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f-x-f-1-1-x-tan-1-x-x-0-Find-f-x-

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Question Number 145190 by qaz last updated on 03/Jul/21
f(x)+f(1−(1/x))=tan^(−1) x,(x≠0) Find f(x)=?
f(x)+f(1−1x)=tan−1x,(x≠0)Findf(x)=?
Answered by EDWIN88 last updated on 03/Jul/21
(1)f(x)+f(((x−1)/x))= tan^(−1) (x) replace x→((x−1)/x) (2)f(((x−1)/x))+f(((((x−1)/x)−1)/((x−1)/x)))=tan^(−1) (((x−1)/x)) ⇒f(((x−1)/x))+f((1/(1−x)))=tan^(−1) (((x−1)/x)) replace x→(1/(1−x)) (3)f((1/(1−x)))+f((((1/(1−x))−1)/(1/(1−x))))=tan^(−1) ((1/(1−x))) ⇒f((1/(1−x)))+f(x)= tan^(−1) ((1/(1−x))) sum equation (1),(2),(3) ⇒2 [f(x)+f((1/(1−x)))+f(((x−1)/x))]=tan^(−1) (x)+tan^(−1) (((x−1)/x))+tan^(−1) ((1/(1−x))) ⇒2 [f(x)+tan^(−1) (((x−1)/x))]=tan^(−1) (x)+tan^(−1) (((x−1)/x))+tan^(−1) ((1/(1−x))) 2f(x)=tan^(−1) (x)+tan^(−1) ((1/(1−x)))−tan^(−1) (((x−1)/x)) f(x)=(1/2)[ tan^(−1) (x)+tan^(−1) ((1/(1−x)))−tan^(−1) (((x−1)/x))] =(1/2)[ tan^(−1) (((x+(1/(1−x)))/(1−(x/(1−x)))))−tan^(−1) (((x−1)/x))] =(1/2)[ tan^(−1) (((x+1−x^2 )/(1−2x)))−tan^(−1) (((x−1)/x))] =(1/2)tan^(−1) (((((x+1−x^2 )/(1−2x))−((x−1)/x))/(1+(((x+1−x^2 )(x−1))/((1−2x)x))))) =(1/2)tan^(−1) (((x^2 +x−x^3 −(1−2x)(x−1))/(1−2x^2 +(x−1)(x+1−x^2 ))))
(1)f(x)+f(x−1x)=tan−1(x)replacex→x−1x(2)f(x−1x)+f(x−1x−1x−1x)=tan−1(x−1x)⇒f(x−1x)+f(11−x)=tan−1(x−1x)replacex→11−x(3)f(11−x)+f(11−x−111−x)=tan−1(11−x)⇒f(11−x)+f(x)=tan−1(11−x)sumequation(1),(2),(3)⇒2[f(x)+f(11−x)+f(x−1x)]=tan−1(x)+tan−1(x−1x)+tan−1(11−x)⇒2[f(x)+tan−1(x−1x)]=tan−1(x)+tan−1(x−1x)+tan−1(11−x)2f(x)=tan−1(x)+tan−1(11−x)−tan−1(x−1x)f(x)=12[tan−1(x)+tan−1(11−x)−tan−1(x−1x)]=12[tan−1(x+11−x1−x1−x)−tan−1(x−1x)]=12[tan−1(x+1−x21−2x)−tan−1(x−1x)]=12tan−1(x+1−x21−2x−x−1x1+(x+1−x2)(x−1)(1−2x)x)=12tan−1(x2+x−x3−(1−2x)(x−1)1−2x2+(x−1)(x+1−x2))
Commented by qaz last updated on 03/Jul/21
thank you
thankyou

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