Question

Explain why an odd-degree function must always have at least one real root.

Answer 1

I explained for general case of odd-degree function

First, degree of a polynomial is the highest degree of variable.

There are several ways to explain it, but I prefer to use the limit tool.

Assume that \[f(x) = a_n x^n + ....\]

In the above polynomial, n is the highest degree. Let n is an odd number (1,3,5, ...., 2k+1, etc.)

Then take the limit when x --> +infinity and -infinity. (actually I am using intermediate theorem to prove also)

\[\lim_{x \rightarrow +\infty} f(x) = \lim_{x \rightarrow +\infty} a_n x^n\]

\[\lim_{x \rightarrow -\infty} f(x) = \lim_{x \rightarrow -\infty} a_n x^n\]

The limit only takes into account the highest degree when x goes to infinity.

In case of n = odd numbers, you can see

\[\lim_{x \rightarrow +\infty} = +\infty\]
and \[\lim_{x \rightarrow -\infty} = -\infty\]

So those 2 limits are not the same sign (>0 and <0) ==> Odd-degree polynomial (graph) must go through the x-axis at least 1 point ==> at least 1 real solution exist. (Intermediate theorem)

Expand to case of n = even number, of course 2 limit is the same sign (+infty, >0) so we can not use limit to say about solution of even-degree polynomials

Answer 2

for a layman complex roots always occur in pairs.