Final answer:
To find the absolute maximum and minimum values of the function on the closed interval, we find the critical points and evaluate the function at those points as well as at the endpoints.
Step-by-step explanation:
To find the absolute maximum and absolute minimum values of the function f(x) = 3x^4-4x^3-12x^2 on the closed interval [-2, 1], we need to identify the critical points and endpoints of the interval.
First, we find the derivative of f(x) which is f'(x) = 12x^3 - 12x^2 - 24x.
Next, we set f'(x) equal to zero and solve for x to find the critical points. By solving the equation 12x^3 - 12x^2 - 24x = 0, we find three critical points: x = -2, x = 0, and x = 2.
Since the interval [-2, 1] includes the critical points -2 and 0, we evaluate the function at these points as well as at the endpoints -2 and 1 to find the absolute maximum and minimum values.
By substituting these values into f(x), we get f(-2) = 76, f(0) = 0, f(1) = -13, and f(2) = -28.
Therefore, the absolute maximum value is 76 and it occurs at x = -2, while the absolute minimum value is -28 and it occurs at x = 2.