Question

How do you find the inverse of `f(x)=(x+1)/(x-2)` and graph both f and `f^-1`?

Answer 1

The answer is `f^-1(x)=(2x+1)/(x-1)`

Explanation:

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

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

`yx-2y=x+1`

`yx-x=2y+1`

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

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

Therefore,

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

Verification,

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

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

`=3x/3=x`

graph{(y-(x+1)/(x-2))(y-x)=0 [-10, 10, -5, 5]}

graph{(y-(2x+1)/(x-1))(y-x)=0 [-10, 10, -5, 5]}

The graphs are symmetric wrt `y=x`