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Question Number 193685 by cortano12 last updated on 18/Jun/23

$${\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2-4\mathrm{xf}\left(\mathrm{x}\right)+3=0$$$${\mathrm{f}}^{-1}\left(3\right)=?$$

Answered by mr W last updated on 18/Jun/23

$$y^2-4xy+3=0$$$$\Rightarrow x=\frac{y^2+3}{4y}$$$$\Rightarrow f^{-1}\left(x\right)=\frac{x^2+3}{4x}$$$$\Rightarrow f^{-1}\left(3\right)=\frac{3^2+3}{4\times 3}=1$$

Answered by MM42 last updated on 18/Jun/23

$$f^{-1}\left(x\right)=3\Rightarrow x=f\left(3\right)$$$$\Rightarrow 9-12x+3=0\Rightarrow x=1=f^{-1}\left(3\right)$$

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