Proof: Suppose has a maximum at ,

Define two sequences , in by

Clearly and , and and for all . We have

By the inequality theorem,

Also,

so

Since , we conclude that . The proof for minimum points is left to you.

be a function that is continuous on and differentiable on . Suppose that . Then there is a number such that .

Proof: We know from the extreme value theorem that has a maximum at some point . If , then the critical point theorem says , and we are finished. Suppose . We know there is a point such that has a minimum at . If we get by the critical point theorem, so suppose . Then since and , we have , and it follows that is a constant function on , and in this case for all .

This theorem says that the tangent to the graph of at some point is parallel to the chord joining to .

Proof: Let

so the equation of the line joining to is , and

Let

Then

and is continuous on and differentiable on . By Rolle's theorem, there is some such that ; i.e., ; i.e.,

Then

so

But

so , and there is no point in with

If is an interval, and are points in with , then every point in is in the interior of .

- a)
- If for all , then is increasing on .
- b)
- If for all , then is strictly increasing on .
- c)
- If for all , then is decreasing on .
- d)
- If for all , then is strictly decreasing on .
- e)
- If for all , then is constant on .

Proof: All five statements have similar proofs. I'll prove only part a).

Suppose for all . Then for all with we have is continuous on and differentiable on , so by the mean value theorem

Hence, is increasing on .

a) If is differentiable on and is strictly increasing on , then for all .

b) If is differentiable on , and has a maximum at , then .

c) If is continuous on and is differentiable on , and for all , then is strictly increasing on .

*Let be a subset of
, let
, and let
be a point such that is differentiable at . Let be a subset
of containing , and let
be the restriction of
to , i.e.
*

Proof:
Let be any sequence in
such that
. Then is a sequence in
,
and hence

It follows that

I've shown that

and the set

is called the

Note that all points of are limit points of . If , then for some real number

and

If we solve equation (10.35) for we get

By using this value for in equation (10.36) we get

Let be defined by

Then is differentiable, and for all . We have

so by the restriction theorem

In general, the real and imaginary parts of a differentiable function are not differentiable.

and

Hence, the sequences and have different limits, so does not exist.

However, we do have the following theorem.

Proof: Since is differentiable at there is a function
on such that is continuous at and

If and , then and , so

and

Since is continuous at , and are continuous at , so equations (10.40) and (10.41) show that and are differentiable and

We have by direct calculation,

so

(This example just illustrates that the theorem is true in a special case.)

Proof: Define a function
by

By the chain rule, is differentiable on and . Since for all , we have for all . Hence

and hence

If and , then for all .

- a)
- Show that if then the segment is a subset of .
- b)
- Let be a function such that for all . Show that is constant on .