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Domain of g(x) = arctan(2x)

  1. Dec 10, 2007 #1
    Question 2.

    I agree that the range of arctan is [-pi/2, pi/2], however due to the fact that we have arctan(2x), not just arctan(x), wouldn't that mean that the range of the g(x) function would be [-pi/4, pi/4] instead? 'cause then when your pi/4 goes into the arctan, it gets multiplied by the 2, making it pi/2, which is the maximum allowed into the arctan function.

    Am I wrong, or is the solution? If it's me, where is my mistake?

    Also, in Question 3, wouldn't the answer be 5pi/6? Don't the arctan and tan cancel each other out, like squaring a root?
  2. jcsd
  3. Dec 10, 2007 #2


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    The upper limit of the range of arctan(x) is pi/2 because limit as x->infinity, arctan(x)->pi/2. Changing x to 2x doesn't change that. For question 3) arctan(tan(x))=x only for x in a certain part of the range of tan. You've already said the domain of arctan is [-pi/2,pi/2]. For x outside that range arctan(tan(x))=x CAN'T be true.
    Last edited: Dec 10, 2007
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