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 $x=0$

With differentiation we get:$$g'(x)=\frac{f'(x)*x-f(x)}{x^2}$$ We want to show that $g'(x)>0 \ \Leftrightarrow \ f'(x)*x-f(x) > 0 \ (1)\ \forall \ x > 0 $
Reforming (1) we got: $$f'(x) > \frac{f(x)}{x} \ (2)$$ Applying the mean value theorem at [0,x], we got $f'(c)=\frac{f(x)}{x}$, $ 0<c<x$. So we want to show that $f'(x) > f'(c) \ \forall x > 0$. But since f' is an increasing function we got: $$ x > c \Leftrightarrow f'(x)>f'(c)=\frac{f(x)}{x}$$. So from (2) and since $g'(x) > 0 \ \forall \ x >0 $, g(x) is an increasing function in $[0,\infty).$
