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Question Number 188407 by HeferH last updated on 01/Mar/23

xf(x) = f(x + 2) f(2) = 2 f(8) = ?

$${xf}\left({x}\right)\:=\:{f}\left({x}\:+\:\mathrm{2}\right) \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{2} \\ $$$${f}\left(\mathrm{8}\right)\:=\:?\: \\ $$

Answered by horsebrand11 last updated on 01/Mar/23

(i) x f(x)=f(x+2) (ii) (x−2)f(x−2)=f(x) x=2⇒f(2)=0?

$$\left({i}\right)\:{x}\:{f}\left({x}\right)={f}\left({x}+\mathrm{2}\right) \\ $$$$\left({ii}\right)\:\left({x}−\mathrm{2}\right){f}\left({x}−\mathrm{2}\right)={f}\left({x}\right) \\ $$$$\:{x}=\mathrm{2}\Rightarrow{f}\left(\mathrm{2}\right)=\mathrm{0}?\:\: \\ $$

Answered by manxsol last updated on 01/Mar/23

xf(x)=f(x+2) x=2 2f(2)=f(4) f(4)=4 x=4 4f(4)=f(6) f(6)=16 x=6 6f(6)=f(8) 6(16)=f(8) f(8)=96

$${xf}\left({x}\right)={f}\left({x}+\mathrm{2}\right) \\ $$$$ \\ $$$${x}=\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{2}{f}\left(\mathrm{2}\right)={f}\left(\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{4}\right)=\mathrm{4} \\ $$$${x}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{f}\left(\mathrm{4}\right)={f}\left(\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{6}\right)=\mathrm{16} \\ $$$${x}=\mathrm{6}\:\:\:\:\:\:\:\mathrm{6}{f}\left(\mathrm{6}\right)={f}\left(\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}\left(\mathrm{16}\right)={f}\left(\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{8}\right)=\mathrm{96} \\ $$$$ \\ $$

Commented by mr W last updated on 01/Mar/23

you just took f(2)=2, but this is not proved, even wrong, since f(2)=0.

$${you}\:{just}\:{took}\:{f}\left(\mathrm{2}\right)=\mathrm{2},\:{but}\:{this}\:{is}\:{not} \\ $$$${proved},\:{even}\:{wrong},\:{since}\:{f}\left(\mathrm{2}\right)=\mathrm{0}. \\ $$

Answered by mr W last updated on 01/Mar/23

xf(x)=f(x+2) 0×f(0)=f(2) ⇒f(2)=0 2×f(2)=f(4) ⇒f(4)=0 4×f(4)=f(6) ⇒f(6)=0 6×f(6)=f(8) ⇒f(8)=0 generally f(2n)=0

$${xf}\left({x}\right)={f}\left({x}+\mathrm{2}\right) \\ $$$$\mathrm{0}×{f}\left(\mathrm{0}\right)={f}\left(\mathrm{2}\right)\:\Rightarrow{f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{2}×{f}\left(\mathrm{2}\right)={f}\left(\mathrm{4}\right)\:\Rightarrow{f}\left(\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{4}×{f}\left(\mathrm{4}\right)={f}\left(\mathrm{6}\right)\:\Rightarrow{f}\left(\mathrm{6}\right)=\mathrm{0} \\ $$$$\mathrm{6}×{f}\left(\mathrm{6}\right)={f}\left(\mathrm{8}\right)\:\Rightarrow{f}\left(\mathrm{8}\right)=\mathrm{0} \\ $$$${generally}\:{f}\left(\mathrm{2}{n}\right)=\mathrm{0} \\ $$

Commented by manxsol last updated on 01/Mar/23

dato f(2)=2 0×f(0)=f(2)⇒f(2)=0 f(2) has two imagen⇒f no is funtion. i am worked with funtion not with relation

$${dato}\:{f}\left(\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{0}×{f}\left(\mathrm{0}\right)={f}\left(\mathrm{2}\right)\Rightarrow{f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)\:{has}\:{two} \\ $$$$\:{imagen}\Rightarrow{f}\:{no}\:{is}\:{funtion}. \\ $$$${i}\:{am}\:{worked}\:{with}\:{funtion} \\ $$$$\:{not}\:{with}\:{relation} \\ $$$$ \\ $$

Commented by manxsol last updated on 01/Mar/23

0 cannot be in the domain of the given funtion.the function goes through (2,2). I dont have to try it .

$$\mathrm{0}\:{cannot}\:{be}\:{in}\:{the}\:{domain}\: \\ $$$${of}\:{the}\:{given}\:{funtion}.{the} \\ $$$$\:{function}\:{goes}\:{through}\:\left(\mathrm{2},\mathrm{2}\right). \\ $$$${I}\:{dont}\:{have}\:{to}\:{try}\:{it}\:. \\ $$

Commented by mr W last updated on 01/Mar/23

f(2)=2 can not be obtained, even when x≠0.

$${f}\left(\mathrm{2}\right)=\mathrm{2}\:{can}\:{not}\:{be}\:{obtained},\:{even}\:{when} \\ $$$${x}\neq\mathrm{0}. \\ $$

Commented by manxsol last updated on 01/Mar/23

that is very good and true answer.Grateful.

$${that}\:{is}\:\:{very}\:{good}\:{and}\:{true} \\ $$$$\:{answer}.{Grateful}. \\ $$

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