$f:\mathbb R \to \mathbb R$ be differentiable such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is $f(x)>0,\forall x>0$? Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is it true that $f(x)>0,\forall x>0$ ?
 A: Because $f'(x)>f(x)$ we have that $ e^{-x} f(x)$ is strictly increasing.That is the conclusion.
A: Yes.  Hint: Consider $x=0$. Since $f'(0)>f(0)=0$, we know
$$
\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0}\frac{f(h)}{h}>0.
$$
Therefore, for $h$ positive and sufficiently close to $0$, the numerator is positive.  Continue with this idea.
A: Let $y(x)=e^{-x}f(x)$. Then $ f$ (strictly) positive $ \iff  y$ (striclty) positive. 
$\forall x $,  $   y'(x)=e^{-x}(f'(x)-f(x)) \ge 0$ and if $x> 0, y'(x)>0$.
Therefore $y$ is positive for $x \ge 0$. 
Now lets suppose it exists $t>0$ such that $y(t)=0$. Then for  $\epsilon >0 $ small enough, because $y'(t) >0$, for $x \in ]t-\epsilon,t[$ , $y(x) < y(t) $, which is in contradiction with the previous point.
A: I think it should work like this:
By definition of limit
$\lim_{h \to 0^+}\frac{f(0+h)-f(0)}{h}=f^\prime(0)$
Since $f^\prime(0)>f(0)$ and $f(0)=0$, $\lim_{h \to 0^+}\frac{f(h)}{h}>0$ then $f(x)>0$ for $x$ in $]0,\delta]$.
$\lim_{h \to \delta^+}\frac{f(\delta+h)-f(\delta)}{h}=f^\prime(\delta)>f(\delta)>0\rightarrow\lim_{h \to \delta^+}{f(\delta+h)-f(\delta)}>0\rightarrow f(x)>0\ for\ x\in(\delta,\delta_1) $ And so on.. So yes it is true
