Problem 37 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The function \(f(x,y) = \frac{2}{x(y^2 + 1)}\) is continuous for all points in \(\mathbb{R}^2\) except those on the vertical line \(x = 0\).

Step 1: Identify the Function and Its Domain

The function we are working with is $$ f(x, y) = \frac{2}{x(y^2 + 1)} $$ Let's first determine the domain of the function by finding where the denominator is not zero:

Step 2: Find the Denominator

The denominator of the function is given by \(x(y^2+1)\). We'll find where the denominator is zero, so we can understand where the function is not defined. Solve the equation \(x(y^2+1) = 0\).

Step 3: Factor and Solve for x and y

To find the zeroes of the denominator, we can factor the expression: $$ x(y^2 + 1) = 0 $$ This expression will be equal to zero if either \(x = 0\), or if \(y^2 + 1 = 0\). However, the latter equation has no real solution for \(y\), as \(y^2\) is always non-negative, and thus \(y^2 + 1 > 0\) for all \(y \in \mathbb{R}\).

Step 4: Identify Points of Discontinuity

The only case where the denominator of the function is zero is when \(x = 0\). Therefore, the function is discontinuous along the line \(x = 0\) in \(\mathbb{R}^2\).

Step 5: Determine Points of Continuity

Since the function is discontinuous along the line \(x = 0\), it is continuous everywhere else in \(\mathbb{R}^2\). The points of continuity of the function are given by the set \(\{(x,y) \in \mathbb{R}^2 : x \neq 0\}\). Overall, the function \(f(x,y) = \frac{2}{x(y^2 + 1)}\) is continuous for all points in \(\mathbb{R}^2\) except those on the vertical line \(x = 0\).