Question

Given the function h(x) = 4x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

Answer 1

Average rate of change in Section A:

$\dfrac{h(1)-h(0)}{1-0}=\dfrac{4-0}{1-0}=4$

Average rate of change in Section B:

$\dfrac{h(3)-h(2)}{3-2}=\dfrac{12-8}{3-2}=4$

As you can see, the average rates of change are the same, as expected. $h(x)=4x$ is linear, which means it has a constant rate of change over any interval in its domain.

Answer 2

Answer:

The rate of change of section A is 3 and rate of change of section B is 48.

Step-by-step explanation:

The given function is

$h(x)=4^x$

Formula for average rate:

$m=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

Section A is from x = 0 to x = 1, the average rate of change of section A is

$m_1=\frac{h(1)-h(0)}{1-0}=\frac{4-1}{1}=3$

Section B is from x = 2 to x = 3, the average rate of change of section B is

$m_2=\frac{h(3)-h(2)}{3-2}=\frac{64-16}{1}=48$

Therefore rate of change of section A is 3 and rate of change of section B is 48.

$m_1\times k=m_2$

$3\times k=48$

$k=12$

Therefore average rate of change of Section B is 12 times of Section A.

The given function is an exponential function and rate of change is not constant.