Question

The maximum value of $x^{1/x}$ is

• A. $\frac{1}{e^e}$
• B. $e$
• C. $e^{1/e}$
• D. $\frac{1}{e}$

Answer 1

Answer: (C)

$e^{1/e}$

Let $f\left(x\right)=x^{1/x}$
On taking log on both sides, we get
$\log f\left(x\right)=\frac{1}{x}\log x$
On differentiating we get
$\frac{1}{f\left(x\right)}{f}^{\prime }\left(x\right)=\frac{1-\log x}{x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\left(\frac{1-\log x}{x^2}\right)f\left(x\right)$
For maxima and minima, put $f^{\prime }\left(x\right)=0$
$\frac{1-\log x}{x^2}=0$
$\Rightarrow 1-\log x=0\Rightarrow x=e$
After differentiating, we get
$f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right)\left(\frac{1-\log x}{x^2}\right)-f\left(x\right)\left[\frac{-x-2(1-\log x)x}{x^4}\right]$
$={\left(\frac{1-\log x}{x^2}\right)}^{2}f\left(x\right)-f\left(x\right)\left(\frac{2\log x-3}{x^3}\right)$
$f^{\prime \prime }\left(x\right)=$-ve$<0$ at $x=e$
$\therefore$ $f\left(x\right)$ is maximum at $x=e$ and maximum value of $f\left(x\right)$ is $e^{1/e}$
f(x)=x1/x