Question Number 42730 by Rio Michael last updated on 01/Sep/18
$${Given}\:{that}\:{f}\left({x}\right)\:=\:\sqrt{\mathrm{1}−{x}}\:{Find} \\ $$$$\left.{a}\right)\:{D}_{{f}} \:{for}\:{the}\:{arranged}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.{b}\right)\:{fg}\:{if}\:{fh}=\:{g}\left({x}\right)\:{and}\:{h}\left({x}\right)=\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4} \\ $$$$\left.{c}\right)\:{A}\left({x}\right)=\:\begin{cases}{\sqrt{\mathrm{1}−{x}}\:,\:{x}\neq\:\mathrm{1}}\\{{x}^{\mathrm{2}} ,{x}\neq\mathrm{0}}\end{cases} \\ $$$${find}\:{A}^{−\mathrm{1}} . \\ $$$$ \\ $$$$ \\ $$
Commented by Joel578 last updated on 03/Sep/18
$$\mathrm{For}\:\left({c}\right) \\ $$$$\mathrm{if}\:{x}\:=\:−\mathrm{1},\:\mathrm{what}\:\mathrm{function}\:{A}\left({x}\right)\:\mathrm{should}\:\mathrm{be}\:\mathrm{defined}? \\ $$$${A}\left({x}\right)\:=\:\sqrt{\mathrm{1}\:−\:{x}}\:\:\mathrm{or}\:{A}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:? \\ $$
Answered by Joel578 last updated on 02/Sep/18
$$\left({a}\right) \\ $$$${D}_{{f}} \:=\:\left\{{x}\:\in\:\mathbb{R}\::\:{x}\:\leqslant\:\mathrm{1}\:\right\} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$$\left({f}\:.\:{h}\right)\left({x}\right)\:=\:\left(\mathrm{3}{x}^{\mathrm{2}} \:−\:\mathrm{4}\right)\sqrt{\mathrm{1}−{x}}\:=\:{g}\left({x}\right) \\ $$$$\:\left({f}\:.\:{g}\right)\left({x}\right)\:=\:\left(\sqrt{\mathrm{1}\:−\:{x}}\right)^{\mathrm{2}} \left(\mathrm{3}{x}^{\mathrm{2}} \:−\:\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{1}\:−\:{x}\right)\left(\mathrm{3}{x}^{\mathrm{2}} \:−\:\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\mathrm{3}{x}^{\mathrm{3}} \:+\:\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:−\:\mathrm{4} \\ $$