let $f :[0,1]→\mathbb R$ be a fixed continous function such that f is differentiable on (0,1) and $ f(0)=f(1)=0$ .then the equation $ f'(x)=f(x)$ admits
1.No solution $x\in (0,1)$ 2. More than one solution $x\in (0,1)$ 3.Exactly one solution $x\in (0,1)$ 4.At least one solution $x\in (0,1)$
Actually mean value theorem doesn't work for $f'(x)$. The function $f$ has a fixed point and at that point $f'(x)=1$ and at some point $x_0 , f'(x)=0$ (by mean value theorem).but it doesn't get me anywhere! What am I missing?
Obviously 1. Is false since f(x)=0 ( also 3.)
- Is also false since $f(x)=sin\pi x$
So 4. Is true but i stuck to prove . Please help