Find the inverse of the function `f(x) = x/(x+1)`

First, we let `y = f(x).`

So we now can say: `y = x/(x+1)`

To find the inverse I can switch x and y and solve for y:

First, `x = y/(y+1)`

Rewrite as: `x = 1 - (1/(y+1))`

Subtract 1: `(x - 1) = - 1/(y + 1)`

Multiply by (y + 1): `(y + 1)(x - 1) = -1`

`yx - y + x - 1 = -1`

`y(x - 1) = -x` Divide by (x - 1)

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

Therefore** inverse is:** `f^(-1)(x) = - x/(x - 1)`

Given

`f(x)=x/(1+x)`

Let `y=f(x)=x/(1+x)` (i)

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

`1/x+1=1/y`

`1/x=1/y-1`

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

`x=y/(1-y)` (ii)

By def. of inverse function

`f^(-1)(f(x))=x`

Thus from (i)

`f^(-1)(y)=x`

substitute x from (ii) ,we have

`f^(-1)(y)=y/(1-y)`

**Thus inverse of function f is defined as**

`f^(-1)(x)=x/(1-x)`