Analysis problem: Show that$ f(x)$ is less or equal than$ g(x)$ Analysis problem:
Let $f$ and $g$ be differentiable on $ \mathbb R$. Suppose that $f(0)=g(0)$ and that $f' (x)$ is less or equal than $g' (x)$ for all $x$ greater or equal than $0$ Show that $f(x)$ is less or equal than$g(x)$ for all $x$ greater or equal than $0$.
Is my proof correct?
I am trying to use the Generalized Mean Value Theorem:

As $f$and g are differentiable on$ \mathbb R$, $f$ and $g$ are continuous on $ \mathbb R$ and we can use the Generalized Value Theorem. Using the starting condition $f(0)=g(0)$, we have that for any b that is greater than$ 0$, exist a $c$ element of $(0,b)$ such that
$f' (c) g(b) =  g' (c) f(b)$
By the starting conditions,
$f' (x) $is less or equal than$g' (x)$ for all $x$ greater or equal than $0$
Therefore, $f(b)$ is less or equal than $g(b)$ for any b element of $(0, b)$
As $b$ is any number bigger than$ 0$
$f(x)$ is less or equal than $g(x)$ for any $x$ greater or bigger than$ 0$. Q.E.D.

Let $h(x)=f(x)-g(x);x\in [0,\infty)$
$h^{'}(x)=f^{'}(x)-g^{'}(x)\le 0\implies h$ is decreasing on $[0,\infty)\implies h(x)\le h(0)\forall x\in [0,\infty)\implies f(x)\le g(x)$
