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Question Number 217541 by hardmath last updated on 15/Mar/25
f:R→Rf(f(x))=x2−x+1f(0)=?
Answered by mr W last updated on 16/Mar/25
sayf(0)=kf(k)=f(f(0))=02−0+1=1replacexwithf(x):f(f(f(x)))=f2(x)−f(x)+1f(x2−x+1)=f2(x)−f(x)+1―setx=0:f(1)=f2(0)−f(0)+1⇒f(1)=k2−k+1setx=1:f(1)=f2(1)−f(1)+1(f(1)−1)2=0⇒f(1)=1k2−k+1=1⇒k(k−1)=0⇒k=0or1ifk=0,i.e.f(0)=0,butf(k)=f(0)=1≠0⇒k=0isnotvalid.sok=f(0)=1istheonlysolution.
Commented by hardmath last updated on 16/Mar/25
thankyoudearprofessor
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