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Describing Graphs and Applications of the Derivative

This chapter covers the concepts of describing graphs of functions, the first and second derivative rules, curve sketching, optimization problems, and applications of derivatives to business and economics.

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Describing Graphs and Applications of the Derivative

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  1. Chapter 2Applications of the Derivative

  2. Chapter Outline • Describing Graphs of Functions • The First and Second Derivative Rules • The First and Second Derivative Tests and Curve Sketching • Curve Sketching (Conclusion) • Optimization Problems • Further Optimization Problems • Applications of Derivatives to Business and Economics

  3. §2.1 Describing Graphs of Functions

  4. Section Outline • Increasing and Decreasing Functions • Relative and Absolute Extrema • Changing Slope • Concavity • Inflection Points • x- and y-Intercepts • Asymptotes • Describing Graphs

  5. Increasing Functions

  6. Decreasing Functions

  7. Relative Maxima & Minima

  8. Absolute Maxima & Minima

  9. Changing Slope EXAMPLE Draw the graph of a function y = f (T) with the stated properties. In certain professions the average annual income has been rising at an increasing rate. Let f (T) denote the average annual income at year T for persons in one of these professions and sketch a graph that could represent f (T). SOLUTION Since f (T) is rising at an increasing rate, this means that the slope of the graph of f (T) will continually increase. The following is a possible example. Notice that the slope becomes continually steeper.

  10. Concavity Concave Up Concave Down

  11. Inflection Points Notice that an inflection point is not where a graph changes from an increasing to a decreasing slope, but where the graph changes its concavity.

  12. Intercepts

  13. Asymptotes

  14. 6-Point Graph Description

  15. Describing Graphs EXAMPLE Use the 6 categories previously mentioned to describe the graph. SOLUTION 1) The function is increasing over the intervals The function is decreasing over the intervals Relative maxima are at x = -1 and at x = 5.5. Relative minima is at x = 3 and at x = -3.

  16. Describing Graphs CONTINUED 2) The function has a (absolute) maximum value at x = -1. The function has a (absolute) minimum value at x = -3. 3) The function is concave up over the interval The function is concave down over the interval This function has exactly one inflection point, located at x =1. 4) The function has three x-intercepts, located at x = -2.5, x = 1.25, and x = 4.5. The function has one y-intercept at y = 3.5. 5) Over the function’s domain, , the function is not undefined for any value of x. 6) The function does not appear to have any asymptotes, horizontal or vertical.

  17. §2.2 The First and Second Derivative Rules

  18. Section Outline • First Derivative Rule • Second Derivative Rule

  19. First Derivative Rule

  20. First Derivative Rule EXAMPLE Sketch the graph of a function that has the properties described. f(-1) = 0; for x < -1; for x > -1. SOLUTION The only specific point that the graph must pass through is (-1, 0). Further, we know that to the left of this point, the graph must be decreasing ( for x < -1) and to the right of this point, the graph must be increasing ( for x > -1). Lastly, the graph must have zero slope at that given point ( ).

  21. Second Derivative Rule

  22. First & Second Derivative Scenarios

  23. First & Second Derivative Rules EXAMPLE Sketch the graph of a function that has the properties described. f(x) defined only for x≥ 0; (0, 0) and (5, 6) are on the graph; for x≥ 0; for x < 5, , for x > 5. SOLUTION The only specific points that the graph must pass through are (0, 0) and (5, 6). Further, we know that to the left of (5, 6), the graph must be concave down ( for x < 5) and to the right of this point, the graph must be concave up ( for x > 5). Also, the graph will only be defined in the first and fourth quadrants (x≥ 0). Lastly, the graph must have positive slope everywhere that it is defined.

  24. First & Second Derivative Rules CONTINUED

  25. First & Second Derivative Rules EXAMPLE Looking at the graphs of and for x close to 10, explain why the graph of f(x) has a relative minimum at x = 10.

  26. First & Second Derivative Rules CONTINUED SOLUTION At x = 10 the first derivative has a value of 0. Therefore, the slope of f(x) at x = 10 is 0. This suggests that either a relative minimum or relative maximum exists on the function f(x) at x = 10. To determine which it is, we will look at the second derivative. At x = 10, the second derivative is above the x-axis, suggesting that the second derivative is positive when x = 10. Therefore, f(x) is concave up when x = 10. Since at x = 10, f(x) has slope 0 and is concave up, this means that the f(x) has a relative minimum at x = 10.

  27. First & Second Derivative Rules EXAMPLE After a drug is taken orally, the amount of the drug in the bloodstream after t hours is f(t) units. The figure below shows partial graphs of the first and second derivatives of the function.

  28. First & Second Derivative Rules CONTINUED (a) Is the amount of the drug in the bloodstream increasing or decreasing at t = 5? (b) Is the graph of f(t) concave up or concave down at t = 5? (c) When is the level of the drug in the bloodstream decreasing the fastest? SOLUTION (a) To determine whether the amount of the drug in the bloodstream is increasing or decreasing at t = 5, we will need to consider the graph of the first derivative since the first derivative of a function tells how the function is increasing or decreasing. At t = 5 the value of the first derivative is -4. Therefore, the value of the first derivative is negative at t = 5. Therefore, the function is decreasing at t = 5.

  29. First & Second Derivative Rules CONTINUED (b) To determine whether the graph of f(t) is concave up or concave down at t = 5, we will need to consider the graph of the second derivative at t = 5. At t = 5, the value of the second derivative is 0.5. Therefore, the value of the second derivative is positive at t = 5. Therefore, the function is concave up at t = 5. (c) To determine when the level of the drug in the bloodstream is decreasing the fastest, we need to determine when the first derivative is the smallest. This occurs when t = 4.

  30. §2.3 The First and Second Derivative Tests and Curve Sketching

  31. Section Outline • Curve Sketching • Critical Values • The First Derivative Test • The Second Derivative Test • Test for Inflection Points

  32. Curve Sketching A General Approach to Curve Sketching 1) Starting with f(x), we compute 2) Next, we locate all relative maximum and relative minimum points and make a partial sketch. 3) We study the concavity of f(x) and locate all inflection points. 4) We consider other properties of the graph, such as the intercepts, and complete the sketch.

  33. Critical Values

  34. First Derivative Test

  35. First Derivative Test EXAMPLE Find the local maximum and minimum points of SOLUTION First we find the critical values and critical points of f: The first derivative if 9x – 3 = 0 or 2x + 1 = 0. Thus the critical values are x = 1/3 and x = -1/2. Substituting the critical values into the expression of f:

  36. First Derivative Test CONTINUED Thus the critical points are (1/3, 43/18) and (-1/2, 33/8). To tell whether we have a relative maximum, minimum, or neither at a critical point we shall apply the first derivative test. This requires a careful study of the sign of , which can be facilitated with the aid of a chart. Here is how we can set up the chart.

  37. First Derivative Test CONTINUED • Divide the real line into intervals with the critical values as endpoints. • Since the sign of depends on the signs of its two factors 9x – 3 and 2x + 1, determine the signs of the factors of over each interval. Usually this is done by testing the sign of a factor at points selected from each interval. • In each interval, use a plus sign if the factor is positive and a minus sign if the factor is negative. Then determine the sign of over each interval by multiplying the signs of the factors and using • A plus sign of corresponds to an increasing portion of the graph f and a minus sign to a decreasing portion. Denote an increasing portion with an upward arrow and a decreasing portion with a downward arrow. The sequence of arrows should convey the general shape of the graph and, in particular, tell you whether or not your critical values correspond to extreme points.

  38. First Derivative Test CONTINUED 1/3 -1/2 Critical Points, Intervals x < -1/2 -1/2 < x < 1/3 x > 1/3 __ __ + 9x - 3 __ + + 2x + 1 __ + + Decreasing on Increasing on Increasing on Local minimum Local maximum

  39. First Derivative Test CONTINUED You can see from the chart that the sign of varies from positive to negative at x = -1/2. Thus, according to the first derivative test, f has a local maximum at x = -1/2. Also, the sign of varies from negative to positive at x = 1/3; and so f has a local minimum at x = 1/3. In conclusion, f has a local maximum at (-1/2, 33/8) and a local minimum at (1/3, 43/18). NOTE: Upon the analyzing the various intervals, had any two consecutive intervals not alternated between “increasing” and “decreasing”, there would not have been a relative maximum or minimum at the value for x separating those two intervals.

  40. Second Derivative Test

  41. Second Derivative Test EXAMPLE Locate all possible relative extreme points on the graph of the function Check the concavity at these points and use this information to sketch the graph of f(x). SOLUTION We have The easiest way to find the critical values is to factor the expression for

  42. Second Derivative Test CONTINUED From this factorization it is clear that will be zero if and only if x = -3 or x = -1. In other words, the graph will have horizontal tangent lines when x = -3 and x = -1, and no where else. To plot the points on the graph where x = -3 and x = -1, we substitute these values back into the original expression for f(x). That is, we compute Therefore, the slope of f(x) is 0 at the points (-3, 0) and (-1, -4). Next, we check the sign of at x = -3 and at x = -1 and apply the second derivative test: (local maximum) (local minimum).

  43. Second Derivative Test CONTINUED The following is a sketch of the function. (-3, 0) (-1, -4)

  44. Test for Inflection Points

  45. Second Derivative Test EXAMPLE Sketch the graph of SOLUTION We have We set and solve for x. (critical values)

  46. Second Derivative Test CONTINUED Substituting these values of x back into f(x), we find that We now compute (local minimum) (local maximum).

  47. Second Derivative Test CONTINUED Since the concavity reverses somewhere between , there must be at least one inflection point. If we set , we find that So the inflection point must occur at x = 0. In order to plot the inflection point, we compute The final sketch of the graph is given below.

  48. Second Derivative Test CONTINUED (0, 2)

  49. §2.4 Curve Sketching (Conclusion)

  50. Section Outline • The Second Derivative Test • Graphs With Asymptotes

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