If $g(x) = \frac{x}{x+1}$, and $f\circ g(x) = x^2$, find $f(x)$?

Question

$$g(x) = \frac{x}{x+1}$$

$$f\circ g(x) = x^2$$

$$f(x) =\text{ ?}$$

Hello, I don't have any clue how to solve that. Do you have any ideas how to solve that? Thanks in advance.

Answer 1

I'll give you a hint: Write $f \circ g(x) =f \circ \dfrac{x}{1+x}=f\big(\dfrac{x}{1+x}\big)=x^2 $
Then plug $t=\dfrac{x}{1+x}$ and solve for $x$.

Can you proceed from here?

Answer 2

HINTS:

First find the inverse function for $g$, so

$$g^{-1}\left(\frac{u}{u+1}\right)=u$$

Then

$$(f\circ g)(x)=f(g(x))=x^2$$ so $$f\left(g\left(g^{-1}\left(\frac{u}{u+1}\right)\right)\right)=\left(g^{-1}\left(\frac{u}{u+1}\right)\right)^2$$

You should be able to continue from there.

Answer 3

It is given:

$$f(\frac{x}{x+1})=x^2$$

If we let:

$$u=\frac{x}{x+1}$$

Then,

$$x=\frac{u}{1-u}$$

Thus,

$$f(u)=(\frac{u}{1-u})^2$$

Which implies ....