Question

Let $f\left(x\right)=1+xfor0\le x\le 2$ $=3-xfor2<x\le 3.$ Determine the form of $g\left(x\right)=f\left(f\left(x\right)\right)$ and hence find the number of points of discontinuity of g

Answer 1

By definitions of the functions $f$ and $g$, we have

$g\left(x\right)=f\left(f\left(x\right)\right)=1+f\left(x\right)$ for $0\le f\left(x\right)\le 2$
$=3-f\left(x\right)$ for $2<f\left(x\right)\le 3.$
Now we find the form of $g\left(x\right)$ as a function of $x$.

We consider two sub-intervals $I_1(0\le x\le 2)$ and $I_2(2<x\le 3)$ of the interval $I(0\le x\le 3).$
(i) On $I_1,$ we have $f\left(x\right)=1+x$
Now $0\le f\left(x\right)\le 2\Rightarrow 0\le 1+x\le 2$ $\Rightarrow -1\le x\le 1\Rightarrow 0\le x\le 1$ on $I_1.$
$\therefore g\left(x\right)=1+f\left(x\right)=1+(1+x)=2+x$ for $0\le x\le 1.$

and $2<f\left(x\right)\le 3\Rightarrow 2<1+x\le 3$

$\Rightarrow 1<x\le 2$ which belong to $I_1$
$\therefore g\left(x\right)=3-(1+x)=2-x,1<x\le 2.$
(ii) On $I_2$,we have $f\left(x\right)=3-x.$

Now $0\le f\left(x\right)\le 2$
$\Rightarrow 0\le 3-x\le 2\Rightarrow -3\le -x\le -1\Rightarrow 3\ge x\ge 1\Rightarrow 1\le x\le 3.\Rightarrow 2\le x\le 3$ on $I_2.$
$\therefore g\left(x\right)=1+(3-x)=4-x$ for $2<x\le 3.$

and $2<f\left(x\right)\le 3\Rightarrow 2<3-x\le 3.$
$\Rightarrow -1<-x\le 0\Rightarrow 0\le x<1<ϵI_2.$
$\therefore g\left(x\right)$ is not defined in this case.
Thus $g$ is defined as follows :$g\left(x\right)=2+x$ for $0\le x\le 1$
$=2-x$ for $1<x\le 2$

$=4-x$ for $2<x\le 3.$
We test the function $g$ for continuity at $x=1$ and $x=2$ only.$g\left(1\right)=3,g(1-0)=\underset{h\to 0}{lim}[2+(1-h)]=3$ and $g(1+0)=\underset{h\to 0}{lim}[2-(1+h)]=1.$
Hence $g$ is discontinuous at $x=1$ $g\left(2\right)=2-2=0,$
$g(2-0)=\underset{h\to \to 0}{lim}[2-(2-h)]=0$ and $g(2+0)=\underset{h\to \to 0}{lim}[4-(2+h)]=2.$
Hence $g$ is also discontinuous at $x=2.$