Math Core

The sign of the derivative tells you which way a graph is heading. A positive slope means the graph is climbing, and a negative slope means it's falling. In this lesson you'll use that idea to find exactly where a function increases and decreases, and to justify your answer the way the AP exam expects.

The derivative sign rule

Recall that $f$ is increasing on an interval if $f(x_1) < f(x_2)$ whenever $x_1 < x_2$ in that interval, and decreasing if $f(x_1) > f(x_2)$ whenever $x_1 < x_2$ .

Increasing/decreasing test

Suppose $f$ is differentiable on an open interval $I$ .

If $f'(x) > 0$ for every $x$ in $I$ , then $f$ is increasing on $I$ .
If $f'(x) < 0$ for every $x$ in $I$ , then $f$ is decreasing on $I$ .

Why is this true? It's the mean value theorem at work. Take any $x_1 < x_2$ in $I$ . The MVT gives a $c$ between them with

f(x_2) - f(x_1) = f'(c)\,(x_2 - x_1).

If $f' > 0$ everywhere on $I$ , the right side is a positive number times a positive number, so $f(x_2) > f(x_1)$ . The function went up. The same argument with $f' < 0$ shows the function goes down.

Sign charts

A derivative can only change sign where it equals zero or is undefined, which means at critical points (or at points where $f$ itself is undefined). So to find where $f$ increases and decreases:

Find $f'(x)$ and factor it as much as possible.
Find the critical points, plus any $x$ -values not in the domain of $f$ . These split the number line into intervals.
Test one value in each interval (or read signs from the factors) to find the sign of $f'$ .
Where $f' > 0$ , $f$ increases. Where $f' < 0$ , $f$ decreases.

Worked example: A cubic

Find the intervals on which $f(x) = x^3 - 3x^2 - 9x + 2$ is increasing and decreasing.

Solution. $f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)$ . The critical points are $x = -1$ and $x = 3$ . Check the sign of each factor on each interval:

interval	$x < -1$	$-1 < x < 3$	$x > 3$
sign of $x + 1$	$-$	$+$	$+$
sign of $x - 3$	$-$	$-$	$+$
sign of $f'(x)$	$+$	$-$	$+$
behavior of $f$	increasing	decreasing	increasing

So $f$ is increasing for $x < -1$ or $x > 3$ , and decreasing on $-1 < x < 3$ .

Justification: $f$ is increasing on $(-\infty, -1)$ and $(3, \infty)$ because $f'(x) > 0$ there, and decreasing on $(-1, 3)$ because $f'(x) < 0$ there.

The picture below shows $f$ and $f'$ together. Wherever the graph of $f'$ is above the $x$ -axis, the graph of $f$ is rising. Wherever $f'$ dips below the axis, $f$ falls. The turning points of $f$ line up exactly with the zeros of $f'$ .

f(x) = x³ − 3x² − 9x + 2 (solid) and f′(x) = 3(x − 3)(x + 1) (dashed). f rises where f′ > 0 and falls where f′ < 0.Open in grapher →

Tip

Instead of plugging a test value into $f'$ , read the sign from the factored form. Factors that are always positive, like $e^x$ or $x^2 + 1$ , never affect the sign, so you can ignore them. This is faster and avoids arithmetic slips.

Reading from the graph of f′

AP questions often show you the graph of $f'$ , not $f$ . Don't confuse the two. You're looking at the slope of $f$ , so the only question is: is this graph above or below the $x$ -axis?

Worked example: Given the graph of the derivative

The graph below is the graph of $f'$ , the derivative of a function $f$ . On what intervals is $f$ decreasing?

The graph of f′ (not f).Open in grapher →

Solution. $f$ decreases where $f'(x) < 0$ , meaning where the graph of $f'$ is below the $x$ -axis. That happens only for $x < -2$ . At $x = 1$ the graph of $f'$ touches the axis but doesn't go below it, so $f$ keeps increasing on both sides of $1$ . So $f$ is decreasing on $(-\infty, -2)$ because $f' < 0$ there, and increasing on $(-2, 1)$ and $(1, \infty)$ .

Common mistake

On a graph of $f'$ , "the graph is going down" does not mean $f$ is decreasing. It means $f'$ is decreasing (the slope of $f$ is getting smaller). For increasing and decreasing $f$ , only the sign of $f'$ matters: above the axis or below it.

Watch the domain

Points where $f$ is undefined also split the number line, even though they aren't critical points. The sign of $f'$ can change across a vertical asymptote, and a function can't be "increasing across" a place where it doesn't exist.

Worked example: A rational function

Find where $f(x) = \dfrac{x^2}{x - 1}$ is increasing and decreasing.

Solution. The domain excludes $x = 1$ . By the quotient rule,

f'(x) = \frac{2x(x - 1) - x^2}{(x - 1)^2} = \frac{x^2 - 2x}{(x - 1)^2} = \frac{x(x - 2)}{(x - 1)^2}.

The critical points are $x = 0$ and $x = 2$ , and $x = 1$ is a break in the domain. The denominator is positive wherever it's defined, so $f'$ has the sign of $x(x - 2)$ .

interval	$x < 0$	$0 < x < 1$	$1 < x < 2$	$x > 2$
sign of $f'$	$+$	$-$	$-$	$+$

$f$ is increasing for $x < 0$ or $x > 2$ , and decreasing on $(0, 1)$ and on $(1, 2)$ . Don't merge the last two into " $0 < x < 2$ ": that interval contains $x = 1$ , where $f$ isn't defined, and in fact $f(0.5) = -0.5$ while $f(1.5) = 4.5$ , so $f$ is not decreasing across the whole stretch.

Transcendental functions

The same method works for any differentiable function. Just remember which factors can never be zero.

Worked example: A trig function on a closed interval

Find where $f(x) = x + 2\cos x$ is increasing on $[0, 2\pi]$ .

Solution. $f'(x) = 1 - 2\sin x$ . Set it to $0$ : $\sin x = \dfrac{1}{2}$ , so $x = \dfrac{\pi}{6}$ or $x = \dfrac{5\pi}{6}$ . Test a value in each piece:

$x = 0$ : $f'(0) = 1 > 0$ .
$x = \dfrac{\pi}{2}$ : $f'\!\left(\dfrac{\pi}{2}\right) = 1 - 2 = -1 < 0$ .
$x = \pi$ : $f'(\pi) = 1 > 0$ .

So $f$ is increasing on $\left[0, \dfrac{\pi}{6}\right]$ and $\left[\dfrac{5\pi}{6}, 2\pi\right]$ , and decreasing on $\left[\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right]$ .

A note on endpoints: a function that increases on $(a, b)$ and is continuous at $a$ and $b$ also increases on $[a, b]$ , so AP graders accept either open or closed intervals. In this course's answer boxes, use strict inequalities (open intervals) unless a problem says otherwise.