Show that the equationx5 − 6x + c = 0

where c is a constant has at most one real root on the interval [−1, 1].

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- Dec 3rd 2007, 06:51 PMsingh1030intermediate value theorem/rolle's theoremShow that the equationx5 − 6x + c = 0

where c is a constant has at most one real root on the interval [−1, 1].

- Dec 3rd 2007, 06:54 PMJhevon
- Dec 3rd 2007, 07:28 PMcolby2152
- Dec 3rd 2007, 07:57 PMThePerfectHacker
- Dec 3rd 2007, 08:05 PMJhevon
I don't know

well, since you refuse to respond, i will assume you meant , 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 . 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 and , with (without loss of generality). Then we have . Thus, by Rolle's theorem, there exists an in the interval such that . Now . But there is no real for which this attains zero in much less . Thus we have a contradiction. Therefore, has at most one root in

QED - Dec 7th 2007, 08:35 AMsingh1030
to find the two inital roots by the IVT don't we need to know what the value of c is?

- Dec 8th 2007, 02:55 PMJhevon