# What are the extrema and saddle points of f(x)=2x^2 lnx?

Mar 21, 2018

The domain of definition of:

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

is the interval $x \in \left(0 , + \infty\right)$.

Evaluate the first and second derivatives of the function:

$\frac{\mathrm{df}}{\mathrm{dx}} = 4 x \ln x + 2 {x}^{2} / x = 2 x \left(1 + 2 \ln x\right)$

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = 2 \left(1 + 2 \ln x\right) + 2 x \cdot \frac{2}{x} = 2 + 4 \ln x + 4 = 6 + \ln x$

The critical points are the solutions of:

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

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

and as $x > 0$:

$1 + 2 \ln x = 0$

$\ln x = - \frac{1}{2}$

$x = \frac{1}{\sqrt{e}}$

In this point:

$f ' ' \left(\frac{1}{\sqrt{e}}\right) = 6 - \frac{1}{2} = \frac{11}{2} > 0$

so the critical point is a local minimum.

The saddle points are the solutions of:

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

$6 + \ln x = 0$

$\ln x = - 6$

$x = \frac{1}{e} ^ 6$

and as $f ' ' \left(x\right)$ is monotone increasing we can conclude that $f \left(x\right)$ is concave down for $x < \frac{1}{e} ^ 6$ and concave up for $x > \frac{1}{e} ^ 6$

graph{2x^2lnx [-0.2943, 0.9557, -0.4625, 0.1625]}