Let $f: N \rightarrow R$ be a function defined as $f(x)=4x^{2}+12x+15$. Show that $f: N \rightarrow S$,where $S$ is the range of $f$,is invertible. Find the inverse of $f$.

(N/A) Let $y$ be an arbitrary element of the range $S$. Then $y = 4x^{2} + 12x + 15$ for some $x \in N$.
This can be written as $y = (2x + 3)^{2} + 6$.
Solving for $x$,we get $(2x + 3)^{2} = y - 6$,which implies $2x + 3 = \sqrt{y - 6}$ (since $x \in N$,$2x + 3 > 0$).
Thus,$x = \frac{\sqrt{y - 6} - 3}{2}$.
Define $g: S \rightarrow N$ by $g(y) = \frac{\sqrt{y - 6} - 3}{2}$.
Now,$g(f(x)) = g((2x + 3)^{2} + 6) = \frac{\sqrt{(2x + 3)^{2} + 6 - 6} - 3}{2} = \frac{(2x + 3) - 3}{2} = x$.
Also,$f(g(y)) = f\left(\frac{\sqrt{y - 6} - 3}{2}\right) = \left(2\left(\frac{\sqrt{y - 6} - 3}{2}\right) + 3\right)^{2} + 6 = (\sqrt{y - 6} - 3 + 3)^{2} + 6 = (\sqrt{y - 6})^{2} + 6 = y - 6 + 6 = y$.
Since $g \circ f = I_{N}$ and $f \circ g = I_{S}$,$f$ is invertible and $f^{-1}(y) = \frac{\sqrt{y - 6} - 3}{2}$.