Suppose that $f: (a, b) → R$ is twice differentiable on $(a, b)$. Suppose $a < x_1 < x_2 < x_3 < b$ and that $f(x_1) = f(x_2) = f(x_3)$. Show that there exists $c \in (x_1, x_3)$ such that $f^{(2)}(c) = 0$.
I understand that I can use Rolle's Theorem on $[x_1,x_2]$ and then to $[x_2,x_3]$ to find two values which I can use Rolle's to find a value such that $f''(c)=0.$
I have attempted to also use the mean value Theorem, but is the problem that $x_1$ is not equal to $x_3$?
$f':[x_1,x_3] \rightarrow \mathbb{R}$ and differentiable on $(x_1,x_3),$ then there exists some $c\in (x_1,x_3)$ with $$f''(c)=\frac{f'(x_3)-f'(x_1)}{x_3-x_1},$$
as $$f'(x_3)=\frac{f(x)-f(x_3)}{x-x_3} = f'(x_1)=\frac{f(x)-f(x_1)}{x-x_1}.$$
you wrote: "use Rolle's Theorem on $[x_1,x_2]$ and then on $[x_2,x_3]$ to find two values..."
That is correct, you may apply Rolle on each interval for the function $f$. Rolle gives us $\alpha \in (x_1,x_2)$ satisfying $f'(\alpha)=0$, and then next gives us $\beta \in (x_2,x_3)$ satisfying $f'(\beta)=0.$
The only step you were missing is to use interval $[\alpha,\beta]$ rather than $[x_1,x_3]$ for the third application of Rolle, this time using $f'(x).$