A topic in the Open Knowledge Graph — a free, open map of 15,495 topics and the order to learn them in.
An absolute maximum of f on an interval is the largest value f takes anywhere on that interval, and an absolute minimum is the smallest. The Extreme Value Theorem guarantees that a function continuous on a closed interval [a, b] has both. Fermat's theorem says that an interior extremum can only occur where f'(x) = 0 or f'(x) does not exist. So the closed interval method is: find the critical numbers inside (a, b), evaluate f at them and at the endpoints a and b, and pick the largest and smallest values.
Sketch graphs on closed intervals and mark where the highest and lowest points occur, so students see that the answer can be at an endpoint or at a critical number. Show counterexamples that break each hypothesis of the Extreme Value Theorem (an open interval, a discontinuity). Then practice the method with a table of candidate x-values and their f-values.
The first and second derivative tests find local extrema: points that are higher or lower than their immediate neighbors. Many problems ask a different question. What is the very largest or smallest value a function takes over a whole interval? These are the absolute, or global, extrema.
The first issue is whether they exist at all. On the open interval (0, 1), the function f(x) = x gets as close to 1 as you like but never reaches it, so it has no largest value. On [-1, 1], a function with a jump can also miss its highest point. The Extreme Value Theorem rules out these problems: if f is continuous on a closed interval [a, b], then f attains an absolute maximum and an absolute minimum somewhere in [a, b]. Both hypotheses matter. The theorem says the extremes exist; it does not say where they are.
To locate them, use Fermat's theorem: if f has a local extremum at an interior point c and f'(c) exists, then f'(c) = 0. An absolute extremum at an interior point is also a local one, so it must occur at a critical number, where f' is zero or undefined. The only other places to look are the endpoints, which Fermat's theorem does not cover. This gives the closed interval method.
1. Check that f is continuous on [a, b].
2. Find the critical numbers of f in (a, b).
3. Evaluate f at each critical number and at a and b.
4. The largest value is the absolute maximum; the smallest is the absolute minimum.
Example: f(x) = x³ - 3x on [0, 2]. Here f'(x) = 3x² - 3, which is zero at x = 1 and x = -1. Only x = 1 is in the interval. Evaluate: f(0) = 0, f(1) = 1 - 3 = -2 and f(2) = 8 - 6 = 2. The absolute maximum is 2, at x = 2, and the absolute minimum is -2, at x = 1. Notice that the maximum sits at an endpoint where the derivative is not zero.
The method does not require classifying critical numbers as maxima or minima. You compare the f-values directly. That makes it the backbone of optimization problems in which the variable is restricted to a closed interval, such as the side length of a box cut from a fixed sheet.