Find out whether the given function is even, odd or neither even nor odd
`" where " f(x)={{:(xabs(x),",",x le -1),([1+x]+[1-x],",",-1 lt x lt 1),(-xabs(x),",",x ge 1 ):}`
where `||` and [] represent then modulus and greater integer functions.

Answer

To determine whether the given function \( f(x) \) is even, odd, or neither, we will analyze the function based on its definition and apply the properties of even and odd functions.

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

### Step 2: Calculate \( f(-x) \)
To check if the function is even or odd, we need to find \( f(-x) \) for each piece of the function.

1. **For \( x \leq -1 \)**:
- Here, \( -x \geq 1 \), so:
\[ f(-x) = -x | -x | = -x(-x) = x^2 \]

2. **For \( -1 < x < 1 \)**:
- Here, \( -x \) will be in the range \( -1 < -x < 1 \), so:
\[ f(-x) = (1 - x) + (1 + x) = 2 \]

3. **For \( x \geq 1 \)**:
- Here, \( -x \leq -1 \), so:
\[ f(-x) = -(-x)|-x| = x|x| = x^2 \]

### Step 3: Compare \( f(-x) \) with \( f(x) \)
Now we will compare \( f(-x) \) with \( f(x) \) for each case:

1. **For \( x \leq -1 \)**:
- \( f(x) = x |x| = x(-x) = -x^2 \)
- \( f(-x) = x^2 \)
- Here, \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \).

2. **For \( -1 < x < 1 \)**:
- \( f(x) = (1+x) + (1-x) = 2 \)
- \( f(-x) = 2 \)
- Here, \( f(-x) = f(x) \), indicating that this part is even.

3. **For \( x \geq 1 \)**:
- \( f(x) = -x |x| = -x^2 \)
- \( f(-x) = x^2 \)
- Here, \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \).

### Conclusion
Since \( f(-x) \) is equal to \( f(x) \) in the interval \( -1 < x < 1 \) but not in the other intervals, the function does not satisfy the conditions for being an even or odd function overall. Therefore, the function is **neither even nor odd**.