Problem 38 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The function \(f(x, y)=\frac{x^{2}+y^{2}}{x\left(y^{2}-1\right)}\) is continuous for all points in \(\mathbb{R}^2\), except for the lines x = 0, y = 1, and y = -1.

Step 1: Determine the discontinuity points

To find the discontinuous points, let's set the denominator of the given function to zero. $$x\left(y^{2}-1\right) = 0$$ Now, we can find the possible values of x and y for which this equation holds true.

Step 2: Find the possible values of x and y

To find the values of x and y, we can consider two cases: when x = 0, and when y² - 1 = 0. Case 1: When x = 0 The function is discontinuous along the line x = 0, regardless of the value of y. Case 2: When y² - 1 = 0 $$y^2 = 1$$ $$y = ±\sqrt{1}$$ We have two values, y = 1, and y = -1. The function is discontinuous along the lines y = 1 and y = -1, regardless of the value of x.

Step 3: Identify the continuous points

Now we have found the functions discontinuous points, which are along the lines x = 0, y = 1, and y = -1. Since these are only lines, we can see that the function will be continuous for all other points in \(\mathbb{R}^{2}\), excluding these lines. In conclusion, the function \(f(x, y)=\frac{x^{2}+y^{2}}{x\left(y^{2}-1\right)}\) is continuous for all points in \(\mathbb{R}^2\), except for the lines x = 0, y = 1, and y = -1.