Question Number 202184 by mr W last updated on 22/Dec/23
$${with}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{30}\:{and}\:{x}\in{R} \\ $$$${solve}\:{f}\left({f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)\right)=\mathrm{0} \\ $$
Answered by cortano12 last updated on 22/Dec/23
$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{6}\right)^{\mathrm{2}} −\mathrm{6}\:\Rightarrow\mathrm{x}=\sqrt{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{6}}−\mathrm{6} \\ $$$$\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\:\sqrt{\mathrm{x}+\mathrm{6}}−\mathrm{6}\: \\ $$$$\:\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\right)\:=\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{0}\right)=\sqrt{\mathrm{6}}−\mathrm{6} \\ $$$$\:\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\:=\:\mathrm{f}^{−\mathrm{1}} \left(\sqrt{\mathrm{6}}−\mathrm{6}\right)=\sqrt{\sqrt{\mathrm{6}}}−\mathrm{6} \\ $$$$\:\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\:\mathrm{f}^{−\mathrm{1}} \left(\sqrt{\sqrt{\mathrm{6}}}−\mathrm{6}\right)=\sqrt{\sqrt{\sqrt{\mathrm{6}}}}−\mathrm{6} \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{f}^{−\mathrm{1}} \left(\sqrt{\sqrt{\sqrt{\mathrm{6}}}}−\mathrm{6}\right)=\:\sqrt{\sqrt{\sqrt{\sqrt{\mathrm{6}}}}}−\mathrm{6} \\ $$$$\:\Rightarrow\:\mathrm{x}\:=\:\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\mathrm{6}}}}}}−\mathrm{6}\:=\:\sqrt[{\mathrm{32}}]{\mathrm{6}}\:−\:\mathrm{6} \\ $$
Commented by mr W last updated on 22/Dec/23
$${thanks}! \\ $$$$−\sqrt[{\mathrm{32}}]{\mathrm{6}}−\mathrm{6}\:{is}\:{also}\:{a}\:{root}. \\ $$
Commented by esmaeil last updated on 22/Dec/23
$${hi} \\ $$$${why}? \\ $$$${f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)={f}^{−\mathrm{1}} \left(\mathrm{0}\right) \\ $$
Commented by mr W last updated on 23/Dec/23
$${f}\left({x}\right)={t}=\mathrm{0}\:\Rightarrow{x}={f}^{−\mathrm{1}} \left({t}\right)={f}^{−\mathrm{1}} \left(\mathrm{0}\right) \\ $$$${f}\left({g}\left({x}\right)\right)={t}=\mathrm{0}\:\Rightarrow{g}\left({x}\right)={f}^{−\mathrm{1}} \left({t}\right)={f}^{−} \left(\mathrm{0}\right) \\ $$$${f}\left({f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)\right)={t}=\mathrm{0}\:\Rightarrow{f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)={f}^{−\mathrm{1}} \left({t}\right)={f}^{−} \left(\mathrm{0}\right) \\ $$
Answered by MATHEMATICSAM last updated on 22/Dec/23
![f(x) = x^2 + 12x + 30 = (x + 6)^2 − 6 ⇒ f(f(x)) = [(x + 6)^2 − 6 + 6]^2 − 6 = [(x + 6)^2 ]^2 − 6 = (x + 6)^4 − 6 ⇒ f(f(f(x))) = [(x + 6)^4 −6 + 6]^2 − 6 = (x + 6)^8 − 6 So we can say f(f(f(f(f(x))))) = (x + 6)^(32) − 6 (x + 6)^(32) − 6 = 0 ⇒ (x + 6)^(32) = 6 ⇒ x + 6 = ± (6)^(1/(32)) ⇒ x = ± (6)^(1/(32)) − 6 (Ans)](https://www.tinkutara.com/question/Q202188.png)
$${f}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{12}{x}\:+\:\mathrm{30}\:=\:\left({x}\:+\:\mathrm{6}\right)^{\mathrm{2}} \:−\:\mathrm{6} \\ $$$$\Rightarrow\:{f}\left({f}\left({x}\right)\right)\:=\:\left[\left({x}\:+\:\mathrm{6}\right)^{\mathrm{2}} \:−\:\mathrm{6}\:+\:\mathrm{6}\right]^{\mathrm{2}} \:−\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\left[\left({x}\:+\:\mathrm{6}\right)^{\mathrm{2}} \right]^{\mathrm{2}} \:−\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({x}\:+\:\mathrm{6}\right)^{\mathrm{4}} \:−\:\mathrm{6} \\ $$$$\Rightarrow\:{f}\left({f}\left({f}\left({x}\right)\right)\right)\:=\:\left[\left({x}\:+\:\mathrm{6}\right)^{\mathrm{4}} \:−\mathrm{6}\:+\:\mathrm{6}\right]^{\mathrm{2}} \:−\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({x}\:+\:\mathrm{6}\right)^{\mathrm{8}} \:−\:\mathrm{6} \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{can}\:\mathrm{say}\: \\ $$$${f}\left({f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)\right)\:=\:\left({x}\:+\:\mathrm{6}\right)^{\mathrm{32}} \:−\:\mathrm{6}\: \\ $$$$ \\ $$$$\left({x}\:+\:\mathrm{6}\right)^{\mathrm{32}} \:−\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\left({x}\:+\:\mathrm{6}\right)^{\mathrm{32}} \:=\:\mathrm{6} \\ $$$$\Rightarrow\:{x}\:+\:\mathrm{6}\:=\:\pm\:\sqrt[{\mathrm{32}}]{\mathrm{6}} \\ $$$$\Rightarrow\:{x}\:=\:\pm\:\sqrt[{\mathrm{32}}]{\mathrm{6}}\:−\:\mathrm{6}\:\left(\boldsymbol{\mathrm{Ans}}\right) \\ $$
Commented by mr W last updated on 22/Dec/23
$${thanks}! \\ $$