Homework answers, week of Oct. 7
#1, p. 75 Because
f
is continuous on
[
a
,
b
]
, the extremevalue theorem implies that
there exist real numbers
m
and
M
such that
m
≤
f
(
x
)
≤
M
for all
x
∈
[
a
,
b
]
,
and, moreover, that there exist numbers
x
m
,
x
M
∈
[
a
,
b
]
such that
f
(
x
m
) =
m
and
f
(
x
M
) =
M
. Since
f
is an increasing function,
f
(
x
)
<
f
(
y
)
if and only
if
x
<
y
. Consequently, if
x
M
<
b
, then
M
=
f
(
x
M
)
<
f
(
b
)
, a contradiction.
Therefore,
M
=
f
(
b
)
and
x
M
=
b
. Similarly, if
x
m
>
a
, then
m
=
f
(
x
m
)
>
f
(
a
)
, which contradicts the deﬁnition of
m
. Hence
m
=
f
(
a
)
and
x
m
=
a
.
Therefore,
f
([
a
,
b
]) = [
f
(
a
)
,
f
(
b
)]
.
#2, p. 75 Let
M
=
sup
f
([
a
,
b
])
. Then there exists a sequence
{
y
n
} ⊂
f
([
a
,
b
])
such
that
{
y
n
} →
M
. Consequently, there exists a sequence
{
x
n
} ⊂
[
a
,
b
]
such
that
f
(
x
n
) =
y
n
for every
n
. The BolzanoWeierstrauss theorem implies that
there is a subsequence
{
x
n
k
}
that converges to a point
p
∈
[
a
,
b
]
. Since
f
is
continuous,
{
f
(
x
n
k
)
} →
f
(
p
)
. Furthermore, since
f
(
x
n
k
) =
y
n
k
for every
k
,
{
f
(
x
n
k
)
} →
M
. The uniqueness theorem for limits implies that
f
(
p
) =
M
.
Hence
f
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 thieme
 Calculus, Real Numbers, Continuous function, Metric space, Uniform continuity, xm

Click to edit the document details