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# Let ${{\text{R}}_{+}}$ be the set of all positive real numbers. Show that the function $f$ : ${{\text{R}}_{+}}\to \left[ -5,\infty \right)$ : $f\left( x \right)=9{{x}^{2}}+6x-5$ is invertible. Find ${{f}^{-1}}$.

Last updated date: 20th Jul 2024
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Hint: We know that any function is invertible, if and only if, the function is one-one and onto. So, we must first prove that the function is invertible. And to find the inverse, we must assume $x=f\left( y \right)$, and then solve the equation for y. The function we get, will be the inverse function.

We are given that $f$ : ${{\text{R}}_{+}}\to \left[ -5,\infty \right)$ : $f\left( x \right)=9{{x}^{2}}+6x-5$.
We can see that the domain is ${{\text{R}}_{+}}$ that is the set of all positive real numbers, codomain is $\left[ -5,\infty \right)$ and the function is given as $f\left( x \right)=9{{x}^{2}}+6x-5$.
First, we need to show that the function is invertible.
Let us assume two values ${{x}_{1}},{{x}_{2}}\in {{\text{R}}_{+}}$ such that $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$.
Using the definition of this function, we can write
$9{{x}_{1}}^{2}+6{{x}_{1}}-5=9{{x}_{2}}^{2}+6{{x}_{2}}-5$
Adding 5 on both sides, we get
$9{{x}_{1}}^{2}+6{{x}_{1}}=9{{x}_{2}}^{2}+6{{x}_{2}}$
We can also write this as
$9\left( {{x}_{1}}^{2}-{{x}_{2}}^{2} \right)+6\left( {{x}_{1}}-{{x}_{2}} \right)=0$
Using the expansion for difference of squares, we can write
$9\left( {{x}_{1}}+{{x}_{2}} \right)\left( {{x}_{1}}-{{x}_{2}} \right)+6\left( {{x}_{1}}-{{x}_{2}} \right)=0$
We can take a factor as common, and so we now have
$\left( {{x}_{1}}-{{x}_{2}} \right)\left[ 9\left( {{x}_{1}}+{{x}_{2}} \right)+6 \right]=0$
So, now either $\left( {{x}_{1}}-{{x}_{2}} \right)=0$ or $\left[ 9\left( {{x}_{1}}+{{x}_{2}} \right)+6 \right]=0$.
We know that since both ${{x}_{1}}\text{ and }{{x}_{2}}$ are positive, the term $\left[ 9\left( {{x}_{1}}+{{x}_{2}} \right)+6 \right]$ can never be equal to 0.
Thus, we have $\left( {{x}_{1}}-{{x}_{2}} \right)=0$. And so, ${{x}_{1}}={{x}_{2}}$.
Hence, the function is one-one. Or, we can say that the function is injective.
Let us now assume a variable ${{y}_{1}}\in \left[ -5,\infty \right)$, such that $f\left( {{x}_{1}} \right)={{y}_{1}}$.
Thus, we get
$9{{x}_{1}}^{2}+6{{x}_{1}}-5={{y}_{1}}$
Or, we can write
$9{{x}_{1}}^{2}+6{{x}_{1}}={{y}_{1}}+5$
Adding 1 on both sies, we get
$9{{x}_{1}}^{2}+6{{x}_{1}}+1={{y}_{1}}+5+1$
This can also be written as
${{\left( 3{{x}_{1}}+1 \right)}^{2}}={{y}_{1}}+6$
Thus, we have
$3{{x}_{1}}+1=\sqrt{{{y}_{1}}+6}$
$\Rightarrow {{x}_{1}}=\dfrac{\sqrt{{{y}_{1}}+6}-1}{3}$
We can see here that for every ${{y}_{1}}\in \left[ -5,\infty \right)$, the corresponding ${{x}_{1}}\in {{\text{R}}_{+}}$.
Thus, the function is onto or surjective.
Hence, we can say that the function is invertible.
To find the inverse function, we can assume a $y={{f}^{-1}}\left( x \right)$. So, $x=f\left( y \right)$.
Thus, we have
$x=9{{y}^{2}}+6y-5$
$\Rightarrow x+5=9{{y}^{2}}+6y$
$\Rightarrow x+6=9{{y}^{2}}+6y+1$
$\Rightarrow x+6={{\left( 3y+1 \right)}^{2}}$
So, we get
$y=\dfrac{\sqrt{x+6}-1}{3}$
Hence, the inverse function is ${{f}^{-1}}\left( x \right)=\dfrac{\sqrt{x+6}-1}{3}$.

Note: We must be careful and check for all the boundary conditions while checking whether the given function satisfies the criteria for being both injective and surjective. Also, we must remember to check whether the function is invertible or not, whenever we have to find the inverse of any function.