Question

Sue interests $800 in an investment that can be represented by the function m=800(1.08)^t where t is the number of years the money is invested for, and m is the value of the investment in dollars. Enter the average rate or change, rounded to the nearest dollar per year, from year 2 to year 4

Answer 1

`\begin{array}{llll} f(x)~from\\\\ x_1 ~~ to ~~ x_2 \end{array}~\hfill slope = m \implies \cfrac{ \stackrel{rise}{f(x_2) - f(x_1)}}{ \underset{run}{x_2 - x_1}}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array} \\\\[-0.35em] ~\dotfill\\\\ m(t)=800(1.08)^t \qquad \begin{cases} t_1=2\\ t_2=4 \end{cases}\implies \cfrac{m(4)-m(2)}{4 - 2} \\\\\\ \cfrac{800(1.08)^4~~ - ~~800(1.08)^2}{2} ~~ \approx ~~ \cfrac{1088.39-933.12}{2}~~ \approx ~~\text{\LARGE 78} ~~ (\$)/(year)`