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Question Number 207879 by efronzo1 last updated on 29/May/24

Answered by mr W last updated on 29/May/24

$$f\left(x\right)=x^2-6x+12={(x-3)}^2+3⩾3$$$$f\left(a\right)=65539=a^2-6a+12$$$$\Rightarrow a^2-6a-65527=0$$$$\Rightarrow a=3+256=259(3-256<3rejected)$$$$f\left(b\right)=a=259=b^2-6b+12$$$$\Rightarrow b^2-6b-247=0$$$$\Rightarrow b=3+16=19(3-16<3rejected)$$$$f\left(c\right)=b=19=c^2-6c+12$$$$\Rightarrow c^2-6c-7=0$$$$\Rightarrow c=3+4=7(3-4<3rejected)$$$$f\left(f\left(f\left(f\left(x\right)\right)\right)\right)=65539$$$$f\left(f\left(f\left(x\right)\right)\right)=a$$$$f\left(f\left(x\right)\right)=b$$$$f\left(x\right)=c=7=x^2-6x+12$$$$\Rightarrow x^2-6x+5=0$$$$\Rightarrow x=1,5✓$$

Commented by efronzo1 last updated on 29/May/24

$${\mathrm{x}}^2-6\mathrm{x}+5=0$$$$(\mathrm{x}-1)(\mathrm{x}-5)=0$$$$\mathrm{x}=1\mathrm{and}\mathrm{x}=5$$

Commented by mr W last updated on 29/May/24

$$yes,thanks!$$

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