Let $f(x)= \begin{cases}1+x, & 0 \leq x \leq 2 \ 3-x, & 2

Question

Let $f(x)= \begin{cases}1+x, & 0 \leq x \leq 2 \ 3-x, & 2<x \leq 3\end{cases}$ Determine the form of $g(x)=f(f(x))$ and hence find the sum of values at which $g$ is discontinuous.

Professor PC Thomas Classes · Answer

$f(x)= \begin{cases}1+x, & 0 \leq x \leq 2 \ 3-x, & 2<x \leq 3\end{cases}$\ $f(f(x))= \begin{cases}1+f(x), & 0 \leq f(x) \leq 2 \ 3-f(x), & 2<f(x) \leq 3\end{cases}$\ Now $0 \leq x<2 \quad \Rightarrow 1 \leq x+1 \leq 3$ $$ \Rightarrow 1 \leq f(x) \leq 3 $$ $2<x \leq 3 \Rightarrow-3 \leq-x \leq-2$\ $\Rightarrow \quad 0 \leq 3-x \leq 1 \Rightarrow 0 \leq f(x) \leq 1$\ $\Rightarrow \quad 0 \leq x \leq 1 \Rightarrow 1 \leq f(x) \leq 2$ $$ \begin{aligned} & 1<x \leq 2 \Rightarrow 2<f(x) \leq 3 \ & 2<x \leq 3 \Rightarrow 0<f(x)<1 \end{aligned} $$ $\Rightarrow f(f(x))= \begin{cases}2+x, & 0 \leq x \leq 1 \ 2-x, & 1<x \leq 2 \ 4-x, & 2<x \leq 3\end{cases}$\ at $x=1$,\ R.H.L. $=\lim _{h ightarrow 0} g(1+h)=\lim _{h ightarrow 0} 2-(1+h)=1$\ $g(1)=3 \quad 	herefore$ Discontinuous at $x=1$\ at $x=2$,\ R.H.L. $=\lim _{h ightarrow 0} g(2+h)=\lim _{h ightarrow 0} 4-(2+h)=2$\ $g(2)=0 \quad 	herefore$ Discontinuous at $x=2$\