I assume by f¹, you actually mean f⁻¹ as in the inverse of f. I also assume you are asked to find f(x) (as in the inverse of f⁻¹) and f⁻¹(4).
Given that
[tex]f^{-1}(x+2) = \dfrac{x-1}{x+1}[/tex]
with x ≠ 1, we can find f⁻¹(x) by replacing x + 2 with x :
[tex]f^{-1}(x + 2) = \dfrac{x-1}{x+1} = \dfrac{(x+2) - 3}{(x + 2) - 1} \implies f^{-1}(x) = \dfrac{x-3}{x-1}[/tex]
Then when x = 4, we have
[tex]f^{-1}(4) = \dfrac{4-3}{4-1} = \dfrac13[/tex]
Of course, we also could have just substituted x = 2 into the definition of f⁻¹(x + 2) :
[tex]f^{-1}(4) = f^{-1}(2+2) = \dfrac{2-1}{2+1} = \dfrac13[/tex]
To find f(x), we fall back to the definition of an inverse function:
[tex]f^{-1}\left(f(x)\right) = x[/tex]
Then by definition of f⁻¹, we have
[tex]f^{-1}\left(f(x)\right) = \dfrac{f(x)-3}{f(x)-1} = x[/tex]
Solve for f :
[tex]f(x) - 3 = x (f(x) - 1)[/tex]
[tex]f(x) - 3 = x f(x) - x[/tex]
[tex]f(x) - x f(x) = 3 - x[/tex]
[tex](1 - x) f(x) = 3-x[/tex]
[tex]f(x) = \dfrac{3-x}{1-x} = \dfrac{x-1}{x-3}[/tex]
