differentiable function I'm trying to solve this problem
Let $f:[a,b] \to \mathbb R$ a differentiable function with continuous derivative.
Suppose further that $f$ is twice differentiable on $(a,b)$.
Prove that if $f(a)=f(b)$ and $f'(a)=f'(b)=0$, then
  exist $x_1, x_2 \in (a,b)$ with $x_1 \neq x_2$ such that $f''(x_1) = f''(x_2)$.
I'm trying to solve this problem graphically intuitively ; I tried, and likely the problem is real: if $f(a) = f(b)$ and $f'(a) = f'(b) = 0$, the points $a$ and $b$ are the points of maximum and minimum for the function.
but how can you prove that there are $x_1, x_2 \in (a,b)$ with $x_1 \neq x_2$
such that $f''(x_1) = f''(x_2)$?
 A: Hint: Apply the Mean Value Theorem three times (or Rolle's Theorem if you prefer).
Applying it once to $f$ over $[a,b]$ gives you a point $c$ with $a<c<b$ where $f'(c)=0$.
Now apply the Mean Value Theorem to $f'$ on each of the intervals $[a,c]$ and $[c,b]$.
(Note that the Mean Value Theorem gives you a point $c$ strictly between the endpoints of the interval that you're working over.)
 It's not true that $a$ and $b$ necessarily give the minimum and maximum of the function. They could in fact give neither. For instance, the graph could resemble a $\sin$ wave over $[0,2\pi]$ that's been "smoothed out" at the endpoints (so that the derivative is zero at the endpoints).
A: If the function is constant, everything is easy.  
If the function is not constant, it attains a maximum and a minimum value in our interval, and these values are distinct.
The max and min cannot both occur at endpoints, since $f(a)=f(b)$. So there is a local extremum in $(a,b)$, and therefore a point $c\in(a,b)$ such that $f'(c)=0$.  Now you should be able to use Rolle's Theorem to show that there exist points $x_1\in(a,c)$ and $x_2\in(c,b)$ such that $f''(x_1)=f''(x_2)=0$.
