The set of values for which `x^(3)+1 ge x^(2)+x` is

Options

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

Answer

To solve the inequality \( x^3 + 1 \geq x^2 + x \), we can follow these steps:

### Step 1: Rearrange the inequality
First, we will rearrange the inequality to bring all terms to one side:
\[ x^3 + 1 - x^2 - x \geq 0 \]
This simplifies to:
\[ x^3 - x^2 - x + 1 \geq 0 \]

### Step 2: Factor the polynomial
Next, we will try to factor the polynomial \( x^3 - x^2 - x + 1 \). We can use the Rational Root Theorem to test possible rational roots. Testing \( x = 1 \):
\[ 1^3 - 1^2 - 1 + 1 = 1 - 1 - 1 + 1 = 0 \]
So, \( x = 1 \) is a root. We can factor \( x - 1 \) out of the polynomial.

Using synthetic division to divide \( x^3 - x^2 - x + 1 \) by \( x - 1 \):
\[ \begin{array}{r|rrrr} 1 & 1 & -1 & -1 & 1 \\ & & 1 & 0 & -1 \\ \hline & 1 & 0 & -1 & 0 \\ \end{array} \]
This gives us:
\[ x^3 - x^2 - x + 1 = (x - 1)(x^2 - 1) = (x - 1)(x - 1)(x + 1) = (x - 1)^2(x + 1) \]

### Step 3: Solve the factored inequality
Now we have:
\[ (x - 1)^2(x + 1) \geq 0 \]
The critical points are \( x = 1 \) and \( x = -1 \).

### Step 4: Test intervals
We will test the intervals determined by the critical points \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \):

1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[ (-2 - 1)^2(-2 + 1) = 1 \cdot (-1) = -1 \quad (\text{not } \geq 0) \]

2. **Interval \( (-1, 1) \)**: Choose \( x = 0 \):
\[ (0 - 1)^2(0 + 1) = 1 \cdot 1 = 1 \quad (\text{is } \geq 0) \]

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

### Step 5: Include critical points
At \( x = -1 \):
\[ (-1 - 1)^2(-1 + 1) = 0 \quad (\text{is } \geq 0) \]
At \( x = 1 \):
\[ (1 - 1)^2(1 + 1) = 0 \quad (\text{is } \geq 0) \]

### Conclusion
The solution to the inequality \( x^3 + 1 \geq x^2 + x \) is:
\[ x \in [-1, 1] \cup (1, \infty) \]
In interval notation, this can be expressed as:
\[ x \geq -1 \]