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Math Help - [SOLVED] Halving the interval (bisection)?

  1. #1
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    [SOLVED] Halving the interval (bisection)?

    can someone help me with halving the interval, i know the theory, method, etc but i dont know how to arrive at the final answer when it asks to solve to two decimal places.
    e.g Question:
    Use the method of bisection to solve the following equation to 2 decimal places given that:
    x^3 - 3x - 20 = 0 has a root between 3 and 3.1

    the answer is 3.08.

    thanks
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  2. #2
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    Quote Originally Posted by noobonastick View Post
    can someone help me with halving the interval, i know the theory, method, etc but i dont know how to arrive at the final answer when it asks to solve to two decimal places.
    e.g Question:
    Use the method of bisection to solve the following equation to 2 decimal places given that:
    x^3 - 3x - 20 = 0 has a root between 3 and 3.1

    the answer is 3.08.

    thanks
    Use a table: f(x)=x^3-3x-20

    1. f(3)=-2~~~~f(3.1)=0.491~~~~\longrightarrow~3<x<3.1

    2. f(3.05)=-0.777...~~~~f(3.1)=0.491~~~~\longrightarrow~3.05<x  <3.1

    3. f(3.075)=-0.1489... ~~~~f(3.1)=0.491~~~~ \longrightarrow~ 3.075<x<3.1


    Because f(3.0875)=\bold{+0.165...} you now have to examine:

    f(3.08125)=\bold{+0.0099...} It is still positive. Therefore you have to examine now:

    4. f(3.078125)=-0.695...~~~~f(3.08125)=0.0099~~~~ \longrightarrow~3.078125<x<3.08125

    Thus the result to 2 dp is 3.08

    By the way: A linear interpolation yields the same result in one step.
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  3. #3
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    Thanks, so basically I have to keep halving until the upper and lower x values round off to the same two decimal places?
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  4. #4
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    Quote Originally Posted by noobonastick View Post
    Thanks, so basically I have to keep halving until the upper and lower x values round off to the same two decimal places?
    Correct!
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