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I'm a little confused by the concept of "piecewise-continuous". My understanding is that if a function $f$ is piecewise-continuous on some domain $[a, b]$, it is not necessarily continuous on $[a, b]$, but it can be cut into pieces which are continuous on subintervals of $[a, b]$. Does this sound correct?

Would the following function be piecewise-continuous?

\begin{cases} -12 & -1 \leq x < 0 \\ \frac{x}{1-x^3} & 0\leq x < 1 \\ 5-x & 1\leq x \leq 2 \end{cases}

lioness99a
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Nażysław Zbyłutowicz
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  • Yep, the qualifier piecewise means having the desired property in subintervals. For instance, a function can be piecewise linear. –  Mar 06 '17 at 08:41

2 Answers2

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Yes this is correct, and the function you described is an example of a function that is piecewise continuous, but not continuous on $[-1,2]$

SquirtleSquad
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Your example is correct. Another way to describe the piecewise-continuous functions is like this: if $f$ is defined in a closed interval $[a,b]$ then for every point in $(a,b)$ both lateral limits exists, and the lateral limits $\lim_{x\to a^+}f(x)$ and $\lim_{x\to b^-}f(x)$ also exists.

Masacroso
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