Final answer:
To find the absolute maximum and minimum, we find the critical points and evaluate the function at these points and the interval's endpoints; the absolute maximum is 216 at x=2, and the absolute minimum is -8 at x=0.
Step-by-step explanation:
To find the absolute maximum and absolute minimum values of the function f(x) = 8(x²-1)³ on the interval [-1, 2], you first need to find the critical points of the function within the interval by differentiating and setting the derivative equal to zero. Then, evaluate the function at the critical points and at the endpoints of the interval. The highest value will be the absolute maximum and the lowest value will be the absolute minimum.
To start, find the derivative of f(x): f'(x) = 8·3(x²-1)²·2x = 48x(x²-1)². The critical points are where f'(x)=0 or undefined. Clearly, f'(x) is never undefined on the given interval, and f'(x)=0 when x=0 or x²-1=0, i.e., x=±1.
Now, evaluate f(x) at x=-1, 0, 1, and 2. We get
- f(-1) = 8((-1)²-1)³ = 0,
- f(0) = 8((0)²-1)³ = -8,
- f(1) = 8((1)²-1)³ = 0,
- f(2) = 8((2)²-1)³ = 8(3)³ = 216.
Therefore, the absolute maximum value is 216 at x=2, and the absolute minimum value is -8 at x=0.