Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

1 Answer 1

A polynomial is continuous and so bounded on any bounded interval.
$$\lim_{x\rightarrow \pm \infty} f(x) = \sigma \cdot \infty$$

where $\sigma$ is the sign of the leading coefficient and so $f(x)$ is bounded above or below accordingly.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.