Consider the function f(x)=x31+x3

Obviously this function is discontinuous at x=−1 therefore discontinuous on 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 x0∈R. Is this reasoning correct? Can an unbounded discontinuous function be locally bounded?
Answer
If I understand your definition correctly, the function f(x)={x−1x<00x=0x+1x>0
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)=p(x)q(x) is continuous if q(x)≠0 and is unbounded around x if q(x)=0.
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