Help pls! (04.03 HC)

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

Part A: Find the average rate of change of each section. (4 points)

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. (6 points)

Substitute the equation and values into the average rate of change formula. 6 part A

Section A:

3*2^1 = 6

3*2^2=12

12-6 = 6

rate of change = 6

SECTION B:

3*2^3 = 24

3*2^4 = 48

48-24 = 24

rate of change = 24

Part B 24/6 = 4 ( Section B is 4 times greater) because the equation is raised to the x power the lager the x value the greater the rate of change would be

h(x) = 3 * (2)^x

Section A is from x = 1 to x = 2

h(1) = 3 * (2)^1 = 3 * 2 = 6

h(2) = 3 * (2)^2 = 3 * 4 = 12

so

the average rate of change = (12 - 6)/(2 - 1) = 6

Section B is from x = 3 to x = 4

h(3) = 3 * (2)^3 = 3 * 8 = 24

h(4) = 3 * (2)^4 = 3 * 16 = 48

so

the average rate of change = (48 - 24)/(4 - 3) = 24

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. (6 points)

the average rate of change of section B is 24 and the average rate of change of section A is 6

So 24/6 = 4

The average rate of change of Section B is 4 times greater than the average rate of change of Section A

It's exponential function, not a linear function; so the rate of change is increasing.

Part a: section a h(x)=3(2)^xh(1)=3(2)^1h(1)=3(2)h(1)=9h(x)=3(2)^2h(2)=3(4)h(2)=8181-9/2-1 = 72/1 = 72section b h(x)=3(2)^xh(3)=3(2)^3h(3)=3(8)h(3)=6561h(x)=3(2)^xh(4)=3(2)^4h(4)=3(16)h(4)=4304672143046721-6561/4-3 = 43040160/1 = 43040160

Part A) The average rate of change of section A is 6 and the average rate of change of section B is 24

Part B) The average rate of change of Section B is 4 times greater than the average rate of change of Section A. see the explanation

Step-by-step explanation:

Part A) Find the average rate of change of each section

we know that

To find the average rate of change, we divide the change in the output value by the change in the input value

the average rate of change is equal to

[tex]\frac{f(b)-f(a)}{b-a}[/tex]

In this problem we have

Section A

[tex]a=1[/tex]

[tex]b=2[/tex]

[tex]f(a)=f(1)=3(2)^1=6[/tex]

[tex]f(b)=f(2)=3(2)^2=12[/tex]

Substitute

[tex]\frac{12-6}{2-1}=6[/tex]

Section B

[tex]a=3[/tex]

[tex]b=4[/tex]

[tex]f(a)=f(3)=3(2)^3=24[/tex]

[tex]f(b)=f(4)=3(2)^4=48[/tex]

Substitute

[tex]\frac{48-24}{4-3}=24[/tex]

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

Divide the average rate of change section B by the average rate of change of Section B

[tex]\frac{Section\ B}{Section\ A}=\frac{24}{6}[/tex]

[tex]\frac{Section\ B}{Section\ A}=4[/tex]

[tex]Section\ B=(4)*Section\ A[/tex]

so

The average rate of change of Section B is 4 times greater than the average rate of change of Section A.

One rate of change is greater than the other, because it is an exponential function and the slope grows bigger.