The function f: R - {-1} → R defined as $f(x)=\frac{x-1}{x+1}$ is increasing in the interval:

(-∞, ∞) - {-1}

Given

$f(x)=\frac{x-1}{x+1}$

Domain $x\ne-1$

Derivative

$f'(x)=\frac{(x+1)-(x-1)}{(x+1)^2}$

$=\frac{x+1-x+1}{(x+1)^2}$

$=\frac{2}{(x+1)^2}$

Since $(x+1)^2>0$ for all $x\ne-1$,

$f'(x)>0$ for all $x\ne-1$

Hence the function is increasing on both intervals separated by $x=-1$.

The function is increasing on $(-\infty,-1)$ and $(-1,\infty)$.