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Find points where tangent line is horizontal

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the points on the graph of x3-y3=3xy-3 where the tangent line is horizontal


    2. Relevant equations

    y = f(x) so implicit differentiation must be used when taking the derivative of y

    (xy)' = xy' + y


    3. The attempt at a solution

    So if the tangent line is horizontal the slope = 0 and the derivative of y' must be equal to zero.

    So first taking the derivative I end up with

    3x2-3y2y' = 3xy' + 3y

    Factor out the three and divide by three on both sides to cancel them out. That leaves me with

    x2-y2y' = xy' + y

    Subtract y on both sides, add y2y' on both sides and get

    x2-y = xy' + y2y'

    Factor out the y' to get

    x2-y = y'(x+y2)

    divide by x+y on both sides and

    y' = (x2-y)/(x+y2)

    So at this point, I set y' equal to zero and solve for x? Don't I remove the y variables or something like that?
     
    Last edited: Jun 9, 2013
  2. jcsd
  3. Jun 9, 2013 #2
    Remove it how?

    You're looking for points (x,y) where dy/dx = 0. You have dy/dx = some fraction. When is a fraction zero?
     
  4. Jun 9, 2013 #3
    Well I understand part of what you're saying.

    I can solve the y' so that x2 = y and then go back into the equation of y' and replace y with x2 and demonstrate that it equals zero. Do I plug x2 into the original y and solve for x? Or take the derivative and solve for x?
     
  5. Jun 9, 2013 #4

    Dick

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    Second choice. Substitute y=x^2 into the original equation and find x. Having a horizontal tangent at (x,y) means it has to satisfy the original equation AND y'=0. Two equations.
     
    Last edited: Jun 9, 2013
  6. Jun 10, 2013 #5
    OK, so x3-y3 = 3xy -3

    Substitute,

    x3 -(x2)3 = 3(x)(x2) - 3

    x3 - x6 = 3x3-3

    Subtract 3x3 both sides and get

    -2x3-x6 = -3

    So at this point, I'm not sure whats factorable and whats not. I could factor out an x3 and get

    x3(-2-x3) = -3 but I'm not sure how much that helps me. Do I need to use the quadratic formula at some point?

    NOPENOPENOPE I think I got it. I believe x = √3 cubed root rather
     
  7. Jun 10, 2013 #6

    ehild

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    Not quite. And what about the other root?

    ehild
     
  8. Jun 10, 2013 #7
    Yeah I'm starting to realize that wasn't right

    x3 - x6 = 3x3-3

    is that at least on the right track?
     
  9. Jun 10, 2013 #8

    Dick

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    Fine so far. Move everything to one side and try to factor if. It's a quadratic in u=x^3.
     
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