This is a question in our book, and the answer in the teacher's book is "No", and a friend of mine says that their teacher also says that a rational function always has a definite $n$ number of docontinuity points.
But I can think of a few functions that have an infinite number of discontinuity points;
1) $f(x)= \frac{1}{\sin x} $ which is discontinuous at every x value that satisfies $x= n\pi$ where $n$ is an integer. The same is also true for this function but with cosine and tan and their respective infinite sets of discontinuity points
2) $f(x) = \frac{1}{\sqrt{r^2-x^2}} $ which is discontinuous over the whole interval $[-r,r]$ and has an infinte number of discontinuity points that belong to this interval
Now I think I'm wrong here because I'm not sure if these count as "rational functions" or not.
Actually, since a rational function is a quotient of two polynomial functions, since polynomial functions are continuous and since the quotient of two continuous functions is continuous, every rational function has zero points of discontinuity.