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Question Number 4671 by 314159 last updated on 20/Feb/16

Define a function f:R→R such that f(f(x))=x^2 −x+1 for all real x.Evaluate f(0).

$${Define}\:{a}\:{function}\:{f}:{R}\rightarrow{R}\:{such}\:{that}\:{f}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$${for}\:{all}\:{real}\:{x}.{Evaluate}\:{f}\left(\mathrm{0}\right). \\ $$$$ \\ $$

Answered by 123456 last updated on 20/Feb/16

f(f(x))=x^2 −x+1 f(f(f(x)))=f(x)^2 −f(x)+1 f(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)^2 −2f(1)+1=0 (f(1)−1)^2 =0 f(1)=1 f(f(f(0)))=f(0)^2 −f(0)+1 f(1)=f(0)^2 −f(0)+1 1=f(0)^2 −f(0)+1 f(0)(f(0)−1)=0 f(0)=0∨f(0)=1 if f(0)=0 f(f(0))=f(0)=0≠1 so f(0)=1 ⋮

$${f}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$${f}\left({f}\left({f}\left({x}\right)\right)\right)={f}\left({x}\right)^{\mathrm{2}} −{f}\left({x}\right)+\mathrm{1} \\ $$$${f}\left({f}\left(\mathrm{0}\right)\right)=\mathrm{1} \\ $$$${f}\left({f}\left(\mathrm{1}\right)\right)=\mathrm{1} \\ $$$${f}\left({f}\left({f}\left(\mathrm{1}\right)\right)\right)={f}\left(\mathrm{1}\right)^{\mathrm{2}} −{f}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)^{\mathrm{2}} −{f}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{f}\left(\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\left({f}\left(\mathrm{1}\right)−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left({f}\left({f}\left(\mathrm{0}\right)\right)\right)={f}\left(\mathrm{0}\right)^{\mathrm{2}} −{f}\left(\mathrm{0}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)={f}\left(\mathrm{0}\right)^{\mathrm{2}} −{f}\left(\mathrm{0}\right)+\mathrm{1} \\ $$$$\mathrm{1}={f}\left(\mathrm{0}\right)^{\mathrm{2}} −{f}\left(\mathrm{0}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)\left({f}\left(\mathrm{0}\right)−\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\vee{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{if}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({f}\left(\mathrm{0}\right)\right)={f}\left(\mathrm{0}\right)=\mathrm{0}\neq\mathrm{1} \\ $$$$\mathrm{so} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\vdots \\ $$

Commented by 123456 last updated on 21/Feb/16

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

$${f}\left(\mathrm{0}\right)={a}\Rightarrow{f}\left({f}\left(\mathrm{0}\right)\right)={f}\left({a}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)={b}\Rightarrow{f}\left({f}\left(\mathrm{1}\right)\right)={f}\left({b}\right)=\mathrm{1} \\ $$$${f}\left({f}\left({a}\right)\right)={f}\left(\mathrm{1}\right)={b} \\ $$$${f}\left({f}\left({b}\right)\right)={f}\left(\mathrm{1}\right)={b} \\ $$$${a}^{\mathrm{2}} −{a}+\mathrm{1}={b}\Rightarrow{a}=\mathrm{0}\vee{a}=\mathrm{1} \\ $$$${b}^{\mathrm{2}} −{b}+\mathrm{1}={b}\Rightarrow{b}=\mathrm{1} \\ $$$${f}\left({x}\right)={x} \\ $$$${f}\left({f}\left({x}\right)\right)={f}\left({x}\right)={x} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}={x} \\ $$$${x}=\mathrm{1} \\ $$

Commented by 314159 last updated on 21/Feb/16

Thanks a lot!

$${Thanks}\:{a}\:{lot}! \\ $$

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