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anonymous

  • 5 years ago

find the domain Q(y)=sqrt (y^2+1) - cuberoot (1-y)

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  1. anonymous
    • 5 years ago
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    Domain of y is all real numbers. Explanation: since the quantity y^2 is always positive ((y^2)+1)^(1/2) is always defined and since the a cube root is defined along all reals y can take any value. There does not exist a value of y that will make the function undefined

  2. anonymous
    • 5 years ago
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    and what about in this case 4 / (x-9) - sqrt (x^3 -36)

  3. anonymous
    • 5 years ago
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    x definitely can't equal 9 (since you would be dividing by 0) and x^3 must be greater than 36 (since you can't have a negative in a square root so x > 36^(1/3) and x \[\neq\] 9

  4. anonymous
    • 5 years ago
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    thanx

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