Let $f$ be a function such that $f'''$ is continuous and $f(0)=f'(0)=f'(1)=0$. Prove that there exists $c \in (0,1)$ such that
$f(1)=-\frac{1}{12} f'''(c)$
Let $f$ be a function such that $f'''$ is continuous and $f(0)=f'(0)=f'(1)=0$. Prove that there exists $c \in (0,1)$ such that
$f(1)=-\frac{1}{12} f'''(c)$
If $f(1)\neq 0$, up to rescaling we can assume $f(1)=1$ and $ f(x)=3x^2-2x^3+g(x), $ where $g\in\mathcal{C}^3([0,1])$ satisfies: $ g(0)=g'(0)=g(1)=g'(1)=0.$ In this setting, the statement is equivalent to: $ \exists\xi\in(0,1):\;g'''(\xi)=0. $ If $g$ is constantly zero on $[0,1]$ there is nothing to prove. Otherwise, $|g|$ takes its maximum positive value in $\eta\in(0,1)$, and, by Fermat's theorem, $g'(\eta)=0.$ By Rolle's theorem, there are at least two distinct zeros of $g''(x)$ in $(0,1)$, the first one in $(0,\eta)$ and the latter in $(\eta,1)$. By Rolle's theorem again, there is a $\xi\in(0,1)$ such that $g'''(\xi)=0$, QED. If $f(1)=0$, it is sufficient to take $g=f$.