Math 2250 Written HW #12 Solutions
1. Consider
f
(
x
) =
1
4
x
2
+ 7
(a) Find all critical points of
f
(
x
). For each one determine whether it is a local maximum,
a local minimum, or neither.
Answer:
To find the critical points, first we need to compute
f
0
(
x
) =

1
(4
x
2
+ 7)
2
·
8
x
=

8
x
(4
x
2
+ 7)
2
.
Since the denominator is always positive,
f
0
(
x
) = 0 exactly when the numerator is zero,
so the only critical point occurs when
x
= 0, so this is the point (0
,
1
/
7).
Now, again since the denominator is always positive,
f
0
(
x
) has the same sign as the
numerator, which is

8
x
.
Therefore,
f
0
(
x
) changes sign from positive to negative at
x
= 0, so the first derivative test implies that (0
,
1
/
7) is a local maximum.
(b) If it exists, what is the absolute maximum of
f
(
x
)?
If it exists, what is the absolute
minimum?
Answer:
Since
f
(
x
) has only one critical point, and that critical point is a local max
imum, it must also be an absolute maximum. Therefore, the absolute maximum value
of
f
(
x
) is 1
/
7. On the other hand, any absolute minimum would also have to be a local
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 Fall '08
 CHESTKOFSKY
 Calculus, Critical Point, Derivative, Cos, lim, Mathematical analysis

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