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.