I wanted to prove the following statement: in case of a polynomial of an even degree, there exists some $y\in\mathbb{R}$ for which there is NO $x\in\mathbb{R}$ such that $f(x)=y$.
The proof I initially came up with was as following. I thought of splitting the proof into two cases: with positive and negative leading coefficient. Then, we can can prove that the polynomial with a positive leading coefficient attains a global minimum, and hence our statement follows for every $y<y_{\min}$. Similarly, we can prove that each polynomial with a negative leading coefficient attains a global maximum, and hence the statement follows for every $y>y_{\max}$.
However, is there an alternative (or perhaps a simpler) way to prove this statement?
Best Answer
I will answer this formally since it hasn't been (rigorously) answered yet, in my opinion. This is likely a duplicate, but I couldn't find a duplicate.
If we're considering polynomials of degree $0$, this is trivial, since such polynomials are constant. If the degree is $\geq 2$, then note
$$p(x) \stackrel{\text{def}}{=}\sum_{1 \leq i \leq 2n} a_ix^i = a_{2n}x^{2n}\left(1 + \sum_{1 \leq i \leq 2n-1} \frac{a_i}{a_{2n}}\frac{1}{x^{2n-i}}\right) $$
The sum inside the brackets clearly goes to $0$ as $|x| \to \infty$, and hence $$\lim_{x \to ±\infty} p(x) = \text{sgn}(a_{2n}) \cdot \infty$$
This implies there's an $M>0$ such that $p(x) > 0$ $\text{XOR}$ $p(x) < 0$ for all $x \in \mathbb{R} \setminus [-M,M]$. Hence, were $p$ surjective onto $\mathbb{R}$, we would need one of $(-\infty, 0]$ or $[0, +\infty)$ to be subsets of the image of $[-M, M]$ under $p$.
However, this leads to a contradiction since the continuous image of a (finite) closed interval is a (finite) closed interval.