Free AP Calculus AB Analytical Applications of Differentiation

AP Calculus AB · Analytical Applications of Differentiation · free lesson

The Candidates Test for Absolute Extrema

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

The highest and lowest values on an interval

A relative maximum is higher than the points near it. An absolute maximum of $f$ on an interval is higher than (or equal to) every value of $f$ on that interval; an absolute minimum is lower than or equal to every value. Together they are the absolute extrema (also called global extrema).

The Extreme Value Theorem says when they are sure to exist: if $f$ is continuous on a closed interval $[a, b]$ , then $f$ has an absolute maximum and an absolute minimum on $[a, b]$ . This topic is about finding them.

Where to look: the candidates

Suppose the absolute maximum happens at a point $c$ inside the interval. Then it is also a relative maximum, so $c$ is a critical point. The only other places it can happen are the two ends. So the absolute extrema can only be at a short list of candidates: the endpoints $a$ and $b$ , and the critical points between them.

Rule.

The Candidates Test (also called the closed interval method). To find the absolute extrema of a continuous function $f$ on $[a, b]$ :

Find the critical points of $f$ in $a < x < b$ : where $f'(x) = 0$ or $f'(x)$ does not exist.
Work out $f$ at each critical point and at both endpoints $a$ and $b$ .
The largest of these values is the absolute maximum; the smallest is the absolute minimum.

No sign chart is needed: the values are simply compared. Take $f(x) = 2x^3 - 3x^2 - 12x + 4$ on $[-2, 4]$ .

Step 1. Differentiate and find where $f' = 0$ :

\begin{aligned} f'(x) &= 6x^{2} - 6x - 12 = 6(x + 1)(x - 2) \end{aligned}

So the critical points are $x = -1$ and $x = 2$ , both inside $[-2, 4]$ . ( $f'$ exists everywhere, so there are no others.)

Step 2. Work out $f$ at the candidates, the two critical points and the two endpoints:

Candidate $x$	$-2$ (endpoint)	$-1$ (critical point)	$2$ (critical point)	$4$ (endpoint)
$f(x)$	$0$	$11$	$-16$	$36$

Step 3. Compare. The largest value is $36$ , at $x = 4$ ; the smallest is $-16$ , at $x = 2$ . So on $[-2, 4]$ the absolute maximum of $f$ is $f(4) = 36$ and the absolute minimum is $f(2) = -16$ .

Notice: the relative maximum at $x = -1$ is not the absolute maximum. The curve climbs higher at the right end. Watch the graph draw itself, with the four candidates marked:

The interval decides

The same function on a different interval can have its absolute extrema somewhere else. Drag the right end $b$ of the interval $[-2, b]$ . While $b$ is small, the hilltop at $x = -1$ is the highest point. Once $b$ passes a certain value, the right end climbs above it and becomes the absolute maximum. A critical point outside the interval (for example $x = 2$ when $b < 2$ ) is not a candidate at all.

Points where f' does not exist

Step 1 asks for both kinds of critical point. For $f(x) = x^{2/3}$ on $[-1, 8]$ , $f'(x) = \frac{2}{3x^{1/3}}$ is never 0, but it does not exist at $x = 0$ . The candidates are $-1$ , $0$ and $8$ :

$f(-1) = 1$ , $f(0) = 0$ , $f(8) = 4$ . The absolute minimum is $f(0) = 0$ , at the cusp, and the absolute maximum is $f(8) = 4$ . Looking only for $f' = 0$ would have missed the minimum.

Only one critical point

When $f$ is continuous on an interval (closed or not) and has only one critical point $c$ there, the first derivative test settles it: if $f'$ changes from positive to negative at $c$ , then $f$ rises up to $c$ and falls all the way after it, so $f(c)$ is the absolute maximum on the interval. A change from negative to positive gives the absolute minimum. This is how the AP exam often wants an absolute extremum on an open interval justified.

Note.

On the AP exam the usual slips are: giving the $x$ -value when the maximum value was asked (or the other way round); leaving out the endpoints; leaving out a critical point where $f'$ does not exist; keeping a critical point that lies outside the interval; and putting the candidates into $f'$ instead of $f$ . A free-response answer lists every candidate with its value of $f$ , then states the conclusion: "the absolute maximum value of $f$ on $[a, b]$ is $f(c) = \ldots$ ".

Worked examples

Each example is one kind of question the AP exam asks about absolute extrema. The method is the Candidates Test: list the endpoints and the critical points inside the interval, work out $f$ at each, and compare. Every number in the working is found by the engine.

Example.

1. A polynomial, with a critical point outside the interval. Find the absolute maximum and minimum values of $f(x) = x^4 - 8x^2 + 3$ on $[-1, 3]$ .

Critical points. $f'$ exists everywhere, and

\begin{aligned} f'(x) &= 4x^{3} - 16x = 4x(x + 2)(x - 2) \end{aligned}

So $f'(x) = 0$ at $x = -2$ , $x = 0$ and $x = 2$ . But $-2$ is not in $[-1, 3]$ , so it is not a candidate. The candidates are the endpoints $-1$ and $3$ and the critical points $0$ and $2$ .

Candidate $x$	$-1$ (endpoint)	$0$	$2$	$3$ (endpoint)
$f(x)$	$-4$	$3$	$-13$	$12$

The absolute maximum value is $f(3) = 12$ , at the right endpoint. The absolute minimum value is $f(2) = -13$ .

Check: the gray part of the curve, outside the interval, does not count, even though it also dips to $-13$ at $x = -2$ .

Example.

2. A trigonometric function. Find the absolute extrema of $f(x) = x + 2\cos x$ on $[0, \pi]$ .

\begin{aligned} f'(x) &= 1 - 2\sin(x) \end{aligned}

$f'(x) = 0$ when $\sin x = \frac{1}{2}$ ; on $[0, \pi]$ that is $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ . Using $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ :

Candidate $x$	$0$	$\frac{\pi}{6}$	$\frac{5\pi}{6}$	$\pi$
$f(x)$ exactly	$2$	$\frac{\pi}{6} + \sqrt{3}$	$\frac{5\pi}{6} - \sqrt{3}$	$\pi - 2$
$f(x)$ as a decimal	$2.000$	$2.256$	$0.886$	$1.142$

The absolute maximum is $f\left({\frac{\pi}{6}}\right) = \frac{\pi}{6} + \sqrt{3} \approx {2.256}$ and the absolute minimum is $f\left({\frac{5\pi}{6}}\right) = \frac{5\pi}{6} - \sqrt{3} \approx {0.886}$ . On a no-calculator question the answer is left exact; the decimals only help to compare. (To compare without them: $\sqrt{3} \approx 1.73$ and $\pi \approx 3.14$ .)

Example.

3. A critical point where f' does not exist. Find the absolute extrema of $f(x) = 3x^{2/3} - 2x$ on $[-1, 8]$ .

\begin{aligned} f'(x) &= 2x^{-1/3} - 2 = \frac{2(1 - x^{1/3})}{x^{1/3}} \end{aligned}

$f'(x) = 0$ when $x^{1/3} = 1$ , that is $x = 1$ ; and $f'(x)$ does not exist at $x = 0$ , where $f(0)$ is defined. Both are inside $[-1, 8]$ .

Candidate $x$	$-1$ (endpoint)	$0$ ( $f'$ does not exist)	$1$ ( $f' = 0$ )	$8$ (endpoint)
$f(x)$	$5$	$0$	$1$	$-4$

The absolute maximum value is $f(-1) = 5$ and the absolute minimum value is $f(8) = -4$ , both at endpoints. The two critical points give a relative minimum (the cusp at 0) and a relative maximum (at 1), but neither is absolute.

Example.

4. From a table. The function $f$ is differentiable on $[0, 6]$ , and its only critical points are $x = 1$ , $x = 3$ and $x = 5$ . Some values of $f$ and $f'$ are in the table. What are the absolute maximum and minimum values of $f$ on $[0, 6]$ ?

$x$	$0$	$1$	$3$	$5$	$6$
$f(x)$	$2$	$7$	$-1$	$6$	$3$
$f'(x)$	$5$	$0$	$0$	$0$	$-4$

$f$ is differentiable, so it is continuous on the closed interval $[0, 6]$ : the Extreme Value Theorem says the absolute extrema exist. The candidates are the endpoints 0 and 6 and the three critical points, and the table gives $f$ at all five. The largest value of $f$ in the table is $7$ and the smallest is $-1$ .

So the absolute maximum value is $f(1) = 7$ and the absolute minimum value is $f(3) = -1$ . The row of $f'$ only says which $x$ are critical points; its numbers (such as $f'(0) = {5}$ ) are slopes, not values of $f$ , and are never compared.

Example.

5. A tank, with a calculator. The amount of water in a storage tank is $A(t) = 2000 + 300\sin\left(\frac{t}{3}\right) - 2t^2$ gallons, $t$ hours after noon, for $0 \le t \le 12$ . At what time is the amount of water greatest, and how much water is in the tank then? Justify your answer.

The candidates are the endpoints and the times where $A'(t) = 0$ . Differentiate:

\begin{aligned} A'(t) &= 100\cos(\frac{1}{3}t) - 4t \end{aligned}

On the calculator, graph $A'$ on $0 \le t \le 12$ and find its zero: $A'(t) = 0$ only at $t \approx {4.205}$ . Now compare the values of $A$ :

Candidate $t$	$0$	$4.205$	$12$
$A(t)$ , gallons	$2000$	$2260.355$	$1484.959$

The amount of water is greatest at $t \approx {4.205}$ hours after noon, when the tank holds about $2260.355$ gallons.

The justification is the comparison: the value at the one critical point is larger than the values at both ends. (Since $A'$ changes from positive to negative at that point and nowhere else, the single-critical-point reasoning gives the same answer.)

Example.

6. An open interval, one critical point. Show that $f(x) = x + \frac{9}{x}$ has an absolute minimum on the interval $x > 0$ , and find it.

There are no endpoints here, so the Candidates Test does not apply. Use the one critical point instead.

\begin{aligned} f'(x) &= 1 - \frac{9}{x^{2}} \end{aligned}

For $x > 0$ , $f'(x) = 0$ only at $x = 3$ ( $x = -3$ is outside the interval). Test values: $f'(1) = {-8}$ and $f'(4) = {\frac{7}{16}}$ . So $f'$ is negative on $0 < x < 3$ and positive on $x > 3$ : $f$ falls all the way to $x = 3$ and rises all the way after it.

Because $x = 3$ is the only critical point on $x > 0$ and $f'$ changes from negative to positive there, $f(3) = 6$ is the absolute minimum of $f$ on $x > 0$ . ( $f$ has no absolute maximum there: it grows without bound near 0 and for large $x$ .)

Practice

Answer AP-style questions on the candidates test for absolute extrema. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise the candidates test for absolute extrema with new numbers