Question

prove that f(x) =ax+b whre a and b are constant and a>0

Answer 1

If a≠0a≠0, then f(x)f(x) will have an inverse function g(x)=1ax−bag(x)=1ax−ba such that for all x∈Rx∈R, f(g(x))=g(f(x))=xf(g(x))=g(f(x))=x.  The existence of such an inverse shows that f(x)f(x) is both one-one and onto;  however, this argument doesn't work if a=0a=0 because it would involve division by zero.

For a=0a=0, f(x)=bf(x)=b, which is quickly seen to be neither onto (it only ever attains one value) or one-one (many different numbers are mapped to the same value).