We encountered this question in the class today. My lecturer asked me to define a function $f(x)=\tan(x) - x$. Then we were asked to find its derivative, $f'(x)$, i.e, $\sec^2(x) - 1$, and solve the inequality $f'(x)>0$. The answer was $(0,π/2)$.
My question is, if $f'(x)>0$, it just proves that $f(x)$ is increasing in the interval, it doesn't prove if $f(x)$ is always positive there. Ok, agreed that since $\tan0=0$, $\tan x>x$ in $(0,π/2)$ since it's increasing. But there is possibility of loosing a solution right? It could be possible that in an interval $\tan x-x$ is decreasing but $\tan(x)$ is still $>x$.