Math 125A, Fall 2009
Solutions for Homework 2
(Prepared by Matt Low)
1. Let
f
(
x
) =
x

cos
x
. By direct computation, we see that
f
(0)
<
0 and
f
(
π
2
)
>
0. Hence, by the
intermediate value theorem (Theorem 18.2), there exists a number
c
∈
(0
,
π
2
) such that
f
(
c
) = 0.
In other words,
c
= cos
c
.
2. (a) True. By Theorem 18.1, there exists a
c
∈
[
a, b
] such that
f
(
x
)
≤
f
(
c
) for all
x
∈
[
a, b
].
(b) False. For example, the function
f
: [0
,
2]
→
R
deﬁned by:
f
(
x
) =
±
x,
if 0
≤
x <
1

x
+ 3
,
if 1
≤
x
≤
2
assumes a maximum value, but is not continuous.
(c) True. By Theorem 18.1, there exist numbers
c, d
∈
[

2
,
2] for which
f
(
c
)
≤
f
(
x
)
≤
f
(
d
)
for all
x
∈
[

2
,
2], and according to the intermediate value theorem,
f
assumes every value
in between, so
f
([

2
,
2]) = [
f
(
c
)
, f
(
d
)].
Remark (by Xia): In the case that
f
(
c
) =
f
(
d
), we get only a single point. So, a more
suitable answer to this question is “False” as a single point is not an interval. The countex