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Math Help - find the domain of the function

  1. #1
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    find the domain of the function

    f(x) = \sqrt{4-x^2} + \sqrt{sinx}

    \sqrt{4-x^2} is defined if  -2 \leq x \leq 2

    \sqrt{sinx} is defined if  2\pi k \leq x \leq \pi + 2 \pi k

    So, to find the domain of the function, am I supposed to determine their union or intersection?

    (I think it's the intersection)

     (-2 \leq x \leq 2) \cap  (2\pi k \leq x \leq \pi + 2 \pi k) = 0 \leq x \leq 2

    Is that the correct way to find the domain or am I supposed to find their union?

    Thanks!
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  2. #2
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    Quote Originally Posted by jayshizwiz View Post
    = 0 \leq x \leq 2
    I agree.
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  3. #3
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    In order to be able to find a value for f(x), you must be able to calculate both \sqrt{4- x^2} and \sqrt{sin(x)}. That is why both " -2\le x\le 2" and " 2k\pi\le x\le \pi+ 2k\pi must be true.


    I think it is amusing that Pickslides says "I agree with you" and just below has his "If I agreed with you we'd both be wrong"!
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    In order to be able to find a value for f(x), you must be able to calculate both \sqrt{4- x^2} and \sqrt{sin(x)}. That is why both " -2\le x\le 2" and " 2k\pi\le x\le \pi+ 2k\pi must be true.


    I think it is amusing that Pickslides says "I agree with you" and just below has his "If I agreed with you we'd both be wrong"!
    Haha!!
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