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

Given function f (x + 1) + f (x - 1) = x^{2}

then f^{- 1} (x) = ?

(A) \pm \sqrt{1 - 2 x}; x ⩽ \frac{1}{2}

(B) \pm \sqrt{x + 2}; x ⩾ - 2

(C) \pm \sqrt{2 x + 1}; x ⩾ - \frac{1}{2}

(D) \pm \sqrt{3 x - 1}; x ⩾ \frac{1}{3}

(E) \pm \sqrt{2 x - 1}; x ⩾ \frac{1}{2}

Answered by liberty last updated on 07/Jan/21

let f (x) = {ax}^{2} + bx + c

\Rightarrow f (x + 1) = {ax}^{2} + 2 ax + bx + a + b + c

\Rightarrow f (x - 1) = {ax}^{2} - 2 ax + bx + a - b + c

____________________________-

\Rightarrow {2 ax}^{2} + 2 bx + 2 (a + c) = x^{2} + 0 x + 0

{\begin{cases} a = \frac{1}{2} \\ b = 0; c = - \frac{1}{2} \end{cases}

∴ f (x) = \frac{1}{2} x^{2} - \frac{1}{2}; x^{2} = 2 y + 1

\Rightarrow x = \pm \sqrt{2 y + 1} \Rightarrow f^{- 1} (x) = \pm \sqrt{2 x + 1}; x ⩾ - \frac{1}{2}

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