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Question Number 171955 by infinityaction last updated on 22/Jun/22

f(x)+ f((1/(1−x))) = 1+(1/(x(1−x))) f(x) = ??

$$\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)+\:\boldsymbol{{f}}\left(\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{{x}}}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{x}}\left(\mathrm{1}−\boldsymbol{{x}}\right)} \\ $$$$\:\:\:\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:=\:\:??\:\:\:\:\:\:\:\:\: \\ $$

Answered by thfchristopher last updated on 22/Jun/22

1+(1/(x(1−x))) =((x−x^2 +1)/(x(1−x))) =((−(1−x)^2 +2−3x)/(x(1−x))) =((2−3x)/(x(1−x)))−((1−x)/x) =(2/x)−(1/(1−x))−((1−x)/x) =((x+1)/x)−(1/(1−x)) Comparison by terms, f(x)=1+(1/x) f((1/(1−x)))=((−1)/(1−x))

$$\mathrm{1}+\frac{\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$=\frac{{x}−{x}^{\mathrm{2}} +\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$=\frac{−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{2}−\mathrm{3}{x}}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$=\frac{\mathrm{2}−\mathrm{3}{x}}{{x}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{1}−{x}}{{x}} \\ $$$$=\frac{\mathrm{2}}{{x}}−\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}−{x}}{{x}} \\ $$$$=\frac{{x}+\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\mathrm{Comparison}\:\mathrm{by}\:\mathrm{terms}, \\ $$$${f}\left({x}\right)=\mathrm{1}+\frac{\mathrm{1}}{{x}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)=\frac{−\mathrm{1}}{\mathrm{1}−{x}} \\ $$

Commented by infinityaction last updated on 22/Jun/22

i know solution of this question but your solution is wrong

$${i}\:{know}\:{solution}\:{of}\:{this}\:{question} \\ $$$${but}\:{your}\:{solution}\:{is}\:{wrong} \\ $$

Commented by infinityaction last updated on 22/Jun/22

i need any other way of this question

$${i}\:{need}\:{any}\:{other}\:{way}\:{of}\:{this} \\ $$$${question} \\ $$

Commented by infinityaction last updated on 22/Jun/22

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

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

Commented by thfchristopher last updated on 22/Jun/22

Haha, I just tried on error. I certainly know my answer should be gussed wrong.

$$\mathrm{Haha},\:\mathrm{I}\:\mathrm{just}\:\mathrm{tried}\:\mathrm{on}\:\mathrm{error}.\:\mathrm{I}\:\mathrm{certainly}\:\mathrm{know} \\ $$$$\mathrm{my}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{gussed}\:\mathrm{wrong}. \\ $$

Answered by mr W last updated on 22/Jun/22

f(x)+f((1/(1−x)))=1+(1/(x(1−x))) ...(i) replace x in (i) with (1/(1−x)): f((1/(1−x)))+f((1/(1−(1/(1−x)))))=1+(1/((1/(1−x))(1−(1/(1−x))))) f((1/(1−x)))+f(((x−1)/x))=1−(((1−x)^2 )/x) ...(ii) replace x in (i) with ((x−1)/x): f(((x−1)/x))+f((1/(1−((x−1)/x))))=1+(1/(((x−1)/x)(1−((x−1)/x)))) f(((x−1)/x))+f(x)=1+(x^2 /(x−1)) ...(iii) (i)+(iii)−(ii): 2f(x)=1+(1/(x(1−x)))+1+(x^2 /(x−1))−1+(((1−x)^2 )/x) 2f(x)=((2(1+x^2 ))/x) ⇒f(x)=((1+x^2 )/x)

$${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)=\mathrm{1}+\frac{\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)}\:\:\:...\left({i}\right) \\ $$$${replace}\:{x}\:{in}\:\left({i}\right)\:{with}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}: \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)=\mathrm{1}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}−{x}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\mathrm{1}−\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{{x}}\:\:\:...\left({ii}\right) \\ $$$${replace}\:{x}\:{in}\:\left({i}\right)\:{with}\:\frac{{x}−\mathrm{1}}{{x}}: \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}−\mathrm{1}}{{x}}}\right)=\mathrm{1}+\frac{\mathrm{1}}{\frac{{x}−\mathrm{1}}{{x}}\left(\mathrm{1}−\frac{{x}−\mathrm{1}}{{x}}\right)} \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+{f}\left({x}\right)=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{1}}\:\:\:...\left({iii}\right) \\ $$$$\left({i}\right)+\left({iii}\right)−\left({ii}\right): \\ $$$$\mathrm{2}{f}\left({x}\right)=\mathrm{1}+\frac{\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)}+\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{x}−\mathrm{1}}−\mathrm{1}+\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{{x}} \\ $$$$\mathrm{2}{f}\left({x}\right)=\frac{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}+{x}^{\mathrm{2}} }{{x}} \\ $$

Commented by infinityaction last updated on 22/Jun/22

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by peter frank last updated on 22/Jun/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by Tawa11 last updated on 25/Jun/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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