f-x-x-2-4x-3-x-2-3-f-1-x-

Question Number 182581 by sciencestudent last updated on 11/Dec/22

f (x) = \frac{x^{2} + 4 x - 3}{x^{2} - 3}

f^{- 1} (x) = ?

Answered by cortano1 last updated on 11/Dec/22

{yx}^{2} - 3 y = x^{2} + 4 x - 3

(y - 1) x^{2} - 4 x = 3 y - 3

x^{2} - (\frac{4}{y - 1}) x = \frac{3 y - 3}{y - 1}, y \neq 1

{(x - (\frac{2}{y - 1}))}^{2} = 3 + \frac{4}{{(y - 1)}^{2}}

\Rightarrow x = \frac{2}{y - 1} \pm \sqrt{\frac{{3 y}^{2} - 6 y + 7}{{(y - 1)}^{2}}}

\Rightarrow f^{- 1} (x) = \frac{2 \pm \sqrt{{3 x}^{2} - 6 x + 7}}{x - 1}

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