Question Number 4261 by 123456 last updated on 06/Jan/16
$${f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({x}\right)=? \\ $$
Commented by Yozzii last updated on 06/Jan/16
$${f}\left({x}\right)=\mathrm{0}\:{for}\:{example}.\: \\ $$$$ \\ $$$${f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)={f}\left(\mathrm{1}\right)={constant}. \\ $$$$\Rightarrow{f}'\left({x}\right){f}\left(\mathrm{1}−{x}\right)−{f}\left({x}\right){f}^{'} \left(\mathrm{1}−{x}\right)=\mathrm{0} \\ $$$$\frac{{f}^{'} \left({x}\right)}{{f}^{'} \left(\mathrm{1}−{x}\right)}=\frac{{f}\left({x}\right)}{{f}\left(\mathrm{1}−{x}\right)}={r}\left({x}\right)\: \\ $$$$ \\ $$$${x}=\mathrm{1}:\:{f}\left(\mathrm{1}\right){f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1}\:{if}\:{f}\left(\mathrm{1}\right)\neq\mathrm{0}.\: \\ $$$$\: \\ $$
Answered by prakash jain last updated on 07/Jan/16
$${f}\left({x}\right)={e}^{{kx}} \\ $$$${f}\left(\mathrm{1}−{x}\right)={e}^{{k}−{kx}} \\ $$$${f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)={e}^{{k}} ={f}\left(\mathrm{1}\right) \\ $$
Commented by prakash jain last updated on 07/Jan/16
$$\mathrm{Thanks}.\:\mathrm{Deleted}\:{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\:\mathrm{part}\:\mathrm{of}\:\mathrm{answer}. \\ $$
Commented by Yozzii last updated on 07/Jan/16
$${f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)={ax}×{a}\left(\mathrm{1}−{x}\right)={a}^{\mathrm{2}} {x}\left(\mathrm{1}−{x}\right) \\ $$$${f}\left(\mathrm{1}\right)={a}\neq{a}^{\mathrm{2}} {x}\left(\mathrm{1}−{x}\right)\Rightarrow{f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)\neq{f}\left(\mathrm{1}\right) \\ $$