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Question Number 194190 by tri26112004 last updated on 29/Jun/23

Explanation Why:  While f(ax+b)+f(cx+d)=ex+g  then f(x)=Ax^2 +Bx+C ¿

$${Explanation}\:{Why}: \\ $$$${While}\:{f}\left({ax}+{b}\right)+{f}\left({cx}+{d}\right)={ex}+{g} \\ $$$${then}\:{f}\left({x}\right)={Ax}^{\mathrm{2}} +{Bx}+{C}\:¿ \\ $$

Commented by Tinku Tara last updated on 30/Jun/23

For the given condition it is not  necessary for f(x) to be quadratic.  f(x)=Ax+B  f(ax+b)=Aax+Ab+B  f(cx+d)=Acx+Ad+B  g=A(b+d)+2B  e=A(a+c)  A=(e/(a+c))  B=g−((e(b+d))/((a+c)))

$$\mathrm{For}\:\mathrm{the}\:\mathrm{given}\:\mathrm{condition}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{necessary}\:\mathrm{for}\:{f}\left({x}\right)\:\mathrm{to}\:\mathrm{be}\:\mathrm{quadratic}. \\ $$$${f}\left({x}\right)={Ax}+{B} \\ $$$${f}\left({ax}+{b}\right)={Aax}+{Ab}+{B} \\ $$$${f}\left({cx}+{d}\right)={Acx}+{Ad}+{B} \\ $$$${g}={A}\left({b}+{d}\right)+\mathrm{2}{B} \\ $$$${e}={A}\left({a}+{c}\right) \\ $$$${A}=\frac{{e}}{{a}+{c}} \\ $$$${B}={g}−\frac{{e}\left({b}+{d}\right)}{\left({a}+{c}\right)} \\ $$

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