Question

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?

Answer 1

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!

Answer 2

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)`.