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Math Help - Proof with Intermediate Value Theorem

  1. #1
    BadAtCalc
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    Unhappy Proof with Intermediate Value Theorem

    Is any real number exactly 1 less than its cube? Answer this question by applying the intermediate value property (you need to determin a suitable finction and a suitable interval)

    I really dont understand, can someone help?

    Thanks
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  2. #2
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    Quote Originally Posted by BadAtCalc View Post
    Is any real number exactly 1 less than its cube? Answer this question by applying the intermediate value property (you need to determin a suitable finction and a suitable interval)

    I really dont understand, can someone help?

    Thanks
    You mean,

    x=x^3-1

    Rewrite as,

    x^3 -x -1=0

    When x=10, the polynomial is positive.

    When x=-10, the polynomial is negative.

    Thus, there exists at least one zero on the interval (-10,10).

    This is Mine 51th Post!!!
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post

    This is Mine 51th Post!!!
    ha, this is my 7th post!!!!
    Last edited by Jhevon; March 29th 2007 at 07:57 AM.
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