16 x 32 9y 2 144
31
y
2
y2 x 32 1. This is a hyperbola with a 4, b 3, and 16 9 c 16 9 5. The center is 3 0, the foci are 3 5, the vertices are
x
1
3 4, and the asymptotes are y 43 x 3 y 43 x 4 and y 4 43 x.
54. 4x 2 4x 8y 9 0 4 x 2 x 8y 9 0
y
2 4 x 2 x 14 8y 9 1 4 x 12 8 y 1
2 x 12 2 y 1. This is a parabola with 4 p 2 p 12 . The vertex is 1 1 , the focus is 1 3 , and the directrix is y 1 1 1 . 2 2 2 2 2
1 x
1
55. x 2 16 4 y 2 2x x 2 8x 4y 2 16 0 x 2 8x 16 4y 2 16 16 4y 2 x 42 y 12 x 4.
In #54, vertex should be at (1/2, 1), not at (1/2, 1/2).
y
1
Thus, the conic is degenerate, and its graph is the pair of lines y 12 x 4 and
x
1
y 12 x 4.
56. x 2 y 2 10 x y 1 x 2 10x y 2 10y 1 x 2 10x 25 y 2 10y 25 1 25 25 x 52 y 52 1. This is a hyperbola with a 1, b 1, and c 1 1 2. The center is 5 5, the foci are 5 2 5 , the vertices are 5 1 5 4 5 and 6 5, and the asymptotes are y 5 x 5 y x and y x 10.
y
1 x
1
57. 3x 2 4y 2 6x 24y 39 0 3 x 2 2x 4 y 2 6y 39 3 x 2 2x 1 4 y 2 6y 9 39 3 36
3 x 12 4 y 32 0 x 1 and y 3. This is a degenerate conic whose
y
(1, 3)
1 1
x
graph is the point 1 3.
58. x 2 4y 2 20x 40y 300 0 x 2 20x 4 y 2 10y 300 x 2 20x 100 4 y 2 10y 25 300 100 100 x 102 4 y 52 100. Since u 2 2 0 for all u, , there is no x y such that x 102 4 y 52 100. So there is no solution, and the graph is empty.
32
CHAPTER 11 Conic Sections
59. 2x 2 4x y 5 0 y 2x 2 4x 5.
-2
2
4
-5 -10
60. 4x 2 9y 2 36y 0 4x 2 9 y 2 4y 0 4x 2 9 y 2 4y 4 36 9 y 2 4y 4 36 4x 2 y 2 4y 4 4 49 x 2 y 2 4 49 x 2 y 2 4 49 x 2
5
-4
61. 9x 2 36 y 2 36x 6y x 2 36x 36 y 2 6y 9x 2 36x 45 y 2 6y 9 9 x 2 4x 5 y 32 y 3 9 x 2 4x 5 y 3 3 x 2 4x 5
-2
2
2
4
4
-10
62. x 2 4y 2 4x 8y 0 x 2 4x 4 y 2 2y 0 x 2 4x 4 4 y 2 2y 1 0 4 y 12 x 22
2
2 y 1 x 2 y 1 12 x 2 y 1 12 x 2.So
y 1 12 x 2 12 x 2 and y 1 12 x 2 12 x. This is a degenerate
-5
conic.
63. 4x 2 y 2 4 x 2y F 0 4 x 2 x y 2 8y F 2 4 x 2 x 14 y 2 8y 16 16 1 F 4 x 12 y 12 17 F (a) For an ellipse, 17 F 0 F 17.
(b) For a single point, 17 F 0 F 17. (c) For the empty set, 17 F 0 F 17.
64. The parabola x 2 y 100 x 2 y 100 has 4 p 1 p 14 . The vertex is 0 100 and so the focus is 399 0 100 14 0 399 4 . Thus, one vertex of the ellipse is 0 100, and one focus is 0 4 . Since the second focus of 401 the ellipse is 0 0, the second vertex is 0 14 . So 2a 100 14 401 4 a 8 . Since 2c is the distance between 2
2
399 401 399 2 2 2 the foci of the ellipse, 2c 399 4 , c 8 , and then b a c 64 64 25. The center of the ellipse is 2 y 399 2 x x2 8y 3992 8 0 399 and so its equation is 1. 2 1, which simplifies to 8 25 25 160,801 401 8
SECTION 11.4 Shifted Conics
65. (a) x 2 4 p y p, for
p 2, 32 , 1, 12 , 12 , 1, 32 , 2. 3
1 2
p= 2
p= 1
2
8
-4
0 -4
4
8
3
1
_2
x 2 4 py vertically p units so that the vertex is at 0 p. The
focus of x 2 4 py is at 0 p, so this point is also shifted p units
vertically to the point 0 p p 0 0. Thus, the focus is located
(c) The parabolas become narrower as the vertex moves toward the origin.
-8 p= _ 2
(b) The graph of x 2 4 p y p is obtained by shifting the graph of
at the origin.
4 -8
33
p= _1 _2
66. Since 0 0 and 1600 0 are both points on the parabola, the x-coordinate of the vertex is 800. And since the highest point it reaches is 3200, the y-coordinate of the vertex is 3200. Thus the vertex is 800 3200, and the equation is of the form x 8002 4 p y 3200. Substituting the point 0 0, we get 0 8002 4 p 0 3200 640,000 12,800 p
p 50. So an equation is x 8002 4 50 y 3200 x 8002 200 y 3200. 67. Since the height of the satellite above the earth varies between 140 and 440, the
y
length of the major axis is 2a 140 2 3960 440 8500 a 4250. Since the center of the earth is at one focus, we have a c earth radius 140 3960 140 4100
c a 4100 4250 4100 150. Thus, the center of the ellipse is 150 0. So b2 a 2 c2 42502 1502 18,062,500 22500 18,040,000. Hence,
an equation is
410
Path of satellite Earth
140
3960
x
Center of ellipse is (_150, 0)
y2 x 1502 1. 18,062,500 18,040,000
68. (a) We assume that 0 1 is the focus closer to the vertex 0 0, as shown in the figure in the text. Then the center of the ellipse is 0 a and 1 a c. So c a 1 and a 12 a 2 b2 a 2 2a 1 a 2 b2 b2 2a 1.
x2 y a2 y 22 x2 1. If we choose 1. If we choose a 2, then we get 2a 1 3 4 a2 x2 y 52 1. (Answers will vary, depending on your choices of a 1.) a 5, then we get 9 25 (b) Since a vertex is at 0 0 and a focus is at 0 1, we must have c a 1 (a 0), and the center of the hyperbola Thus, the equation is
is 0 a. So c a 1 and a 12 a 2 b2 a 2 2a 1 a 2 b2 b2 2a 1. Thus the equation x2 x2 y 22 y a2 1. If we let a 2, then we get 1. If we let a 5, then we get 2 2a 1 4 5 a x2 y 52 1. (Answers will vary, depending on your choices of a.) 25 11
is
(c) Since the vertex is at 0 0 and the focus is at 0 1, we must have
y
p 1. So x 02 4 1 y 0 x 2 4y, and there is no other
possible parabola.
(d) Graphs will vary, depending on the choices of a in parts (a) and (b). (e) The ellipses are inside the parabola and the hyperbolas are outside the parabola. All touch at the origin.
1 1
x
34
CHAPTER 11 Conic Sections
69. (a) Factored form: y 2x 2 8x 6 2 x 2 4x 3 2 x 3 x 1 Vertex form: y 2 x 2 4x 6 2 x 2 4x 4 6 8 2 x 22 2
Standard form: y 2 x 22 2 12 y x 22 1 x 22 12 y 2 x 22 4 18 y 2
(b) To sketch the graph, we use the vertex form or the standard form. The maximum or minimum value is immediately evident from the vertex form. The standard form gives the focus. The factored form gives the real and complex zeroes. 70. Consider the blue circle, centered at P x y with radius r, tangent to the two red circles. Because the blue circle is tangent to the inner red circle, a line from 1 0 to P passes through the point of tangency, and so its length is the sum of the radii of the two circles, that is, 3 r. Similarly, the length of the line segment from 1 0 to P is the radius of the outer red circle minus the radius of the blue circle, that is, 5 r . The sum of these distances—that is, the distance from 1 0 to P to 1 0—is thus 3 r 5 r 8. But by definition, an ellipse is a set of points whose distances from two fixed points (the foci of the ellipse) sum to a constant. To find the equation of this ellipse, noting that it is centered at the origin, we substitute the points 4 0 and 3 0 into the x2 y2 42 32 standard equation 2 2 1, obtaining 2 1 a 4 and 1 b 3. Thus, an equation of the ellipse is a b a b2 y2 x2 1. 16 9
11.5 ROTATION OF AXES 1. If the x- and y-axes are rotated through an acute angle to produce the new X- and Y -axes, then the x y-coordinates x y and the XY -coordinates X Y of a point P in the plane are related by the formulas x X cos Y sin , y X sin Y cos , X x cos y sin , and Y x sin y cos . 2. (a) In general, the graph of the equation Ax 2 Bx y C y 2 Dx E y F 0 is a conic section.
AC . B (c) The discriminant of this equation is B 2 4AC. If the discriminant is 0 the graph is a parabola, if it is positive the graph is an ellipse, and if it is negative the graph is a hyperbola. 2 and 3. x y 1 1, 45 . Then X x cos y sin 1 1 1 1 2 2 Y x sin y cos 1 1 1 1 0. Therefore, the XY -coordinates of the given point are X Y 2 0 . (b) To eliminate the x y-term from this equation we rotate the axes through an angle that satisfies cot 2
2
2
4. x y 2 1, 30 . Then X x cos y sin 2 23 1 12 12 1 2 3 and Y x sin y cos 2 12 1 23 12 2 3 . Therefore, the XY -coordinates of the given point are X Y 12 1 2 3 12 2 3 . 3 3 , 60 . Then X x cos y sin 3 12 3 23 0 and Y x sin y cos 3 23 3 12 2 3. Therefore, the XY -coordinates of the given point are X Y 0 2 3 .
5. x y
6. x y 2 0, 15 . Then X x cos y sin 2 cos 15 0 sin 15 19319 and Y x sin y cos 2 sin 15 0 cos 15 05176. Therefore, the XY -coordinates of the given point are approximately X Y 19319 05176.
35
SECTION 11.5 Rotation of Axes
7. x y 0 2, 55 . Then X x cos y sin 0 cos 55 2 sin 55 16383 and Y x sin y cos 0 sin 55 2 cos 55 11472. Therefore, the XY -coordinates of the given point are approximately X Y 16383 11472. 2 4 2 , 45 . Then X x cos y sin 2 1 4 2 1 5 and 8. x y 2 2 Y x sin y cos 2 1 4 2 1 3. Therefore, the XY -coordinates of the given point are 2
2
X Y 5 3.
9. x 2 3y 2 4, 60 . Then x X cos 60 Y sin 60 12 X 23 Y and y X sin 60 Y cos 60 23 X 12 Y . 2 2 3 23 X 12 Y 4 Substituting these values into the equation, we get 12 X 23 Y 3Y 2 Y2 X2 9 2 3Y 2 3Y 2 3 3XY X2 3XY 3XY 3XY 3X 2 3 4 X 4 4 2 4 4 2 4 4 4 4 4 2 2 2X 2 2 3XY 4 X 2 3XY 2.
10. y x 12 , 45 . Then x X cos 45 Y sin 45 22 X 22 Y and y X sin 45 Y cos 45 22 X 22 Y . Substituting these values into the equation, we get 2 2 X2 2 X 2Y 2 X 2Y 1 2X 2Y 1 2Y 1 2 2 2 2 2 2 2 2 2 Y2 X2 2 22 X 22 Y 1 2Y 1 2 2 Y2 X2 22 X 2X 22 Y 1 2Y 1 22 Y 0 X 2 Y 2 2XY 3 2X 2Y 2 0. 2 2 11. x 2 y 2 2y, cos1 35 . So cos 35 and sin 45 . Then 2 2 X cos Y sin 2 X sin Y cos 2 2 X sin Y cos 35 X 45 Y 45 X 35 Y 2 45 X 35 Y
24XY 16Y 2 16X 2 24XY 9Y 2 8X 6Y 7X 2 48XY 7Y 2 8X 6Y 9X 2 0 25 25 25 25 25 25 5 5 25 25 25 5 5 7Y 2 48XY 7X 2 40X 30Y 0. 12. x 2 2y 2 16, sin1 35 . So sin 35 and cos 45 . Then x 45 X 35 Y and y 35 X 45 Y . Substituting these
values into the equation, we get 4 X 3 Y 2 2 3 X 4 Y 2 16 16 X 2 24 XY 9 Y 2 2 9 X 2 24 XY 16 Y 2 16 5 5 5 5 25 25 25 25 25 25
16 X 2 18 X 2 9 Y 2 32 Y 2 24 XY 48 XY 16 34 X 2 41 Y 2 24 XY 16 34X 2 41Y 2 24XY 400. 25 25 25 25 25 25 25 25 25
3X Y 13. x 2 2 3x y y 2 4, 30 . Then x X cos 30 Y sin 30 23 X 12 Y 12 and y X sin 30 Y cos 30 12 X 23 Y 12 X 3Y . Substituting these values into the 2 1 2 1 X 3Y 3X Y 2 3 12 3X Y 2 X 3Y 4 equation, we get 12 2 2 2 3X Y 2 3 3X Y X 3Y X 3Y 16 3X 2 2 3XY Y 2 6X 2 4 3XY 6Y 2 X 2 2 3XY 3Y 2 16 8X 2 8Y 2 16
X2 Y2 1. This is a hyperbola. 2 2
36
CHAPTER 11 Conic Sections
1 X 1 Y 1 X Y 14. x y x y, 4 . Therefore, x X cos 4 Y sin 4 2
2
2
1 X 1 Y 1 X Y . Substituting into the equation gives and y X sin 4 Y cos 4
1 X Y 2 2
X
2
2
1 X Y 2
2
2
2
1 X Y 1 X Y 12 2 2
Y2 1. This is a hyperbola. 2
X 2 Y 2 2 X X 2 Y 2 2 2X
15. (a) x y 8 0x 2 x y 0y 2 8. So A 0, B 1, and C 0, and so the
2
(c)
discriminant is B 2 4AC 12 4 0 0 1. Since the discriminant is
positive, the equation represents a hyperbola. (b) cot 2
y
1
x
1
AC 0 2 90 45 . Therefore, B
x 22 X 22 Y and y 22 X 22 Y . After substitution, the original 2 2 equation becomes 22 X 22 Y 2 X 2 Y 8
X2 Y2 X Y X Y 8 1. This is a hyperbola with a 4, b 4, and c 4 2. Hence, the vertices 2 16 16 are V 4 0 and the foci are F 4 2 0 .
16. (a) x y 4 0 0x 2 x y 0y 2 4 0. So A 0, B 1, and C 0,
y
(c)
and so the discriminant is B 2 4AC 12 4 0 0 1. Since the discriminant is positive, the equation represents a hyperbola.
1
x
1
AC 0 2 90 45 . Therefore, (b) cot 2 B
x 22 X 22 Y and y 22 X 22 Y . After substitution, the original 2 2 equation becomes 22 X 22 Y 2 X 2 Y 40
Y2 X2 X Y X Y 4 1. This is a hyperbola with a 2 2, b 2 2, and c 4. Hence, the 2 8 8 vertices are V 0 2 2 and the foci are F 0 4.
17. (a) x 2 2 3x y y 2 2 0. So A 1, B 2 3, and C 1, and so the 2 discriminant is B 2 4AC 2 3 4 1 1 0. Since the discriminant is positive, the equation represents a hyperbola.
(c)
y
1 1
11 AC 1 2 60 30 . Therefore, (b) cot 2 B 2 3 3
x 23 X 12 Y and y 12 X 23 Y . After substitution, the original equation becomes 2 2 3 1 1 X 3Y 1 X 3Y 2 3 23 X 12 Y 20 2 X 2Y 2 2 2 2 3 X 2 3 XY 1 Y 2 3 3X 2 2XY 3Y 2 14 X 2 23 XY 34 Y 2 2 0 4 2 4 2 X 2 34 32 14 XY 23 3 23 Y 2 14 32 34 2 2X 2 2Y 2 2 Y 2 X 2 1.
x
37
SECTION 11.5 Rotation of Axes
18. (a) 13x 2 6 3x y 7y 2 16. Then A 13, B 6 3, and C 7, and so 2 the discriminant is B 2 4AC 6 3 4 13 7 256. Since the
(c)
y
1
discriminant is negative, the equation represents an ellipse.
1
13 7 AC 1 2 60 30 . Therefore, (b) cot 2 B 6 3 2
x
x 23 X 12 Y and y 12 X 23 Y . After substitution, the original equation becomes 2 2 1 X 3Y 7 1 X 3Y 13 23 X 12 Y 6 3 23 X 12 Y 16 2 2 2 2 13 3X 2 2 3XY Y 2 3 3 3X 2 2XY 3Y 2 74 X 2 2 3XY 3Y 2 16 4 2 9 7 XY 13 3 6 3 7 3 Y 2 13 9 21 16 16X 2 4Y 2 16 X 2 39 4 2 4 2 2 2 4 2 4
Y2 1. This is an ellipse with a 2, b 1, and c 4 1 3. Thus, the vertices are V 0 2 and the 4 foci are F 0 3 .̀ X2
19. (a) 11x 2 24x y 4y 2 20 0. So A 11, B 24, and C 4, and so the discriminant is B 2 4AC 242 4 11 4 0. Since the
discriminant is positive, the equation represents a hyperbola.
AC 11 4 7 7 . Therefore, cos 2 25 B 24 24 cos 1725 35 and sin 1725 45 . Hence, 2 2
(b) cot 2
(c)
y
1 1
x
7 , we have 3X Since cos 2 25 45 Y and y 45 X 35 Y . After substitution, the original 5 2 10626 , so 53 . equation becomes 2 4 X 3 Y 4 4 X 3 Y 2 20 0 11 35 X 45 Y 24 35 X 45 Y 5 5 5 5 11 9X 2 24XY 16Y 2 24 12X 2 7XY 12Y 2 4 16X 2 24XY 9Y 2 20 0 25 25 25
x
X 2 99 288 64 XY 264 168 96 Y 2 176 288 36 500 125X 2 500Y 2 500 1 X 2 Y 2 1. 4
38
CHAPTER 11 Conic Sections
20. (a) 21x 2 10 3x y 31y 2 144. Then A 21, B 10 3, and C 31, 2 and so the discriminant is B 2 4AC 10 3 4 21 31 2304.
(c)
y
1
Since the discriminant is negative, the equation represents an ellipse.
1
21 31 AC 1 (b) cot 2 2 120 60 . B 10 3 3
x
Therefore, x 12 X 23 Y and y 23 X 12 Y . After substitution, the original equation becomes 2 2 3 3 1 1 21 12 X 23 Y 10 3 12 X 23 Y 144 2 X 2 Y 31 2 X 2 Y 21 X 2 2 3XY 3Y 2 5 3 2 2 144 3X 2 2XY 3Y 2 31 4 2 4 3X 2 3XY Y 15 93 XY 21 3 10 3 31 3 Y 2 63 15 31 144 36X 2 16Y 2 144 X 2 21 4 2 4 2 2 2 4 2 4 1 X 2 1 Y 2 1. This is an ellipse with a 3, b 2, and c 9 4 5. 4 9 21. (a)
2 3x 3x y 3. So A 3, B 3, and C 0, and so the discriminant 3 0 9. Since the discriminant is positive, is B 2 4AC 32 4
(c)
y
1
the equation represents a hyperbola.
x
1
1 AC 2 60 30 . Therefore, (b) cot 2 B 3
x 23 X 12 Y and y 12 X 23 Y . After substitution, the equation 2 1 X 3Y 3 becomes 3 23 X 12 Y 3 23 X 12 Y 2 2 3 3 4 3X 2 2 3XY Y 2 4 3X 2 2XY 3Y 2 3 3 3 3 3 3 3 X2 3 Y 2 3 3 X2 6 Y2 1 Y 2 1. This X 2 3 4 3 3 4 3 XY 6 4 4 4 4 2 2 2 2 3 is a hyperbola with a 2 and b 2 3. 3
22. (a) 153x 2 192x y 97y 2 225. Then A 153, B 192, and C 97, and so the discriminant is B 2 4AC 1922 4 153 97 22500.
Since the discriminant is negative, the equation represents an ellipse.
AC 153 97 56 56 (b) cot 2 cos 2 . Therefore, B 192 192 200 156200 4 3 16 12 cos 156200 2 20 5 and sin 2 20 5
(c)
y
1
cos1 54 369 . Substituting gives 2 3 X 4 Y 97 3 X 4 Y 2 225 153 45 X 35 Y 192 45 X 35 Y 5 5 5 5 153 16X 2 24XY 9Y 2 192 12X 2 7XY 12Y 2 97 9X 2 24XY 16Y 2 225 25 25 25
1
X 2 2448 2304 873 XY 3672 1344 2328 Y 2 1377 2304 1552 5625 5625X 2 625Y 2 5625 X 2 19 Y 2 1. This is an ellipse with a 3, b 1, and c 9 1 2 2.
x
39
SECTION 11.5 Rotation of Axes
23. (a) x 2 2x y y 2 x y 0. So A 1, B 2, and C 1, and so the
y
(c)
discriminant is B 2 4AC 22 4 1 1 0. Since the discriminant is
zero, the equation represents a parabola.
1 1
AC (b) cot 2 0 2 90 45 . Therefore, B
x
x 22 X 22 Y and y 22 X 22 Y . After substitution, the original equation becomes 2 2 X 2Y 2 X 2Y 2 X 2Y 2 2 2 2 2 2 2
2 2 2 22 X 22 Y 22 X 22 Y 0 2 X 2 Y 1 X 2 XY 1 Y 2 X 2 Y 2 1 X 2 XY Y 2 2Y 0 2X 2 2Y 0 X 2 2 Y . This is a 2 2 2 2
1 . parabola with 4 p 1 and hence the focus is F 0 2
4 2
In #24(c), the parabola should appear to be symmetric with respect to the X-axis. 24. (a) 25x 2 120x y 144y 2 156x 65y 0. Then A 25, B 120, and
(c)
y
C 144, and so the discriminant is
B 2 4AC 1202 4 25 144 0. Since the discriminant is zero,
the equation represents a parabola.
AC 25 144 119 cos 2 119 169 . Therefore, B 120 120 1119169 5 . Hence, cos 1119169 12 13 2 13 and sin 2
(b) cot 2
2 2
x
Since cos 2 119 169 , we have
12X 5Y 5X 12Y 2 452 , so 23 . x and y . Substituting gives 13 13 13 13 5Y 2 5Y 12Y 12Y 2 5Y 5X 5X 12X 12X 12X 144 120 156 25 13 13 13 13 13 13 13 13 13 13 Y2 12Y X2 5X 65 0 25 122 120 12 5 144 52 25 52 120 5 12 144 122 13 13 169 169 Y X 156 12 65 5 156 5 65 12 0 169Y 2 169X 0 X Y 2 . This is a parabola with 13 13 4 p 1.
40
CHAPTER 11 Conic Sections
25. (a) 2 3x 2 6x y 3x 3y 0. So A 2 3, B 6, and C 0, and so the discriminant is B 2 4AC 62 4 2 3 0 36. Since the
y
(c)
2
discriminant is positive, the equation represents a hyperbola. AC 2 3 1 (b) cot 2 2 120 60 . Therefore, B 6 3
2
x
x 12 X 23 Y and y 23 X 12 Y , and substituting gives 2 3 3 3 1 1 1 2 3 12 X 23 Y 6 12 X 23 Y 2 X 2Y 3 2 X 2 Y 3 2 X 2Y 0 3 X 2 23XY 3Y 2 3 3X 2 2XY 3Y 2 3 X 3Y 3 3X Y 0 2 2 2 2 X 2 23 3 2 3 X 23 3 2 3 XY 3 3 Y 2 3 2 3 3 2 3 Y 32 32 0 3X 2 2 3X 3 3Y 2 0 X 2 2X 3Y 2 0 3Y 2 X 2 2X 1 1 X 12 3Y 2 1. This is a hyperbola with a 1, b 33 , c 1 13 2 , and C 1 0. 3
26. (a) 9x 2 24x y 16y 2 100 x y 1. Then A 9, B 24, and
C 16, and so the discriminant is B 2 4AC 242 4 9 16 0. Since the discriminant is zero, the equation represents a parabola.
(c)
y 1 1
x
AC 9 16 7 7 369 . Now cos 2 25 B 24 24 4 and sin 1725 3 , and so cos 1725 2 5 2 5
(b) cot 2
x 45 X 35 Y , y 35 X 45 Y . By substitution, 2 3 X 4 Y 16 3 X 4 Y 2 100 4 X 3 Y 3 X 4 Y 1 9 45 X 35 Y 24 45 X 35 Y 5 5 5 5 5 5 5 5 9 2 2 24 12X 2 7XY 12Y 2 16 9X 2 24XY 16Y 2 100 1 X 7 Y 1 25 16X 24XY 9Y 25 25 5 5 625Y 2 500X 3500Y 2500 5Y 2 28Y 4X 20 196 196 96 24 14 2 4 X 24 . This is a 5 Y 2 28 5 Y 25 4X 20 5 4X 5 4 X 5 Y 5 5 5 14 . parabola with 4 p 45 and V 24 5 5
27. (a) 52x 2 72x y 73y 2 40x 30y 75. So A 52, B 72, and C 73, and so the discriminant is
B 2 4AC 722 4 52 73 10,000. Since the discriminant is decidedly negative, the equation represents an ellipse.
SECTION 11.5 Rotation of Axes
41
52 73 7 AC . Therefore, as in Exercise 19(b), we get cos 35 , sin 45 , and B 72 24 x 35 X 45 Y , y 45 X 35 Y . By substitution, 2 4 X 3 Y 73 4 X 3 Y 2 52 35 X 45 Y 72 35 X 45 Y 5 5 5 5 40 35 X 45 Y 30 45 X 35 Y 75 52 9X 2 24XY 16Y 2 72 12X 2 7XY 12Y 2 73 16X 2 24XY 9Y 2 25 25 25
(b) cot 2
24X 32Y 24X 18Y 75 468X 2 832Y 2 864X 2 864Y 2 1168X 2 657Y 2 1250Y 1875 2500X 2 625Y 2 1250Y 1875 100X 2 25Y 2 50Y 75 X 2 14 Y 12 1. This is an ellipse with a 2, b 1, c 4 1 3, and center C 0 1.
(c)
y
1 x
1
7 , we have 2 cos1 7 10626 and so 53 . Since cos 2 25 25
28. (a) 7x 24y2 49x 2 336x y 576y 2 600x 175y 25. So A 49, B 336, and C 576, and so the
discriminant is B 2 4AC 3362 4 49 576 0. Since the discriminant is zero, the equation represents a parabola. AC 49 576 527 1527625 7 and (b) cot 2 . Therefore, cos 25 cos 2 527 625 2 B 336 336 7 24 24 7 24 sin 1527625 2 25 . Substituting x 25 X 25 Y and y 25 X 25 Y gives 7 X 24 Y 2 336 7 X 24 Y 24 X 7 Y 576 24 X 7 Y 2 49 25 25 25 25 25 25 25 25 7 X 24 Y 175 24 X 7 Y 25 600 25 25 25 25 49 2 2 336 168X 2 527XY 168Y 2 576 576X 2 336XY 49Y 2 625 49X 336XY 576Y 625 625
168X 576Y 168X 49Y 25 X 2 2401 56,448 331,776 XY 16,464 177,072 193,536 Y 2 28,224 56,448 28,224
(c)
6252 Y 15,625 1 Y 1 . This is a parabola with 390,625X 2 390,625Y 15,625 25X 2 25Y 1 X 2 Y 25 25 1 4 p 1 and vertex 0 25 . y
1 1
x
1 527 14748 and so 74 . Since cos 2 527 625 , we have 2 cos 625
42
CHAPTER 11 Conic Sections
29. (a) The discriminant is B 2 4AC 42 4 2 2 0. Since the discriminant is 0, the equation represents a parabola. (b) 2x 2 4x y 2y 2 5x 5 0 2y 2 4x y 2x 2 5x 5 2 y 2 2x y 2x 2 5x 5 2 y 2 2x y x 2 2x 2 5x 5 2x 2 2 y x2 5x 5 y x2 52 x 52 y x 52 x 52 y x 52 x 52
5
5
30. (a) The discriminant is B 2 4AC 22 4 1 3 8 0. Since the discriminant is negative, the equation represents an ellipse. (b) x 2 2x y 3y 2 8 3y 2 2x y 8 x 2 3 y 2 23 x y 8 x 2
3 y 2 23 x y 19 x 2 8 x 2 13 x 2 3 y 13 x 83 29 x 2 y 13 x 83 29 x 2 y 13 x 83 29 x 2
y 13 x
2
2
5
8 23 x 2
-5
5 -5
31. (a) The discriminant is B 2 4AC 102 4 6 3 28 0. Since the discriminant is positive, the equation represents a hyperbola. (b) 6x 2 10x y 3y 2 6y 36 3y 2 10x y 6y 36 6x 2 3y 2 2 5x 3 y 36 6x 2 y 2 2 53 x 1 y 12 2x 2 2 2 y 2 2 53 x 1 y 53 x 1 53 x 1 12 2x 2
10
-10
10
2 10 2 2 y 53 x 1 25 9 x 3 x 1 12 2x
-10
2 y 53 x 1 79 x 2 10 3 x 13 y 53 x 1 79 x 2 10 3 x 13 y 53 x 1 79 x 2 10 3 x 13
(or a degenerate conic)
32. (a) The discriminant is B 2 4AC 62 4 9 1 0. Since the discriminant is 0, the equation represents a parabola. (b) 9x 2 6x y y 2 6x 2y 0 y 2 6x y 2y 9x 2 6x
5
y 2 2 3x 1 y 9x 2 6x
y 2 2 3x 1 y 3x 12 3x 12 9x 2 6x 2 y 3x 1 1 y 3x 1 1 y 3x 1 1. So y 3x or y 3x 2. This is a degenerate conic.
Note that the conic can't be both a parabola and a degenerate.
-2
2 -5
SECTION 11.5 Rotation of Axes
43
33. (a) 7x 2 48x y 7y 2 200x 150y 600 0. Then A 7, B 48, and C 7, and so the discriminant is
B 2 4AC 482 4 7 7 0. Since the discriminant is positive, the equation represents a hyperbola. We 7 AC cos 45 and sin 35 . now find the equation in terms of XY -coordinates. We have cot 2 B 24 Therefore, x 45 X 35 Y and y 35 X 45 Y , and substitution gives
2 3 X 4 Y 7 3 X 4 Y 2 200 4 X 3 Y 150 3 X 4 Y 600 0 7 45 X 35 Y 48 45 X 35 Y 5 5 5 5 5 5 5 5 7 16X 2 24XY 9Y 2 48 12X 2 7XY 12Y 2 7 9X 2 24XY 16Y 2 25 25 25
160X 120Y 90X 120Y 600 0 112X 2 168XY 63Y 2 576X 2 336XY 576Y 2 63X 2 168XY 112Y 2 6250X 15,000 0 25X 2 25Y 2 250X 600 0 25 X 2 10X 25 25Y 2 600 625 X 52 Y 2 1. This is a hyperbola with a 1, b 1, c 1 1 2, and center C 5 0.
(b) In the XY -plane, the center is C 5 0, the vertices are V 5 1 0 V1 4 0 and V2 6 0, and the foci are F 5 2 0 . In the x y-plane, the center is C 45 5 35 0 35 5 45 0 C 4 3, the vertices are 12 4 3 3 4 24 18 V1 45 4 35 0 35 4 45 0 V1 16 5 5 and V2 5 6 5 0 5 6 5 0 V2 5 5 , and the foci are F1 4 45 2 3 35 2 and F2 4 45 2 3 35 2 . (c) In the XY -plane, the equations of the asymptotes are Y X 5 and Y X 5. In the x y-plane, these equations
become x 35 y 45 x 45 y 35 5 7x y 25 0. Similarly, x 35 y 45 x 45 y 35 5 x 7y 25 0.
34. (a) 2 2 x y2 7x 9y 2 2x 2 4 2x y 2 2y 2 7x 9y. Therefore, A 2 2, B 4 2, and C 2 2, X Y X Y AC x y 2X. Thus the 0 2 90 45 . Thus x , y and so cot 2 B 2 2 2 16X 2Y 8X 2 16X 2Y 4X 2 8X Y 4 X 2 2X 1 Y 4 equation becomes 2 2 2X 2 X 12 14 Y 4. This is a parabola.
1 . Thus, in XY -coordinates the vertex is V 1 4 and the focus is F 1 63 . In x y-coordinates, (b) 4 p 14 p 16 16 5 2 3 2 79 2 47 2 V and F 32 32 . 2 2 (c) The directrix is Y 65 16 . Thus
65 x y or y x 6516 2 . 16 2
35. We use the hint and eliminate Y by adding: x X cos Y sin x cos X cos2 Y sin cos and
y X sin Y cos y sin X sin2 Y sin cos , and adding these two equations gives x cos y sin X cos2 sin2 x cos y sin X. In a similar manner, we eliminate X by subtracting:
x X cos Y sin x sin X cos sin Y sin2 and y X sin Y cos y cos X sin cos Y cos2 , so x sin y cos Y cos2 sin2 x sin y cos Y . Thus, X x cos y sin and Y x sin y cos .
44
36.
CHAPTER 11 Conic Sections
x
y 1. Squaring both sides gives x 2 x y y 1 2 x y 1 x y, and squaring both sides again
gives 4x y 1 x y2 1 x y x x 2 x y y x y y 2 4x y x 2 y 2 2x y 2x 2y 1 AC x 2 y 2 2x y 2x 2y 1 0. Then A 1, B 2, and C 1, and so cot 2 0 2 90 B
45 . Therefore, x 22 X 22 Y and y 22 X 22 Y , and substituting gives 2 2 X 2Y 2 X 2Y 2 X 2Y 2 2 2 2 2 2 2 2 22 X 22 Y 2 22 X 22 Y 2 22 X 22 Y 1 0 1 X 2 2XY Y 2 X 2 Y 2 1 X 2 2XY Y 2 2 2X 1 0 X 2 1 1 1 XY 1 1 2 2 2 2 1 . This is a parabola with Y 2 12 1 12 2 2X 1 0 2Y 2 2 2X 1 Y 2 2X 12 2 X 2 2 1 4 p 2 and vertex V 0 . However, in the original equation we must have x 0 and y 0, so we get only the 2 2 part of the parabola that lies in the first quadrant.
37. (a) Z
x y
X cos sin , Z , and R . Y
cos x cos sin X X cos Y sin Y . Equating the entries in this Thus Z R Z y sin cos Y X sin Y cos
sin
matrix equation gives the first pair of rotation of axes formulas. Now cos sin cos sin 1 1 and so Z R 1 Z R cos2 sin2 sin cos sin cos X cos sin x x cos y sin . Equating the entries in this matrix equation gives Y sin cos y x sin y cos
the second pair of rotation of axes formulas. cos 1 sin 1 cos 2 sin 2 (b) R1 R2 sin 1 cos 1 sin 2 cos 2 cos 1 cos 2 sin 1 sin 2 cos 1 sin 2 sin 1 cos 2 cos 1 2 sin 1 2 sin 1 2 cos 1 2 sin 1 cos 2 cos 1 sin 2 sin 1 sin 2 cos 1 cos 2 38. (a) Using A A cos2 B sin cos C sin2 , B 2 C a sin cos B cos2 sin2 , and C A sin2 B sin cos C cos2 , we first expand the terms. 2 2 B 2 C a sin cos B cos2 sin2
2 4 C A2 sin2 cos2 4B C A sin cos cos2 sin2 B 2 cos2 sin2 4 C A2 sin2 cos2 4B C A sin cos3 4B C A sin3 cos
B 2 cos4 2B 2 sin2 cos2 B 2 sin4
SECTION 11.5 Rotation of Axes
4A C 4 A cos2 B sin cos C sin2 A sin2 B sin cos C cos2
4A2 sin2 cos2 4AB sin cos3 4AC cos4 4AB sin3 cos 4B 2 sin2 cos2
4BC sin cos3 4AC sin4 4BC sin3 cos 4C 2 sin2 cos2
Since there are many terms in this expansion, we find the coefficients of like trignometric terms. 2 B Term 4A C Sum cos4
B2
sin cos3
4B C A
sin2 cos2 sin3 cos sin4 So 2
B 4A C
4 C A2 2B 2 4B C A B2
4AC
4AB 4BC
B 2 4AC
4A2 4B 2 4C 2
4BC 4AB 4AB 4BC 0 8AC 2B 2 2 B 2 4AC
4AC
B 2 4AC
4AB 4BC
4BC AB 4AB 4BC 0
B 2 4AC cos4 2 B 2 4AC sin2 cos2 B 2 4AC sin4
2 cos4 2 sin2 cos2 sin4 B 2 4AC cos2 sin2 B 2 4AC 12 B 2 4AC
B 2 4AC
So B 2 4AC B 2 4A C .
(b) Using A A cos2 B sin cos C sin2 and C A sin2 B sin cos C cos2 , we have A C A cos2 B sin cos C sin2 A sin2 B sin cos C cos2 A sin2 cos2 C sin2 cos2 AC
(c) Since F F, F is also invariant under rotation. 39. Let P be the point x1 y1 and Q be the point x2 y2 and let P X 1 Y1 and Q X 2 Y2 be the images of P and Q under the rotation of . So X 1 x1 cos y1 sin Y1 x1 sin y1 cos , X 2 x2 cos y2 sin and Y2 x2 sin y2 cos . Thus d P Q X 2 X 1 2 Y2 Y1 2 , where X 2 X 1 2
2 2 x2 cos y2 sin x1 cos y1 sin x2 x1 cos y2 y1 sin
x2 x1 2 cos2 x2 x1 y2 y1 sin cos y2 y1 2 sin2
and Y2 Y1 2
2 2 x2 sin y2 cos x1 sin y1 cos x2 x1 sin y2 y1 cos
x2 x1 2 sin2 x2 x1 y2 y1 sin cos y2 y1 2 cos2
So X 2 X 1 2 Y2 Y1 2 x2 x1 2 cos2 x2 x1 y2 y1 sin cos y2 y1 2 sin2
x2 x1 2 sin2 x2 x1 y2 y1 sin cos y2 y1 2 cos2
x2 x1 2 cos2 y2 y1 2 sin2 x2 x1 2 sin2 y2 y1 2 cos2 x2 x1 2 cos2 sin2 y2 y1 2 sin2 cos2 x2 x1 2 y2 y1 2
45
46
CHAPTER 11 Conic Sections
Putting these equations together gives d P Q X 2 X 1 2 Y2 Y1 2 x2 x1 2 y2 y1 2 d P Q.
11.6 POLAR EQUATIONS OF CONICS 1. All conics can be described geometrically using a fixed point F called the focus and a fixed line called the directrix. For a distance from P to F fixed positive number e the set of all points P satisfying e is a conic section. If e 1 the conic is a distance from P to parabola, if e 1 the conic is an ellipse, and if e 1 the conic is a hyperbola. The number e is called the eccentricity of the conic. ed ed 2. The polar equation of a conic with eccentricity e has one of the following forms: r or r . 1 e cos 1 e sin 3. Substituting e 23 and d 3 into the general equation of a conic with vertical directrix, we get r r
6 . 3 2 cos
4. Substituting e 43 and d 3 into the general equation of a conic with vertical directrix, we get r r
12 . 3 4 cos
2 3 3 1 23 cos
4 3 3 1 43 cos
5. Substituting e 1 and d 2 into the general equation of a conic with horizontal directrix, we get r r
2 . 1 sin
6. Substituting e 12 and d 4 into the general equation of a conic with horizontal directrix, we get r r
4 . 2 sin
12 1 sin 1 4 2 1 12 sin
20 45 r . 1 4 cos 1 4 cos 12 06 2 r . 8. r 2 csc r sin 2 y 2. So d 2 and e 06 gives r 1 06 sin 1 06 sin 9. Since this is a parabola whose focus is at the origin and vertex at 5 2, the directrix must be y 10. So d 10 and 10 1 10 . e 1 gives r 1 sin 1 sin d P F 2 2 10. Since the vertex is at 2 0 we have d P F 2. Now, since e we get 04 d P 5. d P d P 04 04 7 28 The directrix is x 7, so substituting e 04 and d 7 we get r r . 1 04 cos 1 04 cos 6 is Graph II. The eccentricity is 1, so this is a parabola. When 0, we have r 3 and when 11. r 2 , we 1 cos have r 6. 2 1 12. r is Graph III. r , so e 12 and this is an ellipse. When 0, r 2, and when , r 23 . 2 cos 1 1 cos 7. r 5 sec r cos 5 x 5. So d 5 and e 4 gives r
2
3 13. r is Graph VI. e 2, so this is a hyperbola. When 0, r 3, and when , r 3. 1 2 sin 5
14. r
5 3 is Graph I. r , so e 1 and this is a parabola. When 0, r 53 and when , r 53 . 3 3 sin 1 sin
SECTION 11.6 Polar Equations of Conics
47
15. r
4 12 , so e 23 and this is an ellipse. When 0, r 4, and when , r 4. is Graph IV. r 2 3 2 sin 1 sin
16. r
6 12 is Graph V. r , so e 32 and this is a hyperbola. When 0, r 12 5 , and when , 2 3 cos 1 3 cos
3
2
we have r 12.
4 has e 1 and d 4, so 1 sin it represents a parabola.
17. (a) The equation r
3
18. (a) The equation r
3 2 has e 1 2 2 sin 1 sin
and d 32 , so it represents a parabola. V(3/4, ¹/2) _4
_8
_4 y=_4
O
4 V(2, 3¹/2)
ed , 1 e sin the directrix is parallel to the polar axis and has equation y 4. The vertex is 2 32 .
e 1 and d 53 , so it represents a parabola. x=5/3
O
O
4
ed , 1 e sin the directrix is parallel to the polar axis and has equation y d 32 . The vertex is 34 2 .
(b) Because the equation is of the form r
5
5 3 has 3 3 cos 1 cos
1
y=3/2 2
8
(b) Because the equation is of the form r
19. (a) The equation r
_2
2
V(5/6, 0)
3
ed , 1 e cos the directrix is parallel to the polar axis and has equation x d 53 . The vertex is 56 0 .
(b) Because the equation is of the form r
2
20. (a) The equation r
2 5 has 5 5 cos 1 cos
e 1 and d 25 , so it represents a parabola. x=_2/5
1 V(1/5, ¹) O
1
ed , 1 e cos the directrix is parallel to the polar axis and has equation x d 25 . The vertex is 15 .
(b) Because the equation is of the form r
48
CHAPTER 11 Conic Sections
21. (a) The equation r ellipse.
2 4 has e 12 1, so it represents an 1 2 cos 1 cos 2
ed (b) Because the equation is of the form r with d 4, the directrix is 1 e cos vertical and has equation x 4. Thus, the vertices are V1 4 0 and V2 43 .
x=_4 1 Vª(4/3, ¹)
1
O
VÁ(4, 0)
(c) The length of the major axis is 2a V1 V2 4 43 16 3 and the center is at the midpoint of V1 V2 , 43 0 . The minor axis has length 2b where 2 2 b2 a 2 c2 a 2 ae2 83 83 12 16 3 , so 8 3 2b 2 16 3 3 462.
6 2 has e 23 1, so it represents an 2 3 2 sin 1 sin
VÁ(6, ¹/2)
ed with d 3, the directrix is 1 e sin horizontal and has equation y 3. Thus, the vertices are V1 6 2 and V2 65 32 .
O
22. (a) The equation r ellipse.
3
(b) Because the equation is of the form r
1 Vª(6/5, 3¹/2)
1 y=_3
(c) The length of the major axis is 2a V1 V2 6 65 36 5 and the center is at the midpoint of V1 V2 , 12 5 2 . The minor axis has length 2b where 2 18 2 2 36 , so b2 a 2 c2 a 2 ae2 18 5 5 3 5 12 5 2b 2 36 5 5 537.
23. (a) The equation r ellipse.
3 12 has e 34 1, so it represents an 4 3 sin 1 3 sin 4
ed (b) Because the equation is of the form r with d 4, the directrix is 1 e sin horizontal and has equation y 4. Thus, the vertices are V1 12 7 2 and V2 12 32 .
96 (c) The length of the major axis is 2a V1 V2 12 7 12 7 and the center is at 3 the midpoint of V1 V2 , 36 7 2 . The minor axis has length 2b where
2 48 3 2 144 , so b2 a 2 c2 a 2 ae2 48 7 7 4 7 24 7 2b 2 144 7 7 907.
VÁ(12/7, ¹/2) O 1 4 8 Vª(12, 3¹/2)
y=4
SECTION 11.6 Polar Equations of Conics
49
9
24. (a) The equation r ellipse.
18 2 has e 34 1, so it represents an 4 3 cos 1 3 cos
x=6
4
VÁ(18/7, 0)
ed (b) Because the equation is of the form r with d 6, the directrix is 1 e cos 0 and V2 18 . vertical and has equation x 6. Thus, the vertices are V1 18 7
Vª(18, ¹)
2 O
5
144 (c) The length of the major axis is 2a V1 V2 18 7 18 7 and the center is at the midpoint of V1 V2 , 54 7 . The minor axis has length 2b where
2 72 3 2 324 , so b2 a 2 c2 a 2 ae2 72 7 7 4 7 36 7 2b 2 324 7 7 1361.
In #25, indicate vertical axis scale 8 has e 2 1, so it represents a hyperbola. 1 2 cos ed with d 4, the transverse axis is (b) Because the equation has the form r 1 cos horizontal and the directrix has equation x 4. The vertices are V1 83 0 and
x=4
25. (a) The equation r
VÁ(8/3, 0)
Vª(_8, ¹)
O
V2 8 8 0.
(c) The center is the midpoint of V1 V2 , 16 3 0 . To sketch the central box and the
asymptotes, we find a and b. The length of the transverse axis is 2a 16 3 , and so 2 2 a 83 , and b2 c2 a 2 ae2 a 2 83 2 83 64 3 , so 8 3 b 64 3 3 462.
10 has e 4 1, so it represents a hyperbola. 1 4 sin ed with d 52 , the transverse axis (b) Because the equation has the form r 1 cos
In #26, indicate scale in both axes; I believe that ticks are every 2 units for the x-axis, but every 1 unit for the y-axis
26. (a) The equation r
VÁ(2, 3¹/2)
is vertical and the directrix has equation y 52 . The vertices are
10 3 3 V1 10 3 2 3 2 and V2 2 2 . (c) The center is the midpoint of V1 V2 , 83 32 . To sketch the central box and the
asymptotes, we find a and b. The length of the transverse axis is 2a 43 , and so 2 2 a 23 , and b2 c2 a 2 ae2 a 2 23 4 23 20 3 , so b 20 3 258.
y=_5/2 Vª(_10/3, ¹/2)
50
CHAPTER 11 Conic Sections
27. (a) The equation r hyperbola.
10 20 has e 32 1, so it represents a 2 3 sin 1 3 sin
VÁ(4, 3¹/2)
2
(b) Because the equation has the form r
O y=_20/3
ed with d 20 3 , the transverse axis 1 cos
is vertical and the directrix has equation y 20 3 . The vertices are 3 3 V1 20 2 20 2 and V2 4 2 . (c) The center is the midpoint of V1 V2 , 12 32 . To sketch the central box and the
Vª(_20, ¹/2)
In #27, indicate scale in both axes.
asymptotes, we find a and b. The length of the transverse axis is 2a 16, and so 2 a 8, and b2 c2 a 2 ae2 a 2 8 32 82 80, so b 80 4 5 894.
28. (a) The equation r hyperbola.
3 6 has e 72 1, so it represents a 7 2 7 cos 1 cos
x=6/7
2
1
ed with d 67 , the transverse axis (b) Because the equation has the form r 1 cos is horizontal and the directrix has equation x 67 . The vertices are V1 23 0 and V2 65 65 0 . (c) The center is the midpoint of V1 V2 , 14 15 0 . To sketch the central box and the
O VÁ(2/3, 0)
Vª(_6/5, ¹)
8 , and so asymptotes, we find a and b. The length of the transverse axis is 2a 15 4 , and b2 c2 a 2 ae2 a 2 4 7 2 4 2 4 , so a 15 15 2 15 5 b 45 2 5 5 089.
29. (a) r
4 e 3, so the conic is a hyperbola. 1 3 cos
(b) The vertices occur where 0 and . Now 0 r
4 1, 1 3 cos 0
4 4 and r 2. Thus the vertices are 1 0 and 1 3 cos 2 2 .
In #29 indicate scale in both axes.
(_2, ¹)
(1, 0)
8
8 3 r e 1, so the conic is a parabola. 3 3 cos 1 cos 8 8 4 (b) Substituting 0, we have r . Thus the vertex is 43 0 . 3 3 cos 0 6 3
30. (a) r
1
( 43 , 0)
SECTION 11.6 Polar Equations of Conics
2 e 1, so the conic is a parabola. 1 cos 2 (b) Substituting , we have r 22 1. Thus the vertex is 1 . 1 cos
31. (a) r
(1, ¹)
1
10
32. (a) r
10 3 r e 23 , so the conic is an ellipse. 3 2 sin 1 2 sin
(10, ¹2 )
3
3 (b) The vertices occur where 2 and 2 . Now 2
10 10 10 10 10 and 32 r 2. Thus, 3 3 2 sin 1 5 3 2 sin 2 2 3 . the vertices are 10 and 2 2 2
r
33. (a) r
1
(2, 3¹ 2 )
1 6 6 2 r e 12 , so the conic is an ellipse. 2 sin 1 1 sin
(2, ¹2)
2
3 (b) The vertices occur where 2 and 2 . Now 2
1
6 6 6 6 2 and 32 r 6. Thus, the 3 2 sin 2 3 1 2 sin 2 vertices are 2 2 and 6 32 .
r
(6, 3¹ 2 )
5
34. (a) r
5 2 e 32 , so the conic is a hyperbola. r 2 3 sin 1 3 sin
1
2
3 (b) The vertices occur where 2 and 2 . Now 2
(
5 5 5 5 and 32 r 55 1. Thus, 3 2 3 sin 1 2 3 sin 2 2 3 the vertices are 5 2 and 1 2 .
r
¹ _5, 2
)
(1, 3¹ 2)
7
35. (a) r
7 2 r e 52 , so the conic is a hyperbola. 2 5 sin 1 5 sin 2
7 7 7 and 3 3 2 5 sin 2 77 1. Thus, the vertices are 73 2 and 3
3 (b) The vertices occur where 2 and 2 . r
32 r 1 32 . 36. (a) r
7
(_ 73 , ¹2 )
1
(1, 3¹ 2)
2 5 sin 2
8
8 3 r e 13 , so the conic is an ellipse. 3 cos 1 1 cos 3
(b) The vertices occur where 0 and . Now 0 8 8 r 84 2 and r 82 4. Thus, the vertices 3 cos 0 3 cos are 2 0 and 4 .
(4, ¹)
(2, 0) 1
51
52
CHAPTER 11 Conic Sections 1
37. (a) r
1 4 e 34 , so the 4 3 cos 1 3 cos 4
conic is an ellipse. The vertices occur where 0 and . Now 0 r
1 1 and 4 3 cos 0
(b) If the ellipse is rotated through 3 , the equation of the resulting conic is r 1.0
1 17 . Thus, the vertices r 4 3 cos are 1 0 and 17 . We have d 13 , so the
1 . 4 3 cos 3
0.5
directrix is x 13 .
0.5
x=-1/3
O
1
2
38. (a) r
2 5 e 35 , so the 5 3 sin 1 3 sin 5
conic is an ellipse. The vertices occur where 2 and 32 . Now 2 r and 32 r
2
(b) If the ellipse is rotated through 23 , the equation of the resulting conic is r
2 1 5 3 sin 2
2 . 5 3 sin 23 0.5
14 . Thus, the 5 3 sin 32
1 3 2 vertices are 1 2 and 4 2 . We have d 3 , so the directrix is y 23 .
1
O y=_2/3
-1.0
-0.5 -0.5
SECTION 11.6 Polar Equations of Conics
2 e 1, so the conic is a parabola. 1 sin 2 Substituting 2 , we have r t 1 sin 1, 2 so the vertex is 1 . Because d 2, the directrix 2
39. (a) r
In #39 indicate scale in both axes.The ticks are every 2 units for both axes.
is y 2.
(b) If the ellipse is rotated through 4 , the equation of the resulting conic is
r
2 . 1 sin 4 2
y=2
-8
O
-6
-4
-2
2 -2 -4 -6 -8
9 9 2 e 1, so the conic 2 2 cos 1 cos is a parabola. Substituting 0, we have 9 9 , so the vertex is 94 0 . r 2 2 cos 0 4
40. (a) r
Because d 92 , the directrix is x 92 .
(b) If the ellipse is rotated through 56 , the equation of the resulting conic is
r
9 . 2 2 cos 56
x=9/2
O
10
5 1
5
In #40 indicate scale in y-axis .
-5
41. The ellipse is nearly circular when e is close to 0 and becomes more elongated as
4
e 1 . At e 1, the curve becomes a parabola.
e=0.4 e=0.6
2
2 -2
4 e=0.8
e=1
-4
d where d 12 , d 2, and 1 sin d 10. As d increases, the parabolas get flatter while the vertex moves further
42. (a) Shown are the graphs of the conics r
10 d=10
from the focus at the origin.
-10
10 d=2 -10
1
d=2
53
54
CHAPTER 11 Conic Sections
e where e 05, e 1, and 1 e sin e 10. As e increases, the conic changes from an ellipse to a parabola, and
(b) Shown are the graphs of the conics r
finally to a hyperbola. The vertex gets closer to the directrix (shown as a
2 e=10 -2
2 e=0.5 e=1 -2
dashed line in the figure).
ed we need to show that ed a 1 e2 . 1 e cos e2 d 2 From the proof of the Equivalent Description of Conics we have a 2 2 . Since the conic is an ellipse, e 1 1 e2 and so the quantities a, d, and 1 e2 are all positive. Thus we can take the square roots of both sides and maintain 2d 2 a 1 e2 ed e ed equality. Thus a 2 r . ed a 1 e2 . As a result, r 2 a 1 e cos 1 e cos 1 e2 1 e2
43. (a) Since the polar form of an ellipse with directrix x d is r
(b) Since 2a 299 108 we have a 1495 108 , so a polar equation for the earth’s orbit (using e 0017) is 1495 108 1 00172 149 108 . r 1 0017 cos 1 0017 cos
44. (a) Using the form of the equation from Exercise 27, the perihelion distance occurs when and the aphelion distance occurs when 0. Since the focus is at the origin, the perihelion distance is 2 a 1 e a 1 e2 a 1 e 1 e a 1 e 1 e a 1 e and the aphelion distance is a 1 e. 1e 1e 1e 1e (b) Given e 0017 and 2a 299 108 we have a 1495 108 . Thus the perihelion distance is
1495 108 [1 0017] 1468 108 km and the aphelion distance is 1495 108 1 0017 1520 108 km.
45. From Exercise 44, we know that at perihelion r 443 109 a 1 e and at aphelion r 737 109 a 1 e. 737 109 1e a 1 e 1664 1664 1 e 1 e 9 a 1 e 1e 443 10 0664 1664 1 e 1664 0664 2664e e 025. 2664 Dividing these equations gives
46. Since the focus is at the origin, the distance from the focus to any point on the conic is the absolute value of the r-coordinate of that point, which we can obtain from the polar equation. 47. The r-coordinate of the satellite will be its distance from the focus (the center of the earth). From the r-coordinate we can easily calculate the height of the satellite.
CHAPTER 11 REVIEW 1. (a) y 2 4x. This is a parabola with 4 p 4 p 1. The vertex is 0 0, the
(b)
y
focus is 1 0, and the directrix is x 1.
1 1
x
CHAPTER 11 1 y 2 y 2 12x. This is a parabola with 4 p 12 p 3. The vertex 2. (a) x 12
Review
55
1
x
y
(b)
is 0 0, the focus is 3 0, and the directrix is x 3.
2
3. (a) 18 x 2 y x 2 8y. This is a parabola with 4 p 8 p 2. The vertex is
y
(b)
0 0, the focus is 0 2, and the directrix is y 2.
1
4. (a) x 2 8y. This is a parabola with 4 p 8 p 2. The vertex is 0 0,
1
x
1
x
1
y
(b)
vertex is 0 0, the focus is 0 2, and the directrix is y 2.
6. (a) 2x y 2 0 y 2 2x. This is a parabola with 4 p 2 p 12 . The vertex is 0 0, the focus is 12 0 , and the directrix is x 12 .
x
y
(b)
the focus is 0 2, and the directrix is y 2.
5. (a) x 2 8y 0 x 2 8y. This is a parabola with 4 p 8 p 2. The
1
1
(b)
y
1 1
x
56
CHAPTER 11 Conic Sections
7. (a) y 22 4 x 2. This is a parabola with 4 p 4 p 1. The vertex is
y
(b)
2 2, the focus is 1 2, and the directrix is x 3.
1 x
1
8. (a) x 32 20 y 2. This is a parabola with 4 p 20 p 5. The
y
(b)
vertex is 3 2, the focus is 3 7, and the directrix is y 3.
1 x
5
9. (a) 12 y 32 x 0 y 32 2x. This is a parabola with 4 p 2 p 12 . The vertex is 0 3, the focus is 12 3 , and the directrix is x 12 .
y
(b)
1
x
1
10. (a) 2 x 12 y x 12 12 y. This is a parabola with 4 p 12 p 18 . The vertex is 1 0, the focus is 1 18 , and the directrix is y 18 .
y
(b)
1 1 x
11. (a) 12 x 2 2x 2y 4 x 2 4x 4y 8 x 2 4x 4 4y 8
y
(b)
x 22 4 y 3. This is a parabola with 4 p 4 p 1. The vertex is 2 3, the focus is 2 3 1 2 2, and the directrix is y 3 1 4.
12. (a) x 2 3 x y x 2 3x 3y x 2 3x 94 3y 94 2 x 32 3 y 34 . This is a parabola with 4 p 3 p 34 . The vertex is 32 34 , the focus is 32 34 34 32 0 , and the directrix is y 34 34 32 .
1 1
(b)
x
y
2
1
x
CHAPTER 11
13. (a)
y2 x2 1. This is an ellipse with a 5, b 3, and c 25 9 4. The 9 25 center is 0 0, the vertices are 0 5, and the foci are 0 4.
y
(c)
1
(b) The length of the major axis is 2a 10 and the length of the minor axis is
y2 x2 1. This is an ellipse with a 7, b 3, and c 49 9 2 10. 49 9 The center is 0 0, the vertices are 7 0, and the foci are 2 10 0 .
x
1
2b 6.
14. (a)
57
Review
y
(c)
1
(b) The length of the major axis is 2a 14 and the length of the minor axis is
x
1
2b 6.
15. (a)
y2 x2 1. This is an ellipse with a 7, b 2, and c 49 4 3 5. 49 4 The center is 0 0, the vertices are 7 0, and the foci are 3 5 0 .
(c)
y
1
(b) The length of the major axis is 2a 14 and the length of the minor axis is
x
1
2b 4.
16. (a)
y2 x2 1. This is an ellipse with a 6, b 2, and c 36 4 4 2. 4 36 The center is 0 0, the vertices are 0 6, and the foci are 0 4 2 .
(c)
y
1
(b) The length of the major axis is 2a 12 and the length of the minor axis is
x
1
2b 4.
y2 x2 1. This is an ellipse with a 4, b 2, and 17. (a) x 2 4y 2 16 16 4 c 16 4 2 3. The center is 0 0, the vertices are 4 0, and the foci are 2 3 0 .
(c)
y
1 x
1
(b) The length of the major axis is 2a 8 and the length of the minor axis is 2b 4.
y2 x2 1. This is an ellipse with a 12 , b 13 , and 19 14 c 14 19 16 5. The center is 0 0, the vertices are 0 12 , and the foci are 0 16 5 .
18. (a) 9x 2 4y 2 1
(b) The length of the major axis is 2a 1 and the length of the minor axis is 2b 23 .
(c)
y 0.5
0.5 x
58
CHAPTER 11 Conic Sections
19. (a)
y2 x 32 1This is an ellipse with a 4, b 3, and 9 16 c 16 9 7. The center is 3 0, the vertices are 3 4, and the foci are 3 7 .
(c)
y
1 1
x
1
x
(b) The length of the major axis is 2a 8 and the length of the minor axis is 2b 6.
20. (a)
y 32 x 22 1. This is an ellipse with a 5, b 4, and 25 16 c 25 16 3. The center is 2 3, the vertices are
(c)
y 1
2 5 3 3 3 and 7 3, and the foci are 2 3 3 1 3 and 5 3.
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 8.
21. (a)
y 32 x 22 1. This is an ellipse with a 6, b 3, and 9 36 c 36 9 3 3. The center is 2 3, the vertices are 2 3 6 2 9 and 2 3, and the foci are 2 3 3 3 .
(c)
y 1 x
1
(b) The length of the major axis is 2a 12 and the length of the minor axis is 2b 6.
22. (a)
x2 y 52 1. This is an ellipse with a 5, b 3, and 3 25 c 25 3 22. The center is 0 5, the vertices are 0 5 5 0 10 and 0 0, and the foci are 0 5 22 .
(c)
y 1
1
x
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 2 3. 23. (a) 4x 2 9y 2 36y 4x 2 9 y 2 4y 4 36 4x 2 9 y 22 36 x2 y 22 1. This is an ellipse with a 3, b 2, and 9 4 c 9 4 5. The center is 0 2, the vertices are 3 2, and the foci are 5 2 .
(c)
y
(b) The length of the major axis is 2a 6 and the length of the minor axis is 2b 4.
1 1
x
CHAPTER 11
24. (a) 2x 2 y 2 2 4 x y 2x 2 4x y 2 4y 2 2 x 2 2x 1 y 2 4y 4 2 2 4 2 x 12 y 22 8 x 12
(c)
y
1 x
1
y 22
59
Review
1. This is an ellipse with a 2 2, b 2, and 4 8 c 8 4 2. The center is 1 2, the vertices are 1 2 2 2 , and
the foci are 1 2 2 1 0 and 1 4. (b) The length of the major axis is 2a 4 2 and the length of the minor axis is 2b 4.
y2 y2 x2 x2 1 0. This is a hyperbola with a 4, b 3, and 9 16 16 9 c 16 9 25 5. The center is 0 0, the vertices are 0 4, the foci
25. (a)
(b)
1
are 0 5, and the asymptotes are y 43 x.
26. (a)
y2 x2 1. This is a hyperbola with a 7, b 4 2, and 49 32 c 49 32 9. The center is 0 0, the vertices are 7 0, the foci are
(b)
28. (a)
x2 y2 1. This is a hyperbola with a 2, b 7, and 4 49 c 4 49 53. The center is 0 0, the vertices are 2 0, the foci are 53 0 , and the asymptotes are y 72 x.
x2 y2 1. This is a hyperbola with a 5, b 2, and 25 4 c 25 4 29. The center is 0 0, the vertices are 0 5, the foci are 0 29 , and the asymptotes are y 52 x.
1
x
2
x
1
x
1
x
y
2
9 0, and the asymptotes are y 4 7 2 x.
27. (a)
y
(b)
y
2
(b)
y
2
60
CHAPTER 11 Conic Sections
y2 x2 1. This is a hyperbola with a 4, b 2 2, 16 8 and c 16 8 24 2 6. The center is 0 0, the vertices are 4 0, the foci are 2 6 0 , and the asymptotes are y 2 4 2 x y 1 x.
29. (a) x 2 2y 2 16
y
(b)
1
2
x2 y2 1. This is a hyperbola 30. (a) x 2 4y 2 16 0 4y 2 x 2 16 4 16 with a 2, b 4, and c 4 16 2 5. The center is 0 0, the vertices are 0 2, the foci are 0 2 5 , and the asymptotes are y 24 x
1
x
1
x
y
(b)
1
y 12 x.
31. (a)
y2 x 42 1. This is a hyperbola with a 4, b 4 and 16 16 c 16 16 4 2. The center is 4 0, the vertices are 4 4 0 which are 8 0 and 0 0, the foci are 4 4 2 0 , and the asymptotes
y
(b)
1
1 x
are y x 4.
32. (a)
y 22 x 22 1. This is a hyperbola with a 2 2, b 2 2 and 8 8 c 8 8 4. The center is 2 2, the vertices are 2 2 2 2 , the
(b)
y 1
x
1
foci are 2 4 2 2 2 and 6 2, and the asymptotes are y 2 x 2 y x and y x 4.
33. (a)
x 12 y 32 1. This is a hyperbola with a 2, b 6, and 4 36 c 4 36 2 10. The center is 1 3, the vertices are 1 3 2 1 1 and 1 5, the foci are 1 3 2 10 , and the
(b)
1
1 8 asymptotes are y 3 13 x 1 y 13 x 10 3 and y 3 x 3 .
34. (a)
x2 y 32 1. This is a hyperbola with a 3, b 4, and 3 16 c 3 16 19. The center is 0 3, the vertices are 0 3 3 , the foci are 0 3 19 , and the asymptotes are y 3 43 x y 43 x 3 and y 43 x 3.
y
(b)
2
x
2
x
y
1
CHAPTER 11
35. (a) 9y 2 18y x 2 6x 18 9 y 2 2y 1 x 2 6x 9 9 9 18
61
Review y
(b)
x 32 y 12 9 y 12 x 32 18 1. This is a 2 18 hyperbola with a 2, b 3 2, and c 2 18 2 5.The center is 3 1, the vertices are 3 1 2 , the foci are 3 1 2 5 , and
1 1
x
the asymptotes are y 1 13 x 3 y 13 x and y 13 x 2. 36. (a) y 2 x 2 6y y 2 6y 9 x 2 9 y 32 x 2 9
y
(b)
x2 y 32 1. This is a hyperbola with a 3, b 3, and 9 9 c 9 9 3 2. The center is 0 3, the vertices are 0 3 3 0 6 and 0 0, the foci are 0 3 3 2 , and the asymptotes are y 3 x
1 x
1
y x 3 and y x 3.
37. This is a parabola that opens to the right with its vertex at 0 0 and the focus at 2 0. So p 2, and the equation is y 2 4 2 x y 2 8x.
38. This is an ellipse with the center at 0 0, a 12, and b 5. The equation is then
y2 y2 x2 x2 1. 2 1 2 144 25 12 5
39. From the graph, the center is 0 0, and the vertices are 0 4 and 0 4. Since a is the distance from the center to a
vertex, we have a 4. Because one focus is 0 5, we have c 5, and since c2 a 2 b2 , we have 25 16 b2
b2 9. Thus an equation of the hyperbola is
x2 y2 1. 16 9
40. This is a parabola that opens to the left with its vertex at 4 4, so its equation is of the form y 42 4 p x 4 with
p 0. Since 0 0 is a point on this hyperbola, we must have 0 42 4 p 0 4 16 16 p p 1. Thus the
equation is y 42 4 x 4.
41. From the graph, the center of the ellipse is 4 2, and so a 4 and b 2. The equation is
x 42 y 22 1 2 4 22
x 42 y 22 1. 16 4 42. From the graph, the center is at 1 0, and the vertices are 0 0 and 2 0. Since a is the distance form the center to a b vertex, a 1. From the graph, the slope of one of the asymptotes is 1 b 1. Thus an equation of the hyperbola is a x 12 y 2 1. 43.
x2 x2 y 1 y 1 x 2 12 y 1. This is a parabola with 12 12 4 p 12 p 3. The vertex is 0 1, the focus is 0 1 3 0 2, and the directrix is y 1 3 4.
y 1
1
x
62
44.
CHAPTER 11 Conic Sections
y2 y x2 12x 2 y 2 12y 0 12x 2 y 2 12y 36 36 12 144 12 x2 y 62 1. This is an ellipse with a 6, b 3, and 3 36 c 36 3 33. The center is 0 6, the foci are 0 6 33 , and the
y
x2 y2 1. This is a hyperbola with a 12, b 12, 144 144 and c 144 144 12 2. The center is 0 0, the foci are 0 12 2 , the
y
1
vertices are 0 6 6 0 0 and 0 12.
x
1
45. x 2 y 2 144 0
4
vertices are 0 12, and the asymptotes are y x.
x
4
x 32 y 2 1. This is a 46. x 2 6x 9y 2 x 2 6x 9 9y 2 9 9 hyperbola with a 3, b 1, and c 9 1 10. The center is 3 0, the foci are 3 10 0 , the vertices are 3 3 0 6 0 and 0 0, and the
y
1 1
x
asymptotes are y 13 x 3 y 13 x 1 and y 13 x 1.
47. 4x 2 y 2 8 x y 4 x 2 2x y 2 8y 0 4 x 2 2x 1 y 2 8y 16 4 16 4 x 12 y 42 20 x 12 y 42 1. This is an ellipse with a 2 5, b 5, and 5 20 c 20 5 15. The center is 1 4, the foci are 1 4 15 , and the vertices are 1 4 2 5 .
48. 3x 2 6 x y 10 3x 2 6x 6y 10 3 x 2 2x 1 6y 10 3 13 2 3 x 12 6y 13 3 x 12 6 y 13 6 x 1 2 y 6 . This is a parabola with 4 p 2 p 12 . The vertex is 1 13 6 the focus is 1 10 5 13 1 8 1 13 6 2 1 6 1 3 , and the directrix is y 6 2 3 . 49. x y 2 16y x 64 y 2 16y 64 y 82 x 64. This is a parabola with 4 p 1 p 14 . The vertex is 64 8, the focus is 64 14 8 255 8 , and the directrix is x 64 14 257 4 4 .
y
3 3
x
y
1
x
1
y
2 10
x
CHAPTER 11
50. 2x 2 4 4x y 2 y 2 2x 2 4x 4 y 2 2 x 2 2x 1 4 2
y2 x 12 1. This is a hyperbola with a 2, 2 b 1, and c 2 1 3. The center is 1 0, the foci are 1 3 , the vertices are 1 2 , and the asymptotes are y 2 x 1 y 2 x 2 and y 2 x 2. 51. 2x 2 12x y 2 6y 26 0 2 x 2 6x y 2 6y 26 2 x 2 6x 9 y 2 6y 9 26 18 9 2 x 32 y 32 1 x 32 1 2
63
y
1
y 2 2 x 12 2
Review
x
1
y x
1
_1
y 32 1. This is an ellipse with a 1, b 22 , and
c 1 12 22 . The center is 3 3, the foci are 3 3 22 , and the
vertices are 3 3 1 3 4 and 3 2. 52. 36x 2 4y 2 36x 8y 31 36 x 2 x 4 y 2 2y 31 36 x 2 x 14 4 y 2 2y 1 31 9 4
y
1
2 2 y 12 1. This is a 36 x 12 4 y 12 36 x 12 9
1
x
hyperbola with a 1, b 3, and c 1 9 10. The center is 12 1 , the foci are 12 10 1 , the vertices are 12 1 1 12 1 and 32 1 , and the asymptotes are y 1 3 x 12 y 3x 52 and y 3x 12 .
2 2 2 y 1 27 25 2 9 x 5 8 y 1 8 y 53. 9x 2 8y 2 15x8y27 0 9 x 2 53 x 25 75 36 4 4 6 2 4 . However, since the left-hand side of the equation is greater than or equal to 0, there is no point that satisfies this equation. The graph is empty. y 54. x 2 4y 2 4x 8 x 2 4x 4 4y 2 8 4 x 22 4y 2 12 y2 x 22 1. This is an ellipse with a 2 3, b 3, and 12 3 c 12 3 3. The center is 2 0, the foci are 2 3 0 1 0 and 5 0, and the vertices are 2 2 3 0 .
1 1
x
55. The parabola has focus 0 1 and directrix y 1. Therefore, p 1 and so 4 p 4. Since the focus is on the y-axis and the vertex is 0 0, an equation of the parabola is x 2 4y.
56. The parabola with vertex at the origin and focus F 5 0 has p 5, so 4 p 20. The focus is on the x-axis, so an equation is y 2 20x.
57. The ellipse with center at the origin and with x-intercepts 2 and y-intercepts 5 has a vertical major axis, a 5, and b 2, so an equation is
x2 y2 1. 4 25
64
CHAPTER 11 Conic Sections
58. The hyperbola has vertices 0 2 and asymptotes y 12 x. Therefore, a 2, and the foci are on the y-axis. Since the slopes of the asymptotes are 12
a y2 x2 b 2a 4, an equation of the hyperbola is 1. b 4 16
59. The ellipse has center C 0 4, foci F1 0 0 and F2 0 8, and major axis of length 10. Then 2c 8 0 c 4. Also,
since the length of the major axis is 10, 2a 10 a 5. Therefore, b2 a 2 c2 25 16 9. Since the foci are on
the y-axis, the vertices are on the y-axis, and an equation of the ellipse is
x2 y 42 1. 9 25
60. The hyperbola has center C 2 4, foci F1 2 7 and F2 2 1, and vertices V1 2 6 and V2 2 2. Thus, 2a 6 2 4 a 2. Also, 2c 7 1 6 c 3. So b2 9 4 5. Since the hyperbola has center C 2 4, its equation is y 42 x 22 1. 4 5
61. The ellipse has foci F1 1 1 and F2 1 3, and one vertex is on the x-axis. Thus, 2c 3 1 2 c 1, and so the
center of the ellipse is C 1 2. Also, since one vertex is on the x-axis, a 2 0 2, and thus b2 4 1 3. So an
equation of the ellipse is
x 12 y 22 1. 3 4
62. The parabola has vertex V 5 5 and directrix the y-axis. Therefore, p 0 5 p 5 4 p 20. Since the parabola opens to the right, its equation is y 52 20 x 5.
63. The ellipse has vertices V1 7 12 and V2 7 8 and passes through the point P 1 8. Thus, 2a 12 8 20 8 12 x 72 y 22 a 10, and the center is 7 7 2. Thus an equation of the ellipse has the form 1. 2 2 100 b
1 72 8 22 1 3600 36b2 100b2 64b2 3600 b2 225 4 . 100 b2 4 x 72 x 72 y 22 y 22 Therefore, an equation of the ellipse is 1 1. 2254 100 225 100 Since the point P 1 8 is on the ellipse,
64. The parabola has vertex V 1 0, horizontal axis of symmetry, and crosses the y-axis where y 2. Since the parabola has a horizontal axis of symmetry and V 1 0, its equation is of the form y 2 4 p x 1. Also, since the parabola crosses
the y-axis where y 2, it passes through the point 0 2. Substituting this point gives 22 4 p 0 1 4 p 4.
Therefore, an equation of the parabola is y 2 4 x 1.
65. The length of the major axis is 2a 186,000,000 a 93,000,000. The eccentricity is e ca 0017, and so c 0017 93,000,000 1,581,000. (a) The earth is closest to the sun when the distance is a c 93,000,000 1,581,000 91,419,000.
(b) The earth is furthest from the sun when the distance is a c 93,000,000 1,581,000 94,581,000.
66. We sketch the LORAN station on the y-axis and place the x-axis halfway between them as x2 y2 suggested in the exercise. This gives us the general form 2 2 1. Since the ship is 80 miles a b closer to A than to B we have 2a 80 a 40Since the foci are 0 150, we have c 150. Thus b2 c2 a 2 1502 402 20900. So this places the ship on the hyperbola given by the x2 y2 1600 y2 225 y2 1. When x 40, we get 1 1600 20,900 1600 20,900 1600 209 y 415. (Note that y 0, since A is on the positive y-axis.) Thus, the ship’s position is
equation
approximately 40 415.
y A 150
40 x
150 B
CHAPTER 11
67. (a) The graphs of
x2 y2 1 for k 1, 2, 4, and 8 are shown in 16 k 2 k2
the figure. (b) c2 16 k 2 k 2 16
Review
65
y k=8
c 4. Since the center is 0 0,
2
the focus of each of the ellipses is 4 0.
k=4 2 x k=2 k=1
68. (a) The graphs of y kx 2 for k 12 , 1, 2, and 4 are shown in the figure. 1 1 1 (b) y kx 2 x 2 y 4 y. Thus the foci are 0 . k 4k 4k
10
k=4
k=2
8
k=1
6 4
(c) As k increases, the focus gets closer to the vertex.
1
2 -2
0
k=2 2
69. (a) x 2 4x y y 2 1. Then A 1, B 4, and C 1, so the discriminant is 42 4 1 1 12. Since the discriminant is positive, the equation represents a hyperbola. 11 AC 0 2 90 45 . Therefore, x 22 X 22 Y and y 22 X 22 Y . (b) cot 2 B 4 Substituting into the original equation gives 2 2 2 2 2 2 2 2 4 22 X 22 Y 1 2 X 2 Y 2 X 2 Y 2 X 2 Y y 1 X 2 2XY Y 2 2 X 2 XY XY Y 2 1 X 2 2XY Y 2 1 (c) 2 2 3X 2 Y 2 1 3X 2 Y 2 1. This is a hyperbola with a 1 , b 1, and 3 c 13 1 2 . Therefore, the hyperbola has vertices V 1 0 and foci 3 3 F 2 0 , in XY -coordinates. 3
1
1
x
66
CHAPTER 11 Conic Sections
70. (a) 5x 2 6x y 5y 2 8 2x 8 2y 4 0. Then A 5, B 6, and C 5, so the discriminant is
62 4 5 5 64. Since the discriminant is negative, the equation represents an ellipse. AC (b) cot 2 0 2 90 45 . Therefore, x 22 X 22 Y and y 22 X 22 Y . Substituting B into the original equation gives 2 2 2 5 22 X 22 Y 6 22 X 22 Y 2 X 2 Y 2 5 22 X 22 Y 8 2 22 X 22 Y 8 2 22 X 22 Y 4 0 5 X 2 2XY Y 2 3 X 2 Y 2 5 X 2 2XY Y 2 8X 8Y 8X 8Y 4 0 2 2 y 2 2 2 2 2X 8Y 16Y 4 0 X 4 Y 2Y 2 X 2 4 Y 12 6 (c) X2 Y 12 1. This ellipse has a 6, b 26 , and c 6 32 3 2 2 . 6 32 Therefore, the vertices are V 6 1 and the foci are F 3 2 2 1 .
1 1
71. (a) 7x 2 6 3x y 13y 2 4 3x 4y 0. Then A 7, B 6 3, and C 13, so the discriminant is 2 6 3 4 7 13 256. Since the discriminant is negative, the equation represents an ellipse.
x
AC 7 13 1 2 60 30 . Therefore, x 23 X 12 Y and y 12 X 23 Y . B 6 3 3 Substituting into the original equation gives 2 1 X 3Y 7 23 X 12 Y 6 3 23 X 12 Y 2 2 2 13 12 X 23 Y 4 3 23 X 12 Y 4 12 X 23 Y 0 7 3X 2 2 3XY Y 2 3 3 3X 2 3XY XY 3Y 2 4 2 2 2 6X 2 3Y 2X 2 3Y 0 13 4 X 2 3XY 3Y y 9 13 8X Y 2 7 9 39 0 4X 2 8X 16Y 2 0 (c) X 2 21 4 2 4 4 2 4 1 4 X 2 2X 1 16Y 2 4 X 12 4Y 2 1. This ellipse has a 1, x 1 b 12 , and c 1 14 12 3. Therefore, the vertices are V 1 1 0 V1 0 0 and V2 2 0 and the foci are F 1 12 3 0 .
(b) cot 2
Note that the conic can't be both a parabola and a degenerate.
CHAPTER 11
67
Review
72. (a) 9x 2 24x y 16y 2 25. Then A 9, B 24, and C 16, so the discriminant is 242 4 9 16 0. Since the discriminant is zero, the equation represents a parabola.
(b) cot 2
9 16 7 7 AC cos 2 , so cos B 24 24 25
(or a degenerate conic).
1725 35 , sin 2
53 , and thus x 35 X 45 Y and y 45 X 35 Y . Substituting,
1725 45 2 y
(c)
9x 2 24x y 16y 2 25 2 4 X 3 Y 16 4 X 3 Y 2 25 9 35 X 45 Y 24 35 X 45 Y 5 5 5 5
1 x
1
25X 2 25 X 2 1 X 1. Thus the graph is a degenerate conic that consists of two lines. Converting back to x y-coordinates, we see that X 35 x 45 y, so 35 x 45 y 1 3x 4y 5.
73. 5x 2 3y 2 60 3y 2 60 5x 2 y 2 20 53 x 2 . This conic is an ellipse.
5
-5
5 -5
74. 9x 2 12y 2 36 0 12y 2 9x 2 36 y 2 34 x 2 3 y 34 x 2 3.
5
This conic is a hyperbola.
-5
5 -5
75. 6x y 2 12y 30 y 2 12y 30 6x y 2 12y 36 66 6x y 62 66 6x y 6 66 6x y 6 66 6x. This conic is
20
a parabola.
10
-10
76. 52x 2 72x y 73y 2 100 73y 2 72x y 52x 2 100 0. Using the quadratic formula,
72x 72x2 473 52x 2 100 y 146
1
-2
2 ,000x 2 29,200 72x 10146 72x201467325x 10 2 36 73 x 73 73 25x
This conic is an ellipse.
10
2 -1
68
CHAPTER 11 Conic Sections
1 e 1. Therefore, this is a 1 cos parabola.
2 e 23 . Therefore, this is an 3 2 sin ellipse.
77. (a) r
78. (a) r
(b)
(b)
( 25 , ¹2) 1
( 21 , ¹)
_1
1
(2, 3¹ 2 ) 4 e 2. Therefore, this is a 1 2 sin hyperbola.
12 e 4. Therefore, this is a 1 4 cos hyperbola.
79. (a) r
80. (a) r
(b)
(b)
(_4, 3¹ 2 ) ( 43 , ¹2)
(_4, 0)
( 125 , ¹)
1
1
x2 y2 2 0 1 is the standard equation of an ellipse centered at the origin passing through 2 4 and 0 2, corresponding to graph IV.
81. (a) 2x 2 y 2 4
(b) 3x 2 4y 12 y 34 x 2 3 is an equation of a parabola opening downward with vertex 0 3, corresponding to graph III. (c) x 2 y 2 4 is the standard equation of a circle centered at the origin with radius 2, corresponding to graph II.
(d) x 2 2x y y 2 x y 0 has discriminant B 2 4AC 22 4 1 1 0, so it is an equation of a rotated parabola, corresponding to graph VIII. (e) x y 1 y 1x is an equation of a hyperbola centered at the origin, corresponding to graph I. x2 1 is an equation of a (f) 4y 2 x 2 8y 0 4 y 2 2y x 2 0 4 y 12 x 2 4 y 12 4 hyperbola centered at 0 1, corresponding to graph VII. (g) 2y 2 5x 4y 8 5x 2 y 2 2y 8 2 y 12 8 2 x 2 25 y 12 . This is an equation of a parabola opening rightward with vertex 2 1, corresponding to graph VI.
(h) 153x 2 192x y 97y 2 225 has discriminant B 2 4AC 1922 4 153 97 22,500, so it is an equation of a rotated ellipse, corresponding to graph V.
CHAPTER 11
Test
1
x
1
x
1
x
69
CHAPTER 11 TEST 1. x 2 12y. This is a parabola with 4 p 12 p 3. The focus is 0 3 and the directrix is y 3.
2.
_1
y2 x2 1. This is an ellipse with a 4, b 2, and c 16 4 2 3. The 16 4 vertices are 4 0, the foci are 2 3 0 , the length of the major axis is 2a 8, and the length of the minor axis is 2b 4.
3.
y
x2 y2 1. This is a hyperbola with a 3, b 4, and c 9 16 5. The 9 16 vertices are 0 3, the foci are 0 5, and the asymptotes are y 34 x.
y
1
y
1
4. The parabola with vertex 0 0 and focus 4 0 has p 4, so 4 p 16. The focus lies to the right of the vertex, so an equation is y 2 16x.
5. The ellipse with foci 3 0 and vertices 4 0 has a 4 and c 3, so c2 a 2 b2 9 16 b2 b2 16 9 7. Thus, an equation is
x2 y2 1. 16 7
6. The hyperbola has foci 0 5 and asymptotes y 34 x. Since the foci are 0 5, c 5, the foci are on a 3 a a 34 b. Then the y-axis, and the center is 0 0. Also, since y 34 x x, it follows that b b 4 2 3 2 2 c2 52 25 a 2 b2 34 b b2 25 16 b b 16, and by substitution, a 4 4 3. Therefore, an equation of the hyperbola is
x2 y2 1. 9 16
7. This is a parabola that opens to the left with its vertex at 0 0. So its equation is of the form y 2 4 px with p 0. Substituting the point 4 2, we have 22 4 p 4 4 16 p p 14 . So an equation is y 2 4 14 x y 2 x.
8. This is an ellipse tangent to the x-axis at 0 0 and with one vertex at the point 4 3. The center is 0 3, and a 4 and b 3. Thus the equation is
x2 y 32 1. 16 9
70
CHAPTER 11 Conic Sections
9. This a hyperbola with a horizontal transverse axis, vertices at 1 0 and 3 0, and foci at 0 0 and 4 0. Thus the center is 2 0, and a 3 2 1 and c 4 2 2. Thus b2 22 12 3. So an equation is x 22
x 22 y2 1 2 3 1
y2 1. 3
10. 16x 2 36y 2 96x 36y 9 0 16 x 2 6x 36 y 2 y 9 16 x 2 6x 9 36 y 2 y 14 9 144 9 2 x 32 16 x 32 36 y 12 144
y 12
2
y 1 x
1
1. This is an ellipse with a 3, b 2, and c 9 4 5. The center is 3 12 , the vertices are 3 3 12 0 12 and 6 12 , and the foci are h c k 3 5 12 3 5 12 and 3 5 12 . 9
4
11. 9x 2 8y 2 36x 64y 164 9 x 2 4x 8 y 2 8y 164 9 x 2 4x 4 8 y 2 8y 16 164 36 128
y
x 22 y 42 9 x 22 8 y 42 72 1. This is a hyperbola 8 9 with a 2 2, b 3, and c2 a 2 b2 17. The center is 2 4, the foci are 2 17 4 , the vertices are 2 2 2 4 , and the asymptotes are
1 1
x
y 4 3 4 2 x 2 y 3 4 2 x 4 3 2 2 and y 3 4 2 x 4 3 2 2 .
12. 2x y 2 8y 8 0 y 2 8y 16 2x 8 16 y 42 2 x 4. This is a parabola with 4 p 2 p 12 . The vertex is 4 4, the focus is 4 12 4 72 4 , and the directrix is x 4 12 92 .
y 1 1
x
13. The ellipse with center 2 0, foci 2 3 and major axis of length 8 has a horizontal major axis with 2a 8 a 4. Also, c 3, so b2 a 2 c2 16 9 7. Thus, an equation is
y2 x 22 1. 7 16
14. The parabola has focus 2 4 and directrix the x-axis (y 0). Therefore, 2 p 4 0 4 p 2 4 p 8, and the vertex is 2 4 p 2 2. Hence, an equation of the parabola is x 22 8 y 2 x 2 4x 4 8y 16 x 2 4x 8y 20 0.
15. We place the vertex of the parabola at the origin, so the parabola contains the points 3 3, and the equation is of the form y 2 4 px. Substituting the point 3 3, we get 32 4 p 3 9 12 p p 34 . So the focus is 34 0 , and we should place the light bulb 34 inch from the vertex.
CHAPTER 11
Test
16. (a) 5x 2 4x y 2y 2 18. Then A 5, B 4, and C 2, so the discriminant is 42 4 5 2 24. Since the discriminant is negative, the equation represents an ellipse. AC 52 3 (b) cot 2 . Thus, cos 2 35 and so cos 135 255, 2 B 4 4 135 5 sin 5 . It follows that x 2 5 5 X 55 Y and y 55 X 2 5 5 Y . By 2 2 2 5 X 2 5Y 2 5 X 2 5Y substitution, 5 2 5 5 X 55 Y 4 2 5 5 X 55 Y 18 5 5 5 5 4X 2 4XY Y 2 45 2X 2 4XY XY 2Y 2 25 X 2 4XY 4Y 2 18
2 2 8 2 1 8 4 18 6X 2 Y 2 18 X Y 1. This is an X 2 4 85 25 XY 4 12 5 5 Y 5 5 3 18 ellipse with a 3 2 and b 3. y (c) (d) In XY -coordinates, the vertices are V 0 3 2 . Therefore, in
1
x
1
x y-coordinates, the vertices are x 3 2 and y 6 2 5 5 3 2 6 2 3 2 6 2 V1 , and x and y 5 5 5 5 2 . V2 3 2 6 5
Since cos 2 35 we have
5
2 cos1 35 5313 , so 27 . 17. (a) Since the focus of this conic is the origin and the directrix is x 2, the equation has the form
ed . Subsituting e 12 and d 2 we 1 e cos 1 2 get r r . 1 2 cos 1 cos
(b) r
3 3 2 r . So e 12 and the 2 sin 1 1 sin 2
conic is an ellipse.
r
(3, ¹2 )
2
( 32 , ¹) O
1
O
1
(
3¹ 1, 2
( 32 , 0) )
71
72
FOCUS ON MODELING
FOCUS ON MODELING Conics in Architecture 1. Answers will vary. 2. (a) The difference P F2 P A c is a constant (the length of the string). Also, P F1 P A d is constant. Subtracting these two equations, we find that P F2 P F1 c d is a constant. This is the definition of a hyperbola. (b) Reflect the entire apparatus through a line perpendicular to F1 F2 .
4. As the lids are twisted more, the vertices of the hyperbolic cross sections get closer together. 5. (a) The tangent line passes though the point a a 2 , so an equation is y a 2 m x a.
(b) Because the tangent line intersects the parabola at only the one point a a 2 , the system
y a 2 m x a y x2
has
only one solution, namely x a, y a 2 . y a 2 m x a y a 2 m x a (c) a 2 m x a x 2 x 2 mx am a 2 0. This quadratic 2 yx y x2 has discriminant m2 4 1 am a 2 m 2 4am 4a 2 m 2a2 . Setting this equal to 0, we find m 2a.
(d) An equation of the tangent line is y a 2 2a x a y a 2 2ax 2a 2 y 2ax a 2 .
6. (a) F1 and F2 are points of tangency of the spheres to the plane, and the vertical line through Q 1 and Q 2 is also tangent to each sphere. Since P lies outside of both spheres, we see that P F1 P Q 1 and P F2 P Q 2 . (b) Each sphere is tangent to the enclosing cylinder along a circle in a vertical plane. The distance between these two planes is constant, so P Q 1 P Q 2 Q 1 Q 2 is constant for any choice of P.
(c) Combining the results of parts (a) and (b), we see that P F1 P F2 P Q 1 P Q 2 is constant.
(d) By definition, an ellipse is a curve consisting of all points whose distances to two fixed points F1 and F2 add up to a constant. That is the case here.
CORRECTIONS: 2,4,11
CHAPTER 12
SEQUENCES AND SERIES
12.1
Sequences and Summation Notation 1
12.2
Arithmetic Sequences 6
12.3
Geometric Sequences 11
12.4 12.5
Mathematical Induction 20 The Binomial Theorem 29 Chapter 12 Review 33 Chapter 12 Test 39
¥
FOCUS ON MODELING: Modeling with Recursive Sequences 41
1
12 SEQUENCES AND SERIES 12.1 SEQUENCES AND SUMMATION NOTATION 1. A sequence is a function whose domain is the natural numbers. 2. The nth partial sum of a sequence is the sum of the first n terms of the sequence. So for the sequence an n 2 the fourth partial sum is S4 12 22 32 42 30.
3. an n 3. Then a1 1 3 2, a2 2 3 1, a3 3 3 0, a4 4 3 1, and a100 100 3 97.
4. an 2n 1. Then a1 2 1 1 1, a2 2 2 1 3, a3 2 3 1 5, a4 2 4 1 7, and a100 2 100 1 199. 1 1 1 1 1 1 1 1 . Then a1 1, a2 , a3 , a4 , and 5. an 3n 4 3 1 4 3 2 4 2 3 3 4 5 3 4 4 8 1 1 a100 . 3 100 4 296
6. an n 3 2. Then a1 13 2 3, a2 23 2 10, a3 33 2 29, a4 43 2 66, and a100 1003 2 1,000,002.
7. an 3n . Then a1 31 3, a2 32 9, a3 33 27, a4 34 81, and a100 3100 5154 1047 . n1 11 21 31 1 , a 1 41 1 , and . Then a1 15 1, a2 15 15 , a3 15 25 8. an 15 4 5 125 1001 6338 1070 . a100 15
1 1 1 1n 11 12 13 14 . Then a1 1, a2 , a3 , a4 , and 2 2 2 2 4 9 16 n 1 2 3 42 1 1100 . a100 2 10,000 100 1 1 1 1 1 1 1 1 1 1 , and a100 . 1, a2 , a3 , a4 10. an 2 . Then a1 2 2 2 2 2 4 9 16 10,000 n 1 2 3 4 100 9. an
11. an 1 1n . Then a1 1 11 0, a2 1 12 2, a3 1 13 0, a4 1 14 2, and a100 1 1100 2.
1 2 3 4 1n1 n 12 1 13 2 14 3 15 4 . Then a1 , a2 , a3 , a4 , and n1 11 2 21 3 31 4 41 5 100 1101 100 . a100 101 101 13. an n n . Then a1 11 1, a2 22 4, a3 33 27, a4 44 256, and a100 100100 10200 . 12. an
14. an 3. Then a1 3, a2 3, a3 3, a4 3, and a100 3. 15. an 2 an1 3 and a1 4. Then a2 2 4 3 14, a3 2 14 3 34, a4 2 34 3 74, and a5 2 74 3 154.
1 2 2 1 1 a 24 4 4, a3 , a4 3 , and a5 9 . 16. an n1 and a1 24. Then a2 6 6 6 3 6 9 6 54 17. an 2an1 1 and a1 1. Then a2 2 1 1 3, a3 2 3 1 7, a4 2 7 1 15, and a5 2 15 1 31. 1 1 1 1 2 1 3 1 5 , a3 and a1 1. Then a2 , a4 , and a5 . 18. an 1 an1 11 2 3 5 8 1 1 1 2 1 3 2
3
5
1
2
CHAPTER 12 Sequences and Series
19. an an1 an2 , a1 1, and a2 2. Then a3 2 1 3, a4 3 2 5, and a5 5 3 8.
20. an an1 an2 an3 and a1 1, a2 1, and a3 1. Then a4 1 1 1 3 and a5 3 1 1 5. 21. (a) a1 7, a2 11, a3 15, a4 19, a5 23,
a6 27, a7 31, a8 35, a9 39, a10 43
Delete tick marks at 2.5 and 7.5, since the domain is only integers
22. (a) a1 2, a2 6, a3 12, a4 20, a5 30,
a6 42, a7 56, a8 72, a9 90, a10 110
(b)
(b) 40
100
20
50
0
0
5
0
10
12 12 23. (a) a1 12 1 12, a2 2 6, a3 3 4,
12 12 12 a4 12 4 3, a5 5 , a6 6 2, a7 7 ,
0
5
10
24. (a) a1 6, a2 2, a3 6, a4 2, a5 6, a6 2, a7 6, a8 2, a9 6, a10 2
(b)
3 12 4 12 6 a8 12 8 2 , a9 9 3 , a10 10 5
(b)
5 10
0
5 0
0
5
a6 05, a7 2, a8 05, a9 2, a10 05
a6 2, a7 1, a8 3, a9 2, a10 1
(b) 2 1 5
10
26. (a) a1 1, a2 3, a3 2, a4 1, a5 3,
(b)
0
5
10
25. (a) a1 2, a2 05, a3 2, a4 05, a5 2,
0
0
10
4 2 0 -2 -4
5
10
27. 2, 4, 6, 8, . All are multiples of 2, so a1 2, a2 2 2, a3 3 2, a4 4 2, . Thus an 2n.
28. 1, 3, 5, 7, . All are odd numbers, so a1 2 1, a2 2 2 1, a3 3 2 1, a4 4 2 1, . Thus an 2n 1. 29. 3, 9, 27, 81, . All are powers of 3, so a1 3, a2 32 , a3 33 , a4 34 , . Thus an 3n .
1 , 1 , . The denominators are all powers of 3, and the terms alternate in sign. Thus a 30. 13 , 19 , 27 1 81
11 12 , a , 2 31 32
13 14 1n , a4 , . So an . 3 4 3n 3 3 31. 4, 9, 14, 19, . The difference between any two consecutive terms is 5, so a1 5 1 1, a2 5 2 1, a3 5 3 1, a4 5 4 1, . Thus, an 5n 1. a3
32. 10, 3, 4, 11, . The difference between any two consecutive terms is 7, so a1 7 1 17, a2 7 2 17, a3 7 3 17, a4 7 4 17, . Thus, an 7n 17.
33. 5, 25, 125, 625, . These terms are powers of 5, and the terms alternate in sign. So a1 12 51 , a2 13 52 , a3 14 53 , a4 15 54 , . Thus an 1n1 5n .
34. 3, 03, 003, 0003, . The ratio of any two consecutive terms is 10, so a1 3 100 , a2 3 101 , a3 3 102 , n1 1 a4 3 103 , . Thus an 3 10 .
SECTION 12.1 Sequences and Summation Notation
3
7 , 9 , . We consider the numerator separately from the denominator. The numerators of the terms differ 35. 1, 34 , 59 , 16 25
by 2, and the denominators are perfect squares. So a1 a5
2 1 1 2 2 1 2 3 1 2 4 1 , a2 , a3 , a4 , 2 2 2 1 2 3 42
2 5 1 2n 1 , . Thus an . 2 5 n2
36. 34 , 45 , 56 , 67 , . Both the numerator and the denominator increase by 1, so a1 a4
42 n2 , . Thus an . 43 n3
12 22 32 ,a ,a , 13 2 23 3 33
37. 0, 2, 0, 2, 0, 2, . These terms alternate between 0 and 2. So a1 1 1, a2 1 1, a3 1 1, a4 1 1, a5 1 1, a6 1 1, Thus an 1 1n . n1
38. 1, 12 , 3, 14 , 5, 16 , . So a1 1, a2 21 , a3 31 , a4 41 , . Thus an n 1
.
39. a1 2, a2 4, a3 6, a4 8, . Therefore, an 2n. So S1 2, S2 2 4 6, S3 2 4 6 12, S4 2 4 6 8 20, S5 2 4 6 8 10 30, and S6 2 4 6 8 10 12 42. 40. a1 12 , a2 22 , a3 32 , a4 42 , . Therefore, an n 2 . So S1 12 1, S2 122 5, S3 532 59 14, S4 14 42 14 16 30, S5 30 52 30 25 55, and S6 55 62 55 36 91.
1 1 1 1 1 4 1 1 41. a1 13 , a2 2 , a3 3 , a4 4 , . Therefore, an n . So S1 13 , S2 13 2 , S3 13 2 3 13 27 , 3 9 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 121 1 364 S4 13 2 3 4 40 81 , and S5 3 32 33 34 35 243 , S6 3 32 33 34 35 36 729 . 3 3 3 42. a1 4, a2 4, a3 4, a4 4, . Therefore, an 4 1n1 . So S1 4, S2 4 4 0, S3 4 4 4 4, S4 0, S5 4, and S6 0. 2 2 2 2 2 2 2 2 , and S4 23 2 3 4 80 43. an n . So S1 23 , S2 2 89 , S3 23 2 3 26 27 81 . Therefore, 3 3 3 3 3 3 3 3 n 3 1 . Sn 3n 1 1 44. an . So S1 12 13 , S2 12 13 13 14 12 13 13 14 12 14 , n1 n2 1 1 S3 2 3 13 14 14 15 12 13 13 14 14 15 12 15 , and S4 12 13 13 14 14 15 15 16 12 13 13 14 14 15 15 16 12 16 . Therefore, 1 1 1 1 1 1 1 1 1 1 Sn 12 13 13 14 n1 n2 2 3 3 n1 n1 n2 2 n2 . n n 1. So S1 1 2 1 2, S2 1 2 2 3 1 2 2 3 1 3, 1 2 2 3 3 4 1 2 2 3 3 4 1 4, S3 S4 1 2 2 3 3 4 4 5 1 2 2 3 3 4 4 51 5
45. an
Therefore, 1 2 2 3 n n 1 Sn 1 2 2 3 3 n n n 1 1 n 1
4
CHAPTER 12 Sequences and Series
In #46, replace "k" with "n" (4 times)
k log k log k 1. So S1 log 1 log 2 log 2, S2 log 2 log 2 log 3 log 3, k 1 S3 log 2 log 2 log 3 log 3 log 4 log 2 log 2 log 3 log 3 log 4 log 4, and
46. an log
S4 log 2 log 2 log 3 log 3 log 4 log 4 log 5
log 2 log 2 log 3 log 3 log 4 log 4 log 5 log 5
Therefore, Sn log n 1. 4 47. k1 k 1 2 3 4 10 4 48. k1 k 2 1 22 32 42 1 4 9 16 30
49. 50. 51.
52.
1 1 1 1 1 25 k1 3k 3 6 9 12 36
4
51
j 1 2 3 4 50 51 j1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
8
i i 1 1 1
12
02020202 8
i 4 10 10 10 10 10 10 10 10 10 10 90 53. k1 2k1 20 21 22 23 24 1 2 4 8 16 31 54. i31 i2i 1 21 2 22 3 23 2 8 24 34
5
for #58, I get 0.153446
55. 385 56. 15,550 57. 46,438 58. 0153146 59. 22 60. 0688172 4 3 3 3 3 3 61. k1 k 1 2 3 4 1 8 27 64 4 j 1 0 1 2 3 3 2 15 0 62. j1 j 1 2 3 4 5 3 2 5 63. 6k0 k 4 4 5 6 7 8 9 10 64. 9k6 k k 3 6 9 7 10 8 11 9 12 54 70 88 108 k 3 4 5 100 65. 100 k3 x x x x x n j1 x j 12 x 13 x 2 14 x 3 1n1 x n x x 2 x 3 1n1 x n 66. j1 1 67. 4 8 12 16 48 12 k1 4k 10 68. 2 5 8 29 k1 3k 1 2 69. 12 22 32 102 10 k1 k
1 1 1 1 1 1k 100 k2 k ln k 2 ln 2 3 ln 3 4 ln 4 5 ln 5 100 ln 100 999 1 1 1 1 1 k1 71. 12 23 34 999 1000 k k 1 1 2 3 n k 72. 2 2 2 2 nk1 2 1 2 3 n k k 73. 1 x x 2 x 3 x 100 100 k0 x k1 k x k1 74. 1 2x 3x 2 4x 3 5x 4 100x 99 100 k1 1 75. 2, 2 2, 2 2 2, 2 2 2 2, . We simplify each term in an attempt to determine a formula for an . So a1 212 , a2 2 212 232 234 , a3 2 234 274 278 , a4 2 278 2158 21516 , . Thus 70.
n
n
an 22 12 .
76. F1 1, F2 1, F3 2, F4 3, F5 5, F6 8, F7 13, F8 21, F9 34, F10 55
77. (a) A1 $2004, A2 $200801, A3 $201202, A4 $201605, A5 $202008, A6 $202412
SECTION 12.1 Sequences and Summation Notation
5
(b) Since 3 years is 36 months, we get A36 $214916. 78. (a) I1 0, I2 $050, I3 $150, I4 $301, I5 $503, I6 $755 (b) Since 5 years is 60 months, we have I60 $97700.
79. (a) P1 35,700, P2 36,414, P3 37,142, P4 37,885, P5 38,643 (b) Since 2014 is 10 years after 2004, P10 42,665.
80. (a) The amount owed at the end of the month, An , is the amount owed at the beginning of the month, An1 , plus the interest, 0005An1 , minus the $200 monthly repayment. Thus An An1 0005An1 200 An 1005An1 200.
(b) A1 9850, A2 969925, A3 954774, A4 939548, A5 924246, A6 908867. Thus the amount owing after six months is $908867.
81. (a) The number of catfish at the end of the month, Pn , is the population at the start of the month, Pn1 , plus the increase in population, 008Pn1 , minus the 300 catfish harvested. Thus Pn Pn1 008Pn1 300 Pn 108Pn1 300. (b) P1 5100, P2 5208, P3 5325, P4 5451, P5 5587, P6 5734, P7 5892, P8 6064, P9 6249, P10 6449, P11 6665, P12 6898. Thus there should be 6898 catfish in the pond at the end of 12 months.
82. (a) Let n be the number of years since 2022, so P0 $240,000. Each month, the median price of a house increases to 106 times its price the previous month. Thus, Pn 106n P0 . (b) Since 2030 2022 8, we calculate P8 382,52353. In 2030, the median price of a house is estimated to be $382,524.
83. (a) Let An be the salary in the nth year. Then A1 $45,000. Since the salary increases by $2000 each year, An An1 2000. Thus, A1 $45,000 and An An1 2000.
(b) A5 A4 2000 A3 2000 2000 A2 2000 4000 A1 2000 6000 $53,000. 84. (a) Let n be the days after the experiment starts, so C0 4. Each day the concentration increases by 10%, so after n days the concentration of the brine solution is Cn 110Cn1 with C0 4.
(b) C8 110C7 110 110C6 1108 C0 1108 4 86. Thus, the brine solution concentration is 86 g/L of salt.
85. Let Fn be the number of pairs of rabbits in the nth month. Clearly F1 F2 1. In the nth month each pair that is two or more months old (that is, Fn2 pairs) will add a pair of offspring to the Fn1 pairs already present. Thus Fn Fn1 Fn2 . So Fn is the Fibonacci sequence. 86. (a) an n 2 . Then a1 12 1, a2 22 4, a3 32 9, a4 42 16.
(b) an n 2 n 1 n 2 n 3 n 4, a1 12 1 1 1 2 1 3 1 4 1 0 1 2 3 1, a2 22 2 1 2 2 2 3 2 4 4 1 0 1 2 4, a3 32 3 1 3 2 3 3 3 4 9 2 1 0 1 9,
a4 42 4 1 4 2 4 3 4 4 16 3 2 1 0 16.
Hence, the sequences agree in the first four terms. However, for the second sequence, a5 52
5 1 5 2 5 3 5 4 25 4 3 2 1 49, and for the first sequence, a5 52 25, and thus the sequences disagree from the fifth term on. (c) an n 2 n 1 n 2 n 3 n 4 n 5 n 6 agrees with an n 2 in the first six terms only.
(d) an 2n and bn 2n n 1 n 2 n 3 n 4.
6
CHAPTER 12 Sequences and Series
a n if an is even 2 87. an1 3an 1if an is odd
With a1 11, we have a2 34, a3 17, a4 52, a5 26, a6 13, a7 40,
a8 20, a9 10, a10 5, a11 16, a12 8, a13 4, a14 2, a15 1, a16 4, a17 2, a18 1, (with 4, 2, 1 repeating). So a3n1 4, a3n2 2, and a3n 1, for n 5. With a1 25, we have a2 76, a3 38, a4 19, a5 58, a6 29, a7 88, a8 44, a9 22, a10 11, a11 34, a12 17, a13 52, a14 26, a15 13, a16 40, a17 20, a18 10, a19 5, a20 16, a21 8, a22 4, a23 2, a24 1, a25 4, a26 2, a27 1, (with 4, 2, 1 repeating). So a3n1 4, a3n2 2, and a3n3 1 for n 7. We conjecture that the sequence will always return to the numbers 4, 2, 1 repeating.
88. an anan1 anan2 , a1 1, and a2 1. So a3 a31 a31 a2 a2 1 1 2, a4 a42 a41 a2 a3 1 2 3, a5 a53 a52 a2 a3 1 2 3, a6 a63 a63 a3 a3 2 2 4, a7 a74 a73 a3 a4 2 3 5, a8 a85 a84 a3 a4 2 3 5, a9 a95 a95 a4 a4 3 3 6, and a10 a106 a105 a4 a5 3 3 6. The definition of an depends on the values of certain preceding terms. So an is the sum of two preceding terms whose choice depends on the values of an1 and an2 (not on n 1 and n 2).
12.2 ARITHMETIC SEQUENCES 1. An arithmetic sequence is sequence where the difference between successive terms is constant. 2. The sequence an a n 1 d is an arithmetic sequence where a is the first term and d is the common difference. So, for the arithmetic sequence an 2 5 n 1 the first term is 2 and the common difference is 5. 3. True. The nth partial sum of an arithmetic sequence is the average of the first and last terms times n. 4. True. If we know the first and second terms of an arithmetic sequence then we can find any other term.
5. (a) a1 7 3 1 1 7, a2 7 3 2 1 10,
a3 7 3 3 1 13, a4 7 3 4 1 16,
6. (a) a1 10 20 1 1 10, a2 10 20 2 1 10, a3 10 20 3 1 30,
a5 7 3 5 1 19
a4 10 20 4 1 50,
(b) The common difference is 3. (c)
a5 10 20 5 1 70
an 20
(b) The common difference is 20.
15
(c)
10
60
5 0
an
40 1
2
3
4
5 n
20 0
1
2
3
4
5 n
SECTION 12.2 Arithmetic Sequences
7. (a) a1 3 5 1 1 3,
8. (a) a1 7 3 1 1 7, a2 7 3 2 1 4,
a2 3 5 2 1 8,
a3 7 3 3 1 1, a4 7 3 4 1 2,
a3 3 5 3 1 13,
a5 7 3 5 1 5
a4 3 5 4 1 18,
(b) The common difference is 3.
a5 3 5 5 1 23
(c)
(b) The common difference is 5. an 5
(c)
1
2
an 10 5
3
4
4
0 _5
5 n
1
2
3
5 n
_10 _20
10. (a) a1 12 1 1 0, a2 12 2 1 12 ,
9. (a) a1 15 05 1 1 15, a2 15 05 2 1 2,
a3 12 3 1 1, a4 12 4 1 32 ,
a3 15 05 3 1 25,
a5 12 5 1 2
a4 15 05 4 1 3,
(b) The common difference is 12 .
a5 15 05 5 1 35
(c)
(b) The common difference is 05. (c)
2
an 3
1
2 1 0
an
0
1
2
3
4
1
2
3
4
5 n
5 n
11. a 10, d 6, an a d n 1 10 6 n 1. So a10 10 6 10 1 44. 12. a 5, d 2, an a d n 1 5 2 n 1. So a10 5 2 10 1 13.
13. a 06, d 1, an a d n 1 06 n 1. So a10 06 10 1 84.
14. a 18, d 02, an a d n 1 18 02 n 1. So a10 18 02 10 1 0.
15. a 52 , d 12 , an a d n 1 52 12 n 1. So a10 52 12 10 1 2. 16. a 3, d 3, an a d n 1 3 3 n 1. So a10 3 3 10 1 10 3.
17. a4 a3 a3 a2 a2 a1 6. The sequence is arithmetic with common difference 6.
18. a4 a3 a3 a2 a2 a1 12. The sequence is arithmetic with common difference 12.
19. Since a3 a2 7 and a4 a3 6, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 20. a4 a3 a3 a2 a2 a1 32. The sequence is arithmetic with common difference 32.
21. Since a2 a1 4 2 2 and a4 a3 16 8 8, the terms of the sequence do not have a common difference. This sequence is not arithmetic.
22. a4 a3 a3 a2 a2 a1 2. This sequence is arithmetic with common difference 2. 23. a4 a3 32 0 32 , a3 a2 0 difference 32 .
3 32 , a2 a1 32 3 32 . This sequence is arithmetic with common 2
7
8
CHAPTER 12 Sequences and Series
8 4 24. a4 a3 ln 16 ln 8 ln 16 8 ln 2, a3 a2 ln 8 ln 4 ln 4 ln 2, a2 a1 ln 4 ln 2 ln 2 ln 2. This
sequence is arithmetic with common difference ln 2.
25. a4 a3 77 60 17, a3 a2 60 43 17, 4 a1 43 26 17. This sequence is arithmetic with common difference 17. 1 and a a 1 1 1 , the terms of the sequence do not have a common difference. 26. Since a4 a3 15 14 20 3 2 4 3 12
This sequence is not arithmetic.
27. a1 4 7 1 11, a2 4 7 2 18, a3 4 7 3 25, a4 4 7 4 32, a5 4 7 5 39. This sequence is arithmetic, the common difference is d 7 and an 4 7n 4 7n 7 7 11 7 n 1. 28. a1 4 21 6, a2 4 22 8, a3 4 23 12, a4 4 24 20, a5 4 25 36. Since a4 a3 8 and a3 a2 4, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 1 1 1 1 1 1 1 1 1 1 , a2 , a3 , a4 , a5 . Since 29. a1 1 2 1 3 1 2 2 5 1 2 3 7 1 2 4 9 1 2 5 11
2 and a a 1 1 2 , the terms of the sequence do not have a common difference. This a4 a3 19 17 63 3 2 7 3 21
sequence is not arithmetic.
30. a1 1 12 32 , a2 1 22 2, a3 1 32 52 , a4 1 42 3, a5 1 52 72 . This sequence is arithmetic, the common difference is d 12 and an 1 n2 1 12 n 12 12 32 12 n 1.
31. a1 6 1 10 4, a2 6 2 10 2, a3 6 3 10 8, a4 6 4 10 14, a5 6 5 10 20. This sequence is arithmetic, the common difference is d 6 and an 6n 10 6n 6 6 10 4 6 n 1. 32. a1 3 11 1 2, a2 3 12 2 5, a3 3 13 3 0, a4 3 14 4 7,
a5 3 15 5 2. Since a4 a3 7 and a3 a2 5, the terms of the sequence do not have a common difference. This sequence is not arithmetic.
33. 6, 8, 10, 12, . Then d a2 a1 8 6 2, a5 a4 2 12 2 14, an 6 2 n 1, and a100 6 2 99 204.
34. 5, 0, 5, 10, . Then d a2 a1 0 5 5, a5 a4 5 10 5 15, an 5 5 n 1, and a100 5 5 99 490.
35. 29, 11, 7, 25, . Then d a2 a1 11 29 18, a5 a4 18 25 18 43, an 29 18 n 1, and a100 29 18 99 1753.
36. 64, 49, 34, 19, . Then d a2 a1 49 64 15, a5 a4 15 19 15 4, an 64 15 n 1, and a100 64 15 99 1421. 37. 4, 9, 14, 19, . Then d a2 a1 9 4 5, a5 a4 5 19 5 24, an 4 5 n 1, and a100 4 5 99 499.
38. 11, 8, 5, 2, . Then d a2 a1 8 11 3, a5 a4 3 2 3 1, an 11 3 n 1, and a100 11 3 99 286.
39. 12, 8, 4, 0, . Then d a2 a1 8 12 4, a5 a4 4 0 4 4, an 12 4 n 1, and a100 12 4 99 384.
40. 14 , 34 , 54 , 74 , . Then d a2 a1 34 14 12 , a5 a4 12 74 12 94 , an 14 12 n 1, and a100 14 12 99 199 4 .
41. 25, 265, 28, 295, . Then d a2 a1 265 25 15, a5 a4 15 295 15 31, an 25 15 n 1, and a100 25 15 99 1735.
42. 15, 123, 96, 69, . Then d a2 a1 123 15 27, a5 a4 27 69 27 42, an 15 27 n 1, and a100 15 27 99 2523.
43. 2, 2 s, 2 2s, 2 3s, . Then d a2 a1 2 s 2 s, a5 a4 s 2 3s s 2 4s, an 2 n 1 s, and a100 2 99s.
SECTION 12.2 Arithmetic Sequences
9
44. t, t 3, t 6, t 9, . Then d a2 a1 t 3 t 3, a5 a4 3 t 9 3 t 12, an t 3 n 1, and a100 t 3 99 t 297. 45. a50 1000 and d 6. Thus, a50 a1 d 50 1 1000 a1 6 50 1 a1 1000 294 706 and a2 706 6 712.
46. a100 750 and d 20. Thus, 750 a1 20 99 a1 1230 and a5 1230 20 4 1150.
1 . Thus, a 1 8 1 5 and 47. a14 23 and a9 14 , so 23 a1 13d and 14 a1 8d 23 14 5d d 12 1 4 12 12 5 1 n 1. an 12 12
48. a12 118 and a8 146, so 118 a1 11d and 146 a1 7d 118 146 4d d 7. Thus, a1 146 7 7 195 and an 195 7 n 1.
49. a1 25 and d 18, so an 601 25 18 n 1 601 33. Thus, 601 is the 33rd term of the sequence.
50. a1 3500 and d 15, so an 2795 2795 3500 15 n 1 n 48. Thus, 2795 is the 48th term of the sequence. 51. a 3, d 5, n 20. Then S20 20 2 2 3 19 5 1010.
52. a 10, d 8, n 30. Then S30 30 2 [2 10 29 8] 3180.
53. a 40, d 14, n 15. Then S15 15 2 [2 40 14 14] 870. 54. a 2, d 23, n 25. Then S25 25 2 [2 2 24 23] 6850.
55. a a1 4 and a3 2 4 3 1 d d 3, so S15 15 2 [2 4 14 3] 255.
56. a3 45 a 3 1 d 45 a 2d 45 and a7 55 a 7 1 d 55 a 6d 55. Solving the system, we have a 2d a 6d 45 55 4d 10 d 52 and a 45 2 52 40. Therefore, 5 4900. 2 40 48 S49 49 2 2
57. 1 5 9 401 is a partial sum of an arithmetic series with a 1 and d 5 1 4. The last term is
401 an 1 4 n 1, so n 1 100 n 101. So the partial sum is S101 101 2 1 401 101 201 20,301.
58. 5 25 0 25 60 is a partial sum of an arithmetic sequence with a 5 and d 25 5 25. The last
65 n 1 n 27. So the partial sum is S 27 5 60 1485 7425. term is 60 an 5 25 n 1, so 25 27 2 2
59. 250 233 216 97 is a partial sum of an arithmetic sequence with a 250 and d 233 250 17. The last
10 term is 97 an 250 17 n 1, so n 1 97250 17 9 n 10. So the partial sum is S10 2 250 97 1735.
60. 89 85 81 13 is a partial sum of an arithmetic sequence with a 89 and d 85 89 4. The last term is 20 13 an 89 4 n 1, so n 1 1389 4 19 n 20. So the partial sum is S20 2 89 13 1020.
61. 07 27 47 567 is a partial sum of an arithmetic sequence with a 07 and d 27 07 2. The last term is 567 an 07 2 n 1 28 n 1 n 29. So the partial sum is S29 29 2 07 567 8323.
62. 10 99 98 01 is a partial sum of an arithmetic sequence with a 10 and d 01. The last term
is 01 an 10 01 n 1, so 99 n 1 n 100. So the partial sum is S100 100 2 10 01 505. 10 63. k0 3 025k is a partial sum of an arithmetic sequence with a 3 025 0 3 and d 025. The last term is a11 3 025 10 55. So the partial sum is S11 11 2 3 55 4675. 20 64. n0 1 2n is a partial sum of an arithmetic sequence where a 1 2 0 1, d 2, and the last term is a21 1 2 20 39. So the partial sum is S21 21 2 1 39 399.
65. We have an arithmetic sequence with a 5 and d 2. We seek n such that 2700 Sn
n [2a n 1 d]. Solving for 2
n [10 2 n 1] 5400 10n 2n 2 2n n 2 4n 2700 0 n 50 n 54 0 2 n 50 or n 54. Since n is a positive integer, 50 terms of the sequence must be added to get 2700.
n, we have 2700
10
CHAPTER 12 Sequences and Series
n [2 12 n 1 8] 4n 2 8n 2700 0 4 n 27 n 25 0. 2 Thus, 2700 is the sum of the first 25 terms of the sequence.
66. a 12 and d 8, so Sn 2700 2700
67. Let x denote the length of the side between the length of the other two sides. Then the lengths of the three sides of the triangle are x a, x, and x a, for some a 0. Since x a is the longest side, it is the hypotenuse, and by the Pythagorean Theorem, we know that x a2 x 2 x a2 x 2 2ax a 2 x 2 x 2 2ax a 2 x 2 4ax 0 x x 4a 0 x 4a (x 0 is not a possible solution). Thus, the lengths of the three sides are x a 4a a 3a, x 4a, and x a 4a a 5a. The lengths 3a, 4a, 5a are proportional to 3, 4, 5, and so the triangle is similar to a 3-4-5 triangle.
68. P 10110 10210 10310 101910 101231910 . Now, 1 2 3 19 is an arithmetic series with 1 19 a 1, d 1, and n 19. Thus, 1 2 3 19 S19 19 190, and so P 1019010 1019 . 2 69. The sequence 1, 35 , 37 , 13 , is harmonic if 1, 53 , 73 , 3, forms an arithmetic sequence. Since 53 1 73 53 3 73 23 , the sequence of reciprocals is arithmetic and thus the original sequence is harmonic.
70. The two original numbers are 3 and 5. Thus, the reciprocals are 13 and 15 , and their average is 1 1 1 1 5 3 4 . Therefore, the harmonic mean is 15 . 2 3 5 2 15 15 15 4
71. The diminishing values of the computer form an arithmetic sequence with a1 12,500 and common difference d 1875. Thus the value of the computer after 6 years is a7 12,500 7 1 1875 $1250. 72. The number of poles in a layer can be viewed as an arithmetic sequence, where a1 25 and the common difference is 1. The number of poles in the first 12 layers is S12 12 2 [2 25 11 1] 6 39 234.
73. The increasing salary values form an arithmetic sequence with a1 45,000 and common difference d 2000. Then
his total earnings for a ten-year period are S10 10 2 [2 45,000 9 2000] 540,000. Thus, the total earnings for the
10-year period are $540,000.
74. The number of cars that can park in a row can be viewed as an arithmetic sequence, where a1 20 and the common difference is 2. Thus the number of cars that can park in the 21 rows is S21 21 2 [2 20 20 2] 105 80 840.
75. The number of seats in the nth row is given by the nth term of an arithmetic sequence with a1 15 and common n difference d 3. We need to find n such that Sn 870. So we solve 870 Sn [2 15 n 1 3] for n. We have 2 n 870 27 3n 1740 3n 2 27n 3n 2 27n 1740 0 n 2 9n 580 0 x 20 x 29 0 2 n 20 or n 29. Since the number of rows is positive, the theater must have 20 rows. 76. The sequence is 16, 48, 80, . This is an arithmetic sequence with a 16 and d 48 16 32. (a) The total distance after 6 seconds is S6 62 32 5 32 3 192 576 ft. (b) The total distance after n seconds is Sn n2 [32 32 n 1] 16n 2 ft.
77. The number of gifts on the 12th day is 1 2 3 4 12. Since a2 a1 a3 a2 a4 a3 1, the number of gifts on the 12th day is the partial sum of an arithmetic sequence with a 1 and d 1. So the sum is 1 12 6 13 78. S12 12 2
78. (a) We want an arithmetic sequence with 4 terms, so let a1 10 and a4 18. Since the sequence is arithmetic, 8 46 are the a4 a1 3d 18 10 8 3d 8 d 83 . Therefore, a2 10 83 38 and a 10 2 3 3 3 3 two arithmetic means between 10 and 18.
(b) We want an arithmetic sequence with 5 terms, so let a1 10 and a5 18. Since the sequence is arithmetic, a5 a1 4d 18 10 8 4d 8 d 2. Therefore, a2 10 2 12, a3 10 2 2 14, and a4 10 3 2 16 are the three arithmetic means between 10 and 18.
SECTION 12.3 Geometric Sequences
11
(c) We want an arithmetic sequence with 6 terms, with the starting dosage a1 100 and the final dosage a6 300. Since the sequence is arithmetic, a6 a1 5d 300 100 200 5d 200 d 40. Therefore, a2 140, a3 180, a4 220, a5 260, and a6 300. The patient should take 140 mg, then 180 mg, then 220 mg, then 260 mg, and finally arrive at 300 mg.
12.3 GEOMETRIC SEQUENCES 1. A geometric sequence is a sequence where the ratio between successive terms is constant. 2. The sequence an ar n1 is a geometric sequence where a is the first term and r is the common ratio. So, for the geometric sequence an 2 5n1 the first term is 2 and the common ratio is 5. 3. True. If we know the first and second terms of a geometric sequence then we can find all other terms. 4. (a) The nth partial sum of a geometric sequence an ar n1 is given by Sn a
1 rn . 1r
k1 a ar ar 2 ar 3 is is an infinite geometric series. If r 1, then this series (b) The series k1 ar converges and its sum is S a 1 r. If r 1 the series diverges.
5. (a) a1 7 30 7, a2 7 31 21,
6. (a) a1 6 050 6, a2 6 051 3,
a3 7 32 63, a4 7 33 189,
a3 6 052 15, a4 6 053 075,
a5 7 34 567
a5 6 054 0375
(b) The common ratio is 3.
(b) The common ratio is 05.
(c)
(c)
an
an
600
6 4
400
2
200
0 _2
500 300
1
2
3
2 3 a3 8 14 12 , a4 8 14 18 , 4 1 a5 8 14 32
5 n
8. (a) a1 19 30 19 , a2 19 31 13 , a3 19 32 1, a4 19 33 3,
a5 19 34 9
(b) The common ratio is 3.
(b) The common ratio is 14 . an 5
4
3
5 n
4
0 1 7. (a) a1 8 14 8, a2 8 14 2,
(c)
2
_4
100
0
1
(c)
an 0
1
_2
2
3
4
5 n
_10 _20
9. a 7, r 4. So an ar n1 7 4n1 and a4 7 43 448.
10. a 32 , r 3. So an ar n1 32 3n1 and a4 32 33 81 2.
1
2
3
4
5 n
_4 _6 _8 _10
This is the wrong art for #7
12
CHAPTER 12 Sequences and Series
11. a 5, r 3. So an ar n1 5 3n1 and a4 5 33 135. n1 n 3 12. a 3, r 3. So an ar n1 3 3 3 and a4 3 3 9.
a a 6 12 24 a2 2, and 4 2. Since these ratios are the same, the sequence is geometric with common 2, 3 a1 3 a2 6 a3 12 ratio 2. a 31 a 48 93 16 and 3 . Since these ratios are not the same, this is not a geometric sequence. 14. 2 a1 3 a2 48 16 3 a 4 12 23 a and 3 . Since these ratios are not the same, this is not a geometric sequence. 15. 2 a1 13 2 a2 12 3 1 a 1 a 1 144 48 16 a , 3 , and 4 . Since these ratios are the same, the sequence is 16. 2 a1 432 3 a2 144 3 a3 48 3 13.
geometric with common ratio 13 .
17.
a2 32 34 38 1 a 1 a 1 , 3 , and 4 . Since these ratios are the same, the sequence is geometric with a1 3 2 a2 32 2 a3 34 2
common ratio 12 . 18.
a2 103 109 1027 1 a 1 a 1 , 3 , and 4 . Since these ratios are the same, the sequence is geometric with a1 10 3 a2 103 3 a3 109 3 common ratio 13 .
19. 20.
a2 a 14 1 12 1 and 3 . Since these ratios are not the same, this is not a geometric sequence. a1 12 a2 12 2
a2 e4 a e6 a e8 2 e2 , 3 4 e2 , and 4 6 e2 . Since these ratios are the same, the sequence is geometric with a1 a2 a3 e e e
common ratio e2 . a a 11 121 1331 a 11, 3 11, and 4 11. Since these ratios are the same, the sequence is geometric 21. 2 a1 10 a2 11 a3 121 with common ratio 11. 22.
1
1
2
6
a 1 3 a2 41 and 4 81 . Since these ratios are not the same, this is not a geometric sequence. a1 2 a3 4
23. a1 2 31 6, a2 2 32 18, a3 2 33 54, a4 2 34 162, and a5 2 35 486. This sequence is geometric, the common ratio is r 3, and an a1 r n1 6 3n1 .
24. a1 4 31 7, a2 4 32 13, a3 4 33 31, a4 4 34 85, and a5 4 35 247. Since a3 a2 31 13 7 and a 13 are different ratios, this is not a geometric sequence. a1 2 1 1 1 1 1 1 1 1 1 , a3 3 , a4 4 , and a5 5 . This sequence is geometric, the 25. a1 , a2 2 4 16 64 256 1024 4 4 4 4 n1 . common ratio is r 14 and an a1r n1 14 14 26. a1 11 21 2, a2 12 22 4, a3 13 23 8, a4 14 24 16, and
a5 15 25 32. This sequence is geometric, the common ratio is r 2, and an a1r n1 2 2n1 . 27. Since ln a b b ln a, we have a1 ln 50 ln 1 0, a2 ln 51 ln 5, a3 ln 52 2 ln 5, a4 ln 53 3 ln 5, a5 ln 54 4 ln 5. Since a1 0 and a2 0, this sequence is not geometric. a a 4 27 are 28. a1 11 1, a2 22 4, a3 33 27, a4 44 256, and a5 55 3125. Since 2 4 and 3 a1 1 a2 4 different, this is not a geometric sequence.
SECTION 12.3 Geometric Sequences
13
a 29. 2, 6, 18, 54, . Then r 2 62 3, a5 a4 3 54 3 162, and an 2 3n1 . a1 14 n1 a2 3 28 56 2 2 56 2 112 2 30. 7, 14 . 3 , 9 , 27 , . Then r a 7 3 , a5 a4 3 27 3 81 , and an 7 3 1 009 a 03, a5 a4 03 00081 03 000243, and 31. 03, 009, 0027, 00081, . Then r 2 a1 03 an 03 03n1 .
n1 a 2 2, a5 a4 2 2 2 2 4, and an 2, 2, 2 2, . Then r 2 2 . a1 1 1 , . Then r a2 12 1 , a a 1 1 1 1 , a 144 1 n1 . 33. 144, 12, 1, 12 n 5 4 12 12 12 12 144 12 a1 144 a 2 1 2 1 , and a 8 1 n1 . 34. 8, 2, 12 , 18 , . Then r 14 , a5 a4 18 14 32 n 4 a1 8 4 a 353 35. 3, 353 , 373 , 27, . Then r 2 323 , a5 a4 323 27 323 3113 , and a1 3 n1 3 32n23 32n13 . an 3 323
32. 1,
t2 n1 a2 t t4 t t5 t t t2 t3 t4 , and an t 2 , a5 a4 . 36. t, , , , . Then r 2 4 8 a1 t 2 2 8 2 16 2
n1 a s 27 37. 1, s 27 , s 47 , s 67 , . Then r 2 s 2n27 . s 27 , a5 a4 s 27 s 67 s 27 s 87 , and an s 27 a1 1 38. 5, 5c1 , 52c1 , 53c1 , . Then r n1 5 5cnc 5cnc1 . an 5 5c
a2 5c1 5c , a5 a4 5c 53c1 5c 54c1 , and a1 5
3 2 a 4 39. a1 14, a2 4. Thus r 2 and a4 a1r 41 14 27 16 49 . a1 14 7 4 a 3 40. a1 8, a2 6. Thus r 2 , so a5 a1r 51 8 34 81 32 . a1 4 41. a3
1 a a r5 9 and a6 9. Thus, 6 1 2 r 3 r 3 27, so r 3. Therefore, a1 32 13 3 a3 13 a1r
1 and a 1 3 1 . a1 27 2 27 9
a7 8 2 81 32 329 a 12 . Thus, , so r . Therefore, a4 r 3 a1 a1 34 r3 r3 9 a4 12 27 3 827 2 r n1 2 81 . and the nth term is an a1r n 2 3
42. a4 12 and a7
9 a 9216 a 18 512 r 8. Therefore, a1 23 and 43. a3 18 and a6 9216. Thus, r 3 6 a3 18 64 32 r 9 an 8n1 . 32 27 3 a6 729256 a 54 3 44. a3 54 and a6 729 r . Thus, a1 23 384 and 256 . Thus, r a 54 512 8 r 382 3 a2 384 38 144. a 729 1728, a2 1728 075 1296, and a3 1296 075 972. 45. r 075 and a4 729, so a1 34 r 0753
14
CHAPTER 12 Sequences and Series
6 a 18 1 . 46. r 16 and a3 18, so a1 23 648 and a7 648 16 2 72 r 16 n1 47. a 1536 and r 12 , so an ar n1 6 1536 12 1536 21n log2 6 log2 1536 1 n n 1 log2 1536 6 1 log2 256 9. Thus, 6 is the ninth term.
n5 468,750 48. a2 30 and a5 3750, so r 3 3750 30 125 and r 5. Thus, an 468,750 3750 5
,750 n 5 log5 468 3750 log5 125 n 8. Therefore, the eighth term of the sequence is 468,750.
1 26 5 63 315. 12 1 1 4 80 2 3 2 81 80 . 50. a 23 , r 13 , n 4. Then S4 3 81 2 3 1 1
49. a 5, r 2, n 6. Then S6 5
3
3
a ar 5 a 2 51. a3 28, a6 224, n 6. So 6 2 r 3 . So we have r 3 6 224 28 8, and hence r 2. Since a3 a r , we a3 a3 ar a 28 1 26 7 63 441. get a 23 2 7. So S6 7 12 r 2 012 a 000096 a 52. a2 012, a5 000096, n 4. So r 3 5 0008 r 02, and thus a1 2 06. a2 012 r 02 Therefore, S4 06
1 024 07488. 1 02
a 53. 1 3 9 2187 is a partial sum of a geometric sequence, where a 1 and r 2 31 3. Then the last term is a1 2187 an 1 3n1 n 1 log3 2187 7 n 8. So the partial sum is S8 1
1 38 3280. 13 1
1 is a partial sum of a geometric sequence for which a 1 and r a2 2 1 . The last 54. 1 12 14 18 512 2 a1 1 10 n1 1 12 1 1 341 term is an , where an 1 2 , so n 10. So the partial sum is S10 1 512 . 512 1 1 2
30 a 2. The 55. 15 30 60 960 is a partial sum of a geometric sequence for which a 15 and r 2 a1 15 last term is an 960 15 2n1 n 7, so the partial sum is S7 15
1 27 645. 1 2
a 1 2560 56. 5120 2560 1280 20 is a partial sum of a geometric sequence for which a 5120 and r 2 . a1 5120 2 9 n1 1 12 n 9, so the partial sum is S9 5120 10,220. The last term is an 20 5120 12 1 12 a 57. 125 125 125 12,500,000 is a partial sum of a geometric sequence for which a 125 and r 2 10. The a1 1 108 13,888,88875. 1 10 a 1 58. 10,800 1080 108 0000108 is a partial sum of a geometric sequence for which a 10,800 and r 2 . a1 10 9 1 n1 1 10 1 The last term is an 0000108 10,800 10 n 9, so the partial sum is S9 10,800 11,999999988. 1 1 10 last term is an 12,500,000 125 10n1 n 8, so the partial sum is S8 125
SECTION 12.3 Geometric Sequences
59.
5
k1
k1
3 12
3
5 1 12 1 12
5 k1 1 32 55 60. 8 32 8 3 2 k1 1 2 5
93 16
1 26 105 1 2 k1 5 k1 1 23 5 211 63. 3 23 3 2 27 1 k1
61.
6
15
1 56 39,060 15 k1 6 k1 1 32 6 1330 64. 64 32 64 k1 1 32
5 2k1 5
62.
3
6
10 5k1 10
1 is an infinite geometric series with a 1 and r 1 . Therefore, it is convergent with sum 65. 1 13 19 27 3
S
3 a 1 . 1 1r 2 1 3
66. 1 12 14 18 is an infinite geometric series with a 1 and r 12 . Therefore, it is convergent with sum S
2 1 1 . 3 1 3 1 2 2
1 is an infinite geometric series with a 1 and r 1 . Therefore, it is convergent with sum 67. 1 13 19 27 3
S
3 a 1 . 1 1r 4 1 3
4 8 is an infinite geometric series with a 2 and r 2 . Therefore, it is convergent with sum 68. 25 25 125 5 5 2 2 2 S 5 2 53 . 3 1 5 5 2 3 69. 1 32 32 32 is an infinite geometric series with a 1 and r 32 1. Therefore, the series diverges.
70.
1 1 1 1 1 1 1 8 10 12 is an infinite geometric series with a 6 and r 2 . Therefore, it is convergent with 6 9 3 3 3 3 3 3 1
sum S
1 9 1 a 6 3 . 6 1 1r 648 3 8 1 9
71. 3 32 34 38 is an infinite geometric series with a 3 and r 12 . Therefore, it is convergent with sum S
3 2. 1 12
72. 1 1 1 1 is an infinite geometric series with a 1 and r 1. Because r 1 1 1, the series diverges.
73. 3 3 11 3 112 3 113 is an infinite geometric series with a 3 and r 11 1. Therefore, the series diverges. 10 3 74. 100 9 3 1 10 is an infinite geometric series with a
100 100 a 9 10 1000 9 with sum S 100 9 13 117 . 13 1r 1 3 10
10
100 3 3 and r 100 . Therefore, it is convergent 9 10 9
10
1 1 is an infinite geometric series with a 1 and r 1 . Therefore, the sum of the series is 75. 1 12 4 2 2 2 2 2 1 1 2 2 1. S 1 21 1 2
16
CHAPTER 12 Sequences and Series
76. 1 2 2 2 2 4 is an infinite geometric series with a 1 and r 2. Because r 2 1, the series diverges. 9 9 9 is an infinite geometric series with a 9 and r 1 . Thus 77. 0999 10 100 1000 10 10 9
0999
a 10 1 1. 1r 1 10
53 53 53 53 78. 02535353 02 1000 100,000 10,000,000 is an infinite geometric series (after the first term) with a 1000
53 1000 53 100 53 , and so 02535353 2 53 2 99 53 251 . 1000 99 990 1 10 990 990 990 1 100 3 3 3 3 1 79. 0030303 100 10,000 1,000,000 is an infinite geometric series with a 100 and r 100 . Thus 3 1 3 a 1001 . 0030303 1r 99 33 1 100 25 25 80. 211252525 211 1025 ,000 1,000,000 100,000,000 is an infinite geometric series (after 25 25 10,000 1 , and so the first term) with a 1025 ,000 and r 100 . Thus 000252525 1 9900 1 100 1 . Thus 005353 and r 100
211252525
25 211 99 25 20,914 10,457 211 . 100 9900 9900 9900 4950
112 112 112 112 81. 0112 0112112112 1000 1,000,000 1,000,000,000 is an infinite geometric series with a 1000 and 1 . Thus 0112112112 r 1000
112
a 10001 1r 1
1000
112 . 999
123 123 123 123 1 82. 0123123123 1000 1,000,000 1,000,000,000 is an infinite geometric series with a 1000 and r 1000 .
Thus 0123123123
123 123 41 1000 . 1 999 333 1 1000
a 83. Since we have 5 terms, let us denote a1 5 and a5 80. Also, 5 r 4 because the sequence is geometric, and so a1 r 4 80 5 16 r 2 . If r 2, the three geometric means are a2 10, a3 20, and a4 40. (If r 2, the three geometric means are a2 10, a3 20, and a4 40, but these are not between 5 and 80.)
84. The sum is given by a b a 2 2b a 3 3b a 10 10b a a 2 a 3 a 10 b 2b 3b 10b a
1 a 10 10 1 a 10 b 10b a 55b 1a 2 1a
a a3 85. (a) 5 3 5 3 is neither arithmetic nor geometric because a2 a1 a3 a2 and 2 . a1 a2 (b) 13 1 53 73 is arithmetic with d 23 , so the next term is a5 73 23 3. (c) 3 3 3 3 9 is geometric with r 3, so the next term is a5 9 3.
(d) 3 32 0 32 is arithmetic with d 32 , so the next term is a5 32 32 3.
86. (a) 1 1 1 1 is geometric with r 1, so the next term is a5 1. 1 (b) 5 3 5 6 5 1 is geometric with r 516 , so a5 1 516 . 6 5
a a3 (c) 2 1 12 2 is neither arithmetic nor geometric because a2 a1 a3 a2 and 2 . a1 a2
SECTION 12.3 Geometric Sequences
17
(d) x 1 x x 1 x 2 is arithmetic with d 1, so a5 x 2 1 x 3.
87. (a) The value at the end of the year is equal to the value at beginning less the depreciation, so Vn Vn1 02Vn1 08Vn1 with V1 160,000. Thus Vn 160,000 08n1 .
(b) Vn 100,000 08n1 160,000 100,000 08n1 0625 n 1 log 08 log 0625 log 0625 211. Thus it will depreciate to below $100,000 during the fourth year. n1 log 08 88. Let an denote the number of ancestors a person has n generations back. Then a1 2, a2 4, a3 8, . Since 4 8 2, this is a geometric sequence with r 2. Therefore, a 2 214 215 32,768. 15 2 4
89. Since the ball is dropped from a height of 80 feet, a 80 . Also since the ball rebounds three-fourths of the distance n fallen, r 34 . So on the nth bounce, the ball attains a height of an 80 34 . Hence, on the fifth bounce, the ball goes 5 a5 80 34 80243 1024 19 ft high.
90. a 5000, r 108. After 1 hour, there are 5000 108 5400, after 2 hours, 5400 108 5832, after 3 hours, 5832 108 629856, after 4 hours, 629856 108 68024448, and after 5 hours, 68024448 108 7347 bacteria. After n hours, the number of bacteria is an 5000 108n .
91. Let an be the amount of water remaining at the nth stage. We start with 5 gallons, so a 5. When 1 gallon (that is, 15 of the mixture) is removed, 45 of the mixture (and hence 45 of the water in the mixture) remains. Thus, a1 5 45 a2 5 45 45 , n 3 and in general, an 5 45 . The amount of water remaining after 3 repetitions is a3 5 45 64 25 , and after 5 5 repetitions it is a5 5 45 1024 625 .
92. Let aC denote the term of the geometric series that is the frequency of middle C. Then aC 256 and aC1 512. Since aC 256 this is a geometric sequence, r 512 256 2, and so aC2 r 2 22 64. 93. Let an be the height the ball reaches on the nth bounce. From the given information, an is the geometric sequence n an 9 13 . (Notice that the ball hits the ground for the fifth time after the fourth bounce.) 2 3 4 (a) a0 9, a1 9 13 3, a2 9 13 1, a3 9 13 13 , and a4 9 13 19 . The total distance traveled is 8 a0 2a1 2a2 2a3 2a4 9 2 3 2 1 2 13 2 19 161 9 17 9 ft.
(b) The total distance traveled at the instant the ball hits the ground for the nth time is 2 3 4 n1 Dn 9 2 9 13 2 9 13 2 9 13 2 9 13 2 9 13 2 3 4 n1 9 2 9 9 13 9 13 9 13 9 13 9 13 n n3 n 1 13 2 9 9 18 13 9 27 1 13 1 1 3
94. The annual salaries form a geometric sequence with a 45,000 and r 105. Then Sn 45,000
1 105n . At the end of 1 105
1 10510 566,00516, or $566,00516. 1 105 95. Let a1 1 be the man with 7 wives. Also, let a2 7 (the wives), a3 7a2 72 (the sacks), a4 7a3 73 (the cats), 10 years, the total earnings are S10 45,000
and a5 7a4 74 (the kits). The total is a1 a2 a3 a4 a5 1 7 72 73 74 , which is a partial sum of a geometric sequence with a 1 and r 7. Thus, the number in the party is S5 1
1 75 2801. 17
18
CHAPTER 12 Sequences and Series
96. (a) (b)
10
k1
1 k1 50 2
50
k1 1 k1 50 2
10 1 12
50 1 12
1 12
100 1 00009765 999023 mg
100 mg
97. Let an be the height the ball reaches on the nth bounce. We have a0 1 and an 12 an1 . Since the total distance d traveled includes the bounce up as well and the distance down, we have 2 3 4 d a0 2 a1 2 a2 1 2 12 2 12 2 12 2 12 i 2 3 1 1 1 12 12 12 1 1 2 i0
1
1 12
3
Thus the total distance traveled is about 3 m.
2 98. The time required for the ball to stop bouncing is t 1 1 1 which is an infinite geometric series with 2 2 2 2 1 2 2 21 2 2. Thus the a 1 and r 1 . The sum of this series is t 1 2 21 21 21 21 1 2 time required for the ball to stop is 2 2 341 s. 99. (a) If a square has side x, then by the Pythagorean Theorem the length of the side of the square formed by x2 x x2 x 2 x 2 . In our case, x 1 and the side of the joining the midpoints is, 2 2 4 4 2 2 first inscribed square is 1 , the side of the second inscribed square is 1 1 1 , the side of the 2 2 2 2 3 third inscribed square is 1 , and so on. Since this pattern continues, the total area of all the squares is 2 2 4 6 2 3 1 1 1 2 1 1 12 12 12 2. A1 2 2 2 1 12 2 3 (b) As in part (a), the sides of the squares are 1, 1 , 1 , 1 , . Thus the sum of the perimeters is 2 2 2 2 3 1 1 1 4 , which is an infinite geometric series with a 4 and r 1 . Thus S 414 4 2 2 2 2 4 2 424 2 4 2 21 4 8 4 2. the sum of the perimeters is S 21 21 21 21 1 1 2
2 2 100. Let An be the area of the disks of paper placed at the nth stage. Then A1 R 2 , A2 2 12 R 2R , 2 1 1 2 2 2 2 A3 4 14 R 4 R , . We see from this pattern that the total area is A R 2 R 4 R . Thus, the total area, A, is an infinite geometric series with a1 R 2 and r 12 . So A
R2
1 12
2 R 2 .
SECTION 12.3 Geometric Sequences
19
101. Let an denote the area colored blue at nth stage. Since only the middle squares are colored blue, an 19 area remaining yellow at the n 1 th stage. Also, the area remaining yellow at the nth stage is 89 of the area 2 3 remaining yellow at the preceding stage. So a1 19 , a2 19 89 , a3 19 89 , a4 19 89 , . Thus the total area 2 3 colored blue A 19 19 89 19 89 19 89 is an infinite geometric series with a 19 and r 89 . So the total area is A
1 9
1 89
1.
102. Let a1 , a2 , a3 , be a geometric sequence with common ratio r. Thus a2 a1 r, a3 a1 r 2 , , an a1 r n1 . Hence, 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n1 1 1 , , , and so a2 a1 r a1 r a3 a1 r 2 a1 r an a1 r n1 a1 r a1 r 2 a1 r n1 1 1 1 1 , , , is a geometric sequence with common ratio . a1 a2 a3 r 103. a1 a2 a3 is a geometric sequence with common ratio r . Thus a2 a1 r, a3 a1 r 2 , , an a1 r n1 . Hence log a2 log a1r log a1 log r, log a3 log a1 r 2 log a1 log r 2 log a1 2 log r, , log an log a1 r n1 log a1 log r n1 log a1 n 1 log r , and so log a1 log a2 log a3 is an arithmetic sequence with common difference log r.
104. Since a1 a2 a3 is an arithmetic sequence with common difference d, the terms can be expressed as a2 a1 d, 2 a3 a1 2d, , an a1 n 1 d. So 10a2 10a1 d 10a1 10d , 10a3 10a1 2d 10a1 10d , , n1 , and so 10a1 10a2 10a3 is a geometric sequence with common ratio r 10d . 10an 10a1 n1d 10a1 10d 105. By the partial sum formula with n 2k , we write the left-hand side of the equation to be proved as k
k
1 r r 2 r 2 1 S2k By the same formula with n 2k1 , the right-hand side is
1 r2 1r
k1 k1 1 r 2k1 k1 1 k1 2 2 2 2 1 r r r r 1 r2 1 S2k1 r 1 1r k1 2 1 r2 1r k
These are equal, and so we are done.
1 r2 1r
20
CHAPTER 12 Sequences and Series
12.4 MATHEMATICAL INDUCTION 1. Mathematical induction is a method of proving that a statement P n is true for all natural numbers n. In Step 1 we prove that P 1 is true.
2. (b) (ii) We prove “If P kis true then P k 1 is true.” 3. Let P n denote the statement 2 4 6 2n n n 1. Step 1: P 1 is the statement that 2 1 1 1, which is true. Step 2: Assume that P k is true; that is, 2 4 6 2k k k 1. We want to use this to show that P k 1 is true. Now 2 4 6 2k 2 k 1 k k 1 2 k 1
induction hypothesis
k 1 k 2 k 1 [k 1 1]
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
4. Let P n denote the statement 1 4 7 10 3n 2
n 3n 1 . 2
1 [3 1 1] 12 , which is true. 2 2 k 3k 1 Step 2: Assume that P k is true; that is. 1 4 7 3k 2 . We want to use this to show that 2 P k 1 is true. Now k 3k 1 3k 1 induction hypothesis 1 4 7 10 3k 2 [3 k 1 2] 2 3k 2 k 6k 2 3k 2 5k 2 k 3k 1 6k 2 k 1 3k 2 k 1 [3 k 1 1] 2 2 2 2 2 2 Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. Step 1: P 1 is the statement that 1
n 3n 7 . 2 1 3 1 7 Step 1: We need to show that P 1 is true. But P 1 says that 5 , which is true. 2 k 3k 7 Step 2: Assume that P k is true; that is, 5 8 11 3k 2 . We want to use this to show that 2 P k 1 is true. Now
5. Let P n denote the statement 5 8 11 3n 2
5 8 11 3k 2 [3 k 1 2]
k 3k 7 3k 5 2
6k 10 3k 2 13k 10 3k 2 7k 2 2 2 k 1 [3 k 1 7] 3k 10 k 1 2 2
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
induction hypothesis
SECTION 12.4 Mathematical Induction
6. Let P n denote the statement 12 22 32 n 2 Step 1: P 1 is the statement that 12
21
n n 1 2n 1 . 6
123 , which is true. 6
Step 2: Assume that P k is true; that is, 12 22 32 k 2
k k 1 2k 1 . 6
We want to use this to show that P k 1 is true. Now,
k k 1 2k 1 k 12 induction hypothesis 6 2k 2 k 6k 6 k 2k 1 6 k 1 k 1 k 1 6 6 2k 2 7k 6 k 1 k 2 2k 3 k 1 6 6
12 22 32 k 2 k 12
k 1 [k 1 1] [2 k 1 1] 6
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 7. Let P n denote the statement 1 2 2 3 3 4 n n 1 Step 1: P 1 is the statement that 1 2
n n 1 n 2 . 3
1 1 1 1 2 , which is true. 3
Step 2: Assume that P k is true; that is, 1 2 2 3 3 4 k k 1
k k 1 k 2 . We want to use this to 3
show that P k 1 is true. Now 1 2 2 3 3 4 k k 1 k 1 [k 1 1]
k k 1 k 2 k 1 k 2 3 k k 1 k 2 3 k 1 k 2 k 1 k 2 k 3 3 3 3
induction hypothesis
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 8. Let P n denote the statement 1 3 2 4 3 5 n n 2 Step 1: P 1 is the statement that 1 3
n n 1 2n 7 . 6
129 , which is true. 6
Step 2: Assume that P k is true; that is, 1 3 2 4 3 5 k k 2
k k 1 2k 7 . We want to use this to 6
show that P k 1 is true. Now 1 3 2 4 3 5 k k 2 k 1 [k 1 2]
k k 1 2k 7 k 1 k 3 6 2k 2 7k 6k 18 k 2k 7 6 k 3 k 1 k 1 6 6 6
k 1 [k 1 1] [2 k 1 7] 6
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
induction hypothesis
22
CHAPTER 12 Sequences and Series
9. Let P n denote the statement 13 23 33 n 3
n 2 n 12 . 4
12 1 12 , which is clearly true. 4 k 2 k 12 Step 2: Assume that P k is true; that is, 13 23 33 k 3 . We want to use this to show that 4 P k 1 is true. Now
Step 1: P 1 is the statement that 13
13 23 33 k 3 k 13
k 2 k 12 k 13 4 k 12 k 2 4 k 1 4
k 12 k 22
induction hypothesis
k 12 k 2 4k 4 4
k 12 [k 1 1]2
4 4 Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
10. Let P n denote the statement 13 33 53 2n 13 n 2 2n 2 1 . Step 1: P 1 is the statement that 13 12 2 12 1 , which is clearly true. Step 2: Assume that P k is true; that is, 13 33 53 2k 13 k 2 2k 2 1 . We want to use this to show that P k 1 is true. Now
13 33 53 2k 13 2k 13 k 2 2k 2 1 2k 13
induction hypothesis
2k 4 k 2 8k 3 12k 2 6k 1 2k 4 8k 3 11k 2 6k 1 k 2 2k 1 2k 2 4k 1 k 12 2 k 12 1
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
11. Let P n denote the statement 23 43 63 2n3 2n 2 n 12 . Step 1: P 1 is true since 23 2 12 1 12 2 4 8. Step 2: Assume that P k is true; that is, 23 43 63 2k3 2k 2 k 12 . We want to use this to show that P k 1 is true. Now 23 43 63 2k3 [2 k 1]3 2k 2 k 12 [2 k 1]3
induction hypothesis
2k 2 k 12 8 k 1 k 12 k 12 2k 2 8k 8 2 k 12 k 22 2 k 12 [k 1 1]2
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
SECTION 12.4 Mathematical Induction
23
1 1 1 n 1 . 12 23 34 n n 1 n1 1 12 , which is clearly true. Step 1: P 1 is the statement that 12 1 1 1 1 k Step 2: Assume that P k is true; that is . We want to use this to show that 12 23 34 k k 1 k 1 P k 1 is true. Now
12. Let P n denote the statement
1 1 1 1 k 1 12 23 k k 1 k 1 k 2 k 1 k 1 k 2
induction hypothesis
k 2 2k 1 k k 2 1 k 12 k 1 k 2 k 1 k 2 k 1 k 2 k 1 k 1 k 2 k 1 1
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 13. Let P n denote the statement 1 2 2 22 3 23 4 24 n 2n 2 1 n 1 2n . Step 1: P 1 is the statement that 1 2 2 [1 0], which is clearly true.
Step 2: Assume that P k is true; that is, 1 2 2 22 3 23 4 24 k 2k 2 1 k 1 2k . We want to use
this to show that P k 1 is true. Now
1 2 2 22 3 23 4 24 k 2k k 1 2k1 2 1 k 1 2k k 1 2k1
induction hypothesis
2 1 k 1 2k k 1 2k 2 1 2k 2k 2 1 k 2k1 2 1 [k 1 1] 2k1
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 14. Let P n denote the statement 1 2 22 2n1 2n 1.
Step 1: P 1 is the statement that 1 21 1, which is clearly true.
Step 2: Assume that P k is true; that is, 1 2 22 2k1 2k 1. We want to use this to show that P k 1 is true. Now
1 2 22 2k1 2k 2k 1 2k
induction hypothesis
2 2k 1 2k1 1
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 15. Let P n denote the statement n 2 n is divisible by 2.
Step 1: P 1 is the statement that 12 1 2 is divisible by 2, which is clearly true.
Step 2: Assume that P k is true; that is, k 2 k is divisible by 2. Now k 12 k 1 k 2 2k 1 k 1 k 2 k 2k 2 k 2 k 2 k 1. By the induction hypothesis, k 2 k is divisible by 2, and clearly 2 k 1 is divisible by 2. Thus, the sum is divisible by 2, so P k 1 is true. Therefore, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
24
CHAPTER 12 Sequences and Series
16. Let P n denote the statement that 5n 1 is divisible by 4.
Step 1: P 1 is the statement that 51 1 4 is divisible by 4, which is clearly true.
Step 2: Assume that P k is true; that is, 5k 1 is divisible by 4. We want to use this to show that P k 1 is true. Now, 5k1 1 5 5k 1 5 5k 5 4 5 5k 1 4 which is divisible by 4 since 5 5k 1 is divisible by 4 by the induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
17. Let P n denote the statement that n 2 n 41 is odd.
Step 1: P 1 is the statement that 12 1 41 41 is odd, which is clearly true.
Step 2: Assume that P k is true; that is, k 2 k 41 is odd. We want to use this to show that P k 1 is true. Now, k 12 k 1 41 k 2 2k 1 k 1 41 k 2 k 41 2k, which is also odd because k 2 k 41 is odd by the induction hypothesis, 2k is always even, and an odd number plus an even number is always odd. Therefore, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 18. Let P n denote the statement that n 3 n 3 is divisible by 3.
Step 1: P 1 is the statement that 13 1 3 3 is divisible by 3, which is true.
Step 2: Assume that P k is true; that is, k 3 k 3 is divisible by 3. We want to use this to show that P k 1 is true. Now, k 13 k 1 3 k 3 3k 2 3k 1 k 1 3 k 3 k 3 3k 2 3k k 3 k 3 3 k 2 k , which is divisible by 3, since k 3 k 3 is divisible by 3 by the induction hypothesis, and 3 k 2 k is divisible by 3. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
19. Let P n denote the statement that 8n 3n is divisible by 5.
Step 1: P 1 is the statement that 81 31 5 is divisible by 5, which is clearly true.
Step 2: Assume that P k is true; that is, 8k 3k is divisible by 5. We want to use this to show that P k 1 is true. Now, 8k1 3k1 8 8k 3 3k 8 8k 8 5 3k 8 8k 3k 5 3k , which is divisible by 5 because 8k 3k
is divisible by 5 by our induction hypothesis, and 5 3k is divisible by 5. Thus P k 1 follows from P k. So by the
Principle of Mathematical Induction, P n is true for all n.
20. Let P n denote the statement that 32n 1 is divisible by 8.
Step 1: P 1 is the statement that 32 1 8 is divisible by 8, which is clearly true.
Step 2: Assume that P k is true; that is, 32k 1 is divisible by 8. We want to use this to show that P k 1 is true. Now, 32k1 1 9 32k 1 9 32k 9 8 9 32k 1 8, which is divisible by 8, since 32k 1 is divisible by 8 by the induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true
for all n.
21. Let P n denote the statement n 2n . Step 1: P 1 is the statement that 1 21 2, which is clearly true.
Step 2: Assume that P k is true; that is, k 2k . We want to use this to show that P k 1 is true. Adding 1 to both sides
of P k we have k 1 2k 1. Since 1 2k for k 1, we have 2k 1 2k 2k 2 2k 2k1 . Thus k 1 2k1 ,
which is exactly P k 1. Therefore, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
SECTION 12.4 Mathematical Induction
25
22. Let P n denote the statement n 12 2n 2 , for all n 3.
Step 1: P 3 is the statement that 3 12 2 32 or 16 18, which is true.
Step 2: Assume that P k is true; that is, k 12 2k 2 , k 3. We want to use this to show that P k 1 is true. Now k 22 k 2 4k 4 k 2 2k 1 2k 3 k 12 2k 3 2k 2 2k 3
induction hypothesis
2k 2 2k 3 2k 1
because 2k 1 0 for k 3
2k 2 4k 2 2 k 12 Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 3. 23. Let P n denote the statement 1 xn 1 nx, if x 1.
Step 1: P 1 is the statement that 1 x1 1 1x, which is clearly true.
Step 2: Assume that P k is true; that is, 1 xk 1 kx. Now, 1 xk1 1 x 1 xk 1 x 1 kx,
by the induction hypothesis. Since 1 x 1 kx 1 k 1 x kx 2 1 k 1 x (since kx 2 0), we have
1 xk1 1 k 1 x, which is P k 1. Thus P k 1 follows from P k. So the Principle of Mathematical Induction, P n is true for all n. 24. Let P n denote the statement 100n n 2 , for all n 100.
Step 1: P 100 is the statement that 100 100 1002 , which is true.
Step 2: Assume that P k is true; that is, 100k k 2 . We want to use this to show that P k 1 is true. Now 100 k 1 100k 100 k 2 100 k 2 2k 1 k 12
induction hypothesis because 2k 1 100 for k 100
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 100. 25. Let P n be the statement that an 5 3n1 .
Step 1: P 1 is the statement that a1 5 30 5, which is true.
Step 2: Assume that P k is true; that is, ak 5 3k1 . We want to use this to show that P k 1 is true. Now, ak1 3ak 3 5 3k1 , by the induction hypothesis. Therefore, ak1 3 5 3k1 5 3k , which is exactly P k 1. Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
26. an1 3an 8 and a1 4. Then a2 3 4 8 4, a3 3 4 8 4, a4 3 4 8 4, , and the conjecture is that an 4. Let P n denote the statement that an 4. Step 1: P 1 is the statement that a1 4, which is true. Step 2: Assume that P k is true; that is, ak 4. We want to use this to show that P k 1 is true. Now, ak1 3 ak 8 3 4 8 4, by the induction hypothesis. This is exactly P k 1, so by the Principle of Mathematical Induction, P n is true for all n. 27. Let P n be the statement that x y is a factor of x n y n for all natural numbers n. Step 1: P 1 is the statement that x y is a factor of x 1 y 1 , which is clearly true.
Step 2: Assume that P k is true; that is, x y is a factor of x k y k . We want to use this to show that P k 1 is true. Now, x k1 y k1 x k1 x k y x k y y k1 x k x y x k y k y, for which x y is a factor because x y is a factor of x k x y, and x y is a factor of x k y k y, by the induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
26
CHAPTER 12 Sequences and Series
28. Let P n be the statement that x y is a factor of x 2n1 y 2n1 .
Step 1: P 1 is the statement that x y is a factor of x 1 y 1 , which is clearly true.
Step 2: Assume that P k is true; that is, x y is a factor of x 2k1 y 2k1 . We want to use this to show that P k 1 is true. Now x 2k11 y 2k11 x 2k1 y 2k1 x 2k1 x 2k1 y 2 x 2k1 y 2 y 2k1 x 2k1 x 2 y 2 x 2k1 y 2k1 y 2
for which x y is a factor. This is because x y is a factor of x 2 y 2 x y x y and x y is a factor of
x 2k1 y 2k1 by our induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n .
29. Let P n denote the statement that F3n is even for all natural numbers n. Step 1: P 1 is the statement that F3 is even. Since F3 F2 F1 1 1 2, this statement is true. Step 2: Assume that P k is true; that is, F3k is even. We want to use this to show that P k 1 is true. Now, F3k1 F3k3 F3k2 F3k1 F3k1 F3k F3k1 F3k 2 F3k1 , which is even because F3k is even by the induction hypothesis, and 2 F3k1 is even. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
30. Let P n denote the statement that F1 F2 F3 Fn Fn2 1. Step 1: P 1 is the statement that F1 F3 1. But F1 1 2 1 F3 1, which is true. Step 2: Assume that P k is true; that is, F1 F2 F3 Fk Fk2 1. We want to use this to show that P k 1 is true. Now Fk12 1 Fk3 1 Fk2 Fk1 1 Fk2 1 Fk1 F1 F2 F3 Fk Fk1 by the induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
31. Let P n denote the statement that F12 F22 F32 Fn2 Fn Fn1 . Step 1: P 1 is the statement that F12 F1 F2 or 12 1 1, which is true. Step 2: Assume that P k is true, that is, F12 F22 F32 Fk2 Fk Fk1 . We want to use this to show that P k 1 is true. Now 2 2 Fk Fk1 Fk1 F12 F22 F32 Fk2 Fk1
Fk1 Fk Fk1 Fk1 Fk2
induction hypothesis
by definition of the Fibonacci sequence
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
SECTION 12.4 Mathematical Induction
27
32. Let P n denote the statement that F1 F3 F2n1 F2n . Step 1: P 1 is the statement that F1 F2 , which is true since F1 1 and F2 1. Step 2: Assume that P k is true; that is, F1 F3 F2k1 F2k , for some k 1. We want to use this to show that P k 1 is true; that is, F1 F3 F2k11 F2k1 . Now F1 F3 F2k1 F2k1 F2k F2k1
induction hypothesis
F2k2
definition of F2k2
F2k1 Therefore, P k 1 is true. So by the Principle of Mathematical Induction, P n is true for all n. n Fn1 Fn 1 1 . 33. Let P n denote the statement Fn Fn1 1 0
Step 1: Since
1 1 1 0
2
1 1 1 0
2 1
1 1 1 0
k
1 1
k
Fk1 Fk Fk1
1 1 1 0
Step 2: Assume that P k is true; that is,
1 1 1 0
k1
1 1
1 0
F3 F2
F2 F1
Fk1
Fk
1 1 1 0
Fk Fk1
Fk Fk1
Fk
, it follows that P 2 is true.
. We show that P k 1 follows from this. Now,
Fk1
Fk
Fk Fk1
Fk2 Fk1 Fk1
Fk
1 1 1 0
induction hypothesis
by definition of the Fibonacci sequence
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 2. 34. Let a1 1 and an1
1 Fn , for n 1 . Let P n be the statement that an , for all n 1. 1 an Fn1
Step 1: P 1 is the statement that a1
F1 F , which is true since a1 1 and 1 11 1. F2 F2
Step 2: Assume that P k is true; that is, ak ak1
1 (definition of ak1 ) 1 ak
Fk . We want to use this to show that P k 1 is true. Now Fk1
1 Fk 1 Fk1
(induction hypothesis)
Fk1 F k1 (definition of Fk2 ). Fk Fk1 Fk2
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 100.
35. Since F1 1, F2 1, F3 2, F4 3, F5 5, F6 8, F7 13, our conjecture is that Fn n, for all n 5. Let P n denote the statement that Fn n.
Step 1: P 5 is the statement that F5 5 5, which is clearly true.
Step 2: Assume that P k is true; that is, Fk k, for some k 5. We want to use this to show that P k 1 is true. Now, Fk1 Fk Fk1 k Fk1 (by the induction hypothesis) k 1 (because Fk1 1). Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 5.
28
CHAPTER 12 Sequences and Series
36. Since 100 10 103 , 100 11 113 , 100 12 123 , , our conjecture is that 100n n 3 , for all natural numbers n 10. Let P n denote the statement that 100n n 3 , for n 10.
Step 1: P 10 is the statement that 100 10 1,000 103 1,000, which is true.
Step 2: Assume that P k is true; that is, 100k k 3 . We want to use this to show that P k 1 is true. Now 100 k 1 100k 100 k 3 100
induction hypothesis
k3 k2
because k 10
k 3 3k 2 3k 1 k 13 Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 10. 37. (a) P n n 2 n 11 is prime for all n. This is false as the case for n 11 demonstrates: P 11 112 11 11 121, which is not prime since 112 121.
(b) n 2 n, for all n 2. This is true. Let P n denote the statement that n 2 n. Step 1: P 2 is the statement that 22 4 2, which is clearly true.
Step 2: Assume that P k is true; that is, k 2 k. We want to use this to show that P k 1
is true. Now k 12 k 2 2k 1. Using the induction hypothesis (to replace k 2 ), we have
k 2 2k 1 k 2k 1 3k 1 k 1, since k 2. Therefore, k 12 k 1, which is exactly P k 1. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
(c) 22n1 1 is divisible by 3, for all n 1 . This is true. Let P n denote the statement that 22n1 1 is divisible by 3. Step 1: P 1 is the statement that 23 1 9 is divisible by 3, which is clearly true.
Step 2: Assume that P k is true; that is, 22k1 1 is divisible by 3. We want to use this to show that P k 1 is true. Now, 22k11 1 22k3 1 4 22k1 1 3 1 22k1 1 3 22k1 22k1 1 , which is
divisible by 3 since 22k1 1 is divisible by 3 by the induction hypothesis, and 3 22k1 is clearly divisible by 3. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
(d) The statement n 3 n 12 for all n 2 is false. The statement fails when n 2: 23 8 2 12 9.
(e) n 3 n is divisible by 3, for all n 2. This is true. Let P n denote the statement that n 3 n is divisible by 3. Step 1: P 2 is the statement that 23 2 6 is divisible by 3, which is clearly true.
Step 2: Assume that P k is true; that is, k 3 k is divisible by 3. We want to use this to show that P k 1 is true. Now k 13 k 1 k 3 3k 2 3k 1 k 1 k 3 3k 2 2k k 3 k 3k 2 2k k k 3 k 3 k 2 k . The term k 3 k is divisible by 3 by our induction hypothesis, and the term 3 k 2 k is clearly divisible by 3. Thus k 13 k 1 is divisible by 3, which is exactly P k 1. So by the Principle of Mathematical Induction, P n is true for all n.
(f) n 3 6n 2 11n is divisible by 6, for all n 1. This is true. Let P n denote the statement that n 3 6n 2 11n is divisible by 6. Step 1: P 1 is the statement that 13 6 12 11 1 6 is divisible by 6, which is clearly true.
Step 2: Assume that P k is true; that is, k 3 6k 2 11k is divisible by 6. We show that P k 1 is then also true. Now
k 13 6 k 12 11 k 1 k 3 3k 2 3k 1 6k 2 12k 6 11k 11 k 3 3k 2 2k 6 k 3 6k 2 11k 3k 2 9k 6 k 3 6k 2 11k 3 k 2 3k 2 k 3 6k 2 11k 3 k 1 k 2
SECTION 12.5 The Binomial Theorem
29
In this last expression, the first term is divisible by 6 by our induction hypothesis. The second term is also divisible by 6. To see this, notice that k 1 and k 2 are consecutive natural numbers, and so one of them must be even (divisible by 2). Since 3 also appears in this second term, it follows that this term is divisible by 2 and 3 and so is divisible by 6. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
38. The induction step fails when k 2, that is, P 2 does not follow from P 1. If there are only two cats, Tadpole and
Sparky, and we remove Sparky, then only Tadpole remains. So at this point, we still know only that Tadpole is black. Now removing Tadpole and putting Sparky back leaves Sparky (who is not necessarily black) alone. So the induction hypothesis does not allow us to conclude that Sparky is black.
12.5 THE BINOMIAL THEOREM 1. An algebraic expression of the form a b, which consists of a sum of two terms, is called a binomial.
2. We can find the coefficients in the expansion of a bn from the nth row of Pascal’s Triangle. So a b4 1a 4 4a 3 b 6a 2 b2 4ab3 1b4 .
n n! 4 4! 3. The binomial coefficients can be calculated directly using the formula . So 4. k! n k! 3!1! k 3 4. To expand a bn we can use the Binomial Theorem. Using this theorem we find 4 4 4 3 4 2 2 4 4 4 a a b a b ab3 b . a b4 0 1 2 3 4
5. x y6 x 6 6x 5 y 15x 4 y 2 20x 3 y 3 15x 2 y 4 6x y 5 y 6
6. 2x 14 2x4 4 2x3 6 2x2 4 2x 1 16x 4 32x 3 24x 2 8x 1 2 3 4 1 1 1 1 4 1 1 4 x 4 4x 3 6x 2 4x x 4 4x 2 6 2 4 7. x x x x x x x x 8. x y5 x 5 5x 4 y 10x 3 y 2 10x 2 y 3 5x y 4 y 5
9. x 15 x 5 5x 4 10x 3 10x 2 5x 1 6 a b a 3 6a 2 a b 15a 2 b 20a ab b 15ab2 6 ab2 b b3 10. a 3 6a 2 ab 15a 2 b 20ab ab 15ab2 6b2 ab b3
or a 3 6a 52 b12 15a 2 b 20a 32 b32 15ab2 6a 12 b52 b3 . 5 5 4 3 2 11. x 2 y 1 x 2 y 5 x 2 y 10 x 2 y 10 x 2 y 5x 2 y 1 x 10 y 5 5x 8 y 4 10x 6 y 3 10x 4 y 2 5x 2 y 1 6 12. 1 2 16 6 15 2 15 14 2 20 13 2 2 15 12 4 6 1 4 2 23 1 6 2 30 40 2 60 24 2 8 99 70 2 13. 2x 3y3 2x3 3 2x2 3y 3 2x 3y2 3y3 8x 3 36x 2 y 54x y 2 27y 3 3 2 3 14. 1 x 3 13 3 12 x 3 3 1 x 3 x 3 1 3x 3 3x 6 x 9 5 4 3 2 1 5 1 1 1 1 1 x x2 x2 x 15. 5 x 10 x 10 x x 5 x x x x x x 5 10 10 1 5 72 2 12 5x x 52 x x x x x 2 x 3 x 4 x 5 x x 5 1 x5 16. 2 25 5 24 10 23 10 22 5 2 3240x 20x 2 5x 3 58 x 4 32 2 2 2 2 2 2 6 6 5 4! 6! 17. 15 4! 2! 2 1 4! 4
30
CHAPTER 12 Sequences and Series
8 7 6 5! 8! 8 8 7 56 18. 3! 5! 3 2 1 5! 3 100 99 98! 100 100! 19. 4950 98! 2! 98! 2 1 98 10 9 8 7 6 5! 10 10! 3 2 7 6 252 20. 5! 5! 5 4 3 2 1 5! 5 3 2! 4 3 2! 3 4 3! 4! 18 21. 1! 2! 2! 2! 1 2! 2 1 2! 1 2 5! 5 4 3! 5 4 3! 5 5 5! 22. 10 10 100 2! 3! 3! 2! 2 1 3! 3! 2 1 2 3 5 5 5 5 5 5 23. 0 1 2 3 4 5 1 15 25 32 5 5! 5! 5! 5 5 5 5 5 5! 24. 1 0. Notice that the first and sixth 1 1! 4! 2! 3! 3! 2! 4! 1! 0 1 2 3 4 5 terms cancel, as do the second and fifth terms and the third and fourth terms. 25. x 2y4 40 x 4 41 x 3 2y 42 x 2 4y 2 43 x 8y 3 44 16y 4 x 4 8x 3 y 24x 2 y 2 32x y 3 16y 4 26. 1 x5 50 15 51 14 x 52 13 x 2 53 12 x 3 54 1 x 4 55 x 5 1 5x 10x 2 10x 3 5x 4 x 5 2 3 4 5 6 1 6 6 4 1 6 6 6 5 1 6 3 1 6 2 1 6 6 1 1 27. 1 1 1 1 1 1 1 x x x x x x x 2 0 1 3 4 5 6 15 6 20 15 6 1 1 2 3 4 5 6 x x x x x x 4 2 3 4 28. 2A B 2 40 2A4 41 2A3 B 2 42 2A2 B 2 43 2A B 2 44 B 2 16A4 32A3 B 2 24A2 B 4 8AB 6 B 8 20 19 29. The first three terms in the expansion of x 2y20 are 20 x 20 , 20 2y 40x 19 y, and 0 x 1 x 20 18 2 18 2 2 x 2y 760x y . 30 12 30 12 29 x x 30. The first four terms in the expansion of x 12 1 are 30 x 15 , 30 1 30x 292 , 0 1 27 30 12 28 2 14 , and 30 x 12 x 435x 1 13 4060x 272 . 2 3 25 23 13 24 253 are 25 a 25a 263 , and 25 a 253 . 31. The last two terms in the expansion of a 23 a 13 24 a 25 a 40 1 40 40 , 40 x 39 1 32. The first three terms in the expansion of x 40x 38 , and are 40 x x 0 1 x x 40 38 1 2 x 780x 36 . 2 x 18 33. The middle term in the expansion of x 2 1 occurs when both terms are raised to the 9th power. So this term is 18 2 9 9 x 1 48,620x 18 . 9 16 4 16 16 34. The fifth term in the expansion of ab 120 is 20 4 ab 1 4845a b . 2 23 2 23 35. The 24th term in the expansion of a b25 is 25 23 a b 300a b . 3 27 3 27 36. The 28th term in the expansion of A B30 is 30 27 A B 4060A B . 1 99 99 37. The 100th term in the expansion of 1 y100 is 100 99 1 y 100y . 1 1 25 25 2 24 is 1 x 25x 47 . 38. The second term in the expansion of x 2 x x
SECTION 12.5 The Binomial Theorem
31
39. The term that contains x 4 in the expansion of x 2y10 has exponent r 4. So this term is 10 4 104 13,440x 4 y 6 . 4 x 2y 12 r 40. The rth term in the expansion of 2y is 12 2 y 12r . The term that contains y 3 occurs when 12 r 3 r 9 3 2 y 3520 2y 3 . r 9 . Therefore, the term is 12 9
r 2 12r 12 r 242r 41. The rth term is 12 r a b . Thus the term that contains b8 occurs where 24 2r 8 r 8. So r a b 8 8 8 8 the term is 12 8 a b 495a b . 8r 8r 8r xr 1 r8 8r 8r r8 8r x 2r8 . So the term that does not contain x occurs 42. The rth term is r8 8xr 2x 2 x 2 4 8 1 when 2r 8 0 r 4. Thus, the term is 4 8x4 17,920. 2x 43. x 4 4x 3 y 6x 2 y 2 4x y 3 y 4 x y4
44. x 15 5 x 14 10 x 13 10 x 12 5 x 1 1 [x 1 1]5 x 5 45. 8a 3 12a 2 b 6ab2 b3 30 2a3 31 2a2 b 32 2ab2 33 b3 2a b3 4 4 3 2 46. x 8 4x 6 y 6x 4 y 2 4x 2 y 3 y 4 40 x 2 41 x 2 y 42 x 2 y 2 43 x 2 y 3 44 y 4 x 2 y x h3 x 3
x 3 3x 2 h 3xh 2 h 3 x 3
3x 2 h 3xh 2 h 3
h 3x 2 3xh h 2
3x 2 3xh h 2 h h 4 4 4 3 x 1 x h 42 x 2 h 2 43 xh 3 44 h 4 x 4 x 4 4x 3 h 6x 2 h 2 4xh 3 h 4 x 4 x h4 x 4 0 48. h h h 3 2 2 3 h 4x 6x h 4xh h 4x 3 h 6x 2 h 2 4xh 3 h 4 4x 3 6x 2 h 4xh 2 h 3 h h 100 49. 101100 1 001100 . Now the first term in the expansion is 100 1, the second term is 0 1 100 98 100 99 2 1 1 001 1, and the third term is 2 1 001 0495. Now each term is nonnegative, so 47.
h
h
101100 1 001100 1 1 0495 2. Thus 101100 2. n n n n n! n! n! n! 1. 1. Therefore, . 50. 0!n! 1 n! n n!0! n! 1 0 n 0 n n! n n 1! n n n! n n 1! 51. n. n. Therefore, 1 1! n 1! 1 n 1! 1 n1 n 1! 1! n 1! 1 n n n. 1 n1 n n! n n! for 0 r n. 52. r! n r ! n r r n r! r! n! n n n! . 53. (a) r 1 r r 1! [n r 1]! r! n r! n! r n! n! n r 1 n! (b) r r 1! n r 1! r! n r 1 n r ! r 1! [n r 1]! r! n r ! r n! n r 1 n! r! n r 1! r! n r 1! Thus a common denominator is r! n r 1!.
32
CHAPTER 12 Sequences and Series
(c) Therefore, using the results of parts (a) and (b), r n! n! n n n! n r 1 n! r! n r 1! r! n r 1! r 1 r r 1! [n r 1]! r! n r!
n! r n r 1 n! n 1 r n! n r 1 n! n1 n 1! r! n r 1! r! n r 1! r! n 1 r ! r ! n 1 r! r n 54. Let P n be the proposition that is an integer for the number n, 0 r n. r n 0 Step 1: Suppose n 0. If 0 r n, then r 0, and so 1, which is obviously an integer. Therefore, P 0 r 0 is true. k1 Step 2: Suppose that P k is true. We want to use this to show that P k 1 must also be true; that is, is an r k k k 1 by the key property of binomial coefficients integer for 0 r k 1. But we know that r 1 r r k k (see Exercise 49). Furthermore, and are both integers by the induction hypothesis. Since the sum of two r 1 r k 1 integers is always an integer, must be an integer. Thus, P k 1 is true if P k is true. So by the Principal of r n Mathematical induction, is an integer for all n 0, 0 r n. r
55. By the Binomial Theorem, the volume of a cube of side x 2 inches is 3 3 3 2 3 3 3 3 2 x x 2 2 x 3 3 2x 2 3 4x 8 x 3 6x 2 12x 8. The volume x 2 x 2 0 1 3 2
of a cube of side x inches is x 3 , so the difference in volumes is x 3 6x 2 12x 8 x 3 6x 2 12x 8 cubic inches. n pr q nr . In this case, 56. By the Binomial Theorem, the coefficient of pr in the expansion of p qn is n r p 09, q 01, n 5, and r 3, so the probability that the archer hits the target exactly 3 times in 5 attempts is 5 5 P 093 012 00729. 093 0153 2 53 57. Notice that 100!101 100!100 100! and 101!100 101 100!100 101100 100!100 . Now
100! 1 2 3 4 99 100 and 101100 101 101 101 101. Thus each of these last two expressions consists of 100 factors multiplied together, and since each factor in the product for 101100 is larger than each factor in the product for
100!, it follows that 100! 101100 . Thus 100!100 100! 100!100 101100 . So 100!101 101!100 . 58.
11 2
1214
13318
1 4 6 4 1 16
1 5 10 10 5 1 32
Conjecture: The sum is 2n . Proof: 2n 1 1n
n 0
10 1n
n
11 1n1
1 n n n n 0 1 2 n
n 2
12 1n2
n n
1n 10
CHAPTER 12
n
n
n
Review
n
59. 0 0n 1 1n 10 1n 11 1n1 12 1n2 1n 10 n n 1 n 2 n n n 0 1n 1k k n 0 1 2
CHAPTER 12 REVIEW 1. an
1 4 9 16 100 n2 12 22 32 42 102 . Then a1 , a2 , a3 , a4 , and a10 . n1 11 2 21 3 31 4 41 5 10 1 11
2n 21 22 23 24 8 . Then a1 11 2, a2 12 2, a3 13 , a4 14 4, and n 1 2 3 3 4 1024 512 210 . a10 110 10 10 5
2. an 1n
1 2 1n 1 11 1 12 1 13 1 , a3 . Then a1 0, a2 0, 3 3 3 8 4 n 1 2 33 2 1 1 14 1 110 1 a4 , and a . 10 64 32 500 43 103
3. an
n n 1 1 1 1 2 2 1 3 3 1 4 4 1 . Then a1 1, a2 3, a3 6, a4 10, and 2 2 2 2 2 10 10 1 a10 55. 2
4. an
654 2n! 2 1! 2 2! 2 3! . Then a1 1 15, 1, a2 2 3, a3 3 n 2 n! 8 2 1! 2 2! 2 3! 2 4! 2 10! 8765 105, and a10 10 654,729,075. a4 4 16 2 4! 2 10! n1 11 21 3! 31 4! 3, a3 6, 6. an . Then a1 1, a2 2! 1! 2! 2! 2 2 2 2 41 5! 10 1 11! 10, and a10 55. a4 2! 3! 2! 9! 2 2
5. an
7. an an1 2n 1 and a1 1. Then a2 a1 4 1 4, a3 a2 6 1 9, a4 a3 8 1 16, a5 a4 10 1 25, a6 a5 12 1 36, and a7 a6 14 1 49. 1 1 1 1 1 a a a a a a , a5 4 , a6 5 , and 8. an n1 and a1 1. Then a2 1 , a3 2 , a4 3 n 2 2 3 6 4 24 5 120 6 720 1 a . a7 6 7 5040 9. an an1 2an2 , a1 1 and a2 3. Then a3 a2 2a1 5, a4 a3 2a2 11, a5 a4 2a3 21, a6 a5 2a4 43, and a7 a6 2a5 85. 12 3an1 and a1 3 312 . Then a2 3a1 3 3 3 312 334 , a3 3a2 3 334 374 378 , a4 3a3 3 378 3158 31516 , a5 3a4 3 31516 33116 33132 , a6 3a5 3 33132 36332 36364 , a7 3a6 3 36364 312764 3127128 .
10. an
33
34
CHAPTER 12 Sequences and Series
5 5 5 5 5 5 12. (a) a1 1 , a2 2 , a3 3 , 2 4 8 2 2 2 5 5 5 5 , a5 5 a4 4 16 32 2 2
11. (a) a1 2 1 5 7, a2 2 2 5 9,
a3 2 3 5 11, a4 2 4 5 13,
a5 2 5 5 15 (b)
(b)
an
an
14 12 10 8 6 4 2 0
2 1
1
2
3
4
0
5 n
1
2
3
4
5 n
(c) S5 7 9 11 13 15 55
5 5 5 5 5 155 (c) S5 2 4 8 16 32 32
(d) This sequence is arithmetic with common difference
(d) This sequence is geometric with common ratio 12 .
2. 31 3 32 9 33 27 , 13. (a) a1 2 , a2 3 , a3 4 4 8 16 2 2 2
7 1 2 , a2 4 3, 2 2 2 5 3 3 4 5 a3 4 , a4 4 2, a5 4 2 2 2 2 2
14. (a) a1 4
34 81 35 243 , a5 6 a4 5 32 64 2 2 (b)
(b)
an
3
4
2
2
0
(c) S5
an
1
1
2
3
4
0
5 n
1
2
3
4
5 n
5 3 25 7 (c) S5 3 2 2 2 2 2
3 9 27 81 243 633 4 8 16 32 64 64
(d) This sequence is geometric with common ratio 32 .
(d) This sequence is arithmetic with common difference 12 .
15. 5 55 6 65 . Since 55 5 6 55 65 6 05, this is an arithmetic sequence with a1 5 and d 05. Then a5 a4 05 7. 16. 2 2 2 3 2 4 2 . Since 2 2 2 3 2 2 2 4 2 3 2 2, this is an arithmetic sequence with a1 2 and d 2. Then a5 a4 2 4 2 2 5 2.
17. t 3 t 2 t 1 t . Since t 2 t 3 t 1 t 2 t t 1 1, this is an arithmetic sequence with a1 t 3 and d 1. Then a5 a4 1 t 1. 4 2, this is a geometric sequence with a 2 and r 2. Then 18. 2 2 2 2 4 . Since 2 2 2 2 1 2 2 2 a5 a4 r 4 2 4 2.
1 t2 t 1 1 1 19. t 3 t 2 t 1 . Since 3 2 , this is a geometric sequence with a1 t 3 and r . Then a5 a4 r 1 . t t t t t t 20. 1 32 2 52 . Since 32 1 2 32 , and 1 21. 34 12 13 29 . Since 23 4 4 . a5 a4 r 29 23 27
32 2 3 , the series is neither arithmetic nor geometric. 1
2 2 1 3 9 2 , this is a geometric sequence with a 3 and r 2 . Then 1 3 4 3 1 1 2 3
CHAPTER 12
Review
35
1 1 2 1 1 1 1 1 a and a , this is a geometric sequence with a1 a and r . Then 22. a 1 2 . Since 1 a a a 1 a a a 1 1 1 a5 a4 r 2 3 . a a a 12 2 2i 24i 6i 2i, 2 2i, 2i, this is a geometric sequence with common 23. 3 6i 12 24i . Since 3 6i i 12 i ratio r 2i. a 2 2i 24. The sequence 2, 2 2i, 4i, 4 4i, 8, is geometric (where i 2 1), since 2 1 i, a1 2 a3 4i 2 2i 8 8i a 1 1 i i 4i 4 4i 1 i, 4 1 1 2 1 1 i, a2 2 2i 2 2i 2 2i 8 a3 4i i i i i a5 8 4 4i 32 32i 8 1 i. Thus the common ratio is i 1 and the first term is 2. So the a4 4 4i 4 4i 4 4i 32 nth term is an a1r n1 2 1 in1 .
25. a6 17 a 5d and a4 11 a 3d. Then, a6 a4 17 11 a 5d a 3d 6 6 2d d 3. Substituting into 11 a 3d gives 11 a 3 3, and so a 2. Thus a2 a 2 1 d 2 3 5. 26. a20 96 and d 5. Then 96 a20 a 19 5 a 95 a 96 95 1. Therefore, an 1 5 n 1. 2 27. a3 9 and r 32 . Then a5 a3 r 2 9 32 81 4.
3 28. a2 10 and a5 1250 27 . Then r
n1 an ar n1 6 53 .
a5 a2
1250 27 125 r 5 and a a a2 10 6. Therefore, 1 27 3 5 10 r 3
29. (a) An 52,000 104n1
(b) A1 $52,000, A2 52,0001041 $54,080, A3 52,0001042 $56,24320, A4 52,0001043 $58,49293, A5 52,000 1044 $60,83265, A6 52,000 1045 $63,26595
30. (a) An 55,000 1,600 n 1
(b) A1 $55,000, A2 55,000 1600 56,600, A3 55,000 1600 2 58,200, A4 55,000 1600 3 59,800, A5 55,000 1600 4 61,400, A6 55,000 1600 5 63,000. The salary for Position I is higher in the sixth year.
31. Let an be the number of bacteria in the dish at the end of 5n seconds. So a0 3, a1 3 2, a2 3 22 , a3 3 23 , . Then, clearly, an is a geometric sequence with r 2 and a 3. Thus at the end of 60 5 12 seconds, the number of bacteria is a12 3 212 12,288.
32. Let d be the common difference in the arithmetic sequence a1 , a2 , a3 , , so that an a1 n 1 d, n 1, 2, 3, , and let e be the common difference for b1 , b2 , b3 , , so that bn b1 n 1 e. Then an bn a1 n 1 a b1 n 1 e a1 b1 n 1 d e, n 1, 2, 3, . Thus a1 b1 a2 b2 is an arithmetic sequence with first term a1 b1 and common difference d e. 33. Suppose that the common ratio in the sequence a1 a2 a3 is r . Also, suppose that the common ratio in the sequence
b1 b2 b3 is s. Then an a1r n1 and bn b1 s n1 , n 1 2 3 . Thus an bn a1r n1 b1 s n1 a1 b1 rsn1 . So the sequence a1 b1 a2 b2 a3 b3 is geometric with first term a1 b1 and common ratio rs. 34. (a) Yes. If the common difference is d, then an a1 n 1 d. So an 2 a1 2 n 1 d, and thus the sequence a1 2 a2 2 a3 2 is an arithmetic sequence with the common difference d, but with the first term a1 2. (b) Yes. If the common ratio is r, then an a1 r n1 . So 5an 5a1 r n1 , and the sequence 5a1 5a2 5a3 is also geometric, with common ratio r, but with the first term 5a1 .
36
CHAPTER 12 Sequences and Series
35. (a) 6 x 12 is arithmetic if x 6 12 x 2x 18 x 9. 12 x x 2 72 x 6 2. (b) 6 x 12 is geometric if 6 x 36. (a) 2 x y 17 is arithmetic. Therefore, 15 17 2 a4 a1 a 3d a 3d. So d 5, and hence, x a d 2 5 7 and y a 2d 2 2 5 12. 13 a4 ar 3 17 3 . So (b) 2 x y 17 is geometric. Therefore, 17 2 a a r r 2 1 2 13 23 17 13 23 1713 and y a ar 2 2 2 2 17 213 1723 . x a2 ar 2 17 3 2 2 2 37.
38.
6
2 2 2 2 2 k3 k 1 3 1 4 1 5 1 6 1 16 25 36 49 126
2i
4
21
22
23
24
4
6
8
i 1 2i 1 2 1 1 2 2 1 2 3 1 2 4 1 2 3 5 7
6
210 140 126 120 596 105 105
k1 2 20 3 21 4 22 5 23 6 24 7 25 2 6 16 40 96 224 384 k1 k 1 2 40. 5m1 3m2 31 30 31 32 33 13 1 3 9 27 121 3
39.
41. 42. 43. 44.
10
2 2 2 2 2 2 2 2 2 2 2 k1 k 1 0 1 2 3 4 5 6 7 8 9
1 1 1 1 1 1 1 1 j2 j 1 1 2 3 4 5 98 99
100 50
k1
10
3k
3 32 33 34 349 350 2k1 22 23 24 25 250 251
2 n 2 1 2 2 2 3 2 9 2 10 n1 n 2 1 2 2 2 3 2 9 2 10 2
45. 3 6 9 12 99 3 1 3 2 3 3 3 33 2 46. 12 22 33 1002 100 k1 k
33
k1 3k
47. 1 23 2 24 3 25 4 26 100 2102
48.
1 212 2 222 3 232 4 242 100 21002 k2 100 k1 k 2
1 1 1 1 1 999 k1 k k 1 12 23 34 999 1000
49. 1 09 092 095 is a geometric series with a 1 and r
09 09. Thus, the sum of the series is 1
1 0531441 1 096 468559. 1 09 01 50. 3 37 44 10 is an arithmetic series with a 3 and d 07. Then 10 an 3 07 n 1 07 n 1 7 S6
143 n 11. So the sum of the series is S11 11 2 3 10 2 715. 51. 5 2 5 3 5 100 5 is an arithmetic series with a 5 and d 5. Then 100 5 an 5 5 n 1 n 100. So the sum is S100 100 5 100 5 50 101 5 5050 5. 2
52. 13 23 1 43 33 is an arithmetic series with a 13 and d 13 . Then an 33 13 13 n 1 n 99. So the 2 99 1 99 100 1650. sum is S99 99 2 3 2 3 3 6 53. n0 3 4n is a geometric series with a 3, r 4, and n 7. Therefore, the sum of the series is S7 3
1 47 35 1 47 9831. 1 4
CHAPTER 12
54.
Review
37
8
k2 is a geometric series with a 7, r 512 , and n 9. Thus, the sum of the series is k0 7 5 1 592 1 592 1 5 1 592 5 55 7 S9 7 7
15 1 5 1 5 1 5 74 1 625 5 5 3125 5467 1092 5 79088
4 8 is a geometric series with a 1 and r 2 . Therefore, it is convergent with sum 55. 1 25 25 125 5
S
5 1 a . 2 1r 7 1 5
56. 01 001 0001 00001 is an infinite geometric series with a 01 and r 01. Therefore, it is convergent with 1 01 . sum S 1 01 9
57. 5 5 101 5 1012 5 1013 is an infinite geometric series with a 5 and r 101. Because r 101 1, the series diverges. 1 1 1 58. 1 12 13 32 is an infinite geometric series with a 1 and r . Thus, it is convergent with sum 3 3 3 1 3 S 12 3 3 . 31 1 1 3
2
59. 1 98 98 diverges.
3 98 is an infinite geometric series with a 1 and r 98 . Because r 98 1, the series
60. a ab2 ab4 ab6 is an infinite geometric series with first term a and common ratio b2 . Because b 1, b2 1, a . 1 b2 61. We have an arithmetic sequence with a 7 and d 3. Then n n n Sn 325 [2a n 1 d] [14 3 n 1] 11 3n 650 3n 2 11n 3n 50 n 13 0 2 2 2 is inadmissible). Thus, 13 terms must be added. n 13 (because n 50 3 and so the series is convergent with sum S
62. We have a geometric series with S3 52 and r 3. Then 52 S3 a 3a 9a 13a a 4, and so the first term is 4. 1 215 2 215 1 65,534, and so the total 63. This is a geometric sequence with a 2 and r 2. Then S15 2 12 number of ancestors is 65,534. n 3n 1 64. Let P n denote the statement that 1 4 7 3n 2 . 2 1 [3 1 1] 12 Step 1: P 1 is the statement that 1 , which is true. 2 2 k 3k 1 Step 2: Assume that P k is true; that is, 1 4 7 3k 2 . We want to use this to show that 2 P k 1 is true. Now 1 4 7 10 3k 2 [3 k 1 2]
k 3k 1 3k 1 2
3k 2 k 6k 2 k 3k 1 6k 2 2 2 2
3k 2 5k 2 k 1 3k 2 2 2 k 1 [3 k 1 1] 2
induction hypothesis
38
CHAPTER 12 Sequences and Series
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
1 1 1 1 n . 13 35 57 2n 1 2n 1 2n 1 1 1 Step 1: P 1 is the statement that , which is true. 13 211 1 1 1 1 k Step 2: Assume that P k is true; that is, . We want to use this to 13 35 57 2k 1 2k 1 2k 1 show that P k 1 is true. Now 1 1 1 1 1 13 35 57 2k 1 2k 1 2k 1 2k 3
65. Let P n denote the statement that
k 1 2k 1 2k 1 2k 3
induction hypothesis
k 2k 3 1 2k 2 3k 1 2k 1 2k 3 2k 1 2k 3
k 1 k 1 k 1 2k 1 2k 3 2 k 1 1 2k 1 2k 3
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 1 1 1 1 1 1 1 n 1. 66. Let P n denote the statement that 1 1 2 3 n
Step 1: P 1 is the statement that 1 11 1 1, which is clearly true. 1 1 1 1 1 1 1 k 1. We want to use this to Step 2: Assume that P k is true; that is, 1 1 2 3 k show that P k 1 is true. Now 1 1 1 1 1 1 1 1 1 1 1 2 3 k k 1 1 1 1 1 1 1 1 1 1 1 1 2 3 k k 1 1 induction hypothesis k 1 1 k1 k 1 1
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
67. Let P n denote the statement that 7n 1 is divisible by 6.
Step 1: P 1 is the statement that 71 1 6 is divisible by 6, which is clearly true.
Step 2: Assume that P k is true; that is, 7k 1 is divisible by 6. We want to use this to show that P k 1 is true. Now 7k1 1 7 7k 1 7 7k 7 6 7 7k 1 6, which is divisible by 6. This is because 7k 1 is divisible by
6 by the induction hypothesis, and clearly 6 is divisible by 6. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
68. Let P n denote the statement that F4n is divisible by 3. Step 1: Show that P 1 is true, but P 1 is true since F4 3 is divisible by 3. Step 2: Assume that P k is true; that is, F4k is divisible by 3. We want to use this to show that P k 1 is true. Now, F4k1 F4k4 F4k2 F4k3 F4k F4k1 F4k1 F4k2 F4k F4k1 F4k1 F4k F4k1 2 F4k 3 F4k1 , which is divisible by 3 because F4k is divisible by 3 by our induction hypothesis, and 3 F4k1 is clearly divisible by 3. Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
CHAPTER 12
Test
39
69. an1 3an 4 and a1 4. Let P n denote the statement that an 2 3n 2. Step 1: P 1 is the statement that a1 2 31 2 4, which is clearly true.
Step 2: Assume that P k is true; that is, ak 2 3k 2. We want to use this to show that P k 1 is true. Now ak1 3ak 4 3 2 3k 2 4
definition of ak1 induction hypothesis
2 3k1 6 4 2 3k1 2
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 5! 54 54 5! 5 5 10 10 100 70. 2! 3! 3! 2! 2 2 2 3 10 10 10! 10! 10 9 10 9 8 7 71. 45 210 255 2 6 2! 8! 6! 4! 2 432 72. 5k0 5k 50 51 52 53 54 55 2 0!5!5! 1!5!4! 2!5!3! 2 1 5 10 32 8 73. 8k0 8k 8k 2 80 88 2 81 87 2 82 86 2 83 85 84 84 2 2 82 2 282 2 562 702 12,870 74. A B3 30 A3 31 A2 B 32 AB 2 33 B 3 A3 3A2 B 3AB 2 B 3 75. x 25 50 x 5 51 x 4 2 52 x 3 22 53 x 2 23 54 x 24 55 25
x 5 10x 4 40x 3 80x 2 80x 32 6 76. 1 x 2 60 16 61 15 x 2 62 14 x 4 63 13 x 6 64 12 x 8 65 x 10 66 x 12
1 6x 2 15x 4 20x 6 15x 8 6x 10 x 12 77. 2x y4 40 2x4 41 2x3 y 42 2x2 y 2 43 2x y 3 44 y 4 16x 4 32x 3 y 24x 2 y 2 8x y 3 y 4 3 19 3 19 78. The 20th term is 22 19 a b 1540a b . 20 23 20 23 19 79. The first three terms in the expansion of b23 b13 b b are 20 b403 , 20 0 1 23 18 13 2 b13 20b373 , and 20 b b 190b343 . 2 r 10r . The term that contains A6 occurs when r 6. Thus, the 80. The rth term in the expansion of A 3B10 is 10 r A 3B 10 6 term is 6 A 3B4 210A6 81B 4 17,010A6 B 4 .
CHAPTER 12 TEST 1. an 2n 2 n a1 1, a2 6, a3 15, a4 28, a5 45, a6 66, and S6 1 6 15 28 45 66 161. 2. an1 3an n, a1 2 a2 3 2 1 5, a3 3 5 2 13, a4 3 13 3 36, a5 3 36 4 104, and a6 3 104 5 307. 3. (a) The common difference is d 5 2 3. (b) an 2 n 1 3
(c) a35 2 3 35 1 104
3 1. 4. (a) The common ratio is r 12 4 n1 (b) an a1r n1 12 14 101 (c) a10 12 14 38 65,3536 4
40
CHAPTER 12 Sequences and Series 1
1 1. r 15 , so a5 ra4 25 5. (a) a1 25, a4 15 . Then r 3 5 25 125 8 1 15 97,656 58 1 (b) S8 25 1 12,500 3125 1 5
6. (a) a1 10 and a10 2, so 9d 8 d 89 and a100 a1 99d 10 88 78. 10 8 60 (b) S10 10 2 [2a 10 1 d] 2 2 10 9 9
7. Let the common ratio for the geometric series a1 a2 a3 be r, so that an a1r n1 , n 1, 2, 3, . Then 2 n1 an2 a1r n1 a12 r 2 . Therefore, the sequence a12 , a22 , a32 , is geometric with common ratio r 2 . 5 2 1 12 1 22 1 32 1 42 1 52 0 3 8 15 24 50 8. (a) n1 1 n (b) 6n3 1n 2n2 13 232 14 242 15 252 16 262 2 4 8 16 10 2
3
9
9. (a) The geometric sum 13 22 23 24 210 has a 13 , r 23 , and n 10. So 3 3 3 3 12310 58,025 1 1 1024 S10 3 123 3 3 1 59,049 59,049 .
1 1 1 has a 1 and r 212 1 . Thus, (b) The infinite geometric series 1 12 2 2 232 2 2 2 21 1 S 2 2. 11 2
21
21
21
n n 1 2n 1 . 6 123 , which is true. Step 1: Show that P 1 is true. But P 1 says that 12 6 k k 1 2k 1 . We want to use this to show that Step 2: Assume that P k is true; that is, 12 22 32 k 2 6 P k 1 is true. Now
10. Let P n denote the statement that 12 22 32 n 2
12 22 32 k 2 k 12
k k 1 2k 1 k 12 6
induction hypothesis
k 1 2k 2 k 6k 6 k 1
k k 1 2k 1 6 k 12 6 6 2 2 k 1 2k 7k 6 k 1 2k k 6k 6 6 6 k 1 [k 1 1] [2 k 1 1] k 1 k 2 2k 3 6 6
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 5 2 3 4 5 11. 2x y 2 50 2x5 51 2x4 y 2 52 2x3 y 2 53 2x2 y 2 54 2x y 2 55 y 2 12.
10
32x 5 80x 4 y 2 80x 3 y 4 40x 2 y 6 10x y 8 y 10
3 7 3 3 3 3x 2 120 27x 128 414,720x
13. (a) Each week he gains 24% in weight, that is, 024an . Thus, an1 an 024an 124an for n 1.
a0 is given to be 085 lb. Then a0 085, a1 124 085, a2 124 124 085 1242 085, a3 124 1242 085 1243 085, and so on. So we can see that an 085 124n .
(b) a6 124a5 124 124a4 1246 a0 1246 085 31 lb (c) The sequence a1 a2 a3 is geometric with common ratio 124.
Modeling with Recursive Sequences
41
FOCUS ON MODELING Modeling with Recursive Sequences 00365 00001. Thus the amount in the account at 365 the end of the nth day is An 10001An1 with A0 $275,000.
1. (a) Since there are 365 days in a year, the interest earned per day is
(b) A0 $275,000, A1 10001A0 10001 275,000 $275,02750, A2 10001A1 10001 10001A0 100012 A0 $275,05500,
A3 10001A2 100013 A0 $275,08251, A4 100014 A0 $275,11002, A5 100015 A0 $275,13753,
A6 100016 A0 $275,16504, A7 100017 A0 $275,19256
(c) An 10001n 275,000
2. (a) Tn Tn1 15 with T1 5.
(b) T1 5, T2 T1 15 5 15 65, T3 T2 15 5 15 15 5 2 15 80, T4 T3 15 5 2 15 15 5 3 15 95, T5 T4 15 5 3 15 15 5 4 15 110, T6 T5 15 5 4 15 15 5 5 15 125 (c) This is an arithmetic sequence with Tn 5 15 n 1.
(d) Tn 65 5 15n 15 615 15n n 41. So the student swims 65 minutes on the 41st day.
(e) Using the partial sum of an arithmetic sequence, the student swims for 30 2 [2 5 30 1 15] 15 535 8025 minutes 13 hours 225 minutes.
003 00025. Thus the amount in the account at the 12 end of the nth month is An 10025An1 100 with A0 $100.
3. (a) Since there are 12 months in a year, the interest earned per day is
(b) A0 $100, A1 10025A0 100 10025 100 100 $20025,
A2 10025A1 100 10025 10025 100 100 100 100252 100 10025 100 100 $30075, A3 10025A2 100 10025 100252 100 10025 100 100 100 100253 100 100252 100 10025 100 100 $40150, A4 10025A3 100 10025 100253 100 100252 100 10025 100 100 100 100254 100 100253 100 100252 100 10025 100 100 $50251
(c) An 10025n 100 100252 100 10025 100 100, the partial sum of a geometric series, so An 100
10025n1 1 1 10025n1 100 . 1 10025 00025
(d) Since 5 years is 60 months, we have A60 100
1002561 1 $658083. 00025
4. (a) The amount An of pollutants in the lake in the nth year is 30% of the amount from the preceding year (030An1 ) plus the amount discharged that year (2400 tons). Thus An 030An1 2400. (b) A0 2400, A1 030 2400 2400 3120,
A2 030 [030 2400 2400] 2400 0302 2400 2400 2400 2400 3336, A3 030 0302 2400 2400 2400 2400 2400 0033 2400 0302 2400 2400 2400 2400 34008, A4 030 0033 2400 0302 2400 2400 2400 2400 2400
0034 2400 0033 2400 0302 2400 2400 2400 2400 34202
42
FOCUS ON MODELING
(c) An is the partial sum of a geometric series, so An 2400
1 030n1
2400 1 030 n1 34286 1 030
(e)
4000
1 030n1 070
2000
1 0307 0 34278 tons. The sum of a geometric 0 10 20 070 1 series, is A 2400 34286 tons. 070 5. (a) For Plan I, the interest earned in the first year is 005U0 , and so the additional amount invested is 01 105U0 . Thus, U1 105U0 01 105U0 . The interest earned in the second year is 005U1 and the additional amount invested is 01 105U1 , so U2 105U1 01 105U1 . We see that Un 105Un1 01 105Un1 . For Plan 2, V0 earns 5% interest in the first year, and the additional amount invested is 500n, so V1 105V0 500 1. After the second year, V2 105V1 500 2. Thus, Vn 105Vn1 500n. (d) A6 2400
(b)
n
Un
Vn
n
Un
Vn
0
5000
5000
5
10,22732
14,40054
1
5775
5750
10
21,12467
40,21234
2
667013
703750
20
89,25030
160,45901
3
770399
888938
30
377,07656
419,21761
4
889811
11,33384
40
1,593,12324
903,59757
From the tables, we see that Un overtakes Vn somewhere between n 30 and n 40. We find that the smallest value of n for which Un Vn (other than n 1) is n 32 (U32 503,02955 and V32 494,46242).
CORRECTIONS: 2,4,7,18,22,24,32
CHAPTER 13
PROBABILITY AND STATISTICS
13.1
Counting 1
13.2
Probability 9
13.3
Binomial Probability 17
13.4
Expected Value 22 Chapter 13 Review 25 Chapter 13 Test 30
¥
FOCUS ON MODELING: The Monte Carlo Method 32
1
13 COUNTING AND PROBABILITY 13.1 COUNTING 1. The Fundamental Counting Principle says that if one event can occur in m ways and a second event can occurs in n ways, then the two events can occur in order in m n ways. So if you have two choices for shoes and three choices for hats, then the number of different shoehat combinations you can wear is 2 3 6.
2. The number of ways of arranging r objects from n objects in order is called the number of permutations of n objects taken n! r at a time, and is given by the formula P n r . So, if 3 students are chosen from a class of 10 to serve as n r! president, vice president, and treasurer, then the order in which these students are assigned these positions matters. Thus, 10! 720. the number of ways of choosing the 3 class officers is P 10 3 10 3! 3. The number of ways of choosing r objects from n objects is called the number of combinations of n objects taken r at n! . So if 3 students are chosen from a class of 10 to serve as a a time, and is given by the formula Cn r r! n r! committee, then the order in which these students are chosen doesn’t matter. Thus, the number of ways of choosing the 10! 120. 3 class officers is C 10 3 3! 10 3!
4. (a) False. When counting combinations, order does not matter. (b) True. When counting permutations, order matters.
(c) False. For a set of n distinct objects, the number of different combinations of these objects is less than the number of different permutations. (d) True. If we have a set with five distinct objects then the number of different ways of choosing two members of this set is the same as the number of ways of choosing three members. 5. (a) P 8 3
8! 8! 8 7 6 336 5! 8 3!
6. (a) P 11 4 7. (a) C 8 3
8! 8! 876 56 3! 8 3! 3! 5! 321
8. (a) C 11 4 9. (a) P 12 4 10. (a) P 9 6 11. (a) P 8 2
11! 11! 11 10 9 8 7920 7! 11 4!
11 10 9 8 11! 330 4! 7! 4321
(b) P 9 2
9! 9! 9 8 72 7! 9 2!
(b) P 10 5 (b) C 9 2
10! 10! 30,240 5! 10 5!
9! 9! 98 36 2! 9 2! 2! 7! 21
(b) C 10 5
10! 10! 252 5! 10 5! 5! 5!
12! 12! 12! 12! 12 11 10 9 11,880 (b) C 12 4 495 8! 4! 12 4! 4! 8! 12 4!
9! 9! 9! 9! 9 8 7 6 5 4 60,480 (b) C 9 6 84 3! 6! 9 6! 6! 3! 9 6! 8! 8! 8 7 56 6! 8 2!
12. (a) P 10 3
10! 720 10 3!
(b) P 8 6
8! 8! 8 7 6 5 4 3 20,160 2! 8 6!
(b) P 10 7
10! 604,800 10 7!
1
2
CHAPTER 13 Counting and Probability
In Solutions 13–16, we use the fact that C n p C n n p. See Exercise 4(d). 8! 8! 28 (b) C 8 6 C 8 2 28 13. (a) C 8 2 2! 8 2! 2! 6! 14. (a) C 10 3 15. (a) C 50 2
10! 120 3! 10 3!
50! 1225 2! 50 2!
(b) C 10 7 C 10 3 120 (b) C 50 48 C 50 2 1225
100! 4950 (b) C 100 98 C 100 2 4950 2! 100 2! 17. By the Fundamental Counting Principle, the number of possible singlescoop ice cream cones is number of ways to number of ways to 4 3 12. choose the flavor choose the type of cone 16. (a) C 100 2
18. By the Fundamental Counting Principle, the possible number of threeletter words is number of ways to number of ways to number of ways to . choose the first letter choose the second letter choose the third letter
(a) Since repetitions are allowed, we have 26 choices for each letter. Thus, there are 26 26 26 17,576 words.
(b) Since repetitions are not allowed, we have 26 choices for the first letter, 25 choices for the second letter, and 24 choices for the third letter. Thus there are 26 25 24 15,600 words.
19. (a) By the Fundamental Counting Principle, the possible number of ways 8 horses can complete a race, assuming no ties in any position, is number of ways to number of ways to number of ways to 87654321 choose the first finisher choose the second finisher choose the eighth finisher 8! 40,320 (b) By the Fundamental Counting Principle, the possible number of ways the first, second, and third place can be decided, assuming no ties, is number of ways to number of ways to number of ways to 8 7 6 336. choose the first finisher choose the second finisher choose the third finisher
20. Since there are four choices for each of the five questions, by the Fundamental Counting Principle there are 4 4 4 4 4 1024 different ways the test can be completed.
21. The number of possible sevendigit phone numbers is number of ways to number of ways to number of ways to . choose the first digit choose the second digit choose the seventh digit Since the first digit cannot be a 0 or a 1, there are only 8 digits to choose from, while there are 10 digits to choose from for the other 6 digits in the phone number. Thus the number of possible sevendigit phone numbers is 8 10 10 10 10 10 10 8,000,000.
22. Since a runner can only finish once, there are no repetitions. And since we are assuming that there is no tie, the number of different finishes is number of ways to number of ways to number of ways to 5 4 3 2 1 120. choose the first runner choose the second runner choose the fifth runner
23. Since there are 4 main courses, there are 6 ways to choose a main course. Likewise, there are 5 drinks and 3 desserts so there are 5 ways to choose a drink and 3 ways to choose a dessert. So the number of different meals consisting of a main course, a number of ways to number of ways to number of ways to drink, and a dessert is 4 5 3 60. choose the main course choose a drink choose a dessert 24. By the Fundamental Counting Principle, the number of different routes from town A to town D via towns B and C is number of routes number of routes number of routes 4 5 6 120. from A to B from B to C from C to D
4
SECTION 13.1 Counting
3
25. The number of possible sequences of heads and tails when a coin is flipped 5 times is number of possible number of possible number of possible 2 2 2 2 2 outcomes on the first flip outcomes on the second flip outcomes on the fifth flip 25 32 Here there are only two choices, heads or tails, for each flip.
26. Since each die has six different faces, the number of different outcomes when rolling a red and a white die is 6 6 36.
27. Since there are six different faces on each die, the number of possible outcomes when a red die and a blue die and a white die are rolled is number of possible number of possible number of possible 6 6 6 63 216. outcomes on the red die outcomes on the blue die outcomes on the white die
28. The number of possible pantsshirtshoes outfits is number of ways number of ways number of ways 5 8 12 480. to choose pants to choose a shirt to choose shoes
29. The number of different California license plates possible is number of ways to number of ways number of ways 9 263 103 158,184,000. choose a nonzero digit to choose 3 letters to choose 3 digits
30. The number of possible ID numbers consisting of one letter followed by three digits is 26 10 10 10 26,000.
31. Since successive numbers cannot be the same, the number of possible choices for the second number in the combination is only 59. The third number in the combination cannot be the same as the second in the combination, but it can be the same as the first number, so the number of possible choices for the third number in the combination is also 59. So the number of possible combinations consisting of a number in the clockwise direction, a number in the counterclockwise direction, and then a number in the clockwise direction is 60 59 59 208,860. 32. The number of possible license plates of two letters followed by three digits is number of ways to number of ways 262 103 676,000. Since 676,000 8,000,000, there will not be choose 2 letters to choose 3 digits enough different license plates for the state’s 8 million registered cars.
33. Since a student can hold only one office, the number of ways that a president, a vice president, and a secretary can be chosen from a class of 30 students is number of ways number of ways to number of ways 30 29 28 24,360. to choose a president choose a vice president to choose a secretary 34. The number of ways to choose a graduate president, an undergraduate treasurer, and a vice president is number of ways
to choose a president from the 10 graduates
number of ways
to choose a treasurer
from the 7 undergraduates
number of ways
to choose a vice president 10 7 15 1050. from the remaining 15 students
35. We have 7 choices for the first digit and 10 choices for each of the other 8 digits. Thus, the number of Social Security numbers is 7 108 700,000,000.
36. The number of possible ways to arrange three girls and four boys is number of ways to number of ways 3! 4! 144. arrange 3 girls to arrange 4 boys 37. (a) The number of ways to select 5 of the 8 objects is C 8 5
8! 56. 5! 3!
(b) A set with 8 elements has 28 256 subsets.
38. We may choose any subset of the 8 available brochures. There are 28 256 ways to do this.
39. Each subset of toppings constitutes a different way a hamburger can be ordered. Since a set with 10 elements has 210 1024 subsets, there are 1024 different ways to order a hamburger.
4
CHAPTER 13 Counting and Probability
40. We consider a set of 20 objects (the shoppers in the mall) and a subset that corresponds to those shoppers that enter the store. Since a set of 20 objects has 220 1,048,576 subsets, there are 1,048,576 outcomes to their decisions.
41. (a) The number of ways to seat 10 people in a row of 10 chairs is 10! 3,628,800.
10! 10! 6! 151,200. 6! 4! 4! 42. The number of ways of selecting 3 objects in order (a 3letter word) from 6 distinct objects (the 6 letters) assuming that the letters cannot be repeated is P 6 3 6 5 4 120. (b) The number of ways to choose 6 out of 10 people and seat them in 6 chairs is C 10 66!
43. In selecting these officers, order is important and repetition is not allowed, so the number of ways of choosing 3 officers from 15 students is P 15 3 2730.
44. The number of ways of selecting 3 objects in order (a 3digit number) from 4 distinct objects (the 4 digits) with no repetition of the digits is P 4 3 4 3 2 24. 45. Since the order of finish is important, we want the number of permutations of 8 objects (the contestants) taken 3 at a time, 8! 8! which is P 8 3 8 7 6 336. 5! 8 3!
46. The number of ways of ordering 8 pieces in order (without repeats) is P 8 8 8! 40,320.
47. The number of ways of ordering 9 distinct objects (the contestants) is P 9 9 9! 362,880. Here a runner cannot finish more than once, so no repetitions are allowed, and order is important. 48. The number of ways of ordering three of the five distinct flags is P 5 3 60.
49. The number of ways of ordering 1000 distinct objects (the contestants) taking 3 at a time is P 1000 3 1000 999 998 997,002,000. We are assuming that a person cannot win more than once, that is, there are no repetitions. 50. In selecting these officers, order is important and repetition is not allowed, so the number of ways of choosing 4 officers from 30 students is P 30 4 657,720.
51. We first place Kit in the first seat, and then seat the remaining 4 students. Thus the number of these arrangements is number of ways to number of ways to seat P 1 1 P 4 4 1! 4! 24. seat Kit in the first seat the remaining 4 students
52. We start by placing Kit in the middle seat, and then we place the remaining 4 students in the remaining four seats. Thus the number of ways number of ways to seat number of possible arrangements is 1 P 4 4 1! 4! 24. to seat Kit the remaining 4 students
53. Here we have 6 objects, of which 2 are blue marbles and 4 are red marbles. Thus the number of distinguishable permutations 6 5 4! 6! 15. is 2! 4! 2 4! 54. Here we have 14 objects (the 14 balls) of which 5 are red balls, 2 are white balls, and 7 are blue balls. So the number of 14! 72,072. distinguishable permutations is 5! 2! 7! 55. The number of distinguishable permutations of 12 objects (the 12 coins), from like groups of size 4 (the pennies), size 3 (the 12! 277,200. nickels), size 2 (the dimes) and size 3 (the quarters) is 4! 3! 2! 3! 56. The word ELEEMOSYNARY has 12 letters of which 3 are E, 2 are Y, and the rest are distinct. So we wish to find the number of distinguishable permutations of 12 objects (the 12 letters) from like groups of sizes 3 and 2 and 7 like groups of size 1. 12! replace "12" with "14" (2 times) 39,916,800. We get 3! 2! 1! 1! 1! 1! 1! 1! 57. The number of distinguishable permutations of 12 objects (the 12 ice cream cones) from like groups of size 3 (the vanilla cones), size 2 (the chocolate cones), size 4 (the strawberry cones), and size 5 (the butterscotch cones) is 14! 2,522,520. 3! 2! 4! 5!
SECTION 13.1 Counting
5
58. This is the number of distinguishable permutations of 7 objects (the students) from like groups of size 3 (the ones who stay in the threeperson room), size 2 (the ones who stay in the twoperson room), size 1 (the one who stays in the oneperson 7! 420. room), and size 1 (the one who sleeps in the car). This number is 3! 2! 1! 1! 59. The number of distinguishable permutations of 8 objects (the 8 cleaning tasks) from like groups of sizes 5, 2, and 1 workers 8! is 168. 5! 2! 1! 60. The number of distinguishable permutations of 30 objects (the students) from like groups of sizes 8, 11, and 11 is 30! 4,128 ,840,588,600. 8! 11! 11! 61. Here we are interested in the number of ways of choosing three objects (the three members of the committee) from a set of 25! 2300. 25 objects (the 25 members). The number of combinations of 25 objects taken three at a time is C 25 3 3! 22! 6! 20. 62. We want the number of ways of choosing a group of three from a group of six. This number is C 6 3 3! 3! 12! 220. 63. We want the number of ways of choosing a group of three from a group of 12. This number is C 12 3 3! 9! 64. We want the number of ways of choosing a group of 6 people from a group of 10 people. The number of combinations of 10! 210. 10 objects (people) taken 6 at a time is C 10 6 6! 4! 65. We want the number of ways of choosing a group (the 5card hand) where order of selection is not important. The number 52! of combinations of 52 objects (the 52 cards) taken 5 at a time is C 52 5 2,598,960. 5! 47! 66. Since order is not important in a 7card hand, the number of combinations of 52 objects (the 52 cards) taken 7 at a time is 52! 133,784,560. C 52 7 7! 45! 67. The order of selection is not important, hence we must calculate the number of combinations of 10 objects (the 10 questions) 10! taken 7 at a time. This gives C 10 7 120. 7! 3! 68. In this exercise, we assume that the pizza toppings cannot be repeated, so we are interested in the number of ways to select a 16! subset of 3 toppings from a set of 16 toppings. The number of ways this can occur is C 16 3 560. 3! 13! 69. We assume that the order in which the violinist plays the pieces in the recital is not important, so the number of combinations 12! of 12 objects (the 12 pieces) taken 8 at a time is C 12 8 495. 8! 4! 70. The order in which the shirts are selected is not important and no shirt is repeated. So the number of combinations of eight 8! shirts taken five at a time is C 8 5 56. 5! 3! 71. The order in which the pants are selected is not important and no pair is repeated, so the number of combinations of ten 10! 120. pairs of pants taken three at a time is C 10 3 3! 7! 72. (a) Since Kai must go on the field trip, we first pick Kai to go on the field trip, and then select the six other students from 29! the remaining 29 students. Since C 29 6 475,020, there are 475,020 ways to select the students to go on 6! 23! the field trip with Kai. (b) We first take Kai out of the class of 30 students and select the seven students from the remaining 29 students. Thus 29! there are C 29 7 1,560,780 ways to pick the 7 students for the field trip. 7! 22! (c) We are interested only in the group of 7 students taken from the class of 30 students, not the order in which they are 30! 2,035,800. picked. Thus the number is C 30 7 7! 23!
6
CHAPTER 13 Counting and Probability
73. Since the order in which the numbers are selected is not important, the number of combinations of 49 numbers taken 6 at a 49! 13,983,816. time is C 49 6 6! 43! 74. The number of distinguishable permutations of 13 objects (the total number of blocks the jogger must travel) which can be 13! 1287. partitioned into like groups of size 8 (the east blocks) and of size 5 (the north blocks) is 8! 5! 75. (a) The number of ways of choosing 5 students from the 20 students is C 20 5
20! 15,504. 5! 15!
12! 792. 5! 7! (c) We use the Fundamental Counting Principle to count the number of possible committees with 3 math majors and 2 physics majors. Thus, we get number of ways to choose number of ways to choose C 12 3 C 8 2 220 28 6160. 3 of 12 math majors 2 of the 8 physics majors
(b) The number of ways of choosing 5 students for the committee from the 12 math majors is C 12 5
76. The number of ways 2 of 10 Americans and 2 of 10 Canadians can be chosen is number of ways to pick number of ways to pick C 10 2 C 10 2 45 45 2025. 2 of 10 Americans 2 of 10 Canadians 77. The number of ways the committee can be chosen is number of ways to number of ways to number of ways to C 20 1 C 19 1 C 18 4 choose a chair choose a secretary choose four other members 20 19 3060 1,162,800 78. (a) We choose 2 of the 9 children and 4 of the 16 adults: number of ways to choose number of ways to choose C 9 2 C 16 4 36 1820 65,520. 2 of 9 children 4 of 16 adults (b) Method 1: We consider the number of ways of selecting the group of 6 from the 25 campers and subtract the groups that contain no children and those that contain one child. The number of groups that contain at least two children is C 25 6 C 9 0 C 16 6 C 9 1 C 16 5 177,100 1 8008 9 4368 129,780. Method 2: In the method we construct all the groups that are possible: groups with 2 children and 4 adults
groups with 3 children and 3 adults
groups with 4 children
and 2 adults
groups with 5 children and 1 adult
groups with 6 children
C 9 2 C 16 4 C 9 3 C 16 3 C 9 4 C 16 2 C 9 5 C 16 1 C 9 6 C 16 0 36 1820 84 560 126 120 126 16 84 1
65,520 47,040 15,120 2,016 84 129,780.
79. The number of ways the committee can be chosen is number of ways to choose number of ways to choose number of ways to choose number of ways to choose 2 of 6 freshmen 3 of 8 sophomores 4 of 12 juniors 5 of 10 seniors C 6 2 C 8 3 C 12 4 C 10 5 15 56 495 252 104,781,600 80. The leading and supporting roles are different (order counts), while the extra roles are not (order doesn’t count). Also, the instrumental roles must be filled by the instrumentalists and the singing roles by the singers. Thus, number of ways the ensemble can be chosen is
number of ways to choose
the leading and supporting instrumentalists
number of ways to choose
the leading and supporting singers
number of ways to choose 5 extras
number of ways to choose 3 extras
from the 8 remaining instrumentalists
from the 10 remaining singers
P 10 2 P 12 2 C 8 5 C 10 3 90 132 56 120 79,833,600.
SECTION 13.1 Counting
7
81. We choose 3 forwards from the forwards, 2 defensemen from the defensemen, and a goalie from the 2 goalies. Thus the number of ways to pick the 6 starting players is number of ways to number of ways to number of ways to C 12 3 C 6 2 C 2 1 pick 3 of 12 forwards pick 2 of 6 defensemen pick 1 of 2 goalies 220 15 2 6600 82. To order a pizza, we must make several choices. First the size (4 choices), the type of crust (2 choices), and then the toppings. Since there are 14 toppings, the number of possible choices is the number of subsets of the 14 toppings, that is 214 choices. So by the Fundamental Counting Principle, the number of possible pizzas is 4 2 214 131,072. 83. We count the total number of committees and subtract the number that contain both Skyler and Riley. The total number of committees possible is C 10 4 and the number that contain both Skyler and Riley is C 8 2, so the number of possible committees is C 10 4 C 8 2 210 28 182. 84. Method 1: We consider the number of fivemember committees that can be formed from the 26 interested people and subtract the numbers of those that contain no teacher and those that contain no students. Thus the number of committees is C 26 5 C 12 5 C 14 5 65,780 792 2002 62,986. Method 2: Here we construct all the committees that are possible. Thus the number of committees is committees with
1 student and 4 teachers
committees with
2 students and 3 teachers
committees with
3 students and 2 teachers
committees with
4 students and 1 teacher
C 14 1 C 12 4 C 14 2 C 12 3 C 14 3 C 12 2 C 14 4 C 12 1 14 495 91 220 364 66 1001 12 6,930 20,020 24,024 12,012 62,986
85. Since the two algebra books must be next to each other, we first consider them as one object. So we now have four objects to arrange and there are 4! ways to arrange these four objects. Now there are two ways to arrange the two algebra books. Thus the number of ways that five mathematics books may be placed on a shelf if the two algebra books are to be next to each other is 2 4! 48. 86. We treat Fin and Sydney as one object and Riley and Kelly as one object, so we need the number of ways of permuting eight objects. We then multiply this by the number of ways of arranging Fin and Sydney within their group and arranging Riley and Kelly within their group. Thus, the number of possible arrangements is P 8 8 P 2 2 P 2 2 8! 2! 2! 161,280. 87. (a) To find the number of ways the students and teachers can be seated, we first select and place a student in the first seat select 1 of arrange the and then arrange the other 7 people: C 4 1 P 7 7 4 7! 20,160. the 4 students remaining 7 people (b) To find the number of ways the students and teachers can be seated, we first select and place teachers in the first and arrange 2 of arrange the last seats and then arrange the other 6 people: P 4 2 P 6 6 12 the 4 teachers remaining 6 people grey 6! 8,640. 88. (a) We treat the people wearing white jackets as one object and find the number of ways of permuting five objects, then permute the four wearing gray jackets. Thus the number of arrangements is P 5 5 P 4 4 5! 4! 2880.
(b) Method 1: The number of ways the wearers of white and gray jackets can be seated is number of ways the number of ways the number of ways to select white jackets can be arranged gray jackets can be arranged the jacket color of the first seat P 4 4 P 4 4 C 2 1 4! 4! 2 1152 Method 2: There are eight choices for the first seat, four for the second, three each for the third and fourth, two each for the fifth and sixth, and one each for the seventh and eighth. Thus the number of possibilities is 8 4 3 3 2 2 1 1 1152.
8
CHAPTER 13 Counting and Probability
89. The number of ways the winner can be chosen is number of ways number of ways number of ways to choose 6 semifinalists to choose 2 finalists to choose the winner from the 30 contestants from the 6 semifinalists from the 2 finalists
C 30 6 C 6 2 C 2 1 593,775 15 2 17,813,250
90. There are many different possibilities here, so we consider the complement where no professor is chosen for the delegates and subtract this number from the way to select three people from the group of eight people, which is C 8 3. If the professor cannot to be selected, then we must select three people from a group of five, and this can be done in C 5 3 ways. Thus the number of delegations that contain a professor is C 8 3 C 5 3 56 10 46.
91. Since there are 26 letters, the possible number of combinations of the first and the last initials is 26 26 676. Since 677 676, there must be at least two people that have the same first and last initials in any group of 677 people.
92. When 2 objects are chosen from 10 objects, it determines a unique set of 8 objects, those not chosen. So choosing 2 of 10 objects is the same a choosing 8 of the 10. In general, every subset of r objects chosen from a set of n objects determines a corresponding set of n r objects, namely, those not chosen. Therefore, the total number of combinations for each type are equal. 93. We are only interested in selecting a set of 3 marbles to give to Alex and a set of 2 marbles to give to Sasha, not the order in which we hand out the marbles. Since both C 10 3 C 7 2 and C 10 2 C 8 3 count the number of ways this can be done, these numbers must be equal. (Calculating these values shows that they are indeed equal.) In general, if we wish to find two distinct sets of k and r objects selected from n objects (k r n), then we can either first select the k objects from the n objects and then select the r objects from the n k remaining objects, or we can first select the r objects from n n r n nk . the n objects and then the k objects from the n r remaining objects. Thus, r k k r 94. (a) x y5 x y x y x y x y x y
x y x y x y x x x y yx yy
x y x y x x x x x y x yx x yy yx x yx y yyx yyy
x y x x x x x x x y x x yx x x yy x yx x x yx y x yyx x yyy
yx x x yx x y yx yx yx yy yyx x yyx y yyyx yyyy
x x x x x x x x x y x x x yx x x x yy x x yx x x x yx y x x yyx x x yyy
x yx x x x yx x y x yx yx x yx yy x yyx x x yyx y x yyyx x yyyy
yx x x x yx x x y yx x yx yx x yy yx yx x yx yx y yx yyx yx yyy
yyx x x yyx x y yyx yx yyx yy yyyx x yyyx y yyyyx yyyyy
(b) There are ten terms that contain two x’s and three y’s. Their sum is
x x yyy x yx yy x yyx y x yyyx yx x yy yx yx y yx yyx yyx x y yyx yx yyyx x (c) To count the number of terms with two x’s, we must count the number of ways to pick two of the five positions to contain an x. This number is C 5 2. n (d) In the Binomial Theorem, the coefficient is the number of ways of picking r positions in a term with n factors to r contain an x. By definition, this is C n r .
SECTION 13.2 Probability
9
13.2 PROBABILITY 1. The set of all possible outcomes of an experiment is called the sample space. A subset of the sample space is called an event. The sample space for the experiment of tossing two coins is S H H H T T H T T , and the event “getting at least 3 n E . one head” is E H H H T T H. The probability of getting at least one head is P E n S 4 2. (a) The probability of E or F occurring is P E F P E P F P E F.
(b) If the events E and F have no outcomes in common (that is, the intersection of E and F is empty), then the events are called mutually exclusive. So in drawing a card from a deck, the event E, “getting a heart,” and the event F, “getting a spade,” are mutually exclusive.
(c) If E and F are mutually exclusive, then the probability of E or F is PE F P E P F. n E F . So in tossing a die, the conditional 3. The conditional probability of E given that F occurs is P E F n F probability of the event E, “getting a six,” given that that the event F, “getting an even number,” has occurred is P E F 13 .
4. (a) The probability of E and F occurring is PE F P E P F E. So if two marbles are drawn consecutively, without replacement, from a jar that contains six blue and four red marbles, then the probability that the first marble 6 4 4. drawn is blue (E) and the second is red (F) is P E F P E P F E 10 9 15
(b) If the occurrence of E does not affect the probability of the occurrence F, then the events are called independent. If E and F are independent events, then the probability of E and F is PE F P E P F. So if two marbles are drawn consecutively, with replacement, from a jar that contains six blue and four red marbles, then the probability that 6 4 6 . the first marble drawn is blue (E) and the second is red (F) is P E F P E P F 10 10 25
5. (a) S 1 2 3 4 5 6 (b) E 2 4 6 (c) E 5 6
6. (a) There are two possible outcomes of the coin toss and 52 possible outcomes of drawing a card, so n S 2 52 104. (b) H A H A H A H A
(c) T J T Q T K T J T Q T K T J T Q T K T J T Q T K
(d) H 2 H 3 H4 H5 H6 H7 H8 H9 H10 H J H Q H K H A
7. Let H stand for head and T for tails. (a) The sample space is S H H H T T H T T .
1 (b) Let E be the event of getting exactly two heads, so E H H. Then P E nE nS 4 .
3 (c) Let F be the event of getting at least one head. Then F H H H T T H, and P F nF nS 4 .
2 1 (d) Let G be the event of getting exactly one head, that is, G H T T H . Then P G nG nS 4 2 .
8. Let H stand for heads and T for tails; the numbers 1, 2, , 6 are the faces of the die. (a) S H 1 H2 H3 H 4 H5 H6 T 1 T 2 T 3 T 4 T 5 T 6
(b) Let E be the event of getting heads and rolling an even number. Then E H2 H4 H6, and 3 1 P E nE nS 12 4 .
(c) Let F be the event of getting heads and rolling a number greater than 4. Then F H5 H6, and 2 1 P F nF nS 12 6 .
3 1 (d) Let G be the event of getting tails and rolling an odd number. Then G T 1 T 3 T 5, and P G nG nS 12 4 .
10
CHAPTER 13 Counting and Probability
1 9. (a) Let E be the event of rolling a six. Then P E nE nS 6 .
3 1 (b) Let F be the event of rolling an even number. Then F 2 4 6. So P F nF nS 6 2 .
1 (c) Let G be the event of rolling a number greater than 5. Since 6 is the only face greater than 5, P G nG nS 6 . 2 1 10. (a) Let E be the event of rolling a two or a three. Then P E nE nS 6 3 .
3 1 (b) Let F be the event of rolling an odd number. So F 1 3 5, and P F nF nS 6 2 .
1 (c) Let G be the event of rolling a number divisible by 3. Then G 3 6 and P G nG nS 3 .
4 1 11. (a) Let E be the event of choosing a king. Since a deck has four kings, P E nE nS 52 13 .
(b) Let F be the event of choosing a face card. Since there are three face cards per suit and four suits, 12 3 P F nF nS 52 13 .
3 10 . (c) Let F be the event of choosing a face card. Then P F 1 P F 1 13 13
13 1 12. (a) Let E be the event of choosing a heart. Since there are 13 hearts, P E nE nS 52 4 .
26 1 (b) Let F be the event of choosing a heart or a spade. Since there are 13 hearts and 13 spades, P F nE nS 52 2 .
(c) Let G be the event of choosing a heart, a diamond or a spade. Since there are 13 cards in each suit, 39 3 P G nG nS 52 4 .
5 13. (a) Let E be the event of selecting a red ball. Since the jar contains five red balls, P E nE nS 8 .
(b) Let F be the event of selecting a yellow ball. Since there is only one yellow ball, 1 7 P F 1 P F 1 nF nS 1 8 8 .
0 (c) Let G be the event of selecting a black ball. Since there are no black balls in the jar, P G nG nS 8 0.
14. (a) Let E be the event of selecting a white ball or a yellow ball. Since there are two white balls and one yellow ball, 3 5 P E 1 P E 1 nE nS 1 8 8 . (b) Let F be the event of selecting a red ball, a white ball, or a yellow ball. Since all the types of balls are in the jar, P E 1. (c) Let G be the event of selecting a white ball. Since there are two white balls, 2 6 3 P E 1 P E 1 nE nS 1 8 8 4 .
C 13 5 1287 0000495. C 52 5 2,598,960 (b) Let E be the event of choosing five cards of the same suit. Since there are four suits and 13 cards in each suit, 5,148 4 C 13 5 000198. n E 4 C 13 5. Also, n S C 52 5. Therefore, P E C 52 5 2,598,960 (c) Let E be the event of dealing five face cards. Since there are 3 face cards in each suit and 4 suits, 792 C 12 5 0000305. P E C 52 5 2,598,960 (d) Let E be the event of dealing a royal flush (ace, king, queen, jack, and 10 of the same suit). Since there is only one such 4 4 153908 106 . sequence for each suit, there are only 4 royal flushes, so P E C 52 5 2,598,960
15. (a) Let E be the event of dealing five hearts. Since there are 13 hearts, P E
16. (a) Let E be the event of choosing three defective charging devices. Since there are four defective charging devices, C 4 3 4 1 n E 0018. n E C 4 3. Also, n S C 12 3. Therefore, P E n S C 12 3 220 55
SECTION 13.2 Probability
11
(b) Let E be the event of choosing three functioning charging devices. Since there are eight functioning charging devices, C 8 3 56 14 n E 0255. n E C 8 3, so P E n S C 12 3 220 55 17. (a) Let E be the event of choosing two red balls. Since there are three red balls, n E C 3 2. Also, n S C 8 2. C 3 2 3 n E 0107. Therefore, P E n S C 8 2 28 (b) Let E be the event of choosing two white balls. Since there are five white balls, n E C 5 2, so C 5 2 10 5 n E 0357. P E n S C 8 2 28 14 3. 18. (a) Let E be the event of choosing a T. Since 3 of the 16 letters are T’s, P E 16 5. (b) Let F be the event of choosing a vowel. Since there are 5 vowels, P F 16 5 11 . (c) Let F be the event of choosing a vowel. Then P F 1 16 16
19. (a) Let E be the probability that at least one card is a spade. The number of hands that do not contain a spade is the number 7411 C 39 5 0778. of possible fivecard hands using the other three suits, that is, C 39 5. Thus, P E 1 C 52 5 9520 (b) Let E be the probability that at least one card is a face card. The number of hands that do not contain a face card is the number of possible fivecard hands using the cards of the deck that are not face cards, that is, C 40 5. Thus, 6221 C 40 5 0747. P E 1 C 52 5 8330 20. (a) S 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6 2 6 3 6 4 6 5 6 6 6 1. (b) Let E be the event of getting a sum of 7. Then E 1 6 2 5 3 4 4 3 5 2 6 1, and P E 36 6
4 1. (c) Let F be the event of getting a sum of 9. Then F 3 6 4 5 5 4 6 3, and P F 36 9 6 1. (d) Let E be the event that the two dice show the same number. Then P E 36 6
(e) Let E be the event that the two dice show different numbers. Then E is the event that the two dice show the same number. Thus, P E 1 P E 1 16 56 . 5 (f) Let E be the event of getting a sum of 9 or higher. Then P E 10 36 18 .
3 21. (a) Let E be the event that the spinner stops on red. Since 12 of the regions are red, P E 12 16 4 .
(b) Let F be the event that the spinner stops on an even number. Since 8 of the regions are evennumbered, 8 1. P F 16 2
4 1. (c) Since 4 of the evennumbered regions are red, P E F P E P F P E F 34 12 16
4 1. 22. (a) Let E be the event that the spinner stops on blue. Since only 4 of the regions are blue, P E 16 4
8 1. (b) Let F be the event that the spinner stops on an odd number. Since 8 of the regions are oddnumbered, P F 16 2
(c) Since none of the oddnumbered regions are blue, P E F P E P F 14 12 34 .
23. (a) Yes, the events are mutually exclusive since the number cannot be both even and odd. So P E F P E P F 36 36 1.
(b) No, the events are not mutually exclusive since 6 is both even and greater than 4. So P E F P E P F P E F 36 26 16 23 .
24. (a) No, the events are not mutually exclusive since 4 is greater than 3 and also less than 5. So P E F P E P F P E F 36 46 16 1.
12
CHAPTER 13 Counting and Probability
(b) Yes, the events are mutually exclusive since there are only 2 numbers less than 3, namely 1 and 2, but they are not divisible by 3. So P E F P E P F 26 26 23 . 25. (a) No, the events E and F are not mutually exclusive since the jack, queen, and king of spades are both face cards and 12 3 11 spades. So P E F P E P F P E F 13 52 52 52 26 .
(b) Yes, the events E and F are mutually exclusive since the card cannot be both a heart and a spade. So 13 1 P E F P E P F 13 52 52 2 .
26. (a) No, events E and F are not mutually exclusive since a king can be a club. So 4 1 4 P E F P E P F P E F 13 52 52 52 13 .
(b) No, events E and F are not mutually exclusive since an ace can be a spade. So 4 13 1 4 . P E F P E P F P E F 52 52 52 13
27. (a) Let E denote a roll of five and F a roll greater than three. Using the formula for conditional probability, we have 1 n E F . P E F n F 3 (b) Let E denote a roll of three and F an odd roll. Using the formula for conditional probability, we have 1 n E F . P E F n F 3 n E F 4 1 . n F 12 3 n E F 1 (b) Let E denote drawing a king and F drawing a spade. Then P E F . n F 13 n E F 1 (c) Let E denote drawing a spade and F drawing a king. Then P E F . n F 4
28. (a) Let E denote drawing a queen and F drawing a face card. Then P E F
29. Let E denote the spinner stopping on an even number and F the spinner stopping on red. Then n E F 4 1 P E F . n F 12 3 30. Let E denote the spinner stopping on a number divisible by 3 and F the spinner stopping on blue. Then 1 n E F . P E F n F 4 31. (a) There is only one red ball numbered 3 and one green ball numbered 3. If the ball drawn is numbered 3, then the probability it is red is 12 . (b) There is only one ball numbered 7 and it is green, so if the ball drawn is numbered 7, then the probability it is green is 1. (c) There are two evennumbered red balls and three evennumbered green balls, so if the ball is evennumbered, then the probability it is a red ball is 25 . (d) There are five red balls, and two are evennumbered, so if the ball drawn is red, then the probability it is evennumbered is 25 . 32. (a) If the first ball drawn is red, there are four red balls and seven green balls remaining, so the probability that the second 4 . ball is red is 11 5. (b) If the first ball drawn is green, there are five red balls and six green balls remaining, so the required probability is 11
(c) If the first ball drawn is oddnumbered, then there are six oddnumbered and five evennumbered balls remaining, so the 5 . required probability is 11
(d) If the first ball drawn is evennumbered, then there are seven oddnumbered and four evennumbered balls remaining, 4 . so the required probability is 11
SECTION 13.2 Probability
13
33. (a) Let E be the event of drawing a black ball first. Because the jar contains seven black balls and three white balls, the 7 . The probability of the second ball being white is 3 1 , so the probability of the first ball being black is P E 10 9 3
7 1 7. probability of the intersection is 10 3 30
7 2 7 . (b) Here the probability of the second ball being black is 69 23 , so the probability of the intersection is 10 3 15
34. (a) Let E be the event of drawing a red sock. Since three pairs are red, the drawer contains six red socks, and so 6 1 P E nE nS 18 3 .
5 (b) Let F be the event of drawing another red sock. Since there are 17 socks left of which 5 are red, P F nF nS 17 .
5 5 . (c) In this case, the probability is P E F P E P F 13 17 51
4 and the probability of the second being a king is 4 , so the 35. (a) The probability of the first card being an ace is 52 51 4 4 4 . probability of the intersection is 52 51 663
4 and the probability of the second being an ace is 3 , so the (b) The probability of the first card being an ace is 52 51 4 3 1 . probability of the intersection is 52 51 221
36. (a) Yes, the first roll does not influence the outcome of the second roll.
1 1 . (b) The probability of getting a six on both rolls is P E F P E P F 16 6 36
37. Let E be the event of a “one” facing up on the first roll, and let F be the event of an even number facing up on the second 1. roll. Since these events are independent, P E F P E P F 16 36 12
38. (a) Yes, they are independent. The toss of the coin does not influence the roll of the die. (b) The probability of getting a tail is 12 and the probability of getting an even number is 36 12 , so the probability of the intersection is 12 12 14 .
39. (a) Yes. What happens on spinner A does not influence what happens on spinner B.
2 1. (b) The probability that A stops on red and B stops on yellow is P E F P E P F 24 8 8
40. (a) Let E A and E B be the event that the respective spinners stop on a purple region. Since these events are independent, 1 . P E A E B P E A P E B 14 28 16 (b) Let FA and FB be the event that the respective spinners stop on a blue region. Since these events are independent, 1. P FA FB P FA P FB 14 18 32
41. (a) Let M and F stand for “male”and “female”. Then S M M M M F M M M M F M M M M F M M M M F F F M M F M F M F M M F M F F M M F M F M M F F M F F F F M F F F F M F F F F M F F F F 1. (b) Let E be the event that the child gets only male goldfish. Then E M M M M and P E 16
(c) Let F be the event that the child gets two male and two female goldfish. Then
6 3. F F F M M F M F M F M M F M F F M M F M F M M F F, so P F 16 8
(d) Let F be the event that the child gets four goldfish of the same gender. Then F M M M M F F F F, and 2 1. P F 16 8
(e) Let H be the event that the child gets at least two female goldfish. Then H is the event that the child gets fewer than two female goldfish. Thus, H M M M M F M M M M F M M M M F M M M M F, so n H 5, and 5 11 . P H 1 P H 1 16 16
42. Let E be the event that a 13card bridge hand consists of all cards from the same suit. Since there are exactly 4 such hands 4 63 1012 . (one for each suit), P E C 52 13
14
CHAPTER 13 Counting and Probability
43. Let E be the event that the ball lands in an oddnumbered slot. Since there are 18 odd numbers between 1 and 36, 9 P E 18 38 19 .
44. (a) Let E be the event that the toddler arranges the word FRENCH. Since the letters are distinct, there are P 6 6 ways of 1 1 arranging the blocks of which only one spells the word FRENCH. Thus P E 00014. P 6 6 720 (b) Let E be the event that the toddler arranges the letters in alphabetical order. Since there are P 6 6 ways of arranging 1 1 the blocks of which only one is in alphabetical order, P E 00014. P 6 6 720 45. Let E be the event of selecting the 6 winning numbers. Since there is only one way to select them, 1 1 715 108 . P E C 49 6 13,983,816 46. Let E be the event that no marketing employee is chosen. The number of ways that no marketing employee is chosen is the C 11 6 000078. same as the number of ways that only product managers are chosen, which is C 11 6. Thus P E C 30 6 47. The sample space consist of all possible truefalse combinations, so n S 210 . Let E be the event that the student answers 1 1 . all 10 questions correctly. Since there is only one way to answer all 10 questions correctly, P E 10 1024 2 48. Let E be the event that the batch will be discarded. Thus, E is the event that at least one defective bulb is found. It is easier to find E , the event that no defective bulbs are found. Since there are 10 bulbs in the batch of which 8 are nondefective, C 8 3 C 8 3 P E . Thus P E 1 P E 1 1 04667 05333. C 10 3 C 10 3
49. (a) Let E be the event that the monkey types HAMLET as the first word. Since HAMLET contains six letters and there are 1 48 typewriter keys, P E 6 818 1011 . 48 (b) Let F be the event that the monkey types TO BE OR NOT TO BE as the first words. Since this phrase has 18 characters 1 (including spaces), P F 18 547 1031 . 48 1 1 00014. 720 6! (b) The probability that the monkey arranges the six blocks to spell HAMLET three consecutive times is the probability of 1 3 268 109 . three independent events E, and hence is equal to [P E]3 720
50. (a) Let E be the event that the monkey arranges the six blocks to spell HAMLET. Then P E
51. Let E be the event that the toddler will arrange the eight blocks to spell TRIANGLE or INTEGRAL. The number of ways of arranging these blocks is the number of distinguishable permutations of eight blocks. Since no two blocks are the same, the 2 496 105 , number of distinguishable permutations is 8!. Two of these arrangements result in event E, so P E 8! or 00000496. 52. Let E be the event that you predict the correct order for the horses to finish the race. Since there are eight horses, there are P 8 8 8! ways that the horses could finish, of which you predict only one. Thus, 1 1 P E 248 105 . P 8 8 40,320 53. (a) Let E be the event that the pea is tall. Since tall is dominant, E T T T t t T . So P E 34 . (b) E is the event that the pea is short. So P E 1 P E 1 34 14 .
SECTION 13.2 Probability
54.
(a) Let E be the event that the offspring will be tall. Since only offspring
Parent 2
Parent 1
15
t
t
T
Tt
Tt
t
tt
tt
with genotype Tt will be tall, P E 24 12 . (b) E is the event that the offspring will not be tall (thus, the offspring is short). So P E 1 P E 1 12 12 .
55. Let E be the event that the player wins on spin 1, and let F be the event that the player wins on spin 2. What happens on the first spin does not influence what happens on the second spin, so the events are independent. Thus, 1 1 1 . P E F P E P F 38 38 1444
56. Let E be the event that the committee consists entirely of juniors and F the event that it consists entirely of seniors. The sample space is the set of all ways that five people can be chosen from the group of 14. These events are mutually exclusive, C 8 5 6 56 31 C 6 5 . so P E F P E P F C 14 5 C 14 5 2002 1001 57. Let E, F and G denote the events of rolling two ones on the first, second, and third rolls, respectively, of a pair of dice. The 1 1 1 1 214 105 . events are independent, so P E F G P E P F P G 36 3 36 36 36
58. Let E be the event that a player has exactly five winning numbers and F be the event that a player has all six winning numbers. These events are mutually exclusive. For a players to have exactly five winning numbers means that the player has five of the six winning numbers and one number that was not selected in the lottery. So n E C 6 5 C 43 1. C 6 5 C 43 1 1 Thus, P at least five winning numbers P E F P E P F 00000185. C 49 6 C 49 6 59. Let E be the event that the marble is red and F be the event that the number is oddnumbered. Then E is the event that the marble is blue, and F is the event that the marble is evennumbered. 6 3 (a) P E 16 8 8 1 (b) P F 16 2
6 8 3 11 (c) P E F P E P F P E F 16 16 16 16 8 5 13 . (d) P E F P E P F P E F 10 16 16 16 16
60. The number of ways a set of six numbers can be selected from a set of 49 is C 49 6. Since the games are independent, the 2 1 probability of winning the lottery two times in a row is 511 1015 . C 49 6 4 10 . The probability of selecting a red ball from each jar 61. The probability of selecting two red balls from jar B is 57 6 21 8 7 4 is 37 57 15 49 . The probability of selecting two red balls after putting all balls in one jar is 14 13 13 . Hence, picking both balls from jar B gives the greatest probability.
1 , and is the same for the second and third wheels. The events are 62. (a) The probability that the first wheel has a bar is 11 1 1 1 1 . independent, and so the probability of getting 3 bars is 11 11 11 1331
(b) The probability of getting a number on the first wheel is 10 11 , the probability of getting the same number on the second 1 , and the probability of getting the same number on the third wheel is 1 . Thus, the probability of getting wheel is 11 11 1 1 10 the same number on each wheel is 10 11 11 11 1331 .
(c) We use the complement, no bar, to determine the probability of at least one bar. The probability that the first wheel does not have a bar is 10 11 , and is the same for the second and third wheels. Since the events are independent, the probability 3
10 10 10 1000 1000 331 of getting no bar is 10 11 11 11 3 1331 , and so P at least one bar 1 1331 1331 . 11
16
CHAPTER 13 Counting and Probability
63. Let E be the event that the student opens the lock within an hour. The number of combinations that can be tried in one hour is 10 60 600. The number of possible combinations is P 40 3, assuming that no number is repeated. Thus, 600 5 600 0010. P E P 40 3 59,280 494 64. Let E be the event that of the three bulbs selected, two are defective and one is not. Then number of ways to pick two defective and one functional bulb C 6 2 C 18 1 15 18 135 P E . number of ways to select three bulbs C 24 3 2024 1012 65. (a) Let E be the event that the curriculum committee consists of two parttime faculty and four fulltime faculty. So N committees with two parttimers and four fulltimers C 8 2 C 10 4 28 210 490 P E 0317. N ways to select a sixmember committee C 18 6 18564 1547 (b) Let F be the event that curriculum committee consists of two or fewer parttime faculty. Then P F
1 210 8 252 28 210 C 8 0 C 10 6 C 8 1 C 10 5 C 8 2 C 10 4 C 18 6 C 18 6 C 18 6 18,564 193 8106 18,564 442
249 (c) F is the event that the curriculum committee has more that two parttime faculty: P F 1 P F 1 193 442 442 .
138,415 C 95 3 0856. C 100 3 161,700 (b) The probability that at least one of the three fittings inspected is defective is 1 P E 0144.
66. (a) Let E be the event that the plumber inspects three nondefective fittings. Then P E
67. Let E be the event that Alex stands next to Sydney. To find n E we treat Alex and Sydney as one object and find the number of ways to arrange the 19 objects and then multiply the result by the number of ways to arrange Alex and Sydney. 2 19! 2! 01. So n E 19! 2!. The sample space is all the ways that 20 people can be arranged. Thus, P E 20! 20 68. Let E be the event that the monkey arranges the six blocks to spell BUBBLE. The number of ways of arranging these blocks is the number of distinguishable permutations of six blocks. Since there are three blocks labeled B, 6! the number of distinguishable permutations is . Only one of these arrangements spells the word BUBBLE. Thus, 3! 1 1 3! . P E 6! 6! 120 3! 69. Let E be the event that the monkey arranges the 11 blocks to spell PROBABILITY. The number of ways of arranging these blocks is the number of distinguishable permutations of 11 blocks. Since there are two blocks labeled B and two 11! . Only one of these arrangements spells the word blocks labeled I, the number of distinguishable permutations is 2! 2! 1 1 2! 2! . PROBABILITY. Thus P E 11! 11! 9,979,200 2! 2! 70. (a) Because the events are independent, the probability that the family has two boys given that the oldest child is a boy is 12 . (b) There are four equally likely possibilities: boyboy, boygirl, girlboy, and girlgirl. In three cases, at least one of the children is a boy, and in one of those cases both children are boys. Thus, the probability that the family has two boys given that one of the children is a boy is 13 . 71. (a) The sample space S consists of all possible choices of 2 marbles, so n S C 10 2 45. The event E that both marbles chosen are green has n E C 4 2 6. Thus, the probability that 2 green marbles are chosen is n E 6 2 . n S 45 15
SECTION 13.3 Binomial Probability
17
4 2 and (b) Let E be the event that the first marble is green and F that the second marble is green. Then P E 10 5 2 . P F E 39 13 , so the probability that both marbles chosen are green is P E P F E 25 13 15
The answers are the same, but the thought process involved is different. If 3 marbles are chosen, then using the method of part (a), we find that n S C 10 3 120 and n E C 6 1 C 4 2 6 6 36 (where E is the event that 1 red and 2 green marbles are chosen), so 36 3 n E . P E n S 120 10 Using the method of part (b), with E the event that the marbles are chosen greengreenred, F the event that they are 4 3 6 1, chosen greenredgreen, and G the event that they are chosen redgreengreen, we have P E 10 9 8 10
4 6 3 1 , and P G 6 4 3 1 . Thus, the probability that 1 red and 2 green marbles are chosen is P F 10 9 8 10 10 9 8 10 3 , as before. P E P F P G 10
13.3 BINOMIAL PROBABILITY 1. A binomial experiment is one in which there are exactly two outcomes. One outcome is called success and the other is called failure. 2. If a binomial experiment has probability p of success then the probability of failure is 1 p. The probability of getting
exactly r successes in n trials of this experiment is C n r pr 1 pnr . 3. P 2 successes in 5 C 5 2 072 033 013230 4. P 3 successes in 5 C 5 3 073 032 030870 5. P 0 success in 5 C 5 0 070 035 000243 6. P 5 successes in 5 C 5 5 075 030 016807 7. P 1 success in 5 C 5 1 071 034 002835 8. P 1 failure in 5 C 5 1 031 074 036015. Note that exactly one failure is the same as exactly 4 successes, and P 4 successes in 5 C 5 4 074 031 036015. 9. P at least 4 successes P 4 successes P 5 successes C 5 4 074 031 C 5 5 075 030 036015 016807 052822 10. P at least 3 successes P 3 successes P 4 successes P 5 successes
C 5 3 073 032 C 5 4 074 031 C 5 5 075 030 030870 036015 016807 083692
11. P at most 1 failure P 0 failure P 1 failure P 5 successes P 4 successes C 5 5 075 030 C 5 4 074 031
016807 036015 052822
12. P at most 2 failures P 0 failure P 1 failure P 2 failure
P 5 successes P 4 successes P 3 successes
C 5 5 075 030 C 5 4 074 031 C 5 3 073 032
016807 036015 030870 083692
18
CHAPTER 13 Counting and Probability
13. P at least 2 successes P 2 successes P 3 successes P 4 successes P 5 successes 013230 030870 036015 016807 096922
14. P at most 3 failures P 0 failure P 1 failure P 2 failures P 3 failures
P 5 successes P 4 successes P 3 successes P 2 successes 016807 036015 030870 013230 096922 (b)
15. (a) Outcome
Probability
1
02
2
02
3
02
4
02
5
02
0.2
0
1
2
3
4
5
In #16, include tick for 5 16. (a)
(b) Outcome
Probability
1
05
2
03
3
01
4
01
5
0
17. (a)
0.1
0
1
2
3
4
0
1
2
3
4
(b) r
Probability
0
4
1 16 1 4 3 8 1 4 1 16
r
Probability
0
00776
1
02592
2
03456
3
02304
4
00768
5
00102
1 2 3
18. (a)
1 16
(b)
0.5
0
1
2
3
4
5
SECTION 13.3 Binomial Probability
19. (a)
(b) r
Probability
0
02097
1
03670
2
02753
3
01147
4
00287
5
00043
6
000036
7
0000013
r
Probability
0
0000001
1
0000054
2
0001215
3
0014580
4
0098415
5
0354294
6
0531441
20. (a)
0.3
(b)
0
1
2
0
1
2
3
4
5
6
7
5
6
0.5
3
4
2 4 5 21. Here “success” is “face is 4” and P face is 4 16 . Then P 2 successes in 6 C 6 2 16 020094. 6 22. Here P success 08 and P failure 02. (a) P 0 success in 7 C 7 0 080 027 00000128 (b) P 7 successes in 7 C 7 7 087 020 0209715
(c) P the archer hits the target more than once 1 [P 0 success in 7 P 1 success in 7] 1 C 7 0 080 027 C 7 1 081 026 099963
(d) P at least 5 successes P 5 successes P 6 successes P 7 successes C 7 5 085 022 C 7 6 086 021 C 7 7 087 020 085197
23. P 4 successes in 10 C 10 4 044 066 025082
24. The complement is that none of the raccoons had rabies.
P at least one had rabies 1 P none had rabies 1 C 4 0 010 094 1 06561 03439
25. (a) P 5 in 10 C 10 5 0455 0555 023403
(b) P at least 3 1 P at most 2 P 0 in 10 P 1 in 10 P 2 in 10 1 C 10 0 0450 05510 C 10 1 0451 0559 C 10 2 0452 0558 090044
26. (a) P 12 in 15 C 15 12 0112 093 331695 1010
19
20
CHAPTER 13 Counting and Probability
(b) P at least 12 P 12 in 15 P 13 in 15 P 14 in 15 P 15 in 15 C 15 12 0112 093 C 15 13 0113 092 C 15 14 0114 091 C 15 15 0115 090 340336 1010
27. (a) The complement of at least one seed germinating is no seed germinating, so P at least 1 germinates 1 P 0 germinates 1 C 4 0 0750 0254 099609.
(b) P at least 2 germinate P 2 germinates P 3 germinates P 4 germinates C 4 2 0752 0252 C 4 3 0753 0251 C 4 4 0754 0250
094922 (c) P 4 germinates C 4 4 0754 0250 031641
28. (a) P at least 3 boys P 3 boys P 4 boys P 5 boys C 5 3 053 052 C 5 4 054 051 C 5 5 055 050 05
(b) P at least 4 girls P 4 girls P 5 girls P 6 girls P 7 girls C 7 4 054 053 C 7 5 055 052 C 7 6 056 051 C 7 7 057 050 05
29. (a) P all 10 are boys C 10 10 05210 0480 00014456. (b) P all 10 are girls C 10 0 0520 04810 000064925. (c) P 5 in 10 are boys C 10 5 0525 0485 024413. 30. (a) P 2 in 12 C 12 2 022 0810 028347. (b) The complement of “at least 3” is “at most 2”, so
P at least 3 in 12 1 P at most 2 in 12 1 [P 0 in 12 P 1 in 12 P 2 in 12] 1 C 12 0 020 0812 C 12 1 021 0811 C 12 2 022 0810
044165 31. (a) P 3 in 3 C 3 3 00053 09950 0000000125.
(b) The complement of “one or more bulbs is defective” is “none of the bulbs is defective.” So P at least 1 is defective 1 P none is defective 1 C 3 0 00050 09953 0014925.
32. P at least 1 in 10 1 P 0 in 10 1 C 10 0 0050 09510 040126
33. The complement of “2 or more workers call in sick” is “fewer than 2 workers calls in sick.” So P 2 or more 1 [P 0 in 8 P 1 in 8] 1 C 8 0 0040 0968 C 8 1 0041 0967 0038147
34. P 3 in 5 favor C 5 3 063 042 03456
35. (a) P 6 in 6 C 6 6 0756 0250 017798 (b) P 0 in 6 C 6 0 0750 0256 000024414 (c) P 3 in 6 C 6 3 0753 0253 013184
SECTION 13.3 Binomial Probability
21
(d) P at least 2 seasick 1 P at most 1 seasick 1 [P 6 in 6 OK P 5 in 6 OK] 1 C 6 6 0756 0250 C 6 5 0755 0251 046606 36. (a) P machine breaks P at least 1 component fails 1 P no component fails 1 C 4 0 0010 0994 0039404 (b) P no component fails C 4 0 0010 0994 0960596 (c) P 3 components fail C 4 3 0013 0991 000000396 37. (a) The complement of “at least one inherits the disease” is “none inherits the disease.” Then P at least 1 inherits the disease 1 P none inherits the disease 1 C 4 0 0250 0754 068359. (b) P at least 3 inherit the disease P 3 inherit the disease P 4 inherit the disease C 4 3 0253 0751 C 4 4 0254 0750 005078
38. There are 52 cards in the deck, of which to each suit, so P heart P spade P diamond P club 025. 13 belong (a) P 3 in 3 are hearts C 3 3 0253 0750 0015625 (b) P 2 in 3 are spades C 3 2 0252 0751 0140625 (c) P 0 in 3 are diamonds C 3 0 0250 0753 0421875 (d) P at least 1 is a club 1 P none is a club 1 C 3 0 0250 0753 0578125
39. Alex (a nonsmoker) is already in the room, concerns the four other participants assigned to the room. exercise so this (a) P 1 in 4 is a smoker C 4 1 031
073 04116
(b) P at least one smoker 1 P no smoker 1 C 4 0 030 074 07599
40. (a) P 2 or more 1 [P 0 in 100 P 1 in 100] 1 C 100 0 0020 098100 C 100 1 0021 09899 059673
(b) Since P at least 1 interested 1 P 0 interested and 09835 0507, the telephone consultant needs to make at least 35 calls to ensure at least a 05 probability of reaching one or more interested parties.
41. (a) P 8 or more recover P 0 dies P 1 dies P 2 die C 10 0 060 0410 C 10 1 061 049 C 10 2 062 048 00123
(b) Yes, the drug appears to be effective.
42. (a) P 5 or more hits P 5 hits P 6 hits P 7 hits P 8 hits C 8 5 065 043 C 8 6 066 042 C 8 7 067 041 C 8 8 068 040 0594
(b) No, the coaching does not appear to have made any difference.
22
0.273438
CHAPTER 13 Counting and Probability
0.3
43. (a) Number of heads
Probability
0
0003906
1
003125
2
0109375
3
021875
4
0273475
0.1
5
021875
0.05
6
0109375
7
003125
8
0003906
0.25 0.2 Probability
0.15
0
2
4
6
8
Number of heads
If n 8, then 4 heads has the greatest probability of occurring. If the coin is flipped 100 times, then 50 heads has the greatest probability of occurring.
Height of bars is incorrect. Please fix
0.3
(b) Number of heads
Probability
0
0001953
1
0017578
2
0070313
3
0164063
4
0246094
5
0246094
6
0164063
7
0070313
8
0017578
9
0001953
0.25 0.2 Probability
0.15 0.1 0.05 0
2
4
6
8
Number of heads
If n 9, then 4 and 5 heads are the most likely outcomes. If the coin is flipped 101 times, then 50 and 51 heads are the most likely outcomes.
13.4 EXPECTED VALUE 1. If a game gives payoffs of $10 and $100 with probabilities 09 and 01, respectively, then the expected value of this game is E 10 09 100 01 $19. 2. If you played the game in Exercise 1 many times then you would expect your average payoff per game to be about $19. 3. You win $2 with probability 12 and $1 with probability 12 . Thus, E 2 12 1 12 15, and so your expected winnings are $150 per game.
4. The probability that you win $10 is 16 , and the probability that you lose $1 is 56 . Thus E 10 16 1 56 0833, and so your expectation is $0833.
1 , the expected value of this game is 5. Since the probability of drawing the ace of spades is 52 1 1 51 49 094. So your expected winnings are $094 per game. E 100 52 52 52 6. The expected value of this game is E 3 12 2 12 52 25. So your expected winnings are $250 per game.
SECTION 13.4 Expected Value
23
7. Since the probability that you roll a six is 16 , the expected value of this game is E 3 16 050 56 55 6 09167. So you expect to win $092 per game. 2 2 8. The probability that you get two tails is 12 14 , the probability that you get one tail and one head is C 2 1 12 12 , 2 and the probability that you get two heads is 12 14 . If you get two heads, you will receive $4, if you get one head and one tail, you will get $2 $1 $1, and if you get two tails, you will lose $2. Thus the expected value of this game is 1 1 E 4 4 1 2 14 1. So your expected winnings are $1 per game. 2
9. Since the probability that the die shows an even number equals the probability that that die shows an odd number, the expected value of this game is E 2 12 2 12 0. So you should expect to break even after playing this game many times.
10. Since there are 4 aces, 12 face cards, and only one 8 of clubs, the expected value of this game is 4 26 12 13 1 $1425. E 104 52 52 52
11. Since it costs $050 to play, if you get a silver dollar, you win only 1 050 $050. Thus the expected value of this game 2 050 8 030. So your expected winnings are $030 per game. In other words, you is E 050 10 10 should expect to lose $030 per game.
8 7 56 , and the probability of not choosing 12. The probability of choosing two white balls (that is, no black ball) is 10 9 90 56 34 two white balls (that is, at least one black ball) is 1 90 90 . Therefore, the expected value of this game is 34 E 5 56 90 0 90 3111. Thus, your expected winnings are $311 per game. 1 1 37 2 00526. 13. You can either win $35 or lose $1, so the expected value of this game is E 35 38 38 38
Thus, the expected value is $00526 per game.
1 . After the first prize winner is selected, then 2 106 1 1 P winning the second prize . Similarly, P winning the third prize . So the expected 6 2 10 1 2 106 2 1 1 1 105 104 $0555. value of this game is E 106 6 6 2 10 2 10 1 2 106 2 (b) Since we expect to win $0555 on the average per game, if we pay $100, then our net outcome is a loss of $0445 per game. Hence, it is not worth playing, because on average you will lose $0445 per game.
14. (a) We have P winning the first prize
15. By the rules of the game, a player can win $10 or $5, break even, or lose $100. Thus the expected value of this game is 10 5 10 100 2 78 E 10 100 100 100 0 100 050. So the expected winnings per game are $050.
16. Since the safe has a six digit combination, there are 106 possible combinations to the safe, of which only one is correct. The 1 106 1 6 0. expected value of this game is E 10 1 1 106 106 17. If the stock goes up to $20, the investors make $20 $5 $15; if the stock falls to $1, they lose $5 $1 $4. So the expected value of the profit is E 15 01 4 09 21. Thus, the investors’ expected profit per share is $210, that is, they should expect to lose $210 per share. They did not make a wise investment. 3 1 18. Since the wheels of the slot machine are independent, the probability that you get three watermelons is 11 . So the expected value of this game is E 475 13 025 1 13 $0246. 11
11
24
CHAPTER 13 Counting and Probability
19. There are C 49 6 ways to select a set of six numbers from the group of 49 numbers, of which only one is a winning set. 1 1 Thus the expected value of this game is E 106 1 1 1 $093. C 49 6 C 49 6 20. (a) Since the life insurance policy costs $25 per year, we have the expected value E 7500 25 00003 25 09997 2275. (b) The expected yearly income is 450,000 2275 $10,237,500.
21. The expected number is E 2 015 3 045 4 030 5 010 335 hours of TV. 22. The expected number is 005 3 015 2 045 1 035 0 09 foreign languages.
23. The expected number is 3 030 2 045 1 015 0 010 195 times in any given week. 24. (a) Number of girls
Probability
0
1 8 3 8 3 8 1 8
1 2 3
(b) The expected number of girls is 0 18 1 38 2 38 3 18 15.
25. (a) A standard deck contains 1 ace of spades and 48 cards that are not aces. Thus, the expected value is 1 12 48 1 3 $023, and so the game is not fair. 52 52 2 13 1 x 48 1 0 x $2400. (b) The game would be fair with payout x, where 52 52 2
26. (a) The expected value is 13 20 23 10 0, and so the game is fair.
3070 10 27. (a) The expected value is 16 16 30 35 36 2 36 9 , and so the game is not fair. 1 x 35 2 0 x $70. (b) The game would be fair with payout x, where 36 36
1 28. (a) The expected value is 16 12 10 11 12 1 12 , and so the game is not fair.
(b) The game would be fair with payout x, where 16 12 x 11 12 1 0 x $11. 1 1 1 600 1 1 1 1 1 23 , and so the game is not fair. 29. (a) The expected value is 52 6 2 52 6 2 624 1 1 1 1 1 1 0 x $623. (b) The game would be fair with payout x, where 52 6 2 x 1 52 6 2 30. (a) The expected value is 28 1 68 12 18 , and so the game is not fair. (b) The game would be fair with payout x, where 28 x 68 12 0 28 x 38 x $150.
31. If you win, you win $1 million minus the price of the stamp. If you lose, you lose only the price of the stamp (currently
63
44 cents). So the expected value of this game is 999,99956 expect to lose 39 cents on each entry, and so it’s not worth it.
20 106 1 1 039. Thus, you 044 20 106 20 106
-0.63 -0.58 Currently the price of a first class stamp is $0.63.
CHAPTER 13
Review
25
CHAPTER 13 REVIEW 1. The number of possible outcomes is number of outcomes number of outcomes number of ways 2 6 52 624. when a coin is tossed a die is rolled to draw a card 2. (a) The number of 3digit numbers that can be formed using the digits 1–6 if a digit can be used any number of times is 6 6 6 216. (b) The number of 3digit numbers that can be formed using the digits 1–6 if a digit can be used only once is 6 5 4 120.
3. (a) Order is not important, and there are no repetitions, so the number of different twoelement subsets is 54 5! 10. C 5 2 2! 3! 2
5! 20. 3! 4. Since the order in which the people are chosen is not important and a person cannot be bumped more than once (no (b) Order is important, and there are no repetitions, so the number of different twoletter words is P 5 2
repetitions), the number of ways that 7 passengers can be bumped is C 120 7 59488 1010 . 5. You earn a score of 70% by answering exactly 7 of the 10 questions correctly. The number of different ways to answer the 10! 120. questions correctly is C 10 7 7! 3! 6. There 2 ways to answer each of the 10 truefalse questions and 4 ways to answer each of the 5 multiplechoice questions. So the number of ways that this test can be completed is 210 45 1,048,576.
7. You must choose 2 of the 10 questions to omit, and the number of ways of choosing these 2 questions is 10! 45. C 10 2 2! 8! 8. Since the order of the scoops of ice cream is not important and the scoops cannot be repeated, the number of ways to have a banana split is C 15 4 1365. 9. The maximum number of employees using this security system is number of choices number of choices number of choices 26 26 26 17,576. for the first letter for the second letter for the third letter 10. Since there are n! ways to arrange a group of size n and 5! 120, there are 5 students in this class. 11. We could count the number of ways of choosing 7 of the flips to be heads; equivalently we could count the number of ways of 10! 120. choosing 3 of the flips to be tails. Thus, the number of different ways this can occur is C 10 7 C 10 3 3! 7! 12. The number ways to form a license plate consisting of 2 letters followed by 3 numbers is 26 26 10 10 10 676,000. Since there are fewer possible license plates than 700,000, there must be fewer than 700,000 licensed cars in the Yukon. x! x! 13. Let x be the number of people in the group. Then C x 2 10 10 20 x x 1 20 2! x 2! x 2! x 2 x 20 0 x 5 x 4 0 x 5 or x 4. So there are 5 people in this group.
14. Each topping corresponds to a subset of a set with n elements. Since a set with n elements has 2n subsets and 211 2048, the pizza parlor offers 11 toppings. 15. A letter can be represented by a sequence of length 1, a sequence of length 2, or a sequence of length 3. Since each symbol is either a dot or a dash, the possible number of letters is number of letters number of letters number of letters 23 22 2 14. using 3 symbols using 2 symbols using 1 symbol 16. Since the nucleotides can be repeated, the number of possible words of length n is 4n . Since 42 16 20 and 43 64, the minimum length of word needed is 3.
26
CHAPTER 13 Counting and Probability
17. (a) Since we cannot choose a major and a minor in the same subject, the number of ways a student can select a major and a minor is P 16 2 16 15 240. (b) Again, since we cannot have repetitions and the order of selection is important, the number of ways to select a major, a first minor, and a second minor is P 16 3 16 15 14 3360.
(c) When we select a major and 2 minors, the order in which we choose the minors is not important. Thus the number of number of ways number of ways to ways to select a major and 2 minors is 16 C 15 2 16 105 1680. to select a major select two minors
18. (a) Solution 1: Since the leftmost digit of a threedigit number cannot be zero, there are 9 choices for this first digit and 10 choices for each of the other two digits. Thus, the number of threedigit numbers is 9 10 10 900. Solution 2: Since there are 999 numbers between 1 and 999, of which the numbers between 1 and 99 do not have three digits, there are 999 99 900 threedigit numbers. (b) There are 1001 numbers from 0–1000. From part (a), there are 900 threedigit numbers. Therefore the probability that 900 0899. the number chosen is a threedigit number is P E 1001
19. Because the letters are distinct, the number of anagrams of the word RANDOM is 6! 720. 20. Because two letters are the same, the number of anagrams of the word BLOB is
4! 12. 2!
21. Because three letters are the same, the number of anagrams of the word BUBBLE is
6! 120. 3!
22. Because there are two sets of four indistinguishable letters (I, S) and one set of two indistinguishable letters (P), the number 11! of anagrams of the word MISSISSIPPI is 34,650. 4! 4! 2! 23. (a) The possible number of committees is C 18 7 31,824.
(b) Since we must select the 4 graduates from the group of 10 graduates and the 3 undergraduates from the group of 8 undergraduates, the possible number of committees is number of ways to number of ways to C 10 4 C 8 3 210 56 11,760. choose 4 of 10 graduates choose 3 of 8 undergraduates (c) We remove Alex (an undergraduate) from the group of 18, so the possible number of committees is C 17 7 19,448. (d) The possible number of committees is possible number of possible number of possible number of committees with 5 undergraduates committees with 6 undergraduates committees with 7 undergraduates C 8 5 C 10 2 C 8 6 C 10 1 C 8 7 C 10 0 56 45 28 10 8 1 2808
(e) Since the committee is to have 7 members, “at most 2 graduates” is the same as “at least 5 undergraduates,” which we found in part (d). So there are 2808 possibilities. (f) We select the specific offices first, then complete the committee from the remaining members of the group. So the number of possible committees is number of ways to choose number of ways to choose a chairman, a vicechairman, P 18 3 C 15 4 4896 1365 6,683,040. and a secretary four other members
24. Method 1: We choose the 5 states first and then one of the two senators from each state. Thus the number of committees is C 50 5 25 67,800,320. Method 2: We choose one of 100 senators, then choose one of the remaining 98 senators (deleting the chosen senator and the other senator form that state), then choose one of the remaining 96 senators, continuing this way until the 5 senators are chosen. Finally, we need to divide by the number of ways to arrange the 5 senators. Thus the number of committees is 10098969492 67,800,320. 5!
CHAPTER 13
Review
27
2 25. (a) The probability that the marble is red is 10 15 3 .
8 . (b) The probability that the marble is even numbered is 15 2. (c) The probability that the marble is white and an odd number is 15
7 5 12 4 (d) The probability that the marble is red or odd numbered is P red P odd P red odd 10 15 15 15 15 5 .
26. Let Rn denote the event that the nth marble is red and let Wn denote the event that the nth marble is white. 9 3 (a) P both marbles are red P R1 R2 P R1 P R2 R1 10 15 14 7 . (b) Solution 1: The probability that one is white and that the other is red is C 10 1 C 5 1 10 number of ways to select one white and one red . number of ways to select two marbles C 15 2 21 Solution 2:
P one white and one red P W1 R2 P R1 W2 P W1 P R2 W1 P R1 P W2 R1 5 10 10 5 10 15 14 15 14 21
(c) Solution 1: Let E be the event “at least one is red”. Then E is the event “both are white”. 5 4 2 . Thus P E 1 2 19 . P E P W1 W2 P W1 P W2 W1 15 14 21 21 21
9 19 Solution 2: P at least one is red P one red and one white P both red 10 21 21 21 (from (a) and (b)).
(d) Since 5 of the 15 marbles are both red and evennumbered, the probability that both marbles are red and evennumbered 5 4 2 . is 15 14 21
2 1 1 . (e) Since 2 of the 15 marbles are both white and oddnumbered, the probability that both are white is 15 14 105
27. (a) S H H H H H T H T H H T T T H H T H T T T H T T T . (b) P H H H 18
(c) P 2 or more heads P exactly 2 heads P 3 heads 38 18 48 12 (d) P tails on the first toss 48 12
28. The probability that you select a mathematics book is
4 2 number of ways to select a mathematics book 04. number of ways to select a book 10 5
29. Since rolling a die and selecting a card are independent, 4 1. P both show a six P die shows a six P card is a six 16 52 78 4 1 30. (a) P ace 52 13
(b) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a jack. Then 4 4 2. P E F P E P F 52 52 13
(c) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a spade. Then 4 13 1 4 . P E F P E P F P E F 52 52 52 13
(d) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a red card. Then n E F 2 1. P E F 52 26 n S 1 1 1 1 . 31. (a) Since these events are independent, the probability of getting the ace of spades, a six, and a head is 52 6 2 624 1 1 1 (b) The probability of getting a spade, a six, and a head is 13 52 6 2 48 .
3 1 3 (c) The probability of getting a face card, a number greater than 3, and a head is 12 52 6 2 52 .
32. (a) The probability the first die shows some number is 1, and the probability the second die shows the same number is 16 . So the probability each die shows the same number is 1 16 16 .
28
CHAPTER 13 Counting and Probability
(b) By part (a), the event of showing the same number has probability of 16 , and the complement of this event is that the dice show different numbers. Thus the probability that the dice show different numbers is 1 16 56 . 33. (a) Since there are four kings in a standard deck, P 4 kings
1 1 C 4 4 52515049 369 106 . C 52 4 270,725
4321 13121110 C 13 4 4321 11 000264. (b) Since there are 13 spades in a standard deck, P 4 spades 52515049 4165 C 52 4 4321 2 26252423 2 C 26 4 4321 92 011044. (c) Since there are 26 red cards and 26 black cards, P all same color 833 52515049 C 52 4 4321
34. In the numbers game lottery, there are 1000 possible “winning” numbers. 1 . (a) The probability that the player wins $500 is 1000 (b) There are P 3 3 6 ways to arrange the digits 1, 5, 9. However, if the player wins only $50, it means that their 5 1 . number 159 was not the winning number. Thus the probability is 1000 200
35. The contestant knows the first digit and must arrange the other four digits. Since only one of the P 4 4 24 arrangements 1. is correct, the probability that the contestant guesses correctly is 24
36. The number of different pizzas is the number of subsets of the set of 12 toppings, that is, 212 4096. The number of pizzas with anchovies is the number of ways of choosing anchovies and then choosing a subset of the 11 remaining toppings, that 1 is, 1 211 2048. Thus, P getting anchovies 2048 4096 2 .
Note that this makes intuitive sense: for each pizza combination without anchovies there is a corresponding one with anchovies, so half will have anchovies and half will not. 37. (a) Since there are only two sock colors, any three socks must contain a matching pair. (b) Method 1: If the two socks drawn form a matching pair then they are either both red or both blue. So C 20 2 C 30 2 choosing a both red or 051. P P P both red P both blue C 50 2 C 50 2 matching pair both blue Method 2: The complement of choosing a matching pair is choosing one sock of each color. So C 20 1 C 30 1 1 049 051. P choosing a matching pair 1 P different colors 1 C 50 2 38. (a) number of codes choices for 1st digit choices for 2nd digit choices for 5th digit 10 10 10 10 10 105 100,000
(b) Since there are five numbers (0, 1, 6, 8 and 9) that can be read upside down, we have number of codes choices for 1st digit choices for 2nd digit choices for 5th digit 55 3125.
55 n E 1 5 . n S 32 10 (d) Suppose a zip code is turned upside down. Then the middle digit remains the middle digit, so it must be a digit that reads the same when turned upside down, that is, a 0, 1 or 8. Also, the last digit becomes the first digit, and the next to last digit becomes the second digit. Thus, once the first two digits are chosen, the last two are determined. Therefore, the number of zip codes that read the same upside down as right side up is number of codes choices for 1st digit choices for 2nd digit choices for 5th digit 5 5 3 1 1 75. (c) Let E be the event that a zip code can be read upside down. Then by parts (a) and (b), P E
39. (a) Order is important, and repeats are possible. Thus there are 10 choices for each digit. So the number of different Zip+4 codes is 10 10 10 109 .
(b) If a Zip+4 code is to be a palindrome, the first 5 digits can be chosen arbitrarily. But once chosen, the last 4 digits are determined. Since there are 10 ways to choose each of the first 5 digits, there are 105 palindromes. 5
(c) By parts (a) and (b), the probability that a randomly chosen Zip+4 code is a palindrome is 109 104 . 10
CHAPTER 13
Review
29
40. (a) Using the rule for the number of distinguishable combinations, the number of divisors of N is 7 1 2 1 5 1 144.
(b) An even divisor of N must contain 2 as a factor. Thus we place a 2 as one of the factors and count the number of distinguishable combinations of M 26 32 55 . So using the rule for the number of distinguishable combinations, the number of even divisors of N is 6 1 2 1 5 1 126.
(c) A divisor is a multiple of 6 if 2 is a factor and 3 is a factor. Thus we place a 2 as one of the factors and a 3 as one of the factors. Then we count the number of distinguishable combinations of K 26 31 55 . So using the rule for the number of distinguishable combinations, the number of even divisors of N is 6 1 1 1 5 1 84.
126 7 (d) Let E be the event that the divisor is even. Then using parts (a) and (b), P E nE nS 144 8 . 4 1 41. (a) P king 52 13
8 2 (b) P king or ace 52 13
4 1 number of kings . number of face cards 12 3 number of kings 4 1 (d) The probability that the card is a king given that it is not an ace is . number of nonaces 48 12 3 0 1 1 12 . 42. (a) Because each card is replaced, the probability that all three cards are kings is C 3 3 13 13 2197 4 4 4 1 1 Another method:We can calculate the probability as 3 . 52 52 52 2197 13 2 1 1 36 12 (b) The probability that exactly two cards are jacks is C 3 2 . 13 13 2197 0 3 1000 3 10 . (c) The probability that none of the cards is a face card is C 3 0 13 13 2197 (d) “At least one of the cards is a face card” is the complement of the event in part (c), so its probability is 1000 1197 1 . 2197 2197 4 4 5 43. (a) P 4 sixes in 8 rolls C 8 4 16 0026048. 6 (c) The probability that the card is a king given that it is a face card is
(b) There are three even numbers on a die and three odds numbers, so P even P odd 05. Thus
P 2 or more evens in 8 rolls 1 P fewer than 2 evens 1 [P 0 evens P 1 even] 1 C 8 0 050 058 C 8 1 051 057 096484
44. (a) P 5 in 5 are white flesh C 5 5 035 070 000243 (b) P 0 in 5 are white flesh C 5 0 030 075 016807 (c) P 2 in 5 are white flesh C 5 2 032 073 03087
(d) P 3 or more are red flesh P 3 in 5 are red flesh P 4 in 5 are red flesh P 5 in 5 are red flesh C 5 3 032 073 C 5 4 031 074 C 5 5 030 075 083692
45. (a) The probability that 9 or more patients would have recovered without the drug is C 12 9 0659 0353 C 12 10 06510 0352 C 12 11 06511 0351 C 12 12 06512 0350 0347.
(b) No, the drug does not appear to be effective.
30
CHAPTER 13 Counting and Probability
46. The probabilities are as follows: 0 head: C 4 0 070 034 00081 1 head: C 4 1 071 033 00756
2 heads: C 4 2 072 032 02646 3 heads: C 4 3 073 031 04116 4 heads: C 4 4 074 030 02401
Outcome (heads)
Probability
0
00081
1
00756
2
02646
3
04116
4
02401
47. There are 36 possible outcomes in rolling two dice and 6 ways in which both dice show the same number, namely, 1 1, 6 1 30 0. 2 2, 3 3, 4 4, 5 5, and 6 6. So the expected value of this game is E 5 36 36 1 , 48. Using the same logic as in Exercise 32(a), the probability that all three dice show the same number is 1 16 16 36
1 35 . Thus, the expected value of this game is while the probability they are not all the same is 1 36 36 1 35 30 E 5 36 1 36 36 083. So your expected winnings per game are $083; that is, you expect to lose
$083 per game.
49. Since the contestant makes a guess as to the order of ratification of the 13 original states, the number of such guesses is 1 . Thus, the expected value is P 13 13 13!, while the probability that the contestant guesses the correct order is 13! 1 13! 1 E 1,000,000 0 000016. So the contestant’s expected winnings are $000016. 13! 13!
50. The expected number of times the athlete goes jogging in any given week is 04 3 01 2 02 1 03 0 16.
CHAPTER 13 TEST 1. The order is fixed, but for each grandchild they have three choices of pictures. Thus, the number of possibilities is 3 3 3 3 81.
2. There are 4 main courses, 3 desserts, and 6 drinks to choose from, so the total number of possibilities is 4 3 6 72. 3. (a) If repetition is allowed, then each letter can be chosen in 26 ways and each digit in 10 ways, so the number of possible passwords is 264 103 456,976,000.
(b) If repetition is not allowed, then the first letter can be chosen in 26 ways, the second in 25 ways, the third in 24 ways, and the fourth in 23 ways. The first digit can be chosen in 10 ways, the second in 9 ways, and the third in 8 ways. Thus, in this case the total number of possible passwords is 26 25 24 23 10 9 8 258,336,000. 4. (a) Order is important in the arrangement, therefore the number of ways to arrange P 30 4 657,720.
(b) Here we are interested in the group of books to be taken on vacation, so order is not important. Therefore, the number of ways to choose these books is C 30 4 27,405.
5. There are two choices to be made: Choose a road to travel from Ajax to Barrie, and then choose a different road from Barrie to Ajax. Since there are 4 roads joining the two cities, we need the number of permutations of 4 objects (the roads) taken 2 at a time (the road there and the road back). This number is P 4 2 4 3 12. 6. A customer must choose a size of pizza and must make a choice of toppings. There are 4 sizes of pizza, and each choice
of toppings from the 14 available corresponds to a subset of the 14 objects. Since a set with 14 objects has 214 subsets, the number of different pizzas this parlor offers is 4 214 65,536. 7. (a) We want the number of ways of arranging 4 distinct objects (the letters L, O, V, E). This is the number of permutations of 4 objects taken 4 at a time. Therefore, the number of anagrams of the word LOVE is P 4 4 4! 24.
CHAPTER 13
Test
31
(b) We want the number of distinguishable permutations of 6 objects (the letters K, I, S, S, E, S) consisting of three like groups of size 1 and a like group of size 3 (the S’s). Therefore, the number of different anagrams of the word KISSES is 6! 6! 120. 1! 1! 1! 3! 3! 8. We choose the officers first. Here order is important, because the officers are different. Thus there are P 30 3 ways to do this. Next we choose the other 5 members from the remaining 27 members. Here order is not important, so there are C 27 5 ways to do this. Therefore the number of ways that the board of directors can be chosen is 27! 1,966,582,800. P 30 3 C 27 5 30 29 28 5! 22! 9. One card is drawn from a standard 52card deck. 1 (a) Since there are 26 red cards, the probability that the card is red is 26 52 2 . 4 1 . (b) Since there are 4 kings, the probability that the card is a king is 52 13
2 1. (c) Since there are 2 red kings, the probability that the card is a red king is 52 26
10. Let R be the event that the ball chosen is red. Let E be the event that the ball chosen is evennumbered. 5 03846. (a) Since 5 of the 13 balls are red, P R 13 6 04615. (b) Since 6 of the 13 balls are evennumbered, P E 13
5 6 2 9 06923. (c) P R or E P R P E P R E 13 13 13 13
11. Let E be the event of choosing 3 males. Then number of ways to choose 3 males C 5 3 n E 0022. P E n S number of ways to choose 3 goldfish C 15 3
n E 6 1 . n S 36 6 13. There are 4 students and 12 astrological signs. Let E be the event that at least 2 have the same astrological sign. Then E is the event that no 2 have the same astrological sign. It is easier to find E . So number of ways to assign 4 different astrological signs P 12 4 55 12 11 10 9 . P E number of ways to assign 4 astrological signs 12 12 12 12 96 124 41 Therefore, P E 1 P E 1 55 96 96 0427. 14. (a) P 6 heads in 10 tosses C 10 6 0556 0454 023837. 12. Two dice are rolled. Let E be the event of getting doubles. Since a double may occur in 6 ways, P E
(b) “Fewer than 3 heads” is the same as “0, 1, or 2 heads.” So
P fewer than 3 heads P 0 head in 10 P 1 head in 10 P 2 heads in 10 C 10 0 0550 04510 C 10 1 0551 0459 C 10 2 0552 0458 002739
4 1 , the probability 15. A deck of cards contains 4 aces, 12 face cards, and 36 other cards. So the probability of an ace is 52 13
3 , and the probability of a nonace, nonface card is 36 9 . Thus the expected value of this game of a face card is 12 13 52 13 52 1 3 9 85 is E 10 13 1 13 5 13 13 0654, that is, about $065.
32
FOCUS ON MODELING
FOCUS ON MODELING The Monte Carlo Method 1. (a) You should find that with the switching strategy, you win about 90% of the time. The more games you play, the closer to 90% your winning ratio will be. 1 , since there are 10 doors and (b) The probability that the contestant has selected the winning door to begin with is 10 9 . If the contestant switches, only 1 is a winner. So the probability that the contestant has selected a losing door is 10 1 , and the they exchange a losing door for a winning door (and vice versa), so the probability that they lose is now 10 9 . probability that they win is now 10
2. (a) You should find that you get a combination consisting of one head and one tail about 50% of the time. (b) The possible gender combinations are B B BG G B GG. Thus, the probability of having one child of each sex is 2 1. 4 2
3. (a) You should find that player A wins about 78 of the time. That is, if you play this game 80 times, player A should win approximately 70 times. (b) The game will end when either player A gets one more head or player B gets three more tails. Each toss is independent, and both heads and tails have probability 12 , so we obtain the following probabilities. Outcome
Probability
H
1 2 1 1 1 2 2 4 1 1 1 1 2 2 2 8 1 1 1 1 2 2 2 8
TH TTH TTT
Player A
Since Player A wins for any outcome that ends in heads, the probability that he wins is 12 14 18 78 . 4. (a) If you simulate 80 World Series with coin tosses, you should expect the series to end in 4 games about 10 times, in 5 games about 20 times, in 6 games about 25 times, and in 7 games about 25 times. (b) We first calculate the number of ways that the series can end with team A winning. (Note that a team must win the final game plus three of the preceding games to win the series.) To win in 4 games, team A must win 4 games right off the bat, and there is only 1 way this can happen. To win in 5 games, team A must win the final game plus 3 of the first 4 games, so this can happen in C 4 3 4 ways. To win in 6 games, team A must win the final game plus 3 of the first 54 5 games, so this can happen in C 5 3 10 ways. To win in 7 games, team A must win the final game plus 3 21 654 of the first 6 games, so this can happen in C 6 3 20 ways. By symmetry, it is also true for team B that 321 they can win in 4 games 1 way, in 5 games 4 ways, in 6 games 10 ways, and in 7 games 20 ways. The probability that any particular team wins a given game is 12 ; this fact, together with the assumption that the games are independent allows us to calculate the probabilities in the following table. Series
Number of possibilities
4 games
2
5 games
8
6 games
20
7 games
40
Probability 2 12 12 12 12 18 8 12 12 12 12 12 14 5 20 12 12 12 12 12 12 16 5 40 12 12 12 12 12 12 12 16
The Monte Carlo Method
33
5 6 5 7 5 13 58. Thus, on average, we expect a World Series to end in (c) The expected value is 18 4 14 5 16 16 16
about 58 games.
5. With 1000 trials, you are likely to obtain an estimate for that is between 31 and 32. 6. Modify the TI84 program in Problem 5 to the following: PROGRAM:PROB6 :0P :For(N,1,1000) :randX:randY :P+(X2 Y)P :End :Disp “PROBABILITY IS APPROX”,P/1000 You should find that the probability is very close to 13 . 7. (a) We can use the following TI84 program to model this experiment. It is a minor modification of the one given in Problem 5. PROGRAM:PROB7 :0P :For(N,1,1000) :randX:randY :P+((X+Y)1)P :End :Disp “PROBABILITY IS APPROX”,P/1000 You should find that the probability is very close to 12 . (b) Following the hint, the points in the square for which x y 1 are the ones that lie below the line x y 1. This triangle has area 12 (it takes up half the square), so the probability that x y 1 is 12 .