Previous in Relation and Functions Next in Relation and Functions
Question Number 193685 by cortano12 last updated on 18/Jun/23
$$\:\:\:\:\:\:\:\left(\mathrm{f}\left(\mathrm{x}\right)\right)^{\mathrm{2}} −\mathrm{4xf}\left(\mathrm{x}\right)+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{3}\right)=?\: \\ $$
Answered by mr W last updated on 18/Jun/23
$${y}^{\mathrm{2}} −\mathrm{4}{xy}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{y}^{\mathrm{2}} +\mathrm{3}}{\mathrm{4}{y}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\mathrm{4}{x}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left(\mathrm{3}\right)=\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{3}}{\mathrm{4}×\mathrm{3}}=\mathrm{1} \\ $$
Answered by MM42 last updated on 18/Jun/23
$${f}^{−\mathrm{1}} \left({x}\right)=\mathrm{3}\Rightarrow{x}={f}\left(\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{9}−\mathrm{12}{x}+\mathrm{3}=\mathrm{0}\Rightarrow{x}=\mathrm{1}={f}^{−\mathrm{1}} \left(\mathrm{3}\right) \\ $$