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Chrysostomou

Member

Registered: 2009-05-25

Posts: 9

Rolle's theorem

Hello I have some question about the Rolle's theorem

Here is the question:
Use Rolle's theorem to show that f(x)=x^3+3x-5 has exactly one real root.

what is my limit!?I need some limit right!?

Thanks
NaNa

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Rolle's theorem

The contrapositive of Rolle's Theorem is useful here, since f'(x) has no real roots.
You can then deduce that f(x) must be injective and so eliminate the possibility of it having more than one root.

Showing that f(x) has at least one root doesn't require Rolle, it's just automatic from the fact that it's a cubic function.

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JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Rolle's theorem

Showing that f(x) has at least one root doesn't require Rolle, it's just automatic from the fact that it's a cubic function.

No, no. Use the intermediate-value theorem.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Rolle's theorem

Hi;

  Isn't mathsy right. Because it is a cubic it must have at least 1 real root or 3 real roots. Doesn't prove their is 1 real root though.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Rolle's theorem

Did you read what I posted? Use the intermediate-value theorem! The IVT proves that every cubic has at least one real root.

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Rolle's theorem

(of real coeffecients)

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Rolle's theorem

Bobby, we're both right, but Jane didn't like how I used knowledge about cubic functions without proof. She's just being more rigourous than me.

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Why did the vector cross the road?
It wanted to be normal.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Rolle's theorem

Hi mathsy;

I think that your statement, the one in the quotes is rigorous enough, for the reason below.

Hi Jane;

Yes I did read it, thoroughly, the IVT does do that. But since Gauss proved that a cubic (real or complex coefficients) has 3 roots and since complex roots must come in pairs then every cubic (with real coefficients- thanks luca for pointing out complex coeffs don't apply here) must have at least one real root. I know it doesn't prove that it has only one real root.

Glad to be talking to you again, haven't had you reply in a couple of days.

Last edited by bobbym (2009-05-27 10:24:56)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Rolle's theorem

bobbym, surely that only holds for cubic equations of real coeffecients, surely the cubic (x-i)^3 has no real roots for example?

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The Beginning Of All Things To End.
The End Of All Things To Come.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Rolle's theorem

Hi luca;

Your absolutely right but I was referring to the fact that it must have 3 roots which it does.But it does need real coefficients to have at least 1 real root. Will adjust the post to make your point clear. Thanks!

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Rolle's theorem

Also, could we not use the idea of the discriminant
For

If all coefficients are real and If D<0 then we have 1 real root and 2 complex roots.

a=1
b=0
c=3
d= -5

D= -4(27)-27(25) = -783<0

x^3+3x-5 has 1 real root.

Last edited by bobbym (2009-05-27 11:03:04)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Rolle's theorem

That would work, but it's fairly heavy machinery. Rolle gets the same result far less strenuously.

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Why did the vector cross the road?
It wanted to be normal.