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HOMEWORK 7: SOLUTIONS/HINTS
(17.12a) Let
x
2
(
a;b
), then as
Q
\
(
a;b
)
±
(
a;b
) is dense there is a sequence of
rational numbers
r
n
converging to
x
. Since
f
is continuous and
f
(
r
n
) = 0
for all
n
, we now get
f
(
x
) =
f
(lim
r
n
) = lim
f
(
r
n
) = 0
:
(17.13a) Let
x
2
R
,
±
2
(0
;
1) and let
² >
0. If
x
is rational (resp. irrational) then
there is an irrational (resp. rational) number
y
2
R
such that
j
x
²
y
j
< ²
and consequently we get in either case that
j
f
(
x
)
²
f
(
y
)
j
= 1
> ±:
So
f
cannot be continuous at
x:
(18.4) De±ne
f
(
x
) :=
1
x
²
x
0
;
for
x
2
S:
Then
f
is continuous and as the sequence
x
n
2
S
converging to
x
0
, we get
lim
j
f
(
x
n
)
j
= lim
1
j
x
n
²
x
0
j
= +
1
;
which shows that
f
is unbounded on
S:
(18.5) De±ne
h
(
x
) :=
g
(
x
)
²
f
(
x
). Then
h
is a continuous function on [
a;b
] and
h
(
a
) =
g
(
a
)
²
f
(
a
)
³
0 and
h
(
b
) =
g
(
b
)
²
f
(
b
)
´
0. By the Intermediate
Value Theorem there exists an
x
0
2
[
a;b
] such that
h
(
x
0
) = 0 and hence
g
(
x
0
) =
f
(
x
0
)
:
(18.6) Take
f
(
x
) = cos(
x
),
g
(
x
) =
x
,
a
= 0 and
b
=
±
2
in problem (18.5), then
there is an
x
0
2
[0
;
±
2
] such that
x
0
= cos(
x
0
)
:
Since the equation does not
hold for
x
= 0 or
x
=
±
2
, it follows that
x
0
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 Spring '11
 Peter

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