Let $f$ be strictly increasing and such that $f(x)+f^{-1}(x)+1=e^x$. Is it true that $f$ has at most one fixed point?
I am told the answer is yes, but I am having trouble proving it. It's obvious that it must have at most two, but why can it not have two?
If $x$ is a fixed point, then
$$x + x + 1 = e^x,$$
and that equation has only two solutions (in $\mathbb{R}$), one of them is $0$.
Note that since $f^{-1}$ is everywhere defined, $f$ must be surjective, and hence continuous.
Suppose $0$ were a fixed point. Then for small $x > 0$, you have either $0 < f(x) < x$ or $x < f(x)$. Switching the roles of $f$ and $f^{-1}$ if necessary, let's assume that $x < f(x)$ for $0 < x < \varepsilon$.
Then the functional equation implies
$$x < f(x) < e^x - 1$$
for $0 < x < \varepsilon$. Squeezing shows that $f$ has a right derivative in $0$, and
$$D_+ f(0) = \lim_{x\downarrow 0} \frac{f(x)-f(0)}{x-0} = 1.$$
But then $f^{-1}$ also has a right derivative in $0$, and
$$D_+ f^{-1}(0) = \frac{1}{D_+ f(0)} = 1$$
too. But that means the right derivative of $g\colon x\mapsto f(x) + f^{-1}(x)$ in $0$ is $D_+ f(0) + D_+ f^{-1}(0) = 2$, which contradicts $g(x) = e^x - 1$, from which we obtain
$$D_+ g(0) = g'(0) = e^0 = 1.$$
So $0$ cannot be a fixed point of $f$.