# The equation $f'(x)=f(x)$ admits a solution

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.)

1. Is also false since $$f(x)=sin\pi x$$

• $\sin \pi x$ does not satisfy $f'(x) = f(x)$. – GEdgar Dec 6 '18 at 12:10
• @GEdgar you are right!give me an example of function which has excatly give one solution – Cloud JR Dec 6 '18 at 19:58

hint

$$g(x)=f(x)e^{-x}$$

$$g(0)=g(1)$$

Rolle?

• How do you get that idea of defining function like that... Any easier way?plz help – Cloud JR Dec 5 '18 at 19:21
• From the difference $$f'(x)-f(x)=0$$ – hamam_Abdallah Dec 5 '18 at 19:22
• And you think how to make that , also $f'(x)=f(x)$ suggest using e^x ...your answer(hellHound ) is very enlightening – Cloud JR Dec 5 '18 at 19:25
• You mean $e^{-x}$. If we had $$f'(x)+f(x)=0$$ we think $e^x$. – hamam_Abdallah Dec 5 '18 at 19:29

Look at $$g(x)=f(x) \cdot \exp (-x)$$. Then $$g$$ is also differentiable on $$(0,1)$$ with derivative $$g'(x) = (f(x)-f'(x)) \exp (-x)$$. Noting that $$g(0) = g(1) = 0$$, Rolle's theorem applies and you get a $$c \in (0,1)$$ such that $$g'(c) = 0$$. This implies $$f(c) = f'(c)$$.

• Yes I realised and edited it, thank you. It seems like we have posted our answer at almost the same time. – hellHound Dec 5 '18 at 19:21
• Also the same answer lol...idk which one to accept – Cloud JR Dec 5 '18 at 19:22