Process of graphing a function $\arccos(\ln{(\frac{x+1}{x-3})})$

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$$\arccos\left(\log{\left(\frac{x+1}{x-3}\right)}\right)$$

We need to graph this function. I can graph $g(x)=\log{\left(\frac{x+1}{x-3}\right)}$ but I get trouble when I want to graph $\arccos(g(x))$.

What is the easiest thought process for graphing such functions.

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$$f(x)=\arccos\left(\ln{\frac{x+1}{x-3}}\right)$$ domain of $\arccos$ is $[-1,1]$ thus we must have $$-1\le \ln{\frac{x+1}{x-3}}\le 1;\;\frac{x+1}{x-3}> 0$$

$$ x\leq -\frac{3+e}{e-1}\lor x\geq \frac{3 e+1}{e-1}$$

$$\lim_{x\to\infty}\arccos\left(\ln \left(\frac{x+1}{x-3}\right)\right)=\frac{\pi}{2}$$ Horizontal asymptote $y=\frac{\pi}{2}$ $$\lim_{x\to \frac{-3-e}{e-1}}\arccos\left(\log \left(\frac{x+1}{x-3}\right)\right)=\pi$$ $$\lim_{x\to \frac{3 e+1}{e-1}}\arccos\left(\log \left(\frac{x+1}{x-3}\right)\right)=0$$ Derivative

$$y'=\frac{4}{(x-3) (x+1) \sqrt{1-\log ^2\left(\frac{x+1}{x-3}\right)}}$$ $y'>0$ for $x<-1\lor x>3$


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