If I have a function of class $C^2$ and $f(-1)=f(0)=f(1)$ then I know there must be a $c \in {]-1,1[}$ such that $f'(c)=0$ by Rolle's theorem.
BUT is it also true that there must be a $c \in {]-1,1[}$ such that $f''(c)=0$ ?
If I have a function of class $C^2$ and $f(-1)=f(0)=f(1)$ then I know there must be a $c \in {]-1,1[}$ such that $f'(c)=0$ by Rolle's theorem.
BUT is it also true that there must be a $c \in {]-1,1[}$ such that $f''(c)=0$ ?
By Rolle's theorem, there must be a $c_1 \in \left]-1,0\right[$ such that $f'(c_1) = 0$, and there must be a $c_2\in \left]0,1\right[$ such that $f'(c_2) = 0$. If we define the function $g(x) = f'(x)$, we have $g(c_1) = g(c_2) = 0$ What does Rolle's theorem say about $g$?