Question

How do you find the inverse of ${{x}^{2}}-x-2$?

Answer 1

Hint: We first explain the expression of the function as we take $y=f\left( x \right)={{x}^{2}}-x-2$. 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.

Complete step-by-step answer:
We need to find the inverse of the equation of ${{x}^{2}}-x-2$.
The given equation is a function of $x$ where we can write $y=f\left( x \right)={{x}^{2}}-x-2$.
If we take the inverse of the equation, we will get $x={{f}^{-1}}\left( y \right)$.
The given function was of $x$. We convert it to a function of $y$ and that becomes the inverse of $y={{x}^{2}}-x-2$.
We need to express the value of $x$ with respect to $y$.
We first form the square and get
\[\begin{align}
  & y={{x}^{2}}-x-2 \\
 & \Rightarrow y={{\left( x \right)}^{2}}-2\times x\times \dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}-2-{{\left( \dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow y+\dfrac{9}{4}={{\left( x-\dfrac{1}{2} \right)}^{2}} \\
\end{align}\]
Now we take square root on both sides and get
\[\begin{align}
  & y+\dfrac{9}{4}={{\left( x-\dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow x-\dfrac{1}{2}=\pm \sqrt{y+\dfrac{9}{4}} \\
 & \Rightarrow x=\dfrac{1}{2}\pm \sqrt{y+\dfrac{9}{4}} \\
\end{align}\]
Therefore, the expression of $x$ with respect to $y$ is \[x=\dfrac{1}{2}\pm \sqrt{y+\dfrac{9}{4}}\].
Now as we know any general expression of any function is defined by the function of $x$.
We try to interchange the position of $x$ and $y$ in the equation \[x=\dfrac{1}{2}\pm \sqrt{y+\dfrac{9}{4}}\] to form the inverse equation in general form.
From \[x=\dfrac{1}{2}\pm \sqrt{y+\dfrac{9}{4}}\], we get \[y=\dfrac{1}{2}\pm \sqrt{x+\dfrac{9}{4}}\]. So, $y={{f}^{-1}}\left( x \right)=\dfrac{1}{2}\pm \sqrt{x+\dfrac{9}{4}}$
Therefore, the inverse function of ${{x}^{2}}-x-2$ is \[y=\dfrac{1}{2}\pm \sqrt{x+\dfrac{9}{4}}\].

Note: All quadratic equations cannot have an inverse. It’s necessary to understand the concept of domain as the inverse function has to be well-defined. For our inverse function of \[y=\dfrac{1}{2}\pm \sqrt{x+\dfrac{9}{4}}\], the restriction is $x\ge -\dfrac{9}{4}$. Otherwise, the function becomes imaginary.