Mathematics
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OpenStudy (anonymous):
Find f^2(x), f^3(x), and f^n(x):
F(x)=x-3
14 years ago
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OpenStudy (lalaly):
\[f^2(x)=(x-3)^2\]
14 years ago
OpenStudy (lalaly):
and the rest are the same
14 years ago
OpenStudy (anonymous):
I'm not sure about that:
for the example problem:
F(x)=x+1
F^2(x)=x+2
14 years ago
OpenStudy (anonymous):
I'm studying iterated functions, if that helps at all
14 years ago
myininaya (myininaya):
why is one f and the other F?
14 years ago
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OpenStudy (anonymous):
no difference- sorry
14 years ago
myininaya (myininaya):
so you have f^2 not f^(2)
so f^2 means to square f
14 years ago
myininaya (myininaya):
like lala did
14 years ago
OpenStudy (anonymous):
ah, i didnt realize it was different. It is \[f ^{2}(x)\]
14 years ago
OpenStudy (anonymous):
Which is what I put in the problem... lol
14 years ago
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OpenStudy (lalaly):
what is the answer?
14 years ago
myininaya (myininaya):
\[f^n(x)=(x-3)^n\]
14 years ago
OpenStudy (anonymous):
Ok- but then the example makes no sense according to that
14 years ago
OpenStudy (lalaly):
what are u studying about?
14 years ago
OpenStudy (anonymous):
\[f ^{2}(x) =f(fx))\]
=\[f(x)+1\]
=x+3
Studying iterated functions
14 years ago
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OpenStudy (anonymous):
that example above is for f(x) = x+1
14 years ago
OpenStudy (lalaly):
oh ok so it would be\[f(x-3)=(x-3)-3\]
14 years ago
OpenStudy (anonymous):
ah, weird...I guess it makes sense, but I dont know why
14 years ago
OpenStudy (anonymous):
i guess i just keep replacing x with x-3?
14 years ago
OpenStudy (lalaly):
f(f(x)=f(x-3) replace the x with x-3
14 years ago
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OpenStudy (lalaly):
yeah exactly
14 years ago
OpenStudy (anonymous):
yep- then how would i express that with n?
14 years ago
OpenStudy (anonymous):
\[f ^{n}(x)=\]
14 years ago
myininaya (myininaya):
so you are doing composition functions?
14 years ago
myininaya (myininaya):
\[f^3(x)=f \circ f \circ f (x)\]
is this what is meant by your notation?
14 years ago
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OpenStudy (anonymous):
no, or at least we haven't talked about it at all
14 years ago
OpenStudy (anonymous):
it might be what is meant myinanaya
14 years ago
OpenStudy (anonymous):
I'm just putting in what is in the book
14 years ago
myininaya (myininaya):
so no where in the book does it say how f^3 is defined?
14 years ago
myininaya (myininaya):
or f^n or whatever
14 years ago
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OpenStudy (lalaly):
like u said myin
14 years ago
OpenStudy (anonymous):
for f(x) = x+1
\[f ^{3}(x)=f(f ^{2}(x))\]
\[=f ^{2}(x)+1\]
\[=(x+2)+1\]
\[=x+3\]
14 years ago
myininaya (myininaya):
lol lets assume what i thought is true okay?
14 years ago
OpenStudy (lalaly):
u dont have to assume lol it is true
14 years ago
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OpenStudy (anonymous):
Sorry- I wasn't saying you were wrong earlier- I was saying I wasn't really studying this
14 years ago
OpenStudy (lalaly):
it's okay i was wrong xD
14 years ago
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
14 years ago
myininaya (myininaya):
well i'm saying lalay
f^2 can be read as f * f
or f composed with f
14 years ago
myininaya (myininaya):
there might be other notations for all i know
14 years ago
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OpenStudy (anonymous):
ah wow thank you so much! It makes a lot of sense now- sorry the notation was confusing
14 years ago
OpenStudy (lalaly):
yeah :) well found about something new:D
14 years ago