\(x\) years after the year 2015, the population of the city of Fulton is given by the function \(y= f(x) = 35000(1.03^x)\). \(x\) years after the year 2015, the population of the city of Greenville is...\(x\) years after the year 2015, the population of the city of Fulton is given by the function \(y= f(x) = 35000(1.03^x)\). \(x\) years after the year 2015, the population of the city of Greenville is given by the function \(y = g(x) = 80000(0.95^x)\). The functions represent population size as a function of time after the year 2015 . We restrict the domain in this context, using the “practical domain” as the set of all non-negative real numbers: \(x≥0\).
\(x\) years after the year 2015, the population of the city of Fulton is given by the function \(y= f(x) = 35000(1.03^x)\). \(x\) years after the year 2015, the population of the city of Greenville is...\(x\) years after the year 2015, the population of the city of Fulton is given by the function \(y= f(x) = 35000(1.03^x)\). \(x\) years after the year 2015, the population of the city of Greenville is given by the function \(y = g(x) = 80000(0.95^x)\). The functions represent population size as a function of time after the year 2015 . We restrict the domain in this context, using the “practical domain” as the set of all non-negative real numbers: \(x≥0\).
\(x\) years after the year 2015, the population of the city of Fulton is given by the function \(y= f(x) = 35000(1.03^x)\). \(x\) years after the year 2015, the population of the city of Greenville is...\(x\) years after the year 2015, the population of the city of Fulton is given by the function \(y= f(x) = 35000(1.03^x)\). \(x\) years after the year 2015, the population of the city of Greenville is given by the function \(y = g(x) = 80000(0.95^x)\). 6) To set the X values to what you want, hit the 2nd button and the Window button . Set the beginning value of 0, and the amount to increase the x value by 0.25 then hit the 2nd button then the Graph button.
