Question Number 132031 by mathlove last updated on 10/Feb/21
Commented by kaivan.ahmadi last updated on 10/Feb/21
$${p}\left({x}\right)={f}\left({x}\right)\left({x}−\mathrm{1}\right)+\mathrm{3}={g}\left({x}\right)\left({x}+\mathrm{1}\right)+\mathrm{7} \\ $$$${p}\left(\mathrm{1}\right)=\mathrm{3},{p}\left(−\mathrm{1}\right)=\mathrm{7} \\ $$$${p}\left({x}\right)={t}\left({x}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{k}\left({x}\right) \\ $$$${p}\left(\mathrm{1}\right)={k}\left(\mathrm{1}\right)\Rightarrow{k}\left(\mathrm{1}\right)=\mathrm{3}\Rightarrow\left(\mathrm{1},\mathrm{3}\right)\in{k} \\ $$$${p}\left(−\mathrm{1}\right)={k}\left(−\mathrm{1}\right)=\mathrm{7}\Rightarrow\left(−\mathrm{1},\mathrm{7}\right)\in{k} \\ $$$${k}\left({x}\right)−\mathrm{3}=\frac{\mathrm{7}−\mathrm{3}}{−\mathrm{1}−\mathrm{1}}\left({x}−\mathrm{1}\right)\Rightarrow{k}\left({x}\right)−\mathrm{3}=−\mathrm{2}\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow{k}\left({x}\right)=−\mathrm{2}{x}+\mathrm{5} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 10/Feb/21
$${k}\left(\mathrm{1}\right)=\mathrm{3},{k}\left(−\mathrm{1}\right)=\mathrm{7} \\ $$$${k}\left({x}\right)={ax}+{b} \\ $$$$\begin{cases}{{x}=\mathrm{1}\Rightarrow{a}+{b}=\mathrm{3}}\\{{x}=−\mathrm{1}\Rightarrow−{a}+{b}=\mathrm{7}}\end{cases}\:\:\Rightarrow{b}=\mathrm{5}\:\:,{a}=−\mathrm{2} \\ $$$$\Rightarrow{k}\left({x}\right)=−\mathrm{2}{x}+\mathrm{5} \\ $$
Commented by mathlove last updated on 11/Feb/21
$${thanks}\:{sir} \\ $$