Question Number 192049 by sciencestudentW last updated on 06/May/23
$${fog}\left({x}\right)=\mathrm{4}{x}−\mathrm{1} \\ $$$${g}\left({x}\right)={x}−\mathrm{2} \\ $$$${f}\left({x}\right)=? \\ $$
Answered by AST last updated on 06/May/23
$${f}\left({g}\left({x}\right)\right)=\mathrm{4}{x}−\mathrm{1}\Rightarrow{f}\left({x}−\mathrm{2}\right)=\mathrm{4}\left({x}−\mathrm{2}\right)+\mathrm{7} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{4}{x}+\mathrm{7} \\ $$
Answered by qaz last updated on 07/May/23
$${f}\circ{g}={f}\circ\begin{pmatrix}{{x}}\\{{x}−\mathrm{2}}\end{pmatrix}=\begin{pmatrix}{{x}}\\{\mathrm{4}{x}−\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow{f}=\begin{pmatrix}{{x}}\\{\mathrm{4}{x}−\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{x}−\mathrm{2}}\end{pmatrix}^{−\mathrm{1}} =\begin{pmatrix}{{x}}\\{\mathrm{4}{x}−\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{x}−\mathrm{2}}\\{{x}}\end{pmatrix}=\begin{pmatrix}{{x}−\mathrm{2}}\\{\mathrm{4}{x}−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{{x}}\\{\mathrm{4}{x}+\mathrm{7}}\end{pmatrix} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{4}{x}+\mathrm{7} \\ $$