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

Nov 3, 2016

$f \left(x\right)$ has local extrema at $x = 0$ and $x = \sqrt{e}$

#### Explanation:

$f \left(x\right) = {x}^{2} / \ln x$

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

$f \left(x\right)$ will have local extrema where $f ' \left(x\right) = 0$

$\frac{\ln x \cdot 2 x - {x}^{2} \cdot \frac{1}{x}}{\ln x} ^ 2 = 0$

$2 x \ln x - x = 0$

$x \left(2 \ln x - 1\right) = 0$

$x = 0$ or $\ln x = \frac{1}{2} \to x = {e}^{\frac{1}{2}} = \sqrt{e}$

Hence $f \left(x\right)$ has local extrema at $x = 0 , \left(0 , 0\right)$ and $x = \sqrt{e} , \left(\sqrt{e} , 2 e\right)$

This can be seen from the graph of $f \left(x\right)$ below:

graph{x^2/lnx [-11.91, 33.68, -8.58, 14.24]}