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

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### 2 Answers

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)**

Here we are given the function f(x)=(2x-1)/(2x+1) and we have to find the inverse of f(x) .

Let's take y = f(x) = (2x-1)/(2x+1)

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

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

=> 2yx + y = 2x - 1

=> 2yx - 2x = -1 -y

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

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

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

Replace y with x and x with y

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

**Therefore the required inverse function is f(x) = (1+x)/(2-2x)**

To confirm: The inverse of a function f(x), f^-1(x) is one that gives back x.

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

=> [2+ 2x - 2 +2x] / [2 +2x + 2 +2x] = 4x/4 = x