# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for F(x)= (x^2)(lnx)?

Jan 3, 2017

By analysing $F ' \left(x\right)$ over the domain of $F \left(x\right) : x > 0$
$F \left(x\right)$ has an absolute minimum at $x = \frac{1}{\sqrt{e}}$
$F \left(x\right)$ is decreasing for $0 < x < \frac{1}{\sqrt{e}}$ and increasing for $x > \frac{1}{\sqrt{e}}$

#### Explanation:

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

F'(x) = x^2.!/x + lnx*2x (Product Rule)

$= x + 2 x \cdot \ln x$

For local extrema $F ' \left(x\right) = 0$

Thus: $x + 2 x \ln x = 0$

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

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

Since $F \left(x\right)$ is defined for $x > 0$ our local extrema will be at:

$\ln x = - \frac{1}{2} \to x = {e}^{- \frac{1}{2}} = \frac{1}{\sqrt{e}} \cong 0.60653$

Hence:
$F \left(x\right)$ has an absolute minimum at $x = \frac{1}{\sqrt{e}}$
$F \left(x\right)$ is decreasing for $x < \frac{1}{\sqrt{e}}$ and increasing for $x > \frac{1}{\sqrt{e}}$

This is shown by the graph of $F \left(x\right)$ below:

graph{x^2*lnx [-0.542, 2.877, -0.449, 1.26]}