Question Number 226015 by mr W last updated on 18/Nov/25
Answered by fantastic2 last updated on 18/Nov/25
$${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}…\left({i}\right) \\ $$$${put}\:{x}=\mathrm{2}\:{in}\:{i} \\ $$$${f}\left(\mathrm{2}\right)+{f}\left(−\mathrm{1}\right)=\mathrm{2}….\mathrm{1} \\ $$$${put}\:−\mathrm{1}\:{in}\:{i} \\ $$$${f}\left(−\mathrm{1}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}\:…..\mathrm{2} \\ $$$${put}\:\frac{\mathrm{1}}{\mathrm{2}}\:{in}\:{i} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:…..\mathrm{3} \\ $$$$\mathrm{1}−\mathrm{2}\Rightarrow{f}\left(\mathrm{2}\right)−{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{3}….\mathrm{4} \\ $$$$\mathrm{4}+\mathrm{3}\Rightarrow\mathrm{2}{f}\left(\mathrm{2}\right)=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{7}}{\mathrm{4}}\checkmark \\ $$
Commented by mr W last updated on 18/Nov/25
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Commented by mr W last updated on 18/Nov/25
$${we}\:{get}\:{similarly} \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{3}} −{x}+\mathrm{1}}{\mathrm{2}{x}\left({x}−\mathrm{1}\right)} \\ $$
Commented by fantastic2 last updated on 18/Nov/25
$${ye} \\ $$