Identify the given functions whether odd or even or neither: `f(x)={(x|x| ,, xlt=-1),([x+1]+[1-x] ,, -1lt x lt1),(-x|x| ,, xgt=1):}`

Explanation

To determine whether the given function \( f(x) \) is odd, even, or neither, we will analyze the function piece by piece based on the provided definition:

The function is defined as follows:
\[ f(x) = \begin{cases} -x|x| & \text{if } x \leq -1 \\ [x + 1] + [1 - x] & \text{if } -1 < x < 1 \\ -x|x| & \text{if } x \geq 1 \end{cases} \]

### Step 1: Evaluate \( f(-x) \)

We need to evaluate \( f(-x) \) for different intervals of \( x \).

1. **For \( x \leq -1 \)**:
- Here, \( -x \geq 1 \), so we use the third case of the function:
\[ f(-x) = -(-x)|-x| = -x^2 \]

2. **For \( -1 < x < 1 \)**:
- Here, \( -x \) will be in the range \( -1 < -x < 1 \). Thus, we use the second case of the function:
\[ f(-x) = [-x + 1] + [1 - (-x)] = [-x + 1] + [1 + x] \]

3. **For \( x \geq 1 \)**:
- Here, \( -x \leq -1 \), so we use the first case of the function:
\[ f(-x) = -(-x)|-x| = -x^2 \]

### Step 2: Compare \( f(-x) \) with \( f(x) \)

Now, we will compare \( f(-x) \) with \( f(x) \) in each case:

1. **For \( x \leq -1 \)**:
- \( f(x) = -x|x| = -x(-x) = x^2 \)
- \( f(-x) = -x^2 \)
- Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), this case is neither.

2. **For \( -1 < x < 1 \)**:
- We need to calculate \( f(x) \):
\[ f(x) = [x + 1] + [1 - x] \]
- Since \( f(-x) \) is dependent on the values of \( x \), we need to analyze it further. However, we can see that \( f(-x) \) does not equal \( f(x) \) or \( -f(x) \) in general.

3. **For \( x \geq 1 \)**:
- \( f(x) = -x|x| = -x^2 \)
- \( f(-x) = -x^2 \)
- Here, \( f(-x) = f(x) \), which indicates that it is even in this interval.

### Conclusion

Since we have established that:
- For \( x \leq -1 \), it is neither odd nor even.
- For \( -1 < x < 1 \), it is also neither odd nor even.
- For \( x \geq 1 \), it is even.

Thus, the overall function \( f(x) \) is neither odd nor even.