1. ## another domain question

hi
i need a little help with this domain question
let $\displaystyle f(x)=\tan (x)-\sqrt{\tan^{2}(x)-\tan (x)}$
such that $\displaystyle x \in ]\frac{-\pi}{2},\frac{+\pi}{2}[$
this is how i started :
$\displaystyle D = \left \{ x\in ]\frac{-\pi}{2},\frac{+\pi}{2}[\mid tan^{2}(x)-tan(x)\geq 0 \right \}$
next step :
$\displaystyle \forall x\in ]\frac{-\pi}{2},\frac{+\pi}{2}[$ we have $\displaystyle tan^{2}(x)-tan(x)$ $\displaystyle positive$ and so $\displaystyle tan(x) \in ]-\infty ,0]\cup [1,+\infty [$
what should i do next ?
thanks a lot

2. Originally Posted by Raoh
hi

$\displaystyle \forall x\in ]\frac{-\pi}{2},\frac{+\pi}{2}[$ we have $\displaystyle tan^{2}(x)-tan(x)$ $\displaystyle positive$
$\displaystyle \forall x\in ]0,\frac{+\pi}{4}[$ we have $\displaystyle tan^{2}(x)-tan(x)$ $\displaystyle NEGATIVE$

3. $\displaystyle \uparrow$ You've already stated the answer! $\displaystyle \uparrow$

I mean, it would be more explicit if you wrote what values of x are implied by $\displaystyle tan^2(x)-tanx\geq0$.

Do you know how to do that?

4. hi
well ..we have $\displaystyle tan^2(x)-tanx\geq0$ which means $\displaystyle tan(x) \in (- \infty ,0] \cup [1,+ \infty )$,and so $\displaystyle x \in (\frac{-\pi}{2},0]\cup [\frac{\pi}{4},\frac{+\pi}{2})$
am i right ?
thanks