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?

You are almost there. We have
 $$f(b) - f(a) = f'(c)(b-a)$$
Since $b>a$, we have $b-a\gt 0$. The sign on the LHS is solely decided by f$\prime (c)$. If f'(c)>0, f(b)-f(a)>0 or f(b)>f(a). 
