Question Number 151453 by mathdanisur last updated on 21/Aug/21
$$\mathrm{find}\:\mathrm{all}\:\mathrm{continous}\:\mathrm{functions}\:\mathrm{f}\::\:\mathbb{R}\rightarrow\mathbb{R} \\ $$$$\mathrm{such}\:\mathrm{that}: \\ $$$$\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:=\:\mathrm{f}\left(\sqrt{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{4}} }\right)\:;\:\forall\mathrm{x}\in\mathbb{R} \\ $$
Commented by MJS_new last updated on 21/Aug/21
$${f}\left({x}\right)={c}\wedge{c}\in\mathbb{R} \\ $$
Commented by mathdanisur last updated on 21/Aug/21
$$\mathrm{wrong} \\ $$
Commented by MJS_new last updated on 21/Aug/21
$$\mathrm{no}.\:\mathrm{maybe}\:\mathrm{there}\:\mathrm{are}\:\mathrm{other}\:\mathrm{functions}.\:\mathrm{but}\:\mathrm{with} \\ $$$${f}\left({x}\right)={c}:\:{f}\left({x}^{\mathrm{2}} +\mathrm{1}\right)={f}\left(\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\right)={c} \\ $$
Commented by mathdanisur last updated on 22/Aug/21
$$\mathrm{need}\:\mathrm{to}\:\mathrm{prove} \\ $$
Commented by MJS_new last updated on 22/Aug/21
$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{what}\:\mathrm{do}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{prove}? \\ $$$${f}\left({x}\right)=\mathrm{5}\:\Rightarrow\:{f}\left({x}^{\mathrm{2}} +\mathrm{1}\right)={f}\left(\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\right)={f}\left(\frac{{x}^{\mathrm{37}} +\mathrm{3ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}\right)}{\mathrm{5e}^{\mathrm{i}{x}} +\mathrm{9}\zeta\left(\mathrm{33}\right)}\right)=\mathrm{5} \\ $$
Commented by mathdanisur last updated on 22/Aug/21
$${please}\:{solve}\:{again},\:{thanks} \\ $$
Commented by MJS_new last updated on 22/Aug/21
$$\mathrm{this}\:\mathrm{is}\:\mathrm{weird} \\ $$