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Question-83035




Question Number 83035 by bshahid010@gmail.com last updated on 27/Feb/20
Commented by Tony Lin last updated on 27/Feb/20
f(1)=4f(0)⇒f(0)=1  f(2)=((√(f(0)))+(√(f(1))))^2   =f(0)+f(1)+2(√(f(0)f(1)))  =(9/4)f(1)=9f(0)=9  f(3)=4f(1)=16f(0)=16  ⇒f(x)=(x+1)^2
$${f}\left(\mathrm{1}\right)=\mathrm{4}{f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=\left(\sqrt{{f}\left(\mathrm{0}\right)}+\sqrt{{f}\left(\mathrm{1}\right)}\right)^{\mathrm{2}} \\ $$$$={f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)+\mathrm{2}\sqrt{{f}\left(\mathrm{0}\right){f}\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{9}}{\mathrm{4}}{f}\left(\mathrm{1}\right)=\mathrm{9}{f}\left(\mathrm{0}\right)=\mathrm{9} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{4}{f}\left(\mathrm{1}\right)=\mathrm{16}{f}\left(\mathrm{0}\right)=\mathrm{16} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$
Answered by mr W last updated on 27/Feb/20
with x=y=0,  f(1)=(2(√(f(0))))^2 =4f(0)=4  ⇒f(0)=1  with y=0,  f(x+1)=((√(f(x)))+1)^2   let g(x)=(√(f(x)))  (g(x+1))^2 =(g(x)+1)^2   g(x+1)=g(x)+1  let g(x)=Ax+B  Ax+A+B=Ax+B+1  ⇒A=1  g(x)=x+B  g(0)=0+B=1 ⇒B=1  ⇒g(x)=x+1  ⇒f(x)=(g(x))^2 =(x+1)^2
$${with}\:{x}={y}=\mathrm{0}, \\ $$$${f}\left(\mathrm{1}\right)=\left(\mathrm{2}\sqrt{{f}\left(\mathrm{0}\right)}\right)^{\mathrm{2}} =\mathrm{4}{f}\left(\mathrm{0}\right)=\mathrm{4} \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${with}\:{y}=\mathrm{0}, \\ $$$${f}\left({x}+\mathrm{1}\right)=\left(\sqrt{{f}\left({x}\right)}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${let}\:{g}\left({x}\right)=\sqrt{{f}\left({x}\right)} \\ $$$$\left({g}\left({x}+\mathrm{1}\right)\right)^{\mathrm{2}} =\left({g}\left({x}\right)+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${g}\left({x}+\mathrm{1}\right)={g}\left({x}\right)+\mathrm{1} \\ $$$${let}\:{g}\left({x}\right)={Ax}+{B} \\ $$$${Ax}+{A}+{B}={Ax}+{B}+\mathrm{1} \\ $$$$\Rightarrow{A}=\mathrm{1} \\ $$$${g}\left({x}\right)={x}+{B} \\ $$$${g}\left(\mathrm{0}\right)=\mathrm{0}+{B}=\mathrm{1}\:\Rightarrow{B}=\mathrm{1} \\ $$$$\Rightarrow{g}\left({x}\right)={x}+\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({g}\left({x}\right)\right)^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

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