Question

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

Answer 1

$f:R^{+}\to [-5,\infty ]$ given by $f\left(x\right)=9x^2+6x-5$

To show: f is one-one & onto

Let us assume that f is not one-one.

$\therefore$ there exists $2$ as more numbers for which images are same

For $x_1,x_2\in R+$ & $x_1\pm x_2$

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

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

$9x_1^2+6x_2=9x_1^2+6x_2$

$9(x_1^2-x_2^2)+6(x_1-x_2)=0$

$(x_1-x_2)\left[9\right(x_1-x_2)+6]=0$.

Since $x_1$ & $x_2$ are positive

$9(x_1-x_2)+6>0$

$\therefore x_1-x_2=0$ $\Rightarrow x_1=x_2$

Therefore, it contradicts are assumption.

Hence the function is one-one

Now, let as prove that f is onto

A function $f:X\to Y$ is onto it for every y$\in$Y, these exists a pie image in X.

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

$=9x^2+6x+1-6$

$=(3x+1{)}^2-6$

Now, for all $x\in R+$ as $[0,\infty ),f\left(x\right)\in [-5,\infty )$

$\therefore$ Range$=$co-domain

Hence f is onto

We have proved, that function is one-one & onto. Hence in turn it is proved that f is invertible.