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 function $f$, is invertible. Find the inverse of $f$.

The correct answer is $\dfrac{\sqrt{x-6} - 3}{2}$

$f(x) = 4x^2 + 12x + 15$

Using the formula $\text{Last Term}$ = $\dfrac{(\text{Middle Term})^2}{4 \times \text{First Term}}$ = $\dfrac{(12)^2}{4 \times 4}$ = $9$


∴ Adding and subtracting 9 from the equation, we get

$f(x) = 4x^2 + 12x + +9 - 9+ 15$ = $(2x + 3)^2 + 6$ ----> (1)


$y$ = $(2x + 3)^2 + 6$

⇒ $y - 6$ = $(2x + 3)^2$

⇒ $g(y) = x$ = $\dfrac{\sqrt{y-6} - 3}{2}$ ----> (2)


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

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

∴ $fog(y)$ = $y$



$gof(x)$ = $g(f(x))$ = $g((2x + 3)^2 + 6)$

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

∴ $gof(x)$ = $x$


∴ $f$ is invertible


$f^{-1}$ = $\dfrac{\sqrt{x-6} - 3}{2}$