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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