Question

Use the function $f\left(x\right)=x^{1/x},x>0$, determine the bigger of the two numbers $e^{\pi }$ and ${\pi }^e.$

Answer 1

We have $f\left(x\right)=x^{1/x}$
$\therefore \log f\left(x\right)=\frac{1}{x}{\log }_{e}x,$
Differentiating with
respect to x, we get
$\frac{1}{f\left(x\right)}.f^{\prime }\left(x\right)=\frac{1}{x}.\frac{1}{x}-\frac{1}{x^2}{\log }_{e}x$
or $f^{\prime }\left(x\right)=\frac{x^{1/x}}{x^2}[1-{\log }_{e}x]$ ...(1)
We have to
consider the function for $x=e$ to $x=\pi ,\pi >e$
$\therefore$ for $e<x<\pi ,{\log }_{e}x>1$ as $x>e.$
Then it follows from
(1) that $f^{\prime }\left(x\right)<0$ for
$x>e.$
Hence $f\left(x\right)$ is a decreasing function of x for
$x>e.$ Since $\pi >e$,
we
conclude $f\left(\pi \right)<f\left(e\right)\Rightarrow {\pi }^{1/\pi }<e^{1/e}$
$\Rightarrow {\left({\pi }^{1/\pi }\right)}^{e\pi }<{\left(e^{1/e}\right)}^{e\pi }\Rightarrow {\pi }^e<e^{\pi }.$
Thus $e^{\pi }$ is bigger that ${\pi }^e.$