Introduction

This topic focuses on using the First Derivative Test to classify critical points as local extrema. The essential knowledge tells us that the first derivative of a function can determine the location of relative (local) extrema of the function. This test uses sign changes in the derivative to identify whether critical points are local maxima, local minima, or neither.

The First Derivative Test

Statement of the Test

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

Local Maximum: If $f^{'} (x)$ changes from positive to negative at $x = c$ , then $f$ has a local maximum at $x = c$ .

Local Minimum: If $f^{'} (x)$ changes from negative to positive at $x = c$ , then $f$ has a local minimum at $x = c$ .

Neither: If $f^{'} (x)$ does not change sign at $x = c$ , then $f$ has neither a local maximum nor minimum at $x = c$ .

Understanding Sign Changes

The sign change of $f^{'} (x)$ tells us about the behavior of $f (x)$ :

Positive to Negative ( $+ \to -$ ):

Function increases then decreases
Creates a peak (local maximum)

Negative to Positive ( $- \to +$ ):

Function decreases then increases
Creates a valley (local minimum)

No Sign Change:

Function continues in same direction
No local extremum occurs

Applying the First Derivative Test

Complete Procedure

Find all critical points (solve $f^{'} (x) = 0$ and find where $f^{'} (x)$ is undefined)
Create a sign chart for $f^{'} (x)$ using the critical points
Determine the sign of $f^{'} (x)$ on each interval
Classify each critical point based on sign changes
State conclusions clearly

Example 1:

Use the First Derivative Test to find and classify all local extrema of $f (x) = x^{3} - 3 x^{2} - 9 x + 5$ .

Find critical points: $f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1)$ $x = - 1$ and $x = 3$

Sign analysis:

For $x < - 1$ : $f^{'} (x) > 0$ (increasing)
For $- 1 < x < 3$ : $f^{'} (x) < 0$ (decreasing)
For $x > 3$ : $f^{'} (x) > 0$ (increasing)

At $x = - 1$ : $f^{'}$ changes from $+$ to $-$ , so local maximum

At $x = 3$ : $f^{'}$ changes from $-$ to $+$ , so local minimum

When No Extremum Occurs

Example 2:

Use the First Derivative Test on $f (x) = x^{3}$ .

Find critical points: $f^{'} (x) = 3 x^{2}$ $3 x^{2} = 0$ $x = 0$

Sign analysis:

For $x < 0$ : $f^{'} (x) = 3 x^{2} > 0$ (increasing)
For $x > 0$ : $f^{'} (x) = 3 x^{2} > 0$ (increasing)

At $x = 0$ : $f^{'}$ does not change sign (stays positive), so neither a local maximum nor minimum.

This is an inflection point, not an extremum.

Justifying Extrema with Proper Format

AP Exam Justification Standards

A complete justification must include:

Identification of the critical point
Sign of $f^{'} (x)$ before the critical point
Sign of $f^{'} (x)$ after the critical point
Conclusion based on the sign change

Example 3:

Justify that $f (x) = 2 x^{3} - 15 x^{2} + 24 x$ has a local maximum at $x = 1$ .

Find $f^{'} (x)$ : $f^{'} (x) = 6 x^{2} - 30 x + 24 = 6 (x - 1) (x - 4)$

Critical points: $x = 1$ and $x = 4$

Justification for $x = 1$ being a local maximum:

For $x < 1$ (test $x = 0$ ): $f^{'} (0) = 24 > 0$

For $1 < x < 4$ (test $x = 2$ ): $f^{'} (2) = 6 (1) (- 2) = - 12 < 0$

Since $f^{'} (x)$ changes from positive to negative at $x = 1$ , the function has a local maximum at $x = 1$ by the First Derivative Test.

Functions with Undefined Derivatives

Critical Points from Undefined Derivatives

When $f^{'} (x)$ is undefined at a point in the domain, that point is also a critical point and must be tested.

Example 4:

Find and classify local extrema of $f (x) = x^{2/3} (x - 5) = x^{5/3} - 5 x^{2/3}$ .

Find $f^{'} (x)$ : $f^{'} (x) = \frac{5}{3} x^{2/3} - \frac{10}{3} x^{- 1/3} = \frac{5 x ^{2/3}}{3} - \frac{10}{3 x ^{1/3}}$

$f^{'} (x) = \frac{5 x - 10}{3 x ^{1/3}}$

Set numerator equal to zero: $5 x - 10 = 0$ , so $x = 2$

$f^{'} (x)$ is undefined when $x = 0$

Critical points: $x = 0$ and $x = 2$

Sign analysis:

For $x < 0$ : $f^{'} (- 1) = \frac{- 15}{3 ( - 1 )} > 0$ (increasing)
For $0 < x < 2$ : $f^{'} (1) = \frac{- 5}{3} < 0$ (decreasing)
For $x > 2$ : $f^{'} (3) = \frac{5}{3 ( 3 ) ^{1/3}} > 0$ (increasing)

At $x = 0$ : changes from $+$ to $-$ , so local maximum

At $x = 2$ : changes from $-$ to $+$ , so local minimum

Multiple Critical Points

Example 5:

Find all local extrema of $g (x) = x^{4} - 4 x^{3} + 4 x^{2}$ .

Find critical points: $g^{'} (x) = 4 x^{3} - 12 x^{2} + 8 x = 4 x (x^{2} - 3 x + 2) = 4 x (x - 1) (x - 2)$ $x = 0, 1, 2$

Sign analysis:

For $x < 0$ : $g^{'} (- 1) = 4 (- 1) (- 4) (- 3) < 0$ (decreasing)
For $0 < x < 1$ : $g^{'} (0.5) = 4 (0.5) (- 1.25) (- 1.5) > 0$ (increasing)
For $1 < x < 2$ : $g^{'} (1.5) = 4 (1.5) (- 0.75) (- 0.5) > 0$ (increasing)
For $x > 2$ : $g^{'} (3) = 4 (3) (2) (1) > 0$ (increasing)

At $x = 0$ : changes from $-$ to $+$ , so local minimum

At $x = 1$ : no sign change (stays $+$ ), so neither

At $x = 2$ : no sign change (stays $+$ ), so neither

FiveHive

FiveHive

5.4 - Using the First Derivative Test to Determine Relative (Local) Extrema

Introduction

The First Derivative Test

Statement of the Test

Understanding Sign Changes

Applying the First Derivative Test

Complete Procedure

When No Extremum Occurs

Justifying Extrema with Proper Format

AP Exam Justification Standards

Functions with Undefined Derivatives

Critical Points from Undefined Derivatives

Multiple Critical Points

Practice Section