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Domain of a function

  1. Mar 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the domain of the function: f(x,y) = xy*√(x2+y)

    2. Relevant equations
    The domain is the set of all input values for which the function is defined.

    3. The attempt at a solution
    D = {(x,y): xy>0}

    I plugged the function into Wolfram and it gave the result:

    D = {(x,y): x2+y ≥ 0 }

    I can see how that solution is correct, but isn't mine as well? I'm saying that the product of x and y is greater than 0, which implies neither x nor y = 0, hence the > symbol.

    The wolfram solution seems a little less explicit to me.
     
  2. jcsd
  3. Mar 22, 2016 #2

    BvU

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    Your D would be the first and the third quadrant. What is f(-1,-3) according to you ?
    Sketch the two answers in the xy plane and see that you include areas that should be excludede and vice versa.
     
  4. Mar 22, 2016 #3
    f(-1,-3) is in the second quadrant of the domain, and according to my domain it maps to 3.

    I've stated that D=(x,y): xy>0.that whatever the x or y values take on in the domain, the output will always be greater than 0.

    This is incorrect though. The product of the equation states √(x2+y). The point (-3,1) could not be mapped in the reals because √(32+1) = √-8.

    √(x2+y) has to be greater than or equal to 0, therefore x2+y ≥ 0
     
  5. Mar 22, 2016 #4

    BvU

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    Correct. For a point x,y to be in the domain, f(x,y) has to exist. It doesn't have to be > 0. ( [edit] quote added to indicate what precisely is correct. )

    Oops ! I mentioned the point (1, -3) for which x2 + y = -2 < 0 so (-1,3) is not in D.

    The point (-3,1), however, happily resides in D: f(-3,1) = - 3 √8 !!

    -----------------------------------------------------------------------------

    [edit] : added:

    No. f(-1,-3) does not exist.

    You seem to think f(x,y) = x,y but it is not.

    Furthermore, you seem to mix up domain and range. Check your notes/textbook.
     
    Last edited: Mar 22, 2016
  6. Mar 22, 2016 #5

    SammyS

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    Your solution is not correct.

    If y=0, then x can be any number.

    If x=0, then y≥0 .

    The important issue here is that ##\ \sqrt{x^2+y\,}\ ## gives a real number result so long as ##\ x^2+y\ge0\ ##.
     
  7. Mar 22, 2016 #6

    Ray Vickson

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    Draw the curve ##C: \:y = -x^2## in the ##(x,y)## plane. For a point ##p =(x,y)## to be in the domain of ##f##, where must ##p## be located in relation to the curve ##C##?
     
  8. Mar 22, 2016 #7
    I meant to say that the a point (x,y) is in a quadrant on the xy-plane and a function will either map or not map this point. (i.e. it exists or does not exist.)
     
    Last edited by a moderator: Mar 22, 2016
  9. Mar 22, 2016 #8

    Mark44

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    That's pretty vague, to say "a point (x,y) is in a quadrant on the xy-plane". Also, a point in the x-y plane exists whether or not f maps it to some value.

    There's a fair amount of confusion in this thread. The domain is not, as has been mentioned, the first and third quadrants. The domain is ##\{(x, y) | x^2 + y \ge 0\}##. It would be helpful to sketch a graph of the domain, which is a geometric figure you should already know, together with all the interior points of this figure.

    The important thing if for the quantity inside the radical to be >= 0. All you need to say is that ##x^2 + y \ge 0##.
     
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