MAT 3749 2.3 Part III Handout
Maximum Value
A function f has an (absolute) maximum at c if f (c)  f ( x) for all x in D, where D is
the domain of f . The number f (c ) is called the maximum value of f on D.
Local Maximum
A function f has a local maximum at c if f (c)  f ( x) for all x in a neighborhood of c.
The Extreme Value Theorem
If f is continuous on  a, b , then f attains its maximum value f  c  and minimum
value f  d  at some numbers c and d in  a, b .
Lemma (Homework)
(a) If f (c ) exists and is positive, then there is a neighborhood D of c on which
 f  x   f  c  if x  c
.

 f  x   f  c  if x  c
(b) If f (c ) exists and is negative, then there is a neighborhood D of c on which
 f  x   f  c  if x  c
.

 f  x   f  c  if x  c
Analysis
Last Edit: 2/6/2016 4:42 AM
Fermat’s Theorem
If f has a local maximum or minimum at c , and if f (c ) exists, then f (c)  0 .
Analysis
In light of the lemma, it makes sense to try using a contradiction.
Conceptual Diagrams
Case 1 f (c)  0
Then, by lemma, there is a neighborhood D of c such that
(1) f  x   f  c  x  c in D and
(2) f  x   f  c  x  c in D .
We want to show that f does not have a local maximum at c and f does not have a local
minimum at c .
In order to show that f
does not have a local
maximum at c , what
exactly do we need to
demonstrate?
Which of the two
conditions, (1) or (2), is
relevant?
How does this condition
implicate the statement
above?
2
Proof
Suppose f (c)  0 .
Case 1 f (c)  0 .
Then, by lemma, there is a neighborhood D of c such that
(1) f  x   f  c  x  c in D and (2) f  x   f  c  x  c in D .
Let E be any neighborhood of c .
(1)  f  x   f  c  x  c in D  E .
 x  c in E such that f  x   f  c 
 f does not have a local maximum at c .
(2)  f  x   f  c  x  c in D  E .
 x  c in E such that f  x   f  c 
 f does not have a local minimum at c .
This is a contradiction to the fact that f has a local maximum or minimum at c .
Case 2 f (c)  0 . (similar)
Thus, if f has a local maximum or minimum at c , and if f (c ) exists, then f (c)  0 .
3
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 .
Analysis
We have 3 conditions here. They may give us hints of how to approach this problem. On
the other hand, we need to make sure we make use of all of them.
Conceptual Diagrams
After studying the conceptual
diagrams, what do you think
where we should choose c ?
Based on the first condition,
which previous result(s) is/are
relevant?
Do we have enough to get
exactly what we want to
choose c ? Why?
Not the end of the world…at least we get a good starting point.
4
Since f is continuous on  a, b , by EVT, f attains its maximum value f   and
minimum value f    at some  ,    a, b .
If one of them is local, then we are done. If both are not local, we need to think about
what to do.
If both are not local, where are
 and  ?
What is the implication of
these locations?
What does this tell you about
the function f ?
Where to choose c ?
Good Job. We have resolve the case when f has absolute maximum / minimum values
at both endpoints.
Now, let us look carefully how to finish our proof when f has local maximum /
minimum values at either  or  .
In the actual proof, it is cleaner to use the cases f    f    and f    f    , rather
than the equivalent statements above.
Suppose f    f    , how
do we know one of  and 
is not one of the end points?
We can conclude that at least one of  ,  are in  a, b  . Let that point be c .
5
Short stop for reality check
Which of the 3 conditions that
we have not used?
Which previous result(s) is/are
relevant?
Our gut feelings tell us that we need to use Fermat’s Theorem to show that f   c   0 .
We need to check the two hypotheses before we use it.
f has a local maximum or
minimum at c . Why?
f   c  exists. Why?
We can now conclude that by Fermat’s Theorem, f   c   0 .
6
Proof
Since f is continuous on  a, b , by EVT, f attains its maximum value f   and
minimum value f    at some  ,    a, b .
Case 1 f    f   
f is a constant function on  a, b  . Thus, f   x   0, x   a, b  .
Choose any c   a, b  , we have f   c   0 .
Case 2 f    f   
Since f (a )  f (b) ,  ,  are not both the endpoints.
Thus, at least one of  ,  are in  a, b  . Let that point be c .
So, f attains its maximum or minimum value at c   a, b  .
That means f has a local maximum or minimum at c   a, b  .
Since f is differentiable on  a, b  , f   c  exists. By Fermat’s Theorem, f   c   0 .
For all cases, we have proved that c   a, b  such that f   c   0 .
7
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
Analysis
.
8
Proof
f b  f  a 
 x  a  and h  x   f  x   g  x  .
ba
Since g is a polynomial, it is continuous on  a, b and differentiable on  a, b  .
Let g  x   f  a  
So, h is continuous on  a, b and differentiable on  a, b  .
Also,
h  x   f   x   g   x 
f b  f  a 


 f  x   f a 
 x  a 
ba


f b  f  a 
 f  x 
ba
Now,
and
f b  f  a 


h a  f a  g a  f a   f a 
 a  a  
ba


0
f b  f  a 


h b  f b   g b   f b    f  a  
 b  a  
ba


0
By Rolle’s Theorem, c   a, b  such that
f c 
h  c   0
f b  f  a 
0
ba
f b  f  a 
f c 
ba
9
Theorem If f ( x)  0 for all x in an interval  a, b  , then f is constant on  a, b  .
Proof
Corollary If f ( x)  g ( x) for all x in an interval  a, b  , then f ( x)  g ( x)  C on  a, b  .
Proof
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