Question

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

Answer 1

We have,

f : N$\to$N is a function defined as f (x) = 9x² + 6x $-$ 5.

Let y = f (x) = 9x² + 6x $-$ 5

$$\begin{gathered} \Rightarrow y=9x^2+6x-5 \\ \Rightarrow y=9x^2+6x+1-1-5 \\ \Rightarrow y=\left(9x^2+6x+1\right)-6 \\ \Rightarrow y={\left(3x+1\right)}^2-6 \\ \Rightarrow y+6={\left(3x+1\right)}^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow \sqrt{y+6}=3x+1\left(\because y\in N\right) \\ \Rightarrow \sqrt{y+6}-1=3x \\ \Rightarrow x=\frac{\sqrt{y+6}-1}{3} \\ \Rightarrow g\left(y\right)=\frac{\sqrt{y+6}-1}{3}\left[\mathrm{Let}x=g\left(y\right)\right] \end{gathered}$$

Now,

$$\begin{gathered} fog\left(y\right)=f\left[g\left(y\right)\right] \\ =f\left(\frac{\sqrt{y+6}-1}{3}\right) \\ =9{\left(\frac{\sqrt{y+6}-1}{3}\right)}^2+6\left(\frac{\sqrt{y+6}-1}{3}\right)-5 \\ =9\left(\frac{y+6-2\sqrt{y+6}+1}{9}\right)+2\left(\sqrt{y+6}-1\right)-5 \\ =y+6-2\sqrt{y+6}+1+2\sqrt{y+6}-2-5 \\ =y \\ =I_Y,\mathrm{Identity}\mathrm{function} \end{gathered}$$

$$\begin{gathered} gof\left(x\right)=g\left[f\left(x\right)\right] \\ =g\left(9x^2+6x-5\right) \\ =\frac{\sqrt{\left(9x^2+6x-5\right)+6}-1}{3} \\ =\frac{\sqrt{\left(9x^2+6x+1\right)}-1}{3} \\ =\frac{\sqrt{{\left(3x+1\right)}^2}-1}{3} \\ =\frac{\left(3x+1\right)-1}{3} \\ =\frac{3x}{3} \\ =x \\ =I_X,\mathrm{Identity}\mathrm{function} \end{gathered}$$

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

Thus, f is invertible.

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

Now,

f ^$-$1(43) = $\frac{\sqrt{43+6}-1}{3}=\frac{{\displaystyle \sqrt{49}-1}}{3}=\frac{7-1}{3}=\frac{6}{3}=2$

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