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Digiron [165]
3 years ago
7

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

Mathematics
1 answer:
Aleksandr [31]3 years ago
5 0
A)  average rate=Δy/Δx=(y2-y1)/(x2-x1)=(h2-h1)/(x2-x1)
from x1 = 0 to x2= 1, <span>h(x) = 4^x, h(0)=4^0=1, h(1)=4^1=4
</span>average rate=(h2-h1)/(x2-x1)=(4-1)/(1-0)=3  (section A for interval x=0 to x=1)
from x1 = 2 to x2= 3, h(x) = 4^x, h(0)=4^2=16, h(1)=4^3=64
average rate=(h2-h1)/(x2-x1)=(64-16)/(3-2)=48  (section B for interval x=2 to x=3)
B) av.rate section B/ av. rate section A=48/4=12
this is exponential function, and for Δx=1, Δy change more  for bigger numbers

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