# Describing The Graph Of A Function In Two Variables

I am currently reading about an example problem, of which is mentioned in the title of this thread.

The function is $f(x,y)= \sqrt{16-4x^2-y^2}$. The one part of the discussion I don't quite understand is why the range of this function is $0 \le z \le 4$

Also, as a side note, is the domain of a function of two variables, $z= f(x,y)$, the ordered-pairs (x,y), and the range is real numbers? That is, the function maps an ordered-pair to a real number, the real number being the z-value?

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It sometimes helps to do a plot to get your hands around it visually and then work the math. Regards – Amzoti Mar 4 '13 at 21:53

To see that the range of $z=f(x,y)$ is $0 \le z \le 4$, first note that $f(0,0)=4$ and $f(2,0)=0$ so the range is at least that. Since the squares are at least zero, the value under the square root sign cannot be greater than $16$, so the maximum function value cannot be more than $4$. The square root is defined to be greater than or equal to zero, so it cannot be less. This shows the range is as claimed.
Your side not is correct: the domain is the set of allowable inputs to the function, which here are ordered pairs $(x,y)$. The range is the range of outputs of the function, in this case a set of real numbers.
@EliMackenzie: No, it is ordered pairs. In your example, we have $-2 \le x \le 2$ and $-4 \le y \le 4$, but $(1,4)$ is not in the domain. The domain is in fact an ellipse centered on the origin. – Ross Millikan Mar 4 '13 at 22:24
@EliMackenzie: the reason I said the domain is ordered pairs is that $x$ has a certain allowable range (shown in my first comment) and $y$ has a certain allowable range, but you can't pick any $x$ in that range and pair it with any $y$. This is why the domain is a set of ordered pairs. Only pairs with $4x^2+y^2 \le 16$ are in the domain. – Ross Millikan Mar 5 '13 at 1:12