Math 3110 Homework 14 Solutions
Exercise 15.1.3
If
f
(
x
) changes sign from negative to positive on [
a, b
], then there are some
h, k
∈
[
a, b
] with
h < k
,
f
(
h
)
<
0, and
f
(
k
)
>
0. Since
f
(
x
) is diﬀerentiable on [
a, b
] and [
h, k
]
⊆
[
a, b
],
f
(
x
) is diﬀerentiable on [
h, k
]. Then by Theorem 15.1, there is some
c
∈
[
h, k
] such that
f
(
k
)

f
(
h
) =
f
0
(
c
)(
k

h
), from which
f
0
(
c
) =
f
(
k
)

f
(
h
)
k

h
. Since
f
(
k
)
>
0
> f
(
h
) and
k > h
,
f
0
(
c
) is a quotient of two positive numbers, and so
f
0
(
c
)
>
0.
Exercise 15.2.1
(a)
We prove the contrapositive, namely, that if
f
(
x
) is not bounded on
I
, then neither is
f
0
(
x
). Suppose that
f
(
x
) is not bounded on
I
. Pick some
a
∈
I
. Let the length of the
interval
I
be
`
. For any
m >
0, since
f
(
x
) is not bounded, there is some
b
∈
I
such that

f
(
b
)

> m`
+

f
(
a
)

. Hence,

f
(
b
)

f
(
a
)
 ≥ 
f
(
b
)

f
(
a
)

> m`
+

f
(
a
)

f
(
a
)

=
m`
. By
Theorem 15.1, there is some
c
∈
I
(as
I
contains all points between
a
and
b
, regardless of
whether the interval is [
a, b