Concavity and Inflection Points, calculus lesson

Two cars can both be speeding up while one gains speed faster and faster and the other gains it more and more slowly. Their position graphs both rise, yet they bend in opposite directions. The first derivative tells you whether a graph goes up or down. The second derivative tells you which way it bends, and that extra information sharpens both graph sketches and the hunt for maxima and minima.

Key terms

Concave up and concave down

Definition Let $f$ be differentiable on an open interval $I$ . The function $f$ is concave up on $I$ if $f'$ is strictly increasing on $I$ , and concave down on $I$ if $f'$ is strictly decreasing on $I$ . Equivalently, $f$ is concave up on $I$ exactly when its graph lies above every one of its tangent lines on $I$ , touching each only at its point of tangency, so that for all $a \neq x$ in $I$

f(x) > f(a) + f'(a)(x - a),

and concave down when the inequality is reversed.

In plain words A concave up graph bends like a cup that holds water, and its slopes keep getting larger. A concave down graph bends like a cap, and its slopes keep getting smaller.

Example For $f(x) = x^3$ the slope is $f'(x) = 3x^2$ . On $(-\infty, 0)$ the slope decreases from large positive values toward $0$ , so $f$ is concave down there. On $(0, \infty)$ the slope increases, so $f$ is concave up there.

The sign of the second derivative

Definition This is the concavity test. Suppose $f''$ exists at every point of an open interval $I$ .

If $f''(x) > 0$ for all $x$ in $I$ , then $f$ is concave up on $I$ .
If $f''(x) < 0$ for all $x$ in $I$ , then $f$ is concave down on $I$ .

The test follows from the increasing and decreasing test applied to $f'$ . Its converse fails. The function $x^4$ is concave up on all of $\mathbb{R}$ even though its second derivative is $0$ at $x = 0$ .

In plain words A positive second derivative means the slope is growing, so the graph bends up. A negative one means the slope is shrinking, so the graph bends down.

Example Take $f(x) = x^4 - 4x^3$ . Then $f'(x) = 4x^3 - 12x^2$ and

f''(x) = 12x^2 - 24x = 12x(x - 2).

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

Inflection points

Definition A point $P = (c, f(c))$ on the graph of $f$ is an inflection point if $f$ is continuous at $c$ and there is a $\delta > 0$ such that $f$ is concave up on $(c - \delta, c)$ and concave down on $(c, c + \delta)$ , or concave down on $(c - \delta, c)$ and concave up on $(c, c + \delta)$ .

If $f''(c)$ exists at an inflection point, then $f''(c) = 0$ , because $f'$ has a local maximum or minimum at $c$ . The reverse is false. A zero of $f''$ is only a candidate, and the concavity has to actually change there.

In plain words An inflection point is where the graph switches from bending one way to bending the other. A zero second derivative alone does not make one.

Example For $f(x) = x^4 - 4x^3$ the table above shows a change of concavity at $x = 0$ and at $x = 2$ . Since $f(0) = 0$ and $f(2) = 16 - 32 = -16$ , the inflection points are $(0, 0)$ and $(2, -16)$ . By contrast, $g(x) = x^4$ has $g''(0) = 0$ , but $g''(x) = 12x^2 > 0$ on both sides of $0$ , so the origin is not an inflection point of $g$ .

The second derivative test

Definition Let $f$ be differentiable on an open interval containing $c$ , with $f'(c) = 0$ , and suppose $f''(c)$ exists.

If $f''(c) > 0$ , then $f$ has a local minimum at $c$ .
If $f''(c) < 0$ , then $f$ has a local maximum at $c$ .

In plain words At a point where the tangent is flat, a graph that bends up must sit at the bottom of a valley, and a graph that bends down must sit at the top of a hill.

Example Let $f(x) = x^3 - 3x$ . Then $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$ , so the critical points are $x = 1$ and $x = -1$ , and $f''(x) = 6x$ .

f''(1) = 6 > 0, \qquad f''(-1) = -6 < 0.

So $f(1) = -2$ is a local minimum and $f(-1) = 2$ is a local maximum.

When the second derivative test is inconclusive

Definition If $f'(c) = 0$ and $f''(c) = 0$ , or if $f''(c)$ does not exist, the second derivative test gives no conclusion. The point $c$ may be a local minimum, a local maximum or neither. In that case use the first derivative test, which looks at the sign of $f'$ on each side of $c$ .

In plain words A second derivative of zero means the test cannot tell. It does not mean there is no extremum, and it does not mean there is one.

Example Each of these functions has $f'(0) = 0$ and $f''(0) = 0$ .

Function	Sign of $f'$ left of $0$	Sign of $f'$ right of $0$	What happens at $0$
$x^4$	negative	positive	local minimum
$-x^4$	positive	negative	local maximum
$x^3$	positive	positive	neither

The same zero second derivative leads to three different answers, which is why the test stays silent.

Check yourself

Find where $f(x) = x^3 - 6x^2 + 5$ is concave up, where it is concave down, and its inflection point.

Show the answer

The second derivative is $f''(x) = 6x - 12$ , which is negative for $x < 2$ and positive for $x > 2$ . So $f$ is concave down on $(-\infty, 2)$ and concave up on $(2, \infty)$ . The inflection point is $(2, f(2)) = (2, -11)$ .

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

Show the answer

The derivatives are

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

so the critical points are $x = 1$ and $x = -2$ . Since $f''(1) = 18 > 0$ , there is a local minimum $f(1) = -7$ . Since $f''(-2) = -18 < 0$ , there is a local maximum $f(-2) = 20$ .

Does $f(x) = \sqrt[3]{x}$ have an inflection point at the origin, even though $f''(0)$ does not exist?

Show the answer

Yes. For $x \neq 0$ the second derivative is

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

which is positive for $x < 0$ and negative for $x > 0$ . The function is continuous at $0$ and the concavity changes there, so $(0, 0)$ is an inflection point.

Concavity and Inflection Points

Theory, part 73

Practice, part 74

Key terms

Concave up and concave down

The sign of the second derivative

Inflection points

The second derivative test

When the second derivative test is inconclusive

Check yourself

How this lesson connects

Builds on

Used later in