The Candidates Test — AP® Calculus AB Notes

The First Derivative Test finds peaks and valleys, but only compared with their neighbors. Many real questions want more: the hottest moment of the whole day, the lowest level the reservoir reaches all month. That is an absolute extremum, and on a closed interval there is a short, reliable way to find it without any sign analysis at all.

Why a short list is enough

Put together what you already know. If $f$ is continuous on $[a, b]$ :

The Extreme Value Theorem says an absolute maximum and an absolute minimum exist.
Each one is either at an endpoint, or inside the interval.
If it is inside, it is also a relative extremum, so it is at a critical point.

So the absolute extrema are among a short list of candidates: the two endpoints and the critical points in between. Evaluate $f$ at each candidate, and the largest value is the absolute maximum; the smallest is the absolute minimum.

The graph of f(x) = 2x cubed minus 9x squared plus 12x on the closed interval from 0 to 3, with four candidates: endpoints (0, 0) and (3, 9), critical points (1, 5) and (2, 4). A table lists the values; the absolute maximum is 9 at x = 3 and the absolute minimum is 0 at x = 0. The relative maximum (1, 5) and relative minimum (2, 4) are not absolute. — **Figure 1.** Four candidates, four values. The absolute maximum is at an endpoint; the relative maximum inside loses. LevlPrep original diagram.

The Candidates Test, step by step

The Candidates Test (also called the closed interval method) for $f$ continuous on $[a, b]$ :

Find $f'(x)$ .
Find the critical points in the open interval $(a, b)$ : where $f'(x) = 0$ or $f'(x)$ does not exist. Throw away any outside the interval.
Make a table of $x$ and $f(x)$ for the endpoints and each critical point.
The largest $f(x)$ in the table is the absolute maximum value; the smallest is the absolute minimum value.

Worked example 1. Find the absolute extrema of $f(x) = 2x^3 - 9x^2 + 12x$ on $[0, 3]$ .

$f'(x)$ $= 6x^2 - 18x$ $+ 12$ $= 6(x - 1)(x - 2)$ , zero at $x = 1$ and $x = 2$ , both in $(0, 3)$ .

$x$	$0$	$1$	$2$	$3$
$f(x)$	$0$	$5$	$4$	$9$

Absolute maximum: $9$ , at $x = 3$ . Absolute minimum: $0$ , at $x = 0$ . Both are at endpoints; the relative maximum $f(1) = 5$ and relative minimum $f(2) = 4$ are not the absolute ones.

No signs of $f'$ are needed, and nothing has to be said about which critical points are peaks. The comparison does all the work.

Exact answers without a calculator

Worked example 2. Find the absolute minimum of $f(x) = x + 2\cos x$ on $[0, \pi]$ .

$f'(x) = 1 - 2\sin x = 0$ when $\sin x = \frac{1}{2}$ : $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ .

Candidates: $f(0) = 2$ ; $f\left(\frac{\pi}{6}\right)$ $= \frac{\pi}{6} + \sqrt{3}$ $\approx 2.256$ ; $f\left(\frac{5\pi}{6}\right)$ $= \frac{5\pi}{6} - \sqrt{3}$ $\approx 0.886$ ; $f(\pi) = \pi - 2 \approx 1.142$ .

Absolute minimum: $\frac{5\pi}{6} - \sqrt{3}$ , at $x = \frac{5\pi}{6}$ . Keep it exact; the decimals are only for comparing.

With a calculator

On a calculator question the critical points may not have a formula. Graph $f'$ or use the solver to find its zeros, store each one, and evaluate $f$ at the stored values. Then compare with the endpoints.

Worked example 3. A reservoir holds $V(t)$ $= 40 + 12\sin\left(\frac{t}{2}\right) - 0.5t$ thousand gallons, $0 \le t \le 10$ days. Find the least volume.

$V'(t)$ $= 6\cos\left(\frac{t}{2}\right) - 0.5$ $= 0$ at $t \approx 2.975$ and $t \approx 9.592$ (radian mode).

Candidates: $V(0) = 40$ , $V(2.975) \approx 50.471$ , $V(9.592) \approx 23.246$ , $V(10) \approx 23.493$ .

The least volume is about $23.246$ thousand gallons, at $t \approx 9.592$ days. The endpoint value $23.493$ was close: only the comparison shows which is smaller.

Local versus global: the point readers take away

A justification like " $V'$ changes from negative to positive at $t = 9.592$ , so this is the absolute minimum" earns no credit. A sign change is a local fact: it compares $V(9.592)$ with nearby values only. An endpoint, or another critical point far away, could be lower. For an absolute extremum you need a global argument. On a closed interval, the Candidates Test is the simplest one: show the table of candidates and values, then name the largest or smallest.

Question asks for	Tool	Justification
A relative extremum	First Derivative Test	" $f'$ changes from $+$ to $-$ at $x = c$ "
An absolute extremum on $[a, b]$	Candidates Test	A table of endpoints and critical points with their values