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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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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![### Problem Statement
1. Are functions \( f(x) = \frac{x^2 - 1}{x + 1} \) and \( g(x) = x - 1 \) the same? Why? Explain in full sentences.
### Solution Explanation
To determine if the two functions \( f(x) = \frac{x^2 - 1}{x + 1} \) and \( g(x) = x - 1 \) are the same, we need to analyze and compare them carefully.
- The function \( f(x) \) is given by the expression \( \frac{x^2 - 1}{x + 1} \), which can be factored as:
\[
f(x) = \frac{(x - 1)(x + 1)}{x + 1}
\]
Since \( x + 1 \) in the numerator and denominator cancel each other out (for \( x \neq -1 \)):
\[
f(x) = x - 1, \text{ for } x \neq -1
\]
- The function \( g(x) \) is defined as \( g(x) = x - 1 \).
Both \( f(x) \) and \( g(x) \) appear to be identical when \( x \neq -1 \). However, \( f(x) \) is not defined at \( x = -1 \) because it would result in division by zero, while \( g(x) \) is defined for all real numbers.
Therefore, \( f(x) \) and \( g(x) \) are not exactly the same. They are equivalent for \( x \neq -1 \), but \( f(x) \) has a point of discontinuity at \( x = -1 \) where it is not defined.
Thus, your final explanation should note:
1. \( f(x) = g(x) \) for all values of \( x \) except \( x = -1 \).
2. The functions differ because \( f(x) \) is undefined at \( x = -1 \), whereas \( g(x) \) is defined for all \( x \).
This critical difference underscores the importance of domain considerations when evaluating the equivalence of mathematical functions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58deed7b-e5cf-455c-8dcf-5c14f5ab5a29%2F4fc0cfc9-68af-4dea-a8bd-b70c5c38528d%2F03lcljh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
1. Are functions \( f(x) = \frac{x^2 - 1}{x + 1} \) and \( g(x) = x - 1 \) the same? Why? Explain in full sentences.
### Solution Explanation
To determine if the two functions \( f(x) = \frac{x^2 - 1}{x + 1} \) and \( g(x) = x - 1 \) are the same, we need to analyze and compare them carefully.
- The function \( f(x) \) is given by the expression \( \frac{x^2 - 1}{x + 1} \), which can be factored as:
\[
f(x) = \frac{(x - 1)(x + 1)}{x + 1}
\]
Since \( x + 1 \) in the numerator and denominator cancel each other out (for \( x \neq -1 \)):
\[
f(x) = x - 1, \text{ for } x \neq -1
\]
- The function \( g(x) \) is defined as \( g(x) = x - 1 \).
Both \( f(x) \) and \( g(x) \) appear to be identical when \( x \neq -1 \). However, \( f(x) \) is not defined at \( x = -1 \) because it would result in division by zero, while \( g(x) \) is defined for all real numbers.
Therefore, \( f(x) \) and \( g(x) \) are not exactly the same. They are equivalent for \( x \neq -1 \), but \( f(x) \) has a point of discontinuity at \( x = -1 \) where it is not defined.
Thus, your final explanation should note:
1. \( f(x) = g(x) \) for all values of \( x \) except \( x = -1 \).
2. The functions differ because \( f(x) \) is undefined at \( x = -1 \), whereas \( g(x) \) is defined for all \( x \).
This critical difference underscores the importance of domain considerations when evaluating the equivalence of mathematical functions.
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