Problem 62 - Essential Calculus Early Transcendentals

Question

(a) What are the domain and range of \(f ?\) (b) What is the \(x\) -intercept of the graph of \(f ?\) (c) Sketch the graph of \(f .\) $$ f(x)=\ln (x-1)-1 $$

Step-by-Step Solution

Answer

Domain: \((1, \infty)\); Range: \((-\infty, \infty)\); x-intercept: \((e+1, 0)\).

Step 1: Domain of the function

The function is defined as \( f(x) = \ln(x-1) - 1 \). The natural logarithm function, \( \ln(x) \), is only defined for \( x > 0 \). Therefore, \( x - 1 > 0 \) must be true, which implies \( x > 1 \). Hence, the domain of the function is \( x \in (1, \infty) \).

Step 2: Range of the function

Since \( f(x) = \ln(x-1) - 1 \), and the range of the natural logarithm function \( \ln(x-1) \) is \((-\infty, \infty)\), subtracting 1 from \( \ln(x-1) \) doesn't affect the range's infinite nature. Hence, the range of \( f(x) \) is \( (-\infty, \infty) \).

Step 3: Finding the x-intercept

To find the x-intercept, set \( f(x) = 0 \) and solve for \( x \):\[ \ln(x-1) - 1 = 0 \]\[ \ln(x-1) = 1 \]Exponentiating both sides gives \( x - 1 = e \), hence \( x = e + 1 \). Therefore, the x-intercept is at \( (e+1, 0) \).

Step 4: Sketching the graph

To sketch the graph of \( f(x) = \ln(x-1) - 1 \), start by plotting the x-intercept at \((e+1, 0)\). The curve moves upwards passing through this point. The vertical asymptote is at \( x = 1 \), as the function approaches \(-\infty\) as \( x \to 1^+ \). As \( x \to \infty \), \( f(x) \to \infty \). The graph is located entirely to the right of \( x = 1 \) and rises gradually. Also, note the function is defined only for \( x > 1 \).

Key Concepts

Domain and Range

The domain and range of a function are crucial in understanding its behavior. **Domain** refers to all the possible input values (or 'x-values') that will produce a valid output. For the function \( f(x) = \ln(x-1) - 1 \), the natural logarithm is only defined for positive numbers. Hence, we need \( x-1 > 0 \), which simplifies to \( x > 1 \). Therefore, the domain of our function is the interval \( (1, \infty) \).
The **range** describes all the possible output values (or 'y-values'). Since the natural logarithm function itself has a range of \( (-\infty, \infty) \), subtracting 1 doesn't change this infinite nature. Therefore, the range of \( f(x)= \ln(x-1) - 1 \), like the original log function, is also \( (-\infty, \infty) \).
To sum up:

• The Domain is \((1, \infty)\).
• The Range is \((-\infty, \infty)\).

Graphing Functions

Graphing a function helps to visualize its behavior and key attributes. For \( f(x) = \ln(x-1) - 1 \), start by identifying key points, like the **x-intercept**, and any **asymptotes**. There is a **vertical asymptote** at \( x = 1 \) because as \( x \) approaches 1 from the right, the logarithm heads to \(-\infty\), thus the function goes down steeply.
For \( x > 1 \), the graph rises gently to the right as \( x \) increases, reflecting how the logarithm behaves. The intercepts and asymptotic behavior provide a good framework for sketching. Begin by marking the x-intercept and then proceed to draw the graph moving upwards, demonstrating the range from \( 1 \) to \( \infty \).
Remember, the graph doesn’t exist at or to the left of \( x = 1 \) due to the domain restriction.

x-intercepts

A graph's **x-intercepts** are where it crosses the x-axis. For the function \( f(x) = \ln(x-1) - 1 \), this intersection occurs when the function's value equals zero. To solve this, we set \( f(x) = 0 \): \[ \ln(x-1) - 1 = 0 \] Simplifying, we find \( \ln(x-1) = 1 \). To eliminate the logarithm, apply the inverse: exponentiation. This gives \( x - 1 = e \) since the natural exponent of 1 is \( e \), or \( x = e + 1 \). Thus, the x-intercept is \( (e+1, 0) \), meaning at this point, the graph crosses the x-axis.
This intercept provides crucial insight while graphing, helping to shape and guide the function's layout on paper or in digital plots.