Exercise:
Find f(0) such that f(f(x))=x2−x+1 where f:R→R
Solution: Let f(x)=x2−x+1 and we have f(f(x))=x2−x+1
⇔f(x2−x+1)=f2(x)−f(x)+1
For x=0⇔f(1)=f2(0)−f(0)+1
For x=1⇔f(1)=f2(1)−f(1)+1
So f2(0)−f2(1)−f(0)+f(1)=0⇔(f(0)−f(1))(f(0)+f(1))−(f(0)−f(1))=0
⇔(f(0)−f(1))(f(0)+f(1)−1)=0⇔f(0)=f(1)orf(0)+f(1)=1
So we have two cases to be considered in this discussion
If f(0)=f(1)⇔f2(0)−f(0)+1=f(0)⇔f(0)[f(0)−1]=f(0)−1⇔f(0)=1
If f(0)+f(1)=1⇔f(1)=1−f(0)⇔f2(0)−f(0)+1=1−f(0)⇔f(0)=0
Now if f(0)=1 then f(f(0))=f2(0)−f(0)+1=1−1+1=1⇔f(1)=1 true
If f(0)=0⇔f(f(0))=f2(0)−f(0)+1=0−0+1=1⇔f(0)=1 false
Therefore f(0)=1
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the idea of solution credit to Trần Hữu Phúc
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