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: $f'(x)>f(x)$ $=> f'(0)>f(0)$ $=> f'(0)>0$

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
Then $f(x_0) >0$ and $f'(x_0) \le 0$, or $f(t)>0$ and $f'(t)=0$, a contradiction.

Hence solution $A$.