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Let $f:[a,b] \rightarrow \mathbb{R} $ be a continuous function in $[a,b]$ and differentiable in $(a,b)$ such that $f(a)=f(b)=0$ . Show that $7f(c)+cf'(c)=0$ for some $c \in (a,b)$.

I tried to use rolle's theorem and the mean value theorem, but got stuck. Any hint will be appreciated.

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  • $\begingroup$ Are you sure about that $7$ there? $\endgroup$ – José Carlos Santos Jan 30 '18 at 23:30
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    $\begingroup$ Sure. Me too. ;-) $\endgroup$ – José Carlos Santos Jan 30 '18 at 23:35
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    $\begingroup$ I can show it to be valid if $0\not\in[a,b]$, by using function $g(x)=x^7f(x)$ and applying Rolle,s theorem, but the case $a\le 0\le b$ is still escaping me. $\endgroup$ – user491874 Jan 30 '18 at 23:36
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    $\begingroup$ @user8734617 Just work in the interval $[a,0]$ (or $[0,b]$), there should be a zero of the derivative in the interior. $\endgroup$ – orole Jan 30 '18 at 23:41
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    $\begingroup$ @orole Indeed, $g(a)=g(b)=g(0)=0$... I think you should have the honour of writing the answer up in this case! $\endgroup$ – user491874 Jan 31 '18 at 0:01
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Apply the Rolle's theorem to the function $$x^7f(x).$$

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    $\begingroup$ You might want to show you got the factor $x^7$ by using the differential equation $7f+xf'=0$. The integrating factor is $e^{7\log x} =x^7$. $\endgroup$ – Paramanand Singh Jan 31 '18 at 8:14
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Hint :

Take the expression you have and solve it as an ODE :

$$7f(x) + xf'(x) = 0 \Leftrightarrow \frac{f'(x)}{f(x)} = - \frac{7}{x} \Rightarrow f(x) = \frac{c_1}{x^7}$$

For $x=c$ the expressions yields :

$$f(c) = \frac{c_1}{c^7}$$

This can only be equal to $0$ iff $c_1 = 0 \Rightarrow f(x) = 0$.

For $f(x)=0$ though the hypothesis holds, since $f$ will be continuous in $[a,b]$, differentiable in $(a,b)$ and it obviously is $f(a) = f(b) = 0$. This, by Rolle's Theorem implies that $\exists c \in (a,b) : f'(c) =0$, which is also true, since if $f(x)=0$ then also $f'(x) = 0$.

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    $\begingroup$ How exactly is this solving the problem? It is not given that $7f(x)+f’(x)=0$ on $[a,b]$. $\endgroup$ – user491874 Jan 31 '18 at 0:52
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    $\begingroup$ Are you trying to prove that $f(x) $ must be identically $0$ in order for the result to be true? $\endgroup$ – Paramanand Singh Jan 31 '18 at 8:25
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This augments the answer given by @Bumblebee.

Set $g(x)=x^7f(x)$.

We have $g(a)=g(b)=0$ and $g$ is also continuous on $[a,b]$ and differentiable on $(a,b)$. Thus, by Rolle's theorem, there is a point $c\in(a,b)$ such that $0=g'(c)=7c^6f(c)+c^7f'(c)=c^6\left(7f(c)+cf'(c)\right)$. Distinguish two cases:

  • $0\not\in (a,b)$: then $c^6\ne 0$ so we conclude that $7f(c)+cf'(c)=0$
  • $0\in (a,b)$. Note, however, $g(0)=0$, so we can apply Rolle's theorem to $g(x)$ on $(a,0)$ (or $(0,b)$) and conclude that there is a point $c\in(a,0)$ (or $c\in(0,b)$) such that $0=c^6\left(7f(c)+cf'(c)\right)$, like before. However, on $(a,0)$ (or $(0,b)$) we can guarantee $c\ne 0$, so the conclusion $7f(c)+cf'(c)=0$ follows.

The $2^{nd}$ part of this proof is courtesy of @orole.

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