1
$\begingroup$

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 $

  • 0
    I might have a gist of what you've said. Please take a look at my working: Define $F(x) = \int\limits_0^x {f(x)dx} $, then $F(0)=0$ and $F(1)=f(1)$. By MVT, there exists $c \in (0,1)$ s.t. $F'(c)=f(c)=f(1)=f(0)$. By Rolle's Theorem, there exists $x_{1} \in (0,c)$ and $x_{2} \in (c,1)$ s.t. $f'(x_{1})=f'(x_{2})$. Another application of Rolle's Theorem gives $x_{0} \in (x_{1},x_{2}) s.t. f''(x_{0})=0$.2012-11-29

1 Answers 1

2

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)) $