The domain of the function
`f(x) = sqrt(x^(14) - x^(11) + x^(6) - x^(3) + x^(2) + 1)` is

Options

A. `(- infty, infty)`
B. `[0, infty)`
C. `(- infty, 0]`
D. R /[ 0,1]

Answer

To find the domain of the function \( f(x) = \sqrt{x^{14} - x^{11} + x^{6} - x^{3} + x^{2} + 1} \), we need to ensure that the expression inside the square root is non-negative, since the square root of a negative number is not defined in the real number system.

### Step-by-step Solution:

1. **Identify the expression inside the square root**:
\[ g(x) = x^{14} - x^{11} + x^{6} - x^{3} + x^{2} + 1 \]

2. **Analyze the terms**:
- The term \( x^{14} \) is always non-negative for all real \( x \) because any even power of a real number is non-negative.
- The term \( -x^{11} \) can be negative or positive depending on the value of \( x \).
- The term \( x^{6} \) is also non-negative for all real \( x \).
- The term \( -x^{3} \) can be negative or positive depending on the value of \( x \).
- The term \( x^{2} \) is non-negative for all real \( x \).
- The constant term \( +1 \) is always positive.

3. **Consider the overall expression**:
- For large values of \( |x| \) (both positive and negative), the term \( x^{14} \) will dominate the expression, making \( g(x) \) positive.
- For \( x = 0 \), we have:
\[ g(0) = 0^{14} - 0^{11} + 0^{6} - 0^{3} + 0^{2} + 1 = 1 \]
which is positive.
- For \( x = 1 \):
\[ g(1) = 1^{14} - 1^{11} + 1^{6} - 1^{3} + 1^{2} + 1 = 1 - 1 + 1 - 1 + 1 + 1 = 2 \]
which is also positive.
- For \( x = -1 \):
\[ g(-1) = (-1)^{14} - (-1)^{11} + (-1)^{6} - (-1)^{3} + (-1)^{2} + 1 = 1 + 1 + 1 + 1 + 1 + 1 = 6 \]
which is positive.

4. **Conclusion on the domain**:
- Since \( g(x) \) is positive for all tested values and tends to be dominated by the positive \( x^{14} \) term for large \( |x| \), we can conclude that \( g(x) \) is non-negative for all real \( x \).
- Therefore, the function \( f(x) \) is defined for all real numbers.

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