1. Chapter 2 Applications of the Derivative

2. § 2.1 Describing Graphs of Functions

3. Increasing Functions

4. Decreasing Functions

5. Relative Maxima & Minima

6. Absolute Maxima & Minima

7. Changing SlopeEXAMPLE SOLUTION Draw the graph of a function y = f (T) with the stated properties. 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. 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). Notice that the slope becomes continually steeper.

8. Concavity Concave Up Concave Down

9. 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.

10. Intercepts Definition x-Intercept: A point at which a graph crosses the x-axis. y-Intercept: A point at which a graph crosses the y-axis.

11. Asymptotes Definition Horizontal Asymptotes: A straight, horizontal line that a graph follows indefinitely as x increases without bound. Vertical Asymptotes: A straight, vertical line that a graph follows indefinitely as y increases without bound. Horizontal asymptotes occur when exists, in which case the asymptote is: If a function is undefined at x = a, a vertical asymptote occurs when a denominator equals zero, in which case the asymptote is: x = a.

12. 6-Point Graph Description

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

14. Describing Graphs 2) The function has a (absolute) maximum value at x = -1. The function has a (absolute) minimum value at x = -3. CONTINUED 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.

15. § 2.2 The First and Second Derivative Rules

16. First Derivative Rule

17. EXAMPLE SOLUTION Sketch the graph of a function that has the properties described. 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). Lastly, the graph must have zero slope at that given point ( ). f (-1) = 0; for x -1.

18. Second Derivative Rule

19. First & Second Derivative Scenarios

20. First & Second Derivative RulesEXAMPLE SOLUTION Sketch the graph of a function that has the properties described. 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). 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. f (x) defined only for x ≥ 0; (0, 0) and (5, 6) are on the graph; for x ≥ 0; for x 5.

21. First & Second Derivative RulesCONTINUED

22. EXAMPLE Use the given information to draw the function near x = 3: f(3) = 4, f’(3) = -2, f’’(3) = - 2.

23. First & Second Derivative RulesEXAMPLE 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.

24. First & Second Derivative RulesSOLUTION 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. CONTINUED

25. First & Second Derivative RulesEXAMPLE 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.

26. First & Second Derivative Rules (a) Is the amount of the drug in the bloodstream increasing or decreasing at t = 5? CONTINUED (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.

27. First & Second Derivative RulesCONTINUED (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.

28. § 2.3 The First and Second Derivative Tests and Curve Sketching

29. 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. A General Approach to Curve Sketching

30. Critical Values DefinitionExample Critical Values: Given a function f (x), a number a in the domain such that either or is undefined. For the function below, notice that the slope of the function is 0 at x = -2 and the slope is undefined at x = -0.4. Also notice that the function has a relative minimum and a relative maximum at these points, respectively.

31. First Derivative Test

32. EXAMPLE Find the local maximum and minimum points of

33. Second Derivative Test

34. 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).

35. Second Derivative Test The following is a sketch of the function in the previous example. (-3, 0) (-1, -4)

36. Test for Inflection Points

37. Second Derivative TestEXAMPLE (a) Suppose the graph of y = g(x) is given below. If g(x) = f’(x), what is the nature of f(x) when x = -1? (b) Suppose the graph of y = g(x) is given below. If g(x) = f’’(x), what is the nature of f(x) when x = -1?

38. Curve Sketching 1) Find the local minimum and local maximum of y = f(x) and the increasing and decreasing intervals. 2) Find the inflection points (if any) of y = f(x). These inflection points are possible critical values (i.e. possible local minimum or local maximum) on the graph of y = f’(x). Given the graph of y = f(x), how to sketch the graph of y = f’(x)?

39. § 2.5 Optimization Problems

40. Maximizing AreaEXAMPLE Find the dimensions of the rectangular garden of greatest area that can be fenced off (all four sides) with 300 meters of fencing.

41. Minimizing CostEXAMPLE (Cost) A rectangular garden of area 75 square feet is to be surrounded on three sides by a brick wall costing $10 per foot and on one side by a fence costing $5 per foot. Find the dimensions of the garden such that the cost of materials is minimized.

42. Cost, Revenue & ProfitEXAMPLE (Cost and Profit) A one-product firm estimates that its daily total cost function (in suitable units) is C(x) = x 3 – 6x 2 + 13x + 15 and its total revenue function is R(x) = 28x. Find the value of x that maximizes the daily profit.