Part II: The Answers
296
Answers
401–500
To determine whether the function is increasing or decreasing on (0, 1), take a point in
the interval and substitute it into the derivative. If x = 1
2
, then ′ ( )
( )
f 1
2
4 1 1
2
1
2
0
2
=
−
> ,
so the function is increasing on (0, 1). Likewise, if you use x = 2, then ′
f ( )
(
)
2
4 1 4
2
0
=
−
< ,
so the function is decreasing on (1, ∞).
461.
local maximum at (–1, 7); local minimum at (2, –20)
Begin by finding the derivative of the function f
(x) = 2x
3
– 3x
2
– 12x:
′
f x
x
x
( ) =
−
−
6
6 12
2
Then set this derivative equal to zero and solve for x to find the critical numbers:
6
6 12 0
2 0
2
1 0
2 1
2
2
x
x
x
x
x
x
x
−
− =
− − =
−
+ =
= −
(
)(
)
,
Next, determine whether the function is increasing or decreasing on the intervals
(–∞, –1), (–1, 2), and (2, ∞) by picking a point inside each interval and substituting it
into the derivative. Using the values x = –2, x = 0, and x = 3 gives you
′ − = −
− − − >
′
′
=
−
− <
=
−
f
f
f
( )
( )
( )
( )
( )
( )
( )
( )
2 6 2
6 2 12 0
0
0
0
3
3
6
6
12 0
6
6
2
2
3
( ( )
3 12 0
− >
Therefore, f
(x) is increasing on (–∞, –1), decreasing on (–1, 2), and increasing again on
(2, ∞). That means there’s a local maximum at x = –1 and a local minimum at x = 2.
Now enter these values in the original function to find the coordinates of the local maximum and minimum. Because f
(–1) = 2(–1)
3
– 3(–1)
2
– 12(–1) = 7, the local maximum occurs
at (–1, 7). Because f
(2) = 2(2)
3
– 3(2)
2
– 12(2) = –20, the local minimum occurs at (2, –20).
462.
no local maxima; local minimum at (16, –16)
Begin by finding the derivative of the function f x
x
x x
x
( ) = −
= −
8
8
1 2
:
′
( )
f x
x
x
x
x
( ) = −
= −
=
−
−
1 8 1
2
1 4
4
1 2
Then find the critical numbers. Setting the numerator equal to zero gives you x − =
4 0
so that x = 4, or x = 16. Setting the denominator equal to zero gives you x = 0. Next,
determine whether the function is increasing or decreasing on the intervals (0, 16)
and (16, ∞) by taking a point in each interval and substituting it into the derivative.
296
Answers
401–500
To determine whether the function is increasing or decreasing on (0, 1), take a point in
the interval and substitute it into the derivative. If x = 1
2
, then ′ ( )
( )
f 1
2
4 1 1
2
1
2
0
2
=
−
> ,
so the function is increasing on (0, 1). Likewise, if you use x = 2, then ′
f ( )
(
)
2
4 1 4
2
0
=
−
< ,
so the function is decreasing on (1, ∞).
461.
local maximum at (–1, 7); local minimum at (2, –20)
Begin by finding the derivative of the function f
(x) = 2x
3
– 3x
2
– 12x:
′
f x
x
x
( ) =
−
−
6
6 12
2
Then set this derivative equal to zero and solve for x to find the critical numbers:
6
6 12 0
2 0
2
1 0
2 1
2
2
x
x
x
x
x
x
x
−
− =
− − =
−
+ =
= −
(
)(
)
,
Next, determine whether the function is increasing or decreasing on the intervals
(–∞, –1), (–1, 2), and (2, ∞) by picking a point inside each interval and substituting it
into the derivative. Using the values x = –2, x = 0, and x = 3 gives you
′ − = −
− − − >
′
′
=
−
− <
=
−
f
f
f
( )
( )
( )
( )
( )
( )
( )
( )
2 6 2
6 2 12 0
0
0
0
3
3
6
6
12 0
6
6
2
2
3
( ( )
3 12 0
− >
Therefore, f
(x) is increasing on (–∞, –1), decreasing on (–1, 2), and increasing again on
(2, ∞). That means there’s a local maximum at x = –1 and a local minimum at x = 2.
Now enter these values in the original function to find the coordinates of the local maximum and minimum. Because f
(–1) = 2(–1)
3
– 3(–1)
2
– 12(–1) = 7, the local maximum occurs
at (–1, 7). Because f
(2) = 2(2)
3
– 3(2)
2
– 12(2) = –20, the local minimum occurs at (2, –20).
462.
no local maxima; local minimum at (16, –16)
Begin by finding the derivative of the function f x
x
x x
x
( ) = −
= −
8
8
1 2
:
′
( )
f x
x
x
x
x
( ) = −
= −
=
−
−
1 8 1
2
1 4
4
1 2
Then find the critical numbers. Setting the numerator equal to zero gives you x − =
4 0
so that x = 4, or x = 16. Setting the denominator equal to zero gives you x = 0. Next,
determine whether the function is increasing or decreasing on the intervals (0, 16)
and (16, ∞) by taking a point in each interval and substituting it into the derivative.