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Question Number 83035 by bshahid010@gmail.com last updated on 27/Feb/20

Commented by Tony Lin last updated on 27/Feb/20

$$f\left(1\right)=4f\left(0\right)\Rightarrow f\left(0\right)=1$$$$f\left(2\right)={(\sqrt{f\left(0\right)}+\sqrt{f\left(1\right)})}^2$$$$=f\left(0\right)+f\left(1\right)+2\sqrt{f\left(0\right)f\left(1\right)}$$$$=\frac{9}{4}f\left(1\right)=9f\left(0\right)=9$$$$f\left(3\right)=4f\left(1\right)=16f\left(0\right)=16$$$$\Rightarrow f\left(x\right)={(x+1)}^2$$

Answered by mr W last updated on 27/Feb/20

$$withx=y=0,$$$$f\left(1\right)={\left(2\sqrt{f\left(0\right)}\right)}^2=4f\left(0\right)=4$$$$\Rightarrow f\left(0\right)=1$$$$withy=0,$$$$f(x+1)={(\sqrt{f\left(x\right)}+1)}^2$$$$letg\left(x\right)=\sqrt{f\left(x\right)}$$$${\left(g(x+1)\right)}^2={(g\left(x\right)+1)}^2$$$$g(x+1)=g\left(x\right)+1$$$$letg\left(x\right)=Ax+B$$$$Ax+A+B=Ax+B+1$$$$\Rightarrow A=1$$$$g\left(x\right)=x+B$$$$g\left(0\right)=0+B=1\Rightarrow B=1$$$$\Rightarrow g\left(x\right)=x+1$$$$\Rightarrow f\left(x\right)={\left(g\left(x\right)\right)}^2={(x+1)}^2$$

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