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

    If    f(x)= ((x^( 2) − 2x −8)/(x^( 2) −7x +12))     then ,find :           f^( −1) (x)=?

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

Answered by TheSupreme last updated on 07/Feb/22

D=R−{4,3}  f(x)=(((x−4)(x+2))/((x−4)(x−3)))=((x+2)/(x−3))=1+(5/(x−3))  y−1=(5/(f^− (y)−3))  f^(−1) (y)=3+(5/(y−1))

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

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

 f(x)= ((x^( 2) − 2x −8)/(x^( 2) −7x +12))=(((x−4)(x+2))/((x−4)(x−3)))   y= ((x+2)/(x−3))=((x−3+5)/(x−3))=1+(5/(x−3))  y−1=(5/(x−3))  x−3=(5/(y−1))⇒x=(5/(y−1))+3=((3y−3+5)/(y−1))  x=((3y+2)/(y−1))  Exchanging  x & y  y=((3x+2)/(x−1))  ∴ f^(−1) (x)=((3x+2)/(x−1))

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

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