Question Number 32161 by jarjum last updated on 20/Mar/18
$${Let}\:{a}\:{function}\:{F}\::{R}\rightarrow{R}\:{be}\:{defined}\:{by} \\ $$$${f}\left({x}\right)=\mathrm{1}+{ax},\alpha\neq\:\mathrm{0}\:{for}\:{all}\:{X}\:\in\:{R}.\:{Show} \\ $$$${that}\:{f}\:{is}\:{invertible}\:{and}\:{find}\:{its}\:{inverse} \\ $$$${function}.{Also}\:{find}\:{the}\:{value}\:\left({s}\right)\:{of}\:\alpha \\ $$$${if}\:{inverse}\:{of}\:{f}\:{is}\:{itself} \\ $$
Answered by mrW2 last updated on 21/Mar/18
$${f}\left({x}\right)=\mathrm{1}+{ax} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}−\mathrm{1}}{{a}} \\ $$$$ \\ $$$${with}\:{a}=−\mathrm{1}: \\ $$$${f}^{−\mathrm{1}} \left({x}\right)={f}\left({x}\right) \\ $$