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Let $f(x)$ be a continuous function in $[0,1]$ and differentiable in $(0,1)$ such that $f(1) = 0$

Prove that there is $x_0\in(0,1)$ such that $x_0\cdot f'(x_0) + f(x_0) = 0$

My attempt -

I really couldn't make much progress but it does feel to me like some kind of Lagrange Theorem manipulation. i tried some kind of messing around, and could only prove that there exists $x_0$ such that - $$f'(x_0) = -f(0)$$

Any hints ?

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    $\begingroup$ Hint: put $g(x)=xf(x)$. $\endgroup$ – Kelenner Dec 30 '17 at 17:29
  • $\begingroup$ Use an integrating factor.. $\endgroup$ – Cameron Williams Dec 30 '17 at 17:35
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Apply Rolle's theorem to $g(x) = xf(x)$ on the interval $[0 , 1]$ and use product rule to find $g'(x)$.

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  • $\begingroup$ Hey Mohammad! In the future, you should endeavor to use MathJax in your answers to make them easier to read. In your case, converting to MathJax is as simple as enclosing your mathematical expressions in dollars signs (e.g. $g(x) = xf(x)$) $\endgroup$ – eepperly16 Dec 30 '17 at 18:14
  • $\begingroup$ Thanks, I am learning about it. $\endgroup$ – Mohammad Riazi-Kermani Dec 30 '17 at 19:21
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as @Kelenner pointed out in a comment

if you put $g(x) = xf(x)$

what you want to show is equivalent to showing that $\exists x_0 \; g'(x_0)=0$

also you have that $g(0) = 0$ and $g(1) = f(1) = 0$

then according to the Mean value theorem there exists $x_0 \in (0,1)$ such that $ g'(x_0) = x_0\cdot f'(x_0) + f(x_0) = 0$

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