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

First ley us cross multiply:

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

==> 2xy + y = 2x - 1

Now group temrs with x:

==> 2xy - 2x = -y-1

Now factor x:

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

Now divide by 2y-2

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

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

Then the inverse is:

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

We can write y= (2x-1) / (2x+1)

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

=> 2xy +y = 2x-1

=> 2xy - 2x = -1-y

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

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

Now replace x with y:

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

**Therefore the inverse of y= (2x-1) / (2x+1) **

**is (x-1)/ (2-2x).**

To find the inverse of y = (2x-1)/(2x+1).

We know that at 2x+1 = 0 or x =-1/2, y does nt exist.

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

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

2xy-2x = -1-y

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

By swapping x and y we get:

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

So y = -(x+1)/2(x-1) is the inverse.

We'll solve this equation y=(2x-1)/(2x+1) 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 factorizeby 2x:

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

We'll divide by 2(y-1)

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

Now, we'll interchange x and y:

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

So, the inverse function is:

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