Show that a function is increasing. Qusestion: Let f be a continuous and differentiable function on $[0, \infty[$, with $f(0) = 0$ and such that $f'$ is an increasing function on $[0, \infty[$. Show that the function g, defined on $[0, \infty[$ by $$g(x) = \begin{cases} \frac{f(x)}{x},   x\gt0& \text{is an increasing function.}\\ f'(0), x=0 \end{cases}$$
I have tried to solve this problem but I don't know whether I have done it right or not. 
Solution: I have applied mean value theorem on the interval $[0, x]$. Then, $$\frac{f(x)}{x} =f'(c) , 0\lt c \lt  x$$ 
It is given that $f'$ is an increasing function. So I deduce that $\frac{f(x)}{x}$ is also increasing.
Further, $$g(x) = f'(c) \text {such that } 0<c<x$$ Therefore, $$g(0) =f'(c) \text{such that} 0<c<0$$
So,  $c=0$
Thus $g(x) = f'(0)$ at $x=0$

$c$ depends on $x$, and what you did doens't prove that if $x_1\leq x_2$ then $c_{x_1}\leq c_{x_2}$. 
But we can write for $x>0$, since  $f'$ is increasing hence integrable over finite intervals 
$$\frac{f(x)}x=\frac{f(x)-f(0)}x=\frac 1x\int_0^xf'(t)dt=\int_0^1f'(xs)ds$$
by the substitution $t=xs$. This formula also works for $x=0$, and now it's easy to deduce that $g$ is increasing: if $x_1\leq x_2$, for all $0\leq s\leq 1$ we have $sx_1\leq sx_2$ and since $f'$ is increasing...
