Question

If $f:R-\left\{3\right\}\to R-\left\{5\right\}$ is a function defined by $f\left(x\right)=\frac{3x-3}{x-5},$ then show that $f$ is one-one and onto and also find $f^{-1}$

Answer 1

Given $f\left(x\right)=\frac{3x-3}{x-5}$

For one-one, we have to show
If $f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2$
$\Rightarrow \frac{3x_1-3}{x_1-5}=\frac{3x_2-3}{x_2-5}\Rightarrow x_1=x_2$
$\therefore f\left(x\right)$ is one-one.

For onto, let $y=\frac{3x-3}{x-5}$
$\Rightarrow x=\frac{5y-3}{y-3}...\left(1\right)$
The above equation is defined for forall $y\in R-\left\{3\right\}$
$\because \forall x\in R-\left\{5\right\},\forall y\in R-\left\{3\right\}$
$\therefore f\left(x\right)$ is onto.

From equation (1) we get
$f^{-1}\left(y\right)=x{f}^{-1}\left(y\right)=\frac{5y-3}{y-3}\therefore f^{-1}\left(x\right)=\frac{5x-3}{x-3}$