Problem 102 - A Graphical Approach to Precalculus with Limits

Question

Solve the equation \(f(x)=0\) analytically and then use a graph of \(y=f(x)\) to solve the inequalities \(f(x)<0\) and \(f(x) \geq 0\). $$f(x)=\ln (x-1)-\ln (x+1)$$

Step-by-Step Solution

Answer

There is no solution to \(f(x)=0\); \(f(x)<0\) for \(x>1\).

Step 1: Solve the equation analytically

We start by solving the equation \(f(x) = 0\), which is given as \(\ln(x-1) - \ln(x+1) = 0\). Using the property of logarithms, \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\), we can rewrite the equation as \(\ln\left(\frac{x-1}{x+1}\right) = 0\). This implies \(\frac{x-1}{x+1} = 1\). Solve for \(x\) by cross-multiplying to get \(x - 1 = x + 1\), which simplifies to \(-1 = 1\). This is not possible, so there is no solution for \(f(x) = 0\).

Step 2: Investigate the domain of f(x)

The function \(f(x) = \ln(x-1) - \ln(x+1)\) involves logarithms, which are only defined for positive arguments. Thus, \(x-1 > 0\) and \(x+1 > 0\), meaning \(x > 1\). There is no further restriction from \(x+1\), which suggests that the domain of \(f(x)\) is \(x > 1\).

Step 3: Use a graph to solve \\(f(x)<0\\) and \\(f(x)\geq0\\)

Since \(f(x) = \ln\left(\frac{x-1}{x+1}\right)\), we are interested in when this expression is less than zero (\(f(x) < 0\)) and when it is non-negative (\(f(x) \geq 0\)). Calculate \(\frac{x-1}{x+1}<1\): this is always true for \(x > 1\). Thus, \(f(x) < 0\) for all \(x > 1\), and \(f(x) \geq 0\) is never true in this domain.

Key Concepts

Inequalities

Inequalities are mathematical statements indicating that two expressions are not equal in value. They are used to show that one expression is either greater than or less than another. When solving inequalities, you often look for a range of values that satisfy the condition. In the context of the function \( f(x) = \ln(x-1) - \ln(x+1) \), we're tasked with finding when \( f(x) < 0 \) and \( f(x) \geq 0 \).

1. **Understanding the Signs**:
- When \( \ln\left(\frac{x-1}{x+1}\right) < 0 \), it implies that \( \frac{x-1}{x+1} \) is less than 1.
- When \( \ln\left(\frac{x-1}{x+1}\right) \geq 0 \), it would imply that \( \frac{x-1}{x+1} \) is greater than or equal to 1, which, based on our solution, is not possible for any \( x > 1 \).

2. **Graphical Insights**: Inequalities are often easier to understand when visualized on a graph. By plotting \( y = f(x) \), you'll notice how the curve behaves relative to the x-axis. If \( f(x) < 0 \), the graph is below the x-axis, and if \( f(x) \geq 0 \), it's above or on the x-axis. However, in our scenario, it's always below for \( x > 1 \).

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions and are critical in many areas of mathematics and science. They help us to understand exponential growth, decay, and many other kinds of rate changes.

1. **Basic Concepts**:
- A logarithm \( \ln(x) \) asks the question: "To what power must the base (e) be raised, to produce the number x?" Here, \( \ln \) refers specifically to the natural logarithm where the base is Euler's number, \( e \).
- Logarithms are only defined for positive numbers, meaning that their arguments (what you're taking the log of) must be greater than zero.

2. **Properties of Logarithms**: Logarithms have several properties that make them very useful, especially in solving equations. For example:
- \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \), which is used in simplifying expressions.
- This property was crucial in simplifying \( \ln(x-1) - \ln(x+1) \) into \( \ln\left(\frac{x-1}{x+1}\right) \).

Function Domain

Understanding the domain of a function tells us the set of all possible inputs (x-values) for the function. For logarithmic functions, the domain is determined based on where the expression inside the logarithm is greater than zero.

1. **Domain Insights for Logarithms**:
- For the function \( f(x) = \ln(x-1) - \ln(x+1) \), we need both \( x-1 > 0 \) and \( x+1 > 0 \) for the logarithms to be defined.
- From \( x-1 > 0 \), we find that \( x > 1 \). The condition \( x+1 > 0 \) is naturally satisfied when \( x > 1 \), making \( x > 1 \) the entire domain of the function.

2. **Visual Representation**: Graphically, the domain of \( f(x) \) is the x-values where the graph is defined and does not include any breaks or undefined regions. Seeing the plotted graph will show \( f(x) \) only existing right of \( x = 1 \), since any value less than or equal to 1 would make at least one of the logarithms undefined.