Question-199781

Question Number 199781 by cortano12 last updated on 09/Nov/23

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Answered by qaz last updated on 09/Nov/23

f(x)=((2(1−2x))/(x(1−x)))−f((1/(1−x))) =((2(1−2x))/(x(1−x)))−(((2(1−(2/(1−x))))/((1/(1−x))(1−(1/(1−x)))))−f((1/(1−(1/(1−x)))))) =((2(1−2x))/(x(1−x)))−((2(1+x)(1−x))/x)+f(1−(1/x)) =((2(1−2x))/(x(1−x)))−((2(1+x)(1−x))/x)+(((2(1−2(1−(1/x))))/((1−(1/x))(1−(1−(1/x)))))−f((1/(1−(1−(1/x)))))) =((2(1−2x))/(x(1−x)))−((2(1+x)(1−x))/x)+((2x(2−x))/(x−1))−f(x) ⇒f(x)=(((1−2x))/(x(1−x)))−(((1+x)(1−x))/x)+((x(2−x))/(x−1))

$${f}\left({x}\right)=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\left(\frac{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}−{x}}\right)}{\frac{\mathrm{1}}{\mathrm{1}−{x}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{2}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)}{{x}}+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{2}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)}{{x}}+\left(\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)}\right)\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{2}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)}{{x}}+\frac{\mathrm{2}{x}\left(\mathrm{2}−{x}\right)}{{x}−\mathrm{1}}−{f}\left({x}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\frac{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)}{{x}}+\frac{{x}\left(\mathrm{2}−{x}\right)}{{x}−\mathrm{1}} \\ $$

Commented by Frix last updated on 09/Nov/23