babyagelesszj

2022-07-15

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) $f(x)=\frac{1}{\mathrm{sin}x}$ 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) $f(x)=\frac{1}{\sqrt{{r}^{2}-{x}^{2}}}$ 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.

But I can think of a few functions that have an infinite number of discontinuity points;

1) $f(x)=\frac{1}{\mathrm{sin}x}$ 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) $f(x)=\frac{1}{\sqrt{{r}^{2}-{x}^{2}}}$ 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.

Jayvion Mclaughlin

Beginner2022-07-16Added 14 answers

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.

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