Question

If $f\left(x\right)=\frac{x-1}{x+1},x\ne -1$ then show that $f\left(f\right(x\left)\right)=\frac{-1}{x}$, where $x\ne 0)$.

Answer 1

We have, $f\left(x\right)=\frac{x-1}{x+1}$, where $x\ne -1$.

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

$=\frac{\left\{\right(x-1)-(x+1\left)\right\}}{(x+1)}\times \frac{(x+1)}{\left\{\right(x-1)+(x+1\left)\right\}}=\frac{-2}{2x}=\frac{-1}{x}$.

Hence, $f\left\{f\right(x\left)\right\}=\frac{-1}{x}$, where $x\ne 0$.