Steps to find absolute extrema

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

  1. Find all critical numbers $c$ of the function $f(x)$ on the open interval $(a,b)$.
  2. Find the function values $f(c)$ for each critical number $c$ found in step 1.

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

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

DO:  Use the steps above to find the absolute maximum and absolute minimum values of $f(x) = \displaystyle{\frac{2x}{x^2-2x+2}}$ on the interval $[0,3]$.  Do not look at the solution (below) until you have attempted this on your own.

Other examples are in this video.


Solution to the example above: 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 values are at $x = \pm \sqrt2$. However, only the critical value $x=\sqrt2$ 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 $x=\sqrt2$).