# What are the absolute max and min?

Finds the absolute maximum and minimum of a continuous function on a closed interval [a, b], from its critical numbers and end points.

- Page: https://www.acalculator.org/math/absolute-extrema-calculator
- JSON spec: https://www.acalculator.org/math/absolute-extrema-calculator.json
- Version: fb56c220e0ab

## Default answer

Example with the default inputs (Function f(x) -x^2 + 3x - 2, From x = a 1, To x = b 3): On [1, 3], -x^2 + 3x - 2 has absolute maximum 0.25 at x = 3/2 and absolute minimum -2 at x = 3.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| f | Function f(x) | A function continuous on [a, b]. |
| a | From x = a | From x = a: a number or a constant such as pi. |
| b | To x = b | To x = b: a number or a constant such as pi. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| max | Absolute maximum | The largest value of f on [a, b]. |
| maxAt | Maximum at x = | Where f takes its largest value. |
| min | Absolute minimum | The smallest value of f on [a, b]. |
| minAt | Minimum at x = | Where f takes its smallest value. |
| candidates | Values checked | f at each end point and critical number in between. |
| derivative | f′(x) = | The derivative of f. |

## Method

Closed interval method: f at a, at b, and at each critical number in (a, b), where f′ is 0 or does not exist; the largest is the absolute maximum, the smallest the absolute minimum.

## Assumptions

- f is continuous on [a, b]; angles in radians.
- Values within 10⁻¹² of the extreme value count as ties.

## Worked examples

1. f = -x^2 + 3x - 2, a = 1, b = 3 gives max = 0.25, maxAt = 3/2, min = -2, minAt = 3, candidates = f(1) = 0; f(3/2) = 1/4; f(3) = -2. Source: OpenStax, Calculus Volume 1, section 4.3 Maxima and Minima, Example 4.13 (https://openstax.org/books/calculus-volume-1/pages/4-3-maxima-and-minima), part a.
2. f = x^2 - 3x^(2/3), a = 0, b = 2 gives max = 0, maxAt = 0, min = -2, minAt = 1. Source: OpenStax, Calculus Volume 1, section 4.3 Maxima and Minima, Example 4.13 (https://openstax.org/books/calculus-volume-1/pages/4-3-maxima-and-minima), part b.
3. f = x^3 - 3x, a = -2, b = 2 gives max = 2, maxAt = -1, 2, min = -2, minAt = -2, 1.

## FAQ

### What are absolute extrema?

The absolute maximum of f on an interval is its largest value there, and the absolute minimum is its smallest. By the extreme value theorem, a function continuous on a closed interval [a, b] has both.

### How do I find the absolute max and min on a closed interval?

Use the closed interval method: find the critical numbers of f inside (a, b), where f′ is 0 or does not exist; evaluate f there and at a and b; the largest value is the absolute maximum and the smallest the absolute minimum. For f(x) = −x² + 3x − 2 on [1, 3]: f(1) = 0, f(3/2) = 1/4 and f(3) = −2.

### What is the difference between absolute and local extrema?

A local maximum is the largest value near a point; an absolute maximum is the largest on the whole interval. An absolute extremum inside (a, b) is also a local one, but it may also sit at an end point, where it is not a critical number.

### Can the maximum occur at more than one point?

Yes. x³ − 3x on [−2, 2] reaches its maximum 2 at both x = −1 and x = 2, and its minimum −2 at x = −2 and x = 1. The page lists every point where the value ties with the extreme one (within 10⁻¹² of its size).

### Why does the page refuse a function such as 1/x on [−1, 1]?

The function is not continuous on the interval (it has a pole at 0), so the extreme value theorem does not apply and it has no absolute maximum or minimum there.

### How is this page related to the critical number calculator?

The critical number calculator finds every critical number of f on the whole real line and classifies each as a local maximum or minimum. This page uses the same search on (a, b) and adds the end points to find the absolute extrema.

## Sources

- OpenStax, Calculus Volume 1, section 4.3 Maxima and Minima (retrieved 2026-10-03): https://openstax.org/books/calculus-volume-1/pages/4-3-maxima-and-minima