In this case, a reasonable domain to plot the growth function - 6.97, the y-intercept of the graph of the function f(d) represent = 11 (initial numbers), and the average rate of change of the function f(d) from d = 2 to d = 7 = 0.11447.
Given:
f(d) = 11[tex](1.01)^{d}[/tex]
Part A:
f(d) = 11.79
11.79=11[tex](1.01)^{d}[/tex]
[tex](1.01)^{d}[/tex] = 11.79/11
[tex](1.01)^{d}[/tex] = 1.071818
d = [tex]\frac{log 1.071818}{log 1.01}[/tex]
d = 6.97
Part B:
y- intercept is obtained when domain = 0
d = 0 ⇒ f(0)
= 11(1.01)⁰
= 11*1 = 11
f(d) = 11
Part C:
The average rate of change -
[tex]Rate \ of \ change = \frac{\partial }{\partial d} f(d)\\ = \frac{\partial}{\partial d} (1.01)^d .11\\ = 11(1.01)^d \ log (1.01)\\Average \ rate \ of \ change = \frac{\int\limits^7_2 \frac{\partial f(d)}{\partial d} \partial d}{\int\limits^7_2 \partial d } \\= \frac{\int_{2}^{7} 11(1.01)^d log_e 1.01 \ \partial d}{\int_{2}^{7} \partial d}\\= \frac{11 log_2 1.01}{5} \int_{2}^{7} 1.01^d \partial d\\= \frac{11}{5} log_e 1.01\times [\frac{1.01^d}{log_e(1.01)}]_2^7\\[/tex]
[tex]= \frac{11}{5}[1.01^7-1.01^2]\\= \frac{11}{5}\times 0.05203\\= 0.11447 \ mm/day[/tex]
Thus, a reasonable domain to plot the growth function - 6.97, the y-intercept of the graph of the function f(d) represent = 11 (initial numbers), and the average rate of change of the function f(d) from d = 2 to d = 7 = 0.11447.
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https://brainly.com/question/19516348
