Question Number 171583 by mr W last updated on 18/Jun/22
$${if}\:{f}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1},\:{find}\:{f}\left(\mathrm{0}\right)=? \\ $$
Commented by infinityaction last updated on 18/Jun/22
![f(f(f(x))) = f(x)^2 −f(x)+1 f(f(0)) = 1 so f(1) = f(0)^2 −f(0) +1 that is f(0)^2 −f(0)−f(1)+1 = 0 .....1 repeating the process since f(f(1)) = 1 f(1) = f(1)^2 −f(1)+1 [f(1)−1]^2 = 0 f(1) = 1 put this value in eq^n . 1 f(0){f(0)−1} = 0 f(0) =1 or f(0) = 0 f(0)= 1 and f(f(0)) = f(0) f(f(0)) = 0 rejected it is right ????](https://www.tinkutara.com/question/Q171586.png)
$${f}\left({f}\left({f}\left({x}\right)\right)\right)\:=\:{f}\left({x}\right)^{\mathrm{2}} −{f}\left({x}\right)+\mathrm{1} \\ $$$$\:\:\:{f}\left({f}\left(\mathrm{0}\right)\right)\:=\:\mathrm{1} \\ $$$$\:\:\:{so}\:\: \\ $$$$\:\:\:{f}\left(\mathrm{1}\right)\:=\:{f}\left(\mathrm{0}\right)^{\mathrm{2}} \:−{f}\left(\mathrm{0}\right)\:+\mathrm{1} \\ $$$$\:\:\:\:{that}\:{is} \\ $$$$\:\:\:{f}\left(\mathrm{0}\right)^{\mathrm{2}} −{f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}\right)+\mathrm{1}\:=\:\mathrm{0}\:\:…..\mathrm{1} \\ $$$$\:\:\:{repeating}\:{the}\:{process}\:{since} \\ $$$$\:\:\:\:\:{f}\left({f}\left(\mathrm{1}\right)\right)\:=\:\mathrm{1} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{1}\right)\:=\:{f}\left(\mathrm{1}\right)^{\mathrm{2}} −{f}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:\left[{f}\left(\mathrm{1}\right)−\mathrm{1}\right]^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\:\:\:{f}\left(\mathrm{1}\right)\:\:=\:\mathrm{1} \\ $$$$\:\:\:\:{put}\:{this}\:{value}\:{in}\:{eq}^{{n}} .\:\mathrm{1} \\ $$$$\:\:\:\:{f}\left(\mathrm{0}\right)\left\{{f}\left(\mathrm{0}\right)−\mathrm{1}\right\}\:=\:\mathrm{0} \\ $$$$\:\:\:{f}\left(\mathrm{0}\right)\:=\mathrm{1}\:{or}\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{0}\right)=\:\mathrm{1} \\ $$$$\:\:\:\:{and} \\ $$$$\:\:\:\:\:\:\:{f}\left({f}\left(\mathrm{0}\right)\right)\:=\:{f}\left(\mathrm{0}\right) \\ $$$$\:\:\:\:\:{f}\left({f}\left(\mathrm{0}\right)\right)\:\:=\:\:\mathrm{0}\:\:{rejected} \\ $$$$\:\:\:\:{it}\:{is}\:{right}\:????\:\: \\ $$
Commented by mr W last updated on 18/Jun/22
$${very}\:{nice}\:{sir}! \\ $$
Commented by infinityaction last updated on 18/Jun/22
$${thank}\:{you}\:{sir} \\ $$
Answered by mr W last updated on 18/Jun/22
$${f}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1}\:>\mathrm{0} \\ $$$${f}\left({f}\left({f}\left({x}\right)\right)\right)={f}^{\mathrm{2}} \left({x}\right)−{f}\left({x}\right)+\mathrm{1} \\ $$$${f}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)={f}^{\mathrm{2}} \left({x}\right)−{f}\left({x}\right)+\mathrm{1} \\ $$$${with}\:{x}=\mathrm{1}: \\ $$$${f}\left(\mathrm{1}\right)={f}^{\mathrm{2}} \left(\mathrm{1}\right)−{f}\left(\mathrm{1}\right)+\mathrm{1}\:\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${with}\:{x}=\mathrm{0}: \\ $$$${f}\left(\mathrm{1}\right)={f}^{\mathrm{2}} \left(\mathrm{0}\right)−{f}\left(\mathrm{0}\right)+\mathrm{1}=\mathrm{1}\:\Rightarrow{f}\left(\mathrm{0}\right)=\:\mathrm{1}\:{or}\:\mathrm{0}\:\left({rejected}\right) \\ $$
Commented by infinityaction last updated on 18/Jun/22
$${amazing}\:{solution}\:{sir} \\ $$
Answered by cortano1 last updated on 18/Jun/22
