Find the domain and range of the following function:
`f(x)=tan^(-1)(sqrt([x]+[-x]))+sqrt(2-|x|)+1/(x^(2))` (where [ ] denotes greatest integer function)

Explanation

To find the domain and range of the function \( f(x) = \tan^{-1}(\sqrt{[x] + [-x]}) + \sqrt{2 - |x|} + \frac{1}{x^2} \), where \([ ]\) denotes the greatest integer function, we will analyze each component of the function step by step.

### Step 1: Analyze the term \( \tan^{-1}(\sqrt{[x] + [-x]}) \)

The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\), and \([-x]\) gives the largest integer less than or equal to \(-x\).

- For any real number \(x\), \([x] + [-x]\) is always equal to \(-1\) when \(x\) is not an integer (since \([x] = n\) and \([-x] = -n-1\) for \(n \in \mathbb{Z}\)).
- When \(x\) is an integer, \([x] + [-x] = 0\).

Thus, we have:
- If \(x\) is an integer, \(\sqrt{[x] + [-x]} = \sqrt{0} = 0\).
- If \(x\) is not an integer, \(\sqrt{[x] + [-x]} = \sqrt{-1}\), which is not defined in the real numbers.

### Step 2: Determine the domain from the first term

From the analysis above, the first term \( \tan^{-1}(\sqrt{[x] + [-x]}) \) is defined only when \(x\) is an integer. Therefore, we need to restrict \(x\) to integer values.

### Step 3: Analyze the term \( \sqrt{2 - |x|} \)

For the square root to be defined, the expression inside must be non-negative:
\[ 2 - |x| \geq 0 \implies |x| \leq 2. \]
This means:
\[ -2 \leq x \leq 2. \]
Since \(x\) must also be an integer, the possible integer values are \( -2, -1, 0, 1, 2 \).

### Step 4: Analyze the term \( \frac{1}{x^2} \)

The term \( \frac{1}{x^2} \) is defined for all \(x\) except \(x = 0\). Therefore, we must exclude \(x = 0\) from our domain.

### Step 5: Combine the results for the domain

Taking into account all the restrictions:
- The integer values in the interval \([-2, 2]\) are \(-2, -1, 0, 1, 2\).
- Excluding \(0\) (due to the term \( \frac{1}{x^2} \)), the valid integer values are:
\[ \text{Domain: } \{-2, -1, 1, 2\}. \]

### Step 6: Calculate the range of the function

Now we will evaluate \(f(x)\) at the points in the domain:

1. **At \(x = -2\)**:
\[ f(-2) = \tan^{-1}(0) + \sqrt{2 - 2} + \frac{1}{(-2)^2} = 0 + 0 + \frac{1}{4} = \frac{1}{4}. \]

2. **At \(x = -1\)**:
\[ f(-1) = \tan^{-1}(0) + \sqrt{2 - 1} + \frac{1}{(-1)^2} = 0 + 1 + 1 = 2. \]

3. **At \(x = 1\)**:
\[ f(1) = \tan^{-1}(0) + \sqrt{2 - 1} + \frac{1}{(1)^2} = 0 + 1 + 1 = 2. \]

4. **At \(x = 2\)**:
\[ f(2) = \tan^{-1}(0) + \sqrt{2 - 2} + \frac{1}{(2)^2} = 0 + 0 + \frac{1}{4} = \frac{1}{4}. \]

### Step 7: Compile the range

From the evaluations:
- \(f(-2) = \frac{1}{4}\)
- \(f(-1) = 2\)
- \(f(1) = 2\)
- \(f(2) = \frac{1}{4}\)

Thus, the range of \(f(x)\) is:
\[ \text{Range: } \left\{ \frac{1}{4}, 2 \right\}. \]

### Final Answer

- **Domain:** \(\{-2, -1, 1, 2\}\)
- **Range:** \(\left\{ \frac{1}{4}, 2 \right\}\)