Expanding Composite Functions( help needed)

1. Nov 18, 2004

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?

2. Nov 18, 2004

cepheid

Staff Emeritus
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

3. Nov 18, 2004

matrix_204

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

4. Nov 18, 2004

cepheid

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

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

5. Nov 18, 2004

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.

6. Nov 18, 2004

matrix_204

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

7. Nov 18, 2004

Hurkyl

Staff Emeritus
Why do you think that?

Anyways, what about using f^-1 somewhere?

8. Nov 18, 2004

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.

9. Nov 18, 2004

cepheid

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

10. Nov 18, 2004

matrix_204

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

11. Nov 18, 2004

matrix_204

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

12. Nov 18, 2004

Hurkyl

Staff Emeritus
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. Nov 18, 2004

matrix_204

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