How Do I Expand Composite Functions in Mathematics?

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Homework Help Overview

The discussion revolves around expanding composite functions in mathematics, specifically focusing on the inverse of a function and the properties of one-to-one functions. The original poster is seeking assistance with an assignment that involves finding the formula for g^-1 and expanding the function f(g^-1(x)).

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the nature of composite functions and the implications of one-to-one functions. There are attempts to clarify how to treat g^-1(x) within the context of function f, and questions arise about the validity of using specific functions like f(x) = x^2 as examples.

Discussion Status

The conversation is ongoing, with various interpretations of the problem being discussed. Some participants suggest using different functions to illustrate one-to-one properties, while others question the assumptions made about the squaring function. There is no explicit consensus on the correct approach or solution at this stage.

Contextual Notes

Participants note that they have not yet learned about the chain rule or derivatives, which may limit their approaches. There is also a discussion about the restrictions needed for certain functions to be considered one-to-one.

matrix_204
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Expanding Composite Functions(urgent help needed)

I had to solve an assignment question, we are asked to find the formula for g^-1, except once i plug n do everything, i come up with x=f(g^-1(x)+c), which is x=f(g^-1(x)) + f(c), now i m suppose to put the formula in terms of g^-1=..., but i don't know how i can expand the function f(g^-1(x)). So how do i expand composite functions?
 
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g-1(x) is just the argument you are passing into the function f. So treat it no differently than you would if you were passing in x, or t, or y etc:

E.g. if f(x) = x2, then f(g-1(x)) = [g-1(x)]2
 
Also, we haven't learned the chain rule or derivatives yet, so it doesn't involve any of that in the solution.
 
:confused: There was no differentiation of any kind in my last post.

If you could post the expression for f(x), that would be helpful.
 
well that is obvious to me, its just that, i don't know how to expand.
This is the question.
Suppose f is a one-one function. g(x)=f(x + c) for all x s.t. x +c (element)dom f , now prove that g is one-one and find the formula for g^-1.
 
so basically, any one-one function f, can possibly work, don't you think?
 
x=f(g^-1(x)+c), which is x=f(g^-1(x)) + f(c)

Why do you think that?


Anyways, what about using f^-1 somewhere?
 
I realized though, that since the squaring funcion is one-one. That means i can suppose for f that it is f(x)=x^2.
So then, x=g(g^-1(x))=f(g^-1(x)) + f(c)=[g^-1(x)]^2 + c^2.
So x - c^2=[g^-1(x)]^2.
So g^-1(x)=sq.root(x - c^2).
Plz tell me if this is right.
 
f(x) = x^2 is not one to one (unless you restrict the domain of the inverse function appropriately).
 
  • #10
so if i restrict the domain of the inverse, then it becomes one-one and then my solution will be correct right.
 
  • #11
btw can someone give me a proof of the function f(x)=x^2, as being one-one.
 
  • #12
that since the squaring funcion is one-one. That means i can suppose for f that it is f(x)=x^2.

You cannot; you're reversing the statement.

The statement "the squaring function is one-one" does not mean you can say "a one-one function is the squaring function".

(And, of course, you have cephid's observation that the squaring function is not one-one)
 
  • #13
i c, i realized that i should have used the cubic function, since that is one-one.
 

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