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Question Number 75212 by chess1 last updated on 08/Dec/19

Answered by mind is power last updated on 08/Dec/19

x=y=0⇒ f(0)=1 y=−2x f(x)+1−10x^2 =f(5x)+x^2 +1 ⇒f(x)−f(5x)=11x^2 y=3x⇒ f(x)+f(5x)+15x^2 =x^2 +2 f(x)+f(5x)=−14x^2 +2 ⇒2f(x)=−3x^2 +2⇒f(x)=((−3x^2 +2)/2) f(10)=((−298)/2)=−149

$$\mathrm{x}=\mathrm{y}=\mathrm{0}\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{y}=−\mathrm{2x} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{1}−\mathrm{10x}^{\mathrm{2}} =\mathrm{f}\left(\mathrm{5x}\right)+\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{5x}\right)=\mathrm{11x}^{\mathrm{2}} \\ $$$$\mathrm{y}=\mathrm{3x}\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{5x}\right)+\mathrm{15x}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{5x}\right)=−\mathrm{14x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\Rightarrow\mathrm{2f}\left(\mathrm{x}\right)=−\mathrm{3x}^{\mathrm{2}} +\mathrm{2}\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{−\mathrm{3x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{f}\left(\mathrm{10}\right)=\frac{−\mathrm{298}}{\mathrm{2}}=−\mathrm{149} \\ $$

Commented by chess1 last updated on 08/Dec/19

thanks

$$\mathrm{thanks} \\ $$

Commented by mind is power last updated on 08/Dec/19

y′re Welcom

$$\mathrm{y}'\mathrm{re}\:\mathrm{Welcom} \\ $$

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