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Math Help - Bisection, Newton's, and secant method help!

  1. #1
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    Bisection, Newton's, and secant method help!

    Can't seem to find the answer to this one.

    Question:
    Using the bisection method, Newton's method, and the secant method, find the largest positive root correct to three decimal places of x^3-5x+3=0. (Hint: All roots are in [-3, +3]).

    Not really sure what the bisection method is but here's the newton's method and the secant method.

    Bisection method:

    Newton's Method: Xn+1=Xn-(f(xn)/f '(xn)

    Secant Method: Xn+1=Xn-((Xn-Xn-1)/(f(xn)-f(xn-1))) * f(xn)

    Any help would be appreciated!
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  2. #2
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    Quote Originally Posted by Porter1 View Post

    Not really sure what the bisection method is but here's the newton's method and the secant method.

    Bisection method:
    With the bisection method you take the given interval and cut it in half.

    You then decide which half to take given the functions value at the interval fences.

    Your interval is [-3,3] so first of all find f(-3) and f(3). You will notice one result is positive and one negative. This means the root is on the interval.

    Now bisect the interval \frac{-3+3}{2} = 0 and find the function value for that, f(0) , this bisection gives two new intervals to choose from either [-3,0] or [0,3].

    You now have f(-3) = -39, f(3)=45 and f(0)=3 as pick the interval such that either side is still positive and negative. i.e f(-3) = -39 and f(0)=3 means the root is on [-3,0] and not [0,3] , bisect ths new interval again applying the same logic until you converge on the root.
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  3. #3
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    Quote Originally Posted by Porter1 View Post
    Can't seem to find the answer to this one.

    Question:
    Using the bisection method, Newton's method, and the secant method, find the largest positive root correct to three decimal places of x^3-5x+3=0. (Hint: All roots are in [-3, +3]).

    Not really sure what the bisection method is but here's the newton's method and the secant method.

    Bisection method:

    Newton's Method: Xn+1=Xn-(f(xn)/f '(xn)

    Secant Method: Xn+1=Xn-((Xn-Xn-1)/(f(xn)-f(xn-1))) * f(xn)

    Any help would be appreciated!
    Since you have the formulas for Newton's method and the secant method what is the problem?

    Show us what you have done and/or explain what the difficulty is that you have with these.

    CB
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