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I've been trying to figure out how can I solve this exercise but I haven't had much luck so far. Do you think you can help me out a bit? Pointing out what might I possibly be doing wrong?

The exercise is as follows:

Find the coordinates where the tangent to the curve is horizontal

$x^3+3xy+2y^2+4y=1$

Given that it's difficult to solve for either x or y. I decided to differentiate implicitly. And here's what I got:

$- {3x^2+3y\over 4y+3x+4}=0 $

In order to find the horizontal tangents, the first order differential must be zero, and for this case particularly:

$ 3x^2+3y=0 $

Now, solving for x:

$x=\sqrt{-y}$ $x=-\sqrt{-y}$

Which tells me that y must be positive. (Real field)

But now I'm stuck there. Just looking at the answers I can't think of anything else but some numbers that might satisfy the equation; $(1,-1)$, $(-1,-1)$,$(0,0)$ But I wouldn't know how to get there, nor I know if those are the right coordinates. Can you help me out? Thanks in advance.

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    I actually have, and I think I'm really far from finding the correct answer.2012-11-08

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Your first steps seem right, but then it is much more natural to solve for $y$: $~~y = -x^2$. Of course, the points have to lie on the original curve, so substitute this back into the original equation to get $ 2x^4 - 2x^3 - 4x^2 - 1 = 0 ~. $ So now you need the real roots of this quartic. This is slightly tricky though; it certainly has real roots, but I don't think it has any rational roots (assume it does, derive a contradiction), for example...

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    Actually it was one of the problems on my Calculus quiz. And I had asked my teacher about it in class, but he didn't actually answer because he was a bit busy. And I was surprised when I saw it on my quiz, anyway, I talked to him and he agreed on just leaving it till the implicit differentiation.2012-11-08