Assuming that the function $f$ is differentiable in $(0,1)$ and continuous on $[0,1]$. If $f(1) = 0$, show that there exists one $c \in (0,1)$, such that $$f(c) = \frac{c f'(c)}{100}.$$
My Attempt: I want to use Rolle's Theorem by defining $g(x) = xf(x)$. See that $g(0) = 0 = g(1)$. So that there exists $c$ such that $$g'(c) = 0 \Longleftrightarrow f(c) + cf'(c) = 0.$$ But I still don't get the answers. Thanks in advance.
I think $f(x) = 1 - x$ gives a counterexample for your statement.
One has $f'(x) = -1$, and the solution of $$ 1 - c = -\frac{c}{100} $$ is $c = \frac{100}{99}$, which is not in $(0, 1)$.