Problem 16 - Calculus Early Transcendentals

Question

Find the domain of the following functions. $$f(x, y)=\frac{12}{y^{2}-x^{2}}.$$

Step-by-Step Solution

Answer

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Step 1: Set up the inequality to find the domain

As mentioned, we need to find the values of x and y that make the denominator nonzero. So we need to solve the inequality: $$y^2 - x^2 \neq 0$$ This inequality can also be written as: $$(y-x)(y+x) \neq 0$$

Step 2: Find the values where the inequality holds

In order for this inequality to hold, it must have that either \((y-x)\) or \((y+x)\) is nonzero. We will analyze these two cases separately to find the domain of the function. Case 1: \((y-x)\neq0\) This inequality holds when y is not equal to x. In other words, we must have \(y \neq x\). This will be the first condition for the domain. Case 2: \((y+x)\neq0\) This inequality holds when y is not equal to -x. In other words, we must have \(y \neq -x\). This will be the second condition for the domain.

Step 3: Combine the conditions for the domain

Combining the conditions from both cases, we obtain the domain of the function. The function \(f(x, y)\) is defined for all real values of \(x\) and \(y\) with the following conditions: 1. \(y \neq x\) 2. \(y \neq -x\) The domain of the function \(f(x, y) = \frac{12}{y^2 - x^2}\) is all values of \((x, y)\) such that \(y \neq x\) and \(y \neq -x\).

Key Concepts

Function Domain

In multivariable calculus, the domain of a function refers to the set of input values for which the function is defined. For the function \(f(x, y) = \frac{12}{y^2 - x^2}\), we are particularly interested in determining the pairs of \(x\) and \(y\) that allow the function to produce real values. To do this, we need to ensure the denominator \(y^2 - x^2\) is not zero, as division by zero is undefined in mathematics.

To find the domain, we set up an inequality:

• Firstly, ensure that \(y^2 - x^2 eq 0\).
• Factor this inequality as \((y-x)(y+x) eq 0\).

By breaking it into these factors, we can analyze where each separately is non-zero, thereby identifying specific conditions under which the function remains defined. By pursuing each factor, a clearer picture of the domain emerges.

Inequalities

Inequalities are a fundamental part of solving for function domains, especially in cases involving divisions, like our example \(f(x, y) = \frac{12}{y^2 - x^2}\). Inequalities allow us to determine where a function remains valid and help prevent undefined behavior, such as division by zero.

For the function in question, we had:

• \((y-x)(y+x) eq 0\). This converts our domain problem into finding where the product of these two expressions is not equal to zero.

By solving these inequalities:

• \((y-x) eq 0\) means \(y eq x\).
• \((y+x) eq 0\) means \(y eq -x\).

Each inequality gives us a restriction on the values \(y\) can take in relation to \(x\), ultimately combining these restrictions defines the domain.

Real Values

When exploring the domain of functions in multivariable calculus, ensuring the result is a real value is pivotal. A real value implies that the function outputs numbers we are familiar with in everyday mathematics, as opposed to complex values.

For the function \(f(x, y) = \frac{12}{y^2 - x^2}\), we focus on maintaining real outputs by preventing the denominator from equating to zero, which is otherwise undefined. Therefore, in this function:

• The conditions \(y eq x\) and \(y eq -x\) are set to ensure all calculations within the function remain real and do not breach undefined mathematical territory.

Through this approach, we create a valid range of inputs that guarantee the function produces real-valued results, hence making it suitable for further calculations and applications in real-world scenarios.