Question

The function f(x)=1200(1.055)x models the balance of an investment x years after it is made.

How does the average rate of change between years 21 and 25 compare to the average rate of change between years 1 and 5?

Answer 1

B i took the test

Explanation:

Answer 2

option-B

Explanation:

We are given function as

`f(x)=1200(1.055)^x`

Average rate of change between 21 and 25 years:

we can use formula

`A=(f(b)-f(a))/(b-a)`

so, we have

a=21 and b=25

`f(21)=1200(1.055)^(21)=3693.88`

`f(25)=1200(1.055)^(25)=4576.0708`

`A=(f(25)-f(21))/(25-21)`

now, we can plug values

`A=(4576.0708-3693.88)/(25-21)`

`A_1=220.5477`

Average rate of change between 1 and 5 years:

we can use formula

`A=(f(b)-f(a))/(b-a)`

so, we have

a=1 and b=5

`f(1)=1200(1.055)^(1)=1266`

`f(5)=1200(1.055)^(5)=1568.352`

`A=(f(5)-f(1))/(5-1)`

now, we can plug values

`A_2=(1568.352-1266)/(5-1)`

`A_2=75.588`

now, we can find ratio

`(A_1)/(A_2)=(220.5477)/(75.588)`

`(A_1)/(A_2)=3`

`A_1=3A_2`