The limit of $x^x$ goes to $1$ as we approach either $0$ or $1$, but I'm interested in the smallest possible value for $x^x$ in this interval. Is there a way to find this number, and is it rational?

up vote 4 down vote accepted

Well, since $\frac{{\rm d}x^x}{{\rm d}x} = x^x (\ln(x)+1)$, we clearly have a critical point at $x = \frac 1e$. This gives rise to the minimum value $ e^{-\frac 1e}$, which is not rational.

  • I'm sure it is irrational, but is there a proof? – Lord Shark the Unknown Jun 1 at 6:58
  • I was wondering about that as well. It should be possible to prove that but I don't have a proof at hand. Maybe someone else can give a pointer here. – Florian Jun 1 at 6:59
  • According to Wolfram, it is unknown. – Elliot G Jun 1 at 7:03
  • @Florian I guess it is open whether this number is rational. – Peter Jun 1 at 9:20

Let $f(x)=\ln x^x=x\ln x$. Then $f'(x)=1+\ln x$ which vanishes at $x=e^{-1}$. At $x=e^{-1}$, $x^x=e^{-1/e}$. That is the minimum of $x^x$ on $[0,1]$.

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