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Question Number 128401 by bramlexs22 last updated on 07/Jan/21

$$\mathrm{Given}\mathrm{function}\mathrm{f}(x+1)+\mathrm{f}(x-1)=x^2$$$$\mathrm{then}{\mathrm{f}}^{-1}\left(x\right)=?$$$$\left(A\right)\pm \sqrt{1-2x};x⩽\frac{1}{2}$$$$\left(B\right)\pm \sqrt{x+2};x⩾-2$$$$\left(C\right)\pm \sqrt{2x+1};x⩾-\frac{1}{2}$$$$\left(D\right)\pm \sqrt{3x-1};x⩾\frac{1}{3}$$$$\left(E\right)\pm \sqrt{2x-1};x⩾\frac{1}{2}$$

Answered by liberty last updated on 07/Jan/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$$$$\Rightarrow \mathrm{f}(\mathrm{x}+1)={\mathrm{ax}}^2+2\mathrm{ax}+\mathrm{bx}+\mathrm{a}+\mathrm{b}+\mathrm{c}$$$$\Rightarrow \mathrm{f}(\mathrm{x}-1)={\mathrm{ax}}^2-2\mathrm{ax}+\mathrm{bx}+\mathrm{a}-\mathrm{b}+\mathrm{c}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}-$$$$\Rightarrow {2\mathrm{ax}}^2+2\mathrm{bx}+2(\mathrm{a}+\mathrm{c})={\mathrm{x}}^2+0\mathrm{x}+0$$$$\{\begin{array}{l}\mathrm{a}=\frac{1}{2}\\ \mathrm{b}=0;\mathrm{c}=-\frac{1}{2}\end{array}$$$$\therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{x}}^2-\frac{1}{2};{\mathrm{x}}^2=2\mathrm{y}+1$$$$\Rightarrow \mathrm{x}=\pm \sqrt{2\mathrm{y}+1}\Rightarrow {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\pm \sqrt{2\mathrm{x}+1};\mathrm{x}⩾-\frac{1}{2}$$

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