To solve the problem, we need to determine the number of points where the function \( f(x) = \left\lfloor x + \frac{1}{3} \right\rfloor + \left\lfloor x + \frac{2}{3} \right\rfloor \) is discontinuous in the interval \( (0, 3) \). The greatest integer function \( \lfloor x \rfloor \) is known to be discontinuous at integer values.
### Step-by-Step Solution:
1. **Identify the components of the function**:
The function \( f(x) \) consists of two greatest integer functions:
\[
f(x) = \left\lfloor x + \frac{1}{3} \right\rfloor + \left\lfloor x + \frac{2}{3} \right\rfloor
\]
2. **Determine the intervals for discontinuity**:
The greatest integer function \( \lfloor x \rfloor \) is discontinuous at integer values. Therefore, we need to find the points where \( x + \frac{1}{3} \) and \( x + \frac{2}{3} \) are integers.
3. **Set up equations for discontinuity**:
- For \( \left\lfloor x + \frac{1}{3} \right\rfloor \):
\[
x + \frac{1}{3} = n \quad \text{(where \( n \) is an integer)}
\]
This gives:
\[
x = n - \frac{1}{3}
\]
- For \( \left\lfloor x + \frac{2}{3} \right\rfloor \):
\[
x + \frac{2}{3} = m \quad \text{(where \( m \) is an integer)}
\]
This gives:
\[
x = m - \frac{2}{3}
\]
4. **Find integer values of \( n \) and \( m \) in the interval \( (0, 3) \)**:
- For \( n = 1 \):
\[
x = 1 - \frac{1}{3} = \frac{2}{3}
\]
- For \( n = 2 \):
\[
x = 2 - \frac{1}{3} = \frac{5}{3}
\]
- For \( n = 3 \):
\[
x = 3 - \frac{1}{3} = \frac{8}{3}
\]
- For \( m = 1 \):
\[
x = 1 - \frac{2}{3} = \frac{1}{3}
\]
- For \( m = 2 \):
\[
x = 2 - \frac{2}{3} = \frac{4}{3}
\]
- For \( m = 3 \):
\[
x = 3 - \frac{2}{3} = \frac{7}{3}
\]
5. **List all points of discontinuity**:
The points of discontinuity found are:
- From \( n \): \( \frac{2}{3}, \frac{5}{3}, \frac{8}{3} \)
- From \( m \): \( \frac{1}{3}, \frac{4}{3}, \frac{7}{3} \)
Combining these, we have:
\[
\frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{7}{3}, \frac{8}{3}
\]
6. **Count the unique points**:
The unique points of discontinuity in the interval \( (0, 3) \) are:
\[
\frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{7}{3}, \frac{8}{3}
\]
This gives us a total of **6 points**.
### Final Answer:
The number of points where \( f(x) \) is discontinuous for \( x \in (0, 3) \) is **6**.
