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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)=?
$${Given}\:{f}\left({x}\right)=\frac{{px}+{q}}{{x}+\mathrm{2}}\:,\:{q}\neq\:\mathrm{0} \\ $$$${f}^{−\mathrm{1}} \:\left({q}\right)\:=\:−\mathrm{1}\:{then}\:{f}^{−\mathrm{1}} \left(\mathrm{2}{q}\right)=? \\ $$
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)
$${q}={f}\left(−\mathrm{1}\right)\:\:\:\:\:{q}={q}−{p}\:\:{p}=\mathrm{0} \\ $$$${f}\left({x}\right)=\frac{{q}}{{x}+\mathrm{2}}\:\:\Rightarrow\:\:{let}\:{f}\left({x}\right)={y}\: \\ $$$${y}=\frac{{q}}{{x}+\mathrm{2}}\:\:\:\:\:{xy}+\mathrm{2}{y}={q}\:\:\:\:\:{x}=\frac{{q}−\mathrm{2}{y}}{{y}} \\ $$$${hence}\:{f}^{−\mathrm{1}} \left({x}\right)=\frac{{q}−\mathrm{2}{x}}{{x}}\:\:,{x}\neq\mathrm{0}\: \\ $$$$\Rightarrow\:{f}^{−\mathrm{1}} \left(\mathrm{2}{q}\right)=\frac{{q}−\mathrm{4}{q}}{\mathrm{2}{q}}=\frac{−\mathrm{3}{q}}{\mathrm{2}{q}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left(\mathrm{2}{q}\right)=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$
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
$${f}\left({x}\right)=\frac{{px}+{q}}{{x}+\mathrm{2}}\:,\:{q}\neq\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}\left(\mathrm{f}\left(\mathrm{x}\right)−\mathrm{p}\right)=\mathrm{q}−\mathrm{2f}\left(\mathrm{x}\right)\Rightarrow\mathrm{x}=\frac{\mathrm{q}−\mathrm{2f}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)−\mathrm{p}} \\ $$$$\Rightarrow\mathrm{f}\:^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{q}−\mathrm{2x}}{\mathrm{x}−\mathrm{p}}.\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis} \\ $$$$\mathrm{f}\:^{−\mathrm{1}} \left(\mathrm{q}\right)=−\mathrm{1we}\:\mathrm{get}\:\frac{\mathrm{q}−\mathrm{2q}}{\mathrm{q}−\mathrm{p}}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{q}−\mathrm{2q}=\mathrm{p}−\mathrm{q}\Rightarrow\mathrm{p}=\mathrm{0}\Rightarrow\mathrm{f}\:^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{q}−\mathrm{2x}}{\mathrm{x}} \\ $$$$\Rightarrow\mathrm{f}\:^{−\mathrm{1}} \left(\mathrm{2q}\right)=\frac{\mathrm{q}−\mathrm{2}.\mathrm{2q}}{\mathrm{2q}}=\frac{−\mathrm{3q}}{\mathrm{2q}}=−\mathrm{3}/\mathrm{2} \\ $$

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