What is the average rate of change of the function f(n) from n = 2 to n = 6, and what does it represent? f(n) = 8(1.05)^n
@TheSmartOne
@.Sam.
@ParthKohli
@imqwerty
alright so \[ f(n) = 8(1.05)^{n}\]
yes
so what this seems like to me is like we start off with a number 8 right and it's growing by 5%
we can probably find the slope of this. HINT; rate of change is \[m = \frac{ \Delta~y }{ \Delta~x } = \frac{ y_{2}-y_{1} }{ x_{2}-x_{1} }\] so can you first find the y value when n = 2 \[8(1.05)^{2} = ?\] and n = 6 \[8(1.05)^{6} = ?\]
honestly all i need to be able to find out is the average rate of change
we are doing that..and i'm tying to show you how to do that.
ok
so I was asking if you could figure out what 8(1.05)^2 = ? and 8(1.05)^6
8.82 and 10.72
alright so now let's put this together n = 2, f(2) = 8.82 n = 6, f(6) = 10.72 \[\frac{ f(6)-f(2) }{ n_{2}-n_{1} } = m\]
tell me what you get
@PeriodicNinja
-5.04? @Photon336
how did you get that answer?
was i supposed to keep the 6 and 2 inside of f(x)
Here's how you set it up \[\frac{ (10.72)-(0.82) }{ (6-2) } = ?\]
i put it upside down
@PeriodicNinja do you see how I did it?
yeah
take a look. at the graph I'll post it. if we take the slope that we just found right and then make a line in which we put that in for example a line that goes through (2,8.82) we get this
typo above that should be 8.82
here the slope is 0.475 or the average rate of change between n = 2 and n = 6
but like any graph, I believe this is exponential, the graph will get steeper as you increase in x.
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