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 5: Antiderivatives and Initial Conditions**
**Objective:**
Determine the function \( f(x) \) using antiderivatives, given that its derivative is \( f'(x) = \frac{1}{\sqrt{x}} \) and it satisfies the condition \( f(9) = 8 \).
**Solution Outline:**
1. **Find the Antiderivative:**
- Start by integrating the derivative \( f'(x) = \frac{1}{\sqrt{x}} \).
- Rewrite \( \frac{1}{\sqrt{x}} \) as \( x^{-\frac{1}{2}} \).
- Integrate \( x^{-\frac{1}{2}} \) to find \( f(x) \):
\[
\int x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}} + C
\]
- Thus, the general solution is \( f(x) = 2\sqrt{x} + C \).
2. **Apply the Initial Condition:**
- Use the given condition \( f(9) = 8 \) to solve for \( C \).
- Substitute \( x = 9 \) into the general solution:
\[
8 = 2\sqrt{9} + C
\]
- Simplify and solve for \( C \):
\[
8 = 2 \times 3 + C \Rightarrow 8 = 6 + C \Rightarrow C = 2
\]
3. **Final Solution:**
- Substitute \( C = 2 \) back into the general solution:
\[
f(x) = 2\sqrt{x} + 2
\]
**Conclusion:**
The function \( f(x) = 2\sqrt{x} + 2 \) satisfies both the derivative condition and the initial condition given.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F042a0375-4506-4632-a5f0-d842c57b4108%2F4557a13d-5a60-498c-bf74-12fa65c21522%2F3voy5vn_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 5: Antiderivatives and Initial Conditions**
**Objective:**
Determine the function \( f(x) \) using antiderivatives, given that its derivative is \( f'(x) = \frac{1}{\sqrt{x}} \) and it satisfies the condition \( f(9) = 8 \).
**Solution Outline:**
1. **Find the Antiderivative:**
- Start by integrating the derivative \( f'(x) = \frac{1}{\sqrt{x}} \).
- Rewrite \( \frac{1}{\sqrt{x}} \) as \( x^{-\frac{1}{2}} \).
- Integrate \( x^{-\frac{1}{2}} \) to find \( f(x) \):
\[
\int x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}} + C
\]
- Thus, the general solution is \( f(x) = 2\sqrt{x} + C \).
2. **Apply the Initial Condition:**
- Use the given condition \( f(9) = 8 \) to solve for \( C \).
- Substitute \( x = 9 \) into the general solution:
\[
8 = 2\sqrt{9} + C
\]
- Simplify and solve for \( C \):
\[
8 = 2 \times 3 + C \Rightarrow 8 = 6 + C \Rightarrow C = 2
\]
3. **Final Solution:**
- Substitute \( C = 2 \) back into the general solution:
\[
f(x) = 2\sqrt{x} + 2
\]
**Conclusion:**
The function \( f(x) = 2\sqrt{x} + 2 \) satisfies both the derivative condition and the initial condition given.
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