Math Core

Every relative maximum or minimum happens at a critical point, but not every critical point is a maximum or minimum. The first derivative test sorts the candidates: it looks at how the sign of $f'$ changes as you walk across each critical point.

The idea: watch the sign change

Think of hiking along the graph from left to right. At the top of a hill, you stop climbing and start descending: the slope goes from positive to negative. At the bottom of a valley, you stop descending and start climbing: the slope goes from negative to positive. If the slope is positive on both sides, you just paused on a ledge and kept climbing, so it's neither a peak nor a valley.

The first derivative test

Suppose $c$ is a critical point of $f$ , and $f$ is continuous at $c$ .

If $f'$ changes from positive to negative at $c$ , then $f$ has a relative maximum at $c$ .
If $f'$ changes from negative to positive at $c$ , then $f$ has a relative minimum at $c$ .
If $f'$ does not change sign at $c$ , then $f$ has no relative extremum at $c$ .

The test works whether $f'(c) = 0$ or $f'(c)$ doesn't exist. That makes it more powerful than the second derivative test you'll see next lesson, which fails at corners and cusps.

Using the test

The steps are the same as for finding increasing and decreasing intervals, plus one more.

Find $f'(x)$ and the critical points.
Make a sign chart for $f'$ .
At each critical point, read the sign change and classify it.
If asked for the extreme value, plug $c$ into the original function $f$ .

Worked example: A quartic with a non-extremum

Find and classify all relative extrema of $f(x) = x^4 - 4x^3$ .

Solution. $f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$ . The critical points are $x = 0$ and $x = 3$ .

interval	$x < 0$	$0 < x < 3$	$x > 3$
sign of $4x^2$	$+$	$+$	$+$
sign of $x - 3$	$-$	$-$	$+$
sign of $f'(x)$	$-$	$-$	$+$

At $x = 0$ , $f'$ is negative on both sides, so $f$ has no relative extremum at $x = 0$ .
At $x = 3$ , $f'$ changes from negative to positive, so $f$ has a relative minimum at $x = 3$ . Its value is $f(3) = 81 - 108 = -27$ .

f(x) = x⁴ − 4x³ (solid) and f′(x) = 4x²(x − 3) (dashed). f′ touches zero at x = 0 without changing sign, so f has a flat spot there but no extremum.Open in grapher →

The squared factor $x^2$ is the reason nothing happens at $x = 0$ . A factor raised to an even power doesn't change sign, so the derivative's sign doesn't flip there. A factor raised to an odd power does flip. That's a quick way to predict the chart before you build it.

Common mistake

Don't assume that $f'(c) = 0$ means an extremum. You must show the sign change. On the AP exam, " $f'(3) = 0$ , so $f$ has a minimum at $x = 3$ " earns no justification credit. Write instead: " $f$ has a relative minimum at $x = 3$ because $f'$ changes from negative to positive at $x = 3$ ."

Critical points where f′ doesn't exist

Worked example: A cusp

Find and classify the relative extrema of $f(x) = x^{2/3}(x - 5)$ .

Solution. In the previous lesson you found $f'(x) = \dfrac{5(x - 2)}{3\sqrt[3]{x}}$ , with critical points $x = 0$ (where $f'$ is undefined) and $x = 2$ (where $f' = 0$ ). The cube root $\sqrt[3]{x}$ has the same sign as $x$ .

interval	$x < 0$	$0 < x < 2$	$x > 2$
sign of $x - 2$	$-$	$-$	$+$
sign of $\sqrt[3]{x}$	$-$	$+$	$+$
sign of $f'(x)$	$+$	$-$	$+$

At $x = 0$ , $f'$ changes from positive to negative, and $f$ is continuous there, so $f$ has a relative maximum at $x = 0$ . The value is $f(0) = 0$ .
At $x = 2$ , $f'$ changes from negative to positive, so $f$ has a relative minimum at $x = 2$ . The value is $f(2) = 2^{2/3}(-3) = -3\sqrt[3]{4} \approx -4.76$ .

f(x) = x^(2/3)(x − 5). The relative maximum is at a cusp, where f′ does not exist.Open in grapher →

Using the graph of f′

When you're given the graph of $f'$ , the first derivative test becomes a matter of reading where the graph crosses the $x$ -axis:

crosses from above to below: relative maximum of $f$ ;
crosses from below to above: relative minimum of $f$ ;
touches without crossing: no extremum.

Worked example: From a factored derivative

The function $f$ is defined for all real numbers, and its derivative is $f'(x) = (x + 1)^2(x - 4)(x - 7)$ . At which $x$ -values does $f$ have a relative maximum? A relative minimum?

The graph of f′(x) = (x + 1)²(x − 4)(x − 7), drawn at 1/20 vertical scale. It touches the axis at x = −1 and crosses it at x = 4 and x = 7.Open in grapher →

Solution. The critical points are $x = -1$ , $4$ , $7$ . Because $(x + 1)^2 \ge 0$ , the sign of $f'$ is the sign of $(x - 4)(x - 7)$ away from $x = -1$ :

interval	$x < -1$	$-1 < x < 4$	$4 < x < 7$	$x > 7$
sign of $f'$	$+$	$+$	$-$	$+$

$f$ has a relative maximum at $x = 4$ ( $f'$ changes from positive to negative) and a relative minimum at $x = 7$ ( $f'$ changes from negative to positive). At $x = -1$ , $f'$ doesn't change sign, so there is no extremum.

Tip

A relative extremum is a point on the graph of $f$ . If a question asks "at what $x$ ," give the $x$ -value. If it asks for "the relative maximum value," give $f(c)$ , which usually means evaluating the original function, not the derivative.