For what value of k the following piecewise defined function will be continuous? [k - 2 cos x, e4x+1 O f(x) = k = 2 k = 3 k=2+e k = 3 + e k=2-e x ≤ 0 x > 0
For what value of k the following piecewise defined function will be continuous? [k - 2 cos x, e4x+1 O f(x) = k = 2 k = 3 k=2+e k = 3 + e k=2-e x ≤ 0 x > 0
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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![### Problem Statement
For what value of \( k \) the following piecewise defined function will be continuous?
\[
f(x) =
\begin{cases}
k - 2 \cos x, & x \leq 0 \\
e^{4x+1}, & x > 0
\end{cases}
\]
### Answer Choices:
1. \( k = 2 \)
2. \( k = 3 \) (selected)
3. \( k = 2 + e \)
4. \( k = 3 + e \)
5. \( k = 2 - e \)
### Explanation:
To determine the value of \( k \) that makes the function \( f(x) \) continuous, we need to ensure that the limit from the left as \( x \) approaches 0 equals the limit from the right as \( x \) approaches 0, and both limits are equal to the function value at \( x = 0 \).
#### Left-Hand Limit as \( x \) approaches 0:
For \( x \leq 0 \), the function is \( f(x) = k - 2 \cos x \). As \( x \) approaches 0 from the left:
\[
\lim_{{x \to 0^-}} (k - 2 \cos x) = k - 2 \cos(0) = k - 2
\]
#### Right-Hand Limit as \( x \) approaches 0:
For \( x > 0 \), the function is \( f(x) = e^{4x+1} \). As \( x \) approaches 0 from the right:
\[
\lim_{{x \to 0^+}} e^{4x+1} = e^1 = e
\]
#### Equating the Limits:
To ensure continuity at \( x = 0 \):
\[
k - 2 = e
\]
Solving for \( k \):
\[
k = e + 2
\]
### Conclusion:
The piecewise function \( f(x) \) will be continuous if \( k = 2 + e \).
Since the selected answer \( k = 3 \) does not satisfy this condition, the correct answer should be \( k = 2 + e \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49dd86f5-e3e4-495a-9b19-f3d039df6a7c%2F7ae46f95-b49e-4759-bed2-608c6a936746%2Fc92woep_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
For what value of \( k \) the following piecewise defined function will be continuous?
\[
f(x) =
\begin{cases}
k - 2 \cos x, & x \leq 0 \\
e^{4x+1}, & x > 0
\end{cases}
\]
### Answer Choices:
1. \( k = 2 \)
2. \( k = 3 \) (selected)
3. \( k = 2 + e \)
4. \( k = 3 + e \)
5. \( k = 2 - e \)
### Explanation:
To determine the value of \( k \) that makes the function \( f(x) \) continuous, we need to ensure that the limit from the left as \( x \) approaches 0 equals the limit from the right as \( x \) approaches 0, and both limits are equal to the function value at \( x = 0 \).
#### Left-Hand Limit as \( x \) approaches 0:
For \( x \leq 0 \), the function is \( f(x) = k - 2 \cos x \). As \( x \) approaches 0 from the left:
\[
\lim_{{x \to 0^-}} (k - 2 \cos x) = k - 2 \cos(0) = k - 2
\]
#### Right-Hand Limit as \( x \) approaches 0:
For \( x > 0 \), the function is \( f(x) = e^{4x+1} \). As \( x \) approaches 0 from the right:
\[
\lim_{{x \to 0^+}} e^{4x+1} = e^1 = e
\]
#### Equating the Limits:
To ensure continuity at \( x = 0 \):
\[
k - 2 = e
\]
Solving for \( k \):
\[
k = e + 2
\]
### Conclusion:
The piecewise function \( f(x) \) will be continuous if \( k = 2 + e \).
Since the selected answer \( k = 3 \) does not satisfy this condition, the correct answer should be \( k = 2 + e \).
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