Question

The function f(x) = x^x has a stationary point at
(a) x = e (b) x = $\frac{1}{e}$ (c) x = 1 (d) x = $\sqrt{e}$

Answer 1

The values of x for which $f\text{'}\left(x\right)=0$ are called stationary values.

Let $f\left(x\right)=x^x$.

$f\left(x\right)=x^x$

$\Rightarrow \log f\left(x\right)=\log x^x$

$\Rightarrow \log f\left(x\right)=x\log x$

Differentiating both sides with respect to x, we get

$\frac{1}{f\left(x\right)}\times f\text{'}\left(x\right)=x\times \frac{1}{x}+\log x\times 1$

$\Rightarrow f\text{'}\left(x\right)=x^x\left(1+\log x\right)$

For stationary point, we have

$f\text{'}\left(x\right)=0$

$\Rightarrow x^x\left(1+\log x\right)=0$

$\Rightarrow 1+\log x=0\left[x^x>0\right]$

$\Rightarrow \log x=-1$

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

Thus, the function f(x) = x^x has a stationary point at $x=\frac{1}{e}$.

Hence, the correct answer is option (b).