The interval(s) which satisfy `x^(10)-x^(7)+x^(4)-x+1 gt0` is (are)

Options

A. `-1le x le 0`
B. `0 lt x lt 1`
C. `x ge 1`
D. `-oo lt x lt oo`.

Answer

To solve the inequality \( x^{10} - x^{7} + x^{4} - x + 1 > 0 \), we will analyze the function step by step.

### Step 1: Define the function
Let \( f(x) = x^{10} - x^{7} + x^{4} - x + 1 \).

### Step 2: Factor out common terms
We can factor out \( x^{7} \) from the first two terms and \( x \) from the next two terms:
\[ f(x) = x^{7}(x^{3} - 1) + x(x^{3} - 1) + 1 \]
This can be rewritten as:
\[ f(x) = (x^{7} + x)(x^{3} - 1) + 1 \]

### Step 3: Analyze the factors
1. **Factor \( x^{7} + x \)**:
- This factor is always non-negative for all real \( x \) because both \( x^{7} \) and \( x \) are non-negative when \( x \geq 0 \) and non-positive when \( x < 0 \).

2. **Factor \( x^{3} - 1 \)**:
- This factor is zero when \( x = 1 \) and negative for \( x < 1 \) and positive for \( x > 1 \).

### Step 4: Determine the sign of \( f(x) \)
- For \( x < 1 \):
- \( x^{3} - 1 < 0 \)
- \( x^{7} + x \) can be either positive or negative depending on \( x \).
- Therefore, \( f(x) \) can be negative.

- For \( x = 1 \):
- \( f(1) = 1^{10} - 1^{7} + 1^{4} - 1 + 1 = 1 \) (which is positive).

- For \( x > 1 \):
- \( x^{3} - 1 > 0 \)
- \( x^{7} + x > 0 \)
- Therefore, \( f(x) > 0 \).

### Step 5: Conclusion
From the analysis:
- \( f(x) > 0 \) for \( x \in (1, \infty) \).
- \( f(x) = 0 \) at \( x = 1 \).
- \( f(x) < 0 \) for \( x < 1 \).

Thus, the intervals which satisfy \( f(x) > 0 \) are:
\[ \text{The solution is } (1, \infty). \]