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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![### Finding a Function Given its Derivative
**Problem Statement:**
Find \( f \) such that \( f'(x) = 4x - 5 \) and \( f(4) = 0 \).
**Solution:**
To solve for \( f(x) \), we need to find the antiderivative (indefinite integral) of \( f'(x) = 4x - 5 \).
### Step-by-Step Solution:
1. **Integrate \( f'(x) \) to Find \( f(x) \):**
The function \( f(x) \) can be found by integrating the derivative function \( 4x - 5 \):
\[ f(x) = \int (4x - 5) \, dx \]
2. **Calculate the Indefinite Integral:**
\[ \int (4x - 5) \, dx = \int 4x \, dx - \int 5 \, dx = 4 \int x \, dx - 5 \int 1 \, dx \]
Use the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \) for \( n = 1 \):
\[ 4 \int x \, dx = 4 \cdot \frac{x^2}{2} = 2x^2 \]
The integral of a constant is:
\[ \int 5 \, dx = 5x \]
Combining these results gives us:
\[ f(x) = 2x^2 - 5x + C \]
where \( C \) is the constant of integration.
3. **Solve for the Constant \( C \) Using the Initial Condition \( f(4) = 0 \):**
Substitute \( x = 4 \) and \( f(4) = 0 \) into the equation to find \( C \):
\[ 0 = 2(4)^2 - 5(4) + C \]
\[ 0 = 2 \cdot 16 - 20 + C \]
\[ 0 = 32 - 20 + C \]
\[ 0 = 12 + C \]
\[ C = -12 \]
4. **Write the Final Function:**
\[ f(x) = 2x^2 - 5x - 12 \]
Thus, the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19d0c0ee-b2f7-4400-bddc-d84e0eebb818%2Fbf7fa740-eee4-4250-8789-c1c9aa9aaf44%2Fmnex6s6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding a Function Given its Derivative
**Problem Statement:**
Find \( f \) such that \( f'(x) = 4x - 5 \) and \( f(4) = 0 \).
**Solution:**
To solve for \( f(x) \), we need to find the antiderivative (indefinite integral) of \( f'(x) = 4x - 5 \).
### Step-by-Step Solution:
1. **Integrate \( f'(x) \) to Find \( f(x) \):**
The function \( f(x) \) can be found by integrating the derivative function \( 4x - 5 \):
\[ f(x) = \int (4x - 5) \, dx \]
2. **Calculate the Indefinite Integral:**
\[ \int (4x - 5) \, dx = \int 4x \, dx - \int 5 \, dx = 4 \int x \, dx - 5 \int 1 \, dx \]
Use the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \) for \( n = 1 \):
\[ 4 \int x \, dx = 4 \cdot \frac{x^2}{2} = 2x^2 \]
The integral of a constant is:
\[ \int 5 \, dx = 5x \]
Combining these results gives us:
\[ f(x) = 2x^2 - 5x + C \]
where \( C \) is the constant of integration.
3. **Solve for the Constant \( C \) Using the Initial Condition \( f(4) = 0 \):**
Substitute \( x = 4 \) and \( f(4) = 0 \) into the equation to find \( C \):
\[ 0 = 2(4)^2 - 5(4) + C \]
\[ 0 = 2 \cdot 16 - 20 + C \]
\[ 0 = 32 - 20 + C \]
\[ 0 = 12 + C \]
\[ C = -12 \]
4. **Write the Final Function:**
\[ f(x) = 2x^2 - 5x - 12 \]
Thus, the
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