Question Number 1147 by prakash jain last updated on 04/Jul/15

f(f(x))=x^2 −x+1  f(0)=?

Commented byprakash jain last updated on 04/Jul/15

f(f(0))=1

Commented by123456 last updated on 04/Jul/15

f•f=x^2 −x+1         =x^2 −2×(1/2)×x+(1/4)+1−(1/4)         =(x−(1/2))^2 +(3/4)         =(((2x−1)^2 +3)/4)  Δ=(−1)^2 −4(1)(1)=1−4=−3  x=((1±ı(√3))/2)

Answered by 123456 last updated on 04/Jul/15

f(f(x))=x^2 −x+1  f(f(f(x)))=(f(x))^2 −f(x)+1  f(f(0))=1  f(f(f(0)))=(f(0))^2 −f(0)+1⇒f(1)=(f(0))^2 −f(0)+1  f(f(1))=1  f(f(f(1)))=(f(1))^2 −f(1)+1⇒f(1)=(f(1))^2 −f(1)+1  f(1)=1  1=(f(0))^2 −f(0)+1  f(0)=0∨f(0)=1

Commented byprakash jain last updated on 05/Jul/15

If f(0)=0  f(f(0))=1⇒f(0)=1  Is this a contradiction?

Commented by123456 last updated on 05/Jul/15

yes :)  then f(0)=1

Commented byprakash jain last updated on 16/Jul/15

In the question the function is defined as  f(f(x))=x^2 −x+1  So  f(f(f(x)))=(f(x))^2 −f(x)+1

Commented byRasheed Soomro last updated on 12/Aug/15

Dear Parkash  THANKS for guidance.I was wrong. I did not understand  the logic of Mr 123456 at first! But now I do (with your help).  Appreciation for  123456