Candidates Test for Absolute Extrema

When a function is continuous on a closed interval, the Extreme Value Theorem promises an absolute maximum and an absolute minimum. The candidates test (also called the closed interval method) is how you find them: make a short list of every place they could be, evaluate the function at each one, and pick the biggest and smallest. No sign charts needed.

Key ideas

Where can an absolute extremum be?

On a closed interval $[a, b]$, an absolute maximum or minimum can only happen at:

• a critical point inside the interval (where $f'(x) = 0$ or $f'(x)$ doesn’t exist), or
• an endpoint, $x = a$ or $x = b$.

Anywhere else, the graph is going up or down, so a nearby point is higher and another is lower.

The candidates test

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

1. Check the hypothesis: $f$ is continuous on the closed interval $[a, b]$, so the EVT says absolute extrema exist.
2. Find the critical points of $f$ and keep only those inside $(a, b)$.
3. Evaluate $f$ at each critical point and at both endpoints. A table works well.
4. The largest value is the absolute maximum; the smallest is the absolute minimum.

If two candidates tie for the largest value, the absolute maximum happens at both $x$-values (and the same for the minimum).

The four candidates for $f(x) = x^3 - 3x^2 + 1$ on $[-1, 4]$ (Example 2).

Writing the answer

AP questions often say “find the absolute maximum value of $f$ on $[a, b]$. Justify your answer.” A complete answer shows the table of candidates and a conclusion sentence such as “The absolute maximum value of $f$ on $[-1, 4]$ is $17$, at $x = 4$.” The table itself is the justification.

When the interval isn’t closed

The candidates test needs a closed interval. On an open interval or the whole real line, use the first derivative test instead. A handy fact for optimization: if a continuous function has only one critical point on an interval, and it’s a relative maximum, then it’s the absolute maximum on that interval (and the same for a minimum).

Worked examples

Example 1: One critical point

Find the absolute extrema of $f(x) = x^2 - 6x + 2$ on $[0, 5]$.

Solution. $f$ is a polynomial, so it’s continuous on $[0, 5]$.

$f'(x) = 2x - 6 = 0$ gives $x = 3$, which is in $(0, 5)$.

$x$	$0$ (endpoint)	$3$ (critical point)	$5$ (endpoint)
$f(x)$	$2$	$-7$	$-3$

The absolute maximum is $2$, at $x = 0$. The absolute minimum is $-7$, at $x = 3$.

Example 2: A tie

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

Solution. $f$ is continuous on $[-1, 4]$.

$$f'(x) = 3x^2 - 6x = 3x(x - 2)$$

Critical points: $x = 0$ and $x = 2$, both in $(-1, 4)$.

$x$	$-1$	$0$	$2$	$4$
$f(x)$	$-1 - 3 + 1 = -3$	$1$	$8 - 12 + 1 = -3$	$64 - 48 + 1 = 17$

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

Example 3: A trig function (radians)

Find the absolute extrema of $f(x) = x - 2\sin x$ on $[0, \pi]$.

Solution. $f$ is continuous on $[0, \pi]$. Remember: $x$ is in radians.

$f'(x) = 1 - 2\cos x = 0$ gives $\cos x = \dfrac{1}{2}$. In $(0, \pi)$, that’s only $x = \dfrac{\pi}{3}$.

$x$	$0$	$\dfrac{\pi}{3}$	$\pi$
$f(x)$	$0$	$\dfrac{\pi}{3} - \sqrt{3} \approx -0.685$	$\pi \approx 3.142$

The absolute maximum is $\pi$, at $x = \pi$. The absolute minimum is $\dfrac{\pi}{3} - \sqrt{3} \approx -0.685$, at $x = \dfrac{\pi}{3}$.

Example 4: A critical point where f′ is undefined

Find the absolute extrema of $f(x) = x^{2/3}(x - 5)$ on $[-1, 4]$.

Solution. $f$ is continuous everywhere (cube roots are defined for negative numbers too). From critical points and extrema, $f'(x) = \dfrac{5(x - 2)}{3x^{1/3}}$, so the critical points are $x = 0$ (undefined) and $x = 2$ (zero). Both are in $(-1, 4)$.

$x$	$-1$	$0$	$2$	$4$
$f(x)$	$(1)(-6) = -6$	$0$	$-3\sqrt[3]{4} \approx -4.762$	$-\sqrt[3]{16} \approx -2.520$

For $x = -1$: $(-1)^{2/3} = \left(\sqrt[3]{-1}\right)^2 = 1$. For $x = 4$: $4^{2/3}(-1) = -\sqrt[3]{16}$.

The absolute maximum is $0$, at $x = 0$ (the cusp). The absolute minimum is $-6$, at $x = -1$.

Common mistakes

Forgetting the endpoints. In Example 1, the absolute maximum is at an endpoint. If you only check critical points, you’ll miss it.

Including critical points outside the interval. For $f(x) = x^4 - 8x^2$ on $[-1, 3]$, the critical point $x = -2$ isn’t in the interval, so it isn’t a candidate.

Missing critical points where f′ is undefined. In Example 4, the absolute maximum is at $x = 0$, where $f'$ doesn’t exist. Always check the denominator of $f'$.

Answering with the wrong thing. “Absolute maximum value” means the $y$-value ($17$). “Where” or “at what $x$” means the $x$-value ($x = 4$). Give both when you’re not sure.

Using the test on an interval where f isn’t continuous. For $f(x) = \dfrac{1}{x - 2}$ on $[0, 3]$, the function blows up at $x = 2$ and has no absolute maximum or minimum. Check continuity first.

Arithmetic slips with negatives. Write out each substitution, especially powers of negative numbers. $(-1)^3 = -1$, but $(-1)^2 = 1$.

Practice

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

Solution

$f'(x) = 2x - 4 = 0$ gives $x = 2$.

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

Absolute maximum $6$ at $x = 5$; absolute minimum $-3$ at $x = 2$.

2. (Warm-up) Find the absolute extrema of $f(x) = 3x - 5$ on $[1, 4]$.

Solution

$f'(x) = 3$, which is never $0$, so there are no critical points. Only the endpoints are candidates: $f(1) = -2$ and $f(4) = 7$.

Absolute maximum $7$ at $x = 4$; absolute minimum $-2$ at $x = 1$.

3. (Core) Find the absolute extrema of $f(x) = 2x^3 + 3x^2 - 12x$ on $[-3, 2]$.

Solution

$$f'(x) = 6x^2 + 6x - 12 = 6(x + 2)(x - 1)$$

Critical points: $x = -2$ and $x = 1$, both in $(-3, 2)$.

$x$	$-3$	$-2$	$1$	$2$
$f(x)$	$-54 + 27 + 36 = 9$	$-16 + 12 + 24 = 20$	$2 + 3 - 12 = -7$	$16 + 12 - 24 = 4$

Absolute maximum $20$ at $x = -2$; absolute minimum $-7$ at $x = 1$.

4. (Core) Find the absolute extrema of $f(x) = xe^{-x}$ on $[0, 3]$. Give decimals to three places.

Solution

$f'(x) = (1 - x)e^{-x} = 0$ gives $x = 1$.

$x$	$0$	$1$	$3$
$f(x)$	$0$	$e^{-1} \approx 0.368$	$3e^{-3} \approx 0.149$

Absolute maximum $\dfrac{1}{e} \approx 0.368$ at $x = 1$; absolute minimum $0$ at $x = 0$.

5. (Core) Find the absolute extrema of $f(x) = x + 2\cos x$ on $[0, \pi]$ (radians).

Solution

$f'(x) = 1 - 2\sin x = 0$ gives $\sin x = \dfrac{1}{2}$: $x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$, both in $(0, \pi)$.

$x$	$0$	$\dfrac{\pi}{6}$	$\dfrac{5\pi}{6}$	$\pi$
$f(x)$	$2$	$\dfrac{\pi}{6} + \sqrt{3} \approx 2.256$	$\dfrac{5\pi}{6} - \sqrt{3} \approx 0.886$	$\pi - 2 \approx 1.142$

Absolute maximum $\dfrac{\pi}{6} + \sqrt{3} \approx 2.256$ at $x = \dfrac{\pi}{6}$; absolute minimum $\dfrac{5\pi}{6} - \sqrt{3} \approx 0.886$ at $x = \dfrac{5\pi}{6}$.

6. (Core) Find the absolute extrema of $f(x) = (x^2 - 4)^{2/3}$ on $[-1, 3]$. Give decimals to three places.

Solution

$f'(x) = \dfrac{4x}{3\sqrt[3]{x^2 - 4}}$. Critical points: $x = 0$ (where $f' = 0$) and $x = \pm 2$ (where $f'$ is undefined). Only $0$ and $2$ are in $(-1, 3)$.

$x$	$-1$	$0$	$2$	$3$
$f(x)$	$(-3)^{2/3} = \sqrt[3]{9} \approx 2.080$	$4^{2/3} = \sqrt[3]{16} \approx 2.520$	$0$	$5^{2/3} = \sqrt[3]{25} \approx 2.924$

Absolute maximum $\sqrt[3]{25} \approx 2.924$ at $x = 3$; absolute minimum $0$ at $x = 2$.

7. (Core) Find the absolute extrema of $f(x) = \dfrac{\ln x}{x}$ on $[1, e^2]$.

Solution

$$f'(x) = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$$

$f'(x) = 0$ when $\ln x = 1$, so $x = e$, which is in $(1, e^2)$.

$x$	$1$	$e$	$e^2$
$f(x)$	$0$	$\dfrac{1}{e} \approx 0.368$	$\dfrac{2}{e^2} \approx 0.271$

Absolute maximum $\dfrac{1}{e}$ at $x = e$; absolute minimum $0$ at $x = 1$.

8. (Challenge) Find the absolute extrema of $f(x) = x^4 - 8x^2$ on $[-1, 3]$.

Solution

$$f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2)$$

Critical points: $x = -2$, $0$, $2$. Only $0$ and $2$ are in $(-1, 3)$.

$x$	$-1$	$0$	$2$	$3$
$f(x)$	$1 - 8 = -7$	$0$	$16 - 32 = -16$	$81 - 72 = 9$

Absolute maximum $9$ at $x = 3$; absolute minimum $-16$ at $x = 2$.

9. (Challenge) A drone flies straight up and down. Its vertical velocity is $v(t) = t^3 - 9t^2 + 24t$ m/s for $0 \le t \le 5$ seconds. Find its greatest and least velocity during this time, and when they happen.

Solution

$v$ is a polynomial, so it’s continuous on $[0, 5]$.

$$v'(t) = 3t^2 - 18t + 24 = 3(t - 2)(t - 4)$$

Critical points: $t = 2$ and $t = 4$.

$t$ (s)	$0$	$2$	$4$	$5$
$v(t)$ (m/s)	$0$	$8 - 36 + 48 = 20$	$64 - 144 + 96 = 16$	$125 - 225 + 120 = 20$

The greatest velocity is $20$ m/s, at both $t = 2$ s and $t = 5$ s. The least velocity is $0$ m/s, at $t = 0$.