Answers and Explanations 297
Answers
401–500
So if x = 1, you have ′
f ( )
1 1 4
1
0
= −
< so that f
(x) is decreasing on (0, 16). And if x = 25,
you have ′
f ( )
25 1
4
25
0
= −
> , so f
(x) is increasing on (16, ∞). Therefore, the function
has a local minimum at x = 16. Finish by finding the coordinates: f ( )
16 16 8 16
16
= −
= − ,
so the local minimum is at (16, –16).
463.
local maximum at (64, 32); local minimum at (0, 0)
Begin by finding the derivative of the function f
(x) = 6x
2/3
– x:
′
( )
f x
x
x
x
x
x
x
( ) =
−
=
−
= −
−
6 2
3
1
4
4
1 3
3
3
3
3
3
Then find the critical numbers. Setting the numerator equal to zero gives you
4
0
3
−
=
x
so that 4 = x
1/3
, or 64 = x. Setting the denominator equal to zero gives you
x = 0 as a solution.
Next, determine whether the function is increasing or decreasing on the intervals
(–∞, 0), (0, 64), and (64, ∞) by taking a point in each interval and substituting it into
the derivative. Using the test points x = –1, x = 1, and x = 125 gives you the following:
′
′ = −
>
′
= −
= −
− = − −
−
= +
−
<
f
f
f
( )
( )
(
)
1
4
1
1
4 1
1
0
1
125
4 1
1
0
4 125
125
4
3
3
3
3
3
3
5 5
5
0
<
Therefore, f
(x) is decreasing on (–∞, 0), increasing on (0, 64), and decreasing on (64, ∞).
That means f
(x) has a local minimum at x = 0; f
(0) = 0, so a local minimum occurs at (0, 0).
Also, f
(x) has a local maximum at x = 64; f
(64) = 6(64)
2/3
– 64 = 32, so the local maximum
occurs at (64, 32).
464.
local maximum at 2
3
3 3
2
π ,
; local minimum at 4
3
3 3
2
π , −
Begin by finding the derivative of the function f
(x) = 2 sin x – sin 2x:
′
(
)
f x
x
x
x
x
x
x
( )
cos
cos
cos
cos( )
cos
c os
=
−(
) (2)
=
−
=
−
−
= −
2
2
2
2
2
2
2 2
1
4
2
c cos
cos
2
2
2
x
x
+
+
Next, find the critical numbers by setting the derivative equal to zero:
−
+
+
−
− =
+
− =
4
2
2 0
2
1 0
2
1
1 0
2
2
cos
cos
cos
cos
( cos
)(cos
)
x
x
x
x
x
x
=
Answers
401–500
So if x = 1, you have ′
f ( )
1 1 4
1
0
= −
< so that f
(x) is decreasing on (0, 16). And if x = 25,
you have ′
f ( )
25 1
4
25
0
= −
> , so f
(x) is increasing on (16, ∞). Therefore, the function
has a local minimum at x = 16. Finish by finding the coordinates: f ( )
16 16 8 16
16
= −
= − ,
so the local minimum is at (16, –16).
463.
local maximum at (64, 32); local minimum at (0, 0)
Begin by finding the derivative of the function f
(x) = 6x
2/3
– x:
′
( )
f x
x
x
x
x
x
x
( ) =
−
=
−
= −
−
6 2
3
1
4
4
1 3
3
3
3
3
3
Then find the critical numbers. Setting the numerator equal to zero gives you
4
0
3
−
=
x
so that 4 = x
1/3
, or 64 = x. Setting the denominator equal to zero gives you
x = 0 as a solution.
Next, determine whether the function is increasing or decreasing on the intervals
(–∞, 0), (0, 64), and (64, ∞) by taking a point in each interval and substituting it into
the derivative. Using the test points x = –1, x = 1, and x = 125 gives you the following:
′
′ = −
>
′
= −
= −
− = − −
−
= +
−
<
f
f
f
( )
( )
(
)
1
4
1
1
4 1
1
0
1
125
4 1
1
0
4 125
125
4
3
3
3
3
3
3
5 5
5
0
<
Therefore, f
(x) is decreasing on (–∞, 0), increasing on (0, 64), and decreasing on (64, ∞).
That means f
(x) has a local minimum at x = 0; f
(0) = 0, so a local minimum occurs at (0, 0).
Also, f
(x) has a local maximum at x = 64; f
(64) = 6(64)
2/3
– 64 = 32, so the local maximum
occurs at (64, 32).
464.
local maximum at 2
3
3 3
2
π ,
; local minimum at 4
3
3 3
2
π , −
Begin by finding the derivative of the function f
(x) = 2 sin x – sin 2x:
′
(
)
f x
x
x
x
x
x
x
( )
cos
cos
cos
cos( )
cos
c os
=
−(
) (2)
=
−
=
−
−
= −
2
2
2
2
2
2
2 2
1
4
2
c cos
cos
2
2
2
x
x
+
+
Next, find the critical numbers by setting the derivative equal to zero:
−
+
+
−
− =
+
− =
4
2
2 0
2
1 0
2
1
1 0
2
2
cos
cos
cos
cos
( cos
)(cos
)
x
x
x
x
x
x
=