Question

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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)n


Part 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?

Answer 1

Answer:

Part A) The reasonable domain to plot  the growth function is the interval [0,5]

Part B) The average rate of change is $0.21\ \frac{cm}{day}$

see the explanation

Step-by-step explanation:

Part A)

Let

f(n) -----> the height of the plant in cm

n ----> the number of days

we have

$f(n)=10(1.02)^n$

This is a exponential function of the form

$f(x)=a(b)^x$

where

a is the initial value

b is the base

r is the rate of growth

b=(1+r)

In this problem we have

$a=10\ cm$ ----> initial value or y-intercept

$b=1.02$

$r=b-1=1.02-1=0.02$

$r=2\%$

For f(n)=11.04 cm

Find the value of n

substitute in the exponential function

$11.04=10(1.02)^n$

$11.04/10=(1.02)^n$

$1.104=(1.02)^n$

Apply log both sides

$log(1.104)=(n)log(1.02)$

$n=log(1.104)/log(1.02)$

$n=5\ days$

so

The reasonable domain to plot  the growth function is the interval -----> [0,5]

$0 \leq x \leq 5$

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

$\frac{f(b)-f(a)}{b-a}$

In this problem we have

$f(a)=f(1)=10(1.02)^1=10.2\ cm$  

$f(b)=f(5)=10(1.02)^5=11.04\ cm$

$a=1$

$b=5$

Substitute

$\frac{11.04-10.2}{5-1}=0.21\ \frac{cm}{day}$

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

Answer 2

Answer:

A)  0 ≤ n ≤ 5

B)  0.21 cm per day

Step-by-step explanation:

Given:

$f(n)=10(1.02)^n$

where:

• f(n) = height of the plant (in cm)
• n = number of days

Part A

To find the upper limit of the domain, find when the height of the plant was approximately 11.04 cm by substituting f(n) = 11.04 and solving for n:

$\implies 10(1.02)^n=11.04$

$\implies (1.02)^n=\dfrac{11.04}{10}$

$\implies (1.02)^n=1.104$

$\implies \ln (1.02)^n= \ln 1.104$

$\implies n\ln 1.02= \ln 1.104$

$\implies n=\dfrac{ \ln 1.104}{\ln 1.02}$

$\implies n=4.996304095...$

$\implies n=5$

Therefore, a suitable domain to plot the growth function is 0 ≤ n ≤ 5

Part B

The average rate of change of function f(x) over the interval a ≤ x ≤ b is given by:

$\dfrac{f(b)-f(a)}{b-a}$

Therefore, the average rate of change of function f(n) over the interval 1 ≤ n ≤ 5 is:

$\implies \dfrac{f(5)-f(1)}{5-1}$

$\implies \dfrac{10(1.02)^5-10(1.02)^1}{5-1}$

$\implies 0.210202008$

Therefore, the average rate of change is 0.21 cm per day.

This represents the average growth of the plant each day.