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Math Help - Find non-overlapping intervals

  1. #1
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    Find non-overlapping intervals

    I've spent two hours looking through my text book trying to find something that explains what this question is asking but I still have not figured out what to do. Can someone please help? Thanks!

    Find three non-overlapping intervals of length 1 such that each interval contains a solution of the equation. The endpoints of your intervals should be integers.
    2x^3-x^2-4x+2=0
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  2. #2
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    Have you solved the equation yet?

    That will be a good start...
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  3. #3
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    Quote Originally Posted by beanus View Post
    I've spent two hours looking through my text book trying to find something that explains what this question is asking but I still have not figured out what to do. Can someone please help? Thanks!

    Find three non-overlapping intervals of length 1 such that each interval contains a solution of the equation. The endpoints of your intervals should be integers.
    2x^3-x^2-4x+2=0
    Have you found the solutions?

    if not here is a hint:
    2x^3-x^2-4x+2=0 \iff x^2(2x-1)-2(2x-1)=0 \iff (2x-1)(x^2-2)

    Can you find the three solutions from here? These will divide the real line into three parts(the three intervals that you are looking for.)
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  4. #4
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    Yes I found the solutions already, .5, sqrt2, -sqrt2. I'm just confused about where to go from there.
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  5. #5
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    Quote Originally Posted by beanus View Post
    Yes I found the solutions already, .5, sqrt2, -sqrt2. I'm just confused about where to go from there.
    try plotting the solutions on a number line
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  6. #6
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    (-00,-sqrt2) (-sqrt2,sqrt2) (sqrt2,00)?
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  7. #7
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    Quote Originally Posted by beanus View Post
    (-00,-sqrt2) (-sqrt2,sqrt2) (sqrt2,00)?
    No the question asked for three non overlaping interval and they must have length 1. Two of your intervals have infiite length. What can you do to fix this>
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  8. #8
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    I suppose (-2,-1) (0,1) (1,2)?
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    yes those will work
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  10. #10
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    Great thanks guys!! That's so easy now that I know what they were looking for.
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  11. #11
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    Quote Originally Posted by beanus View Post
    I've spent two hours looking through my text book trying to find something that explains what this question is asking but I still have not figured out what to do. Can someone please help? Thanks!

    Find three non-overlapping intervals of length 1 such that each interval contains a solution of the equation. The endpoints of your intervals should be integers.
    2x^3-x^2-4x+2=0
    You do not need to solve the cubic to answer this question.

    From the Cauchy bound on roots we know that the absolute values of all the roots are less than or equal to 3. So make a table:

    Code:
    x          -3    -2    -1     0     1     2     3
    p(x)      -49   -10     3     2    -1     6    35
    Hence as the cubic changes signs between -2 and -1, between 0 and 1 and between 1 and 2 there is a root in [-2,-1], [0,1] and [1,2].

    (To make this calculus, or at least pre-calculus we can quote the Intermediate Value Theorem to justify the conclusion that those intervals contain a root)

    CB
    Last edited by CaptainBlack; February 9th 2011 at 07:41 PM.
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