Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com