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
![**Problem Statement**
Given that the graph shown below represents the function \( f(x) = 3\sin(x) \), determine \( g'(\pi) \) if:
\[
g(x) = \int_{0}^{x} f(t) \, dt
\]
**Graph Description**
The graph is a visual representation of the function \( f(x) = 3\sin(x) \). The graph intersects the x-axis at multiple points and exhibits the typical sinusoidal wave pattern. It oscillates between \( y = 3 \) and \( y = -3 \) with a period characteristic of the sine function, adjusted by the amplitude factor of 3.
**Question Options**
- \( 0 \)
- \( \frac{\pi}{2} \)
- \( \pi \)
- \( 2\pi \)
**Solution Explanation**
To solve for \( g'(\pi) \):
Using the Fundamental Theorem of Calculus, the derivative of \( g(x) \) with respect to \( x \), \( g'(x) \), is given by:
\[
g'(x) = f(x)
\]
Thus, \( g'(\pi) = f(\pi) \). Given \( f(x) = 3\sin(x) \):
\[
f(\pi) = 3\sin(\pi)
\]
Since \(\sin(\pi) = 0\):
\[
f(\pi) = 3 \times 0 = 0
\]
Therefore, \( g'(\pi) = 0 \).
**Correct Answer:**
- \( 0 \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf75a45c-687d-4994-8001-f519eebb3c9c%2F2e6e4149-c328-410d-aa77-39cb71cdf0fb%2Fopz1nw_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Given that the graph shown below represents the function \( f(x) = 3\sin(x) \), determine \( g'(\pi) \) if:
\[
g(x) = \int_{0}^{x} f(t) \, dt
\]
**Graph Description**
The graph is a visual representation of the function \( f(x) = 3\sin(x) \). The graph intersects the x-axis at multiple points and exhibits the typical sinusoidal wave pattern. It oscillates between \( y = 3 \) and \( y = -3 \) with a period characteristic of the sine function, adjusted by the amplitude factor of 3.
**Question Options**
- \( 0 \)
- \( \frac{\pi}{2} \)
- \( \pi \)
- \( 2\pi \)
**Solution Explanation**
To solve for \( g'(\pi) \):
Using the Fundamental Theorem of Calculus, the derivative of \( g(x) \) with respect to \( x \), \( g'(x) \), is given by:
\[
g'(x) = f(x)
\]
Thus, \( g'(\pi) = f(\pi) \). Given \( f(x) = 3\sin(x) \):
\[
f(\pi) = 3\sin(\pi)
\]
Since \(\sin(\pi) = 0\):
\[
f(\pi) = 3 \times 0 = 0
\]
Therefore, \( g'(\pi) = 0 \).
**Correct Answer:**
- \( 0 \)
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