For what value of the number k is the following function continuous at r=2 if x < 2, if x = 2, if x > 2 f(x) = = (A) 3 (B) 2 (C) (D) All values of k work 1-x² k 3r-9 -3 (E) No such value of k exists
For what value of the number k is the following function continuous at r=2 if x < 2, if x = 2, if x > 2 f(x) = = (A) 3 (B) 2 (C) (D) All values of k work 1-x² k 3r-9 -3 (E) No such value of k exists
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 Piecewise Functions
For what value of the number \(k\) is the following function continuous at \(x = 2\)?
\[
f(x) =
\begin{cases}
1 - x^2 & \text{if } x < 2, \\
k & \text{if } x = 2, \\
3x - 9 & \text{if } x > 2
\end{cases}
\]
**Options:**
- (A) 3
- (B) 2
- (C) -3
- (D) All values of \(k\) work
- (E) No such value of \(k\) exists
### Explanation:
To determine the value of \(k\) that makes the function continuous at \(x = 2\), we must ensure that:
\[
\lim_{{x \to 2^-}} f(x) = f(2) = \lim_{{x \to 2^+}} f(x)
\]
1. **Left-hand limit (\(\lim_{{x \to 2^-}} f(x)\))**:
- When \(x < 2\), \(f(x) = 1 - x^2\).
- \(\lim_{{x \to 2^-}} f(x) = 1 - 2^2 = 1 - 4 = -3\).
2. **Right-hand limit (\(\lim_{{x \to 2^+}} f(x)\))**:
- When \(x > 2\), \(f(x) = 3x - 9\).
- \(\lim_{{x \to 2^+}} f(x) = 3(2) - 9 = 6 - 9 = -3\).
3. **Value of the function at \(x = 2\), \(f(2)\)**:
- This is given as \(k\).
### Conclusion:
For the function to be continuous at \(x = 2\):
\[
\lim_{{x \to 2^-}} f(x) = f(2) = \lim_{{x \to 2^+}} f(x)
\]
\[
-3 = k = -3
\]
Hence, the value of \(k\) must be](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbca33481-84f0-4d0e-9458-2fdcb949c0ec%2F271841f9-8e55-486d-97ea-fab0b70ff0b4%2Fkjio3rx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Continuity of Piecewise Functions
For what value of the number \(k\) is the following function continuous at \(x = 2\)?
\[
f(x) =
\begin{cases}
1 - x^2 & \text{if } x < 2, \\
k & \text{if } x = 2, \\
3x - 9 & \text{if } x > 2
\end{cases}
\]
**Options:**
- (A) 3
- (B) 2
- (C) -3
- (D) All values of \(k\) work
- (E) No such value of \(k\) exists
### Explanation:
To determine the value of \(k\) that makes the function continuous at \(x = 2\), we must ensure that:
\[
\lim_{{x \to 2^-}} f(x) = f(2) = \lim_{{x \to 2^+}} f(x)
\]
1. **Left-hand limit (\(\lim_{{x \to 2^-}} f(x)\))**:
- When \(x < 2\), \(f(x) = 1 - x^2\).
- \(\lim_{{x \to 2^-}} f(x) = 1 - 2^2 = 1 - 4 = -3\).
2. **Right-hand limit (\(\lim_{{x \to 2^+}} f(x)\))**:
- When \(x > 2\), \(f(x) = 3x - 9\).
- \(\lim_{{x \to 2^+}} f(x) = 3(2) - 9 = 6 - 9 = -3\).
3. **Value of the function at \(x = 2\), \(f(2)\)**:
- This is given as \(k\).
### Conclusion:
For the function to be continuous at \(x = 2\):
\[
\lim_{{x \to 2^-}} f(x) = f(2) = \lim_{{x \to 2^+}} f(x)
\]
\[
-3 = k = -3
\]
Hence, the value of \(k\) must be
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