Final answer:
The maximum value of the function y = 3x^3 - 3x^2 - 3x occurs at x = 2 and the minimum value occurs at x = 1 on the interval [0, 2].
Step-by-step explanation:
To find the maximum and minimum values of the function y = 3x^3 - 3x^2 - 3x on the interval [0, 2], we first need to take the derivative of the function to find the critical points. The derivative of the function is:
y' = 9x^2 - 6x - 3.
Setting the derivative equal to zero to find the critical points gives us:
9x^2 - 6x - 3 = 0.
Solving this quadratic equation, we find the critical points are:
x = 1 and x = -1/3.
However, since x = -1/3 is not within the interval [0, 2], we only consider x = 1. By evaluating the original function at the endpoints of the interval (x=0 and x=2) and the critical point (x=1), we can determine the maximum and minimum values within the interval. The function values are:
- y(0) = 3(0)^3 - 3(0)^2 - 3(0) = 0,
- y(1) = 3(1)^3 - 3(1)^2 - 3(1) = -3,
- y(2) = 3(2)^3 - 3(2)^2 - 3(2) = 12.
From these values, we can see that the maximum value occurs at x = 2 and the minimum value occurs at x = 1. Therefore, the correct answer is 'c. Max at x = 2, Min at x = 1'.