what is f^-1(x) given f(x)=(2x-1)/(2x+1) ?

First, we'll
write:

f(x)=(2x-1)/(2x+1) as
y=(2x-1)/(2x+1)

Now, we'll solve this equation for x,
multiplying both sides by (2x+1):

2xy+y =
(2x-1)

We'll move all terms containing x, to the left side
and all terms in y, to the right side:

2xy-2x =
-1-y

We'll factorize by
x:

x(2y-2) =
-(1+y)

x=-(1+y)/2(y-1)

We'll multiply
the denominator by -1 and we'll get:

x =
(1+y)/2(1-y)

Now, we'll interchange x and
y.

y = (1+x)/2(1-x)

So, the
inverse function is:

[f(x)]^(-1)
=(1+x)/2(1-x)