Part II: The Answers
310
Answers
401–500
Next, find the second derivative of the function:
′′
(
)
+
(
) (
)
(
)
(
)
(
f x
x
x
x
x
x
x
x x
( )
(
)
( )
(
)
=
+
−
− −
+
+
=
−
+
2
2
2
2
2
4
2
4
2
4 2
4 2
4
2
4 ) )
(
)(
)
(
)
(
) (
) (
)




2
2
2
2
4
2
2
2
4
4
4
4
2
4
4 2
4
−
+
− +
+
=
−
+
+ + − +
x x
x
x
x x
x
x
x
(
)
2 2
4
2
2
3
4
2
12
4
+
=
−
−
+
(
)
(
)
(
)
(
)
x
x
x
Substitute the critical numbers into the second derivative to see whether the second
derivative is positive or negative at those values:
′′
′′
f
f
( )
( )( )
( )
( )
( )( )
( )
− =
>
=
−
<
2
4 8
8
0
2
4 8
8
0
3
3
Because f
 
(x) is concave up when x = –2 and concave down when x = 2, a local minimum
occurs at x = –2, and a local maximum occurs when x = 2. Because f ( ) ( )
− =
−
−
+
= −
2
2
2
4
1
4
2
and f ( )
2
2
2 4
1
4
2
=
+
= , the local minimum occurs at − −
2 1
4
,
( ) , and
the local maximum occurs at 2 1
4
,
( ) .
480.
local maximum at π
π
3
3 3
,
−
(
) ; local minimum at 5
3
3 5
3
π
π
, −
−
(
)
Begin by finding the first derivative of the function f
 
(x) = 2 sin x – x:
′
f x
x
( )
cos
=
−
2
1
Set the first derivative equal to zero to find the critical numbers. From the equation 2
cos x – 1 = 0, you get cos x = 1
2
, so the critical numbers are x = π
3
and x = 5
3
π .
Next, find the second derivative:
′′
f x
x
( )
sin
= −2
Substitute the critical numbers into the second derivative to see whether the second
derivative is positive or negative at these values:
′′ ( ) = − = −


 


 
<
′′ ( ) = −
= − −


 

f
f
π
π
π
π
3
2
3
2
3
2
0
5
3
2
5
3
2
3
2
sin
sin
 
 
> 0