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Finding implied domains and ranges

  1. May 18, 2006 #1

    Could someone please help me out here?

    state the implied domain and range of
    B) tan (2arccos(x))

    ok, the domain of tan for which arctan exists (conventionally) is (-pi/2 , pi/2) -- really the domain of tan is R, but we're using the restricted one.

    therefore, we know that the internal function must be within this:
    -pi/2 < 2arccos(x)) < pi/2
    dividing by two
    -pi/4 < arccos(x)) < pi/4
    Now, to find what x is between, we must take the cos of both sides, BUT WHY DO WE HAVE TO FLIP THE INEQUALITIES? I mean, sure, it makes no sense if one doesn't flip them -- however i presume that in maths one can't just arbitrarily change things around because they dont make sense.

    Oh, crap, i forgot to consider the internal function's domain / range....

    What, exactly, am i supposed to consider here? The domain or the range of the internal function?

    For some questions, all one has to do is find the intersection of the domain of the external function and the range of the internal function, but this logic, i've noticed, doesn't always work. I dont really understand what i'm doing.... i guess i'm just learning it by rote without really understanding it, which irritates me.

    Like with this question: y=arctan (sin x)

    If one uses the Domain intersection Range logic, one gets [-1, 1] for the implied domain, whcih is wrong.

    On the other hand, if one uses the other logic, ie that:
    the dom of arctan is R, domain of restricted sin x is [-pi/2, pi/2], therefore for implied domain for that question is [-pi/2, pi/2].

    I don't understand this. Implied ranges too, but i guess once i understand domains, ranges wont be a problem.

    How does one go about doing questions like this? And could someone please explain what i'm actually doing?

    Last edited: May 18, 2006
  2. jcsd
  3. May 18, 2006 #2


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    Staff Emeritus
    Science Advisor

    You don't "flip the inequalities" here. If you were working on 0< x< pi/2 then, since cosine is a decreasing function, x< y implies cos(x)> cos(y). However, here, "taking the cos of both sides" (you really mean "of all three parts") since cos(-pi/4)= cos(pi/4)= sqrt(2)/2, you would have the nonsensical sqrt(2)/2< x< sqrt(2). Better than just "taking the cos of both sides" is to look at the graph of cosine between -pi/4 and pi/4. cos(-pi/4)= sqrt(2)/2 and cosine rises to 1 at 0 and then goes back down to sqrt(2)/2 at pi/4. x must be between sqrt(2)/2 and 1.

    THINK! You know that the domain of (the principal value of) tan, in this problem, is 0 to pi. Since that is contained in sqrt(2)/2 to 1, what values of x will give arccos(x) between sqrt(2)/2 and 1? Obviously pi/4 to pi/2. The domain of this function is [-pi/2, -pi/4] union [pi/4, pi/2]

    Oh, I see where you are going wrong: [-1, 1] is the "Domain intersection Range" but the domain of the entire function is the subset of the domain of the "internal function" that gives those values.
    Any value of [-pi/2, pi/2] (the domain of the principal value of sine) gives a value of sin x between in [-1, 1] which then gives a valid arctan value. The domain of y= arctan(sin x) is [-pi/2, pi/2], not [-1, 1]. You didn't "go back" to the domain of sine from [-1, 1].

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