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 resonable domain to plot the growth function?
PART B) What does the y-intercept of the graph of the function f(n) represent?
PART C) What is the average rate of change of the function f(n) from n=1 to n= 5, and what does it represent?

PART A)
11.04 = 10(1.02)^n
1.104 = 1.02^n
ln 1.104 = ln 1.02^n
ln 1.104 = n ln 1.02
n = ln 1.104/ ln 1.02
n = 4.99630409516
4.99 can be rounded to 5.
So a reasonable domain would be 0 ≤ x < 5

PART B)
f(0) = 10(1.02)^0
f(0) = 10(1)
f(0) = 10
The y-intercept represents the height of the plant when they began the experiment.

PART C)
f(1) = 10(1.02)^1
f(1) = 10(1.02)
f(1) = 10.2
(1, 10.2)

f(5) = 10(1.02)^5
f(5) = 10(1.1040808)
f(5) = 11.040808
(5, 11.040808)

Find the slope between those two points.

\( \text{Slope } = \frac{ y_2 - y_1 } { x_2 - x_1 } \\ \\= \frac{ \frac{1380101}{125000} - \frac{51}{5}}{ 5 - 1} \\\\ =\frac{ \frac{105101}{125000}}{ 4} \\\\= \frac{ \frac{105101}{125000}\cdot125000}{ 4\cdot125000} \\\\=\frac{ 105101}{ 500000}\\\\\\=\boxed{\bf{0.210202}} \)

The slope represents how much the plant is growing every day.

Let

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

n-> is the number of days

we now that

The equation to show the height of the plant is equal to

\( f(n)= 10*(1.02)^n \)

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?

Find the value of n for \( f(n)=11.04 \)

\( f(n)= 10*(1.02)^n \)

\( 11.04= 10*(1.02)^n \)

\( 1.104=(1.02)^n \)

Applying logarithm both sides

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

\( ln(1.104)=n*ln(1.02)\\\\ n= \frac{ln(1.104)}{ln(1.02)} \\ \\ n=5\ days \)

So a reasonable domain would be

\( 0 \leq n \leq 5 \)

therefore

the answer Part a) is

\( 0 \leq n \leq 5 \)

see the attached figure N 1

Part b) What does the y-intercept of the graph of the function f(n) represent?

we know that

the y-intercept is when the value of n is equal to zero

so

\( n=0 \)

\( f(0)= 10*(1.02)^0 \)

\( f(0)= 10\ cm \)

The y-intercept represents the height of the plant when they began the experiment (\( 10\ cm \))

see the attached figure N 1

therefore

the answer Part b) is

The y-intercept represents the height of the plant when they began the experiment

Part c) What is the average rate of change of the function f(n) from n=1 to n= 5, and what does it represent?

\( For \ n=1\ find\ f(1) \)

\( f(1)= 10*(1.02)^1 \)

\( f(1)= 10.2\ cm \)

\( For \ n=5\ find\ f(5) \)

\( f(5)= 10*(1.02)^5 \)

\( f(5)= 11.04\ cm \)

Find the slope m

\( m=\frac{f(5)-f(1)}{(5-1)} \\ \\ m=\frac{(11.04-10.2)}{4} \\ \\ m=0.21\ \frac{cm}{day} \)

The slope represents how much the plant is growing every day (\( 0.21\ \frac{cm}{day} \))

therefore

the answer part c) is

The slope represents how much the plant is growing every day

PART A)  11.04 = 10(1.02)^n  1.104 = 1.02^n  ln 1.104 = ln 1.02^n  ln 1.104 = n

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