The number of points of discontinuity for f(x) = sgn(sin x), `x in [0,4pi] ` is ___________.

Answer

To determine the number of points of discontinuity for the function \( f(x) = \text{sgn}(\sin x) \) over the interval \( [0, 4\pi] \), we can follow these steps:

### Step 1: Understand the Signum Function
The signum function, denoted as \( \text{sgn}(x) \), is defined as:
- \( \text{sgn}(x) = 1 \) if \( x > 0 \)
- \( \text{sgn}(x) = 0 \) if \( x = 0 \)
- \( \text{sgn}(x) = -1 \) if \( x < 0 \)

### Step 2: Identify Points of Discontinuity
The function \( f(x) = \text{sgn}(\sin x) \) will be discontinuous at points where \( \sin x = 0 \) because at these points, the value of the signum function changes.

### Step 3: Find Where \( \sin x = 0 \)
The sine function is equal to zero at integer multiples of \( \pi \):
\[ \sin x = 0 \quad \Rightarrow \quad x = n\pi \quad \text{for } n \in \mathbb{Z} \]
Within the interval \( [0, 4\pi] \), we can find the specific values:
- \( n = 0 \): \( x = 0 \)
- \( n = 1 \): \( x = \pi \)
- \( n = 2 \): \( x = 2\pi \)
- \( n = 3 \): \( x = 3\pi \)
- \( n = 4 \): \( x = 4\pi \)

### Step 4: Count the Points of Discontinuity
The points where \( \sin x = 0 \) in the interval \( [0, 4\pi] \) are:
- \( 0 \)
- \( \pi \)
- \( 2\pi \)
- \( 3\pi \)
- \( 4\pi \)

Thus, there are **5 points** of discontinuity.

### Final Answer
The number of points of discontinuity for \( f(x) = \text{sgn}(\sin x) \) in the interval \( [0, 4\pi] \) is **5**.

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