Find the domain? `f(x)=sqrt((x^2+4x)C_(2x^2+3))`

Answer

To find the domain of the function \( f(x) = \sqrt{(x^2 + 4x) C_{(2x^2 + 3)}} \), we need to ensure that the expression inside the square root is non-negative and that the parameters for the combination \( C \) are valid.

### Step-by-Step Solution:

1. **Understanding the Combination**:
The expression \( C_{(n, r)} \) is defined as \( \frac{n!}{r!(n-r)!} \) and is valid when \( n \geq r \) and both \( n \) and \( r \) are non-negative integers. Here, we have:
- \( n = x^2 + 4x \)
- \( r = 2x^2 + 3 \)

2. **Setting Up the Inequality**:
For the combination to be defined, we need:
\[ x^2 + 4x \geq 2x^2 + 3 \]

3. **Rearranging the Inequality**:
Rearranging gives:
\[ x^2 - 4x + 3 \leq 0 \]
This can be rewritten as:
\[ -x^2 + 4x - 3 \geq 0 \]

4. **Factoring the Quadratic**:
We can factor the quadratic:
\[ x^2 - 4x + 3 = (x - 1)(x - 3) \leq 0 \]

5. **Finding the Critical Points**:
The critical points from the factors are \( x = 1 \) and \( x = 3 \).

6. **Testing Intervals**:
We test the intervals determined by these critical points:
- For \( x < 1 \): Choose \( x = 0 \), \( (0 - 1)(0 - 3) = 3 > 0 \)
- For \( 1 < x < 3 \): Choose \( x = 2 \), \( (2 - 1)(2 - 3) = -1 < 0 \)
- For \( x > 3 \): Choose \( x = 4 \), \( (4 - 1)(4 - 3) = 3 > 0 \)

The expression \( (x - 1)(x - 3) \leq 0 \) holds for \( 1 \leq x \leq 3 \).

7. **Considering Non-Negative Integers**:
Since \( n \) and \( r \) must be non-negative integers, we check the integers in the interval \( [1, 3] \):
- The integers are \( 1, 2, 3 \).

8. **Final Domain**:
Therefore, the domain of the function \( f(x) \) is:
\[ \{1, 2, 3\} \]