Proof:
Let
a >
0. Set
p
(
x
) =
x
2

a
on
[0
, b
]
where
b
2
> a
. For example,
let
b
=
a
+ 1. Then,
p
(0) =

a <
0 and
p
(
b
) =
b
2

a >
0. Therefore, by the
IntermediateValue Theorem, there exists at least one number
c
∈
(0
, b
)
such
that
p
(
c
) = 0.
That is,
c
2

a
= 0,
which implies
c
=
√
a
.
5. Fix a positive number
P
.
Let
R
be the set of all rectangles with perimeter
P
= 4. Prove that there is a member of
R
that has maximum area.
Proof:
Let
R
be a member of
R
,
and let
R
have length
x
and width
y
.
Since the perimeter is
4,
2
x
+ 2
y
= 4
so
y
= 2

x
.
The area of
R
is
A
(
x
) =
x
(2

x
) = 2
x

x
2
,
0
≤
x
≤
2
(since
x
and
y
are each
nonnegative). Now
A
is a continuous function on the compact set
D
= [0
,
2].
Therefore
A
has a maximum value (and a minimum value which is 0). That
is, there is a number
c
∈
[0
,
2]
such that
A
(
x
)
≤
A
(
c
)
for all
x
∈
[0
,
2].
6. Let
f
: [
a, b
]
→
R
and
g
: [
a, b
]
→
R
be continuous functions. Prove that
if
f
(
a
)
< g
(
a
)
and
g
(
b
)
< f
(
b
),
then there is at least one number
c
∈
(
a, b
)
such that
f
(
c
) =
g
(
c
).
Proof:
Set
h
(
x
) =
f
(
x
)

g
(
x
). Then
h
: [
a, b
]
→
R
is continuous,
h
(
a
) =
f
(
a
)

g
(
a
)
<
0
and
h
(
b
) =
f
(
b
)

g
(
b
)
>
0. Therefore, by the Intermediate
Value Theorem, there is a number
c
∈
(
a, b
) such that
h
(
c
) =
f
(
c
)

g
(
c
) = 0
which implies
f
(
c
) =
g
(
c
).
7. Let
f
:
D
→
R
. Prove that if
f
is bounded on a neighborhood of each
x
∈
D
and
D
is compact, then
f
is bounded on
D
.
Proof:
For each
x
∈
D
there is a neighborhood
N
(
x
)
of
x
and a
number
M
x
such that

f
(
x
)
 ≤
M
x
for all
x
∈
N
(
x
).
The collection of
neighborhoods
N
(
x
)
, x
∈
D
is an open cover of
D
. Since
D
is compact,
this open cover has a finite subcover,
N
(
x
1
)
, N
(
x
2
)
,
· · ·
, N
(
x
k
). Associated
with each neighborhood
N
(
x
p
)
,
1
≤
p
≤
k
there is a number
M
x
p
.
Let
M
= max
M
x
1
, M
x
2
,
· · ·
, M
x
p
.
Now, choose any point
z
∈
D
. Then
z
∈
N
(
x
p
)
for some
p
∈ {
1
,
2
,
· · ·
, k
}
.
This implies that,

f
(
z
)
 ≤
M
x
p
≤
M
.
Therefore we can conclude that

f
(
z
)
 ≤
M
for all
z
∈
D
which means
f
is bounded on
D
.
8. Use the definition of the derivative to find the derivative of
f
(
x
) =
√
4 +
x
at
x
= 5.
METHOD 1:
lim
x
→
5
f
(
x
)

f
(5)
x

5
= lim
x
→
5
√
4 +
x

3
x

5
= lim
x
→
5
√
4 +
x

3
x

5
·
√
4 +
x
+ 3
√
4 +
x
+ 3
= lim
x
→
5
x

5
(
x

5)(
√
4 +
x
+ 3)
= lim
x
→
5
1
√
4 +
x
+ 3
=
1
6
Therefore,
f
0
(5) =
1
6
.
METHOD 2:
Alternatively, in terms of
h
, we have
lim
h
→
0
f
(5 +
h
)

f
(5)
h
= lim
h
→
0
p
4 + (5 +
h
)

3
h
= lim
h
→
0
√
9 +
h

3
h
·
√
9 +
h
+ 3
√
9 +
h
+ 3
= lim
h
→
0
h
h
(
√
9 +
h
+ 3)
= lim
h
→
0
1
√
9 +
h
+ 3
=
1
6
Therefore,
f
0
(5) =
1
6
.
9. Use the definition of the derivative to find the derivative of
f
(
x
) =
1
2
x
+ 1
.
METHOD 1:
For any number
c
6
=

1
/
2, we have
lim
x
→
c
f
(
x
)

f
(
c
)
x

c
= lim
x
→
c
1
2
x
+ 1

1
2
c
+ 1
x

c
= lim
x
→
c
(2
c
+ 1)

(2
x
+ 1)
(2
x
+ 1)(2
c
+ 1)
x

c
= lim
x
→
c
(2
c
+ 1)

(2
x
+ 1)
(
x

c
)(2
x
+ 1)(2
c
+ 1)
= lim
x
→
c

2
(2
x
+ 1)(2
c
+ 1)
=

2
(2
c
+ 1)
2
Therefore,
f
0
(
x
) =

2
(2
x
+ 1)
2
,
for
x
6
=

1
2
.
METHOD 2:
For any number
c
6
=

1
/
2, we have
lim
h
→
0
f
(
c
+
h
)

f
(
c
)
h
= lim
h
→
0
1
2(
c
+
h
) + 1

1
2
c
+ 1
h
= lim
h
→
0
(2
c
+ 1)

(2
c
+ 2
h
+ 1)
(2
c
+ 2
h
+ 1)(2
c
+ 1)
h
= lim
h
→
0
(2
c
+ 1)

(2
c
+ 2
h
+ 1)
h
(2
c
+ 2
h
+ 1)(2
c
+ 1)
= lim
h
→
0

2
h
h
(2
c
+ 2
h
+ 1)(2
c
+ 1)
= lim
h
→
0

2
(2
c
+ 2
h
+ 1)(2
c
+ 1)
=

2
(2
c
+ 1)
2
Therefore,
f
0
(
x
) =

2
(2
x
+ 1)
2
,
for
x
6
=

1
2
.
10. Set
f
(
x
) =
(
x
3
+ 3
x
+ 2
,
x
≤
1
3
x
2
+ 3
,
x >
1
.
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 Fall '08
 Staff
 Math, Calculus, Derivative, lim, Continuous function