I with 1 question. VERY PLEASE

[STAR3URY]

I need help with 1 question. VERY URGENT PLEASE!

Homework Statement

if F[g(x)] = R(x) where g(x) and r(x) is known find f(x).

Homework Equations

none

The Attempt at a Solution

I did it but i don't know if it is right:

(g^-1 of x) = x

so...

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

 

[Numzie]

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

 

[STAR3URY]

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

how did u get that?

 

[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

 

[Dick]

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

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.

 

[EnumaElish]

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

 

[Dick]

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

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.