$f(x)$ is continuous on $[0,\,1]$, differentiable on $(0,\,1)$, and satisfies $$ f(1)=\frac{1}{2}\int_{0}^{\frac{1}{2}}e^{1-x^2}f(x)dx. $$ Prove that there exists $\xi\in(0,\,1)$ such that $f'(\xi)=2\xi f(\xi)$.
Let $F(x)=e^{1-x^2}f(x)$, then it's equivalent to prove that $F(\xi)'=0$. Also, we have $F(1)=\frac{1}{2}\int_{0}^{\frac{1}{2}}F(x)\,dx$. But I don't know how to continue.
I think there is a mistype. The correct equation is $ f(1)=2\int_0^{1/2} e^{1-x^{2}}f(x)\, dx$. I will answer the question with this correction. I would like to thank Martin R for suggesting this correction.
Suppose $F'$ does not vanish at any point. Then $F'(x) >0$ for all $x$ or $F'(x) <0$ for all $x$. This is because any derivative has IVP. In the first case $F$ is strictly increasing so $2 \int_0^{\frac 1 2} F(x)\, dx < 2 \int_0^{\frac 1 2} F(1)\, dx =F(1)$, a contradiction. Similarly in the second case we get $ 2 \int_0^{\frac 1 2} F(x)\, dx > 2 \int_0^{\frac 1 2} F(1)\, dx =F(1)$.
The statement is wrong, a counterexample is $F(x) = 1-\frac 67 x$. $$ F(1)=\frac{1}{2}\int_{0}^{\frac{1}{2}}F(x)\,dx = \frac 17 \, , $$ but $F'$ is nowhere zero.
If you actually meant the equation $$ f(1)= 2\int_{0}^{\frac{1}{2}}e^{1-x^2}f(x)dx $$ then the conclusion follows from the mean-value theorem for integrals, as demonstrated in Rolle's Theorem related exercise.
