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Question Number 118740 by Jamshidbek2311 last updated on 19/Oct/20

$$f(x+2)+f(x-1)=2x^2+14$$$$f\left(x\right)=?$$

Commented by PRITHWISH SEN 2 last updated on 19/Oct/20

$$\mathrm{x}=\mathrm{x}+1$$$$\mathrm{f}(\mathrm{x}+3)+\mathrm{f}\left(\mathrm{x}\right)={2\mathrm{x}}^2+4\mathrm{x}+16.....\left(\mathrm{i}\right)$$$$\mathrm{x}=\mathrm{x}-2$$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(\mathrm{x}-3)={2\mathrm{x}}^2-8\mathrm{x}+22...\left(\mathrm{ii}\right)$$$$\mathrm{adding}\left(\mathrm{i}\right)\mathrm{\&}\left(\mathrm{ii}\right)$$$$\mathrm{f}(\mathrm{x}+3)+\mathrm{f}(\mathrm{x}-3)+2\mathrm{f}\left(\mathrm{x}\right)=4\mathrm{f}\left(\mathrm{x}\right)+2\times 3^2={4\mathrm{x}}^2-4\mathrm{x}+38$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{{4\mathrm{x}}^2-4\mathrm{x}+38-2\times 3^2}{4}={\mathrm{x}}^2-\mathrm{x}+5$$

Answered by bemath last updated on 19/Oct/20

$$letf\left(x\right)={px}^2+qx+r$$$$f(x+2)={px}^2+4px+4p+qx+2q+r$$$$={px}^2+(4p+q)x+4p+2q+r$$$$f(x-1)={px}^2-2px+p+qx-q+r$$$$={px}^2+(-2p+q)x+p-q+r$$$$\Rightarrow f(x+2)+f(x-1)=$$$$2{px}^2+(2p+2q)x+5p+q+2r=2x^2+14$$$$\{\begin{array}{l}2p=2\Rightarrow p=1\\ 2p+2q=0;q=-1\\ 5p+q+2r=14;4+2r=14;r=5\end{array}$$$$\therefore f\left(x\right)=x^2-x+5$$$$$$

Commented by Jamshidbek2311 last updated on 19/Oct/20

$$wrong.$$

Commented by benjo_mathlover last updated on 19/Oct/20

$$checking$$$$f\left(x\right)=x^2-x+5\to \{\begin{array}{l}f(x+2)=x^2+4x+4-x-2+5=x^2+3x+7\\ f(x-1)=x^2-2x+1-x+1+5=x^2-3x+7\end{array}$$$$thenf(x+2)+f(x-1)=2x^2+14$$$$\mathit{C}o\mathit{r}\mathit{r}\mathit{e}\mathit{c}\mathit{t}.$$

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