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Chapter 12: Problem 48

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\sqrt{4-x^{2}-y^{2}}$$

Short Answer

Expert verified
Answer: The function \(f(x, y) = \sqrt{4-x^2-y^2}\) is continuous at all points \((x, y)\) in \(\mathbb{R}^2\) that satisfy \(x^2+y^2 \le 4\). This includes all points inside or on the boundary of the closed circle with center \((0, 0)\) and radius 2.

Step by step solution

01

Find the domain of the function

The domain of a function is the set of all possible input values (in this case, \((x, y)\)) for which the function is defined. Since the function involves a square root, we know that the expression under the square root must be non-negative. Therefore, we have: $$4-x^2-y^2 \ge 0$$ Rearranging, we get: $$x^2+y^2 \le 4$$ This inequality represents a closed circle in the plane with center \((0, 0)\) and radius 2. So the domain of the function is all points \((x, y)\) inside or on the boundary of this circle.
02

Analyze the continuity within the domain

We know that the square root is continuous for non-negative input values. Since we have already restricted the domain to only include points \((x, y)\) where the input to the square root, \(4-x^2-y^2\), is non-negative, our function is continuous over this domain.
03

State the answer

The function \(f(x, y) = \sqrt{4-x^2-y^2}\) is continuous at all points \((x, y)\) in \(\mathbb{R}^2\) that satisfy \(x^2+y^2 \le 4\). This includes all points inside or on the boundary of the closed circle with center \((0, 0)\) and radius 2.

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