# Expanding Composite Functions( help needed)

#### matrix_204

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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#### cepheid

Staff Emeritus
Gold Member
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

#### matrix_204

Also, we haven't learned the chain rule or derivatives yet, so it doesn't involve any of that in the solution.

#### cepheid

Staff Emeritus
Gold Member
There was no differentiation of any kind in my last post.

If you could post the expression for f(x), that would be helpful.

#### matrix_204

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.

#### matrix_204

so basically, any one-one function f, can possibly work, dont you think?

#### Hurkyl

Staff Emeritus
Gold Member
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?

#### matrix_204

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

Staff Emeritus
Gold Member
f(x) = x^2 is not one to one (unless you restrict the domain of the inverse function appropriately).

#### matrix_204

so if i restrict the domain of the inverse, then it becomes one-one and then my solution will be correct right.

#### matrix_204

btw can someone give me a proof of the function f(x)=x^2, as being one-one.

#### Hurkyl

Staff Emeritus
Gold Member
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)

#### matrix_204

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

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