Solutions, Assignment #6
Section 21.
3.
lim
x
→
5
x
2

4
x

5
x

5
= lim
x
→
5
(
x
+ 1)(
x

5)
x

5
= lim
x
→
5
(
x
+ 1) = 6. Set
f
(5) = 6.
6.
(a)
True. Set
g
(
x
)
≡
k
,
constant.
Then
g
is continuous on
R
. By Theorem 21.6 (a)
g
·
f
=
kf
is continuous on
D
.
(b)
True.
g
= (
f
+
g
)

f
.
(c)
False.
f
(
x
) =
x
2
, g
(
x
) =
1
/x
x
negationslash
= 0
0
x
= 0
on
[

1
,
1].
f
is continuous,
g
is not
continuous at
x
= 0,
f
(
x
)
g
(
x
) =
x
is continuous.
(d)
False.
Set
f
(
x
) =
1
if
x
is a rational number

1
if
x
is an irrational number
on
[

1
,
1].
f
2
≡
1
is
continuous,
f
is everywhere discontinuous.
(e)
False. Set
f
(
x
) = 1
/x
on
D
= (0
,
1].
(f)
False. Let
f
be the example in (d) and let
g
=

f
;
f
+
g
≡
0 which is continuous on
[

1
,
1].
(g)
False. Let
f
be the example in (d) and let
g
=

f
;
f
·
g
≡
1
which is continuous on
[

1
,
1].
(h)
False. Let
f
and
g
be the function in (d). Then
(
g
◦
f
)(
x
) = 1
for all
x
∈
[

1
,
1];
f
and
g
are not continuous,
g
◦
f
is continuous.
13.
f
is continuous at
c
and
f
(
c
)
>
0. Let
epsilon1
=
f
(
c
)
/
2
>
0. Since
lim
x
→
c
f
(
x
) =
f
(
c
),
there is a
neighborhood
N
(
c, δ
) =
U
such that

f
(
x
)

f
(
c
)

< epsilon1
for all
x
∈
U
∩
D
. Now

f
(
x
)

f
(
c
)

< epsilon1
implies

epsilon1 < f
(
x
)

f
(
c
)
< epsilon1.
It now follows that
f
(
c
)

epsilon1 < f
(
x
)
< epsilon1
+
f
(
c
)
or
f
(
c
)
2
< f
(
x
)
<
3
f
(
c
)
2
.
for all
x
∈
U
∩
D
. In particular,
f
(
x
)
> f
(
c
)
/
2 =
α
for all
x
∈
U
∩
D
.
14.
Let
epsilon1
= 1. There exists a neighborhood
U
of
c
such that

f
(
x
)

f
(
c
)

<
1
for all
x
∈
U
∩
D.
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 Fall '08
 Staff
 Intermediate Value Theorem, Continuous function, intermediatevalue theorem