Determine whether the following functions are even or odd or neither even nor odd:
`f(x)={(|lne^(x)|,,,xle-1),([2+x]+[2-x],,,-1ltxlt1),(e^(lnx),,,xge1):}`,where [.] is GIF

Explanation

To determine whether the given function \( f(x) \) is even, odd, or neither, we will analyze each piece of the piecewise function separately.

The function is defined as follows:

\[ f(x) = \begin{cases} | \ln e^x | & \text{if } x \leq -1 \\ [2+x] + [2-x] & \text{if } -1 < x < 1 \\ e^{\ln x} & \text{if } x \geq 1 \end{cases} \]

### Step 1: Analyze \( f(x) \) for \( x \leq -1 \)

For \( x \leq -1 \):
\[ f(x) = | \ln e^x | = | x | \quad \text{(since } \ln e^x = x\text{)} \]
Since \( x \leq -1 \), \( |x| = -x \). Therefore:
\[ f(x) = -x \]

### Step 2: Analyze \( f(x) \) for \( -1 < x < 1 \)

For \( -1 < x < 1 \):
\[ f(x) = [2+x] + [2-x] \]
Here, \( [.] \) denotes the greatest integer function (GIF). We will evaluate this expression for specific values of \( x \) in the interval \( -1 < x < 1 \).

### Step 3: Analyze \( f(x) \) for \( x \geq 1 \)

For \( x \geq 1 \):
\[ f(x) = e^{\ln x} = x \]

### Step 4: Find \( f(-x) \)

Now, we need to find \( f(-x) \) for each case:

1. **For \( x \leq -1 \)**:
\[ f(-x) = f(x) = -x \quad \text{(since } -x \geq 1\text{)} \]

2. **For \( -1 < x < 1 \)**:
\[ f(-x) = [2 - x] + [2 + x] \]

3. **For \( x \geq 1 \)**:
\[ f(-x) = f(-x) \quad \text{(since } -x \leq -1\text{)} \]

### Step 5: Compare \( f(x) \) and \( f(-x) \)

- For \( x \leq -1 \):
\[ f(x) = -x \quad \text{and} \quad f(-x) = x \]
Here, \( f(-x) = -f(x) \) implies that the function is odd.

- For \( -1 < x < 1 \):
We need to evaluate \( [2+x] + [2-x] \) and \( [2-x] + [2+x] \). After simplification, we find that \( f(-x) \) will not equal \( f(x) \) or \( -f(x) \).

- For \( x \geq 1 \):
\[ f(x) = x \quad \text{and} \quad f(-x) = -x \]
Again, \( f(-x) = -f(x) \) implies that the function is odd.

### Conclusion

Since we have established that \( f(x) \) is odd for the intervals \( x \leq -1 \) and \( x \geq 1 \), and neither even nor odd for the interval \( -1 < x < 1 \), we conclude that the function \( f(x) \) is neither even nor odd overall.