Math 117, F 09, HW#9
(sample solutions)
P.105, #15.
Let
x
0
∈
D
, and let
± >
0. Since
f
is continuous at
x
0
, there exists
δ
1
>
0
such that

f
(
x
)

f
(
x
0
)

< ±
for all
x
∈
D
with

x

x
0

< δ
1
. Likewise, there exists
δ
2
>
0
such that

g
(
x
)

g
(
x
0
)

< ±
for all
x
∈
D
with

x

x
0

< δ
2
. Set
δ
= min
{
δ
1
, δ
2
}
>
0.
If
x
∈
D
with

x

x
0

< δ
, then

x

x
0

< δ
1
and

x

x
0

< δ
2
, so

f
(
x
)

f
(
x
0
)

< ±
and

g
(
x
)

g
(
x
0
)

< ±
. Then
f
(
x
0
)

± < f
(
x
)
< f
(
x
0
) +
±
and
g
(
x
0
)

± < g
(
x
)
< g
(
x
0
) +
±
,
and hence
h
(
x
0
)

±
= max
{
f
(
x
0
)

±, g
(
x
0
)

±
}
<
max
{
f
(
x
)
, g
(
x
)
}
=
h
(
x
)
h
(
x
) = max
{
f
(
x
)
, g
(
x
)
}
<
max
{
f
(
x
0
) +
±, g
(
x
0
) +
±
}
=
h
(
x
0
) +
±,
which shows that

h
(
x
)

h
(
x
0
)

< ±
. Therefore
h
is continuous at
x
0
.
±
P.105, #17.
Deﬁne
g
: [0
,
∞
)
→
R
to be the square root function:
g
(
x
) =
√
x
for
x
∈
[0
,
∞
). By Exercise 6, p. 104,
g
is continuous. The function
√
f
is just the composition
g
◦
f
, and we are given that
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 08/27/2011 for the course MATH 117 taught by Professor Akhmedov,a during the Fall '08 term at UCSB.
 Fall '08
 Akhmedov,A
 Math

Click to edit the document details