37. f x) = f is discontinuous at x) = f x) = - x = -2 is a...

EXERCISE 3-3 103

37. f(x) =

!

2x

x + 2; f is discontinuous at x = -2

!

limx"#2#

f(x) = ∞,

!

limx"#2+

f(x) = -∞; x = -2 is a vertical asymptote.

!

limx"#

!

2x

x + 2 =

!

limx"#

!

2x

x = 2; y = 2 is a horizontal asymptote.

39. f(x) =

!

x2 + 1

x2 " 1 =

!

x2 + 1

(x " 1)(x + 1); f is discontinuous at x = -1, x = 1.

At x = -1:

!

limx"#1#

f(x) = ∞,

!

limx"#1+

= -∞; x = -1 is a vertical asymptote.

At x = 1:

!

limx"1#

f(x) = -∞,

!

limx"1+

f(x) = ∞; x = 1 is a vertical asymptote.

!

limx"#

!

x2 + 1

x2 " 1 =

!

limx"#

!

x2

x2 = 1; y = 1 is a horizontal asymptote.

41. f(x) =

!

x3

x2 + 6. Since x2 + 6 ≠ 0 for all x, f is continuous for all x;

there are no vertical asymptotes.

!

limx"#

!

x3

x2 + 6 =

!

limx"#

!

x3

x2 =

!

limx"#

x = ∞; there are no horizontal asymptotes.

43. f(x) =

!

x

x2 + 4. Since x2 + 4 ≠ 0 for all x, f is continuous for all x;

there are no vertical asymptotes.

!

limx"#

!

x

x2 + 4 =

!

limx"#

!

x

x2 =

!

limx"#

!

1

x = 0; y = 0 is a horizontal asymptote.

45. f(x) =

!

x2

x " 3; f is discontinuous at x = 3.

At x = 3:

!

limx"3#

f(x) = -∞,

!

limx"3+


!

limx"#

!

x2

x " 3 =

!

limx"#

!

x2

x =

!

limx"#

x = ∞; there are no horizontal asymptotes.

47. f(x) =

!

2x2 + 3x " 2

x2 " x " 2 =

!

(x + 2)(2x " 1)

(x " 2)(x + 1); f is discontinuous at x = -1,

x = 2.

At x = -1:

!

limx"#1#

f(x) = -∞,

!

limx"#1+

f(x) = ∞; x = -1 is a vertical asymptote.

104 CHAPTER 3 LIMITS AND THE DERIVATIVE

At x = 2:

!

limx"2#

f(x) = -∞,

!

limx"2+


!

limx"#

!

2x2 + 3x " 2

x2 " x " 2 =

!

limx"#

!

2x2


49. f(x) =

!

2x2 " 5x + 2

x2 " x " 2 =

!

(x " 2)(2x " 1)

(x " 2)(x + 1); f is discontinuous at x = -1, x = 2.

At x = -1:

!

limx"#1#

f(x) = ∞,

!

limx"#1+

f(x) = -∞; x = -1 is a vertical asymptote.

At x = 2:

!

limx"2

f(x) =

!

limx"2

!

2x " 1

x + 1

#

$ %

&

' ( = 1

!

limx"#

!

2x2 " 5x + 2

x2 " x " 2 =

!

limx"#

!

2x2


51.

!

x2 + 2x " 3

x " 1 =

!

(x " 1)(x + 3)

x " 1 = x + 3 provided x ≠ 1

!

limx"1

!

x2 + 2x " 3

x " 1 =

!

limx"1

(x + 3) = 4; set a = 4.

53.

!

x2 + 4x + 3

x + 3 =

!

(x + 3)(x + 1)

x + 3 = x + 1, provided x ≠ -3

!

limx"#3

!

x2 + 4x + 3

x + 3 =

!

limx"#3

(x + 1) = -2; set a = -2.

55. If n ≥ 1 and an > 0, then

!

limn"#

(anxn + an-1x

n-1 + … + a0) = ∞

If n ≥ 1 and an < 0, then

!

limn"#

(anxn + an-1x

n-1 + … + a0) = -∞

57. f(x) = 2x4 – 5x + 11

!

limx"#

f(x) =

!

limx"#

2x4 = ∞;

!

limx"#$

f(x) =

!

limx"#$

2x4 = ∞

59. f(x) = 7x2 + 9x3 + 5x

!

limx"#

f(x) =

!

limx"#

9x3 = ∞;

!

limx"#$

f(x) =

!

limx"#$

9x3 = -∞

EXERCISE 3-3 105

61. f(x) =

!

x2 " 5x " 7

x + 11

!

limx"#

f(x) =

!

limx"#

!

x2

x =

!

limx"#

x = ∞;

!

limx"#$

f(x) =

!

limx"#$

!

x2

x =

!

limx"#$

x = -∞

63. f(x) =

!

5x5 + 7x4 " 10

"x3 + 6x2 + 3

!

limx"#

f(x) =

!

limx"#

!

5x5

"x3 =

!

limx"#

-5x2 = -∞

!

limx"#$

f(x) =

!

limx"#$

!

5x5

"x3 =

!

limx"#$

-5x2 = -∞

65. (A) Since C(x) is a linear function of x, it can be written in the form C(x) = mx + b Since the fixed costs are $200, b = 200. Also, C(20) = 3800, so 3800 = m(20) + 200

20m = 3600 m = 180 Therefore, C(x) = 180x + 200

(B)

!

C(x) =

!

C(x)

x =

!

180x + 200

x

(D)

!

C(x) =

!

180x + 200

x =

!

180 +200

x1

(C)

C(x)

As x increases, the numerator tends to 180 and the denominator is 1. Therefore,

!

C(x) tends to 180 or $180 per board.

67. (A) Ce(x) = 950 + 56x;

!

Ce(x) =

!

Ce(x)

x =

!

950 + 56x

x =

!

950

x + 56

(B) Cc(x) = 900 + 66x;

!

Cc(x) =

!

Cc(x)

x =

!

900 + 66x

x =

!

900

x + 66

(C) Set Cc(x) = Ce(x) and solve for x: 900 + 66x = 950 + 56x 10x = 50 x = 5 The total costs for the two models are equal at x = 5 years.


(D) Set

!

Cc(x) =

!

Ce(x) and solve for x.

!

900

x + 66 =

!

950

x + 56

!

900 " 950

x = -10

-

!

50

x = -10

-10x = -50 x = 5 The average costs for the two models are equal at x = 5 years.

(E)

!

limx"#

!

Ce(x) =

!

limx"#

!

950

x+ 56

"

# $

%

& ' = 56

!

limx"#

!

Cc(x) =

!

limx"#

!

900

x+ 66

"

# $

%

& ' = 66

For large x, the energy efficient model is approximately $10 per year cheaper to operate than the conventional model.

69. C(t) =

!

5t2(t + 50)

t3 + 100 =

!

5t3 + 250t2

t3 + 100

!

limt"#

C(t) =

!

limt"#

!

5t3

t3 = 5; the long-term drug concentration is 5 mg/ml.

71. P(x) =

!

2x

1 " x, 0 ≤ x < 1

(A) P(0.9) =

!

2(0.9)

1 " 0.9 =

!

1.8

0.1 = 18; $18 million

(B) P(0.95) =

!

2(0.95)

1 " 0.95 =

!

1.9

0.05 = 38; $38 million

(C)

!

limx"1#

P(x) =

!

limx"1#

!

2x

1 " x = ∞; removal of 100% of the contamination

would require an infinite amount of money; impossible.

73. v(s) =

!

vmaxs

KM + s

(A)

!

lims"#

v(s) =

!

lims"#

!

vmaxs

KM + s =

!

lims"#

!

vmaxs

s = vmax

(B) v(KM) =

!

vmax " KM

KM + KM =

!

vmaxKM

2KM =

!

vmax

2

EXERCISE 3-3 107

(C)

0

0.5

1

1.5

2

2.5

3

3.5

4

0 50 100 150 200 250 300 350

Concentration

Velocity

!

Vmax

2= 2

v

!

s

!

Vmax = 4

!

KM =20

(D) v(s) =

!

4s

20 + s

(E) v(15) =

!

4(15)

20 + 15 =

!

60

35 =

!

12

7

Set v = 3 and solve for s:

3 =

!

4s

20 + s

60 + 3s = 4s s = 60 Thus, s = 60 when v = 3.

75. (A) Cmax = 18, M = 150

0

2

4

6

8

10

12

14

16

18

20

0 200 400 600 800 1000 1200

°K

CTE

!

Cmax

2= 9

!

Cmax = 18

M = 150

C

K

Kelvins

!!

T

(B) C(T) =

!

18T

150 + T

(C)

(D) C(600) =

!

18(600)

150 + 600 = 14.4

To find T when C(T) = 12, solve

!

18T

150 + T = 12 for T:

18T = 1800 + 12T 6T = 1800 T = 300 Thus, C(T) = 12 at T = 300K

500 1000

5

10

15

T

C

100 200 300

2

4

S

v


EXERCISE 3-4 Things to remember: 1. AVERAGE RATE OF CHANGE

For y = f(x), the AVERAGE RATE OF CHANGE FROM x = a TO x = a + h is

!

f(a + h) " f(a)

(a + h) " a =

!

f(a + h) " f(a)

h h ≠ 0

The expression

!

f(a + h) " f(a)

h is called the DIFFERENCE

QUOTIENT. 2. INSTANTANEOUS RATE OF CHANGE For y = f(x), the INSTANTANEOUS RATE OF CHANGE AT x = a is

!

limh"0

!

f(a + h) " f(a)

h

if the limit exists.

3. SECANT LINE

A line through two points on the graph of a function is called a SECANT LINE. If (a, f(a)) and ((a + h), f(a + h)) are two points on the graph of y = f(x), then

Slope of secant line =

!

f(a + h) " f(a)

h [Difference quotient]

4. SLOPE OF A GRAPH For y = f(x), the SLOPE OF THE GRAPH at the point (a, f(a))

is given by

!

limh"0

!

f(a + h) " f(a)

h

provided the limit exists. The slope of the graph is also the SLOPE OF THE TANGENT LINE at the point (a, f(a)).

5. THE DERIVATIVE For y = f(x), we define THE DERIVATIVE OF f AT x, denoted by

f'(x), to be

f'(x) =

!

limh"0

!

f(x + h) " f(x)

h if the limit exists.

If f'(x) exists for each x in the open interval (a, b), then f is said to be DIFFERENTIABLE OVER (a, b).

EXERCISE 3-4 109

6. INTERPRETATIONS OF THE DERIVATIVE The derivative of a function f is a new function f'. The

domain of f' is a subset of the domain of f. Interpretations of the derivative are:

a. Slope of the tangent line. For each x in the domain of f', f'(x) is the slope of the line tangent to the graph of f at the point (x, f(x)).

b. Instantaneous rate of change. For each x in the domain of f', f'(x) is the instantaneous rate of change of y = f(x) with respect to x.

c. Velocity. If f(x) is the position of a moving object at time x, then v = f'(x) is the velocity of the object at that time.

7. THE FOUR STEP PROCESS FOR FINDING THE DERIVATIVE OF A FUNCTION f. Step 1. Find f(x + h). Step 2. Find f(x + h) - f(x).

Step 3. Find

!

f(x + h) " f(x)

h.

Step 4. Find limh!0

!

f(x + h) " f(x)

h.

1. (A)

!

f(2) " f(1)

2 " 1 =

!

1 " 4

1 = -3 is the slope of the secant line through

(1, f(1)) and (2, f(2)).

(B)

!

f(1 + h) " f(1)

h =

!

5 " (1 + h)2 " 4

h =

!

5 " [1 + 2h + h2] " 4

h

=

!

"2h " h2

h = -2 - h;

slope of the secant line through (1, f(1)) and (1 + h, f(1 + h))

(C)

!

f(1 + h) " f(1)

h =

!

limh"0

(-2 - h) = -2;

slope of the tangent line at (1, f(1))

3. f(x) = 3x2

(A) Average rate of change:

!

f(4) " f(1)

4 " 1 =

!

3(4)2 " 3(1)2

3 =

!

48 " 3

3 = 15.

(B) Slope of the secant line:

!

f(4) " f(1)

4 " 1 = 15.


(C)

!

f(1 + h) " f(1)

1 + h " 1 =

!

3(1 + h)2 " 3(1)2

h =

!

3(1 + 2h + h2) " 3

h

=

!

6h + 3h2

h = 6 + 3h.

(D) limh!0

!

f(1 + h) " f(1)

h =

limh!0

(6 + 3h) = 6.

(E) 6. (F) 6. (G) f(1) = 3, f'(1) = 6; tangent line: y - 3 = 6(x - 1) or y = 6x - 3.

5. f(x) = -5

Step 1. Find f(x + h): f(x + h) = -5

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = -5 - (-5) = 0

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

0

h = 0

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

0 = 0

Thus, f'(x) = 0. f'(1) = 0, f'(2) = 0, f'(3) = 0

7. f(x) = 3x - 7

Step 1. Find f(x + h): f(x + h) = 3(x + h) - 7 = 3x + 3h - 7

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = 3x + 3h - 7 - (3x - 7) = 3h

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

3h

h = 3

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

3 = 3

Thus, f'(x) = 3. f'(1) = 3, f'(2) = 3, f'(3) = 3

EXERCISE 3-4 111

9. f(x) = 2 - 3x2

Step 1. Find f(x + h): f(x + h) = 2 - 3(x + h)2 = 2 - 3(x2 + 2xh + h2) = 2 - 3x2 - 6xh - 3h2

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = 2 - 3x2 - 6xh - 3h2 - (2 - 3x2) = -6xh - 3h2

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

"6xh " 3h2

h = -6x - 3h

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

(-6x - 3h) = -6x

Thus, f'(x) = -6x. f'(1) = -6, f'(2) = -12, f'(3) = -18

11. f(x) = x2 + 6x - 10

Step 1. Find f(x + h): f(x + h) = (x + h)2 + 6(x + h) - 10 = x2 + 2xh + h2 + 6x + 6h - 10

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = x2 + 2xh + h2 + 6x + 6h - 10 - (x2 + 6x - 10) = 2xh + h2 + 6h

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

2xh + h2 + 6h

h = 2x + h + 6

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

(2x + h + 6) = 2x + 6

Thus, f'(x) = 2x + 6. f'(1) = 8, f'(2) = 10, f'(3) = 12

13. f(x) = 2x2 - 7x + 3 Step 1. Find f(x + h):

f(x + h) = 2(x + h)2 - 7(x + h) + 3 = 2(x2 + 2xh + h2) - 7x - 7h + 3 = 2x2 + 4xh + 2h2 - 7x - 7h + 3


Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = 2x2 + 4xh + 2h2 - 7x - 7h + 3 - (2x2 - 7x + 3) = 4xh + 2h2 - 7h

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

4xh + 2h2 " 7h

h = 4x + 2h - 7

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

(4x + 2h - 7) = 4x - 7

Thus, f'(x) = 4x - 7. f'(1) = -3, f'(2) = 1, f'(3) = 5

15. f(x) = -x2 + 4x - 9

Step 1. Find f(x + h): f(x + h) = -(x + h)2 + 4(x + h) - 9 = -(x2 + 2xh + h2) + 4x + 4h - 9 = -x2 - 2xh - h2 + 4x + 4h - 9

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = -x2 - 2xh - h2 + 4x + 4h - 9 - (-x2 + 4x - 9) = -2xh - h2 + 4h

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

"2xh " h2 + 4h

h = -2x - h + 4

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

(-2x - h + 4) = -2x + 4

Thus, f'(x) = -2x + 4. f'(1) = 2, f'(2) = 0, f'(3) = -2

17. f(x) = 2x3 + 1

Step 1. Find f(x + h): f(x + h) = 2(x + h)3 + 1 = 2(x3 + 3x2h + 3xh2 + h3) + 1 = 2x3 + 6x2h + 6xh2 + 2h3 + 1

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = 2x3 + 6x2h + 6xh2 + 2h3 + 1 - (2x3 + 1) = 6x2h + 6xh2 + 2h3

EXERCISE 3-4 113

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

6x2h + 6xh2 + 2h3

h = 6x2 + 6xh + 2h2

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

(6x2 + 6xh + 2h2) = 6x2

Thus, f'(x) = 6x2. f'(1) = 6, f'(2) = 24, f'(3) = 54

19. f(x) = 4 +

!

4

x

Step 1. Find f(x + h):

f(x + h) = 4 +

!

4

x + h

Step 2. Find f(x + h) - f(x):

f(x + h) - f(x) = 4 +

!

4

x + h -

!

4 +4

x

"

# $

%

& ' =

!

4

x + h -

!

4

x

=

!

4x " 4(x + h)

x(x + h) = -

!

4h

x(x + h)

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

"4h

x(x + h)

h = -

!

4

x(x + h)

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

-

!

4

x(x + h) = -

!

4

x2

Thus, f'(x) = -

!

4

x2.

f'(1) = -4, f'(2) = -1, f'(3) = -

!

4

9

21. f(x) = 5 + 3

!

x

Step 1. Find f(x + h): f(x + h) = 5 + 3

!

x + h

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = 5 + 3

!

x + h - (5 + 3

!

x ) = 3(

!

x + h -

!

x )


Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

3( x + h " x)

h =

!

3( x + h " x)

h ·

!

( x + h + x)

( x + h + x)

=

!

3(x + h " x)

h( x + h + x) =

!

3h

h( x + h + x) =

!

3

x + h + x

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

!

3

x + h + x =

!

3

2 x

Thus, f'(x) =

!

3

2 x.

f'(1) =

!

3

2, f'(2) =

!

3

2 2=3 2

4, f'(3) =

!

3

2 3=

3

2

23. f(x) = 10

!

x + 5 Step 1. Find f(x + h):

f(x + h) = 10

!

x + h + 5

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = 10

!

x + h + 5 - 10

!

x + 5

= 10

!

x + h + 5 " x + 5( )

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

10 x + h + 5 " x + 5( )h

=

!

10 x + h + 5 " x + 5( )h

·

!

x + h + 5 + x + 5( )x + h + 5 + x + 5( )

=

!

10[x + h + 5 " (x + 5)]

h x + h + 5 + x + 5( ) =

!

10h

h x + h + 5 + x + 5( )

=

!

10

x + h + 5 + x + 5

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

!

10

x + h + 5 + x + 5 =

!

10

2 x + 5 =

!

5

x + 5

Thus, f'(x) =

!

5

x + 5.

f'(1) =

!

5

6=5 6

6, f'(2) =

!

5

7=5 7

7, f'(3) =

!

5

8=

5

2 2=5 2

4

EXERCISE 3-4 115

25. f(x) =

!

3x

x + 2

Step 1. Find f(x + h):

f(x + h) =

!

3(x + h)

x + h + 2

Step 2. Find f(x + h) - f(x):

f(x + h) - f(x) =

!

3(x + h)

x + h + 2 -

!

3x

x + 2

=

!

3(x + h)(x + 2) " 3x(x + h + 2)

(x + h + 2)(x + 2)

=

!

3x2 + 3xh + 6x + 6h " 3x2 " 3xh " 6x

(x + h + 2)(x + 2)

=

!

6h

(x + h + 2)(x + 2)

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

6h

(x + h + 2)(x + 2)

h =

!

6

(x + h + 2)(x + 2)

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

!

6

(x + h + 2)(x + 2) =

!

6

(x + 2)2

Thus, f'(x) =

!

6

(x + 2)2.

f'(1) =

!

2

3, f'(2) =

!

3

8, f'(3) =

!

6

25

27. y = f(x) = x2 + x (A) f(1) = 12 + 1 = 2, f(3) = 32 + 3 = 12

Slope of secant line:

!

f(3) " f(1)

3 " 1=12 " 2

2 = 5

(B) f(1) = 2, f(1 + h) = (1 + h)2 + (1 + h) = 1 + 2h + h2 + 1 + h = 2 + 3h + h2

Slope of secant line:

!

f(1 + h) " f(1)

h =

!

2 + 3h + h2 " 2

h = 3 + h

(C) Slope of tangent line at (1, f(1)):

limh!0

!

f(1 + h) " f(1)

h =

limh!0

(3 + h) = 3

(D) Equation of tangent line at (1, f(1)): y - f(1) = f'(1)(x - 1) or y - 2 = 3(x - 1) and y = 3x - 1.


29. f(x) = x2 + x

(A) Average velocity:

!

f(3) " f(1)

3 " 1 =

!

32 + 3 " (12 + 1)

2=12 " 2

2

= 5 meters/sec.

(B) Average velocity:

!

f(1 + h) " f(1)

h =

!

(1 + h)2 + (1 + h) " (12 + 1)

h

=

!

1 + 2h + h2 + 1 + h " 2

h

=

!

3h + h2

h = 3 + h meters/sec.

(C) Instantaneous velocity: limh!0

!

f(1 + h) " f(1)

h =

limh!0

(3 + h) = 3 m/sec.

31. F'(x) does exist at x = a.

33. F'(x) does not exist at x = c; the graph has a vertical tangent line at (c, F(c)).

35. F'(x) does exist at x = e; F'(e) = 0.

37. F'(x) does exist at x = g.

39. f(x) = x2 - 4x (A) Step 1. Find f(x + h):

f(x + h) = (x + h)2 - 4(x + h) = x2 + 2xh + h2 - 4x - 4h Step 2. Find f(x + h) - f(x):

f(x + h) - f(x) = x2 + 2xh + h2 - 4x - 4h - (x2 - 4x) = 2xh + h2 - 4h

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

2xh + h2 " 4h

h = 2x + h - 4

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

(2x + h - 4) = 2x - 4

Thus, f'(x) = 2x - 4.

EXERCISE 3-4 117

(B) f'(0) = -4, f'(2) = 0, f'(4) = 4

(C) Since f is a quadratic function, the graph of f is a parabola.

y intercept: y = 0 x intercepts: x = 0, x = 4 Vertex: (2, -4)

41. To find v = f'(x), use the four-step process on the position function

f(x) = 4x2 - 2x. Step 1. Find f(x + h):

f(x + h) = 4(x + h)2 - 2(x + h) = 4(x2 + 2xh + h2) - 2x - 2h

= 4x2 + 8xh + 4h2 - 2x - 2h Step 2. Find f(x + h) - f(x):

f(x + h) - f(x) = 4x2 + 8xh + 4h2 - 2x - 2h - (4x2 - 2x) = 8xh + 4h2 - 2h

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

!

8xh + 4h2 " 2h

h = 8x + 4h - 2

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

(8x + 4h - 2) = 8x - 2

Thus, the velocity, v(x) = f'(x) = 8x - 2 f'(1) = 8 · 1 - 2 = 6 ft/sec

f'(3) = 8 · 3 - 2 = 22 ft/sec f'(5) = 8 · 5 - 2 = 38 ft/sec

43. (A) The graphs of g and h are vertical translations of the graph of f. All three functions should have the same derivative.

(B) m(x) = x2 + C Step 1. Find m(x + h): m(x + h) = (x + h)2 + C

Step 2. Find m(x + h) - m(x): m(x + h) - m(x) = (x + h)2 + C - (x2 + C) = x2 + 2xh + h2 + C - x2 + C = 2xh + h2

Step 3. Find

!

m(x + h) " m(x)

h:

!

m(x + h) " m(x)

h =

!

2xh + h2

h = 2x + h


Step 4. limh!0

!

m(x + h) " m(x)

h:

limh!0

!

m(x + h) " m(x)

h =

limh!0

(2x + h) = 2x

Thus, m'(x) = 2x.

45. (A) The graph of f(x) = C, C a constant, is a horizontal line C units above or below the x axis depending on the sign of C. At any given point on the graph, the slope of the tangent line is 0.

(B) f(x) = C Step 1. Find f(x + h): f(x + h) = C

Step 2. Find f(x + h) - f(x): f(x + h) - f(x) = C - C = 0

Step 3. Find

!

f(x + h) " f(x)

h:

!

f(x + h) " f(x)

h =

0

h = 0

Step 4. Find limh!0

!

f(x + h) " f(x)

h:

limh!0

!

f(x + h) " f(x)

h =

limh!0

0 = 0

Thus, f'(x) = 0.

47. The graph of f(x) =

!

2x,x < 12,x " 1

# $ %

& ' ( is:

f is not differentiable at x = 1 because the graph of f has a sharp corner at this point.

49. f(x) =

!

x2 + 1 if x < 01 if x " 0

# $ %

It is clear that f'(x) =

!

2x if x < 00 if x > 0

" # $

Thus, the only question is f'(0). Since

y

x

1

limx!0

"f'(x) =

limx!0

"2x = 0 and

limx!0

+f'(x) =

limx!0

+0 = 0

f is differentiable at 0 as well; f is differentiable for real numbers.

EXERCISE 3-4 119

51. f(x) = |x|

limh!0

!

f(0 + h) " f(0)

h =

limh!0

!

0 + h " 0

h =

limh!0

!

h

h

The limit does not exist. Thus, f is not differentiable at x = 0.

53. f(x) =

!

x3 = x1/3

limh!0

!

f(0 + h) " f(0)

h =

limh!0

!

(0 + h)1 3 " 01 3

h =

limh!0

!

h1 3

h =

limh!0

!

1

h2 3

The limit does not exist. Thus, f is not differentiable at x = 0.

55. f(x) =

!

1 " x2

!

f(0 + h) " f(0)

h =

!

1 " h2 " 1

h =

!

1 " h2 " 1

h ·

!

1 " h2 + 1

1 " h2 + 1

=

!

1 " h2 " 1

h 1 " h2 + 1#

$ %

&

' ( =

!

"h

1 " h2 + 1

limh!0

!

f(0 + h) " f(0)

h =

limh!0

!

"h

1 " h2 + 1

= 0

f is differentiable at 0; f'(0) = 0.

57. The height of the ball at x seconds is h(x) = 576 - 16x2. To find when the ball hits the ground, we solve: 576 - 16x2 = 0 16x2 = 576 x2 = 36 x = 6 seconds

The velocity of the ball is given by h'(x) = -32x. The velocity at impact is h'(6) = -32(6) = -192; the ball hits the ground at 192 ft/sec.

59. R(x) = 60x - 0.025x2 0 ≤ x ≤ 2,400. (A) Average rate of change:

!

R(1,050) " R(1,000)

1,050 " 1,000

=

!

60(1,050) " 0.025(1,050)2 " [60(1,000) " 0.025(1,000)2]

50

=

!

35,437.50 " 35,000

50 = $8.75


(B) Step 1. Find R(x + h): R(x + h) = 60(x + h) - 0.025(x + h)2 = 60x + 60h - 0.025(x2 + 2xh + h2) = 60x + 60h - 0.025x2 - 0.050xh - 0.025h2

Step 2. Find R(x + h) - R(x): R(x + h) - R(x) = 60x + 60h - 0.025x2 - 0.050xh - 0.025h2 - (60x - 0.025x2) = 60h - 0.050xh - 0.025h2

Step 3. Find

!

R(x + h) " R(x)

h:

!

R(x + h) " R(x)

h =

!

60h " 0.050xh " 0.025h2

h

= 60 - 0.050x - 0.025h

Step 4. Find limh!0

!

R(x + h) " R(x)

h:

limh!0

!

R(x + h) " R(x)

h =

limh!0

(60 - 0.050x - 0.025h)

= 60 - 0.050x Thus, R'(x) = 60 - 0.050x.

(C) R(1,000) = 60(1,000) - 0.025(1,000)2 = $35,000; R'(1,000) = 60 - 0.05(1,000) = $10; at a production level of 1,000 car seats, the revenue is $35,000 and is increasing at the rate of $10 per seat.

61. (A) S(t) = 2

!

t + 10

Step 1. Find S(t + h): S(t + h) = 2

!

t + h + 10

Step 2. Find S(t + h) - S(t): S(t + h) - S(t) = 2

!

t + h + 10 - 2

!

t + 10 = 2(

!

t + h + 10 -

!

t + 10)

Step 3. Find

!

S(t + h) " S(t)

h:

!

S(t + h) " S(t)

h =

!

2 t + h + 10 " t + 10( )h

=

!

2 t + h + 10 " t + 10( )h

·

!

t + h + 10 + t + 10( )t + h + 10 + t + 10( )

=

!

2[t + h + 10 " (t + 10)]

h t + h + 10 + t + 10( ) =

!

2h

h t + h + 10 + t + 10( )

=

!

2

t + h + 10 + t + 10

EXERCISE 3-4 121

Step 4. Find limh!0

!

S(t + h) " S(t)

h:

limh!0

!

S(t + h) " S(t)

h =

limh!0

!

2

t + h + 10 + t + 10 =

!

1

t + 10

Thus, S'(t) =

!

1

t + 10.

(B) S(15) = 2

!

15 + 10 = 2

!

25 = 10;

S'(15) =

!

1

15 + 10 =

!

1

25=1

5 = 0.2

After 15 months, the total sales are $10 million and are INCREASING at the rate of $0.2 million = $200,000 per month.

(C) The estimated total sales are $10.2 million after 16 months and $10.4 million after 17 months.

63. p(t) = 14t2 - 6.6t + 602.4 (A) Step 1. Find p(t + h):

p(t + h) = 14(t + h)2 - 6.6(t + h) + 602.4

Step 2. Find p(t + h) - p(t): p(t + h) - p(t) = 14(t + h)2 - 6.6(t + h) + 602.4 - (14t2 - 6.6t + 602.4) = 28th + 14h2 - 6.6h

Step 3. Find

!

p(t + h) " p(t)

h:

!

p(t + h) " p(t)

h =

!

28th + 14h2 " 6.6h

h = 28t + 14h - 6.6

Step 4. Find limh!0

!

p(t + h) " p(t)

h:

limh!0

!

p(t + h) " p(t)

h =

limh!0

(28t + 14h - 6.6) = 28t - 6.6

Thus, p'(t) = 28t - 6.6. (B) The year 2010 corresponds to t = 15.

p(15) = 14(15)2 - 6.6(15) + 602.4 = 3,653.4 thousand tons; p'(t) = 28t - 6.6, p'(15) = 28(15) - 6.6 = 413.4; in 2010 the US will produce 3,653.4 thousand tons of zinc and this quantity is increasing at the rate of 413.4 thousand tons/year.

65. (A)

R(x) ≈ 0.6666666667x2 + 24.88484848x + 1071.018182; R(20) ≈ 1,835.4; R'(x) = 1.3333333334x + 24.88484848; R’(20) ≈ 51.6 In 2016, 1,835.4 billion kilowatts will be sold and the amount sold is increasing at the rate of 51.6 billion kilowatts per year.


67. (A) P(t) = 80 + 12t - t2 Step 1. Find P(t + h):

P(t + h) = 80 + 12(t + h) - (t + h)2 = 80 + 12t + 12h - (t + h)2

Step 2. Find P(t + h) - P(t): P(t + h) - P(t) = 80 + 12t + 12h - (t + h)2 - (80 + 12t - t2) = 12h - 2th - h2

Step 3. Find

!

P(t + h) " P(t)

h:

!

P(t + h) " P(t)

h =

!

12h " 2th " h2

h = 12 - 2t - h

Step 4. Find limh!0

!

P(t + h) " P(t)

h:

limh!0

!

P(t + h) " P(t)

h =

limh!0

(12 - 2t - h) = 12 - 2t

Thus, P'(t) = 12 - 2t.

(B) P(3) = 80 + 12(3) - (3)2 = 107; P'(3) = 12 - 2(3) = 6

After 3 hours, the ozone level is 107 ppb and is INCREASING at the rate of 6 ppb per hour.

69. f(t) = 0.008t2 – 0.5t + 14.4 (A) Step 1. Find f(t + h):

f(t + h) = 0.008(t + h)2 – 0.5(t + h) + 14.4 Step 2. Find f(t + h) - f(t):

f(t + h) – f(t) = 0.008(t + h)2 - 0.5(t + h) + 14.4 - (0.008t2 -0.5t + 14.4) = 0.016th + 0.008h2 – 0.5h

Step 3. Find

!

f(t + h) " f(t)

h:

!

f(t + h) " f(t)

h =

!

0.016th + 0.008h2 " 0.5h

h

= 0.016t + 0.008h – 0.5

Step 4. Find limh!0

!

f(t + h) " f(t)

h:

limh!0

!

f(t + h) " f(t)

h =

limh!0

(0.016t + 0.008h – 0.5)

= 0.016t – 0.5 Thus, f’(t) = 0.016t – 0.5.

EXERCISE 3-5 123

(B) The year 2010 corresponds to t = 30; f(30) = 0.008(30)2 – 0.5(30) + 14.4 = 6.6, f’(30) = -0.02.

In 2010, the number of male infant deaths per 100,000 births will be 6.6 and will be decreasing at the rate of 0.02 male infant deaths per 100,00 births.

EXERCISE 3-5 Things to remember:

1. DERIVATIVE NOTATION Given y = f(x), then

f'(x), y',

!

dy

dx

all represent the derivative of f at x.

2. CONSTANT FUNCTION RULE If f(x) = C, C a constant, then f'(x) = 0. Also

y' = 0 and

!

dy

dx = 0.

3. POWER RULE If f(x) = xn, n any real number, then

f'(x) = nxn-1.

Also, y' = nxn-1 and

!

dy

dx = nxn-1

4. CONSTANT MULTIPLE PROPERTY If y = f(x) = ku(x), where k is a constant, then

f'(x) = ku'(x). Also,

y' = ku' and

!

dy

dx = k

!

du

dx.

5. SUM AND DIFFERENCE PROPERTY If y = f(x) = u(x) ± v(x), then

f'(x) = u'(x) ± v'(x). Also,

y' = u' ± v' and

!

dy

dx =

!

du

dx ±

!

dv

dx

[Note: This rule generalizes to the sum and difference of any given number of functions.]


1. f(x) = 7; f'(x) = 0 (using 2) 3. y = x9;

!

dy

dx = 9x8 (using 3)

5.

!

d

dxx3 = 3x2 (using 3) 7. y = x-4; y' = -4x-5 (using 3)

9. g(x) = x8/3; g'(x) =

8

3x5/3

(using 3)

11. y =

1

x10;

!

dy

dx = -10x-11 =

!10

x11

13. f(x) = 5x2; f'(x) = 5(2x) = 10x (using 4)

15. y = 0.4x7; y' = 0.4(7x6) = 2.8x6

17.

d

dx

x3

18

!

" #

$

% & =

1

18(3x2) =

1

6x2

19. h(x) = 4f(x); h'(2) = 4 · f'(2) = 4(3) = 12

21. h(x) = f(x) + g(x); h'(2) = f'(2) + g'(2) = 3 + (-1) = 2

23. h(x) = 2f(x) - 3g(x) + 7; h'(2) = 2f'(2) - 3g'(2) = 2(3) - 3(-1) = 9

25.

d

dx(2x - 5) =

d

dx(2x) -

d

dx(5) = 2

27. f(t) = 2t2 - 3t + 1; f'(t) = (2t2)' - (3t)' + (1)' = 4t - 3

29. y = 5x-2 + 9x-1; y' = -10x-3 - 9x-2

31.

d

du(5u0.3 - 4u2.2) =

d

du(5u0.3) -

d

du(4u2.2) = 1.5u-0.7 - 8.8u1.2

33. h(t) = 2.1 + 0.5t - 1.1t3; h'(t) = 0.5 - (1.1)3t2 = 0.5 - 3.3t2

35. y =

2

5x4 =

2

5x-4; y' =

2

5(-4x-5) = -

8

5x-5 =

!8

5x5

37.

d

dx

!

3x2

2"

7

5x2

#

$ % %

&

' ( ( =

d

dx

!

3

2x2

"

# $

%

& ' -

d

dx

!

7

5x"2

#

$ %

&

' ( = 3x +

14

5x-3 = 3x +

14

5x3

39. G(w) =

5

9w4 + 5

!

w3 =

5

9w-4 + 5w1/3;

G'(w) = -

20

9w-5 +

5

3w-2/3 =

!20

9w5 +

5

3w2 3

EXERCISE 3-5 125

41.

d

du(3u2/3 - 5u1/3) =

d

du(3u2/3) -

d

du(5u1/3)

= 2u-1/3 -

5

3u-2/3 =

2

u1 3!

5

3u2 3

43. h(t) =

3

t3 5 -

6

t1 2 = 3t-3/5 - 6t-1/2;

h'(t) = 3

!

"3

5t"8 5

#

$ %

&

' ( - 6

!

"1

2t"3 2

#

$ %

&

' ( = -

9

5t-8/5 + 3t-3/2 =

!9

5t8 5 +

3

t3 2

45. y =

!

1

x3 =

1

x1 3 = x-1/3; y' = -

1

3x-4/3 =

!1

3x4 3

47.

d

dx

!

1.2

x" 3.2x"2 + x

#

$ %

&

' ( =

d

dx(1.2x-1/2 - 3.2x-2 + x)

=

d

dx(1.2x-1/2) -

d

dx(3.2x-2) +

d

dx(x)

= -0.6x-3/2 + 6.4x-3 + 1 =

!0.6

x3 2+6.4

x3 + 1

49. f(x) = 6x - x2

(A) f'(x) = 6 - 2x

(B) Slope of the graph of f at x = 2: f'(2) = 6 - 2(2) = 2 Slope of the graph of f at x = 4: f'(4) = 6 - 2(4) = -2

(C) Tangent line at x = 2: y - y1 = m(x - x1) x1 = 2

y1 = f(2) = 6(2) - 22 = 8

m = f'(2) = 2 Thus, y - 8 = 2(x - 2) or y = 2x + 4.

Tangent line at x = 4: y - y1 = m(x - x1) x1 = 4

y1 = f(4) = 6(4) - 42 = 8

m = f'(4) = -2 Thus, y - 8 = -2(x - 4) or y = -2x + 16

(D) The tangent line is horizontal at the values x = c such that f'(c) = 0. Thus, we must solve the following: f'(x) = 6 - 2x = 0 2x = 6 x = 3

37. f x) = f is discontinuous at x) = f x) = - x = -2 is a...

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