1. Answers to even-numbered HW problems S-2 42,943,441.08 Ex 6 a) C(98,230) =$7.04 b) C(1.03, 172) means the cost of operating a car when gas is $1.03 per gallon and you travel 172 miles. It’s value is $5.54. Ex 8 a) 45 degrees b) R(30) = 84 degrees c) 13.66 degrees d) 10.12 degrees Ex 16 a).355 dynes b) 199.61 dynes Section 1.1

2. WebAssign Homework

3. WebAssign Homework – Section 1-1

4. Last time, we found a formula for the area of a garden surrounded by 20 yards of chicken wire on three sides. We said the formula is an example of a function and we used the functional notation A(W) instead of just A. A(W) = 20W – 2W 2 Today, we will refine the concept of a FUNCTION.

5. Consider the function A(W) = 20W – 2W 2 Warm-up Use a calculator to compute the values of A(4), A( ), A(10), and A(-7). Round answers to the nearest hundredth (two decimal places). 7 8

6. Consider the function A(W) = 20W – 2W 2 Warm-up A(4) = 48 Use a calculator to compute the values of A(4), A( ), A(10), and A(-7). Round answers to the nearest hundredth (two decimal places). 7 8

7. Consider the function A(W) = 20W – 2W 2 Warm-up A( ) = 15.97A(4) = 48 7 8 Use a calculator to compute the values of A(4), A( ), A(10), and A(-7). Round answers to the nearest hundredth (two decimal places). 7 8

8. Consider the function A(W) = 20W – 2W 2 Warm-up A( ) = 15.97A(4) = 48A(10) = 0 7 8 Use a calculator to compute the values of A(4), A( ), A(10), and A(-7). Round answers to the nearest hundredth (two decimal places). 7 8

9. Consider the function A(W) = 20W – 2W 2 Warm-up A( ) = 15.97A(4) = 48A(10) = 0A(-7) = -238 7 8 Use a calculator to compute the values of A(4), A( ), A(10), and A(-7). Round answers to the nearest hundredth (two decimal places). 7 8

10. Consider the function A(W) = 20W – 2W 2 Warm-up A( ) = 15.97A(4) = 48A(10) = 0A(-7) = -238 Input value Output or Function value 7 8 Use a calculator to compute the values of A(4), A( ), A(10), and A(-7). Round answers to the nearest hundredth (two decimal places). 7 8

11. Consider the function A(W) = 20W – 2W 2 Warm-up A( ) = 15.97A(4) = 48A(10) = 0A(-7) = -238 We could also write this in table form 7 8 Use a calculator to compute the values of A(4), A( ), A(10), and A(-7). Round answers to the nearest hundredth (two decimal places). 7 8

12. Consider the function A(W) = 20W – 2W 2 Warm-up A( ) = 15.97A(4) = 48A(10) = 0A(-7) = -238 We could also write this in table form W410-7 A(W)4815.970-238 7 8 7 8 Input values Output or Function values Use a calculator to compute the values of A(4), A( ), A(10), and A(-7). Round answers to the nearest hundredth (two decimal places). 7 8

13. A function can be thought of as a pairing of the numbers in one set with the numbers in a second set. Sometimes the rule for the pairs is given as an equation or is easy to spot – sometimes it isn’t. Definition of a function: A function is a correspondence between two sets that assigns to each element of the first set exactly one element of the second set. The first set of numbers is called the domain. domain range. WA(W) 448 15.97 100 -7-238 7 8 The second set of numbers is called the range.

14. Definition of a function: A function is a correspondence between two sets that assigns to each element of the domain exactly one element of the range. We usually refer to the function by the letter name for the range. In this case, we would call it function A. In this example, the domain is {4,, 10, -7} and the range is {48, 15.97, 0, -238} 7 8 Note: the table for the function says the same thing as A(4) = 48, A( ) = 15.97, A(10) = 0, A(-7) = -238. 7 8 WA(W) 448 15.97 100 -7-238 7 8

15. Summary: In a function, the first components (domain) never repeat. The second components (range) may repeat. Example: A(W) = 20W – 2W 2 W47/810-7 A(W)4815.970-238 A function can be represented in several ways, including:  As a formula Example:  As a table  As an arrow diagram 4 48 10 0 15.97 -7 -238 Example: 7 8  As a set of number pairs Example: {(4, 48), (, 15.97), (10, 0), (-7, -238)} 7 8 For example, W47/810-7 A(W)4815.970-238 6 48

16. Using functions given as tables of values

17. The population of the United States has increased steadily over the years. Thus, the population (N) of the U.S. is a function of the year (d). We can use N(d) in place of N.

18. N = N(d) We can still use function notation in this situation. For example: What is meant by N(1980)? “the population of the United States in 1980” How can we use the table to approximate the population of the United States in 1985? Since 1985 is midway between 1980 and 1990, both of which are in the table, one method is to average the populations for 1980 and 1990. N(1980) = 226.5 million

19. N = N(d) N(1980) = N(1990) = 226.5 million 248.7 million = 237.6 million Average population = A reasonable estimate for the population of the U.S. in 1985 is 237.6 million. Therefore, N(1985)  237.6 million United States Population July 1, 1987242,288,9182,156,0310.89 July 1, 1986240,132,8872,209,0920.92 July 1, 1985237,923,7952,098,8930.89 July 1, 1984235,824,9022,032,9080.87 July 1, 1983233,791,9942,127,536 Source: http://www.npg.org/facts/us_historical_pops.htm

20. N = N(d) How would you estimate the population of the United States in 1945? Averaging assumes that values of the function (in this case, population) increase at a constant (uniform) rate.

21. United States Population July 1, 1950152,271,4173,083,2872.05 July 1, 1949149,188,1302,556,8281.73 July 1, 1948146,631,3022,505,2311.72 July 1, 1947144,126,0712,737,5051.92 July 1, 1946141,388,5661,460,4011.04 July 1, 1945139,928,1651,530,8201.10 July 1, 1944138,397,3451,657,9921.21 July 1, 1943136,739,3531,879,8001.38 July 1, 1942134,859,5531,457,0821.09 July 1, 1941133,402,4711,280,0250.96 July 1, 1940132,122,4461,242,7280.95 The U.S. population grew by less than 8 million from 1940-1945. The U.S. population grew by more than 12 million from 1945-1950. Source: http://www.npg.org/facts/us_historical_pops.htm The rate of growth was faster in the second half of the decade. If we had averaged, million people

22. N = N(d) How would you estimate the population of the United States in 1945? Averaging assumes that values of the function (in this case, population) increase at a constant (uniform) rate. Averaging would not be an appropriate method for estimating the United States population in 1945 because historical evidence suggests the population did not rise at a uniform rate between 1940 and 1950. The population increased far more rapidly from 1945 to 1950 because of the post World War II Baby Boom.

23. N = N(d) What if we wanted an estimate of the population in 1983? To find the average rate of change in population over an interval, divide the change (difference) in population by the number of years in the interval (change in years). In our example: So the average rate of change for the ten year period from 1980 to 1990 is 2.22 million people per year. To do this, we will find the average yearly change in population between 1980 and 1990 (called the average rate of change). This means that, on average, the population increased by 2.22 million per year from 1980 to 1990. Therefore, our estimate for the population in 1983 is 226.5 + 3(2.22) = 233.16 million

24. Therefore, our estimate for the population in 1983 is 226.5 + 3(2.22) = 233.16 million United States Population Jul 1, 1990249.62 million Jul 1, 1989246.82 million Jul 1, 1988244.50 million Jul 1, 1987242.29 million Jul 1, 1986240.13 million Jul 1, 1985237.92 million Jul 1, 1984235.83 million Jul 1, 1983233.79 million Jul 1, 1982231.66 million Jul 1, 1981229.47 million Jul 1, 1980227.23 million http://www.multpl.com/united-states-population/table

25. United States Population July 1, 1950152,271,4173,083,2872.05 July 1, 1949149,188,1302,556,8281.73 July 1, 1948146,631,3022,505,2311.72 July 1, 1947144,126,0712,737,5051.92 July 1, 1946141,388,5661,460,4011.04 July 1, 1945139,928,1651,530,8201.10 July 1, 1944138,397,3451,657,9921.21 July 1, 1943136,739,3531,879,8001.38 July 1, 1942134,859,5531,457,0821.09 July 1, 1941133,402,4711,280,0250.96 July 1, 1940132,122,4461,242,7280.95 The U.S. population grew by less than 8 million from 1940-1945. The U.S. population grew by more than 12 million from 1945-1950. Source: http://www.npg.org/facts/us_historical_pops.htm Average rate of change from 1940-1945: 1.56 million people per year Average rate of change from 1945-1950: 2.48 million people per year

26. N = N(d) To find the average rate of change in population over an interval, divide the change (difference) in population by the number of years in the interval (change in years). So the average rate of change for the ten year period from 1980 to 1990 is 2.22 million people per year. This means that, on average, the population increased by 2.22 million per year from 1980 to 1990. Therefore, our estimate for the population in 1983 is 226.5 + 3(2.22) = 233.16 million

27. P = Number of Breeding cccPairs of Bald Eagles Bald Eagles were once very common throughout most of the United States. Their population numbers have been estimated at 300,000 to 500,000 birds in the early 1700s. Their population fell to threatened levels in the continental U.S. of less than 10,000 nesting pairs by the 1950s. Bald eagles were officially declared an endangered species in 1967 in all areas of the United States Shown is a table indicating the estimated number of breeding pairs of bald eagles in the lower 48 states in various years. Answer each of the following questions about the function P = P(y), where y is the year and P is the number of breeding pairs of bald eagles. 1940195219561981198619922000 9,5514,7503,2101,1881,8753,3996,471 y = Year Practice

28. 1.Explain the meaning of P(1956) and give its value. 2.Estimate the value of P(1989). 3.Find the average yearly rate of change from 1992 to 2000. 4.Use your answer to question 3 to estimate the number of breeding pairs of bald eagles in 1994. P = Number of Breeding cccPairs of Bald Eagles 1940195219561981198619922000 9,5514,7503,2101,1881,8753,3996,471 y = Year P(1989) is approximately 2,637. The average yearly rate of change from 1992 to 2000 was 384 breeding pairs per year. P(1956) represents the number of breeding pairs of bald eagles in the U.S. in 1956. P(1956) = 3,210 There were approximately 4,167 breeding pairs of bald eagles in the U.S. in 1994. [3,399 + 2(384)]

29. Homework: Read section 1.2 (omit pages 46–51) Do the following problems: Page 52 # S-1, S-3, S-8, S-11, S-15 Pages 53-56 # 1, 7, 8, 10 Not part c Not parts g and h