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Question Number 32640 by Rio Mike last updated on 30/Mar/18

Given the function f:x→ ((x +1)/(3x)) and g : x → x−1.Find a) fg b)f°g c) gf^(−1) (x)

$${Given}\:{the}\:{function}\:{f}:{x}\rightarrow\:\frac{{x}\:+\mathrm{1}}{\mathrm{3}{x}} \\ $$$${and}\:{g}\::\:{x}\:\rightarrow\:{x}−\mathrm{1}.\mathrm{Find}\: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{fg} \\ $$$$\left.\mathrm{b}\right)\mathrm{f}°\mathrm{g} \\ $$$$\left.\mathrm{c}\right)\:{gf}^{−\mathrm{1}} \left({x}\right) \\ $$

Answered by Joel578 last updated on 30/Mar/18

f(x) = ((x + 1)/(3x)), g(x) = x − 1 (f ○ g)(x) = f(x − 1) = (((x − 1) + 1)/(3(x − 1))) = (x/(3x − 3)) (g ○ f)(x) = g(((x + 1)/(3x))) = ((x + 1)/(3x)) − ((3x)/(3x)) = ((1 − 2x)/(3x)) f^(−1) (x) = (1/(3x − 1)) (g ○ f^(−1) )(x) = g((1/(3x − 1))) = (1/(3x − 1)) − ((3x − 1)/(3x − 1)) = ((2 − 3x)/(3x − 1))

$${f}\left({x}\right)\:=\:\frac{{x}\:+\:\mathrm{1}}{\mathrm{3}{x}},\:\:{g}\left({x}\right)\:=\:{x}\:−\:\mathrm{1} \\ $$$$\left({f}\:\circ\:{g}\right)\left({x}\right)\:=\:{f}\left({x}\:−\:\mathrm{1}\right)\:=\:\frac{\left({x}\:−\:\mathrm{1}\right)\:+\:\mathrm{1}}{\mathrm{3}\left({x}\:−\:\mathrm{1}\right)}\:=\:\frac{{x}}{\mathrm{3}{x}\:−\:\mathrm{3}} \\ $$$$ \\ $$$$\left({g}\:\circ\:{f}\right)\left({x}\right)\:=\:{g}\left(\frac{{x}\:+\:\mathrm{1}}{\mathrm{3}{x}}\right)\:=\:\frac{{x}\:+\:\mathrm{1}}{\mathrm{3}{x}}\:−\:\frac{\mathrm{3}{x}}{\mathrm{3}{x}}\:=\:\frac{\mathrm{1}\:−\:\mathrm{2}{x}}{\mathrm{3}{x}} \\ $$$$ \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}{x}\:−\:\mathrm{1}} \\ $$$$\left({g}\:\circ\:{f}^{−\mathrm{1}} \right)\left({x}\right)\:=\:{g}\left(\frac{\mathrm{1}}{\mathrm{3}{x}\:−\:\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}{x}\:−\:\mathrm{1}}\:−\:\frac{\mathrm{3}{x}\:−\:\mathrm{1}}{\mathrm{3}{x}\:−\:\mathrm{1}}\:=\:\frac{\mathrm{2}\:−\:\mathrm{3}{x}}{\mathrm{3}{x}\:−\:\mathrm{1}} \\ $$

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