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March 27, 2017

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find the coordinates of the point where the tangent to the curve y=x^3 +x +2 at the point (1,4) meets the curve again. [ans:-2,-8]
pls help me i don't understand the question....

  • math - ,

    dy/dx = 3x^2 + 1
    at (1,4) , dy/dx = 4
    tangent equation
    4 = 4(1) + b
    b = 0
    the tangent equation at (1,4) is y = 4x
    where does this cross or touch y = x^3 + x + 2 ?
    x^3 + x + 2 = 4x
    x^3 - 3x + 2 = 0
    let f(x) = x^3 - 3x + 2
    f(1) = 1 - 3 + 2 = 0 , so x-1) is a factor

    by synthetic division ...
    x^3 - 3x + 2 = (x-1)(x^2 + x - 2)
    = (x-1)(x-1)(x+2)

    then the roots of x^3 - 3x + 2 = 0 are
    x = 1 and x = -1 and x = -2
    we knew x = 1 from the tangent part, so the new x = -2 is the other intersection point value
    if x = -2
    y = -8 - 2 + 2 = -8

    the other point is (-2,-8)

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