Question:
If $f(x)=\frac{x-1}{x+1}$, then $f(x) f\left(-\frac{1}{x}\right)$ is equal to _________ .
Solution:
If $f(x)=\frac{x-1}{x+1}$
$f\left(\frac{-1}{x}\right)=\frac{\frac{-1}{x}-1}{\frac{-1}{x}+1}$
$=\frac{\frac{(-1-x)}{x}}{\frac{(-1+x)}{x}}$
$=-\frac{(1+x)}{x-1}$
$f\left(\frac{-1}{x}\right)=\frac{1+x}{1-x}$
$\therefore f(x) f\left(\frac{-1}{x}\right)=\frac{x-1}{x+1} \times \frac{1+x}{1-x}$
$\therefore f(x) f\left(\frac{-1}{x}\right)=-1$