Question

in 2012, the population of city had been growing steadily, and the exponential function defined by $f(t)=42.4cdot1.06^{t}$ models the population from 2005 to 2012, where t represents the number of years since 2005 and f(t) represents the population of the city, measured in thousands of people. use the given information to answer parts a through f.
a. it is a growth function with growth rate, $r = 6%$.
(type an integer or a decimal.)
b. it is a decay function with decay rate, $r=square%$.
(type an integer or a decimal.)
c. according to the model, what is the initial value? what does the value mean in this situation? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the initial value is 42.4. this means that the population of the city in 2005 was approximately 42.4 thousand people. (type an integer or a decimal.)
b. the initial value is $square$. this means that the population of the city increased by approximately $square$ thousand people per year. (type an integer or a decimal.)
d. the equation for continuous growth is $y = ae^{kt}$. set the value of b in your model equal to $e^{k}$, and use your graphing calculator to determine the value for k graphically. this will be the continuous growth or decay model.
$kapproxsquare$
(round to three decimal places as needed.)

Explanation:

Step1: Recall the relationship between \(b\) and \(k\)

We know that for the exponential functions \(y = a\cdot b^{t}\) and \(y=a\cdot e^{kt}\), if \(b = e^{k}\). In the given function \(f(t)=42.4\cdot1.06^{t}\), we have \(b = 1.06\).

Step2: Solve for \(k\)

Since \(b = e^{k}\), then \(k=\ln(b)\). Substitute \(b = 1.06\) into the formula \(k=\ln(1.06)\).
Using a calculator, \(\ln(1.06)\approx0.058\) (rounded to three decimal places).

Answer:

\(k\approx0.058\)