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Question Number 23833 by Joel577 last updated on 07/Nov/17

f(x^2 + x) + 2f(x^2 − 3x + 2) = 9x^2 − 15x then what is the value of f(10) ?

$${f}\left({x}^{\mathrm{2}} \:+\:{x}\right)\:+\:\mathrm{2}{f}\left({x}^{\mathrm{2}} \:−\:\mathrm{3}{x}\:+\:\mathrm{2}\right)\:=\:\mathrm{9}{x}^{\mathrm{2}} \:−\:\mathrm{15}{x} \\ $$$$\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{f}\left(\mathrm{10}\right)\:? \\ $$

Commented by prakash jain last updated on 07/Nov/17

x=0 f(0)+2f(2)=0 x=1 f(2)+2f(0)=−6 f(2)=2 f(0)=−4 f(x) is linear function f(x)=ax+b ⇒f(x)=3x−4 f(x^2 +x)+2f(x^2 −3x+2)= =3(x^2 +x)−4+2[3(x^2 −3x+2)−4] =3x^2 +3x−4+2(3x^2 −9x+2) =9x^2 −15x f(x)=3x−4 f(10) =30−4=26

$${x}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)+\mathrm{2}{f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)+\mathrm{2}{f}\left(\mathrm{0}\right)=−\mathrm{6} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{2} \\ $$$${f}\left(\mathrm{0}\right)=−\mathrm{4} \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{linear}\:\mathrm{function} \\ $$$${f}\left({x}\right)={ax}+{b} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{3}{x}−\mathrm{4} \\ $$$${f}\left({x}^{\mathrm{2}} +{x}\right)+\mathrm{2}{f}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}\right)= \\ $$$$=\mathrm{3}\left({x}^{\mathrm{2}} +{x}\right)−\mathrm{4}+\mathrm{2}\left[\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}\right)−\mathrm{4}\right] \\ $$$$=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{4}+\mathrm{2}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{2}\right) \\ $$$$=\mathrm{9}{x}^{\mathrm{2}} −\mathrm{15}{x} \\ $$$${f}\left({x}\right)=\mathrm{3}{x}−\mathrm{4} \\ $$$${f}\left(\mathrm{10}\right) \\ $$$$=\mathrm{30}−\mathrm{4}=\mathrm{26} \\ $$

Commented by ajfour last updated on 07/Nov/17

you make it easy enough sir!

$${you}\:{make}\:{it}\:{easy}\:{enough}\:{sir}! \\ $$

Commented by Joel577 last updated on 08/Nov/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by kaivan.ahmadi last updated on 13/Dec/17

f(0)=−4 az the Mr parkash slution if we set x^2 +x=10⇒ x^2 −3x+2=−4x+12 and 9x^2 −15x=−24x+90 Now replace f(10)+2f(−4x+12)=−24x+90 and so for x=3 we have f(10)+2f(0)=18⇒f(10)=26 in this way we didnt use the linearly of f.

$$\mathrm{f}\left(\mathrm{0}\right)=−\mathrm{4}\:\:\:\mathrm{az}\:\mathrm{the}\:\mathrm{Mr}\:\mathrm{parkash}\:\mathrm{slution} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{set}\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}=\mathrm{10}\Rightarrow \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2}=−\mathrm{4x}+\mathrm{12} \\ $$$$\mathrm{and} \\ $$$$\mathrm{9x}^{\mathrm{2}} −\mathrm{15x}=−\mathrm{24x}+\mathrm{90} \\ $$$$\mathrm{Now}\:\mathrm{replace} \\ $$$$\mathrm{f}\left(\mathrm{10}\right)+\mathrm{2f}\left(−\mathrm{4x}+\mathrm{12}\right)=−\mathrm{24x}+\mathrm{90} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{for}\:\mathrm{x}=\mathrm{3}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}\left(\mathrm{10}\right)+\mathrm{2f}\left(\mathrm{0}\right)=\mathrm{18}\Rightarrow\mathrm{f}\left(\mathrm{10}\right)=\mathrm{26} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{way}\:\mathrm{we}\:\mathrm{didnt}\:\mathrm{use}\:\mathrm{the} \\ $$$$\mathrm{linearly}\:\mathrm{of}\:\:\:\mathrm{f}. \\ $$$$ \\ $$

Commented by prakash jain last updated on 13/Dec/17

Thanks. It is a good approach. Very useful when it is difficult to evaluate f(x) in terms of x alone.

$$\mathrm{Thanks}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{good}\:\mathrm{approach}. \\ $$$$\mathrm{Very}\:\mathrm{useful}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{difficult} \\ $$$$\mathrm{to}\:\mathrm{evaluate}\:{f}\left({x}\right)\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{x}\:\mathrm{alone}. \\ $$

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