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Question Number 1125 by 123456 last updated on 18/Jun/15

$$f:\mathbb{R}\to \mathbb{R}$$$$g:\mathbb{R}\to \mathbb{R}$$$$f(x+y)=f\left(x\right)+g\left(y\right)$$$$g\left(xy\right)=f\left(x\right)g\left(y\right)$$$$f\left(x\right)=?$$$$g\left(x\right)=?$$

Commented by 123456 last updated on 18/Jun/15

$$f\left(0\right)=f\left(0\right)+g\left(0\right)\Rightarrow g\left(0\right)=0$$$$f\left(x\right)=f\left(x\right)+g\left(0\right)\Rightarrow g\left(0\right)=0$$$$g\left(0\right)=f\left(0\right)g\left(0\right)\Rightarrow f\left(0\right)=1\vee g\left(0\right)=0$$$$g\left(0\right)=f\left(x\right)g\left(0\right)\Rightarrow f\left(x\right)=1\vee g\left(0\right)=0$$$$g\left(1\right)=f\left(1\right)g\left(1\right)\Rightarrow f\left(1\right)=1\vee g\left(1\right)=0$$$$g\left(x\right)=f\left(1\right)g\left(x\right)\Rightarrow f\left(1\right)=1\vee g\left(x\right)=0$$$$g(-1)=f\left(1\right)g(-1)\Rightarrow f\left(1\right)=1\vee g(-1)=0$$

Answered by prakash jain last updated on 20/Jun/15

$$g\left(x\right)=f\left(x\right)g\left(1\right)$$$$g\left(1\right)=1/k$$$$f\left(x\right)=kg\left(x\right)$$$$g\left(xy\right)=kg\left(x\right)g\left(y\right)$$$$g\left(x\right)=x,k=1$$$$f\left(x\right)=kg\left(x\right)=x$$

Commented by prakash jain last updated on 20/Jun/15

$$g\left(xy\right)=\frac{g\left(x\right)g\left(y\right)}{g\left(1\right)}\Rightarrow g\left(x\right)=x^r,g\left(1\right)=1$$$$f\left(x\right)=x^r$$$$f(x+y)=f\left(x\right)+g\left(y\right)$$$${(x+y)}^r=x^r+y^r\Rightarrow r=1$$$$g\left(x\right)=f\left(x\right)=x$$

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