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 { as } y \geq-5 \Rightarrow y+6\gt0] \\\\ &\Rightarrow x=\frac{\sqrt{y+6}-1}{3} \\ \end{aligned} $

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

Let us define, $ g:[-5, \infty) \rightarrow \mathbf{R}_{+} \\ $

as

$ g(y)=\frac{\sqrt{y+6}-1}{3} \\ $

We now have:

$$ \begin{aligned} (g \circ f)(x)=g(f(x)) &=g\left(9 x^{2}+6 x-5\right) \\\\ &=g\left((3 x+1)^{2}-6\right) \\\\ &=\frac{\sqrt{(3 x+1)^{2}-6+6}-1}{3} \\\\ &=\frac{3 x+1-1}{3}=x \\ \end{aligned} $$

And, $ (f \circ g)(y)=f(g(y))=f\left(\frac{\sqrt{y+6}-1}{3}\right) \\ $

$$ \begin{aligned} &=\left[3\left(\frac{\sqrt{y+6}-1}{3}\right)+1\right]^{2}-6 \\\\ &=(\sqrt{y+6})^{2}-6=y+6-6=y \\ \end{aligned} $$

$ \therefore g \circ f=\mathbf{I}_{\mathbf{R}_{+}} \\ $ and

$ f \circ g=\mathbf{I}_{[-5, \infty)} \\ $

Hence, f is invertible and the inverse of f is given by $ f^{-1}(y)=g(y)=\frac{\sqrt{y+6}-1}{3} \\ $.