Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch a graph of the given logarithmic function. $$ f(x)=\log _{2}(x-1) $$

Step-by-Step Solution

Answer

Sketch the graph with a vertical asymptote at \( x = 1 \), it passes through points \( (2, 0) \) and \( (3, 1) \), and increases for \( x > 1 \).

Step 1: Identify the Function Type

The given function is a logarithmic function in the form of \( f(x) = \log_{2}(x-1) \). This indicates a log base 2 function shifted to the right by 1 unit.

Step 2: Determine the Domain

Since \( \log_{2}(x-1) \) is only defined for positive arguments, the inside of the log must be greater than zero: \( x-1 > 0 \). Solving this inequality, we find that the domain of the function is \( x > 1 \).

Step 3: Find the Vertical Asymptote

Logarithmic functions have a vertical asymptote where the argument of the log equals zero. Setting \( x-1 = 0 \) gives \( x = 1 \). Therefore, there is a vertical asymptote at \( x = 1 \).

Step 4: Identify Key Points

Find key points that will help in sketching the graph:1. At \( x = 2 \), \( f(x) = \log_{2}(2-1) = \log_{2}(1) = 0 \).2. At \( x = 3 \), \( f(x) = \log_{2}(3-1) = \log_{2}(2) = 1 \).These points help define the shape of the graph as it increases to the right of the asymptote.

Step 5: Sketch the Graph

Using the details from previous steps:1. Draw a vertical line (asymptote) at \( x = 1 \).2. Plot the key points: \( (2, 0) \) and \( (3, 1) \).3. Sketch a curve starting from close to the asymptote at \( (1, -\infty) \), passing through the key points, and increasing to the right.

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial, as it tells us all the possible inputs (x-values) that the function can accept. For the logarithmic function given, \( f(x) = \log_{2}(x-1) \), the domain is determined by the expression inside the logarithm. Because the logarithm of a non-positive number is undefined, the argument \( x-1 \) must be greater than zero. To find this, simply solve the inequality:

• \( x - 1 > 0 \)
• \( x > 1 \)

This means the domain of \( f(x) \) is all real numbers greater than 1, or in interval notation, it is \( (1, \infty) \). In simpler terms, any x-value greater than 1 is acceptable, and nothing less or equal to 1 can be used.

Vertical Asymptote

A vertical asymptote in a graph signifies a line that the graph approaches but never touches or crosses. For logarithmic functions, vertical asymptotes occur where the argument inside the logarithm equals zero. In our function \( f(x) = \log_{2}(x-1) \), this happens when \( x-1 = 0 \):

• Solve for x: \( x = 1 \)

Therefore, \( x = 1 \) is the vertical asymptote for this function. This asymptote helps define the boundary of the function's behavior. As the input values (x) approach 1 from the right, the function's output will head towards negative infinity, sharply decreasing near this vertical line.

Key Points in Graphing

Key points on a graph can help you outline the function's shape and see how it behaves around important x-values. For \( f(x) = \log_{2}(x-1) \), finding specific points for x that are easy to calculate provides a clear guide in drawing the graph.Some useful points to consider are:

• At \( x = 2 \): \( f(x) = \log_{2}(2-1) = \log_{2}(1) = 0 \). This results in the point (2, 0) on the graph.
• At \( x = 3 \): \( f(x) = \log_{2}(3-1) = \log_{2}(2) = 1 \). This gives us the point (3, 1).

These points are crucial as they show how the function starts from a vertical asymptote and then increases to the right, depicting its growth pattern.

Transformation of Functions

Transformations shift, stretch, or compress the graph of a function in different ways. In the function \( f(x) = \log_{2}(x-1) \), the formula itself signals a transformation:

• The \( -1 \) within the log's argument shifts the graph 1 unit to the right.
• There are no vertical shifts or reflections as there are no added or subtracted terms outside the logarithm, or negative signs altering the function's direction.

Understanding these transformations lets you quickly determine how the graph of the function differs from its basic form \( g(x) = \log_{2}(x) \). Such graphical manipulations are key in predicting the function's shape by simple equation analysis, without detailed plotting.