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Math Help - Tangent to a curve

  1. #1
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    Tangent to a curve

    For curve  x^3 + y^3 - 9xy = 0 find x co-ordinates where the tangent to the curve is horizontal.

    I differentiated:  \frac{d}{dx}[x^3 + y^3 - 9xy] = \frac{3y - x^2}{y^2 - 3x}

    Let this equal zero for horizontal tangents, but I don't know how to solve from here, the answer is  54^{\frac{1}{3}} but my answers always seem to have a y term in them.

    Thanks in advance,
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  2. #2
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    horizontal tangent ...

    3y - x^2 = 0

    y = \frac{x^2}{3}

    x^3 + \left(\frac{x^2}{3}\right)^3 - 9x\left(\frac{x^2}{3}\right) = 0

    solve for x
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