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Math Help - gradient

  1. #1
    Junior Member
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    gradient

    find the gradient to the curve y = x^3 - 3x^2 at the point x = 3
    then find the equations of the tangent and the normal to the curve at the point where x = 3


    thanks
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  2. #2
    Member Henderson's Avatar
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    f(x) = x^3 - 3x^2
    f'(x) = 3x^2 - 6x
    f'(3) = 3(3)^2 - 6(3) = 9

    So the slope of your tangent is 9, and the slope of your normal is -\frac{1}{9}.

    Also, since
    f(3) = (3)^3 - 3(3)^2 = 0,
    both equations will pass through the point (3, 0).

    Let us know if you need more help from there.
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