Question

Let $f\left(x\right)=-1+|x-1|,1\le x\le 3$ and $g\left(x\right)=2-|x+1|,-2\le x\le 2$, then (fog)(x) is equal to

• A. $\{\begin{array}{c}x+1-2\le x\le 0\\ x-10<x\le 2\end{array}$
• B. $\{\begin{array}{c}x-1-2\le x\le 0\\ x+0<x\le 2\end{array}$
• C. $\{\begin{array}{c}-1-x-2\le x\le 0\\ x-10<x\le 2\end{array}$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$\left(fog\right)\left(x\right)=f(x+3),-2\le x+3\le -1f(-x+1),-1\le -x+2\le 2$
$=\{\begin{array}{cc}f(x+3),& -2\le x+3\le -1\\ f(-x+1),& -1\le -x+1\le 1\\ f(-x+1),& 1\le -x+1\le 2\end{array}$
We can see that f(x) will not be defined when $1\le -x+1\le 2.$
$=\{\begin{array}{cc}f(x+3),& -2\le x+3\le -1\\ f(-x+1),& -1\le -x+1\le 1\end{array}$
$=\{\begin{array}{cc}x+1,& -2\le x\le -1\\ -x-1,& -1\le x\le 0\\ x-1,& 0\le x\le 1\end{array}$