Find the maximum and minimum values of the function `f(x) = x+ sin 2x, (0 lt x lt pi)`.

Answer

To find the maximum and minimum values of the function \( f(x) = x + \sin(2x) \) in the interval \( (0, \pi) \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sin(2x)) = 1 + 2\cos(2x) \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ 1 + 2\cos(2x) = 0 \]
This simplifies to:
\[ 2\cos(2x) = -1 \quad \Rightarrow \quad \cos(2x) = -\frac{1}{2} \]

### Step 3: Solve for \( x \)
The cosine function equals \(-\frac{1}{2}\) at specific angles:
\[ 2x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad 2x = \frac{4\pi}{3} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]
Dividing by 2 gives us:
\[ x = \frac{\pi}{3} + k\pi \quad \text{or} \quad x = \frac{2\pi}{3} + k\pi \]
Considering the interval \( (0, \pi) \), we take:
\[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{2\pi}{3} \]

### Step 4: Evaluate the second derivative
To determine whether these critical points are maxima or minima, we find the second derivative:
\[ f''(x) = \frac{d}{dx}(f'(x)) = \frac{d}{dx}(1 + 2\cos(2x)) = -4\sin(2x) \]

### Step 5: Test the critical points
1. For \( x = \frac{\pi}{3} \):
\[ f''\left(\frac{\pi}{3}\right) = -4\sin\left(2 \cdot \frac{\pi}{3}\right) = -4\sin\left(\frac{2\pi}{3}\right) = -4 \cdot \frac{\sqrt{3}}{2} = -2\sqrt{3} < 0 \]
This indicates a local maximum.

2. For \( x = \frac{2\pi}{3} \):
\[ f''\left(\frac{2\pi}{3}\right) = -4\sin\left(2 \cdot \frac{2\pi}{3}\right) = -4\sin\left(\frac{4\pi}{3}\right) = -4 \cdot \left(-\frac{\sqrt{3}}{2}\right) = 2\sqrt{3} > 0 \]
This indicates a local minimum.

### Step 6: Evaluate the function at critical points and endpoints
Now we evaluate \( f(x) \) at the critical points and the endpoints of the interval:
1. At \( x = \frac{\pi}{3} \):
\[ f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \]

2. At \( x = \frac{2\pi}{3} \):
\[ f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} + \sin\left(\frac{4\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \]

3. At the endpoints \( x = 0 \) and \( x = \pi \):
\[ f(0) = 0 + \sin(0) = 0 \]
\[ f(\pi) = \pi + \sin(2\pi) = \pi \]

### Step 7: Compare values
Now we compare the values:
- \( f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \)
- \( f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \)
- \( f(0) = 0 \)
- \( f(\pi) = \pi \)

### Conclusion
The maximum value occurs at \( x = \frac{\pi}{3} \) and the minimum value occurs at \( x = \frac{2\pi}{3} \).

### Final Answer
- Maximum value: \( f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \)
- Minimum value: \( f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \)