Skill: Classify local extrema using the first derivative

A critical number is an interior point of the domain where $f'(x)=0$ or where $f'(x)$ is undefined. To classify a critical number:

• If $f'(x)$ changes from positive to negative, $f$ changes from increasing to decreasing, so there is a local maximum.
• If $f'(x)$ changes from negative to positive, $f$ changes from decreasing to increasing, so there is a local minimum.
• If the sign of $f'(x)$ does not change, there is no local extremum.

Worked example

Classify the critical numbers of

$$f(x)=\frac{x^5}{5}-\frac{x^4}{4}-\frac{2x^3}{3}.$$

1. Find the derivative

$$f'(x)=x^4-x^3-2x^2.$$

Factor:

$$f'(x)=x^2(x^2-x-2)$$ $$f'(x)=x^2(x+1)(x-2).$$

2. Find the critical numbers

Set the derivative equal to zero:

$$x^2(x+1)(x-2)=0.$$

Thus,

$$x=-1,\qquad x=0,\qquad x=2.$$

The derivative is defined for every real number, so these are all the critical numbers.

3. Make a sign chart

The critical numbers divide the number line into four intervals. Test one value in each interval.

Interval	Test value	Sign of $f'(x)$	Behavior of $f$
$(-\infty,-1)$	$-2$	$+$	Increasing
$(-1,0)$	$-0.5$	$-$	Decreasing
$(0,2)$	$1$	$-$	Decreasing
$(2,\infty)$	$3$	$+$	Increasing

4. Classify each critical number

• At $x=-1$, the sign changes from $+$ to $-$. Therefore, $f$ has a local maximum at $x=-1$.
• At $x=0$, the sign remains negative on both sides. Therefore, $x=0$ is not a local extremum.
• At $x=2$, the sign changes from $-$ to $+$. Therefore, $f$ has a local minimum at $x=2$.

The important point is that $f'(x)=0$ alone does not guarantee a maximum or minimum. The sign change in the derivative determines the classification.