Find the domain of the following function: `f(x)=5/([(x-1)/2])-3^(sin^(-1)x^(2))+((7x+1)!)/(sqrt(x+1))`, where [.] denotes greatest integer function.

Explanation

To find the domain of the function \( f(x) = \frac{5}{\left(\frac{x-1}{2}\right)} - 3^{\sin^{-1}(x^2)} + \frac{(7x+1)!}{\sqrt{x+1}} \), we need to analyze each component of the function for restrictions.

### Step 1: Analyze the first term \(\frac{5}{\left(\frac{x-1}{2}\right)}\)

The denominator cannot be zero:
\[ \frac{x-1}{2} \neq 0 \implies x - 1 \neq 0 \implies x \neq 1 \]

**Hint:** The denominator of a fraction cannot be zero.

### Step 2: Analyze the second term \(-3^{\sin^{-1}(x^2)}\)

The function \(\sin^{-1}(x^2)\) is defined only for \(x^2\) in the interval \([-1, 1]\). Since \(x^2 \geq 0\), we have:
\[ 0 \leq x^2 \leq 1 \implies -1 \leq x \leq 1 \]

**Hint:** The inverse sine function is defined for inputs between -1 and 1.

### Step 3: Analyze the third term \(\frac{(7x+1)!}{\sqrt{x+1}}\)

1. The factorial \((7x+1)!\) is defined only for non-negative integers:
\[ 7x + 1 \geq 0 \implies 7x \geq -1 \implies x \geq -\frac{1}{7} \]

2. The square root \(\sqrt{x+1}\) requires:
\[ x + 1 > 0 \implies x > -1 \]

**Hint:** Factorials are defined for non-negative integers, and square roots are defined for non-negative values.

### Step 4: Combine the conditions

From the analysis, we have the following conditions:
1. \(x \neq 1\)
2. \(-1 < x \leq 1\)
3. \(x \geq -\frac{1}{7}\)

Now, we need to combine these intervals:
- The condition \(-1 < x \leq 1\) gives us the interval \((-1, 1]\).
- The condition \(x \geq -\frac{1}{7}\) gives us the interval \([- \frac{1}{7}, \infty)\).

### Step 5: Find the intersection of the intervals

The intersection of the intervals \((-1, 1]\) and \([- \frac{1}{7}, \infty)\) is:
\[ [-\frac{1}{7}, 1] \]
However, since \(x \neq 1\), we exclude \(1\):
\[ [-\frac{1}{7}, 1) \]

### Final Domain

Thus, the domain of the function \(f(x)\) is:
\[ \boxed{[-\frac{1}{7}, 1)} \]