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Question Number 120300 by bemath last updated on 30/Oct/20
$${Determine}\:{all}\:{function}\:{f}:{R}\rightarrow{R} \\ $$$${which}\:{satisfy}\:{f}\left({a}+{x}\right)−{f}\left({a}−{x}\right)=\mathrm{4}{ax} \\ $$$${for}\:{all}\:{real}\:{a}\:{and}\:{x}. \\ $$
Commented by benjo_mathlover last updated on 30/Oct/20
$${let}\:{f}\left({x}\right)={px}^{\mathrm{2}} +{qx}+{r} \\ $$$${f}\left({a}+{x}\right)={p}\left({a}^{\mathrm{2}} +\mathrm{2}{ax}+{x}^{\mathrm{2}} \right)+{qx}+{aq}+{r} \\ $$$${f}\left({a}+{x}\right)={px}^{\mathrm{2}} +\left(\mathrm{2}{ap}+{q}\right){x}+{a}^{\mathrm{2}} {p}+{aq}+{r} \\ $$$$ \\ $$$${f}\left({a}−{x}\right)={p}\left({a}^{\mathrm{2}} −\mathrm{2}{ax}+{x}^{\mathrm{2}} \right)+{aq}−{qx}+{r} \\ $$$${f}\left({a}−{x}\right)={px}^{\mathrm{2}} −\left(\mathrm{2}{ap}+{q}\right){x}+{a}^{\mathrm{2}} {p}+{aq}+{r} \\ $$$$ \\ $$$${f}\left({a}+{x}\right)−{f}\left({a}−{x}\right)=\left(\mathrm{4}{ap}+\mathrm{2}{q}\right){x}\:\equiv\:\mathrm{4}{ax} \\ $$$$\rightarrow\:\mathrm{4}{ap}+\mathrm{2}{q}\:=\:\mathrm{4}{a}\:;\:\mathrm{4}{ap}−\mathrm{4}{a}=−\mathrm{2}{q} \\ $$$$\mathrm{4}{a}\left(\mathrm{1}−{p}\right)=\:\mathrm{2}{q}\:\Rightarrow\:\mathrm{1}−{p}=\frac{{q}}{\mathrm{2}{a}} \\ $$$$\Rightarrow{p}=\mathrm{1}−\frac{{q}}{\mathrm{2}{a}}\:=\:\frac{\mathrm{2}{a}−{q}}{\mathrm{2}{a}} \\ $$$${f}\left({x}\right)=\:\left(\frac{\mathrm{2}{a}−{q}}{\mathrm{2}{a}}\right){x}^{\mathrm{2}} +{qx}+{r}\: \\ $$