Assume that $f(x)$ is continuous in $[a,b]$ and differential in $(a,b)$, and $f(a)=f(b)=0$, prove that there exists a $\xi\in(a,b)$, s.t $f'(\xi)=f(\xi)$.
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1What have you tried so far? – Giorgos Giapitzakis Apr 29 '21 at 01:10
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I get an anwser that let $F(x)=e^{-x}f(x)$, then $F(b)=0,F(a)=0$,so we just apply the Roll theorem to complete this proof.But I don't know how I get $F(x)$? – Wang Aliber Apr 29 '21 at 01:23
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2When you come across such equalities it's always a good practice to multiply with something that is always nonzero in order to make the expression equal the derivative of some function. The exponential function $e^{\pm x}$ is a very good candidate since it is always positive and is ($\pm$) the derivative of itself. – Giorgos Giapitzakis Apr 29 '21 at 01:25
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@WangAliber FYI, your question is basically Prove there exists $\xi \in (a,b)$ such that $f(\xi)=f^{(n+1)}(\xi)$. with $n = 0$. However, since that other problem involves a more general, and likely harder to solve, problem than yours, I'm not considering it to be a duplicate. Nonetheless, although I couldn't offhand find any, I suspect some do exist. Regardless, I hope you find my linked post to be of some help. Good luck with solving your problem. – John Omielan Apr 29 '21 at 02:33
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I assume you mean "s.t. $f'(\xi) = 0$", since as stated you are trying to prove something false. – Mark Saving Apr 29 '21 at 02:52
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Thanks for your help! – Wang Aliber May 08 '21 at 12:08