If-2f-x-f-x-1-x-2-determine-f-x-

Question Number 96826 by bobhans last updated on 05/Jun/20

$$\mathrm{If}\:\mathrm{2f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)\:=\:\mathrm{x}^{\mathrm{2}} \:.\:\mathrm{determine}\:\mathrm{f}\left(\mathrm{x}\right)\: \\ $$

Answered by john santu last updated on 05/Jun/20

let f(x) = px^2 +qx+r f(x−1)= px^2 −2px+p+qx−q+r f(x−1)= px^2 +(q−2p)x+p−q+r ⇒2f(x)+f(x−1)= x^2 ⇒3px^2 +(3q−2p)x+3r+p−q = x^2 3p=1 ⇒p = (1/3) 3q−2p = 0 ⇒q = (2/3)p = (2/9) 3r+p−q=0 ⇒r = ((q−p)/3) = (((2/9)−(1/3))/3) r = −(1/(27)) ∴ f(x) = (1/3)x^2 +(2/9)x−(1/(27))

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:{px}^{\mathrm{2}} +{qx}+{r}\: \\ $$$$\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\:{px}^{\mathrm{2}} −\mathrm{2}{px}+{p}+{qx}−{q}+{r} \\ $$$$\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\:{px}^{\mathrm{2}} +\left({q}−\mathrm{2}{p}\right){x}+{p}−{q}+{r} \\ $$$$\Rightarrow\mathrm{2f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\:\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{px}^{\mathrm{2}} +\left(\mathrm{3}{q}−\mathrm{2}{p}\right){x}+\mathrm{3}{r}+{p}−{q}\:=\:{x}^{\mathrm{2}} \\ $$$$\mathrm{3}{p}=\mathrm{1}\:\Rightarrow{p}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{3}{q}−\mathrm{2}{p}\:=\:\mathrm{0}\:\Rightarrow{q}\:=\:\frac{\mathrm{2}}{\mathrm{3}}{p}\:=\:\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\mathrm{3}{r}+{p}−{q}=\mathrm{0}\:\Rightarrow{r}\:=\:\frac{{q}−{p}}{\mathrm{3}}\:=\:\frac{\frac{\mathrm{2}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{3}} \\ $$$${r}\:=\:−\frac{\mathrm{1}}{\mathrm{27}} \\ $$$$\therefore\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{9}}{x}−\frac{\mathrm{1}}{\mathrm{27}}\: \\ $$$$ \\ $$

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