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Question Number 135797 by benjo_mathlover last updated on 16/Mar/21
$$ \\ $$How do you solve f(3) when f(-1) = -1 and f(0) =1 and f(x) =f(x-1)-2f(x-2)?
Answered by liberty last updated on 17/Mar/21
$${f}\left({x}\right)={f}\left({x}−\mathrm{1}\right)−\mathrm{2}{f}\left({x}−\mathrm{2}\right) \\ $$$${replace}\:{x}−\mathrm{2}\:{by}\:{x}\:{we}\:{get} \\ $$$$\mathrm{2}{f}\left({x}\right)={f}\left({x}+\mathrm{1}\right)−{f}\left({x}+\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{x}=−\mathrm{1}\Rightarrow\mathrm{2}{f}\left(−\mathrm{1}\right)={f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}\right) \\ $$$$\Rightarrow−\mathrm{2}=\mathrm{1}−{f}\left(\mathrm{1}\right);\:{f}\left(\mathrm{1}\right)=\:\mathrm{3} \\ $$$$ \\ $$$$\left(\mathrm{2}\right){x}=\mathrm{0}\Rightarrow\mathrm{2}{f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)−{f}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{2}\:=\:\mathrm{3}−{f}\left(\mathrm{2}\right);\:{f}\left(\mathrm{2}\right)=\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{3}\right){x}=\mathrm{1}\Rightarrow\mathrm{2}{f}\left(\mathrm{1}\right)={f}\left(\mathrm{2}\right)−{f}\left(\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{6}\:=\:\mathrm{1}−{f}\left(\mathrm{3}\right);\:{f}\left(\mathrm{3}\right)=−\mathrm{5} \\ $$