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M.Qayyum
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Here is my question, that what is the domain of f, while
f(x,y)=1/(x-y2)
f(x,y)=1/(x-y2)
What Values of (x, y) Make f(x, y) = 1/(x - y^2) Undefined?
- Thread starter M.Qayyum
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SUMMARY
The function f(x, y) = 1/(x - y^2) is undefined when the denominator equals zero. This occurs specifically when x = y^2. Therefore, the values of (x, y) that make f(x, y) undefined are all pairs where x is equal to the square of y. This relationship defines the domain restrictions for the function.
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