TITLE: Let f be a continuous and differentiable function such that f(a)=f(b)=0
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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.