1 votes 1 votes Show that the function f(x) = ax + b from R->R is invertible, where a and b are constants, with a$\neq$0, and find the inverse of f How to check whether this function is onto? pls give a detailed solution Set Theory & Algebra kenneth-rosen discrete-mathematics functions set-theory&algebra + – aditi19 asked Nov 26, 2018 • edited Mar 4, 2019 by Pooja Khatri aditi19 1.3k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Magma commented Nov 26, 2018 reply Follow Share y = ax+b x = ay+b x-b/a = y f-(x) = x -b /a 0 votes 0 votes Magma commented Nov 26, 2018 reply Follow Share onto mean : for every element in co-domain there should be a pre-image in domain right ?? now assume there's is an element in co-domain 'c' c is an arbitrary element now definitely element 'c' have pre-image in domain right ?? x = b - c / a now a and b are const it's given in the question and a ! =0 for any value of c we get the pre-image in domain you can check it now C can be any real number for every value in c we get pre-image in domain of the function 0 votes 0 votes aditi19 commented Nov 26, 2018 reply Follow Share oh! got it now... thanks 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes To find inverse of a function if inverse exits.. do this Put. f(x) = y ..........(1) f^-1(y)=x .......(2) In our case f(x) =ax+b =y ....... (3) Now put , x= f^-1(y) in (3) We have y = a.f^-1(y)+b (y-b)/a =f^-1(y) Puy x in place of y f^-1(x)=(x-b)/a this is the inverse of f(x) Kingdarab answered Aug 29, 2020 Kingdarab comment Share Follow See all 0 reply Please log in or register to add a comment.