Question

If $f:R-\left\{3\right\}\to R$ is defined by $f\left(x\right)=\frac{x+3}{x-3}$, show that $f\left[\frac{3x+3}{x-1}\right]=x$ for $x\ne 1$.

Answer 1

Given that:

$f\left(x\right)=\frac{x+3}{x-3}$ where, $f:R\left\{3\right\}\to R$

To Prove: $f\left[\frac{3x+3}{x-1}\right]=x$ for $x\ne 1$

Solution:

$f\left[\frac{3x+3}{x-1}\right]=\frac{\frac{3x+3}{x-1}+3}{\frac{3x+3}{x-1}-3}$

$=\frac{3x+3+3x-3}{3x+3-3x+3}$

$=\frac{6x}{6}$

$=x$

Hence, $f\left[\frac{3x+3}{x-1}\right]=x$