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

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

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

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

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

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

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