# analysis question

Suppose $f(x) \in \mathbb{R}[x]$ is such that $\operatorname{deg}{f(x)} = 2011$, then $\exists \: c \in \mathbb{R}$ such that $f(c) = f'(c)$. How can I prove/disprove the above statement. Any hints?

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Consider $g(x) = f(x)-f'(x)$. Now apply the fact that an odd degree polynomial has at-least one real root.

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+1 Painfully and beautifully simple... –  DonAntonio May 30 '12 at 2:39

Chandrasekhar's answer is boss, I wondered what would happen if we replace 2011 say with 2010.

But note that any polynomial $f = \sum b_i z^i \in \mathbb{R}[x]$ can be written as $p - p', p = \sum a_i x^i \in \mathbb{R}[x]$ (set $a_n$ to be $b_n$, where $f$ is degree $n$, then just inductively solve $a_i - (i+1)a_{i+1} = b_i$ for $a_{i}$), so this being true for $f$ of even degree is equivalent to $\mathbb{R}$ being algebraically closed o_O!

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