For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. There are a couple of key points to note about the statement of this theorem. Typically, it is proved in a course on real analysis. The proof of the extreme value theorem is beyond the scope of this text. If f f is a continuous function over the closed, bounded interval, , then there is a point in at which f f has an absolute maximum over and there is a point in at which f f has an absolute minimum over. We say that f ( x ) = x 2 + 1 f ( x ) = x 2 + 1 does not have an absolute maximum (see the following figure). We say that 1 is the absolute minimum of f ( x ) = x 2 + 1 f ( x ) = x 2 + 1 and it occurs at x = 0. However, since x 2 + 1 ≥ 1 x 2 + 1 ≥ 1 for all real numbers x x and x 2 + 1 = 1 x 2 + 1 = 1 when x = 0, x = 0, the function has a smallest value, 1, when x = 0. Therefore, the function does not have a largest value. Absolute ExtremaĬonsider the function f ( x ) = x 2 + 1 f ( x ) = x 2 + 1 over the interval ( − ∞, ∞ ). In this section, we look at how to use derivatives to find the largest and smallest values for a function. Finding the maximum and minimum values of a function also has practical significance because we can use this method to solve optimization problems, such as maximizing profit, minimizing the amount of material used in manufacturing an aluminum can, or finding the maximum height a rocket can reach. This information is important in creating accurate graphs. Given a particular function, we are often interested in determining the largest and smallest values of the function.
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