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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 th...