Theorem about bounded functions.

Question

Is there a theorem that says that if $f:[a,b]\rightarrow \mathbb{R} $ is a piecewise function, then it is bounded?

Answer 1

Not without more restrictions, like continuity (which is enough).

For example, consider $$f(x) = \begin{cases} 0,& x=a \\ \dfrac{1}{x-a},& a<x\leq b\\ \end{cases} $$

If your function is continuous, then it is bounded since the continuous image of a compact set is compact (in $\mathbb{R}$, this means it is closed and bounded).