bounded on (
a, b
). Prove that
f
is uniformly continuous on (
a, b
).
Solution:
By assumption
∃
M >
0 such that

f
(
x
)
 ≤
M
, for any
x
∈
(
a, b
).
Let
>
0 and take
δ
=
/M
. Let
y, z
∈
(
a, b
) such that

z

y

< δ
. By the mean value theorem applied
to
f
on the interval between
y
and
z
, there exists
c
such that
f
(
z
)

f
(
y
) =
f
(
c
)(
z

y
). Thus

f
(
z
)

f
(
y
)

=

f
(
c
)

z

y
 ≤
M

z

y

< Mδ
=
.
Since this is true for any
y, z
such that

z

y

< δ
, it follows that
f
is uniformly continuous on (
a, b
).
(b) (10 pts) Give an example to show that if the hypothesis
f
bounded on (
a, b
) is omitted, then the
statement of (a) is no longer true.
Solution:
Let
f
: (0
,
1)
→
R
, defined by
f
(
x
) = ln
x
.
f
is differentiable on (0
,
1), and its derivative
f
(
x
) = 1
/x
is clearly not bounded on (0
,
1).
The function
f
is not uniformly continuous because it
cannot be extended by continuity at 0 (the limit of
f
as
x
→
0
+
is
∞
).
4.
(15 pts) Show that ln(
x
+ 1)
≤
x
, for all
x
≥
0.
Solution:
Let
f
(
x
) =
x

ln(
x
+ 1).
Note that for any
x
≥
0,
f
(
x
) = 1

1
/
(
x
+ 1)
≥
0.
Thus
f
is
increasing on the interval [0
,
∞
). Thus if
x
≥
0, then
f
(
x
)
≥
f
(0). Noting that
f
(0) = 0, this is exactly
the inequality we had to prove.
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5.
(15 pts) Suppose [
a, b
] is a closed, bounded, nondegenerate interval. Is the following statement true?
For any continuous function
f
: [
a, b
]
→
R
, the function

f

, defined by

f

(
x
) =

f
(
x
)

, is integrable on
[
a, b
]
.
Briefly justify your answer.
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 Fall '12
 Draghic
 Calculus, Topology, Continuous function, 10 pts, 15 pts

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