Problem 265 - Calculus Volume 1

Question

For the following exercises, sketch the graph of the logarithmic function. Determine the domain, range, and vertical asymptote. $$ f(x)=\ln (x-1) $$

Step-by-Step Solution

Answer

Domain: \( (1, \infty) \), Range: \( (-\infty, \infty) \), Vertical Asymptote: \( x=1 \).

Step 1: Identify the Domain

The domain of the function involves finding the set of all possible input values (x-values) for which the function is defined. For \( f(x) = \ln(x-1) \), the argument of the logarithm (i.e., \( x-1 \)) must be positive. Thus, we set the inequality \( x-1 > 0 \), which simplifies to \( x > 1 \). So the domain is \( (1, \infty) \).

Step 2: Determine the Range

The range of a logarithmic function is all real numbers. The function \( f(x) = \ln(x-1) \) can take any real value from negative infinity to positive infinity, as the values of \( x-1 \) range from very small positive numbers to very large positive numbers.

Step 3: Find the Vertical Asymptote

A vertical asymptote occurs where the function is undefined. Since \( \ln(x-1) \) is undefined at \( x=1 \), there is a vertical asymptote at \( x=1 \). The graph of the function approaches but never touches or crosses this vertical line.

Step 4: Sketch the Graph

To sketch the graph, note that it approaches the vertical asymptote at \( x=1 \) and increases towards the right. Starting just to the right of \( x=1 \), plot the curve that moves steeply upward to the left and increases gradually as \( x \) goes to positive infinity. The graph is always above the x-axis, as \( \ln \) of a positive number is positive.

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values that the function can accept. For a logarithmic function like \( f(x) = \ln(x-1) \), the focus is on ensuring that the argument of the logarithm, \( x-1 \), is positive. This is because the logarithm of zero or a negative number is not defined in the realm of real numbers.
The process involves setting up an inequality for the expression within the logarithm. Here, \( x-1 > 0 \). Solving this inequality gives us \( x > 1 \). Thus, the domain of \( f(x) = \ln(x-1) \) is all values of \( x \) greater than 1. In interval notation, this is shown as \( (1, \infty) \).
This means you can plug in any value of \( x \) into the function, as long as it is greater than 1, and the function will yield a valid real number output.

Range of a Function

The range of a function consists of all possible output values. For logarithmic functions, such as \( f(x) = \ln(x-1) \), the range is particularly interesting. Regardless of the input values, as long as they fall within the domain, the logarithmic function can produce any real number.
This happens because as \( x \) gets very close to the vertical asymptote (but never reaches it), the function \( f(x) \) plummets downwards towards negative infinity. Conversely, as \( x \) increases significantly, the output of the function can climb as high as it needs, tending towards positive infinity.
The result is that the range of \( f(x) = \ln(x-1) \) spans all real numbers, represented in interval notation as \( (-\infty, \infty) \). This characteristic is a hallmark of logarithmic functions.

Vertical Asymptotes

A vertical asymptote is a line where a function approaches but never actually touches or crosses. The logarithmic function \( f(x) = \ln(x-1) \) has a vertical asymptote because the function is undefined for values where the logarithmic expression is zero or negative.
For \( f(x) = \ln(x-1) \), notice the function becomes undefined at \( x = 1 \). This is because at \( x=1 \), the term inside the logarithm becomes zero \( \ln(0) \), which is not a valid operation in the real number system. Therefore, there is a vertical asymptote at \( x=1 \).
Graphically, this asymptote appears as a vertical line on the graph, which the curve of the function will approach but never cross. It serves as an important boundary for the behavior of the function, indicating that as \( x \) inches closer to 1 from the right, the output of \( f(x) \) will trend towards negative infinity.