First Derivative Test

The derivative tells you the slope of a graph at every point. Positive slope means the graph is going up; negative slope means it’s going down. So just from the sign of $f'$ you can say where a function rises, where it falls, and where it turns around. The first derivative test turns this into a reliable way to find relative maximums and minimums, and it’s the justification AP graders expect most often.

Key ideas

Increasing and decreasing

On an interval:

• If $f'(x) \gt 0$, then $f$ is increasing.
• If $f'(x) \lt 0$, then $f$ is decreasing.

This follows from the Mean Value Theorem: if $f'$ is positive everywhere between two points, the average rate of change between them is positive too, so the right-hand point is higher.

Sign charts

$f'$ can only change sign at a critical point (where $f' = 0$ or is undefined) or where $f$ itself is undefined. So:

1. Find the critical points, plus any $x$-values not in the domain of $f$.
2. Mark them on a number line. They split it into intervals.
3. Pick a test value in each interval and find the sign of $f'(x)$ there. (Only the sign matters, so the factored form of $f'$ is the easiest to use.)
4. Read off: $+$ means increasing, $-$ means decreasing.

The sign chart for $f'(x) = 3(x + 1)(x - 3)$ matches the shape of $f(x) = x^3 - 3x^2 - 9x + 2$ (Example 1).

The first derivative test

Let $c$ be a critical point of a continuous function $f$.

As $x$ passes $c$ from left to right, $f'$ changes from…	At $x = c$, $f$ has a…
positive to negative	relative maximum
negative to positive	relative minimum
doesn’t change sign	neither

Picture it: up then down is a hilltop; down then up is a valley; up then up (or down then down) is just a pause.

Justification sentences for the AP exam

AP graders want the reason tied to $f'$, at the specific $x$-value:

• ”$f'(x)$ changes from positive to negative at $x = -1$, so $f$ has a relative maximum at $x = -1$.”
• ”$f'(x) \gt 0$ on $(3, \infty)$, so $f$ is increasing on $(3, \infty)$.”
• ”$f'(x)$ does not change sign at $x = 0$, so $f$ has no relative extremum there.”

Avoid vague phrases like “the slope goes up and down” or “it” without naming $f$ or $f'$. Write about the function you’re given: if the question gives the graph of $f'$, talk about the sign of $f'$, not the shape of a graph of $f$ you haven’t been shown.

Worked examples

Example 1: A cubic

Let $f(x) = x^3 - 3x^2 - 9x + 2$. Find the intervals where $f$ is increasing and decreasing, and find all relative extrema.

Solution.

$$f'(x) = 3x^2 - 6x - 9 = 3(x + 1)(x - 3)$$

Critical points: $x = -1$ and $x = 3$. Test each interval:

Interval	Test value	$3(x + 1)(x - 3)$	$f$ is
$(-\infty, -1)$	$x = -2$	$3(-)(-) = +$	increasing
$(-1, 3)$	$x = 0$	$3(+)(-) = -$	decreasing
$(3, \infty)$	$x = 4$	$3(+)(+) = +$	increasing

$f$ is increasing on $(-\infty, -1)$ and $(3, \infty)$, and decreasing on $(-1, 3)$.

$f'$ changes from positive to negative at $x = -1$, so $f$ has a relative maximum there: $f(-1) = -1 - 3 + 9 + 2 = 7$.

$f'$ changes from negative to positive at $x = 3$, so $f$ has a relative minimum there: $f(3) = 27 - 27 - 27 + 2 = -25$.

Example 2: A critical point that isn’t an extremum

Find the relative extrema of $f(x) = x^4 - 4x^3$.

Solution.

$$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$$

Critical points: $x = 0$ and $x = 3$. The factor $x^2$ is never negative, so the sign of $f'$ depends only on $x - 3$:

Interval	$(-\infty, 0)$	$(0, 3)$	$(3, \infty)$
Sign of $f'$	$-$	$-$	$+$

$f'$ does not change sign at $x = 0$, so there’s no extremum there (the graph flattens out, then keeps falling). $f'$ changes from negative to positive at $x = 3$, so $f$ has a relative minimum: $f(3) = 81 - 108 = -27$.

Example 3: A break in the domain

Let $f(x) = x + \dfrac{4}{x}$. Find the intervals of increase and decrease and the relative extrema.

Solution. The domain is all $x \ne 0$.

$$f'(x) = 1 - \frac{4}{x^2} = \frac{x^2 - 4}{x^2} = \frac{(x - 2)(x + 2)}{x^2}$$

Critical points: $x = \pm 2$. Also put $x = 0$ on the chart, since $f$ is undefined there (it is not a critical point).

Interval	$(-\infty, -2)$	$(-2, 0)$	$(0, 2)$	$(2, \infty)$
Sign of $f'$	$+$	$-$	$-$	$+$

$f$ is increasing on $(-\infty, -2)$ and $(2, \infty)$, and decreasing on $(-2, 0)$ and $(0, 2)$. Write these as two separate intervals: $f$ isn’t defined at $0$, so you can’t say “decreasing on $(-2, 2)$”.

Relative maximum at $x = -2$ ($f'$ changes from $+$ to $-$): $f(-2) = -2 - 2 = -4$.

Relative minimum at $x = 2$ ($f'$ changes from $-$ to $+$): $f(2) = 2 + 2 = 4$.

(Yes, the relative maximum is lower than the relative minimum. That’s possible because the graph breaks apart at $x = 0$.)

Example 4: Given only f′

A function $f$ is differentiable, with $f'(x) = (x + 2)(x - 1)^2(x - 4)$. Find the $x$-values of all relative extrema of $f$, and justify your answers.

Solution. Critical points: $x = -2$, $1$, $4$. The factor $(x - 1)^2$ is never negative, so it doesn’t affect the sign.

Interval	$(-\infty, -2)$	$(-2, 1)$	$(1, 4)$	$(4, \infty)$
$(x + 2)$	$-$	$+$	$+$	$+$
$(x - 4)$	$-$	$-$	$-$	$+$
Sign of $f'$	$+$	$-$	$-$	$+$

• $f'$ changes from positive to negative at $x = -2$, so $f$ has a relative maximum at $x = -2$.
• $f'$ changes from negative to positive at $x = 4$, so $f$ has a relative minimum at $x = 4$.
• $f'$ does not change sign at $x = 1$, so $f$ has no relative extremum at $x = 1$.

You don’t know $f$ itself, so you can’t find the $y$-values, and the question doesn’t ask for them.

Common mistakes

Saying “f′ = 0, so it’s a maximum.” A zero derivative only makes $c$ a candidate. You need the sign change to classify it, as Example 2 shows.

Leaving undefined points off the sign chart. In Example 3, skipping $x = 0$ would give the wrong picture. Put every critical point and every point where $f$ is undefined on the number line.

Joining intervals across a break. “Decreasing on $(-2, 2)$” is wrong in Example 3, because $0$ isn’t in the domain. List the intervals separately: $(-2, 0)$ and $(0, 2)$.

Using the sign of f instead of f′. Where the graph of $f$ is above the axis tells you nothing about increasing or decreasing. It’s the sign of $f'$ that matters. This trips people up most when a question gives the graph of $f'$.

Weak justifications. “Because the graph goes up and then down” earns no credit on the AP exam. Name the derivative, the sign change, and the $x$-value.

Giving a point when an x-value is asked for (or vice versa). “At what $x$” wants $x = -1$. “Find the relative maximum value” wants $f(-1) = 7$.

Practice

1. (Warm-up) Let $f(x) = x^2 - 10x$. Find where $f$ is increasing and decreasing, and find its relative extremum.

Solution

$f'(x) = 2x - 10 = 2(x - 5)$, which is negative for $x \lt 5$ and positive for $x \gt 5$.

$f$ is decreasing on $(-\infty, 5)$ and increasing on $(5, \infty)$. $f'$ changes from negative to positive at $x = 5$, so there’s a relative minimum: $f(5) = 25 - 50 = -25$.

2. (Warm-up) A function $g$ is differentiable, and $g'(x)$ changes from negative to positive at $x = 3$. What does $g$ have at $x = 3$? Write the justification sentence.

Solution

A relative minimum. ”$g'(x)$ changes from negative to positive at $x = 3$, so $g$ has a relative minimum at $x = 3$.”

3. (Warm-up) Let $f(x) = 12x - x^3$. Find the intervals of increase and decrease, and the relative extrema.

Solution

$f'(x) = 12 - 3x^2 = 3(2 - x)(2 + x)$.

Interval	$(-\infty, -2)$	$(-2, 2)$	$(2, \infty)$
Sign of $f'$	$-$	$+$	$-$

$f$ is decreasing on $(-\infty, -2)$ and $(2, \infty)$, and increasing on $(-2, 2)$.

Relative minimum at $x = -2$ ($-$ to $+$): $f(-2) = -24 + 8 = -16$. Relative maximum at $x = 2$ ($+$ to $-$): $f(2) = 24 - 8 = 16$.

4. (Core) Let $f(x) = xe^{-x}$. Find the intervals of increase and decrease and the relative extremum.

Solution

By the product rule:

$$f'(x) = e^{-x} + x(-e^{-x}) = (1 - x)e^{-x}$$

$e^{-x} \gt 0$ always, so the sign comes from $1 - x$: positive for $x \lt 1$, negative for $x \gt 1$.

$f$ is increasing on $(-\infty, 1)$ and decreasing on $(1, \infty)$. $f'$ changes from positive to negative at $x = 1$, so $f$ has a relative maximum: $f(1) = e^{-1} = \dfrac{1}{e} \approx 0.368$.

5. (Core) Let $f(x) = x^{2/3}(x - 5)$, so $f'(x) = \dfrac{5(x - 2)}{3x^{1/3}}$. Find the relative extrema.

Solution

Critical points: $x = 0$ (where $f'$ is undefined but $f(0) = 0$ exists) and $x = 2$.

Interval	$(-\infty, 0)$	$(0, 2)$	$(2, \infty)$
$x - 2$	$-$	$-$	$+$
$x^{1/3}$	$-$	$+$	$+$
Sign of $f'$	$+$	$-$	$+$

$f'$ changes from positive to negative at $x = 0$: relative maximum, $f(0) = 0$.

$f'$ changes from negative to positive at $x = 2$: relative minimum, $f(2) = 2^{2/3}(-3) = -3\sqrt[3]{4} \approx -4.762$.

6. (Core) Let $f(x) = \dfrac{x^2}{x - 1}$. Find the intervals of increase and decrease and the relative extrema.

Solution

The domain is $x \ne 1$.

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

Critical points: $x = 0$ and $x = 2$. Also mark $x = 1$ (not in the domain). The denominator is positive, so the sign comes from $x(x - 2)$:

Interval	$(-\infty, 0)$	$(0, 1)$	$(1, 2)$	$(2, \infty)$
Sign of $f'$	$+$	$-$	$-$	$+$

Increasing on $(-\infty, 0)$ and $(2, \infty)$; decreasing on $(0, 1)$ and $(1, 2)$.

Relative maximum at $x = 0$: $f(0) = 0$. Relative minimum at $x = 2$: $f(2) = \dfrac{4}{1} = 4$.

7. (Core) Let $f(x) = x + 2\cos x$ on $[0, 2\pi]$ (radians). Find the $x$-values of the relative extrema and the relative extreme values.

Solution

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

Test values: at $x = 0.1$, $f' \approx 1 - 0.2 \gt 0$; at $x = \frac{\pi}{2}$, $f' = 1 - 2 = -1 \lt 0$; at $x = \pi$, $f' = 1 - 0 = 1 \gt 0$.

$f'$ changes from positive to negative at $x = \dfrac{\pi}{6}$: relative maximum,

$$f\left(\frac{\pi}{6}\right) = \frac{\pi}{6} + 2 \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{6} + \sqrt{3} \approx 2.256$$

$f'$ changes from negative to positive at $x = \dfrac{5\pi}{6}$: relative minimum,

$$f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} + 2\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} - \sqrt{3} \approx 0.886$$

8. (Core) A function $f$ has derivative $f'(x) = x^2(x + 1)(x - 4)^3$. At what $x$-values does $f$ have a relative maximum or minimum? Justify your answer.

Solution

Critical points: $x = -1$, $0$, $4$. The factor $x^2$ doesn’t affect the sign; $(x - 4)^3$ has the same sign as $x - 4$.

Interval	$(-\infty, -1)$	$(-1, 0)$	$(0, 4)$	$(4, \infty)$
Sign of $f'$	$(-)(-) = +$	$(+)(-) = -$	$(+)(-) = -$	$(+)(+) = +$

• $f'$ changes from positive to negative at $x = -1$, so $f$ has a relative maximum at $x = -1$.
• $f'$ changes from negative to positive at $x = 4$, so $f$ has a relative minimum at $x = 4$.
• $f'$ does not change sign at $x = 0$, so there is no relative extremum there.

9. (Challenge) Show that $f(x) = \ln(x^2 + 1) - x$ is decreasing on its whole domain, and explain why it has no relative extrema.

Solution

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

The numerator $-(x - 1)^2$ is never positive, and the denominator is always positive, so $f'(x) \le 0$ for all $x$, with $f'(x) = 0$ only at $x = 1$. A single point where $f' = 0$ doesn’t stop $f$ from decreasing, so $f$ is decreasing on $(-\infty, \infty)$.

The only critical point is $x = 1$, and $f'$ doesn’t change sign there (negative on both sides), so $f$ has no relative extrema.

10. (Challenge) Find the value of $k$ so that $f(x) = x^3 - kx$ has a relative maximum at $x = -2$. Then find the relative maximum value.

Solution

$f'(x) = 3x^2 - k$. A relative maximum at $x = -2$ needs $f'(-2) = 0$: $12 - k = 0$, so $k = 12$.

Check with the first derivative test: $f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$ is positive for $x \lt -2$ and negative on $(-2, 2)$. It changes from positive to negative at $x = -2$, so it really is a relative maximum.

$f(-2) = -8 + 24 = 16$.