If p(x) is a polynomial of odd degree, show that the equation p(x)=0 has at least one solution. how would I do this?

2 Answers
Feb 20, 2018

Odd degree functions come in two cases, both depending on their highest degree terms.

When y = +ax^b

We know that b is odd and a is positive. Thus, lim_(x-> -oo) = -oo because a negative value of x paired with an odd exponent will give a negative number. Meanwhile, lim_(x-> oo) = oo, because a positive value of x paired with an odd exponent will still give a positive number. Thus in this case, the function must cross the x-axis at least once, or else the limits as x-> +- oo would be on the same side (either positive or negative infinity). We also note that polynomials are continuous for all x.

When y= -ax^b

Everything is reversed from example one. The limit as x ->-oo = oo because the negative cancels the positive in example 1. On the same vein, we see that limit as x-> oo = -oo. Therefore, as above, the function must cross the x-axis at least once otherwise the limits couldn't have +oo on one side and -oo on the other.

Hopefully this helps!

Feb 20, 2018

Doesn't quite fit in a sentence, check below.

Explanation:

Since polynomial functions are continuous everywhere, it would be enough to find two points "at" which the polynomial has a different sign (one at which it's positive, one at which it's negative, and let's call them a and b with a < b), then we can use Bolzano's theorem to show that there will be a point c in (a,b) such that the polynomial at c is 0, or c is a root.

Our polynomial of odd degree has the general form

p(x)=a_k x^k+a_(k-1)x^(k-1)+...+a_1x+a_0

where k is an odd natural number and a_k != 0.

Let's take the limits at positive and negative infinity of p(x), and without any loss of generality* we will assume that a_k > 0.

lim_(x to +infty) p(x) = lim_(x to +infty) a_k x^k = a_k lim_(x to +infty) x^k =+infty

lim_(x to -infty) p(x) = lim_(x to -infty) a_k x^k = a_k lim_(x to -infty) x^k= -infty

The second line is explained because k is odd, any negative raised to an odd power remains negative.

From the above limits we can see that there will be a value for x, let's call it a, sufficiently large and negative so that p(a)<0 and there will be a value for x, we'll call it b, sufficiently large and positive so that p(b)>0. Therefore there exist a c in (a,b) such that p(c)=0 and there's our root.

*Now if you were to consider the case where a_k<0, we'd just see that the limit at +infty becomes -infty and the limit at -infty becomes +infty. We'd just have to switch places for our a and b and we'd still be fine. Here when using Bolzano's theorem, if a < b, it doesn't matter if p(a) < 0 < p(b) or p(b) < 0 < p(a).