Consider $f: \mathbf{R}_{+} \rightarrow[-5, \infty)$ given by $f(x)=9 x^{2}+6 x-5 .$ Show that $f$ is invertible with $f^{-1}(y)=\left(\frac{(\sqrt{y+6})-1}{3}\right)$

Solution:

$f: \mathbf{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 \\ $

$ \begin{aligned} &\Rightarrow y=(3 x+1)^{2}-1-5=(3 x+1)^{2}-6 \\\\ &\Rightarrow(3 x+1)^{2}=y+6 \\\\ &\Rightarrow 3 x+1=\sqrt{y+6} \quad[\text …