The exhaustive domain of `f(x)=sqrt(x^(12)-x^9+x^4-x+1)` is

Answer

To find the exhaustive domain of the function \( f(x) = \sqrt{x^{12} - x^9 + x^4 - x + 1} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

\[ x^{12} - x^9 + x^4 - x + 1 \geq 0 \]

### Step 1: Analyze the Expression
We start by rewriting the expression:

\[ x^{12} - x^9 + x^4 - x + 1 \]

### Step 2: Factor Out Common Terms
Let's factor out \( x^4 \) from the first two terms and \( -x \) from the next two terms:

\[ = x^4(x^8 + 1) - x(x^8 + 1) + 1 \]

This can be rearranged as:

\[ = (x^4 - x)(x^8 + 1) + 1 \]

### Step 3: Analyze Each Factor
1. **The term \( x^8 + 1 \)**: This term is always positive for all real \( x \) because \( x^8 \) is non-negative and adding 1 makes it strictly positive.

2. **The term \( x^4 - x \)**: We need to analyze when this term is non-negative.

### Step 4: Solve \( x^4 - x \geq 0 \)
Factoring gives:

\[ x(x^3 - 1) \geq 0 \]

This can be solved by finding the roots:

- \( x = 0 \)
- \( x^3 - 1 = 0 \) gives \( x = 1 \)

### Step 5: Test Intervals
Now we test the intervals defined by the roots \( x = 0 \) and \( x = 1 \):

1. **Interval \( (-\infty, 0) \)**: Choose \( x = -1 \):
\[ -1(-1^3 - 1) = -1(-2) = 2 \geq 0 \quad \text{(True)} \]

2. **Interval \( (0, 1) \)**: Choose \( x = 0.5 \):
\[ 0.5(0.5^3 - 1) = 0.5(0.125 - 1) = 0.5(-0.875) = -0.4375 < 0 \quad \text{(False)} \]

3. **Interval \( (1, \infty) \)**: Choose \( x = 2 \):
\[ 2(2^3 - 1) = 2(8 - 1) = 2(7) = 14 \geq 0 \quad \text{(True)} \]

### Step 6: Conclusion
The expression \( x^4 - x \) is non-negative for \( x \in (-\infty, 0] \) and \( x \in [1, \infty) \). Since \( x^8 + 1 \) is always positive, the entire expression \( x^{12} - x^9 + x^4 - x + 1 \) is non-negative in these intervals.

Thus, the exhaustive domain of \( f(x) \) is:

\[ \text{Domain} = (-\infty, 0] \cup [1, \infty) \]