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Expanding Composite Functions( help needed)

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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?
 

cepheid

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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.
 

cepheid

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: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 dont 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, dont you think?
 

Hurkyl

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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.
 

cepheid

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f(x) = x^2 is not one to one (unless you restrict the domain of the inverse function appropriately).
 
so if i restrict the domain of the inverse, then it becomes one-one and then my solution will be correct right.
 
btw can someone give me a proof of the function f(x)=x^2, as being one-one.
 

Hurkyl

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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)
 
i c, i realized that i should have used the cubic function, since that is one-one.
 

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