To solve the inequality $3x^3 > 6x^2$, let's proceed step-by-step:

Step 1: Simplify the inequality

$3x^3 > 6x^2$ Divide both sides by 3: $x^3 > 2x^2$ Next, subtract $2x^2$ from both sides: $x^3 - 2x^2 > 0$ Factor out the common term $x^2$: $x^2(x - 2) > 0$

Step 2: Identify critical points

The critical points are obtained by setting $x^2(x - 2) = 0$: $x^2 = 0 \quad \Rightarrow \quad x = 0$ $x - 2 = 0 \quad \Rightarrow \quad x = 2$ Thus, the critical points are $x = 0$ and $x = 2$.

Step 3: Analyze the sign changes

We will test the sign of $x^2(x - 2)$ in the intervals determined by the critical points: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.

1. For $x \in (-\infty, 0)$:

   • $x^2 > 0$ (since the square of any real number is non-negative)
   • $x - 2 < 0$ (since $x < 2$)
   • Thus, $x^2(x - 2) < 0$.

2. For $x \in (0, 2)$:

   • $x^2 > 0$
   • $x - 2 < 0$
   • Thus, $x^2(x - 2) < 0$.

3. For $x \in (2, \infty)$:

   • $x^2 > 0$
   • $x - 2 > 0$
   • Thus, $x^2(x - 2) > 0$.

Step 4: Determine the solution set

The inequality $x^2(x - 2) > 0$ holds in the interval $(2, \infty)$. At $x = 0$ and $x = 2$, the expression equals zero, which does not satisfy the strict inequality.

Final answer in interval notation:

$(2, \infty)$

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

Related Questions:

1. How do you solve $x^3 < 2x^2$ in interval notation?
2. What are the steps to solve a rational inequality like $\frac{x^2 - 4}{x + 1} > 0$?
3. How do you graphically represent the solution to the inequality $3x^3 > 6x^2$?
4. What changes if the inequality was $3x^3 \geq 6x^2$?
5. How do you test intervals when solving inequalities?

Tip:

Always factor out common terms first when solving polynomial inequalities; it simplifies finding critical points and analyzing the intervals.