Answer

Verified

373.5k+ views

**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

$x={{f}^{-1}}\left( y \right)=\dfrac{-3-\sqrt{54+5y}}{5}$.

Recently Updated Pages

How do you find slope point slope slope intercept standard class 12 maths CBSE

How do you find B1 We know that B2B+2I3 class 12 maths CBSE

How do you integrate int dfracxsqrt x2 + 9 dx class 12 maths CBSE

How do you integrate int left dfracx2 1x + 1 right class 12 maths CBSE

How do you find the critical points of yx2sin x on class 12 maths CBSE

How do you find the general solution to dfracdydx class 12 maths CBSE

Trending doubts

The provincial president of the constituent assembly class 11 social science CBSE

Gersoppa waterfall is located in AGuyana BUganda C class 9 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

The hundru falls is in A Chota Nagpur Plateau B Calcutta class 8 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE