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)
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:
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