Question

The maximum value of $\left(\frac{1}{x}\right)^{x}$ is

• A. $e$
• B. $e^e$
• C. $e^{\frac{1}{e}}$
• D. $\left(\frac{1}{e}\right)^{\frac{1}{e}}$

Answer 1

Answer: (C)

Let $y = \left(\frac{1}{x}\right)^{x}$
$\Rightarrow log \,y = x\,log \left(\frac{1}{x}\right)$
$\Rightarrow log \,y = -x\,log\,x$
Differentiating w.r.t. $x$., we get
$\Rightarrow \frac{1}{y} \frac{dy}{dx} = \left(1+log\,x\right)$
$\Rightarrow \frac{dy}{dx} = -y\left(1+log\,x\right)$
$\Rightarrow \frac{d^{2}y}{dx^{2}} = - \frac{dy}{dx}\left(1+log\,x\right) - \frac{y}{x}$
$\Rightarrow \frac{d^{2}y}{dx^{2}} =y\left(1+log\,x\right)^{2} - \frac{y}{x}$
$\Rightarrow \frac{d^{2}y}{dx^{2}} = \left(\frac{1}{x}\right)^{x} \left(1+log\,x\right)^{2} -\frac{1}{x^{\left(x+1\right)}}$
For maximum value, $\frac{dy}{dx} = 0$
$-y\left(1+log\,x\right) = 0$
$\Rightarrow 1+log\,x = 0\left(\because y \ne 0\right) $
$\Rightarrow log\,x = -1$
$\Rightarrow x = e^{-1}$
$\Rightarrow x= 1/e$
Also $\left[\frac{d^{2}y}{dx^{2}}\right]_{x = \frac{1}{e}} = e^{1/e}\left(1+log \frac{1}{e}\right)^{2} - e ^{\left(1/e+1\right)}$
$= e^{1/e} \left(1-log\,e\right)^{2} - e^{1/e+1}$
$= - e ^{1/e+1} < 0$
So, $x = 1/e$ is a point of local maxima.
Hence, local maximum value $y = \left(e\right)^{1/e}$.