Let f be a continuous and differentiable function such that f(a)=f(b)=0 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.
 A: Apply the Rolle's theorem to the function $$x^7f(x).$$
A: 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.
A: 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 $f'(x) = 0$.
