8. Given the function $f(x)=\frac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}$. If $g(x)=$ $f^{-1}(-x)$, then $g(x)$ is an increasing function on the interval $(\quad$.
(A) $(-1,+\infty)$ 
(B) $(-1,+\infty)$ is a decreasing function
(C) $(-\infty,-1)$ is an increasing function
(D) $(-\infty,-1)$ is a decreasing function
8.C.

Let $y=\frac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}$. Solving for $x$ yields $x=\ln \frac{y+1}{y-1}$.
Therefore, $f^{-1}(x)=\ln \frac{x+1}{x-1}$.
Thus, $g(x)=f^{-1}(-x)=\ln \frac{-x+1}{-x-1}=\ln \frac{x-1}{1+x}$.
From $\frac{x-1}{1+x}>0$, we get $x1$.
Hence, the domain of $g(x)$ is $(-\infty,-1) \cup(1,+\infty)$.
Options (A) and (B) are incorrect.
By $g(x)=\ln \frac{x-1}{x+1}=\ln \left(1-\frac{2}{x+1}\right)$, on the interval $(-\infty, -1)$, $\frac{2}{x+1}$ decreases as $x$ increases, so $1-\frac{2}{x+1}$ increases as $x$ increases. Since $\ln x$ is an increasing function, $g(x)$ increases as $x$ increases, meaning $g(x)$ is an increasing function on the interval $(-\infty, -1)$.