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Determine the Set of Points at Which the Function is Continu

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  1. Feb 16, 2016 #1
    1. The problem statement, all variables and given/known data
    "Determine the set of points at which the function is continuous"
    ƒ(x,y,z) = √y/(√(x2-y2-z2))

    2. Relevant equations
    ƒ(x,y,z) = √y/(√(x2-y2-z2))

    3. The attempt at a solution
    If y is zero this function would be discontinuous or if the denominator became zero.
    As much detail as you could provide in your explanation would be greatly appreciated, thank you.
     
  2. jcsd
  3. Feb 16, 2016 #2

    RUber

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    Why would you have a discontinuous function at y = 0?
    First, you need to define the domain for the function. The denominator being zero would be inadmissible, values of x, y, z such that ##x^2 -y^2 -z^2 =0## would be outside the domain.
    Secondly, if you are only dealing with real numbers, you would also want to remove any negatives under the square roots.
    Other than that, rational expressions like this are generally continuous everywhere they are defined.
     
  4. Feb 16, 2016 #3
    You're right, the numerator could be zero but the denominator should not ever be zero so, would the domain then be {(x,y,z) ∈ ℝ3 : x2+y2+z2>0}?
     
  5. Feb 16, 2016 #4

    RUber

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    Double check your signs.
    And there is also a requirement for y to be *edit* non-negative (from the numerator).
    If you draw the region, it should look like a half-cone.
     
  6. Feb 18, 2016 #5

    HallsofIvy

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    This is not a "rational expression".
     
  7. Feb 18, 2016 #6

    RUber

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    Thank you, I misused the term. Is there a general term for functions of this type?

    What I meant to say was: f(x,y,z), as a ratio of continuous functions, g(y) and h(x,y,z) will be continuous anywhere in the intersection of the domains of g(y) and h(x,y,z) with the exception of those points where h(x,y,z) = 0.
     
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