If $f'(x)\le g'(x)$, prove $f(x)\le g(x)$ I have to do the following exercise:

Let $f$ and $g$ two differentiable functions such that $f(0)=g(0)$ and $f'(x)\leq g'(x)$ for all $x$ in $\mathbb{R}$. Prove that $f(x)\leq g(x)$ for any $x\geq0$.

Now, I know this is true because the first derivative of a function is the angular coefficient of the function in a point $x$. So, $f'(x)\leq g'(x)$ means, in other words, that the function $g(x)$ grows faster than $f(x)$. I think this is the base for a more formal proof, could someone help me to figure out a more formal proof?   

Hint: consider the function $h(x) := g(x) - f(x)$, so that $h(0) = 0$ and $h'(x) \geq 0$ for every $x\geq 0$.
Now, for a fixed $x> 0$, it is enough to apply the mean value theorem to $h$ in the interval $[0, x]$.
