Let f : N→N be a function defined as f (x)


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


We have,

$f: N \rightarrow N$ is a function defined as $f(x)=9 x^{2}+6 x-5$

Let $y=f(x)=9 x^{2}+6 x-5$

$\Rightarrow y=9 x^{2}+6 x-5$

$\Rightarrow y=9 x^{2}+6 x+1-1-5$

$\Rightarrow y=\left(9 x^{2}+6 x+1\right)-6$

$\Rightarrow y=(3 x+1)^{2}-6$

$\Rightarrow y+6=(3 x+1)^{2}$

$\Rightarrow \sqrt{y+6}=3 x+1 \quad(\because y \in N)$

$\Rightarrow \sqrt{y+6}-1=3 x$

$\Rightarrow x=\frac{\sqrt{y+6}-1}{3}$

$\Rightarrow g(y)=\frac{\sqrt{y+6}-1}{3} \quad$ [Let $\left.x=g(y)\right]$


$f o g(y)=f[g(y)]$



$=9\left(\frac{y+6-2 \sqrt{y+6}+1}{9}\right)+2(\sqrt{y+6}-1)-5$

$=y+6-2 \sqrt{y+6}+1+2 \sqrt{y+6}-2-5$


$=I_{Y}$, Identity function

$g o f(x)=g[f(x)]$

$=g\left(9 x^{2}+6 x-5\right)$

$=\frac{\sqrt{\left(9 x^{2}+6 x-5\right)+6}-1}{3}$

$=\frac{\sqrt{\left(9 x^{2}+6 x+1\right)-1}}{3}$

$=\frac{\sqrt{(3 x+1)^{2}-1}}{3}$

$=\frac{(3 x+1)-1}{3}$

$=\frac{3 x}{3}$


$=I_{X}$, Identity function

Since, fog(y) and gof(x) are identity function.

Thus, f is invertible.

So, $f^{-1}(x)=g(x)=\frac{\sqrt{x+6}-1}{3}$.



And $f^{-1}(163)=\frac{\sqrt{163+6}-1}{3}=\frac{\sqrt{169}-1}{3}=\frac{13-1}{3}=\frac{12}{3}=4$

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