The curve $$x^2y^3-3xy^2-9y+9=0$$ is given. I want to determine what points on the curve, for a neighbourhood to said points, $y$ can safely be expressed as a function of $x$. I guess what this means is that I need to find sets $A$ such that for $x_i \in A$ yields one and only one corresponding y-term $y_i$. Furthermore, this means that the y-term in the gradient in every such point is separate from zero, i.e. $$\frac{\partial f}{\partial y} (x_i, y_i) \neq 0$$ where $f$ is the the function $f : x \rightarrow y$. Would this be accurate?
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There are two related questions, and it's not entirely clear to me which you're asking: 1. Find the set of $x$ such that if $(x, y_{1})$ and $(x, y_{2})$ are on your curve, then $y_{1} = y_{2}$. 2. Find the set of $(x, y)$ on the curve such that in some open neighborhood $V$ of $(x, y)$, the intersection of $V$ with your curve may be expressed as the graph of a function $y = \phi(x)$. Interpretation 2. seems likelier; the implicit function theorem (checking that the partial with respect to $y$ is non-zero) gives a sufficient condition. Was the exact wording everything before "I guess..."? – Andrew D. Hwang May 20 '15 at 21:12
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Lozansky, why do you want to know? – Will Jagy May 20 '15 at 21:42
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@user86418 Yes I agree that 2) is a better description of the problem. I haven't heard about the implicit function theorem but seems like what I implied. The exact wording is not in English but it is very close to what I wrote before "I guess...". – Lozansky May 20 '15 at 21:49
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@WillJagy Because it is an exercise in my book. – Lozansky May 20 '15 at 21:50
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You can get a pretty good picture of the solution set online. To verify what that shows, and get specifics, you need to find the points where the graph is vertical, which happens twice, both at nice points. There are also asymptotes and so on. I made a small error the first time and the whole thing became impossible. Anyway, pictures first, always. Best if you can draw it yourself, but a computer picture will do here. – Will Jagy May 20 '15 at 21:54
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An elementary way: assume $(x,y_1)$ and $(x,y_2)$ lie on the curve. Then
$$x^2y_1^3-3xy_1^2-9y_1+9=x^2y_2^3-3xy_2^2-9y_2+9$$ $$0=x(y_1^3-y_2^3)-3x(y_1^2-y_2^2)-9(y_1-y_2)$$ $$0=x(y_1^2+y_1y_2+y_2^2)-3x(y_1+y_2)+9$$
Now find when a real solution $y_2$ exists by taking a discriminant. As I'm working it out, it appears that we need to use the quadratic formula to solve for $x$ in terms of $y_1$, so it's not at all a pleasant bash...
Another approach - that for which you were probably looking - is finding the $x$-extrema by finding the critical points of the function.
Cody Johnson
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picture https://www.wolframalpha.com/input/?i=plot+x^2+y^3%E2%88%923xy^2%E2%88%929y%2B9%3D0 which can probably be verified by hand, as far as number of (real) components, asymptotes, and so on.... The link seems to have split... – Will Jagy May 20 '15 at 21:02
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Yes, agreed. Overall a somewhat bashy problem, but hey, it answers the question. – Cody Johnson May 20 '15 at 21:04
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Understood (and expected, else the problem would be /complete/ trash), but how could one elegantly synthesize that? – Cody Johnson May 20 '15 at 22:29
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Implicitly solve $x' = 0.$ Not $y',$ rather $x',$ or $\partial x / \partial y.$ I posted a picture as an answer. The book with the question was *extremely* careful about making the task of setting $x'=0$ computationally, um, self-contained, entirely do-able by hand. OHHH: or find the points where the gradient is horizontal! – Will Jagy May 20 '15 at 22:32
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next day; the point where the gradient is vertical also comes out nice. In all cases with the gradient, we are able to solve for the product $xy,$ which tells us $x^2 y^2$ for free. Plugging that back into the original gives a single $y$ value. Anyway, two values for $xy$ with horizontal gradient, one for vertical. Extremely careful construction of the problem, I would not know how to make one of these come out this well. – Will Jagy May 21 '15 at 18:24
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I think I will post this picture, I think students here undervalue the skill of drawing; in this case, I am putting a computer picture, but it can be substantiated. The calculation part, for you to finish, is to correctly identify the two points where the graph is perfectly vertical, one in the first quadrant, one in the second. Also verify that the apparent asymptotes make sense, and so on. If you make no mistakes, the two vertical points come out with nice coordinates. If, as I did the first time, you make any small error, it comes out garbage and confusing. EDIT: for many of you, we are looking for the exact points where the gradient of the two-variable function is horizontal. This is the easiest case of Lagrange multipliers: to find the local extrema of the function $x$ constrained to the curve(s) $f(x,y) = x^2 y^3 -3x y^2 - 9 y + 9$ is set to $0,$ find the points on the curve where $\nabla f$ is a scalar multiple of $(1,0),$ which is to say, the points on the curve(s) where $$ \frac{\partial f}{ \partial y} = 0. $$
Will Jagy
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