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Math Help - help!

  1. #1
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    help!

    show that the cubic y=x^3+2x^2+5 has zero somewhere between x=-2.5 and x=-3. pick a suitable starting estimate of the root, x0, and perfrom 3 iterations of newtons method. test your final estimate to see how good it is.

    how do i go about this?
    thanks
    jimmy
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  2. #2
    Super Member Showcase_22's Avatar
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    The first part is the intermediate value theorem.

    The second bit is newton Rhapson method.

    Newton's Method -- from Wolfram MathWorld
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  3. #3
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    i dont understand the intermediate value therom, do you know any good references for this?
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  4. #4
    Super Member Showcase_22's Avatar
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    y=x^3+2x^2+5
    f(x)=x^3+2x^2+5

    f(-3)=(-3)^3+2(-3)^2+5=-27+18+5=-4<0

    f(-2.5)>0 (I can't work this out in my head!)

    f(x) is continous since it's a polynomial (and all polynomials are continuous).

    By the Intermediate Value Theorem f(x) must assume every value between f(-3) and f(-2.5) since it's continuous. Therefore \exists \ x_0 \in \mathbb{R} \ s.t \ f(x_0)=0. x_0 is a solution to f(x).
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