$f: [0, 1) \to \mathbb{R}$ is a function such that $f(x)-x^3$ and $f(x)-3x$ are both increasing functions. i.e., $f'(x)-3x^2 > 0$ and $f'(x)-3 > 0$. Therefore, $f'(x) > 3$. Now, differentiating the function $f(x)-x^2-x$ w.r.t $x$ we get $f'(x)-2x-1$. Now, $f'(x)-2x-1 > 3-2x-1 = 2-2x$. Since, $x \in [0, 1)$, $2-2x > 0$. i.e. the derivative of the given function is always positive. So, the function is increasing.

parmenides51 wrote: Assume that $f : [0,1)\to R$ is a function such that $f(x)-x^3$ and $f(x)-3x$ are both increasing functions. .... cygnusx251 wrote: i.e., $f'(x)-3x^2 > 0$ .... Nobody said derivative do exist. Nobody even said that $f(x)$ is continuous .... .

pco wrote: parmenides51 wrote: Assume that $f : [0,1)\to R$ is a function such that $f(x)-x^3$ and $f(x)-3x$ are both increasing functions. .... cygnusx251 wrote: i.e., $f'(x)-3x^2 > 0$ .... Nobody said derivative do exist. Nobody even said that $f(x)$ is continuous .... . Thanks for your point. I did not notice that. Sorry for this foolish mistake

Posted here in 2006, but not sure what the source is.