If `f(x)=[x]`, where [x] is the greatest integer not greater than x, in (-4, 4), then `f(x)` is

Options

A. discontinuous at `x=0`, only in (-4, 4)
B. continuous at `x=0` only in (-4,4)
C. discontinuous at every integral point of (-4,4)
D. continuous at every integral point of (-4,4)

Answer

To solve the problem, we need to analyze the function \( f(x) = [x] \), where \([x]\) is the greatest integer function, defined as the greatest integer not greater than \( x \). We will determine the continuity of this function in the interval \((-4, 4)\).

### Step-by-Step Solution:

1. **Understanding the Greatest Integer Function**:
The greatest integer function \([x]\) gives the largest integer less than or equal to \( x \). For example:
- \([2.3] = 2\)
- \([-1.7] = -2\)
- \([3] = 3\)

2. **Identifying the Points of Interest**:
The function \([x]\) is piecewise constant, meaning it remains constant between integers and jumps at integer values. The integers in the interval \((-4, 4)\) are \(-4, -3, -2, -1, 0, 1, 2, 3\).

3. **Analyzing Continuity**:
A function is continuous at a point \( c \) if:
- \( f(c) \) is defined.
- The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
- The limit equals \( f(c) \).

We will check the continuity at each integer point:

- **At \( x = -4 \)**:
- \( f(-4) = -4 \)
- \( \lim_{x \to -4^-} f(x) = -4 \)
- \( \lim_{x \to -4^+} f(x) = -4 \)
- Continuous at \( x = -4 \).

- **At \( x = -3 \)**:
- \( f(-3) = -3 \)
- \( \lim_{x \to -3^-} f(x) = -3 \)
- \( \lim_{x \to -3^+} f(x) = -3 \)
- Continuous at \( x = -3 \).

- **At \( x = -2 \)**:
- \( f(-2) = -2 \)
- \( \lim_{x \to -2^-} f(x) = -2 \)
- \( \lim_{x \to -2^+} f(x) = -2 \)
- Continuous at \( x = -2 \).

- **At \( x = -1 \)**:
- \( f(-1) = -1 \)
- \( \lim_{x \to -1^-} f(x) = -1 \)
- \( \lim_{x \to -1^+} f(x) = 0 \)
- Discontinuous at \( x = -1 \).

- **At \( x = 0 \)**:
- \( f(0) = 0 \)
- \( \lim_{x \to 0^-} f(x) = -1 \)
- \( \lim_{x \to 0^+} f(x) = 0 \)
- Discontinuous at \( x = 0 \).

- **At \( x = 1 \)**:
- \( f(1) = 1 \)
- \( \lim_{x \to 1^-} f(x) = 0 \)
- \( \lim_{x \to 1^+} f(x) = 1 \)
- Discontinuous at \( x = 1 \).

- **At \( x = 2 \)**:
- \( f(2) = 2 \)
- \( \lim_{x \to 2^-} f(x) = 1 \)
- \( \lim_{x \to 2^+} f(x) = 2 \)
- Discontinuous at \( x = 2 \).

- **At \( x = 3 \)**:
- \( f(3) = 3 \)
- \( \lim_{x \to 3^-} f(x) = 2 \)
- \( \lim_{x \to 3^+} f(x) = 3 \)
- Discontinuous at \( x = 3 \).

4. **Conclusion**:
The function \( f(x) = [x] \) is discontinuous at every integer point in the interval \((-4, 4)\) except at the endpoints \(-4\), \(-3\), and \(-2\). Therefore, the correct answer is that \( f(x) \) is discontinuous at every integral point.

### Final Answer:
Discontinuous at every integral point in the interval \((-4, 4)\).