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Question Number 202184 by mr W last updated on 22/Dec/23

$$withf\left(x\right)=x^2+12x+30andx\in R$$$$solvef\left(f\left(f\left(f\left(f\left(x\right)\right)\right)\right)\right)=0$$

Answered by cortano12 last updated on 22/Dec/23

$$\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}+6)}^2-6\Rightarrow \mathrm{x}=\sqrt{\mathrm{f}\left(\mathrm{x}\right)+6}-6$$$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}+6}-6$$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\right)={\mathrm{f}}^{-1}\left(0\right)=\sqrt{6}-6$$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)={\mathrm{f}}^{-1}(\sqrt{6}-6)=\sqrt{\sqrt{6}}-6$$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)={\mathrm{f}}^{-1}(\sqrt{\sqrt{6}}-6)=\sqrt{\sqrt{\sqrt{6}}}-6$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{f}}^{-1}(\sqrt{\sqrt{\sqrt{6}}}-6)=\sqrt{\sqrt{\sqrt{\sqrt{6}}}}-6$$$$\Rightarrow \mathrm{x}=\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{6}}}}}-6=\sqrt[32]{6}-6$$

Commented by mr W last updated on 22/Dec/23

$$thanks!$$$$-\sqrt[32]{6}-6isalsoaroot.$$

Commented by esmaeil last updated on 22/Dec/23

$$hi$$$$why?$$$$f\left(f\left(f\left(f\left(x\right)\right)\right)\right)=f^{-1}\left(0\right)$$

Commented by mr W last updated on 23/Dec/23

$$f\left(x\right)=t=0\Rightarrow x=f^{-1}\left(t\right)=f^{-1}\left(0\right)$$$$f\left(g\left(x\right)\right)=t=0\Rightarrow g\left(x\right)=f^{-1}\left(t\right)=f^{-}\left(0\right)$$$$f\left(f\left(f\left(f\left(f\left(x\right)\right)\right)\right)\right)=t=0\Rightarrow f\left(f\left(f\left(f\left(x\right)\right)\right)\right)=f^{-1}\left(t\right)=f^{-}\left(0\right)$$

Answered by MATHEMATICSAM last updated on 22/Dec/23

$$f\left(x\right)=x^2+12x+30={(x+6)}^2-6$$$$\Rightarrow f\left(f\left(x\right)\right)={[{(x+6)}^2-6+6]}^2-6$$$$={\left[{(x+6)}^2\right]}^2-6$$$$={(x+6)}^4-6$$$$\Rightarrow f\left(f\left(f\left(x\right)\right)\right)={[{(x+6)}^4-6+6]}^2-6$$$$={(x+6)}^8-6$$$$\mathrm{So}\mathrm{we}\mathrm{can}\mathrm{say}$$$$f\left(f\left(f\left(f\left(f\left(x\right)\right)\right)\right)\right)={(x+6)}^{32}-6$$$$$$$${(x+6)}^{32}-6=0$$$$\Rightarrow {(x+6)}^{32}=6$$$$\Rightarrow x+6=\pm \sqrt[32]{6}$$$$\Rightarrow x=\pm \sqrt[32]{6}-6\left(\mathbf{Ans}\right)$$

Commented by mr W last updated on 22/Dec/23

$$thanks!$$

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