Engineering High School

Find the absolute maximum and minimum values on the closed interval [-1,8] for the function below. If a maximum or minimum value does not exist, enter NONE. f(x) = 1 − x2/3

Answers

Answer 1

The absolute maximum value on the closed interval [-1,8] for the function f(x) = 1 − x^(2/3) is f(1) = 0. The absolute minimum value does not exist.

What is the process for finding the absolute maximum and minimum values on a closed interval?

To find the absolute maximum and minimum values on a closed interval, we need to follow these steps:

1. Find the critical points of the function within the interval by taking its derivative and solving for x. In this case, the derivative of f(x) = 1 - x^(2/3) is f'(x) = -2x^(-1/3)/3. Setting f'(x) equal to zero, we get -2x^(-1/3)/3 = 0. This equation has no solution since x^(-1/3) is undefined for x = 0.

2. Evaluate the function at the endpoints of the interval. In this case, we need to calculate f(-1) and f(8). Evaluating the function at these points, we get f(-1) = 2 and f(8) = -7.

3. Compare the values obtained in steps 1 and 2 to determine the absolute maximum and minimum. Since there are no critical points within the interval, we compare the function values at the endpoints. We find that f(-1) = 2 is the maximum value, and f(8) = -7 is the minimum value.

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