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Question Number 187988 by cortano12 last updated on 24/Feb/23

find function f(x) and g(x) such that { ((f(2x−1)+g(1−x)=x+1)),((f((x/(x+1)))+2g((1/(2x+2)))=3)) :}

$$\:\mathrm{find}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{such}\:\mathrm{that}\:\begin{cases}{\mathrm{f}\left(\mathrm{2x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x}+\mathrm{1}}\\{\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)+\mathrm{2g}\left(\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{2}}\right)=\mathrm{3}}\end{cases} \\ $$

Answered by MathGuy last updated on 24/Feb/23

Answer :− do x→x+1 in eq.1 & x→x−1 in eq.2 you will get f(2x+1)+g(−x)=x+2 as new eq.1. call it eq.3 & f(((x−1)/x))+2g((1/(2x)))=3 as new eq.2. call it eq.4 do x→−x in eq.3 & x→(1/(2x)) in eq.4 you will get f(1−2x)+g(x)=2−x ; call it eq.5 & f(1−2x)+2g(x)=3 ; call it eq.6 doing [eq.6−eq.5] gives g(x)=x+1 put value of g(x) in eq.5 you will get f(1−2x)=1−2x now putting x→((1−x)/2) , gives f(x)=x so we have f(x)=x & g(x)=x+1

$${Answer}\::−\:{do}\:{x}\rightarrow{x}+\mathrm{1}\:{in}\:{eq}.\mathrm{1}\:\&\:{x}\rightarrow{x}−\mathrm{1}\:{in}\:{eq}.\mathrm{2} \\ $$$${you}\:{will}\:{get} \\ $$$${f}\left(\mathrm{2}{x}+\mathrm{1}\right)+{g}\left(−{x}\right)={x}+\mathrm{2}\:{as}\:{new}\:{eq}.\mathrm{1}.\:{call}\:{it}\:{eq}.\mathrm{3} \\ $$$$\& \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}}{\mathrm{2}{x}}\right)=\mathrm{3}\:{as}\:{new}\:{eq}.\mathrm{2}.\:{call}\:{it}\:{eq}.\mathrm{4} \\ $$$${do}\:{x}\rightarrow−{x}\:{in}\:{eq}.\mathrm{3}\:\&\:{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}{x}}\:{in}\:{eq}.\mathrm{4} \\ $$$${you}\:{will}\:{get} \\ $$$${f}\left(\mathrm{1}−\mathrm{2}{x}\right)+{g}\left({x}\right)=\mathrm{2}−{x}\:\:;\:{call}\:{it}\:{eq}.\mathrm{5} \\ $$$$\& \\ $$$${f}\left(\mathrm{1}−\mathrm{2}{x}\right)+\mathrm{2}{g}\left({x}\right)=\mathrm{3}\:\:;\:{call}\:{it}\:{eq}.\mathrm{6} \\ $$$$ \\ $$$${doing}\:\left[{eq}.\mathrm{6}−{eq}.\mathrm{5}\right]\:{gives} \\ $$$$ \\ $$$${g}\left({x}\right)={x}+\mathrm{1} \\ $$$$ \\ $$$${put}\:{value}\:{of}\:{g}\left({x}\right)\:{in}\:{eq}.\mathrm{5} \\ $$$${you}\:{will}\:{get} \\ $$$${f}\left(\mathrm{1}−\mathrm{2}{x}\right)=\mathrm{1}−\mathrm{2}{x} \\ $$$${now}\:{putting}\:{x}\rightarrow\frac{\mathrm{1}−{x}}{\mathrm{2}}\:,\:{gives} \\ $$$${f}\left({x}\right)={x} \\ $$$$ \\ $$$${so}\:{we}\:{have}\:{f}\left({x}\right)={x}\:\&\:{g}\left({x}\right)={x}+\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 24/Feb/23

such that { ((f(2x−1)+g(1−x)=x+1...(i))),((f((x/(x+1)))+2g((1/(2x+2)))=3....(ii))) :} (i):_(−) 2x−1=y⇒x=((y+1)/2) (i)⇒f(y)+g(1−((y+1)/2))=((y+1)/2)+1 f(y)+g(((1−y)/2))=((y+3)/2)....(iii) (ii):_(−) (x/(x+1))=y x−yx=y x=(y/(1−y)) (ii)⇒f(y)+2g((1/(2((y/(1−y)))+2)))=3 f(y)+2g((1/((2y+2−2y)/(1−y))))=3 f(y)+2g(((1−y)/2))=3......(iv) (iv)−(iii):g(((1−y)/2))=3−((y+3)/2)=((3−y)/2) Let ((1−y)/2)=x g(x)=((1−y+2)/2)=((1−y)/2)+1=x+1 2(iii)−(iv): (vi): f(y)+2g(((1−y)/2))=3 2(iii): 2f(y)+2g(((1−y)/2))=y+3 2(iii)−(iv): f(y)=y+3−3 f(y)=y f(x)=x

$$\:\mathrm{such}\:\mathrm{that}\:\begin{cases}{\mathrm{f}\left(\mathrm{2x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x}+\mathrm{1}...\left({i}\right)}\\{\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)+\mathrm{2g}\left(\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{2}}\right)=\mathrm{3}....\left({ii}\right)}\end{cases} \\ $$$$\underset{−} {\left({i}\right):} \\ $$$$\mathrm{2}{x}−\mathrm{1}={y}\Rightarrow{x}=\frac{{y}+\mathrm{1}}{\mathrm{2}} \\ $$$$\left({i}\right)\Rightarrow{f}\left({y}\right)+{g}\left(\mathrm{1}−\frac{{y}+\mathrm{1}}{\mathrm{2}}\right)=\frac{{y}+\mathrm{1}}{\mathrm{2}}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{f}\left({y}\right)+{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)=\frac{{y}+\mathrm{3}}{\mathrm{2}}....\left({iii}\right) \\ $$$$\underset{−} {\left({ii}\right):} \\ $$$$\frac{{x}}{{x}+\mathrm{1}}={y} \\ $$$${x}−{yx}={y} \\ $$$${x}=\frac{{y}}{\mathrm{1}−{y}} \\ $$$$\left({ii}\right)\Rightarrow{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}}{\mathrm{2}\left(\frac{{y}}{\mathrm{1}−{y}}\right)+\mathrm{2}}\right)=\mathrm{3} \\ $$$$\:\:\:\:\:{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}}{\frac{\mathrm{2}{y}+\mathrm{2}−\mathrm{2}{y}}{\mathrm{1}−{y}}}\right)=\mathrm{3} \\ $$$$\:\:\:\:\:{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)=\mathrm{3}......\left({iv}\right) \\ $$$$\left({iv}\right)−\left({iii}\right):{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)=\mathrm{3}−\frac{{y}+\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}−{y}}{\mathrm{2}} \\ $$$${Let}\:\frac{\mathrm{1}−{y}}{\mathrm{2}}={x} \\ $$$$\:\:\:\:\:\:\:\:\:\:{g}\left({x}\right)=\frac{\mathrm{1}−{y}+\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{1}−{y}}{\mathrm{2}}+\mathrm{1}={x}+\mathrm{1} \\ $$$$\mathrm{2}\left({iii}\right)−\left({iv}\right): \\ $$$$\left({vi}\right):\:\:\:\:\:\:{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)=\mathrm{3} \\ $$$$\mathrm{2}\left({iii}\right):\:\:\mathrm{2}{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)={y}+\mathrm{3} \\ $$$$\:\:\mathrm{2}\left({iii}\right)−\left({iv}\right):\:\:{f}\left({y}\right)={y}+\mathrm{3}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({y}\right)={y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)={x} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Feb/23

find function f(x) and g(x) such that { ((f(2x−1)+g(1−x)=x+1...(i))),((f((x/(x+1)))+2g((1/(2x+2)))=3.....(ii))) :} (i): 1−x=y⇒ x=1−y (i)⇒f( 2(1−y)−1 )+g(y)=1−y+1 f(1−2y)+g(y)=2−y.....(iii) (ii): (1/(2x+2))=y⇒2xy+2y=1⇒x=((1−2y)/(2y)) (ii)⇒f((((1−2y)/(2y))/(((1−2y)/(2y))+1)))+2g(y)=3 (((1−2y)/(2y))/(((1−2y)/(2y))+1))=1−2y f(1−2y)+2g(y)=3......(iv) (iv)−(iii): g(y)=3−(2−y)=y+1 g(x)=x+1 2(iii)−(iv): 2(iii)⇒ 2f(1−2y)+2g(y)=4−2y (iv)⇒ f(1−2y)+2g(y)=3 f(1−2y)=4−2y−3 f(1−2y)=1−2y f(x)=x

$$\:\mathrm{find}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{such}\:\mathrm{that}\:\begin{cases}{\mathrm{f}\left(\mathrm{2x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x}+\mathrm{1}...\left({i}\right)}\\{\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)+\mathrm{2g}\left(\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{2}}\right)=\mathrm{3}.....\left({ii}\right)}\end{cases}\: \\ $$$$\left(\mathrm{i}\right):\:\:\mathrm{1}−\mathrm{x}=\mathrm{y}\Rightarrow\:\mathrm{x}=\mathrm{1}−\mathrm{y} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{i}\right)\Rightarrow\mathrm{f}\left(\:\mathrm{2}\left(\mathrm{1}−\mathrm{y}\right)−\mathrm{1}\:\right)+\mathrm{g}\left(\mathrm{y}\right)=\mathrm{1}−\mathrm{y}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)+\mathrm{g}\left(\mathrm{y}\right)=\mathrm{2}−\mathrm{y}.....\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{ii}\right):\:\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{2}}=\mathrm{y}\Rightarrow\mathrm{2xy}+\mathrm{2y}=\mathrm{1}\Rightarrow\mathrm{x}=\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}} \\ $$$$\:\:\:\:\:\:\left(\mathrm{ii}\right)\Rightarrow\mathrm{f}\left(\frac{\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}}}{\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}}+\mathrm{1}}\right)+\mathrm{2g}\left(\mathrm{y}\right)=\mathrm{3} \\ $$$$\frac{\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}}}{\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}}+\mathrm{1}}=\mathrm{1}−\mathrm{2y} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)+\mathrm{2g}\left(\mathrm{y}\right)=\mathrm{3}......\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{iv}\right)−\left(\mathrm{iii}\right):\:\:\:\mathrm{g}\left(\mathrm{y}\right)=\mathrm{3}−\left(\mathrm{2}−\mathrm{y}\right)=\mathrm{y}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{2}\left(\mathrm{iii}\right)−\left(\mathrm{iv}\right): \\ $$$$\:\:\:\:\mathrm{2}\left(\mathrm{iii}\right)\Rightarrow\:\mathrm{2f}\left(\mathrm{1}−\mathrm{2y}\right)+\mathrm{2g}\left(\mathrm{y}\right)=\mathrm{4}−\mathrm{2y} \\ $$$$\:\:\:\:\:\:\left(\mathrm{iv}\right)\Rightarrow\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)+\mathrm{2g}\left(\mathrm{y}\right)=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)=\mathrm{4}−\mathrm{2y}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)=\mathrm{1}−\mathrm{2y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x} \\ $$

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