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So I have this question for AP Calculus. It doesn't seem to be that difficult, but I just can't wrap my head around it. I believe that I have answered (a) and (b) correctly, but I have no clue as to what to do for (c). I know that for a tangent line to be horizontal the slope has to be $0$, so the numerator of the derivative has to be $0$, but I don't know the process in which to get that.

A curve in the $xy$ plane is defined by $xy^2-x^3y=2$.

(a) Find $\frac{dy}{dx}$

(b) Find an equation for the tangent line at each point on the curve with $x$-coordinate $1$.

(c) Find the $x$-coordinate of each point in the curve where the tangent line is horizontal.

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So you say you have computed $dy/dx$ correctly, if so, then in the form $"dy/dx="$ gives you the gradient function. What do you know about gradient functions or, more simply just gradients, and horizontal lines?

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  • $\begingroup$ I haven't learned gradient functions yet. All I know is that the numerator of the derivative is suppose to equal 0. $\endgroup$ – SiDiX Dec 11 '13 at 11:29
  • $\begingroup$ Exactly. This is because a horizontal line has zero gradient, for this to be true, your complicated $dy/dx$ must be zero, meaning (if it is a fraction) the numerator is zero. Hence, you must solve this numerator equal to zero (solve in $x$) and hey presto, you should have your $x$ co-ordinate. $\endgroup$ – Autolatry Dec 11 '13 at 11:32
  • $\begingroup$ Yes, I know that. However, I get that x=sqrt(y/3) Is that the answer? $\endgroup$ – SiDiX Dec 11 '13 at 11:35
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    $\begingroup$ Try solving for y and plugging that into x = sqrt(y/3) $\endgroup$ – mdenton8 Dec 11 '13 at 11:36
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    $\begingroup$ Yes, so you should end up with (after solving for your numerator in $dy/dx$ that $y=3x^2$. This tells you that this relationship says something about where the minimum or maximum occurs. Plug $y=3x^2$ into your initial (implicit) relationship between $x$ and $y$ in the very first line in your jpeg. You will then have only $x$ to solve for, you should get two values. $\endgroup$ – Autolatry Dec 11 '13 at 11:42
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What do you know about horizontal lines? What is their slope? You learned that in Algebra 1.

And we know that the slope of the tangent line of a function is equal to the derivative of the function right? That's elementary calculus.

Now that you know the slope of what you are looking for, you can just set the derivative of the function equal to the slope.

Make sure you solve for y in your original equation so that when you get your derivative equation, you only have one variable (x). This is a standard way to solve implicit functions and differential equations in AP Calc BC.

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  • $\begingroup$ Yes, I know all of that, I have set the numerator of the derivative to 0 and got x=sqrt(y/3) However, I believe that the problem was asking for the specific points. $\endgroup$ – SiDiX Dec 11 '13 at 11:36
  • $\begingroup$ Right, see my comment on the other guy's answer $\endgroup$ – mdenton8 Dec 11 '13 at 11:36
  • $\begingroup$ Just solve for y and solve for x $\endgroup$ – mdenton8 Dec 11 '13 at 11:37
  • $\begingroup$ Solve for y in the regular (non-derivative) equation? $\endgroup$ – SiDiX Dec 11 '13 at 11:37
  • $\begingroup$ Yup, still commenting on the other guy's question, sorry about that. $\endgroup$ – mdenton8 Dec 11 '13 at 11:38

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