Math Core

Two graphs can both be increasing yet look completely different: one bends upward like the start of a rocket launch, the other flattens out like a runner getting tired. The first derivative can't tell them apart, but the second derivative can. In this lesson you'll use $f''$ to describe how a graph bends and to classify critical points in a single step.

Concavity

Definition

Concavity

Let $f$ be differentiable on an open interval $I$ .

$f$ is concave up on $I$ if $f'$ is increasing on $I$ . The graph bends upward, like a cup, and lies above its tangent lines.
$f$ is concave down on $I$ if $f'$ is decreasing on $I$ . The graph bends downward, like a frown, and lies below its tangent lines.

Concavity is about the slope changing, not about the graph going up or down. A concave up graph can be falling, as long as it is falling less and less steeply (think of the left half of $y = x^2$ ).

Since $f''$ is the derivative of $f'$ , the increasing/decreasing test applied to $f'$ gives a test for concavity.

Concavity test

If $f''(x) > 0$ on an interval, then $f$ is concave up there.
If $f''(x) < 0$ on an interval, then $f$ is concave down there.

Points of inflection

Definition

Point of inflection

A point of inflection is a point on the graph of $f$ where $f$ is continuous and the concavity changes (from up to down or from down to up).

The concavity can only change where $f''(x) = 0$ or $f''(x)$ is undefined. So those are the candidates for inflection points, and you confirm them with a sign chart for $f''$ , just like you confirmed extrema with a sign chart for $f'$ .

Worked example: Concavity of a cubic

Find the intervals of concavity and the inflection point of $f(x) = x^3 - 3x^2 - 9x + 2$ .

Solution. $f'(x) = 3x^2 - 6x - 9$ and $f''(x) = 6x - 6 = 6(x - 1)$ . $f''(x) = 0$ at $x = 1$ .

For $x < 1$ , $f''(x) < 0$ , so $f$ is concave down.
For $x > 1$ , $f''(x) > 0$ , so $f$ is concave up.

The concavity changes at $x = 1$ , so there is a point of inflection at $(1, f(1)) = (1, -9)$ .

In the graph below, compare $f$ with $f'$ . The inflection point of $f$ lines up with the lowest point of $f'$ : that's where the slope of $f$ stops decreasing and starts increasing.

f(x) = x³ − 3x² − 9x + 2 (solid) and f′ (dashed). f is concave down where f′ decreases and concave up where f′ increases; the switch is at x = 1.Open in grapher →

Common mistake

$f''(c) = 0$ does not guarantee an inflection point. For $f(x) = x^4$ , $f''(x) = 12x^2$ is $0$ at $x = 0$ , but $f''$ is positive on both sides, so the graph is concave up everywhere and $(0, 0)$ is not an inflection point. Always justify with a sign change of $f''$ (or, equivalently, a change in whether $f'$ is increasing or decreasing).

The second derivative test

At a critical point where $f'(c) = 0$ , the tangent line is horizontal. If the graph is concave up there, it must be the bottom of a cup: a minimum. If it's concave down, it's the top of a frown: a maximum.

The second derivative test

Suppose $f'(c) = 0$ and $f''$ exists near $c$ .

If $f''(c) > 0$ , then $f$ has a relative minimum at $x = c$ .
If $f''(c) < 0$ , then $f$ has a relative maximum at $x = c$ .
If $f''(c) = 0$ , the test is inconclusive. Use the first derivative test instead.

For example, $x^4$ , $-x^4$ and $x^3$ all have $f'(0) = 0$ and $f''(0) = 0$ , yet they have a minimum, a maximum and neither at $x = 0$ . That's why a zero second derivative tells you nothing on its own.

Worked example: Classifying with f″

Use the second derivative test to classify the critical points of $f(x) = 2x^3 - 9x^2 + 12x$ .

Solution. $f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)$ , so the critical points are $x = 1$ and $x = 2$ . Then $f''(x) = 12x - 18$ .

$f''(1) = -6 < 0$ , so $f$ has a relative maximum at $x = 1$ (value $f(1) = 5$ ).
$f''(2) = 6 > 0$ , so $f$ has a relative minimum at $x = 2$ (value $f(2) = 4$ ).

Worked example: Inflection points of a quartic

Find the inflection points of $f(x) = x^4 - 6x^2$ .

Solution. $f'(x) = 4x^3 - 12x$ and $f''(x) = 12x^2 - 12 = 12(x - 1)(x + 1)$ .

interval	$x < -1$	$-1 < x < 1$	$x > 1$
sign of $f''$	$+$	$-$	$+$
concavity	up	down	up

The concavity changes at both $x = -1$ and $x = 1$ . Since $f(\pm 1) = 1 - 6 = -5$ , the inflection points are $(-1, -5)$ and $(1, -5)$ .

f(x) = x⁴ − 6x². It is concave up outside [−1, 1] and concave down between the inflection points (±1, −5).Open in grapher →

Justifying on the AP exam

Justifications must cite the right function:

claim about $f$	justify with
increasing / decreasing	sign of $f'$
relative max / min	sign change of $f'$ , or sign of $f''$ at a point where $f' = 0$
concave up / down	sign of $f''$ , or $f'$ increasing / decreasing
inflection point	sign change of $f''$ , or $f'$ changing from increasing to decreasing (or vice versa)

Tip

If you're given the graph of $f'$ , you don't need $f''$ at all. $f$ is concave up where the graph of $f'$ is rising, concave down where it's falling, and inflection points of $f$ are at the relative extrema (the peaks and valleys) of $f'$ .

Practice

Practice 1

Find the $x$ -coordinate of the inflection point of $f(x) = x^3 - 6x^2 + 5$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

On what interval is $f(x) = x^3 - 6x^2 + 5$ concave up?

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 3

On what interval is $f(x) = x^4 - 24x^2$ concave down? Use strict inequalities.

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 4

Find the $x$ -coordinates of all inflection points of $f(x) = x^4 - 6x^3 + 12x^2 + 1$ .

Separate answers with commas, e.g. 2, -5

Practice 5

A twice-differentiable function $g$ has $g'(4) = 0$ and $g''(4) = -3$ . Which statement is true?

Practice 6

Find the $x$ -coordinate of the inflection point of $f(x) = xe^{-x}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Use the second derivative test to find the $x$ -value at which $f(x) = x^3 - 3x^2 - 9x + 20$ has a relative minimum.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

For $f(x) = x^{5/3}$ , which statement about the point $(0, 0)$ is true?