Use Rolle's Theorem and argue the case that $\displaystyle f(x)=x^5-7x+c$ has at most one real root in the interval [-1,1]

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- Oct 21st 2010, 07:26 PMdrewbearRolle's Theorem!
Use Rolle's Theorem and argue the case that $\displaystyle f(x)=x^5-7x+c$ has at most one real root in the interval [-1,1]

- Oct 21st 2010, 07:44 PMJhevon
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 $\displaystyle \displaystyle x = x_1$ and $\displaystyle \displaystyle x = x_2$, both in the interval you are considering. and you may also assume that $\displaystyle x_1 < x_2$. then that means

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

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

Where can you get with that? - Oct 21st 2010, 08:12 PMdrewbear
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?

- Oct 21st 2010, 08:50 PMTheEmptySet
c is a constant so when you take the derivative you get

$\displaystyle f'(x)=5x^4-7$

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