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Math Help - How do I choose the correct range to find zeros

  1. #1
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    How do I choose the correct range to find zeros

    How do I choose the correct range of numbers to find the zeros...
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  2. #2
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    Quote Originally Posted by coopsterdude View Post
    How do I choose the correct range of numbers to find the zeros...
    Trial and error, We know from Descartes rule of signs that there is exactly one positive root, and 0 or 2 negative roots. Also we know that for large positive x f(x) is positive, and "large" negative x f(x) is negative from the sign of the x^3 term.

    So evaluate f(0) it should be -ve, take positive steps untill f changes sign (in this case between 1 and 2). Then as the sum of the roots is equal to -the coefficient of x^2, we conclude that the negative roots if they exist, are >=-5.

    There are almost certainly other methods of establishing a suitable range.

    RonL
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    Forum Admin topsquark's Avatar
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    A simple way, if you are allowed to use it, is to graph the function (as on a calculator or a computer). Then you can "eye-ball" where the solutions are and use a table to compute them to get the exact values.

    -Dan
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    Thanks for your help. I used the trial and error method but it's so time consuming.

    Thanks again,

    coopsterdude
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    Quote Originally Posted by coopsterdude View Post
    Thanks for your help. I used the trial and error method but it's so time consuming.

    Thanks again,

    coopsterdude
    A rather simple method (that computer programs use) is the "bisection method".

    First find an interval that has only one zero and work in that interval.

    Example,

    y=x^3+x-1
    We can to find the zero of this. That is to solve,
    x^3+x-1=0
    You can draw a graph and see where it intersects only once and work in that interval. Look below.
    So work on interval,
    [0,2]

    Now, chose the midpoint x=1 does it work?
    1^3+1-1=1>0 too large.
    Meaning the solution must be in the interval.
    [0,1].
    Now chose the midpoint there x=.5 does it work?
    (.5)^3+(.5)-1=-.375 too little.
    Meaning the solution must be in the interval.
    [.5,1]
    Choose the midpoint there x=.75 does it work?
    (.75)^3+(.75)-1=.1719 too large.
    Thus, the solution is on the interval
    [.75,1]
    And so on.
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