(x + 1, x < 0 f(x) = 0, x = 0 (x² + 1, x>0 Can also be displayed as: x + 1 when x < 0, 0 when x = 0, x² + 1 when x > 0 f(x) = Is f(x) continuous at x = 0? Why or why not? You must show all of your work and reasoning to receive credit. Use the rules of continuity.
(x + 1, x < 0 f(x) = 0, x = 0 (x² + 1, x>0 Can also be displayed as: x + 1 when x < 0, 0 when x = 0, x² + 1 when x > 0 f(x) = Is f(x) continuous at x = 0? Why or why not? You must show all of your work and reasoning to receive credit. Use the rules of continuity.
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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![### Continuity of a Piecewise Function
Given the piecewise function:
\[
f(x) =
\begin{cases}
x + 1, & \text{if } x < 0 \\
0, & \text{if } x = 0 \\
x^2 + 1, & \text{if } x > 0
\end{cases}
\]
Which can also be displayed as:
\[
f(x) =
\begin{cases}
x + 1 \text{ when } x < 0 \\
0 \text{ when } x = 0 \\
x^2 + 1 \text{ when } x > 0
\end{cases}
\]
### Question:
Is \( f(x) \) continuous at \( x = 0 \)? Why or why not?
### Instructions:
You must show all of your work and reasoning to receive credit. Use the rules of continuity.
### Solution Approach:
To determine if \( f(x) \) is continuous at \( x = 0 \), you must check the following conditions:
1. \( f(0) \) is defined.
2. The limit of \( f(x) \) as \( x \) approaches 0 exists.
3. The limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).
By evaluating the piecewise function at \( x = 0 \):
- \( f(0) = 0 \) (condition 1 is satisfied).
Next, calculate the left-hand limit as \( x \) approaches 0:
- For \( x < 0 \), \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + 1) = 1 \).
Then, calculate the right-hand limit as \( x \) approaches 0:
- For \( x > 0 \), \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + 1) = 1 \).
Since both the left-hand limit and the right-hand limit as \( x \) approaches 0 are equal to 1:
- \( \lim_{x \to 0} f(x) = 1 \) (condition 2 satisfied).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F89d20bf7-5284-426d-a0f1-0ea4150b8e2e%2Ff1989dd5-1b7b-4e8a-923e-725a3ea2f0c7%2F7qt94b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Continuity of a Piecewise Function
Given the piecewise function:
\[
f(x) =
\begin{cases}
x + 1, & \text{if } x < 0 \\
0, & \text{if } x = 0 \\
x^2 + 1, & \text{if } x > 0
\end{cases}
\]
Which can also be displayed as:
\[
f(x) =
\begin{cases}
x + 1 \text{ when } x < 0 \\
0 \text{ when } x = 0 \\
x^2 + 1 \text{ when } x > 0
\end{cases}
\]
### Question:
Is \( f(x) \) continuous at \( x = 0 \)? Why or why not?
### Instructions:
You must show all of your work and reasoning to receive credit. Use the rules of continuity.
### Solution Approach:
To determine if \( f(x) \) is continuous at \( x = 0 \), you must check the following conditions:
1. \( f(0) \) is defined.
2. The limit of \( f(x) \) as \( x \) approaches 0 exists.
3. The limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).
By evaluating the piecewise function at \( x = 0 \):
- \( f(0) = 0 \) (condition 1 is satisfied).
Next, calculate the left-hand limit as \( x \) approaches 0:
- For \( x < 0 \), \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + 1) = 1 \).
Then, calculate the right-hand limit as \( x \) approaches 0:
- For \( x > 0 \), \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + 1) = 1 \).
Since both the left-hand limit and the right-hand limit as \( x \) approaches 0 are equal to 1:
- \( \lim_{x \to 0} f(x) = 1 \) (condition 2 satisfied).
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