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Consider $f:{{R}^{+}}\to \left[ -9,\infty \right)$ given by $f\left( x \right)=5{{x}^{2}}+6x-9$. Prove that $f$ is invertible with ${{f}^{-1}}\left( y \right)=\dfrac{\sqrt{54+5y}-3}{5}$.

Last updated date: 21st Jun 2024
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Hint: We first explain the expression of the function. We convert the function from $y$ of $x$ to $x$ of $y$. The inverse function on being conjugated gives the value of $x$. At the end we interchange the terms to make it a general equation. We also need to eliminate the wrong function depending on the domain of the main function.

Complete step-by-step solution:
We need to find the inverse of the equation of $f\left( x \right)=5{{x}^{2}}+6x-9$.
The given equation is a function of $x$ where we can write $y=f\left( x \right)=5{{x}^{2}}+6x-9$.
This gives the quadratic equation as $5{{x}^{2}}+6x-9-y=0$.
If we take the inverse of the equation, we will get $x={{f}^{-1}}\left( y \right)$.
We need to express the value of $x$ with respect to $y$. We find the roots for the quadratic equation.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part $\sqrt{{{b}^{2}}-4ac}$ of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation.
In the given equation we have $5{{x}^{2}}+6x-9-y=0$. The values of a, b, c is $5,6,-\left( 9+y \right)$ respectively.
We put the values and get $x$ as \[x=\dfrac{-6\pm \sqrt{{{6}^{2}}-4\times \left( -9-y \right)\times 5}}{2\times 5}=\dfrac{-3\pm \sqrt{54+5y}}{5}\].
This gives two inverse function $x={{f}^{-1}}\left( y \right)=\dfrac{-3+\sqrt{54+5y}}{5},\dfrac{-3-\sqrt{54+5y}}{5}$.
It’s given for $f:{{R}^{+}}\to \left[ -9,\infty \right)$, $y\in \left[ -9,\infty \right)$. This gives $\sqrt{54+5y}\in \left[ 3,\infty \right)$.
Now the domain for $f:{{R}^{+}}\to \left[ -9,\infty \right)$ is the positive real line. The function $x=\dfrac{-3-\sqrt{54+5y}}{5}$ becomes invalid for the range of positive real line.
Therefore, the inverse function of $f\left( x \right)=5{{x}^{2}}+6x-9$ is ${{f}^{-1}}\left( y \right)=\dfrac{\sqrt{54+5y}-3}{5}$.

Note: For the function $x={{f}^{-1}}\left( y \right)=\dfrac{-3-\sqrt{54+5y}}{5}$, the domain is $\left[ -9,\infty \right)$. Putting the values, we get $\dfrac{-3-\sqrt{54+5y}}{5}\in \left( -\infty ,-\dfrac{6}{5} \right]$. It is out of range for the inverse function where
$x={{f}^{-1}}\left( y \right)=\dfrac{-3-\sqrt{54+5y}}{5}$.