The natural domain of the function f(x,y) = 1/sqrt(x^2-y) is defined as D = {(x,y) | x^2 - y > 0}, indicating that y must be less than x^2. The range of the function is R = {(x,Y) | f > 0}, which specifies that the output values of f are positive real numbers. It is crucial to note that the range should consist of real numbers rather than ordered pairs. Misinterpretations regarding the domain and range were clarified during the discussion.
PREREQUISITESMathematics students, educators, and anyone involved in teaching or learning about multivariable functions and their properties.
No, it's not. That would be [itex]x^2- y\ne 0[/itex]. You are requiring that y be less than x2.Derill03 said:define the natural domain and range for f(x,y) = 1/sqrt(x^2-y)
I get D= {(x,y)|x^2-y>0} same as saying y can not equal x^2
And, as Dick said, the range of f is a set of numbers, not of pairs of numbers.R= {(x,Y)|f>0}
can someone tell me if I've approached this correctly?