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Question: Let the functions \(f\left( x \right)=-1+\left| x-1 \right|,-1\le x\le 3\) and \(g\left( x \right)=2...

Let the functions f(x)=−1+∣x−1∣,−1≤x≤3f\left( x \right)=-1+\left| x-1 \right|,-1\le x\le 3 and g(x)=2−∣x+1∣,−2≤x≤2g\left( x \right)=2-\left| x+1 \right|,-2\le x\le 2 . Then (fog)(x)=\left( fog \right)\left( x \right)= :

  1. x+1x+1
  2. −1−x-1-x
  3. x−1x-1

x+1 if −2≤x≤−1 −1−x if −1≤x≤0 x−1 if 0≤x≤2 \begin{aligned} & 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{aligned}

Explanation

Solution

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

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