### Question 1 **True or False:** Evaluate the statement: \[ \frac{d}{dx} \left[ 2 + \frac{1}{x} \right] = -\frac{1}{x^2} \] - **False** (selected) - True ### Explanation: This is a calculus question asking if the derivative of the function \( f(x) = 2 + \frac{1}{x} \) with respect to \( x \) is equal to \( -\frac{1}{x^2} \). To solve this, use the power rule for derivatives. The derivative of \( 2 \) is 0, and the derivative of \( \frac{1}{x} \) which is \( x^{-1} \) is calculated as follows: \[ \frac{d}{dx} \left( x^{-1} \right) = -1 \cdot x^{-2} = -\frac{1}{x^2} \] So, the derivative \(\frac{d}{dx} \left[ 2 + \frac{1}{x} \right]\) is indeed \(-\frac{1}{x^2}\). Therefore, the answer should be **True** instead of False.
### Question 1 **True or False:** Evaluate the statement: \[ \frac{d}{dx} \left[ 2 + \frac{1}{x} \right] = -\frac{1}{x^2} \] - **False** (selected) - True ### Explanation: This is a calculus question asking if the derivative of the function \( f(x) = 2 + \frac{1}{x} \) with respect to \( x \) is equal to \( -\frac{1}{x^2} \). To solve this, use the power rule for derivatives. The derivative of \( 2 \) is 0, and the derivative of \( \frac{1}{x} \) which is \( x^{-1} \) is calculated as follows: \[ \frac{d}{dx} \left( x^{-1} \right) = -1 \cdot x^{-2} = -\frac{1}{x^2} \] So, the derivative \(\frac{d}{dx} \left[ 2 + \frac{1}{x} \right]\) is indeed \(-\frac{1}{x^2}\). Therefore, the answer should be **True** instead of False.
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...
Related questions
Question
![### Question 1
**True or False:**
Evaluate the statement:
\[
\frac{d}{dx} \left[ 2 + \frac{1}{x} \right] = -\frac{1}{x^2}
\]
- **False** (selected)
- True
### Explanation:
This is a calculus question asking if the derivative of the function \( f(x) = 2 + \frac{1}{x} \) with respect to \( x \) is equal to \( -\frac{1}{x^2} \).
To solve this, use the power rule for derivatives. The derivative of \( 2 \) is 0, and the derivative of \( \frac{1}{x} \) which is \( x^{-1} \) is calculated as follows:
\[
\frac{d}{dx} \left( x^{-1} \right) = -1 \cdot x^{-2} = -\frac{1}{x^2}
\]
So, the derivative \(\frac{d}{dx} \left[ 2 + \frac{1}{x} \right]\) is indeed \(-\frac{1}{x^2}\). Therefore, the answer should be **True** instead of False.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc7eb3f3-8ec3-4594-8f18-33cddb117e48%2F5474106b-aa9b-45af-90ff-e721d7044663%2Fc5yqxxg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 1
**True or False:**
Evaluate the statement:
\[
\frac{d}{dx} \left[ 2 + \frac{1}{x} \right] = -\frac{1}{x^2}
\]
- **False** (selected)
- True
### Explanation:
This is a calculus question asking if the derivative of the function \( f(x) = 2 + \frac{1}{x} \) with respect to \( x \) is equal to \( -\frac{1}{x^2} \).
To solve this, use the power rule for derivatives. The derivative of \( 2 \) is 0, and the derivative of \( \frac{1}{x} \) which is \( x^{-1} \) is calculated as follows:
\[
\frac{d}{dx} \left( x^{-1} \right) = -1 \cdot x^{-2} = -\frac{1}{x^2}
\]
So, the derivative \(\frac{d}{dx} \left[ 2 + \frac{1}{x} \right]\) is indeed \(-\frac{1}{x^2}\). Therefore, the answer should be **True** instead of False.
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