Question

Show that the function $x^x$ is minimum at $x=\frac{1}{e}$.

Answer 1

Let $y=x^x$.

Taking natural logarithms on both sides, we get

$lny=ln{x}^x$

$\therefore lny=xlnx$

Differentiating wrt $x$, we have

$\frac{1}{y}\frac{dy}{dx}=\frac{dx}{dx}lnx+x\frac{lnx}{dx}$

$\therefore \frac{1}{y}\frac{dy}{dx}=lnx+x\frac{1}{x}$

$\therefore \frac{1}{y}\frac{dy}{dx}=lnx+1$

$\therefore \frac{dy}{dx}=x^x(1+lnx)$

For minima, the derivative is 0. Hence,

$x^x(1+lnx)=0$

$\therefore (1+lnx)=0\text{or}{x}^x=0$

Also, $x^x$ can not be 0. Hence, only feasible solution is

$(1+lnx)=0$

$\therefore lnx=-1$

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

To ensure that we have a minima, we look at the nature of the function. It is strictly increasing after $x=e^{-1}$ as the function is only defined for $x>0$ and strictly decreasing for $x\in (0,e^{-1})$.

Hence, proved.