Question

$\text{If}f\left(x\right)=\frac{x-1}{x+1},\text{then find f(f(x)).}$

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

Answer 1

Answer: (D)

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

$\text{Given,}f\left(x\right)=\frac{x-1}{x+1}$
$\therefore f\left(f\right(x\left)\right)=\frac{\left(\frac{x-1}{x+1}\right)-1}{\left(\frac{x-1}{x+1}\right)+1}$
$=\frac{\frac{(x-1)-(x+1)}{x+1}}{\frac{(x-1)+(x+1)}{x+1}}$
$=\frac{-2}{2x}=\frac{-1}{x}$