Answers and Explanations 301
Answers
401–500
′′
(
)
f x
x
x
( ) = −
= −
−
6 1
3
2
4 3
4 3
Next, set the denominator of the second derivative equal to zero to get x
4/3
= 0 so that x = 0.
To determine the concavity on the intervals (–∞, 0) and (0, ∞), pick a point from each
interval and substitute it into the second derivative to see whether it’s positive or
negative. Using the values x = –1 and x = 1, you have the following:
′′ − = −
−
<
′′ = −
<
f
f
( ) ( )
( ) ( )
1
2
1
0
1
2
1
0
4 3
4 3
Therefore, the f
 
(x) is concave down on the intervals (–∞, 0) and (0, ∞).
468.
concave up on (–∞, 0), 1
2
, ∞
( ) ; concave down on 0 1
2
,
( )
Begin by finding the first derivative of the function f
 
(x) = x
1/3
(x + 1) = x
4/3
+ x
1/3
:
′
f x
x
x
( ) =
+
−
4
3
1
3
1 3
2 3
Next, use the power rule to find the second derivative:
′′
f x
x
x
x
x
x
x
( )
(
)
(
)
=
−
=
−
=
−
−
−
−
4
9
2
9
2
9
2 1
2 2 1
9
2 3
5 3
5 3
5 3
Then find where the second derivative is equal to zero or undefined. Setting the numerator
equal to zero gives you 2x – 1 = 0 so that x = 1
2
, and setting the denominator equal to
zero gives you 9x
5/3
= 0 so that x = 0.
To determine the concavity on the intervals (–∞, 0), 0 1
2
,
( ) , and 1
2
,∞
( ) , pick a point
from each interval and substitute it into the second derivative to see whether it’s positive
or negative. Using the values x = –1, x = 1
4
, and x = 1 gives you the following:
′′ − =
− −
−
>
′′ ( ) = ( ) −
( )
<
′′ =
f
f
f
( )
( )
( )
( )
(
1
4 1 2
9 1
0
1
4
4 1
4
2
9 1
4
0
1
4 1
5 3
5 3
) )
( )
− >
2
9 1
0
5 3
Therefore, f
 
(x) is concave up on the intervals (–∞, 0) and 1
2
, ∞
( ) , and f
 
(x) is concave
down on the interval 0 1
2
,
( ) .