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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 $f(x)=sin\pi x$

So 4. Is true but i stuck to prove . Please help

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

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

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

Rolle?

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

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

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