Let $f: [0,6] \rightarrow \mathbb R$ be a three times differentiable function such that $f(0)=f(1.4)=f(3.9)=f(5.2)$. Prove that there is c $\epsilon$ (0,6) such that $f^{'''}(c)=0$.
Now using Rolle Theorem I know that from (0,5.2) I have that $f^{'}(x)=0$. But I am struggling with what to do from (5.2,6). I figure I can use the fact that we are looking for $f^{'''}(c)=0$, but I don't know what to do.
Suppose by contradiction $f''(c)\ne 0$ for any $c\in (0,6)$.In that case suppose $f''(x)>0$ for all $x\in(0,6)$.This means the function is concave upwards and can only hit the $x$ axis at atmost $2$ points which is not the case.Similarly in the case when $f''(x)<0$ for all $x\in (0,6)$ it will be concave downwards and meet the x axis at atmost $2$ points.Again a contradiction
From Rolle's theorem, you know that there is a $c_1 \in (0, 1.4)$ such that $f'(c_1) = 0$ and that there is a $c_2 \in (1.4, 3.9)$ such that $f'(c_2) = 0$ and that there is a $c_3 \in (3.9, 5.2)$ such that $f'(c_3) = 0$. Then since you know that $f'(c_1) = f'(c_2) = f'(c_3) = 0$, you can apply Rolle's theorem again to show that there is $c_4, c_5$ such that $c_1<c_4<c_2<c_5<c_3$ and $f''(c_4) = f''(c_5) = 0$. Finally, apply it again to show that there is a $c$ such that $f'''(c) = 0$.
