Q. 82 - Calculus

Question

determine the global extrema of  function $f\left(x\right)=x^2\ln \left(0·2x\right) , I=\left(0, 4\right) , J=\left(0,\infty \right)$

Step-by-Step Solution

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 has no global maximum 

Step 1. Given information

The given function is $f\left(x\right)=x^2\ln \left(0·2x\right)$

Step 2. Calculate global maximum and minimum

$$\begin{gathered} f\left(x\right)=x^2\ln \left(0·2x\right) \\ f\text{'}\left(x\right)=0 \\ \frac{d\left(x^2\ln \left(0·2x\right)\right)}{dx}=0 \\ {x}^2·\frac{1}{0·2x}\times 0·2+\ln \left(0·2x\right)\times 2x=0 \\ x\left(1+2\ln \left(0·2x\right)\right)=0 \\ x=0; x=\frac{e^{{\displaystyle \frac{-1}{2}}}}{0·2} \\ x=0; x=\frac{5}{\sqrt{e}} \\ Again, \\ \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }{x}^2\ln \left(0·2x\right) \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{0·2x}}\times 0·2}{{\displaystyle \frac{-2}{x^3}}} \\ =\underset{x\to 0}{\lim }\left(\frac{-1}{x}·\frac{x^3}{2}\right) \\ =\underset{x\to 0}{\lim }\frac{-x^2}{2} \\ Again \\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }{x}^2\ln \left(0·2x\right) \\ =\infty \end{gathered}$$

Step 3. The solution

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 has no global maximum