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.
Best Answer
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...