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Math 2413
Homework Set 8 – Solutions
10 Points
1.
(2 pts)
()
() ( )( )
43
2
32
2
3
1
5
1
2
42
3
0
2
325
fx
xx
x
f
x
x
x
¢
¢¢
=+

=+=+
So, it looks like possible inflection points are :
5
2
3,0,
x
=
.
Here is a number line for the 2
nd
derivative.
From this we get the following concavity information.
55
22
Concave Up :
3
0
,
Concave Down :
3
,0
x
x
<
<
< <
¥
¥

<<
The inflection points are then :
5
2
3,0,
x
.
4. (2 pts)
This is a polynomial and so is continuous and differentiable everywhere, in particular, on the
given interval and so the Mean Value Theorem can be used.
() ( ) ()
2
6
1
4
1
2
3
6
16
g
x
gg
¢
+

=
=
Applying the Mean Value Theorem and solving gives,
() ( )
( )
2
2
1
6
12
42
6
1
4
1
14
1
23
6
1
4
1
3
0
7
12
7
0.7116
, 3.0449
ff
cc
fc
c

¢

+
=
=
==
 
= &
=
–
The second solution is not in the interval and the first solution is in the interval and the only value that
satisfies the Mean Value Theorem is 0.7116.
8. (2 pts)
A quick sketch is to the right.
Here are the equations we need.
Maximize :
Constraint : 10
02
V
r
h
r
hr
p
pp
=
=+
Solve the constraint for
h
and plug into the volume equation.
( ) ()
2
223
1
11
2
10
0
10
0
50
rr
h
r
V
r
r
r
Øø
=

=

ºû
2
3
2
5
03
V
r
r
V
¢
=

From the first derivative we get the critical points :
100
3
3.2574
r
=
–
=–
.
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 Spring '07
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 Calculus, Inflection Points

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