Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Understanding the Logarithmic Function**
Consider the function \( g(x) = \log (x + 1) \).
### Analyzing the Function:
1. **Domain**:
To determine the domain of \( g(x) \), we need to find the values of \( x \) for which the argument of the logarithm is positive. This is because the logarithm of a non-positive number is not defined. Hence, we have:
\[
x + 1 > 0 \implies x > -1
\]
Therefore, the domain of \( g \) is the interval from \( -1 \) (not inclusive) to infinity. Formally, we can write this as:
\[
\text{The domain of } g \text{ is the interval from } (-1, \infty).
\]
2. **x-intercept**:
To find the x-intercept, we need to solve for \( x \) when \( g(x) = 0 \). Therefore, we set up the equation:
\[
\log (x + 1) = 0
\]
Recall that \( \log a = b \) is equivalent to \( a = 10^b \). So in this case:
\[
x + 1 = 10^0 \implies x + 1 = 1 \implies x = 0
\]
Thus, the x-intercept is at \( x = 0 \).
---------------------------
**Summary:**
- The domain of \( g \) is the interval from \( (-1, \infty) \).
- The x-intercept of the function \( g(x) \) is at \( x = 0 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a4215dc-32d4-4c07-b1ef-f2c19838cd70%2Fbe8aa483-f467-438a-be6e-08c05468d41d%2Fqxflpja_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding the Logarithmic Function**
Consider the function \( g(x) = \log (x + 1) \).
### Analyzing the Function:
1. **Domain**:
To determine the domain of \( g(x) \), we need to find the values of \( x \) for which the argument of the logarithm is positive. This is because the logarithm of a non-positive number is not defined. Hence, we have:
\[
x + 1 > 0 \implies x > -1
\]
Therefore, the domain of \( g \) is the interval from \( -1 \) (not inclusive) to infinity. Formally, we can write this as:
\[
\text{The domain of } g \text{ is the interval from } (-1, \infty).
\]
2. **x-intercept**:
To find the x-intercept, we need to solve for \( x \) when \( g(x) = 0 \). Therefore, we set up the equation:
\[
\log (x + 1) = 0
\]
Recall that \( \log a = b \) is equivalent to \( a = 10^b \). So in this case:
\[
x + 1 = 10^0 \implies x + 1 = 1 \implies x = 0
\]
Thus, the x-intercept is at \( x = 0 \).
---------------------------
**Summary:**
- The domain of \( g \) is the interval from \( (-1, \infty) \).
- The x-intercept of the function \( g(x) \) is at \( x = 0 \).
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