Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)f(x) = 3x4 + 8x3 − 48x2

To find the critical numbers of the function f(x) = 3x^4 + 8x^3 - 48x^2, we first need to find the derivative of the function.

The derivative of f(x) is f'(x) = 12x^3 + 24x^2 - 96x.

Next, we set the derivative equal to zero and solve for x to find the critical points.

So, 12x^3 + 24x^2 - 96x = 0.

We can factor out a 12x from each term to simplify this equation:

12x(x^2 + 2x - 8) = 0.

Setting each factor equal to zero gives us the solutions x = 0 and x^2 + 2x - 8 = 0.

The quadratic equation x^2 + 2x - 8 = 0 can be factored to (x - 2)(x + 4) = 0, which gives us the solutions x = 2 and x = -4.

So, the critical numbers of the function are x = 0, x = 2, and x = -4.