Question Number 192076 by sciencestudentW last updated on 07/May/23
$${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{6}{x}\:\:\:\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$
Answered by AST last updated on 07/May/23
$${y}={f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{6}{x}=\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$${x}=\underset{−} {+}\sqrt{{y}+\mathrm{9}}−\mathrm{3} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\underset{−} {+}\sqrt{{x}+\mathrm{9}}−\mathrm{3} \\ $$