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Question Number 15463 by myintkhaing last updated on 10/Jun/17
$${Find}\:{the}\:{domain}\:{and}\:{range}\:{of}\:{a}\:{function}\:{for}\:{which}\:{f}\left({x}\right)=\frac{\mathrm{1}+\mathrm{2}{x}}{{x}}. \\ $$
Answered by Tinkutara last updated on 10/Jun/17
$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\forall\:{x}\:\in\:{R}\:−\:\left\{\mathrm{0}\right\}. \\ $$$$\mathrm{Domain}\:=\:{R}\:−\:\left\{\mathrm{0}\right\} \\ $$$$\mathrm{Let}\:{y}\:=\:\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{{x}} \\ $$$${xy}\:=\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$$${x}\left({y}\:−\:\mathrm{2}\right)\:=\:\mathrm{1} \\ $$$${x}\:=\:\frac{\mathrm{1}}{{y}\:−\:\mathrm{2}} \\ $$$$\therefore\:{y}\:−\:\mathrm{2}\:\neq\:\mathrm{0}\:\mathrm{which}\:\mathrm{gives}\:{y}\:\neq\:\mathrm{2} \\ $$$$\mathrm{Hence}\:\mathrm{Range}\:=\:{R}\:−\:\left\{\mathrm{2}\right\} \\ $$
Commented by myintkhaing last updated on 11/Jun/17
$${thank}\:{you} \\ $$