Question Number 1068 by 123456 last updated on 01/Jun/15 | ||
$${f}:\mathbb{R}\rightarrow\mathbb{R}_{+} ,{g}:\mathbb{R}\rightarrow\mathbb{R}_{+} \\ $$$$\mathrm{2}{f}\left({x}\right)={f}\left({x}−\mathrm{1}\right)+{g}\left({x}+\mathrm{1}\right) \\ $$$$\left[{g}\left({x}\right)\right]^{\mathrm{2}} ={f}\left({x}−\mathrm{1}\right){g}\left({x}+\mathrm{1}\right) \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{1},{g}\left(\mathrm{1}\right)=\mathrm{2},{f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)=? \\ $$ | ||
Answered by prakash jain last updated on 02/Jun/15 | ||
$$\mathrm{2}{f}\left(\mathrm{0}\right)={f}\left(−\mathrm{1}\right){g}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\left[{g}\left(\mathrm{0}\right)\right]^{\mathrm{2}} ={f}\left(−\mathrm{1}\right){g}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{g}\left(\mathrm{0}\right)=\sqrt{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)=\sqrt{\mathrm{2}} \\ $$ | ||
Commented by navajyoti.tamuli.tamuli@gmail. last updated on 13/Jun/15 | ||
$$ \\ $$$$\mathrm{2}{f}\left(\mathrm{0}\right)={f}\left(−\mathrm{1}\right)+{g}\left(\mathrm{1}\right)=\mathrm{3}=>{f}\left(\mathrm{0}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$\left[{g}\left(\mathrm{0}\right)\right]^{\mathrm{2}} ={f}\left(−\mathrm{1}\right){g}\left(\mathrm{1}\right)=\mathrm{2}=>{g}\left(\mathrm{0}\right)=\sqrt{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)=\frac{\mathrm{3}}{\sqrt{\mathrm{2}}} \\ $$ | ||