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Mathematics 18 Online
OpenStudy (anonymous):

Find f^2(x), f^3(x), and f^n(x): F(x)=x-3

OpenStudy (lalaly):

\[f^2(x)=(x-3)^2\]

OpenStudy (lalaly):

and the rest are the same

OpenStudy (anonymous):

I'm not sure about that: for the example problem: F(x)=x+1 F^2(x)=x+2

OpenStudy (anonymous):

I'm studying iterated functions, if that helps at all

myininaya (myininaya):

why is one f and the other F?

OpenStudy (anonymous):

no difference- sorry

myininaya (myininaya):

so you have f^2 not f^(2) so f^2 means to square f

myininaya (myininaya):

like lala did

OpenStudy (anonymous):

ah, i didnt realize it was different. It is \[f ^{2}(x)\]

OpenStudy (anonymous):

Which is what I put in the problem... lol

OpenStudy (lalaly):

what is the answer?

myininaya (myininaya):

\[f^n(x)=(x-3)^n\]

OpenStudy (anonymous):

Ok- but then the example makes no sense according to that

OpenStudy (lalaly):

what are u studying about?

OpenStudy (anonymous):

\[f ^{2}(x) =f(fx))\] =\[f(x)+1\] =x+3 Studying iterated functions

OpenStudy (anonymous):

that example above is for f(x) = x+1

OpenStudy (lalaly):

oh ok so it would be\[f(x-3)=(x-3)-3\]

OpenStudy (anonymous):

ah, weird...I guess it makes sense, but I dont know why

OpenStudy (anonymous):

i guess i just keep replacing x with x-3?

OpenStudy (lalaly):

f(f(x)=f(x-3) replace the x with x-3

OpenStudy (lalaly):

yeah exactly

OpenStudy (anonymous):

yep- then how would i express that with n?

OpenStudy (anonymous):

\[f ^{n}(x)=\]

myininaya (myininaya):

so you are doing composition functions?

myininaya (myininaya):

\[f^3(x)=f \circ f \circ f (x)\] is this what is meant by your notation?

OpenStudy (anonymous):

no, or at least we haven't talked about it at all

OpenStudy (anonymous):

it might be what is meant myinanaya

OpenStudy (anonymous):

I'm just putting in what is in the book

myininaya (myininaya):

so no where in the book does it say how f^3 is defined?

myininaya (myininaya):

or f^n or whatever

OpenStudy (lalaly):

like u said myin

OpenStudy (anonymous):

for f(x) = x+1 \[f ^{3}(x)=f(f ^{2}(x))\] \[=f ^{2}(x)+1\] \[=(x+2)+1\] \[=x+3\]

myininaya (myininaya):

lol lets assume what i thought is true okay?

OpenStudy (lalaly):

u dont have to assume lol it is true

OpenStudy (anonymous):

Sorry- I wasn't saying you were wrong earlier- I was saying I wasn't really studying this

OpenStudy (lalaly):

it's okay i was wrong xD

myininaya (myininaya):

\[f(x)=x-3\] \[f^2=f \circ f(x)=f(x-3)=(x-3)-3=x-6\] \[f^3=f \circ f \circ f(x)=f(x-6)=(x-6)-3=x-9\] \[f^4=f \circ f \circ f \circ f(x)=f(x-9)=(x-9)-3=x-12\] so what is the pattern f=x-3 f^2=x-3(2) f^3=x-3(3) f^4=x-3(4) .... f^n=x-3(n)=x-3n

myininaya (myininaya):

well i'm saying lalay f^2 can be read as f * f or f composed with f

myininaya (myininaya):

there might be other notations for all i know

OpenStudy (anonymous):

ah wow thank you so much! It makes a lot of sense now- sorry the notation was confusing

OpenStudy (lalaly):

yeah :) well found about something new:D

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