Proof needed: polynomial of odd degree

[John O' Meara]

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.

 

[statdad]

first think about this: if $$p(x)$$ has odd degree, what do you know about

$$\lim_{x \to \infty} p(x)$$

and

$$\lim_{x \to -\infty} p(x)$$

 

[John O' Meara]

I guess they go to +infinity and -infinity, respectively.

 

[statdad]

I guess they go to +infinity and -infinity, respectively.

Correct. First, concentrate on $$\lim_{x \to \infty} p(x)$$.

Since this limit is positive infinity, you know that for $$x$$ 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?

 

[John O' Meara]

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.

 

[Office_Shredder]

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

 

[John O' Meara]

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?