Inverse of $x^x$ in $(0, \infty)$

Question

A question I came across read something along the lines of:
"$f(x) = x^x$, defined on $x \in (0, \infty)$ and $g(x)$ is the inverse of $f$. Find the derivative of $g$, in terms of $g$."
I did get the right answer, according to the solutions; but then, I tried graphing $f$, and to me, it seems to be a non-injective function. I have read, however, that a function must be bijective for it to be invertible. Am I missing something? Can $f$ really have an inverse in the given range?

Answer 1

You are right, the function $f(x)=x^x$ is not bijective in $x \in (0, \infty)$, it has a minimum at $x_0 = 1/e \approx 0.37$. Hence, to invert it, you should restrict to one of the two "branches": $(0,x_0)$ and $(x_0, \infty)$.

Still, the derivative you found is valid for the inverse of each branch.

For a simpler example: $f(x)=x^2$ is not bijective on $(-\infty, \infty)$, it has two inverses for its two branches: $g_1(y) = -\sqrt{y}$ and $g_2(y) = \sqrt{y}$

The rule of inverse derivative $g' = \frac{1}{f'}$ gives

$$g' = \frac{1}{2x}= \frac{1}{2g} \tag{1}$$

since $g(y)=x$. And you can verify that $(1)$ is indeed valid for both $g_1 $ and $g_2$