Answers, Assignment #13
Exercises 3.3
1. Set
f
(
x
)=
x
3

4
x
+ 2. Then
f
(

3) =

13
,f
(

2) = 2;
f
has a zero in (

3
,

2);
f
(

1)=5
(0) = 2
(1) =

1;
f
has a zero in (0
,
1);
f
(2) = 2;
f
has a zero in (1
,
2).
2. Set
f
(
x
) = sin
x
+ 2 cos
x

x
2
. Then
f
(0) = 1
(
π/
2)=1

π
2
4
<
0. Therefore,
f
has a
zero in (0
,π/
2) =
⇒
sin
x
+ 2 cos
x
=
x
2
for some
x
in (0
2).
3. Set
f
(
x
x
2

2. Then
f
(0) =

2
(2) = 2. Therefore,
f
has a zero in (0
,
2) =
⇒
x
2
=2
for some
x
∈
(0
,
2).
4. (a) False. The statement says that a continuous function on a domain
D
has a maximum
value. There are many counterexamples: (1)
f
(
x
1
x
on
D
=(0
,
∞
). (2)
f
(
x
x
on (0
,
1). A continuous function on a FINITE CLOSED INTERVAL (i.e., compact)
DOES have a maximum value (and a minimum value).
(b) False. Let
f
(
x
1
x
on
D
,
1].
D
is bounded,
f
(
D
)=[1
,
∞
) is not.
(c) True. If
k
=
f
(
a
) or if
k
=
f
(
b
), then we are done. If
f
(
a
)
<k<f
(
b
), then apply
Thoeorem 13.
(d) False.
f
(
x
x
on (0
,
1).
(e) False.
f
(
x
x
on (0
,
1).
f
has neither a maximum nor a minimum value.
5. (a) False.
f
(
x
)
≡
1 on (0
,
1); (0
,
1) is open,
f
[(0
,
1)] =
{
1
}
is closed.
(b) False. Take
D
=
R
and
f
(
x
e
x
.
D
is closed and
f
(
D
)=(0
,
∞
) is open.
(c) True.
f

1
[
f
(
D
)] =
D
. By Theorem 10 and its Corollary, the inverse image of an open
set is open.
(d) False. Take
D
=(

π,π
) and
f
(
x
) = sin
x
.
D
is not closed and
f
(
D
)=[

1
,
1] is
closed. The example in (a) will also work.
(e) False. The example in (a) works here; (0
,
1) is not compact,
f
[(0
,
1)] =
{
1
}
is compact.
(f) False.
f
(
x
) = sin
x
on (
∞
,
∞
); (
∞
,
∞
) is not bounded,
f
[(
∞
,
∞
)] = [

1
,
1] is
bounded.