Extreme Values, calculus lesson

The highest point of a mountain range and the top of a small hill on its slope are both peaks, but in different senses. One beats every other point. The other only beats the points near it. Calculus keeps these two ideas apart and asks where such peaks and valleys can hide on the graph of a function.

Key terms

Global maxima and minima

Definition Let $f$ have domain $D$ and let $c$ be in $D$ . Then $f$ has a global (or absolute) maximum at $c$ if $f(c) \ge f(x)$ for every $x$ in $D$ , and a global (or absolute) minimum at $c$ if $f(c) \le f(x)$ for every $x$ in $D$ .

In plain words A global maximum is the highest point of the whole graph and a global minimum is the lowest. A function may have neither.

Example For $f(x) = x^2$ on $D = [-1, 2]$ , the global maximum is $f(2) = 4$ and the global minimum is $f(0) = 0$ . On $D = \mathbb{R}$ the same formula has the global minimum $0$ and no global maximum.

Local maxima and minima

Definition Let $f$ have domain $D$ and let $c$ be in $D$ . Then $f$ has a local (or relative) maximum at $c$ if there is a $\delta > 0$ such that $f(c) \ge f(x)$ for every $x$ in $D$ with $\lvert x - c \rvert < \delta$ . A local minimum is defined the same way with $f(c) \le f(x)$ . Some books also require an open interval around $c$ to lie inside $D$ , which rules out local extrema at endpoints. Here endpoints are allowed.

In plain words A local maximum only has to beat its neighbors. Every global maximum is also a local one, but not the other way around.

Example For $f(x) = x^3 - 3x$ on $\mathbb{R}$ , $f(-1) = 2$ is a local maximum and $f(1) = -2$ is a local minimum. Neither is global, since $f(3) = 18 > 2$ and $f(-3) = -18 < -2$ .

Extreme values

Definition A maximum or minimum value of $f$ , global or local, is called an extreme value or extremum (plural extrema). The extreme value is the number $f(c)$ , and $c$ is a point where it is attained. A global maximum value, when it exists, is unique, but it may be attained at many points.

In plain words Keep the value and the location apart. “The maximum is 1” names a height, while “the maximum occurs at $x = \frac{\pi}{2}$ ” names a place.

Example On $\mathbb{R}$ , $\sin x$ has global maximum value $1$ , attained at every $x = \frac{\pi}{2} + 2k\pi$ with $k$ an integer, and global minimum value $-1$ .

Domain points and endpoints

Definition A point $c$ of the domain $D$ is an interior point if some open interval $(c - \delta, c + \delta)$ with $\delta > 0$ lies inside $D$ . For a domain that is an interval, an endpoint is a point of $D$ at the end of the interval, such as $a$ or $b$ in $[a, b]$ . An endpoint is not an interior point.

In plain words At an interior point you can step a little left and a little right and stay in the domain. At an endpoint you can only go one way.

Example For $f(x) = \sqrt{x}$ with domain $[0, \infty)$ , the point $0$ is an endpoint and every $c > 0$ is an interior point. The global minimum of $\sqrt{x}$ is at the endpoint $0$ , where the graph simply starts.

Extrema at endpoints

Definition An extreme value may occur at an endpoint of the domain, and there the derivative of $f$ (or its one-sided derivative) need not be $0$ . Endpoints must therefore always be checked as separate candidates.

In plain words A graph can reach its highest point simply because the domain ends there, while still climbing.

Example For $f(x) = 2x + 1$ on $[0, 1]$ , the minimum $1$ is at $x = 0$ and the maximum $3$ is at $x = 1$ . The derivative is $2$ everywhere, never $0$ .

Critical points

Definition A critical point (or critical number) of $f$ is an interior point $c$ of the domain such that either $f'(c) = 0$ or $f'(c)$ does not exist.

In plain words These are the interior places where the graph is flat or where it has a corner, cusp or vertical tangent. As the next lesson proves, an interior local extremum can occur only at a critical point.

Example For $f(x) = x^3 - 12x + 1$ the derivative $f'(x) = 3x^2 - 12$ is zero at $x = 2$ and $x = -2$ , and these are the critical points. For $f(x) = \lvert x \rvert$ , the derivative does not exist at $0$ , so $0$ is a critical point.

Stationary and singular points

Definition A critical point $c$ is a stationary point if $f'(c) = 0$ , and a singular point if $f$ is not differentiable at $c$ .

In plain words At a stationary point the tangent line is horizontal. At a singular point there is no tangent slope at all. Either kind may or may not be an extremum.

Example For $f(x) = x^{2/3}$ the derivative is $f'(x) = \frac{2}{3} x^{-1/3}$ for $x \neq 0$ , and $f'(0)$ does not exist, so $0$ is a singular point. It is a minimum, since $x^{2/3} \ge 0 = f(0)$ for every $x$ . For $f(x) = x^3$ the point $0$ is stationary, yet $f$ has no extremum there because $f(x) < 0$ for $x < 0$ and $f(x) > 0$ for $x > 0$ .

Extrema on open intervals

Definition A function defined on an open interval $(a, b)$ need not have a global maximum or minimum, even if it is continuous and bounded. If it does have a global extremum on $(a, b)$ , that point is an interior point, so by Fermat’s theorem in the next lesson it is a critical point.

In plain words On an open interval the function can get closer and closer to a value without ever reaching it, because the point that would reach it has been left out.

Example On $(-1, 2)$ , $f(x) = x^2$ has global minimum $0$ at the critical point $0$ . It has no global maximum, since its values approach $4$ as $x \to 2^{-}$ but $x = 2$ is not in the domain. On $(0, 1)$ , $g(x) = x$ has neither a maximum nor a minimum.

Check yourself

Find all critical points of $f(x) = x^4 - 8x^2$ .

Show the answer

The derivative is $f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2)$ , which exists everywhere and is zero at $x = -2$ , $0$ and $2$ . These are all stationary points.

Find the global maximum and minimum of $f(x) = \lvert x - 2 \rvert$ on $[0, 5]$ , and say whether each occurs at an endpoint, a stationary point or a singular point.

Show the answer

The minimum is $0$ at $x = 2$ , a singular point. The maximum is $3$ at the endpoint $x = 5$ , since $f(0) = 2 < 3$ .

Does $f(x) = \frac{1}{1 + x^2}$ have a global maximum on $\mathbb{R}$ ? A global minimum?

Show the answer

The global maximum is $1$ at $x = 0$ , because $1 + x^2 \ge 1$ . There is no global minimum, since the values approach $0$ as $\lvert x \rvert \to \infty$ but are always positive.

Extreme Values

Theory, part 65

Practice, part 66

Key terms

Global maxima and minima

Local maxima and minima

Extreme values

Domain points and endpoints

Extrema at endpoints

Critical points

Stationary and singular points

Extrema on open intervals

Check yourself

How this lesson connects

Builds on

Used later in