Let $A = R - \{ 3\} ,B = R - \{ 1\}$.
If $f:A \to B$ be defined by $f(x) = \frac{{x - 2}}{{x - 3}}$,$\forall x \in A$.
Then, show that $f$ is bijective.

It is given that,, $A = R - \{ 3\} ,B = R - \{ 1\}$.

$f:A \to B$ is defined by $f(x) = \frac{{x - 2}}{{x - 3}},\forall x \in A$

For injectivity
Let $f\left( {{x_1}} \right) = f\left( {{x_2}} \right) \Rightarrow \frac{{{x_1} - 2}}{{{x_1} - 3}} = \frac{{{x_2} - 2}}{{{x_2} - 3}}$

$\Rightarrow$ $\left( {{x_1} - 2} \right)\left( {{x_2} - 3} \right) = \left( {{x_2} - 2} \right)\left( {{x_1} - 3} \right)$

$\Rightarrow$ ${x_1}{x_2} - 3{x_1} - 2{x_2} + 6 = {x_1}{x_2} - 3{x_2} - 2{x_1} + 6$

$\Rightarrow$ $- 3{x_1} - 2{x_2} = - 3{x_2} - 2{x_1}$

$\Rightarrow$ $- {x_1} = - {x_2} \Rightarrow {x_1} = {x_2}$

So, $f(x)$ is an injective function.

For surjectivity

Let $y = \frac{{x - 2}}{{r - 3}} \Rightarrow x - 2 = xy - 3y$

$\Rightarrow$ $x(1 - y) = 2 - 3y \Rightarrow x = \frac{{2 - 3y}}{{1 - y}}$

$\Rightarrow$ $x = \frac{{3y - 2}}{{y - 1}} \in A,\forall y \in B$

[codomain]

So, $f(x)$ is surjective function.

Hence we can say that, $f(x)$ is a bijective function.