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Thread: 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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    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
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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 $\displaystyle x^5 - 6x + c = 0$??
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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 $\displaystyle g(x) = (1/6)x^6 - 3x^2 + c$ on $\displaystyle [-1,1]$ then $\displaystyle g(-1)=g(1)$ thus there is a $\displaystyle k\in (-1,1)$ such that $\displaystyle g'(k)=0 \implies f(k)=0$.
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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 $\displaystyle x^5 - 6x + c = 0$??
    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 $\displaystyle x^5 - 6x + c = 0$, 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 $\displaystyle [-1,1]$. 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 $\displaystyle r_1$ and $\displaystyle r_2$, with $\displaystyle r_1 < r_2$ (without loss of generality). Then we have $\displaystyle f(r_1) = f(r_2) = 0$. Thus, by Rolle's theorem, there exists an $\displaystyle r$ in the interval $\displaystyle (r_1,r_2)$ such that $\displaystyle f'(r) = 0$. Now $\displaystyle f'(x) = 5x^4 - 6$. But there is no real $\displaystyle x$ for which this attains zero in $\displaystyle [-1,1]$ much less $\displaystyle (r_1,r_2) \subset [-1,1]$. Thus we have a contradiction. Therefore, $\displaystyle x^5 - 6x + c = 0$ has at most one root in $\displaystyle [-1,1]$

    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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