Let $ f $ be a function continuous on $ [0,1] $ and twice differentiable on $ (0,1) $ . Suppose that
$ \int_{0}^{1}f(x) dx = f(0)=f(1) $
Prove that there exists a number $ x_{0} \in (0,1) $ such that $ f''(x_{0})=0 $
Let $ f $ be a function continuous on $ [0,1] $ and twice differentiable on $ (0,1) $ . Suppose that
$ \int_{0}^{1}f(x) dx = f(0)=f(1) $
Prove that there exists a number $ x_{0} \in (0,1) $ such that $ f''(x_{0})=0 $
Define $ F(x)=\int_{0}^{x} f(x)\ dx $ By Integral Mean Value Theorem, $ \int_{0}^{1}f(x)\ dx =f(x_1)\int_{0}^{1}dx=f(x_1) \hspace{1 in} (for\ x_1\in(0,1)) $ This should gives you a great hint in relating F(x) to f(x).
Remarks: The Full Version of Integral Mean Value Theorem is: $ Suppose\ f(x)\ and \ g(x)\ are\ real\ valued\ function\ defined \ on\ [a,b]\ such\ that \ f(x)\ is\ continuous\ and \ g(x)\geqslant 0\ for \ every \ x \in[a,b].\\Hence \ there\ is \ a \ number\ x_1 \in[a,b]\ such\ that\\ \int_{a}^{b} f(x)g(x)\ dx=f(x_1)\int_{a}^{b}g(x)\ dx \hspace{1in} (for\ x_1\in(a,b)) $