Problem 7 - Calculus with Concepts in Calculus

Question

Find the domain of the function. \(f(x, y)=\frac{1}{x+y}\)

Step-by-Step Solution

Answer

The domain of \( f(x, y) = \frac{1}{x+y} \) is all \( (x, y) \) except where \( x + y = 0 \).

Step 1: Understanding the Function

We start with the function given by \( f(x, y) = \frac{1}{x+y} \). This is a rational function, where the numerator is constant and the denominator is \( x + y \).

Step 2: Identifying Invalid Conditions

The domain of a function includes all possible input values (\(x, y\) pairs) that do not make the function undefined. For a rational function, the denominator must not be zero.

Step 3: Set Denominator Not Equal to Zero

To find where \( f(x, y) \) is undefined, set the denominator equal to zero and solve for \( x + y \): \( x + y = 0 \). Solving this equation gives \( x = -y \).

Step 4: Domain Description

The domain consists of all \( (x, y) \) pairs except those that satisfy \( x + y = 0 \). Therefore, the domain is all numbers \( \mathbb{R}^2 \) except for the line where \( x + y = 0 \).

Key Concepts

Understanding Rational Functions

A rational function is a type of function that is expressed as a fraction where both the numerator and the denominator are polynomials. For example, in the given function \( f(x, y) = \frac{1}{x+y} \), the numerator is a constant, while the denominator is a linear expression \( x + y \). The nature of rational functions implies that they can behave differently from polynomials because they can involve division.

• Rational functions can have restrictions in their domains due to the division by a variable expression.
• These functions can present vertical asymptotes or holes in their graphs, where they are not defined.

Typically, when working with rational functions, we must be mindful of these restrictions to determine where the function will be defined and where it will not be.

Detecting Undefined Conditions in Functions

For any function, particularly rational functions, the domain refers to the set of all input values that keep the function well-defined. Undefined conditions occur whenever there is a division by zero. This happens because division by zero is undefined in mathematics.To find undefined conditions in \( f(x, y) = \frac{1}{x+y} \), we need to:

• Identify when the denominator, here \( x + y \), equals zero.
• Exclude these values from the domain.

Detecting these points of undefined conditions helps in mapping the exact domain of the function. It ensures the function remains valid and does not include points where it cannot operate successfully.

Denominator Not Equal to Zero: Ensuring a Defined Function

In any rational function, such as \( f(x, y) = \frac{1}{x+y} \), it is crucial to ensure the denominator does not equal zero. By setting the denominator \( x+y \) not equal to zero, we can find the restricted values that cannot be part of the domain.

• Set \( x+y eq 0 \).
• Solve for the exclusion condition, specifically \( x + y = 0 \), which can be rewritten as \( x = -y \).

Through this process, any pair \((x, y)\) satisfying \( x = -y \) will make the denominator zero and the function undefined. Thus, these values must be excluded from the domain. The domain of the function is all real \((x, y)\) pairs except those where \( x + y = 0 \), ensuring the function remains continuous and well-defined.