Question

Let $A=R-\left\{3\right\}$ and $B=R-\left\{1\right\}$. consider the function $f:A$$\to$ $B$ defined by $f\left(x\right)=\left(\frac{x-2}{x-3}\right)$. Show that f is one-one and onto and hence find $f^{-1}$.

Answer 1

$A=R-\left\{3\right\}$

$B=R-\left\{1\right\}$

$f:A\to B$

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

$f\left(x_1\right)=f\left(x_2\right)$

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

$(x_2-3)(x_1-2)=(x_2-2)(x_1-3)$

$x_1{x}_2-3x_1-2x_2+6=x_1{x}_2-3x_2-2x_1+6$

$-3x_1-2x_2=-3x_2-2x_1$

$-x_1=-x_2$

$x_1=x_2$

So, $f\left(x\right)$ is one-one

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

$y=\frac{x-2}{x-3}$

$y(x-3)=x-2$

$yx-3y=x-2$

$yx-x=3y-2$

$x(y-1)=3y-2$

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

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

$=\frac{\frac{3y-2}{y-1}-2}{\frac{3y-2}{y-1}-3}$

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

$=\frac{3y-2-2y+2}{3y-2-3y+3}$

$=\frac{3y-2y}{-2+3}$

$=y$

$f\left(x\right)=y$

$f\left(x\right)$ is onto.

So $f\left(x\right)$ is bijective and invertible

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

$y=\frac{x-2}{x-3}$

$x=\frac{y-2}{y-3}$

$x(y-3)=y-2$

$xy-3x=y-2$

$xy-y=3x-2$

$y(x-1)=3x-2$

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

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