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 countable, 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 finite in number or as equivalent classes (for analytic functions) but I am not able to prove that.
I can define the equivalence relation $ℜ$ by $$cℜd⇔f^{(1)}(c)=((1-t)/(1-u)).f^{(1)}(d)$$
where $u∈ℝ$ such that $g(u)=0$, $c∈(u,v)$ and $t∈ℝ$ such that $g(t)=0$, $d∈(t,v)$