Absolute Extrema on Finite Closed Intervals

Question

In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

f(t) = 2 − |t|, −1 ≤ t ≤ 3

Accepted Answer

Determine the absolute minimum and maximum values of the function F of X equals 5 minus absolute value of X on the interval -4 to 2, and identify where they occur. Now our function has an absolute value, so we'll have to split this function up into two piecewise functions. So, Or X is less than 0, we will let the absolute value of X equals negative X. And for X is greater than the equals 0, we'll let the absolute value. B equals to positive X. We didn't have F of X equals 5 minus X or 5 + X. This means a derivative is equals to one. We'll do the same for the other segment. F of X equals 5 minus X, which means our derivative is -1. Now, We noticed that we have constants for both of our derivatives. This means that there are no critical points. For either segment. So, because there are no critical points, we have to check something else. Let's use our first derivative test to determine if X equals 0 is a max or a mint or neither. Now, the first derivative test. Tells us That there is a derivative if there's a sign change between. The left part of the function and the right part of the function. Around that point. So, we're going to check. For sign changes. Well, we noticed that if X is less than 0, character derivative is positive. And as X is squared than 0, our derivative is negative. So we do have a sign change. Which means we have a local max or min at the value X equals 0. And in particular, we go from positive to negative. Because this is the order that we're changing the signs, we have a local maximum. No. Let's check the end points. We have F of X equals 5 minus the opposite value of X. And if we check F of -4. We have 5 minus the absolute value of -4, which is 5 minus 4, or just 1. We'll also check F2. F of 2 will be 5 minus the absolute value of 2, which is just 3. And if we check f of 0. We get 5 minus the absolute value of 0, which is just 5. We can then see that we have A maximum At X equals 0. And the minimum At X equals. -4. This means in the absolute maximum at X equals 0. And the absolute minimum. Is that X equals -4. The values of these. are F of X Equals 5. And F of X equals. One for Max and Min respectively. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Absolute Extrema

Critical Points

Graphing Absolute Value Functions

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