# 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 $2x^5-4x^4+ax$

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 $\boxed{\Sigma}$ 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.

Now did i read your question correctly?
• Feb 9th 2013, 08:39 AM
johng
Re: Use Rolle's theorem to prove function cannot have 5 zeros.
If $g$ is any differentiable function with zeros $x_1, Rolle's Theorem guarantees the derivative $g^\prime$ has 4 zeros. (Don't you see why?) Similarly, if $g$ has n zeros, $g^\prime$ has n-1 zeros.

Now let $f(x)=2x^5-5x^4+ax$. If $f$ had 5 real zeros, then $f^\prime$ has 4 zeros and so $f^{\prime\prime}$ would have 3 zeros. But $f^{\prime\prime}(x)=40x^3-60x^2=x^2(40x-60)$ clearly has only 2 zeros. So $f$ can not have 5 zeros.

I think using Latex with this editor (or any editor which requires HTML tags) is a real pain. Notice I was "lazy" and didn't put tags around some of the numbers, n and n-1, so the fonts are different. I do agree for short expressions (a polynomial for example), it's pretty easy. So I'd say use enough Latex so that the text is readable, but don't worry about trying to produce "professional" copy.
• Feb 9th 2013, 01:59 PM
mcleja
Re: Use Rolle's theorem to prove function cannot 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, $thanks!$