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 equal 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 solutions to \( \cos(2x) = -\frac{1}{2} \) are:
\[ 2x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad 2x = \frac{4\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
Dividing by 2 gives:
\[ x = \frac{\pi}{3} + k\pi \quad \text{or} \quad x = \frac{2\pi}{3} + k\pi \]
Considering the interval \( (0, \pi) \):
- For \( k = 0 \): \( x = \frac{\pi}{3} \) and \( x = \frac{2\pi}{3} \)

### Step 4: Evaluate the function at critical points and endpoints
Next, we evaluate \( f(x) \) at the critical points and the endpoints of the interval:
- At \( x = \frac{\pi}{3} \):
\[ f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(2 \cdot \frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \]

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

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

### Step 5: Compare the values
Now we compare the values obtained:
- \( f(0) = 0 \)
- \( f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \)
- \( f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \)
- \( f(\pi) = \pi \)

### Step 6: Determine maximum and minimum
To determine which is maximum and minimum, we need to evaluate the numerical values of \( f\left(\frac{\pi}{3}\right) \) and \( f\left(\frac{2\pi}{3}\right) \):
- \( f\left(\frac{\pi}{3}\right) \approx 1.047 + 0.866 \approx 1.913 \)
- \( f\left(\frac{2\pi}{3}\right) \approx 2.094 - 0.866 \approx 1.228 \)

Thus, we have:
- Minimum value: \( f(0) = 0 \)
- Maximum value: \( f(\pi) = \pi \)

### Final Result
The maximum value of \( f(x) \) is \( \pi \) and the minimum value is \( 0 \).