If `f : [0, oo) rarr [2, oo)` be defined by `f(x) = x^(2) + 2, AA xx in R`. Then find `f^(-1)`.

Explanation

`f(x) = x^(2) + 2 f : [0, oo) rarr [2, oo)`
`f(x_(1)) = f(x_(2))`
`rArr x_(1)^(2) + 2 = x_(2)^(2) + 2`
`rArr (x_(1) - x_(2))(x_(1)+x_(2)) = 0`
`rArr x_(1) = x_(2)` Hence one-one
`y = x^(2) + 2`
`rArr x = +- sqrt(y-2)`
`:. x - sqrt(y-2)` is onto `{ because f : [0, oo) rarr [2, oo)`
`x = -sqrt(y-2)` is not onto
`:.` f is invertible. Hence `f^(-1)` exists
`y = x^(2) + 2`
`x = sqrt(y-2)`
`f^(-1)(y) = sqrt(y-2)`
`f^(-1(x) = sqrt(x-2)`