Question

Show that $f\left(x\right)=x^x$ is minimum when $x=\frac{1}{e}$.

Answer 1

Let $y=x^x$

$\log y=x\log x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\log x+x\times \frac{1}{x}$

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

$\Rightarrow \frac{d^{2}y}{d{x}^2}=x^x\frac{d}{dx}(\log x+1)+(\log x+1)\frac{d}{dx}\left(x^x\right)$

$=x^x\left(\frac{1}{x}\right)+{(\log x+1)}^2$ where $\frac{d}{dx}\left(x^x\right)=\log x+1$

$=x^{x-1}+{(\log x+1)}^2$

For maximum or minimum value $\frac{dy}{dx}=0$

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

$\therefore x^x=0$ or $\log x+1=0$

$\Rightarrow x=0$ is not possible.

Hence $\log x=-1$

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

$\frac{d^{2}y}{d{x}^2}=x^{x-1}+{(\log x+1)}^2$

$={\frac{1}{e}}^{\frac{1}{e}-1}+{(\log \frac{1}{e}+1)}^2$

$=e+0=e>0$

Hence the function is increasing at $x=\frac{1}{e}$ and has minimum value at $x=\frac{1}{e}$