DEFINITION OF RELATIVE MAXIMUM. A function
is said to have a relative maximum at a point
if there is some
for all satisfying
The concept of relative minimum is similarly defined by reversing the inequality.
DEFINITION OF EXTREMUM. A number which is either a relative maximum or a relative minimum of a function is called an extreme value or an extremum of
THEOREM (VANISHING OF THE DERIVATIVE AT AN INTERIOR EXTREMUM.) Let be defined on an open interval , and assume that has a relative maximum or a relative minimum at an interior point of . If the derivative exists, then
Proof Define a function
on I as follows:
Since exists, as , so is continuous at we wish to prove that . We shall do this by showing that each of the inequalities and leads to a contradiction.
Assume . By the sign-preserving property of continuous functions, there is an interval about in which is positive. Therefore the numerator of the quotient has the same sign as the denominator for all in this interval. In other words, when , and when . This contradicts the assumption that has an extremum at . Hence, the inequality is impossible. A similar argument shows that we cannot have . Therefore , as asserted. Since , this proves the theorem.
be a function which is continuous everywhere on a closed interval
and has a derivative at each point of the open interval
Also, assume that
Then there is at least one point in the open interval such that .
Proof. We assume that for every in the open interval and we arrive at a contradiction as follows: By the extreme-value theorem for continuous functions, must take on its absolute maximum and its absolute minimum somewhere in the closed interval The previous theorem tells us that neither extreme value can be taken at any interior point (otherwise the derivative would vanish there). Hence, both extreme values are taken on at the endpoints and . But since this means that , and hence is constant on . This contradicts the fact that for all in It follows that for at least one satisfying which proves the theorem.
MEAN-VALUE THEOREM FOR DERIVATIVES. Assume that
is continuous everywhere on a closed interval
and has a derivative at each point of the open interval
Then there is at least one interior point
Proof. To apply Rolle's theorem we need a function which has equal values at
To construct such a function, we modify
as follows. Let
is continuous on
and has a derivative in the open interval
Applying Rolle's theorem to
we find that
When , this gives us the desired result.
be two functions continuous on a closed interval
and having derivatives in the open interval
Proof. We let
Applying Rolle's theorem to , we find that for some in . Computing from the formula defining , we obtain Cauchy's mean-value formula. Note that the mean-value theorem is the special case obtained by taking
The mean-value theorem may be used to deduce properties of a function from a knowledge of the algebraic sign of its derivative. This is illustrated by the following theorem.
Theorem. Let be a function which is continuous on a closed interval and assume has a derivative at each point of the open interval Then we have:
(a) If for every in , is strictly increasing on ;
(b) If for every in , is strictly decreasing on ;
(c) If for every in , is constant throughout
Proof. To prove (a) we must show that
and apply the mean-value theorem to the closed subinterval
Since both and are positive, so is , and this means , as asserted. This proves (a), and the proof of (b) is similar. To prove (c), we again use the mean-value theorem:
We have for every in , so is constant
Theorem. Assume is continuous on a closed interval and assume that the derivative exists everywhere in the open interval except possibly at a point
(a) If is positive for all and negative for all , then has a relative maximum at
(b) If, on the other hand, is negative for all and positive for all , then has a relative minimum at
Proof. In case (a), the previous theorem tells us that is strictly increasing on and strictly decreasing on . Hence for all in , so has a relative maximum at This proves (a) and the proof of (b) is entirely analogous.
If a function is continuous on a closed interval , the extreme-value theorem tells that it has an absolute maximum and an absolute minimum somewhere in . If has a derivative at each interior point, then the only places where extrema can occur are:
1) at the endpoints and ;
2) at those interior points where .
Points of type (2) are often called critical points of . To decide whether there is a maximum or minimum (or neither) at a critical point , we need more information about Usually the behavior of at a critical point can be determined from the algebraic sign of the derivative near . The next theorem shows that a study of the sign of the second derivative near can also be helpful.
THEOREM. SECOND-DERIVATIVE TEST FOR AN EXTREMUM AT A CRITICAL POINT. Let be a critical point of in an open interval ; that is, assume and 0. Assume also that the second derivative exists in Then we have the following:
(a) If is negative in , has a relative maximum at .
(b) If is positive in , has a relative minimum at .
Proof. Consider case (a), in The function is strictly decreasing in : But , so changes its sign from positive to negative at . Hence, has a relative maximum at . The proof in case (b) is entirely analogous.
DEFINITION OF A CONVEX FUNCTION. A function
is said to be convex on an interval
if, for all
and for every
We say is concave on if the reverse inequality holds,
The sign of the second derivative also governs the convexity or the concavity of . The next theorem shows that the function is convex in intervals where is positive. It suffices to discuss only the convex case, because if is convex, then is concave.
THEOREM. DERIVATIVE TEST FOR CONVEXITY. Assume is continuous on and has a derivative in the open interval . If is increasing on , then is convex on . In particular, is convex if exists and is nonnegative in
We wish to prove that
this is the same as proving that
By the mean-value theorem (applied twice), there exist points and satisfying and such that
Since is increasing; we have . Also, we have , so we may write
which proves the required inequality for convexity.