and we say that has a

Points where has a maximum or a minimum are called

If is the function whose graph is shown, then has a maximum at , and has minimums at and .

The extreme value property is another assumption that is really a theorem,
(although the proof requires yet another assumption, namely *completeness*
of the real numbers.)

The following exercise shows that all of the hypotheses of the extreme value property are necessary.

- a)
- Give an example of a continuous function on such that has no maximum on .
- b)
- Give an example of a bounded continuous function on the closed interval , such that has no maximum on
- c)
- Give an example of a function on such that has no maximum on .
- d)
- Give an example of a continuous function on that has neither a maximum nor a minimum on , or else explain why no such function exists.

- a)
- Show that every continuous function from an interval to is bounded. (Hint: Use the extreme value property,)
- b)
- Is it true that every continuous function from an open interval to is bounded?
- c)
- Give an example of a function from to that is not bounded.

Proof: We will consider only the case where has a maximum. Suppose
has a
maximum at and is differentiable at . Then is an interior point
of
so we can find sequences and in
such that , , for all
,
and for all
.

Hence by the inequality theorem for limits,

It follows that .

and we say that has a

Sometimes we say that has a

Proof: The proof is the same as the proof of theorem 12.17.

The function whose graph is shown in the figure has local maxima at and local minima at , and . It has a global maximum at , and it has no global minimum.

From the critical point theorem, it follows that to investigate the extreme points of , we should look at critical points, or at points where is not differentiable (including endpoints of domain ).

From this we see that the critical set for is . Since is a continuous function on a closed interval we know that has a maximum and a minimum on . Now

Hence has global maxima at and , and has global minima at and . The graph of is shown.

Here and clearly for all . I can see by inspection that has a maximum at ; i.e.,

I also see that , and that is strictly decreasing on

thus has no local extreme points other than . Also is very small when is large. There is no point in calculating the critical points here because all the information about the extreme points is apparent without the calculation.

- a)
- for .
- b)
- for .