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

f : ${R_ + } \to [ - 5,\infty )$ and $f(x) = 9{x^2} + 6x - 5$

Injectivity

Let ${x_1},{x_2} \in R$
$\to f({x_1}) = 9x_1^2 + 6{x_1} - 5$and $f({x_2}) = 9x_2^2 + 6{x_2} - 5$

If$f({x_1}) = f({x_2})$

$\Rightarrow$ $9x_1^2 + 6{x_1} - 5 = 9x_2^2 + 6{x_2} - 5 \Rightarrow 9x_1^2 + 6{x_1} = 9x_2^2 + 6{x_2}$

$\Rightarrow$ $9(x_1^2 - x_2^2) + 6({x_1} - {x_2}) = 0 \Rightarrow ({x_1} - {x_2})[9({x_1} + {x_2}) + 6] = 0$

$\Rightarrow$ ${x_1} - {x_2} = 0 \Rightarrow {x_1} = {x_2}$ $\therefore f(x)$ is one-one.

Surjectivity

$\Rightarrow$ $f:{R_ + } \to [ - 5,\infty )$ is given, lety $y \in [ - 5,\infty )$ (co-domain off ) then there exist an element x$\in$R+ (domain of f), such that f(x) $=$ y

$\Rightarrow$ $y = 9{x^2} + 6x - 5 \Rightarrow 9{x^2} + 6x - (5 + y) = 0$
$\Rightarrow$ $x = \cfrac{{ - (6) \pm \sqrt {{{(6)}^2} + 4 \times 9(5 + y)} }}{{18}}$

$\Rightarrow$ $x = \cfrac{{ - 6 \pm 6\sqrt {1 + (5 + y)} }}{{18}}$

$\Rightarrow$ $x = \cfrac{{ - 6 + 6\sqrt {y + 6} }}{{18}}$ $\Rightarrow$ $x = \cfrac{{(\sqrt {y + 6)} - 1}}{3}$
$\Rightarrow$ $f(x) = f\left( {\cfrac{{\sqrt {y + 6} - 1}}{3}} \right)$

$= 9{\left[ {\cfrac{{\sqrt {y + 6} - 1}}{3}} \right]^2} + 6\left[ {\cfrac{{\sqrt {y + 6} - 1}}{3}} \right] - 5$

$=$ $9\left[ {\cfrac{{y + 6 + 1 - 2\sqrt {y + 6} }}{6}} \right] + 2(\sqrt {y + 6} - 1) - 5$

$= y + 7 - 2\sqrt {y + 6} + 2\sqrt {y + 6} - 2 - 5$

$\Rightarrow$ $f(x) = y$ . $\Rightarrow$ $f(x)$ is onto.

Thus, f(x) is bijective and hence invertible.

Now, we show that inverse of f is $\cfrac{{(\sqrt {y + 6} ) - 1}}{3}$

Wehave, $f(x) = y \Rightarrow {f^{ - 1}}(y) = x$

$\Rightarrow$ ${f^{ - 1}}(y) = \cfrac{{(\sqrt {y + 6} ) - 1}}{3}$ or ${f^{ - 1}}(x) = \cfrac{{\sqrt {x + 6} - 1}}{3}$