h
=
f
(
g
(
x
0
) +
Δ
g
)

f
(
g
(
x
0
)) =
f
(
g
+
Δ
g
)

f
(
g
)
.
Thus we apply the fundamental lemma of di
ff
erentiation,
h
= [
f
(
g
) +
η
(
Δ
g
)]
Δ
g,
1
f
(
g
) +
η
(
Δ
g
)
=
Δ
g
h
Note that
f
(
g
(
x
))
>
0 for all
x
∈
(
a, b
) and
η
(
Δ
g
)
→
0 as
h
→
0, thus,
lim
h
→
0
Δ
g/h
= lim
h
→
0
1
f
(
g
) +
η
(
Δ
g
)
=
1
f
(
g
(
x
))
.
Thus
g
(
x
) =
1
f
(
g
(
x
))
,
g
(
f
(
x
)) =
1
f
(
x
)
.
3. Suppose
g
is a real function on
R
1
, with bounded derivative (say

g

≤
M
).
Fix
>
0, and define
f
(
x
) =
x
+
g
(
x
).
Prove that
f
is onetoone if
is small enough. (A set of admissible values of
can
be determined which depends only on
M
.)
Proof:
For every
x < y
, and
x, y
∈
R
, we will show that
f
(
x
) =
f
(
y
).
By using MeanValue Theorem:
g
(
x
)

g
(
y
) =
g
(
c
)(
x

y
)
where x < c < y,
(
x

y
) + ((
x
)

g
(
y
)) = (
g
(
c
) + 1)(
x

y
)
,
that is,
f
(
x
)

f
(
y
) = (
g
(
c
) + 1)(
x

y
)
.
(
*
)
Since

g
(
x
)

≤
M
,

M
≤
g
(
x
)
≤
M
for all
x
∈
R
. Thus 1

M
≤
g
(
c
) + 1
≤
1 +
M
, where
x < c < y
. Take
c
=
1
2
M
, and
g
(
c
) + 1
>
0
where
x < c < y
for all
x, y
. Take into equation (*), and
f
(
x
)

f
(
y
)
<
0
since
x

y <
0, that is,
f
(
x
) =
f
(
y
), that is, f is onetoone (injective).
2