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Question Number 165687 by mnjuly1970 last updated on 06/Feb/22

$$$$$$\mathrm{I}ff\left(x\right)=\frac{x^2-2x-8}{x^2-7x+12}$$$$then,find:f^{-1}\left(x\right)=?$$$$$$

Answered by TheSupreme last updated on 07/Feb/22

$$D=\mathbb{R}-\{4,3\}$$$$f\left(x\right)=\frac{(x-4)(x+2)}{(x-4)(x-3)}=\frac{x+2}{x-3}=1+\frac{5}{x-3}$$$$y-1=\frac{5}{f^{-}\left(y\right)-3}$$$$f^{-1}\left(y\right)=3+\frac{5}{y-1}$$$$$$

Answered by Rasheed.Sindhi last updated on 06/Feb/22

$$f\left(x\right)=\frac{x^2-2x-8}{x^2-7x+12}=\frac{\cancel{(x-4)}(x+2)}{\cancel{(x-4)}(x-3)}$$$$y=\frac{x+2}{x-3}=\frac{x-3+5}{x-3}=1+\frac{5}{x-3}$$$$y-1=\frac{5}{x-3}$$$$x-3=\frac{5}{y-1}\Rightarrow x=\frac{5}{y-1}+3=\frac{3y-3+5}{y-1}$$$$x=\frac{3y+2}{y-1}$$$$\mathcal{E}xchangingx\mathrm{\&}y$$$$y=\frac{3x+2}{x-1}$$$$\therefore f^{-1}\left(x\right)=\frac{3x+2}{x-1}$$

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