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Question Number 164770 by cortano1 last updated on 21/Jan/22

{ ((f(3x−1)+g(6x−1)=3x)),((f(x+1)+x g(2x+3)=2x^2 +x)) :} f(x)=?

$$\begin{cases}{{f}\left(\mathrm{3}{x}−\mathrm{1}\right)+{g}\left(\mathrm{6}{x}−\mathrm{1}\right)=\mathrm{3}{x}}\\{{f}\left({x}+\mathrm{1}\right)+{x}\:{g}\left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{2}{x}^{\mathrm{2}} +{x}}\end{cases} \\ $$$$\:{f}\left({x}\right)=? \\ $$

Answered by blackmamba last updated on 21/Jan/22

let { ((f(x)=px+q)),((g(x)=ax+b)) :} { ((f(3x−1)=3px+q−p)),((g(6x−1)=6ax+b−a)) :} ⇒(3p+6a)x+b+q−(a+p)=3x { ((3p+6a=3)),((b+q=a+p )) :} { ((f(x+1)=px+q+p)),((g(2x+3)=2ax+b+3a)) :} ⇒px+q+p+2ax^2 +(b+3a)x=2x^2 +x ⇒2ax^2 +(b+3a+p)x+p+q=2x^2 +x { ((p+q=0⇒p=−q)),((2a=2⇒a=1 ; b+3a+p=1 ; b+p=−2)) :} { ((3p+6 =3⇒p=−1 ∧ q=1)),((b+1=1+(−1)⇒b=−1)) :} ∴ { ((f(x)=−x+1)),((g(x)= x−1)) :}

$$\:\mathrm{let}\:\begin{cases}{{f}\left({x}\right)={px}+{q}}\\{{g}\left({x}\right)={ax}+{b}}\end{cases} \\ $$$$\:\begin{cases}{{f}\left(\mathrm{3}{x}−\mathrm{1}\right)=\mathrm{3}{px}+{q}−{p}}\\{{g}\left(\mathrm{6}{x}−\mathrm{1}\right)=\mathrm{6}{ax}+{b}−{a}}\end{cases}\:\Rightarrow\left(\mathrm{3}{p}+\mathrm{6}{a}\right){x}+{b}+{q}−\left({a}+{p}\right)=\mathrm{3}{x} \\ $$$$\:\begin{cases}{\mathrm{3}{p}+\mathrm{6}{a}=\mathrm{3}}\\{{b}+{q}={a}+{p}\:}\end{cases} \\ $$$$\begin{cases}{{f}\left({x}+\mathrm{1}\right)={px}+{q}+{p}}\\{{g}\left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{2}{ax}+{b}+\mathrm{3}{a}}\end{cases} \\ $$$$\:\Rightarrow{px}+{q}+{p}+\mathrm{2}{ax}^{\mathrm{2}} +\left({b}+\mathrm{3}{a}\right){x}=\mathrm{2}{x}^{\mathrm{2}} +{x} \\ $$$$\Rightarrow\mathrm{2}{ax}^{\mathrm{2}} +\left({b}+\mathrm{3}{a}+{p}\right){x}+{p}+{q}=\mathrm{2}{x}^{\mathrm{2}} +{x} \\ $$$$\:\begin{cases}{{p}+{q}=\mathrm{0}\Rightarrow{p}=−{q}}\\{\mathrm{2}{a}=\mathrm{2}\Rightarrow{a}=\mathrm{1}\:;\:{b}+\mathrm{3}{a}+{p}=\mathrm{1}\:;\:{b}+{p}=−\mathrm{2}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{3}{p}+\mathrm{6}\:=\mathrm{3}\Rightarrow{p}=−\mathrm{1}\:\wedge\:{q}=\mathrm{1}}\\{{b}+\mathrm{1}=\mathrm{1}+\left(−\mathrm{1}\right)\Rightarrow{b}=−\mathrm{1}}\end{cases} \\ $$$$\:\therefore\:\begin{cases}{{f}\left({x}\right)=−{x}+\mathrm{1}}\\{{g}\left({x}\right)=\:{x}−\mathrm{1}}\end{cases} \\ $$

Answered by TheSupreme last updated on 21/Jan/22

6x−1=t f(((t−1)/2))+g(t)=((t+1)/2) 2x+3=t f(((t−1)/2))+((t−3)/2)g(t)=(((t−3)/2))(t−2) g(t)(1−((t−3)/2))=((t−3)/2)(t−2)−((t+1)/2) g(t)=(((((t−3)(t−2))/2)−((t+1)/2))/(1−((t−3)/2)))=((t^2 −6t+5)/(5−t))=(((t−5)(t−1))/(−(t−5)))=1−t f(3x+1)+2−6x=3x f(3x+1)=9x−2 f(u)=3u+1

$$\mathrm{6}{x}−\mathrm{1}={t} \\ $$$${f}\left(\frac{{t}−\mathrm{1}}{\mathrm{2}}\right)+{g}\left({t}\right)=\frac{{t}+\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{x}+\mathrm{3}={t} \\ $$$${f}\left(\frac{{t}−\mathrm{1}}{\mathrm{2}}\right)+\frac{{t}−\mathrm{3}}{\mathrm{2}}{g}\left({t}\right)=\left(\frac{{t}−\mathrm{3}}{\mathrm{2}}\right)\left({t}−\mathrm{2}\right) \\ $$$${g}\left({t}\right)\left(\mathrm{1}−\frac{{t}−\mathrm{3}}{\mathrm{2}}\right)=\frac{{t}−\mathrm{3}}{\mathrm{2}}\left({t}−\mathrm{2}\right)−\frac{{t}+\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left({t}\right)=\frac{\frac{\left({t}−\mathrm{3}\right)\left({t}−\mathrm{2}\right)}{\mathrm{2}}−\frac{{t}+\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{{t}−\mathrm{3}}{\mathrm{2}}}=\frac{{t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{5}}{\mathrm{5}−{t}}=\frac{\left({t}−\mathrm{5}\right)\left({t}−\mathrm{1}\right)}{−\left({t}−\mathrm{5}\right)}=\mathrm{1}−{t} \\ $$$${f}\left(\mathrm{3}{x}+\mathrm{1}\right)+\mathrm{2}−\mathrm{6}{x}=\mathrm{3}{x} \\ $$$${f}\left(\mathrm{3}{x}+\mathrm{1}\right)=\mathrm{9}{x}−\mathrm{2} \\ $$$${f}\left({u}\right)=\mathrm{3}{u}+\mathrm{1} \\ $$

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