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Math Help - intermediate value theorem/rolle's theorem

  1. #1
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    intermediate value theorem/rolle's theorem

    Show that the equation
    x
    5 6x + c = 0

    where
    c is a constant has at most one real root on the interval [1, 1].
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by singh1030 View Post
    Show that the equation
    x
    5 6x + c = 0

    where
    c is a constant has at most one real root on the interval [1, 1].
    i think you need to clarify the problem. why is there an x vertically above the 5?
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  3. #3
    GAMMA Mathematics
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    Quote Originally Posted by Jhevon View Post
    i think you need to clarify the problem. why is there an x vertically above the 5?
    Is that supposed to be x^5 - 6x + c = 0??
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    Quote Originally Posted by singh1030 View Post
    Show that the equation
    x
    5 6x + c = 0

    where
    c is a constant has at most one real root on the interval [1, 1].
    Let g(x) = (1/6)x^6 - 3x^2 + c on [-1,1] then g(-1)=g(1) thus there is a k\in (-1,1) such that g'(k)=0 \implies f(k)=0.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    Is that supposed to be x^5 - 6x + c = 0??
    I don't know

    Quote Originally Posted by singh1030 View Post
    Show that the equation
    x
    5 6x + c = 0

    where
    c is a constant has at most one real root on the interval [1, 1].
    well, since you refuse to respond, i will assume you meant x^5 - 6x + c = 0, if not. use this as a guide to do the right problem.

    Proof

    Assume to the contrary that there are more than one root of the equation in the interval [-1,1]. That is, there are at least two roots in this interval. Pick any two roots, if there are only two, pick both. Let the roots be r_1 and r_2, with r_1 < r_2 (without loss of generality). Then we have f(r_1) = f(r_2) = 0. Thus, by Rolle's theorem, there exists an r in the interval (r_1,r_2) such that f'(r) = 0. Now f'(x) = 5x^4 - 6. But there is no real x for which this attains zero in [-1,1] much less (r_1,r_2) \subset [-1,1]. Thus we have a contradiction. Therefore, x^5 - 6x + c = 0 has at most one root in [-1,1]

    QED
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  6. #6
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    to find the two inital roots by the IVT don't we need to know what the value of c is?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by singh1030 View Post
    to find the two inital roots by the IVT don't we need to know what the value of c is?
    no. notice that the punch line of my proof was using the derivative. it does not matter what c is here, since it would not matter in the derivative.
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