f(x) = 5 sin(x) + 5 cos(x),    0 ≤ x ≤ 2𝜋, find the interval on which f is concave down and up

To determine the intervals where \( f(x) = 5 \sin(x) + 5 \cos(x) \) is concave up or down, we'll follow these steps:

1. **Rewrite the function**: Simplify \( f(x) \) using a trigonometric identity.
   
   Since \( f(x) = 5 \sin(x) + 5 \cos(x) \), we can factor out the 5:
   \[
   f(x) = 5(\sin(x) + \cos(x))
   \]
   
   Using the identity \( \sin(x) + \cos(x) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \), we rewrite \( f(x) \) as:
   \[
   f(x) = 5\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)
   \]

2. **Find \( f'(x) \)**: Differentiate \( f(x) \) with respect to \( x \) to find the first derivative.
   \[
   f'(x) = 5\sqrt{2} \cos\left(x + \frac{\pi}{4}\right)
   \]

3. **Find \( f''(x) \)**: Differentiate \( f'(x) \) to get the second derivative, which will help identify concavity.
   \[
   f''(x) = -5\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)
   \]

4. **Determine concavity**: Analyze \( f''(x) \) for sign changes.
   
   - \( f(x) \) is **concave up** where \( f''(x) > 0 \).
   - \( f(x) \) is **concave down** where \( f''(x) < 0 \).
   
   Since \( f''(x) = -5\sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \), the sign of \( f''(x) \) depends on \( \sin\left(x + \frac{\pi}{4}\right) \).

5. **Solve \( \sin\left(x + \frac{\pi}{4}\right) = 0 \)**: To find the points where concavity changes, solve for:
   \[
   x + \frac{\pi}{4} = n\pi, \quad n \in \mathbb{Z}
   \]
   which gives:
   \[
   x = n\pi - \frac{\pi}{4}
   \]

   For \( 0 \leq x \leq 2\pi \), we get critical points at:
   \[
   x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4}, \quad x = \frac{9\pi}{4} \quad (\text{exclude } \frac{9\pi}{4} \text{ as it is out of range})
   \]

6. **Determine intervals**:
   - For \( 0 \leq x < \frac{\pi}{4} \) and \( \frac{5\pi}{4} < x \leq 2\pi \), \( f''(x) < 0 \), so \( f(x) \) is **concave down**.
   - For \( \frac{\pi}{4} < x < \frac{5\pi}{4} \), \( f''(x) > 0 \), so \( f(x) \) is **concave up**.

### Final Answer
- **Concave down** on \( \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \).
- **Concave up** on \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \).

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Here are five related questions:
1. How would the concavity intervals change if the function were \( f(x) = 5 \sin(x) - 5 \cos(x) \)?
2. What is the significance of the concavity in relation to the function’s maximum and minimum points?
3. How could the concavity intervals be used to sketch an accurate graph of \( f(x) \)?
4. What is the role of trigonometric identities in simplifying functions like \( f(x) \)?
5. How does changing the amplitude of \( \sin(x) \) and \( \cos(x) \) affect concavity?

**Tip**: When analyzing concavity, remember that positive values of the second derivative indicate an upward curvature, while negative values indicate a downward curvature.

Learn how to determine the intervals where f(x) = 5 sin(x) + 5 cos(x) is concave up and concave down on [0, 2π]. This solution uses trigonometric identities and the second derivative test to identify concavity changes. Suitable for students in Grades 11-12.

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