Timeline for Is the reciprocal function continuous?
Current License: CC BY-SA 3.0
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| Jul 5, 2017 at 16:36 | comment | added | Daniel R. Collins | Every freshman calculus text that I've seen explicitly restricts the definition of a continuous function to considering only points in the domain, such that the answer is "yes", $f(x) = 1/x$ is a continuous function. E.g., in Stein/Barcellos Sec. 2.8 that is the very first example. | |
| Aug 10, 2014 at 21:34 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
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| Aug 10, 2014 at 16:38 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
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| Aug 10, 2014 at 16:28 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
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| Aug 10, 2014 at 16:20 | comment | added | Pete L. Clark | @Darksonn: The Fundamental Theorem of Calculus applies separately on each interval. Thus the most general antiderivative of $g$ is $F(x) = \log x + C_1$ for $x > 0$ and $\log (-x) + C_2$ for $x < 0$, where $C_1,C_2$ are two arbitrary constants. (metacompactness and I had some previous exchanges about this which we both deleted.) | |
| Aug 10, 2014 at 9:04 | comment | added | user5402 | @Darksonn Of course. | |
| Aug 10, 2014 at 9:02 | comment | added | Alice Ryhl | @metacompactness By the fundamental theorem of calculus if $F(x)$ is the antiderivative of $f(x)$ then any other antiderivative can only differ by a constant. So no, there's no other antiderivatives. | |
| Aug 9, 2014 at 23:40 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
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| Aug 9, 2014 at 22:27 | comment | added | djechlin | +1 especially for how a mathematician would salvage the first question. | |
| Aug 9, 2014 at 14:25 | vote | accept | James S. Cook | ||
| Aug 9, 2014 at 4:16 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
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| Aug 9, 2014 at 4:10 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
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| Aug 9, 2014 at 4:03 | history | answered | Pete L. Clark | CC BY-SA 3.0 |