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Interval halving

  1. Oct 24, 2005 #1
    could someone give me a hand with this please?

    i need to use the interval halving method to show that the function f has a root in the interval [a,b]. I need to approximate that root and determine a bound on the error of my estimate.

    f(x)=x^3+2x^2+pi(x)-(square root of 2)

    I have determined that f(-1)= -3.55581 and that

    f(1)= 4.72738



    the actual answer to the problem in the back of the book is: root is approx=0.25 and error at most 1/8

    I'm getting confused because wouldn't the error be (-1+1)/2? This would equal zero.

    Any help would be great.
  2. jcsd
  3. Nov 2, 2005 #2


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    The interval halving method requires you to keep narrowing an interval to find a smaller and smaller interval that contains a desired root. In this case, the intervale halving method goes like this:

    -1) -3.55581
    +1) +4.72738

    So there must be a root between -1 and +1. Try the midpoint of the interval, i.e the midpoint of [-1,+1] or zero. This gives:

    -1) -3.55581
    0) -1.414
    +1) +4.72738

    So now you know that the solution is in the interval [0,+1]. You begin the problem again with:

    0) -1.414
    +1) +4.72738

    So there must be a root somewhere between 0 and 1. Try 0.5

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