Function problem?

Question

We know that: f(x+1/x-1)=x. Find out f(-1), f(3), f(0).

Answer 1

put t = (x+1)/(x-1)
tx - t = x + 1
x = (t+1)/(t-1)
therefore,
f(x) = (x+1)/(x-1)

f(-1) = 0, f(3) = 2, f(0) = -1

Answer 2

f(-1) For this we must find x so x+1/x-1=-1 so x+1/x=0 x^2+1=0 impossible
f(3) x+1/x -1= 3 x^2 -4x+1 =0 x=((4+-sqrt(12))/2 To have a function we must have further restrictions because there cant be two values.Lets say x>1 Then f(3) =2+sqrt(3)
f(0) x+1/x-1=0 x^2-x+1=0 which has no real roots.So f(0) does not exist

Answer 3

f(y) = x where y=(x+1)/(x-1), so

xy - y = x+1
→ x(y-1) = y+1
→ x = (y+1)/(y-1)
→ f(y) = x = (y+1)/(y-1).

Plug y=-1, y=3, y=0 in f(y) = (y+1)/(y-1).

Answer 4

Let (x+1)/(x-1) = y
x+1 = yx - y
x(y-1) = y+1
x = (y+1)/(y-1)
Thus, f(y) = (y+1)/(y-1)
Substituting, f(-1) = 0, f(3) = 2, f(0) = -1