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If-f-x-f-x-1-f-x-1-where-f-10-6-and-f-20-2f-21-then-f-16-




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

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