Part II: The Answers
302
Answers
401–500
469.
concave up on (–∞, –2), −2
5
2
5
,
, (2, ∞); concave down on − −
2 2
5
,
, 2
5
2
,
Begin by finding the first and second derivatives of the function f
(x) = (x
2
– 4)
3
:
′
(
)
(
)
f x
x
x
x x
( )
( )
=
−
=
−
3
4 2
6
4
2
2
2
2
′′
(
)
(
)
(
)
(
)
f x
x
x
x
x
x
x x
x
( )
( )
=
−
+ ( )( )
−
=
−
+
−
=
−
6
4
6 2
4 2
6
4
24
4
6
2
2
2
2
2
2
2
2
4 4
4 4
6
4 5
4
2
2
2
2
(
) (
)
(
)
(
)(
)
x
x
x
x
− +
=
−
−
Next, set each factor in the second derivative equal to zero and solve for x: x
2
– 4 = 0
gives you the solutions x = 2 and x = –2, and 5x
2
– 4 = 0, or x
2
4
5
= , has the solutions
x = 2
5
and x = −2
5
.
To determine the concavity on the intervals (–∞, –2), − −
2
2
5
,
, − 2
5
2
5
,
,
2
5
2
,
, and (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 = –3, x = –1, x = 0,
x = 1, and x = 3, you have the following:
′′ −
( )= −
( ) −
(
) −
( ) −
(
) >
′′ −
( )= −
( ) −
(
) −
( ) −
(
)
f
f
3 6
3
4 5 3
4
0
1 6
1
4 5 1
4
2
2
2
2
< <
′′ ( ) = −
( ) −
( )>
′′ ( ) = ( ) −
(
) ( ) −
(
) <
′′ ( ) = (
0
0 6 4
4
0
1 6 1
4 5 1
4 0
3 6 3
2
2
f
f
f
) ) −
(
) ( ) −
(
) >
2
2
4 5 3
4
0
Therefore, f
(x) is concave up on the intervals (–∞, –2), −2
5
2
5
,
, and (2, ∞), and f
(x) is
concave down on the intervals − −
2 2
5
,
and 2
5
2
,
.
470.
concave up on 0 253 2
.
, π
(
) , π
π
− 0 253 3
2
.
,
(
) ; concave down on the intervals (0, 0.253),
π π
2
0 253
,
.
−
(
) ,
Begin by finding the first and second derivatives of the function f
(x) = 2 cos x – sin(2x):
′
(
)
f x
x
x
x
x
( )
sin
cos( )
sin
cos( )
= −
−
()
= −
−
2
2
2
2
2
2
3
2
2
π π
,
( )
302
Answers
401–500
469.
concave up on (–∞, –2), −2
5
2
5
,
, (2, ∞); concave down on − −
2 2
5
,
, 2
5
2
,
Begin by finding the first and second derivatives of the function f
(x) = (x
2
– 4)
3
:
′
(
)
(
)
f x
x
x
x x
( )
( )
=
−
=
−
3
4 2
6
4
2
2
2
2
′′
(
)
(
)
(
)
(
)
f x
x
x
x
x
x
x x
x
( )
( )
=
−
+ ( )( )
−
=
−
+
−
=
−
6
4
6 2
4 2
6
4
24
4
6
2
2
2
2
2
2
2
2
4 4
4 4
6
4 5
4
2
2
2
2
(
) (
)
(
)
(
)(
)
x
x
x
x
− +
=
−
−
Next, set each factor in the second derivative equal to zero and solve for x: x
2
– 4 = 0
gives you the solutions x = 2 and x = –2, and 5x
2
– 4 = 0, or x
2
4
5
= , has the solutions
x = 2
5
and x = −2
5
.
To determine the concavity on the intervals (–∞, –2), − −
2
2
5
,
, − 2
5
2
5
,
,
2
5
2
,
, and (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 = –3, x = –1, x = 0,
x = 1, and x = 3, you have the following:
′′ −
( )= −
( ) −
(
) −
( ) −
(
) >
′′ −
( )= −
( ) −
(
) −
( ) −
(
)
f
f
3 6
3
4 5 3
4
0
1 6
1
4 5 1
4
2
2
2
2
< <
′′ ( ) = −
( ) −
( )>
′′ ( ) = ( ) −
(
) ( ) −
(
) <
′′ ( ) = (
0
0 6 4
4
0
1 6 1
4 5 1
4 0
3 6 3
2
2
f
f
f
) ) −
(
) ( ) −
(
) >
2
2
4 5 3
4
0
Therefore, f
(x) is concave up on the intervals (–∞, –2), −2
5
2
5
,
, and (2, ∞), and f
(x) is
concave down on the intervals − −
2 2
5
,
and 2
5
2
,
.
470.
concave up on 0 253 2
.
, π
(
) , π
π
− 0 253 3
2
.
,
(
) ; concave down on the intervals (0, 0.253),
π π
2
0 253
,
.
−
(
) ,
Begin by finding the first and second derivatives of the function f
(x) = 2 cos x – sin(2x):
′
(
)
f x
x
x
x
x
( )
sin
cos( )
sin
cos( )
= −
−
()
= −
−
2
2
2
2
2
2
3
2
2
π π
,
( )