Analytic Properties of $f(x) = x^x$

calculusderivativesoptimization

The function $f(x) = x^x, x > 0$ can be plotted on graphing software and inspected to see a local minimum around .367. The function is convex, decreasing from 0 to its minimum, and increasing thereafter. The derivative of the function can be found by implicit differentiation to be $f'(x) = x^x(\ln(x)+1)$.

  1. Can the exact value of the local minimum be found?
  2. Can someone explain intuitively why the function first decreases and then increases?
  3. Is there anything interesting about the class of functions $\{ \, f(x) \, | \, f'(x) = f(x) \cdot g(x) \, \}$

Edit: I forgot, I'll be subject to a firing squad if I don't explain what I have tried!

  1. Setting the derivative equal to 0 doesn't do much for me. If this can be solved exactly without optimization algorithms, I suspect something really clever will need to happen.

  2. I tried thinking about what is happing for different types of inputs (irrational $x$, rational $x$, natural $x$). Still didn't get anywhere.

  3. Nothing to try really, does this class of functions come up anywhere in math?

Best Answer

The minimum is attained at $x=\frac1e$ as can be seen by setting the derivative equal to $0$ and solving for $x$.

Your statement in a comment that it is not always true that if $x>y>0$ then $x^x>y^x$ is false. We have $$x>y\implies \log x > \log x\implies x\log x > x\log y \implies x^x>y^x $$

Perhaps you meant to say it is not always true that $x>y>0$ implies $x^x>y^y$. This is true. Taking logarithms, there's no particular reason to believe that $x\log x > y\log y$ and indeed, it isn't always true.

To try to explain it intuitively, note first that we're concerned with numbers $0<x<y<1$. When you raise such a number to a power, the larger the power, the smaller the result. When we compare $x^x$ and $y^y$, for some choices of $x$ and $y$, this effect dominates, and $y^y$ is the smaller of the two numbers. For other choices of $x$ and $y$, the fact that $x$ is smaller dominates.