Where maxima are not located

Theorem 7887

Suppose that I is an interval and $c \in I$, and $f:I\rightarrow(-\infty,\infty)$. If $f'(c) \in (-\infty,0)\cup(0,\infty)$then f(c) is neither the maximum nor the minimum value of f.

At first glance this theorem does not appear very useful, and in some sense you are correct. In fact, in the grand scheme of things, most functions are not differentiable anywhere, so the theorem tells us nothing. However, in general the functions we wish to study have intervals for their domains and have derivatives at most points. For such functions, this theorem is invaluable. Let us see why it is true.

There are four cases to consider, based on what the sign of f'(c) is and whether we wish to rule f(c) out as a maximum or a minimum. We will present two cases here, and the other two cases will be left as homework or test questions, or both.

Case 1:

If f'(c) > 0 then f(c) is not the maximum value of f. Suppose that f(c) were the maximum value of f. Then for x > c we have

$$\frac{f(x) - f(c)}{x-c} \leq 0$$

since we have a fraction whose denominator is positive and whose numerator is not positive. This is the source the following contradiction:

$$\begin{array} {rcl} f'(c) & = & {\displaystyle \lim_{x\rig... ...x-c}}\\ & \leq & \lim_{x\rightarrow c^+} 0\\ & = & 0\end{array}$$

which says that $f'(c) \leq 0$, the opposite of what we assumed to start with. Therefore, if f'(c) > 0 then f(c) cannot be the maximum value of f.

Case 2:

If f'(c) < 0 then f(c) is not the maximum value of f. Suppose that f(c) were the maximum value of f. Then for x < c we have

$$\frac{f(x) - f(c)}{x-c} \geq 0$$

since we have a fraction whose denominator is negative and whose numerator is not positive. This is the source the following contradiction:

$$\begin{array} {rcl} f'(c) & = & {\displaystyle \lim_{x\rig... ...x-c}}\\ & \geq & \lim_{x\rightarrow c^-} 0\\ & = & 0\end{array}$$

which says that $f'(c) \geq 0$, the opposite of what we assumed to start with. Therefore, if f'(c) > 0 then f(c) cannot be the maximum value of f.

Case 3:

If f'(c) > 0 then f(c) is not the minimum value of f.

Case 4:

If f'(c) < 0 then f(c) is not the minimum value of f.

What does the theorem allow us to conclude about the location of the maximum and minimum values of f? First of all, if f is not a continuous function whose domain is an interval, there may not be any such values. Secondly, if f is a continuous function whose domain is a closed interval, then the maximum and minimum values can only be attained at

• The endpoints of the domain;
• Points where the derivative is not defined;
• Points where the derivative is .

This is what we call a sieve for location of maximum and minimum values. It captures a (hopefully) small set of numbers for further testing as location of the maximum and minimum values of our function if it is a continuous function defined on a closed interval! Here is the most famous example of the use of this sieve, Rolle's Theorem.

Theorem 7931 (M. Rolle, 1691)

Suppose a < b, $f:[a,b]\rightarrow (-\infty,\infty)$ is continuous, f'(x) exists for every $x\in(a,b)$, and f(a) = f(b) = 0. Then for some $c\in(a,b)$ we have f'(c) = 0.

It is easy to see why this is true. If the conclusion were false, that is that f'(x) is equal to nowhere in (a,b), then both the maximum and the minimum of f would occur at the endpoints of [a,b]. Since f has the value at both a and b then f(x) = 0 for every x in its domain, and it derivative is everywhere in (a,b), contradicting our assumption that the derivative was equal to nowhere in (a,b). QED

The geometric interpretation of Rolle's Theorem is that if f is a continuous function whose domain is a closed interval which has tangent lines at every point of its graph except possibly the endpoints, then at least one of those tangent lines is horizontal.