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Domain of a function

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Use set or interval notation to give the domain of the function:

    g(t)=(x-5)/sinx


    2. Relevant equations



    3. The attempt at a solution

    So looking at this, I recognize that x cannot be 0,180,360, etc... it's just putting this into set or interval notation that confuses me. Also, I notice that g is a function of t which is not in the equation, would that therefore mean that the domain can be all real numbers?
     
  2. jcsd
  3. Nov 28, 2011 #2

    Mark44

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    Staff: Mentor

    Your function should be g(x), not g(t), since t doesn't appear in the formula on the right side.

    Instead of 0, 180, 360, etc. degrees, you should be thinking in terms of radians. The sine function is zero at integer multiples of π, so these will not be in the domain.

    Mod note: As this does not appear to be a calculus-related question, I am moving it to the Precalc section.
     
  4. Nov 28, 2011 #3
    Hmm, I wonder if I can argue for the marks back then if it's a typo.

    Anyways, thanks for the help Mark. Substituting 0, 180, and 360 for 0, pi, 2pi, etc...
    I do understand what it cannot be, just putting it into interval notation is confusing me. I guess I'm thinking as x can increase without bounds, but 2pi essentially = 0pi.

    Would it be something like: (2pi, pi)U(pi, 0)?
     
  5. Nov 29, 2011 #4

    Mark44

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    You have your intervals backwards - they should be (0, [itex]\pi[/itex]) U ([itex]\pi[/itex], 2[itex]\pi[/itex]) U ...

    But this doesn't include the intervals on the left side of 0. A better way is to write it {x [itex]\in[/itex] R | x [itex]\neq[/itex] k[itex]\pi[/itex], k an integer}.
     
  6. Nov 29, 2011 #5
    Ah yes, that makes perfect sense. Thanks again for the help Mark.
     
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