# What are the extrema of  f(x)=x/(x^2+9) on the interval [0,5]?

Dec 18, 2015

Find the critical values of $f \left(x\right)$ on the interval $\left[0 , 5\right]$.

$f ' \left(x\right) = \frac{\left({x}^{2} + 9\right) \frac{d}{\mathrm{dx}} \left[x\right] - x \frac{d}{\mathrm{dx}} \left[{x}^{2} + 9\right]}{{x}^{2} + 9} ^ 2$

$f ' \left(x\right) = \frac{{x}^{2} + 9 - 2 {x}^{2}}{{x}^{2} + 9} ^ 2$

$f ' \left(x\right) = - \frac{{x}^{2} - 9}{{x}^{2} + 9} ^ 2$

$f ' \left(x\right) = 0$ when $x = \pm 3$.
$f ' \left(x\right)$ is never undefined.

To find the extrema, plug in the endpoints of the interval and any critical numbers inside the interval into $f \left(x\right)$, which, in this case, is only $3$.

$f \left(0\right) = 0 \leftarrow \text{absolute minimum}$

$f \left(3\right) = \frac{1}{6} \leftarrow \text{absolute maximum}$

$f \left(5\right) = \frac{5}{36}$

Check a graph:

graph{x/(x^2+9) [-0.02, 5, -0.02, 0.2]}