My task is to find the domain of inverse of : $f(x) = \dfrac{\cos e^x}{1-\cos e^x}$ .
Now, if I am correct the inverse should be : $f^{-1}(x) = \ln\Bigl(\arccos\dfrac {x}{x+1}\Bigr)$ .
If I solve the ln part I get : $\langle -\infty,-1\rangle$
And for the arccos part I get : $\langle -\infty,-1\rangle$ and $\langle -\infty,-1\rangle \cup [ -\frac{1}{2},\infty\rangle $ giving the final result for arccos part: $ [-\frac{1}{2},\infty\rangle$ .
And the final domain does not make sense cause the final intersection would be an empty set. Could you please help me solve my problem.
The function has no global inverse, because it assumes each of its value infinitely many times.
However, we can invert a branch of it, say over an interval of the form $(-\infty,a]$.
We have $\lim_{x\to-\infty}f(x)=\infty$. Moreover, writing the function as $$ f(x)=\frac{1}{1-\cos e^x}-1 $$ we can easily compute its derivative as $$ f'(x)=-\frac{e^x\sin e^x}{(1-\cos e^x)^2} $$ showing that the function is decreasing on the interval $(-\infty,\log\pi]$. The function has a minimum at $\log\pi$ and $$ f(\log\pi)=\frac{\cos\pi}{1-\cos\pi}=-\frac{1}{2} $$ Thus we can invert the function over the interval $(-\infty,\log\pi]$ and the domain of the inverse is $[-1/2,\infty)$.
An analytic expression of this inverse can be obtained as follows: write $$ y=\frac{1}{1-\cos e^x}-1 $$ with $y\ge-1/2$, so $$ \cos e^x=\frac{y}{y+1} $$ For $y\ge-1/2$ we have $-1\le y/(y+1)<1$, so we can use the standard branch of the arccosine and $$ x=\log\arccos\frac{y}{y+1} $$ so the inverse we found can be expressed as $$ g(x)=\log\arccos\frac{x}{x+1} $$
There are other intervals where the function is invertible, for instance $[\log\pi,\log(2\pi))$, $\log(2\pi),\log(3\pi)]$ and so on. Pick your preferred one.