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$xy^2-x^3y = 2$

$y' = \frac{3x^2y-y^2}{2xy-x^3}$

find the x-coordinate of each point on the curve where the tangent line is horizontal.

I know that $3x^2y-y^2 = 0 \Rightarrow$ then $y=3x^2.$

If I plug y into the original I get

$x(3x^2)^2-x^3(3x^2)=2 \Rightarrow x(9x^4)-3x^5 = 2 \Rightarrow 9x^5 - 3x^5 = 2 -> x^5 = \frac{1}{3} \Rightarrow x = (\frac{1}{3})^{(\frac{1}{5})}$

Is this all I have to do?

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  • $\begingroup$ I think you answered the question. $\endgroup$ – Claude Leibovici Dec 18 '13 at 12:31
  • $\begingroup$ So there's only one solution right? $\endgroup$ – SiDiX Dec 18 '13 at 12:51
  • $\begingroup$ @SiDix.Take care : you and I missed one solution since y' can be zero also if y = 0 (the term you factored out). But, if y=0, we should have 0 = 2 from the equation of the curve. Then, only one root (the one you found). $\endgroup$ – Claude Leibovici Dec 18 '13 at 13:06
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Yes, you've found the unique solution: $\quad x = \left(\frac 13\right)^{1/5}$.

Note: It also happens to be true that at $y = 0$, then $y' = 0$. But it turns out that $y = 0$ does not satisfy the original equation, and so does not lie on the curve. Hence, we have only the solution that you found.

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  • $\begingroup$ @amWhy: Needs another UV +1 $\endgroup$ – Amzoti Dec 19 '13 at 2:44
  • $\begingroup$ @amWhy: Needs another $\color{red}{\mathbf{+1}}$. $\endgroup$ – user93957 Dec 26 '13 at 12:54

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