Proof needed: polynomial of odd degree

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SUMMARY

If p(x) is a polynomial of odd degree, then the equation p(x)=0 has at least one real solution. This conclusion is derived from the Intermediate Value Theorem, which states that if a continuous function f is defined on a closed interval [a,b] and takes on opposite signs at the endpoints, then there exists at least one c in (a,b) such that f(c)=0. By analyzing the limits of p(x) as x approaches positive and negative infinity, it is established that p(x) transitions from negative to positive values, confirming the existence of a real root.

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  • Understanding of polynomials and their properties
  • Familiarity with the Intermediate Value Theorem
  • Knowledge of limits in calculus
  • Basic proof techniques, including mathematical induction
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  • Learn about polynomial behavior at infinity
  • Explore proof techniques using mathematical induction
  • Investigate examples of odd degree polynomials and their roots
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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 p(x) has odd degree, what do you know about

<br /> \lim_{x \to \infty} p(x)<br />

and

<br /> \lim_{x \to -\infty} p(x)<br />
 
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 \lim_{x \to \infty} p(x).

Since this limit is positive infinity, you know that for x large enough, p(x) &gt; 0 has to be true, right? You can make a similar statement about the other limit.&lt;br /&gt; &lt;br /&gt; You don&amp;#039;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?
 
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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