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Question-207879




Question Number 207879 by efronzo1 last updated on 29/May/24
Answered by mr W last updated on 29/May/24
f(x)=x^2 −6x+12=(x−3)^2 +3≥3  f(a)=65539=a^2 −6a+12  ⇒a^2 −6a−65527=0  ⇒a=3+256=259     (3−256<3 rejected)  f(b)=a=259=b^2 −6b+12  ⇒b^2 −6b−247=0  ⇒b=3+16=19     (3−16<3 rejected)  f(c)=b=19=c^2 −6c+12  ⇒c^2 −6c−7=0  ⇒c=3+4=7      (3−4<3 rejected)  f(f(f(f(x))))=65539  f(f(f(x)))=a  f(f(x))=b  f(x)=c=7=x^2 −6x+12  ⇒x^2 −6x+5=0  ⇒x=1, 5 ✓
$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{12}=\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}\geqslant\mathrm{3} \\ $$$${f}\left({a}\right)=\mathrm{65539}={a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{12} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −\mathrm{6}{a}−\mathrm{65527}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{3}+\mathrm{256}=\mathrm{259}\:\:\:\:\:\left(\mathrm{3}−\mathrm{256}<\mathrm{3}\:{rejected}\right) \\ $$$${f}\left({b}\right)={a}=\mathrm{259}={b}^{\mathrm{2}} −\mathrm{6}{b}+\mathrm{12} \\ $$$$\Rightarrow{b}^{\mathrm{2}} −\mathrm{6}{b}−\mathrm{247}=\mathrm{0} \\ $$$$\Rightarrow{b}=\mathrm{3}+\mathrm{16}=\mathrm{19}\:\:\:\:\:\left(\mathrm{3}−\mathrm{16}<\mathrm{3}\:{rejected}\right) \\ $$$${f}\left({c}\right)={b}=\mathrm{19}={c}^{\mathrm{2}} −\mathrm{6}{c}+\mathrm{12} \\ $$$$\Rightarrow{c}^{\mathrm{2}} −\mathrm{6}{c}−\mathrm{7}=\mathrm{0} \\ $$$$\Rightarrow{c}=\mathrm{3}+\mathrm{4}=\mathrm{7}\:\:\:\:\:\:\left(\mathrm{3}−\mathrm{4}<\mathrm{3}\:{rejected}\right) \\ $$$${f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)=\mathrm{65539} \\ $$$${f}\left({f}\left({f}\left({x}\right)\right)\right)={a} \\ $$$${f}\left({f}\left({x}\right)\right)={b} \\ $$$${f}\left({x}\right)={c}=\mathrm{7}={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{12} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1},\:\mathrm{5}\:\checkmark \\ $$
Commented by efronzo1 last updated on 29/May/24
  x^2 −6x+5 = 0    (x−1)(x−5)=0     x=1 and x=5
$$\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{5}\:=\:\mathrm{0} \\ $$$$\:\:\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\:\:\:\mathrm{x}=\mathrm{1}\:\mathrm{and}\:\mathrm{x}=\mathrm{5}\: \\ $$
Commented by mr W last updated on 29/May/24
yes, thanks!
$${yes},\:{thanks}! \\ $$

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