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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Prove that this polynomial has at least 1 root between 0 and 1.

*Last edited by bobbym (2009-08-09 11:10:48)*

**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.**

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

Assume that there is a triplet (a, b, c) such that the polynomial has no roots between 0 and 1. By the IVT we know that f(0) and f(1) must either both be greater than 0 or less than 0. Since every term of the polynomial has a parameter we only need to consider the case where both are greater than 0.

Now solve for b:

Substitute these values into our previous inequality:

These inequality signs are strict, so we have a contradiction. There is no triplet (a, b, c) where both f(0) and f(1) are greater than 0 (or less than 0), and so by the IVT there must be at least one root between 0 and 1 for all possible triplets.

*Last edited by TheDude (2009-08-13 05:14:44)*

Wrap it in bacon

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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

Im not following. You cant just substitute expressions for inequalities in other inequalities

TheDude wrote:

\\

Substitute these values into our previous inequality:

The Right hand side is ok, but not the left side. You have:

if im not missing anything.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi TheDude;

There is also a much easier way to attack this problem then the IVT.

**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.**

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

Kurre wrote:

Im not following. You cant just substitute expressions for inequalities in other inequalities

The Right hand side is ok, but not the left side. You have:

if im not missing anything.

Right?

bobbym wrote:

Hi TheDude;

There is also a much easier way to attack this problem then the IVT.

Probably true, but I'm something of a 1-trick pony

*Last edited by TheDude (2009-08-13 23:57:53)*

Wrap it in bacon

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi TheDude;

How about when b= -4 and c=1.

**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.**

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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

TheDude wrote:

Right?

No. You forget that there is a minus sign. We have 3c<-b. But:

there is a minus sign infront of the 3c/2, so we cant apply the inequality there.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I thought that someone would get this. I even thought that I knew who it would be.

Here is the solution.

Now F(0) = F(1) = 0. Now by Rolle's theorem

must have some point d in (0,1) where f(d) =0. So d is a root in (0,1).

*Last edited by bobbym (2009-08-24 01:20:52)*

**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.**

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