Question

If $f\left(x\right)=\frac{x+1}{x-1},$ show that $f\left[f\right(x\left)\right]=x.$

Answer 1

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

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

$\Rightarrow f\left[f\right(x\left)\right]=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}$

$\Rightarrow f\left[f\right(x\left)\right]=\frac{x+1+x-1}{x+1-x+1}$

$\Rightarrow f\left[f\right(x\left)\right]=\frac{2x}{2}=x$

Hence $f\left[f\right(x\left)\right]=x$