Section 1.3Exponential Functions25EXAMPLE 5Predicting the U.S. PopulationUse the population data in Table 1.8 to estimate the population for the year 2000. Com-pare the result with the actual 2000 population of approximately 281.4 million.SOLUTIONModelLet x0 represent 1880,x1 represent 1890, and so on. We enter the datainto the grapher and find the exponential regression equation to bef(x)(56.4696)(1.1519)x.Figure 1.26 shows the graph of fsuperimposed on the scatter plot of the data.Solve GraphicallyThe year 2000 is represented by x12. Reading from the curve,we findf12308.2.The exponential model estimates the 2000 population to be 308.2 million, an overesti-mate of approximately 26.8 million, or about 9.5%.Now try Exercise 39(a, b).EXAMPLE 6Interpreting Exponential RegressionWhat annualrate of growth can we infer from the exponential regression equation inExample 5?SOLUTIONLet rbe the annual rate of growth of the U.S. population, expressed as a decimal. Be-cause the time increments we used were 10-year intervals, we have(1r)101.1519r101.15191r0.014The annual rate of growth is about 1.4%.Now try Exercise 39(c).The Number eMany natural, physical, and economic phenomena are best modeled by an exponentialfunction whose base is the famous number e, which is 2.718281828 to nine decimalplaces. We can define eto be the number that the function f(x) (1 1x)xapproachesas xapproaches infinity. The graph and table in Figure 1.27 strongly suggest that such anumber exists.The exponential functions yexand yexare frequently used as models of expo-nential growth or decay. For example, interest compounded continuouslyuses the modelyP•ert, where Pis the initial investment,ris the interest rate as a decimal, and tistime in years.[–1, 15] by [–50, 350]X = 121Y = 308.1712Figure 1.26(Example 5)XY1 = (1+1/X)^X10002000300040005000600070002.71692.71762.71782.71792.7182.71812.7181Y1Figure 1.27A graph and table of valuesfor f(x) (1 1x)xboth suggest that asx→,f x→e2.718.