The maximum value of $(\frac{1}{x})^x$ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $e^{\frac{1}{e}}$
$\frac{1}{x^x}$
$\log y=-x\log x$ differentiating wrt x
$\frac{1}{y}\frac{dy}{dx}=-\log x-1⇒\frac{dy}{dx}=-\frac{(\log x+1)}{x^x}$
so at optimize point
$\frac{dy}{dx}=0⇒\log x=-1$
$⇒x=\frac{1}{e}$
max. value → $y(\frac{1}{e})=e^{\frac{1}{e}}$