If $f$ is differentiable infinite number of times, then there exist $x\in (0,1)$ such that
$$\frac{f(1)-f(x)}{x}=f'(x)$$.
I have tried to use Lagranges Mean Value Theorem .
It lands me with $f(1)-f'(x)=(1-x)f'(x+\theta(1-x))$ where $\theta\in(0,1)$.
Then I thought that I could differentiate both sides again and again to get some form of $\theta^{n}$ but to no avail.
Even the differential equation $xf'(x)+f(x)=f(1)$ admits a solution but I don't think that it will help me.
Can anyone tell me how to proceed with this?
Best Answer
We can assume without loss of generality that $f(1)=0$. Now let $g(x) = x f(x)$. We have $g(0)=g(1)=0$. Hence there is a point $c$ in $(0,1)$ where $g'(c) = 0$. It follows that $c f'(c) + f(c) = 0$, QED.
More generally if $\alpha>0$ and we choose $g(x)=x^\alpha (f(x)-f(1))$, the same argument applies, we have $g(0)=g(1)=0$ and there is a point $c\in (0, 1)$ such that $\alpha c^{\alpha-1}(f(c)-f(1))+ c^\alpha f'(c)=0$, i.e. \begin{equation} f'(c) = \alpha \frac{f(1)-f(c)}{c} \end{equation}