- #1
zeraoulia
- 12
- 0
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.