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Question Number 122128 by bemath last updated on 14/Nov/20

For which numbers a,b,c,d will the function ψ(x)= ((ax+b)/(cx+d)) satisfy ψ(ψ(x))= x for all x.

Forwhichnumbersa,b,c,dwillthefunctionψ(x)=ax+bcx+dsatisfyψ(ψ(x))=xforallx.

Commented by liberty last updated on 14/Nov/20

from ψ(ψ(x)) = x , we get ψ(x)=ψ^(−1) (x) first, we find ψ^(−1) (x). ⇒ψ^(−1) (x) = ((−dx+b)/(cx−a)) = ((ax+b)/(cx+d)) ⇒(−dx+b)(cx+d)=(ax+b)(cx−a) ⇒ −cd x^2 +(bc−d^2 )x+bd = ac x^2 +(bc−a^2 )x−ba → { ((−cd = ac , a=−d)),((bc−d^2 =bc−a^2 ; a^2 = d^2 )),((bd = −ba ; a = −d)) :} therefore we get a = −d and { ((b is arbitary constant)),((c is arbitary constant)) :}

fromψ(ψ(x))=x,wegetψ(x)=ψ−1(x)first,wefindψ−1(x).⇒ψ−1(x)=−dx+bcx−a=ax+bcx+d⇒(−dx+b)(cx+d)=(ax+b)(cx−a)⇒−cdx2+(bc−d2)x+bd=acx2+(bc−a2)x−ba→{−cd=ac,a=−dbc−d2=bc−a2;a2=d2bd=−ba;a=−dthereforewegeta=−dand{bisarbitaryconstantcisarbitaryconstant

Answered by TANMAY PANACEA last updated on 14/Nov/20

ψ(ψ(x))=((a(((ax+b)/(cx+d)))+b)/(c(((ax+b)/(cx+d)))+d))=x ((a^2 x+ab+bcx+bd)/(acx+bc+cdx+d^2 ))=x ((x(a^2 +bc)+ab+bd)/(x(ac+cd)+bc+d^2 ))= a^2 +bc=1 ab+bd=0→b(a+d)=0 ac+cd=0→c(a+d)=0 bc+d^2 =1 now if a+d≠0 b=0 and c=0 a^2 =1 a=1 and d=1 a=1 b=0 c=0 d=1

ψ(ψ(x))=a(ax+bcx+d)+bc(ax+bcx+d)+d=xa2x+ab+bcx+bdacx+bc+cdx+d2=xx(a2+bc)+ab+bdx(ac+cd)+bc+d2=a2+bc=1ab+bd=0→b(a+d)=0ac+cd=0→c(a+d)=0bc+d2=1nowifa+d≠0b=0andc=0a2=1a=1andd=1a=1b=0c=0d=1

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