Question Number 6946 by 314159 last updated on 03/Aug/16
$${Find}\:{all}\:{continuous}\:{functions}\:{f}\left({x}\right)\:{such}\: \\ $$$${that}\:{f}\left(\mathrm{2}{x}+\mathrm{1}\right)={f}\left({x}\right)\:{for}\:{all}\:{real}\:{x}. \\ $$
Commented by Yozzii last updated on 04/Aug/16
$${f}\left(\mathrm{1}\right)={f}\left(\mathrm{0}\right) \\ $$$${f}'\left({x}\right)=\mathrm{2}{f}'\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${f}\left({x}\right)={c}\in\mathbb{R}\Rightarrow\:{f}\left(\mathrm{2}{x}+\mathrm{1}\right)={c}={f}\left({x}\right)\:\forall{x}\in\mathbb{R} \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 04/Aug/16
$${Same}\:{answer}\:{as}\:{above}\:{but}\:{with}\:{different}\:{process}\: \\ $$$${Suppose}\:{that}\:{f}\left({x}\right)\:{is}\:{a}\:{linear}\:{polynomial}\:{ax}+{b} \\ $$$$\:\:\:\:\:\:{f}\left({x}\right)={ax}+{b} \\ $$$$\:\:\:\:\:\:{f}\left(\mathrm{2}{x}+\mathrm{1}\right)={a}\left(\mathrm{2}{x}+\mathrm{1}\right)+{b}=\mathrm{2}{ax}+{a}+{b} \\ $$$${As}\:\:\:\:{f}\left(\mathrm{2}{x}+\mathrm{1}\right)={f}\left({x}\right) \\ $$$${So}\:\:\:\:\mathrm{2}{ax}+{a}+{b}={ax}+{b} \\ $$$${Comparing}\:{coefficients}: \\ $$$$\:\:\:\:\:\:\:{a}=\mathrm{2}{a}\:\:\wedge\:\:{a}+{b}={b} \\ $$$$\:\:\:\:\:\:\:\:{a}=\mathrm{0},\:{b}={b} \\ $$$${Hence}\:{f}\left({x}\right)={b} \\ $$$${Same}\:{result}\:{if}\:{f}\left({x}\right)\:{be}\:{supposed}?\:{ax}^{\mathrm{2}} +{bx}+{c} \\ $$