There is something I am trying to prove:
Let $f:\mathbb R\to\mathbb R$ be a nonconstant polynomial. Show that the image of the function is either the real line, $[a, \infty)$, or $(-\infty,a]$ for some $a\in\mathbb R$.
Here I tried to separate them to odd and even powers of polynomials, and intuitively, I see why it will be true. I am unable to make it rigorous (I thought about using the max/min value theorem for the even cases but not sure exactly how it will work). We probably use the intermediate value theorem for the odd case?
Hint If the degree is odd, prove that
$$\lim_{x \to \infty} P(x)= \pm \infty \\ \lim_{x \to -\infty} P(x)= \mp \infty$$
Now, for each real value, by the above $P(x)$ takes a larger and smaller value, thus by IVT it takes the value.
If the degree is even,
$$\lim_{x \to \infty} P(x)= \lim_{x \to -\infty} P(x)= \pm \infty$$
Now, deending on the sign, you can find a $c$ such that outside $[-c,c]$ the Polynomial is always larger/smaller than $P(0)$.
Use the fact that on $[-c,c]$ the polynomial attains his absolute min and max.