Chapter 6: Problem 22
Find the domain of each logarithmic function analytically. You may wish to support your answer graphically. $$f(x)=\log \left(\frac{x+1}{x-5}\right)$$
Short Answer
Expert verified
The domain of \(f(x)\) is \((-1, 5) \cup (5, \infty)\).
Step by step solution
01
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.
02
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.
03
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 \(x - 5 < 0\)
04
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 \(x > 5\).
05
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 \(-1 < x < 5\).
06
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 \(x = 5\).
07
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 \((-1, 5) \cup (5, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
Understanding domain helps in identifying where a function is legitimate and can be used in calculations.
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.
- 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 \( -1 < x < 5 \).
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:
To solve such inequalities effectively:
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 \(-1 < x < 5 \).
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.
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 \( \frac{x+1}{x-5} \):
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.
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.
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.