Answer:
Given function:
[tex]f(x)=2(3)^x[/tex]
The average rate of change of function f(x) over the interval a ≤ x ≤ b is given by:
[tex]\dfrac{f(b)-f(a)}{b-a}[/tex]
[tex]\begin{aligned}\textsf{Average rate of change - Section A} & =\dfrac{f(1)-f(0)}{1-0}\\\\& =\dfrac{2(3)^1-2(3)^0}{1-0}\\\\& =\dfrac{6-2}{1}\\\\& =4 \end{aligned}[/tex]
[tex]\begin{aligned}\textsf{Average rate of change - Section B} & =\dfrac{f(3)-f(2)}{3-2}\\\\& =\dfrac{2(3)^3-2(3)^2}{3-2}\\\\& =\dfrac{54-18}{1}\\\\& =36 \end{aligned}[/tex]
[tex]\dfrac{\textsf{Average rate of change of Section B}}{\textsf{Average rate of change of Section A}}=\dfrac{36}{4}=9[/tex]
Therefore, the average rate of change of Section B is 9 times greater than Section A.
The function is an exponential function (with 3 as the growth factor) so the rate of change increases over time.
