Proof needed: polynomial of odd degree

  • Context: Undergrad 
  • Thread starter Thread starter John O' Meara
  • Start date Start date
  • Tags Tags
    Degree Polynomial Proof
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 8K views
John O' Meara
Messages
325
Reaction score
0
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.
 
Physics news on Phys.org
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?