Let $f(x) = \frac{x+1}{x-1}$. Then for $x^2 \neq 1, f(-x)$ is
$\textbf{(A)}\ \frac{1}{f(x)}\qquad \textbf{(B)}\ -f(x)\qquad \textbf{(C)}\ \frac{1}{f(-x)}\qquad \textbf{(D)}\ -f(-x)\qquad \textbf{(E)}\ f(x)$
We find $f(-x) = \frac{-x+1}{-x-1} = \frac{x-1}{x+1} = \frac{1}{f(x)}$, so the answer is $\boxed{\textbf{(A)}}$.