Question Number 105593 by Study last updated on 30/Jul/20 | ||
$${f}\left({x}\right)=\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}×\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{2}{x}+\mathrm{1}}\:\:\:\:\:\:{Dom}_{{f}} =? \\ $$ | ||
Answered by mathmax by abdo last updated on 30/Jul/20 | ||
$$\left.\mathrm{D}_{\mathrm{f}} =\right]−\frac{\mathrm{1}}{\mathrm{2}},+\infty\left[\right. \\ $$ | ||
Commented by Study last updated on 30/Jul/20 | ||
$${why}\:{show}\:{u}\:{practice} \\ $$ | ||
Answered by mathmax by abdo last updated on 30/Jul/20 | ||
$$\mathrm{what}\:\mathrm{practice}\:\mathrm{here}\:?\:\:\:\mathrm{x}\in\:\mathrm{Df}\:\Leftrightarrow\:\mathrm{2x}+\mathrm{1}\:\geqslant\mathrm{0}\:\mathrm{and}\:\mathrm{2x}+\mathrm{1}\:\neq\mathrm{0}\:\Rightarrow\mathrm{2x}+\mathrm{1}>\mathrm{0}\:\Rightarrow \\ $$$$\left.\mathrm{x}>−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{D}_{\mathrm{f}} =\right]−\frac{\mathrm{1}}{\mathrm{2}},+\infty\left[\right. \\ $$ | ||