Problem 49 - Calculus Early Transcendentals

Question

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$g(x, y)=\sqrt[3]{x^{2}+y^{2}-9}$$

Step-by-Step Solution

Answer

Question: For the given function \(g(x, y)=\sqrt[3]{x^{2}+y^{2}-9}\), state the region in \(\mathbb{R}^2\) at which the function is continuous. Answer: The given function \(g(x, y)=\sqrt[3]{x^{2}+y^{2}-9}\) is continuous at all points in \(\mathbb{R}^{2}\) for which \(x^2 + y^2 \ge 9\), i.e., the points on and outside the circle of radius 3 centered at the origin.

Step 1: Identify the Discontinuities

To determine at which points the function is not continuous, we need to find out where the cube root is not defined. This occurs when the expression inside the cube root is negative. So, we must identify the region in \(\mathbb{R}^2\) where \(x^2 + y^2 - 9 < 0\): $$x^2 + y^2 - 9 < 0$$

Step 2: Simplify Inequality

Now, rearranging the inequality, we have: $$x^2 + y^2 < 9$$ This represents all the points inside a circle with a radius of 3 centered at the origin (since 9 is the square of the radius).

Step 3: Determine Continuous Points

Since we have found that the function is not continuous for all points within a circle of radius 3 around the origin (\((x^2 + y^2 < 9\))), we will now identify the region of \(\mathbb{R}^2\) where the given function is continuous. This happens when the expression inside the cube root is non-negative: $$x^2 + y^2 - 9 \ge 0$$ Rearranging, we get \(x^2 + y^2 \ge 9\). This inequality represents the points on the circle and all the points outside the circle with a radius of 3 centered at the origin. So, the given function \(g(x, y)=\sqrt[3]{x^{2}+y^{2}-9}\) is continuous at all points in \(\mathbb{R}^{2}\) except for those inside the circle of radius 3 centered at the origin (i.e., for \(x^2 + y^2 \ge 9\)).