$f : R \rightarrow R$ is given by, $f ( x )=4 x +3$

For one – one

Let $f(x)=f(y)$

$\Rightarrow 4 x+3=4 y+3$

$\Rightarrow 4 x=4 y$

$\Rightarrow x=y$

$\therefore f$ is a one – one function

For onto

For $y \in R,$ let $y=4 x+3$

$\Rightarrow x=\frac{y-3}{4} \in R$

Therefore, for any $y \in R ,$ there exists $x =\frac{y-3}{4} \in R ,$ such that

$f(x)=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y$

$\therefore f$ is onto.

Thus, $f$ is one $-$ one and onto and therefore, $f^{-1}$ exists.

Let us define $g:$ $R \rightarrow R$ by $g(x)=\frac{y-3}{4}$

Now,

$(gof)(x)=g(f(x))=g(4 x+3)=\frac{(4 x+3)-3}{4}=\frac{4 x}{4}=x$

and

$(fog)(y)=f(g(y))=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y-3+3=y$

$\therefore $     $gof= fog = I _{ R }$

Hence, $f$ is invertible and the inverse of $f$ is given by $f^{-1}(y)=g(y)=\frac{y-3}{4}$