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Functions from a Calculus PerspectiveCHAPTER 1CHCHAPAPTETERR 11

Chapter 1 Get Ready

1.

0-2-43.

-4 -2 0 2 4-6-85.

6420 8 10-27. y = 2 + 3x 9. y = ± √ �� 2x - 7 11. y =

3 √ �� -x - 9

13. D = n _ 12

; 25.5 dozen donuts 15. 1

17. -50 x 2 + 25x + 1 19. 12 n 2 - 60n + 77

Lesson 1-1

1. {x | x > 50, x ∈ �}; (50, ∞) 3. {x | x ≤ -4, x ∈ �}; (-∞, -4] 5. {x | 8 < x < 99, x ∈ �}; (8, 99) 7. {x | x < -19 or x > 21, x ∈ �}; (-∞, -19) ∪ (21, ∞) 9. {x | 0.25n = x, n ≥ -1, n ∈ } 11. {x | x ≤ -45 or x > 86, x ∈ �}; (-∞, -45] ∪ (86, ∞) 13. {x | x = 5n, n ∈ } 15. function 17. function 19. function 21. function 23. function 25. function 27. not a function 29a. {(1, 70), (2, 75), (3, 70), (4, 62), (5, 65)} 29b. Yes; there is exactly one estimated high temperature each day. Yes; there is exactly one low temperature each day. 31a. -207 31b. 24 y 3 + 12y + 9 31c. -375 b 3 - 675 b 2 - 435b - 90

33 a. g(-2) = 3 (-2) 3 __

(-2) 2 + (-2) - 4

= 3(-8) __

4 + (-2) - 4

= -24 _

-2

= 12

b. g(5x) = 3 (5x) 3 __

(5x) 2 + (5x) - 4

= 3(125 x 3 )

__ 25 x 2 + 5x - 4

= 375 x 3 __

25 x 2 + 54 - 4

c. g(8 - 4b) = 3 (8 - 4b) 3

__ (8 - 4b) 2 + (8 - 4b) - 4

= 3(512 - 768b + 384b 2 - 64b 3 )

___ 64 - 64b + 16 b 2 + 8 - 4b - 4

= -192 b 3 + 1152 b 2 - 2304b + 1536

___ 16 b 2 - 68b + 68

= -48 b 3 + 288 b 2 - 576b + 384

___ 4 b 2 - 17b + 17

35a. -0.8 35b. -1 -

1 _

8x 35c.

-48 b 3 + 288 b 2 - 576b + 384 ___

4 b 2 - 17b + 17

37a. 20 √ � 6 37b. 10|x| √ � 6 37c. 5 |7 + n| √ � 6 39. (-∞, -4) ∪ (-4, -1) ∪ (-1, ∞) 41. (-∞, ∞) 43. (0.25, ∞) 45. (-∞, -1) ∪ (-1, 0) ∪ (0, ∞)

47. Yes; sample answer: Because length must be positive, the

domain of the function is (0, ∞). 49. 21; 157 51. 1; 8 1 _ 6

53a. 8.65 million; 9.36 million 53b. integral values inside the interval [0, 10]55. No; a vertical line would pass through infinitely many points. 57. No; the y-axis is a vertical line that passes through two points on the graph, (0, 0) and (0, -4).

59 Sample answer: Because presidential elections are held every 4 years, and do not have a finite end, it is impractical to display the set in interval notation. If set-builder notation is used, the interval can be taken into account and a finite interval is not necessary. Therefore, the set of presidential election years beginning in 1792 can be described in set-builder notation as {x | x = 4n + 1792, n ∈ �}.

61. D = {x | x ≥ 1874, x ∈ �} 63. -5; -5; 0

65. 1 _

a + 4 ; 1

_ a + h + 4

; -1 __

a 2 + ah + 8a + 4h + 16 67. a 2 - 6a + 8;

a 2 + 2ah + h 2 - 6a - 6h + 8; 2a – 6 + h 69. - a 5 ; - a 5 - 5 a 4 h - 10 a 3 h 2 - 10 a 2 h 3 - 5a h 4 - h 5 ; -5 a 4 - 10 a 3 h - 10 a 2 h 2 - 5a h 3 - h 4 71. 7a - 3; 7a + 7h - 3; 7 73. a 3 ; a 3 +

3 a 2 h + 3a h 2 + h 3 ; 3 a 2 + 3ah + h 2 75a. A(ℓ) = ℓ 2 _

1.8 ; [5, 11.5]

75b. A(h) = 2.1 h 2 ; [2.4, 5.5] 75c. 52.9 in 2 77. No; sample answer: Most nonnegative x-values are paired with two y-values because it is necessary to take both the positive and negative values of the absolute value of x when solving the equation for y. 79a.

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x) = x

[-10, 10] scl: 1 by [-10, 10] scl: 1

= x2f (x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

= x3f (x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

= x 4f (x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

= x 5f (x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

= x 6f (x)

R1connectED.mcgraw-hill.com

R01_R50_EM_SelAns_664293.indd R1R01_R50_EM_SelAns_664293.indd R1 05/10/12 6:59 PM05/10/12 6:59 PM

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79b. n Range

1 (-∞, ∞)

2 [0, ∞)

3 (-∞, ∞)

4 [0, ∞)

5 (-∞, ∞)

6 [0, ∞)

79c. Sample answer: When n is even in f (x) = x n , the range is [0, ∞). 79d. Sample answer: When n is odd in f (x) = x n , the range is (-∞, ∞).

81 When the denominator of 1 __

(x + 3)(x + 1)(x - 5) is zero, the

expression is undefined. Therefore, f(x) is undefined for x = -3, x = -1, and x = 5. In interval notation the domain can be described as (-∞, -3) ∪ (-3, -1) ∪ (-1, 5) ∪ (5, ∞), and in set-builder notation the domain can be described as {x | x ≠ -3, x ≠ -1, x ≠ -5, x ∈ �}. I prefer set-builder notation because instead of listing the four intervals in which x can be an element, it lists the three real numbers that x cannot be equal to.

83. true 85. False; sample answer: Two or more elements in X may be matched with the same element in Y. 87. Sample answer: If each of the possible inputs is assigned to exactly one output in the verbal description, the relation is a function. 89. Sample answer: If each input value in the table is paired with a unique output value, the relation is a function. 91. Sample answer: If each x-value can be paired with exactly one y-value after the equation is solved for y, then the relation is a function. 93. 0.73 95. 216,216 97. (-3.5, 2) 99. (9, 6); consistent and independent 101. (2, 2); consistent and independent 103. {1, 2, 4, 5, 8, 9, 11, 12} 105. � 107. B 109. A

Lesson 1-2

1a. 37.75 1b. 32.68 1c. 28.21

3 The function value at x = -8 appears to be about 10. Find f (-8). f (x) = |x| + 2f (-8) = |-8| + 2 = 8 + 2 or 10The function value at x = -3 appears to be about 5. Find f(-3).f (-3) = |-3| + 2 = 3 + 2 or 5The function value at x = 0 appears to be about 2. Find f (0).f (0) = |0| + 2 = 2

5a. 4 _ 3 5b. -1 5c. undefined 7a. 36,000 tons, 46,000 tons,

54,000 tons; 35,750 tons, 46,082 tons, 53,113 tons 7b. about 2004 or x = 11; f (10) = -0.0013 (10) 4 + 0.0513 (10) 3 - 0.662 (10) 2 + 4.128(10) + 35.75 ≈ 49; f (12) = -0.0013 (12) 4 +

0.0513 (12) 3 - 0.662 (12) 2 + 4.128(12) + 35.75 ≈ 52 9. D = (-∞, ∞), R = [2, ∞) 11. D = (-4, 4], R = [-1, 6] 13. D = [-5, ∞), R = [-2, ∞) 15a. Sample answer: copper: D = [-150, 150], R = [1.75]; aluminum: D = [-150, 150], R = [0.6, 1.5]; zinc: D = [-150, 150], R = [0.5, 1.3]; steel: D = [-150, 150], R = [0.2, 1.75] 15b. Sample answer: copper ≈ 1.75 J, aluminum ≈ 1.2 J, zinc ≈ 0.5 J, steel ≈ 1.5 J

17 From the graph, it appears that f (x) intersects the y-axis at approximately (0, 0), so the y-intercept is 0. Find f (0).

f (0) = 2(0 ) 3 - (0 ) 2 - 3(0)f (0) = 0Therefore, the y-intercept is 0.From the graph, the x-intercepts appear to be at about

-1 and 3 _ 2 . Let f (x) = 0 and solve for x.

2 x 3 - x 2 - 3x = 0x(2x - 3)(x + 1) = 0x = 0 or 2x - 3 = 0 or x + 1 = 0

x = 3 _ 2 x = -1

Therefore, the zeros of f are 0, 3 _ 2 , and -1.

19. y-intercept: 3; no zeros; √

x + 3 = 0

√

x ≠ -321. y-intercept: -2; zeros: -

1 _

2 , 2 _

3 ;

6 x 2 - x - 2 = 0(2x + 1)(3x - 2) = 02x + 1 = 0 or 3x - 2 = 0

x = -

1 _

2 x = 2 _

3

23. y-intercept: 6; zeros: -2, -3; x 2 + 5x + 6 = 0(x + 2)(x + 3) = 0x + 2 = 0 or x + 3 = 0 x = -2 x = -3

25. x-axis;

x y (x, y )

1 2 (1, 2)

1 -2 (1, -2)

2 √ 5 (2, √ 5 )

2 - √ 5 (2, - √ 5 )

3 √ 6 (3, √ 6 )

3 - √ 6 (3, - √ 6 )

Because x = (-y) 2 - 3 is equivalent to x = y 2 - 3, the graph is symmetric with respect to the x-axis.

27. x-axis, y-axis, and origin;

x y (x, y )

1 2 √ 2 _

5 (1, 2 √ 2

_ 5 )

1 -

2 √ 2 _

5 (1, -

2 √ 2 _

5 )

2 √ 35

_ 5 (2,

√ 35 _

5 )

2 -

√ 35 _

5 (2, -

√ 35 _

5 )

3 4 √ 5 _

5 (3, 4 √ 5

_ 5 )

3 -

4 √ 5 _

5 (3, -

4 √ 5 _

5 )

x y (x, y )

-3 4 √ 5 _

5 (-3, 4 √ 5

_ 5 )

-2 √ 35

_ 5 (-2,

√ 35 _

5 )

-1 2 √ 2 _

5 (-1, 2 √ 2

_ 5 )

1 2 √ 2 _

5 (1, 2 √ 2

_ 5 )

2 √ 35

_ 5 (2,

√ 35 _

5 )

3 4 √ 5 _

5 (3, 4 √ 5

_ 5 )

Because 9 x 2 - 25 (-y) 2 = 1 is equivalent to 9 x 2 - 25 y 2 = 1,the graph is symmetric with respect to the x-axis.

Because 9 (-x) 2 - 25 y 2 = 1 is equivalent to 9 x 2 - 25 y 2 = 1, the graph is symmetric with respect to the y-axis.

(Continued on the next page)

R2 | Selected Answers


Selected A

nswers and S

olutions

x y (x, y )

-3 -

4 √ � 5 _

5 (-3, -

4 √ � 5 _

5 )

-2 -

√ � 35 _

5 (-2, -

√ � 35 _

5 )

-1 -

2 √ � 2 _

5 (-1, -

2 √ � 2 _

5 )

1 2 √ � 2 _

5 (1, 2 √ � 2

_ 5 )

2 √ � 35

_ 5 (2,

√ � 35 _

5 )

3 4 √ � 5 _

5 (3, 4 √ � 5

_ 5 )

Because 9(- x 2 ) - 25 (-y) 2 = 1 is equivalent to 9 x 2 - 25 y 2 = 1, the graph is symmetric with respect to the origin.

29. origin; 31. y-axis

x y (x, y )

-10 1 (-10, 1)

-5 2 (-5, 2)

-1 10 (-1, 10)

1 -10 (1, -10)

5 -2 (5, -2)

10 -1 (10, -1)

x y (x, y )

-3 9 (-3, 9)

-2 -16 (-2, -16)

-1 -7 (-1, -7)

1 -7 (1, -7)

2 -16 (2, 16)

3 9 (3, 9)

Because -y = -

10 _

(-x) is

equivalent to y = - 10 _ x , the

graph is symmetric with respect to the origin.

Because y = (-x) 4 - 8 (-x) 2 is equivalent to y = x 4 - 8 x 2 , the graph is symmetric with respect to the y-axis.

33. y-axis;

x y (x, y )

-2 √ � 2 6 (-2 √ � 2 , 6)

-

√ � 14 _

2 12 (-

√ � 14 _

2 , 12)

-1 2 √ � 14 + 6 (-1, 2 √ � 14 + 6)

√ � 14

_ 2 12 (

√ � 14 _

2 , 12)

1 2 √ � 14 + 6 (1, 2 √ � 14 + 6)

2 √ � 2 6 (2 √ � 2 , 6)

Because (y - 6) 2 + 8 (-x) 2 = 64 is equivalent to (y - 6) 2 + 8 x 2 = 64, the graph is symmetric with respect to the y-axis.

35.

[-5, 5] scl: 1 by [-9, 1] scl: 1

neither;f (-x) = -2(-x ) 3 + 5(-x) - 4 = -2 x 3 - 5x - 4

37.

[-5, 5] scl: 1 by [-5, 5] scl: 1

Even; the graph of h(x) is symmetric with respect to the y-axis.

h(-x) = √ ��

(-x ) 2 - 9

= √ ��

x 2 - 9 = h(x)

39.

[-5, 5] scl: 1 by [-3, 7] scl: 1

Even; the graph of f (x) is symmetric with respect to the y-axis.f (-x) = |(-x ) 3 |

= |- x 3 |

= f(x)

41.

[-6, 4] scl: 1 by [-12, 8] scl: 2

neither;

g(-x) = (-x ) 2

_ -x + 1

= x 2 _

-x + 1

43a. undefined 43b. 0 43c. 1.5

45 a. Because x represents years since 2001 and the model is only valid from 2001 to 2005, the relevant domain of the function is {x | 0 ≤ x ≤ 4, x ∈ �}. The graph does not appear to extend below f(0) = 1.2 or above f (4) ≈ 11.2. Therefore, the range of the function is approximately R = {y | 1.2 million ≤ y ≤ 11.2 million, y ∈ �}.

b. Because 2003 is 2 years after 2001, x = 2. Sample answer: From the graph, it appears that the function value when x = 2 is about 4.1. Find h(2).

h(2) = 0.5(2 ) 2 + 0.5(2) + 1.2= 2 + 1 + 1.2 or 4.2

c. Sample answer: From the graph, it appears that the y-intercept is about 1.1. Find h(0).

h(0) = 0.5(0 ) 2 + 0.5(0) + 1.2= 1.2

The y-intercept represents the number of millions of households with only wireless service in 2001.

d. There are no zeros because there were more than zero households with only wireless phone service for all of the years in the domain.

47a.

[-5, 15] scl: 2 by [-100, 400] scl: 50

47b. [0, 6]; The relevant domain represents the interval of time beginning when the dose was first taken and ending when the pain reliever left the bloodstream. Because time cannot be negative, x ≥ 0. The amount of pain reliever in the bloodstream is zero when the dose is first taken at x = 0, and is zero again at x = 6. Therefore, x ≤ 6. Combining these restrictions, the relevant domain is {x | 0 ≤ x ≤ 6, x �} or [0, 6]. 47c. about 346 milligrams49.

[-30, 30] scl: 6 by [-36, 24] scl: 6

no zeros

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51.

[-25, 15] scl: 5 by [-20, 20] scl: 5

2 √ �� x + 12 - 8 = 0

2 √ �� x + 12 = 8

√ �� x + 12 = 4

x + 12 = 16 x = 4

53.

[-20, 20] scl: 5 by [-20, 20] scl: 5

6 _ x + 3 = 0

6 + 3x = 0 3x = -6 x = -2

55. D = (-8, -4] � (-2, ∞), R = (-6, ∞) 57. D = (-∞, -6] � (0, 4) � [7, ∞), R = [-8, ∞) 59a. D = [0, 70], R ≈ [-19.19, 12.58]59b. Sample answers: 12, 12.58; The y-intercept represents the percent population change in 1930.59c. 15.4; The zeros represent the times at which the percent population change was and will be 0.59d. 203.83%; Sample answer: This value does not seem realistic because it would mean that the population increased by about 204%, which does not seem likely in comparison to the population trends from 1930 to 2000.61a.

x 1.99 1.999 2 2.001 2.01

f (x ) -100 -1000 undefi ned 1000 100

61b. Sample answer: As x approaches 2 from the left, the value of f (x) becomes greater and greater. As x approaches 2 from the right, the value of f (x) becomes more and more negative.61c.

[-10, 10] scl: 1 by [-10, 10] scl: 1

Sample answer: As x approaches 2 from the left, the graph decreases without bound. As x approaches 2 from the right, the graph increases without bound. 61d. Sample answer: As the value of x becomes increasingly large, or when x > 3, the denominator of the fraction will also become increasingly large. This creates smaller fractions that will continue to decrease but will never reach zero or cross the x-axis. This also applies as x decreases at an increasing rate, or when x < -1. However, since x is negative, the value will also be negative but it too will never reach 0. As the value of x approaches 2, the difference of x and 2 becomes smaller and smaller. When the difference d is -1 < d < 1, the denominator is smaller than the numerator, thus producing a larger number. If the difference is positive, the fraction will approach infinity. If the difference is negative, the fraction will approach negative infinity. Only at 2 will the fraction fail to exist because the difference in the denominator cannot be 0.

63.

[-10, 10] scl: 2 by [-40, 20] scl: 6

Even; the graph of g(n) is symmetric with respect to the y-axis.g(-n) = n 2 - 37

= (-n ) 2 - 37

= n 2 - 37

= g(n)

65.

[-5, 5] scl: 1 by [-5, 5] scl: 1

Odd; the graph of f (g) is symmetric with respect to the origin.f (-g) = (-g ) 9

= - g 9

= -f ( g)

67.

[-5, 5] scl: 1 by [-200, 200] scl: 50

Odd; the graph of h(y) is symmetric with respect to the origin.h(-y) = (-y ) 5 - 17(-y ) 3 +

16(-y)

= - y 5 + 17 y 3 - 16y = -h( y)

69. Sample answer: y

x

−4

−8

−4−8

8

4

4 8

71 The graph of a relation is symmetric with respect to the origin if, for every point (x, y) on the graph, the point (-x, -y) is also on the graph. Therefore, the points (3, 18), (2, 9), and (1, 3) are also on the graph.

73. Sample answer: For a relation to be a function, each x-value must map to exactly one y-value. A relation with two y-intercepts located at (0, y 1 ) and (0, y 2 ) cannot be a function because two different y-values cannot map to the same x-value. A relation with two x-intercepts located at ( x 1 , 0) and ( x 2 , 0) can be a function because two different x-values can map to the same y-value. 75. False; sample answer: If n is 0, the range is {y | y = 0}. If n is negative, the range is {y | y ≤ 0, y ∈ }. The range will only be {y | y ≥ 0, y ∈ } if n is positive. 77. False; sample answer: Consider the odd function y = x 3 and the point (2, 8) on its graph. Reflecting the point (2, 8) in the line y = -x produces the point (-8, -2) which is not on the graph of y = x 3 .79. odd; Sample answer: Proof: Given: a is an odd function and b(x) = a(-x) Prove: b is an odd function 1. b(x) = a(-x) (Given) 2. a is an odd function (Given) 3. a(–x) = -a(x) (Def. odd function) 4. b(x) = -a(x) (Trans. Prop. of Equality using 1 and 3) 5. –b(x) = a(x) (Div. Prop. of Equality) 6. a(x) = -b(x) (Reflexive Prop. of Equality) 7. b (-x) = a(-(-x)) (Substitute –x for x in 1)

y

x

−8

−16

−2−4

16

8

2 4



Selected A

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olutions

8. b (-x) = a(x) (Substitute x for -(-x).) 9. b (-x) = -b(x) (Trans. Prop. of Equality using

8 and 6) 10. b is an odd function. (Def. odd function) 81. even; Sample answer: Proof: Given: a is an odd function and b(x) = [a(x) ] 2 Prove: b is an even function 1. b(x) = [a(x) ] 2 (Given) 2. a is an odd function (Given) 3. a(–x) = -a(x) (Def. odd function) 4. b(–x) = [a(-x) ] 2 (Substitute -x for x in 1) 5. b(–x) = [-a(x) ] 2 (Trans. Prop. of Equality using

4 and 3) 6. b(-x) = [a(x) ] 2 (Mult. Prop.) 7. b(-x) = b(x) (Trans. Prop. of Equality using

6 and 1) 8. b is an even function. (Def. even function)83. odd; Sample answer: Proof: Given: a is an odd function and b(x) = [a(x) ] 3 Prove: b is an odd function 1. b(x) = [a(x) ] 3 (Given) 2. a is an odd function (Given) 3. a(-x) = -a(x) (Def. odd function) 4. b(-x) = [a(-x) ] 3 (Substitute -x for x in 1) 5. b(-x) = [-a(x) ] 3 (Trans. Prop. of Equality using

4 and 3) 6. b(-x) = –[a(x) ] 3 (Mult. Prop.) 7. –b(-x) = [a(x) ] 3 (Div. Prop. of Equality) 8. –b(-x) = b(x) (Trans. Prop. of Equality using

7 and 1) 9. b(-x) = -b(x) (Div. Prop. of Equality) 10. b is an odd function. (Def. odd function)85. Sometimes; sample answer: The relation described by x 2 +(y - 2 ) 2 = 4 is symmetric with respect to the line y = 2, but is not a function. A relation whose points all lie on the line y = 2 is symmetric with respect to the line y = 2 and is a function. 87. Sometimes; sample answer: The relation described by x 2 + y 2 = 25 is symmetric with respect to the x- and y-axes but is not a function. The relation {(x, 0) | - 1 ≤ x ≤ 1, x � �} is symmetric with respect to the x- and y-axes and is a function. 89a. -13 89b. 16 x 2 + 40x + 3

89c. 9 n 2 - 24n - 6 91a. 8 91b. 2 x 6 + 2

_ x 4 - 2

91c. 2 x 3 + 6 x 2 + 6x + 4

__ x 2 + 2x - 1

93a. 22,308 93b. 144 93c. 792

95. no solution 97. 3 99. 1 _ 7 101. 125

103. R 2 + 2RW + W 2 105. -15 + 8i 107. B 109. B

Lesson 1-3

1. Continuous; f (-5) = √ � 21 or about 4.58, lim x→-5

≈ 4.58, and

lim x→-5 = f (-5). 3. Discontinuous at x = -6; h (-6) is undefined

and lim x→-6 = -12, so h(x) has a removable discontinuity at x = -6.

Continuous at x = 6. h (6) = 0, lim x→6 h (x) = 0, and lim x→6

h (x) = h(6).

5. Discontinuous; g(1) is undefined and g(x) approaches -∞ as x approaches 1 from the left and ∞ as x approaches 1 from the right, so g(x) has an infinite discontinuity at x = 1. 7. Discontinuous at x = 1; h (1) is undefined and h(x) approaches -∞ as x approaches 1 from the left and ∞ as x approaches 1 from the right, so h(x) has an infinite discontinuity at x = 1. Discontinuous at x = 4; h (4) is undefined and lim x→4

h(x) = 1 _ 3 , so h(x)

has a removable discontinuity at x = 4. 9. Discontinuous at x =

-6; f (x) approaches -25 as x approaches -6 from the left and 8 as

x approaches -6 from the right, so f (x) has a jump discontinuity at x = -6.

11 a. Find f (0.4).

f (w) =

7.4 _ w

f (0.4) =

7.4

_

(0.4)

= 18.5

The function is defined at w = 0.4. Find lim w→0.4

f (w). Construct a table that shows values of f (w)

for w-values approaching 0.4 from the left and from the right.

w 0.39 0.399 0.3999 0.4 0.4001 0.401 0.41

f (w) 18.974 18.546 18.505 18.495 18.454 18.049

lim w→0.4 f (w) = 18.5.

Because lim w→0.4 = f (0.4), f (w) is continuous as w = 0.4.

b. The function is discontinuous at w = 0. Construct a table that shows values of f (w) for w-values approaching 0 from the right. The function does not have to be examined for negative values of w since a wall cannot have a negative thickness.

w 0 0.001 0.01 0.1

f (w) 7400 740 74

Because f (0) is undefined and f (w) approaches ∞ as w approaches 0 from the right, f (w) is discontinuous at w = 0 and has an infinite discontinuity at w = 0.

c. Evaluate the function for several w-values in its domain. Use these points to construct a graph.

y

w

f (w) = 7.4w

13. 1 and 2 15. -1 and 0, 1 and 2 17. -1 and 0, 0 and 1, 2 and 3 19. no zeros on the interval 21. 2 and 3 23. From the graph, it appears that f (x) → ∞ as x → -∞ and f(x) → -∞ as x → ∞.

x -10,000 -1000 0 1000 10,000

f (x ) 5 · 10 12 5 · 10 9 -1 -5 · 10 9 -5 · 10 12

25. From the graph, it appears that f (x) → -4 as x → -∞ and f (x) → -4 as x → ∞.

x -10,000 -1000 0 1000 10,000

f (x ) -3.9981 -3.9811 –0.8333 –4.0191 –4.0019

27. From the graph, it appears that f (x) → 0 as x → -∞ and f (x) → 0 as x → ∞.

x -10,000 -1000 0 1000 10,000

f (x ) -8 · 1 0 -4 -8 · 1 0 -3 0 8 · 10 -3 8 · 10 -4



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29. From the graph, it appears that f (x) → -2 as x → -∞ and f (x) →-2 as x → ∞.

x -1000 -100 0 100 1000

f (x ) -1.99999 -1.9999 –5 -1.9999 -1.99999

31a.

[0, 200] scl: 20 by [0, 0.5] scl: 0.05

31b. Sample answer: The end behavior of the graph indicates that as the concentration of the catalyst solution is increased, the chemical reaction rate approaches 0.5.

x 0 10 100 1000 10,000

R(x ) 0 0.2273 0.4464 0.4941 0.4994

33. Sample answer: As x → ∞, the fraction will decrease, and q(x) will approach 0. 35. Sample answer: As x → ∞, the fraction will approach x _ x , so p(x) will approach 1. 37. Sample answer: As x → ∞, the fraction will decrease, and c(x) will approach 0. 39. Sample answer: As x → ∞, h(x) will increase without bound and approach ∞. 41. Sample answer: As the mass of the railway car continues to increase, the railway car’s kinetic energy will approach 0.

43 h (x) has an infinite discontinuity at x = 0 because h (0) is undefined and h (x) approaches ∞ as x approaches 0 from the left and the right.

w -0.1 -0.01 -0.001 0 0.001 0.01 0.1

h (x ) 136.4 13,636 1.36 × 1 0 6 1.36 × 1 0 6 13,636 136.4

From the graph, it appears that as x → -∞ and ∞, h (x) → 0.

x h (x )

-10,000 1.36 × 1 0 -8

-1000 1.36 × 1 0 -6

0 undefi ned

1000 1.36 × 1 0 -6

10,000 1.36 × 1 0 -8

The table supports this conjecture.

45.

[-5, 5] scl: 1 by [-10, 10] scl: 1discontinuity: infinite at x = -1, x = 2, and x = 3; end behavior: lim x→-∞

f(x) = 0, lim x→∞

f (x) = 0; zeros: x = 0

47.

[-20, 20] scl: 2 by [-20, 20] scl: 2

discontinuity: infinite at x = 3 and x = -6; end behavior:

lim x→-∞

h(x) = 4, lim x→∞

h(x) = 4; zeros: x = -3 and 1 _ 4

49.

[-20, 20] scl: 2 by [-20, 20] scl: 2

discontinuity: infinite at x = -4 and x = 3; end behavior: lim x→-∞

h(x) = -∞, lim x→∞

h(x) = ∞; zeros: x = -5, 4, and 6

51.

[-20, 20] scl: 2 by [-2500, 2500] scl: 500

x -10,000 -1000 -100 0 100 1000 10,000

f (x ) -1 · 10 16 -1 · 10 12 -1 · 10 8 -4 -9 · 10 7 -1 · 10 12 -1 · 10 16

53.

[-80, 80] scl: 8 by [-80, 80] scl: 8

end behavior: lim x→-∞

f (x) = 16,

lim x→∞

f (x) = 16

x -10,000 -1000 -100 0 100 1000 10,000

f (x ) 16.024 16.244 18.824 – 13.913 15.764 15.976

55a. f (n) = 3n + 4000

_ n 55b.

[0, 1000] scl: 100 by [0, 400] scl: 50

55c. $3; Sample answer: As n approaches ∞, f (n) approaches 3. Therefore, the average cost approaches $3 as the number of shirts sold increases.57a. Sample answer: If y

xx−4−8

8

12

4

4 8

f (x) = x 4+ 3x 3

+ 2x 2

n is even, f (x) approaches ∞ as x approaches -∞ and ∞.

57b. Sample answer: If n is y

xx

f (x) = x 5+ 2x 4

+ 3x 3

odd, f (x) approaches -∞ as x approaches -∞ and ∞ as x approaches ∞.

end behavior: lim x→-∞

f (x) = -∞,

lim x→∞

f (x) = -∞



Selected A

nswers and S

olutions

59. Infinite; f (0) is undefined, and f (x) approaches -∞ as x approaches 0 from the left and ∞ as x approaches 0 from the right.

61 The graph is continuous at x = 3, then by substitution: (3) 2 + a = (3)b + a x = 3

9 + a = 3b + a Simplify.

9 = 3b Subtract.

b = 3 Divide.

The graph is continuous at x = -3, then by substitution:

(-3)b + a = √ �� -b + 3 x = -3

(-3)3 + a = √ �� -3 + 3 b = 3

-9 + a = 0 Simplify.

a = 9 Subtract.

63. lim x→-∞

f (x) = ∞; because f is odd, f (-x) = -f (x). 65. lim x→-∞

f (x)

= ∞; with y-axis symmetry, f (x) = f (-x).

67.

[-10, 10] scl: 2 by [-10, 10] scl: 2

Even; the graph of h(x) is symmetric with respect to the y-axis.

h(-x) = √ ��

(-x ) 2 - 16

=

√ ��

x 2 - 16 = h(x)

69. no zeros

[-10, 10] scl: 2 by [-10, 10] scl: 2

71. D = (-∞, 1 - √ � 11 ) ∪

(1 - √ � 11 , 1 + √ � 11 ) ∪

(1 + √ � 11 , ∞)

73a. 100,000 73b. 7290 73c. 999,900,000

75. ⎡

⎢

⎣

-32

4

50

-30

-16

10 ⎤

�

⎦

77. infinite solutions 79. (2, 3, -1)

81. J 83. D

Lesson 1-4

1. f is increasing on (-∞, -0.5), decreasing on (-0.5, 1), and increasing on (1, ∞). 3. f is decreasing on (-∞, 2.5) and increasing on (2.5, ∞). 5. f is increasing on (-∞, 0) and increasing on (0, ∞). 7. f is increasing on (-∞, -2), decreasing on (0, 4), and increasing on (4, ∞). 9. f is constant on (-∞, -5), increasing on (-5, -3.5), and decreasing on (-3.5, ∞).

11a.

y

t0.5

4

8

12

1 1.5

( ) = -16 2+ 23.8 + 5 11b. about 13.9 ft

13–21. Your extrema values should be close to the actual extrema values given.

13 It appears that f (x) has an absolute maximum at x = -1.5, a relative minimum at x = 0, and an absolute maximum at

x = 1.5. It also appears that lim x→-∞

f (x) = -∞ and lim x→∞

f (x) =-∞, so we conjecture that this function has no absolute

minimum.

x -100 -2 -1.5 -1

f (x ) -1 · 1 0 9 -1 2.94 2

Because f (-1.5) > f (-2) and f (-1.5) > f (-1), there is an absolute maximum in the interval (-2, -1). The approximate value of the absolute maximum is 2.94.

x -0.5 0 0.5

f (x ) -0.063 -1 -0.063

Because f (0) < f (-0.5) and f (0) < f (0.5), there is a relative maximum in the interval (-0.5, 0.5). The approximate value of this relative minimum is –1.

x 1 1.5 2 100

f (x ) 2 2.94 -1 -1 · 1 0 9

Because f (1.5) > f (1) and f (1.5) > f (2), there is also an absolute maximum in the interval (1, 2). The approximate value of the absolute maximum is 2.94.

15. abs. min: (-3.71, -1334.6); rel. max: (0.11, 0.001); rel. min: (3.59, -1042.52) 17. rel. min: (-1, -4); rel. max: (1.08, 4.18) 19. rel. min: (1.2, -1.11); rel. max: (2, 0) 21. rel. max: (-1.14, -1.58); rel. min: (1.14, -8.42) 23. rel. max: (1.08, 0.04); rel. min: (-1.08, -10.04) 25. abs. min: (-1.38, -7.08) 27. rel. max: (1.11, 2.12); rel. min: (-0.17, -1.08) 29. rel. max: (-1.11, 1.32); rel. min: (0, -4) 31. rel. max: (-1.66, 3.43); rel. min: (0.93, -3.82) 33. r = 1.85 in.; h = 3.70 in. 35. 28 37. -16 39. -309 41. 472 43. -0.45 45. ≈ 0.219

47 a. f (0) = 12 and f (10) = 15.6

m = 15.6 - 12 _

10 - 0 Rate of change

= 3.6 _

10 or 0.36 million pounds more per year

b. f (5) = 11.925 and f (10) = 15.6

m = 15.6 - 11.925 __

10 - 5 Rate of change

= 3.675 _

5 or 0.735 million pounds more per year

49a. [5, 15]: 5; [15, 20]: 3; [25, 45]: 0.5 49b. The object is increasing in speed, or accelerating, along all three intervals. It is accelerating at the fastest rate for the interval [5, 15]. While it is very slowly accelerating for the interval [25, 45], it is still increasing its speed. 49c. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.

51a. y

2 4 6 8

10,000

20,000

30,000

40,000

xO

I(x) = -1.465x 5+ 35.51x 4

- 277.99x 3+ 741.06x 2 + 847.8x + 25362



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51b. about $1280.93; Sample answer: This value represents the average change in net personal income in the U.S. from 2000 to 2007. 51c. The rate of change was lowest from 2000 to 2004 at about $826.43 and highest from 2003 to 2007 at about $1711.44. 53. 14.46 in. × 14.46 in.

[0, 100] scl: 10 by [0, 5000] scl: 500

× 14.46 in.; Sample answer: The function for the surface area of the box in terms of the length of one side of the

base is f(w) = 2 w 2 + 12,096

_ w .

Graph the function. The absolute minimum occurred when w = 14.46 in.

55. Sample answer: 57. Sample answer:

y

x

f(x)

y

x

f(x)

59. Sample answer: 61. (-5, -1); maximum

63. (2.3, 8); minimum

65. (0, -5); minimum

y

xf (x) = ⎪x⎥

67. Sample answer: One reason for the variation in average rate of change may be that Simeon’s family encountered traffic later in their trip, thus slowing them down. Another may be that Simeon’s family traveled on residential roads for the first hour before entering a highway for the next three hours. The two instances on the graph where Simeon’s family appeared to not travel any distance may be as a result of the family needing to stop to eat or take a break. They may have also encountered an accident where traffic was completely stopped.

69. Sample answer: 31 0; Sample answer: When a function is constant on an interval, the y-coordinates of each point in the interval are the same. Because f (x) is constant on the interval [a, b], f (a) = f (b).

m = f (a) - f (b)

_ a - b

= 0 _

a - b or 0

So, the slope of the secant line is zero.

73.

[-1080, 1080] scl: 90 by [-1, 1] scl: 0.1

There are an infinite number of relative maxima and minima. The relative maximum is 1 and occurs at {x | x = 90 + 360n, n � �}. The relative minimum is -1 and occurs at {x | x = 270 + 360n, n � �}.

75. Sample answer: When a function is increasing on an interval, the average rate of change is positive. When a function is decreasing on an interval, the average rate of change is negative. When a function is constant on an interval, the average rate of change is zero.

77. Continuous; f (3) = 2, lim x→3 f (x) = 2, and lim x→3

f (x) = f (3).

79.

[-10, 10] scl: 1 by [-10, 10] scl: 1

Even; the graph of f (x) is symmetric with respect to the y-axis.f (-x) = |(-x ) 5 | = | x 5 |

= f (x)

81.

[-16, 16] scl: 2 by [-22, 18] scl: 2

neither; g(-x) = (-x ) 2

_ -x + 3

= x 2 _

-x + 3

83. (-∞, -3] � [3, ∞) 85. (5, 3, 2) 87. -6, 3 89. 7, -3 91. 9 - 7i 93. 200 or 400 amps 95. G 97. G

Lesson 1-5

1. D = {x | x ∈ �}, R = {y | y ∈ �}. The graph has a y-intercept at (0, 0) and x-intercepts for {x | 0 ≤ x ≤ 1, x ∈ �}. The graph has no symmetry. The graph has a jump discontinuity for {x | x ∈ �}. lim x→-∞


f (x) = ∞. The graph is

constant for {x | x ∉ �}. The graph increases for {x | x ∈ �}. 3. D = {x | x ∈ �}, R = {y | y ∈ �}. The graph has an intercept at (0, 0). The graph is symmetric with respect to the origin. The graph is continuous. lim x→-∞


f (x) = ∞. The graph is

increasing on (-∞, ∞).

5. D = {x|x ∈ �}, R = {y|y = c, c ∈ �}. If c = 0, all real numbers are x-intercepts. If c ≠ 0, there are no x-intercepts. The graph has a y-intercept at (0, c). If c ≠ 0, the graph is symmetric with respect to the y-axis. If c = 0, the graph is symmetric with respect to the x-axis, y-axis, and origin. The graph is continuous. lim x→-∞

f (x) = c

and lim x→∞

f (x) = c. The graph is constant on (-∞, ∞).

7. y

−8

−4

−4−8

8

4

4 8 x

f (x) g(x)

9. y

−8

−4

−4−8

8

4

4 8 x

f (x)

g(x)

y

−4

−8

−4−8

8

4

4 8 x



Selected A

nswers and S

olutions

11. y

−8

−4

8

4

−4−8 4 8 x

f (x)

g(x)

13. y

−16

−8

16

8

−8−16 8 16x

f (x)

g(x)

15. The graph of g(x) is the graph of f (x) translated 5 units to the right and g(x) = �x - 5�, or translated 5 units down and g(x) = �x� - 5. 17. The graph of g(x) is the graph of f (x) reflected in the y-axis and translated 5 units right when g(x) = �5 - x�, or reflected in the y-axis and translated 5 units up when g(x) = �-x� + 5.

19 Because there is a two-month delay, d(x) is the graph of p(x) translated two units to the right. Therefore, d(x) = 10 (x - 2) 3 - 70 (x - 2) 2 + 150(x - 2) + 2

21. The graph of g(x) is the graph of f (x) translated 5 units down; g(x) = |x| - 5.

23. The graph of g(x) is the graph of f (x) translated 1 unit to the right and 2 units down; g(x) = |x - 1| - 2.

25. f (x) = √ �

x ; the graph of g(x) y

−8

−4

8

4

4 8−4−8 x

g(x)

f (x)

is the graph of f (x) translated 8 units to the left and expanded vertically.

27. f (x) = �x�; the graph of g(x) y

−8

−4

8

4

4 8−8 −4 x

f (x)

g(x)

is the graph of f (x) translated 6 units to the right and expanded vertically.

29. f (x) = |x|; g(x) is the graph y

−8

−4

8

4

4 8−4−8 x

g(x)

f (x)

of f (x) translated 5 units to the left, expanded vertically, and reflected in the x-axis.

31. f (x) = √ �

x ; g(x) is the graph of y

x

f (x)

g(x)

f (x) translated 3 units to the left and compressed vertically.

33. y

−8

−4

8

4

4 8−8 −4 x

35. y

−4

8

4

4 8−4−8 x

37. y

x

39a. The graph of c (x) is the graph of f (x) compressed vertically and translated 1.99 units up. 39b. c (x) = 2.49 + 0.05�x�

39c.

[0, 17] scl: 4 by [0, 5] scl: 1

39d. Yes; the plans will equal each other at 10 minutes.

41. y

−8

4

8

−4

−4−8 4 8 x

f (x)

y

−8

4

8

−4

−4−8 4 8 x

g(x)

y

−8

4

8

−4

−4−8 4 8 x

h(x)

43. y

−8

4

8

−4

−4−8 4 8 x

f (x)

y

−8

4

8

−4

−4−8 4 8 x

g(x)

y

−8

4

8

−4

−4−8 4 8 x

h(x)



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45. y

−32

16

32

−16

−4−8 4 8 x

f (x)

y

−32

16

32

−16

−4−8 4 8 x

g(x)

y

−32

16

32

−16

−4−8 4 8 x

h(x)

47a. f (x) = ⎧

⎨

⎩

0.4�x� + 2.50 if �x� = x 0.4�x+ 1� + 2.50 if �x� < x

47b. Taxi Cost

Pric

e (d

olla

rs)

1.702.503.304.104.905.706.507.30

Fare Units1 2 3 4 5 6 7 8 9 10

y

x

47c. The graph of f (x) is translated 0.5 unit up.

49. A vertical shift of 5 is f (x) + 5 or 1 _ x + 5.

A horizontal shift of -7 is f (x - (-7)) or 1 _

x + 7 .

A vertical expansion by a factor of 2 is 2f (x) or 2 ( 1 _ x ) = 2 _ x .

Combining the attributes, g(x) = 2 f (x - (-7)) + 5 or 2 _

x + 7 + 5.

y

−8

4

8

−4

−4−8 4 8 x

g(x) =2

x + 7+ 5

51. translated one unit to the left; translated one unit down 53. translated 2 units to the left; expanded vertically; translated 7

units down 55. g(x) = 2 _

x - 3 + 4 57. g(x) = 4 √ �� x + 4 - 6

59a. g(x) = 1.12f (x) 59b. There is no affect on f (x). While the opening is delayed, the number of days after the opening, which determines the domain of the function, is unaffected. 59c. g(x) = f (x) - 0.45 61. f (x) = x 3 ; Sample answer: The graph of g(x) is the graph of f (x) translated 4 units to the right, expanded vertically, reflected in the x-axis, and translated 2 units up. 63. f(x) =

√

� x ;

Sample answer: The graph of g(x) is the graph of f (x) translated 3 units to the right, reflected in the x-axis, and translated 5 units up.

65. y

−16

8

16

−8

−8−16 8 16x

67. y

4

8

-4

-8

4-8 8 x

69. g(x) = -

24 _

√ �� x + 6 + 18 71. g(x) = 8

_ √ �� 2x + 7

+ 4

y

−16

8

16

−8

−8−16 8 16x

g(x) = 24_√x + 6

18+- y

−16

8

16

−8

−8−16 8 16x

g(x) = 8√2x + 7

4+

73. Sample answer: Both; for the greatest integer function, a shift of a units left is identical to a shift of a units up. 75. Sample answer: Order is important because different graphs can be obtained depending on the order the transformations are performed. For example, if (a, b) is on the original graph and there is a translation 6 units up and then a reflection in the x-axis, the resulting point on the transformation will be (a, -b - 6). However, if (a, b) is reflected in the x-axis first and then translated 6 units up, the resulting point will be (a, -b + 6). 77. Sometimes; sample answer: f (x) = x 3 is an odd function and f (-x) ≠ -| f (x)| when x = -1. However, f (x) = 0 is an odd function and f (-x) = -| f (x)| for all x.

79 Sample answer: There are an infinite number of functions that meet these criteria. Any values of a, b, and c for g(x) = a √ �� x + b + c are valid such that f (-2) = -6. To simplify this process, let a = 1.

f (x) = a √ �� x + b - c -6 = √ �� -2 + b - c c - 6 = √ �� -2 + b (c - 6 ) 2 = -2 + b (c - 6 ) 2 + 2 = b Any values of b and c that satisfy this equation will be valid. If

c = 8, b = 6 and g(x) = √ �� x + 6 - 8.83. 6 85. 0; Sample answer: As x → ∞, the denominator of the fraction will increase and the value of the fraction will approach 0, so g(x) will approach 0. 87. 1; Sample answer: As x → ∞, the

fraction will get closer and closer to x _ x , so p(x) will approach 1.

Taxi Cost

Pric

e (d

olla

rs)

2.503.304.104.905.706.507.308.10

Fare Units1 2 3 4 5 6 7 8 9 10

y

x



Selected A

nswers and S

olutions

89. y-intercept = 0; zeros: -1, 0, 2;

x 3 - x 2 - 2x = 0 x( x 2 - x - 2) = 0 x(x - 2)(x + 1) = 0x = 0 or x - 2 = 0 or x + 1 = 0 x = 2 x = -191. about 118.60 93. B 95. G

Lesson 1-6

1. ( f + g)(x) = x 2 + √ �

x + 4; D = [0, ∞); ( f - g)(x) = x 2 - √ �

x + 4; D

= [0, ∞); ( f · g)(x) = x 5 _ 2 + 4 x

1 _ 2 ; D = [0, ∞); (

f _ g ) (x) =

x 2 + 4 _

√

� x ; D = (0, ∞)

3. ( f + g)(x) = x 2 + 6x + 8; D = (-∞, ∞); ( f - g)(x) = x 2 + 4x + 4;

D = (-∞, ∞); ( f · g)(x) = x 3 + 7 x 2 + 16x + 12; D = (-∞, ∞); ( f _ g ) (x)

= x + 3; D (-∞,-2) ∪ (-2,∞) 5. ( f + g)(x) = x 2 + 10x; D = (-∞, ∞); ( f - g)(x) = x 2 - 8x; D = (-∞, ∞); ( f · g)(x) = 9 x 3 + 9 x 2 ; D =

(-∞, ∞); ( f _ g ) (x) =

x + 1 _

9 ; D = (-∞,0) ∪ (0,∞) 7. ( f + g)(x) = x 3 +

x + 6 _ x ; D = (-∞,0) ∪ (0,∞); ( f - g)(x) = - x 3 - x + 6 _ x ;

D = (-∞,0) ∪ (0,∞); ( f · g)(x) = 6 x 2 + 6; D = (-∞,0) ∪ (0,∞);

( f _ g ) (x) = 6

_ x 4 + x 2

; D = (-∞,0) ∪ (0,∞) 9. ( f + g)(x) = 1 _

√ �

x - 4 √

� x ;

D = (0, ∞); ( f - g)(x) = 1 _

√ �

x - 4 √

� x ; D = (0, ∞); ( f · g)(x) = 4; D = (0, ∞);

( f _ g ) (x) = 1 _

4x ; D = (0, ∞) 11. ( f + g)(x) = √ �� x + 8 + √ �� x + 5 - 3;

D = [-5, ∞); (f - g)(x) = √ �� x + 8 - √ �� x + 5 + 3; D = [-5, ∞);

( f · g)(x) = √ ��

x 2 + 13x + 40 - 3 √ �� x + 8 ; D = [-5, ∞); ( f _ g ) (x)

=

√ ��

x 2 + 13x + 40 + 3 √ �� x + 8 ___

x - 4 ; D = [-5,4) ∪ (4,∞)13a. ( f + g)(x) = 40x

+ 550; x ≥ 0 13b. ( f + g)(x) represents all factors that influence the budget. 13c. $710; The budget for 4 weeks.

15. [ f ◦ g](x) = 8x - 19; [ g ◦ f ](x) = 8x - 20; [ f ◦ g](6) = 29

17. [ f ◦ g](x) = - x 4 - 2 x 3 - 3 x 2 - 2x + 7; [ g ◦ f ](x) = x 4 -

17 x 2 + 73; [ f ◦ g](6) = -1841 19. [ f ◦ g](x) = - x 6 - 2 x 3 + 2;

[ g ◦ f ](x) = - x 6 + 9 x 4 - 27 x 2 + 28; [ f ◦ g](6) = -47,086

21. [ f ◦ g](x) = 1 _

x 2 - 3 for x ≠ ± √ � 3 23. [ f ◦ g](x) = |x|

25. [ f ◦ g](x) =

5 √ �� 6 - x _

6 - x for x < 6 27. [ f ◦ g](x) = |x + 2|

29 a. Since the denominator cannot be zero, 1 - c 2 _

v 2 ≠ 0. Thus,

v ≠ c. Because velocity cannot be negative, D = {v | 0 ≤ v < c, v ∈ �}.

b. m(10) = 100 __

√

��

1 -

10 2 _

(3 × 10 8 ) 2

= 100

m(10,000) = 100 __

√

��

1 -

10,000 2 _

(3 × 10 8 ) 2

= 100.000000056

m(1,000,000) = 100 __

√

��

1 -

(1 × 10 6 ) 2 _

(3 × 10 8 ) 2

= 100.00055556

c. lim v→c m(v) = ∞, since as m approaches c, v 2

_ c 2

approaches 1,

making √

��

1 - v 2 _

c 2 very small. 100 divided by a very small

number results in a very large quotient. d. Sample answer: Let g(v) = 1 - v 2

_ c 2

and f (v) = 100 _

√ �

v .

m(v) = f [g(v)]

= 100 _

√

��

1- v 2 _

c 2

31. Sample answer: f (x) = 6 _ x - 8; g(x) = x + 5 33. Sample answer: f (x) = �-3x�; g(x) = x - 9 35. Sample answer: f (x) = x 3 ; g(x) =

√ �

x + 4 37. Sample answer: f (x) = 8 _ x 2

; g(x) = x - 5

39. Sample answer: f (x) = x + 6

_ √

� x ; g(x) = x - 1

41a. h[ f (x)]; The bonus is found after subtracting 300,000 from the

total sales. 41b. $6000 43. Sample answer: f (x) = x - 4 _

x 2 + 1 ;

g(x) = √ �� x - 1 45. Sample answer: f (x) = 8 _

x + 2 + 5 √

� x ; g(x) = x 2

47. Sample answer: f (x) = x − 4 _

2x − 9 +

√ ��

4 _

x - 4 ; g(x) = x + 4

49. f (0.5) = -0.75, f (-6) = 22, f (x + 1) =

x 2 + 4x + 1 51. f (0.5) = -3.6, f (-6) = -300, f (x + 1) =

-9 x 2 - 22x + x + 1

_ 10

- 62 _

5

53 The domain of h(x) = √ �

x + 3 is {x | x ≥ 0, x��} and of g(x) = x 2 - 6 is all real numbers. Therefore the domain of g ◦ h is {x | x ≥ 0, x��}.

[g ◦ h](x) = g[(h(x)]

= g( √ �

x + 3)

= [( √ �

x + 3 ) 2 - 6]

= [x + 6 √ �

x + 9 - 6]

= x + 6 √ �

x + 3

Since the domain of f (x) = x + 8 is all real numbers, the domain of f ◦ g ◦ h is {x | x ≥ 0, x��}.

[ f ◦ g ◦ h](x) = f ( g[(h(x)])

= f (x + 6 √ �

x + 3)

= (x + 6 √ �

x + 3) + 8

= x + 6 √ �

x + 11

Therefore, [f ◦ g ◦ h](x) = x + 6 √ �

x + 11 for x ≥ 0.

55. [ f ◦ g ◦ h](x) = √

��

1 _ x 2

+ 2 for x ≠ 0 57a. g(x) = x 2 + 4

57b. g(x) = 4x + 8 59a. g(x) = 1 _ 4x

59b. g(x) =

1 _

32 x

1 _ 3

61. 1 63. 0 65. 0

67a. {m | m > 0, m ∈ �}; The molar mass of the gas cannot be negative or zero. 67b. about 7.22 m/s 67c. The velocity will decrease. 67d. Sample answer: v(m) = f [ g(x)];

f (m) = √ �

m ; g(m) = (24.9435)(303)

__ m 69. Sample answer:

f (x) = √ �

x ; g(x) = x 2 + 8; h(x) = x - 5 71. Sample answer: f (x) = 4 _ x ;

g(x) = x 2 + 1; h(x) = √ �

x + 3 73. [ f ◦ g](x) = x for

x ≥ - 19; [ g ◦ f ](x) = | x + 4 | - 4 75. [ f ◦ g](x) = √

��

√ ��

9 - x 2

for - 3 ≤ x ≤ 3; [g ◦ f ](x) = √ �� 9 - x for 0 ≤ x ≤ 9

77. [ f ◦ g](x) = 24 - 6x _ 12 - x

for x ≠ 4 and x ≠ 12; [g ◦ f ](x) = 4x + 2

_ 4x - 1

for x

≠ -

1 _

2 and x ≠ 1 _

4

79. y

xO

h(x) (h - f )(x)

f (x)

81. y

x

(h + g)(x)

h(x)

g(x)



Sel

ecte

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83. {x | -10 ≤ x ≤ -4 or 2 ≤ x ≤ 8, x ∈ �} 85. {x | 8 ≤ x ≤ 10, x ∈

�} 87. odd 89. even

91. Sample answer: f (x) = 1 _ x , x ≠ 0

93. Sample answer: This can occur when the range of g(x) is a subset of the domain of f (x). For example, when f (x) = √

� x and

g(x) = x 2 + 4, the range of g(x) is all real numbers greater than or equal to 4. This range is a subset of the domain of f (x), which is all real numbers greater than or equal to 0. Thus, f [ g(x)] exists for all real numbers.

95 The domain of h(x) = 4 _ x is {x | x ≠ 0, x � �}. To evaluate g ◦ h,

you must be able to evaluate g(x) = √ �� x + 1 for each of these h(x)-values, which is only possible when h(x) ≥ -1. That is,

when 4 _ x ≥ -1. Since we do not know if x is positive or

negative, we cannot multiply both sides of this inequality by

x to solve for x. Instead, rewrite the inequality as 4 _ x - 1 ≥ 0.

When x = -4, 4 _ x + 1 = 0 and when x = 0, 4 _ x + 1 is undefined.

Test intervals to determine the solutions of 4 _ x + 1 ≥ 0 are

therefore x ≤ -4, -4 ≤ x < 0, and x > 0. Test points in these

intervals to determine the solutions of the inequality. Since

4 _ x + 1 ≥ 0 is true for x = -5 and x = 1, but not for x = -1, the solution is x ≤ -4 or x > 0. Therefore the domain of g ◦ h is {x | x ≤ -4 or x ≥ 0, x � �}. To find f ◦ g ◦ h, you must be able

to evaluate f(x) = 1 _

x - 2 , for each of these g[h(x)]-values,

which can only be done for g[h(x)] ≠ -2. This means that we must exclude from the domain those values for

which √ ��

4 + x _ x = - 2.

√ ��

4 + x _ x = -2

4 + x _ x = 4

4x = 4 + x 3x = 4

x =

4 _

3

Therefore, the domain of f ◦ g ◦ h is {x | x ≤ -4 or 0 < x < 4 _

3 or x > 4 _

3 � �}. In interval notation, the domain is

(-∞, -4] ∪ (0, 4 _ 3 ) ∪ ( 4 _

3 , ∞) .

97. Sample answer: Order is important because there usually will be different outputs for f [ g(x)] and g[ f (x)]. For example, if f (x) = x 2 and g(x) = x + 2, f [g(x)] = x 2 + 4x + 4 and g[ f (x)] = x 2 + 2.

99.

[-10, 10] scl: 1 by [-10, 10] scl: 1

rel. max: (0, 4); rel. min: (1, 3)

101.

[-10, 10] scl: 1 by [-10, 10] scl: 1

abs. min: (-0.75, -2.11)

103. -1.73, 1.73105a.

RBI

135

0

140

145

150

155

160

HR40 45 50 55 60

105b. D = {43, 48, 54}, R = {137, 146, 148, 150, 156} 105c. No; each of the domain values 48 and 54 is paired with two different range values. 107. G

Lesson 1-7

1. no 3. no 5. yes 7. yes 9. yes 11. yes 13. no 15. no

17. no 19. yes; f -1 (x) = 4 _

x + 1 ; x ≠ -1 21. yes; f -1 (x) =

8 − 36 _

x 2 ; x > 0 23. yes; f -1 (x) =

8x + 3 _

x − 6 ; x ≠ 6 25. no

27. f [ g(x)] = -6(3 - x)

_ 6 + 3 =

-18 + 6x _ 6 + 3 = -3 + x + 3 = x;

g[ f (x)] = 3 - (-6x + 3)

__ 6 = 6x _

6 = x 29. f [ g(x)] = -3 (

√ ��

5 - x _ 3 )

2

+ 5

= -3(5 - x)

_ 3 + 5 = x - 5 + 5 = x; g[ f (x)]= √

��

5 - (-3 x 2 + 5)

__ 3 = √

��

3 x 2 _

3

= √ �

x 2 = x 31. f [ g(x)] = 2 ( 3

√ ��

x + 6 _

2 )

3

- 6 = 2(x + 6)

_ 2 - 6 = x + 6 - 6

= x; g[ f (x)] = 3

√

��

2 x 3 - 6 + 6

_ 2 =

3

√ ��

2 x 3 _

2 =

3

√ �

x 3 = x

33. f [ g(x)] = ( √ �� x + 8 - 4) 2 + 8 √ �� x + 8 - 32 + 8 = x + 8 - 8 √ �� x + 8

+ 16 + 8 √ �� x + 8 - 24 = x + 24 - 24 = x; g[ f (x)] =

√ ��

x 2 + 8x + 8 + 8 - 4 = √ ��

x 2 + 8x + 16 - 4 = √ ��

(x + 4) 2 -

4 = x + 4 − 4 = x 35. f [ g(x)] = 4

_ x − 1

+ 4 _

4 _

x − 1 =

4 + 4x − 4 _

x − 1 ÷

4 _

x − 1 =

4 + 4x − 4 _

4 = 4x _

4 = x; g[ f (x)] = 4

_ x + 4

_ x − 1 = 4

_ x + 4 − x _ x

=

4 ÷ 4 _ x = x

37 a. If y = 0.5m x 2 , the inverses comes from x = 1 _ 2 m y 2 .

2x = m y 2 Original equation.

y 2 = 2x _ m Divide.

y = √ �

2x _ m Take the square root and reject the negative root.

g(x) = √ �

2x _ m Replace y with g ( x ).

g(x) = velocity in m/s, x = kinetic energy in J and m = mass in kg.

b. f [ g(x)] = f ( √ �

2x _ m ) = 1 _ 2 m ( √

�

2x _ m ) 2

=

1 _

2 m ( 2x _ m ) = x; g[f(x)] = g

( 1 _ 2 m x 2 ) = ( √

��

2 ( 1 _

2 m x 2 ) _ m ) = √

� x 2 = x; because f [g(x)] = g [ f (x)]

= x, the functions are inverses.However they are not inverse functions until the domain of f(x) is restricted to D = [0, ∞).




Selected A

nswers and S

olutions

c.

[0, 10] scl: 1 by [0, 10] scl: 1

39. y

x

−4

−8

−4−8

8

4

4 8

y = (x)g-1

y = g(x)

41. y

x

−4

−8

−4−8

8

4

4 8

y = (x)g-1

y = g(x)

43. y

x

−4

−8

−4−8

8

4

4 8

y = (x)g-1

y = g(x)

45a. Sample answer: The graph of the function is linear, so it passes the horizontal line test. Therefore, it is a one-to-one function and it has an inverse; f −1 (x) = x _

0.66 .

45b. x represents the value of the currency in U.S. dollars, and f −1 (x) represents the value of the currency in Euros. 45c. x ≥ 0; You cannot exchange negative money. 45d. 151.52 47. f -1 exists. 49. f -1 exists. 51. f -1 does not exist. 53. f -1 exists.

x 8 7 6 5 4 3

f -1 (x) -10 -9 -8 -7 -6 -5

55 Since the vertex of the parabola is at (5, 0), either x ≤ 5 or x ≥

5 will make the inverse one-to-one. Choose x ≥ 5.

x = (y - 5) 2 Original function.

√ �

x = y - 5 Interchange x and y.

√ �

x + 5 = y Add 5 to each side.

Thus, f -1 (x) = √ �

x + 5.

57. Sample answer: x ≥ 0, f -1 (x) = √ �� x + 6 59. f: D = {x | x ≥ 6, x ∈ �}, R = {y | y ≥ 0, y ∈ �}; f -1 : D = {x | x ≥ 0, x ∈ �}, R = {y | y ≥ 6, y ∈ �} 61. f: D = {x | x ≠ 4, x ∈ �}, R = {y | y ≠ 3, y ∈ �}; f -1 : D = {x | x ≠ 3, x ∈ �}, R = {y | y ≠ 4, y ∈ �}63a. Sample answer: f (x) = 272.6x + 1710.4

63b. f −1 (x) = x − 1710.4 _

272.6 ; x represents the number of nesting pairs

and f −1 (x) represents the number of years after 1984. 63c. 1996

65. f −1 (x) =

⎧

⎨

⎩

- √ �

x if 16 ≤ x2.5 - 0.5x if 13 > x

y

x

−16

−32

−16−32

32

16

16 32

f (x)

(x)f -1

67a. v -1 (x) = 3x; represents the formula for the flow rate of the

gas. 67b. 5000 ft _ s 67c. ≈ 35.3 ft 3 _ s 69. [ f -1 ◦ g -1 ](x) =

x + 2

_ 16

71. [f ◦ g ] -1 (x) = x − 44 _

16 73. (f · g) -1 (x)

√ �� x + 49 - 5 __

4 ,

x ≥ -1.25 75. [ f -1 ◦ g -1 ](x) = √ ��

x 2 + 3 for x ≥ 0 77. [ f ◦ g ] -1 (x) = x + 3 for x ≥ 1 79. ( f · g -1 )(x) = x 4 + 5 x 2 + 4 for x ≥ 0

81a. 81b. D = {x | x ∈ �}, R = { y | y ∈ positive multiples of 0.4}

Cost

(dol

lars

)

0.40

0.81.21.6

22.42.83.23.6

y

Time (minutes)1 2 3 4 5 6 7 8 9 x

C(x) =

�0.4x�

81c.

Tim

e (m

inut

es)

10

23456789y

Cost (dollars)0.8 1.6 2.4 3.2 x

81d. D = {x | x ∈ positive multiples of 0.4}; R = {y | y ∈ �} 81e. The inverse gives the number of possible minutes spent using the scanner that costs x dollars.

83 The inverse of (x, y) is (y, x). Thus, if f contains a zero at 6, (6, 0) is a point on the graph. By definition of an inverse, the graph of the inverse must contain (0, 6), which is its y-intercept.

85. Sample answer: False; constant functions are linear, but they do not pass the horizontal line test. Therefore, constant functions are not one-to-one functions and do not have inverse functions.

87. Sample answer: Yes; one function that does this is f (x) = 1 _ x .

Even though both limits approach 0, they do it from opposite sides of 0 and no x-values ever share a corresponding y-value. Therefore, the function passes the horizontal line test. 89. Sample

answer: If f (x) = x 2 , f does not have an inverse because it is not one-to-one. If the domain is restricted to x ≥ 0, then the function is

now one-to-one and f -1 exists; f -1 (x) = √ �

x . 91. [ f ◦ g](x) = 1 _ 2 x -

4 for {x | x ∈ �}; [g ◦ f ](x) = 1 _ 2 x - 1 for {x | x ∈ �} 93a. expanded

horizontally 93b. translated 5 units to the right and 2 units down 93c. expanded vertically, translated 6 units up 95a. compressed horizontally 95b. translated 5 units to the right 95c. compressed horizontally, translated 4 units down 97. (1, -4, 4) 99. (10, -7, 1) 101. D 103. A

Chapter 1 Study Guide and Review

1. true 3. true 5. false; end behavior 7. true 9. true 11. function 13. function 15. 14 17. D = {x | x ∈ �} 19. D = {a | a ≠ -5, a ∈ �} 21. D = [-8, 8], R = [0, 8]

23. -9; 9 _ 4 25. 0; 0, 4, -4 27. continuous at x = 4; The function is

defined when x = 4. The function approaches 4 when x approaches 4 from both sides, and f (4) = 4. 29. continuous at x = 0; The function is defined when x = 0.



Sel

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and

Sol

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ns

The function approaches 0 when x approaches 0 from both sides, and f (0) = 0. continuous at x = 7; The function is defined when x = 7. The function approaches 0.5 when x approaches 7 from both sides, and f (7) = 0.5. 31. continuous at x = 1; The function is defined when x = 1. The function approaches 2 when x approaches 1 from both sides, and f (1) = 2. 33. From the graph, it appears that as x → ∞, f (x) → 0; as x → -∞, f (x) → 0. 35. f is decreasing on (-∞, -3), increasing on (-3, -1.5), decreasing on (-1.5, 0.5), and increasing on (0.5, ∞); rel. min. at (-3, 3), rel. max. at (-1.5, 6) and rel. min. at (0.5, -7). 37. 0

39. f (x) = x 2 ; g(x) is the 41. f (x) = �x�; g(x) is the graph of f (x) reflected in the x-axis and translated 6 units to the right and 5 units down.

graph of f (x) compressed vertically by a factor

of 1 _ 4 and translated

3 units up.

y

x

−4

−8

−4−8

8

4

4 8

f (x) = x 2

g(x) = -(x - 6)2- 5

y

x

f (x) = �x�

1_4

x + 3

43. The graph is reflected in the x-axis and translated 4 units right

and 1 unit up; g(x) = - √ �� x - 4 + 1. 45. ( f + g)(x) =

4 x 2 + 5x - 2; D = (-∞, ∞); (f - g)(x) = 4 x 2 - 5x; D =

(-∞, ∞); (f � g)(x) = 20 x 3 - 4 x 2 - 5x + 1; D = (-∞, ∞); ( f _ g ) (x)

= 4 x 2 - 1 _

5x - 1 ; D = (-∞, 1 _

5 ) ∪ ( 1 _

5 , ∞) 47. (f + g)(x) =

x + 1 _

x 2 ;

D = (-∞, 0) ∪ (0, ∞); (f - g)(x) = x - 1 _

x 2 ; D = (-∞, 0) ∪ (0, ∞);

( f � g)(x) = 1 _ x 3

; D = (-∞, 0) ∪ (0, ∞); ( f _ g ) (x) = x; D = (-∞, 0) ∪ (0, ∞)

49. x 2 - 8x + 23; x 2 + 2x + 3; 11

51. [f ◦ g](x) = 1 _

2x - 9 for x ≠ 9 _

2 53. no 55. yes

57. f -1 (x) = 3 √ �� x + 2 59. h -1 (x) = 1 _

4 x 2 - 3, x ≥ 0

61a. f(x) = ⎧

⎨

⎩

39.99 if 0 ≤ x ≤ 500

39.99 + 0.2(x - 500) if x > 500

61b. $39.99; $49.99

61c. y

x120

20

40

60

80

240 360 480

63. No; sample answer: At the time of her promotion, her income had a jump discontinuity.

65. 67a. A(x) = 6.4516x cm 2

Spee

d (m

/s)

0

12

24

36

48

60

72

Time (s)1 2 3 4 5 6

v(t) = √(9.8t)2+ 49

67b. A -1 (x) = 1 _

6.4516 x in 2

Chapter 1 Connect to AP Calculus

1. 5 3. Sample answer: 5 5. Sample answer: The secant line appears to become a line tangent to the graph of f. 7. Sample answer: The exact rate of change is 5 which is equivalent to what the other values appeared to be approaching. 9. Sample answer: To find the rate of change of a function f (x) at the point x, find the difference quotient, and simplify the expression. Substitute h = 0 into the simplified expression, and the result is the rate of change of the function at x.

Polynomial and Rational Functions CHAPTER 2CHCHAPAPTETERR 22

Chapter 2 Get Ready

1. (x - 4)(x + 5) 3. (2x - 3)(x - 7) 5. (4x - 5)(3x + 7) 7. 4x + 79. y

x

−4

−8

−4−8

8

4

4 8

f (x) = -2

11. y

x

−4

−8

−12

−4−8 4 8f (x) = -x 2

+ x - 6

13. y

x

−4

−8

−4−8

8

4

4 8

f (x) = 3x 2- x - 2

15. {x | x ≤ 6, x ∈ �}; (-∞, 6] 17. {x | -2 < x < 9, x ∈ �}; (-2, 9) 19. {x | x < -4 or x > 5, x ∈

�}; (−∞, -4) ∪ (5, ∞) 21. {x | 9.99 < x < 19.99, x ∈ �}; (9.99, 19.99)

Lesson 2-1

1. y

xO−2−4

8

12

16

4

2 4

f (x) = 5x 2

D = (-∞, ∞), R = [0, ∞); intercept: 0; lim x→-∞

f(x) = ∞ and lim x→∞

f(x) = ∞;

continuous for all real numbers; decreasing: (-∞, 0); increasing: (0, ∞)



Selected A

nswers and S

olutions

3. y

x

h (x) = -x 3

D = (-∞, ∞), R = (-∞, ∞); intercept: 0; lim x→-∞

f (x) = ∞ and lim x→∞

f (x) = -∞;

continuous for all real numbers; decreasing: (-∞, ∞)

5. y

x

g (x) = x 91_3



f (x) = ∞;

continuous for all real numbers; increasing: (-∞, ∞)

7. y

x

f (x) = - x71_2



f (x) = -∞;

continuous for all real numbers; decreasing: (-∞, ∞)

9. y

x−4−8

8

12

16

4

4 8

f (x) = 2x -4

D = (-∞, 0) ∪ (0, ∞), R = (0, ∞); no intercepts; lim x→-∞

f (x) = 0 and lim x→∞

f(x)

= 0; infinite discontinuity at x = 0; increasing: (-∞, 0); decreasing: (0, ∞)

11. y

x

−4

−8

−4−8

8

4

4 8

f (x) = -8x -5

D = (-∞, 0) ∪ (0, ∞), R = (-∞, 0) ∪ (0, ∞); no

intercepts; lim x→-∞


f (x) = 0; infinite discontinuity at x

= 0; increasing: (-∞, 0) and (0, ∞)

13. y

x

−4

−8

−4−8

8

4

4 8

f (x) = x -92_5

D = (-∞, 0) ∪ (0, ∞), R = (-∞, 0) ∪ (0, ∞); no intercepts; lim x→-∞

f (x) = 0

and lim x→∞

f (x) = 0; infinite

discontinuity at x = 0; increasing: (-∞, 0) and (0, ∞)

15. y

x

−4

−8

−4−8

8

4

4 8

h(x) = 3_4

x -3

D = (-∞, 0) ∪ (0, ∞),

R = (-∞, 0) ∪ (0, ∞); no intercepts; lim x→-∞


f (x) = 0; infinite discontinuity at x

= 0; decreasing: (-∞, 0) and (0, ∞)

17a. D = (0, ∞), R = (0, ∞)

17b. V(r)

r2 4 6 8

20

40

60

80

V (r) =4_3

r 3π

19. y

x

−4

−8

−4−8

8

4

4 8

f (x) = -6x 1_5


f (x) = ∞

and lim x→∞

f (x) = -∞; continuous for all real numbers; decreasing: (-∞, ∞)

21. y

x4 8 12 16

4

8

12

16

f (x) = 10x-

1_6

D = (0, ∞), R = (0, ∞); no intercepts; lim x→∞

f (x) = 0; continuous

on (0, ∞); decreasing: (0, ∞)

23. y

x

h(x) =3_4

x3_5


f (x) = -∞ and

lim x→∞

f (x) = ∞; continuous for all

real numbers; increasing: (-∞, ∞)

25. y

x

f (x) = x- 2_3


f (x) = 0 and

lim x→∞

f (x) = 0; infinite discontinuity

at x = 0; increasing: (-∞, 0); decreasing: (0, ∞)

27. yx

h(x) = -4x 7_4

D = [0, ∞), R = (-∞, 0]; intercept: 0;

lim x→∞

f (x) = -∞; continuous on

[0, ∞); decreasing: (0, ∞)



Sel

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29. y

x

h(x) = 2_3

x- 8_5


f (x) = 0 and

lim x→∞

f (x) = 0; infinite discontinuity

at x = 0; increasing: (-∞, 0); decreasing: (0, ∞)

31a.

[0, 10] scl: 1 by [0, 150000] scl: 10000

31b. y = 0.77 x 5.75 31c. about 235,906,039

33a.

[0, 45] scl: 5 by [40, 50] scl: 1

33b. f (x) = 55.14 x -0.0797 33c. about 39.54°F

35. y

x

−6

−12

12

6

200 400−200−400

g(x) = -2 5√ 1024 + 8x

D = (-∞, ∞), R = (-∞, ∞); x-intercept: -128, y-intercept: -8;

lim x→-∞


f (x) = -∞; continuous for all real

numbers; decreasing: (-∞, ∞)

37 Evaluate the function for several x-values in its domain.

x 12 _ 7 2 3 4 5 6 7

f (x ) 4 5.4 7 8 8.8 9.5 10.1

Use these points to construct a graph.

y

x

h(x) = 4 + √7x - 12

Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 7x = 12. Solve for x when the radicand is 0 to find the restriction on the domain.

0 = 7x - 12

12 = 7x

12 _

7 = x

D = ⎡

⎢

⎣

12 _

7 , ∞

⎞

⎥

⎠

, R = [4, ∞); no intercepts; lim x→∞

f (x) = ∞;

continuous on ⎡

⎢

⎣

12 _

7 , ∞

⎞

⎥

⎠

; increasing: ⎛

⎪

⎝

12 _

7 , ∞

⎞

⎥

⎠

39. yx

−20

−40

−60

−80

−4−8 4 8

f (x) = - 3√(25x - 7) 2 - 49

D = (-∞, ∞), R = (-∞, -49.00]; y-intercept: -52.66; lim x→-∞

f (x) = -∞

and lim x→∞

f (x) = -∞; continuous for all real numbers; increasing: (-∞, 0.28); decreasing: (0.28, ∞)

41. y

x

−4

−8

8

4

6 12 18 24

g (x) = √22 - x - √3x - 3

D = [1, 22], R = [- √ � 63 , √ � 21 ]; x-intercept: 6.25; continuous on [1, 22]; decreasing: (1, 22)

43a. about 6.89 Mcal 43b. about 7.18 lb 45. no solution 47. 7 49. 2, 4 51. 8 53. no solution 55. 10 57. Yes; sample answer: The function follows the form f (x) = a x n , where n is a positive integer. In this case, a = –2a and n = 4. 59. Yes; sample answer: The function follows the form f (x) = a x n , where n is a

positive integer. In this case, a = 7 _ 3 and n = ab. 61. No; sample

answer: The function is not a monomial function because the exponent for x is negative. 63. x ≥ −2 65. x ≤ −6 67. x ≥ 5

69 a. Volume vs. Pressure

Pres

sure

(atm

osph

eres

)

0

1

2

3

4

Volume (liters)1 2 3 4 v

P(v)

.

b. Use the power regression function on the graphing calculator to find values for a and n.

P(v) = 3.62 v -1

c. Sample answer: Yes; the problem states that the volume and pressure are inversely proportional, and in the power function, the exponent of the volume variable is -1.



Selected A

nswers and S

olutions

d. Graph the regression

[0, 5] scl: 0.5 by [0, 5] scl: 0.5

equation using a graphing calculator. To predict the pressure of the gas if the volume is 3.25 liters, use the CALC function on the graphing calculator. Let v = 3.25.

The pressure of 3.25 liters of the gas is about 1.12 atmospheres.

e. Graph the regression

[0, 6] scl: 0.5 by [0, 5] scl: 0.5

equation using a graphing calculator. To predict the pressure of the gas if the volume is 6 liters, use the CALC function on the graphing calculator. Let v = 6.

The pressure of 6 liters of the gas is about 0.60 atmospheres.71. c 73. d

75 The function is undefined at x = 0, therefore; it is a power function of the form f (x) = a x -n or f (x) = a 1 _

x n , where a is a real

constant and n is a natural number. Start by assuming n = 1 and solve for a by substituting a set of points (2, 2) for x and y.

f (x) = a 1 _ x n

2 = a 1 _ 2 1

2 = a 1 _ 2

4 = a When n = 1, f (x) = 4 · 1 _ x or 4 _ x for the point (2, 2). If this power

function is true for the second point (-1, -4), then f (x) can represent the graph.

f (x) = 4 _ x

-4 = 4 _

-1

-4 = -4 Since f (x) is true for (-1, -4), f (x) = 4 _ x or 4 x -1 is a power

function for the graph.

77. f (x) = 3 x 1 _ 2 79a. 7.98 mm 79b. 7.35 mm 79c. increase

81. √

��

8 n · 2 7 _

4 -n =

√ ��

( 2 3 ) n · 2 7

_ ( 2 2 ) -n

= √

��

2 3n · 2 7 _

2 -2n

= √ ��

2 (3n + 7) - (-2n)

= √ ��

2 5n + 7

= √ ��

2 4n + 6 · 2 n + 1

= √ ��

2 4n + 6 · √

�� 2 n + 1

= 2 2n + 3 √ ��

2 n + 1

85. Sample answer: As n increases, the value of 1 _ n approaches 0.

This means that the value of 1 _ x n

will approach 1 when x is positive

and -1 when x is negative. Therefore, for positive values of x, f (x) will approach 1 + 5 or 6 and will resemble the line y = 6. For negative values of x, f (x) will approach

-1 + 5 or 4 and will resemble the line y = 4. 87a. r = -1 + 3 √ � B _

10

87b. 3.23% 89. ( f + g)(x) = x 3 + x 2 - 1

_ x + 1

, D = (-∞, -1) ∪

(-1, ∞); ( f - g)(x) = - x 3 - x 2 + 2x + 1

__ x + 1

, D = (-∞, -1) ∪

(-1, ∞); ( f · g)(x) = x 2 - x; D = (-∞, -1) ∪ (-1, ∞);

( f _ g ) (x) = x __

x 3 + x 2 - x - 1 , D = (-∞, -1) ∪ (-1, 1) ∪ (1, ∞)

91. y

xO

−4

−8

−4−8

8

4

4 8

f (x)

y

xO

−4

−8

−4−8

8

4

4 8

g (x)

y

xO

−4

−8

−4−8

8

4

4 8

h (x)

93. y

xO

−4

−8

−12

−4−8 4 8

f (x)

y

xO−4−8 4 8

4

8

12

g (x)

y

xO

−4

−8

−12

−4−8 4 8

h (x)

95. f is increasing on (-∞, -4) and increasing on (-4, ∞).

97. ( 1 _ 6 -

√ � 6 _

3 ) + (

√ � 3 _

3 +

√ � 2 _

6 ) i 99. -

24 _

169 + 10

_ 169

i 101. J

103. H

Lesson 2-2

1. y

x

−4

−8

−4−8

8

4

4 8

(x + 5) 2f (x) =

3. y

x

−4

−8

−4−8

8

4

4 8

x 4- 6f (x) =

5.

x

−4

−8

−4−8

8

4

4 8

(2x) 4f (x) =

y



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7. 9.y

x

−8

−16

−8−16

16

8

8 16

(x - 3) 4 + 6f (x) =

y

x

−8

−16

−8−16

16

8

8 16

(x - 9) 5_3

f (x) =

11.

Volu

me

(gal

lons

)

10

23456789

Time (minutes)0.2 0.4 0.6 0.8 t

v (t)Draining Water

v (t) = 10(1 t) 2

13. The degree is 6, and the leading coefficient is 2. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x) = ∞. 15. The degree is 6, and the

leading coefficient is -6. Because the degree is even and the leading coefficient is negative, lim x→-∞


f (x) = -∞. 17. The degree is 5, and the leading coefficient is -2. Because the degree is odd and the leading coefficient is negative, lim x→-∞


f (x) = -∞.

19. The degree is 4, and the leading coefficient is -2. Because the degree is even and the leading coefficient is negative, lim x→-∞


f (x) = -∞. 21. The degree is 6, and the

leading coefficient is 1. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x) = ∞.

23 The degree of f (x) is 5, so it will have at most five real zeros and four turning points.

0 = x 5 + 3 x 4 + 2 x 3

0 = x 3 ( x 2 + 3x + 2)

0 = x 3 (x + 2)(x + 1) So, the zeros are -1, -2, and 0. 25. 4 real zeros and 3 turning points; ± √ � 3 27. 6 real zeros and 5

turning points; 2 and 3 √ �� -2 29. 6 real zeros and

5 turning points; 0, -2, and 2 31. 4 real zeros and 3 turning

points; 0, -

1 _

2 , and 3 _

2 33a. The degree is 4, and the leading

coefficient is 1. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x) = ∞.

33b. 0, -4, 1 (multiplicity: 2)

33c. Sample answer: (-5, 180), (-2, -36), (0.5, 0.5625), (2, 12)

33d. y

x

−20

−40

−4−8

40

20

4 8 f (x) = x(x + 4)(x - 1 ) 2

35a. The degree is 4, and the leading coefficient is -1. Because the degree is even and the leading coefficient is negative, lim x→-∞


f (x) = -∞.

35b. 0, -3 (multiplicity: -2), 5

35c. Sample answer: (-4, -36), (-1, -24), (2, 150), (6, -486)

35d. y

x

−100

−200

−4−8

200

100

4 8

f (x) = x(x + 3 ) 2 (x 5)

37a. The degree is 5, and the leading coefficient is -1. Because the degree is odd and the leading coefficient is negative, lim x→-∞

f (x) = ∞

and lim x→∞

f (x) = -∞.

37b. 0, 3, -2 (multiplicity: 3) 37d.

y

x

−60

−120

−4−8

120

60

4 8

f (x) = x(x 3)(x + 2 ) 337c. Sample answer: (-3, 18),

(-1, -4), (1, 54), (4, -864)

39a. The degree is 3, and the leading coefficient is 3. Because the degree is odd and the leading coefficient is positive,

lim x→-∞


f (x) = ∞. 39b. 0, 4, -3

39c. Sample answer: (-4, -96), (-2, 36), (2, -60), (5, 120)

39d. y

x

−40

−80

−4−8

80

40

4 8

f (x) = 3 x 3 3 x 2 36x

41a. The degree is 4, and the leading coefficient is 1. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x) = ∞. 41b. 0 (multiplicity: 2),

4, -5 41c. Sample answer: (-6, 360), (-2, -72), (2, -56), (5, 250)

41d. y

x

−60

−120

−4−8

120

60

4 8

f (x) = x 4 + x 3 20 x 2

43a. Sample answer: f (x) = -0.0175 x 4 + 0.355 x 3 - 2.276 x 2 + 5.72x + 3.5 43b. 14.8 ft 45. Sample answer: f (x) = 0.09 x 3 - 2.70 x 2 +

24.63x - 65.21 47. Sample answer: f (x) = 4.05 x 4 - 0.09 x 3 + 6.69 x 2 - 222.03x + 2697.74 49a. According to the data, the values are increasing as x increases, therefore, lim x→∞

f (x) = ∞.



Selected A

nswers and S

olutions

49b.

[0, 12] scl: 1 by [400, 1400] scl: 200

Sample answer: f (x) = 0.146 x 4 - 3.526 x 3 + 32.406 x 2 - 63.374x + 473.255; The line is not a good fit. There are many outlying points.

49c. The leading coefficient is 0.146, so lim x→∞

f (x) = ∞. Sample

answer: The prediction was accurate because the leading coefficient is positive, as x → ∞, f (x) → ∞. 51. yes 53. No; undefined at x = 0 55. Sample answer: f (x) = x 3 - 9 x 2 + 27x - 27 57. Sample answer: f (x) = x 4 + 11 x 3 + 15 x 2 - 175x - 500 59. Sample answer: f (x) = x 5 + 4 x 4 61. Sample answer: f (x) = x 5

- x 4 - 6 x 3 63. Sample answer: f (x) = x 6 + 3 x 4 + 3 x 2 + 1 65. n is odd; a n is positive. 67. n is even; a n is negative. 69. f (x) = x 3 - 8 x 2 - 3x + 90 71. f (x) = x 5 - 9 x 4 + 23 x 3 - 3 x 2 - 36x 73. f (x) = 12 x 4 + 83 x 3 + 131 x 2 - 54x - 72

75 a. Sample answer: Enter (0, 56.1), (10, 63), (20, 68.6),

(30, 74.8), (40, 74.9), (50, 79.2) into STAT EDIT; the linear regression (r = 0.97) is f(x) = 0.45x + 58.19

b. Find f (x) for x = 65 since 2015 is 65 years since 1950. Graph the regression equation using a graphing calculator. Use the CALC feature on the calculator to find f (65).

[0, 70] scl: 5 by [0, 100] scl: 10

f (65) = 87.4%.

c. Sample answer: If f (x) = 85, then

85 = 0.45x + 58.226.8 = 0.45x59.

−

5 = x. So 59.

−

5 years after 1950, about 2010, 85% of the population will live in metropolitan areas.

77. Sample answer: f (x) = -(x + 2)(x + 3)(x - 4)(x - 1) ·

( x 2 + 6)

y

x

−320

−2−4

640

320

2 4

f (x) = (x + 2)(x + 3)(x 4)(x 1)(x 2+ 6)

79. Sample answer: f (x) = -(x - 1)(x - 2)(x + 1 ) 2 ( x 2 + 2)

y

x

−8

−16

−4−8

8

4 8

f (x) = (x 1)(x 2)(x + 1)2(x 2+ 2)

81a. degree = 4 81b. -6, -2 (multiplicity: 2), 4 81c. f (x) = 0.5(x + 2 ) 2 (x + 6)(x - 4) 83a. degree = 4 83b. -4, 3 (multiplicity: 2), -1

83c. f (x) = 1 _ 8 (x - 3 ) 2 (x + 1)(x + 4)

85. 4 real zeros and 3 turning points; no real zeros

87 The degree of the function is 4, so f has at most 4 distinct real zeros and at most 4 - 1 or 3 turning points. To find the real zeros, solve the related equation f (x) = 0 by factoring.

-24 x 4 + 24 x 3 - 6 x 2 = 0 -6 x 2 (4 x 2 - 4x + 1) = 0 -6 x 2 (2x - 1)(2x - 1) = 0 -6 x 2 (2x - 1) 2 = 0

The expression above has 4 factors, but solving for

x yields only 2 distinct real zeros: x = 0 and x = 1 _ 2 .

Both of the zeros have multiplicity 2.

89a.

f (x)g(x)

h(x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x) g(x)

h(x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x)

g(x)

h(x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x) g(x)

h(x)

[-1, 1] scl: 0.1 by [-1, 1] scl: 0.1

f (x) g(x)

h(x)

[-10, 10] scl: 1 by [-10, 10] scl: 1

f (x)

g(x)

h(x)

[-1, 1] scl: 0.1 by [-1, 1] scl: 0.1

89b. Sample answer: Very close to the origin, the function approximates the behavior of the term of lower degree, or h(x). 89c. Sample answer: As x → ∞ and −∞, the function approximates the behavior of the term of higher degree, or g(x). 89d. Sample answer: As x → ∞ and −∞, the function approximates the behavior of function a and, very close to the origin, the function approximates the behavior of function b. 91. Sample answer: No; a polynomial function f (x) cannot have both an absolute maximum and absolute minimum because as x → ∞, f (x) will approach either ∞ or -∞. If f (x)→ ∞ as x → ∞, then an absolute maximum is impossible. If f (x) → -∞ as x → ∞, then an absolute minimum is impossible. 93. Rearranging the terms gives f (x) = x 3 - x 2 - 12x + 5 x 2 - 5x - 60. Notice how the first set of three terms has a common factor of x and the second set of three terms has a



Sel

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nsw

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and

Sol

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ns

common factor of 5. After factoring using the Distributive Property, f (x) = x( x 2 - x - 12) + 5( x 2 - x - 12). Now notice how the factors inside the parentheses are identical. Using the Commutative Property, f (x) = (x + 5) · ( x 2 - x - 12). After factoring the second factor, f (x) = (x + 5) · (x - 4)(x + 3). The function is now completely factored and the zeros of the function are -5, 4, and -3. These were determined by setting each factor equal to zero and solving for x. 95. Sample answer: There is one turning point at the absolute maximum, one at the relative minimum, and one at the relative maximum. Therefore, the minimum degree is 3 + 1 or 4. 97. 46 99. no solution 101. 2x 2 + x + 7 103. 12 x 2 - 30x + 22 105. The graph of g(x) is the graph of f (x) reflected in the x-axis and translated 2 units left and 4 units down; g(x) = -(x + 2 ) 2 - 4. 107a. $50 107b. Sample answer: The company’s competition might offer a similar product at a lower cost. 109. ≈12.54 111. F 113. H

Lesson 2-3

1. (x + 3)(x - 5)(x + 4) 3. (x + 5)(x + 2)(x - 4) 5. -3(x - 5)(x + 6)(x - 6) 7. (x + 7)(x - 1)(x + 3 ) 2 9. 5 x 3 + 17 x 2

+ 74x + 295 + 1192 _

x - 4 11. 2 x 3 - 8 x 2 + 22x - 47 + 100

_ x + 2

13. 3 x 5 + 3 x 3 - 6 x 2 - 2x + 4 - 2 _

2x - 1 15. x + 4

17. 2 x 2 - 4x + 2 + 2 __

3 x 3 + 2x + 3 19. x 3 + x 2 + 5x + 4 + 2

_ x - 2

21 Because x - 4, c = 4. Set up the synthetic division as follows, using a zero placeholder for the missing x 2 -term in the dividend. Then follow the synthetic division procedure.

4 3 -9 0 -24 -48

12 12 48 96

3 3 12 24 48

The quotient is 3 x 3 + 3 x 2 + 12x + 24 +

48 _

x - 4 .

23. 6 x 4 + 14 x 3 + 12 x 2 + 12x + 18 + 46 _

2x - 3 25. 15 x 4 +

12 x 3 + 9 x 2 + 6x + 20 _

3 + 76

_ 3(3x − 2)

27. 12 x 5 + 6 x 4 − 3 x 3 − 5

29. 2,151,000 students 31. 711 33. 45,536 35. -67,978 37. -125,184 39. no, no 41. yes, yes; f (x) = (3x - 1)(x - 5) · (x - 4)(x + 2) 43. no, no 45. yes, no; f (x) = (4x + 3)( x 4 - 3 x 3 + 12 x 2 - 3x + 21) 47. 5 seconds 49. ( x 2n + 2 x n - 10) ·

( x n - 1) 51. 3( x n - 3)( x n + 6)(3 x n - 1) 53. 2 55. -4 57a. v(x) = - x 3 + 12 x 2 - 47x + 60

57b. Volume of Clay

Volu

me

( ft3 )

0

20

40

60

Removed Amount (ft)1 2 3

= -x3+ 12x2

- 47x + 60v(x)

x

v(x)

57c. 36 = - x 3 + 12 x 2 - 47x + 60 57d. about 0.60 ft 59. f (x) = (x + 6)(x - 2 ) 2 (3x – 7)(2x + 5)

61 Using the Factor Theorem, (x - 1) is a factor if f (1) = 0.

f (1) = 18(1 ) 165 - 15(1 ) 135 + 8(1 ) 105 - 15(1 ) 55 + 4= 18(1) - 15(1) + 8(1) - 15(1) + 4

= 18 - 15 + 8 - 15 + 4= 0

Since f (1) = 0, (x - 1) is a factor of the polynomial. 63. True; sample answer: The Remainder Theorem states that if h( y) is divided by y - (-2), then the remainder is r = h(-2), which is -1. 65. −30 67. -5 69. even; negative

71. odd; negative 73a. h = v 2 _

64 73b. Yes; the pump can propel

water to a height of about 88 ft.

75. (-1, 1) 77. ( 1 _ 3 , 2 _

3 ) 79. (-

6 _

43 , -

64 _

43 ) 81. K 83. H

Lesson 2-4

1. ±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±30, ±36, ±45, ±60, ±90, ±180; 6, 5, 1, -6 3. ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30; -5, 6, -1, 1 5. ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20,

±30, ±60, ±

1 _

2 , ±

3 _

2 , ±

5 _

2 , ±

15 _

2 , ±

1 _

3 , ±

2 _

3 , ±

4 _

3 , ±

5 _

3 , ±

10 _

3 , ±

20 _

3 , ±

1 _

6 , ±

5 _

6 ;

−

5 _

3 , −

1 _

2 7. ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±16, ±18, ±24, ±27,

±36, ±48, ±54, ±72, ±108, ±144, ±216, ±432; 4 (multiplicity: 2), 3 9. 1 in., 5 in., 9 in. 11. -4, 1, -3 (multiplicity: 2) 13. 4, -5, 2

(multiplicity: 2) 15. 1 _ 2 , 2 _

3 , -4 (multiplicity: 2) 17. 2 days

19. Sample answer: [-6, 5]; 3, 2, -4, −

1 _

2 21. Sample answer: [-4,

7]; -2, 5, 1, 3 _ 2 23. Sample answer: [-2, 7]; 3 (multiplicity: 2),

-1, 3 _ 2 25. Sample answer: [-3, 5]; 1 _

2 (multiplicity: 2), -2, 3

(multiplicity: 2) 27. 3 or 1 positive zeros, 1 negative zero 29. 2 or 0 positive zeros, 2 or 0 negative zeros

31. 4, 2 or 0 positive zeros, 1 negative zero 33. Sample answer: f (x) = x 4 + 4 x 3 - 19 x 2 - 106x - 120 35. Sample answer: f (x) = x 4 - 19 x 3 + 113 x 2 - 163x - 296 37. Sample answer: f (x) = x 4 - 5 x 3 - 21 x 2 + 119x - 130 39. Sample answer: f (x) = x 4 - 6 x 3 + 19 x 2 + 36x − 150 41. Sample answer: f (x) = x 4 - 28 x 3 + 296 x 2 - 1372x + 2263

43 a. g(x) has possible rational zeros of ±1, ±2, ±4, ±8. By using synthetic division, it can be determined that x = -2 is a rational zero.

-2 1 -3 -12 0 8

-2 10 4 -8

1 -5 -2 4 0

By using synthetic division on the depressed polynomial, it can be determined that x = -1 is a rational zero.

-1 1 -5 -2 4

-1 6 -4

1 -6 4 0

The remaining quadratic factor ( x 2 - 6x + 4) yields no rational zeros. Use the quadratic formula to find the zeros.

-b ± √

�� b 2 - 4ac __

2a =

-(-6) ± √ ��

(-6 ) 2 - 4(1)(4) ___

2(1)

= 6 ± √ � 20

_ 2

= 3 ± √ � 5




Selected A

nswers and S

olutions

So, g(x) written as a product of linear and irreducible quadratic factors is g(x) = (x + 2)(x + 1)(x – 3 + √ � 5 )(x - 3 - √ � 5 ).

b. g(x) written as a product of linear factors is g(x) = (x + 2)(x + 1)(x - 3 + √ � 5 )(x - 3 - √ � 5 ).

c. The zeros are -2, -1, 3 - √ � 5 , 3 + √ � 5 .

45a. f(x) = ( x 2 – 4x + 13)(4x - 3)(x - 4) 45b. f(x) = (4x - 3) ·

(x - 4)(x - 2 + 3i)(x - 2 - 3i) 45c. 3 _ 4 , 4, 2 - 3i, 2 + 3i

47a. h(x) = (x + 3)(x - 5)(x + √ � 2 )(x - √ � 2 ) 47b. h(x) = (x + 3)(x - 5) (x + √ � 2 ) (x - √ � 2 ) 47c. -3, 5, √ � 2 , - √ � 2

49. 3, -4, 1 _ 2 , 3i, -3i; h(x) = (x - 3) ·(x + 4)(2x - 1)(x + 3i) ·

(x - 3i) 51. -3, -2, 1, 3 + i, 3 - i; g(x) = (x + 3)(x + 2) ·

(x - 1)(x - 3 + i)(x - 3 - i) 53. 3, 2 √ � 2 , -2 √ � 2 , 2i, -2i; f(x) =

(x - 3) (x + 2 √ � 2 ) (x - 2 √ � 2 ) (x + 2i)(x - 2i) 55a. V(�) = 1 _ 3 ·

�

3 - 3 �

2 55b. 6300 = 1 _ 3 �

3 - 3 �

2 55c. base: 30 in. by 30 in.; height:

21 in. 57. Sample answer: f (x) = x 3 - 6 59. Sample answer: f (x) = x 3 + 2 61. g(x) = 3(x - 4)(x - 1) ·(x + 5i) (x - 5i); 4, 1, ±5i 63. -1 65. no rational zeros

67 Solve d = 0.0000008 x 2 (200 − x) for d = 0.8.0.8 = 0.0000008 x 2 (200 − x) = 0.0000008 x 2 (200 − x) − 0.8

Use a graphing calculator to graph y = 0.0000008 x 2 · (200 − x) − 0.8 and solve for y = 0. Using the CALC menu, find the zeros of the function.

[0, 200] scl: 10 by [-1, 1] scl: 0.5

The zeros of the function occur at x = 100 and x = 161.8. A vertical deflection of 0.8 feet will occur when a weight is placed 100 feet or about 161.8 feet from the left piling.

69. (x + 2 3 √ � 2 ) ( x 2 - 2x 3

√ � 2 + 4 3 √ � 4 ) 71. (3 x 2 +

3 √ � 4 ) ·

(9 x 4 - 3 x 2 3 √ � 4 + 2

3 √ � 2 ) 73. Julius; sample answer: Angie

did not divide by the factors of the leading coefficient. 75a. The graphs of -f (x) are the graphs of f (x) reflected in the x-axis. The zeros are the same. 75b. The graphs of f (-x) are the graphs of f (x) reflected in the y-axis. The zeros are opposites.

77. False; sample answer: The third-degree polynomial x 3 + x 2 - 17x + 15 has three real zeros and no nonreal zeros.

79. x 1

_ 3 ,

x 2 _

3 ,

x 3 _

3 81. - x 1 , - x 2 , - x 3

83. Sample answer: If a polynomial has an imaginary zero, then its complex conjugate is also a zero of the polynomial. Therefore, any polynomial that has one imaginary zero has at least two imaginary

zeros. 85. x 3 + 3 x 2 + 6x + 12 + 23 _

x − 2 87. 2 x 2 + 2x − 3

_ x − 1

89. The degree is 6, and the leading coefficient is 4. Because the degree is even and the leading coefficient is positive, lim x→-∞


f (x) = ∞.

91. It appears that f (x) has a relative minimum of -3 at x = 0 and a relative maximum of 3 at x = -2. 93. It appears that f (x) has a relative maximum of 0 at x = 0 and x = -4 and a relative minimum of -80 at x = -2 and -150 at x = 2. 95. E 97. A

Lesson 2-5

1. D = {x | x ≠ 2, -2, x ∈ }; x = 2, x = -2, y = 1 3. D = {x | x ≠ 4, -3, x ∈ }; x = 4, x = -3 5. D = {x | x ≠ 0, -2, x ∈ }; x = 0, x = -2, y = 2 7. D = {x | x ≠ 2, -4, x ∈ }; x = 2, x = -4, y = 0 9. asymptotes: x = -4, x = 5, y = 1; x-intercepts: -2, 3; y-intercept:

3 _ 10

; D = {x | x ≠ -4, 5, x ∈ }

y

−4

−8

−4−8

8

4

4 8 x

y = 1

x = 5x = -4

f (x) = (x + 2)(x - 3)__(x + 4)(x - 5)

11. asymptotes: x = -2, x = 2, y = 0; y-intercept: -2; D = {x | x ≠ -2, 2, x ∈ }

13. asymptotes: x = -5, x = 6, y = 0; x-intercept: -2;

y-intercept: -

1 _

15 ;

D = {x | x ≠ -5, 6, x ∈ }

15. asymptotes: x = -3, x = -1; x-intercepts: -5, 2, 0; y-intercept: 0; D = {x | x ≠ -3, -1, x ∈ }

17. asymptote: y = 0; x-intercept: 8; y-intercept:

-

8 _

5 ; D = {x | x ∈ }

19a. D = {z | z ≥ 0, z ∈ } 19b. horizontal asymptote:

y = 3 _ 2 ; x-intercept: 1; y-intercept: -

3 _

5

y

−4

−8

−4−8

8

4

4 8 x

y = 0

x = 2x = -2

f (x) = 8__(x - 2)(x + 2)

y

−4

−8

−4−8

8

4

4 8 x

y = 0

x = -5 x = 6

g(x) = (x + 2) (x + 5)__(x + 5)2 (x - 6)

y

−80

−160

−4−8

160

80

4 8 x

x = -3

x = -1

h(x) = x2(x - 2)(x + 5)

x 2+ 4x + 3

__

y

−4

−8

−4−8

8

4

4 8 x

y = 0

f (x) = x - 8__x2

+ 4x + 5



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19c. Car Wash

Prof

it (th

ousa

nds)

0

1

-1

2

Week

p (z)

z10 20 30 40

p(z) =3z 2 3

2z 2+ 7z + 5

__

21. asymptotes: x = 3

√ ��

- 4 _ 3 , y = 0; y-intercept: 1 _

4 ;

D = ⎧

⎨

⎩

x | x ≠ 3

√ ��

- 4 _ 3 , x ∈

⎫

⎬

⎭

y

−4

−8

−4−8

8

4

4 8 x

y = 0

x =

3√

-4_3

h(x) =

4x2- 2x + 1

3x 3+ 4

__

23. asymptotes: x = 4, y = 1; x-intercept: -7; y-intercept: -

7 _

4 ;

D = {x | x ≠ 4, x ∈ }

y

−8

−16

−8−16

16

8

8 16x

y = 1

x = 4g(x) =

x + 7_x - 4

25. asymptote: x = 4; x-intercepts: -2, -1, 0; y-intercept: 0; D = {x | x ≠ 4, x ∈ }

y

−240

−8−16

240

120

8 16xx = 4

g (x) =

x 3+ 3x2

+ 2xx - 4

__

−120

27 g(x) can be written as g(x) = (x - 2)(x + 2)

__ (x - 2)(x + 2)(x + 1)

or 1 _

x + 1 .

The function is undefined at b(x) = 0, so D = {x | x ≠ -2, -1, 2, x ∈ }. There is a vertical asymptote at x = -1, the real zero of the simplified denominator.There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.Since the simplified numerator has no real zeros, there are no x-intercepts. The y-intercept is 1 because g(0) =1.

There are holes at (-2, -1) and (2, 1 _ 3 ) because the original

function is undefined when x = -2 and x = 2. Graph the asymptotes and intercept. Then find and plot points.

y

−2

−4

−2−4

4

2

2 4 x

x = -1g (x) =

x2- 4

x 3+ x 2

- 4x - 4__

y = 0

29. asymptotes: x = -4; y = 0; hole: (5, 11 _

81 ) ; x-intercept: -

1 _

2 ;

y-intercept: 1 _ 16

; D = {x | x ≠ -4, 5, x ∈ }

y

−4

−8

−4−8

8

4

4 8 x

y = 0x = -4

g (x) = (2x + 1)(x - 5)__(x - 5)(x + 4) 2

31a. 1 _ 8 = 1 _

d i + 1 _

32 31b. 10 2 _

3 cm 33. ±2 √ � 3 − 2 35. no

solution 37. 5 39. -6 41. -1 43. Sample answer:

f(x) = x(x - 4)

__ (x - 1)(x - 6 ) 2

45 a. There is a vertical asymptote at the real zero of the denominator r 1 = 30. There is a horizontal asymptote at r 2

= 30 _

1 or r 2 = 30, the ratio of the leading coefficients of the

numerator and denominator, because the degrees of the polynomials are equal.

[0, 240] scl: 30 by [0, 100] scl: 10

b. Evaluate the function for each value of r 1 to determine r 2 .

r 2 = 30(45)

_ 45 − 30

r 2 = 30(60)

_ 60 − 30

= 90 = 60

r 2 = 30(50)

_ 50 − 30

r 2 = 30(65)

_ 65 − 30

= 75 = 55.7

r 2 = 30(55)

_ 55 − 30

r 2 = 30(70)

_ 70 − 30

= 66 = 52.5

r 1 45 50 55 60 65 70

r 2 90 75 66 60 55.7 52.5

c. No; sample answer: As r 1 approaches infinity, r 2 approaches 30. This suggests that the average speeds reached during the first leg of the trip have no bounds. Being able to reach an infinite speed is not reasonable. The same holds true about r 2 as r 1 approaches 30 from the right.

47. Sample answer: f (x) = 2 x 2 − 14x + 20

__ x 2 + x − 12

49. ≈-2.65, ≈2.65

51. ≈-3.87, ≈1.20, ≈3.70



Selected A

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olutions

53a.

FunctionHorizontal

Asymptote

f (x) = x 2 − 5x + 4 _

x 3 + 2 y = 0

h (x) = x 3 − 3 x 2 + 4x − 12 __

x 4 − 4 y = 0

g (x) = x 4 − 1 _ x 5 + 3

y = 0

53b.

[-5, 10] scl: 1 by [-1, 1] scl: 0.1

[-5, 10] scl: 1 by [-1, 1] scl: 0.1

[-5, 10] scl: 1 by [-1, 1] scl: 0.1

53c.

FunctionReal Zeros of

Numerator

f(x) = x 2 − 5x + 4 _

x 3 + 2 1, 4

h(x) = x 3 − 3 x 2 + 4x − 12 __

x 4 − 4 3

g(x) = x 4 − 1 _ x 5 + 3

-1, 1

53d. Sample answer: When the degree of the numerator is less than the degree of the denominator and the numerator has at least one real zero, the graph of the function will have y = 0 as an asymptote and will intersect the asymptote at the real zeros of the numerator. 59. Sample answer: The test intervals are used to determine the location of points on the graph. Because many rational functions are not continuous, one interval may include y-values that are vastly different than the next interval. Therefore, at least one, and preferably more than one, point is needed for every interval in order to sketch a reasonably accurate graph of the function. 61. ±1, ±2, ±3, ±6, ±9, ±18; -2 63. no; yes; f (x) = (x - 2) · ( x 3 - 13x - 12) 65. yes; no; f (x) = (3x − 2) · (2 x 3 + 21 x 2 + 60x + 25) or f (x) = (3x − 2)(2x + 1)(x + 5 ) 2 67. yes; no; f (x) = (x + 2)(4 x 4 + 7 x 3 - 2 x 2 ) or f (x) = x 2 (x + 2 ) 2 (4x − 1)

69. 71.

y

−4−8

8

12

4

x−12

f(x) = (x + 7) 2

y

−4

−8

−4−8

8

4

4 8 x

f (x) = x 4- 5

73a. (12 - 2x)(16 - 2x) = 96

73b. 73c. 8 ft by 12 ft 75. -4 + 4i

y

x

−40

−80

80

40

4 8 12

( ) = 4 2- 56 + 96 77. C 79. C

Lesson 2-6

1. [-4, 2] 3. ⎛

⎪

⎝

-∞, -

1 _

3 ⎤

�

⎦

∪ [8, ∞) 5. (-∞, -

1 _

2 ) ∪ ( 2 _

3 , ∞)

7. (-3, -

3 _

4 ) ∪ (0, ∞) 9. (-∞, ∞) 11. ∅ 13. {4} 15. {-2}

17. -0.0004 x 2 + 80x - 1,000,000 ≥ 2,000,000; [50,000, 150,000] or 50,000 ≤ x ≤ 150,000 19. (-∞, 5)

21 3x - 2 _

x + 3 < 6

3x - 2 _

x + 3 - 6 < 0

3x - 2 _

x + 3 - 6 (

x + 3 _

x + 3 ) < 0

3x - 2 _

x + 3 -

6x + 18 _

x + 3 <0

-3x - 20 _

x + 3 < 0

Let f (x) = -3x - 20 _

x + 3 . The zeros and undefined points of the

inequality are the zeros of the numerator, x = - 20 _

3 , and

denominator, x = -3. Create a sign chart. Then choose and test x-values in each interval to determine if f (x) is positive or negative.

x203

(-) (+) (-)

--3

The solutions of -3x - 20 _

x + 3 < 0 are x-values such that f (x) is

negative. From the sign chart, you can see that the solution set

is (-∞, -

20 _

3 ) ∪ (-3, ∞).

23. ⎛

⎪

⎝

-∞, 14 _

13 ⎤

�

⎦

∪ ( 5 _ 3 , ∞) 25.

⎛

⎪

⎝

-3, 2 _ 3 ⎤

�

⎦

∪ (1, ∞) 27. (-∞, 13 _

6 ) ∪

(4, ∞) 29. 750 + 25x _ x < 120; 8 to 14 people 31. (–∞, –5] ∪

[8, ∞) 33. (-∞, -3] ∪ [3, ∞) 35. (–∞, -6) ∪ (-6, 6) ∪ (6, ∞) 37. [-4, 3] 39a. (56 - ) ≥ 588 or -

2 + 56 - 588 ≥ 0 39b. [14, 42]; The length of the playing field is at least 14 and at most 42 feet. 39c. 0 < (56 - ) ≤ 588; (0, 14] ∪ [42, 56); Sample answer: The area of the playing field must be greater than 0 square feet but at most 588 square feet, so

0 < ≤ 14 or 42 ≤ < 56. 41. ⎡

⎢

⎣

40 _

9 , 8

⎤

�

⎦

43. ⎛

⎪

⎝

-12, 25 _

12 ⎤

�

⎦

45. x 2 +2x - 8 < 0 47. (-5, -

4 _

3 ) ∪ (0, 4) 49. (-∞, -8] ∪

⎡

⎢

⎣

-3, -

1 _

3 ⎤

�

⎦

∪ [1, 6] 51. (-∞, -3) ∪ (-3, -

1 _

2 ) ∪ (-

1 _

2 , 4)

53. (-∞, -2) ∪ (3, 6) ∪ (6, ∞) 55a. 75 + t _ 5 ≥ 89.5

55b. 75; Sample answer: Since t ≥ 72.5, Jarrick will have to spend 75 minutes studying for the test. 57. 36 unit s 2 59. (-k, k) 61. (-∞, ∞) 63. Neither; sample answer: The expression is undefined at 3, but it is positive everywhere that it is defined, so the solution set is (−∞, 3) ∪ (3, ∞). 65. a < 0 and |a + b| > |c + d|



Sel

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67 Because (x - a ) 2n is always an even degree, by definition, the end behavior is to approach ∞ as x → ± ∞. The absolute minimum occurs at x - a = 0 or x = a. Since any other value

of (x - a) to an even power must be positive in the reals,

(x - a ) 2n > 0 for all values of x except when x = a, since (0 ) 2n > 0 is false. Thus, the solution set is (-∞, a) ∪ (a, ∞).

69. Sample answer: There will be instances when x - 2 is a negative value. When this occurs, we are multiplying an inequality by a negative and the inequality symbol will need to be reversed. The problem occurs here because x - 2 could be positive or negative. 71. D = {x|x ≠ -6, x ∈ �}; x = -6 73. about 0.37 cm or 9 cm 75. 3a 3 – 9a 2 + 7a - 6 77. x 2 - xy + y 2 79a. 100,000 79b. 30,240 79c. 8000; 2352 81. F


1. leading coefficient 3. repeated zeros 5. Factor Theorem 7. Vertical 9. Oblique

11. y

x

f (x) = 5x 6

D = (-∞, ∞), R = [0, ∞); intercept:

0; lim x→-∞


f (x) = ∞;

continuous for all real numbers; decreasing: (-∞, 0); increasing: (0, ∞)

13.

x

−4

−8

−4−8

8

4

4 8

y

f (x) = x -9

D = (-∞, 0) ∪ (0, ∞), R = (-∞, 0) ∪ (0, ∞); no intercepts; lim x→-∞


f (x)

= 0; infinite discontinuity at x = 0; decreasing: (-∞, 0) and (0, ∞)

15. yx

−4

−8

−12

−16

4 8 12 16

√5x 6 11f (x) =

D = [1.2, ∞), R = [-11, ∞); x-intercept: 25.4; lim x→∞

f (x) = ∞; continuous on [1.2,

∞); increasing: (1.2, ∞)

17. 4 19. 4 21. lim x→∞

f (x) = -∞; lim x→-∞

f (x) = -∞;

The degree is even and the leading coefficient is negative.

23. lim x→∞

f (x) = ∞ and lim x→-∞

f (x) = ∞; The degree is even and the

leading coefficient is positive. 25. 3 real zeros and 2 turning points; 0, 3, and 4 27. 4 real zeros and 3 turning points; -3, -1, 1, and 3 29a. lim x→∞

f (x) = ∞; lim x→-∞

f(x) = ∞

29b. 0 (multiplicity 3), 3, -4 (multiplicity 2)

29c. Sample answer: (-1, 36), (0, 0), (1, -50), (2, -288)

29d. y

x

−160

−320

−4−8

320

160

4 8

x 3(x 3)(x + 4)2f (x) =

31. x 2 + 10x + 20 + 35 _

x - 2 33. x 4 + 3 x 3 - x 2 − 9 + 1

_ 2x − 1

35. x 3 + x 2 + 9x + 9 37. yes; f(x) = (x + 3)( x 2 - 8) 39. yes; yes;

f(x) = (x - 2 ) 2 (x + 1 ) 2 41. ±1, ±3, ±5, ±15; -3 43. ±1, ±2, ±5,

±10, ±

1 _

3 , ±

2 _

3 , ±

5 _

3 , ±

10 _

3 ; 2, -

1 _

3 45. 2, 1 _

2 , 4 _

3 47. 3 _

2 , -2, 2, 4 49. 1, 3,

2i, -2i; f(x) = (x - 3)(x - 1)(x - 2i)(x + 2i) 51. D = {x | x ≠ 5, -5, x ∈ �}; x = 5, x = -5, y = 1 53. D = {x | x ≠ -3, -9, x ∈ �}; x = -3, x = -9, y = 1

55. asymptotes: x = -4, y = 1;

x-intercept: 2; y- intercept: -

1 _

2 ;

D = {x | x ≠ -4, x ∈ �}

57. asymptotes: y

x

−4

−8

−4−8

8

4

4 8

f (x) =

x(x + 7) __

(x + 6)(x 3)

x = -6y = 1

x = 3

x = -6, x = 3, y = 1;x-intercepts: 0 and -7; y-intercept: 0; D = {x | x ≠ –6, 3, x ∈ �}

59. x = 0, x = 1, x = 5, y = 0; x-intercepts: 4 and -4; D = {x | x ≠ 0, 1, 5, x ∈ �}

y

x

−8

−16

−4−8

16

8

4 8

f (x) = x 2 16__

x 3 6 x 2 + 5x

x = 5

x = 1

x = 0

y = 0

61. -3 63. 7 _

3 65. (-∞, -2) ∪ (8, ∞) 67. (-5, -

3 _

2 ) 69. ∅

71. (0, 5) 73. (3, 26 _

7 ) ∪ (4, ∞)

75a.

[0, 3.5] scl: 0.5 by [0, 100] scl: 10

75b. v(t) = 43.63 t -1.12 75c. about 35.6 mi/h 75d. about 0.94 second

77. 6.13 f t 2 79. 15 m by 50 m or 25 m by 30 m 81. 16 cakes


1. 21.86 unit s 2 3. 25.13 unit s 2 ; Sample answer: The approximation is less than the actual area. As more rectangles are used, the approximation approaches the actual area. 5. Sample answer: 48 unit s 2 7. Yes; sample answer: For this example, using right endpoints for the rectangles would produce rectangles that overlap the curve and include area that lies above the curve. This would produce a higher approximation. However, this is not always the case. If the curve is symmetrical, like the previous example, the approximations will be the same regardless of the

y

x

−4

−8

−4−8

8

4

4 8

x = -4f(x) = x 2_

x + 4

y = 1



Selected A

nswers and S

olutions

endpoint used. 9. Sample answer: The more rectangles that are used, the better the approximation of the area. Smaller rectangles fit the desired region better than larger rectangles, thus producing more accurate approximations. 11a. 4 rectangles: 14 unit s 2 ; 8 rectangles: 13.75 11b. Sample answer: Approximating the area using 4 or 8 rectangles probably does not give a good representation of the actual area. The nature of the curve prevents rectangles of widths of 1 and 0.5 units from really fitting well beneath it. 11c. Yes; sample answer: The graph is not symmetrical. Using right endpoints to determine the height of the rectangles would produce different rectangles with different areas.

Trigonometric Functions 3CHAPTER 3CHCHAPAPTETERR 33

Chapter 3 Get Ready

1. 6 √ � 2 3. 3.52 5. no 7. 7 9. vertical asymptote at x = -5 11. x = 3, x = 5, y = 0 13. no asymptotes

Lesson 3-1

1. sin θ = 4 √ � 2

_ 9 , cos θ = 7 _

9 , tan θ =

4 √ � 2 _

7 , csc θ =

9 √ � 2 _

8 ,

sec θ = 9 _ 7 , cot θ =

7 √ � 2 _

8 3. sin θ =

9 √ � 97 _

97 , cos θ =

4 √ � 97 _

97 ,

tan θ = 9 _ 4 , csc θ =

√ � 97 _

9 , sec θ =

√ � 97 _

4 , cot θ = 4 _

9

5. sin θ = √ �� 165

_ 29

, cos θ = 26 _

29 , tan θ =

√ �� 165 _

26 , csc θ =

29 √ �� 165 _

165 ,

sec θ = 29 _

26 , cot θ =

26 √ �� 165 _

165 7. sin θ = 3 _

5 , cos θ = 4 _

5 , tan θ = 3 _

4 ,

csc θ = 5 _ 3 , sec θ = 5 _

4 , cot θ = 4 _

3 9. cos θ = 3 _

5 , tan θ = 4 _

3 , csc θ = 5 _

4 ,

sec θ = 5 _ 3 , cot θ = 3 _

4 11. sin θ =

3 √ � 10 _

10 , cos θ =

√ � 10 _

10 , csc θ =

√ � 10 _

3 ,

sec θ = √ � 10 , cot θ = 1 _ 3 13. sin θ =

2 √ � 14 _

9 , tan θ =

2 √ � 14 _

5 ,

csc θ = 9 √ � 14

_ 28

, sec θ = 9 _ 5 , cot θ =

5 √ � 14 _

28 15. sin θ =

√ � 26 _

26 ,

cos θ = 5 √ � 26

_ 26

, tan θ = 1 _ 5 , csc θ = √ � 26 , sec θ =

√ � 26 _

5

17. sin θ = √ � 77

_ 9 , cos θ = 2 _

9 , tan θ =

√ � 77 _

2 , csc θ =

9 √ � 77 _

77 ,

cot θ = 2 √ � 77

_ 77

19. 3.2 21. 4.1 23. 43.0 25. 7.5

27

35°

25 ft An acute angle measure and the adjacent side length are given,

so the tangent function can be used to find the length of the opposite side.

tan θ = opp

_ adj

Tangent function

tan 35° = x _ 25

θ = 35°, opp = x, and adj = 25

25 tan 35° = x Multiply each side by 25.

17.5 = x Use a calculator.

So, the ravine is about 17.5 feet wide.

29a.

Elwood Ave.

Oak St.

Maple St.

x

32°

0.8 mi

29b. 1.3 mi 31. 14° 33. 35° 35. 59° 37. 53° 39. 424 ft

41a.

55°

375 ft

41b. 307 ft 43. about 3.1 ft

45 a. Because the angle of depression from the lighthouse to a ship and the angle of elevation from that ship to the lighthouse are congruent, you can label the angles of elevation at each ship as shown. Label the distance from the lighthouse to the first ship y and the distance between the two ships x + y.

156 ft

Ship 1 yx Ship 2

7°

7°

27°

27°

b. From the smaller right triangle, you can use the tangent function to find the value of y.

tan θ = opp

_ adj

Tangent function

tan 27° = 156 _ y θ = 27°, opp = 156, and adj = y

y tan 27° = 156 Multiply each side by y.

y = 156 _

tan 27° Divide each side by tan 27°.

y ≈ 306.167 Use a calculator.

From the larger right triangle, you can use the tangent function to find x.

tan 7° =

156 _ x + y θ = 7°, opp = 156, and adj = x + y

(x + y) tan 7° = 156 Multiply each side by x + y.

x + y = 156 _

tan 7° Divide each side by tan 7°.

x = 156 _

tan 7° - y Subtract y from each side.

x = 156 _

tan 7° - 306.167 y = 306.167

x ≈ 964.35 Use a calculator.

Therefore, the ships are about 964 feet apart.

47. B = 70°, b ≈ 16.5, c ≈ 17.5 49. P ≈ 43°, Q ≈ 47°, r ≈ 34.0 51. K ≈ 19°, j ≈ 18.0, k ≈ 6.2 53. H = 41°, f ≈ 19.6, h ≈ 17.1

55a.

65 ft

10 ft 55b. 9°

57. √ � 3

_ 2 59.

2 √ � 3 _

3 61. √ � 3 63. 45° 65. 60° 67. 30° 69. 2



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45 ft

shipwreck

20 ft

70°

57°

73. about 0.92 75. > 77. < 79. > 81. about 17.9 in.

about 100 ft

83. Sample answer: For an acute angle θ of a right triangle,

sin θ = opp

_ hyp

, cos θ = adj

_ hyp

, tan θ = opp

_ adj

, and cot θ = adj

_ opp .

Using these definitions, sin θ

_ cos θ

=

opp

_ hyp

_ adj

_ hyp

= opp

_ adj

= tan θ.

Similarly, cos θ

_ sin θ

=

adj

_ hyp

_ opp

_ hyp

=

adj _ opp = cot θ.

85. The trigonometric functions are transcendental functions because they cannot be expressed in terms of algebraic operations. For example, there is no way to find the value of θ in y = cos θ by adding, subtracting, multiplying, or dividing a constant and θ or raising θ to a rational power.

87 Sample answer: Given a triangle with hypotenuse c and legs a

and b, let sin θ = a _ c and cos θ = b _ c . By the Pythagorean Theorem, c 2 = a 2 + b 2 .

(sin θ ) 2 + (cos θ) 2 = ( a _ c ) 2 + ( b _

c )

2

= a 2 _

c 2 + b 2

_ c 2

= a 2 + b 2

_ c 2

= c 2 _

c 2 or 1

Therefore, (sin θ ) 2 + (cos θ ) 2 = 1.

89. False; sample answer: In �ABC, the measure of angle B is less than the measure of angle A, cos B ≈ 0.7986, and cos A ≈ 0.6018. Therefore, cos B > cos A, and thus the statement is false.

53°

37°

3

4

5

91. Sample answer: Since the cosine function is the reciprocal of the secant function, the sine function is the reciprocal of the cosecant function, and the tangent function is the reciprocal of the cotangent function, you can find the value of a secant, cosecant, or cotangent function on a graphing calculator by finding one divided by the reciprocal of the function. 93. 0.6495. y

x

f (x) = - 3x - 2

D = (-∞, ∞), R = (-∞, 0); no

x-intercept, y-intercept: -

1 _

9 ;

horizontal asymptote at y = 0; lim x→-∞

= 0, lim x→∞

= -∞;

decreasing on (-∞, ∞)

97. y

−4

−8

−12

−16

4 8 12 16 x

f (x) = - 4-x + 6

D = (-∞, ∞), R = (-∞, 0); no x-intercept, y-intercept: -4096; horizontal asymptote at y = 0; lim x→-∞

= -∞, lim x→∞

= 0;

increasing on (-∞, ∞)

99. 4

101a.

[0, 10] scl: 1 by [0, 1000] scl: 100

101b. y = 904.254x-0.149

101c. 611,068103. G 105. G

Lesson 3-2

1. 11° 46′ 23� 3. 141° 32′ 56� 5. 87.886° 7. 45.357° 9. 17.63°

11. 5π

_ 4 13. -

π

_ 4 15. 450° 17. -210° 19–25. n is an integer.

19. -75° + 360n°; Sample answer: 285°, -435°

y

-75°

285°

x

y

-75°

-435°

x

21. -150° + 360n°; Sample answer: 210°, -510°

y

-150°

210°

x

y

-150°

-510°

x

23. -

3π

_

4 + 2nπ; Sample answer: 5π

_

4 , -

11π

_ 4

y

5π

4

3π

4-

x

y

x

11π_

4

3π_4

25. 3π

_

2 + 2nπ; Sample Answer: 7π

_

2 , -

π

_

2

y

3π

27π

2

x

y

3π

2

π

2-

x



Selected A

nswers and S

olutions

27. 1.3 m 29. 5.2 yd 31. 3.9 mi

33 a. Convert 3024° to radian measure, and then use the arc

length formula s = rθ.

3024° = 3024° ( π radians _

180° )

= 16.8π radians

Substitute r = 13 and θ = 16.8π.s = rθ

= 13(16.8π)

≈ 686.124 or about 686 feet

b. Let r 1 represent the distance the first rider is from the center of the carousel and r 2 represent the distance the second rider is from the center.s = r 2 θ - r 1 θ

= 18(16.8π) - 13(16.8π) ≈ 263.894 or about 264 feet

35. 1.125 rev _

min 37. 6 m 39. 16.3 in. 41a. 1292π rad

_ min

to

1680π rad _

min 41b. 50 mi/h to 65 mi/h 43. 2.0 i n 2 45. 90.5 y d 2

47. 500.5 f t 2 49. 12.7 i n 2

51 First, convert 68° to radians.

θ = 68° ( π

_

180 )

= 17 _

45 π radians

Use the area formula for a sector to find the radius.

A = 1 _ 2 r 2 θ

29 = 1 _ 2 r 2 ( 17

_ 45

π) A = 29 and θ = 17

_ 45

π

29 = 17 _

90 π r 2 Multiply.

r 2 = 2610 _

17π

Solve for r 2 .

r = √

��

2610 _

17π

Take the positive square root of each side.

r ≈ 6.99 or about 7 ft

53. 12 in. 55a. Quadrant I: 0 < θ < π

_

2 55b. Quadrant II:

π

_

2 < θ < π 55c. Quadrant III: π < θ < 3π

_

2

55d. Quadrant IV: 3π

_

2 < θ < 2π 57. about 537 mi

59. 3.3 radians or about 191° 61. 1.5 radians or about 85.9° 63a. about 18.8 in. 63b. about 5.2 ft

65 Let θ 1 represent the sector with a central angle of 53° and θ 2 represent the sector with a central angle of 72°. First, convert each angle to radian measure.

53° = 53° ( π radians _

180° ) or 0.29π

72° = 72° ( π radians _

180° ) or 0.40π

Then use the area formula for a sector to find the area of the entire shaded region.

A = 1 _ 2 r 2 θ 1 + 1 _

2 r 2 θ 2

= 1 _ 2 (11 ) 2 (0.29π) + 1 _

2 (11 ) 2 (0.40π)

≈ 132 i n 2

67a. about 0.43 mi/h 67b. about 41 mi/h 69. The angle does not have a complement since it is greater than π

_

2 or 90°;

supplement: π

_ 6 71. Complements and supplements are not

defined for negative angles. 73. Sarah; sample answer: The formula for the length of an intercepted arc is s = rθ. Therefore,

the perimeter of the sector is the sum of the length of the intercepted arc and twice the radius, or P = rθ + 2r. Because P = 10r, using substitution, 10r = rθ + 2r. Upon simplifying, θ = 8 radians. 75. Decrease; sample answer: If the radius decreased, then the linear speed would also decrease because the linear speed is directly proportional to the radius. 77. Increase; sample answer: The equation for linear speed can also be written as v = rω. If the angular speed increased, then the linear speed would also increase because the linear speed is directly proportional to the angular speed. 79a. The perimeter would double; sample answer: The perimeter of a sector of a circle P is equal to the sum of the arc length s and two times the radius r, so P = s + 2r. Because s = rθ, if the radius doubled, the arc length would become a new arc length s’ = (2r)θ, which is equal to s’ = 2(rθ) or s’ = 2s. So, the perimeter would be P = 2s + 2(2r) or P = 2(s + 2r). Therefore, the perimeter would double. 79b. The area would quadruple; sample

answer: Because A = 1 _ 2 r 2 θ, if the radius doubled, the area would

become a new area A’ = 1 _ 2 (2r ) 2 θ, which is equal to A’ = 1 _

2 4 r 2 θ or A’

= 4 (

1

_

2

r 2 θ) . Therefore, the area would quadruple.

81. cos θ = √ �� 161

_ 15

, tan θ = 8 √ �� 161

_ 161

, csc θ = 15 _

8 , sec θ =

15 √ �� 161 _

161 , cot θ

= √ �� 161

_ 8 83. sin θ =

19 √ � 26 _

130 , cos θ =

17 √ � 26 _

130 , tan θ = 19

_ 17

,

csc θ = 5 √ � 26

_ 19

, sec θ = 5 √ � 26

_ 17

85. 2 ln 5 - 4 ln 2 87. ln 2 - ln 5

89. ±1, ±2, ±4, ±5, ±10, ±20; -2, 2, 5 91. f (x) → -∞ as x → -∞, f (x) → ∞ as x → ∞ 93. h (x) → 2 as x → -∞, h(x) → 2 as x → ∞ 95. {x|-4 ≤ x < 10, x ∈ }; [-4, 10) 97. B 99. C

Lesson 3-3

1. sin θ = 4 _ 5 , cos θ = 3 _

5 , tan θ = 4 _

3 , csc θ = 5 _

4 , sec θ = 5 _

3 ,

cot θ = 3 _ 4 3. sin θ = -

3 _

5 , cos θ = -

4 _

5 , tan θ = 3 _

4 , csc θ = -

5 _

3 ,

sec θ = -

5 _

4 , cot θ = 4 _

3 5. sin θ = -

8 √ � 65 _

65 , cos θ =

√ � 65 _

65 ,

tan θ = -8, csc θ = -

√ � 65 _

8 , sec θ = √ � 65 , cot θ = -

1 _

8

7. sin θ = 15 _

17 , cos θ = -

8 _

17 , tan θ = -

15 _

8 , csc θ = 17

_ 15

,

sec θ = -

17 _

8 , cot θ = -

8 _

15 9. 1 11. undefined 13. 0 15. 0

17. 45° 19. 5π

_

12

y

x

135°

y

x

7π

12

21. 45° 23. π

_

6

O

y

x

-405°

y

x

5π

6



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25. -

1 _

2 27.

√ � 2 _

2 29. 2 31. -

√ � 3 _

3 33. sin θ =

2 √ � 5 _

5 ,

cos θ = √ � 5

_ 5 , csc θ =

√ � 5 _

2 , sec θ = √ � 5 , cot θ = 1 _

2

35. cos θ = 2 √ � 6

_ 5 , tan θ = -

√ � 6 _

12 , csc θ = -5, sec θ =

5 √ � 6 _

12 ,

cot θ = -2 √ � 6 37. sin θ = -

√ � 6 _

3 , cos θ =

√ � 3 _

3 , tan θ = - √ � 2 ,

csc θ = -

√ � 6 _

2 , cot θ = -

√ � 2 _

2 39. sin θ = -

√ � 2 _

2 , cos θ =

√ � 2 _

2 ,

csc θ = - √ � 2 , sec θ = √ � 2 , cot θ = -1

41 Let the center of the carousel represent the origin on the coordinate plane and Zoe’s position after the 210° rotation have coordinates (x, y). The seat rotates 210°, so the reference angle is 210° - 180° or 30°. Because the final position of the seat corresponds to Quadrant III, the sine and cosine of 210° are negative.

y

x

210°

40

cos θ = x _ r sin θ = y _ r

cos 210° = x _ 40

sin 210° = y _

40

-cos 30° = x _ 12

sin 30° =

y

_ 12

-

√ � 3 _

2 = x _

40 -

1 _

2 =

y _

40

-20 √ � 3 = x -20 = yTherefore, the coordinates of Zoe’s seat after a rotation of 210°

are (-20 √ � 3 , -20) or (-34.6, -20).

43. -2 45. 1 _ 2 47. 2 49. undefined 51. √ � 3 53. - √ � 3

55. √ � 3

_ 2 57. - √ � 3

59. t 0 0.5 1 1.5 2 2.5

θ 22 0 -22 0 22 0

61. π

_

2 63. 4π

_ 3 or 5π

_ 3 65. π

_

6 67. 8 _

11 , 11

_ 8 69. 12

_ 13

, 12 _

13

71. sin θ = -

2 √ � 5 _

5 , cos θ = -

√ � 5 _

5 , tan θ = 2, csc θ = -

√ � 5 _

2 ,

sec θ = - √ � 5 , cot θ = 1 _ 2 73. (-

5 _

2 ,

5 √ � 3 _

2 ) 75. (-4, -4 √ � 3 )

77a. 11 m 77b. 7:00 a.m. and 7:00 p.m.

79 a. On the unit circle, cos θ = 0 at π

_

2 and 3π

_

2 . Since the cosine

function has a period of 2π, cos θ will also equal 0 at 2π

multiples of π

_

2 and 3π

_

2 . Therefore,

cos (n · π

_ 2 ) = 0 when n is an odd integer.

b. The cosecant function is the reciprocal of the sine function. So, csc θ will be undefined when sin θ = 0. On the unit circle, sin θ = 0 at 0, π, and 2π multiples of those angles. Therefore, csc

(n · π

_ 2 ) is undefined when n is an even

integer.81. False; sample answer: The expression tan (-t) = -tan t means that tangent is an odd function. The tangent of an angle depends on what quadrant the terminal side of the angle lies in.

83. Sample answer: The sine function is represented by the y-coordinate on the unit circle. Comparing sin t and sin (-t) for different values of t, notice that the y-coordinate is positive for sin t and is negative for sin (-t). For instance, on the first unit circle, sin t = b and sin (-t) = -b. Now find -(sin t) to verify the relationship. -(sin t) = -(b) or -b, which is equivalent to sin (-t). Thus, -sin t = sin (-t).

y

x

b

-b

sin t = b

sin (-t) = -b

-t0

π

2

t

y

x

d

-d

sin t = d

sin (-t) = -d

-tt 0

π

2

85. Sample answer: Since tan t = sin t _ cos t , we can analyze -tan t and

tan (-t) by first looking at tan t and tan (-t) on the unit circle for a given value of t. Regardless of the sign of t, the value of cosine remains the same. However, the value of sine is positive for t but negative for -t. This results in tan t = a _

b , but tan (-t) = -

a _ b . Now

find -tan t to verify the relationship. -tan t = - (

a _ b ) or -

a _ b which

is equivalent to tan (-t). Thus, -tan t = tan (-t).

y

x

tan t = a_b

tan (-t) = -a_b

-t0

π

2

t

y

x

sin (-t) = -a

cos (-t) = b

-t0

π

2

cos t = b sin t = a

t

87. 168° 21′ 89. 14.089° 91a. 6° 91b. about 3.2 in. 93. 1 _ 3

95. 3.5 97. ±1, ±3; -3 99. ±1, ±3, ±

1 _

2 , ±

3 _

2 ; -3, 1 _

2 , 1

101. ±1, ±2, ±

1 _

4 , ±

1 _

2 ; -

1 _

4 103. 45° 105. J

Lesson 3-4

1. The graph of g(x) is the graph of f (x) compressed vertically.

The amplitude of g(x) is 1 _ 2 .

f (x) = sin x

-1

1

x

y

π 2π 3π 4π

g(x) = sin x12

3. The graph of g(x) is the graph of f (x) expanded vertically. The amplitude of g(x) is 6.

g(x) = 6 cos x

x

y

-2

-4

-6

2

4

6

2π 4π

f (x) = cos x



Selected A

nswers and S

olutions

5. The graph of g(x) is the graph of f (x) compressed horizontally.

The period of g(x) is π

_

2 .

-1

1

x

y

O

g(x) = sin 4x

f(x) = sin x

2ππ

7. The graph of g(x) is the graph of f (x) expanded horizontally. The period of g(x) is 10π.

f (x) = cos x

x

y

-1

1

2π 4π 6π 8π 10π 12π 14π

g(x) = cos 1_5

x

9. Sample answer: y = 0.15 sin 330πt 11. Sample answer: y = 0.25 sin 1864πt 13. Sample answer: y = 0.2 sin 1246πt 15. amplitude = 1; period = 6π; frequency = 1

_ 6π

;

phase shift = -

3π

_

2 ; vertical shift = 0

-1

1

x

y

3π

29π

215π

2-3π_2

y = cos ( x_3

+ π_2 )

17. amplitude = 1; period = 2π

_

3 ; frequency = 3

_ 2π

;

phase shift = 0; vertical shift = -2

x

y

Oπ

2π

32π

3

5π

6π 7π

6

y = sin 3x - 2

-1

-2

-3

π

6

19. amplitude = 1; period = 2π; frequency = 1 _

2π

;

phase shift = -

5π

_


x

y

π-π

y = sin (x + 5π_6 ) + 42

4

21a. amplitude = 5.465; period = 13; phase shift = 4.417; vertical

shift = 7.485 21b. Sample answer: y = 5.465 cos ( π

_ 6.5

x - π

_ 1.47

) +

7.485 21c. about 7.28 ft 23. x = -

π

_

4 and 3π

_

4 25. x = -

π

_

4

and 3π

_

4 27. none

29 First, find the period of y = 1.5 sin (2t + c).

Period = 2π

_

|b|

Period formula

= 2π

_

|2|

or π b = 2

Since the period is π, the function will reach a maximum

height at π

_

4 radians. The phase shift is the difference between

the horizontal position of the function at π

_

4 and 7π

_

12 radians,

which is π

_

3 radians. Substitute π

_

3 and b into the phase shift

formula to find c.

Phase shift = -

c _ |b|

Phase shift formula

π

_

3 = -

c _ |2|

phase shift = π

_

3 and b = 2

-

2π

_

3 = c Multiply each side by -2.

Therefore, the equation is y = 1.5 sin (2t -

2π

_ 3 ) .

31. Sample answer: y = 3 sin 2x 33. Sample answer: y = 2 cos 4x + 1 35. Sample answer: y = 5 cos 2x

37. Sample answer: y = 3 _ 2 cos 4x

39a.

[-20, 20] scl: 5 by [-40, 40] scl: 5

39b. The graph of y = 2x cos x oscillates between the graphs of y = 2x and y = -2x.

39c.

[-20, 20] scl: 5 by [-200, 200] scl: 50

39d. The graph of y = x 2 sin x oscillates between the graphs of y = x 2 and y = - x 2 . 39e. The graph of y = f (x) sin x or y = f (x) cos x will oscillate between the graphs of y = f (x) and y = -f (x).

41. True; sample answer: The graph of y = cos x is a horizontal translation of graph of y = sin x. Therefore, a cosine function can be written from any sine function using the same amplitude and period by applying the necessary phase shift.

43 The graph of y = cos x completes one cycle on the interval [0, 2π]. Therefore, the graph of y = cos 1500x will complete 1500 cycles on [0, 2π]. Since cos 1500x has two zeros per cycle and there are 1500 cycles, there will be 1500(2) or 3000 zeros.

45. Sample answer: Although the pulse of light can be represented as a function with a period, it is not a sinusoidal function because the distance the pulse of light is from the ground changes at a constant rate. As a result, the graph of this function would resemble the graph below.



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47. sin θ = -

√ � 17 _

17 , cos θ =

4 √ � 17 _

17 , tan θ = -

1 _

4 , csc θ = - √ � 17 , sec θ

= √ � 17

_ 4 , cot θ = -4 49. sin θ =

5 √ � 41 _

41 , cos θ =

4 √ � 41 _

41 , tan θ = 5 _

4 , csc

θ = √ � 41

_ 5 , sec θ =

√ � 41 _

4 , cot θ = 4 _

5 51. -

7π

_

3 53. 480° 55. 3 real

zeros and 2 turning points; -4, 0, and 2 57. 5 real zeros and 4 turning points; -2, 0, and 2 59. yes; f –1 (x) = -x - 2 61. no 63. C 65. C

Lesson 3-5

1.

π

y

-2

-1

-3

3

2

1

-π

y = 2 tan x

x

3. y

-2

2

2π

3

y = cot (x - )

π_6

x 2π_3

5.

π

2

y

-2

-1

-3

3

2

1

y = - cot x 1_4

x

π_2

7. y

-2

-1

-3

3

2

1

π

4

y = -2 tan (6x - π)

x π_4

9. y

-1

-0.5

-1.5

1.5

1

0.5

π-π

x

y = csc 2x 1_5

11.

π

y

-2

-1

-3

3

2

-π 2π-2π

y = sec (x + π)

x

13.

π

y

-4

-2

-6

6

4

2

-π 2π-2π

y = 4 sec (x - 3π_4 )

x

15. y

-2

-1

-3

3

2

4π

32π

3-4π

3-2π

3

y = csc (x - 2π_3 )

3_2

x

17. The damping factor is 3 _ 5 x;

the amplitude of the function is decreasing as x approaches 0 from both directions.

[-5π, 5π] scl: by [-10, 10] scl: 1π

2

19. The damping factor is 2 x 2 ; the amplitude of the function is decreasing as x approaches 0 from both directions.

[-2π, 2π] scl: by [-50, 50] scl: 5π

2

21. The damping factor is 1 _ 3 x;

the amplitude of the function is decreasing as x approaches 0 from both directions.

[-5π, 5π] scl: by [-10, 10] scl: 1π

2

23. The damping factor is e 0.5x ; the amplitude of the function is decreasing as x approaches -∞.

[-π, 3π

] scl: by [-50, 50] scl: 10π

2

25. The damping factor is |x|; the amplitude of the function is decreasing as x approaches 0 from both directions.

[-π, π] scl: by [-2.5, 2.5] scl: 0.5π

4

27a. Sample answer: y = 8 e -3t cos 5πt 27b. about 0.06 second

29. y

-8

-4

-12

12

8

4

2ππ-2π -π

y = sec x + 3

x

31. y

2π-2π

6

4

2

x

y = csc x_3

2

-2

-4

-6



Selected A

nswers and S

olutions

33. y

-π π

-4

-2

-6

6

4

2

x

y = cot (2x + π) 3

35 a.

θ

150 ft

Since d is the length of the hypotenuse and you are given the length of the side opposite θ, you can use the sine function to write d as a function of θ.

sin θ = opp

_ hyp

sin θ = 150 _

d

d sin θ = 150

d = 150 _

sin θ or d = 150 csc θ

b. d

θ

160

240

80 d = 150 csc θ

π

4π

23π

4

c. Substitute θ = 45° into the function you found in part a.d = 150 csc θ = 150 csc 45° = 150 √ � 2 ≈ 212.1 ft

37. 0.427 and 2.715 39. -2.880, -1.833, 0.262, and 1.309 41. -2.356 and 0.785 43a. 5331.2 N

43b. T = 2665.6 sec θ _ 2

43c.

[0, 180°] scl: 45° by [0, 8000] scl: 1000

43d. total rope: about 12.9 m; tension: about 3769.7 N 43e. θ ≈ 97.0°; about 4023 N

45. b 47. a 49. 51.

[-2π, 2π] scl: by [-2, 2] scl: 0.5π

2 [-2π, 2π] scl: by [-2, 2] scl: 0.5π

2

The expressions are The expressions are notequivalent. equivalent for all real numbers.

53 The general form of the tangent function is y = a tan (bx + c) + d,

where the amplitude is equal to |a|, the period is equal to π

_

b ,

the phase shift is equal to -

c _ |b|

, and the vertical shift is equal to

d. Since we are given the period, phase shift, and vertical shift,

we can find the values of b, c, and d and substitute them into the general form of the tangent function.

period = π

_

2 phase shift = -

c _ |b|

π

_

|b|

= π

_

2 π

_

4 = -

c _ ±2

π = π|b|

_ 2 π

_

4 = -

c _ 2

2 = |b| -

π

_

2 = c

±2 = bvertical shift = -1Therefore, a tangent function with a period π

_

2 , a phase shift

of π

_

4 units to the right, and a vertical shift of 1 unit down is

given by y = tan (2x - π

_ 2 ) - 1 or

y = tan (-2x - π

_ 2 ) - 1.

55. y = cot ( x _ 3 - π

_ 6 ) + 4, y = cot

(-

x _ 3 - π

_ 6 ) + 4

57. To find the y-intercept, let t = 0.y = k e -ct cos wt Original equation

= k e -c(0) cos (w · 0) Substitute t = 0.

= k e 0 cos 0 Multiply.

= k(1)(1) e 0 = 1, cos 0 = 1

= k Simplify.

59. True; sample answer: Since y = csc x = 1 _

sin x , asymptotes will

occur for values of x when sin x = 0. Since y = cot x = cos x _ sin x

,

asymptotes will also occur for values of x when sin x = 0.

61. Sample answer: y = csc (x + π

_

2 ) ; y = -cot

(x - π

_

2 )

63. amplitude = 3; period = π; frequency = 1 _ π

; phase shift = π

_


y

π-π

20

10

15

5

-10

-5

+ 10y = 3 sin (2x - )

π

3

xπ_2

π_2

65. amplitude = 1 _ 2 ; period = π

_

2 ; frequency = 2 _

π

;

phase shift = π

_


y2

1

-1

π

4π

2 π_2

x π_4

y = 1_2

cos (4x ) + 1

67. sin θ = √ � 37

_ 37

, tan θ = 1 _ 6 , csc θ = √ � 37 , sec θ =

√ � 37 _

6 ,

cot θ = 6 69. about 3.6% 71. (x + 1)(x + 2)(x - 1) 73. (x - 4)(x + 2)(x + 1) 75. C 77. B



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Lesson 3-6

1. 0 3. π

_

4 5. -

π

_

4 7. π

_

4 9. π

_

6 11. π

_

4 13. π

_

6 15. π

_

6

17.

y = arcsin x

x

y

π

2

π

2

1-1

-

19.

x

y

π

2

π

2

π

4

1-1-2-3-4

-

π

4-

y = sin 1(x + 3)

21.

x

y

y = arccos x

-1-2 21

π

π

2

π

2

3π

2

-

23.

x

yπ

2

π

2

y = arctan x

-1-2 21

-

25.

x-1-2 21

y = tan 1 (x + 1)

yπ

π

2

-π

π

2-

27 a. θ

30 m

Since x is the length of the side opposite θ and you are given the length of the side adjacent to θ, you can use the tangent function to write θ as a function of x.

tan θ = opp

_ adj

tan θ = x _ 30

θ = ta n -1 x _ 30

or arctan x _ 30

b. Substitute x = 6 and x = 14 into the function you found in part a.

θ = ta n -1 x _ 30

θ = ta n -1 x _ 30

θ = ta n -1 6 _ 30

θ = ta n -1 14 _

30

≈ 11.3° ≈ 25.0°

29. 3 _ 4 31. 2 _

9 33. π

_

4 35.

√ � 2 _

2 37. 1 39.

√ � 2 _

2

41 Let u = arccos x, so cos u = x. You can draw a right triangle, where the side adjacent to u is equal to x and the hypotenuse is equal to 1.

1

The domain of the inverse cosine function is restricted to Quadrants I and II, so u must lie in one of those quadrants. Using the Pythagorean Theorem, you can find the length of

the side opposite u to be √ ��

1 2 - x 2 . Now, find tan u.

tan u = opp

_ adj

= √

�� 1 2 - x 2 _ x or

√ ��

1 - x 2 _ x

So, tan (arccos x) = √

�� 1 - x 2 _ x

43. √ ��

1 - x 2 45. 1 _ x 47. x √

�� 1 − x 2 _

1 − x 2 49. The graph of g(x) is the

graph of f (x) translated 1 unit to the right and 2 units down. 51. The graph of g(x) is the graph of f (x) expanded vertically and translated 6 units down 53. The graph of g(x) is the graph of f (x) compressed horizontally and translated 5 units up. 55a. about 36° 55b. about 11 ft

57. D = {x|-1 ≤ x ≤ 1}; 59. D = {x|x ∈ �};

R = {y|0 ≤ y ≤ 1} R = {y|-

π

_ 2 ≤ y ≤

π

_ 2 }

[-2, 2] scl: 1 by [-2, 2] scl: 1

[-5, 5] scl: 1 by [- , ] scl: π

2π π

61. D = {x|-1 ≤ x ≤ 1, x ≠ 0}; R = {y|y ≠ 0}

[-2, 2] scl: 1 by [-10, 10] scl: 2



Selected A

nswers and S

olutions

63. tan (si n -1 x) or cot (co s -1 x) 65a. Sample answer: For f ◦ f −1 , the domain is [−1, 1] and the range is [−1, 1]. For

f −1 ◦ f, the domain is (−∞, ∞) and the range is ⎡

⎣

-

π

_ 2 , π

_ 2 ⎤

⎦

.

65b. Sample answer:

x -2.0 -1.5 -1.0 -0.75 -0.5 -0.25 0

f ◦ f -1 -1.0 -0.75 -0.5 -0.25 0

x 0.25 0.5 0.75 1.0 1.5 2.0

f ◦ f -1 0.25 0.5 0.75 1.0

x -2.0 -1.5 -1.0 -0.75 -0.5 -0.25 0

f -1 ◦ f -1.142 -1.5 -1.0 -0.75 -0.5 -0.25 0

x 0.25 0.50 0.75 1.0 1.5 2.0

f -1 ◦ f 0.25 0.50 0.75 1.0 1.5 1.142

f ◦ f −1 f −1 ◦ f

[-2, 2] scl: 1 by [-2, 2] scl: 1 [-2, 2] scl: 1 by [-2, 2] scl: 1

65c. Sample answer: For g ◦ g -1 , the domain is [-1, 1] and the range is [-1, 1]. For g -1 ◦ g, the domain is (-∞, ∞) and the range is [0, π].The graph of g ◦ g -1 should be the line y = x for -1 ≤ x ≤ 1. The inverse property of trigonometric functions states that on the closed interval [-1, 1], cos (co s -1 x) = x. The graph of g -1 ◦ g should be the line y = x for 0 ≤ x ≤ π. Once the graph reaches π, it will turn and decrease until it reaches the x-axis at the same rate. When it reaches the x-axis, it will turn again and increase until it reaches π. It will continue to do this as x approaches infinity.

65d. g ◦ g −1 g −1 ◦ g


65e. Sample answer: For f (x) = tan (ta n -1 x), due to the inverse property of trigonometric functions, for all values of x, f (x) = x. This should result in the line y = x for all real numbers. The graph of g(x) = ta n -1 (tan x) will be different because tan x is undefined for multiples of π. As a result, asymptotes for multiples of π can

be expected. We can also expect a range of ⎡

⎣

-

π

_ 2 , π

_ 2 ⎤

⎦

due to the definition of arctan.

f (x ) = tan(ta n -1 x ) g (x ) = ta n -1 (tan x )

[-10, 10] scl: 1 by [-10, 10] scl: 1 [-5, 5] scl: 0.5 by [-π, π] scl: π

2

67.

x

y

π

2

π

2

(0.5, )

y = nis -1 x

π

6

-1 1

-

x

yπ

1-1

(0.5, )

y = soc -1 x

π

3

Notice from the graphs of y = si n -1 x and y = co s -1 x that when

x = 0.5, si n -1 x = π

_

6 and co s -1 x = π

_

3 . So, si n -1 x + co s -1 x is π

_

6

+ π

_

3 or π

_

2 . When x = 1, si n -1 x = π

_

2 and co s -1 x = 0, so si n -1

x + co s -1 x is π

_

2 + 0 or π

_

2 . When x = -1, si n -1 x = -

π

_ 2 and co s -1 x

= π, so si n -1 x + co s -1 x is -

π

_

2 + π or π

_

2 . Therefore, on the

interval [-1, 1] it appears that si n -1 x + co s -1 x = π

_

2 . The graph of

y = si n -1 x + co s -1 x supports this conjecture.

y

x−0.5 0.5

y = sin-1x + cos-1x

π

4

3π

4

3π

4-

π

4-

69. Sample answer: The function is odd. The definition of an odd function states for every x in the domain of f, f (-x) = -f (x). If we let si n -1 x = u, we have x = sin u. From Lesson 3-3, we know that the sine function is odd, so -x = sin (-u). From here, we can get si n -1 (-x) = -u. Graphically, it can be seen that for every x in the domain of f, f (-x) = -f (x). 71. Sample answer: The function is odd. The definition of an odd function states for every x in the domain of f, f (-x) = -f (x). If we let ta n -1 x = u, we have x = tan u. From Lesson 3-3, we know that the tangent function is odd, so -x = tan (-u). From here, we can get ta n -1 (-x) = -u. Graphically, it can be seen that for every x in the domain of f, f (-x) = -f (x).

73.

xO

2

4

-4

-2

y = 3 tan θ

π

2π

2-

y

75. y

-8

-4

8

4

-3π -π π 3π

θ

y = 3 csc 1_2 θ

77. 5.8 79. 6 81. [ f ◦ g](x) = 6 - 5 _ x ; [ g ◦ f ](x) = 1 _

6 - 5x ;

[ f ◦ g](4) = 4.75 83. 10 questions 85. H 87. F



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Lesson 3-7

1. A = 32°, a = 11.2, b = 19.8 3. K = 82°, j = 16.2, � = 21.4 5. T = 20°, t = 29.0, u = 17.7 7. about 0.85 ft

9 a. Let B = 8°, C = 157°, and a = 90.

8°90 mi

157°

First, find A.m ∠A = 180° - 8° - 157° or 15° Since the measures of A, B, and a are known, you can use the Law of Sines to find b.

sin 15° _

90 = sin 8°

_ b

b = 90 sin 8° _

sin 15° or 48.4 mi

So, the entire trip is 90 + 48.4 or 138.4 miles.

b. If the direct flight distance is c,

sin 15° _

90 = sin 157°

_ c

c = 90 sin 157° _

sin 15°

= 135.9Therefore, the distance of a direct flight to the destination is about 135.9 miles.

11. no solution 13. B = 43°, C = 45°, c = 24.7 15. B = 90°, C = 60°, c = 32.9 17. B = 90°, C = 45°, c = 10 19. B = 63°, C = 78°, c = 18.7 and B = 117°, C = 24°, c = 7.8 21. B = 70°, C = 49°, c = 12.2 and B = 110°, C = 9°, c = 2.5 23. B = 70°, C = 56°, c = 31.8 and B = 110°, C = 16°, c = 10.6 25. 33.4 mi 27. B = 39°, C = 99°, a = 12.9 29. Q = 27°, R = 80°, p = 14.6 31. R = 96°, S = 39°, T = 45° 33. C = 162°, D = 2°, b = 24.1 35. 15.6°; 71.5°; 92.9° 37. 47.6 c m 2 39. 1133.0 f t 2

41. 20.9 y d 2 43a. 4511.5 step s 2 43b. 14,617 f t 2 45. 51.5 m m 2 47. 313.2 f t 2 49. 259.8 i n 2 51. about 22.3 ft or about 26.1 ft

53 To find the total area of the figure, find the sum of the areas of

�FJE and �FGH.

For �FJE: s = 1 _ 2 (a + b + c)

= 1 _ 2 (12 + 20 + 27)

= 29.5

Area = √ �� s(s - a)(s - b)(s - c)

= √ �� 29.5(29.5 - 12)(29.5 - 20)(29.5 - 27)

≈ 110.7 m m 2

For �FGH: s = 1 _ 2 (17 + 21 + 28)

= 33

Area = √ �� 33(33 - 17)(33 - 21)(33 - 28)

≈ 178.0 m m 2

So, the total area is 110.7 + 178.0 or 288.7 m m 2 .

55. 1062.6 f t 2 57. South Bay is about 25.72 mi from the boat and Steep Rock is about 31.92 mi from the boat. 59. 994.2 m m 2 61. 884.2 i n 2 63. ≈ 40.9 m 65. Neither; for the acute case, h = 21 sin 34° or 11.7. Because a < b and h < a, there are two solutions.

67. Sample answer: The domain of inverse cosine includes angles measured 0 to 180 degrees. The domain of inverse sine includes angles measured -90 to 90 degrees. 69.

BB

A A

C C

h1h1

h2

h2

b baa

c

c

A is acute. A is obtuse.

Sample answer: Let h 1 be an altitude of either triangle shown

above. From the definition of the sine function, sin A = h 1

_ b

or h 1 = b sin A, and sin B = h 1

_ a or h 1 = a sin B. Equating the two

values of h 1 , b sin A = a sin B, or sin A _ a = sin B _ b , where sin A ≠ 0

and sin B ≠ 0. When an altitude h 2 is drawn from vertex B to side

AC (extended in the obtuse triangle), sin A = h 2

_ c or h 2 = c sin A,

and sin C = h 2

_ a or h 2 = a sin C. Equating the two values of h 2 , c sin

A = a sin C, or sin A _ a = sin C _ c . By the Transitive Property of equality,

sin A _ a = sin B _ b = sin C _ c .

71 Draw a diagram to represent the situation. 35°850 mi. XB

C

A2110 mi.

To find the amount of time that it takes to complete one full orbit, we need to find the measure of the arc intercepted by points B and X. From the information given, m∠XAC = 90° + 35° or 125° and the length of XC is 2110 + 850 or 2960 miles. Now that we have two side lengths and an angle measure, we can find m∠C in �XAC.

X

A

C

125°

2960 mi.2110 mi.

First, use the Law of Sines to find ∠X.

sin 125° _

2960 = sin X _

2110

2110 sin 125° = 2960 sin X

2110 sin 125° __

2960 = sin X

si n -1 ( 2110 sin 125° __

2960 ) = X

35.73° ≈ X

So, ∠ACX is 180° - (125° + 35.73°) or about 19.27°. Because point B is directly above point A, ∠ACX ∠BCX. From

geometry, m∠BCX = m XB . Therefore, m XB ≈ 19.27°. Find the amount of time that it takes to complete one full orbit or 360°.

19.27° _

14 = 360°

_ x

19.27°x = (360°)(14) x ≈ 261.55Convert 261.55 minutes to hours.

261.55 min · 1 h _

60 min ≈ 4.36 h



Selected A

nswers and S

olutions

Therefore, one full orbit takes approximately 4.36 hours or 4 hours and 22 minutes.

73. 2π

_

3 75. π

_

4

77. f (x) = -2x; the amplitude of the function is decreasing as x approaches 0 from both sides.

[ 4π, 4π] scl: π by [ 20, 20] scl: 4

79. f (x) = (x - 1 ) 2 ; the amplitude of the function is decreasing as x approaches 0 from both sides.

[ 8π, 8π] scl: 2π by [ 150, 150] scl: 50

81. Latitude Distance

0° 24,881.4

30° 21,547.9

45° 17,593.8

60° 12,440.7

90° 0

The distances range from about 24,881 mi to 0 mi.

83. Sample answer: f (x) = x 3 - 5 x 2 - x + 5 85. Sample answer:

f (x) = x 3 + 3 x 2 + 4x + 12 87. E

89a. θ

θ =

1_4

cos 12t

-0.25

0.25

t1 2

89b. The period is π

_

6 or about 0.524 seconds. This represents the

amount of time it takes for the pendulum to complete one full

swing or cycle. The amplitude is 1 _ 4 . This represents the maximum

angular displacement of the pendulum. The frequency is 6 _ π

or

about 1.91. This represents the number of swings the pendulum completes per second. 89c. The maximum angular displacement is 14.3°. 89d. The midline represents when the pendulum is vertical and there is no angular displacement. 89e. about 0.101 + 0.524n seconds and 0.161 + 0.524n seconds


1. true 3. false, angle of depression 5. true 7. false, amplitude

9. false, arccosine 11. sin θ =

12 _

13 , cos θ = 5 _

13 ,

tan θ =

12 _

5 , csc θ =

13 _

12 , sec θ = 13

_ 5 , cot θ = 5 _

12 13. 5.0

15. 19° 17. 3π

_

4 19. 315°

21. 342° + 360n°, n is an integer; Sample answer: 702°, -18°y

x

342°702°

342°

-18°

y

x

23. 53.0 i n 2 25. 60° 27. π

_

4

y

x

240°

y

x

-3π

4

29. sin θ = √ � 21

_ 5 , tan θ =

√ � 21 _

2 , csc θ =

5 √ � 21 _

21 , sec θ = 5 _

2 ,

cot θ = 2 √ � 21

_ 21

31. cos θ = 12 _

13 , csc θ = – 13

_ 5 , sec θ = 13

_ 12

,

tan θ = -

5 _

12 , cot θ = -

12 _

5 33. 0 35. undefined

37. The graph of g(x) is the graph of f (x) stretched vertically. The amplitude of g(x) is 5, and the period is 2π.

-2

2

4

x

y

-4

g(x) = 5 sin x

f (x) = sin x

3ππ

39. The graph of g(x) is the graph of f (x) compressed vertically. The amplitude

of g(x) is 1 _ 2 , and the period

is 2π.

-0.5

0.5

1

x

y

-1

f (x) = sin x

π 3π-π

g(x) = 1_2

sin x

41. amplitude = 2; period = 2π; frequency = 1 _

2π

; phase shift = π; vertical shift = 0

3π

29π

2-1

1

2

y

-2

-3

3

3π3π

2

x-

y = 2 cos (x π)



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43. amplitude = 1 _ 2 ; period = 2π; frequency = 1

_ 2π

;

phase shift = -

π

_


-0.5

0.5

1

x

y

-1

3ππ-π

y = 1_2

cos (x +

π_2 )

45.

π

y

x

-4

-2

-6

6

4

2

-π-2π 2π

y = 3 tan x

47.

π

y

x-π-2π 2π

-2

2

y = cot (x +

π_3 )

49.

2π

y

x

-4

-2

4

2

-2π-4π 4π

y = 2 sec (

x_2 )

51.

π

y

x

-2

-1

-3

-4

4

2

1

3

-π

y = sec (x )

53. -

π

_

2 55. -

π

_

6

57. -

π

_

4 59. -

π

_

3

61. B = 12°, C = 146°, c = 16.4 63. B = 29°, C = 73°, c = 19.5 65. A = 78°, B = 65°, C = 37° 67a. 43 ft 67b. 0.07 69a. 150π/min or 2.5π/s 69b. 27,000°/min 71a. 60° 71b. 550 ft 71c. 2.5 min

73a. 12 h 30 min 73b. 11:15 p.m. 75. 6.3 ft 77a. 63° 77b. 84 unit s 2


1. d

t

70

140

210

280

1 2 3 4

3. Sample answer: If one of the cars slowed down, the distance between the two cars would grow at a slower rate. Therefore, the average rate of change of the distance between the two cars would be smaller. If one of the cars sped up, the distance between the two cars would grow at a faster rate. Therefore, the average rate of change of the distance between the two cars would be larger. 5. For t = 1, A ≈ 79 c m 2 . For t = 2, A ≈ 314 c m 2 . For t = 3, A ≈ 707 c m 2 . For t = 4, A ≈ 1257 c m 2 . For t = 5, A ≈ 1963 c m 2 . 7. For t = 2, m approaches 314 c m 2 /sec. For t = 3, m approaches 471 c m 2 /sec. For t = 4, m approaches 628 c m 2 /sec. 9. Sample answer: In the first activity, the graph resembled a linear function. Therefore, the average rate of change could be found by using any two points since the rate of change remained constant. In the second activity, the graph resembled a quadratic function. The difference quotient had to be used to approximate the rate of change at each individual value of t.

Appendix: Trigonometry and the Unit Circle

Guided Example 3 √ � 2

_ 2 , -

√ � 2 _

2 ,

-

√ � 2 _

2 _

√ � 2

_ 2

, -1

Guided Example 4 1 _

sin (-

7π

_ 6 )

; 1 _

2 ; 2;

cos ( π _ 2 ) _

sin ( π _ 2 ) ; 0 _

1 ; 0

Guided Example 5 √ � 85 ; √ � 85

_ 11

; 11 _

√ � 85 ; 6;

6 _

11 ;

11 _

6 ; opposite A,

adjacent to A, √ � 85

_ 6 ;

6 _

√ � 85

Questions

1a. 7π

_ 9 1b. (-0.766, 0.643)

3.

5. 1 _ 2 ,

√ � 3 _

2 ,

√ � 3 _

3 , √ � 3 ,

2 √ � 3 _

3 , 2 7. -

√ � 2 _

2 , -

√ � 2 _

2 , 1, 1, - √ � 2 , - √ � 2

9a. no 9b. These values do not satisfy the Pythagorean Identify: 0.7 2 + 0.3 2 = 0.58 ≠ 1 11a. (cos 3, sin 3) 11b. (-0.990, 0.141)13a. positive 13b. positive 15a. undefined 15b. zero17a. 15° 17b. 2520° 17c. about 1037.5 mph

19. -

41 _

40 21. By the first part the Logarithmic Graph Similarity

Theorem, the graphs of y = 2 · 3 x and y = log 3 x are congruent. By the second part, the graphs of y = log 3 x and y + log 5 x are similar. Therefore the graph of y = log 5 x is similar to the graph of y = 2 · 3 x . 23a. 1995 to 1997, 1998 to 1999, 2000 to 2002, 2004 to 2005

23b. 1995 to 1997, 2004 to 2005 23c. 7.6 mWh d.38.5 mWh



Selected A

nswers and S

olutions

Trigonometric Identities and Equations CHAPTER 4CHCHAPAPTETERR 44

Chapter 4 Get Ready

1. -8, 3 3. -

1 _

2 , 5 5. -2, 0, 3 7. 12 s 9. a ≈ 13.62,

b ≈ 23.30, A = 32° 11. k ≈ 9.13, J ≈ 42.63°, L ≈ 75.37°

13. √ � 2

_ 2 15. - √ � 3 17. 1 _

2

Lesson 4-1

1. 7 _ 5 3. 5 5.

√ � 35 _

35 7.

7 √ � 10 _

20 9. sec θ = √ � 26 , cos θ = 1

_ √ � 26

or √ � 26

_ 26

11. tan θ = √ � 15 , sin θ = √ � 15

_ 4 13. cos θ =

√ � 55 _

8 ,

tan θ = 3 _

√ � 55 or

3 √ � 55 _

55 15. cot θ = -

2 _

√ � 77 or -

2 √ � 77 _

77 ,

sin θ = √ � 77

_ 9 17. -1.24 19. 1.52 21. -1.35 23. sin x

25. 2 sin x 27. 2 cos x 29. ta n 2 x 31. sin x cos x 33. sin x 35. cot x - cos x + tan x + sin x tan x

37 a. I = I 0 - I 0 _

cs c 2 θ

= I 0 (1 - 1

_ c sc 2 θ

)

= I 0 (1 - si n 2 θ)

= I 0 c os 2 θ

b. I = I 0 co s 2 θ

= I 0 ( √ � 3

_ 2 )

2

= I 0 ( 3 _ 4 )

= 3 I 0

_ 4

Thus, 3 _ 4 of the original intensity emerges.

39. se c 2 x (csc x + 1) 41. sec x (csc x - 1) 43. 2 - 2 cos x 45. cos x (csc x + 1) 47. sec x - 1 49. Even; the graph shown is f (x) = sec x. f (-x) = sec (-x) = sec x = f (x). Since f (-x) = f (x), f (x) = sec x is an even function. 51. -cot x 53. -csc x 55a.

x -2π -π 0 π 2π

y 1 1 1 1 1 1

y 2 1 1 1 1 1

y 3 0 0 0 0 0

y 4 0 0 0 0 0

55b.

[-2π, 2π] scl: by [-4, 4] scl: 1

y1 = tan x + 1

π_2

[-2π, 2π] scl: by [-4, 4] scl: 1

y2 = sec x cos x - sin x sec x

π_2

[-2π, 2π] scl: by [-4, 4] scl: 1

y3 = tan x sec x - sin x

π_2

[-2π, 2π] scl: by [-4, 4] scl: 1

y4 = sin x tan2 x

π_2

55c. y 1 ≠ y 2 , y 3 = y 4 55d. Sample answer: The graphs of y 1 and y 2 support the conjecture that y 1 ≠ y 2 . Although the table from part a and the graphs of y 3 and y 4 support the conjecture that y 3 = y 4 , it is impossible to determine whether the two equations are equal over the entire domain unless the equations are verified algebraically. 57. -2 ln |cos x|

59. ln |cs c 2 x| 61a. csc θ = d _ mλ

61b. θ = 0.11 radian

63. Jenelle; 1 - si n 2 x __ si n 2 x - co s 2 x

= 1 - si n 2 x __ (1 - co s 2 x) - co s 2 x

= co s 2 x _ 1 - 2 co s 2 x

65. sin x = ± √ ��

1 - co s 2 x ,

csc x = ±

√

�� 1 - co s 2 x _

1 - co s 2 x , sec x = 1

_ cos x , tan x = ±

√

�� 1 - co s 2 x _ cos x ,

cot x = ±

cos x √ ��

1 -co s 2 x __

1 - co s 2 x 67. False; sample answer: The domain

does not include values of x where csc x is undefined, such as nπ.

69 According to the Pythagorean Theorem, x 2 + y 2 = r 2 . When each side of the equation is divided by x 2 ,

1 + y 2

_ x 2

=

r 2 _

x 2 . Since tan =

y _ x and sec = r _ x , 1 +

y 2 _

x 2 = r 2

_ x 2

is

equivalent to 1 + ta n 2 x = se c 2 x or ta n 2 x + 1 = se c 2 x.73. C = 73°, a ≈ 55.6, b ≈ 48.2 75. B ≈ 46°, C ≈ 69°, c ≈ 5.1

77. A ≈ 40°, B ≈ 65°, b ≈ 2.8 79. 3 81. √ � 2

_ 2 83. 4 _

5

85. True; all of the elements of A are also elements of B. 87. D 89. B

Lesson 4-2

1. (se c 2 θ - 1) co s 2 θ = (ta n 2 θ) co s 2 θ

= ( si n 2 θ

_ co s 2 θ

) co s 2 θ

= si n 2 θ

3. sin θ - sin θ co s 2 θ = sin θ (1 - co s 2 θ) = sin θ si n 2 θ = si n 3 θ

5. co t 2 θ cs c 2 θ - co t 2 θ = co t 2 θ (cs c 2 θ - 1) = co t 2 θ co t 2 θ = co t 4 θ



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7. sec θ

_ sin θ

- sin θ

_ cos θ

= 1

_ cos θ

_

sin θ - sin θ

_ cos θ

= 1 _

sin θ cos θ - si n 2 θ

_ sin θ cos θ

= 1 - si n 2 θ

_ sin θ cos θ

= co s 2 θ

_ sin θ cos θ

= cos θ

_ sin θ

= cot θ9. cos θ

_ 1 + sin θ

+ tan θ

= cos θ

_ 1 + sin θ

+ sin θ

_ cos θ

= cos θ

_ cos θ

· cos θ

_ 1 + sin θ

+ 1 + sin θ

_ 1 + sin θ

· sin θ

_ cos θ

= co s 2 θ + sin θ + si n 2 θ

__

cos θ (1 + sin θ)

= 1 + sin θ

__ cos θ (1 + sin θ)

= 1 _

cos θ

= sec θ

11. 1 _

1 - tan 2 θ + 1

_ 1 - co t 2 θ

= 1 _

1 - si n 2 θ

_ co s 2 θ

+ 1 _

1 - co s 2 θ

_ si n 2 θ

= 1 __

co s 2 θ

_ co s 2 θ

- si n 2 θ

_ co s 2 θ

+ 1 __

si n 2 θ

_ si n 2 θ

- co s 2 θ

_ si n 2 θ

= 1 __

co s 2 θ - si n 2 θ

__ co s 2 θ

+ 1 __

si n 2 θ - co s 2 θ

__ si n 2 θ

= co s 2 θ

__ co s 2 θ - si n 2 θ

+ si n 2 θ

__ si n 2 θ - co s 2 θ

= co s 2 θ

__ co s 2 θ - si n 2 θ

+ -1 _

-1 · si n 2 θ

__ si n 2 θ - co s 2 θ

= co s 2 θ

__ co s 2 θ - si n 2 θ

+ -si n 2 θ

__ co s 2 θ - si n 2 θ

= c os 2 θ - si n 2 θ

__ co s 2 θ - si n 2 θ

= 1

13. (csc θ - cot θ)(csc θ + cot θ) = csc 2 θ - cot 2 θ = 1

15. 1 _

1 - sin θ + 1

_ 1 + sin θ

= 1 + sin θ

_ 1 + sin θ

· 1 _

1 - sin θ + 1 - sin θ

_ 1 - sin θ

· 1 _

1 + sin θ

= 1 + sin θ

_ 1 - si n 2 θ

+ 1 - sin θ

_ 1 - si n 2 θ

= 2 _

1 - si n 2 θ

= 2 _

co s 2 θ

= 2 sec 2 θ

17. csc 4 θ - c ot 4 θ = (cs c 2 θ - co t 2 θ)(cs c 2 θ + co t 2 θ) = [cs c 2 θ - (cs c 2 θ - 1)][cs c 2 θ + (cs c 2 θ - 1)] = 2 cs c 2 θ - 1 = 2(co t 2 θ + 1) - 1 = 2 co t 2 θ + 2 - 1 = 2 co t 2 θ + 1

19 a. v 2 ta n 2 θ

_ 2g se c 2 θ

=

v 2 ( sin 2 θ

_ cos 2 θ

)

_ 2g

( 1

_ cos 2 θ

)

= v 2 sin 2 θ

_ 2g

b. h = v 2 si n 2 θ

_ 2g

= (110 ) 2 (sin 80 ) 2

__ 2(9.8)

≈ 598.7 m

21. ta n 2 θ - si n 2 θ = si n 2 θ

_ co s 2 θ

- si n 2 θ

= si n 2 θ

_ co s 2 θ

- si n 2 θ ( co s 2 θ

_ co s 2 θ

)

= si n 2 θ

_ co s 2 θ

- si n 2 θ co s 2 θ

_ co s 2 θ

= si n 2 θ - si n 2 θ co s 2 θ

__ co s 2 θ

= si n 2 θ (1 - co s 2 θ )

__ co s 2 θ

= si n 2 θ si n 2 θ

_ co s 2 θ

= si n 2 θ ( si n 2 θ

_ co s 2 θ

)

= si n 2 θ ta n 2 θ

23. 1 + csc θ

_

sec θ =

1 + 1 _

sin θ _

1 _

cos θ

= sin θ

_ sin θ

+ 1 _

sin θ

__

1 _

cos θ

= sin θ + 1

_ sin θ

· cos θ _

1

= sin θ cos θ + cos θ

__

sin θ

= sin θ cos θ

_ sin θ

+ cos θ

_ sin θ

= cos θ + cot θ

25. 1 + ta n 2 θ

_ 1 - ta n 2 θ

= 1 + si n 2 θ

_ co s 2 θ

_ 1 - si n 2 θ

_ co s 2 θ

= co s 2 θ

_ co s 2 θ

+ si n 2 θ

_ co s 2 θ

__ co s 2 θ

_ co s 2 θ

- si n 2 θ

_ co s 2 θ

= co s 2 θ + si n 2 θ

__

co s 2 θ

__ co s 2 θ - si n 2 θ

__

co s 2 θ

= co s 2 θ + si n 2 θ

__

co s 2 θ · co s 2 θ

__ co s 2 θ - si n 2 θ

= co s 2 θ + si n 2 θ

__ co s 2 θ - si n 2 θ

= 1 __

co s 2 θ - si n 2 θ

= 1 __

co s 2 θ - (1 - co s 2 θ)

= 1 _

2 co s 2 θ - 1

27. sec θ - cos θ = 1 _

cos θ - cos θ

= 1 _

cos θ - co s 2 θ

_ cos θ



Selected A

nswers and S

olutions

= 1 - co s 2 θ

_ cos θ

= si n 2 θ

_ cos θ

= tan θ sin θ

29. (csc θ - cot θ ) 2 = ( 1 _

sin θ -

cos θ

_ sin θ

) 2

= ( 1 - cos θ

_ sin θ

) 2

= (1 - cos θ ) 2

_ si n 2 θ

= (1 - cos θ ) 2

_ 1 - co s 2 θ

= (1 - cos θ) (1 - cos θ)

__ (1 - cos θ) (1 + cos θ)

= 1 - cos θ

_ 1 + cos θ

31. 2 + csc θ sec θ

__

csc θ sec θ = 2

_ csc θ sec θ

+ 1

= 2 · 1 _

csc θ · 1

_ sec θ

+ 1

= 2 sin θ cos θ + (si n 2 + co s 2 θ)

= si n 2 θ + 2sin θ cos θ + co s 2 θ

= (sin θ + cos θ ) 2

33. I m - I m _

co t 2 θ + 1 = I m

(1 - 1

_ co t 2 θ + 1

)

= I m (1 - 1

_ cs c 2 θ

)

= I m (1 - si n 2 θ)

= I m co s 2 θ

35.

[-2π, 2π

] scl: by [-4, 4] scl: 1π

2

Y1 = sec x + tan x

[-2π, 2π

] scl: by [-4, 4] scl: 1π

2

Y2 = 1 sec x - tan x

1 __

sec x - tan x = 1 __

1 _ cos x - sin x _ cos x

= 1 _

1 - sin x _ cos x

= cos x _ 1 - sin x

� 1 + sin x _ 1 + sin x

= cos x + sin x cos x __

1 - si n 2 x

= cos x + sin x cos x __

co s 2 x

= 1 _ cos x + sin x _ cos x

= sec x + tan x

37.

[-2π, 2π

] scl: by [-4, 4] scl: 1π

2

Y1 = cot2 x - 11 + cot2 x

[-2π, 2π

] scl: by [-4, 4] scl: 1π

2

Y2 = 1 - 2 sin2 x

co t 2 x - 1 _

1 + co t 2 x = co t 2 x - 1

_ cs c 2 x

= co s 2 x _ si n 2 x

- 1

_ 1

_ si n 2 x

= co s 2 x - si n 2 x = (1 - si n 2 x) - si n 2 x = 1 - 2 si n 2 x

39.

[-2π, 2π

] scl: by [-4, 4] scl: 1π

2

Y1 = cos2 - sin2

[-2π, 2π

] scl: by [-4, 4] scl: 1π

2

Y2 = cot x - tan xtan x + cot x

cot x - tan x __ tan x + cot x =

cos x _ sin x

- sin x _ cos x __

sin x _ cos x + cos x _ sin x

= co s 2 x _ sin x cos x

- si n 2 x _ sin x cos x

__

si n 2 x _ sin x cos x

+ co s 2 x _ sin x cos x

= co s 2 x - si n 2 x __

sin x cos x __

si n 2 x + co s 2 x __

sin x cos x

= co s 2 x - si n 2 x __ si n 2 x + co s 2 x

= co s 2 x - si n 2 x

41. √

��

sec x - 1 _

sec x + 1 = √

��

sec x - 1 _

sec x + 1 · √

��

sec x - 1 _

sec x - 1

= √

��

(sec x - 1 ) 2

_ se c 2 x - 1

= √

��

(sec x - 1 ) 2

_ ta n 2 x

= ⎪ sec x - 1 _

tan x ⎥

43. ln |cot x| + ln |tan x cos x|

= ln ⎪ cos x _ sin x

⎥ + ln ⎪ sin x _ cos x · cos x⎥

= ln ⎪ cos x _ sin x

· sin x⎥

= ln |cos x|



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45. si n 4 θ - co s 4 θ - 1 = (si n 2 θ + co s 2 θ)(si n 2 θ - co s 2 θ) - 1 = (si n 2 θ - co s 2 θ) - 1 = 1 - co s 2 θ - co s 2 θ - 1 = -2 co s 2 θ

47. se c 6 θ - ta n 6 θ = (se c 3 θ - ta n 3 θ)(se c 3 θ + ta n 3 θ) = (sec θ - tan θ)(se c 2 θ + sec θ tan θ + ta n 2 θ) ·

(sec θ + tan θ)(se c 2 θ - sec θ tan θ + ta n 2 θ) = (se c 2 θ - ta n 2 θ)[(1 + ta n 2 θ) + sec θ tan θ + ta n 2 θ] ·

[(1 + ta n 2 θ) - sec θ tan θ + ta n 2 θ] = (1 + 2 ta n 2 θ + sec θ tan θ)(1 + 2 ta n 2 θ - sec θ tan θ) = (1 + 2 ta n 2 θ ) 2 - (sec θ tan θ ) 2 = 1 + 4 ta n 2 θ + 4 ta n 4 θ - se c 2 θ ta n 2 θ = 1 + ta n 2 θ (4 + 4 ta n 2 θ - se c 2 θ) = 1 + ta n 2 θ [4 + 4(se c 2 θ - 1) - se c 2 θ] = 1 + ta n 2 θ (4 + 4 se c 2 θ - 4 - se c 2 θ) = 1 + ta n 2 θ (3 se c 2 θ) = 1 + 3 ta n 2 θ se c 2 θ = 3 se c 2 θ ta n 2 θ + 1

49. se c 2 x cs c 2 x = (ta n 2 x + 1)cs c 2 x

= ( si n 2 x _ co s 2 x

+ 1) ( 1

_ si n 2 x

)

= 1 _

co s 2 x + 1

_ si n 2 x

= se c 2 x + cs c 2 x

51. cos h 2 x - sin h 2 x

= 1 _ 4 ( e x + e -x ) 2 - 1 _

4 ( e x - e -x ) 2

= 1 _ 4 [ e 2x + 2 + e -2x - ( e 2x - 2 + e -2x )]

= 1 _ 4 (4)

= 1

53. 1 - tan h 2 x = 1 - sin h 2 x _ cos h 2 x

= cos h 2 x _ cos h 2 x

- sin h 2 x _ cos h 2 x

= cos h 2 x - sin h 2 x __ cos h 2 x

= 1 _

cos h 2 x

= sec h 2 x55.

[-2π, 2π] scl: by [-4, 4] scl: 1

y =sec x_cos x -

tan x sec x__csc x

π_2

[-2π, 2π] scl: by [-4, 4] scl: 1π

2

y = 1

sec x _ cos x - tan x sec x _ csc x = 1 _ cos x · 1

_ cos x - sin x _ cos x · 1

_ cos x _

1 _

sin x

= 1 _

co s 2 x - si n 2 x _

co s 2 x

= se c 2 x - ta n 2 x = 1

57.

[-2π, 2π] scl: by [-4, 4] scl: 1

y = (tan x + sec x)(1 - sin x)

π_2

[-2π, 2π] scl: by [-4, 4] scl: 1

y = cos x

π_2

(tan x + sec x)(1 - sin x) = tan x - tan x sin x + sec x - sec x sin x = sin x _ cos x - sin x _ cos x · sin x + 1

_ cos x - 1 _ cos x · sin x

= sin x _ cos x - sin x _ cos x · sin x + 1 _ cos x - 1

_ cos x · sin x

= sin x _ cos x - sin 2 x _ cos x + 1 _ cos x - sin x _ cos x

= si n 2 x _ cos x + 1 _ cos x

= 1 - si n 2 x _ cos x

= co s 2 x _ cos x = cos x

59a. sin x = 1 _ 2

59b. The graphs of y = sin x 59c.

and y = 1 _ 2 intersect

y

x

π

65π

6

(-

√3_2

, 1_2 ) (

√3_2

, 1_2 )at π

_ 6 and 5π

_ 6 over [0, 2π).

[0, 2π] scl: by [-2, 2] scl: 1

y =1_2

y = sin x

π_2

59d. The graphs of y = sin x

and y = 1 _ 2 intersect at -

11π

_ 6 ,

-

7π _

6 , π

_ 6 , and 5π

_ 6 over

(-2π, 2π).

[-2π, 2π] scl: by [-2, 2] scl: 1

y =1_2

y = sin x

π_2

59e. Sample answer: Since sine is a periodic function, the solutions

of sin x = 1 _ 2 are x = π

_ 6 + 2nπ and x = 5π

_ 6 + 2nπ, where n is an

integer.

61 Using the Law of Sines, sin β

_

b = sin α

_ a , so b =

a sin β

_ sin α

.

A = 1 _ 2 ab sin γ

A = 1 _ 2 a (

a sin β

_ sin α

) sin γ

A = a 2 sin β sin γ

__

2 sin α


__ 2 sin (180° - (β + γ))


__ 2 sin (β + γ)



Selected A

nswers and S

olutions

63. Sample answers: tan x sin x + cos x = sec x and sin x + cot x cos x = csc x;

tan x sin x + cos x = sin x _ cos x · sin x + cos x

= si n 2 x _ cos x + cos x

= 1 - co s 2 x _ cos x + cos x

= 1 _ cos x - cos x + cos x

= 1 _ cos x

= sec x

sin x + cot x cos x = sin x + cos x _ sin x

· cos x

= sin x + co s 2 x _ sin x

= sin x + 1 - si n 2 x _ sin x

= sin x + 1 _

sin x - sin x

= 1 _

sin x

= csc x

65. Sample answer: You could start on the left side of the identity and simplify it as much as possible. Then, you could move to the right side and simplify until it matches the left side. 67. 1 69. co t 2 θ 71. 1

73.

π-π

-1

-2

1

2

x

y

y =1_4

tan x

3π

2π

2-

π

2-3π

2

75.

-1

-2

1

2

x

y

π

32π

3-

π

3π

6π

25π

6-

π

6

y =1_2

sec 3x

77. 19π

_ 6 79. 1305° 81. 1620

_ π

≈ 515.7° 83. (-7, 4)

85. (-∞, -4] ∪ [4, ∞) 87. (-∞, -7] ∪ [1, ∞) 89. B 91. D

Lesson 4-3

1. 7π

_ 6 + 2nπ, 11π

_ 6 + 2nπ, n � � 3. π

_ 3 + 2nπ, 2π

_ 3 + 2nπ,

4π

_ 3 + 2nπ, 5π

_ 3 + 2nπ, n � � 5. π

_ 6 + nπ, 5π

_ 6 + nπ, n � �

7. π

_ 4 + 2nπ, 3π

_ 4 + 2nπ, n � � 9. π

_ 4 + nπ, 3π

_ 4 + nπ, n � �

11. π

_ 3 + nπ, n � � 13. π

_ 2 , 3π

_ 2 15. π

_ 2 , 3π

_ 2 17. 0, π

19 a. d = 1 _ 32

v 0 2 sin 2θ

40 = 1 _ 32

(50 ) 2 sin 2θ

1280 _

2500 = sin 2θ

2θ = 30.8° θ = 15.4°Sine is also positive in Quadrant II.So, 180 - 30.8 = 149.2°.So if 2θ = 149.2°, θ = 74.6°.Hence, the possible interval is [15.4°, 74.6°].

b. 50 = 1 _ 32

(50 ) 2 sin 2θ

1600 _

2500 = sin 2θ

2θ = 39.8 or 2θ = 140.2

θ = 19.9° θ = 70.1°

21. π

_ 2 , 7π

_ 6 , 11π

_ 6 23. 0, 2π

_ 3 , 4π

_ 3 25. π

_ 2 , 7π

_ 6 , 11π

_ 6 27. π

_ 2 , 0

29. π

_ 3 , 7π

_ 4 31. π

_ 4 + 2nπ, 5π

_ 4 + 2nπ, n � � 33. 2π

_ 3 , 4π

_ 3 35. π

_ 3 ,

5π

_ 3 37. 0.41 39. 1.84, 4.49 41. π

_ 2 , 3π

_ 2 43. π

_ 4 , 5π

_ 4 45. π

_ 4 , 5π

_ 4 ,

9π

_ 4 , 13π

_ 4 47. π

_ 4 , 3π

_ 4 , 5π

_ 4 , 7π

_ 4 , 9π

_ 4 , 11π

_ 4 , 13π

_ 4 , 15π

_ 4 49a. 2.39 radians

49b. 3.01 radians 51. 0 ≤ x < π

_ 4 or 7π

_ 4 < x < 2π

53. 0 < x ≤ 2π 55. 0 ≤ x < π

_ 4 or 3π

_ 4 < x < 2π 57. Both; sample

answer: Vijay’s solutions are correct; however, they are not stated in the simplest form. For example, his solutions of x = π

_ 4 + nπ and

x = 5π

_ 4 + nπ could simply be stated as x = π

_ 4 + nπ because when

n = 1, π

_ 4 + nπ is equivalent to 5π

_ 4 .

59 4 co s 2 x - 4 si n 2 co s 2 x + 3 si n 2 x - 3 = 0 4 co s 2 x (1 - si n 2 x) + 3(si n 2 x - 1) = 0 4 co s 2 x (1 - si n 2 x) - 3(1 - si n 2 x) = 0 (1 - si n 2 x)(4 co s 2 x - 3) = 0 1 - si n 2 x = 0 or 4 co s 2 x - 3 = 0

si n 2 x = 1 co s 2 x = 3 _ 4

sin x = ±1 cos x = ±

√ 3 _

2

x = π

_ 2 , 3π

_ 2 x = π

_ 6 , 5π

_ 6 , 7π

_ 6 , 11π

_ 6

61. Sample answer: 2 si n 2 x = √ 2 sin x 63. Sample answer: When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps.

65. 1 + tan θ

_ 1 + cot θ

= 1 + sin θ

_ cos θ

_ 1 + cos θ

_ sin θ

= sin θ + cos θ

__

cos θ

_ sin θ + cos θ

__

sin θ

= sin θ + cos θ

__

cos θ · sin θ

__ sin θ + cos θ

= sin θ

_ cos θ

67. √ 3

_ 3 69. √ 2 71. ( f - g)(x) = 2 x 2 - 11x - 1

73. ( f _ g ) (x) =

2 x 2 - 5x + 3 __

6x + 4 , x ≠ -

2 _

3 75. D 77. D



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Lesson 4-4

1. √ � 6 - √ � 2

_ 4 3.

√ � 6 - √ � 2 _

4 5. -2 + √ � 3 7. -cos nwt

9a. y = 31.65 sin ( π

_ 6 x - 2.09

) + 42.65 9b. The new function h(x)

represents the average of the high and low temperatures for each

month. 11. √ � 3

_ 3 13. 1 15. 1 _

2 17. tan θ

19. sin 4y 21. cos x

23 V L = IwL cos (wt + π

_ 2 )

= IwL ⎡

⎢

⎣

cos (wt) cos π

_ 2 - sin (wt) sin π

_ 2 ⎤

⎦

= IwL [cos (wt) · 0 - sin (wt) · 1] = IwL [-sin (wt)] = -IwL sin wt

25. 2x ( √ ��

1 - x 2 ) - x ( √ ��

1 - 4 x 2 ) 27. ( √ � 3 - x) √

�� x 2 + 1 __

2 x 2 + 2

29. - x 4 + x 3 + x + ( x 3 + x)

√ ��

1 - x 2 ___

x 4

31. - x 2 + 1 + (2x - 2 x 3 ) √

�� 1 - x 2 ___

2 x 4 - 3 x 2 + 1

33. sec ( π

_ 2 - x

) = 1

_ cos

( π

_ 2 - x

)

= 1 __

cos π

_ 2 cos x + sin π

_ 2 sin x

= 1 __

0(cos x) + 1(sin x)

= 1 _

sin x

= csc x35. cos (π - θ) = cos π cos θ + sin π sin θ = -1(cos θ) + 0(sin θ) = -cos θ

37. sin (π - θ) = sin π cos θ - cos π sin θ = 0(cos θ) - (-1)(sin θ) = sin θ

39. cos (270° - θ) = cos 270° cos θ + sin 270° sin θ = 0(cos θ) + (-1)(sin θ) = -sin θ

41. 2π

_ 3 ; 4π

_ 3 43. π

_ 3 , 5π

_ 3 45. π

_ 4 , 5π

_ 4

47. cos (α + β)

_ sin α sin β

= cos α cos β - sin α sin β

__

sin α cos β

= cos α cos β

_ sin α cos β

- sin α sin β

_ sin α cos β

= cos α

_ sin α

- sin β

_ cos β

= cot α - tan β

49. sin (a + b) + sin (a - b) = sin a cos b + cos a sin b + sin a cos b - cos a sin b = 2 sin a cos b

51. The function is equivalent to y = 1.

co s 2 (x + π

_ 4 ) + co s 2

(x - π

_ 4 )

= (cos x cos π

_ 4 - sin x sin π

_ 4 ) 2 +

(cos x cos π

_ 4 + sin x sin π

_ 4 ) 2

= ( √ � 2

_ 2 cos x -

√ � 2 _

2 sin x)

2

+ (

√ � 2 _

2 cos x +

√ � 2 _

2 sin x)

2

= 1 _ 2 co s 2 x - sin x cos x + 1 _

2 si n 2 x + 1 _

2 co s 2 x +

sin x cos x + 1 _ 2 si n 2 x

= co s 2 x + si n 2 x = 1

[-2π, 2π] scl: by [-2, 2] scl: 1π_2

53. sin z = sin [π - (x + y)] = sin π cos (x + y) - cos π sin (x + y) = 0 · cos (x + y) - [(-1) sin (x + y)] = 0 - [(-1)(sin x cos y + cos x sin y)] = sin x cos y + cos x sin y

55a. sin x cos h + cos x sin h - sin x ___

h

55b.

[-2π, 2π

] scl: by [-2, 2] scl: 1π

2

[-2π, 2π

] scl: by [-2, 2] scl: 1π

2

[-2π, 2π

] scl: by [-2, 2] scl: 1π

2

[-2π, 2π

] scl: by [-2, 2] scl: 1π

255c. cos x57. tan (α + β) =

sin (α + β) _

cos (α + β)

= sin α cos β + cos α sin β

__ cos α cos β - sin α sin β

= sin α cos β

_ cos α cos β

+ cos α sin β

_ cos α cos β

__ cos α cos β

_ cos α cos β

- sin α sin β

_ cos α cos β

= sin α

_ cos α +

sin β

_ cos β

__ 1 -

sin α sin β

_ cos α cos β

= tan α + tan β

__ 1 - tan α tan β

59. sin (α + β) = cos [90° - (α + β)] = cos [(90° - α) - β)] = cos (90° - α) cos β + sin (90° - α) sin β = sin α cos β + cos α sin β

61. sin (x + y + z) = sin [(x + y) + z] = sin (x + y) cos z + cos (x + y) sin z = (sin x cos y + cos x sin y) cos z +

(cos x cos y - sin x sin y) sin z = sin x cos y cos z + cos x sin y cos z +

cos x cos y sin z - sin x sin y sin z



Selected A

nswers and S

olutions

63 sin (x - y) = sin x cos y - cos x sin y

= (-

2 _

3 ) ( 1 _

3 ) - (

√ � 5 _

3 ) (

2 √ � 2 _

3 )

= -

2 _

9 -

2 √ � 10 _

9

= -2 - 2 √ � 10

_ 9

65a. sin 3x cos 2x = cos 3x sin 2xsin 3x cos 2x - cos 3x sin 2x = 0

sin (3x - 2x) = 0 sin x = 0 x = 0 and x = π

65b.

[-0, 2π] scl: by [-2, 2] scl: 1π_2

67. cos (x + h) - cos x

__ h

= cos x cos h - sin x sin h - cos x ___ h

= cos x (cos h - 1) - sin x sin h

___ h

= cos x ( cos h - 1 _

h ) - sin x ( sin h _

h )

69. ≈ 31.3°

71. sec θ

_ sin θ

- sin θ

_ cos θ

= 1 _

sin θ cos θ - sin θ

_ cos θ

= 1 _

sin θ cos θ - si n 2 θ

_ sin θ cos θ

= co s 2 θ

_ sin θ cos θ

= cos θ

_ sin θ

= cot θ

73. π

_ 3 75a. Exponential; the base, 1 + r _ n , is fixed, but the exponent,

nt, is variable since the time t can vary. 75b. A(t) = 1000(1.01 ) 4t 75c. $2216.72 77. ±1, ±2, ±3, ±5, ±6, ±10, ±15 and ±30; -3, 2 79. D 81. A

Lesson 4-5

1. -

24 _

25 ; -

7 _

25 ; 24

_ 7 3. -

720 _

1681 ; -

1519 _

1681 ; 720

_ 1519

5. -

4 _

5 ; 3 _

5 ; -

4 _

3

7. -

24 _

25 ; -

7 _

25 ; 24

_ 7 9. π

_ 6 , π

_ 2 , 5π

_ 6 , 3π

_ 2 11. π

_ 6 , 5π

_ 6 , 3π

_ 2 13. π

_ 3 , 5π

_ 3

15 a. d =

v 0 2 sin 2θ

_ 32

Distance equation

242 = 88 2 sin 2θ

_ 32

d = 242 and v 0 2 = 88

(32)(242)

_ 8 8 2

= sin 2θ Solve for sin 2θ.

1 = sin 2θ Simplify.

90° = 2θ si n -1 1 = 90°

45° = θ Divide each side by 2.

b. d = v 0 2 sin 2θ

_

32

= 2 v 0 2 sin θ cos θ

__

32 sin 2θ = 2 sin θ cos θ

= v 0 2 sin θ cos θ

__

16 Simplify.

17. tan θ - tan θ cos 2θ

__

1 + cos 2θ 19.

cos θ + cos θ cos 2θ

__ 1 - cos 2θ

21. cos θ - cos θ cos 4θ

__ 8 23.

3 - 4 cos 2θ + cos 4θ

__ 4 + 4 cos 2θ

25. π

_ 6 + nπ, 5π

_ 6 + nπ, n � � 27. nπ, 3π

_ 4 + 2nπ, n � �

29.

√ ��

2 +

√ � 2 _

2 31. 1 -

√ � 2 33. 0, π

_ 3 , 5π

_ 3 35. 0

37. 1 _ 2 cos 2θ + 1 _

2 cos 4θ 39. 1 _

2 sin 5x + 1 _

2 sin x 41.

1 + √ � 3 _

2

43. √ � 2 - 1

_ 6 45.

√ � 2 _

2 47. -

3 √ � 6 _

2 49. nπ, π

_ 2 + nπ, n � �

51. π

_ 2 n, π

_ 4 + π

_ 2 n, n � � 53. 2πn _

5 , π

_ 5 + 2πn _

5 , nπ, n � �

55. ±cos 3x 57. 1 _ 2 (cos 2a + cos 2b) 59. 1 _

2 [sin (2b + θ + π) +

sin (θ - π)] 61a. 1 + cos � + sin �

__ 1 + cos � - sin �

61b. 2 + √ � 3

63. cos 2θ = cos (θ + θ) = cos θ cos θ - sin θ sin θ = co s 2 θ - si n 2 θ

65. tan 2θ = tan (θ + θ)

= tan θ + tan θ

__ 1 - tan θ tan θ

= 2 tan θ

_ 1 - ta n 2 θ

67. ta n 2 θ = 1 - cos 2θ

_ 1 + cos 2θ

tan θ = ±

√

��

1 - cos 2θ

_ 1 + cos 2θ

tan θ _ 2 = ±

√

��

1 - cos θ

_ 1 + cos θ

69. i. 2 co s 2 5θ - 1 = 2 ⎡

⎢

⎣

1 _

2 (1 + cos 10θ)

⎤

⎦

- 1

= 1 + cos 10θ - 1 = cos 10θ

ii. 2 co s 2 5θ - 1 = 2 cos 5θ cos 5θ - 1

= 2 ⎧

⎨

⎩

1 _

2 [cos (5θ - 5θ) + cos (5θ + 5θ)]

⎫

⎬

⎭

- 1

= (cos 0 + cos 10θ) - 1 = 1 + cos 10θ - 1 = cos 10θ

71. 1 _ 16

[5 - 7 cos 2θ + 3 cos 4θ - cos 2θ cos 4θ]

73. 1 _ 16

(5 cos θ + 7 cos θ cos 2θ + 3 cos θ cos 4θ +

cos θ cos 2θ cos 4θ)

75a.

[-2π, 2π

] scl: by [-4, 4] scl: 1π

2

f(x) = 4(sin θ cos - cos θ sin )π

4π

4



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

75b. 4 sin (θ - π

_ 4 ) ;

4 sin (θ - π

_ 4 ) = 4

(sin θ cos π

_ 4 - cos θ sin π

_ 4 ) , difference of two

angles identity

75c.

[-2π, 2π

] scl: by [-2, 2] scl: 1π

2

g(x) = cos2 (θ - ) - sin2

(θ - )π

3π

3

75d. cos (2θ - 2π

_ 3 ) ;

cos (2θ - 2π

_ 3 ) = cos

⎡

⎢

⎣

2 (θ - π

_ 3 ) ⎤

�

⎦

= co s 2 (θ - π

_ 3 ) - si n 2

(θ - π

_ 3 )

77. I, II, or between I and II; I 79. I, II, or between I and II; II

81 cos 4θ = cos 2(2θ)

= 1 - 2 si n 2 2θ

= 1 - 2 (sin 2θ)(sin 2θ)

= 1 - 2 (2 sin θ cos θ)(2 sin θ cos θ)

= 1 - 8 si n 2 θ co s 2 θ

83. ta n 2 θ = si n 2 θ

_ co s 2 θ

= 1 - cos 2θ

_

2 _

1 + cos 2θ

_

2

= 1 - cos 2θ

_ 1 + cos 2θ

85. 1 _ 2 [sin (α + β) + sin (α - β)]

=

1 _

2 (sin α cos β + cos α sin β +

sin α cos β - cos α sin β)

= 1 _ 2 (2 sin α cos β)

= sin α cos β

87. Let x = α + β

_

2 and let y =

α - β

_ 2 .

2 cos ( α + β

_

2 ) cos (

α - β

_ 2 )

= 2 cos x cos y

= 2 ⎧

⎨

⎩

1 _

2 [cos (x + y) + cos (x - y)]

⎫

⎬

⎭

= cos (x + y) + cos (x - y)

= cos ( α + β

_

2 +

α - β

_ 2 ) + cos (

α + β

_ 2 -

α - β

_ 2 )

= cos α + cos β

89. Let x = α + β

_

2 and let y =

α - β

_ 2 .

-2 sin ( α + β

_

2 ) sin (

α - β

_ 2 )

= -2 sin x sin y

= -2 ⎧

⎨

⎩

1 _

2 [cos (x - y) - cos (x + y)]

⎫

⎬

⎭

= -cos (x - y) + cos (x + y)

= -cos ( α + β

_

2 -

α - β

_ 2 ) + cos (

α + β

_ 2 +

α - β

_ 2 )

= -cos β + cos α = cos α - cos β

91. √ � 6 + √ � 2

_ 4 93.

1 _

2 95. -

√ � 3 _

2 97. the 104th day of the year, or

around April 14

99. √ � 3 _

3 101. - 1

103. y

x

5_4

= 4xf (x)

D = [ 0, ∞), R = [ 0, ∞); intercept:

(0, 0); lim x→∞

f (x) = ∞; continuous on

[ 0, ∞); increasing on (0, ∞)

105. y

x

4

8

−4

−8

−4−8 4 8

= 4x 5f (x)

D = (-∞, ∞), R = (-∞, ∞);

intercept: (0, 0); lim x→-∞

f (x) =

-∞ and lim x→∞

f (x) = ∞; continuous

for all real numbers; increasing on

(-∞, ∞)

107. J


1. 1 _

cos θ ; reciprocal identity 3. tan 2 θ; pythagorean identity

5. -tan θ; odd-even identity 7. cos 2α; double-angle identity 9. si n 2 θ; power-reducing identity

11. sec θ = √ � 10 , cos θ = 1 _

√ � 10 or

√ � 10 _

10 13. csc θ = -

5 _

4 ,

tan θ = -

4 _

3 15. sec θ = 5

_ 2 √ � 5

or √ � 5

_ 2 , sin θ = -

1 _

√ � 5 or -

√ � 5 _

5 17. 1

19. sec x 21. s ec 2 x + tan x sec x

23. sin θ

_ 1 - cos θ

+ sin θ

_ 1 + cos θ

= sin θ (1 + cos θ)

__ 1 - co s 2 θ

+ sin θ (1 - cos θ)

__ 1 - cos 2 θ

= sin θ (1 + cos θ)

__ si n 2 θ

+ sin θ (1 - cos θ)

__ s in 2 θ

= 1 + cos θ

_

sin θ + 1 - cos θ

_

sin θ

= 2 _

sin θ

= 2 csc θ

25. cot θ

_ 1 + csc θ

+ 1 + csc θ

_

cot θ

= cot 2 θ

__ cot θ (1 + csc θ)

+ (1 + csc θ) 2


= co t 2 θ + (1 + csc θ) 2


= cs c 2 θ - 1 + 1 + 2 csc θ + cs c 2 θ

___

cot θ (1 + csc θ)

= 2 c sc 2 θ + 2 csc θ


= 2 csc θ (csc θ + 1)


= 2 _

sin θ · sin θ

_ cos θ

= 2 _

cos θ

= 2 sec θ



Selected A

nswers and S

olutions

27. c ot 2 θ

_ 1 + csc θ

= co t 2 θ (1 - csc θ)

__ (1 + csc θ)(1 - csc θ)

= co t 2 θ (1 - csc θ)

__ 1 - csc 2 θ

= c ot 2 θ (1 - csc θ)

__ -c ot 2 θ

= -(1 - csc θ) = csc θ - 1

29. sec θ + csc θ

__

1 + tan θ =

1 _

cos θ +

1 _

sin θ

__ cos θ

_ cos θ

+ sin θ

_ cos θ

= sin θ + cos θ

__

sin θ cos θ

_ sin θ + cos θ

__

cos θ

= sin θ + cos θ

__

sin θ cos θ · cos θ

__ sin θ + cos θ

= 1 _

sin θ

= csc θ

31. sin θ

__ sin θ + cos θ

= sin θ

_ cos θ

__ sin θ

_ cos θ

+ cos θ

_ cos θ

= tan θ

_ 1 + tan θ

33. π

_

4 , 3π

_

4 35. π

_

3 , 2π

_

3 , 4π

_

3 , 5π

_

3 37. 0, π, π

_

6 , 5π

_

6

39. nπ, π

_

2 + 2nπ, n � � 41. 2π

_

3 + 2nπ, 4π

_

3 + 2nπ, n � �

43. 2nπ, π

_

2 + nπ, n � � 45.

√ � 6 + √ � 2 _

4 47. 2 - √ � 3

49. - √ � 6 - √ � 2

_ 4 51. 0 53.

√ � 2 _

2

55. cos (θ + 30°) - sin (θ + 60°) = cos θ cos 30° - sin θ sin 30° -

(sin θ cos 60° + cos θ sin 60°)

=

√ � 3 _

2 cos θ - 1 _

2 sin θ - 1 _

2 sin θ -

√ � 3 _

2 cos θ

= -sin θ

57. cos (θ -

π

_ 3 ) + cos

(θ +

π

_ 3 )

= cos θ cos π

_

3 + sin θ sin π

_

3 + cos θ cos π

_

3 - sin θ sin π

_

3

= 1 _ 2 cos θ +

√ � 3 _

2 sin θ + 1 _

2 cos θ -

√ � 3 _

2 sin θ

= cos θ

59. 4 √ � 2

_ 9 , -

7 _

9 , -

4 √ � 2 _

7 61. -

24 _

25 , -

7 _

25 , 24

_ 7 63.

√ ��

2 +

√ � 3 _

2

65. √ � 2 + 1 67. -

√ ��

2 - √ � 2 _

2 69.

√ �� 145 _

145 ,

12 √ �� 145 _

145

71. sin α

_ 1 - cos α

= sin α

_ 1 - cos α

· 1 + cos α

_ 1 + cos α

73. yes

= sin α (1 + cos α)

__ 1 - co s 2 α

= sin α (1 + cos α)

__ si n 2 α

= 1 + cos α

_

sin α


1. Sample answer: The values state that the rate of change for f (x)

at π

_

2 is 0, at 0 is 1, and at π is -1. This should be the slope of the

lines tangent to f (x) at those values of x. The graph of f (x) and the tangent lines verify this conclusion.

1y

1π 2π

3. For x = 3π

_

2 , the rate of change is 0. For x = 2π, the rate of change

is 1. For x = 5π

_

2 , the rate of change is 0.

5a. m = cos (x + h) - cos x

__ h

5b. m = cos x ( cos h - 1 _

h ) - sin x ( sin h _

h )

5c. lim h→o cos h - 1

_ h = 0; lim h→o

sin h _ h = 1

5d. m = cos x (0) - sin x (1) 5e. m = -sin x

5f. For x = 0, the rate of change is 0. For x = π

_

2 , the rate of change

is -1. For x = π, the rate of change is 0. For x = 3π

_

2 , the rate of

change is 1.

Applications of Trigonometry5CHAPTER 5CHCHAPAPTETERR 55

Lesson 5-1

1. a ≈ 8.98

3. C ≈ 49.2° 5. C ≈ 21.4° 7. ∠C = 78°, b ≈ 109.5 cm, c ≈ 119.2 cm 9. ∠C = 90°, a = 10 in., c = 20 in.

11.

19 in.in.

33�

102� 45�

13. ∠C = 90°, a = 15 mi, b = 15 mi

15. ∠A = 57°, b ≈ 49.5 km, c ≈ 17.1 km

17.

0.8 cm

3.6 cm56�

112� 12�

19a. 10 cm 19b. 0 19c. 2 19d. 1

21. not possible 23. B = 60°, C = 90°, b = 12.9 √ � 3 mi

25 A ≈ 39°, B ≈ 82°, a ≈ 42.6 mi or A ≈ 23°, B ≈ 98°, a ≈ 26.4 mi

27. A ≈ 39°, B ≈ 82°, a ≈ 42.6 ft or A ≈ 23°, B ≈ 98°, a ≈ 26.4 ft

29. not possible 31. A ≈ 80.0°, B ≈ 38.0°, b ≈ 1.8 × 10 25 mi

33. A 1 ≈ 19.3°, A 2 ≈ 160.7°, 48° + 160.7° > 180°; no second solution possible 35. C 1 ≈ 71.3°, C 2 ≈ 108.7°, 57° + 108.7° 6 180°; two

solutions possible 37. not possible, sin A > 1 39. √ � 2

_ 2 41. 34.6

million miles or 119.7 million miles 43a. No 43b. ≈ 3.9 mi

45. V ↔ S = 41.7 km, V ↔ P = 80.8 km

47a. No 47b. about 201.5 ft 47c. ≈15 sec

49. Two triangles

Angles Sides

A 1 ≈ 41.1° a = 12 cm

B = 26° b = 8 cm

C 1 ≈ 112.9° c 1 ≈ 16.8 cm

Angles Sides

A 2 ≈ 138.9° a = 12 cm

B = 26° b = 8 cm

C 2 ≈ 15.1° c 2 ≈ 4.8 cm



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

51. Angles Sides

A 1 ≈ 47.0° a = 9 cm

B 1 ≈ 109.0° b 1 ≈ 11.6 cm

C ≈ 24° c = 5 cm

Angles Sides

A 2 ≈ 133.0° a = 9 cm

B 2 = 23.0° b 2 ≈ 4.8 cm

C ≈ 24° c = 5 cm

53. a ≈ 33.7 ft, c ≈ 22.3 ft 55. Rhymes to Tarryson: 61.7 km, Sexton to Tarryson: 52.6 km 57. ≈ 3.2 mi 59. h ≈ 161.9 yd

61. angle = 90°; sides ≈ 9.8 cm, 11 cm; diameter ≈ 11 cm; it is a right triangle 63a. about 3187 m 63b. about 2613 m 63c. about 2368 m

65.

30�

60�

10.2 cm

10.2(√3) cm

20.4 cm √ � 3 =

sin 60° _ sin 30°

; √ � 2 =

sin 90° _

sin 45°

67. A = 19°, B = 31°, C = 130°, a = 45 cm, b ≈ 71.2 cm, c ≈ 105.8 cm

69. ≈12,564 mph

71. tan 2 x - sin 2 x =

sin 2 x _ cos 2 x

- sin 2 x

= sin 2 x _ cos 2 x

-

sin 2 x cos 2 x __ cos 2 x

=

sin 2 x - sin 2 x cos 2 x __ cos 2 x

=

sin 2 x(1 - cos 2 x) __

cos 2 x

=

sin 2 x sin 2 x _ cos 2 x

= sin 2 x sin 2 x _ cos 2 x

= sin 2 x tan 2 x

73a. y =

5 _ 9 x -

2 _ 9 73b. √ �� 106 units

Lesson 5-2

1. yes 3. no 5. yes 7. verified 9. B ≈ 41.4°11. a ≈ 7.24 13. A ≈ 41.6° 15. A ≈ 120.4°, B ≈ 20.6°, c ≈ 53.5 cm17. A ≈ 23.8°, C ≈ 126.2°, b ≈ 16 mi

19.

A

B

C

538 mm

465 mm

260.9 mm59.8�

91.2� 29�

21. A ≈ 137.9°, B ≈ 15.6°, C ≈ 26.5°

23. A ≈ 119.3°, B ≈ 41.5°, C ≈ 19.2°

25.

B

A

C

2.9 � 1025 mi

4.1 � 1025 mi

2.3

� 1

025

mi

43.5�

103.3� 33.2�

27. A ≈ 139.7°, B ≈ 23.7°, C ≈ 16.6° 29. C ≈ 86.3°

31. about 1668 mi 33. P ≈ 27.7°; heading 297.7° 35. It cannot be constructed (available length ≈ 10,703.6 ft) 37. 1678.2 mi39. P ≈ 22.4 cm, A ≈ 135°, B ≈ 23.2°, C ≈ 21.8°

41. A ≈ 20.6°, B ≈ 15.3°, C ≈ 144.1° 43. 58.78 cm

45. a = 13 A ≈ 132.2°

b.= 5 B ≈ 16.3°

c = √ � 82 C ≈ 30.5°

47. 33.7°; 150 ft 2 49a. 0.65 = 65 %, 49b. $1,950,000

51. about 453,529 km 2 53. 357 + 502 = 889 < 902

55. (1) a 2 = b 2 + c 2 - 2bc cos A (2) b 2 = a 2 + c 2 - 2ac cos B, use substitutino for a 2 and (2) becomes b 2 = ( b 2 + c 2 - 2bc cos A) + c 2 - 2ab cos B. Then 0 = 2c 2 - 2bc cos A - 2ac cos B, 2bc cos A + 2ac

cosB = 2c 2 , b cos A + a cos B = c 57. 2

59. sin x = -5

_ 13

, csc x = -13

_ 5 cos x =

12 _ 13

, sec x = 13

_ 12

,

tan x = -5

_ 12

, cot x = -12 _

5

Lesson 5-3

1. 3. 5.

7. 9.

11. Terminal point: (5, -1) , magnitude: √ � 53 13. Terminal point: (-1,1) , magnitude: √ � 34

15a. 17a.

15b. √ � 73 21c. 20.6° 17b. √ � 29 17c. 68.2°

19. ⟨-10.9, 5.1⟩ 21. ⟨106, -92.2⟩ 21. ⟨-9.7, -2.6⟩

25a. ⟨-1, 9⟩ 25b. ⟨5, -3⟩

25c. ⟨-0.5, 15⟩ 25d. ⟨8, -9⟩

27a. ⟨8, 4⟩ 27b. ⟨6, -8⟩

12 knots

9 knots

6 knots

V1

V2

V3 V4

25�210 N

250 N

J

M

x

y

54 76321�3�2�1�1�2�3�4�5

2345

1(�3, 2)

(4, 5)

H7, 3I

4 � (�3) � 75 � 2 � 3

x

y

84�8 �4�4

�8

4

8

(5, �3)

�1 �5 � �62� (� 3) � 5

(�1, 2)

H�6, 5I

t

y

84�8 �4�4

�8

4

8

H8, 3I QI

x

y

42�4 �2�2

�4

2

4

H�2, �5I

QIII

x

y

5�5

�55

u

r

v

(�1, 9)

x

y

5�5

5

�5

u

r

�v

(5, �3)

x

y

5�5

10

20

2u

r

1.5v

(�0.5, 15)x

y

5

�10

6

u

r�2v

(8, 9)

x

y

5�2

5

�5

u

rv

(8, 4)x

y

5�2

�5

u

r �v

(6, �8)



Selected A

nswers and S

olutions

27c. ⟨15.5, 5⟩ 27d. ⟨5, -14⟩

29a. ⟨-3, 6⟩ 29b. ⟨-5, -2⟩

29c. ⟨-6.5, 10⟩ 29d. ⟨-6, -6⟩

31. True 33. False 35. True37. u + v = ⟨8, 6⟩ 39. u + v = ⟨-9, -6⟩

u + v = ⟨-6, 2⟩ u + v = ⟨7, 4⟩

x

y

5�5

5

�5

u � v

u � v

x

y

5�5

5

�5u � v

u � v

41. u + v = ⟨-3, -6⟩ 43. u = 8i + 15j u - v = ⟨-7, 0⟩ ⎪u⎥ = 17

45.p = -3.2i - 5.7j 47a.

⎪p

⎥ ≈ 6. 54

47b. v = ⟨-11.5, -3.3⟩

47c. v = -11.5i, -3.3j49a. 49b. w = ⟨2.5, 9.2⟩ 49c. w = 2.5i + 9.2j

51a. p = -2i + 2j; ⎪p

⎥ = 2 √ � 2 , θ = 135°

51b. q = 6i - 8j; ⎪q

⎥ = 10, θ = 306.9°

51c. r = -2i + 1.5j; ⎪r⎥ = 2.5, θ = 143.1°51d. s = 10i - 13j; ⎪s⎥ ≈ 16.4, θ = 307.6°53a. p = 2 √ � 2 i + 2j;

⎪p

⎥ ≈ 3.5, θ = 35.3°

53b. q = 8 √ � 2 i + 12j; ⎪q

⎥ ≈ 16.5, θ ≈ 46.7°

53c. r = 5.5 √ � 2 i + 6.5j; ⎪r⎥ ≈ 10.1, θ ≈ 39.9°

53d. s = 11 √ � 2 i + 17j; ⎪s⎥ ≈ 23.0, θ ≈ 47.5°55a. p = 8i + 4j;

⎪p

⎥ ≈ 8.9, θ ≈ 26.6°

55b. q = 16i + 4j; ⎪q

⎥ ≈ 16.5, θ ≈ 14.0°

55c. r = 18i + 8j; ⎪r⎥ ≈ 19.7, θ ≈ 24.0°55d. s = 20i + 4j; ⎪s⎥ ≈ 20.4, θ ≈ 11.3°

57. ⟨ 7 _ 25

, 24

_ 25

⟩ , verified 59. ⟨ -20

_ 29

, 21

_ 29

⟩ , verified

61. 20

_ 29

i - 21

_ 29

j, verified 63. 7 _

25 i -

24 _

25 j, verified

65. ⟨-

13 _

√ �� 178 ,

3 _

√ �� 178 ⟩ , verified 67.

6 _

√ �� 157 i +

11 _

√ �� 157 j, verified

69. ≈ 4.48 ⟨

5 _

√ � 29 ,

2 _

√ � 29 ⟩ ≈ ⟨4.16, 1.66⟩

71. ≈ 5.83 ⟨

8 _

√ � 73 ,

-3 _

√ � 29 ⟩ ≈ ⟨5.46, -2.05⟩ 73. ≈ 14.4 75. ≈ 24.3°

77. hor. comp. ≈ 79.9 ft/sec; vert. comp. ≈ 60.2 ft/sec79. heading 68.2° at 266.7 mph 81. ≈ (82.10 cm, 22.00 cm) 83. 1⟨a, b⟩ = ⟨1a, 1b⟩ = ⟨a, b⟩

85. ⟨a, b⟩ = - ⟨c, d⟩ = ⟨a - c, b - d⟩ = ⟨a + (-c), b + (-d)⟩ = ⟨a, b⟩ + ⟨-c, -d⟩ = ⟨a, b⟩ + -1⟨c, d⟩ = u + (-1v)87. (ck)u = ⟨cka, ckb⟩ = c⟨ka, kb⟩ = c (ku) c(ku) = ⟨cka, ckb⟩ = ⟨kca, kcb⟩ = k⟨ca, cb⟩ = k (cu) 89. u + (-u) = ⟨a, b⟩ + ⟨-a, -b⟩ = ⟨a - a, b - b⟩ = ⟨0, 0⟩

91. (c + k)u = (c + k)⟨a, b⟩ = ⟨(c + k)a, (c + k)b⟩ =

⟨ca + ka, cb + kb⟩ = ⟨ca, cb⟩ + ⟨ka, kb⟩ = cu + ku

93. ⟨1, 3⟩ + ⟨3, 3⟩ + ⟨4, -1⟩ + ⟨2, -4⟩ + ⟨-4, -3⟩ + ⟨-6, 2⟩ = ⟨0, 0⟩

95. Answers will vary, one possibility: 0°, 81.4°, -34°

99. x = 0, ± √ � 7 ; see graph

Mid-Chapter Check

1. sin B = b sin a _ a 2. cos B =

a 2 + c 2 - b 2 __

2ac

3. a ≈ 129 m, B ≈ 86.5°, C ≈ 62.5°4. A ≈ 42.3°, b ≈ 81.5°, c ≈ 56.2°

5. A = 44° a = 2.1 km 6. A = 18.5° a = 70 yd

B ≈ 68.1° b ≈ 2.8 km B.≈ 134.5° b ≈ 157.1 yd

C ≈ 67.9° c = 2.8 km C ≈ 27° c = 100 yd

or

A = 44° a = 2.1 km

B ≈ 23.9° b ≈ 1.2 km

C ≈ 112.1° c = 2.8 km

7. about 60.7 ft 8. 169 m 9. α ≈ 49.6°; β ≈ 92.2°; γ ≈ 38.2°

9. 9.4 mi

Reinforcing Basic Concepts

1. 2. For ∠A = 35°, a ≈ 10.3

For ∠A = 50°, a ≈ 14.2

For ∠A = 70°, a ≈ 19.1;

yes, very close

Very close.

x

y

2010

5

�5

2u

1.5v

r

(15.5, 5)

x

y

�3

�10

u

r�2v

(5, �14)

x

y

5�5

5

�3

u

rv

(�3, 6)

x

y

3�5

5

�5

u

r

�v

(�5, �2)

x

y

�5

10

5

2u

r

1.5v

(�6.5, 10)

x

y

�5 3

�5

3

u

rv

(�6, �6)

x

y

�2 8

16

�4

u

x

y

5�5

5

�5

u � v

u � v

x

y

�10 8

4

�5

v

�r � 16�

x

y

5�5

4

�5

p

x

y

5�5

10

5 w

�r � 74.5�

x

y

54321�5�4�3�2�1�2�4�6�8

�10

468

10

2

Latitude Distance

A = 35° a = 11.6 cm

B ≈ 81.5° a = 20 cm

C ≈ 63.5° a = 18 cm



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

Lesson 5-4

1. ⟨6, 8⟩ 3. ⟨-5, 10⟩ 5. -6i - 8j 7. -2.2i + 0.4j

9. ⟨-11.48, -9.16⟩ 11. ⟨-24, -27⟩ 13. ⎪ F 3 ⎥ ≈ 3336.8; θ ≈ 268.5°

15. 37.16 kg 17. 644.49 lb 19. 2606.74 kg 21. approx. 286.79 lb

23. approx. 43.8° 25. 1125 N-m 27. approx. 957.0 ft

29. approx. 64,951.90 ft-lb 31. approx. 451.72 lb

33. approx. 2819.08 N-m 35. 800 ft-lb 37. 118 ft-lb

39. verified 41. verified 43a. 29 43b. 45° 45a. 0 45b. 90°

47a. 1 47b. 89.4° 49. yes 51. no 53. yes 55. 3.68 57. -4

59. 3.17 61a. ⟨3.73, 1.40⟩ 61b. u 1 = ⟨3.73, 1.40⟩, u 2 = ⟨-1.73, 4.60⟩

63a. ⟨-0.65, 0.11⟩ 63b. u 1 = ⟨-0.65, 0.11⟩, u 2 = ⟨-1.35, -8.11⟩

65a. 10.54i + 1.76j 65b. u 1 = 10.54i + 1.76j, u 2 = -0.54i + 3.24j

67a. projectile is about 375 ft away, and 505.52 ft high

67b. approx. 1.27 sec and 12.26 sec

69a. projectile is about 424.26 ft away, and 280.26 ft high

69b. approx. 2.44 sec and 6.40 sec

71. about 74.84 ft; t ≈ 3.9 - 1.2 = 2.7 sec

73. w · (u + v) = ⟨e, f ⟩ · ⟨a + c, b + d⟩

= e(a + c) + f (b + d) = ea + ec + fb + fd

= (ea + fb) + (ec + fd)

= ⟨e, f ⟩ · ⟨a, b⟩ + ⟨e, f ⟩ · ⟨c, d⟩

= w · u + w · v

75. 0 · u = ⟨0, 0⟩ · ⟨a, b⟩ = 0(a) + 0(b) = 0

u · 0 = ⟨a, b⟩ · ⟨0, 0⟩ = a(0) + b(0) = 0

77. θ ≈ 56.9°; answers will vary. 79. x ≈ -20

81. a ≈ 138.4, B ≈ 106.8° C ≈ 41.2°; P ≈ 560.4 m, A ≈ 11,394.3 m 2

Lesson 5-5

1. z 2 = z 1 + z 3 3. z 2 = z 1 + z 3

5. 2 √ � 2 (cos 225° + i sin 225°) 7. 10(cos 210° + i sin 210°)

9. 6 ⎡

⎢

⎣

cos ( 3π

_ 4 ) + i sin ( 3π

_ 4 )

⎤

⎦

11. 8 ⎡

⎢

⎣

cos ( 11π

_ 6 ) + i sin ( 11π

_

6 )

⎤

⎦

13. 10 cis ⎡

⎢

⎣

tan -1 ( 6 _ 8 )

⎤

⎦

; 10 cis 36.9°

15. 13 cis ⎡

⎢

⎣

180° + tan -1 ( 12 _

5 )

⎤

⎦

; 13 cis 247.4°

17. 18.5 cis ⎡

⎢

⎣

tan -1 ( 17.5 _

6 )

⎤

⎦

; 18.5 cis 1.2405

19. 2 √ � 34 cis ⎡

⎢

⎣

π + tan -1 (-

5 _

3 )

⎤

⎦

; 2 √ � 34 cis 2.1112

21. r = 2, θ =

π

_ 4

z = 2 cis (

π

_ 4 )

=

√ � 2 +

√ � 2 i

23. r = 4

√ � 3 , θ =

π

_ 3

z = 4

√ � 3 cis

(

π

_ 3 )

= 2 √ � 3 + 6i

25. r = 17 , θ = tan -1 ( 15 _

8 )

z = 17 cis ⎡

⎢

⎣

tan -1 ( 15 _

8 )

⎤

⎦

= 17 (

8 _

17 +

15 _

17 i) = 8 + 15i

27. r = 6, θ = π - tan -1 ( 5

_ √ � 11

)

z = 6 cis ⎡

⎢

⎣

π - tan -1 5 _

√ � 11

⎤

⎦

= 6 (-

√ � 11

_ 6 +

5 _

6 i) = -

√ � 11 + 5i

29. r 1 = 2 √ � 2 , r 2 = 3

√ � 2 , θ 1 = 135°, θ 2 = 45°;

z = z 1 z 2 = -12 + 0i� r = 12, θ = 180°;

r 1 r 2 = 2 √ � 2 (3 √ � 2 ) = 12�

θ 1 + θ 2 = 135° + 45°= 180°�

31. r 1 = 2, r 2 = 2, θ 1 = 30°, θ 2 = 60°;

z =

z 1 _ z 2 =

√ � 3 _

2 -

1 _

2 i � r = 1, θ = -30°;

r 1 _ r 2 =

2 _

2 = 1�

θ 1 - θ 2 = 30° - 60° = -30°�

33. z 1 z 2 = -24 + 0i, z 1

_ z 2 = -

4 _

3 +

4

√ � 3 _

3 i

35. z 1 z 2 = 21 √ � 3 - 21i,

z 1 _ z 2 =

√ � 3

_ 7 +

1 _

7 i

37. z 1 z 2 = -10.84 + 12.04i, z 1

_ z 2 = -1.55 - 4.76i

39. z 1 z 2 = 0 + 40i, z 1

_ z 2 =

5

√ � 3 _

4 +

5 _

4 i

41. z 1 z 2 = -10 - 10 √ � 3 i,

z 1 _ z 2 =

-5 _

2 + 0i

43. z 1 z 2 = -2,93 + 8.5i, z 1

_ z 2 = 2.29 + 3.28i

45. verified; verified, u 2 + v 2 + w 2 = uv + uw + vw

(1 + 4 √ � 3 i) + (97 + 20

√ � 3 i) + (-39 + 60

√ � 3 i)

= (17 + 12 √ � 3 i) + (-3 + 16

√ � 3 i) + (45 + 56

√ � 3 i),

59 + 84 √ � 3 i = 59 + 84

√ � 3 i

47a. V(t) = 170 sin(120πt)

47b.

47c. t ≈ 0.00257 sec 49a. 17 cis 28.1° 49b. 51 V

yi

x

z3

z1

z2

2

7 8

4

6yi

xz3

z2

z1

x

yi

�1 1 2�2

�1

1

2

2

3

4

(√2, √2)

�4

x

yi

5�5

�2

5

(2√3, 6)

4√3

�3

x

yi

5

10

�2

20

17

(8, 15)

� � tan�1� �158

x

yi

3

�3

�5

5

6

� � tan�1� �5√11

(�√11, 5)

t V(t)

0 0

0.001 62.6

0.002 116.4

0.003 153.8

0.004 169.7

0.005 161.7

0.006 131.0

0.007 81.9

0.008 21.3



Selected A

nswers and S

olutions

51a. 8.60 cis 324.5° 51b. 15.48 V 53a. 13 cis 22.6° 53b. 22.1 V

55. I = 2 cis 30°; Z = 5 √ � 2 cis 45°; V = 10

√ � 2 cis 75°

57. I =

√ � 13 cis 326.3°; Z =

17 _

4 cis 61.9°; V =

17 √ � 13 _

4 cis 28.2°

59. V = 4 cis 60°; Z = 4 √ � 2 cis 315°; I =

√ � 2 _

2 cis 105°

61. V = 5 cis 306.9°; Z = 8.5 cis 61.9°; I =

10 _

17 cis 245°

63. √ � 65 cis 29.7°

__ 4 65. verified 67. z 2 =

24 _

5 -

7 _

5 i, z 3 =-

24 _

5 +

7 _

5 i

69. 5π

_

24 , 13π

_

24 , 29π

_

24 , 37π

_

24

71.

Lesson 5-6

1. r = 3 √ � 2 ; n = 4; θ = 45°; -324 3. r = 2; n = 3; θ = 120°; 8

5. r = 1; n = 5; θ = -60°; 1 _

2 +

√ � 3 _

2 i 7. r = 1; n = 6; θ = -45°; i

9. r = 4; n = 3; θ = 330°; -64i

11. r =

√ � 2 _

2 ; n = 5; θ = 135°;

1 _

8 -

1 _ 8 i

13. verified 15. verified 17. verified 19. verified

21. r = 1; n = 5; θ = 0°; roots: 1, 0.3090 ± 0.9511i, -0.8090 ± 0.5878i

23. r = 243; n = 5; θ = 0°; roots: 3, 0.9271 ± 2.8532i, -2.4271 ±1.7634i

25. r = 27; n = 3; θ = 270°; roots: 3i, -3 √ � 3

_ 2 -

3 _

2 i,

3 √ � 3 _

2 -

3 _

2 i

27. 2, 0.6180 ± 1.9021i, -1.6180 ± 1.1756i

29. 3 √ � 3

_ 2 +

3 _

2 i, -

3 √ � 3 _

2 +

3 _

2 i, -3i

31. 1.1346 + 0.1797i, 0.1797 + 1,1346i, -1.0235 + 0.5215i, -0.8123 - 0.8123i, 0.5215 - 1.0235i

33. x = 1, -

1 _ 2 ±

√ � 3 _

2 i. These are the same results as in Example 3.

35. r = 16; n = 4; θ = 120°; roots: √ � 3 + i, -1 + √ � 3 i, - √ � 3 -i, 1 - √ � 3 i

37. r = 7 √ � 2 ; n = 4; θ = 225°; roots: 0.9855 + 1.4749i,

-1.4749 + 0.9855i -0.9855 - 1.4749i, 1.4749i - 0.9855i

39. D = -4, z 0 = 8 1 _

6 cis 45°, z 1 = 8

1 _

6 cis 165°, z 2 = 8

1 _

6 cis 285°,

z 0 = 8 1 _

6 cis 75°, z 1 = 8

1 _

6 cis 195°, z 2 = 8

1 _

6 cis 315° 41. verified

43a. numerator: -117 + 44j, denominator: -21 + 72j

43b. 1 +

4 _ 3 j 43c. verified 45. Answers will vary. 47. -7 - 24i

49. z ≈ -2.7320, z ≈ 0.7320, z ≈ 2.

Note:Using sum and difference identities, all three solutions can actually be given in exact form: -1 - √ � 3 , -1 + √ � 3 , 2.

51. tan 2 x _

sec x + 1 =

sec 2 x -1 _ sec x + 1

=

(sec x + 1)(sec x - 1) __

sec x + 1

= sec x - 1

=

1 _ cos x -

cos x _ cos x

=

1 - cos x _ cos x

53. y = -

4 _ 5 x +

12 _ 5

Summary and concept review

1.

2. 3. approx. 41.84 ft

4. approx. 20.2° and 159.8°

5.

6. no; 36° 7. apporx. 36.9° 8. apporx. 385.5 m

9. 133. 2°, 30.1°, and 16.7° 10. 85,570.7 m 2

11. 12. -8i + 3j; ⎪u⎥ ≈ 8.54; θ ≈ 159.4°

13. horiz. comp. ≈ 11.08, vertical comp.14.18

14. ⟨-4, -2⟩; ⎪2u + v⎥ ≈ 4.47, θ 206.6° 15. 7 _

√ �� 193 i +

12

_ √ �� 193

j

16. QII; since the x-compotent is negative and the y-component is positive.

17. 1 _ 6 mi 18. approx. 19.7° 19. ⟨-25, -123⟩ 20. approx. -0.87

21. 4 22. p · q = -6; θ ≈ 97.9° 23. 4340 ft-lb

24. approx. 417.81 lb 25. approx. 8156.77 ft-lb

26a. x ≈ 269.97 ft; y ≈ 285.74 ft 26b. approx. 0.74 sec

27. 2(cos 240° + i sin 240°) 28. 3 + 3i

1�1�3 2 3 4

y

x

Angles Sides

A = 36° a ≈ 205.35 cm

B = 21° b ≈ 125.20 cm

C = 123° a ≈ 293 cm

Angles Sides

A = 28° a ≈ 140.59 yd

B = 10° b ≈ 52 yd

C = 142° c ≈ 184.36 yd

Angles Sides

A = 35° a = 67 cm

B 1 ≈ 64.0° b = 105 cm

C 1 ≈ 81.0° c 1 ≈ 115.37 cm

Angles Sides

A = 35° a ≈ 67 cm

B 2 = 116.0° b ≈ 105 cm

C 2 ≈ 29.0° c 2 ≈ 56.63 cm

y

x9

5

29.1�

10.30



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

29. 30. z 1 z 2 = 16 cis ( 5π

_ 12

) ; z 1

_ z 2 = 4 cis ( π

_ 12

)

31. 2 √ � 3 + 2j 32. ⎪Z⎥ ≈ 10.44 θ ≈ 16.7, 10.44 cis 16.7°

33. - 16 - 16 √ � 3i 34. verified 35. 5 √ � 3

_ 2 +

5 _

2 i, -

5 √ � 3 _

2 +

5 _

2 i,- 5i

36. 6, -3 ± 3i √ � 3 37. 2 - 2i, -2 ± 2i 38. 1 ± 2i, -1 ± 2i

39. verified

Mixed Review

1. Angles sides

A = 41° a ≈ 13.44 in

B = 27° b ≈ 9.30 in

C = 112° c = 19 in

, Area ≈ 57.9 in 2

3. x 16.09, y 13.50 5. approx. 176.15 ft 7. approx. 793.70 mph; heading 28.2°

9. One solution possible since side a > side b

Angles sides

A = 31° a = 36 m

B ≈ 20.1° b = 24 m

C ≈ 128.9° c ≈ 54.4 m

11. No; barely touches (“tangent”) at 30°

13a. 4 √ � 2 (cos 315° + i sin 315°) 13b. -3 + 3 √ � 3 iyi

x

�4

445�

yi

x

120�

6

15. ≈ 13.1° 17. comp v u ≈ -0.87, proj v u ≈ -

38 _

53 +

26 _

53 i

19. z 0 =

√ � 6 _

2 +

√ � 2 _

2 i, z 1 = -

√ � 2 _

2 +

√ � 6 _

2 i,

z 2 = -

√ � 6 _

2 -

√ � 2 _

2 i, z 3 =

√ � 2 _

2 -

√ � 6 _

2 i

Practice Test

1. 6.58 mi 2. 137.18 ft

3. Angles Sides (in.)

A 1 ≈ 58.8° a = 15

B = 20° b = 6

C 1 ≈ 101.2° c 1 ≈ 17.21

Angles Sides (in.)

A 2 ≈ 121.2° a = 15

B = 20° b = 6

C 2 ≈ 38.8° c 2 ≈ 11.0

4a. No 4b. 2.66 mi 5a. No 5b.1 5c. 8.43 sec

6a. 2.30 mi 6b. 7516.5 ft

7. A ≈ 438,795 mi 2 , P ≈ 61.7°, B ≈ 61.2°, M ≈ 57.1°

8. speed ≈ 73.36 mph bearing ≈ 47.8° 9. θ ≈ 36.5°

10. 63.48 cm to the right and 130.05 cm down from the initial point on the ceiling

11. ⎪F3⎥ ≈ 212.94 N, θ ≈ 251.2°

12a. θ ≈ 42.5° 12b. proj v u = ⟨-2.4, 7.2⟩ 12c. u 1 = ⟨-2.4, 7.2⟩,

u 2 = ⟨-6.6, -2.2⟩

13. 104.53 ft; 3.27 sec 14. 2 cis ( π _ 4 ) 15. 48 √ � 2 cis 75°; verified

16. -8 - 8 √ � 3 i 17. verified 18. 5 √ � 3

_ 2 + 5 _

2 i, -

5 √ � 3 _

2 + 5 _

2 i, -5i

19. 2.3039 ± 1.5192i, -2.3039 ± 1.5192i 20. ≈ 2, 414,300 mi 2

Strengthening Core Skills

Exercise 1: 664.46 lb, 640.86 lb

Exercise 2: 106.07 lb, 106.07 lb

Exercise 3: yes

Study Guide and Review

Lesson 5-1

1. ambiguous 3. I; II 5. Answers will vary.

Lesson 5-2

1. cosines 3. pythagorean 5. B ≈ 33.1°, C ≈ 129.9°, a ≈ 19.8; law of sines

Lesson 5-3

1. scalar 3. directed; line 5. Answers will vary.

Lesson 5-4

1. equilibrium; zero 3. orthogonal 5. Answers will vary.

Lesson 5-5

1. modulus; argument 3. multiply; add

5. 2(cos 240° + i sin 240°), z is in QIII

Lesson 5-6

1. r 5 [cos (5θ) + i sin (5θ) ]; De Moivre’s 3. complex

5. z 5 = 2 cis 366° = 2 cis 6°, z 6 = 2 cis 438° = 2 cis 78°, z 7 = 2 cis 510° = 2 cis 150°; Answers will vary

yi

x

30�

2

4 5

3

5√3�\, e�



Selected A

nswers and S

olutions

For Homework Help, go to Hotmath.com

System of Equations and Matrices6CHAPTER 6CHCHAPAPTETERR 66

Chapter 6 Get Ready

1. (4, 1) 3. (-2, 4) 5. (-3, -2) 7. dog: $15, cat: $12

9. ⎡

⎢

⎣

-10

34

-8

4

0

-10 ⎤

�

⎦

11. ⎡

⎢

⎣

6

13

23

3

-7

23 ⎤

�

⎦

13. ⎡

⎢

⎣

-7

45

9

4

-5

4 ⎤

�

⎦

15. 2 x 3 - x + 4 + -12 _

x + 2

17. 2 x 4 + 4 x 3 + 7 x 2 + 16x + 41 + 88 _

x - 2

Lesson 6-1

1. (-2, -7) 3. ( 1 _ 2 , 5) 5. (8, 9, 5) 7. (-6, 4, 1 _

6 )

9. ⎡

⎢

⎣

12

-3

-5

8

-9

10 ⎤

�

⎦

11.

⎡

⎢

⎣

3

-10

4

- 5

1

0

7

8

-15

9

6

-8

⎤

�

⎦

13.

⎡

⎢

⎣

1

7

6

0

-8

2

0

3

5

-3

12

4

0

9

-15

-8

11

-5

4

-13

⎤

�

⎦

15 a. The total amount of money raised can be represented by 30c + 40p + 200g = 684.50. A linear equation that relates the cost of a pie to the cost of a cake is p - c = -2, and a linear equation that relates the cost of a cake to a giant cookie is c - 5g = 0.

b. Let the first column of the augmented matrix represent c, the second column represent p, the third column represent g, and the fourth column represent the cost in dollars.

⎡

⎢

⎣

30

-1

1

40

1

0

200

0

-5

684.50

-2

0

⎤

�

⎦

c. Perform elementary row operations to obtain a row-echelon form of the matrix. Begin by switching the third and first rows so that the first row has a 1 in the first entry.

⎡

⎢

⎣

30

-1

1

10

1

0

200

0

-5

684.5

-2

0

⎤

�

⎦

interchange rows

⎡

⎢

⎣

1

-1

30

0

1

40

-5

0

200

0

-2

684.5

⎤

�

⎦

R 1 + R 2

⎡

⎢

⎣

1

0

30

0

1

40

-5

-5

200

0

-2

684.5

⎤

�

⎦

-30 R 1 + R 3

⎡

⎢

⎣

1

0

0

0

1

40

-5

-5

350

0

-2

684.5

⎤

�

⎦

-40 R 2 + R 3

⎡

⎢

⎣

1

0

0

0

1

0

-5

-5

550

0

-2

764.5

⎤

�

⎦

1 _

550 R 3

⎡

⎢

⎣

1

0

0

0

1

0

-5

-5

1

0

-2

1.39

⎤

�

⎦

You can use substitution to find that c = 6.95 and p = 4.95. Therefore, the solution is (6.95, 4.95, 1.39). So, a cake costs $6.95, a pie costs $4.95, and a giant cookie costs $1.39.17. yes 19. yes 21. yes 23. (-3, 1) 25. (-1, 2) 27. (10, -2, 1) 29. (52, 10, 24) 31. Sample answer: There are an infinite number of solutions to the system. One of the rows in the matrix is eliminated, so the solution set is (-2.75s + 8, 0.5s + 0.25, s). More information is needed to determine which of these solutions is valid. 33. (z + 5, 9 - z, z) 35. (3z + 1, 2z + 9, z) 37. (5z - 3, 4z + 1, z) 39. (w - 2, 3w - 5, 2w - 4, w)

41. ⎡

⎢

⎣

1

0

5 _ 6

1

10 _

3

-8 ⎤

�

⎦

;

⎡

⎢

⎣

1

0

0

1

10

-8 ⎤

�

⎦

43.

⎡

⎢

⎣

1

0 0

-

10 _

159

1

0

43 _

159

-0.16…

1

148 _

159

3535 _

2144

-7

⎤

�

⎦

;

⎡

⎢

⎣

1

0

0

0

1

0

0

0

1

3

2.8

-7

⎤

�

⎦

45. R 2 + R 3

47 The fourth row of the second matrix is different from the fourth row in the first matrix. Therefore, an operation involving the fourth row was performed. The first entry in the fourth row was -3 in the first matrix but is 0 in the second matrix. Thus, 3 was added to this entry. Since there is no 3 in the first column for any of the three other rows, one of the rows must have been multplied by a factor. Starting with the first row, if the row is multiplied by 3, the entries will become 3, 45, 6, 8, and 42. Adding this new row to the fourth row results in the fourth row for the second matrix. Thus, the row operation performed is 3 R 1 + R 4 .

49a. x + y = 10; 0.20x + 0.40y = 2.5 49b. (7.5, 2.5); The nurses will

need to mix 7.5 L of 20% and 2.5 L of 40% saline. 51. 3 or 27 _

34

53. False; Sample answer: Because the equation in the last row is 0 = 0, the value of one of the variables cannot be determined. Therefore, the corresponding system of equations may have infinitely many solutions. 55. Sample answer: In Gaussian elimination, matrices are in row-echelon form. In Gauss-Jordan elimination, matrices are in reduced row-echelon form. Matrices in row-echelon form and reduced row-echelon form may have rows consisting entirely of 0s. If they do, these rows appear at the bottom of the matrices. In both forms of matrices, the first entry in any row with nonzero entries is 1. This is called a leading 1. In both forms of matrices, if there are two successive rows with nonzero entries, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row. For a system with a unique solution, when the matrix is in reduced row-echelon form, every column that has a leading 1 has 0s above and/or below that 1.

57. ta n 2 x _ 2 =

si n 2 x _ 2 _

co s 2 x _ 2

= 1 - cos

⎡

⎣

2 ( x _ 2 )

⎤

⎦

__ 1 + cos

⎡

⎣

2 ( x _ 2 )

⎤

⎦

= 1 - cos x _ 1 + cos x



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

59. √ � 2 - √ � 6

_ 4 61.

√ � 2 - √ � 6 _

4 63. 2 - √ � 3 65. about 46.1 ft

67a. y = 65,000(6.2 ) x 67b. about 22,890,495,000 69. 2, 9 71. B 73. C

Lesson 6-2

1. AB = [19 -54]; BA is undefined. 3. AB = [7 15 -16]; BA is

undefined. 5. AB is undefined; BA = ⎡

⎢

⎣

18

-11 ⎤

�

⎦

.

7. AB = ⎡

⎢

⎣

-9

-41

6

-14

12

65 ⎤

�

⎦

; BA is undefined.

9 Let a 3 × 3 matrix represent the number of free throws, 2-pointers, and 3-pointers that each player made, and let a 3 × 1 matrix represent the point value of each shot.

FT 2-pt 3-pt

Rey

Chris

Jerry

⎡

⎢

⎣

44

37

35

32

24

39

25

31

29

⎤

�

⎦

FT

2-pt

3-pt

⎡

⎢

⎣

1

2

3

⎤

�

⎦

Perform matrix multiplication to determine the total amount of points scored by each player.

⎡

⎢

⎣

44

37

35

32

24

39

25

31

29

⎤

�

⎦

·

⎡

⎢

⎣

1

2

3

⎤

�

⎦

=

⎡

⎢

⎣

183

178

200

⎤

�

⎦

So, Rey scored 183 points, Chris scored 178 points, and Jerry scored 200 points.

11. ⎡

⎢

⎣

2

4

-3

-5

1

9

3

-6

-7

⎤

�

⎦

·

⎡

⎢

⎣

x 1

x 2

x 3

⎤

�

⎦

=

⎡

⎢

⎣

9

35

-6

⎤

�

⎦

; (1, -5, -6)

13.

⎡

⎢

⎣

2

6

-4

-10

-1

8

7

5

-3

⎤

�

⎦

·

⎡

⎢

⎣

x 1

x 2

x 3

⎤

�

⎦

=

⎡

⎢

⎣

7

-2

-22

⎤

�

⎦

; (2.5, -3, -4)

15. ⎡

⎢

⎣

2

8

-4

6

-12

10

-5

7

-1

⎤

�

⎦

·

⎡

⎢

⎣

x 1

x 2

x 3

⎤

�

⎦

=

⎡

⎢

⎣

-20

28

7

⎤

�

⎦

; (0.5, 1.5, 6)

17. ⎡

⎢

⎣

-1

-5

2

-3

11

1

9

8

-13

⎤

�

⎦

·

⎡

⎢

⎣

x 1

x 2

x 3

⎤

�

⎦

=

⎡

⎢

⎣

25

33

-45

⎤

�

⎦

; (14, 5, 6) 19. yes

21. no 23. yes 25. yes 27. singular 29. A -1 = ⎡

⎢

⎣

-3

2

-5

3 ⎤

�

⎦

31. A -1 =

⎡

⎢

⎣

-44

16

9

-5

2

1

-14

5

3

⎤

�

⎦

33. singular 35. 3; ⎡

⎢

⎣

-

2 _

3

-1

5 _ 3

2 ⎤

�

⎦

37. 6;

⎡

⎢

⎣

3 _

2

-1

7 _ 6

-

2 _

3

⎤

�

⎦

39. -11;

⎡

⎢

⎣

1 _

11

0

-

4 _

11

5 _ 11

1

13 _

11

8 _ 11

2

23 _

11

⎤

�

⎦

41. 0; singular

43. -14;

⎡

⎢

⎣

12 _

7

15 _

7

-

13 _

7

1

1

-1

-

1 _

14

-

3 _

14

2 _ 7

⎤

�

⎦

45 Find the determinate of the matrix formed by the three vertices.

⎪

1

4

4

1

-1

3

1

1

1

⎥ = 1

⎪ -1

3

1

1 ⎥ - 1

⎪ 4

4

1

1 ⎥ + 1

⎪ 4

4

-1

3 ⎥

= -4 - 0 + 16 or 12

Since the area of triangle is 1 _ 2 of the determinate of the

coordinates, A = 1 _ 2 (12) or 6 unit s 2 .

47. 13 1 _ 2 uni ts 2 49.

⎡

⎢

⎣

2

-5

8

4 ⎤

�

⎦

51. x = 0.5, y = -3 53. c 3

55a. A 2 = ⎡

⎢

⎣

1

0

2

1 ⎤

�

⎦

, A 3 = ⎡

⎢

⎣

1

0

3

1 ⎤

�

⎦

, A 4 = ⎡

⎢

⎣

1

0

4

1 ⎤

�

⎦

, A n = ⎡

⎢

⎣

1

0

n

1 ⎤

�

⎦

55b. B 2 = ⎡

⎢

⎣

2

2

2

2 ⎤

�

⎦

, B 3 = ⎡

⎢

⎣

4

4

4

4 ⎤

�

⎦

= ⎡

⎢

⎣

2 2

2 2

2 2

2 2 ⎤

�

⎦

, B 4 = ⎡

⎢

⎣

8

8

8

8 ⎤

�

⎦

= ⎡

⎢

⎣

2 3

2 3

2 3

2 3 ⎤

�

⎦

,

B 5 = ⎡

⎢

⎣

16

16

16

16 ⎤

�

⎦

= ⎡

⎢

⎣

2 4

2 4

2 4

2 4 ⎤

�

⎦

, B n = ⎡

⎢

⎣

2 n - 1

2 n - 1

2 n - 1

2 n - 1 ⎤

�

⎦

or B n = 2 n - 1 ⎡

⎢

⎣

1

1

1

1 ⎤

�

⎦

55c. C 2 = ⎡

⎢

⎣

4

8

0

4 ⎤

�

⎦

= ⎡

⎢

⎣

2 2

2 3

0

2 2 ⎤

�

⎦

,

C 3 = ⎡

⎢

⎣

8

24

0

8 ⎤

�

⎦

= ⎡

⎢

⎣

2 3 3 · 2 3

0

2 3 ⎤

�

⎦

, C 4 = ⎡

⎢

⎣

16

64

0

16 ⎤

�

⎦

= ⎡

⎢

⎣

2 4 4 · 2 4

0

2 4 ⎤

�

⎦

,

C 5 = ⎡

⎢

⎣

32

160

0

32 ⎤

�

⎦

=

⎡

⎢

⎣

2 5 5 · 2 5

0

2 5 ⎤

�

⎦

, C n =

⎡

⎢

⎣

2 n n · 2 n

0

2 n ⎤

�

⎦

55d. ⎡

⎢

⎣

128

896

0

128 ⎤

�

⎦

57.

⎡

⎢

⎣

12

-4

32

4

0

6

⎤

�

⎦

59.

⎡

⎢

⎣

13

7

-5

27

7

16

-10

8

-43

⎤

�

⎦

61. ⎡

⎢

⎣

3

-2

5

-1 ⎤

�

⎦

63.

⎡

⎢

⎣

2

1 _ 3

1

1

-

4 _

3

3

⎤

�

⎦

65a. 4 65b. 4

65c. They are the same.

65d. ⎪

a d

g

b e

h

c

f

i

⎥ =

⎪ a d

g

b e

h

c

f

i

⎥ a d

g

b e

h

= (aei + bfg + cdh) - (gec + hfa + idb)= aei + bfg + cdh - gec - hfa - idb= a(ei - hf ) + b( fg - id) + c(dh - ge)= a(ei - hf ) - b(id - fg) + c(dh - ge)= a(ei - hf ) - b(di - gf ) + c(dh - ge)

= a ⎪ e

h

f

i ⎥ - b ⎪

d g

f

i ⎥ + c

⎪ d g

e

h ⎥ 65e. No; sample answer:

If A =

⎡

⎢

⎣

1

0

2

3

2

1

3

1

3

2

1

2

0

3

1

1

⎤

�

⎦

, det(A) = -60.



Selected A

nswers and S

olutions


⎪

1

0

2

3

2

1

3

1

3

2

1

2

0

3

1

1

⎥ =

⎪

1

0

2

3

2

1

3

1

3

2

1

2

0

3

1

1

⎥

1

0

2

3

2

1

3

1

3

2

1

2

= 1(1)(1)(1) + 2(2)(1)(3) + 3(3)(2)(1) + 0(0)(3)(2) - [3(3)(2)(0) + 1(1)(3)(1) + 2(1)(0)(2) + 1(2)(1)(3)]

� 1 + 12 + 18 + 0 - (0 + 3 + 0 + 6) = 22Since 22 ≠ -60, this method does not work for a 4 × 4 matrix. 67. Suppose B and C are both inverses of square matrix A. Then by definition, AB = BA = I and AC = CA = I. Then B(AC) = BI = B and (BA)C = IC = C. But B(AC) = (BA)C by the Associative Property of Matrix Multiplication. Therefore, B = C.

69 Let B = ⎡

⎢

⎣

r

u

x

s v

y

t

w

z ⎤

⎦

.

Set up three systems of equations.AB = A · B

⎡

⎢

⎣

14

4

1

6

4

18

33

13

12

⎤

⎦

= ⎡

⎢

⎣

3

1

6

-1

0

4

5

2

1

⎤

⎦

· ⎡

⎢

⎣

r

u

x

s v

y

t

w

z ⎤

⎦

3r - u + 5x = 14r + 0u + 2x = 46r + 4u + x = 1

3s - v + 5y = 6s + 0v + 2y = 46s + 4v + y = 18

3t - w + 5z = 33t + 0w + 2z = 136t + 4w + z = 12

Solve the first system.r + 2x = 4r = 4 - 2x

(4)(3r - u + 5x = 14)(+)6r + 4u + x = 1

12r - 4u + 20x = 56 ____________________ (+)6r + 4u + x = 1 18r + 21x = 57

18(4 - 2x) + 21x = 57 72 - 36x + 21x = 57 -15x = -15 x = 1 r = 4 - 2(1) r = 2

3r - u + 5x = 143(2) - u + 5(1) = 14 6 - u + 5 = 14 -u = 3 u = -3

Solve the next system.s + 2y = 4s = 4 - 2y

(4)(3s - v + 5y = 6)(+)6s + 4v + y = 18

12s - 4v + 20y = 24 ____________________ (+)6s + 4v + y = 18 18s + 21y = 42

18s + 21y = 4218(4 - 2y) + 21y = 42 72 - 36y + 21y = 42 -15y = -30 y = 2

s = 4 - 2(2) s = 0

3s - v + 5y = 63(0) - v + 5(2) = 6 -v + 10 = 6 -v = -4 v = 4

Solve the last system.t + 2z = 13t = 13 - 2z

4(3t - w + 5z = 33)(+)6t + 4w + z = 12

12t - 4w + 20z = 132 _____________________ (+)6t + 4w + z = 12 18t + 21z = 144

18t + 21z = 14418(13 - 2z) + 21z = 144 234 - 36z + 21z = 144 -15z = -90 z = 6 t = 13 - 2(6) t = 1

3t - w + 5z = 333(1) - w + 5(6) = 33 3 - w + 30 = 33 -w = 0 w = 0

B = ⎡

⎢

⎣

r

u x

s v

y

t

w

z ⎤

⎦

= ⎡

⎢

⎣

2

-3

1

0

4

2

1

0

6

⎤

⎦

71. Sample answer: A = ⎡

⎢

⎣

2

2

2

2 ⎤

⎦

, B =

⎡

⎢

⎣

3

3

3

3 ⎤

⎦

73. C(A + B) =

⎡

⎢

⎣

c 11

c 21

c 12

c 22 ⎤

⎦

(

⎡

⎢

⎣

a 11

a 21

a 12

a 22 ⎤

⎦

+ ⎡

⎢

⎣

b 11

b 21

b 12

b 22 ⎤

⎦

)

= ⎡

⎢

⎣

c 11

c 21

c 12

c 22 ⎤

⎦

⎡

⎢

⎣

a 11 + b 11

a 21 + b 21

a 12 + b 12

a 22 + b 22 ⎤

⎦

= ⎡

⎢

⎣

c 11 ( a 11 + b 11 ) + c 12 ( a 21 + b 21 )

c 21 ( a 11 + b 11 ) + c 22 ( a 21 + b 21 )

c 11 ( a 12 + b 12 ) + c 12 ( a 22 + b 22 )

c 21 ( a 12 + b 12 ) + c 22 ( a 22 + b 22 )

⎤

⎦

= ⎡

⎢

⎣

c 11 a 11 + c 11 b 11 + c 12 a 21 + c 12 b 21

c 21 a 11 + c 21 b 11 + c 22 a 21 + c 22 b 21

c 11 a 12 + c 11 b 12 + c 12 a 22 + c 12 b 22

c 21 a 12 + c 21 b 12 + c 22 a 22 + c 22 b 22

⎤

⎦

= ⎡

⎢

⎣

c 11 a 11 + c 12 a 21 + c 11 b 11 + c 12 b 21

c 21 a 11 + c 22 a 21 + c 21 b 11 + c 22 b 21

c 11 a 12 + c 12 a 22 + c 11 b 12 + c 12 b 22

c 21 a 12 + c 22 a 22 + c 21 b 12 + c 22 b 22

⎤

⎦



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

= ⎡

⎢

⎣

c 11 a 11 + c 12 a 21

c 21 a 11 + a 22 a 21

c 11 a 12 + c 12 a 22

c 21 a 12 + c 22 a 22 ⎤

�

⎦

+

⎡

⎢

⎣

c 11 b 11 + c 12 b 21

c 21 b 11 + c 22 b 21

c 11 b 12 + c 12 b 22

c 21 b 12 + c 22 b 22 ⎤

�

⎦

= CA + CB

75. c(AB) = c ( ⎡

⎢

⎣

a 11

a 21

a 12

a 22 ⎤

�

⎦

⎡

⎢

⎣

b 11

b 21

b 12

b 22 ⎤

�

⎦

) Subst.

= c ⎡

⎢

⎣

a 11 b 11 + a 12 b 21

a 21 b 11 + a 22 b 21

a 11 b 12 + a 12 b 22

a 21 b 12 + a 22 b 22 ⎤

�

⎦

Def. Matrix Mult.

= ⎡

⎢

⎣

c( a 11 b 11 + a 12 b 21 )

c( a 21 b 11 + a 22 b 21 )

c( a 11 b 12 + a 12 b 22 )

c( a 21 b 12 + a 22 b 22 ) ⎤

�

⎦

Def. Scalar Mult.

= ⎡

⎢

⎣

c a 11 b 11 + ca 12 b 21

ca 21 b 11 + ca 22 b 21

ca 11 b 12 + ca 12 b 22

ca 21 b 12 + ca 22 b 22 ⎤

�

⎦

Dist. Prop.

= ⎡

⎢

⎣

ca 11

ca 21

ca 12

ca 22 ⎤

�

⎦

⎡

⎢

⎣

b 11

b 21

b 12

b 22 ⎤

�

⎦

Def. Matrix Mult.

= (c ⎡

⎢

⎣

a 11

a 21

a 12

a 22 ⎤

�

⎦

) ⎡

⎢

⎣

b 11

b 21

b 12

b 22 ⎤

�

⎦

Def. Scalar Mult.

= (cA)B Subst.

= c(AB) = c ( ⎡

⎢

⎣

a 11

a 21

a 12

a 22 ⎤

�

⎦

⎡

⎢

⎣

b 11

b 21

b 12

b 22 ⎤

�

⎦

) Subst.

= c ⎡

⎢

⎣

a 11 b 11 + a 12 b 21

a 21 b 11 + a 22 b 21

a 11 b 12 + a 12 b 22

a 21 b 12 + a 22 b 22 ⎤

�

⎦

Def. Matrix Mult.

= ⎡

⎢

⎣

c( a 11 b 11 + a 12 b 21 )

c( a 21 b 11 + a 22 b 21 )

c( a 11 b 12 + a 12 b 22 )

c( a 21 b 12 + a 22 b 22 ) ⎤

�

⎦

Def. Scalar Mult.

= ⎡

⎢

⎣

c a 11 b 11 + c a 12 b 21

c a 21 b 11 + c a 22 b 21

c a 11 b 12 + c a 12 b 22

c a 21 b 12 + c a 22 b 22 ⎤

�

⎦

Dist. Prop.

= ⎡

⎢

⎣

a 11 c b 11 + a 12 c b 21

a 21 c b 11 + a 22 c b 21

a 11 c b 12 + a 12 c b 22

a 21 c b 12 + a 22 c b 22 ⎤

�

⎦

Comm. Prop.

= ⎡

⎢

⎣

a 11 a 12

a 21 a 22 ⎤

�

⎦

⎡

⎢

⎣

cb 11 cb 12

cb 21 cb 22 ⎤

�

⎦

Def. Matrix Mult.

= A(cB) Subst.

77. 6 × 8; sample answer: Since the number of columns of matrix A must be equal to the number of rows of matrix B, matrix B must have 6 rows. The product matrix AB has the same number of columns as matrix B. So, matrix B must have 8 columns.

79. ⎡

⎢

⎣

10

-6

-3

4

-12

20 ⎤

�

⎦

81.

⎡

⎢

⎣

1

3

9

0

-6

2

0

5

14

-4

18

10

0

8

-12

-16

19

-2

3

-26

⎤

�

⎦

83. - π

_ 18

85. 135° 87. 8 89. 5 91. 9 93. C 95. G

Lesson 6-3

1. (3, 2) 3. (-1, 6) 5. (-6, 7, 8) 7. no unique solution 9. 4 sitcoms, 3 talk shows, and 2 movies 11. (-2, -2)

13. (1, 1 _ 2 ) 15. no unique solution 17. (6, 3, -4)

19. 15.5 gallons

21 Let x = the cost of each extra gaming minute, y = the cost of each extra call minute, and z = the cost of each text message. The corresponding system of equations is shown below.

30x + 12y + 40z = 52.9018x + 15y + 55z = 48.076x + 7y = 13.64

The coefficient matrix is

⎡

⎢

⎣

30

18

6

12

15

7

40

55

0

⎤

�

⎦

. Calculate the

determinant of the matrix.

⎪

30

18

6

12

15

7

40

55

0

⎥ = 30

⎪ 15

7

55

0 ⎥ - 12

⎪ 18

6

55

0 ⎥ + 40

⎪ 18

6

15

7 ⎥

= 30(-385) - 12(-330) + 40(36) or -6150

Since the determinant does not equal 0, you can apply Cramer’s Rule to find x, y, and z.

x = | A x |

_ |A|

=

⎪ 52.90

48.07

13.64

12

15

7

40

55

0

⎥

__ -6150

= 0.99

y = | A y |

_

|A|

=

⎪

30

18

6

52.90

48.07

13.64

40

55

0

⎥

__ -6150

= 1.10

z = | A z |

_ |A|

=

⎪

30

18

6

12

15

7

52.90

48.07

13.64

⎥

__ -6150

= 0.6

Each extra gaming minute is $0.99, each extra call minute is $1.10, and each text message is $0.25.

23. ⎡

⎢

⎣

1

2

-2

1 ⎤

�

⎦

25. ⎡

⎢

⎣

-1

-4

2

3 ⎤

�

⎦

27. (-6, 2, 5) 29. (-2, 4, 1)

31 When the determinant of a 2 × 2 matrix is 0, the matrix is

singular and has no inverse. Let A = ⎡

⎢

⎣

n 1

-8

2 ⎤

�

⎦

.

Find an equation for the determinant of A and solve for n when the determinant is 0.

⎪ n

1

-8

2 ⎥ = n(2) - 1(-8) or 2n + 8

2n + 8 = 0 2n = -8 n = -4Therefore, when n = -4, the system cannot be solved using an inverse matrix.

33. 0 35. 0.4 kg of 35% alloy, 0.8 kg of 55% alloy, and 1.3 kg of 60% alloy 37. AB = 32 mm, BC = 36 mm, AC = 21 mm

39.

⎡

⎢

⎣

2x _ 2 - 3x

- x 2 _

2 - 3x

-

3 _

2 - 3x

1 _

2 - 3x

⎤

�

⎦

41. ⎡

⎢

⎣

-2i -1

-3

i ⎤

�

⎦

43. X = (A + B ) -1 (D)

45. X = (-C - D)(I - A ) -1 47. X = (B - A ) -1 (AD)



Selected A

nswers and S

olutions


49. Neither; if the determinant of the coefficient matrix is 0, then there is no unique solution. The system may have no solution, or it may have an infinite number of solutions. The system of equations shown has an infinite number of solutions. 51. Yes; sample answer:

A 2 = ⎡

⎢

⎣

a 2 + bc

ac + cd

ab + bd

bc + d 2 ⎤

�

⎦

. Therefore,

( A 2 ) -1 = 1 __

a 2 d 2 - 2abcd + b 2 c 2 ·

⎡

⎢

⎣

bc + d 2

-ac - cd

-ab - bd

a 2 + bc ⎤

�

⎦

and

A -1 = 1 _

ad - bc ·

⎡

⎢

⎣

d -c

-b a ⎤

�

⎦

=

⎡

⎢

⎣

d _ ad - bc

-c _ ad - bc

-b _ ad - bc

a _ ad - bc

⎤

�

⎦

, so

( A -1 ) 2 = 1 __

a 2 d 2 - 2abcd + b 2 c 2 ·

⎡

⎢

⎣

bc + d 2

-ac - cd -ab - bd

a 2 + bc ⎤

�

⎦

.

Thus, ( A 2 ) -1 = ( A -1 ) 2 . 53a. Sample answer: Gauss-Jordan elimination can be used to solve any system of linear equations. It is possible to perform row operations on any matrix. 53b. Sample answer: Inverse matrices can only be used to solve systems with square coefficient matrices because matrix multiplication can only be done if the number of columns of the first matrix is equal to the number of rows of the second matrix. 53c. Sample answer: Cramer’s Rule uses determinants to solve systems, and because it is only possible to find the determinant of a square matrix, this method can only be used to solve square systems.

55. AB =

⎡

⎢

⎣

54

202

136

26

-58

67

-76

-4

-66

⎤

�

⎦

, BA =

⎡

⎢

⎣

38

43

-84

2

-122

38

112

27

14

⎤

�

⎦

57. yes 59a. Sirius 59b. Sirius: 1.45, Vega: 0.58 59c. Vega 61. F 63. H

Lesson 6-4

1. 2 _

x + 3 + -1

_ x + 2

3. 10 _

x + 3 + -9

_ x + 4

5. 3 _

-2x - 9 + 1

_ x + 5

7. 3 + 2 _ x + 5 _

x - 2 9. -2 + 4 _ x + 1

_ x - 2

+ 3 _

x + 4

11. x + 4 + -8 _

x + 3 + -6

_ x + 5

13. -3 _ x + 4

_ x + 1

+ 2 _

(x + 1 ) 2

15. -2 _ x + 1

_ x + 5

+ -7 _

(x + 5 ) 2 17. 4 _ x + -4

_ x - 8

+ 49 _

(x - 8 ) 2

19 a. Decompose s(x) = 20x + 10πx + 20

__ π x 3 + π x 2

. First, divide the numerator by π.

s(x) = 20x + 10πx + 20

__ π x 3 + π x 2

= 20

_ π

x + 10x + 20 _

π

__ x 3 + x 2

The fraction is proper, so we can rewrite the expression as partial fractions with constant numerators, A, B, and C, and denominators that are the linear factors of the original denominator. This time, however, we have repeated factors in the denominator. x 3 + x 2 = ( x 2 )(x + 1)

Because the factor x 2 has a multiplicity of 2, include partial fractions with denominators of x, x 2 , and x + 1.

20

_ π

x + 10x + 20 _

π

__ x 3 + x 2

= A _ x + B _

x 2 + C _

x + 1

Multiply each side by the LCD, x 3 + x 2 .

20

_ π

x + 10x + 20 _

π

__ x 3 + x 2

= A _ x + B _

x 2 + C _

x + 1

20

_ π

x + 10x + 20 _

π

__ x 3 + x 2

= A(x)(x + 1)

_ x 3 + x 2

+ B(x + 1)

_ x 3 + x 2

+ C x 2 _

x 3 + x 2

20 _

π

x + 10x + 20 _

π

= A(x)(x + 1) + B(x + 1) + C x 2

( 20 _

π

+ 10) x + 20 _

π

= A x 2 + Ax + Bx + B + C x 2

( 20 _

π

+ 10) x + 20 _

π

= A x 2 + C x 2 + Ax + Bx + B

( 20 _

π

+ 10) x + 20 _

π

= (A + C) x 2 + (A+ B)x + B

Equate the coefficients on the left and right side of the equation to form a system of equations. In other words, the coefficients of the x-terms on the left side of the equation must equal the coefficients of the x-terms on the right side.A + C = 0

A + B = 20 _

π

+ 10

B = 20 _

π

Use any method to solve the system.

B = 20 _

π

A + B = 20 _

π

+ 10

A + 20 _

π

= 20 _

π

+ 10

A = 10 A + C = 0 10 + C = 0 C = -10Replace A, B, and C with 10, 20

_ π

, and -10 in the partial fraction decomposition.

20

_ π

x + 10x + 20 _

π

__ x 3 + x 2

= A _ x + B _

x 2 + C _

x + 1

= 10 _

x + π

_ x 2

+ -10 _

x + 1

b.

[-10, 10] scl: 1 by [-100, 100] scl: 10

21. 2 _ x + x _ x 2 + 3

+ -11x + 8

_ ( x 2 + 3 ) 2

23. 8x _ x 2 - 6

+ 7 _

( x 2 - 6 ) 2

25. 4x _ x 2 - 3x + 3

+ -17x + 20

__ ( x 2 - 3x + 3 ) 2

27. 3.4x + 0.6

_ 4 x 2 + 1

+ -0.6 _

x + 1

29. -2 _ x + 3

_ x - 2

+ 4 _

x + 2 31. 1

_ x 2 + 3

+ x + 2

_ ( x 2 + 3 ) 2

33. 2x + -17 _

x + 5

+ 4 _

x + 1 35. 6 _ x , -6

_ x + 1

, -7 _

(x + 1 ) 2 37. A =

2 _

3 t, B = r - 2 _

3 t - 12

39. A = -t, B = 2, C = t + 3, D = 5r - 2

41 Rewrite the expression as partial fractions with constant numerators A, B, C, and D and denominators that are linear factors of the original denominator.

x 3 + 4 __

( x 2 - 1)( x 2 + 3x + 2) = A _

(x - 1) + B _

(x + 1) + C _

(x + 1 ) 2 + D _

(x + 2)



Sel

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Sol

utio

ns x 3 + 4 = A(x + 1 ) 2 (x + 2) + B(x + 1)(x + 2)(x - 1) +

C(x - 1)(x + 2) + D(x + 1 ) 2 (x - 1)

x 3 + 4 = A( x 3 + 4 x 2 + 5x + 2) + B( x 3 + 2 x 2 - x - 2) + C( x 2 + x - 2) + D( x 3 + x 2 - x - 1)

x 3 + 4 = x 3 (A + B + D) + x 2 (4A + 2B + C + D) + x(5A - B + C - D) + 2A - 2B - 2C - D

Equate the coefficients on the left and right sides of the equation.A + B + D = 14A + 2B + C + D = 05A - B + C - D = 02A - 2B - 2C - D = 0To solve the system, you can write it in matrix form EX = F and solve for X. E X F

⎡

⎢

⎣

1

4

5

2

1

2

-1

-2

0

1

1

-2

1

1

-1

-1

⎤

�

⎦

·

⎡

⎢

⎣

A

B C

D

⎤

�

⎦

=

⎡

⎢

⎣

1

0

0

4

⎤

�

⎦

You can use a graphing calculator to find X = E -1 F.

So, A = 5 _ 12

, B = -

3 _

4 , C = -

3 _

2 , and D = 4 _

3 . Therefore,

x 3 + 4

__ ( x 2 - 1)( x 2 + 3x + 2)

= -

3 _

4 _

x + 1 +

-

3 _

2 _

(x + 1 ) 2 +

5 _ 12

_

x - 1 +

4 _ 3 _

x + 2

43. 7x + 2 + 1 _ x + 8 _

x - 1 + 2

_ (x - 1 ) 2

+ -4 _

x + 2 +

-4x + 4 _

x 2 + 1

45. I 47. 0; The partial fraction decomposition of f (x) is

1 _

x + 1 + -3

_ (x + 1 ) 2

+ 3 _

(x + 1 ) 3 . As x approaches infinity, the values of

the three terms approach 0. So, the limit of the expression is 0.49. b 51. d

53. True; -5 _ x +

4 + x _ x 2 - 3

+ x 2 + 1

_ ( x 2 - 3 ) 2

=

-5( x 2 - 3) 2 + (4 + x)(x)( x 2 - 3) + x( x 2 + 1) ____

x( x 2 - 3 ) 2

=

-5 x 4 + 30 x 2 - 45 + 4 x 3 - 12x + x 4 - 3 x 2 + x 3 + x _____ x( x 2 - 3 ) 2

=

-4 x 4 + 5 x 3 + 27 x 2 - 11x - 45 ___

x( x 2 - 3 ) 2

55. Sample answer: First, factor the denominator, if necessary, and rewrite the equation with constant numerators, A, B, C, …, and denominators that are the linear or prime quadratic factors of the original denominator. Write the original expression equal to the new expression and multiply each side by the LCD. Use the Distributive Property to group like terms. Equate the coefficients on both sides of the equation to get a system of two equations. Then solve the system by writing it in matrix form CX = D and solving for X. Finally, use substitution to find the partial fraction decomposition. 57. (2, -1, 3) 59a. $24,900 59b. $26,36059c. She made $1460. 61. t an 2 θ 63a. y = 0.25 sin 128 πt, y = 0.25 sin 512 πt, y = 0.25 sin 1024 πt 63b. Sample answer: As the frequency increases, the period decreases. 65. F 67. H

Lesson 6-5

1. 31 at (10, 1); 1 at (0, 1) 3. 5 at (2, -3); -6 at (-2, 4) 5. 11 at (5, 2); -5 at (1, 4) 7. 0 at (4, 1); -12 at (4, 4)

9 a. The objective function is f (x, y) = 125x + 55y, where x is the number of physicals and y is the number of checkups. The constraints are 0 ≤ x ≤ 6, y ≥ 0, and 20y + 40x ≤ 420 (7 hours = 420 minutes).

b. y

x

x = 0(6, 9)

(6, 0)(0, 0)

(0, 21)

40x + 20y = 420

x = 6

y = 0

Chec

kups

Physicalsc. The polygonal region of feasible solutions has four vertices

located at (0, 21), (6, 9), (6, 0), and (0, 0). Find the value of the objective function at each of the vertices.

f (0, 0) = 0 f (6, 9) = 1245f (6, 0) = 750 f (0, 21) = 1155

So, the maximum income is $1245 when Olivia schedules 6 checkups and 9 physicals.

11a. f (w, a) = 600w + 700a, w ≥ 0, a ≥ 0, 10w + 15a ≤ 70, 4.5w + 9a ≤ 36

11b. y

x

E-al

bum

s

Web Sites

11c. The company should create 7 Web sites and no E-albums for a maximum profit of $4200.

13. The problem has multiple maxima of 15 from (1, 2) to (3, 1), or along the line y = -0.5x + 2.5 in [-1, 3]. There is a minimum of 0 at (0, 0). 15. The problem has multiple minima of -8 from (0, 2) to (2, 5), or along the line y = 1.5x + 2 in [0, 2]. There is a maximum of 18 at (3, 0). 17. The problem has multiple maxima of 40 from (0, 4) to (5, 0), or along the line y = -0.8x + 4 in [0, 5]. There is a minimum of 0 at (0, 0). 19. Sample answer: 1 21. Sample answer: -3 23. Sample answer: 5 25. Sample answer: f (x, y) = 4x - y 27. Sample answer: f (x, y) = -2x + 5y 29a. f (x, y) = 16x + 12y, 3x + 2y ≤ 40, x + y ≤ 15, x ≥ 0, y ≥ 0 29b. 10 batches of Mango Sunrise and 5 batches of Oasis Dream; $220 31. 48 at (1, 10); -24 at (-3, -6) 33. 40 unit s 2 35. 46.5 unit s 2 37. No; sample answer: If a and b are negative, then the maximum value of P, if it exists, will be negative. 39a. Sample answer: A bakery makes up to 100 apple and lemon meringue pies each week. Due to labor costs, the number of lemon meringue pies that are made needs to be less than or equal to 25 plus two times the number of apple pies. The customer demand for lemon meringue pies is at least twice that of apple pies. The bakery makes a profit of $8 per apple pie and $10 per lemon meringue pie. How many of each type of pie should be made to maximize profit? 39b. The constraints are y ≤ 25 + 2x, y ≥ 2x, x + y ≤ 100, x ≥ 0, and y ≥ 0. The objective function is given by f(x, y) = 8x + 10y.



Selected A

nswers and S

olutions


39c.

Lem

on M

erin

gue

Pies

10

0

20

30

40

50

60

80

90

100

Apple Pies10 20 30 40 50 60 70 80 90 100

70

(0, 0)

(0, 25)

(33.3, 66.7)(25, 75)

The graph of the system of inequalities forms a polygonal region with vertices at (0, 0), (0, 25), (25, 75), and (33.3, 66.7). 39d. Since f is greatest at (25, 75), the bakery should make 25 apple pies and 75 lemon meringue pies to earn a

maximum profit of $950. 41. 5 _

y - 1 + 3

_ y + 2

43. 3.5 _

m + 2 + 1.5

_ m - 2

45. race car, 6.5 points; snowboard,

5.25 points 47. tan θ sin θ cos θ cs c 2 θ

= sin θ

_ cos θ

� sin θ � cos θ � 1 _

si n 2 θ

= 1 49. 19.5 y d 2 51. 6.4 c m 2 53. G


1. augmented matrix 3. reduced row-echelon 5. inverse matrix 7. determinant 9. constraints 11. (1, 1)

13. (2, 3, -1) 15. (33, 15, -10) 17. (1, 3) 19. (2, 2, 3)

21. (1, 0, 5) 23. AB is undefined; BA = ⎡

⎢

⎣

-13

15

15

-3

-39

14 ⎤

�

⎦

.

25. AB = [-11 23]; BA is undefined. 27.

⎡

⎢

⎣

8 _

19

3 _ 19

-

1 _

19

2 _ 19

⎤

�

⎦

29. singular 31. (-4, 5) 33. (0, 1, -1) 35. (2, -5, 8) 37. (9, -3) 39. (3, -3, -4) 41. (0, -2, 7)

43. 1 _

x - 2 + 1

_ x + 2

45. 1 _

x - 5 - 3

_ x - 6

47. 2x - 1 _ x + 3 _

x - 7

49. 2 - 2 _ x + 6 _

x - 2 51. 1 _

2 + 2 _ x - 1.5

_ 2x - 3

53. max at (3, 4) = 2, min

at (0, 7) = -7 55. max at (10, 8) = 18, min at (0, 4) = 4 57. max at (-4, 20) = 44, min at (2, 2) = 14 59. plain hamburgers = $3.00, cheeseburgers = $4.50, veggie burgers = $5.50 61. 11 strawberry, 13 pineapple, 24 cherry 63a. f(p, g) = 25p + 20g, p + g ≥ 60, 0 ≤ p ≤ 50, 0 ≤ g ≤ 30 63b. 50 lb of paper and 10 lb of glass 63c. $1450


1. 0 < x < 8; Sample answer: x must be greater than 0 for the box to have any height, but it must also be less than 8 or the value for the width of the box would be negative or zero. 3. height ≈ 2.94 in., length ≈ 14.12 in., width ≈ 10.12 in., maximum volume ≈ 420.11 i n 3 5. radius = 1.55 in., height = 3.09 in. minimum surface area = 45.19 i n 2

7. (6.67, 1.83), Sample answer: A minimum time of 1.83 hours will be obtained if the racer takes a path that will have him/her at the sidewalk 3.33 miles before the pier. 9. m = 0; Sample answer: The slope of the tangent line is 0. The tangent line is a horizontal line.

Appendix: Nested Quantifiers

1a. For every real number x there exists a real number y such that x is less than y. 1b. For every real number x and real number y, if x and y are both nonnegative, then their product is nonnegative. 1c. For every real number x and real number y, there exists a real number z such that xy = z. 3a. There is some student in your class who has sent a message to some student in your class. 3b. There is some student in your class who has sent a message to every student in your class. 3c. Every student in your class has sent a message to at least one student in your class. 3d. There is a student in your class who has been sent a message by every student in your class. 3e. Every student in your class has been sent a message from at least one student in your class. 3f. Every student in the class has sent a message to every student in the class. 5a. Sarah Smith has visited www.att.com. 5b. At least one person has visited www.imdb.org. 5c. Jose Orez has visited at least one website. 5d. There is a website that both Ashok Puri and CindyYoon have visited. 5e. There is a person besides David Belcher who has visited all the websites that David Belcher has visited. 5f. There are two different people who have visited exactly the same websites. 7a. Abdallah Hussein does not like Japanese cuisine. 7b. Some student at your school likes Korean cuisine, and everyone at your school likes Mexican cuisine.7c. There is some cuisine that either Monique Arsenault or Jay Johnson likes. 7d. For every pair of distinct students at your school, there is some cuisine that at least one them does not like. 7e. There are two students at your school who like exactly the same set of cuisines. 7f. For every pair of students at your school, there is some cuisine about which they have the same opinion (either they both like it or they both do not like it). 9a. ∀xL(x, Jerry) 9b. ∀x∃yL(x, y) 9c. ∃y∀xL(x, y) 9d. ∀x∃y¬L(x, y) 9e. ∃x¬L(Lydia, x) 9f. ∃x∀y¬L(y, x) 9g. ∃x(∀yL(y, x) ∧ ∀z((∀wL(w, z)) → z = x)) 9h. ∃x∃y(x ≠ y ∧ L(Lynn, x) ∧ L(Lynn, y) ∧ ∀z(L(Lynn, z) → (z = x ∨ z=y))) 9i. ∀xL(x, x) 9j. ∃x ∀ y (L(x, y) ↔ x = y) 11a. A(Lois, Professor Michaels) 11b. ∀x(S(x) → A(x, Professor Gross)) 11c. ∀x(F(x) → (A(x, Professor Miller) ∨ A(Professor Miller, x))) 11d. ∃x(S(x) ∧ ∀y(F(y) → ¬A(x, y))) 11e. ∃x(F(x) ∧ ∀y(S(y) → ¬A(y, x))) 11f. ∀y(F(y) → ∃x(S(x) ∨ A(x, y))) 11g. ∃x(F(x) ∧ ∀y((F(y) ∧ (y ≠ x)) → A(x, y))) 11h. ∃x(S(x) ∧

∀y(F(y) → ¬A(y, x))) 13a. ¬M(Chou, Koko) 13b. ¬M(Arlene, Sarah) ∧ ¬T(Arlene, Sarah) 13c. ¬M (Deborah, Jose) 13d. ∀xM(x, Ken) 13e. ∀x¬T(x, Nina) 13f. ∀x(T -x, Avi) ∨ M(x, Avi)) 13g. ∃x∀y(y ≠ x → M(x, y)) 13h. ∃x∀y(y ≠ x → (M(x, y) ∨ T(x, y))) 13i. ∃x∃y(x ≠ y ∧ M(x, y) ∧ M(y, x)) 13j. ∃xM(x, x)13k. ∃x∀y(x ≠ y → (¬M(x, y) ∧ ¬T(y, x))) 13l. ∀x(∃y(x ≠ y ∧ (M(y, x) ∨ T(y, x)))) 13m. ∃x∃y(x ≠ y ∧ M(x, y) ∧ T(y, x))13n. ∃x∃y(x ≠ y ∧ ∀z((z ≠ x ∧ z ≠ y) → (M(x, z) ∨ M(y, z) ∨ T(x, z) ∨ T(y, z)))) 15a. ∀xP(x), where P(x) is “x needs a course in discrete mathematics” and the domain consists of all computer science students 15b. ∃xP(x), where P(x) is “x owns a personal computer” and the domain consists of all students in this class15c. ∀x∃yP(x, y), where P(x, y) is “x has taken y,” the domain for x consists of all students in this class, and the domain for y consists of all computer science classes 15d. ∃x∃yP(x, y), where P(x, y) and domains are the same as in part (c) 15e. ∀x∀yP(x, y), where



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utio

ns

P(x, y) is “x has been in y,” the domain for x consists of all students in this class, and the domain for y consists of all buildings on campus 15f. ∃x∃y∀z(P(z, y) → Q(x, z)), where P(z, y) is “z is in y” and Q(x, z) is “x has been in z”; the domain for x consists of all students in the class, the domain for y consists of all buildings on campus, and the domain of z consists of all rooms. 15g. ∀x∀y∃z(P(z, y) ∧ Q(x, z)), with same environment as in part (f) 17a. ∀u∃m(A(u, m) ∧ ∀n(n ≠ m → ¬A(u, n))), where A(u, m) means that user u has access to mailbox m 17b. ∃p∀e(H(e) ∧ S(p, running)) → S (kernel, working correctly), where H(e) means that error condition e is in effect and S(x, y) means that the status of x is y 17c. ∀u∀s(E(s, .edu) → A(u, s)), where E(s, x) means that website s has extension x, and A(u, s) means that user u can access website s 17d. ∃x∃y(x ≠ y ∧ ∀z((∀s M(z, s)) ↔ (z = x ∨ z = y))), where M(a, b) means that system a monitors remote server b 19a. ∀x∀y((x < 0) ∧ (y < 0) → (x + y < 0)) 19b. ¬∀x∀y ((x > 0) ∧ (y > 0) → (x − y > 0)) 19c. ∀x∀y (x2 + y2 ≥ (x + y ) 2 ) 19d. ∀x∀y

( ⎪xy

⎥ = ⎪x⎥

⎪y

⎥ ) 21. ∀x∃a∃b∃c∃d ((x > 0) → x = a2 + b2 + c2 + d2),

where the domain consists of all integers 23a. ∀x ∀y ((x < 0) ∧ (y < 0) → (xy >0)) 23b. ∀x(x − x = 0) 23c. ∀x∃a∃b(a ≠ b ∧ ∀c(c2 = x ↔ (c = a ∨ c = b))) d) ∀x((x < 0) → ¬∃y(x = y2)) 25a. There is a multiplicative identity for the real numbers. 25b. The product of two negative real numbers is always a positive real number. 25c. There exist real numbers x and y such that x2 exceeds y but x is less than y. 25d. The real numbers are closed under the operation of addition. 27a. True 27b. True 27c. True 27d. True 27e. True 27f. False 27g. False 27h. True 27i. False 29a. P(1,1) ∧ P(1,2) ∧ P(1,3) ∧ P(2,1) ∧ P(2,2) ∧ P(2, 3) ∧ P(3, 1) ∧ P(3, 2) ∧ P(3, 3) 29b. P(1, 1) ∨ P(1, 2) ∨ P(1, 3) ∨ P(2, 1) ∨ P(2, 2) ∨ P(2, 3) ∨ P(3,1) ∨ P(3, 2) ∨ P(3, 3) 29c. (P (1, 1) ∧ P(1, 2) ∧ P(1, 3)) ∨ (P (2 , 1) ∧ P(2, 2) ∧ P(2, 3)) ∨ (P (3, 1) ∧ P(3, 2) ∧ P(3, 3)) 29d. (P (1, 1) ∨ P(2, 1) ∨ P(3, 1)) ∧ (P (1, 2) ∨ P(2, 2) ∨ P(3, 2)) ∧ (P (1, 3) ∨ P(2, 3) ∨

P(3, 3)) 31a. ∃x∀y∃z¬T (x, y, z) 31b. ∃x∀y¬P(x, y) ∧ ∃x∀y ¬ Q(x, y) 31c. ∃x∀y (¬P(x, y) ∨ ∀z¬R(x, y, z)) 31d. ∃x∀y(P(x, y)∧¬Q(x, y)) 33a. ∃x∃y¬P(x, y) 33b. ∃y∀x¬P(x, y) 33c. ∃y∃x(¬P(x, y) ∧ ¬Q(x, y)) 33d. (∀x∀yP(x, y)) ∨ (∃x∃y¬Q(x, y)) 33e. ∃x(∀y∃z¬P(x,y,z) ∨ ∀z∃y¬P(x, y, z)) 35. Any domain with four or more members makes the statement true; any domain with three or fewer members makes the statement false. 37a. There is someone in this class such that for every two different math courses, these are not the two and only two math courses this person has taken. 37b. Every person has either visited Libya or has not visited a country other than Libya. 37c. Someone has climbed every mountain in the Himalayas. 37d. There is someone who has neither been in a movie with Kevin Bacon nor has been in a movie with someone who has been in a movie with Kevin Bacon. 39a. x = 2, y = −2 39b. x = −4 39c. x = 17, y = −1 41. ∀x ∀y ∀z((x · y) · z = x · (y · z)) 43. ∀m∀b(m = 0 → ∃x(mx + b = 0 ∧ ∀w(mw + b = 0 → w = x))) 45a. True 45b. False 45c. True 47. ¬(∃x∀yP(x, y)) ↔ ∀x(¬∀yP(x, y)) ↔ ∀x∃y¬P(x,y) 49a. Suppose that ∀xP(x) ∧ ∃xQ(x) is true. Then P(x) is true for all x and there is an element y for which Q(y) is true. Because P(x) ∧

Q(y) is true for all x and there is a y for which Q(y) is true, ∀x∃y(P(x) ∧ Q(y)) is true. Conversely, suppose that the second proposition is true. Let x be an element in the domain. There is a y such that Q(y) is true, so ∃xQ(x) is true. Because ∀xP(x) is also true, it follows that the first proposition is true. 49b. Suppose that ∀xP(x) ∨ ∃xQ(x) is true. Then either P(x) is true for all x, or there exists a y for which Q(y) is true. In the former case, P(x) ∨ Q(y) is true for all x, so ∀x∃y(P(x) ∨ Q(y)) is true. In the latter case, Q(y) is true for a particular y, so P(x) ∨ Q(y) is true for all x and consequently ∀x∃y(P(x) ∨ Q(y)) is true. Conversely, suppose that the second proposition is true. If P(x) is true for all x, then the first proposition is true. If not, P(x) is false for some x, and for this x

there must be a y such that P(x) ∨ Q(y) is true. Hence, Q(y) must be true, so ∃yQ(y) is true. It follows that the first proposition must hold. 51. We will show how an expression can be put into prenex normal form (PNF) if subexpressions in it can be put into PNF. Then, working from the inside out, any expression can be put in PNF. (To formalize the argument, it is necessary to use the method of structural induction that will be discussed in Section 5.3.) By Exercise 45 of Section 1.4, we can assume that the proposition uses only ∨ and ¬ as logical connectives. Now note that any proposition with no quantifiers is already in PNF. (This is the basis case of the argument.) Now suppose that the proposition is of the form QxP(x), where Q is a quantifier. Because P(x) is a shorter expression than the original proposition, we can put it into PNF. Then Qx followed by this PNF is again in PNF and is equivalent to the original proposition. Next, suppose that the proposition is of the form ¬P. If P is already in PNF, we slide the negation sign past all the quantifiers using the equivalences in Table 2 in Section 1.4. Finally, assume that proposition is of the form P ∨ Q, where each of P and Q is in PNF. If only one of P and Q has quantifiers, then we can use Exercise 46 in Section 1.4 to bring the quantifier in front of both. If both P and Q have quantifiers, we can use Exercise 45 in Section 1.4, Exercise 48, or part (b) of Exercise 49 to rewrite P ∨ Q with two quantifiers preceding the disjunction of a proposition of the form R ∨ S, and then put R ∨ S into PNF.

Conic Sections and Parametric EquationsCHAPTER 7CHCHAPAPTETERR 77

Chapter 7 Get Ready

1. x = 1; (0, -12); (1, -13) 3. x = -1; (0, -8); (-1, -10) 5. x = 2; (0, -4); (2, -16) 7. x = 25; (0, 550); (25, 543.75) 9. 108 11. 100 13. -172 15. vertical asymptote at x = -5 17. vertical asymptotes at x = -5 and x = 1; horizontal asymptote at y = 0 19. vertical asymptote at x = 4.

Lesson 7-1

1. vertex: (3, 7);

x

8

4

12

−4

y

4 8

16(x - 3)

2= 12(y - 7)focus: (3, 10);

directrix: y = 4; axis of symmetry: x = 3

3. vertex: (-2, 4);

x

4

-4

8

−8 −4

y

4

12

(y - 4)2

= 20(x + 2)

focus: (3, 4); directrix: x = -7; axis of symmetry: y = 4



Selected A

nswers and S

olutions


5. vertex: (-8, 3);

x

4

8

12

16

−4−12

y

(x + 8)2

= 8(y - 3)

focus: (-8, 5); directrix: y = 1; axis of symmetry: x = -8

7. vertex: (1, -5);

x

4

-8

-12

−8 4 8

y

(y + 5)2

= 24(x - 1)

focus: (7, -5); directrix: x = -5; axis of symmetry: y = -5

9. vertex: (-8, -2);

x−4y

-4(y + 2) = (x + 8)2

-2

-4

-6

-8

focus: (-8, -3); directrix: y = -1; axis of symmetry: x = -8

11. Sample answer: The focus is 4 feet above the ground.

13 a. y 2 - 180x + 10y + 565 = 0

y 2 + 10y = 180x - 565

y 2 + 10y + 25 = 180x - 565 + 25

y 2 + 10y + 25 = 180x - 540

(y + 5) 2 = 180(x - 3)

b. The equation in standard form has y as the squared term, which means that the parabola opens horizontally. Because 4p = 180, p = 45 and the graph opens to the right. The equation is in the form (y - k) 2 = 4p(x - h), so h = 3 and k = -5. Since the stern is located at the vertex of the parabola formed, it is at the point (h, k) or (3, -5). The swimmer is at the focus, located at (h + p, k), which is (3 + 45, -5) or (48, -5). The distance the swimmer is from the stern of the boat represents the length of rope needed to attach the swimmer to the stern. Using the distance formula, the distance between these two points is

√ ��

(48 - 3) 2 + (-5 − (-5)) 2 or 45. Therefore, the length of rope attaching the swimmer to the stern of the boat is 45 feet.

15. x 2 = 8(y + 7);

x

-8

-4

−4−8 4 8

x 2- 17 = 8y + 39

yvertex: (0, -7); focus: (0, -5); directrix: y = -9; axis of symmetry: x = 0

17. x 2 = -24(y + 1);

-20

-30

-10

10

−20 10 20

y

x

3x 2+ 72 = -72y

vertex: (0, -1); focus: (0, -7); directrix: y = 5; axis of symmetry: x = 0

19. y 2 = 20(x – 3);

x

-4

4

−4−8 4 8

y60x - 80 = 3y 2

+ 100vertex: (3, 0); focus: (8, 0); directrix: x = -2; axis of symmetry: y = 0

21. (y - 4) 2 = 10(x - 2);

x

8

12

−8 −4 8

y-72 = 2y 2

- 16y - 20xvertex: (2, 4); focus: (4.5, 4); directrix: x = -0.5; axis of symmetry: y = 4

23. (x + 6) 2 = 18(y + 9);

x

-20

10

−10 10

yx 2- 18y + 12x = 126

vertex: (-6, -9); focus: (-6, -4.5); directrix: y = -13.5; axis of symmetry: x = -6

25. 9 in. 27. (y + 1) 2 = 24(x + 4)

xO

-2

-4

−4−8−12

y

2

4

(y + 1)2

= 24(x + 4)

29. (x + 3) 2 = 8(y - 2) 31. (y - 4) 2 = -32(x - 7)

x

-4

−8

y

4

8

4(x + 3)

2= 8(y - 2)

xO

-10

−10−20

y

10

20

10 20

(y - 4)2 = -32(x - 7)



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns33. (x - 1) 2 = -12(y - 6) 35. (x - 8) 2 = 16(y + 7)

xO

-4

−4

y

4

8

12

4 8

(x - 1)2 = -12(y - 6)

xO

y4

4 8 12

(x - 8)2

= 16(y + 7)

37. (x - 1) 2 = -12(y - 5) 39. (x + 4) 2 = -4(y - 1)

−8

−4

−4

8

4

4 8

y

x

(x - 1)2= -12(y - 5)

y

−4

−8

−8−12

8

4

4 x

(x + 4)2= -4(y - 1)

41. (y + 9) 2 = 4(x + 6) 43. (x + 5) 2 = 16(y + 6)y

O

−4

−8

−12

−16

−4−8 4 8 x

(y + 9)2= 4(x + 6)

y

xO−8

4

−4

−8

−12

(x + 5)2= 16(y + 6)

45. y = -8x - 45 47. y = 4x + 14

49 The graph opens vertically. Determine the vertex and focus. -0.25 (x - 6) 2 = y - 9

(x - 6) 2 = -4(y - 9) Because 4p = -4, p = -1. The vertex is (6, 9) and the focus is

(6, 8). We need to determine d, the distance between the focus and the point of tangency, C. This is one leg of the isosceles triangle.

d = √ ��

( x 2 − x 1 ) 2 + ( y 2 - y 1 ) 2

= √ ��

(10 - 6) 2 + (5 - 8) 2

= 5 Use d to find A, the endpoint of the other leg of the isosceles

triangle. Since p is negative, the parabola opens down and A will be above the focus.

A = (6, 8 + 5) or (6, 13) Points A and C both lie on the line tangent to the parabola. To

find an equation for this line, first use two points to find the slope m.

m = 5 - 13 _

10 − 6 or -2

Then use a point on the line and the slope to write an equation for the line.

y - y 1 = m(x - x 1 ) Point-slope formula

y - 5 = -2(x - 10) ( x 1 , y 1 ) = (10, 5) and m = -2

y - 5 = -2x + 20 Simplify.

y = -2x + 25 Solve for y.

51. opens downward 53. opens upward 55. (x - 3) 2 = -4(y - 5) 57. (y - 1) 2 = -16(x + 5) 59a. Sample answer:

x 2 = -180.27(y + 20) 59b. about 30.35 m 61. (y − 4) 2 = −4(x + 6) 63. (y + 5) 2 = 8(x + 3)

y

x−4−8

4

8

12

−4

(−10, 8)

(−6, 4)

y

−12

6

6 12 18 x(−1, −1)

(5, 3) (15, 7)

65. (2, 8) 67. (0, -1) 69a. y 2 = 2 _ 3 x 69b. 1.53 ft

71a. (y - 2) 2 = 8(x + 3) or (y - 2) 2 = -8(x + 3)

71b. y

x

2p

pX F W

D

p

Sample answer: To prove that the endpoints of the latus rectum X and W and the point of intersection of the axis and directrix D are the vertices of a right isosceles triangle �XDW, we need to show that ∠XDW is a right angle and that

−−−

XD � −−−

WD . Since −−

XF � −−−

FW , −−

FD � −−

FD , and ∠XFD � ∠WFD, �XFD � �FWD by SAS. Thus, −−−

XD � −−−

WD . To prove that ∠XDW is a right angle, we can first prove

that ∠XDF and ∠FDW are both 45° angles. Since −−

XF � −−

FD and ∠XFD is a right angle, �XFD is an isosceles right triangle. Therefore, ∠XDF is 45°. This also means that ∠FDW is 45°. Likewise, ∠XDW is a right angle. Thus, �XDW is an isosceles right triangle. 73. Sample answer: x 2 = 4(y - 2) 75. Sample answer: y 2 = 12x 77a. i. 1 unit; ii. 2 units; iii. 4 units

77b.

(3, 0) (6, 0)

(4, 0) x

y 77c. As the focus is moved farther away from the vertex, the parabolas become wider.77d. Sample answer: (x + 1) 2 = 4(y + 7)

77e. Sample answer: The parabolas all have a vertex of (0, -1) and open downward. The first equation produces the narrowest parabola and the second equation produces the widest parabola.

79 The width of the sector is 3 units, so y = 1.5. A = 4 _ 3 xy and the

area of the sector is 2.4 square units. Substitute the values into the equation and solve for x.

2.4 = 4 _ 3 x(1.5)

2.4 = 2x1.2 = xTherefore, the vertex of the parabola is (0, 0), and two other points on the parabola are (1.2, 1.5) and (1.2, -1.5).Set up and solve a system of equations using y as the



Selected A

nswers and S

olutions


independent variable.a y 2 + by + c = x

a (0) 2 + b(0) + c = 0 a (1.5) 2 + b(1.5) + c = 1.2 a (-1.5) 2 + b(-1.5) + c = 1.2 c = 0 2.25a + 1.5b = 1.2 2.25a - 1.5b = 1.2 4.5a = 2.4

a = 8 _ 15

2.25 · 8 _ 15

+ 1.5b = 1.2

1.2 + 1.5b = 1.2 1.5b = 0 b = 0 The equation of the parabola is 8 _

15 y 2 = x or y 2 = 15

_ 8 x.

81. Quadrants I and IV; the vertex is (-2, 5) and p = -2. Since the vertex is to the left of the y-axis, and the parabola opens to the left, no points will be to the right of the y-axis, or in Quadrants I and IV.

85. max at (7, 8.5) = 81.5, min at (2, 0) = 16 87. -1 _

y + 2 + 3

_ y + 1

89.

cot θ = 7 _ 6 , csc θ =

√ � 85 _

6

91.

2

4

6

x

y

y = tan x + 4

π

4π

43π

43π

4- -

93.

-4

2

4

x

y

-3_4

y = csc x

3π

2π

23π

2-

π

2-

95. 2x

97. y

xO

−4

−8

8

4

= x 3- x 2

- 4x + 4f (x)

99. y

x

= x 4- 4x 2

+ 2h (x)

101. G 103. G

Lesson 7-2

1. y

x

−4

−8

−4−8

8

4

4 8

y 2_49

+ (x + 2)2_

9= 1

3. 5.

y

x

−3

−6

−9

4 8

x 2+ 9y 2

- 14x + 36y + 49 = 0

y

x

9x 2+ y 2

+ 126x + 2y + 433 = 0

7. (x - 3) 2

_ 100

+ (y + 3) 2

_ 36

= 1 9. (x - 2) 2

_ 25

+ (y - 2) 2

_ 9 = 1

11. (x + 6) 2

_ 64

+ (y - 3) 2

_ 100

= 1 13. (x − 2) 2

_ 225

+ (y - 8) 2

_ 81

= 1

15. 0.837 17. 0.426 19. 0.511 21. 0.447 23a. 28.8 in.

23b. x 2 _

324 +

y 2 _

207 = 1 25.

(x - 1) 2 _

8 +

(y + 3) 2 _

4 = 1; ellipse

27. (y + 9) 2 = 12(x - 6); parabola 29. (x - 7) 2

_ 3 +

(y + 2) 2 _

9 = 1;

ellipse

31 2 x 2 + 7 y 2 + 24x + 84y + 310 = 0

2 x 2 + 24x + 7 y 2 + 84y + 310 = 0

2 ( x 2 + 12x + 36) + 7 ( y 2 + 12y + 36) + 310 = 0 + 72 + 252

2 (x + 6) 2 + 7 (y + 6) 2 = 14

(x + 6) 2

_ 7 +

(y + 6) 2 _

2 = 1

The related conic is an ellipse because a ≠ b and the graph is of the

form (x − h) 2

_ a 2

+ (y - k) 2

_ b 2

= 1.

33. (x - 3) 2 + y 2 = 4 35. (x + 4) 2 + (y + 3) 2 = 36

y

x

(3, 0)

y

x

−12

−8−12 4

(−4, −3)

4

37. y

x

P(x, y)V 2(b, 0)

(0, 0)

F 1(0, c)

F 2(0, -c)

V 1(-b, 0)

V 3(0, a)

V 4(0, -a)

PF 1 + PF 2 = 2a

√ ��

(x - 0) 2 + (y - c) 2 + √ ��

(x - 0) 2 + (y - (-c)) 2 = 2a

√ ��

x 2 + (y - c) 2 + √ ��

x 2 + (y + c) 2 = 2a

√ ��

x 2 + (y - c) 2 = 2a - √ ��

x 2 + (y + c) 2



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

x 2 + y 2 - 2cy + c 2 = 4 a 2 − 4a √ ��

x 2 + (y + c) 2 +

x 2 + y 2 + 2cy + c 2

4a √ ��

x 2 + (y + c) 2 = 4 a 2 + 4cy

a √ ��

x 2 + (y + c) 2 = a 2 + cy

a 2 ( x 2 + y 2 + 2cy + c 2 ) = a 4 + 2 a 2 cy + c 2 y 2

a 2 x 2 + a 2 y 2 + 2 a 2 cy + a 2 c 2 = a 4 + 2 a 2 cy + c 2 y 2

a 2 y 2 - c 2 y 2 + a 2 x 2 = a 4 - a 2 c 2

y 2 ( a 2 - c 2 ) + a 2 x 2 = a 2 ( a 2 - c 2 )

y 2 b 2 + a 2 x 2 = a 2 b 2

y 2

_ a 2

+ x 2 _

b 2 = 1

39. (x - 1) 2

_ 16

+ (y - 3) 2

_ 1 = 1 41.

(x - 2) 2 _

16 +

(y - 2) 2 _

49 = 1

43 a. The length of the major axis is 43.4 + 28.6 + 0.87 or 72.87.

The value of a is 72.87 _

2 or 36.435. The distance from the focus

(the sun) to the vertex is 28.6 + 0.87 _

2 or 29.035. Therefore, the

value of c, the focus to the center, is 36.435 - 29.035 or 7.4. In an ellipse,

c 2 = a 2 - b 2 . The value of b is √ ��

a 2 - c 2 =

√ ��

36.435 2 - 7.4 2 ≈ 35.676. The value of 2b, the length of the minor axis, is about 35.676 · 2 or about 71.35 million mi.

b. The eccentricity e is equal to c _ a . e = 7.4 _

36.435 ≈ 0.203

45. center: (0, -6); foci: (±5 √ � 3 , -6) ; vertices: (±10, -6)

47. center: (-1, 0); foci: (-1, ±7); vertices: (-1, ± √ � 65 )

49. x 2 _

100 +

y 2 _

64 = 1 51.

(y + 4) 2 _

36 +

(x - 2) 2 _

16 = 1

53a. x 2 + y 2 = 64, x 2 + y 2 = 78453b. y

x

−15

−30

−15

30

15

15

53c. 5 days

55. (7, -1), (3, -5) 57. (-2, 3), (6, -5) 59. x 2 _

16 +

y 2 _

12 = 1

61. (x - 6.5) 2 + (y - 4.5) 2 = 32.5 63. (x - 1) 2 + (y + 7) 2 = 16; center: (1, -7), radius: 4 65. (x + 1) 2 + (y - 4) 2 = 64; center: (-1, 4), radius: 8 67. Sample answer: No; if a 2 = p and b 2 = p + r, then c = √

�� -r . If a 2 = p + r and b 2 = p, then

c = √ � r and the foci are (0, ± √

� r ).

69 The area of the ellipse A = 24π. Substitute into the area formula to obtain an equation with a and b. 24π = πab 24 = abSubstitute a = b + 5 into the equation and solve for b. 24 = (b + 5)b 24 = b 2 + 5b

0 = b 2 + 5b - 24 0 = (b + 8)(b - 3)Since b cannot be negative, b = 3 and a = 3 + 5 or 8.Therefore, an equation for the ellipse is

x 2 _

8 2 +

y 2 _

3 2 = 1 or x 2

_ 64

+ y 2

_ 9 = 1.

71. Yes; sample answer: If (x, y) is a point on the ellipse, then (-x, -y) must also be on the ellipse.

x 2

_ a 2

+ y 2

_ b 2

= 1

(-x) 2

_ a 2

+ (-y) 2

_ b 2

= 1

x 2

_ a 2

+ y 2

_ b 2

= 1

Thus, (-x, -y) is also a point on the ellipse and the ellipse is symmetric with respect to the origin.

73. Sample answer: When a is much greater than c, c _ a is close to 0. Since e = c _ a , the value of e is close to zero and the foci are near the center of the ellipse. So, the ellipse is nearly circular.

75. vertex: ( 5 _ 4 , -

55 _

8 ) ; focus: ( 5 _

4 , -7) ; directrix: y = -

27 _

4 ;

axis of symmetry: x = 5 _ 4

yx

−4

−8

−12

−16

−2 2 4

y = -2x 2+ 5x - 10

77. 160 My Real Babies and 320 My First Babies

79. sin (θ + π

_ 3 ) - cos

(θ + π

_ 6 )

= sin θ cos π

_

3 + cos θ sin π

_

3 - cos θ cos π

_

6 + sin θ sin π

_

6

= 1 _ 2 sin θ +

√ � 3 _

2 cos θ -

√ � 3 _

2 cos θ + 1 _

2 sin θ

= 1 _ 2 sin θ + 1 _

2 sin θ

= sin θ

81. π

_

4 , 5π

_

4 83. 3π

_

2 , 7π

_

6 , 11π

_

6 85. [-7, 5] 87. 4 real zeros and

3 turning points; 0, -4, and -2 89. 5 real zeros and 4 turning points; 0, 5, and -2 91. 3 + 4i 93. B 95. C

Lesson 7-3

1. y

−8

−8

8

4

8 x

(−5, 0) (5, 0)(−4, 0)(4, 0)

x 2_16

- y2_

9= 1

3.

−8

y

−4

−4

8

4

4 x−8

(−7, 0) (7, 0)(−8.89, 0) (8.89, 0)

x 2_49

- y2_

30= 1



Selected A

nswers and S

olutions


5. y

−8

−8

8

8 x

(−5.48, 0) (5.48, 0)(3, 0)

(−3, 0) y 2_9

- x2_

21= 1

7. y

−8

16

8 16 x−16

(0, 9)

(0, −9.43)

(0, 9.43)

(0, −9) y 2_81

- x2_

8= 1

9. y

x

(−2, 0)(−3.16, 0)

(2, 0)

(3.16, 0)

3x 2- 2y 2

= 12

11. y

−20

−10−20

20

10

10 20xy 2_

225- x

2_81

= 1

13. y

x

8

4

−4−8

16

12

4 8

(0, 13.1)

(0, 0.92)

(y - 7)2_4

+x 2_

33 = 1

15. y

x

−16

−8

16

16 24(13.12, 1)

(-3.12, 1)

(5, 1)

(x - 5 )2_49

-

(y - 1) 2_17

= 1

17. y

xO

−16

−24

8

8(−17.05, −5) (5.05, −5)(−6, −5)

(x + 6 ) 2 _ 64

- (y + 5) 2

_ 58

= 1

19. y

x

−8

−4−8

8

4

4

(−2, 5)

(−2, -7)

- + 3 - 4x + 6y = 28x 2 y 221. y

x

−4

−12

8

4

(-7, 7.92)

(-7, -3.92)

-5 x 2 + 2 y 2 - 70x - 8y = 287

23. (y - 1) 2

_ 15

- (x + 1) 2

_ 49

= 1 25. (x - 3) 2

_ 27

- (y + 1) 2

_ 9 = 1

27. (y + 8) 2

_ 16

- (x + 3) 2

_ 33

= 1 29. (x + 7) 2

_ 25

- (y - 2) 2

_ 49

= 1

31 The vertices are 4 units apart, so 2a = 4, a = 2, and a 2 = 4. The center is the midpoint of the vertices, or (-2, -3). Therefore, h = -2 and k = -3. The conjugate axis is of length 12, so 2b = 12, b = 6, and b 2 = 36. The x-coordinates of the vertices are different, so the transverse axis is horizontal and the a 2 -term goes with the x 2 -term. The equation for the hyperbola is

(x + 2) 2

_ 4 -

(y + 3) 2 _

36 = 1.

33a. (y - 4) 2

_ 9 -

(x - 5) 2 _

225 _

7 = 1 33b. 90 ft 35. 1.27 37. 1.33

39. 1.68 41. 1.32 43. 2.83 45. hyperbola 47. parabola 49. ellipse 51. hyperbola 53. circle

55 a. The common difference is 18 kilometers and the absolute value of the difference of the distances from any point on a hyperbola to the foci is 2a, so 2a = 18, a = 9, and a 2 = 81. The two airports are the foci of the hyperbola and are 72 kilometers apart, so 2c = 72, c = 36, and c 2 = 1296. c 2 = a 2 + b 2 , so b 2 = 1296 - 81 or 1215. Airport B is due south of airport A, so the transverse axis is vertical and a 2 -term goes with the y 2 -term. The equation for the hyperbola is

y 2

_ 81

- x 2 _

1215 = 1.

b. y

x

−20

−40

−20−40

20

40

20 40

(0, 36) (Airport A)

(Airport B)(0, −36)

The plane is on the top branch because it is closer to airport A.

c. Because the highway is the transverse axis and the plane is 40 kilometers from the highway, x = 40. Substitute 40 for x in the equation.

y 2

_ 81

- 40 2 _

1215 = 1

y 2

_ 81

- 1.317 = 1

y 2

_ 81

= 2.317

y 2 = 187.67 y ≈ 13.7 The coordinates of the plane are (40, 13.7).

57. y

x

F 1 (0, c)

P(x, y)

F 2 (0, -c)

| PF 1 - PF 2 | = 2a

⎪ √ ��

(x - 0) 2 + (y − c) 2 - √ ��

(x - 0) 2 + [y - (-c)] 2 ⎥ = 2a

√ ��

x 2 + (y - c) 2 - √ ��

x 2 + (y + c) 2 = 2a

√ ��

x 2 + (y - c) 2 = 2a + √ ��

x 2 + (y + c) 2

x 2 + (y - c) 2 = 4 a 2 + 4a √ ��

x 2 + (y + c) 2 + x 2 + (y + c) 2



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

-4cy - 4 a 2 = 4a √ ��

x 2 + (y + c) 2

-cy - a 2 = a √ ��

x 2 + (y + c) 2

c 2 y 2 + 2 a 2 cy + a 4 = a 2 x 2 + 2 a 2 cy + a 2 c 2 + a 2 y 2

c 2 y 2 - a 2 y 2 - a 2 x 2 = a 2 c 2 - a 4

( c 2 - a 2 ) y 2 - a 2 x 2 = a 2 ( c 2 - a 2 )

y 2

_ a 2

- x 2 _

c 2 - a 2 = 1

y 2

_ a 2

- x 2 _

b 2 = 1

By the Pythagorean Theorem, c 2 - a 2 = b 2 .

59. (-1.3, -5.7), (7, -1.5) 61. (-5, -10), (6.9, 13.7)

63. (0, 6), (0, -6) 65a. x 2 _

550 2 -

y 2 _

7901 2 = 1 65b. on the right

branch 67. x 2 _

25 -

y 2 _

9 = 1 69. (641.2, 3178.3), (1212.8, 640.6),

(331.6, 4454.4), (2351.0, 100.8) 71. (x + 4) 2

_ 25

- (y - 3) 2

_ 49

= 1

73 The foci are 14 units apart, so c = 7 and c 2 = 49.The center is the midpoint of the foci, (6, -2), so h = 6 and k =

-2. The eccentricity c _ a is 7 _ 6 , so a = 6 and a 2 = 36.

b 2 = c 2 - a 2 = 49 - 36 or 13. The x-coordinates of the vertices are different, so the transverse axis is horizontal and the a 2 -term goes with the x 2 -term. The equation for the hyperbola is

(x - 6) 2

_ 36

- (y + 2) 2

_ 13

= 1.

75. 2 x 2 _

121 -

2 y 2 _

121 = 1 77. Sample answer: x 2

_ 16

- y 2

_ 48

= 1

79. Sample answer: If the transverse is horizontal, then a is a horizontal distance and b is a vertical distance. So, the slopes of

the asymptotes are ±

b _ a . If the transverse is vertical, then a is a

vertical distance and b is a horizontal distance. So, the slopes of the asymptotes are ±

a _ b .

81. y 2

_ 9 - x 2

_ 72

= 1 83. Sample answer: First, determine whether the orientation of the hyperbola is vertical or horizontal. Then use the foci to locate the center of the hyperbola and determine the values of h and k. Use the transverse axis length to find a 2 . Find c, the distance from the center to a focus. Use the equation b 2 = c 2 - a 2 to find b 2 . Finally, use the correct standard form to write the equation, depending on whether the transverse axis is parallel to the x-axis or to the y-axis.

85. y

x−4−8

−4

−8

4 8

x 2_64

+

(y + 5) 2_49

= 1

87a. 105 ft87b. 5 seconds

89.

⎡

⎢

⎣

1

6

3

-7

5

-4

8

-2

9

⎤

�

⎦

·

⎡

⎢

⎣

x 1

x 2

x 3

⎤

�

⎦

=

⎡

⎢

⎣

-3

2

26

⎤

�

⎦

; (-5, 10, 9) 91. 2nπ; n ∈ �

93. no solution 95. sin θ = 24 _

25 , cos θ = 7 _

25 , tan θ = 24

_ 7 ,

csc θ = 25 _

24 , sec θ = 25

_ 7 , cot θ = 7 _

24 97. -

3 _

2 , -1 (multiplicity: 2),

4 + 9i, 4 - 9i; (2x + 3) (x + 1) 2 (x - 4 + 9i)(x - 4 - 9i) 99. H 101. G

Lesson 7-4

1. hyperbola; (x′) 2 + 2 √ � 3 x′y′ - (y′) 2 + 18 = 0 3. parabola; (y′) 2 -

8x′ = 0 5. parabola; (x′) 2 + 2 √ � 3 x′y′ + 3(y′) 2 +

16 √ � 3 x′ - 16y′ = 0 7. ellipse; 2 (x′) 2 + 2 (y′) 2 - 5 √ � 2 x +

5 √ � 2 y - 6 = 0 9. hyperbola; 25 (x′) 2 - 4(y′) 2 + 64 = 0

11. y′ 2

_ 8 - x′ 2

_ 8 = 1 13. x′ 2

_ 4 +

y′ 2 _

9 = 1 15.

(y′ - 1) 2 _

1 _ 2 -

(x′ - 1) _

1 = 1

17. (x′) 2

_ 1 _ 6 -

(y′ ) 2 _

2 = 1 19. x 2 - xy + y 2 - 4 = 0

21. x 2 + 10 √ � 3 xy + 11 y 2 - 144 = 0 23. 13 x 2 + 6 √ � 3 xy +

7 y 2 - 112 = 0 25. 3 x 2 - 10xy + 3 y 2 + 128 = 0

27. 23 x 2 + 26 √ � 3 xy - 3 y 2 - 144 = 0

29 a. Find the equations for x and y for θ = 30°.x′ = x cos θ + y sin θ

= √ � 3

_ 2 x + 1 _

2 y

y′ = y cos θ - x sin θ

= √ � 3

_ 2 y - 1 _

2 x

Substitute into 144 (x′) 2 + 64 (y′) 2 = 576

144 ( √ � 3

_ 2 x + 1 _

2 y)

2

+ 64 ( √ � 3

_ 2 y - 1 _

2 x)

2

= 576

144 ( 3 _ 4 x 2 + 1 _

4 y 2 +

√ � 3 _

2 xy) + 64 ( 3 _

4 y 2 + 1 _

4 x 2 -

√ � 3 _

2 xy) = 576

108 x 2 + 36 y 2 + 72 √ � 3 xy + 48 y 2 + 16 x 2 - 32 √ � 3 xy = 576

124 x 2 + 40 √ � 3 xy + 84 y 2 - 576 = 0

31 x 2 + 10 √ � 3 xy + 21 y 2 - 144 = 0

b. Graph the equation by solving for y.

21 y 2 + (10 √ � 3 x) y + (31 x 2 - 144) = 0Use the quadratic formula.

y = -10 √ � 3 x ± √

��

(10 √ � 3 x) 2 - 4(21)(31 x 2 - 144) ____

2(21)

y = -10 √ � 3 x ± √

�� 12096 - 2304 x 2 ___

42

Graph the conic.

y

x

31x 2- 10 √3 xy + 21y 2

= 144

31.

x

−4

−8

−4−8

8

4

84

(x')2_25

- (y')2_36

= 1

y



Selected A

nswers and S

olutions


33. y

x

+x')) 2 )) 2 6 y' = 248

35. y

−2−4

4

6

8

2

2 4 x

y = 3(x')2- 2x' + 5

37.

[-6.61, 14.6] scl: 1by [-2, 12] scl: 1

x 2- 2xy + y 2

- 5x - 5y = 0 39.

[-7.58, 7.58] scl: 1by [-5, 5] scl: 1

8x 2+ 5xy - 4y 2

= -2

41.

[-10.58, 4.58] scl: 1 by[-2, 8] scl: 1

2x 2+ 4xy + 2y 2

+ 2 √2 x - 2 √2 y = -12

43.

[-10, 10] scl: 1 by [-10, 10] scl: 1

x 2+ 10 √3 xy + 11y 2

- 64 = 0

45.

[-10, 10] scl: 1 by [-10, 10] scl: 1

x 2- 2 √3 xy - y 2

+ 18 = 0 47. intersecting lines y = 4x and y = -4x 49. point at (0, 0) 51. b 53. c

55a. A x 2 + Bxy + C y 2 + Dx + Ey + F = 0 is equivalent to A′ (x′) 2 + B′x′y′ + C′ (y′) 2 + D′x′ + E′y′ + F′ = 0 by rotating the conic

section through θ such that cot 2θ = A - C _ B

. F is unaffected by this rotation, so F = F′.

55b. A x 2 + Bxy + C y 2 + Dx + Ey + F = A′ (x′) 2 + B′x′y′ + C′ (y′) 2 + D′x′ + E′y′ + F′; by subtracting out the xy, x, y, and F terms, the remaining statement is true. A x 2 + C y 2 = A′ (x′) 2 + C′

(y′) 2 ; A must equal A′ and B must equal B′ in order for the statement to be true. 55c. A x 2 + Bxy + C y 2 + Dx + Ey + F = A′ (x′) 2 + B′x′y′ + C′(y′) 2 + D′x′ + E′y′ + F′; by subtracting out the x, y, and F terms, the remaining statement is true. A x 2 + Bxy + C y 2 = A′ (x′) 2 + B′x′y′ + C ′(y′) 2 ; A must equal A′, B must equal B′, and C must equal C′ in order for the statement to be true. Therefore, B 2 - 4AC = (B′) 2 - 4(A′C′);

A (x cos θ + y sin θ) 2 + B(x cos θ + y sin θ)(y cos θ - x sin θ).

57 For each equation, solve for y using the quadratic formula. Then graph using a graphing calculator.

9 x 2 + 4xy + 5 y 2 - 40 = 0

5 y 2 + 4x(y) + (9 x 2 - 40) = 0

y = -4x ± √

�� (4x) 2 - 4(5)( 9x 2 - 40) ___

2(5)

= -4x ± √

�� 800 - 164 x 2 __

10

x 2 - xy - 2 y 2 - x - y + 2 = 0 -2 y 2 + (-x - 1)y + ( x 2 - x + 2) = 0

y = x + 1 ± √

�� (-x - 1) 2 - 4(-2)( x 2 - x + 2) ____

2(-2)

= x + 1 ± √

�� 9 x 2 - 6x + 17 __

-4

[-10, 10] scl: 1 by [-10, 10] scl: 1

The graphs intersect at four points: (-1.9, 2.2), (-1.5, -1.5), (1.9, -2.2), (1.9, 0.8).

59a.Equation Graph

Minimum Angle of

Rotation

x 2 - 5x + 3 - y = 0 parabola 360°

6 x 2 + 10 y 2 = 15 ellipse 180°

2xy = 9 hyperbola 180°

59b. Sample answer: A parabola has 1 line of symmetry and the minimum angle of rotation is a complete circle. An ellipse and a hyperbola have 2 lines of symmetry and the minimum angle of rotation is a half circle. 59c. 130° 61. Let x = x′ cos θ + y′ sin θ and y = -x′ sin θ + y′ cos θ.

r 2 = x 2 + y 2 = (x ′ cos θ + y ′ sin θ) 2 + (-x ′ sin θ + y′ cos θ) 2 = (x ′) 2 cos 2 θ + 2x ′y ′cos θ sin θ + (y ′) 2 sin 2 θ + (x ′) sin 2 θ

- 2x ′y ′cos θ sin θ + (y ′) 2 cos 2 θ = [ (x ′) 2 + (y ′) 2 ] cos 2 θ + [ (x ′) 2 + (y ′) 2 ] sin 2 θ = [ (x ′) 2 + (y ′) 2 ]( cos 2 θ + sin 2 θ) = [ (x ′) 2 + (y ′) 2 ](1) = (x ′) 2 + (y ′) 2

63. cos θ (x = x ′ cos θ - y ′ sin θ) sin θ (y = x ′ sin θ + y ′ cos θ) x cos θ = x ′ cos 2 θ - y ′ sin θ cos θ _______________________________ + y sin θ = x ′ sin 2 θ + y ′ sin θ cos θ

x cos θ + y sin θ = x ′ cos 2 θ + x ′ sin 2 θ x cos θ + y sin θ = x ′ ( cos 2 θ + sin 2 θ) x cos θ + y sin θ = x ′

sin θ (x = x ′ cos θ − y ′ sin θ) cos θ (y = x ′ sin θ + y ′ cos θ) x sin θ = x ′ cos θ sin θ − y ′ sin 2 θ ________________________________ − y cos θ = x′ cos θ sin θ + y ′ cos 2 θ

x sin θ − y cos θ = − y ′ sin 2 θ − y ′ cos 2 θ x sin θ − y cos θ = −y ′ ( sin 2 θ + cos 2 θ) y cos θ − x sin θ = y′



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns65. Sample answer: The discriminant is defined as B 2 - 4AC, or in this instance, (B′ ) 2 − 4A′C′. Since a conic that is rotated has no B′ term, the discriminant reduces to −4A′C′. Thus, only the A′ and C′ terms determine the type of conic. Therefore, −4A′C′ < 0 would be an ellipse or a circle, −4A′C′ = 0 would be a parabola, and -4A′C′ > 0 would be a hyperbola. For a circle or an ellipse, A′ and C′ need to share the same sign. For a parabola, either A′ or C′ has to be equal to 0. For a hyperbola, A′ and C′ need to have opposite signs.

67. y

−4

−8

−4−8

8

4

4 8 x

x 2_9

- y2_

64= 1

69. y

−16

−8

16

8

8x

(x - 3) 2_64

- (y - 7) 2_

25= 1

71. 0.447 73a. s + d = 5000, 0.035s + 0.05d = 227.50 73b. savings account: $1500; certificate of deposit: $3500 75. 2 77. yes, yes; (x - 4)(x - 2)(x + 2)(x + 3) 79. D 81. A

Lesson 7-5

1. yx

−2

−8

6 12 18 24

t = 0

t = 5t = 4t = 3t = 2

t = 1

t = -1t = -2 t = -3 t = -4

t = -5

x = t 2+ 3

y = t_4

- 5

3. y

x−16

16

24

8

16

t = 0

t = 4

t = 5

t = 6

t = 2t = 1 t = -1

t = -2

t = -3

t = -4

t = -5

t = -6

x = -5t_2

+ 4

y = t 2- 8 t = 3

5. y

x

−4

−8

−4

4

4 8

t = 0

t = 4

t = 3t = 2

t = 1t = -1t = -2

t = -3

t = -4x = 2t - 1

y = -t 2_2

+ 7

7. y

x

4

6

8

2

1 2 3 4

t = 1

t = 4 t = 6 t = 8

t = 7

t = 0t = 5t = 3

t = 2

x = t_2

y = -√t + 5

9. y = 0.25 x 2 + 2.5x + 10.25 11. y = 5 √ �� x + 2 y

x−4−8−12

4

8

12

y

x

−4

−8

−4

4

4 8

13. y = 3 x 2 + 24x + 48 15. y = 3 √

� x _

5 + 9

y

x

−4

−8

−4−8

8

4

4 8

y

x

−8

−16

−8−16

16

8

8 16

17. y = - x 2 + 15 _

4 x + 100 19.

x 2 _

49 +

y 2 _

4 = 1

y

xO

40

80

120

124 8

t = 0t = 1

t = 2

t = 3

y

x

−4

−4−8

4

4

θ = π

2θ = 3π

2

θ = 0

θ = π

x = 7 sin θy = 2 cos θ

21. x 2

_ 9 +

y 2 _

9 = 1 23.

x 2 _

25 +

y 2 _

36 = 1

x

θ = π

2

θ = 3π

2

θ = 0θ = π

x = 3 cos θy = 3 sin θ

y y

xO

−4

−8

−8

4

4 8

θ = π

2

θ = 3π

2

θ = 0θ = π

x = 5 cos θy = 6 sin θ

25. x 2 + y 2

_ 49

= 1 27. x = t _ 8 and y = √

��

9 - t 2 _

64

x

−8

−4−8

8

4 8

θ = π

2θ = 3π

2

θ = 0

θ = π

y

x = sin θy = 7 cos θ

y

x

t = 0

t = 24t = -24

t = 8xy2

= 9 - x2



Selected A

nswers and S

olutions


29. x = 5t - 20 and y = -25 t 2 + 200t - 390

y

x

−80

−120

8 16

t = 2

t = 3

t = 6

t = 5t = 4

y = 10 - x2

t = + 4x5

31. x = 1 - 2t and y = - t 2 + t + 1 _

2

y

O−4−8

−4

−8

4

t = -2

t = 0

t = 2

t = 1

t = 3

t = 4 t = -3

8t = -1

x

y =3 - x2_

4

t = 1 - x2

33. yes

35 Solve the first equation for t. x = log t

10 x = t

Substitute into second equation.

y = t + 3

y = 10 x + 3

ln x = log t, x must be > 0, so the domain restriction is x > 0.

37. y = 10 x + 4 39. y = 2 · 10 1 _ x - 8, x ≠ 0 41. x = t + 4 and y = 7

+ 3t 43. x = t - 2 and y = -6 + 4t, 0 ≤ t ≤ 4 45. b 47. a

49 a. The position equations are x = t v 0 cos θ and

y = t v 0 sin θ - 1 _ 2 g t 2 + h 0 . The initial velocity v 0 is

0.75 and θ is 45°. The surface of the water is at y = 0, so the value of the initial height h 0 is 0.3. By substitution, the position equations are x = t · 0.75 cos 45° and y = t · 0.75 sin 45° - 4.9 t 2 + 0.3.

b. Find the value of t for which y = 0 in the vertical position equation.0 = t · 0.75 sin 45° − 4.9 t 2 + 0.3

0 = 3 √ � 2

_ 8 t - 4.9 t 2 + 0.3.

The positive zero in the graph of the equation is about 0.3074, so t ≈ 0.3074 seconds. Substitute this for t in x = t · 0.75 cos 45°, the horizontal position equation.

x = 0.3074(0.75) · √ � 2

_ 2 ≈ 0.16.

This means that when the frog reaches the surface of the water, he is only about 0.16 m from where he jumped. Therefore, he is about 0.5 - 0.16 or 0.34 meters from the other bank.

c. The 0.4 meter distance is a horizontal distance, so substitute the values into the horizontal position equation and solve for v 0 . x = t v 0 cos θ

0.4 = 0.38 · v 0 · √ � 2

_ 2

1.49 = v 0

51a. about 27.2 ft/s 51b. about 1 second 51c. about 1.84 seconds 53. Sample answer: x = 4 + t, y = 10 + 3t 55. Sample answer: The horizontal distance is modeled by the cosine function, which is 0 at 90°. This would imply that the projectile has no horizontal movement. The corresponding parametric equation would be x = 0. 57. Sample answer: Parametric equations show both horizontal

and vertical positions of an object over time, while rectangular equations can only show one or the other.

59. y

x

(x′) 2 (y′ ) 2 = 1-

61. (y - 2) 2

_ 4 -

(x - 3) 2 _

5 = 1

63. 2 tan 2 x 65. 3 ln 3 + ln 2; 3.99 67. 3 ln 2 - ln 3; 0.97

69. horizontal asymptote at y = 1; y

x−8

16

8

8 16

vertical asymptote at x = -6; D = {x | x ≠ -6, x ∈ �}

71. vertical asymptotes at x = -5; y

x

16

8

8 16

oblique asymptote at y = x + 3; x-intercepts 0 and -8; y-intercept 0; D = (-∞, -5) ∪ (-5, ∞)

73. 7 75. 23 77. E 79a. 21 x 2 + 31 y 2 − 10 √ � 3 xy = 4

79b. ellipse 79c. 4 (x′ ) 2 + 9 (y′ ) 2 = 1 79d. ≈ 0.745

79e. ( √ � 5

_ 6 , 0) , (-

√ � 5 _

6 , 0)


1. conic section 3. directrix 5. foci 7. center 9. parametric

11. vertex: (-3, -2), y

x

−4

−8

−8

8

4

4 8(−3, −2)

focus: (-3, 1); axis of symmetry: x = -3; directrix: y = -5

13. vertex: (2, -1), focus: (2, -2); y

x

−8

−4−8

8

4

4 8

(2, -1)

axis of symmetry: x = 2; directrix: y = 0



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y

x

−4

−8

−16

8

8 16

(1, 5) y

x

−4

−8

−4

8

4

4 8

(−2, 1)

19. (y + 4) 2 = -4(x + 3) 21. (x - 3) 2 = 4(y + 7)

y

x

−8

−8

8

4

4 8

(−3, −4)

y

x

−8

−4−8

8

4

4

(3, −7)

23. y

xO

−4

−8

−12

−4−8 4 8

(x - 3 ) ( )2_16

+

y + 6 2_25

= 1 25.

(x - 5) 2 _

25 +

(y - 2) 2 _

9 = 1

27. (x - 1) 2 + (y - 2) 2 = 30; circle 29. (x - 2) 2 = -4(y + 5); parabola

31. y

xO

−8

−16

−8

16

8 16

(x + 7 )2_18

-

(y - 6) 2_36

= 1

33. y

x

−4

−8

−4−8

8

4

4 8

(−1.83, 2) (3.83, 2)

x 2 - y 2 - 2x + 4y - 7 = 0

35. y 2

_ 9 - x 2

_ 16

= 1

37. x 2 _

4 -

y 2 _

9 = 1 39. parabola

41. 43.


45.

[-5, 5] scl: 1 by [-5, 5] scl: 1

47. (x ′ ) 2 - 2 √ � 3 x ′y ′ + 3(y ′ ) 2 + (2 √ � 3 - 4)x ′ + (4 √ � 3 + 2)y ′ = 20; parabola 49. 9 (y ′) 2 + 4(x ′) 2 = 36; ellipse

51. y

x−2

8

12

4

-4

2 4 6

t = 0t = 1t = -1

t = -2

t = -3

t = -4

t = 2

t = 3

t = 4

53. y = x 2 _

4 - 2 55. (y - 1) 2 = 4(x + 1)

y

x

−4

−8

−4−8

8

4

4 8

y

x

−4

−8

−4−8

8

4

4 8

57a. x 2 + y 2 = 900 59b. y

x

−4

−8

−4−8

8

4

4 8

57b. 5 seconds 59a. 3 x 2 + 2 √ � 3 xy + y 2 + 8x - 8 √ � 3 y = 0


1. 376.99 units 3 3. 100.53 units 3 5. Sample answer: The approximation will get closer to the actual volume of the solid because as the cylinders decrease in height, they will better fill the volume of the solid. 7. 188.5 units 3



Selected A

nswers and S

olutions


Vectors8CHAPTER 8CHCHAPAPTETERR 88

Chapter 8 Get Ready

1. 3; (-

1 _

2 , 4) 3. √ � 29 ; (-

1 _

2 , -8) 5. 5.4 7. 4.0 9. 22.8 ft

11. no solution 13. B ≈ 39°, C ≈ 50°, c ≈ 23.0

Lesson 8-1

1. scalar 3. vector 5. vector

7. Sample answer: 9. Sample answer:

EW

N

S

h

1 cm : 3 in./s

205°

EW

N

S

j

1 cm : 1 ft/min

300°

11. Sample answer:

EW

N

S

m

1 cm : 10 m

55°

13. 1.4 cm, 50° 15. 1.0 cm, 46° 17. 2.3 cm, 188° 19. 45 mph, 31° 21. 2.6 km due north 23. 250 m due south

25 Let a = 17 miles east and b = 16 miles south. Draw a diagram to represent a and b using a scale of 1 cm : 4 mi.

4.25 cm

1 cm : 4 mi

4.0 cm

a

E

N

b

E

N

Translate b so that its tail touches the tip of a. Then draw the resultant vector a + b.

5.9 cm

a

ba + b

S47°E

E

N

Measure the length of a + b and then measure the angle this vector makes with the horizontal. The length of the vector is approximately 5.9 centimeters, which is 5.9 × 4 or 23.6 miles. Therefore, the resultant is about 23.6 miles at a bearing of S47°E.

27. -2n

m - 2nm

29.

1_2

p

3n

3n + 1_2

p

31. 2n

-m p

p + 2n -m



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33.

3n

m

3n - 1_2

p + m

1_2

p

35. 7 mi/h 37. about 1.49 m/s at a bearing of S51°W

39. about 1.13 cm; 41. about 0.72 in./min; about 0.98 cm about 0.19 in./min

1.5 cm

x

y49°

255°

3_4

in./min

xy

43a.

33°

190 N

1 cm : 50 Nx

y

43b. about 159.3 N; about 103.5 N 45. 84 lb at 162° 47. 11.7 lb at 215°

49 a. Let v represent the vertical component of the force and r represent the magnitude of the resultant force.

r

86 newtons

v

31°

Use the tangent ratio to find v.

tan 31° = |v|

_ 86

|x| = 86 tan 31°

≈ 51.674

The vertical component of the force is about 52 newtons.

b. Use the cosine ratio to find r.

cos 31° = 86 _

|r|

|r| cos 31° = 86

|r| =

86 _

cos 31°

≈ 100.33

The magnitude of the resultant force is about 100 newtons.

51 Draw a diagram to represent a and b using a scale of 1 cm : 4 mi/h.

a b

35°

35° 45°100°

125°

E

N

1 cm : 4 mi/h

The angle created by a and the x-axis is 125° - 90° or 35°. Draw a horizontal where the tip of a and the tail of b meet, as shown. b makes a 45° angle and a makes a 35° angle with the horizontal. Thus, the angle created by a and b is 180° - 35° - 45° or 100°. Draw the resultant a + b. The three vectors form a triangle.

a + b

100°

E

N

12 mi/h15 mi/h

Use the Law of Cosines to find the magnitude of a + b. c 2 = a 2 + 2 b 2 - 2ab cos θ |a + b | 2 = 1 5 2 + 1 2 2 -2(15)(12) cos 100°

|a + b| =

√ ��

1 5 2 + 1 2 2 -2(15)(12) cos 100°

≈ 20.77Use the Law of Sines to find the angle opposite of b.

sin 100° _

20.77 = sin B _

12

12 sin 100° _

20.77 = sin B

si n -1

12 sin 100° _

20.77 = B

34.68 ≈ BThe angle opposite b is about 35°. To find the bearing of a + b, subtract 35° from 125°. Thus, the direction of a + b is a bearing of 90°. Since the equilibrant vector is the opposite of the resultant vector, it will have a magnitude of about 20.77 mi/h at a bearing of about 270°.

53. about 14.87 ft/s at a bearing of 69° 55a. Sample answers: T 1x = T 1 cos 65°; T 1y = T 1 sin 65°; T 2x = T 2 cos 35°; T 2y = T 2 sin 35°55b. T 1 ≈ 2079.5 lb; T 2 ≈ 1072.8 lb 55c. T 1x ≈ 878.8 lb; T 1y ≈ 1884.7 lb; T 2x ≈ 878.8 lb; T 2y ≈ 615.3lb 57. about 5.2 ft at 54° 59. about 3.4 yd at 212°61. Sample answer:

(a + b) + c

a + (b + c)

a

a

b

b

c

c

63. Sample answer:

3 units 4 units

5 unitsθ

65. Sample answer: In order for the direction to have a consistent meaning, it must be measured using a common reference. Lack of a common reference would cause ambiguity in the reporting of the direction of the vector. 67a. The magnitude of a added to the magnitude of b is greater than or equal to the magnitude of the vector created by a + b. 67b. True; sample answer: The vector created by a + b has to account for the direction of both vectors. This can result in a very small magnitude, |a + b|, if a and b have opposite directions. Calculating the sum of the magnitudes, |a| + |b|, will result in the greatest possible value because direction is not being considered. This value can only be achieved by |a + b| if a and b are parallel vectors with the same direction. 69. Yes; sample answer: It is possible for the sum of two vectors to be equal to one of the components only when one of the vectors is the zero vector. 71. 36.7 m



Selected A

nswers and S

olutions


73. (x - 4 ) 2 + (y - 5 ) 2 = 16 75. (x - 2 ) 2 = 4(y - 3)y

x

−4

−4

8

4

4 8

(4, 5)

(x - 4)2+ (y - 5)2

= 16

y

xO

−4

−4−8

8

12

4

4 8(2, 3)

(x - 2)2= 4(y - 3)

77a. Sample answer: [12 25 45 50],

⎡

⎢

⎣

60

40

25

5

⎤

�

⎦

77b. $3095 79. π

_ 6 + 2nπ, 5π

_ 6 + 2nπ,

3π

_ 2 + 2nπ; n ∈ 81. G 83. H

Lesson 8-2

1. ⟨7, 4⟩; √ � 65 ≈ 8.1 3. ⟨-7, -3⟩; √ � 58 ≈ 7.6 5. ⟨13, 2⟩;

√ �� 173 ≈ 13.2 7. ⟨-6.5, 4.5⟩; √ �� 62.5 ≈ 7.9 9. 11 _

2 , 23

_ 2 ;

√

��

325 _

2 ≈ 12.7 11. ⟨-21, 13⟩ 13. ⟨-55, -13⟩ 15. ⟨26, 6⟩

17. ⟨-42, -18⟩ 19a. f = ⟨-9, 10⟩, n = ⟨76, 84⟩, w = ⟨0, -170⟩ 19b. ⟨67, -76⟩

21. u = 3 √ � 10

_ 10

, -

√ � 10 _

10 23. u =

2 √ � 5 _

5 ,

√ � 5 _

5

25. u = -

√ � 26 _

26 , -

5 √ � 26 _

26 27. u = 3 _

5 , -

4 _

5

29. -16i + 8j 31. -9.5i - 8.3j 33. 13i + 11j 35. 0

37 a. Nadia’s rowing can be represented by the vector r = ⟨0, 5⟩ and the current can be represented by the vector c = ⟨3, 0⟩.

3 mph

5 mph

shore

r + c

Add the vectors representing r and c to find the resultant vector, v.

v = r + c = ⟨0, 5⟩ + ⟨3, 0⟩ or ⟨3, 5⟩

The speed at which Nadia is traveling is the magnitude of v.

|v| = √ ��

3 2 + 5 2 = √ � 34 or about 5.8Thus, she is traveling at about 5.8 miles per hour.

b. Find θ by constructing a right triangle with the given vectors and using the tangent ratio.

3 mph

3 mph

5 mph

shore

5 mph

θ

tan θ = 5 _ 3

θ = tan -1 5 _ 3

θ ≈ 59°Thus, Nadia is traveling at an angle of about 59° with respect to the shore.

39. ⟨-2 √ � 2 , 2 √ � 2 ⟩ 41. ⟨8 √ � 3 , -8⟩ 43. ⟨-8.6, 12.29⟩ 45. 111.8° 47. 216.9° 49. 45° 51. 290.6° 53a. about 674.3 mph 53b. about S86°E 55. No; sample answer: The magnitude and direction are not the same for both vectors, so they are not equivalent. 57. Yes; sample answer: The magnitude and direction are the same for both vectors, so they are equivalent.

59a. Sample answer: 59b. about 46.1 mph;

Groundspeed:518 mph

Air speed:480 mph

wind3° about N46°E

59c. about 538 mph; about N79°E

61. Sample answer: (0, -2) 63a. wire 1: ⟨-755.2, 932.6⟩; wire 2: ⟨898.8, 438.4⟩; wire 3: ⟨544.0, -440.5⟩ 63b. ⟨687.6, 930.5⟩ 63c. about 1157 N; about 53.5° 65. Sample answer: a = kb, where k is any real number scalar. 67. Sample answer: If the initial point of a vector is (a, b) and the vector has magnitude m, then the locus of terminal points (x, y) are

the points that satisfy the equation √ ��

(x - a ) 2 + (y - b ) 2 = m.

69 tan θ = b _ a

tan 4y = y _ x

x tan 4y = y

x = y _

tan 4y

71. (a + b) + c = (⟨ x 1 , y 1 ⟩ + ⟨ x 2 , y 2 ⟩) + ⟨ x 3 , y 3 ⟩

= ⟨ x 1 + x 2 , y 1 + y 2 ⟩ + ⟨ x 3 , y 3 ⟩ = ⟨ x 1 + x 2 + x 3 , y 1 + y 2 + y 3 ⟩

= ⟨ x 1 , y 1 ⟩ + ⟨ x 2 + x 3 , y 2 + y 3 ⟩ = ⟨ x 1 , y 1 ⟩ + (⟨ x 2 , y 2 ⟩ + ⟨ x 3 , y 3 ⟩)

= a + (b + c)

73. |ka| = |k⟨ x 1 , y 1 ⟩|

= |⟨k x 1 , k y 1 ⟩|

= √ ��

(k x 1 ) 2 + (k y 1 ) 2

= √ ��

k 2 x 1 2 + k 2 y 1 2

= √ ��

k 2 ( x 1 2 + y 1 2 )

= √ �

k 2 √ ��

x 1 2 + y 1 2

= |k| |⟨ x 1 , y 1 ⟩|

= |k| |a|

75. y = x 2 _

9 + 4x _

3 + 6 77. y = 7 √ �� x + 1 79. 3

_ z + 2

- 1 _

2z - 3

81. -6 _

x + 3 + -3

_ x - 3

83. sin (60° + θ) + sin (60° - θ)

= sin 60° cos θ + cos 60° sin θ + sin 60° cos θ - cos 60° sin θ

= √ � 3

_ 2 cos θ + 1 _

2 sin θ +

√ � 3 _

2 cos θ - 1 _

2 sin θ

= 2 √ � 3

_ 2 cos θ

= √ � 3 cos θ

85. 2 ln 7 - 2 ln 3 87. 2 ln 3 - 3 ln 7 89. 5984 91. H 93. H



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1. 8; not orthogonal 3. 0; orthogonal 5. -30; not orthogonal 7. 8; not orthogonal 9a. 15,6209b. The total revenue that can be made by selling all of the

basketballs is $15,620. 11. √ � 97 ≈ 9.8 13. 5 √ � 13 ≈ 18.0

15. √ �� 785 ≈ 28.0 17. 100.0° 19. 82.9° 21. 48.4° 23. 159.1°

25. pro j v u = 15 _

29 , -

6 _

29 ; u = 15

_ 29

, -

6 _

29 + 72

_ 29

, 180 _

29

27. pro j v u = 120 _

17 , -

30 _

17 ; u = 120

_ 17

, -

30 _

17 + 16

_ 17

, 64 _

17

29. pro j v u = - 78 _

73 , 208

_ 73

; u = - 78 _

73 , 208

_ 73

+ 224 _

73 , 84

_ 73

31. pro j v u = 93 _

13 , - 62

_ 13

; u = 93 _

13 , -

62 _

13 + -

28 _

13 , -

42 _

13

33 The combined weight of Malcolm’s sister and the wagon is the force exerted due to gravity, F = ⟨0, -78⟩. To find the force -

w 1 required to keep her from sliding down the slope, project F onto a unit vector v in the direction of the side of the hill.

wagon

v

F

w115°

Step 1 Find a unit vector v in the direction of the side of the hill.v = ⟨|v| cos θ, |v| sin θ⟩

= ⟨1 � cos 15°, 1 � sin 15°⟩

= ⟨0.966, 0.259⟩

Step 2 Find w 1 , the projection of F onto the unit vector v, pro j v F.

pro j v F = ( F · v

_ |v | 2

) v

= (F � v)v = (⟨0, -78⟩ � ⟨0.966, 0.259⟩)v ≈ -20.2v

Since w 1 points down the hill, the force required is - w 1 = -(-20.2v) or 20.2v. Since v is a unit vector, 20.2 pounds represents the force required to keep Malcolm’s sister from sliding down the hill.

35. 801 J 37. Sample answer: ⟨-12, 3⟩ 39. Sample answer: ⟨8, 14⟩ 41a. ⟨16.38, 11.47⟩, ⟨22.94, -32.77⟩

41b. dot product; (20 cos 35°)(40 cos 55°) + (20 sin 35°) � (-40 sin 55°) = 0 43. Sample answer: u = ⟨-1, 7⟩ 45. Sample answer: u = ⟨4, -5⟩ 47. Perpendicular; sample answer: Since u � v = 0, the vectors are perpendicular.

49 Find the angle between u and v.

cos θ =

u · v _

|u||v|

cos θ =

⟨5, 7⟩ · ⟨-15, -21⟩

__

|⟨5, 7⟩||⟨-15, -21⟩|

cos θ =

-75 + (-147) __

√ � 74

√ �� 666

cos θ =

-222 _

222

cos θ = -1 θ = co s -1 (-1) or 180°

Parallel; sample answer: The angle between the two vectors is 180°. They are headed in opposite directions.

51. 5π

_ 12

53. π

_ 2 55. about 141.1 J 57. 39.8°, 50.2°, 90°

59. 48.9°, 65.6°, 65.6° 61. ⟨-5.36, 0.55⟩ or ⟨2.03, -4.99⟩ 63. ⟨-9.03, 7.90⟩ or ⟨1.09, 11.95⟩ 65. False; sample answer: All three vectors may originate at the same point and not form a triangle at all. If so, the angle between d and f may be acute, right, or obtuse. 67. False; sample answer: If either a or b is a zero vector, it would be orthogonal to v, but could not be parallel to the other vector.

69. cos 90° = u · v _

|u| · |v|

Angle between u and v where θ = 90°.

0 = u · v _

|u| · |v|

cos 90° = 0

0 = u · v Multiply each side by |u| � |v|.

71. u · (v + w) = u · v + u · w ⟨ u 1 , u 2 ⟩ · (⟨ v 1 , v 2 ⟩ + ⟨ w 1 , w 2 ⟩) ⟨ u 1 , u 2 ⟩ · ⟨ v 1 , v 2 ⟩ +

⟨ u 1 , u 2 ⟩ · ⟨ w 1 , w 2 ⟩

⟨ u 1 , u 2 ⟩ · ⟨ v 1 + w 1 , v 2 + w 2 ⟩ ( u 1 v 1 + u 2 v 2 ) + ( u 1 w 1 + u 2 w 2 )

u 1 ( v 1 + w 1 ) + u 2 ( v 2 + w 2 ) u 1 v 1 + u 1 w 1 + u 2 v 2 + u 2 w 2

u 1 v 1 + u 1 w 1 + u 2 v 2 + u 2 w 2 = u 1 v 1 + u 1 w 1 + u 2 v 2 + u 2 w 2

73. Sample answer: For two nonzero vectors ⟨a, b⟩ and ⟨c, d ⟩, the dot product ⟨a, b⟩ � ⟨c, d⟩ is the sum of the products of the x-coordinates and the y-coordinates, or ac + bd.

75. -

137 _

4 , -9.2 77. Jada: about 173.8 ft/s and about

108.6 ft/s; James: about 143.4 ft/s and about 124.7 ft/s

79. y

−8

−16

8

8 16x−8

x 2_81

- y2_

49= 1

−16

81. π

_ 6 83.

√ � 7 _

4 85. 6

87a. t = 4 87b. when t = 4 87c. 10 amps 89. F 91. H

Lesson 8-4

1.

(1, -2, -4)

z

y

x



Selected A

nswers and S

olutions


3. z

y

x

(-5, -4, -2)

5. z

y

x

(-5, 3, 1)

7.

4

8

z

y

x

(4, -10, -2)

4 8-8

4

-8

-4

-4

-8

8

9. 5 √ � 6 ≈ 12.2; (-

3 _

2 , 5, 13

_ 2 ) 11. √ �� 542 ≈ 23.3; (-

3 _

2 , -

9 _

2 , 3)

13. 4 √ � 14 ≈ 15.0; (3, 4, 4) 15. longitude: 38.55°, latitude: 101°, altitude: 1207 m 17.

(0, -4, 4)

z

y

x

19.

(-1, 3, -4)

z

y

x

21. z

y

x

(6, 8, -2)

8-4-8 4

4

8

-4

-84

-8

8

23. z

y

x

(7, -6, 6)

8-4-8 4

-4

4

-8

8

4

-8

8

25. ⟨16, 2, 8⟩; 18; 8 _ 9 , 1 _

9 , 4 _

9 27. ⟨-3, -5, -10⟩; √ �� 134 ;

-

3 √ �� 134 _

134 , -

5 √ �� 134 _

134 , -

5 √ �� 134 _

67

29. ⟨-1, 8, -10⟩; √ �� 165 ; -

√ �� 165 _

165 ,

8 √ �� 165 _

165 , -

2 √ �� 165 _

33

31. ⟨4, -15, 5⟩; √ �� 266 ; 2 √ �� 266

_ 133

, -

15 √ �� 266 _

266 ,

5 √ �� 266 _

266

33. ⟨11, -23, -13⟩; 3 √ � 91 ; 11 √ � 91

_ 273

, -

23 √ � 91 _

273 , -

√ � 91 _

21

35 Let P represent the point at which the rope and the pole are connected and let B represent the position of the ball as it is

being served. Then ⎯⎯⎯� PB represents the vector from the pole to

the ball.



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⎯⎯⎯� PB = ⟨5 - 0, 3.6 - 0, 4.7 - 9.8⟩

= ⟨5, 3.6, -5.1⟩

The magnitude of ⎯⎯⎯� PB is

√ ��

5 2 + 3. 6 2 + (-5.1 ) 2 ≈ 8.0.

Thus, the rope is about 8 feet long.

37. ⟨-65, -18, 56⟩ 39. ⟨48, 12, -38⟩ 41. ⟨22, 36, 3⟩ 43. -63i + 28j + 56k 45. 50i - 18j + 10k 47. -13i + 2j + 21k 49. 3.1i - 23.6j + 13.4k

51 Let P = ( x 2 , y 2 , z 2 ). Use the Midpoint Formula and the points M and N to solve for P.

N = ( x 1 + x 2

_ 2 ,

y 1 + y 2 _

2 ,

z 1 + z 2 _

2 )

( 7 _ 2 , 1, 2) = (

3 + x 2 _

2 ,

4 + y 2 _

2 ,

5 + z 2 _

2 )

Solve for x 2 , y 2 , and z 2 .

3 + x 2

_ 2 = 7 _

2

4 + y 2 _

2 = 1

5 + z 2 _

2 = 2

3 + x 2 = 7 4 + y 2 = 2 5 + z 2 = 4

x 2 = 4 y 2 = -2 z 2 = -1

So, P = (4, -2, -1).

53. (3, -2, 7) 55. 34 ft 57. scalene 59. isosceles

61. d = √ ��

( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2 + ( z 2 - z 1 ) 2

r = √ ��

(x - h ) 2 + (y - k ) 2 + (z - ) 2

r 2 = (x - h ) 2 + (y - k ) 2 + (z - ) 2

63. (x - 6 ) 2 + y 2 + (z + 1 ) 2 = 1 _ 4 65. x 2 + (y - 7 ) 2 +

(z + 1 ) 2 = 144 67a. about 199.2 mi/h 67b. N4.1°E 67c. about 75.9° 69. Sample answer: It is more reasonable to use two dimensions when describing position on a map since the map itself is two-dimensional. When the description of the position of an object requires the added dimension of altitude, such as the position of an airplane, it is more reasonable to use three dimensions. 71. ⟨-0.19, -0.04⟩; u = ⟨-0.19, -0.04⟩ + ⟨-0.81, 4.04⟩

73. ⟨-13, -3⟩, √ �� 178 ≈ 13.3 75. ⟨6, 18⟩, √ �� 360 ≈ 19.0

77a. about 2.6 feet to the left and to the right of the center of the arch 77b. 8 ft; Sample answer: With the string anchored by thumbtacks at the foci of the arch and held taut by a pencil, the sum of the distances from each thumbtack to the pencil will remain constant. When the pencil is at one of the bottom corners of the arch, exactly 4 feet from the center, the distance from the far thumbtack to the bottom corner will be 4 + 2.6 or about 6.6 feet and the distance from the other thumbtack to the bottom corner will be 4 - 2.6 or about 1.4 feet. Thus, the total length of string needed is 6.6 + 1.4 or 8 feet.

79. cos -1 ( 1 _ 3 ) + nπ, c os -1 (-

1 _

3 ) + nπ, n ∈ �

81.

x

y

π

π

2

y = cos -1 (x - 2)

π

2

2 3 41-1-

83.

x

y

-0.8 0.8 1.6-1.6

π

4

y = sin -1 3x

π

4-

85. F 87. J

Lesson 8-5

1. 0; orthogonal 3. -13; not orthogonal 5. -15; not orthogonal 7. 0; orthogonal

9 Let u = ⟨55.5, 55.5, -55.5⟩ and v = ⟨-55.5, -55.5, -55.5⟩. Use the formula for the angle between two vectors.

cos θ = u · v _

|u||v|

cos θ = ⟨55.5, 55.5, -55.5⟩ ⟨-55.5, -55.5, -55.5⟩

____ |⟨55.5, 55.5, -55.5⟩||⟨-55.5, -55.5, -55.5⟩|

cos θ = -3080.25 + (-3080.25) + 3080.25

___ √ �� 9240.75

√ �� 9240.75

cos θ = -3080.25 _

9240.75

θ = co s -1 (-

1 _

3 ) or about 109.5°

The angle between the vectors formed by the oxygen-hydrogen bonds is about 109.5°.

11. 88.9° 13. 37.5° 15. 16.6°17. ⟨25, 6, 71⟩

(u × v) · u (u × v) · v

= ⟨25, 6, 71⟩ · ⟨4, 7, -2⟩ = ⟨25, 6, 71⟩ · ⟨-5, 9, 1⟩

= 25(4) + 6(7) + 71(-2) = 25(-5) + 6(9) + 71(1) = 0 = 0

19. ⟨-56, -35, -42⟩

(u × v) · u

= ⟨-56, -35, -42⟩ · ⟨5, -8, 0⟩

= (-56)(5) + (-35)(-8) + (-42)(0) = 0

(u × v) · v

= ⟨-56, -35, -42⟩ · ⟨-4, -2, 7⟩

= (-56)(-4) + (-35)(-2) + (-42)(7) = 021. ⟨29, 12, 13⟩

(u × v) · u (u × v) · v

= ⟨29, 12, 13⟩ · ⟨-4, 1, 8⟩ = ⟨29, 12, 13⟩ · ⟨3, -4, -3⟩

= 29(-4) + 12(1) + 13(8) = 29(3) + 12(-4) + 13(-3) = 0 = 0

23a. ⟨0, -72.08, 0⟩ 23b. about 72.08 N · m parallel to the negative

y-axis 25. 13 √ � 19 ≈ 56.7 unit s 2 27. √ �� 6821 ≈

82.6 unit s 2 29. √ �� 2158 ≈ 46.5 unit s 2 31. 206 unit s 3 33. 102 unit s 3 35. 69 unit s 3 37. Sample answer: ⟨5, 5, 3⟩

39. Sample answer: ⟨-8, 0, 7⟩ 41. Sample answer: ⟨-1, 0, 4⟩ 43. not collinear 45. parallel 47. parallel

49. -

11 √ � 3 _

2 , -

11 _

2 , 0 51. parallelogram; about 9.4 unit s 2 ;

rectangle 53. 0 55. ⟨0, 0, 0⟩ 57a. about 2.1 N 57b. about 98.8° 59. 5 61. Always; sample answer: The cross product of



Selected A

nswers and S

olutions


two three-dimensional vectors results in a vector that is perpendicular to both original vectors.

63 Calculate u × v.

u × v = ⎪

i

4

-3

j

6

-2

k

c

5

⎥

= ⎪

6

-2

c

5 ⎥ i -

⎪

4

-3

c

5 ⎥ j +

⎪

4

-3

6

-2 ⎥ k

= (30 + 2c)i - (20 + 3c)j + 10k

For u × v to equal 34i - 26j + 10k, set the components equal and solve for c. 30 + 2c = 34, so c = 2. -(20 + 3c) = -26, so c = 2. This implies the system is consistent. Thus, c = 2.

65. Sample answer: To determine if two vectors are parallel or perpendicular, you can use the formula for the angle between two vectors. If the angle measures 0 or 180, they are parallel. If the angle measures 90, they are perpendicular. You can also find the component form of the two vectors and use the ratios of the coordinates to determine if two vectors are parallel. If the ratios of all three coordinates of the component vectors are the same, then the vectors are parallel. To determine if two vectors are perpendicular, you can find the dot product of the vectors. If the dot product is 0, then the vectors are perpendicular.

67. √ �� 562 ≈ 23.7; ( 33

_ 2 , 9, -

37 _

2 ) 69. -22; not orthogonal 71. -33;

not orthogonal 73. 3 _

m - 4 + -1

_ m + 4

75. s ec 2 θ cot 2 θ - cot 2 θ = c ot 2 θ (se c 2 θ - 1) = c ot 2 θ ta n 2 θ = 1

77. A ≈ 55°, C ≈ 78°, b ≈ 17.9 79. A ≈ 103°, B ≈ 49°, C ≈ 28° 81. 29.102° 83. D 85a. horizontal: 45 √ � 3 ft/s; vertical: 45 ft/s 85b. scalars 85c. about 226.0 ft 85d. ⟨-77.3, 212.4⟩ 85e. ⟨-0.342, 0.940⟩ 85f. about N51.1°W


1. false; ends 3. true 5. true 7. false; projection 9. true 11. vector 13. 4.0 cm, 11° 15. 2.8 cm, 297º

17. 80 m due east 19. ⟨6, 1⟩; √ � 37 ≈ 6.1 21. ⟨14, 5⟩;

√ �� 221 ≈ 14.9 23. ⟨-8, -6⟩ 25. ⟨-18, -1⟩

27. -

7 √ � 53 _

53 ,

2 √ � 53 _

53 29. -

5 √ � 89 _

89 , -

8 √ � 89 _

89

31. -1; not orthogonal 33. 0; orthogonal 35. 135.0°

37. z

y

x (1, 2, -4)

39. z

y

x

(5, -3, -2)

41. 2 √ � 38 ≈ 12.3; (-1, 5, 6) 43. 4 √ � 19 ≈ 17.4; (-3, -4, 2)

45. z

y

x

(0, -3, 4)

47.

z

y

x

(-2, -3, 5)

49. 0; orthogonal 51. ⟨17, -1, 10⟩; ⟨17, -1, 10⟩ · ⟨1, -3, -2⟩ = 0 and ⟨17, -1, 10⟩ · ⟨2, 4, -3⟩ = 0 53. about 49.8 ft/s; about 23.2 ft/s 55. about 1081.8 lb 57a. about 5.3 mph 57b. about 40.9° 59. 800 J 61a. ⟨-2.3, 0, 0⟩ 61b. about 2.3 N · m parallel to the negative x-axis


1. different 3. No; sample answer: Every point (x, y) in a plane has a vector associated to it, and there are infinitely many points in a given plane. 5. Sample answer: Representing a vector field in only two dimensions ignores the third dimension and assumes that the force created by wind would remain constant along the z-plane.



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1. ⟨9, -26⟩and ⟨25, 16⟩ 3. [63.31, -0.50°] 5a. the second quadrant because the resultant’s x-compenent is negative and the y-component is positive. 5b. No, the magnitude of the resultant vector is

√ ��

(12 cos 62° + 31 cos 170°)2 + (12 sin 62° + 31 sin 170°)2

≈ 29.6 < 31.

7a. u

v

u + v

7b.

u

v

v + u

7c.

v

v

v + v

7d.

w

v

v + w

7e.

w

-w

7f.

v

v - w

-w

9a. ⟨-1, 3⟩ 9b. ⟨-7, 19⟩ 11a. The angle is 0. 11b. 4.19 knots 11c. 16.05 knots 13. The fourth force that puts the other three in equilibrium has magnitude of 12.44 and is 65.34 degrees above the negative x-axis. 15. Answers vary. Sample: Here is a counterexample. Let r = s = 1 and α = β = 45°. Then

�

u + �

v = [1, 45°] + [1, 45°] = [2, 45°]. But [r + s, α + β] = [2, 90°].

17. x √ �� 2 + 2cos θ = 2x cos θ _ 2 19a.

19b. y = √ �� x + 4 + 1 21. (-1, 13), (-5, -1), (5, 1)

Polar Coordinates and Complex Numbers9CHAPTER 9CHCHAPAPTETERR 99

Chapter 9 Get Ready

1.

[-10, 10] scl: 2 by [-2, 18] scl: 2

even; symmetric with respect to the y-axisf (-x) = (-x ) 2 + 10 = x 2 + 10 = f (x)

3.

[-10, 10] scl: 1 by [-10, 10] scl: 1

neitherg(-x) = √ �� -x + 9 which is not equal to g(x) or -g(x).

5.

[-10, 10] scl: 1 by [-10, 10] scl: 1

odd; symmetric with respect to the origing(-x) = 3(-x ) 5 - 7(-x) = 3(- x 5 ) + 7x = -3 x 5 + 7x = -g(x)

7. even 9. abs. max.: (2.25, 9.13) 11. real. max.: (-1.67, 6.48); rel. min.: (1, -3)



Selected A

nswers and S

olutions


13. 165° + 360k°; 525°; -195°y

x

165°525°

y

x

165°

-195°

15. -10° + 360k°; 350°; -370°

-370°

-10°

y

x

350°

-10°

y

x

17. 4π

_ 3 + 2kπ; 10π

_ 3 ; -

2π

_ 3

10π

3

4π

3

y

x

-2π

3

4π

3

y

x

Lesson 9-1

1.

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°54321

(1, 120°)

3. 5.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

(-2, )2π

3

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

(4, - )

5π

6

7. 9.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

(-1, - )

5π

3

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321(-4, π)

11. 13a.

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°54321

(4.5, -210°) 30°150°

300°240°

330°210°

90°60°120°

270°

0°180°603612

(57, 45°)

(15, 240°)(41, 315°)

13b. 13 points 15. (2, 120°), (2, -240°), (-2, -60°)

17. (3, 5π

_ 3 ) ,

(3, -

π

_ 3 ) , (- 3, -

4π

_ 3 ) 19. (5, 5π

_ 3 ) ,

(5, -

π

_ 3 ) ,

(-5, 2π

_ 3 ) 21. (1, 300°), (1, -60°), (-1, 120°)

23. 25.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

5321

(4, )

5π

3

(4, π)(4, )

π

6r = 4

4

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

(1.5, )

3π

2

r = 1.5(1.5, )

π

3(1.5, π)

27. 29a. r = 6.3, r = 16, r = 225

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°54321

(2, -15°) (-1, -15°)

(-4, -15°) θ = -15°

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°25015050

r = 6.3r = 16 mm

r = 225 mm

29b. ≈0.5% 31. ≈10.70 33. 8 35. ≈3.05 37. 5

39. ≈6.08 41. ≈5.35

43 a. The parameters -60° ≤ θ ≤ 150° and 0 ≤ r ≤ 40 describe a sector of a circle with radius 40 units from -60° to 150°.

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°5010 30

b. The area of a sector of a circle with radius r and central

angle θ is A = 1 _ 2 r 2 θ, where θ is measured in radians. The angle

of the sector is 150° - (-60°) or 210°, which is 210° × π

_

180°

or 7π

_

6 radians. Therefore, the area of the sector is 1 _

2 (40 ) 2 ( 7π

_ 6 )

or about 2932.2 m 2 .

45. (-2.5, π

_ 2 ) 47. (1.25, 160°) 49. (6, 160°) 51. ≈10 ft

53. ≈174.46°



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55 Substitute the given information into the Polar Distance Formula.

P 1 P 2 = √ ��

r 1 2 + r 2 2 -2 r 1 r 2 cos ( θ 2 - θ 1 )

3.297 = √ ��

r 2 + 4 2 - 2r(4) cos (160° - 120°)

3.29 7 2 = r 2 + 16 - 8r cos 40°

= r 2 − 8r cos 40°r + (16 - 3.29 7 2 )

Use the Quadratic Formula to solve for r.

r = 8 cos 40° ± √

�� (-8 cos 40° ) 2 - 4(1)(16 - 3.29 7 2 ) ____

2(1)

≈ 1 or 5.1357. Sample answer: θ = π

_ 12

59. r = 4 or r = -4 61. Sample

answer: The sum and product of the r-values are calculated. Both of these operations are commutative. Due to the odd-even identities of trigonometric functions, cos (-θ) = cos θ. Therefore, the order of the angles does not matter.63.

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°54321

( r 1, 1 ) θ

-

2( r 2, 2) θ

2θ 1θr 1

1r 2

The triangle created by the two points and the origin results in a triangle with two known sides and an included angle. Therefore, by the Law of Cosines, ( P 1 P 2 ) 2 = r 1 2 + r 2 2 − 2 r 1 r 2 cos ( θ 2 − θ 1 ) or

P 1 P 2 = √ ��

r 1 2 + r 2 2 - 2 r 1 r 2 cos ( θ 2 − θ 1 ) .

65. Erina; sample answer: Sona plotted a point where the distance between the polar axis and the ray equals 5 units. She should have measured 5 units along the terminal side of the angle. 67. -3; not orthogonal 69. -14; not orthogonal 71. ⟨1, 44, -17⟩

73. vertex: (2, 7); y

x−4−8

12

8

4

-14(x - 2) = (y- 7)2

focus: (-1.5, 7); directrix: x = 5.5; axis of symmetry: y = 7

75. vertex: (3, 5); y

x

−4

−4

8

12

4

4 8

y = 1 _ 2 x 2 - 3x + 19_

2

focus: (3, 5.5); directrix: y = 4.5; axis of symmetry: x = 3

77.

⎡

⎢

⎣

12

4

11

0

14

0

-13

19

-10

-5

0

-6

0

6

2

7

23

33

-19

-25

⎤

�

⎦

79.

⎡

⎢

⎣

1

2

-5

0

8

-5

0

1

-3

11

8

-4

25

13

26

17

⎤

�

⎦

81. 3π

_ 4 +πn; n � � 83. A 85. B

Lesson 9-2

1. 3.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

10.5

r = -cos θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

10.5

r = 1_2

cos θ

5. 7.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = -sec θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = -4 cos θ

9. 11.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

106

r = 3 + 3 cos θ

2 84

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

106

r = 4 - 3 cos θ

842

13. 15.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = 2 - 2 sin θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

108642

r = 5 + 4 sin θ

17. 19.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

10.5

r = sin 4θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

42

r = 5 cos 3θ

31



Selected A

nswers and S

olutions


21. 23.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

10.5

r = 1_2

sin 3θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

21

r = 2 sin 5θ

25a.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6r = 3 cos 5θ

54321

symmetry with respect to the

polar axis; |r| = 3 when θ = 0, π

_ 5 , 2π

_ 5 , 3π

_ 5 ,

4π

_ 5 , and π; zeros for r when θ

=

π

_ 10

, 3π

_ 10

, π

_ 2 , 7π

_ 10

, and 9π

_ 10

25b.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

2010r = 20 cos 8θ

symmetry with respect to the polar axis, the

line θ =

π

_ 2 , and the

pole; |r| = 20 when

θ = 0, π

_ 8 , π

_ 4 , 3π

_ 8 , π

_ 2 , 5π

_ 8 ,

3π

_ 4 , 7π

_ 8 , and π; zeros for

r when θ =

π

_ 16

, 3π

_ 16

, 5π

_ 16

,

7π

_ 16

, 9π

_ 16

, 11π

_ 16

, 13π

_ 16

,

and 15π

_ 16

27. Spiral of Archimedes; 29. Cardioid;

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

4020

r = 4θ + 1

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6r = 6 + 6 cos θ

126

31. Spiral of Archimedes; 32. Lemniscate;

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

8040

r = 5θ + 2

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r 2= 9 sin 2θ

35. r = 3 sin θ 37. r = 3 cos 4θ 39. r = -2 cos θ

41 a. Since the fan is in the shape of a rose curve, the equation is of the form r = a cos nθ or r = a sin nθ. Since the fan has 5 blades and 5 is odd, n = 5. The length of each blade is 4 ft, so a

= 4. Thus, two equations that can represent the blades are r = 4 cos 5θ and r = 4 sin 5θ.

b.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = 4 cos 5θ

r = 4 cos 5θ is symmetric with respect to the polar axis, so you can find points on the interval [0, π] and then use polar axis symmetry to complete the graph.

θ 0

π

_ 6

π

_ 4

π

_ 3

π

_ 2 2π

_ 3 3π

_ 4 5π

_ 6 π

r 4 -3.5 -2.8 2 0 -2 2.8 3.5 -4

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = 4 sin 5θ

r = 4 sin 5θ is symmetric with respect to the line

θ = π

_ 2 , so you can find points

on the interval

⎡

⎣

-

π

_ 2 , π

_ 2 ⎤

⎦

and then use

symmetry to complete the graph.

θ -

π

_ 2 -

π

_ 3 -

π

_ 4 -

π

_ 6 0

π

_ 6

π

_ 4

π

_ 3

π

_ 2

r -4 3.5 2.8 -2 0 2 -2.8 -3.5 4

43. Substitute (-r, θ ) for (r, θ ). r 2 = 4 sin 2θ

(-r ) 2 = 4 sin 2θ

r 2 = 4 sin 2θ

Since this substitution produces an equivalent equation, r 2 = 4 sin 2θ is symmetric with respect to the pole.45. Substitute (r, π - θ ) for (r, θ ).r = 5 cos 8θ

r = 5 cos 8(π - θ )r = 5 cos (8π - 8θ )r = 5(cos 8π cos 8θ + sin 8π sin 8θ )r = 5[(1) cos 8θ - (0) sin 8θ ]r = 5 cos 8θ

Since this substitution produces an equivalent equation, r = 5 cos 8θ is symmetric with respect to the line θ = π

_ 2 . Substitute

π - θ for θ.47a. r = 2 sin 5θ or r = 2 cos 5θ 47b. r = 7 sin 2θ or r = 7 cos 2θ 47c. r = 6 sin 4θ or r = 6 cos 4θ 49. r = 2 + 2 sin θ 51. a 53. b

55 Make a table of values for r = 2 − sin 2θ.

θ 0

π

_ 6

π

_ 4

π

_ 3

π

_ 2 2π

_ 3 3π

_ 4 5π

_ 6

r 2 1.13 1 1.13 2 2.86 3 2.86

θπ 7π

_ 6 5π

_ 4 4π

_ 3 3π

_ 2 5π

_ 3 7π

_ 4 11π

_ 6

r 2 1.13 1 1.13 2 2.86 3 2.86

θ 2π 13π

_ 6 9π

_ 4 7π

_ 3 5π

_ 2 8π

_ 3 11π

_ 4 17π

_ 6

r 2 1.13 1 1.13 2 2.86 3 2.86



Sel

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nsNotice that the points in the first and third tables are equivalent polar coordinates. Therefore, for a complete polar graph, the function must be graphed on the interval 0 ≤ θ ≤ 2π. So x = 2π.

57. b 59. d 61a.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

84

0 ≤ θ ≤ 3π

1062

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

84

-3π ≤ θ ≤ 0

1062

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

84

-3π ≤ θ ≤ 3π

1062

61b. Sample answer: The equation r = θ is symmetric with respect to the line θ = π

_

2

when the interval for θ is -a ≤ θ ≤ a.

61c. Substitute (-r, -θ) for (r, θ). r = θ-r = -θ

r = θThis substitution produces an equivalent equation, therefore r = θ

is symmetric with respect to the line θ = π

_

2 .

61d. Sample answer: It does not affect the other classic curves. The classic curves all consist of either a sine or cosine function. Therefore, to achieve a complete graph, they just need to be graphed for all values of θ within their period. Extending the interval of θ to include additional values outside of the period will result in the graph repeating itself. Because the spiral of Archimedes does not contain a trigonometric function, additional values of θ result in different values of r. 63. Sample answer: r 2 and r 3

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

1

r 1 = cos θ r 2 = cos (θ -

π_2 )

r 3 = cos(θ - π)

are the graphs of r 1 after a

rotation about the pole of π

_ 2

and π, respectively. The graph of r = cos (θ − d) will be the graph of r = cos θ after a rotation of d about the pole.

65. To test for symmetry with respect to the line θ = π

_ 2 ,

substitute (r, π − θ) for (r, θ).r = a + b cos 2(π − θ) r = a + b cos (2π − 2θ) r = a + b (cos 2π cos 2θ + sin 2π sin 2θ) r = a + b [(1) cos 2θ + (0) sin 2θ] r = a + b cos 2θ

Since this substitution produces an equivalent equation, the graph

of r = a + b cos 2θ is symmetric with respect to the line θ = π

_ 2 .

67. Sample answer: The value of a determines the diameter of the circle. If a > 0, the graph is located in Quadrants I and IV of the plane with the polar axis containing a diameter. If a < 0, the graph is located in Quadrants II and III of the plane with the polar axis containing a diameter.

69. 71.

54321

r = 3.5

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

4321

θ = 225° 30°150°

300°240°

330°210°

60°120° 90°

270°

0°180°

73. 144.3° 75. 21 _

5 i - 2 _

3 j 77. -15.8i - 6.1j

79.

(3, 0)(-3, 0)

y

x

x 2_9

- y2_

25= 1

(-5.83, 0)

(5.83, 0)

81. y

x

(x + 1)2_4

-

(y + 3)2_9

= 1

(-3, -3) (1, -3)

(-4.61, -3) (2.61, -3)

83. (y + 5) 2 = -20(x - 4) 85. G 87. Hy

x

−4

−8

−12

−4 4 8

4

(y + 5)2= -20(x - 4)

Lesson 9-3

1. ( √ � 2 , √ � 2 ) 3. (-

5 _

2 , -

5 √ � 3 _

2 ) 5. (1, √ � 3 ) 7. (0, 3) 9. (0, 2)

11. (-

√ � 3 _

2 , 1 _

2 ) 13. (12.21, 0.96) and (-12.2, 4.10) 15. (13.42, 4.25)

and (-13.42, 1.11) 17. (3.61, 5.30) and (-3.61, 2.16) 19. (3.16a, 1.25) and (-3.16a, 4.39) 21. (60.54, 5.74) and

(-60.54, 2.61) 23. ( √ � 2 , 7π

_ 4 ) and (- √ � 2 , 3π

_ 4 )



Selected A

nswers and S

olutions


25a. about 0.90 mi north and 27. circle; r = -10 cos θ about 1.20 mi east 25b. Sample answer: (2.06, 194.04°)

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

108642

r = -10 cos θ

29. parabola; r = cot θ csc θ 31. hyperbola; r = 2 sec 2θ cos θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = cot θ csc θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321r = 2 sec 2θ cos θ

33. line; θ = π

_ 3 35. circle; r = 16 sin θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

θ = _π3

54321

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

20124

r = 16 sin θ

37. y = - √ � 3 x; line 39. x 2 − 4x + y 2 = 0; circle

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

θ = -_3π

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = 4 cos θ

41. y = 8; line 43. x = -7y or -

1 _

7 x = y; line

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

108642

r = 8 csc θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

cot θ = -7

45. x = 1; line

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = sec θ

47 a.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = 2 + 2 cos θ

This graph is symmetric with respect to the polar axis, so you can find points on the interval [0, π] and then use polar axis symmetry to complete the graph.

θ 0

π

_ 6

π

_ 4

π

_ 3

π

_ 2 2π

_ 3 3π

_ 4 5π

_ 6 π

r 4 3.7 3.4 3 2 1 0.6 0.3 0

b. No; sample answer: The given point is on the negative x-axis, directly behind the microphone. The polar pattern indicates that the microphone does not pick up any sound from this direction.

49.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

126

θ +(

7π_4 )r = 10 csc

√ � 2

_ 2 y −

√ � 2 _

2 x = 10 or

y = x + 10 √ � 2 ; line

51.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

θ -(

11π

6 )r = -2 sec _

√ � 3

_ 2 x − 1 _

2 y = -2 or

y = √ � 3 x + 4; line

53.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

108642

r = 5 cos + 5 sin cos 2 - sin 2

θ

θ θ

θ___

x − y = 5 or y = x − 5; line



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns55.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

θ +(

π

2 )r = 4 cos _

x 2 + y 2 + 4y = 0 or x 2 + (y + 2) 2 = 4; circle

57.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r =

4___ 6 cos θ - 3 sin θ

line; r = 4 __

6 cos θ - 3 sin θ

59.

0π

π

2

3π

2

4π

35π

3

π

32π

3

5π

6π

6

20161284

r = 12 cos θ + 16 sin θ

circle; r = 12 cos θ + 16 sin θ

61. Sample answer: y = -4; r = -4 csc θ 63. Sample answer: x 2 + (y − 2) 2 = 4; r = 4 sin θ 65. 39π yd 2 or about 122.52 yd 2

67a. i

R

−4

−8

−4−8

8

4

4 8

(6, 8)

67b.

0°108642

(10, 53.13°) 30°150°

300°240°

330°210°

90°60°120°

270°

180°

(10, 53.13°)

67c.

(-3, 3)

R

i

67d.

0°54321

(4.24, 135°)30°150°

300°240°

330°210°

90°60°120°

270°

180°

(4.24, 135°)

67e. r = √ ��

a 2 + b 2 , θ = tan -1 b _ a , when a is positive;

θ = tan -1 b _ a + 180° when a is negative. 69. (x − a) 2 + y 2 = a 2 , radius = a, center = (a, 0) 71. Sample answer: Rectangular equations that are not functions, such as equations representing ellipses or circles, are easier to graph in polar form. Equations that represent functions, such as linear functions, are easier to graph in rectangular form.

73 r 2 (4 cos 2 θ + 3 sin 2 θ) + r(−8a cos θ + 6b sin θ) = 12 − 4 a 2 − 3 b 2

4 r 2 cos 2 θ + 3 r 2 si n 2 θ − 8ar cos θ + 6br sin θ = 12 − 4 a 2 − 3 b 2

4 (r cos θ) 2 + 3 (r sin θ) 2 − 8a(r cos θ) + 6b(r sin θ) = 12 − 4 a 2 − 3 b 2

4 x 2 + 3 y 2 − 8ax + 6by = 12 − 4 a 2 − 3 b 2

4 x 2 − 8ax + 4 a 2 + 3 y 2 + 6by + 3 b 2 = 12

4( x 2 −2ax + a 2 ) + 3( y 2 + 2by + b 2 ) = 12

4(x − a ) 2 + 3(y + b ) 2 = 12

(x - a) 2

_

3 +

(y + b) 2 _

4 = 1

75. 77.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = 1 - 2 sin θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

4321

r = 2 sin 3θ

79. (1, -

2π

_ 3 ) , (1, 4π

_ 3 ) , (-1, -

5π

_ 3 )

81. 91.8°, not orthogonal 83. 98.3°, not orthogonal

85. (y - 1) 2 = x 87a. x 2 _

2916 -

y 2 _

49,984 = 1

y

x

−4

−8

−4−8

8

4

4 8

(y - 1) = x2

87b. y

x

−80

−160

−80−160

160

80

80 160

(230, 0)(−230, 0)

ship onthis branch

87c. (-60.2, 110)



Selected A

nswers and S

olutions


89.

⎡

⎢

⎣

1

0

0

0

1

0

0

0

1

-4

-1

7

⎤

�

⎦

91.

⎡

⎢

⎣

1

0

0

0

1

0

0

0

1

2

-3

4

⎤

�

⎦

93. G 95. J

Lesson 9-4

1. e = 1; parabola; y = 5 3. e = 3; hyperbola; x = 7 5. e = 1; parabola; x = -2 7. e = 4; hyperbola; y = -8

9 a. Factor 2 from the denominator, to rewrite the equation as r

= 7 _ 2 _

1 + sin θ . Since the coefficient of sin θ is 1, e = 1, which

means the conic is a parabola. Then 7 _ 2 = ed, but since e = 1, d

= 7 _ 2 , so the directrix is y = 7 _

2 or y = 3.5.

b. Factor 5 from the denominator, to rewrite the equation as r

=

28 _

5 __

1 + 2.5 cos θ . The coefficient of cos θ is 2.5, so e = 2.5 > 1,

which means the conic is a hyperbola. Then 28 _

5 = ed = 2.5d.

Dividing by 2.5, d = 28 _

12.5 = 2.24, so the directrix is x = 2.24.

11.

0

π

2

30186

x = -8

x

y

r = 6

1 - 0.75 cos θ __

r = 6 __

1 - 0.75 cos θ

13. π

2

focusy = 8

4 x

y

r =0.8

1 + 0.1 sin θ __

08

r = 0.8 __

1 + 0.1 sin θ

15.

0

π

2

54321 x

y

x = -1.5

r =1.5

1 - cos θ __ r = 1.5

_ 1 - cos θ

17.

0

π

2

4020

x = -5

focusx

y

r =7.5

1 - 1.5 cos θ __

r = 7.5 __

1 - 1.5 cos θ

19. (x + 2) 2

_ 64

+ y 2

_ 60

= 1 21. x 2 _

51 +

(y + 7) 2 _

100 = 1

23. (y + 9) 2

_ 9 -

x 2 _

72 = 1 25.

(x - 4 ) 2 _

100 +

y 2 _

84 = 1

27. parabola 29. hyperbola

[0, 2π] scl: by [-14, 6] scl: 2

by [-14, 6] scl: 2

π_24

[0, 2π] scl: by [-3, 7] scl: 1

by [-7, 3] scl: 1

π_24

31. a 33. c 35. e = 1; parabola; y = -5 37. e = 1 _

5 ; ellipse; x = 8

0

π

2

108642 x

y

r =1

0.2 - 0.2 sin θ __

y = -5

0

π

2

108642x = 8

x

y

r =8

cos θ + 5__

39. b 2 = a 2 - c 2 b 2 = a 2 - (ae ) 2

b 2 = a 2 - a 2 e 2 b 2 = a 2 (1 - e 2 )

b = a √

1- e 2

41. PF = ePQ c - a = e[a - (c - d)] c - a = e(a - c + d) ae - a = e(a - ae + d) ae - a = ae - a e 2 + de -a = -a e 2 + de a e 2 - a = de

a e 2 - a _ e = d

a( e 2 - 1)

_ e = d

43. r = 5 _

1 + sin θ 45. r = 6

__ 1 + 0.5 cos θ

47. r’ = 4.5 __

1 + 1.25 cos θ’

49a. Sample answer: r = 1.2 __

1 + 0.4 cos θ , ellipse; r = 1.8

__ 1 + 0.6 cos θ

,

ellipse; r = 3 _

1 + cos θ , parabola; r = 4.8

__ 1 + 1.6 cos θ

, hyperbola;

r = 6 _

1 + 2 cos θ , hyperbola

49b.

-4

-8

4

4

0

8-4-8

8e = 1

e = 2

e = 1.6

y

x

e = 0.6

e = 0.4

π

2

49c. Sample answer: The ellipse with the smaller eccentricity is much more circular than the ellipse with greater eccentricity. As e



Sel

ecte

d A

nsw

ers

and

Sol

utio

nsapproaches 1, the ellipses open farther to the left, approaching the parabola with e = 1. For values where e > 1, the hyperbola with the greater eccentricity approaches its asymptotes much more rapidly than the hyperbola with the smaller eccentricity. Also, as e approaches 1, one branch of the hyperbola moves farther and farther away from its asymptotes, approaching the parabola with e = 1.

49d. Sample answer: r = 0.125 __

1 + 0.5 cos θ ; r = 0.5

__ 1 + 0.5 cos θ

;

r = 2 __

1 + 0.5 cos θ

49e.

-2

-4

2

2 4-2-4

4

y

x

x = 0.25

x = 1

x = 4

r =0.125__

1 + 0.5 cos θr =

2__1 + 0.5 cos θ

r =0.5__

1 + 0.5 cos θ

0

π

2

49f. Sample answer: As the value of the directrix increases from 0.25 to 4, the distances between the vertices and the foci of the ellipses increase.51.

(x, y)

x

y

y = d

Statements (Reasons)

1. PF = ePQ (Definition of a conic section)

2. √ ��

x 2 + y 2 = e(d - y) (PF = √ ��

x 2 + y 2 and PQ = d - y)

3. r = e(d - r sin θ) (r = √ ��

x 2 + y 2 and y = r sin θ)

4. r = ed - er sin θ (Distributive Property)5. r + er sin θ = ed (Isolate the r-terms.)6. r(1 + e sin θ) = ed (Factor.)

7. r = ed _ 1 + e sin θ

(Solve for r.)

53. Sample answer: A conic section can be defined as a figure that is formed when a plane intersects a double-napped right cone or as the locus of points such that the distance from a point to the focus and the distance from the point to the directrix is a constant ratio.

55 Since one focus is at the pole and the other focus appears to be to the left of the pole, the equation is of the form

r = ed _ 1 + e cos θ

. Substitute the points (2, 0) and (8, π) and solve

for e in the resulting system of equations.

2 =

ed _ 1 + e cos 0

8 = ed _ 1 + e cos π

2 = ed _ 1 + e

8 = ed _ 1 - e

2(1 + e) = ed 8(1 − e) = ed

2(1 + e) = 8(1 − e) ed = 8(1 - e )

2 + 2e = 8 − 8e Multiply.

10e = 6 Add 8e - 2 to each side.

e = 6 _ 10

or 3 _ 5 Solve for e.

Substitute e = 3 _ 5 into 2(1 + e) = ed and solve for d.

2 (1 + 3 _ 5 ) = 3 _

5 d e =

3 _

5

16 _

5 = 3 _

5 d Simplify.

d = 16 _

3 Solve for d.

Therefore, r = 3 _ 5 · 16

_

3 _

1 + 3 _ 5 cos θ

or r = 16 _

5 + 3 cos θ .

57. Sample answer: Let P = (r, θ) be a point on the conic. Then the distance from P to the focus located at (0, 0) is |r|. In terms of θ,

depending on the conic, |r| = | |

ed _ 1 + e cos θ

| | ,

|r| = | |

ed _ 1 - e cos θ

| | , |r| =

| |

ed _ 1 + e sin θ

| | , or |r| =

| |

ed _ 1 - e sin θ

| | .

59. (5.39, 4.33), (-5.39, 1.19) 61. cardioid 63. spiral of Archimedes

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

8642

r = 3 + 3 cos θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

161284

r = 5_2

θ

65. (x - 3 ) 2

_ 25

+ (y - 4 ) 2

_ 9 = 1 67. 1996 Olympics

69. -

336 _

625 ; -

527 _

625 ; 336

_ 527

71. asymptotes at x = π

_

6 + nπ

-2

-4

y

-π ππ

23π

2

2

4

x +

x

(

π

3 )y = sec _

3π

2-

π

2-

73. asymptotes at x = π

_

2 + 3π

_

2 n

-2

-4

y

2

4

6

[ ](

2_3 (

x

x - + 0.75

3π

2-

π

2-

π

2-π

3π

2π

y = 2 cot π

2



Selected A

nswers and S

olutions


75. sin θ = √ � 5

_ 5 , cos θ =

2 √ � 5 _

5 , tan θ = 1 _

2 , sec θ =

√ � 5 _

2 ,

cot θ = 2 77. G 79. H

Lesson 9-5

1. 4 √ � 2 ≈ 5.66 3. 2 √ � 13 ≈ 7.21

(4, 4)

R

i

(−4, −6)

R

i

5. 5 7. √ � 58 ≈ 7.62(3, 4)

R

i

−4−8 4 8

−4

−8

4

8

(−3, −7)

R

i

9a. i

8

12

16

4

4 8 12 16R

(10, 15)

9b. 18.03 N; about 56.31°11. √ � 5 (cos 2.68 + i sin 2.68)

13. 2 √ � 2 (cos 7π

_ 4 + i sin 7π

_ 4 )

15. 2 √ � 5 (cos 2.03 + i sin 2.03)

17. 3 √ � 2 (cos π

_ 4 + i sin π

_ 4 )

19. -1.98 + 0.28i 21. 3 √ � 2

_ 2 +

3 √ � 2 _

2 i

0

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6(-2, 3)

54321π

π

2

3π

2

R

i

0

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

42 54321

π_4 )(3,

π

π

2

3π

2

R

i

23. -1 - √ � 3 i 25. 3 _ 2

0

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

(2, 4π_3 )

54321π

π

2

3π

2

R

i

R

i

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°54321

, 360°32 )(

27. 10(cos 180° + i sin 180°); -10 29. 4(cos 360° +

i sin 360°); 4 31. 2 (cos 3π

_ 4 + i sin 3π

_ 4 ) ; - √ � 2 + √ � 2 i

33. 3 (cos π

_ 2 + i sin π

_ 2 ) ; 3i 35. 1 _

6 (cos π

_ 6 + i sin π

_ 6 ) ;

√ � 3 _

12 + 1 _

12 i

37. 2035 - 828i 39. -8i 41. -112 - 384i 43. -5 + 12i 45. -16 47. 0.97 + 0.26i, 0.26 + 0.97i, -0.71 + 0.71i, -0.97 - 0.26i, -0.26 - 0.97i, 0.71 - 0.71i 49. 0.22 + 1.67i, -1.67 + 0.22i, -0.22 - 1.67i, 1.67 - 0.22i 51. 1.64 + 0.55i, -0.02 + 1.73i, -1.65 + 0.52i, -1.00 - 1.41i, 1.03 - 1.39i 53. 1, -1

55 a. Evaluate the trigonometric values and simplify.5(cos 0.9 + j sin 0.9) = 5 cos 0.9 + 5j sin 0.9 ≈ 3.11 + 3.92 j ohms8(cos 0.4 + j sin 0.4) = 8 cos 0.4 + 8j sin 0.4 ≈ 7.37 + 3.12 j ohms

b. Find the sum.(3.11 + 3.92j) + (7.37 + 3.12j) = 10.48 + 7.04j ohms

c. Find the modulus r and argument θ.

r = √ ��

a 2 + b 2 θ = ta n -1 b _ a

= √ ��

10.4 8 2 + 7.0 4 2 = ta n -1 7.04 _

10.48

≈ 12.63 ≈ 0.59 The polar form of 10.48 + 7.04j is 12.63(cos 0.59 +

j sin 0.59) ohms.

57. (3 + i)(3 - i) = 10; √ � 10 (cos 0.3218 + i sin 0.3218) �

√ � 10 [cos (-0.3218) + i sin (-0.3218)] = 10 59. (-6 + 5i) �(2 - 3i) = 3 + 28i; √ � 61 (cos 2.4469 + i sin 2.4469) �

√ � 13 [cos (-0.9828) + i sin (-0.9828)] ≈ 3 + 28i 61. (3 - 2i)(1 + √ � 3 i) ≈ 6.464 + 3.196i; √ � 13 [cos (-0.5880) +

i sin (-0.5880)] � 2(cos 1.0472 + i sin 1.0472) ≈ 6.464 + 3.196i 63a. a translation 3 units to the right and 4 units down 63b. a reflection in the real axis 63c. a rotation of 90° counterclockwise about the origin 63d. a dilation by a factor of 0.25 65. -4; 1 + i, -1 + i, -1 - i, 1 - i

67. √ � 3

_ 2 + 1 _

2 i, -

√ � 3 _

2 + 1 _

2 i, i

69. 2.77 + 1.15i, -1.15 + 2.77i, -2.77 - 1.15i, 1.15 - 2.77i 71. 0.79 + 0.79i, -1.08 + 0.29i, 0.29 - 1.08i 73. Blake; sample answer: Alma only converted the expression into polar form. She needed to use De Moivre’s Theorem to find the fifth power.

75 Let r be any of the roots depicted. Since r lies on a circle of

radius 3, |r| = 3. The root r 1 at π

_ 6 can be represented by r 1 = 3

(cos π

_ 6 + i sin π

_ 6 ) . The root r 2 at 5π

_ 6 can be represented by r 2 =

3 (cos 5π

_ 6 + i sin 5π

_ 6 ) . The root r 3 at 3π

_ 2 can be represented by r 3

= 3 (cos 3π

_ 2 + i sin 3π

_ 2 ) . To determine the number whose roots

are r 1 , r 2 , and r 3 , use De Moivre’s Theorem to cube any one of them.

r 1 3 = ⎡

⎢

⎣

3 (cos π

_ 6 + i sin π

_ 6 ) ⎤

�

⎦

3

= 27 (cos π

_ 2 + i sin π

_ 2 ) = 27i

77. Given: z 1 = r 1 (cos θ 1 + i sin θ 1 ) and z 2 = r 2 (cos θ 2 +

i sin θ 2 ); Prove: z 1

_ z 2 = r 1

_ r 2 [cos ( θ 1 − θ 2 ) + i sin ( θ 1 − θ 2 )]

z 1

_ z 2 = r 1 (cos θ

1

+ i sin θ 1 )

__ r 2 (cos θ 2 + i sin θ 2 )

= r 1

_ r 2 ( cos θ 1 + i sin θ 1

__ cos θ 2 + i sin θ 2

) ·

( cos θ 2 - i sin θ 2

__ cos θ 2 - i sin θ 2

)

= r 1

_ r 2 ·



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

( cos θ 1 cos θ 2 - i sin θ 2 cos θ 1 + i sin θ 1 cos θ 2 - i 2 sin θ 1 sin θ 2

_____ cos 2 θ 2 - i sin θ 2 cos θ 2 + i sin θ 2 cos θ 2 - i 2 sin 2 θ 2

)

= r 1

_ r 2 ·

( cos θ 1 cos θ 2 - i sin θ 2 cos θ 1 + i sin θ 1 cos θ 2 - i 2 sin θ 1 sin θ 2

_____ cos 2 θ 2 + sin 2 θ 2

)

= r 1

_ r 2 (cos θ 1 cos θ 2 - i sin θ 2 cos θ 1 + i sin θ 1 cos θ 2 +

sin θ 1 sin θ 2 )

= r 1

_ r 2 [(cos θ 1 cos θ 2 + sin θ 1 sin θ 2 ) + i(sin θ 1 cos θ 2 -

sin θ 2 cos θ 1 )]

= r 1

_ r 2 [cos ( θ 1 - θ 2 ) + i sin ( θ 1 − θ 2 )]

79. Always; sample answer: If z = a + bi and − z = a - bi, z + − z = a + bi + a - bi or 2a and z · − z = (a + bi)(a - bi) or a 2 + b 2 .81. Sample answer: Since one of the roots must be a positive real number, a vertex of the polygon lies on the positive real axis and the polygon is symmetric about the real axis. This means that the non-real complex roots occur in conjugate pairs. Since the imaginary part of the sum of two complex conjugates is 0, the imaginary part of the sum of all of the roots must be 0. 83. y 2 = -14(x - 3.5)

85. circle; r = 6 cos θ 87. circle; r = 2 sin θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6r = 6 cos θ

108642

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

21r = 2 sin θ

89. y

x

−4

−8

−4−8

8

4

4 8

y 2 2x 2 16 = 0

91. center: (-8, 7); foci: (-8, 7 ± 6 √ � 2 ); vertices: (-8, 16), (-8, -2) 93. center: (7, -6); foci: (7 ± √ � 5 , -6); vertices: (10, -6), (4, -6) 95. (-1, -3, 7) 97a. about 7.94 billion 97b. about 51 years 99. F


1. polar graph 3. polar coordinate system 5. argument 7. modulus 9. lemniscate

11. 13.

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°54321

(-0.5, -210°)

30°150°

300°240°

330°210°

90°60°120°

270°

0°180°54321

(4, -120°)

15. 17.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6(-2, -60°)

(4, -60°)

(1, -60°)54321

θ = -60°

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

(7, )

5π

6

(7, )

π

2

r = 7(7, )

11π

6

108642

19. ≈4.36 21. ≈6.74

23. 25.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

21

r = sin 3θ

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

6

5π

6π

6

r = 5 cos 2θ

54321

27. 29.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

r = 2 + 2 cos θ

54321

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6r = 2 - sin θ

54321

31. 33.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6r = 3 - 2 cos θ

54321

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

r = 3 sin θ

35. (-5.10, 4.91) and (5.10, 1.77) 37. (2a, 0) and (-2a, π)

39. x 2 + y 2 = 25; circle 41. x = 6; line

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6r = 5

54321

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

r = 6 sec θ

108642

43. e = 1; parabola; y = 3.5 45. e = 2; hyperbola; y = -7



Selected A

nswers and S

olutions


47. r = 3 __

1 - 0.5 sin θ ; y = -6 49.

(y - 1 _ 3 )

2 _

25 _

9 + x 2

_ 8 _ 3 = 1

0

π

2

r =1 - 0.5 sin θ

3

y = -6

108642

y

x

51. √ � 10 ≈ 3.16 53. 2 √ � 5 ≈ 4.47

y

x(3, -1)

y

x

(−4, 2)

55. 3.317(cos 0.441 + i sin 0.441)57. 4.359(cos 3.550 + i sin 3.550)59. 3i 61. - √ � 2 − √ � 2 i

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6(3, )

π

2

54321

�

�

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

(-2, )

π

4

54321

�

�

63. -4 √ � 3 - 4i 65. 15 _

2 +

15 √ � 3 _

2 i 67. 404 - 1121i 69. 1.895 -

0.376i, -0.622 + 1.829i, -1.273 - 1.453i 71a. 20 71b. Sample answer: (2, 0°) or (2, 180°)73. spiral of Archimedes

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6r = θ +

31

21

54321

75. 8605 ft 77a. 15.2 - 37.9j ohms 77b. 21.9 + 66.0j amps


1. Less; sample answer: The shortest distance between two points is a straight line, which is given by our line segments. Since the graph connects the points with a curve, it is going to be longer than the measurements given by the line segments. 3. Sample answer: The approximation approaches the actual arc length. 5. Sample answer: The graph is symmetric with respect to the y-axis.

If the lengths of three line segments that are on either side of the y-axis are found, the result can be doubled to account for the line segments on the other side. 7. Sample answer: As n 9. Sample answer: 20.6 unitsincreases, Δθ decreases. The approximation approaches the actual arc length.

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

8642

Sequences and Series0CHAPTER 100CHCHAPAPTETERR 1010

Chapter 10 Get Ready

1. x 3 + 9 x 2 + 27x + 27 3. As x → ∞, f(x) → 0; as x → -∞, f(x) → 0.

x -10,000 -1000 0 1000 10,000

f (x ) -4 · 10 -4 -0.004 undefi ned 0.004 4 · 10 -4

5. D = (-∞, ∞); R = (0, ∞); y

x

f(x) = 3 x

y-intercept: 1; asymptote: x-axis; lim x→-∞

f (x) = ∞, lim x→∞

f (x) = 0;


7. D = (-∞, ∞); R = (0, ∞); y

x

h(x) = 0.1 x + 2

y-intercept: 0.01; asymptote: x-axis; lim x→-∞

h(x) = ∞, lim x→∞

h(x) = 0;


9. 4 11. -3

Lesson 10-1

1. Sample answer: 29, 36, 43, 50 3. Sample answer: 1, 1 _ 3 , 1 _

9 ,

1 _ 27

5. Sample answer: -54, -67, -80, -93 7. 0, 3, 8, 15

9. 1, 9 _ 7 , 5 _

3 , 11

_ 5 11a. 16,350; 31,350; 46,350 11b. 2098, 2497, 2896,

3295 11c. a n = 1699 + 399n 11d. $16,063 13. 6.125 15. 18

17 a. In the family tree, each male will have only a female parent, and each female will have one of each parent. Starting from the left, the male has a mother. His mother has both a mother and father, who have two and one parent, respectively.

M F

M

F

F

MF

b. Let B n be the number of bees in the nth previous level of the tree, with level 0 corresponding to the original male bee. Each of the B n − 1 bees at level n − 1 will have a mother, so there are B n − 1 female bees at level n. Each female at level n − 1 will also have a father and each of these is a mother to exactly one of the bees at level n − 2, so there are B n − 2 male

bees at level n. Thus, B n = B n − 1 + B n − 2 with B 0 = 1 and



Sel

ecte

d A

nsw

ers

and

Sol

utio

nsB 1 = 1. This is the Fibonacci sequence, in which the previous two terms are added to get the next term. The 11th term in this sequence is 144, so there will be 144 parent bees in the 11th previous generation.

19. convergent 21. convergent 23. divergent 25. convergent 27. convergent 29. -22.65 31. 25 33. -27

35. 6 37. 0 39. -17 41. 300 43. 70,707

_ 1,000,000

45. 8 _ 9

47. $1750.93 49. Sample answer: a 1 = 6, a n = a n - 1 + 2, n ≥ 2; a n = 2n + 4 51. Sample answer: a 1 = 7, a n = 3 a n - 1 , n ≥ 2; a n = 7 (3) n - 1 53. a 1 = 5, a n = a n - 1 + 8, n ≥ 2; 5, 13, 21, 29,

… 55. 3, 6, 11, 18, … ; a n = n 2 + 2 57. ∑

n=1

8

(n - 3)

59 In the sum 8 + 27 + 64 + … + 1000, each term is a perfect cube. The first term is 8 = 2 3 and the last term is 1000 = 10 3 .

Thus, 8 + 27 + 64 + … + 1000 = ∑

n=2

10

n 3 .

61. ∑

n=3

9

(-2 ) n 63. divergent; 1 65. convergent; ≈-0.4

67. b 69. c 71. e 73a. 1 _ 1 , 2 _

1 , 3 _

2 , 5 _

3 , 8 _

5 , 13

_ 8 , 21

_ 13

, 34 _

21 , 55

_ 34

, 89 _

55

73b. y

x

0.250.5

0.751

1.251.5

1.752

1 2 3 4 5 6 7 8 9 10

73c. yes; 1.618 73d. The two ratios are equivalent to three decimal places. 73e. Sample answer: the Mona Lisa and the Taj Mahal 75. divergent 77. convergent 79. a n = -0.5n + 1.75 81a. 0.4, 0.04, 0.004, 0.0004, 0.00004

81b.

0.2

0.4

2 4 6

81c. As n → ∞, the terms approach a value of 0. 81d. 0.44444; 0.4444444; 0.444444444 81e. Sample answer: The sum approaches 4 _

9 .

81f. Sample answer: The sum will be the decimal 0.44…, with n number of 4s. The sum of the first five terms has 5 fours, the sum of the first seven terms has 7 fours, and sum of the first nine terms has 9 fours. 83a. Sample answer: a n = 1 _ n 83b. Sample answer: a n = 1 _ n + 3 83c. Sample answer: a n = 2n 85. True; the left side is equal to 4 + 10 + 18 + 28 + 40 or 100. The right side is equal to (1 + 4 + 9 + 16 + 25) + 3(1 + 2 + 3 + 4 + 5) or 100.

87 Write out the next several terms of the sequence. a 6 = a 5 - a 4 = -17 - (-15) = -2 a 7 = a 6 - a 5 = -2 - (-17) = 15 a 8 = a 7 - a 6 = 15 - (-2) = 17Since a 7 = a 1 and a 8 = a 2 , the sequence will repeat the first six terms. The sum of these six terms is 0, so the sum of the first 60 terms will also be zero.

89. - √ � 2 - √ � 2 i 91. 5

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

54321

5π

4(2, )

0π

π

2

3π

2

4π

35π

3

π

32π

3

7π

611π

6

5π

6π

6

108642

(5, 0)

93. e = 4; hyperbola; y = 5 95. no 97. no 99. √ � 26 ≈ 5.10; (1.5, -13, 13.5) 101. √ �� 373 ≈ 19.31; (2.5, -2, -3)

103. 2 - √ � 3 105. -

√ � 2 + √ � 6 _

4 107. 1 _

x - 2

_ 2x + 1

109. B 111. C

Lesson 10-2

1. -3; 11, 8, 5, 2 3. -9; 90, 81, 72, 63 5. 4; 9, 13, 17, 21 7. -5; -19.5, -24.5, -29.5, -34.5

9 a. In this problem, a 1 = 1, a 2 = 3, and a 3 = 5. Find the common difference.a 2 - a 1 = 3 - 1 or 2a 3 - a 2 = 5 - 3 or 2Add 2 to the third term to find the fourth term, and so on to find the eighth term. a 4 = 5 + 2 or 7 a 5 = 7 + 2 or 9 a 6 = 9 + 2 or 11 a 7 = 11 + 2 or 13 a 8 = 13 + 2 or 15So, there are 15 band members in the 8th row.b. For an explicit formula, substitute a 1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence. a n = a 1 + (n - 1)d a n = 1 + (n - 1)2 a n = 1 + 2(n - 1) or 2n - 1For the recursive formula, state the first term a 1 and then indicate that the next term is the sum of the previous term a n - 1 and d. a 1 = 1, a n = a n - 1 + 2

11. a n = -6 + 11(n - 1); a 1 = -6, a n = a n - 1 + 11 13. a n = 4 + 15(n - 1); a 1 = 4, a n = a n - 1 + 15 15. a n = 7 + (-10.5) ·(n - 1); a 1 = 7, a n = a n - 1 - 10.5 17. a n = 1 + 36(n - 1); a 1 = 1, a n = a n - 1 + 36 19. d = -7 21. a 12 = -8 23. a 6 = -311 25. d = -3 27. 13, 7, 1 29. 20, 37, 54, 71 31. -3, -1.5, 0, 1.5, 3, 4.5, 6 33. a n = n 2 + 4n + 7 35. a n = -2 n 2 + n + 9 37. a n = - n 2 - 5n + 12 39. 3978 41. -126 43. 12,008 45a. 3.75 mi 45b. day 25 47. 1176 49. 9588 51. 728 53. -756 55. 3792 ft 57. 14.8 59. 8

61. 6 63. ∑

n=1

11

6n 65. ∑

n=4

15

(4n + 1) 67. ∑

n=2

12

( n _ 5 - 3

)

69a. Sample answer: ∑

n=1

35

(5n + 19) 69b. 3815 seats

69c. 2560 seats 71. quadratic; a n = n 2 + 2n + 5 73 . cubic; a n = 3 n 3 + 2 n 2 - n + 1 75. quartic; a n = 2 n 4 - n 3 - 1

77. -

2 _

3 79. 1 _

3 81. Both; sample answer: Candace’s formula

could be simplified to Peter’s formula. 83. Sample answer: We know that a n = a 1 + (n - 1)d. Similarly, a k = a 1 + (k - 1)d. Solving the second equation for a 1 and substituting into the first yields:



Selected A

nswers and S

olutions


a n = a k - (k - 1)d + (n - 1)d = a k - kd + d + nd - d = a k + nd - kd = a k + (n - k)d

85. False; sample answer: you must also know n.

87 a. The arithmetic sequence of odd natural numbers is a 1 = 1, a 2 = 3, a 3 = 5, a 5 = 7, a 6 = 9, a 7 = 11, a 8 = 13, a 9 = 15… . Find the common difference. a 2 - a 1 = 3 - 1 or 2 a 3 - a 2 = 5 - 3 or 2

Find S 7 . Find S 9 .

S n = n _ 2 [2 a 1 + (n - 1)d] S n = n _

2 [2 a 1 + (n - 1)d]

S 7 = 7 _ 2 [2(1) + (7 - 1)(2)] S 9 = 9 _

2 [2(1) + (9 - 1)(2)]

= 7 _ 2 (14) or 49 = 9 _

2 (18) or 81

b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum of the first n terms of the sequence of odd natural numbers is n 2 .

c. Write an explicit formula for the arithmetic sequence of odd natural numbers. a n = a 1 + (n - 1)d = 1 + (n - 1)(2) = 1 + 2n - 2 or 2n - 1Substitute a n = 2n - 1 and a 1 = 1 into the formula for the sum of a finite arithmetic series.

S n = n _ 2 ( a 1 + a n )

= n _ 2 [1 + (2n - 1)]

= n _ 2 (2n) or n 2

89. Sample answer: 24, 28, 32, 36 91. Sample answer: 10, 3, -4,

-11 93. 3 (cos 11π

_ 6 + i sin 11π

_ 6 ) ;

3 √ � 3 _

2 - 3 _

2 i 95. 26; not

orthogonal 97. 251.6° 99. 135°

101. y

x

−4

−8

8

4

4 8 12

f (x) = ln x

g (x) = 3 ln (x - 1)

The graph of g(x) is the graph of f(x) shifted 1 unit to the right and expanded vertically.

103. J 105. J

Lesson 10-3

1. -2; 2, -4, 8 3. 1.5; 1.6875, 2.53125, 3.796875 5. 5; 250x, 1250x,

6250x 7. 3; 27x + 135, 81x + 405, 243x + 1215 9a. Sample answer: The sequence of circumferences is π d 1 ,

π d 2 , π d 3 , π d 4 , π d 5 . The common ratio is π d 2

_ πd 1

or d 2

_ d 1

.

9b. Sample answer: The sequence of areas is π (

d 1 _

2 )

2

, π (

d 2 _

2 )

2

,

π (

d 3 _

2 )

2

, π (

d 4 _

2 )

2

, π (

d 5 _

2 )

2

. The common ratio is π (

d 2 _

2 )

2

_

π (

d 1 _

2 )

2

= d 2 2

_ d 1 2

or

( d 2

_ d 1

) 2

. 11. a n = 64 ( 1 _ 4 )

n - 1 ; a 1 = 64, a n = 1 _

4 a n - 1 13. a n = 4

(-3) n - 1 ; a 1 = 4, a n = -3 a n - 1 15. a n = 20 (1.5) n - 1 ; a 1 = 20, a n = 1.5 a n - 1 17. a n = 1 _

32 (2) n - 1 ; a 1 = 1 _

32 , a n = 2 a n - 1

19 a. The number of bacteria doubles every 15 minutes, so in the course of an hour, it increases by a factor of 2 4 or 16, since there are 4 15-minute intervals. Thus, the number of bacteria in the dish after t hours is b t = b 0 · 16 t .

b. Substituting b 0 = 12 and t = 4, the number of bacteria after 4 hours is b 4 = 12 · 16 4 or 786,432.

21. 56 23. 5 _ 72

25. -4 27. 1 _ 4 29. 288 31. ≈$360.97 million

33. -192, 144, -108 or 192, 144, 108 35. 6, -18 37. -120, 300 39. t 5 , t 2 , t -1 , t -4 41. 85.5 43. 1501 45. 39,063

47. 1140 49. -484 51. 32.64768 53. 4921 _

27 55. 3,145,725

57. does not exist 59. 7.5 61. 100

63 The common ratio r is ⎪-

3 _

4 ⎥ < 1. Therefore, the infinite

geometric series has a sum. Find a 1 .

a 1 = 35 (−

3 _

4 )

1 - 1 or 35

Use the formula for the sum of an infinite geometric series to find the sum.

S = a 1 _

1 - r

= 35 _

1 - (-

3 _

4 ) or 20

65. 1 _ 3 67. -3 69. 31.104 71. 3 73a. $5, 1.1 73b. 52

75. 2 77. -0.2 79. 6, 3, 3 _ 2 81. 56, -14, 3.5 83. -36, -14.4,

-5.76 85. -138, 27.6, -5.52 87. 12, -3, 3 _ 4 89. neither; 5 _

12 ,

6 _ 14

, 7 _ 16

91. arithmetic; 60, 72, 84 93. neither; 100k, 121k, 144k

95. geometric; 75 √ � 5 , 375, 375 √ � 5 97. ∑

n=1

6

3 (4) n - 1

99. ∑

n=1

5

50 (1.7) n - 1 101. ∑

n=1

6

0.2(-5) a n - 1 103a. 1 _ 16

103b. d,

17 _

16 d, 273

_ 256

d, 4369 _

4096 d 103c. a 1 = d, a n = a n - 1 + ( 1 _

16 )

n - 1 d

105 Find the common ratio. a 2 ÷ a 1 = 4 ÷ 16 or 0.25 a 3 ÷ a 2 = 1 ÷ 4 or 0.25Sample answer: The common ratio is less than 1, so the sequence is converging to 0, and the sum of the series can be calculated. Therefore, Emilio is correct.

107. If |r| > 1, then |S n | increases without limit. Therefore, the corresponding sequence will be divergent, and the sum of the series cannot be calculated. 109. False; you must also know the value of the first term. 111. Sometimes; if |r| < 1, then the series has a sum, and the sum is a negative number. If |r| > 1, then the corresponding sequence is divergent, and the series has no sum. 113. 63 115. 24,300 117a. r 2 = 9 cos 2θ or r 2 = 9 sin 2θ 117b. r 2 = 16 cos 2θ or r 2 = 16 sin 2θ

119.

30°150°

300°240°

330°210°

60°120° 90°

270°

0°180°4321



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns121. ⟨-30, 6, 24⟩

⟨-30, 6, 24⟩ · ⟨1, 9, -1⟩ = -30(1) + 6(9) + 24(-1) = 0⟨-30, 6, 24⟩ · ⟨-2, 6, -4⟩ = -30(-2) + 6(6) + 24(-4) = 0

123. ⟨8, -21, 18⟩

⟨8, -21, 18⟩ · ⟨9, 0, -4⟩ = 8(9) + (-21)0 + 18(-4) = 0⟨8, -21, 18⟩ · ⟨-6, 2, 5⟩ = 8(-6) + (-21)2 + 18(5) = 0

125. ⟨-16, 18, -20⟩; 14 √ � 5 ; -

8 √ � 5 _

35 ,

9 √ � 5 _

35 , -

2 √ � 5 _

7 127. C

129. A

Lesson 10-4

1. Let P n be the statement 3 + 5 + 7 + � + (2n + 1) = n(n + 2). Because 3 = 1(1 + 2) is a true statement, P n is true for n = 1. Assume that P k : 3 + 5 + 7 + � + (2k + 1) = k(k + 2) is true for a positive integer k. Show that P k + 1 must be true. 3 + 5 + 7 + � + (2k + 1) = k(k + 2)3 + 5 + 7 + � + (2k + 1) + [2(k + 1) + 1] = k(k + 2) +

[2(k + 1) + 1]3 + 5 + 7 + � + (2k + 1) + [2(k + 1) + 1] = k 2 + 4k + 33 + 5 + 7 + � + (2k + 1) + [2(k + 1) + 1] = (k + 1)(k + 3)3 + 5 + 7 + � + (2k + 1) + [2(k + 1) + 1] = (k + 1) ·

[(k + 1) + 2]This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 3 + 5 + 7 + � + (2n + 1) = n(n + 2) is true for all positive integers n.3. Let P n be the statement 2 + 2 2 + 2 3 + � + 2 n = 2( 2 n - 1). Because 2 = 2( 2 1 - 1) is a true statement, P n is true for n = 1. Assume that 2 + 2 2 + 2 3 + � + 2 k = 2( 2 k - 1) is true for a positive integer k. Show that P k + 1 must be true.

2 + 2 2 + 2 3 + � + 2 k = 2( 2 k - 1)2 + 2 2 + 2 3 + � + 2 k + 2 k + 1 = 2( 2 k - 1) + 2 k + 1 2 + 2 2 + 2 3 + � + 2 k + 2 k + 1 = 2 · 2 k + 1 - 22 + 2 2 + 2 3 + � + 2 k + 2 k + 1 = 2( 2 k + 1 - 1)This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 2 + 2 2 + 2 3 + � + 2 n = 2( 2 n - 1) is true for all positive integers n.5. Let P n be the statement 1 + 4 + 7 + � + (3n - 2) =

n(3n - 1)

_ 2 . Because 1 =

1 · [3(1) - 1] __

2 is a true statement, P n is true

for n = 1. Assume that 1 + 4 + 7 + � + (3k - 2) =

k(3k - 1)

_ 2 is true for a positive integer k. Show that P k + 1 must be

true.

1 + 4 + 7 + � + (3k - 2) = k(3k - 1)

_ 2

1 + 4 + 7 + � + (3k - 2) + [3(k + 1) - 2] = k(3k - 1)

_ 2 +

[3(k + 1) - 2]

1 + 4 + 7 + � + (3k - 2) + [3(k + 1) - 2] = k(3k - 1) + 2(3k + 1)

__ 2

1 + 4 + 7 + � + (3k - 2) + [3(k + 1) - 2] = 3 k 2 + 5k + 2

__ 2

1 + 4 + 7 + � + (3k - 2) + [3(k + 1) - 2] = (k + 1)(3k + 2)

__ 2

1 + 4 + 7 + � + (3k - 2) + [3(k + 1) - 2] = (k + 1)[3(k + 1) - 1]

__ 2

This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 1 + 4 +

7 + � + (3n - 2) = n(3n - 1)

_ 2 is true for all positive integers n.

7. Let P n be the statement 1 + 2 + 4 + � + 2 n - 1 = 2 n - 1. Because 1 = 2 1 - 1 is a true statement, P n is true for n = 1. Assume that 1 + 2 + 4 + � + 2 k - 1 = 2 k - 1 is true for a positive integer k. Show that P k + 1 must be true.

1 + 2 + 4 + � + 2 k - 1 = 2 k - 11 + 2 + 4 + � + 2 k - 1 + 2 (k + 1) - 1 = 2 k - 1 + 2 (k + 1) - 1 1 + 2 + 4 + � + 2 k - 1 + 2 (k + 1) - 1 = 2( 2 k ) - 11 + 2 + 4 + � + 2 k - 1 + 2 (k + 1) - 1 = 2 k + 1 - 1This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 1 + 2 + 4 + � + 2 n - 1 = 2 n - 1 is true for all positive integers n.

9. Let P n be the statement 1 _ 2 + 1 _

2 2 + 1 _

2 3 + � + 1 _

2 n = 1 - 1 _

2 n .

Because 1 _ 2 = 1 - 1 _

2 1 is a true statement, P n is true for n = 1.

Assume that 1 _ 2 + 1 _

2 2 + 1 _

2 3 + � + 1 _

2 k = 1 - 1 _

2 k is true for a positive

integer k. Show that P k + 1 must be true.

1 _ 2 + 1 _

2 2 + 1 _

2 3 + � + 1 _

2 k = 1 - 1 _

2 k

1 _ 2 + 1 _

2 2 + 1 _

2 3 + � + 1 _

2 k +

( 1

_ 2 k + 1

) = 1 - 1 _

2 k +

( 1

_ 2 k + 1

)

1 _ 2 + 1 _

2 2 + 1 _

2 3 + � + 1 _

2 k +

( 1

_ 2 k + 1

) = 1 - 2

_ 2 · 2 k

+ 1 _

2 k + 1

1 _ 2 + 1 _

2 2 + 1 _

2 3 + � + 1 _

2 k +

( 1

_ 2 k + 1

) = 1 - 2

_ 2 k + 1

+ 1 _

2 k + 1

1 _ 2 + 1 _

2 2 + 1 _

2 3 + � + 1 _

2 k +

( 1

_ 2 k + 1

) = 1 - 1

_ 2 k + 1

This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 1 _

2 + 1 _

2 2

+ 1 _ 2 3

+ � 1 _ 2 n

= 1 - 1 _ 2 n

is true for all positive integers n.

11 a. From the pattern of dots, we can see that each triangle contains one greater number of dots than the previous row.

4th row: 1 + 2 + 3 + 4 = 105th row: 1 + 2 + 3 + 4 + 5 = 156th row: 1 + 2 + 3 + 4 + 5 + 6 = 217th row: 1 + 2 + 3 + 4 + 5 + 6 + 7 = 288th row: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36b. The sequence of numbers shows a pattern of common second differences.

1 3 6 10 15 21 28 362 3 4 5 6 7 8

1 1 1 1 1 1This means there exists a quadratic expression of the form a n 2 + bn + c to model this sequence. Write and solve a system of three equations to find this expression.Equation 1: a(1 ) 2 + b(1) + c = 1Equation 2: a(2 ) 2 + b(2) + c = 3Equation 3: a(3 ) 2 + b(3) + c = 6

Simplify.Equation 1: a + b + c = 1Equation 2: 4a + 2b + c = 3Equation 3: 9a + 3b + c = 6



Selected A

nswers and S

olutions


Add (-1)Equation 1 to Equation 2 and (-1)Equation 1 to Equation 3 to get the following system and solve for a.3a + b = 2 -6a - 2b = -48a + 2b = 5 (+) 8a + 2b = 5 2a = 1

a = 1 _ 2

Then solve for b and then c.

3 ( 1 _ 2 ) + b = 2

3 _ 2 + b = 2 1 _

2 + 1 _

2 + c = 1

b = 2 - 3 _ 2 1 + c = 1

b = 1 _ 2 c = 0

Therefore, the expression that models this sequence is 1 _

2 n 2 + 1 _

2 n, which is

n 2 + n _ 2 or

n(n + 1) _

2 . So, a n =

n(n + 1) _

2 .

c. Let P n be the statement 1 + 3 + 6 + � + n(n + 1)

_ 2 =

n(n + 1)(n + 2)

__ 6 . Because 1 =

(1)[(1) + 1][(1) + 2] __

6 is a true

statement, P n is true for n = 1. Assume that 1 + 3 + 6 + � + k(k + 1)

_ 2

= k(k + 1)(k + 2)

__ 6 is true for a positive integer k. Show that P k + 1

must be true.

1 + 3 + 6 + � + k(k + 1)

_ 2 =

k(k + 1)(k + 2) __

6

1 + 3 + 6 + � + k(k + 1)

_ 2 +

(k + 1)[(k + 1) + 1] __

2 =

k(k + 1)(k + 2) __

6

+ (k + 1)[(k + 1) + 1]

__ 2

1 + 3 + 6 + � + k(k + 1)

_ 2 +

(k + 1)[(k + 1) + 1] __

2 =

k(k + 1)(k + 2) + 3(k + 1)(k + 2)

___ 6

1 + 3 + 6 + � + k(k + 1)

_ 2 +

(k + 1)[(k + 1) + 1] __

2 =

(k + 1)(k + 2)(k + 3)

__ 6

1 + 3 + 6 + � + k(k + 1)

_ 2 +

(k + 1)[(k + 1) + 1] __

2 =

(k + 1)[(k + 1) + 1][(k + 1) + 2]

___ 6


6 + � + n(n + 1)

_ 2 =

n(n + 1)(n + 2) __

6 is true

for all positive integers n.

13. Let P n be the statement 9 n - 1 is divisible by 8. P 1 is the statement that 9 1 - 1 is divisible by 8. P 1 is true because 9 1 -1 is 8, which is divisible by 8. Assume P k is true, where k is a positive integer, and show that P k + 1 must be true. That is, show that 9 k - 1 = 8r for some integer r implies that 9 k + 1 - 1 is divisible by 8. 9 k - 1 = 8r 9 k = 8r + 1 9 � 9 k = 9(8r + 1) 9 k + 1 = 72r + 9 9 k + 1 - 1 = 72r + 8 9 k + 1 - 1 = 8(9r + 1)Because r is an integer, 9r + 1 is an integer and 8(9r + 1) is divisible by 8. Therefore, 9 k + 1 - 1 is divisible by 8. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 9 n - 1 is divisible by 8 for all positive integers n.

15. Let P n be the statement 2 3n - 1 is divisible by 7. P 1 is true because 2 3(1) - 1 is 7, which is divisible by 7. Assume P k is true, where k is a positive integer, and show that P k + 1 must be true. That is, show that 2 3k - 1 = 7r for some integer r implies that 2 3(k + 1) - 1 is divisible by 7. 2 3k - 1 = 7r 2 3k = 7r + 1 2 3 � 2 3k = 2 3 (7r + 1) 2 3(k + 1) = 56r + 8 2 3(k + 1) - 1 = 56r + 7 2 3(k + 1) - 1 = 7(8r + 1)Because r is an integer, 8r + 1 is an integer and 7(8r + 1) is divisible by 7. Therefore, 2 3(k + 1) - 1 is divisible by 7. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 2 3n - 1 is divisible by 7 for all positive integers n.

17. Let P n be the statement 3 n ≥ 3n. P 1 is true, since 3 1 ≥ 3(1) is a true inequality. Assume P k is true, that 3 k ≥ 3k if k is a positive integer k ≥ 1, and show that P k + 1 must be true. That is, show that

3 k ≥ 3k implies 3 k + 1 ≥ 3(k + 1). Use both parts of the inductive hypothesis. 3 k ≥ 3k3 · 3 k ≥ 3 · 3k 3 k + 1 ≥ 9k

k ≥ 1 6k ≥ 6If 6k ≥ 6 and 6 ≥ 3, then by the Transitive Property of Inequality, 6k ≥ 3. 6k ≥ 33k + 6k ≥ 3k + 3 9k ≥ 3(k + 1)

By the Transitive Property of Inequality, if 3 k + 1 ≥ 9k and 9k ≥ 3(k + 1), then 3 k + 1 ≥ 3(k + 1). Since P n is true for n = 1 and P k implies P k + 1 for k ≥ 1, P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 3 n ≥ 3n is true for integer values n ≥ 1.19. Let P n be the statement 2 n > 2n. P 3 is true, since 2 3 ≥ 2(3) is a true statement. Assume P k is true, that 2 k > 2k if k is a positive integer k ≥ 3, and show that P k + 1 must be true. That is, show that 2 k > 2k implies 2 k + 1 > 2(k + 1). Use both parts of the inductive hypothesis. 2 k > 2k2 · 2 k > 2 · 2k 2 k + 1 > 4k

k ≥ 3 2k ≥ 6If 2k ≥ 6 and 6 ≥ 2, then by the Transitive Property of Inequality, 2k ≥ 2. 2k ≥ 22k + 2k ≥ 2k + 2 4k ≥ 2(k + 1)

By the Transitive Property of Inequality, if 2 k + 1 > 4k and 4k ≥ 2(k + 1), then 2 k + 1 > 2(k + 1). Since P n is true for n = 3 and P k implies P k + 1 for k ≥ 3, P n is true for n = 4, n = 5, and so on. By the extended principle of mathematical induction, 2 n > 2n is true for integer values n ≥ 3. 21. Let P n be the statement 3n < 4 n . P 1 is true, since 3(1) < 4 1 is a true statement. Assume P k is true, that 3k < 4 k if k is a positive integer k ≥ 1, and show that P k + 1 must be true. That is, show that 3k < 4 k implies 3(k + 1) < 4 k + 1 . Use both parts of the inductive hypothesis. 3k < 4 k 4 � 3k < 4 � 4 k 12k < 4 k + 1

k ≥ 1 3k ≥ 3 3k - 3 ≥ 0 6k - 3k - 3 ≥ 06k - 3(k + 1) ≥ 0 6k ≥ 3(k + 1) 3(k + 1) ≤ 6k Since k is a positive integer, 6k ≤ 12k. If 3(k + 1) ≤ 6k and 6k ≤ 12k then 3(k + 1) ≤ 12k.



Sel

ecte

d A

nsw

ers

and

Sol

utio

nsBy the Transitive Property of Inequality, if 3(k + 1) ≤ 12k and 12k < 4 k + 1 , then 3(k + 1) < 4 k + 1 . Since P n is true for n = 1 and P k implies P k + 1 for k ≥ 1, P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 3n < 4 n is true for integer values n ≥ 1.

23. Let P n be the statement 2n < 1. 5 n . P 7 is true, since 2(7) < 1. 5 7 is a true statement. Assume P k is true, that 2k < 1. 5 k if k is a positive integer k ≥ 7, and show that P k + 1 must be true. That is, show that 2k < 1. 5 k implies 2(k + 1) < 1. 5 k + 1 . Use both parts of the inductive hypothesis.

2k < 1. 5 k 1.5 � 2k < 1.5 � 1. 5 k 3k < 1. 5 k + 1

k ≥ 7If k ≥ 7 and 7 ≥ 2, then by the Transitive Property of Inequality, k ≥ 2. k ≥ 2 2 ≤ k 2k + 2 ≤ 2k + k2(k + 1) ≤ 3k

By the Transitive Property of Inequality, if 2(k + 1) ≤ 3k and 3k < 1. 5 k + 1 , then 2(k + 1) < 1. 5 k + 1 . Since P n is true for n = 7 and P k implies P k + 1 for k ≥ 7, P n is true for n = 8, n = 9, and so on. By the extended principle of mathematical induction, 2n < 1. 5 n is true for all integers n ≥ 7.

25. Let P n be the statement that there exists a set of $0.05 and $0.06 stamps that adds to $0.01n for n > 20. For n = 21, the conjecture is true, because $0.01(21) = $0.05(3) + $0.06(1). Assume that there exists a set of $0.05 and/or $0.06 stamps that adds to $0.01k for k > 20. Show that this implies the existence of a set of $0.05 and/or $0.06 stamps that adds to $0.01(k + 1). Case 1 The set contains at least one $0.05 stamp. Replace it with a

$0.06 stamp and the set’s value is increased to $0.01k + 0.01 or $0.01(k + 1), which is exactly P k + 1 .

Case 2 The set contains no $0.05 stamps. The set must contain at least four $0.06 stamps, because the set’s value must be greater than $0.20. Replace four of the $0.06 stamps with five $0.05 stamps and the set’s value is increased by $0.01, to $0.01k + $0.01 or $0.01(k + 1), which is exactly P k + 1 .

In both cases, P n is true for n = k + 1. Because P n is true for n = 21 and P k implies P k + 1 for k > 20, P n is true for n = 22, n = 23, and so on. That is, by the extended principle of mathematical induction, all stamps greater than $0.20 can be formed using just $0.05 and $0.06 stamps.

27. Since each oblong number has one more column than row, the sequence of oblong numbers is defined explicitly by the formula a n = n(n + 1). To prove that the sum of the first n oblong numbers

is given by S n = n 3 + 3 n 2 + 2n __

3 , you must prove that 2 + 6 + � +

n(n + 1) = n 3 + 3 n 2 + 2n __

3 is true for all positive integers n. Let this

statement be P n . Because

2 = 1 3 + 3(1 ) 2 + 2(1)

__ 3 is a true statement, P n is true for n = 1.

Assume that 2 + 6 + � + k(k + 1) = k 3 + 3 k 2 + 2k __

3 is true for a

positive integer k. Show that P k + 1 must be true.

2 + 6 + � + k(k + 1) = k 3 + 3 k 2 + 2k __

3

2 + 6 + � + k(k + 1) + (k + 1)[(k + 1) + 1] = k 3 + 3 k 2 + 2k __

3 +

(k + 1)[(k + 1) + 1]

2 + 6 + � + k(k + 1) + (k + 1)[(k + 1) + 1] = k 3 + 3 k 2 + 2k __

3 +

( k 2 + 3k + 2)

2 + 6 + � + k(k + 1) + (k + 1)[(k + 1) + 1] =

k 3 + 3 k 2 + 2k + 3 k 2 + 9k + 6 ___

3

2 + 6 + � + k(k + 1) + (k + 1)[(k + 1) + 1] =

k 3 + 3 k 2 + 2k + 3 k 2 + (3k + 6k) + (3 + 1 + 2)

____ 3

2 + 6 + � + k(k + 1) + (k + 1)[(k + 1) + 1] =

( k 3 + 3 k 2 + 3k + 1) + (3 k 2 + 6k + 3) + (2k + 2)

____ 3

2 + 6 + � + k(k + 1) + (k + 1)[(k + 1) + 1] =

(k + 1 ) 3 + 3(k + 1 ) 2 + 2(k + 1)

___ 3


� + n(n + 1) = n 3 + 3 n 2 + 2n __

3 is true for all positive integers n.

29. Let P n be 1 + 4 + � + (3n - 2) = n _ 2 (3n - 1). Because

1 = 1 _ 2 [3(1) - 1] is a true statement, P n is true for n = 1. Assume

that 1 + 4 + � + (3k - 2) = k _ 2 (3k - 1) is true for a positive integer

k. Show that P k + 1 must be true.

1 + 4 + � + (3k - 2) = k _ 2 (3k - 1)

1 + 4 + � + (3k - 2) + (3k + 1) = k _ 2 (3k - 1) + (3k + 1)

1 + 4 + � + (3k - 2) + (3k + 1) = k(3k - 1)

_ 2 +

2(3k + 1) _

2

1 + 4 + � + (3k - 2) + (3k + 1) = 3 k 2 + 5k + 2

__ 2

1 + 4 + � + (3k - 2) + (3k + 1) = (k + 1)(3k + 2)

__ 2

1 + 4 + � + (3k - 2) + (3k + 1) = (k + 1)

_ 2 [3(k + 1) - 1]


� + (3n - 2) = n _ 2 (3n - 1) or ∑

a=1

n

(3a - 2) = n _ 2 (3n - 1) is true for

all positive integers n.

31. Let P n be 1 _ 3 + 1 _

15 + � + 1

_ 4 n 2 - 1

= n _ 2n + 1

. Because 1 _ 3 =

1 _

2(1) + 1 is a true statement, P n is true for n = 1. Assume that 1 _

3

+ 1 _ 15

+ � + 1 _

4 n 2 - 1 = k _

2k + 1 is true for a positive integer k. Show

that P k + 1 must be true.

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

= k _ 2k + 1

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

+ 1 __

4(k + 1 ) 2 - 1 = k _

2k + 1 + 1

__ 4(k + 1 ) 2 - 1

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

+ 1 __

4 k 2 + 8k + 3 = k _

2k + 1 + 1

__ 4 k 2 + 8k + 3

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

+ 1 __

4 k 2 + 8k + 3 = k _

2k + 1 +

1 __

(2k + 1)(2k + 3)

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

+ 1 __

4 k 2 + 8k + 3 =

k(2k + 3) + 1 __

(2k + 1)(2k + 3)

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

+ 1 __

4 k 2 + 8k + 3 =

2 k 2 + 3k + 1 __

(2k + 1)(2k + 3)

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

+ 1 __

4 k 2 + 8k + 3 =

(2k + 1)(k + 1) __

(2k + 1)(2k + 3)

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

+ 1 __

4 k 2 + 8k + 3 =

k + 1 _

2k + 3

1 _ 3 + 1 _

15 + � + 1

_ 4 k 2 - 1

+ 1 __

4 k 2 + 8k + 3 =

k + 1 _

2(k + 1) + 1



Selected A

nswers and S

olutions



3 + 1 _

15

+ � + 1 _

4 n 2 - 1 = n _

2n + 1 or ∑

a=1

n

1 _

4 a 2 - 1 =

n _ 2n + 1


33. Let P n be 1 _ 6 + 1 _

12 + � + 1

__ (n + 1)(n + 2)

= n _ 2(n + 2)

. Because 1 _ 6

= 1 __

(1 + 1)(1 + 2) is a true statement, P n is true for

n = 1. Assume that 1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

=

k _ 2(k + 2)

is true for a positive integer k. Show that P k + 1 must be

true.

1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

= k _ 2(k + 2)

1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

+ 1 __

[(k + 1) + 1][(k + 1) + 2] =

k _ 2(k + 2)

+ 1 __

[(k + 1) + 1][(k + 1) + 2]

1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

+ 1 __

(k + 2)(k + 3) = k _

2(k + 2) +

1 __

(k + 2)(k + 3)

1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

+ 1 __

(k + 2)(k + 3) =

k(k + 3) + 2 __

2(k + 2)(k + 3)

1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

+ 1 __

(k + 2)(k + 3) =

k 2 + 3k + 2 __

2(k + 2)(k + 3)

1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

+ 1 __

(k + 2)(k + 3) =

(k + 2)(k + 1) __

2(k + 2)(k + 3)

1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

+ 1 __

(k + 2)(k + 3) =

k + 1 _

2(k + 3)

1 _ 6 + 1 _

12 + � + 1

__ (k + 1)(k + 2)

+ 1 __

(k + 2)(k + 3) =

k + 1 __

2[(k + 1) + 2]


6 + 1 _

12

+ � + 1 __

(n + 1)(n + 2) = n _

2(n + 2) or ∑

a=1

n

1 __

(a + 1)(a + 2) = n _

2(n + 2) is

true for all positive integers n.

35. Let P n be 1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2n)(2n + 2)

=

n _ 2(2n + 2)

. Because 1 _

2 � 4 = 1

_ 2(2 + 2)

is a true statement, P n is true for

n = 1. Assume that 1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � +

1 _

(2k)(2k + 2) = k _

2(2k + 2) is true for a positive integer k. Show that

P k + 1 must be true.

1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2k)(2k + 2)

= k _ 2(2k + 2)

1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2k)(2k + 2)

+ 1 __

[2(k + 1)][2(k + 1) + 2]

= k _ 2(2k + 2)

+ 1 __

[2(k + 1)][2(k + 1) + 2]

1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2k)(2k + 2)

+ 1 __

(2k + 2)(2k + 4) =

k _ 2(2k + 2)

+ 1 __

(2k + 2)(2k + 4)

1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2k)(2k + 2)

+ 1 __

(2k + 2)(2k + 4) =

k(2k + 4) + 1(2)

__ 2(2k + 2)(2k + 4)

1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2k)(2k + 2)

+ 1 __

(2k + 2)(2k + 4) =

2 k 2 + 4k + 2

__ 2(2k + 2)(2k + 4)

1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2k)(2k + 2)

+ 1 __

(2k + 2)(2k + 4) =

2(k + 1 ) 2

__ 4(k + 1)(2k + 4)

1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2k)(2k + 2)

+ 1 __

(2k + 2)(2k + 4) =

k + 1 _

4k + 8

1 _

2 � 4 + 1

_ 4 � 6

+ 1 _

6 � 8 + � + 1

_ (2k)(2k + 2)

+ 1 __

(2k + 2)(2k + 4) =

k + 1 __

2[2(k + 1) + 2]

This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and

so on. That is, by the principle of mathematical induction, 1 _

2 � 4

+ 1 _

4 � 6 + 1

_ 6 � 8

+ � + 1 _

(2n)(2n + 2) = n _

2(2n + 2) is true for all positive

integers n.

37. false; n = 2

39. Let P n be the statement that 2 2n + 1 + 3 2n + 1 is divisible by 5. P 1 is the statement that 2 2(1) + 1 + 3 2(1) + 1 is divisible by 5. P 1 is true, since 2 2(1) + 1 + 3 2(1) + 1 = 35, which is divisible by 5. Assume P k is true, where k is a positive integer, and show that P k + 1 must be true. That is, show that 2 2k + 1 + 3 2k + 1 = 5r for some integer r implies that 2 2(k + 1) + 1 + 3 2(k + 1) + 1 is divisible by 5. 2 2k + 1 + 3 2k + 1 = 5r 2 2k + 1 = 5r - 3 2k + 1 2 2k + 1 � 2 2 = 5r - 3 2k + 1 � 2 2 2 2k + 1 � 2 2 = (5r - 3 2k + 1 )( 3 2 - 5) 2 2k + 3 = 45r - 25r - 3 2k + 3 + 5( 3 2k + 1 ) 2 2k + 3 + 3 2k + 3 = 45r - 25r - 3 2k + 3 + 5( 3 2k + 1 ) + 3 2k + 3 2 2k + 3 + 3 2k + 3 = 20r + 5( 3 2k + 1 ) 2 2(k + 1) + 1 + 3 2(k + 1) + 1 = 5(4r + 3 2k + 1 )

Because r and k are integers, 4r + 3 2k + 1 is an integer and 5(4r + 3 2k + 1 ) is divisible by 5. Therefore, 2 2k + 3 + 3 2k + 3 is divisible by 5. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 2 2n + 1 + 3 2n + 1 is divisible by 5 for all positive integers n.

41. Let P n be the statement (x + 1 ) n ≥ nx, for integer values n ≥ 1 and x ≥ 1. P 1 is true, since when n = 1 and x ≥ 1, (x + 1 ) 1 ≥ 1x is a true statement. Assume that (x + 1 ) k ≥ kx is true for an integer k ≥ 1. Show that (x + 1 ) k + 1 ≥ (k + 1)x must be true. (x + 1 ) k ≥ kx(x + 1)(x + 1 ) k ≥ (x + 1)kx (x + 1 ) k + 1 ≥ k x 2 + kxNow we need to prove that k x 2 + kx ≥ (k + 1)x. First note that since k ≥ 1 and x ≥ 1, both kx and x - 1 are nonnegative. Hence their product kx(x - 1) is nonnegative. Thus k x 2 - kx ≥ 0, which implies k x 2 ≥ kx. But kx ≥ x, so by the Transitive Property of Inequality, k x 2 ≥ x. Add kx to each side of k x 2 ≥ x and factor to obtain k x 2 + kx ≥ (k + 1)x. Since (x + 1 ) k + 1 ≥ k x 2 + kx and k x 2 + kx ≥ (k + 1)x, then (x + 1 ) k + 1 ≥ (k + 1)x. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, (x + 1 ) n ≥ nx is true for all integers n ≥ 1 and for x ≥ 1.

43. S n = 2 n 2 ; Let P n be the statement 2 + 6 + 10 + 14 + � + (4n - 2) = 2 n 2 . Because 2 = 2(1 ) 2 is a true statement, P n is true for n = 1. Assume that 2 + 6 + 10 + 14 + � + (4k - 2) = 2 k 2 is true for a positive integer k. Show that P k + 1 must be true.



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns 2 + 6 + 10 + 14 + � + (4k - 2) = 2 k 2 2 + 6 + 10 + 14 + � + (4k - 2) + [4(k + 1) - 2] = 2 k 2 +

[4(k + 1) - 2]2 + 6 + 10 + 14 + � + (4k - 2) + (4k + 2) = 2 k 2 + (4k + 2)2 + 6 + 10 + 14 + � + (4k - 2) + (4k + 2) = 2( k 2 + 2k + 1)2 + 6 + 10 + 14 + � + (4k - 2) + (4k + 2) = 2(k + 1 ) 2 This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, 2 + 6 + 10 + 14 + � + (4n - 2) = 2 n 2 is true for all positive integers n.

45. S n = 1 _ 2 n

- 1; Let P n be -

1 _

2 + (-

1 _

4 ) + (-

1 _

8 ) + (-

1 _

16 ) + � +

(-

1 _

2 n ) = 1 _

2 n - 1. Because -

1 _

2 = 1 _

2 1 - 1 is a true statement, P n is

true for n = 1. Assume that -

1 _

2 + (-

1 _

4 ) + (-

1 _

8 ) + (-

1 _

16 ) + � +

(-

1 _

2 k ) = 1 _

2 k - 1 is true for a positive integer k. Show that P k + 1

must be true. -

1 _

2 + (-

1 _

4 ) + (-

1 _

8 ) + (-

1 _

16 ) + � +

(-

1 _

2 k ) =

1 _ 2 k

- 1

-

1 _

2 + (-

1 _

4 ) + (-

1 _

8 ) + (-

1 _

16 ) + � +

(-

1 _

2 k ) +

(-

1 _

2 k + 1 ) =

1 _ 2 k

- 1 - 1 _

2 k + 1

-

1 _

2 + (-

1 _

4 ) + (-

1 _

8 ) + (-

1 _

16 ) + � +

(-

1 _

2 k ) +

(-

1 _

2 k + 1 ) =

2 _ 2 ( 1 _ 2 k

) - 1 - 1

_ 2 k + 1

-

1 _

2 + (-

1 _

4 ) + (-

1 _

8 ) + (-

1 _

16 ) + � +

(-

1 _

2 k ) +

(-

1 _

2 k + 1 ) =

2 _

2 k + 1 - 1 - 1

_ 2 k + 1

-

1 _

2 + (-

1 _

4 ) + (-

1 _

8 ) + (-

1 _

16 ) + � +

(-

1 _

2 k ) +

(-

1 _

2 k + 1 ) =

1 _

2 k + 1 - 1


so on. That is, by the principle of mathematical induction, -

1 _

2 +

(-

1 _

4 ) + (-

1 _

8 ) + (-

1 _

16 ) + � + (-

1 _

2 n ) = 1 _

2 n - 1 is true for all

positive integers n.

47. Let P n be the statement that f 1 + f 2 + � + f n = f n + 2 - 1. Because f 1 = 1 and f 1 + 2 - 1 = f 3 - 1 = 2 - 1 or 1, then f 1 = f 1 + 2 - 1 is true. So, P n is true for n = 1. Assume that f 1 + f 2 + � + f k = f k + 2 - 1 is true for a positive integer k. Show that P k + 1 must be true.

f 1 + f 2 + � + f k = f k + 2 - 1

f 1 + f 2 + � + f k + f k + 1 = f k + 2 - 1 + f k + 1

f 1 + f 2 + � + f k + f k + 1 = f k + 3 - 1

This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. So, f 1 + f 2 + � + f n = f n + 2 - 1 is true for all positive integers n.

49. Let P n be the statement that if a polygon has n sides, the sum of the interior angles is 180(n - 2). P 3 is true, since when n = 3, the figure is a triangle and the sum of its interior angles is 180 and 180(3 - 2) = 180. Assume that if a polygon has k sides, the sum of the interior angles is 180(k - 2) for a positive integer k ≥ 3. Show that P k + 1 must be true. That is, show that if a polygon has k + 1

sides, the sum of the interior angles is 180(k + 1 - 2). Use the induction hypothesis and its restriction that k ≥ 3. Consider a convex polygon with k + 1 vertices. Because k ≥ 3, it follows that k + 1 ≥ 3. If we take one vertex x, there is another vertex y such that there is one vertex between x and y in one direction, and k - 2 vertices in the other direction. Join x and y by a new edge, dividing the original polygon into two polygons. The sum of the new polygons’ interior angle sums is the sum of the original polygon’s interior angles. One of the new polygons is a triangle and the other a polygon has k vertices (all but the one isolated between x and y). The triangle has interior angle sum of 180, and by the inductive hypothesis, the other polygon has an interior angle sum of (k - 2)180. Summing these, we get (k + 1 - 2)180, and the theorem is proved. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the extended principle of mathematical induction, if a polygon has n sides, the sum of the interior angles is 180(n - 2) is true for integer values of n ≥ 3.

51. Let P n be the statement ( x _ y

) n = x n _

y n . Because

( x _ y

) 1 = x 1

_ y 1

is a true

statement, P n is true for n = 1. Assume that ( x _ y

) k = x k _

y k is true for a

positive integer k. Show that P k + 1 must be true.

( x _ y

) k = x k _

y k

x _ y � ( x _ y

) k =

x _ y � x k _ y k

( x _ y

) k + 1

=

x k + 1 _

y k + 1

This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction,

( x _ y

) n

= x n _ y n


53. Let P n be the statement cos nπ = (-1 ) n for any positive integer n. Because cos π = (-1 ) 1 is a true statement, P n is true for n = 1. Assume that cos kπ = (-1 ) k is true for a positive integer k. Show that P k + 1 must be true. cos kπ = (-1 ) k cos kπ · cos π = (-1 ) k · cos πcos kπ cos π + (-sin kπ sin π + sin kπ sin π) = (-1 ) k · (-1) (cos kπ cos π - sin kπ sin π) + sin kπ sin π = (-1 ) k + 1 cos (kπ + π) + sin kπ sin π = (-1 ) k + 1 cos (kπ + π) + sin kπ · 0 = (-1 ) k + 1 cos (kπ + π) + 0 = (-1 ) k + 1 cos [(k + 1)π] = (-1 ) k + 1 This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, cos nπ = (-1 ) n is true for all positive integers n.55. Let S n = a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (n - 1)d] and P n be the statement a 1 + ( a 1 + d) + ( a 1 + 2d) + � +

[ a 1 + (n - 1)d] = n _ 2 [2 a 1 + (n - 1)d]. Because a 1 = 1 _

2 [2 a 1 +

(1 - 1)d] is a true statement, P n is true for n = 1. Assume

a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] = k _ 2 [2 a 1 +

(k - 1)d] is true for a positive integer k. Show that P k + 1 must be true. a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] =

k _ 2 [2 a 1 + (k - 1)d]

a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] + ( a 1 + kd) =

k _ 2 [2 a 1 + (k - 1)d] + ( a 1 + kd)



Selected A

nswers and S

olutions


a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] + ( a 1 + kd) =

k[2 a 1 + (k - 1)d] + 2( a 1 + kd)

___ 2

a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] + ( a 1 + kd) =

2 a 1 k + k(k - 1)d + 2 a 1 + 2kd

___ 2

a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] + ( a 1 + kd) =

2 a 1 k + 2 a 1 + [k(k - 1) + 2k]d

___ 2

a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] + ( a 1 + kd) =

2 a 1 (k + 1) + ( k 2 + k)d

__ 2

a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] + ( a 1 + kd) =

2 a 1 (k + 1) + (k + 1)kd

__ 2

a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (k - 1)d] + ( a 1 + kd) =

k + 1 _

2 (2 a 1 + kd)

a 1 + ( a 1 + d) + � + [ a 1 + (k - 1)d] + { a 1 + [(k + 1) - 1]d} =

k + 1 _

2 {2 a 1 + [(k + 1) - 1]d}

This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, a 1 + ( a 1 + d) + ( a 1 + 2d) + � + [ a 1 + (n - 1)d] =

n _ 2 [2 a 1 + (n - 1)d] is true for all positive integers n.

57 a. Sample answer: The conjecture: 2 n + 1 is divisible by 3 is false.

b. counterexample: n = 259. Let P n be the statement a n < 3. Because √ � 6 < 3 is a true statement, P n is true for n = 1. Assume that a k < 3 is true for a positive integer k. Show that P k + 1 must be true.

a k + 1 = √

�� 6 + a k Each successive term is the square root of 6 more

than the previous term.

( a k + 1 ) 2 = 6 + a k Square both sides.

( a k + 1 ) 2 - 6 = a k Solve for a k . a k < 3 Inductive hypothesis

( a k + 1 ) 2 - 6 < 3 Substitution

( a k + 1 ) 2 < 9 Add 6 to each side.

a k + 1 < 3 Take the square root of each side.

This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. That is, by the principle of mathematical induction, P n is true for all positive integers n.61. 78,125 63. 192 65. 5.83 67. 6.32

−4

−8

−4−8

8

4

4 8

(5, −3)

i

R

−4

−8

−4−8

8

4

4 8

(2, 6)i

R

69. (- √ � 3 , -1) 71. ⟨- √ � 2 , 0, √ � 2 ⟩

73. vertex: (4, -7); focus:

(4, -7 1 _ 4 ) ; axis of symmetry:

x = 4; directrix: y = -6 3 _ 4

y

x

−4

−8

−12

−4−8 4 8

(4, −7)

75. vertex: (0, 6); focus: (1, 6); axis of symmetry: y = 6; directrix: x = -1

y

x

8

12

16

4

4 8 12 16

2(y - 6) = 4x

77. C 79. B

Lesson 10-5

1. 16 + 32x + 24x 2 + 8 x 3 + x 4 3. 64 a 3 - 48 a 2 b + 12a b 2 - b 3 5. 2187 x 7 + 10,206 x 6 y + 20,412 x 5 y 2 + 22,680 x 4 y 3 +

15,120 x 3 y 4 + 6048 x 2 y 5 + 1344x y 6 + 128 y 7 7. 81 c 4 - 108 c 3 d + 54 c 2 d 2 - 12c d 3 + d 4 9. a 3 - 3 a 2 b + 3a b 2 - b 3 11. 3360 13. 262,440 15. 56 17. -126 19. 280 21. 1,088,640 23. 35

_ 27

25. 21,504 27a. 1820 27b. 12,870 29a. 0.12% 29b. 3.37% 29c. 15.38% 31. 1024 t 5 + 3840 t 4 + 5760 t 3 + 4320 t 2 + 1620t + 24333. 64 m 6 - 192 m 5 n + 240 m 4 n 2 - 160 m 3 n 3 + 60 m 2 n 4 - 12m n 5 + n 6 35. 2187 p 7 + 5103 p 6 q + 5103 p 5 q 2 + 2835 p 4 q 3 + 945 p 3 q 4 + 189 p 2 q 5 + 21p q 6 + q 7 37. 16,807 c 10 + 36,015 c 8 d + 30,870 c 6 d 2 + 13,230 c 4 d 3 + 2835 c 2 d 4 + 243 d 5

39. ∑

r = 0

15

( 15

r ) (2q) 15 - r (3) r 41. ∑

r = 0

31

( 31

r ) (11x) 31 - r (y) r

43. ∑

r = 0

22

( 22

r ) (3f ) 22 - r (-

3 _

4 g)

r

45 a. Each opening of a treasure chest represents a trial, so n = 15. You want to find how many different ways is it possible to open the chest and find gold coins exactly 9 times, so let x = 9. Find the value of the term n C x . n C x = 15 C 9

= 15! _

(15 - 9)!9!

= 5005b. A success in this situation is finding gold coins, so

p = 5 _ 6 and q = 1 - 5 _

6 or 1 _

6 . Each opening of a treasure chest

represents a trial, so n = 15. You want to find the probability of finding gold more than 12 times out of those 15 trials, so find the sum of x = 13, x = 14, and x = 15. To find these probabilities, find the value of the terms n C x p x q n - x in the expansions of (p + q ) n .

n C x p x q n - x = 15 C 13 ( 5 _ 6 )

13 ( 1 _

6 )

15 - 13

= 15! __

(15 - 13)!13! ( 5 _

6 )

13 ( 1 _

6 )

2

= 0.2726

C x p x q n - x = 15 C 14 ( 5 _ 6 )

14 ( 1 _

6 )

= 15! __

(15 - 14)!14! ( 5 _

6 )

14 ( 1 _

6 )

1

= 0.1947



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns C x p x q n - x = 15 C 15 ( 5 _

6 )

15 ( 1 _

6 )

15 - 15

= 15! __

(15 - 15)!15! ( 5 _

6 )

15 ( 1 _

6 )

0

= 0.0649

The probability of finding gold more than 12 times is 0.2726 + 0.1947 + 0.0649 or 0.5322 or 53.22%.

47. 16 d 4 + 32 √ � 5 d 3 + 120 d 2 + 40 √ � 5 d + 25 49. 1024 s 5 +

640 s 4 t + 160 s 3 t 2 + 20 s 2 t 3 + 5 _ 4 s t 4 + 1 _

32 t 5 51. 3150 53. 231

_ 512

55. -7 + 24i 57. 1121 - 404i 59. 5 + √ � 2 i

_ 8 61. g(x) = x 4 + 12x 3 +

54x 2 + 113x + 96 63. g(x) = x 6 - 6 x 5 + 15 x 4 - 18 x 3 + 9 x 2 - 1 65a. f(x) = x 3 : 3x 2 + 3xh + h 2 , h ≠ 0 f(x) = x 4 : 4 x 3 + 6 x 2 h + 4x h 2 + h 3 , h ≠ 0 f(x) = x 5 : 5 x 4 + 10 x 3 h + 10 x 2 h 2 + 5x h 3 + h 4 , h ≠ 0 f(x) = x 6 : 6 x 5 + 15 x 4 h + 20 x 3 h 2 + 15 x 2 h 3 + 6x h 4 + h 5 , h ≠ 0 f(x) = x 7 : 7 x 6 + 21 x 5 h + 35 x 4 h 2 + 35 x 3 h 3 + 21 x 2 h 4 + 7x h 5 + h 6 , h ≠ 0 f(x) = x n : n x n - 1 + n C 2 x n - 2 h + n C 3 x n - 3 h 2 + …

+ n C r x n - r h r - 1 + … + h n - 1 , h ≠ 0

65b.

f (x ) = x 3 f (x ) = x 4 f (x ) = x 5

h = 0.1

3 x 2 + 0.3x + 0.01 4 x 3 + 0.6 x 2 + 0.04x + 0.001

5 x 4 + 1 x 3 + 0.1 x 2 + 0.005x + 0.0001

h = 0.01

3 x 2 + 0.03x + 0.0001

4 x 3 + 0.06 x 2 + 0.0004x +

0.000001

5 x 4 + 0.1 x 3 + 0.001 x 2 + 0.000005x + (1 × 10 -8 )

h = 0.001

3 x 2 + 0.003x + 0.000001

4 x 3 + 0.006 x 2 + 0.000004x + (1 × 10 -9 )

5 x 4 + 0.01 x 3 + 0.00001 x 2 + (5 × 10 -9 )x + (1 × 10 -12 )

h = 0.0001

3 x 2 + 0.0003x + 0.00000001

4 x 3 + 0.0006 x 2 + 0.00000004x + (1 × 10 -12 )

5 x 4 + 0.001 x 3 + (1 × 10 -7 ) x 2 + (5 × 10 -12 )x + (1 × 10 -16 )

f (x ) = x 6 f (x ) = x 7

h = 0.1

6 x 5 + 1.5 x 4 + 0.2 x 3 + 0.015 x 2 + 0.0006x + 0.00001

7 x 6 + 2.1 x 5 + 0.35 x 4 + 0.035 x 3 + 0.0021 x 2 + (7 × 10 -5 )x + (1 × 10 -6 )

h = 0.01

6 x 5 + 0.15 x 4 + 0.002 x 3 + 0.000015 x 2 + (6 × 10 -8 )x + (1 × 10 -10 )

7 x 6 + 0.21 x 5 + 0.0035 x 4 + (3.5 × 10 -5 ) x 3 + (2.1 × 10 -7 ) x 2 + (7 × 10 -10 )x + (1 × 10 -12 )

h = 0.001

6 x 5 + 0.015 x 4 + 0.00002 x 3 + (1.5 × 10 -8 ) x 2 + (6 × 10 -12 )x + (1 × 10 -15 )

7 x 6 + 0.021 x 5 + 0.000035 x 4 + (3.5 × 10 -8 ) x 3 + (2.1 × 10 -11 ) x 2 + (7 × 10 -15 )x + (1 × 10 -18 )

h = 0.0001

6 x 5 + 0.0015 x 4 + 0.0000002 x 3 + (1.5 × 10 -11 ) x 2 + (6 × 10 -16 )x + (1 × 10 -20 )

7 x 6 + 0.0021 x 5 + 0.00000035 x 4 + (3.5 × 10 -11 ) x 3 + (2.1 × 10 -15 ) x 2 + (7 × 10 -20 )x + (1 × 10 -24 )

Sample answer: As h decreases, all the terms except the first in each expression get closer to 0.

65c. y

x

−0.4

−0.4−0.8

0.8

1.2

0.4

0.4 0.8

Sample answer: The graphs of the functions are almost exactly the same. 65d. n x n - 1

67. Sample answer: Chinese mathematician Chu Shih-chieh used the triangle in 1303. If the second number of any row is prime, all other numbers in the row (excluding the 1s) are divisible by the prime. The second diagonal of the triangle is the set of natural numbers. The third diagonal is the set of triangular numbers. 69. Sample answer: Treat x + y as a single term and expand [(x + y) + z] n ; (x + y) n + n (x + y) n - 1 z + n C 2 (x + y) n - 2 z 2 + n C 3

(x + y) n - 3 z 3 + … + n C r (x + y) n - r z r + … + z n . 71. Sample

answer: The first and last numbers in each row are 1. Every other number is formed by adding the two numbers just above that number in the previous row. The expansion of (a + b) n - 1 has n terms and the expansion of (a + b) n has n + 1 terms. The coefficients of the terms of the expansion of (a + b) n are all positive, whereas every other coefficient in the expansion of (a - b) n is negative.

73 Each term in the expansion of ( 1 _ 2v

+ 6 v 7 ) 8 will be of the form 8

C r ( 1 _ 2v

) 8 - r

(6 v 7 ) r . The term 8 C r ( 1 _ 2v

) 8 - r

(6 v 7 ) r can be written as 8

C r ( 1 _

2 8 - r v 8 - r ) ( 6 r v 7r ) or 8 C r ( 6 r v 7r _

2 8 - r v 8 - r ) . For v to be completely

divided out of the term, the exponent of the v in the numerator must be equal to the exponent of the v in the denominator. So, 7r = 8 - r or r = 1. Since r = 1, this will be the second term of the expansion.

8 C r ( 1 _ 2v

) 8 - r

(6 v 7 ) r = 8 C 1 ( 1 _ 2v

) 8 - 1

(6 v 7 ) 1

= 8 C 1 ( 1

_ 128 v 7

) (6 v 7 )

= 8 ( 6 v 7

_ 128 v 7

)

= 3 _ 8

75. Sample answer: n = 3 77. 65,534 79. circle: (x - 8) 2 +

(y + 5) 2 = 25 81. circle: (x + 1) 2 + (y + 12) 2 = 4

83. x + 2 _

x - 3 + 4

_ x - 5

85. no 87. yes 89. √ � 6 - √ � 2

91a. y

x

80

120

160

40

10 20

91b. y = 30(1.065 ) x 91c. ≈ $396.70 93. F 95. H



Selected A

nswers and S

olutions


Lesson 10-6

1. g(x) = 4 _

1 - x =

∑

n=0

∞

(

x + 3 _

4 )

n

for -7 < x < 1

[-7, 1] scl: 1 by [-5, 5] scl: 1

3. g(x) = 2 _

1 - x 2 =

∑

n=0

∞

(

x 2 + 1 _

2 )

n

for -1 < x < 1

[-1, 1] scl: 0.1 by [-5, 5] scl: 1

5. g(x) = 2 _

5 - 3x =

∑

n=0

∞

(

3x - 3 _

2 )

n

for 1 _ 3 < x < 5 _

3

scl: 1 by [-5, 5] scl: 1⎡

⎣

1_3

, 5_3

⎤

⎦

7. 1.648 9. 0.648 11. 0.741

13 Use the power series representation of e x to find the fifth

partial sum of P = 3.5 e 0.08t and evaluate the expression for t = 4, t = 12, and t = 52.

3.5e 0.08t ≈ 3.5 (1 + 0.08t + (0.08t) 2

_ 2!

+ (0.08t) 3

_ 3!

+ (0.08t) 4

_ 4!

)

3.5 e 0.08(4)

≈ 3.5 (1 + 0.08(4) + [0.08(4) ] 2

_ 2!

+ [0.08(4) ] 3

_ 3!

+ [0.08(4) ] 4

_ 4!

)

≈ 4.8 mussels/ m 2

3.5 e 0.08(12)

≈ 3.5 (1 + 0.08(12) + [0.08(12) ] 2

_ 2!

+ [0.08(12) ] 3

_ 3!

+ [0.08(12) ] 4

_ 4!

)


3.5 e 0.08(52)

≈ 3.5 (1 + 0.08(52) + [0.08(52) ] 2

_ 2!

+ [0.08(52) ] 3

_ 3!

+ [0.08(52) ] 4

_ 4!

)


15. 0.886 17. 0.588 19. 0.476 21. 2e i π

_ 6 23. 2e

i 7π

_ 4 25. 2 e

i 5π

_ 3

27. 2 e i 3π

_ 4 29. ≈1.792 + iπ 31. ≈0.896 + iπ 33. ≈1.472 + iπ

35a. y

x

−1

−2

−4

2

1

2−2 4

y = sin x

y = S 3(x)

y

x

−1

−2

−2−4

2

1

2 4

y = sin x

y = S 4(x)

Sample answer: Sample answer:convergence, (-1.5, 1.5) convergence, (-2.5, 2.5)

y

x

−1

−2

−2−4

2

1

2 4

y = sin x

y = S 5(x)

Sample answer: convergence, (-3.5, 3.5)

35b.

y

x

−1

−2

−2−4

2

1

2 4

y = cos x

y = S 3(x)y

x

−1

−2

−2−4

2

1

2 4

y = cos x

y = S 4(x)

Sample answer: Sample answer: convergence, (-1.5, 1.5) convergence, (-2.5, 2.5)

y

x

−1

−2

−2−4

2

1

2 4

y = cos x

y = S 5(x)

Sample answer: convergence, (-3, 3)

35c. Sample answer: The interval of convergence widens as n increases. As n approaches infinity, the power series equals the trigonometric function that it represents.

37. ≈1.243 + iπ

_ 2 39. ≈0.288 + iπ 41. ≈0.406 + iπ, ≈-1.609 + iπ

43. two terms: | e 2.1 - (1 + 2.1)|

__ e 2.1

= 0.62 or 62%;

three terms: ⎪ e 2.1 - (1 + 2.1 +

2.1 2 _

2 ) ⎥

__ e 2.1

= 0.35 or 35%;

six terms: ⎪ e 2.1 - (1 + 2.1 +

2.1 2 _

2 + 2.1 3

_ 6 + 2.1 4

_ 24

+ 2.1 5 _

120 ) ⎥

____ e 2.1

= 0.02 or 2%

45. sin 1 ≈ 1 -

1 3 _

3 ! + 1 5

_ 5 ! - 1 7

_ 7!

or 0.8415

cos 1 ≈ 1 -

1 2 _

2! + 1 4

_ 4!

- 1 6 _

6! or 0.5403

tan 1 = sin 1 _

cos 1 ≈ 0.8415

_ 0.5403

or 1.5575

sec 1 = 1 _

cos 1 = 1

_ 0.5403

or 1.8508

1.850 8 2 - 1.557 5 2 ≈ 0.9997The expected value is exactly 1 because sec 2 x - tan 2 x = 1.

47. ∑

n=0

∞

( 4kqx - x 4 + 2 x 2 - 1

__ 4kqx

) n

49 sin x = x - x 3 _

3! + x 5

_ 5!

- x 7 _

7! + …

sin (-x) = -x - (-x) 3

_ 3!

+ (-x) 5

_ 5!

- (-x) 7

_ 7!

+ …

= -x + x 3 _

3! - x 5

_ 5!

+ x 7 _

7! - …

= - (x -

x 3 _

3! + x 5

_ 5!

- x 7 _

7! + …)

= -sin x



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

51a. tan -1 x ≈ x - x 3 _

3 + x 5

_ 5 - x 7

_ 7 + x 9

_ 9 51b. 0.0997

51c. y

x

−1

−2

−2−4

2

1

2 4

y = tan -1x

y = S 3(x)

y

x

−1

−2

−2−4

2

1

2 4

y = tan -1x

y = S 4(x)

y

x

−1

−2

−2−4

2

1

2 4

y = tan -1x

y = S 5(x)

51d. As n increases, the graphs of the partial sums more closely resemble the graph of f(x) = tan -1 x on the interval (-1, 1). Outside of the interval (-1, 1), the end behavior of the polynomial approximations causes the graphs of the partial sums to diverge from the graph of f(x) = tan -1 x.

53. Sample answer: The power series representation for sine is

given by sin x ≈ x - x 3 _

3! + x 5

_ 5!

- x 7 _

7! + x 9

_ 9!

- …. For x-values on the

interval [-0.1, 0.1], the cubic term and those of higher degree have

values less than or equal to ⎪ 1 _

6000 ⎥ . This represents less than one-

tenth of 1 percent of the value. Therefore, for the values of x on the interval [-0.1, 0.1], the first term of the series, x, is a close approximation of sin x. 55. If α and β differ by an integer multiple of 2π, then the complex numbers represented by e iα and e iβ will be the same.

57. e ix + e -ix = (cos x + i sin x) + (cos x - i sin x) e ix + e -ix = 2 cos x

cos x = e ix + e -ix _

2

59. 81m 4 + 108 √ � 2 m 3 + 108 m 2 + 24 √ � 2 m + 4 61. p 16 + 8 p 14 q + 28

p 12 q 2 + 56 p 10 q 3 + 70 p 8 q 4 + 56 p 6 q 5 + 28 p 4 q 6 + 8 p 2 q 7 + q 8

63. 16 + 16i 65. -

1 _

4 i 67. (9, 2) 69. (7, 1, 3) 71. yes

73a. 54.41% 73b. Sample answer: women who participated in hiking or quilting but not sculpting 75. F


1. true 3. false; explicitly 5. false; recursively 7. false; common difference 9. false; diverge 11. Sample answer: 33, 41, 49, 57 13.

-20 15. 42 2 _ 3 17. a n = -6 + 5(n - 1); a 1 = -6, a n = a n - 1 + 5

19. 11,325 21. -1194 23. 1.5; 16.875, 25.3125, 37.96875 25. a n = 10 (-2) n – 1 ; a 1 = 10, a n = (–2) a n - 1 27. 133.1129. Let P n be the statement 6 n - 9 is divisible by 3. P 1 is the statement that 6 1 - 9 is divisible by 3. P 1 is true because 6 1 - 9 is -3, which is divisible by 3. Assume P k is true, where k is a positive integer, and show that P k + 1 must be true. That is, show that 6 k - 9 = 3r for some integer r implies that 6 k + 1 - 9 is divisible by 3.

6 k - 9 = 3r 6 k = 3r + 9 6 · 6 k = 6(3r + 9) 6 k + 1 = 18r + 54 6 k + 1 - 9 = 18r + 45 6 k + 1 - 9 = 3(6r + 15)Because r is an integer, 6r + 15 is an integer, and 3(6r + 15) is divisible by 3. Therefore, 6 k + 1 - 9 is divisible by 3. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 6 n - 9 is divisible by 3 for all positive integers n.31. Let P n be the statement 5 n + 3 is divisible by 4. P 1 is the statement that 5 1 + 3 is divisible by 4. P 1 is true because 5 1 + 3 is 8, which is divisible by 4. Assume P k is true, where k is a positive integer, and show that P k + 1 must be true. That is, show that 5 n + 3 = 4r for some integer r implies that 5 k + 1 + 3 is divisible by 4. 5 k + 3 = 4r 5 k = 4r - 3 5 · 5 k = 5(4r - 3) 5 k + 1 = 20r - 15 5 k + 1 + 3 = 20r - 12 5 k + 1 + 3 = 4(5r - 3)Because r is an integer, 5r - 3 is an integer, and 4(5r - 3) is divisible by 4. Therefore, 5 k + 1 + 3 is divisible by 4. Because P n is true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 5 n + 3 is divisible by 4 for all positive integers n.

33. Let P n be the statement 1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ 1 _

3 · 4 · 5 + …

+ 1 __

n(n + 1)(n + 2) =

n(n + 3) __

4(n + 1)(n + 2) . Because 1

__ (1 + 1)(1 + 2)

=

(1 + 3) __

4(1 + 1)(1 + 2) is a true statement, P n is true for n = 1. Assume that

1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ 1 _

3 · 4 · 5 + … + 1

__ k(k + 1)(k + 2)

= k(k + 3)

__ 4(k + 1)(k + 2)

is true for a positive integer k. Show that P k + 1 must be true.

1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ … + 1 __

k(k + 1)(k + 2) =

k(k + 3) __

4(k + 1)(k + 2)

1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ … + 1 __

k(k + 1)(k + 2) +

1 ___

(k + 1)[(k + 1) + 1][(k + 1) + 2] =

k(k + 3) __

4(k + 1)(k + 2) +

1 ___

(k + 1)[(k + 1) + 1][(k + 1) + 2]

1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ … + 1 __

k(k + 1)(k + 2) +

1 ___

(k + 1)[(k + 1) + 1][(k + 1) + 2] =

k(k + 3) __

4(k + 1)(k + 2) +

1 __

(k + 1)(k + 2)(k + 3)

1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ … + 1 __

k(k + 1)(k + 2) +

1 ___

(k + 1)[(k + 1) + 1][(k + 1) + 2] =

k(k + 3 ) 2 + 4 __

4(k + 1)(k + 2)(k + 3)

1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ … + 1 __

k(k + 1)(k + 2) +

1 ___

(k + 1)[(k + 1) + 1][(k + 1) + 2] =

k 3 + 6 k 2 + 9k + 4 __

4(k + 1)(k + 2)(k + 3)

1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ … + 1 __

k(k + 1)(k + 2) +

1 ___

(k + 1)[(k + 1) + 1][(k + 1) + 2] =

(k + 1 ) 2 + (k + 4) __

4(k + 1)(k + 2)(k + 3)



Selected A

nswers and S

olutions


1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ … + 1 __

k(k + 1)(k + 2) +

1 ___

(k + 1)[(k + 1) + 1][(k + 1) + 2] =

(k + 1)(k + 4) __

4(k + 2)(k + 3)

1 _

1 · 2 · 3 + 1

_ 2 · 3 · 4

+ … + 1 __

k(k + 1)(k + 2) +

1 ___

(k + 1)[(k + 1) + 1][(k + 1) + 2] =

(k + 1)[(k + 1) + 3] ___

4[(k + 1) + 1][(k + 1) + 2]


so on. That is, by the principle of mathematical induction, 1 _

1 · 2 · 3

+ 1 _

2 · 3 · 4 + 1

_ 3 · 4 · 5

+ … + 1 __

n(n + 1)(n + 2)

= n(n + 3)

__ 4(n + 1)(n + 2)


35. Let P n be the statement 5n < 6 n . P 1 is true, since 5(1) < 6 1 is a true statement. Assume P k is true, that 5k < 6 k if k is a positive integer, and show that P k + 1 must be true. That is, show that 5k < 6 k implies 5(k + 1) < 6 k + 1 . Use both parts of the inductive hypothesis.

5k < 6 k 6 � 5k < 6 � 6 k 30k < 6 k + 1

k ≥ 1 1 ≤ k 25 � 1 ≤ 25 � k 25 ≤ 25kIf 5 ≤ 25 and 25 ≤ 25k, then by the Transitive Property of Inequality, 5 ≤ 25k. 5 ≤ 25k 5k + 5 ≤ 25k + 5k5(k + 1) ≤ 30k

By the Transitive Property of Inequality, if 5(k + 1) ≤ 30k and 30k < 6 k + 1 , then 5(k + 1) < 6 k + 1 . Since P n is true for k ≥ 1 and P k implies P k + 1 for k ≥ 1, P n is true for n = 2, n = 3, and so on. By the principle of mathematical induction, 5n < 6 n is true for all positive integers n.

37. m 6 - 30 m 5 n + 375 m 4 n 2 - 2500 m 3 n 3 + 9375 m 2 n 4 - 18,750m n 5 + 15,625 n 6 39. 240,185,088 41. 16,384 m 7 + 86,016 m 6 n + 193,536 m 5 n 2 + 241,920 m 4 n 3 + 181,440 m 3 n 4 + 81,648 m 2 n 5 + 20,412m n 6 + 2187 n 7

43. ∑

n=0

∞

( 2x + 2

_ 3 )

n ;

[-5, 5] scl: 1 by [-2, 8] scl: 1

y = g(x)

y = S 5(x)- 5 _

2 < x < 1 _

2

45. 0.273 47. 1.967 + iπ 49a. 9 49b. 80

51. Let P n be the set of $10 and $25 bills that adds to $10n for n ≥ 6. For n = 6, the first possible case, the conjecture is valid, since $10(6) = $10(1) + $25(2). Assume that for n = k, there exists a set of $10 and/or $25 bills that adds to $10k. Show that P n is valid for n = k + 1. There are two cases to consider.

Case 1: P k contains at least three $25 bills. Replace two $25 bills in P k with six $10 bills and the value of P k is increased by $10, to $10k + 10 or $10(k + 1), which is P k + 1 .

Case 2: P k contains no $25 bills. P k must contain at least six $10 bills, since the value of P k must be greater than or equal to $60. Replace four of the $10 bills with two $25 bills and the value of P k is increased by $10, to $10k + 10 or $10(k + 1), which is P k + 1 .

In both cases, P n is valid for n = k + 1. Since P n is valid for n = 6 and for n = k + 1, it is valid for n = 7, n = 8, and so on. That is, by the extended principle of mathematical induction, all multiples of $5 greater than $60 can be formed using just $10 and $25 bills. 53. 31.173 ft


1. Sample answer: Δx is decreasing. As Δx becomes smaller, the rectangles will better approximate the area of the region. 3. Sample answer: Δx approaches 0. 5. Sample answer: Δx approaches 0, the number of terms approaches ∞, and the area approaches 16.5.

7. ∑

i=1

10

f( x i ) · 0.4; 63.36 units 2

Limits and Derivatives1CHAPTER 111CHCHAPAPTETERR 1111

Chapter 11 Get Ready

1. From the graph, it appears that f (x) → 0 as x → -∞ and f (x) → 0 as x → ∞. 3. From the graph, it appears that f (x) → 1 as x → -∞ and f (x) → 1 as x → ∞. 5. 1200 7. -17 9. $290 11. D = {x | x ≠ 10, x ∈ �}; x = 10 13. D = {x | x ≠ 2, x ≠ -4, x ∈ �}; x = 2, y = 1 15. -12, -17, -22, -27 17. -324, 972, -2916, 8748 19. 0, 7, 14, 21

Lesson 11-1

1. 10

[-2, 8] scl: 1 by [-5, 15] scl: 2

x 4.99 4.999 5 5.001 5.01

f (x ) 9.96 9.996 10.004 10.04

3. -15

[-8, 2] scl: 1 by [-19, 1] scl: 2

x -2.01 -2.001 -2 -1.999 -1.99

f (x ) -14.98 -14.998 -15.002 -15.02

5. 25

[-5, 5] scl: 1 by [-10, 30] scl: 5

x 2.99 2.999 3 3.001 3.01

f (x ) 24.56 24.956 25.044 25.44



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns7. 0

[-5, 5] scl: 1 by [-5, 15] scl: 2

x -0.01 -0.001 0 0.001 0.01

f (x ) -0.0002 -0.000002 -0.000002 -0.0002

9. 5.72

[-2, 8] scl: 1 by [-2, 8] scl: 1

x 5.99 5.999 6 6.001 6.01

f (x ) 5.70 5.719 5.723 5.74

11. 0 13. 0 15. 1 17. does not exist 19. -7 21. 7 23. does not exist 25. 0 27. does not exist 29. 4 31. ∞

33. -∞ 35. does not exist 37. ∞ 39. 0 41. 1 _ 3 43. -1

45. 0 47a. lim w→1 f (w) = 250; lim w→3

f (w) = 100 47b. 0; Sample answer:

Eventually, the vaccine will eliminate all cases of the infection.

49 a. Using a graphing calculator, graph

Y 1 = 12(1.25012 ) X - 12 in the window [0, 20] × [0, 1100].

[0, 20] scl: 2 by [0, 1100] scl: 100

b. Use the TRACE feature. Move the cursor until X is close to 5, 10, and 20. The y-values near these values are about 25, 100, and 1031. Therefore, the number of people who have viewed the video after 60 days is p(60) = 12(1.25012 ) 60 or about 7,880,000.

c. ∞; Sample answer: The graph appears to increase without bound, which suggests that an infinite amount of people may view the video.

51 a. Using a graphing calculator, graph

Y 1 = 2000(0.7 ) (X - 1) in the window [1, 15] × [0, 2000].

[1, 15] scl: 1 by [0, 2000] scl: 200

b. Use the TRACE feature. Move the cursor until X is close to 5, 10, and 15. The y-values near these values are about 480.2, 80.71, and 13.56.

c. The graph approaches the x-axis, so lim t→∞

= 0.

d. No; the sum of the infinite series is approximately 6666.67 meters, which is less than the 7000 meters required to reach the hospital.

53. -1 55. does not exist 57. 6 59. no; vertical asymptote 61. no; oscillations 63. no; vertical asymptote 65. Neither;

sample answer: If f (x) approaches a different value from the left than from the right, then the limit at that point does not exist.

67. lim x→1 f (x) does not exist; lim x→1

g(x) does not exist; lim x→a f (x)

_ g (x)

does not exist; sample answer: If the denominator of a rational function is equal to zero for a given point, the limit at this point does not exist.

69. Sample answer: 71. Sample answer: If f (x) is y

−2

−2−4

4

2

2 4 x

continuous at x = a, you can substitute a into the function. If the function is not continuous, you may be able to simplify it and then substitute in a. If neither of these methods works, you must estimate the limit graphically.

73. H 0 : μ = 4 (claim), H a : μ ≠ 4 75. H 0 : μ ≥ 10, H a : μ < 10 (claim)

77. r = 4 _

1 - sin θ 79. 63.0°

0

π

2

84

r = 4__1 - sin θ

x

y 81.

[-10, 10] scl: 1 by [-10, 10] scl: 1

83. A 85. C

Lesson 11-2

1. -25 3. 10 5. 21 1 _ 9 7. 6 9. 20 11. Not possible; when x = 16,

the denominator is 0. 13. 2 15. -9216 17. -62 1 _ 2

19. Not possible; when x = 5, the denominator is 0. 21a. lim v→0

m = m 0 . As the velocity of the object approaches 0, the

mass of the object will approach its initial, or rest mass. 21b. The mass of the object would begin to increase without bound.

23. 11 25. 3 27. 1 6 _ 13

29. 4 31. -12 33. 1 _ 6 35. ∞

37. 0 39. 1 _ 3 41. ∞ 43. -∞ 45. 2

47 a. The length of the capsule before it is submerged is its

length at t = 0.

�(0) = 105(0 ) 2

_ 10 + 0 2

+ 25 = 25 mm

b. lim t→∞

105 t 2 _

10 + t 2 + 25 = lim t→∞

105 t 2

_ t 2

_ 10

_ t 2

+ t 2 _

t 2 + 25

= lim t→∞

105 _

10 _

t 2 + 1

+ 25

= lim t→∞

105 __

lim t→∞

10 _

t 2 + lim t→∞

1 + lim t→∞

25

= 105 _

0 + 1 + 25

= 130

c. The sponge animal will not grow more than 130 mm in length.

49. ∞ 51. -4 53. 2 55. 5 _ 2 57. 1 _

4



Selected A

nswers and S

olutions


59a. Years Since 2006 Rise in

Population

1 398

2 2430

3 5550

59b. approximately 9576 people 59c. 12,000 people

59d. Sample answer: A city’s borders may limit the amount of possible growth and building opportunities. 61. 2 63. No limit exists. 65. 1 67. 4.5 69. -

1 _

2 71. 2

73 lim h→0

f (x + h) - f (x) __

h = lim h→0

√ �� x + h - √

� x __

h

= lim h→0

√ �� x + h - √ �

x __

h ·

√ �� x + h + √ �

x __

√ �� x + h + √ �

x

= lim h→0

(x + h) - x __

h ( √ �� x + h + √ � x )

= lim h→0 h __

h ( √ �� x + h + √ � x )

= lim h→0 1

__

√ �� x + h + √ � x

= 1 __

lim h→0 √ �� x + h + lim h→0

√ � x

= 1 _

2 √ �

x or

√ �

x _

2x

75. 2x 77. 0.0000125

79. Let P n be the statement if lim x→c f (x) = L, then lim x→c

⎡ ⎣

f(x) ⎤ ⎦

n =

⎡ ⎣

lim x→c f (x)

⎤ ⎦

n or L n for any integer n.

Because lim x→c ⎡ ⎣

f (x) ⎤ ⎦

1 = L 1 is a true statement, P 1 is true. Assume that lim x→c

⎡ ⎣

f (x) ⎤ ⎦

k = L k is true for an integer k. Show that P k + 1 must be

true.

lim x→c ⎡ ⎣

f (x) ⎤ ⎦

k = L k

lim x→c ⎡ ⎣

f(x) ⎤ ⎦

k · lim x→c ⎡ ⎣

f (x) ⎤ ⎦

1 = L k · lim x→c ⎡ ⎣

f (x) ⎤ ⎦

1

lim x→c ⎡ ⎣

f (x) ⎤ ⎦

k + 1 = L k · L

lim x→c ⎡ ⎣

f (x) ⎤ ⎦

k + 1 = L k + 1

This final statement is exactly P k + 1 , so P k + 1 is true. Because P n is

true for n = 1 and P k implies P k + 1 , P n is true for n = 2, n = 3, and

so on. That is, by the principle of mathematical induction, if lim x→c

f (x) = L, then lim x→c ⎡ ⎣

f (x) ⎤ ⎦

n =

⎡ ⎣

lim x→c f (x)

⎤ ⎦

n or L n for any integer n.

81. Sample answer: Sometimes; if the limit in question is not at a vertical asymptote, then it is true; if the limit in question is at a vertical asymptote, then it is not true. 83. Sample answer: A limit

of ∞

_ ∞

does not translate to 1 because infinity is not a real number; it is more of a concept. Further analyze this problem by graphing the original rational function and observing the behavior of the graph around the limit. 85. 0; undefined

87a.

[0, 105] scl: 10 by [40, 100] scl: 5

The data appear to have a positive linear correlation.

87b. r ≈ 0.975; The correlation coefficient indicates that the data have a strong positive linear correlation. Since t ≈ 2.306 and 12.41

> 2.306, the statistic falls within the critical region and the null hypothesis is rejected. Therefore, the correlation is significant at the 5% level. 87c. y = 0.295x + 49.927; A slope a = 0.295 indicates that for each additional year, the life expectancy will increase by a rate of 0.295 per year. The y-intercept b = 49.927 indicates that in 1900, the average life expectancy was approximately 50 years.

87d. Using this model, the average life expectancy in 2080 will be about 103 years, which is probably not reasonable.

89. x 2 _

4 =

y 2 _

25 = 1 y

x

−4

−8

−4−8

8

4

4 8

x 2_4

+ y2_

25= 1

91. H 93. G

Lesson 11-3

1. -3; 5 3. 6; 12 5. 12; 3 7. m = -2 9. m = 2x

11. m = -2x 13. m = -6 x 2 15. m = -

√ �

x _

2 x 2

17a. m = 0.18 x 2 - 2.16x 17b. -3.6, -6.3, -6.3 19. 64.8 mi/h21. 49.2 mi/h 23. 36 mi/h 25. -96 ft/s 27. -95 ft/s 29. -512 ft/s 31. -121.6 ft/s 33. v(t) = 28t 35. v(t) = 5

37. v(t) = 3 t 2 − 2t + 1 39. v(t) = √ � t

_ 2t

- 6t

41 a. Determine the height of the skydiver at times t = 2 and t = 5.

h(2) = 15,000 - 16 · 2 2 = 14,936 h(5) = 15,000 - 16 · 5 2 = 14,600 The average velocity between the 2 nd and

5 th seconds is

h(5) - h(2)

_ 5 - 2

= 14600 - 14936 __

3 = -112 ft/s

b. Calculate v(2) and v(5).

v(2) = lim h→0 h(2 + h) - h(2)

__ h

= lim h→0 15000 - 16(2 + h ) 2 - ⎡

⎣

15000 - 16 · 2 2 ⎤ ⎦

____ h

= lim h→0 15000 - 16(4 + 4h + h 2 ) -15000 + 64

___ h

= lim h→0 -64 - 64h - 16 h 2 + 64

__ h

= lim h→0 h(-64 -16h)

__ h

= lim h→0 (-64 - 16h)

= -64

v(5) = lim h→0 h(5 + h) - h(5)

__ h

= lim h→0 15000 - 16(5 + h) 2 - ⎡

⎣

15000 - 16 · 5 2 ⎤ ⎦

____ h

= lim h→0 15000 - 16(25 + 10h + h 2 ) -15000 + 400

____ h

= lim h→0 -400 - 160h - 16 h 2 + 400

___ h

= lim h→0 h(-160 -16h)

__ h



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

= lim h→0 (-160 - 16h)

= -160 The velocities at times t = 2 and t = 5 are -64 ft/s and

-160 ft/s.

c. v(t) = lim h→0 h(t + h) - h(t)

__ h

= lim h→0 15000 - 16(t + h ) 2 -

⎡ ⎣

15000 - 16 t 2 ⎤ ⎦

___ h

= lim h→0 15000 - 16( t 2 + 2th + h 2 ) - 15000 + 16 t 2

____ h

= lim h→0 -16 t 2 - 32th - 16 h 2 + 16 t 2

___ h

= lim h→0 h(-32t - 16h)

__ h

= lim h→0 (-32t - 16h)

= -32t43a. v(t) = -32t + 75 43b. 59 ft/s 43c. t ≈2.344 seconds 43d. ≈90.39 ft 45. y = -4x + 1

[10, -10] scl: 1 by [10, -10] scl: 1

47 Find an equation for the slope of g(x) and determine the x-value for which it equals the slope of the line perpendicular to y = -

1 _

6 x + 9, namely m = 6.

m = lim h→0 g(x + h) - g(x)

__ h

= lim h→0 1 _ 2 (x + h ) 2 + 4(x + h) - (

1 _

2 x 2 + 4x)

___ h

= lim h→0 xh + 1 _

2 h 2 + 4h

__ h

= lim h→0 (x + 1 _

2 h + 4)

= x + 1 _ 2 (0) + 4 or x + 4

Set this expression equal to the desired slope of 6 and solve. x + 4 = 6 x = 2 The y-value for g(x) at x = 2 is g(2) = 1 _

2 · 2 2 + 4 · 2 or 10. Use

the point-slope formula to find the equation of the line with slope m = 6 that passes through (2, 10).

y - y 1 = m(x - x 1 ) y - 10 = 6(x - 2) y - 10 = 6x - 12 y = 6x - 2

[10, -10] scl: 1 by [10, -10] scl: 1

49. c 51. b 53a. -40 ft/s 53b. -136 ft/s 53c. -232 ft/s 55. Jillian; sample answer: The graph of f (x) has a slope of -1 when x < 0 and a slope of 1 when x > 0. Therefore, the graph of this equation would be two horizontal lines

y =

⎧

⎨

⎩

-1 if x < 0

1 if x > 0 and would not be continuous.

57. False; sample answer: If a graph is not circular, a tangent line can intersect a graph at more than one point. For example, the graph of y = sin x. 63a. The average growth of the investment for the first 4 years is about $41.20 per year. 63b. After exactly 4 years, the investment is growing at a rate of about $42.90 per year. 61. 1

63a.

0.04

0.08

0.12

0.16

00.2 0.4 0.6 0.8 r

v (r)v (r) = 0.65(0.5 2 - r 2)

63b. 0 in./s 65a. 72

65b. 58 65c. 68–71

67. 729 69. 891

71. 23.3, 29.4, 35.5, 41.6

73. -29, -13, 3, 19, 35, 51, 67, 83, 99 75. G

77. G

Lesson 12-4

1. f ′(x) = 8x; f ′(2) = 16, f ′(-1) = -8 3. m′(j) = 14; m′(-7) = 14, m′(-4) = 14 5. f ′(c) = 3 c 2 + 4c - 1; f ′(-2) = 3, f ′(1) = 6

7. y′(f ) = -11 9. p′(v) = 7 11. b′(m) = 2 m -

1 _

3 - 3 m

1 _ 2

13. f ′(x) =

3 _

2 x

-

1 _

2 - 3 _

2 x

1 _ 2 - x

-

3 _

2 15. p′(k) = 5.2 k 4.2 - 38.4 k 3.8 + 3

17 a. Find f ′(h).

f ′(h) = -0.0036 · 3 h 2 - 0.01 · 2h + 2.04 = -0.0108 h 2 - 0.02h + 2.04

b. Evaluate f ′(h) at h = 2, h = 14, and h = 20.

f ′(2) = -0.0108 · 2 2 - 0.02 · 2 + 2.04 ≈ 1.96 deg/hr

f ′(14) = -0.0108 · 1 4 2 - 0.02 · 14 + 2.04 ≈ -0.36 deg/hr

f ′(20) = -0.0108 · 2 0 2 - 0.02 · 20 + 2.04 ≈ -2.68 deg/hr

c. Determine when f ′(h) = 0.

−0.0108 h 2 - 0.02h + 2.04 = 0

h = -(-0.02) ± √

(-0.02 ) 2 - 4(-0.0108)(2.04) ____

2(-0.0108)

h ≈ −14.70 or h ≈ 12.85

Since 0 ≤ h ≤ 24, choose only the positive value. Evaluate f at h = 0, h = 12.85 and h = 24. f (0) = -0.0036 · 0 3 − 0.01 · 0 2 + 2.04 · 0 + 52 = 52 f (12.85) = -0.0036 · 12.8 5 3 - 0.01 · 12.8 5 2 + 2.04 · 12.85 + 52 ≈ 68.92 f (24) = −0.0036 · 2 4 3 - 0.01 · 2 4 2 + 2.04 · 24 + 52 ≈ 43.43 Thus, the maximum temperature is 68.92°F.

19. critical points: ( 2 √ 3

_ 3 , 6.92) and (-

2 √ 3 _

3 , 13.08) ; max: 25, min: -5

21. critical point: (-5, -10); max: -2, min: -11 23. critical point: (-9, 405); max: 405, min: 385 25. critical points: (0, 2) and (2.25, -6.54); max: 66, min: -6.54



Selected A

nswers and S

olutions


27a. h′(t) = 65 - 32t 27b. (2.03, 71.02); (0, 5) 27c. Yes; the maximum height that Gabe can throw the ball is approximately 71 ft. This is more than the 70 ft needed to reach Zach’s window. 29. g′(x) = -45 x 4 + 60 x 3 - 12x + 10

31. s′(t) = 69 _

2 t

21 _

2 + 66 t 10 - 6 t

1 _ 2 - 8 33. c′(t) = 13 t 12 - 33 t 10 + 9 t 8 -

132 t 7 + 35 t 6 + 30 t 4 - 88t 35. q′(a) = 19 _

8 a

11 _

8 - 221

_ 8 a

9 _ 8 +

a - 39 _

4 a

-

1 _

4 37. h ′(x) =

19 _

48 x

13 _

6 + 37

_ 192

x 13

_ 24

+ 14

_ 15

x 4 _ 3 + 17

_ 60

x -

7 _

24

39. f ′(m) = -12 _

(3 + 2m ) 2 41. r′(t) = 10t _

(3 - t 2 ) 2

43. v′(t) = - t 4 + 10 t 3 - 13 t 2 + 12

__ ( t 3 - 4t ) 2

45. f ′(x) = - x 4 + 11 x 2 + 6

__ (- x 2 + 3 ) 2

47. t′(w) =

2 w 3 - 1 _

w 2

49a. r′(x) = 0.375 x 2 - 22.5x + 250 49b. 14.72, 45.28 49c. Sample answer: The solution of 14.72 represents the price that Nayla and Deirdre should charge per sweatshirt to make the maximum possible revenue. The solution of 45.28 is not relevant because the revenue is 0 when x = 40.

51. y = 10x + 45

x−2−4

10

20

30

2 4

y

f (x) = -5x2 - 10x + 25

y = 10x + 45

53. y = -21.6x - 40.76 x

−24

−12

−36

−2−4 2 4

y

y = -21.6x - 40.76

f (x) = 4x2 - 12x - 35

55a. f ′′(x) = 80 x 3 - 12x 55b. g′′′(x) = -420 x 4 + 96x - 42 55c. h (4) (x) = 1080 x −7 + 240 x -6 57. Sample answer: 59. Sample answer:

y

x

y

x

−4

−8

−4−8

8

4

4 8

61a. A′ = 2πr; V′ = 4π r 2 61b. Sample answer: The derivative of the formula for the area of a circle is the formula for the circumference of a circle. The derivative of the formula for the volume of a sphere is the formula for the surface area of a sphere.

61c.

a a

61d. A = 4 a 2 ; A′ = 8a; V = 8 a 3 ; V′ = 24 a 2 61e. Sample answer: When the area of a square is written in terms of its apothem, the derivative is the formula for the perimeter of the square. When the volume of a cube is written in terms of the apothems of its faces, the derivative is the formula for the surface area of the cube.

63 Treat x and z as constants.

f ′(y) = 10 x 2 · 3 y 2 + 0 - 6x · 2y + 0 - 11 x 8 · 1 · z 7 = 30 x 2 y 2 − 12xy - 11 x 8 z 7 65. True; sample answer: The power for f (x) is 5n + 3. Due to the Power Rule, this will become the coefficient of the derivative. The power of the derivative will be one less than the original power. It will then be (5n + 3) - 1 or 5n + 2. 67. Sample answer:

f ′(x)g (x) - f (x)g′(x)

__ [g(x) ] 2

= lim h→0 f (x + h)g(x) - f (x)g(x + h)

___ hg(x + h)g(x)

= lim h→0 f (x + h)g(x) - f (x)g(x) + f (x)g(x) - f (x)g(x + h)

____ hg(x + h)g(x)

= lim h→0 [f (x + h) - f (x)]g(x) - [g(x + h) - g(x)]f (x)

____ hg(x + h)g(x)

= lim h→0 f(x + h) - f(x)

__ h g(x) -

g(x + h) - g(x) __

h f(x)

___ g(x + h)g(x)

=

g(x) lim h→0 f (x + h) - f (x)

__ h - f(x) lim h→0

g(x + h) - g(x)

__ h ____

g(x) lim h→0 g(x + h)

= f ′(x)g(x) - f (x)g′(x)

__ [g(x) ] 2

69. -3; 3 71. 6; 12 73. -

1 _

3

75. Gym Class Exercise Amountfor a Week

Gym Class Exercise Amountfor a Week

Days of Exercise

Prob

abili

ty

0.15

0.10

0.05

0

0.20

0.25

0.30

0 1 2 3 4 5 X

P (X)

Days, X 0 1 2 3 4 5

P(x ) 0.10 0.20 0.23 0.27 0.13 0.07

77. 0.841 79. 0.239 81. a n = 1.4(-2.5 ) n - 1 ; a 1 = 1.4, a n = -2.5 a n - 1 83. D 85. B

Lesson 11-5

1. 15 unit s 2 3. 18.29 unit s 2 5. 16.23 unit s 2

7 a. The graph of the semicircle is on the interval [1, 10]. Form rectangles with left endpoints and a width of 1.

R 1 = 1 · [−(1 ) 2 + 10(1) ] 0.5 = 3

R 2 = 1 · [−(2 ) 2 + 10(2) ] 0.5 = 4

R 3 = 1 · [−(3 ) 2 + 10(3) ] 0.5 ≈ 4.58



Sel

ecte

d A

nsw

ers

and

Sol

utio

ns

R 4 = 1 · [−(4 ) 2 + 10(4) ] 0.5 ≈ 4.90 R 5 = 1 · [−(5 ) 2 + 10(5) ] 0.5 = 5

R 6 = 1 · [−(6 ) 2 + 10(6) ] 0.5 ≈ 4.90

R 7 = 1 · [−(7 ) 2 + 10(7) ] 0.5 ≈ 4.58

R 8 = 1 · [−(8 ) 2 + 10(8) ] 0.5 = 4

R 9 = 1 · [−(9 ) 2 + 10(9) ] 0.5 = 3 Total area ≈ 37.96

b. For this section, some rectangles use right endpoints and some use left endpoints.

R 1 = 1 · [−(1 ) 2 + 10(1) ] 0.5 = 3

R 2 = 1 · [−(2 ) 2 + 10(2) ] 0.5 = 4

R 3 = 1 · [−(3 ) 2 + 10(3) ] 0.5 ≈ 4.58

R 4 = 1 · [−(4 ) 2 + 10(4) ] 0.5 ≈ 4.90

R 5 = 1 · [−(4 ) 2 + 10(4) ] 0.5 ≈ 4.90

R 6 = 1 · [−(7 ) 2 + 10(7) ] 0.5 ≈ 4.58

R 7 = 1 · [−(8 ) 2 + 10(8) ] 0.5 = 4

R 8 = 1 · [−(9 ) 2 + 10(9) ] 0.5 = 3 Total area ≈ 32.96

c. The radius is 5.

Area = 1 _ 2 π r 2

= 1 _ 2 π 5 2

= 12.5π

≈ 39.27The first estimate is closer. Sample answer: The additional area outside of the semicircle that is included in the first estimate helps account for the area in the region that is not being included by the rectangles9. right endpoints: 12.6 unit s 2 ; left endpoints: 9.4 unit s 2 ; average: 11 unit s 2 11. right endpoints: 18.91 unit s 2 ; left endpoints: 19.66 unit s 2 ; average: 19.285 unit s 2 13. right endpoints: 12.75 unit s 2 ; left endpoints: 12.25 unit s 2 ;

average: 12.5 unit s 2 15. 52 unit s 2 17. 23 1 _ 3 unit s 2

19. 30 unit s 2 21. 8 2 _ 3 unit s 2 23. 12 unit s 2

25. 48 unit s 2 27. 80 5 _ 6 unit s 2 29. 67.2 f t 2

31a. height: 4 units, base: 4 units; 8 unit s 2 31b. 8 unit s 2

33 Find Δx and x i .

Δx = b - a _ n = 0-(-1)

_ n = 1 _ n

x i = a + iΔx = −1 + i _ n .Calculate the definite integral that gives the area.

∫ -1

0

( x 3 + 2)dx = lim n→∞

∑

i=1

n

f( x i )Δx

= lim n→∞

∑

i=1

n

⎡

⎢

⎣

(-1 + i _ n ) 3 + 2

⎤

⎦

1 _ n

= lim n→∞

1 _ n ∑

i=1

n

⎡

⎢

⎣

-1 + 3 ( i _ n ) - 3 ( i _ n ) 2 + ( i _ n )

3 + 2

⎤

⎦

= lim n→∞

1 _ n ∑

i=1

n

⎡

⎢

⎣

1 + 3i _ n - 3 i 2 _

n 2 + i

3 _

n 3 ⎤

⎦

= lim n→∞

1 _ n ⎡

⎢

⎣

∑

i=1

n

1 + ∑

i=1

n

3i _ n - ∑

i=1

n

3 i 2 _

n 2 + ∑

i=1

n

i 3 _

n 3 ⎤

⎦

= lim n→∞

1 _ n ⎡

⎢

⎣

∑

i=1

n

1 +

3 _ n ∑

i=1

n

i -

3 _

n 2 ∑

i=1

n

i 2 +

1 _

n 3 ∑

i=1

n

i 3 ⎤

⎦

= lim n→∞

1 _ n (n +

3 _ n

n(n + 1) _

2 - 3 _

n 2 n(n + 1)(2n + 1)

__ 6 + 1 _

n 3 n 2 (n + 1 ) 2

_ 4 )

= lim n→∞

( n _ n +

3n(n + 1) _

2 n 2 -

3n(2 n 2 + 3n + 1) __

6 n 3 +

n 2 ( n 2 + 2n + 1) __

4 n 4 )

= lim n→∞

(1 +

3(n + 1) _

2n -

2 n 2 + 3n + 1 __

2 n 2 +

n 2 + 2n + 1 _

4 n 2 )

= lim n→∞

(1 + 3 ( 1 _

2 + 1 _

2n ) -

(1 + 3 _

2n + 1

_ 2 n 2

) +

( 1 _ 4 + 1 _

2n + 1

_ 4 n 2

) )

= lim n→∞

(1) + 3 lim n→∞

(

1 _

2 + 1 _

2n ) - lim n→∞

(1 + 3 _

2n + 1

_ 2 n 2

) +

lim n→∞

( 1 _ 4 + 1 _

2n + 1

_ 4 n 2

)

= 1 + 3 ( 1 _ 2 + 0) - (1 + 0 + 0) + ( 1 _

4 + 0 + 0) = 1 3 _

4 or 1.75

35. 12 1 _ 2 unit s 2 37. 3 _

4 unit s 2 39. 4 unit s 2 41. 3 3 _

4 unit s 2

43. Neither; sample answer: If the function is increasing, then using the right endpoints will cause the area of the rectangles to be greater than the actual area while using the left endpoints causes the area of the rectangles to be less. But, if the function is decreasing, then using the left endpoints will cause the area of the rectangles to be greater than the actual area while using the right endpoints causes the area of the rectangles to be less. 45. Sample answer: The integral gives the area of each cross section. Multiplying this area by the total length of the tunnel would compute the volume of the tunnel.

47. t 3 _

3 + 2t unit s 2 49. j′(x) = 44 x 10 + 198 x 8 - 120 x 4 - 396 x 2

51. s′(t) = 51 _

2 t

15 _

2 - 168 t 7 - 5 _

2 t

1 _ 2 + 35 53. -7 55. 3

57. 2 _ 9 59. E = 3.16; 33.34 < μ < 39.66 61. Discrete; the number of

mobile phone calls is countable and therefore discrete. 63. D 65. B

Lesson 11-6

1. F(x) = 1 _ 6 x 6 + C 3. F(z) = 3 _

4 z

4 _ 3 + C 5. Q(r) = 15

_ 28

r 7 _ 5 +

15 _

32 r

4 _ 3 + 2 _

3 r

3 _ 2 + C 7. G(a) = 2 a 4 + 5 _

3 a 3 − 9 _

2 a 2 + 3a + C

9. M(t) = 4 t 4 − 4 t 3 + 10 t 2 − 11t + C 11a. s(t) = -16 t 2 + C

11b. 64 11c. 28 ft 13. 5 n 4 − 3 n 3 − 9 n 2 + 4n + C 15. 46.5

17. 22.37 19. 18.93 21. 2 w 7.1 - 3 w 6.7 + 4 w 3.3 + 3w + C

23 a. Find the antiderivative of v(t) to find s(t).

s(t) = −32 t 1 + 1

_ 1 + 1

+ 34t + C = −16 t 2 + 34t + C

Substitute t = 0 and s(t) = 0 and solve for C.

0 = −16 · 0 2 + 34 · 0 + C 0 = C So, s(t) = -16 t 2 + 34t.

b. Solve s(t) = 0 to determine when the flea lands. −16 t 2 + 34t = 0 (-16t + 34)t = 0 -16t + 34 = 0 or t = 0

Since t = 0 is when the flea jumps, solve −16t + 34 = 0.

−16t = −34. So, t = -34 _

-16 or 17

_ 8 or 2.125 seconds.



Selected A

nswers and S

olutions


25a.

[0, 12] scl: 1 by [0, 24] scl: 2

25b. s 1 (t) = 3.25 t 2 _

2 - 0.2 t 3

_ 3 ;

s 2 (t) = 1.2 t 2 _

2 + 0.03 t 3

_ 3

25c. 10.34 seconds for strategy one and 11.80 seconds for strategy two 27. 27 29. 5.33 31. 16.4 33. - x 3 - 4 x 2 + 24 35. 16 x 4 + 20 x 2 + 4x - 132

37 ∫ x x 2

(16 t 3 - 15 t 2 + 7) dt

= 16 t 3 + 1

_ 3 + 1

- 15 t 2 + 1

_ 2 + 1

+ 7t �

�

�

x

x 2

= 4 t 4 - 5 t 3 + 7t �

�

�

x

x 2

= ⎡

⎣

4 ( x 2 ) 4 - 5 ( x 2 )

3 + 7 x 2

⎤

⎦

- ⎡

⎣

4 x 4 - 5 x 3 + 7x ⎤

⎦

= 4 x 8 - 5 x 6 - 4 x 4 + 5 x 3 + 7 x 2 - 7x

39. 4 _ 3 π R 3 41. 44,152.52; 44,100 43. 2,156,407.8; 2,153,645

45a. y

x

4

4

4

f (x) = x 3 - 6 x 2 + 8x

45b. 4; -4 45c. The two areas are equivalent; however, the value of the integral of f (x) that corresponds to the area above the x-axis is positive, and the value of the integral of f(x) that corresponds to the area below the x-axis is negative. 45d. 0; 8 45e. Signed area is the difference between the absolute values of the areas found above and below the x-axis. Total area is the sum of the absolute values of the areas found above and below the x-axis. 47. Sometimes; sample answer: Changing the order of the limits will change the sign of the original answer unless the answer is 0. 49. Sometimes; sample answer: If f(x) is an even function, a ≥ 0, and b ≥ 0, the identity will hold true. If f(x) is an odd function, a ≥ 0 and b ≥ 0, or a ≤ 0 and b ≤ 0, the identity will hold true. 51. Because the graph is below the x-axis, f(x) is negative. Each f(x)

is negative and Δx is positive, so each term in the sum ∑

i=1

n

f( x i )Δx

is negative. Therefore, the sum is negative. Because ∫ a b f(x)dx is a

limit of negative sums, it is also negative.53. Sample answer: 1) Decide which rule to apply to find the antiderivative of the

function: 2 x 3 + C. a) Power rule. b) Constant Multiple of a Power rule (applicable here). c) Sum and Difference rules. 2) Remove/ignore the constant C since this is a definite

integral. 3) Evaluate the antiderivative at the upper and lower limits

and find the difference.

2 x 3 �

�

�

0

2

= 2(2 ) 3 - 2(0 ) 3 = 16 - 0 = 16

4) The area under the graph on the interval [0, 2] is 16 square units.

55. 30 57. g′(n) = 2 n 4 + 2 n 2 + 4

__ ( n 2 + 1 ) 2

59a. 16% 59b. 24 61. B

63a. 4 m 63b. 1 m/s 63c. s′(t) = 2t - 3 63d. -1 m/s; 5 m/s 63e. 1.5 s 63f. Sample answer: After 1.5 s, the particle changes direction, moving to the right instead of to the left.


1. instantaneous rate of change 3. direct substitution 5. indeterminate form 7. differentiation 9. instantaneous velocity 11. -1 y

xO

f(x) = 2x - 7

x 2.99 2.999 3 3.001 3.01

f (x ) -1.02 -1.002 -0.998 -0.98

13. 5 15. ∞ 17. 9 19. Not possible; when x = 25, the denominator is 0. 21. -

1 _

6 23. -1; -1 25. m = -2x + 3

27. 31 ft/s 29. v(t) = 24t 31. g′(t) = -2t + 5;

g′(-4) = 13 and g′(1) = 3 33. p′(v) = -9 35. t′(x) = -

18 _

5 x

1 _ 5

37. f ′(m) = -25 _

(5 + 2m ) 2 39. 5.16 unit s 2 41. 4 2 _

3 unit s 2

43. 4 2 _ 3 unit s 2 45. G(n) = 5 _

2 n 2 - 2n + C

47. M(t) = 3 _ 2 t 4 − 4 t 3 + t 2 − 11t + C 49. 8 _

3 x 3 + C

51. 2466.53 unit s 2 53a.

t 0 1 2 3

v 15 30 50 68.2

53b. v (t)

t

20

40

60

80

2 4 6 8

v (t) = 450__5 + 25(0.4)t

53c. 90 53d. Sample answer: The stamp will peak in value atabout $90. 55a. 200 55b. The maximum value of Eldin’s coin collection is equal to lim t→∞

v(t). 55c. Yes; the offer is

more than the collection is worth. 57a. v(t) = -32t + 35 57b. 19 ft/s 57c. ≈1.09 s 57d. ≈20.64 ft 59a. s(t) = -16 t 2 + 26t 59b. 1.63 s


1. Sample answer: 4 x 3 is the derivative of the expression within the parentheses. 3. Sample answers: 4( x 4 - 3 ) 3 4 x 3 or 16 x 3 ( x 4 - 3 ) 3 5. s′(t) = 8t( t 2 - 1 ) 3

7. c′(r) = 3(3r − 2 r 2 ) 2 (3 − 4r) 9. f ′(x) = √ �� 100 + 8x

_ 25 + 2x

or 2 √ �� 25 + 2x

_ 25 + 2x



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Logic2CHAPTER 122CHCHAPAPTETERR 1212

Lesson 12-1

5. Not a statement 6. Statement 7. Statement 8. Statement9. Not a statement 10. Statement 11. Not a statement12. Statement 13. Not a statement 14. Not a statement15. Compound statement 16. Simple statement17. Compound statement 18. Compound statement19. Simple statement 20. Simple statement21. Compound statement 22. Compound statement23. Simple statement 24. Compound statement 25. Conjunction26. Disjunction 27. Biconditional 28. Conjunction29. Disjunction 30. Conditional 31. Biconditional32. Conditional 33. The sky is not blue.34. Your computer has a virus. 35. The dorm room is large.36. The class is full. 37. You will fail this class.38. He does not have large biceps. 39. Universal 40. Universal41. Existential 42. Existential 43. Universal 44. Universal45. Existential 46. Universal 47. Universal 48. Universal49. Existential 50. Universal51. Not all fish swim in water; Some fish do not swim in water.52. Someone who passes algebra has not studied; Not everyone who passes algebra has studied.53. No people who live in glass houses throw stones.54. There is no one in this class that won’t pass; Everyone in this class will pass.55. Not every happy dog wags its tail; Some happy dog does not wag its tail.56. Some men can join a sorority. 57. There is no four leaf clover.58. Some person who participated in the study will not get a hundred dollars; Not every person who participated in the study will get a hundred dollars.59. Someone with green eyes wears glasses.60. Not everyone in the class was bored by the professor’s lecture; Someone in the class was not bored by the professor’s lecture.61. None of my friends have an iPhone.62. Someone gets out of here alive. 63. p ∧ q 64. ∼p 65. ∼q → p66. ∼(q ∨ p) 67. ∼q 68. ∼p 69. q ∨ ∼p 70. ∼q ∨ p 71. q ↔ p72. p → q 73. ∼q 74. ∼p ∧ ∼q 75. q → p 76. ∼p 77. ∼q ∨ p78. ∼(∼q ∧ p) 79. q ↔ p 80. ∼(q ∨ p) or ∼q ∧ ∼p 81. ∼q → p82. ∼p ↔ ∼q 83. The plane is on time and the sky is clear.84. The plane is not on time or the sky is clear.85. If the sky is clear, then the plane is on time.86. If the sky is clear, then the plane is not on time.87. The plane is not on time and the sky is not clear.88. The sky is clear if and only if the plane is on time.89. The plane is on time or the sky is not clear.90. The plane is not on time if and only if the sky is not clear.91. If the sky is clear, then the plane is or is not on time.92. If the plane is on time, then the sky is clear; or the plane is not on time.93. Trudy does not live off campus.94. If Mark lives on campus, then Trudy lives off campus.95. Mark lives on campus or Trudy does not live off campus.96. Trudy lives off campus if and only if Mark lives on campus.97. If Mark does not live on campus, then Trudy does not live off campus.98. Mark does not live on campus.99. Mark lives on campus or Trudy lives off campus.100. Mark does not live on campus or Trudy lives off campus; or Trudy does not live off campus.

101. Trudy lives off campus or Mark lives on campus.102. If Mark lives on campus or Trudy lives off campus, then it is not the case that Trudy does not live off campus.103. It cannot be classified as true or false.104. It cannot be classified as true or false.

Lesson 12-2

5. FFFT 6. TTFT 7. FFTF 8. TFTT 9. FTTF 10. FFTT11. TTFF 12. TFFF 13. TFTT 14. FFTF 15. TTFF16. TFTT 17. TTFF 18. TTTTFTTT 19. TTFFTFFF20. TTTTTTTT 21. TTTTTTTT 22. TTFTFTFT23. TFFFTFTF 24. TTFTFFFF 25. TTTFFFTF26. TFTFTFFF 27. FTFTFTTT 28. TTTTTTTT29. FFFTTTTT 30. FFFTFFTF 31. FFFFTFFF32. TFTTFFTT 33. TFTFFFFF 34. FTTFTFTF 35. False36. False 37. True 38. True 39. True 40. True41. Let p be “if you take their daily product,” q be “you cut your calories intake by 10%, and r be “you lose at least 10 pounds in the next 4 months.”42. (p ∧ q) → r 43. TFTTTTTT 44. True 45. True 46. True47. Let p be “the attendance for the following season is over 2 million,” q be “he will add 20 million dollars to the payroll,” and r be “the team will make the playoffs the following year.”48. p → (q ∧ r) 49. TFFFTTTT 50. False 51. True 52. False53. The statements are equivalent. 54. Order does matter.55. False

Lesson 12-3

7. Tautology 8. Neither 9. Self-contradiction 10. Neither11. Tautology 12. Self-contradiction 13. Neither 14. Neither 15. Neither 16. Neither 17. Equivalent 18. Negations19. Neither 20. Neither 21. Neither 22. Neither 23. Negations 24. Equivalent 25. Neither 26. Equivalent27. q → p; ~p → ~q; ~q → ~p 28. ~q → ~p; p → q; q → p29. p → ~q; q → ~p; ~p → q 30. q → ~p; p → ~q; ~q → p31. ~q → p; ~p → q; q → ~p 32. p → q; ~q → ~p; ~p → ~q33. The concert is not long and it is not fun.34. The soda is not sweet and it is carbonated.35. It is cold or I am not soaked.36. I will not walk in the Race for the Cure walkathon or I will not be tired.37. I will not go to the beach or I will get sunburned.38. The coffee is not a latte and it is not an espresso.39. The student is not a girl and the professor is a man.40. I will not go to college or I will not get a degree.41. It is not right and it is not wrong.42. Our school colors are blue and green. 43. p → q 44. ~q → ~p45. p → q 46. p → q 47. p → q 48. p → q 49. ~p → ~q50. The statements in exercises 43, 44, 45, 46, 47, and 48 are all equivalent.51. Converse: If he did get a good job, then he graduated with a Bachelor’s degree in Management Information Systems. Inverse: If he did not graduate with a Bachelor’s degree in Management Information Systems, then he will not get a good job. Contrapositive: If he did not get a good job, he did not graduate with a Bachelor’s degree in Management Information Systems.52. Converse: If she cannot buy the green Ford Focus, she did not earn $5,000 this summer as a barista at the coffee house. Inverse: If she does earn $5,000 this summer as a barista at the coffee house, she can buy the green Ford Focus. Contrapositive: If she can buy the green Ford Focus, then she did earn $5,000 this summer as a barista at the coffee house.


R51_R113_EM_SelAns_664293.indd R106R51_R113_EM_SelAns_664293.indd R106 09/10/12 9:32 AM09/10/12 9:32 AM

Selected A

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53. Converse: If I host a party in my dorm room, then the American Idol finale is today.Inverse: If the American Idol finale is not today, then I will not host a party in my dorm room.Contrapositive: If I do not host a party in my dorm room, then the American Idol finale is not today.54. Converse: If I replace the battery, then my cell phone will not charge.Inverse: If my cell phone will charge, then I will not replace the battery.Contrapositive: If I do not replace the battery then my cell phone will charge.55. Converse: If I go to Nassau for spring break then I will lose 10 pounds by March 1.Inverse: If I do not lose 10 pounds by March 1 then I will not go to Nassau for spring break.Contrapositive: If I do not go to Nassau for spring break then I did not lose 10 pounds by March 1.56. Converse: If the politician goes to jail then he got caught taking kickbacks.Inverse: If the politician does not get caught taking kickbacks, then he will not go to jail.Contrapositive: If the politician does not go to jail, then he did not get caught taking kickbacks.57. ~(p → q) ≡ p ∧ ~q 58. ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)59. Answers vary 60. Answers vary 61. Equivalent62. Equivalent

Lesson 12-4

7. Valid 8. Valid 9. Invalid 10. Valid 11. Invalid12. Valid 13. Valid 14. Valid 15. Invalid 16. Invalid17. Valid 18. Valid 19. Valid 20. Valid 21. Invalid22. Invalid 23. Valid 24. Invalid 25. Invalid 26. Valid27. Valid 28. Invalid 29. Invalid 30. Valid 31. Invalid32. Valid 33. Invalid 34. Valid 35. Invalid 36. Invalid37. Valid 38. Invalid 39. Invalid 40. Valid 41. Valid42. Invalid 43. Invalid 44. Valid 45. Valid 46. Invalid47. Invalid 48. Valid 49. Valid 50. Invalid 51. Valid52. Invalid 53. Answers vary 54. Answers vary

Lesson 12-5

5.

Computers

Calculators

6.

Real thingsUnicorns

7.

�

Going tocollege

People

8.

�

DVDburners

CDburners

9. Easy thingsMath courses

10.

�

Resultsin weight

loss

Fad diet

11.

�

U.S. laws Mexicanlaws

12. Smart people

Mensamembers

13. Low fatthings

Cheese-burgers

14.

�

Politicians Crooks

15. Invalid 16. Invalid 17. Valid 18. Invalid 19. Invalid20. Invalid 21. Valid 22. Valid 23. Invalid 24. Invalid25. Invalid 26. Valid 27. Invalid 28. Invalid 29. Valid30. Invalid 31. Invalid 32. Invalid 33. Invalid34. Invalid 35. Valid 36. Valid 37. Invalid 38. Valid39. All A is C. 40. No S is P.41. No calculators can make breakfast.42. Some things with brains are prejudiced.



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1. Not a statement 2. Statement 3. Not a statement 4. Statement 5. Not a Statement 6. Compound 7. Simple8. Compound 9. Compound 10. Simple 11. It is not scary.12. The cell phone is not out of juice. 13. The popsicle is not green.14. Some people who live in glass houses throw stones.15. No failing students can learn new study methods.16. Not everyone will pass the test on logic; Someone will not pass the test on logic.17. All printers have ink. 18. Some of these links are broken.19. None of the contestants will be voted off the island.20. Not all SUVs are gas guzzlers; Some SUVs are not gas guzzlers.21. Conjunction 22. Conditional 23. Disjunction24. Biconditional 25. Conditional 26. p ∧ q 27. q → p28. q ↔ p 29. q ∧ ~p 30. ~p → ~q 31. ~(q ∧ p) 32. ~(p → q)33. ~q ↔ ~p 34. ~(~q) 35. ~p ∧ ~q 36. It is cool or it is not cloudy.37. If it is cloudy, then it is cool.38. It is cool if and only if it is cloudy.39. If it is cool or cloudy, then it is cool.40. It is not true that it is not cool or it is cloudy.41. FTTT 42. TTFT 43. FFFT 44. TTTT 45. FFFT46. TFTTTTTT 47. TFTFFFTF 48. TTFTTTTT49. False 50. True 51. True 52. True 53. Tautology54. Neither 55. Neither 56. Tautology 57. Neither58. Not equivalent 59. Not equivalent 60. Equivalent61. The internet connection is not dial-up and it’s not DSL.62. We will not increase sales and our profit margin will not go down.63. The signature is authentic or the check is valid.64. It is strenuous or I am not tired.65. Let p be the statement “I will be happy” and q be the statement “I get rich.” The compound statement is p → q.66. Let p be the statement “having a good career” and q be the statement “having a fulfilling life.” The compound statement is p → q.67. Converse: If I start riding my bike to work, gas prices will go higher.Inverse: If gas prices do not go any higher, I will not start riding my bike to work.Contrapositive: If I do not start riding my bike to work, then gas prices will not go any higher.68. Converse: If my parents kill me, then I didn’t pass this class.Inverse: If I pass this class, then my parents won’t kill me.Contrapositive: If my parents don’t kill me, then I did pass this class.69. Converse: If it rains, the festival will move inside the student center.Inverse: If the festival did not move inside the student center, then it did not rain.Contrapositive: If it does not rain, then the festival will not move inside the student center.70. Invalid 71. Invalid 72. Invalid 73. Invalid 74. Valid75. Invalid 76. Invalid 77. Invalid 78. Invalid 79. Invalid80. Valid 81. Valid 82. Invalid 83. Invalid 84. Invalid85. Invalid 86. Valid

Chapter 12 Test

1. Statement 2. Statement 3. Not a statement 4. Not a statement5. The image is not uploading to my online bio.6. Not all men have goatees; Some men do not have goatees.7. No students ride a bike to school.8. Some short people can dunk a basketball. 9. p ∧ q10. q → p 11. p ↔ q 12. p ∨ q 13. ~(~p ∧ q) 14. ~q ∧ ~p

15. It is sunny or not warm. 16. If it is warm, then it is sunny.17. It is sunny if and only if it is warm.18. If it is sunny or warm, then it is sunny.19. It is not the case that it is not sunny or it is warm. 20. FTTT21. FFTFTFTF 22. FTTTFTFT 23. TTFF 24. TFTTTTTT25. Self-contradiction 26. Neither 27. Neither 28. Neither29. Tautology 30. Equivalent 31. Equivalent32. Converse: If I am healthy, then I exercise regularly.Inverse: If I do not exercise regularly, then I will not be healthy.Contrapositive: If I am not healthy, then I do not exercise regularly.33. It is cold or it is not snowing.34. I am not hungry and I am not thirsty. 35. Invalid 36. Invalid37. Valid 38. Invalid 39. Invalid 40. Invalid 41. Valid42. Invalid 43. Invalid 44. Valid 45. Valid

Boolean Algebra3CHAPTER 133CHCHAPAPTETERR 1313

Lesson 13-1

1a. 1 1b. 1 1c. 0 1d. 0 3a. (1 · 1) + ( −−−

0 · 1 + 0) = 1 + ( −

0 + 0) = 1 + (1 + 0) = 1 + 1 = 1 3b. (T ∧ T) ∨ (⌐(F ∧ T) ∨ F) ≡ T

5a. x y z − x y

1 1 1 0

1 1 0 0

1 0 1 0

1 0 0 0

0 1 1 1

0 1 0 1

0 0 1 0

0 0 0 0

5b. x y z x +y z

1 1 1 1

1 1 0 1

1 0 1 1

1 0 0 1

0 1 1 1

0 1 0 0

0 0 1 0

0 0 0 0

5c. x y z x − y + −−

xyz

1 1 1 0

1 1 0 1

1 0 1 1

1 0 0 1

0 1 1 1

0 1 0 1

0 0 1 1

0 0 0 1

5d. x y z x (y z +

−

y − z )

1 1 1 1

1 1 0 0

1 0 1 0

1 0 0 1

0 1 1 0

0 1 0 0

0 0 1 0

0 0 0 0

7a. 110 111

011

101

010

100

000 001

7b. 110 111

011

101

010

100

000 001


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7c. 110 111

011

101

010

100

000 001

7d. 110 111

011

101

010

100

000 001

9. (0, 0) and (1, 1) 11. x + xy = x · 1 + xy = x(1 + y) = x(y + 1) = x · 1 = x13.

x y z x − y − y z − x zx − y +

−

y

z + − x z − x y − y z x − z

− x y +

−

y

z + x − z

1 1 1 0 0 0 0 0 0 0 0

1 1 0 0 1 0 1 0 0 1 1

1 0 1 1 0 0 1 0 1 0 1

1 0 0 1 0 0 1 0 0 1 1

0 1 1 0 0 1 1 1 0 0 1

0 1 0 0 1 0 1 1 0 0 1

0 0 1 0 0 1 1 0 1 0 1

0 0 0 0 0 0 0 0 0 0 0

15. x x + x x · x

0 0 0

1 1 1

17. x x + 1 x · 0

0 1 0

1 1 0

19.

x y z y + zx +

(y + z)x + y

(x + y)

+ zyz x(yz) xy (xy)z

1 1 1 1 1 1 1 1 1 1 1

1 1 0 1 1 1 1 0 0 1 0

1 0 1 1 1 1 1 0 0 0 0

1 0 0 0 1 1 1 0 0 0 0

0 1 1 1 1 1 1 1 0 0 0

0 1 0 1 1 1 1 0 0 0 0

0 0 1 1 1 0 1 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

21.

x y xy −−

(xy) − x − y − x + − y x + y −−−−

(x + y) − x − y

1 1 1 0 0 0 0 1 0 0

1 0 0 1 0 1 1 1 0 0

0 1 0 1 1 0 1 1 0 0

0 0 0 1 1 1 1 0 1 1

23. 0 . −

0 = 0 . 1 = 0; 1 . −

1 = 1 . 0 = 025.

x y x ⊕ y x + y xy −−

(xy) (x + y) −−

(xy) x − y − x y x − y + − x y

1 1 0 1 1 0 0 0 0 0

1 0 1 1 0 1 1 1 0 1

0 1 1 1 0 1 1 0 1 1

0 0 0 0 0 1 0 0 0 0

27a. True, as a table of values can show 27b. False; take x = 1, y = 1, z = 1, for instance 27c. False; take x = 1, y = 1, z = 0, for instance 29. By De Morgan’s laws, the complement of an expression is like the dual except that the complement of each variable has been taken. 31. 16 33. If we replace each 0 by F, 1 by T, Boolean sum by ∨, Boolean product by ∧, and

_ by ¬ (and x

by p and y by q so that the variables look like they represent propositions, and the equals sign by the logical equivalence symbol), then −−

xy = − x + − y becomes ¬(p ∧ q) ≡ ¬p ∨ ¬q and −−−

x + y = − x − y becomes ¬(p ∨ q) ≡ ¬p∧¬q. 35. By the domination, distributive, and identity laws, x ∨ x = (x ∨ x)∧1 = (x ∨ x)∧(x ∨

−

x ) = x ∨ (x ∧ − x ) = x ∨ 0 = x. Similarly, x ∧ x =(x ∧ x) ∨ 0=(x ∧ x) ∨

(x ∧

−

x ) = x ∧ (x ∨

−

x ) = x ∧ 1 = x. 37. Because 0 ∨ 1 = 1 and 0 ∧ 1 = 0 by the identity and commutative laws, it follows that

−

0 = 1. Similarly, because 1 ∨ 0 = 1 and 1 ∧ 0 = 1, it follows that

−

1 = 0.39. First, note that x∧0 = 0 and x ∨ 1 = 1 for all x, as can easily be proved. To prove the first identity, it is sufficient to show that (x ∨

y)∨(x ∧ y) = 1 and (x ∨ y) ∧ ( − x ∧

−

y ) = 0. By the associative, commutative, distributive, domination, and identity laws, (x ∨ y) ∨ ( − x ∧

−

y ) = y ∨ [x ∨ ( − x ∧ − y )] = y ∨[(x ∨

−

x ) ∧ (x ∨ − y )] = y ∨ [1 ∧ (x ∨ − y )] = y ∨ (x ∨ − y ) = (y ∨ − y ) ∨ x = 1 ∨ x = 1 and (x ∨ y)∧( − x ∧

−

y ) = − y ∧[ − x ∧ (x∨y)] = − y ∧[( − x ∧ x) ∨ ( − x ∧y)] = − y ∧ [0 ∨ ( − x ∧ y)] = − y ∧( − x ∧ y) = − x ∧ (y ∧ − y ) = − x ∧ 0 = 0. The second identity is proved in a similar way. 41. Using the hypotheses, Exercise 35, and the distributive law it follows that x = x ∨ 0 = x ∨ (x ∨ y) = (x ∨ x) ∨ y = x ∨ y = 0. Similarly, y = 0. To prove the second statement, note that x = x ∧ 1 = x ∧ (x ∧ y) = (x ∧ x)∧y = x ∧ y = 1. Similarly, y = 1. 43. Use Exercises 39 and 41 in the Supplementary Exercises in Chapter 9 and the definition of a complemented, distributed lattice to establish the five pairs of laws in the definition.

Lesson 13-2

1a. − x − y z 1b. − x y − z 1c. − x yz 1d. − x − y − z 3a. xyz + xy − z + x − y z + x − y − z + − x yz + − x y − z + − x − y z 3b. xyz + xy − z + − x yz 3c. xyz + xy − z + x − y z + x − y − z 3d. x − y z + x − y − z 5. w x y − z + w x − y z + w − x y z +

−−

w x y z +

−−

w x − y − z +

−−

w − x − y z +

−−

w − x y − z + w − x − y − z 7a. − x + − y + z 7b. x + y + z 7c. x + − y + z9. y1 + y2 + … + y n = 0 if and only if yi = 0 for i = 1, 2, … , n. This holds if and only if x i = 0 when y i = x i and x i = 1 when y i = − x i . 11a. (x + y + z) 11b. (x + y + z) (x + y + − z ) (x + − y + z) ( − x + y + z) ( − x + y + − z ) 11c. (x + y + z) (x + y + − z ) (x + − y + z) (x + − y + − z ) 11d.(x + y + z) (x + y + − z ) (x + − y + z) (x + − y + − z )( − x + − y + z) ( − x + − y + − z ) 13a. x + y + z 13b. x +

−−−−−−−−

[y + −−−−−

( − x + z) ]

13c. −−−−−

(x + − y ) 13d. −−−−−−−−−−

[x + −−−−−−−

(x +

−

y + − z ) ] 15a.

x − x x ↓ x

1 0 0

0 1 1

15b. x y xy x ↓ x y ↓ y (x ↓ x) ↓ (y ↓ y)

1 1 1 0 0 1

1 0 0 0 1 0

0 1 0 1 0 0

0 0 0 1 1 0



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1 1 1 0 1

1 0 1 0 1

0 1 1 0 1

0 0 0 1 1

17a. {[(x | x) | (y | y)] | [(x | x) | (y | y)]} |(z | z) 17b. {[(x | x) | (z | z)] | y} | {[(x | x) | (z | z)] | y} 17c. x 17d. [x | (y | y)] | [x | (y | y)]

19. It is impossible to represent − x using + and · because there is no way to get the value 0 if the input is 1.

Lesson 13.3

1. (x + y) − y 3.

−−−

(xy) + ( − z + x) 5. (x + y + z) + ( − x + y + z) + ( − x + − y + − z )7. vwx

vwy

vwz

vxy

vxz

wxy

wxz

wyz

xyz

vwx + vwy + vwz + vxy + vxz +vyz + wxy + wxz + wyz + xyz

vwx

vwy

vwz

vxy

vxz

wxy

wxz

wyz

xyz

vyz

vyz

9. HA

FA

FA

FA

FA

FA

s0

s1

s2

s3

s4

s5

s6

FA = full adderHA = half adder

x0

y0

x1y1

x2y2

x3y3

x4y4

x5y5

11. xy

bi

xy

bi

xy

bi

xy

bi

d

xy

bi

xy

bi

xy

bi

xy

bi

bi+1

13. x1

y1

x1

y1

x1

y1

x0y0

x1y1 + x0 y0(x1y1 + x1y1)

15a. x

xx

15b. x

x

x + yy

y

15c. x

y

xyx

y

15d. x

x

y

y

x y

y

x



Selected A

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17.

Sum = x y

x

y

Carry = xy

Circuitfrom15 (d)

Circuitfrom15 (c)

19. c0c1x3

c0c1x2

c0c1x1

c0c1x0

Lesson 13.4

1a. y y

x 1

x

1b. xy and − x − y

3a.

x

x

y y

1

3b.

x

x

y y

1

1

3c.

x

x

y y

1

11

1

5a. yz yz

x 1

yz yz

x

5b. − x yz, − x − y − z , xy − z 7a. yz yz yz yz

x

x

1

7b. yz yz yz yz

x

x 11

7c. yz yz yz yz

x

x 1

111

9. Implicants: xyz, xy − z , x − y − z , − x y − z , xy, x − z , y − z ; prime implicants: xy, x − z , y − z ; essential prime implicants: xy, x − z , y − z

yz yz yz yz

x

x 1

111

11. The 3-cube on the right corresponds to w; the 3-cube given by the top surface of the whole figure represents x; the 3-cube given by the back surface of the whole figure represents y; the 3-cube given by the right surfaces of both the left and the right 3-cube represents z. In each case, the opposite 3-face represents the complemented literal. The 2-cube that represents wz is the right face of the 3-cube on the right; the 2-cube that represents − x y is bottom rear; the 2-cube that represents − y − z is front left.

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

wxyz

13a. yz yz

wx

yz yz

wx

wx 1

wx

13b. −−

w xyz, −−

w − x y − z , −−

w x − y − z , wxy − z 15a.

x1x2

x3x4x5 x3x4x5 x3x4x5 x3x4x5 x3x4x5 x3x4x5 x3x4x5 x3x4x5

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15c.

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x1x2

x1x2 1 1

1 1

1 1

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15d.

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x1x2 1 1

1 1

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15e.

x1x2

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15f.

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17a. 64 17b. 6 19. Rows 1 and 4 are considered adjacent. The pairs of columns considered adjacent are: columns 1 and 4, 1 and 12, 1 and 16, 2 and 11, 2 and 15, 3 and 6, 3 and 10, 4 and 9, 5 and 8, 5 and 16, 6 and 15, 7 and 10, 7 and 14, 8 and 13, 9 and 12, 11 and 14, 13 and 16.

21. SmithJones

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23a. − x z 23b. y 23c. x − z + − x z + − y z 23d. xz + − x y + − y − z 25a. wxz + wx − y + w − y z + w − x y − z 25b. x − y z + −−

w − y z + wxy − z + w − x yz + −−

w − x y − z 25c. − y z + wxy + w − x − y + −−

w − x y − z 25d. wy + yz + − x y + wxz + −−

w − x z 27. x(y + z)29. z

yx

zyx

zyx

zw

w

31. − x − z + xz 33. We use induction on n. If n = 1, then we are looking at a line segment, labeled 0 at one end and 1 at the other end. The only possible value of k is also 1, and if the literal is x 1 , then the subcube we have is the 0-dimensional subcube consisting of the endpoint labeled 1, and if the literal is −−

x 1 , then the subcube we have is the 0-dimensional subcube consisting of the endpoint labeled 0. Now assume that the statement is true for n; we must show that it is true for n + 1. If the literal x n + 1 (or its complement) is not part of the product, then by the inductive hypothesis, the product when viewed in the setting of n variables corresponds to an (n − k)-dimensional subcube of the n-dimensional cube, and the Cartesian product of that subcube with the line segment [0, 1] gives us a subcube one dimension higher in our given (n + 1)-dimensional cube, namely having dimension (n + 1)−k, as desired. On the other hand, if the literal x n + 1 (or its complement) is part of the product, then the product of the remaining k − 1 literals corresponds to a subcube of dimension n−(k−1) = (n + 1)−k in the n-dimensional cube, and that slice, at either the 1-end or the 0-end in the last variable, is the desired subcube.

Supplementary Exercises

1a. x = 0, y = 0, z = 0; x = 1, y = 1, z = 1 1b. x = 0, y = 0, z = 0; x = 0, y = 0, z = 1; x = 0, y = 1, z = 0; x = 1, y = 0, z = 1; x = 1, y = 1, z = 0; x = 1, y = 1, z = 1 1c. No values 3a. Yes 3b. No 3c. No 3d. Yes 5. 2 2 n−1 7a. If F( x 1 , . . . , x n ) = 1, then (F + G)( x 1 , . . . , x n ) = F( x 1 , . . . , x n ) + G( x 1 , . . . , x n ) = 1 by the dominance law. Hence, F ≤ F + G. 7b. If (F G)( x 1 , . . . , x n ) = 1, then F( x 1 , . . . , x n ) · G( x 1 , . . . , x n ) = 1. Hence, F( x 1 , . . . , x n ) = 1. It follows that FG ≤ F. 9. Because F( x 1 , . . . , x n ) = 1 implies that F( x 1 , . . . , x n ) = 1, ≤ is reflexive. Suppose that F ≤ G and G ≤ F. Then F( x 1 , . . . , x n ) = 1 if and only if G( x 1 , . . . , x n ) = 1. This implies that F = G. Hence, ≤ is antisymmetric. Suppose that F ≤ G ≤ H. Then if



Selected A

nswers and S

olutions


F( x 1 , . . . , x n ) = 1, it follows that G( x 1 , . . . , x n ) = 1, which implies that H( x 1 , . . . , x n ) = 1. Hence, F ≤ H, so ≤ is transitive. 11a. x = 1, y = 0, z = 0 11b. x = 1, y = 0, z = 011c. x = 1, y = 0, z = 0 13.

x y x � y x ⊕ y −−−−

(x ⊕ y)

1 1 1 0 1

1 0 0 1 0

0 1 0 1 0

0 0 1 0 1

15.Yes, as a truth table shows 17a. 6 17b. 5 17c. 5 17d. 619. x

ySum = x � y

Carry = xy

21. x 3 + x 2 − x 1 23. Suppose it were with weights a and b. Then there would be a real number T such that xa + yb ≥ T for (1,0) and (0,1), but with xa + yb < T for (0,0) and (1,1). Hence, a ≥ T, b ≥ T, 0 < T, and a + b < T. Thus, a and b are positive, which implies that a + b > a ≥ T , a contradiction.



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