The domain of definiton of the function `f(x)=(1)/(sqrt(x^(12)-x^(9)+x^(4)-x+1))` , is

Options

A. `(-oo, -1)`
B. `(1, oo)`
C. `(-1, 1)`
D. `(-oo, oo)`

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{x^{12} - x^{9} + x^{4} - x + 1}} \), we need to ensure that the expression inside the square root is positive, as the square root of a non-positive number is not defined in the real number system.

### Step-by-Step Solution:

1. **Identify the condition for the function to be defined**:
\[ x^{12} - x^{9} + x^{4} - x + 1 > 0 \]
This inequality must hold true for \( f(x) \) to be defined.

2. **Analyze the polynomial**:
Let \( g(x) = x^{12} - x^{9} + x^{4} - x + 1 \). We need to determine where \( g(x) > 0 \).

3. **Evaluate \( g(x) \) for various intervals**:
- **For \( x < 1 \)**:
- Each term in \( g(x) \) can be analyzed:
- \( x^{12} \) is positive.
- \( -x^{9} \) is negative but less than \( x^{12} \).
- \( x^{4} \) is positive.
- \( -x \) is negative.
- The constant \( +1 \) ensures positivity.
- Thus, \( g(x) \) remains positive for \( x < 1 \).

- **For \( x = 1 \)**:
\[ g(1) = 1^{12} - 1^{9} + 1^{4} - 1 + 1 = 1 - 1 + 1 - 1 + 1 = 1 > 0 \]

- **For \( x > 1 \)**:
- Each term in \( g(x) \) remains positive:
- \( x^{12} \) is much larger than \( x^{9} \).
- \( x^{4} \) is positive.
- \( -x \) is negative but does not outweigh the positive contributions.
- The constant \( +1 \) ensures positivity.
- Thus, \( g(x) \) remains positive for \( x > 1 \).

4. **Conclusion**:
Since \( g(x) > 0 \) for all \( x \) in the real number line, the function \( f(x) \) is defined for all real numbers.

### Domain of the function:
\[ \text{Domain of } f(x) = (-\infty, \infty) \]