Question

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

Answer 1

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.

Answer 2

Just took the test, the answer was "A) The average rate of change between years 11 and 15 is about $\frac{1}{3}$ the rate between years 1 and 5."