The Extreme Value Theorem

The Extreme Value Theorem

Theorem 1 (The Extreme Value Theorem): If $f$ is a continuous function on the closed interval $[a, b]$, then $f$ contains both an absolute maximum and absolute minimum on $[a, b]$.

Example 1

Find the absolute maximum values of $f(x) = \sin x + \cos x$ on the interval $[0, 2\pi]$.

Note that $\sin x + \cos x$ is continuous on the interval $[0, 2\pi]$, thus there must be an absolute maximum and and absolute minimum on the interval $[0, 2\pi]$. Let's differentiate to find these values:

\begin{align} f(x) = \sin x + \cos x \\ f'(x) = \cos x - \sin x \\ 0 = \cos x - \sin x \\ \sin x = \cos x \\ x = \frac{\pi}{4}, \frac{5 \pi}{4}\\ \\ f(\frac{\pi}{4}) = \sqrt{2} \\ f(\frac{5 \pi}{4}) = - \sqrt{2} \end{align}

Of course, we must also check the endpoints of our domain. $f(0) = 1$, and $f(2\pi) = 1$, neither of which value is greater than $\sqrt{2}$ or less than $-\sqrt{2}$. Thus there is an absolute max at $(\frac{\pi}{4}, \sqrt{2})$ and an absolute minimum at $(\frac{5\pi}{4}, -\sqrt{2})$ on the interval $[0, 2\pi]$.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License