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:
Find \( f \) such that \( f'(x) = 8x^2 + 3x - 4 \) and \( f(0) = 9 \).
**Solution:**
To find \( f(x) \) given its derivative, we need to integrate the function \( f'(x) \).
1. Write the given derivative:
\[ f'(x) = 8x^2 + 3x - 4 \]
2. Integrate with respect to \( x \):
\[
f(x) = \int (8x^2 + 3x - 4) \, dx
\]
3. Perform the integration term by term:
\[
f(x) = \int 8x^2 \, dx + \int 3x \, dx - \int 4 \, dx
\]
4. Evaluate each integral:
\[
\int 8x^2 \, dx = 8 \cdot \frac{x^3}{3} = \frac{8}{3} x^3
\]
\[
\int 3x \, dx = 3 \cdot \frac{x^2}{2} = \frac{3}{2} x^2
\]
\[
\int 4 \, dx = 4x
\]
Combine these results:
\[
f(x) = \frac{8}{3} x^3 + \frac{3}{2} x^2 - 4x + C
\]
5. Use the initial condition \( f(0) = 9 \) to solve for \( C \):
\[
f(0) = \frac{8}{3}(0)^3 + \frac{3}{2}(0)^2 - 4(0) + C = 9
\]
\[
C = 9
\]
Therefore, the function \( f(x) \) is:
\[
f(x) = \frac{8}{3} x^3 + \frac{3}{2} x^2 - 4x + 9
\]
**Final Answer:**
\[
f(x) = \frac{8}{3} x^3 + \frac{3}{2} x^2 - 4x + 9
\]
### Explanation:
This process involved integrating the given derivative and then using the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19d0c0ee-b2f7-4400-bddc-d84e0eebb818%2Fe11da377-f804-41b5-b0ee-f0fad5ca1862%2Fhsoinw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Find \( f \) such that \( f'(x) = 8x^2 + 3x - 4 \) and \( f(0) = 9 \).
**Solution:**
To find \( f(x) \) given its derivative, we need to integrate the function \( f'(x) \).
1. Write the given derivative:
\[ f'(x) = 8x^2 + 3x - 4 \]
2. Integrate with respect to \( x \):
\[
f(x) = \int (8x^2 + 3x - 4) \, dx
\]
3. Perform the integration term by term:
\[
f(x) = \int 8x^2 \, dx + \int 3x \, dx - \int 4 \, dx
\]
4. Evaluate each integral:
\[
\int 8x^2 \, dx = 8 \cdot \frac{x^3}{3} = \frac{8}{3} x^3
\]
\[
\int 3x \, dx = 3 \cdot \frac{x^2}{2} = \frac{3}{2} x^2
\]
\[
\int 4 \, dx = 4x
\]
Combine these results:
\[
f(x) = \frac{8}{3} x^3 + \frac{3}{2} x^2 - 4x + C
\]
5. Use the initial condition \( f(0) = 9 \) to solve for \( C \):
\[
f(0) = \frac{8}{3}(0)^3 + \frac{3}{2}(0)^2 - 4(0) + C = 9
\]
\[
C = 9
\]
Therefore, the function \( f(x) \) is:
\[
f(x) = \frac{8}{3} x^3 + \frac{3}{2} x^2 - 4x + 9
\]
**Final Answer:**
\[
f(x) = \frac{8}{3} x^3 + \frac{3}{2} x^2 - 4x + 9
\]
### Explanation:
This process involved integrating the given derivative and then using the
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