# `f :[0,1]->R ` `int_0^1 f(x)dx = f(1)-f (0)` Show `f'(c )=f(c), 0<c <1`

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### 1 Answer

You need to consider the function `g(x) = f(x) - f'(x)` and you need to prove that there exists a value` c in (0,1)` such that `f'(c) - f(c) = 0` .

You need to use the mean value theorem, such that:

`g(c) = (int_0^1 g(x)dx)/(1-0)`

`g(c) = int_0^1 (f(x) - f'(x)) dx`

Using the property of linearity of integral yields:

`g(c) = int_0^1 f(x)dx - int_0^1 f'(x)dx`

The problem provides the information that` int_0^1 f(x)dx = f(1) - f(0)` , such that:

`g(c) = f(1) - f(0) - f(x)|_0^1`

Using the fundamental theorem of calculus yields:

`g(c) = f(1) - f(0) - (f(1) - f(0)) => g(c) = f(1) - f(0) - f(1) + f(0) = 0`

Since `g(c) = f(c) - f'(c)` yields that `f(c) - f'(c) = 0.`

**Hence, testing if `f(c) = f'(c)` holds, `c in (0,1)` , using the mean value theorem, yields ` f(c) - f'(c) = 0` , thus `f(c) = f'(c).` **