About the Critical Points Calculator
This critical points calculator finds every x where the derivative of your function is zero or undefined on the interval you choose, then classifies each one. Type a function such as x^3 − 3x, x^4 − 4x^3 or sin(x) + cos(x), set the interval, and it differentiates the function symbolically, locates the zeros of f′(x) numerically to high precision, and applies the second derivative test — falling back to the first derivative test when f″ is zero.
It is built for calculus students working on optimisation and curve-sketching problems, and for anyone who needs the highest or lowest value of a function over a range, such as maximum profit, minimum cost or peak height. The closed-interval method is included: the calculator compares the function at the critical points and at both endpoints to report the absolute maximum and minimum.
Because roots are found numerically on the interval you give, choose an interval wide enough to contain the behaviour you care about. Points where f′ does not exist are found too — corners such as |x − 2|, cusps such as x^(2/3) and vertical tangents such as x^(1/3) (fractional powers with odd denominators are evaluated as real roots, so they work for negative x). Trigonometric functions use radians.
How to use the critical points calculator
- 1Type the function using x as the variable, e.g. x^3 - 3x.
- 2Set the interval start a and end b to search.
- 3Read the list of critical points and whether each is a local max, min or neither.
- 4Use the absolute maximum and minimum for closed-interval optimisation problems.
- 5Check the chart: critical points are where the f′ curve crosses or touches zero.
Formula and method
A critical point (critical number) of f is an x-value c in the domain of f where the derivative is zero or does not exist. Local maxima and minima of a function can only occur at critical points, which is why they are the first step in curve sketching and optimisation.
The second derivative test classifies a point with f′(c) = 0: if f″(c) < 0 the graph is concave down, so c is a local maximum; if f″(c) > 0 it is a local minimum. When f″(c) = 0 or does not exist, the first derivative test is used: f′ changing from + to − means a maximum, − to + a minimum, and no sign change means neither (for example x³ at 0). On a closed interval [a, b], the absolute maximum and minimum are the largest and smallest values of f among the critical points and the two endpoints.
- f′(x)
- First derivative — zero at smooth peaks and valleys
- f″(x)
- Second derivative — the concavity used to classify the point
- [a, b]
- Interval searched for critical points
Worked examples
f(x) = x³ − 3x on [−3, 3]
f′(x) = 3x² − 3 = 0 at x = ±1. f″(x) = 6x is −6 at x = −1 (local max, f = 2) and 6 at x = 1 (local min, f = −2). The endpoints give f(−3) = −18 and f(3) = 18, which are the absolute extremes on this interval.
Flat inflection: f(x) = x⁴ − 4x³ on [−2, 5]
f′(x) = 4x³ − 12x² = 4x²(x − 3), zero at x = 0 and x = 3. At x = 0, f″ = 0 and f′ is negative on both sides, so it is neither a max nor a min. At x = 3, f″ = 36 > 0, a local minimum of −27. The absolute maximum on [−2, 5] is f(5) = 125.
sin(x) on [0, 7]
cos(x) = 0 at π/2 ≈ 1.5708 and 3π/2 ≈ 4.7124 within [0, 7]. sin has a local (and absolute) maximum of 1 at π/2 and a minimum of −1 at 3π/2.
A corner: f(x) = |x − 2| on [0, 5]
The derivative is −1 left of 2 and +1 right of 2, so it does not exist at x = 2 — a critical point where f has a sharp minimum of 0. On [0, 5] the absolute maximum is f(5) = 3.
Frequently asked questions
What is a critical point?+
A critical point is a value c in the domain of f where f′(c) = 0 or f′(c) does not exist. Graphically these are flat tangents (peaks, valleys, flat inflections) or sharp corners and cusps.
Is every critical point a maximum or minimum?+
No. f(x) = x³ has f′(0) = 0, but the graph keeps rising through x = 0 — it is a flat inflection point. Use the first or second derivative test to classify each critical point.
What is the second derivative test?+
If f′(c) = 0 and f″(c) > 0, f has a local minimum at c; if f″(c) < 0, a local maximum. If f″(c) = 0 the test is inconclusive and you fall back on the sign of f′ on either side.
How do I find the absolute maximum and minimum on a closed interval?+
Evaluate f at every critical point inside [a, b] and at both endpoints a and b. The largest value is the absolute maximum and the smallest is the absolute minimum (the closed-interval method).
Why does the calculator need an interval?+
Many functions, such as sin(x), have infinitely many critical points. Searching a finite interval keeps the answer finite and lets the calculator report absolute extrema. Widen the interval if you expect more points.
Are points where f is undefined critical points?+
No — a critical point must be in the domain of f. For f(x) = 1/x, f′ is undefined at 0, but so is f, so x = 0 is not a critical point (it is a vertical asymptote).