(N/A) To show that $f$ is invertible,we must prove that $f$ is both one-one and onto.
$1$. One-one:
Let $f(x_1) = f(x_2)$.
$4x_1 + 3 = 4x_2 + 3$
$4x_1 = 4x_2$
$x_1 = x_2$.
Thus,$f$ is one-one.
$2$. Onto:
For any $y \in R$,let $y = 4x + 3$.
Then $4x = y - 3$,which gives $x = \frac{y-3}{4}$.
Since $y \in R$,$x = \frac{y-3}{4} \in R$.
For every $y \in R$,there exists $x = \frac{y-3}{4}$ such that $f(x) = 4(\frac{y-3}{4}) + 3 = y - 3 + 3 = y$.
Thus,$f$ is onto.
Since $f$ is both one-one and onto,it is invertible.
To find $f^{-1}$,we use the relation $f(x) = y \implies x = f^{-1}(y)$.
From $y = 4x + 3$,we have $x = \frac{y-3}{4}$.
Therefore,$f^{-1}(y) = \frac{y-3}{4}$,or $f^{-1}(x) = \frac{x-3}{4}$.