If $\tan (x)>x$. Find interval of $x$.

Question

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 $>x$.

I'm confused.

Answer 1

Note that the derivative is positive everywhere it is defined, i.e. for all $x\ne(2k+1)\dfrac\pi2$.

For a continuous function and $x_1>x_0$, if you know $f(x_0)>0$ and $f$ increasing, then $f(x_1)>0$. But if $f(x_0)>0$ and $f$ decreasing, you can't conclude about the sign of $f(x_1)$.

Answer 2

$f'(x)>0$ in the interval $(0, \frac{\pi}{2})$, so $f$ is increasing on this interval. Since $f(0)=0$ and $f$ is continuous on the interval, $f$ must be positive on the interval. Note that $f(\frac{\pi}{2})$ is undefined, and then $f(\frac{\pi}{2}+ \epsilon)$ for some small $\epsilon>0$ is negative. So we can say that $f$ is always positive on the interval you've specified.

Answer 3

No. $\tan x -x$ is always increasing on each interval on which it is defined, since it is differentiable, and its derivative has a second form: $$(\tan x-x)'=1+\tan^2x-1=\tan^2x>0$$ except at the isolated points $\;\Bigl\{\dfrac\pi2+k\pi\:\Big\vert\: k\in\mathbf Z\Bigr\}$, where it is equal to $0$.