Part II: The Answers
292
Answers
401–500
Setting the derivative equal to zero and solving gives you the critical numbers x = 0
and x = ±1, all of which fall within the given interval. Next, substitute the endpoints of
the interval along with the critical numbers into the original function and pick the
largest and smallest values:
f
f
f
( ) ( )
( )
( ) ( )
( )
( )
( )
− = −
− −
+ =
− = −
− −
+ =
= −
+
2
2
2 2
4 12
1
1
2 1
4 3
0 0 2 0
4
2
4
2
4
4 4 4
1 1 2 1
4 3
3 3 2 3
4 67
4
2
4
2
=
= −
+ =
= −
+ =
f
f
( )
( )
( )
( )
Therefore, the absolute maximum is 67, and the absolute minimum is 3.
453.
absolute maximum: 1
2
; absolute minimum: 0
Begin by finding the derivative of the function; then find any critical numbers on the
given interval. The derivative of f x
x
x
( ) = +
2
1
is
′
(
)
(
)
(
)
f x
x
x x
x
x
x
( )
( )
( )
=
+
−
+
=
−
+
2
2
2
2
2
2
1 1
2
1
1
1
Next, find the critical numbers by setting the numerator equal to zero and solving
for x (note that the denominator will never be zero):
1
0
1
1
2
2
−
=
=
± =
x
x
x
Only x = 1 is in the given interval, so don’t use x = –1. Next, substitute the endpoints
of the interval along with the critical number into the original function and pick the
largest and smallest values:
f
f
f
( )
( )
( )
0 0
1
1
2
3
3
10
=
=
=
Therefore, the absolute maximum is 1
2
, and the absolute minimum is 0.
454.
absolute maximum: 2; absolute minimum: − 3
Begin by finding the derivative of the function; then find any critical numbers on the
given interval. The derivative of f t
t
t
t
t
( ) =
− =
−
4
4
2
2 1 2
( ) is
′
( )
( )
( )
( )
−
f t
t
t
t
t
t
t
t
t
( )
(
)
=
−
+
−
−
= −
−
−
=
−
1 4
1
2
4
2
4
4
4 2
2 1 2
2
1 2
2
1 2
2
2 1 2
2 2
2
1 2
4 − t
( )