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f-1-x-1-f-x-1-x-2x-1-x-f-x-Stepwise-process-is-required-

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Question Number 2272 by Rasheed Soomro last updated on 12/Nov/15
f(((−1)/(x−1)))+f(((x−1)/x))=((2x−1)/x) f(x)=? Stepwise process is required.
f(−1x−1)+f(x−1x)=2x−1xf(x)=?Stepwiseprocessisrequired.
Answered by Rasheed Soomro last updated on 16/Nov/15
f(((−1)/(x−1)))+f(((x−1)/x))=((2x−1)/x) In order to get equation involving f(x), we have to replace ((−1)/(x−1)) (or ((x−1)/x) ) by x : Let ((−1)/(x−1))=y⇒x=((y−1)/y) f(((−1)/(x−1)))+f(((x−1)/x))=((2x−1)/x) ⇒f(y)+f(((((y−1)/y)−1)/((y−1)/y)))=((2(((y−1)/y))−1)/((y−1)/y)) ⇒f(y)+f(((−1)/(y−1))) =((y−2)/(y−1)) ⇒f(x)+f(((−1)/(x−1))) =((x−2)/(x−1)) [replacing y by x ] ⇒f(x)=((x−2)/(x−1))−f(((−1)/(x−1)))........................(1) f(((−1)/(x−1)))=((((−1)/(x−1))−2)/(((−1)/(x−1))−1))−f(((−1)/(((−1)/(x−1))−1))) [ replace x by ((−1)/(x−1)) in (1)] =((2x−1)/x)−f(((x−1)/x)) [ replace x by ((−1)/(x−1)) in (1)] f(((x−1)/x))=((((x−1)/x)−2)/(((x−1)/x)−1))−f(((−1)/(((x−1)/x)−1))) [replace x by ((x−1)/x) in (1) ] =x+1−f(x) [replace x by ((x−1)/x) in (1) ] −−−−−−−−−−−−−−−−−−−−− From (1) f(x)=((x−2)/(x−1))−f(((−1)/(x−1))) =((x−2)/(x−1))−f(((−1)/(x−1))) =((x−2)/(x−1))−(((2x−1)/x)−f(((x−1)/x))) =((x−2)/(x−1))−((2x−1)/x)+x+1−f(x) 2f(x)=((x−2)/(x−1))−((2x−1)/x)+x+1 f(x)=(1/2)(((x−2)/(x−1))−((2x−1)/x)+x+1)
f(−1x−1)+f(x−1x)=2x−1xInordertogetequationinvolvingf(x),wehavetoreplace−1x−1(orx−1x)byx:Let−1x−1=y⇒x=y−1yf(−1x−1)+f(x−1x)=2x−1x⇒f(y)+f(y−1y−1y−1y)=2(y−1y)−1y−1y⇒f(y)+f(−1y−1)=y−2y−1⇒f(x)+f(−1x−1)=x−2x−1[replacingybyx]⇒f(x)=x−2x−1−f(−1x−1)……………………(1)f(−1x−1)=−1x−1−2−1x−1−1−f(−1−1x−1−1)[replacexby−1x−1in(1)]=2x−1x−f(x−1x)[replacexby−1x−1in(1)]f(x−1x)=x−1x−2x−1x−1−f(−1x−1x−1)[replacexbyx−1xin(1)]=x+1−f(x)[replacexbyx−1xin(1)]−−−−−−−−−−−−−−−−−−−−−From(1)f(x)=x−2x−1−f(−1x−1)=x−2x−1−f(−1x−1)=x−2x−1−(2x−1x−f(x−1x))=x−2x−1−2x−1x+x+1−f(x)2f(x)=x−2x−1−2x−1x+x+1f(x)=12(x−2x−1−2x−1x+x+1)

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