Consider the function $$f(x) = \frac{x^3}{1+x^3}$$
Obviously this function is discontinuous at $x = -1$ therefore discontinuous on $\mathbb{R}$. Moreover, it is unbounded at the same point. Now, I would not say that this function is locally bounded either, as not all sets $f(A)$ are bounded for a neighbourhood $A$ about any $x_0 \in \mathbb{R}$. Is this reasoning correct? Can an unbounded discontinuous function be locally bounded?
Answer
If I understand your definition correctly, the function $$f(x) = \begin{cases}x - 1 & x < 0\\
0 & x = 0\\
x+1 & x>0\end{cases}$$
fits the bill.
If you want a rational function, then no can do, because if $p, q$ are polynomials with no common factors, then $f(x)=\frac{p(x)}{q(x)}$ is continuous if $q(x)\neq 0$ and is unbounded around $x$ if $q(x)=0$.
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