Number of solutions of the equation $e^{f(x)}=f(x)+2$

Question

Let $f$ be an everywhere differentiable function, and suppose that $f(x)=0$ has a unique solution, and suppose that $f$ has no local extreme points.

What is the number of solutions of the equation $$ e^{f(x)}=f(x)+2. $$ Thanks!

Answer 1

As was noted in the comments, $x$ satisfies $e^{f(x)}=f(x)+2$ if and only if $f(x)=-W\left(-\frac{1}{e^2}\right)-2$ or $f(x)=-W_{-1}\left(-\frac{1}{e^2}\right)-2$, where $W$ and $W_{-1}$ are two branches of the Lambert $W$ function. Importantly, one of these values is positive and one of them is negative (they are about $1.4619$ and $-1.84141$ according to Wolfram Alpha).

So the question becomes: how many solutions can there be to $f(x)=1.4619$ and $f(x)=-1.84141$. Now use the fact that $f$ is differentiable everywhere and has no local extrema to prove that either $f$ is strictly increasing or $f$ is strictly decreasing. Now show that a strictly increasing or strictly decreasing function with a unique solution to $f(x)=0$ can take on any given $y$-value at most once. This proves that there are at most two solutions. You can easily come up with examples of $f(x)$ that take on both values, only one value, or neither value.