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 at a Point
Consider the following piecewise function \( f(x) \):
\[
f(x) =
\begin{cases}
kx + 5 & \text{for } x \le 2 \\
x - 1 & \text{for } x > 2
\end{cases}
\]
We need to determine the value of \( k \) for which the function \( f(x) \) is continuous at \( x = 2 \).
### Steps to Determine Continuity:
1. **Evaluate the Function from the Left at \( x = 2 \):**
\[
\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} (kx + 5) = k \cdot 2 + 5 = 2k + 5
\]
2. **Evaluate the Function from the Right at \( x = 2 \):**
\[
\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} (x - 1) = 2 - 1 = 1
\]
3. **Set the Left and Right Limits Equal for Continuity at \( x = 2 \):**
\[
2k + 5 = 1
\]
4. **Solve for \( k \):**
\[
2k + 5 = 1 \\
2k = 1 - 5 \\
2k = -4 \\
k = -2
\]
Therefore, the value of \( k \) that makes \( f(x) \) continuous at \( x = 2 \) is \( -2 \).
### Answer Options:
- A) -2
- B) -1
- C) 0
- D) 1
- E) 7
The correct answer is **A) -2**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffffffe81-8b7d-45ab-b8d5-e6ff78c2fa3b%2F2257c6f6-ef97-4e93-9451-dbecbac44bba%2Fgoj9xbg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Continuity of a Piecewise Function at a Point
Consider the following piecewise function \( f(x) \):
\[
f(x) =
\begin{cases}
kx + 5 & \text{for } x \le 2 \\
x - 1 & \text{for } x > 2
\end{cases}
\]
We need to determine the value of \( k \) for which the function \( f(x) \) is continuous at \( x = 2 \).
### Steps to Determine Continuity:
1. **Evaluate the Function from the Left at \( x = 2 \):**
\[
\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} (kx + 5) = k \cdot 2 + 5 = 2k + 5
\]
2. **Evaluate the Function from the Right at \( x = 2 \):**
\[
\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} (x - 1) = 2 - 1 = 1
\]
3. **Set the Left and Right Limits Equal for Continuity at \( x = 2 \):**
\[
2k + 5 = 1
\]
4. **Solve for \( k \):**
\[
2k + 5 = 1 \\
2k = 1 - 5 \\
2k = -4 \\
k = -2
\]
Therefore, the value of \( k \) that makes \( f(x) \) continuous at \( x = 2 \) is \( -2 \).
### Answer Options:
- A) -2
- B) -1
- C) 0
- D) 1
- E) 7
The correct answer is **A) -2**.
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