Find the domain of the function `f(x)=3/([x/2])-5^(cos^(-1)x^(2))+( (2x+1)!)/(sqrt(x+1))` [.] denotes the greatest integar function.

Answer

To find the domain of the function \( f(x) = \frac{3}{\lfloor \frac{x}{2} \rfloor} - 5^{\cos^{-1}(x^2)} + \frac{(2x+1)!}{\sqrt{x+1}} \), we need to analyze each component of the function to determine the values of \( x \) for which the function is defined.

### Step 1: Analyze the term \( \frac{3}{\lfloor \frac{x}{2} \rfloor} \)

The term \( \lfloor \frac{x}{2} \rfloor \) (the greatest integer function) must not be equal to zero because division by zero is undefined.

- The greatest integer function \( \lfloor \frac{x}{2} \rfloor = 0 \) when \( 0 \leq \frac{x}{2} < 1 \), which translates to \( 0 \leq x < 2 \).
- Therefore, \( x \) must not be in the interval \( [0, 2) \).

### Step 2: Analyze the term \( -5^{\cos^{-1}(x^2)} \)

The term \( \cos^{-1}(x^2) \) is defined when \( x^2 \) is in the interval \( [-1, 1] \). Since \( x^2 \) is always non-negative, we only need to consider \( 0 \leq x^2 \leq 1 \).

- This gives us the condition \( -1 \leq x \leq 1 \).

### Step 3: Analyze the term \( \frac{(2x+1)!}{\sqrt{x+1}} \)

For the factorial \( (2x + 1)! \) to be defined, \( 2x + 1 \) must be a non-negative integer. Therefore, we require:

- \( 2x + 1 \geq 0 \) which simplifies to \( x \geq -\frac{1}{2} \).

Additionally, the term \( \sqrt{x+1} \) must be defined and non-zero:

- This means \( x + 1 > 0 \) or \( x > -1 \).

### Step 4: Combine the conditions

Now we have the following conditions:
1. \( x \notin [0, 2) \)
2. \( -1 \leq x \leq 1 \)
3. \( x \geq -\frac{1}{2} \)

From condition 2, we know \( x \) must be between -1 and 1. From condition 3, we have \( x \geq -\frac{1}{2} \). Therefore, the valid range for \( x \) from these two conditions is:

- \( -\frac{1}{2} \leq x \leq 1 \).

Now, we must exclude the interval \( [0, 2) \) from this range. The overlap with the interval \( [0, 2) \) means we must exclude \( [0, 1] \) from \( [-\frac{1}{2}, 1] \).

### Final Domain

Thus, the domain of the function is:

\[ x \in [-\frac{1}{2}, 0) \cup (1, \infty) \]