Let $y$ be an arbitrary element of range $f$. Then $y=4 x^{2}+12 x+15,$ for some $x$ in $N ,$ which implies that $y=(2 x+3)^{2}+6 .$ This gives $x=\frac{((\sqrt{y-6})-3)}{2},$ as $y \geq 6$

Let us define $g: S \rightarrow N$ by $g(y)=\frac{((\sqrt{y-6})-3)}{2}$

Now          $gof\,(x)=g(f(x))=g\left(4 x^{2}+12 x+15\right)$ $=g\left((2 x+3)^{2}+6\right)$

$=\frac{((\sqrt{(2 x+3)^{2}+6-6})-3)}{2}=\frac{(2 x+3-3)}{2}=x$

and       $fog (y)=f\left(\frac{((\sqrt{y-6})-3)}{2}\right)=\left(\frac{2((\sqrt{y-6})-3)}{2}+3\right)^{2}+6$

$=((\sqrt{y-6})-3+3))^{2}+6=(\sqrt{y-6})^{2}+6=y-6+6=y$

Hence, $gof=I_{ N }$ and $f o g=I_{s} .$ This implies that $f$ is invertible with $f^{-1}=g$.