Critical Points Calculator
- Free
- No sign-up
- Works on phones
- Shows the working
This critical points calculator finds every point where \( f'(x) = 0 \) on an interval you choose, classifies each one as a local maximum, local minimum or neither, and reports the absolute maximum and minimum of the function on that interval. It is built 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
- Enter \( f(x) \). Use
^for powers,sqrt(x),sin(x),ln(x),e^(x)andpi; implicit multiplication like9xorx e^(-x)works. - Enter the interval endpoints \( a \) and \( b \) (with \( a < b \)). The calculator searches for critical points only inside \( [a, b] \), and those endpoints are also used for the absolute extrema.
- 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:
- differentiates \( f \) symbolically;
- solves \( f'(x) = 0 \) numerically on \( [a, b] \), showing exact values such as \( \sqrt2 \) or \( \frac{\pi}{2} \) when it recognizes them;
- 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 \( f''(c) = 0 \);
- 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. Only roots inside \( [a, b] \) are found, and the absolute extrema are only valid for that interval. For periodic functions such as
sin(x), pick a window like0to2pi. - 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.
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.
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.
