Q:

PLEASE HELP 100PTS!!!!!!!! I SOLVED IT BUT I WANT TO CHECK MY ANSWERS!!DONT COPY OTHER ANSWERS I CAN TELL AND REPORT YOU!!!!A scientist is studying the growth of a particular species of plant. He writes the following equation to show the height of the plant f(n), in cm, after n days:f(n) = 10(1.02)nPart A: When the scientist concluded his study, the height of the plant was approximately 11.04 cm. What is a reasonable domain to plot the growth function?Part B: What is the average rate of change of the function f(n) from n = 1 to n = 5, and what does it represent?

Accepted Solution

A:
Answer:Part A) The reasonable domain to plot  the growth function is the interval [0,5]Part B) The average rate of change is [tex]0.21\ \frac{cm}{day}[/tex] see the explanationStep-by-step explanation:Part A) Letf(n) -----> the height of the plant in cmn ----> the number of dayswe have[tex]f(n)=10(1.02)^n[/tex]This is a exponential function of the form[tex]f(x)=a(b)^x[/tex]wherea is the initial valueb is the baser is the rate of growthb=(1+r)In this problem we have[tex]a=10\ cm[/tex] ----> initial value or y-intercept[tex]b=1.02[/tex][tex]r=b-1=1.02-1=0.02[/tex][tex]r=2\%[/tex]For f(n)=11.04 cmFind the value of nsubstitute in the exponential function[tex]11.04=10(1.02)^n[/tex][tex]11.04/10=(1.02)^n[/tex][tex]1.104=(1.02)^n[/tex]Apply log both sides[tex]log(1.104)=(n)log(1.02)[/tex][tex]n=log(1.104)/log(1.02)[/tex][tex]n=5\ days[/tex]soThe reasonable domain to plot  the growth function is the interval -----> [0,5][tex]0 \leq x \leq 5[/tex]Part B) What is the average rate of change of the function f(n) from n = 1 to n = 5, and what does it represent?the average rate of change is equal to [tex]\frac{f(b)-f(a)}{b-a}[/tex] In this problem we have [tex]f(a)=f(1)=10(1.02)^1=10.2\ cm[/tex]  [tex]f(b)=f(5)=10(1.02)^5=11.04\ cm[/tex] [tex]a=1[/tex] [tex]b=5[/tex] Substitute [tex]\frac{11.04-10.2}{5-1}=0.21\ \frac{cm}{day}[/tex] The average rate of change is the change of the function values (output values) divided by the change of the input values.That represent ----> The plant grew an average of 0.21 cm per day during that time interval