Question
Let $\text{f(x)=}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}.$ Find fof.

Answer

$\text{f(x)=}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}$
$\therefore$ Range of $\text{f}=[0,3]\subseteq$ Domain of f.
$\therefore$ fof(x) = f(f(x))
$=\text{f}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}$
$\text{fof(x)}=\begin{cases}2+\text{x},&0\leq\text{x}\leq1\\2-\text{x},&1<\text{x}\leq2\\4-\text{x},&2<\text{x}\leq3\end{cases}$

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