The number of points where `f(x)= sgn (x^(2)-3x+2)+[x-3], x in [0,4],` is discontinuous is (where [.] denotes the greatest integer function )_____.

Answer

To determine the number of points where the function \( f(x) = \text{sgn}(x^2 - 3x + 2) + [x - 3] \) is discontinuous for \( x \in [0, 4] \), we will analyze the two components of the function separately: the signum function and the greatest integer function.

### Step 1: Analyze the Signum Function
The signum function \( \text{sgn}(x^2 - 3x + 2) \) is defined as follows:
- \( \text{sgn}(x) = 1 \) if \( x > 0 \)
- \( \text{sgn}(x) = 0 \) if \( x = 0 \)
- \( \text{sgn}(x) = -1 \) if \( x < 0 \)

We need to find where \( x^2 - 3x + 2 = 0 \):
\[ x^2 - 3x + 2 = (x - 1)(x - 2) = 0 \]
Thus, the roots are \( x = 1 \) and \( x = 2 \). The signum function will change its value at these points, indicating discontinuity.

### Step 2: Identify Points of Discontinuity for the Signum Function
The function \( \text{sgn}(x^2 - 3x + 2) \) will be discontinuous at:
- \( x = 1 \)
- \( x = 2 \)

### Step 3: Analyze the Greatest Integer Function
The greatest integer function \( [x - 3] \) is discontinuous at integer values of \( x - 3 \), which means it is discontinuous at:
- \( x = 3 \) (since \( [3 - 3] = [0] \))
- \( x = 4 \) (since \( [4 - 3] = [1] \))

Additionally, we need to check the integer points in the interval \( [0, 4] \):
- \( x = 0 \) gives \( [0 - 3] = [-3] \)
- \( x = 1 \) gives \( [1 - 3] = [-2] \)
- \( x = 2 \) gives \( [2 - 3] = [-1] \)
- \( x = 3 \) gives \( [3 - 3] = [0] \)
- \( x = 4 \) gives \( [4 - 3] = [1] \)

The greatest integer function is discontinuous at all integer points in the interval \( [0, 4] \), which are:
- \( x = 0 \)
- \( x = 1 \)
- \( x = 2 \)
- \( x = 3 \)
- \( x = 4 \)

### Step 4: Combine Points of Discontinuity
Now, we combine the points of discontinuity from both components:
- From the signum function: \( x = 1, 2 \)
- From the greatest integer function: \( x = 0, 1, 2, 3, 4 \)

### Step 5: List Unique Points of Discontinuity
The unique points of discontinuity are:
- \( x = 0 \)
- \( x = 1 \)
- \( x = 2 \)
- \( x = 3 \)
- \( x = 4 \)

### Conclusion
Thus, the total number of points where \( f(x) \) is discontinuous is:
\[ \text{Total points of discontinuity} = 5 \]

### Final Answer
The number of points where \( f(x) \) is discontinuous is **5**.

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