To draw the graph of the function \( |f(x)| = 2 + \sin x \) for \( x \) in the interval \([0, 2\pi]\), we will follow these steps:
### Step 1: Understand the function \( f(x) \)
The function given is \( |f(x)| = 2 + \sin x \). This means that \( f(x) \) can take both positive and negative values, but we will focus on the absolute value.
### Step 2: Analyze the sine function
The sine function, \( \sin x \), oscillates between -1 and 1. Therefore, \( 2 + \sin x \) will oscillate between:
- Minimum value: \( 2 - 1 = 1 \)
- Maximum value: \( 2 + 1 = 3 \)
### Step 3: Sketch the graph of \( 2 + \sin x \)
1. **Identify key points**:
- At \( x = 0 \): \( 2 + \sin(0) = 2 + 0 = 2 \)
- At \( x = \frac{\pi}{2} \): \( 2 + \sin\left(\frac{\pi}{2}\right) = 2 + 1 = 3 \)
- At \( x = \pi \): \( 2 + \sin(\pi) = 2 + 0 = 2 \)
- At \( x = \frac{3\pi}{2} \): \( 2 + \sin\left(\frac{3\pi}{2}\right) = 2 - 1 = 1 \)
- At \( x = 2\pi \): \( 2 + \sin(2\pi) = 2 + 0 = 2 \)
2. **Plot these points**:
- \( (0, 2) \)
- \( \left(\frac{\pi}{2}, 3\right) \)
- \( (\pi, 2) \)
- \( \left(\frac{3\pi}{2}, 1\right) \)
- \( (2\pi, 2) \)
3. **Draw the curve**:
- Connect the points smoothly, creating a wave-like graph that peaks at \( ( \frac{\pi}{2}, 3 ) \) and dips to \( ( \frac{3\pi}{2}, 1 ) \).
### Step 4: Reflect the graph across the x-axis
Since we are dealing with the absolute value, we need to consider the reflection of the graph of \( 2 + \sin x \) across the x-axis for any negative values. However, since \( 2 + \sin x \) never goes below 1, we will not have any negative values to reflect.
### Step 5: Final graph of \( |f(x)| \)
The graph of \( |f(x)| = 2 + \sin x \) will remain the same as the graph of \( 2 + \sin x \) since it does not dip below zero.
### Conclusion
The final graph of \( |f(x)| = 2 + \sin x \) will oscillate between 1 and 3, with the following key points:
- Maximum at \( ( \frac{\pi}{2}, 3 ) \)
- Minimum at \( ( \frac{3\pi}{2}, 1 ) \)
