How do I finish this? **Problem Statement:**Determine where the function \( f(x) = \frac{1}{2} x^2 + \cos(x) \) is concave up or concave down on the interval \([0, 2\pi]\). Also, find the x and y coordinates of all inflection points within this interval.**Calculations:**1. **First Derivative:** \[ f'(x) = x - \sin(x) \]2. **Second Derivative:** \[ f''(x) = 1 - \cos(x) \]3. **Finding Inflection Points:** Set \( f''(x) = 0 \): \[ 1 - \cos(x) = 0 \] \[ \cos(x) = 1 \] This occurs at: \[ x = 0, 2\pi \; \text{(within the interval [0, 2π])} \]4. **Test Intervals for Concavity:** - Interval \( (0, 2\pi) \) Use test points within the intervals to determine concavity: - Test point: \(\pi\) \[ f''(\pi) = 1 - \cos(\pi) = 1 - (-1) = 2 \quad \text{(concave up)} \] - Therefore, \( f(x) \) is concave up on the interval \( (0, 2\pi) \) except at the boundaries \( x = 0 \) and \( x = 2\pi \).**Graph Explanation:**The graph on the left-hand side is a plot on a coordinate system with labeled intervals and markings that illustrate the concavity of the function. Critical points such as \( 0, \pi, 2\pi \) are marked on the x-axis. The areas labeled with ‘+’ and ‘-’ indicate regions where the function is concave up and concave down, respectively:- At \( x = 0 \) and \( x = 2\pi \), the concavity changes, indicating potential inflection points.- The concavity between these points is evaluated based on \( f''(x) \).**Conclusion:**- The inflection points

Name: How do I finish this? **Problem Statement:**Determine where the functi
Uploaded: 2024-03-05T11:52:21.000Z
Channel: Bartleby
Description: How do I finish this? **Problem Statement:**Determine where the function \( f(x) = \frac{1}{2} x^2 + \cos(x) \) is concave up or concave down on the interval \([0, 2\pi]\). Also, find the x and y coordinates of all inflection points within this inter