Question Number 213291 by MathematicalUser2357 last updated on 02/Nov/24
$${Find}\:{domain}\:{of}\:{y}_{\mathrm{213291}} : \\ $$$${y}_{\mathrm{213291}} =\frac{\mathrm{3}+{e}^{\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{{x}−\mathrm{6}}} }{\mathrm{log}_{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$
Answered by MrGaster last updated on 02/Nov/24
$${D}_{{y}\mathrm{213291}} :\begin{cases}{{x}\neq\mathrm{6}}\\{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}>\mathrm{0}}\\{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\neq\mathrm{1}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{{D}_{{y}\mathrm{213291}} =\left(−\infty,−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\cup\left(−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\cup\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\cup\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}},\mathrm{6}\right)\cup\left(\mathrm{6},+\infty\right)}\\\hline\end{array} \\ $$