Problem 2 - Calculus with Concepts in Calculus

Question

Find the domain of the function. \(f(x, y)=\sqrt{x+y}\)

Step-by-Step Solution

Answer

The domain is \( \{ (x, y) \mid x+y \geq 0 \} \).

Step 1: Understanding the Function

The function given is a two-variable function: \( f(x, y) = \sqrt{x+y} \). We need to find out which values of \( x \) and \( y \) make the expression under the square root a non-negative number. This is because the square root is only defined for non-negative values.

Step 2: Identifying the Condition for the Square Root

The function \( \sqrt{u} \) is defined for \( u \geq 0 \). In our case, \( u = x + y \). Therefore, the expression \( x + y \) must be greater than or equal to zero for the square root to be defined. Hence, the condition is \( x+y \geq 0 \).

Step 3: Finding the Domain

The domain of \( f(x, y) \) is the set of all \( (x, y) \) pairs that satisfy the condition \( x+y \geq 0 \). This can be represented in the coordinate plane as the region that lies on or above the line \( x + y = 0 \), which is equivalent to \( y = -x \). Therefore, the domain includes all pairs \( (x, y) \) where the sum of \( x \) and \( y \) is non-negative.

Step 4: Expressing the Domain in Set Notation

In set notation, the domain is expressed as: \( \{ (x, y) \mid x+y \geq 0 \} \). This notation succinctly captures all the pairs \( (x, y) \) such that their sum is non-negative.

Key Concepts

Understanding Two-Variable Functions

A two-variable function is a mathematical expression involving two independent variables, often denoted as \( x \) and \( y \). Unlike single-variable functions that map an input to one output, two-variable functions consider pairs of inputs. Think of it as handing the function two numbers, and it returns a value based on these. In this exercise, we're looking at \( f(x, y) = \sqrt{x+y} \), which uses both \( x \) and \( y \) together inside a square root. This means the function's behavior relies on both variables simultaneously. To understand and find the domain, which is all the permissible inputs, we consider all pairs \( (x, y) \) that make \( f(x, y) \) a real number. This requires understanding under what conditions this square root exists and is well-defined.

Conditions for Square Roots

Square roots are only defined for non-negative numbers. This is because you can't take the square root of a negative number without moving into complex numbers, which we generally want to avoid in standard real-number analysis. In the function \( f(x, y) = \sqrt{x+y} \), what's under the square root is the sum of \( x \) and \( y \). To ensure the square root is valid, \( x + y \) must be 0 or positive, hence \( x + y \geq 0 \). This forms the basic condition we need to satisfy for every point \( (x, y) \) that our function \( f \) can accept. Simply put, any values \( x \) and \( y \) that don't meet this condition aren't in the domain of the function.

Using Set Notation to Express Domains

Set notation is a powerful and concise way to describe domains of functions. For our function \( f(x, y) = \sqrt{x+y} \), we need to express all \( (x, y) \) pairs that meet the condition \( x+y \geq 0 \). In set notation, this is represented as \( \{ (x, y) \mid x+y \geq 0 \} \). The curly braces \( \{ \} \) indicate a set, and the vertical bar \( \mid \) stands for 'such that.' Ultimately, what this tells us is we're including every combination of \( x \) and \( y \) where their sum is not negative. Providing the domain in this format allows us to quickly understand which inputs the function can handle, and it captures the information cleanly and efficiently.