Question

Find the domain of each function. $$ h(x)=\log _{3}\left(\frac{x}{x-1}\right) $$

Short answer

Domain: \( (-\infty, 0) \cup (1, +\infty) \).

Step-by-step solution

Identify domain restrictions for the logarithm

The argument of the logarithm function must be greater than zero. Therefore, we need to solve the inequality: \( \frac{x}{x-1} > 0 \).

Analyze the inequality \( \frac{x}{x-1} > 0 \)

To solve \( \frac{x}{x-1} > 0 \), determine where the function changes sign. This depends on the critical points: \( x = 0 \) and \( x = 1 \). These are the points where the numerator and denominator are zero, respectively.

Test intervals around critical points

Divide the number line into three intervals based on the critical points: 1. \( (-\infty, 0) \)2. \( (0, 1) \)3. \( (1, +\infty) \). Determine the sign of \( \frac{x}{x-1} \) at a test point within each interval.

Evaluate each interval

For \( x \in (-\infty, 0) \), choose \( x = -1 \): \( \frac{-1}{-1-1} = \frac{-1}{-2} > 0 \). The interval \( (-\infty, 0) \) satisfies the inequality.For \( x \in (0, 1) \), choose \( x = 0.5\): \( \frac{0.5}{0.5-1} = \frac{0.5}{-0.5} < 0 \). The interval \( (0, 1) \) does not satisfy the inequality. For \( x \in (1, +\infty) \), choose \( x = 2 \): \( \frac{2}{2-1} = \frac{2}{1} > 0 \). The interval \( (1, +\infty) \) satisfies the inequality.

Combine the valid intervals

The valid intervals from our analysis are \( (-\infty, 0) \) and \( (1, +\infty) \). Thus, the domain of the function is the union of these intervals: \( (-\infty, 0) \cup (1, +\infty) \).

Key concepts

Understanding Logarithmic Functions

A logarithmic function is a mathematical function involving an expression of a number as the power to which a fixed base must be raised to produce that number. For example, in \(log_{3}(x)\), 3 is the base, and the equation solves for the power to get x. Logarithmic functions have several important properties:

• The argument (the value inside the log function) must always be greater than zero.
• The base of the logarithm is a positive real number not equal to 1.

In our exercise with \(h(x)=log_{3}\left(\frac{x}{x-1}\right)\), the function \(\frac{x}{x-1}\) needs to be greater than zero to be defined. This is crucial because the log function is only defined for positive real numbers.

Solving Inequalities

Inequalities represent a range of values instead of a single number. In our problem, we face the inequality \(\frac{x}{x-1} > 0\). This tells us that the fraction \(\frac{x}{x-1}\) needs to be positive.
To solve this:

• Identify where the function equals zero or is undefined by setting the numerator and denominator to zero:
• 
  **Critical Point 1:** \(x = 0\).
  **Critical Point 2:** \(x = 1\).

These points divide the number line, helping us test which intervals satisfy the inequality.

• For \(-\infty < x < 0\): The function is positive.
• For \(0 < x < 1\): The function is negative.
• For \(x > 1\): The function is positive.

Identifying Critical Points

Critical points in an equation are values that make the numerator or the denominator zero. Finding these helps in determining the intervals to test for inequalities. Here, we have two critical points:

• \(x = 0\) (numerator is zero)
• \(x = 1\) (denominator is zero)

Using these critical points, we divided the number line into intervals for testing the sign of \(\frac{x}{x-1}\).
After evaluating the sign of the function in each interval, we found:

• The interval \((-\infty, 0)\) is valid.
• The interval \((0, 1)\) is not valid.
• The interval \((1, \infty)\) is valid.

Therefore, the domain of the function \(h(x)=log_{3}\left(\frac{x}{x-1}\right)\) is \(-\infty < x < 0\) and \(x > 1\).