chapter 1: functions - cengagecollege.cengage.com/.../3e/students/ssm/ch01_1.pdf · 2008-06-27 ·...

Copyright Houghton Mifflin Company. All rights reserved. 1

Chapter 1: Functions EXERCISES 1.1 1. {x|0 ≤ x < 6}

0 6 3. {x|x ≤ 2}

2 5. a. Since ∆x = 3 and m = 5, then ∆y, the

change in y, is ∆y = 3 · m = 3 · 5 = 15 b. Since ∆x = –4 and m = 5, then ∆y, the

change in y, is ∆y = –4 · m = –4 · 5 = –20

7. For (2, 3) and (4, –1), the slope is −1− 34 − 2 = −4

2 = −2

9. For (–4, 0) and (2, 2), the slope is

2 − 02 − (−4) = 2

2 + 4 = 26 = 1

3 11. For (0, –1) and (4, –1), the slope is

−1− (−1)4 − 0 = −1+ 1

4 = 04 = 0

13. For (2, –1) and (2, 5), the slope is

5 − (−1)2 − 2 = 5 +1

0 undefined 15. Since y = 3x – 4 is in slope-intercept form,

m = 3 and the y-intercept is (0, –4). Using the slope m = 3, we see that the point 1 unit over and 3 units up is also on the line

y

x–1 2 3

1

–4

34

1

2

5

17. Since y = − 1

2 x is in slope-intercept form, m = − 1

2 and the y-intercept is (0, 0). Using m = − 1

2 , we see that the point 1 unit over and 12 unit down is also on the line.

2

–2–4

–2

2 4

y

x

19. The equation y = 4 is the equation of the horizontal line through the points with y-coordinate 4. Thus, m = 0 and the y-intercept is (0, 4).

2

–1

1

2 3

y

x4

4

3

5

1

2 Chapter 1: Functions

Copyright Houghton Mifflin Company. All rights reserved.

21. The equation x = 4 is the equation of the vertical line through the points with x-coordinate 4. Thus, m is not defined and there is no y-intercept.

2

–2–4–2

2 4

y

x

4

–4

23. First, solve for y: 2x − 3y = 12

−3y = −2x + 12y = 2

3 x− 4

Therefore, m = 23 and the y-intercept is (0, –4).

2

–4

–22 4

y

x6

4

25. First, solve for y:

x + y = 0y = − x

Therefore, m = –1 and the y-intercept is (0, 0).

2

–2–4–2

2 4

y

x

4

–4

27. First, solve for y: x − y = 0

−y = − xy = x

Therefore, m = 1 and the y-intercept is (0, 0).

2

–2–4–2

2 4

y

x

4

–4

29. First, put the equation in slope-intercept form:

y = x+ 23

y = 13 x + 2

3

Therefore, m = 13 and the y-intercept is 0, 2

3( ).

2

1

2

y

x4

3

6

31. First, solve for y: 2 x3 − y = 1

−y = − 23 x + 1

y = 23 x − 1

Therefore, m = 23 and the y-intercept is (0, –1).

2

–1–1

–2

2 4

y

x5

1

1 3

33.

on [–10, 10] by [–10, 10]

35.

on [–10, 10] by [–10, 10]

Exercises 1.1


3

37.

on [–75, 75] by [–75, 75]

39. y = –2.25x + 3

41. y − −2( ) = 5 x − −1( )[ ]

y + 2 = 5x + 5y = 5x + 3

43. y = –4

45. x = 1.5 47. First, find the slope.

m = −1− 37− 5 = −4

2 = −2 Then use the point-slope formula with this slope and the point (5, 3). y − 3 = −2 x − 5( )y− 3 = −2x + 10y = −2x + 13

49. First, find the slope.

m = −1− −1( )5−1 = −1+1

4 = 0 Then use the point-slope formula with this slope and the point (1,–1). y − −1( ) = 0 x − 1( )y+ 1 = 0y = −1

51. The y-intercept of the line is (0, 1), and ∆y = –2 for ∆x = 1. Thus, m = ∆y

∆x = −21 = −2 . Now,

use the slope-intercept form of the line: y = –2x + 1.

53. The y-intercept is (0, –2), and ∆y = 3 for

∆x = 2. Thus, m = ∆y∆x = 3

2 . Now, use the slope-intercept form of the line: y = 3

2 x − 2

55. First, consider the line through the points (0, 5) and (5, 0). The slope of this line is m = 0−5

5−0 = −55 = −1. Since (0, 5)

is the y-intercept of this line, use the slope-intercept form of the line: y = –1 · x + 5 or y = –x + 5. Now consider the line through the points (5, 0) and (0, –5). The slope of this line is m = −5−0

0−5 = −5−5 = 1. Since (0,–5)

is the y-intercept of the line, use the slope-intercept form of the line: y = 1 · x – 5 or y = x – 5. Next, consider the line through the points (0, –5) and (–5, 0). The slope of this line is m = 0− −5( )

−5−0 = 5−5 = −1. Since

(0, –5) is the y-intercept, use the slope-intercept form of the line: y = –1 · x – 5 or y = –x – 5. Finally, consider the line through the points (–5, 0) and (0, 5). The slope of this line is m = 5−0

0− −5( ) = 55 = 1. Since (0,

5) is the y-intercept, use the slope-intercept form of the line: y = 1 · x + 5 or y = x + 5. 57. If the point x1 , y1( ) is the y-intercept (0, b),

then substituting into the point-slope form of the line gives y − y1( )=m x− x1( )y− b( ) =m x− 0( )y −b =mx

y =mx+ b

59. To find the x-intercept, substitute y = 0 into the equation and solve for x:

y =mx+ b0 =mx+ b

−mx = b

x = b−m

If m ≠ 0, then a single x-intercept exists.



61. a.

on [–5, 5] by [–5, 5] b.

on [–5, 5] by [–5, 5]

63. Low demand: [0, 8); average demand: [8, 20); high demand: [20, 40); critical demand: [40, ∞)

65. a. The value of x corresponding to the year 2010 is x = 2010 – 1900 = 110. Substituting x = 110 into the

equation for the regression line gives

y = −0.356x + 257.44y = −0.356(110) + 257.44 = 218.28 second

Since 3 minutes = 180 seconds, 218.28 = 3 minutes 38.28 seconds. Thus the world record in the year 2010 will be 3 minutes 38.28 seconds.

b. To find the year when the record will be 3 minutes 30 seconds, first convert 3 minutes 30 seconds to seconds: 3 minutes 30 seconds = 3 minutes · 60 sec

min + 30 seconds = 210 seconds. Now substitute 210 seconds into the equation for the regression line and solve for x.

y = −0.356x + 257.44210= −0.356x + 257.44

0.356x = 257.44 − 210x = 47.44

0.356 ≈ 133.26

Since x represents the number of years after 1900, the year corresponding to this value of x is 1900 + 133.26 = 2033.26 ≈ 2033. The world record will be 3 minutes 30 seconds in 2033.

67. a. To find the linear equation, first find the

slope of the line containing these points. m = 14−6

3−1 = 82 = 4

Next, use the point-slope form with the point (1, 6): y − y1 = m x − x1( )y− 6 = 4 x −1( )y− 6 = 4x − 4y = 4x + 2

69. a. First, find the slope of the line containing the points. m = 212−32

100− 0 = 180100 = 9

5 Next, use the point-slope form with the point (0, 32): y − y1 = m x − x1( )y− 32= 9

5 x − 0( )y = 9

5 x + 32

b. c.

To find the profit at the end of 2 years, substitute 2 into the equation y = 4x + 2. y = 4(2) + 2 = $10 million The profit at the end of 5 years is y = 4(5) + 2 = $22 million.

b. Substitute 20 into the equation. y = 9

5 x+ 32

y = 95 20( ) + 32 = 36+ 32 = 68°F

Exercises 1.1


5

a.

Price = $50,000; useful lifetime = 20 years; scrap value =

$6000V = 50,000− 50, 000−6000

20( )t 0 ≤ t ≤ 20

= 50,000− 2200t 0 ≤ t ≤ 20

71.

b. Substitute t = 5 into the equation. V = 50,000− 2200t

= 50,000− 22005( )= 50,000− 11,000= $39,000

73. y = Rx

1+ R−1K

x

= RKxK + R− 1( )x

.

RK = K + (R− 1)K.x < K⇒RK > K + (R− 1)x⇒

RKK + R−1( )x

> 1.

y = RKK + R− 1( )x

x

so y > x.

Similarly,

y = RKxK + R− 1( )x

.

Rx= x + (R− 1)x.K > x⇒Rx< K + (R −1)x⇒

RxK + R−1( )x

< 1.

y = RxK + R− 1( )x

K

so y < K.Thusx < y < K.

c.

on [0, 20] by [0, 50,000]

This means that if the parent population is less thanthe carrying capacity, each subsequent generation will be greater and approach the carrying capacity without meeting or exceeding it.

75. a.

on [0, 30] by [20, 30]

77. a.

on [10, 16] by [0, 100]

b. The value 2010 – 1980 = 30 corresponds to the year 2010, so use x = 30. From the graph in part (a), when x = 30, y = 28.7 on line y1. So the median age at first marriage formen in 2010 is 28.7 years.

When x = 30, y = 27.2 on line y2 . So the median age at first marriage forwomen in 2010 is 27.2 years.

b. From the graph in (a), when x = 12, y = 39.7. So the probability that a high school graduate smoker will quit is 39.7%.



c. The value 2020 – 1980 = 40 corresponds to the year 2020. Substitute 30 into each equation. y1 = 24.8 + 0.13x = 24.8 + 0.13(40) = 30 yearsfor men

y2 = 22.1 + 0.17x = 22.1 + 0.17(40) = 28.9 yearsforwomen

c.

on [10, 16] by [0, 100] When x = 16, y = 60.4. So the probability that a college graduate smoker will quit is 60.4%.

79. a.

b.

on [0, 50] by [60, 90]

on [0, 50] by [60, 90] y1 = 0.203x + 65.92

c. The value x = 2025 – 1960 = 65 corresponds to the year 2025. Substitute 65 into the equation. y = 0.203x + 65.92y = 0.203(65) +65.92

≈ 79.1

EXERCISES 1.2

1. 22 ⋅ 2( )2= 22 ⋅21( )2

= 23( )2= 26 = 64

3. 2 −4 = 124 = 1

16

5. 12( )−3

= 2−1( )−3= 23 = 8

7. 58( )−1

= 85

9. 4 −2 ⋅ 2−1 = 22( )−2 ⋅ 2−1

= 2−4 ⋅ 2−1 = 2−5 = 125 = 1

32

11. 32( )−3

= 23( )3

= 23

33 = 827

13. 13( )−2

− 12( )−3

= 31( )2

− 21( )3

= 32

12 − 23

13

= 9 − 8 = 1

15.

23( )−2[ ]−1

= 32( )2[ ]−1

= 32

22( )−1

= 94( )−1

= 49

17. 251 2 = 25 = 5

19. 253 2 = 25( )3

= 53 = 125

21. 16 3 4 = 164( )3= 23 = 8

23. −8( )2 3 = −83( )2

= −2( )2 = 4

Exercises 1.2


7

25. −8( )5 3 = −83( )5= −2( )5 = −32

27. 2536( )3 2

= 2536( )3

= 56( )3

= 53

63 = 125216

29. 27

125( )2 3= 27

1253( )2 = 3

5( )2= 32

52 = 925

31.

132( )2 5

= 132

5( )2

= 12( )2

= 14

33. 4 −1 2 = 1

41 2 = 14

= 12

35. 4 −3 2 = 1

4 3 2 = 14( )3 = 1

23 = 18

37. 8−2 3 = 1

82 3 = 183( )2

= 122 = 1

4

39. −8( )−13 = 1−8( )1 3 = 1

−83 = 1−2 = − 1

2

41. −8( )−2 3 = 1

−8( )2 3 = 1−8( )23

= 1643 = 1

4

43. 2516( )−1 2

= 1625( )1 2

= 1625( )= 4

5

45. 2516( )−3 2

= 1625( )3 2

= 1625( )3

= 45( )3

= 43

53 = 64125 47. − 1

27( )−5 3= −27( )5 3 = −273( )5

= −3( )5 = −243

49. 7 0.39 ≈ 2.14

51. 82.7 ≈ 274.37

53. −8( )7 3 = −128

55. 52( )−1[ ]−2

= 6.25

57. 4( )−1[ ]0.5= 0.5

59. 0.4 −7( )−1 7= 0.4

61. 0.1( )0.1[ ]0.1≈ 0.977

63. 1− 1100( )−1000

≈ 2.720

65. x3 ⋅ x2( )2= x5( )2

= x10

67. z 2 z ⋅z2( )2 z[ ]3= z2 z3( )2z[ ]3

= z2 ⋅ z6 ⋅ z( )3

= z9( )3= z27

69. x2( )2[ ]2

= x 4( )2= x

8

71.

ww2( )3

w 3w= w3w 6

w3w= w5

73. 5xy 4( )2

25x3y3 = 25x2y8

25x3y3 = y5

x

75.

9xy 3z( )2

3 xyz( )2 = 81x2y6z2

3x2y2 z2= 27y4

77. 2u2vw3( )2

4 uw2( )2 = 4u 4v2w6

4u2w 4 = u2v2w2

79. Average body thickness

= 0.4(hip-to-shoulder length)3/ 2

= 0.4(16)3 /2

= 0.4( 16)3 = 0.4(64)= 25.6 ft



81. ′ C = x 0.6C= 40.6C ≈ 2.3C

To quadruple the capacity costs about 2.3 times as much.

83.

on [0, 5] by [0, 3] Capacity can be multiplied by about 3.2

85. Heart rate( ) = 250 weight( )−1 4

= 25016( )−1 4

= 125 beats per minut

87.

on [0, 200] by [0, 150] Heart rate decreases more slowly as body weight increases.

89. Time to build the 50th Boeing 707( )

= 15050−0.322( )≈ 42.6 thousand work-hours

It took approximately 42,600 work-hours to build the 50th Boeing 707.

91. a. Increase in ground motion = 10B−A

= 108.3−6.8

= 101.5 ≈ 32

The 1906 San Francisco earthquake had about 32 times the ground motion of the 1994 Northridge, California, earthquake.

b. Increase in ground motion = 10B−A

= 108.1−7.2

= 100.9 ≈ 8

The 1933 Miyagi earthquake had about 8 times the grand motion of the Kobe (Japan) earthquake.

93. S = 60

11 x0.5

= 6011 3281( )0.5 ≈ 312 mph

95.

on [0,100] by[0,4] x ≈ 18.2. Therefore, the land area must be increased by a factor of more than 18 to double the number of species.

Exercises 1.3


9

97. a.

on [–2, 32] by [1000, 3500]

b. y1 = 3261x−0.267 c. For x = 50,

y1 = 3261x−0.267

= 326150( )−0.267 ≈ 1147 work-hours

It will take approximately 1147 work-hours to build the fiftieth supercomputer.

EXERCISES 1.3 1. Yes 3. No 5. No 7. No 9. Domain = {x | x ≤ 0 or x ≥ 1} 11. a. f x( ) = x − 1

f 10( ) = 10− 1 = 9 = 3

13. a. h z( ) = 1z+ 4

h −5( ) = 1−5+ 4 = −1

b. Domain = {x | x ≥ 1} since f x( ) = x− 1 is defined for all values of x ≥ 1.

b. Domain ={z | z ≠ –4} since h z( ) = 1z+4

is defined for all values of z except z = –4.

c. Range = {y | y ≥ 0} c. Range = {y | y ≠ 0} 15. a. h x( ) = x1 4

h 81( ) = 811 4 = 814 = 3

17. a. f x( ) = x2 3

f −8( ) = −8( )2 3 = −8( )23 = 643 = 4

b. Domain = {x | x ≥ 0} since h x( ) = x1 4 is defined for nonnegative values of x.

b. Domain = ℜ

c. Range = {y | y ≥ 0} c. Range = {y | y ≥ 0} 19. a. f x( ) = 4 − x 2

f 0( ) = 4 − 02 = 2

21. a. f x( ) = − x

f −25( ) = − −25( ) = 5

b. f x( ) = 4 − x 2 is defined for values of x such that 4 − x2 ≥ 0 . Thus, 4 − x2 ≥ 0

−x2 ≥ −4x2 ≤ 4

–2 ≤ x ≤ 2 Domain = {x | –2 ≤ x ≤ 2}

b.

c.

Domain = {x | x ≤ 0} since f x( ) = −x is defined only for values of x ≤ 0. Range = {y | y ≥ 0}

c. Range = {y | 0 ≤ y ≤ 2}



23.

2

–1

–4

–21 2

y

x

4

25.

1

–1

–2

–11 2

y

x

2

3

27.

x

y

2–2

–10

10

–4

29.

x

y

–2 –1

–10

10

1 2 3 4

31. a. x = −b2a = − −40( )

2 1( ) = 402 = 20

To find the y-coordinate, evaluate f at x = 20. f 20( ) = 20( )2 − 40 20( )+ 500 = 100

The vertex is (20, 100).

33. a. x = −b

2a = − −80( )2 −1( ) = 80

−2 = −40 To find the y-coordinate, evaluate f at x = –40. f −40( )= − −40( )2 − 80 −40( )− 1800= −200

The vertex is (–40, –200). b.

on [15, 25] by [100, 120]

b.

on [–45, –35] by [–220, –200] 35. x 2 − 6x− 7 = 0

x − 7( ) x +1( )= 0

Equals 0 Equals 0

at x = 7 at x = −1

x = 7, x = −1

37. x 2 + 2x = 15

x2 + 2x − 15= 0x + 5( ) x− 3( )= 0

Equals 0 Equals 0

at x = −5 at x = 3

x = −5, x = 3

39. 2x 2 + 40 = 18x

2x2 − 18x + 40 = 0x 2 − 9x + 20 = 0x − 4( ) x − 5( ) = 0

Equals 0 Equals 0

at x = 4 at x = 5

x = 4, x = 5

41. 5x2 − 50x = 0

x 2 −10x = 0x x − 10( ) = 0

Equals 0

at x = 10

x = 0, x = 10

43. 2x 2 − 50= 0

x 2 − 25= 0x − 5( ) x + 5( )= 0

Equals 0 Equals 0

at x = 5 at x = −5

x = 5, x = −5

45. 4x 2 + 24x + 40 = 44x2 + 24x + 36= 0

x2 + 6x + 9 = 0

x + 3( )2 = 0x = −3

Exercises 1.3


11

47. −4x2 + 12x = 8

−4x2 + 12x − 8 = 0x2 − 3x + 2 = 0x − 2( ) x − 1( ) = 0

Equals 0 Equals 0

at x = 2 at x = 1

x = 2, x = 1

49. 2x2 − 12x + 20 = 0x 2 − 6x+ 10 = 0

Use the quadratic formula with a = 1, b = –6, and c = 10.

x =− −6( ) ± −6( )2 − 4 1( ) 10( )

2 1( )

= 6 ± 36 − 402

= 6 ± −42 Not a real numbe

2x2 − 12x + 20 = 0

has no real solutions.

51. 3x 2 + 12 = 0

x2 + 4 = 0

x2 = −4x = ± −4 Not a real numbe

3x 2 +12 = 0

has no real solutions.

53.

on [–5, 6] by [–22, 6] x = –4, x = 5

55.

on [–1, 9] by [–10, 40]

x = 4, x = 5

57.

on [–7, 1] by [–2, 16]

x = –3 59.

on [–5, 3] by [–5, 30]

No real solutions

61.

on [–4, 3] by [–9, 15]

x = –2.64, x = 1.14 63.

on [–10, 10] by [–10, 10] a. Their slopes are all 2, but they have

different y-intercepts. b. The line 2 units below the line of the

equation y = 2x – 6 must have y-intercept –8. Thus, the equation of this line is y = 2x – 8.

65. Let x = the number of board feet of wood. Then C(x) = 4x + 20



67. Let x = the number of hours of overtime. Then P(x) = 15x + 500

69. a. p d( )= 0.45d+ 15p 6( )= 0.45 6( )+ 15

= 17.7 pounds per square in

b. p d( ) = 0.45d + 15p 35,000( ) = 0.45 35,000( )+ 15

= 15,765pounds per square in

71. D v( )= 0.055v2 + 1.1v

D 40( )= 0.05540( )2 + 1.1 40( ) = 132 ft

73. a. N t( ) = 200+ 50t 2

N 2( ) = 200+ 50 2( )2

= 400cells

b. N t( ) = 200+ 50t2

N 10( ) = 200+ 50 10( )2

= 5200cells 75. v x( )= 60

11 x

v 1454( )= 6011 1454 ≈ 208 mph

77.

on [0, 5] by [0, 50] The object hits the ground in about 2.92 seconds

79. a. The break-even points occur when

C(x) = R(x). 180x + 16,000= −2x2 + 660x

2x2 − 480x +16,000 = 0

x =480 ± (−480)2 − 4(2)(16,000)

2(2)

=480 ± 102,400

4=

480± 3204

= 8004

or 1604

= 200or 40

The company will break even when it makes either 40 devices or 200 devices.

81. a. The break-even points occur when C(x) = R(x). 100x + 3200= −2x2 + 300x2x2 − 200x + 3200= 0

x =200± (−200)2 − 4(2)(3200)

2(2)

=200± 14,400

4=

200 ±1204

= 3204

or 804

= 80or 20

The store will break even when it sells 20 machines or 80 machines.

Exercises 1.4


13

b. The profit function, P(x), is the revenue function minus the cost function. P(x) = −2x2 + 660x− (180x + 16,000)

= −2x2 + 480x −16,000

The profit is maximized at the vertex, x = − b

2a .

x = −480

2(−2)=

−480−4

= 120

The profit is maximized when 120 devices are made. The maximum profit is P(120) = −2(120)2 + 480(120) −16,000

= 12,800

The maximum profit is $12,800.

b. The profit function, P(x), is the revenue function minus the cost function. P(x) = −2x2 + 300x − (100x + 3200)

= −2x2 + 200x− 3200


2a .

x = −200

2(−2)=

−200−4

= 50

The profit is maximized when 50 machines are sold. The maximum profit is P(150) = −2(50)2 + 200(50) − 3200

= 1800

The maximum profit is $1800. 83. (w +a)(v+ b) = c

v+ b = cw + a

v =c

w + a−b

85. a.

on [0, 5] by [0, 4.5]

b.

on [0.5, 4.5] by [3.5, 4.5] The quadratic regression formula for this data is y = 0.125x2 − 0.555x+ 4.225.

c. y = 0.125x2 − 0.555x + 4.225

= 0.125 5( )2 − 0.5555( )+ 4.225≈ $4.6 million

EXERCISES 1.4 1. Domain = {x | x < –4 or x >0} Range = {y | y < –2 or y > 0} 3. a. f (x) = 1

x+ 4

f (23) = 123+ 4 = 1

27

5. a. f x( ) = x2

x−1

f −1( ) = −1( )2

−1−1 = − 12

b. Domain = {x | x ≠ –4} b. Domain = {x | x ≠ 1} c. Range = {y | y ≠ 0} c. Range = {y | y ≤ 0 or y ≥ 4} 7. a. f x( ) = 12

x x+ 4( )

f 2( ) = 122 2+ 4( ) = 1

9. a. g x( ) = 4x

g − 12( )= 4−1 2 = 1

41 2 = 12

b. Domain = {x | x ≠ 0, x ≠ –4} b. Domain = ℜ c. Range = {y | y ≤ –3 or y > 0} c. Range = {y | y > 0}



11.

x5 + 2x 4 − 3x3 = 0

x 3 x 2 + 2x − 3( )= 0

x3 x+ 3( ) x −1( ) = 0

Equals 0 Equals 0 Equals 0

at x = 0 atx = −3 at x = 1

x = 0, x = −3, x = 1

13.

5x3 − 20x = 0

5x x2 − 4( )= 0

5x x − 2( ) x + 2( ) = 0


at x = 0 atx = 2 at x = −2

x = 0, x = 2, x = −2

15. 2x3 +18x = 12x 2

2x3 − 12x 2 +18x = 02x x2 − 6x + 9( )= 0

2x x − 3( )2 = 0

Equals 0 Equals 0

at x = 0 atx = 3

x = 0, x = 3

17.

6x 5 = 30x 4

6x5 − 30x 4 = 06x4 x − 5( ) = 0

Equals 0 Equals 0

at x = 0 atx = 5

x = 0, x = 5

19. 3x 5 2 − 6x3 2 = 9x1 2

3x 5 2 − 6x3 2 − 9x1 2 = 03x12 x 2 − 2x − 3( )= 0

3x1 2 x− 3( ) x +1( )= 0


at x = 0 atx = 3 at x = −1

x = 0, x = 3, x = −1

21.

on [–5, 5] by [–20, 20] x = –2, x = 0, and x = 4

23.

on [–4, 5] by [–30, 5] x = –1, x = 0, and x = 3

25.

on [–2, 5] by [–50, 50] x = 0 and x = 3

27.

on [–6, 3] by [–100, 1600] x = –5 and x = 0

29.

on [0, 4] by [–5,25] x = 0 and x = 1

31.

on [–3, 3] by [–25, 10] x ≈ –1.79, x = 0, and x ≈ 2.79

Exercises 1.4


15

33.

1 2 3–1–2

3

6

9

x

y

35.

37.

4 62x

2

22

4

24

y

39. Polynomial 41. Exponential function 43. Polynomial 45. Rational function 47. Piecewise linear function 49. Polynomial 51. None of these

53. a. y4 = 3x , because it is the exponential function with the largest base.

b. y1 = 13( )x , because it is the exponential

function with the smallest base.

c.

d. (0, 1), because any nonzero number raised to the 0 is equal to 1.

55. a.

on [–5, 5] by [–5, 5]

57. a. f g x( )( )= g x( )[ ]5= 7x −1( )5

b. Domain = ℜ; range = the set of integers b. g f x( )( )= 7 f x( )[ ]− 1 = 7 x5( )− 1 = 7x 5 − 1

59. a. f g x( )( )= 1

g x( ) = 1x2 +1

61. a. f g x( )( )= g x( )[ ]3− g x( )[ ]2

= x −1( )3− x − 1( )2

b. g f x( )( )= f x( )[ ]2+1 = 1

x2 + 1 b. g f x( )( )= f x( ) − 1 = x 3 − x 2 − 1

y

x

–24

2

4



63. a. f g x( )( )=

g x( )[ ]3 −1

g x( )[ ]3 +1=

x2 − x( )3−1

x2 − x( )3 +1

65. a. f g x( )( )= a g x( )[ ]+b = a cx+ d( ) +b= acx+ad + b

b. g f x( )( )= f x( )[ ]2− f x( )

= x3 −1x3 +1( )2

− x3 −1x3 +1

b. Yes

67. f x + h( ) = 5 x +h( )2 = 5 x2 + 2hx+ h2( )= 5x 2 +10hx+ 5h2

69. f x + h( ) = 2 x + h( )2 − 5 x + h( )+ 1

= 2x2 + 4xh+ 2h2 − 5x − 5h+ 1

71. f x+ h( )− f x( )

h = 5 x+h( )2 − 5x2

h

= 5x2 +10hx+ 5h2 − 5x2

h

= 10x + 5h

73. f x+ h( )− f x( )

h =2 x+h( )2 −5 x+h( )+1− 2x2 − 5x+1( )

h

= 2x2 +4hx+2h 2 −5x− 5h+1−2x2 +5x−1h

= 4x + 2h− 5

75. f x+ h( )− f x( )

h =7 x+h( )2 −3 x +h( )+2 − 7x2 − 3x+2( )

h

= 7x2 +14hx+7h2 − 3x−3h+ 2− 7x2 +3x −2h

= 14x + 7h− 3

77. f x+ h( )− f x( )h = x+h( )3 − x3

h = x3 +3x2h+3xh2 +h3 − x3

h = 3x 2 + 3xh+ h2

79. f x + h( )− f x( )h =

2x+h

− 2x

h = 2h x + h( ) − 2

hx = 2xhx x + h( ) −

2 x + h( )hx x + h( ) = 2x − 2x − 2h

hx x + h( ) = − 2x x + h( )

81. f x+ h( )− f x( )

h =1

x+h( )2− 1x2

h = 1h x+h( )2 − 1

hx2 =x2 − x2 +2hx+h 2( )hx2 x+h( )2 = x2 − x2 − 2hx−h2

hx2 x+ h( )2 = −2x−hx2 x+h( )2

83. a. 2.70481

b. 2.71815| c. 2.71828 d. Yes, 2.71828

85. The graph of y = (x + 3)3 + 6 will be the graph of y = x3 shifted 3 units to the left and 6 units up.

on [10, 5] by [8, 12]

87. P x( )= 5221.0053( )x

P 50( ) = 5221.0053( )50 ≈ 680 million people in 1750

89. a. For x = 3000, use ƒ(x) = 0.10x. f x( ) = 0.10x

f 3000( ) = 0.10 3000( ) = $300

Exercises 1.4


17

b. For x = 5000, use ƒ(x) = 500 + 0.30(x – 5000).

f x( ) = 500+ 0.30 x − 5000( )f 5000( ) = 500+ 0.30 5000− 5000( )

= $500

c. For x = 10,000, use ƒ(x) = 500 + 0.30(x – 5000).

f x( ) = 500+ 0.30 x − 5000( )f 10,000( ) = 500+ 0.30 10,000− 5000( )

= 500+ 0.30 5000( )= $2000

d.

5000 10000

1000

2000

x

y

91. a. For x = 0.66,

use ƒ(x) = 10.5x. f x( )= 10.5x

f 0.66( )= 10.5 0.66( )= 6.93 years

For x = 1.33, use ƒ(x) = 10.5x.

f x( ) = 10.5xf 1.33( ) = 10.5 1.33( )

= 13.965 years For x = 10,

use ƒ(x) = 21 + 4(x – 2). f x( ) = 21+ 4 x − 2( )f 10( ) = 21+ 4 10− 2( )

= 21+ 4 8( )= 53 years

93. We must find the composition R(v(t)). R v t( )( )= 2 v t( )[ ]0.3

= 2 60+ 3t( )0.3

R(10) = 2(60+ 3⋅10)0.3 = 2(90)0.3

≈ $7.7 million



b.

10x

y

305070

95. a. f (10) = 410 = 1,048,576 cells 97. a.

on [21.6, 40] by [0, 2000]

b. f (15) = 415 = 1,073,741,824 cells No the mouse will not survive beyond day 15.

b. At about x = 27.9 mpg

REVIEW EXERCISES FOR CHAPTER 1 1. {x | 2 < x ≤ 5}

2 5 2. {x | –2 ≤ x < 0}

22 0 3. {x | x ≥ 100}

100 4. {x | x ≤ 6}

6 5. Hurricane: [74, ∞); storm: [55, 74);

gale: [38, 55); small craft warning: [21,38) 6. a.

b. c. d.

(0, ∞) (–∞, 0) [0, ∞) (–∞, 0]

7. y − −3( )= 2 x − 1( )

y + 3 = 2x − 2y = 2x − 5

8. y − 6 = −3 x − −1( )[ ]

y− 6 = −3x − 3y = −3x + 3

9. Since the vertical line passes through the

point with x-coordinate 2, the equation of the line is x = 2.

10. Since the horizontal line passes through the point with y-coordinate 3, the equation of the line is y = 3.

11. First, calculate the slope from the two points.

m = −3−32 − −1( ) = −6

3 = −2 Now use the point-slope formula with this slope and the point (–1, 3). y − 3 = −2 x − −1( )[ ]y− 3 = −2x − 2y = −2x + 1

12. First, calculate the slope from the points. m = 4− −2( )

3−1 = 62 = 3

Now, use the point-slope formula with this slope and the point (1, –2). y − −2( ) = 3 x − 1( )y + 2 = 3x − 3y = 3x − 5

Review Exercises for Chapter 1


19

13. Since the y-intercept is (0, –1), b = –1. To find the slope, use the slope formula with the points (0, –1) and (1, 1). m = 1− −1( )

1−0 = 2 Thus the equation of the line is y = 2x – 1.

14. Since the y-intercept is (0, 1), b = 1. To find the slope, use the slope formula with the points (0, 1) and (2, 0). m = 0−1

2 −0 = − 12

The equation of the line is y = − 12 x + 1

15. a. Use the straight-line depreciation formula

with price = 25,000, useful lifetime = 8, and scrap value = 1000. Value = price− price – scrap value

useful lifetime( )t= 25,000− 25,000−1000

8( )t= 25,000− 24, 000

8( )t= 25,000− 3000t

16. a. Use the straight-line depreciation formula with price = 78,000, useful lifetime = 15, and scrap value = 3000. Value = price− price – scrap value

useful lifetime( )t= 78,000− 78, 000−3000

15( )t= 78,000− 75, 000

15( )t= 78,000− 5000t

b. Value after 4 years = 25,000− 3000 4( )

= 25,000− 12,000= $13,000

b. Value after 8 years = 78, 000− 5000 8( )

= 78, 000− 40,000

= $38, 000

17. a.

b.

on [0,30] by [0,30]

on [0,30] by [0,30] The regression line y1 = –0.33x + 25.24 fits the data reasonably well.

18. a. b.

on [0,50] by [0,100]

on [0,50] by [0,100] The regression line y1 = 1.12x + 25.4 fits the data reasonably well.

c. For the year 2010, y = –0.33(35) + 25.24 = 13.69 million tons

c. For the year 2010, y = 1.12(50) + 25.4 = 81.4 to 1

19. 16( )−2

= 6−1( )−2= 62 = 36

20. 43( )−1

= 34

21. 641 2 = 64 = 8 22. 1000 13 = 10003 = 10

23. 81( )−3 4 = 1

813 4 = 1814( )3 = 1

33 = 127

24. 100 −3 2 = 1

1003 2 = 1100( )3 = 1

103 = 11000

25. f (11) = 11− 7 = 2

26. 9

16( )−32 = 16

9( )3

2 = 169( )3 = 4

3( )3= 64

27



27. 13.97

28. 112.32

29. a. f (11) = 11− 7 = 2

30. a. g(−1) = 1

−1+3 = 12

b. Domain = {x ≥ 7} because is defined only for all values of x ≥ 7.

b. Domain = {t | t ≠ –3}

c. Range = {y | y > 0} c. Range = {y | y ≠ 0}

31. a. h 16( ) = 16−3 4 = 116( )3 4

= 116

4( )3

= 12( )3

= 18

32. a. w 8( ) = 8−4 3 = 18( )4 3

= 18

3( )4

= 12( )4

= 116

b. Domain = {w | w > 0} because the fourth root is defined only for nonnegative numbers and division by 0 is not defined.

b. Domain = {z | z ≠ 0}

c. Range = {y | y > 0} c. Range = {y | y > 0} 33. Yes 34. No

35.

8

28

4222x

y

36.

22

2

21

4

6

54321

y

x

37.

2122 212324x

210

86

y

38.

21 321x

23

9

y

6

39. a.

3x 2 + 9x = 03x x + 3( ) = 0

Equals 0 Equals 0 at x = 0 at x = –3

x = 0 and x = –3

40. a. 2x2 − 8x −10 = 0

2 x 2 − 4x − 5( )= 0

x − 5( ) x + 1( ) = 0

Equals 0 Equals 0

at x = 5 at x = –1

x = 5 and x = –1 b. Use the quadratic formula with a = 3,

b = 9, and c = 0 −9± 92 − 4 3( ) 0( )

2 3( ) = −9± 816

= −9±96

= 0,−3

x = 0 and x = –3

b. Use the quadratic formula with a = 2, b = –8, and c = –10 − −8( )± −8( )2 −4 2( ) −10( )

2 2( ) = 8± 64+804

= 8±124

= 5,−1

x = 5 and x = –1



21

41. a.

3x2 + 3x + 5 = 113x2 + 3x − 6 = 03 x2 + x − 2( )= 0

x + 2( ) x − 1( ) = 0

Equals 0 Equals 0 at x = –2 at x = 1

x = –2 and x = 1

42. a. 4x 2 − 2 = 24x2 = 4x2 = 1x = ±1

x = 1 and x = –1

b. Use the quadratic formula with a = 3, b = 3, and c = –6 −3± 32 −4 3( ) −6( )

2 3( ) = −3± 9+ 726

= −3±96

= −2,1

x = –2 and x = 1

b. Use the quadratic formula with a = 4, b = 0, c = –4 −0 ± 02 −4 4( ) −4( )

2 4( ) = ± 648

= ±88

= ±1

x = 1 and x = –1 43. a. Use the vertex formula with a = 1 and

b = –10. x = −b

2a = − −10( )2 1( ) = 10

2 = 5 To find y, evaluate f(5). f 5( )= 52 − 10 5( )− 25 = −50

The vertex is (5, –50).

44. a. Use the vertex formula with a = 1 and b = 14. x = −b

2a = −142 1( ) = −7

To find y, evaluate f(–7). f −7( )= −7( )2 +14 −7( ) −15 = −64

The vertex is (–7, –64) b.

on [2, 8] by [250, 240]

b.

on [210, 24] by [264, 254] 45. Let x = number of miles per day.

C(x) = 0.12x + 45 46. Use the interest formula with P = 10,000 and

r = 0.08. I(t) = 10,000(0.08)t = 800t

47. Let x = the altitude in feet.

T x( )= 70 − x300

48. Let t = the number of years after 2000. C t( )= 0.58t + 27

30= 0.58t + 27t ≈ 5 years after 200

49. a. The break-even points occur when

C(x) = R(x). 80x + 1950= −2x2 + 240x

2x2 − 160x +1950= 0

x =160± (−160)2 − 4(2)(1950)

2(2)

=160± 10,000

4

= 160±1004

= 2604

or 604

= 65 or 15

The store will break even when it installs 15 receivers or 65 receivers.

50. a. The break-even points occur when C(x) = R(x). 220x + 202,500= −3x2 + 2020x

3x2 − 1800x + 202,500= 0

x =1800± (−1800)2 − 4(3)(202,500)

2(3)

=1800± 810,000

6=

1800± 9006

=2700

6 or

9006

= 450 or 150

The outlet will break even when it sells 150 air conditioners or 450 air conditioners.



b. The profit function, P(x), is the revenue

function minus the cost function. P(x) = −2x2 + 240x− (80x + 1950)

= −2x2 + 160x −1950


2a .

x = −160

2(−2)=

−160−4

= 40

The profit is maximized when 40 receivers are installed. The maximum profit is P(40) = −2(40)2 +160(40) −1950

= 1250

The maximum profit is $1250.

b. The profit function, P(x), is the revenue function minus the cost function. P(x) = −3x2 + 2020x − (220x + 202,500)

= −3x2 +1800x − 202,500


2a .

x = −18002(−3)

=−1800

−6= 300

The profit is maximized when 300 air conditioners are sold. The maximum profit is P(300) = −3(300)2 + 1800(300) − 202,500

= 67,500The maximum profit is $67,500.

51. a. f −1( ) = 3

−1( ) −1− 2( ) = 33 = 1 52. a. f −8( ) = 16

−8( ) −8+ 4( ) = 1632 = 1

2

b. Domain = {x | x ≠ 0, x ≠ 2} b. Domain = {x | x ≠ 0, x ≠ –4} c. Range = {y | y > 0 or y ≤ –3} c. Range = {y | y > 0 or y ≤ –4}

53. a. g32

= 93/ 2 = ( 9)3 = 27 54. a. g

53

= 85 /3 = ( 83 )5 = 25 = 32

b. Domain = ℜ b. Domain = ℜ c. Range = {y | y > 0} c. Range = {y | y > 0}

55. 5x4 + 10x3 = 15x2

5x4 + 10x 3 − 15x 2 = 05x2 x2 + 2x − 3( )= 0

x2 x + 3( ) x − 1( ) = 0

Equals 0 Equals 0 Equals 0at x = 0 at x = −3 at x = 1x = 0, x = −3, and x = 1

56. 4x 5 + 8x4 = 32x 3

4x 5 + 8x4 − 32x 3 = 04x3 x 2 + 2x − 8( )= 0

x3 x + 4( ) x − 2( ) = 0


57. 2x 5 2 − 8x 3 2 = 10x1 2

2x5 2 − 8x3 2 − 10x1 2 = 02x1 2 x 2 − 4x − 5( )= 0

x1 2 x − 5( ) x + 1( ) = 0

Equals 0 Equals 0 Equals 0at x = 0 at x = 5 at x = −1x = 0, x = 5, and x = −1

58. 3x 5 2 + 3x 3 2 = 18x1 2

3x 5 2 + 3x 3 2 −18x1 2 = 03x1 2 x2 + x − 6( )= 0

x1 2 x + 3( ) x − 2( ) = 0




23

59.

22 221 1

16

1

4

y

x

60.

4 4 6 82

2x

2

2

6

8

4

y

61.

3

2

1

22

21

23

22 321 21x

y

62. 3

2

1

22

21

23

22 321 21x

y

63. a. f g x( )( )= g x( )[ ]2 +1

= 1x( )2

+ 1

= 1x2 + 1

64. a. b.

f g x( )( )= g x( ) = 5x − 4 g f x( )( )= 5 f x( )[ ]− 4 = 5 x − 4

b. g f x( )( )= 1f x( ) = 1

x2 +1

65. a. f g x( )( )= g x( )+1g x( )−1 = x3 +1

x3 −1

66. a. f g x( )( )= 2g x( ) = 2x2

b. g f x( )( )= f x( )[ ]3= x +1

x−1( )3 b. g f x( )( )= f x( )[ ]2

= 2x( )2= 22x

67. f x+ h( )− f x( )

h =2 x+h( )2 −3 x+h( )+1− 2x2 − 3x+1( )

h

=2 x2 + 2hx+h2( )−3x− 3h+1− 2x2 +3x−1

h

= 2x2 +4hx+2h 2 −3h− 2x2

h = 4x + 2h− 3

68. f x+ h( )− f x( )

h =5x+h − 5

xh = 5

h x+h( ) − 5hx

= 5xhx x+h( ) − 5 x+h( )

hx x+ h( ) = 5x−5x− 5hhx x +h( ) = − 5

x x+ h( )

69. The advertising budget A as a function of t is

the composition of A(p) and p(t). A p t( )( )= 2 p t( )[ ]0.15 = 2 18+ 2t( )0.15

A 4( ) = 2 18+ 2 4( )[ ]0.15 = 2 26( )0.15

≈ $3.26 million

70. a.

x 4 − 2x3 − 3x 2 = 0

x 2 x2 − 2x − 3( )= 0

x2 x − 3( ) x + 1( ) = 0


at x = 0 at x = 3 at x = −1

x = 0, x = 3, and x = −1



b.

on [25, 5] by [25, 5] 71. a.

x3 + 2x 2 − 3x = 0

x x 2 + 2x − 3( )= 0

x x + 3( ) x − 1( ) = 0


72. a. b.

The points suggest a parabolic (quadratic) curve.

on [0.5, 5.5] by [1.5, 3] b.

on [25, 5] by [25, 5]

c. For year 6, y ≈ 0.136 6( )2 − 0.624 6( )+ 2.5 ≈ $3.6 millionFor year 7, y = 0.136 7( )2 − 0.6247( )+ 2.5 ≈ $4.8 million

chapter 1: functions - cengagecollege.cengage.com/.../3e/students/ssm/ch01_1.pdf · 2008-06-27 ·...

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