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

Jul 21, 2017

${f}_{\min} = f \left(1\right) = 0$
${f}_{\max} = f \left({e}^{- 2}\right) \approx 0.541$

#### Explanation:

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

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

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

Applying the product rule

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

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

For local maxima or minima: $f ' \left(x\right) = 0$

Let $z = \ln x$

$\therefore {z}^{2} + 2 z = 0$

$z \left(z + 2\right) = 0 \to z = 0 \mathmr{and} z = - 2$

Hence for local maximum or minimum:

$\ln x = 0 \mathmr{and} \ln x = - 2$

$\therefore x = 1 \mathmr{and} x = {e}^{-} 2 \approx 0.135$

Now examine the graph of $x {\left(\ln x\right)}^{2}$ below.

graph{x(lnx)^2 [-2.566, 5.23, -1.028, 2.87]}

We can observe that simplified $f \left(x\right)$ has a local minimum at $x = 1$ and a local maximum at $x \in \left(0 , 0.25\right)$

Hence: ${f}_{\min} = f \left(1\right) = 0$ and ${f}_{\max} = f \left({e}^{- 2}\right) \approx 0.541$