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Question Number 860 by 123456 last updated on 28/Mar/15

f(x^2 )=[f(x)]^2 f(1)=1

$${f}\left({x}^{\mathrm{2}} \right)=\left[{f}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$

Commented by prakash jain last updated on 28/Mar/15

f(x)=0 or f(x)=1 f(x)=x^k , k∈R

$${f}\left({x}\right)=\mathrm{0}\:\mathrm{or}\:{f}\left({x}\right)=\mathrm{1} \\ $$$${f}\left({x}\right)={x}^{{k}} ,\:\:{k}\in\mathrm{R} \\ $$

Answered by prakash jain last updated on 28/Mar/15

f(x^2 )=f(x)∙f(x) divide by x^2 ((f(x∙x))/(x∙x))=((f(x))/x)∙((f(x))/x) g(x)=((f(x))/x) If g(xy)=g(x)g(y)⇒g(x)=x^k , k∈R hence f(x)=x^k , k∈R

$${f}\left({x}^{\mathrm{2}} \right)={f}\left({x}\right)\centerdot{f}\left({x}\right) \\ $$$$\mathrm{divide}\:\mathrm{by}\:{x}^{\mathrm{2}} \\ $$$$\frac{{f}\left({x}\centerdot{x}\right)}{{x}\centerdot{x}}=\frac{{f}\left({x}\right)}{{x}}\centerdot\frac{{f}\left({x}\right)}{{x}} \\ $$$${g}\left({x}\right)=\frac{{f}\left({x}\right)}{{x}} \\ $$$$\mathrm{If}\:{g}\left({xy}\right)={g}\left({x}\right){g}\left({y}\right)\Rightarrow{g}\left({x}\right)={x}^{{k}} ,\:{k}\in\mathrm{R} \\ $$$$\mathrm{hence} \\ $$$${f}\left({x}\right)={x}^{{k}} ,\:{k}\in\mathrm{R} \\ $$

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