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Thread: Find the domain range and inverse

  1. July 28th 2010, 01:58 PM #1
    dat1611
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    Find the domain range and inverse

    f(x)=arcsin(sqrt(x))

    i think the domain is x>=1
    inverse is sin(sqrt(x))
    not sure about the range can someone help
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  2. July 28th 2010, 02:13 PM #2
    pickslides
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    I think x \in [0,1]

    and f^{-1}= \sin^2{x}

    Now I have given you the domain, what do you think the range could be?
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  3. July 28th 2010, 02:34 PM #3
    dat1611
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    [0,pi/2]
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  4. July 28th 2010, 03:30 PM #4
    pickslides
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    Yep.
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  5. July 28th 2010, 03:38 PM #5
    dat1611
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    how did you come up with the domain
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  6. July 28th 2010, 03:47 PM #6
    pickslides
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    I asked myself 3 questions, what is the domain of \sqrt{x} ? what is the domain of \arcsin{x} and what is there intersection?
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