Find the maximum and minimum values of the function `f(x) = sin x + cos 2x`.

Answer

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

### Step 1: Differentiate the function
We start by finding the first derivative of the function \( f(x) \).

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

Using the derivatives of sine and cosine, we have:

\[ f'(x) = \cos x - 2\sin 2x \]

### Step 2: Set the derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ \cos x - 2\sin 2x = 0 \]

Using the identity \( \sin 2x = 2\sin x \cos x \), we can rewrite the equation:

\[ \cos x - 4\sin x \cos x = 0 \]

Factoring out \( \cos x \):

\[ \cos x(1 - 4\sin x) = 0 \]

### Step 3: Solve for critical points
This gives us two cases to consider:

1. \( \cos x = 0 \)
2. \( 1 - 4\sin x = 0 \)

For \( \cos x = 0 \):

\[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

For \( 1 - 4\sin x = 0 \):

\[ \sin x = \frac{1}{4} \]

### Step 4: Find the second derivative
Next, we find the second derivative to determine the nature of the critical points:

\[ f''(x) = -\sin x - 4\cos 2x \]

### Step 5: Evaluate critical points
1. **At \( x = \frac{\pi}{2} \)**:

\[ f''\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) - 4\cos(\pi) = -1 + 4 = 3 \]

Since \( f''\left(\frac{\pi}{2}\right) > 0 \), this point is a local minimum.

Calculating \( f\left(\frac{\pi}{2}\right) \):

\[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos(\pi) = 1 - 1 = 0 \]

2. **At \( \sin x = \frac{1}{4} \)**:

Let \( x = \sin^{-1}\left(\frac{1}{4}\right) \). We need to evaluate \( f''(x) \):

Calculating \( f''(x) \):

\[ f''(x) = -\sin\left(\sin^{-1}\left(\frac{1}{4}\right)\right) - 4\cos\left(2\sin^{-1}\left(\frac{1}{4}\right)\right) \]

Using \( \sin\left(\sin^{-1}\left(\frac{1}{4}\right)\right) = \frac{1}{4} \) and \( \cos(2\theta) = 1 - 2\sin^2(\theta) \):

\[ \cos\left(2\sin^{-1}\left(\frac{1}{4}\right)\right) = 1 - 2\left(\frac{1}{4}\right)^2 = 1 - 2 \cdot \frac{1}{16} = 1 - \frac{1}{8} = \frac{7}{8} \]

Thus,

\[ f''(x) = -\frac{1}{4} - 4 \cdot \frac{7}{8} = -\frac{1}{4} - \frac{28}{8} = -\frac{1}{4} - \frac{7}{2} = -\frac{1 + 14}{4} = -\frac{15}{4} \]

Since \( f''(x) < 0 \), this point is a local maximum.

Calculating \( f(x) \):

\[ f\left(\sin^{-1}\left(\frac{1}{4}\right)\right) = \frac{1}{4} + \cos(2\sin^{-1}\left(\frac{1}{4}\right)) = \frac{1}{4} + \frac{7}{8} = \frac{2}{8} + \frac{7}{8} = \frac{9}{8} \]

### Conclusion
The maximum value of the function is \( \frac{9}{8} \) and the minimum value is \( 0 \).