# What are the local extrema, if any, of f (x) =(lnx)^2/x?

Jan 6, 2016

There is a local minimum of $0$ at $1$. (Which is also global.) and a local maximum of $\frac{4}{e} ^ 2$ at ${e}^{2}$.

#### Explanation:

For $f \left(x\right) = {\left(\ln x\right)}^{2} / x$, note first that the domain of $f$ is the positive real numbers, $\left(0 , \infty\right)$.

Then find

$f ' \left(x\right) = \frac{\left[2 \left(\ln x\right) \left(\frac{1}{x}\right)\right] \cdot x - {\left(\ln x\right)}^{2} \left[1\right]}{x} ^ 2$

$= \frac{\ln x \left(2 - \ln x\right)}{x} ^ 2$.

$f '$ is undefined at $x = 0$ which is not in the domain of $f$, so it is not a critical number for $f$.

$f ' \left(x\right) = 0$ where

$\ln x = 0$   or   $2 - \ln x = 0$

$x = 1$   or   $x = {e}^{2}$

Test the intervals $\left(0 , 1\right)$, $\left(1 , {e}^{2}\right)$, and $\left({e}^{2} , \infty\right)$.

(For test numbers, I suggest ${e}^{-} 1 , {e}^{1} , {e}^{3}$ -- recall $1 = {e}^{0}$ and ${e}^{x}$ is increasing.)

We find that $f '$ changes from negative to positive as we pass $1$, so $f \left(1\right) = 0$ is a local minimum,

and that $f '$ changes from positive to negative as we pass ${e}^{2}$, so $f \left({e}^{2}\right) = \frac{4}{e} ^ 2$ is a local maximum.