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Question Number 95150 by bobhans last updated on 23/May/20

how do you solve f(x) +2 f((1/(1−x))) = x for f ?

$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{f}\left(\mathrm{x}\right)\:+\mathrm{2}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\mathrm{x}\: \\ $$$$\mathrm{for}\:\mathrm{f}\:?\: \\ $$

Answered by mr W last updated on 23/May/20

f(x)+2f((1/(1−x)))=x ...(i) f((1/(1−x)))+2f(((x−1)/x))=(1/(1−x)) ...(ii) f(((x−1)/x))+2f(x)=((x−1)/x) ...(iii) (i)−2(ii): f(x)−4f(((x−1)/x))=x−(2/(1−x)) ...(iv) 4(iii)+(iv): 9f(x)=((4(x−1))/x)+x−(2/(1−x)) ⇒f(x)=((x^3 +3x^2 −6x+4)/(9x(x−1)))

$${f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}\:\:\:...\left({i}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+\mathrm{2}{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:...\left({ii}\right) \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+\mathrm{2}{f}\left({x}\right)=\frac{{x}−\mathrm{1}}{{x}}\:\:\:...\left({iii}\right) \\ $$$$\left({i}\right)−\mathrm{2}\left({ii}\right): \\ $$$${f}\left({x}\right)−\mathrm{4}{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)={x}−\frac{\mathrm{2}}{\mathrm{1}−{x}}\:\:\:...\left({iv}\right) \\ $$$$\mathrm{4}\left({iii}\right)+\left({iv}\right): \\ $$$$\mathrm{9}{f}\left({x}\right)=\frac{\mathrm{4}\left({x}−\mathrm{1}\right)}{{x}}+{x}−\frac{\mathrm{2}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{4}}{\mathrm{9}{x}\left({x}−\mathrm{1}\right)} \\ $$

Commented by bobhans last updated on 23/May/20

ok. i compare to my way f(x)+ 2f((1/(1−x))) = x f((1/(1−x)))+2f((1/(1−(1/(1−x))))) = (1/(1−x)) f((1/(1−x)))+2 f(((x−1)/x)) = (1/(1−x)) ...(2) f((1/(1−(1/(1−x))))) + 2f ((((1/(1−x))−1)/(1/(1−x)))) = (1/(1−(1/(1−x)))) f(((x−1)/x)) + 2f(x) = ((x−1)/x) ...(3) (1)−2×(2)+4×(3) 9f(x) = ((x^3 +3x^2 −6x+4)/(x(x−1))) ⇒ f(x) = ((x^3 +3x^2 −6x+4)/(9x(x−1)))

$$\mathrm{ok}.\:\mathrm{i}\:\mathrm{compare}\:\mathrm{to}\:\mathrm{my}\:\mathrm{way} \\ $$$${f}\left({x}\right)+\:\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:=\:{x}\: \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\: \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+\mathrm{2}\:{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:...\left(\mathrm{2}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)\:+\:\mathrm{2}{f}\:\left(\frac{\frac{\mathrm{1}}{\mathrm{1}−{x}}−\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}} \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:+\:\mathrm{2}{f}\left({x}\right)\:=\:\frac{{x}−\mathrm{1}}{{x}}\:...\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)−\mathrm{2}×\left(\mathrm{2}\right)+\mathrm{4}×\left(\mathrm{3}\right)\: \\ $$$$\mathrm{9}{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{4}}{{x}\left({x}−\mathrm{1}\right)}\:\Rightarrow\:{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{4}}{\mathrm{9}{x}\left({x}−\mathrm{1}\right)} \\ $$

Commented by I want to learn more last updated on 23/May/20

Weldone sir

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

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