Rolle's Theorem!

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October 21st 2010, 07:26 PM
drewbear

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]
October 21st 2010, 07:44 PM
Jhevon

Quote:

Originally Posted by drewbear

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?
October 21st 2010, 08:12 PM
drewbear

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?
October 21st 2010, 08:50 PM
TheEmptySet

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 $f'(x)$ ?