To find the critical numbers of the function $f(x) = 3x^4 + 4x^3 - 36x^2$, follow these steps:

1. Find the derivative of the function: $f'(x) = \frac{d}{dx}(3x^4 + 4x^3 - 36x^2) = 12x^3 + 12x^2 - 72x.$

2. Set the derivative equal to zero to find the critical points: $12x^3 + 12x^2 - 72x = 0.$

3. Factor the derivative equation: Factor out the common term: $12x(x^2 + x - 6) = 0.$ Now, factor the quadratic: $12x(x + 3)(x - 2) = 0.$

4. Solve for the values of $x$: The solutions are: $x = 0, \quad x = -3, \quad x = 2.$

Therefore, the critical numbers of the function are $0, -3,$ and $2$.

Would you like more details or have any questions?

Here are five related questions:

1. How do you determine if a critical point is a local minimum or maximum?
2. Can there be critical points at which the derivative does not exist?
3. What role do critical numbers play in curve sketching?
4. How do you find inflection points of a function?
5. How does factoring help in solving polynomial equations?

Tip: Always check the second derivative to determine the concavity and nature (minimum/maximum) of critical points.