Question

$f\left(x\right)=x^x$ has a stationary point at

• A. $x=e$
• B. $x=\frac{1}{e}$
• C. $x=\sqrt{e}$
• D. $x=1$

Answer 1

The correct option is B $x=\frac{1}{e}$

Given : $f\left(x\right)=x^x$
Let $y=x^x$
$\Rightarrow \log y=x\log x$
Differentiating both sides w.r.t $x$ , we get
$\Rightarrow \frac{1}{y}.\frac{dy}{dx}=x.\frac{1}{x}+\log x$
$\Rightarrow \frac{dy}{dx}=x^x[1+\log x]$
Now,
$\frac{dy}{dx}=0$
$\Rightarrow x^x(1+\log x)=0$
$\Rightarrow \log x=-1$
$\Rightarrow x=e^{-1}=\frac{1}{e}$
$f\left(x\right)=x^x$ has a stationary point at $x=\frac{1}{e}$
Hence, option B is correct.