I with 1 question. VERY PLEASE

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Homework Help Overview

The problem involves finding the function f(x) given the relationship F[g(x)] = R(x), where g(x) and R(x) are known functions. The context suggests a focus on function composition and inverses.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various interpretations of the relationship between the functions, with some suggesting that f(x) could be expressed as R(x)/g(x). Others question the validity of this approach and the assumptions behind it.

Discussion Status

The discussion is ongoing, with multiple interpretations being explored. Some participants have offered reasoning based on the properties of function composition and inverses, while others are questioning the assumptions made in the original poster's approach.

Contextual Notes

There is a mention of the need for g(x) to be invertible for certain statements to hold true. Additionally, participants are clarifying the distinction between multiplication and function composition in their reasoning.

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I need help with 1 question. VERY URGENT PLEASE!

Homework Statement


if F[g(x)] = R(x) where g(x) and r(x) is known find f(x).


Homework Equations


none


The Attempt at a Solution


I did it but i don't know if it is right:

(g^-1 of x) = x

so...

F(x) = r(g^-1 of x)
 
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f(x)=R(x)/g(x)
 
Numzie said:
f(x)=R(x)/g(x)


how did u get that?
 
Well I'm not 100% certain its that but I just used logic. Division is the inverse of multiplication so if f[g(x)]=R(x) then f(x)=R(x)/g(x). i.e. f[g(2)]=12 then f(x)=12/2. 12/2=6 and 6(2)=12
 
Numzie said:
Well I'm not 100% certain its that but I just used logic. Division is the inverse of multiplication so if f[g(x)]=R(x) then f(x)=R(x)/g(x). i.e. f[g(2)]=12 then f(x)=12/2. 12/2=6 and 6(2)=12

It's not multiplication. It's composition of functions. And assuming g is invertible then the OP is correct. Except don't say g^(-1)(x)=x. That's ridiculous. g(g^(-1)(x))=x.
 
Last edited:
Isn't the OP correct only if F(g) = g(F) and R(g) = g(R) ?
 
EnumaElish said:
Isn't the OP correct only if F(g) = g(F) and R(g) = g(R) ?

No. The OP is ok with the answer. What the OP means to say is substitute y=g(x), so x=g^(-1)(y). So F(y)=R(g^(-1)(y)). Now just replace the dummy y with x.
 

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