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
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 Spring '08
 UPPAL
 Derivative, Sin, Cos, Mathematical analysis, Convex function

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