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Question-187027




Question Number 187027 by yaslm last updated on 12/Feb/23
Answered by Farhadazizi last updated on 12/Feb/23
fog(x)=f(g(x))=((2g(x)−4)/(g(x)))=((2(√(x−1))−4)/( (√(x−1))))  gof(x)=g(f(x))=(√(f(x)−1))=(√((x−4)/x))  f(x)=((2x−4)/x)=2−(4/x)⇛f(x)−2=−(4/x)  −(1/(f(x)−2))=(x/4)⇛x=((−4)/(f(x)−2))⇛f^(−1) (x)=((−4)/(x−2))    g(x)=(√(x−1))⇒(g(x))^2 +1=x      ⇒g^(−1) (x)=x^2 +1
$${fog}\left({x}\right)={f}\left({g}\left({x}\right)\right)=\frac{\mathrm{2}{g}\left({x}\right)−\mathrm{4}}{{g}\left({x}\right)}=\frac{\mathrm{2}\sqrt{{x}−\mathrm{1}}−\mathrm{4}}{\:\sqrt{{x}−\mathrm{1}}} \\ $$$${gof}\left({x}\right)={g}\left({f}\left({x}\right)\right)=\sqrt{{f}\left({x}\right)−\mathrm{1}}=\sqrt{\frac{{x}−\mathrm{4}}{{x}}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{4}}{{x}}=\mathrm{2}−\frac{\mathrm{4}}{{x}}\Rrightarrow{f}\left({x}\right)−\mathrm{2}=−\frac{\mathrm{4}}{{x}} \\ $$$$−\frac{\mathrm{1}}{{f}\left({x}\right)−\mathrm{2}}=\frac{{x}}{\mathrm{4}}\Rrightarrow{x}=\frac{−\mathrm{4}}{{f}\left({x}\right)−\mathrm{2}}\Rrightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{−\mathrm{4}}{{x}−\mathrm{2}} \\ $$$$ \\ $$$${g}\left({x}\right)=\sqrt{{x}−\mathrm{1}}\Rightarrow\left({g}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{1}={x} \\ $$$$\:\:\:\:\Rightarrow{g}^{−\mathrm{1}} \left({x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$
Commented by yaslm last updated on 12/Feb/23
thanks sir,

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