Absolute Maximum and Minimum Values

Quick answer

An absolute maximum is the largest value a function takes on its whole domain or interval, not only near a point. For a continuous function on a closed interval [a, b], it is guaranteed to exist, and it can occur only at a critical point inside or at an endpoint. So evaluate f at those few candidates and compare. On an open or unbounded interval the extremes may not exist; a single critical point that is a local extremum is then absolute.

What you'll learn

Tell absolute extrema from local extrema
Find absolute extrema on a closed interval by comparing candidates
Explain why only critical points and endpoints need checking
Decide absolute extrema on open intervals with one critical point

Absolute versus local

A local maximum is higher than everything nearby. An absolute maximum is higher than, or equal to, every value on the whole interval. The same goes for minima.

A local maximum need not be absolute — there may be a higher peak elsewhere — and an absolute maximum can sit at an endpoint, where it is not a local peak at all.

The closed interval method

For $f$ continuous on $[a, b]$ :

Find the critical points of $f$ in the open interval $(a, b)$ .
Evaluate $f$ at each of them and at the endpoints $a$ and $b$ .
The largest value is the absolute maximum; the smallest is the absolute minimum.

f(x) = x³ − 3x² + 1

Four candidates on [−1, 4]

Why a few candidates are enough

The Extreme Value Theorem guarantees that the absolute maximum exists. Suppose it occurs at some point $c$ .

If $c$ is inside the interval, it is also a local maximum. At a local maximum of a differentiable function the tangent is horizontal, so $f'(c) = 0$ — or $f$ is not differentiable at $c$ . Either way, $c$ is a critical point. If $c$ is not inside, it is an endpoint.

Every absolute extremum is at a critical point or an endpoint, so comparing those finitely many values cannot miss it.

Example 1 — Compare the candidates

Find the absolute extrema of $f(x) = x^3 - 3x^2 + 1$ on $[-1, 4]$ .

$f'(x) = 3x^2 - 6x = 3x(x - 2)$ , so the critical points are $0$ and $2$ , both in the interval.

$x$	$-1$	$0$	$2$	$4$
$f(x)$	$-3$	$1$	$-3$	$17$

The absolute maximum is $17$ , at $x = 4$ . The absolute minimum is $-3$ , reached at both $x = -1$ and $x = 2$ .

Open and unbounded intervals

Without a closed interval, there may be no absolute extremum: $f(x) = x^2$ on $(0, 1)$ has none. Analyze instead with the sign of $f'$ and the behavior near the ends.

One situation settles it quickly. If $f$ is continuous on an interval and has exactly one critical point there, and that point is a local minimum, it is the absolute minimum. The function falls all the way to it and rises all the way after, so no other value can be lower. The same holds for a single local maximum.

Worked examples

Common mistakes

Practice problems

Find the absolute extrema of $f(x) = x^2 - 4x + 1$ on $[0, 5]$ .

Answer

Max $6$ at $x = 5$ ; min $-3$ at $x = 2$

Full solution

$f'(x) = 2x - 4 = 0$ at $x = 2$ . $f(0) = 1$ , $f(2) = -3$ , $f(5) = 6$ .
Find the absolute extrema of $f(x) = x^3 - 12x$ on $[-3, 3]$ .

Answer

Max $16$ at $x = -2$ ; min $-16$ at $x = 2$

Full solution

Critical points $\pm 2$ . $f(-3) = 9$ , $f(-2) = 16$ , $f(2) = -16$ , $f(3) = -9$ .
Find the absolute extrema of $f(x) = \sin x + \cos x$ on $[0, \pi]$ .

Answer

Max $\sqrt{2}$ at $x = \tfrac{\pi}{4}$ ; min $-1$ at $x = \pi$

Full solution

$f'(x) = \cos x - \sin x = 0$ at $x = \tfrac{\pi}{4}$ . $f(0) = 1$ , $f\!\left(\tfrac{\pi}{4}\right) = \sqrt{2}$ , $f(\pi) = -1$ .
Find the absolute extrema of $f(x) = x e^{-x}$ on $[0, 3]$ .

Answer

Max $\tfrac{1}{e}$ at $x = 1$ ; min $0$ at $x = 0$

Full solution

$f'(x) = e^{-x}(1 - x) = 0$ at $x = 1$ . $f(0) = 0$ , $f(1) = e^{-1} \approx 0.368$ , $f(3) = 3e^{-3} \approx 0.149$ .
Find the absolute extrema of $f(x) = |x - 2|$ on $[0, 5]$ .

Answer

Max $3$ at $x = 5$ ; min $0$ at $x = 2$

Full solution

The critical point is $x = 2$ , where $f'$ does not exist. $f(0) = 2$ , $f(2) = 0$ , $f(5) = 3$ .
Does $f(x) = \tfrac{1}{x}$ have an absolute maximum on $(0, 2]$ ? An absolute minimum?

Answer

No maximum; minimum $\tfrac{1}{2}$ at $x = 2$

Full solution

$f$ decreases on the interval. As $x \to 0^+$ it grows without bound, so there is no maximum. The smallest value is at the closed end: $f(2) = \tfrac{1}{2}$ .
Find the absolute maximum of $f(x) = \tfrac{x}{x^2 + 1}$ on $[0, \infty)$ .

Answer

$\tfrac{1}{2}$ , at $x = 1$

Full solution

$f'(x) = \tfrac{1 - x^2}{(x^2 + 1)^2}$ , zero at $x = 1$ in the interval, positive before and negative after. The only critical point is a local max, so it is absolute: $f(1) = \tfrac{1}{2}$ .
Find the absolute extrema of $f(x) = x^{2/3}$ on $[-1, 8]$ .

Answer

Max $4$ at $x = 8$ ; min $0$ at $x = 0$

Full solution

$f'(x) = \tfrac{2}{3}x^{-1/3}$ is never $0$ and undefined at $0$ . $f(-1) = 1$ , $f(0) = 0$ , $f(8) = 4$ .
Find the absolute extrema of $f(x) = 2x^3 - 9x^2 + 12x$ on $[0, 3]$ .

Answer

Max $9$ at $x = 3$ ; min $0$ at $x = 0$

Full solution

$f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)$ . $f(0) = 0$ , $f(1) = 5$ , $f(2) = 4$ , $f(3) = 9$ .
A student finds the absolute maximum of $f(x) = x^2$ on $[-1, 3]$ as $f(0) = 0$ , the only critical point. What went wrong?

Hint

Is $x = 0$ a maximum at all?

Answer

The endpoints were skipped, and $x = 0$ is a minimum. The maximum is $f(3) = 9$ .

Full solution

Candidates: $f(-1) = 1$ , $f(0) = 0$ , $f(3) = 9$ .

The absolute maximum is $9$ at the endpoint $x = 3$ ; the absolute minimum is $0$ at $x = 0$ .

Frequently asked questions

What is the difference between an absolute and a local maximum?

A local maximum is the highest value near a point. An absolute maximum is the highest value on the whole interval under consideration.

What is the closed interval method?

For f continuous on [a, b]: find the critical points in (a, b), evaluate f at them and at a and b, and pick the largest and smallest values.

Why are the endpoints included?

The largest value can sit at an end of the interval, where the function is still rising or falling and f′ need not be 0.

Does every function have an absolute maximum?

No. On an open or unbounded interval, or with a discontinuity, it may not. Continuity on a closed interval guarantees one.

Can an absolute extreme value occur at more than one point?

Yes. The value is unique, but it can be attained at several x-values.

What to learn next

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