Consider the function $f(x)=x^{\frac{1}{x}}$. Its

maximum value is $e^{\frac{1}{e}}$

The correct answer is Option (2) → maximum value is $e^{\frac{1}{e}}$

$f(x)=x^{\frac{1}{x}}⇒x∈(0,∞)$

let,

$y=x^{\frac{1}{x}}⇒\log y=\frac{1}{x}\log x$

$⇒g'(x)=\frac{1-\log x}{x^2}$

and,

$g'(x)=0$

$⇒1-\log x=0$

$⇒x=e$ and, $g''(x)<0$

∴ $f(x=e)=e^{\frac{1}{e}}$ is maximum value of f(x).