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

$$f\left(x\right)=\frac{x^2+4x-3}{x^2-3}$$$$f^{-1}\left(x\right)=?$$

Answered by cortano1 last updated on 11/Dec/22

$${\mathrm{yx}}^2-3\mathrm{y}={\mathrm{x}}^2+4\mathrm{x}-3$$$$(\mathrm{y}-1){\mathrm{x}}^2-4\mathrm{x}=3\mathrm{y}-3$$$${\mathrm{x}}^2-\left(\frac{4}{\mathrm{y}-1}\right)\mathrm{x}=\frac{3\mathrm{y}-3}{\mathrm{y}-1},\mathrm{y}\ne 1$$$${(\mathrm{x}-\left(\frac{2}{\mathrm{y}-1}\right))}^2=3+\frac{4}{{(\mathrm{y}-1)}^2}$$$$\Rightarrow \mathrm{x}=\frac{2}{\mathrm{y}-1}\pm \sqrt{\frac{{3\mathrm{y}}^2-6\mathrm{y}+7}{{(\mathrm{y}-1)}^2}}$$$$\Rightarrow {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{2\pm \sqrt{{3\mathrm{x}}^2-6\mathrm{x}+7}}{\mathrm{x}-1}$$

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