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Key Concept 1

Key Concept 1. Analyze Increasing and Decreasing Behavior. A. Use the graph of the function f ( x ) = x 2 – 4 to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically. Example 1. Analyze Graphically

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Key Concept 1

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  1. Key Concept 1

  2. Analyze Increasing and Decreasing Behavior A. Use the graph of the function f(x) = x2 – 4 to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically. Example 1

  3. Analyze Graphically From the graph, we can estimate that f is decreasing on and increasing on . Analyze Increasing and Decreasing Behavior Support Numerically Create a table using x-values in each interval. The table shows that as x increases from negative values to 0, f(x) decreases; as x increases from 0 to positive values, f(x) increases. This supports the conjecture. Example 1

  4. Answer:f(x) is decreasing on and increasing on . Analyze Increasing and Decreasing Behavior Example 1

  5. Analyze Increasing and Decreasing Behavior B. Use the graph of the function f(x) = –x3+ x to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically. Example 1

  6. Analyze Graphically From the graph, we can estimate that f is decreasing on , increasing on , and decreasing on . Analyze Increasing and Decreasing Behavior Support Numerically Create a table using x-values in each interval. Example 1

  7. 0.5 1 2 2.5 3 –13.125 –24 –6 Analyze Increasing and Decreasing Behavior Example 1

  8. The table shows that as x increases to , f(x) decreases; as x increases from , f(x) increases; as x increases from , f(x) decreases. This supports the conjecture. Answer:f(x) is decreasing on and and increasing on Analyze Increasing and Decreasing Behavior Example 1

  9. Use the graph of the function f(x) =2x2 + 3x – 1 to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically. A. f(x) is increasing on (–∞, –1) and (–1, ∞). B. f(x) is increasing on (–∞, –1) and decreasing on (–1, ∞). C. f(x) is decreasing on (–∞, –1) and increasing on (–1, ∞). D. f(x) is decreasing on (–∞, –1) and decreasing on (–1, ∞). Example 1

  10. Key Concept 2

  11. Estimate and Identify Extrema of a Function Estimate and classify the extrema to the nearest 0.5 unit for the graph of f(x). Support the answers numerically. Example 2

  12. Analyze Graphically It appears that f(x) has a relative minimum at x = –1 and a relative maximum at x = 2. It also appears that so we conjecture that this function has no absolute extrema. Estimate and Identify Extrema of a Function Example 2

  13. Estimate and Identify Extrema of a Function Support Numerically Choose x-values in half unit intervals on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x. Because f(–1.5) > f(–1) and f(–0.5) > f(–1), there is a relative minimum in the interval (–1.5, –0.5) near –1. The approximate value of this relative minimum is f(–1) or –7.0. Example 2

  14. Estimate and Identify Extrema of a Function Likewise, because f(1.5) < f(2) and f(2.5) < f(2), there is a relative maximum in the interval (1.5, 2.5) near 2. The approximate value of this relative maximum is f(2) or 14. f(100) < f(2) and f(–100) > f(–1), which supports our conjecture that f has no absolute extrema. Answer:To the nearest 0.5 unit, there is a relative minimum at x = –1 and a relative maximum at x = 2. There are no absolute extrema. Example 2

  15. Estimate and classify the extrema to the nearest 0.5 unit for the graph of f(x). Support the answers numerically. A. There is a relative minimum of 2 at x = –1 and a relative maximum of 1 at x = 0. There are no absolute extrema. B. There is a relative maximum of 2 at x = –1 and a relative minimum of 1 at x = 0. There are no absolute extrema. C. There is a relative maximum of 2 at x = –1 and no relative minimum. There are no absolute extrema. D. There is no relative maximum and there is a relative minimum of 1 at x = 0. There are no absolute extrema. Example 2

  16. Use a Graphing Calculator to Approximate Extrema GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of f(x) = x4 – 5x2 – 2x + 4. State the x-value(s) where they occur. f(x) = x4 – 5x2 – 2x + 4 Graph the function and adjust the window as needed so that all of the graph’s behavior is visible. Example 3

  17. Use a Graphing Calculator to Approximate Extrema From the graph of f, it appears that the function has one relative minimum in the interval (–2, –1), an absolute minimum in the interval (1, 2), and one relative maximum in the interval (–1, 0) of the domain. The end behavior of the graph suggests that this function has no absolute extrema. Example 3

  18. Use a Graphing Calculator to Approximate Extrema Using the minimum and maximum selection from the CALC menu of your graphing calculator, you can estimate that f(x) has a relative minimum of 0.80 at x = –1.47, an absolute minimum of –5.51 at x = 1.67, and a relative maximum of 4.20 at x = –0.20. Example 3

  19. Use a Graphing Calculator to Approximate Extrema Answer:relative minimum: (–1.47, 0.80); relative maximum: (–0.20, 4.20);absolute minimum: (1.67, –5.51) Example 3

  20. GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of f(x) = x3 + 2x2 – x – 1. State the x-value(s) where they occur. A. relative minimum: (0.22, –1.11);relative maximum: (–1.55, 1.63) B. relative minimum: (–1.55, 1.63);relative maximum: (0.22, –1.11) C. relative minimum: (0.22, –1.11);relative maximum: none D. relative minimum: (0.22, 0); relative minimum: (–0.55, 0)relative maximum: (–1.55, 1.63) Example 3

  21. Use Extrema for Optimization FUEL ECONOMY Advertisements for a new car claim that a tank of gas will take a driver and three passengers about 360 miles. After researching on the Internet, you find the function for miles per tank of gas for the car is f(x) = 0.025x2 + 3.5x + 240, where x is the speed in miles per hour . What speed optimizes the distance the car can travel on a tank of gas? How far will the car travel at that optimum speed? Example 4

  22. Use Extrema for Optimization We want to maximize the distance a car can travel on a tank of gas. Graph the function f(x) = –0.025x2 + 3.5x + 240 using a graphing calculator. Then use the maximum selection from the CALC menu to approximate the x-value that will produce the greatest value for f(x). Example 4

  23. Use Extrema for Optimization The graph has a maximum of 362.5 for x ≈ 70. So the speed that optimizes the distance the car can travel on a tank of gas is 70 miles per hour. The distance the car travels at that speed is 362.5 miles. Answer:The optimal speed is about 70 miles per hour. The car will travel 362.5 miles when traveling at the optimum speed. Example 4

  24. VOLUME A square with side length x is cut from each corner of a rectangle with dimensions 8 inches by 12 inches. Then the figure is folded to form an open box, as shown in the diagram. Determine the length and width of the box that will allow the maximum volume. A. 6.43 in. by 10.43 in. B. 4.86 in. by 8.86 in. C. 3 in. by 7 in. D. 1.57 in. by 67.6 in. Example 4

  25. Key Concept3

  26. Find Average Rates of Change A. Find the average rate of change of f(x) = –2x2 + 4x + 6 on the interval [–3, –1]. Use the Slope Formula to find the average rate of change of f on the interval [–3, –1]. Substitute –3 for x1 and –1 for x2. Evaluate f(–1) and f(–3). Example 5

  27. Find Average Rates of Change Simplify. The average rate of change on the interval [–3, –1] is 12. The graph of the secant line supports this conclusion. Answer:12 Example 5

  28. Find Average Rates of Change B. Find the average rate of change of f(x) = –2x2 + 4x + 6 on the interval [2, 5]. Use the Slope Formula to find the average rate of change of f on the interval [2, 5]. Substitute 2 for x1 and 5 for x2. Evaluate f(5) and f(2). Example 5

  29. Find Average Rates of Change Simplify. The average rate of change on the interval [2, 5] is –10. The graph of the secant line supports this conclusion. Answer:–10 Example 5

  30. A. 27 B. 11 C. D. –19 Find the average rate of change of f(x) = –3x3+ 2x + 3 on the interval [–2, –1]. Example 5

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