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Question Number 119909 by bramlexs22 last updated on 28/Oct/20

$$Givenf\left(x\right)=\frac{px+q}{x+2},q\ne 0$$$$f^{-1}\left(q\right)=-1then{f}^{-1}\left(2q\right)=?$$

Answered by TITA last updated on 28/Oct/20

$$q=f(-1)q=q-pp=0$$$$f\left(x\right)=\frac{q}{x+2}\Rightarrow letf\left(x\right)=y$$$$y=\frac{q}{x+2}xy+2y=qx=\frac{q-2y}{y}$$$$hence{f}^{-1}\left(x\right)=\frac{q-2x}{x},x\ne 0$$$$\Rightarrow f^{-1}\left(2q\right)=\frac{q-4q}{2q}=\frac{-3q}{2q}$$$$\Rightarrow f^{-1}\left(2q\right)=\frac{-3}{2}$$$$$$

Answered by 1549442205PVT last updated on 28/Oct/20

$$f\left(x\right)=\frac{px+q}{x+2},q\ne 0$$$$\Rightarrow \mathrm{x}(\mathrm{f}\left(\mathrm{x}\right)-\mathrm{p})=\mathrm{q}-2\mathrm{f}\left(\mathrm{x}\right)\Rightarrow \mathrm{x}=\frac{\mathrm{q}-2\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{p}}$$$$\Rightarrow \mathrm{f}{}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{q}-2\mathrm{x}}{\mathrm{x}-\mathrm{p}}.\mathrm{From}\mathrm{the}\mathrm{hypothesis}$$$$\mathrm{f}{}^{-1}\left(\mathrm{q}\right)=-1\mathrm{we}\mathrm{get}\frac{\mathrm{q}-2\mathrm{q}}{\mathrm{q}-\mathrm{p}}=-1$$$$\Rightarrow \mathrm{q}-2\mathrm{q}=\mathrm{p}-\mathrm{q}\Rightarrow \mathrm{p}=0\Rightarrow \mathrm{f}{}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{q}-2\mathrm{x}}{\mathrm{x}}$$$$\Rightarrow \mathrm{f}{}^{-1}\left(2\mathrm{q}\right)=\frac{\mathrm{q}-2.2\mathrm{q}}{2\mathrm{q}}=\frac{-3\mathrm{q}}{2\mathrm{q}}=-3/2$$

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