Is this a correct way to show that $f''(c)=0$?

Question

I am not sure if my solution is correct, so any feedback would be appreciated.

Suppose:

1. $f$ is twice differentiable in $(a,b)$
2. $a<x_1<x_2<x_3<b$
3. $f(x_1)=f(x_2)=f(x_3)$

Show that there exist a point $c\in(a,b)$ such that $f''(c)=0$

My proposed solution:

For $x_4\in (x_1,x_2)$ and $x_5\in (x_2,x_3)$ Lagrange's mean value theorem yields:

$f'(x_4)= \frac{f(x_2)-f(x_1)}{x_2-x_1}=0$ and $f'(x_5)= \frac{f'(x_3)-f'(x_2)}{x_3-x_2}=0$

Now let $c\in (x_4,x_5)$ and once again the mean value theorem:

$f''(c)= \frac{f'(x_5)-f'(x_4)}{x_5-x_4}=0$