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?
Asked
Active
Viewed 149 times
5
User
- 1,671
Monochrome
- 83
-
2If $y=x^x \tag 1$ then $\ln y = x\ln x$, then taking Lambert-$W$ at both sides: $W(\ln y) = W(x\ln x) = \ln x$ and thus $$f(x) = \exp(W(\ln x)) = \frac{\ln x}{W(\ln x)}\tag 2$$ is the inverse of (1). Notice that $W$ has two branches. The (real) inverse is only defined if $x\geqslant e^{-1/e}$. – emacs drives me nuts Aug 16 '22 at 13:30
-
1The derivative of $g$ in terms of $g$. What's that supposed to mean? – emacs drives me nuts Aug 16 '22 at 13:35
-
1I meant that the answer I got was: $\frac{1}{x(1+\ln(g(x))}$ – Monochrome Aug 16 '22 at 13:36
-
For any differentiable function $F$ with inverse $F^{-1}$, $$(F^{-1})'=\frac{1}{F'\circ F^{-1}}$$ – K.defaoite Aug 16 '22 at 13:43
-
Note that $f(x)=x^x$ could also be written as $f(x)={^2x}$ in tetration form. So, its inverse is the "super square root" of $x$. – Geoffrey Trang Aug 16 '22 at 13:51
-
Sorry, but does this answer your question? Inverse of $x^x$ – Тyma Gaidash Aug 16 '22 at 14:38
1 Answers
5
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 $g_2$
leonbloy
- 63,430