Given the function g(x) = 6(4)x, section a is from x = 0 to x = 1 and section b is from x = 2 to x = 3.

part a: find the average rate of change of each section.

part b: how many times greater is the average rate of change of section b than section a? explain why one rate of change is greater than the other

We are given the function:

g(x) = 6 (4)^x

Part A.

To get the average rate of change, we use the formula:

average rate of change = [g(x2) – g(x1)] / (x2 – x1)

Section A:

average rate of change = [6 (4)^1 – 6 (4)^0] / (1 – 0) = 18

Section B:

average rate of change = [6 (4)^3 – 6 (4)^2] / (3 – 2) = 288

Part B.

288 / 18 = 16

Therefore the average rate of change of Section B is 16 times greater than in Section A.

The average rate of change is greater between x = 2 to x = 3 than between x = 1 and x = 0 because an exponential function's rate of change increases with increasing x (not constant).

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: