Absolute and Local Extrema

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Absolute and Local Extrema

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How-to

To find the absolute extrema of a continuous function on a closed interval $[a,b]$ :

Find all critical numbers $c$ of the function $f(x)$ on the open interval $(a,b)$ .

Find the function values $f(c)$ for each critical number $c$ found in step 1.

Evaluate the function at the endpoints. That is, find $f(a)$ and $f(b)$ .

The largest value found in steps 2 and 3 above will be the absolute maximum and the smallest value will be the absolute minimum.

Example: Find the absolute maximum and absolute minimum values of

$f(x) = \displaystyle{\frac{2x}{x^2-2x+2}}$ on the interval

$[0,3]$ .

Solution: The denominator is always positive (its roots are complex), so the function is continuous everywhere and the Extreme Value Theorem applies. Using the quotient rule, the derivative works out to

$f'(x) = \frac{2(-x^2 +2)}{(x^2-2x+2)^2}$

The critical points are at

$x = \pm \sqrt2$ . However, only the critical point

$x=\sqrt2$ is in our interval. So the three points we need to consider are the endpoints

$x=0$ and

$x=3$ , and the critical point

$x=\sqrt2$ .

Since

$\begin{eqnarray*} f(0) & = & 0, \cr f(\sqrt2) & = & 1+\sqrt2, \cr f(3) & = & 6/5, \end{eqnarray*}$ the largest value is

$1+\sqrt2$ , which is the absolute maximum (achieved at

$x=\sqrt2$ ).

This method, and this example, are worked out in detail in the following video. However, beware of a mistake towards the end. The critical number $x=-1$ should not be considered, since it isn't between 0 and 3.