Question

Let $f:N\to N$ be a function defined as $f\left(x\right)=9x^2+6x-5$. Show that $f:N\to S$, where $S$ is the range of $f$, is invertible. Find the inverse of $f$ and hence find $f^{-1}\left(43\right)$ and $f^{-1}\left(163\right)$

Answer 1

$f\left(x\right)=9x^2+6x-5$

$x=\frac{-6\pm \sqrt{36+180}}{18}=\frac{-1\pm \sqrt{6}}{3}$

Approximate value of these values is $0.466,-1.15$

Thus, for all the natural numbers, the value of the function will be positive and hence, its bijective. Its inverse is thus possible.

Let $f\left(x\right)$ be $y.$

We get $9x^2+6x-5-y=0$

$x=\frac{-6\pm \sqrt{36+180+36y}}{18}=\frac{-1\pm \sqrt{y+6}}{3}$

Or, $f^{-1}x=\frac{-1\pm \sqrt{x+6}}{3}$

$f^{-1}\left(43\right)=\frac{-1\pm 7}{3}=2$ (negative value discarded)

$f^{-1}\left(163\right)=\frac{-1\pm 13}{3}=4$ (negative value discarded)