Question Number 154668 by EDWIN88 last updated on 20/Sep/21
$$\:{If}\:{f}\left({x}\right)={f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)\:{where} \\ $$$${f}\left(\mathrm{10}\right)=\mathrm{6}\:{and}\:{f}\left(\mathrm{20}\right)=\mathrm{2}{f}\left(\mathrm{21}\right) \\ $$$${then}\:{f}\left(\mathrm{16}\right)=\ldots? \\ $$
Commented by mathdanisur last updated on 20/Sep/21
$$\mathrm{f}\left(\mathrm{16}\right)\:=\:-\:\mathrm{f}\left(\mathrm{21}\right) \\ $$$$\mathrm{f}\left(\mathrm{21}\right)\:=\:-\:\mathrm{6}\:\Rightarrow\:\mathrm{f}\left(\mathrm{16}\right)\:=\:\mathrm{6} \\ $$
Commented by Rasheed.Sindhi last updated on 20/Sep/21
$${More}\:{details}\:{please}! \\ $$
Answered by Rasheed.Sindhi last updated on 20/Sep/21
$${f}\left({x}\right)={f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right) \\ $$$${f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)={f}\left({x}+\mathrm{1}\right) \\ $$$${Applying}\:{definition}\:{on}\:{left}\:{side}: \\ $$$${f}\left({x}\right)−{f}\left({x}−\mathrm{1}\right)={f}\left({x}\right)+{f}\left({x}+\mathrm{2}\right) \\ $$$${f}\left({x}−\mathrm{1}\right)=−{f}\left({x}+\mathrm{2}\right) \\ $$$${x}−\mathrm{1}\rightarrow{x} \\ $$$${f}\left({x}\right)=−{f}\left({x}+\mathrm{3}\right) \\ $$$$\blacktriangleright{f}\left(\mathrm{10}\right)=−{f}\left(\mathrm{13}\right)=\mathrm{6}\Rightarrow{f}\left(\mathrm{13}\right)=−\mathrm{6} \\ $$$$\blacktriangleright{f}\left(\mathrm{13}\right)=−{f}\left(\mathrm{13}+\mathrm{3}\right)=−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{16}\right)=\mathrm{6} \\ $$
Commented by mathdanisur last updated on 20/Sep/21
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{Ser} \\ $$
Commented by Rasheed.Sindhi last updated on 20/Sep/21
$$\mathcal{T}{hanks}\:{ser}! \\ $$