Question

On the interval [0, 2], find the max and min of the function y = 3x^3 - 3x^2 - 3x.

A) Max at x = 2, Min at x = 0
B) Max at x = 0, Min at x = 2
C) Max at x = 1, Min at x = 2
D) Max at x = 2, Min at x = 1

Answer 1

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 find the critical points of the function. The minimum value occurs at x = 0 and the maximum value occurs at x = 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 find the critical points of the function. Critical points occur when the derivative is equal to zero or does not exist. Taking the derivative of y with respect to x, we get y' = 9x^2 - 6x - 3. Setting y' equal to zero and solving for x gives us:

9x^2 - 6x - 3 = 0

Using the quadratic formula, we find that x = (-(-6) ± √((-6)^2 - 4(9)(-3))) / (2(9))

Simplifying further, we get x = (6 ± √(216)) / 18

x = (6 ± 6√6) / 18

The critical points are x = (1 ± √6) / 3. Since the interval is [0, 2], we need to check the function at the endpoints as well. Evaluating the function at x = 0 and x = 2, we get:

y(0) = 0

y(2) = 24

Therefore, the minimum value occurs at x = 0 and the maximum value occurs at x = 2.