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Proof needed: polynomial of odd degree

  1. Sep 29, 2009 #1
    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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  3. Sep 29, 2009 #2

    statdad

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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]
     
  4. Sep 29, 2009 #3
    I guess they go to +infinity and -infinity, respectively.
     
  5. Sep 29, 2009 #4

    statdad

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    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.

    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?
     
  6. Sep 29, 2009 #5
    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.
     
  7. Sep 29, 2009 #6

    Office_Shredder

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    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
     
  8. Sep 29, 2009 #7
    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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