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

(A) $f: R_{+} \rightarrow [-5, \infty)$ is given by $f(x) = 9x^{2} + 6x - 5$.
Let $y$ be an arbitrary element of $[-5, \infty)$.
Set $y = 9x^{2} + 6x - 5$.
$y = (3x + 1)^{2} - 1 - 5 = (3x + 1)^{2} - 6$.
$y + 6 = (3x + 1)^{2}$.
Since $x \in R_{+}$,$x > 0$,so $3x + 1 > 1$. Thus,$3x + 1 = \sqrt{y+6}$.
$x = \frac{\sqrt{y+6}-1}{3}$.
Define $g: [-5, \infty) \rightarrow R_{+}$ by $g(y) = \frac{\sqrt{y+6}-1}{3}$.
$(g \circ f)(x) = g(f(x)) = g((3x+1)^{2}-6) = \frac{\sqrt{(3x+1)^{2}-6+6}-1}{3} = \frac{3x+1-1}{3} = x$.
$(f \circ g)(y) = f(g(y)) = 9\left(\frac{\sqrt{y+6}-1}{3}\right)^{2} + 6\left(\frac{\sqrt{y+6}-1}{3}\right) - 5 = (\sqrt{y+6}-1)^{2} + 2(\sqrt{y+6}-1) - 5 = (y+6 - 2\sqrt{y+6} + 1) + 2\sqrt{y+6} - 2 - 5 = y + 7 - 7 = y$.
Since $g \circ f = I_{R_{+}}$ and $f \circ g = I_{[-5, \infty)}$,$f$ is invertible and $f^{-1}(y) = \frac{\sqrt{y+6}-1}{3}$.