Problem: Let $f(x)$ be an n-th degree real polynomial. Show that $\exists N \in \mathbb{R}$ such that $f(x)$ is either strictly increasing or decreasing on $[N, \infty)$.
I know that by the Fundamental Theorem of Algebra, this polynomial has at most $n$ roots, so past the rightmost root $k$, the function is either strictly positive or strictly negative, but I have no clue how to prove it is strictly increasing or decreasing past a certain $N$. I know I could try to prove that the function is one to one past $N$ as well, but I have no clue how to do that either. Any tips/help would be greatly appreciated.
Here is a more analytic proof.
Let $$f(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$$ so the derivative is $$f'(x) = na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + \dots + 2a_2 x + a_1.$$ We can factor out the first term to obtain $$f'(x) = na_n x^{n-1} \left( 1 + \frac{n-1}{n} \frac{a_{n-1}}{a_n} \frac{1}{x} + \dots + \frac{2}{n} \frac{a_2}{a_{n}} \frac{1}{x^{n-1}} + \frac{1}{n}\frac{a_1}{a_n}\frac{1}{x^{n-1}} \right) =na_n x^{n-1} \left( 1 + g(x) \right), $$ where $g(x) \to 0$ as $x \to \pm \infty$. So, there exists $N$ such that $|x| > N$ implies $g(x) \in (-1,1)$ and thus $1 + g(x) \in (0,2)$, which is positive. This allows us to fix the sign of the derivative for $|x|$ large enough.