91Ó°ÊÓ

We can notice immediately that the function is quite simple, given by the exponential function: $$p(x, y) = e^{x - y}$$ This function is composed of basic algebraic operations (addition and subtraction) and the exponential function. We know that these basic operations and the exponential function are individually continuous over their respective domains.

Now, let's verify that each component function is continuous at any given point \((a, b)\) in \(\mathbb{R}^2\). First, let's look at the addition and subtraction operations: $$x - y$$ The operations of addition and subtraction are continuous over their domains, which in this case are the real numbers. Since we're considering all points in \(\mathbb{R}^2\), the domain of this operation is valid, and thus this part of the function is continuous everywhere. Next, let's check the continuity of the exponential function: $$e^z$$ where \(z = x - y\). The exponential function, \(e^z\), is also continuous over its valid domain, which in this case is again the real numbers. Since we're considering all points in \(\mathbb{R}^2\), the domain of the exponential function is also valid.

We have now verified that both the algebraic operation (addition and subtraction) and the exponential function are continuous on their respective domains. Since the function \(p(x, y) = e^{x - y}\) is composed entirely of these continuous component functions, it inherits their properties and must also be continuous over the entire domain of \(\mathbb{R}^2\).

We found that the function \(p(x, y) = e^{x - y}\) is continuous at all points \((x, y)\) in \(\mathbb{R}^2\).