This preview shows pages 1–2. Sign up to view the full content.
MAT 137Y 200809 Winter Session, Solutions to Problem Set 6
1
(SHE 4.1)
12. We have
f
(
x
) =
x
2
/
3

1.
(a)
f
0
(
x
) =
2
3
x

1
/
3
has no zeroes in
(

1
,
1
)
.
(b)
f
0
(
0
)
clearly does not exist, so
f
is not differentiable at 0, and as such we can not apply
Rolle’s Theorem since we require differentiability for all
x
∈
(

1
,
1
)
. Hence this does not
contradict Rolle’s Theorem.
24. Consider the function
f
(
x
) =
6
x
5
+
13
x
+
1, which is continuous and differentiable for all
x
since
it is a polynomial. Since
f
(

1
) =

18
<
0
<
1
=
f
(
0
)
, by the Intermediate Value Theorem there
exists
c
∈
(

1
,
0
)
such that
f
(
c
) =
0, so
f
(
x
)
has at least one root. Now suppose
c
1
and
c
2
are
distinct real roots (where
c
1
<
c
2
). Then
f
(
c
1
) =
f
(
c
2
)
, so by Rolle’s Theorem there exists
γ
∈
(
c
1
,
c
2
)
such that
f
0
(
γ
) =
0. But
f
0
(
x
) =
30
x
4
+
13
>
0 for all
x
∈
R
, so this contradicts the
assumption that there exist distinct real roots. Hence
f
(
x
)
has exactly one real root.
26. For part (a), suppose that
f
has two zeroes
x
1
,
x
2
∈
(
a
,
b
)
.
f
is differentiable on
(
x
1
,
x
2
)
and
continuous on
[
x
1
,
x
2
]
. By Rolle’s Theorem,
f
0
has a zero in
(
x
1
,
x
2
)
which contradicts the hy
pothesis.
For part (b), if
f
had three zeroes in
(
a
,
b
)
, then by Rolle’s Theorem,
f
0
would have at least two
zeroes in
(
a
,
b
)
and
f
00
would have at least one zero in
(
a
,
b
)
which contradicts the hypothesis.
38. We prove for all
a
,
b
∈
R
that

cos
b

cos
a
 ≤ 
b

a

. If
a
=
b
the proof is obvious. Otherwise,
assume without loss of generality that
a
<
b
. As
f
(
x
) =
cos
x
is continuous and differentiable
for all
x
, then by the Mean Value Theorem, there exists
c
∈
(
a
,
b
)
such that
f
0
(
c
) =
f
(
b
)

f
(
a
)
b

a
=
⇒ 
sin
c
=
cos
b

cos
a
b

a
=
⇒ 
sin
c

=
±
±
±
±
cos
b

cos
a
b

a
±
±
±
±
.
But

sin
c
 ≤
1 for all
c
, so
±
±
±
±
cos
b

cos
a
b

a
±
±
±
±
≤
1
=
⇒ 
cos
b

cos
a
 ≤ 
b

a

.
4
(SHE 4.2)
30. Since
d
dx
(

1
4
x

4

25
4
x
4
/
5
) =
f
0
(
x
)
, it follows that
f
(
x
) =

1
4
x

4

25
4
x
4
/
5
+
C
. But
f
(
1
) =
0, so
0
=

1
4

25
2
+
C
, so
C
=
13
2
. Thus,
f
(
x
) =

1
4
x

4

25
4
x
4
/
5
+
13
2
,
x
>
0.
56. (a) Suppose
f
0
(
x
)
>
g
0
(
x
)
for all
x
∈
(
0
,
c
)
and
f
(
0
) =
g
(
0
)
. Now consider the function
F
(
x
) =
f
(
x
)

g
(
x
)
. Then
F
0
(
x
) =
f
0
(
x
)

g
0
(
x
)
>
0 for all
x
∈
(
0
,
c
)
, so
F
is increasing on
(
0
,
c
)
.
Since
F
(
0
) =
f
(
0
)

g
(
0
) =
0, it follows that
F
(
x
)
>
0 on
(
0
,
c
)
. Therefore
f
(
x
)

g
(
x
)
>
0
or
f
(
x
)
>
g
(
x
)
on
(
0
,
c
)
.
(b) Let
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 UPPAL

Click to edit the document details