Critical Points Calculator

This critical points calculator finds every point where \( f'(x) = 0 \) on an interval you choose. It classifies each one as a local maximum, local minimum or neither, and reports the absolute maximum and minimum of the function on that interval. We built it for the classic “find and classify the critical points” and “closed interval method” problems in Calculus 1.

It’s free, runs instantly in your browser and needs no sign-up.

How to use the critical points calculator

1. Enter \( f(x) \). Use ^ for powers, sqrt(x), sin(x), ln(x), e^(x) and pi; implicit multiplication like 9x or x e^(-x) works.
2. Enter the interval endpoints \( a \) and \( b \) (with \( a < b \)). The calculator searches for critical points only inside \( [a, b] \), and it also uses those endpoints for the absolute extrema.
3. Press Find extrema. You get \( f'(x) \), a table of critical points with \( x \), \( f(x) \) and the type, and the absolute max and min on the interval.

How it works

A critical point is a value \( c \) in the domain where \( f'(c) = 0 \) or \( f'(c) \) does not exist. The calculator:

1. differentiates \( f \) symbolically;
2. solves \( f'(x) = 0 \) numerically on \( [a, b] \), showing exact values such as \( \sqrt2 \) or \( \frac{\pi}{2} \) when it recognizes them;
3. classifies each root with the second derivative test, and falls back to the first derivative test (the sign of \( f' \) just left and right of \( c \)) when \( f''(c) = 0 \);
4. compares \( f \) at the critical points and both endpoints to pick the absolute extrema.

$$f''(c) > 0 \Rightarrow \text{local min}, \qquad f''(c) < 0 \Rightarrow \text{local max}$$

Worked example

Find and classify the critical points of \( f(x) = x^3 - 6x^2 + 9x + 1 \), then find its absolute extrema on \( [-1, 5] \).

Step 1: derivative. \( f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3) \).

Step 2: solve \( f'(x) = 0 \). The critical points are \( x = 1 \) and \( x = 3 \).

Step 3: classify. \( f''(x) = 6x - 12 \) (the second derivative calculator gives it directly). Since \( f''(1) = -6 < 0 \), \( x = 1 \) is a local maximum with \( f(1) = 5 \). Since \( f''(3) = 6 > 0 \), \( x = 3 \) is a local minimum with \( f(3) = 1 \).

Step 4: absolute extrema. Compare all candidates:

\( x \)	\( -1 \)	\( 1 \)	\( 3 \)	\( 5 \)
\( f(x) \)	\( -15 \)	\( 5 \)	\( 1 \)	\( 21 \)

The absolute maximum is 21 at \( x = 5 \) and the absolute minimum is \( -15 \) at \( x = -1 \). Both occur at endpoints, not at the critical points, which is why you always check the endpoints. The calculator above is preloaded with this example.

Limits of the calculator (and what to check by hand)

• Points where \( f' \) is undefined. The calculator solves \( f'(x) = 0 \) only. A cusp or corner, like \( x = 0 \) for \( f(x) = x^{2/3} \), has \( f'(x) = \frac{2}{3x^{1/3}} \), which is undefined at 0 and never zero, so it won’t be listed. Check the zeros of the denominator of \( f' \) yourself.
• “Neither (stationary point)”. This label means \( f' = 0 \) but \( f' \) does not change sign, as for \( f(x) = x^3 \) at 0, where \( f'(x) = 3x^2 \geq 0 \) on both sides.
• Interval choice matters. The calculator finds only roots inside \( [a, b] \), and the absolute extrema are only valid for that interval. For periodic functions such as sin(x), pick a window like 0 to 2pi.
• Long lists. The table shows up to 12 critical points.

For the full method with more examples, see how to find critical points. Word problems that ask you to maximize area or minimize cost reduce to the same steps; see optimization problems. To find where the concavity changes instead, use the inflection point calculator.

Absolute extrema on a closed interval

On \( [a, b] \), the largest and smallest values occur at critical points or at the endpoints, so compare them all.

Example: \( f(x) = x^3 - 3x \) on \( [0, 3] \). The critical points are \( x = \pm 1 \); only \( x = 1 \) is in the interval. Values: \( f(0) = 0 \), \( f(1) = -2 \), \( f(3) = 18 \). The absolute maximum is 18 at \( x = 3 \) and the absolute minimum is −2 at \( x = 1 \). See the absolute extrema calculator.

Critical points where the derivative does not exist

\( f(x) = x^{2/3} \) has \( f'(x) = \frac{2}{3x^{1/3}} \), undefined at \( x = 0 \). Since \( f(0) = 0 \) is defined, \( x = 0 \) is a critical point, and it is a minimum with a sharp corner (a cusp). Always check where \( f' \) is undefined as well as where it is zero.

The first derivative test

Check the sign of \( f' \) on each side of a critical point: from + to − is a local maximum, from − to + a local minimum, no change means neither (as for \( x^3 \) at 0).

Practice problems

1. Find and classify the critical points of \( f(x) = x^4 - 8x^2 \).
2. Find the absolute extrema of \( f(x) = x^2 - 4x \) on \( [0, 5] \).

Answers: 1. \( f'(x) = 4x(x^2 - 4) \): minima at \( x = \pm 2 \) (value −16), maximum at \( x = 0 \) (value 0). 2. Minimum −4 at \( x = 2 \), maximum 5 at \( x = 5 \).

Reading and Checking the Results

After using the critical points calculator, review the output carefully. First, check the reported derivative, f'(x). Ensure it matches your own calculation. Then, examine the table of critical points. Each row lists an x-value, its corresponding f(x) value, and the classification (local maximum, local minimum, or neither).

Next, verify the classification. A local maximum means the function value is higher than nearby points. A local minimum means it is lower. 'Neither' indicates a saddle point or an inflection point where the derivative is zero, like for f(x) = x^3 at x=0. Finally, locate the absolute maximum and minimum. These are the highest and lowest function values on your chosen interval, including endpoints.

Common Mistakes and How to Avoid Them

One common mistake is entering an incorrect function or interval. Double-check your input for typos, especially with exponents or parentheses. For example, x^2+1 is different from x^(2+1). Also, ensure your interval [a, b] has 'a' less than 'b'. The calculator expects a valid range.

Another error is forgetting to consider points where the derivative does not exist. The critical points calculator primarily finds where f'(x) = 0. If your function has sharp corners or vertical tangents, these critical points will not appear in the results. Always manually inspect the derivative for undefined points, such as denominators that become zero.

Sometimes users misinterpret the 'neither' classification. This simply means the critical point is not a local extremum. It is still a point where the slope is zero, but the function does not change direction. Always consider the full context of the problem when interpreting these results.

Related guides

• How to Find Critical Points (and Classify Them)
• Optimization Problems in Calculus
• How to Find Inflection Points
• Power Rule for Derivatives

Further reading

• Critical Points (Paul’s Online Math Notes) — the definition plus examples where \( f' \) is zero or undefined.
• The Shape of a Graph, Part I (Paul’s Online Math Notes) — increasing and decreasing intervals and the first derivative test.

When the derivative is a quadratic, the quadratic formula calculator gives its roots exactly.

FAQ

How do you find critical points of a function?

Differentiate, then solve \( f'(x) = 0 \) and find where \( f'(x) \) is undefined, keeping only values in the domain of \( f \). Each solution is a critical point.

Is every critical point a maximum or minimum?

No. A critical point can be neither, like \( x = 0 \) for \( x^3 \). Use the first or second derivative test to classify it; the calculator does this for you.

What is the difference between local and absolute extrema?

A local extremum is the highest or lowest value near a point. The absolute extremum is the highest or lowest value on the whole interval, and it can occur at an endpoint.

Why do I need to enter an interval?

The interval tells the calculator where to search for solutions of \( f'(x) = 0 \) and which endpoints to compare for the absolute maximum and minimum.