Question

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

• A. ${\left(1/e\right)}^e$
• B. $e^{1/e}$
• C. 1
• D. none of these

Answer 1

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

$y=(\frac{1}{x}{)}^x$
Or
$lny=-xln\left(x\right)$
Differentiating y with respect to x, we get
$y^{\prime }\frac{1}{y}=-\left[ln\right(x)+1]$
Or
$y^{\prime }=-\left(\frac{1}{x}{)}^x\right[ln\left(x\right)+1]$
Now for the critical points
$-\left(\frac{1}{x}{)}^x\right[ln\left(x\right)+1]=0$
Since
$(\frac{1}{x}{)}^x$ cannot be equal to 0, we get
$ln\left(x\right)+1=0$ or
$x=e^{-1}$
Hence
$f\left(e^{-1}\right)$
$=e^{\frac{1}{e}}$