Question Number 74265 by Maclaurin Stickker last updated on 21/Nov/19
$${Determine}\:{the}\:{values}\:{of}\:{m}\in\mathbb{R}\:{for}\: \\ $$$${which}\:{the}\:{function}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}{x}^{\mathrm{2}} −{mx}+{m}}} \\ $$$${is}\:{the}\:{set}\:{of}\:{real}\:{numbers}. \\ $$
Answered by MJS last updated on 21/Nov/19
$$\mathrm{2}{x}^{\mathrm{2}} −{mx}+{m}=\mathrm{0} \\ $$$${x}=\frac{{m}\pm\sqrt{{m}\left({m}−\mathrm{8}\right)}}{\mathrm{4}} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{no}\:\mathrm{zeros}: \\ $$$${m}\left({m}−\mathrm{8}\right)<\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{0}<{m}<\mathrm{8} \\ $$$$\mathrm{for}\:\mathrm{0}<{m}<\mathrm{8}:\:\forall{x}\in\mathbb{R}:\:\mathrm{2}{x}^{\mathrm{2}} −{mx}+{m}>\mathrm{0} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{range}\:\mathrm{is}\:\mathrm{0}<{f}\left({x}\right)\leqslant\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{8}{m}−{m}^{\mathrm{2}} }} \\ $$