Question: > 
> Show that if $f$ is continuous in $[a,b]$ and differentiable in $(a,b)$, and $f(a)=f(b)=0$, then for a real $\alpha$ there is an $x \in (a,b)$ such that $\alpha f(x)+f'(x)=0$.
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I believe that I am doing a logical mistake in my following solution of this problem or I am not interpreting correctly the request of the problem (English is not my first language). My solution: since the hypotheses of the mean value theorem are satisfied and it is $f(b)=f(a)=0$, there exists $x \in (a,b)$ such that $0=f(b)-f(a)=(b-a)f'(x)$. This implies $0=(b-a)f'(x)$ and, since for definition of interval $[a,b]$ it is $a<b$, hence $b-a>0$ and so $0=(b-a)f'(x)$ implies $f'(x)=0$. So $\alpha f(x)+f'(x)=0$ is true taking $\alpha=0$ for the $x \in (a,b)$ given by the mean value theorem.

I am suspicious that the problem doesn't ask for the existence of a particular real $\alpha$ that makes the identity $\alpha f(x)+f'(x)=0$ true for a $x \in (a,b)$, then I am not understanding what the problem means with "for a real $\alpha$". Can someone help explain me this better or, if I correctly understood the request but my proof is wrong, explain me why it is wrong?

Answer: 
You are right, the formulation is indeed a little imprecise and "for any real $\alpha$" would be the better formulation. "...for a real $\alpha$ ..." could mean "$\exists \alpha \in \mathbb{R}$.." or "$\forall \alpha\in \mathbb{R}$...". In the first interpretation your solution is correct. In the second interpretation for a given $\alpha$ set $g(x):=\exp(\alpha x)f(x)$ on $[a,b]$ and apply the mean value theorem to $g$.
Indeed, this is true for al $\alpha$. We may assume that $f'(a)\not=0$, as well as $f'(b)$, otherwise $a$ (or $b$) does the job. Changing $f$ to $-f$, we may even assume that $f'(a)>0$. Let $b\_1$ be the first $0$ of $f$, $b\_1=Min (x/f(x)=0$. Note that $b\_1>a$, and $f$ is $>0$ on $[a,b\_1]$, so that $f'(b\_1)\leq 0$. Now, if $g(x)=\alpha f(x)+f'(x)$, $g(a)>0,g(b\_1)\leq0$, so somewhere inbetween, $g(x)=0$.