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 the function \( f(x) = \int_{0}^{x} (u^2 - 6u + 5) \, du \), determine the y-coordinate of its inflection point.
### Solution
The selected answer is: \(-3\).
### Explanation
1. **Understanding the Function**: The given function \( f(x) \) is defined as an integral function:
\[
f(x) = \int_{0}^{x} (u^2 - 6u + 5) \, du
\]
2. **Step-by-Step Approach**:
- **Determine \( f(x) \)**: Integrate \( u^2 - 6u + 5 \) with respect to \( u \) from \( 0 \) to \( x \):
\[
\int_{0}^{x} (u^2 - 6u + 5) \, du = \left[ \frac{u^3}{3} - 3u^2 + 5u \right]_{0}^{x} = \frac{x^3}{3} - 3x^2 + 5x
\]
So, \( f(x) = \frac{x^3}{3} - 3x^2 + 5x \).
- **Find the First Derivative \( f'(x) \)**: Differentiate \( f(x) \) with respect to \( x \):
\[
f'(x) = x^2 - 6x + 5
\]
- **Find the Second Derivative \( f''(x) \)**: Differentiate \( f'(x) \) with respect to \( x \):
\[
f''(x) = 2x - 6
\]
- **Determine Inflection Point**: An inflection point occurs where \( f''(x) = 0 \):
\[
2x - 6 = 0 \quad \Rightarrow \quad x = 3
\]
- **Find the y-coordinate**:
\[
f(3) = \frac{3^3}{3} - 3(3^2) + 5(3) = 9 - 27 + 15 = -3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5cf73431-844e-4c23-9b17-47dbe521c237%2Fd372aa5b-159c-4dda-9179-ec234d04c032%2Fs9e67oe_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Given the function \( f(x) = \int_{0}^{x} (u^2 - 6u + 5) \, du \), determine the y-coordinate of its inflection point.
### Solution
The selected answer is: \(-3\).
### Explanation
1. **Understanding the Function**: The given function \( f(x) \) is defined as an integral function:
\[
f(x) = \int_{0}^{x} (u^2 - 6u + 5) \, du
\]
2. **Step-by-Step Approach**:
- **Determine \( f(x) \)**: Integrate \( u^2 - 6u + 5 \) with respect to \( u \) from \( 0 \) to \( x \):
\[
\int_{0}^{x} (u^2 - 6u + 5) \, du = \left[ \frac{u^3}{3} - 3u^2 + 5u \right]_{0}^{x} = \frac{x^3}{3} - 3x^2 + 5x
\]
So, \( f(x) = \frac{x^3}{3} - 3x^2 + 5x \).
- **Find the First Derivative \( f'(x) \)**: Differentiate \( f(x) \) with respect to \( x \):
\[
f'(x) = x^2 - 6x + 5
\]
- **Find the Second Derivative \( f''(x) \)**: Differentiate \( f'(x) \) with respect to \( x \):
\[
f''(x) = 2x - 6
\]
- **Determine Inflection Point**: An inflection point occurs where \( f''(x) = 0 \):
\[
2x - 6 = 0 \quad \Rightarrow \quad x = 3
\]
- **Find the y-coordinate**:
\[
f(3) = \frac{3^3}{3} - 3(3^2) + 5(3) = 9 - 27 + 15 = -3
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