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Domain of a function 
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#1
Nov2811, 08:21 PM

P: 96

1. The problem statement, all variables and given/known data
Use set or interval notation to give the domain of the function: g(t)=(x5)/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
Nov2811, 08:27 PM

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P: 21,314

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 calculusrelated question, I am moving it to the Precalc section. 


#3
Nov2811, 09:10 PM

P: 96

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)? 


#4
Nov2911, 12:36 AM

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P: 21,314

Domain of a function
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}. 


#5
Nov2911, 08:43 AM

P: 96

Ah yes, that makes perfect sense. Thanks again for the help Mark.



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