Question

Determine the local extrema of a function.

$f\left(x\right)=x^2\ln \left(0.2x\right),I=(0,4],J=(0,\infty )$

Short answer

On $I$, $f$has a global maximum at $x=0$and a global minimum at $x=\frac{5}{\sqrt{e}}$. On $J$, $f$has no global minimum at$x=\frac{5}{\sqrt{e}}$and it has no global maximum.

Step-by-step solution

Step 1. Given Information.

The function,

$f\left(x\right)=x^2\ln \left(0.2x\right),I=(0,4],J=(0,\infty )$.

Step 2. The critical point.

For critical points, we consider,

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \frac{d}{dx}\left(x^2\ln \right(0.2x\left)\right)=0 \\ {x}^2\times \frac{1}{0.2x}\times 0.2+\ln \left(0.2x\right)2x=0 \\ x+2x\ln \left(0.2x\right)=0 \\ x(1+2\ln \left(0.2x\right))=0 \end{gathered}$$

We consider,

either $x=0$

or, $$\begin{gathered} 1+2\ln \left(0.2x\right)=0 \\ 2\ln \left(0.2x\right)=-1 \\ \ln \left(0.2x\right)=-\frac{1}{2} \\ 0.2x=e^{-\frac{1}{2}} \\ x=\frac{e^{-{\displaystyle \frac{1}{2}}}}{0.2} \\ x=\frac{1}{0.2\sqrt{e}} \\ x=\frac{5}{\sqrt{e}} \end{gathered}$$

Again,

$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(x^2\ln \left(0.2x\right)\right) \\ =\underset{x\to 0}{\lim }\frac{\ln \left(0.2x\right)}{{\displaystyle \frac{1}{x^2}}}\left[\frac{-\infty }{\infty }\right] \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{0.2x}}\times 0.2}{-{\displaystyle \frac{2}{x^3}}} \\ =\underset{x\to 0}{\lim }(-\frac{1}{x}\times \frac{x^3}{2}) \\ =\underset{x\to 0}{\lim }(-\frac{x^2}{2}) \end{gathered}$$

The graph of the function with limit $I=(0,1)$is shown below,

Step 3. The values on I and J.

On I, $f$has a global maximum at $x=0$and a global minimum at $x=\frac{5}{\sqrt{e}}$.

Now,

$$\begin{gathered} \underset{f\left(x\right)\to \infty }{\lim }=\underset{x\to \infty }{\lim }{x}^2\ln \left(0.2x\right) \\ =\infty \end{gathered}$$

The graph of the function for $J=(0,\infty )$is shown below,

On $J$, $f$has no global minimum and no global maximum.