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Thread: another domain question

  1. #1
    Super Member
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    Smile 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
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Raoh View Post
    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 $\displaystyle tan^{2}(x)-tan(x)$ $\displaystyle NEGATIVE$
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  3. #3
    No one in Particular VonNemo19's Avatar
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    $\displaystyle \uparrow$ 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?
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  4. #4
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    Smile

    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
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