calculus

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determine the point of intersection of the tangents at the points of inflection to the curve f(x)= x^4 – 24x^2 – 2

  • calculus -

    y' = 4x^3 - 48x
    y '' = 12x^2 - 48
    at points of inflection, y'' = 0
    12x^2 = 48
    x^2 = 4
    x = ±2

    when x=2 , f(2) = 16 - 96 - 2 = -82 , slope = 32 - 96 = -64
    when x = -2, f(-2) = -82 , slope = -32 + 96 = 64

    1st tangent: slope = -64, point is (2,-82)
    -82 = -64(2) + b
    b=46
    first tangent equation: y = -64x + 46

    2nd tangent: slope = 64 , point is (-2,-82)
    -82 = 64(-2) + b
    b = 46
    second tangent is y = 64x + 46

    intersection of y = 64x + 46 and y = -64x + 46
    64x + 46 = -64x + 46
    128x = 0
    x = 0, then y = 46

    the two tangents intersect at (0,46)

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