Question

$f\left(x\right)=-1+|x-2|,0\le x\le 4$
$g\left(x\right)=2-|x|,-1\le x\le 3$
Which of the following is true

• A. $fog\left(x\right)=\{\begin{array}{ccc}-(1+x),& & -1\le x\le 0\\ x+1& & 0<x\le 2\end{array}$
• B. $gof\left(x\right)=\{\begin{array}{cc}x+1,& 0\le x<1\\ 3-x,& 1\le x\le 2\\ x-1,& 2<x\le 3\\ 5-x,& 3<x\le 4\end{array}$
• C. $fog\left(x\right)=\{\begin{array}{ccc}-(1+2x),& & -1\le x\le 0\\ x-1& & 0<x\le 2\end{array}$
• D. $gof\left(x\right)=\{\begin{array}{cc}x+1,& 0\le x<1\\ 3-x,& 1\le x\le 2\\ x+1,& 2<x\le 3\\ 5-x,& 3<x\le 4\end{array}$

Answer 1

The correct option is B $gof\left(x\right)=\{\begin{array}{cc}x+1,& 0\le x<1\\ 3-x,& 1\le x\le 2\\ x-1,& 2<x\le 3\\ 5-x,& 3<x\le 4\end{array}$

$f\left(x\right)=-1+|x-2|,0\le x\le 4$
$g\left(x\right)=2-|x|,-1\le x\le 3$

$g\left(f\right(x\left)\right)=2-|-1+|x-2||,$

$gof\left(x\right)=\{\begin{array}{cc}2-[2-x-1]& 0\le x<1\\ 2+[2-x-1],& 1\le x\le 2\\ 2+[x-2-1],& 2<x\le 3\\ 2-[x-2-1],& 3<x\le 4\end{array}gof\left(x\right)=\{\begin{array}{cc}x+1,& 0\le x<1\\ 3-x,& 1\le x\le 2\\ x-1,& 2<x\le 3\\ 5-x,& 3<x\le 4\end{array}$