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CONTENTS Chapter 0

Functions .............................................................................................1

Chapter 1

The Derivative ..................................................................................26

Chapter 2

Applications of the Derivative ..........................................................73

Chapter 3

Techniques of Differentiation .........................................................121

Chapter 4

The Exponential and Natural Logarithmic Functions .....................147

Chapter 5

Applications of the Exponential and Natural Logarithm Functions .........................................................................................177

Chapter 6

The Definite Integral .......................................................................196

Chapter 7

Functions of Several Variables .......................................................230

Chapter 8

The Trigonometric Functions .........................................................267

Chapter 9

Techniques of Integration ...............................................................286

Chapter 0 Functions 0.1

Functions and Their Graphs

16. h( s ) =

(

)

− 32

−3 ⎛ 3⎞ h ⎜− ⎟ = = 12 = 3 ⎝ 2 ⎠ 1+ − 3 −2

( 2)

h(a + 1) =

5. 17.

6. 7. [2, 3)

3⎞ ⎛ 8. ⎜ −1, ⎟ ⎝ 2⎠

9. [–1, 0)

10. [–1, 8)

11.

(−∞, 3)

12. ⎡⎣ 2, ∞

13.

f ( x) = x 2 − 3 x

)

f ( x) = 3 x + 2, h ≠ 0

f (3 + h ) − f (3) (3h + 11) − 11 3h = = =3 h h h

18.

f ( x) = x 2 , h ≠ 0 f (1 + h ) = (1 + h ) = 1 + 2h + h 2 2

f (1) = 12 = 1

f (3) = 3 2 − 3(3) = 9 − 9 = 0 f (−7) = (−7) 2 − 3(−7) = 49 + 21 = 70 f ( x) = x 3 + x 2 − x − 1

19. a.

f (1) = 1 + 1 − 1 − 1 = 0 f (−1) = (−1) 3 + (−1) 2 − (−1) − 1 = 0 3

9 ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ f ⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ −1 = − ⎝2⎠ ⎝2⎠ ⎝2⎠ ⎝2⎠ 8 f (a) = a 3 + a 2 − a − 1 f ( x) = x 2 − 2 x

(

)

2 f (1 + h ) − f (1) 1 + 2h + h − 1 = h h 2h + h 2 = = 2+h h

f (5) = 5 2 − 3(5) = 25 − 15 = 10

a +1 a +1 = 1 + (a + 1) a + 2

f (3 + h ) = 3 (3 + h ) + 2 = 9 + 3h + 2 = 3h + 11 f (3) = 3 (3) + 2 = 11

f (0) = 0 2 − 3(0) = 0

15.

1 ⎛1⎞ h ⎜ ⎟ = 2 = 32 = ⎝ 2 ⎠ 1+ 1 3

14.

s (1 + s )

20. a. 2

f (a + 1) = (a + 1) − 2(a + 1) = (a 2 + 2a + 1) − 2a − 2 = a 2 − 1

f (a + 2) = (a + 2) 2 − 2(a + 2) = (a 2 + 4a + 4) − 2a − 4 = a 2 + 2a

21.

k ( x ) = x + 273 5933 = x + 273 ⇒ x = 5660 The boiling point of tungsten is 5660°C. 9 x + 32 5 9 f ( x ) = (5660) + 32 = 10220 5 The boiling point of tungsten is 10220°F. f ( x) =

f (0) represents the number of laptops sold in 2015.

f (5) = 150 + 2(5) + 5 2 = 150 + 10 + 25 = 185 In 2020, the company will sell 185 laptops.

8x ( x − 1)( x − 2) all real numbers such that x ≠ 1, 2 or (−∞, −1)  (−1, 2)  (2, ∞ ) f ( x) =

Chapter 0 Functions

22.

1 t all real numbers such that t > 0 or (0, ∞ ) f (t ) =

1 3− x all real numbers such that x < 3 or ( −∞, −3)

23. g ( x ) =

4 x ( x + 2) all real numbers such that x ≠ 0, –2 or (−∞, −2)  (−2, 0)  (0, ∞ )

24. g ( x) =

37. positive

38. negative

39. [−1, 3]

40. −1, 5, 9

41.

(−∞, − 1]  [5, 9]

43.

f (1) ≈ .03; f (5) ≈ .037

44.

f (6) ≈ .03

45.

[0, .05]

47.

1⎞ ⎛ f ( x) = ⎜ x − ⎟ ( x + 2) ⎝ 2⎠

42.

[ −1, 5]  [9, ∞ ]

46. t ≈ 3

1⎞ 25 ⎛ f (3) = ⎜ 3 − ⎟ (3 + 2) = ⎝ ⎠ 2 2 No, (3, 12) is not on the graph.

25.

48. f(x) = x(5 + x)(4 – x) f(–2) = –2(5 + (–2))(4 – (–2)) = –36 No, (–2, 12) is not on the graph. 49. g ( x) =

26.

g (1) =

3x − 1 x2 + 1 3 (1) − 1

2 =1 2

(1) + 1 Yes, (1, 1) is on the graph. 50. g ( x) = 27.

g ( 4) =

x2 + 4 x+2

( 4) 2 + 4

20 10 = 6 3

4+2 ⎛ 1⎞ No, ⎜ 4, ⎟ is not on the graph. ⎝ 4⎠

51. 28.

f ( x) = x 3 f (a + 1) = (a + 1) 3

52.

⎛5⎞ f ( x) = ⎜ ⎟ − x ⎝x⎠

5 − (2 + h) (2 + h) 5 − (2 + h) 2 1 − 4h − h 2 = = (2 + h) 2+h

f (2 + h) =

29. function

30. not a function

31. not a function

32. not a function

33. not a function

34. function

35.

f (0) = 1; f (7 ) = −1

36.

f ( 2) = 3; f ( −1) = 0

53.

for 0 ≤ x < 2 ⎪⎧ x f ( x) = ⎨ ⎪⎩1 + x for 2 ≤ x ≤ 5 f (1) = 1 = 1 f (2) = 1 + 2 = 3 f (3) = 1 + 3 = 4

Section 0.1 Functions and Their Graphs

54.

⎧1 for 1 ≤ x ≤ 2 ⎪ f ( x) = ⎨ x ⎪ x 2 for 2 < x ⎩ 1 f (1) = = 1 1 1 f (2) = 2

60. P ( x ) = a.

⎧π x 2 for x < 2 ⎪ f ( x) = ⎨1 + x for 2 ≤ x ≤ 2.5 ⎪4 x for 2.5 < x ⎩ f (1) = π (1) 2 = π f(2) = 1 + 2 = 3 f(3) = 4(3) = 12

⎧ 3 for x < 2 ⎪ 4− x ⎪ 56. f ( x) = ⎨ 2 x for 2 ≤ x < 3 ⎪ 2 ⎪ x − 5 for 3 ≤ x ⎩ 3 f (1) = =1 4 −1 f(2) = 2(2) = 4 f (3) = 3 2 − 5 = 4 = 2

57. a. b. 58.

110 x − 25 10 x + n

P (30) =

110 (30) − 25 10 (30) + 5

3275 ≈ 10.738 305

Each partner’s profit was be approximately, $10.738 thousand or $10,738.

f (3) = 3 2 = 9

55.

110 (30) − 25

⇒ n + 300 = 10 (30) + n 3275 n= − 300 = 355 5

61. Entering Y1 = 1/X + 1 will graph the function 1 f ( x) = + 1 . In order to graph the function x 1 , you need to include parentheses f ( x) = x +1 in the denominator: Y1 = 1/(X + 1). 62. Entering Y1 = X ^ 3 / 4 will graph the function x3 f ( x) = . In order to graph the function 4 y = x 3 4 , you need to include parentheses in the exponent: Y1 = X ^ (3/4).

63.

f ( x) = − x2 + 2 x + 2

64.

f ( x) =

for 50 ≤ x ≤ 3000 ⎧0.06 x f ( x) = ⎨ ⎩0.02 x + 15 for 3000 < x f (3000) = 0.06 (3000) = 180 f ( 4500) = 0.02 ( 4500) + 15 = 105

3275 ⇒ 5

59.

1 x +1 2

Chapter 0 Functions

0.2

Some Important Functions

5. y = −2 x + 3

1. y = 2 x − 1

−1 5

0 3

0 –1

1 1

–1 –3 f ( x) = 2 x − 1

6. y = 0 2.

y=3

7. x − y = 0

3. y = 3 x + 1

f ( x) = 3x + 1

–1 –1

–1 –2

4. y = −

1 x−4 2

2 –5 0 –4 –2 –3

8. 3x + 2 y = −1

3 −5 1 –2 −3

Section 0.2 Some Important Functions

9. x = 2 y − 1

16. 2 + 3x = 2 y 2 + 3 (0 ) = 2 y ⇒ y = 1

The y-intercept is (0, 1). 2 + 3x = 2 (0) ⇒ 3x = −2 ⇒ x = −

2 3

⎛ 2 ⎞ The x-intercept is ⎜ − , 0 ⎟ . ⎝ 3 ⎠

–3 –1

17. a.

Cost is $(24 + 200(.45)) = $114.

b. f(x) = .45x + 24 18. Let x be the volume of gas (in thousands of cubic feet) extracted. f(x) = 5000 + .10x

10.

19. Let x be the number of days of hospital confinement. f(x) = 700x + 1900 20. a.

11.

total cost = 84, 000 (10) = 840, 000

f ( x) = 9x + 3 f ( 0 ) = 9 (9 ) + 3 = 3 The y-intercept is (0, 3). 1 3

9 x + 3 = 0 ⇒ 9 x = −3 ⇒ x = − ⎛ 1 ⎞ The x-intercept is ⎜ − , 0 ⎟ . ⎝ 3 ⎠

12.

cost per empl = 12 (1000) + 12 ( 25)( 240) = 12, 000 + 72, 000 = 84, 000

21.

1 x −1 2 1 f (0) = − (0) − 1 = −1 2 The y-intercept is (0, –1). 1 1 − x − 1 = 0 ⇒ − x = 1 ⇒ x = −2 2 2 The x-intercept is (–2, 0). f ( x) = −

13. f(x) = 5 The y-intercept is (0, 5). There is no x-intercept.

c( x) = 84, 000 x

c(25) = 840, 000 ( 25) = 21, 000, 000

50 x , 0 ≤ x ≤ 100 105 − x From example 6, we know that f(70) = 100. The cost to remove 75% of the pollutant is 50 ⋅ 75 f (75) = = 125. 105 − 75 The cost of removing an extra 5% is $125 − $100 = $25 million. To remove the final 5% the cost is f(100) – f(95) = 1000 – 475 = $525 million. This costs 21 times as much as the cost to remove the next 5% after the first 70% is removed. f ( x) =

22. a.

14. f(x) = 14 The y-intercept is (0, 14). There is no x-intercept. 15. x − 5 y = 0 0 − 5y = 0 ⇒ y = 0 The x- and y-intercept is (0, 0).

f (85) =

20(85) = $100 million 102 − 85

b. f(100) – f(95) = 1000 – 271.43 ≈ $728.57 million 23.

1 ⎛K⎞ f ( x) = ⎜ ⎟ x + ⎝V ⎠ V

Chapter 0 Functions

f(x) = .2x + 50 K 1 1 = .2 and = 50. If = 50, We have V V V 1 K then V = . Now, = .2 implies V 50 K 1 1 1 = .2, so K = ⋅ = . 1 5 50 250 50

1 2 x + 3x−π 2 1 a = , b = 3, c = −π 2

30. y =

31.

a = 2, b = –4, c = 0 vertex: ⎛ − ( −4) ⎛ − ( −4) ⎞ ⎞ , f⎜ ⎜ ⎟ ⎟ = (1, f (1)) = (1, − 2) ⎝ 2 ( 2) ⎠ ⎠ ⎝ 2 ( 2)

1 ⎛K ⎞ 1 1 ⎛K⎞ y = ⎜ ⎟ x + , ⎜ ⎟ ⋅ 0 + = , so the ⎝V ⎠ ⎝ ⎠ V V V V ⎛ 1⎞ y-intercept is ⎜ 0, ⎟ . ⎝ V⎠ 1 ⎛K⎞ Solving ⎜ ⎟ x + = 0, we get ⎝V ⎠ V K 1 1 x = − ⇒ x = − , so the x-intercept V V K ⎛ 1 ⎞ is ⎜ − , 0 ⎟ . ⎝ K ⎠

⎛ 1 ⎞ 24. From 17(b), ⎜ − , 0 ⎟ is the x-intercept. From ⎝ K ⎠ the experimental data, (–500, 0) is also the 1 1 x-intercept. Thus − = −500 ⇒ K = . 500 K ⎛ 1⎞ Again from 17(b), ⎜ 0, ⎟ is the y-intercept. ⎝ V⎠ From the experimental data, (0, 60) is also the 1 1 y-intercept. Thus = 60 ⇒ V = . 60 V

f ( x) = 2 x 2 − 4 x

32. g (t ) = −t 2 + 4t − 3 a = –1, b = 4, c = –3 vertex: ⎛ −4 ⎛ −4 ⎞ ⎞ , g⎜ ⎜ ⎟ ⎟ = ( 2, g ( 2)) = ( 2, 1) ⎝ 2 ( −1) ⎠ ⎠ ⎝ 2 ( −1)

25. y = 3x − 4 x a = 3, b = –4, c = 0 x 2 − 6x + 2 1 2 2 26. y = = x − 2x + 3 3 3 1 2 a = , b = –2, c = 3 3

–3

27. y = 3 x − 2 x 2 + 1 a = –2, b = 3, c = 1 28. y = 3 − 2 x + 4 x 2 a = 4, b = –2, c = 3 29. y = 1 − x 2 a = –1, b = 0, c = 1

33.

f ( x) =

{

3 2x + 1

for x < 2 for x ≥ 2

Section 0.2 Some Important Functions

x<2

x≥2

x≥3

f ( x) = 3

f ( x) = 2x + 1

f ( x) = x + 1

36.

34.

for 0 ≤ x < 1 ⎧4 x ⎪ f ( x) = ⎨8 − 4 x for 1 ≤ x < 2 ⎪2 x − 4 for x ≥ 2 ⎩

0≤x<1

⎧1 for 0 ≤ x < 4 ⎪ x f ( x) = ⎨ 2 ⎪⎩2 x − 3 for 4 ≤ x ≤ 5

0≤x<4

4≤x≤5

1 f ( x) = x 2

f ( x) = 2x − 3

3 2

1≤x<2

f ( x) = 4 x

f ( x) = 8 − 4 x

1 2

3 2

x≥2

37.

f ( x) = 2x − 4

f ( x) = x100 , x = −1 f ( −1) = ( −1)

f ( x) = x 5 , x =

1 2

100

35.

38.

⎧4 − x for 0 ≤ x < 2 ⎪ f ( x) = ⎨2 x − 2 for 2 ≤ x < 3 ⎪x + 1 for x ≥ 3 ⎩

1 ⎛1⎞ ⎛1⎞ f ⎜ ⎟=⎜ ⎟ = ⎝2⎠ ⎝2⎠ 32

39.

0≤x<2

f (10 −2 ) = 10 −2 = 10 −2

2≤x<3

f ( x) = 4 − x

f ( x) = 2x − 2

5 2

f ( x) = x , x = 10 −2

40.

f ( x) = x , x = π f (π ) = π = π

41.

f ( x) = x , x = –2.5 f (–2.5) = −2.5 = 2.5

Chapter 0 Functions

42.

2 3 2 2 ⎛ 2⎞ f ⎜− ⎟ = − = ⎝ 3⎠ 3 3

f ( x) = x , x = −

9. 43.

10. 44.

11. 45.

12. 46. 13.

0.3

The Algebra of Functions

f ( x) + g ( x) = ( x 2 + 1) + 9 x = x 2 + 9 x + 1

f ( x) − h( x) = ( x 2 + 1) − (5 − 2 x 2 ) = 3 x 2 − 4

f ( x) g ( x) = ( x 2 + 1)(9 x) = 9 x 3 + 9 x 2

4. g ( x)h( x ) = (9 x)(5 − 2 x ) = 45 x − 18 x

f (t ) t 2 + 1 t 2 1 t 1 t 2 + 1 = = + = + = g (t ) 9t 9t 9t 9 9t 9t

g (t ) 9t = h(t ) 5 − 2t 2

2 1 2( x + 2) + ( x − 3) + = x−3 x+2 ( x − 3)( x + 2) 3x + 1 = 2 x − x−6

14.

3 −2 3( x − 2) + (−2)( x − 6) + = x−6 x−2 ( x − 6)( x − 2) x+6 = 2 x − 8 x + 12 x −x x( x − 4) + (− x)( x − 8) + = x−8 x−4 ( x − 8)( x − 4) 4x = 2 x − 12 x + 32 −x x (− x)( x + 5) + x( x + 3) + = x+3 x+5 ( x + 3)( x + 5) −2 x = 2 x + 8 x + 15 ( x + 5)( x + 10) + x( x − 10) x+5 x + = ( x − 10)( x + 10) x − 10 x + 10 2 x 2 + 5 x + 50 = x 2 − 100 x + 6 x − 6 ( x + 6)( x + 6) + ( x − 6)( x − 6) + = ( x − 6)( x + 6) x−6 x+6 2 2 x + 72 = 2 x − 36 5 − x x(5 + x ) − (5 − x)( x − 2) x − = ( x − 2)(5 + x) x−2 5+ x 2 2 x − 2 x + 10 = 2 x + 3x − 10 t t + 1 t (3t − 1) − (t − 2)(t + 1) − = (t − 2)(3t − 1) t − 2 3t − 1 2 2t + 2 = 2 3t − 7t + 2

15.

− x 2 + 5x x 5− x ⋅ = 2 x − 2 5 + x x + 3 x − 10

16.

5 − x x + 1 − x 2 + 4x + 5 ⋅ = 5 + x 3 x − 1 3x 2 + 14 x − 5

x x 2 + 5x − x 2 = x ⋅5+ x = 17. 5 − x x − 2 5 − x − x 2 + 7 x − 10 5+ x s +1 s +1 s − 2 s2 − s − 2 18. 3s − 1 = ⋅ = s 3s − 1 s 3s 2 − s s−2

Section 0.3 The Algebra of Functions

19.

20.

5 − ( x + 1) x + 1 − x + 4 x +1 ⋅ = ⋅ ( x + 1) − 2 5 + ( x + 1) x − 1 6 + x − x 2 + 3x + 4 = 2 x + 5x − 6

30.

x+2 5 − ( x + 2) + ( x + 2) − 2 5 + ( x + 2) x + 2 3− x = + x x+7 ( x + 2)( x + 7) + (3 − x)( x) 12 x + 14 = = 2 x( x + 7) x + 7x

32.

5 − ( x + 5) 5 + ( x + 5) 5 − ( x + 5) ( x + 5) − 2 21. = ⋅ x+5 5 + ( x + 5) x+5 ( x + 5) − 2 −x x + 3 = ⋅ 10 + x x + 5 2 − x − 3x = 2 x + 15 x + 50

22.

1 t

1 −2 t

1 t 1 = ⋅ = ,t≠0 t 1 − 2t 1 − 2t

1 u = 5u − 1 ⋅ u = 5u − 1 , u ≠ 0 23. 1 u 5u + 1 5u + 1 5+ u

f ( x 3 − 5 x 2 + 1) = ( x 3 − 5 x 2 + 1) 6

31. ( x + h) 2 − x 2 = x 2 + 2 xh + h 2 − x 2 = 2 xh + h 2 −h 1 1 x− x−h − = = x + h x x ( x + h) x ( x + h)

(

24.

+1 1+ x2 x2 1+ x2 x2 = ⋅ = ,x ≠ 0 2 2 2 ⎛ 1 ⎞ 3 3 x − x − x 3⎜ 2 ⎟ −1 ⎝x ⎠

25.

⎛ x ⎞ ⎛ x ⎞ = f⎜ ⎝ 1 − x ⎟⎠ ⎜⎝ 1 − x ⎟⎠

(

⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ 27. h ⎜ = − 5⎜ +1 ⎝ 1 − x ⎟⎠ ⎜⎝ 1 − x ⎟⎠ ⎝ 1 − x ⎟⎠

)

⎡(t + h )3 + 5⎤ − t 3 + 5 ⎦ 34. ⎣ h t 3 + 3t 2 h + 3th 2 + h 3 + 5 − t 3 − 5 = h 3t 2 h + 3th 2 + h 3 h(3t 2 + 3th + h 2 ) = = h h = 3t 2 + 3th + h 2

35. a.

1 ⎞ ⎛ C ( A(t ) ) = 3000 + 80 ⎜ 20t − t 2 ⎟ ⎝ 2 ⎠ = 3000 + 1600t − 40t 2

C (2) = 3000 + 1600(2) − 40(2) 2 = 3000 + 3200 − 160 = $6040

36. a.

C ( f (t )) = .1(10t − 5) 2 + 25(10t − 5) + 200 = .1(100t 2 − 100t + 25) + 250t − 125 + 200 = 10t 2 + 240t + 77.5

C (4) = 10(4) 2 + 240(4) + 77.5 = $1197.50

( ) ( ) − 5 (t 6 ) + 1 = t18 − 5t12 + 1

26. h t 6 = t 6

)

⎡ 4 (t + h ) − (t + h ) 2 ⎤ − 4t − t 2 ⎣ ⎦ 33. h 4t + 4h − (t 2 + 2th + h 2 ) − 4t + t 2 = h 4h − 2th − h 2 h(4 − 2t − h) = = h h = 4 − 2t − h

5−

1 ⎛1⎞ 37. h( x ) = f (8 x + 1) = ⎜ ⎟ (8 x + 1) = x + ⎝8⎠ 8 h(x) converts from British to U.S. sizes.

38. f(x + 1):

( ) 1 −x x 6

28. g x 6 =

29. g (t 3 − 5t 2 + 1) = =

t 3 − 5t 2 + 1 1 − (t 3 − 5t 2 + 1) t 3 − 5t 2 + 1

[−10, 10] by [0, 20]

−t 3 + 5t 2

(continued on next page)

Chapter 0 Functions

f(x) – 2:

(continued) f(x – 1):

[−5, 5] by [−5, 15] The graph of f(x) + c is the graph of f(x) shifted up (if c > 0) or down (if c < 0) by c

[−10, 10] by [0, 20] f(x + 2):

units. 40. This is the graph of f ( x) = x 2 shifted 1 unit to the right and 2 units up.

[−10, 10] by [0, 20] f(x – 2): [−5, 5] by [−5, 15]

[−10, 10] by [0, 20] The graph of f(x + a) is the graph of f(x) shifted to the left (if a > 0) or to the right (if a < 0) by a units.

41. This is the graph of f ( x) = x 2 shifted 2 units to the left and 1 unit down.

39. f(x) + 1:

[−5, 5] by [−5, 15] 42.

[−5, 5] by [−5, 15] f(x) – 1:

[−4, 4] by [−10, 10] They are not the same function. 43.

[−5, 5] by [−5, 15] f(x) + 2: [−15, 15] by [−10, 10] ⎛ x ⎞ f ( f ( x )) = f ⎜ = ⎝ x − 1 ⎟⎠

[−5, 5] by [−5, 15]

x x −1 x −1 x −1

x = x, x ≠ 1 x − ( x − 1)

Section 0.4 Zeros of Functions—The Quadratic Formula and Factoring

0.4 1.

Zeros of Functions—The Quadratic Formula and Factoring

−b ± b 2 − 4ac 7 ± (−7) 2 − 4(11)(1) = 2a 2(11) 7± 5 7+ 5 7− 5 = = , 22 22 22

2x 2 − 7x + 6 = 0 a = 2, b = –7, c = 6 b 2 − 4ac = 49 − 4(2)(6) = 1 = 1

7. 5 x 2 − 4 x − 1 = 0

−b ± b 2 − 4ac 7 ± 1 3 = = 2, 2a 4 2

−b ± b 2 − 4ac 4 ± (−4) 2 − 4(5)(−1) = 2a 2(5) 4 ± 36 4 ± 6 1 = = = 1, − 10 10 5

f ( x) = 3 x 2 + 2 x − 1 3x 2 + 2 x − 1 = 0 a = 3, b = 2, c = –1 b 2 − 4ac = 4 2 − 4(3)(−1) = 16 = 4 x=

8. x 2 − 4 x + 5 = 0 −b ± b 2 − 4ac 4 ± (−4) 2 − 4(1)(5) = 2a 2(1) 4 ± −4 = 2 −4 is undefined, so there is no real solution.

−b ± b 2 − 4ac −2 ± 4 1 = = , −1 2a 6 3

f (t ) = 4t 2 − 12t + 9

4t 2 − 12t + 9 = 0 −b ± b 2 − 4ac 12 ± (−12) 2 − 4(4)(9) = 2a 2(4) 12 ± 0 3 = = 8 2

f ( x) =

()

2 1 −b ± b 2 − 4ac −1 ± 1 − 4 4 (1) x= = 2a 2 14

()

9. 15 x 2 − 135 x + 300 = 0 −b ± b 2 − 4ac 2a 135 ± (−135) 2 − 4(15)(300) = 2(15) 135 ± 225 135 ± 15 = = = 5, 4 30 30

1 2 x + x +1 4

1 2 x + x +1 = 0 4

−1 ± 0 1 2

10. z 2 − 2 z − z=

= −2

f ( x) = −2 x 2 + 3 x − 4

−2 x 2 + 3 x − 4 = 0 2

−b ± b − 4ac −3 ± 3 − 4(–2)(–4) = 2a 2(–2) −3 ± −23 = −4 −23 is undefined, so f(x) has no real zeros.

f (a) = 11a 2 − 7 a + 1 11a 2 − 7a + 1 = 0

f ( x) = 2 x 2 − 7 x + 6

11.

5 =0 4

−b ± b 2 − 4ac 2a 2±

(− 2 ) − 4(1) (− 54 ) = 2 ± 7 2

2(1) 2+ 7 2− 7 , 2 2

3 2 x − 6x + 5 = 0 2

()

2 3 −b ± b 2 − 4ac 6 ± (−6) − 4 2 (5) x= = 2a 2 32

()

6± 6 6 6 = 2+ , 2− 3 3 3

Chapter 0 Functions

12. 9 x 2 − 12 x + 4 = 0 −b ± b 2 − 4ac 12 ± (−12) 2 − 4(9)(4) = 2a 2(9) 12 ± 0 2 = = 18 3

30. x 2 + x +

31.

14. x 2 − 10 x + 16 = ( x − 2)( x − 8) 15. x 2 − 16 = ( x − 4)( x + 4) 16. x 2 − 1 = ( x + 1)( x − 1)

(

)

(

)

17. 3x 2 + 12 x + 12 = 3 x 2 + 4 x + 4

32.

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

18. 2 x 2 − 12 x + 18 = 2 x 2 − 6 x + 9

= 2( x − 3)( x − 3) = 2( x − 3) 2

(

19. 30 − 4 x − 2 x 2 = −2 −15 + 2 x + x 2 = −2( x − 3)( x + 5)

20. 15 + 12 x − 3 x 2 = −3(−5 − 4 x + x 2 ) = −3( x − 5)( x + 1) 21. 3x − x 2 = x (3 − x) 22. 4 x 2 − 1 = (2 x + 1)(2 x − 1)

)

x 2 − 10 x + 9 = x − 9 x 2 − 11x + 18 = 0 (x – 9)(x – 2) = 0 x = 9, 2 y=x–9=9–9=0 y = 2 – 9 = –7 Points of intersection: (9, 0), (2, –7)

33. y = x 2 − 4 x + 4 y = 12 + 2 x − x 2 x 2 − 4 x + 4 = 12 + 2 x − x 2 2x 2 − 6x − 8 = 0 2( x 2 − 3x − 4) = 0 2( x − 4)( x + 1) = 0 x = 4, −1 y = x 2 − 4 x + 4 = 4 2 − 4(4) + 4 = 4

23. 6 x − 2 x 3 = −2 x( x 2 − 3) = −2 x x − 3 x + 3

)(

(

24. 16 x + 6 x 2 − x 3 = x 16 + 6 x − x 2

)

= x(8 − x)( x + 2) = − x( x − 8)( x + 2)

)

25. x3 − 1 = ( x − 1) x 2 + x + 1

(

26. x 3 + 125 = ( x + 5) x 2 − 5 x + 25

(

)

27. 8 x 3 + 27 = ( 2 x + 3) 4 x 2 − 6 x + 9 28. x 3 −

2 x 2 − 5 x − 6 = 3x + 4 2 x 2 − 8 x − 10 = 0 −b ± b 2 − 4ac 8 ± (−8) 2 − 4(2)(–10) = 2a 2(2) 8 ± 144 8 ± 12 = = = 5, − 1 4 4 y = 3x + 4 = 15 + 4 = 19 y = –3 + 4 = 1 Points of intersection: (5, 19), (–1, 1)

13. x + 8 x + 15 = ( x + 5)( x + 3)

(

y = (−1) 2 − 4(−1) + 4 = 9 Points of intersection: (4, 4), (–1, 9)

34. y = 3 x 2 + 9 y = 2x 2 − 5x + 3 3x 2 + 9 = 2 x 2 − 5 x + 3 x 2 + 5x + 6 = 0 ( x + 3)( x + 2) = 0 x = −3, −2 y = 3 x 2 + 9 = 3(−3) 2 + 9 = 36

)

y = 3(−2) 2 + 9 = 21 Points of intersection: (–3, 36), (–2, 21)

1 ⎛ 1⎞⎛ x 1⎞ = ⎜ x − ⎟ ⎜ x2 + + ⎟ 8 ⎝ 2⎠⎝ 2 4⎠

29. x 2 − 14 x + 49 = ( x − 7 )

(

1 ⎛ 1⎞ = ⎜x + ⎟ 4 ⎝ 2⎠

Section 0.4 Zeros of Functions—The Quadratic Formula and Factoring

35. y = x 3 − 3x 2 + x

−b ± b 2 − 4ac 2 ± x= = 2a

y = x 2 − 3x x 3 − 3x 2 + x = x 2 − 3x 3 x − 4x 2 + 4x = 0

y = x − 3x = 0 − 3(0) = 0 y = 2 2 − 3(2) = 4 − 6 = −2 Points of intersection: (0, 0), (2, –2)

(

) − 12 (2 − 3 ) + 5 = 25 − 232 3

y = 16 x 3 + 25 x 2 30 x 3 − 3 x 2 = 16 x 3 + 25 x 2 14 x 3 − 28 x 2 = 0 14 x 2 ( x − 2) = 0 x = 0 or x = 2 y = 30(0) 3 − 3(0) 2 = 0

()

2 ± (−2) 2 − 4 12 ( −2) 2

()

y = 30(2) 3 − 3(2) 2 = 30(8) – 3(4) = 228 Points of intersection: (0, 0), (2, 228)

1 2

2± 8 = 2 + 2 2, 2 − 2 2 1 y = 2x = 2(0) = 0 =

39.

( ) y = 2 (2 − 2 2 ) = 4 − 4 2 y = 2 2+2 2 = 4+4 2

Points of intersection: (0, 0),

(2 + 2 2, 4 + 4 2 ) , (2 − 2 2, 4 − 4 2 ) 1 3 x + x2 + 5 2 1 y = 3x 2 − x + 5 2 1 3 1 x + x 2 + 5 = 3x 2 − x + 5 2 2 1 3 1 2 x − 2x + x = 0 2 2 1⎞ ⎛1 2 x ⎜ x − 2x + ⎟ = 0 ⎝2 2⎠

40.

37. y =

x = 0 or

) − 12 (2 + 3 ) + 5 = 25 + 232 3

38. y = 30 x 3 − 3x 2

1 3 x − 2x 2 − 2x = 0 2 ⎛1 ⎞ x ⎜ x 2 − 2x − 2⎟ = 0 ⎝2 ⎠ 1 x = 0 or x 2 − 2 x − 2 = 0 2

−b ± b − 4ac = 2a

(

Points of intersection: (0, 5), ⎛ 23 3 ⎞ ⎛ 23 3 ⎞ ⎜ 2 − 3, 25 − 2 ⎟ , ⎜ 2 + 3, 25 + 2 ⎟ ⎝ ⎠ ⎝ ⎠

1 3 x − 2x 2 = 2x 2

()

2 12

y =3 2+ 3 y =3 2− 3

1 36. y = x 3 − 2 x 2 2 y = 2x

(−2)2 − 4 ( 12 )( 12 )

= 2± 3 1 1 y = 3 x 2 − x + 5 = 3(0) 2 − (0) + 5 = 5 2 2

x ( x 2 − 4 x + 4) = 0 x( x − 2)( x − 2) = 0 ⇒ x = 0, 2 2

1 2 1 x − 2x + = 0 2 2

41.

21 −x=4 x 21 − x 2 = 4 x x 2 + 4 x − 21 = 0 ( x + 7)( x − 3) = 0 ⇒ x = −7, 3 2 =3 x−6 x 2 − 6 x + 2 = 3x − 18 x 2 − 9 x + 20 = 0 ( x − 4)( x − 5) = 0 ⇒ x = 4, 5 x+

14 =5 x+4 2 x + 4 x + 14 = 5 x + 20 x2 − x − 6 = 0 ( x − 3)( x + 2) = 0 ⇒ x = 3, − 2 x+

42. 1 = 1=

5 6 + x x2 5x + 6 x2

x 2 − 5x − 6 = 0 ( x − 6)( x + 1) = 0 ⇒ x = 6, − 1

43.

Chapter 0 Functions

x 2 + 14 x + 49 2

50.

x +1 x + 14 x + 49 = 0 ( x + 7) 2 = 0 ⇒ x = −7 2

[−1.5, 2] by [−2, 3] The zeros are approximately −.689 and 1.170.

x 2 − 8 x + 16 =0 44. 1+ x x 2 − 8 x + 16 = 0 ( x − 4) 2 = 0 ⇒ x = 4

51.

45. C(x) = 275 + 12x R ( x ) = 32 x − .21x 2 C ( x) = R ( x) 275 + 12 x = 32 x − .21x 2 2 .21x − 20 x + 275 = 0 Thus

[−4, 4] by [−6, 10] Approximate points of intersection: (–0.41, –1.83) and (2.41, 3.83) 52.

20 ± (−20) 2 − 4(.21)275 .42 = 16, 667 or 78, 571 subscribers

46.

[−2, 2] by [−5, 2] Approximate points of intersection: (–.65, –1.35) and (1.15, –3.15)

⎛ 1 ⎞ x + ⎜ ⎟ x 2 = 175 ⎝ 20 ⎠ x 2 + 20 x − 3500 = 0 ( x − 50)( x + 70) = 0 x = 50 mph

53.

47.

[−3, 5] by [−80, 30] Approximate points of intersection: (2.14, –25.73) and (4.10, –21.80) [−4, 5] by [−4, 10] The zeros are –1 and 2.

54.

48.

[0, 4] by [−1, 3] Approximate point of intersection: (1.27, .79)

[−4, 5] by [−4, 10] The zeros are –2 and 1.

Answers may vary for exercises 55−58. 55.

49.

[−2, 7] by [−2, 4] The zero is approximately 4.56.

[−5, 22] by [−1400, 100]

Section 0.5 Exponents and Power Functions

( ) =8

56.

19. (25) 3 / 2 =

( 25 ) = 125 3

(

20. (27) 2 / 3 = 3 27

[−1, 1] by [−10, 10] 57.

) =9 2

21. (1.8) 0 = 1 22. 91.5 = 9 3 / 2 =

( 9 ) = 27 3

23. 16 0.5 = 161/ 2 = 4

[−20, 4] by [−500, 2500]

24. 810.75 = 813/ 4 = 27

58.

25. 4 −1/ 2 =

0.5

18. 16 3 / 4 = 4 16

−2 / 3

1 1 = 4 2

( ) =4

[−5, 15] by [−100, 100]

⎛1⎞ 26. ⎜ ⎟ ⎝8⎠

Exponents and Power Functions

27. (.01) −1.5 =

1. 33 = 27

2. (−2) 3 = −8

100

3. 1

4. 0

= 82 / 3 = 3 8 1 (.01)

3/ 2

1 = 1000 .001

1 28. 1−1.2 = 1.2 = 1 1

5. (.1) 4 = (.1)(.1)(.1)(.1) = .0001

29. 51/ 3 ⋅ 2001/ 3 = 10001/ 3 = 10

6. (100) 4 = (100)(100)(100)(100) = 100, 000, 000

30. (31/ 3 ⋅ 31/ 6 ) 6 = (31/ 2 ) 6 = 27

7. −4 2 = −16

31. 61/ 3 ⋅ 6 2 / 3 = 61 = 6

8. (.01) 3 = .000001

9. (16)1/ 2 = 16 = 4

32. (9 4 / 5 ) 5 / 8 = 91/ 2 = 3

10. (27)1/ 3 = 3 27 = 3

33.

11. (.000001)1/3 = 3 .000001 = .01 ⎛ 1 ⎞ 12. ⎜ ⎝ 125 ⎟⎠

1/ 3

13. 6 −1 =

1 6

34. =3

15. (.01) −1 =

1 1 = 125 5 ⎛1⎞ 14. ⎜ ⎟ ⎝2⎠

10 4 54

= 2 4 = 16

35 / 2 31/ 2

= 3(5 / 2) − (1/ 2) = 34 / 2 = 9

(

35. (21/ 3 ⋅ 32 / 3 ) 3 = 3 2 3 9 −1

1 = 1 =2 2

) = ( 3 18 ) = 18 3

36. 20 0.5 ⋅ 5 0.5 = (100)1/ 2 = 10 2/3

82 / 3

1 = 100 .01

⎛ 8 ⎞ 37. ⎜ ⎟ ⎝ 27 ⎠

1 5

38. (125 ⋅ 27)1/ 3 = 1251/ 3 ⋅ 271/ 3 = 15

16. (−5) −1 = −

( ) = 16

17. 8 4 / 3 = 3 8

39.

74 / 3 71/ 3

2/3

4 9

= 7 (4 / 3) − (1/ 3) = 7 3 / 3 = 7

Chapter 0 Functions

40. (61/ 2 ) 0 = 6 (1/ 2)(0) = 6 0 = 1 41. ( xy ) 6 = x 6 y 6

⎛ 3x 2 ⎞ 33 ⋅ x 6 27 x 6 61. ⎜ = = ⎟ 23 ⋅ y 3 8y3 ⎝ 2y ⎠

42. ( x1/ 3 ) 6 = x (1/ 3)(6) = x 2 43. 44.

x4 ⋅ y5

= x 4 ⋅ y 5 ⋅ x −1 ⋅ y −2 = x 3 y 3

xy 2

= x3

x −3

1/2

64.

46. ( x 3 ⋅ y 6 )1/ 3 = x 3(1/ 3) ⋅ y 6(1/ 3) = xy 2 4 ⎞3

⎛x x x 47. ⎜ 2 ⎟ = 2(3) = 6 y y ⎝y ⎠

⎛x⎞ 48. ⎜ ⎟ ⎝ y⎠

4(3)

−2

62. 63.

45. x −1/2 =

1 1 = 9x 3 x

60. (9 x) −1/ 2 =

⋅ y2 = 2

x5 y

x2 1 1 ⋅ = 3 5 y x x y

2x = 2 x ⋅ x −1/ 2 = 2 x x 1 yx

−5

x5 y

65. (16 x 8 ) −3 / 4 = 16 −3 / 4 ⋅ x −6 =

66. (−8 y 9 ) 2 / 3 = (−8) 2 / 3 y 9(2 / 3) = 4 y 6 67.

⎛ 1 ⎞ x⎜ ⎟ ⎝ 4x ⎠

5/ 2

= =

49. ( x 3 y 5 ) 4 = x 3(4) ⋅ y 5(4) = x12 y 20 1 + x (1 + x ) 3 / 2 = (1 + x )1/ 2 (1 + x) 3 / 2 = (1 + x ) (1/ 2) + (3 / 2) = (1 + x) 2 = x 2 + 2x + 1

50.

1 8x 6

2 ⎞3

⎛y x 5 ⋅ y 2(3) = x 5 ⋅ y 6 ⋅ x −3 = x 2 y 6 51. x 5 ⋅ ⎜ ⎟ = x3 ⎝ x ⎠

68.

69.

(25 xy ) 3 / 2 2

x y

x1/ 2

= 45 / 2 x 5 / 2 1

x1/ 2 ⋅ x −5 / 2 32

32 x 2

(25) 3 / 2 x 3 / 2 y 3 / 2 2

x y

15 x

=−

55.

56.

57.

3 x 1 ⋅ 4 =− 3 15 x 5x

70. (−32 y −5 ) 3 / 5 = (−32) 3 / 5 y −5(3 / 5) = −

y −2 x −4 x

72.

= x3 y 2 =

⋅

= (−3) 3 ⋅ x 3 =

58. (−3 x) 3 = −27 x 3 59.

71.

−x y x y = ⋅ = x2 − xy x y x3

8 y3

For exercises 71−82, f ( x ) = 3 x and g ( x ) =

53. (2 x) 4 = 2 4 ⋅ x 4 = 16 x 4

−3x

125 y

(−27 x 5 ) 2 / 3 (−27) 2 / 3 x 5(2 / 3) = = 9x3 1/ 3 3 x x

52. x −3 ⋅ x 7 = x 7 − 3 = x 4

54.

f ( x) g ( x) = 3 x ⋅

1 x

1 x2

= x1 3 ⋅ x −2 = x − 5 3 =

. 1 x

f ( x) 3 x = = x1 3 ⋅ x 2 = x 7 3 1 g ( x) x2

1 g ( x) x 2 1 73. = = x −2 ⋅ x −1 3 = x −7 3 = 7 3 f ( x) 3 x x 3

( ) ⋅ x12 = x ⋅ x −2 = x −1 = 1x

74. ⎡⎣ f ( x )⎤⎦ g ( x ) = 3 x

x ⋅ 3 x 2 = x1/ 3 ⋅ x 2 / 3 = x

Section 0.5 Exponents and Power Functions 3

(

)

3 1 ⎞ 3 ⎛ 75. ⎡⎣ f ( x ) g ( x )⎤⎦ = ⎜ 3 x ⋅ 2 ⎟ = x1 3 ⋅ x −2 ⎝ ⎠ x 1 −5 3 3 = x = x −5 = 5 x

(

76.

f ( x) ⎛ 3 x ⎞ = g ( x) ⎜ 1 ⎟ ⎜ 2⎟ ⎝x ⎠

)

(

= x1 3 ⋅ x 2

) = (x7 3 ) 12

= x7 6

77.

1 ⎞ ⎛ f ( x) g ( x) = ⎜ 3 x ⋅ 2 ⎟ ⎝ x ⎠

(

= x

78.

3 f

(

= x1 3 ⋅ x −2

)

(

= x1 3 ⋅ x − 2

)

) = x −5 9 = x 51 9

( )

⎛ 1 ⎞ f ( g ( x )) = f ⎜ 2 ⎟ = f x −2 = 3 x −2 ⎝x ⎠ 13 1 = x −2 = x−2 3 = 2 3 x

( 3 x ) = f ( x1 3 ) = 3 x1 3 13 = ( x1 3 ) = x 1 9

f ( g ( x )) = f

( )

⎛ 1 ⎞ ⎛ 1 ⎞ 82. g ( g ( x )) = g ⎜ 2 ⎟ = g x −2 = ⎜ −2 ⎟ ⎝x ⎠ ⎝x ⎠ 1 = −4 = x 4 x 1 1 = ( x − 1) x x

(

85. x −1/ 4 + 6 x1/ 4 = x −1/ 4 1 + 6 x 86. 87.

(Law 6)

89.

f ( x) = x 2 ⇒ f (4) = (4) 2 = 16

90.

f ( x) = x 3 ⇒ f (4) = (4) 3 = 64

91.

f ( x) = x −1 ⇒ f (4) = (4) −1 =

92.

f ( x) = x1/ 2 ⇒ f (4) = (4)1/ 2 = 2

93.

f ( x) = x 3 / 2 ⇒ f (4) = (4) 3 / 2 = 8

94.

f ( x) = x −1/ 2 ⇒ f (4) = (4) −1/ 2 =

1 2

95.

f ( x) = x −5 / 2 ⇒ f (4) = (4) −5 / 2 =

1 32

96.

f ( x) = x 0 ⇒ f (4) = 4 0 = 1

1 4

r⎞ ⎛ formula A = P ⎜1 + ⎟ , where P is the principal, ⎝ m⎠ r is the annual interest rate, m is the number of interest periods per year, and t is the number of years. ⎛ .06 ⎞ 97. A = 500 ⎜1 + ⎟ ⎝ 1 ⎠

1(6)

⎛ .08 ⎞ 98. A = 700 ⎜1 + ⎟ ⎝ 1 ⎠

1(8)

≈ $709.26 ≈ $1295.65

⎛ .095 ⎞ 99. A = 50, 000 ⎜1 + ⎟ ⎝ 4 ⎠ ⎛ .12 ⎞ 100. A = 20, 000 ⎜1 + ⎟ ⎝ 4 ⎠

84. 2 x 2 / 3 − x −1/ 3 = x −1/ 3 (2 x − 1)

x − y

1/ 2

( ) ( )

x−

⎛a⎞ =⎜ ⎟ 1/ 2 ⎝b⎠ b

In exercises 97−104, use the compound interest

1 ⎛ 1 ⎞ 80. g ( f ( x )) = g 3 x = g x1 3 = ⎜ 1 3 ⎟ = 2 3 ⎝x ⎠ x

83.

a b

( )

81.

a = b a1/ 2

) = x −5 6 = x 5 6 1

= x −5 3

79.

−5 3 1 2

( x ) g ( x ) = ⎛⎜⎝ 3 x ⋅ 2 ⎞⎟⎠

(

88.

⎛1 1⎞ y = xy ⎜ − ⎟ x ⎝ y x⎠

a ⋅ b = ab 1/ 2 1/ 2 a ⋅ b = (ab)1/ 2 (Law 5)

)

⎛ .05 ⎞ 101. A = 100 ⎜1 + ⎝ 12 ⎟⎠

≈ $127,857.61

4(3)

≈ $28, 515.22

12(10)

⎛ .045 ⎞ 102. A = 500 ⎜1 + ⎟ ⎝ 12 ⎠

≈ $164.70 12(1)

.06 ⎞ ⎛ 103. A = 1500 ⎜1 + ⎝ 365 ⎟⎠

4(10)

≈ $522.97 365(1)

≈ $1592.75

Chapter 0 Functions

.06 ⎞ ⎛ 104. A = 1500 ⎜1 + ⎝ 365 ⎟⎠

0.6

365(3)

⎛ .068 ⎞ 105. A = 1000 ⎜1 + ⎟ ⎝ 1 ⎠

≈ $1795.80

Functions and Graphs in Applications

1(18)

≈ $3268.00

106. At the end of the first year, there will be A1 = A0 (1 + .08) = 4000(1.08) = $4320 in the

account. At the end of the second year, there will be A2 = A1 (1 + .08) = ( 4320 + 4000)(1.08) = $8985.60 in the account. At the end of the third year, there will be A3 = A2 (1 + 0.8) = (8985.60 + 4000)(1.08) = 14, 024.448 in the account. (Note that we hold the decimals since this is a partial answer. We will round at the end of the calculations.) At the end of the fourth year, there will be A4 = A3 (1 + .08) = (14, 024.448 + 4000)(1.08) ≈ 19, 466.40384 in the account. No additional deposits are made, so use the compound interest formula to compute the amount in the account after another four years: ⎛ .08 ⎞ A = 19, 466.40384 ⎜1 + ⎟ ⎝ 1 ⎠ ≈ $26, 483.83.

107. A = 500 + 500r + =

(

375 2 125 3 125 4 r + r + r 2 4 64

500 256 + 256r + 96r 2 + 16r 3 + r 4 256

(

125 16 + 32r + 24r 2 + 8r 3 + r 4 2

)

125 4 r 2

109. If the speed is 2x, then 1 1 1 (2 x )2 = 4 x 2 = 4 ⎛⎝⎜ x 2 ⎞⎠⎟ . 20 20 20

( )

110. 5E–5 = 5 ⋅ 10 −5 = .00005 111. 8.103E–4 = 8.103 ⋅ 10 −4 = .0008103 112. 1.35E13 = 1.35 ⋅ 1013 = 13, 500, 000, 000, 000 113. 8.23E–6 = 8.23 ⋅ 10 −6 = .00000823

1(4)

108. A = 1000 + 2000r + 1500r 2 + 500r 3 + =

7. P = 2 ( x + 3 x ) = 8 x 3x 2 = 25

8. A = 3x 2 8 x = 30 9. A = π r 2 2π r = 15 10. P = 2r + 2h + π r The area of the window is represented by 1 A = 2rh + π r 2 . 2 1 ⎛ ⎞ 2rh + ⎜ ⎟ π r 2 = 2.5 ⎝2⎠

Section 0.6 Functions and Graphs in Applications

11. V = x 2 h The surface area of the box is represented by

20. V = 2π r 3 = 54π ⇒ r 3 = 27 ⇒ r = 3 From exercise 14, we know that the surface

S = x 2 + 4 xh.

area is equal to 6 πr 2 . Thus, in this example

x 2 + 4 xh = 65

S = 6π 3 2 = 54π in.2

( )

⎛x⎞ ⎛x⎞ 12. SA = 2 xw + 2 x ⎜ ⎟ + 2 w ⎜ ⎟ = 3 xw + x 2 ⎝2⎠ ⎝2⎠ The volume is represented by ⎛x⎞ 1 xw ⎜ ⎟ = x 2 w. ⎝2⎠ 2 ⎛1⎞ 2 ⎜⎝ ⎟⎠ wx = 10 2

21. a.

C (50) − C (40) = (73 + 4 (50)) − (73 + 4 ( 40)) = 273 − 233 = $40 The cost will rise $40.

13. π r 2 h = 100

22. a.

P(x) = 4x – C(x) P(100) = 400 – (10 + 75) = $315

Cost = 5π r 2 + 6π r 2 + 7(2π rh) = 11π r 2 + 14π rh

b. P(101) = 404 – (10.1 + 75) = $318.9 Increase is $3.90.

π h2 ⎛h⎞ ⎛h⎞ + π h2 14. 2π ⎜ ⎟ + 2π ⎜ ⎟ h = ⎝2⎠ ⎝2⎠ 2 3π h 2 = = 30π 2 2

πh ⎛h⎞ V =π ⎜ ⎟ h= ⎝2⎠ 4

15. 2x + 3h = 5000 A = xh

16. h = 2500 f = 4 + 2 h

73 + 4 x = 225 ⇒ x = 38 When 38 T-shirts are sold, the cost will be $225.

23. a.

80 = 200 .4 Sales will break-even when 200 scoops are sold. .4 x − 80 = 0 ⇒ x =

30 = .4 x − 80 ⇒ x = 275 Sales of 275 scoops will generate a daily profit of $30.

40 = .4 x − 80 ⇒ x = 300 To raise the daily profit to $40, 300 – 275 = 25 more scoops will have to be sold.

24. a.

160 = 12 x − 200 ⇒ x = 30 30 thousand subscribers are needed for a monthly profit of $160 thousand

166 = 12 x − 200 ⇒ x = 30.5 thousand There will need to be 30,500 – 30,000 = 500 new subscribers.

25. a.

P ( x ) = R( x) − C ( x) = 21x − 9 x − 800 = 12 x − 800

b. P(120) = 1440 – 800 = $640 c. 17. C = 10 (2 + 2h ) + 8 ( 2 ) = 36 + 20h 2

26. a.

18. 5 x + 4(4 xh) = 5 x + 16 xh = 150 19. 8x = 40  x = 5 A = 3 x 2 = 3(25) = 75 cm2

1000 = 12 x − 800 ⇒ x = 150 R (150) = 21(150) = $3150 P ( x) = R( x) − C ( x) = 1200 x − (550 x + 6500) = 650 x − 6500 P (12) = 650(12) − 6500 = $1300 The company will earn $1300. C ( x) = 14, 750 = 550 x + 6500 ⇒ x = 15 P(15) = 650(15) – 6500 = $3250

Chapter 0 Functions

27. f(6) = 270 cents

50. Find h(0). Find the y-intercept of the graph.

28. From the graph, f(r) = 330 for r = 1 and r = 6.87.

51. a.

29. A 100-inch3 cylinder with radius 3 inches costs $1.62 to construct. 30. The least expensive cylinder has radius 3 inches and costs $1.62 to construct. The cost drops until the radius is 3 in. and then increases.

[0, 6] by [−30, 120] b. Using the Trace command or the Value command, the height is 96 feet.

31. f(3) = $1.62; f(6) = $2.70, so the additional cost = 2.70 – 1.62 = $1.08 32. f(1) = $3.30; f(3) = $1.62, so the amount saved is 3.30 – 1.62 = $1.68 33. From the graph, we see that revenue = $1800 and cost = $1200.

34. The revenue is $1400 when production is 20 units.

Graphing Y2 = 64 and using the Intersect command, the height is 64 feet when x = 1 and x = 4 seconds.

35. The cost is $1400 when production is 40 units. 36. 1800 – 1200 = $600 37. C(1000) = $4000 38. Find the x-coordinate of the point on the graph whose y-coordinate is 3500. 39. Find the y-coordinate of the point on the graph whose x-coordinate is 400.

d. Using the Trace command or the Zero command, the ball hits the ground when x = 5 seconds.

40. C (600) − C (500) = 3136 − 2875 = $261 41. The greatest profit, $52,500, occurs when 2500 units of goods are produced. 42. P(1500) = $42,500 43. Find the x-coordinate of the point on the graph whose y-coordinate is 30,000.

44. Find the y-coordinate of the point on the graph whose x-coordinate is 2000.

Using the Trace command or the Maximum command, the maximum height is reached when x = 2.5 seconds. The maximum height is 100 feet.

45. Find h(3). Find the y-coordinate of the point on the graph whose t-coordinate is 3. 46. Find t such that h(t) is as large as possible. Find the t-coordinate of the highest point of the graph.

52. a.

47. Find the maximum value of h(t). Find the y-coordinate of the highest point of the graph. 48. Solve h(t) = 0. Find the t-intercept of the graph. 49. Solve h(t) = 100. Find the t-coordinates of the points whose y-coordinate is 100.

[0, 70] by [−400, 2000]

Chapter 0 Fundamental Concept Check Exercises

b. Using the Trace command or the Value command, the cost is $1050.

d. c.

R(400) – R(350) = 5000 Revenue would decrease by $5000. e.

The additional cost is $22.11. d. Graphing Y2 = 510 and using the Intersect command, the daily cost is $510 when 10 units are produced.

R(450) – R(400) = 4000 No, the store should not spend $5000 on advertising, since the revenues would only increase by $4000.

Chapter 0 Fundamental Concept Check Exercises 1. Real numbers can be thought of as points on a number line, where each number corresponds to one point on the line, and each point determines one real number. Every real number has a decimal representation. A rational number is a real number with a finite or infinite repeating decimal, such as − 52 = −2.5, 1, 133 = 4.333. An irrational

53. a.

[200, 500] by [42000, 75000] b. Graphing Y2 = 63, 000 and using the Intersect command, the revenue is $63,000 when sales are 350 bicycles per year.

number is a real number with an infinite, nonrepeating decimal representation, such as − 2 = −1.414213 or π = 3.14159 . 2. x < y means x is less than y; x ≤ y means x is less than or equal to y; x > y means x is greater than y; x ≥ y means x is greater than or equal to y. 3. An open interval (a, b) does not contain its endpoints a and b but a closed interval [a, b] does not contain a and b.

Using the Trace command or the Value command, the revenue is $68,000 when 400 bicycles are sold per year.

4. A function of a variable x is a rule f that assigns a unique number f ( x ) to each value

of x. 5. The value of a function at x is the unique number f ( x ) .

Chapter 0 Functions

6. The domain of a function is the set of values that the independent variable x is allowed to assume. The range of a function is the set of values that the function assumes.

13. Sum: f ( x ) + g ( x )

Difference: f ( x ) − g ( x ) Product: f ( x ) g ( x )

7. The graph of a function f ( x ) is the curve that

consists of the set of all points ( x, f ( x )) in

Composition: f ( g ( x ))

the xy-plane. A curve is the graph of a function if and only if each vertical line cuts or touches the curve at no more than one point.

If f ( x ) = 3x 2 and g ( x ) = 3x + 1, then f ( x ) + g ( x ) = 3x 2 + 3x + 1

8. A linear function has the form f ( x ) = mx + b.

f ( x ) − g ( x ) = 3x 2 − (3x + 1) = 3 x 2 − 3 x − 1

When m = 0, the function is a constant function. f ( x ) = 3 x − .5 is a linear function.

f ( x ) g ( x ) = 3 x 2 (3x + 1) = 9 x3 + 3 x 2 f ( x) 3x 2 = g ( x ) 3x + 1

f = −2 is a constant function.

9. An x-intercept is a point at which the graph of a function intersects the x-axis. A y-intercept is a point at which the graph intersects the y-axis. To find the x-intercept, set f ( x ) = 0 and solve

for x, if possible. The y-intercept is the point (0, f (0)) . 10. A quadratic function has the form f ( x ) = ax 2 + bx + c, where a ≠ 0. The graph

is a parabola. 11. a.

Quadratic function: f ( x ) = ax 2 + bx + c, where a ≠ 0; f ( x ) = −2 x 2 + 4 x + 9

b. Polynomial function: p ( x ) = an x n + an −1 x n −1 +  + a0 , where n

is a nonnegative integer and a0 , a1 ,  , an are real numbers, an ≠ 0, and n is a nonnegative integer; f ( x ) = x5 + 3x3 − 7 x + 3 c.

f ( x) Rational function: h ( x ) = , where f g ( x)

and g are polynomials; h ( x ) =

2x − 3 x2 + 1

d. Power function: f ( x ) = x r , where r is a

real number; f ( x ) = x = x1 2 12.

f ( x ) = x is defined as

{

x f ( x) = −x

if x ≥ 0 . if x < 0

f ( x) g ( x)

Quotient:

(

)

f ( g ( x )) = 3 (3x + 1) = 3 9 x 2 + 6 x + 1 2

= 27 x + 18 x + 3

14. x = a is a zero of f ( x ) if f ( a ) = 0. 15. Two methods for finding the zeros of a quadratic function are using factoring or using the quadratic equation. 16.

1 br

br b s = br + s

b− r =

br = br − s s b

(b ) = b

(ab) = a r br

ar ⎛a⎞ ⎜⎝ ⎟⎠ = r b b

r s

17. In the formula A = P (1 + i ) , A represents the n

compound amount, P represents the principal amount, i represents the interest rate, and n represents the number of interest periods. 18. To solve f ( x ) = b geometrically from the

graph of y = f ( x ) , draw the horizontal line y = b. The line intersects the graph at a point

(a, b) if and only if f (a ) = b. Thus, x = a is a solution of f ( x ) = b. 19. To find f ( a ) geometrically from the graph of y = f ( x ) , draw the vertical line x = a. This

line intersects the graph at the point ( a , f ( a )) .

Chapter 0 Review Exercises

Chapter 0 Review Exercises 1.

2 x 2 2 k (1) = 1 + = 3 1 No, the point (1, –2) is not on the graph.

10. k ( x) = x 2 +

1 x 3 1 f (1) = 1 + = 2 1 82 1 3 1 f (3) = 3 + = = 27 3 3 3 1 3 f (−1) = (−1) + = −2 (−1) f ( x) = x 3 +

11. 5 x 3 + 15 x 2 − 20 x = 5 x( x 2 + 3x − 4) = 5 x ( x − 1)( x + 4)

17 1 ⎛ 1⎞ ⎛ 1⎞ f ⎜ − ⎟ = ⎜ − ⎟ − 2 = − = −2 ⎝ 2⎠ ⎝ 2⎠ 8 8 3 1 1 5 2 f 2 = 2 + =2 2+ = 2 2 2

13. 18 + 3 x − x 2 = (− x − 3)( x − 6) = (−1)( x − 6)( x + 3)

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

14. x 5 − x 4 − 2 x 3 = x 3 ( x 2 − x − 2) = x 3 ( x − 2)( x + 1) 2

3+ 2 2 ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ f⎜ = 2⎜ + 3⎜ = ⎟ ⎟ ⎟ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 2

12. 3x 2 − 3x − 60 = 3( x 2 − x − 20) = 3 ( x − 5)( x + 4)

( ) ( )

5 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ f ⎜− ⎟ = 2 ⎜− ⎟ + 3 ⎜− ⎟ = − ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ 16

f ( x) = x 2 − 2 f (a − 2) = (a − 2) 2 − 2 = a 2 − 4a + 2 1 − x2 x +1 1 f (a + 1) = − (a + 1) 2 (a + 1) + 1 1 = − (a + 1) 2 a+2

16. y = −2 x 2 − x + 2 ⇒ −2 x 2 − x + 2 = 0.

17. Substitute 2x − 1 for y in the quadratic equation, then find the zeros:

1 ⇒ x ≠ 0, −3 x( x + 3)

5 x 2 − 3x − 2 = 2 x − 1 ⇒ 5 x 2 − 5 x − 1 = 0. −b ± b 2 − 4ac 5 ± (−5) 2 − 4(5)(–1) = 2a 2(5) 5±3 5 = 10 Now find the y-values for each x value: ⎛5+ 3 5 ⎞ 3 5 y = 2x − 1 = 2 ⎜ −1 = ⎝ 10 ⎟⎠ 5 x=

f ( x) = x − 1 ⇒ x ≥ 1

f ( x) = x 2 + 1 , all values of x

f ( x) =

1 , x>0 3x

⎛5− 3 5 ⎞ −3 5 −1 = y = 2x − 1 = 2 ⎜ ⎝ 10 ⎟⎠ 5 Points of intersection: ⎛5+ 3 5 3 5 ⎞ ⎛5− 3 5 3 5⎞ , ,− ⎟ , ⎝⎜ ⎟ ⎝⎜ 10 ⎠ 5 10 5 ⎠

x −1 x2 +1

⎛1⎞ h⎜ ⎟ = ⎝2⎠

−b ± b 2 − 4ac 3 ± (−3) 2 − 4(5)(−2) = 2a 2(5) 3± 7 2 = ⇒ x = 1 or x = − 10 5

9. h( x ) =

15. y = 5 x 2 − 3 x − 2 ⇒ 5 x 2 − 3 x − 2 = 0.

−b ± b 2 − 4ac 1 ± (−1) 2 − 4(−2)(2) = 2a 2(−2) 1 ± 17 −1 + 17 −1 − 17 or x = = ⇒x= 4 4 −4

f ( x) =

( 12 ) − 1 = − 3 2 ( 12 ) + 1 5 2

3⎞ ⎛1 Yes, the point ⎜ , − ⎟ is on the graph. ⎝2 5⎠

Chapter 0 Functions

18. Substitute x − 5 for y in the quadratic equation, then find the zeros:

1− x 2 − 1 + x 3x + 1 (1 − x )(3x + 1) − 2(1 + x ) = (1 + x )(3x + 1) 3x 2 + 1 =− (1 + x)(3 x + 1) 3x 2 + 1 =− 2 3x + 4 x + 1

27. g ( x) − h( x) =

−x2 + x + 1 = x − 5 ⇒ x2 − 6 = 0 ⇒ x = ± 6 Now find the y-values for each x value: y = x−5= 6 −5 y = − 6 −5 Points of intersection:

( 6, 6 − 5) , (− 6, – 6 − 5)

( ) 20. f ( x) − g ( x) = ( x 2 − 2 x ) − (3 x − 1) 19.

f ( x) + g ( x) = x 2 − 2 x + (3 x − 1) = x 2 + x − 1

28.

f ( x ) + h( x ) = =

2 +1 3 x x −1 x(3 x + 1) + 2 x 2 − 1 2

21.

(

f ( x ) h( x ) = x − 2 x 2

1/ 2

)( x )

= x ⋅ x − 2x ⋅ x = x5 / 2 − 2x3 / 2

22.

23.

= 1/ 2

f ( x) x 2 − 2 x = = x 3 / 2 − 2 x1/ 2 h( x ) x

30.

1− x x −1 1+ x x − ( x − 1)(1 − x) x

f ( x) − g ( x + 1) = = =

f ( x) + g ( x) = =

−

g ( x) =

1 − ( x + 1) 2 x − 1 1 + ( x + 1) x ( x + 2) − (− x) x 2 − 1 x

−

(

( x − 1) ( x + 2) 2

( x 2 − 1) (3x + 1)

x + x2 + x

( x 2 − 1) ( x + 2)

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

For exercises 31−36, f ( x ) = x 2 − 2 x + 4,

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

26.

5x 2 + x − 2

1− x 2 − 1 + x 3( x − 3) + 1 (1 − x)(3x − 8) − 2(1 + x) = (1 + x)(3x − 8) 2 −3 x + 9 x − 10 = (1 + x )(3x − 8) −3 x 2 + 9 x − 10 = 3x 2 − 5 x − 8

f ( x) g ( x) = ( x 2 − 2 x)(3 x − 1) = 3x 3 − x 2 − 6 x 2 + 2 x = 3x 3 − 7 x 2 + 2 x

f ( x) − g ( x) =

)

29. g ( x) − h( x − 3) =

24. g ( x)h( x) = (3 x − 1) x = 3 x ⋅ x1/ 2 − x1/ 2 = 3 x 3 / 2 − x1/ 2 25.

(

( x 2 − 1) (3x + 1)

= x 2 − 5x + 1 2

)

1 x

and h ( x ) =

1 . x −1 2

31.

⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ f ( g ( x )) = f ⎜ 2 ⎟ = ⎜ 2 ⎟ − 2 ⎜ 2 ⎟ + 4 ⎝x ⎠ ⎝x ⎠ ⎝x ⎠ 1 2 = 4 − 2 +4 x x

32. g ( f ( x )) = g ( x 2 − 2 x + 4) = ⎛ 1 ⎞ 33. g ( h ( x )) = g ⎜ = ⎝ x − 1 ⎟⎠ = x − 2 x +1 =

( x 2 − 2 x + 4) 1

( ) 1 x −1

( x − 1)

= 2

1 1 x − 2 x +1

Chapter 0 Review Exercises

⎛ 1 ⎞ 34. h ( g ( x )) = h ⎜ 2 ⎟ = ⎝x ⎠

35.

1 1 −1 x2

x 1 = −1 1− x x

= 1

42.

⎛ 1 ⎞ f ( h ( x )) = f ⎜ ⎝ x − 1 ⎟⎠ 2

⎛ 1 ⎞ ⎛ 1 ⎞ =⎜ − 2⎜ +4 ⎝ x − 1 ⎟⎠ ⎝ x − 1 ⎟⎠ 1 2 = − +4 2 x −1 x −1

(

37.

44.

45. a.

⎛1⎞ (0.25) −1 = ⎜ ⎟ ⎝4⎠

12 ⋅ 2

b. −1

≈ 18314.94

At the end of 2 years, the account balance is about $16,247. At the end of 5 years, the account balance is about $18,315. 46. a.

P = 7000, r = .09, m = 2 ⎛ .09 ⎞ A (t ) = 7000 ⎜1 + ⎟ ⎝ 2 ⎠

P (t ) = 750 + 25t + .1t 2 C ( P (t )) = 1 + .4 750 + 25t + .1t = 1 + 300 + 10t + .04t 2 = .04t 2 + 10t + 301

A (10) = 7000 (1.045)

= 7000 (1.045)

2 ⋅10

= 16882.00

A ( 20) = 7000 (1.045) = 40714.55 At the end of 10 years, the account balance is about $16,882. At the end of 20 years, the account balance is about $40,715.

47. a.

)

2 ⋅ 20

P = 15000, m = 1, t = 10 A ( r ) = 15000 (1 + r )

A (.04) = 15000 (1 + .04) = 22203.66 10

A (.06) = 15000 (1 + .06) = 26862.72 10

R( x) = 5 x − x 2 200 ⎞ ⎛ f (d ) = 6 ⎜1 − ⎝ d + 200 ⎟⎠ 200 ⎞ ⎛ R ( f ( d )) = 5 ⋅ 6 ⎜1 − ⎝ d + 200 ⎟⎠

48. a.

P = 7000, m = 1, t = 20 A ( r ) = 7000 (1 + r )

b. 2

⎡ ⎛ 200 ⎞ ⎤ − ⎢6 ⎜1 − ⎟ ⎝ d + 200 ⎠ ⎥⎦ ⎣ 2 200 ⎞ 200 ⎞ ⎛ ⎛ = 30 ⎜1 − − − 36 1 ⎝ d + 200 ⎠⎟ ⎝⎜ d + 200 ⎠⎟

41.

≈ 16247.14

12 ⋅5

(

⎛ .04 ⎞ A ( 2) = 15000 ⎜1 + ⎝ 12 ⎟⎠ ⎛ .04 ⎞ A (5) = 15000 ⎜1 + ⎝ 12 ⎟⎠

39. C(x) = carbon monoxide level corresponding to population x P(t) = population of the city in t years C(x) = 1 + .4x

40.

P = 15000, r = .04, m = 12

12t

( 100 ) = 1000 1/3 (.001) = ( 3 .001) = .1

38. (100) 3/ 2 =

x (8 x 2 / 3 ) = x1/ 3 ⋅ 8 x 2 / 3 = 8 x

= 15000 (1.00333)

−1

12t

( ) = 27 5 85/3 = ( 3 8 ) = 2 5 = 32

(81) 3/ 4 = 4 81

⎛ .04 ⎞ A (t ) = 15000 ⎜1 + ⎝ 12 ⎟⎠

( x − 2x + 4 − 1)

= x ⋅ x 5 ⋅ y 3 ⋅ y −6 =

x3 / 2 = x 3 / 2 ⋅ x −1/ 2 = x x

)

x − 2x + 4 − 1

−5 6

43.

)

36. h ( f ( x )) = h x 2 − 2 x + 4 =

xy 3

A (.07 ) = 7000 (1 + .07 ) A (.12) = 7000 (1 + .12)

( x + 1) = ( x + 1) 4 / 2 = ( x + 1) 2 = x 2 + 2 x + 1 4

= 27087.79

= 67524.05

Chapter 1 The Derivative 1.1

The Slope of a Straight Line

1. y = 3 − 7 x ; y-intercept: (0, 3), slope: −7 3x + 1 3 1 ⎛ 1⎞ 2. y = = x + ; y-intercept: ⎜ 0, ⎟ , ⎝ 5⎠ 5 5 5 3 slope: 5

x+3 1 3 ⇒ y = x+ ; 2 2 2 3 1 ⎛ ⎞ y-intercept: ⎜ 0, ⎟ , slope: ⎝ 2⎠ 2

3. x = 2 y − 3 ⇒ y =

4. y = 6 ⇒ y = 0 x + 6; y-intercept: (0, 6), slope: 0 x 1 5. y = − 5 ⇒ y = x − 5; y-intercept: (0, −5), 7 7 1 slope: 7 −4 x − 1 4 1 ⇒ y=− x− ; 6. 4 x + 9 y = −1 ⇒ y = 9 9 9 1⎞ 4 ⎛ y-intercept: ⎜ 0, − ⎟ , slope = − ⎝ 9⎠ 9

7. slope = –1, (7, 1) on line. Let (x, y) = (7, 1), m = –1. y − y1 = m ( x − x1 ) ⇒ y − 1 = −( x − 7) ⇒ y = −x + 8 8. slope = 2; (1, −2) on line. Let (x, y) = (1, −2), m = 2. y − y1 = m ( x − x1 ) ⇒ y + 2 = 2( x − 1) ⇒ y = 2x − 4 1 9. slope = ; (2, 1) on line. 2

1 y − y1 = m ( x − x1 ) ⇒ y − 1 = ( x − 2) ⇒ 2 1 y= x 2 7 ⎛1 2⎞ ; ⎜ , − ⎟ on line. 3 ⎝4 5⎠ 7 ⎛1 2⎞ Let ( x1 , y1 ) = ⎜ , − ⎟ ; m = . ⎝4 5⎠ 3

2 7⎛ 1⎞ = ⎜x − ⎟ ⇒ ⎝ 5 3 4⎠

7 59 x− 3 60

⎛5 ⎞ ⎛ 5 ⎞ 11. ⎜ , 5 ⎟ and ⎜ − , −4 ⎟ on line. ⎝7 ⎠ ⎝ 7 ⎠ slope =

y 2 − y1 −4 − 5 −9 63 = 5 5 = 10 = x 2 − x1 − − 10 − 7

63 ⎛5 ⎞ Let ( x1 , y1 ) = ⎜ , 5 ⎟ , m = . ⎝7 ⎠ 10 y − y1 = m ( x − x1 ) ⇒ y − 5 =

63 ⎛ 5⎞ ⎜⎝ x − ⎟⎠ 10 7

⎛1 ⎞ 12. ⎜ ,1⎟ and (1, 4) on line. ⎝2 ⎠

slope=

y 2 − y1 4 − 1 3 = = =6 x 2 − x1 1 − 12 12

Let ( x1 , y1 ) = (1, 4), m = 6.

y − y1 = m ( x − x1 ) ⇒ y − 4 = 6 ( x − 1) ⇒ y = 6x − 2

13. (0, 0) and (1, 0) on line. y − y1 0 − 0 slope = 2 = =0 x 2 − x1 1 − 0 y − 0 = 0( x − 0) ⇒ y = 0 ⎛ 1 1⎞ ⎛2 ⎞ 14. ⎜ − , − ⎟ and ⎜ ,1⎟ on line. ⎝ 2 7⎠ ⎝3 ⎠

slope =

( ) 3 ( 2)

1 8 y 2 − y1 1 − − 7 48 7 = 2 = = =m 7 x 2 − x1 49 − −1 6

⎛2 ⎞ Let ( x1 , y1 ) = ⎜ ,1⎟ . ⎝3 ⎠

y − y1 = m ( x − x1 ) ⇒ y − 1 =

1 Let ( x1 , y1 ) = ( 2,1) ; m = . 2

10. slope =

y − y1 = m ( x − x1 ) ⇒ y +

48 ⎛ 2⎞ ⎜⎝ x − ⎟⎠ ⇒ 49 3

48 17 x+ 49 49

15. Horizontal through (2, 9). Let ( x1 , y1 ) = (2, 9), m = 0 (horizontal line). y − y1 = m ( x − x1 ) ⇒ y − 9 = 0 ( x − 2) ⇒ y=9

Section 1.1 The Slope of a Straight Line

16. x-intercept is 1; y-intercept is –3. The intercepts (1, 0) and (0, –3) are on the line. y − y1 −3 − 0 slope = 2 = =3=m x 2 − x1 0 −1 y-intercept (0, b) = (0, –3) y = mx + b ⇒ y = 3 x − 3 17. x-intercept is −π ; y-intercept is 1. The intercepts (−π , 0) and (0, 1) are on the line. y 2 − y1 1− 0 1 = = x 2 − x1 0 − (−π ) π y-intercept (0, b) = (0, 1) x y = mx + b ⇒ y = + 1 slope=

23. Parallel to y = 3 x + 7 ; x-intercept is 2. slope = m = 3. Let ( x1 , y1 ) = (2, 0).

y − y1 = m ( x − x1 ) ⇒ y − 0 = 3 ( x − 2) ⇒ y = 3x − 6

24. Parallel to y − x = 13 ; y-intercept is 0. y = x + 13 , slope = m = 1, b = 0. y = mx + b ⇒ y = x 25. Perpendicular to y + x = 0; (2, 0) on line. y + x = 0 ⇒ y = − x ⇒ slope = m1 = −1 m1 ⋅ m2 = −1 ⇒ −1 ⋅ m2 = −1 ⇒ m2 = 1 Let ( x 2 , y 2 ) = (2, 0). y − y 2 = m2 ( x − x 2 ) ⇒ y − 0 = x − 2 ⇒ y = x−2

18. Slope = 2; x-intercept is –3. The x-intercept (–3, 0) is on the line. Let ( x1 , y1 ) = (−3, 0), m = 2.

26. Perpendicular to y = −5 x + 1; (1, 5) on line. slope = m1 = −5 m1 ⋅ m2 = −1 ⇒ −5m2 = −1 ⇒ m2 =

y − y1 = m ( x − x1 ) ⇒ y − 0 = 2 ( x + 3) ⇒ y = 2x + 6

y − y1 = m ( x − x1 ) ⇒ y − 0 = −2 ( x + 2) ⇒ y = −2 x − 4

( 7, 2). Let ( x1 , y1 ) = ( 7, 2) , m = 0 (horizontal line). y − y1 = m ( x − x1 ) ⇒ y − 2 = 0 ( x − 7 ) ⇒

y − y 2 = m2 ( x − x 2 ) ⇒ y − 5 = y=

21. Parallel to y = x; (2, 0) on line. Let ( x1 , y1 ) = (2, 0); slope = m = 1.

y − y1 = m ( x − x1 ) ⇒ y − 0 = 1( x − 2) ⇒ y = x−2

22. Parallel to x + 2y = 0; (1, 2) on line. 1 1 x + 2 y = 0 ⇒ y = − x; m = − 2 2 Let ( x1 , y1 ) = (1, 2).

y=−

1 5 x+ 2 2

1 24 x+ 5 5

27.

Start at (1, 0), then move one unit right and one unit up to (2, 1).

28.

Start at (−1, 1), then move one unit up and two units to the right.

20. Horizontal through

y=2

1 5

Let ( x 2 , y 2 ) = (1, 5).

19. Slope = −2; x-intercept is –2. The x-intercept (–2, 0) is on the line. Let ( x1 , y1 ) = (−2, 0), m = −2.

y − y1 = m ( x − x1 ) ⇒ y − 2 = −

1 ( x − 1) ⇒ 5

29. Start at (1, −1), then move one unit up and three units to the left. Alternatively, move one unit down and three units to the right.

1 ( x − 1) ⇒ 2

Chapter 1 The Derivative

30. Start at (0, 2), then move zero units up and any distance (for example, one unit) right.

38. Slope = –3, (2, 2) on line. x1 = 2, y1 = 2 If x = 3, then y − 2 = −3(3 − 2) ⇒ y = −1. If x = 4, then y − 2 = −3(4 − 2) ⇒ y = −4. If x = 1, then y − 2 = −3(1 − 2) ⇒ y = 5. The points are (3, −1), (4, −4), and (1, 5). 39.

f ( 2) = 1 ⇒ (2,1) lies on the line. Thus, the

31. (a)–(C) x- and y-intercepts are 1. (b)–(B) x-intercept is 1, y-intercept is –1. (c)–(D) x- and y-intercepts are –1. (d)–(A) x-intercept is –1, y-intercept is 1. 1 1 x⇒m=− 2 2 The slope of the line through (−1, 2) and 1 (3, b) is also − . . 2 1 b−2 m=− = ⇒ −4 = 2b − 4 ⇒ b = 0 2 3 − (−1)

1− 0 = 1 . If x = 3 and 2 −1 y −1 y = f (3) , then 1 = ⇒ 1 = y − 1 ⇒ y = 2. 3− 2 Thus f (3) = 2.

slope of the line is

32. x + 2 y = 0 ⇒ y = −

40. First find the slope of 2x + 3y = 0. 2 x + 3 y = 0 ⇒ 3 y = −2 x ⇒ y = −

1 2

2 3 Now find the slope of the line through (3, 4) and (–1, 2). y − y1 2 − 4 −2 1 m2 = 2 = = = x 2 − x1 −1 − 3 −4 2 Since the slopes are not equal, the lines are not parallel.

41. l1

1 unit in the x-direction, then you 2 1 must move ⋅ 2 = 1 unit in the y-direction. 2

If you move

35. m = −3, h = .25 If you move .25 unit in the x-direction, then you must move −3 ⋅ .25 = −.75 unit in the y-direction. 2 1 , h= 3 2 1 If you move unit in the x-direction, then you 2 1 2 1 must move ⋅ = unit in the y-direction. 2 3 3

36. m =

37. Slope = 2, (1, 3) on line. x1 = 1, y1 = 3 If x = 2, then y − 3 = 2(2 − 1) ⇒ y = 5. If x = 3, then y − 3 = 2(3 − 1) ⇒ y = 7. If x = 0, then y − 3 = 2(0 − 1) ⇒ y = 1. The points are (2, 5), (3, 7), and (0, 1).

2 x⇒ 3

m1 = −

1 33. m = , h = 3 3 If you move 3 units in the x-direction, then you must move 1 unit in the y-direction to return to the line.

34. m = 2, h =

f (1) = 0 ⇒ (1, 0) lies on the line.

43.

Slope = m = –2 y-intercept: (0, –1) y = mx + b y = –2x – 1

44.

1 3 y-intercept: (0, 1) y = mx + b 1 y = x +1 3

42. l 2

Slope = m =

45. a is the x-coordinate of the point of intersection of y = –x + 4 and y = 2. Use substitution to find the x-coordinate. 2 = −x + 4 ⇒ x = 2 So a = 2. f(a) is the y-coordinate of the intersection point. So f(a) = 2.

Section 1.1 The Slope of a Straight Line

46. a is the x-coordinate of the point of 1 intersection of y = x and y = x + 1. Use 2 substitution to find the x-coordinate. 1 1 x = x +1 ⇒ x = 1 ⇒ x = 2 2 2 So a = 2. f(a) is the y-coordinate of the intersection point. Substituting x = 2 into y = x gives y = 2. So f(a) = 2. 47. C ( x) = 12 x + 1100 a.

C (10) = 12(10) + 1100 = $1220

b. The marginal cost is the slope of line. Marginal cost = m = $12/unit c.

It would cost an additional $12 to raise the daily production level from 10 units to 11 units.

48. C ( x + 1) − C ( x) = (12 ( x + 1) + 1100) − (12 x + 1100) = 12 x + 12 + 1100 − 12 x − 1100 = 12 $12 is the marginal cost. It is the additional cost incurred when the production level of this commodity is increased one unit, from x to x + 1, per day. 49. Let x be the number of months since January 1, 2020. Then (0, 3.19) is one point on the line. The slope is –.04 since the price fell $.04 per month. Therefore, P ( x) = −.04 x + 3.19 gives the price of gasoline x months after January 1, 2020. On April 1, 2020, 3 months later, the cost of one gallon of gasoline is: P (3) = −.04(3) + 3.19 = $3.07 / gallon. So, 15 gallons cost 15 ⋅ 3.07 = $46.05. On September 1, 2020, 8 months after January 1, the cost of one gallon of gasoline is: P (8) = −.04(8) + 3.19 = $2.87 / gallon. So, 15 gallons cost 15 ⋅ 2.87 = $43.05. 50. Let y be the value of monthly exports in millions of dollars. Let x be the number of months since Sept 1, 2003. Since the rate of change of y is constant, we conclude that y is a linear function of x whose slope is equal to its rate of change, m = 42.5. On September 1 (when x = 0), the value of monthly exports was 0 dollars, since the ban had just ended. So the point (0, 0) is on the graph of y. Using the point-slope form, we have y − 0 = 42.5 ( x − 0) ⇒ y = 42.5 x.

The end of December 2003 corresponds to x = 4, at which time the exports had reached the value y = 42.5 ⋅ 4 = 170 million dollars 51. Let x = the cost of order. Then C ( x ) = .03 x + 5. 52. a.

The points (7.25, .2) and (8, .18) are on the line. The slope of the line is y − y1 .18 − .2 2 = =− m= 2 x2 − x1 8 − 7.25 75 Let ( x1 , y1 ) = (8,.18). Then, the equation of the line is y − y1 = m ( x − x1 ) 2 y − .18 = − ( x − 8) 75 2 59 y=− x+ 75 150 2 59 Thus, Q( x) = − x + . 75 150

b. Let Q(x) = .1 (10 employees per 100) and solve for x. 2 59 .1 = − x + 75 150 22 2 − = − x ⇒ x = 11 75 75 The hourly wage should be $11 in order for the quit ratio to drop to 10 employees per 100. 53. The points (3.10, 1500) and (3.25, 1250) are on the line. The slope of the line is y − y1 1500 − 1250 5000 m= 2 = =− . x2 − x1 3.10 − 3.25 3 Let ( x1 , y1 ) = (3.10,1500). The equation of the line is y − y1 = m ( x − x1 ) 5000 y − 1500 = − ( x − 3.10) 3 5000 20, 000 y=− x+ 3 3 5000 20, 000 G ( x) = − x+ 3 3 Now find G(3.34): 5000 20, 000 G (3.34) = − = 1100 (3.34) + 3 3 gallons.

Chapter 1 The Derivative

54. Solve for x: 5000 20, 000 G ( x) = 2200 = − x+ 3 3 3 ⎞ ⎛ 13, 400 ⎞ ⎛ x = ⎜− ⎟ ⎜− ⎟ = 2.68 ⎝ 3 ⎠ ⎝ 5000 ⎠ The owner should set the price at $2.68 in order to sell 2200 gallons per day.

55. a.

C ( x) = mx + b

b = $1500 (fixed costs) Total cost of producing 100 rods is $2200. C (100) = m(100) + 1500 = $2200 ⇒ m = 7 Thus, C ( x) = 7 x + 1500.

b. Marginal cost at x = 100 is m = $7/rod c.

Since the marginal cost = $7, the cost of raising the daily production level form 100 to 101 rods is $7. Alternatively, C (101) − C (100) = 2207 − 2200 = $7.

56. Each unit sold increases the pay by 5 dollars. Thus, the slope is her pay per unit sold. The weekly pay is 60 dollars if no units are sold. Thus, the y-intercept is her base pay. 57. If the monopolist wants to sell one more unit of goods, then the price per unit must be lowered by 2 cents. No one will pay 7 dollars or more for a unit of goods. 58. x = degrees Fahrenheit, y = degrees Celsius, so the points (32, 0) and (212, 100) lie on the line. 100 − 0 100 5 = = m= 212 − 32 180 9 Now find b: 5 160 . y = mx + b ⇒ 32 = (0) + b ⇒ b = − 9 9 9 5 160 = 37 Thus, y = x + 32. y = (98.6) − 5 9 9 98.6°F corresponds to 37°C. 59. The point (0, 1.5) is on the line and the slope is 6 (ml/min). Let y be the amount of drug in the body x minutes from the start of the infusion. Then y − 1.5 = 6( x − 0) ⇒ y = 6 x + 1.5. 60. Eliminating 2 ml/hour means that the rate is 1 − ml/min. (The rate given in exercise 59 is 30 1 179 in ml/min.) y = 6 x + 1.5 − x= x + 1.5 30 30

61. The diver starts at a depth of 212 ft, which is represented as −212. Thus, the function is y (t ) = 2t − 212. 62. First we must determine how long it will take the diver to reach 150 feet depth. −150 = 2t − 212 ⇒ 62 = 2t ⇒ t = 31 sec The diver must then rest for 5 minutes or 5 ⋅ 60 = 300 sec, which is 331 sec after she started ascending. The remaining depth can be determined by y = 2 (t − 331) − 150 = 2t − 812.

Thus, the function giving depth as a function of time is ⎧2t − 212 0 ≤ t ≤ 31 ⎪ y (t ) = ⎨−150 31 ≤ t ≤ 331 ⎪2t − 812 t ≥ 331 ⎩ The first 62 ft will take the diver 31 sec to ascend. To determine how long it will take the diver to ascend final 150 ft, solve 150 = 2t ⇒ t = 75 sec. Therefore, it will take the diver 31 + 300 + 75 = 406 sec to reach the surface. 63. a. b.

C ( x ) = 7 x + 230 R ( x ) = 12 x

64. C ( x ) = R ( x ) ⇒ 7 x + 230 = 12 x ⇒ 230 = 5 x ⇒ x = 46 The business will break even when 46 t-shirts are sold. 65. Using

f ( x 2 ) − f ( x1 ) = m and the hint, x 2 − x1

f ( x) − f ( x1 ) = m ⇒ f ( x) − f ( x1 ) = m( x − x1 ) x − x1 f ( x) = m( x − x1 ) + f ( x1 ) = mx + (− mx1 + f ( x1 )) Let b = − mx1 + f ( x1 ) . Then f ( x) = mx + b .

66. a–c.

f (3 + h) − f (3) f (3 + h) − 2 = 3+ h −3 h

Section 1.2 The Slope of a Curve at a Point

67. a.

The points (0, 54) and (36, 66) lie on the line. y − y1 66 − 54 1 slope = 2 = = x 2 − x1 36 − 0 3 1 1 y − 54 = ( x − 0) ⇒ y = x + 54 3 3

Graphing the line y = 5000 and using the INTERSECT command, the point ($60,138.25, 1600) is on both lines. An average itemized deduction of $1600 corresponds to a reported income of $60,138.25.

An increase of $15,000 in income level will correspond to an increase of $15,000(.0217) = $325.50 in itemized deductions.

b. Every year since 2014, 13 % = .33% more

of the world population becomes urban. c.

68. a.

The year 2020 is represented by x = 6. 1 f (6) = (6) + 54 = 56 3 Thus, in 2020, 56% of the world’s population will be urban. 1 1 72 = x + 54 ⇒ 18 = x ⇒ x = 54 3 3 72% of the world’s population will be urban 54 years after 2014, or in 2068.

(20,000, 729) and (50,000, 1380) on line. y − y1 1380 − 729 slope = 2 = x2 − x1 50, 000 − 20, 000 651 = = .0217 30, 000 y − 1380 = .0217 ( x − 50, 000) ⇒ y = .0217 x + 295

1.2

The Slope of a Curve at a Point

1. −

4 3

2. 0

3. 1

4. 1

5. 1

7. –2

8. −

1 2 1 3

9. Small positive slope; large positive slope 10. Zero slope; large negative slope 11. Zero slope; small negative slope 12. Let m P = slope at point P. Then, m A = 1, m B = 8, 1 mC = 0, mD = −6, mE = 0, mF = − . 2

For 13–24, note that the slope of the line tangent to the graph of y = x 2 at the point (x, y) is 2x. 13. The slope at (–.4, .16) is 2(–.4) = –.8. Let ( x1 , y1 ) = (−.4, .16), m = –.8.

[0, 75000] by [0, 2000] c.

For every increase of $1 in reported income, the average itemized deductions increase by $.0217. (Alternatively, an increase of $100 in reported income corresponds to an average increase of $2.17 in itemized deductions.)

d. Using the TRACE or VALUE feature on a graphing calculator, the point (75,000, 1992.5) is on the line. Thus, the average amount of itemized deductions on a return reporting income of $75,000 is $1922.50.

y − .16 = −.8 ( x − ( −.4)) y − .16 = −.8 ( x + .4) y = −.8 x − .16

14. The slope at (–2, 4) is 2x = 2(–2) = –4. Let ( x1 , y1 ) = (–2, 4), m = –4 .

y − 4 = −4 ( x − ( −2)) ⇒ y − 4 = −4 ( x + 2) ⇒ y = −4 x − 4