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The curve y=x^3-3x^2-8x+4 has tangent L at point P (1,-8). Given that the Line M is parallel to L and is also a tangent to Q show that the shortest distance between L and M is 16 root 2

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1 answer

  1. y' = 3x^2 - 6x - 8
    at (1,-8), y' = 3 - 6 - 8 = -11
    So at Q, the slope of the tangent must also be -11
    3x^2 - 6x -8 = -11
    3x^2 - 6x + 3 = 0
    x^2 - 2x + 1 = 0
    (x-1)^2 = 0
    x = 1
    mmmhhh, expecting two values of x

    y'' = 6x-6
    = 0 for point of inflection
    x = 1
    if x = 1, y = 1-3-8+4 = -6

    ahhh, found the problem.
    you said that there was a tangent at (1, -8), but that point does not lie on the curve

    There is a typo, or the question is bogus.

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