Problem 61 - Algebra and Trigonometry

Question

\(59-64\) Find the domain of the function. $$ g(x)=\log _{3}\left(x^{2}-1\right) $$

Step-by-Step Solution

Answer

The domain of \(g(x)\) is \((-\infty, -1) \cup (1, \infty)\).

Step 1: Identify the Condition for Logarithmic Functions

The function involves a logarithm of the expression \(x^2 - 1\). A logarithmic function is defined only when the argument (the part inside the logarithm) is greater than zero. Thus, the condition for the domain is: \(x^2 - 1 > 0\).

Step 2: Solve the Inequality

To find \(x\) values that satisfy \(x^2 - 1 > 0\), solve the inequality: \(x^2 - 1 = 0\) to find the values where the expression equals zero. This simplifies to \(x = \pm 1\). Now, consider the inequality in interval form: the expression \(x^2 - 1\) is greater than zero for \(x < -1\) and \(x > 1\).

Step 3: Express the Domain

Based on the solution of the inequality, the domain of \(g(x)\) in interval notation is \((-\infty, -1) \cup (1, \infty)\). This represents all \(x\) values where the logarithmic function is defined.

Key Concepts

Logarithmic Functions

Logarithmic functions are mathematical functions that help us solve problems involving exponents. The function \( g(x) = \log_{3}(x^2 - 1) \) is an example of a logarithmic function. Here, \( \log_{3} \) indicates a logarithm with a base of 3. The term inside, or the **argument**, of the logarithm is \( x^2 - 1 \).

For any logarithmic function, its argument must be positive. This means \( x^2 - 1 \) should be greater than 0 for the function to be valid. If the inside of the logarithm isn't positive, the function becomes undefined. This is key when considering the domain of a function, as we'll only include x-values in our domain where the argument stays positive.

Inequalities

Inequalities are mathematical expressions involving symbols like greater than \( (>) \), less than \( (<) \), or equals \( (=) \). Solving inequalities helps us find out which values of \( x \) are permissible.

In our function \( g(x) = \log_{3}(x^2 - 1) \), the inequality we need to focus on is \( x^2 - 1 > 0 \).

• First, equate \( x^2 - 1 \) to zero: \( x^2 - 1 = 0 \).
• Solve this equation to find the critical points \( x = \pm 1 \).

These critical points are where the expression shifts from positive to negative or vice versa.

Evaluating the inequality, we find \( x^2 - 1 > 0 \) holds for \( x < -1 \) and \( x > 1 \). This solution helps us define the domain of the function later.

Interval Notation

Interval notation is a method of writing sets of numbers, particularly to describe the domain of functions. In the context of our logarithmic function \( g(x) = \log_{3}(x^2 - 1) \), interval notation helps us express all possible values of \( x \) based on the conditions we've derived from the inequality.

After solving \( x^2 - 1 > 0 \), we find that values of \( x \) satisfying this inequality are in the two intervals \( (\infty, -1) \) and \( (1, \infty) \).

• \( (-\infty, -1) \) covers all values less than \(-1\).
• \( (1, \infty) \) covers all values greater than \(1\).

The use of round brackets indicates that the endpoints \(-1\) and \(1\) are not included, as the inequality is strictly "greater than" and not "greater than or equal to." Therefore, the domain in interval notation is denoted as \( (-\infty, -1) \cup (1, \infty) \), clearly showing which values make the logarithmic function valid.