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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![### Problem Statement
**Find where the function is increasing:**
\[ y = x^2 - 3 \ln x, \quad (x > 0) \]
### Explanation
To determine where the function \( y = x^2 - 3 \ln x \) is increasing, we will need to find its first derivative and analyze its critical points. A function is increasing where its derivative is positive.
### Steps to Solve
1. **Calculate the first derivative of the function:**
\[ \frac{dy}{dx} = \frac{d}{dx} (x^2 - 3 \ln x) \]
2. **Apply the power rule and the derivative of the natural logarithm:**
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( -3 \ln x \) is \( -\frac{3}{x} \).
\[ \frac{dy}{dx} = 2x - \frac{3}{x} \]
3. **Find the critical points by setting the first derivative equal to zero:**
\[ 2x - \frac{3}{x} = 0 \]
4. **Solve for \( x \):**
\[ 2x = \frac{3}{x} \]
\[ 2x^2 = 3 \]
\[ x^2 = \frac{3}{2} \]
\[ x = \sqrt{\frac{3}{2}}, \quad x = -\sqrt{\frac{3}{2}} \]
Since \( x > 0 \), we only consider the positive root:
\[ x = \sqrt{\frac{3}{2}} \]
5. **Analyze the intervals around the critical point to determine where the derivative is positive:**
- For \( 0 < x < \sqrt{\frac{3}{2}} \):
Choose a test point \( x = 1 \):
\[ 2(1) - \frac{3}{1} = 2 - 3 = -1 \] (derivative is negative)
- For \( x > \sqrt{\frac{3}{2}} \):
Choose a test point \( x = 2 \):
\[ 2(2) - \frac{3}{2} = 4 - 1.5 = 2.5 \] (derivative](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9a5f5a5-4eae-476c-bcab-4a001ac27552%2F26e8ef59-06ac-4c1b-a572-af350461bfff%2Fbjbrg5s_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Find where the function is increasing:**
\[ y = x^2 - 3 \ln x, \quad (x > 0) \]
### Explanation
To determine where the function \( y = x^2 - 3 \ln x \) is increasing, we will need to find its first derivative and analyze its critical points. A function is increasing where its derivative is positive.
### Steps to Solve
1. **Calculate the first derivative of the function:**
\[ \frac{dy}{dx} = \frac{d}{dx} (x^2 - 3 \ln x) \]
2. **Apply the power rule and the derivative of the natural logarithm:**
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( -3 \ln x \) is \( -\frac{3}{x} \).
\[ \frac{dy}{dx} = 2x - \frac{3}{x} \]
3. **Find the critical points by setting the first derivative equal to zero:**
\[ 2x - \frac{3}{x} = 0 \]
4. **Solve for \( x \):**
\[ 2x = \frac{3}{x} \]
\[ 2x^2 = 3 \]
\[ x^2 = \frac{3}{2} \]
\[ x = \sqrt{\frac{3}{2}}, \quad x = -\sqrt{\frac{3}{2}} \]
Since \( x > 0 \), we only consider the positive root:
\[ x = \sqrt{\frac{3}{2}} \]
5. **Analyze the intervals around the critical point to determine where the derivative is positive:**
- For \( 0 < x < \sqrt{\frac{3}{2}} \):
Choose a test point \( x = 1 \):
\[ 2(1) - \frac{3}{1} = 2 - 3 = -1 \] (derivative is negative)
- For \( x > \sqrt{\frac{3}{2}} \):
Choose a test point \( x = 2 \):
\[ 2(2) - \frac{3}{2} = 4 - 1.5 = 2.5 \] (derivative
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