Answer
(a) The average rate of growth
(i) from 2006 to 2008: $2120lo/ye$
(ii) from 2008 to 2010: $89lo/ye$
b) The instantaneous rate of growth is $346.5lo/ye$
c) The instantaneous rate of growth is $346.5lo/ye$
Work Step by Step
(a) The average rate of growth
(i) from 2006 to 2008: $$\frac{N(2008)-N(2006)}{2008-2006}=\frac{16680-12440}{2008-2006}=2120(lo/ye)$$
(ii) from 2008 to 2010: $$\frac{N(2010)-N(2008)}{2010-2008}=\frac{16858-16680}{2010-2008}=89(lo/ye)$$
The unit of $N$ is $lo$ (locations), and the unit of the year is $ye$ (year). Therefore, the unit of the average rate of growth is $lo/ye$.
We can conclude that there is a steep drop in the growth of locations from the period 2006-2008 to the period 2008-2010.
(b) The average rate of growth
(i) from 2010 to 2012: $$\frac{N(2012)-N(2010)}{2012-2010}=\frac{18066-16858}{2012-2010}=604(lo/ye)$$
(ii) from 2008 to 2010: $89(lo/ye)$
The instantaneous rate of growth in 2010 would be the average of 2 average rates of growth from 2008 to 2010 and from 2010 to 2012: $$\frac{89+604}{2}=346.5(lo/ye)$$
The unit is also $lo/ye$.
(c) To measure the slope of the tangent at 2010, we would calculate the average rate of growth from 2008 to 2012: $$m=\frac{N(2012)-N(2008)}{2012-2008}=\frac{18066-16680}{4}=346.5$$
The slope of the above tangent is also the instantaneous rate of growth in 2010, which is $346.5lo/ye$
