If f(x)=(x-1)/(x+1) then prove that f(2x)=3f(x)+1/f(x)+3

We have, f (x) = \frac{(x - 1)}{(x + 1)} Now, \frac{3 f (x) + 1}{f (x) + 3} = \frac{\frac{3 (x - 1)}{(x + 1)} + 1}{\frac{(x - 1)}{(x + 1)} + 3} = \frac{\frac{3 (x - 1) + (x + 1)}{(x + 1)}}{\frac{(x - 1) + 3 (x + 1)}{(x + 1)}} = \frac{3 (x - 1) + (x + 1)}{(x - 1) + 3 (x + 1)} = \frac{3 x - 3 + x + 1}{x - 1 + 3 x + 3} = \frac{4 x - 2}{4 x + 2} = \frac{2 (2 x - 1)}{2 (2 x + 1)} = \frac{2 x - 1}{2 x + 1} = f (2 x) Hence, proved .

If f(x)=(x-1)/(x​+1) then prove that f(2x)=3f(x)+1/f(x)+3