વિધેય $f: R _{+} \rightarrow[-5, \infty)$, $f(x)=9 x^{2}+6 x-5$ દ્વારા વ્યાખ્યાયિત છે. સાબિત કરો કે $f$ વ્યસ્તસંપન્ન છે અને $f^{-1}(y)=\left(\frac{(\sqrt{y+6})-1}{3}\right)$

$f : R _+\rightarrow[-5, \infty)$ is given as $f ( x )=9 x ^{2}+6 x -5$

Let $y$ be an arbitrary element of $[-5, \infty)$

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

$\Rightarrow y=(3 x+1)^{2}-1-5=(3 x+1)^{2}-6$

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

$\Rightarrow 3 x+1=\sqrt{Y+6}$         $[$ as $y \geq-5 \Rightarrow y+6>0]$

$\Rightarrow x =\left(\frac{(\sqrt{ y +6})-1}{3}\right)$

$\therefore f$ is onto, thereby range $f =[-5, \infty)$

Let us define $g:[-5, \infty) \rightarrow R_+$ as $g(y)=\left(\frac{(\sqrt{y+6})-1}{3}\right)$

Now, $(gof)(x)=g(f(x))=g\left(9 x^{2}+6 x-5\right)$ $=g\left((3 x+1)^{2}-6\right)$

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

$=\frac{3 x+1-1}{3}=\frac{3 x}{3}=X$

and

$(\text { fog })(y)=f(g(y))=\left(\frac{\sqrt{Y+6}-1}{3}\right)$ $=\left[3\left(\frac{\sqrt{Y+6}-1}{3}\right)+1^{2}\right]-6$

$=(\sqrt{Y+6})^{2}-6=+6-6=y$

$\therefore gof=x=I_{R}$ and $fog$ $=y=I_{R a n g e} f$

Hence, $f$ is invertible and the inverse of $f$ is given by

$f^{-1}(y)=g(y)=\left(\frac{\sqrt{Y+6}-1}{3}\right)$