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

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!

## closed as off-topic by Martin R, DMcMor, Nosrati, user10354138, SilJun 21 at 17:01

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• It is $$f(x)=-W\left(-\frac{1}{e^2}\right)-2$$ – Dr. Sonnhard Graubner Jun 21 at 15:36
• Yes, but I asked about the number of solutions, not the solutions them self – boaz Jun 21 at 15:44
• @Dr.SonnhardGraubner That is not correct, as there are not necessarily values of $x$ satisfying the two equations you wrote. There are either zero, one, or two solutions depending on the function $f(x)$. – kccu Jun 21 at 15:49
• @AjayMishra I'm not sure what you're getting at. If $f(x)=x$, then there are two solutions to $e^{f(x)}=f(x)+2$, i.e., $e^x=x+2$. But I already stated in my comment that one solution is possible for certain choices of $f(x)$, for instance $f(x)=e^x-1$. – kccu Jun 21 at 16:15
• Is $e^{f(x)}=f(x)+2$ supposed to hold for all $x$? Is $f(x)$ real or complex function? – Sil Jun 21 at 16:35

## 1 Answer

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.