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?

---

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.