The **mean values theorem** says that there exists a $c∈(u,v)$ such that $$f(v)-f(u)=f′(c)(v-u)$$ My question is: Assume that $u$ is a root of $f$, hence we obtain $$f(v)=f′(c)(v-u)$$ Assume that $f$ is a non-zero analytic function in the whole real line. My interest is about the real $c∈(u,v)$. I believe that the set of those $c$ is contable, otherwise we conclud that $f′(c)$ is constant (in an open set containing $c$) since $v$ is a constant and hence $f$ is identically zero. Also, I think that the set of those $c$ is countable (for analytic functions) but I am not able to prove that.(adsbygoogle = window.adsbygoogle || []).push({});

**Physics Forums - The Fusion of Science and Community**

# Prove that the set of those $c$ is countable

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Prove that the set of those $c$ is countable

Loading...

**Physics Forums - The Fusion of Science and Community**