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Question Number 10734 by ABD last updated on 23/Feb/17

f(x−1)+f(x+1)=2x^2 +6 ⇒f(x)=?

$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}\:\Rightarrow{f}\left({x}\right)=? \\ $$

Answered by bar Jesús last updated on 23/Feb/17

y1=f(x−1)=2(x−1)^2 +6 y2=f(x+1)=2(x+1)^2 +6 y2−y1=y=2x^2 −4x+2−2x^2 −4x−2 y= −8x

$${y}\mathrm{1}={f}\left({x}−\mathrm{1}\right)=\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{6} \\ $$$${y}\mathrm{2}={f}\left({x}+\mathrm{1}\right)=\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{6} \\ $$$$ \\ $$$${y}\mathrm{2}−{y}\mathrm{1}={y}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2} \\ $$$${y}=\:−\mathrm{8}{x} \\ $$$$ \\ $$$$ \\ $$

Answered by sandy_suhendra last updated on 24/Feb/17

(x−1)^2 +2+(x+1)^2 +2=2x^2 +6 so f(x−1)=(x−1)^2 +2 and f(x+1)=(x+1)^2 +2 f(x)=x^2 +2

$$\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}+\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}=\mathrm{2x}^{\mathrm{2}} +\mathrm{6} \\ $$$$\mathrm{so}\:\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\:\:\mathrm{and}\:\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{2} \\ $$

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