Use Rolle's theorem to prove function cannot have 5 zeros.

• Feb 8th 2013, 09:54 PM
mcleja
Use Rolle's theorem to prove function cannot have 5 zeros.
I have to show that Rolle's theorem can be used to prove that the function 2x5 − 5x4 + ax cannot have five distinct real zeros. I know that Rolle's theorem means that if f(a)=f(b)=0 there must be some x value c between a and b such that f'(c)=0. I am unsure of how to do this, would i show that the derivative does not equal zero 4 times and then use Rolle's theorem to say therefore there must be less than 5 zeros for the function.

Would this be a way to do it? If not how would i do it?

Thanks.
• Feb 9th 2013, 04:25 AM
Plato
Re: Use Rolle's theorem to prove function cannot have 5 zeros.
Quote:

Originally Posted by mcleja
I have to show that Rolle's theorem can be used to prove that the function 2x5 − 5x4 + ax cannot have five distinct real zeros. I know that Rolle's theorem means that if f(a)=f(b)=0 there must be some x value c between a and b such that f'(c)=0. I am unsure of how to do this, would i show that the derivative does not equal zero 4 times and then use Rolle's theorem to say therefore there must be less than 5 zeros for the function.

Look at the following>
$$2x^5-4x^4+ax$$ gives $\displaystyle 2x^5-4x^4+ax$

Isn't that a lot easier to read than "2x5 − 5x4 + ax". It takes so little effort to learn to do the LaTeX code.
Click on the “go advanced” tab. On the toolbar you will see $\displaystyle \boxed{\Sigma}$ clicking on that give the LaTeX wraps, . The code goes between them.

You will find that helpers are more likely to reply to well formatted questions.

If $\displaystyle g$ is any differentiable function with zeros $\displaystyle x_1<x_2<x_3<x_4<x_5$, Rolle's Theorem guarantees the derivative $\displaystyle g^\prime$ has 4 zeros. (Don't you see why?) Similarly, if $\displaystyle g$ has n zeros, $\displaystyle g^\prime$ has n-1 zeros.
Now let $\displaystyle f(x)=2x^5-5x^4+ax$. If $\displaystyle f$ had 5 real zeros, then $\displaystyle f^\prime$ has 4 zeros and so $\displaystyle f^{\prime\prime}$ would have 3 zeros. But $\displaystyle f^{\prime\prime}(x)=40x^3-60x^2=x^2(40x-60)$ clearly has only 2 zeros. So $\displaystyle f$ can not have 5 zeros.
Thanks Plato and johng, i will try to use latex from now on. And thanks to you johng the problem seems crystal clear to me now, $\displaystyle thanks!$