# Proving a statement about polynomials.

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?

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I can imagine a proof that any polynomial which yields all $y \in \mathbb{R}$ must be of odd degree, but this is only a slight variation on the approach you've outlined. –  hardmath Feb 27 '13 at 19:08
Your argument looks pretty simple to me. I suppose you can say that without loss of generality the leading coefficient is positive as otherwise you could apply the proof to $- f(x)$. –  JavaMan Feb 27 '13 at 19:14
@hardmath - well, I think the proof of such a statement would be a bit easier (we could resort to intermediate value theorem), but I don't see how we could move from that to the statement above. If you see how this could be done, it would be great if you could share your thoughts :) –  Johnny Westerling Feb 27 '13 at 19:17
@JavaMan - it just seemed to me that such a statement shouldn't require the hassle involved with proving that a function attains an extremum (derivatives, etc) –  Johnny Westerling Feb 27 '13 at 19:19
@JohnnyWesterling I think that's overkill showing it attains its extrema, all you really needed is boundedness. –  muzzlator Feb 27 '13 at 19:22

$$\lim_{x\rightarrow \pm \infty} f(x) = \sigma \cdot \infty$$
where $\sigma$ is the sign of the leading coefficient and so $f(x)$ is bounded above or below accordingly.