Question

Let $f\left(x\right)=\left\{\begin{array}{ll}1+x,& 0\le x\le 2\boxed{}\\ 3-x,& 2<x\le 3\end{array}\right.$. Find fof.

Answer 1

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}1+x,& 0\le x\le 2\\ 3-x,& 2<x\le 3\end{array}\right. \\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}, \\ f\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}1+x,& 0\le x\le 1\end{array} \\ \begin{array}{ll}1+x,& 1<x\le 2\end{array} \\ \begin{array}{ll}3-x,& 2<x\le 3\end{array}\\ \end{array}\right. \\ \mathrm{When}\begin{array}{ll},& 0\le x\le 1\end{array} \\ \mathrm{Then},f\left(x\right)=1+x \\ \mathrm{Now}\mathrm{when}\begin{array}{ll},& 0\le x\le 1\end{array}\mathrm{then}\begin{array}{ll},& 1\le x+1\le 2\end{array} \\ \mathrm{Then},f\left(f\right(x\left)\right)=1+\left(1+x\right)=2+x\left[\because 1\le f\left(x\right)<2\right] \\ \mathrm{When}\begin{array}{ll},& 1<x\le 2\end{array} \\ \mathrm{Then},f\left(x\right)=1+x \\ \mathrm{Now}\mathrm{when}\begin{array}{ll},& 1<x\le 2\end{array}\mathrm{then}\begin{array}{ll},& 2<x+1\le 3\end{array} \\ \mathrm{Then},f\left(f\right(x\left)\right)=3-\left(1+x\right)=2-x\left[\because 2\le f\left(x\right)<3\right] \\ \mathrm{When}\begin{array}{ll},& 2<x\le 3\end{array} \\ \mathrm{Then},f\left(x\right)=3-x \\ \mathrm{Now}\mathrm{when}\begin{array}{ll},& 2<x\le 3\end{array}\mathrm{then}\begin{array}{ll},& 0\le 3-x<1\end{array} \\ \mathrm{Then},f\left(f\right(x\left)\right)=1+\left(3-x\right)=4-x\left[\because 0\le f\left(x\right)<1\right] \\ f\left(f\left(x\right)\right)=\left\{\begin{array}{l}\begin{array}{ll}2+x,& 0\le x\le 1\end{array} \\ \begin{array}{ll}2-x,& 1<x\le 2\end{array} \\ \begin{array}{ll}4-x,& 2<x\le 3\end{array}\\ \end{array}\right. \end{gathered}$$