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with-f-x-x-2-12x-30-and-x-R-solve-f-f-f-f-f-x-0-




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

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