ANALYSIS
EXERCISE 11–SOLUTIONS
1. By the definition of continuity for every
x
0
∈
[
a, b
]
lim
x
→
x
0
f
(
x
) =
f
(
x
0
)
.
Using the Heine definition one can compute the above limit along
the sequences
x
n
→
x
0
,
x
n
6
=
x
0
. Choose
x
n
∈
Q
. Then
f
(
x
n
) = 0
for all
n
∈
N
, and it is clear that
f
(
x
0
) = 0.
2.
(a) Let
f
:
R
→
R
be defined by
f
(
x
) =
x
3

x
+ 3. Then
f
(

2) =

3
<
0 and
f
(0) = 3
>
0. By the intermediate value theorem
(
∃
x
∈
[

2
,
0]) (
f
(
x
) = 0).
(b) Let
f
:
R
→
R
be defined by
f
(
x
) =
x
5
+
x
+ 1. Then
f
(

2) =

33
<
0 and
f
(0) = 1
>
0. By the intermediate value theorem
(
∃
x
∈
[

2
,
0]) (
f
(
x
) = 0).
3. Let
ψ
=
f

g
. Then
ψ
is continuous.
ψ
(
a
) =
f
(
a
)

g
(
a
)
<
0, and
ψ
(
b
) =
f
(
b
)

g
(
b
)
>
0. Therefore by the intermediate value theorem
(
∃
x
∈
(
a, b
)) (
ψ
(
x
) = 0).
4. Let
f
:
R
→
R
be defined by
f
(
x
) =
x
4

3
x

9.
f
(

2) = 13
>
0,
f
(2) = 1
>
0.
But
f
(0) =

9
<
0.
So by the intermediate value
theorem there is an
x
1
∈
(

2
,
0) such that
f
(
x
1
) = 0, and there is
an
x
2
∈
(0
,
2) such that
f
(
x
2
) = 0.
5. Let
g
: [0
,
1]
→
R
defined by
g
(
x
) =
f
(
x
)

x
. Then
g
(0) =
f
(0)
≥
0.
If
g
(0) = 0 then
x
= 0 is the desired solution.
Now assume that
g
(0)
>
0.
g
(1) =
f
(1)

1
≤
0. If
g
(1) = 0 then
x
= 1 is the desired
solution.
Next assume that
g
(1)
<
0.
Then by the intermediate
value theorem
(
∃
x
∈
(0
,
1)) (
g
(
x
) = 0).
6.
(a) Assume for a contradiction that
α
6∈
I
. There are two possibili
ties: either
α < a
or
α > b
. Consider the first case,
α < a
. Put
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 Winter '10
 Spyros
 Continuity, Intermediate Value Theorem, Continuous function, Uniform continuity

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