Question

1. Show that the function f(x)= | cos 2x| is a periodic function

2. Find the local extreme value of (1/x)^x , if it exists.

Answer 1

1)

Consider the graph of f(x)=cos 2x below

From which the graph of f(x)=|cos 2x| is

Hence, as shown in the figure above the function f(x)=|cos 2x| is a periodic function with period $\frac{π}{2}$

2)

Given f(x)=$(\frac{1}{x})^x$ ,x>0

$ln(f(x))=ln((\frac{1}{x})^x)$

$ln(f(x))=xln(\frac{1}{x})$

$ln(f(x))=-xln(\frac{1}{x})$

Differentiate to get;

$\frac{1}{f(x)}\cdot f'(x)=-(x\cdot\frac{1}{x}+ln\ x)$

$\implies f'(x)=-(\frac{1}{x})^x(1+ln\ x)$

For critical points f'(x)=0

$\implies -(\frac{1}{x})^x(1+ln\ x)=0$

ln x=-1

$x=e^{-1}$

Hence local maximum exists and occurs at $x=e^{-1}$

Value $f(e^{-1})=(\frac{1}{e^{-1}})^{e^{-1}}$

=$e^{e^{-1}}$