Part II: The Answers
300
Answers
401–500
f π
π
π
π
π
6
6
2
6
6
2
3
2
6
3
( )
( )
= +
= +
= +
cos
f 5
6
5
6
2
5
6
5
6
2
3
2
5
6
3
π
π
π
π
π
( )
( )
=
+
=
+
−
=
−
cos
Therefore, the local maxima are −
−
+
11
6
11
6
3
π
π
,
(
) and π π
6 6
3
, +
(
) , and the local
minima are −
− −
7
6
7
6
3
π
π
,
(
) and 5
6
5
6
3
π π
,
−
(
) .
466.
concave up on (1, ∞); concave down on (–∞, 1)
To determine concavity, examine the second derivative. Because a derivative measures
a “rate of change” and the first derivative of a function gives the slopes of tangent
lines, the derivative of the derivative (the second derivative) measures the rate of
change of the slopes of the tangent lines. If the second derivative is positive on an
interval, the slopes of the tangent lines are increasing, so the function is bending
upward, or is concave up. Likewise, if the second derivative is negative on an interval,
the slopes of the tangent lines are decreasing, so the function is bending downward, or
is concave down.
Begin by finding the first and second derivatives of the function f
(x) = x
3
– 3x
2
+ 4:
′
=
−
′′
=
−
f x
x
x
f x
x
( )
( )
3
6
6
6
2
Setting the second derivative equal to zero gives you 6x – 6 = 0 so that x = 1.
To determine the concavity on the intervals (–∞, 1) and (1, ∞), 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 = 0 and x = 2, you have f
"(0) = 6(0) – 6 < 0 and f
"(2) = 6(2) – 6 > 0.
Therefore, f
(x) is concave up on the interval (1, ∞) and concave down on the
interval (–∞, 1).
467.
concave up nowhere; concave down on (–∞, 0), (0, ∞)
Begin by finding the first and second derivatives of the function f
(x) = 9x
2/3
– x:
′
( )
f x
x
x
( ) =
−
=
−
−
−
9 2
3
1
6
1
1 3
1 3
300
Answers
401–500
f π
π
π
π
π
6
6
2
6
6
2
3
2
6
3
( )
( )
= +
= +
= +
cos
f 5
6
5
6
2
5
6
5
6
2
3
2
5
6
3
π
π
π
π
π
( )
( )
=
+
=
+
−
=
−
cos
Therefore, the local maxima are −
−
+
11
6
11
6
3
π
π
,
(
) and π π
6 6
3
, +
(
) , and the local
minima are −
− −
7
6
7
6
3
π
π
,
(
) and 5
6
5
6
3
π π
,
−
(
) .
466.
concave up on (1, ∞); concave down on (–∞, 1)
To determine concavity, examine the second derivative. Because a derivative measures
a “rate of change” and the first derivative of a function gives the slopes of tangent
lines, the derivative of the derivative (the second derivative) measures the rate of
change of the slopes of the tangent lines. If the second derivative is positive on an
interval, the slopes of the tangent lines are increasing, so the function is bending
upward, or is concave up. Likewise, if the second derivative is negative on an interval,
the slopes of the tangent lines are decreasing, so the function is bending downward, or
is concave down.
Begin by finding the first and second derivatives of the function f
(x) = x
3
– 3x
2
+ 4:
′
=
−
′′
=
−
f x
x
x
f x
x
( )
( )
3
6
6
6
2
Setting the second derivative equal to zero gives you 6x – 6 = 0 so that x = 1.
To determine the concavity on the intervals (–∞, 1) and (1, ∞), 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 = 0 and x = 2, you have f
"(0) = 6(0) – 6 < 0 and f
"(2) = 6(2) – 6 > 0.
Therefore, f
(x) is concave up on the interval (1, ∞) and concave down on the
interval (–∞, 1).
467.
concave up nowhere; concave down on (–∞, 0), (0, ∞)
Begin by finding the first and second derivatives of the function f
(x) = 9x
2/3
– x:
′
( )
f x
x
x
( ) =
−
=
−
−
−
9 2
3
1
6
1
1 3
1 3