# The function f(x)=(2x+1)/(2x-1) has an inverse. Which is the inverse?

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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 = 1-2y

We'll factorize:

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

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

Now, we'll interchange x and y:

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

So, the inverse function is:

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

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

Multiply by (2x-1

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

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

Move x terms to the left side:

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

Factorize x:

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

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

Then the inverse for the function f(x) is:

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

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

Solution:

Let y = (2x+1)/(2x-1). To get the inverse make x the subject>

Multiplying by (2x-1), we get:

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

2xy-2x = y+1. Or

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

x = (y-1)/[2(y+1)]. Or sapping x and y,

y = (x-1)/{2(x+1)] is the inverse of the function.