Final answer:
The absolute minimum value of f(x) = x^3 - 6x^2 + 12x + 6 on the closed interval [0,4] is 6.
Step-by-step explanation:
To determine the absolute minimum value of the function f(x) = x^3 - 6x^2 + 12x + 6 on the closed interval [0,4], we need to find the critical points and evaluate the function at these points. To find the critical points, we take the derivative of the function and set it equal to 0. The derivative of f(x) is f'(x) = 3x^2 - 12x + 12. Setting f'(x) = 0, we have 3x^2 - 12x + 12 = 0. Solving this quadratic equation, we get x = 2. Plugging x = 2 back into f(x), we get f(2) = 2^3 - 6(2)^2 + 12(2) + 6 = 8 - 24 + 24 + 6 = 14.
Now, we need to check the endpoints of the interval [0,4]. Evaluating f(0), we get f(0) = 0^3 - 6(0)^2 + 12(0) + 6 = 6. Evaluating f(4), we get f(4) = 4^3 - 6(4)^2 + 12(4) + 6 = 64 - 96 + 48 + 6 = 22.
Comparing the values f(2), f(0), and f(4), we can see that the absolute minimum value of f on the closed interval [0,4] is f(0) = 6.