The First Derivative Test — AP® Calculus AB Notes

You now know where a function rises and where it falls. Put two intervals side by side and a peak or valley appears between them. The First Derivative Test is that observation, stated carefully enough to use as a justification. It is one of the most scored sentences on the whole exam.

The test

Let $c$ be a critical point of $f$ , and let $f$ be continuous at $c$ .

If $f'$ changes from positive to negative at $c$ , then $f$ has a relative maximum at $c$ . The graph rises up to $c$ , then falls.
If $f'$ changes from negative to positive at $c$ , then $f$ has a relative minimum at $c$ . The graph falls, then rises.
If $f'$ does not change sign at $c$ , then $f$ has no relative extremum at $c$ . The graph keeps going the same way.

Three graphs, each with a critical point at x = c. Left: f' changes from positive to negative, the graph rises then falls, a relative maximum. Middle: f' changes from negative to positive, a relative minimum. Right: f' is positive on both sides, the graph rises through a flat point, no relative extremum. — **Figure 1.** Only the change in the sign of $f'$ decides. A flat point with the same sign on both sides is not an extremum. LevlPrep original diagram.

Notice what the test does not use: the value $f'(c)$ . It works the same at a point where $f'(c) = 0$ and at a corner or cusp where $f'(c)$ does not exist. What matters is the sign just before and just after.

Sign charts

A sign chart is a number line marked with the critical points and the sign of $f'$ on each interval between them. It is a good way to organize the work:

Worked example 1. Find the relative extrema of $f(x) = x^3 - 3x^2 - 9x + 4$ .

$f'(x)$ $= 3x^2 - 6x - 9$ $= 3(x - 3)(x + 1)$ , zero at $x = -1$ and $x = 3$ .

Test values: $f'(-2) = 15 > 0$ , $f'(0) = -9 < 0$ , $f'(4) = 15 > 0$ . Sign chart: $+$ , then $-$ after $-1$ , then $+$ after $3$ .

At $x = -1$ , $f'$ changes from positive to negative: relative maximum, $f(-1)$ $= -1 - 3 + 9 + 4$ $= 9$ .

At $x = 3$ , $f'$ changes from negative to positive: relative minimum, $f(3)$ $= 27 - 27 - 27 + 4$ $= -23$ .

But the chart is scratch work. The scoring guidelines do not accept a sign chart by itself as a justification. Write the sentence: " $f'$ changes from positive to negative at $x = -1$ , so $f$ has a relative maximum there."

Points where $f'$ does not exist

Worked example 2. $f(x)$ $= x^{5/3} - 5x^{2/3}$ . Find its relative extrema.

$f'(x)$ $= \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3}$ $= \frac{5x - 10}{3x^{1/3}}$ $= \frac{5(x - 2)}{3x^{1/3}}$ .

Critical points: $x = 2$ (numerator zero) and $x = 0$ ( $f'$ undefined; $f$ is defined and continuous there).

Signs: for $x < 0$ , both $x - 2$ and $x^{1/3}$ are negative, so $f' > 0$ . For $0 < x < 2$ , $f' < 0$ . For $x > 2$ , $f' > 0$ .

So $f$ has a relative maximum at $x = 0$ ( $f'$ changes from positive to negative at a cusp) and a relative minimum at $x = 2$ .

Zeros that are not extrema

A squared factor in $f'$ is the usual reason a critical point is not an extremum. For $f(x) = x^4 - 4x^3$ :

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

The factor $4x^2$ is never negative, so the sign of $f'$ is the sign of $x - 3$ . At $x = 0$ , $f'$ is negative on both sides: no extremum, even though $f'(0) = 0$ . At $x = 3$ , $f'$ changes from negative to positive: a relative minimum.

Reading the test from a graph of $f'$

What the graph of $f'$ does at $x = c$	What $f$ has at $x = c$
Crosses the axis going down ( $+$ to $-$ )	relative maximum
Crosses the axis going up ( $-$ to $+$ )	relative minimum
Touches the axis and turns back (no sign change)	no extremum
Has a peak or valley above or below the axis	no extremum (that is about concavity)

The sentence to write

Name the function, name the point, give the direction of the change:

" $f'$ changes from negative to positive at $x = 5$ , so $f$ has a relative minimum at $x = 5$ ."

Answers that lose the point: " $f'(5) = 0$ " (no sign change), "it changes sign" (which function? which direction?), or "the graph crosses the axis" (which graph?). And remember this test is local: it finds relative extrema, not absolute ones. For the absolute maximum on an interval, you need the next topic.