# Is this proof correct/complete?

Let $f(x)=\sum\limits_{k=0}^{n}c_kx^k$ be a polynomial where $c_0$ and $c_n$ have different sings. Show $\exists x_0 \in \mathbb{R}$ such that $f(x_0)=0$.

My workings so far: Lets assume $c_0>0$ and thus $c_n<0$. If this is not the case we can simply look at $f^*(x)=-f(x)$ and $f^*(x)$ will satisfy this condition where $f^*(x_0)=0$ implies $f(x_0)=0$. By this assumption we know $f(0)>0$. Then for sufficiently large $x_l$ we have: $$|c_nx_l^n|>\left|\sum\limits_{k=0}^{n-1}c_kx_l^k\right|$$ Therefore, as $c_n<0$, it follows that: $$f(x_l)=\sum\limits_{k=0}^{n-1}c_kx_l^k-|c_n|x_l^n<0$$ Now because $f(x)$ is a polynomial and therefore continuous, we can apply the intermediate value theorem to conclude that $\exists x_0 \in [0,x_l]$ such that $f(x_0)=0$.

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Yes. Your instructor might want more details for the part "For sufficiently large $x_l$ we have...", but it looks fine to me. – N. S. Feb 3 '13 at 17:02
Thanks, that is indeed the part that I was worrying about. How do I make a statement like that more rigorous? – Slugger Feb 3 '13 at 17:03
Intermediate value theorem, not mean value theorem. – 1015 Feb 3 '13 at 17:05
Note that if $x_l >1$ you have $\left|\sum\limits_{k=0}^{n-1}c_kx_l^k\right| \leq \sum\limits_{k=0}^{n-1} \left|c_kx_l^k\right|< \sum\limits_{k=0}^{n-1} \left|c_kx_l^{n-1}\right| < [\sum\limits_{k=0}^{n-1} \left|c_k\right| ]x_l^{n-1}$. It is easy to make $|c_l|x_l^n$ larger than $[\sum\limits_{k=0}^{n-1} \left|c_k\right| ]x_l^{n-1}$. – N. S. Feb 3 '13 at 17:06
You don't show that the polynomial is $> 0$ at any point. – vonbrand Feb 3 '13 at 17:32

It is easier if you take the limit at $+\infty$.

Assume $c_0<0$ and $c_n>0$ first.

Then $f(0)=c_0<0$, while $\lim_{x\rightarrow +\infty} f(x)=\lim_{x\rightarrow +\infty}c_nx^n=+\infty$.

So we can find $x>0$ such that $f(x)>0$.

By the intermediate value theorem, there exists $x_0>0$ such that $f(x_0)=0$.

I let you do the case $c_0>0$ and $c_n<0$.

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