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
![Title: Calculating the Derivative of an Integral Function
*Mathematical Concepts Covered: Derivatives, Integral Functions, Fundamental Theorem of Calculus*
---
### Problem Statement
The graph shown below is \( f(x) = \sqrt{2x} - 2 \).
\[ g(x) = \int_{0}^{x} f(t) \, dt \]
What is \( g'(2) \)?
---
### Graph Explanation
The graph depicts the function \( f(x) = \sqrt{2x} - 2 \) plotted on the Cartesian plane with the X-axis ranging from -5 to 5 and the Y-axis ranging from -5 to 5. The function starts at \( x = 2 \) and proceeds linearly upwards, cutting the Y-axis at -2.
---
### Solution Options
- \( 0 \)
- \( \sqrt{2} \)
- \( 2 \)
- \( 4 \)
---
### Solution Explanation
To solve for \( g'(2) \), we utilize the Fundamental Theorem of Calculus, which states that if \( G(x) = \int_{a}^{x} f(t) \, dt \), then:
\[ G'(x) = f(x) \]
Therefore,
\[ g'(x) = f(x) \]
Substituting \( x = 2 \) into the function \( f(x) \):
\[ g'(2) = f(2) = \sqrt{2 \cdot 2} - 2 \]
\[ g'(2) = \sqrt{4} - 2 \]
\[ g'(2) = 2 - 2 \]
\[ g'(2) = 0 \]
Hence, the value of \( g'(2) \) is \( 0 \).
---
### Correct Answer
- \( 0 \)
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb401025-c6b7-46e8-825c-86d3393fab5e%2F59c89b07-e551-436c-b94f-570fb8572c57%2F76btzba_processed.png&w=3840&q=75)
Transcribed Image Text:Title: Calculating the Derivative of an Integral Function
*Mathematical Concepts Covered: Derivatives, Integral Functions, Fundamental Theorem of Calculus*
---
### Problem Statement
The graph shown below is \( f(x) = \sqrt{2x} - 2 \).
\[ g(x) = \int_{0}^{x} f(t) \, dt \]
What is \( g'(2) \)?
---
### Graph Explanation
The graph depicts the function \( f(x) = \sqrt{2x} - 2 \) plotted on the Cartesian plane with the X-axis ranging from -5 to 5 and the Y-axis ranging from -5 to 5. The function starts at \( x = 2 \) and proceeds linearly upwards, cutting the Y-axis at -2.
---
### Solution Options
- \( 0 \)
- \( \sqrt{2} \)
- \( 2 \)
- \( 4 \)
---
### Solution Explanation
To solve for \( g'(2) \), we utilize the Fundamental Theorem of Calculus, which states that if \( G(x) = \int_{a}^{x} f(t) \, dt \), then:
\[ G'(x) = f(x) \]
Therefore,
\[ g'(x) = f(x) \]
Substituting \( x = 2 \) into the function \( f(x) \):
\[ g'(2) = f(2) = \sqrt{2 \cdot 2} - 2 \]
\[ g'(2) = \sqrt{4} - 2 \]
\[ g'(2) = 2 - 2 \]
\[ g'(2) = 0 \]
Hence, the value of \( g'(2) \) is \( 0 \).
---
### Correct Answer
- \( 0 \)
---
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