What is a critical number?

A number c in the domain of f where f'(c)=0 or f'(c) doesn't exist. Local maximums and minimums can only happen at critical numbers.

Can the absolute maximum be at an endpoint?

Yes, and it often is. That's why the closed interval method evaluates f at both endpoints as well as at the critical numbers.

Can a function have more than one absolute maximum?

It has only one absolute maximum value, but that value can be reached at more than one x. In x^3-3x^2+1 on [-1,4], the minimum value -3 happens at both x=-1 and x=2.

Is every critical number a maximum or minimum?

No. f(x)=x^3 has a critical number at 0 but keeps increasing through it. Critical numbers are candidates that must be checked.

What is the difference between absolute and local extrema?

An absolute extreme is the highest or lowest value over the whole interval. A local extreme is highest or lowest only compared with nearby points. Every absolute extreme inside the interval is also a local one, but not the other way round.

How to Find Absolute Maximum and Minimum Values

What is the difference between absolute and local maximum and minimum?

Many practical questions ask for the largest or smallest value of something: the highest point of a path, the lowest cost, the greatest profit. Calculus answers them by looking at the slope, but first we need precise names for the two kinds of "highest point".

Definition: absolute maximum and minimum

$f (c)$ is the absolute maximum of $f$ on an interval if $f (c) \geq f (x)$ for every $x$ in the interval, and the absolute minimum if $f (c) \leq f (x)$ for every $x$ in the interval. Together they are called the absolute extreme values (or absolute extrema).

Definition: local maximum and minimum

$f (c)$ is a local maximum if $f (c) \geq f (x)$ for all $x$ near $c$ , like the top of a hill that isn't necessarily the highest hill. A local minimum is the bottom of a valley in the same sense.

f (x) = x^{3} - 3 x^{2} + 1

[- 1, 4]

: a local maximum at

x = 0

, the absolute maximum

17

at the endpoint

x = 4

, and the absolute minimum

- 3

, reached at both

x = - 1

and

x = 2

The graph shows why the difference matters: the hilltop at $x = 0$ is a local maximum, but the function is much higher at the right-hand end, so the hilltop isn't the absolute maximum.

Is there always a maximum and minimum?

Not always. $f (x) = \frac{1}{x}$ on $(0, 1]$ grows without bound near $0$ , so it has no absolute maximum. And $f (x) = x$ on the open interval $(0, 1)$ gets as close to $1$ as we like but never reaches it, so it has no absolute maximum either. Both problems disappear for continuous functions on closed intervals (intervals that include their endpoints):

The Extreme Value Theorem

f

is continuous on a closed interval

[a, b]

, then

f

has an absolute maximum and an absolute minimum on

[a, b]

What are critical numbers?

At the top of a smooth hill, the tangent line is horizontal, so the derivative is $0$ . The reason: just before the top the graph rises (slope $\geq 0$ ), and just after it falls (slope $\leq 0$ ), so at the top itself the slope can only be $0$ . The other way a high or low point can happen is at a sharp corner or cusp, where there's no derivative at all. These are the only candidates inside an interval, so they get a name:

Definition: critical number

A critical number of $f$ is a number $c$ in the domain of $f$ where $f^{'} (c) = 0$ or $f^{'} (c)$ does not exist.

Where local extremes can happen

f

has a local maximum or minimum at

c

, then

c

is a critical number of

f

. (This is often called Fermat's theorem.)

Common mistake

Assuming every critical number is a maximum or minimum. $f (x) = x^{3}$ has $f^{'} (0) = 0$ , but the graph keeps rising through $x = 0$ : it levels off for an instant and carries on. Critical numbers are only candidates; you still have to check.

y = x^{3}

has a horizontal tangent at

x = 0

but no maximum or minimum there.

How do you find the absolute maximum and minimum on a closed interval?

Since the absolute extremes exist (for a continuous function on $[a, b]$ ) and must happen either at a critical number or at an endpoint, we only need to compare a short list of values:

Check that $f$ is continuous on the closed interval $[a, b]$ .
Find the critical numbers of $f$ that lie inside $(a, b)$ : solve $f^{'} (x) = 0$ , and find where $f^{'} (x)$ doesn't exist.
Evaluate $f$ at each critical number and at both endpoints $a$ and $b$ .
Compare: the largest value is the absolute maximum and the smallest is the absolute minimum.

Worked example: A polynomial

Find the absolute maximum and minimum of $f (x) = x^{3} - 3 x^{2} + 1$ on $[- 1, 4]$ .

$f$ is a polynomial, so it is continuous on $[- 1, 4]$ .
$f^{'} (x) =$ $3 x^{2} - 6 x =$ $3 x (x - 2)$ , which is $0$ at $x = 0$ and $x = 2$ . Both are inside the interval, and $f^{'}$ exists everywhere.
Evaluate: $f (- 1) = - 1 - 3 + 1 = - 3$ ; $f (0) = 1$ ; $f (2) = 8 - 12 + 1 = - 3$ ; $f (4) = 64 - 48 + 1 = 17$ .
The largest value is $17$ and the smallest is $- 3$ .

Answer: Absolute maximum $17$ at $x = 4$ ; absolute minimum $- 3$ , at both $x = - 1$ and $x = 2$ .

Worked example: A critical number where the derivative doesn't exist

Find the absolute maximum and minimum of $f (x) = x^{2 / 3}$ on $[- 1, 8]$ .

$f (x) =$ $x^{2 / 3} =$ ${(\sqrt[3]{x})}^{2}$ is continuous everywhere, including on $[- 1, 8]$ .
$f^{'} (x) =$ $\frac{2}{3} x^{- 1 / 3} =$ $\frac{2}{3 \sqrt[3]{x}}$ . This is never $0$ (the top is $2$ ), but it doesn't exist at $x = 0$ , where the bottom is $0$ . So $x = 0$ is a critical number.
Evaluate: $f (- 1) =$ ${(\sqrt[3]{- 1})}^{2} =$ $(- 1)^{2} =$ $1$ ; $f (0) = 0$ ; $f (8) =$ ${(\sqrt[3]{8})}^{2} =$ $2^{2} =$ $4$ .
Compare: the largest is $4$ and the smallest is $0$ .

Answer: Absolute maximum $4$ at $x = 8$ ; absolute minimum $0$ at $x = 0$ (a sharp point, or cusp, on the graph).

Worked example: A trig function

Find the absolute maximum and minimum of $f (x) = \sin x + \cos x$ on $[0, π]$ .

$f$ is continuous everywhere.
$f^{'} (x) = \cos x - \sin x$ . Setting it to $0$ gives $\cos x = \sin x$ ; on $[0, π]$ that happens only at $x = \frac{π}{4}$ .
Evaluate: $f (0) = 0 + 1 = 1$ ; $f (\frac{π}{4}) =$ $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} =$ $\sqrt{2} \approx 1.414$ ; $f (π) = 0 + (- 1) = - 1$ .
Compare the three values.

Answer: Absolute maximum $\sqrt{2}$ at $x = \frac{π}{4}$ ; absolute minimum $- 1$ at $x = π$ .

Common mistake

Forgetting the endpoints. In the first example the absolute maximum is at the endpoint $x = 4$ , where $f^{'}$ isn't $0$ at all. Critical numbers find extremes inside the interval; endpoints must always be checked separately.

How do you find local maximums and minimums?

For local extremes (on an open interval or the whole real line), find the critical numbers and then test each one. The first derivative test checks whether $f^{'}$ changes from positive to negative (a local max) or negative to positive (a local min); the second derivative test checks the sign of $f^{″}$ at the critical number. Both are explained in increasing, decreasing and concavity.

Word problems that ask for the largest or smallest possible value of something, such as the most area for a given fence, use the same ideas: see optimization problems.

How to find absolute maximum and minimum values