Question

Let $f:N\to R$ be defined by $f\left(x\right)=4x^2+12x+15$. show that $f:N\to S$ where S is the range of fuction $f$, is invertible. Also find the inverse of $f$.

Answer 1

Given : $f\left(x\right)=4x^2+12x+15$

Let $x,y\in N$ such that

$f\left(x\right)=f\left(y\right)$

$⟹4x^2+12x+15=4y^2+12y+15$

$⟹4x^2-4y^2+12x-12y=0$

$⟹4(x^2-y^2)+12(x-y)=0$

$⟹(x^2-y^2)+3(x-y)=0$

$⟹(x-y)(x+y+3)=0$

$⟹(x-y)=0$ .......... $\because (x+y+3)\ne 0$ as $x,y\in N$

$⟹x=y$

Thus, $f$ is one-one

$f$ is onto if there exist $x\in N$ such that $f\left(x\right)=y$

$⟹4x^2+12x+15=y$

$⟹4x^2+12x+15-y=0$

$⟹x=\frac{-12\pm \sqrt{{12}^2-4(4\times (15-y\left)\right)}}{8}$

$⟹x=\frac{-12\pm \sqrt{144-16(15-y)}}{8}$

$⟹x=\frac{-12\pm \sqrt{16y-96}}{8}$

$⟹x=\frac{-12\pm 4\sqrt{y-6}}{8}$

$⟹x=\frac{-3\pm \sqrt{y-6}}{2}$

$⟹x=\frac{-3+\sqrt{y-6}}{2}$ ......... As $x\in N$

Hence, $f^{-1}\left(x\right)=\frac{-3+\sqrt{x-6}}{2}$