To identify where the function $f(x) = x + 2 \sin(x)$ is increasing or decreasing, we need to:

1. Find the derivative of $f(x)$:

   • $f'(x) = \frac{d}{dx} \left( x + 2 \sin(x) \right)$
   • This gives: $f'(x) = 1 + 2 \cos(x)$

2. Determine where $f'(x) > 0$ and $f'(x) < 0$:

   • To find when the function is increasing, we solve $f'(x) > 0$.
   • To find when the function is decreasing, we solve $f'(x) < 0$.

Let's start by analyzing the critical points:

• $f'(x) = 0 \Rightarrow 1 + 2 \cos(x) = 0 \Rightarrow \cos(x) = -\frac{1}{2}$
• This occurs at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ in the interval $(0, 2\pi)$.

Now, we check the sign of $f'(x)$ in each of the three intervals formed by these critical points:

• $(0, \frac{2\pi}{3})$
• $(\frac{2\pi}{3}, \frac{4\pi}{3})$
• $(\frac{4\pi}{3}, 2\pi)$

Next, we will test the sign of $f'(x)$ in these intervals:

1. For $0 < x < \frac{2\pi}{3}$, we choose a test point like $x = \frac{\pi}{2}$:

   • $f'(\frac{\pi}{2}) = 1 + 2 \cdot 0 = 1$ (positive, so increasing in this interval).

2. For $\frac{2\pi}{3} < x < \frac{4\pi}{3}$, we choose a test point like $x = \pi$:

   • $f'(\pi) = 1 + 2(-1) = -1$ (negative, so decreasing in this interval).

3. For $\frac{4\pi}{3} < x < 2\pi$, we choose a test point like $x = \frac{3\pi}{2}$:

   • $f'(\frac{3\pi}{2}) = 1 + 2 \cdot 0 = 1$ (positive, so increasing in this interval).

Final Answer:

• Increasing intervals: $(0, \frac{2\pi}{3}) \cup (\frac{4\pi}{3}, 2\pi)$
• Decreasing intervals: $(\frac{2\pi}{3}, \frac{4\pi}{3})$

Would you like more details or have any questions?

Here are 5 related questions to extend your understanding:

1. How would the behavior change if the function was $f(x) = x + \sin(x)$?
2. What role does the amplitude of the sine function play in increasing or decreasing intervals?
3. Can you find the points where the function has a local maximum or minimum?
4. How would you determine concavity and points of inflection for this function?
5. What happens to the increasing and decreasing intervals if the domain changes?

Tip: When solving for critical points, always check both the derivative's sign changes and any potential boundary points in the given domain!