Find the value(s) of k that makes the given function continuous. 4x 5 if x < k f(x) = { = x++1 ifk ≤x≤6 (Use a comma to separate multiple answers. Enter an exact value or values.) Provide your answer below:
Find the value(s) of k that makes the given function continuous. 4x 5 if x < k f(x) = { = x++1 ifk ≤x≤6 (Use a comma to separate multiple answers. Enter an exact value or values.) Provide your answer below:
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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Question
![**Problem Statement:**
Find the value(s) of \( k \) that makes the given function continuous.
\[
f(x) =
\begin{cases}
4x + 5 & \text{if } x < k \\
-4x + 1 & \text{if } k \le x \le 6
\end{cases}
\]
(Use a comma to separate multiple answers. Enter an exact value or values.)
**Solution Workspace:**
Provide your answer below:
\( k = \_\_\_\_\_ \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fffcf92b5-4154-452a-af6b-b4e696e552b0%2F30e516b7-a437-4b62-b1ef-5a0accae87f6%2Fsbz6aqt_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the value(s) of \( k \) that makes the given function continuous.
\[
f(x) =
\begin{cases}
4x + 5 & \text{if } x < k \\
-4x + 1 & \text{if } k \le x \le 6
\end{cases}
\]
(Use a comma to separate multiple answers. Enter an exact value or values.)
**Solution Workspace:**
Provide your answer below:
\( k = \_\_\_\_\_ \)
![### Application of the Intermediate Value Theorem
**Problem Statement:**
If \( f(x) \) is continuous in the interval \([-2, 1]\), \( f(-2) = -4 \), and \( f(1) = -1 \), can we use the Intermediate Value Theorem to conclude that \( f(c) = 0 \) where \(-2 < c < 1\)?
**Options to Select From:**
1. Yes, the Intermediate Value Theorem guarantees that \( f(c) = 0 \).
2. No, the Intermediate Value Theorem does not guarantee this conclusion, but \( f(c) = 0 \) could be true and cannot be ruled out.
**Correct Answer:**
- Yes, the Intermediate Value Theorem guarantees that \( f(c) = 0 \).
**Explanation:**
The Intermediate Value Theorem (IVT) states that for any continuous function \( f(x) \) on a closed interval \([a, b]\), if \( N \) is any number between \( f(a) \) and \( f(b) \) (inclusive), then there exists at least one \( c \) in the open interval \((a, b)\) such that \( f(c) = N \).
Given:
- \( f(x) \) is continuous on \([-2, 1]\),
- \( f(-2) = -4 \),
- \( f(1) = -1 \).
Since 0 is a number between \( f(-2) = -4 \) and \( f(1) = -1 \), by the IVT, there must exist a \( c \) in the interval \((-2, 1)\) such that \( f(c) = 0 \).
Thus, the correct answer is the first option:
- **Yes, the Intermediate Value Theorem guarantees that \( f(c) = 0 \).**
### Visualization:
There are no graphs or diagrams provided in this question. However, to understand it better, you can imagine a continuous curve that starts at \(-4\) when \( x = -2 \) and ends at \(-1\) when \( x = 1 \). The curve must pass through every value between \(-4\) and \(-1\) at least once, including 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fffcf92b5-4154-452a-af6b-b4e696e552b0%2F30e516b7-a437-4b62-b1ef-5a0accae87f6%2Fwg2gqp_processed.png&w=3840&q=75)
Transcribed Image Text:### Application of the Intermediate Value Theorem
**Problem Statement:**
If \( f(x) \) is continuous in the interval \([-2, 1]\), \( f(-2) = -4 \), and \( f(1) = -1 \), can we use the Intermediate Value Theorem to conclude that \( f(c) = 0 \) where \(-2 < c < 1\)?
**Options to Select From:**
1. Yes, the Intermediate Value Theorem guarantees that \( f(c) = 0 \).
2. No, the Intermediate Value Theorem does not guarantee this conclusion, but \( f(c) = 0 \) could be true and cannot be ruled out.
**Correct Answer:**
- Yes, the Intermediate Value Theorem guarantees that \( f(c) = 0 \).
**Explanation:**
The Intermediate Value Theorem (IVT) states that for any continuous function \( f(x) \) on a closed interval \([a, b]\), if \( N \) is any number between \( f(a) \) and \( f(b) \) (inclusive), then there exists at least one \( c \) in the open interval \((a, b)\) such that \( f(c) = N \).
Given:
- \( f(x) \) is continuous on \([-2, 1]\),
- \( f(-2) = -4 \),
- \( f(1) = -1 \).
Since 0 is a number between \( f(-2) = -4 \) and \( f(1) = -1 \), by the IVT, there must exist a \( c \) in the interval \((-2, 1)\) such that \( f(c) = 0 \).
Thus, the correct answer is the first option:
- **Yes, the Intermediate Value Theorem guarantees that \( f(c) = 0 \).**
### Visualization:
There are no graphs or diagrams provided in this question. However, to understand it better, you can imagine a continuous curve that starts at \(-4\) when \( x = -2 \) and ends at \(-1\) when \( x = 1 \). The curve must pass through every value between \(-4\) and \(-1\) at least once, including 0.
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