What are the absolute max and min?

Type a function of x and a closed interval [a, b]. The page checks f at both end points and at every critical number in between, and gives the absolute maximum and minimum and where they occur.

Absolute maximum: 0.25

On [1, 3], -x^2 + 3x - 2 has absolute maximum 0.25 at x = 3/2 and absolute minimum -2 at x = 3.

Maximum at x =: 3/2
Absolute minimum: -2
Minimum at x =: 3
Values checked: f(1) = 0; f(3/2) = 1/4; f(3) = -2
f′(x) =: 3 - 2x

Absolute maximum: 0.25. On [1, 3], -x^2 + 3x - 2 has absolute maximum 0.25 at x = 3/2 and absolute minimum -2 at x = 3.

How to calculate

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

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.

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.

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

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

Function f(x) -x^2 + 3x - 2, From x = a 1, To x = b 3 gives Absolute maximum 0.25, Maximum at x = 3/2, Absolute minimum -2, Minimum at x = 3, Values checked 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
Function f(x) x^2 - 3x^(2/3), From x = a 0, To x = b 2 gives Absolute maximum 0, Maximum at x = 0, Absolute minimum -2, Minimum at x = 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
Function f(x) x^3 - 3x, From x = a -2, To x = b 2 gives Absolute maximum 2, Maximum at x = -1, 2, Absolute minimum -2, Minimum at x = -2, 1.

How it works

For f(x) and a < b (numbers or constants such as pi):

Continuity. f must be a real number at a and at b, and have no break inside (a, b): no point where f jumps, goes to ±∞, or stops being a real number, found by a sign and turn scan of f on a grid of 8,001 points (as on the critical number calculator), and no zero of a denominator in f where f is not a real number (sin(x)/x at 0). Otherwise there is no answer.
Critical numbers. A computer algebra system (nerdamer, open source) finds f′, checked against a numeric difference quotient. The zeros and breaks of f′ strictly between a and b are found by the same grid scan and bisection, with each zero exact where the algebra solves f′(x) = 0 and the solution checks; a point that is a fraction p/q with q ≤ 12 to 10⁻⁹ is written as that fraction. A break of f′ (f′ does not exist) where f is defined is also a critical number. If f′ is 0 everywhere, only the end points are checked.
Values. f is evaluated at a, at each critical number, and at b. At an exact point whose value is within 10⁻⁹ × max(1, |value|) of the largest or the smallest, or within 10⁻¹² of 0 (sin(π) is 1.2 × 10⁻¹⁶ in double precision), the algebra also simplifies f there; that exact value is shown when it equals the decimal value to 10⁻⁹, and its value is then used. Other values show to 10 significant figures.
Answer. The absolute maximum is the largest value and the absolute minimum the smallest. Every point whose value is within 10⁻¹² × max(1, |extreme|) of it is listed.

Angles are in radians and ln is the natural logarithm.

Worked examples by hand

f(x) = −x² + 3x − 2 on [1, 3] (OpenStax Calculus Volume 1, Example 4.13a). f′(x) = −2x + 3 = 0 at x = 3/2. f(1) = 0, f(3/2) = −9/4 + 9/2 − 2 = 1/4, f(3) = −9 + 9 − 2 = −2. Absolute maximum 1/4 at x = 3/2; absolute minimum −2 at x = 3.

f(x) = x² − 3x^(2/3) on [0, 2] (Example 4.13b). f′(x) = 2x − 2x^(−1/3) = 0 at x = 1 (and does not exist at the end point 0). f(0) = 0, f(1) = −2, f(2) = 4 − 3 × 2^(2/3) ≈ −0.762. Absolute maximum 0 at x = 0; absolute minimum −2 at x = 1.

f(x) = x³ − 3x on [−2, 2]. f′(x) = 3x² − 3 = 0 at x = ±1. f(−2) = −2, f(−1) = 2, f(1) = −2, f(2) = 2. Absolute maximum 2 at x = −1 and x = 2; absolute minimum −2 at x = −2 and x = 1.

Other questions people ask

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.