# This is a question in our book, and the answer in the teacher's book is "No", and a friend of mine s

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) $f\left(x\right)=\frac{1}{\mathrm{sin}x}$ 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) $f\left(x\right)=\frac{1}{\sqrt{{r}^{2}-{x}^{2}}}$ which is discontinuous over the whole interval $\left[-r,r\right]$ 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.
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Jayvion Mclaughlin
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.