Question

Consider the function $f:R\to [-5,\infty )$ defined by $f\left(x\right)=9x^2+6x-5$, where R is negative real numbers. Show that $f$ is invertible and find its inverse.

Answer 1

Solution:-

$f:R\to [-5,\infty )$

$f\left(x\right)=9x^2+6x-5$

A function is invertible if the function is one-one and onto.

Let $x_1,x_2\in R^{-}$, such that

$f\left(x_1\right)=f\left(x_2\right)$

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

$\Rightarrow 3({x_1}^2-{x_2}^2)+2(x_1-x_2)=0$

$\Rightarrow (x_1-x_2)(3(x_1+x_2)+2)=0$

$\because x_1,x_2\in R^{-}$

$\Rightarrow (3(x_1+x_2)+2)\ne 0$

$\therefore x_1=x_2$

Thus, $f\left(x\right)$ is one-one.

The function $f:X\to Y$ is onto if for every $y\in Y$, there exist a perimage in X, such that $y=f\left(x\right)$

$\therefore y=9x^2+6x-5$

$\Rightarrow 9x^2+6x-(5+y)=0$

Here,

$a=9$

$b=6$

$c=-(5+y)$

From quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we have

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

$\Rightarrow x=\frac{-1\pm \sqrt{y+6}}{3}$

$\because x\in R^{-}$

$\therefore x=\frac{-1-\sqrt{y+6}}{3}$

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