###### Example 10.3.1

Find the maximum and minimum values of \(\displaystyle f(x)=\frac{x-1}{x^2+1}\) on the interval \([-2,2]\text{.}\)

Because the denominator of \(f(x)\) is never zero, \(x^2+1 \ne 0\text{,}\) \(f(x)\) is continuous everywhere. Consequently, the Extreme Value Theorem guarantees that \(f\) will have a maximum and minimum value on the closed interval \([-2,2]\text{.}\)

First, we compute \(f'(x)\text{,}\) which involves the quotient rule.

Next, we find critical points. Again, the denominator is nonzero, so \(f'(x)\) will be defined and continuous for every value \(x\text{.}\) We only need to find solutions to \(f'(x)=0\text{,}\) which are solutions to \(-x^2+2x+1=0\text{.}\) The quadratic formula gives

The critical values are \(x=1-\sqrt{2}\approx 0.414\) and \(1+\sqrt{2} \approx 2.414\text{.}\)

If we were to test these critical values as turning points, we would look at the signs of \(f'(x)\) on the intervals formed by the critical points. The sign analysis is summarized with the number line below, showing that \(x=1-\sqrt{2}\) is a local minimum and \(x=1+\sqrt{2}\) is a local maximum.

To find the extreme values on the interval, we really just need to compare the values of \(f(x)\) at the end points of the interval with the critical points that are inside the interval. Because \(x=1+\sqrt{2}\) is outside \([-2,2]\text{,}\) we do not include that point.

We finish by interpreting our results. The maximum value of \(f\) on the interval \([-2,2]\) is \(\frac{1}{5}\text{,}\) occurring at \(x=2\text{.}\) The minimum value of \(f\) on the interval is \(\frac{\sqrt{2}}{4-2\sqrt{2}} \approx -1.207\text{,}\) occurring at \(x=1-\sqrt{2} \approx 0.414\text{.}\) A graph of the function showing these extremes is given below.