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f-x-2-x-2f-x-2-3x-2-9x-2-15x-then-what-is-the-value-of-f-10-




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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