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

Understanding the First Derivative Test

The first derivative test uses the sign of

f'(x)

around a critical point to figure out whether it’s a local max or a local min.

For a minimum:

f'(x)

goes negative

\rightarrow

\rightarrow

positive

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

For a maximum:

f'(x)

goes positive

\rightarrow

\rightarrow

negative

Formal Justification Statements

Let

c

and

d

be critical numbers.

Minimum: There is a minimum value at

x = c

because

f'(x)

changes sign from negative to positive.

Maximum: There is a maximum value at

x = d

because

f'(x)

changes sign from positive to negative.

Neither: There is no maximum or minimum at a critical point where

f'(x)

does not change sign.

For example,

f(x) = x^3 - 6x^2 + 12x

has

f'(x) = 3x^2 - 12x + 12 = 3(x - 2)^2

. So

x = 2

is a critical point, but

f'(x) > 0

on both sides of it. The function keeps increasing through

x = 2

, so there’s no relative extremum there.

The Procedure

Find $f'(x)$

Find all critical points (where $f'(x) = 0$ or $f'(x)$ doesn’t exist)

Make a sign chart with intervals between the critical points

Test the sign of $f'(x)$ in each interval

Look at the sign changes to classify each critical point

Write justification statements

Example 1

Given

g(x) = (x^2 - 9)^{2/3}

Step 1: Find the derivative. Chain rule:

g'(x) = \frac{2}{3}(x^2 - 9)^{-1/3} \cdot 2x = \frac{4x}{3(x^2 - 9)^{1/3}}

Step 2: Find critical points.

g'(x) = 0

when the numerator is zero:

4x = 0 \Rightarrow x = 0

g'(x)

is undefined when the denominator is zero:

x^2 - 9 = 0 \Rightarrow x = \pm 3

Critical points:

x = -3, 0, 3

Step 3: Sign chart.

Interval	$(-\infty, -3)$	$x = -3$	$(-3, 0)$	$x = 0$	$(0, 3)$	$x = 3$	$(3, \infty)$
Sign of $g'(x)$	$(-)$	Undef	$(+)$	$0$	$(-)$	Undef	$(+)$

Step 4: Test points.

$x = -4$ : $\frac{4(-4)}{3(7)^{1/3}} < 0$

$x = -1$ : $\frac{4(-1)}{3(-8)^{1/3}} = \frac{-4}{3(-2)} > 0$

$x = 1$ : $\frac{4(1)}{3(-8)^{1/3}} = \frac{4}{3(-2)} < 0$

$x = 4$ : $\frac{4(4)}{3(7)^{1/3}} > 0$

Step 5: Classify.

Minimum at

x = -3

and

x = 3

because

g'(x)

changes sign from negative to positive.

Maximum at

x = 0

because

g'(x)

changes sign from positive to negative.

Example 2: Domain Restrictions

Given

h(x) = \dfrac{x^2}{x - 4}

, where

x \neq 4

Find the derivative with the quotient rule:

h'(x) = \frac{2x(x-4) - x^2(1)}{(x-4)^2} = \frac{x^2 - 8x}{(x-4)^2} = \frac{x(x - 8)}{(x-4)^2}

Potential critical points:

From

h'(x) = 0

x(x - 8) = 0 \Rightarrow x = 0

x = 8

From

h'(x)

undefined:

x = 4

Check the domain.

x = 4

is not in the domain of

h(x)

, it’s a vertical asymptote. So

x = 4

is NOT a critical point and cannot be a max or min. You still put it in the sign chart though, since the behavior can change there.

Sign testing. Since

(x-4)^2

is always positive, you only need to look at the numerator

x(x - 8)

$x < 0$ : $(-)(-) = (+)$

$0 < x < 8$ : $(+)(-) = (-)$

$x > 8$ : $(+)(+) = (+)$

Sign chart:

Interval	$(-\infty, 0)$	$x = 0$	$(0, 4)$	$x = 4$	$(4, 8)$	$x = 8$	$(8, \infty)$
Sign of $h'(x)$	$(+)$	$0$	$(-)$	Undef (not a CP)	$(-)$	$0$	$(+)$

Classification:

Maximum at

x = 0

because

h'(x)

changes sign from positive to negative.

Minimum at

x = 8

because

h'(x)

changes sign from negative to positive.

Again,

x = 4

creates a vertical asymptote but it’s not a critical point since it’s not in the domain.

Maximum/Minimum VALUE vs. LOCATION

This distinction matters a lot on exams.

“Where is the maximum?” asks for the x-coordinate:

x = c

“What is the maximum value?” asks for the y-coordinate:

f(c)

So if

f

has a maximum at

x = 3

Location of the maximum: $x = 3$

Maximum value: $f(3)$ , which you get by plugging 3 into the original function

Both of these often show up as multiple choice options, so read the question.

Key Takeaways

The first derivative test looks at sign changes of $f'(x)$ at critical points

Minimum: $f'(x)$ changes from $(-)$ to $(+)$

Maximum: $f'(x)$ changes from $(+)$ to $(-)$

Justification is required: say “ $f'(x)$ changes sign from ___ to ___”

The sign chart organizes your info but is not justification by itself

Points not in the domain cannot be critical points or extrema

Still include domain restrictions in your sign chart to get the intervals right

“Where” asks for the $x$ -value, “what value” asks for the $y$ -value

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