Final answer:
The absolute maximum of the function occurs at x = −1 with a value of 5, and the absolute minimum occurs at x = 1 with a value of 1, within the interval [−1, 1].
Step-by-step explanation:
To find the absolute minimum and maximum of the function y = 3x2/3 − 2x over the interval [−1, 1], we need to evaluate the function at its critical points and at the endpoints of the interval. We'll first find the derivative to locate the critical points.
The derivative of the function y = 3x2/3 − 2x is y' = 2x∑3/3 − 2. Setting the derivative equal to zero, we solve for the critical points. Yet there's no critical point satisfying the equation within the interval [−1, 1].
Hence, we evaluate the function at the endpoints, x = −1 and x = 1.
For x = −1: y = 3(−1)2/3 − 2(−1) = 3 + 2 = 5
For x = 1: y = 3(1)2/3 − 2(1) = 3 − 2 = 1
Comparing these values, we see that the absolute maximum value of y on the interval [−1, 1] is 5 when x = −1, and the absolute minimum value is 1 when x = 1.