Let `A=R-{3},B=R-{1},` and let `f: AvecB` be defined by `f(x)=(x-2)/(x-3)` is `f` invertible? Explain.

Explanation

Let `x_(1),x_(2) in A`
and let `f(x_(1))=f(x_(2 ))`
`implies (x_(1)-2)/(x_(1)-3)=(x_(2)-2)/(x_(2)-3)`
`implies x_(1)x_(2)-3x_(1)-2x_(2)+6=x_(1)x_(2)-3x_(2)-2x_(1)+6`
So, `f` is one-one.
Now `y=(x-2)/(x-3)`.
Clearly for any value of y other than 'I' there exists unique x.
So, range of function is `R-{1}.`
Thus, `f(x)` is onto.
Therefore, `f(x)` is invertible.
Now `y=(x-2)/(x-3)`
`implies xy-3y=x-2`
`implies x(y-1)=3y-2`
`implies x=(3y-2)/(y-1)`
Therefore, `f^(-1): A to B,f^(-1)(x)=(3x-2)/(x-1)`