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Question Number 194688 by CrispyXYZ last updated on 13/Jul/23
$$\mathrm{find}\:\mathrm{all}\:\mathrm{function}\:{f}:\:\mathbb{R}\:\rightarrow\:\mathbb{R}\:\mathrm{such}\:\mathrm{that}\:\forall{x},\:{y}\in\mathbb{R}, \\ $$$${f}\left({x}−{f}\left({y}\right)\right)={f}\left({f}\left({y}\right)\right)+{xf}\left({y}\right)+{f}\left({x}\right)−\mathrm{1}. \\ $$
Answered by Tinku Tara last updated on 13/Jul/23
$${put}\:{x}={f}\left({y}\right) \\ $$$${f}\left(\mathrm{0}\right)={f}\left({x}\right)+{x}^{\mathrm{2}} +{f}\left({x}\right)−\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)={c} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}+{c}−{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${put}\:{y}=\mathrm{0} \\ $$$${f}\left({x}−{c}\right)={f}\left({c}\right)+{cx}+{f}\left({x}\right)−\mathrm{1} \\ $$$$\frac{\mathrm{1}+{c}−\left({x}−{c}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{1}+{c}−{c}^{\mathrm{2}} }{\mathrm{2}}+{cx}+\frac{\mathrm{1}+{c}−{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1} \\ $$$$\frac{\mathrm{1}+{c}−{x}^{\mathrm{2}} }{\mathrm{2}}+{cx}−\frac{{c}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{1}+{c}−{c}^{\mathrm{2}} }{\mathrm{2}}+{cx}+\frac{\mathrm{1}+{c}−{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1} \\ $$$$\frac{\mathrm{1}+{c}}{\mathrm{2}}−\mathrm{1}=\mathrm{0}\Rightarrow{c}=\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by CrispyXYZ last updated on 14/Jul/23
$$\mathrm{Thank}\:\mathrm{you}! \\ $$