Question

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

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

Answer 1

Answer: (C)

$e^{1/e}$

Answer 2

Let $y=\left(\frac{1}{x}\right)^{x}$
$\Rightarrow y=x^{-x}$
$\therefore \frac{d y}{d x}=x^{-x}(-1-\log x)$
$\Rightarrow \frac{d y}{d x}=-x^{-x}(1+\log x)$
$\left[\because \frac{d}{d x} f(x)^{d(x)}=f(x)^{g(x)}\right.$
\left.\left\\{g(x) \cdot \frac{1}{f(x)} \cdot f'(x)+g'(x) \log f(x)\right\\}\right]
For maxima,
$\frac{dy}{dx} = 0$
$\Rightarrow 1 + \log\,x = 0\,\,[\because x^{-x} \ne0 ]$
$\Rightarrow \log\,x = -1$
$\Rightarrow x = e^{-1}$
Hence, the maximum value of $\left(\frac{1}{x}\right)^x$ is $\left(\frac{1}{e^{-1}}\right)^{e^{-1}}$
i.e., $(e)^{1/e}$ at $x = e^{-1}$