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When trying to find the absolute extrema of a function over a specified range, the Closed Interval Method is your go-to strategy. This method is specifically useful for intervals that include both their endpoints, which is typical in practical scenarios. Here's how it works:

• First, determine any critical points where the function's derivative is zero or does not exist within the interval.
• Next, evaluate the function at these critical points and at the endpoints of the interval.
• Finally, compare the function values obtained from these evaluations.

The highest value you find is the absolute maximum, and the lowest is the absolute minimum, ensuring that you have accounted for all possible locations where extrema might occur in the closed interval.

Critical points are values within the domain of a function where its derivative either equals zero or does not exist. These points are crucial because they indicate potential peaks and valleys in the function—points where the function could reach a local maximum or minimum. In our case with the function \(f(x) = 2(3-x)\), the derivative is given by \(f'(x) = -2\).

Upon examining \(f'(x) = -2\), we see that it’s a constant value and never equals zero nor is it undefined. Therefore, for the interval \([-1,2]\), there are no critical points to evaluate within, simplifying your task to just checking the endpoints, since any extrema would necessarily be at the endpoints in this scenario.

Evaluating a function at its endpoints is a straightforward process, yet fundamental in applying the Closed Interval Method. It involves plugging in the endpoint values of the interval directly into the original function to compute their respective outputs. This not only helps verify potential extrema but is mandatory when critical points are absent.

Taking our function \(f(x) = 2(3-x)\) and evaluating at the interval endpoints \([-1,2]\):

• For \(x = -1\): Substitute into the function to find \(f(-1) = 2(3 - (-1)) = 8\).
• For \(x = 2\): Substitute into the function to find \(f(2) = 2(3 - 2) = 2\).

By comparing the function values at \(x=-1\) and \(x=2\), we determine that the absolute maximum value of 8 occurs at \(x=-1\), and the absolute minimum value of 2 occurs at \(x=2\). These results emphasize the importance of endpoints when critical points are not present in the function's derivative.