# I need help with 1 question. VERY PLEASE!

1. Oct 16, 2007

### STAR3URY

I need help with 1 question. VERY URGENT PLEASE!

1. The problem statement, all variables and given/known data
if F[g(x)] = R(x) where g(x) and r(x) is known find f(x).

2. Relevant equations
none

3. The attempt at a solution
I did it but i dont know if it is right:

(g^-1 of x) = x

so...

F(x) = r(g^-1 of x)

2. Oct 16, 2007

### Numzie

f(x)=R(x)/g(x)

3. Oct 16, 2007

### STAR3URY

how did u get that?

4. Oct 16, 2007

### Numzie

Well I'm not 100% certain its that but I just used logic. Division is the inverse of multiplication so if f[g(x)]=R(x) then f(x)=R(x)/g(x). i.e. f[g(2)]=12 then f(x)=12/2. 12/2=6 and 6(2)=12

5. Oct 16, 2007

### Dick

It's not multiplication. It's composition of functions. And assuming g is invertible then the OP is correct. Except don't say g^(-1)(x)=x. That's ridiculous. g(g^(-1)(x))=x.

Last edited: Oct 16, 2007
6. Oct 17, 2007

### EnumaElish

Isn't the OP correct only if F(g) = g(F) and R(g) = g(R) ?

7. Oct 17, 2007

### Dick

No. The OP is ok with the answer. What the OP means to say is substitute y=g(x), so x=g^(-1)(y). So F(y)=R(g^(-1)(y)). Now just replace the dummy y with x.