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Math Help - application of differentiation

  1. #1
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    application of differentiation

    the equation of a curve is y = 2 x^3 -21 x^2 +78x-98
    Find the gradient at the point (3,1). Find the x-coordinate of the point at which the tangent of the curve is parallel to the tangent at (3,1).

    ------------

    i can solve the first part of the question.....answer is 6....
    the second part of the question is the one i have problem with.....it sounds very confusing....
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Here
    f '(x)=6x^2-42x+78
    Now for tangent to be parrallel to yours
    6(x^2-7x+13)=6
    x^2-7x+12=0
    Hence
    (x-3)(x-4)=0
    So the x coordinate of other point is 4
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