What if $f'(c) = 0$ but the sign of $f'(x)$ does not change around $c$?

The point $c$ is not a local extremum. It is typically a point of inflection where the tangent is horizontal, such as $x=0$ for $f(x)=x^3$.

Are endpoints of a closed interval considered critical points?

By standard definition, critical points must lie in the interior of the domain. Endpoints are tested separately when finding absolute extrema on a closed interval.

What if $f'(c)$ is undefined but $f(c)$ is defined?

This is a valid critical point. It often indicates a cusp, corner, or vertical tangent on the graph. You must use the First Derivative Test to classify it.

How to find critical points and classify them

Critical points occur where the first derivative of a function is zero or undefined. These points correspond to potential local maxima, local minima, or inflection points on the graph of the function.

This method applies to any continuous function of a single variable. Classification is done using either the First Derivative Test or the Second Derivative Test to determine the behavior of the function at these specific points.

The setup

Identify the function $f(x)$ and explicitly state its domain. Ensure that the rules of differentiation (product, quotient, chain rule) can be applied to compute the first derivative $f'(x)$ and the second derivative $f''(x)$ .

The steps

Compute the first derivative $f'(x)$ .
Set $f'(x) = 0$ and solve for $x$ to find stationary points.
Identify any $x$ -values in the domain of $f(x)$ where $f'(x)$ is undefined.
Collect all $x$ -values from steps 2 and 3; these are the critical points $c$ .
Compute the second derivative $f''(x)$ .
Evaluate $f''(c)$ for each critical point (Second Derivative Test):

If $f''(c) > 0$ , $f(c)$ is a local minimum.
If $f''(c) < 0$ , $f(c)$ is a local maximum.
If $f''(c) = 0$ or is undefined, use the First Derivative Test by checking the sign of $f'(x)$ immediately to the left and right of $c$ .

Checking the result

Verify that every critical point $c$ is actually within the original domain of $f(x)$ . Confirm your classification by checking the sign changes of $f'(x)$ on the intervals bounded by the critical points.

Common errors

A frequent mistake is including points where the original function $f(x)$ is undefined (like vertical asymptotes) as critical points. Another common error is assuming that $f''(c) = 0$ implies there is no local extremum; the test is simply inconclusive, and the First Derivative Test must be used.

Worked example

Find and classify the critical points of $f(x) = x^3 - 3x^2 - 9x + 5$ .

$f'(x) = 3x^2 - 6x - 9$ Set the first derivative to zero: $3x^2 - 6x - 9 = 0$ $3(x^2 - 2x - 3) = 0$ $3(x - 3)(x + 1) = 0$ The critical points are $x = 3$ and $x = -1$ . There are no points where $f'(x)$ is undefined. Compute the second derivative: $f''(x) = 6x - 6$ Evaluate at $x = -1$ : $f''(-1) = 6(-1) - 6 = -12$ Since $f''(-1) < 0$ , there is a local maximum at $x = -1$ . Evaluate at $x = 3$ : $f''(3) = 6(3) - 6 = 12$ Since $f''(3) > 0$ , there is a local minimum at $x = 3$ .

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References: Stewart, Calculus: Early Transcendentals, Chapter 4 · OpenStax Calculus Volume 1, Applications of Derivatives