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 term in the function separately.

### 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 zero because it is in the denominator.

1. \(\lfloor \frac{x}{2} \rfloor = 0\) when \(0 \leq x < 2\).
2. Therefore, we need \(x < 0\) or \(x \geq 2\).

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

The term \(\cos^{-1}(x^2)\) is defined for \(x^2\) in the range \([-1, 1]\). Since \(x^2 \geq 0\), we have:

1. \(0 \leq x^2 \leq 1\) implies \(-1 \leq x \leq 1\).

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

1. The factorial \((2x+1)!\) is defined for \(2x + 1 \geq 0\), which gives \(x \geq -\frac{1}{2}\).
2. The term \(\sqrt{x+1}\) requires \(x + 1 > 0\), which gives \(x > -1\).

### Step 4: Combine the conditions

Now we have the following conditions from the three terms:

1. From \(\frac{3}{\lfloor \frac{x}{2} \rfloor}\): \(x < 0\) or \(x \geq 2\).
2. From \(-5^{\cos^{-1}(x^2)}\): \(-1 \leq x \leq 1\).
3. From \(\frac{(2x+1)!}{\sqrt{x+1}}\): \(x \geq -\frac{1}{2}\) and \(x > -1\).

### Step 5: Find the intersection of all conditions

- For \(x < 0\) or \(x \geq 2\):
- If \(x < 0\), the intersection with \([-1, 1]\) gives \([-1, 0)\).
- If \(x \geq 2\), this does not intersect with \([-1, 1]\).

- For \(x \geq -\frac{1}{2}\) and \(x > -1\):
- The relevant range is \([-1/2, 1]\).

### Step 6: Final Domain

Combining these intervals, we find that the valid values for \(x\) are:

- From the first condition: \([-1, 0)\)
- From the second condition: \([-1/2, 1]\)

The intersection of these intervals gives us:

\[ [-\frac{1}{2}, 0) \]

Thus, the domain of the function \(f(x)\) is:

\[ \text{Domain of } f(x) = [-\frac{1}{2}, 0) \]