$h(x)={f(x)\over x}$ is decreasing or increasing or both over $[0,\infty)$ $f$ is real valued function on $[0,\infty)$ such that $f''(x)>0$ for all $x$ and $f(0)=0$
Then $h(x)={f(x)\over x}$ is decreasing or increasing or both over $[0,\infty)$
$h'(x)={xf'(x)-f(x)\over x^2}$
What I can conclude from here? 
 A: Since $f(0)=0$ then for each $x>0$ there is $c\in(0,x)$ such that $f'(c)=\frac{f(x)}{x}$ by mean value theorem.
Let $x>0$ then we have that:
$h'(x)=\frac{xf'(x)-f(x)}{x^{2}}=\frac{f'(x)-\frac{f(x)}{x}}{x}=\frac{f'(x)-f'(c)}{x}=\frac{f'(x)-f'(c)}{x-c}\cdot\frac{x-c}{x}=f''(d)\frac{x-c}{x}>0$
since $c<x$, $x>0$, and $f''(d)>0$. So $h$ is increasing on $[0,\infty)$.
A: Since f''(x) > 0  on [0,∞] we have that $\int f''(x)\,dx$  = f'(x) > 0 for all x ≥0.  So f is increasing everywhere (we know that anyway since f''(x) > 0 means f is concave upwards, but maybe this is a little more precise).  Further it is increasing faster than x, because f'(x) = 1 and f''(x) = 0. So the numerator of f(x)/x is increasing faster than the denominator, which suggests intuitively that h should be increasing.
We can be a little more mathematical about it.  By the mean value theorem f($x_1$)/x$_1$ = f'(c) for some c such that 0 ≤ c ≤ x$_1$ .  If x$_2$ > x$_1$ then  f($x_2$)/x$_2$ = f'(d) for some d such that 0 ≤ d ≤ x$_2$ .  Because f is concave upwards, d > c, so f'(d) > f'(c) (since f' is monotonic).
Thus h($x_1$) < h($x_2$) and is also monotonic.
A note about h'(x).  We are taught in elementary calculus that if f'(x) > 0 on an interval then f is monotonic there.  But as you see from this example (and there are worse ones -- much worse), it may not be easy to prove that f'(x) > 0.  However, sometimes you can think yourself through this kind of problem with more elementary methods.
