Question Number 183556 by Shrinava last updated on 26/Dec/22 | ||
Answered by Frix last updated on 26/Dec/22 | ||
$${f}^{−\mathrm{1}} \left({x}\right)=\frac{{bx}+\mathrm{8}}{\left({a}−\mathrm{2}{b}\right){x}+{a}+\mathrm{2}} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{8}\right)=\frac{\mathrm{8}\left({b}+\mathrm{1}\right)}{\mathrm{9}{a}−\mathrm{2}\left(\mathrm{8}{b}−\mathrm{1}\right)} \\ $$ | ||