Classify local extrema using the first derivative

The derivative’s sign determines whether a function is increasing or decreasing, allowing critical numbers—interior domain points where $f'(x)=0$ or $f'(x)$ is undefined—to be classified with a sign chart. A change from positive to negative indicates a local maximum, while a change from negative to positive indicates a local minimum; no sign change means the point is not a local extremum, so $f'(x)=0$ alone is insufficient. The standard test applies to single-variable functions and does not include more advanced generalized extrema.

Detailed Explanation: 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.

Learn by doing: Classify local extrema using the first derivative

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Critical Points - Derivative Chart and Maxima or Minima to True or False

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