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State the range for the function f(x) = -In(x − 1) + 2 : O (-00,00) O (-∞, -1) O (0,00) O (-00, 0)

State the range for the function f(x) = -In(x − 1) + 2 : O (-00,00) O (-∞, -1) O (0,00) O (-00, 0)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Determining the Range of the Function In this task, you are asked to determine the range for the function \( f(x) = -\ln(x - 1) + 2 \). Below the problem statement, there are four multiple-choice options. #### Problem Statement: **State the range for the function \( f(x) = -\ln(x - 1) + 2 \):** #### Options: - \(\circ\) \((-\infty, \infty)\) - \(\circ\) \((-\infty, -1)\) - \(\circ\) \((0, \infty)\) - \(\circ\) \((-\infty, 0)\) ### Explanation of the Function's Behavior The given function is \( f(x) = -\ln(x - 1) + 2 \). 1. **Domain**: First, we identify the domain of the function \( \ln(x - 1) \), which is valid for \( x - 1 > 0 \) or \( x > 1 \). 2. **Analysis of \( -\ln(x - 1) \)**: - As \( x \) approaches \( 1 \) from the right (\( x \to 1^+ \)), \( \ln(x - 1) \to -\infty \), so \( -\ln(x - 1) \to \infty \). - As \( x \) increases, \( \ln(x - 1) \) becomes larger, and nearing infinity, \( -\ln(x - 1) \to -\infty \). 3. **Shifting by 2**: - \( f(x) \) shifts the \( -\ln(x - 1) \) function up by 2 units. - Therefore, the range of \( f(x) \) is moved up by 2 units along the y-axis. ### Determining the Range: From the behavior of \( -\ln(x - 1) \): - When \( x = 1 \), \( f(x) \) tends towards \( \infty \). - As \( x \to \infty \), \( f(x) \to -\infty + 2 = -\infty \). Thus, the range of the function \( f(x) = -\
Transcribed Image Text:### Determining the Range of the Function In this task, you are asked to determine the range for the function \( f(x) = -\ln(x - 1) + 2 \). Below the problem statement, there are four multiple-choice options. #### Problem Statement: **State the range for the function \( f(x) = -\ln(x - 1) + 2 \):** #### Options: - \(\circ\) \((-\infty, \infty)\) - \(\circ\) \((-\infty, -1)\) - \(\circ\) \((0, \infty)\) - \(\circ\) \((-\infty, 0)\) ### Explanation of the Function's Behavior The given function is \( f(x) = -\ln(x - 1) + 2 \). 1. **Domain**: First, we identify the domain of the function \( \ln(x - 1) \), which is valid for \( x - 1 > 0 \) or \( x > 1 \). 2. **Analysis of \( -\ln(x - 1) \)**: - As \( x \) approaches \( 1 \) from the right (\( x \to 1^+ \)), \( \ln(x - 1) \to -\infty \), so \( -\ln(x - 1) \to \infty \). - As \( x \) increases, \( \ln(x - 1) \) becomes larger, and nearing infinity, \( -\ln(x - 1) \to -\infty \). 3. **Shifting by 2**: - \( f(x) \) shifts the \( -\ln(x - 1) \) function up by 2 units. - Therefore, the range of \( f(x) \) is moved up by 2 units along the y-axis. ### Determining the Range: From the behavior of \( -\ln(x - 1) \): - When \( x = 1 \), \( f(x) \) tends towards \( \infty \). - As \( x \to \infty \), \( f(x) \to -\infty + 2 = -\infty \). Thus, the range of the function \( f(x) = -\
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