Let `x_(1), x_(2) in` and f `(x_(1)) = f (x_(2))`
`rArr 3x_(1) + 4 = 3x_(2) +4`
`rArr 3x_(1) = 3x_(2)`
`rArr x_(1) = x_(2)`
`:.` f is one-one.
Again , let f (x) = y when y `in` R (co-domain)
`rArr 3x + 4 = y`
`rArr 3x = y-4`
`rArrx= (y-4)/(3) in` (domain)`AAy in R`
Therefore f is onto.
`because` f is one-one onto. Hence Proved.
`:.` Inverse of exists
`:. f^(-1) : R rarr R` is defined as
`f^(-1)(y) = (y-4)/(3)` Ans.