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Let $f$ be a continuous function on [0,1] differentiable on (0,1) such that $f(1)=0$ then prove that for some $c$ $$cf'(c)+ f(c)=0$$

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Hint: Consider the function $g(x)=xf(x)$, and apply Rolle's theorem.

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  • $\begingroup$ Wow ,i feel like a fool now :p, $\endgroup$ – avz2611 May 17 '14 at 15:44
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    $\begingroup$ @user142634 You shouldn't. One has to come up with the function which I think is hard without prior experience. $\endgroup$ – BCLC May 17 '14 at 15:55
  • $\begingroup$ @user1337 How do you know that if f(x) is cont in [0, 1] and diff in (0,1), then g(x) is cont in [0, 1] and diff in (0,1)? $\endgroup$ – BCLC May 17 '14 at 15:56
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    $\begingroup$ @BCLC $g(x)$ is the product of two functions which are continuous in $[0,1]$, and is thus continuous there. Similarly $g(x)$ is the product of two functions which are differentiable in $(0,1)$, and is thus differentiable there. $\endgroup$ – user1337 May 17 '14 at 15:58
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    $\begingroup$ @user142634 You shouldn't. The key here is to recognize the expression in your question as the product rule applied to $xf(x)$ at the point $x=c$. This might require some experience. $\endgroup$ – user1337 May 17 '14 at 16:00

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