Absolute extrema on a closed interval

Theorem: a function continuous on a closed interval [a, b] has an absolute maximum and an absolute minimum there. Each is reached at a critical point inside or at an end point.

Method: list the critical points in (a, b), add a and b, evaluate f at all of them, and pick the largest and the smallest values.

Worked example: f(x) = x + 2 on [0, 1] has no critical point, so the extremes sit at the ends: maximum 3 at x = 1, minimum 2 at x = 0.

Worked example: f(x) = 2x³ − 15x² + 36x + 1 on [1, 5]. f′ = 6(x − 2)(x − 3). The values f(1) = 24, f(2) = 29, f(3) = 28 and f(5) = 56 give absolute maximum 56 at x = 5 and absolute minimum 24 at x = 1.

In that example x = 2 is a local maximum and x = 3 a local minimum, yet neither is an absolute extreme; the end points win.

Worked example: f(x) = 12x^(4/3) − 6x^(1/3) on [−1, 1]. f′ = 2(8x − 1)/x^(2/3), so the critical points are x = 1/8 (f′ = 0) and x = 0 (f′ undefined). The values f(−1) = 18, f(0) = 0, f(1/8) = −9/4 and f(1) = 6 give maximum 18 and minimum −9/4.

No second derivative test is needed here: comparing the values settles which is largest.