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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![## Redefining the Function for Continuity
To ensure the function \( f \) is continuous everywhere, we need to redefine it at \( x = 2 \).
This is our function:
\[ f(x) = \begin{cases}
x^3 - 2x + 1 & \text{if } x \neq 2 \\
7 & \text{if } x = 2
\end{cases} \]
### Explanation
1. **For \( x \neq 2 \)**: The function is defined as \( f(x) = x^3 - 2x + 1 \).
2. **For \( x = 2 \)**: The function is defined as \( f(x) = 7 \).
### Steps to Redefine
To redefine \( f \) so that it is continuous everywhere, including at \( x = 2 \), we perform the following steps:
1. **Find the limit of \( f(x) \) as \( x \) approaches 2**:
\[ \lim_{x \to 2} (x^3 - 2x + 1) \]
2. **Substitute \( x = 2 \) into \( x^3 - 2x + 1 \) to find the limit value**:
\[ 2^3 - 2(2) + 1 = 8 - 4 + 1 = 5 \]
3. **Redefine \( f(2) \) to match this limit value**:
\[ f(2) = 5 \]
### Final Defined Function
The redefined function to ensure it is continuous everywhere would be:
\[ f(x) = \begin{cases}
x^3 - 2x + 1 & \text{if } x \neq 2 \\
5 & \text{if } x = 2
\end{cases} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b45534a-2d1e-4c33-8e4b-f4b4691d8c9d%2Fda95802f-5ef8-4487-8585-ee62d4eb48fb%2Fu0ozpq8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Redefining the Function for Continuity
To ensure the function \( f \) is continuous everywhere, we need to redefine it at \( x = 2 \).
This is our function:
\[ f(x) = \begin{cases}
x^3 - 2x + 1 & \text{if } x \neq 2 \\
7 & \text{if } x = 2
\end{cases} \]
### Explanation
1. **For \( x \neq 2 \)**: The function is defined as \( f(x) = x^3 - 2x + 1 \).
2. **For \( x = 2 \)**: The function is defined as \( f(x) = 7 \).
### Steps to Redefine
To redefine \( f \) so that it is continuous everywhere, including at \( x = 2 \), we perform the following steps:
1. **Find the limit of \( f(x) \) as \( x \) approaches 2**:
\[ \lim_{x \to 2} (x^3 - 2x + 1) \]
2. **Substitute \( x = 2 \) into \( x^3 - 2x + 1 \) to find the limit value**:
\[ 2^3 - 2(2) + 1 = 8 - 4 + 1 = 5 \]
3. **Redefine \( f(2) \) to match this limit value**:
\[ f(2) = 5 \]
### Final Defined Function
The redefined function to ensure it is continuous everywhere would be:
\[ f(x) = \begin{cases}
x^3 - 2x + 1 & \text{if } x \neq 2 \\
5 & \text{if } x = 2
\end{cases} \]
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