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Question Number 217541 by hardmath last updated on 15/Mar/25

f : R → R  f(f(x)) = x^2  − x + 1  f(0) = ?

f:RRf(f(x))=x2x+1f(0)=?

Answered by mr W last updated on 16/Mar/25

say f(0)=k  f(k)=f(f(0))=0^2 −0+1=1    replace x with f(x):  f(f(f(x)))=f^2 (x)−f(x)+1  f(x^2 −x+1)=f^2 (x)−f(x)+1  set x=0:  f(1)=f^2 (0)−f(0)+1  ⇒f(1)=k^2 −k+1    set x=1:  f(1)=f^2 (1)−f(1)+1  (f(1)−1)^2 =0  ⇒f(1)=1    k^2 −k+1=1  ⇒k(k−1)=0  ⇒k=0 or 1  if k=0, i.e. f(0)=0,   but f(k)=f(0)=1≠0  ⇒k=0 is not valid.  so k=f(0)=1 is the only solution.

sayf(0)=kf(k)=f(f(0))=020+1=1replacexwithf(x):f(f(f(x)))=f2(x)f(x)+1f(x2x+1)=f2(x)f(x)+1setx=0:f(1)=f2(0)f(0)+1f(1)=k2k+1setx=1:f(1)=f2(1)f(1)+1(f(1)1)2=0f(1)=1k2k+1=1k(k1)=0k=0or1ifk=0,i.e.f(0)=0,butf(k)=f(0)=10k=0isnotvalid.sok=f(0)=1istheonlysolution.

Commented by hardmath last updated on 16/Mar/25

thankyou dear professor

thankyoudearprofessor

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