Question

Let $f:N\to N$ be a function defined as $f\left(x\right)=4x^2+12x+15$. Show that $f:N\to S$ is invertible. Find the inverse of $f$ and hence find $f^{-1}\left(31\right)$ and $f^{-1}\left(87\right)$ .

Answer 1

$y={4x}^2+12x+15$

$y-15={4x}^2+12x$

${4x}^2+12x+15-y=0$

$x=\frac{-12\pm \sqrt{144-4(15-y)}\times 4}{8}$

$=\frac{-12\pm \sqrt{144-(60-4y)\times 4}}{8}$

$=\frac{-12\pm \sqrt{144-240-{16y}^2}}{8}$

$=\frac{-12\pm \sqrt{{16y}^2-6}}{8}$

$=\frac{-3\pm \sqrt{y^2-6}}{2}$

$f^{-1}\left(31\right)$ $=\frac{-3\pm \sqrt{{31}^2-6}}{2}=\frac{-3\pm 30.90}{2}=\frac{27.9}{2}\text{or}\frac{-33.90}{2}=13.95\text{or}-16.95$

$f^{-1}\left(87\right)$ $=\frac{-3\pm \sqrt{{87}^2-6}}{2}=\frac{-3\pm 86.96}{2}=41.98\text{or}-44.98$

$\therefore$ $f^{-1}\left(31\right)=13.95$ or $-16.95,$ $f^{-1}\left(87\right)=41.98$ or $-44.98$.