Let f:[-1,3]R be defined as f(x)=cases|x|+[x], & -1 x

Question

Let $f:[-1,3]	o\mathbb{R}$ be defined as $f(x)=\begin{cases}|x|+[x], & -1\le x<1\x+|x|, & 1\le x<2\x+[x], & 2\le x\le3\end{cases}$, where $[t]$ denotes the greatest integer $\le t$. Then, $f$ is discontinuous at:

Accepted Answer

$$f(x)=\begin{cases}-x-1, & -1\le x<0\x, & 0\le x<1\2x, & 1\le x<2\x+2, & 2\le x<3\6, & x=3\end{cases}$$
Checking each join: $f(0^-)=-1\ne f(0)=0$; $f(1^-)=1\ne f(1)=2$; $f(3^-)=5\ne f(3)=6$. So $f$ is discontinuous at exactly three points ($x=0,1,3$).

Solutions · 1