Answers and Explanations 309
Answers
401–500
Solving 4x(1 – 2x
2
) = 0 gives you x = 0, x = 1
2
, and x = − 1
2
as the critical numbers.
Next, find the second derivative:
′′
f x
x
( ) = −
4 24
2
Substitute the critical numbers into the second derivative to see whether the second
derivative is positive or negative at those values:
′′ −
= −
−
= − <
′′ ( ) = − ( ) = >
′
f
f
1
2
4 24
1
2
4 12 0
0 4 24 0
4 0
2
2
′ ′
= −
= − <
f
1
2
4 24 1
2
4 12 0
2
So the original function has local maxima when x = 1
2
and x = − 1
2
and a local minimum
when x = 0. Finding the corresponding y values on the original function gives you the
following:
f
f
1
2
2 1
2
2 1
2
2
2
2
4
1
2
0 2 0
2
2
4
2
=
−
= − =
( ) = ( ) − 0 0 0
1
2
2
1
2
2
1
2
2
2
2
4
1
2
4
2
4
( ) =
−
= −
− −
= − =
f
Therefore, the local maxima are at − 1
2
1
2
,
and 1
2
1
2
,
, and the local minimum is
at (0, 0).
479.
local maximum at 2 1
4
,
( ) ; local minimum at − −
2 1
4
,
( )
Begin by finding the first derivative of the function f x
x
x
( ) =
+
2
4
:
′
(
)
(
)
(
)
f x
x
x x
x
x
x
( )
( )
( )
=
+
−
+
= − +
+
2
2
2
2
2
2
4 1
2
4
4
4
Then set the numerator and denominator of the first derivative equal to zero to find
the critical numbers. Setting the numerator equal to zero gives you –x
2
+ 4 = 0, which
has the solutions x = 2 and x = –2. Notice that the denominator of the derivative
doesn’t equal zero for any value of x.
Answers
401–500
Solving 4x(1 – 2x
2
) = 0 gives you x = 0, x = 1
2
, and x = − 1
2
as the critical numbers.
Next, find the second derivative:
′′
f x
x
( ) = −
4 24
2
Substitute the critical numbers into the second derivative to see whether the second
derivative is positive or negative at those values:
′′ −
= −
−
= − <
′′ ( ) = − ( ) = >
′
f
f
1
2
4 24
1
2
4 12 0
0 4 24 0
4 0
2
2
′ ′
= −
= − <
f
1
2
4 24 1
2
4 12 0
2
So the original function has local maxima when x = 1
2
and x = − 1
2
and a local minimum
when x = 0. Finding the corresponding y values on the original function gives you the
following:
f
f
1
2
2 1
2
2 1
2
2
2
2
4
1
2
0 2 0
2
2
4
2
=
−
= − =
( ) = ( ) − 0 0 0
1
2
2
1
2
2
1
2
2
2
2
4
1
2
4
2
4
( ) =
−
= −
− −
= − =
f
Therefore, the local maxima are at − 1
2
1
2
,
and 1
2
1
2
,
, and the local minimum is
at (0, 0).
479.
local maximum at 2 1
4
,
( ) ; local minimum at − −
2 1
4
,
( )
Begin by finding the first derivative of the function f x
x
x
( ) =
+
2
4
:
′
(
)
(
)
(
)
f x
x
x x
x
x
x
( )
( )
( )
=
+
−
+
= − +
+
2
2
2
2
2
2
4 1
2
4
4
4
Then set the numerator and denominator of the first derivative equal to zero to find
the critical numbers. Setting the numerator equal to zero gives you –x
2
+ 4 = 0, which
has the solutions x = 2 and x = –2. Notice that the denominator of the derivative
doesn’t equal zero for any value of x.