Local maximum and local minimum

Derivatives can be used to find a function's local extremal values (i.e., local maximum or minimum values).

If $f(x)$ is a continuous function defined on a closed interval $[a,b]$, then it always has a maximum value and a minimum value (c.f. the Extreme Value Theorem). Finding a maximum or minimum of a function on its entire domain is a global problem.

On the other hand, examining continuity or differentiability around $x=a$ is a local problem in the sense that it only concerns the neighbor of the point $x=a$. We may also define the notions of local maxima and local minima.

Definition (Local maximum and local minimum)

Let $f(x)$ be a function defined on an interval $I$ and let $a\in I$.

1. $f(a)$ is said to be a local maximum value of the function $f(x)$ if there exists  $\delta >0$ such that, for all $x\in (a-\delta ,a+\delta )\cap I$, $x\ne a$ implies $f(x)<f(a)$. $x=a$ is said to be a local maximum point if $f(a)$ is a local maximum value.
2. $f(a)$ is said to be a local minimum value of the function $f(x)$ if there exists  $\delta >0$ such that, for all $x\in (a-\delta ,a+\delta )\cap I$, $x\ne a$ implies $f(x)>f(a)$. $x=a$ is said to be a local minimum point if $f(a)$ is a local minimum value.

We use the term extreme value to mean either a local maximum value or a local minimum value.

Remark. If a local maximum value is actually a maximum value, then the maximum value is called a global maximum value. A global minimum value is similarly defined. Note that a global maximum (minimum) value is a local maximum (minimum) value, but not vice versa. □

Example. The constant function $f(x)=c$ ($c$ is a constant) has the maximum value $c$ and the minimum value $c$, but it does not have a local maximum value or local minimum value. □

Example. Let $f(x)=(x^2-1)(x^2-4)$ (draw the graph!). $f(0)=4$ is a local maximum value. In fact, take $\delta =\sqrt{5}>0$, then for all $x$ such that $0<|x|<\delta$, we have

$$0<x^2<{\delta }^2$$

(i.e., $x^2$ is strictly positive) and

$$-5<x^2-5<{\delta }^2-5=0$$

(i.e., $x^2-5$ is strictly negative) so that

$$f(x)=x^2(x^2-5)+4<4=f(0).$$

This means that $f(0)>f(x)$ for all $x\in (-\delta ,+\delta ),x\ne 0$. That is, $f(0)$ is a local maximum value (check the above definition again!). □

Definition (Stationary/critical point)

Let $f(x)$ be a differentiable function. $x=a$ is said to be a critical point or stationary point if $f^{\prime }(a)=0$. In this case, $f(a)$ is called a critical value or stationary value.

Theorem

If a differentiable function $f(x)$ has a local extremum value (i.e., either a local maximum or a local minimum value) at $x=a$, then $x=a$ is a stationary point.

Proof. We prove only for the local maximum case (the local minimum case is similar).

Suppose $f(x)$ takes a local maximum value at $x=a$. Then there exists $\delta >0$ such that, for all $x\in (a-\delta ,a+\delta )$, $f(x)\le f(a)$. Since $f(x)$ is differentiable at $x=a$ (by assumption), we have

$$\begin{array}{rcl}{f}^{\prime }(a)& =& \underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}\\ & =& \underset{x\to a+0}{lim}\frac{f(x)-f(a)}{x-a}\\ & =& \underset{x\to a-0}{lim}\frac{f(x)-f(a)}{x-a}.\end{array}$$

• If $0<a-x<\delta$ (i.e., $x\in (a-\delta ,a)$), then $\frac{f(x)-f(a)}{x-a}\ge 0$ so, by taking the limit as $x\to a-0$, we have $f^{\prime }(a)\ge 0$. 
• If $0<x-a<\delta$ (i.e., $x\in (a,a+\delta )$), then $\frac{f(x)-f(a)}{x-a}\le 0$ so, by taking the limit as $x\to a+0$, we have $f^{\prime }(a)\le 0$. 

Therefore, we have $f^{\prime }(a)=0$. ■

Remark. The converse of this theorem is not true. That is, $f^{\prime }(a)=0$ does not imply $f(a)$ is an extremum. For example, $f(x)=x^3$ has $f^{\prime }(0)=0$, but $f(0)$ is neither a local maximum value nor a local minimum value (draw the graph!). □