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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 at 21:53

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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.

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So, the domain is x and y values, which we just so happen to represent by ordered-pairs, and and not the ordered-pairs themselves? – Mack Mar 4 at 22:02
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@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 at 22:24
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@EliMackenzie: no, the domain is the set of values that can be input to the function. The range is the set of output values, which is the set of values the function can take on. – Ross Millikan Mar 5 at 0:53
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@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 at 1:12
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@EliMackenzie: that is correct, because then the number under the square root will not be negative. – Ross Millikan Mar 5 at 1:29
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