The extreme value theorem states that a continuous real-valued function defined over a closed interval attains its minimum and maximum at some points in the closed interval. In other words, if \(f\) is a real-valued function continuous on a closed interval \([a,b]\), then by the extreme value theorem, there are two points \(l\) and \(u\) in the interval \([a,b]\) such that for all \(x \in [a,b]\), we have $$f(l) \leq f(x) \leq f(u).$$

The extreme value theorem is more specific than the boundedness theorem (see the post on the Bolzano-Weierstrass theorem) which only states that any real-valued continuous function over a closed interval is bounded in that interval.

#### A proof of the extreme value theorem

By definition, the maximum and minimum of a function, known generically as extremum, are the largest and smallest value taken by the function, within a given set (usually an interval or the whole domain of the function).

**Result**. A real-valued continuous function \(f\) on \([a,b]\) achieves its extrema (i.e., its minima and maxima).

The proof is as follows:

By the boundedness theorem – see Bolzano-Weierstrass theorem – the range of \(f\) over the interval \([a,b]\), i.e., the following set $$\{f(x): a \leq x \leq b\}$$ is bounded. Set $$M = \sup \{f(x) : x \in [a,b]\}.$$ Suppose \(f\) never achieves the value \(M\) and define $$g(x) = M – f(x).$$ It is obvious that \(g\) is positive over \([a,b]\), and so, the function \(1/g\) is continuous. Therefore, by the boundedness theorem, \(1/g\) is bounded above by a positive real number \(N\). This implies that $$M-f(x) > 1/N,$$ and so, $$f(x) < M – 1/N,$$ contradicting the definition of \(M\). Thus we have proved that \(f\) achieves its maxima. In a similar way, one may prove that \(f\) achieves its minima and the proof is complete.

#### Some examples related to the extreme value theorem

In the following, we give some examples related to the extreme value theorem

- The function \(f(x) = \sin x^3\) is continuous over the closed interval \([-\pi, \pi]\). So, \(f\) must attain its extrema by the extreme value theorem. On the other hand, from elementary trigonometry we know that no matter what \(x \in \mathbb{R}\) is, we have $$ -1 \leq \sin x^3 \leq 1.$$ Now, observe that $$f(\sqrt[3]{\pi/2}) = 1$$ and $$f(-\sqrt[3]{\pi/2}) = -1.$$
- The function \(g(x) = 2-x\) is continuous and bounded over the interval \((0,2]\) but it does not attain its least upper bound.
- The function \(h(x) = 1/x\) is continuous over the interval \((0,2]\) but it is not even bounded.

#### How to find the extreme values of functions

For a continuous function defined in a closed interval, the extreme value theorem guarantees it will always reach some absolute maximum value. It will also attain some absolute minimum value at least once. However, it does not tell us exactly where these extremes occur, or what their actual values are.

We therefore need stronger results to show how to find the extreme values of continuous functions. For example, the first derivative test is an effective result that helps us to find the extreme values of differentiable functions.

**Remark**. See Wiktionary, for a lexical discussion of the word extremum.