To solve the inequality \(x^3 - 36x \leq 0\), we can follow these steps:

### Step 1: Factor the expression
The expression \(x^3 - 36x\) can be factored as follows:
\[
x^3 - 36x = x(x^2 - 36)
\]
Notice that \(x^2 - 36\) is a difference of squares, which can be factored further:
\[
x^2 - 36 = (x - 6)(x + 6)
\]
So the complete factorization is:
\[
x(x - 6)(x + 6) \leq 0
\]

### Step 2: Determine the critical points
The critical points occur where each factor is equal to zero:
\[
x = 0, \quad x - 6 = 0 \quad \Rightarrow \quad x = 6, \quad x + 6 = 0 \quad \Rightarrow \quad x = -6
\]
So, the critical points are \(x = -6\), \(x = 0\), and \(x = 6\).

### Step 3: Analyze the sign of the expression in each interval
The real line is divided into the intervals by the critical points: \((-\infty, -6)\), \((-6, 0)\), \((0, 6)\), and \((6, \infty)\).

We need to determine the sign of the product \(x(x - 6)(x + 6)\) in each of these intervals.

1. **For \(x \in (-\infty, -6)\)**:
   - \(x\) is negative,
   - \(x - 6\) is negative,
   - \(x + 6\) is negative.
   - Product: Negative × Negative × Negative = Negative.

2. **For \(x \in (-6, 0)\)**:
   - \(x\) is negative,
   - \(x - 6\) is negative,
   - \(x + 6\) is positive.
   - Product: Negative × Negative × Positive = Positive.

3. **For \(x \in (0, 6)\)**:
   - \(x\) is positive,
   - \(x - 6\) is negative,
   - \(x + 6\) is positive.
   - Product: Positive × Negative × Positive = Negative.

4. **For \(x \in (6, \infty)\)**:
   - \(x\) is positive,
   - \(x - 6\) is positive,
   - \(x + 6\) is positive.
   - Product: Positive × Positive × Positive = Positive.

### Step 4: Include the critical points
The inequality is \(x(x - 6)(x + 6) \leq 0\), so we include the points where the product is zero:
- The expression equals zero at \(x = -6\), \(x = 0\), and \(x = 6\).

### Final Solution
The solution to the inequality is:
\[
x \in [-6, 0] \cup [6, \infty)
\]

Would you like more details on this solution or have any other questions?

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Here are some related questions for practice:
1. Solve the inequality \(x^3 - 27x \geq 0\).
2. Solve the inequality \(x^2 - 4x \leq 0\).
3. Factor the expression \(x^4 - 16x^2\) and solve \(x^4 - 16x^2 \geq 0\).
4. Solve the inequality \((x-1)(x+2)(x-3) \leq 0\).
5. Determine the intervals where the function \(f(x) = x^3 - 9x\) is non-negative.

**Tip:** When solving inequalities, factor the expression if possible, and test intervals to determine where the inequality holds.

Learn how to solve the inequality x^3 - 36x ≤ 0 step-by-step with detailed explanations. This problem involves inequalities, polynomial functions, and factoring methods. Suitable for students in Grades 10-12.

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