How do you find the local maximum and minimum values of  f(x) = 2 x + 3 x ^{ -1 }  using both the First and Second Derivative Tests?

Dec 14, 2016

Find critical numbers:

$f ' \left(x\right) = 2 - \frac{3}{x} ^ 2 = \frac{2 {x}^{2} - 3}{x} ^ 2$

$f ' \left(x\right) = 0$ at $x = \pm \sqrt{\frac{3}{2}}$ and $f ' \left(x\right)$ DNE at $x = 0$.

$0$ is not in the domain of $f$, so the only critical numbers ar $\pm \sqrt{\frac{3}{2}}$

First Derivative Test for Local Extrema

On $\left(- \infty , - \sqrt{\frac{3}{2}}\right)$, $f '$ is positive
On $\left(- \sqrt{\frac{3}{2}} , 0\right)$, $f '$ is negative $\text{ }$ so $f \left(- \sqrt{\frac{3}{2}}\right)$ is a relative maximum.
On $\left(0 , \sqrt{\frac{3}{2}}\right)$, $f '$ is negative
On $\left(\sqrt{\frac{3}{2}} , \infty\right)$, $f '$ is positive $\text{ }$ so $f \left(\sqrt{\frac{3}{2}}\right)$ is a relative minimum.

Second Derivative Test for Local Extrema

$f ' ' \left(x\right) = \frac{6}{x} ^ 3$

$f ' ' \left(- \sqrt{\frac{3}{2}}\right)$ is negative, so $f \left(- \sqrt{\frac{3}{2}}\right)$ is a relative maximum.

$f ' ' \left(- \sqrt{\frac{3}{2}}\right)$ is positive, so $f \left(- \sqrt{\frac{3}{2}}\right)$ is a relative minimum.

Finish by doing arithmetic

Find the values by doing the arithmetic $f \left(- \sqrt{\frac{3}{2}}\right)$ and $f \left(\sqrt{\frac{3}{2}}\right)$. (Actually, we only need to do one of these, because $f$ is an odd function.)