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Question Number 136740 by bramlexs22 last updated on 25/Mar/21
Given f(2f^(−1) (x))= (x/(2−x)) . what is f(x) ?
$$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{2f}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)=\:\frac{\mathrm{x}}{\mathrm{2}−\mathrm{x}}\:. \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{f}\left(\mathrm{x}\right)\:? \\ $$
Answered by Olaf last updated on 25/Mar/21
f(2f^(−1) (x)) = (x/(2−x)) (1) Let y = 2f^(−1) (x) ⇒ x = f((y/2)) (1) : f(y) = ((f((y/2)))/(2−f((y/2)))) (1/(f(y))) = ((2−f((y/2)))/(f((y/2)))) = (2/(f((y/2))))−1 (2) Let g(y) = (1/(f(y))) (2) : g(y) = 2g((y/2))−1 (3) ⇒ g′(y) = 2×(1/2)g′((y/2)) = g′((y/2)) g′(y) = g′((y/2)) = ... = g′((y/2^n )) g′(y) = lim_(n→∞) g′((y/2^n )) = g′(0) = cst a ⇒g(y) = ay+b (3) : ay+b = 2(a(y/2)+b)−1 b = 1, g(y) = ay+1 f(y) = (1/(g(y))) = (1/(ay+1)) Let x = f(y) ⇒ y = f^(−1) (x) x = (1/(ay+1)) y = (1/a)((1/x)−1) = f^(−1) (x)
$$ \\ $$$${f}\left(\mathrm{2}{f}^{−\mathrm{1}} \left({x}\right)\right)\:=\:\frac{{x}}{\mathrm{2}−{x}}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{y}\:=\:\mathrm{2}{f}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow\:{x}\:=\:{f}\left(\frac{{y}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{1}\right)\::\:{f}\left({y}\right)\:=\:\frac{{f}\left(\frac{{y}}{\mathrm{2}}\right)}{\mathrm{2}−{f}\left(\frac{{y}}{\mathrm{2}}\right)}\: \\ $$$$\frac{\mathrm{1}}{{f}\left({y}\right)}\:=\:\frac{\mathrm{2}−{f}\left(\frac{{y}}{\mathrm{2}}\right)}{{f}\left(\frac{{y}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{2}}{{f}\left(\frac{{y}}{\mathrm{2}}\right)}−\mathrm{1}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{Let}\:{g}\left({y}\right)\:=\:\frac{\mathrm{1}}{{f}\left({y}\right)} \\ $$$$\left(\mathrm{2}\right)\::\:{g}\left({y}\right)\:=\:\mathrm{2}{g}\left(\frac{{y}}{\mathrm{2}}\right)−\mathrm{1}\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\Rightarrow\:{g}'\left({y}\right)\:=\:\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}{g}'\left(\frac{{y}}{\mathrm{2}}\right)\:=\:{g}'\left(\frac{{y}}{\mathrm{2}}\right) \\ $$$${g}'\left({y}\right)\:=\:{g}'\left(\frac{{y}}{\mathrm{2}}\right)\:=\:…\:=\:\:{g}'\left(\frac{{y}}{\mathrm{2}^{{n}} }\right) \\ $$$${g}'\left({y}\right)\:=\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{g}'\left(\frac{{y}}{\mathrm{2}^{{n}} }\right)\:=\:{g}'\left(\mathrm{0}\right)\:=\:\mathrm{cst}\:{a} \\ $$$$\Rightarrow{g}\left({y}\right)\:=\:{ay}+{b} \\ $$$$\left(\mathrm{3}\right)\::\:{ay}+{b}\:=\:\mathrm{2}\left({a}\frac{{y}}{\mathrm{2}}+{b}\right)−\mathrm{1}\: \\ $$$${b}\:=\:\mathrm{1},\:{g}\left({y}\right)\:=\:{ay}+\mathrm{1} \\ $$$${f}\left({y}\right)\:=\:\frac{\mathrm{1}}{{g}\left({y}\right)}\:=\:\frac{\mathrm{1}}{{ay}+\mathrm{1}} \\ $$$$\mathrm{Let}\:{x}\:=\:{f}\left({y}\right)\:\Rightarrow\:{y}\:=\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$${x}\:=\:\frac{\mathrm{1}}{{ay}+\mathrm{1}} \\ $$$${y}\:=\:\frac{\mathrm{1}}{{a}}\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)\:=\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$

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