Concavity and the Second Derivative Test

Knowing that a graph is going up isn’t the whole story: is it curving upward like a smile, or bending over like a frown? That bending is called concavity, and the second derivative measures it. Concavity helps you sketch accurate graphs, find where a rate of change is greatest, and gives a quick second way to classify maximums and minimums.

Key ideas

Concave up and concave down

On an interval:

• If $f''(x) \gt 0$, then $f$ is concave up: the graph bends upward like a cup, and the slopes $f'$ are increasing.
• If $f''(x) \lt 0$, then $f$ is concave down: the graph bends downward like a cap, and the slopes $f'$ are decreasing.

Another way to see it: a concave-up graph lies above its tangent lines, and a concave-down graph lies below them.

$f(x) = x^3 - 6x^2 + 9x + 1$ has $f''(x) = 6x - 12$: concave down for $x \lt 2$, concave up for $x \gt 2$.

Concavity and increasing/decreasing are separate ideas. A graph can be increasing and concave down (rising, but more and more slowly) or decreasing and concave up (falling, but levelling off).

Points of inflection

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

To find them, make a sign chart for $f''$, just like the one you make for $f'$:

1. Find where $f''(x) = 0$ or $f''(x)$ doesn’t exist (with $f$ defined there).
2. Test the sign of $f''$ on each interval.
3. A point of inflection is where the sign of $f''$ actually changes.

$f''(c) = 0$ on its own is not enough. For $f(x) = x^4$, $f''(x) = 12x^2$ is $0$ at $x = 0$, but it’s positive on both sides, so the graph is concave up everywhere and $(0, 0)$ is not a point of inflection.

Justification (the wording AP graders expect): ”$f''(x)$ changes from negative to positive at $x = 2$, so the graph of $f$ has a point of inflection at $x = 2$.”

The second derivative test

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

$f''(c)$	Conclusion	Picture
$f''(c) \gt 0$	relative minimum at $x = c$	flat tangent at the bottom of a cup
$f''(c) \lt 0$	relative maximum at $x = c$	flat tangent at the top of a cap
$f''(c) = 0$	inconclusive: could be max, min, or neither	use the first derivative test

The test is quick when $f''$ is easy to compute, because you only evaluate at the critical point instead of building a sign chart.

Inconclusive really means inconclusive. All three functions $x^4$, $-x^4$, and $x^3$ have $f'(0) = 0$ and $f''(0) = 0$. The first has a minimum at $0$, the second has a maximum, and the third has neither.

Justification: ”$f'(3) = 0$ and $f''(3) \gt 0$, so $f$ has a relative minimum at $x = 3$.”

Worked examples

Example 1: Concavity and an inflection point

Find the intervals where $f(x) = x^3 - 6x^2 + 5$ is concave up and concave down, and any points of inflection.

Solution.

$$f'(x) = 3x^2 - 12x, \qquad f''(x) = 6x - 12 = 6(x - 2)$$

$f''(x) = 0$ at $x = 2$. For $x \lt 2$, $f''(x) \lt 0$; for $x \gt 2$, $f''(x) \gt 0$.

$f$ is concave down on $(-\infty, 2)$ and concave up on $(2, \infty)$. $f''$ changes sign at $x = 2$, and $f(2) = 8 - 24 + 5 = -11$, so the point of inflection is $(2, -11)$.

Example 2: Two inflection points

Find the intervals of concavity and the points of inflection of $f(x) = x^4 - 4x^3$.

Solution.

$$f'(x) = 4x^3 - 12x^2, \qquad f''(x) = 12x^2 - 24x = 12x(x - 2)$$

Interval	$(-\infty, 0)$	$(0, 2)$	$(2, \infty)$
Sign of $f''$	$(-)(-) = +$	$(+)(-) = -$	$(+)(+) = +$
Concavity	up	down	up

$f''$ changes sign at both $x = 0$ and $x = 2$. Points of inflection: $(0, 0)$ and $(2, 16 - 32) = (2, -16)$.

Example 3: The second derivative test

Use the second derivative test to find the relative extrema of $f(x) = 2x^3 - 9x^2 + 12x$.

Solution.

$$f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2), \qquad f''(x) = 12x - 18$$

Critical points: $x = 1$ and $x = 2$.

• $f''(1) = 12 - 18 = -6 \lt 0$, so $f$ has a relative maximum at $x = 1$: $f(1) = 2 - 9 + 12 = 5$.
• $f''(2) = 24 - 18 = 6 \gt 0$, so $f$ has a relative minimum at $x = 2$: $f(2) = 16 - 36 + 24 = 4$.

Example 4: When the test is inconclusive

Classify the critical points of $f(x) = x^4 - 4x^3$.

Solution. From Example 2, $f'(x) = 4x^2(x - 3)$, so the critical points are $x = 0$ and $x = 3$, and $f''(x) = 12x^2 - 24x$.

• $f''(3) = 108 - 72 = 36 \gt 0$, so there’s a relative minimum at $x = 3$: $f(3) = 81 - 108 = -27$.
• $f''(0) = 0$: the second derivative test is inconclusive. Switch to the first derivative test. $f'(x) = 4x^2(x - 3)$ is negative just left of $0$ and just right of $0$ (since $x^2 \gt 0$ and $x - 3 \lt 0$). $f'$ doesn’t change sign, so there’s no extremum at $x = 0$.

In fact, $(0, 0)$ is a point of inflection (Example 2) with a horizontal tangent.

Common mistakes

Saying f″(c) = 0 means a point of inflection. It only makes $c$ a candidate. Check that $f''$ actually changes sign there. $f(x) = x^4$ is the classic counterexample.

Saying f″(c) = 0 means “no extremum”. It means the second derivative test can’t decide. Use the first derivative test, which always gives an answer.

Mixing up which sign means max or min. $f''(c) \gt 0$ means concave up, a cup, so the flat point is at the bottom: a minimum. Picture the cup, don’t memorize the signs.

Using the second derivative test at a point where f′(c) ≠ 0. The test only classifies critical points where $f'(c) = 0$. A positive $f''$ at an ordinary point just means concave up there.

Forgetting points where f″ doesn’t exist. For $f(x) = \sqrt[3]{x}$, $f''$ is undefined at $0$, but the concavity still changes there, so $(0, 0)$ is a point of inflection.

Giving only the x-value of a point of inflection when the question asks for the point. “Find the point of inflection” wants both coordinates: $(2, -11)$, not just $x = 2$.

Practice

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

Solution

$f'(x) = 3x^2 + 6x$ and $f''(x) = 6x + 6 = 6(x + 1)$.

$f''(x) \lt 0$ for $x \lt -1$ and $f''(x) \gt 0$ for $x \gt -1$. So $f$ is concave down on $(-\infty, -1)$ and concave up on $(-1, \infty)$.

Point of inflection: $f(-1) = -1 + 3 - 2 = 0$, so $(-1, 0)$.

2. (Warm-up) A function $g$ has $g'(4) = 0$ and $g''(4) = -3$. What does $g$ have at $x = 4$? Justify your answer.

Solution

A relative maximum. ”$g'(4) = 0$ and $g''(4) \lt 0$, so by the second derivative test, $g$ has a relative maximum at $x = 4$.”

3. (Warm-up) For $f(x) = x^4$, $f''(0) = 0$. Does the graph of $f$ have a point of inflection at $x = 0$? Explain.

Solution

No. $f''(x) = 12x^2$, which is positive for every $x \ne 0$. $f''$ doesn’t change sign at $x = 0$, so the graph is concave up on both sides and there’s no point of inflection.

4. (Core) Find the intervals of concavity and the point of inflection of $f(x) = xe^x$.

Solution

$$f'(x) = e^x + xe^x = (x + 1)e^x, \qquad f''(x) = e^x + (x + 1)e^x = (x + 2)e^x$$

$e^x \gt 0$, so the sign of $f''$ is the sign of $x + 2$. Concave down on $(-\infty, -2)$, concave up on $(-2, \infty)$.

Point of inflection: $\left(-2, -2e^{-2}\right) \approx (-2, -0.271)$.

5. (Core) Find the intervals of concavity and the points of inflection of $f(x) = \sin x + \cos x$ on $(0, 2\pi)$ (radians).

Solution

$f'(x) = \cos x - \sin x$ and $f''(x) = -\sin x - \cos x$.

$f''(x) = 0$ when $\sin x = -\cos x$, so $\tan x = -1$: $x = \dfrac{3\pi}{4}$ or $x = \dfrac{7\pi}{4}$.

Test values: $f''\left(\frac{\pi}{2}\right) = -1 - 0 = -1 \lt 0$; $f''(\pi) = 0 + 1 = 1 \gt 0$; $f''\left(\frac{15\pi}{8}\right) \approx 0.383 - 0.924 \lt 0$.

Concave down on $\left(0, \dfrac{3\pi}{4}\right)$ and $\left(\dfrac{7\pi}{4}, 2\pi\right)$; concave up on $\left(\dfrac{3\pi}{4}, \dfrac{7\pi}{4}\right)$.

$f\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$ and $f\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0$. Points of inflection: $\left(\dfrac{3\pi}{4}, 0\right)$ and $\left(\dfrac{7\pi}{4}, 0\right)$.

6. (Core) Use the second derivative test to find the relative extrema of $f(x) = x^3 - 12x + 1$.

Solution

$f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$, so the critical points are $x = \pm 2$. $f''(x) = 6x$.

• $f''(-2) = -12 \lt 0$: relative maximum, $f(-2) = -8 + 24 + 1 = 17$.
• $f''(2) = 12 \gt 0$: relative minimum, $f(2) = 8 - 24 + 1 = -15$.

7. (Core) Show that $f(x) = \sqrt[3]{x}$ has a point of inflection at $(0, 0)$, even though $f''(0)$ doesn’t exist.

Solution

$f(x) = x^{1/3}$, $f'(x) = \dfrac{1}{3}x^{-2/3}$, and

$$f''(x) = -\frac{2}{9}x^{-5/3} = -\frac{2}{9x^{5/3}}$$

$f''(0)$ is undefined, but $f$ is continuous at $0$. For $x \lt 0$, $x^{5/3} \lt 0$, so $f''(x) \gt 0$ (concave up). For $x \gt 0$, $f''(x) \lt 0$ (concave down). The concavity changes at $x = 0$, so $(0, 0)$ is a point of inflection.

8. (Core) Let $f(x) = x + \dfrac{4}{x}$ for $x \gt 0$. Use the second derivative test to classify the critical point.

Solution

$f'(x) = 1 - \dfrac{4}{x^2} = 0$ gives $x^2 = 4$, so $x = 2$ (since $x \gt 0$).

$f''(x) = \dfrac{8}{x^3}$, so $f''(2) = 1 \gt 0$. $f$ has a relative minimum at $x = 2$: $f(2) = 2 + 2 = 4$.

9. (Challenge) Find constants $a$ and $b$ so that $f(x) = ax^3 + bx^2$ has a point of inflection at $(1, 2)$.

Solution

$f''(x) = 6ax + 2b$. For an inflection at $x = 1$ we need $f''(1) = 0$: $6a + 2b = 0$, so $b = -3a$.

The point $(1, 2)$ is on the graph: $f(1) = a + b = 2$. Substitute: $a - 3a = 2$, so $a = -1$ and $b = 3$.

Check: $f(x) = -x^3 + 3x^2$ has $f''(x) = -6x + 6$, which changes from positive to negative at $x = 1$, and $f(1) = -1 + 3 = 2$. ✓

10. (Challenge) A function $f$ has derivative $f'(x) = (x - 1)^2(x + 3)$. Find the $x$-values of all relative extrema and points of inflection of $f$, and the intervals of concavity.

Solution

Extrema: $f' = 0$ at $x = 1$ and $x = -3$. $(x - 1)^2 \ge 0$, so the sign of $f'$ is the sign of $x + 3$: negative for $x \lt -3$, positive for $x \gt -3$ (except $0$ at $x = 1$). $f'$ changes from negative to positive at $x = -3$: relative minimum. No sign change at $x = 1$: no extremum.

Concavity: by the product rule,

$$\begin{aligned} f''(x) &= 2(x - 1)(x + 3) + (x - 1)^2 \\ &= (x - 1)\big(2(x + 3) + (x - 1)\big) \\ &= (x - 1)(3x + 5) \end{aligned}$$

$f''(x) = 0$ at $x = -\dfrac{5}{3}$ and $x = 1$.

Interval	$\left(-\infty, -\frac{5}{3}\right)$	$\left(-\frac{5}{3}, 1\right)$	$(1, \infty)$
Sign of $f''$	$(-)(-) = +$	$(-)(+) = -$	$(+)(+) = +$

Concave up on $\left(-\infty, -\frac{5}{3}\right)$ and $(1, \infty)$; concave down on $\left(-\frac{5}{3}, 1\right)$. Points of inflection at $x = -\dfrac{5}{3}$ and $x = 1$.