7.4 The second derivative, convexity and points of inflection — Learn

Prerequisites: differentiating powers, exponentials, trigonometric functions and logarithms from Section 7.3; reading the sign of an expression.

By the end of this section you can: find and use the second derivative; interpret it as the rate of change of the gradient; decide where a curve is convex or concave; and find a point of inflection, checking that the second derivative changes sign.

1. The second derivative

Differentiate a function once and you have its gradient. Differentiate the gradient and you have the second derivative. It measures how fast the gradient itself is changing.

The notation is

f''(x) \quad \text{or} \quad \frac{d^2y}{dx^2}.

The square sits on the $d$ in the numerator and on the $x$ in the denominator. It is not a power to apply to the function.

Example 1. Find $f''(x)$ for $f(x) = x^4 - 6x^2$ .

Differentiate once:

f'(x) = 4x^3 - 12x.

Differentiate again:

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

Example 2. Find $\dfrac{d^2y}{dx^2}$ for $y = \sin 2x$ .

\frac{dy}{dx} = 2\cos 2x, \qquad \frac{d^2y}{dx^2} = -4\sin 2x.

2. Convex and concave sections

The sign of the second derivative tells you how the curve bends.

Where $f''(x) > 0$ , the gradient is increasing. The curve turns upwards, like the inside of a cup. It is convex there.
Where $f''(x) < 0$ , the gradient is decreasing. The curve turns downwards, like a frown. It is concave there.

Read the sign of $f''(x)$ , not the sign of $f(x)$ . A curve can be above the $x$ -axis and still be concave.

Example 3. For $f(x) = x^3 - 3x^2 + 2x$ , find where the curve is convex and where it is concave.

Differentiate twice:

f'(x) = 3x^2 - 6x + 2, \qquad f''(x) = 6x - 6.

Set $f''(x) = 0$ : $6x - 6 = 0$ , so $x = 1$ .

For $x < 1$ , $f''(x) < 0$ , so the curve is concave.
For $x > 1$ , $f''(x) > 0$ , so the curve is convex.

The curve changes from concave to convex at $x = 1$ .

3. Points of inflection

A point of inflection is a point where the curve changes from convex to concave, or from concave to convex. At such a point the second derivative is zero. The reverse is not always true: $f''(x) = 0$ is necessary, but it is not enough.

To prove that a point is a point of inflection, check that $f''(x)$ changes sign as $x$ passes through it. Find $f''$ to the left and to the right.

Example 4. For $f(x) = x^4$ , the second derivative is $f''(x) = 12x^2$ . At $x = 0$ this is zero. Is there a point of inflection?

Look at the sign of $f''$ on each side. For every $x \neq 0$ , $12x^2 > 0$ , so the curve is convex on both sides. The second derivative does not change sign. There is no point of inflection at $x = 0$ ; the curve flattens and turns up again.

Example 5. For $f(x) = x^3 - 3x^2 + 2x$ from Example 3, find the point of inflection and its coordinates.

We already found $f''(x) = 6x - 6$ , which is zero at $x = 1$ . Use the sign table:

$x$	below $1$ , say $x = 0$	$x = 1$	above $1$ , say $x = 2$
$f''(x) = 6x - 6$	$-6$ , concave	$0$	$6$ , convex

The second derivative changes from negative to positive, so $x = 1$ is a point of inflection. The $y$ -coordinate is

f(1) = 1 - 3 + 2 = 0.

The point of inflection is $(1, 0)$ .

4. Using the second derivative to classify a turning point

The second derivative also distinguishes a maximum from a minimum. At a turning point, $f'(x) = 0$ . Then:

if $f''(x) > 0$ , the curve is convex, so the point is a minimum;
if $f''(x) < 0$ , the curve is concave, so the point is a maximum;
if $f''(x) = 0$ , the test gives no answer, and you must use a sign table for $f'(x)$ .

Section 7.5 uses this test to solve the box problem.

Watch out. Three mistakes are common with the second derivative.

Writing $f''(x)$ as $f'(x^2)$ . The double prime marks a second differentiation, not a square. To find $f''(x)$ , differentiate $f'(x)$ once more.

Assuming that $f''(x) = 0$ always means a point of inflection. The curve $y = x^4$ has $f''(0) = 0$ but bends upwards on both sides. Always check the sign of $f''$ either side of the point.

Confusing convex with above the axis. Convex and concave describe the bending, not the height. The curve $y = x^2 - 4$ is convex everywhere, even where it dips below the $x$ -axis. Re-check: pick one point each side of a suspected inflection and compare the signs of the second derivative.

You should now be able to

find a second derivative and use the notation $f''(x)$ and $\dfrac{d^2y}{dx^2}$ ;
explain the second derivative as the rate of change of the gradient;
decide where a curve is convex or concave from the sign of $f''$ ;
find a candidate point of inflection and test it with a sign table;
use the second derivative to classify a turning point.