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How would I find all the discontinuities in the following function.

$$f(x)=\begin{cases} x+2,&\text{if }x<-2\\ 2,&\text{if }x=-2\\ -x^2+4,&\text{if }-2<x\le 1\\ x+1,&\text{if }1<x\;. \end{cases}$$

I have found jump discontinuities in $x=-2$ and $x=1$ but I am unsure if those are all the ones there are.

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These are the only two, but the discontinuity at $-2$ is a removable discontinuity, not a jump discontinuity: the limits from each side are $0$.

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Oh I did not see that thanks. – Fernando Martinez Jan 26 '13 at 23:16
@Fernando: You’re welcome. – Brian M. Scott Jan 26 '13 at 23:17

You have found all. Note that on the intervals: $(-\infty, -2)$, $(-2,1)$, and $(1,\infty)$ the function is given by continuous functions. So the only place that the function could be discontinuous is at $-2$ and at $1$.

Note however, that at $-2$ you have $$ \lim_{x\to -2^-}f(x) = 0\quad\text{and}\quad \lim_{x\to -2^+} f(x) = 0. $$ So here the graph doesn't jump.

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