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Math 2413
Homework Set 7 – Solutions
10 Points
1. (2 pts)
() ( )
( ) ( )( )
( )
( ) ( ) ( ) ( )
2
43
22
33
2
2
4
4
1
4
1
6
4
14
4
4
1
4
1
4
4
4
4
4
1
4
8
16
g
z z
z
z
zz
z
z
z
¢
=
 ++
+
Øø
=+
++ =+
 +
ºû
The five critical points are then,
( )
2
1
4
2
1
16
4
02
1
40
8
1
60
1
51
3
1.4781
, 1.3531
z

=
=–
+
=&
=
+=
=

–
4. (2 pts)
()
( ) ( )
2
12
1
3
2
2
3
2
2
2
3
6
87
2
4
28
2
8
7
2
4
2
8
3
2
5
4
0
53
3
t
t
tt
P
t
t
t
t t tt


+
¢
=


++



+=
=
Critical points are then where the derivative is zero (
i.e.
where the numerator is zero) and where the
derivative doesn’t exist (
i.e.
where the denominator is zero). For the critical points we need to recall
that the function also has to exist at these points (this is especially true for the second set in this case).
From the numerator we get the two critical points :
1
5
2
0 3 1
5
1.6762
, 6.3238
t
=–=
.
From the denominator we get the two critical points :
1
,7
t
=
(function exists at these two points!).
7. (2 pts)
From
#1
we know that the critical points are :
1
4
2
, ,
1.4781
z
=
–
.
However the
only ones that are in the given interval are :
1
4
2
1.4781
z
=

.
So evaluating the function at these
three critical points and the endpoints of the interval gives,
( ) ( ) ( ) ()
1
4
2
0
0
1.4781
4
1918.817
2
2.
1
504.074
3 1
5625
gg
g

=

=

=

==
So the absolute max is 5625 which occurs at
1
z
=
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This note was uploaded on 11/13/2010 for the course MA 141 taught by Professor Wears during the Spring '07 term at N.C. State.
 Spring '07
 WEARS
 Math, Calculus

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