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Question Number 11238 by Joel576 last updated on 18/Mar/17

f(1 − 2x) = g(x + 3)  f^(−1) (x) = ?    (A) 7 − 2g^(−1) (x)  (B) 2g^(−1) (x) + 7  (C) 2g^(−1) (x) − 5  (D) 5 − 2g^(−1) (x)

$${f}\left(\mathrm{1}\:−\:\mathrm{2}{x}\right)\:=\:{g}\left({x}\:+\:\mathrm{3}\right) \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:? \\ $$$$ \\ $$$$\left(\mathrm{A}\right)\:\mathrm{7}\:−\:\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left(\mathrm{B}\right)\:\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{7} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right)\:−\:\mathrm{5} \\ $$$$\left(\mathrm{D}\right)\:\mathrm{5}\:−\:\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right) \\ $$

Answered by mrW1 last updated on 18/Mar/17

u=f(1−2x)=g(x+3)  ⇒1−2x=f^(−1) (u)     ...(i)  ⇒x+3=g^(−1) (u)      ...(ii)  2×(ii)+(i) ⇒7=f^(−1) (u)+2g^(−1) (u)  ⇒f^(−1) (u)=7−2g^(−1) (u)  ⇒f^(−1) (x)=7−2g^(−1) (x)

$${u}={f}\left(\mathrm{1}−\mathrm{2}{x}\right)={g}\left({x}+\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}{x}={f}^{−\mathrm{1}} \left({u}\right)\:\:\:\:\:...\left({i}\right) \\ $$$$\Rightarrow{x}+\mathrm{3}={g}^{−\mathrm{1}} \left({u}\right)\:\:\:\:\:\:...\left({ii}\right) \\ $$$$\mathrm{2}×\left({ii}\right)+\left({i}\right)\:\Rightarrow\mathrm{7}={f}^{−\mathrm{1}} \left({u}\right)+\mathrm{2}{g}^{−\mathrm{1}} \left({u}\right) \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({u}\right)=\mathrm{7}−\mathrm{2}{g}^{−\mathrm{1}} \left({u}\right) \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\mathrm{7}−\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right) \\ $$

Commented by Joel576 last updated on 18/Mar/17

thank you very much

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

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