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Chapter 12: Problem 47

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\ln \left(x^{2}+y^{2}\right)$$

Short Answer

Expert verified
Answer: The function \(f(x, y)=\ln \left(x^{2}+y^{2}\right)\) is continuous everywhere in \(\mathbb{R}^{2}\) except at the origin, \((0, 0)\).

Step by step solution

01

Check the domain of the function

To ensure that \(f(x, y)=\ln \left(x^{2}+y^{2}\right)\) is continuous, we must first make sure it's defined. The natural logarithm function is defined for positive arguments, so we want \(x^{2}+y^{2}>0\). This is true for all points in \(\mathbb{R}^{2}\) except for the origin, \((0, 0)\).
02

Analyze the continuity of the function

The function \(f(x, y)\) is a composition of functions \(g(x, y) = x^2 + y^2\) and \(h(z) = \ln{z}\). As both \(g(x, y)\) and \(h(z)\) are continuous functions (with the exception of the origin for \(h(z)\)), their composition \(f(x, y) = h(g(x, y))\) will also be continuous wherever it is defined.
03

Conclude where the function is continuous

Since \(x^2+y^2 > 0\) for all points in \(\mathbb{R}^{2}\) except for the origin, the function \(f(x, y) = \ln \left(x^{2}+y^{2}\right)\) is continuous everywhere in \(\mathbb{R}^{2}\) except at \((0, 0)\). Therefore, the given function is continuous in \(\mathbb{R}^{2} - \{(0, 0)\}\).

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