Let f(x) be a derivable function, $f'(x) > f(x)$ and $f(0) = 0$. What can be said about the sign of $f(x)?$ 
Let $f(x)$ be a derivable function, $f'(x) > f(x)$ and $f(0) = 0$. Then 
(A) $f(x) > 0$ for all $x > 0$
(B) $f(x) < 0$ for all $x > 0$
(C) no sign of $f(x)$ can be ascertained
(D) $f(x)$ is a constant function 

Since the mean value theorems are not applicable where do I start?
I tried this:
$f'(x)>f(x)$ 
$=> f'(0)>f(0)$ 
$=> f'(0)>0$
 A: Let $x\ge 0$,then
$$f'(x)>f(x) \implies e^{-x} f'(x) > e^{-x} f(x) \implies \frac{d}{dx} [e^{-x}f(x)]>0.$$ This means that $g(x)=e^{-x}f(x)$ is an increasing function. So $g(x) \ge g(0), for x \ge 0$. Then we have $g(x)>0 \implies f(x) \ge 0 ~\text{for}~ x\ge 0.$
A: The answer is A.
Let $F(x)=f(x)\cdot e^{-x}$ ,so
$$F(0)=0,\,F’(x)=\left(f’(x)-f(x)\right)\cdot e^{-x}>0$$
So
$$\forall x>0,F(x)>0,f(x)=F(x)\cdot e^x>0$$
So why I multiply $e^{-x}$?It just need some ways to solve it. I want to find a helpful function $F(x)=f(x)g(x)$ to use $f’(x)>f(x)$, we know $F’(x)=f’(x)g(x)+f(x)g’(x)$, so if $g’(x)=-g(x)$, we can use it.
Which $g(x)$ is $g’(x)=-g(x)$ ？ 
$e^{-x}$!
A: Attempt:
Let $x >0$;
You got $f'(0)>0$.
$\dfrac{f(x)-f(0)}{x}=\dfrac{f(x)}{x}$;
$\lim_{x \rightarrow 0^+} f(x)/x= f'(0)>0$.
For $x_0$ small enough $f(x_0)/x_0 >0$, i.e. $f(x_0)>0$.
Assume there is an $x_1 >x_0$ s.t. $f(x_0)>f(x_1)$.
The continuos function $f$ attains its maximum on $[x_0,x_1]$.
$f'(x_0) \le 0$, if maximum occurs at $x_0$, or $f'(t)=0$ if maximum occurs at $x_0<t <x_1$.
Then $f(x_0) >0$ and $f'(x_0) \le 0$, or $f(t)>0$ and $f'(t)=0$, a contradiction.
Hence solution $A$.
