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Question-186870

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Question Number 186870 by 073 last updated on 11/Feb/23
Commented by 073 last updated on 11/Feb/23
solution please??
$$\mathrm{solution}\:\mathrm{please}?? \\ $$
Answered by Ar Brandon last updated on 11/Feb/23
{ ((fog^(−1) (x)=3x+2),(...(i))),((gof(x)=2x−1),(...(ii))) :} (ii) ⇒f(x)=g^(−1) (2x−1) ⇒f(3)=g^(−1) (5) ⇒fof(3)=fog^(−1) (5)=3(5)+2=17
$$\begin{cases}{{f}\mathrm{o}{g}^{−\mathrm{1}} \left({x}\right)=\mathrm{3}{x}+\mathrm{2}}&{…\left({i}\right)}\\{{g}\mathrm{o}{f}\left({x}\right)=\mathrm{2}{x}−\mathrm{1}}&{…\left({ii}\right)}\end{cases} \\ $$$$\left({ii}\right)\:\Rightarrow{f}\left({x}\right)={g}^{−\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{f}\left(\mathrm{3}\right)={g}^{−\mathrm{1}} \left(\mathrm{5}\right) \\ $$$$\Rightarrow{f}\mathrm{o}{f}\left(\mathrm{3}\right)={f}\mathrm{o}{g}^{−\mathrm{1}} \left(\mathrm{5}\right)=\mathrm{3}\left(\mathrm{5}\right)+\mathrm{2}=\mathrm{17} \\ $$
Commented by AROUNAMoussa last updated on 12/Feb/23
Toutes mes excuses. C'était une erreur de manipulation
Commented by ARUNG_Brandon_MBU last updated on 12/Feb/23
Pas de quoi, l'ami.
Answered by ARUNG_Brandon_MBU last updated on 11/Feb/23
{ ((fog^(−1) (x)=3x+2),(...(i))),((gof(x)=2x−1),(...(ii))) :} (ii) ⇒f(x)=g^(−1) (2x−1) ⇒g^(−1) (x)=f(((x+1)/2)) ...(iii) (iii) in (i) ⇒fof(((x+1)/2))=3x+2 ⇒fof(x)=6x−1 ⇒f(3)=6(3)−1=17
$$\begin{cases}{{f}\mathrm{o}{g}^{−\mathrm{1}} \left({x}\right)=\mathrm{3}{x}+\mathrm{2}}&{…\left({i}\right)}\\{{g}\mathrm{o}{f}\left({x}\right)=\mathrm{2}{x}−\mathrm{1}}&{…\left({ii}\right)}\end{cases} \\ $$$$\left({ii}\right)\:\Rightarrow{f}\left({x}\right)={g}^{−\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{g}^{−\mathrm{1}} \left({x}\right)={f}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)\:\:…\left({iii}\right) \\ $$$$\left({iii}\right)\:\mathrm{in}\:\left({i}\right)\:\Rightarrow{f}\mathrm{o}{f}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)=\mathrm{3}{x}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{f}\mathrm{o}{f}\left({x}\right)=\mathrm{6}{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{f}\left(\mathrm{3}\right)=\mathrm{6}\left(\mathrm{3}\right)−\mathrm{1}=\mathrm{17} \\ $$
Answered by aba last updated on 11/Feb/23
fof(3)=f(f(3))=f(g^(−1) (g(f(3))))=f(g^(−1) (5))=3×5+2=17
$$ \\ $$$${fof}\left(\mathrm{3}\right)={f}\left({f}\left(\mathrm{3}\right)\right)={f}\left({g}^{−\mathrm{1}} \left({g}\left({f}\left(\mathrm{3}\right)\right)\right)\right)={f}\left({g}^{−\mathrm{1}} \left(\mathrm{5}\right)\right)=\mathrm{3}×\mathrm{5}+\mathrm{2}=\mathrm{17} \\ $$
Answered by horsebrand11 last updated on 12/Feb/23
{ ((f(g^(−1) (x))=3x+2)),((g(f(x))=2x−1)) :} ⇒g^(−1) (x)=f^(−1) (3x+2) ⇒f(x)=g^(−1) (2x−1) ⇒f(f(x))=f(g^(−1) (2x−1)) ⇒f(f(3))=f(g^(−1) (5)) ⇒f(f(3))=3.5 + 2 = 17
$$\:\begin{cases}{{f}\left({g}^{−\mathrm{1}} \left({x}\right)\right)=\mathrm{3}{x}+\mathrm{2}}\\{{g}\left({f}\left({x}\right)\right)=\mathrm{2}{x}−\mathrm{1}}\end{cases} \\ $$$$\:\Rightarrow{g}^{−\mathrm{1}} \left({x}\right)={f}^{−\mathrm{1}} \left(\mathrm{3}{x}+\mathrm{2}\right) \\ $$$$\Rightarrow{f}\left({x}\right)={g}^{−\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left({f}\left({x}\right)\right)={f}\left({g}^{−\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right)\right) \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{3}\right)\right)={f}\left({g}^{−\mathrm{1}} \left(\mathrm{5}\right)\right) \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{3}\right)\right)=\mathrm{3}.\mathrm{5}\:+\:\mathrm{2}\:=\:\mathrm{17} \\ $$

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