Find the inflection points for the function given. f(x) = 6x + 2 minus sin(x), 0 less than x less than 3pi

To find the inflection points of the function \( f(x) = 6x + 2 - \sin(x) \) on the interval \( 0 < x < 3\pi \), we need to follow these steps:

1. **Find the second derivative of \( f(x) \).**

   The first derivative \( f'(x) \) is given by:

   \[
   f'(x) = \frac{d}{dx}(6x + 2 - \sin(x)) = 6 - \cos(x)
   \]

   The second derivative \( f''(x) \) is:

   \[
   f''(x) = \frac{d}{dx}(6 - \cos(x)) = \sin(x)
   \]

2. **Determine where the second derivative is zero or undefined.**

   Set the second derivative equal to zero to find potential inflection points:

   \[
   \sin(x) = 0
   \]

   The solutions for \( \sin(x) = 0 \) within the interval \( 0 < x < 3\pi \) are:

   \[
   x = \pi, \quad x = 2\pi
   \]

3. **Verify that these points are inflection points.**

   To confirm that \( x = \pi \) and \( x = 2\pi \) are inflection points, check if the concavity changes around these points. We do this by evaluating \( f''(x) \) just before and after each point.

   - For \( x = \pi \):
     - \( f''(x) < 0 \) for \( x \) slightly less than \( \pi \) (e.g., \( x = \pi - 0.1 \))
     - \( f''(x) > 0 \) for \( x \) slightly greater than \( \pi \) (e.g., \( x = \pi + 0.1 \))
     - Thus, there is a change in concavity at \( x = \pi \).

   - For \( x = 2\pi \):
     - \( f''(x) > 0 \) for \( x \) slightly less than \( 2\pi \) (e.g., \( x = 2\pi - 0.1 \))
     - \( f''(x) < 0 \) for \( x \) slightly greater than \( 2\pi \) (e.g., \( x = 2\pi + 0.1 \))
     - Thus, there is a change in concavity at \( x = 2\pi \).

Therefore, the inflection points of the function \( f(x) = 6x + 2 - \sin(x) \) on the interval \( 0 < x < 3\pi \) are at \( x = \pi \) and \( x = 2\pi \).

If you have any questions or want more details about this solution, let me know!

Here are some related questions you might be interested in:

1. How do you determine the concavity of a function?
2. What is the significance of inflection points in a graph?
3. How do you find the first derivative of a trigonometric function?
4. What is the difference between critical points and inflection points?
5. Can inflection points occur at endpoints of an interval?

**Tip:** When looking for inflection points, always check the sign of the second derivative on either side of the critical point to confirm a change in concavity.

Learn how to find inflection points for the function f(x) = 6x + 2 - sin(x) on the interval 0 < x < 3π. This problem involves calculus concepts such as derivatives and inflection points, and includes step-by-step solutions and verifications. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation