Question Number 208370 by lmcp1203 last updated on 14/Jun/24
$${if}\:\:\:\left({fof}\right)\left({x}\right)={f}\left({x}\right)+{x}\:\:{and}\:{f}\left(\mathrm{1}\right)=\mathrm{1}\:\:\: \\ $$$${find}\:\:{fofofofofofofofofof}\left(\mathrm{1}\right) \\ $$
Commented by A5T last updated on 14/Jun/24
$${f}\left({f}\left({x}\right)\right)={f}\left({x}\right)+{x}\Rightarrow\:{f}\left({f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)+\mathrm{1}\Rightarrow\mathrm{1}\overset{?} {=}\mathrm{0} \\ $$
Commented by mr W last updated on 14/Jun/24
$${f}\left({x}\right)=\mathrm{1}={constant}? \\ $$
Commented by lmcp1203 last updated on 14/Jun/24
$${sorry}.\:{its}\:{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$
Answered by MM42 last updated on 14/Jun/24
$$\bigstar \\ $$$$\left({fof}\right)\left(\mathrm{1}\right)={f}\left({f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\neq{f}\left(\mathrm{1}\right)+\mathrm{1}\:\:\:\:\:\:?!!! \\ $$$$ \\ $$
Commented by lmcp1203 last updated on 14/Jun/24
$${this}\:{problem}\:{is}\:{wrong}.\:{thanks} \\ $$
Answered by mathzup last updated on 14/Jun/24
$${oooooof}.... \\ $$