The second derivative test and inflection points

Warm-up

Three quick questions from earlier lessons.

Discover

At a critical point the graph is flat. If the graph is concave up there, like a cup, is the point a maximum or a minimum?

A minimum
A maximum

Key idea

The second derivative can classify a critical point, and locates changes in concavity.

Second derivative test: if

f'(c) = 0

and

f''(c) > 0

, local minimum;

f''(c) < 0

, local maximum;

f''(c) = 0

, no conclusion. An inflection point is where

f''

changes sign.

Worked example

$f(x) = x^3 - 6x^2$ . Classify the critical points and find the inflection point.

Critical numbers

0

and

4

f' = 3x² − 12x = 3x(x − 4).

Local max at

0

, local min at

4

f'' = 6x − 12: f''(0) = −12, f''(4) = 12.

Inflection point at

x = 2

f'' changes sign at x = 2.

Your turn, guided

Your turn: $f(x) = x^3 - 9x^2$ . Where is the inflection point?

x = 3

f''(x) = 6x − 18.

Practice ★

$f(x) = x^3 - 3x^2$ . At which $x$ does $f$ have an inflection point?
$f(x) = x^3 - 12x^2$ . At which $x$ does $f$ have an inflection point?

$f'(8) = 0$ and $f''(8) = 3$ . What happens at $x = 8$ ?
1. A local minimum
2. A local maximum
3. An inflection point

Watch out!

Practice ★★

$f'(1) = 0$ and $f''(1) = 0$ . What can you conclude about $x = 1$ ?
1. Nothing yet: it could be a maximum, a minimum or neither
2. It is an inflection point
3. It is a local maximum

Exit ticket

Three questions to check you have got it.

$f(x) = x^3 - 3x^2$ . At which $x$ does $f$ have an inflection point?
$f(x) = x^3 - 12x^2$ . At which $x$ does $f$ have an inflection point?
$f(x) = x^3 - 15x^2$ . At which $x$ does $f$ have an inflection point?

Summary card

Second derivative test

$f' = 0$ , $f'' > 0$ : min. $f' = 0$ , $f'' < 0$ : max. $f'' = 0$ : test fails.
Inflection: $f''$ changes sign.

Watch out: $f''(c) = 0$ is not enough for an inflection point.

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