Use the function $x^{1/x}, x > 0$, to determine the bigger of $e^\pi$ and $\pi^e$.

$e^π>π^e$

The correct answer is Option (1) → $e^π>π^e$

Given $f(x) = x^{1/x}, x > 0$.

Taking logarithm of both sides, we get

$\log f(x)=\frac{1}{x}\log x=\frac{\log x}{x}$, dfff. w.r.t. x, we get

$\frac{1}{f(x)}.f'(x)=\frac{x.\frac{1}{x}-\log x.1}{x^2}⇒f'(x)x^{1/x}.\frac{1-\log x}{x^2},x>0$.

Now $f'(x)=0⇒ 1-\log x=0⇒ \log x = 1⇒x= e$.

Therefore, $x = e$ is the only point where extremum may occur.

When $x <e$, slightly, $f'(x) > 0$ and when $x >e$ slightly $f'(x) < 0$

$⇒ f$ is maximum at $x = e$

$⇒ f(e) > f(x)$ for all $x>0⇒  f(e) > f(\pi)$

$⇒e^{1/e}>\pi^{1/\pi}⇒(e^{1/e})^{\pi e}>(\pi^{1/\pi})^{\pi e}⇒e^π>π^e$