Answers and Explanations 307
Answers
401–500
To determine the concavity on the intervals (–∞, –1), (1, 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 = –2, x = − 1
2
, and x = 1 gives you
′′
′′ ( ) ( )
( )
′′
f
f
f
( )
( )
( )
− =
− +
−
<
−
=
− +
−
>
2
10 2 10
9 2
0
1
2
10 1
2
10
9 1
2
0
4 3
4 3
( ( )
( )
( )
1
10 1 10
9 1
0
4 3
=
+
>
Because the concavity changes at x = –1, the original function has an inflection point
there. The corresponding y value on f
 
(x) is
f ( ) ( )
( )
− = −
− −
= − −
= −
1
1
5 1
1 5
6
5 3
2 3
Therefore, the inflection point is (–1, –6).
476.
no local maxima; local minimum at (0, 1)
Begin by finding the first derivative of the function f x
x
( ) =
+
2
2
3
1
(
) :
′
(
)
(
)
(
)
f x
x
x
x x
x
x
( )
( )
/
=
+
=
+
=
+
−
−
2
3
1
2
4
3
1
4
3
1
2
1 3
2
1 3
2
1 3
Next, find any critical numbers of the function. Setting the numerator of the derivative
equal to zero gives you the solution x = 0. Note that no real values can make the
denominator equal to zero.
Then find the second derivative:
′′
(
)
(
)
(
)
(
)
(
)
f x
x
x
x
x
x
x
( )
( ) ( )
( )
=
+
−
+
+
=
−
3
1
4
4
3 1
3
1
2
3
1
12
2
1 3
2
2 3
2
1 3
2
2 2
1 3
2
2
2 3
2
2 3
2
2 3
2
2
2
1
8
1
9
1
4
1
3
1 2
9
+
−
+
+
=
+
+ −
+
−
−
(
)
(
)
(
)
(
)
(
)
(
)
x x
x
x
x
x
x 1 1
4 3
3 2
9
1
4
12
9
1
2 3
2
2
2
4 3
2
2
4 3
(
)
(
)
(
)
(
)
=
+ −
+
=
+
+
x
x
x
x
x