To solve the inequality \( x^3 - x \geq 0 \), let's follow these steps:

### Step 1: Factor the inequality
First, rewrite the inequality:

\[
x^3 - x \geq 0
\]

Factor the expression on the left-hand side:

\[
x(x^2 - 1) \geq 0
\]

Now, notice that \( x^2 - 1 \) is a difference of squares:

\[
x(x - 1)(x + 1) \geq 0
\]

### Step 2: Find the critical points
The critical points occur when any of the factors are equal to zero. Set each factor to zero:

\[
x = 0, \quad x - 1 = 0 \implies x = 1, \quad x + 1 = 0 \implies x = -1
\]

So, the critical points are \( x = -1 \), \( x = 0 \), and \( x = 1 \).

### Step 3: Analyze the sign of the expression
We now analyze the sign of the expression \( x(x - 1)(x + 1) \) in the intervals determined by the critical points: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 1) \), and \( (1, \infty) \).

1. **For \( x \in (-\infty, -1) \):**  
   Each factor is negative, so the product is negative.
   
   \[
   (-)(-)(-) = -
   \]
   Therefore, \( x(x - 1)(x + 1) < 0 \).

2. **For \( x \in (-1, 0) \):**  
   \( x + 1 \) is positive, while \( x \) and \( x - 1 \) are negative, so the product is positive.
   
   \[
   (-)(-)(+) = +
   \]
   Therefore, \( x(x - 1)(x + 1) > 0 \).

3. **For \( x \in (0, 1) \):**  
   \( x \) is positive, but \( x - 1 \) is negative and \( x + 1 \) is positive, so the product is negative.
   
   \[
   (+)(-)(+) = -
   \]
   Therefore, \( x(x - 1)(x + 1) < 0 \).

4. **For \( x \in (1, \infty) \):**  
   All the factors are positive, so the product is positive.
   
   \[
   (+)(+)(+) = +
   \]
   Therefore, \( x(x - 1)(x + 1) > 0 \).

### Step 4: Include critical points
At \( x = -1 \), \( x = 0 \), and \( x = 1 \), the expression equals zero:

\[
x(x - 1)(x + 1) = 0
\]

### Step 5: Write the solution
The inequality \( x(x - 1)(x + 1) \geq 0 \) holds when the expression is either positive or zero. From our sign analysis, this happens in the intervals:

\[
x \in [-1, 0] \cup [1, \infty)
\]

Thus, the solution to the inequality is:

\[
x \in [-1, 0] \cup [1, \infty)
\]

Would you like further details or have any questions?

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Here are five related questions to explore further:
1. How would the solution change if the inequality were strict (\( x^3 - x > 0 \))?
2. What is the solution to the inequality \( x^2 - x \geq 0 \)?
3. Can you solve a similar inequality for \( x^4 - x \geq 0 \)?
4. How would you solve \( x^3 - 2x \geq 0 \)?
5. Can you solve the equation \( x^3 - x = 0 \) for its exact roots?

**Tip:** When factoring polynomials, always check for differences of squares and common factors!

Learn how to solve the inequality x^3 - x >= 0 by factoring the cubic equation into x(x - 1)(x + 1). This solution involves key algebraic concepts such as analyzing the sign of the product and identifying critical points, suitable for students in Grades 9-11.

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