calculus

Find the points of intersection of the curves y=2sin(x-3) and y=-4x^2 + 2

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  1. tough question!

    I used Newton's Method
    and I let
    y = 2sin(x-3) + 4x^2 - 2
    then dy/dx = 2cos(x-3) + 8x

    xnew = x - (2sin(x-3) + 4x^2 - 2)/(2cos(x-3) + 8x)

    I started with x = 1
    and my first xnew was.974691
    make that your current x and find the next xnew
    I quickly converged to
    x = 0.974237

    set your calculator to radians and test my answer, it works making both equations equal.
    (error was .000000418 between LS and RS

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  2. Reiny found a positive root but there is a negative root x=-0.55 (approximately)
    We can started with x=-1

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