How to Find Critical Numbers of a Function in 5 Steps

To find critical numbers of a function, take its derivative, set it equal to zero and solve, then find every x-value where the derivative is undefined; the x-values from both cases are the critical numbers. These points are the only candidates for local maxima, local minima, and other turning behavior, so the process reduces to one derivative and two conditions.

What Are Critical Numbers?

A critical number is an x-value in the domain of a function where the derivative is either zero or does not exist. That definition matters because of Fermat’s theorem: if a differentiable function has a local extremum at an interior point, the derivative there must be zero. So every local extremum that isn’t at an endpoint has to appear on the critical number list.

The list is a set of candidates, not confirmed answers. A critical number can also be a flat spot with no extremum, such as x = 0 on the graph of f(x) = x³.

Two practical reminders:

• The x-value must lie in the function’s domain. A point where the function itself is undefined is never a critical number.
• Endpoints of a closed interval are not critical numbers under this definition, though they still must be checked when hunting for absolute extrema.

Conditions for Critical Numbers

A number c qualifies if it satisfies at least one of two conditions, with c in the domain of f:

1. f′© = 0. The tangent line is horizontal. This is the classic case for smooth curves like polynomials.
2. f′© is undefined. The derivative fails to exist — a corner, a cusp, or a vertical tangent.

Condition 2 is the one students skip, and it is often where the interesting behavior hides. Consider f(x) = |x|. The derivative is −1 for x < 0 and +1 for x > 0, so f′(0) does not exist, yet x = 0 is where the minimum sits. Similarly, f(x) = x^(1/3) has a vertical tangent at x = 0.

Notice what is not on the list: points where the function is undefined. If f has a vertical asymptote at x = 2, then x = 2 is not a critical number, no matter how strange the graph looks there.

Why the undefined case still counts

A local extremum requires only that the function change direction — not that it be smooth. Any sharp turn in a continuous graph forces the derivative to fail, which makes those x-values legitimate candidates. Skipping them can cause you to miss an extremum entirely.

The 5-Step Method

Step 1: Write down the domain. Before differentiating, note where f is defined. Rational functions lose points where the denominator is zero; radical functions with even roots require a nonnegative radicand.

Step 2: Find the derivative f′(x). Use the power, product, quotient, and chain rules as needed. Simplify the result — a single factored fraction is far easier to analyze than a mess of terms.

Step 3: Solve f′(x) = 0. Set the numerator of a simplified derivative to zero (for rational derivatives) or the whole expression to zero, and solve for x. Keep only solutions that lie in the domain from Step 1.

Step 4: Find where f′(x) is undefined. For a rational derivative, these are the zeros of the denominator. For radicals, look for values that make an even root’s radicand negative. Again, discard anything outside the domain of f.

Step 5: Combine and list. Merge the results of Steps 3 and 4 into one list of x-values. If the problem also asks for the critical points, plug each x back into f to get the y-coordinate.

A quick check on your answer

Verify each candidate by substituting it back into the original derivative. If f′© is neither zero nor undefined, it is not a critical number, and you can drop it.

Worked Example

Let f(x) = x³ − 3x. The domain is all real numbers. Differentiating gives f′(x) = 3x² − 3. Setting it to zero: 3x² − 3 = 0, so x² = 1 and x = ±1. The derivative is a polynomial, so it is defined everywhere — no Step 4 candidates. The critical numbers are x = −1 and x = 1.

Key Takeaways

• A critical number is an x-value in the domain of f where f′(x) = 0 or f′(x) is undefined.
• Both conditions matter. Corners, cusps, and vertical tangents produce critical numbers that the zero-solving step will never find.
• Points outside the domain are never critical numbers — check the domain first and filter at the end.
• Critical numbers are candidates for local extrema, not guarantees; use the first or second derivative test to classify them.
• The five steps are: find the domain, differentiate, solve f′(x) = 0, find where f′ is undefined, then combine and keep only valid x-values.