Please help I will medal/fan (NO DIRECT ANSWERS) :) A scientist is studying the growth of a particular species of plant. He writes the following equation to show the height of the plant f(n), in cm, after n days: f(n) = 8(1.05)^n Part C: What is the average rate of change of the function f(n) from n = 2 to n = 6, and what does it represent?

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Please help I will medal/fan (NO DIRECT ANSWERS) :) A scientist is studying the growth of a particular species of plant. He writes the following equation to show the height of the plant f(n), in cm, after n days: f(n) = 8(1.05)^n Part C: What is the average rate of change of the function f(n) from n = 2 to n = 6, and what does it represent?

Mathematics
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I know that aroc=\[\frac{ f(b)-f(a) }{b-a }\]
so you could try \(\frac{f(6)-f(2)}{6-2}\) ?

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the average is then 1. I know that the f(b) and f(a) are the y values, and b and a are the x values. @IrishBoy123
\[\frac{ y2-y1 }{ x2-x1 }\]
\(\large \frac{8(1.05)^6 - 8(1.05)^2}{6-2}\) \(\large = 8\frac{(1.05)^6 - (1.05)^2}{4}\) \(\large = 2[(1.05)^6 - (1.05)^2]\) that's shouldn't give you 1
okay then... this is a different equation. Could you help me solve it?
do you see where it came from? the rest is just pumping it into a calculator
\(f(n) = 8(1.05)^n\) so \(f(6) = 8(1.05)^6\) \(f(2) = 8(1.05)^2\)
okay what exactly would I type into my calculator, it is not hard to get the numbers messed up. Sorry for being such a bish I woke up early.. :( @IrishBoy123
The average rate of change is (change in height)/(change in n) The change in height is 10.72 - 8.82 = 1.9 The change in n is 6 - 2 = 4 So the average rate of change is 1.9/4 = 0.475 But what is the average rate of change? It is the average daily (because a single change in n corresponds to a day) growth of the plant (because f(n) represents height)
so you got the 0.475. good. and yes it is the average change per day. it is a simplification - ie assuming the change per day is the same - becauseit isn't this is an exponential so the change per day increases...
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So everything I said was correct? @IrishBoy123 thanks!!
sounded good to me

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