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
Thus g(x) = f'(0) at $x=0$