Question Number 194282 by alcohol last updated on 02/Jul/23
$${f}\left({f}\left({x}\right)\right)\:=\:{ax}\:+\:{b} \\ $$$$\mathrm{1}.\:{show}\:{that}\:{f}\left({ax}+{b}\right)\:=\:{af}\left({x}\right)\:+\:{b} \\ $$$${deduce}\:{f}\:'\left({ax}\:+\:{b}\right) \\ $$$$\mathrm{2}.\:{Show}\:{that}\:{f}\:'\left({x}\right)\:{is}\:{a}\:{constant}\: \\ $$$${hence}\:{deduce}\:{f} \\ $$
Answered by Frix last updated on 02/Jul/23
![f(x)=αx+β ⇒ f(f(x))=α^2 x+(α+1)β ⇒ a=α^2 _([⇒ a>0]) ∧b=(α+1)β ⇔ α=±(√a)∧β=(b/(1±(√a))) ⇒ f(x)=±(√a)x+(b/(1±(√a))) f(ax+b)=f(α^2 x+(α+1)β)= =α^3 x+(α^2 +α+1)β af(x)+b=α^2 (αx+β)+(α+1)β= =α^3 x+(α^2 +α+1)β f′(ax+b)=α^3 =±a^(3/2) f′(x)=±(√a)](https://www.tinkutara.com/question/Q194287.png)
$${f}\left({x}\right)=\alpha{x}+\beta\:\Rightarrow \\ $$$${f}\left({f}\left({x}\right)\right)=\alpha^{\mathrm{2}} {x}+\left(\alpha+\mathrm{1}\right)\beta\:\Rightarrow \\ $$$$\underset{\left[\Rightarrow\:{a}>\mathrm{0}\right]} {{a}=\alpha^{\mathrm{2}} }\wedge{b}=\left(\alpha+\mathrm{1}\right)\beta\:\Leftrightarrow\:\alpha=\pm\sqrt{{a}}\wedge\beta=\frac{{b}}{\mathrm{1}\pm\sqrt{{a}}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\pm\sqrt{{a}}{x}+\frac{{b}}{\mathrm{1}\pm\sqrt{{a}}} \\ $$$${f}\left({ax}+{b}\right)={f}\left(\alpha^{\mathrm{2}} {x}+\left(\alpha+\mathrm{1}\right)\beta\right)= \\ $$$$=\alpha^{\mathrm{3}} {x}+\left(\alpha^{\mathrm{2}} +\alpha+\mathrm{1}\right)\beta \\ $$$${af}\left({x}\right)+{b}=\alpha^{\mathrm{2}} \left(\alpha{x}+\beta\right)+\left(\alpha+\mathrm{1}\right)\beta= \\ $$$$=\alpha^{\mathrm{3}} {x}+\left(\alpha^{\mathrm{2}} +\alpha+\mathrm{1}\right)\beta \\ $$$${f}'\left({ax}+{b}\right)=\alpha^{\mathrm{3}} =\pm{a}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${f}'\left({x}\right)=\pm\sqrt{{a}} \\ $$