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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![**The graph shown below is \( f(x) = 2x^2 - 1 \). If \( g(x) = \int_{0}^{x} f(t) \, dt \), find \( g'(3) \).**
The image depicts a graph of the quadratic function \( f(x) = 2x^2 - 1 \). The function shows a parabola opening upwards with the vertex at the point (0, -1).
### Explanation of Graph
- The x-axis ranges from -5 to 5.
- The y-axis ranges from -5 to 5.
- The parabola intersects the y-axis at \( -1 \) (the vertex) and opens upwards, reflecting the positive coefficient of \( x^2 \).
- The graph indicates it is the plot of the function \( f(x) \).
### Problem Statement
We are given \( g(x) \), which is defined as the integral of the function \( f(t) \) from \( 0 \) to \( x \):
\[ g(x) = \int_{0}^{x} f(t) \, dt \]
We need to find \( g'(3) \).
### Solution Approach
By the Fundamental Theorem of Calculus, Part 1, if \( G(x) \) is an antiderivative of \( f(x) \), then:
\[ g'(x) = f(x) \]
So, to find \( g'(3) \), we simply need to evaluate \( f(3) \):
\[ f(x) = 2x^2 - 1 \]
Substitute \( x = 3 \):
\[ f(3) = 2(3)^2 - 1 = 2 \cdot 9 - 1 = 18 - 1 = 17 \]
Thus:
\[ g'(3) = 17 \]
Therefore, \( g'(3) \) is \( 17 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F883933e7-6864-477d-8bb2-712659e12b95%2Fc0960064-a1ab-4cd3-a2d7-7eb5ac08bb1d%2Fpniujcy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**The graph shown below is \( f(x) = 2x^2 - 1 \). If \( g(x) = \int_{0}^{x} f(t) \, dt \), find \( g'(3) \).**
The image depicts a graph of the quadratic function \( f(x) = 2x^2 - 1 \). The function shows a parabola opening upwards with the vertex at the point (0, -1).
### Explanation of Graph
- The x-axis ranges from -5 to 5.
- The y-axis ranges from -5 to 5.
- The parabola intersects the y-axis at \( -1 \) (the vertex) and opens upwards, reflecting the positive coefficient of \( x^2 \).
- The graph indicates it is the plot of the function \( f(x) \).
### Problem Statement
We are given \( g(x) \), which is defined as the integral of the function \( f(t) \) from \( 0 \) to \( x \):
\[ g(x) = \int_{0}^{x} f(t) \, dt \]
We need to find \( g'(3) \).
### Solution Approach
By the Fundamental Theorem of Calculus, Part 1, if \( G(x) \) is an antiderivative of \( f(x) \), then:
\[ g'(x) = f(x) \]
So, to find \( g'(3) \), we simply need to evaluate \( f(3) \):
\[ f(x) = 2x^2 - 1 \]
Substitute \( x = 3 \):
\[ f(3) = 2(3)^2 - 1 = 2 \cdot 9 - 1 = 18 - 1 = 17 \]
Thus:
\[ g'(3) = 17 \]
Therefore, \( g'(3) \) is \( 17 \).
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