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Chapter 12: Problem 37

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\frac{2}{x\left(y^{2}+1\right)}$$

Short Answer

Expert verified
The function f(x, y) = 2 / (x(y^2 + 1)) is continuous for all points (x, y) in 鈩澛� except when x = 0. Thus, the set of continuous points is (x, y) 鈭� 鈩澛� : x 鈮� 0.

Step by step solution

01

1. Identify the domain of the function

The domain of the function \(f(x, y)\) is the set of all \((x, y)\) values for which the function is defined. The function will be undefined when the denominator is equal to zero: $$x(y^2+1)=0$$ This occurs when either \(x=0\) or \(y^2+1=0\). The term \(y^2+1\) is always positive, and adding 1 only makes it larger, so \(y^2+1\) can never be zero. Thus, the only restriction on the domain of the function is that \(x\) cannot be 0: $$(x, y) \in \mathbb{R}^2, x \neq 0$$
02

2. Determine if there are any discontinuities

A function is discontinuous at a point if there is a "hole" or "jump" in the graph. In our case, the function is undefined only when \(x=0\), which means there is a discontinuity along the line \(x=0\). For all other points in the domain, the function is continuous. We can also observe that the function is a rational function, and rational functions are continuous where they are defined.
03

3. Report the continuous points of the function

Having determined that the function is continuous for all points where it is defined, we conclude that the function is continuous for all points \((x, y) \in \mathbb{R}^2\) except when \(x=0\). Thus, the set of continuous points of the function is: $$(x, y) \in \mathbb{R}^2 : x \neq 0$$

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