Rolle's Theorem and the Mean Value Theorem Let $f\colon \mathbb{R} \to \mathbb{R}$ have derivatives of all orders. Suppose that $a < b$ and that $f(a)=f(b)=f'(a)=f'(b) =0$. Prove that $f'''(c) = 0$ for some $c$ in $(a,b)$.
 A: Apply the mean value theorem to the function $f$ on the interval $[a,b]$: there's $c_1\in(a,b)$ such that:
$$f(b)-f(a)=0=(b-a)f'(c_1)\iff f'(c_1)=0$$
Now apply the mean value theorem to the function $f'$ on the interval $[a,c_1]$ and $[c_1,b]$ we find $c_2\in(a,c_1)$ and $c_3\in(c_1,b)$ such that
$$f''(c_2)=f(c_3)=0$$
and finaly apply for the last time the mean value theorem to the function $f''$ on the interval $[c_2,c_3]$ we find $c\in(c_2,c_3)\subset(a,b)$ such that
$$f'''(c)=0$$
A: As $f(a)=f(b)=0$, there exists a $c_1\in (a,b)$, with $f'(c_1)=0$, due to Rolle's Theorem.
Since $$f'(a)=f'(c_1)=f'(b)=0,$$
using Rolle's Theorem for $f'$ now, we get $c_2,\in(a,c_1)$ and $c_3\in(c_1,b)$, such that $$f''(c_2)=f''(c_3)=0.$$ 
Finally, using once again Rolle's Theorem for $f''$, we get $c,\in(c_2,c_3)$, such that $$f'''(c)=0.$$  
A: We may assume the function is not identically $0$ in our interval. Without loss of generality we may assume that it reaches a maximum $\gt 0$, say at $c$ strictly between $a$ and $b$. Then $f'(c)=0$. 
It follows that $f''(p)=0$ for some $p$ strictly between $a$ and $c$, and $f''(q)=0$ for some $q$ strictly between $c$ and $b$. 
