Let f,g:RR be functions defined by f(x)=cases[x], & x

Question

Let $f,g:\mathbb{R}	o\mathbb{R}$ be functions defined by $f(x)=\begin{cases}[x], & x<0\ |1-x|, & x\ge0\end{cases}$ and $g(x)=\begin{cases}e^x-x, & x<0\ (x-1)^2-1, & x\ge0\end{cases}$, where $[x]$ denotes the greatest integer $\le x$. Then, the function $f\circ g$ is discontinuous at exactly:

Accepted Answer

$$f\circ g(x)=\begin{cases}[g(x)], & g(x)<0\ |1-g(x)|, & g(x)\ge0\end{cases}=\begin{cases}|1+x-e^x|, & x<0\ 1, & x=0\ [(x-1)^2-1], & 0<x<2\ |2-(x-1)^2|, & x\ge2\end{cases}$$

Checking the boundary points $x=0$ and $x=2$ (where $g$ changes branch, and where $g(x)$ crosses $0$), the composite function $f\circ g$ is found to be discontinuous at exactly $x=0$ and $x=2$: two points.

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