‡
Solutions to Practice Problems for Test 2, Math 2015
1)
Ÿ

1
5
H
3
f
H
x
L

g
H
x
LL
„
x
=
3
Ÿ

1
5
f
H
x
L
„
x

Ÿ

1
5
g
H
x
L
„
x
=
3
I
Ÿ

1
2
f
H
x
L
„
x
+
Ÿ
2
5
f
H
x
L
„
x
M

I
Ÿ

1
2
g
H
x
L
„
x
+
Ÿ
2
5
g
H
x
L
„
x
M
=
3
H
3

1
L

H
5
+
2
L
=
1
There is no rule that would allow us to calculate
Ÿ

1
2
f
H
x
L
g
H
x
L
„
x
.
2)
M
£
=
28

7
M
can be rewritten as
M
£
=
7
H
M

4
L
, so the equilibrium solution is
M
= 4 and the general solution is
M
=
4
+
Ce

7
t
. Solving
M
(0)=5 gives 5 = 4+
C
, or
C
= 1.
So the solution is
M
=
4
+
e

7
t
.
3)
F
is increasing where
f
is positive, so from about
x
= 0 to
x
ª
1.6.
F
is then decreasing for
x
greater than about 1.6.
The
largest value for
F
occurs at the local maximum at
x
ª
1.6, and the smallest value is at the end of the interval (after
F
has
been decreasing) at
x
= 4.
A graph of the antiderivative
F
is shown below:
1.5
2
2.5
3
3.5
4
6
4
2
4) a) Avg Value =
1
ÅÅÅÅÅÅÅÅ
6

0
Ÿ
0
6
1
ÅÅÅÅ
3
H
30
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 Spring '06
 AWKohler
 Math, Order theory, Y1 f HxL, Y0 ÅÅÅÅ H30, ÅÅÅÅÅÅÅÅÅ7 ÅÅÅ +c

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