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Given-f-x-px-q-x-2-q-0-f-1-q-1-then-f-1-2q-

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Question Number 119909 by bramlexs22 last updated on 28/Oct/20
Given f(x)=((px+q)/(x+2)) , q≠ 0 f^(−1) (q) = −1 then f^(−1) (2q)=?
Givenf(x)=px+qx+2,q≠0f−1(q)=−1thenf−1(2q)=?
Answered by TITA last updated on 28/Oct/20
q=f(−1) q=q−p p=0 f(x)=(q/(x+2)) ⇒ let f(x)=y y=(q/(x+2)) xy+2y=q x=((q−2y)/y) hence f^(−1) (x)=((q−2x)/x) ,x≠0 ⇒ f^(−1) (2q)=((q−4q)/(2q))=((−3q)/(2q)) ⇒f^(−1) (2q)=((−3)/2)
q=f(−1)q=q−pp=0f(x)=qx+2⇒letf(x)=yy=qx+2xy+2y=qx=q−2yyhencef−1(x)=q−2xx,x≠0⇒f−1(2q)=q−4q2q=−3q2q⇒f−1(2q)=−32
Answered by 1549442205PVT last updated on 28/Oct/20
f(x)=((px+q)/(x+2)) , q≠ 0 ⇒x(f(x)−p)=q−2f(x)⇒x=((q−2f(x))/(f(x)−p)) ⇒f^(−1) (x)=((q−2x)/(x−p)).From the hypothesis f^(−1) (q)=−1we get ((q−2q)/(q−p))=−1 ⇒q−2q=p−q⇒p=0⇒f^(−1) (x)=((q−2x)/x) ⇒f^(−1) (2q)=((q−2.2q)/(2q))=((−3q)/(2q))=−3/2
f(x)=px+qx+2,q≠0⇒x(f(x)−p)=q−2f(x)⇒x=q−2f(x)f(x)−p⇒f−1(x)=q−2xx−p.Fromthehypothesisf−1(q)=−1wegetq−2qq−p=−1⇒q−2q=p−q⇒p=0⇒f−1(x)=q−2xx⇒f−1(2q)=q−2.2q2q=−3q2q=−3/2

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