Question

(04.03 MC)

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) = 8(1.05)n

Part A: When the scientist concluded his study, the height of the plant was approximately 11.26 cm. What is a reasonable domain to plot the growth function? (4 points)

Part B: What does the y-intercept of the graph of the function f(n) represent? (2 points)

Part C: What is the average rate of change of the function f(n) from n = 2 to n = 6, and what does it represent? (4 points)

(10 points)

Answer 1

A) 0 ≤ n ≤ 7

B) initial height of plant = 8 cm

C) average rate of change = 0.475 cm/day.

Plant grew an average of 0.475 cm per day from day 2 to day 6

Explanation:

`f(n)=8(1.05)^n`

Part A

Given height = 11.26 cm

`\implies 8(1.05)^n=11.26`

`\implies(1.05)^n=(11.26)/(8)`

`\implies(1.05)^n=1.4075`

Taking natural logs:

`\implies \ln(1.05)^n=\ln(1.4075)`

`\implies n\ln(1.05)=\ln(1.4075)`

`\implies n=(\ln(1.4075))/(\ln(1.05))`

`\implies n=7.005819448`

`\implies n \approx7`

Therefore, the scientist concluded his study after 7 days, so the domain is 0 ≤ n ≤ 7

Part B

y-intercept is when the curve intersects the y-axis (when n = 0):

`\implies f(0)=8(1.05)&^0=8*1=8`

The y-intercept represents the initial height of the plant, which is 8 cm

Part C

Rate of change = change in y ÷ change in x

⇒ rate of change = change in f(n) ÷ change in n

`\implies f(2)=8(1.05)^2=8.82`

`\implies f(6)=8(1.05)^6=10.72076513...`

`\implies \textsf{rate of change} \ =(f(6)-f(2))/(6-2)=0.4751912813...`

So the average rate of change was 0.475 cm/day. This indicates that the plant grew an average of 0.475 cm per day from day 2 to day 6 (over a period of 4 days).

ETA: please see attached graph. The gray dotted curve is the function, and the black solid curve is the function with the restricted domain of 0≤n≤7 since the study took 7 days.