To find the domain of the functions \( f(\lfloor |x| \rfloor) \) and \( f(\lfloor 2x + 3 \rfloor) \) given that \( f(x) \) is defined in the interval \([-3, 2]\), we will analyze each function step by step.
### Step 1: Analyze \( f(\lfloor |x| \rfloor) \)
1. **Understanding the function**: The function \( \lfloor |x| \rfloor \) represents the greatest integer less than or equal to the absolute value of \( x \).
2. **Determine the range of \( \lfloor |x| \rfloor \)**: Since \( f(x) \) is defined for \( x \) in \([-3, 2]\), we need \( \lfloor |x| \rfloor \) to also fall within this interval.
3. **Finding the values of \( |x| \)**:
- The absolute value \( |x| \) is always non-negative.
- The greatest integer function \( \lfloor |x| \rfloor \) can take values \( 0, 1, 2, \) or \( 3 \) (since \( |x| \) can be less than 3).
4. **Setting the inequalities**:
- We need \( \lfloor |x| \rfloor \) to be in the interval \([-3, 2]\).
- This means \( \lfloor |x| \rfloor \) can take values \( 0, 1, 2 \) (since it cannot be negative).
5. **Finding the corresponding \( x \)**:
- For \( \lfloor |x| \rfloor = 0 \): \( |x| \) is in \([0, 1)\) which gives \( x \in [-1, 1] \).
- For \( \lfloor |x| \rfloor = 1 \): \( |x| \) is in \([1, 2)\) which gives \( x \in [-2, -1) \cup [1, 2) \).
- For \( \lfloor |x| \rfloor = 2 \): \( |x| \) is in \([2, 3)\) which gives \( x \in [-3, -2) \cup [2, 3) \).
6. **Combining the intervals**:
- The valid intervals for \( x \) are \( [-3, -2) \cup [-2, -1) \cup [1, 2) \cup [-1, 1] \).
- Thus, the overall domain for \( f(\lfloor |x| \rfloor) \) is \( [-3, 3) \).
### Step 2: Analyze \( f(\lfloor 2x + 3 \rfloor) \)
1. **Understanding the function**: The function \( \lfloor 2x + 3 \rfloor \) represents the greatest integer less than or equal to \( 2x + 3 \).
2. **Determine the range of \( \lfloor 2x + 3 \rfloor \)**: We need \( \lfloor 2x + 3 \rfloor \) to fall within the interval \([-3, 2]\).
3. **Setting the inequalities**:
- We have \( -3 \leq 2x + 3 \leq 2 \).
- Rearranging gives:
- From \( -3 \leq 2x + 3 \): \( 2x \geq -6 \) → \( x \geq -3 \).
- From \( 2x + 3 \leq 2 \): \( 2x \leq -1 \) → \( x \leq -\frac{1}{2} \).
4. **Combining the intervals**:
- The valid interval for \( x \) is \( [-3, -\frac{1}{2}] \).
### Final Answer
Thus, the domains of the functions are:
- For \( f(\lfloor |x| \rfloor) \): \( [-3, 3) \)
- For \( f(\lfloor 2x + 3 \rfloor) \): \( [-3, -\frac{1}{2}] \)
