Problem 33 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Provide a brief explanation. Answer: Yes, the function \(f(x, y) = x^2 + 2xy - y^3\) is continuous at all points in \(\mathbb{R}^2\). This is because it is a polynomial function in two variables, and polynomial functions are continuous everywhere. Since it is defined for all points in \(\mathbb{R}^2\), the limit exists for all points, and the limit equals the function value, it satisfies the conditions for continuity.

Step 1: Function definition

Given the function: $$f(x, y) = x^2 + 2xy - y^3$$ This is a polynomial function in two variables, \(x\) and \(y\). Since both \(x^2, 2xy,\) and \(y^3\) are defined for all \((x, y) \in \mathbb{R}^2\), \(f(x, y)\) is defined for all points in \(\mathbb{R}^2\).

Step 2: Existence of limit

To show that the limit exists for all points \((x_0, y_0) \in \mathbb{R}^2\), we need to prove that: $$\lim_{(x,y) \to (x_0, y_0)} f(x, y) = L$$ for some value of \(L\). The function is given by the polynomial: $$f(x, y) = x^2 + 2xy - y^3$$ Since \(x^2\), \(2xy\), and \(y^3\) are continuous everywhere individually, their sum is also continuous everywhere. Therefore, the limit \(\lim_{(x,y) \to (x_0, y_0)} f(x, y)\) exists for all points in \(\mathbb{R}^2\).

Step 3: Evaluating limit and comparing to function value

Now we need to evaluate the limit as \((x, y) \to (x_0, y_0)\) and compare it to \(f(x_0, y_0)\). The limit for the given function is as follows: $$\lim_{(x,y) \to (x_0, y_0)} f(x, y) = \lim_{(x,y) \to (x_0, y_0)} (x^2 + 2xy - y^3)$$ Since the function is continuous everywhere, it follows that: $$\lim_{(x,y) \to (x_0, y_0)} (x^2 +2xy - y^3) = (x_0^2 + 2x_0y_0 - y_0^3) = f(x_0, y_0)$$ Since all three conditions for continuity are satisfied (function defined, limit exists, and limit equals function value), the given function \(f(x, y)=x^2+2xy-y^3\) is continuous at all points in \(\mathbb{R}^2\).