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I'm working though (on my own; and unfortunately I don't live near a university) a text on Real Analysis: Bartle and Sherbert, 4th Ed. Chapter 5 is "Continuous Functions". In Section 5.2 ("Combinations of Continuous Functions") one of the exercises (#3, pp. 133) asks for functions f and g which are discontinuous at some point c in their shared domain, but when the functions are multiplied that function is continuous at the point c.

One of the difficulties in studying on my own is that -- for open-ended questions like this -- it is difficult to know if my own ideas are correct. I understand the answer suggested in the book, and I think I understand the question.

However, I have no clear way of determining if MY idea is (also) correct. I came up with the two functions:

f: = (e^(1/(1-x)))/(1-x)

g: = the reciprocal of f

Thus, when multiplied together, their product-function is the constant 1, and continuous at 1 (and elsewhere). But each is discontinuous at the point x = 1.

Thus, I believe this answer is also correct? Is it?

Thanks for taking the time to help me! (I'm actively looking for a local tutor, as we're a large research-base oriented community and there are at least some mathematicians working here; but no luck so far...)

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    $\begingroup$ By "inverse" do you mean "reciprocal"? $\endgroup$ – lulu Aug 13 at 22:49
  • $\begingroup$ Yes, sorry. Corrected. Thanks! $\endgroup$ – Larry Cosner Aug 13 at 22:50
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Actually, your $f$ is undefined at $1$, so it's neither continuous nor discontinuous there.

You can take$$f(x)=\begin{cases}1&\text{ if }x\in\mathbb Q\\2&\text{ otherwise}\end{cases}$$and$$g(x)=\begin{cases}2&\text{ if }x\in\mathbb Q\\1&\text{ otherwise.}\end{cases}$$Then $f$ and $g$ are discontinuous everywhere, but $f\times g=2$.

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  • $\begingroup$ Indeed, this was the answer the book had (or very similar). Thanks! $\endgroup$ – Larry Cosner Aug 13 at 22:52
  • $\begingroup$ And got it. Thanks also. I'll keep looking...! $\endgroup$ – Larry Cosner Aug 13 at 22:53
  • $\begingroup$ The exponential, when in the denominator, accomplishes what I want, as it approaches the limiting value of 0 (at x = 1), but remains defined, right? So I need another function that has this property, as the simple (1-x) is undefined at x=1. Is that correct? $\endgroup$ – Larry Cosner Aug 13 at 22:55
  • $\begingroup$ I cannot answer, because I do not understand what you are saying. $\endgroup$ – José Carlos Santos Aug 13 at 23:00
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    $\begingroup$ The expression $1-x$ is defined at $x=1$. But it is $0$ there, and therefore any expression of the type $\frac{f(x)}{1-x}$ is undefined at $x=1$. $\endgroup$ – José Carlos Santos Aug 14 at 7:26

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