Introduction

This topic introduces the Second Derivative Test as an alternative method for classifying critical points. The essential knowledge tells us that the second derivative of a function may determine whether a critical point is the location of a relative (local) maximum or minimum and when a continuous function has only one critical point on an interval on its domain and the critical point corresponds to a relative (local) extremum of the function on the interval, then that critical point also corresponds to the absolute (global) extremum of the function on the interval. This test provides a quick way to classify critical points when it applies.

The Second Derivative Test

Statement of the Test

Let $c$ be a critical point where $f^{'} (c) = 0$ .

If $f^{''} (c) > 0$ , then $f$ has a local minimum at $x = c$ .

If $f^{''} (c) < 0$ , then $f$ has a local maximum at $x = c$ .

If $f^{''} (c) = 0$ , the test is inconclusive (use the First Derivative Test instead).

Why the Test Works

The second derivative tells us about concavity:

$f^{''} (c) > 0$ means concave up at $c$ → critical point is at the bottom of a "cup" → local minimum
$f^{''} (c) < 0$ means concave down at $c$ → critical point is at the top of a "cap" → local maximum

Example 1:

Use the Second Derivative Test to classify the critical points of $f (x) = x^{3} - 6 x^{2} + 9 x$ .

Find critical points: $f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$

$x = 1$ and $x = 3$

Find $f^{''} (x)$ : $f^{''} (x) = 6 x - 12$

Test at $x = 1$ : $f^{''} (1) = 6 - 12 = - 6 < 0$

Local maximum at $x = 1$

Test at $x = 3$ : $f^{''} (3) = 18 - 12 = 6 > 0$

Local minimum at $x = 3$

Comparing First and Second Derivative Tests

Advantages and Limitations

Second Derivative Test:

Advantage: Often quicker - just evaluate $f^{''}$ at the critical point
Limitation: Fails when $f^{''} (c) = 0$ or $f^{''} (c)$ is undefined

First Derivative Test:

Advantage: Always works if $f^{'}$ exists near the critical point
Limitation: Requires testing intervals around the critical point

Example 2:

Use the Second Derivative Test on $g (x) = 2 x^{3} - 9 x^{2} + 12 x - 3$ .

Find critical points: $g^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$

Therefore the critical points are $x = 1$ and $x = 2$

Find $g^{''} (x)$ : $g^{''} (x) = 12 x - 18$

Test at $x = 1$ : $g^{''} (1) = 12 - 18 = - 6 < 0$

Local maximum at $x = 1$

Test at $x = 2$ : $g^{''} (2) = 24 - 18 = 6 > 0$

Local minimum at $x = 2$

When the Second Derivative Test Fails

The Inconclusive Case

When $f^{''} (c) = 0$ , the Second Derivative Test provides no information. We must use the First Derivative Test instead.

Example 3:

Apply the Second Derivative Test to $h (x) = x^{4}$ .

Find critical points: $h^{'} (x) = 4 x^{3}$

$x = 0$

Find $h^{''} (x)$ : $h^{''} (x) = 12 x^{2}$

Test at $x = 0$ : $h^{''} (0) = 0$

The test is inconclusive. We must use the First Derivative Test:

For $x < 0$ : $h^{'} (x) < 0$ (decreasing)
For $x > 0$ : $h^{'} (x) > 0$ (increasing)

Since $h^{'}$ changes from negative to positive, $x = 0$ is a local minimum.

Another Inconclusive Example

Example 4:

Use the Second Derivative Test on $f (x) = x^{5} - 5 x$ .

Find critical points:

$f^{'} (x) = 5 x^{4} - 5 = 5 (x^{4} - 1) = 5 (x - 1) (x + 1) (x^{2} + 1)$

$x = - 1$ and $x = 1$

Find $f^{''} (x)$ : $f^{''} (x) = 20 x^{3}$

Test at $x = - 1$ : $f^{''} (- 1) = - 20 < 0$

Local maximum at $x = - 1$

Test at $x = 1$ : $f^{''} (1) = 20 > 0$

Local minimum at $x = 1$

The One-Critical-Point Theorem

Global Extrema from Single Critical Points

Theorem: If a continuous function has only one critical point on an interval, and that critical point is a local extremum, then it is also the absolute (global) extremum on that interval.

This powerful result means we don't need to check endpoints when there's only one critical point that we've identified as a local extremum.

Example 5:

Show that $f (x) = x^{2} + \frac{4}{x}$ has an absolute minimum on $(0, \infty)$ .

Find critical points:

$f^{'} (x) = 2 x - \frac{4}{x ^{2}}$

$2 x - \frac{4}{x ^{2}} = 0$

$2 x = \frac{4}{x ^{2}}$

$2 x^{3} = 4$

$x^{3} = 2$

$x = 32$ (only critical point on $(0, \infty)$ )

Use Second Derivative Test:

$f^{''} (x) = 2 + \frac{8}{x ^{3}}$

$f^{''} (32) = 2 + \frac{8}{2} = 6 > 0$

Since $f^{''} (32) > 0$ , there is a local minimum at $x = 32$ .

By the One-Critical-Point Theorem, this is also the absolute minimum on $(0, \infty)$ .

Choosing the Appropriate Test

Decision Guide

Use the Second Derivative Test when:

You need to classify one or more critical points where $f^{'} (c) = 0$
Computing $f^{''} (c)$ is straightforward
You want a quick answer

Use the First Derivative Test when:

$f^{''} (c) = 0$ (Second Derivative Test fails)
$f^{'} (c)$ is undefined
You want guaranteed results
You're already analyzing increasing/decreasing intervals

Example 6:

For $g (x) = x^{4} - 8 x^{2} + 3$ , classify all critical points.

Find critical points:

$g^{'} (x) = 4 x^{3} - 16 x = 4 x (x^{2} - 4) = 4 x (x - 2) (x + 2)$

$x = - 2, 0, 2$

Find $g^{''} (x)$ : $g^{''} (x) = 12 x^{2} - 16$

Test each critical point:

$g^{''} (- 2) = 48 - 16 = 32 > 0$ → local minimum

$g^{''} (0) = - 16 < 0$ → local maximum

$g^{''} (2) = 48 - 16 = 32 > 0$ → local minimum

FiveHive

FiveHive

5.7 - Using the Second Derivative Test to Determine Extrema

Introduction

The Second Derivative Test

Statement of the Test

Why the Test Works

Comparing First and Second Derivative Tests

Advantages and Limitations

When the Second Derivative Test Fails

The Inconclusive Case

Another Inconclusive Example

The One-Critical-Point Theorem

Global Extrema from Single Critical Points

Choosing the Appropriate Test

Decision Guide

Practice Section