Section 5.6
#1)
First suppose
f
is increasing. Then for any
x
∈
[
a, b
],
a
≤
x
≤
b
, and hence
f
(
a
)
≤
f
(
x
)
≤
f
(
b
). Hence,
f
(
a
) is an absolute minimum and
f
(
b
) is an absolute maximum.
If
f
is strictly increasing, then
x
∈
(
a, b
)
=
⇒
a < x < b
=
⇒
f
(
a
)
< f
(
x
)
< f
(
b
), and so
f
(
a
) and
f
(
b
) are the unique absolute minimum and maximum.
#5)
First suppose that
f
is continuous at
a
. By Problem 1),
f
(
a
) is a lower bound for
{
f
(
x
) :
x
∈
(
a, b
]
}
.
We must show it is the greatest lower bound, or equivalently, show that for all
>
0, there exists
x
0
∈
(
a, b
] such that
f
(
x
0
)
< f
(
a
) +
.
Fix
>
0.
Since
f
is continuous at
a
, we can find
δ >
0
such that if
x
∈
[
a, b
] and

x

a

< δ
, then
f
(
a
)

< f
(
x
)
< f
(
a
) +
.
Therefore, choosing any
a < x
0
< a
+
δ
will work.
Conversely, suppose that
f
(
a
) = inf
{
f
(
x
) :
x
∈
(
a, b
]
}
.
Fix
>
0.
We can find
x
0
∈
(
a, b
] such
that
f
(
a
)
≤
f
(
x
0
)
< f
(
a
) +
.
Set
δ
=
x
0

a >
0.
Suppose
x
∈
[
a, b
], and

x

a

< δ
.
Then
a
≤
x < a
+
δ
=
x
0
. Therefore, since
f
is increasing,
f
(
a
)
≤
f
(
x
)
≤
f
(
x
0
)
< f
(
a
) + , by the definition
of
x
0
. Ergo,

f
(
x
)

f
(
a
)

<
, and
f
is continuous at
a
.
#8)
Suppose
f
(
x
0
) =
y
=
g
(
x
1
) for some
x
0
, x
1
∈
I
.
If
x
0
≥
x
1
, then
y
=
f
(
x
0
)
≥
f
(
x
1
) since
f
is
increasing. But
f
(
x
1
)
> g
(
x
1
) =
y
, so we get
y > y
, contradiction. Hences,
x
0
< x
1
.
#9)
If
x
is rational, then
f
(
x
) =
x
is rational. If
x
is irrational, then
f
(
x
) = 1

x
is irrational. Suppose
therefore, that
f
(
x
1
) =
f
(
x
2
) =
y
.
If
y
is rational, then
x
1
, x
2
must both be rational and hence
y
=
x
1
=
x
2
. If
y
is irrational, then
x
1
, x
2
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 Spring '08
 JUNGE
 Intermediate Value Theorem, Continuous function, Order theory, absolute maximum, xc

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