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Diferentiation
Written by
MenGen Tsai
email: [email protected]
1. Let
f
be defned For all real
x
, and suppose that

f
(
x
)

f
(
y
)
 ≤
(
x

y
)
2
For all real
x
and
y
. Prove that
f
is constant.
Proof:

f
(
x
)

f
(
y
)
(
x

y
)
2
For all real
x
and
y
. ±ix
y
,

f
(
x
)

f
(
y
)
x

y

x

y

. Let
x
→
y
, thereFore,
0
≤
lim
x
→
y
f
(
x
)

f
(
y
)
x

y
≤
lim
x
→
y

x

y

=0
It implies that (
f
(
x
)

f
(
y
))
/
(
x

y
)
→
0 as
x
→
y
. Hence
f
±
(
y
) = 0,
f
=
const
.
2. Suppose
f
±
(
x
)
>
0 in (
a,b
). Prove that
f
is strictly increasing in (
),
and let
g
be its inverse Function. Prove that
g
is di²erentible, and that
g
±
(
f
(
x
)) =
1
f
±
(
x
)
(
a < x < b
)
.
Proof:
±or every pair
x>y
in (
),
f
(
x
)

f
(
y
)=
f
±
(
c
)(
x

y
) where
y < c < x
by MeanValue Theorem. Note that
c
∈
(
) and
f
±
(
x
)
>
0
in (
), hence
f
±
(
c
)
>
0.
f
(
x
)

f
(
y
)
>
0,
f
(
x
)
>f
(
y
) iF
x > y
,
f
is
strictly increasing in (
).
Let Δ
g
=
g
(
x
0
+
h
)

g
(
x
0
). Note that
x
0
=
f
(
g
(
x
0
)), and thus,
(
x
0
+
h
)

x
0
=
f
(
g
(
x
0
+
h
))

f
(
g
(
x
0
))
,
1
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View Full Document h
=
f
(
g
(
x
0
)+ Δ
g
)

f
(
g
(
x
0
)) =
f
(
g
+ Δ
g
)

f
(
g
)
.
Thus we apply the fundamental lemma of diFerentiation,
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
). ±ix
± >
0, and de²ne
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:
±or 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
>
0
where
x < c < y
for all
. 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
4. If
C
0
+
C
1
2
+
...
+
C
n

1
n
+
C
n
n
+1
=0
,
where
C
0
, ...,
C
n
are real constants, prove that the equation
C
0
+
C
1
x
+
...
+
C
n

1
x
n

1
+
C
n
x
n
has at least one real root between 0 and 1.
Proof:
Let
f
(
x
)=
C
0
x
+
...
+
C
n
n
+1
x
n
+1
.
f
is diFerentiable in
R
1
and
f
(0) =
f
(1) = 0. Thus,
f
(1)

f
(0) =
f
±
(
c
) where
c
∈
(0
,
1) by
MeanValue Theorem. Note that
f
±
(
x
C
0
+
C
1
x
+
...
+
C
n

1
x
n

1
+
C
n
x
n
.
Thus,
c
∈
(0
,
1) is one real root between 0 and 1 of that equation.
5. Suppose
f
is de±ned and diFerentiable for every
x>
0, and
f
±
(
x
)
→
0
as
x
→
+
∞
. Put
g
(
x
f
(
x
+ 1)

f
(
x
). Prove that
g
(
x
)
→
0 as
x
→
+
∞
.
Proof:
f
(
x
+ 1)

f
(
x
f
±
(
c
)(
x

x
) where
x < c < x
by MeanValue Theorem. Thus,
g
(
x
f
±
(
c
) where
x < c < x
+ 1,
that is,
lim
x
→
+
∞
g
(
x
) = lim
x
→
+
∞
f
±
(
c
) = lim
c
→
+
∞
f
±
(
c
) = 0
.
6. Suppose
(a)
f
is continuous for
x
≥
0,
(b)
f
±
(
x
) exists for
0,
(c)
f
(0) = 0,
3
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f
±
is monotonically increasing.
Put
g
(
x
)=
f
(
x
)
x
(
x>
0)
and prove that
g
is monotonically increasing.
Proof:
Our goal is to show
g
±
(
x
)
>
0 for all
0
⇔
g
±
(
x
xf
±
(
x
)

f
(
x
)
x
2
>
0
⇔
f
±
(
x
)
>
f
(
x
)
x
.
Since
f
±
(
x
) exists,
f
(
x
)

f
(0) =
f
±
(
c
)(
x

0) where 0
<c<x
by
MeanValue Theorem.
⇒
f
±
(
c
f
(
x
)
x
where 0
. Since
f
±
is
monotonically increasing,
f
±
(
x
)
>f
±
(
c
), that is,
f
±
(
x
)
>
f
(
x
)
x
for all
0.
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This note was uploaded on 07/15/2008 for the course MATH 118 taught by Professor Ponce during the Spring '08 term at UCSB.
 Spring '08
 Ponce

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