Counter-Example

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This question is a true and false question.

A rational function can have infinitely many x-values at which it is not continuous.

The textbook has the answer as:

false, because a rational function can be written as $\frac{P(x)}{Q(x)}$ where $P$ and Q are polynomials of degree m and n, respectively. It can at most have $n$ discontinuities.

I thought it could be

true, because I can write it as $$\frac{x}{(x-\infty)...(x-1)(x)(x+1)...(x+\infty)}$$

Is my thinking wrong?

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$$(x-\infty)\ldots(x-1)(x)(x+1)\ldots(x+\infty)$$ is not a meaningfull expression. There is no number $\infty$. The expression $$\cdots(x-1)(x)(x+1)\cdots$$ isn't meaningfull, too.

There are functions that have a zero at any $z \in \mathbb{Z}$ but these functions are not polynomials.

A polynomial has only finitely many zeroes. So you are wrong and the book is right.