Math 310 
hw
8 solutions
18.4, 18.6, 18.10;
Friday, 6. Nov 2009
18.4 Let
S
⊂
R
and suppose there is a sequence (
x
n
)
n
in
S
that converges to a number
x
0
/
∈
S
. Show that there exists
an unbounded continuous function on
S
.
We define
f
:
S
→
R
by
f
(
x
) :=
1
x

x
0
for
x
∈
S.
Then
f
is continuous, because it is the composition of the two continuous functions:
x
7→
x

x
0
and
y
7→
1
/y
.
We now verify that
f
is unbounded. Let
M >
0 be given. Then since
x
n
→
x
0
there is
N
such that for all
n
∈
N
,
n > N
implies that
0
<

x
n

x
0

<
1
M
.
Let
n
∈
N
and suppose that
n > N
. Then since

f
(
x
n
)

=
1

x
n

x
0

>
1
1
/M
=
M,
f
is unbounded on
S
.
18.6 Prove that
x
= cos
x
for some
x
in (0
,
π
2
).
We suppose that the function
x
7→
cos
x
is continuous on
R
. Let
f
be the
R
valued function defined by
f
(
x
) =
x

cos
x
for
x
∈
R
. Then
f
is continuous since it is the difference of continuous functions. Observe that
f
(0) =

1
<
0
<
π
2
=
f
π
2
.
Since
f
is continuous on [0
,
π
2
] the Intermediate Value Theorem (18.2) applies and we conclude that there is
x
∈
(0
,
π
2
) such that
x

cos
x
=
f
(
x
) = 0. Hence,
x
= cos
x
as desired.
18.10 Suppose that
f
is continuous on [0
,
2] and that
f
(0) =
f
(2). Prove that there exist
x, y
in [0
,
2] such that

y

x

= 1
and
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 janeday
 Intermediate Value Theorem, Continuous function

Click to edit the document details