To solve the problem, we need to analyze the given piecewise function \( f(x) \) and determine the values of \( x \) for which \( f(x) \geq 0 \).
### Step 1: Define the function based on the intervals
The function \( f(x) \) is defined as follows:
- For \( -3 < x \leq -1 \), \( f(x) = \lfloor x \rfloor \) (greatest integer function).
- For \( -1 < x < 1 \), \( f(x) = |\lfloor x \rfloor| \).
- For \( 1 \leq x < 3 \), \( f(x) = |-\lfloor x \rfloor| \).
### Step 2: Analyze each interval
#### Interval 1: \( -3 < x \leq -1 \)
In this interval:
- For \( -3 < x < -2 \), \( \lfloor x \rfloor = -3 \) so \( f(x) = -3 \) (which is negative).
- For \( -2 \leq x \leq -1 \), \( \lfloor x \rfloor = -2 \) so \( f(x) = -2 \) (which is also negative).
Thus, in the interval \( -3 < x \leq -1 \), \( f(x) < 0 \).
#### Interval 2: \( -1 < x < 1 \)
In this interval:
- For \( -1 < x < 0 \), \( \lfloor x \rfloor = -1 \) so \( f(x) = |-1| = 1 \) (which is positive).
- For \( 0 \leq x < 1 \), \( \lfloor x \rfloor = 0 \) so \( f(x) = |0| = 0 \) (which is non-negative).
Thus, in the interval \( -1 < x < 1 \), \( f(x) \geq 0 \).
#### Interval 3: \( 1 \leq x < 3 \)
In this interval:
- For \( 1 \leq x < 2 \), \( \lfloor x \rfloor = 1 \) so \( f(x) = | -1 | = 1 \) (which is positive).
- For \( 2 \leq x < 3 \), \( \lfloor x \rfloor = 2 \) so \( f(x) = | -2 | = 2 \) (which is also positive).
Thus, in the interval \( 1 \leq x < 3 \), \( f(x) \geq 0 \).
### Step 3: Combine the intervals where \( f(x) \geq 0 \)
From our analysis:
- \( f(x) < 0 \) for \( -3 < x \leq -1 \).
- \( f(x) \geq 0 \) for \( -1 < x < 1 \).
- \( f(x) \geq 0 \) for \( 1 \leq x < 3 \).
Therefore, the combined intervals where \( f(x) \geq 0 \) are:
- \( (-1, 1) \) and \( [1, 3) \).
### Final Result
Thus, the solution set for \( x \) such that \( f(x) \geq 0 \) is:
\[
(-1, 3)
\]
Since \( -1 \) is not included in the interval and \( 3 \) is also not included, the correct answer is:
\[
(-1, 3)
\]
