1. Copyright © Cengage Learning. All rights reserved. Combinations of Functions SECTION 7.1

2. 2 Learning Objectives 1 Compute the sum, difference, product, or quotient of functions to model a real-world situation 2 Determine the practical and theoretical domain of the combination of functions

3. 3 Combining Functions Graphically and Numerically

4. 4 The graph in Figure 7.1 shows the life expectancy of men and women born between 1980 and 2004. What is the trend? Figure 7.1 Life Expectancy

5. 5 Combining Functions Graphically and Numerically The graph in Figure 7.1 shows the life expectancy of men and women born between 1980 and 2004. Looking at the overall trend in life expectancies in Figure 7.2, we see that the difference in life expectancy between women and men (D) decreases as time increases. Figure 7.1Figure 7.2 Life Expectancy

6. 6 Combining Functions Graphically and Numerically To see this more clearly, we calculate the differences throughout the data set, as shown in Table 7.1. Table 7.1

7. 7 Combining Functions Graphically and Numerically The table confirms the overall trend that the life expectancy gap is decreasing as time increases. Table 7.1 (continued)

8. 8 Combining Functions Graphically and Numerically The gap in life expectancy generally decreases from 7.4 years in 1980 to 5.2 years in 2004. We create a scatter plot of the differences in Figure 7.3 and observe the same trend. Figure 7.3 Difference in Life Expectancy

9. 9 Combining Functions Symbolically

10. 10 Example 1 – Combining Functions Symbolically Using linear regression, we determine that the life expectancy for women can be modeled by and the life expectancy for men by, where t is the number of years since 1980. a. Determine a function that models the difference in life expectancy between men and women. Name this function D(t).

11. 11 Example 1 – Solution We combine the like terms to simplify the function. b. Interpret the meaning of the slope and vertical intercept of D(t) in the context of this situation. cont’d

12. 12 Example 1 – Solution The slope of –0.098 means the difference in life expectancy between men and women is decreasing by 0.098 years of age per year. The vertical intercept of (0, 7.6) means that the difference in 1980 is 7.6 years. (The actual difference is 7.4 but the model predicts 7.6.) c. Graph D(t) along with the scatter plot shown in Figure 7.3. Comment on the accuracy of the model. cont’d

13. 13 Example 1 – Solution c. Figure 7.4 shows the graph of along with the scatter plot of the data. We see that this function models the data reasonably well. cont’d Figure 7.4 Difference in Life Expectancy

14. 14 Example 1 – Combining Functions Symbolically d. Describe the practical domain and range of D(t). cont’d

15. 15 Example 1 – Solution The practical domain for the function is the years for which the model can reasonably represent the situation. We choose the interval 0  t  30, which means we reserve the use of our model for years between 1980 (t = 0) and 2010 (t = 30). The practical range for the function is the differences in life expectancy for which the model can reasonably represent the situation. Based on the practical domain, we determine that the interval 4.66  D  7.60 is the practical range. This corresponds to the difference in life expectancy in 2010 (4.66) and in 1980 (7.60). cont’d

16. 16 Dividing Functions

17. 17 Dividing Functions The U.S. federal government brings in money (revenue) primarily through taxes paid by its citizens. It spends money (expenses) on military, social programs, and education, among other programs. Currently, the government spends more money than it receives creating a “national debt.” To make up for this deficit, it borrows money by selling securities like Treasury bills, notes, bonds, and savings bonds to the public.

18. 18 Dividing Functions Based on data from 1940 to 2004, the national debt can be modeled by the exponential function, where t is measured in years since 1940 and D is measured in trillions of dollars. (Modeled from www.whitehouse.gov data) The population of the United States can be modeled by the linear function, where t is years since 1940 and P is millions of people. We use these models in the next example as we combine functions using division.

19. 19 Example 3 – Combining Functions Using Division Suppose every person (including children) in the United States was to pay an equal amount of money to eliminate the national debt. Using the functions D and P just defined, create a function A that would give the amount of money each person in the United States would need to pay. Then evaluate and interpret A(70). Solution: The function D gives the total national debt (in trillions of dollars) for a given year (since 1940).

20. 20 Example 3 – Solution If we divide this value by the population for that same year, we get an average amount of debt per person in the United States. Note the units used in this combination of functions is trillion dollars per million people. cont’d

21. 21 Example 3 – Solution Evaluating A(70) will provide us the amount of money required per person to completely pay for the national debt in 2010, since 2010 is 70 years after 1940. We express this result as a fraction and simplify. cont’d

22. 22 Example 3 – Solution If everyone paid an equal amount, every man, woman, and child in the United States would have to pay $32,100 in 2010 in order to pay off the national debt. cont’d

23. 23 Multiplying Functions

24. 24 Multiplying Functions The National Automobile Dealers Association (NADA) reports statistics on the sales of new vehicles in the United States. The number of new vehicles sold per dealership in the United States is given in Table 7.2 and Figure 7.6. Table 7.2 Figure 7.6 New Vehicles Sold

25. 25 Multiplying Functions We model these data using the quartic function, as seen in Figure 7.7. Figure 7.7 New Vehicles Sold

26. 26 Multiplying Functions The NADA also reports the average retail selling price of these new vehicles. Table 7.3 and Figure 7.8 provide this information. Figure 7.8 Table 7.3 New Vehicle Average Selling Price

27. 27 Multiplying Functions We model these data using the linear function, as seen in Figure 7.9. In the next example, we use these models to investigate the product of two functions. Figure 7.9 New Vehicle Average Selling Price

28. 28 Example 5 – Interpreting the Product of Two Functions The function models the number of new vehicles sold, where t is measured in years since 1995 and V is the number of new vehicles per dealership in the United States. The function models the average retail selling price, P, of a new vehicle t years after 1995. a. Find and interpret f (15) given. b. Express the function symbolically. c. Interpret the meaning of the function.

29. 29 Example 5(a) – Solution We first evaluate and interpret V(15). This tells us that 15 years after 1995 (2010), the number of new vehicles sold per dealership will be 2180 vehicles. Next we evaluate and interpret P(15).

30. 30 Example 5(a) – Solution This tells us that 15 years after 1995 (2010), the average retail selling price of a new vehicle will be approximately $32,700. To evaluate f (15), we calculate. cont’d

31. 31 Example 5(a) – Solution This tells us that 15 years after 1995 (2010), a dealership will generate approximately $71 million in income from the sales of new vehicles. That is, if 2180 new vehicles are sold at an average cost of $32,700 each, roughly $71 million in income is generated. cont’d

32. 32 Example 5(b) – Solution We now express symbolically. cont’d

33. 33 Example 5(c) – Solution Since V(t) is the number of new vehicles sold per dealership and P(t) is the average retail selling price per vehicle (in dollars), we interpret to be the total income from new vehicle sales per dealership t years after 1995. That is, if each dealership sold V(t) vehicles at a price of P(t) each, the income for the dealership would be f (t) dollars. cont’d

34. 34 Multiplying Functions The following box summarizes our work so far in this section and introduces formal notation for the combination of functions.