Question

The function f(x)=125(0.9)x models the population of a species of fly in millions after x years.

How does the average rate of change between years 11 and 15 compare to the average rate of change between years 1 and 5?

A) The average rate of change between years 11 and 15 is about 1/3 the rate between years 1 and 5.

B) The average rate of change between years 11 and 15 is about 1/2 the rate between years 1 and 5 .

C) The average rate of change between years 11 and 15 is about 3 times the rate between years 1 and 5.

D) The average rate of change between years 11 and 15 is about 2 times the rate between years 1 and 5 .

Answer 1

Just took the test, the answer was "A) The average rate of change between years 11 and 15 is about
`(1)/(3)` the rate between years 1 and 5."

Answer 2

A word to the wise: It's f(x)=125(0.9)^x, where ^ represents exponentiation.

In this case the ave. value over the interval [11, 15] is

125(0.9)^15 - 125(0.9)^11
------------------------------------- = (125/4) [ 0.9^15 - 0.9^11)
15 - 11 = (31.25) [ 0.2059 - 0.3138 ] = a negative result
= (31.25)(-0.1079) = -3.372 (av. r. of c.
over the interval [11,15] )

Do the same thing for the time interval [1,5]. Then compare the two rates of change.