Question

Let the functions $f\left( x \right)=-1+\left| x-1 \right|,-1\le x\le 3$ and $g\left( x \right)=2-\left| x+1 \right|,-2\le x\le 2$ . Then $\left( fog \right)\left( x \right)=$ :
1) $x+1$
2) $-1-x$
3) $x-1$
4)
$\begin{align}
  & x+1\text{ if }-2\le x\le -1 \\
 & -1-x\text{ if }-1\le x\le 0 \\
 & x-1\text{ if }0\le x\le 2 \\
\end{align}$

Answer 1

Hint: Here in this question we have been given that $f\left( x \right)=-1+\left| x-1 \right|,-1\le x\le 3$ and $g\left( x \right)=2-\left| x+1 \right|,-2\le x\le 2$ . We have been asked to evaluate the value of $\left( fog \right)\left( x \right)$ which can be also represented as $f\left( g\left( x \right) \right)$ .

Complete step-by-step solution:
Now we need to say that $g\left( x \right)$ should lie between $\left[ -1,3 \right]$ in the expression $f\left( g\left( x \right) \right)$ .
Therefore $-1\le g\left( x \right)\le 3$ can be further simplified and given as
$\begin{align}
  & -1\le 2-\left| x+1 \right|\le 3 \\
 & \Rightarrow -3\le -\left| x+1 \right|\le 1 \\
 & \Rightarrow -1\le \left| x+1 \right|\le 3 \\
\end{align}$ .
Now by simplifying this further with two conditions that is when $x<-1$ and when $x\ge -1$ .
When $x\ge -1$ we will have $-1\le x+1\le 3$ this can be further simplified as $-2\le x\le 2$ . Hence we can say that one condition is $-1\le x\le 2$ .
When $x<-1$ we will have $-1\le -x-1\le 3$ this can be further simplified as $-4\le x\le 0$ . Hence we can say that the other condition is $-4\le x<-1$ .
Therefore the valid domain is $x\in \left[ -4,2 \right]$ .
Now by using definitions of both the functions we have$f\left( g\left( x \right) \right)=-1+\left| 1-\left| x+1 \right| \right|$ .
Here initially we have 2 conditions when $x\ge -1$ then $f\left( g\left( x \right) \right)=-1+\left| -x \right|$ . Therefore here for $x\ge -1$ we have $f\left( g\left( x \right) \right)=-1-x$ . By verifying the valid domain we can say that $f\left( g\left( x \right) \right)=-1-x$ for $x\in \left[ -1,0 \right)$and $f\left( g\left( x \right) \right)=-1+x$ for $x\in \left[ 0,2 \right]$ .
Now when we have $x<-1$ then $f\left( g\left( x \right) \right)=-1+\left| 2+x \right|$ . Therefore when $-2\le x<-1$ we will have $f\left( g\left( x \right) \right)=1+x$ .
Therefore we can conclude that when it is given that $f\left( x \right)=-1+\left| x-1 \right|,-1\le x\le 3$ and $g\left( x \right)=2-\left| x+1 \right|,-2\le x\le 2$ then the value of $\left( fog \right)\left( x \right)$ is given as
$\begin{align}
  & x+1\text{ if }-2\le x\le -1 \\
 & -1-x\text{ if }-1\le x\le 0 \\
 & x-1\text{ if }0\le x\le 2 \\
\end{align}$ .
Hence we will mark the option “4” as correct.

Note: While answering questions of this type we should be sure with the valid domain. We should carefully evaluate the domain of the functions otherwise we may end up having a wrong conclusion. If someone had considered any one condition only in a hurry then they would end up marking the wrong options.