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Question Number 43353 by pieroo last updated on 10/Sep/18

Given the functions f(x)=2x−1 and f•g(x)=x^2 −x+2, find g(x)

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{f}\bullet\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}, \\ $$$$\mathrm{find}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$

Commented by maxmathsup by imad last updated on 10/Sep/18

we have f(x)=y ⇒2x−1=y ⇒x =((y+1)/2) ⇒f^(−1) (x) =((x+1)/2) but fog(x)=x^2 −x+2 ⇒f^(−1) of og(x) =f^(−1) (x^2 −x+2) ⇒ g(x) =f^(−1) (x^2 −x+2) =((x^2 −x+2+1)/2) =((x^2 −x+3)/2) .

$${we}\:{have}\:{f}\left({x}\right)={y}\:\Rightarrow\mathrm{2}{x}−\mathrm{1}={y}\:\Rightarrow{x}\:=\frac{{y}+\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)\:=\frac{{x}+\mathrm{1}}{\mathrm{2}} \\ $$$${but}\:{fog}\left({x}\right)={x}^{\mathrm{2}} −{x}+\mathrm{2}\:\Rightarrow{f}^{−\mathrm{1}} {of}\:{og}\left({x}\right)\:={f}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{2}\right)\:\Rightarrow \\ $$$${g}\left({x}\right)\:={f}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{2}\right)\:=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{2}+\mathrm{1}}{\mathrm{2}}\:=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{3}}{\mathrm{2}}\:. \\ $$

Answered by Joel578 last updated on 10/Sep/18

f(x) = 2x − 1 f(g(x)) = x^2 − x + 2 2g(x) − 1 = x^2 − x + 2 g(x) = ((x^2 − x + 3)/2) = (1/2)x^2 − (1/2)x + (3/2)

$${f}\left({x}\right)\:=\:\mathrm{2}{x}\:−\:\mathrm{1} \\ $$$$\:\:\:\:\:\:{f}\left({g}\left({x}\right)\right)\:=\:{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{2} \\ $$$$\mathrm{2}{g}\left({x}\right)\:−\:\mathrm{1}\:=\:{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

2g(x)−1=x^2 −x+2 2g(x)=x^2 −x+3 g(x)=((x^2 −x+3)/2)

$$\mathrm{2}{g}\left({x}\right)−\mathrm{1}={x}^{\mathrm{2}} −{x}+\mathrm{2} \\ $$$$\mathrm{2}{g}\left({x}\right)={x}^{\mathrm{2}} −{x}+\mathrm{3} \\ $$$${g}\left({x}\right)=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

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