Question Number 1068 by 123456 last updated on 01/Jun/15
![f:R→R_+ ,g:R→R_+ 2f(x)=f(x−1)+g(x+1) [g(x)]^2 =f(x−1)g(x+1) f(−1)=1,g(1)=2,f(0)g(0)=?](https://www.tinkutara.com/question/Q1068.png)
$${f}:\mathbb{R}\rightarrow\mathbb{R}_{+} ,{g}:\mathbb{R}\rightarrow\mathbb{R}_{+} \\ $$$$\mathrm{2}{f}\left({x}\right)={f}\left({x}−\mathrm{1}\right)+{g}\left({x}+\mathrm{1}\right) \\ $$$$\left[{g}\left({x}\right)\right]^{\mathrm{2}} ={f}\left({x}−\mathrm{1}\right){g}\left({x}+\mathrm{1}\right) \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{1},{g}\left(\mathrm{1}\right)=\mathrm{2},{f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)=? \\ $$
Answered by prakash jain last updated on 02/Jun/15
![2f(0)=f(−1)g(1)=2⇒f(0)=1 [g(0)]^2 =f(−1)g(1)=2⇒g(0)=(√2) f(0)g(0)=(√2)](https://www.tinkutara.com/question/Q1072.png)
$$\mathrm{2}{f}\left(\mathrm{0}\right)={f}\left(−\mathrm{1}\right){g}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\left[{g}\left(\mathrm{0}\right)\right]^{\mathrm{2}} ={f}\left(−\mathrm{1}\right){g}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{g}\left(\mathrm{0}\right)=\sqrt{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)=\sqrt{\mathrm{2}} \\ $$
Commented by navajyoti.tamuli.tamuli@gmail. last updated on 13/Jun/15
![2f(0)=f(−1)+g(1)=3=>f(0)=(3/2) [g(0)]^2 =f(−1)g(1)=2=>g(0)=(√2) f(0)g(0)=(3/( (√2)))](https://www.tinkutara.com/question/Q1099.png)
$$ \\ $$$$\mathrm{2}{f}\left(\mathrm{0}\right)={f}\left(−\mathrm{1}\right)+{g}\left(\mathrm{1}\right)=\mathrm{3}=>{f}\left(\mathrm{0}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$\left[{g}\left(\mathrm{0}\right)\right]^{\mathrm{2}} ={f}\left(−\mathrm{1}\right){g}\left(\mathrm{1}\right)=\mathrm{2}=>{g}\left(\mathrm{0}\right)=\sqrt{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)=\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}} \\ $$