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x
f
(
x
) which is continuous, by qn 6.23. Then
20 Choose
g
(
x
)
g
(0)
f
(0)
0 and
g
(1)
1
f
(1)
0. So
g
(
c
)
0 for some
c
[0, 1], and
f
(
c
)
c
.
21 If
a
and
b
areany two points in an intrval domain of
f
, then
f
(
a
) and
f
(
b
) aretwo points in therangeof
f
. The Intermediate Value Theorem
says that, for any
k
,
f
(
a
)
k
f
(
b
)or
f
(
a
)
k
f
(
b
), there is a
c
[
a
,
b
] such that
f
(
c
)
k
. So the range is connected and is therefore
an interval.
22 If
a
x
b
, then
f
(
a
)
f
(
x
)
f
(
b
): because if
f
(
a
)
f
(
x
) then, for some
c
[
x
,
b
),
f
(
c
)
f
(
a
) and
f
is not one
—
one; and if
f
(
x
)
f
(
b
) then, for
some
c
f
(
b
) and, again,
f
is not one
—
one. Applying this
result to the interval [
x
,
b
] wehave
a
x
y
b
implies
f
(
a
)
(
a
,
x
],
f
(
c
)
f
(
x
)
f
(
y
)
f
(
b
), and
f
is strictly monotonic increasing.
If
f
(
a
)
f
(
b
) and
f
is continuous and one
—
one, then
f
is strictly
monotonic decreasing.
23
(i) Yes, immediate consequence of being strictly monotone.
(ii) Yes, consequence of (i) from qns 6.5 and 6.6.
(iii) No. For example, consider
f
: [0, 2]
[0, 3] given by
f
(
x
)
x
on
[0, 1] and
f
(
x
)
x
1 on (1, 2].
24 Yes, from qn 22.
25
(i) Therangeis an intrval from qn 21. But, for
x
[
a
,
b
],
f
(
a
)
f
(
x
)
f
(
b
), taking
f
as increasing. So
c
f
(
a
) and
d
f
(
b
).
f
(
a
).
(ii) Strictly monotonic functions areone
—
one and so invertible.
(iii)
x
If
f
is decreasing then
c
f
(
b
) and
d
y
f
(
x
)
f
(
y
)
g
(
f
(
x
))
g
(
f
(
y
))
f
(
x
)
f
(
y
).
(iv) From qn 5, using completeness.
(v) The sequences are convergent from the continuity of
f
. They are
increasing and decreasing because
f
is strictly monotone.
(vi) Let
t
f
(
x
) from below and
t
f
(
s
) from above.
(vii)
lim
g
(
t
) exists from (iv) and equals
g
(
y
) from (vi).
Similarly for thelimit from abov, so lim
g
(
t
)
g
(
y
) and
g
is
continuous at
y
by qn 6.89.
26
f
only has an inverse on
R
0
or on
R
R
.On
R
0
,
x
x
is
continuous by qn 6.29, and strictly monotonic by qn 2.14. Therangeis
an interval by the Intermediate Value Theorem and is unbounded
above, so
f
is a bijection.
f
:
x
x
is continuous and monotonic on
R
0
, by qn 25.
27
f
is continuous by qn 6.28, strictly monotonic by qn 2.20, and
unbounded above, so
f
is a bijection. Then
f
has a continuous inverse
by qn 25.
