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View Full Document5. Suppose that the function
f
(
x
) is continuous on the
open
interval (0
,
1), and we also have
lim
x
→
0
+
f
(
x
) = 0, lim
x
→
1

f
(
x
) = 1.
(a) (Show that there is a continuous function
g
(
x
) on [0
,
1] for which
g
(
x
) =
f
(
x
) for
every
x
∈
(0
,
1).
Deﬁne the function
g
(
x
) on [0
,
1] as follows:
g
(
x
) =
0
if
x
= 0
f
(
x
)
if 0
< x <
1
1
if
x
= 1
This is continuous for every
x
∈
(0
,
1) (since it’s equal to
f
which is continuous by
assumption). It is also rightcontinuous at
x
= 0 (since lim
x
→
0
+
f
(
x
) = 0 =
g
(0)) and
the same is true for
x
= 1. But this means that
g
is continuous on [0
,
1].
(b) Show that
f
is bounded in (0
,
1).
The function
g
is continuous on [0
,
1] so by the Extreme Value Theorem it is bounded
there:

g
(
x
)

< M
for some
M
for
x
∈
[0
,
1]. But then

f
(
x
)

< M
is true for all
x
∈
(0
,
1) since
g
(
x
) =
f
(
x
) there.
(c) Is it always true that
f
has an absolute maximum in (0
,
1)? If yes, prove it, if not
then give a counterexample.
No. E.g.
f
(
x
) =
x
is a function like this and it does not have an absolute maximum
in (0
,
1). (Since it’s strictly less than 1 and it can be bigger than 1

ε
for any
ε >
0.)
6. Show that the equation
√
x
= cos(
x
3
) has at least one real solution.
The function
f
(
x
) =
√
x

cos(
x
3
) is continuous for every
x
≥
0 (rightcontinuous at
x
= 0).
We have
f
(0) =

1 and
f
(4) =
√
4

cos(64)
≥
2

1 = 1. Thus
f
(0)
<
0
,f
(4)
>
1 and
by Bolzano’s theorem there must be an
x
∈
(0
,
4)
3
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 Fall '08
 Staff
 Math, Calculus, Derivative, Sin, lim, Continuous function

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