Question

If $f\left( x \right)$ is defined on $\left[ -2,2 \right]$ and given by f\left( x \right)=\left\\{ \begin{matrix} -1,\text{ }-2\le x\le 0 \\\ x-1,\text{ }0\le x\le 2 \\\ \end{matrix} \right. and $g\left( x \right)=f\left( \left| x \right| \right)+\left| f\left( x \right) \right|$. Find $g\left( x \right)$

Answer 1

We will find the values of $f\left( \left| x \right| \right)$ and $\left| f\left( x \right) \right|$ by using the function $f\left( x \right)$ individually and adding them up to get the value of $g\left( x \right)$.

Complete step by step solution:
Given that,
f\left( x \right)=\left\\{ \begin{matrix} -1,\text{ }-2\le x\le 0 \\\ x-1,\text{ }0\le x\le 2 \\\ \end{matrix} \right.
The function $\left| x \right|$ convert the $-ve$ values into $+ve$ values i.e. the result of the function $\left| x \right|$ should be a $+ve$ value. We get the $+ve$ value from the function $\left| x \right|$ by substituting the appropriate value either $+x$ or $-x$ to get the result of function as $+ve$. Mathematically
\left| x \right|=\left\\{ \begin{matrix} +x,\text{ for }x\ge 0 \\\ -x,\text{ for }x<0 \\\ \end{matrix} \right.
Here we have $x\in \left[ -2,2 \right]$ then $\left| x \right|\in \left[ 0,2 \right]$, so
$f\left( \left| x \right| \right)=\left| x \right|-1$ for all $x\in \left[ -2,2 \right]$
Substitute the function $\left| x \right|$ in above equation,
f\left( \left| x \right| \right)=\left\\{ \begin{matrix} x-1,\text{ }0\le x\le 2 \\\ -x-1,\text{ }-2\le x<0 \\\ \end{matrix} \right.
Now the value of $\left| f\left( x \right) \right|$ is
\begin{aligned} & \left| f\left( x \right) \right|=\left\\{ \begin{matrix} \left| -1 \right|,\text{ }-2\le x<0 \\\ \left| x-1 \right|,\text{ }0\le x\le 2 \\\ \end{matrix} \right. \\\ & =\left\\{ \begin{matrix} 1,\text{ }-2\le x<0 \\\ \left| x-1 \right|,\text{ }0\le x\le 2 \\\ \end{matrix} \right. \end{aligned}
Now we have to change the value of $\left| x-1 \right|$ in the range of $\left[ 0,2 \right]$ to get a $+ve$. Hence we will take $+\left( x-1 \right)$ for $1\le x\le 2$ and we will take $-\left( x-1 \right)$ for $0\le x<1$. Mathematically we can write it as
\left| x-1 \right|=\left\\{ \begin{matrix} x-1,\text{ }1\le x\le 2 \\\ 1-x,\text{ }0\le x<1 \\\ \end{matrix} \right.
Hence the value of $\left| f\left( x \right) \right|$ is
\left| f\left( x \right) \right|=\left\\{ \begin{matrix} 1,\text{ }-2\le x<0 \\\ 1-x,\text{ }0\le x<1 \\\ x-1,\text{ }1\le x\le 2 \\\ \end{matrix} \right.
Now we have given that $g\left( x \right)=f\left( \left| x \right| \right)+\left| f\left( x \right) \right|$, so
\begin{aligned} & g\left( x \right)=\left\\{ \begin{matrix} -x-1,\text{ }-2\le x\le 0 \\\ x-1,\text{ }0\le x<1 \\\ x-1,\text{ }1\le x\le 2 \\\ \end{matrix} \right.+\left\\{ \begin{matrix} 1,\text{ }-2\le x<0 \\\ 1-x,\text{ }0\le x<1 \\\ x-1,\text{ }1\le x\le 2 \\\ \end{matrix} \right. \\\ & =\left\\{ \begin{matrix} -x-1+1,\text{ }-2\le x\le 0\text{ } \\\ x-1+1-x,\text{ }0\le x<1 \\\ x-1+x-1,\text{ }1\le x\le 2 \\\ \end{matrix} \right. \\\ & =\left\\{ \begin{matrix} -x,\text{ }-2\le x\le 0 \\\ 0,\text{ }0\le x<1 \\\ 2\left( x-1 \right),\text{ }1\le x\le 2 \\\ \end{matrix} \right. \end{aligned}

Note: The value of $\left| x \right|$ in the function $f\left( \left| x \right| \right)$ is modified as $+x$ and $-x$ in the range of $x\in \left[ 0,2 \right]$ and $-2\le x<0$ respectively. While finding the value of $\left| f\left( x \right) \right|$ first apply mode function to the variables in the function $f\left( x \right)$ and then modify the variables based on the range of that variable in order to get a $+ve$ value.