Consider an arbitrary element $y$ of $Y$. By the definition of $Y, y=4 x+3$ for some $x$ in the domain $N$. This shows that $x=\frac{(y-3)}{4} .$ Define $g: Y \rightarrow N$ by $g(y)=\frac{(y-3)}{4} .$ Now, $gof\,(x)=g(f(x))=g(4 x+3)$ $=\frac{(4 x+3-3)}{4}=x$ and $fog\,(y)=f(g(y))=f\left(\frac{(y-3)}{4}\right)$ $=\frac{4(y-3)}{4}+3=y-3+3=y .$ This shows that $gof= I _{ N }$ and $f o g=I_{Y}$, which implies that $f$ is invertible and $g$ is the inverse of $f$