Show that the function `f: Rvec` given by `f(x)=x a+b ,` where `a , b in R ,a!=0` is a bijection.

Explanation

Given function `f: R -> R` and `a x+b, a equiv 0`


`f^{-1}` exist if `f: R -> R` be a bijection function, before this we have to prove `f` be a bijection function.


One-one/many-one:


Let `p, q in R`


`f(p)=f(q)` `Rightarrow a p+b=a q+b` `Rightarrow a p-a q` `-> p=q` So, `f(p)=f(q), forall p, q in R`


`therefore` f is a bijection function.


Onto/into:


Let `f(x)=y, y in R`


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