8.2.3 homework answers

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Core Connections Algebra 2

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Core Connections Algebra 2

Lesson 1.1.1

1-4. a: 1 2 b: 3

1-5. a: h(x) then g(x) b: Yes, g(x) then h(x).

1-6. See graph above right.

# of Buses

4

3

2

1

45 90 135 180

# of Students

1-7. a: y

b: c: y

d:

y

x

1-8. a: Not linear. b: The exponent. c: A parabola.

1-9. Answers will vary.

Lesson 1.1.2 (Day 1)

1-12. y = 2x + 10

See graph and table at right.

x 0 1 2 3 4

y 10 12 14 16 18

1-13. a: x = –13 or 17 b: x = – 3 2 or 7 3

c: x = 0 or 3 d: x = 0 or 5

e: x = 7 or –5 f: x = 1 3

or –5

1-14. a: 14, –4, 3x – 1 b: f(x) = 3x – 1

1-15. a: y = 5x – 2 b: x = 2 5

1-16. a: 21, 15, (0, 15) b: –3, 3, (0, 3)

1-17. a: 16 b: 9 c: 478.38

Temperature

1-18. a: y depends on x; x is independent. Explanations will vary.

b: Temperature is dependent; time is independent.

c: See graph above right.

Time

Lesson 1.1.2 (Day 2)

1-19. y = 30 – x

Graph and table shown at right.

Answers will vary.

x 0 1 6 20

y 30 29 24 10

1-20. See graph below. Possible inputs: x –4 –2 0 1 6

all real numbers; possible outputs: y 8 2 0 0.5 18

any number greater than or equal to zero.

1-21. a: 1 b: x = 12

c: 13 d: no solution

e: x = ± 13 2

≈ ± 2.55 f: x = ± 7 ≈ ± 2.65

1-22. Cube each input: f (x) = x 3

1-23. a: The more gas you buy, the more money you spend. I: gallons, D: dollars

b: People grow a lot in their early years and then their growing slows down.

I: age, D: height

c: As time goes by, the ozone concentration goes down, although the effect is slowing.

I: year, D: ozone

d: As the number of students grows, more classrooms are used and each classroom holds

30 students. I: students, D: classrooms

e: Possible inputs: x can be any number between and including 0 and 120,

possible outputs: y = 1, 2, 3, 4

1-24. They are similar by AA.

a: n m b: m x

1-25. Error in line 2: It should be –14, not +14; x = –37.

Lesson 1.1.3

1-34. a: The numbers between –2 and 4 inclusive.

b: The numbers between –1 and 3 inclusive.

c: No. He is missing all the values between those

numbers. The curve is continuous, so the description

needs to include all real numbers, not just integers.

(–2,3)

(4,–1)

d: See graph at right.

1-35. a: 70 b: 2 c: 43 d: undefined

e: 3x 2 = x – 5 – 3 f: 3x 2 = x – 5 + 7

g: all real numbers h: all real numbers greater than or equal to 5.

i: They are different because the square root of a negative is undefined, whereas any real

number can be squared.

1-36. Chelita is correct about how to find the intercepts, but she makes an error with signs

while factoring. The correct equation is (x − 7)(x − 3) = 0 and the x-intercepts are

7 and 3.

1-37. a: y = x–6

b: y =

x+10

5

c: y = ± x

d: y = ± x+4

e: y = ± x + 5

1-38. a: –7 b: 3.5 c: The x- and y-intercepts.

1-39. a: y = 3x + 24; Table and graph shown at right.

1-40. a: x = 13 b: x = 8

0

24

27

30

33

36

39

Height (in inches)

Time (in weeks)

Lesson 1.1.4

1-46. (2, 1)

1-47. a: 2 b: 10 c: 100 d: ≈ 142.86

1-48. a: x = 5, 3 b: x ≈ 3.39, –0.89 or x =

5± 73

1-49. a: 34 ≈ 5.83 units b: 3 5

1-50. a: 1 52

b:

51

52

1-51. The error is in line 3. It should be: 0 = 5.4x + 23.7, x ≈ –4.39

1-52. a: x ≈ –7.37 b: x = 2.8

Lesson 1.2.1 (Day 1)

1-59. Table and graph shown below right.

D: −∞ < x < ∞ ; R: −∞ < y < ∞

intercepts (0, –4) and

4, 0

1-60. a: ≈ 5.18 b: ≈18.66

( ) or (≈ 1.59, 0)

–3

–2

–1

h(x)

–31

–12

–5

–4

23

c: ≈ 24.62º d: 180 ≈ 13.42

1-61. a: A line, no variables are raised to a power.

b: y = 2 3

x – 2 , graph shown at right.

c: Substitute x = 0 and solve for y, substitute

y = 0 and solve for x. (3, 0) and (0, –2)

d: Answers will vary.

e: The intercepts are (–9, 0) and (0, 6),

graph shown at right.

1-62. a: D: x = –1, 1, 2 b: D: –1 ≤ x < 1 c: D: x ≥ –1 d: D: −∞ < x < ∞

R: y = –2, 1, 2 R: –1 ≤ y < 2 R: y ≥ –1 R: y ≥ –2

1-63. There is an error in line 2. Both sides need to be multiplied by x: 5 = x 2 – 4x,

0 = x 2 –4 –5 = (x –5)(x + 1), x = –1, 5.

1-64. a: x = 3, –2 b: x = 3, –3

Lesson 1.2.1 (Day 2)

1-65. a: 2 b: –4 c: 1 0

is undefined d: Answers will vary.

1-66. a: (0, 3) and ( –

2 3 , 0 ) , see graph at right.

b: These equations are equivalent, they just have different

notation.

1-67. x ≈ 2.72 feet, y ≈ 1.27 feet

1-68. a: D: –2, –1, 2 b: D: –1< x ≤ 1 c: D: x > –1 d: D: −∞ < x < ∞

R: –1, 0, 1 R: –1< y ≤ 2 R: y > –1 R: −∞ < y < ∞

1-69. l = 4w and l + w = 22 or w + 4w = 22

The length is 17.6 cm, and the width is 4.4 cm.

1-70. a: x = −

17 1 ≈ −0.059 b: x =

66

13

≈ 5.08 c: x = –1, 3

1-71. a: (–1, 9) and (5, 21) b: x 2 +17 c: x 2 – 4x – 5

Lesson 1.2.1 (Day 3)

1-72. a: x = 5(y–1)

b: x = –2y+6

c: x = ± y d: x = ± y +100

1-73. y = π x 2 , table and graph shown at right.

1-74. a: 58 ≈ 7.62

y 0 π 4π 9π 16π radius

b: –

7

1-75. Solve x 2 + 2x +1 = 1 ; 0 or –2.

1-76. a: (0, 6) b: (0, 2) c: (0, 0)

d: (0, –4) e: (0, 25) f: (0, 13)

-3

-1 -1

area

Lesson 1.2.2 (Day 1)

1-84. (1, 3) and (7, 81)

1-85. a: x = –6 b: x = 38

13 ≈ 2.92

1-86. Graph shown at right. intercepts: (0, –2) and (4, 0),

domain: x ≥ 0, range: y ≥ –2

f(x)

1-87. x +(x + 18) or x + y = 84 and y = x +18; 33 and 51 meters long.

1-88. a: Table and graph shown at right, y = 2x + 26.

b: 37 weeks after his birthday.

1-89. y = 0

a: (–2, 0) b: (–10, 0) c: (0, 0)

( ) e: (5, 0) f:

( 13, 0)

d: ± 2, 0

1-90. Graph shown at right. domain: −∞ < x < ∞ , range: y ≥ –8

26

28

32

34

-2

4 x

Lesson 1.2.2 (Day 2)

1-91. a: x = y–b

m b: r = ± A π

c: W = V

LH d: y = 1

3–2x

1-92. See table and graph at right. Answers will vary.

1-93. a: Answers will vary.

b: When the y-values are the same, they must be equal.

c: 3x + 15 = 3 – 3x, x = –2

d: y = 9

–2 –1

–1 –2

–0.5 –4

0 undef.

0.5 4

1 2

2 1

e: They cross at the point (–2, 9).

1-94. 7.5 feet

1-95. ( ± 5, 0) ; Graph shown at right.

1-96. a: y

b: y c: y-intercept (0, 3) for both, x-intercept

1-97. a: 4 b: 2 c: 3 d: 1

( –

2 3 , 0 ) for (a) and none for (b).

d: (0, 3) and (2, 7), solve 2x + 3 = x 2 + 3 to

get x = 0

or x = 2

Lesson 1.2.3

1-104. m = –

4 3 , (4, 0), (0, 3), graph shown at right.

1-105. y = 3 2 x – 3

1-106. x =

−3± 21

7± 193

≈ –3.79, 0.79 b: x =

6

≈ 3.48, –1.15

1-107. $12.00

(Schools with an open campus)

1-108. Sample graphs.

# of People

or

= 2

1-110. a: 1, 2, 3, 4, 5 or 6 b: 1 6 c: 4 6

1-109. a: D: –3≤ x ≤ 3 b: D: x = 2 c: D: x ≥ −2

R: y = –2, 1, 3 R: −∞ < y < ∞ R: −∞ < y < ∞

Lesson 1.2.4

1-112. a: A portion of the trip at a specific speed.

b: About 400 miles. It is the total distance on the graph.

c: Graph shown below – a speed of approximately 30 mph for 1 hour, approximately

80 mph for the next 3 hours, 0 mph for 2 hours, approximately 40 mph for 2 hours,

and then approximately 20 mph for the last 2 hours. Note that the step graph assumes

instantaneous change of speed, which is not technically possible.

1-113. a: x = 2 b: x = 4

1-114. mB = 39.8° , 244 ≈ 15.62

1-115. 56 inches

Speed (mph)

Time (hours)

1-116. The independent variable is the volume of

water; the dependent variable is the height

C

of the liquid. The graph is 3 line segments

A

starting at the origin. C is the steepest, and

B is the least steep.

Volume of Water

Added

1-117. Diagrams vary; graph and table below, y = 3x.

1-118. a:

26 1 b: 25 1 x y

1 3

Height of Liquid

2 6

3 9

B

Lesson 2.1.1

2-4. a: See graph at right.

b: Yes, for every possible amount of water usage,

there is only one possible cost.

c: Domain: 0 to 1,000 cubic feet; range: discrete

values including: $12.70, $16.60, $20.50, $24.40,

$29.60, $34.80, $40, $45.20, $50.40, $55.60,

$60.80

Cost ($)

Water Used (ft 3 )

2-5. Smallest: a: 2; b: 0; c: –3; d: none.

Largest: a: none. b: none. c: none. d: 0. e: At the vertex.

2-6. The negative coefficient causes parabolas to open

downward, without changing the vertex. See graph at right.

2-7. a: Parabola with vertex (3, 0), see graph at right.

b: Shifted to the right three units.

2-8. a: 4, 1, 0.25; t(n) = 256(0.25) n – 1

b: They get smaller, but are never negative.

c: See graph at right. They get very close to zero.

d: The domain is n integers greater than or equal to zero.

The domain of the function is all real numbers.

t(n)

2-9. a: y = − 2 3 x − 4 b: y = 2

n

c: x = 2 d: y = 2 3 x − 8 3

2-10. n = 24; 650 = 5 26

Lesson 2.1.2

2-16. Answers will vary.

2-17. a: (0, –6)

b: (–6, 0) and (1, 0)

c: x-intercepts at (0, 0) and (–5, 0) and y-intercept at (0, 0); the graph of p(x) is 6 units

lower than q(x)

d: –6

2-18. a: z = 1.5 b: z = – 18 5

c: z = 8 d: z = –3, 2

2-19. a: 3 b:

x 2 y 4 c: y

2-20. a: 3p + 3d = 22.50 and p + 3d + 3(8) = 37.5, so popcorn costs $4.50 and a soft drink costs

$3.00.

b: Answers will vary.

2-21. a: 146 ≈ 12.1 b: 145 ≈ 12.0 c: 50 ≈ 7.1 d: 5 2

2-22. Maximum profit is $25 million when n = 5 million.

2-23. a: vertex at (–3, –8), opens up, vertically stretched.

b: x-intercepts (–5, 0) and (–1, 0); y-intercept (0, 10)

2-24. a, b, and c: Answers will vary.

2-25. a: y = (x – 8) 2 – 5 b: y = 10(x + 6) 2 c: y = –0.6x(x + 7) 2 – 2

2-26. Answers will vary.

2-27. a: 5 2 b: 6 2 c: 3 5

2-28. a: x = 46.71 b: x = 8.19

2-29. About $ 365.00. b: y = 300(1.04) x

Lesson 2.1.3

2-35. a: y = 0 or 6 b: n = 0 or –5 c: t = 0 or 7 d: x = 0 or –9

2-36. a: (7, −16), y = (x − 7) 2 −16 b: (2, −16), y = (x − 2) 2 −16

c: (7, −9), y = (x − 7) 2 − 9 d: (2, –1)

2-37. a: (2, –1)

b: When x = 2, (x – 2) 2 will equal zero and y = –1, the smallest possible value for y in

the equation. So the y-value of the vertex is the minimum value in the range of the

function.

2-38. a: 9.015 gigatons

b: C(x) = 8(1.01) (x+2) if x represents years since 2000 or 8.16(1.01) x .

2-39. a: 2 b: 1 c: 1 d: 2 e: 2 f: 1

h: If the factored version includes a perfect-square binomial factor, the parabola will

touch at one point only.

2-40. a: 4 b:

16x 4 y 10

c: 6xy3

2-41. a: 8 27

12

27 c: 6

27 d: 1

Lesson 2.1.4

2-50. a: f (x) = (x + 3) 2 + 6 (–3, 6), x = –3

b: y = (x − 2) 2 + 5 , (2, 5) x = 2

c: f (x) = (x − 4) 2 −16 , (4, –16), x = 4

d: y = (x + 3.5) 2 −14.25 , (–3.5, –14.25), x = –13.5

2-51.

b 2 a

2-52. The second graph is a reflection of the first

across the x-axis. See graph at right.

2-53. a: 45 = 3 5 ≈ 6.71; y = 1 2

x + 5 b: 5; x = 3

c: 725 ≈ 26.93; y = − 5 2 x + 5 2

d: 4; y = –2

2-54. After x is factored out, the other factor is a quadratic equation. After using the

Quadratic Formula the solutions are x =

−23± 561

8

or 0.

2-55. a: x = 21 b: x = 10 5 ≈ 22.4 c: x = 50

2-56. a: 1 4 b: 1 3

2-57. B

2-58. a: A cylinder b: 45π = 141.37 cubic units

2-59. a and b: Answers will vary. c: A circle.

2-60. (5, 14)

2-61. a: 0.625 hours or about 37.5 minutes.

b: 0.77 hours or about 46.2 minutes.

c: About $22.99 per minute.

2-62. a: 61 b: 30º c: tan −1 ( 4 5 ) d: 5 3

2-63. a: Years; 1.06; 120,000; 120000(1.06) x

b: Hours; 1.22; 180; 180(1.22) x

Lesson 2.1.5

2-69. Answers will vary.

2-70. See graph at right.

a: It is the slope.

b: No, because only lines have (constant) slopes.

This 2 is the stretch factor.

2-71. a and b: No. Answers will vary.

2-72. a: y = 0.25 ⋅6 x b: y = 12 ⋅0.3 x

2-73. a: x: (1, 0), ( – 5 2 , 0 ), y : (0, –5) b: x: (2, 0), y: none

2-74. See graphs at right.

a: stretched parabola, vertex (0, 5)

b: inverted parabola, vertex (3, –7)

2-75. a: x = ± 5 b: x = ± 11

Lesson 2.2.1 (Day 1)

2-81. Possible equation: y = − 4 25 (x − 5)2 + 8 , standing at (0, 0)

domain: 0 ≤ x ≤ 10; range: 4 ≤ y ≤ 8

2-82. a: x : (− 1 2 , 0), (−1, 0); y : (0, 1) b: x = – 3 4

( ) or (–0.75, –0.125)

c: – 3 4 , – 1 8

2-83. Move it up 0.125 units: y = 2x 2 + 3x +1.125

2-84. a: 2 6 b: 3 2 c: 2 3 d: 5 3

2-85. a: Years; 0.89; 12250; 12250(0.89) x b: Months; 1.005; 1000; 1000(1.005) x

2-86. a: 32 b: x 2 y 2 x c: x2 y

2-87. c + m = 18 and $4.89c + $5.43m = $92.07

10.5 lbs. of Colombian and 7.5 lbs. of Mocha Java.

Lesson 2.2.1 (Day 2)

2-88. a: 15 ft

b: Surface area of concrete: 793.14 sq. ft.; 528.76 cu. ft.; $1,263.74

2-89. a: See graph at right. b: y = 3x +2 c: 2, 5, 8, 11

d: One is continuous and one is discrete. They have

the same slope so the “lines” are parallel, but they

have different intercepts.

2-90. a: 4.116 ⋅10 12 b: y = 1.665(10 12 )(1.0317) t

c: Answers will vary.

2-91. a: 6 x + 3 y b: 32 c: 5 d:

2-92. a: 6x 3 + 8x 4 y b: x 14 y 9

2-93. See graph at right. line of symmetry x = 4

2-94. a: 4π + 4 3 π ≈ 16.755 m 3 b: No; r, r 2 , r 3 relationship; V = 80π

3 ≈ 83.776 m 3

c: V = 4 3 π r3 + 4π r 2

Lesson 2.2.1 (Day 3)

2-95. a: y = 1

x+2 b: y = x2 – 5 c: y = (x – 3) 3 d: y = 2 x – 3

e: y = 3x – 6 f: y = (x + 2) 3 + 3 g: y = (x + 3) 2 – 6 h: y = –(x – 3) 2 + 6

i: y = (x + 3) 3 – 2

2-96. He should move it up 6 units or redraw the axes 6 units lower.

2-97. a: 18 b: 3 2 c: 1 3 or 3

d: 11+ 6 2

2-98. a: (2x – 3y)(2x + 3y) b: 2x 3 (2 + x 2 )(2 − x 2 )

c: (x 2 + 9y 2 )(x − 3y)(x + 3y) d: 2x 3 (4 + x 4 )

2-99. x = −by3 +c+7

a

2-100. a: t(n) = –6n + 26 b: t(n) = −1.5(4) n or − 6(4) n−1

2-101. a: See graph at right. b: 2

c: –1 d:

–13

e: no solution f: Three because the graphs cross three times.

g: x 3 – x 2 – 2x

Lesson 2.2.2 (Day 1)

2-107. a: y = (x – 2) 2 + 3 b: y = (x – 2) 3 + 3 c: y = –2(x + 6) 2

2-108. a: D: all real numbers, R: y ≥ 3 b: D and R: all real numbers

c: D: all real numbers, R: y ≤ 0

2-109. a: compresses or stretches b: shifts up or down

c: shifts left or right d: shifts up or down

2-110. a: y = 0.4 ⋅0.5 x b: y = 8 ⋅2 x

2-111. a: 2 25 b: 3x2 y 3

z 4 c: 54m 4 n d: y 5x 2 z

2-112. a: See table and graph at right.

b: He had 28,900 miles in May.

c: 5600 miles

d: No, he will not be able to go in

December, he will only have

24,200 miles.

Miles

2-113. a: x = ± y 2 +17 b: x = (y + 7)3 − 5

Month

Month Miles

1 15,000

2 18,000

3 22,900

4 25,900

5 28,900

6 8,800

7 11,800

8 14,800

9 19,700

10 22,700

11 25,700

12 5,600

Lesson 2.2.2 (Day 2)

2-114. a: (10, 48) b:

29

( 5

, 9 5 )

2-115. a: 8 3 b: 3 x c: 12 d: 108

2-116. a: g ( 1 2 = –4.75 ) b: g(h +1) = h 2 + 2h − 4

2-117. See graph at right.

a: y = 2x : (0, 0), y = – 1 2

x + 6 : (0, 6), (12, 0)

b: It should be a triangle with vertices

(0, 0), (12, 0), and (2.4, 4.8).

c: Domain 0 ≤ x < 12, Range 0 ≤ y ≤ 4.8

d: A = 1 2

(12)(4.8) = 28.8 square units

2-118. y ≈ 2(x − 5) 2 + 2 and y ≈ − 1 2 (x − 5)2 + 2

2-119. See graph at right.

y = (x +1) 2 − 81; x-intercepts: (–10, 0), (8, 0),

y-intercept: (0, –80); vertex: (–1, –81)

2-120. Yes, when n = 117.

Lesson 2.2.3

2-125. a and : Neither c: Even

2-126. a: b: c: Neither function is odd nor even.

2-127. y = − 3 4 (x − 2)2 + 3

2-128. a: x: (–1, 0), y: (0, 2) b: x: (0, 0), (2, 0), y: (0, 0)

V: (–1, 0), y = 2(x +1) 2 V: (1, 1), y = –(x –1) 2 +1

2-129. a: y = x b:

( 2

, 1 3 ) c:

, 1 3 )

d: The solution to the system is the point at which the lines intersect.

2-130. a: t(n) = 40( 1 4 )n or 10( 1 4 )n−1 b: t(n) = −6n + 4

2-131. a: x: (2, 0), (6, 0) y: (0, 2) vertex (4, –2), D: all real numbers; R: y ≥ –2

b: x: (–4, 0), (2, 0) y: (0, 2) vertex (–1, 3), D: all real numbers; R: y ≤ 3

Lesson 2.2.4

2-139. y = (x + 3.5) 2 − 20.25

2-140. a: See graph at right.

b: Loudness depends on distance.

2-141. See graph at right. The domain is all positive

numbers (or d > 0). The range is all real

numbers greater than or equal to 3 and that

are multiples of 0.25.

$5.00

$4.00

Loudness

Distance

2-142. Answers will vary.

$3.00

2-143. The second graph shifts the first 5 units left and

7 units up and stretches it by a factor of 4.

2-144. a: x 2 –1 b: 2x 3 + 4x 2 + 2x

c: x 3 − 2x 2 − x + 2 d: y: (0, 2), x: (1, 0), (–1, 0), (2, 0)

2-145. a: (a, b) = ( 2, ± 1 2 ) b: (a, b) = ( 1 2 , ± 2 )

2-146. a: y = − 5 9 (x − 3)2 + 5 b: x = − 3

25 (y − 5)2 + 3

0.2 0.4 0.6 0.8

Distance (miles)

2-147. See graph at right.

a: y = 2x 2 − 4x + 6

b: There is no difference, but the explanations vary.

c: y = x 2

d: y = x 2

2-148. a: The graph will be a circle with a center at (5, 8) and a radius of 7.

b: See graph at right.

2-149. a: –2 b: –2 c: 1 2

d: –1

e: The product of the slopes of any two perpendicular lines is –1.

2-150. Answers will vary.

2-151. a: (0, 0), (–24, 0), and (0, 0) b: (6, 0), (10, 0), and (0, 60)

2-152. (3, 2)

Lesson 2.2.5

2-162. x < 2, y = –(x – 2) 2 ; x ≥ 2, y = x + 2

2-163. Any function for which f (x) = f (−x). On a graph, the function will have the y-axis as its

line of symmetry.

2-164. y = −2 x + 3 + 4

2-165. a: (x + 2) 2 + (y − 3) 2 = 4 b: (x −12) 2 + (y +15) 2 = 81

2-166. y = (x − 2.5) 2 + 0.75 , vertex (2.5, 0.75)

2-167. He is incorrect. Answers will vary.

2-168. f (x) = x 2 +1

2-169. ± 11, ± 9, ±19

Lesson 3.1.1

3-5. a: 4x 2 –12x +14 b: 81y4

x 4

3-6. a: 3 b: 4 c: 1 d: 5 e: 2

3-7. They are both correct: x12 y 3

64

.

3-8. a: Horizontal line through (0, 3), domain: all real numbers, range: 3

b: Vertical line through (–2, 0), domain: –2, range: all real numbers

c: (–2, 3)

3-9. m = 15, b = –3

3-10. a: (4, 8, 4 3 ), (5, 10, 5 3 )

b: The long leg is n 3 units long, and the hypotenuse is 2n units long.

3-11. a: 15, 21, 27, 33, t(n) = 6n –3

b: 27, 81, 243, 729, t(n) = 3 n

3-12. a: 1 5 b: 3 c: 27 d: 1 8

Lesson 3.1.2

3-23. a: not equivalent b: equivalent c: equivalent

d: equivalent e: not equivalent f: not equivalent

3-24. a: equal if x = 0 e: equal if x = 0 or x = 1 f: equal if a = 1 or a = 0

3-25. a: Possibilities include x – 2 = 4 or 2x – 4 = 8.

b: They have the solution x = 6.

c: 3 – x = 7, x = –4

3-26. a: t (n) = –3n + 17, points along a line with y-intercept (0, 17) and slope –3;

b: t(n) = 50(0.8) n , points along a decreasing exponential curve with

y-intercept (0, 50)

3-27. a: 4 b: –30 c: 12 d: –2 1 4 e: x = –4, 1 3

3-28. (0, 0) and (–6, 0)

3-29. a: 2x 2 + 6x b: x 2 – 2x –15 c: 2x 2 – 5x – 3 d: x 2 + 6x + 9

3-30. The first graph opens downward, is stretched, and has its vertex at (–1, –3).

The second is the parent graph.

3-31. a: (1, –4) b: (1, –4) c: (–2.5, –4.25)

3-32. a: y8

x 12

d: Domain: −∞ < x < ∞ , Range: y ≥ –4.25

b: −18x3 y + 6x 5 y 2 z

3-33. a: odd b: even c: even

3-34. a: $4.00 b: $4.00

c: $4.00. $5.00 d: See graph above right.

e: No, it is a step function. f: The graph will shift (translate) upward by $2.00.

3-35. a: (x + 2) 2 + (y −13) 2 = 144 b: (x +1) 2 + (y + 4) 2 = 1

c: (x − 3) 2 + (y + 8) 2 = 16

-1 -1 1 2 3 4 5 6 7 8

Hours

3-36. a: 24 blocks per hr. b: 18 blocks per hr. c: 11.08 blocks per hr.

Lesson 3.1.3

3-45. a: n = –2 b: x = –4, 1

3-46. a: equivalent b: equivalent c: equivalent

d: not equivalent e: not equivalent f: not equivalent

3-47. d: equal if a = 0 or b = 0 e: equal if x = 1 f: equal if x = 5 and y = 2

3-48. 10 = 15m + b and 106 = 63m + b; m = 2, b = –20, t (n) = 2n –20

3-49. a: t(n) = 450000 (1.03) n

b: They will make $154,762.37 or 34.39% profit.

3-50. 5xy(x + 2)(x + 5)

3-51. a: They both have the solution x = 2.

b: She divided both sides of the equation by 150 and used the Distributive Property.

3-52. a: –6, –14, –22, –30, t(n) = 18 – 8n

b: 2 5 , 25 2 , 2

125 , 2

625 , t(n) = 50 ( 1 5 ) n

3-53. a: 5 1/2 b: 9 1/3 or 3 2/3 c: 17 x/8 d: 7x 3/4

3-54. a: x 2 + y 2 = 36

b: (x − 2) 2 + (y + 3) 2 = 36

c: (x − 4) 2 + (y + 5) 2 = 36

3-55.

741.8−25

1800−0

= 0.4 ºF/sec

3-56. a: b: Shift the graph up $11.

Price ($)

100

50

Duration (days)

Lesson 3.2.1

3-63. odd numbers; 46 th term: 91; n th term: 2n – 1

3-64. after 44 minutes

3-65. a: 1.03 b: f (n) = 10.25(1.03) n c: $13.78

3-66. (y –2)(y – 2)(y –2)

3-67. a: x 1/5 b: x –3 3

c: x 2

d: x –1/2

e:

xy 8 f: 1

m 3 g: xy3 x h:

81x 6 y 12

3-68. Yes, he can. a: x = 2 b: Divide both sides by 100.

3-69. a: 5m 2 + 9m – 2 b: –x 2 + 4x +12

c: 25x 2 –10xy + y 2 d: 6x 2 –15xy +12x

Lesson 3.2.2

3-78. a: x–4

3x+2 b: 5

x–3

c: 2

3-79. a: 1 b: none c: 2 d: 1

3-80. a: x – 2 = 4 b: For each, x = 6. c: x + 3 = 8, x = 5

3-81. a: x < 0 b: x ≤ –4

3-82. a: 3 7 b: 5 4

3-83. Graph shown at right.

a: y = x 3 ; The vertex has been shifted up 4 and left 2.

b: It would not differ.

3-84. See graph at right.

a: all real numbers

b: See graph at far right.

c: f(x) is a continuous function with

range y > 0 while t(n) is a discrete

sequence with positive integer inputs.

15

10

Lesson 3.2.3

3-90.

2x

3(2x–1) = 2x

6x–3

3-91. a: x ≠ –4 or 2, x+4

x–2

x–4

x+4

b: x ≠ –2 or 3,

2(x+2)

(x–3) 2

3-92. Answers will vary.

3-93. a:

( 3

, –2) b: (4, –9)

3-94. n ⋅ 3 15 = 72 million, n = 5; There were 5 bacteria at first.

3-95. The function is even. A reflection across the y-axis results in the same graph.

3-96. a: m = 6 b: x = 5.5 c: k = 4 d: x = 90

Lesson 3.2.4

3-102. a: Because if x = 4, then the denominator is zero. Since dividing by zero makes the

expression undefined, x ≠ 4.

b: a: x ≠ – 1 3

and x ≠ 5; b: x ≠ 3 or –3

3-103. a:

8x+8

(x–4)(x+2) b: 1

x+2

3-104. a: all real numbers b: –5 < x < 4

c: no solution d: x = 1 3

3-105. a: x – 4 b: 7m–1

3m+2

c:

(4z–1)2

z+2

d: x–3

3-106. a: 1722 b: 1368 c: y = 1500(1.047) n+3

3-107. a: 5(3x–1)

2(4x+1)

b: 1 c: 3 d: –m2

3-108. See graph at right; x-intercept: (–2, 0), y-intercept: (0, –2);

there is no value for f(1), which creates a break in the graph.

3-109. a: –15 b: –4 c: 3 d: –m 2

Lesson 3.2.5

3-113. a:

3x+1

x–7

x–3 c: x–2

2x+12 d: 1

3-114. a: 1 b: 4 c: 2 d: 5

3-115. a: x = 3 b: 0 ≤ x ≤ 6 c: x = 1 or 5 d: x < 2 or x > 4

3-116. Domain: all real numbers; Range y ≥ 0; g(–5) = 8

g(a + 1) = 2a 2 +16a + 32 , x = 1 or x = 7, x = –3

3-117. x = –3 or –11

3-118. a: 1 b: 3 c: 2

3-119. (–3, 8) and (1, –12)

3-120. a:

x+1

x 2 −4

x+6

2(x+2) 2 c: 1 x d: – 1 2

3-121. x = 62

3-122. a: y = – 1 2 x +12 b: y = 2 3 x –15

3-123. The width is 1.5 meters, and the outer dimensions are 8 m by 5 m.

3-124. 80x + 0.5 = 100x, so x = 1 40

of an hour or 1.5 minutes.

3-125. 6 7

3-126. a: (5x –1)(5x +1) b: 5x(x + 5)(x − 5)

c: (x +9)(x –8) d: x(x –6)(x +3)

Lesson 4.1.1

4-7. See graph at right. x = 0 and x = 4

4-8. a: x = 5 or x = –3 b: m = 35

c: no solution d: x = 7

4-9. a: y = 0 b: x = 0

4-10. a: Combining the equations leads to an impossible result,

so there is no solution.

c: There can be no intersection because the lines are parallel.

When assuming there is an intersection, students will find that their work results in a

false statement.

4-11. This is a scalene triangle, because the sides have lengths 29 , 17 and 20 .

4-12. a: 63 b: 0 c: n 3 –1 d: Neither; answers will vary.

4-13. a: (x−2)(x+6)

(x+4)(2x+3)

2x+1

x−5

9m+27

m+3

= 9 d:

n+3

n–1

4-14. a: 0-2 times

b: 0-4 times

c: 0-4 times

d: 1-3 times if you consider parabolas that open up or down. 0-4 times if you consider

rotated parabolas.

Lesson 4.1.2

4-22. Graph y = (x − 3) 2 − 2 and y = x +1 and find the x-values of the points of intersection.

Or, graph y = x 2 − 7x + 6 and find the x-intercepts. x = 1 or x = 6

4-23. See graph at right. x = 3 or x = 6

4-24. a: x = 15 b: x = 7 3

or x = –5

4-25. The lines intersect at the point (2, 6).

Ted will solve the system algebraically by setting 18x – 30 = –22x + 50.

4-26. a = 18.5, b = 5.5

4-27. x = 13, x = 5 is extraneous

4-28. x = 36 b: x = 20 2 or x ≈ 28.28

4-29. See graph at right. x = 1 and x = 3; No.

4-30. a: 1 2 (x − 2)3 +1 = 2x 2 − 6x − 3, x = 0 or x = 4

b: x = 6 is also a solution.

c: 1 2 (x − 2)3 +1 = 0 , x ≈ 0.74

d: Domain and range of f(x): all real numbers, domain of g(x): all real numbers,

range of g(x): y ≥ –7.5

4-31. a: x = –3 b: x = 1 or x = 3 c: x = –8 or x = 13 d: x = 1.2

4-32. a: y = 5 3 x – 4 b: m 2 = Fr2

Gm 1

c: m = 2E

v 2 d: y = ± 10 − (x − 4)2 +1

4-33. (a + b) 2 = a 2 + 2ab + b 2 , substitute numbers, etc.

4-34. a: See graph at top right.

b: See graph at bottom right.

c: Graph (b) is similar to graph (a), but is rotated

90º clockwise.

d: (a) D: all real numbers, R: y ≥ 0; (b) D: x ≥ 0, R: all real numbers

4-35. a: 21.00 b: 117.58

Lesson 4.1.3

4-40. a: (–2, –11); The lines intersect at one point.

b: infinite solutions; The equations are equivalent.

c: (2, 45), and(–1, 3); The line and parabola intersect twice.

d: (3, 6); The line is tangent to the parabola.

4-41. a: y = 3 or y = –5 b: x = – 99 4

c: y = 1 d: x = –13

4-42. a: E t(n) = −2 + 3n ; R t(0) = −2, t(n +1) = t(n) + 3

b: E t(n) = 6( 1 2 )n ; R t(0) = 6, t(n +1) = 1 2 t(n)

c: t(n) = 10 – 7 d: t(n) = 5(1.2) n e: t(n) = 1620

4-43. 19.79 feet

4-44. a: m = – 6 5 , b = (0, –7) b: m = 2 3 , b = (0, –5) c: m = 2, b = (0, –12)

4-45. a: not a function; D: –3 ≤ x ≤ 3; R: –3 ≤ y ≤ 3

b: a function; D: –2 ≤ x ≤ 3; R: –2 ≤ y ≤ 2

4-46. (–7, 11)

Lesson 4.1.4

4-51. 4c + 5p = 32, c + 8p = 35, cylinders weigh 3 oz. and prisms weigh 4 oz.

4-52. Yes. No. Any value of x such that –3 ≤ x ≤ 2 is a solution.

4-53. a: x = 4 b: x = 6 c: x = 6 d: x = 3 2

4-54. a: (4, –6) b: (4, –6) c:

, – 9 4 )

4-55. a: b:

4-56. B

4-57. a: See graph at right. b: x ≈ 0.71

Lesson 4.2.1

4-65. a: boundary point x = –4 b: boundary points x = 4, – 3 2

4-66. a: –4 < x < 1 b: x ≤ –4 or x ≥ 1 c: –1 < x < 4

d: x ≤ –1 or x ≥ 4 e: –1 < x < 4 f: x ≤ –1 or x ≥ 4

4-67. a: y = –3x + 8 b: y = –x – 1 2

4-68. a: No real solutions b: y = 7, y = 13 3

is extraneous

4-69. a:

3x 2 +x−3

2x 3 +9x 2 −5x

b: 3x−5

2x+3

4-70. x = −6 + 4 6 or x = –6 – 4 6

4 x−3

m+5

m+4

4-71. a: x(b + a) b: x(1 + a) c:

d: x–b

4-72. See graph at right.

a: Rectangle; perpendicular lines or slopes.

b: (1, 4), (–3, –3), (–5, 1), (3, 0)

4-78. a: y = 1 2 x – 2 b: y = 2x + 2

4-73. a: –5 < x < 13 b: x ≥ 250 or x ≤ –70 c:

2 3 ≤ x ≤ 2

4-74. a: C = 800 + 60m b: C = 1200 + 40m c: 20 months d: 5 years

4-75. a: input x, output x

b: Replace x with c in first function machine resulting in c – 5 , then substitute this

expression for x in the second function machine, yielding 6(c−5)+8

= 3c −11 .

Substitute this a third time in the final machine, giving (3c−11)+11

= c .

4-76. a:

3x–14

2x–1

4-77. (1, 12) and (–5, 42)

Lesson 4.2.2

4-83. x = –2, y = 3, z = –5; Solve the system to two equations with x and y, then substitute these

values into the third equation to find z.

4-84. a: x ≤ 4 b: x < –6 or x > 6

4-85. red = 10 cm, blue = 14 cm

–5 0 5 –6

0 6

4-86. The points on the line y = 2x – 2 are excluded from the solution region of y < 2x –2.

4-87. a: y = 1 3 x – 4 b: y = 6 5 x – 1 5 c: y = (x + 1) 2 + 4 d: y = x 2 + 4x

4-88. y = 0, x = 0

4-89. 2.11 feet

Lesson 4.2.3

4-92. There is no solution, so the lines are parallel.

4-93. See graph at right.

a: A square, justifications will vary.

b: (0, –3), (4, 1), (–4, 1), (0, 5)

c: 32 square units.

4-94. a: x < 13 b:

5− 57 5+ 57

≤ x ≤

or −0.637 ≤ x ≤ 3.137

13 –1 0 3

4-95. a: no solution b: y ≈ 4.3 or 10.7

4-96. (25, –3)

a: x 2 + 3y = 16 and x 2 − 2y = 31

b: The solutions to the new system are (5, –3) and (–5, –3).

4-97. a: See graph at right; y = –2x +8

b: 63.43º or 116.57º

Lesson 4.2.4

4-99. a: y = 1 2 (x + 3)2 − 2 b: y = x +5 c: x = 1 or x = –5

d: (1, 6) and (–5, 0) e: x < –5 and x > 1 f: x = –1 or x = –5

g: x = –1 h: Answers will vary.

4-100. y ≤ 3x + 3, y ≥ 0.5x − 2 , y ≤ −0.75x + 3

4-101. a: x ≤ 1 or x ≥ 7 3 b: x < 3

0 1 2 x 1 2 3 x

4-102. a: y b: y c:

4-103. 60º

4-104. a: y > 3x –3 b: y < 3 c: y ≥ 3 2 x – 3 d: y ≥ x2 – 9

4-105. a: w = 0 or w = –4 b: w = 0 or w = 2 5

c: w = 0 or w = 6

Lesson 5.1.1

5-8. a: y = 2(x +3) b: Yes, y = x. See graph at right.

5-9. a: 9 b: 4 c: x ≈ 1.89

5-10. x = sin −1 (0.75) ≈ 48.59° ; to check: sin(48.59 ) ≈ 0.75

5-11. x must equal y.

5-12. a: x = 12 5 b: x = 5 2 c: x = 8 d: x = 80 3

5-13. The area between an upward parabola with vertex (0, –5)

and the downward parabola with vertex (1, 7). See graph at right.

5-14. a and b: See graph at right.

c: Possible equation: y = 10x – 5

d: For this equation, approximately $495.

5-15. ≈ 17.74 feet

Weight (ounces)

Cost (dollars)

Lesson 5.1.2

5-26. See graph at right.

5-27. a: y = 1 3

(x + 8) b: y = 2(x – 6) c: y = 2x – 6

5-28. x ≈ 0.53

5-29. a: x 2 – 5x –14 b: 6m 2 +11m – 7 c: x 2 – 6x + 9 d: 4y 2 – 9

5-30. (x + 3) 2 + (y − 5) 2 = 9 . See graph at right.

5-31. a:

x−3

x(x−4) b: 4

c: 2 d: x–1

5-32. a: f (x) ≈ 1.5(1.048) x

b: ≈ $425.04

5-33. See graph at right.

For f(x), domain: –2≤ x ≤ 5, range: –3 ≤ y ≤ 3

For f –1 (x), domain: –3 ≤ x ≤ 3, range: –2≤ y ≤ 5

-4

5-34. a: L(x) = x 2 −1, R(x) = 3(x + 2)

b: 30

c: Order does matter – show by substituting numbers; output is 224 if x = 3 for L(R(x)).

5-35. a: The system has no solution.

b: The graphs do not intersect, they are parallel lines.

5-36. If she adds nothing else to the account and it just sits there making interest, she will have

$440.13 on her eighteenth birthday.

5-37. a: x 2 –10x – 56 b: 4m 2 + 8m – 5

c: x 2 – 81 d: 9y 2 +12y + 4

5-38. a: (2, 0), (–1, 0) b: (–5, 0), (–3, 0)

5-39. x = 2.5

Lesson 5.1.3

5-48. 121 b: 17

5-49. a: 2x 3 + 2x 2 − 3x − 3 b: x 3 − x 2 + x + 3

c: 2x 2 +12x +18 d: 4x 3 − 8x 2 − 3x + 9

5-50. a: x = – 10 7 b: x = 1 3 or x = 1

c: x = 115 d: x = 0 or x = 4

5-51. a: y = ± x − 3 b:

y = 4 ( x − 6)

c: y = x2 +6

5-52. (x − 2) 2 + y 2 = 20 ; circle, x 2 + y 2 = r 2 , center (2, 0)

and radius ≈ 4.5; See graph at right.

5-53. 70

5-54. a: 3 b: y – 4 c: 1 3x d: x

Lesson 5.2.1

5-60. Domain: x > 0; Range: −∞ < y < ∞ ; x-intercept: (1, 0) no y-intercept;

asymptote at x = 0

5-61. a: undefined b: x ≠ 7 c: g(3) = 11 d: f (g(3)) = − 1 2

5-62. a: e(x) = (x −1) 2 − 5

b: One machine undoes the other so e( f (−4)) = −4.

c: They would be reflections of each other across the line y = x.

5-63. See graph at right.

a: Domain: all real numbers, range: y > –3

b: No

c: (0, –2), (1.585, 0)

d: Sample: y +a = 2 x , where a ≤ 0.

5-64. a: x ≈ 36.78 b: x ≈ 31.43

5-65. a: B = 0.07(0.3x) or B = 0.021x

b: S = 0.09(0.7x) or S = 0.063x

c: 0.084x = 5000; $59,523.81

5-66. a: (x + 7)(x – 7) b: 6x(x + 8)

c: (x – 9)(x + 8) d: 2x(x + 2)(x – 2)

5-67. The region between the two parabolas, see graph at right.

-5

Lesson 5.2.2

5-73. x = 2 y , no, yes, yes; They have the same graph or give the same table of (x, y) values,

or one is just a rewritten equation of the other.

5-74. a: x = log 5 (y) b: x = 7 y c: x = log 8 (y)

d: K = log A (C) e: C = A K f: K = ( 1 2 ) N

5-75. a: $1.90, 1.38, 0.96, 0.94, 0.90, 0.88

b: decrease

c: Smaller size. Note: Sketching a graph of rate with respect to

bag size like the one at right may help here.

Rate

5-76. Answers will vary. Possible answers:

5-77. x = –4

a: Factor and use the Zero b: Take the square root (undo)

Product Property (rewrite), (–8, 0) and (1, 0)

c: Quadratic Formula d: Complete the square (rewrite)

Bag size (pounds)

5-78. a: x = 17 3 ≈ 29.44 b: x = 4 2 ≈ 5.66

5-79. See graph at right. domain: x ≥ 0, range: y ≥ 0, x- and y-intercept: (0, 0),

no asymptotes, half of parabola: y = πx 2

5-80. a: A good sketch would be a parabola opening upwards with a locator

point at (–6, –7).

b: Shift the graph up 9 units.

c: The graph is the same except the region below the x-axis is reflected across the axis so

that the graph is entirely above the x-axis.

e: y = x + 7 − 6

Lesson 5.2.3

5-84. Possible answer: y = 2 x +15 5-85. y = log 7 x

5-86. n ≈ 3.66 5-87. (x + 2) 2 + (y − 3) 2 = 4r 2

5-88. $0.66

5-89. See graphs at right.

a: The second is just the first shifted up ten units.

b: y = km x + b

5-90. a: x = 10 or x = –8 b: x = 2 or x –4

c: –2 < x < 4 d: x ≥ 3 or x ≤ –13

5-91. a: x(x + 8) b: (xy + 9z)(xy – 9z)

c: 2(x +8)(x –1) d: (3x + 1)(x – 4)

5-92. a: 2 b:

x+2 c: x−4

(x−2)(x−1)

4 x+16

x(x+2)

Lesson 5.2.4

5-96. a: 12 because 12 .926628408 = 10 b: Answers will vary

5-97. a: x = 25 b: x = 2 c: x = 343 d: x = 3 e: x = 3 f: x = 4

5-98. Less than one; Answers will vary.

5-99. x ≈ 17.973

5-100. a: (2x + 1)(2x – 1) b: (2x +1) 2 c: (2y + 1)(y + 2) d: (3m + 1)(m – 2)

5-101. a: –1 < x < 3 b: x ≤ 1 or x ≥ 2

5-102. No; log 3 2 < 1and log 2 3 > 1

5-103. a: a = y

b x

b: b is the x th root of y a , or b = x y a

5-104. See graphs at right.

Lesson 5.2.5

5-112. f (g(x)) = g( f (x)) = x ; They are inverses.

5-113. No. For f (x) = mx + b, f (a) + f (b) = ma + b + mb + b = m(a + b) + 2b and

f (a + b) = m(a + b) + b .

5-114. x ≈ 1.585

5-115. a: t(n) is arithmetic, h(n) is geometric, q(n) is neither

b: No, because h(n) is increasing much faster than the other two.

c: h(1) = q(1) = 12 and t(2) = h(2) = 36 ; continuous graphs for t(n) and q(n) intersect but

not for an integer n. h(n) is increasing much faster than q(n).

5-116. s(n) = (50 + 7n) 2 − 6(50 − 7n) +17 , neither, it is quadratic and there is no common

difference or multiplier.

5-117. a: 1

5-118. See graph at right.

b: 10x+m

5-119. (–3, 0, 5)

5-120. m ≈ 2.19

-5 5

5-121. a and b: g( f (x)) = log x or f (g(x)) = log x ,

see graphs at right.

c: The log of an absolute value is very different

from the absolute value of a log.

d: See graph at right. Note that x = 0 is an asymptote

5-122. 1 2

no matter where X is placed.

5-123. x ≈ 1.68

5-124. a:

6x−21

(x−4)(x+1)

5+6x

2(x−5) c: 1

x 2 −9

5-125. a: b + a b: 3d + 2c 2 c: x – 1 d: xy

Lesson 6.1.1

6-8. a: Their y- and z-coordinates are zero. b: Answers will vary.

6-9. x = –2, y = 5

6-10. a: 9 b: 4N – 3, arithmetic

6-11. a: x ≈ 1.204 b: x ≈ 1.613 c: x = 6 d: x ≈ 2.004

6-12. a:

25 1 b: x

y 2 c: 1

x 2 y 2 d: b10

6-13. a: x b:

x 2 −3x+2

6-14. a: 1 2

b: –2

c: The product of the slopes is –1, or they are negative reciprocals of each other.

6-15. Heather is correct, because a 4% decrease does not “undo” a 4% increase.

Lesson 6.1.2

6-21. a: (0, 10, 0), (0, 0, 4) b: (8, 0, 0), (0, 6, 0), (0, 0, 12)

c: (0, 0, 4), (0, 0, –4), (2, 0, 0) d: (0, 0, 6)

6-22. A line b: They do not intersect. c: They do not intersect.

6-23. a: y = −2(x + 4) 2 + 2 b: y = 1

x−2 c: y = −x3 + 3

6-24. It is not the parent. The second equation does not have a vertical asymptote, and it has a

maximum value, while y = 1 x

does not.

6-25. a: x = b 3 b: x = b

5a

c: x = b

1+a

6-26. a: No, input equals output only if x ≥ 0.

b: The output is the absolute value of the input value.

c: n + 2, n 2 − 4 , n

d: Because x 2 = x .

6-27. It is the log 5 (x) graph shifted 2 units to the right.

See graph at right.

6-28. a: 254,000 people/year b: 1,574,000 people/year c: 1960 to 2010

6-29. a: –7 b: –102 c: –102 d: –132

Lesson 6.1.3

6-35. See graph at right.

6-36. Yes.

6-37. Answers will vary.

6-38. y ≤ −x + 4, y ≥ 1 3 x

6-39. a: x+3

2x−1 b: 1

(x−3)

• • • • • • • • • • • • • • • •

• • • • • • • • • • • • • • • • •

z

6-40. a: Most solving strategies will yield x = 8 or x = 1.

b: x = 1 does not check, so it is extraneous.

6-41. a: x = –4 or x = 5 2

b: x = –4, 2, or 3

6-42. a: Neither b: Even

6-43. x = 3, y = 1, z = 3

Lesson 6.1.4

6-51. (1, –2, 4)

6-52. a: ≈ $140,809.30 b: ≈ 24.2 years c: ≈ $164,706.25

6-53. x = 7

6-54. a: They both equal 16, but this is a special case (for example, 5 3 ≠ 3 5 ).

6-55. a: x = 6.5 b: x = –3.75 or x = 5

6-56. a: y = 1 3 x + 5 b: y = 2x + 5 c: y = − 1 2 x + 15 2

d: y = 2x

6-57. a: y = −x 2 + 4x b: y = 5 ± x − 3

6-58. a: See graph right.

b: See graph far right.

6-59. 384 feet

Lesson 6.1.5

6-71. x = –1, y = 3, z = 5

6-72. y = 3x 2 − 5x + 7

6-73. a: x+3

x−4 b: 1

6-74. a: y + x 2 b: 2b + 4a2 c: 6x – 1 d: xy

6-75. a: x = 12 y b: y x = 17 c: 2x = log 1.75 y d: 7 = log x 3y

6-76. x = 14

6-77. a: ≈ 0.0488 grams

b: Roughly between 4600 and 6700 depending on how the base is rounded.

c: Never

6-78. a: See graph at right.

b: x > −2; y = x + 2 and x ≤ −2; y = (x + 2) 3

6-79. a: 2 4 b: 2 –3 c: 2 1/2 d: 2 2/3

6-80. x = –1, y = 3, z = 6

6-81. y = 2x 2 − 3x + 5

6-82. a: 24 = b a b: 7 = (2y) 3x c: 5x = log 2 3y d: 6 = log 2q 4 p

6-83. a:

x−4

6-84. Yes, Hannah is correct; 4(x − 3) 2 − 29 = 4x 2 − 24x + 7 and 4(x − 3) 2 − 2 = 4x 2 − 24 + 34

6-85. a: y = 2(x − 2) 2 −1, vertex (2, –1), axis of symmetry x = 2

b: y = 5(x −1) 2 −12 , vertex (1, –12), axis of symmetry x = 1

6-86. See graph at right. y = log(x − 6) + 3

6-87. a: 2a 2 − 4 b: 18a 2 − 4 c: 2a 2 + 4ab + 2b 2 − 4

d: 2x 2 + 28x + 94 e: 50x 2 + 60x +14 f: 10x 2 −17

Lesson 6.2.1

6-95. y = 3 x

6-96. In 2 = 1.04 x the variable is the exponent, but in 56 = x 8 the exponent is known so

you can take the 8 th root.

6-97. x > 100, because 10 2 = 100

6-98. Answers will vary.

6-99. a: 1 8 b: 1 x

c: m ≈ 1.586 d: n ≈ 2.587

e: Answers will vary. x = b 1/a

6-100. 2 1/2 = 2 and 2 −1 = 1 2

6-101. a: –3 < x < 3 b: –2 < x < 1 c: x ≤ –2 or x ≥ 1

6-102. a: Yes

b: See graph at right, (it is not a function).

c: Not necessarily.

d: Functions that have inverse functions have no

repeated outputs; a horizontal line can intersect

the graph in no more than one place.

e: Yes; for example, a sleeping parabola is not a function,

but its inverse is a function.

6-103. a: x = –3, y = 5, z = 10

b: There are infinitely many solutions.

c: The planes intersect in a line.

Lesson 6.2.2

6-113. a: 5.717 b: 11.228

6-114. a: x2

x−1

b: b+a

a−a 2 b

6-115. log 5 7

log5 2

6-116. It is the log 3 (x) graph shifted 4 units to the left. See graph at right.

6-117. 16.5 months; 99.2 months

6-118. They are correct. Vertex: (2.5, –23.75), line of symmetry: x = 2.5.

6-119. a: f (x) = 4(x −1.5) 2 − 3, vertex (1.5, –3), line of symmetry x = 1.5

b: g(x) = 2(x + 3.5) 2 − 20.5 , vertex (–3.5, –20.5), line of symmetry x = –3.5

6-120. a: Consider only x ≥ –2 or x ≤ –2.

b: Depending on the original domain restriction, y = x+7

− 2 or y = − x+7

− 2 .

c: x ≥ –7 and y ≥ –2 or x ≥ –7 and y ≤ –2.

6-121. a:

x 2 −3x−4

6-122. a: 20, 100, 500 b: n = 7

c: No, because there are no terms between the 6 th term (62,500) and the 7 th term

(312,500).

Lesson 6.2.3

6-127. a: y = 40(1.5) x

b: When x = –9, or 9 days before the last day of October (October 22).

6-128. Possible answer: 4 (x+1) = 6

6-129. Answers will vary.

6-130. The graph should show a decreasing exponential function which will have an asymptote

at room temperature.

6-131. y = x 2 − 6x + 8

6-132. a: x ≥ 1 2 and y ≥ 3 b: g(x) = (x−3)2 +1

c: x ≥ 3 and y ≥ 1 2

d: x e: x (They are the same, because f and g are inverses.)

6-133. a: x ≈ 6.24 b: x = 5

6-134. a: (x −1) 2 + y 2 = 9 b: (x + 3) 2 + (y − 4) 2 = 4

6-135. a: x + 5 b: a + 5 c: x – y d: x2 +1

x 2 −1

6-136. a: p −1 (x) = 3 ( x 3 − 6) b: k −1 (x) = 3 ( x−6

3 )

c: h −1 (x) = x+1

d: j−1 (x) = 3x−2

= − 2 x + 3

Lesson 6.2.4

6-138. a: Decreasing by 20% means you multiply by 0.8 each time, and the presence of a

multiplier implies exponential.

b: y = 23500(0.8 x ) c: $9625.60

d: ≈ 6.12 years e: $42,926.44

6-139. a: x = 1 2

b: x > 0 c: x = 1023

6-140. a: x = 2.236 b: x = 4.230 c: x = 0.316

d: x = 2.021 e: x = 3.673

6-141. a: 16 b: 12 c: 12 4 = 20736 d: 54

e: No, they are not inverses (if they were, then the answers to parts (c) and (d) would

have to be 2).

6-143. c(x) = x 2 − 5

6-144. x = 17

6-145. a: 2(x+1)

x+3 b: 3x2 −5x−3

(2x+1) 2

6-146. a: 30° b: 22.6°

6-147. y ≤ −

4 3 x + 3, y ≥ − 4 3 x − 3, x ≤ 3, x ≥ −3

Lesson 7.1.1

7-3. a: The shape would be stretched vertically. In other words, there would be a larger

distance between the lowest and highest points of each cycle.

b: Each cycle would be longer horizontally. Fewer cycles would fit on a page of the

same length.

7-4. See graph at right. domain: x ≠ 3, range: y ≠ 0,

asymptotes at x = 3 and y = 0 f −1 (x) = 2 x + 3

7-5. a: ≈ 27.04 feet b: ≈ 176.88 cm c: ≈ 28.94 meters

7-6. 30º - 60º: 1 2 , 3

2 ; 45º - 45º: 1 , 1 or 2

2 2 2 , 2

7-7. y = 6x − x 2

7-8. x = 5, x ≈ 19.69 does not check.

7-9. a: y = ( x + 5 2 ) 2 +

4 3 , vertex ( − 5 2 , 4

3 ) b: (0, 7)

c: (–5, 7); See graph at right.

7-10. No x-intercepts, y-intercept: (0, 88)

7-11. (x −1) 2 + y 2 = 30 ; See graph at right.

center: (1, 0), intercepts: (± 30 +1, 0) and (0, ± 29)

Lesson 7.1.2 (Day 1)

7-15. a: 30 – 60: hypotenuse: 2, leg: 3 ; isosceles: hypotenuse: 2 , leg: 1

b: See diagram at right.

7-16. ≈ 17.46°

60°

60° 60°

7-17. x = 2 , − 5 2 , y = –10

7-18. a: 0 b: 3 c: 4 d: 64

7-19. y : 3; 4; 5; undefined; 7; 8

a: See graph at right. It is linear. The data does not all

connect because f (1) is undefined.

b: y = x + 5, f (0.9) = 5.9, f (1.1) = 6.1, no asymptote.

c: The complete graph is the line y = x +5 with a hole at (1, 6).

7-20. a: An exponential function b: y = 60000 +12000(0.93) t

7-21. If he drives down the center of the road, the height of the tunnel at the edge of the house

is only approximately 23.56 feet. The house will not fit.

7-22. a: x ≈ 33.752 b: x ≈ 9.663

7-23. x = 18, y = 13, z = 9

Lesson 7.1.2 (Day 2)

7-24. −∞ < θ < ∞

7-25. ≈ 40.5º or 139.5º

7-26. She should have subtracted 3⋅16 = 48 to

account for the factor of three. The vertex is (4, 7).

7-27.

7-28. See graph above right.

7-29. a: b: c: d:

7-30. x = 3 2 or − 1 4 , y = –3

W -1 (x)

7-31. a: See graph at right.

b: No; when the points are interchanged, the

input x = 0 has two outputs.

7-32. R + B + G = 40, R = B + 5, R = 2G ; 18 red, 13 blue and 9 green

Lesson 7.1.3

7-36. See graph at right.

7-37. (a): Above ground just past the highest point.

(b): Just below ground.

(c): Back to the starting point.

See diagram at right.

7-38. ≈ 82.4 feet

(a)

-1

(c)

90 270 450 630

(b)

7-39. a: log 6 = log 3+ log 2 ≈ 0.7781 b: log15 = log 3+ log 5 ≈ 1.1761

c: log 9 = 2 log 3 ≈ 0.9542 d: log 50 = 2 log 5 + log 2 ≈ 1.6990

7-40. x = −3± 6

, y = 1

7-41. y = 3(x +1) 2 − 2 ; See graph at right.

7-42. x ≤ –5

7-43. No real solution.

7-44. C + W + P = 40 , W = C − 5 , C = 2P ; 18 from California, 13 from Washington, and

9 from Pennsylvania

Lesson 7.1.4 (Day 1)

7-53.

( 4 , 1 4 )

or ( −

4 , 1 4 )

7-54. P: (cos 50º , sin 50º ) or (~0.643, ~0.766); Q: (cos110º , sin110º ) or (~ –0.342, ~0.940)

( 2 )

7-55. a: 300° b: 1 2 , 3

2 c: 1

, − 3

7-56. a: 30° b: 60 ° c: 67° d: 23 °

7-57. x = 11

7-58. a: downward parabola, vertex (2, 3), see graph above right.

b: cubic, point of inflection (1, 3), see graph below right.

7-59. Solving graphically: x ≈ –3.2

7-60. a: y = 25d + 0.50m and y = 0.03(2) m−1

b: R vs. T: $55 vs. $15.36, $60 vs. $15,728.64, $100 vs. ~ $1.901×10 28

7-61. All of these problems could be solved using the same system of equations.

Lesson 7.1.4 (Day 2)

7-62. 58º, 122º, 238º, or 302º

7-63. a: An angle in the 4 th quadrant. b: 270 ° or –90°

c: An angle in the 3 rd quadrant. d: Approximately 160°

e: No, an angle with sine equal to 0.9 has cosine equal to ±0.4359, so the point (0.8, 0.9)

is not on the unit circle.

7-64. a: (0.3420, 0.9397) b: (cos70° , sin70 ° )

c: (cos 70°) 2 + (sin 70°) 2 = 0.1170 + 0.8830 = 1

7-65. Graph 2 is sine, while graph 1 is cosine. Answers will vary.

7-66. a: All yes.

c: x = (−180° + 360°n) for all integers n

7-67. y-intercept: (0, –17), x-intercepts: (−2 + 21, 0) and (−2 − 21, 0)

7-68. a: x = –4 b: x =

7-69. 7.07 '

5± 57

c: no solution

d: If a =

x+2 3 , then a + 5 ≠ a. Or, solving yields x = –2, but when substituted, –2 gives a

zero denominator.

7-70. Tess is correct: A sequence has no more than one output for each input. A sequence is a

function with domain limited to positive integers.

Lesson 7.1.5

7-77. a: Same; π 3

and 60° are measures of the same angle.

b: 45º, 135º, 405º, etc.

7-78. a:

2 ≈ 0.707 b: 3

2 ≈ 0.866

7-79. a: Set each factor equal to zero to get x = 0, 1 2

, or 3.

b: Factor to get x(x – 1)(2x + 3) = 0. x = 0, 1, – 3 2

7-80. a: x ≈ 2.657 b: x ≈ 0.936 c: x ≈ –0.711

7-81. He should have subtracted 2 ⋅ 9 4 = 9 2 to account for the factor of 2. The vertex is 3

, − 5 2 )

7-82. a: y = 3(x − 3) 2 −1, vertex: (3, –1), axis of symmetry x = 3

b: y = 3( x − 2 3 ) 2 − 37 3 , vertex: 2

, − 37 3 ) , axis of symmetry: x = 2 3

7-83. a: x = 2.5121 b: x = 5 57y

7-84. See graph at right.

a: No

b: –10 ≤ x ≤ 10, –10 ≤ y ≤ 10

c: 200π

≈ 209.44 sq. units

7-85. f −1 (x) = (x −1) 2 + 3 for x ≥ 1; See graph at right.

Lesson 7.1.6

7-90. a: –0.76 b: – 3

7-91.

π

, 5π 6

7-92.

, π 3 , π 2 , 2π 3 , 3π 4 , 5π 6 , π, 7π 6 , 5π 4 , 4π 3 , 3π 2 , 5π 3 , 7π 4 , 11π

6 , 2π

7-93. See diagram at right.

a: A little less than 360° (almost 344° ).

b: sin6 ≈ –0.3

7-94. a: 1 b: 1 2

c: undefined d: 9

7-95. ~ 4.73% annual interest

7-96.

sin A

cos A = 3

91

≈ −0.3145

7-97. a: f −1 (x) = x3 +1

b: g −1 (x) = 7 x

7-98. a: x = 4 or x = –2 b: x ≈ 2.81

Lesson 7.1.7

7-104. 420°

a: π 3 ± 2π n

2 , 1 2 , 3

7-105. a: 0 b: 0 c: –1

d: 0.5 e: 0 f: undefined

0 1 x

7-106. Some may set up a proportion, others may use

180

7-107. a: 210° b: 300 ° c: π 4 radians

d: 5π 9 radians e: 9π 2

radians f: 630 °

7-108. See graph at right.

7-109. f (x) = 2(x − 4) 2 + 2

f -1 (x)

7-110. a: – 5

13 b: 5

7-111. a: a + b b: 2c c: a + 2b d: 3a + c

7-112. a: See graph at right.

b: Yes, the pizza will never get below room temperature.

temperature

time

Lesson 7.2.1

7-116. a: See graph at right.

b: y = 1+ sin x

c: y : (0, 1), x : ( −

2 π , 0 3π

), ( 2

, 0

7π

, 0), ...

d: Yes, there are infinitely many, at intervals of 2π .

7-117. a: π b: y = sin(x + π )

7-118. a: This may go up and down, but the cycles are probably of differing length.

b: This may or may not be periodic.

c: This is probably approximately periodic.

7-119. y = 100 sin ( x +

π ) − 50 or y = 100 cos x − 50

7-120. Only one needs to be a parent, since y = sin(x + 90°) is the same as y = cos x.

7-121. a: y = 3⋅6 x b: y = −2(0.5) x

7-122. a: x = ± 3 5 = ± 15

b: x = 4, –1 c: x = 4

7-123. a: – 3 b:

7-124. a = −

3125 3 = −0.00096 , possible equation: y = −

3125 (x −125)2 +15

Lesson 7.2.2

7-129. a: y = sin x − π 4

( ) + 2

( ) + 0.5

( ) + 2 or y = sin ( x + 5π 6 ) + 2

( ) −1 or y = −3sin ( x +

π ) −1

b: y = 1.5 sin x + π 2

c: y = − sin x − π 6

d: y = 3sin x − 2π 3

7-130. 360° is the period of y = cosθ , so shifting it 360° left lines up the cycles perfectly.

7-131. Graphing form: y = 2(x −1) 2 + 3 ; vertex (1, 3);

7-132. a: x = (0, 0),

5±3 3

, 0) and y = (0, 0)

7-133. 17.67 years

7-134. a: y = −2 ( x + 1 4 ) 2 + 105

8 , x = all real numbers, y = −∞ < y < 25 8

; Yes it is a function.

b: y = −3(x +1) 2 +15 , domain: all real numbers, range: −∞ < y < 15 ; Yes it is a function.

7-135. 64.16 ° , unsafe

7-136. a: 5,000,000 bytes b: ≈ 12.3 minutes

c: According to the equation, technically never, but for all practical purposes, after

23 minutes.

7-137. See graph at right.

a: The vertex of the graph is at (6, –4) with

two rays emanating at slopes of ±1.

b: See graph at right. Flip all parts of the graph

that are below the x-axis above the x-axis.

Lesson 7.2.3

7-144. a: Amplitude 3, period 4π

c: The differences are the period and

amplitude, and therefore some of the

x-intercepts. They have the same basic shape.

7-145. 1, 2π

2π

= 1 or 2π (1) = 2π

7-146. Colleen’s calculator was in radian mode, while Jolleen’s calculator was in degree mode.

Colleen’s calculation is wrong.

7-147. y = sin 2(x −1) is correct. To shift the graph one unit to the right, subtract 1 from x

before multiplying by anything.

7-148. They are both wrong. The equation needs to be set equal to zero before the Zero Product

Property can be applied. 2x 2 + 5x − 3 = 4 is equivalent to (2x + 7)(x −1) = 0 .

x = 1 or x = − 7 2

7-149. a: 3 b: 1.5 c: 2 d: 12

7-150. a: y = 20 ( 1 2 ) x + 5 b: w = 5.078

7-151. a: Answers vary, if g(x) is linear, tangent lines only.

b: Any line y = b such that b < –8.

Lesson 7.2.4

7-158. a: Yes b: y = cos ( x +

π ) c: y = –sin x

7-159. 6 cycles, period: π 3

7-160. Answers will vary.

7-161. a: 180º b: 540º c: π 6 radians

d: 45º e: 5π 4

radians f: 270º

7-162. a: − 2

b: 3 c: – 1 2 d: 2

e: 1 f: − 1 3

g: π 4 or 5π 4 h: 3π 4 or 7π 4

7-163. ( –1, 1 2 , 2 )

7-164. a: x = 0, x = – 1 2 , or x = 5 3

b: x = 6 or x = –1

7-165. a, b, and c: Answers will vary.

7-166. a: About $564,240 b: In 2025 c: About $36,585

Lesson 8.1.1 (Day 1)

8-8. See graphs and tables below. Parent functions:

a: y = x 3 b: y = x 4 c: y = x 3

x y

–2 –9

–1 0

0 1

1 0

2 3

–2 9

2 9

–2 0

–1 3

0 0

1 –3

2 0

8-9. Functions in parts (a), (b), and (e) are polynomial functions; explanations vary.

8-10. Graphs will vary.

a: 0, 1, or ∞ b: 0. 1, or 2 c: 0, 1, 2, 3

d: 0, 1, 2, 3, or 4 (1 and 3 require the parabola to be tangent to the circle.)

8-11. (–2, –1) and (3, 4)

8-12. a: adds 2; multiplies by 3; ; subtracts 1

b: f −1 (x) = ( x−3

2 )2 +1, g −1 (x) = 3(x + 2) −1

8-13. The second graph is shifted up 5 from the first.

8-14. a: b: c:

8-15. a: 4n– 27 b: At least 2507 times.

8-16. a: 60º, 300º b: 135º, 315º c: 60º, 120º d: 150º, 210º

Lesson 8.1.1 (Day 2)

8-17. The functions in parts (a), (b), (d), (e), (h), (i), and (j) are polynomial functions.

8-18. They are not equivalent. Explanations vary. Students may substitute numbers to check.

Also, the second equation can be written y = – x + 12, which is a line, not a circle.

8-19. a: x = 2 or x = 4 b: x = 3 c: x = –2, x = 0, or x = 2

8-20. See graph at right.

a: 2 b: x = 7, – 7

8-21. x = –1± 6

a: 2 b: At x ≈ 1.45 and x ≈ –3.45

8-22. See graph at right.

8-23. x = –1 or 5

8-24. a: y = ( 3 x ) – 4 b: y = 3 (x–7)

8-25. y: –21.2' ; 0'; 21.2'; 30'; 21.2'; 0'; …; –30'

a: Repeat the pattern for several cycles.

b: 30'

c: y = 30 sin x

Lesson 8.1.2

8-36. At (−3 ± 5, 0) .

8-37. At (74, 0), a double root, and at (–29, 0).

8-38. Possible Answers: a: y = x 2 + x – 6 b: y = 2x 2 + 5x – 3

8-39. a: 2 b: 5 c: 3 d: 6

8-40. Lines, parabolas (vertically oriented), and cubics are polynomial functions because they

can be written in the form y = ax n . Exponentials are not polynomial functions because

“x” is the exponent. Circles are not functions.

8-41. a: y

8-42. a: (x − 2) 2 + (y − 6) 2 = 4 b: (x − 3) 2 + (y − 9) 2 = 9

8-43. See graph at right.

8-44. a: 30º, 150º b: 60º, 240º

c: 30º, 330º d: 225º, 315º

Lesson 8.1.3

8-54. Stretch factor is –2. f (x) = −2(x + 2) 2 (x −1)

8-55. a: degree 4, a 4 = 6 , a 3 = –3 , a 2 = 5 , a 1 = 1, a 0 = 8

b: degree 3, a 3 = −5 , a 2 = 10 , a 1 = 0 , a 0 = 8 ,

c: degree 2, a 2 = –1, a 1 = 1, a 0 = 0

d: degree 3, a 3 = 1, a 2 = –8 , a 1 = 15 , a 0 = 0

e: degree 1, a 1 = 1

f: degree 0, a 0 = 10

8-56. Possible equation: p(x) = 2.5(x + 4)(x −1)(x − 3)

8-57. a: y = 4x 2 + 5x − 6 b: y = x 2 − 5

8-58. There is no real solution, because a radical cannot be equal to a negative value.

8-59. a: C: (3, 7), r: 5 b: C: (0, –5), r: 4

c: x = 4 d: x = 7

8-60. a: x = log17

log 2

b: x = 242 c: x = 4 d: x = 7

8-61. a: –3 < x < 2 b: x ≤ –1 or x ≥ 7 3

8-62. y = 2 + 4sinx

Lesson 8.2.1

8-70. a: –18 – 5i b: 1± 2i c: 5 + i 6

8-71. i 3 = i 2 i = −1i = −i ; 1

8-72. a: –21 b: –10 + 7i c: –22 + i

8-73. Yes, substitute it into the equation to check.

8-74. x = –8

8-75. Yes; both are equivalent to x 2 –10x + 25 .

8-76. a: 7i b: 2i or i 2 c: –16 d: –27i

8-77. a: x+3

2 b: x − 2 + 3

8-78. a: x ≈ 2.24 b: x ≈ ± 2.25

Lesson 8.2.2

8-87. Possible Functions:

a: f (x) = x 2 + 6x +10 b: g(x) = x 2 −10x + 22

c: h(x) = x 3 + 2x 2 − 7x −14 d: p(x) = x 3 + 2x 2 −14x − 40

8-88. a: b 2 − 4ac = −7 , complex b: b 2 − 4ac = 49 , real

8-89. See graph at right. Area = 25 sq. units

8-90. a: Repeat 1, i, –1, –i, etc. b: 1, i, –i, 1

c: 1 d: i, –1, –i

e: 1, i, –1, –i

8-91. a: 1 b: i c: –1

8-92. If n is a multiple of 4, the value is 1; if it is 1 more than a multiple of 4, the value is i; if it

is 2 more than a multiple of 4, the value is –1; if it 3 more than a multiple of 4, the value

is –i.

8-93. a: x =

log 17

log 3

x = 17

8-94. a: 2 b: 4 c: 5 d: 3 e: 1

8-95. a: Standard form for y-intercept at (0, 400) and graphing form for vertex at (0.5, 404).

b: 400 ft; 404 ft

8-96. a: y = log x b: x = 2 c: y = log 2 (x − 2) is one possibility.

Lesson 8.2.3

8-104. a: three real linear factors (one repeated), therefore two real (one single, one double) and

zero complex (non-real) roots

b: one linear and one quadratic factor, therefore one real and two complex (non-real)

roots

c: four linear factors, therefore four real and zero complex (non-real) roots

d: two linear and one quadratic factor, therefore two real and two complex (non-real)

8-105. a: b: c: d:

8-106. a: (3, 0), (0, 0), and (–3, 0) b: See graph at right.

8-107. See graphs below.

a: x-intercepts ( − 5 2 , 0 ), (0, 0) , and

, 0), y-intercept (0, 0)

b: x-intercepts (−3, 0) and

a: b:

( ) (double root), y-intercept (0, 675)

8-108. See graph at far right.

8-109. a: Platform is 11.27 meters off the ground. h = −4.9(t − 5) 2 +133.77 ;

Therefore, the maximum height is 133.77 meters. Time when h = 0 is 10.22 sec.

b: h ≈ −4.9(t −10.22)(t + 0.22) ; Factored form reveals the intercepts, or how long it took

the firework to reach the ground.)

8-110. b ≥ 20 or b ≤ –20

8-111. a: (i − 3) 2 = i 2 − 6i + 9 = −1− 6i + 9 = 8 − 6i

b: (2i −1)(3i +1) = 6i 2 − 3i + 2i −1 = −6 − i −1 = −7 − i

c: (3− 2i)(2i + 3) = 6i − 4i 2 − 6i + 9 = 4 + 9 = 13

8-112. ( ±6, 1 2 )

Lesson 8.3.1

8-120. a: –7 c: (x + 7) d: (x 2 − 2x − 2) f: −7, 1± 3

8-122. Part (c), because (–2)(3)(–5) = 30 and (x)(x)(x) = x 3 not 2x 3 .

8-123. Part (b), because 5 is a factor of the last term, but 2 and 3 are not.

8-124. (x − 5)(x 2 − 4x −1); zeros: 5, 2 ± 5

8-125. a: (x − 2)(5x + 3) b: −

5 3 , 2 c: Explanations will vary.

d: 3 and 2 are factors of 6, while 5 is a factor of the lead coefficient.

8-126. a: See the combination histogram boxplot at right.

The five number summary (for the box plot) is

0, 2.75, 8, 15.7, 36.5.

b: The distribution has a right skew and an outlier at

36.5 pounds so the center is best described by the

median of 8.0 pounds and the spread by the IQR

of 12.95 pounds.

c: The median is better in this case because it is not

affected by skewing and outliers.

d: The IQR is better in this case because it is less affected by skewing and outliers than

the standard deviation.

e: If you remove the outlier from the data the mean drops to 8.7 pounds which is below

the profitable minimum. You could suggest running the test a few more weeks

because perhaps as people get used to the composting program they will participate

even more.

0 6 12 18 24 32 36 42

8-127. a: b:

8-128. a: See graph below.

b: Yes, it is a solution to the equation.

Lesson 8.3.2 (Day 1)

8-138. a: It shows that (x – 3) is a double factor and 3 is a double root.

b: p(x) = (x − 3) 2 (x 2 + 2x −1) ; x = 3, −1± 2

8-139. a: x 2 − 6x + 25 = 0 b: x 2 − 6x + 25 = 0 c: Answers will vary.

8-140. a: 3+2i

−4+7i ⋅ −4−7i

−4−7i = 2−29i

65

b: 2

65 − 29

65 i

8-141. a: 12 5 − 1 5 i b: − 3

13 + 11

13 i

8-142. See graph at right; x + 3 −1; x ≥ −3, y ≥ −1

8-143. a: See graph below left. −1, 1± 3 i

b: See graph below right. 2, −1± 3 i

8-144. (3, 4, –1)

8-145. x = 1 2

8-146. a: See graph below left, locator( −

2 π , 0 ) , period = 2π, amplitude: 3

b: See graph below right, locator (0, 0), period:

2 π , amplitude: 2, inverted

Lesson 8.3.2 (Day 2)

8-147. p(x) = x 3 + 5x 2 + 33x + 29

8-148. a: p(2) = 0 b: (x – 2) c: (x 2 − 4x −1) d: 2, 2 ± 5

8-149. a:

17 8 + 15

17

i b: 2 + 5i

8-150. a: Regular: (361, 367, 369 373, 380 grams); Diet (349, 354, 356.5, 361, 366 grams)

b: See histograms at right.

Regular

c: Regular: The mean is 369.6 grams, which falls at the middle

of the distribution on the histogram. The shape is singlepeaked

and symmetric, so the mean should be a good

measure of the center. There are no outliers, so the standard

deviation of 4.34 grams could be used to describe spread.

Diet: The mean is 357.5 grams; this mean also falls at the

center of the data on the histogram. The data is doublepeaked

but still fairly symmetric so the mean could be used

to represent the center. There are no outliers so the standard

deviation of 5.12 grams could be used to describe spread.

d: The regular cola cans are noticeably heavier (or had more mass) than the diet cans.

The lightest regular can is at the third quartile of the diet sample and the median of

the regular cans is heavier than the most massive diet can. The spread of each

distribution is similar and they are both reasonably symmetric but the diet cans have a

double peaked distribution.

e: Answers will vary.

348 360 372 384

Diet

8-151. See graph at right. a: 4 b: (±4, 3) and (±3, −4)

8-152. a: x = 4 ( 1 is extraneous) b: x = 1 4

8-153. a: b:

c: d:

8-154. a: x = 2, (x – 2) b: x = 2, −3 ± 2i ; (x − 2) ( x − (−3+ 2i) )( x − (−3− 2i) )

8-155. a: x = 2π 3 , 4π 3 b: x = π 6 , 7π 6 c: x = 0, π d: x = π 4 , 7π 4

Lesson 8.3.2 (Day 3)

8-156. a: − 1 5 − 5 7 i b: 1 – 2i

8-157. At 6 years, it will be worth $23,803.11. At 7 years it will be worth $25,707.36.

8-158. a: x = 5 9

b: x = 3 c: x = 48 d: x ≈ 1.46

8-159. Students should show the substitution of the coordinates of the point into both equations

to verify.

8-160. x = 2 or x ≈ 1.1187

8-161. a: x ≈ 781.36 b: x = 6 c: x = 1, 1 5

d: x = 0, 1, 2

8-162. When you find the complement of the angle, the x and y values reverse.

8-163. a: −7 ⋅ −7 = i 7 ⋅i 7 = i 2 49 = −7

b: She multiplied −7 ⋅ −7 to get 49 = 7 .

c: −7 is undefined in relation to real numbers, and is only defined as the imaginary

number 7i , so it must be written in its imaginary form before operations such as

addition or multiplication can be performed.

d: a and b must be non-negative real numbers.

8-164. a: π 3

5π

12 c: 7π 6 d: 5π 4

Lesson 8.3.3

8-169. (0, 0), (3, 0), and (−0.5, 0)

8-170. See graph at right.

8-171. a: (x + 10)(x − 10) b: x −

3+ 37

3− 37

( x −

2 )

c: (x + 2i)(x − 2i) d: (x − (1+ i))(x − (1− i))

8-172. a: real b: complex c: complex

d: real e: real f: complex

8-173. It is not; 16 + 8 ≠ 32 − 40

8-174. a: x = 5 or 1 b: x = 4 or 0 c: x = 7 d: x = 1

8-175. a: 24

b: (x 3 − 3x 2 − 7x + 9) ÷ (x − 5) = (x 2 + 2x + 3) with a remainder of 24.

8-176. a: y = x 2 +1 b: y = x 2 – 2x –1

8-177. a: b:

Lesson 9.1.1

9-7. a: Population: U.S. employees; the population is too large to conveniently measure so

sampling should be used.

b: Population: students in the class. A census can be taken.

c: Population: All carrots. To measure the Vitamin A in a carrot, it must be destroyed, so

even if it were possible to measure all carrots, it would not be wise. Sampling must be

used.

d: Population: The public (the media does not generally make this population very clear).

It could be all voting adults, all adults, or all people in the state. In any case, the

population is too large, so sampling must be used.

e: Population: Elevator cables. To find this, elevator cables must hold greater and greater

weight until they break. If all elevator cables were tested, there would be none left to

use for elevators. Sampling must be used.

f: Population: Your friends. A census can be taken.

9-8. a: The five-number summary is (1, 19.5, 29, 40.5, 76 cups of coffee per hour).

b: The typical number of cups sold in an hour is 29 as determined by the median.

Looking at the shape of the distribution the median is a satisfactory representation of

the distribution. The distribution has a skew. There is a gap between 60 and 70 cups.

The IQR is 21 cups. Seventy-six cups of coffee in one hour is an apparent outlier.

9-9. a: (x ±1), (x ± 7)

9-10. a ≤ 25

b: Neither are factors. Use substitution to determine whether x = –1 and x = 1 are zeros.

Or you could use the Remainder Theorem and divide to see that neither are factors

because there is a remainder.

9-11. a: 30º or 150º b: 120º or 240º c: 45º or 225º

d: 35.26º, 144.74º, 215.26º, or 324.76º

9-12. f (g(x)) = g( f (x)) = x

9-13. a: 10 times stronger b: 10 0.8 = 6.3 times stronger

c: log(0.5) ≈ –0.3, so 6.2 – 0.3 ≈ 5.9 on the Richter scale

9-14. a: (x + 2 − 3)(x + 2 + 3) = x 2 + 4x +1

b: (x + 2 − i)(x + 2 + i) = x 2 + 4x + 5

9-15. posts: $3, boards: $2, piers: $10

Lesson 9.1.2

9-22. a: The question implies that the questioner holds this opinion, thus biasing results.

b: The question assumes that the respondent believes that the climate is changing and

will think that one of the given factors is important, and that it is important to slow

global climate change, biasing results.

c: The question implies that teacher salaries should be raised.

9-23. Sample questions given:

a: Are a majority of Americans in favor of replacing the Electoral College with a

popular vote?

b: Do consumers prefer the taste of the new “improved” cookie recipe?

c: What was the class average on the semester final exam?

d: What was the average for high school students taking the state math proficiency

examination last year?

Hush Puppy

9-24. a: See plots at right. Hush Puppy: min = 19.7,

Q1 = 44.5, med = 58.3, Q3 = 70.1, max = 79.5;

Quiet Down: min = 14.2, Q1 = 37.4, med = 54.85,

Q3 = 63.3, max = 102.1

b: Hush Puppy: The distribution is left skewed so its

center and spread are best described by the median

of 58.3 dB and IQR of 25.6 dB there are no apparent

outliers. Quiet Down: Has some potential outliers

over 100 dB or is perhaps dual-peaked. The main

body of data has a left skew. The center and spread

are best described by the median of 54.85 dB and IQR

of 25.9 dB.

d: The Hush Puppy looks better now because those three

high readings from the Quiet Down model are a lot

more significant. Or perhaps the Quiet Down could

be redesigned to eliminate those high readings.

12 24 36 48 60 72 84 96 10

Quiet Down

9-25. y = 2(x + 2) 2 − 3 , (–2, –3). See graph at right.

9-26. y = 6, z = 2

9-27. a: x = 4 b: x = 200

9-28. a: π 6 b: π 12

c: –

12 d: 7π 2

9-29. y = − 1 4

(x − 2)(x + 2)2

9-30. a: 160π

, about 167.6 cubic feet b: 6 feet

c: It is not changing, the angle is ≈ 68.20°

Lesson 9.1.3

9-36. a: This information could be found on the web for all American League players.

It would be a census and the answer would be a parameter.

b: An experiment would need to be conducted on a sample of eggs. The findings would

be a statistic.

c: Random high school students could be surveyed, possibly from different high schools

in different parts of the country. Surveying every high school student would be almost

impossible, so this survey would be a sample and the answer would be a statistic.

9-37. a: closed b: open c: open d: closed

9-38. Answers will vary.

9-39.

9

; k = 7, 8 are factorable.

9-40. blue block: 8 grams, red block: 16 grams

9-41. a: The more rabbits you have, the more new ones you get, a linear model would grow by

the same number each year. A sine function would be better if the population rises

and falls, but more data would be needed to apply this model.

9-42.

x+5

9-43.

b: R = 80, 000(5.4772...) t

c: ≈ 394 million

d: 1859, it seems okay that they grew to 80,000 in 7 years, if they are growing

exponentially.

e: No, since it would predict a huge number of rabbits now. The population probably

leveled off at some point or dropped drastically and rebuilt periodically.

9-44. a: 4π 5

15π

c: 100º d: 255º e: 1710º f:

11π

Lesson 9.2.1

9-50. a: When asked to choose between an “m” and a “q,” most people prefer “m,” regardless

of the Cola taste.

b: The “to protect” interpretation is not part of the Bill of Rights; it will bias the results.

c: The statistics chosen for the lead-in will bias the results.

9-51. a: Only certain types of people typically respond.

b-d: Answers will vary.

e: Students should observe some clear preferences for some numbers, letters, and colors.

This would provide evidence that people cannot behave or choose randomly.

9-52. Mean: 7.6 g, mean distance-squared: 2.56+0.16+0.16+1.96+0.36

5−1

= 1.3 ,

sample standard deviation≈ 1.14 g (and not ≈ 1.02).

9-53.

, 6, –3)

9-54. x 2 + 25

9-55. 2, ±5i

9-56. ± 15

9-57. a: 3 + 2i b: 1 + 4i c: 5 + i d: − 1 2 + 5 2 i

9-58. a: x = 32 b: x = 1 6

Lesson 9.2.2

9-62. a: Not likely; this samples the population of people with phone numbers listed online that

are home midday. In each of these, we could also notice that we only get responses

from those who agree to participate in our polling activities—already a very

unrepresentative sample.

b: Not likely; this samples the population of people who shop at this particular grocery

store.

c: Not likely; this samples the population of people who attend movies.

d: Not likely: this samples the population of people who go downtown at the time you are

there.

e: This sample is likely to be more representative, as it is closer to random.

9-63. Sample diagram:

30 get

classroom

SAT prep

90 student

volunteers

take SAT

random assignment

on-line

30 get no

additional

instruction

All Students retake

the SAT. Compare

average change in

scores between the

groups

9-64. Children would need to be randomly assigned to treatment groups, one that gets spanked

and another where there is no spanking. After a period of years the IQ of both groups can

be tested and compared. Any variable that has the suggestion or potential to lower the IQ

of a human does not belong in a clinical experiment. Who would decide who, when, and

how the spankings would be administered? Would you spank kids randomly?

9-65. Mean: 52 g, mean distance-squared: 64+64+4+144+4

≈ 8.37g (and not ≈ 7.48g).

9-66. one point of intersection: (2, 2)

= 70 , sample standard deviation

9-67. See graph at right; a sphere, V = 32π

cubic units

2 x

9-68. x 3 + 8 = (x + 2)(x 2 − 2x + 4)

9-69. a: b:

Lesson 9.3.1

9-75. a: They will not show the people who named other stations or people saw the station

logo and knew what station the interviewer was from.

b: Surveying outside the gym does not give you a random sample.

c: No, more people drive during the day. You should look at the probability of being in

an accident.

d: About half of all power plants are below average. It does not mean that it is unsafe.

9-76. Answers will vary.

9-77. a: 3 2 = 9 b: 3 0 = 1 c: 3 3 + 3 1 = 27 + 3 = 30 d: – 9 2

9-78. a: n = 2; 1 7 b: n = 0, 1; 2 7 c: n = 3, 4, 5, 6; 4 7

9-79. – 1 5

9-80. a: x = 4 b: x = 9

9-81. ≈ 278 months or 23 years

9-82. a: y

9-83. a: No

b: No, no number of trials will assure there are no red ones.

c: Not possible.

Lesson 9.3.2

9-88. a: The typical number of tardy students (center) is the median, 3 students, because 50%

of the students fall below 3. The shape is single-peaked and symmetric with no gaps

or clusters. The IQR (spread) is 1 student, because Q1 is 2 and Q3 is 3. There are no

apparent outliers.

b: (0.43 + 0.13 + 0.07)(30) = 19 days

c: ≈ 30%

9-89. a: Answers will vary.

b: Yes, assuming a sufficient sample size, a controlled randomized experiment can show

cause and effect because of the random assignment of subjects to the groups.

9-90. Experiments can be very expensive, time consuming, and in some cases involving

humans or animals, unethical.

9-91. 6 sq. units; see graph at right.

9-92. a: f −1 3

(x) = x +1 − 3

b: See graph below right.

c: Yes, each x is paired with no more than one y.

9-93. a: double roots at –1, 2, 5 b: Same as the previous except

reflected over the x-axis

9-94. a: (– 4, 0), (–2, 0) and (0, –16) b: domain: all real numbers, range: y ≤ 2

9-95. a: y = 1 4 (x +1)2 +

8 3 , vertex = ( −1, 8

3 ) , x = –1

b: y = 1 4 (x +10)2 +16 , vertex = (–10, 16), x = –10

9-96. a: x + 3 b: (x + 3) 2 + (y − 2) 2

Lesson 9.3.3

9-103. a: 667.87 lunches; sample standard deviation is 56.17 lunches.

b: (576, 624, 665, 700.5, 785)

c: See histogram at right.

d: Since the shape is fairly symmetric, we’ll use

mean as the measure of center; the typical number

of lunches served is 668. The shape is single

peaked and fairly symmetric with no gaps or

clusters, the standard deviation is about

56 lunches, and there are no apparent outliers.

e: 10.8%

f: Use half of the 680 – 720 bin. 0.216 + 0.351 + (0.5)(0.108) ≈ 62.1%

9-104. a: See graph at right.

b: 2.7333 tardy students, 1.1427

c: See graph middle right.

d: See graph middle right. 11.0%

e: See graph far right.

normalcdf(4, 10^99, 2.7333, 1.1427) = 0.1338. 0.1338(180) ≈ 24 days

9-105. a: See graph at right.

b: 50 %

c: See graph at right.

normalcdf(11.5, 12.5, 12, 0.33) = 87%

9-106. a: log 3 (5m) b: log 6 ( p m

) c: not possible d: log(10) = 1

0.108

0.216

0.351

0.135

0.081

9-107. a:

2x 2 −2x−7

(x−3)(x+1)(x−2)

b: −x2 +2x−4

x(x−2) x+2

x−5 d: x(x2 − 2x + 4)

9-108. a: x = ± 26 b: x =

9-109. See graphs at far right.

−2± 10

9-110. a: 1224 ≈ 34.99 b: (x +1) 2 + (y +1) 2

9-111. a: f −1 (x) = 1 3 ( x−5

2 )2 +1 = 1

12 (x − 5)2 +1 for x ≥ 5

Lesson 10.1.1

10-7. a: $165 b: t(n) = 50 + 5(n −1) c: $930

10-8. a: 4050 b: 300 + 550 + 800 + … + 4050; t(n) = 300 + 250(n −1)

10-9. a: t(n) = 3+ 7(n −1) b: t(n) = 20 − 9(n −1)

10-10. –2

10-11. a: x-intercepts (2.71, 0) and (5.29, 0), y-intercept (0, 43)

b: x-intercepts (–1, 0) and (2.5, 0), y-intercept (0, –5)

10-12. a: normcdf(70, 79, 74, 5) = 0.629, About 63% would be considered average.

b: normcdf(–10^99, 66, 74, 5) = 0.055, Between 5 and 6% of would be in excellent

shape.

c: normcdf(–10^99, 66, 70, 5) – 0.0548 from part (b) = 0.157; There would be a nearly

16% increase in young women classified as being in excellent shape.

10-13. a: 233 units b: (x − 5) 2 + (y − 2) 2 units

10-14. $20.14

10-15. 34,800 people

10-16. Yes, because the sum of the diameters is 830 mm.

10-17. 235

10-18. (−6) + (−3) + 0 + 3+ 6 + 9 +12 +15 +18

10-19. It is the 55 th term.

10-20. 220

10-21. a: 10 P 5 = 30, 240 b: 10 ⋅9 4 = 65, 610

10-22. n = ±6 2

10-23. a: 2 b: 3 4

10-24. a: It looks like an endless wave repeating the original cycle over and over again.

b: A polynomial of degree n has at most n roots, but f(x) = sin(x) has infinitely many

roots. Also, every polynomial eventually heads away from the x-axis.

Lesson 10.1.2

10-34. a: odds: t(n) = 1+ 2(n −1) , evens: t(n) = 2 + 2(n −1)

b: odds: 5625, evens: 5700

10-35. a: t(n) = 21− 4(n −1)

b: 31; You can solve the equation 25 − 4n = −99 .

c: –1209

10-36. 15

10-37. 6 P 4 = 360

10-38. Degree 4; Graph shown at right.

10-39. f (x) = 1 4

(x − 3)(x − 2)(x +1)

10-40. a: f (x) = (x + 3) 2 − 2 , vertex (–3, –2) b: f (x) = (x − 5) 2 − 25 , vertex (5, –25)

10-41. For David: normcdf(122, 10^99,149,13.6) = 0.976

For Regina: normcdf(130, 10^99,145,8.2) = 0.966

For now David is relatively faster.

10-42. a: π 4

12 c: − π 12

d: 5π 2

Lesson 10.1.3

10-49. a: 16,200 b: 16,040 c: 564

10-50. 11+ 22 + 33+ ...+ 99 = 495

10-51. a: Sample response: The terms decrease by two, then add seven, then decrease by two,

and then add seven continually.

b: It is not arithmetic because the difference from one term to the next is not constant.

c: Find the sum of each “unzipped” series and then add these sums together.

The sum is 32,240.

10-52. a: 12 C 10 = 66 b: 9 C 7 = 36

10-53. Some may substitute for x, others may set x equal to 3+ i 2 and work back to the

equation, others may write the two factors and multiply to get the original equation, and

others may solve by completing the square.

10-54. The graphs of y = 2 x and y = 5 − x intersect at only one point.

10-55. h = $2.50, m = $1.75

10-56. a: 17 b: 5

10-57.

Lesson 10.1.4

11

10-62. a: ∑ (60 −13k) = −198 b: ∑

k=1

10-63. 495,550

(3+ 7(k −1)) =

n[3+(3+7(n−1))]

10-64. It works for the integers from 1 through 39.

10-65. a: 7 ⋅ 3 n b: 10(0.6) n−1

10-66. 9!= 362,880

10-67. a: normalcdf(–10^99, 59, 63.8, 2.7) = 0.0377; 3.77%

b: (0.0377)(324)(half girls) = 6 girls

= n(−1+7n)

c: normalcdf(72,10^99, 63.8, 2.7) = 0.00119. (0.00119)(324)(half) = 0.19 girls.

We would not expect to see any girls over 6ft tall.

10-68. a: b:

10-69. cos( 2π 3 ) = – 1 2 , cos( 4π 3 ) = – 1 2 , cos( 5π 3 ) = 1 2

Lesson 10.2.1 (Day 1)

10-87. a: 3+ 30 + 300 + 3,000 + 30,000 + 300,000 = 333,333

b: Write the series 3+ 30 + ...+ 300,000 = S(6) twice. Multiply one of them by 10.

Subtract 10S(6) − S(6) = 2,999,997 = 9S(6) . Divide by 9 to get 333,333.

c: ∑ 3⋅10 n−1 = 3⋅10n −3

i=1

10-88. a: A sequence would represent the list of the class sizes of the graduating classes as the

number of years since the school opened increased. The corresponding series would

represent the growing number of alumni.

b: t(10) = 150 ; total = 960

c: n(36 + 6n) = 36n + 6n 2

10-89. a: 15 b: –615

10-90. 210, arithmetic

10-91. a: 23 P 3 = 10, 626 b: 23 C 3 = 1771 c: 1⋅22 ⋅22 = 484 d: 4 ⋅22 ⋅22 = 1848

10-92. See graph at right.

10-93. a: x−2

c: Use the Distributive Property to factor and the multiplicative property of 1 to reduce.

10-94. a and b: no amplitude, period = π , LP = (0, 0).

10-95. a: x = 125

2 b: x = − 4 5

c: x = 0.04 d: y = 9 4

Lesson 10.2.1 (Day 2)

10-96. Calculate the sums of two geometric series, the first with 25 terms, the second with 15.

Retirement at age 55: $1,093,777; at age 65: $1,115,934

10-97. $20,000 at 8% and $30,000 at 6.5%

10-98. a: 8!= 40320 b: 1⋅ 7!= 5040 c: 1⋅ 7! + 7!⋅1 = 10080

10-99. a: 272 = 4 17 units b: (x + 3) 2 + (y + 5) 2 units

10-100. a: x = 5 2 b: y =10 c: x = –3, 2 d: y = − 15 4

10-101. a: (2, 8) and (4, 4) b: (3+ i,6 − 2i) and (3− i,6 + 2i)

c: In system (a), the solutions are the points of intersection. In system (b), the solutions

show that they do not intersect.

10-102. tan(160°) = –0.3640 , tan(200°) = 0.3640 , tan(340°) = –0.3640

10-103. a: x = 4 b: x = 48 c: x = 3 d: x = 6

10-104. 26 + i

Lesson 10.2.2

10-117. 500 miles

10-118. a: 7 C 2 = 21 b: 7 C 3 = 35 c: 7 C 4 = 35

d: Choosing three points to form a triangle is the same as choosing four points to not be

part of the triangle. Those four points form a quadrilateral, 7 C 3 =

4!3! 7! =

3!4! 7! = 7 C 4 .

10-119. a: 10t + u − (10u + t) = 27 b: x = –13

10-120. $1157

10-121. a, d

10-122. a: (x − 3)(x 2 − 2x + 5) b: 3, 1± 2i

10-123.

10-124. (−1+ i, 3), (−1− i, 3)

10-125. When r ≥ 1, r n increases in size as n increases, so the expression 1− r n does not get

close to 1, and being able to replace that expression with 1 is a key part of the

derivation of the formula.

10-126.

121

10-127. a: 45 b: 792 c: 7

10-128. a: x = –3, 4 b: x = –1.5, 3 c: x =

d: Never e: x = 0, –2, 4 3 f: x = 0

10-129. a: −16x 5/2 y 4 z 2 b: 3 1/2 x 3/2 y 8/3

−1 ± 57

10-130. a: 0, 5 seconds b: 0 ≤ t ≤ 5 c: 5 seconds d: 1 < t < 4

10-131. Yes; use the Quadratic Formula or direct substitution.

10-132. a: x = –1, 4 b: x ≤ –1 or x ≥ 4 c: –1 ≤ x ≤ 4

Lesson 10.3.1 (Day 1)

10-145. x 7 + 7x 6 y + 21x 5 y 2 + 35x 4 y 3 + 35x 3 y 4 + 21x 2 y 5 + 7xy 6 + y 7

10-146. −640w 2 z 3

10-147. 1365

10-148. a: 16 b: Not possible. r > 1, and the terms keep increasing.

10-149. 4 C 0 = 1, 4 C 1 = 4, 4 C 2 = 6, 4 C 3 = 4, 4 C 4 = 1

a: The number of possibilities are the elements of the 4 th row of the triangle.

b: 1, 6, 15, 20, 15, 6, 1; Use the 6 th row of the triangle.

10-150. a: 9 C 3 = 84 b: 9 C 2 = 36 c: 10 C 3 = 9 C 3 + 9 C 2

d: The tenth row entry of Pascal’s Triangle is the sum of the two ninth row entries

above it and these numbers correspond to the total number of combinations when

one more choice is added.

10-151. See graph at right.

10-152. a: See graph below right.

b: Answers will vary. See graph below right.

normalcdf (6.71, 13.29, 10, 2) = 0.9000

10-153. (−2, 3, − 1 2 )

10-154. a: y = x 2 − 4x + 5 = (x − 2) 2 +1

b: (2, 1)

Lesson 10.3.1 (Day 2)

10-155. 42x 3

10-156. 728

10-157. 81x 4 +108x 3 + 54x 2 +12x +1

10-158. a: f (x) = (x + 2) 2 + 2 b: (x + 3) 2 + (y − 4) 2 = 25

10-159. 32.9 mm

10-160. y = −2x + 34

10-161. a: See graph at right. b: f −1 (x) = [2(x +1)] 2 – 4

c: x ≥ –1, y ≥ –4 d: 5

10-162. a: x = 7 b: x = 1.5 c: x ≈ 1.75 d: x ≈ 1.87

10-163. a: (0, –5), (4, 3), (8, 3) b: See graph at right.

c: y = − 1 4 x2 + 3x − 5 d: 10 seconds

height

e: 0 ≤ x ≤ 10 f: 0 ≤ x < 2

Lesson 10.3.2

10-169. Robin: $11,887.58; Tyrell: $11,808.38; difference: $79.20

10-170. a: $10,304.55; it rounds off to the same amount.

b: $10,832,870, 680 and $10,832,775,720, a difference of $94,960.

c: Maybe billionaires, or other investors of large amounts.

10-171. a: 1+ 3 n + 3

n 2 + 1

n 3

b: 1+ 5 n + 10

n 2 + 10

n 3 + 5

n 4 + 1

n 5

10-172. a: 14.7 lbs./sq. in. b: ≈ 12.55 lbs./sq. in. c: ≈ 14.83 lbs./sq. in.

10-173. a: 2 = (1.015) 4t , 2 = e 0.06t

b: Quarterly, 11.64 years; Continuously, 11.55 years

c: The difference is about one month, so probably not.

10-174. (−5, 0), ( 2 3 , 0), (− 1 4 , 0)

10-175. 8x 3 − 36x 2 + 54x − 27

10-176. a: log 2 (5x) b: log 2 (5x 2 ) c: x = 17

d: x = −

20 9 = −0.45 e: x = 15 f: x = 4

10-177. a: 10t + u

b: 10u + t

c: t + u = 11 and 10t + u − (10u + t) = 27

d: 74 and 47

Lesson 11.1.1

11-5. a: preface b: biased wording c: desire to please d: fair question

11-6. a: Using the normal model here is not a good idea because the data is not symmetric,

single-peaked, and bell-shaped. A different model would represent the data better.

b: 9/48 ≈ 19 th percentile; In 19% of the hours over the two day period the coffee shop

was not profitable.

c: 41/48 = 85 th percentile. In 85% of the hours over the two-day period, the coffee shop

would not have been profitable. If this data represents a typical 48-hour period, now

would not be the time to expand.

11-7. If the nickel is tossed and the die rolled there are 12 equally likely outcomes. He can

make a list of each of the possible outcomes, assign one player to each outcome

(H1-Juan, H2-Rolf, … T6-Jordan), and then toss the coin and roll the die to select.

11-8. a: No, by observation, a curved regression line may be

better. See graph at right.

b: Exponential growth.

4 5

c: m = 8.187 ⋅1.338 d , where m is the percentage of mold,

and d is the number of days. Hannah predicted the

1 2 3

mold covered 20% of a sandwich on Wednesday.

Day

Hannah measured to the nearest percent.

b: 0.60

0.80 = 75% y

11-9. a: f −1 (x) = ln x

b: f(x): domain is all real numbers, range is y > 0, y-intercept is (0, 1),

asymptote is y = 0. f –1 (x): domain is x > 0, range is all real numbers,

x-intercept is (1, 0), asymptote is x = 0. See graphs at right.

c: 2 and 3

d: Above log base 2 for x > 1 and below it for 0 < x < 1, below

log base 3 for x > 1 and above for 0 < x < 1.

11-10. a: 9.00646832 for both b: 3.10628372 for both

c: It will take about 9 years to double.

d: After about three years and one month, the car

will be worth less than half of the original price.

11-11. 765

Garage?

yes no

0.80 0.20

11-12. a: 2i b: –2 + 2i

yes

0.60 0.15

0.75

11-13. a: 5%. See the table at right.

no

0.20 0.05

0.25

% Mold

40

20

Large

Backyard?

Lesson 11.1.2

11-18. Answers should be close to P(sum 6 or less) = 15

≈ 0.42 , so 0.42(7) = 2.94 or about

3 days per week.

11-19. Theoretical probabilities: P(sum 6 or less) = 15

36 ≈ 0.42 ,

P(sum 7 or more) =

36 21 ≈ 0.58 , so about3 days a week.

11-20. a: F 2%, D 16 – 2 = 14%, C 84 – 16 = 68%, B 98 – 84 = 14%, A 100 – 98 = 2%

b: F 0, D 76.5 – (2)17.4 = 41.7, C 75.5 – 17.4 = 58.1, A 76.5 – (2)17.4 = 111.3

c: The normal model is not a good idea because the distribution of scores is strongly

skewed left. Also, the minimum grade required for an A would be 111.3, which is

probably not possible.

11-21. Answers vary.

11-22. a:

, 11 5 ) b: ( −5, 1 2 )

11-23. a: 12 b: 1 2

11-24. Both equal 3 8 .

11-25. The base of the natural logs is e; e is between 2 and 3, and ln e = 1; x = e

11-26. a: 1.79175 b: 2.4849 c: 2.7726 d: –1.0986

Lesson 11.1.3

11-28. Theoretically, the series is expected to last for 5.8125 games.

11-29. Theoretically, a streak of 4 or more has probability of about 48%, a streak of five or more

about 25%, a streak of six or more 12%, and a streak of seven or more about 6%.

11-30. The survey is not random. They will get a lot more representation from adults. People

may be influenced by the responses of others. An individual may have several favorites

but only states one.

11-31. a: x = 1 4 y2 −1, parabola b: x 2 + y 2 = 49 , circle

11-32. a: 51 b: 64.77

11-33. They are equivalent and simplify to x 2 .

11-34. a: 1

b: 580 c: (–9, 1) d: y − 2 =

(x − 3)

11-35. a: ≈ 266.67 b: 37.5% c: 27% d: 135

11-36. a: See diagram at right.

b: In the RR rectangle, 18

38 ⋅ 18

38 = 1444 324 ≈ 22.44% .

c: Add the RR, BR, and GR rectangles,

324

1444 + 1444 324 + 1444 36 ≈ 47.37%.

d: Considering only the column in which the

second spin is red (RR, BR, GR), the probability

the first spin is red (RR)

is

1444

1444 + 1444 324 + 1444

≈ 47.37%.

R

G

e: They are the same.

Lesson 11.2.1

11-41. Theoretically, there will be 7.39 streaks of three or more in 60 days.

11-42. a: 7.83%

b: 5% and 11%

c: About 7.83% ± 3% def. flashlights

11-43. Each used a convenience sample. Each sample came from a distinct population with

people who have many things in common, including attitudes and beliefs about the

subject matter of murals.

11-44. a: –344 b: –6740

11-45. a: x = 10p b: x = 3q – 2p

11-46. y = 5, z = 3

11-47. See table at right. 0.6 ⋅0.5 = 0.3 = 30%

11-48. (–1, –3) and (3, 5)

11-49. a: 10 log 2 ≈ 3

b: 20 ⋅10 6

c: Two sounds have equivalent pressures, or one sound has a pressure of 20 micropascals.

d: 100

Second Jar

yellow

0.30

orange

0.50

white

0.20

black

0.60

First Jar

purple

0.40

0.18 0.12

0.30 0.20

0.12 0.08

Lesson 11.2.2

11-52. a: 0.56 and 0.74 b: ≈ 65% ± 9%

11-53. 3% ± 0.85%. Answers vary, but should be 3% plus/minus a number smaller than 1.7%

11-54. The energy usage will change by –1.3% to 1.7%. Indiana could save up to $11.7 million,

but could end up spending $15.3 million.

11-55. a: Answers will vary. Like a tournament, the students could be paired and the coin

tossed for each pair. The “winners” are paired and the coin tossed again, repeating

until there is just one student.

b: 3 tosses. Tossing a coin 3 times has 8 equally likely outcomes: HHH, HHT, HTH, HTT,

THH, THT, TTH, TTT. Assign each student an outcome and toss the coin 3 times.

c: Yes. Use the method outlined in part b repeatedly until a baseball player is not

selected.

11-56. a: y = 2(x +

4 7 )2 − 105

, graph shown at right,

vertex (−

4 7 , − 105

8 ) , axis of symmetry x = − 4

b: y = 3(x − 1 6 )2 −

12 97 , graph shown at far right,

vertex( 1 6 , − 12 97 ) , axis of symmetry x = 1 6

11-57. a: b:

c: For part (a), the parts above the x-axis stay the same, the parts below the x-axis are

reflected upward across the axis. For part (b), the part of the graph to the right of the

y-axis remains the same, and the part to the left of the y-axis is replaced by a reflection

of the part on the right of the axis.

11-58. a: Some may predict the amount due will be far too much for a state to pay.

b: ≈ 1.126 ⋅ 10 15 dollars

c: A ≈ $2.791⋅10 16 , or about 2.68 ⋅10 16 dollars more.

11-59. a: x = log(3)

log(2)

c: x = log(12)

log(7)

log(8)

b: x =

log(5)

log(b)

d: x =

log(a)

11-60. Both 31.5%. Neither 16.5%. See table at right.

Dimples

0.45

Widow’s Peak

0.70 0.30

0.315

0.55 0.165

Lesson 11.2.3

11-64. Upper bound: 25145 lower bound: 24563; Students should report that 90% of the time we

can expect that the copy machine will need maintenance after 24854 ± 291 copies. The

interval of confidence is from 24563 to 25145. 25000 copies is within the interval of

confidence, so the research company may support the copy machine company’s claim.

They may also state that the copy machine company is pushing the limits with a claim of

“at least 25000 copies” since the range within the margin of error is from 24563 to 25145.

11-65. a: 0.12

b: A difference of zero is not within the margin of error, so it is not a plausible result. A

difference of zero means that there is no difference in the percentage of food removed

with detergent compared with the percentage of food cleaned off with plain water.

c: Yes. Because a difference of zero is not within the margin of error, a difference of

zero is not a plausible result for the population of all food cleaned. We are convinced

there is between 3.5% and 20.5% (12% ± 8.5%) more food removed with detergent

than with plain water.

11-66. a: 1/8 or 12.5%, yes

b: Her study provides no reliable evidence of her conclusion. She used a convenience

sample of only 4 people. Her question introduced bias with the preface about violence

and crime. There may also have been a desire to please the interviewer. She used a

closed question forcing romantic comedies as the only alternative to action movies.

Even if there is no real preference among moviegoers, it is plausible that 4 people will

have the same opinion between any two types of movies.

11-67. a: The number of cards on the field is 768 ×1029 = 790, 272 cards. The probability is

or 0.000 001265 .

790272

790,272 cards

52 cards

1 pack

= 15,198 packs of cards.

The maximum loss is if the first player chooses a card and wins:

−$1, 000, 000 prize − ($0.99)(15,198) cost of packs + $5 from the player

= −$1, 015, 041.02 . (If nobody plays, then the million dollars is not paid out, and the

boosters do not have the maximum possible loss.)

c: If all of the chances were purchased,

−1, 000, 000 prize − ($0.99)(15,198) cost of packs + ($5)(790, 272) from players

= $2, 936, 313.98

d: On average half the cards would be sold before there was a winner,

−1, 000, 000 prize − ($0.99)(15,198) cost of packs + ($5)(395,136) from players

= $960, 633.98 .

e: 176,000,000

790,272

≈ 223 football fields would have to be covered to give the same odds as

winning the state lottery!

11-68. (0, 0)

11-69. a: 5 + i b: –1 + 9i c: 26 + 7i d: –0.56 + 0.92i

11-70. a: 4 ≤ y ≤ 10 b: m > 1 or m < –2

11-71. a: x = –3 or x = 2 b: x < –3 or x > 2

Lesson 11.2.4

11-75. Yes, 20% is within the margin of error of 13% ± 10%.

11-76. a: 0.10

b: i: 25% of 40 + 15% of 40 = 16 putts went into the hole

ii: 1 to 16 will represent a putt that went into the hole; 17 to 80 will represent a putt

that missed.

c: A difference of zero means that there is no difference between the proportion of putts

that went in the hole with the new club and the proportion of putts that went in with

the old club. A difference of zero is within the margin of error, so it is a plausible

result.

d: No. A difference of zero is within the margin of error, and a difference of zero is a

plausible result for the population of all putts. We are not convinced there is a true

difference in the number of putts that go in with the new club compared to the old

club.

11-77. Yes. As little as 11.75 ounces is still within specifications.

11-78. The first process is wildly out of control; systems wildly out of control are often caused

by inexperienced operators. The second process is fully in control. The third process is

technically out of control at only one point, but the cyclical nature of the process is

disconcerting. Any explanation that is cyclical over 20 hours is acceptable.

11-79. a: y

11-80. a: x = a+b

c

b: x = ab + ac c: x = a, b

d: x = 0, c e: x = a+b

f: x =

b−a

11-81. a: x = −26 b: x = 10 or x = 3

11-82. a: x = 1 b: x = ±2

parents

niece

boyfriend

11-83. a: See diagram at right.

b: 1 4 / 1 2 = 0.5

c: 1 9 + 1 6 + 1 6 + 1 4 = 25

36 ≈ 69%

d: 1 6 / 25

36 = 6 25 = 24%

1 1

9 18

18

Lesson 11.3.1

11-87. This is a “Gambler’s Ruin” problem. The player with more coins always has a better

chance of winning in the long run. P(Jill) = 4 6 ≈ 67%, P(Jack) = 2 6 ≈ 33% .

11-88. a: See possible diagrams at right and answers below.

Cell A is the proportion of times the system correctly

activated the alarm.

Cell B is the proportion of times the alarm was

correctly not activated.

Cell C is the proportion of times A happened and the

alarm was incorrectly not activated.

Cell D is the proportion of times A did not happen and

the alarm was activated.

b: 0.03966/(0.03966 + 0.00096) = 97.6%

c: Yes, it is independent because the accuracy of

alarm is the same regardless of whether event A

occurs or not.

Detection System

Event A

Yes No

Correct

Incorrect

A B

C D

correct

0.00096

incorrect

0.00004

11-89. y = 1 5 x + 27 5

11-90. (± 7, 3), (0, −4)

Not Event A

11-91. a: 10 + 11i b: 13 c: 29 d: a 2 + b 2

0.95904

0.03996

11-92. a: b:

11-93. a: –35 b: 123

11-94. a: x = ±2 3 b: x = 2 c: x = 2 9

−1± 13

or x ≈ 0.434 or –0.768

11-95. a: any polynomial with 5 x-intercepts; b: a polynomial graph with 3 x-intercepts and

another ‘bend’; c: no x-intercepts, could have two ‘humps’; d: 2 x-intercepts and up to

two ‘humps.’

Lesson 12.1.1

12-5. a: always b: never

c: always d: True for x = π 4 + 2π n and x = 5π 4 + 2π n

12-6. a: (x − 2)(x + 2) b: (y − 9)(y + 9)

c: (1− x)(1+ x) d: (1− sin x)(1+ sin x)

12-7. a: ≈ 80.86 b: ≈ 24.05 c: ≈ 15.50º

trigonometric ratios, Law of Sines, Law of Cosines, Pythagorean theorem

12-8. The graphs of y = sin(2x) and y = 2sin(x) intersect at integer multiples of π. Solving the

equation algebraically yields x = 0 + πn.

12-9. a: x ≥ 2, f (x) ≥ 2 b: f −1 (x) = (x−2)2

+ 2 c: x ≥ 2, f −1 (x) ≥ 2

12-10. a: Stretched (amplitude = 3), shifted left

2 π , and shifted down 4

12-11. The first process is fully in control. The second process is wildly

out of control. The third process is out of control; beginning at the

9 th hour, there are 12 consecutive points above the centerline.

12-12. a: 90 b: 190 c: 35 d: 405,150

12-13. a: 12 P 5 = 95, 040 b: 12 C 5 = 792

12-14. Sample answers: h = π 2 , 5π 2 , − 3π 2

12-15. a: negative b: negative c: positive d: negative

12-16. a ≤ 25

12-17. See graph at right.

12-18.

12-19. a: x = 9 b: x = –9

12-20. a: 3π 5

12-21. a: 5 C 2 i 4 C 1

12 C 3

d: 5 C 3 + 4 C 3 + 3 C 3

16π

c: 140º d: 285º e: 1530º f:

13π

=

11 2 b: 4 C 3

12 C = 1

3 55 c: 5 C 1 i 4 C 1 i 3 C 1

12 C = 3

3 11

44 3 e: 5 C 1 i 4 C 2

3 22

12-22. a: a 3 + 3a 2 b + 3ab 2 + b 3 b: 8m 3 + 60m 2 +150m +125

f: 1−

( ) = 29

11 + 44

Lesson 12.1.2

12-29. a: x = π 6 , 5π 6 b: x = 5π 6 , 7π 6 c: x = π 4 , 3π 4 d: x = 0

12-30. No, 52º and 308º have the same value for the cosine, while 128º

has the exact opposite cosine value. See diagram at right.

128º

52º

12-31. See graph at right.

–x

12-37. a: 24

12-32. f −1 (x) = −x + 6

12-33.

12-34. a: yes b: x 4 + x 3 + x 2 + x +1; yes c: x n + x n−1 + x n−2 +…+ x +1

Recipient

12-35. a: See possible diagrams at right and answers below.

Cell A is the proportion of people correctly

identified as drug users.

Cell B is the proportion correctly identified as

not drug users.

Cell C is the proportion the test failed to identify

as drug users but who are.

Cell D is the proportion identified as drug users

who are not.

This can also be modeled as a tree diagram.

b: 0.02 are actually using; 0.01 are told they are

User

Not Using

D

0.02277

using, but are actually not.

Drug User

c: 0.00977/(0.00977 + 0.02277) ≈ 30%

0.00023

d: From part (b), about 1 out of 100 people

receiving assistance will lose their assistance

correct 0.98703

because they have been falsely accused of

Not Drug User

using drugs. That seems high considering that

only 2 out of 100 are actually using drugs.

From part (c), 30% of the people identified as

using drugs will be falsely accused and unfairly

lose their money.

incorrect 0.00977

e: Yes, they are independent because the accuracy of the test stays the same whether or

not a person uses drugs. To test, check whether P(A)⋅ P(B) = P(A and B) , for

example, P(drug user)⋅ P(test correct) = P(drug user and test correct) .

12-36. a: 0.0253 b: 26 C5

52 C 5

c: 0.000495 d: 13 C5

e: 0.0019808

Drug Test

360º – 52º

=308º

Lesson 12.1.3 (Day 1)

12-43. a: 30º, 50º or

6 π , 5π 6 b: 120º, 140º or 2π 3 , 4π 3

c: 45º, 225º or

4 π , 5π 4

d: ≈ 35.26º, 144.74º, 215.26º, 324.74º or 0.62, 2.53, 3.76, 5.67

12-44. a: domain: –3 ≤ x ≤ 3, range: –3 ≤ y ≤ 3, not a function

b: domain: –3 ≤ x ≤ 4, range: –2 ≤ y ≤ 4, not a function

c: domain: x ≤ 3, range: y ≤ 4, yes a function

d: domain: −∞ < x < ∞ , range: y ≥ –2, yes a function

12-45. a: x = 2 b: no solution

12-46. a: b:

–2 –2

12-47. a: (4, 8) b: (0, –2, 3)

Suspect

Person

Not Suspect

12-48. a: See possible diagrams at right and answers below.

Cell A is the proportion correctly identified as

suspects.

Cell B is the proportion correctly identified as not

being suspects.

Cell C is the proportion the software failed to

identify but who actually are suspects.

Cell D is the proportion the software identified as

suspects who are not.

b: 0.000999995/(0.000004995 + 0.000999995)

≈ 99.5%

12-49. 12 C 5 + 12 C 4 + 12 C 3 + 12 C 2 + 12 C 1 + 12 C 0 = 1586

12-50. 2x 4 − 2x + −1

Facial ID Software

Not a Suspect

0.000004995

0.000000005

0.99899501

0.000999995

12-51. a: See graph at right.

b: The number of defects seems to be staying at or within

control limits, but there is a cycle apparent every eight

hours. Perhaps as the inspectors work through their

shift, they get tired and catch fewer errors.

Lesson 12.1.3 (Day 2)

12-52. The restrictions are needed so that the inverses will be functions. The domain of the sine

function is restricted to −

2 π ≤ x ≤ 2 π , the domain of the cosine function is restricted to

0 ≤ x ≤ π , and the domain of the tangent function is restricted to −

2 π < x < 2 π .

12-53. a: x = 7π 6

+ 2π n,

6 + 2π n b: x = 6 π + 2π n,

6 + 2π n

c: x = 3π 4 + π n d: x = π n

12-54. a: shifted up 1 unit b: shifted left

c: reflected over the x-axis d: vertically stretched by a factor of 4

12-55. a: y

c: g(x) = − f (x)

12-56. a: 5 6

3x+8

2x 2

c: x2 +2x+3

(x+1)(x−1)

12-57. f (x) = 2(x −1) 2 −1; domain: all real numbers; range: f (x) ≥ −1;

vertex: (1, –1); line of symmetry: x = 1; See graph at right.

d: sin2 θ+cosθ

sinθ cosθ

12-58. a: x ≈ 1.356 b: x ≈ 2.112 c: x ≈ 1.792

12-59. a: a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4

b: 81m 4 − 216m 3 + 216m 2 − 96m +16

12-60. a: 8 C 3 + 8 C 4 = 56 + 70 = 126

b: If mushrooms are a known topping, then choose the rest of the toppings from only

7 remaining toppings so 7 C3+7C2

126

126 56 = 4 9

Lesson 12.1.4

12-67. a: a c b: c b c: c b d: b a

12-68. a and b: ± π 6 , ± 5π 6 , ± 7π 6 , ± 11π

12-69. The solution is equivalent to the solutions of cos(x) = 1 2

300°; or 0, π,

3 π , 5π 3

and sin(x) = 0 . 0°, 180°, 60°,

12-70. 9.10

12-71. 1.083, 8.3%

12-72. a: a 3 + b 3 b: x 3 − 8 c: y 2 +125 d: x 3 − y 3

e: They consist only of two terms; they are sums or differences of cubes.

12-73. a: (x + y)(x 2 − xy + y 2 ) b: (x − 3)(x 2 + 3x + 9)

c: (2x − y)(4x 2 + 2xy + y 2 ) d: (x +1)(x 2 − x +1)

12-74. y = 10

216 (x + 6)3 −10

12-75. a: 3C 1 i 10 C 3

13C4

= 360

715 ≈ 0.503 b: 11

13C4 = 1 65 = 0.015

12-76. a: c a b: a b c: b a d: c a

12-77. The solution is equivalent to the solutions of sin(x) = − 1 2

and sin(x) = 0 . 90° + 180°n,

210° + 360°n, 330° + 360°n; or

2 π + π n, 7π 6

12-78. 2x 2 + 4x −1

12-79. a: x 2 (x + 2y)(x 2 − 2xy + 4y 2 )

b: (2y 2 − 5x)(4y 2 +10xy 2 + 25x 2 )

c: (x + y)(x 2 − xy + y 2 )(x − y)(x 2 + xy + y 2 )

12-80. Possible equation: y = 1 8 (x − 3)3 + 3 Inverse: y = 3 8(x − 3) + 3

12-81. y = 4(0.4) x + 5

12-82. a: 126a 5 b 4 b: 1120x 4 y 4

12-83. x 2 − 6x + 34 = 0

12-84. P(3 or 4 or 5) ≈ 0.99

Lesson 12.2.1

12-90. a: 1 b: cos(4w) c: tan(θ)

12-91. ≈ 75.52°, 75.52°, and 28.96°

12-92. a: x = 30º + 360ºn or x = 150º + 360ºn b: no solution

12-93. 3x 2 − x + 2

12-94. x 3 − 2x 2 − 3x + 9

12-95.

10!

12-96. (5 – 2)! because 3! > 2!

12-97. a: 18 b: –12 c: –1 + 7i d: –14.5 e: x = 0, –7

12-98. a: p = 44, a = 20 b: y = 20 cos

( 22

(x −15) ) + 3

12-99. Possibilities include: sin 2 x = 1− cos 2 x , sin x = ± 1− cos 2 x ,

sin 2 x = (1− cos x)(1+ cos x)

12-100. − 4 5

12-101. a: sinθ

cosθ b: 1

sinθ

cosθ

sinθ d: 1

12-102. a: y − 9 = 315

(x − 2) or y =

315

12-103. They intersect at

, 0) and (3, 10).

12-104. a:

2(x−1) b: 4x

3x 2 +10x+3

12-105. a: 41.41° b: 28.30°

x − 306 b: y = 0.25(6)x

12-106. roots: −0.4 ± 0.8i 6 ; vertex: (–0.4, 19.2); f (x) = 5(x + 0.4) 2 +19.2

12-107. a: (0.9) 5 ≈ 0.59

b: 10(0.9) 2 (0.1) 3 + 5(0.9)(0.1) 4 + (0.1) 5 ≈ 0.00856

Lesson 12.2.2 (Day 1)

12-109. 90.21 feet or 4.71 feet

12-110. a: 2x 4 − x 2 + 3x + 5 = (x −1)(2x 3 + 2x 2 + x + 4) + 9

b: x 5 − 2x 3 +1 = (x − 3)(x 4 + 3x 3 + 7x 2 + 21x + 63) +190

12-111. a: x = 3π 2 b: x = π 3 , 5π 3 c: x = π 4 , 5π 4 d: x = 7π 6 , 11π

12-112. a: x ≈ 69.34 b: x ≈ 5.35

12-113. a: 10 P 5 = 30, 240 b: 10 i 9 4 = 65, 610

12-114. ( ±

5 3 , 4 5 )

12-115. (x + 4) 2 + (y − 6) 2 = 64 ; circle, center: (–4, 6), r = 8

12-116. a: See graph at right below.

b: The process has a lot of variability for the first 14 hours.

There are two out-of-control points (one upper and one

lower). Apparently an adjustment was made at hour 14

because the process is much less variable, but now there are

nine consecutive points above the centerline of 0.075.

Apparently another adjustment was made at hour 22, but

this apparently swung the process to the very low end.

12-117. a: 2x+1

3x−2

b: x2 +2x+4

x(4x+5)

Lesson 12.2.2 (Day 2)

12-118. a: (sinθ + cosθ) 2 =

sin 2 θ + 2 sinθ cosθ + cos 2 θ =

1+ 2 sinθ cosθ

( ) =

c: (tanθ cosθ) sin 2 θ + 1

sec 2 θ

( ) (sin 2 θ + cos 2 θ) =

cosθ cosθ

(sinθ)(1) = sinθ

b: tanθ + cotθ =

cosθ + cosθ

sinθ =

sin 2 θ+cos 2 θ

= cscθ secθ

12-119. See unit circle at right. θ = π 6 , 5π 6 , 7π 6 , 11π

12-120. m∠B = 86.17º or 1.5 radians

12-121. The equation of the parabola is y = 1 8 (x − 3)2 + 3 .

It is a function; every input has only one output.

12-122. a: x ≈ 1.839 b: x ≈ –1.839 c: x ≈ 1.839

12-123. a: The two lines intersect at (8, 17).

b: No solution; the lines are parallel.

12-124. a: There is one way to choose all five.

5!

5!0!

= 1 . In order to have the formula give a

reasonable result for all situations, it is necessary to define 0! as equal to 1.

b: There is one way to choose nothing.

0!5! = 1

12-125. AC = 10 inches

12-126. Possible answer: f (x) = x 3 − 5x 2 + 8x − 6

Lesson 12.2.3

12-130. sin π 12 = sin π 3 − π 4

12-131. a: sin π 3 + π 4

( ) = 6− 2

( ) = 6+ 2

( ) = cos ( 7π 6 − 4

π ) = − 6+ 2

b: cos 3π 4 + π 6

; cos π 12 = cos π 3 − π 4

12-132. sin(x + x) = sin x cos x + cos x sin x = 2 sin x cos x

( ) = 2+ 6

cos(x + x) = cos x cos x − sin x sin x = cos 2 x − sin 2 x

12-133. a: See graph at right.

b: Possible answer: f (x) = − sin x

c: cos x + π 2

12-134. sin x + cos x

( ) = cos x cos π 2 − sin x sin π 2 = − sin x

12-135. Divide by x – 3, then solve the resulting quadratic; x = 1 ± i.

12-136. a: 2 b: a – 2

12-137. 3 C 1 ( 1 4 ) 2 3

( 4 ) = 9

64 ≈ 0.141 x

Recommendations

Selected Answers for Core Connections Algebra 2

Page 2 and 3: Lesson 1.1.1 1-4. a: 1 2 b: 3 1-5.
Page 4 and 5: Lesson 1.1.3 1-34. a: The numbers b
Page 6 and 7: Lesson 1.2.1 (Day 2) 1-65. a: 2 b:
Page 8 and 9: Lesson 1.2.2 (Day 2) 1-91. a: x = y
Page 10 and 11: Lesson 1.2.4 1-112. a: A portion of
Page 12 and 13: Lesson 2.1.2 2-16. Answers will var
Page 14 and 15: Lesson 2.1.4 2-50. a: f (x) = (x +
Page 16 and 17: Lesson 2.2.1 (Day 1) 2-81. Possible
Page 18 and 19: Lesson 2.2.2 (Day 1) 2-107. a: y =
Page 20 and 21: Lesson 2.2.3 2-125. a and : Neither
Page 22 and 23: Lesson 2.2.5 2-162. x < 2, y = -(x
Page 24 and 25: Lesson 3.1.3 3-45. a: n = -2 b: x =
Page 26 and 27: Lesson 3.2.3 3-90. 2x 3(2x-1) = 2x
Page 28 and 29: Lesson 4.1.1 4-7. See graph at righ
Page 30 and 31: Lesson 4.1.3 4-40. a: (-2, -11); Th
Page 32 and 33: Lesson 4.2.2 4-83. x = -2, y = 3, z
Page 34 and 35: Lesson 5.1.2 5-26. See graph at rig
Page 36 and 37: Lesson 5.2.1 5-60. Domain: x > 0; R
Page 38 and 39: Lesson 5.2.3 5-84. Possible answer:
Page 40 and 41: Lesson 6.1.1 6-8. a: Their y- and z
Page 42 and 43: Lesson 6.1.3 6-35. See graph at rig
Page 44 and 45: Lesson 6.2.1 6-95. y = 3 x 6-96. In
Page 46 and 47: Lesson 6.2.3 6-127. a: y = 40(1.5)
Page 48 and 49: Lesson 7.1.1 7-3. a: The shape woul
Page 50 and 51: Lesson 7.1.2 (Day 2) 7-24. −∞ <

Lesson 7.1.4 (Day 1) 7-53. 15 ( 4 ,

Lesson 7.1.5 7-77. a: Same; π 3 an

Lesson 7.1.7 7-104. 420° a: π 3

Lesson 7.2.2 7-129. a: y = sin x

Lesson 7.2.4 7-158. a: Yes b: y = c

Lesson 8.1.1 (Day 2) 8-17. The func

Lesson 8.1.3 8-54. Stretch factor i

Lesson 8.2.2 8-87. Possible Functio

Lesson 8.3.1 8-120. a: -7 c: (x + 7

Lesson 8.3.2 (Day 2) 8-147. p(x) =

Lesson 8.3.3 8-169. (0, 0), (3, 0),

Lesson 9.1.2 9-22. a: The question

Lesson 9.2.1 9-50. a: When asked to

Lesson 9.3.1 9-75. a: They will not

Lesson 9.3.3 9-103. a: 667.87 lunch

Lesson 10.1.2 10-34. a: odds: t(n)

Lesson 10.1.4 11 10-62. a: ∑ (60

Lesson 10.2.1 (Day 2) 10-96. Calcul

Lesson 10.3.1 (Day 1) 10-145. x 7 +

Lesson 10.3.2 10-169. Robin: $11,88

Lesson 11.1.2 11-18. Answers should

Lesson 11.2.1 11-41. Theoretically,

Lesson 11.2.3 11-64. Upper bound: 2

Lesson 11.2.4 11-75. Yes, 20% is wi

Lesson 12.1.1 12-5. a: always b: ne

Lesson 12.1.3 (Day 1) 12-43. a: 30

Lesson 12.1.4 12-67. a: a c b: c b

Lesson 12.2.2 (Day 1) 12-109. 90.21

Lesson 12.2.3 12-130. sin π 12 = s

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Active Calculus

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Appendix B Answers to Activities

1 understanding the derivative 1.1 how do we measure velocity 1.1.1 position and average velocity, activity 1.1.2 ..

$AV_{[0.4,0.8]} = 12.8$ ft/sec; $AV_{[0.7,0.8]} = 8$ ft/sec; the other average velocities are, respectively, 6.56, 6.416, 0, 4.8, 6.24, 6.384, all in ft/sec.
Like a straight line with slope about 6.4.
About 6.4 feet per second.

1.1.2 Instantaneous Velocity

Activity 1.1.3 ..

$AV_{[1.5,2]} = -24$ ft/sec, which is negative.
The instantaneous velocity at $t = 1.5$ is approximately $-16$ ft/sec; at $t = 2\text{,}$ the instantaneous velocity is about $-32$ ft/sec, and $-16>-32\text{.}$
When the ball is rising, its instantaneous velocity is positive, while when the ball is falling, its instantaneous velocity is negative.

Activity 1.1.4 .

1.2 the notion of limit 1.2.1 the notion of limit, activity 1.2.2 ..

$2\text{.}$
$12\text{.}$
$\frac{1}{2}\text{.}$

1.2.2 Instantaneous Velocity

Activity 1.2.3 ..

$6 + h\text{.}$
$6.2$ meters/min.
$6$ meters per minute.

Activity 1.2.4 .

$AV_{[0.5,1]} = \frac{1-1}{1-0.5} = 0\text{,}$ $AV_{[1.5,2.5]} = \frac{3-1}{2.5-1.5} = 2\text{,}$ and $AV_{[0,5]} = \frac{5-0}{5-0} = 1\text{.}$
Take shorter and shorter time intervals and draw the lines whose slopes represent average velocity. If those lines’ slopes are approaching a single number, that number represents the instantaneous velocity.
The instantaneous velocity at $t = 2$ is greater than the average velocity on $[1.5,2.5]\text{.}$

1.3 The derivative of a function at a point 1.3.1 The Derivative of a Function at a Point

Activity 1.3.2 ..

$f$ is linear.
The average rate of change on $[1,4]\text{,}$ $[3,7]\text{,}$ and $[5,5+h]$ is $-2\text{.}$
$f'(1)=-2\text{.}$
$f'(2)=-2\text{,}$ $f'(\pi)=-2\text{,}$ and $f'(-\sqrt{2})=-2\text{,}$ since the slope of a linear function is the same at every point.

Activity 1.3.3 .

The vertex is $(\frac{1}{2},36)\text{.}$
$\frac{s(2)-s(1)}{2-1} = -32$ feet per second.
$s'(1) = -16\text{.}$
$s'(a)$ is positive whenever $0 \le a \lt \frac{1}{2}\text{;}$ $s'(a)$ to be negative whenever $\frac{1}{2} \lt a \lt 2\text{;}$ $s'(\frac{1}{2}) = 0\text{.}$

Activity 1.3.4 .

$AV_{[2,4]} \approx 9171$ people per decade is expected to be the average rate of change of the city’s population over the two decades from 2030 to 2050.
\begin{equation*} P'(2) = \lim_{h \to 0} \frac{P(2+h)-P(2)}{h} = \lim_{h \to 0} 25000e^{2/5}\left( \frac{e^{h/5} - 1}{h}\right) \end{equation*} Because there is no way to remove a factor of $h$ from the numerator, we cannot eliminate the $h$ that is making the denominator go to zero.
\begin{equation*} P'(2) = \lim_{h \to 0} \frac{P(2+h)-P(2)}{h} \approx 7458.5 \end{equation*} which is measured in people per decade.
See the graph provided in (a) above. The magenta line has slope equal to the average rate of change of $P$ on $[2,4]\text{,}$ while the green line is the tangent line at $(2,P(2))$ with slope $P'(2)\text{.}$
It appears that the tangent line’s slope at the point $(a,P(a))$ will increase as $a$ increases.

1.4 The derivative function 1.4.1 How the derivative is itself a function

Activity 1.4.2 ., activity 1.4.3 ..

$f'(x) = 0\text{.}$
$g'(t) = 1\text{.}$
$p'(z) = 2z\text{.}$
$q'(s) = 3s^2\text{.}$
$F'(t) = \frac{-1}{t^2}\text{.}$
$G'(y) = \frac{1}{2\sqrt{y}}\text{.}$

1.5 Interpreting, estimating, and using the derivative 1.5.2 Toward more accurate derivative estimates

Activity 1.5.2 ..

$F'(30) \approx = 3.85$ degrees per minute.
$F'(60) \approx = 1.56$ degrees per minute.
$F'(75) \gt F'(90)\text{.}$
The value $F(64) = 330.28$ is the temperature of the potato in degrees Fahrenheit at time 64, while $F'(64) = 1.341$ measures the instantaneous rate of change of the potato’s temperature with respect to time at the instant $t = 64\text{,}$ and its units are degrees per minute. Because at time $t = 64$ the potato’s temperature is increasing at 1.341 degrees per minute, we expect that at $t = 65\text{,}$ the temperature will be about 1.341 degrees greater than at $t = 64\text{,}$ or in other words $F(65) \approx 330.28 + 1.341 = 331.621\text{.}$ Similarly, at $t = 66\text{,}$ two minutes have elapsed from $t = 64\text{,}$ so we expect an increase of $2 \cdot 1.341$ degrees: $F(66) \approx 330.28 + 2 \cdot 1.341 = 332.962\text{.}$
Throughout the time interval $[0,90]\text{,}$ the temperature $F$ of the potato is increasing. But as time goes on, the rate at which the temperature is rising appears to be decreasing. That is, while the values of $F$ continue to get larger as time progresses, the values of $F'$ are getting smaller (while still remaining positive). We thus might say that “the temperature of the potato is increasing, but at a decreasing rate.”

Activity 1.5.3 .

It costs $800 to make 2000 feet of rope.
“dollars per foot.”
$C(2100) \approx = 835\text{,.}$
Either $C'(2000) = C'(3000)$ or $C'(2000) > C'(3000)\text{.}$
Impossible. The total cost function $C(r)$ can never decrease.

Activity 1.5.4 .

$f'(90) \approx 0.0006$ liters per kilometer per kilometer per hour.
At 80 kilometers per hour, the car is using fuel at a rate of 0.015 liters per kilometer.
When the car is traveling at 90 kilometers per hour, its rate of fuel consumption per kilometer is increasing at a rate of 0.0006 liters per kilometer per kilometer per hour.

1.6 The second derivative 1.6.3 Concavity

Activity 1.6.2 ..

Increasing: $0\lt t\lt 2\text{,}$ $3\lt t\lt 5\text{,}$ $7\lt t\lt 9\text{,}$ and $10\lt t\lt 12\text{.}$ Decreasing: never.
Velocity is increasing on $0\lt t\lt 1\text{,}$ $3\lt t\lt 4\text{,}$ $7\lt t\lt 8\text{,}$ and $10\lt t\lt 11\text{;}$ $y = v(t)$ is decreasing on $1\lt t\lt 2\text{,}$ $4\lt t\lt 5\text{,}$ $8\lt t\lt 9\text{,}$ and $11\lt t\lt 12\text{.}$ Velocity is constant on $2\lt t\lt 3\text{,}$ $5\lt t\lt 7\text{,}$ and $9\lt t\lt 10\text{.}$
$a(t) = v'(t)$ and $a(t) = s''(t)\text{.}$
$s''(t)$ is positive since $s'(t)$ is increasing.
increasing .
decreasing .
concave up .
concave down .

Activity 1.6.3 .

Degrees Fahrenheit per minute.
$F''(30) \approx -0.119\text{.}$
At the moment $t = 30\text{,}$ the temperature of the potato is 251 degrees; its temperature is rising at a rate of 3.85 degrees per minute; and the rate at which the temperature is rising is falling at a rate of 0.119 degrees per minute per minute.
Increasing at a decreasing rate.

Activity 1.6.4 .

1.7 limits, continuity, and differentiability 1.7.1 having a limit at a point, activity 1.7.2 ..

$f(-2) = 1\text{;}$ $f(-1)$ is not defined; $f(0) = \frac{7}{3}\text{;}$ $f(1) = 2\text{;}$ $f(2) = 2\text{.}$
\begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ \text{and} \lim_{x \to -2^+} f(x) = 1 \end{equation*} \begin{equation*} \lim_{x \to -1^-} f(x) = \frac{5}{3} \ \text{and} \lim_{x \to -1^+} f(x) = \frac{5}{3} \end{equation*} \begin{equation*} \lim_{x \to 0^-} f(x) = \frac{7}{3} \ \text{and} \lim_{x \to 0^+} f(x) = \frac{7}{3} \end{equation*} \begin{equation*} \lim_{x \to 1^-} f(x) = 3 \ \text{and} \lim_{x \to 1^+} f(x) = 3 \end{equation*} \begin{equation*} \lim_{x \to 2^-} f(x) = 2 \ \text{and} \lim_{x \to 2^+} f(x) = 2 \end{equation*}
$\lim_{x \to -2} f(x)$ does not exist. The values of the limits as $x \to a$ for $a = -1, 0, 1, 2$ are $\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}$
$a = -2\text{,}$ $a = -1\text{,}$ and $a = 1\text{.}$

1.7.2 Being continuous at a point

Activity 1.7.3 ..

$a = -2\text{;}$ $a = +2\text{.}$
$a = 3\text{.}$
$a = -1\text{;}$ $a = 3\text{.}$
$a=-2\text{;}$ $a = 2\text{;}$ $a = 3\text{;}$ $a = -1\text{.}$
“If $f$ is continuous at $x = a\text{,}$ then $f$ has a limit at $x = a\text{.}$ ”

1.7.3 Being differentiable at a point

Activity 1.7.4 ..

$g$ is piecewise linear.
\begin{align*} g'(0) =\mathstrut \amp \lim_{h \to 0} \frac{g(0+h)-g(0)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|0+h|-|0|}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|h|}{h} \end{align*}
$\lim_{h \to 0^+} \frac{|h|}{h} = 1 \text{,}$ but $\lim_{h \to 0^-} \frac{|h|}{h} = -1 \text{.}$
$a = -3, -2, -1, 1, 2, 3\text{.}$

1.8 The Tangent Line Approximation 1.8.2 The local linearization

Activity 1.8.2 ..

$L(-1) = -2\text{;}$ $L'(-1) = 3\text{.}$
$g(-1) = -2\text{;}$ $g'(-1) = 3\text{.}$
$g(-1.03) \approx L(-1.03) = -2.09\text{.}$
Concave up.

Activity 1.8.3 .

$L(x) = -1 + 2(x-2)\text{.}$
$f(2.07) \approx L(2.07) = -0.86\text{.}$
See the image in part e .

2 Computing Derivatives 2.1 Elementary derivative rules 2.1.2 Constant, Power, and Exponential Functions

Activity 2.1.2 ..

$f'(t) = 0\text{.}$
$g'(z) = 7^z \ln(7)\text{.}$
$h'(w) = \frac{3}{4} w^{-1/4}\text{.}$
$\frac{dp}{dx} = 0\text{.}$
$r'(t) = (\sqrt{2})^t \ln (\sqrt{2})\text{.}$
$\frac{d}{dq}[q^{-1}] = -q^{-2}\text{.}$
$\frac{dm}{dt} = -3t^{-4} = -\frac{3}{t^4}\text{.}$

2.1.3 Constant Multiples and Sums of Functions

Activity 2.1.3 ..

$f'(x) = \frac{5}{3}x^{2/3} - 4 x^3 + 2^x \ln(2)\text{.}$
$g'(x) = 14e^x + 3 \cdot 5x^4 - 1\text{.}$
$h'(z) = \frac{1}{2}z^{-1/2} - 4z^{-5} + 5^z \ln(5)\text{.}$
$\frac{dr}{dt} = \sqrt{53} \cdot 7 t^6 - \pi e^t\text{.}$
$\frac{ds}{dy} = 4y^3\text{.}$
$q'(x) = 2x - 2x^{-2}\text{.}$
$p'(a) = 12a^3 - 6 a^2 + 14a - 1\text{.}$

Activity 2.1.4 .

$h'(4) = \frac{3}{16}\text{.}$
(i.) $P'(4) = 2(1.37)^4 \ln(1.37) \approx 2.218$ million cells per day; (ii.) the population is growing at an increasing rate.
$y - 25 = -33(a+1)\text{.}$
The slope is a number, while the equation is, well, an equation.

2.2 The sine and cosine functions 2.2.1 The sine and cosine functions

Activity 2.2.2 ..

Figure 2.2.3 .
$1,0,-1,0,1,0,-1,0,1\text{.}$
$f'(0) = f'(-2\pi) = f'(2\pi) = 1\text{.}$
$\frac{d}{dx}[\sin(x)] = \cos(x)\text{.}$

Activity 2.2.3 .

Figure 2.2.6 .
$0,-1,0,1,0,-1,0,1,0\text{.}$
$g'(\frac{\pi}{2})=g'(-\frac{3\pi}{2})=-1\text{.}$
$\frac{d}{dx}[\cos(x)] = -\sin(x)\text{.}$

Activity 2.2.4 .

$\frac{dh}{dt} = -3\sin(t) - 4\cos(t)\text{.}$
$f'(\frac{\pi}{6}) = 2 + \frac{\sqrt{3}}{4}\text{.}$
$y - \frac{\pi^2}{4} = (\pi-2)(x-\frac{\pi}{2})\text{.}$
$p'(z) = 4z^3 + 4^z \ln(4) - 4\sin(z)\text{.}$
$P'(2) = 8\cos(2) \approx -3.329$ hundred animals per decade.

2.3 The product and quotient rules 2.3.1 The product rule

Activity 2.3.2 ..

$m'(w) = 3w^{17} \cdot 4^w \ln(4) + 4^w \cdot 51w^{16}\text{.}$
$h'(t) = (\sin(t) + \cos(t)) \cdot 4t^3 + t^4 \cdot (\cos(t) - \sin(t))\text{.}$
$f'(1) = e(\cos(1) + \sin(1)) \approx 3.756\text{.}$
$L(x) = -\frac{1}{2}(x+1)\text{.}$

2.3.2 The quotient rule

Activity 2.3.3 ..

$r'(z)=\frac{(z^4+1) 3^z \ln(3) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.}$
$v'(t) = \frac{(\cos(t) + t^2)\cos(t) - \sin(t)(-\sin(t) + 2t)}{(\cos(t) + t^2)^2}\text{.}$
$R'(0) = \frac{2}{9}\text{.}$
$I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}$ $I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}$ and $I'(5) = \frac{-400}{e^5} \approx -2.695\text{,}$ each in candles per millisecond.

2.3.3 Combining rules

Activity 2.3.4 ..

$f'(r) = (5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)] + (4^r - 2\cos(r))[15r^2 + \cos(r)]\text{.}$
$p'(t) = \frac{t^6 \cdot 6^t [-\sin(t)] - \cos(t) [t^6 \cdot 6^t \ln(6) + 6^t \cdot 6t^5]}{(t^6 \cdot 6^t)^2}\text{.}$
$g'(z) = 3 [z^7 e^z + 7z^6e^z] - 2[z^2 \cos(z) + 2z\sin(z)] + \frac{(z^2+1) 1 - z(2z)}{(z^2 + 1)^2}\text{.}$
$s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414$ feet per second.
$p'(3) = 30$ and $q'(3) = \frac{13}{8}\text{.}$

2.4 Derivatives of other trigonometric functions 2.4.1 Derivatives of the cotangent, secant, and cosecant functions

Activity 2.4.2 ..

All real numbers $x$ such that $x \ne \frac{\pi}{2} + k\pi\text{,}$ where $k = \pm 1, \pm 2, \ldots\text{.}$
$h'(x) = \frac{\sin(x)}{\cos^2(x)}\text{.}$
$h'(x) = \sec(x) \tan(x)\text{.}$
$h$ and $h'$ have the same domain: all real numbers $x$ such that $x \ne \frac{\pi}{2}+k\pi\text{,}$ where $k = 0, \pm 1, \pm 2, \ldots\text{.}$

Activity 2.4.3 .

All real numbers $x$ such that $x \ne k\pi\text{,}$ where $k = 0, \pm 1, \pm 2, \ldots\text{.}$
$h'(x) = -\frac{\cos(x)}{\sin^2(x)}\text{.}$
$h'(x) = -\csc(x) \cot(x)\text{.}$
$p$ and $p'$ have the same domain: all real numbers $x$ such that $x \ne k\pi\text{,}$ where $k = 0, \pm 1, \pm 2, \ldots\text{.}$

Activity 2.4.4 .

$m = f'(\frac{\pi}{3}) =10\sqrt{3} + \frac{4}{3}\text{.}$
$p'(\frac{\pi}{4}) = \frac{\pi^2}{16} \sqrt{2} + \frac{\sqrt{2}\pi}{2} + \frac{\pi}{2} - 1\text{.}$
$h'(t) = \frac{(t^2+1) \sec^2(t) - 2t \tan(t)}{(t^2 + 1)^2} + 2e^t \sin(t) - 2 e^t\cos(t)\text{.}$
$g'(r) = \frac{r \sec(r) \tan(r) + \sec(r) - r ln(5) \sec(r)}{5^r}\text{.}$
$s'(2) = \frac{15\cos(2) - 15\sin(2)}{e^2} \approx -2.69$ inches per second.

2.5 The chain rule 2.5.1 The chain rule

Activity 2.5.2 ..

$h'(x) = -4x^3\sin(x^4)\text{.}$
$h'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}\text{.}$
$h'(x) = 2^{\sin(x)}\ln(2)\cos(x)\text{.}$
$h'(x) = -5\cot^4(x) \csc^2(x)\text{.}$
$h'(x) = 9(\sec(x)+e^x)^8 (\sec(x)\tan(x) + e^x)\text{.}$

2.5.2 Using multiple rules simultaneously

Activity 2.5.3 ..

$p'(r) = \frac{4(6r^5 + 2e^r)}{2\sqrt{r^6 + 2e^r}}\text{.}$
$m'(v) = -3v^2 \sin(v^2)\sin(v^3) + 2v \cos(v^3)\cos(v^2)\text{.}$
$h'(y) = \frac{(e^{4y}+1) [-10\sin(10y)] - \cos(10y) [4e^{4y}]}{(e^{4y}+1)^2}\text{.}$
$s'(z) = 2^{z^2\sec(z)} \ln(2) [z^2 \sec(z)\tan(z) + \sec(z) \cdot 2z]\text{.}$
$c'(x) = \cos(e^{x^2}) [e^{x^2}\cdot 2x]\text{.}$

Activity 2.5.4 .

$y - 2 = \frac{1}{4}(x-0)\text{.}$
$v(1) = s'(1) = -\frac{3}{8}$ inches per second; the particle is moving left at the instant $t = 1\text{.}$
$P'(1000) = 30 e^{-0.0323} (-0.0000323) \approx -0.000938$ inches of mercury per foot.
$C'(2) = -10 \text{;}$ $D'(-1) = -20\text{.}$

2.6 Derivatives of Inverse Functions 2.6.2 The derivative of the natural logarithm function

Activity 2.6.2 ..

$h'(x) = x + 2x\ln(x)\text{.}$
$p'(t) = \frac{(e^t + 1) \frac{1}{t} - \ln(t) \cdot e^t}{(e^t + 1)^2}\text{.}$
$s'(y) = \frac{1}{\cos(y) + 2} \cdot (-\sin(y))\text{.}$
$z'(x) = \sec^2(\ln(x)) \cdot \frac{1}{x}\text{.}$
$m'(z) = \frac{1}{\ln(z)} \cdot \frac{1}{z}\text{.}$

2.6.3 Inverse trigonometric functions and their derivatives

Activity 2.6.3 ..

$\tan(r(x)) = x\text{.}$
$r'(x) = \cos^2(r(x))\text{.}$
$r'(x) = \cos^2(\arctan(x))\text{.}$
$\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}\text{.}$
$r'(x) = \frac{1}{1+x^2}\text{.}$

Activity 2.6.4 .

$f'(x) = \left[x^3 \cdot \frac{1}{1+x^2} + \arctan(x) \cdot 3x^2 \right] + \left[e^x \cdot \frac{1}{x} + \ln(x) \cdot e^x\right]\text{.}$
$p'(t) = 2^{t\arcsin(t)} \ln(2) [t \cdot \frac{1}{\sqrt{1-t^2}} + \arcsin(t) \cdot 1]\text{.}$
$h'(z) = 27(\arcsin(5z) + \arctan(4-z))^{26} \left[\frac{1}{\sqrt{1-(5z)^2}} \cdot 5 + \frac{1}{1+(4-z)^2} \cdot (-1) \right]\text{.}$
$s'(y) = -\frac{1}{y^2}\text{.}$
$m'(v) = \frac{1}{\sin^2(v)+1} \cdot \left[ 2\sin(v)\cos(v) \right]\text{.}$
$\displaystyle g'(w) = \frac{1}{1+ \left( \frac{\ln(w)}{1+w^2} \right)^2} \cdot \left[ \frac{(1+w^2) \frac{1}{w} - \ln(w) \cdot 2w}{(1+w^2)^2} \right] $

2.7 Derivatives of Functions Given Implicitly 2.7.1 Implicit Differentiation

Activity 2.7.2 ..

The graph of the curve fails the vertical line test.
$\frac{dy}{dx} = \frac{1}{5y^4 - 15y^2 + 4}\text{.}$
$y = -\frac{1}{6}x + 1\text{.}$
$(1.418697,0.543912)\text{,}$ $(-1.418697,-0.543912)\text{,}$ $(-3.63143, 1.64443)\text{,}$ and $(3.63143, -1.64443)\text{.}$

Activity 2.7.3 .

Horizontal at $x \approx 0.42265\text{,}$ thus $(0.42265, -1.05782); (0.42265, 0.229478); (0.42265, 0.770522); (0.42265, 2.05782)\text{.}$ There are four more points where $x \approx 1.57735\text{.}$
When $y = \frac{1}{2}, \frac{1 \pm \sqrt{5}}{2}\text{,}$ so one point is $(2.21028, \frac{1}{2})\text{.}$
$y - 1 = \frac{1}{2}(x-1)\text{.}$

Activity 2.7.4 .

$\frac{dy}{dx}(-3y^2 - 6x) = 6y-3x^2 $ and the tangent line has equation $y - 3 = 1(x+3)\text{.}$
$\frac{dy}{dx} = \frac{3x^2 + 1}{\cos(y) + 1}$ and the tangent line has equation $y = \frac{1}{2}x\text{.}$
$\frac{dy}{dx} = \frac{3e^{-xy} - 3xye^{-xy}}{3x^2e^{-xy}+2y}$ and the tangent line is $y - 1 = 0.234950(x - 0.619061)\text{.}$

2.8 Using Derivatives to Evaluate Limits 2.8.1 Using derivatives to evaluate indeterminate limits of the form $\frac{0}{0}\text{.}$

Activity 2.8.2 ..

$\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1\text{.}$
$\lim_{x \to \pi} \frac{\cos(x)}{x} = -\frac{1}{\pi}\text{.}$
$\lim_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}} = -2\text{.}$
$\lim_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1} = 0\text{.}$

Activity 2.8.3 .

$\lim_{x \to 2} \frac{f(x)}{g(x)} = \frac{1}{8}\text{.}$
$\lim_{x \to 2} \frac{p(x)}{q(x)} = 1\text{.}$
$\lim_{x \to 2} \frac{r(x)}{s(x)} \lt 0\text{.}$

2.8.2 Limits involving $\infty$

Activity 2.8.4 ..

$\lim_{x \to \infty} \frac{x}{\ln(x)} = \infty\text{.}$
$\lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \frac{1}{2}\text{.}$
$\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = 0\text{.}$
$\lim_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}} = -\infty\text{.}$
$\lim_{x \to \infty} xe^{-x} = 0\text{.}$

3 Using Derivatives 3.1 Using derivatives to identify extreme values 3.1.1 Critical numbers and the first derivative test

Activity 3.1.2 ..

$x = -4$ or $x = 1\text{.}$
$g$ has a local maximum at $x = -4$ and neither a max nor min at $x = 1\text{.}$
$g$ does not have a global minimum; it is unclear (at this point in our work) if $g$ increases without bound, so we can’t say for certain whether or not $g$ has a global maximum.
$\lim_{x \to \infty} g'(x) = \infty\text{.}$

3.1.2 The second derivative test

Activity 3.1.3 ..

$x = -1$ is an inflection point of $g\text{.}$
$g$ is concave up for $x \lt -1\text{,}$ concave down for $-1 \lt x \lt 2\text{,}$ and concave down for $x \gt 2\text{.}$
$g$ has a local minimum at $x = -1.67857351\text{.}$
$g$ is a degree 5 polynomial.

Activity 3.1.4 .

If $\frac{2}{k^2} \gt 1\text{,}$ then the equation $\cos(kx) = \frac{2}{k^2}$ has no solution. Hence, whenever $k^2 \lt 2\text{,}$ or $k \lt \sqrt{2} \approx 1.414\text{,}$ it follows that the equation $\cos(kx) = \frac{2}{k^2}$ has no solutions $x\text{,}$ which means that $h''(x)$ is never zero (indeed, for these $k$ -values, $h''(x)$ is always positive so that $h$ is always concave up). On the other hand, if $k \ge \sqrt{2}\text{,}$ then $\frac{2}{k^2} \le 1\text{,}$ which guarantees that $\cos(kx) = \frac{2}{k^2}$ has infinitely many solutions, due to the periodicity of the cosine function. At each such point, $h''(x) = 2 - k^2 \cos(kx)$ changes sign, and therefore $h$ has infinitely many inflection points whenever $k \ge \sqrt{2}\text{.}$
To see why $h$ can only have a finite number of critical numbers regardless of the value of $k\text{,}$ consider the equation \begin{equation*} 0 = h'(x) = 2x - k\sin(kx)\text{,} \end{equation*} which implies that $2x = k\sin(kx)\text{.}$ Since $-1 \le \sin(kx) \le 1\text{,}$ we know that $-k \le k\sin(kx) \le k\text{.}$ Once $|x|$ is sufficiently large, we are guaranteed that $|2x| \gt k\text{,}$ which means that for large $x\text{,}$ $2x$ and $k\sin(kx)$ cannot intersect. Moreover, for relatively small values of $x\text{,}$ the functions $2x$ and $k\sin(kx)$ can only intersect finitely many times since $k\sin(kx)$ oscillates a finite number of times. This is why $h$ can only have a finite number of critical numbers, regardless of the value of $k\text{.}$

3.2 Using derivatives to describe families of functions 3.2.1 Describing families of functions in terms of parameters

Activity 3.2.2 ..

$p$ has two critical numbers ( $x = \pm \sqrt{\frac{a}{3}}$ ) whenever $a \gt 0$ and no critical numbers when $a \lt 0\text{.}$
When $a \lt 0\text{,}$ $p$ is always increasing and has no relative extreme values. When $a\gt 0\text{,}$ $p$ has a relative maximum at $x = -\sqrt{\frac{a}{3}}$ and a relative minimum at $x = +\sqrt{\frac{a}{3}}\text{.}$
$p$ is CCD for $x \lt 0$ and $p$ is CCU for $x\gt 0\text{,}$ making $x = 0$ an inflection point.

Activity 3.2.3 .

$h$ is an always increasing function.
$h$ is always concave down.
$\lim_{x \to \infty} a(1-e^{-bx}) = a\text{,}$ and $\lim_{x \to \infty} a(1-e^{-bx}) = -\infty\text{.}$
If $b$ is large and $x$ is close to zero, $h'(x)$ is relatively large near $x = 0\text{,}$ and the curve’s slope will quickly approach zero as $x$ increases. If $b$ is small, the graph is less steep near $x = 0$ and its slope goes to zero less quickly as $x$ increases.

Activity 3.2.4 .

$L$ is an always increasing function.
$L$ is concave up for all $t \lt -\frac{1}{k} \ln \left(\frac{1}{c}\right)$ and concave up for all other values of $t\text{.}$
$\lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = A\text{,}$ and \begin{equation*} \lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = 0\text{.} \end{equation*}
The inflection point on the graph of $L$ is $( -\frac{1}{k} \ln \left(\frac{1}{c}\right), \frac{A}{2})\text{.}$

3.3 Global Optimization 3.3.1 Global Optimization

Activity 3.3.2 ..

$x = \pm \sqrt{2} \approx \pm 1.414\text{.}$
On $[-2,3]\text{,}$ $g$ has a global maximum at $x = 3$ and a global minimum at $x = \sqrt{2}\text{.}$
On $[-2,2]\text{,}$ $g$ has a global maximum at $x = -\sqrt{2}$ and a global minimum at $x = \sqrt{2}\text{.}$
On $[-2,1]\text{,}$ $g$ has a global maximum at $x = -\sqrt{2}$ and a global minimum at $x = 1\text{.}$

Activity 3.3.3 .

Absolute maximum: $e^{-1}\text{;}$ absolute minimum: $0\text{.}$
Absolute maximum: $\sqrt{2}\text{;}$ absolute minimum: $-1\text{.}$
Absolute maximum: 9.8; absolute minimum: 8.
Absolute minimum 3; no absolute maximum.
Absolute minimum $0\text{;}$ absolute maximum $\frac{1}{a}e^{-1}\text{.}$
Absolute minimum $b-1\text{;}$ no absolute maximum.

3.3.2 Moving toward applications

Activity 3.3.4 ..

$V(x) = x (10-2x) (15-2x) = 4x^3 - 50x^2 + 150x\text{.}$
$1 \le x \le 3\text{.}$
$x = \frac{25 \pm 5\sqrt{7}}{6} \approx 6.371459426, 1.961873908\text{.}$
$\displaystyle V(1.961873908) = 132.0382370$
$\displaystyle V(1) = 104$
$\displaystyle V(3) = 108$
Absolute maximum: 132.0382370; absolute minimum: 104.

3.4 Applied Optimization 3.4.1 More applied optimization problems

Activity 3.4.2 ..

Let the can have radius $r$ and height $h\text{.}$
$V = \pi r^2 h\text{;}$ $S = 2 \pi r^2 + 2 \pi r h\text{;}$ $C = 2 \pi r^2 \cdot 0.027 + 2 \pi r h \cdot 0.015\text{.}$
$C(r) = 0.054 \pi r^2 + 0.48 \frac{1}{r}\text{,}$ $r \gt 0\text{.}$
$r = \sqrt[3]{ \frac{0.48}{0.108 \pi} } \approx 1.12259\text{;}$ $h \approx 4.041337\text{;}$ minimum cost $C(1.12259) \approx 0.64137\text{.}$

Activity 3.4.3 .

Activity 3.4.4 ., activity 3.4.5 ., 3.5 related rates 3.5.1 related rates problems, activity 3.5.2 ..

$r = \frac{3}{4}h\text{.}$
$V = \frac{3}{16} \pi h^3\text{.}$
$\frac{dV}{dt} = \frac{9}{16} \pi h^2 \frac{dh}{dt} \text{.}$
$\left. \frac{dh}{dt} \right|_{h=3} = \frac{64}{81\pi} \approx 0.2515$ feet per minute.
Most rapidly when $h = 3\text{.}$

Activity 3.5.3 .

$\frac{dh}{dt} = 4000 \sec^2 (\theta) \frac{d\theta}{dt}\text{.}$
$h \frac{dh}{dt} = z \frac{dz}{dt}\text{.}$
$\left. \frac{dz}{dt} \right|_{h=3000} = 360 \ \text{feet/sec}; $ $\left. \frac{d\theta}{dt} \right|_{h=3000} = \frac{12}{125} $ radians per second.

Activity 3.5.4 .

$3s = 2x\text{.}$
$3 \frac{ds}{dt} = 2\frac{dx}{dt}\text{.}$
$\left. \frac{ds}{dt} \right|_{x=8} = 2$ feet per second.
at a constant rate.
Let $y$ represent the location of the tip of the shadow; $\frac{dy}{dt} = 5$ feet/sec.

Activity 3.5.5 .

4 the definite integral 4.1 determining distance traveled from velocity 4.1.1 area under the graph of the velocity function, activity 4.1.2 ..

Using 8 rectangles of width $0.25\text{,}$ $D \approx 3.875\text{.}$
$s(t) = \frac{1}{8}t^4 - \frac{1}{2} t^3 + \frac{3}{4} t^2 + \frac{3}{2}t\text{.}$
$s(2) - s(0) = \frac{1}{8}2^4 - \frac{1}{2}2^3 + \frac{3}{4}2^2 + \frac{3}{2} 2 = 4\text{.}$

4.1.2 Two approaches: area and antidifferentiation

Activity 4.1.3 ..

On $(0,1)\text{,}$ $s$ is increasing because velocity is positive.
$s(t) = 32t - 16t^2\text{.}$
$s(1) - s(\frac{1}{2}) = 4\text{.}$
$A = 4$ feet is the total distance the ball traveled vertically on $[\frac{1}{2},1]\text{.}$
$s(1) - s(0) = 16$ is the vertical distance the ball traveled on the interval $[0,1]\text{.}$ Equivalently, the area between the velocity curve and the $t$ -axis on $[0,1]$ is $A = 16$ feet.
$s(2) - s(0) = 0\text{,}$ so the ball has zero change in position on the interval $[0,2]\text{.}$

4.1.3 When velocity is negative

Activity 4.1.4 ..

Total distance traveled is $2\text{;}$ change in position is $0\text{.}$
$0 \lt t \lt 1$ and $4 \lt t \lt 8 \text{.}$
$s(8) - s(0) = 5 \ \mbox{m} \text{,}$ while the distance traveled on $[0,8]$ is $D = 13\text{,}$ and thus these two quantities are different.
See the figure below.

4.2 Riemann Sums 4.2.1 Sigma Notation

Activity 4.2.2 ..

$\displaystyle 65 $
$\displaystyle 32 $
\begin{equation*} 3 + 7 + 11 + 15 + \cdots + 27 = \sum_{k=1}^{7} 4k-1\text{.} \end{equation*}
\begin{equation*} 4 + 8 + 16 + 32 + \cdots + 256 = \sum_{i=2}^{8} 2^i\text{.} \end{equation*}
\begin{equation*} \sum_{i=1}^{6} \frac{1}{2^i} = \frac{63}{64}\text{.} \end{equation*}

4.2.2 Riemann Sums

Activity 4.2.3 ..

$L_4 = \frac{311}{48} \approx 6.47917\text{,}$ $R_4 = \frac{335}{48} \approx 6.97917\text{,}$ and $M_4 = \frac{637}{96} \approx 6.63542\text{.}$
\begin{equation*} \frac{L_4 + M_4}{2} = \frac{646}{96} \ne \frac{637}{96} = M_4\text{.} \end{equation*}
$L_n$ is an under-estimate; $R_n$ is an over-estimate.

4.2.3 When the function is sometimes negative

Activity 4.2.4 ..

The change in position is approximately $-1.44$ feet.
$D \approx 2.336\text{.}$
$-\frac{4}{3}$ is the object’s total change in position on $[1,5]\text{.}$

4.3 The Definite Integral 4.3.1 The definition of the definite integral

Activity 4.3.2 ..

$\int_0^1 3x \, dx = \frac{3}{2}\text{.}$
$\int_{-1}^4 (2-2x) \, dx = -5\text{.}$
$\int_{-1}^1 \sqrt{1-x^2} \, dx = \frac{\pi}{2}\text{.}$
$\int_{-3}^4 g(x) \, dx = \frac{3\pi}{4} - \frac{3}{2}\text{.}$

4.3.2 Some properties of the definite integral

Activity 4.3.3 ..

$\int_5^2 f(x) \,dx = -2\text{.}$
$\int_0^5 g(x) \,dx = 3\text{.}$
$\int_0^5 (f(x) + g(x))\, dx = 2\text{.}$
$\int_2^5 (3x^2 - 4x^3) \, dx = -492\text{.}$
$\int_5^0 (2x^3 - 7g(x)) \, dx = -\frac{583}{2}\text{.}$

4.3.3 How the definite integral is connected to a function’s average value

Activity 4.3.4 ..

$y = v(t) = \sqrt{4-(t-2)^2}$ is the top half of the circle $(t-2)^2 + y^2 = 4\text{,}$ which has radius 2 and is centered at $(2,0)\text{.}$
$\int_0^4 v(t) \, dt = 2\pi\text{.}$
The object moved $2 \pi$ meters in 4 minutes.
$v_{\text{AVG} }[0,4] = \frac{\pi}{2}\text{,}$ meters per minute,.
$D = 2\pi\text{.}$

4.4 The Fundamental Theorem of Calculus 4.4.1 The Fundamental Theorem of Calculus

Activity 4.4.2 ..

$\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = 1\text{.}$
$\int_0^1 e^x \, dx = e-1\text{.}$
$\int_{-1}^{1} x^5 \, dx = 0\text{.}$
$\int_0^2 (3x^3 - 2x^2 - e^x) \, dx = \frac{23}{3} - e^2\text{.}$

4.4.2 Basic antiderivatives

Activity 4.4.3 ..

$\int_0^1 \left(x^3 - x - e^x + 2\right) \,dx = \frac{11}{4} - e\text{.}$
$\int_0^{\pi/3} (2\sin (t) - 4\cos(t) + \sec^2(t) - \pi) \, dt = 1 - \sqrt{3} - \frac{\pi^2}{3}\text{.}$
$\int_0^1 (\sqrt{x} - x^2) \, dx = \frac{1}{3}\text{.}$

4.4.3 The total change theorem

Activity 4.4.4 ..

The person burned exactly $\frac{400}{3}$ calories in the first 10 minutes of the workout.
$C(40) - C(0) = \int_0^{40} C'(t) \, dt = \int_0^{40} c(t) \, dt$ is the total calories burned on $[0,40]\text{.}$
The exact average rate at which the person burned calories on $0 \le t \le 40$ is \begin{equation*} c_{\operatorname{AVG} [0,40]} = \frac{1}{40-0} \int_0^{40} c(t) \, dt = \frac{1}{40} \cdot \frac{1700}{3} = \frac{1700}{120} \approx 14.17 \ \text{cal/min}\text{.} \end{equation*}
One time at which the instantaneous rate at which calories are burned equals the average rate on $[0,40]$ is $t = \frac{5}{3}(6 - \sqrt{6}) \approx 5.918\text{.}$

5 Evaluating Integrals 5.1 Constructing Accurate Graphs of Antiderivatives 5.1.1 Constructing the graph of an antiderivative

Activity 5.1.2 ..

$F$ is increasing on $(0,2)$ and $(5,7)\text{;}$ $F$ is decreasing on $(2,5)\text{.}$
$F$ is concave up on $(0,1)\text{,}$ $(4,6)\text{;}$ concave down on $(1,3)\text{,}$ $(6,7)\text{;}$ neither on $(3,4)\text{.}$
A relative maximum at $x = 2\text{;}$ a relative minimum at $x = 5\text{.}$
$F(1) = -\frac{1}{2}\text{;}$ $F(2) = \frac{\pi}{4} - \frac{1}{2}\text{;}$ $F(3) = \frac{\pi}{4} - 1\text{;}$ $F(4) = \frac{\pi}{4}-2\text{;}$ $F(5) = \frac{\pi}{4} - \frac{5}{2}\text{;}$ $F(6) = \frac{\pi}{2} - \frac{5}{2}\text{;}$ $F(7) = \frac{3\pi}{4} - \frac{5}{2}\text{;}$ $F(8) = \frac{3\pi}{4} - \frac{5}{2}\text{;}$ and $F(-1) = -1\text{.}$
Use the function values found in (d) and the earlier information regarding the shape of $F\text{.}$
$G(x) = F(x) + 1\text{.}$

5.1.2 Multiple antiderivatives of a single function

Activity 5.1.3 ..

$H(x) = -\cos(x) + 2\text{.}$

5.1.3 Functions defined by integrals

Activity 5.1.4 ..

$A$ is increasing on $(0,1.5)\text{,}$ $(4,6)\text{;}$ $A$ is decreasing on $(1.5,4)\text{.}$
$A$ is concave up on $(0,1)$ and $(3,5)\text{;}$ $A$ is concave down on $(1,3)$ and $(5,6)\text{.}$
At $x = 1.5\text{,}$ $A$ has a relative maximum; $A$ has a relative minimum at $x = 4\text{.}$
$A(0) = -\frac{1}{2}\text{;}$ $A(1) = 0\text{;}$ $A(2) = 0\text{;}$ $A(3) = -2\text{;}$ $A(4) = -3.5\text{,}$ $A(5) = -2\text{,}$ $A(6) = -0.5\text{.}$
Use your work in (a)-(d) appropriately.
$B(x) = A(x) + \frac{1}{2}\text{.}$

5.2 The Second Fundamental Theorem of Calculus 5.2.1 The Second Fundamental Theorem of Calculus

Activity 5.2.2 ..

$A'(x) = f(x)\text{.}$
$A(1) = -\frac{\pi}{4}\text{.}$
$A$ is increasing wherever $f$ is positive; $A$ is CCU wherever $f$ is increasing. $A(2) = 0\text{,}$ $A(3) = -0.5\text{,}$ $A(4) = -1.5\text{,}$ $A(5) = -2\text{,}$ $A(6) = -2 + \frac{\pi}{4}\text{,}$ and $A(7) = -2 + \frac{\pi}{2}\text{.}$
$F$ and $A$ differ by the constant $\frac{\pi}{4} - \frac{1}{2}\text{.}$
$B$ and $C$ have the same shape as $A$ and $F\text{,}$ and differ from $A$ by a constant. Observe that $B(3) = 0$ and $C(1) = 0\text{.}$

5.2.2 Understanding Integral Functions

Activity 5.2.3 ..

See the plot at below left.
$F' = f\text{.}$
$F$ is increasing for all $x \gt 0\text{;}$ $F$ is decreasing for $x \lt 0$
$F$ is CCU on $-1 \lt x \lt 1$ and CCD for $x \lt -1$ and $x \gt 1\text{.}$
$F(5) \approx 1.64038\text{;}$ $F(10) \approx 2.35973\text{.}$
See the graph at below right.

5.2.3 Differentiating an Integral Function

Activity 5.2.4 ..

$\frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right] = e^{x^2}\text{.}$
$\int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt = \frac{x^4}{1+x^4} - \frac{16}{17}\text{.}$
$\frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right] = -\cos(x^3)\text{.}$
$\int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt = \ln(1+x^2)-\ln(10)\text{.}$
$\frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right] = \sin(x^6) \cdot 3x^2\text{.}$

5.3 Integration by Substitution 5.3.1 Reversing the Chain Rule: First Steps

Activity 5.3.2 ..

$\int \sin(8-3x) \, dx = -\frac{1}{3} (-\cos(8-3x)) + C\text{.}$
$\int \sec^2 (4x) \, dx = \frac{1}{4} \tan(4x) + C\text{.}$
$\int \frac{1}{11x - 9} \, dx = \frac{1}{11} \ln|11x - 9| + C\text{.}$
$\int \csc(2x+1) \cot(2x+1) \, dx = -\frac{1}{2}\csc(2x+1) + C\text{.}$
$\displaystyle \int \frac{1}{\sqrt{1-16x^2}}\, dx = \frac{1}{4} \arcsin(4x) + C$
$\int 5^{-x}\, dx = -\frac{1}{\ln(5)}5^{-x} + C\text{.}$

5.3.2 Reversing the Chain Rule: $u$ -substitution

Activity 5.3.3 ..

$\int \frac{x^2}{5x^3+1} \, dx = \frac{1}{15} \ln(5x^3 + 1) + C\text{.}$
$\int e^x \sin(e^x) \, dx = -\cos(e^x) + C\text{.}$
$\int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx = 2\sin(\sqrt{x}) + C\text{.}$

5.3.3 Evaluating Definite Integrals via $u$ -substitution

Activity 5.3.4 ..

$\int_{x=1}^{x=2} \frac{x}{1 + 4x^2} \, dx = \frac{1}{8} (\ln(17) - \ln(5))\text{.}$
$\int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx = -\frac{1}{20}(2e^{-1}+3)^{10} + \frac{1}{20}(2e^{0}+3)^{10}\text{.}$
$\int_{2/\pi}^{4/\pi} \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx = 1 - \frac{\sqrt{2}}{2}\text{.}$

5.4 Integration by Parts 5.4.1 Reversing the Product Rule: Integration by Parts

Activity 5.4.2 ..

$\int t e^{-t} dt = -te^{-t} - e^{-t} + c\text{.}$
$\int 4x \sin(3x) dx = -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + c \text{.}$
$\int z \sec^2(z) dz = z \tan(z) + \ln |\cos(z)| + c \text{.}$
$\int x\ln(x) dx = \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + c \text{.}$

5.4.2 Some Subtleties with Integration by Parts

Activity 5.4.3 ..

$\int{\arctan(x) dx} = x\arctan(x) - \frac{1}{2} \ln \left( | 1 + x^2 | \right) + c \text{.}$
$\int \ln(z) dz = z \ln(z) - z + c \text{.}$
$\int t^3 \sin(t^2) dt = \frac{1}{2} \left( -t^2 \cos\left(t^2 \right) + \sin\left(t^2\right) \right) \text{.}$
$\int s^5 e^{s^3} ds = \frac{1}{3} \left( s^3 e^{s^3} - e^{s^3} \right) + c \text{.}$
$\int e^{2t} \cos\left( e^t \right) dt = e^t \sin \left( e^t \right) + \cos \left( e^t \right) + c \text{.}$

5.4.3 Using Integration by Parts Multiple Times

Activity 5.4.4 ..

$\int x^2 \sin(x) dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + c \text{.}$
$\int t^3 \ln(t) dt = \frac{1}{4} t^4 \ln(t) - \frac{1}{16} t^4 + c \text{.}$
$\int e^z \sin(z) dz = -\frac{1}{2}e^z \cos(z) + \frac{1}{2}e^z \sin(z) + c \text{.}$
$\int s^2 e^{3s} ds = \frac{1}{3}s^2 e^{3s} - \frac{2}{9}s e^{3s} + \frac{2}{27} e^{3s} + c \text{.}$
$\int t \arctan(t) dt = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2}t - \frac{1}{2} \arctan(t) + c \text{.}$

5.5 Other Options for Finding Algebraic Antiderivatives 5.5.1 The Method of Partial Fractions

Activity 5.5.2 ..

$\int \frac{1}{x^2 - 2x - 3} \, dx = \frac{1}{4}\ln|x-3| - \frac{1}{4}\ln|x+1| + C\text{.}$
$\int \frac{x^2+1}{x^3 - x^2} \, dx = -\ln|x| + x^{-1} + 2\ln|x-1| + C\text{.}$
$\int \frac{x-2}{x^4 + x^2}\, dx = \ln|x| + 2x^{-1} - \frac{1}{2} \ln|1+x^2| + 2\arctan(x) + C\text{.}$

5.5.2 Using an Integral Table

Activity 5.5.3 ..

$\int \sqrt{x^2 + 4} \, dx = \frac{x}{2} \sqrt{x^2+4} + 2 \ln | x + \sqrt{x^2+4}| + C\text{.}$
$\int \frac{x}{\sqrt{x^2 +4}} \, dx = \sqrt{x^2 + 4} + C\text{.}$
$\int \frac{2}{\sqrt{16+25x^2}}\, dx = \frac{2}{5} \ln| 5x + \sqrt{16+25x^2} | + C\text{.}$
$\int \frac{1}{x^2 \sqrt{49-36x^2}} \, dx = - \frac{\sqrt{49-36x^2}}{49x} + C\text{.}$

5.6 Numerical Integration 5.6.1 The Trapezoid Rule

Activity 5.6.2 ..

$\int_1^2 \dfrac{1}{x^2} dx = \dfrac{1}{2}\text{.}$
The trapezoid rule overestimates; the midpoint rule underestimates.
$f(x) = \dfrac{1}{x^2}$ is concave up on $[1, 2]\text{.}$

5.6.3 Simpson’s Rule

Activity 5.6.3 ..

Plot the data.
$\int_0^{1.8} v(t) dt\text{.}$
\begin{align*} L_3 \amp = 165.6 \text{ ft } \amp R_3 \amp = 105.6 \text{ ft } \amp T_3 \amp = 135.6 \text{ ft}\text{.} \end{align*} $R_3$ and $T_3$ are underestimates.
$M_3 = 143.4 \text{ ft}\text{;}$ overestimate.
$S_6 = 140.8 \text{ ft}\text{.}$
Simpson’s rule gives the best approximation of the distance traveled, $\int_0^{1.8} v(t) dt \approx 140.8 \text{ ft}\text{,}$ which leads to $AV_{[0,1.8]} \approx \frac{140.8}{1.8} \approx 78.22 \text{ ft/sec}\text{.}$

5.6.4 Overall observations regarding $L_n\text{,}$ $R_n\text{,}$ $T_n\text{,}$ $M_n\text{,}$ and $S_{2n}\text{.}$

Activity 5.6.4 ..

\begin{align*} \int_0^1 f(x) dx \amp = \frac{5}{3} \amp \int_0^1 g(x) dx \amp = \frac{7}{4} \amp \int_0^1 h(x) dx \amp = \frac{9}{5} \end{align*}
Left endpoint rule results are overestimates; right endpoint rules are underestimates; midpoint rules are overestimates; trapezoid rules are underestimates. Simpson’s rule is exact for both $f$ and $g\text{,}$ while a slight underestimate of $\int_0^1 h(x) dx\text{.}$

6 Using Definite Integrals 6.1 Using Definite Integrals to Find Area and Length 6.1.1 The Area Between Two Curves

Activity 6.1.2 ..

$A = \int_{0}^{16} (\sqrt{x} - \frac{1}{4}x) \, dx = \frac{32}{3}\text{.}$
$A = \int_{-\sqrt{20}/3}^{\sqrt{20}/3} ((12-2x^2)-(x^2-8)) \, dx \frac{160 \sqrt{\frac{5}{3}}}{3} \approx 68.853\text{.}$
$A = \int_0^\frac{\pi}{4} \left( \cos(x) - \sin(x) \right) dx = \sqrt{2} - 1\text{.}$
The left-hand region has area \begin{equation*} A_1 = \int_{\frac{1 - \sqrt{5}}{2}}^0 \left( \left(x^3 - x \right) - x^2\right) dx = \dfrac{13 - 5\sqrt{5}}{24} \approx 0.075819\text{.} \end{equation*} The right-hand region has area \begin{equation*} A_2 = \int_0^{\frac{1 + \sqrt{5}}{2}} \left( x^2 - \left(x^3 - x \right) \right) dx = \dfrac{13 + 5\sqrt{5}}{24} \approx 1.007514 \text{.} \end{equation*}

6.1.2 Finding Area with Horizontal Slices

Activity 6.1.3 ..

$A = \int_{y=-\sqrt{2}}^{y=\sqrt{2}} (6-2y^2 - y^2) \, dy = 8\sqrt{2} \approx 11.314\text{.}$
$A = \int_{y=-1}^{y=1} (2-2y^2-(1-y^2)) \, dy = \frac{4}{3}\text{.}$
$\displaystyle A = \int_{y=0}^{y=1} \left(2-y - \sqrt{y} \right) \, dy = \frac{5}{6} $
$A = \int_{0}^{3} (y - (y^2 - 2y)) \, dy = \frac{9}{2}\text{.}$

6.1.3 Finding the length of a curve

Activity 6.1.4 ..

$L \approx 2.95789\text{.}$
$L = \int_{-2}^{2} \sqrt{\frac{4}{4-x^2}} \, dx = 2\pi\text{.}$
$L = \int_0^1 \sqrt{1 + e^{6x}(9x^2 + 6x + 1)} \, dx \approx 20.1773\text{.}$
We will usually have to estimate the value of $\int_a^b \sqrt{1+f'(x)^2} \, dx$ using computational technology.
Approximately $(14.9165,f(14.9165)) = (14.9165, 23.2502)\text{.}$

6.2 Using Definite Integrals to Find Volume 6.2.1 The Volume of a Solid of Revolution

Activity 6.2.2 ..

$V = \int_0^4 \pi (\sqrt{x})^2 \, dx = \int_0^4 \pi x \, dx = 8\pi\text{.}$
$V = \int_0^4 \pi (4-(\sqrt{x})^2) \, dx = \int_0^4 \pi (4-x) \, dx = 8\pi\text{.}$
$V = \int_0^1 \pi(x - x^6) \, dx = \frac{5}{14}\pi\text{.}$
$V = \int_{-\sqrt{3}}^{\sqrt{3}} \pi( (x^2 + 4)^2 - (2x^2 + 1)^2) \, dx = \frac{136\sqrt{3}}{5}\pi\text{.}$
$V = \int_0^2 \pi y^4 \, dy = \frac{32}{5}\pi\text{.}$

6.2.2 Revolving about the $y$ -axis

Activity 6.2.3 ..

$V = \int_0^2 \pi y^4 \Delta \, dy\text{.}$
$V = \int_0^2 \pi (16 - y^4) \, dy\text{.}$
$V = int_0^{\sqrt{2}} \pi ( 4x^2 - x^6 ) \, dx\text{.}$
$V = \int_0^{2\sqrt{2}} \pi( y^{2/3} - y^2/4 ) \, dy\text{.}$
$V = \int_0^3 \pi( (y+1)^2 - (y-1)^4 ) \, dy\text{.}$

6.2.3 Revolving about horizontal and vertical lines other than the coordinate axes

Activity 6.2.4 ..

\begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (2x+2)^2 - (x^3 + 2)^2 ) \, dx = \frac{4}{21}(21+8\sqrt{2}) \pi \approx 19.336\text{.} \end{equation*}
\begin{equation*} V = \int_{0}^{\sqrt{2}} \pi ( (4 - x^3)^2 - (4-2x)^2 ) \, dx = \left( 8-\frac{32\sqrt{2}}{21} \right)\pi \approx 18.3626\text{.} \end{equation*}
\begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (y^{1/3} + 1)^2 - (\frac{1}{2}y + 1)^2 ) \, dy = \frac{2}{15}(15 + 8\sqrt{2}) \pi \approx 11.022\text{.} \end{equation*}
\begin{equation*} V = \int_{0}^{2\sqrt{2}} \pi( (5 - \frac{1}{2}y)^2 - (5 - y^{1/3})^2 ) \, dy = \frac{2}{15}(75-8\sqrt{2})\pi \approx 26.677\text{.} \end{equation*}

6.3 Density, Mass, and Center of Mass 6.3.1 Density

Activity 6.3.2 ..

$M = 10 - 10e^{-2} \approx 8.64665$ grams.
$V = \int_{0}^{5} \pi (4 - \frac{4}{5}x)^2 \, dx = \frac{80\pi}{3} \approx 83.7758 \mbox{m}^3\text{.}$
$M = \frac{64000\pi}{3} \approx 67020.6433 \mbox{kg} \text{.}$
$\displaystyle M = \int_{0}^{5} (400 + \frac{200}{1+x^2}) \cdot \pi (4-\frac{4}{5}x)^2 \, dx = 128 \pi (\frac{265}{3} + 24 \arctan(5) - 5 \ln(26)) \approx 42224.8024 \mbox{kg}$
$b \approx 3.0652\text{.}$

6.3.2 Weighted Averages

Activity 6.3.3 ..

$\overline{x} = \frac{x_1 + x_2}{2} = 3\text{.}$
$\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = 3\text{.}$
$\overline{x} = \frac{x_1 + x_2 + x_3 + x_4}{4} = 2.75\text{.}$
$\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{16}{7}\text{.}$
$\overline{x} = \frac{2x_1 + 3x_2 + 1x_3 + 1x_4}{7} = \frac{17}{7}\text{.}$
$\overline{x} = \frac{2x_1 + 3x_2 + 2x_3 + 1x_4}{8} = \frac{20}{8}\text{.}$
Answers will vary.
If we have an existing arrangement and balancing point, moving one of the locations to the left will move the balancing point to the left; similarly, moving one of the locations to the right will move the balancing point to the right. If instead we add weight to an existing location, if that location is left of the balancing point, the balancing point will move left; the behavior is similar if on the right.

6.3.3 Center of Mass

Activity 6.3.4 ..

$M = \int_{0}^{20} 4 + 0.1x \, dx = 100$ g.
Greater than 10.
$\overline{x} = \frac{\int_{0}^{20} x (4 + 0.1x)) \, dx}{\int_{0}^{20} 4 + 0.1x \, dx} = \frac{32}{3}\text{.}$
Slightly to the right of the center of mass for $\rho(x)\text{.}$
$\overline{x} = \frac{\int_{0}^{20} x 4e^{0.020732x} \, dx}{\int_{0}^{20} 4e^{0.020732x} \, dx} \approx 10.6891\text{,}$

6.4 Physics Applications: Work, Force, and Pressure 6.4.1 Work

Activity 6.4.2 ..

$W = \int_0^{200} 0.3(200-h) \, dh = 6000 \text{ foot-pounds}\text{.}$
$W = \int_0^{100} (40-0.1h) \, dh = 3500 \text{foot-pounds}\text{.}$
$B_{\text{AVG} [0,100]} \approx 25.9798 \text{ pounds}\text{.}$
$k = 15\text{.}$
$W = \int_0^1 15x \, dx = \frac{15}{2} \text{ foot-pounds}\text{.}$
$W = \int_1^{1.5} 15x \, dx = 9.375 \text{ foot-pounds}\text{.}$

6.4.2 Work: Pumping Liquid from a Tank

Activity 6.4.3 ..

\begin{equation*} W = \int_{2}^{3} 9.81 \cdot 4000\pi \cdot x \, dx = 308~190 \, \text{newton-meters}\text{.} \end{equation*}
\begin{equation*} W = \int_{3}^{8} 62.4 \pi (100-x^2)(x+5) \, dx \approx 673593 \, \text{foot-pounds}\text{.} \end{equation*}
\begin{equation*} W = \int_{1}^{3} 62.4 (50 - \frac{25}{2}x) x \, dx = 5720 \, \text{foot-pounds}\text{.} \end{equation*}

6.4.3 Force due to Hydrostatic Pressure

Activity 6.4.4 ..

$F = \int_{x = 0}^{x=50} (6240 x) dx = 7~800~000 \text{ pounds } \text{.}$
$F = \int_{x=10}^{x=30} 124.8 (x - 10)\sqrt{900 - x^2} dx = 800~244 \text{ pounds } \text{.}$
$F = \int_{x=1}^{x=4} 62.4 (x - 1)(5 - 1.25x) dx = 351 \text{ pounds } \text{.}$

6.5 Improper Integrals 6.5.1 Improper Integrals Involving Unbounded Intervals

Activity 6.5.2 ..

$\int_1^{10} \frac{1}{x} dx = \ln(10)$ $\int_1^{1000} \frac{1}{x} dx = \ln(1000)$ $\int_1^{100000} \frac{1}{x} dx = \ln(100000)$
$\int_1^b \frac{1}{x} dx = \ln(b)\text{.}$
$\displaystyle \lim_{b \to \infty} \int_1^b \frac{1}{x} dx = \lim_{b \to \infty} \ln(b) = \infty$
$\int_1^{10} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{10}}$ $\int_1^{1000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{1000}}$ $\int_1^{100000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{100000}}$
$\int_1^b \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{b}}\text{.}$
$\displaystyle \lim_{b \to \infty} \int_1^b \frac{1}{x^{3/2}} dx = \lim_{b \to \infty} \left( 2 - \frac{2}{\sqrt{b}} \right) = 2$
Both graphs have a vertical asymptote at $x = 0$ and for both graphs, the $x$ -axis is a horizontal asymptote. However, the graph of $y = \frac{1}{x^{3/2}}$ will ’’approach the $x$ -axis faster’’ than the graph of $y = \frac{1}{x}\text{.}$
The area bounded by the graph of $y = \frac{1}{x}\text{,}$ the $x$ -axis, and the vertical line $x = 1$ is infinite or unbounded. However, The area bounded by the graph of $y = \frac{1}{x^{3/2}}\text{,}$ the $x$ -axis, and the vertical line $x = 1$ is equal to 2.

6.5.2 Convergence and Divergence

Activity 6.5.3 ..

$\displaystyle \int_1^\infty \frac{1}{x^2} dx = 1 $
$\displaystyle \int_0^\infty e^{-x/4} dx = 4 $
$\displaystyle \int_2^\infty \frac{9}{(x+5)^{2/3}} dx = \infty $
$\displaystyle \int_4^\infty \frac{3}{(x+2)^{5/4}} dx = \frac{12}{6^{1/4}} $
$\displaystyle \int_0^\infty x e^{-x/4} dx = 16 $
If $0 \lt p \lt 1\text{,}$ $\int_1^\infty \frac{1}{x^p} dx$ diverges, while if $p \gt 1\text{,}$ the integral converges.

6.5.3 Improper Integrals Involving Unbounded Integrands

Activity 6.5.4 ..

$\displaystyle \int_0^1 \frac{1}{x^{1/3}}dx = \frac{3}{2} $
$\displaystyle \int_0^2 e^{-x} dx = 1 - e^{-2} $
$\displaystyle \int_1^4 \frac{1}{\sqrt{4-x}} dx = 2\sqrt{3} $
$\int_{-2}^2 \frac{1}{x^2} \, dx$ diverges.
$\displaystyle \int_0^{\pi/2} \tan(x) dx = \infty $
$\displaystyle \int_0^1 \frac{1}{\sqrt{1-x^2}} dx = \frac{\pi}{2} $

7 Differential Equations 7.1 An Introduction to Differential Equations 7.1.1 What is a differential equation?

Activity 7.1.2 ..

Let $P$ be the population $t$ the time in years; $\frac{dP}{dt} = 0.0125P\text{.}$
Let $m$ be the mass $t$ the time in days; $\frac{dm}{dt} = -0.056m\text{.}$
Let $B$ be the balance $t$ be time in years; $\frac{dB}{dt} = 0.04B - 1000\text{.}$
Let $t$ be time in minutes $H$ the temperature of the hot chocolate; $\frac{dH}{dt} = -0.1(H - 70)\text{.}$
Let $t$ be time in minutes and $H$ the temperature of the soda; \begin{equation*} \frac{dH}{dt} = 0.1(70 - H) = -0.1(H - 70)\text{.} \end{equation*}

7.1.2 Differential equations in the world around us

Activity 7.1.3 ..

For the skydiver: \begin{align*} \left. \frac{dv}{dt}\right|_{(v = 0.5)} \amp \approx 1.5 \amp \left. \frac{dv}{dt}\right|_{(v = 1)} \amp \approx 1.2 \amp \left. \frac{dv}{dt}\right|_{(v = 1.5 )} \amp \approx 0.9\\ \left. \frac{dv}{dt}\right|_{(v = 2)} \amp \approx 0.6 \amp \left. \frac{dv}{dt}\right|_{(v = 2.5)} \amp \approx 0.3 \end{align*}
$\frac{dv}{dt} = -0.6v + 1.8\text{.}$
The rate of change of velocity with respect to time is a linear function of velocity.
$0 \lt v \lt 3\text{.}$
$3 \lt v \lt 5\text{.}$
$v = 3\text{.}$

7.1.3 Solving a differential equation

Activity 7.1.4 ..

$v(t) = 1.5t - 0.25t^2$ is not a solution to the given DE.
$v(t) = 3 + 2e^{-0.5t}$ is a solution to the given DE.
$v(t) = 3$ is a solution to the given DE.
$v(t) = 3 + Ce^{-0.5t}$ is a solution to the given DE for any choice of $C\text{.}$

7.2 Qualitative behavior of solutions to DEs 7.2.1 Slope fields

Activity 7.2.2 ..

\begin{equation*} \frac{dy}{dt} = 2 \left( -\frac{1}{2} e^{-t/2} \right) = -e^{-t/2} \end{equation*} and \begin{equation*} -\frac{1}{2}( y - 4 ) = -\frac{1}{2} \left( 4 + 2e^{-t/2} \right) = -e^{-t/2} \end{equation*} In addition, $y(0) = 4 + 2e^0 = 6\text{.}$
A constant function.

7.2.2 Equilibrium solutions and stability

Activity 7.2.3 ..

$y = 0$ and $y = 4\text{.}$
$y = 4$ is stable; $y = 0$ is unstable.
Figure 7.2.12 is for an ustable equilibrium; Figure 7.2.13 is for a stable equilibrium.

7.3 Euler’s method 7.3.1 Euler’s Method

Activity 7.3.2 ..

$y = t^2 - t\text{,}$ with errors $e_1 = 0.04\text{,}$ $e_2 = 0.08\text{,}$ $e_3 = 0.12\text{,}$ $e_4 = 0.16\text{,}$ $e_5 = 0.2\text{.}$
If we first think about how $y_1$ is generated for the initial value problem $\frac{dy}{dt} = f(t) = 2t-1, \ y(0) = 0\text{,}$ we see that $y_1 = y_0 + \Delta t \cdot f(t_0)\text{.}$ Since $y_0 = 0\text{,}$ we have $y_1 = \Delta t \cdot f(t_0)\text{.}$ From there, we know that $y_2$ is given by $y_2 = y_1 + \Delta t f(t_1)\text{.}$ Substituting our earlier result for $y_1\text{,}$ we see that $y_2 = \Delta t \cdot f(t_0) + \Delta t f(t_1)\text{.}$ Continuing this process up to $y_5\text{,}$ we get \begin{equation*} y_5 = \Delta t \cdot f(t_0) + \Delta t f(t_1) + \Delta t f(t_2) + \Delta t f(t_3) + \Delta t f(t_4) \end{equation*} This is precisely the left Riemann sum with five subintervals for the definite integral $\int_0^1 (2t-1)~dt\text{.}$
Solutions to this differential equation all differ by only a constant.

Activity 7.3.3 .

$y = 0$ or $y = 6\text{;}$ $y = 0$ is unstable, $y = 6$ is stable.
The solution will tend to $y = 6\text{.}$
The value of $y_i = 6$ for every value of $i\text{.}$

7.4 Separable differential equations 7.4.1 Solving separable differential equations

Activity 7.4.2 ..

$\displaystyle \frac{dP}{dt} = 0.03 P$
$P = Ce^{0.03t}\text{.}$
$P = 10000 e^{0.03t}\text{.}$
The doubling time is $t = \frac{\ln(2)}{0.03} \approx 23.105$ years.
The doubling time is $t = \frac{1}{k} \ln(2)\text{.}$

Activity 7.4.3 .

$\displaystyle k = \frac{1}{30}$
$\displaystyle T = 75 + Ce^{-t/30}$
The temperature of the coffee tends to 75 degrees.
$T(20) = 75 + 30e^{-2/3} \approx 90.4^\circ$ F.
$t = -30 \ln \left( \frac{1}{6} \right) \approx 53.75$ minutes.

Activity 7.4.4 .

$y = -1 + C e^{\left(2t - \frac{t^2}{2} \right)}\text{.}$
$y = \frac{1}{2} \ln \left( e^{t^2} + C \right)\text{.}$
$y = -1 + 3 e^{2t}\text{.}$
$y = -\frac{1}{2t + \frac{3}{2}} = -\frac{2}{4t + 3}\text{.}$
$y = \frac{4}{t^2 + 1}\text{.}$

7.5 Modeling with differential equations 7.5.1 Developing a differential equation

Activity 7.5.2 ..

$\frac{dA}{dt} = 0.05A\text{.}$
$\frac{dA}{dt} = 0.05A - 10000\text{.}$
$t = 20 \ln(2) \approx 13.86$ years.
At least $200000.
Up to $15000 every year.

Activity 7.5.3 .

$\frac{dM}{dt} = -kM\text{,}$ where $k$ is a positive constant.
$k = -\frac{1}{2} \ln \left( \frac{1}{2} \right) \approx 0.34657\text{.}$
$\frac{dM}{dt} = 3 - kM\text{,}$ where $k$ is a positive constant.
The equilibrium solution $mM = \frac{3}{k}$ is stable.
$M = \frac{3}{k} \left( 1 - e^{-kt} \right)\text{.}$
About 2.426 milligrams per hour.

7.6 Population Growth and the Logistic Equation 7.6.1 The earth’s population

Activity 7.6.2 ..

$P'(0) \approx 0.0755\text{.}$
$P(0) = 6.084\text{.}$
$k \approx 0.012041\text{.}$
$P(t) = 6.084 e^{0.012041t}\text{.}$
$P(10) \approx 6.8878\text{.}$
$t = \frac{1}{0.012041} \ln \left( \frac{12}{6.084}\right) \approx 56.41\text{,}$ or in the year 2056.
$P(500) \approx 3012.3$ billion.

7.6.2 Solving the logistic differential equation

Activity 7.6.3 ..

When $P = \frac{N}{2}\text{.}$
When the population is 6.125 billion.
$P = \frac{12.5}{1.0546e^{-0.025t} + 1} \text{;}$ $P(100) = 11.504$ billion.
$t = \frac{1}{-0.025} \ln \left( \frac{\left( \frac{12.5}{9} - 1 \right)}{1.0546}\right) \approx 39.9049$ (so in about year $2040$ ).
$\lim_{t \to \infty} P(t) = N\text{.}$

8 Sequences and Series 8.1 Sequences 8.1.1 Sequences

Activity 8.1.2 ., activity 8.1.4 ..

The sequence $\left\{\frac{1+2n}{3n-2}\right\}$ converges to $\frac{2}{3}\text{.}$
The sequence $\left\{\frac{5+3^n}{10+2^n}\right\}$ diverges to infinity.
$\frac{10^n}{n!} \to 0$ as $n \to \infty\text{.}$

8.2 Geometric Series 8.2.1 Geometric Series

Activity 8.2.2 ..

$rS_n = ar+ar^2+ar^3 + \cdots + ar^n \text{.}$
$S_n - rS_n = a - ar^n\text{.}$
$S_n = a\frac{1-r^n}{1-r}\text{.}$

Activity 8.2.3 .

Observe that \begin{equation*} S = \lim_{n \to \infty} S_n\text{.} \end{equation*}
If $r \gt 1\text{,}$ then $\lim_{n \to \infty} r^n = \infty\text{.}$ If $0 \lt r \lt 1\text{,}$ then $\lim_{n \to \infty} r^n = 0\text{.}$
Since \begin{equation*} S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1-r^n}{1-r} \end{equation*} and \begin{equation*} \lim_{n \to \infty} r^n = 0 \end{equation*} for $0 \lt r \lt 1\text{,}$ we conclude that \begin{equation*} S = \lim_{n \to \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r} \end{equation*} when $0 \lt r \lt 1\text{.}$

Activity 8.2.4 .

$(2)\left(\frac{1}{3}\right) \left[1 + \left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)^2 + \cdots \right] \text{.}$
$ar^3+ar^4+ar^5 + \cdots = \frac{ar^3}{1-r}\text{.}$
$r^n+ar^{n+1}+ar^{n+2} + \cdots = \frac{ar^n}{1-r}\text{.}$

8.3 Series of Real Numbers 8.3.1 Infinite Series

Activity 8.3.2 ..

See the table in part c .
$\displaystyle \displaystyle\sum_{k=1}^{1} \frac{1}{k^2}=1$
$\displaystyle \displaystyle\sum_{k=1}^{2} \frac{1}{k^2}=1.25$
$\displaystyle \displaystyle\sum_{k=1}^{3} \frac{1}{k^2}=1.361111111$
$\displaystyle \displaystyle\sum_{k=1}^{4} \frac{1}{k^2}=1.423611111$
$\displaystyle \displaystyle\sum_{k=1}^{5} \frac{1}{k^2}=1.463611111$
$\displaystyle \displaystyle\sum_{k=1}^{6} \frac{1}{k^2}=1.491388889$
$\displaystyle \displaystyle\sum_{k=1}^{7} \frac{1}{k^2}=1.511797052$
$\displaystyle \displaystyle\sum_{k=1}^{8} \frac{1}{k^2}=1.527422052$
$\displaystyle \displaystyle\sum_{k=1}^{9} \frac{1}{k^2}=1.539767731$
$\displaystyle \displaystyle\sum_{k=1}^{10} \frac{1}{k^2} =1.549767731 \phantom{1.54976773}$
It appears $\{S_n\}$ converges to something a bit larger than 1.5.

8.3.2 The Divergence Test

Activity 8.3.3 ..

$S_n = \sum_{k=1}^n a_k \text{.}$
$S_{n-1} = \sum_{k=1}^{n-1} a_k \text{.}$
$\sum_{k=1}^{n} a_k - \sum_{k=1}^{n-1} a_k = a_{n} \text{.}$
$\lim_{n \to \infty} S_n = L$ and $\lim_{n \to \infty} S_{n-1} = L\text{.}$
We have $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(S_n - S_{n-1}\right) = 0 \text{.}$

Activity 8.3.4 .

$\sum \frac{k}{k+1}$ diverges.
$\sum (-1)^k$ diverges.
The Divergence Test does not apply.

8.3.3 The Integral Test

Activity 8.3.5 ..

The $n$ th partial sum of the series $\sum_{k=1}^{\infty} \frac{1}{k}$ is the left hand Riemann sum of $f(x)$ on the interval $[1,n]\text{.}$
$\sum_{k=1}^{n} \frac{1}{k} \gt \int_{1}^{n} \frac{1}{x} \ dx \text{.}$
$\sum_{k=1}^{\infty} \frac{1}{k} \gt \int_{1}^{\infty} \frac{1}{x} \ dx \text{.}$
$\int_{1}^{\infty} f(x) \ dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x} \ dx \infty $ so the series $\sum_{k=1}^{\infty} \frac{1}{k}$ diverges.

Activity 8.3.6 .

$\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges.
$\sum_{k=1}^{\infty} \frac{1}{k^p}$ converges when $p \gt 1\text{.}$
$\sum_{k=1}^{\infty} \frac{1}{k^p}$ diverges when $p \lt 1\text{.}$

The $p$ -series $\sum_{k=1}^{\infty} \frac{1}{k^p}$ converges if and only if $p \gt 1\text{.}$

8.3.4 The Limit Comparison Test

Activity 8.3.7 ..

$\frac{k+1}{k^3+2}$ looks like $\frac{k}{k^3}$ when $k$ is large.
$\lim_{k \to \infty} \frac{a_k}{b_k} = 1$ so $a_k \approx b_k$ for large values of $k\text{.}$
$\sum \frac{k+1}{k^3+2}$ converges.

Activity 8.3.8 .

8.3.5 the ratio test, activity 8.3.9 ..

$\frac{a_{k+1}}{a_k} \approx \frac{2}{3}$ when $k$ is large.
$\sum \frac{2^k}{3^k-k}$ converges.

Activity 8.3.10 .

$\sum \frac{n}{2^n}$ converges.
$\sum \frac{k^3+2}{k^2+1}$ diverges.
$\sum \frac{10^k}{k!}$ converges.
$\sum \frac{k^3-2k^2+1}{k^6+4}$ converges.

8.4 Alternating Series 8.4.1 The Alternating Series Test

Activity 8.4.2 ..

There appears to be a limit for the sequence of partial sums.

Activity 8.4.3 .

$\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+2}$ converges.
$\sum_{k=1}^{\infty} \frac{2(-1)^{k+1}k}{k+5}$ diverges.
$\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln(k)}$ converges.

8.4.2 Estimating Alternating Sums

Activity 8.4.4 ., 8.4.3 absolute and conditional convergence, activity 8.4.5 ..

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \lt \sum_{k=1}^{\infty} \frac{1}{k^2} \text{.}$
$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \gt \sum_{k=1}^{\infty} -\frac{1}{k^2} \text{.}$
We expect this series to converge to some finite number between $\sum_{k=1}^{\infty} -\frac{1}{k^2}$ and $\sum_{k=1}^{\infty} \frac{1}{k^2}\text{.}$

Activity 8.4.6 .

$\sum (-1)^k \frac{\ln(k)}{k}$ converges.
$\sum (-1)^k \frac{\ln(k)}{k}$ converges conditionally.
$\sum (-1)^k \frac{\ln(k)}{k^2}$ converges.
$\sum (-1)^k \frac{\ln(k)}{k^2}$ converges absolutely.

8.4.4 Summary of Tests for Convergence of Series

Activity 8.4.7 ..

$\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k-2}}$ diverges.
$\sum_{k=1}^{\infty} \ \frac{k}{1+2k}$ diverges.
$\sum_{k=0}^{\infty} \ \frac{2k^2+1}{k^3+k+1}$ diverges.
$\sum_{k=0}^{\infty} \ \frac{100^k}{k!}$ converges.
$\sum_{k=1}^{\infty} \ \frac{2^k}{5^k}$ is a geometric series with ratio $\frac{2}{5}$ and sum $\frac{2}{3}\text{.}$
$\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1}$ converges.
$\sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k}$ is a convergent geometric series with $a = \frac{3}{49}$ and $r = \frac{3}{7}$ and sum $\frac{3}{28}\text{.}$
$\sum_{k=2}^{\infty} \ \frac{1}{k^k}$ converges.
$\sum_{k=1}^{\infty} \ \frac{(-1)^{k+1}}{\sqrt{k+1}}$ converges.
$\sum_{k=2}^{\infty} \ \frac{1}{k \ln(k)}$ diverges.
$\sum_{k=2}^{\infty} \frac{(-1)^k}{\ln(k)}$ converges very slowly.

8.5 Taylor Polynomials and Taylor Series 8.5.1 Taylor Polynomials

Activity 8.5.2 ..

$f^{(k)}(0) = k! \text{.}$
\begin{equation*} P_n(x) = \sum_{k=0}^n x^k\text{.} \end{equation*}
$f^{(k)}(0) = 0 \text{ if } k \text{ is odd, and } f^{(2k)}(0) = (-1)^k \text{.}$
$P_n(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots + (-1)^{n/2}\frac{x^n}{n!}$ if $n$ is even and $P_n(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots + (-1)^{(n-1)/2}\frac{x^(n-1)}{(n-1)!}$ if $n$ is odd.
$f^{(k)}(0) = 0 \text{ if } k \text{ is even } \ \ \ \text{ and } \ \ \ f^{(2k+1)}(0) = (-1)^k \text{.}$
$P_n(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots + (-1)^{(n-1)/2}\frac{x^n}{n!}$ if $n$ is odd and $P_n(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots + (-1)^{n/2+1}\frac{x^{n-1}}{(n-1)!}$ if $n$ is even.

8.5.2 Taylor Series

Activity 8.5.3 ..

$\displaystyle P(x) = 1 + x + x^2 + x^3 + \cdots + x^n + \cdots$
$P(x) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \cdots + (-1)^{n}\frac{1}{(2n)!}x^{2n} + \cdots \text{.}$
$P(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots + (-1)^{n}\frac{1}{(2n+1)!}x^{2n+1} + \cdots \text{.}$
$\displaystyle P_n(x) = 1 + x + x^2 + x^3 + \cdots + x^n$

Activity 8.5.4 .

It appears that as we increase the order of the Taylor polynomials, they fit the graph of $f$ better and better over larger intervals.
The Taylor polynomials converge to $\frac{1}{1-x}$ only on the interval $(-1,1)\text{.}$

8.5.3 The Interval of Convergence of a Taylor Series

Activity 8.5.5 ..

The interval $(-1,1)\text{.}$
$(-\infty, \infty)\text{.}$

8.5.4 Error Approximations for Taylor Polynomials

Activity 8.5.6 ., activity 8.5.7 ..

Compare Example 8.5.6 .
Use the fact that that $|f^{(n)}(x)| \le e^c$ on the interval $[0,c]$ for any fixed positive value of $c\text{.}$
Repeat the argument in (a) but replace $e^c$ with $1\text{,}$ and everything else holds in the same way.
Combine the results of (a) and (b)
$n = 28\text{.}$

8.6 Power Series 8.6.1 Power Series

Activity 8.6.2 ..

$[0,2)\text{.}$
$(-1,1)\text{.}$
$(-5,3)\text{.}$
$\{0\}\text{.}$

8.6.2 Manipulating Power Series

Activity 8.6.3 ..

$\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k$ for $-1 \lt x \lt 1\text{.}$
$f(x) = g(-x^2) = \sum_{k=0}^{\infty} (-1)^k x^{2k}\text{.}$
$-1 \lt x \lt 1\text{.}$

Activity 8.6.4 .

$f'(x) = - x + \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots + (-1)^k \frac{x^{k-1}}{(k-1)!} + \cdots \text{.}$
$\frac{d}{dx} \cos(x) = -\sin(x) \text{.}$
$f''(x) = - 1 + \frac{x^2}{2!} - \frac{x^4}{4!} + \cdots \text{.}$

Activity 8.6.5 .

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Appendix B Answers to Activities

1.1.2 Instantaneous Velocity

Activity 1.1.4 .

1.2.2 Instantaneous Velocity

Activity 1.2.4 .

1.3 The derivative of a function at a point 1.3.1 The Derivative of a Function at a Point

Activity 1.3.3 .

Activity 1.3.4 .

1.4 The derivative function 1.4.1 How the derivative is itself a function

1.5 Interpreting, estimating, and using the derivative 1.5.2 Toward more accurate derivative estimates

Activity 1.5.3 .

Activity 1.5.4 .

1.6 The second derivative 1.6.3 Concavity

Activity 1.6.3 .

Activity 1.6.4 .

1.7.2 Being continuous at a point

1.7.3 Being differentiable at a point

1.8 The Tangent Line Approximation 1.8.2 The local linearization

Activity 1.8.3 .

2 Computing Derivatives 2.1 Elementary derivative rules 2.1.2 Constant, Power, and Exponential Functions

2.1.3 Constant Multiples and Sums of Functions

Activity 2.1.4 .

2.2 The sine and cosine functions 2.2.1 The sine and cosine functions

Activity 2.2.3 .

Activity 2.2.4 .

2.3 The product and quotient rules 2.3.1 The product rule

2.3.2 The quotient rule

2.3.3 Combining rules

2.4 Derivatives of other trigonometric functions 2.4.1 Derivatives of the cotangent, secant, and cosecant functions

Activity 2.4.3 .

Activity 2.4.4 .

2.5 The chain rule 2.5.1 The chain rule

2.5.2 Using multiple rules simultaneously

Activity 2.5.4 .

2.6 Derivatives of Inverse Functions 2.6.2 The derivative of the natural logarithm function

2.6.3 Inverse trigonometric functions and their derivatives

Activity 2.6.4 .

2.7 Derivatives of Functions Given Implicitly 2.7.1 Implicit Differentiation

Activity 2.7.3 .

Activity 2.7.4 .

2.8 Using Derivatives to Evaluate Limits 2.8.1 Using derivatives to evaluate indeterminate limits of the form \(\frac{0}{0}\text{.}\)

Activity 2.8.3 .

2.8.2 Limits involving \(\infty\)

3 Using Derivatives 3.1 Using derivatives to identify extreme values 3.1.1 Critical numbers and the first derivative test

3.1.2 The second derivative test

Activity 3.1.4 .

3.2 Using derivatives to describe families of functions 3.2.1 Describing families of functions in terms of parameters

Activity 3.2.3 .

Activity 3.2.4 .

3.3 Global Optimization 3.3.1 Global Optimization

Activity 3.3.3 .

3.3.2 Moving toward applications

3.4 Applied Optimization 3.4.1 More applied optimization problems

Activity 3.4.3 .

Activity 3.5.3 .

Activity 3.5.4 .

Activity 3.5.5 .

4.1.2 Two approaches: area and antidifferentiation

4.1.3 When velocity is negative

4.2 Riemann Sums 4.2.1 Sigma Notation

4.2.2 Riemann Sums

4.2.3 When the function is sometimes negative

4.3 The Definite Integral 4.3.1 The definition of the definite integral

4.3.2 Some properties of the definite integral