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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. (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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sqdancefan

Answer:

Part A:

  • section A: 18
  • section B: 288

Part B: rate of change in section B is 16 times as great

Explanation:

Part A:

The rate of change is computed as ...

... rate of change = (change in g(x)) / (change in x)

In section A, this is ...

... (change in g(x)) / (change in x) = (24 - 6)/(1 - 0) = 18

In section B, this is ...

... (change in g(x)) / (change in x) = (384 -96)/(3 -2) = 288

Part B:

The function, like any exponential function, increases at an increasing rate. The increase in the rate of change reflects the fact that as x increases, the rate of change of g(x) increases. (See the attachment for a graph.)

Ver imagen sqdancefan
apologiabiology

average rate of change of f(x) from x=a to x=b is the slope from (a,f(a)) to (b,f(b))


A. [tex]g(x)=6(4)^x[/tex]


section A:

x=0 to x=1

find f(0) and f(1)

[tex]f(0)=6(4)^0=6(1)=6[/tex]

[tex]f(1)=6(4)^1=6(4)=24[/tex]

average rate of change is slope from (0,6) to (1,24) or (24-6)/(1-0)=18/1=18


section B:

x=2 to x=3

find f(2) and f(3)

[tex]f(2)=6(4)^2=6(16)=96[/tex]

[tex]f(3)=6(4)^3=6(64)=384[/tex]

average rate of change is slope from (2,96) to (3,384) or (384-96)/(3-2)=288/1=288



B.

avg rate of change of A=18

avg rate of change of B=288

B/A=288/18=16 times greater


it's greater because an exponential function increases at an increasing rate

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