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Homework Help: Write an equation of each horizontal tangent line to the curve.

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Write an equation of each horizontal tangent line to the curve.

    2. Relevant equations

    y = 2y^3 + 6x^2y - 12x + 6y = 1
    y' = (4x - 2xy) / x^2 + y^2 + 1)

    3. The attempt at a solution

    Well, horizontal tangent line means the derivative equals zero. Thus,

    4x - 2xy = 0
    2x(2-y) = 0
    x = 0, y = 2​

    Since I need an equation, I plug x = 0 back into the original function and end up with

    2y^3 + 6y = 1​

    Then I need to solve for y. I guess this is more of an algebra question than a calculus question? But nevertheless, I don't know how to solve for y.. I could do 2y(y^2 + 3) = 1 but that doesn't help me.
     
  2. jcsd
  3. May 2, 2010 #2

    tiny-tim

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    Hi lude1! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Where does the 4x come from? :confused:

    (Or should the 12x be 12x2 ?)
     
  4. May 2, 2010 #3
    Ah, yes, you are correct! It should be 12x2

    Thus:

    y = 2y3 + 6x2y - 12x2 + 6y = 1

    and

    y' = (4x - 2xy) / (x2 + y2 + 1)
     
  5. May 2, 2010 #4

    tiny-tim

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    (why do you keep starting with "y =" ? :frown:)

    hmm … then, sorry, I don't think your 2y3 + 6y = 1 has a straightforward answer

    (unless the = 1 should be = 8)

    oh, and don't forget you still have to deal with the other case, y = 2. :wink:
     
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