Let $f(x) = \left\{\begin{matrix} e^{\frac{\ln x}{x}} &, & x>0 \\ 0 & , & x = 0 \end{matrix}\right.$. If $F$ is a primitive of $f$ then prove that there exists $\xi \in (2, 4)$ such that
$$f'(\xi) \left ( F(\xi) - F(2) \right ) = f(\xi) \left ( f \left ( 2 \right ) - f(\xi) \right )$$
This calls for Rolle's theorem but I do not see the way to apply it. Any hints?
Let $g : (2,4) \rightarrow \mathbb{R}$ defined by $$g(x)=F(x)f(x)-F(2)f(x)-f(2)F(x)$$
Then $g(2)=-f(2)F(2)$. But by direct computation, one has $f(2)=f(4)$, so one deduces \begin{align*} g(4)& =F(4)f(4)-F(2)f(4)-f(2)F(4) \\ &=F(4)f(2)-F(2)f(2)-f(2)F(4) \\&=-f(2)F(2) \end{align*}
so $g(2)=g(4)$, so by Rolle's theorem, there exists $\xi \in (2,4)$ with $g'(\xi)=0$, i.e. with $$f^2(\xi)+F(\xi)f'(\xi)-F(2)f'(\xi) - f(2) f(\xi)=0$$
which is equivalent to $$\boxed{f'(\xi) \left ( F(\xi) - F(2) \right ) = f(\xi) \left ( f \left ( 2 \right ) - f(\xi) \right )}$$