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CONTINUITY
Problem 17.4: Prove that the function
√
x
is continuous on its domain [0
,
∞
)
.
Proof.
Given
± >
0, we need to ﬁnd a
δ >
0 such that

x

x
o

< δ
implies

f
(
x
)

f
(
x
o
)

< ±.

f
(
x
)

f
(
x
o
)

=

√
x

√
x
o

=
±
±
±
±
x

x
o
√
x
+
√
x
o
±
±
±
±
<

x

x
o

√
x
o
<
δ
√
x
o
.
(
?
)
We want

f
(
x
)

f
(
x
o
)

< ±
. The only problem that we might run into is if
x
o
= 0. So if we have
two cases:
Case 1:
x
o
>
0. We set
δ
=
±
√
x
o
and

f
(
x
)

f
(
x
o
)

< ±
from (
*
).
Case 2:
x
o
= 0. This is slightly diﬀerent. Since

x

x
o

< δ
, and in this case
x
o
= 0,

x

< δ
. Then
√
x <
√
δ
. Thus if we set set
δ
=
±
2
, we have

f
(
x
)

f
(
x
o
)

=

√
x

√
x
o

=

√
x

0

<

√
x

<
√
δ
=
±.
Thus, let
δ
= min
{
±
√
x
o
,±
2
}
, then

f
(
x
)

f
(
x
o
)

< ±
for all
x
o
∈
[0
,
∞
). Thus,
f
is continuous.
±
Problem 17.6: A
rational function
is a function
f
of the form
p/q
where
p
and
q
are polynomial
functions. The domain of
f
is
{
x
∈
R
:
q
(
x
)
6
= 0
}
. Prove that every rational function is continuous.
d
We state problem 17.5 here without proofs (can easily proved using induction):
If
m
∈
N
, then the function
f
(
x
) =
x
m
is continuous on
R
.
Every polynomial function
p
(
x
) =
a
o
+
a
1
x
+
...
+
a
n
x
n
is continuous on
R
.
c
Proof.
Suppose
f
=
p/q
for some polynomial functions
p,q
. Since both
p
and
q
are polynomial
functions, they are continuous in
R
. By theorem 17.4,
p/q
is continuous provided
q
(
x
)
6
= 0. Since
the domain of
f
is
{
x
∈
R
:
q
(
x
)
6
= 0
}
,
f
is continuous.
±
Problem 17.8: Let
f
and
g
be realvalued functions.
(a) Show that min(
f,g
) =
1
2
(
f
+
g
)

1
2

f

g

.
(b) Show that min(
f,g
) =

max(

f,

g
).
(c) Use (a) or (b) to prove that if
f
and
g
are continuous at
x
o
∈
R
, then min(
f,g
) is
continuous at
x
o
.
Proof.
Note that the real hard part is to come up with max(
f,g
) =
1
2
(
f
+
g
) +
1
2

f

g

.
(a) This should be an easy exercise to verify the following statement is true:
min(
f,g
) =
1
2
(
f
+
g
)

1
2

f

g

.
(
?
)
There are two possibilities here: whether
f
≥
g
or
f < g
. Recall that

a

b

=
²
a

b
if
a
≥
b

(
a

b
)
if
a < b
giamath@gmail.com.
1
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View Full Documenti. If
f
≥
g
, then

f

g

=
f

g
and min(
f,g
) =
g
. While
1
2
(
f
+
g
)

1
2

f

g

=
1
2
(
f
+
g
)

1
2
(
f

g
)
=
1
2
f
+
1
2
g

1
2
f
+
1
2
g
=
g
= min(
f,g
)
so (
?
) holds.
ii. If
f < g
, then

f

g

=

(
f

g
) =
g

f
and min(
f,g
) =
f
. While
1
2
(
f
+
g
)

1
2

f

g

=
1
2
(
f
+
g
)

1
2
(
g

f
)
=
1
2
f
+
1
2
g

1
2
g
+
1
2
f
=
f
= min(
f,g
)
so (
?
) holds as well.
(b) Since we have max(
f,g
) =
1
2
(
f
+
g
) +
1
2

f

g

. Observe that
max(

f,

g
) =
1
2
(

f
+ (

g
)) +
1
2
 
f

(

g
)

=

1
2
(
f
+
g
) +
1
2
 
(
f

g
)

=

±
1
2
(
f
+
g
)

1
2

(
f

g
)

²
=

min(
f,g
)
Thus, min(
f,g
) =

max(

f,

g
)
.
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 Spring '09
 Continuity

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