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, Sil Jun 21 at 17:01

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  • $\begingroup$ It is $$f(x)=-W\left(-\frac{1}{e^2}\right)-2$$ $\endgroup$ – Dr. Sonnhard Graubner Jun 21 at 15:36
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    $\begingroup$ Yes, but I asked about the number of solutions, not the solutions them self $\endgroup$ – boaz Jun 21 at 15:44
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    $\begingroup$ @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 $f(x)$. $\endgroup$ – kccu Jun 21 at 15:49
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    $\begingroup$ @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 $f(x)=e^x-1$. $\endgroup$ – kccu Jun 21 at 16:15
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    $\begingroup$ Is $e^{f(x)}=f(x)+2$ supposed to hold for all $x$? Is $f(x)$ real or complex function? $\endgroup$ – Sil Jun 21 at 16:35

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.


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