Question

How do you write \[\left| {{x}^{2}}+x-12 \right|\] as a piecewise function?

Answer 1

Hint: In this problem, we have to find the piecewise function for the given equation. We can first find the points where the quadratic is zero to simplify the restrictions and add pieces as needed. We can then form the functions using those points to get the final answer. We can also draw the graph for better understanding.

Complete step by step solution:
We know that the given equation is,
\[\left| {{x}^{2}}+x-12 \right|\]…….. (1)
We can now use the definition,
\[\left| A \right|=\left\{ \begin{matrix}
   A; & A\ge 0 \\
   -A; & A < 0 \\
\end{matrix} \right.\] ……. (2)
We can now find the points where the quadratic is zero to simplify the restrictions and add pieces as needed.
We can now use the definition (2) for the given equation (1), we get
\[y=\left\{ \begin{matrix}
   {{x}^{2}}+x-12; & {{x}^{2}}+x-12\ge 0 \\
   -\left( {{x}^{2}}+x-12 \right); & {{x}^{2}}+x-12 < 0 \\
\end{matrix} \right\}\]
Now we can find the value of x from the equation \[{{x}^{2}}+x-12\], we get
We can now factorize the equation, we get
\[\left( x-3 \right)\left( x+4 \right)=0\]
Where
\[\begin{align}
  & -3+4=1 \\
 & -3\times 4=-12 \\
\end{align}\]
Therefore, the values of x are -4 and 3.
This means that \[{{x}^{2}}+x-12\] for \[x\le -4\] and \[x\ge 3\].
We can now modify the restriction for the first piece to be \[x\le -4\] and add a third piece with the restriction \[x\ge 3\], we get
\[y=\left\{ \begin{matrix}
   {{x}^{2}}+x-12; & x\le -4 \\
   -\left( {{x}^{2}}+x-12 \right); & {{x}^{2}}+x-12 < 0 \\
   {{x}^{2}}+x-12; & x\ge 3 \\
\end{matrix} \right\}\]
 Now we can modify the restriction for the middle piece to be \[-4 < x < 3\] and distribute the -1.
\[y=\left\{ \begin{matrix}
   {{x}^{2}}+x-12; & x\le -4 \\
   -\left( {{x}^{2}}+x-12 \right); & -4 < x < 3 \\
   {{x}^{2}}+x-12; & x\ge 3 \\
\end{matrix} \right\}\]
Now we can draw the graph for piecewise function with each piece in different colour.
\[y=\left\{ \begin{matrix}
   {{x}^{2}}+x-12; & x\le -4 \\
   -\left( {{x}^{2}}+x-12 \right); & -4 < x < 3 \\
   {{x}^{2}}+x-12; & x\ge 3 \\
\end{matrix} \right\}\]

Note: Students make mistakes while finding the restrictions and while adding the pieces as needed. We should also know how to handle the limit values. We should know to draw the graph in a piecewise method with different colours for better understanding.