Show that $f:Q\to Q$ defined by $f\left(x\right)=5 x+4$ is a bijection and $\text{ find }{f}^{-1}\text{. }$ 

Given $f\left(x\right)=5 x+4$ 

Let $x_1,x_2\in Q$ 

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

$\Rightarrow 5x_1+4=5x_2+4\Rightarrow 5x_1=5x_2\Rightarrow x_1=x_2$

$\therefore$ ’ f ’ is one-one 

Let $y\in Q\text{ and }f\left(x\right)=y$

$\Rightarrow 5x+4=y \Rightarrow 5x=y-4$

$\Rightarrow x=\frac{y-4}{5}\in Q\text{ then }f\left(x\right)=f\left(\frac{y-4}{5}\right)$

$5\left(\frac{y-4}{5}\right)+4=y-4+4=y$

$\therefore f$ is onto

Hence f  is bijection

To find $f^{-1}:$

Let $f\left(x\right)=y\Rightarrow f^{-1}\left(y\right)=x$

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

$\Rightarrow x=\frac{y-4}{5}\in Q$

$\therefore f^{-1}\left(y\right)=\frac{y-4}{5}$

$\Rightarrow f^{-1}\left(x\right)=\frac{x-4}{5},\forall x\in Q$