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Question Number 143987 by liberty last updated on 20/Jun/21

If f(x^2 −6x+6)+f(x^2 −4x+4)=2x  ∀x∈R then f(−3)+f(9)−5f(1)=?

$${If}\:{f}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}\right)+{f}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\right)=\mathrm{2}{x} \\ $$$$\forall{x}\in{R}\:{th}\mathrm{e}{n}\:{f}\left(−\mathrm{3}\right)+{f}\left(\mathrm{9}\right)−\mathrm{5}{f}\left(\mathrm{1}\right)=? \\ $$

Answered by mitica last updated on 20/Jun/21

x=1⇒f(1)+f(1)=2⇒f(1)=1  x=3⇒f(−3)+f(1)=6⇒f(−3)=5  x=5⇒f(1)+f(9)=10⇒f(9)=9  ⇒f(−3)+f(9)−5f(1)=9

$${x}=\mathrm{1}\Rightarrow{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${x}=\mathrm{3}\Rightarrow{f}\left(−\mathrm{3}\right)+{f}\left(\mathrm{1}\right)=\mathrm{6}\Rightarrow{f}\left(−\mathrm{3}\right)=\mathrm{5} \\ $$$${x}=\mathrm{5}\Rightarrow{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{9}\right)=\mathrm{10}\Rightarrow{f}\left(\mathrm{9}\right)=\mathrm{9} \\ $$$$\Rightarrow{f}\left(−\mathrm{3}\right)+{f}\left(\mathrm{9}\right)−\mathrm{5}{f}\left(\mathrm{1}\right)=\mathrm{9} \\ $$

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