How to we get the inverse function?

Question

Let $f(x)=\sqrt{\frac{e^x-1}{x}}, x \neq 0$, $f(0)=1$.

It is given that $g=f^2$.

I want to show that the function $f^{-1}:(0,+\infty) \to \mathbb{R}$ is a function and that the equation $f(x)=f^{-1}(x)$ has at least 2 solutions, $x_2>x_1>1$.

Also, show that $x_1, x_2$ are the only solutions of the above equation and that there is a unique $\xi>1$ such that $g(\xi)+2\xi f(\xi)=e^{\xi}$. (we suppose that $f$ is convex)

According to wolframalpha, the inverse function is the following:

What exactly is $W$ ? And how to we come to the inverse of the function?

EDIT: I have thought the following:

WE have that $f(f^{-1}(x))=x$. Thus we get that $\frac{e^{f^{-1}(x)-1}}{f^{-1}(x)}=x^2$.

Does this help ? Can we find like that the formula of the function $f^{-1}$ ?

Answer 1

Lambert's $W$ is defined as the function $W(x)$ that satisfies the relationship

$$W(x)e^{W(x)}=x\tag1$$

Let $y(x)=f^{-1}(x)$ denote the inverse function of the function

$$f(x)=\begin{cases}\sqrt{\frac{e^x-1}{x}}&,x\ne 0\\\\1&,x=0\end{cases}\tag2$$

Then it is easy to see from $(2)$ that $y(x)$ satisfies the equation

$$e^{y(x)}=x^2\left(y(x)+\frac1{x^2}\right)\tag3$$

Substituting $s(x)=-\left(y(x)+\frac1{x^2}\right)$ into $(3)$ yields

$$s(x)e^{s(x)}=-\frac{e^{-1/x^2}}{x^2}\tag4$$

Comparing $(4)$ with $(1)$ reveals

$$s(x)=W\left(-\frac{e^{-1/x^2}}{x^2}\right)$$

whereupon we find that

$$f^{-1}(x)=-W\left(-\frac{e^{-1/x^2}}{x^2}\right)-\frac1{x^2}$$

which agrees with the result from Wolframalpha that the OP reported.

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Now, let's take a step backward. First, the function $f(x)$ as given in $(2)$ is differentiable with $f'(x)>0$ on $[0,\infty)$. Therefore, $f(x)$ has a unique inverse function for $x\ge 0$.

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Next, let's show that $f(x)=x$ has two roots. Note that $f(x)>x$ for $x=0$.

Then, using $e<3$ it is easy to see that $f(2)=\sqrt{\frac{e^2-1}{2}}<2$. So, $f(x)<x$ at $x=2$.

Finally, since $e^x-1\ge \frac{x^4}{24}$, we find that $f(x)\ge \frac{x^{3/2}}{24}$ and hence $f(x)>x$ for $x>24$.

Putting this together, we see that $f(x)=x$ for two distinct values ox $x$, $x_1$ and $x_2$ such that $0<x_1<2$ and $2< x_2<24$. A numerical root search finds that $x_1 \approx 1.545$ and $x_2\approx 4.567$.

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Note that at any fixed point $x_0$ of $f$, $f(x_0)=x_0\implies x_0=f^{-1}(x_0)$. Therefore, $x_1$ and $x_2$ are also solutions of the equation $f(x)=f^{-1}(x)$.

Answer 2

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

The wolframalpha solution is incomplete.

It should be \begin{align} f^{-1}(x) &= \begin{cases} f_0^{-1}(x)= -\Wp(-\tfrac1{x^2}\exp(-\tfrac1{x^2}))- \tfrac1{x^2} ,\quad & x\in(0,1) ,\\ f_1^{-1}(x)= -\Wm(-\tfrac1{x^2}\exp(-\tfrac1{x^2}))- \tfrac1{x^2} ,\quad & x\in[1,\infty) , \end{cases} \end{align}

where $\Wp$ is the principal branch and $\Wm$ is the other real branch of the Lambert $\W$ function.

And indeed the equation $f(x)=f^{-1}(x)$ has two real solutions, which can be fouund numerically as

\begin{align} x_1&\approx 1.5450 ,\\ x_2&\approx 4.5670 . \end{align}

$\endgroup$