Question

Given the function h(x) = 3(5)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

Answer 1

We can write the function in terms of y rather than h(x) so that:

y = 3 (5)^x

 

A. The rate of change is simply calculated as:

r = (y2 – y1) / (x2 – x1) where r stands for rate

 

Section A:

rA = [3 (5)^1 – 3 (5)^0] / (1 – 0)

rA = 12

 

Section B:

rB = [3 (5)^3 – 3 (5)^2] / (3 – 2)

rB = 300

 

B. We take the ratio of rB / rA:

rB/rA = 300 / 12

rB/rA = 25

 

So we see that the rate of change of section B is 25 times greater than A