Question

Let $f:R\to R$ be defined by $f\left(x\right)=3x+4$, $xϵR$. Is $f$ invertible? If so, give a formula for $f^{-1}$.

Answer 1

$f\left(x\right)=3x+4=ax+b$

If $a\ne 0$, then f(x) will have an inverse function $g\left(x\right)=\frac{1}{a}x-\frac{b}{a}$ such that for all $x\in R,f\left(g\right(x\left)\right)=g\left(f\right(x\left)\right)=x$. The existence of such an inverse shows that f(x) is both one-one and onto; however, this argument doesn't work if a=0 because it would involve division by zero.

For $a=0,f\left(x\right)=b$, 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).