Timeline for answer to Why is an ellipse, hyperbola, and circle not a function? by David K
Current License: CC BY-SA 3.0
Post Revisions
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2014 at 19:07 | audit | First posts | |||
| Dec 6, 2014 at 19:09 | |||||
| Dec 6, 2014 at 17:39 | audit | First posts | |||
| Dec 6, 2014 at 17:40 | |||||
| Nov 11, 2014 at 19:01 | history | edited | David K | CC BY-SA 3.0 |
thanks
|
| Nov 11, 2014 at 13:50 | comment | added | David K | @DavidZ Goos had already convinced me that I used the word "valid" poorly. In fact the test is always valid in the plane; one just has to understand what question it answers. And of course it is relevant as well; thanks for pointing out that misstatement. | |
| Nov 11, 2014 at 13:42 | history | edited | David K | CC BY-SA 3.0 |
corrections as discussed in comments
|
| Nov 11, 2014 at 12:44 | comment | added | David Z | I like most of this answer, but @Goos has a point. If I may make a suggestion: "valid only for functions of one variable...output of the function" to "valid only for relationships between exactly one input variable and one output variable, and only when you have used the horizontal axis to plot the input and the vertical axis to plot the output." (or something like that) And delete the very last sentence; that's just not correct. | |
| Nov 11, 2014 at 10:36 | comment | added | Jeppe Stig Nielsen | To state the same in another way. The graph of the function $f(x,y)=\frac{x^2}9 - \frac{y^2}4$ would be a surface in space. Every vertical line in space (that is a line parallel to the $z$ axis) intersects the surface exactly once. The hyperbola mentioned is the inverse image $f^{-1}( \{ 0 \} )$. That is the intersection of the graph with the $xy$ plane (given by $z=0$). | |
| Nov 10, 2014 at 22:48 | comment | added | David K | @Goos I suppose I could change "is only valid for" to something like "will tell you only whether something is a ..." (and slight modifications to make the words after that fit grammatically). That might be a bit less ambiguous. | |
| Nov 10, 2014 at 22:43 | comment | added | Caleb Stanford | @DavidK I suppose in the context of the original question your answer makes more sense than I'm giving it credit for. | |
| Nov 10, 2014 at 22:42 | comment | added | Caleb Stanford | @DavidK Okay, I think I understand what you're saying. I just think your wording is poor. The vertical line test certainly applies to a set like $x^2/9 - y^2/4 = 0$. When you say it is "not relevant" I can't see how you don't really mean, it fails in this case. | |
| Nov 10, 2014 at 22:38 | comment | added | David K | @Goos: By "valid for functions of one variable, etc." I meant that this test will determine whether something is exactly the kind of function I described, plotted the way I described. If its answer is "no" then you know the graph is not a single-variable function plotted that way. It's not the right tool for other kinds of functions, nor for single-variable functions plotted in other ways. | |
| Nov 10, 2014 at 22:20 | comment | added | Caleb Stanford | According to this answer, the vertical line test is only relevant to things which already, by definition, pass the test. What a useless test... | |
| Nov 10, 2014 at 22:17 | comment | added | Caleb Stanford | I don't know why this answer has so many upvotes, as it seems to entirely disregard what the vertical line test actually is. The vertical line test is absolutely applicable to any set of points in $\mathbb{R}^2$. It tells you, given such a set of points, whether that set is the output of a function $y = f(x)$. | |
| Nov 10, 2014 at 22:14 | comment | added | Caleb Stanford |
What in the world do you mean by The "vertical line" test is valid only for functions of exactly one variable? Surely this test would be completely useless if we could only apply it to things which were already functions--then it would always return true!
|
|
| Nov 10, 2014 at 12:36 | comment | added | David K | @Conclusio If $G_f$ means "the graph of $y=f(x)$ on a plane with Cartesian coordinates $x$ and $y$", then yes, $G_f$ for $f(x)=x^2$ is a parabola, as I stated. Usually this is what we would understand "the graph of $x^2$" to mean, because it's such a well-used convention. If things start to get complicated, however, sometimes we can clear up confusion by being very explicit about exactly what we mean. | |
| Nov 10, 2014 at 8:58 | comment | added | Fytch | But wouldn't the graph $G_f$ be a parabola? I reckon that calling a function a certain shape basically expresses the same as calling the function's graph $G$ that shape. (I'm sorry for my bad English) | |
| Nov 10, 2014 at 5:58 | vote | accept | JohnMerlino | ||
| Nov 10, 2014 at 5:56 | comment | added | yshavit | +1. I would add that $f(x, y)$ can be plotted onto $z$ to make a three-dimensional image. You can see it on wolfram alpha. | |
| Nov 10, 2014 at 5:13 | history | answered | David K | CC BY-SA 3.0 |