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$f : (0,1)$ to $R$ differentiable in $(0,1)$ and $f$ defined by $f(x)=f'(x) + \int_{0}^{1} f(x) dx$ for all $x$ element of $(0,1)$ . If there exist $a,b$ in $(0,1)$ such that $f(a)=f(b)=\frac{a+b}{2}$ . Find the value of $f(\frac{a+b}{2})$

I get the fact that $f'(a)=f'(b)$ and using Mean Value Theorem, but I did not know how to get the answer. Maybe someone can help me to solve this question

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    Any constant function $f(x) = C$ is a solution to $f(x) = f'(x)+\int_0^1 f(x)\,dx$. Are you sure the equation isn't $f'(x) = f(x)+\int_0^1 f(x)\,dx$? – 2017-01-22
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    the question isn't wrong – 2017-01-22

1 Answers 1

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It should read $f(x)=f'(x) + \int_{0}^{1} f(t) dt$.

If yes, then we have: $f$ is twice differentiable and $f'(x)=f''(x)$. This gives, with some constants $c,d$:

$f(x)=ce^x+d$.

If $c \ne 0$, then $f$ is injective, hence $a=b$.

If $c=0$, the $f(a)=f(b)=d$ for all $a,b$.

So, something went wrong !