Problem 41 - College Algebra with Corequisite Support

Question

For the following exercises, use a graph to help determine the domain of the functions. $$ f(x)=\sqrt{\frac{(x+1)(x-1)}{x}} $$

Step-by-Step Solution

Answer

The domain is \((-1, 0) \cup (0, 1) \cup (1, \infty)\).

Step 1: Identify the Function

The function given is \( f(x) = \sqrt{\frac{(x+1)(x-1)}{x}} \). We need to determine where this function is defined.

Step 2: Determine Radical Expression Validity

For the expression inside the square root to be valid, it must be non-negative. Thus, \( \frac{(x+1)(x-1)}{x} \geq 0 \). This inequality will help us find the domain by identifying where the expression inside the square root is greater than or equal to zero.

Step 3: Solve for Critical Points

The critical points occur where the expression is equal to zero or undefined. We solve for when \( (x+1)(x-1) = 0 \) or \( x = 0 \). These critical points are \( x = -1, x = 1, \) and \( x = 0 \).

Step 4: Test Intervals Between Critical Points

We must test the sign of \( \frac{(x+1)(x-1)}{x} \) between the intervals defined by our critical points: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 1) \), and \( (1, \infty) \). Choose test points like \( x = -2, -0.5, 0.5, \) and \( 2 \).

Step 5: Evaluate Interval Signs

- For \( x = -2: \frac{(-2+1)(-2-1)}{-2} < 0 \).- For \( x = -0.5: \frac{(-0.5+1)(-0.5-1)}{-0.5} > 0 \).- For \( x = 0.5: \frac{(0.5+1)(0.5-1)}{0.5} < 0 \).- For \( x = 2: \frac{(2+1)(2-1)}{2} > 0 \).

Step 6: Include Valid Intervals in Domain

Based on the sign evaluation, the function \( f(x) \) is non-negative, and thus valid for \( x \in (-1, 0) \cup (1, \infty) \). We omit \( x = 0 \) because the original expression is undefined at \( x = 0 \) (division by zero).

Step 7: Write the Domain

Combining all findings, the domain of \( f(x) = \sqrt{\frac{(x+1)(x-1)}{x}} \) is \((-1, 0) \cup (0, 1) \cup (1, \infty)\).

Key Concepts

Interval Notation

Interval notation is a convenient mathematical notation used to describe the domain or range of a function. It is a way of writing subsets of the real number line, allowing us to easily interpret which parts of the domain are included.

• Round brackets ( ) indicate intervals that do not include the endpoints, known as open intervals.
• Square brackets [ ] imply intervals that do include the endpoints, referred to as closed intervals.

For example, if we say "from -1 to 0" and "from 1 to infinity," not including -1 and 1, we write it as \((-1, 0)\cup (1, \infty)\).
This notation shows where a function can "live," providing a clear guide to its domain. In our case, it's helpful to know that the function is undefined at certain points due to division by zero or negative values under a square root.

Radical Expression

Radical expressions involve a root, such as a square root or cube root. In the case of functions with radical expressions, the expression inside the radical sign dictates if and where the function is defined.
To determine the domain of a radical expression:

• The expression inside the square root sign must be non-negative (\(\geq 0\)).
• Values leading to negative outputs in a square root are omitted from the domain as they lead to non-real numbers.

In our function, \(f(x) = \sqrt{\frac{(x+1)(x-1)}{x}}\), the terms inside the root, \(\frac{(x+1)(x-1)}{x}\), must be non-negative, meaning we only consider intervals where this condition is satisfied. Solving these inequalities ensures that no complex numbers pop up within the evaluated domain.

Functions With Fractions

Functions with fractions often lead to particular restrictions in determining their domains. Fractions require careful assessment because division by zero makes them undefined, which affects the domain directly.

• Avoid x-values that make the denominator zero, as they will cause the function to be undefined.
• Expressions need to meet conditions for both numerator and denominator to stay within the real number system.

For our specific function, \(f(x) = \sqrt{\frac{(x+1)(x-1)}{x}}\), it's necessary to ensure \(x eq 0\) because it would cause division by zero. This recognition helps filter the valid sections where the function can operate, affecting both the logics in our inequality checks and ultimately the domain.

Inequality Solutions

Solving inequalities is a key skill for determining domains, especially for functions with complex expressions. It involves determining where certain expressions are true (positive, in this case).
To solve inequalities:

• Establish critical points, where the expression is zero or undefined.
• Divide the number line into intervals based on these points and test values within each interval to determine where the inequality holds.

In our provided steps, testing intervals with selected values (e.g., \(x=-2, -0.5, 0.5, 2\)) helped identify which intervals satisfied the inequality \(\frac{(x+1)(x-1)}{x} \geq 0\). The use of test values and assessments make solving inequalities both straightforward and practical to find valid domains, influencing how intervals—such as the domain \((-1, 0)\cup (1, \infty)\)—are formed.