The function $f(x) = \frac{x-2}{x+1}; x≠-1$ is increasing when (Where R is a set of real numbers)

$x∈R-\{-1\}$

The correct answer is Option (2) → $x∈R-\{-1\}$

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

$f'(x)=\frac{(x+1)\cdot1-(x-2)\cdot1}{(x+1)^2} =\frac{x+1-x+2}{(x+1)^2} =\frac{3}{(x+1)^2}$

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

$f'(x)=\frac{3}{(x+1)^2}>0$ for all $x\neq -1$.

Therefore, the function is increasing for all real numbers except $x=-1$.

The function is increasing on: $\;R-\{-1\}$.