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Question Number 152769 by liberty last updated on 01/Sep/21

Commented by mathdanisur last updated on 01/Sep/21

f(x)=(x/2)+((x^2 +3)/(x^2 −1))

f(x)=x2+x2+3x2−1

Answered by tabata last updated on 01/Sep/21

let: y = ((x−3)/(x+1)) ⇒ xy + y = x−3 ⇒ x (y−1)=−3−y ⇒ x = ((3+y)/(1−y)) f(y) + f ( (((3+y+3−3y)/(1−y))/((1−y−3−y)/(1−y))) ) = ((3+y)/(1−y)) f(y) + f (((−3+y)/(1+y))) = ((3+y)/(1−y)) → (1) Replase y to x ⇒ f(x)+f (((−3+x)/(1+x)))=((3+x)/(1−x)) → (1) let: w = ((x+3)/(1−x)) ⇒ w − wx = x + 3 ⇒ w−3 = x(1+w)⇒ x = ((w−3)/(1+w)) f ( (((w−3−3−3w)/(1+w))/((w−3+1+w)/(1+w))) ) + f (w ) = ((w−3)/(1+w)) f ( ((3+w)/(1−w))) + f (w ) = ((w−3)/(1+w)) → (2) Replase w to x : f ( ((3+x)/(1−x))) + f (x) = ((x −3)/(1+x)) → (2) (1) + (2) ⇒ 2 f(x) + x = ((3+x)/(1−x)) + ((x −3)/(1+x)) 2 f (x) + x = (((3+x)(1+x)+(x−3)(1−x))/((1−x)(1+x))) 2 f (x) + x = ((8x)/(1−x^2 )) ⇒ f (x)= (1/2) ( ((8x−x+x^3 )/(1−x^2 )) ) ⟨ M . T ⟩

let:y=x−3x+1⇒xy+y=x−3⇒x(y−1)=−3−y⇒x=3+y1−yf(y)+f(3+y+3−3y1−y1−y−3−y1−y)=3+y1−yf(y)+f(−3+y1+y)=3+y1−y→(1)Replaseytox⇒f(x)+f(−3+x1+x)=3+x1−x→(1)let:w=x+31−x⇒w−wx=x+3⇒w−3=x(1+w)⇒x=w−31+wf(w−3−3−3w1+ww−3+1+w1+w)+f(w)=w−31+wf(3+w1−w)+f(w)=w−31+w→(2)Replasewtox:f(3+x1−x)+f(x)=x−31+x→(2)(1)+(2)⇒2f(x)+x=3+x1−x+x−31+x2f(x)+x=(3+x)(1+x)+(x−3)(1−x)(1−x)(1+x)2f(x)+x=8x1−x2⇒f(x)=12(8x−x+x31−x2)⟨M.T⟩

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