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Find where the function is increasing. 2 y = x² - 3lnx, (x > 0)

Find where the function is increasing. 2 y = x² - 3lnx, (x > 0)

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.7: Applications
Problem 16EQ
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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
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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