MAT 3749 2.4 Handout
Rolle’s Theorem
Let f be continuous on  a, b and differentiable on  a, b  such that f (a )  f (b) . Then
c   a, b  such that f   c   0 .
The Mean Value Theorem
Let f be continuous on  a, b and differentiable on  a, b  . Then c   a, b  such that
f (c) 
f (b)  f (a)
.
ba
Cauchy’s Mean Value Theorem
Let f and g be continuous on  a, b and differentiable on  a, b  . Then c   a, b  such
that
f (c)  g  b   g  a    g   c   f (b)  f (a)  .
In particular, if g  b   g  a  and g   c   0 , then
f (c) f (b)  f (a)
.

g  c  g b   g  a 
Remarks
1. The Mean Value Theorem is a special case of Cauchy’s Mean Value Theorem.
Last Edit: 7/14/2017 2:18 AM
2. Cauchy’s Mean Value Theorem is NOT a consequence of the Mean Value Theorem.
If we apply the MVT to f and g , c1 , c2   a, b  such that
f (c1 ) 
f (b)  f (a )
g (b)  g (a )
and g (c2 ) 
ba
ba
Cauchy’s Mean Value Theorem
Let f and g be continuous on  a, b and differentiable on  a, b  . Then c   a, b  such
that
f (c)  g  b   g  a    g   c   f (b)  f (a)  .
Analysis
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Proof
Let h  x   f ( x)  g  b   g  a    g  x   f (b)  f (a)  on  a, b .
Since f and g are continuous on  a, b and differentiable on  a, b  , h is continuous on
a, b and differentiable on  a, b  .
Then h  x   f ( x)  g  b   g  a    g   x   f (b)  f (a )  on  a, b  .
Also,

h  a   f (a)  g  b   g  a   g  a   f (b)  f (a)  f (a) g  b   g  a  f (b)


h  b   f (b)  g  b   g  a    g  b   f (b)  f (a)  f (a) g  b   g  a  f (b)
That is, h(a)  h(b) .
By Rolle’s Theorem, c   a, b  such that
h  c   0
 f (c)  g  b   g  a    g   c   f (b)  f (a )   0
 f (c)  g  b   g  a    g   c   f (b)  f (a ) 
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l’Hospital’s Rule Version 1
Let f and g be differentiable on a neighborhood of a , at which f  a   g  a   0 .
If lim
xa
f  x
f  x
f  x
exists, then lim
exists and equal to lim
.
xa g  x 
xa g   x 
g x
Analysis
1. Let us look at how Cauchy’s MVT can help us to prove l’Hospital’s Rule.
2. We will need to check all the hypotheses of Cauchy’s MVT.
Conceptual Diagram
Cauchy’s MVT requires the existence of intervals  a, b and  a, b  . In order to make it
work, let us create another point, say d on the right side of a . Now we have the
intervals  a, d  and  a, d  .
By Cauchy’s MVT,
c   a, d  such that…
f (c)

g c
How is this similar to what
we want to prove?
From our diagram, if we
take the limit as d  a ,
what kind of limit is this?
When d  a , what is c
approaching?
4
Show that
f ( x)
f ( x)
lim
 lim
x a g  x 
x a g   x 
From symmetry, it is not difficult to prove the other side of the limit:
f ( x)
f ( x)
.
lim
 lim
x a g  x 
x a g   x 
There are 4 conditions that 1.
we need to check before we
can use Cauchy’s MVT.
2.
They are…
3.
4.
What conditions do we
have to prove the 4
conditions above?
A.
B.
C.
We can use (A) to prove (1) and (2).
We have already used (B).
So, (C) must be involved in proving (3) and (4). It turns out that we need to use both (A)
and (C) to show that g   x   0 which implies (3) and (4).
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Let   1 . Write down the
definition of
f  x
lim
 L.
x a g   x 
Why the definition above
f
implies that
is defined
g
on
N   a   , a    a, a    .
Why does this imply
1. g   x   0 on N and
2. f and g are
differentiable on N ?
What is the difference
between N and
a  , a   ?
Why f and g are also
differentiable at a ?
So, f and g are differentiable and therefore continuous on  a   , a    .
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Proof
Suppose lim
x a
f  x
 L.
g  x
Let   1 . Then   0 such that if 0  x  a   , then
In particular,
f  x
 L  1.
g x
f
is defined on N   a   , a    a, a    .
g
This implies
1. g   x   0 on N .
2. f and g are differentiable on N .
Also, f and g are differentiable on a neighborhood of a
 f and g are differentiable at a
So, f and g are differentiable and therefore continuous on  a   , a    .
Let d   a, a    . Then f and g are continuous on  a, d  and differentiable on  a, d  .
By Cauchy’s MVT, Then c   a, d  such that
f (c)  g  d   g  a    g   c   f (d )  f (a) 
Now,
1. g   c   0 since c  N .
2. g  a   g  d  . Otherwise, by Rolle’s Theorem, g   e   0 for some e   a, d   N
which is a contradiction to g   x   0 on N .
Thus,
f (c) f (d )  f (a) f (d )  0 f (d )



g c  g  d   g  a  g  d   0 g  d 
As d  a  , c  a  . So,
lim
d a
f (d )
f (c)
 lim
g  d  d a g   c 
 lim
d a
f (d )
f (c)
 lim
g  d  ca g   c 
f ( x)
f ( x)
 lim
xa g  x 
xa g   x 
f ( x)
f ( x)
Similarly, we can prove that lim
.
 lim
x a g  x 
x a g   x 
f ( x)
f ( x)
Thus, lim
.
 lim
x a g  x 
x a g   x 
 lim
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