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Question Number 13099 by Joel577 last updated on 14/May/17

If f(x + 5) = g(2x −1) Find 2f^(−1) (x) (A) g^(−1) (x) + 11 (D) g^(−1) (x/2) + 6 (B) g^(−1) (x) + 9 (E) g^(−1) (2x) + 6 (C) g^(−1) (x) + 6

$$\mathrm{If}\:{f}\left({x}\:+\:\mathrm{5}\right)\:=\:{g}\left(\mathrm{2}{x}\:−\mathrm{1}\right) \\ $$$$\mathrm{Find}\:\mathrm{2}{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$ \\ $$$$\left(\mathrm{A}\right)\:{g}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{11}\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:{g}^{−\mathrm{1}} \left({x}/\mathrm{2}\right)\:+\:\mathrm{6} \\ $$$$\left(\mathrm{B}\right)\:{g}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{E}\right)\:{g}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\:+\:\mathrm{6} \\ $$$$\left(\mathrm{C}\right)\:{g}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{6} \\ $$

Answered by 433 last updated on 14/May/17

h(x)=x+5 & k(x)=2x−1 foh=gok f=gokoh^(−1) f^(−1) =hok^(−1) g^(−1) 2f^(−1) =2(hok^(−1) g^(−1) ) k^(−1) (x)=((x+1)/2) 2f^(−1) (x)=2h(k^(−1) (g^(−1) )(x) =2h(((g^(−1) (x)+1)/2)) =2(((g^(−1) (x)+1)/2)+5)=g^(−1) (x)+1+10 =g^(−1) (x)+11

$${h}\left({x}\right)={x}+\mathrm{5}\:\&\:{k}\left({x}\right)=\mathrm{2}{x}−\mathrm{1} \\ $$$${foh}={gok} \\ $$$${f}={gokoh}^{−\mathrm{1}} \\ $$$${f}^{−\mathrm{1}} ={hok}^{−\mathrm{1}} {g}^{−\mathrm{1}} \\ $$$$\mathrm{2}{f}^{−\mathrm{1}} =\mathrm{2}\left({hok}^{−\mathrm{1}} {g}^{−\mathrm{1}} \right) \\ $$$${k}^{−\mathrm{1}} \left({x}\right)=\frac{{x}+\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{f}^{−\mathrm{1}} \left({x}\right)=\mathrm{2}{h}\left({k}^{−\mathrm{1}} \left({g}^{−\mathrm{1}} \right)\left({x}\right)\right. \\ $$$$=\mathrm{2}{h}\left(\frac{{g}^{−\mathrm{1}} \left({x}\right)+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}\left(\frac{{g}^{−\mathrm{1}} \left({x}\right)+\mathrm{1}}{\mathrm{2}}+\mathrm{5}\right)={g}^{−\mathrm{1}} \left({x}\right)+\mathrm{1}+\mathrm{10} \\ $$$$={g}^{−\mathrm{1}} \left({x}\right)+\mathrm{11} \\ $$

Commented by Joel577 last updated on 15/May/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by mrW1 last updated on 14/May/17

f(x + 5) = g(2x −1)=u ⇒x+5=f^(−1) (u) ...(i) ⇒2x−1=g^(−1) (u) ...(ii) 2×(i)−(ii): ⇒11=2f^(−1) (u)−g^(−1) (u) ⇒2f^(−1) (u)=g^(−1) (u)+11 ⇒2f^(−1) (x)=g^(−1) (x)+11 ⇒Answer (A)

$${f}\left({x}\:+\:\mathrm{5}\right)\:=\:{g}\left(\mathrm{2}{x}\:−\mathrm{1}\right)={u} \\ $$$$\Rightarrow{x}+\mathrm{5}={f}^{−\mathrm{1}} \left({u}\right)\:\:\:\:\:...\left({i}\right) \\ $$$$\Rightarrow\mathrm{2}{x}−\mathrm{1}={g}^{−\mathrm{1}} \left({u}\right)\:\:\:...\left({ii}\right) \\ $$$$\mathrm{2}×\left({i}\right)−\left({ii}\right): \\ $$$$\Rightarrow\mathrm{11}=\mathrm{2}{f}^{−\mathrm{1}} \left({u}\right)−{g}^{−\mathrm{1}} \left({u}\right) \\ $$$$\Rightarrow\mathrm{2}{f}^{−\mathrm{1}} \left({u}\right)={g}^{−\mathrm{1}} \left({u}\right)+\mathrm{11} \\ $$$$\Rightarrow\mathrm{2}{f}^{−\mathrm{1}} \left({x}\right)={g}^{−\mathrm{1}} \left({x}\right)+\mathrm{11} \\ $$$$ \\ $$$$\Rightarrow{Answer}\:\left({A}\right) \\ $$

Commented by RasheedSindhi last updated on 14/May/17

Easy n simple!

$$\mathcal{E}\mathrm{asy}\:\mathrm{n}\:\mathrm{simple}! \\ $$

Commented by Joel577 last updated on 15/May/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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