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

Consider f:R+→[−9,∞)f:{{R}^{+}}\to \left[ -9,\infty \right) given by f(x)=5x2+6x−9f\left( x \right)=5{{x}^{2}}+6x-9. Prove that ff is invertible with f−1(y)=54+5y−35{{f}^{-1}}\left( y \right)=\dfrac{\sqrt{54+5y}-3}{5}.

Explanation

Solution

We first explain the expression of the function. We convert the function from yy of xx to xx of yy. The inverse function on being conjugated gives the value of xx. 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(x)=5x2+6x−9f\left( x \right)=5{{x}^{2}}+6x-9.
The given equation is a function of xx where we can write y=f(x)=5x2+6x−9y=f\left( x \right)=5{{x}^{2}}+6x-9.
This gives the quadratic equation as 5x2+6x−9−y=05{{x}^{2}}+6x-9-y=0.
If we take the inverse of the equation, we will get x=f−1(y)x={{f}^{-1}}\left( y \right).
We need to express the value of xx with respect to yy. We find the roots for the quadratic equation.
We know for a general equation of quadratic ax2+bx+c=0a{{x}^{2}}+bx+c=0, the value of the roots of xx will be x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}. This is the quadratic equation solving method. The root part b2−4ac\sqrt{{{b}^{2}}-4ac} of x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} is called the discriminant of the equation.
In the given equation we have 5x2+6x−9−y=05{{x}^{2}}+6x-9-y=0. The values of a, b, c is 5,6,−(9+y)5,6,-\left( 9+y \right) respectively.
We put the values and get xx as x=−6±62−4×(−9−y)×52×5=−3±54+5y5x=\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(y)=−3+54+5y5,−3−54+5y5x={{f}^{-1}}\left( y \right)=\dfrac{-3+\sqrt{54+5y}}{5},\dfrac{-3-\sqrt{54+5y}}{5}.
It’s given for f:R+→[−9,∞)f:{{R}^{+}}\to \left[ -9,\infty \right), y∈[−9,∞)y\in \left[ -9,\infty \right). This gives 54+5y∈[3,∞)\sqrt{54+5y}\in \left[ 3,\infty \right).
Now the domain for f:R+→[−9,∞)f:{{R}^{+}}\to \left[ -9,\infty \right) is the positive real line. The function x=−3−54+5y5x=\dfrac{-3-\sqrt{54+5y}}{5} becomes invalid for the range of positive real line.
Therefore, the inverse function of f(x)=5x2+6x−9f\left( x \right)=5{{x}^{2}}+6x-9 is f−1(y)=54+5y−35{{f}^{-1}}\left( y \right)=\dfrac{\sqrt{54+5y}-3}{5}.

Note: For the function x=f−1(y)=−3−54+5y5x={{f}^{-1}}\left( y \right)=\dfrac{-3-\sqrt{54+5y}}{5}, the domain is [−9,∞)\left[ -9,\infty \right). Putting the values, we get −3−54+5y5∈(−∞,−65]\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(y)=−3−54+5y5x={{f}^{-1}}\left( y \right)=\dfrac{-3-\sqrt{54+5y}}{5}.