Question

If f is a real function defined by $f\left(x\right)=\frac{x-1}{x+1}$, then prove that
$f\left(2x\right)=\frac{3f\left(x\right)+1}{f\left(x\right)+3}$

Answer 1

$3f\left(x\right)+1=\frac{3(x-1)}{x+1}+1=\frac{4x-2}{x+1}$

$f\left(x\right)+3=\frac{x-1}{x+1}+3=\frac{4x+2}{x+1}$

Dividing we get

$\frac{4x-2}{4x+2}=\frac{2x-1}{2x+1}=f\left(2x\right)$