$f:\mathbb{R}\rightarrow \mathbb{R}$ differentiable, $f'(x)>f(x)$ for all $x$, and $f(0)=0$, then $f(x)>0$ for all $x>0$ Suppose $f:\mathbb{R}\rightarrow \mathbb{R}$ is differentiable everywhere, $f'(x)>f(x)$ for all $x\in \mathbb{R}$, and $f(0)=0$, then $f(x)>0$ for all $x>0$.
I tried using the proof given in this post: Show that if $f( 0 ) = 0$ and $f'( x ) > f( x )$ for all $x \in \mathbb{ R }$ then $f( x ) > 0$ for all $x > 0$.
but I realized that I am not given that $f'$ is continuous. I have been playing with the following:
Since $f'(x)>f(x)$ for all $x$ then $f'(0)=\lim_{h\rightarrow 0} \frac{f(h)}{h}>f(0)=0$. I want to say that this means that $f$ must be positive over $N_{\epsilon}(0)$ for some $\epsilon>0$. Otherwise, this would somehow contradict that $f$ is differentiable. 
If I know that $f$ is positive over some epsilon neighborhood around zero, I know that the function has "lift off" after zero. From there I can argue that the function can't come back down to zero because it would contradict that $f'(x)>f(x)$. Can someone help me formalize this hand wavy argument? I would greatly appreciate it!

Suppose that on the contrary there is a positive number $a$ such that $f(a) \leq 0$. Now that $f'(0) > f(0) = 0$ and hence there is an interval of type $(0, h]$ such that $f$ is positive in it and clearly we can take $h < a$. The condition $f(a) \leq 0$ and the fact that $f(x) > 0$ for all $x \in (0, h]$ allows us to infer the existence of a $b \in (0, a]$ such that $f(b) = 0$ (via intermediate value property).
Now $f$ is continuous so there is a first value of $x$ in $[h, b]$ such that $f(x) = 0$ and let's call it $c$. Then $f(0) = 0 = f(c)$ and $f(x) > 0$ for all $x \in (0, c)$. It now follows by Rolle's theorem that there is a $d\in (0, c)$ for which $f'(d) = 0$. But $f'(d) > f(d) > 0$ so we arrive at a contradiction.
