Math 508
Exam 2
Jerry L. Kazdan
December 8, 2006
12:00 – 1:20
Directions
This exam has two parts, Part A has 3 shorter problems (8 points each, so 24 points),
Part B has 5 traditional problems (15 points each, so 75 points).
Closed book, no calculators – but you may use one 3
00
×
5
00
card with notes.
Part A: Short Problems
(3 problems, 8 points each).
A–1. A continuous function
f
:
R
→
R
has the property that
Z
x
0
f
(
t
)
dt
= cos(
x
)
e

x
+
C,
where
C
is some constant. Find both
f
(
x
) and the constant
C
.
Solution:
Letting
x
= 0, we find that 0 = 1 +
C
, so
C
=

1.
To compute
f
, use the
fundamental theorem of calculus. Thus take the derivative of both sides
f
(
x
) =
d
dx
[cos(
x
)
e

x
+
C
] =

sin(
x
)
e

x

cos(
x
)
e

x
.
A–2. A function
h
:
R
→
R
with two continuous derivatives has the property that
h
(0) = 2,
h
(1) = 0, and h(3)=1. Prove there is at least one point
c
in the interval 0
< x <
3 where
h
00
(
c
)
>
0 by finding some
explicit
m >
0 (such as
m
= 3
/
2) with
h
00
(
c
)
≥
m
.
Solution:
By the mean value theorem applied twice, there is some
a
∈
(0
,
1) and some
b
∈
(1
,
3) so that
h
0
(
a
) =
h
(1)

h
(0)
1

0
=

2
,
h
0
(
b
) =
h
(3)

h
(1)
3

1
=
1
2
.
Thus, by the mean value theorem again there is some
c
∈
(
a, b
) so that
h
00
(
c
) =
h
0
(
b
)

h
0
(
a
)
b

a
=
1
2
+ 2
b

a
>
5
/
2
3
=
5
6
.
A–3. Say a smooth function
u
(
x
) satisfies
u
00

c
(
x
)
u
= 0
for 0
≤
x
≤
1 (here
c
(
x
) is some given
contunuous function).
If
c
(
x
)
>
0 everywhere, show that there is
no
point where
u
(
x
) is both positive
and
has a
local maximum.
If we also knew that
u
(0) = 0 and
u
(1) = 0, why can we conclude that
u
(
x
) = 0 for all
0
≤
x
≤
1?
Solution:
If
u
has a positive maximum at some point
p
, then
u
00
(
p
)
≤
0 and
u
(
p
)
>
0.
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 Fall '10
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 Math, Topology, Continuous function, Metric space, t dt

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