Suppose that the function $f$ has second derivative in the interval $(a,b)$ and that $f(x_1) = f(x_2)=f(x_3)$, with $a < x_1<x_2<x_3<b$. Then, there exist a $\epsilon \in (a,b)$, such that $f''(\epsilon) = 0.$
My Attempt:
I am going to use Rolle Theorem on that problem. Basically, Rolle Theorem gives us, exist a $\delta_1, \delta _2$ such that $$f'(\delta_1) = \frac{f(x_2) - f(x_1)}{x_2 - x_1} = 0$$ $$f'(\delta_2) = \frac{f(x_2) - f(x_3)}{x_2 - x_3} = 0.$$ So, if $g(x) = f'(x)$, then $$g'(\epsilon) = \frac{g(\delta_2) - g(\delta_1)}{\delta_2 - \delta_1} = 0.$$ Is it true?