Let $f:\Bbb R \to \Bbb R$ be a differentiable function such that $\exists x $ satisfying $f(x)>0$ and $f'(y) <0$ whenever $f(y)>0$. I want to show that $f$ is always positive before $x$.
I claim that since $f'(x) <0$ then there is an open set $(\delta,x)$ such that $f(z)>f(x)>0$ for all $z \in (\delta,x)$. Apply the same thing to $z$ etc and have that $\forall z \in (-\infty, 0]$, $f(z) >0$. However, I wanted a more formal argument to finish the proof or make it more elegant. I thought of using the mean value theorem, but I cannot guarantee that it is known that the point it gives has a positive value.