Proof needed: polynomial of odd degree

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    Degree Polynomial Proof
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Discussion Overview

The discussion centers on proving that if p(x) is a polynomial of odd degree, then the equation p(x)=0 has at least one real solution. The conversation explores the application of the intermediate value theorem and the behavior of polynomials at infinity.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant proposes that the proof could start with the specific case of p(x)=x^3 and then use induction for general odd degree polynomials.
  • Another participant suggests considering the limits of p(x) as x approaches positive and negative infinity to understand its behavior.
  • It is noted that as x approaches positive infinity, p(x) tends to positive infinity, and as x approaches negative infinity, p(x) tends to negative infinity.
  • A participant expresses uncertainty about how to select the closed interval [a,b] for applying the intermediate value theorem.
  • Another participant clarifies that for very negative x, p(x) is less than zero, and for very positive x, p(x) is greater than zero, allowing for the selection of appropriate a and b.
  • One participant concludes that since the conditions of the theorem are satisfied (f(a)<0 and f(b)>0), there is at least one real solution to p(x). However, it is unclear if this is considered a complete proof.

Areas of Agreement / Disagreement

Participants generally agree on the application of the intermediate value theorem and the behavior of odd degree polynomials at infinity, but there is no consensus on whether the discussion has reached a complete proof.

Contextual Notes

The discussion does not resolve the specifics of the proof structure or the completeness of the argument presented.

John O' Meara
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Prove: If p(x) is a polynomial of odd degree then the equation p(x)=0 has at least one real solution. I know the following theorem is required: If f is continuous on [a,b] and if f(a) and f(b) are nonzero and have opposite signs, then there is at least one solution of the equation f(x)=0 in the interval (a,b). Which is a conquence of the intermediate value theorem. I am studying this on my own and I do not know how one would go about this proof. Does one first prove it for say p(x)=x^3, then use induction to prove it for any odd degree polynomial? Please help. Thanks.
 
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first think about this: if [tex]p(x)[/tex] has odd degree, what do you know about

[tex] \lim_{x \to \infty} p(x)[/tex]

and

[tex] \lim_{x \to -\infty} p(x)[/tex]
 
I guess they go to +infinity and -infinity, respectively.
 
John O' Meara said:
I guess they go to +infinity and -infinity, respectively.

Correct. First, concentrate on [tex]\lim_{x \to \infty} p(x)[/tex].

Since this limit is positive infinity, you know that for [tex]x[/tex] large enough, [tex]p(x) > 0 [tex]has to be true, right? You can make a similar statement about the other limit.<br /> <br /> You don't need to know where these things happen; just that there are real numbers where p is becomes negative and becomes positive, and stays with those signs. How does this help you with your theorem?[/tex][/tex]
 
I think I need a little more information, as I think f(a) < 0 and f(b) > 0 how do I select my closed interval [a,b]? Sorry for the long delay in getting back to you, I had to go away quickly.
 
If x is very negative, f(x)<0. So you can find a such that f(a)<0. If x is very positive, f(x)>0. So you can find b such that f(b)>0. Just use those a and b; you don't need to specify them any further
 
Since, we have shown that f(a)<0 and f(b)>0 that is the hypothesis of the theorem is satisfied, the conclusion of the theorem implies that there is at least one real solution to P(x). Is that it finished?
 

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