Problem 22 - A Graphical Approach to Precalculus with Limits

Question

Find the domain of each logarithmic function analytically. You may wish to support your answer graphically. $$f(x)=\log \left(\frac{x+1}{x-5}\right)$$

Step-by-Step Solution

Answer

The domain of \(f(x)\) is \((-1, 5) \cup (5, \infty)\).

Step 1: Understand the Logarithmic Function

The function is given as \(f(x) = \log \left(\frac{x+1}{x-5}\right)\). For this function to be defined, the expression inside the logarithm, \(\frac{x+1}{x-5}\), must be positive because the logarithm of a non-positive number is undefined in the real number system.

Step 2: Set the Expression Greater Than Zero

To find where \(f(x)\) is defined, set the expression \(\frac{x+1}{x-5} > 0\). This inequality will help us find the values of \(x\) for which the function is defined.

Step 3: Solve the Inequality

The inequality \(\frac{x+1}{x-5} > 0\) implies that both numerator \((x+1)\) and denominator \((x-5)\) have the same sign. Solve this by considering two cases:1. \(x + 1 > 0\) and \(x - 5 > 0\)2. \(x + 1 < 0\) and \(x - 5 < 0\)

Step 4: Analyze Case 1: Both Numerator and Denominator Positive

For \(x + 1 > 0\), we have \(x > -1\). For \(x - 5 > 0\), we have \(x > 5\). The overlap of these conditions is \(x > 5\).

Step 5: Analyze Case 2: Both Numerator and Denominator Negative

For \(x + 1 < 0\), we have \(x < -1\). For \(x - 5 < 0\), we have \(x < 5\). The overlap of these conditions is \(-1 < x < 5\).

Step 6: Combine the Results

Combine both cases’ solutions, the valid intervals are \(-1 < x < 5\) and \(x > 5\). Thus, the domain is the union of these intervals: \(x > -1\), excluding \(x = 5\).

Step 7: Exclude Undefined Point

The function is undefined at \(x = 5\) because it makes the denominator zero, which is not allowed in a fraction. Therefore, the final domain of \(f(x)\) is \((-1, 5) \cup (5, \infty)\).

Key Concepts

Domain of a Function

In mathematics, the domain of a function refers to the complete set of possible values of the independent variable, often denoted as \( x \). For the function \( f(x) = \log \left(\frac{x+1}{x-5}\right) \), the concept of domain is crucial because logarithms are only defined for positive numbers. This means that the expression \( \frac{x+1}{x-5} \) must be greater than zero.

To find the domain analytically:

• Identify when the numerator, \( x+1 \), and the denominator, \( x-5 \), are greater than zero simultaneously.
• Check when both are negative at the same time.

This problem requires two cases:

• Case 1: Both \( x+1 \) and \( x-5 \) are positive. Solve \( x+1 > 0 \) and \( x-5 > 0 \), leading to \( x > 5 \).
• Case 2: Both \( x+1 \) and \( x-5 \) are negative. Solve \( x+1 < 0 \) and \( x-5 < 0 \), leading to \( -1 < x < 5 \).

The final domain combines these intervals, except for \( x = 5 \), which would make the denominator zero. Therefore, the domain is \((-1, 5) \cup (5, \infty) \).

Understanding domain helps in identifying where a function is legitimate and can be used in calculations.

Solving Inequalities

Solving inequalities is a vital skill when working with logarithmic functions since it helps determine the range of \( x \) values that make the function defined. In the function \( f(x) = \log \left(\frac{x+1}{x-5}\right) \), the inequality \( \frac{x+1}{x-5} > 0 \) is crucial to identify where the logarithm is valid.

The inequality \( \frac{x+1}{x-5} > 0 \) implies the numerator and the denominator must have the same sign:

• If \( x+1 > 0 \) and \( x-5 > 0 \), then \( x > 5 \).
• If \( x+1 < 0 \) and \( x-5 < 0 \), then \(-1 < x < 5 \).

These inequalities result from analyzing when the quotient of two expressions remains positive.

To solve such inequalities effectively:

• Break the problem into cases to determine when the signs of the expressions align.
• Consider critical points where the numerator or denominator becomes zero, as these are key boundaries.

Practicing this method enhances problem-solving and critical thinking skills, especially in mathematics.

Graphical Analysis

Analyzing functions graphically provides an intuitive understanding of where a function is defined and helps validate analytical solutions. For \( f(x) = \log \left(\frac{x+1}{x-5}\right) \), a graphical representation can illustrate why the domain excludes certain values, such as when \( x = 5 \).

By plotting \( \frac{x+1}{x-5} \):

• You can observe the behavior near \( x = 5 \), where the function becomes undefined as the graph shoots towards infinity or negative infinity (asymptotic behavior).
• The sections \(-1 < x < 5\) and \(x > 5\) are where the graph remains positive, fitting into the identified domain.

Visualization helps bridge the gap between rote calculation and conceptual understanding. It answers "why" the solutions are so instead of just "how" to compute them.

Tools like graphing calculators or software can be invaluable for students to cross-check and deepen their understanding of mathematical concepts. Graphical analysis is thus a powerful complement to algebraic solutions in learning mathematics.