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 the function \( f \).
Given:
- \( f''(x) = 8x^3 + 5 \)
- \( f(1) = 2 \)
- \( f'(1) = 3 \)
**Solution:**
We need to find \( f(x) \).
Begin by integrating \( f''(x) = 8x^3 + 5 \) to find \( f'(x) \).
1. Integrate \( f''(x) \):
\[
f'(x) = \int (8x^3 + 5) \, dx = 2x^4 + 5x + C
\]
where \( C \) is a constant.
2. Use the condition \( f'(1) = 3 \) to find \( C \):
\[
f'(1) = 2(1)^4 + 5(1) + C = 3
\]
\[
2 + 5 + C = 3
\]
\[
C = -4
\]
So, \( f'(x) = 2x^4 + 5x - 4 \).
3. Integrate \( f'(x) \) to find \( f(x) \):
\[
f(x) = \int (2x^4 + 5x - 4) \, dx = \frac{2}{5}x^5 + \frac{5}{2}x^2 - 4x + D
\]
where \( D \) is another constant.
4. Use the condition \( f(1) = 2 \) to find \( D \):
\[
f(1) = \frac{2}{5}(1)^5 + \frac{5}{2}(1)^2 - 4(1) + D = 2
\]
\[
\frac{2}{5} + \frac{5}{2} - 4 + D = 2
\]
Simplifying:
\[
\frac{2}{5} + \frac{25}{10} - \frac{40}{10} + D = 2
\]
\[
\frac{2}{5} + \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6252efdc-f0a9-41e1-9e81-932de2dbfbdd%2F22036da4-1f10-4ec4-90d3-9043c4b15f16%2Fxibeawe_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the function \( f \).
Given:
- \( f''(x) = 8x^3 + 5 \)
- \( f(1) = 2 \)
- \( f'(1) = 3 \)
**Solution:**
We need to find \( f(x) \).
Begin by integrating \( f''(x) = 8x^3 + 5 \) to find \( f'(x) \).
1. Integrate \( f''(x) \):
\[
f'(x) = \int (8x^3 + 5) \, dx = 2x^4 + 5x + C
\]
where \( C \) is a constant.
2. Use the condition \( f'(1) = 3 \) to find \( C \):
\[
f'(1) = 2(1)^4 + 5(1) + C = 3
\]
\[
2 + 5 + C = 3
\]
\[
C = -4
\]
So, \( f'(x) = 2x^4 + 5x - 4 \).
3. Integrate \( f'(x) \) to find \( f(x) \):
\[
f(x) = \int (2x^4 + 5x - 4) \, dx = \frac{2}{5}x^5 + \frac{5}{2}x^2 - 4x + D
\]
where \( D \) is another constant.
4. Use the condition \( f(1) = 2 \) to find \( D \):
\[
f(1) = \frac{2}{5}(1)^5 + \frac{5}{2}(1)^2 - 4(1) + D = 2
\]
\[
\frac{2}{5} + \frac{5}{2} - 4 + D = 2
\]
Simplifying:
\[
\frac{2}{5} + \frac{25}{10} - \frac{40}{10} + D = 2
\]
\[
\frac{2}{5} + \frac
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