Extreme Value Theorems and the Closed Interval Method, calculus lesson

If you drive from one town to another along a road, at some moment of the trip you were at your highest altitude and at some moment at your lowest. That seems obvious, but it rests on two facts. The road has no jumps, and the trip starts and ends. The extreme value theorem makes this precise, and Fermat’s theorem tells you where to look for those highest and lowest moments.

Key terms

Extreme value theorem

Definition If $f$ is continuous on a closed bounded interval $[a, b]$ , then $f$ attains a global maximum and a global minimum on $[a, b]$ . That is, there are numbers $c$ and $d$ in $[a, b]$ with

f(c) \le f(x) \le f(d) \quad \text{for every } x \text{ in } [a, b].

In plain words A continuous function on a closed interval always has a highest and a lowest point. The theorem says they exist but not where they are.

Example For $f(x) = x^3 - 3x$ on $[0, 2]$ , the theorem promises a maximum and a minimum. They turn out to be $f(2) = 2$ and $f(1) = -2$ .

Continuity on a compact interval

Definition A closed bounded interval $[a, b]$ is called compact. Both hypotheses of the extreme value theorem, continuity at every point of $[a, b]$ and the interval being closed and bounded, are needed. Dropping either one allows functions with no maximum or no minimum. The hypotheses are sufficient, not necessary, so a function that fails them may still happen to have extrema.

In plain words Each condition blocks a different way of escaping. Without the endpoints the function can creep toward a value it never reaches. Without boundedness it can climb forever. With a jump or a blow up it can skip its top value.

Example Each of these misses one hypothesis and has no global maximum.

Function	Interval	What fails
$f(x) = x$	$(0, 1)$	not closed
$f(x) = x$	$[0, \infty)$	not bounded
$f(x) = \frac{1}{x}$ for $x > 0$ , $f(0) = 0$	$[0, 1]$	not continuous at $0$

On the other hand, the step function equal to $0$ on $[0, 1)$ and $1$ on $[1, 2]$ is discontinuous, yet it has maximum $1$ and minimum $0$ .

Fermat’s theorem

Definition If $f$ has a local maximum or local minimum at an interior point $c$ of its domain, and $f'(c)$ exists, then $f'(c) = 0$ .

Here is why for a local maximum. Since $c$ is interior, $f(c + h) - f(c) \le 0$ for all small $h \neq 0$ . So the difference quotient

\frac{f(c + h) - f(c)}{h}

is $\le 0$ for small $h > 0$ and $\ge 0$ for small $h < 0$ . Both one-sided limits equal $f'(c)$ , so $f'(c) \le 0$ and $f'(c) \ge 0$ , which forces $f'(c) = 0$ . For a local minimum the inequalities are reversed.

In plain words At the top of a smooth hill the tangent line is horizontal. The converse is false. A zero derivative does not guarantee a peak or a valley.

Example For $f(x) = x^2 - 4x$ , the local minimum is at $x = 2$ , and indeed $f'(2) = 2 \cdot 2 - 4 = 0$ . For $g(x) = x^3$ , $g'(0) = 0$ but $0$ is not an extremum, which shows the converse fails.

Closed interval method

Definition Let $f$ be continuous on $[a, b]$ . To find its global maximum and minimum values,

find the values of $f$ at the critical points of $f$ in $(a, b)$ ,
find $f(a)$ and $f(b)$ ,
the largest value from steps 1 and 2 is the global maximum and the smallest is the global minimum.

This works because the extrema exist by the extreme value theorem, and by Fermat’s theorem each one sits either at an endpoint or at an interior critical point.

In plain words List every place the answer could be, evaluate $f$ there, and compare the numbers. No sign charts are needed.

Example Take $f(x) = x^3 - 3x^2 + 1$ on $\left[-\frac{1}{2}, 4\right]$ . Then $f'(x) = 3x^2 - 6x = 3x(x - 2)$ , so the critical points are $0$ and $2$ .

$x$	$-\frac{1}{2}$	$0$	$2$	$4$
$f(x)$	$\frac{1}{8}$	$1$	$-3$	$17$

The global maximum is $17$ at $x = 4$ and the global minimum is $-3$ at $x = 2$ .

Endpoints and nondifferentiable candidates

Definition In the closed interval method the candidate list must contain both endpoints and every interior point where $f'$ does not exist, not only the points where $f'(x) = 0$ . Leaving any of them out can miss the answer.

In plain words The derivative equation finds only flat spots. Corners, cusps and the two ends of the interval have to be added by hand.

Example Let $f(x) = x^{2/3}$ on $[-1, 8]$ . Its derivative $\frac{2}{3} x^{-1/3}$ is never $0$ and does not exist at $x = 0$ . The candidates are $f(-1) = 1$ , $f(0) = 0$ and $f(8) = 4$ . The minimum $0$ is at the singular point $0$ and the maximum $4$ is at the endpoint $8$ . Solving $f'(x) = 0$ alone would have found nothing.

Check yourself

Find the global maximum and minimum of $f(x) = 2x^3 - 3x^2 - 12x$ on $[-2, 3]$ .

Show the answer

The derivative is $f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1)$ , so the critical points are $-1$ and $2$ . The values are $f(-2) = -4$ , $f(-1) = 7$ , $f(2) = -20$ and $f(3) = -9$ . The maximum is $7$ at $x = -1$ and the minimum is $-20$ at $x = 2$ .

Find the global maximum and minimum of $f(x) = \lvert x - 1 \rvert$ on $[0, 3]$ .

Show the answer

The only critical point is the singular point $1$ . The values are $f(0) = 1$ , $f(1) = 0$ and $f(3) = 2$ . The maximum is $2$ at $x = 3$ and the minimum is $0$ at $x = 1$ .

Why does the extreme value theorem not apply to $\tan x$ on $\left[0, \frac{\pi}{2}\right)$ , and does $\tan x$ have a maximum there?

Show the answer

The interval is not closed, and $\tan$ is not defined at $\frac{\pi}{2}$ . There is no maximum, since $\tan x \to \infty$ as $x \to \frac{\pi}{2}^{-}$ . The minimum $0$ at $x = 0$ does exist.

Extreme Value Theorems and the Closed Interval Method

Theory, part 67

Practice, part 68

Key terms