The domain of `f(x)=(x^(2)+x+1)^(-3//2)` is :

Options

A. `(-oo, oo)`
B. `(0, oo)`
C. `((3)/(2),oo)`
D. none of these

Answer

To find the domain of the function \( f(x) = (x^2 + x + 1)^{-\frac{3}{2}} \), we need to determine for which values of \( x \) the function is defined.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is given as \( f(x) = (x^2 + x + 1)^{-\frac{3}{2}} \). The expression inside the parentheses, \( x^2 + x + 1 \), is raised to the power of \(-\frac{3}{2}\).

2. **Identifying Conditions for the Domain**:
- The expression \( (x^2 + x + 1)^{-\frac{3}{2}} \) is defined if \( x^2 + x + 1 > 0 \) because we cannot have a negative number raised to a fractional power that results in a real number.
- Additionally, since the expression is in the denominator, \( x^2 + x + 1 \) must not equal zero.

3. **Finding the Roots**:
To analyze when \( x^2 + x + 1 = 0 \), we can use the discriminant method. The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( D = b^2 - 4ac \).
- Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \).
- Calculating the discriminant:
\[ D = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative (\( D < 0 \)), the quadratic equation \( x^2 + x + 1 = 0 \) has no real roots.

4. **Analyzing the Quadratic**:
A quadratic with a negative discriminant does not cross the x-axis, which means it is always positive. Since \( a = 1 \) (the coefficient of \( x^2 \)) is positive, the parabola opens upwards.

5. **Conclusion about the Domain**:
Since \( x^2 + x + 1 \) is always greater than zero for all real \( x \), the function \( f(x) \) is defined for all real numbers. Therefore, the domain of \( f(x) \) is:
\[ \text{Domain of } f(x) = (-\infty, \infty) \]

### Final Answer:
The domain of \( f(x) = (x^2 + x + 1)^{-\frac{3}{2}} \) is \( (-\infty, \infty) \).