Problem 46 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The function \(p(x, y)=e^{x-y}\) is continuous at every point \((x, y)\) in its domain, which is \(\mathbb{R}^{2}\).

Step 1: Break down the function

Given the function \(p(x, y) = e^{x - y}\), we can break it down into its constituent operations: addition, subtraction and exponentiation. The function can be rewritten as $$p(x, y) = e^{f(x, y)}$$ where \(f(x, y) = x - y\).

Step 2: Determine the continuity of the constituent functions

We have the following elementary functions and their corresponding continuity properties: 1. The addition and subtraction functions, \(f(x, y) = x \pm y\), are continuous in \(\mathbb{R}^{2}\). 2. The exponential function, \(g(x) = e^x\), is continuous in \(\mathbb{R}\).

Step 3: Use rules of continuity for composite functions

Now that we know our constituent functions are continuous, we must determine whether the composite function, \(p(x, y)\), is continuous. If a function \(g \circ f(x, y)\) is the composition of two functions \(f(x, y)\) and \(g(u)\), the following must hold for it to be continuous: - \(f\) is continuous at each point \((x, y)\) in its domain. - \(g\) is continuous at each point \(u = f(x, y)\) in its domain. If these conditions are satisfied, then the composite function \(g \circ f(x, y)\) will also be continuous at each point \((x, y)\) in its domain.

Step 4: Analyze continuity based on rules

We know that both constituent functions are continuous individually, so let's check the conditions mentioned in Step 3: 1. The function \(f(x, y) = x - y\) is continuous in \(\mathbb{R}^{2}\). 2. The function \(g(u) = e^u\) is continuous in \(\mathbb{R}\), and since the range of \(f(x, y)\), which is \(\mathbb{R}\), is contained in the domain of \(g(u)\), we can conclude that \(g(f(x, y))\) is continuous. Therefore, the composite function \(p(x, y) = e^{x - y}\) is continuous at every point \((x, y)\) in its domain, which is \(\mathbb{R}^{2}\).