Question

Find the absolute maximum and minimum values of the function below. f(x) = x^3 - 3x^2 + 3, -1/2 ≤ x ≤ 4 SOLUTION: Since f is continuous on [-1/2, 4], we can use the Closed Interval Method: f(x) = x^3 - 3x^2 + 3 f'(x) = 3x^2 - 6x Since f'(x) exists for all x, the only critical numbers of f occur when f'(x) = 0, that is, x = 0 or x = 2. Notice that each of these critical numbers lies in [-1/2, 4]. The values of f at these critical numbers are f(0) = 3 and f(2) = -1. The values of f at the endpoints of the interval are f(-1/2) = 2.125 and f(4) = 19. Comparing these four numbers, we see that the absolute maximum value is f(4) = 19 and the absolute minimum value is f(2) = -1. Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in the figure.

          Find the absolute maximum and minimum values of the function below.
f(x) = x^3 - 3x^2 + 3, -1/2 ≤ x ≤ 4

SOLUTION: Since f is continuous on [-1/2, 4], we can use the Closed Interval Method:
f(x) = x^3 - 3x^2 + 3
f'(x) = 3x^2 - 6x
Since f'(x) exists for all x, the only critical numbers of f occur when f'(x) = 0, that is, x = 0 or x = 2. Notice that each of these critical numbers lies in [-1/2, 4]. The values of f at these critical numbers are f(0) = 3 and f(2) = -1. 
The values of f at the endpoints of the interval are f(-1/2) = 2.125 and f(4) = 19. 
Comparing these four numbers, we see that the absolute maximum value is f(4) = 19 and the absolute minimum value is f(2) = -1. 
Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in the figure.

Added by Sherri J.

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