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Math Help - Rolle's Theorem!

  1. #1
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    Rolle's Theorem!

    Use Rolle's Theorem and argue the case that f(x)=x^5-7x+c has at most one real root in 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 drewbear View Post
    Use Rolle's Theorem and argue the case that f(x)=x^5-7x+c has at most one real root in the interval [-1,1]
    You can show it is possible to have a root using the intermediate value theorem.

    To use Rolle's theorem to argue there is at most one, you can proceed thusly:

    Assume, to the contrary, there are two roots (or more, but at least 2 is fine), say for \displaystyle x = x_1 and \displaystyle x = x_2, both in the interval you are considering. and you may also assume that x_1 < x_2. then that means

    \displaystyle f(x_1) = f(x_2) = 0

    and so according to Rolle's theorem, there must be a point \displaystyle x = x_3, such that \displaystyle x_1 < x_3 < x_2 and \displaystyle f'(x_3) = 0.

    Where can you get with that?
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  3. #3
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    i am sorry but i am still a little confused. i understand the IVT and the proving by contradiction, but the c variable is throwing me off. do i have to solve for that ever or is that just a constant of no importance?
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  4. #4
    Behold, the power of SARDINES!
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    c is a constant so when you take the derivative you get

    f'(x)=5x^4-7

    As Jhevon suggested where are the zero's of f'(x)?
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