Question

If $f\left(x\right)=\frac{x-1}{x+1}$, then $f\left[f\right\{f\left(x\right)\left\}\right]$ is equal to

• A. $-x$
• B. $-\frac{1}{x}$
• C. $-f\left(x\right)$
• D. $-\frac{1}{f\left(x\right)}$

Answer 1

The correct option is D $-\frac{1}{f\left(x\right)}$

If $f\left(x\right)=\frac{x-1}{x+1}$, then
$f\left[f\right\{f\left(x\right)\left\}\right]=f\left[f\left(\frac{x-1}{x+1}\right)\right]$
$=f\left[\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}\right]$
$=f\left[\frac{\frac{x-1-x-1}{x+1}}{\frac{x-1+x+1}{x+1}}\right]$
$=f[-\frac{2}{2x}]=f(-\frac{1}{x})$
Now, $f(-\frac{1}{x})=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}$
$=\frac{\frac{-1-x}{x}}{\frac{-1+x}{x}}=-\frac{x+1}{x-1}=-\frac{1}{f\left(x\right)}$
$\therefore f\left[f\right\{f\left(x\right)\left\}\right]=-\frac{1}{f\left(x\right)}$