1-4 Extrema and Average Rates of Change

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1 Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (, 2.5) and increasing on (2.5, ). Create a table of values using x-values in each interval. (, 2.5): 12 25, , (2.5, ): ,541 The tables support the conjecture that the graph of f is decreasing on (, 2.5) and increasing on (2.5, ). When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (, 6), decreasing on ( 6, 3), decreasing on ( 3, 0), and increasing on (0, ). Create a table of values using x-values in each interval. (, 6): ( 6, 3): ( 3, 0): (0, ): The tables support the conjecture that the graph of f is increasing on (, 6), decreasing on ( 6, 3), decreasing on ( 3, 0), and increasing on (0, ). esolutions Manual - Powered by Cognero Page 1

2 9. When the graph is viewed from left to right, we can estimate that the graph of f is constant on (, 5), increasing on ( 5, 3.5), and decreasing on ( 3.5, ). Create a table of values using x-values in each interval. (, 5): ( 5, 3.5): ( 3.5, ): The tables support the conjecture that the graph of f is constant on (, 5), increasing on ( 5, 3.5), and decreasing on ( 3.5, ). 12. Sample answer: It appears that f (x) has relative maxima at x = 0.5 and x = 0.75 and relative minima at x = 0 and x = It also appears that and, so we conjecture that this function has no absolute extrema Because f ( 0.5) > f ( 1) and f ( 0.5) > f (0), there is a relative maximum in the interval ( 1, 0). The approximate value of this relative maximum is Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is Estimate and classify the extrema for the graph of each function. Support the answers numerically. Because f (0) < f ( 0.5) and f (0) < f (0.5), there is a relative minimum in the interval ( 0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimum in the interval (2, 2.5). esolutions Manual - Powered by Cognero Page 2

3 The approximate value of this relative minimum is 9.1. f( 100) < f ( 0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema. 15. Sample answer: It appears that f (x) has an absolute minimum at x = 3.75, a relative maximum at x = 0, and a relative minimum at x = 3.5. It also appears that = and =, so we conjecture that this function has no absolute maximum Because f ( 3.75) < f ( 4) and f ( 3.75) < f ( 3.5), there is an absolute minimum in the interval ( 4, 3.5). The approximate value of the absolute minimum is Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is Because f (0) > f ( 0.5) and f (0) > f (0.5), there is a relative maximum in the interval ( 0.5, 0.5). The approximate value of this relative minimum is 0. f( 100) > f ( 3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum. esolutions Manual - Powered by Cognero Page 3

4 18. Sample answer: It appears that f (x) has a relative maximum at x = 1.5 and a relative minimum at x = 1.5. It also appears that and, so we conjecture that this function has no absolute extrema Because f ( 1.5) > f ( 2) and f ( 1.5) > f ( 1), there is a relative maximum in the interval ( 2, 1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is 5.7. f( 100) < f ( 1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema. 21. Sample answer: It appears that f (x) has a relative maximum at x = 1.25 and a relative minimum at x = It also appears that and, so we conjecture that this function has no absolute extrema Because f ( 1.25) > f ( 1.5) and f ( 1.25) > f ( 1), there is a relative maximum in the interval ( 1.5, 1). The approximate value of this relative maximum is Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is f( 100) < f ( 1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema. esolutions Manual - Powered by Cognero Page 4

5 GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur. 24. f (x) = x 4 + 3x f (x) = x 5 + 3x 2 + x 1 The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum. Use the maximum tool from the CALC menu to locate the absolute maximum. The absolute maximum is at (2.25, 6.54). The relative maximum is (1.11, 2.12) and the relative minimum is ( 0.17, 1.08). 30. f (x) = 0.008x x 4 0.2x x 2 0.7x esolutions Manual - Powered by Cognero Page 5

6 The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima. The relative maximums are (2.49, 1.45) and ( 3.72, 14.23). The relative minimums are (0.32, 0.11) and (5.90, 6.83). 33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round to the nearest hundredth of an inch, if necessary. The formula for the surface area of a cylinder is SA = 2πr 2 +2πrh. Substitute 20.5π for SA and get h in terms of r. We need to maximize the volume of the cylinder. The formula for the volume is V = πr 2 h. Substitute for h. esolutions Manual - Powered by Cognero Page 6

7 Graph the equation and locate the maximum. 39. f (x) = 2x 4 5x 3 + 4x 6; [ 1, 5] f(5) = 2(5) 4 5(5) 3 + 4(5) 6 f(5) = f(5) = 1861 The maximum occurs when x equals about This is the radius of the cylinder. Substitute this answer and solve for h. f( 1) = 2( 1) 4 5( 1) 3 + 4( 1) 6 f( 1) = f( 1) = 7 Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a surface area of in 2. Find the average rate of change of each function on the given interval. 36. f (x) = 3x 3 2x 2 + 6; [2, 6] f(x) = 3x 3 2x 2 + 6; [2, 6] f(6) = 3(6) 3 2(6) f(6) = f(6) = 582 f(3) = 3(3) 3 2(3) f(3) = f(3) = 22 esolutions Manual - Powered by Cognero Page 7