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

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)

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  1. 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

  2. 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.

  3. 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

  4. 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.

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