Two functions f(x)=x^x and f(x)=x^(-x)

Question

I was playing with desmos. Then I found a pair of interesting graphs, namely $f(x)=x^x$(the purple one) and $f(x)=x^{-x}$.(the green one) They both have their global maximum/minimum at a point about $x=0.368$, which I found out is very close to $\frac1e$. Is there some kind of explanation about why this is happening?

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Answer 1

$$f_1(x)=\frac 1 {f_2(x)}$$ so you should expect that where one achieves its minimum, the other one achieves its maximum.

A closer inspection shows that $$\ln f_1(x)=x\ln x=-f_2(x)$$ and $$\frac{\partial (\ln f_1)}{\partial x}(x)=\ln(x)-1$$ So the minimum/maximum are indeed achieved exactly at $\frac 1 e$.

Answer 2

That minimum of $x^x$ happens at where it’s derivaive is zero. $\frac{dx^x}{dx}=x^x(log(x)+1)=0$ gives $x=1/e$ for positive x. Similar argument can be made for $x^{-x}$.

Answer 3

Let's say $y = x^x$. It would have extrema when $\frac{dy}{dx} = 0$.

To differentiate this function, first take $ln()$ on both sides. We get $ln(y) = xln(x)$. Now differentiate w.r.t x. $\frac{dy/dx}{y} = 1 + ln(x)$. So if $dy/dx = 0$, then $ln(x) = -1$ which means that there is an extrema at $\frac{1}{e}$ whose nature can be determined by the second derivative test.

So it's no surprise that you see a global maxima/minima at $\frac{1}{e}$.