Question

if $f\left(x\right)=\{\begin{array}{cc}1+x;& 0\le x\le 2\\ 3-x;& 2<x\le 3\end{array}$ then determine $\left(fof\right)\left(2\right).$

Answer 1

Given $f\left(x\right)=\{\begin{array}{cc}1+x;& 0\le x\le 2...\left(A\right)\\ 3-x;& 2<x\le 3...\left(B\right)\end{array}$ Now $\left(fof\right)x=f\left[f\left(x\right)\right]=\{\begin{array}{cc}f(1+x);& 0\le x\le 2\\ f(3-x);& 2<x\le 3\end{array}$ We will redefine above keeping in view the definitions given in (A) and (B).Now $0\le x\le 2\Rightarrow 1\le 1+x\le 3...\left(1\right)$ $\Rightarrow 0<1+x\le 2and2<1+x\le 3$ and $2<x\le 3\Rightarrow -3\le -x<-2$ $\Rightarrow 3-3\le 3-x<3-2or0<3-x<1...\left(2\right)$ $\left(fof\right)x=\{\begin{array}{ccc}f\left(t\right),& 1\le t\le 3,& t=1+xby\left(1\right)\\ f\left(z\right)& 0<z<1,& z=3-xby\left(2\right)\end{array}$ Again we write above as $\{\begin{array}{cc}f\left(t\right)& 0\le t\le 2\\ & 2<t\le 3\\ f\left(z\right)& 0<z<1\end{array}$ $1+t=1+1x=2+xby\left(A\right)$ $3-t=\beta -(1+x)=2-xby\left(B\right)$ $1+z=1+3-x=4-xby\left(A\right)$