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E-2 Determining Extremes To determine extremes on a closed interval, we must identify any peaks and valleys that create a local maximum value or a local minimum value respectively. To explore the development of determining extremes, go to the concept attainment activity Extremes take two forms, absolute and relative An absolute maximum (global maximum) occurs at the highest yvalue on a graph, in the example to the left it occurs at (3,4). An absolute minimum (global minimum) occurs at the lowest yvalue on a graph, in the example it occurs at (7,-3). A relative maximum (local maximum) occurs if you can move to the right and left of a point and in both cases decrease the value. In other words, a hill is created. Examples occur at (3,4) and (1,3). A relative minimum (local minimum) occurs if you can move to the right and left of a point and in both cases increase the value. In other words, a valley is created. Examples occur at (1,0) and (7,-3). Extreme Value Theorem If f(x) is a continuous function on a closed interval, then f(x) has both an absolute minimum and an absolute maximum on that closed interval. This theorem means that if a function has distinct endpoints (closed) and has no breaks (continuous), it MUST have an absolute maximum and an absolute minimum value. Clarification This function is continuous on a This function is continuous on This function is discontinuous closed interval an open interval on a closed interval There is an absolute maximum and minimum as this example fits the Extreme Value Theorem perfectly. There is no absolute minimum as there is not a defined point that we can label as the minimum. The following diagrams display derivative that DO NOT EXIST. There is no absolute maximum as there is not a defined point that we can label as the maximum. f '(a) does not exist because there is a corner at x = c. f '(a) does not exist at x = a f '(a) does not exist because because the tangent line is vertical. f(x) is not continuous at x = a. (The slope of the tangent line is 1/0) Critical Number x = c is a critical number for the function f(x) if f(c) is defined and f '(c) = 0 or f '(c) does not exist. Another way of explaining a critical number, is that it occurs anywhere that the slope on the graph is either horizontal or vertical. Example 1 Determine any critical points for f (x) = 2x2 + 10x + 12 Example 2 Determine any critical points for Example 3 Determine any critical points for Determining Absolute Extremes Complete the following steps to determine extremes 1. Determine f '(x). 2. Determine any critical numbers. 3. Determine if any corners exist. 4. Calculate the values of f(x) at all endpoints, critical numbers, and corners and assign a maximum and minimum value. Example 4 Determine the absolute extremes for f(x) = x3 + 3x2 + 3x + 1 on the interval [-2,3] Step One - Determine f '(x) f '(x) = 3x2 + 6x + 3 Step Two - Determine any critical numbers. 0 = 3x2 + 6x + 3 0 = 3(x2 + 2x + 1) 0 = x2 + 2x + 1 0 = (x + 1)(x + 1) The critical value occurs at x = -1 Step Three - Determine if any corners exist. No corners exist on a polynomial function, only on an absolute value or a split function. Step Four - Calculate the values of f(x) at all endpoints, critical numbers, and corners and assign a maximum and minimum value. The easiest way to do this is to list all endpoints and the critical value in a chart to make your conclusion. Item x f(x) left endpoint -2 (-2)3 + 3(-2)2 + 3(-2) + 1 = -1 critical number -1 (-1)3 + 3(-1)2 + 3(-1) + 1 = 0 right endpoint 3 (3)3 + 3(3)2 + 3(3) + 1 = 61 The minimum value occurs at (-2, -1). The maximum value occurs at (3, 61). Example 5 Determine the absolute extremes for on the interval [0,8] Item x f(x) left endpoint 0 6(0)2/3 - 4(0) + 3 = 3 critical number 1 6(1)2/3 - 4(1) + 3 = 5 right endpoint 8 6(8)2/3 - 4(8) + 3 = -21 The minimum value occurs at (8,-21). The maximum value occurs at (1,5). Many additional examples are found in the question and answer section, including absolute value and split functions.