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compositon function exercise asked by Mayssoune Zein EL Dine in the math group


Exercise:

Find f(0) such that f(f(x))=x2x+1 where f:RR

Solution: Let f(x)=x2x+1 and we have f(f(x))=x2x+1

f(x2x+1)=f2(x)f(x)+1

For x=0f(1)=f2(0)f(0)+1

For x=1f(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)=0f(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)1f(0)=1

If f(0)+f(1)=1f(1)=1f(0)f2(0)f(0)+1=1f(0)f(0)=0

Now if f(0)=1 then f(f(0))=f2(0)f(0)+1=11+1=1f(1)=1 true

If f(0)=0f(f(0))=f2(0)f(0)+1=00+1=1f(0)=1 false

Therefore f(0)=1


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the idea of solution credit to Trần Hữu Phúc

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