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

The correct option is C $e^{1/e}$

Let $y={\left(\frac{1}{x}\right)}^x=(x^{-1}{)}^x=x^{-x}$

Take natural log both sides,

$\ln y=\ln x^{-x}=-x\ln x$

Differentiate both sides w.r.t. x

$\frac{1}{y}\frac{dy}{dx}-\ln x-1$, use product rule

$\Rightarrow \frac{dy}{dx}=-y(\ln x+1)=-x^{-x}(1+\ln x)$

For max/min of y, substitute $\frac{dy}{dx}=0$

$\Rightarrow \ln x+1=0$

$\Rightarrow \ln x=-1\Rightarrow x=e^{-1}=\frac{1}{e}$

Therefore maximum value of y will be $y\left(1/e\right)=e^{1/e}$