Question involving the application of the mean value theorem (MVT) 
Suppose that $f(0)=0$ and $f'$ is increasing. Prove that the function $g(x) = \frac{f(x)}{x}$ is increasing on $(0,\infty)$.

My attempt: We want to show that $g(a)<g(b)$. So first, we use the MVT on $f$. Suppose $a<b$ are any two points on $I$. Since $f'$ is increasing we know that $f(a)\leq f(b) \Rightarrow f(a) -f(b) \leq 0 $. By the MVT, there is some point $c\in(a,b)$ so that $f'(c) = \frac{f(b)-f(a)}{b-a}$. So $f(b) - f(a) = f'(c)(b-a) = xg(b)-xg(a) = f'(c)(b-a)$.
Then where do I go from here?

Let $a<b$.  $g(a) = f^\prime(c)$ for some $c \in (0,a)$ by the mean value theorem for $f$ on $[0,a]$.  As you've observed, $bg(b) - ag(a) = f^\prime(d)(b-a)$ for some $d \in (a, b)$.  As $d>c$, we have $$g(b) = \frac{1}{b}((b-a)f^\prime(d) + af^\prime(c)) > \frac{1}{b}((b-a)f^\prime(c) + af^\prime(c)) = f^\prime(c) = g(a)$$ since $f^\prime$ is increasing. 
