51. f(x) = 3x4 - 6x2 - 7
(A) f'(x) = 12x3 - 12x
(B) Slope of the graph of x = 2: f'(2) = 12(2)3 - 12(2) = 72
Slope of the graph of x = 4: f'(4) = 12(4)3 - 12(4) = 720
(C) Tangent line at x = 2: y - y1 = m(x - x1), where x1 = 2,
y1 = f(2) = 3(2)4 - 6(2)2 - 7 = 17, m = 72.
y - 17 = 72(x - 2) or y = 72x - 127
Tangent line at x = 4: y - y1 = m(x - x1), where x1 = 4,
y1 = f(4) = 3(4)4 - 6(4)2 - 7 = 665, m = 720.
y - 665 = 720(x - 4) or y = 720x - 2215
(D) Solve f'(x) = 0 for
12x3 - 12x =
12x(x2 - 1) =
12x(x - 1)(x + 1) =
x=
x:
0
0
0
-1, x = 0, x = 1
53. f(x) = 176x - 16x2
(A) v = f'(x) = 176 - 32x
(C) Solve v =
for x:
176 - 32x
32x
x
(B) v x =0 = f'(0) = 176 ft/sec.
f'(x) = 0
v x =3 = f'(3) = 176 - 32(3) = 80 ft/sec.
= 0
= 176
= 5.5 sec.
55. f(x) = x3 - 9x2 + 15x
(A) v = f'(x) = 3x2 - 18x + 15
(B) v x =0 = f'(0) = 15 feet/sec.
v x =3 = f'(3) =
(C) Solve v = f'(x)
3x2 - 18x + 15
3(x2 - 6x + 5)
3(x - 5)(x - 1)
x
3(3)2 - 18(3) + 15 = -12 feet/sec.
=
=
=
=
=
0 for x:
0
0
0
1, x = 5
57. f(x) = x2 - 3x - 4 x = x2 - 3x - 4x1/2
f'(x) = 2x - 3 - 2x-1/2
The graph of f has a horizontal tangent line at the value(s) of x
where f'(x) = 0. Thus, we need to solve the equation
!
2x - 3 - 2x-1/2 = 0
By graphing the function y = 2x - 3 - 2x-1/2, we see that there is one
zero. To four decimal places, it is x = 2.1777.
126
CHAPTER 3
LIMITS AND THE DERIVATIVE
3
59. f(x) = 3 x 4 - 1.5x2 - 3x = 3x4/3 - 1.5x2 - 3x
f'(x) = 4x1/3 - 3x - 3
The graph of f has a horizontal tangent line at the value(s) of x
where
f'(x) = 0. Thus, we need to solve the equation
!
4x1/3 - 3x - 3 = 0
Graphing the function y = 4x1/3 - 3x - 3, we see that there is one
zero. To four decimal places, it is x = -2.9018.
61. f(x) = 0.05x4 - 0.1x3 - 1.5x2 - 1.6x + 3
f'(x) = 0.2x3 + 0.3x2 - 3x - 1.6
The graph of f has a horizontal tangent line at the value(s) of x
where f'(x) = 0. Thus, we need to solve the equation
0.2x3 + 0.3x2 - 3x - 1.6 = 0
By graphing the function y = 0.2x3 + 0.3x2 - 3x - 1.6 we see that
there are three zeros. To four decimal places, they are
x1 = -4.4607, x2 = -0.5159, x3 = 3.4765
63. f(x) = 0.2x4 - 3.12x3 + 16.25x2 - 28.25x + 7.5
f'(x) = 0.8x3 - 9.36x2 + 32.5x - 28.25
The graph of f has a horizontal tangent line at the value(s) of x
where f'(x) = 0. Thus, we need to solve the equation
0.8x3 - 9.36x2 + 32.5x - 28.25 = 0
Graphing the function y = 0.8x3 - 9.36x2 + 32.5x - 28.25, we see that
there is one zero. To four decimal places, it is x = 1.3050.
65. f(x) = ax2 + bx + c; f'(x) = 2ax + b.
The derivative is 0 at the vertex of the parabola:
2ax + b = 0
b
x = 2a
67. (A) f(x) = x 3 + x
(B) f(x) = x 3
(C) f(x) = x 3 - x
69. f(x) = (2x - 1)2 = 4x2 - 4x + 1
f'(x) = 8x - 4
71.
20
d "10x + 20%
d "
d
d
20%
(10) +
(20x-1) = -20x-2 = - 2
$
' =
$10 +
' =
&
x
dx #
dx #
dx
dx
x
x&
3x ! 4
3x
4
1
1 -2
!
73. y =
=
= x-1 x
2
2
2
12x
12x
12x
4
3
!
!
2
1
dy
1
2 -3
= - x-2 +
x
= - 2 +
3x 3
4x
dx
4
3
75. y =
10
x 4 ! 3x 3 + 5
= x2 - 3x + 5x-2; y' = 2x - 3 - 10x-3 = 2x - 3 - 3
2
x
x
EXERCISE 3-5
127
77. f(x) = x3
Step 1. Find f(x + h):
f(x + h) = (x + h)3 = x3 + 3x2h + 3xh2 + h3
Step 2. Find f(x + h) - f(x):
f(x + h) - f(x + h) = x3 + 3x3h + 3xh2 + h3 - x3
= 3x2h + 3xh2 + h3
f(x + h) ! f(x)
Step 3. Find
:
h
3x 2h + 3xh2 + h3
f(x + h) ! f(x)
=
= 3x2 + 3xh + h2
h
h
f(x + h) ! f(x)
Step 4. Find lim
:
h
h!0
f(x + h) ! f(x)
lim
= lim (3x2 + 3xh + h2) = 3x2
h
h!0
h!0
d
3
2
Thus,
(x ) = 3x .
dx
1
1 -2/3
79. f(x) = x1/3; f'(x) =
x
=
3x2/3
3
The domain of f' is the set of all real numbers except x = 0. The
graph of f is smooth, but it has a vertical tangent line at (0, 0).
81. (A) S(t) = 0.03t3 + 0.5t2 + 2t + 3
S'(t) = 0.09t2 + t + 2
(B) S(5) = 0.03(5)3 + 0.5(5)2 + 2(5) + 3 = 29.25
S'(5) = 0.09(5)2 + 5 + 2 = 9.25
After 5 months, sales are $29.25 million and are increasing
at the rate of $9.25 million per month.
(C) S(10) = 0.03(10)3 + 0.5(10)2 + 2(10) + 3 = 103
S'(10) = 0.09(10)2 + 10 + 2 = 21
After 10 months, sales are $103 million and are increasing
at the rate of $21 million per month.
3,780
= 1,000 - 3,780x-1
x
3,780
-2
N'(x) = 3,780x
=
x2
3,780
(B) N'(10) =
= 37.8
(10)2
At the $10,000 level of advertising, sales are INCREASING at the
rate of 37.8 boats per $1000 spent on advertising.
3,780
N'(20) =
= 9.45
(20)2
83. (A) N(x) = 1,000 -
At the $20,000 level of advertising, sales are INCREASING at the
rate of 9.45 boats per $1000 spent on advertising.
128
CHAPTER 3
LIMITS AND THE DERIVATIVE
(B) L(12) ≈ 3,919.09 or 3,900 rounded to the
nearest hundred; L'(12) = -417.75 or -400.
Interpretation: In 1992, 3,900 limousines
were produced and limousine production was
DECREASING at the rate of 400 limousines per
year.
85. (A)
(C) L(18) ≈ 3,883.2 or 3,900 rounded to the nearest hundred;
L'(18) ≈ 631.39 or 600.
Interpretation: In 1998, 3,900 limousines were produced and limousine
production was INCREASING at the rate of 600 limousines per year.
87. y = 590x-1/2, 30 ≤
dy
First, find
=
dx
rate of change of
x ≤ 75
!295
d
590x-1/2 = -295x-3/2 =
, the instantaneous
x3 2
dx
pulse when a person is x inches tall.
(A) The instantaneous rate of change of pulse rate at x = 36 is:
!295
!295
=
= -1.37 (1.37 decrease in pulse rate)
3
2
(36)
216
(B) The instantaneous rate of change of pulse rate at x = 64 is:
!295
!295
=
= -0.58 (0.58 decrease in pulse rate)
3
2
(64)
512
89. y = 50 x , 0 ≤ x ≤ 9
First, find y' = (50 x ) ' = (50x1/2) ' = 25x-1/2
25
=
, the rate of learning at the end of x hours.
x
!
25
!
(A) Rate of learning
at the end of 1 hour:
= 25 items/hr
1
! learning at the end of 9 hours: 25 = 25 = 8.33 items/hr
(B) Rate of
3
9
!
EXERCISE 3-6
Things to remember:
1.
!
INCREMENTS
For y = f(x),
∆x = x2 – x1
x2 = x1 + ∆x
∆y = y2 – y1
= f(x2) – f(x1)
= f(x1 + ∆x) – f(x1)
∆y represents the change in y corresponding to a ∆x change in
x. ∆x can be either positive or negative.
[Note: ∆y depends on the function f, the input x, and the
increment ∆x.]
EXERCISE 3-6
129
2.
DIFFERENTIALS
If y = f(x) defines a differential function, then the
differential dy, or df, is defined as the product of f’(x)
and dx, where dx = ∆x. Symbolically,
dy = f’(x)dx
or
df = f’(x)dx
where
dx = ∆x
[Note: The differential dy (or df) is actually a function
involving two independent variables, x and dx; a change in
either one or both will affect dy (or df).]
1. ∆x = x2 - x1 = 4 - 1 = 3
∆y = f(x2) - f(x1) = 3·42 - 3·12 = 48 - 3 = 45
∆y
45
∆x = 3 = 15
f(x1 + ∆x) - f(x1)
3.
∆x
f(1 + 2) - f(1)
=
2
3·32 - 3·12
=
2
24
= 2
= 12
5. ∆y = f(x2) - f(x1) = f(3) - f(1) = 3·32 - 3·12 = 27 - 3 = 24
∆x = x2 - x1 = 3 - 1 = 2
∆y
24
∆x = 2 = 12
7. y = 30 + 12x2 - x3
dy = (30 + 12x2 - x3)'dx = (24x - 3x2)dx
x

9. y = x21 - 9 
x 
x 
1 
 
  1


dy = x21 - 9  'dx = x2- 9 + 1 - 9 2x dx = 2x - 3 x2 dx
590
11. y =
= 590x-1/2
x
-295
 1
dy = 590- 2 x-3/2dx = 3/2 dx
x
3(2 + ∆x)2 - 3·22
f(2 + ! x) " f(2)
13. (A)
=
∆x
!x
3(22 + 4∆x + ∆x2) - 12
=
∆x
∆x(12 + 3∆x)
=
∆x
= 12 + 3∆x, ∆x ≠ 0
15. y = (2x + 1)3
dy = 3(2x + 1)2(2)dx
= 6(2x + 1)2dx
130
CHAPTER 3
LIMITS AND THE DERIVATIVE
(B) As ∆x tends to zero,
then, clearly, 12 + 3∆x
tends to 12. Note the
values in the following
table:
∆x
1
0.1
0.01
0.001
12 + 3∆x
15
12.3
12.03
12.003
x
x + 9
(x2 + 9)(1) - x(2x)
9 - x2
dy =
dx =
dx
2
2
(x + 9)
(x2 + 9)2
17. y =
2
19. y = f(x) = x2 - 3x + 2
∆y = f(5 + 0.2) - f(5)
(using 1)
= f(5.2) - f(5)
= (5.2)2 - 3(5.2) + 2 - (52 - 3·5 + 2) = 1.44
dy = (x2 - 3x)'
x =5
dx = (2x - 3)
x =5
(0.2)
#
2&
21. y = f(x) = 75 %1 " (
$
x'
∆y = f[5 + (-0.5)] - f(5)
= f(4.5) - f(5)
#
#
2 &
2&
= 75!%1 "
( - 75 %1 " (
$
$
4.5'
5'
= 41.67 - 45 = -3.33
) #
150
2 &, /
dy = +75%1 " (. x =5 dx =
(-0.5)
x2 x =5
x '!
* $
!
= 1.4
#
= 6 %"
$
1&
( = -3
2'
23. A cube with sides of length x has volume V(x) = x3. If we increase
! the length of each side by an amount !∆x = dx, then the approximate
change in volume is:
dV = 3x2dx
Therefore, letting x = 10 and dx = 0.4 [= 2(0.2), since each face has
a 0.2 inch coating], we have
dV = 3(10)2(0.4) = 120 cubic inches,
which is the approximate volume of the fiberglass shell.
25. f(x) = x2 + 2x + 3; f'(x) = 2x + 2; x = -0.5; ∆x = dx
(A) ∆y = f(-0.5 + ∆x) - f(–0.5)
= (-0.5 + ∆x)2 + 2(-0.5 + ∆x) + 3 - [(-0.5)2 + 2(-0.5) + 3]
= (-0.5 + ∆x)2 + 2(-0.5 + ∆x) + 0.75 = ∆x + (∆x)2
dy = f'(-0.5)dx = 1 · dx = dx = ∆x
1
(B)
-1
1
-1
EXERCISE 3-6
131
(C) ∆y(0.1) = (-0.5 + 0.1)2 + 2(-0.5 + 0.1) + 0.75 = 0.11
dy(0.1) = 0.1
∆y(0.2) = (-0.5 + 0.2)2 + 2(-0.5 + 0.2) + 0.75 = 0.24
dy(0.2) = 0.2
∆y(0.3) = (-0.5 + 0.3)2 + 2(-0.5 + 0.3) + 0.75 = 0.39
dy(0.3) = 0.3
27. f(x) = x3 - 2x2; f'(x) = 3x2 - 4x; x = 1; ∆x = dx
(A) ∆y = f(1 + ∆x) - f(1) = (1 + ∆x)3 - 2(1 + ∆x)2 - [13 - 2(1)2]
= (1 + ∆x)3 - 2(1 + ∆x)2 + 1
= -∆x + (∆x)2 + (∆x)3
dy = f'(1)dx = (-1)dx = -dx
0.5
(B)
-0.5
0.5
-0.5
(C) ∆y(0.05) = (1 + 0.05)3 - 2(1 + 0.05)2 + 1 ≈ -0.0474
dy(0.05) = -0.05
∆y(0.10) = (1 + 0.10)3 - 2(1 + 0.10)2 + 1 = -0.089
dy(0.10) = -0.10
∆y(0.15) = (1 + 0.15)3 - 2(1 + 0.15)2 + 1 ≈ -0.1241
dy(0.15) = -0.15
29. True
f(x) = mx + b; f'(x) = m; x = 3; ∆x = dx
∆y = f(3 + ∆x) - f(3) = m(3 + ∆x) + b - (3m + b)
= m ∆x
dy = f'(3)dx = m dx
Thus, ∆y = dy
31. False.
At x = 2, dy = f'(2)dx. dy = 0 for all dx implies only that
f'(2) = 0.
Example. Let f(x) = x2 - 4x
3
3x 2 " 2x + 1 = (3x2 - 2x + 1)1/3
(6x - 2)dx
1
dy = (3x 2 ! 2x + 1)-2/3(6x - 2)dx =
3(3x2 - 2x + 1)2/3
3
33. y =
!
132
CHAPTER 3
LIMITS AND THE DERIVATIVE
35. y = f(x) = 52 x
= 52x1/2; x = 4, ∆x = dx = 0.3
∆y = f(x + ∆x) - f(x)
Let x = 4, ∆x = 0.3.
Then:
∆y = 52(4 + 0.3)1/2 - 52(4)1/2
= 52(4.3)1/2 - 104
≈ 107.83 - 104
≈ 3.83
26
dx
x1/2
Let x = 4, dx = 0.3. Then:
26
dy = 1/2 (0.3)
4
26
= 2 (0.3) = 3.9
dy = f'(x)dx =
37. Given N(x) = 60x - x2, 5 ≤ x ≤ 30. Then N'(x) = 60 - 2x. The
approximate change in the sales dN corresponding to a change ∆x = dx
in the amount x (in thousands of dollars) spent on advertising is:
dN = N'(x)dx
Thus, letting x = 10 and dx = 1, we get:
dN = N'(10)·1 = (60 - 2·10) = 40
There will be a 40-unit increase in sales (approximately) when the
advertising budget is increased from $10,000 to $11,000.
Similarly, letting x = 20 and dx = 1, we get:
dN = N'(20)·1 = (60 - 2·20) = 20-unit increase.
_
39. C(x)
400
x
1
x, x ≥ 1.
2
If we increase production per hour by an amount ∆x = dx, then the
approximate change in average cost is:
_
_
dC = C '(x)dx
# 400
1&
= %" 2 + ( dx
2'
$ x
=
+ 5 +
Thus, letting x = 20 and dx = 5, we have
# 400
#
1&
1&
! _
+
dC = %"
( 5 = %"1 + ( 5 = -2.50,
2
$
2'
2'
$ (20)
that is, the average cost per racket will decrease $2.50.
Letting x = 40 and dx = 5, we have
# 400 !1 &
# 1
1&
! _
+
"
+
dC = %"
5
=
(
%
( 5 = 1.25,
2
$ 4
2'
2'
$ (40)
that is, the average cost per racket will increase $1.25.
!
!
EXERCISE 3-6
133
41. y =
590
, 30 ≤ x ≤ 75.
x
Thus, y = 590x-1/2 and y' = -295x-3/2 =
-295
.
x3/2
The approximate change in pulse rate for a height change from 36 to
37 inches is given by:
-295
dy = 3/2
[Note: ∆x = dx = 1.]
x
x =36
= -1.37 per minute
Similarly, the approximate change in pulse rate for a height change
from 64 to 65 inches is given by:
-295
dy = 3/2
x
x =64
= -0.58 per minute
43. Area: A(r) = πr2 ≈ 3.14r2; A'(r) = 6.28r.
An approximate increase in cross-sectional area when the radius of an
artery is increased from 2 mm to 2.1 mm is given by:
dA = A'(r) r =2 × 0.1
[Note: ∆r = 0.1.]
(
= 6.28r r =2
)
× 0.1 = 12.56 × 0.1 = 1.256 mm2 ≈ 1.26 mm2
150
#
2&
45. N(t) = 75 %1 " ( , 3 ≤ t ≤ 20; N'(t) =
.
t2
$
t'
The approximate improvement from 5 to 5.5 weeks of practice is given
by:
dN != N'(t) t = 5 × 0.5
[Note: ∆t = 0.5.]
150
=
× 0.5 = 6 × 0.5 = 3 words per minute
t2 t = 5
47. N(t) = 30 + 12t2 - t3, 0 ≤ t ≤ 8; N'(t) = 24t - 3t2.
(A) The approximate change in votes when time changes from 1 to 1.1 years
= N'(t) t = 1 × 0.1
[Note: ∆t = 0.1.]
=
(24t ! 3t2
)
t= 1
× 0.1 = 21 × 0.1 = 2.1 thousand or 2100 increase.
(B) The approximate change in votes when time changes from 4 to 4.1 years
= N'(t) t = 4 × 0.1
[Note: ∆t = 0.1.]
=
(24t ! 3t2
)
t= 4
× 0.1 = 48 × 0.1 = 4.8 thousand or 4800 increase.
(C) The approximate change in votes when time changes from 7 to 7.1
years
= N'(t) t = 7 × 0.1
[Note: ∆t = 0.1.]
=
134
(24t ! 3t2
CHAPTER 3
)
t= 7
× 0.1 = 2.1 thousand or 2100 increase.
LIMITS AND THE DERIVATIVE