SOLUTIONS CHAPTER 14.2
MATH 132 WI01
2.
f
(
x
) =

2
x
2

6
x
+5, and the interval of definition is [

2
,
3]. To find
absolute
extrema we are intersted in finding a few good numbers, as few as possible, and
plug those numbers in
f
, and compare the results  by picking the largest result we
want to get the
absolute maximum
, and by picking the smallest result we want
to have the
absolute minimum
. How do we find those numbers? Two candidates
are the definition interval’s
endpoints
 in our case

2 and 3. The others are going
to be the
critical numbers
, that is, zeroes and DNEs
1
of the derivative of
f
,
f
0
.
Let’s compute
f
0
:
f
0
(
x
) =

2
*
2
x

6 =

4
x

6.
f
0
(
x
) = 0
→ 
4
x

6 = 0
→

4
x
= 6
→
x
=
6
(

4)
→
x
=

3
2
.
f
0
exists for every
x
, so there’s no case of DNE,
so

3
2
is the only critical number. The last thing we must make sure is whether
the critical number is INSIDE the interval [

2
,
3] ... well, in our case it is! Let’s
start pluging in numbers:
f
(

2) =

2
*
(

2)
2

6
*
(

2) + 5 =

2
*
4 + 12 + 5 =

8 + 12 + 5 = 9
f
(3) =

2
*
3
2

6
*
3 + 5 =

2
*
9

18 + 5 =

18

18 + 5 =

27
f
(

3
2
) =

2
*
(

3
2
)
2

6
*
(

3
2
)+5 =

2
*
9
4
+3
*
3+5 =

9
2
+9+5 =

4
.
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 Spring '08
 Staff
 Math, Calculus, Critical Point, absolute maximum, absolute minimum

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